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

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By AM-GM,
 
By AM-GM,
  
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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>
+
<cmath>\begin{align*}
−
*<math>\frac {1 + y_i}{n}\ge \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}</math>
+
\frac {1}{n}\sum_{j\neq i}{(1 - y_j)} &\geq \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}\\
−
*<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>
+
\frac {1 + y_i}{n} &\geq \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}\\
−
*<math>= \prod_{i = 0}^n{(1 - y_i)}</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>\prod_{i = 0}^n{\frac {1 + y_i}{1 - y_i}}\ge \prod_{i = 0}^n{n} = n^{n + 1}</math>
+
\prod_{i = 0}^n\frac {1 + y_i}{1 - y_i} &\geq \prod_{i = 0}^n{(1 - y_i)}\\
 +
\prod_{i = 0}^n\frac {1 + y_i}{1 - y_i} &\geq \prod_{i = 0}^n{n}\\
 +
\prod_{i = 0}^n\frac {1 + y_i}{1 - y_i} &\geq n^{n + 1}
 +
\end{align*}</cmath>
  
−
Note that by the addition formula for tangents, <math>\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}</math>.
+
Note that by the addition formula for tangents, <cmath>\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}</cmath>.
  
−
So <math>\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}</math>, as desired.
+
So <math>\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}</math>, as desired. <math>\blacksquare</math>
−
 
 
−
<math>\text{QED}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 17:54, 10 May 2023

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,

\begin{align*} \frac {1}{n}\sum_{j\neq i}{(1 - y_j)} &\geq \prod_{j\neq i}{(1 - y_j)^{\frac {1}{n}}}\\ \frac {1 + y_i}{n} &\geq \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\frac {1 + y_i}{1 - y_i} &\geq \prod_{i = 0}^n{(1 - y_i)}\\ \prod_{i = 0}^n\frac {1 + y_i}{1 - y_i} &\geq \prod_{i = 0}^n{n}\\ \prod_{i = 0}^n\frac {1 + y_i}{1 - y_i} &\geq n^{n + 1} \end{align*}

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. $\blacksquare$

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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