Difference between revisions of "1981 IMO Problems/Problem 1"
(→Solution) |
m (→Solution) |
||
Line 15: | Line 15: | ||
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]], | 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]], | ||
− | + | ||
− | < | + | <cmath> |
{(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} | {(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} | ||
− | </ | + | </cmath>, |
− | + | ||
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> | 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> |
Latest revision as of 12:06, 26 August 2024
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 or one of the three excenters. But since we know is inside , we can say is the incenter.
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 |