# Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 29"

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== Solutions == | == Solutions == | ||

=== Solution 1 === | === 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 <math>\sqrt{\frac92 \cdot \frac 52 \cdot \frac 32 \cdot \frac 12} = \frac34 \sqrt{15}</math>. But it also has area <math>\frac{abc}{4R}</math> (where <math>R</math> is the [[circumradius]]) so <math>R = \frac{2\cdot3\cdot4}{4 | + | One simple solution is using [[area]] formulas: by [[Heron's formula]], a [[triangle]] with sides of length 2, 3 and 4 has area <math>\sqrt{\frac92 \cdot \frac 52 \cdot \frac 32 \cdot \frac 12} = \frac34 \sqrt{15}</math>. But it also has area <math>\frac{abc}{4R}</math> (where <math>R</math> is the [[circumradius]]) so <math>R = \frac{2\cdot3\cdot4}{4 (\frac{3}{4}\sqrt{15})} = \frac8{\sqrt{15}} \Longrightarrow \mathrm{(B)}</math>. |

=== Solution 2 === | === Solution 2 === |

## Revision as of 08:54, 31 August 2012

## Contents

## Problem

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

## 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, .