Difference between revisions of "2014 AIME II Problems/Problem 12"
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Simplification leads to <math>x=-y</math> and <math>x+y=0</math>. | Simplification leads to <math>x=-y</math> and <math>x+y=0</math>. | ||
− | Therefore, <math>cos(3C)=1</math>. So <math>\angle C</math> could be <math>0^\circ</math> or <math>120^\circ</math>. We eliminate <math>0^\circ</math> and use law of cosines to get our answer: | + | Therefore, <math>\cos(3C)=1</math>. So <math>\angle C</math> could be <math>0^\circ</math> or <math>120^\circ</math>. We eliminate <math>0^\circ</math> and use law of cosines to get our answer: |
<cmath>m=10^2+13^2-2\cdot 10\cdot 13\cos\angle C</cmath> | <cmath>m=10^2+13^2-2\cdot 10\cdot 13\cos\angle C</cmath> |
Revision as of 15:04, 9 February 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 .
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.