An octagon which has side lengths 3, 3, 11, 11, 15, 15, 15 and 15 is inscribed in a circle. What is the area of the octagon?

I tried using the cosine law on the triangles made when connected with the center but the numbers became really hard to use when I solved for the sine of the angles to easily get the triangle area.

This is a problem from BIMC 2017 individual question 6, where the answer is 567. ;)

  • $\begingroup$ Are those side lengths in order? $\endgroup$ – Arthur Jul 16 '18 at 9:04
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    $\begingroup$ @Arthur I don't think it matters. For a given radius, the area of the triangles depends only on the sides. But since the central angles have to add up to $2\pi,$ you should be able to solve for the radius. $\endgroup$ – saulspatz Jul 16 '18 at 9:07
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    $\begingroup$ Where does this problem come from? So far, I can't see any way to do it except numerically. Is there reason to believe it has some clever exact solution? $\endgroup$ – saulspatz Jul 16 '18 at 9:09
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    $\begingroup$ Given that the order does not matter, I'd put them in the order $3, 15, 11, 15, 3, 15, 11, 15$. The symmetry not only makes opposite sides parallel, the sides of length $3$ will be perpendicular to the sides of length $11$. That has to help a lot. $\endgroup$ – Jaap Scherphuis Jul 16 '18 at 9:43
  • $\begingroup$ This problem from BIMC 2017 individual question 6 where the answer is 567 ;) $\endgroup$ – SuperMage1 Jul 16 '18 at 9:52

Arranging the sides in the order $3,15,11,15,3,15,11,15$, we obtain two overlapping, perpendicular inscribed rectangles of sides $3×h$ and $11×k$ respectively, where $h,k$ are to be determined. The Pythagorean theorem gives $$3^2+h^2=11^2+k^2$$ $$h^2-k^2=112$$ Furthermore, the rectangles are aligned and have a common centroid. This gives $$\left(\frac{h-11}2\right)^2+\left(\frac{k-3}2\right)^2=15^2$$ $$(h-11)^2+(k-3)^2=30^2$$ This immediately suggests finding Pythagorean triples $(a,b,30)$, of which there is only one: $18^2+24^2=30^2$. If we let $h-11=18$ and $k-3=24$, we see that the other equation $h^2-k^2=112$ is "miraculously" satisfied.

Thus $h=29$ and $k=27$. The octagon's area can then be calculated easily by adding the rectangles' areas, subtracting their intersection and adding the triangles' areas: $$3h+11k-33+2\left(\frac{h-11}2\right)\left(\frac{k-3}2\right)=87+297-33+2\cdot9\cdot12=567$$

  • $\begingroup$ I've neatened up your figure with Geogebra, if you'd like i.imgur.com/ztpQFxc.png $\endgroup$ – Jam Jul 17 '18 at 12:22
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    $\begingroup$ @Jam one of the traditions I've had with my answers here is hand-drawn diagrams, particularly because I answer most questions on my mobile phone. But a neat diagram is always good to have. $\endgroup$ – Parcly Taxel Jul 17 '18 at 12:25

Arrange the triangles so there is a $15$ in each quadrant; the $3$ sides are vertical and the $11$ sides are horizontal. Then the distance between $(\sqrt{r^2-9/4},3/2)$ and $(11/2,\sqrt{r^2-121/4})$ is 15.


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