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Saudi Arabian Mathematical Competitions

Geometry Difficulty 7.5 National Olympiad, round 2 Prove it Saudi Arabia

Let $BC$ be a chord of a circle $(O)$ such that $BC$ is not a diameter. Let $AE$ be a diameter perpendicular to $BC$ such that $A$ belongs to the larger $\operatorname{arc} BC$ of $(O)$ . Let $D$ be a point on the larger $\operatorname{arc} BC$ of $(O)$ which is different from $A$ . Suppose that $AD$ intersects $BC$ at $S$ , $ED$ intersects $BC$ at $T$ . Let $F$ be the midpoint of $ST$ and $I$ be the second intersection of the circle $(ODF)$ with $BC$ .
1. Let the line passing $I$ and parallel to $OD$ intersect $AD$ and $ED$ at $M$ and $N$ respectively. Find the maximum value of the area of triangle $MDN$ when $D$ moves on the larger $\operatorname{arc} BC$ of $(O)$ (such that $D \neq A$ ).
2. Prove that the perpendicular from $D$ to $ST$ passes through the midpoint of $MN$ .

Solution

1) First, note that $\angle ADE = 90^\circ$ then $DO$ , $DF$ are two medians of the triangles $ADE$ , $SDT$ .
Then $\angle ODF = \angle ODT + \angle FDT = \angle OET + \angle FTD = 90^\circ$ . Hence, $\angle OIF = 180^\circ - \angle ODF = 90^\circ$ which implies that $I$ is the midpoint of $BC$ .
Since $ODE$ is isosceles triangle then $INE$ , $IMA$ are also isosceles which implies that $IE = IN$ , $IM = IA$ . Hence $MN = IM - IN = IA - IE = \text{const}$ .

Two triangles $DMN$ and $DAE$ are similar with the constant ratio, then to maximize the area of $DMN$ , we have to maximize the area of triangle $ADE$ . We have
$[ADE] = \frac{1}{2} DA \cdot DE \leq \frac{DA^2 + DE^2}{2} = \frac{AE^2}{2}.$
The equality occurs when $DA = DE$ or $D$ lies on circle such that $ADE$ is isosceles right triangle.

2) Denote $P$ as the midpoint of $MN$ then
$\angle PDN = \angle PND = \angle INE = \angle IEN,$
thus $DP \parallel AE$ or the perpendicular line from $D$ to $ST$ passes through the midpoint of $MN$ . $\square$

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