The value of (x-y) of quadrilateral inscribed in circle A quadrilateral that has consecutive sides of length $70,90,110,130$ is inscribed in a circle and also has a circle inscribed in it . The point of tangency of the inscribed circle to the side of length $130$ divides that side into segments of $x$ and $y$ . If $y\geq x$ then what is the value of $(y-x)$ 
 A: Call the quadrilateral $ABCD$, with sides $AB=70, BC=90, CD=130, DA=110$, i.e. in the order 
$$\begin{array}{ccc}
A&\leftarrow70\rightarrow&B\\
\begin{array}{c}\uparrow\\110\\\downarrow\end{array}&O&\begin{array}{c}\uparrow\\90\\\downarrow\end{array}\\
D&E&C\\
&\leftarrow130\rightarrow&\\
\end{array}$$
For $ABCD$ to be cyclic, $\angle B + \angle D = 180^\circ$, so $\cos\angle B = -\cos\angle D$. Consider square of length of $AC$:
$$\begin{align*}
AC^2 = 130^2+110^2 -2\times130\times110\cos\angle D =& 70^2+90^2-2\times70\times90\cos\angle B\\
\cos\angle D =& \frac{16000}{41200}\\
=& \frac{40}{103}
\end{align*}$$
Similarly,
$$\begin{align*}
BD^2 = 110^2+70^2 -2\times110\times70\cos\angle A =& 130^2+90^2-2\times130\times90\cos\angle C\\
\cos\angle C =& \frac{8000}{38800}\\
=& \frac{20}{97}
\end{align*}$$
Let the centre of the inscribed circle be $O$, and the tangent point of the inscribed circle on $CD$ be $E$. The line $OC$ bisects $\angle BCD$, and the line $OD$ bisects $\angle CDA$. $OE$ is a line perpendicular with $CD$. Consider $\triangle OCD$,
$$OD\sin\angle ODE = OC\sin\angle OCE\\
OD\cos\angle ODE + OC\cos\angle OCE = 130$$
With half-angle formulae,
$$\sin\angle ODE = \sqrt{\frac{1-\cos\angle CDA}{2}}\\
\cos\angle ODE = \sqrt{\frac{1+\cos\angle CDA}{2}}\\
\sin\angle OCE = \sqrt{\frac{1-\cos\angle BCD}{2}}\\
\cos\angle OCE = \sqrt{\frac{1+\cos\angle BCD}{2}}$$
These should be enough to find the ratio $OD:OC$ and hence find out both of $DE$ and $CE$.
A: First of all, sorry for reviving this two-years-old thread. I searched this problem on this StackExchange and hoped to find a reasonable answer but failed to do so. However, with some careful work, I have found a definite answer to the problem and wish to place it here. I'd also like to point out that @peterwhy is correct in noticing that the lengths must be $70, 90, 130, 110$ consecutively.
We approach the problem with algebra and similar triangles. Let $O$ be the center of the inscribed circle, $AB = 70,$ $BC = 90,$ $CD = 130,$ and $AD = 110.$ From $O,$ we draw perpendicular radii to the tangent segments. Let $P, Q, R, S$ be the tangent points on $AB, BC, CD, AD,$ respectively. Further, let
$$AS = AP = w,$$
$$BP = BQ = x,$$
$$CQ = CR = y,$$ and
$$DR = DS = z.$$
It follows that
$$w + z = 70,$$
$$z + y = 90,$$
$$y + x = 130,$$ and
$$x + w = 110.$$
The system above does not yield a unique solution, which is expected. Now we use the fact that the quadrilateral is cyclic (meaning it can be inscribed in a circle). The solution given by @peterwhy uses this but takes a wrong turn by using extensive trigonometry. Let us see how we can use this fact to our advantage.
Because $ABCD$ is cyclic, angles $QCR$ and $SAP$ must add to $180$ degrees. Because $OC$ and $OA$ each bisect $QCR$ and $SAP,$ respectively, it quickly follows that triangles $COR$ and $OAS$ are similar right triangles. Since $OR = OS = r,$ the radius of the inscribed circle, then we have
$$\frac{w}{r} = \frac{y}{x}$$
$$wy = r^{2}.$$
Using a similar argument for triangles $DOR$ and $OBQ,$ we have
$$xz = r^{2},$$
and we combine the two to get
$$wy = xz.$$
We can now find a unique solution. Using the substitutions $w = 110 - x$ and $z = 90 - y,$ we have
$$x(90 - y) = (110 - x)y$$
$$90x - xy = 110y - xy$$
$$90x = 110y$$
$$x = \frac{11}{9}y.$$
We can substitute this into $x + y = 130$ and find that $y = \frac{117}{2}.$ Since the positive difference between $x$ and $y$ is $x - y = \frac{11}{9}y - y = \frac{2}{9}y,$ we have our final answer of $\frac{2}{9} \times \frac{117}{2} = \boxed{13}.$
This is the correct answer.
