# Be kind for 2020 (bkf2020)'s Math log

## What is this?

These are math problems I am trying out this week. I'll include notes. My goal is to seriously attempt/solve ALL of the problems (as of Jan 02, 2022). This starts the week of Jan 03, 2022 to Jan 09, 2022. It's fine if I can't solve all the problems.

Possible status for each problem: Not started, Started, Attempted Seriously, Solved

## Week of Apr 25, 2022 to May 01, 2022

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2022 AIME II Problem 13 Solved My solution was similar to Solution 2 on the AoPS wiki.
2020 AIME I Problem 12 Solved My solution was similar to the MAA solution.

I managed to find that 32 and 75 divide the final number, but only knew that 54, 2 × 54, or 4 × 54 divided the final number. I didn't notice to take 149n - 2n mod 5 to then prove that 4 divides the final number.

Another mistake I made was writing 76 in the final number instead of 75 for some reason...

2020 AIME II Problem 12 Not started

## Week of Apr 18, 2022 to Apr 24, 2022

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2022 AIME II Problem 12 Solved My solution was almost the same as the one on AoPS wiki. I made the mistake of writing 2a + 2b as the answer and NOT a + b.
2022 AIME II Problem 13 Not started

## Week of Apr 11, 2022 to Apr 17, 2022

No new AIME problems. Did some medium/hard AIME problems for Semicolon math.

## Week of Apr 04, 2022 to Apr 10, 2022

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2022 AIME II Problem 9 Solved Got 245 originally because my method counted B5A7 as creating a new region when it doesn't.
2022 AIME II Problem 10 Solved
2022 AIME II Problem 11 Solved Didn't compute `(6sqrt(5))^2` correctly the first try.
2022 AIME II Problem 12 Not started

## Week of Mar 28, 2022 to Apr 03, 2022

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2022 AIME II Problem 9 Not started
2022 AIME II Problem 10 Not started
2022 AIME II Problem 11 Not started
2022 AIME II Problem 12 Not started

## Week of Mar 07, 2022 to Mar 13, 2022

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2022 AIME I Problem 12 Not started
2022 AIME I Problem 13 Not started
2022 AIME I Problem 14 Not started
2022 AIME II Problem 6 Solved
2022 AIME II Problem 7 Solved
2022 AIME II Problem 8 Solved I made the mistake of assuming that you can write n as `4 * floor(n / 4) + k1`. The solution I used for this problem isn't too similar to any of the solutions on the AoPS wiki, but I don't feel like adding commentary here currently.

## Week of Feb 28, 2022 to March 06, 2022

Sorry that I didn't make any progress this week.

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2022 AIME I Problem 12 Not started
2022 AIME I Problem 13 Not started
2022 AIME I Problem 14 Not started
2022 AIME II Problem 6 Not started
2022 AIME II Problem 7 Not started
2022 AIME II Problem 8 Not started

## Week of Feb 21, 2022 to Feb 27, 2022

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2020 AIME II Problem 12 Started

Let O be the center of the square with label 2000 and ABCD be the square with label 1099. Let P be the center of the square with label 200. Then one of these two cases can occur:

``````
P
A--B
|  |
C--D
O
``````
``````
P
A--B
|  |
C--D
O
``````
In the first case, we need the slope of OP to be between the slopes of OA and OD in order for OP to intersect the square. Similarly, in the second case, we need the slope of OP to be between the slopes of OB and OC. This is one fact I noticed.

Another fact I noticed was that ceil(1099/n) gave the row the square with label 1099 should be in, and this may be used with the first fact to calculate the slopes, but that would be very messy. Need to try more.

2020 AIME II Problem 13 Solved Solved with a bashy solution. Got 2arctan(2/r) + arctan(3/r) + 2arctan(4/r) = 180. Then, found this in terms of r using tangent sum of angle formulas and solved for the value or r. Might spend some time finding cleaner solutions.
2022 AIME I Problem 4 Solved See LaTeX commentary here.
2022 AIME I Problem 5 Solved My solution was similar to Solution 2 on the AoPS wiki. I calculated D/2 using similar triangles, but forgot to multiply by 2 to get D.
2022 AIME I Problem 6 Solved On the test, I forgot the case when a = 20. When redoing this problem, I forgot the case where a = 7 and b = 9 resulting the arithmetic sequence 3, 5, 7, 9. I need to be more organized with casework.
2022 AIME I Problem 7 Solved On the test, I believed abc - def = 2 was the smallest possible value for the numerator, but abc - def = 1 is possible. The two numbers were abc = 36 and def = 35. This solution makes it more convincing that abc = 36 and def = 35 result in the best answer.
2022 AIME I Problem 10 Solved My solution was similar to Solution 1 on the AoPS wiki, but I made a lot of computational mistakes until I got the right answer.

## Week of Feb 14, 2022 to Feb 20, 2022

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2020 AIME I Problem 11 Solved My solution was similar to the official MAA solution, but I made a mistake by forgetting to add the case where a = 1 (even though I counted it). In the future, I need to organize casework in a more organized fashion, so I don't miss cases...
2020 AIME I Problem 12 Started

I thought of n being the multiples of φ(3^3), φ(5^5), and φ(7^7). (See Euler's Totient function.) However, n can be smaller, as shown by manual calculation for 3^3.

To find the value of n so 2^n was congruent to 149^n mod 3^3 = 27, I found the value of n so 2^n was congruent to 14^n mod 27. I then multiplied by the inverse of 2^n (since the inverse of 2 exists mod 27) to get 7^n congruent to 1 mod 27. The smallest value of n should divide φ(3^3) = 18, and I found it to be 9. So n should be a multiple of 9.

For 5^5 and 7^7, I thought of finding the inverses, but that isn't as easy.

2020 AIME I Problem 13 Solved My solution was similar to MAA solution 2. One mistake I made was believing EFI and IBC were congruent, but checking the wrong angles: FEI and IBC. Another mistake I made was calculating EF as IM/r (see the solution text for context) instead of BC * IM/r. One thing I learned from this problem was to consider the ratio of similar triangles as the ratio of heights (it took me a while to realize that I should use this fact).
2020 AIME II Problem 12 Not started
2020 AIME II Problem 13 Not started

## Week of Feb 07, 2022 to Feb 13, 2022

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2020 AIME II Problem 9 Started My original solution was wrong because
1. it assumed the people sitting together have to be adjacent, and
2. the people sitting together have a fixed position, but they can move around.
I used the official MAA solution and I completed it by hand as well. But, I think I could have tried harder, so I'm just marking this as 'started'.
2020 AIME II Problem 10 Started

My original solution was wrong, because it assumed 4 would always have a modulo inverse mod `n + 5`, but this may not always be true. So, I glanced at MAA solution 1 and completed a similiar solution.

But, this new approach was wrong because it forgot to test the values of n. Just because `n^2(n+1)^2` leaves a remainder of 68 mod `n + 5` does NOT mean `n^2(n+1)^2/4` leaves a remainder of 17 mod ` n + 5`. Also, I forgot that we needed `n + 5 > 17`, but this did not matter here, and checking the values of n would be an additional measure against this mistake.

Lastly, checking the values of n wasn't as tedious as it looked, but I should be careful here, since the computation isn't super trivial.

Leaving it as 'started' because I could have thought more.

2020 AIME II Problem 11 Solved

Avoided getting stuck on one approach for this problem. I believe I noticed that I wasn't using the fact that the entire P(x) polynomial was given, so I wrote

• `P(x) = x^2 - 3x - 7`
• `Q(x) = x^2 - ax + 2`
• `R(x) = x^2 - bx + c`
and found the corresponding values of `P + Q`, `P + R`, and `Q + R`.

Then, I used this fact from the problem:

• `P + Q = 2(x - p)(x - q)`
• `P + R = 2(x - p)(x - r)`
• `Q + R = 2(x - q)(x - r).`
This is because pairs of `P + Q`, `P + R`, and `Q + R` share the same roots, but these common roots are distinct.

After that, I found `p + q, pq, p + r, pr, q + r, qr` in terms of `a, b, c`, and so on...

The resulting solution was similiar to Solution 1 on the AoPS wiki.

2020 AIME I Problem 10 Started Originally, I assumed that n and m just couldn't contain factors of 2, 3, 5, or 7. So, I set n = 121 and m = 143, but gcd(n + m, 210) was not 1. I checked the AoPS wiki solution and believe I got the hint for what went wrong simply by glancing over solution 1. Then, I fixed my solution by making m even: m = 2 * 143 = 286. Solution 3 (official MAA) is the one is used to get a more rigorous solution to this problem. Marking as 'started' since part of the solution got spoiled.
2020 AIME I Problem 11 Not started

## Week of Jan 31, 2022 to Feb 06, 2022

Only did one problem this week, will try to do more next week.

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2021 AIME II #12 Solved Likely got motivated by the solution of 2021 AIME I #11 (similiar problem)

## Week of Jan 24, 2022 to Jan 30, 2022

Note: I have problems to do for a computer science class, and so school work, so I may have less time...

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2021 AIME II Problem 10 Started Assumed the planes must intersect the top of the spheres, but this would mean only one plane could exist. This led to the wrong answer...
2020 AIME I Problem 9 Solved My solution was similiar to solution 2 on the AoPS wiki.

## Week of Jan 17, 2022 to Jan 23, 2022

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2021 AIME II Problem 10 Not started
2020 AIME I Problem 6 Solved
2020 AIME I Problem 7 Solved
2020 AIME I Problem 8 Solved Originally, my solution was similar to Solution 4 (Official MAA 1) on the AoPS wiki, but I made many mistakes, so it took a while. I think I didn't compute everything in the most efficient way possible. Thus, I looked at Solution 5.
2020 AIME I Problem 9 Not started

## Week of Jan 10, 2022 to Jan 16, 2022

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2021 AIME II Problem 8 Solved

My solution let Pi, T be the probability of getting the desired result if the ant is at a vertex of the top face after i moves AND i - 1 moves. Similarly, let Pi, B be the probability of getting the desired result if the ant is at a vertex of the bottom face after i moves AND i - 1 moves.

This meant Pi, T = ½ Pi + 1, T + ½ Pi + 2, B, because the ant can either stay on the top face, or it moves to the bottom face and stays at the bottom face for an additional move. Similarly, Pi, B = ½ Pi + 1, B + ½ Pi + 2, T.

Using this, I calculated the values of Pi,B and Pi,T for 1 ≤ i ≤ 8. Note P8, B = 0 and P8, T = 1.

The answer was 2/3 * P1, B + 1/3 * P2, T, but I got 1/2 * P1, B + 1/2 * P2, T, since I thought the ant only had 2 moves when it started, but it actually has 3 moves.

I realized that Pi, T + Pi, B = 1 and this can be proved using recursion. This may motivate the idea behind solution 3 on the AoPS wiki. That solution is pretty similar to the solution I outlined here.

2021 AIME II Problem 10 Not started
2021 AIME II Problem 11 Solved My solution was similar to solution 2 on the AoPS wiki.

## Week of Jan 03, 2022 to Jan 09, 2022

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Reflection: I tried doing all these problems on the weekend, and I think that is ineffective. I should have spaced out some time over the week to think about these problems.

2021 AIME II Problem 8 Not started
2021 AIME I Problem 10 Solved My solution was similiar to Solution 1.
2021 AIME I Problem 11 Solved

First, I was skeptical that ABA1B1 was a convex quadrilateral. However, let the intersection of AC and BD be E. Then, angle AEB is (arc AB + arc CD) / 2 while angle BEC is (arc BC + arc AD) / 2. Arc AB is less than arc BC, and arc CD is less than arc AD as AB < BC and CD < AD. This means angle AEB is less than angle BEC. That makes angle AEB acute, and it seems arc AD is less than 180, so triangle AEB is also acute. As a result, ABA1B1 is a convex quadrilateral.

In the future, I should draw the circle first and then draw ABCD.

After that, I wasn't so sure on where to continue, so I looked at Solution 1 and Solution 1.1. I didn't know how to come up with such a solution, so I tried coming up with my own solution. I did some angle chasing and got where you would come up with solution 1. The mistake I made was getting triangle EC1D1 similiar to triangle EDC instead of triangle EC1D1 similiar to triangle ECD. Then, I understood where you would come up with solution 1.1 by eliminating A1, B1, C1, and D1 and trying law of cosines since the absolute values of the cosines of angles AEB, BEC, CED, and AED are the same.

## Week of Dec 27, 2021 to Jan 02, 2022

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2021 AIME II Problem 6 Solved
2021 AIME II Problem 7 Solved
2021 AIME II Problem 8 Not started
2021 AIME II Problem 9 Solved Had to use the solution for this problem. I'm not fully sure how one would come up with this solution. Additionally, `gcd(2^a - 1, 2^b - 1) = 2^(gcd(a, b)) - 1` seemed to be true, but I wasn't sure why, so I found this proof on Stackexchange.
2021 AIME I Problem 10 Not started
2021 AIME I Problem 11 Not started

## Week of Dec 20, 2021 to Dec 26, 2021

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2021 AIME I Problem 6 Solved My solution was similar to Solution 1 on the AoPS wiki. I think setting PA = x^2 + y^2 + z^2 would make computation easier, and dividing everything by 12 is important.
2021 AIME I Problem 7 Solved My solution ended up being similar to Solution 2. The mistake I made was forgetting to multiply m and n by powers of 2. I got that m and n should be in the sets {1, 5, 9, 13, 17, 21, 25, 29} and {3, 7, 11, 15, 19, 23, 27}, but I didn't include the sets {2, 10, 18, 26} and so on.
2021 AIME I Problem 8 Solved I used solution 1 for this problem. Originally, I tried finding the number of solutions total for these four equations: x^2 - 20|x| ± c ± 21 = 0. The problem was that I thought if there were two real solutions to something like x^2 - 20x - c - 21 = 0, then there will be four real solutions to x^2 - 20|x| - c - 21.
2021 AIME I Problem 9 Solved Couldn't come up with a clean solution so I used the diagram on the AoPS wiki and Solution 2 on the AoPS wiki. I didn't notice to find the area of triangle ABC or triangle ABD to get a clean equation.
2021 AIME I Problem 10 Not started
2021 AIME I Problem 11 Not started

## Week of Dec 13, 2021 to Dec 19, 2021

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2021 AMC 10A #23 Not started
2021 AMC 10A #25 Not started
2021 AMC 10B Fall #19 Solved The method I used was similiar Solution 1 on Art Of Problem Solving. I think I came up with this because I wanted to express N in a nicer form, and using 7 * (10³¹³-1)/9 is what I came up with.
2021 AMC 10B Fall #20 Solved Tried thinking of something similiar to Solution 2, but I forgot to realize that the probability a player wins is 1 over the number of players. I like Solution 1 because it is simplier, but I didn't realize that the answer is simply the probability the highest roll in the first round is 5.

## Week of Dec 06, 2021 to Dec 12, 2021

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2021 AMC 10B Fall Problem 18 Solved

Previously: Tried solution 2 as it was similiar to what I did on the best, but needs lots of computation. Will try to find a cleaner way online. Also may add old problems again as practice.

Currently: Found clean and simple solution. I think the solutions on the wiki are much more complicated than this.

2021 AMC 10A Problem 23 Not started
2021 AMC 10A Problem 25 Not started
2021 AMC 10A Fall Problem 22 Solved I used this solution on the AoPS wiki. I did not realize to take cross sections using triangles. I only tried doing cross sections using circles. Additionally, when trying this solution out for my own, I didn't realize to use angle bisector on triangle AEF to find OE and OF in terms of r.

## Week of Nov 29, 2021 - Dec 05, 2021

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2021 AMC 10A Problem 23 Not started
2021 AMC 10A Problem 25 Not started
2021 AMC 10A Fall Problem 22 Not started
2021 AMC 10B Fall Problem 15 Solved Tried using similiar triangles but didn't notice that APB is congruent to BQC. Used this solution
2021 AMC 10B Fall Problem 17 Solved Originally didn't do this problem during the test. Solution I came up with is pretty simple, and I need more practice.
Click to view solution The angle between (-1, 4) and (4, 1) is 90. Let the angle between P(-1, 4) and P' be 2θ. Then, the angle between P' and P'' is 90 - 2θ. The angle between P and l is θ and the angle between P'' and m is 45 - θ This makes the angle between l and m 45. Note that the midpoint of the points (1, 5) and (5, -1) lies on m. Note (5, -1) lies on l rotated 90 degrees CLOCKWISE.
2021 AMC 10B Fall Problem 18 Seriously attempted Tried solution 2 as it was similiar to what I did on the best, but needs lots of computation. Will try to find a cleaner way online.

## Week of Nov 22, 2021 - Nov 28, 2021

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2021 AMC 10A Fall Problem 18 Solved Once picking the crop in the top left square, I did casework on what the remaining crops could be. This made my answer 4(4+4+3+6+4) and I added it wrong. Got 4(19) and NOT 4(21). Took a while to find this mistake...
2021 AMC 10A Problem 23 Not started
2021 AMC 10A Problem 24 Solved Used Solution 1.1 for this problem. Took a while to solve because I was trying to find a more clean way, but I should have just used this method.
2021 AMC 10A Problem 25 Not started

## Week of Nov 15, 2021 - Nov 21, 2021

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2021 AMC 10A Fall Problem 17 Solved Found the equation of the solar panel as a plane. See this clean solution. However, if A'B'C'D'E'F' represents the new hexagon, I'm not sure if A'D', B'E' and C'F' will intersect at one point, so the solution isn't fully clear to me.
2021 AMC 10A Fall Problem 18 Not started
2021 AMC 10A Fall Problem 19 Solved Took a while to find the area of 2A in terms of s. Also solved the equation wrong. At this step: 4pi + 8s = 2pi + 16s - 40, I wrote 2pi + 40 = 4s INSTEAD of 2pi + 40 = 80s.
2021 AMC 10A Fall Problem 20 Solved
2021 AMC 10A Fall Problem 21 Solved When computing p, I got C(20,3) * C(17,5) * C(12,4) * C(8,4) * 10 for the numerator. It should be 20 = 4 * 5 not 10, since we choose the positions for the bin with 5 balls and the bin with 4 balls and the order matters.
2021 AMC 10A Problem 23 Not started
2021 AMC 10A Problem 24 Not started
2021 AMC 10A Problem 25 Not started

## Week of Nov 08, 2021 - Nov 14, 2021

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2021 AMC 10A Problem 20 Solved

Using a hint from someone, I found a1 > a2 < a3 > a4 < a5 or a1 < a2 > a3 < a4 > a5. However, I got the wrong answer 24. In the first case, I thought that a1, a3, and a5 must be only 3, 4 or 5, and a2 and a4 must be only 1 or 2. In the second case, I thought that a1, a3, and a5 must be only 1, 2, or 3, and a2 and a4 must be only 4 or 5.

However, I did not consider in the first case, a1 = 2, a2 = 1 OR a4 = 1, a5 = 2. In the second case, a1 = 4, a2 = 5 OR a4 = 5, a5 = 4. This adds 8 possible cases.

2021 AMC 10A Problem 21 Solved
2021 AMC 10A Problem 22 Solved Took 13 minutes to solve; maybe will come up with a faster solution.
2021 AMC 10A Problem 23 Not started
2021 AMC 10A Problem 24 Not started
2021 AMC 10A Problem 25 Not started
2019 AMC 10B Problem 17 Solved
2019 AMC 10B Problem 18 Solved
2019 AMC 10B Problem 19 Solved Got the wrong answer 119. Knew all possible numbers are in form 2^a * 5^b where a and b are between 0 and 10 (inclusive). Subtracted cases where (a, b) = (0, 0) and (10, 10). Forgot (a, b) = (0, 10) and (10, 0).
2019 AMC 10B Problem 20 Solved
2019 AMC 10B Problem 21 Solved
2019 AMC 10B Problem 22 Solved Originally, I had the idea doing this problem by considering the probablility of getting the 2019 ring (where everyone gets \$1) from certain states and a certain number of moves. The two states were 1-1-1 or 2-1-0. Reordering is included. When I tried getting the correct solution (#4 on AoPS wiki), I forgot to make sure all possible ways to transfer money were considered.
2021 AMC 10A Fall Problem 16 Solved I wrote the commentary in a pdf LaTeX file. I think I will do this for SOME problems in the future. View the file here. You can see the source code at `latex/2021_AMC10A_Fall_16.latex` in the website repo.
2021 AMC 10A Fall Problem 17 Not started