Nice one!

Let $X=Y_aY_c\cap Z_aZ_b$ and cyclically define $Y$ and $Z$. By the cyclicities, we have $$\measuredangle Y_aZ_aX=\measuredangle BZ_aZ_b=\measuredangle BAZ_b=\measuredangle Y_cAC=\measuredangle Y_cY_aC=\measuredangle XY_aZ_a$$Therefore, $XZ_aY_a$ is isosceles, meaning that the axis of $Z_aY_a$ is also the internal bisector of $\angle Z_aXY_a$, which is also the internal angle bisector of $\angle YXZ$. Therefore, the three axes must concur at the inventer of $XYZ$.

It was Exam 6, Problem 1, by the way.

MathSaiyan wrote: $BM_A-CM_A = 2(Z_AC-Y_AB)$ How are you getting this? (The first solution I discovered was very close to yours, but the formula for $BM_A - CM_A$ is a bit longer, and the formula for $BM_A^2 - CM_A^2$ does not have the factors of $2$ that yours has.) cadaeibf wrote: Let $X=Y_aY_c\cap Z_aZ_b$ and cyclically define $Y$ and $Z$. By the cyclicities, we have $$\measuredangle Y_aZ_aX=\measuredangle BZ_aZ_b=\measuredangle BAZ_b=\measuredangle Y_cAC=\measuredangle Y_cY_aC=\measuredangle XY_aZ_a$$Therefore, $XZ_aY_a$ is isosceles, meaning that the axis of $Z_aY_a$ is also the internal bisector of $\angle Z_aXY_a$, which is also the internal angle bisector of $\angle YXZ$. Therefore, the three axes must concur at the inventer of $XYZ$. Be careful -- the bisector isn't always internal. This is the shortest solution if one assumes that the bisectors come out "right" (i.e., either all three are internal, or exactly two are external), but it takes a bit of angle chasing to confirm this rigorously (the proper statement to prove is that $\measuredangle\left(p, Z_a Z_b\right) + \measuredangle\left(q, X_b X_c\right) + \measuredangle\left(r, Y_c Y_a\right) = 90^\circ$, where $p, q, r$ are the perpendicular bisectors of $Y_a Z_a, Z_b X_b, X_c Y_c$, respectively). Note: Your solution is the one Georg Schröter found as well. Tintarn wrote: It was Exam 6, Problem 1, by the way. Thanks -- Georg never told me what exam he used this in.

And here is my favorite solution (since it shows something extra): Let $X, Y, Z$ be the centers of the circles $x, y, z$. Let $X', Y', Z'$ be the reflections of the points $X, Y, Z$ through the midpoints of the segments $YZ, ZX, XY$. Thus, triangle $X'Y'Z'$ is the antimedial triangle of triangle $XYZ$. Hence, the homothety with center in the centroid of $\triangle XYZ$ and ratio $-2$ sends $X, Y, Z$ to $X', Y', Z'$. Let $h$ denote this homothety. Thus, $X' = h\left(X\right)$. Let $O$ the circumcenter of $\triangle ABC$. We shall prove that the point $h\left(O\right)$ belongs to the perpendicular bisectors of $Y_a Z_a, Z_b X_b, X_c Y_c$. This will clearly solve the exercise (and yield an Euler-line-like collinearity to boot). We let $M_{UV}$ denote the midpoint of any given segment $UV$. For any point $P$, we let $\overline{P}$ denote the projection of $P$ onto the line $BC$. It is well-known (and follows from the theorem about the midline in a trapezoid) that $\overline{M_{UV}} = M_{\overline{U}\overline{V}}$ for any segment $UV$. We shall refer to this equality as the midpoint-projection identity. The quadrilateral $XYX'Z$ is a parallelogram (by the construction of $X'$). Hence, the midpoints of its diagonals $XX'$ and $YZ$ coincide. In other words, $M_{XX'} = M_{YZ}$. Hence, $\overline{M_{XX'}} = \overline{M_{YZ}}$. However, the midpoint-projection identity yields $\overline{M_{XX'}} = M_{\overline{X}\overline{X'}}$ and $\overline{M_{YZ}} = M_{\overline{Y}\overline{Z}}$. Thus, $M_{\overline{X}\overline{X'}} = \overline{M_{XX'}} = \overline{M_{YZ}} = M_{\overline{Y}\overline{Z}}$. However, $Y$ is the center of the circle $y$, and $\overline{Y}$ is the projection of this center onto the chord $C Y_a$ of this circle. Therefore, $\overline{Y}$ is the midpoint of this chord $C Y_a$. In other words, $\overline{Y} = M_{C Y_a}$. Similarly, $\overline{Z} = M_{B Z_a}$. Thus, $M_{\overline{Y}\overline{Z}} = M_{M_{C Y_a} M_{B Z_a}} = M_{M_{BC} M_{Y_a Z_a}}$. (Here, we have used the fact that $M_{PQ} M_{UV} = M_{UP} M_{QV}$ for any four points $P, Q, U, V$. This fact is clear from vector algebra or from the Varignon parallelogram, although the latter parallelogram is degenerate in the case where we are using this fact.) So we now know that $M_{\overline{X}\overline{X'}} = M_{M_{BC} M_{Y_a Z_a}}$. On the other hand, $X$ is the center of the circle $x$, and $\overline{X}$ is the projection of this center onto the chord $BC$ of this circle. Therefore, $\overline{X}$ is the midpoint of this chord $BC$. In other words, $\overline{X} = M_{BC}$. In view of this, we can rewrite $M_{\overline{X}\overline{X'}} = M_{M_{BC} M_{Y_a Z_a}}$ as $M_{M_{BC} \overline{X'}} = M_{M_{BC} M_{Y_a Z_a}}$. From this, we obtain $\overline{X'} = M_{Y_a Z_a}$ (because if three points $P, U, V$ satisfy $M_{PU} = M_{PV}$, then $U = V$). The perpendicular bisector of the segment $Y_a Z_a$ passes through the midpoint $M_{Y_a Z_a}$ (by definition), i.e., through the point $\overline{X'}$ (since $\overline{X'} = M_{Y_a Z_a}$). Furthermore, this bisector is orthogonal to the line $BC$ (since $Y_a$ and $Z_a$ lie on $BC$). Hence, this bisector is the perpendicular to $BC$ through $\overline{X'}$. But this latter perpendicular clearly passes through $X'$ (since the construction of $\overline{X'}$ entails $X'\overline{X'} \perp BC$). As an upshot, we see that the perpendicular bisector of the segment $Y_a Z_a$ passes through $X' = h\left(X\right)$. Now, let $\mathbf{m}\left(UV\right)$ denote the perpendicular bisector of any segment $UV$. We claim that $\mathbf{m}\left(Y_a Z_a\right)$ is the image of $\mathbf{m}\left(BC\right)$ under the homothety $h$. Indeed, the perpendicular bisectors $\mathbf{m}\left(BC\right)$ and $\mathbf{m}\left(Y_a Z_a\right)$ are both orthogonal to $BC$, and thus are parallel to one another. Moreover, the former bisector passes through $X$ (since $X$ is the center of the circle $x$, while $BC$ is a chord of this circle), while the latter passes through $h\left(X\right)$ (as we saw in the previous paragraph). Hence, the homothety $h$ sends the former bisector to the latter (since a homothety sends any line to a parallel line). Since the former bisector contains $O$ (because $O$ is the circumcenter of $\triangle ABC$), we thus conclude that the latter bisector contains $h\left(O\right)$. In other words, the point $h\left(O\right)$ belongs to the perpendicular bisector of $Y_a Z_a$. Similarly, it belongs to the perpendicular bisectors of $Z_b X_b$ and $X_c Y_c$ as well. This proves our claim and solves the exercise.