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考题24-25 2015 AMC 10A

2018-08-07 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 24

For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ , $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?

$考题24-25 2015 AMC 10A$

Solution

Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that $x^2 + (y - 2)^2 = y^2.$ Simplifying yields $x^2 - 4y + 4 = 0$ , so $x^2 = 4(y - 1)$ . Thus, $y$ is one more than a perfect square.

The perimeter $考题24-25 2015 AMC 10A$ must be less than 2015. Simple calculations demonstrate that $y = 31^2 + 1 = 962$ is valid, but $y = 32^2 + 1 = 1025$ is not. On the lower side, $y = 1$ does not work (because $x > 0$ ), but $y = 1^2 + 1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y = n^2 + 1$ for $1 \le n \le 31$ ), and so our answer is $\boxed{\textbf{(B) } 31}$

Problem 25

Let $S$ be a square of side length $1$ . Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\dfrac{1}{2}$ is $\dfrac{a-b\pi}{c}$ , where $a$ , $b$ , and $c$ are positive integers with $\gcd(a,b,c)=1$ . What is $a+b+c$ ?

$考题24-25 2015 AMC 10A$

Solution 1

Divide the boundary of the square into halves, thereby forming $8$ segments. Without loss of generality, let the first point $A$ be in the bottom-left segment. Then, it is easy to see that any point in the $5$ segments not bordering the bottom-left segment will be distance at least $\dfrac{1}{2}$ apart from $A$ . Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least $0.5$ apart from $A$ is $\dfrac{0 + 1}{2} = \dfrac{1}{2}$ because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)

If the second point $B$ is on the left-bottom segment, then if $A$ is distance $x$ away from the left-bottom vertex, then $B$ must be up to $\dfrac{1}{2} - \sqrt{0.25 - x^2}$ away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is $考题24-25 2015 AMC 10A$

(Alternatively, one can equate the problem to finding all valid $(x, y)$ with $0 < x, y < \dfrac{1}{2}$ such that $x^2 + y^2 \ge \dfrac{1}{4}$ , i.e. $(x, y)$ is outside the unit circle with radius $0.5.$ )

Thus, averaging the probabilities gives $P = \frac{1}{8} \left( 5 + \frac{1}{2} + 1 - \frac{\pi}{4} \right) = \frac{1}{32} \left( 26 - \pi \right).$

Our answer is $\boxed{\textbf{(A) } 59}$ .

Solution 2

Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the same side is $\frac{1}{4}$ , on an adjacent side is $\frac{1}{2}$ , and on the opposite side is $\frac{1}{4}$ . We discuss these three cases.

Case 1: Two points are on the same side. Let the first point be $a$ and the second point be $b$ in the $x$ -axis with $0\le a, b\le 1$ . Consider $(a, b)$ a point on the unit square $[0,1]\times [0,1]$ on the Cartesian plane. The region $\{(a,b): |b-a|> \frac{1}{2}\}$ has the area of $(\frac{1}{2})^2$ . Therefore, the probability that $|b-a|> \frac{1}{2}$ is $\frac{1}{4}$ .

Case 2: Two points are on two adjacent sides. Let the two sides be $[0,1]$ on the x-axis and $[0,1]$ on the y-axis and let one point be $(a, 0)$ and the other point be $(0, b)$ . Then $0\le a, b\le 1$ and the distance between the two points is $\sqrt{a^2+b^2}$ . As in Case 1, $(a, b)$ is a point on the unit square $[0,1]\times [0,1]$ . The area of the region $\{(a,b): \sqrt{a^2+b^2} \le \frac{1}{2}, 0\le a, b\le 1\}$ is $\frac{\pi}{16}$ and the area of its complementary set inside the square (i.e. $AMC10数学竞赛真题$ ) is $1-\frac{\pi}{16}$ . Therefore, the probability that the distance between $(a, 0)$ and $(0, b)$ is at least $\frac{1}{2}$ is $1-\frac{\pi}{16}$ .

Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least $1/2$ is obviously $1$ .

Thus the probability that the probability that the distance between the two points is at least $1/2$ is given by $AMC10数学竞赛真题$ Therefore $a=26$ , $b=1$ , and $c=32$ . Thus, $a+b+c=59$ and the answer is $\textbf{(A).}$