Saudi Booklet

Geometry Difficulty 6.2 National Olympiad Prove it Saudi Arabia

Let $ABC$ be a triangle with $\angle BAC = 90^\circ$ with the altitude $AH$ ( $H \in BC$ ). A circle $(\omega)$ passes through $B, C$ and cuts the segments $AB, AC$ at $M, N$ respectively. Circle $(\omega)$ also cuts the line $AH$ at $D, E$ ( $D$ lies between $A, H$ ). Suppose that $DE = AH\sqrt{5}$ , prove that the circumcircle of triangle $HMN$ is tangent to $BC$ .

Solution

Base on the power from $H$ to the circle $(\omega)$ , $HD \cdot HE = HB \cdot HC = AH^2$ . Moreover, $HD + HE = DE = AH\sqrt{5}$ . Thus, the lengths $HD, HE$ will be the solutions of the quadratic equation
$x^2 - AH\sqrt{5} \cdot x + AH^2 = 0.$
Since $D$ is inside triangle $ABC$ , $HD < AH$ , entails $HE > AH$ and will have $HE > HD$ . From there, solving the above equation, we get
$HD = \frac{-1 + \sqrt{5}}{2}AH \text{ and } HE = \frac{1 + \sqrt{5}}{2}AH.$
Hence,
$AD = AH - HD = \frac{3 - \sqrt{5}}{2}AH \text{ and } AE = AH + HE = \frac{3 + \sqrt{5}}{2}AH.$
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It follows that
$AD \cdot AE = \left(\frac{3 - \sqrt{5}}{2}\right) \left(\frac{3 + \sqrt{5}}{2}\right) AH^2 = AH^2.$
Furthermore, according to the power from $A$ to the circle $(\omega)$ , $AD \cdot AE = AM \cdot AB = AN \cdot AC$ . Therefore $AH^2 = AM \cdot AB = AN \cdot AC$ , which implies that $HM \perp AB$ , $HN \perp AC$ so the quadrilateral $AMHN$ is a rectangle. Therefore, the circle circumscribing triangle $HMN$ is also the circle with diameter $AH$ so it will be tangent to $BC$ . $\square$

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