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

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<math>\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}=1</math>
 
<math>\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}=1</math>
  
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Using <math>AM-GM</math> we get:
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Using AM-GM we get:
  
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<math>\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\le \sqrt[3]{\prod_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}}</math>
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<math>\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})}}</math>
  
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<math>\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\le 1</math>
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<math>\frac{1}{3}\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 1</math>
  
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<math>\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\le 3</math>
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<math>\sum_{i=1}^{3}\frac{sin(A_{i}-\alpha_{i})}{sin(\alpha_{i})}\ge 3</math>
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 11:31, 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 $\measuredangle CAB$, $\measuredangle ABC$, $\measuredangle BCA$, respcetively.

Let $\alpha_{1}$ , $\alpha_{2}$, and $\alpha_{3}$ be $\measuredangle PAB$, $\measuredangle PBC$, $\measuredangle PCA$, 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:

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

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

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

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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