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 2008 AIME I Problem #2

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PostSubject: 2008 AIME I Problem #2   Sun Aug 24, 2008 12:33 pm

Quite easy for AIME, see if you can figure it out.

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 GEM.

-AIME15, Moderator of Algebra Forum
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PostSubject: Re: 2008 AIME I Problem #2   Mon Aug 25, 2008 11:37 am

meh wish this forum had asy.

Let the intersection point of AE and GE be X, and let the intersection point of AE and GM be Y.

(XY+EM)*EM/2=80, (XY+10)*10/2=80, XY=6

Let the foot of perpendicular of G to AE be Z.

proportion - triangle GXY and triangle GEM

GZ : 6 = (GZ+10) : 10, 6GZ+60=10GZ, GZ=15.

therefore, altitude to EM in GEM is 10+15=25.

answer : 25
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