Three circles with two common points Let $ABC$ be a triangle of any type and $A_1,B_1,C_1$ the feet of the heights. Denote $M,N,P$ the orthogonal projections of the point $A$ onto the lines $B_1C_1,C_1A_1$ and $A_1B_1$.
The circes $AMA_1,ANB_1$ and $APC_1$ have two common points from my figure.
The proof seems not too easy, any help would be appreciated!

 A: I'll leave it as a fairly-straightforward exercise for the reader to show that $\bigcirc MNP$ is an excircle of $\triangle A_1B_1C_1$, and that it has center $A$. The result in the question then follows from the claim below, which one would apply to excircle $\bigcirc MNP$.
Note that the notation in the claim is independent of that in the question. Also, the claim is proven for distinct points on a circle; this corresponds to the requirement of distinct $M$, $N$, $P$ in the original question.



Claim. Given distinct $A$, $B$, $C$ on $\bigcirc O$, let $D$ be the point where the tangent lines to $\bigcirc O$ at $B$ and $C$ meet; likewise, let $E$ be the point there the tangents at $C$ and $A$ meet, and let $F$ be the point where the tangents at $A$ and $B$ meet. Then $\bigcirc OAD$, $\bigcirc OBE$, $\bigcirc OCF$ share two points ($O$ and, say, $P$).

For proof, we may take $\bigcirc O$ to be the unit circle, and give our points coordinates
$$A = \operatorname{cis}(2\alpha) \qquad B = \operatorname{cis}(2\beta) \qquad C = \operatorname{cis}(2\gamma)$$
where I'm abusing notation to write "$\operatorname{cis}\theta$" for "$(\cos\theta, \sin\theta)$". Then, since $\angle DOB = \angle DOC = |\beta-\gamma|$ and $|OB|/|OD| = \cos(\beta-\gamma)$, we have
$$D = \frac{\operatorname{cis}(\beta+\gamma)}{\cos(\beta-\gamma)} \quad
E = \frac{\operatorname{cis}(\gamma+\alpha)}{\cos(\gamma-\alpha)} \quad
F = \frac{\operatorname{cis}(\alpha+\beta)}{\cos(\alpha-\beta)}$$
As $\bigcirc OAD$ passes through the origin, its equation has form $x^2+y^2-2 h x - 2 k y = 0$, where $(h,k)$ is its center. Substituting $A$ and $D$ gives a linear system in $h$ and $k$ that we can solve, giving the circle equation:
$$x^2 + y^2 - x\frac{2\sin2\alpha-\sin2\beta-\sin2\gamma}{2\sin(2\alpha-\beta-\gamma)\cos(\beta-\gamma)} + y \frac{2\cos2\alpha-\cos2\beta-\cos2\gamma}{2\sin(2\alpha-\beta-\gamma)\cos(\beta-\gamma)} = 0$$
The equations for $\bigcirc OBE$ and $\bigcirc OCF$ arise from cycling the parameters $\alpha \to \beta \to \gamma \to \alpha$. Taking any two of these circles, we can determine their points of intersection; one of them is, of course, $O$. The other is
$$P := \frac{3 \; \left( \; \cos2\alpha + \cos2\beta + \cos2\gamma \;,\; \sin2\alpha + \sin2\beta + \sin2\gamma \; \right)}{3+2\cos2(\beta-\gamma)+2\cos2(\gamma-\alpha)+2\cos2(\alpha-\beta)}$$
Because the coordinates are symmetric in $\alpha$, $\beta$, $\gamma$, we are assured that $P$ lies on the third circle, as well, proving the claim.

Interestingly, as the coordinates of $P$ have the form $m \;(A + B + C )$, we find that $O$, $P$, and the centroid of $\triangle ABC$ are collinear.
