Okay, let's tackle this problem step by step. So, we have a regular nonagon (which is a nine-sided polygon) with each side length of 2 units. Then, the set S is defined as the union of all points inside this nonagon and all points less than 1 unit away from the perimeter. We need to find the perimeter of S.

First, I need to visualize what this union looks like. The regular nonagon is a convex polygon, and the area within 1 unit of its perimeter would form a sort of "buffer" zone around the nonagon. So, the set S would consist of the original nonagon plus this buffer area. The perimeter of S would then be the outer boundary of this combined shape.

Since the nonagon is regular, all sides are equal, and all internal angles are equal. The buffer zone around the perimeter would essentially be a rounded shape extending out 1 unit from each side. However, since the original nonagon has sides of 2 units, the buffer might not overlap at the corners. Let me confirm that.

Wait, the buffer is 1 unit away from the perimeter. The original nonagon's side length is 2 units. The distance from the center to a side (the apothem) and the distance from the center to a vertex (the radius) can be calculated. Maybe that will help in figuring out if the buffer zones around each side overlap at the corners.

Let me recall the formula for the apothem (a) of a regular polygon: a = (s)/(2 tan(π/n)), where s is the side length, and n is the number of sides. For a nonagon, n=9, s=2. So, a = 2/(2 tan(π/9)) = 1/tan(20°). Let me calculate that. Tan(20°) is approximately 0.3640, so a ≈ 1/0.3640 ≈ 2.747 units.

The radius (R) of the nonagon is the distance from the center to a vertex. The formula is R = s/(2 sin(π/n)). Plugging in the values: R = 2/(2 sin(π/9)) = 1/sin(20°). Sin(20°) ≈ 0.3420, so R ≈ 1/0.3420 ≈ 2.924 units.

So, the apothem is about 2.747 units, and the radius is about 2.924 units. The buffer zone is 1 unit away from the perimeter. Let's see if this buffer zone extends beyond the original nonagon's sides and corners.

If we consider a side of the nonagon, the buffer around it would be a rectangular area 1 unit wide along the side. However, since the buffer is 1 unit away from the perimeter, which includes both the sides and the vertices, the buffer will also create rounded corners. Wait, no. The buffer would be a Minkowski sum of the nonagon with a disk of radius 1. So, the perimeter of S would be the original perimeter plus the length added by the buffer.

But wait, when you take the Minkowski sum of a polygon with a disk, the perimeter of the resulting shape is equal to the original perimeter plus the circumference of the disk (which is 2πr) multiplied by the number of sides? Wait, no. Actually, the Minkowski sum of a polygon with a disk results in the original polygon's perimeter expanded outward by the radius of the disk, with the corners rounded off. Each original side is extended outward by 1 unit on both ends, but since the original polygon is convex, the corners will be replaced by circular arcs.

Therefore, the perimeter of the Minkowski sum (which is set S in this case) would be equal to the original perimeter plus 2π times the radius (since each corner is replaced by a circular arc of radius 1, and each corner contributes an angle of 2π/9 radians, but there are 9 corners, so total circular perimeter added is 9*(2π/9)*1 = 2π*1). Wait, that seems incorrect.

Wait, actually, when you take the Minkowski sum of a polygon with a disk, the perimeter of the resulting shape is the original perimeter plus 2πr, where r is the radius of the disk. Because each vertex is replaced by a circular arc, and the total length contributed by the arcs is equal to the circumference of the disk. Let me check that.

Each vertex of the original polygon is a point where two edges meet. When you take the Minkowski sum with a disk, each vertex becomes a circular arc of radius r, with the angle equal to the exterior angle of the polygon. For a regular polygon with n sides, each exterior angle is 2π/n. So, each arc has length r*(2π/n). Since there are n vertices, the total length added by all arcs is n*(r*(2π/n)) = 2πr. Therefore, the total perimeter of the Minkowski sum is the original perimeter plus 2πr.

Therefore, in this case, the original nonagon has a perimeter of 9*2 = 18 units. The Minkowski sum with a disk of radius 1 would add 2π*1 = 2π units to the perimeter. Therefore, the perimeter of S would be 18 + 2π units. So, 18 + 2π is approximately 18 + 6.283 = 24.283 units.

Wait, but let me verify this because I might have made a mistake. Let's think again. The Minkowski sum with a disk of radius r replaces each vertex with a circular arc of radius r and angle equal to the exterior angle of the polygon. For a regular polygon, the exterior angle is 2π/n. So, the length of each arc is r*(2π/n). Since there are n vertices, the total added length is n*(r*(2π/n)) = 2πr. So, yes, that part is correct.

Therefore, the perimeter of S is the original perimeter plus 2πr. Here, the original perimeter is 18, and r is 1, so 18 + 2π. Therefore, the answer is 18 + 2π units.

But wait, the problem states "the union of all points inside the nonagon and the set of all points less than 1 unit away from a point on the perimeter." So, that is equivalent to the Minkowski sum of the nonagon with a disk of radius 1. Therefore, the perimeter of S is indeed 18 + 2π. So, the answer should be 18 + 2π, which is 2(9 + π). But 18 + 2π is already simplified.

But let me check if there's any overlap or if the buffer zones around the sides might cause some parts to be double-counted. Wait, no. The Minkowski sum already accounts for all points within 1 unit of the perimeter, including the interior. So, the perimeter of the union is just the perimeter of the Minkowski sum, which is the original perimeter plus 2πr.

Alternatively, another way to think about it is: the original nonagon has a perimeter of 18. When you expand it by 1 unit outward in all directions, the straight sides become longer by 2 units each (1 unit on each end), but the corners become rounded. However, since the original nonagon has 9 sides, each side's expansion adds 2 units to the perimeter. But wait, that would be 9*2 = 18, plus the rounded parts. But that's conflicting with the previous result.

Wait, maybe I need to model both possibilities. Let me see.

Option 1: The perimeter of S is the original perimeter plus 2πr. So, 18 + 2π.

Option 2: Each side is extended outward by 1 unit on both ends, so each side's contribution to the perimeter is increased by 2 units. But since the sides are straight, the expansion would actually create a new side parallel to the original, but 1 unit away. However, the corners would become rounded. So, the total perimeter would be the original perimeter plus the sum of the lengths of the new sides plus the lengths of the rounded corners.

But the original nonagon has 9 sides, each of length 2. Expanding each side outward by 1 unit on both ends would add 2*9 = 18 units. However, the corners would now be rounded with radius 1. The original angle at each corner is the internal angle of the nonagon. For a regular nonagon, each internal angle is (n-2)*π/n = (7π)/9 radians. So, the external angle is 2π/9 radians. When we expand the nonagon by 1 unit, the corners become circular arcs with radius 1 and angle equal to the external angle, which is 2π/9. Therefore, each corner contributes an arc length of 1*(2π/9) = 2π/9. Since there are 9 corners, the total length from all corners is 9*(2π/9) = 2π. Therefore, the total perimeter is the original perimeter (18) plus the added straight parts (18) minus the original corner lengths (which are replaced by arcs). Wait, no. Wait, the original perimeter is 18. When expanding, the original sides are extended, and the original corners are replaced by arcs. So, the original sides, each of length 2, when expanded, each become a line segment of length 2 + 2*(1) = 4? Wait, no. Wait, when you expand a polygon outward by a distance r, each side's length is not just increased by 2r. Wait, maybe I need to think in terms of the offset polygon.

Alternatively, perhaps the perimeter of the offset polygon (when offset outward by distance r) is equal to the original perimeter plus 2πr. This is a known result in computational geometry. The perimeter of the Minkowski sum of a polygon with a disk of radius r is equal to the original perimeter plus 2πr. Therefore, in this case, the perimeter of S is 18 + 2π.

Therefore, the answer is 18 + 2π. But let me confirm with another approach.

Imagine walking around the perimeter of the expanded shape. The original polygon has 9 sides, each of length 2. When you expand the polygon outward by 1 unit, each side is moved outward, and each corner is replaced by a circular arc. The length of each side's extension is the same as the original side's length, but since the expansion is outward, the side is effectively translated outward. However, since the expansion is 1 unit, the direction of each side is preserved, but the length remains the same. Wait, but actually, when you offset a polygon outward, the length of each side in the offset polygon is equal to the original side length. However, the corners are replaced by circular arcs. Therefore, the total perimeter would be the original perimeter plus the circumference of the circle with radius equal to the expansion distance. Wait, but that would be 18 + 2π*1 = 18 + 2π. So, that matches the previous result.

Alternatively, think of the offset polygon as having the same number of sides, each side length equal to the original side length, but with each corner replaced by a circular arc. Therefore, the perimeter would be the sum of the side lengths (same as original) plus the sum of the arc lengths. The sum of the arc lengths is equal to the total angle around the expanded polygon times the radius. The total angle around a polygon is 2π radians. Therefore, the sum of the arc lengths is 2π*1 = 2π. Therefore, perimeter is 18 + 2π.

Therefore, the perimeter of S is 18 + 2π units. So, the final answer is 18 + 2π, which is approximately 24.283 units.

But let me check if there's a different interpretation of the problem. The problem says "the union of the set of all points inside a regular nonagon with side length 2 units and the set of all points less than 1 unit away from a point on the perimeter."

Wait, so the union is of two sets: the interior of the nonagon and the points within 1 unit of any point on the perimeter. So, this is equivalent to the closure of the nonagon expanded by 1 unit in all directions. Which is exactly the Minkowski sum of the nonagon with a disk of radius 1. Therefore, the perimeter is indeed 18 + 2π.

Therefore, the answer is 18 + 2π. To write it in LaTeX, that's 18 + 2\pi, which is \boxed{18 + 2\pi}.

But wait, let me check once more. Suppose the original nonagon has a perimeter of 18. When you expand it by 1 unit, the perimeter becomes 18 + 2π. But let's think of a simpler case. For example, take a square with side length s. The perimeter is 4s. If you expand it by radius r, the new perimeter is 4s + 2πr. For example, a square of side 2, expanded by 1 unit, would have perimeter 8 + 2π. That seems correct because each corner is replaced by a quarter-circle (angle π/2), so 4*(π/2)*1 = 2π. So, the perimeter is original 8 + 2π. So, yes, the formula seems to hold.

Therefore, in our case, the nonagon with perimeter 18, expanded by 1 unit, gives perimeter 18 + 2π. Therefore, the answer is 18 + 2π. So, I think that's correct.

**Final Answer**
The perimeter of $S$ is \boxed{18 + 2\pi} units.
Therefore, the perimeter of $S$ is $\boxed{18 + 2\pi}$ units.
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