AIME 2008 I · 第 2 题

AIME 2008 I — Problem 2

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$ , and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$ .

解析

Solution 1

Note that if the altitude of the triangle is at most $10$ , then the maximum area of the intersection of the triangle and the square is $5\cdot10=50$ . This implies that vertex G must be located outside of square $AIME$ .

$AIME diagram$

Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$ . Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$ . Also, $\triangle GXY \sim \triangle GEM$ .

Let the height of $GXY$ be $h$ . By the similarity, $\dfrac{h}{6} = \dfrac{h + 10}{10}$ , we get $h = 15$ . Thus, the height of $GEM$ is $h + 10 = \boxed{025}$ .