1963 AHSME Problems/Problem 10

Problem

Point $P$ is taken interior to a square with side-length $a$ and such that is it equally distant from two consecutive vertices and from the side opposite these vertices. If $d$ represents the common distance, then $d$ equals:

$\textbf{(A)}\ \frac{3a}{5}\qquad \textbf{(B)}\ \frac{5a}{8}\qquad \textbf{(C)}\ \frac{3a}{8}\qquad \textbf{(D)}\ \frac{a\sqrt{2}}{2}\qquad \textbf{(E)}\ \frac{a}{2}$

Solution

[asy] draw((0,0)--(16,0)--(16,16)--(0,16)--(0,0)); draw((0,0)--(8,6)); draw((16,0)--(8,6)); draw((8,16)--(8,6)); draw((8,0)--(8,6),dotted); label("$d$",(4,3),NW); label("$d$",(12,3),NE); label("$d$",(8,11),W); label("$\frac{a}{2}$",(4,0),S); label("$\frac{a}{2}$",(12,0),S); label("$a-d$",(8,2),W); [/asy] Draw a diagram and label it as shown. Because of SSS Congruency, the two bottom triangles are right triangles. By the Pythagorean Theorem, \[d^2 = (a-d)^2 + (\frac{a}{2})^2\] \[d^2 = a^2 - 2ad + d^2 + \frac{a^2}{4}\] \[2ad = \frac{5a^2}{4}\] \[d = \frac{5a}{8}\] Thus, the answer is $\boxed{\textbf{(B)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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