I recently came across a Stackexchange post that asks for the probability that a point chosen randomly inside an equilateral triangle is closer to the center than to any of its edges. The square case was Problem B1 in 50th Putnam 1989.
I thought what about a general n-gon. Obviously for a circle, the answer should be 1/4. To tackle, the general case, we can consider the center of the polygon to be the origin and orient the polygon in such a way that one of its edges is perpendicular to the positive y-axis.
We can scale the polygon such that this edge passes through (0,1). In other words, we scale the polygon such that the inradius is 1. With the polygon scaled to a unit-inradius, the side of the polygon becomes 2\tan(\pi/n).
Now, we know that the locus of the points that are equidistant from the origin and to the line y=1 is a parabola. Using the formulae for distance between two points and distance between a point and a line, we have,
x^2+y^2=\displaystyle\frac{(y-1)^2}{1^2+0^2}
The above formulae is the same one given in this Wikipedia page. Simplifying the above we find the equation of the parabola as y=(1-x^2)/2.
Using the symmetry of the n-gon, we only focus only on the triagulation of the polygon formed by the origin and the edge that lies along the line y=1. The area of one such triangulation can be easily seen as \tan(\frac{\pi}{n}).
The lines through the origin bounding this edge are y=\pm x \tan(\pi/2-\pi/n)=\pm x \cot(\frac{\pi}{n}) . From now on, we restrict ourselves to the positive sign again using symmetry to our advantage.
The point of intersection of the line y=\cot(\frac{\pi}{n})x and the parabola y=(1-x^2)/2. Solving the resulting quadratic we can simplify the positive root as x=\tan(\frac{\pi}{2n}). Therefore the point of intersection is (\tan(\frac{\pi}{2n}),\tan(\frac{\pi}{2n})\cot(\frac{\pi}{n})).
To simplify notation, let t=\tan(\frac{\pi}{2n}) from now on. Therefore, the line becomes y=\pm \frac{1-t^2}{2t}x, the point of intersection becomes (t,\frac{1-t^2}{2}) and the area of the traingulation becomes \frac{2t}{1-t^2}.
Now, the area bounded by the parabola and the x-axis, is given by,
A=\displaystyle\int_0^t \frac{1-x^2}{2}\,dx=\frac{t}{6}(3-t^2)
and
the area of the triangle formed by origin, the point x=t and the point of intersection of the parabola and the line is T=\frac{1}{2}\cdot t \cdot \frac{1-t^2}{2}.
In each of the triangulations of the n-gon, the radom point will be closer to the center than to the edge if it falls within the region bounded by the lines y=\pm \frac{1-t^2}{2t}x and the parabola. If we denote the probability of such an event by p_n, then
p_n=\displaystyle\frac{2(A-T)}{\tan(\frac{\pi}{n})}=\frac{(t^2+3)(1-t^2)}{12}
Substituting for t and simplifying, we have p_n=\frac{1}{12}(4-\sec^4(\frac{\pi}{2n}))
Until then
Yours Aye
Me
I thought what about a general n-gon. Obviously for a circle, the answer should be 1/4. To tackle, the general case, we can consider the center of the polygon to be the origin and orient the polygon in such a way that one of its edges is perpendicular to the positive y-axis.
We can scale the polygon such that this edge passes through (0,1). In other words, we scale the polygon such that the inradius is 1. With the polygon scaled to a unit-inradius, the side of the polygon becomes 2\tan(\pi/n).
Now, we know that the locus of the points that are equidistant from the origin and to the line y=1 is a parabola. Using the formulae for distance between two points and distance between a point and a line, we have,
x^2+y^2=\displaystyle\frac{(y-1)^2}{1^2+0^2}
The above formulae is the same one given in this Wikipedia page. Simplifying the above we find the equation of the parabola as y=(1-x^2)/2.
Using the symmetry of the n-gon, we only focus only on the triagulation of the polygon formed by the origin and the edge that lies along the line y=1. The area of one such triangulation can be easily seen as \tan(\frac{\pi}{n}).
The lines through the origin bounding this edge are y=\pm x \tan(\pi/2-\pi/n)=\pm x \cot(\frac{\pi}{n}) . From now on, we restrict ourselves to the positive sign again using symmetry to our advantage.
The point of intersection of the line y=\cot(\frac{\pi}{n})x and the parabola y=(1-x^2)/2. Solving the resulting quadratic we can simplify the positive root as x=\tan(\frac{\pi}{2n}). Therefore the point of intersection is (\tan(\frac{\pi}{2n}),\tan(\frac{\pi}{2n})\cot(\frac{\pi}{n})).
To simplify notation, let t=\tan(\frac{\pi}{2n}) from now on. Therefore, the line becomes y=\pm \frac{1-t^2}{2t}x, the point of intersection becomes (t,\frac{1-t^2}{2}) and the area of the traingulation becomes \frac{2t}{1-t^2}.
Now, the area bounded by the parabola and the x-axis, is given by,
A=\displaystyle\int_0^t \frac{1-x^2}{2}\,dx=\frac{t}{6}(3-t^2)
and
the area of the triangle formed by origin, the point x=t and the point of intersection of the parabola and the line is T=\frac{1}{2}\cdot t \cdot \frac{1-t^2}{2}.
In each of the triangulations of the n-gon, the radom point will be closer to the center than to the edge if it falls within the region bounded by the lines y=\pm \frac{1-t^2}{2t}x and the parabola. If we denote the probability of such an event by p_n, then
p_n=\displaystyle\frac{2(A-T)}{\tan(\frac{\pi}{n})}=\frac{(t^2+3)(1-t^2)}{12}
Substituting for t and simplifying, we have p_n=\frac{1}{12}(4-\sec^4(\frac{\pi}{2n}))
Until then
Yours Aye
Me
