Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.8 National Olympiad Prove it Italy

Problem:

Given a circle $\omega$ of diameter $AB$ and $P$ a point interior to the segment $AB$ , let $M$ be the midpoint of $PB$ . Let $r, s$ be two parallel lines passing respectively through $M, P$ , neither coincident with the line $AB$ nor orthogonal to it. Let $H$ be the orthogonal projection of $A$ onto $s$ and let $K$ be the intersection point (distinct from $A$ ) between $\omega$ and the line $AH$ . Finally, let $X, Y$ be the intersections of $r$ with $\omega$ , where $X$ is on the opposite side of $H$ with respect to $AB$ .

a. Prove that the triangle $HYK$ is isosceles.

b. Prove that $BXHY$ is a parallelogram.

Solution

Solution:

a.
Since $AB$ is a diameter of $\omega$ , the angle subtended by the corresponding arc is right, and thus $\widehat{AKB} = 90^{\circ} = \widehat{AHP}$ ; we conclude that $BK$ is parallel to $r$ and to $s$ . Then we can apply Thales' theorem to these three parallels and to the transversals $AB, AK$ : $r$ passes through the midpoint of $BP$ and therefore must also pass through the midpoint of $HK$ . Noting that $r$ is perpendicular to the segment $HK$ , we conclude that it must be its perpendicular bisector, whence the thesis (since $Y$ lies on $r$ ).

b.
Recalling that the lines $XY, BK$ are parallel, we note that $\widehat{KXY} = \widehat{XKB} = \widehat{XYB}$ and therefore the segments $BX, YK$ are congruent (since congruent angles subtend the respective arcs). But then $BX = YK = YH$ ; in a completely analogous way one sees that $X$ also lies on the perpendicular bisector of $HK$ and that $BY = XK$ , whence $XH = XK = BY$ . Therefore the quadrilateral $BXHY$ has pairs of opposite sides respectively congruent and is a parallelogram.

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