Problem 1527

1. Given Information and Setup:
- We are given a triangle $\triangle ABC$ with $AB = 10$ and $AC = 11$.
- $I$ is the incenter of $\triangle ABC$.
- $M$ is the midpoint of $AI$.
- We are given that $BM = BC$ and $CM = 7$.

2. Using the Given Conditions:
- We need to find $BC$ in the form $\frac{\sqrt{a} - b}{c}$ where $a$, $b$, and $c$ are positive integers.
- We start by using the given conditions to set up the equation involving $BC$.

3. Using the Distance Formula:
- We use the distance formula and properties of the triangle to derive the necessary equations.
- We know that $BM = BC$ and $CM = 7$.

4. Deriving the Equation:
- We use the formula $2(BM^2 - CM^2) = AB^2 - AC^2 + BI^2 - CI^2$.
- Substituting the given values, we get:
$$2BC^2 - 98 = 100 - 121 + (s - 11)^2 - (s - 10)^2$$
- Simplifying the right-hand side:
$$2BC^2 - 98 = -21 + (s^2 - 22s + 121) - (s^2 - 20s + 100)$$
$$2BC^2 - 98 = -21 + s^2 - 22s + 121 - s^2 + 20s - 100$$
$$2BC^2 - 98 = -21 - 2s + 21$$
$$2BC^2 - 98 = -2s$$

5. **Solving for $BC$:**
- We know that $s$ is the semi-perimeter of $\triangle ABC$, so:
$$s = \frac{AB + AC + BC}{2}$$
- Substituting $s$ into the equation:
$$2BC^2 - 98 = -BC - 21$$
$$2BC^2 + BC - 77 = 0$$

6. Solving the Quadratic Equation:
- We solve the quadratic equation $2BC^2 + BC - 77 = 0$ using the quadratic formula $BC = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$BC = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-77)}}{2 \cdot 2}$$
$$BC = \frac{-1 \pm \sqrt{1 + 616}}{4}$$
$$BC = \frac{-1 \pm \sqrt{617}}{4}$$
- Since $BC$ must be positive, we take the positive root:
$$BC = \frac{\sqrt{617} - 1}{4}$$

7. **Identifying $a$, $b$, and $c$:**
- From the expression $\frac{\sqrt{617} - 1}{4}$, we identify $a = 617$, $b = 1$, and $c = 4$.

8. **Summing $a$, $b$, and $c$:**
- Therefore, $a + b + c = 617 + 1 + 4 = 622$.

The final answer is $\boxed{622}$.