1963 AHSME Problems/Problem 15

Problem

A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle. The ratio of the area of the triangle to the area of the square is:

$\textbf{(A)}\ \sqrt{3}:1\qquad \textbf{(B)}\ \sqrt{3}:\sqrt{2}\qquad \textbf{(C)}\ 3\sqrt{3}:2\qquad \textbf{(D)}\ 3:\sqrt{2}\qquad \textbf{(E)}\ 3:2\sqrt{2}$

Solution

$[asy] draw(circle((0,0),100)); draw((173.205,-100)--(-173.205,-100)--(0,200)--(173.205,-100)); draw((-100,0)--(0,100)--(100,0)--(0,-100)--(-100,0)); draw((0,0)--(0,-100),dotted); draw((0,0)--(-173.205,-100),dotted); [/asy]$ Let the radius of the circle be $r$ . That means the diameter of the circle is $2r$ , so the side length of the square is $r\sqrt{2}$ . Therefore, the area of the square is $2r^2$ .

By using 30-60-90 triangles, half of the side length of an equilateral triangle is $r\sqrt{3}$ , so each side is $2r\sqrt{3}$ units long. Thus, the area of the equilateral triangle is $3r^2\sqrt{3}$ .

The ratio of the area of the equilateral triangle to the area of the square is $\frac{3r^2 \sqrt{3}}{2r^2} = \frac{3 \sqrt{3}}{2}$ , so the answer is $\boxed{\textbf{(C)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1963 AHSME Problems/Problem 15

Problem

Solution

See Also