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2002 JBMO Problems/Problem 2

Problem

Two circles with centers $O_{1}$ and $O_{2}$ meet at two points $A$ and $B$ such that the centers of the circles are on opposite sides of the line $AB$. The lines $BO_{1}$ and $BO_{2}$ meet their respective circles again at $B_{1}$ and $B_{2}$. Let $M$ be the midpoint of $B_{1}B_{2}$. Let $M_{1}$, $M_{2}$ be points on the circles of centers $O_{1}$ and $O_{2}$ respectively, such that $\angle AO_{1}M_{1}= \angle AO_{2}M_{2}$, and $B_{1}$ lies on the minor arc $AM_{1}$ while $B$ lies on the minor arc $AM_{2}$. Show that $\angle MM_{1}B = \angle MM_{2}B$.


Solution

It's easy to see that $B_{1}AB_{2}$ forms a straight line and is parallel to line $O_{1}O_{2}$.

Let us define $\angle AO_{1}M_{1}= \angle AO_{2}M_{2} = \theta$.

Let circumradii of the 2 circles be $R_{1}$ and $R_{2}$ respectively.

Now $\angle ABM_{2} = 180^{\circ} - \theta/2$ and $\angle ABM_{1} = \theta/2$, this implies that:

$\angle ABM_{2} + \angle ABM_{1} = 180^{\circ}$. So $M_{1}BM_{2}$ forms a straight line.

Now since $M$ is the midpoint of $B_{1}B_{2}$, $MO_{2}$ is parallel to $BB_{1}$ and its length is equal to $R_{1}$.

Similarly, we see that, $MO_{1}$ is parallel to $BB_{2}$ and it's length is equal to $R_{2}$.

So $\angle AMO_{1} = \angle AB_{2}B = \angle AO_{2}O_{1}$ (since $O_{2}$ is the circumcenter).

So $AMO_{2}O_{1}$ forms a cyclic quadrilateral.

Thus, we have $\angle MO_{2}A = \angle MO_{1}A$.

Adding $\theta$ to both sides we have: $\angle MO_{2}A + \theta = \angle MO_{1}A + \theta$ or, $\angle MO_{2}M_{2} = \angle MO_{1}M_{1}$

Thus by SAS, $\triangle MO_{2}M_{2}$ is congruent to $\triangle MO_{1}M_{1}$

So, we have $MM_{1} = MM_{2}$, hence $\triangle MM_{1}M_{2}$ is an isoceles triangle.

So, we get $\angle MM_{1}M_{2} = MM_{2}M_{1}$ -- (1)

Now $\angle B_{1}M_{1}B = 90^{\circ} = B_{2}M_{2}B$ So, $\angle B_{1}M_{1}B - \angle MM_{1}M_{2} = \angle B_{2}M_{2}B - \angle MM_{2}M_{1}$

giving $\angle B_{1}M_{1}M = \angle B_{2}M_{2}M$.


$Kris17$

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