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

Revision as of 19:45, 25 October 2007

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, 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

@@ Line 25: / Line 25: @@
 {{alternate solutions}}
-{{IMO box|before=First question|num-a=2|year=1980}}
+{{IMO box|before=First question|num-a=2|year=1981}}
 [[Category:Olympiad Geometry Problems]]

Page

Toolbox

Search

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

Revision as of 19:45, 25 October 2007

Problem

Solution