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