Let $A_0BC_0D$ be a convex quadrilateral inscribed in a circle $\omega$. For all integers $i\geqslant 0$, let $P_i$ be the intersection of lines $A_iB$ and $C_iD$, let $Q_i$ be the intersection of lines $A_iD$ and $BC_i$, let $M _i$ be the midpoint of segment $P _i Q _i$ , and let lines $M _i A _i$ and $M _i C _i$ intersect $\omega$ again at $A _{i+1}$ and $C _{i + 1}$ , respectively. The circumcircles of $\triangle A _3 M _3 C _3$ and $\triangle A _4 M _4 C _4$ intersect at two points $P$ and $M$ . If $A _0 B$ = $3$, $B C _0$ = $4$, $C _0 D$ = $6$, $D A _0 $= $7$, then $PM$ can be expressed in the form $\frac{a \sqrt{b}}{c}$ for positive integers $a,b,c$ such that $\gcd(a,c)=1$ and $b$ is squarefree. Compute $100a +10b+c$ .
Note: a square-free integer (or squarefree integer) is an integer which is divisible by no perfect square other than 1
It's a problem from https://gonitzoggo.com/archive/problem/433/english I'm taking preparations for junior math olympiad contest and recently I came across this problem.
I've done some calculations but I'm not sure about it. By drawing the figures I've figured out that PM is diameter of the circle. Then I calculated the diameter and that is $\frac{3\sqrt{4830}}{28}$ . so $a=3, b=4830$ and $c=28$ where $\gcd(a,c)=1$ . And So, $100a+10b+c=48628$.
This problem is from the Online Math Open Spring 2020 Competition. It was a late problem, #28, and it's very difficult. There are some solutions on the AoPS thread here: https://artofproblemsolving.com/community/c487h2055659. The idea is to first show that all $P, Q$ lie on $P_0Q_0$ (easiest way is projective). You can also show that $UV$ has the same length as the common chord of $(A_0M_0C_0), (A_1M_1C_1)$ (again, projective and basic radical axis facts help here). These are a few basic observations, the rest should be a length bash (although I think you might need a few more easier facts).