Difference between revisions of "1991 IMO Problems/Problem 5"
(→Solution) |
(→Solution) |
||
Line 8: | Line 8: | ||
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> | ||
+ | |||
+ | Using law of sines on <math>\Delta PBC</math> we get: <math>\frac{\left| PB \right|}{sin(A_{3}-\alpha_{3})}=\frac{\left| PC \right|}{sin(\alpha_{2})}</math>, therefore, <math>\frac{\left| PB \right|}{\left| PC \right|}=\frac{sin(A_{3}-\alpha_{3})}{sin(\alpha_{2})}</math> | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 11:21, 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,
Using law of sines on we get: , therefore,
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.