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. | ||
− | 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> | + | 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> |
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 11:19, 12 November 2023
Problem
Let be a triangle and
an interior point of
. Show that at least one of the angles
is less than or equal to
.
Solution
Let ,
, and
be
,
,
, respcetively.
Let ,
, and
be
,
,
, respcetively.
Using law of sines on we get:
, therefore we get:
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.