1959 AHSME Problems/Problem 21: Difference between revisions
Clarkculus (talk | contribs) Created page with "== Problem 21 == If<math> p</math> is the perimeter of an equilateral <math>\triangle</math> inscribed in a circle, the area of the circle is: <math>\textbf{(A)}\ \frac{\pi p^..." |
Clarkculus (talk | contribs) |
||
| Line 3: | Line 3: | ||
<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>\pi\frac{p^2}{27}\rightarrow\boxed{textbf{C}}</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>\pi\frac{p^2}{27}\rightarrow\boxed{\textbf{C}}</math>. | ||
== See also == | == See also == | ||
Revision as of 11:51, 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 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
