Distance Between Intersection Points

We join $X,Y$ . Note that $\angle AXB = \angle AYB = 90^{\circ}$ and $XO$ , $YO$ are the medians of the right angled triangles $AXB$ and $AYB$ . Let $\angle XBO = \alpha$ and $\angle YAO = \beta$ . Thus $\angle OXB = \alpha$ and $\angle XYA = \alpha$ . Similarly $\angle YXB = \beta = \angle YAB$ . Thus $\alpha + \beta = 30^{\circ}$ .

Now if we assume that $AO=XO=YO=BO=x$ , then law of cosines in $\triangle XOY$ yields $XY = x\sqrt{3}$ .

Now note that $\angle WXZ = \angle WYZ = 90^{\circ}$ and hence $WXZY$ is cyclic, $WZ$ being the diameter of the circumscribing circle. . Thus $\angle XZY = \angle XZW + \angle WZY = \angle XYW + \angle WXY= \alpha + \beta = 30^{\circ}$ . Therefore $\angle XWY = 150^{\circ}$ . Now Sine rule in $\triangle XWY$ yields

$\dfrac{XY}{\sin \angle XWY} = 2R = WZ$

or, $\dfrac{x \sqrt{3}}{\sin 150^{\circ}}$ = $WZ$

or, $2 \times x \sqrt{3}= WZ$

or, $2x \times \sqrt{3} = WZ$

or, $WZ = \sqrt{3} AB = \sqrt{3} \times 5\sqrt{3} = \boxed{15}$

WZ=OZ -OW =OA tanOAZ - OA tanOAW =OA(tan1/2BOX - tan 1/2BOY) =OA(tan 0.5 150 - tan0.5 30) =5*root(3)/2( tan 75 - tan 15 ) =15 Answer

Considering symetrical segment..

It is clear that angle ZXW = angle ZYW = 90 (since AB is diameter) therefore, XWYZ is cyclic

Let angle AYO be x and BXO be y. Therefore, OAY=x(isoceles triangle AOY) and OBX=y(isoceles triangle OBX)

angle BOY=2*BAY=2x and XOA=2y

Thus, x+y=30 (since 2x+2y+120=180)

Now, angle XWY=120+x+y=150

Applying sine law in XOY(an isoceles triangle since OX=OY=r), we get XY/sin120 = OX/sin30 which gives XY=7.5

Further, Applying sine law in ZXY, we get XY/sin30=2R(where R is the radius of the circumcircle) Thus 2R=15 and since the circumcircle is also the circumcircle of WXZ(since WXZY is cyclic), therefore 2R=WZ=15.

Without loss of generality, assume that X and Y are equidistant from AB. Then angle YOB=30 and $OY=\frac{5\sqrt{3}}{2}$ . Let the altitude from Y to AB intersect AB at point G. Notice that $\triangle YOG$ is a 30-60-90 right triangle. Therefore, $YG=\frac{5\sqrt{3}}{4}$ and $GO=\frac{15}{4}$ . From this, we deduce that $GB=\frac{5\sqrt{3}}{2}-\frac{15}{4}$ . Then, by similarity of triangles, we have $\frac{ZO}{\frac{5\sqrt{3}}{4}}=\frac{\frac{5\sqrt{3}}{2}}{\frac{5\sqrt{3}}{2}-\frac{15}{4}}$ $=\frac{\frac{75}{8}}{\frac{5\sqrt{3}}{2}-\frac{15}{4}}$ $=\frac{75}{20\sqrt{3}-30}$ $=\frac{15}{4\sqrt{3}-6}$ $=\frac{60\sqrt{3}+90}{12}$ $=\frac{10\sqrt{3}+15}{2}$ .

We now need to subtract WO in order to obtain WZ. Again, we use similar triangles. $\frac{YG}{AG}=\frac{WO}{AO}$ $\frac{WO}{\frac{5\sqrt{3}}{4}}=\frac{\frac{5\sqrt{3}}{2}}{\frac{5\sqrt{3}}{2}+\frac{15}{4}}$ $=\frac{\frac{75}{8}}{\frac{5\sqrt{3}}{2}+\frac{15}{4}}$ $=\frac{75}{20\sqrt{3}+30}$ $=\frac{15}{4\sqrt{3}+6}$ $=\frac{60\sqrt{3}-90}{12}$ $=\frac{10\sqrt{3}-15}{2}$ .

When we subtract WO from ZO, we have $\frac{10\sqrt{3}+15-(10\sqrt{3}-15)}{2}=\frac{30}{2}=\boxed{15}$