Problem 1778

1. Define the Excircle Tangency Points:
Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Similarly, define the points $B_1$ on $CA$ and $C_1$ on $AB$ using the excircles opposite $B$ and $C$, respectively.

2. **Circumcenter of $\triangle A_1B_1C_1$:**
Suppose the circumcenter of triangle $A_1B_1C_1$, denoted as $O$, lies on the circumcircle of triangle $ABC$.

3. Midpoint of Excenters:
Let $I_A, I_B, I_C$ be the excenters of $\triangle ABC$ opposite vertices $A, B, C$, respectively. It is known that the midpoint of $I_BI_C$ lies on the circumcircle of $\triangle ABC$. This midpoint is also the circumcenter of $\triangle I_AI_BI_C$.

4. Claim 1:
Claim: The point $O$ coincides with the midpoint of $I_BI_C$.
Proof: The midpoint of $I_BI_C$ lies on the circumcircle of $\triangle ABC$ and is the center of the nine-point circle of $\triangle I_AI_BI_C$. This nine-point circle passes through $A, B, C$. Let the midpoint of $I_BI_C$ be $M$. Since $M$ lies on the circumcircle of $\triangle ABC$ and $I_BI_C$ is inscribed in a circle with center $M$, we have $MB = MC$. Therefore, triangles $C_1MB$ and $B_1MC$ are congruent (they have $BM = CM$, $BC_1 = B_1C$, and $\angle C_1BM = \angle MB_1C$). Thus, $MC_1 = MB_1$. Hence, $M$ is the intersection point of the perpendicular bisector of $B_1C_1$ with the circumcircle of $\triangle ABC$. Therefore, $O \equiv M$. $\blacksquare$

5. Collinearity of Points:
Let $R$ be the circumcenter of $\triangle I_AI_BI_C$.
Claim 2: $I_C, C_1, R$ are collinear, as are $I_B, B_1, R$ and $I_A, R, A_1$.
Proof: By Nagel's theorem, $I_CR \perp AB$, and since $I_CC_1 \perp AB$, we have that $I_C, C_1, R$ are collinear. Similarly, we obtain the other two collinearities. $\blacksquare$

6. Inscribed Quadrilaterals:
By the previous claim, $\angle BC_1R = \angle BA_1R$, so quadrilateral $BC_1A_1R$ is inscribed in a circle. Similarly, $A_1B_1CR$ is also inscribed in a circle.

7. Angle Calculation:
We have:
$$\angle BRC = \angle BRA_1 + \angle A_1RC = \angle AC_1A_1 + \angle AB_1A_1 = 360^\circ - \angle A - \angle A_1B_1C_1 = 360^\circ - \angle A - (180^\circ - \angle C_1OB_1/2) = 360^\circ - \angle A + \angle C_1OB_1/2 = 180^\circ - \angle A/2$$
where the latter equality follows from:
$$\angle C_1OB_1 = \angle C_1OB + \angle BOB_1 = \angle B_1OC + \angle BOB_1 = \angle BOC = \angle A$$

8. Conclusion:
We have thus obtained that $\angle BRC = 180^\circ - \angle A/2$. This, combined with $BO = CO$ and $\angle BOC = \angle A$, yields that $O$ is the circumcenter of triangle $\triangle BRC$, so:
$$OB = OR = OC = OB_1 = OC_1$$
This gives that in triangle $\triangle I_CRI_B$ that $I_CO = OI_B = OR$, which implies that $\angle I_CRI_B = 90^\circ$.

9. Final Angle Calculation:
Lastly,
$$\angle A = 2\pi - \angle AC_1R - \angle AB_1R - \angle C_1RB_1 = 90^\circ$$
and the proof is complete. $\blacksquare$