1978 AHSME Problems/Problem 29

Problem

Sides and , respectively, of convex quadrilateral are extended past and to points and . Also, and ; and the area of is . The area of is

Solution

Diagram:

[asy] unitsize(1 cm);

pair[] A, B, C, D;

A[0] = (0,0); B[0] = (0.6,1.2); C[0] = (-0.3,2.5); D[0] = (-1.5,0.7); B[1] = interp(A[0],B[0],2); C[1] = interp(B[0],C[0],2); D[1] = interp(C[0],D[0],2); A[1] = interp(D[0],A[0],2);

draw(A[1]--B[1]--C[1]--D[1]--cycle); draw(A[0]--B[1]); draw(B[0]--C[1]); draw(C[0]--D[1]); draw(D[0]--A[1]);

label("", A[0], SW); label("", B[0], SE); label("", C[0], NE); label("", D[0], NW); label("", A[1], SE); label("", B[1], NE); label("", C[1], N); label("", D[1], SW); [/asy]

Notice that the area of \triangle is the same as that of \triangle (same base, same height). Thus, the area of \triangle is twice that (same height, twice the base). Similarly, [\triangle ] = 2 \cdot [\triangle ], and so on.

Adding all of these, we see that the area the four triangles around is twice [\triangle ] + [\triangle ] + [\triangle ] + [\triangle ], which is itself twice the area of the quadrilateral . Finally, [] = [] + 4 \cdot [] = 5 \cdot [] = \fbox{50}.

~ Mathavi