## Tuesday, January 3, 2006

### Low as You Can Go. Topic: Geometry/Inequalities. Level: Olympiad.

SPECIAL POST! This problem was too cool and I couldn't wait another day to post it, so here you go.

Problem: (1981 IMO - #1) Consider a variable point $P$ inside a given triangle $ABC$. Let $D, E, F$ be the feet of the perpendiculars from the point $P$ to the lines $BC, CA, AB$, respectively. Find all points $P$ which minimize the sum

$\frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}$. Solution: So let's try to find some relationship with $PD, BC, PE, CA, PF, AB$. Area is a good bet. Let $S$ be the area of $\triangle ABC$.

We have $S = \frac{1}{2}(PD \cdot BC+PE \cdot CA+PF \cdot AB)$.

We're looking for the minimum $K = \frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}$, which makes us think inequalities and the fractions make us think Cauchy.

Well, amazingly enough, by Cauchy we have

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

So $K \ge \frac{(AB+BC+CA)^2}{2S}$, but the RHS is constant for a given triangle. Therefore the minimum is the equality condition on Cauchy, or

$\frac{PD \cdot BC}{\frac{BC}{PD}} = \frac{PE \cdot CA}{\frac{CA}{PE}} = \frac{PF \cdot AB}{\frac{AB}{PF}}$,

which simplifies to

$PD^2 = PE^2 = PF^2 \Rightarrow PD = PE = PF$.

Hence $K$ is minimized when $P$ is the incenter of $\triangle ABC$. QED.

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Comment: If that's not a cool geometric inequality IMO problem, I don't know what is!

#### 1 comment:

1. I still like my solution by Jensen's better. =P