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Romanian Mathematical Olympiad - Final Round

Geometry Difficulty 6.9 National Olympiad Prove it Romania

The triangle $ABC$ has $\angle BAC = 90^\circ$ and $\angle ABC = 60^\circ$ . The points $D$ and $E$ are taken on the sides $AC$ , respectively $AB$ , so that $CD = 2 \cdot DA$ and $DE$ is the bisector of the angle $\angle ADB$ . Denote $M$ the intersection of the lines $CE$ and $BD$ , and $P$ the intersection of the lines $DE$ and $AM$ . Prove that:

a) the lines $AM$ and $BD$ are perpendicular;

b) $3 \cdot PB = 2 \cdot CM$ .

Solution

a) Let $AB = a$ . Then $BC = 2a$ , $AC = \sqrt{BC^2 - AB^2} = a\sqrt{3}$ , $AD = \frac{a}{3}\sqrt{3}$ , $BD = \sqrt{AB^2 + AD^2} = \frac{2a}{3}\sqrt{3} = 2AD$ . This yields $\angle ABD = 30^\circ$ , hence $\angle ADB = 60^\circ$ .

This gives $\angle ADE = 30^\circ = \angle ACB$ , therefore $DE \parallel BC$ . So $\triangle DME \sim \triangle BMC$ and $\angle ADE \sim \angle ACB$ , whence $\frac{DM}{MB} = \frac{DE}{BC} = \frac{AD}{AC} = \frac{1}{3}$ . This implies $\frac{DM}{DB} = \frac{1}{4}$ , hence $\frac{DM}{DA} = \frac{DA}{DB} = \frac{1}{2}$ , so $\triangle DMA \sim \triangle DAB$ (S.A.S.), which implies $\angle AMD = \angle BAD = 90^\circ$ .

b) Construct $DF \parallel AM$ , $F \in CE$ . Then $\frac{CF}{CM} = \frac{CD}{CA} = \frac{2}{3}$ ; we have to prove that $CF = PB$ .

We have $\angle DAM = 90^\circ - \angle ADB = 30^\circ$ and $\angle ADE = 30^\circ$ , hence $DP = AP$ . In triangle $DPM$ , $PM = \frac{1}{2}DP = \frac{1}{2}AP$ , whence $\frac{MP}{MA} = \frac{MF}{MC} = \frac{1}{3} = \frac{MF}{MC}$ , showing that $PF \parallel AC$ . It follows that $APFD$ is a parallelogram, so $DF = AP = DP$ . Since $\angle CDF = \angle DAM = 30^\circ = \angle BDP$ and $BD = \frac{2a}{3}\sqrt{3} = CD$ , $\triangle CDF \equiv \triangle BDP$ (S.A.S.), therefore $CF = BP$ .

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