2014 AIME II Problems/Problem 12

Problem

Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$

Solution

Note that $\cos{3C}=-\cos{(3A+3B)}$ . Thus, our expression is of the form $\cos{3A}+\cos{3B}-\cos{(3A+3B)}=1$ . Let $\cos{3A}=x$ and $\cos{3B}=y$ . Expanding, we get $x+y-xy+\sqrt{1-x^2}\sqrt{1-y^2}=1$ , or $-\sqrt{1-x^2}\sqrt{1-y^2}=(x-1)(y-1)$ . (why the minus sign at the left side ???) Since the right side of our equation must be non-positive, $x$ or $y$ is $\leq{1}$ , and the other variable must take on a value of 1. WLOG, we can set $x=\cos{3A}=1$ , or $A=0,120$ ; however, only 120 is valid, as we want a nondegenerate triangle. Since $B+C=60$ , the maximum angle in the triangle is 120 degrees. Using Law of Cosines, we get $S^{2}=169+100+2(10)(13)(\frac{1}{2})=\framebox{399}$ , and we're done.

I believe there's a mistake in the solution. There shouldn't be a negative at the left hand side of the equation. We should get: $\sqrt{1-x^2}\sqrt{1-y^2}=(x-1)(y-1)$

squaring both sides we get: $(1-x^2)(1-y^2) = [(x-1)(y-2)]^2$

factoring: $(1+x)(1+y) = (1-x)(1-y)$ $xy-x-y+1 = 1+x+y+xy$ $2x+2y=0$ $x=-y$

Therefore, cos(3C) must equal 1. So C could be 0 or 120 degrees. We eliminate 0 and use law of cosines to get our answer: $\framebox{399}$

2014 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2014 AIME II Problems/Problem 12

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