# 数学竞赛代考|AIME代考美国数学邀请赛|PROBLEM

AIME资格认证的变化

AMC办公室将从2月下旬开始向学校邮寄2012年AMC 10和2012年AMC 12报告，并持续到3月初至3月中旬。 在该AMC 10和AMC 12报告中，将列出学校的AIME合格者名单。

AIME的目的是在AMC10或AMC12之外，为北美许多具有特殊数学能力的高中生提供进一步的挑战和认可。得分最高的美国公民和合法居住在美国和加拿大的学生（根据加权平均分，获得合格分数）被邀请参加美国数学竞赛。
AIME（美国数学邀请考试）是介于AMC10或AMC12和USAMO之间的考试。所有参加AMC 12的学生，如果在可能的150分中取得100分或以上的成绩，或在前5%的学生被邀请参加AIME考试。所有参加AMC 10的学生，在可能的150分中取得120分或以上，或进入前2.5%的学生也有资格参加AIME。本学年AIME I的日期为 ，AIME II的日期为 ， 。美国数学邀请考试没有额外的注册费，除非你选择参加第二次考试。额外的管理/运输费是要收取的，前10名学生的最低费用为，超过10名学生的最低费用为。这在AMC 10/12和AIME教师手册中有更详细的解释。

## 数学竞赛代考|AIME代考美国数学邀请赛|2022 AIME II Problems

Problem
Adults made up $\frac{5}{12}$ of the crowd of people at a concert. After a bus carrying 50 more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.
Solution 1
Let $x$ be the number of people at the party before the bus arrives. We know that $x \equiv 0(\bmod 12)$, as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x+50 \equiv 0(\bmod 25)$, as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x+50 \equiv 0(\bmod 25)$ can be reduced to $x \equiv 0(\bmod 25)$, and since we are looking for the minimum amount of people, $x$ is 300 . That means there are 350 people at the party after the bus arrives, and thus there are $350 \cdot \frac{11}{25}=154$ adults at the party.
reamo
Solution 2 (Kind of lame)
Since at the beginning, adults make up $\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12 .
Call the amount of people in the beginning $x$.Then $x$ must be divisible by 12 , in other words: $x$ must be a multiple of 12 . Since after 50 more people arrived, adults make up $\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25 . This means $x+50$ must be a multiple of 5 .
Notice that if a number is divisible by 5 , it must end with a 0 or 5 . Since 5 is impossible (obviously, since multiples of 12 end in $2,4,6,8,0, \ldots$ ), $x$ must end in 0.

Notice that the multiples of 12 that end in 0 are: $60,120,180$, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert.
So therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350 * \frac{11}{25}=154$. Therefore the answer is 154 .
I know this solution is kind of lame, but this is still pretty straightforward. This solution is very similar to the first one, though.
hastapasta

## 数学竞赛代考|AIME代考美国数学邀请赛|2022 AIME II Problems

Problem
Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac{2}{3}$. When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac{3}{4}$. Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
Solution
Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The 4 circles represent the 4 players, and the arrow is from the winner to the loser with the winning probability as the label.

This problem can be solved by using 2 cases.
Case 1: $C$ ‘s opponent for the semifinal is $A$
The probability $C$ ‘s opponent is $A$ is $\frac{1}{3}$. Therefore the probability $C$ wins the semifinal in this case is $\frac{1}{3} \cdot \frac{1}{3}$. The other semifinal game is played between $J$ and $S$, it doesn’t matter who wins because $C$ has the same probability of winning either one. The probability of $C$ winning in the final $\frac{3}{4}$, so the probability of $C$ winning the tournament in case 1 is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{3}{4}$
Case 2: $C$ ‘s opponent for the semifinal is $J$ or $S$
It doesn’t matter if $C$ s s opponent is $J$ or $S$ because $C$ has the same probability of winning either one. The probability $C$ ‘s opponent is $J$ or $S$ is $\frac{2}{3}$. Therefore the probability $C$ wins the semifinal in this case is $\frac{2}{3} \cdot \frac{3}{4}$. The other semifinal game is played between $A$ and $J$ or $S$. In this case it matters who wins in the other semifinal game because the probability of $C$ winning $A$ and $J$ or $S$ is different.
Case 2.1: $C$ ‘s opponent for the final is $A$
For this to happen, $A$ must have won $J$ or $S$ in the semifinal, the probability is $\frac{3}{4}$. Therefore, the probability that $C$ won $A$ in the final is $\frac{3}{4} \cdot \frac{1}{3}$.
Case 2.1: $C$ ‘s opponent for the final is $J$ or $S$
For this to happen, $J$ or $S$ must have won $A$ in the semifinal, the probability is $\frac{1}{4}$. Therefore, the probability that $C$ won $J$ or $S$ in the final is $\frac{1}{4} \cdot \frac{3}{4}$.
In Case 2 the probability of $C$ winning the tournament is $\frac{2}{3} \cdot \frac{3}{4} \cdot\left(\frac{3}{4} \cdot \frac{1}{3}+\frac{1}{4} \cdot \frac{3}{4}\right)$
Adding case 1 and case 2 together we get $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{3}{4}+\frac{2}{3} \cdot \frac{3}{4} \cdot\left(\frac{3}{4} \cdot \frac{1}{3}+\frac{1}{4} \cdot \frac{3}{4}\right)=\frac{29}{96}$, so the answer is $29+96=125$.

# 美国数学竞赛代考

## 数学竞赛代考|AIME代考美国数学邀请赛|2022 AIME II Problems

reamo
Solution 2 (有点憋㑢)

## 数学竞赛代考|AIME代考美国数学邀请赛|2022 AIME II Problems

Azar、Carl、Jon 和 Sergey 是单打网球锦标寒中剩下的四名选手。他们在半决赛中随机分配对手，这些 比褰的获胜者在决赛中相互交手，以确定锦标赛的获胜者。当阿扎尔打卡尔时，阿扎尔有概率赢得比褰 $\frac{2}{3}$. 当 Azar 或 Carl 扮演 Jon 或 Sergey 时，Azar 或 Carl 有概率赢得比寨 $\frac{3}{4}$. 假设不同比塞的结果是独立 的。卡尔赢得比寨的概率是 $\frac{p}{q}$ ，在哪里 $p$ 和 $q$ 是相对质数的正整数。寻找 $p+q$.

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