## Thursday, January 12, 2017

### 2017 Mathcounts Competition Preparation Strategies

It's the Mathcounts chapter/state/AMCs season. Kids are more motivated.
YEAH !!!!! :)

Hi, Thanks for visiting my blog.

E-mail me at thelinscorner@gmail.com if you want to join my groups.

Currently I'm running different levels of problem solving group lessons, and it's lots of fun learning along with students from different states.

My most advanced group of  students are just AMAZING !! to say the least.
Ha ha, we are using AMC-10, 12 questions as countdown round practices and some can solve the first few AIME problems in less than a minute. Oh dear !!

I can also evaluate your current level, give you suggestions on how to improve on your learning.

So many students are not learning smart.

Problem solving is really fun (and a lot of the times very hard, yes).

Good questions are intriguing and delicious, so come join our vibrant community and have the pleasure of finding things out on your own.

There is no overnight success.

My other blogs :

thelinscorner  : Standardized test preps, books, links/videos for life-time learning

Take care and have fun learning.

Don't forget other equally interesting activities/contests, which engage your creativity  and imagination.

Some also require team work. Go for those and have fun !!

Don't just do math.

Before going full throttle mode for competition math, please spend some time reading this
well- thought-out article from BOGTRO at AoPS "Learn How to Learn".

It will save you tons of time and numerous, unnecessary hours without a clear goal, better method in mind.

Less is more. My best students make steady, very satisfactory progress in much less time than those
counterparts who spent double, triple, or even more multiple times of prep with little to show.

It's all about "deliberate practices", "tenacity", and most of all, "the pleasure of finding things out on your own".

Take care and have fun problem solving.

I have been coaching students for many years. By now, I know to achieve stellar performance you need :
Grit (from TED talk), 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.)

Thanks a lot !!  Mrs. Lin

"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."

"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 !!"

"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."

Now, here are the links to get you started:

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.

Last year's Mathcounts competition problems and answer key

This year's handbook questions.
Near the end of the handbook, there is a page called  problem index (page 82 and 83 for 2013-2014 handbook).
For state/national prep, find your weakness and work on the problems backwards, from the hardest to the easiest.

Here are some other links/sites that are the best.

Mathcounts Mini : 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.

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.

Art of Problem Solving

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.

Register for Alcumus and start using the great tool to practice deliberately.
Change the setting based on the levels of your proficiency of different topics.                                                                     Do Not Rush !!

Awesome site!!

For concepts reviewing, try the following three links.

Mathcounts Bible

Mathcounts Toolbox

Coach Monks's Mathcounts Playbook

You really need to understand how each concept works for the review sheets to be useful.

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.

You don't need a lot of formulas, handbook questions, or test questions to excel.

You simply need to know how the concepts work and apply that knowledge to different problems/situations.

## Thursday, December 29, 2016

### 2013 Mathcounts School and Chapter Harder Problems

Here is the link to the official Mathcounts website.

Some more challenging problems from this year's Mathcounts school/or chapter problems.

2013 school team #10 : Three concepts are testing here :
Hint:
a. If you get rid of the remainder, the numbers will be evenly divided into 192, so you are looking at
those factors of 192 - 12 = 180

b. To leave a remainder of 12, those factors of 180 that are included in the Set must be smaller than 12, otherwise, you can further divide it.

c. To find the median, make sure to line up the numbers from the smallest to the largest and find the middle numbers. If there are even numbers of factors larger than 12, average the middle two. Otherwise, the middle number is the answer.

$$180=2^{2}\times 3^{2}\times 5$$ so there are (2 + 1) (2 + 1) (1 + 1) = 18 factors

The list on the left side gives you the first 9 and if you times those numbers with "5", you get 9 other factors,which are 5, 10, 20, 15, 18, 60, 45, 90 and 180.

Discard the factors that are smaller or equal to 12 and list all the other factors in order and find the median.

2013 Chapter Sprint:
#21: Dimensional change problem : The height of the top pyramid is $$\dfrac {2} {3}$$ of the larger
pyramid so its volume is $$\left( \dfrac {2} {3}\right) ^{3}$$ of the larger pyramid.

$$\left( \dfrac {2} {3}\right) ^{3}\times \dfrac {1} {3}\times \left( \dfrac {36} {4}\right) ^{2}\times 12 =$$ $$96 cm^{3}$$

# 24:  According to the given:   $$xyz=720$$   and   $$2( xy+yz+zx)= 484$$ so
$$( xy+yz+zx )= 242$$

Since x, y and z are all integers, you factor 720 and see if it will come up with the same x, y and z values
for the second condition.

Problem writer(s) are very smart using this number because the numbers "6", "10", "12" would give you
a surface area of 252. (not right)

The three corrrect numbers are "8", "9", and "10" so the answer is $$\sqrt {8^{2}+9^{2}+10^{2}}=$$ $$7\sqrt {5}$$

#25: Geometric probability: Explanations to similar questions and more practices below.

Probability with geometry representations form Aops.

Geometric probability from "Cut the Knots".

#26: This one is similar to 2002 AMC-10B #21, so try that question to get more practices.

#27:
$$\dfrac {1} {A}+\dfrac {1} {B}=\dfrac {1} {2}$$
$$\dfrac {1} {B}+\dfrac {1} {C}=\dfrac {1} {3}$$
$$\dfrac {1} {C}+\dfrac {1} {A}=\dfrac {1} {4}$$
Add them up and you have  $$2 * (\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {12}$$

$$(\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {24}$$

$$\dfrac {1} {\dfrac {1} {A}+\dfrac {1} {B}+\dfrac {1} {C}} =$$$$\dfrac {{24}} {13}$$ hours

#28: Hint : the nth triangular number is the sum of the first "n" natural numbers and $$\dfrac {n\left( n+1\right) } {2}$$ is how you use to find the sum.
From there, you should be able to find how many numbers will be evenly divided by "7".

#29 : Circle questions are very tricky so make sure to find more problems to practice accuracy.

#30 :
Solution I:
Read the solution that is provided by Mathcounts.org here.
Solution II:
Case 1 : $$x-1 > 0\rightarrow x > 1$$ Times ( x - 1) on both sides and you have
$$x^{2}-1>8$$ so x > 3 or x < -3 (discard)

Case 2: $$x-1 < 0$$ so $$x < 1$$ $$\rightarrow x^{2}-1 < 8$$ [You need to change the sign since it's negative.]-3 < x < 3. Combined with x < 1 you have the range as -3 < x < 1

2013 Mathcounts Target #7 and 8:

Target question #8 is very similar to 2011 chapter team #10