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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> = <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> = <math>\measuredangle CAB</math>, <math>\measuredangle ABC</math>, <math>\measuredangle BCA</math>, respcetively.
  
 
Let <math>\alpha_{1}</math> , <math>\alpha_{2}</math>, and <math>\alpha_{3}</math> = <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> = <math>\measuredangle PAB</math>, <math>\measuredangle PBC</math>, <math>\measuredangle PCA</math>, respcetively.

Revision as of 11:12, 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 $A_{3}$ = $\measuredangle CAB$, $\measuredangle ABC$, $\measuredangle BCA$, respcetively.

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


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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