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

(Created page with "Obviously, <math>A+B+C=180</math> in degrees. 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>. ...")
 
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== Problem ==
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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>
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== Solution ==
 
Obviously, <math>A+B+C=180</math> in degrees. 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>, which means that the right side must be non-positive and exactly one of <math>x</math> and <math>y</math> is <math>\leq{1}</math>. However, this means that the other variable must take on a value of 1. WLOG, we can set <math>\cos{3A}=1</math>, or <math>A=0,120</math>; however, only 120 is valid, as we want a degenerate triangle. Since <math>B+C=60</math>, the maximum angle between the sides is clearly 120 degrees. Using Law of Cosines on the triangle, we get <math>S^{2}=169+100+2(10)(13)(\frac{1}{2})=\framebox{399}</math>, and we're done.
 
Obviously, <math>A+B+C=180</math> in degrees. 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>, which means that the right side must be non-positive and exactly one of <math>x</math> and <math>y</math> is <math>\leq{1}</math>. However, this means that the other variable must take on a value of 1. WLOG, we can set <math>\cos{3A}=1</math>, or <math>A=0,120</math>; however, only 120 is valid, as we want a degenerate triangle. Since <math>B+C=60</math>, the maximum angle between the sides is clearly 120 degrees. Using Law of Cosines on the triangle, we get <math>S^{2}=169+100+2(10)(13)(\frac{1}{2})=\framebox{399}</math>, and we're done.
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== See also ==
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{{AIME box|year=2014|n=II|num-b=11|num-a=13}}
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[[Category:Intermediate Trigonometry Problems]]

Revision as of 01:08, 21 May 2014

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

Obviously, $A+B+C=180$ in degrees. 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 $-\sqrt{1-x^2}\sqrt{1-y^2}=(x-1)(y-1)$, which means that the right side must be non-positive and exactly one of $x$ and $y$ is $\leq{1}$. However, this means that the other variable must take on a value of 1. WLOG, we can set $\cos{3A}=1$, or $A=0,120$; however, only 120 is valid, as we want a degenerate triangle. Since $B+C=60$, the maximum angle between the sides is clearly 120 degrees. Using Law of Cosines on the triangle, we get $S^{2}=169+100+2(10)(13)(\frac{1}{2})=\framebox{399}$, and we're done.

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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