, 2009

Solution 1. First, we claim that, if $I$ is the incenter of triangle $ABC$, then $AD$ trisects $\angle HAI$ with $\angle IAH = 3\angle IAD$. Extend rays $AI$ and $AH$ to meet side $BC$ at $I_A$ and $H_A$, respectively. We want to show that $D$ lies on segment $I_A H_A$ with $\angle I_A A H_A = 3\angle I_A A D$. Since $\angle A H_A C = 90^\circ < \angle ADB$, $D$ lies on segment $B H_A$. Since $B O_B O_C C$ is cyclic, we have
$$\begin{aligned} 180^\circ &= \angle CBO_B + \angle O_B O_C C = (\angle ABO_B + \angle CBA) + (360^\circ - \angle AOCO_B - \angle CO_C A) \\ &= \angle ACO_C + \angle CBA + 360^\circ - \angle ACB - 2\angle CDA \\ &= (90^\circ - \angle CDA) + \angle CBA + 360^\circ - \angle ACB - 2\angle CDA, \end{aligned}$$
or $\angle ACB - \angle CBA = 3(90^\circ - \angle CDA)$. Let $H_1$ be the point on side $BC$ such that $\angle BAH_1 = \angle CAH_A$. It follows that
$$\begin{aligned} \angle H_1 A H_A &= \angle BAH_A - \angle H_A A C = (90^\circ - \angle CBA) - (90^\circ - \angle ACB) \\ &= \angle BCA - \angle ABC = 3(90^\circ - \angle ADC) = 3\angle DAH_A. \end{aligned}$$
It is also clear that $AI_A$ bisects $\angle H_1 A H_A$. It is not difficult to see that we have the left configuration shown below and that $\angle D A H_A = 2\angle I_A A D$, as claimed. To complete the proof of our main result, it suffices to show that $AI_A$ bisects $\angle X A D$; that is,
$$\angle BAX = \angle DAC. \qquad (1)$$

We claim that $ABXO_C$ is cyclic. Indeed, by noting that triangles $BXC$ and $O_CXC$ are both isosceles with $BX = O_CX = CX$, we have
$$\begin{align*} \angle XBA + \angle AOCX &= (\angle CBA + \angle CBX) + (\angle AOCO_B + \angle O_BOCX) \\ &= \angle CBA + \angle BCX + \angle ACB + \angle O_BOCX \\ &= \angle CBA + \angle BCX + (\angle O_CCB + \angle ACO_C) + \angle O_BOCX \\ &= (\angle CBA + \angle ACO_C) + \angle O_BOCX + (\angle BCX + \angle O_CCB) \\ &= (\angle CBA + \angle ABO_B) + \angle O_BOCX + \angle O_CCX \\ &= \angle CBO_B + \angle O_BOCX + \angle XOC_C = \angle CBO_B + \angle O_BOC_C; \end{align*}$$
that is, $ABXO_C$ is cyclic if and only if $BO_BOC_C$ is cyclic; this means $ABXO_C$ is cyclic. (This calculation applies when $X$ and $A$ are on opposite sides of line $BC$; but when they are on the same side of $BC$, the calculation is similar.)
Because $ABXO_C$ is cyclic and $O_C$ is the circumcenter of $ACD$, it follows that
$$\angle BAX = \angle BO_CX = 90^\circ - \frac{\angle O_CXB}{2} = 90^\circ - \angle O_CCB = 90^\circ - \angle O_CCD = \angle CAD,$$
which is (1). Our proof is complete.

Solution 2 (Based on work by Delong Meng). (We maintain the same notations as in the first solution.) Let $O$ be the circumcenter of triangle $ABC$. It is well known that $O$ lies on line $AH_1$ (because $\angle BAO = 90^\circ - \angle ACB = \angle H_AHC$). As shown in the first solution, $AD$ trisects $\angle OAH$. Let the other trisector of $\angle OAH$ meet the perpendicular bisector of segment $BC$ at $X_1$. It suffices to show that $X = X_1$. Because $BCO_CO_B$ is cyclic, it suffices to show that $X_1$ lies on the perpendicular bisector of segment $O_BO_C$.
Let $Y$ be the intersection of line $AD$ with this perpendicular bisector. By our angle trisector construction and by the fact that $O_C$ is the circumcenter of triangle $ADC$, we may set
$$\alpha = \angle OAX_1 = \angle X_1AY = \angle YAH = 90^\circ - \angle CDA = \angle O_CAC = \angle O_BAB.$$
Because $AH \parallel OY$, $\angle AYX_1 = \angle YAH = \alpha$. Hence $AX_1Y$ is an isosceles triangle and is similar to both $AO_BB$ and $AO_CC$. Thus the spiral similarity $S_2$ (see (c) in the comment before the first solution) sends $X_1$ to $Y$, because it sends $O_B$ to $B$, and $O_C$ to $C$. Because $Y$ lies on the perpendicular bisector of segment $BC$, $X_1$ lies on the perpendicular bisector of segment $O_BO_C$, completing our proof.

Solution 3 (By Gabriel Carroll). Note that $AO_B = DO_B$ and $AO_C = DO_C$; that is, $D$ and $A$ are reflections of each other across line $O_BO_C$. Let $P$ be the reflection image of $X$, so $APXD$ is an isosceles trapezoid (with bases AD and XP). In particular, we have $\angle ADP = \angle DAX$, and it suffices to show that $DP \perp BC$. To do this, set up a coordinate system with D at the origin, $B = (2b, 0)$, $C = (2c, 0)$. From isosceles triangles $O_BBD, O_CCD, XBC$ we get that $O_B, O_C, X$ have x-coordinates $b, c, b+c$, respectively. But $XO_BPO_C$ is a rhombus (because $XO_B = XO_C$), hence P has x-coordinate zero, and the result follows.