Problem 6 : Strange Fractions

What a funny problem.

Under the given conditions, it turns out that $\frac{AO}{OD}-\frac{BC}{AT}$ is a CONSTANT. Furthermore, the value of this constant is $4$ . In other words, "maximum possible" and the $\frac{a\sqrt{b}}{c}$ form of the answer are both COMPLETELY unnecessary. I suspect the problem is written in this way to prevent fakesolving.

Anyways, here is the solution. Let $H$ be the orthocenter of triangle $ABC$ . Furthermore, let $O'$ be the reflection of $O$ over line $BC$ . Since triangle $DEF$ is similar to triangle $ABC$ with similarity ratio $\frac{1}{2}$ and $O$ is the orthocenter of triangle $DEF$ , it follows that $OD=\frac{AH}{2}$ . Therefore, $OO'=AH$ . Now since $P$ is the nine-point center of triangle $ABC$ , it follows that $P$ is the midpoint of segment $OH$ . In conjunction with $AH\parallel OO'$ , we deduce that $A, P, O'$ are collinear. Since $A, P, Q$ are collinear, it follows that $P, Q, O'$ are collinear.

Now let $A'\not\equiv A$ be the second intersection of line $AI$ with the circumcircle of triangle $ABC$ . $O, D, O', A'$ are collinear on the perpendicular bisector of segment $BC$ . By Menelaus's Theorem, $\frac{AA'}{AI}\cdot\frac{QI}{QO}\cdot\frac{O'O}{O'A}=1$ . Since $Q$ is the midpoint of segment $OI$ , it follows that $\frac{AA'}{AI}=\frac{O'A'}{O'O}$ . Now notice that $\frac{AA'}{AI}=\frac{O'A'}{O'O}=\frac{A'O-O'O}{O'O}=\frac{AO-2\cdot OD}{2\cdot OD}$ Rearranging yields $\frac{AO}{OD}=\frac{2AA'}{AI}+2$ .

The problem is now equivalent to showing that $\frac{2AA'}{AI}-2=\frac{BC}{AT}$ , which we'll show with some computation. Let $a=BC$ , $b=CA$ , $c=AB$ , $s=\frac{a+b+c}{2}$ . We immediately deduce that $\frac{BC}{AT}=\frac{a}{s-a}$ and $AI=\frac{s-a}{\cos\frac{A}{2}}$ . As for the $AA'$ , notice that $A'B=A'C=\frac{a}{2\cos\frac{A}{2}}$ . By Ptolemy's Theorem, $CA\cdot A'B+AB\cdot A'C=AA'\cdot BC$ . Substitution and rearrangement yield $AA'=\frac{b+c}{2\cos\frac{A}{2}}$ . Therefore, $\frac{2AA'}{AI}-2=\frac{\left(2\cdot\frac{b+c}{2\cos\frac{A}{2}}\right)}{\left(\frac{s-a}{\cos\frac{A}{2}}\right)}-2=\frac{b+c}{s-a}-2=\frac{b+c-2(s-a)}{s-a}=\frac{a}{s-a}=\frac{BC}{AT}$ It follows that $\frac{AO}{OD}-\frac{BC}{AT}=4=\frac{4\sqrt{1}}{1}$ , so the answer is $\boxed{6}$ .