Explanation:
Okay, so I need to find the perimeter of this set S. Let me try to visualize what S is. The problem says S is the union of two things: first, all the points inside a regular nonagon with side length 2 units, and second, all the points less than 1 unit away from a point on the perimeter of the nonagon.

So, if I imagine the nonagon, which is a nine-sided polygon, with each side 2 units long. Then, the first part of S is just the interior of this nonagon. The second part is like a buffer zone around the perimeter, extending 1 unit outwards from every point on the perimeter. So, combining these two, S would look like the original nonagon with a sort of "thickened" perimeter around it, right?

Wait, but when they say "the union of the set of all points inside the nonagon and the set of all points less than 1 unit away from the perimeter," does that mean that S is the Minkowski sum of the nonagon with a closed disk of radius 1? Or is it something else?

Hmm. Let me think. If you take all points less than 1 unit away from the perimeter of the nonagon, that's like taking the original nonagon and adding a 1-unit wide area around it. However, the original nonagon is already included in S because S is the union of the interior of the nonagon and that buffer zone. So, in effect, S is the original nonagon expanded outward by 1 unit in all directions, but keeping the interior.

But wait, actually, the union of the interior and the buffer zone would just be the closure of the buffer zone? Or maybe not? Let me clarify.

Suppose you have a set A, which is the interior of the nonagon. Then the set B is all points less than 1 unit away from the perimeter of the nonagon. So B includes points inside the nonagon? Wait, not necessarily. Wait, points less than 1 unit away from the perimeter would be points that are within 1 unit of the boundary of the nonagon. But the interior of the nonagon is points that are more than 0 units away from the perimeter? Wait, no. Actually, the interior points are those that are within the nonagon. The perimeter is the boundary. So points less than 1 unit away from the perimeter would include the interior points near the boundary, but also points outside the nonagon within 1 unit of the perimeter. Wait, so S is the union of the interior of the nonagon and the set of points within 1 unit of the perimeter. But if you take the union of the interior and points within 1 unit of the perimeter, that would actually be the same as the set of points within 1 unit of the nonagon's perimeter or inside it. Which is like taking the original nonagon and expanding it outward by 1 unit. But since the interior is already included, the union would just be the closed disk around the nonagon with radius 1? No, not exactly. Because expanding a polygon by 1 unit outward would result in a sort of rounded polygon, with the original sides extended out and the corners rounded with quarter-circles? Wait, actually, when you take all points within 1 unit of a polygon, it's called an offset polygon or a parallel curve. For convex polygons, offsetting outward would create a new polygon with the same number of sides, but with each side moved outward by 1 unit, and with rounded corners where the sides meet. But the exact shape depends on the original polygon. However, in this case, since the original nonagon is convex, yes, the set S would be the union of the original nonagon and the buffer around it, which would be like the original nonagon expanded by 1 unit in all directions. However, since the buffer zone includes points within 1 unit of the perimeter, which would overlap with the original nonagon. Therefore, S is the original nonagon expanded outward by 1 unit. But the perimeter of S would then consist of the expanded sides and the rounded corners?

Wait, maybe I need to think in terms of the boundary of S. The original perimeter of the nonagon is being expanded outward by 1 unit, but also, where the original sides meet at the corners, the buffer zone would create quarter-circles? Wait, no. If you take all points within 1 unit of the perimeter, then near the corners, you would have a circular arc with radius 1 unit, but how much of an angle does it cover?

Let me recall. If you have a polygon with an interior angle θ at each vertex, then when you offset outward by distance r, the original corner is replaced by a circular arc with angle π - θ. Wait, maybe? Let me think. For a polygon, when you offset outward, each original vertex is replaced by a circular arc. The angle of the arc would depend on the internal angle of the polygon.

For a regular nonagon, each internal angle can be calculated. Let me recall the formula for internal angles of a regular n-gon: it's ((n-2)*180)/n degrees. So for a nonagon, n=9, so internal angle is ((9-2)*180)/9 = (7*180)/9 = 140 degrees. So each internal angle is 140 degrees, which is 140π/180 = 7π/9 radians.

When you offset a polygon outward by distance r, the corner is replaced by a circular arc. The angle of the arc is equal to π - internal angle. Wait, why?

Imagine standing at the corner of the polygon. The original corner is a vertex where two edges meet. When you offset each edge outward, the offset edges will move away from the original polygon. The corner is replaced by an arc that turns around the original corner. The angle of the arc is equal to the angle that the two offset edges would form if they were extended. Wait, perhaps the angle of the arc is π minus the original internal angle?

Wait, for a square, each internal angle is 90 degrees. When you offset a square outward by a distance r, each corner becomes a quarter-circle. The angle of the arc is 90 degrees, which is π/2 radians, which is π - π/2 (since the internal angle is π/2). Wait, π - π/2 is π/2, which matches. Wait, but for a square, the arc is 90 degrees. So if the internal angle is θ, then the arc angle is π - θ? Wait, π - θ would be π - π/2 = π/2, which is correct. For the nonagon, internal angle is 7π/9 radians. Therefore, the arc angle would be π - 7π/9 = 2π/9 radians. Which is 40 degrees. So each corner of the nonagon is replaced by an arc of 40 degrees with radius 1 unit. Therefore, the perimeter of S would consist of the original sides, offset outward by 1 unit, and the arcs replacing the corners.

Wait, but hold on. Since S is the union of the interior of the nonagon and the points within 1 unit of the perimeter. Therefore, the boundary of S would be the original perimeter offset outward by 1 unit, but with the corners replaced by the arcs. Therefore, the perimeter of S is equal to the length of the offset polygon's perimeter.

Therefore, to calculate the perimeter of S, we need to calculate the perimeter of this expanded shape. The perimeter would consist of the lengths of the offset sides plus the lengths of the arcs.

So first, let me figure out the length of the offset sides. For each original side of the nonagon, when we offset it outward by 1 unit, the length of the side remains the same? Wait, no. Wait, when you offset a straight side outward, the length of the side increases? Wait, no. Wait, if you take a straight line segment and offset it outward, keeping it parallel, the length of the segment remains the same, but the offset creates a sort of "buffer" around it. However, when you offset all sides of the polygon, the original sides are extended outward, but the corners are rounded.

But actually, when considering the perimeter of S, which is the union of the original nonagon and the buffer, the perimeter of S is the boundary of the buffer zone. Wait, but since the buffer zone includes the interior of the nonagon, the perimeter of S is the same as the perimeter of the buffer zone. Wait, no. If you take the union, the perimeter of S would consist of the parts of the buffer zone that are not overlapping with the original nonagon. However, since the buffer zone includes all points within 1 unit of the perimeter, and the original nonagon is included, so the union would actually be the closure of the buffer zone. But the boundary of S would be the boundary of the buffer zone. Wait, but how?

Alternatively, perhaps S is the union of two sets: the original nonagon (interior) and the buffer around the perimeter. The union would have a perimeter that is the same as the perimeter of the buffer zone. Because the original nonagon's interior is entirely inside the buffer zone? Wait, no. The buffer zone is all points within 1 unit of the perimeter. The original nonagon's interior is points that are inside the nonagon. However, points inside the nonagon are at distances from the perimeter. The maximum distance from the perimeter inside the nonagon is the inradius. So if the inradius is larger than 1 unit, then the buffer zone would extend beyond the original nonagon, but the union would be the original nonagon plus the buffer zone. But if the inradius is less than 1, then the buffer zone would extend into the original nonagon.

Wait, perhaps I need to calculate the inradius of the nonagon. Let me recall. For a regular polygon with n sides of length s, the inradius r is given by r = (s/2) * cot(π/n). So here, n=9, s=2. Therefore, inradius is (2/2)*cot(π/9) = cot(π/9). Let me compute cot(π/9). π/9 is 20 degrees. Cot(20 degrees) is approximately 2.747, but I don't need the exact value. However, the problem is whether the inradius is less than 1 or not. If the inradius is greater than 1, then the buffer zone would not reach the center of the nonagon. If the inradius is less than 1, the buffer zone would extend into the original nonagon. Wait, but inradius is the distance from the center to the sides. So points within 1 unit of the perimeter would include points inside the nonagon near the perimeter, but depending on how far the inradius is.

Wait, but the inradius is the distance from the center to the sides. So if the inradius is greater than 1, then the buffer zone (which is points within 1 unit of the perimeter) would not reach the center. However, the original nonagon is included in S. Wait, no, S is the union of the interior of the nonagon and points within 1 unit of the perimeter. So even if the buffer zone doesn't reach the center, the interior is included. Therefore, the union is the entire nonagon plus the buffer zone. But in that case, perhaps the perimeter of S is the perimeter of the buffer zone, but since the buffer zone includes parts of the original nonagon, does that affect the perimeter?

Wait, maybe not. Because the buffer zone is all points within 1 unit of the perimeter. So when you take the union with the interior, you are just saying that S includes the interior and the buffer, so S is the union, which would be the closure of the buffer zone. However, the boundary of S is the boundary of the buffer zone. Wait, maybe the buffer zone's boundary is the set of points that are exactly 1 unit away from the perimeter. So the perimeter of S would be the set of points that are 1 unit away from the original perimeter. Which, for a polygon, is called the offset curve.

But as we discussed earlier, for a convex polygon, the offset curve is formed by moving each side outward by 1 unit, and rounding the corners with circular arcs. Therefore, the perimeter of S would be equal to the sum of the lengths of the offset sides plus the sum of the lengths of the arcs.

So to compute this perimeter, we need to know two things:

1. The length of each offset side. Since the original sides are of length 2 units, and when offsetting, each side is moved outward by 1 unit. However, does the length of the side remain the same? Wait, actually, when you offset a straight segment outward, the length of the segment remains the same, but the offset is in the direction perpendicular to the segment. Therefore, if you imagine moving each side outward by 1 unit, then the distance between the original side and the offset side is 1 unit. But since the polygon is regular, all sides are moved outward by the same distance, so the new polygon (if we were to ignore the rounded corners) would have the same side lengths as the original nonagon. However, in reality, the offset polygon is not a polygon anymore, because the corners are rounded. Wait, but if we think about the perimeter contributing from the offset sides, each original side of length 2 units would contribute a length of 2 units to the perimeter? No, wait, because when you offset the sides, they are moved outward, but the direction of the offset is perpendicular to each side. However, when you offset all sides, the corners are replaced by arcs. Therefore, the original sides, when offset, are now sides of a larger polygon, but the total length of those sides would be the same as the original sides? But that can't be, because if you move each side outward, the distance between adjacent sides is increased. Wait, perhaps the length of each offset side is actually longer?

Wait, no. Wait, if you take a straight side of length L, and offset it outward by distance r, keeping it parallel, then the length of the side doesn't change. However, when you do this for all sides of a polygon, the offset polygon's sides are longer because they are "pushed out" and the corners are rounded. Wait, no. Wait, let me think of a square. If you have a square with side length 2, and you offset each side outward by 1 unit, forming a larger square with rounded corners. The sides of the larger square, if you were to make them straight, would have length increased by 2*r, but since we are rounding the corners, the sides are actually the same as the original sides? No. Wait, in reality, the perimeter of the offset square would be the original perimeter plus 2πr. Wait, for a square, when you offset it outward by radius r, the perimeter becomes original perimeter + 2πr. Wait, original perimeter is 4*2=8. If you offset outward, each corner becomes a quarter-circle, so four quarter-circles make a full circle, circumference 2πr. So total perimeter is 8 + 2πr. However, if you think of the straight sides, they have length same as original sides? Wait, but in reality, when you offset a square, the straight sides are not of the same length. Wait, no. If you have a square, and you offset each side outward by distance r, keeping them parallel, then the distance between the original side and the offset side is r. However, if you consider the length of the offset side, since it's parallel, the length remains the same. However, when you offset all sides, the problem is that adjacent sides are offset outward, and the corners are rounded. So the length of the straight parts of the offset shape is same as original? Wait, no. Wait, imagine a square with side length 2. Offset each side outward by 1 unit. The sides are each extended outward, but the corners are replaced by quarter circles. However, when you offset each side, the new side is not longer than original. Therefore, the straight parts of the offset shape have the same length as the original sides, but the total perimeter is original perimeter plus 2πr.

But wait, that seems contradictory, because the offset square's perimeter would be longer than the original. Let me check with numbers. If original square has side length 2, perimeter 8. After offsetting by 1 unit, the perimeter becomes 8 + 2π*1 ≈ 8 + 6.28 ≈ 14.28. Alternatively, if you think of the offset square as having sides that are offset outward, but with rounded corners. If you didn't have the rounded corners, the sides would be longer. Wait, but if you have a square and you just move each side outward, the distance between the original side and the new side is 1 unit. However, if you imagine moving each side outward, the new square would have side length longer. Wait, the original square has side length 2. The distance from the original side to the offset side is 1 unit, but the side length of the new square would be 2 + 2*r*(cot(π/n))? Wait, maybe I need to think in terms of the offsetting process.

Alternatively, perhaps the perimeter of the offset shape is equal to the original perimeter plus 2πr. For a polygon with m sides, the perimeter of the offset shape is original perimeter + 2πr. Wait, that seems to be the case for the square. For a polygon with n sides, if you offset it outward by r, then the perimeter becomes original perimeter + 2πr? That can't be right because for a polygon with many sides, the rounded corners would add up to a full circle. So regardless of the number of sides, as long as you offset outward by r, the perimeter would be original perimeter plus 2πr?

Wait, yes, actually. Because when you offset each side outward by r, the corners are replaced by circular arcs. Each corner contributes an arc of angle equal to the exterior angle of the polygon. For a regular polygon, the sum of the exterior angles is 2π radians. Therefore, when you offset, each exterior angle becomes an arc of angle equal to the exterior angle. Therefore, the total length added by the arcs is r times the total angle, which is r*2π. Therefore, the perimeter of the offset shape is original perimeter + 2πr.

Therefore, for any polygon, when you offset it outward by radius r, the perimeter of the offset shape is original perimeter + 2πr. Therefore, regardless of the number of sides, this formula holds.

Wait, so for the square, original perimeter 8, offset by 1, perimeter becomes 8 + 2π*1 ≈ 14.28. Which seems correct. For a polygon with more sides, like a regular nonagon, the same applies. Therefore, the perimeter of S, which is the set of points within 1 unit of the original nonagon's perimeter, would be equal to the original perimeter of the nonagon plus 2π*1. But wait, but the original problem says S is the union of the interior of the nonagon and points within 1 unit of the perimeter. But if the offset shape is points within 1 unit of the original polygon's perimeter, then the perimeter of S is the perimeter of this offset shape, which is original perimeter + 2πr. Therefore, the perimeter of S would be original perimeter of nonagon + 2π(1).

But wait, let me check again. Suppose that S is the union of the interior and the buffer. However, if the buffer zone is points within 1 unit of the perimeter, then the union would be the closure of the buffer zone, which is the same as the set of points within 1 unit of the perimeter or inside the polygon. However, the boundary of this set is the set of points at exactly 1 unit away from the original perimeter. Which is the offset curve we discussed. Therefore, the perimeter of S is equal to the length of this offset curve. And according to the earlier logic, that is equal to the original perimeter plus 2πr. So if that is the case, then the perimeter of S is equal to the original perimeter of the nonagon plus 2π*1.

But wait, the original perimeter of the nonagon is 9 sides each of length 2 units, so 9*2=18 units. Therefore, perimeter of S would be 18 + 2π(1) ≈ 18 + 6.28 ≈ 24.28. However, I need to confirm this logic. Because earlier, I was confused about the offset sides. But according to the general formula, for any polygon, the perimeter of the offset curve is original perimeter + 2πr.

But why is that? Let me think again. When offsetting a polygon, each side contributes a length equal to its original length, but each corner is replaced by an arc. The total angle of all arcs is equal to 2π, because the sum of the exterior angles of any polygon is 2π. Therefore, the total length added by the arcs is r*(sum of angles) = r*2π. Therefore, the total perimeter is original perimeter + 2πr. Therefore, regardless of the original polygon's shape, as long as it's convex, this formula holds.

Therefore, in this problem, since the nonagon is convex, and S is the set of points within 1 unit of its perimeter (which is equivalent to the offset polygon with radius 1), the perimeter of S is the original perimeter + 2π*1. Therefore, the answer is 9*2 + 2π*1 = 18 + 2π.

Wait, but hold on. The problem says S is the union of the interior of the nonagon and the set of points less than 1 unit away from the perimeter. So is the perimeter of S the same as the perimeter of the offset set?

But if you think of S as the union of the interior and the buffer, then S is a closed set, and its boundary is the set of points that are exactly 1 unit away from the original perimeter. Therefore, the perimeter of S is the length of this boundary, which is the offset curve. Therefore, the perimeter of S is indeed original perimeter + 2π*1.

However, let me verify with a simpler case. Let me take a square with side length 2 units. Its perimeter is 8 units. S would be the union of the interior and points within 1 unit of the perimeter. The perimeter of S would be the boundary of this region. If I imagine this, the boundary is the original square offset outward by 1 unit, with rounded corners. The perimeter of this offset shape is original perimeter + 2π*1 = 8 + 2π, which is correct. Therefore, the same applies to the nonagon. Hence, the perimeter of S is 18 + 2π. But wait, the answer is supposed to be in units. Wait, but the problem says "the perimeter of S". Is that correct?

But let me think again. Wait, if the original nonagon has a perimeter of 18 units, and we add 2π units from the rounded corners. Therefore, the answer is 18 + 2π. However, I need to check if this is correct.

Wait, but maybe there's a misunderstanding here. Let me think again. Suppose that S is the union of the interior of the nonagon and the set of points within 1 unit of the perimeter. However, when you take the union, you include all points inside the nonagon and all points within 1 unit outside of the nonagon. Therefore, the perimeter of S would be the perimeter of the nonagon expanded outward by 1 unit, which would be the original perimeter plus the circumference of a circle with radius 1? Wait, no, it's not a full circle. As we discussed, for a polygon, the perimeter of the offset is original perimeter + 2πr.

Alternatively, another way to think about it: the perimeter of S is the set of points that are exactly 1 unit away from the original nonagon's perimeter. Which, as we said, is the offset curve, which has perimeter original perimeter + 2πr. Therefore, the answer is 18 + 2π.

But wait, the problem says "the set of all points less than 1 unit away from a point on the perimeter of the nonagon". So does that include points inside the nonagon? Because the points inside the nonagon are less than 1 unit away from some point on the perimeter (specifically, points inside are less than the inradius away from the perimeter). However, the original nonagon is included in S. So perhaps S is the Minkowski sum of the nonagon with a closed disk of radius 1.

But the Minkowski sum of a polygon with a disk would be the original polygon expanded by radius 1, with rounded corners, and the perimeter of that shape is original perimeter + 2π*1. So yes, same result. Therefore, the perimeter is 18 + 2π.

But wait, let me check if the formula is correct. If you have a polygon, and you take its Minkowski sum with a disk of radius r, then the perimeter of the resulting shape is equal to the original perimeter plus 2πr.

Let me verify with a simple case. Take a line segment of length L. The Minkowski sum with a disk of radius r is a rectangle with semicircular ends. The perimeter would be 2L (the two straight sides) plus 2πr (the two semicircles). But original perimeter of the line segment is 0 (as a 1-dimensional object), but maybe this is not helpful.

Another example: take a circle of radius R. Its perimeter is 2πR. The Minkowski sum with a disk of radius r is a circle of radius R + r, perimeter 2π(R + r) = 2πR + 2πr. Which matches the formula. Therefore, for the circle, the perimeter increases by 2πr.

For a square with side length L, original perimeter 4L. The Minkowski sum with a disk of radius r is a square with rounded corners. The perimeter is 4L + 2πr. Because each side contributes L, and the corners contribute quarter-circles, totaling a full circle circumference 2πr. Hence, perimeter becomes 4L + 2πr.

Therefore, yes, the formula holds for polygons. Therefore, for the nonagon, original perimeter is 9*2 = 18, so the perimeter of S is 18 + 2π*1 = 18 + 2π. Therefore, the answer is 18 + 2π. But let me check again.

Wait, but the problem says "the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter of the nonagon". Therefore, the union of the interior and the points within 1 unit of the perimeter.

But the Minkowski sum of the nonagon with a disk of radius 1 includes the interior of the nonagon and the points within 1 unit of the perimeter. So yes, that is exactly S. Therefore, the perimeter of S is the perimeter of the Minkowski sum, which is original perimeter + 2π*1.

Therefore, answer is 18 + 2π. However, the problem might expect a numerical value, but since it says "in units", and π is involved, likely wants an exact answer in terms of π. Therefore, 18 + 2π. But wait, let me check once more.

Wait, but hold on. Another thought: when you take the union of the interior and the buffer zone, maybe the perimeter of S is just the perimeter of the buffer zone, which is the same as the offset curve. But the buffer zone itself is the set of points within 1 unit of the perimeter. However, the buffer zone includes the interior of the nonagon. Therefore, the boundary of S is the set of points that are exactly 1 unit away from the perimeter of the nonagon. Which is the offset curve, as discussed. Therefore, the perimeter of S is indeed original perimeter + 2π*1.

But wait, another way to think: suppose that the buffer zone is like a "thickened" perimeter. When you have a polygon with perimeter P, and you create a buffer of width r around it, the area is P*r + π*r². But the perimeter of the buffer zone is P + 2πr. But does this hold?

Wait, actually, the area of the buffer zone (the set of points within r of the polygon) is equal to the area of the polygon plus the perimeter times r plus πr². But the perimeter of the buffer zone is equal to the original perimeter plus 2πr.

But let me check for a square. Area of buffer zone would be area of square + perimeter*r + πr². For a square with side length 2, area is 4. Perimeter is 8. So buffer area would be 4 + 8*1 + π*1² = 4 + 8 + π = 12 + π. But the buffer zone is the square expanded by 1 unit with rounded corners. The area can also be calculated as area of the expanded square (which would have side length 2 + 2*1 = 4) but subtracting the corner parts that are overlapped. Wait, no. Actually, if you expand a square by 1 unit with rounded corners, the area is equal to the area of the square plus the area of the buffer. Alternatively, since the buffer zone is all points within 1 unit of the original square, which can be thought of as the original square plus a rectangle of width 1 on each side, plus quarter circles at each corner.

The area would be original area + perimeter * r + πr². Which is 4 + 8*1 + π*1² = 12 + π, as above. Therefore, the area is correct.

But what about perimeter? The perimeter of the buffer zone is original perimeter + 2πr. For the square, that would be 8 + 2π*1 ≈ 8 + 6.28 ≈ 14.28. Which is correct, as the buffer zone's perimeter is the original perimeter (the "outer" edges) plus the circumference from the rounded corners.

Therefore, the formula seems to hold. Therefore, applying this to the regular nonagon, the perimeter of S is original perimeter + 2π*1. Original perimeter is 9*2 = 18. Therefore, the answer is 18 + 2π.

But the problem says "the union of the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter of the nonagon". Is there a possibility that the perimeter of S is different?

Wait, suppose that the original nonagon's perimeter is entirely inside S. But since S includes the points less than 1 unit away from the perimeter, which includes points outside the nonagon. But the original nonagon is also included. However, the perimeter of S is not just the perimeter of the nonagon. Because when you include the buffer, you are adding new edges. However, the original perimeter is not entirely part of the boundary of S. Because points on the original perimeter are within 1 unit of themselves, so they are included in S. But the boundary of S is the set of points that are exactly 1 unit away from the original perimeter. Therefore, the original perimeter is inside S, not on its boundary. Therefore, the perimeter of S is only the boundary of the buffer zone, which is the offset curve, with perimeter original perimeter + 2πr.

Therefore, I think the answer is 18 + 2π. But let me check if this is correct by thinking about another way.

Alternatively, think about the perimeter of S. S consists of two parts: the interior of the nonagon and the buffer. However, the perimeter of S would be the outer edge of the buffer, because the interior has no perimeter. Wait, no, the union would have a perimeter that is the boundary of the buffer zone, which is what we discussed. Therefore, I think the answer is 18 + 2π.

But let me check with another example. Suppose the nonagon is replaced by a circle. Let's say a circle with radius R. Then the set S would be the union of the interior of the circle and the set of points less than 1 unit away from the perimeter. But the interior of the circle already includes all points within R of the center. The set of points less than 1 unit away from the perimeter would be the annulus between R - 1 and R + 1. Wait, but if R is greater than 1, then the union would be the circle expanded outward by 1 unit. But the perimeter of S would be the circumference of the expanded circle, which is 2π(R + 1). However, original perimeter is 2πR. So perimeter of S is 2πR + 2π*1, which matches our formula: original perimeter + 2πr. Therefore, same result. Therefore, formula holds.

Therefore, I think confident that the perimeter of S is 18 + 2π units. However, maybe there is a trick here. Let me check if the problem says "the union of the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter of the nonagon". Wait, is the buffer zone overlapping with the nonagon? Because points inside the nonagon are less than 1 unit away from the perimeter? Not necessarily. Points inside the nonagon can be more than 1 unit away from the perimeter. For example, the center of the nonagon is at distance equal to the inradius from the perimeter. If the inradius is greater than 1, then the buffer zone (points within 1 unit of the perimeter) does not include the center. Wait, so is S the union of two regions: the interior of the nonagon and the buffer zone. Therefore, S is the set of points either inside the nonagon or within 1 unit of the perimeter.

But in that case, the boundary of S would be the boundary of the buffer zone plus the boundary of the nonagon? No, because the interior of the nonagon is connected to the buffer zone. Wait, but points inside the nonagon are already included in the buffer zone if they are within 1 unit of the perimeter. However, if the inradius is larger than 1, then the buffer zone doesn't reach the center. Therefore, the union would have a boundary that is the union of the original nonagon's boundary and the buffer zone's boundary? But no, because the buffer zone is points within 1 unit of the perimeter, so near the original perimeter, the buffer zone overlaps with the interior of the nonagon. Therefore, the union S would have a boundary that is the set of points that are either on the original perimeter or exactly 1 unit away from the original perimeter. Wait, no.

Wait, suppose you have a point on the original perimeter. It's included in S because it is in the buffer zone (distance 0 from perimeter). The points inside the nonagon are included because of the union. The buffer zone includes points outside of the nonagon within 1 unit of the perimeter. Therefore, the boundary of S is made up of two parts:

1. The parts of the original nonagon's perimeter that are not within 1 unit of any other part of the buffer zone. Wait, this is getting confusing. Maybe it's better to think of S as the union of the nonagon and the buffer zone. The buffer zone is like a 1-unit thick region around the nonagon. However, the union would be the nonagon with a 1-unit thick "border" around it. Therefore, the perimeter of S would consist of two parts:

- The outer edge of the buffer zone (which is the offset curve, with perimeter original perimeter + 2πr)

- And the inner edge of the buffer zone, which is the original perimeter of the nonagon.

But wait, but the original perimeter is entirely inside S. Because S includes the interior of the nonagon and the buffer zone. However, the original perimeter is on the boundary between the interior of the nonagon and the buffer zone. Therefore, the perimeter of S would consist of the outer part of the buffer zone (which is the offset curve) and the original perimeter? No, that can't be. Because when you take the union, the original perimeter is an internal boundary of S.

Wait, actually, if you have two regions, one inside the other, their union has a boundary equal to the boundary of the larger region. But in this case, the buffer zone extends both into the original nonagon and outside. Therefore, the original nonagon's perimeter is completely inside S, but it is not part of the boundary of S. The boundary of S is the outer edge of the buffer zone, which is the offset curve. Because the original perimeter is inside S, not on its boundary.

Let me think of a simple case again. Suppose you have a square, and you take the union of the square and the points within 1 unit of its perimeter. The boundary of this union is the outer edge of the buffer zone, which is the offset curve. The original square's perimeter is not on the boundary of S because the buffer zone overlaps with it. So yes, the boundary is only the offset curve.

Therefore, going back, the perimeter of S is original perimeter + 2πr. Therefore, 18 + 2π.

However, let me check the problem statement once more to be sure.

"The union of the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter of the nonagon."

Wait, the set of all points inside the nonagon is the interior, and the set of all points less than 1 unit away from the perimeter. So the union is the interior plus the buffer. If the buffer extends beyond the nonagon, then the union is the nonagon with a buffer, so the perimeter is the outer edge of the buffer. If the buffer also extends into the nonagon, then the union is the same as the buffer.

But in this case, the buffer includes the interior of the nonagon if the inradius is less than 1. Wait, if the inradius is less than 1, then points inside the nonagon are within 1 unit of the perimeter. Wait, the inradius is the maximum distance from the center to the sides. So the distance from a point inside the nonagon to the perimeter depends on the point. The center is at inradius distance from the perimeter. If the inradius is less than 1, then the entire nonagon is within 1 unit of the perimeter. Therefore, the union of the interior of the nonagon and the buffer zone is just the buffer zone. Therefore, S is the buffer zone, which has perimeter original perimeter + 2π*1.

But if the inradius is greater than 1, then the buffer zone would not include the entire interior of the nonagon. Therefore, S is the union of the interior and the buffer, which is the buffer zone plus the interior, so the perimeter would still be just the outer edge of the buffer zone.

Therefore, regardless of the inradius, the perimeter of S is original perimeter + 2π*1.

Therefore, the answer is 18 + 2π.

Wait, but the problem says "the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter". So the union includes both the interior and the buffer. If the inradius is less than 1, then the buffer already includes the interior, so S is just the buffer. If the inradius is more than 1, then S is the union of the interior and the buffer, which would be the buffer plus the interior. But in this case, the interior is not overlapping with the buffer? Wait, no. Wait, if the inradius is more than 1, then the buffer zone (points within 1 unit of the perimeter) does not include the entire interior. For example, the center is at inradius distance from the perimeter, which is more than 1, so the center is not in the buffer. Therefore, the union of the interior and buffer is the buffer plus the interior. However, the boundary of this union is still the outer edge of the buffer. Because the interior is inside the buffer? No. Wait, the interior is entirely inside the nonagon, which is inside the buffer zone? No. Because the buffer zone is points within 1 unit of the perimeter. So if the inradius is more than 1, then the interior is not entirely within the buffer zone. Therefore, the union would be the buffer zone and the interior. The boundary of the union is the boundary of the buffer zone (the outer edge) and the boundary of the interior (the original perimeter). But since the buffer zone overlaps with the interior near the perimeter, the original perimeter is already included in the buffer zone? No. Wait, the buffer zone is points within 1 unit of the perimeter. The original perimeter is part of the buffer zone. Therefore, the union of the interior and buffer zone would be the buffer zone union the interior. However, the interior is already partly overlapping with the buffer zone. Therefore, the boundary of the union is the outer edge of the buffer zone (the offset curve) and the original perimeter? No, because the original perimeter is on the edge between the interior and buffer zone. Wait, actually, the boundary of S would be the outer edge of the buffer zone (the offset curve) because the interior of the nonagon is connected to the buffer zone. The original perimeter is not a boundary of S.

This is getting too confusing. Maybe a better approach is to think of S as the Minkowski sum of the nonagon and a disk of radius 1. Then, as we discussed, the perimeter of the Minkowski sum is the original perimeter plus 2πr.

Therefore, regardless of the inradius, the perimeter of S is 18 + 2π. Therefore, I think this is the correct answer.

But to be thorough, let me compute the inradius of the nonagon and see if it's less than 1 or not.

As mentioned earlier, the inradius r = (s/2) * cot(π/n), where s is the side length, n is the number of sides.

Given that n = 9, s = 2. Therefore:

r = (2/2) * cot(π/9) = cot(π/9) ≈ cot(20°).

The cotangent of 20 degrees is 1/tan(20°).

tan(20°) ≈ 0.3640, so cot(20°) ≈ 2.747. Therefore, the inradius is approximately 2.747 units.

Which is greater than 1. Therefore, the buffer zone (points within 1 unit of the perimeter) does not include the entire interior of the nonagon. Therefore, S is the union of the interior of the nonagon and the buffer zone.

But then, what does the boundary of S look like?

The boundary of S would consist of two parts:

1. The outer boundary of the buffer zone (the offset curve)

2. The inner boundary of the buffer zone, which is the original perimeter of the nonagon.

Wait, no. Because the buffer zone is points within 1 unit of the perimeter, which includes the perimeter itself. Therefore, the union S is the interior of the nonagon plus the buffer zone. However, since the buffer zone includes the original perimeter, the union S is just the buffer zone union the interior of the nonagon. But since the buffer zone already includes the original perimeter, and the interior is the set of points inside the nonagon. Therefore, the union is all points that are either inside the nonagon or within 1 unit of the perimeter.

But the buffer zone is points within 1 unit of the perimeter. The interior is points inside the nonagon. Therefore, S is the set of points that are inside the nonagon or within 1 unit of the perimeter.

But since the interior of the nonagon is only within 1 unit of the perimeter if they are near the perimeter. However, the center is not within 1 unit of the perimeter (since inradius is ~2.747). Therefore, the union S is the buffer zone (which includes the original perimeter and the points within 1 unit outside the nonagon) plus the interior of the nonagon.

Therefore, the boundary of S is composed of:

- The outer part of the buffer zone (the offset curve)

- The inner part of the buffer zone, which is the original perimeter, but only where the buffer zone starts. Wait, no.

Actually, the boundary of S would consist of:

- The outer boundary of the buffer zone (the offset curve)

- The parts of the original perimeter that are not covered by the buffer zone. Wait, no. The original perimeter is entirely within the buffer zone (because points on the perimeter are 0 units away from the perimeter). Therefore, the original perimeter is inside S, but it is not part of the boundary.

Wait, confusion arises because the buffer zone includes the original perimeter. So the union S includes the interior of the nonagon and the buffer zone. However, the buffer zone is already including the original perimeter and the area within 1 unit of it. Therefore, the union is just the buffer zone union the interior. But since the buffer zone already includes the original perimeter and the area around it, including the interior near the perimeter. However, the interior of the nonagon is not entirely included in the buffer zone. For example, the center of the nonagon is not in the buffer zone. Therefore, S is the buffer zone plus the interior of the nonagon. Therefore, the boundary of S consists of:

- The outer part of the buffer zone (the offset curve)

- The inner part of the nonagon that is not covered by the buffer zone. Wait, but how?

Wait, the boundary of S would be the set of points where S changes from being inside to outside. Since S includes all points inside the nonagon and all points within 1 unit of the perimeter. Therefore, imagine the region S is the nonagon's interior plus a 1-unit wide region around it. So it's like the nonagon with a "border" of width 1 unit.

Therefore, the boundary of S would be:

- The outer edge of the border (the offset curve)

- The inner edge of the border, which is the original perimeter of the nonagon.

But is the original perimeter part of the boundary of S?

If you take a point on the original perimeter, it is in S, but the points inside the nonagon on one side of the perimeter and outside the nonagon on the other side. Wait, the original perimeter is the boundary between the interior of the nonagon and the exterior. However, since S includes the interior of the nonagon and the buffer zone, which includes the original perimeter and the exterior around the nonagon. So the original perimeter is entirely within S, but not part of its boundary. Because points just inside the nonagon are also in S, and points just outside are in the buffer zone. Therefore, the original perimeter is like a line between two regions, but since both sides are included in S (interior and buffer), the original perimeter is not part of the boundary.

Therefore, the boundary of S is only the outer edge of the buffer zone, which is the offset curve. Therefore, the perimeter of S is original perimeter + 2πr, which is 18 + 2π.

But to confirm, consider a simpler case where the nonagon is replaced with a circle. If the circle has radius R, then S is the union of the interior of the circle and the buffer zone of radius 1. The buffer zone would be the annulus between R - 1 and R + 1, but since the interior of the circle is included, S is the union of the circle and the annulus, which is just the annulus from R - 1 to R + 1. However, if R > 1, then the buffer zone includes the interior of the circle (from R - 1 to R). Therefore, the perimeter of S is the outer edge (circle of radius R + 1) and the inner edge (circle of radius R - 1). Wait, but this contradicts our previous conclusion.

Wait, here's the problem. If S is the union of the interior of the circle and the buffer zone (points within 1 unit of perimeter), then if the original circle has radius R > 1, the buffer zone extends from the original circle's perimeter inward by 1 unit to R - 1. Therefore, the union S is the region from R - 1 to R + 1. Therefore, the boundary of S consists of two circles: the inner circle of radius R - 1 and the outer circle of radius R + 1. Therefore, the perimeter of S is the sum of the circumferences of these two circles.

But according to our previous formula, it would have been original perimeter + 2πr. Original perimeter is 2πR, and adding 2π*1 would give 2π(R + 1), which is only the outer circle. But in reality, the perimeter of S would be the outer circle (perimeter 2π(R + 1)) and the inner circle (perimeter 2π(R - 1)), totaling 2π(R + 1 + R - 1) = 4πR. However, this contradicts the previous formula.

Wait, so there must be an error in my previous reasoning.

This suggests that when the buffer zone includes parts of the original shape (interior), the perimeter of S is not just the offset curve, but also includes the inner boundary. Therefore, my earlier conclusion was incorrect.

So, this is a critical mistake. Let me re-examine.

If S is the union of the interior of the circle and the buffer zone (points within 1 unit of the perimeter), then S is the set of points within 1 unit of the perimeter or inside the circle.

But if the circle has radius R > 1, then the set of points within 1 unit of the perimeter includes the annulus between R - 1 and R + 1. Therefore, the union of the interior (points within R) and the buffer zone (annulus between R - 1 and R + 1) is the union of the annulus between R - 1 and R + 1 and the circle of radius R. Therefore, the union is the annulus between R - 1 and R + 1. Therefore, the boundary of S is the two circles: R - 1 and R + 1. Therefore, the perimeter is 2π(R + 1) + 2π(R - 1) = 4πR.

However, if we applied our previous formula, original perimeter (2πR) + 2π*1 = 2π(R + 1), which only accounts for the outer boundary. But clearly, in this case, the perimeter is double that. So where is the error?

Ah, because in the case of the circle, S is not just the buffer zone, but the buffer zone plus the interior. However, the buffer zone already includes the interior of the circle up to a certain point. So the perimeter should account for both the inner and outer edges. Therefore, the formula original perimeter + 2πr is only applicable when the buffer zone does not include the interior of the original shape.

Wait, so when does the buffer zone include the interior? It includes the interior of the original shape if the original shape's inradius is less than r. For a circle, the inradius is equal to the radius. So if the original circle has radius R, and the buffer zone is points within 1 unit of the perimeter (which is the circle), then the buffer zone includes all points within R + 1 of the center (the annulus from R - 1 to R + 1). But the interior of the circle is points within R of the center. Therefore, the union is the annulus from R - 1 to R + 1. So the perimeter is two circles.

But in the previous case of the nonagon, since the inradius is about 2.747, which is greater than 1, the buffer zone (points within 1 unit of the perimeter) does not include the entire interior of the nonagon. Therefore, the union S is the buffer zone (which is the region within 1 unit of the perimeter) plus the interior of the nonagon. Therefore, S is the interior of the nonagon union the buffer zone, which is the region:

- Inside the nonagon (which is not entirely in the buffer zone)

- Plus the buffer zone around the nonagon.

Therefore, the boundary of S would consist of:

- The outer edge of the buffer zone (the offset curve)

- The inner edge of the nonagon where the buffer zone starts.

Wait, but the buffer zone is points within 1 unit of the perimeter. So the inner edge of the buffer zone is the original perimeter. However, since the interior of the nonagon is also included in S, the boundary of S is:

- The outer edge of the buffer zone (offset curve)

- The inner edge where the buffer zone meets the interior of the nonagon.

But this inner edge is a closed curve, which is the original perimeter of the nonagon. But the original perimeter is a boundary between the interior and the buffer zone. Wait, no. The original perimeter is part of the buffer zone. Therefore, if you have the interior of the nonagon and the buffer zone, the boundary of S would be the outer part of the buffer zone and the inner part of the nonagon?

This is getting too tangled. Let me think for the circle case.

If S is the union of the interior of the circle and the buffer zone (points within 1 unit of the perimeter), then for R > 1, S is the annulus from R - 1 to R + 1. The boundary of S is the two circles: R - 1 and R + 1.

But in this case, the "inner" boundary (R - 1) is a part of the buffer zone. Because the buffer zone is points within 1 unit of the perimeter (the original circle), which for points inside, is from R - 1 to R. So the inner boundary of the buffer zone is R - 1, while the outer boundary is R + 1.

However, if we also include the interior of the circle, which is from 0 to R, then the union is from R - 1 to R + 1. Hence the perimeter is two circles.

But if R < 1, then the buffer zone includes the entire interior of the circle (since R - 1 would be negative, so the buffer zone includes everything from 0 to R + 1). Hence, S is from 0 to R + 1, and the perimeter is just the outer circle.

Therefore, in general, for a circle, the perimeter of S is:

- 2π(R + 1) if R ≤ 1 (buffer zone includes the interior)

- 2π(R - 1) + 2π(R + 1) = 4πR if R > 1 (perimeter includes both inner and outer boundaries)

Therefore, for the nonagon, since its inradius is ~2.747 > 1, then S is the union of interior and buffer zone, which would be similar to the circle case, where the perimeter of S would include both the inner boundary (from the original nonagon's perimeter) and the outer boundary (the offset curve).

But how does that work for the nonagon?

If we think of S as the interior of the nonagon union the buffer zone, which is points within 1 unit of the perimeter. For points inside the nonagon, they are in S if they are within 1 unit of the perimeter, or else they are in the interior.

But for the inradius being greater than 1, the interior of the nonagon is not entirely within the buffer zone. Therefore, S is the set of points inside the nonagon or within 1 unit of the perimeter. So this is the union of two regions:

1. The interior of the nonagon (which extends from the center out to the perimeter)

2. The buffer zone around the perimeter, extending out 1 unit

Therefore, the boundary of S would consist of:

- The outer edge of the buffer zone (the offset curve)

- The inner edge of the buffer zone, which is the original perimeter of the nonagon

Wait, no. The original perimeter is part of the buffer zone. But the interior of the nonagon is also part of S. Therefore, the boundary of S would be:

- The outer edge of the buffer zone (offset curve)

- The inner edge of the buffer zone, which is the original perimeter, but only where the buffer zone starts. But since the interior is also part of S, the original perimeter is separating the interior from the buffer zone? Not really, because the buffer zone includes the original perimeter.

This is very confusing. Maybe the safest way is to think in terms of the Minkowski sum.

The set S is the union of the interior of the nonagon and the buffer zone. However, if the buffer zone is the Minkowski sum of the perimeter with a disk of radius 1. Wait, but no. The buffer zone is all points within 1 unit of the perimeter.

But the union of the interior and the buffer zone is the Minkowski sum of the nonagon with a disk of radius 1, because the Minkowski sum includes all points within 1 unit of any point in the nonagon. Which is the same as the interior of the nonagon union the points within 1 unit of the perimeter. Wait, no. The Minkowski sum of the nonagon with a disk of radius 1 is all points within 1 unit of the nonagon. This includes the interior of the nonagon and the buffer zone around it. Therefore, S is precisely the Minkowski sum, so its perimeter is the original perimeter + 2πr.

But in the circle example, if the original shape is a circle of radius R, then the Minkowski sum with a disk of radius 1 is a circle of radius R + 1, with perimeter 2π(R + 1). However, earlier I thought that when R > 1, S would have two boundaries, but that was a mistake.

Wait, no! If S is the Minkowski sum of the original shape with a disk, then it is the set of all points within 1 unit of the original shape. Which for a circle of radius R, would be a circle of radius R + 1. The perimeter is 2π(R + 1). But earlier, I thought that if you take the union of the interior of the circle and the buffer zone, it would be the annulus, but that's incorrect.

Let me clarify. The Minkowski sum of the original circle with a disk of radius 1 is all points within 1 unit of the original circle. Which is the region between R - 1 and R + 1 if R > 1. But this is the same as the set of points within 1 unit of the original circle. However, the interior of the original circle is already within this region. Therefore, the Minkowski sum is exactly the set of points within 1 unit of the original circle, which is the annulus between R - 1 and R + 1. But the perimeter of this region is the outer circle (R + 1) and the inner circle (R - 1). However, the Minkowski sum should be a single region, and its perimeter should be the outer boundary.

But this contradicts. Where is the mistake?

Wait, no. The Minkowski sum of a circle with a disk is another circle, not an annulus. Wait, no. If you take all points within 1 unit of the original circle, it's not just the Minkowski sum. Wait, the Minkowski sum of the original circle (which is a set) with a disk is the set of all points p + q where p is in the circle and q is in the disk. This results in a larger circle with radius R + 1. However, the set of points within 1 unit of the original circle is different.

Wait, the set of points within 1 unit of the original circle is the union of the original circle and the buffer zone around it. This is different from the Minkowski sum. The Minkowski sum is all points within 1 unit of the original circle's interior. Wait, no, the Minkowski sum is all points within 1 unit of the original circle.

Actually, the Minkowski sum of the original shape with a disk is the same as the set of points within 1 unit of the original shape. So for a circle, it would be an annulus if the original circle is considered as a 1-dimensional curve, but if the original shape is the 2-dimensional disk, then the Minkowski sum is the disk with radius R + 1.

But the problem says S is the union of the interior of the nonagon and points within 1 unit of the perimeter. This is different from the Minkowski sum.

The Minkowski sum of the nonagon with a disk is all points within 1 unit of any point in the nonagon, which includes the interior of the nonagon and the buffer zone around it. Therefore, S is the Minkowski sum. Therefore, the perimeter of S is the original perimeter + 2πr.

But in the circle example, if the original shape is a disk (2D), then the Minkowski sum with a disk is another disk, hence perimeter is 2π(R + 1). However, if the original shape is just the 1D perimeter (the circle), then the set of points within 1 unit of the perimeter is the annulus.

But in our problem, the original shape is the interior of the nonagon, and S is the union of the interior and the buffer. Therefore, it's like taking the interior (which is a 2D region) and the buffer (which is a 2D region), and taking their union. This is equivalent to taking the Minkowski sum of the closure of the nonagon (which is the nonagon itself) with a closed disk.

Wait, the closure of the nonagon is the nonagon itself, including the perimeter. Therefore, the Minkowski sum of the nonagon with a closed disk of radius 1 is all points within 1 unit of the nonagon, which includes the interior of the nonagon and the buffer zone. Therefore, S is the Minkowski sum.

Therefore, for the circle example, if the original shape is a closed disk (2D), the Minkowski sum with a disk is another closed disk, perimeter 2π(R + 1). If the original shape is just the 1D perimeter (circle), then the set of points within 1 unit of the perimeter is an annulus, with perimeter 2π(R + 1) + 2π(R - 1) = 4πR.

Therefore, the key is whether the original shape is 1D or 2D. In our problem, the original shape is the interior of the nonagon, which is a 2D region. Therefore, the Minkowski sum is a 2D region, and its perimeter is original perimeter + 2πr.

Therefore, returning to the nonagon: original perimeter is 18, r is 1, so perimeter of S is 18 + 2π.

But in the circle example, if the original shape is the 2D disk, the perimeter of S is 2π(R + 1), which is original perimeter (2πR) + 2πr.

Therefore, this supports the formula.

Therefore, I think the correct answer is 18 + 2π.

However, to ensure, let me think about the circle case again with the formula. If the original shape is the disk, which has perimeter 2πR, and the Minkowski sum with a disk of radius 1 has perimeter 2π(R + 1) = 2πR + 2π*1, which matches the formula.

Therefore, this confirms the formula. Therefore, the answer for the nonagon is 18 + 2π.

But let me check the problem statement once again to ensure I haven't misread.

"Let S be 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 of the nonagon. What, in units, is the perimeter of S?"

Yes, so S is the union of the interior and the buffer. Since the buffer is all points within 1 unit of the perimeter, and the interior is the set of points inside the nonagon. This is equivalent to the Minkowski sum of the nonagon (as a 2D region) with a disk of radius 1. Therefore, the perimeter of S is original perimeter + 2π*1 = 18 + 2π.

Therefore, the answer is 18 + 2π.

But the problem might expect a numerical value, but since π is involved, it's likely to leave it in terms of π.

Therefore, the perimeter of S is 18 + 2π units.

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

To determine the **perimeter of** the region $ S $, we are given a regular nonagon of side length 2 units, and we are to consider the union of:

1. **All points inside the nonagon**, and
2. **All points less than 1 unit away from the perimeter of the nonagon**.

Let's unpack these two components geometrically.

---

### Understanding the Shape of $ S $

- The **interior** of the nonagon is simply its **internal region**.
- The region **less than 1 unit away from the perimeter** forms a "buffer" zone (also known as an *offset region*) around the nonagon's boundary.
- The union of these two sets is exactly the **Minkowski sum** of the nonagon with a closed disk of radius 1.

This gives us the set $ S $ as the region consisting of:
- All points **within 1 unit** of the nonagon's perimeter (the buffer), and
- All points **inside** the nonagon (the interior itself).

Hence, **$ S $ is the union of the interior and the offset buffer** around the nonagon. Importantly, this is **equivalent** to all points within 1 unit of the **nonagon as a whole** (as a 2D object), which makes $ S $ the **inflation** (offset) of the nonagon by 1 unit in all directions.

---

### Calculating the Perimeter of $ S $

For any convex polygon, when you take the **Minkowski sum** with a closed disk of radius $ r $, the resulting shape has:
- The same number of sides (in this case, 9) as the original polygon,
- Its **original perimeter**,
- Plus a **circular component** that replaces the corners.

The key geometric result is that the **new perimeter** equals the **original perimeter** of the polygon *plus* the **circumference of a circle with radius $ r $** (i.e., $ 2\pi r $).

For a regular nonagon (9 sides), with side length 2 units:
- The original perimeter is:
  $$
  P = 9 \times 2 = 18 \text{ units}
  $$
- The added "circular" component from the buffer zone is:
  $$
  \text{Added perimeter} = 2\pi \times 1 = 2\pi \text{ units}
  $$

Thus, the **total perimeter of $ S $** is:
$$
\text{Perimeter of } S = 18 + 2\pi
$$

---

### Final Answer

$$
\boxed{18 + 2\pi}
$$

[Assistant]
Explanation:
Okay, so I need to find the perimeter of this set S. Let me try to visualize what S is. The problem says S is the union of two things: first, all the points inside a regular nonagon with side length 2 units, and second, all the points less than 1 unit away from a point on the perimeter of the nonagon.

So, if I imagine the nonagon, which is a nine-sided polygon, with each side 2 units long. Then, the first part of S is just the interior of this nonagon. The second part is like a buffer zone around the perimeter, extending 1 unit outwards from every point on the perimeter. So, combining these two, S would look like the original nonagon with a sort of "thickened" perimeter around it, right?

Wait, but when they say "the union of the set of all points inside the nonagon and the set of all points less than 1 unit away from the perimeter," does that mean that S is the Minkowski sum of the nonagon with a closed disk of radius 1? Or is it something else?

Hmm. Let me think. If you take all points less than 1 unit away from the perimeter of the nonagon, that's like taking the original nonagon and adding a 1-unit wide area around it. However, the original nonagon is already included in S because S is the union of the interior of the nonagon and that buffer zone. So, in effect, S is the original nonagon expanded outward by 1 unit in all directions, but keeping the interior.

But wait, actually, the union of the interior and the buffer zone would just be the closure of the buffer zone? Or maybe not? Let me clarify.

Suppose you have a set A, which is the interior of the nonagon. Then the set B is all points less than 1 unit away from the perimeter of the nonagon. So B includes points inside the nonagon? Wait, not necessarily. Wait, points less than 1 unit away from the perimeter would be points that are within 1 unit of the boundary of the nonagon. But the interior of the nonagon is points that are more than 0 units away from the perimeter? Wait, no. Actually, the interior points are those that are within the nonagon. The perimeter is the boundary. So points less than 1 unit away from the perimeter would include the interior points near the boundary, but also points outside the nonagon within 1 unit of the perimeter. Wait, so S is the union of the interior of the nonagon and the set of points within 1 unit of the perimeter. But if you take the union of the interior and points within 1 unit of the perimeter, that would actually be the same as the set of points within 1 unit of the nonagon's perimeter or inside it. Which is like taking the original nonagon and expanding it outward by 1 unit. But since the interior is already included, the union would just be the closed disk around the nonagon with radius 1? No, not exactly. Because expanding a polygon by 1 unit outward would result in a sort of rounded polygon, with the original sides extended out and the corners rounded with quarter-circles? Wait, actually, when you take all points within 1 unit of a polygon, it's called an offset polygon or a parallel curve. For convex polygons, offsetting outward would create a new polygon with the same number of sides, but with each side moved outward by 1 unit, and with rounded corners where the sides meet. But the exact shape depends on the original polygon. However, in this case, since the original nonagon is convex, yes, the set S would be the union of the original nonagon and the buffer around it, which would be like the original nonagon expanded by 1 unit in all directions. However, since the buffer zone includes points within 1 unit of the perimeter, which would overlap with the original nonagon. Therefore, S is the original nonagon expanded outward by 1 unit. But the perimeter of S would then consist of the expanded sides and the rounded corners?

Wait, maybe I need to think in terms of the boundary of S. The original perimeter of the nonagon is being expanded outward by 1 unit, but also, where the original sides meet at the corners, the buffer zone would create quarter-circles? Wait, no. If you take all points within 1 unit of the perimeter, then near the corners, you would have a circular arc with radius 1 unit, but how much of an angle does it cover?

Let me recall. If you have a polygon with an interior angle θ at each vertex, then when you offset outward by distance r, the original corner is replaced by a circular arc with angle π - θ. Wait, maybe? Let me think. For a polygon, when you offset outward, each original vertex is replaced by a circular arc. The angle of the arc would depend on the internal angle of the polygon.

For a regular nonagon, each internal angle can be calculated. Let me recall the formula for internal angles of a regular n-gon: it's ((n-2)*180)/n degrees. So for a nonagon, n=9, so internal angle is ((9-2)*180)/9 = (7*180)/9 = 140 degrees. So each internal angle is 140 degrees, which is 140π/180 = 7π/9 radians.

When you offset a polygon outward by distance r, the corner is replaced by a circular arc. The angle of the arc is equal to π - internal angle. Wait, why?

Imagine standing at the corner of the polygon. The original corner is a vertex where two edges meet. When you offset each edge outward, the offset edges will move away from the original polygon. The corner is replaced by an arc that turns around the original corner. The angle of the arc is equal to the angle that the two offset edges would form if they were extended. Wait, perhaps the angle of the arc is π minus the original internal angle?

Wait, for a square, each internal angle is 90 degrees. When you offset a square outward by a distance r, each corner becomes a quarter-circle. The angle of the arc is 90 degrees, which is π/2 radians, which is π - π/2 (since the internal angle is π/2). Wait, π - π/2 is π/2, which matches. Wait, but for a square, the arc is 90 degrees. So if the internal angle is θ, then the arc angle is π - θ? Wait, π - θ would be π - π/2 = π/2, which is correct. For the nonagon, internal angle is 7π/9 radians. Therefore, the arc angle would be π - 7π/9 = 2π/9 radians. Which is 40 degrees. So each corner of the nonagon is replaced by an arc of 40 degrees with radius 1 unit. Therefore, the perimeter of S would consist of the original sides, offset outward by 1 unit, and the arcs replacing the corners.

Wait, but hold on. Since S is the union of the interior of the nonagon and the points within 1 unit of the perimeter. Therefore, the boundary of S would be the original perimeter offset outward by 1 unit, but with the corners replaced by the arcs. Therefore, the perimeter of S is equal to the length of the offset polygon's perimeter.

Therefore, to calculate the perimeter of S, we need to calculate the perimeter of this expanded shape. The perimeter would consist of the lengths of the offset sides plus the lengths of the arcs.

So first, let me figure out the length of the offset sides. For each original side of the nonagon, when we offset it outward by 1 unit, the length of the side remains the same? Wait, no. Wait, when you offset a straight side outward, the length of the side increases? Wait, no. Wait, if you take a straight line segment and offset it outward, keeping it parallel, the length of the segment remains the same, but the offset creates a sort of "buffer" around it. However, when you offset all sides of the polygon, the original sides are extended outward, but the corners are rounded.

But actually, when considering the perimeter of S, which is the union of the original nonagon and the buffer, the perimeter of S is the boundary of the buffer zone. Wait, but since the buffer zone includes the interior of the nonagon, the perimeter of S is the same as the perimeter of the buffer zone. Wait, no. If you take the union, the perimeter of S would consist of the parts of the buffer zone that are not overlapping with the original nonagon. However, since the buffer zone includes all points within 1 unit of the perimeter, and the original nonagon is included, so the union would actually be the closure of the buffer zone. But the boundary of S would be the boundary of the buffer zone. Wait, but how?

Alternatively, perhaps S is the union of two sets: the original nonagon (interior) and the buffer around the perimeter. The union would have a perimeter that is the same as the perimeter of the buffer zone. Because the original nonagon's interior is entirely inside the buffer zone? Wait, no. The buffer zone is all points within 1 unit of the perimeter. The original nonagon's interior is points that are inside the nonagon. However, points inside the nonagon are at distances from the perimeter. The maximum distance from the perimeter inside the nonagon is the inradius. So if the inradius is larger than 1 unit, then the buffer zone would extend beyond the original nonagon, but the union would be the original nonagon plus the buffer zone. But if the inradius is less than 1, then the buffer zone would extend into the original nonagon.

Wait, perhaps I need to calculate the inradius of the nonagon. Let me recall. For a regular polygon with n sides of length s, the inradius r is given by r = (s/2) * cot(π/n). So here, n=9, s=2. Therefore, inradius is (2/2)*cot(π/9) = cot(π/9). Let me compute cot(π/9). π/9 is 20 degrees. Cot(20 degrees) is approximately 2.747, but I don't need the exact value. However, the problem is whether the inradius is less than 1 or not. If the inradius is greater than 1, then the buffer zone would not reach the center of the nonagon. If the inradius is less than 1, the buffer zone would extend into the original nonagon. Wait, but inradius is the distance from the center to the sides. So points within 1 unit of the perimeter would include points inside the nonagon near the perimeter, but depending on how far the inradius is.

Wait, but the inradius is the distance from the center to the sides. So if the inradius is greater than 1, then the buffer zone (which is points within 1 unit of the perimeter) would not reach the center. However, the original nonagon is included in S. Wait, no, S is the union of the interior of the nonagon and points within 1 unit of the perimeter. So even if the buffer zone doesn't reach the center, the interior is included. Therefore, the union is the entire nonagon plus the buffer zone. But in that case, perhaps the perimeter of S is the perimeter of the buffer zone, but since the buffer zone includes parts of the original nonagon, does that affect the perimeter?

Wait, maybe not. Because the buffer zone is all points within 1 unit of the perimeter. So when you take the union with the interior, you are just saying that S includes the interior and the buffer, so S is the union, which would be the closure of the buffer zone. However, the boundary of S is the boundary of the buffer zone. Wait, maybe the buffer zone's boundary is the set of points that are exactly 1 unit away from the perimeter. So the perimeter of S would be the set of points that are 1 unit away from the original perimeter. Which, for a polygon, is called the offset curve.

But as we discussed earlier, for a convex polygon, the offset curve is formed by moving each side outward by 1 unit, and rounding the corners with circular arcs. Therefore, the perimeter of S would be equal to the sum of the lengths of the offset sides plus the sum of the lengths of the arcs.

So to compute this perimeter, we need to know two things:

1. The length of each offset side. Since the original sides are of length 2 units, and when offsetting, each side is moved outward by 1 unit. However, does the length of the side remain the same? Wait, actually, when you offset a straight segment outward, the length of the segment remains the same, but the offset is in the direction perpendicular to the segment. Therefore, if you imagine moving each side outward by 1 unit, then the distance between the original side and the offset side is 1 unit. But since the polygon is regular, all sides are moved outward by the same distance, so the new polygon (if we were to ignore the rounded corners) would have the same side lengths as the original nonagon. However, in reality, the offset polygon is not a polygon anymore, because the corners are rounded. Wait, but if we think about the perimeter contributing from the offset sides, each original side of length 2 units would contribute a length of 2 units to the perimeter? No, wait, because when you offset the sides, they are moved outward, but the direction of the offset is perpendicular to each side. However, when you offset all sides, the corners are replaced by arcs. Therefore, the original sides, when offset, are now sides of a larger polygon, but the total length of those sides would be the same as the original sides? But that can't be, because if you move each side outward, the distance between adjacent sides is increased. Wait, perhaps the length of each offset side is actually longer?

Wait, no. Wait, if you take a straight side of length L, and offset it outward by distance r, keeping it parallel, then the length of the side doesn't change. However, when you do this for all sides of a polygon, the offset polygon's sides are longer because they are "pushed out" and the corners are rounded. Wait, no. Wait, let me think of a square. If you have a square with side length 2, and you offset each side outward by 1 unit, forming a larger square with rounded corners. The sides of the larger square, if you were to make them straight, would have length increased by 2*r, but since we are rounding the corners, the sides are actually the same as the original sides? No. Wait, in reality, the perimeter of the offset square would be the original perimeter plus 2πr. Wait, for a square, when you offset it outward by radius r, the perimeter becomes original perimeter + 2πr. Wait, original perimeter is 4*2=8. If you offset outward, each corner becomes a quarter-circle, so four quarter-circles make a full circle, circumference 2πr. So total perimeter is 8 + 2πr. However, if you think of the straight sides, they have length same as original sides? Wait, but in reality, when you offset a square, the straight sides are not of the same length. Wait, no. If you have a square, and you offset each side outward by distance r, keeping them parallel, then the distance between the original side and the offset side is r. However, if you consider the length of the offset side, since it's parallel, the length remains the same. However, when you offset all sides, the problem is that adjacent sides are offset outward, and the corners are rounded. So the length of the straight parts of the offset shape is same as original? Wait, no. Wait, imagine a square with side length 2. Offset each side outward by 1 unit. The sides are each extended outward, but the corners are replaced by quarter circles. However, when you offset each side, the new side is not longer than original. Therefore, the straight parts of the offset shape have the same length as the original sides, but the total perimeter is original perimeter plus 2πr.

But wait, that seems contradictory, because the offset square's perimeter would be longer than the original. Let me check with numbers. If original square has side length 2, perimeter 8. After offsetting by 1 unit, the perimeter becomes 8 + 2π*1 ≈ 8 + 6.28 ≈ 14.28. Alternatively, if you think of the offset square as having sides that are offset outward, but with rounded corners. If you didn't have the rounded corners, the sides would be longer. Wait, but if you have a square and you just move each side outward, the distance between the original side and the new side is 1 unit. However, if you imagine moving each side outward, the new square would have side length longer. Wait, the original square has side length 2. The distance from the original side to the offset side is 1 unit, but the side length of the new square would be 2 + 2*r*(cot(π/n))? Wait, maybe I need to think in terms of the offsetting process.

Alternatively, perhaps the perimeter of the offset shape is equal to the original perimeter plus 2πr. For a polygon with m sides, the perimeter of the offset shape is original perimeter + 2πr. Wait, that seems to be the case for the square. For a polygon with n sides, if you offset it outward by r, then the perimeter becomes original perimeter + 2πr? That can't be right because for a polygon with many sides, the rounded corners would add up to a full circle. So regardless of the number of sides, as long as you offset outward by r, the perimeter would be original perimeter plus 2πr?

Wait, yes, actually. Because when you offset each side outward by r, the corners are replaced by circular arcs. Each corner contributes an arc of angle equal to the exterior angle of the polygon. For a regular polygon, the sum of the exterior angles is 2π radians. Therefore, when you offset, each exterior angle becomes an arc of angle equal to the exterior angle. Therefore, the total length added by the arcs is r times the total angle, which is r*2π. Therefore, the perimeter of the offset shape is original perimeter + 2πr.

Therefore, for any polygon, when you offset it outward by radius r, the perimeter of the offset shape is original perimeter + 2πr. Therefore, regardless of the number of sides, this formula holds.

Wait, so for the square, original perimeter 8, offset by 1, perimeter becomes 8 + 2π*1 ≈ 14.28. Which seems correct. For a polygon with more sides, like a regular nonagon, the same applies. Therefore, the perimeter of S, which is the set of points within 1 unit of the original nonagon's perimeter, would be equal to the original perimeter of the nonagon plus 2π*1. But wait, but the original problem says S is the union of the interior of the nonagon and points within 1 unit of the perimeter. But if the offset shape is points within 1 unit of the original polygon's perimeter, then the perimeter of S is the perimeter of this offset shape, which is original perimeter + 2πr. Therefore, the perimeter of S would be original perimeter of nonagon + 2π(1).

But wait, let me check again. Suppose that S is the union of the interior and the buffer. However, if the buffer zone is points within 1 unit of the perimeter, then the union would be the closure of the buffer zone, which is the same as the set of points within 1 unit of the perimeter or inside the polygon. However, the boundary of this set is the set of points at exactly 1 unit away from the original perimeter. Which is the offset curve we discussed. Therefore, the perimeter of S is equal to the length of this offset curve. And according to the earlier logic, that is equal to the original perimeter plus 2πr. So if that is the case, then the perimeter of S is equal to the original perimeter of the nonagon plus 2π*1.

But wait, the original perimeter of the nonagon is 9 sides each of length 2 units, so 9*2=18 units. Therefore, perimeter of S would be 18 + 2π(1) ≈ 18 + 6.28 ≈ 24.28. However, I need to confirm this logic. Because earlier, I was confused about the offset sides. But according to the general formula, for any polygon, the perimeter of the offset curve is original perimeter + 2πr.

But why is that? Let me think again. When offsetting a polygon, each side contributes a length equal to its original length, but each corner is replaced by an arc. The total angle of all arcs is equal to 2π, because the sum of the exterior angles of any polygon is 2π. Therefore, the total length added by the arcs is r*(sum of angles) = r*2π. Therefore, the total perimeter is original perimeter + 2πr. Therefore, regardless of the original polygon's shape, as long as it's convex, this formula holds.

Therefore, in this problem, since the nonagon is convex, and S is the set of points within 1 unit of its perimeter (which is equivalent to the offset polygon with radius 1), the perimeter of S is the original perimeter + 2π*1. Therefore, the answer is 9*2 + 2π*1 = 18 + 2π.

Wait, but hold on. The problem says S is the union of the interior of the nonagon and the set of points less than 1 unit away from the perimeter. So is the perimeter of S the same as the perimeter of the offset set?

But if you think of S as the union of the interior and the buffer, then S is a closed set, and its boundary is the set of points that are exactly 1 unit away from the original perimeter. Therefore, the perimeter of S is the length of this boundary, which is the offset curve. Therefore, the perimeter of S is indeed original perimeter + 2π*1.

However, let me verify with a simpler case. Let me take a square with side length 2 units. Its perimeter is 8 units. S would be the union of the interior and points within 1 unit of the perimeter. The perimeter of S would be the boundary of this region. If I imagine this, the boundary is the original square offset outward by 1 unit, with rounded corners. The perimeter of this offset shape is original perimeter + 2π*1 = 8 + 2π, which is correct. Therefore, the same applies to the nonagon. Hence, the perimeter of S is 18 + 2π. But wait, the answer is supposed to be in units. Wait, but the problem says "the perimeter of S". Is that correct?

But let me think again. Wait, if the original nonagon has a perimeter of 18 units, and we add 2π units from the rounded corners. Therefore, the answer is 18 + 2π. However, I need to check if this is correct.

Wait, but maybe there's a misunderstanding here. Let me think again. Suppose that S is the union of the interior of the nonagon and the set of points within 1 unit of the perimeter. However, when you take the union, you include all points inside the nonagon and all points within 1 unit outside of the nonagon. Therefore, the perimeter of S would be the perimeter of the nonagon expanded outward by 1 unit, which would be the original perimeter plus the circumference of a circle with radius 1? Wait, no, it's not a full circle. As we discussed, for a polygon, the perimeter of the offset is original perimeter + 2πr.

Alternatively, another way to think about it: the perimeter of S is the set of points that are exactly 1 unit away from the original nonagon's perimeter. Which, as we said, is the offset curve, which has perimeter original perimeter + 2πr. Therefore, the answer is 18 + 2π.

But wait, the problem says "the set of all points less than 1 unit away from a point on the perimeter of the nonagon". So does that include points inside the nonagon? Because the points inside the nonagon are less than 1 unit away from some point on the perimeter (specifically, points inside are less than the inradius away from the perimeter). However, the original nonagon is included in S. So perhaps S is the Minkowski sum of the nonagon with a closed disk of radius 1.

But the Minkowski sum of a polygon with a disk would be the original polygon expanded by radius 1, with rounded corners, and the perimeter of that shape is original perimeter + 2π*1. So yes, same result. Therefore, the perimeter is 18 + 2π.

But wait, let me check if the formula is correct. If you have a polygon, and you take its Minkowski sum with a disk of radius r, then the perimeter of the resulting shape is equal to the original perimeter plus 2πr.

Let me verify with a simple case. Take a line segment of length L. The Minkowski sum with a disk of radius r is a rectangle with semicircular ends. The perimeter would be 2L (the two straight sides) plus 2πr (the two semicircles). But original perimeter of the line segment is 0 (as a 1-dimensional object), but maybe this is not helpful.

Another example: take a circle of radius R. Its perimeter is 2πR. The Minkowski sum with a disk of radius r is a circle of radius R + r, perimeter 2π(R + r) = 2πR + 2πr. Which matches the formula. Therefore, for the circle, the perimeter increases by 2πr.

For a square with side length L, original perimeter 4L. The Minkowski sum with a disk of radius r is a square with rounded corners. The perimeter is 4L + 2πr. Because each side contributes L, and the corners contribute quarter-circles, totaling a full circle circumference 2πr. Hence, perimeter becomes 4L + 2πr.

Therefore, yes, the formula holds for polygons. Therefore, for the nonagon, original perimeter is 9*2 = 18, so the perimeter of S is 18 + 2π*1 = 18 + 2π. Therefore, the answer is 18 + 2π. But let me check again.

Wait, but the problem says "the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter of the nonagon". Therefore, the union of the interior and the points within 1 unit of the perimeter.

But the Minkowski sum of the nonagon with a disk of radius 1 includes the interior of the nonagon and the points within 1 unit of the perimeter. So yes, that is exactly S. Therefore, the perimeter of S is the perimeter of the Minkowski sum, which is original perimeter + 2π*1.

Therefore, answer is 18 + 2π. However, the problem might expect a numerical value, but since it says "in units", and π is involved, likely wants an exact answer in terms of π. Therefore, 18 + 2π. But wait, let me check once more.

Wait, but hold on. Another thought: when you take the union of the interior and the buffer zone, maybe the perimeter of S is just the perimeter of the buffer zone, which is the same as the offset curve. But the buffer zone itself is the set of points within 1 unit of the perimeter. However, the buffer zone includes the interior of the nonagon. Therefore, the boundary of S is the set of points that are exactly 1 unit away from the perimeter of the nonagon. Which is the offset curve, as discussed. Therefore, the perimeter of S is indeed original perimeter + 2π*1.

But wait, another way to think: suppose that the buffer zone is like a "thickened" perimeter. When you have a polygon with perimeter P, and you create a buffer of width r around it, the area is P*r + π*r². But the perimeter of the buffer zone is P + 2πr. But does this hold?

Wait, actually, the area of the buffer zone (the set of points within r of the polygon) is equal to the area of the polygon plus the perimeter times r plus πr². But the perimeter of the buffer zone is equal to the original perimeter plus 2πr.

But let me check for a square. Area of buffer zone would be area of square + perimeter*r + πr². For a square with side length 2, area is 4. Perimeter is 8. So buffer area would be 4 + 8*1 + π*1² = 4 + 8 + π = 12 + π. But the buffer zone is the square expanded by 1 unit with rounded corners. The area can also be calculated as area of the expanded square (which would have side length 2 + 2*1 = 4) but subtracting the corner parts that are overlapped. Wait, no. Actually, if you expand a square by 1 unit with rounded corners, the area is equal to the area of the square plus the area of the buffer. Alternatively, since the buffer zone is all points within 1 unit of the original square, which can be thought of as the original square plus a rectangle of width 1 on each side, plus quarter circles at each corner.

The area would be original area + perimeter * r + πr². Which is 4 + 8*1 + π*1² = 12 + π, as above. Therefore, the area is correct.

But what about perimeter? The perimeter of the buffer zone is original perimeter + 2πr. For the square, that would be 8 + 2π*1 ≈ 8 + 6.28 ≈ 14.28. Which is correct, as the buffer zone's perimeter is the original perimeter (the "outer" edges) plus the circumference from the rounded corners.

Therefore, the formula seems to hold. Therefore, applying this to the regular nonagon, the perimeter of S is original perimeter + 2π*1. Original perimeter is 9*2 = 18. Therefore, the answer is 18 + 2π.

But the problem says "the union of the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter of the nonagon". Is there a possibility that the perimeter of S is different?

Wait, suppose that the original nonagon's perimeter is entirely inside S. But since S includes the points less than 1 unit away from the perimeter, which includes points outside the nonagon. But the original nonagon is also included. However, the perimeter of S is not just the perimeter of the nonagon. Because when you include the buffer, you are adding new edges. However, the original perimeter is not entirely part of the boundary of S. Because points on the original perimeter are within 1 unit of themselves, so they are included in S. But the boundary of S is the set of points that are exactly 1 unit away from the original perimeter. Therefore, the original perimeter is inside S, not on its boundary. Therefore, the perimeter of S is only the boundary of the buffer zone, which is the offset curve, with perimeter original perimeter + 2πr.

Therefore, I think the answer is 18 + 2π. But let me check if this is correct by thinking about another way.

Alternatively, think about the perimeter of S. S consists of two parts: the interior of the nonagon and the buffer. However, the perimeter of S would be the outer edge of the buffer, because the interior has no perimeter. Wait, no, the union would have a perimeter that is the boundary of the buffer zone, which is what we discussed. Therefore, I think the answer is 18 + 2π.

But let me check with another example. Suppose the nonagon is replaced by a circle. Let's say a circle with radius R. Then the set S would be the union of the interior of the circle and the set of points less than 1 unit away from the perimeter. But the interior of the circle already includes all points within R of the center. The set of points less than 1 unit away from the perimeter would be the annulus between R - 1 and R + 1. Wait, but if R is greater than 1, then the union would be the circle expanded outward by 1 unit. But the perimeter of S would be the circumference of the expanded circle, which is 2π(R + 1). However, original perimeter is 2πR. So perimeter of S is 2πR + 2π*1, which matches our formula: original perimeter + 2πr. Therefore, same result. Therefore, formula holds.

Therefore, I think confident that the perimeter of S is 18 + 2π units. However, maybe there is a trick here. Let me check if the problem says "the union of the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter of the nonagon". Wait, is the buffer zone overlapping with the nonagon? Because points inside the nonagon are less than 1 unit away from the perimeter? Not necessarily. Points inside the nonagon can be more than 1 unit away from the perimeter. For example, the center of the nonagon is at distance equal to the inradius from the perimeter. If the inradius is greater than 1, then the buffer zone (points within 1 unit of the perimeter) does not include the center. Wait, so is S the union of two regions: the interior of the nonagon and the buffer zone. Therefore, S is the set of points either inside the nonagon or within 1 unit of the perimeter.

But in that case, the boundary of S would be the boundary of the buffer zone plus the boundary of the nonagon? No, because the interior of the nonagon is connected to the buffer zone. Wait, but points inside the nonagon are already included in the buffer zone if they are within 1 unit of the perimeter. However, if the inradius is larger than 1, then the buffer zone doesn't reach the center. Therefore, the union would have a boundary that is the union of the original nonagon's boundary and the buffer zone's boundary? But no, because the buffer zone is points within 1 unit of the perimeter, so near the original perimeter, the buffer zone overlaps with the interior of the nonagon. Therefore, the union S would have a boundary that is the set of points that are either on the original perimeter or exactly 1 unit away from the original perimeter. Wait, no.

Wait, suppose you have a point on the original perimeter. It's included in S because it is in the buffer zone (distance 0 from perimeter). The points inside the nonagon are included because of the union. The buffer zone includes points outside of the nonagon within 1 unit of the perimeter. Therefore, the boundary of S is made up of two parts:

1. The parts of the original nonagon's perimeter that are not within 1 unit of any other part of the buffer zone. Wait, this is getting confusing. Maybe it's better to think of S as the union of the nonagon and the buffer zone. The buffer zone is like a 1-unit thick region around the nonagon. However, the union would be the nonagon with a 1-unit thick "border" around it. Therefore, the perimeter of S would consist of two parts:

- The outer edge of the buffer zone (which is the offset curve, with perimeter original perimeter + 2πr)

- And the inner edge of the buffer zone, which is the original perimeter of the nonagon.

But wait, but the original perimeter is entirely inside S. Because S includes the interior of the nonagon and the buffer zone. However, the original perimeter is on the boundary between the interior of the nonagon and the buffer zone. Therefore, the perimeter of S would consist of the outer part of the buffer zone (which is the offset curve) and the original perimeter? No, that can't be. Because when you take the union, the original perimeter is an internal boundary of S.

Wait, actually, if you have two regions, one inside the other, their union has a boundary equal to the boundary of the larger region. But in this case, the buffer zone extends both into the original nonagon and outside. Therefore, the original nonagon's perimeter is completely inside S, but it is not part of the boundary of S. The boundary of S is the outer edge of the buffer zone, which is the offset curve. Because the original perimeter is inside S, not on its boundary.

Let me think of a simple case again. Suppose you have a square, and you take the union of the square and the points within 1 unit of its perimeter. The boundary of this union is the outer edge of the buffer zone, which is the offset curve. The original square's perimeter is not on the boundary of S because the buffer zone overlaps with it. So yes, the boundary is only the offset curve.

Therefore, going back, the perimeter of S is original perimeter + 2πr. Therefore, 18 + 2π.

However, let me check the problem statement once more to be sure.

"The union of the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter of the nonagon."

Wait, the set of all points inside the nonagon is the interior, and the set of all points less than 1 unit away from the perimeter. So the union is the interior plus the buffer. If the buffer extends beyond the nonagon, then the union is the nonagon with a buffer, so the perimeter is the outer edge of the buffer. If the buffer also extends into the nonagon, then the union is the same as the buffer.

But in this case, the buffer includes the interior of the nonagon if the inradius is less than 1. Wait, if the inradius is less than 1, then points inside the nonagon are within 1 unit of the perimeter. Wait, the inradius is the maximum distance from the center to the sides. So the distance from a point inside the nonagon to the perimeter depends on the point. The center is at inradius distance from the perimeter. If the inradius is less than 1, then the entire nonagon is within 1 unit of the perimeter. Therefore, the union of the interior of the nonagon and the buffer zone is just the buffer zone. Therefore, S is the buffer zone, which has perimeter original perimeter + 2π*1.

But if the inradius is greater than 1, then the buffer zone would not include the entire interior of the nonagon. Therefore, S is the union of the interior and the buffer, which is the buffer zone plus the interior, so the perimeter would still be just the outer edge of the buffer zone.

Therefore, regardless of the inradius, the perimeter of S is original perimeter + 2π*1.

Therefore, the answer is 18 + 2π.

Wait, but the problem says "the set of all points inside a regular nonagon... and the set of all points less than 1 unit away from a point on the perimeter". So the union includes both the interior and the buffer. If the inradius is less than 1, then the buffer already includes the interior, so S is just the buffer. If the inradius is more than 1, then S is the union of the interior and the buffer, which would be the buffer plus the interior. But in this case, the interior is not overlapping with the buffer? Wait, no. Wait, if the inradius is more than 1, then the buffer zone (points within 1 unit of the perimeter) does not include the entire interior. For example, the center is at inradius distance from the perimeter, which is more than 1, so the center is not in the buffer. Therefore, the union of the interior and buffer is the buffer plus the interior. However, the boundary of this union is still the outer edge of the buffer. Because the interior is inside the buffer? No. Wait, the interior is entirely inside the nonagon, which is inside the buffer zone? No. Because the buffer zone is points within 1 unit of the perimeter. So if the inradius is more than 1, then the interior is not entirely within the buffer zone. Therefore, the union would be the buffer zone and the interior. The boundary of the union is the boundary of the buffer zone (the outer edge) and the boundary of the interior (the original perimeter). But since the buffer zone overlaps with the interior near the perimeter, the original perimeter is already included in the buffer zone? No. Wait, the buffer zone is points within 1 unit of the perimeter. The original perimeter is part of the buffer zone. Therefore, the union of the interior and buffer zone would be the buffer zone union the interior. However, the interior is already partly overlapping with the buffer zone. Therefore, the boundary of the union is the outer edge of the buffer zone (the offset curve) and the original perimeter? No, because the original perimeter is on the edge between the interior and buffer zone. Wait, actually, the boundary of S would be the outer edge of the buffer zone (the offset curve) because the interior of the nonagon is connected to the buffer zone. The original perimeter is not a boundary of S.

This is getting too confusing. Maybe a better approach is to think of S as the Minkowski sum of the nonagon and a disk of radius 1. Then, as we discussed, the perimeter of the Minkowski sum is the original perimeter plus 2πr.

Therefore, regardless of the inradius, the perimeter of S is 18 + 2π. Therefore, I think this is the correct answer.

But to be thorough, let me compute the inradius of the nonagon and see if it's less than 1 or not.

As mentioned earlier, the inradius r = (s/2) * cot(π/n), where s is the side length, n is the number of sides.

Given that n = 9, s = 2. Therefore:

r = (2/2) * cot(π/9) = cot(π/9) ≈ cot(20°).

The cotangent of 20 degrees is 1/tan(20°).

tan(20°) ≈ 0.3640, so cot(20°) ≈ 2.747. Therefore, the inradius is approximately 2.747 units.

Which is greater than 1. Therefore, the buffer zone (points within 1 unit of the perimeter) does not include the entire interior of the nonagon. Therefore, S is the union of the interior of the nonagon and the buffer zone.

But then, what does the boundary of S look like?

The boundary of S would consist of two parts:

1. The outer boundary of the buffer zone (the offset curve)

2. The inner boundary of the buffer zone, which is the original perimeter of the nonagon.

Wait, no. Because the buffer zone is points within 1 unit of the perimeter, which includes the perimeter itself. Therefore, the union S is the interior of the nonagon plus the buffer zone. However, since the buffer zone includes the original perimeter, the union S is just the buffer zone union the interior of the nonagon. But since the buffer zone already includes the original perimeter, and the interior is the set of points inside the nonagon. Therefore, the union is all points that are either inside the nonagon or within 1 unit of the perimeter.

But the buffer zone is points within 1 unit of the perimeter. The interior is points inside the nonagon. Therefore, S is the set of points that are inside the nonagon or within 1 unit of the perimeter.

But since the interior of the nonagon is only within 1 unit of the perimeter if they are near the perimeter. However, the center is not within 1 unit of the perimeter (since inradius is ~2.747). Therefore, the union S is the buffer zone (which includes the original perimeter and the points within 1 unit outside the nonagon) plus the interior of the nonagon.

Therefore, the boundary of S is composed of:

- The outer part of the buffer zone (the offset curve)

- The inner part of the buffer zone, which is the original perimeter, but only where the buffer zone starts. Wait, no.

Actually, the boundary of S would consist of:

- The outer boundary of the buffer zone (the offset curve)

- The parts of the original perimeter that are not covered by the buffer zone. Wait, no. The original perimeter is entirely within the buffer zone (because points on the perimeter are 0 units away from the perimeter). Therefore, the original perimeter is inside S, but it is not part of the boundary.

Wait, confusion arises because the buffer zone includes the original perimeter. So the union S includes the interior of the nonagon and the buffer zone. However, the buffer zone is already including the original perimeter and the area within 1 unit of it. Therefore, the union is just the buffer zone union the interior. But since the buffer zone already includes the original perimeter and the area around it, including the interior near the perimeter. However, the interior of the nonagon is not entirely included in the buffer zone. For example, the center of the nonagon is not in the buffer zone. Therefore, S is the buffer zone plus the interior of the nonagon. Therefore, the boundary of S consists of:

- The outer part of the buffer zone (the offset curve)

- The inner part of the nonagon that is not covered by the buffer zone. Wait, but how?

Wait, the boundary of S would be the set of points where S changes from being inside to outside. Since S includes all points inside the nonagon and all points within 1 unit of the perimeter. Therefore, imagine the region S is the nonagon's interior plus a 1-unit wide region around it. So it's like the nonagon with a "border" of width 1 unit.

Therefore, the boundary of S would be:

- The outer edge of the border (the offset curve)

- The inner edge of the border, which is the original perimeter of the nonagon.

But is the original perimeter part of the boundary of S?

If you take a point on the original perimeter, it is in S, but the points inside the nonagon on one side of the perimeter and outside the nonagon on the other side. Wait, the original perimeter is the boundary between the interior of the nonagon and the exterior. However, since S includes the interior of the nonagon and the buffer zone, which includes the original perimeter and the exterior around the nonagon. So the original perimeter is entirely within S, but not part of its boundary. Because points just inside the nonagon are also in S, and points just outside are in the buffer zone. Therefore, the original perimeter is like a line between two regions, but since both sides are included in S (interior and buffer), the original perimeter is not part of the boundary.

Therefore, the boundary of S is only the outer edge of the buffer zone, which is the offset curve. Therefore, the perimeter of S is original perimeter + 2πr, which is 18 + 2π.

But to confirm, consider a simpler case where the nonagon is replaced with a circle. If the circle has radius R, then S is the union of the interior of the circle and the buffer zone of radius 1. The buffer zone would be the annulus between R - 1 and R + 1, but since the interior of the circle is included, S is the union of the circle and the annulus, which is just the annulus from R - 1 to R + 1. However, if R > 1, then the buffer zone includes the interior of the circle (from R - 1 to R). Therefore, the perimeter of S is the outer edge (circle of radius R + 1) and the inner edge (circle of radius R - 1). Wait, but this contradicts our previous conclusion.

Wait, here's the problem. If S is the union of the interior of the circle and the buffer zone (points within 1 unit of perimeter), then if the original circle has radius R > 1, the buffer zone extends from the original circle's perimeter inward by 1 unit to R - 1. Therefore, the union S is the region from R - 1 to R + 1. Therefore, the boundary of S consists of two circles: the inner circle of radius R - 1 and the outer circle of radius R + 1. Therefore, the perimeter of S is the sum of the circumferences of these two circles.

But according to our previous formula, it would have been original perimeter + 2πr. Original perimeter is 2πR, and adding 2π*1 would give 2π(R + 1), which is only the outer circle. But in reality, the perimeter of S would be the outer circle (perimeter 2π(R + 1)) and the inner circle (perimeter 2π(R - 1)), totaling 2π(R + 1 + R - 1) = 4πR. However, this contradicts the previous formula.

Wait, so there must be an error in my previous reasoning.

This suggests that when the buffer zone includes parts of the original shape (interior), the perimeter of S is not just the offset curve, but also includes the inner boundary. Therefore, my earlier conclusion was incorrect.

So, this is a critical mistake. Let me re-examine.

If S is the union of the interior of the circle and the buffer zone (points within 1 unit of the perimeter), then S is the set of points within 1 unit of the perimeter or inside the circle.

But if the circle has radius R > 1, then the set of points within 1 unit of the perimeter includes the annulus between R - 1 and R + 1. Therefore, the union of the interior (points within R) and the buffer zone (annulus between R - 1 and R + 1) is the union of the annulus between R - 1 and R + 1 and the circle of radius R. Therefore, the union is the annulus between R - 1 and R + 1. Therefore, the boundary of S is the two circles: R - 1 and R + 1. Therefore, the perimeter is 2π(R + 1) + 2π(R - 1) = 4πR.

However, if we applied our previous formula, original perimeter (2πR) + 2π*1 = 2π(R + 1), which only accounts for the outer boundary. But clearly, in this case, the perimeter is double that. So where is the error?

Ah, because in the case of the circle, S is not just the buffer zone, but the buffer zone plus the interior. However, the buffer zone already includes the interior of the circle up to a certain point. So the perimeter should account for both the inner and outer edges. Therefore, the formula original perimeter + 2πr is only applicable when the buffer zone does not include the interior of the original shape.

Wait, so when does the buffer zone include the interior? It includes the interior of the original shape if the original shape's inradius is less than r. For a circle, the inradius is equal to the radius. So if the original circle has radius R, and the buffer zone is points within 1 unit of the perimeter (which is the circle), then the buffer zone includes all points within R + 1 of the center (the annulus from R - 1 to R + 1). But the interior of the circle is points within R of the center. Therefore, the union is the annulus from R - 1 to R + 1. So the perimeter is two circles.

But in the previous case of the nonagon, since the inradius is about 2.747, which is greater than 1, the buffer zone (points within 1 unit of the perimeter) does not include the entire interior of the nonagon. Therefore, the union S is the buffer zone (which is the region within 1 unit of the perimeter) plus the interior of the nonagon. Therefore, S is the interior of the nonagon union the buffer zone, which is the region:

- Inside the nonagon (which is not entirely in the buffer zone)

- Plus the buffer zone around the nonagon.

Therefore, the boundary of S would consist of:

- The outer edge of the buffer zone (the offset curve)

- The inner edge of the nonagon where the buffer zone starts.

Wait, but the buffer zone is points within 1 unit of the perimeter. So the inner edge of the buffer zone is the original perimeter. However, since the interior of the nonagon is also included in S, the boundary of S is:

- The outer edge of the buffer zone (offset curve)

- The inner edge where the buffer zone meets the interior of the nonagon.

But this inner edge is a closed curve, which is the original perimeter of the nonagon. But the original perimeter is a boundary between the interior and the buffer zone. Wait, no. The original perimeter is part of the buffer zone. Therefore, if you have the interior of the nonagon and the buffer zone, the boundary of S would be the outer part of the buffer zone and the inner part of the nonagon?

This is getting too tangled. Let me think for the circle case.

If S is the union of the interior of the circle and the buffer zone (points within 1 unit of the perimeter), then for R > 1, S is the annulus from R - 1 to R + 1. The boundary of S is the two circles: R - 1 and R + 1.

But in this case, the "inner" boundary (R - 1) is a part of the buffer zone. Because the buffer zone is points within 1 unit of the perimeter (the original circle), which for points inside, is from R - 1 to R. So the inner boundary of the buffer zone is R - 1, while the outer boundary is R + 1.

However, if we also include the interior of the circle, which is from 0 to R, then the union is from R - 1 to R + 1. Hence the perimeter is two circles.

But if R < 1, then the buffer zone includes the entire interior of the circle (since R - 1 would be negative, so the buffer zone includes everything from 0 to R + 1). Hence, S is from 0 to R + 1, and the perimeter is just the outer circle.

Therefore, in general, for a circle, the perimeter of S is:

- 2π(R + 1) if R ≤ 1 (buffer zone includes the interior)

- 2π(R - 1) + 2π(R + 1) = 4πR if R > 1 (perimeter includes both inner and outer boundaries)

Therefore, for the nonagon, since its inradius is ~2.747 > 1, then S is the union of interior and buffer zone, which would be similar to the circle case, where the perimeter of S would include both the inner boundary (from the original nonagon's perimeter) and the outer boundary (the offset curve).

But how does that work for the nonagon?

If we think of S as the interior of the nonagon union the buffer zone, which is points within 1 unit of the perimeter. For points inside the nonagon, they are in S if they are within 1 unit of the perimeter, or else they are in the interior.

But for the inradius being greater than 1, the interior of the nonagon is not entirely within the buffer zone. Therefore, S is the set of points inside the nonagon or within 1 unit of the perimeter. So this is the union of two regions:

1. The interior of the nonagon (which extends from the center out to the perimeter)

2. The buffer zone around the perimeter, extending out 1 unit

Therefore, the boundary of S would consist of:

- The outer edge of the buffer zone (the offset curve)

- The inner edge of the buffer zone, which is the original perimeter of the nonagon

Wait, no. The original perimeter is part of the buffer zone. But the interior of the nonagon is also part of S. Therefore, the boundary of S would be:

- The outer edge of the buffer zone (offset curve)

- The inner edge of the buffer zone, which is the original perimeter, but only where the buffer zone starts. But since the interior is also part of S, the original perimeter is separating the interior from the buffer zone? Not really, because the buffer zone includes the original perimeter.

This is very confusing. Maybe the safest way is to think in terms of the Minkowski sum.

The set S is the union of the interior of the nonagon and the buffer zone. However, if the buffer zone is the Minkowski sum of the perimeter with a disk of radius 1. Wait, but no. The buffer zone is all points within 1 unit of the perimeter.

But the union of the interior and the buffer zone is the Minkowski sum of the nonagon with a disk of radius 1, because the Minkowski sum includes all points within 1 unit of any point in the nonagon. Which is the same as the interior of the nonagon union the points within 1 unit of the perimeter. Wait, no. The Minkowski sum of the nonagon with a disk of radius 1 is all points within 1 unit of the nonagon. This includes the interior of the nonagon and the buffer zone around it. Therefore, S is precisely the Minkowski sum, so its perimeter is the original perimeter + 2πr.

But in the circle example, if the original shape is a circle of radius R, then the Minkowski sum with a disk of radius 1 is a circle of radius R + 1, with perimeter 2π(R + 1). However, earlier I thought that when R > 1, S would have two boundaries, but that was a mistake.

Wait, no! If S is the Minkowski sum of the original shape with a disk, then it is the set of all points within 1 unit of the original shape. Which for a circle of radius R, would be a circle of radius R + 1. The perimeter is 2π(R + 1). But earlier, I thought that if you take the union of the interior of the circle and the buffer zone, it would be the annulus, but that's incorrect.

Let me clarify. The Minkowski sum of the original circle with a disk of radius 1 is all points within 1 unit of the original circle. Which is the region between R - 1 and R + 1 if R > 1. But this is the same as the set of points within 1 unit of the original circle. However, the interior of the original circle is already within this region. Therefore, the Minkowski sum is exactly the set of points within 1 unit of the original circle, which is the annulus between R - 1 and R + 1. But the perimeter of this region is the outer circle (R + 1) and the inner circle (R - 1). However, the Minkowski sum should be a single region, and its perimeter should be the outer boundary.

But this contradicts. Where is the mistake?

Wait, no. The Minkowski sum of a circle with a disk is another circle, not an annulus. Wait, no. If you take all points within 1 unit of the original circle, it's not just the Minkowski sum. Wait, the Minkowski sum of the original circle (which is a set) with a disk is the set of all points p + q where p is in the circle and q is in the disk. This results in a larger circle with radius R + 1. However, the set of points within 1 unit of the original circle is different.

Wait, the set of points within 1 unit of the original circle is the union of the original circle and the buffer zone around it. This is different from the Minkowski sum. The Minkowski sum is all points within 1 unit of the original circle's interior. Wait, no, the Minkowski sum is all points within 1 unit of the original circle.

Actually, the Minkowski sum of the original shape with a disk is the same as the set of points within 1 unit of the original shape. So for a circle, it would be an annulus if the original circle is considered as a 1-dimensional curve, but if the original shape is the 2-dimensional disk, then the Minkowski sum is the disk with radius R + 1.

But the problem says S is the union of the interior of the nonagon and points within 1 unit of the perimeter. This is different from the Minkowski sum.

The Minkowski sum of the nonagon with a disk is all points within 1 unit of any point in the nonagon, which includes the interior of the nonagon and the buffer zone around it. Therefore, S is the Minkowski sum. Therefore, the perimeter of S is the original perimeter + 2πr.

But in the circle example, if the original shape is a disk (2D), then the Minkowski sum with a disk is another disk, hence perimeter is 2π(R + 1). However, if the original shape is just the 1D perimeter (the circle), then the set of points within 1 unit of the perimeter is the annulus.

But in our problem, the original shape is the interior of the nonagon, and S is the union of the interior and the buffer. Therefore, it's like taking the interior (which is a 2D region) and the buffer (which is a 2D region), and taking their union. This is equivalent to taking the Minkowski sum of the closure of the nonagon (which is the nonagon itself) with a closed disk.

Wait, the closure of the nonagon is the nonagon itself, including the perimeter. Therefore, the Minkowski sum of the nonagon with a closed disk of radius 1 is all points within 1 unit of the nonagon, which includes the interior of the nonagon and the buffer zone. Therefore, S is the Minkowski sum.

Therefore, for the circle example, if the original shape is a closed disk (2D), the Minkowski sum with a disk is another closed disk, perimeter 2π(R + 1). If the original shape is just the 1D perimeter (circle), then the set of points within 1 unit of the perimeter is an annulus, with perimeter 2π(R + 1) + 2π(R - 1) = 4πR.

Therefore, the key is whether the original shape is 1D or 2D. In our problem, the original shape is the interior of the nonagon, which is a 2D region. Therefore, the Minkowski sum is a 2D region, and its perimeter is original perimeter + 2πr.

Therefore, returning to the nonagon: original perimeter is 18, r is 1, so perimeter of S is 18 + 2π.

But in the circle example, if the original shape is the 2D disk, the perimeter of S is 2π(R + 1), which is original perimeter (2πR) + 2πr.

Therefore, this supports the formula.

Therefore, I think the correct answer is 18 + 2π.

However, to ensure, let me think about the circle case again with the formula. If the original shape is the disk, which has perimeter 2πR, and the Minkowski sum with a disk of radius 1 has perimeter 2π(R + 1) = 2πR + 2π*1, which matches the formula.

Therefore, this confirms the formula. Therefore, the answer for the nonagon is 18 + 2π.

But let me check the problem statement once again to ensure I haven't misread.

"Let S be 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 of the nonagon. What, in units, is the perimeter of S?"

Yes, so S is the union of the interior and the buffer. Since the buffer is all points within 1 unit of the perimeter, and the interior is the set of points inside the nonagon. This is equivalent to the Minkowski sum of the nonagon (as a 2D region) with a disk of radius 1. Therefore, the perimeter of S is original perimeter + 2π*1 = 18 + 2π.

Therefore, the answer is 18 + 2π.

But the problem might expect a numerical value, but since π is involved, it's likely to leave it in terms of π.

Therefore, the perimeter of S is 18 + 2π units.

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

To determine the **perimeter of** the region $ S $, we are given a regular nonagon of side length 2 units, and we are to consider the union of:

1. **All points inside the nonagon**, and
2. **All points less than 1 unit away from the perimeter of the nonagon**.

Let's unpack these two components geometrically.

---

### Understanding the Shape of $ S $

- The **interior** of the nonagon is simply its **internal region**.
- The region **less than 1 unit away from the perimeter** forms a "buffer" zone (also known as an *offset region*) around the nonagon's boundary.
- The union of these two sets is exactly the **Minkowski sum** of the nonagon with a closed disk of radius 1.

This gives us the set $ S $ as the region consisting of:
- All points **within 1 unit** of the nonagon's perimeter (the buffer), and
- All points **inside** the nonagon (the interior itself).

Hence, **$ S $ is the union of the interior and the offset buffer** around the nonagon. Importantly, this is **equivalent** to all points within 1 unit of the **nonagon as a whole** (as a 2D object), which makes $ S $ the **inflation** (offset) of the nonagon by 1 unit in all directions.

---

### Calculating the Perimeter of $ S $

For any convex polygon, when you take the **Minkowski sum** with a closed disk of radius $ r $, the resulting shape has:
- The same number of sides (in this case, 9) as the original polygon,
- Its **original perimeter**,
- Plus a **circular component** that replaces the corners.

The key geometric result is that the **new perimeter** equals the **original perimeter** of the polygon *plus* the **circumference of a circle with radius $ r $** (i.e., $ 2\pi r $).

For a regular nonagon (9 sides), with side length 2 units:
- The original perimeter is:
  $$
  P = 9 \times 2 = 18 \text{ units}
  $$
- The added "circular" component from the buffer zone is:
  $$
  \text{Added perimeter} = 2\pi \times 1 = 2\pi \text{ units}
  $$

Thus, the **total perimeter of $ S $** is:
$$
\text{Perimeter of } S = 18 + 2\pi
$$

---

### Final Answer

$$
\boxed{18 + 2\pi}
$$

Solution:
$S$ looks like a nonagon with slightly rounded corners.  We draw adjacent sides of the nonagon and look at the boundary of $S$:

[asy]
size(200);
draw((-7.66,-6.43)--(0,0)--(10,0)--(17.66,-6.43));
draw((0,5)--(10,5),blue); draw((13.21,3.83)--(20.87,-2.60),blue);
draw(Arc((10,0),5,50,90),red); draw(Arc((0,0),5,90,130),red);
draw((10,0)--(10,5),dashed); draw((0,0)--(0,5),dashed);
draw((10,0)--(13.21,3.83),dashed);
label("2",(5,0),S); label("1",(10,2.5),W);
draw((-3.21,3.83)--(-10.87,-2.60),blue);
draw((-3.21,3.83)--(0,0),dashed);
[/asy] We can split the portion of $S$ that is outside the nonagon into 9 rectangles and 9 circle sectors, thereby breaking the perimeter of $S$ into alternating straight lines (colored blue above) and curved arcs (colored red above).  The perimeter of $S$ is comprised of nine blue lines and nine red arcs.

Each rectangle has side lengths 1 and 2, so each blue line is 2 units long and the total length of the blue portion of the perimeter is $2\cdot 9 = 18$ units.

Around each vertex of the nonagon, an interior angle, two right angles, and an angle of the circular sector add up to 360 degrees.  The angles inside a nonagon each measure $180(9-2)/9=140$ degrees.  Thus, each circular sector angle measures $360-90-90-140=40$ degrees.  Each sector has radius 1 and arc length $\frac{40^\circ}{360^\circ}(2)(\pi)(1)=\frac{1}{9}(2\pi)$, so nine of these sectors have total arc length $2\pi$.  Thus the total length of the red portion of the perimeter is $2\pi$ units.  (Notice that this is equal to the perimeter of a circle with radius 1, which is what the nine sectors add up to.)

Finally, the perimeter of $S$ is $\boxed{18+2\pi}$ units.