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> | ||
− | == Solution == | + | == 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== | ||
+ | As above, we can see that <math>\cos3A+\cos3B-\cos(3A+3B)=1</math> | ||
+ | |||
+ | Expanding, we get | ||
+ | |||
+ | <math>\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1</math> | ||
+ | |||
+ | <math>\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B</math> | ||
+ | |||
+ | <math>(\cos3A-1)(\cos3B-1)=\sin3A\sin3B</math> | ||
+ | |||
+ | <math>\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1</math> | ||
+ | |||
+ | <math>\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1</math> | ||
+ | |||
+ | Note that <math>\tan{x}=\frac{1}{\tan(90-x)}</math>, or <math>\tan{x}\tan(90-x)=1</math> | ||
+ | |||
+ | Thus <math>\frac{3A}{2}+\frac{3B}{2}=90</math>, or <math>A+B=60</math>. | ||
+ | |||
+ | 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
Contents
[hide]Problem
Suppose that the angles of satisfy Two sides of the triangle have lengths 10 and 13. There is a positive integer so that the maximum possible length for the remaining side of is Find
Solution 1
Note that . Thus, our expression is of the form . Let and . Expanding, we get , or . (why the minus sign at the left side ???) Since the right side of our equation must be non-positive, or is , and the other variable must take on a value of 1. WLOG, we can set , or ; however, only 120 is valid, as we want a nondegenerate triangle. Since , the maximum angle in the triangle is 120 degrees. Using Law of Cosines, we get , 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:
squaring both sides we get:
factoring: Then: Then: Then: 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:
Solution 2
As above, we can see that
Expanding, we get
Note that , or
Thus , or .
Now we know that , so we can just use Law or Cosines to get
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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