1960 AHSME Problems/Problem 6

Problem

The circumference of a circle is $100$ inches. The side of a square inscribed in this circle, expressed in inches, is:

$\textbf{(A) }\frac{25\sqrt{2}}{\pi}\qquad \textbf{(B) }\frac{50\sqrt{2}}{\pi}\qquad \textbf{(C) }\frac{100}{\pi}\qquad \textbf{(D) }\frac{100\sqrt{2}}{\pi}\qquad \textbf{(E) }50\sqrt{2}$

Solution

First, find the diameter of the circle. Plug in the circumference for the formula $C = \pi d$ to solve for $d$ . $100 = \pi d$ $d = \frac{100}{\pi}$ Since all of an inscribed square's vertices touch the circle, and one of the angles of the circle is $90^{\circ}$ , the square's diagonal is the diameter of the circle.

$[asy] draw(circle((0,0),50)); draw((-50,0)--(0,50)--(50,0)--(0,-50)--cycle); draw((-50,0)--(50,0)); draw((-3,47)--(0,44)--(3,47)); label("$\frac{100}{\pi}$",(0,-10)); [/asy]$

From the Pythagorean theorem (or 45-45-90 triangles), the side length is $\frac{50\sqrt{2}}{\pi}$ inches, which is answer choice $\boxed{\textbf{(B)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
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1960 AHSME Problems/Problem 6

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