# What radius of circle intersects these points inside a regular pentagon?

I am looking for the radius of circle OP/OQ. P lies on the altitude of one of the sides of the pentagon, OT. Therefore the angle SOR is $$\frac{\pi}{5}$$. PQ is the chord subtending the arc $$\frac{\pi}{5}$$ which goes through the vertex of the pentagon V. I believe there is only one radius of circle where V, P, and Q are colinear, but if I am wrong please answer with all such radii. Feel free to define any length as the unit as is convenient. I can find the lengths of VR, VO, and VT easily enough (given one as the unit), but the others escape me. Any help is appreciated, thanks.

• Hint…can you show$\angle QVR=\angle RVO=\frac{\pi}{10}$? Commented Apr 12 at 8:27

$$\angle VOP = \angle OVP = \angle Q'PV = 36^\circ$$
so that $$\triangle OPV$$ and $$\triangle VQ'P$$ are similar isosceles triangles. Defining $$p:=|OP|$$ and $$v:=|OV|$$, we have (ignoring an extraneous negative root) ... $$\frac{p}{v} = \frac{v-p}{p} \quad\to\quad p^2+pv-v^2=0 \quad\to\quad p = \frac12v\left(-1+\sqrt{5}\right)=\frac{v}{\phi}$$ where $$\phi=1.618\ldots$$ is the Golden Ratio.
Note that by symmetry, $$B, P, M$$ must also be collinear.
We know that $$\angle TOB = \frac{\pi}{5}$$, hence $$\angle OBT = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}$$. We also know $$\angle BOV = \frac{2\pi}{5}$$, hence $$\angle OBS = \frac{\pi}{2} - \frac{2\pi}{5} = \frac{\pi}{10}.$$ Next, because $$\triangle POQ$$ is isosceles with radii $$OP = OQ$$, and $$\angle POQ \cong \angle TOB = \frac{\pi}{5}$$, then $$\angle QPO = \angle PQO = \frac{1}{2} \left(\pi - \frac{\pi}{5}\right) = \frac{2\pi}{5}.$$ Hence in right $$\triangle VRQ$$, we have $$\angle QVR = \frac{\pi}{2} - \frac{2\pi}{5} = \frac{\pi}{10}.$$ So again by symmetry about the line $$OT$$, it follows that $$\angle SBP \cong \angle SVP \cong \angle QVR = \frac{\pi}{10}$$; therefore, $$\angle OBT$$ is trisected by $$BP$$ and $$BS$$ and $$\angle TBP \cong \angle PBS \cong \angle SBO = \frac{\pi}{10}.$$ This uniquely determines the radius up to a scaling factor. So for instance, $$\frac{OP}{OT} = \frac{OT - PT}{OT} = 1 - \frac{PT/BT}{OT/BT} = 1 - \frac{\tan \frac{\pi}{10}}{\tan \frac{3\pi}{10}} = 3 - \sqrt{5}.$$
• @AnthonyKhodanian $TP < PS$ because if they were equal, then the areas $|\triangle BTP| = |\triangle BTS|$. But $$|\triangle BTP| = \frac{1}{2} BT \cdot BP \sin \angle TBP$$ and similarly $$|\triangle BTS| = \frac{1}{2} BP \cdot BS \sin \angle PBS,$$ and since $\angle TBP \cong \angle PBS$ and $BT < BS$, we have $|\triangle BTP| < |\triangle BTS|$. Commented Apr 12 at 20:21