University of South Carolina High School Math Contest/1993 Exam/Problem 29
Contents
Problem
If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
![$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 8/\sqrt{15} \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2$](http://latex.artofproblemsolving.com/a/d/3/ad39e455fe156ef0c687ce7c7d910ebd4f9c9f27.png)
Solutions
Solution 1
One simple solution is using area formulas: by Heron's formula, a triangle with sides of length 2, 3 and 4 has area . But it also has area
(where
is the circumradius) so
.
Solution 2
Alternatively, let vertex be opposide the side of length 2. Then by the Law of Cosines,
so
. Thus
. Then by the extended Law of Sines,
.