IMO 1996 SL G9

Origin: UKR | Category: Geometry

Problem

In the plane are given a point $O$ and a polygon $F$ (not necessarily convex). Let $P$ denote the perimeter of $F$, $D$ the sum of the distances from $O$ to the vertices of $F$, and $H$ the sum of the distances from $O$ to the lines containing the sides of $F$. Prove that

$$D^2 - H^2 \geq \frac{P^2}{4}.$$

Solution

We first prove the result in the simplest case. Given a 2-gon $ABA$ and a point $O$, let $a, b, c, h$ denote $OA, OB, AB$, and the distance of $O$ from $AB$, respectively. Then $D = a + b$, $P = 2c$, and $H = 2h$, so we should show that

$$(a+b)^2 \geq 4h^2 + c^2. \tag{1}$$

Indeed, let $l$ be the line through $O$ parallel to $AB$, and $D$ the point symmetric to $B$ with respect to $l$. Then

$$(a+b)^2 = (OA + OB)^2 = (OA + OD)^2 \geq AD^2 = c^2 + 4h^2.$$

Now we pass to the general case. Let $A_1A_2 \dots A_n$ be the polygon $F$ and denote by $d_i, p_i,$ and $h_i$ respectively $OA_i$, $A_iA_{i+1}$, and the distance of $O$ from $A_iA_{i+1}$ (where $A_{n+1} = A_1$). By the case proved above, we have for each $i$,

$$d_i + d_{i+1} \geq \sqrt{4h_i^2 + p_i^2}.$$

Summing these inequalities for $i = 1, \dots, n$ and squaring, we obtain

$$4D^2 \geq \left( \sum_{i=1}^n \sqrt{4h_i^2 + p_i^2} \right)^2.$$

It remains only to prove that

$$\sum_{i=1}^n \sqrt{4h_i^2 + p_i^2} \geq \sqrt{4H^2 + P^2}.$$

But this follows immediately from the Minkowski inequality. Equality holds if and only if it holds in (1) and in the Minkowski inequality, i.e., if and only if

$$d_1 = \cdots = d_n \quad \text{and} \quad \frac{h_1}{p_1} = \cdots = \frac{h_n}{p_n}.$$

This means that $F$ is inscribed in a circle with center at $O$ and $p_1 = \cdots = p_n$, so $F$ is a regular polygon and $O$ its center.