Problem I'm trying to solve:
Attempt 1:
Using the cosine rule we find that
$$(IT)^2=14^2+10^2-2(14)(10)\cos(F)\Rightarrow \arccos\bigg(\frac{14^2+10^2-(IT)^2}{2(14)(10)}\bigg)=F$$
Analogously we have
$$\arccos\bigg(\frac{3^2+10^2-(MF)^2}{2(3)(10)}\bigg)=T$$ $$\arccos\bigg(\frac{3^2+10^2-(ST)^2}{2(3)(10)}\bigg)=M$$ $$\arccos\bigg(\frac{3^2+10^2-(IM)^2}{2(3)(10)}\bigg)=S$$ $$\arccos\bigg(\frac{14^2+3^2-(SF)^2}{2(3)(10)}\bigg)=I$$
Adding everything up you can see that the sum of angles $I+S+M+T+F=540$, and the you can work your way back by taking $\cos$ of the sum of the $\arccos$ and using compound angle formulae but I feel like this is way too complicated.
Attempt 2:
Let $O$ be be the center of the circle with radius $x$ and $a,b,c,d,e$ be the angles (in any arbitrary order) between two diagonals meeting at some arbitrary vertex. In the same way label all diagonals $p,q,r,s,t$ (in any arbitrary order). Then by the cosine rule we have $$14^2=p^2+q^2-2pq\cos(a)$$ $$3^2=r^2+s^2-2rs\cos(b)$$ $$10^2=t^2+p^2-2tp\cos(c)$$ $$3^2=q^2+r^2-2qr\cos(d)$$ $$10^2=s^2+t^2-2st\cos(e)$$
as well as
$$14^2=2x^2-2x^2\cos(2a)$$ $$3^2=2x^2-2x^2\cos(2b)$$ $$10^2=2x^2-2x^2\cos(2c)$$ $$3^2=2x^2-2x^2\cos(2d)$$ $$10^2=2x^2-2x^2\cos(2e)$$
and
$$\sin(a)=\frac{7}{x}$$ $$\sin(b)=\frac{3}{2x}$$ $$\sin(c)=\frac{10}{x}$$ $$\sin(d)=\frac{3}{2x}$$ $$\sin(e)=\frac{10}{x}$$
However trying to solve for any $x$ with a corresponding angle gives an infinites number of solutions which doesn't lead me anywhere. On top of that, I'm not sure how that would lead me to the sum of the diagonals.
Other Thoughts:
- Intersecting chord theorem
- Ptolemy's theorem:
$$14(ST)+(3)(10)=(SF)(IT)$$ $$10(IM)+(3)(14)=(MF)(IT)$$ $$10(IT)+(3)(3)=(ST)(IM)$$ $$3(SF)+(10)(10)=(MF)(ST)$$ $$3(MF)+(14)(10)=(SF)(IM)$$
- Brahmagupta's formula for the area of cyclic quadrilaterals (not sure how viable this is though)
The thing is, all the equations above tend to suggest some form of completing the square factorisation, or summing a bunch equations however I don't exactly see where to go from there either. Any suggestions?
So I decided to try my idea with Ptolemy's theorem and I think I have an answer (feels weird and satisfying answering my own question). Let $ST=a$, $SF=b$, $IT=c$, $MF=d$, $IM=e$. Then we can rewrite the equations as
$$14a+30=bc$$ $$10e+42=cd$$ $$10c+9=ae$$ $$3b+100=ad$$ $$3d+140=be$$
Note from the diagram that $d=e=a$ because they are of the same arclength so the equations become:
$$14a+30=bc$$ $$10a+42=ac$$ $$10c+9=a^2$$ $$3b+100=a^2$$ $$3a+140=ab$$
If we consider the bottom two equations then $$b=\frac{a^2-100}{3}$$
and then
\begin{align} 3a+140&=ab \\ 3a+140&=a\bigg(\frac{a^2-100}{3}\bigg) \\ 9a+420&=a^3-100a \\ \end{align}
From there you have
\begin{align} 0&=a^3-109a-420 \\ 0&=(a-12)(a^2+12a+35) \\ 0&=(a-12)(a+7)(a+5) \\ \end{align} So $a=12$ from which follows that $b=\frac{44}{3}$ and $c=\frac{27}{2}$. Then the sum of the diagonals is $a+b+c+d+e=\frac{385}{6}$ so $x+y=391$ in the question.