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 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 $AIME$ .

$[asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label("$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 $h + 10 = \boxed{025}$ .

2008 AIME I (Problems • Answer Key • Resources)
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2008 AIME I Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

See also