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

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Suppose that the angles of <math>\triangle ABC</math> satisfy <math>\cos(3A)+\cos(3B)+\cos(3C)=1.</math> Two sides of the triangle have lengths 10 and 13. There is a positive integer <math>m</math> so that the maximum possible length for the remaining side of <math>\triangle ABC</math> is <math>\sqrt{m}.</math> Find <math>m.</math>
 
Suppose that the angles of <math>\triangle ABC</math> satisfy <math>\cos(3A)+\cos(3B)+\cos(3C)=1.</math> Two sides of the triangle have lengths 10 and 13. There is a positive integer <math>m</math> so that the maximum possible length for the remaining side of <math>\triangle ABC</math> is <math>\sqrt{m}.</math> Find <math>m.</math>
  
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== Solution ==
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== Solution 1 ==
 
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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<math>x=-y</math>
 
<math>x=-y</math>
 
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}</math>
 
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}</math>
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==Solution 2==
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As above, we can see that <math>\cos3A+\cos3B-\cos(3A+3B)=1</math>
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 +
Expanding, we get
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<math>\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1</math>
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<math>\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B</math>
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<math>(\cos3A-1)(\cos3B-1)=\sin3A\sin3B</math>
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<math>\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1</math>
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<math>\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1</math>
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Note that <math>\tan{x}=\frac{1}{\tan(90-x)}</math>, or <math>\tan{x}\tan(90-x)=1</math>
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Thus <math>\frac{3A}{2}+\frac{3B}{2}=90</math>, or <math>A+B=60</math>.
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Now we know that <math>C=120</math>, so we can just use Law or Cosines to get <math>\boxed{399}</math>
  
 
== See also ==
 
== See also ==

Revision as of 18:52, 28 March 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 1

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: $\framebox{399}$

Solution 2

As above, we can see that $\cos3A+\cos3B-\cos(3A+3B)=1$

Expanding, we get

$\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1$

$\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B$

$(\cos3A-1)(\cos3B-1)=\sin3A\sin3B$

$\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1$

$\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1$

Note that $\tan{x}=\frac{1}{\tan(90-x)}$, or $\tan{x}\tan(90-x)=1$

Thus $\frac{3A}{2}+\frac{3B}{2}=90$, or $A+B=60$.

Now we know that $C=120$, so we can just use Law or Cosines to get $\boxed{399}$

See also

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