Difference between revisions of "1957 AHSME Problems/Problem 7"
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==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1957|num-b=6|num-a=8}} | + | {{AHSME 50p box|year=1957|num-b=6|num-a=8}} |
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 08:07, 25 July 2024
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
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
