The problem was as follows:

Regular hexagon $HEXAGN$ is inscribed in the circle $O$, and $R$ is a point on minor arc $HN$ of circle $O$. If $RE=10$ and $RG=8$, then $RN$ can be expressed in the form $a\sqrt b+c$, where $a$, $b$, and $c$ are integers. Compute $RN$ in this form.

No calculator was allowed, and a 10 minute time limit was in place. I found out that the length of a side of the hexagon was $\sqrt {41}$ but I wasn't sure how to find that particular line length.


Use Ptolemy's Theorem.

Ptolemy's Theorem: For a cyclic quadrilateral $ABCD$, $$AC \cdot BD = AB \cdot CD + AD \cdot BC$$ (if you haven't already, try proving this as an exercise; use Law of Cosines).

Let $s$ be the side length of the hexagon. Applying Ptolemy's to quadrilaterals $RHEN$ and $NRHG$, we get the system of equations $$RE \cdot HN = EH \cdot RN + EN \cdot RH \implies 10s = s \cdot RN + s\sqrt{3} \cdot RH$$ $$RG \cdot HN = HG \cdot RN + NG \cdot RH \implies 8s = s\sqrt{3} \cdot RN + s \cdot RH$$

Dividing by $s$ and solving the system gives $RN = 4\sqrt{3} - 5$ and $RH = 5\sqrt{3} -4$.


Since $E$ and $G$ are diametrically opposite, we conclude that $\angle GRE=90^\circ$ and consequently $EG^2=100+64=164$. It follows that the radius of the circumscribed circle is $r\sqrt{41}$.

If $\theta=\frac{1}{2}\angle ROE$, then $RE=2r\sin\theta$, or $$ \sin\theta=\frac{5}{\sqrt{41}}, \qquad\cos\theta=\frac{4}{\sqrt{41}}. $$ But, since $\angle NOE=120^\circ$ we have $\angle NOR =120^\circ-2\theta$, and consequently $$\eqalign{RN&=2r\sin(60^\circ-\theta)=2r\left(\frac{\sqrt{3}}{2}\cos\theta-\frac{1}{2}\sin\theta\right)\cr &= \sqrt{3}\sqrt{41}\cos\theta- \sqrt{41}\sin\theta =4\sqrt{3}-5 } $$


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