Problem — Equilateral triangle from the tangents

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From a point $A$ outside a circle with centre $O$ and radius $r = 4$, with $\overline{OA} = 2r$, draw the two tangents, which touch the circle at $B$ and $C$. Prove that triangle $ABC$ is equilateral, and compute the measure of the central angle $\widehat{BOC}$ and the length of the tangent segment $AB$.

Solution

Triangle $OBA$ is right-angled at $B$, because the radius $OB$ is perpendicular to the tangent. We have $\sin\widehat{BAO} = \frac{\overline{OB}}{\overline{OA}} = \frac{r}{2r} = \frac{1}{2} \implies \widehat{BAO} = 30^\circ.$ By symmetry $\widehat{CAO} = 30^\circ$ as well, so $\widehat{BAC} = 60^\circ$. Moreover $\overline{AB} = \overline{AC}$ (tangent segments from the same point): triangle $ABC$ is isosceles with a vertex angle of $60^\circ$, hence it is equilateral. In the quadrilateral $OBAC$ the angles at $B$ and $C$ are right angles and $\widehat{BAC} = 60^\circ$; since the interior angles sum to $360^\circ$, $\widehat{BOC} = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ.$ The length of the tangent segment is $\overline{AB} = \sqrt{\overline{OA}^{\,2} - r^2} = \sqrt{8^2 - 4^2} = \sqrt{48} = 4\sqrt{3} \approx 6.93.$ $\boxed{\widehat{BOC} = 120^\circ, \qquad \overline{AB} = 4\sqrt{3} \approx 6.93}$