1952 AHSME Problems/Problem 27: Difference between revisions
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== Problem== | |||
The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is: | |||
<math> \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3</math> | |||
==Solution== | |||
If the radius of the circle is <math>r</math>, then the perimeter of the first triangle is <math>3\left(\frac{2r}{\sqrt3}\right)=2r\sqrt3</math>, and the perimeter of the second is <math>3r\sqrt3</math>. So the ratio is <math>\boxed{\frac23{\textbf{ (E)}}}</math>. | |||
==See also== | |||
{{AHSME 50p box|year=1952|num-b=26|num-a=28}} | |||
{{MAA Notice}} | |||
Latest revision as of 20:13, 19 April 2014
Problem
The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:
Solution
If the radius of the circle is 
, then the perimeter of the first triangle is 
, and the perimeter of the second is 
. So the ratio is 
.
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
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