HMMT November, 2024

Solution:

Let the line tangent to $\odot(ABC)$ at $A$ intersect line $BC$ at $P$.

Proof. Note that
$$\angle PAX = \angle PAB + \angle BAX = \angle PCA + \angle CAY = \angle PYA$$

This proves the desired tangency.
Now, by power of point, we have,
$$\begin{aligned} PB \cdot PC & = PX \cdot PY = PA^2 \\ PB(PB+11) & = (PB+2)(PB+8) \\ PB & = 16 \\ PA^2 & = PB \cdot PC = 16 \cdot 27 \\ PA & = 12 \sqrt{3} \end{aligned}$$

By Pythagorean theorem, $AD^2 + DP^2 = PA^2$. Therefore, $PD^2 = PA^2 - AD^2 = 432 - 144 = 288$, so $PD = 12 \sqrt{2}$. Finally, $BD = PD - PB = 12 \sqrt{2} - 16$.

Solution 2:

Since $\angle BAX = \angle CAY$, by Steiner ratio theorem, we get that
$$\frac{AB^2}{AC^2} = \frac{BX}{CX} \cdot \frac{BY}{CY} = \frac{2}{9} \cdot \frac{8}{3} = \frac{16}{27}$$

Thus, if $BD = x$, then by Pythagorean theorem, we get that $AB^2 = 144 + x^2$ and $AC^2 = 144 + (11 - x)^2$. Combining with the displayed equations gives
$$\begin{aligned} 27\left(144 + x^2\right) & = 16\left(144 + (11 - x)^2\right) \\ 27\left(144 + x^2\right) & = 16\left(144 + x^2\right) + 16(-22x + 121) \\ 11\left(144 + x^2\right) & = 16 \cdot 11 \cdot (-2x + 11) \\ x^2 + 144 + 16(2x - 11) & = 0 \\ x^2 + 32x - 32 & = 0 \\ x & = 12 \sqrt{2} - 16, \end{aligned}$$
(where we only take the positive solution since $x > 0$).

Solution 3:

Suppose that $BD = c$. From the length conditions, we get that
$$\tan \angle BAD = \frac{c}{12}, \quad \tan \angle DAX = \frac{2 - c}{12}, \quad \tan \angle DAC = \frac{11 - c}{12}, \quad \tan \angle DAY = \frac{8 - c}{12}$$

Thus, using tangent addition formula, we get that
$$\begin{aligned} & \tan \angle BAX = \tan (\angle BAD + \angle DAX) = \frac{\frac{c}{12} + \frac{2 - c}{12}}{1 - \frac{c}{12} \cdot \frac{2 - c}{12}} = \frac{\frac{2}{12}}{\frac{144 - 2c + c^2}{144}} = \frac{24}{144 - 2c + c^2} \\ & \tan \angle YAC = \tan (\angle DAC - \angle DAY) = \frac{\frac{11 - c}{12} - \frac{8 - c}{12}}{1 + \frac{11 - c}{12} \cdot \frac{8 - c}{12}} = \frac{\frac{3}{12}}{\frac{c^2 - 19c + 232}{144}} = \frac{36}{c^2 - 19c + 232} \end{aligned}$$

Hence, the condition $\angle BAX = \angle CAY$ translates to
$$\begin{aligned} \frac{24}{144 - 2c + c^2} & = \frac{36}{c^2 - 19c + 232} \\ 36\left(c^2 - 2c + 144\right) - 24\left(c^2 - 19c + 232\right) & = 0 \\ 3\left(c^2 - 2c + 144\right) - 2\left(c^2 - 19c + 232\right) & = 0 \\ c^2 + 32c - 32 & = 0 \\ c & = -16 \pm 12 \sqrt{2} \end{aligned}$$
Since $c$ is positive, the answer is $BD = 12 \sqrt{2} - 16$.