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Difference between revisions of "1998 USAMO Problems/Problem 3"

m (moved 1998 USAMO Problems/Problem 1 to 1998 USAMO Problems/Problem 3: Wrong problem, I fix'd it)
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1998 USAMO Problem #1
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== Problem ==
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Problem:
 
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Let <math>a_0,\cdots a_n</math> be real numbers in the interval <math>(0,\frac {\pi}{2})</math> such that
 
Let <math>a_0,\cdots a_n</math> be real numbers in the interval <math>(0,\frac {\pi}{2})</math> such that
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<math>\tan{(a_0 - \frac {\pi}{4})} + \tan{(a_1 - \frac {\pi}{4})} + \cdots + \tan{(a_n - \frac {\pi}{4})}\ge n - 1</math>.
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<cmath>\tan{(a_0 - \frac {\pi}{4})} + \tan{(a_1 - \frac {\pi}{4})} + \cdots + \tan{(a_n - \frac {\pi}{4})}\ge n - 1</cmath>
 
Prove that <math>\tan{(a_0)}\tan{(a_1)}\cdots \tan{(a_n)}\ge n^{n + 1}</math>.
 
Prove that <math>\tan{(a_0)}\tan{(a_1)}\cdots \tan{(a_n)}\ge n^{n + 1}</math>.
  
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Solution:
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== Solution ==
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Let <math>y_i = \tan{(a_i - \frac {\pi}{4})}</math>, where <math>0\le i\le n</math>. Then we have
 
Let <math>y_i = \tan{(a_i - \frac {\pi}{4})}</math>, where <math>0\le i\le n</math>. Then we have
  

Revision as of 10:54, 16 April 2011

Problem

Let $a_0,\cdots a_n$ be real numbers in the interval $(0,\frac {\pi}{2})$ such that \[\tan{(a_0 - \frac {\pi}{4})} + \tan{(a_1 - \frac {\pi}{4})} + \cdots + \tan{(a_n - \frac {\pi}{4})}\ge n - 1\] Prove that $\tan{(a_0)}\tan{(a_1)}\cdots \tan{(a_n)}\ge n^{n + 1}$.

Solution

Let $y_i = \tan{(a_i - \frac {\pi}{4})}$, where $0\le i\le n$. Then we have

  • $y_0 + y_1 + \cdots + y_n\ge n - 1$
  • $1 + y_i\ge \sum_{j\neq i}{(1 - y_j)}$
  • $\frac {1 + y_i}{n}\ge \frac {1}{n}\sum_{j\neq i}{(1 - y_j)}$

By AM-GM,

  • $\frac {1}{n}\sum_{j\neq i}{(1 - y_j)}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$
  • $\frac {1 + y_i}{n}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$
  • $\prod_{i = 0}^n{\frac {1 + y_i}{n}}\ge \prod_{i = 0}^n{\prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$ (Error compiling LaTeX. Unknown error_msg)
  • $= \prod_{i = 0}^n{(1 - y_i)}$
  • $\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}}\ge \prod_{i = 0}^n{n} = n^{n + 1}$

Note that by the addition formula for tangents, $\tan{(a_i)} = \tan{[(a_i - \frac {\pi}{4}) + \frac {\pi}{4}]} = \frac {1 + \tan{(a_i - \frac {\pi}{4})}}{1 - \tan{(a_i - \frac {\pi}{4})}} = \frac {1 + y_i}{1 - y_i}$.

So $\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}} = \tan{(a_0)}\tan{(a_1)}\cdots \tan{(a_n)}\ge n^{n + 1}$, as desired.

$\text{QED}$

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