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Problem 1745

National Olympiad, first round

Geometry Difficulty 6.1 Prove it Romanian Mathematical Olympiad · Romania

Given a triangle $ABC$ with $m(\angle A) = 90^\circ$ , $AC < AB$ , consider on the rays $BA$ and $AC$ points $E$ and $D$ respectively, such that $A \in (BE)$ , $C \in (AD)$ , $AE = AC$ and $AD = AB$ . Denote by $M$ and $N$ the midpoints of $[BC]$ and $[DE]$ respectively, and let $\{R\} = EC \cap BD$ . Show that $MN = RA$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

The hypothesis implies that triangles $\triangle ACE$ and $\triangle ABD$ are right angled and isosceles, so the triangle $\triangle RBE$ is also right angled and isosceles, that is $m(\widehat{RBE}) = m(\widehat{REB}) = 45^\circ$ . By the equality of triangles $\triangle ABC$ and $\triangle ADE$ (CC) we get $BC = DE$ .

In the triangles $BRC$ , $ABC$ , $RDE$ , $ADE$ lines $RM$ , $AM$ , $RN$ , $AN$ are medians corresponding to right angles, so $RM = \frac{1}{2}BC$ , $AM = \frac{1}{2}BC$ , $RN = \frac{1}{2}DE$ , $AN = \frac{1}{2}DE$ , implying $RM = AM = RN = AN$ , that is the quadrilateral $AMRN$ is a rhombus. (*)

In the right angled triangle $ABC$ , as $AM$ is a median, the triangle $\triangle MAC$ is isosceles, so $m(\widehat{MAC}) = m(\widehat{MCA})$ . In the same way in $\triangle ADE$ , $m(\widehat{NAD}) = m(\widehat{NDA})$ . This gives $m(\widehat{MAN}) = m(\widehat{MAC}) + m(\widehat{NAC}) = m(\widehat{MCA}) + m(\widehat{NDA}) = m(\widehat{MCA}) + m(\widehat{ABC}) = 90^\circ$ . (**)

By () and (*), the quadrilateral $AMRN$ is a square, and, as a conclusion $MN = RA$ .

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