Talk:2013 AMC 12A Problems/Problem 12

2013 AMC 12A Problems/Problem 12Vt 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 $x^2 = 5^2 + 4^2 - 40cos 60^{\circ}$ , 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 $A$ .

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!

Page

Toolbox

Search

Talk:2013 AMC 12A Problems/Problem 12