Difference between revisions of "2014 AIME II Problems/Problem 12"
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== Solution == | == Solution == | ||
− | 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>. 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 | + | 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>. 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. |
== See also == | == See also == |
Revision as of 13:32, 27 September 2014
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
. 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.
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.