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A hexagon is inscribed in a circle of radius $r$. Find $r$ if two sides of the hexagon are $7$ units long,while the other four sides are $20$ units long.

Efforts made: I've tried to construct right triangles but i couldn't get anything usefull so far.

Note: This problem is meant to be solved by geometric methods.

Thanks in advance for every answer.

Nameless
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2 Answers2

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Look at one half of the hexagon, and consider the angles subtended by the chords of length $20, 20$ and $7$.

We have $$4\arcsin\left(\frac{10}{r}\right)+2\arcsin\left(\frac{3.5}{r}\right)=\pi$$ Simplifying, this gives $$2\arcsin\left(\frac{10}{r}\right)=\arccos\left(\frac{3.5}{r}\right)$$

Now take the sine of both sides, together with a double angle formula and we get $$2\times\frac{10}{r}\sqrt{1-\frac{100}{r^2}}=\sqrt{1-\frac{3.5^2}{r^2}}$$

This is a quadratic equation in $r^2$ from which we can obtain the values $r=16, r=12.5$. I haven't checked if they're both valid, but can I leave that to you?

David Quinn
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    Could there be an even simpler approach without involving trigonometric relations ?thanks btw for the answer – Nameless Sep 12 '15 at 16:25
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Angle say $\alpha$ subtended by each of two sides having length 7 units at the center of the circle
$$=2\sin^{-1}\left(\frac{\frac{7}{2}}{r}\right)=2\sin^{-1}\left(\frac{7}{2r}\right)$$ Similarly, angle say $\beta$ subtended by each of four sides having length 20 units at the center of the circle
$$=2\sin^{-1}\left(\frac{\frac{20}{2}}{r}\right)=2\sin^{-1}\left(\frac{10}{r}\right)$$

Now, the total angle subtended by all six sides at the center of the circle is $2\pi$ hence, we have $$2\alpha+4\beta=2\pi$$ $$4\sin^{-1}\left(\frac{7}{2r}\right)+8\sin^{-1}\left(\frac{10}{r}\right)=2\pi$$ $$\sin^{-1}\left(\frac{7}{2r}\right)+2\sin^{-1}\left(\frac{10}{r}\right)=\frac{\pi}{2}$$ $$2\sin^{-1}\left(\frac{10}{r}\right)=\frac{\pi}{2}-\sin^{-1}\left(\frac{7}{2r}\right)=\cos^{-1}\left(\frac{7}{2r}\right)$$

$$\cos^{-1}\left(1-2\left(\frac{10}{r}\right)^2\right)=\cos^{-1}\left(\frac{7}{2r}\right)$$ $$1-2\left(\frac{10}{r}\right)^2=\frac{7}{2r}$$ $$2r^2-7r-400=0$$ On solving above quadratic equation we get $r=\frac{7\pm\sqrt{(-7)^2-4(2)(-400)}}{2(2)}=\frac{7\pm{57}}{4}$ $$r=\frac{7+57}{4}=\color{}{16}$$ or $$r=\frac{7-57}{4}=-\frac{25}{2}<0$$ But radius $r>0$ Hence, we get $$\color{red}{r=16\ units}$$

Harish Chandra Rajpoot
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