Difference between revisions of "1959 AHSME Problems/Problem 21"
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<math>\textbf{(A)}\ \frac{\pi p^2}{3} \qquad\textbf{(B)}\ \frac{\pi p^2}{9}\qquad\textbf{(C)}\ \frac{\pi p^2}{27}\qquad\textbf{(D)}\ \frac{\pi p^2}{81}\qquad\textbf{(E)}\ \frac{\pi p^2\sqrt3}{27} </math> | <math>\textbf{(A)}\ \frac{\pi p^2}{3} \qquad\textbf{(B)}\ \frac{\pi p^2}{9}\qquad\textbf{(C)}\ \frac{\pi p^2}{27}\qquad\textbf{(D)}\ \frac{\pi p^2}{81}\qquad\textbf{(E)}\ \frac{\pi p^2\sqrt3}{27} </math> | ||
== Solution == | == Solution == | ||
| − | A side length of the triangle is <math>\frac{p}3</math>. An altitude of the triangle, by 30-60-90 triangles, is <math>\frac{p\sqrt{3}}{6}</math>. Because all classical triangle centers coincide on an equilateral triangle, by centroid properties a circumradius is <math>\frac{p\sqrt{3}}{9}</math>. Finally, the area of the circumcircle is <math> | + | A side length of the triangle is <math>\frac{p}3</math>. An altitude of the triangle, by 30-60-90 triangles, is <math>\frac{p\sqrt{3}}{6}</math>. Because all classical triangle centers coincide on an equilateral triangle, by centroid properties a circumradius is <math>\frac{p\sqrt{3}}{9}</math>. Finally, the area of the circumcircle is <math>\frac{\pi p^2}{27}\rightarrow\boxed{\textbf{C}}</math>. |
== See also == | == See also == | ||
Latest revision as of 12:52, 9 April 2024
Problem 21
If
is the perimeter of an equilateral 
inscribed in a circle, the area of the circle is:
Solution
A side length of the triangle is 
. An altitude of the triangle, by 30-60-90 triangles, is 
. Because all classical triangle centers coincide on an equilateral triangle, by centroid properties a circumradius is 
. Finally, the area of the circumcircle is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
