Which elements are fixed? :

Of course, the locus of $ A$ is a circle with radius $ k$ centered at $ M$. Let's see what happens if we move $ \Delta$. Let $ O_1$ and $ O_2$ be the centers of $ C_1$ and $ C_2$, respectively. Some investigation in the picture (I did it using Geogebra) leads to the conjecture that $ M$ lie on a circle with center $ O$, with $ O$ the midpoint of $ O_1O_2$, and radius $ OP$. I tried to prove this with vectors, but couldn't finish the proof. $ \vec{O}=\frac{\vec{O_1}+\vec{O_2}}{2}$ $ \vec{M}=\frac{\vec{B}+\vec{C}}{2}$ $ \vec{O}-\vec{M}=\frac{1}{2}(\vec{O_1}-\vec{B}+\vec{O_1}-\vec{C})$ $ \vec{MO}=\frac{1}{2}(\vec{BO_1}+\vec{CO_2})$ So we need to prove that the absolute value of this vector is a constant. Denote $ \theta_1$ and $ \theta_2$ the angles that $ BO_1$ and $ CO_2$ make with $ O_1O_2$, respectively. Let $ r_1$ and $ r_2$ be the radii of $ C_1$ and $ C_2$, respectively. By the Law of Cosines $ |\vec{MO}|^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_1+\theta_2)$ So we need to prove that $ \theta_1+\theta_2$ is constant, that is, the angle between the lines $ BO_1$ and $ CO_2$ is constant. From here I couldn't finish. Can someone fill in the rest of the proof? Can someone find a purely synthetic solution?

Yes, M moves on the circle with center O (midpoint of O1O2) and through P. And that's very easy. However, when we move $ \Delta$ , finding the locus of A is really hard.

If we prove that M moves on (O,OP) circle, the problem is easy: the locus of A as we move M is a "ring" between two concentric circles, with center O and radii OP-k and OP+k. But we didn't prove that M actually moves on (O,OP).

Sashsiam_2 wrote: If we prove that M moves on (O,OP) circle, the problem is easy: the locus of A as we move M is a "ring" between two concentric circles, with center O and radii OP-k and OP+k. But we didn't prove that M actually moves on (O,OP).

Kostas Vittas.

[geogebra]1226a16e0a6eb3eb96fe9bee452c3ca0b0d06100[/geogebra] But $ O_1B$ is not parallel to $ O_2C$.

Sashsiam_2 wrote: [geogebra]1226a16e0a6eb3eb96fe9bee452c3ca0b0d06100[/geogebra] But $ O_1B$ is not parallel to $ O_2C$. Sorry dear all my friends because of I misunderstood the problem. I thought that the circles $ (C_{1}),\ (C_{2}),$ tangent (externally, in my drawing) each other at point $ P,$ because of the problem states that they are intersected at one only point. Kostas Vittas. PS. Unfortunately, I can't view your attachment dear Sashiam_2 and I don't understand why.

Hello! Problem: $ C_{1}\left(O_{1}\right)$, $ C_{2}\left(O_{2}\right)$ are two circles and $ \left\{P,Q\right\} = C_{1}\cap C_{2}$. A line $ \Delta$ through the point $ P$ intersects the circles $ C_{1},C_{2}$ at the points $ B,C$ and $ M,O$ are the midpoints of $ BC$ and $ O_{1}O_{2}$. Prove that $ OM = OP$. Solution: Construct the points $ B'\in C_{1}$ and $ C'\in C_{2}$ such that $ B'B\bot BC$ and $ C'C\bot BC$. So $ O_{1}$, $ O_{2}$ are the midpoints of $ PB'$ and $ PC'$. It's easy to prove that the points $ B',Q,C'$ are collinear and the quadrilateral $ BB'C'C$ is a right trapezoid ($ BB'||CC'$ and $ \angle B'BC = \angle C'CB = 90^{\circ}$). Let $ N$ be the midpoint of $ B'C'$. It's easy to prove that $ O$ is the midpoint of $ PN$ and $ NM\bot BC$ ($ MN$ is the central median of the right trapezoid $ BB'C'C$). So $ OM = OP$. P.S. So, when the line $ \Delta$ is drawn arbitrarily, the point $ M$ moves on the circle $ \left(O,OP\right)$. Best regards, Petrisor Neagoe