Harvard-MIT Mathematics Tournament

Solution:

Answer: $2\sqrt{3}$

Let $T$ be the point where the incircle intersects $AD$, and let $r$ be the inradius and $R$ be the circumradius of $\triangle ABD$. Since $BC = CD = 2$, $C$ is on the midpoint of arc $BD$ on the opposite side of $BD$ as $A$, and hence on the angle bisector of $A$. Thus $A$, $I$, and $C$ are collinear. We have the following formulas:
$$\begin{aligned} AI &= \frac{IM}{\sin \angle IAM} = \frac{r}{\sin \frac{A}{2}} \\ BC &= 2R \sin \frac{A}{2} \\ BD &= 2R \sin A \end{aligned}$$
The last two equations follow from the extended law of sines on $\triangle ABC$ and $\triangle ABD$, respectively.

Using $AI = 2 = BC$ gives $\sin^2 \frac{A}{2} = \frac{r}{2R}$. However, it is well-known that $R \geq 2r$ with equality for an equilateral triangle (one way to see this is the identity $1 + \frac{r}{R} = \cos A + \cos B + \cos D$). Hence $\sin^2 \frac{A}{2} \leq \frac{1}{4}$ and $\frac{A}{2} \leq 30^\circ$. Then
$$BD = 2R \left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right) = BC \cdot 2 \cos \frac{A}{2} \geq 2\left(2 \cdot \frac{\sqrt{3}}{2}\right) = 2\sqrt{3}$$
with equality when $\triangle ABD$ is equilateral.

Solution:

Let $T$ be the point where the incircle intersects $AD$, and let $r$ be the inradius and $R$ be the circumradius of $\triangle ABD$. Since $BC = CD = 2$, $C$ is on the midpoint of arc $BD$ on the opposite side of $BD$ as $A$, and hence on the angle bisector of $A$. Thus $A$, $I$, and $C$ are collinear. We have the following formulas:
$$\begin{aligned} AI & = \frac{IM}{\sin \angle IAM} = \frac{r}{\sin \frac{A}{2}} \\ BC & = 2R \sin \frac{A}{2} \\ BD & = 2R \sin A \end{aligned}$$
The last two equations follow from the extended law of sines on $\triangle ABC$ and $\triangle ABD$, respectively.

Using $AI = 2 = BC$ gives $\sin^2 \frac{A}{2} = \frac{r}{2R}$. However, it is well-known that $R \geq 2r$ with equality for an equilateral triangle (one way to see this is the identity $1 + \frac{r}{R} = \cos A + \cos B + \cos D$). Hence $\sin^2 \frac{A}{2} \leq \frac{1}{4}$ and $\frac{A}{2} \leq 30^\circ$. Then
$$BD = 2R \left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right) = BC \cdot 2 \cos \frac{A}{2} \geq 2 \left(2 \cdot \frac{\sqrt{3}}{2}\right) = 2\sqrt{3}$$
with equality when $\triangle ABD$ is equilateral.