Difference between revisions of "1981 IMO Problems/Problem 1"
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== Problem == | == Problem == | ||
− | <math> | + | <math>P</math> is a point inside a given triangle <math>ABC</math>. <math>D, E, F</math> are the feet of the perpendiculars from <math>P</math> to the lines <math>BC, CA, AB</math>, respectively. Find all <math>P</math> for which |
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== Solution == | == Solution == | ||
− | We note that <math> | + | We note that <math>BC \cdot PD + CA \cdot PE + AB \cdot PF</math> is twice the triangle's area, i.e., constant. By the [[Cauchy-Schwarz Inequality]], |
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− | with equality exactly when <math> | + | with equality exactly when <math>PD = PE = PF </math>, which occurs when <math>P </math> is the triangle's incenter, Q.E.D. |
{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | {{IMO box|before=First question|num-a=2|year=1981}} |
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 20:08, 1 February 2021
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, Q.E.D.
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 |