Problem (Geometry):
In square $A B C D$ with diagonal $1, E$ is on $\overline{A B}$ and $F$ is on $\overline{B C}$ with $\mathrm{m} \angle B C E=\mathrm{m} \angle B A F=$ $30^{\circ}$. If $\overline{C E}$ and $\overline{A F}$ intersect at $G$, compute the distance between the incenters of triangles $A G E$ and $C G F$.

Solution:
Let $M$ be the midpoint of $\overline{A G}$, and $I$ the incenter of $\triangle A G E$ as shown below.



Because $\frac{A B}{A C}=\sin 45^{\circ}$ and $\frac{E B}{A B}=\frac{E B}{B C}=\tan 30^{\circ}$,

$$
\begin{aligned}
A E & =A B-E B=A B\left(1-\tan 30^{\circ}\right) \\
& =\sin 45^{\circ}\left(1-\tan 30^{\circ}\right) \\
& =\frac{\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}}{\cos 30^{\circ}} \\
& =\frac{\sin \left(45^{\circ}-30^{\circ}\right)}{\cos 30^{\circ}} \\
& =\frac{\sin 15^{\circ}}{\cos 30^{\circ}} .
\end{aligned}
$$



Note that $\frac{A M}{A E}=\cos 30^{\circ}$ and $\frac{A M}{A I}=\cos 15^{\circ}$. Therefore

$$
\begin{aligned}
\frac{A I}{A E} & =\frac{\cos 30^{\circ}}{\cos 15^{\circ}} \\
& =\frac{\sin 60^{\circ}}{\cos 15^{\circ}} \\
& =\frac{2 \sin 30^{\circ} \cos 30^{\circ}}{\cos 15^{\circ}} \\
& =\frac{2\left(2 \sin 15^{\circ} \cos 15^{\circ}\right) \cos 30^{\circ}}{\cos 15^{\circ}} \\
& =4 \sin 15^{\circ} \cos 30^{\circ} .
\end{aligned}
$$

Thus $A I=\left(4 \sin 15^{\circ} \cos 30^{\circ}\right)\left(\frac{\sin 15^{\circ}}{\cos 30^{\circ}}\right)=4 \sin ^{2} 15^{\circ}=4 \cdot\left(\frac{1-\cos 30^{\circ}}{2}\right)=2-\sqrt{3}$. Finally, the desired distance is $2 I G=2 A I=4-2 \sqrt{3}$.

Final answer: $4-2 \sqrt{3}$