Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National olympiad Prove it Bulgaria

The quadrilateral $ABCD$ is inscribed. The point $H_1$ is the orthocenter of $\triangle ABC$ and the points $A_1$ and $B_1$ are symmetric to the points $A$ and $B$ with respect to the lines $BH_1$ and $AH_1$ , respectively. The point $O_1$ is a center of the circumscribed circle of $\triangle A_1B_1H_1$ . The point $H_2$ is the orthocenter of $\triangle ABD$ and the points $A_2$ and $B_2$ are symmetric to the points $A$ and $B$ with respect to the lines $BH_2$ and $AH_2$ , respectively. The point $O_2$ is a center of the circumscribed circle of $\triangle A_2B_2H_2$ . Denote the line $O_1O_2$ by $l_{AB}$ . The lines $l_{BC}$ , $l_{CD}$ , and $l_{DA}$ are defined analogously. Let $l_{AB} \cap l_{BC} = M$ , $l_{BC} \cap l_{CD} = N$ , $l_{CD} \cap l_{DA} = P$ , and $l_{DA} \cap l_{AB} = Q$ . Prove that the points $M, N, P$ , and $Q$ are concyclic.

Solution

We will need the following lemma.

Lemma. The point $O_1$ lies on the line $H_1O$ , where $O$ is the center of the circumcircle of $\triangle ABC$ . The ratio $H_1O_1 : O_1O$ depends on $\angle ACB$ only.

Proof. Let $K$ and $L$ be the intersection points of $AH_1$ and $BH_1$ with the circumcircle of $\triangle ABC$ . Since $H_1$ and $L$ are symmetric with respect to $AC$ , the quadrilateral $H_1ALA_1$ is a rhombus, as $\angle AH_1A_1 = 2 \angle AH_1L = 2\gamma$ . Analogously, $H_1BKB_1$ is a rhombus with $\angle BH_1B_1 = 2\gamma$ . Therefore $H_1ALA_1$ and $H_1BKB_1$ are similar, whence
$\frac{H_1T}{H_1X} = \frac{H_1Q}{H_1Y} = \frac{1}{1 + 2 \cos 2\gamma}$
This means that $O_1$ belongs to the line $H_1O$ and the ratio $H_1O_1 : O_1O$ depends on $\angle ACB$ only.

It is clear that $CH_1H_2D$ is a parallelogram (it follows from $CH_1 = CH_2 = 2R \cos \gamma$ ) and the points $O_1$ and $O_2$ are homothetic to $H_1$ and $H_2$ , respectively, with respect to $O$ . It follows from the lemma that the ratios of these two homotheties are equal (since they depend on $\gamma$ only). Therefore $O_1O_2 \parallel H_1H_2 \parallel CD$ , i.e. the line $l_{AB}$ is parallel to $CD$ .

We obtain that the sides of $MNPQ$ are parallel to the corresponding sides of $ABCD$ . This means that $MNPQ$ is inscribed.

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