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How would you find the vertices (corners' position) of a shape that inscribes a circle of adjustable radius, given a set of edges? Angles of polygon are not fixed, but edges are.

A few examples:

Finding the vertices of a pentagon with edges 1, 2, 3, 4, and 5 units in length, inscribed on a circle that will fit it.

Finding vertices on a circle of any radii that are the vertices of a kerpleckagon with edges 1, 56, 31, 300, 62, 421, etc.

Update: Just to be clear, the shape made with the fixed edges consists of only convex angles, and is positioned inside of the circle. The output of the function or equation should be the polar coordinate or angle of the vertex.

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  • $\begingroup$ Is the given sequence of edge lengths to "go around cyclically" e.g. in the 1,56,31 example the 56 units length is between the 1 and 31 lengths? [Another issue: Is it easy to show any radius-sequence may be done in some circle of appropriate radius?] $\endgroup$ – coffeemath Jun 11 '16 at 3:00
  • $\begingroup$ One simple restriction on the set of edge lengths is that there not be one length which exceeds the sum of the remaining lengths, otherwise a realization of that would go against the triangle inequality. $\endgroup$ – coffeemath Jun 11 '16 at 3:27
  • $\begingroup$ Logan: I have deleted a previous comment in which I gave the opinion that the circle center would not be inside the pentagon. I now think it is, based on my answer below, which finds a radius of about 2.7175 for the circle. $\endgroup$ – coffeemath Jun 11 '16 at 8:25
  • $\begingroup$ After trying to apply it, I saw that you were answering for the circle inscribed in the shape, when I was curious about the shape inside the circle. Sorry for the confusion. :| $\endgroup$ – ilcj Jun 23 '16 at 22:03
  • $\begingroup$ No, my answer was indeed for the shape inscribed in a circle. The chord lengths $1,2,3,4,5$ are each connecting two points on the circumference of a circle of radius $r,$ and given a chord of length $d$, and the (unknown) $r,$ formula $(1)$ of my answer for $\theta_d$ gives the value of the central angle made by that chord. As said, in this we're assuming the center of the circle lies inside the inscribed shape, so the sum of the five values of $\theta_d$ must be $2\pi.$ For your numbers $1,2,3,4,5$ for chord lengths, it turns out there is a choice $r=r_0$ which works. $\endgroup$ – coffeemath Jun 24 '16 at 4:19
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We show how to numerically obtain the value $r$ for the proper radius so that the pentagonal shape with sides $1,2,3,4,5$ can be inscribed in a circle of radius $r,$ with the added assumption that the center of the circle is interior to the pentagon. Under this assumption $r$ appears to be unique, however to really prove that would involve investigating a complicated sum of inverse cosines. A bit more experimentation led to the guess that there is not a solution in which the circle center was not in the interior of the pentagon.

From the law of cosines, if $\theta_d$ is the angle subtended by a chord of length $d$ in a circle of radius $r,$ [so in particular $\theta_d \le \pi,$ the upper value corresponding to the chord being a diameter] then we have $$\theta_d=\arccos \frac{2r^2-d^2}{2r^2}. \tag{1}$$ In order that the pentagon with sides $1,2,3,4,5$ can be built by putting the triangles each formed by the circle center and one of the chords, we at least require $r>5/2=2.5$ since one of the triangles has sides $r,r,5.$

Now let $g(r)$ denote the result of summing the $\theta_d$ as given in (1) for $d$ from $1$ to $5.$ We find that $g(2.5)=7.5089>2\pi=6.2831.$ But luckily (in a way obviously) the value of $g(r)$ decreases as $r$ increases beyond $2.5,$ and a numerical root-finder shows that when we take $r=r_0=2.717567...$ we get $g(r_0)=2\pi$ which means our desired radius for the pentagon of sides $1,2,3,4,5$ to inscribe in a circle is this $r_0.$ I see no hope of getting an analytical expression for this $r_0,$ since the objective function is defined as a sum of seemingly unrelated inverse cosines.

As for the uniqueness, I admit I only checked that the graph of $g(r)$ is decreasing (or is on my graphing calculator); I didn't feel like tackling $g'(r)$ to really show this. There may be a simple geometric argument to see why $g(r)$ should be decreasing: As we increase $r,$ the angle opposite the chord of length $d$ decreases, and we're adding five such terms.

An example: Take the first vertex at $V_1=(r_0,0)$ and use the chords in the order $5,4,3,2,1.$ Then the second vertex is at $V_2=(r_0 \cos(\theta_5),r_0 \sin(\theta_5))=(-1.88,1.96).$ [Note I'm only rounding to two decimals, also always rounding down, more accurate versions could be obtained.] For the next vertex, we need to use the central angle $\theta_5+\theta_4,$ and the third vertex then becomes $V_3=(-1.79,-2.03).$ Proceeding, we find the last two $V_4=(1.17,-2.45)$ and $V_5=(2.53,-.98).$

A numerical check now shows each vertex is at distance $r_0$ from the origin, thus this five-sided figure is indeed inscribed in the circle as desired in the posted question. And also one can check the chord lengths $V_1V_2$ is 5, $V_2V_3$ is 4, etc, so we do have chords of length $5,4,3,2,1$ in that order going around the circle. Naturally, using only an approximation to $r_0$ our points will not lie exactly on the circle, and the chord distances won't be exactly $5,4,3,2,1.$ But this should convince one that if we somehow could access the exact value of $r_0$, those exact distances would follow.

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