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Problem 2125

National Olympiad second round; IMO P1/P4

Geometry Difficulty 7.7 Prove it IMO Hk TST · Hong Kong

Let $ABCD$ be a convex quadrilateral with $AB = 5$ , $AD = 17$ , and $CD = 6$ . If the angle bisectors of $\angle BAD$ and $\angle ADC$ intersect at the midpoint of $BC$ , find the area of $ABCD$ .

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

Let $M$ be the midpoint of $BC$ . Let $B'$ and $C'$ be points on $AD$ such that $AB' = AB = 5$ and $DC' = DC = 6$ . Then $B'C' = 17 - 5 - 6 = 6$ . Note that $\triangle ABM \cong \triangle AB'M$ and $\triangle DCM \cong \triangle DC'M$ . Note also that $\triangle MB'C'$ is isosceles as $MB' = MB = MC = MC'$ . Let $N$ be the midpoint of $B'C'$ . Then we have $B'N = C'N = 3$ and $\angle MB'C' = \angle MC'B'$ . Observe that $MCDB'$ is cyclic since $MD$ bisects $\angle B'DC$ and $MB' = MC$ (while $DB' \ne DC$ ). Thus,
$2 \angle B'MA = \angle B'MB = \angle C'DC = 2 \angle MDC'.$

Finally we have $MN = \sqrt{MC'^2 - NC'^2} = \sqrt{30-9} = \sqrt{21}$ . Therefore,
$\begin{align*} [ABCD] &= 2([AB'M] + [DC'M] + [MB'N]) \\ &= (AB' + C'D + B'N) \times MN \\ &= (5+6+3)\sqrt{21} = 14\sqrt{21}. \end{align*}$

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.