HMMT-Feb 2011 gen, Problem 31

Let $A B C D$ be a cyclic quadrilateral, and suppose that $B C=C D=2$. Let $I$ be the incenter of triangle $A B D$. If $A I=2$ as well, find the minimum value of the length of diagonal $B D$.

Solution: Algebra \& Geometry Individual Test Let $T$ be the point where the incircle intersects $A D$, and let $r$ be the inradius and $R$ be the circumradius of $\triangle A B D$. Since $B C=C D=2, C$ is on the midpoint of $\operatorname{arc} B D$ on the opposite side of $B D$ as $A$, and hence on the angle bisector of $A$. Thus $A, I$, and $C$ are collinear. We have the following formulas: \[ \begin{aligned} A I & =\frac{I M}{\sin \angle I A M}=\frac{r}{\sin \frac{A}{2}} \\ B C & =2 R \sin \frac{A}{2} \\ B D & =2 R \sin A \end{aligned} \] The last two equations follow from the extended law of sines on $\triangle A B C$ and $\triangle A B D$, respectively. Using $A I=2=B C$ gives $\sin ^{2} \frac{A}{2}=\frac{r}{2 R}$. However, it is well-known that $R \geq 2 r$ 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 \[ B D=2 R\left(2 \sin \frac{A}{2} \cos \frac{A}{2}\right)=B C \cdot 2 \cos \frac{A}{2} \geq 2\left(2 \cdot \frac{\sqrt{3}}{2}\right)=2 \sqrt{3} \] with equality when $\triangle A B D$ is equilateral. Remark: Similar but perhaps simpler computations can be made by noting that if $A C$ intersects $B D$ at $X$, then $A B / B X=A D / D X=2$, which follows from the exterior angle bisector theorem; if $I_{A}$ is the $A$-excenter of triangle $A B C$, then $A I_{A} / X I_{A}=2$ since it is well-known that $C$ is the circumcenter of cyclic quadrilateral $B I D I_{A}$. $\fbox{2 \sqrt{3}}$.

Answer: 2 \sqrt{3}