tag:blogger.com,1999:blog-26684924101901769212024-03-06T00:53:52.696-05:00mathcounts notesThe best math program for middle school studentssomeone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.comBlogger135125tag:blogger.com,1999:blog-2668492410190176921.post-86111924132800050252023-12-10T15:34:00.001-05:002023-12-10T15:34:37.860-05:00Sequences and Series -- Arithmetic and Geometric Sequences <b>Sequences are fun to learn and not really that difficult. </b><br />
<b>There are many similarities between arithmetic and geometric sequences, so </b><br />
<b>learn both together. </b><br />
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<b>Enjoy !!!!! </b><br />
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<b>From Mthcounts Mini: Sequences and Series</b><br />
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<span style="color: blue;"><b>Easier concepts:</b></span><br />
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<a href="http://mathcounts.org/resources/video-library/mathcounts-minis/mathcounts-mini-25-sequences"><span style="color: #274e13;">Sequences</span></a><br />
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<span style="color: #274e13;"><a href="http://www.mathcounts.org/resources/video-library/mathcounts-minis/mini-3-arithmetic-sequencesdetermining-nth-term" target="_blank">Arithmetic sequence/determine the nth term </a></span><br />
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<span style="color: #274e13;"><a href="http://mathcounts.org/resources/video-library/mathcounts-minis/mathcounts-mini-12-arithmetic-and-geometric-sequences">Arithmetic and geometric sequences</a></span><br />
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<a href="http://mathcountsnotes.blogspot.com/2012/05/mathcounts-strategies-some-sums.html" target="_blank">Mathcounts strategies : review some sums </a><br />
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<span style="color: #660000;"><b>Note : Don't just memorize, but really understand the concepts.</b></span><br />
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<span style="color: blue;"><b>Harder concepts:</b></span><br />
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<a href="http://www.youtube.com/watch?v=GS0x9JxVYzI">Sum and Average of An Evenly Space</a><br />
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<a href="http://mathcounts.org/resources/video-library/mathcounts-minis/mathcounts-mini-17-relationships-between-arithmetic"><span style="color: #274e13;">Relationship between arithmetic sequences, mean and median</span></a><br />
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<a href="http://mathcounts.org/resources/video-library/mathcounts-minis/mathcounts-mini-29-sequences-series-and-patterns"><span style="color: #274e13;">Sequences, series and patterns</span></a><br />
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<a href="http://mathcountsnotes.blogspot.com/2013/01/2013-mathcounts-basic-concept-review.html">Some Common Sums</a><br />
<br />someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-32967075666481927392023-11-23T17:35:00.000-05:002023-11-27T09:27:32.910-05:00A Skill for the 21st Century: Problem Solving by Richard Rusczyk<a href="https://www.greatschools.org/gk/articles/why-americas-smartest-students-fail-math/" target="_blank">Does our approach to teaching math fail even the smartest kids ? </a><br />
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Quotes from that article "<span style="background-color: white; color: #333333; font-size: 16px; letter-spacing: 0.32px; line-height: 25px;">According to research from the University of California, Los Angeles, as many as 60 percent of all college students who intend to study a STEM (science, technology, engineering, math) subject end up transferring out. In an era when politicians and educators are beside themselves with worry over American students’ lagging math and science scores compared to the whiz kids of Shanghai and Japan, this attrition trend so troubles experts it has spawned an entire field of research on “STEM drop-out,” citing reasons from gender and race to GPAs and peer relationships."</span><br />
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<a href="http://www.nytimes.com/2011/11/06/education/edlife/why-science-majors-change-their-mind-its-just-so-darn-hard.html" target="_blank">Why Science Majors Change Their Minds (It's Just So Darn Hard) from the New York Times </a></div>
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<a href="http://www.nxtbook.com/nxtbooks/imagine/20110304_MIV/index.php?startid=9" target="_blank">A Skill for the 21st Century: Problem Solving by Richard Rusczyk,</a> founder of "Art of Problem Solving".<br />
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<a href="http://lifehacker.com/top-10-skills-we-wish-were-taught-in-school-but-usuall-1622414148">Top 10 Skills We Wish Were Taught at School, But Usually Aren't </a><br />
from Lifehackersomeone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com3tag:blogger.com,1999:blog-2668492410190176921.post-39558576371815171642023-10-11T10:59:00.000-04:002023-10-11T10:59:49.804-04:002023/ 2024 Mathcounts, AMCs, AIMEs Competition Preparation Strategies <b>Hi, Thanks for visiting my blog.</b><br />
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E-mail me at <b>thelinscorner@gmail.com</b> if you want to learn with me. :) :) :) </div>
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Currently I'm running different levels of problem solving lessons, and it's lots of fun learning along with students from different states/countries. </div>
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So many students are not learning smart.<br />
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Problem solving is really fun (and a lot of the times very hard, yes).<br />
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Good questions are intriguing and delicious, so come join our vibrant community and have the pleasure of finding things out on your own.</div>
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<a href="https://www.ted.com/talks/bel_pesce_5_ways_to_kill_your_dreams" target="_blank">5 ways to kill your dreams from TED Talk</a></div>
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<b><span style="color: #660000;">There is no overnight success.</span></b></div>
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<b><span style="color: blue;"><a href="http://mathcountsnotes.blogspot.com/2015/05/testemonials-of-my-services-so-far-most.html" target="_blank">Testimonials from my students/parents. </a> :) </span></b></div>
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<b>My other blogs :</b></div>
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<a href="https://studentsspeakmagazine.wordpress.com/" target="_blank">Studentsspeak Magazine (featuring students' independent projects)</a></div>
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<a href="http://thelinscorner.blogspot.com/" target="_blank">thelinscorner </a> : Standardized test preps, books, links/videos for life-time learning</div>
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Take care and have fun learning.</div>
<span style="font-family: "georgia" , "times new roman" , serif; font-size: small;"><b style="background-color: white;"><br /></b><b style="background-color: white;">Don't forget other equally interesting activities/contests, which engage your creativity and imagination. </b></span><br />
<span style="font-family: "georgia" , "times new roman" , serif; font-size: small;"><b style="background-color: white;">Some also require team work. Go for those and have fun !! </b></span><br />
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<b><span style="color: #660000; font-family: "georgia" , "times new roman" , serif; font-size: small;">Don't just do math. </span></b></div>
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<span style="color: #1155cc; font-family: "georgia" , "times new roman" , serif; font-size: small;"><a href="http://www.businessinsider.com/qualities-google-looks-for-in-job-candidates-2014-4" style="color: #1155cc;" target="_blank">11 Qualities Google Looks For In Job Candidates </a></span><br />
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<span style="font-family: "georgia" , "times new roman" , serif; font-size: small;"><b>In Chinese :</b> <span style="color: #0066cc; font-weight: bold;"><span style="color: #1155cc;"><a href="http://bbs.wenxuecity.com/znjy/2658101.html" target="_blank">《專題報導》亞裔尖子生 職場難出頭?</a></span></span></span></div>
<br />
Before going full throttle mode for competition math, please spend some time reading this<br />
well- thought-out article from BOGTRO at AoPS <a href="http://mathcountsnotes.blogspot.com/2014/06/learn-how-to-learn-by-bogtro-from-aops.html" target="_blank">"Learn How to Learn"</a>.<br />
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It will save you tons of time and numerous, unnecessary hours without a clear goal, better method in mind.<br />
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Less is more. My best students make steady, very satisfactory progress in much less time than those<br />
counterparts who spent double, triple, or even more multiple times of prep with little to show.<br />
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It's all about "deliberate practices", "tenacity", and most of all, "the pleasure of finding things out on your own". <br />
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Take care and have fun problem solving.<br />
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I have been coaching students for many years. By now, I know to achieve stellar performance you need :<br />
<a href="https://www.ted.com/talks/angela_lee_duckworth_grit_the_power_of_passion_and_perseverance?language=en">Grit (from TED talk)</a>, not only that but self-awareness (so you can fairly evaluate your own progress) and a nurturing-caring environment. (Parents need to be engaged as well.)<br />
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Thanks a lot !! Mrs. Lin<br />
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"Work Smart !!" , "Deliberate practices that target your weakness ", " Relax and get fully rested.", "Pace your time well", "Every point is the same so let go of some questions first; you can always go back to them if time permits."<br />
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"It's tremendous efforts preparing for a major event on top of mounting homework and if you are the ones who want to try that, not your parents and you work diligently towards your goal, good for you !!" <br />
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"Have fun, Mathcounts changes lives, because at middle school level at least, it's one of those rare occasions that the challenges are hard, especially at the state and national level."<br />
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<b>Now, here are the links to get you started: </b><br />
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Of course use my blog. Whenever I have time I analyze students' errors and try to find better ways (the most elegant solutions or the Harvey method I hope) to tackle a problem. Use the search button to help you target your weakness area.<br />
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<a href="http://mathcounts.org/pastcompetition">Last year's Mathcounts competition problems and answer key </a><br />
<br /><a href="https://www.mathcounts.org/sites/default/files/2021-08/2122%20%5BC2C%5D%20HB%20ALL%20250%20Problems%20NO%20SOLUTIONS.pdf" target="_blank">Last year's handbook questions.</a><br />
Near the end of the handbook, there is a page called <b><span style="color: purple;"> problem index (page 7 to page 31 for 2021-2022 handbook)</span></b>.<br />
For state/national prep, find your weakness and work on the problems backwards, from the hardest to the easiest. <br />
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Here are some other links/sites that are <b><span style="color: #660000;">the best</span>.</b><br />
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<a href="https://www.mathcounts.org/resources/mathcounts-minis" rel="nofollow">Mathcounts Mini : </a>At the very least, finish watching and understanding most of the questions from 2010 till now and work on the follow-up sheets, since detailed solutions are provided along with some more challenging problems.<br />
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For those who are aiming for the state/national competition, you can skip the warm-up and go directly to "The Problems" used on the video as well as work on the harder problems afterward.<br />
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<a href="http://www.artofproblemsolving.com/">Art of Problem Solving </a><br />
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The best place to ask for help on challenging math problems. Some of the best students/coaches/teachers are there to help you better your problem solving skills. <br />
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Register for <a href="http://www.artofproblemsolving.com/liz/Alcumus/index.php" target="_blank">Alcumus</a> and start using the great tool to practice deliberately.<br />
Change the setting based on the levels of your proficiency of different topics. Do Not Rush !!<br />
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<b><span style="color: #660000;">Awesome site!! </span></b><br />
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For concepts reviewing, try the following three links.<br />
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<a href="https://blgjh.sharylandisd.org/common/pages/UserFile.aspx?fileId=2549395" target="_blank">Mathcounts Toolbox</a><br />
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<a href="https://monks.scranton.edu/files/courses/ProblemSolving/MathCountsPlaybookBW.pdf" target="_blank">Coach Monks's Mathcounts Playbook</a><br />
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You really need to understand how each concept works for the review sheets to be useful.<br />
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To my exasperation, I have kids who mix up the formulas without gaining a true understanding and appreciation of how an elegant, seemingly simple formula can answer myriads of questions.<br />
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You don't need a lot of formulas, handbook questions, or test questions to excel.<br />
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You simply need to know how the concepts work and apply that knowledge to different problems/situations.<br />
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Hope this is helpful!!<br />
<br />someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com60tag:blogger.com,1999:blog-2668492410190176921.post-27621920741257463022023-08-11T14:06:00.003-04:002023-11-06T07:22:38.201-05:00Som sums' formula:<div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhT13cQErkxg5bJo5MbJxdQC-FCcPLNpYETQK9-mURAbuqztlqhKPAh144x9kJ1FIn3MN7ITOwgV2odoki-fmnD8EHcZd4ifjD-ZwY0OQJre_2oEQKYMIvp8QAYHgkoDMaHIVin5qrDf5U/s1600-h/some+sums+formula.gif"><img alt="" border="0" height="353" id="BLOGGER_PHOTO_ID_5287950702874327170" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhT13cQErkxg5bJo5MbJxdQC-FCcPLNpYETQK9-mURAbuqztlqhKPAh144x9kJ1FIn3MN7ITOwgV2odoki-fmnD8EHcZd4ifjD-ZwY0OQJre_2oEQKYMIvp8QAYHgkoDMaHIVin5qrDf5U/w400-h353/some+sums+formula.gif" style="cursor: hand; float: left; height: 298px; margin: 0px 10px 10px 0px; width: 400px;" width="400" /></a> </div>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-34847802774343808412023-08-04T17:01:00.000-04:002023-08-04T17:01:49.842-04:00V's record <p> V's record </p><p>weekly homework about 30 to 45 minutes </p><p>extra videos, links optional </p><p><b>First lesson:</b></p><p><b>2010 chapter sprint: </b></p><p><span face="Arial, Helvetica, sans-serif" style="background-color: white; color: #222222; font-size: small;">Hi, I got a score of 22 and got questions </span><span face="Arial, Helvetica, sans-serif" style="background-color: white; font-size: small;"><span style="color: #990000;"><b>16, 22, 23, 24,25,27,29, and 30 wrong. </b></span></span></p><p><span face="Arial, Helvetica, sans-serif" style="background-color: white; color: #222222; font-size: small;">I just guessed these questions because I didn't really find a way to do them.</span></p><p><b>Second meet: </b></p><p><b>2011-12 Mathcounts handbook (40 questions total) </b></p><div style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: small;"><span style="color: #222222;">Warm Up </span><b>1:<span style="color: #990000;"> 8</span></b><br /><span style="color: #222222;">Warm Up 2: </span><span style="color: #990000;"><b>19,20</b></span></div><div style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: small;"><span style="color: #222222;">Warm Up 3: </span><span style="color: #990000;"><b>32, 39 </b></span></div><div style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: small;"><span style="color: #222222;">Workout 1: </span><b><span style="color: #990000;">23, 24, 30</span></b></div><div style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: small;"><b><span style="color: #990000;"><br /></span></b></div><div style="background-color: white; font-family: Arial, Helvetica, sans-serif; font-size: small;"><div dir="ltr" style="color: #222222;"><b>Third meet: </b></div><div dir="ltr" style="color: #222222;"><b><br /></b></div><div dir="ltr" style="color: #222222;"><b>2010 Mathcounts school test : </b></div><div dir="ltr" style="color: #222222;"><br /></div><div dir="ltr" style="color: #222222;">Hi, I finished trying the Sprint and Target questions:<div> these are the problems that I got wrong <div>Sprint: I got 7/15 correct.</div><div> 15,16: Attempted but answer was wrong</div><div> 19,20,26,27,29: Didn't attempt</div><div> 30: Couldn't find a good method to do, but was able to solve it by listing out all the possibilities.</div><div>Target: I got 6/8 correct.</div><div> The first 6 were relatively easy and I could find a clear way to do them</div></div><div> The last 2, I couldn't find a way to approach the problem.</div><div><br /></div><div><div dir="ltr">Fourth meet : </div><div dir="ltr"><b>2011 Mathcounts school test :</b> last 15 sprint and last 4 target </div><div dir="ltr"><br /></div><div dir="ltr">Sprint: Out of the last 15, </div><div dir="ltr">I got 7 correct.</div><div dir="ltr">The questions that I got wrong were<b> 20,21,22,24,27,28,29,30.</b> </div><div dir="ltr"><br /></div><div dir="ltr">I tried 21,22, and 27 but the answers were incorrect. I wasn't able to attempt the rest. </div><div dir="ltr"><br /><div>Target: I got every question <b>other than #5.</b> </div><div>However, questions 1 and 2, </div><div>I got wrong at first, but when I retried them, </div><div>I was able to get them. I read the problems wrong and didn't fully understand them the first time.</div><div><br /></div><div><br />Fifth meet : review </div><div><br /></div><div>Sixth meet : </div><div><b>2011 chapter </b></div><div>Hi Mrs Lin! I was able to try both the target and the sprint round questions and here are my results:<div>Target: I missed 3 & 8, and I didn't attempt them.</div><div>Sprint: I got 17,21,23,24,25, 27,29,30.</div><div>I didn't attempt 27,29, and 30,</div></div><div><br /></div><div><b>2012 Mathcounts school </b></div><div>Hi Mrs Lin! I was able to finish both the chapter and target, and here are my results:<div>Target: I got 5,6,and 8 wrong. I didn't try any of them because I didn't know how to do them.</div><div>Sprint: I got 18,23,25,26, and 30</div></div><div><br /></div><div><b>Apr. 30, 2023 </b></div><div><b>2012 Mathcounts chapter </b></div><div>Hi Mrs. Lin! I hope your having a very good day and week! I tried both the target and the sprint. <div>Target: I got only number 8 wrong, but I didn't know how to do it.</div><div>Sprint: I got number 22, 24, 26,28,29, and 30 wrong. I tried to do 22, but I got it wrong. The rest I didn't know how to do them..</div></div><div><br /></div><div><b>May 7th, review </b></div><div><br /></div><div><b>May 13th, 2023 </b></div><div><b>2013 Mathcounts School </b></div><div>Hi Mrs.Lin, I was able to try both the Target and the Sprint and here are my results!<div>Target (5/8) I got 6,7, and 8 wrong, however, I was able to figure out the answer to problem 6 when I reviewed it. </div><div>Sprint: I got 30,29,28, and 24 wrong. </div></div><div><br /></div><div>May 20th, 2023</div><div>Hi Mrs Lin!</div><div><b>2013 Mathcounts Chapter </b><br /><div>I was able to try both the target and sprint round tests and here are my results:</div><div>Target: 6/8. I attempted the first 6 problems and got all of them right. I understood the problems fairly well and was able to do all of them on the first try.</div><div>sprint: I got 10/15 right. I got all problems I attempted right, and didn't attempt <b>30,28,25,24,21, and 19.</b></div></div><div><b><br /></b></div><div>2014 Mathcounts school </div><div>Hi Mrs Lin, I was able to finish both the target and the sprint, and here were my results!<div>Sprint: 22/30, I didn't get 18,19,24,25,26,28,29, and 30. I attempted number 18, but didn't get it.</div><div>Target: I got 5 out of 8 on the target, but after reviewing my answers, I was able to figure out number 5.</div></div><div><br /></div><div>2015 Mathcounts school </div><div>Hi Mrs Lin, I was able to try both the sprint and target. For the sprint, I got problems 20 and 27 wrong, and I didn't attempt any problems past 24 other than 27.<div>For my Target I got numbers 6,7, and 8 wrong, and I didn't know how to approach any of them. </div></div></div><div class="yj6qo ajU" style="cursor: pointer; margin: 2px 0px 0px; outline: none; padding: 10px 0px; width: 22px;"></div></div></div><div class="yj6qo ajU" style="color: #222222; cursor: pointer; margin: 2px 0px 0px; outline: none; padding: 10px 0px; width: 22px;"></div></div>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-13360381970900611142023-05-05T08:07:00.000-04:002023-05-05T08:07:36.037-04:00Pathfinder From Mathcounts Mini :<br />
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<a href="http://mathcounts.org/resources/video-library/mathcounts-minis/mathcounts-mini-7-countingpaths-along-grid">Counting/Paths Along a Grid</a> <br />
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From Art of Problem Solving <br />
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<a href="http://www.youtube.com/watch?v=M8BYckxI8_U">Counting Paths on a Grid</a> <br />
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<a href="https://www.youtube.com/watch?v=trIkPmwTSyA" target="_blank">Math Principles : Paths on a Grid : Two Approaches </a><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhTlCdAh5X7gCki2Zt6Rt9uHutms3GM-QM-RKhVJVRW86BCCotp0uP65khPfIDJ0IKMMwwlr5jtsVzDLE7QmQVmLpaTMKNkMPs_7hHb2-pHUaCMgUUFsqLmcOaIk0lwGt39APJGs2fuT1U/s1600/how+many+ways+to+move+the+dominos.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="180" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhTlCdAh5X7gCki2Zt6Rt9uHutms3GM-QM-RKhVJVRW86BCCotp0uP65khPfIDJ0IKMMwwlr5jtsVzDLE7QmQVmLpaTMKNkMPs_7hHb2-pHUaCMgUUFsqLmcOaIk0lwGt39APJGs2fuT1U/s200/how+many+ways+to+move+the+dominos.jpg" width="200" /></a></div>
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<b><span style="color: #274e13;">Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally? </span></b><br />
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Solution :<br />
You can move the dominoes 5 times to the right at most and 5 down to<br />
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = <b><span style="color: #660000;">252 ways</span></b><br />
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<b><span style="color: #660000;"><span style="color: #274e13;">Question # 2: How many ways can you move from A to B if you can only move downward and to right? </span></span></b><br />
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<span style="color: blue;">Solution :</span>
<span style="color: black;">There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\)</span> * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = <b><span style="color: #660000;">9800 ways</span></b> from A to B<br />
<br />someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-57482589095920948052023-03-03T18:42:00.003-05:002023-03-04T10:53:22.502-05:00Sum of All the Possible Arrangements of Some Numbers<span style="color: black;">Check out <a href="https://mathcounts.org/">Mathcounts</a>, the best middle school competition math program up to the national level. </span><br />
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<div style="color: #274e13;"><b>Questions to ponder: (detailed solutions below) </b></div><div><b><span style="color: #4c1130;">It's extremely important for you to spend some time pondering on </span></b><b><span style="color: #4c1130;">these questions first without peeking on the solutions.</span><span style="color: #134f5c;"> </span></b></div><div style="color: #274e13;">
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<b>#1:
Camy made a list of every possible distinct four-digit positive integer
that can be formed using each of the digits 1, 2 , 3 and 4 exactly once
in each integer. What is the sum of the integers on Camy's list?</b></div>
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<b><span style="color: #006600;"><span style="color: #274e13;">#2: Camy made a list of every possible distinct five-digit positive </span><span style="color: #660000;">even</span><span style="color: #274e13;">
integer that can be formed using each of the digits 1, 3, 4, 5 and 9
exactly once in each integer. What is the sum of the integers on Camy's
list?</span> <span style="color: purple;">(2004 Mathcounts Chapter Sprint #29)</span></span></b><div><span style="color: purple;"><b><br /></b></span></div><div><b><span style="color: #006600;"><span style="color: #274e13;">#3: 2020 Mathcounts state sprint #24 </span></span></b><span style="color: purple;"><b><br /></b></span>
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<span style="color: #006600;"><span style="color: black;">Solutions:</span> </span><br />
<span style="color: #006600;"><span style="color: black;"><b><span style="color: #274e13;">#1: </span></b><b><span style="color: blue;"> </span></b></span></span><br />
<span style="color: #006600;"><span style="color: black;"><b><span style="color: blue;">Solution I:</span></b> </span></span><br />
<span style="color: #006600;"><span style="color: black;">There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times. </span></span><br />
<span style="color: black;">1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 =</span> <b><span style="color: #990000;">66660</span></b><span style="color: black;">, which is the answer.</span><br />
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<b>Solution II: </b></div>
<span style="color: black;">The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrange</span><br />
<span style="color: black;">the four numbers. </span><br />
<span style="color: black;">2.5 (1000 + 100 + 10 + 1) x 24 = <b style="color: #990000;">66660 </b></span><br />
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<span style="color: black;"><b><span style="color: #274e13;">#2:</span></b> <b><span style="color: blue;"> </span></b></span><br />
<span style="color: black;"><b><span style="color: blue;">Solution I:</span></b> </span><br />
<span style="color: black;">Since this time Camy wants five-digit</span> <b><span style="color: #3333ff;">even</span></b> <span style="color: black;">integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely. </span><br />
<span style="color: black;">Again
there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18
and 18 x 6 = 108 (Each number that can be moved freely appears 6 times
evenly.)</span><span style="color: black;">108 x 11110 + 4 x 24 =</span> <b style="color: #990000;">1199976</b><br />
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<b><span style="color: #990000;"><span style="color: blue;">Solution II: </span></span></b><b style="color: #990000;"> <br />
</b><br />
<span style="color: black;">Since this time Camy wants five-digit</span> <b><span style="color: #3333ff;">even</span></b> <span style="color: black;">integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely. </span><br />
<span style="color: black;">There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96</span><br />
<span style="color: black;">As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\) </span> <br />
4.5<i> *</i> ( 10000 + 1000 + 100 + 10) * 24 (arrangements) + 96 = 4.5 * 11110 * 24 + 96 = <b><span style="color: #990000;">1199976</span></b></div><div><b><span style="color: #990000;"><br /></span></b></div><div><b><span style="color: #274e13;">#3: The answer is </span><span style="color: #990000;"><span>101</span><span>.</span></span><span style="color: #274e13;"> </span></b></div><div><b><span style="color: #990000;"><br /></span></b></div><div><b><span style="color: #274e13;">Other applicable problems: (answers below)</span><br />
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<b>#1: What is the sum of all the four-digit positive integers that
can be written with the digits 1, 2, 3, 4 if each digit must be used
exactly once in each four-digit positive integer? <span style="color: purple;">(2003 Mathcounts Sprint #30)</span></b></div>
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<b>#2: What is the average (mean) of all 5-digit numbers that
can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly
once? (You can use a calculator for this question.) <span style="color: purple;">(2005 AMC-10 B) </span><br />
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<b>#3: What is
the sum of all the four-digit positive integers that can be written with
the digits 2, 4, 6, 8 if each digit must be used exactly once in each
four-digit positive integer?</b></div>
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<b>#4: What is
the sum of all the 5-digit positive odd integers that can be written
with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly
once in each five-digit positive integer? </b></div>
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<b>#5:</b><b>What is
the sum of all the four-digit positive integers that can be written with
the digits 2, 3, 4, 5 if each digit must be used exactly once in each
four-digit positive integer?</b></div>
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<b>Answer key: </b><br />
<b>#1: <span style="color: #cc0000;">66660</span></b><br />
<b>#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)</b><br />
<b>4.8 * 11111 =\(\color{red}{53332.8}\) </b><br />
<b>#3: <span style="color: #cc0000;">133320</span></b><br />
<b>#4: <span style="color: #cc0000;">1333272</span></b><br />
<b>#5: <span style="color: #cc0000;">93324</span> <br />
</b></div>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-48839255930950307422023-01-20T09:51:00.001-05:002023-01-20T09:51:05.017-05:002015 Mathcounts State Prep: Mathcounts State Harder QuestionsCheck out <a href="http://mathcounts.org/">Mathcounts</a> here, the best competition math program for middle school students. <br />
Download this year's <a href="http://mathcounts.org/handbook">Mathcounts handbook</a> here.<br />
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<b><span style="color: #274e13;">Question:</span></b> <b><span style="color: #660000;">2010 Mathcounts State Team Round #10</span></b>: <b><span style="color: #274e13;">A square and isosceles triangle of equal height are side-by-side, as shown, with both bases on the x-axis. The lower right vertex of the square and the lower left vertex of the triangle are at (10, 0). The side of the square and the base of the triangle on the x-axis each equal 10 units. A segment is drawn from the top left vertex of the square to the farthest vertex of the triangle, as shown. What is the area of the shaded region?</span></b><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPL_XgDJNN1urya-GKZ8SJbnhkkSBqs0DBRlsQ0dgGhAltMK-XCKtABK_VM6Iq3FWFN3J5p_9BTfMvXH7DkJyHu1FKq6GjDgnDXCLVi2XzSICF_-vzcDVbfONG9eG5whgdv897RwyirLk/s1600/2010+team+%2310+state.gif" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjPL_XgDJNN1urya-GKZ8SJbnhkkSBqs0DBRlsQ0dgGhAltMK-XCKtABK_VM6Iq3FWFN3J5p_9BTfMvXH7DkJyHu1FKq6GjDgnDXCLVi2XzSICF_-vzcDVbfONG9eG5whgdv897RwyirLk/s1600/2010+team+%2310+state.gif" height="480" width="640" /></a></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7JvOy3NSMOrnijfaLk7cKs2x1QNCxKo17ghzEJFGwMs0sxWDBevx240PeRak7-t22x-pwOwa-q1sK2tB55YYd1OxdBUV2k1yz22pcljk_AiBfanBi5n17daRTtioskd11oRBnZE6I7z0/s1600/2010+Mathcounts+Team+%2310.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><br /></a> There are lots of similar triangles for this question, but I think this is the fastest way to find the area.<br />
\(\Delta \)<span style="color: #274e13;"><span style="color: black;">A<span style="color: #274e13;">GB <span style="color: black;">is similar to </span></span></span></span><span style="color: #274e13;"><span style="color: black;"><span style="color: #274e13;"><span style="color: black;">\(\Delta \)<span style="color: #274e13;"><span style="color: black;"><span style="color: #274e13;">DGC and their line ratio is 15 to 10 or 3 : 2.</span></span></span></span></span></span></span><br />
<span style="color: #274e13;"><span style="color: black;"><span style="color: #274e13;"><span style="color: black;"><span style="color: #274e13;"><span style="color: black;"><span style="color: #274e13;"> </span></span></span></span></span></span></span>\(\Delta \)<span style="color: #274e13;"><span style="color: black;"><span style="color: #274e13;">CGF <span style="color: black;">is similar to </span></span></span></span><span style="color: #274e13;"><span style="color: black;"><span style="color: #274e13;"><span style="color: black;">\(\Delta \)<span style="color: #274e13;"><span style="color: black;"><span style="color: #274e13;">CBE. </span></span></span></span></span></span></span><br />
<span style="color: #274e13;"> </span>\(\dfrac {CG} {CB}=\dfrac {GF} {BE}\)<br />
\(\dfrac {2} {5}=\dfrac {GF} {10}\)\(\rightarrow GF=4\)
From there you get the area =<br />
\(\dfrac{10\times 4} {2}=20\)<br />
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<b><span style="color: #274e13;">Question:</span></b> <b><span style="color: #660000;">2010 Mathcounts State #30</span></b>: <b><span style="color: #274e13;">Point D lies on side AC of equilateral triangle ABC such that the measure of angle DBC is 45 degrees. What is the ratio of the area of triangle ADB to the area of triangle CDB? Express your answer as a common fraction in simplest radical form.</span></b><br />
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Since each side is the same for equilateral triangle ABC, once you use the 30-60-90 degree angle ratio and 45-45-90 degree angle ratio, you'll get the side.<br />
Since area ratio stays constant, you can plug in any numbers and it's much easier to use integer first so I use 2 for \(\overline {CD}\).<br />
From there you get the side length for each side is \(\sqrt {3}+1\).<br />
\(\overline {AC}-C\overline {D}=\sqrt {3}+1-2\) = \(\overline {AD} = \sqrt {3}-1\)<br />
\(\Delta ABD\) and \(\Delta CBD\) share the same vertex, so their area ratio is just the side ratio, which is <b><span style="color: #660000;">\(\dfrac {\sqrt {3}-1} {2}\).</span></b><br />
<br />someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-60058488483098867502022-10-15T08:39:00.001-04:002022-10-15T08:39:53.738-04:0016 17 Mathcounts handbook more interesting questions that have nicer solutions <div><b><span style="font-size: medium;">Thanks to Achuth for trying out these questions and time them as an actual Mathcounts test. :) </span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;">First week : warm up 1, 4, 7. (time for 40 mins. like sprint)</span></b></div><div><b><span style="font-size: medium;">Second week : warm up 2, 5, 8.</span></b></div><div><b><span style="font-size: medium;">third week : workout 3 --> all right. (pair 1 to 6, 2 to 7, each time for 6 mins. as </span></b></div><div><b><span style="font-size: medium;">target) </span></b></div><div><b><span style="font-size: medium;">fourth week: workout 4 --> #95, then self correct. </span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;">At lesson: workout 5 and other harder problems. </span></b></div><div><br /></div><div><span style="font-size: medium;"><b>These are nice questions that have various solutions, so it’s better to slow down and try them as puzzles.</b></span></div><div><span style="font-size: medium;"><b><b><br /></b></b></span></div><div><span style="font-size: medium;"><b><b>Less is more and slow is fast.</b></b></span></div><div><span style="font-size: medium;"><b><br /></b></span></div><div><span style="font-size: medium;"><b>If you are new to problem solving, one nice strategy is to make the question much simpler and explore ideas that come to your mind. </b></span></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;">Answer key down below. </span></b></div><b><span style="font-size: medium;"><div><b><span style="font-size: medium;"><br /></span></b></div>#66: A school of 100 fish swims in the ocean and comes to a very wide horizontal pipe. The fish have three choices to get to the food on the other side: swim above the pipe, through the pipe or below the pipe. If we do not consider the fish individually, in how many ways can the entire school of fish be partitioned into three groups with each group choosing a different one of the three options and with at least one fish in each group? </span></b><div><span style="font-size: medium;"><b><br /></b></span><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"> #105
When fully matured, a grape contains 80% water. After the drying process, called dehydration, the resulting raisin is only 20% water. What fraction of the original water in the grape remains after dehydration? Express your answer as a common fraction. </span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"> #112: Cora has five balls—two red, two blue and one yellow—which are indistinguishable except for their color. She has two containers—one red and one green. If the balls are randomly distributed between the two containers, what is the probability that the two red balls will be alone in the red container? Express your answer as a common fraction? </span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"> #116: A 12-foot by 12-foot square bathroom needs to be tiled with 1-foot square tiles. Two of the tiles are the wrong color. If the tiles are placed randomly, what is the probability that the two wrong-colored tiles share an edge? Express your answer as a common fraction.
</span></b></div></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;">#66: <span style="color: #990000;">4851</span></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;">#105: <span style="color: #990000;">1/16</span></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;">#112: <span style="color: #990000;">1/32</span></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;">#116: <span style="color: #990000;">1/39 </span></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div><div><b><span style="font-size: medium;"><br /></span></b></div>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-36776188159631974932022-03-23T15:33:00.000-04:002022-03-23T15:33:55.519-04:00Dimensional Change <div class="separator" style="clear: both; text-align: center;">
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There are lots of questions on dimensional change and this is a very common one.<br />
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Make sure you understand the relationship among linear, 2-D (area) and 3-D (volume) ratio.<br />
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There are many similar triangles featured in the image on the left.<br />
Each of the two legs of the largest triangles is split into 4 equal side lengths.<br />
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<b><span style="color: #274e13;"> </span></b><br />
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<b><span style="color: #274e13;">Question : What is the area ratio of the sum of the two white trapezoids to the largest triangle? </span></b><br />
\(\dfrac {\left( 3+7\right) } {16}=\dfrac {10} {16}=\dfrac{5}{8}\)
<b><span style="color: #660000;"><span style="color: #660000;"><span style="color: #274e13;"> </span></span></span></b><br />
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<b><span style="color: #660000;"><span style="color: #660000;"><span style="color: #274e13;">Question: If the area of the largest triangles are 400 square units, what is the area of the blue-colored trapezoid?</span></span></span></b><br />
\(\dfrac {5} {16}\times 400\) =<b><span style="color: #660000;">125 square units </span></b><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjlNIPTxgGszqukwiP2Sc6T87VnD0j9ZI-mTVraEuTFoJc1Cw42CDTUsWdeVR_i1BMCJ72UPKrmYIYAbSf39SUgNSxpaJr1elfahaexYJS9npYosYkxM3Cj-_yWJalWUgZh3G6zUYjOuyM/s1600/dimentional+change+similar+cones+%2528volume%2529.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjlNIPTxgGszqukwiP2Sc6T87VnD0j9ZI-mTVraEuTFoJc1Cw42CDTUsWdeVR_i1BMCJ72UPKrmYIYAbSf39SUgNSxpaJr1elfahaexYJS9npYosYkxM3Cj-_yWJalWUgZh3G6zUYjOuyM/s320/dimentional+change+similar+cones+%2528volume%2529.jpg" width="210" /></a></div>
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<span style="color: #660000;"><span style="color: black;">Again, each of the two legs are split into three equal segments. </span></span><br />
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<span style="color: #660000;"><span style="color: black;">The volume ration of the cone on the top to the middle frustum to the </span></span><br />
<span style="color: #660000;"><span style="color: black;">bottom frustum is <b><span style="color: #660000;">1 : 7 : 19.</span></b> </span></span><br />
<span style="color: #660000;"><span style="color: black;"> </span></span><br />
<span style="color: #660000;"><span style="color: black;">Make sure you understand why.</span></span><br />
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<span style="color: #660000;"><span style="color: black;"> </span></span><b><span style="color: #660000;"> </span></b>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-49022631309378233192022-03-06T18:19:00.000-05:002022-03-06T18:19:29.274-05:00Mass Points Geometry Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry<br />
so spend some time understanding it.<br />
<br />
<span style="color: #0000ee; text-decoration-line: underline;"><a href="https://www.parkrapids.k12.mn.us/cms/lib/MN02207558/Centricity/Domain/486/Mass_Point.pdf" target="_blank">Basics </a></span><br />
<br />
2014-15 Mathcounts handbook Mass Point Geometry Stretch<br />
<a href="http://academicinnovation.weebly.com/uploads/7/6/3/6/7636030/2014-2015_mathcounts_school_handbook_updated.pdf" target="_blank">from page 39 to page 40</a><br />
<br />
(Talking about motivation, yes, there are students already almost finish<br />
this year's Mathcounts' handbook harder problems.)<br />
<br />
<a href="http://en.wikipedia.org/wiki/Mass_point_geometry">From Wikipedia</a><br />
<br />
<a href="https://www.artofproblemsolving.com/wiki/index.php?title=Mass_points" target="_blank">From AoPS</a><br />
<a href="http://www.sanjosemathcircle.org/handouts/2007-2008/20071114.pdf" target="_blank"><br /></a>
<a href="https://mathcircle.berkeley.edu/sites/default/files/archivedocs/2015/lecture/MassPointGeometry_8Sep2015.pdf" target="_blank">Mass Point Geometry by Tom Rike</a><br />
<br />
<a href="http://www.aquatutoring.org/TYMCM%20Mass%20Points.pdf" target="_blank">Another useful notes </a><br />
<br />
Videos on Mass Point :<br />
<br />
<a href="https://www.youtube.com/watch?v=llX7s1UHdaA" target="_blank">Mass Points Geometry Part I </a><br />
<a href="https://www.youtube.com/watch?v=_5IrxEJADQ0" target="_blank"><br /></a>
<a href="https://www.youtube.com/watch?v=_5IrxEJADQ0" target="_blank">Mass Points Geometry : Split Masses Part II </a><br />
<br />
<a href="https://www.youtube.com/watch?v=CxndN_7Oqvo" target="_blank">Mass Points Geometry : Part III </a><br />
<a href="https://www.youtube.com/results?search_query=mass+points+" target="_blank"><br /></a>
<a href="https://www.youtube.com/results?search_query=mass+points+" target="_blank">other videos from Youtube on Mass Points</a><br />
<br />
It's much more important to fully understand how it works, the easier questions the weights align<br />
very nicely.<br />
<br />
The harder problems the weights are messier, not aligning nicely, so you need to find ways to may them integers (LCM) for easier solving.<br />
<br />
Let me know if you have questions. I love to help (:D) if you've tried.<br />
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<br />someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com1tag:blogger.com,1999:blog-2668492410190176921.post-77174500075449056142021-12-25T17:50:00.000-05:002021-12-25T17:50:59.124-05:00Face Diagonal and Space Diagonal of a Rectangular Prism<div style="color: #660000;">
<b>Face diagonal and space diagonal of a cube </b></div>
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Ways to calculate face and space diagonal.<br />
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Each side of the cube is x units long.<br />
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Use 45-45-90 degree angle ratio<br />
( 1 - 1 -<span style="font-size: larger; white-space: nowrap;">
√<span style="text-decoration: overline;"> 2 </span>
</span> ) or Pythagorean theorem to get the face diagonal.<br />
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Using Pythagorean theorem twice and you'll get the space diagonal.<br />
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<b><span style="color: #660000;">Face diagonal and Space diagonal of a rectangular prism. </span></b><br />
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<br />
Same way to figure out the face<br />
<br />
diagonal of a rectangle as well as<br />
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space diagonal of a rectangular prism. <br />
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Use Pythagorean theorem or<br />
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30-60-90 degree angle ratio<br />
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(1 -<span style="font-size: larger; white-space: nowrap;">√<span style="text-decoration: overline;"> 3 </span></span>- 2) to figure out the face<br />
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diagonal and Pythagorean theorem<br />
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twice to figure out the space diagonal. <br />
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</div>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com2tag:blogger.com,1999:blog-2668492410190176921.post-6309357599040801372019-12-02T18:48:00.000-05:002019-12-02T18:48:46.596-05:002020 Mathcounts State Prep: Simon's Favorite Factoring Trick Check out <a href="http://mathcounts.org/">Mathcounts</a> here, the best competition math program for middle school students. <br />
Download this year's <a href="http://mathcounts.org/handbook">Mathcounts handbook</a> here.<br />
<br />
The most common cases of Simon's Favorite Factoring Trick are:<br />
<br />
I: \(xy+x+y+1=\left( x+1\right) \left( y+1\right)\)<br />
<br />
II: \(xy-x-y+1=\left( x-1\right) \left( y-1\right)\) <br />
<br />
It's easy to learn. <a href="http://www.youtube.com/watch?v=0nN3H7w2LnI">Here is the best tutorial online, by none other than Richard Rusczyk.</a><br />
The method Rusczyk uses at the second half is very nifty. Thanks!!<br />
<br />
Questions to ponder:(answer key below)<br />
<b><span style="color: #274e13;">#1: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that
\(xy+x+y=13\) </span></b><br />
<b><span style="color: #274e13;">#2: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that
\(2xy+2x-3y=18\) </span></b><br />
<b><span style="color: #274e13;">#3: Find the length and the width of a rectangle whose area is equal to its perimeter.</span></b><br />
<b><span style="color: #274e13;">#4: Twice the area of a non-square rectangle equals triple it's perimeter, what is the area of the rectangle? </span></b><br />
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Answer key:<br />
#1: <span style="color: #660000;"> </span><b><span style="color: #274e13;"><span style="color: #660000;">x = 6 and y = 1</span></span></b><br />
#2: <b><span style="color: #660000;">( x, y ) = (4, 2) </span></b><br />
#3: <span style="color: #274e13;"><span style="color: black;">Don't forget square is a kind of rectangle (but not the other way around) so there are two answers: </span></span><br />
<span style="color: #660000;"><b>4 by 4 and 3 by 6 units. </b></span><br />
<span style="color: #274e13;"><span style="color: black;">#4: One side is 4 units and the other 12 units so the answer is 4 x 12 or <b><span style="color: #660000;">48 square units.</span></b> </span></span><br />
<span style="color: #274e13;"><span style="color: black;">There is another one, 6 by 6 that would fit if the question doesn't specify non-square rectangle. </span></span>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com1tag:blogger.com,1999:blog-2668492410190176921.post-11005583159642304002019-11-21T12:00:00.000-05:002019-11-22T08:49:22.640-05:002019 AMC 8 problems, solutions and some thoughts <a href="https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems" target="_blank"><b>2019 AMC 8 problems and solutions, for students, by students</b></a><br />
<br />
<span style="color: #660000;"><b>A student's reflection on this year's test : </b></span><br />
<br />
<b><span style="background-color: white; color: #222222; font-family: "arial" , "helvetica" , sans-serif; font-size: x-small;">Mrs. Lin,</span><br style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: small;" /><span style="background-color: white; font-family: "arial" , "helvetica" , sans-serif;"><span style="color: #222222; font-size: x-small;">I did the AMC 8 yesterday, and it was actually quite easier than last year. I was reviewing my answers, and I believe I only got the last two wrong. I used stars and bars for the last one, but did 21C2 instead of 20C2. I could’ve done number 24, because geometry is really my best subject in math. I realized that I should’ve used mass points later on. It’s fine, though, because it’s still a good score. I think that many people could solve this test pretty well because in many of the last questions you could plug in the multiple choice answers and get the right answer. Also, a lot of it was just plain algebra. Question 20 was actually just an equation, which anybody who knows basic algebra can solve.</span><span style="color: #660000;"> I thought that I would never say this, but I honestly wish that it was harder, because I was hoping for some interesting problems.</span><span style="color: #222222;"> <span style="font-size: x-small;">Those are the problems that get people’s gears turning; this year the problems were quite basic. I think many people will get really good scores on this test, which, along with a good thing, is also not so good because it brings down the credibility of the test.</span></span></span><br style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: small;" /><br style="background-color: white; color: #222222; font-family: Arial, Helvetica, sans-serif; font-size: small;" /><span style="background-color: white; color: #222222; font-family: "arial" , "helvetica" , sans-serif; font-size: x-small;">Thanks,</span></b><br />
<br />
<b>some links that you can review those very basic, but extremely useful strategies on this </b><br />
<b>year's seemingly harder, but not really last two questions. </b><br />
<b><br /></b>
<b><a href="http://mathcountsnotes.blogspot.com/2013/11/mass-points-geometry.html" target="_blank">mass points </a> learn together with <a href="http://mathcountsnotes.blogspot.com/2012/11/trianges-that-share-same-vertexsimilar.html" target="_blank">triangles sharing the same vertex </a></b><br />
<b><br /></b>
<a href="https://mathcountsnotes.blogspot.com/search?q=dimensional+change+" target="_blank"><b>dimensional change / scaling </b></a><br />
<b><br /></b>
<b><a href="https://artofproblemsolving.com/wiki/index.php/Ball-and-urn" target="_blank">balls and urns, stars and bars </a> (lots of variations or twists on this one, so </b><br />
<b>you need to fully understand the concept so to use it well. Be patient !!!!!) </b><br />
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<br />someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-38295841970383566842019-07-07T11:03:00.000-04:002023-03-04T11:04:15.278-05:002012 Harder Mathcounts State Target Questions<span style="color: #274e13;">Check out</span><b><span style="color: #274e13;"> </span></b><span style="color: #274e13;"><a href="http://mathcounts.org/" target="_blank">Mathcounts here</a></span><b><span style="color: #274e13;"> --</span></b><span style="color: #274e13;"><span style="color: black;"> the best competition math program for middle schoolers up to the </span></span><br />
<span style="color: #274e13;"><span style="color: black;">state and national level. </span></span><br />
<br />
<b><span style="color: #274e13;"># 6: A semicircle and a circle are placed inside a square with sides of length 4 cm, as shown. The circle is tangent to two adjacent sides of the square and to the semicircle. The diameter of the semicircle is a side of the square. In centimeters, what is the radius of the circle? Express your answer as a decimal</span><span style="color: #274e13;"> to the nearest hundredth.</span></b> <span style="color: #660000;">[2012 Mathcounts State Target #6]</span><br />
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjA_a-hEqBj4oczuwtmrRd5kLifqCK3JyBuCwz2A52cbKh9JDfKWcgE8hnaan5EmxsB5FcWKjfhrOMyEmddw-WTzIq1wnPoD4gkGqbWmuXimMr63qhdIUkZO63OvkTqDiuvj_Wh-VhHp4/s1600/2012+state+target+%23+6.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="480" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhjA_a-hEqBj4oczuwtmrRd5kLifqCK3JyBuCwz2A52cbKh9JDfKWcgE8hnaan5EmxsB5FcWKjfhrOMyEmddw-WTzIq1wnPoD4gkGqbWmuXimMr63qhdIUkZO63OvkTqDiuvj_Wh-VhHp4/s640/2012+state+target+%23+6.jpg" width="640" /></a><br />
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#6: <span style="color: blue;"> <b>Solution: </b></span><br />
Using Pythagorean theory: (2 + r)<sup>2</sup> = (4-r)<sup>2</sup> + ( 2- r)<sup>2</sup><br />
4 + 4r + r<sup>2</sup> = 16 - 8r + r<sup>2</sup> + 4 - 4r + r<sup>2</sup><br />
r<sup>2</sup> - 16 r + 16 = 0<br />
Using the quadratic formula
You have 8 ±<span style="font-size: larger; white-space: nowrap;">
4√<span style="text-decoration: overline;"> 3 </span></span><br />
Only 8 - 4<span style="font-size: larger; white-space: nowrap;">√<span style="text-decoration: overline;"> 3 </span></span>= <span style="color: #660000;"><b>1.07</b> </span>works<br />
<br />
<a href="http://mathcounts.org/resources/video-library/mathcounts-minis/mathcounts-mini-34-geometry-circles-and-right-triangles" target="_blank">There is a Mathcounts Mini #34 on the same question.</a> Check that out !! <br />
<br />
The above question looks very similar to this year's AMC-10 B #22, so try that one.<br />
(cover the answer choices so it's more like Mathcounts)<br />
<a href="https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_22" target="_blank"><br /></a>
<a href="https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_22" target="_blank">2014 AMC-10 B problem #22 </a><br />
<br />
<div class="MsoNormal" style="color: #274e13;">
<b>#8: In one roll of four standard, six-sided dice, what is the
probability of rolling exactly three different numbers? Express your answer as
a common fraction. </b><span style="color: #660000;"><b>[2012 Mathcounts State Target #8]</b></span><b></b></div>
<div class="MsoNormal" style="color: #274e13;">
<b><br /></b></div>
<div style="color: blue;">
<b>Solution I : Permutation method </b><br />
<span style="color: black;">If order matters, there are 6 * 5 * 4 * 1 ways to choose the number, 1 being the same number as one of the previous one. </span><br />
<span style="color: black;">Let's say if you choose 3 1 4 1.</span><br />
<span style="color: black;"></span><br />
<span style="color: black;">Now for the placement of those 4 numbers on the 4 different dice. There are 4C2 ways to place where the two "1" will </span><br />
<span style="color: black;">be positioned so the answer is :</span> <span style="color: black;">
\(\dfrac {6\times 5\times 4\times 1\times 4C2} {6^{4}}\)
= \(\dfrac{5}{9}\)</span></div>
<div class="postbody" style="max-width: 899px;">
</div>
<b><span style="color: blue;">Solution II: Combination method</span></b><br />
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<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">There
are 6C3 = 20 ways to choose the three numbers. </span><br />
<div class="MsoNormal" style="line-height: normal; margin-bottom: 0in;">
<br />
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">There
are 3 ways that the number can be repeated. [For example: If you choose 1, 2,
and 3, the fourth number could be 1, 2 or 3.]<br />
<br />
There are \(\dfrac {4!} {2!}\) =12</span>ways to arrange the chosen
4 numbers.[same method when you arrange AABC]</div>
So the answer is\(\dfrac{20* 3 *12}{6^4}\) = \(\dfrac{5}{9}\) someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com2tag:blogger.com,1999:blog-2668492410190176921.post-70138951286871087582019-03-11T18:28:00.000-04:002019-03-11T18:28:50.828-04:00Hints/links or Solutions to 2014 Harder Mathcounts State Sprint and Target question <div class="separator" style="clear: both; text-align: center;">
Links, notes, Hints or/and solutions to 2014 Mathcounts state harder problems.</div>
<div class="separator" style="clear: both; text-align: center;">
2014, 2015 Mathcounts state are harder </div>
<div style="margin: 0px; padding: 0px; vertical-align: middle;">
<br />
<b>Sprint round:</b><br />
<b><br /></b>
<b>#14 :</b><br />
<span style="color: #0b5394;"><b>Solution I :</b></span></div>
<div style="margin: 0px; padding: 0px; vertical-align: middle;">
(7 + 8 + 9) + (x + y + z) is divisible by 9, so the sum of the three variables could be 3, 12, or 21.<br />
789120 (sum of 3 for the last three digits) works for 8 but not for 7.<br />
21 is too big to distribute among x, y and z (all numbers are district),<br />
thus only x + y + z = 12 works and z is an even number<br />
__ __ 0 does't work (can't have 6 6 0 and the other pairs all have 7, 8 or 9)<br />
<span style="color: #660000;"><b>264</b></span> works (789264 is the number)<br />
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<span style="color: #0b5394;"><b>Solution II : </b></span><br />
789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...<br />
Try 504 * 1566 = 789264 (it works)<br />
The answer is <span style="color: #660000;"><b>264</b></span>.<br />
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<strong>#18:</strong> <br />
Watch <a href="http://mathcounts.org/resources/video-library/mathcounts-minis/mathcounts-mini-42-more-constructive-counting" target="_blank">this video from Mathcounts mini</a> and use the same method for the first question, <br />
you'll be able to get the answer. It's still tricky, though. <br />
<br />
<strong>#23 :</strong> Drop the heights of the two isosceles triangles and use similar triangles to get the length of FC. <br />
Then solve. <br />
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<b>#24: </b><br />
The key is to see 2<sup>10 </sup> is 1024 or about 10<sup>3</sup>
<br />
<br />
2<sup>30</sup> = ( 2<sup>10 </sup>)<sup>3 </sup> or about (10<sup>3 </sup>)<sup>3</sup>about 10<sup>9</sup> so the
answer is <span style="color: #990000;"><strong>10 digit</strong></span>. <br />
<br />
<strong>#25:</strong> <br />
As you can see, there are two Pythagorean Triples : 9-12-15 and 9-40-41. <br />
Base (40-12) = 28 gives you the smallest area. <br />
The answer is 28 * 18 = <span style="color: #990000;"><strong>504 </strong></span><br />
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<b>#26 :</b>
Let there be A, B, C three winners.
There are 4 cases to distribute the prizes. <br />
A B C<br />
1 1 5 There are 7C1 * 6C1 * \( \dfrac {3!} {2!}\) = 126 ways -- [you can skip the last part for C <br />
because it's 5C5 = 1]<br />
<br />
1 2 4 There are 7C1* 6C2 * 3! = 630<br />
<br />
1 3 3 There are 7C1 * 6C3 * \( \dfrac {3!} {2!}\) = 420<br />
<br />
2 2 3 There are 7C2 * 5C2 * 3 (same as above)<br />
<br />
Add them up and the answer is <span style="color: #990000;"><strong>1806.</strong></span> <br />
<br />
If you can't see why it's \( \dfrac {3!} {2!}\) when there is one repeat, try using easier case to help you understand. <br />
<br />
What about A, B two winners and 4 prizes ? <br />
There are 2 cases, 1 3 or 2 2, and you'll see how it's done.<br />
<br />
<strong>#27 :</strong> Read this and you'll be able to solve this question at ease, just be careful with the sign change. <br />
<a href="http://www.math.washington.edu/~mathcircle/circle/hw_2013_winter/VietasFormulas.pdf" target="_blank">Vieta's Formula and the Identity Theory</a> <br />
<br />
<strong>#28: </strong>There are various methods to solve this question.<br />
I use binomial expansion :<br />
\(11^{12}=\left( 13-2\right)^{12}=12C0*13^{12}\)+ \(12C1*13^{11}*2^{1}\)+...
\(12C11*13^{1}*2^{11}\)+ \(12C12*2^{12}\)
Most of the terms will be evenly divided by 13 except the last term, which is \(2^{12}\) or
4096, which, when divided by 13, leaves a remainder of 1. <br />
<br />
<strong><span style="color: #134f5c;">Solution II :</span></strong> <br />
\(11\equiv -2\left ( mod13\right)\) ; \((-2)^{12}\equiv 4096\equiv 1\left ( mod13\right)\) <br />
<br />
<strong><span style="color: #0b5394;">Solution III :</span></strong>
<br />
Or use Fermat's Little Theorem (Thanks, Spencer !!) <br />
\(11^{13-1}\equiv 11^{12}\equiv1 (mod 13)\)<br />
<br />
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<b>Target Round : </b><br />
<b><br /></b>
<b>#3:</b> Lune of Hippocrates : in seconds solved question.<br />
<b>^__^</b><br />
<br />
<b>#6:</b> This question is very similar to <a href="http://mathcounts.org/resources/video-library/mathcounts-minis/mathcounts-mini-26-probability-and-counting" target="_blank">this Mathcounts Mini.</a><br />
My students should get a virtual bump if they got this question wrong.<br />
<b><br /></b>
<b>#8: </b><span style="color: #0b5394;"><b>Solution I </b></span>: by TMM (Thanks a bunch !!)<br />
Using similar triangles and Pythagorean Theorem.<br />
<span class="cmty-highlight" style="background-color: white; box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span style="font-family: "times"; font-size: small; line-height: normal;"><br /></span></span></span></span></span>
<span style="background-color: white;"><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">The</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> height </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> cone, which can be found us</span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">in</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">g </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> Pythagorean </span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> is </span><span style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; white-space: nowrap;"><img alt="$\sqrt{10^2-5^2}=5\sqrt{3}$" class="latex" height="19" src="https://latex.artofproblemsolving.com/5/0/9/5094916befec088b7631b5e36d1957aa6e9de11b.png" style="border: 0px; box-sizing: border-box; max-height: 600px; max-width: 100%; vertical-align: -2px;" width="137" />.</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span></span><br />
<span style="background-color: white;"><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">Us</span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">in<span style="color: #444444;">g</span></span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">diagram</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> below, let </span><img alt="$r$" class="latex" height="8" src="https://latex.artofproblemsolving.com/b/5/5/b55ca7a0aa88ab7d58f4fc035317fdac39b17861.png" style="border: 0px; box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; max-height: 600px; max-width: 100%; vertical-align: 0px;" width="8" /><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span style="color: #444444;"><span style="font-size: 15px; line-height: 19px;">be the </span></span></span></span></span></span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">radius</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">top</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> cone and let </span><img alt="$h$" class="latex" height="12" src="https://latex.artofproblemsolving.com/8/1/8/8189a5b5a0917b8c93350827be4038af1839139d.png" style="border: 0px; box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; max-height: 600px; max-width: 100%; vertical-align: 0px;" width="10" /><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> be </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> height </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">top</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">cone. </span></span><br />
<span style="background-color: white;"><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">Let </span><img alt="$s=\sqrt{r^2+h^2}$" class="latex" height="18" src="https://latex.artofproblemsolving.com/f/9/d/f9debd9cfbf7003ad15edd1919b61ac37b4fd5c2.png" style="border: 0px; box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; max-height: 600px; max-width: 100%; vertical-align: -2px;" width="104" /><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> be </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> slant height </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">top</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> cone.</span><br style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;" /><br style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;" /><img alt="//cdn.artofproblemsolving.com/images/ad1f21b9f50ef27201faea84feca6f2e6e305786.png" class="bbcode_img" src="https://cdn.artofproblemsolving.com/images/ad1f21b9f50ef27201faea84feca6f2e6e305786.png" style="border: 0px; box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; max-height: 600px; max-width: 100%; vertical-align: middle;" /></span><br />
<span style="background-color: white;"><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span style="background-color: transparent;">Drawing the radius as shown in the diagram, we have two right triangles. Since the bases of the top cone and the original cone are parallel, the two right triangles are similar. So we have the proportion</span></span></span><img alt="\[\dfrac{r}{5}=\dfrac{s}{10}=\dfrac{\sqrt{r^2+h^2}}{10}.\]" class="latexcenter" height="41" src="https://latex.artofproblemsolving.com/2/0/5/205a8d45594c47d92393e07cffca02d3d3a5b415.png" style="border: 0px; box-sizing: content-box; color: #444444; display: block; font-family: sans-serif; font-size: 15px; line-height: 19px; margin: auto; max-height: 600px; max-width: 100%; padding: 1em 0px; vertical-align: middle;" width="160" /><span style="background-color: white;"><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">Cross multiply</span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">in</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">g yields </span></span><img alt="\[10r=5\sqrt{r^2+h^2}\implies 100r^2=25r^2+25h^2\implies 75r^2=25h^2\implies 3r^2=h^2\implies h=r\sqrt{3}.\]" class="latexcenter" height="19" src="https://latex.artofproblemsolving.com/c/3/4/c34a4033b4515ae88c64fda9f900d84924094aa6.png" style="border: 0px; box-sizing: content-box; color: #444444; display: block; font-family: sans-serif; font-size: 15px; line-height: 19px; margin: auto; max-height: 600px; max-width: 100%; padding: 1em 0px; vertical-align: middle;" width="713" /><span style="background-color: white;"><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">This is what </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">we</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> need.</span><br style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;" /><br style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;" /><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">Next, </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> volume </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> orig</span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">in</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">al cone is simply </span><span style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; white-space: nowrap;"><img alt="$\dfrac{\pi\times 25\times 5\sqrt{3}}{3}=\dfrac{125\sqrt{3}}{3}$" class="latex" height="41" src="https://latex.artofproblemsolving.com/6/8/f/68f3a3a8626d7d71bee8898de9db9f7dd3ea353e.png" style="border: 0px; box-sizing: border-box; max-height: 600px; max-width: 100%; vertical-align: -12px;" width="189" />.</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span></span><br />
<span class="cmty-highlight" style="background-color: white; box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">The</span></span></span></span><span style="background-color: white; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> volume </span><span class="cmty-highlight" style="background-color: white; box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="background-color: white; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="background-color: white; box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="background-color: white; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="background-color: white; box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">top</span><span style="background-color: white; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> cone is </span><span style="background-color: white; box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; white-space: nowrap;"><img alt="$\dfrac{\pi\times r^2h}{3}$" class="latex" height="39" src="https://latex.artofproblemsolving.com/0/e/2/0e2e8a88b364f565cd79b16ce27212ad5f245fe3.png" style="border: 0px; box-sizing: border-box; max-height: 600px; max-width: 100%; vertical-align: -12px;" width="62" />.</span><br />
<span style="background-color: white;"><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">From </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> given </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">in</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">formation, </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">we</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> know that </span></span><img alt="\[\dfrac{125\sqrt{3}}{3}-\dfrac{\pi\times r^2h}{3}=\dfrac{125\sqrt{3}}{9}\implies 125\sqrt{3}-r^2h=\dfrac{125\sqrt{3}}{3}\implies r^2h=\dfrac{250\sqrt{3}}{3}.\]" class="latexcenter" height="40" src="https://latex.artofproblemsolving.com/d/2/e/d2eb3f35bb46a600fd5db854308e1f7844964947.png" style="border: 0px; box-sizing: content-box; color: #444444; display: block; font-family: sans-serif; font-size: 15px; line-height: 19px; margin: auto; max-height: 600px; max-width: 100%; padding: 1em 0px; vertical-align: middle;" width="592" /><span style="background-color: white;"><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">We</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> simply substitute </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> value </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><img alt="$h=r\sqrt{3}$" class="latex" height="18" src="https://latex.artofproblemsolving.com/5/2/e/52e93e7c0a83de8727aa2c1631e534f0184b2760.png" style="border: 0px; box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; max-height: 600px; max-width: 100%; vertical-align: -1px;" width="68" /><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> from above to yield </span></span><img alt="\[r^3\sqrt{3}=\dfrac{250\sqrt{3}}{3}\implies r=\sqrt[3]{\frac{250}{3}}.\]" class="latexcenter" height="43" src="https://latex.artofproblemsolving.com/a/d/9/ad9d39850b5a5aa905a1e879a6b5bfebc1804496.png" style="border: 0px; box-sizing: content-box; color: #444444; display: block; font-family: sans-serif; font-size: 15px; line-height: 19px; margin: auto; max-height: 600px; max-width: 100%; padding: 1em 0px; vertical-align: middle;" width="255" /><span style="background-color: white;"><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">We</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> will leave it </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">as</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> is for now so </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> decimals don't get messy.</span><br style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;" /><br style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;" /><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">We</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> get </span><img alt="$h=r\sqrt{3}\approx 7.56543$" class="latex" height="18" src="https://latex.artofproblemsolving.com/2/3/f/23f06646eb4253302beccfec51d656184f53895f.png" style="border: 0px; box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; max-height: 600px; max-width: 100%; vertical-align: -1px;" width="150" /><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> and </span><span style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; white-space: nowrap;"><img alt="$s=\sqrt{r^2+h^2}\approx 8.7358$" class="latex" height="18" src="https://latex.artofproblemsolving.com/3/3/7/33722f797d44f3f1f317e43e3e4d2436af89b132.png" style="border: 0px; box-sizing: border-box; max-height: 600px; max-width: 100%; vertical-align: -2px;" width="178" />.</span></span><br />
<span style="background-color: white;"><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span style="font-family: "times"; font-size: small; line-height: normal;"><br /></span></span></span></span></span></span>
<span style="background-color: white;"><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span style="font-family: "times"; font-size: small; line-height: normal;">The </span><span style="font-family: "times"; font-weight: bold; line-height: normal; margin: 0px; padding: 0px;">lateral</span><span style="font-family: "times"; font-size: small; line-height: normal;"> surface area of the frustum is equal to the </span><span style="font-family: "times"; font-weight: bold; line-height: normal; margin: 0px; padding: 0px;">lateral</span><span style="font-family: "times"; font-size: small; line-height: normal;"> surface area of the original cone minus the lateral surface area of the top cone. The surface area of the original cone is simply </span></span></span></span></span></span><br />
<span style="background-color: white;"><span style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; white-space: nowrap;"><img alt="$5\times 10\times \pi=50\pi$" class="latex" height="19" src="https://latex.artofproblemsolving.com/e/c/0/ec091d6dc38532a76c9e07e627045a4aace18349.png" style="border: 0px; box-sizing: border-box; max-height: 600px; max-width: 100%; vertical-align: 0px;" width="200" />.</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span></span><br />
<span style="background-color: white;"><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">The</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> surface area </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">top</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> cone is </span><span style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; white-space: nowrap;"><img alt="$\pi\times r\times s\approx 119.874$" class="latex" height="13" src="https://latex.artofproblemsolving.com/f/4/1/f41e76261a72913875e72613e28beeb35d933a7d.png" style="border: 0px; box-sizing: border-box; max-height: 600px; max-width: 100%; vertical-align: 0px;" width="156" />.</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span></span><br />
<span style="background-color: white;"><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">So our </span><b style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">lateral</b><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> surface area is </span></span><br />
<span style="color: #444444; font-family: sans-serif;"><span style="background-color: white; font-size: 15px; line-height: 19px;"><br /></span></span><span style="background-color: white;"><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">All </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">we</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">have</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> left is to add </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">two</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> b</span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">as</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">es. </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">The</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> total area </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">of</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> </span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;"><span class="cmty-highlight" style="box-sizing: border-box; padding: 0px 3px;">the</span></span></span></span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">b</span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">as</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">es is </span><span style="box-sizing: border-box; color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px; white-space: nowrap;"><img alt="$25\pi+\pi\cdot r^2\approx 138.477$" class="latex" height="18" src="https://latex.artofproblemsolving.com/1/0/a/10a0a10c8e7ab060fa27ca75ab45f31f374908e2.png" style="border: 0px; box-sizing: border-box; max-height: 600px; max-width: 100%; vertical-align: -1px;" width="200" />.</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;"> So our f</span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">in</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">al ans</span><span class="cmty-highlight" style="box-sizing: border-box; font-family: sans-serif; font-size: 15px; line-height: 19px; padding: 0px 3px;">we</span><span style="color: #444444; font-family: sans-serif; font-size: 15px; line-height: 19px;">r is </span></span><img alt="\[37.207+138.477=175.684\approx\boxed{176}.\]" class="latexcenter" height="23" src="https://latex.artofproblemsolving.com/9/b/9/9b9d7bd69023b24516b90a9c51dbce0f9fffbe08.png" style="border: 0px; box-sizing: content-box; color: #444444; display: block; font-family: sans-serif; font-size: 15px; line-height: 19px; margin: auto; max-height: 600px; max-width: 100%; padding: 1em 0px; vertical-align: middle;" width="282" /></div>
<b><span style="color: #0b5394; font-family: inherit; line-height: 22px;">Solution II</span><span style="font-family: inherit; line-height: 22px;"> </span><span style="font-family: inherit; line-height: 22px;">: </span></b><br />
<span class="postbody" style="font-family: inherit; line-height: 22px; margin: 0px; overflow: hidden; padding: 0px; vertical-align: baseline;">Using dimensional change and ratio, proportion.</span><br />
<span style="font-family: inherit;"><br /></span>
<span style="font-family: inherit;">Cut the cone and observe the shape.</span><br />
<span style="font-family: inherit;"><br />The circumference of the larger circle is 20pi (10 is the radius) and the base of</span><br />
<span style="font-family: inherit;">the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)</span><br />
<span style="font-family: inherit;"><br />To find the part that is the area of the frustum not including the top and bottom circles,</span><br />
<span style="font-family: inherit;">you use the area of the half circle minus the area of the smaller half circle.</span><br />
<span style="font-family: inherit;"><br />Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the</span><br />
<span style="font-family: inherit;">two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).</span><br />
<span style="font-family: inherit;"><br />Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).</span><br />
<span style="font-family: inherit;"><br />Now we can solve this :</span><br />
<span class="postbody" style="font-family: inherit; line-height: 22px; margin: 0px; overflow: hidden; padding: 0px; vertical-align: baseline;"><span style="font-family: inherit;"> \(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about</span><span style="font-family: inherit;"> </span><span style="color: #990000; font-family: inherit;">176 (</span><span style="font-family: inherit;">after you round up)</span>ional change and ratio, proportion.</span><br />
<span style="font-family: inherit;"><br /></span>
<span style="font-family: inherit;">Cut the cone and observe the shape. </span><br />
<span style="font-family: inherit;"><br />
The circumference of the larger circle is 20pi (10 is the radius) and the base of </span><br />
<span style="font-family: inherit;">the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)</span><br />
<span style="font-family: inherit;"><br />
To find the part that is the area of the frustum not including the top and bottom circles, </span><br />
<span style="font-family: inherit;">you use the area of the half circle minus the area of the smaller half circle. </span><br />
<span style="font-family: inherit;"><br />
Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the</span><br />
<span style="font-family: inherit;">two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).</span><br />
<span style="font-family: inherit;"><br />
Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).</span><br />
<span style="font-family: inherit;"><br />
Now we can solve this : </span><br />
<span style="font-family: inherit;"> \(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about <span style="color: #990000;">176 (</span>after you round up)</span><br />
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<span style="color: #0b5394; text-align: center;"><b>Solution III</b></span><span style="text-align: center;"> : Another way to find the surface area of the Frustum is : </span><br />
<span style="text-align: center;">median of the two half circle [same as median of the two bases] * the height [difference of the two radius]</span><br />
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)
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someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-5183277992274721042019-01-05T08:32:00.000-05:002019-01-05T08:32:44.951-05:00Notes to 2018 Mathcounts chapter more interesting problems <b><span style="color: #660000;">2018 Mathcounts Chapter Spring problems : solutions down below </span></b><br />
<b><span style="color: #660000;">Thanks to a boy mathlete who tried these problems and e-mail me for feedback. </span></b><br />
<b><span style="color: #660000;"><br /></span></b>
<b><span style="color: blue;">Please try these problems first before reading the explanations. :D</span></b><br />
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<b>#25 : Three employees split a bonus valued at some number of dollars. Arman first
receives $10 more than one third of the total amount. Bernardo then receives $3
more than one half of what was left. Carson receives the remaining $25. What is
the total dollar value of the bonus?</b><br />
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<b>#27: For a particular list of four distinct integers the mean, median and range have
the same value. If the least integer in the list is 10, what is the greatest value for
an integer in the list?</b><br />
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<b>#29: There are two values of x such that \( |\dfrac {x-2018} {x-2019}|=\dfrac {1} {6}\). . What is the absolute
difference between these two values of x? Express your answer as a common
fraction.</b><br />
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<b><span style="color: #660000;">Target #8</span> : Four congruent circles of radius 2 cm intersect with their centers at intersection
points as shown. What is the area of the shaded region? Express your answer in
terms of π.</b><br />
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<b>#25 : It's easier if you go backward and use inverse operations to solve this question. </b><br />
<b>(25 + 3) *2 = 56 and 56 + 10 = 66, which is </b><b>\( \dfrac {2} {3}\) of the original bonus value, or </b><br />
<b>what is left after </b><b>\( \dfrac {1} {3}\) was given out. </b><br />
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<b>\( \dfrac {2} {3}\) of bonus is 66 dollars, so the answer is 99. </b><br />
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<b>#27: Let the average be x and the three other numbers be a, b, c and </b><b>\( a < b < c \).</b><br />
<b>The least number is 10 (given), so </b><br />
<b>\( 10+a+b+c \)</b><b> = </b><b>\( 4x \)</b><b>---> <span style="color: #660000;">equation 1 </span></b><br />
<b>\( \dfrac {a+b} {2}\) = </b><b>\( x \) </b><b>(how to find the median), so </b><b>\( a+b\)</b><b> = </b><b>\( 2x \)</b><b> <span style="color: #660000;">--- (2)</span></b><br />
<b>\(c-10 = x\) (given because it's the range), \(c= x + 10\) <span style="color: #660000;">---(3)</span></b><br />
<b>Substitute (2) and (3) to equation one and you have \(10 + 2x + x + 10 = 4x\), so \(x = 20 \)</b><br />
<b>\(C = 20 + 10 = 30\), the answer </b><br />
<b><br /></b>
<b>#28 : Let \(x - 2018 = y\) , then \(x - 2019 = y -1\)</b><br />
<b>We then have either </b><br />
<b>\( |\dfrac {y} {y -1}|=\dfrac {1} {6}\) \(\rightarrow\) \(y\) = </b><b>\( \dfrac {-1} {5}\)</b><br />
<b><br /></b>
<b>or </b><b>\( |\dfrac {y} {y-1}|=\dfrac {-1} {6}\) </b><b>\(\rightarrow\) \(y\) = </b><b>\( \dfrac {1} {7}\)</b><br />
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<b>Their positive difference is </b><b>\( \dfrac {12} {35}\)</b><b> </b><b>, the answer. </b><br />
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<b>Target #8 : Thanks to a 5th grader girl mathlete's solution: </b><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEid8elBL7ma2Pr23Hl5VSGlnMcIoeByrFNCK1_e7ijhEPIZPI5X94GqH4IBRWaglFvQvcGQLIh_GlX-5emNycTE31LEly87RSIy0CnQ5AEowG8dBA4ifrbfREipASKS2sdbCjhimUik-bI/s1600/Circle+Problem+Colored++from+R..png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="691" data-original-width="1228" height="225" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEid8elBL7ma2Pr23Hl5VSGlnMcIoeByrFNCK1_e7ijhEPIZPI5X94GqH4IBRWaglFvQvcGQLIh_GlX-5emNycTE31LEly87RSIy0CnQ5AEowG8dBA4ifrbfREipASKS2sdbCjhimUik-bI/s400/Circle+Problem+Colored++from+R..png" width="400" /></a></div>
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<b>If you move parts around, you'll see the answer is exactly a semi-circle with a radius 2, so the answer is \(2\pi\), the answer. </b><br />
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<b>or check out another solution from me: </b><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJ9Y69K4hEJGf9afvluWSmHli9G8zKXh6LwPxSKr6LfdmTRvq6YFgJ4qO8YbkfI8R0E02aZA5JkhCcCvRrRjlRvWlKPXBVlQ47aY74Hb1pW_Lbjw9Ouy2PqvXKUIjH_-qTcizI9AHxrHw/s1600/4+congruent+circles+from+Mrs.+Lin.PNG" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="446" data-original-width="550" height="161" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJ9Y69K4hEJGf9afvluWSmHli9G8zKXh6LwPxSKr6LfdmTRvq6YFgJ4qO8YbkfI8R0E02aZA5JkhCcCvRrRjlRvWlKPXBVlQ47aY74Hb1pW_Lbjw9Ouy2PqvXKUIjH_-qTcizI9AHxrHw/s200/4+congruent+circles+from+Mrs.+Lin.PNG" width="200" /></a></div>
<b><br /></b>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-77234722164112109352018-10-10T13:31:00.001-04:002018-10-10T13:31:45.190-04:00Geometric mean of a right triangle Geometric mean of a right triangle :<br />
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<a href="https://www.youtube.com/watch?v=k8ss6RC752A&t=166s" target="_blank">Video tutorial </a><br />
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<a href="https://mathbitsnotebook.com/Geometry/RightTriangles/RTmeanRight.html" target="_blank">Notes</a><br />
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<a href="https://mathbitsnotebook.com/Geometry/RightTriangles/RTMeanPractice.html" target="_blank">Practice I </a><br />
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<a href="http://www.jmap.org/Worksheets/G.SRT.B.5.Similarity4.pdf" target="_blank">Practice II : focus on the height (or altitude)</a><br />
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<a href="http://www.jmap.org/Worksheets/G.SRT.B.5.Similarity5.pdf" target="_blank">Practice III : focus on the legs </a><br />
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<br />someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-83576301197696538022018-10-01T09:57:00.000-04:002018-10-01T10:01:42.448-04:00The Largest Rectangle Inscribed in Any TriangleFrom Mathcounts Mini : <a href="http://www.mathcounts.org/resources/video-library/mathcounts-minis/mini-45-maximum-area-inscribed-rectangles-triangles" target="_blank">Maximum area of inscribed rectangles and triangles</a>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhA_zN11j0bRsMleq7WVP5-iBvIghWVg0FkC-u9LfKCKyAtHS3Tm7dDQGFy81P0LPVuTavl07HM7YKjK_hOkjYwRzJnBqf2eTuhrv4XeeXfwV9rmGizvecpvnjOwyqXLfz5tO0ymIoh-pQ/s1600/proof+largest+rectangle+in+any+triangle.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="265" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhA_zN11j0bRsMleq7WVP5-iBvIghWVg0FkC-u9LfKCKyAtHS3Tm7dDQGFy81P0LPVuTavl07HM7YKjK_hOkjYwRzJnBqf2eTuhrv4XeeXfwV9rmGizvecpvnjOwyqXLfz5tO0ymIoh-pQ/s640/proof+largest+rectangle+in+any+triangle.jpg" width="640" /></a></div>
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\(\Delta EHI\sim\Delta EFG\) \(\rightarrow\) \(\dfrac {a} {c}=\dfrac {d-b} {d}\)\(\rightarrow\) \(a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}\)<br />
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We are going to find out what the largest area of a rectangle is with the side length a and b.<br />
It can be shown that by substituting the side length "a" with the previous equation + completing the square that
the largest area is half of the area of the triangle the rectangle is embedded.<br />
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\(a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}\).<br />
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From there, you know that when \(b= \dfrac {1} {2}d\), it will give you the largest area, which is \(\dfrac {1} {4}dc\).<br />
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\(a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c\).<br />
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Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.
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<a href="https://www.mathcounts.org/resources/video-library/mathcounts-minis/mini-45-maximum-area-inscribed-rectangles-triangles" target="_blank">Proof without words from Mr. Rusczyk </a><br />
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Try using different types of triangles to experiment and see for yourself.<br />
Paper folding is fun !!!!!<br />
It's very cool :Dsomeone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com2tag:blogger.com,1999:blog-2668492410190176921.post-70054055792103200182018-09-18T16:15:00.000-04:002018-09-18T16:15:50.150-04:00Dimensional Change questions I:Questions written by Willie, a volunteer. Answer key and detailed solutions below.<br />
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1a. There is a regular cylinder, which has a height equal to
its radius. If the radius and height are both increased by 50%, by what % does
the total volume of the cylinder increase?</div>
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1b. If the radius and height are both decreased by 10%, by
what % does the total volume of the cylinder decrease?</div>
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1c. If the radius is increased by 20% and the height is
decreased by 40%, what % of the volume of the original cylinder does the volume
of the new cylinder represent?</div>
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1d. If the radius is increased by 40% and the height is
decreased by 20%, what % of the volume of the original cylinder does the volume
of the new cylinder represent?</div>
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1e. If the height is increased by 125%, what % does the
radius need to be decreased by for the volume to remain the same?</div>
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2. If the side of a cube is increased by 50%, by what % does
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3a. If the volume of a cube increases by 72.8%, by what %
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3b. By what % did the side length of the cube increase?</div>
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4. You have a collection of cylinders, all having a radius
of 5. The first cylinder has a height of 2, the second has a height of 4, the
third a height of 6, etc. The last cylinder has a height of 50. What is the sum
of the volumes of all the cylinders (express your answer in terms of pi)?<br />
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Answer key: (Each question should not take you more than 30 seconds to solve if you really understand the concepts involved.)</div>
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1a. The volume of a cylinder is πr<sup>2</sup>x h (height). The radius itself will be squared and the height stays at
constant ratio. The volume will increased thus (1.5)<sup>3</sup> - 1<sup>3</sup> -- the original 100% of the volume = 2.375 </div>
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<b>=237.5%</b></div>
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<span style="color: #274e13;">1b. Like the previous question: 1</span><sup style="color: #274e13;">3</sup><span style="color: #274e13;"> - 0.9</span><sup style="color: #274e13;">3</sup><span style="color: #274e13;"> [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 = <b><span style="color: #990000;">27.1% decrease</span></b></span><br />
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1c. 1.2<sup>2</sup> [100% + 20% increase = 1.2] x 0.6 [100% -40% = 0.6] = 0.864 or <b><span style="color: #990000;"> </span></b></div>
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<b><span style="color: #990000;">86.4% of the original volume</span></b></div>
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<span style="color: #274e13;">1d. 1.4</span><sup style="color: #274e13;">2</sup><span style="color: #274e13;"> [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = </span><b><span style="color: #990000;">156.8% of the original volume</span><span style="color: #c00000;"> </span></b><br />
<span style="color: #c00000;"><br /></span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="color: #c00000;"> </span><br />
<div style="color: #274e13;">
1e. When the height of a cylinder is increased 125%, the total volume is is 225% of the original cylinder, or 9/4.</div>
<div style="color: #274e13;">
Since the radius is used two times (or squared), it has to decrease 4/9<sup>1/2</sup> = 2/3 for the new cylinder to have the same volume as the old one. [9/4 times 4/9 = 1 or the original volume.]</div>
<span style="color: #274e13;">1 - (2/3) = 1/3<b> =</b></span><b> </b><b style="color: #274e13;">0.<span style="text-decoration: overline;">3</span></b><b><span style="color: #660000;"><span style="color: #274e13;"> </span>= <span style="color: #990000;">33</span>.</span></b><b><span style="color: #990000; text-decoration: overline;">3</span><span style="color: #990000;">%</span></b><br />
<br />
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="color: #274e13;">2. Surface area is 2-D so 1.5</span><sup style="color: #274e13;">2</sup><span style="color: #274e13;"> - 1 = 1.25 =</span> <b><span style="color: #990000;">125% increase</span></b><br />
<br />
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="color: #c00000;"> </span></div>
<div style="color: #274e13;">
3a. If a volume of a cube is increased by 72.8 percent, it's 172.8% or 1.728 of the original volume. Now you are going from 3-D (volume) to 2-D (surface area). 1.728<sup>2/3</sup> = 1.44 or <b><span style="color: #990000;">44% increase</span></b>. [Don't forget to minus 1 (the original volume) since it is asking you the percentage increase.]</div>
<br />
<div style="color: #274e13;">
3b. From surface area, you can get the side increase by using 1.44<sup>1/2</sup> = 1.2, so 20% increase.</div>
<span style="color: #274e13;">Or you can also use 1.728</span><sup style="color: #274e13;">1/3</sup><span style="color: #274e13;"> = 1.2; 1.2 - 1 =</span> <b><span style="color: #990000;">20%</span></b><br />
<br />
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="color: #274e13;">4. The volume of a cylinder is πr</span><sup style="color: #274e13;">2</sup><span style="color: #274e13;">x h . (2 + 4 + 6 + ...50) x 5</span><sup style="color: #274e13;">2</sup><span style="color: #274e13;">π = (25 x 26) x 25π =</span><b><span style="color: #990000;">16250π</span></b></div>
<br /></div>
</div>
</div>
</div>
someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-4301126442209407542018-09-01T09:57:00.000-04:002018-10-03T10:43:07.813-04:002011 Mathcounts Chapter Sprint Round solutions<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhV1WhN7GmZSSsHLG4NO8Xg0TrSVh9CnRWiWC6xaivwHcFtY1Y9bF8LQ-xdJ2Ab0aNvMYFvTaOelS2dTwjcb4DX-m_goF33t4TIapkHkolxLWax3sm3GhOZGRLri7EsInwCIBVfeXVX38c/s1600/2011+chapter+sprint+%252322.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="274" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhV1WhN7GmZSSsHLG4NO8Xg0TrSVh9CnRWiWC6xaivwHcFtY1Y9bF8LQ-xdJ2Ab0aNvMYFvTaOelS2dTwjcb4DX-m_goF33t4TIapkHkolxLWax3sm3GhOZGRLri7EsInwCIBVfeXVX38c/s320/2011+chapter+sprint+%252322.jpg" width="320" /></a><b>#22: The answer is <span style="color: #990000;">2674</span>.</b><br />
<b> See left for explanations.</b><br />
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<b>#23: Let the two consecutive positive integers be x and x + 1.</b><br />
<b>( x + 1 ) / x = 1.02, x + 1 = 1.02x, 0.02x = 1, x = 1 divided by 0.02 = 1 times 100/2 = 50</b><br />
<b>The two numbers are 50 and 51 and their sum is 50 + 51 =<span style="color: #990000;"> 101</span>.</b><br />
<b><br /></b>
<b>#24: The two x-intercepts when y is "0" are 10 or -10; the two y-intercepts when x is "0" are 5 or -5.</b><br />
<b>Area of a rhombus is D1 x D2 / 2 so the answer is [10-(-10)] x [5 -(-5)] = <span style="color: #990000;">100 square units</span>.</b><br />
<b><br /></b>
<b>#25: The area ratio of the two similar triangle is 150/6 so the line ratio is<span style="white-space: nowrap;"> √<span style="text-decoration: overline;"> 150/6 </span> or 5:1.<br />
the length of the hypotenuse of the smaller triangle is 5 inches, so the other two legs are 3 and 4. </span></b><br />
<span style="white-space: nowrap;"><b>(a Pythagorean triple)<br />
The sum of the lengths of the legs of the larger triangle is (3 + 4) * 5 = <span style="color: #990000;">35</span>.</b></span><br />
<span style="white-space: nowrap;"><b><br />
#26: To have same number of boys and girls, the committee needs to consist of 3 boys and 3 girls. </b></span><br />
<span style="white-space: nowrap;"><b>(6C3 x 4C3)/ 10C6 = 80/210 = <span style="color: #990000;">8/21</span></b></span><br />
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#27: When the point (3, 4) is reflected over the x-axis to B, B would = (3, -4). </b></span><br />
<span style="white-space: nowrap;"><b>When B is reflected over the line y = x to C, C would = (-4, 3).</b></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3c0cnT1bN6ZO-YGV9q5YgtXibRaGQf8n6J0r_EHTByGMmIhc5A0aOKmIdO4abOwfaCr5m1yv0Z3e9pjKOr6OflGcLLlhn7goJ5g4MY-K89eizxlMneNawlU2-oPgXuqtOec1ZFOZWvjs/s1600/2011+Mathcounts+sprint+%2523+27.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><b><img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh3c0cnT1bN6ZO-YGV9q5YgtXibRaGQf8n6J0r_EHTByGMmIhc5A0aOKmIdO4abOwfaCr5m1yv0Z3e9pjKOr6OflGcLLlhn7goJ5g4MY-K89eizxlMneNawlU2-oPgXuqtOec1ZFOZWvjs/s320/2011+Mathcounts+sprint+%2523+27.jpg" width="312" /></b></a></div>
<span style="white-space: nowrap;"><b><br />
The area of the triangle is [4 -(-4)] x [ 3 - (-4)]/ 2 =<span style="color: #990000;"> 28 square units</span>. </b></span><br />
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<span style="white-space: nowrap;"><b>#28: Tonisha is 45 miles ahead Sheila when Sheila leaves Maryville at 8: 15 a.m. </b></span><br />
<b><span style="white-space: nowrap;">Each hour Sheila </span><span style="white-space: nowrap;">will be 15 miles closer to Tonisha. 45/15 = 3, which means that 3 hours </span></b><br />
<b><span style="white-space: nowrap;">later Sheila </span></b><b style="white-space: nowrap;">will pass Tonisha. </b><br />
<span style="white-space: nowrap;"><b>8:15 + 3 hours = <span style="color: #990000;">11: 15 a.m</span>.</b></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6r0JKxxDTpqg5exeIJOIYBxfKmFGjJYO9VI_qQPQ1nwXmgpADJTIiZd0kqGm1FfT0BnauvuMkyghluwCi3OIaxXsBkrQa_XmPVo8udq3pRkowJN2Usd6gyr7uiX0SLkGUNOZQ3VlfcgQ/s1600/%252329+2011+Mathcounts+chapter+sprint.jpg" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><b><img border="0" height="253" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6r0JKxxDTpqg5exeIJOIYBxfKmFGjJYO9VI_qQPQ1nwXmgpADJTIiZd0kqGm1FfT0BnauvuMkyghluwCi3OIaxXsBkrQa_XmPVo8udq3pRkowJN2Usd6gyr7uiX0SLkGUNOZQ3VlfcgQ/s320/%252329+2011+Mathcounts+chapter+sprint.jpg" width="320" /></b></a><b><span style="white-space: nowrap;"> </span>#29: Using 30-60-90 degree angle ratio, you can </b><b>make the radius be <span style="font-size: larger; white-space: nowrap;">√<span style="text-decoration-line: overline;"> 3 </span></span>and half of the side of the hexagon would be 1 so each side of the hexagon</b><br />
<b>is 2.</b><br />
<b><br /></b>
<b>The area of the hexagon is <span style="font-size: larger; white-space: nowrap;">(√<span style="text-decoration: overline;"> 3</span></span>/4) times 2<sup>2</sup> times 6 = 6<span style="font-size: larger; white-space: nowrap;">√<span style="text-decoration: overline;"> 3.</span></span></b><br />
<b>The area of the circle is 3Π.</b><br />
<b>The fraction is 3Π/6<span style="font-size: larger; white-space: nowrap;">√<span style="text-decoration: overline;"> 3</span></span> = <span style="font-size: larger; white-space: nowrap;">√<span style="text-decoration: overline;"> 3</span></span> / 6</b><br />
<b>a = 3 and b = 6, ab = <span style="color: #990000;">18</span></b><br />
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<b>#30: Area of triangle KDC is easy to find once you realize the height is just the right triangle with a hypotenuse 6 and a leg 4. (half of the length of CD),</b><br />
<b><br /></b>
<b>Using Pythagorean theorem, you get the height to be 2<span style="font-size: larger; white-space: nowrap;">√<span style="text-decoration: overline;"> 5 </span>.<br />
<br />
8 x </span>2<span style="font-size: larger; white-space: nowrap;">√<span style="text-decoration: overline;"> 5 </span></span>/2 =<span style="color: #990000;"> 8 </span><span style="color: #990000; font-size: larger; white-space: nowrap;">√<span style="text-decoration: overline;"> 5 </span></span>.</b>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-87237705512311439962018-08-01T21:44:00.000-04:002018-08-10T12:59:07.435-04:00Show Your Work, Or, How My Math Abilities Started to Decline<a href="http://everything2.com/index.pl?node_id=903457"><span style="color: #6666cc;">Show your work, or, how my math abilities started to decline</span></a><br />
<br />
I
think it's problematic the way schools teach Algebra. Those meaningless
show-your-work approaches, without knowing what Algebra is truly about.
The overuse of calculators and the piecemeal way of teaching without
the unification of the math concepts are detrimental to our children's
ability to think critically and logically.<br />
<br />
Of course
eventually, it would be beneficial to students if they show their work
with the much more challenging word problems (harder Mathcounts state
team round, counting and probability questions, etc...), but it's totally different from what some schools ask of our capable students.<br />
<br />
How
do you improve problem solving skills with tons of worksheets by going
through 50 to 100 problems all look very much the same? It's called busy
work. <br />
<br />
Quote from <span style="color: #3333ff;">Einstein</span>. <span style="color: #006600;">"Insanity: doing the same thing over and over again and expecting different results."</span><br />
<br />
Quotes from <span style="color: #3333ff;">Richard Feynman</span>, the famous late Nobel-laureate physicist. <span style="color: #990000;">Feynman relates his cousin's unhappy experience with algebra: </span><br />
<br />
<span style="color: #006600;">My
cousin at that time—who was three years older—was in high school and
was having considerable difficulty with his algebra. I was allowed to
sit in the corner while the tutor tried to teach my cousin algebra. I
said to my cousin then, "What are you trying to do?" I hear him talking
about x, you know."Well, you know, 2x + 7 is equal to 15," he said, "and
I'm trying to figure out what x is," and I say, "You mean 4." He says,
"Yeah, but you did it by arithmetic. You have to do it by algebra."And
that's why my cousin was never able to do algebra, because he didn't
understand how he was supposed to do it.<span style="color: #990000;"> I learned
algebra, fortunately, by—not going to school—by knowing the whole idea
was to find out what x was and it didn't make any difference how you did
it.</span> There's no such a thing as, you know, do it by arithmetic, or you do it by algebra. It was a false thing that they had invented in
school, so that the children who have to study algebra can all pass it.
They had invented a set of rules, which if you followed them without
thinking, could produce the answer. Subtract 7 from both sides. If you
have a multiplier, divide both sides by the multiplier. And so on. A
series of steps by which you could get the answer if you didn't
understand what you were trying to do.<br />So I was lucky. </span><span style="color: #990000;">I always learnt things by myself.</span>someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-83665470186578025112018-04-15T09:13:00.000-04:002018-04-15T09:13:03.583-04:00Original Problem from a Student Problem Solver on Similar TrianglesAn original problem on similar triangles from Varun (from FL)<br />
<br />
<iframe class="scribd_iframe_embed" data-aspect-ratio="0.772922022279349" data-auto-height="false" frameborder="0" height="600" id="doc_16130" scrolling="no" src="https://www.scribd.com/embeds/140789100/content?start_page=1&view_mode=scroll&access_key=key-1dg9iwbmt6n152yin1q" width="100%"></iframe>
someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-28112060893386295522018-04-15T09:09:00.000-04:002018-04-15T09:09:50.231-04:00The Grid Technique in Solving Harder Mathcounts Counting Problems : from Vinjai<!--[if !mso]>
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<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">The following notes are from Vinjai, a student I met online. He graciously shares and offers the tips here on how to tackle those harder Mathcounts counting problems. </span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";"><br /></span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">The
point of the grid is to create a bijection in a problem that makes it easier to
solve. Since the grid just represents a combination, it can be adapted to work
with any problem whose answer is a combination.</span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<br /></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">For
example, take an instance of the classic 'stars and bars' problem (also known
as 'balls and urns', 'sticks and stones', etc.):</span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<b><span style="color: #274e13;"><span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">Q: </span><span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">How
many ways are there to pick an ordered triple (a, b, c) of nonnegative integers
such that a+b+c = 8? (The answer is 10C2 or <span style="color: #660000;">45 </span>ways.)</span></span></b></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
</div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="color: blue;"><span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">Solution I: </span></span> </div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";"></span><span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">This
problem is traditionally solved by thinking of ordering 8 stars and 2 bars. An
example is:</span>
</div>
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<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">*
* * | | * * * * *</span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";"> ^
^ ^</span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";"> a
b c</span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";"></span><span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">This
corresponds to a = 3, b = 0, c = 5.</span>
</div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<br /></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";"><span style="color: blue;">Solution II: </span></span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">But
this can also be done using the grid technique:</span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDYnVUHTQ0nJQyhEdLWq8XcHokrkWh__RzXFvFHJO1Nm9akukjdCVUiB7U9bmWLRhqm4hgCM3rOTKel8G5I2t1bsVDBpZQvzsEiN97KvOwDq1-zS3Sd2KBUw1PK2Y1m4BPwvyWddjAa0A/s1600/Vinjai+-+Grid+technique.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDYnVUHTQ0nJQyhEdLWq8XcHokrkWh__RzXFvFHJO1Nm9akukjdCVUiB7U9bmWLRhqm4hgCM3rOTKel8G5I2t1bsVDBpZQvzsEiN97KvOwDq1-zS3Sd2KBUw1PK2Y1m4BPwvyWddjAa0A/s320/Vinjai+-+Grid+technique.png" width="131" /></a></div>
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<br /></div>
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<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman"; mso-no-proof: yes;">
</span><span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";"></span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<br /></div>
<br />
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">The
red path corresponds to the same arrangement: a = 3, b = 0, c = 5. The increase
corresponds to the value: a goes from 0 to 3 (that is an increase of 3), b goes
from 3 to 3 (that is an increase of 0), and c goes from 3 to 8 (that is an
increase of 5). So a = 3, b = 0, c = 5.</span>
<br />
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<br /></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0in;">
<span style="mso-bidi-font-size: 12.0pt; mso-fareast-font-family: "Times New Roman";">Likewise,
using a clever 1-1 correspondence, you can map practically any problem with an
answer of nCk to fit the grid method. The major advantage of this is that it is
an easier way to think about the problem (just like the example I gave may be
easier to follow than the original stars and bars approach, and the example I
gave in class with the dice can also be thought of in a more numerical sense).</span></div>
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<br /></div>
someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com0tag:blogger.com,1999:blog-2668492410190176921.post-18937951202915451112018-04-08T09:15:00.000-04:002018-04-08T09:15:39.527-04:00Learn How to Learn by BOGTRO from AoPS forum -- Thanks a bunch !! <b style="font-family: verdana, arial, helvetica, sans-serif; font-size: 11.5px; line-height: 22px;">I love the following quotes :</b><span class="postbody" style="color: black; font-family: "verdana" , "arial" , "helvetica" , sans-serif; font-size: 11.5px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: 22px; margin: 0px; overflow: hidden; padding: 0px; text-indent: 0px; text-transform: none; vertical-align: baseline; white-space: normal; word-spacing: 0px;"></span><br />
<div class="postbody" style="-webkit-text-stroke-width: 0px; color: black; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 11.5px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: 22px; margin: 0px; orphans: auto; overflow: hidden; padding: 0px; text-align: start; text-indent: 0px; text-transform: none; vertical-align: baseline; white-space: normal; widows: auto; word-spacing: 0px;">
<br /></div>
<div class="postbody" style="-webkit-text-stroke-width: 0px; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 11.5px; font-style: normal; font-variant: normal; letter-spacing: normal; line-height: 22px; margin: 0px; orphans: auto; overflow: hidden; padding: 0px; text-align: start; text-indent: 0px; text-transform: none; vertical-align: baseline; white-space: normal; widows: auto; word-spacing: 0px;">
<span style="background-color: white; font-family: "times new roman" , "times" , serif; font-size: 18px; line-height: 19px;"><span style="color: #38761d;"><b>Insanity: doing the same thing over and over again and expecting different results.</b></span></span></div>
<div class="postbody" style="-webkit-text-stroke-width: 0px; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 11.5px; font-style: normal; font-variant: normal; letter-spacing: normal; line-height: 22px; margin: 0px; orphans: auto; overflow: hidden; padding: 0px; text-align: start; text-indent: 0px; text-transform: none; vertical-align: baseline; white-space: normal; widows: auto; word-spacing: 0px;">
<b>I previously thought it's from Albert Einstein, but it's not. I love it anyway.</b> </div>
<div class="postbody" style="-webkit-text-stroke-width: 0px; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 11.5px; font-style: normal; font-variant: normal; letter-spacing: normal; line-height: 22px; margin: 0px; orphans: auto; overflow: hidden; padding: 0px; text-align: start; text-indent: 0px; text-transform: none; vertical-align: baseline; white-space: normal; widows: auto; word-spacing: 0px;">
<br /></div>
<div class="postbody" style="-webkit-text-stroke-width: 0px; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 11.5px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: 22px; margin: 0px; orphans: auto; overflow: hidden; padding: 0px; text-align: start; text-indent: 0px; text-transform: none; vertical-align: baseline; white-space: normal; widows: auto; word-spacing: 0px;">
<span style="background-color: white; font-family: "open sans" , "helveticaneue" , "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 13px; font-style: italic; font-weight: bold; line-height: 21px;"><span style="color: #990000;">You can practice shooting eight hours a day, but if your technique is wrong, then all you become is very good at shooting the wrong way. Get the fundamentals down and the level of everything you do will rise.”</span></span></div>
<div class="postbody" style="margin: 0px; orphans: auto; overflow: hidden; padding: 0px; text-align: start; text-indent: 0px; vertical-align: baseline; widows: auto;">
<div style="-webkit-text-stroke-width: 0px; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 11.5px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: 22px; text-transform: none; white-space: normal; word-spacing: 0px;">
<span style="background-color: white; font-family: "open sans" , "helveticaneue" , "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 13px; font-style: italic; font-weight: bold; line-height: 21px;">from Michael Jordan </span></div>
<div style="-webkit-text-stroke-width: 0px; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 11.5px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: 22px; text-transform: none; white-space: normal; word-spacing: 0px;">
<span style="background-color: white; font-family: "open sans" , "helveticaneue" , "helvetica neue" , "helvetica" , "arial" , sans-serif; font-size: 13px; font-style: italic; font-weight: bold; line-height: 21px;"><br /></span></div>
<span style="background-color: white; color: #0b5394; font-family: "georgia" , "times new roman" , serif;"><b><span style="line-height: 21px;"><i>"When you first start off trying to solve a problem, the first solutions you come up with are very complex, and most people stop there. But if you keep going, and live with the problem and peel more layer of the onion off, you can often times arrive at some very elegant and simple </i></span><span style="line-height: 21px;"><i>solutions."</i></span></b></span><br />
<span style="font-family: "georgia" , "times new roman" , serif;"><b>- Steve Jobs, 2006 </b></span><br />
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<span style="font-family: "georgia" , "times new roman" , serif;"><b><a href="https://www.ted.com/talks/bel_pesce_5_ways_to_kill_your_dreams?language=en" target="_blank">5 ways to Kill your dreams from TED talk </a></b></span></div>
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Below,<span style="color: #990000; font-weight: bold;"> BOGTRO from AoPS</span><span style="color: black; font-weight: normal;"> has graciously allow me to post his well-thoughtout article on </span><span style="color: black;"><b>"Learn How to Learn".</b></span></div>
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I wish more students will read it , and don't just read it once, but many times at different intervals and really internalize the method. It will help you not just with problem solving/competition math, but learning in general. </div>
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<b style="background-color: white;"><span style="color: purple;">Learn How to Learn </span></b></div>
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<span style="background-color: #f9f9f9; font-family: "verdana" , "arial" , "helvetica" , sans-serif; font-size: 12px; line-height: 22px;">About a month ago I was PMed by a member, asking for advice as to how to prepare for MATHCOUNTS. I (strangely) get a lot of these types of PMs, but this one was slightly different. Whereas normally I could answer something along the lines of "read Volume 1, do practice tests, profit", this user was complaining that despite having rigorously worked through Volume 1 and CMMS (I still don't know what this is, but it's implied to be a book), he was still scoring only in the low 20s on sprints. </span></div>
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To some extent, I was able to relate. Back in my MATHCOUNTS days, I was doing loads of practice tests, learning new techniques to shave off precious seconds, and even practicing hitting a buzzer quickly. But my results only marginally improved. Gradually I understood that he was facing the exact same problem I was - although we were doing plenty of work, <span style="font-weight: bold; margin: 0px; padding: 0px;">we were doing it in the wrong way</span>.</div>
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After some thought, I formulated a long but fairly detailed response. Given that state-national season is rolling around, and with it the usual abundance of "<a class="postlink" href="http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=573751" style="color: #2470ac; margin: 0px; padding: 0px; text-decoration: none;">how do I prepare</a>" threads, I'm reproducing it below (with some minor edits). I referenced sprint several times because that was the specific complaint by the user, but obviously you can replace "sprint" with "target" or even "countdown", or any combination thereof.</div>
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You first need to determine <span style="font-style: italic; font-weight: normal; margin: 0px; padding: 0px;">why</span> it is that you're getting low scores on sprint.<br />
<b>Are you running out of time? </b><br />
<b>Making stupid mistakes? </b><br />
<b>Bad at computation? Or</b><br />
<b> do you honestly not know how to do the problems? </b><br />
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The former three are rectified with simply a lot of (effective) practice, where I say "effective" because simply blazing through problems, checking your score, and moving on is not going to help you very much. You need to be critically analyzing almost every problem - not just the ones you got wrong. Sure, you don't need to think too hard about your process on #2, but questions that take you longer than you would like, you get wrong, or you do in a "bashy" way need to be reviewed.</div>
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Essentially, you should be following something similar to the following process. Of course, this is not something that is going to work for 100% of people. The point here is not that you should be following these guidelines like a bible, but that you need to think about how to get the maximum benefit out of each practice test you take. You may very well find that the below system doesn't work for you (though you should at least give it a chance - it may seem "boring" at first, but after some time you'll be going through it like it's second nature and learning excellent habits along the way), in which case you should come up with an alteration that works for you. If (or more likely when) you choose to develop your own preparation system, keep in mind that the basic elements should be present - rigorous review of problems you got wrong, self-reflection on <span style="font-style: italic; margin: 0px; padding: 0px;">why</span> you got them wrong, and so on.</div>
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<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">Take any MATHCOUNTS sprint round under contest conditions. It doesn't really matter which one you take, though it should be fairly recent for best results. When you're done, score with a simple checkmark or X system - don't look through the solutions immediately. Make a note of the problems that took you a long time, even if you got them correct.</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">Without timing yourself (though you shouldn't spend more than 15 minutes or so), solve the problems that you either got wrong or didn't answer during the test. This will partially tell you if you're getting questions wrong because of time constraints or because you don't know the material.</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">At this point you should have 4 separate categories of problems:<br style="margin: 0px; padding: 0px;" /><ul style="margin: 0px; padding: 0px 0px 0px 2em;">
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">Completely correct - don't worry about these at all. Though there is some benefit to looking these over, they are significantly less important than all the other questions.</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">Correct, but took you a long time. Identify <span style="font-style: italic; margin: 0px; padding: 0px;">why</span> it took you a long time - and if it matters. A problem taking you 2-3 minutes may sound like a killer, but in general if you only have a couple of these questions that's completely fine. Even if there's only one "timesink", you should be looking through alternate solutions to doing these problems. I find that problems that usually cause timesinks are either geometry problems that are semi-direct applications of similar triangles (which are naturally fairly easy to coordinate bash or something similarly slow, but this may take a while) or counting problems where you just listed out the possibilities and counted them up. Unfortunately, many MATHCOUNTS problems have this as their intended solution, so there's not a great deal you can do about those. However, even though there may not be a cleaner solution, minute steps during your bashing may prove important. And in the event that even with optimizations the problem will still take 2-3 minutes, you may want to just skip it altogether even if you know exactly how to do it.</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">Incorrect (or blank), but you solved it after the test. These are questions that you know how to do, but you ran out of time doing. Important is to determine how long it took you to solve these questions. If you solved 2 questions in 30 seconds each after the test, clearly that's worse than solving one problem in the second category. These second and third categories are quite similar and should be evaluated against each other (a quite reasonable rule of thumb is to save any counting question that you don't see how to do within ~10 seconds for later).</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">Incorrect, and you couldn't solve it after the test. Look up the solution, searching (or even posting) on AoPS if necessary (which you should likely do anyway, as MATHCOUNTS official solutions are often horrendous). If it's a situation where you just forgot something that you really knew, it's easy to pass this off as a fluke and move on. However, this is a grave mistake. Perhaps if it happens once or twice in an otherwise good practice, you can kind of gloss over it. But make a note of it anyway. Whenever you hit two problems in the same general category that you didn't solve (keep your categories broad, but not too broad. "Geometry" is too broad a category, while "trignometric relations in geometric models of algebraic inequalities" is too specific to be helpful. Something like "similar triangles" or "factoring" is a much better type of category), you should immediately stop your practicing and look up the relevant sections in whatever book you have (e.g. Volume 1, or whatever CMMS is, or even just an internet search, etc.). Don't move on until you are confident in that area. By "confident", I don't mean that you can approach these kinds of problems once in a while. I mean that once you identify a question as being in your category, you should be able to solve it relatively quickly at least 75% of the time.</li>
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<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">File away every single problem that you got wrong. Categorize these as either "I solved this afterwards" (include the time it took you to solve it - approximate is fine) or "I didn't solve this afterwards". You will need these later. Take a break - read a book, play some FTW, go outside, play League of Legends, whatever floats your boat. There's not much value in overloading yourself, especially so close to chapter. If you're feeling particularly ambitious, review a chapter on a topic that you have trouble with.<span style="font-weight: bold; margin: 0px; padding: 0px;">There is no point to reviewing topics you already can solve problems in regularly</span>.</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">At the end of the week, collect every single problem on your "incorrect problems list". If you're going through a test a day, these shouldn't number more than 50. <span style="font-weight: bold; margin: 0px; padding: 0px;">Do these like you would a test under contest conditions</span>. Compare your results to your incorrect problems paper (how long it took you to solve the problems, and whether you got them correct). The fact that you've seen the problems already should compensate for the fact that you need to work quicker. If you get a problem wrong, do the same process - don't time yourself while solving all of the remaining problems.</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">If you got the same problem wrong twice, there are 3 scenarios:<br style="margin: 0px; padding: 0px;" /><ul style="margin: 0px; padding: 0px 0px 0px 2em;">
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">You got it wrong both times, but finished it after the test both times. This speaks to your (lack of?) time management, something that comes much more naturally with practice. Keep in mind that MATHCOUNTS really only tests a very small amount of concepts (relatively speaking), so working through old problems virtually guarantees that almost all MATHCOUNTS problems will already be more or less familiar to you on test day.</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">You couldn't solve it at all the first time, but solved it after the test the second time. This is improvement, so it's perfectly fine.</li>
<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">You didn't solve it the second time around. This means that you don't understand the concept - back to the books.</li>
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<li style="display: list-item; list-style-image: none !important; list-style-type: square !important; margin: 0px 0px 0px 1.5em; padding: 0px;">Take all the problems you got correct (during the test) off your "incorrect problems" sheet, and continue to repeat the process from the top.</li>
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This may seem like quite a bit of work when typed up here, but in reality it's not. Instead of perpetuating the cycle of "do a practice test, score it, move on, read some books in some disorganized fashion, take another practice test, hope for improvement" (not even necessarily in that order, which is even more problematic), instead we optimize this routine by taking a <span style="font-weight: bold; margin: 0px; padding: 0px;">single practice test</span> a day and making sure that we <span style="font-weight: bold; margin: 0px; padding: 0px;">get everything possible out of it</span>. There are only so many tests, and a frequent complaint is that people have run out of old contests to do. While this may be true, this most likely means that they're <span style="font-weight: bold; margin: 0px; padding: 0px;">not doing the tests properly</span>. A single test with the time taken to reflect, organize, and perform a <span style="font-weight: bold; margin: 0px; padding: 0px;">targeted</span> review is significantly more beneficial than 5 tests taken without a goal in mind.</div>
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All in all, this should take at most a little over an hour per day (a little more at the end of the week). You are, of course, welcome to do more, but there's a sort of diminishing returns law past a certain point. Devoting a great deal of time to MATHCOUNTS is going to seem like a serious mistake in hindsight (I was among the most guilty of this), especially if you realize you were spending time incredibly inefficiently. I won't give an exact quote here (simply because I don't remember it and a quick search doesn't turn it up), but one MATHCOUNTS winner (Albert Ni?) said something along the lines of</div>
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<b style="margin: 0px; padding: 0px;">Quote:</b></div>
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I knew that I wouldn't be the smartest mathlete competing. But I could, quite realistically, be the hardest working one [...]</div>
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In MATHCOUNTS, that's all that's required. But quantifying the term "hard work" is necessary - someone who is pushing a boulder from point A 25% of the way to point B is doing a lot of work for very little benefit, while someone who uses a truck to carry the same boulder to point B is doing significantly less work for significantly more benefit. Perhaps as a more accurate analogy, take two people in a shooting contest. As soon as the whistle blows, person A starts shooting haphazardly at his target, hitting it once in a while but constantly having to reload. Person B, on the other hand, takes his time, lines up his shots, and hits the target with deadly accuracy. This is very similar to MATHCOUNTS. Person A is blowing through his material quickly, getting little benefit overall, but naturally with the experience of shooting comes some slight improvement. On the other side of things, person B is taking the time to think about how best to use his limited resources to improve as best he can. Sure, he starts off a bit slower, and at the end of the day he might still have some ammunition left unused, but overall he hits the target more. The first approach is popular because it's very easy to feel like you're doing something - after all, if you're spending 4 hours a day on practice MATHCOUNTS tests, you're outworking everyone else, right? Don't fall into this trap. Line up your shots.</div>
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someone opposite of Pierrehttp://www.blogger.com/profile/05344831478820677026noreply@blogger.com4