Difference between revisions of "1981 IMO Problems/Problem 1"

Revision as of 12:06, 26 August 2024

Problem

$P$ is a point inside a given triangle $ABC$ . $D, E, F$ are the feet of the perpendiculars from $P$ to the lines $BC, CA, AB$ , respectively. Find all $P$ for which

$\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$

is least.

Solution

We note that $BC \cdot PD + CA \cdot PE + AB \cdot PF$ is twice the triangle's area, i.e., constant. By the Cauchy-Schwarz Inequality,

${(BC \cdot PD + CA \cdot PE + AB \cdot PF) \left(\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} \right) \ge ( BC + CA + AB )^2}$ ,

with equality exactly when $PD = PE = PF$ , which occurs when $P$ is the triangle's incenter or one of the three excenters. But since we know $P$ is inside $\triangle ABC$ , we can say $P$ is the incenter. $\square$

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

@@ Line 21: / Line 21: @@
 </center>
-with equality exactly when <math>PD = PE = PF </math>, which occurs when <math>P </math> is the triangle's incenter, Q.E.D.
+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}}

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Difference between revisions of "1981 IMO Problems/Problem 1"

Revision as of 12:06, 26 August 2024

Problem

Solution