# 1988 AHSME Problems/Problem 20

## Problem

In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label("A", A, NW); label("B", B, SW); label("C", C, SE); label("D", D, NE); label("E", E, S); label("F", F, N); label("G", G, E); label("X", X, NW); label("Y", Y, SW); label("Z", Z, SE); label("W", W, NE); [/asy]$ $\textbf{(A)}\ 4\qquad \textbf{(B)}\ 1+2\sqrt{3}\qquad \textbf{(C)}\ 2\sqrt{5}\qquad \textbf{(D)}\ \frac{8+4\sqrt{3}}{3}\qquad \textbf{(E)}\ 5$

## Solution

Within $WXYZ$, the parallelogram piece has vertical side $BF = \sqrt{1^2 + 2^2} = \sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\triangle AFG$, so we deduce $YZ = AG = \frac{\frac{1}{2} \times 1 \times 2}{\frac{1}{2} \times \sqrt{5}} = \frac{2}{\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put together without gaps or overlaps, so its area is $2^2 = 4$, and thus the vertical side $WZ$ is $\frac{4}{\frac{2}{\sqrt{5}}} = 2\sqrt{5}$, so the required ratio is $\frac{2\sqrt{5}}{\frac{2}{\sqrt{5}}} = \sqrt{5} \times \sqrt{5} = 5$, which is $\boxed{\text{E}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 