Problem 1862

We will construct two parallelograms $R_{1}$ and $R_{3}$, each of them containing $P$, and prove that at least one of the inequalities $|R_{1}| \leq \sqrt{2}|P|$ and $|R_{3}| \leq \sqrt{2}|P|$ 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 $OAB$ has maximal area. Let $a$ be the line through $A$ parallel to $OB$ and $b$ the line through $B$ parallel to $OA$. Let $A'$, $B'$, $a'$ and $b'$ 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 $a$, $b$, $a'$ and $b'$.

Figure 1
Obviously, $A$ and $B$ are located on the boundary of the polygon $P$, and $A$, $B$, $A'$ and $B'$ 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 $a$, $b$, $a'$ or $b'$ were between $O$ and $Z$. If it is $a$, we have $|OZB| > |OAB|$, which is contradictory to the choice of $A$ and $B$. If it is one of the lines $b$, $a'$ or $b'$ almost identical arguments lead to a similar contradiction.
Let $R_{2}$ be the parallelogram $ABA'B'$. Since $A$ and $B$ are points of $P$, segment $AB \subset P$ and so $R_{2} \subset R_{1}$. Since $A$, $B$, $A'$ and $B'$ are midpoints of the sides of $R_{1}$, an easy argument yields
$$|R_{1}| = 2 \cdot |R_{2}| . \tag{1}$$
Let $R_{3}$ be the smallest parallelogram enclosing $P$ defined by lines parallel to $AB$ and $BA'$. 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 $AB$ and $OC$, and by $X'$ the intersection point of $XC$ and the boundary of $R_{3}$. In a similar way denote by $D$ the intersection point of $b$ and $a'$, by $Y$ the intersection point of $A'B$ and $OD$, and by $Y'$ the intersection point of $YD$ and the boundary of $R_{3}$.
Note that $OC = 2 \cdot OX$ and $OD = 2 \cdot OY$, so there exist real numbers $x$ and $y$ with $1 \leq x, y \leq 2$ and $OX' = x \cdot OX$ and $OY' = y \cdot OY$. Corresponding sides of $R_{3}$ and $R_{2}$ are parallel which yields
$$|R_{3}| = x y \cdot |R_{2}| . \tag{2}$$
The side of $R_{3}$ containing $X'$ contains at least one point $X^*$ of $P$; due to the convexity of $P$ we have $AX^*B \subset P$. Since this side of the parallelogram $R_{3}$ is parallel to $AB$ we have $|AX^*B| = |AX'B|$, so $|OAX'B|$ does not exceed the area of $P$ confined to the sector defined by the rays $OB$ and $OA$. In a similar way we conclude that $|OB'Y'A'|$ does not exceed the area of $P$ confined to the sector defined by the rays $OB$ and $OA'$. Putting things together we have $|OAX'B| = x \cdot |OAB|$, $|OBDA'| = y \cdot |OBA'|$. Since $|OAB| = |OBA'|$, we conclude that $|P| \geq 2 \cdot |AX'BY'A'| = 2 \cdot (x \cdot |OAB| + y \cdot |OBA'|) = 4 \cdot \frac{x+y}{2} \cdot |OAB| = \frac{x+y}{2} \cdot |R_{2}|$; this is in short
$$\frac{x+y}{2} \cdot |R_{2}| \leq |P| . \tag{3}$$
Since all numbers concerned are positive, we can combine (1)-(3). Using the arithmetic-geometric-mean inequality we obtain
$$|R_{1}| \cdot |R_{3}| = 2 \cdot |R_{2}| \cdot x y \cdot |R_{2}| \leq 2 \cdot |R_{2}|^{2} \left(\frac{x+y}{2}\right)^{2} \leq 2 \cdot |P|^{2} .$$
This implies immediately the desired result $|R_{1}| \leq \sqrt{2} \cdot |P|$ or $|R_{3}| \leq \sqrt{2} \cdot |P|$.

We construct the parallelograms $R_{1}$, $R_{2}$ and $R_{3}$ in the same way as in Solution 1 and will show that $\frac{|R_{1}|}{|P|} \leq \sqrt{2}$ or $\frac{|R_{3}|}{|P|} \leq \sqrt{2}$.

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 $|R_{1}| = 2a^{2}$ and $|R_{3}| = (a + 2b)(a + 2c)$.
Points $A$, $A'$, $B$ and $B'$ are in the convex polygon $P$. Hence the square $ABA'B'$ 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
$$|P| \geq a^{2} + 2 \cdot \frac{ab}{2} + 2 \cdot \frac{ac}{2} = a(a + b + c)$$
Now assume that both $\frac{|R_{1}|}{|P|} > \sqrt{2}$ and $\frac{|R_{3}|}{|P|} > \sqrt{2}$, then
$$2a^{2} = |R_{1}| > \sqrt{2} \cdot |P| \geq \sqrt{2} \cdot a(a + b + c)$$
and
$$(a + 2b)(a + 2c) = |R_{3}| > \sqrt{2} \cdot |P| \geq \sqrt{2} \cdot a(a + b + c) .$$
All numbers concerned are positive, so after multiplying these inequalities we get
$$2a^{2}(a + 2b)(a + 2c) > 2a^{2}(a + b + c)^{2}$$
But the arithmetic-geometric-mean inequality implies the contradictory result
$$2a^{2}(a + 2b)(a + 2c) \leq 2a^{2}\left(\frac{(a + 2b) + (a + 2c)}{2}\right)^{2} = 2a^{2}(a + b + c)^{2} .$$
Hence $\frac{|R_{1}|}{|P|} \leq \sqrt{2}$ or $\frac{|R_{3}|}{|P|} \leq \sqrt{2}$, as desired.