Difference between revisions of "2014 AIME II Problems/Problem 12"
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squaring both sides we get: | squaring both sides we get: | ||
− | <math>(1-x^2)(1-y^2) = [(x-1)(y- | + | <math>(1-x^2)(1-y^2) = [(x-1)(y-1)]^2</math> |
factoring: | factoring: |
Revision as of 18:19, 9 February 2015
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
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:
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.