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

Revision as of 18:13, 1 February 2021

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

wait can u actually edit this..... if a mod sees this then it is public :O, mod delete it if you can see... I think this just shows up for me

Revision as of 19:45, 25 October 2007 (view source) Temperal (talk \| contribs) m (year) ← Older edit		Revision as of 18:13, 1 February 2021 (view source) V v (talk \| contribs) Newer edit →
Line 28:		Line 28:

	[[Category:Olympiad Geometry Problems]]		[[Category:Olympiad Geometry Problems]]
		+
		+
		+	wait can u actually edit this..... if a mod sees this then it is public :O, mod delete it if you can see... I think this just shows up for me

Page

Toolbox

Search

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

Revision as of 18:13, 1 February 2021

Problem

Solution