Subfield: Geometry


Question: Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P \subset R$ we have

$$
\frac{|R|}{|P|} \leq \sqrt{2}
$$

where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively.

Solution:
We will construct two parallelograms $R_{1}$ and $R_{3}$, each of them containing $P$, and prove that at least one of the inequalities $\left|R_{1}\right| \leq \sqrt{2}|P|$ and $\left|R_{3}\right| \leq \sqrt{2}|P|$ holds (see Figure 1).

First we will construct a parallelogram $R_{1} \supseteq P$ with the property that the midpoints of the sides of $R_{1}$ are points of the boundary of $P$.

Choose two points $A$ and $B$ of $P$ such that the triangle $O A B$ has maximal area. Let $a$ be the line through $A$ parallel to $O B$ and $b$ the line through $B$ parallel to $O A$. Let $A^{\prime}, B^{\prime}, a^{\prime}$ and $b^{\prime}$ be the points or lines, that are symmetric to $A, B, a$ and $b$, respectively, with respect to $O$. Now let $R_{1}$ be the parallelogram defined by $a, b, a^{\prime}$ and $b^{\prime}$.

<img_3903>

Figure 1

Obviously, $A$ and $B$ are located on the boundary of the polygon $P$, and $A, B, A^{\prime}$ and $B^{\prime}$ are midpoints of the sides of $R_{1}$. We note that $P \subseteq R_{1}$. Otherwise, there would be a point $Z \in P$ but $Z \notin R_{1}$, i.e., one of the lines $a, b, a^{\prime}$ or $b^{\prime}$ were between $O$ and $Z$. If it is $a$, we have $|O Z B|>|O A B|$, which is contradictory to the choice of $A$ and $B$. If it is one of the lines $b, a^{\prime}$ or $b^{\prime}$ almost identical arguments lead to a similar contradiction.

Let $R_{2}$ be the parallelogram $A B A^{\prime} B^{\prime}$. Since $A$ and $B$ are points of $P$, segment $A B \subset P$ and so $R_{2} \subset R_{1}$. Since $A, B, A^{\prime}$ and $B^{\prime}$ are midpoints of the sides of $R_{1}$, an easy argument yields

$$
\left|R_{1}\right|=2 \cdot\left|R_{2}\right|
\tag{1}
$$

Let $R_{3}$ be the smallest parallelogram enclosing $P$ defined by lines parallel to $A B$ and $B A^{\prime}$. Obviously $R_{2} \subset R_{3}$ and every side of $R_{3}$ contains at least one point of the boundary of $P$. Denote by $C$ the intersection point of $a$ and $b$, by $X$ the intersection point of $A B$ and $O C$, and by $X^{\prime}$ the intersection point of $X C$ and the boundary of $R_{3}$. In a similar way denote by $D$



the intersection point of $b$ and $a^{\prime}$, by $Y$ the intersection point of $A^{\prime} B$ and $O D$, and by $Y^{\prime}$ the intersection point of $Y D$ and the boundary of $R_{3}$.

Note that $O C=2 \cdot O X$ and $O D=2 \cdot O Y$, so there exist real numbers $x$ and $y$ with $1 \leq x, y \leq 2$ and $O X^{\prime}=x \cdot O X$ and $O Y^{\prime}=y \cdot O Y$. Corresponding sides of $R_{3}$ and $R_{2}$ are parallel which yields

$$
\left|R_{3}\right|=x y \cdot\left|R_{2}\right| .
\tag{2}
$$

The side of $R_{3}$ containing $X^{\prime}$ contains at least one point $X^{*}$ of $P$; due to the convexity of $P$ we have $A X^{*} B \subset P$. Since this side of the parallelogram $R_{3}$ is parallel to $A B$ we have $\left|A X^{*} B\right|=\left|A X^{\prime} B\right|$, so $\left|O A X^{\prime} B\right|$ does not exceed the area of $P$ confined to the sector defined by the rays $O B$ and $O A$. In a similar way we conclude that $\left|O B^{\prime} Y^{\prime} A^{\prime}\right|$ does not exceed the area of $P$ confined to the sector defined by the rays $O B$ and $O A^{\prime}$. Putting things together we have $\left|O A X^{\prime} B\right|=x \cdot|O A B|,\left|O B D A^{\prime}\right|=y \cdot\left|O B A^{\prime}\right|$. Since $|O A B|=\left|O B A^{\prime}\right|$, we conclude that $|P| \geq 2 \cdot\left|A X^{\prime} B Y^{\prime} A^{\prime}\right|=2 \cdot\left(x \cdot|O A B|+y \cdot\left|O B A^{\prime}\right|\right)=4 \cdot \frac{x+y}{2} \cdot|O A B|=\frac{x+y}{2} \cdot R_{2}$; this is in short

$$
\frac{x+y}{2} \cdot\left|R_{2}\right| \leq|P| \text {. }
\tag{3}
$$

Since all numbers concerned are positive, we can combine (1)-(3). Using the arithmeticgeometric-mean inequality we obtain

$$
\left|R_{1}\right| \cdot\left|R_{3}\right|=2 \cdot\left|R_{2}\right| \cdot x y \cdot\left|R_{2}\right| \leq 2 \cdot\left|R_{2}\right|^{2}\left(\frac{x+y}{2}\right)^{2} \leq 2 \cdot|P|^{2}
$$

This implies immediately the desired result $\left|R_{1}\right| \leq \sqrt{2} \cdot|P|$ or $\left|R_{3}\right| \leq \sqrt{2} \cdot|P|$.
We construct the parallelograms $R_{1}, R_{2}$ and $R_{3}$ in the same way as in Solution 1 and will show that $\frac{\left|R_{1}\right|}{|P|} \leq \sqrt{2}$ or $\frac{\left|R_{3}\right|}{|P|} \leq \sqrt{2}$.

<img_3598>

Figure 2

Recall that affine one-to-one maps of the plane preserve the ratio of areas of subsets of the plane. On the other hand, every parallelogram can be transformed with an affine map onto a square. It follows that without loss of generality we may assume that $R_{1}$ is a square (see Figure 2).

Then $R_{2}$, whose vertices are the midpoints of the sides of $R_{1}$, is a square too, and $R_{3}$, whose sides are parallel to the diagonals of $R_{1}$, is a rectangle.

Let $a>0, b \geq 0$ and $c \geq 0$ be the distances introduced in Figure 2. Then $\left|R_{1}\right|=2 a^{2}$ and



$\left|R_{3}\right|=(a+2 b)(a+2 c)$.

Points $A, A^{\prime}, B$ and $B^{\prime}$ are in the convex polygon $P$. Hence the square $A B A^{\prime} B^{\prime}$ is a subset of $P$. Moreover, each of the sides of the rectangle $R_{3}$ contains a point of $P$, otherwise $R_{3}$ would not be minimal. It follows that

$$
|P| \geq a^{2}+2 \cdot \frac{a b}{2}+2 \cdot \frac{a c}{2}=a(a+b+c)
$$

Now assume that both $\frac{\left|R_{1}\right|}{|P|}>\sqrt{2}$ and $\frac{\left|R_{3}\right|}{|P|}>\sqrt{2}$, then

$$
2 a^{2}=\left|R_{1}\right|>\sqrt{2} \cdot|P| \geq \sqrt{2} \cdot a(a+b+c)
$$

and

$$
(a+2 b)(a+2 c)=\left|R_{3}\right|>\sqrt{2} \cdot|P| \geq \sqrt{2} \cdot a(a+b+c)
$$

All numbers concerned are positive, so after multiplying these inequalities we get

$$
2 a^{2}(a+2 b)(a+2 c)>2 a^{2}(a+b+c)^{2} .
$$

But the arithmetic-geometric-mean inequality implies the contradictory result

$$
2 a^{2}(a+2 b)(a+2 c) \leq 2 a^{2}\left(\frac{(a+2 b)+(a+2 c)}{2}\right)^{2}=2 a^{2}(a+b+c)^{2}
$$

Hence $\frac{\left|R_{1}\right|}{|P|} \leq \sqrt{2}$ or $\frac{\left|R_{3}\right|}{|P|} \leq \sqrt{2}$, as desired.