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

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1998 USAMO Problem #1
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== Problem ==
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Let <math>a_0,\cdots a_n</math> be real numbers in the interval <math>\left(0,\frac {\pi}{2}\right)</math> such that
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Problem:
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<cmath>\tan{\left(a_0 - \frac {\pi}{4}\right)} + \tan{\left(a_1 - \frac {\pi}{4}\right)} + \cdots + \tan{\left(a_n - \frac {\pi}{4}\right)}\ge n - 1</cmath>
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Prove that <math>\tan{\left(a_0\right)}\tan{\left(a_1\right)}\cdots \tan{\left(a_n\right)}\ge n^{n + 1}</math>.
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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
 
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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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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 ==
 
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
  
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*<math>\frac {1}{n}\sum_{j\neq i}{(1 - y_j)}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}</math>
 
*<math>\frac {1}{n}\sum_{j\neq i}{(1 - y_j)}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}</math>
 
*<math>\frac {1 + y_i}{n}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}</math>
 
*<math>\frac {1 + y_i}{n}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}</math>
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*<math>\prod_{i = 0}^n{\frac {1 + y_i}{n}}\ge \prod_{i = 0}^n{\prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}</math>
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*<math>\prod_{i = 0}^{n} {\frac{1 + y_i}{n}}\geq \prod_{i = 0}^{n} {\prod_{j\neq i} {(1 - y_j)}^{\frac {1}{n}}}</math>
 
*<math>= \prod_{i = 0}^n{(1 - y_i)}</math>
 
*<math>= \prod_{i = 0}^n{(1 - y_i)}</math>
 
*<math>\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}}\ge \prod_{i = 0}^n{n} = n^{n + 1}</math>
 
*<math>\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}}\ge \prod_{i = 0}^n{n} = n^{n + 1}</math>
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<math>\text{QED}</math>
 
<math>\text{QED}</math>
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==See Also==
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{{USAMO newbox|year=1998|num-b=2|num-a=4}}
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[[Category:Olympiad Trigonometry Problems]]
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[[Category:Olympiad Algebra Problems]]
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[[Category:Olympiad Inequality Problems]]
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{{MAA Notice}}

Revision as of 09:44, 20 July 2016

Problem

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

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,

  • $\frac {1}{n}\sum_{j\neq i}{(1 - y_j)}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$
  • $\frac {1 + y_i}{n}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}$
  • $\prod_{i = 0}^{n} {\frac{1 + y_i}{n}}\geq \prod_{i = 0}^{n} {\prod_{j\neq i} {(1 - y_j)}^{\frac {1}{n}}}$
  • $= \prod_{i = 0}^n{(1 - y_i)}$
  • $\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}}\ge \prod_{i = 0}^n{n} = n^{n + 1}$

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.

$\text{QED}$

See Also

1998 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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