NMO Selection Tests

Let $\ell_D, \ell_E, \ell_F$ be the Simson lines of $D, E, F$ with respect to triangle $ABC$ and let $X = \ell_E \cap \ell_F$, $Y = \ell_F \cap \ell_D$ and $Z = \ell_D \cap \ell_E$. Let $D', E', F'$ be the midpoints of $HD, HE, HF$, respectively, and let $H'$ be the orthocentre of $\triangle$.

Note that $\ell_D$ passes through $D'$. This is a well-known fact about Simson lines. For the sake of completeness we provide a proof: Let $D_B$ and $D_C$ be the reflections of $D$ in $AB$ and $AC$, respectively. As the quadrangles $AHBD_B$ and $AHCD_C$ are both cyclic, $\angle AHD_B + \angle AHD_C = \angle ABD_B + \angle ACD_C = \angle ABD + \angle ACD = 180^\circ$, so $D_B, H, D_C$ are collinear. A factor $\frac{1}{2}$ homothety from $D$ settles the case.

Claim 1. The quadrangles $HE'XF', HF'YD', HD'ZE'$ are cyclic.
Proof. Let $r_E$ and $r_F$ be the rays from $A$ isogonal in $\angle BAC$ to rays $AE$ and $AF$, respectively. Note that $r_E \perp \ell_E$ and $r_F \perp \ell_F$ to write $\angle E'XF' = \angle (r_E, r_F) = 180^\circ - \angle EAF = \angle EDF = 180^\circ - \angle E'HF'$. Hence $HE'XF'$ is cyclic. Similarly, $HF'YD'$ and $HD'ZE'$ are both cyclic.

Claim 2. $H$ is the circumcentre of $\triangle$.
Proof. As $H$ is also the orthocentre of triangle $D'E'F'$, it follows that $\angle HYZ = \angle HF'D' = \angle HE'D' = \angle HZY$, so $HY = HZ$. Similarly, $HX = HY$ and the claim follows.

As $D', E', F'$ lie on the nine-point circle of triangle $DEF$, the midpoint $O'$ of $OH$ is the circumcentre of triangle $D'E'F'$. Proving that $H'$ lies on the circle on diameter $OH$ is equivalent to proving $O'H' = O'H$. The conclusion is then a consequence of Claim 3 below applied to $\triangle$ and points $D', E', F'$.

Claim 3. Let $ABC$ be a triangle with orthocentre $H$ and circumcentre $O \neq H$. Let $X, Y, Z$ be points on the sides $BC, CA, AB$, respectively, such that circles $AYZ, BZX, CXY$ are concurrent at $O$. Then the circumcentre of triangle $XYZ$ lies on the perpendicular bisectrix of $OH$.

Proof. Note that triangles $YOZ$ and $BHC$ are similar and the like. Then so are $XYZO$ and $ABCH$. Letting $O'$ be the circumcentre of triangle $XYZ$, it follows that $XYZOO'$ and $ABCHO$ are also similar. In particular, triangles $XOO'$ and $AHO$ are similar. Varying $X$ along the line $BC$ shows that $O'$ is the image of $X$ under a spiral-similarity from $O$. Consequently, $O'$ lies on some fixed line $\ell$.

As $AH$ and $BC$ are perpendicular, the angle formed by $\ell$ and $OH$ is equal to the angle formed by $OH$ and the image of $BC$ under a rotation through $\angle AHO = 90^\circ$. Letting $X, Y, Z$ be the midpoints of $BC, CA, AB$, respectively, it follows that $\ell$ passes through the midpoint of $OH$. Combining these two observations, it follows that $\ell$ is perpendicular bisectrix of $OH$, as desired. This ends the proof and completes the solution.

As usual, the complex coordinate of a point in the plane is denoted by the corresponding lower case letter. Use the notation in Solution 1 and let the circumcircle of triangles $ABC$ and $DEF$ be centred at the origin and have unit radius. Thus, $a, b, c, d, e, f$ all have a unit absolute value, i.e., $a\bar{a} = b\bar{b} = \ldots = f\bar{f} = 1$ and
$$h = a + b + c = d + e + f = \frac{1}{2}(a + b + c + d + e + f).$$