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

Problem

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

$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 8/\sqrt{15} \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2$

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 $R = \frac{2\cdot3\cdot4}{4 (\frac{3}{4}\sqrt{15})} = \frac8{\sqrt{15}} \Longrightarrow \mathrm{(B)}$ .

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, $R = \frac12 \frac a{\sin A} = \frac12 \cdot \frac{2}{\sqrt{15}/8} = \frac{8}{\sqrt{15}}$ .

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

Contents

Problem

Solutions

Solution 1

Solution 2