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Difference between revisions of "1991 IMO Problems/Problem 5"

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== Solution ==
 
== Solution ==
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Let <math>A_{1}</math> , <math>A_{2}</math>, and <math>A_{3}</math> be <math>\measuredangle CAB</math>, <math>\measuredangle ABC</math>, <math>\measuredangle BCA</math>, respcetively.
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Let <math>A_{1}</math> , <math>A_{2}</math>, and <math>A_{3}</math> be <math>\angle CAB</math>, <math>\angle ABC</math>, <math>\angle BCA</math>, respectively.
  
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Let <math>\alpha_{1}</math> , <math>\alpha_{2}</math>, and <math>\alpha_{3}</math> be <math>\measuredangle PAB</math>, <math>\measuredangle PBC</math>, <math>\measuredangle PCA</math>, respcetively.
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Let <math>\alpha_{1}</math> , <math>\alpha_{2}</math>, and <math>\alpha_{3}</math> be <math>\angle PAB</math>, <math>\angle PBC</math>, <math>\angle PCA</math>, respcetively.
  
 
Using law of sines on <math>\Delta PAB</math> we get: <math>\frac{\left| PA \right|}{sin(A_{2}-\alpha_{2})}=\frac{\left| PB \right|}{sin(\alpha_{1})}</math>, therefore, <math>\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}</math>
 
Using law of sines on <math>\Delta PAB</math> we get: <math>\frac{\left| PA \right|}{sin(A_{2}-\alpha_{2})}=\frac{\left| PB \right|}{sin(\alpha_{1})}</math>, therefore, <math>\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}</math>

Revision as of 12:29, 12 November 2023

Problem

Let $\,ABC\,$ be a triangle and $\,P\,$ an interior point of $\,ABC\,$. Show that at least one of the angles $\,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $30^{\circ }$.

Solution

Let $A_{1}$ , $A_{2}$, and $A_{3}$ be $\angle CAB$, $\angle ABC$, $\angle BCA$, respectively.

Let $\alpha_{1}$ , $\alpha_{2}$, and $\alpha_{3}$ be $\angle PAB$, $\angle PBC$, $\angle PCA$, respcetively.

Using law of sines on $\Delta PAB$ we get: $\frac{\left| PA \right|}{sin(A_{2}-\alpha_{2})}=\frac{\left| PB \right|}{sin(\alpha_{1})}$, therefore, $\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}$

Using law of sines on $\Delta PBC$ we get: $\frac{\left| PB \right|}{sin(A_{3}-\alpha_{3})}=\frac{\left| PC \right|}{sin(\alpha_{2})}$, therefore, $\frac{\left| PB \right|}{\left| PC \right|}=\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}$

Using law of sines on $\Delta PCA$ we get: $\frac{\left| PC \right|}{sin(A_{1}-\alpha_{1})}=\frac{\left| PA \right|}{sin(\alpha_{3})}$, therefore, $\frac{\left| PC \right|}{\left| PA \right|}=\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$

Multiply all three equations we get: $\frac{\left| PA \right|}{\left| PB \right|}\frac{\left| PB \right|}{\left| PC \right|}\frac{\left| PC \right|}{\left| PA \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$

$1=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}\frac{sin(A_{1}-\alpha_{1})}{sin(\alpha_{3})}$

$\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}=1$

Using AM-GM we get:

$\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge \sqrt[3]{\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}}$

$\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 1$

$\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 3$. [Inequality 1]

$\sum_{i=1}^{3}\frac{sin(A_{i})cos(\alpha_{i})-cos(A_{i})sin(\alpha_{i})}{sin(\alpha_{i})}\ge 3$

$\sum_{i=1}^{3}\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]\ge 3$

Note that for $0<\alpha_{i}<180^{\circ}$, $cot(\alpha_{i})$ decreases with increasing $\alpha_{i}$ and fixed $A_{i}$

Therefore, $\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]$ decreases with increasing $\alpha_{i}$ and fixed $A_{i}$

From trigonometric identity:

$sin(x)+sin(y)=2sin\left( \frac{x+y}{2} \right)cos\left( \frac{x-y}{2} \right)$,

since $-1\le cos\left( \frac{x-y}{2} \right) \le 1$, then:

$sin(x)+sin(y) \le 2sin\left( \frac{x+y}{2} \right)$

Therefore,

$sin(A_{1}-30^{\circ})+sin(A_{2}-30^{\circ}) \le 2sin\left( \frac{A_{1}+A_{2}-60^{\circ}}{2} \right)$

and also,

$sin(A_{3}-30^{\circ})+sin(30^{\circ}) \le 2sin\left( \frac{A_{3}}{2} \right)$

Adding these two inequalities we get:

$sin(30^{\circ})+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ sin\left( \frac{A_{1}+A_{2}-60^{\circ}}{2} \right)+sin\left( \frac{A_{3}}{2} \right) \right]$

$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ 2sin\left( \frac{A_{1}+A_{2}+A_{3}-60^{\circ}}{4} \right) \right]$

$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 2\left[ 2sin\left( \frac{180^{\circ}-60^{\circ}}{4} \right) \right]$

$\frac{1}{2}+\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 4sin\left( 30^{\circ} \right)$

$\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le \frac{3}{2}$

$2\sum_{i=1}^{3}sin(A_{i}-30^{\circ})\le 3$.

$\sum_{i=1}^{3}\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le 3$. [Inequality 2]

Combining [Inequality 1] and [Inequality 2] we see the following:

$\sum_{i=1}^{3}\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le \sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}$

This implies that for at least one of the values of $i=1$,$2$,or $3$, the following is true:

$\frac{sin(A_{i}-30^{\circ})}{sin(30^{\circ})}\le \frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}$

or

$\frac{sin(\alpha_{i})}{sin(A_{i}-\alpha_{i})}\le \frac{sin(30^{\circ})}{sin(A_{i}-30^{\circ})}$

Which means that for at least one of the values of $i=1$,$2$,or $3$, the following is true:

$\alpha_{i} \le 30^{\circ}$

Therefore, at least one of the angles $\,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $30^{\circ }$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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