Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.4 Shortlist Prove it China

Find the smallest real number $\alpha$ , such that for any convex polygon $P$ of area $1$ , there exists a point $M$ in the plane, such that the area of the convex hull of $P \cup Q$ is at most $\alpha$ , where $Q$ is the central-symmetric figure of $P$ about $M$ .
(Contributed by Qu Zhenhua and Wu Yuchi)

Solution

(i) If $M$ is outside $P$ or on the boundary. Draw a line $l$ through $M$ such that $P$ , $Q$ lie on different sides of $l$ (they have no common interior points). Then $S(P \cup Q) \ge S(P) + S(Q) \ge 2$ .

(ii) If $M$ is inside $P$ . Let $A', B', C'$ be the respective symmetric points of $A, B, C$ about $M$ . Then $R = P \cup Q = \{A, B, C, A', B', C'\}$ , $R$ has a centre of symmetry, $R$ is a parallelogram or a hexagon. If $R$ is a parallelogram, say $R = ABA'B'$ , then $C$ is inside or on the boundary, and
$S(R) = S(ABA'B') \ge 2S(ABC) = 2.$
If $R$ is a hexagon, $R = AC'BA'CB'$ , then
$\begin{aligned} S(R) &= S(AC'BM) + S(BA'CM) + S(CB'AM) \\ &= (S(AMC) + S(BMC)) + (S(BMA) + S(CMA)) \\ &\quad + (S(CMB) + S(AMB)) = 2. \end{aligned}$
Next, we prove $\alpha = 2$ satisfies the problem statement in two ways. $\square$

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