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, 2011

Geometry Difficulty 5.4 AIME, harder Prove it Singapore

In the acute-angled non-isosceles triangle $ABC$ , $O$ is its circumcentre, $H$ is its orthocentre and $AB > AC$ . Let $Q$ be a point on $AC$ such that the extension of $HQ$ meets the extension of $BC$ at the point $P$ . Suppose $BD = DP$ , where $D$ is the foot of the perpendicular from $A$ onto $BC$ . Prove that $\angle ODQ = 90^\circ$ .

Solution

Drop perpendiculars $OM$ and $QX$ onto $BC$ , and $QY$ from $Q$ onto $AD$ . First $2DM = DM + BD - BM = BD - (BM - DM) = PD - (CM - DM) = PD - CD = PC$ . It is a well-known fact that $2OM = AH$ .

Next $\angle CPQ = \angle DBH = \angle HAQ$ so that the triangles $CPQ$ and $HAQ$ are similar. Thus the triangles $XPQ$ and $YAQ$ are similar. Therefore
$\frac{QX}{DX} = \frac{QX}{QY} = \frac{PC}{AH} = \frac{DM}{OM}.$
Hence the triangles $DXQ$ and $OMD$ are similar. It follows that $\angle ODQ = 90^\circ$ .

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