AoPSWiki

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

Personal tools

Search

Toolbox

University of South Carolina High School Math Contest/1993 Exam/Problem 29

From AoPSWiki

Contents

1 Problem
2 Solutions
- 2.1 Solution 1
- 2.2 Solution 2

Problem

If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 8/\sqrt{15} \qquad \mathrm{(C) \ } 5/2\qquad \mathrm{(D) \ } \sqrt{6}\qquad \mathrm{(...$

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 $\sqrt{\frac92 \cdot \frac 52 \cdot \frac 32 \cdot \frac 12} = \frac34 \sqrt{15}$ . But it also has area $\frac{abc}{4R}$ (where $R$ is the circumradius) so $R = \frac{2\cdot3\cdot4}{4 A} = \frac8{\sqrt{15}} \Longrightarrow \mathrm{(B)}$ .

Solution 2

Alternatively, let vertex $A$ be opposide the side of length 2. Then by the Law of Cosines, $2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos A$ so $\cos A = \frac{3^2 + 4^2 - 2^2}{2\cdot3\cdot 4} = \frac78$ . Thus $\sin A = \sqrt{1 - \left(\frac78\right)^2} = \frac{\sqrt{15}}8$ . Then by the extended Law of Sines, $R = \frac12 \frac a{\sin A} = \frac12 \cdot \frac{2}{\sqrt{15}/8} = \frac{8}{\sqrt{15}}$ .

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/University_of_South_Carolina_High_School_Math_Contest/1993_Exam/Problem_29"

Category: Intermediate Geometry Problems

Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us