Difference between revisions of "1957 AHSME Problems/Problem 7"
Angrybird029 (talk | contribs) (→Solution) |
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| Line 9: | Line 9: | ||
draw(circle((0,0),sqrt(3))); | draw(circle((0,0),sqrt(3))); | ||
dot((0,0)); | dot((0,0)); | ||
| + | draw((0,0)--(0,-sqrt(3))); | ||
</asy> | </asy> | ||
| + | We can see that the radius of the circle is <math>4\sqrt{3}</math>. We know that the radius is <math>\frac{1}{3}</math> of each median line of the triangle; each median line is therefore <math>12\sqrt{3}</math>. Since the median line completes a <math>30</math>-<math>60</math>-<math>90</math> triangle, we can conclude that one of the sides of the triangle is <math>24</math>. Triple the side length and we get our answer, <math>\boxed{72}</math>. | ||
==See Also== | ==See Also== | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:18, 12 October 2020
Problem 7
The area of a circle inscribed in an equilateral triangle is 
. The perimeter of this triangle is:
Solution
![[asy] draw((-3,-sqrt(3))--(3,-sqrt(3))--(0,2sqrt(3))--cycle); draw(circle((0,0),sqrt(3))); dot((0,0)); draw((0,0)--(0,-sqrt(3))); [/asy]](http://latex.artofproblemsolving.com/1/0/4/104af9dce61eb4c467e4ae03dc64a9d78c07c8bc.png)
We can see that the radius of the circle is 
. We know that the radius is 
of each median line of the triangle; each median line is therefore 
. Since the median line completes a 
-
-
triangle, we can conclude that one of the sides of the triangle is 
. Triple the side length and we get our answer, 
.
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
