Interesting math problem that I saw online

[bob012345]

I understand there is no general formula for higher order figures inscribed in a circle in terms of the radius as a function of side lengths. At least not without high powered math and special functions. But still we can use the equation above to numerically solve it for higher order with the caveat that for an $n$-sided figure, the longest side is less than the sum of the other sides. $c_{n}<\sum _{i=1}^{n-1}c_{i}$

We can start with the general formula under the assumption the given sides make a closed figure inscribed by a circle.

$\sum _{i=1}^{n}\arcsin \left(\frac{c_{i}}{2R}\right)=\pi$


Depending on the side lengths, there are two possible cases for the center of the circumscribed circle:

1. Polygon includes the center of the circle (Center Inside)
2. Center of the circle lies outside the polygon (Center Outside)

We can check this by calculating the Critical Radius $R_{\text{crit}}=\frac{L_{n}}{2}$ (Where $L_{n}$ is the longest side). Then we calculate the angle_sum defined as

$A=\sum _{i=1}^{n-1}\arcsin \left(\frac{c_{i}}{2R_{\text{crit}}}\right)$

Then,

• If $A >\frac{\pi}{2}$: The center is inside the polygon.
• If $A < \frac{\pi}{2}$: The center is outside the polygon.
• If $A = \frac{\pi}{2}$: The longest side $L$ is a literal diameter of the circle.

We can then use a solver (I used Desmos) to compute the intersection of two functions as follows;

• If Case 1 (Inside):
  $y=\sum _{i=1}^{n}\arcsin \left(\frac{c_{i}}{2R}\right)\quad \text{and}\quad y=\pi$
• If Case 2 (Outside):
  $y=\sum _{i=1}^{n-1}\arcsin \left(\frac{c_{i}}{2R}\right)\quad \text{and}\quad y=\arcsin \left(\frac{c_{n}}{2R}\right)$
  (Where $c_{n}$is the longest side)

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[bob012345]

Here is an example for the case (2,3,4,5,6). In this case the sum angle is 2.578 rads which is greater than $\frac{\pi}{2}$ so the circle center is inside the polygon.

We use the equations from case 1 above to calculate the intersection.

And here is a case for (2,3,4,5,11). In this case the sum angle is 1.063 rads which is less than $\frac{\pi}{2}$ so the circle center is outside the polygon.

We use the equations from case 2 above to calculate the intersection.

 

[kuruman]

There are m rods length of each does not surpass sum of the others. With these rods we can make m polygons which has a circumcircle. It seems to hold.

Sure. Shown below is a quadrilateral in which one side (green) is equal to the sum of the other three.

There is no circle that passes through all 4 points where the color changes.

If you shorten the green segment by $\epsilon$ keeping the other three at the same length, there will be a closed quadrilateral and a circle passing through all 4 points.

If you lengthen the green segment by $\epsilon$ keeping the other three at the same length, you will no longer have a closed polygon.

Example
The segments shown above are
Red = 1.0 u
Blue =1.6 u
Magenta = 2.2 u
Green = 4.8 u

Using the formula for the radius of the circumscribed circle, I found that if I shorten the green segment by ε = 1×10^-6 u,
$R\approx2,000~$u and that the angle between green and red is 0.04° and between blue and magenta is 179.96°.

 

[anuttarasammyak]

In generalc case of m polygon
∑i=1msin−1⁡Lid=π
seems to hold with choice of parameter d for any L1,L2,..,Lm. So there seems to exist a circumcircle.

I find the above is for polygon which includes centern of circumcircle in it. As shown in #61 including polygon which does not include the center, the condition is
$$(\sum_{i=1}^m \sin^{-1}\frac{f(L_i)L_i}{d})(\sum_{i=1}^m \sin^{-1}\frac{L_i}{d}-\pi)=0$$
defined for d > the longest $L_i$ and $f(L_i) = -1$ for the longest rod, 1 otherwise. When we have plural longest rods, any only one of them should be chosen.

As far as {Li} can form m-polygons, there exist some of them which have circumcircle whose diameter d would satisfy above equation,i.e., have circumicircle.

Proof
Say $L_1 = max\{L_i\}$
case
$$\sum_{i=1}^m\sin^{-1}\frac{L_i}{L_1}-\pi>0$$
$$f(d):=\sum_{i=1}^m\sin^{-1}\frac{L_i}{d}-\pi$$ is continuous monotonous decreasing function to $-\pi$ as d increase from $L_1$. Thus there exists unike value d which satisfies f(d)=0,i.e.,
$$\sum_{i=1}^m\sin^{-1}\frac{L_i}{d}-\pi=0$$

case
$$\sum_{i=1}^m\sin^{-1}\frac{L_i}{L_1}-\pi<0$$
$$\sum_{i=2}^m\sin^{-1}\frac{L_i}{L_1}-\sin^{-1}\frac{L_1}{L_1}<0$$
$$g(d):=\sum_{i=2}^m\sin^{-1}\frac{L_i}{d}-sin^{-1}\frac{L_1}{d}<0$$
for $d = L_1+0$. By expanding g(d) by 1/d
$$g(d)=\frac{\sum_{i=2}^m L_i -L_1}{d}+o(1/d^2) > 0 $$
Thus there exists value d which satisfies g(d)=0,i.e..
$$\sum_{i=2}^m\sin^{-1}\frac{L_i}{d}-\sin^{-1}\frac{L_1}{d}=0$$

QED