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 "1989 AHSME Problems/Problem 19"
m
Line 1: Line 1:
 +
== Problem ==
 +
 
A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?
 
A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?
  
Line 9: Line 11:
  
 
-Solution by '''thecmd999'''
 
-Solution by '''thecmd999'''
 +
 +
== See also ==
 +
{{AHSME box|year=1989|num-b=18|num-a=20}} 
 +
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 06:57, 22 October 2014

Problem

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?

$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(E) \frac{9}{\pi^2}(\sqrt{3}+3) }$

Solution

The three arcs make up the entire circle, so the circumference of the circle is $3+4+5=12$ and the radius is $\frac{12}{2\pi}=\frac{6}{\pi}$. Also, the lengths of the arcs are proportional to their corresponding central angles. Thus, we can write the values of the arcs as $3\theta$, $4\theta$, and $5\theta$ for some $\theta$. By Circle Angle Sum, we obtain $3\theta+4\theta+5\theta=360$. Solving yields $\theta=30$. Thus, the angles of the triangle are $90$, $120$, and $150$. Using $[ABC]=\frac{1}{2}ab\sin{C}$, we obtain $\frac{r^2}{2}(\sin{90}+\sin{120}+\sin{150})$. Substituting $\frac{6}{\pi}$ for $r$ and evaluating yields $\frac{9}{\pi^2}(\sqrt{3}+3)\implies{\boxed{E}}$.


-Solution by thecmd999

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_19&oldid=65426"