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| − | 2013 AMC 12A Problems/Problem 12Vt
| + | Here's what I want to know: what about a plain jane 3-4-5 triangle? Why wouldn't that work here? |
| − | Problem
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| − | The angles in a particular triangle are in arithmetic progression, and the side lengths are <math>4,5,x</math>. The sum of the possible values of x equals <math>a+\sqrt{b}+\sqrt{c}</math> where <math>a, b</math>, and <math>c</math> are positive integers. What is <math>a+b+c</math>?
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| − | <math>\textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44</math>
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| − | Solution
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| − | Because the angles are in an arithmetic progression, and the angles add up to <math>180^{\circ}</math>, the second largest angle in the triangle must be <math>60^{\circ}</math>. Also, the side opposite of that angle must be the second longest because of the angle-side relationship. Any of the three sides, <math>4</math>, <math>5</math>, or <math>x</math>, could be the second longest side of the triangle.
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| − | The law of cosines can be applied to solve for <math>x</math> in all three cases.
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| − | When the second longest side is <math>5</math>, we get that <math>5^2 = 4^2 + x^2 - 2(4)(x)cos 60^{\circ}</math>, therefore <math>x^2 - 4x - 9 = 0</math>. By using the quadratic formula, <math>x = \frac {4 + \sqrt{16 + 36}}{2}</math>, therefore <math>x = 2 + \sqrt{13}</math>.
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| − | When the second longest side is <math>x</math>, we get that <math>x^2 = 5^2 + 4^2 - 40cos 60^{\circ}</math>, therefore <math>x = \sqrt{21}</math>.
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| − | When the second longest side is <math>4</math>, we get that <math>4^2 = 5^2 + x^2 - 2(5)(x)cos 60^{\circ}</math>, therefore <math>x^2 - 5x + 9 = 0</math>. Using the quadratic formula, <math>x = \frac {5 + \sqrt{25 - 36}}{2}</math>. However, <math>\sqrt{-11}</math> is not real, therefore the second longest side cannot equal <math>4</math>.
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| − | Adding the two other possibilities gets <math>2 + \sqrt{13} + \sqrt{21}</math>, with <math>a = 2, b=13</math>, and <math>c=21</math>. <math>a + b + c = 36</math>, which is answer choice <math>A</math>.
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| − | What about a plain jane 3-4-5 triangle, where x would equal 3? Is there something obvious about this that I'm missing? Help!
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