, 2020

Geometry Difficulty 6.1 National Olympiad Prove it Romania

Two congruent squares $ABCD$ and $EFGH$ are placed such that they have disjoint interiors, but $C$ is the midpoint of the line segment $EF$ and the points $B$ , $F$ , $G$ are collinear. The line $BC$ intersects $EH$ at $K$ and the line $AC$ intersects $GH$ at $M$ . Let $L$ be the midpoint of $GH$ , and let the parallel through $K$ to $GH$ intersect $FG$ at $N$ .

a) Prove that $CK = CL = CN$ .
b) Prove that $LM = 2HK$ .

Solution

a) Clearly, triangle $CFB$ is a right triangle, with $\angle F = 90^\circ$ , hence it is similar to $CEK$ . Moreover, since $CF = CE$ the two triangles are congruent, therefore $CK = CB = CL$ . Using the symmetry of $EFGH$ across $CL$ it also follows that $CK = CN$ .

b) In the right triangle $CBF$ we have $CB = 2CF$ , therefore $\angle CBF = 30^\circ$ . But $\angle CBF = \angle CKE = \angle KCL$ so the base angles in the isosceles triangle $CKL$ are both $75^\circ$ . It follows that $HKL$ is a right triangle with a $15^\circ$ angle. In the triangle $LMC$ we also have $\angle LCM = \angle KCM - \angle KCL = \angle ACB - 30^\circ = 15^\circ$ .

Thus, $HKL$ and $LMC$ are similar, and since $LC = 2HL$ , it follows that $LM = 2HK$ , as desired.

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