Welcome 2016! Part 13

By cylic quad angles, $\angle OAB = \angle OCD$ , and so, $\Delta OAB \sim \Delta OCD$ .

Then, it follows that: $\dfrac{OA}{OC}=\dfrac{OB}{OD}=\sqrt{\dfrac{72}{288}}$

$\dfrac{18}{16+x}=\dfrac{16}{18+y}=\dfrac{1}{2}$

Simple algebra yields $x=20, y= 14$ , so $x+y=34$ .

$CO*OB=DO*OA~~\implies~(x+16)*16=(y+18)*18............(A)~~~~~\therefore~y+18=(x+16)*\dfrac 8 9.\\ \dfrac{Area~\Delta~OAB}{ Area~\Delta~ODC}=\dfrac{16*18}{(x+16)(y+18)}=\dfrac{72}{288}=\frac 1 4..........(B)~~~~~\therefore~y+18=4*16*18*\dfrac 1 {x+16}.\\ \implies~ (x+16)*\dfrac 8 9 =4*16*18*\dfrac 1 {x+16},~~\implies~\sqrt{(x+16)^2}=\sqrt{\dfrac{4*16*18*9} 8}=\sqrt{16*81}.\\ \therefore~~x+16=4*9,~~\implies~~x=20.\\ Substituting~in~(A)~~36*16=(y+18)*18,~~\implies~y=14.\\ So~x+y=\Large~~~~\color{#D61F06}{34}.$

$A_{OBA}=\dfrac{1}{2}(16)(18)(\sin \angle COD)$

$72=144 \sin \angle COD$

$\angle COD=0.5$

$A_{ODC}=\dfrac{1}{2}(16+x)(18+y)(0.5)$

$1152=(16+x)(18+y)$

$18+y=\dfrac{1152}{16+x}$ $\color{#D61F06}(1)$

By the power of a point, we have

$16(16+x)=18(18+y)$

Substituting $\color{#D61F06}(1)$ , we have

$16(16+x)=18\left(\dfrac{1152}{16+x}\right)$

$(256+16x)(16+x)=20736$

$4096+256x+256x+16x^2=20736$

$16x^2+512x-16640=0$

$x^2+32x-1040=0$

$(X+52)(x-20)=0$

$x=-52$ or $x=20$ reject the negative value

Solving for $y$ , we have

$18+y=\dfrac{1152}{16+20}$

$18+y=32$

$y=14$

Finally,

$x+y=20+14=$ $\boxed{34}$