Difference between revisions of "1952 AHSME Problems/Problem 27"
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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 21:13, 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 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
