参加加拿大数学竞赛 Canadian Mathematical Olympiad的好处
加拿大奥林匹克数学竞赛
加拿大数学奥林匹克（CMO），顾名思义，是加拿大的国家奥林匹克竞赛。它由加拿大数学协会举办，由永明金融公司赞助。[[http://www.math.ca/Competitions/CMO/ 加拿大数学奥林匹克（CMO）]] 奥林匹克竞赛在加拿大的数学竞赛中扮演着多种角色，最引人注目的是它是加拿大参加国际数学奥林匹克竞赛的主要队伍选拔过程。
资格审查
参加CMO的资格通常是通过加拿大公开数学挑战赛，但只有该比赛的 “官方 “参与者才能参加。一般来说，该比赛的CMO的分数线在70分左右。
奖励
在CMO中表现出色的人有几种不同类型的奖励。
*每年有6人被选入加拿大国际数学奥林匹克队，这是高中数学竞赛的顶峰。
*金、银、铜牌是根据你的分数百分位数颁发的。
SUMaC的故事
加拿大数学竞赛 Canadian Mathematical Olympiad的回报
在CMO中表现出色的学生有几种不同类型的奖励。
每年有6到8名优秀的CMO参与者被选入加拿大数学队，他们接受培训并被派去代表加拿大参加国际数学奥林匹克竞赛，这是高中数学竞赛的巅峰。
对于在加拿大写作的学生，表现最好的学生会获得现金奖励。如果没有 “并列”，那么第一名奖励2000美元，第二名奖励1500美元，第三名奖励1000美元，荣誉奖奖励500美元（通常最多有6人获得荣誉奖）。当出现并列时，奖金将向上集中（例如，并列第一意味着共同获奖者分享3500美元，没有人获得 “第二”）。
加拿大排名第一的学生被认定为CMO冠军，他或她的名字被刻在CMO冠军杯上。
最优秀的学生将在官方网站上被确认，以证明他们的优秀表现。
并且Canadian Mathematical Olympiad所取得的的优异成绩可以极大提升申请大学的竞争力。
下面是几道典型的SUMaC测试题目
Let five points on a circle be labelled $A, B, C, D$, and $E$ in clockwise order. Assume $A E=D E$ and let $P$ be the intersection of $A C$ and $B D$. Let $Q$ be the point on the line through $A$ and $B$ such that $A$ is between $B$ and $Q$ and $A Q=D P$. Similarly, let $R$ be the point on the line through $C$ and $D$ such that $D$ is between $C$ and $R$ and $D R=A P$. Prove that $P E$ is perpendicular to $Q R$.
Solution. We are given $A Q=D P$ and $A P=D R$. Additionally $\angle Q A P=180^{\circ}-\angle B A C=180^{\circ}-\angle B D C=\angle R D P$, and so triangles $A Q P$ and $D P R$ are congruent. Therefore $P Q=P R$. It follows that $P$ is on the perpendicular bisector of $Q R$.
We are also given $A P=D R$ and $A E=D E$. Additionally
$\angle P A E=\angle C A E=180^{\circ}-\angle C D E=\angle R D E$, and so triangles $P A E$ and $R D E$ are congruent.
Therefore $P E=R E$, and similarly $P E=Q E$. It follows that $E$ is on the perpendicular bisector of $P Q$.
Since both $P$ and $E$ are on the perpendicular bisector of $Q R$, the result follows.
Let $k$ be a given even positive integer. Sarah first picks a positive integer $N$ greater than 1 and proceeds to alter it as follows: every minute, she chooses a prime divisor $p$ of the current value of $N$, and multiplies the current $N$ by $p^{k}-p^{-1}$ to produce the next value of $N$. Prove that there are infinitely many even positive integers $k$ such that, no matter what choices Sarah makes, her number $N$ will at some point be divisible by $2018 .$
Solution: Note that 1009 is prime. We will show that if $k=1009^{m}-1$ for some positive integer $m$, then Sarah’s number must at some point be divisible by 2018 . Let $P$ be the largest divisor of $N$ not divisible by a prime congruent to 1 modulo 1009. Assume for contradiction that $N$ is never divisible by 2018. We will show that $P$ decreases each minute. Suppose that in the $t^{\text {th }}$ minute, Sarah chooses the prime divisor $p$ of $N$. First note that $N$ is replaced with $\frac{p^{k+1}-1}{p} \cdot N$ where
$$
p^{k+1}-1=p^{1009^{m}}-1=(p-1)\left(p^{1009^{m}-1}+p^{1009^{m}-2}+\cdots+1\right)
$$
Suppose that $q$ is a prime number dividing the second factor. Since $q$ divides $p^{1009^{m}}-1$, it follows that $q \neq p$ and the order of $p$ modulo $q$ must divide $1009^{m}$ and hence is either divisible by 1009 or is equal to 1 . If it is equal to 1 then $p \equiv 1(\bmod q)$, which implies that
$$
0 \equiv p^{1009^{m}-1}+p^{1009^{m}-2}+\cdots+1 \equiv 1009^{m} \quad(\bmod q)
$$
and thus $q=1009$. However, if $q=1009$ then $p \geq 1010$ and $p$ must be odd. Since $p-1$ now divides $N$, it follows that $N$ is divisible by 2018 in the $(t+1)^{\text {th }}$ minute, which is a contradiction. Therefore the order of $p$ modulo $q$ is divisible by 1009 and hence 1009 divides $q-1$. Therefore all of the prime divisors of the second factor are congruent to 1 modulo 1009. This implies that $P$ is replaced by a divisor of $\frac{p-1}{p} \cdot P$ in the $(t+1)^{\text {th }}$ minute and therefore decreases. Since $P \geq 1$ must always hold, $P$ cannot decrease forever. Therefore $N$ must at some point be divisible by 2018 .
Remark (no credit). If $k$ is allowed to be odd, then choosing $k+1$ to be divisible by $\phi(1009)=1008$ guarantees that Sarah’s number will be divisible by 2018 the first time it is even, which is after either the first or second minute.
加拿大数学奥林匹克CMO辅导|加拿大数学奥林匹克CMO代考
加拿大数学奥林匹克（CMO）是加拿大最重要的国家高级数学竞赛。
参赛者需要得到加拿大数学协会的邀请才能参加。
加拿大青少年数学奥林匹克竞赛（CJMO）也是通过邀请，同时为十年级以下的学生举办。 CJMO不像CMO那样具有挑战性，但有些问题可能会出现在这两项比赛中。
这些3小时的比赛在每年3月的选定时间和日期举行。 所有正式参赛者在同一时间写作。
参加CMO的优秀学生可以随时期待大学的巨大奖学金和招聘通知。他们还将被列入国际数学奥林匹克竞赛加拿大队的选拔名单!
数学代考|加拿大数学竞赛 Canadian Mathematical Olympiad辅导 请认准assignmentutor™. assignmentutor™为您的留学生涯保驾护航。
