Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.8 National Olympiad Prove it Ukraine

Let $ABCD$ be a square with the side-length $10$ . Let $E$ and $F$ be the points on the sides $BC$ and $AD$ respectively, so that $ABEF$ is a rectangle. The rectangle $KLMN$ is such that its vertices $K$ , $L$ , $M$ and $N$ belong to the segments $CD$ , $DF$ , $FE$ and $EC$ respectively. It turns out that the rectangles $ABEF$ and $KLMN$ are equal, and $AB = MN$ . Determine the length of the segment $AL$ .

Fig. 1

Solution

Consider the diagonals $AE$ and $LN$ of the equal rectangles $ABEF$ and $KLMN$ . Clearly, $AE = LN$ . Let $NX$ be the perpendicular to the segment $AD$ (Fig. 1). Clearly, $NX = AB = 10$ . Then $\triangle AEF = \triangle NXL$ , since they are right triangles with equal hypotenuses and sides. Thus, $AF = LX$ . Let $LY \perp BC$ , then the rectangle $LXNY$ is equal to the rectangles $ABEF$ and $KLMN$ . Thus, $\triangle LNY = \triangle NXL$ , since they are right triangles with equal hypotenuses and sides. Hence
$\angle MNE = \angle LNY - \angle LNM = \angle NLX - \angle NLK = \angle KLD,$
so $\triangle MNE = \triangle KLD$ , since they are right triangles with equal hypotenuses and angles. Therefore, $LD = NE = BY = AL$ , hence $L$ is a midpoint of $AD$ , thus, $AL = 5$ .

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