# 2013 AMC 12 B #18

Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove 2 or 4 coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove 1 or 3 coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with 2013 coins and when the game starts with 2014 coins?

# 2021 AIME I #10

I can tell when students are ramping up their preparation for AMC because my solution videos get a lot more views. This one has been receiving a lot of views lately.

Consider the sequence $(a_k)_{k \ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k \ge 1$, if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$, then $a_{k+1} = \frac{m+18}{n+19}$. Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$.

# Introduction to the AMC

Here’s a quick introduction to the American Mathematics Competitions, with particular attention to qualifying for AIME. (Can you spot my acronym error?)

# Get ready for the AMC 10/12: What you need to know now

1. Date November 10 and 16. (The AMC 8 is pushed back to January.) The MAA is offering paper and online formats.
2. Location Check with your school. If you are homeschooled or your school doesn’t host the AMC, check the MAA zip code search for a site near you. You’ll need to register with the local contact, not with the MAA.
3. Prep Don’t overthink this. Take old exams and study the problems you couldn’t solve.
4. Your goal Always to get to the next level. If you are new to the AMC, then your ultimate goal is to qualify for AIME. If you’re already AIME-qualified, then look to USAMO. Don’t bother with trying to get a perfect score. Move forward to more difficult problems. Each tier means greater achievement and greater prestige.
5. Strategy This isn’t the exam where with 5 minutes left you bubble in all the remaining answers. You’ll lose points you could have earned for each blank answer. Write a clean exam: all answers are correct or left blank.

# My Interview with AoPS

AoPS recently interviewed me about my thoughts on math education and math contests. I shared a number of unconventional ideas around math education, including:

• Math isn’t always fun. Learning often feels awkward and uncomfortable. If our mantra is “Learning is Fun!” what message do we send, when it clearly isn’t?
• Using math creatively means students need to know why their algorithms work. Otherwise, it’s like telling a student how use red paint or blue paint, but never showing them how they combine to make purple.
• Studying a math curriculum without participating in a contest is like going to soccer practice every day without ever playing a game.
• Math contest rules are not the boss of you.
• Normalize frustration. Did I mention that learning isn’t fun?

# 2021 AIME I #9

Let $ABCD$ be an isosceles trapezoid with $AD = BC$ and $AB. Suppose that the distances from $A$ to lines $BC, CD$, and $BD$ are $15, 18,$ and $10$, respectively. Let $K$ be the area of $ABCD$. Find $\sqrt{2} \cdot K$.

# 2021 AIME I #8

(Note: I have the wrong problem number written in the video whiteboard.)

Find the number of integers $c$ such that the equation $||20|x|-x^2|-c|=21$ has $12$ distinct solutions.

# 2021 AIME I #7

Find the number of pairs $(m,n)$ of positive integers with $1 \le m < n \le 30$ such that there exists a real number $x$ satisfying $\sin(mx)+\sin(nx)=2$.

# 2021 AIME #5

Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

# 2021 AIME I #4

Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than the third pile.