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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Problem Link  Status  Additional Notes 

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 3^{2} and 7^{5} divide the final number, but only knew that 5^{4}, 2 × 5^{4}, or 4 × 5^{4} divided the final number. I didn't notice to take 149^{n}  2^{n} mod 5 to then prove that 4 divides the final number. Another mistake I made was writing 7^{6} in the final number instead of 7^{5} for some reason... 
2020 AIME II Problem 12  Not started 
Week of Apr 18, 2022 to Apr 24, 2022
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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

2022 AIME II Problem 9  Solved  Got 245 originally because my method counted B_{5}A_{7} 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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

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) + k_{1} .
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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

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:
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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

2020 AIME II Problem 9  Started 
My original solution was wrong because

2020 AIME II Problem 10  Started 
My original solution was wrong, because it assumed 4 would always have a modulo inverse
mod
But, this new approach was wrong because it forgot to test the values of n.
Just because 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 + Q , P + R ,
and Q + R .
Then, I used this fact from the problem:
P + Q , P + R ,
and Q + R share the same roots, but these common roots are distinct.
After that, I found 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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

2021 AIME II Problem 8  Solved 
My solution let P_{i, 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 P_{i, 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 P_{i, T} = ½ P_{i + 1, T} + ½ P_{i + 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, P_{i, B} = ½ P_{i + 1, B} + ½ P_{i + 2, T}. Using this, I calculated the values of P_{i,B} and P_{i,T} for 1 ≤ i ≤ 8. Note P_{8, B} = 0 and P_{8, T} = 1. The answer was 2/3 * P_{1, B} + 1/3 * P_{2, T}, but I got 1/2 * P_{1, B} + 1/2 * P_{2, T}, since I thought the ant only had 2 moves when it started, but it actually has 3 moves. I realized that P_{i, T} + P_{i, 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.
Problem Link  Status  Additional Notes 

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 ABA_{1}B_{1} 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, ABA_{1}B_{1} 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 EC_{1}D_{1} similiar to triangle EDC instead of triangle EC_{1}D_{1} similiar to triangle ECD. Then, I understood where you would come up with solution 1.1 by eliminating A_{1}, B_{1}, C_{1}, and D_{1} 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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

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  20x ± 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  20x  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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Problem Link  Status  Additional Notes 

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 * (10³¹³1)/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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

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 solutionThe 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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

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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Problem Link  Status  Additional Notes 

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 111 or 210. 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 