这是一份美国数学邀请赛代考的成功案例，我们从2020年起就为全世界各地优秀的高中生提供AIME美国数学邀请赛等数学竞赛的代考服务，帮助一系列优秀的热爱数学的高中生取得了满意的成绩，在大学申请的竞争中赢在起跑线上。
一般我们可以保进前百分之一或者前百分之五两个档位，具体情况请在网站上找到我们的联系方式和我们进行沟通。
AIME美国数学邀请赛介绍
AIME初始于1983年。每年举办一次，通常是在3月底或4月初的一个星期二或星期四举行。从2000年开始，AIME每年举行两次，第二次是 “备用 “测试，以满足那些因春假、疾病或任何其他原因而无法参加第一轮测试的学生。然而，在任何情况下，一个学生都不能正式参加两个比赛。备选比赛，通常被称为 “AIME2 “或 “AIME-II”，通常在第一次考试后两周进行，在4月初的一个星期二。
然而，与AMC一样，AIME最近也是在3月初的一个星期二，以及15天后的星期三进行，例如2019年3月13日和20日。2020年，COVID-19疫情的迅速蔓延导致该年的AIME II被取消。取而代之的是，符合条件的学生可以参加美国在线数学邀请考试，其中包含了原本要在AIME II中出现的问题。2021年的AIME I和II也被搬到了网上
报名方式
AMC的成绩结果会在考试结束后6-8周左右公布AMC10/12AMC10/12 主要面向10年级(高一)和12年级(高三)以下的高中生。
AMC10通常涵盖初三和高一数学课程内容，包括等代数、基础几何学（勾股定理、面积体积公示等）、初等数论和概率问题，不包括三角函数、高等代数和高等几何学知识。AMC12涵盖以上全部内容，但不包括微积分。AIME难度高于AMC10/12，并且在AMC10/12中成绩优异的选手可晋级AIME美国数学邀请活动。
AIME是邀请赛，在当年的AMC 10竞赛中排名前2.5%左右或AMC 12竞赛中排名前5%左右才能获邀参赛。
考试时间
美国官网(MAA)公布的2021年AIME考试时间为:
AIME I(主赛): 3月10日(周三)
AIME II(代替赛)：3月18日(周四)
ASDAN中国公布的2021年AIME考试时间为：
AIME I(主赛): 3月11日(周四) 16:00-19:00
AIME II(代替赛)：3月19日(周五) 16:00-19:00
AIME难度设置
通常前几题的难度大致相当于AMC 12的水平，而越往后题目难度越大。通常多数学生能做出第1-5题；到了第6-10题则是区分度最大的题，经过专门的训练、在AMC 12或AMC 10中排名前1%左右的选手一般能做对一部分题；而一般在考场上能做出第11-15题的都是极其顶尖的选手。
从历年受邀参加AIME并获奖的中国学生和所在学校的分布情况来看，不少都是参加过国家队数学奥林匹克集训的选手和长期培养国际数学奥林匹克竞赛(IMO)选手的学校。
AIME往年题目
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.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.
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 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 is $\frac{3}{4}$, 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 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 2.2: $C$ ‘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 $29+96=125$.
