Difference between revisions of "1991 IMO Problems/Problem 5"
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Note that for <math>0<\alpha_{i}<180^{\circ}</math>, <math>cot(\alpha_{i})</math> decreases with increasing <math>\alpha_{i}</math> and fixed <math>A_{i}</math> | Note that for <math>0<\alpha_{i}<180^{\circ}</math>, <math>cot(\alpha_{i})</math> decreases with increasing <math>\alpha_{i}</math> and fixed <math>A_{i}</math> | ||
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+ | Therefore, <math>\left[ sin(A_{i})cot(\alpha_{i})-cos(A_{i})\right]$ decreases with increasing </math>\alpha_{i}<math> and fixed </math>A_{i}$ | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 11:39, 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,
Using law of sines on we get: , therefore,
Multiply all three equations we get:
Using AM-GM we get:
Note that for , decreases with increasing and fixed
Therefore, \alpha_{i}A_{i}$
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.