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

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== Solution ==
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== Solutions ==
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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 \sqrt15</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 A} = \frac8{\sqrt{15}} \Longrightarrow \mathrm{(B)}</math>.
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=== Solution 1 ===
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+
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 A} = \frac8{\sqrt{15}} \Longrightarrow \mathrm{(B)}</math>.
  
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=== Solution 2 ===
 
Alternatively, let [[vertex]] <math>A</math> be opposide the side of length 2.  Then by the [[Law of Cosines]], <math>2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos A</math> so <math>\cos A = \frac{3^2 + 4^2 - 2^2}{2\cdot3\cdot 4} = \frac78</math>.  Thus <math>\sin A = \sqrt{1 - \left(\frac78\right)^2} = \frac{\sqrt{15}}8</math>.  Then by the extended [[Law of Sines]], <math>R = \frac12 \frac a{\sin A} = \frac12 \cdot \frac{2}{\sqrt{15}/8} = \frac{8}{\sqrt{15}}</math>.
 
Alternatively, let [[vertex]] <math>A</math> be opposide the side of length 2.  Then by the [[Law of Cosines]], <math>2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos A</math> so <math>\cos A = \frac{3^2 + 4^2 - 2^2}{2\cdot3\cdot 4} = \frac78</math>.  Thus <math>\sin A = \sqrt{1 - \left(\frac78\right)^2} = \frac{\sqrt{15}}8</math>.  Then by the extended [[Law of Sines]], <math>R = \frac12 \frac a{\sin A} = \frac12 \cdot \frac{2}{\sqrt{15}/8} = \frac{8}{\sqrt{15}}</math>.
  

Revision as of 16:41, 1 April 2009

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

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


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