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Mongolian Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Mongolia

Ordered four points $A$ , $B$ , $C$ , $D$ lie on a given circle. Let $AC$ and $BD$ segments meet at $M$ . A line passing through the point $M$ and given circle meets at $M_1$ and $M_2$ , which line and the circles $\omega(ABM)$ and $\omega(CDM)$ meets at $N_1$ and $N_2$ respectively. Show that $M_1N_1 = M_2N_2$ .

(proposed by B. Ganbileg and U. Batzorig)

Solution

Let $O_1$ and $O_2$ be circumcentres of $\triangle ABM$ and $\triangle CDM$ respectively. Then first, we shall prove that $OO_1MO_2$ is parallelogram. Denote by $\angle ABM = \angle DCM = \varphi$ ; $\Rightarrow \angle AMO_1 = 90^\circ - \varphi$ and $\angle MCD = \varphi \Rightarrow O_1M \perp CD$ ; $OO_2 \perp CD \Rightarrow O_2M \parallel OO_2 \Rightarrow O_1MO_2O$ is parallelogram. So it is sufficient to show that
$\angle OO_2N_2 = \angle OO_2N. \qquad (1)$
Indeed, from (1) $\triangle OO_1N_1 = \triangle N_2O_2O \Rightarrow ON_1 = ON_2$ and $OM_1 = OM_2 \Rightarrow M_1N_1 = M_2N_2$ .

Denote $\alpha, \beta$ by $\angle BMN_1$ ; $\angle BAM$ respectively. Then
$\angle O_1MO_2 = \angle O_1MA + \angle AMD + \angle O_2MD = 90^\circ + \beta + \varphi \Rightarrow$

$\Rightarrow \angle MO_1O = \angle MO_2) = 90^\circ - \beta - \varphi \Rightarrow \\ \rightarrow OO_1N_1 = 90^\circ - \beta - \varphi + 2\alpha + 2\beta = 90 + \beta + 2\alpha - \varphi$
and after same computing we can see $\angle OO_2N_2 = 90^\circ + \beta + 2\alpha - \varphi = \angle OO_1N_1$ .

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