2013 AMC 12A Problem 12

Attempt Problem 12 of the 2013 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

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12.

The angles in a particular triangle are in arithmetic progression, and the side lengths are 4,5,4, 5, and x.x. The sum of the possible values of xx equals a+b+c,a + \sqrt{b} + \sqrt{c}, where a,b,a, b, and cc are positive integers. What is a+b+c?a + b + c?

3636

3838

4040

4242

4444

Answer: A
Concepts:arithmetic sequencelaw of cosinescasework
Difficulty rating: 1740
Solution:

If the angles are αδ,α,α+δ,\alpha - \delta, \alpha, \alpha + \delta, their sum 3α=1803\alpha = 180^\circ gives α=60,\alpha = 60^\circ, so one angle is 60.60^\circ.

If xx is opposite the 6060^\circ angle, the Law of Cosines gives x2=42+52245cos60=21, \begin{gathered} x^2 = 4^2 + 5^2 - 2\cdot 4\cdot 5\cos 60^\circ \\ = 21, \end{gathered} so x=21.x = \sqrt{21}.

If 55 is opposite the 6060^\circ angle, then 25=x24x+16,25 = x^2 - 4x + 16, whose positive solution is x=2+13.x = 2 + \sqrt{13}. If 44 is opposite, then 16=x25x+2516 = x^2 - 5x + 25 has no real solution.

The sum of the possible values is 2+13+21,2 + \sqrt{13} + \sqrt{21}, so a+b+c=2+13+21=36.a + b + c = 2 + 13 + 21 = 36.

Thus, the correct answer is A.

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