Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 29"
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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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+ | 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>. | ||
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Revision as of 14:46, 19 August 2006
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)
Solution
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
.
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,
.