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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.

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

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A little angle-chasing shows that

$$\angle VOP = \angle OVP = \angle Q'PV = 36^\circ $$

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.

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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, $$\frac{OP}{OT} = \frac{OT - PT}{OT} = 1 - \frac{PT/BT}{OT/BT} = 1 - \frac{\tan \frac{\pi}{10}}{\tan \frac{3\pi}{10}} = 3 - \sqrt{5}.$$

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  • $\begingroup$ Thanks! Just curious, is it easy to show from what you've already done whether or not the lenght TP is equal to PS? They look to be so from a glance, but unsure if that is exact. $\endgroup$ Commented Apr 12 at 20:08
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    $\begingroup$ @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 $|\triangle BTP| < |\triangle BTS|$. $\endgroup$
    – heropup
    Commented Apr 12 at 20:21

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