Rude quadrilateral

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Source: Serbian Mathematical Olympiad 2021, P4

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#1 h May 14, 2021, 11:18 AM

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A convex quadrilateral $ABCD$ will be called rude if there exists a convex quadrilateral $PQRS$ whose points are all in the interior or on the sides of quadrilateral $ABCD$ such that the sum of diagonals of $PQRS$ is larger than the sum of diagonals of $ABCD$ .

Let $r>0$ be a real number. Let us assume that a convex quadrilateral $ABCD$ is not rude, but every quadrilateral $A'BCD$ such that $A'\neq A$ and $A'A\leq r$ is rude. Find all possible values of the largest angle of $ABCD$ .

This post has been edited 1 time. Last edited by NZP_IMOCOMP4, May 14, 2021, 11:22 AM

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#2 h Nov 18, 2021, 3:12 PM

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Let $d = \text{max}\{AB, BC, CD, DA, AC, BD\}$ . Note that $d$ is the length of the longest possible segment whose vertices are in the interior or on the sides of quadrilateral $ABCD$ , so $PR + QS \le 2d$ . If segment $MN = d$ ( $M, N \in \{A, B, C, D\}$ ), then we can let $P$ and $Q$ be as close as possible to $M$ and $R$ and $S$ as close as possible to $N$ , making $PR + QS$ as close as possible to $2d$ . Thus, $ABCD$ is not rude if and only if $AC + BD = 2d$ which is equivalent to $\text{max}\{AB, BC, CD, DA\} \le AC = BD = d$ .

If $AB < d$ , then let $A’$ be a point on the same side of line $AC$ as $D$ such that $CA = CA’$ and $AA’ > 0$ is as small as possible. Since $\angle A’CD < \angle ACD$ , $A’D < AD < d$ . Note that $A’B > AB$ , however, as $A’$ gets closer to $A$ , $A’B$ gets closer to $AB$ , so there exists such $A’$ such that $A’B < d$ . Since $\text{max}\{A’B, BC, CD, DA’\} \le A’C = BD = d$ , $A’BCD$ is not rude, contradiction. Therefore, $AB = d$ . Similarly, $AD = d$ .

From this, $AB = AC = AD = BD = d$ , so $\triangle ABD$ is equilateral and $\angle BAD = 60^{\circ}$ . Also, $A$ is the circumcenter of $\triangle BCD$ , so $\angle BCD = 180^{\circ} - \frac{\angle BAD}{2} = 150^{\circ}$ . Note that $\angle ABC = 90^{\circ} - \frac{\angle BAC}{2} < 90^{\circ}$ , and $\angle CDA = 90^{\circ} - \frac{\angle CAD}{2} < 90^{\circ}$ , hence, the only possible value of the largest angle of $ABCD$ is $\boxed{150^{\circ}}$ .

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