AIME 2020 II · 第 2 题

AIME 2020 II — Problem 2

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$ , and $(0,1)$ . The probability that the slope of the line determined by $P$ and the point $\left(\frac58, \frac38 \right)$ is greater than or equal to $\frac12$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

解析

Solution

The areas bounded by the unit square and alternately bounded by the lines through $\left(\frac{5}{8},\frac{3}{8}\right)$ that are vertical or have a slope of $1/2$ show where $P$ can be placed to satisfy the condition. One of the areas is a trapezoid with bases $1/16$ and $3/8$ and height $5/8$ . The other area is a trapezoid with bases $7/16$ and $5/8$ and height $3/8$ . Then,

$\frac{\frac{1}{16}+\frac{3}{8}}{2}\cdot\frac{5}{8}+\frac{\frac{7}{16}+\frac{5}{8}}{2}\cdot\frac{3}{8}=\frac{86}{256}=\frac{43}{128}\implies43+128=\boxed{171}$ ~mn28407

Solution 2 (Official MAA)

The line through the fixed point $\left(\frac58,\frac38\right)$ with slope $\frac12$ has equation $y=\frac12 x + \frac1{16}$ . The slope between $P$ and the fixed point exceeds $\frac12$ if $P$ falls within the shaded region in the diagram below consisting of two trapezoids with area

$\frac{\frac1{16}+\frac38}2\cdot\frac58 + \frac{\frac58+\frac7{16}}2\cdot\frac38 = \frac{43}{128}.$ Because the entire square has area $1,$ the required probability is $\frac{43}{128}$ . The requested sum is $43+128 = 171$ .

$AIME diagram$

Video Solution

https://youtu.be/x0QznvXcwHY?t=190

~IceMatrix

Video Solution 2

https://youtu.be/VBwlVbM0GRw

~avn