Problem 1403

1. Labeling Points of Tangency:
- Let $P$ and $R$ be the points of tangency of the incircle of $\triangle AMN$ with $AN$ and $AM$ respectively. Since $AP = AR$, let $AP = AR = a$.
- Let $Y$ be the point where the incircle of $\triangle AMN$ touches $MN$. Since $MR = MY$, let $MR = MY = b$.
- Let $NP = NY$. Since $XY = 1$, let $NP = NY = c + 1$ and $NL = c$, where $L$ is the point of tangency of the larger circle on $AC$.

2. Tangency Points for Larger Circle:
- Let $Q$ be the point of tangency of the larger circle on $BC$. Since $CL = CQ$, let $CL = CQ = d$.
- Let $T$ be the point of tangency of the larger circle on $AB$. Since $MT = b + 1$ and $TB = QB$, let $TB = QB = e$.

3. Equations from Tangency:
- Since $AN = NC$, we have:
$$a + c + 1 = c + d \implies a + 1 = d$$
Therefore, $d = a + 1$.

- Since $AM = MB$, we have:
$$a + b = b + 1 + e \implies e = a - 1$$
Therefore, $e = a - 1$.

4. **Using $BC = 12$:**
- Since $BC = 12$, we have:
$$d + e = 12 \implies (a + 1) + (a - 1) = 12 \implies 2a = 12 \implies a = 6$$

5. **Substituting $a = 6$:**
- Now, substituting $a = 6$:
$$d = a + 1 = 6 + 1 = 7$$
$$e = a - 1 = 6 - 1 = 5$$

6. **Finding $c$ and $b$:**
- Since $AL = AT$, we have $b = c$.

7. Proportionality in Similar Triangles:
- Since $\triangle ANM$ is similar to $\triangle ACB$, we have:
$$\frac{2c + 1}{12} = \frac{6 + c}{12 + 2c}$$

8. Solving the Proportion:
- Cross-multiplying the proportion:
$$(2c + 1)(12 + 2c) = 12(6 + c)$$
$$24c + 2c^2 + 12 + 2c = 72 + 12c$$
$$2c^2 + 12c + 12 = 72$$
$$2c^2 + 12c - 60 = 0$$
$$c^2 + 6c - 30 = 0$$

9. Solving the Quadratic Equation:
- Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$c = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1}$$
$$c = \frac{-6 \pm \sqrt{36 + 120}}{2}$$
$$c = \frac{-6 \pm \sqrt{156}}{2}$$
$$c = \frac{-6 \pm 2\sqrt{39}}{2}$$
$$c = -3 \pm \sqrt{39}$$

10. Selecting the Positive Root:
- Since $c$ must be positive:
$$c = -3 + \sqrt{39}$$

11. Calculating Side Lengths:
- $AB = 2a = 2 \times 6 = 12$
- $AC = 2c + 1 = 2(-3 + \sqrt{39}) + 1 = -6 + 2\sqrt{39} + 1 = -5 + 2\sqrt{39}$

The final answer is $\boxed{ AB = 12 }$ and $AC = -5 + 2\sqrt{39}$