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Italian Mathematical Olympiad

Geometry Difficulty 4.7 AIME Prove it Italy

Problem:

Let $ABC$ be an acute triangle, and let $D, E$ be the feet of the altitudes from $A, B$ . Let $A'$ be the midpoint of $AD$ , $B'$ the midpoint of $BE$ . $CA'$ intersects $BE$ at $X$ , $CB'$ intersects $AD$ at $Y$ . Prove that there exists a circle passing through the points $A', B', X, Y$ .

Solution

Solution:

The triangles $ADC$ and $BEC$ are similar, being right triangles with the same angle at $C$ : it follows that the triangles $BB'C$ and $AA'C$ are also similar, and, in particular, that $\widehat{BB'C} = \widehat{CA'A}$ holds. There are now two cases: either the quadrilateral $A'XB'Y$ is crossed, or it is not.

In the first case, $A'$ and $B'$ see the segment $XY$ under the same angle.

In the second case, the quadrilateral $A'XB'Y$ has two opposite angles that are supplementary. In both cases, what has been proved is sufficient to establish the cyclicity of the quadrilateral $A'XB'Y$ , that is, that the vertices $A', X, B', Y$ belong to one and the same circle.

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