# 1964 AHSME Problems/Problem 33

## Problem

$P$ is a point interior to rectangle $ABCD$ and such that $PA=3$ inches, $PD=4$ inches, and $PC=5$ inches. Then $PB$, in inches, equals:

$\textbf{(A) }2\sqrt{3}\qquad\textbf{(B) }3\sqrt{2}\qquad\textbf{(C) }3\sqrt{3}\qquad\textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }2$

$[asy] draw((0,0)--(6.5,0)--(6.5,4.5)--(0,4.5)--cycle); draw((2.5,1.5)--(0,0)); draw((2.5,1.5)--(0,4.5)); draw((2.5,1.5)--(6.5,4.5)); draw((2.5,1.5)--(6.5,0),linetype("8 8")); label("A",(0,0),dir(-135)); label("B",(6.5,0),dir(-45)); label("C",(6.5,4.5),dir(45)); label("D",(0,4.5),dir(135)); label("P",(2.5,1.5),dir(-90)); label("3",(1.25,0.75),dir(120)); label("4",(1.25,3),dir(35)); label("5",(4.5,3),dir(120)); [/asy]$

## Solution

From point $P$, create perpendiculars to all four sides, labeling them $a, b, c, d$ starting from going north and continuing clockwise. Label the length $PB$ as $x$.

We have $a^2 + b^2 = 5^2$ and $c^2 + d^2 = 3^2$, leading to $a^2 + b^2 + c^2 + d^2 = 34$.

We also have $a^2 + d^2 = 4^2$ and $b^2 + c^2 = x^2$, leading to $a^2 + b^2 + c^2 + d^2 = 16 + x^2$.

Thus, $34 = 16 + x^2$, or $x = \sqrt{18} = 3\sqrt{2}$, which is option $\boxed{\textbb{(B)}$ (Error compiling LaTeX. ! File ended while scanning use of \boxed.)