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Problem 1349

AIME late

Geometry Difficulty 5.5 Prove it Ukrainian National Mathematical Olympiad, 3rd Round · Ukraine

Let $ABCD$ be a cyclic quadruple. Let us denote the midpoints of $AB$ , $BC$ , $CD$ and $DA$ by $M$ , $L$ , $N$ and $K$ respectively. It is known, that $\angle BMN = \angle MNC$ . Prove that:

a) $\angle DKL = \angle CLK$ ;

b) $ABCD$ has a pair of parallel sides.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

a) Using the properties of inscribed angles we get $KM \perp BD$ , $KN \perp AC$ , $\angle ABD = \angle ACD \Rightarrow \angle AMK = \angle ABD = \angle ACD = \angle KND$ . Thus $\angle KMN = \pi - \angle AMK - \angle BMN = \pi - \angle KND - \angle MNC = \angle KNM$ , hence $\square KMN$ is isosceles and $KM = KN$ , which implies that $KMLN$ is rhombus and $\angle NKL = \angle NLK$ (fig. 8). Since by analogy we have that $\angle AMK = \angle KND$ , $\angle DKN = \angle NLC$ , then $\angle KLD = \angle DKN + \angle NKL = \angle NLC + \angle LNK = \angle CLK$ , which proves part a).

b) Since $KM = KN$ , then $DB = 2KM = 2KN = AC$ and $\angle ABC + \angle DAB = \pi$ . We have $DA \perp BC$ , or $\angle ABC = \angle DAB$ . But since $\angle DAC = \angle DBC$ , then $\angle CAB = \angle DAB - \angle DAC = \angle ABC - \angle DBC = \angle ABD = \angle ACD$ , and $AB \perp BC$ .

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