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

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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.
 
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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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 we get: <math>\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}</math>
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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>
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 11:19, 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, $\frac{\left| PA \right|}{\left| PB \right|}=\frac{sin(A_{2}-\alpha_{2})}{sin(\alpha_{1})}$

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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