How many sides of a regular polygon are visible from a random point?

Maybe I'm jumping the gun, but it seems to me that when R -> ∞, then the polygon becomes a point, and the lines separating the areas of visibility become rays from this point at equal angles. Then your contention is proven upon inspection.

To help visualize, starting with the diagrams provided, just move all the lines towards the center of the polygon. So, for example in the case of n is even, pairs of lines come together, vanishing the probability of seeing just one face. Etc.

As far as I'm concerned, Scythian has a perfectly rigorous proof; everything else is bookkeeping. If you're worried about whether your picture really converges to his picture and want lim r→ ∞s involved, here's a more exhaustive explanation. In the odd case, notice that each gray cone (corresponding to a region with (n+1)/2 sides visible) is paired with the green cone on the opposite side. Obviously the green side of the cone contains some blue and yellow regions at its tip, but those are of a fixed size, and won't affect the limit.

The only tricky point is that the center of your circle is not on a point where the green and gray cones intersect, but rather a fixed distance D away from all of them. To handle this, start running the limit when the radius r is large enough that D/r < ε. To be more concrete, think about your circle being of radius one, and the vertical angle I'm examining being distance ε from the center. Draw another circle of radius 1 centered at the vertical angle. The error term added from shifting the center to the vertical angle (at which the two cones would be exactly equal) is contained in the two lunes you'll see in this picture. These are contained in an annulus of radii 1+ε and 1-ε, which has total area 4πε.

If you want me to translate back, a gray cone does not have the exact same area as the cone opposite it inside the circle, but the difference is no more than 4πD/r, and thus the total difference over the disk is bounded above by 4πnD/r. This obviously goes to zero in the limit, so the green and gray regions are asymptotically 1/2 of the circle.

For the even case, there's even less to worry about. P(yellow) grows linearly with radius, while the total area enclosed by your circle grows quadratically. I don't need any computer simulations to know that for any 2n-gon, the probability of n-1 sides being visible will be O(1/r).

The vertices of the n-gon in standard position are going to be

V(k) = (cos kϑ, sin kϑ) where ϑ = 2π/n

and the k's can be chosen from any set of n consecutive integers.

My plan is to put the observer at infinity in the positive x direction; then rotate the polygon and record the number of sides that are visible. To rotate the vertices through an angle of α is simple:

V(k, α) = (cos (kϑ + α), sin (kϑ + α)).

So let F(α) be the number of vertices, V(k,α), that are visible from an observer at (infinity, 0). First we list some properties of F.

(1) F(α) = F(α + ϑ) for any integer z.

(2) F(α) = F(-α)

(3) F(π - α) + F(α) = n + 2

To prove (3) note that F(π - α) is the number of vertices that an observer at (-infinity, 0) sees. So F(π - α) + F(α) is the total number of vertices seen from both sides of the polygon. This is n+2 because the highest and lowest vertex get seen from both sides.

From (1) and (2) we conclude that we only need to observe F(α) for α ∈ [0, ϑ/2]

CASE 1: n = 4N ( ϑ = π/(2N))

Then V(N, 0) = (cos π/2, sin π/2) = (0, 1)

and V(-N, 0) = (cos -π/2, sin -π/2) = (0, -1)

It follows that V(-N, 0) through V(N, 0) are visible.

So F(0) = 2N+1 = n/2 + 1

Notice that

V(N, ϑ/2) = (cos (π/2 + ϑ/2), sin (π/2 + ϑ/2)) = (-sin ϑ/2, cos ϑ/2)

V(N-1, ϑ/2) = (cos (π/2 - ϑ/2), sin (π/2 - ϑ/2)) = (sin ϑ/2, cos ϑ/2)

which both have the same y-coordinate.

Too,

V(-N, ϑ/2) = (cos (-π/2 + ϑ/2), sin (-π/2 + ϑ/2)) = (sin ϑ/2, -cos ϑ/2)

V(-N-1, ϑ/2) = (cos (-π/2 - ϑ/2), sin (-π/2 - ϑ/2)) = (-sin ϑ/2, -cos ϑ/2)

which both have the same y-coordinate.

As α gets larger, V(N, α) moves into the second quadrant and its y-coordinate gets smaller. V(N, α) will be visible as long as its y-coordinate is greater than that of V(N-1, α). As we have shown, the y-coordinates are equal when α = ϑ/2. So V(N, α) is the highest visible vertex for all α in [0, ϑ/2). When α = ϑ/2, V(N-1, α) is the highest visible vertex.

We conclude that V(-N, α) and V(N, α) will stay visible for all 0 ≤ α < ϑ/2

So F(α) = 2N+1 for all α ∈ [0, ϑ/2)

Hence

F(α) = 2N+1 = n/2 + 1 for all α in [0, ϑ/2)

F(ϑ/2) = 2N = n/2

It follows that P(n/2) = 1

CASE 2. n = 4N + 2

Then V(k, 0) = (cos kϑ, sin kϑ) where ϑ = π/(2N+1)

If we solve kϑ = π/2 - ϑ/2 for k, we get

k = (2N+1)/2 - 1/2 = N

We see Nϑ is in the first quadrant, (N+1)ϑ is in the second quadrant, and they both have the same y-coordinate.

It follows that all points from V(-N) to V(N) will be visible.

So F(0) = 2N+1 = n/2

As α gets larger, V(N, α) moves towards the positive y-axis and its y-coordinate gets larger while the y-coordinate of V(N+1, α) gets smaller. Hence V(N, α) is visible and V(N+1, α) is not visible. On the other hand, as α get larger, V(-N, α) moves counter-clockwise away from the negative y-axis and the magnitude of its y-coordinate gets smaller while the y-coordinates of V(-N-1, α) moves closer to the negative y-axis and the magnitude of its y-coordinate gets larger. Hence V(-N-1) now becomes visible.

It follows that, for α ∈ (0, ϑ/2], the points V(-N-1, α) through V(N, α) are visible. Hence F(α) = 2N+2 = n/2 + 1

We conclude that P(n/2 + 1) = 1.

I concede that this is too much work. I can formalize Scythian's solution using my notation but that's about it.

On the unit circle, halfway between V[k, α) and V(k+1, α) is the point

(cos ((k + 1/2)ϑ + α), sin ((k + 1/2)ϑ + α)).

It's pretty easy to see that a face is visible if and only if -π/2 < kϑ + ϑ/2 + α < π/2; which simplifies to

-n/4 < k + (1/2 + α/ϑ) < n/4 where 0 ≤ α ≤ ϑ/2

and then counting the k's that make this true.

CASE 1 n = 4N

-N < k + (1/2 + α/ϑ) < N where 0 ≤ α ≤ ϑ/2

In the case α = ϑ/2, we get -N ≤ k ≤ N-2. So F(ϑ/2) = 2N-1 = n/2 - 1

In the case 0 ≤ α < ϑ/2, we get -N ≤ k ≤ N-1. So F(α) = 2N = n/2

Hence P(n/2) = 1

CASE 2. n = 4N + 2

-N - 1/2 < k + (1/2 + α/ϑ) < N + 1/2 where 0 ≤ α ≤ ϑ/2

In the case α = 0, we get -N ≤ k ≤ N-1. So F(0) = 2N = n/2 - 1

In the case 0 < α ≤ ϑ/2, we get -N - 1 ≤ k ≤ N-1. So F(α) = 2N+1 = n/2

So P(n/2) = 1

CASE 3:

So, to seek out complete sum of the angles in a ordinary polygon, the equation is: a hundred and eighty(n-2) = A the place n is the # of aspects, and A is the sum of all inside angles if the internal angle measure is a hundred and forty four, A could be 144n, so, one hundred eighty(n-2) = 144n 180n-360 = 144n 36n = 360 so, n = 10 so there are 10 sides in that polygon.

A regular polygon is a polygon whose sides are all the same length, and whose angles are all the same. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180(degrees) * (n - 2) degrees.