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While going through my phone's old pictures, I stumbled on a problem someone had sent me a while back. I remember trying hard, giving up, and then forgetting about it.
I tried to see how I can solve it via trigonometry or properties of isosceles triangles, but to no avail.
How is this problem to be solved?
This is straightforward using trig form of Ceva,
$$\sin \angle OAC\sin \angle OCB\sin \angle OBA = \sin \angle OAB\sin \angle OBC\sin \angle OCA$$
Substituting $\angle OBC = x$ and $\angle OBA = 100-x$ and solving $$x=80^{\circ}$$
Here's a way to do it without Ceva.
From the diagram, we have $$ \begin{align} \angle POC&=30^\circ,\\ \angle AOC &= 150^\circ,\\ \angle B &=100^\circ,\\ \angle POC&=30^\circ. \end{align}$$ Since $\triangle CPO$ is isosceles, we may take $\overline{OP}=\overline{CP}=1.$ Applying the law of cosines to $\triangle CPO$ gives$$\overline{OC}^2=1+1-2\cos(120^\circ)\implies\overline{OC}=\sqrt3$$
Applying the law of sines in $\triangle ACO$ gives $$\frac{\overline{AO}}{\sqrt3}=\frac{\sin10^\circ}{\sin20^\circ}$$ so that $$\overline{AO}=\frac{\sqrt3}{2\cos10^\circ}$$
In $\triangle BOP,$ the law of sines gives$$\frac{\overline{OB}}{\overline{OP}}=\frac{\sin60^\circ}{\sin x},$$ or $$\overline{OB}=\frac{\sqrt3}{2\sin x}\tag1$$
In $\triangle BOA,$ the law of sines gives $$ \frac{\overline{OB}}{\overline{OA}} =\frac{\sin20^\circ}{\sin(100^\circ-x)}$$ so that $$\overline{OB}=\frac{\sqrt3}{2\cos10^\circ}\cdot\frac{2\sin10^\circ\cos10^\circ}{\sin(100^\circ-x)}=\frac{\sqrt3\sin10^\circ}{\sin{(100^\circ-x)}}\tag2$$
Combining $(1)$ and $(2)$ gives $$2\sin10^\circ\sin x=\sin(100^\circ-x)$$ and we see that $$\boxed{x=80^\circ}$$ since $\sin80^\circ=\cos10^\circ.$
I did it wrong half a dozen times before getting it right, so Ceva is better.
