IMO Hk TST

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Hong Kong

Let $ABCD$ be a cyclic quadrilateral. Show that the orthocentres of the triangles $\triangle ABC$ , $\triangle BCD$ , $\triangle CDA$ and $\triangle DAB$ are the vertices of a quadrilateral congruent to $ABCD$ and show that the centroids of the same triangles are the vertices of a cyclic quadrilateral.

Solution

Let $O$ be the centre of $(ABCD)$ . Let $H_A, H_B, H_C, H_D$ and $G_A, G_B, G_C, G_D$ be the orthocentres and centroids of $\triangle DAB$ , $\triangle ABC$ , $\triangle BCD$ and $\triangle CDA$ respectively. Let $M$ be the midpoint of $AB$ . Recall that $O$ lies on $H_A G_A$ , etc., with $OG_A : G_A H_A = 1 : 2$ . Also, we have $DG_A : G_A M = 2 : 1 = CG_B : G_B M$ . From these, we know that $H_A H_B \parallel G_A G_B \parallel DC$ , and
$H_A H_B = 3 G_A G_B = CD.$

By symmetry, we have $H_B H_C = DA$ , $H_C H_D = AB$ , $H_D H_A = BC$ , $H_A H_C = CA$ and $H_B H_D = DB$ . Thus, $H_A H_B H_C H_D$ is congruent to $CDAB$ . It follows

that $H_A H_B H_C H_D$ is concyclic. Now, note that $G_A G_B G_C G_D$ is the image of $H_A H_B H_C H_D$ under the homothety with centre $O$ and ratio $\frac{1}{3}$ . Therefore, $G_A G_B G_C G_D$ is cyclic.

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