orl wrote: In $\bigtriangleup ABC$, $D$ is a point on side $BC$. $O_1$ and $I_1$ are the circumcenter and incenter of $\bigtriangleup ABD$ respectively, and $O_2$ and $I_2$ are the circumcenter and incenter of $\bigtriangleup ADC$ respectively. $O_1I_1$ intersects $O_2I_2$ at $P$. Find the locus of point $P$ as $D$ moves along $BC$. My dynamic sketch shows that the locus is an algebraic curve of degree $\geq 3$. Thus, the problem cannot be an olympiad problem. Was it translated correctly? Darij

It should be equilateral $\triangle ABC$. Note that now the answer is still an algebraic curve of degree 2. (the locus is the curve $y^2-\frac{x^2}{3}=1$, where BC=2 is the x-axis with the midpoint as the origin)

As the above post notes, the locus is the curve $y = - \sqrt{\frac{x^2 +3}{3}}$ where $x \in (-1, 1)$ and $A$, $B$ and $C$ lie at $(0, \sqrt{3})$, $(-1,0)$ and $(1,0)$, respectively.. Without loss of generality, assume $BD \geq CD$ and let $\angle ADB = \alpha$. Claim. Points $A, B, D, O_2, I_2$ are all concyclic. Proof. First, note that $\angle AO_2D = 2 \angle ACD = 120^{\circ}$. On the other hand $\angle A I_2D = 90^{\circ} + \frac{1}{2} \angle ACD = 120^{\circ}$. Hence $A,O_2, I_2, D$ are concyclic. Next, we show that $O_2$ lies on the circumcircle of $ADC$ centered at $O_1$. Note that $\angle AO_1B = 2\alpha = \angle AO_2C$. It follows that there is a spiral similarity $\triangle AO_1B \sim AO_2C$. This implies that $\triangle AO_1O_2 \sim ABC$, so $O_2O_1 = AO_1 = DO_1 = BO_1$. Hence $A, D, C, O_2$ are concyclic. Combining both these steps proves the claim. By symmetry, it follows that $A, C, D, O_1, I_1$ are also concyclic. Next we show that $D$ is the center of the circumcircle of $PO_1O_2$. First observe that $PO_1 \parallel AI_2$ since \[ \angle AO_1P = \angle AO_1I_1 = 180^{\circ} - \angle ADI_1 = 180^{\circ} - \frac{1}{2} \alpha \]and \[ \angle I_2 A O_1 = 90 - \angle I_2 D A = \frac{1}{2} \alpha . \]Thus \[ \angle O_2 P O_1 = \angle O_2 P O_1 = 180^{\circ} - \angle AI_2P = \angle AI_2O_2 = \angle ADO_2 = 30^{\circ} . \]Since $DO_1O_2$ is equilateral, that is precisely half $\angle O_2DO_1$, so the claim is proven. Now $\angle PDO_1 = 180^{\circ} - 2\angle DOP_1$. We can very easily compute \[ \angle DO_1P = \angle DO_1I_1 = \angle DAI_1 = \frac{1}{2}DAB, \]so $\angle PDO_1 = 180^{\circ} - \angle DAB$. However, $\angle O_1DB = 90^{\circ} - \angle DAB$, so $\angle BDP = 90^{\circ}$. This leaves us with two crucial facts: 1. $PD \times BC$ 2. $PD = DO_1$. These are enough for us to determine the precise coordinates of $P$. Let $D = (x,0)$. By the Law of Sines \[2 DO_1 = \frac{AD}{\sin \angle ABD} = \frac{\sqrt{3 + x^2}}{\frac{\sqrt 3}{2}} \implies PD = \sqrt{\frac{x^2+3}{3}} \]Clearly $P$ is on the opposite side of $BC$ from $A$, so the conclusion is immediate.