2025/2026 Mathcounts, AMCs, AIMEs Competition Preparation Strategies

Hi, Thanks for visiting my blog.

E-mail me at thelinscorner@gmail.com if you want to learn with me.  :) :) :) 

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

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.

5 ways to kill your dreams from TED Talk

There is no overnight success.

Testimonials from my students/parents.   :) 

My other blogs :

thelinscorner  : Standardized test preps (ONLY the hardest problems), books, links/videos for life-time learning

Studentsspeak Magazine  (featuring students' independent projects)

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.  

11 Qualities Google Looks For In Job Candidates 

In Chinese : 《專題報導》亞裔尖子生 職場難出頭？

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.

Newest Mathcounts' competition problems and answer key

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.

                                                             Do Not Rush !!

Awesome site!!
       
For concepts reviewing, try the following three links.
 
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.

Hope this is helpful!!

The Grid Technique in Solving Harder Mathcounts Counting Problems : from Vinjai

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. 

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.

For example, take an instance of the classic 'stars and bars' problem (also known as 'balls and urns', 'sticks and stones', etc.):

Q: 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 45 ways.)

Solution I: 

This problem is traditionally solved by thinking of ordering 8 stars and 2 bars. An example is:

* * * |    | * * * * *

  ^       ^       ^

  a       b       c

This corresponds to a = 3, b = 0, c = 5.

Solution II: 

But this can also be done using the grid technique:

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.

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).

Mathcounts prep

 Hi, Thanks for visiting my blog.

E-mail me at thelinscorner@gmail.com if you want to learn with me.  :) :) :) 

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

Sum of All the Possible Arrangements of Some Numbers

Check out Mathcounts, the best middle school competition math program up to the national level.

Questions to ponder: (detailed solutions below) 

It's extremely important for you to spend some time pondering on these questions first without peeking on the solutions. 

#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?

#2: Camy made a list of every possible distinct five-digit positive even 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? (2004 Mathcounts Chapter Sprint #29)

#3: 2020 Mathcounts state sprint #24 













Solutions:
#1:  
Solution I: 
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.

Solution II: 

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
the four numbers. 
2.5 (1000 + 100 + 10 + 1) x 24 = 66660 

#2:  
Solution I: 
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
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.)108 x 11110 + 4 x 24 = 1199976

Solution II: 

Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely. 
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\)                               
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements)  + 96 = 4.5 * 11110 * 24 + 96 = 1199976

#3: The answer is 101. 

Other applicable problems: (answers below)

#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? (2003 Mathcounts Sprint #30)

#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.) (2005 AMC-10 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?

#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?  

#5: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?

  





Answer key: 
#1: 66660
#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)
4.8 * 11111 =\(\color{red}{53332.8}\) 
#3: 133320
#4: 1333272
#5: 93324

2015 Mathcounts State Prep: Mathcounts State Harder Questions

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

Question: 2010 Mathcounts State Team Round #10: 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?

 There are lots of similar triangles for this question, but I think this is the fastest way to find the area.
  \(\Delta \)AGB is similar to \(\Delta \)DGC and their line ratio is 15 to 10 or 3 : 2.
     \(\Delta \)CGF is similar to \(\Delta \)CBE. 
     \(\dfrac {CG} {CB}=\dfrac {GF} {BE}\)
     \(\dfrac {2} {5}=\dfrac {GF} {10}\)\(\rightarrow GF=4\) From there you get the area =
      \(\dfrac{10\times 4} {2}=20\)

Question: 2010 Mathcounts State #30:  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.

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.
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 \(\overline {CD}\).
From there you get the side length for each side is \(\sqrt {3}+1\).
\(\overline {AC}-C\overline {D}=\sqrt {3}+1-2\) = \(\overline {AD} = \sqrt {3}-1\)
\(\Delta ABD\) and \(\Delta CBD\) share the same vertex, so their area ratio is just the side ratio, which is \(\dfrac {\sqrt {3}-1} {2}\).

2020 Mathcounts State Prep: Simon's Favorite Factoring Trick

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

The most common cases of Simon's Favorite Factoring Trick are:

I:  \(xy+x+y+1=\left( x+1\right) \left( y+1\right)\)

II:  \(xy-x-y+1=\left( x-1\right) \left( y-1\right)\)

It's easy to learn. Here is the best tutorial online, by none other than Richard Rusczyk.
The method Rusczyk uses at the second half is very nifty. Thanks!!

Questions to ponder:(answer key below)
#1: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that \(xy+x+y=13\) 
#2: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that \(2xy+2x-3y=18\)
#3: Find the length and the width of a rectangle whose area is equal to its perimeter.
#4: Twice the area of a non-square rectangle equals triple it's perimeter, what is the area of the rectangle? 













Answer key:
#1:  x = 6 and y = 1
#2: ( x, y ) = (4, 2) 
#3: Don't forget square is a kind of rectangle (but not the other way around) so there are two answers: 
4 by 4 and 3 by 6 units. 
#4: One side is 4 units and the other 12 units so the answer is 4 x 12 or 48 square units. 
There is another one, 6 by 6 that would fit if the question doesn't specify non-square rectangle. 

Geometry : Harder Chapter Level Quesitons

Question #1 : The area ratio of two equilateral triangles are 4 to 9 and the sum of their perimeter is 30 √ 3 . What is the area of the a. smaller triangle   b. larger triangle?  

Solution: 
If the area ratio of two similar polygons is 4 to 9, their corresponding line ratio would be √ 4  to √ 9  
or 2 to 3.[Make sure you know why.]
The perimeter of the two equilateral triangles is 30 √ 3 so the smaller triangle has a perimeter of
2/5 *  30√ 3 or 12 √ 3. One side is 4√ 3 . Using the formula of finding the area of an equilateral triangle \(\dfrac{\sqrt3*s^2}{4}\) , you get the area to be 12√ 3.

Use the same method to get the area of the larger triangle as 27√ 3.
You can also use ratio relationship to get the area of the larger triangle by 
multiply 12√ 3 by 9/4.


2007 Mathcounts Chapter Sprint #30: In parallelogram ABCD, AB = 16 cm, DA = 32 cm, and sides AB and DA form a 45-degree interior angle. In isosceles trapezoid WXYZ with WX ≠ YZ, segment WX is the longer parallel side and has length 16 cm, and two interior angles each have a measure of 45 degrees. Trapezoid WXYZ has the same area as parallelogram ABCD. What is the length of segment YZ?

Solution I:
Make sure you know how to get the unknown leg fast. The height of the parallelogram is 8√2, so the area of the parallelogram is 48 square units. [Check out the special right triangle section here if you can't get the height fast.]

Let YZ of the trapezoid be x and draw the height. Using 45-45-90 degree angle ratio, you'll get the height. (See image above.)
Area of the trapezoid is average of the two bases time height. WX = 16 (given)
\(\dfrac{(16+x)* (16-x)}{4}\) = 48 ; 256 - x² = 192 ;  - x² = - 64;  x = 8 = YZ

Solution II: 

Make the y be the height of the trapezoid. YZ = 16 - 2y.  \(\dfrac{(16-2y + 16)}{2}\) * y = 48
\({(16 -y)* y = 48}\)\(\rightarrow\) \({16y -y^2 = 48}\) \(\rightarrow\) \({y^2 - 16y + 48 = 0}\) \(\rightarrow\) \({(y -12)(y -4)=0}\) \(\rightarrow\) \({y = 4}\) or \({y = 12}\)(doesn't work)
 YZ = 16 - 2y. Plug in y = 4 and you have  YZ = 8

Solution III: Let the height be y and you have \(\dfrac{(\overline{YZ}+ 16)* y}{2}"\) = 48 ; ( YZ + 16) * y = 96
When there are some numbers multiply together equal another number, it's a factoring question.
32 * 3 = 96, YZ = 16 (doesn't work)
24 * 4 = 96, YZ =8

2017 Mathcounts State Prep: Volume of a Regular Tetrahedron and Its Relationship with the Cube it's Embedded

How to find volume of a tetrahedron (right pyramid) with side length one.

The above link gives you a visual interpretation of the relationship of a regular tetrahedron, its
relationship with the cube that it is embedded and the other kind of tetrahedron (right angle pyramid).

The side of the cube is \(\dfrac {S} {\sqrt {2}}\) so the volume of the regular embedded tetrahedron is
\(\dfrac {1} {3}\times \left( \dfrac {S} {\sqrt {2}}\right) ^{3}\)=\(\dfrac {1} {3}*\dfrac {s^{3}} {2\sqrt {2}}\)= \(\dfrac {\sqrt {2}S^{3}} {12}\).

You can also fine the height of the tetrahedron and then \(\dfrac {1} {3}\)*base*space height to get the volume.
Using Pythagorean theory, the hypotenuse S and one leg which is \(\dfrac {2} {3}\) of the height of the equilateral triangle base, you'll get the space height.

First week's review reminders

For 16-17 Mathcounts handbook :

Warm-up 1 : review #3, 7, 9.
Think of ways to get these questions in seconds.

Warm-up 2 : review 17,18,19,20 -- slow down on #18, similar types are very easy to
get wrong the first try

direct and inverse relationship 

From AoPS videos :

7.1 Direct Proportion

7.2 Inverse Proportion

7.3 Joint Proportion

Instructions for reviewing 

2014 School Rounds link here 

Sprint : 

#18 (trial and error are faster), #20 (loose pennies out first)
#22, #26, #27, #28, #29, #30 -- Read the solutions and make sure to fully understand
what major concept(s) is (are) tested -- these are not hard problems.

Target : 

#1 : Calendar questions could be tricky, so slow down a bit.
       Check what stage it starts first.

#2 : WOW, this one is a killer, so tedious -- hard to get it right the first time. 

# 3 and 4 are in seconds questions.

# 5: One line, two numbers and you are done.
       With a calculator, you don't even need to write anything down, right?  :)

#6 : more interesting question

#7 and 8 : standard questions, not hard, just need to be careful and thorough.

Team : 

# 2, 3, 5, 6 (very tedious, I suggest skipping first)
8, 9 (more tedious than 8 and 10, so don't work on problems in order), and
#10 (very easy if you've noticed what is actually tested) It's in seconds question. 

Agai, scan those questions and only try those that are your weak spots or marked red.

Please check the solution files I'd sent you to learn the better methods.

Whenever you have extra time, use the other links (individually based) to keep learning.

Keep me posted. Have fun at problem solving.

Don't just do math.  :) 

Math related video : Making Stuff Faster, 

which includes "The Travelling Salesman Problem" and a competition between an astrophysicist and a paleontologist on how to move passengers boarding the planes faster 

Good luck, from Mrs. Lin 

Testimonials for my services. So far, all through words of mouth locally or chance meets online and it's great.

Testimonials from my students/parents  :) 

Dear Mrs Lin,

I just did not get the opportunity earlier to thank you for all your help. _____ was able to make to the National MathCounts largely because of your excellent guidance and coaching. I do not know how to thank you. He had a great once-in-a-lifetime experience there and he really loved the competition as well as meeting other people.

If you can please provide me your mailing address, he wants to send you a gift as a token of thanks for your guidance and tutoring. 

Sincerely,

Dear Mrs. Lin,

I have been meaning to tell you, but just didn't get the time. He just 'LOVES' your sessions. He said he is learning so much and gets to do lots of problems and he likes all the tips/shortcuts you are teaching. He looks forward to your session - he was so upset when we couldn't get back on time from _____ because he didn't want to miss your session. Honestly we came back Tue night only because he cried so much that he didn't want to miss your class:) 

In school, his teacher focuses more on details like, put all the steps, write neatly, don't disturb the class by asking unnecessary questions, don't ask for more work, behave properly etc....so he is not too happy with math in school.

Thank you so much for making such a big difference in his life. He  enjoys doing the homework you are assigning him and has not complained at all. He said "Mrs Lin is so smart and I love her classes...wish she lived next to our house...so I can go to her and have live classes' [disclaimer : I'm not ; my students are much smarter than I and I learn along with them and it's exhilarating ]   Thank you so much!!

One thing he said was it would be helpful if he can have targeted practice worksheets on the tips covered during the class, after that class, so he can practice those shortcuts/tips more.

Hello Mrs. Lin,

This is ______ 's mom. We greatly appreciate your help in working with
______, keeping him motivated and providing him wth constant
encouragement. We are fortunate to have great mentors like you who

are so selfless in their services to our younger generation.

We have a request, can you please share your address. ______ wanted
to send you a card to express his thanks.

BTW, _______ has his chapter level competition tomorrow.

Warm Regards

Mrs. Lin,

Good Afternoon.

Me and my son, ___ and I, have used your blog pages and got a great insight into several of the techniques that you use to solve the problems. I am glad to inform you that _____ was placed 14th in the State Mathcounts for ____. Your blog information has helped us a lot in this preparation and we really want to appreciate all that you do in sharing the information. 

_____ is completing Geometry this year and he hopes to have a exciting next year for AMC 8 and Mathcounts. We hope to learn a lot more from your blog pages.

Regards,

Hi Mrs Lin,

I just finished The One World School House by Salman Khan and I have been thinking of you and the online community you created. Even though _____ and ____ have taken web based courses before, what makes your class different is your inspiration and enthusiasm - you really care about them as unique individuals and you sincerely expect the best from them, more than they (and I sometimes) think they can achieve. Thank you for encouraging them to become a more responsible and self driven learner.

Dear Mrs.Lin,

             I went to NSF tests today and got first place in the Math Bee III. The problems didn't seem that bad although I'm waiting to see my score online. I will most likely be going to nationals which will be held in Ohio this year so that's convenient. Lastly I would like to thank you for all you have done for me. I believe I've grown much more as a student under you and really appreciate everything you have shown me and taught me. I wouldn't be succeeding now if it wasn't for you. :)

Sincerely,

So far,in January state, in algebra 2 he came individual 2nd and one regional February he came 1st (competing with same state high school students while as an 8th grader) 

He could not participate in one regional bec of conflict with MathCounts chapter. We also want to share with you that he got admission into a private school for 9th gr with 100% scholarship. [more than 50 students in that high school are national merit semifinalists, so highly competitive]
Thank you,

An elementary whiz kid, national winners at Math Kangaroo and Math Olympiad. 

_____ will be off to summer sleep away camp where he is not allowed to have any electronics starting _____.  

_________ will be his last class with you for over a month.  He really has been enjoying it and  just so you know I never  have to ask him twice to do the work- crazy because everything else I ask him to do takes at least 5 tries :) 

He will be back online with you at _______ . 

There are many more but I'll take my time to update/upload these infor. sheets. 

To be continued ... 

I'm quite busy these days with resuming our Math Circle + many other projects (my students don't just excel at math, but many other areas equally fun and challenging + most Asian students have much bigger problems with critical reading/not to mention writing, taking initiatives and being strong leaders, and those are my other projects. 

Work-Life balance is utmost important. Less is more. 

Notes to Sunday Nights' Problem Solving Group Lessons

This week's work : Please review the last 8 hardest AMC-8 problems from 2005 to 2011 + finish the last 8 hardest AMC-8 problems from 2012 and 2013 if you haven't done that.

Some notes and questions.

Question #1 : How many ways can you climb up a ten-step staircase if you climb only one or two steps at a time ?

Solutions I :
Make a chart, starting with a smaller case.
one-step staircase -- 1 way, which is 1.
two-step staircase -- 2 ways, which is 1, 1 or 2.
three-step staircase -- 3 ways, which is 1 1 1, 1 2 or 2 1.
four-step staircase -- 5 ways, which is 1111, 211, 121, 112, 2222.

Notice the pattern - - 1 , 2, 3, 5, 8, 13, 21, 34, 55, 89, which is the answer; it also happens that it's
part of the Fibonacci numbers.

Why does it work that way ?

Well, if you climb the 3-step staircase, there are two cases :
You either take one step at first, and there are 2-step left, which leaves you two ways to climb the remaining staircase.
Or you take two step at first, and there are 1 step left, which leaves you one way to climb the remaining staircase.
If you climb the 4-step staircase, again, there are two similar cases :
You either take one step at first, and  then there are 3-step left, which leaves you 3 ways to climb the remaining staircase.
Or you take two step at first, and there are 2 step left, which leaves you 2 way to climb the remaining staircase.

Thus, it's always the sum of the previous two terms for the next staircase steps.
This concept is called recursion.

Now try another question :
2010 AMC-8 #25 : Every day at school, Jo climbs a flight of 6 stairs. Joe can take the stairs, 1, 2 or 3 at a time. For example, Jo could climb, 3, then 1, then 2. In how many ways can Jo climb the stairs ?

Solution I :
List out all the possible ways.
111111  -- 1 way
21111 -- 5C1 = 5 ways to arrange the steps
2211 -- 4!/2! x 2! = 6 ways to arrange the steps
222-- 1 way
3111 -- 4C1 or 4 ways to arrange the steps
123 -- 3! or 6 ways
33 -- 1 way
Sum them up and the answer is 24.

Solution II :
Using recursion, starting with the smallest case.
1 step -- 1 way
2 steps-- 2 ways (11, or 2)
3 steps -- 4 ways (111, 21, 12, or 3)
4 steps -- 1 + 2 + 4 = 7 ways (why??)
5 steps -- 2 + 4 + 7 = 13 ways.
6 steps -- 4 + 7 + 13 = 24 ways, which is the answer.

Besides these, please review the following questions.

Answer key below.

Q #1  1988 AMC-8 #25 : A palindrome is a whole number that reads the same forwards as backwards. If one neglects the colon, certain times displayed on a digital watch are palindromes. Three examples are 1: 01, 4: 44 and 12: 21. How many times during a 12-hour period will be palindromes?

Q#2  2004 Mathcounts sprint  #21: If |-2a + 1| < 13, what is the sum of the distinct possible integer values of a?

Q#3  2004 Mathcounts sprint #30 : A particular right square-based pyramid has a volume of 63,960 cubic meters and a height of 30 meters. What is the number of meters in the length of the lateral height (AB) of the pyramid? Express your answer to the nearest whole number.










Answer key :

#1 : 57

#2 : 6

#3 : 50

Analytical Geometry : Circle Equations

Circle Equations from Math is Fun 

How to Find Equation of a Circle Passing 3 Given Points 

7 methods included ; Amazing !!

Practice finding the equation of a Circle given 3 points -- 

Q #1 : (1, 3), (7, 3) and (1, -3)

Answer : (x -4)² + y² = 18

Q #2  : (3, 4), (3, -4), (0, 5)

Answer : x² + y² = 25

Q #3 : A (1, 1), B (2, 4), C (5, 3)

Answer : (x-3)² + (y -2)² = 5

Solution : 

The midpoint of line AB on the Cartesian plane is \((\frac{3}{2}, \frac{5}{2})\) and the slope is \((\frac{3}{1})\) so the slope of the perpendicular bisector of line AB is \((\frac{-1}{3})\).

The equation of the line bisect line AB and perpendicular to line AB is thus :  

y - \((\frac{5}{2}\)) =\(\frac{-1}{3}\) [x - \((\frac{3}{2})\)] --- equation 1

The midpoint of line BC on the Cartesian plane is \((\frac{7}{2}, \frac{7}{2})\) and the slope is \((\frac{-1}{3})\) so the slope of the perpendicular bisector of line BC is 3.

And the equation of the line bisect line BC and perpendicular to line BC is 

y - \((\frac{7}{2})\) = 3 [x - \((\frac{7}{2})\)] --- equation 2

Solve the two equations for x and y and you have the center of the circle being (3, 2)

Use distance formula from the center circle to any point to get the radius = 

\(\sqrt{5}\).

The answer is : (x - 3)² + (x - 2)² = 5

More practices on similar questions :  (Answers below for self check)

Q #1 : A (2, 5) , B (2, 13) ,  and C (-6, 5 )

Q #2 : A (0, 7), B ( 6, 5 ), and C (-6, -11 )

Q #3 : A (3, -5) , B (-4, 2) and C (1, 7 )








Answer key :

#1 :  (x - 2) ² + ( y - 9 ) ² = 32

#2 :   x² + (y + 3)² = 100

#3 :  (x - 2) ² + ( y -1) ² = 37