IMO Problem Shortlist

We will construct two parallelograms $R_1$ and $R_3$, each of them containing $P$, and prove that at least one of the inequalities $|R_1|\le \sqrt{2}|P|$ and $|R_3|\le \sqrt{2}|P|$ 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 $A^{\prime }$, $B^{\prime }$, $a^{\prime }$ 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 $b^{\prime }$. 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 $|OZB|>|OAB|$, 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 $AB{A}^{\prime }{B}^{\prime }$. Since $A$ and $B$ are points of $P$, segment $AB\subset P$ 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 $$|R_1|=2\cdot |R_2|.\text{\hspace{1em}(1)}$$ Let $R_3$ be the smallest parallelogram enclosing $P$ defined by lines parallel to $AB$ 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 $AB$ and $OC$, and by $X^{\prime }$ the intersection point of $XC$ and the boundary of $R_3$. In a similar way denote by $D$ the intersection point of $b$ and $a^{\prime }$, by $Y$ the intersection point of $A^{\prime }B$ and $OD$, and by $Y^{\prime }$ 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 $1\le x,y\le 2$ and $O{X}^{\prime }=x\cdot OX$ and $O{Y}^{\prime }=y\cdot OY$. Corresponding sides of $R_3$ and $R_2$ are parallel which yields $$|R_3|=xy\cdot |R_2|.\text{\hspace{1em}(2)}$$ The side of $R_3$ containing $X^{\prime }$ contains at least one point $X^{\ast }$ of $P$; due to the convexity of $P$ we have $A{X}^{\ast }B\subset P$. Since this side of the parallelogram $R_3$ is parallel to $AB$ we have $|A{X}^{\ast }B|=|A{X}^{\prime }B|$, so $|OA{X}^{\prime }B|$ 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 $|O{B}^{\prime }{Y}^{\prime }{A}^{\prime }|$ does not exceed the area of $P$ confined to the sector defined by the rays $OB$ and $O{A}^{\prime }$. Putting things together we have $|OA{X}^{\prime }B|=x\cdot |OAB|$, $|OBD{A}^{\prime }|=y\cdot |OB{A}^{\prime }|$. Since $|OAB|=|OB{A}^{\prime }|$, we conclude that $|P|\ge 2\cdot |A{X}^{\prime }B{Y}^{\prime }{A}^{\prime }|=2\cdot (x\cdot |OAB|+y\cdot |OB{A}^{\prime }|)=4\cdot \frac{x+y}{2}\cdot |OAB|=\frac{x+y}{2}\cdot |R_2|$; this is in short $$\frac{x+y}{2}\cdot |R_2|\le |P|.\text{\hspace{1em}(3)}$$ Since all numbers concerned are positive, we can combine (1)-(3). Using the arithmetic-geometric-mean inequality we obtain $$|R_1|\cdot |R_3|=2\cdot |R_2|\cdot xy\cdot |R_2|\le 2\cdot |R_2{|}^2{\left(\frac{x+y}{2}\right)}^2\le 2\cdot |P{|}^2.$$ This implies immediately the desired result $|R_1|\le \sqrt{2}\cdot |P|$ or $|R_3|\le \sqrt{2}\cdot |P|$.

---

Alternative solution.

We construct the parallelograms $R_1$, $R_2$ and $R_3$ in the same way as in Solution 1 and will show that $\frac{|R_1|}{|P|}\le \sqrt{2}$ or $\frac{|R_3|}{|P|}\le \sqrt{2}$. 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 $a>0$, $b\ge 0$ and $c\ge 0$ be the distances introduced in Figure 2. Then $|R_1|=2a^2$ and $|R_3|=(a+2b)(a+2c)$. Points $A$, $A^{\prime }$, $B$ and $B^{\prime }$ are in the convex polygon $P$. Hence the square $AB{A}^{\prime }{B}^{\prime }$ 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 $$|P|\ge a^2+2\cdot \frac{ab}{2}+2\cdot \frac{ac}{2}=a(a+b+c)$$ Now assume that both $\frac{|R_1|}{|P|}>\sqrt{2}$ and $\frac{|R_3|}{|P|}>\sqrt{2}$, then $$2a^2=|R_1|>\sqrt{2}\cdot |P|\ge \sqrt{2}\cdot a(a+b+c)$$ and $$(a+2b)(a+2c)=|R_3|>\sqrt{2}\cdot |P|\ge \sqrt{2}\cdot a(a+b+c).$$ 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 $$2a^2(a+2b)(a+2c)\le 2a^2{\left(\frac{(a+2b)+(a+2c)}{2}\right)}^2=2a^2(a+b+c{)}^2.$$ Hence $\frac{|R_1|}{|P|}\le \sqrt{2}$ or $\frac{|R_3|}{|P|}\le \sqrt{2}$, as desired.