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Difference between revisions of "2014 AIME II Problems/Problem 12"

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squaring both sides we get:
 
squaring both sides we get:
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<math>(1-x^2)(1-y^2) = [(x-1)(y-2)]^2</math>
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<math>(1-x^2)(1-y^2) = [(x-1)(y-1)]^2</math>
  
 
factoring:
 
factoring:

Revision as of 17:19, 9 February 2015

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: $(1-x^2)(1-y^2) = [(x-1)(y-1)]^2$

factoring: $(1+x)(1+y) = (1-x)(1-y).$ Then: $xy-x-y+1 = 1+x+y+xy.$ Then: $2x+2y=0.$ Then: $x=-y$ 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: $\framebox{399}$

See also

2014 AIME II (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
All AIME Problems and Solutions

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