2008 AIME I Problems/Problem 2

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 $\triangle GEM$ .

Solution

$[asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label("$G$",G,N); label("$A$",A,NW); label("$I$",I,NE); label("$M$",M,NE); label("$E$",E,NW); label("$10$",(M+E)/2,S); [/asy]$

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 $h + 10 = \boxed{025}$ .

Note that we know that the altitude of the triangle is greater than ten because if it's not, the maximum intersection of the areas of the triangle and the square is $5\cdot 10=50$ .

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2008 AIME I Problems/Problem 2

Problem

Solution

See also