Bay Area Mathematical Olympiad

Solution:

Let $D$ be the reflection of $A$ across side $B C$, which clearly lies on $k$. Let $K'$ be the point where lines $B D$ and $M O$ intersect. We will eventually show that $K' = K$. Then $K$ lies on line $B D$, which is therefore the same as line $B K$. Since $B$ and $D$ don't depend on the choice of $M$, this proves that the line $B K$ does not depend on $M$, if all of the constructions are well-defined. One possible configuration is illustrated below. Notice that we cannot have $M = B$, because then the circumcenter $O$ of triangle $\triangle A B M$ would not be well-defined (we will need this later).

We first show that $\angle A M O = \angle B D A$, by focusing on the circumcircle of $\triangle A B M$, as illustrated below. Let $L$ be the point diametrically opposite $M$. Then $\angle L A M$ is right and thus $\angle A M O = 90^{\circ} - \angle A L M$. Since $A D \perp B M$, we have $\angle B D A = \angle B A D = 90^{\circ} - \angle A B M$. But the angles $\angle A B M$ and $\angle A L M$ subtend the same arc, so $\angle A B M = \angle A L M$ and thus $\angle B D A = \angle A M O$, as desired.

We will now show that the points $A, D, M$, and $K'$ are concyclic. Recall the following theorem.
A convex quadrilateral $E F G H$ is cyclic if either of the following are true:
(1) $\angle E F G + \angle E H G = 180^{\circ}$.
(2) $\angle E G F = \angle E H F$.
We will apply this theorem to four separate cases, illustrated below.

(a.i)

(b.i)

(b.ii)

(a) Suppose that $M$ lies to the right of $A D$. Then $O$ and $K'$ are on the same side of $M$.

i. If $B$ and $K'$ are on the same side of $D$, then $\angle A M K' = \angle A M O = \angle A D B = \angle A D K'$. Therefore, quadrilateral $A M D K'$ is cyclic by condition (2).

ii. If $B$ and $K'$ are on opposite sides of $D$, then $\angle A M K' = \angle A M O = \angle A D B = 180^{\circ} - \angle A D K'$, so $\angle A M K' + \angle A D K' = 180^{\circ}$. Therefore, quadrilateral $A M K' D$ is cyclic by condition (1).

(b) Now suppose that $M$ lies to the left of $A D$. Then $O$ and $K'$ are on opposite sides of $M$.

i. If $B$ and $K'$ are on the same side of $D$, then $\angle A D K' = \angle A D B = \angle A M O = 180^{\circ} - \angle A M K'$, so $\angle A D K' + \angle A M K' = 180^{\circ}$. Therefore, quadrilateral $A D K' M$ is cyclic by condition (1).

ii. If $B$ and $K'$ are on opposite sides of $D$, then $\angle A D K' = 180^{\circ} - \angle A D B = 180^{\circ} - \angle A M O = \angle A M K'$. Therefore, quadrilateral $A K' D M$ is cyclic by condition (2).

This proves that the points $A, D, M$, and $K'$ are concyclic. Since $k$ is the unique circle going through points $A, D$, and $M$, it follows that $K'$ lies on $k$. Thus $K'$ lies on line $M O$ and circle $k$, so we have $K' = K$ or $K' = M$. But if $K' = M$, then $M$ lies on line $B D$ and thus $M = B$. We noted above that this cannot be the case (or $O$ will not be well-defined), so we have $K = K'$, as desired.