Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.3 Shortlist Prove it Turkey

In an acute triangle $ABC$ , let $D$ be point on the side $[BC]$ different than the vertices. Let $M_1, M_2, M_3, M_4, M_5$ be the midpoints of the line segments $[AD], [AB], [AC], [BD], [CD]$ , respectively; $O_1, O_2, O_3, O_4$ be the circumcenters of the triangles $ABD, ACD, M_1M_2M_4, M_1M_3M_5$ , respectively; $S$ and $T$ be the midpoints of the line segments $AO_1$ and $AO_2$ , respectively. Prove that $SO_3O_4T$ is an isosceles trapezoid.

Solution

As $\angle O_1M_1A = \angle O_1M_2A = 90^\circ$ , we have that $O_1M_1AM_2$ is a cyclic quadrilateral and the point $S$ is its circumcenter. Hence $S$ is the circumcenter of the triangle $AM_1M_2$ .

Next we observe that the triangles $AM_1M_2$ and $M_4M_2M_1$ are congruent since $AM_1 = M_4M_2 = \frac{AD}{2}$ and $AM_2 = M_4M_1 = \frac{AB}{2}$ . Therefore we obtain that the quadrilateral $SM_1O_3M_2$ is a rhombus and hence the line $M_1M_2$ is the perpendicular bisector of the line segment $[SO_3]$ .

Similarly we can get that $M_1M_3$ is the perpendicular bisector of the line segment $[TO_1]$ . As the points $M_1, M_2$ and $M_3$ are collinear, the result follows.

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