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.

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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!!

2022 AIME I reflection notes from H

2022 AIME I — Reflection Notes from H

Attempted: #4–12

Correct: #4–9, #11–12

Had trouble / wrong: #10

Need Review / Unclear:

#8 – Diagram got way messy. Tried to use side-length scaling to get the box/answer, but it didn’t work.

#10 – Not sure how to visualize; was difficult with a plane not totally perpendicular.

2021 AIME 1 reflection notes from H

AIME — Reflection Notes from H

Did 8 to 15 Incorrect / wrong: #9, #13, #15

Need Review / Unclear:

#9 – Not able to use similar triangles.
#13 – Totally confused; I tried PUP.
#15 – Not sure how to solve a quartic equation (x⁴).

2020 AIME I reflection notes from H

2020 AIME I - Reflection Notes from H

Did: #8–15

Had trouble / wrong: #12, #15

Need Review / Unclear:

#12 – Don’t know LTE, but I was very close to solving this using the binomial theorem.
#15 – Wasn’t able to see the symmetry: the circumcircle of IBC (with the orthocenters) is a reflection of the circumcircle of ABC.

2019 AIME I reflection notes from H

2019 AIME I — Reflection Notes from H

9 out of 15, 2 didn't finished were very close. like to know if there is better strategies for some problems.

Quick stats
• #1–5 took about 22.5 min total.
• #6 took about ~20 min (time sink).

My answers (as written)
#1: 342
#2: 029
#3: 120
#4: 122
#5: 252
#6: 090
#7: 880
#8: (very close — not finished)
#9: (wrong / crossed out)
#10: 357 (marked “silly”)
#11: ✓ (got it, but messy)
#12: (very close — basically solved but error)
#13: (basically solved, but used the wrong operation / wrong term; didn’t finish)
#14: 097
#15: (did not attempt)

Had trouble / wrong

#6 – Tried using binomial expansion, but it became too much computation (huge time sink).
#9 – Went nowhere / ended up crossing it out.
#10 – Silly mistake (need to slow down + verify).
#13 – Basically solved, but multiplied/used the wrong thing (wrong term like a_i) and didn’t finish.

Very close / almost

#8 – Very close (need to push to the finish next time).
#12 – Basically solved, but made an operation mistake (multiplied when I shouldn’t / wrong step).

Need review / improve

#11 – Got it, but I couldn’t visualize well, and the computations got messy.
#14 – Some algebra at the end felt annoying; practice finishing cleanly under time.

Not attempted

#15 – Did not attempt.

2018 AIME I reflection notes from H

2018 AIME I — Reflection Notes from H

Had trouble / wrong: #9, #10, #12, #13

Need Review / Unclear:

#9 – I got the first two terms of the final sum, but I didn’t understand / account for the double counting. a + b = 16 abd b + c = 24.

#10 – Did casework and isolated the 3’s, but wasn’t able to finish.

#12 – I got ±5 (and split into ~15 cases), but couldn’t compute after that.

#13 – Very close, but couldn’t simplify the trig at the end in the expression for the area.

2017 AIME II reflection notes from H

AIME 2017 II — Reflection Notes from H

Correct: #8, #10, #11, #12, #13

Had trouble / wrong: #9, #14, #15

Need Review / Unclear:

#9 – The “given that” statement kind of confused me. I wasn’t sure how to use it / what it was trying to tell me, so I got stuck early.

#14 – No idea how to start. I couldn’t visualize the setup, so I didn’t know what to do next.

#15 – Tried using generating functions, but it became too complicated (too many moving parts). Need a cleaner setup / approach.

2017 AIME I reflection notes from H

Correct: #8, #9, #11, #12, #14

Had trouble / wrong: #10, #13, #15

Need Review / Unclear:

#10 – Tried looking at this geometrically, but couldn’t find the right cyclic quadrilateral / setup (kept thinking about points like z₁, z₂, z₃, etc.). Need to learn how to “see” the key circle configuration faster.

#13 – I knew Q(m) would eventually become 1, but I didn’t realize it would happen at such a small value of m. This could have been done with quicker mental casework / checking small m first.

#15 – Tried using trig, but it got very complicated (too many variables). Need to practice spotting a cleaner non-trig / inequality setup sooner.

2016 AIME II reflection from H

Correct: #8, #9, #11, #12, #13, #14

Had trouble / wrong: #10, #15

Reflection / What happened:

#10 – Tried using no trig (just similar triangles + Ptolemy). Computations got messy, and I wasn’t fully confident the setup was correct.

#15 – I got pretty far into the official solution, but the step right before multiplying by Σ_i=1ⁿ(1-a_i) didn’t feel motivated to me at first (why multiply by that?). It makes more sense now, but I need to see this trick more often.

Notes:

• Cauchy–Schwarz is the right tool here and the method is clean, but it’s very tricky in execution.
• I understand the solution after reading it, but I need more problems like this to become consistent.

2016 AIME I reflection notes from H

2016 AIME I — Reflection Notes from H

Correct: #1, #2, #3, #4, #5, #7, #8, #11, #13, #14

Wrong (with notes):

#6 – My solution attempt did not include angle-chasing — that was fun/confusing for me.

#9 – Seems very tricky with a lot of variables — not sure how to even create an expression for the area using 3 variables.

#10 – Again, computation seems difficult/complex — how to simplify such problems?

#12 – Diagram seemed very complex. Additionally, the solution didn’t fully make sense. One thing I didn’t know is the radical axis theorem, which seems important.

#15 – I figured out the importance of multiple II’s and made some substitutions as such, but solutions seem to be lucky through a lot of assumptions.

2015 AIME I reflection notes from H

2015 AIME I — Reflection Notes from H

Had trouble / wrong: #9, #13, #14

Need Review / Unclear:

#9 – Got lost in casework (time sink).
#13 – Could not simplify easily; forgot identities.
#14 – Wasn’t able to visualize graph / trapezoids; need to practice these weird function graphing problems.

2011 AIME II reflection notes from H

2014 AIME I — Reflection Notes

Worked on: #8–#15

Had trouble / wrong: #8, #10, #13, #15

Need Review / Unclear:

#8 – Struggled using the Chinese Remainder Theorem.

#10 – Could not relate the turn angle to the number of rotations; needed to remember to add an extra rotation in the setup.

#13 – Was able to figure out that the center lies on EG, but was not able to use perpendicular lines correctly.

#15 – Angle chasing was difficult.

2011 AIME II reflection notes from H

2011 AIME II — Reflection Notes

Correct: 8, 9, 10, 11, 14, 15

We did #10 during our lesson already.
#15 – made a calculation error but corrected it.

Need Review / Unclear:

#12 – did not understand PIE solution (Principle of Inclusion–Exclusion).
#13 – understood the solution, but it was slightly confusing for me.

Hints/links or Solutions to 2014 Harder Mathcounts State Sprint and Target question

Links, notes, Hints or/and solutions to 2014 Mathcounts state harder problems.

2014, 2015 Mathcounts state are harder 

Sprint round:

#14 :
Solution I :

(7 + 8 + 9)  + (x + y + z)  is divisible by 9, so the sum of the three variables could be 3, 12, or 21.
789120 (sum of 3 for the last three digits) works for 8 but not for 7.
21 is too big to distribute among x, y and z (all numbers are district),
thus only x + y + z = 12 works and z is an even number
__ __ 0 does't work (can't have 6 6 0 and the other pairs all have 7, 8 or 9)
264 works (789264 is the number)

Solution II : 
789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...
Try 504 * 1566 = 789264 (it works)
The answer is 264.

#18:
Watch this video from Mathcounts mini and use the same method for the first question,
you'll be able to get the answer. It's still tricky, though.

#23 : Drop the heights of the two isosceles triangles and use similar triangles to get the length of FC.
Then solve.

#24: 
The key is to see 2¹⁰ is 1024 or about 10³

2³⁰ = ( 2¹⁰ )³  or about (10³  )³about 10⁹ so the answer is 10 digit.

#25:
As you can see, there are two Pythagorean Triples : 9-12-15 and 9-40-41.
Base (40-12) = 28 gives you the smallest area.
The answer is 28 * 18 = 504

#26 : Let there be A, B, C three winners. There are 4 cases to distribute the prizes.
A     B    C
1      1     5    There are 7C1 * 6C1 * \( \dfrac {3!} {2!}\) = 126 ways -- [you can skip the last part for C
because it's 5C5 = 1]

1       2    4    There are 7C1* 6C2 * 3! = 630

1      3     3    There are 7C1 * 6C3 * \( \dfrac {3!} {2!}\) = 420

2      2    3     There are 7C2 * 5C2 * 3 (same as above)

Add them up and the answer is 1806.

If you can't see why it's \( \dfrac {3!} {2!}\) when there is one repeat, try using easier case to help you understand.

What about A, B two winners and 4 prizes ?
There are 2 cases, 1 3 or 2 2, and you'll see how it's done.

#27 : Read this and you'll be able to solve this question at ease, just be careful with the sign change.
Vieta's Formula and the Identity Theory

#28: There are various methods to solve this question.
I use binomial expansion :
\(11^{12}=\left( 13-2\right)^{12}=12C0*13^{12}\)+ \(12C1*13^{11}*2^{1}\)+... \(12C11*13^{1}*2^{11}\)+ \(12C12*2^{12}\) 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.

Solution II :
\(11\equiv -2\left ( mod13\right)\) ; \((-2)^{12}\equiv 4096\equiv 1\left ( mod13\right)\)

Solution III :  
Or use Fermat's Little Theorem (Thanks, Spencer !!)
\(11^{13-1}\equiv 11^{12}\equiv1 (mod 13)\)

Target Round : 

#3: Lune of Hippocrates : in seconds solved question.
^__^

#6: This question is very similar to this Mathcounts Mini.
My students should get a virtual bump if they got this question wrong.

#8: Solution I : by TMM (Thanks a bunch !!)
Using similar triangles and Pythagorean Theorem.

The height of the cone, which can be found using the Pythagorean  is . 
Usingthediagram below, let  be the radius of the top cone and let  be the height of the topcone. 
Let  be the slant height of the top cone.


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 proportionCross multiplying yields This is what we need.

Next, the volume of the original cone is simply . 
The volume of the top cone is .
From the given information, we know that We simply substitute the value of  from above to yield We will leave it as is for now so the decimals don't get messy.

We get  and .

The lateral surface area of the frustum is equal to the lateral surface area of the original cone minus the lateral surface area of the top cone. The surface area of the original cone is simply 
. 
The surface area of the top cone is . 
So our lateral surface area is 

All we have left is to add the two bases. The total area of thebases is . So our final answer is 

Solution II : 
Using dimensional change and ratio, proportion.

Cut the cone and observe the shape.

The circumference of the larger circle is 20pi (10 is the radius) and the base of
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)

To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.

Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).

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 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).

Now we can solve this :
 \(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about 176 (after you round up)ional change and ratio, proportion.

Cut the cone and observe the shape.

The circumference of the larger circle is 20pi (10 is the radius) and the base of
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)

To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.

Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).

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 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).

Now we can solve this :
 \(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about 176 (after you round up)

Solution III : Another way to find the surface area of the Frustum is : 
median of the two half circle [same as median of the two bases] * the height [difference of the two radius]
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)

2023 Mathcounts school sprint more challenging questions -- Thanks to Ani for trying these problems.

#24 – Probability (Unfair Coin)

Gloria has an unfair coin. Let \( p \) be the probability of heads, with \( \tfrac{1}{2} < p < 1 \). When the coin is flipped three times, the probability of getting exactly two heads is twice the probability of getting exactly two tails. Find the probability of tails when the coin is flipped.

Show Solution

Let \( q = 1 - p \).

\[ P(\text{2 heads}) = 3p^2q, \quad P(\text{2 tails}) = 3q^2p. \] Given \( 3p^2q = 2(3q^2p) \Rightarrow p^2q = 2q^2p. \)
Divide by \( pq \) (nonzero) → \( p = 2q \).

Then \(q = 1-2q \)
Hence \( q = = \boxed{\tfrac{1}{3}}. \)

---

#26 – Expected Value (Sum of Cubes)

Johnny rolls a fair six-sided die six times and sums the cubes of all results. What is the expected value of this total?

Show Solution

Let \( X_i \) be the result of the \( i^\text{th} \) roll. We want \( E[T] = E[X_1^3 + X_2^3 + \cdots + X_6^3] \).

By linearity of expectation: \[ E[T] = 6 \, E[X^3]. \]

\[ \displaystyle E[X^3] = \frac{1^3+2^3+3^3+4^3+5^3+6^3}{6} \]
\[ = \frac{(1+2+3+4+5+6)^2}{6} = \frac{21^2}{6} = \frac{441}{6} = 73.5. \] Therefore E[T] = 6 times 73.5 = 441 -- the answer.

(This uses the identity \( \sum_{k=1}^{n} k^3 = \big(\sum_{k=1}^{n} k\big)^2 \).)

---

#30 – Powers and Estimation

Find the integer \( n \) such that \[ n^7 = 44{,}231{,}334{,}895{,}529. \]

Show Solution

Since \(80^7 \approx 2.1\times10^{13}\) and \(90^7 \approx 4.78\times10^{13}\), \(n\) is between 80 and 90. The last digit of \(n^7\) matches the last digit of \(n\), which is 9. Modulo-9 check also gives \(n \equiv 8 \pmod9\), so the only candidate is \(n=89\).

Indeed, \(89^7 = 44{,}231{,}334{,}895{,}529\). Thus \(n = \boxed{89}.\)

2021 spring AMC 12 B Reflection Notes from H

✅ Correct: 1–21

❌ Wrong / Unanswered:
#22 – Not sure of optimal strategy
#23 – I got stuck with cases
#24–25 – Didn’t attempt

2021 spring AMC 12 A Reflection Notes from H

2021 Spring AMC 12A — Reflection Notes

1–16 right, 15 time sink.
18, 19, 23 also right — review #19 solution.

Incorrect / Unanswered:
17, 20, 21, 22

#24, #25 → did not even look at (during the test)

At our lesson, we talked about solutions on all wrong, left blank except #25.

2016 AMC 12 B Reflection Notes from H

Wrong

#20

• no idea how to approach

#23

• Right approach
• 2 step away from right answer
• right answer → understood

#24

• review using video

#25

• tried to solve
• I liked linear recurrence / characteristic polynomial approach

2025 Mathcounts state more interested questions notes

target #4 : level 1.5 question, not hard at all. 

#5: number theory, learn mod or remainder (pattern) 

# 6 and #8 Good for AMC/ AIME preps as well. 

team notes : 

 #1: easy, just one line 

#2: symmetry and infinite geometric sequence

#3: mental math 

#4: more tedious, remainders (take more time) 

#5: there is a very fast method 

#6: elementary 

#7: use balanced method (similar to mass point) easy to make sillies if your answer is 82, not "84". CAREFUL ! ! 

 #8 : two methods 

#9 : You need to practice and see if you can solve it at a

timely manner, even if you have the right idea (for AMC

as well) 

 #10: more ambiguous question, harder to tell if you are "Really" right.

2016 AMC 12 A Reflection Notes from H

2016 AMC 12A Log

✅ Correct

Problems 1 → 23

❌ Wrong

Problems 24, 25

• Problem 24: Had the right idea but didn’t continue far enough.
• Problem 25: Didn’t understand the problem even after a video.

📘 Embedded Problems

Problem 24 (paraphrase)

There is a smallest positive real number a such that one can choose a positive real b making all roots of the cubic \(x^3 - a x^2 + b x - a\) real. For this minimal a, the corresponding b is unique. What is that value of b?

AoPS #24 (official)    AoPS video #24    Extra discussion

Problem 25 (paraphrase)

Let k be a positive integer. Bernardo writes perfect squares starting with the smallest having k + 1 digits; after each square, Silvia erases the last k digits of it. They continue until the final two numbers left on the board differ by at least 2. Let f(k) be the smallest positive integer that never appears on the board. Find the sum of the digits of \(f(2)+f(4)+f(6)+\cdots+f(2016)\).

Note from Mrs. Lin :  To understand this question more in details, try 

this video, starting at 24: 11. 

AoPS #25 (official)   AoPS video #25