Saudi Booklet

Geometry Difficulty 5.6 AIME, harder Prove it Saudi Arabia

Let $ABC$ be a triangle inscribed in circle $(O)$ with angle $\angle A = 45^\circ$ and $AB < AC$ . Let $AD, AH$ be the angle bisector and altitude of triangle $ABC$ with $D, H$ are on $BC$ . Suppose that $OD$ intersects $AH$ at $E$ and $K$ is the circumcenter of triangle $EBC$ . Prove that $HK \parallel AD$ .

Solution

Let $P, N$ be the projections of $O, K$ on $AE$ and $M, T$ be the midpoints of $BC$ and the minor arc $BC$ of $(O)$ . Let $AH = h$ and $R, R'$ be the radii of $(O), (K)$ respectively.

By Thales' theorem, we have
$\frac{HE}{HA} = \frac{MO}{MT} = \frac{MO}{OT - OM} = \sqrt{2} + 1 \implies HE = (\sqrt{2} + 1)h.$
According to the Pythagorean theorem, $BK^2 - BO^2 = KM^2 - OM^2$ so
$R'^2 - R^2 = KM^2 - OM^2 \implies KM^2 = R'^2 - \frac{R^2}{2}.$
Similarly,
$\begin{align*} AO^2 - (AH - OM)^2 &= OP^2 = KN^2 = KE^2 - (EH - MK)^2 \\ \Leftrightarrow R^2 - \left(h - \frac{R}{\sqrt{2}}\right)^2 &= R'^2 - \left((\sqrt{2} + 1)h - MK\right)^2 \\ \Leftrightarrow MK^2 + h \cdot R\sqrt{2} &= R'^2 - \frac{R^2}{2} - (\sqrt{2} + 1)^2 h^2 + (2 + 2\sqrt{2})h \cdot MK \\ \Leftrightarrow R\sqrt{2} &= (2 + 2\sqrt{2})h + (2 + 2\sqrt{2})MK \\ \Leftrightarrow MK &= h + \frac{R\sqrt{2}}{2 + 2\sqrt{2}} = h + MT. \end{align*}$
Thus $KT = MK - MT = h = AH$ , proving that $AHKT$ is a parallelogram. Therefore, $AD \parallel HK$ . $\square$

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