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>. | 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>. | ||
− | Using the fact that <math>\cos(3A+3B)=\cos 3A\cos 3B-\sin 3A\sin 3B=xy-\sqrt{1-x^2}\sqrt{1-y^2}</math>, 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}=xy-x-y | + | Using the fact that <math>\cos(3A+3B)=\cos 3A\cos 3B-\sin 3A\sin 3B=xy-\sqrt{1-x^2}\sqrt{1-y^2}</math>, 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}=xy-x-y+1=(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>. | 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>. |
Revision as of 17:57, 18 March 2017
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 .
Using the fact that , we get , or .
Squaring both sides, we get . Cancelling factors, .
Expanding, .
Simplification leads to and .
Therefore, . So could be or . We eliminate 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 the Law of 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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