Difference between revisions of "1981 IMO Problems/Problem 1"
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− | with equality exactly when <math>PD = PE = PF </math>, which occurs when <math>P </math> is the triangle's incenter, | + | with equality exactly when <math>PD = PE = PF </math>, which occurs when <math>P </math> is the triangle's incenter or one of the three excenters. But since we know <math>P </math> is inside <math>\triangle ABC </math>, we can say <math>P </math> is the incenter. <math>\square </math> |
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 12:06, 26 August 2024
Problem
is a point inside a given triangle . are the feet of the perpendiculars from to the lines , respectively. Find all for which
is least.
Solution
We note that is twice the triangle's area, i.e., constant. By the Cauchy-Schwarz Inequality,
,
with equality exactly when , which occurs when is the triangle's incenter or one of the three excenters. But since we know is inside , we can say is the incenter.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1981 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |