USA IMO TST

First solution using spiral similarity (Ankan Bhattacharya) First, we prove the part of the problem which does not depend on the condition $AB_1A_1C_1$ is cyclic.

Lemma
Let $ABC$ be a triangle and define $I, D, B_1, C_1$ as in the problem. Moreover, let $M$ denote the midpoint of $\overline{AD}$. Then $\overline{AD}$ is tangent to $(AB_1C_1)$, and moreover $\overline{B_1C_1} \parallel \overline{IM}$.

Proof. Let $E$ and $F$ be the tangency points of the incircle. Denote by $Z$ the Miquel point of $BFEC$, i.e. the second intersection of the circle with diameter $\overline{AI}$ and the circumcircle. Note that $A, Z, D$ are collinear, by radical axis on $(ABC), (AFIE), (BIC)$.

Then the spiral similarity gives us
$$\frac{ZF}{ZE} = \frac{BF}{CE} = \frac{AC_1}{AB_1}$$
which together with $\angle FZE = \angle FAE = \angle BAC$ implies that $\triangle ZFE$ and $\triangle AC_1B_1$ are (directly) similar. (See IMO Shortlist 2006 G9 for a similar application of spiral similarity.)

Now the remainder of the proof is just angle chasing. First, since
$$\angle DAC_1 = \angle ZAF = \angle ZEF = \angle AB_1C_1$$
we have $\overline{AD}$ is tangent to $(AB_1C_1)$. Moreover, to see that $\overline{IM} \parallel \overline{B_1C_1}$, write
$$\begin{aligned} \angle(\overline{AI}, \overline{B_1C_1}) &= \angle IAC + \angle AB_1C_1 = \angle BAI + \angle ZEF = \angle FAI + \angle ZAF \\ &= \angle ZAI = \angle MAI = \angle AIM \end{aligned}$$
the last step since $\triangle AID$ is right with hypotenuse $\overline{AD}$, and median $\overline{IM}$. $\square$

Now we return to the present problem with the additional condition.

Claim. Given the condition, we actually have $\angle AB_1A_1 = \angle AC_1A_1 = 90^\circ$.

Proof. Let $I_A, I_B$ and $I_C$ be the excenters of $\triangle ABC$. Then the perpendiculars to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ from $A_1, B_1, C_1$ respectively meet at the so-called Bevan point $V$ (which is the circumcenter of $\triangle I_A I_B I_C$).

Now $\triangle AB_1C_1$ has circumdiameter $\overline{AV}$. We are given $A_1$ lies on this circle, so if $V \neq A_1$ then $\overline{AA_1} \perp \overline{A_1V}$. But $\overline{A_1V} \perp \overline{BC}$ by definition, which would imply $\overline{AA_1} \parallel \overline{BC}$, which is absurd. $\square$

Claim. Given the condition the points $B_1, I, C_1$ are collinear (hence with $M$).

Proof. By Pappus theorem on $\overline{IB}A_1I_C$ and $\overline{BA_1C}$ after the previous claim. $\square$

To finish, since $\overline{DMA}$ was tangent to the circumcircle of $\triangle AB_1C_1$, we have $MD^2 = MA^2 = MC_1 \cdot MB_1$, implying the required tangency.

Second solution by inversion and mixtilinears (Anant Mudgal) As in the end of the preceding solution, we have $\angle AB_1A_1 = \angle AC_1A_1 = 90^\circ$ and $I \in \overline{B_1C_1}$. Let $M$ be the midpoint of minor arc $BC$ and $N$ be the midpoint of arc $\widehat{BAC}$. Let $L$ be the intouch point on $\overline{BC}$. Let $O$ be the circumcenter of $\triangle ABC$. Let $K = \overline{AI} \cap \overline{BC}$.

Claim. We have $\angle(\overline{AI}, \overline{B_1C_1}) = \angle IAD$.

Proof. Let $Z$ lie on $(ABC)$ with $\angle AZI = 90^\circ$. By radical axis theorem on $(AIZ), (BIC)$, and $(ABC)$, we conclude that $D$ lies on $\overline{AZ}$. Let $\overline{NI}$ meet $(ABC)$ again at $T \neq N$.

Inversion in $(BIC)$ maps $\overline{AI}$ to $\overline{KI}$ and $(ABC)$ to $\overline{BC}$. Thus, $Z$ maps to $L$, so $Z, L, M$ are collinear. Since $BL = CV$ and $OI = OV$, we see that $MLIN$ is a trapezoid with $\overline{IL} \parallel \overline{MN}$. Thus, $\overline{ZT} \parallel \overline{MN}$.

It is known that $\overline{AT}$ and $\overline{AA_1}$ are isogonal in angle $BAC$. Since $\overline{AV}$ is a circumdiameter in $(AB_1C_1)$, so $\overline{AT} \perp \overline{B_1C_1}$. So $\angle ZAI = \angle NMT = 90^\circ - \angle TAI = \angle(\overline{AI}, \overline{B_1C_1})$. $\square$

Let $X$ be the midpoint of $\overline{AD}$ and $G$ be the reflection of $I$ in $X$. Since $AIDG$ is a rectangle, we have $\angle AIG = \angle ZAI = \angle (AI, \overline{B_1C_1})$, by the previous claim. So $\overline{IG}$ coincides with $\overline{B_1C_1}$. Now $AI$ bisects $\angle B_1AC_1$ and $\angle IAG = 90^\circ$, so $(\overline{IG}; \overline{B_1C_1}) = -1$.

Since $\angle IDG = 90^\circ$, we see that $\overline{DI}$ and $\overline{DG}$ are bisectors of angle $B_1DC_1$. Now $\angle XDI = \angle XID \implies \angle XDC_1 = \angle XID - \angle IDB_1 = \angle DB_1C_1$, so $\overline{XD}$ is tangent to $(DB_1C_1)$.