HMMT February

Quadrilaterals $A M I D$ and $D I C T$ are cyclic, having right angles $\angle A M D, \angle A I D$, and $\angle C I D, \angle C T D$ respectively. We see that $M, I$, and $T$ are collinear. For, $m \angle M I D=\pi-m \angle D A M=$ $\pi-m \angle D A B=m \angle B C D=\pi-m \angle D C T=\pi-m \angle D I T$. Therefore, Menelaus' theorem applied to triangle MTB and line $I C A$ gives $$\frac{M I}{I T} \cdot \frac{T C}{C B} \cdot \frac{B A}{A M}=1$$ On the other hand, triangle $A D M$ is similar to triangle $C D T$ since $\angle A M D \cong \angle C T D$ and $\angle D A M \cong$ $\angle D C T$ and thus $A M / C T=A D / C D$. It follows that $$\frac{M I}{I T}=\frac{B C \cdot A M}{A B \cdot C T}=\frac{B C \cdot A D}{A B \cdot C D}=\frac{5 \cdot 10}{3 \cdot 6}=\frac{25}{9}$$ Remarks. The line $M I T$, constructed in this problem by taking perpendiculars from a point on the circumcircle of $A B C$, is known as the Simson line. It is often helpful for us to use directed angles while angle chasing to avoid supplementary configuration issues, such as those arising while establishing the collinearity of $M, I$, and $T$.