Difference between revisions of "1959 AHSME Problems/Problem 21"

(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^...")
 
m (Solution)
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$p$ is the perimeter of an equilateral $\triangle$ inscribed in a circle, the area of the circle is: $\textbf{(A)}\ \frac{\pi p^2}{3} \qquad\textbf{(B)}\ \frac{\pi p^2}{9}\qquad\textbf{(C)}\ \frac{\pi p^2}{27}\qquad\textbf{(D)}\ \frac{\pi p^2}{81}\qquad\textbf{(E)}\ \frac{\pi p^2\sqrt3}{27}$

Solution

A side length of the triangle is $\frac{p}3$. An altitude of the triangle, by 30-60-90 triangles, is $\frac{p\sqrt{3}}{6}$. Because all classical triangle centers coincide on an equilateral triangle, by centroid properties a circumradius is $\frac{p\sqrt{3}}{9}$. Finally, the area of the circumcircle is $\pi\frac{p^2}{27}\rightarrow\boxed{\textbf{C}}$.

See also

1959 AHSC (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png