HMMT-Nov 2022 team, Problem 6

A triangle $X Y Z$ and a circle $\omega$ of radius 2 are given in a plane, such that $\omega$ intersects segment $\overline{X Y}$ at the points $A, B$, segment $\overline{Y Z}$ at the points $C, D$, and segment $\overline{Z X}$ at the points $E, F$. Suppose that $X B>X A, Y D>Y C$, and $Z F>Z E$. In addition, $X A=1, Y C=2, Z E=3$, and $A B=C D=E F$. Compute $A B$.

Solution: Solution: Let $d=A B$ and $x=d / 2$ for ease of notation. Let the center of $(A B C D E F)$ be $I$. Because $A B=C D=E F$, the distance from $I$ to $A B, C D$, and $E F$ are the same, so $I$ is the incenter of $\triangle X Y Z$. Let $\triangle X Y Z$ have inradius $r$. By symmetry, we have $X F=1, Y B=2$, and $Z D=3$. Thus, $\triangle X Y Z$ has side lengths $d+3, d+4$, and $d+5$. Heron's Formula gives the area of $\triangle X Y Z$ is \[ K=\sqrt{(3 x+6)(x+2)(x+1)(x+3)}=(x+2) \sqrt{3(x+1)(x+3)}, \] while $K=r s$ gives the area of $\triangle X Y Z$ as \[ K=(3 x+6) r \] Equating our two expressions for $K$, we have \[ (x+2) \sqrt{3(x+1)(x+3)}=(3 x+6) r \Longrightarrow \sqrt{3(x+1)(x+3)}=3 r \Longrightarrow(x+1)(x+3)=3 r^{2} \] The Pythagorean Theorem gives $x^{2}+r^{2}=4$, so $r^{2}=4-x^{2}$. Plugging this in and expanding gives \[ (x+1)(x+3)=3\left(4-x^{2}\right) \Longrightarrow 4 x^{2}+4 x-9=0 \] This has roots $x=\frac{-1 \pm \sqrt{10}}{2}$, and because $x>0$, we conclude that $d=\sqrt{10}-1$. $\fbox{\sqrt{10}-1}$.

Answer: \sqrt{10}-1