Difference between revisions of "2014 AIME II Problems/Problem 12"
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− | 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> | + | 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>. |
− | + | Squaring both sides, we get <math>(1-x^2)(1-y^2) = [(x-1)(y-1)]^2</math>. Cancelling factors, <math>(1+x)(1+y) = (1-x)(1-y).</math> | |
− | <math> | + | Expanding, <math>xy+x+y+1 = 1-x-y+xy</math>. Simplification leads to <math>x=-y</math>. |
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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}</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> | ||
Revision as of 00:37, 30 January 2016
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 .
Squaring both sides, we get . Cancelling factors, Expanding, . Simplification leads to .
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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