Difference between revisions of "2001 USAMO Problems/Problem 3"
5849206328x (talk | contribs) (→Solution: official solutions) |
Mathgeek2006 (talk | contribs) m (→Solution 3) |
||
Line 63: | Line 63: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
ab + bc + ca &\leq (\sin A + \sin B)\cot\frac{A + B}{2} + (\sin B + \sin C)\cot\frac{B + C}{2} + (\sin C + \sin A)\cot\frac{C + A}{2} \\ | ab + bc + ca &\leq (\sin A + \sin B)\cot\frac{A + B}{2} + (\sin B + \sin C)\cot\frac{B + C}{2} + (\sin C + \sin A)\cot\frac{C + A}{2} \\ | ||
− | &= 2\left(\cos\frac{A - B}{2}\cos\frac{A + B}{2} + \cos\frac{B - C}{2}\cos\frac{B + C}{2} + \cos\frac{C - A}{2}\cos\frac{C + A}{2} \\ | + | &= 2\left(\cos\frac{A - B}{2}\cos\frac{A + B}{2} + \cos\frac{B - C}{2}\cos\frac{B + C}{2} + \cos\frac{C - A}{2}\cos\frac{C + A}{2} \right)\\ |
&= 2(\cos A + \cos B + \cos C) \\ | &= 2(\cos A + \cos B + \cos C) \\ | ||
&= 6 - 4\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right) \\ | &= 6 - 4\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right) \\ |
Latest revision as of 18:28, 1 July 2015
Problem
Let and satisfy
Show that
Solution
Solution 1
First we prove the lower bound.
Note that we cannot have all greater than 1. Therefore, suppose . Then
Now, without loss of generality, we assume that and are either both greater than 1 or both less than one, so . From the given equation, we can express in terms of and as
Thus,
From the Cauchy-Schwarz Inequality,
This completes the proof.
Solution 2
The proof for the lower bound is the same as in the first solution.
Now we prove the upper bound. Let us note that at least two of the three numbers , , and are both greater than or equal to 1 or less than or equal to 1. Without loss of generality, we assume that the numbers with this property are and . Then we have The given equality and the inequality imply that or Dividing both sides of the last inequality by yields Thus, as desired.
The last equality holds if and only if and . Hence equality for the upper bound holds if and only if is one of the triples , , , and . Equality for the lower bound holds if and only if is one of the triples , and .
Solution 3
The proof for the lower bound is the same as in the first solution.
Now we prove the upper bound. It is clear that . If , then the result is trivial. Suppose that . Solving for yields This asks for the trigonometric substitution and , where . Then and if we set and , then , , and , where , , and are the angles of a triangle. We have where the inequality step follows from AM-GM. Likewise, Therefore as desired.
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.