1963 AHSME Problems/Problem 15

Revision as of 15:52, 7 June 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 15)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\boxed{\textbf{(C)}}$.


See Also

1963 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png