Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National Olympiad Prove it Italy

A triangle $ABC$ is given, right-angled at $A$ and with $AC$ the longer leg; let $M$ be the midpoint of $BC$ , $N$ the reflection of $A$ with respect to $BC$ , $O$ the intersection between the perpendicular to $MN$ passing through $N$ and the line containing $BC$ .

a) Prove that angle $OMN$ is twice angle $ACB$ .

b) Prove that the ratio between the areas of $MNO$ and $ABC$ equals one quarter of the ratio between the lengths of $BC$ and $HM$ , where $H$ is the foot of the altitude relative to the hypotenuse of $ABC$ .

Solution

The midpoint $M$ of the hypotenuse $BC$ is also the center of the circle circumscribed about triangle $ABC$ ; the central angle $A\widehat{M}B$ and the inscribed angle $A\widehat{C}B$ subtend the same arc, hence $A\widehat{M}B = 2 A\widehat{C}B$ . Moreover $A\widehat{M}B = O\widehat{M}N$ , given the symmetry of $A$ and $N$ with respect to $BC$ ; from this we get precisely that $O\widehat{M}N = 2 A\widehat{C}B$ .

The respective areas of triangle $MNO$ and triangle $ABC$ , calculated by choosing the hypotenuse as the base in both cases, are equal to $\frac{MO \cdot NH}{2}$ and $\frac{BC \cdot AH}{2}$ ; but, since by symmetry $NH = AH$ , the ratio between the areas is exactly equal to the ratio between $MO$ and $BC$ . Euclid's first theorem applied to the right triangle $MNO$ gives the relation $\frac{MO}{MN} = \frac{MN}{HM}$ , that is $MO = \frac{MN^{2}}{HM}$ . On the other hand we also have $MN = AM$

therefore $\frac{MO}{BC} = \frac{BC^{2}}{4 BC \cdot HM} = \frac{BC}{4 HM}$ , from which the thesis follows.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.