Problem:
Triangle AMC is isosceles with AM=AC. Medians MV and CU are perpendicular to each other, and MV=CU=12. What is the area of â–³AMC?
Answer Choices:
A. 48
B. 72
C. 96
D. 144
E. 192
Solution:
Since quadrilateral UVCM has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that △AUV has 41​ the area of triangle AMC by similarity, so
[UVCM]=43​⋅[AMC]
Thus,
21​⋅12⋅1272[AMC]​=43​⋅[AMC]=43​⋅[AMC]=(C) 96​.​
OR
We know that △AUV∼△AMC, and since the ratios of its sides are 21​, the ratio of of their areas is (21​)2=41​. If △AUV is 41​ the area of △AMC, then trapezoid MUVC is 43​ the area of △AMC.
Let's call the intersection of UC and MVP.
Let UP=x
Then PC=12−x
Since UC⊥MV,UP and CP are heights of triangles △MUV and △MCV, respectively. Both of these triangles have base 12 .
Area of △MUV=2x⋅12​=6x
Area of △MCV=2(12−x)⋅12​=72−6x
Adding these two gives us the area of trapezoid MUVC, which is
6x+(72−6x)=72
This is 43​ of the triangle, so the area of the triangle is
34​⋅72=(C) 96​.
The problems on this page are the property of the MAA's American Mathematics Competitions
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