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Difference between revisions of "2013 AMC 12A Problems/Problem 12"

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NOTE: Problem 12 starts at 8:50 in the youtube video
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(NOTE: Problem 12 starts at 8:50 in the youtube video ~jyang66)
  
 
~sugar_rush
 
~sugar_rush

Latest revision as of 18:06, 25 December 2020

Problem

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

$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44$

Solution

Because the angles are in an arithmetic progression, and the angles add up to $180^{\circ}$, the second largest angle in the triangle must be $60^{\circ}$. 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 $5$, we get that $5^2 = 4^2 + x^2 - 2(4)(x)\cos 60^{\circ}$, therefore $x^2 - 4x - 9 = 0$. By using the quadratic formula, $x = \frac {4 + \sqrt{16 + 36}}{2}$, therefore $x = 2 + \sqrt{13}$.

When the second longest side is $x$, we get that $x^2 = 5^2 + 4^2 - 40\cos 60^{\circ}$, therefore $x = \sqrt{21}$.

When the second longest side is $4$, we get that $4^2 = 5^2 + x^2 - 2(5)(x)\cos 60^{\circ}$, therefore $x^2 - 5x + 9 = 0$. Using the quadratic formula, $x = \frac {5 + \sqrt{25 - 36}}{2}$. However, $\sqrt{-11}$ is not real, therefore the second longest side cannot equal $4$.

Adding the two other possibilities gets $2 + \sqrt{13} + \sqrt{21}$, with $a = 2, b=13$, and $c=21$. $a + b + c = 36$, which is answer choice $\text{(A)}$.

Video Solution

https://www.youtube.com/watch?v=XQpQaomC2tA

(NOTE: Problem 12 starts at 8:50 in the youtube video ~jyang66)

~sugar_rush

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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