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AMC数学竞赛真题2017年10A 17-18

2018-08-27 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

Problem 17

Distinct points $P$ , $Q$ , $R$ , $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$ ?

$\mathrm{\textbf{(A)}}\ 3\qquad\mathrm{\textbf{(B)}}\ 5\qquad\mathrm{\textbf{(C)}}\ 3\sqrt{5}\qquad\mathrm{\textbf{(D)}}\ 7\qquad\mathrm{\textbf{(E)}}\ 5\sqrt{2}$

Solution

Because $P$ , $Q$ , $R$ , and $S$ are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are $(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),$ and $(\pm 5,0).$ We want to maximize $PQ$ and minimize $RS.$ They also have to be the square root of something, because they are both irrational. The greatest value of $PQ$ happens when it $P$ and $Q$ are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be $(-4,3)$ and $(3,-4)$ because the two points are almost across from each other. The least value of $RS$ is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, $R$ is $(3,4)$ and $S$ is $(4,3).$ They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point $(3,4)$ than $(4,3).$ Using the distance formula, we get that $PQ$ is $\sqrt{98}$ and that $RS$ is $\sqrt{2}.$ $美国数学竞赛amc$

Problem 18

Amelia has a coin that lands heads with probability $\frac{1}{3}$ , and Blaine has a coin that lands on heads with probability $\frac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$ ?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P$ , as if she gets to her turn again, she is back where she started with probability of winning $P$ . The chance she wins on her first turn is $\frac{1}{3}$ . The chance she makes it to her turn again is a combination of her failing to win the first turn - $\frac{2}{3}$ and Blaine failing to win - $\frac{3}{5}$ . Multiplying gives us $\frac{2}{5}$ . Thus, $P = \frac{1}{3} + \frac{2}{5}P$ Therefore, $P = \frac{5}{9}$ , so the answer is $美国数学竞赛$ .

Solution 2

Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} ...$ This can be represented by an infinite geometric series, $美国数学竞赛amc10$ . Therefore, $P = \frac{5}{9}$ , so the answer is $9-5 = \boxed{\textbf{(D)}\ 4}$