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

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

Revision as of 01:22, 19 January 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-2)]^2

factoring:$ (Error compiling LaTeX. Unknown error_msg)(1+x)(1+y) = (1-x)(1-y)$$ (Error compiling LaTeX. Unknown error_msg)xy-x-y+1 = 1+x+y+xy$$ (Error compiling LaTeX. Unknown error_msg)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:$ (Error compiling LaTeX. Unknown error_msg)\framebox{399}$

See also

2014 AIME II (ProblemsAnswer KeyResources)
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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