I am solving a the following problem.

Let $P$ be a convex octagon which can be inscribed in a circle such that four of its sides have length $2$ and the other four sides have length $3$. Find all possible values of the area of $P$.

Actually this was an problem of a graduate school admission test. If the sides are $2,3,2,3,2,3,2,3$, we can make a square a around the boundary of the octagon and also we can calculate the area of the square.

But if the $2$ and $3$'s are not consecutive, then how to proceed?

Please help me.

  • 2
    $\begingroup$ Some hints: a) Fix the radius $r$ of the circumscribing circle, can you relate the length of a chord (either 2 or 3 in this case) to the angle subtended by that chord at the centre of the circle? b) Use this to conclude that for any ordering of edges, the circumscribing circle has the same radius. c) Finally relate the area of the octagon to the area of the circle, and the areas of the segments cut off by the edges of the octagon. $\endgroup$ Aug 26, 2021 at 16:11
  • $\begingroup$ You can get from any given octagon to another one by switching between adjacent sides. This leads to the conclusion that there is only one possible area - from the area of the circle you subtract two groups of four identical sections from the circle. $\endgroup$
    – Moti
    Aug 26, 2021 at 20:22
  • $\begingroup$ You have calculated it for one octagon. Now cut it into triangles from the center of the circle. You can rearrange these triangles in any order to create a new octagon of the same area inscribed in the same circle. $\endgroup$
    – Doug M
    Aug 27, 2021 at 17:34

1 Answer 1


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Comment: As you see in figure in all cases we have four triangles like OAB with base $AB=3$ and four triangle like OBC with base $BC=2$. So doing as shown in figure a will give the solution for all.You can check this using the measures shown on figures.


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