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.

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!!

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 

2017 AMC 12 B Reflection Notes from H

2017 AMC 12B — Practice Log

Quick reflections and timing notes.

Q1–Q23: all correct ✅ Q22: time sink ⏳ Q24–Q25: not answered ❌

Notable Questions

• #13: Took me a while; need to keep practicing Burnside's lemma/technique. (note to self: revisit topic & drill)
• #22: Became a time sink. Pause sooner; sketch structure, estimate difficulty, and decide quickly whether to skip.
• #24: Didn’t understand the question—unclear how to set up the average. Re-read carefully; translate wording to variables first.
• #25: Ran out of time.

Process & Timing Notes

• Not enough time at the end—was able to draw the diagram but didn’t complete the setup.
• For average/setup questions: define variables immediately, write the equation before computing.
• When a problem starts ballooning (>3–4 minutes without structure), mark and move.

Follow-Up Plan

Thursday, September 11, 2025

Dimentional Change Questions III: Similar Shapes

There are numerous similar triangle questions on Mathcounts.

Here are the basics:











If two triangles are similar, their corresponding angles are congruent and their corresponding sides will have the same ratio or proportion.




Δ ABC and ΔDEF are similar. \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) = \(\frac{BC}{EF}\)= their height ratio = their perimeter ratio.







Once you know the linear ratio, you can just square the linear ratio to get the area ratio and cube the linear ratio to get the volume ratio. 




Practice Similarity of Triangles here.  Read the notes as well as work on the practice problems.  There is instant feedback online. 


Other practice sheets on Similar Triangles                                                        







Many students have trouble solving this problem when the two similar triangles are superimposed. 




Just make sure you are comparing smaller triangular base with larger triangular base and smaller triangular side with corresponding larger triangular side, etc... In this case:

\(\frac{BC}{DE}\)= \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)




Questions to ponder (Solutions below)




#1: Find the area ratio of Δ ABC to trapezoid BCDE to DEGF to FGIH. You can easily get those ratios using similar triangle properties. All the points are equally spaced and line \(\overline{BC}\)// \(\overline{DE}\) // \(\overline{FG}\) // \(\overline{HI}\). 






#2: Find the volume of the cone ABC to Frustum BCDE to DEGF to FGIH. Again, you can use the similar cone, dimensional change property to easily get those ratios.Same conditions as the previous question.



 










Answer key: 




#1:






 #2:

Monday, September 1, 2025

9/1/2025 Student Reflection Note from H

2018 AMC 12A Notes 🎉

😊 Only 2 Wrong — Great Job!

Wrong

• 22: Not sure how to express √abi cleanly.
      Couldn’t manage complex numbers or split the area into 4 pieces.
• 25: Was able to get the powers of 10 and simplify, but not the final casework step.

Right

15, 16, 17, 18, 19, 20, 21, 23, 24

8/31/2015 Student reflection notes from H

2019 AMC 12B Notes

Problems 20–25 → Wrong

• 20: Tried coordinate bash but didn’t realize AOBX is cyclic.
• 21: Very close, but made a mistake in casework.
• 22: No idea how to estimate.
• 24: Confusing — unsure how to approach.
• 25: Imaginary numbers solution was smart.
      Also homothety solution was smart too.

Problems 15–19, 23 → Right

Saturday, August 23, 2025

8/23/2025 student reflection notes from H ～ Welcome back to the States.

2019 AMC 12A – Reflection Log

Quick notes on misses, themes, and time sinks.

Incorrect

• #22. Problem and Solutions
• #24. Found all prime exceptions but missed 4 — Problem and Solutions
• #25. Got after hint → need work on angle chasing in cyclic quadrilaterals — Problem and Solutions 

Correct

Problems solved:

#15 #16 #17 #18 #19 #20 #21 #23

• #17 → learn Newton's Sums
• #23 → algebra took a long time.

Next Steps

• Drill Newton’s Sums identities; derive first 3–4 power sums using Vieta.
• Re-check prime-exception logic; include 2 and 4 in prime-exception checklist.
• Daily 10-min cyclic quadrilateral angle-chasing (use Ptolemy, equal angles, arc marks).
• Practice pacing to avoid #17-style time sinks; enforce a 2-minute “move on” rule.
• Algebra endurance reps (substitution, factoring patterns), especially for de-jangling #23-type problems.

Tuesday, August 19, 2025

8/19/2025 student reflection notes from H

2020 AMC 12B Reflections

Wrong Questions

• Q19: Not sure how to approach. My casework method was too complicated.
• Q21: Solution made sense. I made the substitution u = 70n + 50, but couldn’t expand it properly.
• Q23: Saw that n = 2, n = 3 worked, but I couldn’t eliminate later cases of n ≥ 4.
• Q25: Solution seemed easy, but I didn’t see the graphical method earlier. I have to get better at identifying that.

Correct Questions

• Q15: Solved correctly.
• Q16: Solved correctly.
• Q17: Solved correctly.
• Q18: Solved correctly.
• Q20: Solved correctly.
• Q22: Solved correctly.
• Q24: Solved correctly.

Friday, August 15, 2025

8/15/2025 2020 AMC 12 A student reflection notes from H while in India

2020 AMC 12A Log

Wrong

• Q18 – Tried other equation → could not solve
• Q23 – Hard casework (I could guess if correct, more of a check)
• Q24 – Weird set, specialized translations in geometric problem
• Q25 – Really hard Q. Required graphing + adaptation for 2π trap but not confident/solid

Right

• Q15
• Q16
• Q17
• Q19
• Q20
• Q21
• Q22

Thursday, August 14, 2025

8/14/2025 SAT/AMC 12 student reflection notes from Ay

1) Reviewed the attached problems and solutions 2) Reviewed the SAT vocabulary words and tried the SAT questions. I was able to answer all three questions correctly. 3) Attempted the 2023 AMC 12 A. I skipped these questions- #15, #21, #22, #24 and #25 Rest all questions were answered correctly.

Thursday, August 7, 2025

8/6/2025 2014 AMC 12 B student reflection notes from H while in India :)

2014 AMC 12B Log

Correct:

• 15
• 16
• 17
• 18 – took time
• 19
• 20
• 22 – time-sink
• 24
• 25

Wrong:

• 21 – struggled with too many variables
• 23 – no idea how to approach

Sunday, August 3, 2025

7/27/2025 2023 Mathcounts state sprint student reflection notes An

2023 Mathcounts State Sprint: 

Q13 Rate silly mistake made miscalculation when I was trying to find the rates always double check if the rates if it make sense with the question 

Q21 number theory had an idea but it didn't work because I didn't count all the ordered triplets
Learn how to do these problems fast and accurately 

Q22 probability didn't know how to do it 

Learn how to do these problems fast and accurately 

Q23 Algebra didn't know how to do it
Learn how to do these problems fast and accurately 

Q25 number Theory didn't know how to do it

Learn how to do these problems fast and accurately 

Q26 number Theory didn't know how to do it

Learn how to do these problems fast and accurately 

Q27 Algebra didn't know how to do it

Learn how to do these problems fast and accurately 

Q28 Geometry didn't know how to do it

Learn how to do these problems fast and accurately 

Q29 Algebra didn't know how to do it

Learn how to do these problems fast and accurately 

Q30 Geometry didn't know how to do it

Learn how to do these problems fast and accurately

Friday, August 1, 2025

8/1/2025 2014 AMC 12 A student reflection notes from H

2014 AMC 12A — Log

Incorrect Problems

Problem 20

• The lengths are totally random
• Why are they rotated?

Problem 23

• No comment

Problem 25

• Weird parabola
• Not sure how to solve

Correct Problems

• Problem 15
• Problem 16
• Problem 17
• Problem 18 — Took time
• Problem 19
• Problem 21 — Hard b/c of endpoint → took a while; Quick inside a hard transformation
• Problem 22 — Took time
• Problem 24 — Graphed it to solve

Thursday, July 31, 2025

7/31/2025 2024 AMC 12B student reflection notes from Ay

1) Reviewed the SAT vocabulary words 2) Reviewed the attached problems and solutions 3) Attempted the 2024 AMC 12 B. I attempted 20 questions out of 25 and skipped these: #15, #18( very easy and I could solve it later), #21, #23, #24 Out of the questions attempted, I got #22 wrong but later realized it was a careless mistake and I could solve it. Rest all questions were answered correctly- #1 to #14, #16, #17, #19 , #20 and #25

Monday, July 28, 2025

7/28/2025 2013 AMC 12 B student reflection notes from H

2013 AMC 12B LOG

Incorrect Problems

• Problem 23 – No idea how to start
• Problem 24 – Almost got it through angle chasing
• Problem 25 – Very close, right idea but missed a few cases

Correct Problems

Problems solved correctly: 15, 16, 17, 18, 19, 20, 21, 22

• Problem 17 – Should learn Cauchy-Schwarz
• Problem 20 – Ran out of time

Friday, July 25, 2025

7/25/2025 H 2013 12 A reflection notes

2013 AMC 12A Log

Incorrect Problems

• Problem 20 – Totally confused
• Problem 24 – Right idea but casework had too many cases
• Problem 25 – No idea how to start

Correct Problems

Problems solved correctly: 15, 16, 17, 18, 19, 21, 22, 23

• Problem 16 – Took too long
• Problem 23 – Angle chasing was hard
• Had to look at answer choices

Friday, July 18, 2025

7/18/25 H 2012 AMC 12 B reflection notes

17 wrong confused

20 wrong, time sink

21 wrong 

22 right of hands

23 wrong understood solution

24 wrong time sink casework

25 easy casework, right 

Wednesday, July 16, 2025

Similar to 2023 Mathcounts chapter sprint #30, but harder (level 2)

This question is similar to, but more difficult than the 2023 Mathcounts Chapter Sprint #30, which is as follows:

Consider the seven points on the circle shown. If George draws line segments connecting pairs of points so that each point is connected to exactly two other points, what is the probability that the resulting figure is a convex heptagon? Express your answer as a common fraction.

Try the question first before you read the solution down below.

Convex Heptagon Probability (7 points on a circle)

Label the points 1 – 7 clockwise around the circle. Each point must be joined to exactly two others, so the drawing is a 2-regular graph (a disjoint union of cycles).

1 . Count all edge–sets (2-regular graphs) on 7 labelled points

• One 7-cycle.
  Fix the cyclic order: the edges of a 7-cycle correspond to a permutation of the 6 points after 1, with direction ignored. Hence
  \( \dfrac{6!}{2}=360 \) distinct 7-cycles.
• One 3-cycle + one 4-cycle.
  
  1. Choose the 3-cycle: \( \binom{7}{3}=35 \).
  2. Orient the 3-cycle: \( (3-1)!/2 = 1 \) way.
  3. Orient the remaining 4-cycle: \( (4-1)!/2 = 3 \) ways.
  Total  \( 35 \times 1 \times 3 = 105 \) edge–sets.

Adding the two cases gives the total number of admissible drawings: \[ N_{\text{total}} = 360 + 105 = 465 . \]

2 . Count the favourable edge–set

Exactly one of those drawings is the perimeter \(1\!-\!2\!-\!3\!-\!4\!-\!5\!-\!6\!-\!7\!-\!1\), which yields a convex heptagon.

3 . Probability

\[ \boxed{\displaystyle \Pr(\text{convex heptagon})=\frac{1}{465}} \]

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Checks: the same counting method gives \(70\) total for 6 points (hexagon) and \(3507\) total for 8 points (octagon), agreeing with \(1/70\) and \(1/3507\) respectively.

Saturday, July 12, 2025

7/12/2025 2012 AMC 12A H reflection notes

2012 AMC 12 A Reflection Notes

Correct Answers: Questions 15–21

• Questions: 15, 16, 17, 18, 19, 20, 21
• Note on #16: Took time on one question using the Law of Cosines

Unanswered but Attempted: Questions 22, 23

• Q22 & Q23:
• Able to solve with help from solution
• Need to improve:
  • Organized counting of objects
  • 3D visualization skills
• Q23 Additional Note:
• Not able to solve during the test
• Watched a video solution afterward—it made sense, but I wouldn’t have made that connection under time pressure

Unanswered: Questions 24, 25

• Q24:
• Not able to compare exponents correctly
• Had to look at the solution to understand
• It was a complicated question
• Q25:
• Unable to solve even after reviewing the solution
• Gaps in understanding—don’t think I could have arrived at that solution myself
• Struggled with 3D visualization and drawing the necessary graph

Tuesday, July 8, 2025

7/8/2025 2011 AMC 12B H reflection notes

Q16 → Wrong due to silly mistake

Skipped Q20

• Time-sink solution requires heavy observation and quick identification.
• Involves quick use of inscribed angles and arcs.

Q25 → Not enough time

Was not able to solve; did not understand substitution of
n → n-k in the solution + solution video.
Redo!

Q24 Notes:

• Easy algebra but looked difficult at first.
• Just required extra time to complete — able to solve afterwards.

Q15, Q17, Q18, Q19, Q21, Q22, Q23 → Correct

• Q17: Online solution took a long time
• I had the fastest solution : the solution was first finding g(f(x)) = 10x-1, then subbing in the 1 to get h_1(1)=9, then continuously subbing 9 back into 10x-1 so it becomes 9, 89, 899, 8999….
• Q19: Took a long time
• Found faster solution but took time to get to it: this was a least upper bound question for the slope which hits the next lattice point. If I had the answers, it would have been easier to substitute them back into and find which one would work, but I had hidden the answers. What I did next was find a few of the closest points to 102, 50 which the slope would first intersect and the slope would be as close to 1/2 as possible. After testing a few values in the form of (n/1)/(2n+1) and (n+1)/(2n) which give values close to 1/2, 50/99 was the least of these.

Expected Value Question : 2024 Mathcounts State Sprint #24 level 1.5

Problem. Dennis rolls three fair six-sided dice, obtaining a, b, c ∈ {1,…,6}. Find \[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr]. \]

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Try the question first before scrolling down to read the solution. 

Solution.

Step 1 — Linearity of expectation.

\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] \;=\; \mathbb{E}[|a-b|]+\mathbb{E}[|b-c|]+\mathbb{E}[|c-a|] \;=\; 3\,\mathbb{E}[|a-b|]. \]

Step 2 — Expected absolute difference of two dice.

Let \(X = |a-b|\). Its distribution is

\[ \begin{array}{c|cccccc} d & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \Pr(X=d) & \tfrac{6}{36} & \tfrac{10}{36} & \tfrac{8}{36} & \tfrac{6}{36} & \tfrac{4}{36} & \tfrac{2}{36} \end{array} \] \[ \mathbb{E}[|a-b|] =\frac{1}{36}\bigl(0\cdot6 + 1\cdot10 + 2\cdot8 + 3\cdot6 + 4\cdot4 + 5\cdot2\bigr) =\frac{70}{36} =\frac{35}{18}. \]

Step 3 — Final answer.

\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] = 3 \times \frac{35}{18} = \boxed{\tfrac{35}{6}}. \]