AMC10 2025 A

AMC10 2025 A · Q25

AMC10 2025 A · Q25. It mainly tests Probability (basic), Coordinate geometry.

A point $P$ is chosen at random inside square $ABCD$. The probability that $\overline{AP}$ is neither the shortest nor the longest side of $\triangle APB$ can be written as $\frac{a + b \pi - c \sqrt{d}}{e}$, where $a, b, c, d,$ and $e$ are positive integers, $\text{gcd}(a, b, c, e) = 1$, and $d$ is not divisible by the square of a prime. What is $a+b+c+d+e$?

在正方形 $ABCD$ 内随机选择一点 $P$。直线 $\overline{AP}$ 既不是 $\triangle APB$ 的最短边也不是最长边的概率可以写成 $\frac{a + b \pi - c \sqrt{d}}{e}$，其中 $a, b, c, d,$ 和 $e$ 是正整数，$\text{gcd}(a, b, c, e) = 1$，且 $d$ 不可被任一质数的平方整除。求 $a+b+c+d+e$？

(A) 25 25

(B) 26 26

(D) 28 28

(E) 29 \qquad 29 \qquad

Answer

Correct choice: (A)

正确答案：（A）

Solution

Say WLOG that $AB$ is the top side of the square, and the square is of side length 1. Let us say that the midpoint of $AB$ is $M$, while the midpoint of $CD$ is $Q$. Drawing a vertical line to split the square in half, we notice that if $P$ is to the left of the line, $AP < BP$, and if P is to the right of the line, $AP > BP$. Also, drawing a quarter circle of radius 1 from point $A$, we can split the area into points P for which $AP < AB$ and $AP > AB$. Because of our constraints, there are 2 cases: Case 1: $AB > AP > BP$ In this case, $P$ will be to the right of the vertical line and inside of the quarter circle. Let us say that the intersection of the vertical line and quarter circle is $N$. The distance from $N$ to $AD$ is 1/2, and we can say that $\angle BAN$ is $60^\circ$. Sector $BAN$ of circle $A$ would therefore have an area of $\frac{\pi}{6}$. Because $\triangle AMN$ is a 30-60-90 triangle, the area of $AMN$ is $\frac{\sqrt{3}}{8}$. The probability of case 1 happening should then be $\frac{\pi}{6}-\frac{\sqrt{3}}{8}$. Case 2: $AB < AP < BP$ In this case, $P$ will be to the left of the vertical line and outside of the quarter circle. Knowing that the quarter circle's area is $\frac{\pi}{4}$, we can subtract the probability of Case 1 happening to get the chance that $P$ is on the left of the vertical line and in circle $A$. Doing this would give $\frac{\pi}{12}+\frac{\sqrt{3}}{8}$. To get the probability of Case 2 happening, we can subtract this from the area of rectangle $AMQD$. This would give us $\frac{1}{2}-\frac{\pi}{12}-\frac{\sqrt{3}}{8}$. Adding both cases, we get the total probability as $\frac{1}{2}+\frac{\pi}{12}-\frac{\sqrt{3}}{4} = \frac{6+\pi-3\sqrt{3}}{12}$. Formatting this gives us $6+1+3+3+12 = \boxed{\textbf{(A) } 25}$.

不失一般性，假设 $AB$ 是正方形的上边，正方形边长为 1。设 $AB$ 中点为 $M$，$CD$ 中点为 $Q$。画一条垂直线将正方形分成两半，我们注意到如果 $P$ 在线的左侧，则 $AP < BP$，如果在右侧，则 $AP > BP$。另外，从点 $A$ 画一个半径为 1 的四分之一圆，可以将区域分为 $AP < AB$ 和 $AP > AB$ 的点 $P$。由于我们的约束，有 2 种情况：情况 1: $AB > AP > BP$ 在这种情况下，$P$ 在垂直线的右侧且在四分之一圆内。设垂直线与四分之一圆的交点为 $N$。$N$ 到 $AD$ 的距离为 1/2，我们可以说 $\angle BAN = 60^\circ$。因此圆 $A$ 的扇形 $BAN$ 面积为 $\frac{\pi}{6}$。因为 $\triangle AMN$ 是 30-60-90 三角形，其面积为 $\frac{\sqrt{3}}{8}$。情况 1 发生的概率为 $\frac{\pi}{6}-\frac{\sqrt{3}}{8}$。情况 2: $AB < AP < BP$ 在这种情况下，$P$ 在垂直线的左侧且在四分之一圆外。知道四分之一圆面积为 $\frac{\pi}{4}$，我们可以减去情况 1 的概率，得到 $P$ 在垂直线左侧且在圆 $A$ 内的概率。这样得到 $\frac{\pi}{12}+\frac{\sqrt{3}}{8}$。要得到情况 2 的概率，我们从矩形 $AMQD$ 的面积中减去这个值，得到 $\frac{1}{2}-\frac{\pi}{12}-\frac{\sqrt{3}}{8}$。将两情况相加，总概率为 $\frac{1}{2}+\frac{\pi}{12}-\frac{\sqrt{3}}{4} = \frac{6+\pi-3\sqrt{3}}{12}$。格式化为 $6+1+3+3+12 = \boxed{\textbf{(A) } 25}$。

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