Difference between revisions of "2013 AMC 12A Problems/Problem 12"
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When the second longest side is 4, 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, | When the second longest side is 4, 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, | + | <math> x = \frac {5 + \sqrt{25 - 36}}{2} </math>. However, <math> \sqrt{-11} </math> is not real, therefore the second longest side cannot equal 4. |
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>. | 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>. |
Revision as of 12:30, 7 February 2013
Because the angles are in an arithmetic progression, and the angles add up to , the second largest angle in the triangle must be . Also, the side opposite of that angle must be the second longest because of the angle-side relationship. Any of the three sides, 4, 5, or x, could be the second longest side of the triangle.
The law of cosines can be applied to solve for x in all three cases.
When the second longest side is five, we get that , therefore . By using the quadratic formula, , therefore .
When the second longest side is x, we get that , therefore .
When the second longest side is 4, we get that , therefore . Using the quadratic formula, . However, is not real, therefore the second longest side cannot equal 4.
Adding the two other possibilities gets , with , and . , which is answer choice .