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

Revision as of 20:34, 24 March 2007

Problem

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

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

is least.

Solution

We note that $\displaystyle BC \cdot PD + CA \cdot PE + AB \cdot PF$ 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 $\displaystyle PD = PE = PF$ , which occurs when $\displaystyle P$ 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.

Resources

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

Page

Toolbox

Search

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

Revision as of 20:34, 24 March 2007

Problem

Solution

Resources