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Two regular polygons are inscribed in the same circle of radius $r$. First one has $k$ sides and second has $p$ sides. We are given that their areas have a ratio of $1.5$. Calculate the area of a regular polygon inscribed in the same circle, having number of sides the sum of the other two numbers.

Area of the first polygon: $$A_k = \frac{1}{2}\cdot k \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{k}\bigg)$$ and area of the second: $$A_p = \frac{1}{2} \cdot p \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{p}\bigg)$$ Also $\frac{A_k}{A_p} = 1.5$ (assuming $k>p$ WLOG) So

$$\frac{A_k}{A_p} = \frac{k}{p} \cdot \frac{\sin\bigg(\dfrac{2\pi}{k}\bigg)}{\sin\bigg(\dfrac{2\pi}{p}\bigg)} = 1.5$$

Now obviously: $$A_{k+p} = \frac{1}{2} \cdot (k+p) \cdot r^2 \cdot \sin\bigg(\dfrac{2\pi}{k+p}\bigg)$$ But I don't know how to continue, i.e. how to find a relation between the sins.

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  • $\begingroup$ Where did the ratio $1.5$ go when you went from the line starting "Also" to the formula following that line? $\endgroup$ – coffeemath Feb 4 '18 at 10:07
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    $\begingroup$ @coffeemath, edited, thanks. $\endgroup$ – Sal.Cognato Feb 4 '18 at 10:25
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There are only limited possibilities for $n$ because we require $n\ge3$ to get any area and also $\frac32\frac n2\sin\frac{2\pi}n<\pi$ because the bigger polygon has to be smaller than the circumcircle. So we enumerate the possibilities: $$\begin{array}{r|l}n&\text{Area}\\\hline 3&1.299\\ 4&2.000\\ 5&2.378\\ 11&2.974\\ 12&3.000\\ 13&3.021\\ \end{array}$$ So $n=5$ is too big because $1.5\times2.378>\pi$. $3$ won't work because $1.5\times1.299$ doesn't match anything. Thus we are left with $n=4$ for the smaller polygon, so the area of the bigger polygon is $1.5\times2.000=3.000$, and this matches $n=12$. The final polygon must have $n=4+12=16$ sides with area $$A=\frac{16}2\sin\frac{2\pi}{16}=8\sin\frac{\pi}8=4\sqrt{2-\sqrt2}$$ So actually @Toby Mak's answer turned out to be correct in that the areas were in fact rational.

EDIT: Given that only one possibility is feasible it is easy to check that for $n=4$ $$A=\frac42\sin\frac{2\pi}4=2\sin\frac{\pi}2=2$$ While for $n=12$ $$A=\frac{12}2\sin\frac{2\pi}{12}=6\sin\frac{\pi}6=3$$ So the identity is exactly satisfied.

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  • $\begingroup$ His answer wasn't wrong but his reasoning was (your numerical method is better but still not completely rigorous) $\endgroup$ – Akababa Feb 4 '18 at 10:41
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    $\begingroup$ @Akababa It's a simple exercise to test that the identity is exactly satisfied for the one remaining feasible solution. See my edit. $\endgroup$ – user5713492 Feb 4 '18 at 10:50
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    $\begingroup$ So its a 16-gon, not 15-gon, right? $\endgroup$ – Sal.Cognato Feb 4 '18 at 11:18
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    $\begingroup$ @Sal.Cognato There are $3$ kinds of mathematicians: those who can count and those who can't. Thanks. $\endgroup$ – user5713492 Feb 4 '18 at 12:01
  • $\begingroup$ @user5713492 humor appreciated... $\endgroup$ – Jean Marie Feb 4 '18 at 17:14

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