<s>[INST] Cyclic Quadrilaterals and Pentagons Let ABCDE be a pentagon with side lengths $AB=9$, $BC=10$, $CD=10$, $DE=13$, and $EA=20$. We're given that $AC=17$ and ABCD is a cyclic quadlirateral, and let $O_1$ be the circumcenter of $ABCD$ and $O_2$ be the circumcenter of DEA. Find the length of segment $O_1O_2$.
We cannot solve this problem without a diagram. I need help with this problem.
 [/INST] The hint.
By law of cosines for $\Delta ABC$ we obtain
$$\cos\measuredangle ABC=\frac{9^2+10^2-17^2}{2\cdot9\cdot10}=-\frac{3}{5},$$
which gives $\cos\measuredangle ADC=\frac{3}{5}$ and by law of cosines for $\Delta ACD$ we obtain:
$$AD^2+100-2\cdot AD\cdot10\cdot\frac{3}{5}=17^2,$$ which gives $AD=21.$
Now, $$p_{\Delta ABC}=\frac{9+10+17}{2}=18,$$
which gives
$$S_{\Delta ABC}=\sqrt{18(18-17)(18-9)(18-10)}=24.$$
Thus, $$R_1=\frac{9\cdot10\cdot17}{4\cdot24}=\frac{255}{16}.$$
Also,$$p_{\Delta ADE}=\frac{13+20+21}{2}=27,$$
which gives
$$S_{\Delta ADE}=\sqrt{27(27-13)(27-20)(27-21)}=126$$ and
 $$R_2=\frac{13\cdot20\cdot21}{4\cdot126}=\frac{65}{6}.$$
Now, by law of cosines for $\Delta O_1O_2D$ calculate $O_1O_2$.
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