# 1952 AHSME Problems/Problem 39

## Problem

If the perimeter of a rectangle is $p$ and its diagonal is $d$, the difference between the length and width of the rectangle is: $\textbf{(A)}\ \frac {\sqrt {8d^2 - p^2}}{2} \qquad \textbf{(B)}\ \frac {\sqrt {8d^2 + p^2}}{2} \qquad \textbf{(C)}\ \frac{\sqrt{6d^2-p^2}}{2}\qquad\\ \textbf{(D)}\ \frac {\sqrt {6d^2 + p^2}}{2} \qquad \textbf{(E)}\ \frac {8d^2 - p^2}{4}$

## Solution $[asy] pair A,B,C,D,E,F,G,H; A=(0,0); B=(10,0); C=(10,5); D=(0,5); E=(5,5.5); F=(5,-0.5); G=(-0.5,2.75); H=(10.5,2.75); draw(A--B--C--D--cycle); draw (B--D); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",(-0.5,5),N); label("x",E); label("x",F); label("y",G); label("y",H); [/asy]$ Let the sides of the rectangle be x and y. WLOG, assume x>y. Then, $2x+2y=p \Rightarrow x+y=\frac{p}{2}$.

By pythagorean theorem, $x^2 + y^2 =d^2$.

Since $x+y=\frac{p}{2}$, $(x+y)^2=\frac{p^2}{4} \Rightarrow x^2+2xy+y^2=\frac{p^2}{4}$.

Rearranging to solve for $2xy$ gives $2xy = \frac{p^2}{4}-d^2$.

Rearranging $(x-y)^2$ in terms of the defined variables becomes $(x-y)^2 = d^2 - (\frac{p^2}{4}-d^2)$.

In order to get (x-y), we have to take the square root of the expression and simplify. $(x-y)=\sqrt{2d^2-\frac{p^2}{4}} \Rightarrow (x-y)=\sqrt{\frac{8d^2-p^2}{4}}$ $\Rightarrow$ $(x-y)=\frac{\sqrt{8d^2+p^2}}{2}$ $\Rightarrow$ $\fbox{A}$.

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