Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1957 AHSME Problems/Problem 7"

(→‎Solution)
(→‎Solution)
Line 11: Line 11:
 
draw((0,0)--(0,-sqrt(3)));
 
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>.
+
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{\textbf{(E)}72}</math>.
  
 
==See Also==
 
==See Also==
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:19, 12 October 2020

Problem 7

The area of a circle inscribed in an equilateral triangle is $48\pi$. The perimeter of this triangle is:

$\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72$

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] We can see that the radius of the circle is $4\sqrt{3}$. We know that the radius is $\frac{1}{3}$ of each median line of the triangle; each median line is therefore $12\sqrt{3}$. Since the median line completes a $30$-$60$-$90$ triangle, we can conclude that one of the sides of the triangle is $24$. Triple the side length and we get our answer, $\boxed{\textbf{(E)}72}$.

See Also

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_7&oldid=134968"