Explanation:
Okay, so I need to find the perimeter of this set S. Let me try to understand what S is. It 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, essentially, S is the original nonagon plus a sort of "buffer" zone around it, right? Like, if you imagine taking the nonagon and then expanding it outward by 1 unit in all directions, but only up to the points that are within 1 unit of the perimeter. Wait, but actually, since it's the union of the interior of the nonagon and all points within 1 unit of the perimeter, that would be like taking the original nonagon and then adding a 1-unit wide strip around it. But actually, when you take all points within 1 unit of the perimeter, that's like taking the Minkowski sum of the perimeter with a disk of radius 1. But since the original nonagon is already included, maybe S is the original nonagon expanded by 1 unit? Hmm, but maybe not exactly, because when you take all points within 1 unit of the perimeter, that includes the original nonagon? Wait, no. If you take all points within 1 unit of the perimeter, that would include the original perimeter and the area inside the nonagon? Wait, no. Wait, if you take all points less than 1 unit away from the perimeter, that would include points inside the nonagon? Let me think.

Suppose you have a point inside the nonagon. Is it less than 1 unit away from the perimeter? Well, if the nonagon has a certain radius, then points near the center would be more than 1 unit away from the perimeter. So actually, the set of points less than 1 unit away from the perimeter would be the area between the original nonagon and a larger nonagon that's expanded outward by 1 unit. But also, the original nonagon is included in S because S is the union of the interior of the nonagon and the points within 1 unit of the perimeter. Wait, but if you take the union, then the points within 1 unit of the perimeter would include the original perimeter and the area near it. However, the original interior of the nonagon is already included in the union. Wait, but actually, if you take all points within 1 unit of the perimeter, that would include points inside the nonagon that are within 1 unit of the perimeter. So, for example, if you have a point near the edge of the nonagon, it's within 1 unit of the perimeter. But points near the center might not be. Therefore, the union of the interior of the nonagon and the points within 1 unit of the perimeter would be the same as the set of points within 1 unit of the perimeter union the interior. Wait, but the interior is already included in the points within 1 unit of the perimeter? No, not necessarily. If the nonagon is large enough, then points near the center are more than 1 unit away from the perimeter. Therefore, the union would be the original nonagon plus a 1-unit buffer around it. Therefore, S is like a larger nonagon with rounded edges? Or is it?

Wait, actually, when you take all points within 1 unit of the perimeter of the nonagon, that's called an offset curve or a parallel curve. For a polygon, offsetting it outward by 1 unit would create a new polygon with the same number of sides, but each side is offset outward, and the corners are rounded with circular arcs. However, if you take the union of the original polygon and the offset region, then S would be the original polygon plus the offset region. But since the original polygon is already inside the offset region? Wait, no. If you take all points within 1 unit of the perimeter, that includes points inside the original polygon that are within 1 unit of the perimeter. So, actually, the union of the original polygon and the points within 1 unit of the perimeter is just the set of points within 1 unit of the perimeter. Wait, no. Because the original polygon is entirely within the set of points within 1 unit of its perimeter? No, that's not true. For example, take a point in the center of the polygon. If the polygon is large enough, that point is more than 1 unit away from the perimeter. Therefore, the union of the original polygon and the points within 1 unit of the perimeter is actually the original polygon expanded by 1 unit in all directions, but with the original polygon's interior included. Wait, but if you take the union, then it's the same as the set of points within 1 unit of the perimeter union the original polygon. But since the original polygon is entirely within the set of points within 1 unit of the perimeter? No, that's not true. Wait, if you take a point inside the polygon, is it within 1 unit of the perimeter? Only if it's within 1 unit of the perimeter. So, the union would be the original polygon plus the points within 1 unit of the perimeter. But actually, the points within 1 unit of the perimeter include the original perimeter and the area near it. However, the original polygon's interior is not entirely included. So, the union would be the original polygon plus the buffer zone. Therefore, S is like the original nonagon with a 1-unit wide strip around it. But how does that look?

Alternatively, maybe S is the Minkowski sum of the nonagon with a closed disk of radius 1. The Minkowski sum would be all points that are within 1 unit of the nonagon. Which would be the original nonagon expanded by 1 unit in all directions. However, the perimeter of this Minkowski sum would be the original perimeter plus twice the length of the perimeter? Wait, no. Wait, when you take the Minkowski sum of a polygon with a disk, the perimeter becomes the original perimeter plus twice the length of the perimeter? No, that doesn't make sense. Wait, actually, when you take the Minkowski sum of a polygon with a disk, the resulting shape is called an offset polygon. The perimeter of the offset polygon is equal to the original perimeter plus 2πr, where r is the radius of the disk. Wait, is that true? Let me think. If you offset a polygon outward by radius r, then each side is extended outward by r, but the corners become rounded with quarter-circles? Wait, no. For a polygon with n sides, when you offset it outward by radius r, each original side becomes a side offset outward, and each original vertex becomes a circular arc with radius r. The angle of each arc is equal to the exterior angle of the polygon. For a regular polygon, the exterior angles sum up to 2π. Therefore, the total length added by the arcs would be r times the sum of the exterior angles, which is 2πr. Therefore, the perimeter of the offset polygon would be the original perimeter plus 2πr. Wait, but the original sides are also extended. Wait, no. Wait, when you offset a polygon outward by radius r, the length of each side is increased by 2r * tan(π/n), or something like that? Wait, maybe not. Let me think again.

Imagine a polygon with side length L. When you offset it outward by radius r, each side is moved outward, but the distance between the original side and the offset side is r. However, the length of the offset side is not necessarily the same as the original side. Wait, actually, if you offset a straight line segment outward by distance r, keeping it parallel, then the length of the offset segment is the same as the original. However, when you offset a polygon, the sides are offset outward, but the corners are replaced by circular arcs. Therefore, the perimeter of the offset polygon would be equal to the original perimeter plus the circumference of a circle with radius r. Wait, because each corner is replaced by a circular arc. For a regular polygon, the sum of the exterior angles is 2π. Therefore, if each corner is replaced by an arc of angle equal to the exterior angle, then the total length added by the arcs is r * sum(exterior angles) = r * 2π. Therefore, the total perimeter of the offset polygon is original perimeter + 2πr. However, the original sides are still there? Wait, no. Wait, if you offset the polygon outward, the original sides are moved outward, but the sides themselves are still straight. However, the corners are replaced by circular arcs. Therefore, the perimeter of the offset polygon is equal to the original perimeter plus 2πr. Wait, but the original sides are not changed in length? Wait, no. If you offset a polygon outward, the length of each side remains the same? Wait, no. If you have a polygon and you offset each side outward by distance r, keeping them parallel, then the length of each side remains the same. However, the corners are replaced by circular arcs. Therefore, the total perimeter would be original perimeter + 2πr. Therefore, if S is the Minkowski sum of the nonagon with a disk of radius 1, then its perimeter would be original perimeter + 2π*1. But wait, the problem says S is the union of the interior of the nonagon and the points within 1 unit of the perimeter. Which is exactly the Minkowski sum of the nonagon with a disk of radius 1. Therefore, the perimeter of S would be original perimeter + 2π*1. Therefore, if I can compute the original perimeter of the nonagon, then add 2π, that would be the answer. But let me check if this is correct.

Wait, but the original nonagon has side length 2. Therefore, its perimeter is 9*2=18 units. Then, adding 2π*1, the perimeter of S would be 18 + 2π? But wait, that seems too straightforward. However, I need to make sure that the Minkowski sum is indeed the same as the union of the interior and the points within 1 unit of the perimeter. Let me think again. The Minkowski sum of the nonagon with a disk of radius 1 is all points that are within 1 unit of the nonagon. Which includes the interior of the nonagon and all points within 1 unit of the perimeter. Therefore, yes, S is exactly that. Therefore, the perimeter of S is original perimeter + 2π*1. Therefore, 18 + 2π. But wait, the problem says "the perimeter of S". However, I need to make sure that when you take the Minkowski sum, the perimeter is indeed original perimeter + 2πr. Let me think of a simple case. Take a square with side length 2. Its perimeter is 8. If we take the Minkowski sum with a disk of radius 1, the resulting shape is a square with rounded corners. The perimeter would be the original perimeter (8) plus the circumference of a circle with radius 1, but wait, no. Wait, each corner is replaced by a quarter-circle. There are four corners, so total circumference added is 4*(1/4)*2π*1 = 2π. Therefore, total perimeter is 8 + 2π. Which matches the formula original perimeter + 2πr. Therefore, same for a polygon. Therefore, for a regular nonagon, the perimeter of S would be original perimeter + 2π*1. Therefore, 9*2 + 2π = 18 + 2π. Therefore, the answer is 18 + 2π. But wait, the problem says "the perimeter of S". However, let me check again.

Wait, but maybe there's a mistake here. Because when you take the union of the interior of the nonagon and the points within 1 unit of the perimeter, is that the same as the Minkowski sum? Let me think. The Minkowski sum of the nonagon with a disk of radius 1 is all points that are within 1 unit of the nonagon. Which includes the interior of the nonagon and the points within 1 unit of the perimeter. Therefore, yes, S is exactly that. Therefore, the perimeter of S is the perimeter of the Minkowski sum, which is original perimeter + 2π*1. Therefore, 18 + 2π. However, let me check with another example. Suppose the nonagon is very small, such that the points within 1 unit of the perimeter include the entire original nonagon. Wait, but if the nonagon is small, then the points within 1 unit of the perimeter would include the original nonagon and a buffer around it. However, if the nonagon is so small that its radius is less than 1, then the Minkowski sum would be a larger shape. But regardless, the perimeter formula still holds. Therefore, maybe the answer is 18 + 2π. However, the problem says "the perimeter of S". Wait, but maybe I need to think about the shape again. When you take the Minkowski sum, the perimeter is the original perimeter plus 2πr. But in the case of the square, we saw that. However, in the case of a polygon, when you offset it, the sides are extended outward, but the corners are replaced by circular arcs. Therefore, the perimeter is original perimeter + 2πr. Therefore, in this case, 18 + 2π. However, let me check if the original perimeter is 18. The nonagon has 9 sides, each of length 2, so yes, 9*2=18. Therefore, the perimeter of S is 18 + 2π. But wait, 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 of these two sets. However, the set of all points inside the nonagon is already included in the set of points less than 1 unit away from the perimeter? No, not necessarily. For example, if the nonagon is large enough, then points near the center are more than 1 unit away from the perimeter. Therefore, the union is the set of points inside the nonagon plus the points within 1 unit of the perimeter. However, the points within 1 unit of the perimeter include the perimeter itself and the area near it. Therefore, the union is the set of points within 1 unit of the perimeter union the interior of the nonagon. But if the nonagon is large enough, then the interior of the nonagon is not entirely within the points within 1 unit of the perimeter. Therefore, the union is the Minkowski sum of the nonagon with a disk of radius 1. Therefore, the perimeter is original perimeter + 2π*1. Therefore, 18 + 2π. However, maybe I need to check if the original nonagon's perimeter is entirely included in the Minkowski sum. Yes, because the perimeter of the nonagon is the set of points at distance 0 from the nonagon, which is included in the set of points within 1 unit of the nonagon. Therefore, the union is indeed the Minkowski sum. Therefore, the perimeter is 18 + 2π. However, let me check again with the square example. Suppose the square has side length 2, perimeter 8. The Minkowski sum with radius 1 would have perimeter 8 + 2π. But if you take the union of the interior of the square and the points within 1 unit of the perimeter, that is the same as the Minkowski sum. Therefore, the perimeter is 8 + 2π. Therefore, same logic applies here. Therefore, answer is 18 + 2π. However, the problem says "the perimeter of S". Wait, but maybe there's a mistake here. Because when you take the Minkowski sum, the perimeter is not just original perimeter plus 2πr. Wait, let me think again. For the square, when you offset it by radius r, the perimeter becomes original perimeter + 2πr. But in reality, when you offset a square, each side is extended outward, but the corners are replaced by quarter-circles. Therefore, the total perimeter is original perimeter (which is 4*side length) plus the circumference of a circle with radius r. Wait, because each corner contributes a quarter-circle, so four corners contribute a full circle. Therefore, total perimeter is original perimeter + 2πr. Therefore, same as before. Therefore, for the nonagon, same thing. Therefore, perimeter of S is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, answer is 18 + 2π. However, the problem says "the perimeter of S". Wait, but maybe I need to check if the original nonagon's perimeter is entirely on the boundary of S. Because when you take the union of the interior and the points within 1 unit of the perimeter, the boundary of S would consist of the original perimeter offset outward by 1 unit, plus the circular arcs at the corners. Therefore, the perimeter of S is the same as the perimeter of the Minkowski sum, which is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, the answer is 18 + 2π. However, let me check if 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 of these two sets. However, the set of points less than 1 unit away from the perimeter includes the perimeter itself and the area within 1 unit of it. The set of points inside the nonagon includes the interior. Therefore, the union is the set of points that are either inside the nonagon or within 1 unit of the perimeter. However, the points within 1 unit of the perimeter include the perimeter and the area near it. However, the interior of the nonagon is not entirely included in the points within 1 unit of the perimeter. Therefore, the union is the set of points within 1 unit of the perimeter union the interior of the nonagon. Which is the same as the set of points within 1 unit of the nonagon. Therefore, S is the Minkowski sum. Therefore, perimeter is original perimeter + 2π*1. Therefore, 18 + 2π. However, maybe I need to think about the actual shape. If you have a nonagon, and you take all points within 1 unit of its perimeter, that would create a sort of "thickened" nonagon. The perimeter of this thickened nonagon would consist of the original sides offset outward by 1 unit, connected by circular arcs at each corner. Therefore, the length contributed by the sides would be the same as the original perimeter, but each side is offset outward. Wait, but if you offset each side outward by 1 unit, the length of each side remains the same? Wait, no. Wait, if you have a straight side of length L, and you offset it outward by distance r, keeping it parallel, then the length of the offset side is still L. However, the distance between the original side and the offset side is r. However, when you do this for all sides, the corners become rounded. Therefore, the total perimeter would be original perimeter + 2πr. Therefore, same as before. Therefore, the perimeter of S is 18 + 2π. Therefore, the answer is 18 + 2π. However, let me check if there's a different interpretation. Maybe the problem is not about the Minkowski sum, but something else. For example, maybe the set S is the union of the interior of the nonagon and the set of points within 1 unit of the perimeter. However, the set of points within 1 unit of the perimeter is a region that includes the perimeter and a buffer around it. However, the union with the interior would be the same as the set of points within 1 unit of the nonagon. Therefore, same as before. Therefore, perimeter is 18 + 2π. However, maybe the problem is in 3D? No, it's in 2D. Therefore, I think my reasoning is correct. Therefore, the perimeter of S is 18 + 2π. However, let me check if the original nonagon's side length is 2. Wait, the problem says "a regular nonagon with side length 2 units". Therefore, perimeter is 9*2=18. Therefore, adding 2π*1, the perimeter of S is 18 + 2π. Therefore, the answer is 18 + 2π. However, let me check if there's a different approach. For example, maybe the perimeter of S is just the perimeter of the original nonagon plus the perimeter of the buffer. But the buffer is a 1-unit wide strip around the nonagon. However, the perimeter of the buffer would be... Wait, if you have a buffer around a polygon, the perimeter is not straightforward. However, as we discussed earlier, it's original perimeter + 2πr. Therefore, same result. Therefore, I think the answer is 18 + 2π. However, let me check if the problem says "the set of all points inside... and the set of all points less than 1 unit away from a point on the perimeter". So, the union. Therefore, the boundary of S would be the boundary of the buffer region. Because the interior of the nonagon is already included in the buffer region? Wait, no. Wait, the buffer region is all points within 1 unit of the perimeter. The interior of the nonagon is not entirely within the buffer region. Therefore, the union is the buffer region plus the interior. Therefore, the boundary of S would be the boundary of the buffer region, but with the parts that are inside the original nonagon? Wait, no. Wait, the buffer region is all points within 1 unit of the perimeter. The interior of the nonagon is a separate set. Therefore, the union would have a boundary that is the boundary of the buffer region, but also the original perimeter? Wait, no. Wait, the buffer region includes the original perimeter. Therefore, the union of the interior of the nonagon and the buffer region is the same as the buffer region union the interior. But since the buffer region already includes the perimeter and the area near it, and the interior is the area inside the nonagon. However, the buffer region includes points inside the nonagon that are within 1 unit of the perimeter. Therefore, the union is the set of points within 1 unit of the nonagon. Therefore, the boundary of S is the same as the boundary of the buffer region, which is the original perimeter offset outward by 1 unit, with rounded corners. Therefore, the perimeter of S is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, the answer is 18 + 2π. However, let me check if there's a different way to compute it. For example, if I think about the perimeter of S as the original perimeter plus the perimeter contributed by the buffer. However, the buffer is a 1-unit wide region around the nonagon. However, the perimeter of the buffer would be... Well, if you imagine the buffer as a sort of "annular" region around the nonagon, but since the nonagon is a polygon, the buffer has rounded corners. Therefore, the perimeter of the buffer is equal to the original perimeter plus 2π*1. Therefore, same result. Therefore, I think the answer is 18 + 2π. Therefore, I think that is the answer. However, let me check if the problem says "the set of all points inside... and the set of all points less than 1 unit away from a point on the perimeter". Therefore, the union. So, if you imagine the original nonagon, and then you take all points within 1 unit of the perimeter, which is like a 1-unit wide strip around the nonagon. However, the union of the original nonagon and this strip would be the original nonagon plus the strip. However, the strip is already overlapping with the original nonagon near the edges. Therefore, the perimeter of S would be the perimeter of the strip, but with the parts that are inside the original nonagon removed? Wait, no. Wait, the perimeter of S is the boundary of the union. The union is the original nonagon plus the buffer. However, the buffer includes the original perimeter. Therefore, the boundary of the union would be the boundary of the buffer, because the original nonagon is entirely inside the buffer? No, because the buffer is all points within 1 unit of the perimeter. Therefore, the original nonagon is not entirely within the buffer. For example, the center of the nonagon is not within 1 unit of the perimeter if the nonagon is large enough. Therefore, the union is the original nonagon plus the buffer. Therefore, the boundary of the union would consist of two parts: the parts of the buffer that are outside the original nonagon, and the parts of the original nonagon that are not overlapped by the buffer. However, the original nonagon's perimeter is entirely overlapped by the buffer, because the buffer includes all points within 1 unit of the perimeter. Therefore, the boundary of the union would be the boundary of the buffer. Therefore, the perimeter of S is the same as the perimeter of the buffer, which is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, the answer is 18 + 2π. However, let me check if there's a different interpretation. For example, maybe the problem is considering the set S as the union of the interior of the nonagon and the set of points within 1 unit of the perimeter. Therefore, the boundary of S would be the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, because the buffer region is points within 1 unit of the perimeter. The interior of the nonagon is not entirely within the buffer region. Therefore, the union is the buffer region union the interior. Therefore, the boundary of the union is the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, that's not true. Wait, if you take the union of the interior of the nonagon and the buffer region, then the buffer region includes the perimeter and the area near it. The interior of the nonagon is the area inside the nonagon. Therefore, the union is the set of points that are either inside the nonagon or within 1 unit of the perimeter. Therefore, the boundary of this union would be the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, because the buffer region is points within 1 unit of the perimeter. The interior of the nonagon is points that are inside the nonagon. Therefore, the union is the set of points that are inside the nonagon or within 1 unit of the perimeter. Therefore, the boundary of this union would be the boundary of the buffer region, but also the parts of the original nonagon that are not overlapped by the buffer. Wait, but the buffer region includes the perimeter of the nonagon. Therefore, the union is the buffer region union the interior of the nonagon. However, the buffer region already includes the perimeter and the area near it. Therefore, the union is the buffer region union the interior. But since the buffer region includes the perimeter and the area near it, and the interior is the area inside the nonagon. Therefore, the union is the set of points within 1 unit of the perimeter union the interior. Therefore, the boundary of the union is the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, that's not true. For example, take a point in the center of the nonagon. If the nonagon is large enough, that point is more than 1 unit away from the perimeter, so it's not in the buffer region. Therefore, the union is the buffer region union the interior. Therefore, the boundary of the union is the boundary of the buffer region, plus the boundary of the interior? No, because the interior is entirely inside the buffer region? No, the interior is not entirely inside the buffer region. Therefore, the union is the buffer region union the interior. Therefore, the boundary of the union is the boundary of the buffer region, but with the parts that are inside the interior of the nonagon removed? Wait, this is getting confusing. Maybe I need to think of it as the union of two regions: the interior of the nonagon and the buffer region. The buffer region is all points within 1 unit of the perimeter. The interior of the nonagon is all points inside the nonagon. Therefore, the union is the set of points that are either inside the nonagon or within 1 unit of the perimeter. Therefore, the boundary of this union would be the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, because the buffer region is points within 1 unit of the perimeter. The interior of the nonagon is points inside the nonagon. Therefore, the union is the set of points that are inside the nonagon or within 1 unit of the perimeter. Therefore, the boundary of the union is the set of points that are either on the boundary of the buffer region or on the boundary of the interior of the nonagon. However, the boundary of the interior of the nonagon is the perimeter of the nonagon. But the buffer region includes the perimeter of the nonagon. Therefore, the union's boundary is the boundary of the buffer region. Because the perimeter of the nonagon is already included in the buffer region. Therefore, the boundary of the union is the same as the boundary of the buffer region. Therefore, the perimeter of S is the same as the perimeter of the buffer region, which is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, the answer is 18 + 2π. However, let me check if there's a different way to think about it. For example, if I imagine the perimeter of S as the original perimeter plus the perimeter contributed by the buffer. However, the buffer is a 1-unit wide region around the nonagon. However, the perimeter of the buffer is not just the original perimeter plus something. Wait, but earlier analysis with the square shows that it's original perimeter + 2πr. Therefore, same here. Therefore, answer is 18 + 2π. Therefore, I think this is correct. Therefore, I think the answer is 18 + 2π. However, let me check if the problem says "the set of all points inside... and the set of all points less than 1 unit away from a point on the perimeter". Therefore, the union. Therefore, the perimeter of S is the same as the perimeter of the Minkowski sum, which is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, the answer is 18 + 2π. However, let me check if the problem is from an Olympiad or something, and maybe the answer is different. Wait, but I don't have that context. Alternatively, maybe I need to calculate the perimeter of S by considering the original nonagon's perimeter and adding the perimeter of the buffer. However, the buffer is a 1-unit wide region around the nonagon. However, the perimeter of the buffer is not straightforward. However, as we discussed, it's original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, I think the answer is 18 + 2π. Therefore, I will go with that.

**Final Answer**
\boxed{18 + 2\pi}

To determine the perimeter of the set $ S $, we begin by understanding its definition:

> $ S $ is 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.

---

### Step 1: Understanding the Shape of $ S $

This union can be interpreted geometrically as the **Minkowski sum** of the nonagon with a closed disk of radius 1. In other words, $ S $ is the set of all points that are within 1 unit of the nonagon, including:
- The interior of the nonagon,
- The perimeter of the nonagon,
- A 1-unit wide "buffer" zone around the nonagon.

This shape is known as the **offset polygon** or **buffered polygon**. It has:
- Straight sides that are offset outward from the original nonagon,
- Rounded corners formed by circular arcs of radius 1.

---

### Step 2: Perimeter of the Offset Polygon

For a regular polygon with $ n $ sides, when you offset it outward by a distance $ r $, the resulting shape has:
- The same number of sides as the original polygon,
- Each side is offset outward, but the **length of each side remains unchanged** (since the offset is perpendicular to the side),
- The **corners are replaced by circular arcs** of radius $ r $, with each arc corresponding to the exterior angle of the polygon.

The total length of the arcs is:
$$
\text{Total arc length} = r \times \text{sum of exterior angles}
$$
For any convex polygon, the sum of the exterior angles is $ 2\pi $, so:
$$
\text{Total arc length} = r \times 2\pi
$$

Thus, the **perimeter of the offset polygon** is:
$$
\text{Perimeter of } S = \text{Original perimeter} + 2\pi r
$$

---

### Step 3: Apply to the Given Nonagon

- The nonagon is regular with side length 2.
- It has 9 sides, so its original perimeter is:
  $$
  9 \times 2 = 18
  $$
- The offset is 1 unit, so $ r = 1 $.

Therefore, the perimeter of $ S $ is:
$$
18 + 2\pi
$$

---

### Final Answer

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

[Assistant]
Explanation:
Okay, so I need to find the perimeter of this set S. Let me try to understand what S is. It 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, essentially, S is the original nonagon plus a sort of "buffer" zone around it, right? Like, if you imagine taking the nonagon and then expanding it outward by 1 unit in all directions, but only up to the points that are within 1 unit of the perimeter. Wait, but actually, since it's the union of the interior of the nonagon and all points within 1 unit of the perimeter, that would be like taking the original nonagon and then adding a 1-unit wide strip around it. But actually, when you take all points within 1 unit of the perimeter, that's like taking the Minkowski sum of the perimeter with a disk of radius 1. But since the original nonagon is already included, maybe S is the original nonagon expanded by 1 unit? Hmm, but maybe not exactly, because when you take all points within 1 unit of the perimeter, that includes the original nonagon? Wait, no. If you take all points within 1 unit of the perimeter, that would include the original perimeter and the area inside the nonagon? Wait, no. Wait, if you take all points less than 1 unit away from the perimeter, that would include points inside the nonagon? Let me think.

Suppose you have a point inside the nonagon. Is it less than 1 unit away from the perimeter? Well, if the nonagon has a certain radius, then points near the center would be more than 1 unit away from the perimeter. So actually, the set of points less than 1 unit away from the perimeter would be the area between the original nonagon and a larger nonagon that's expanded outward by 1 unit. But also, the original nonagon is included in S because S is the union of the interior of the nonagon and the points within 1 unit of the perimeter. Wait, but if you take the union, then the points within 1 unit of the perimeter would include the original perimeter and the area near it. However, the original interior of the nonagon is already included in the union. Wait, but actually, if you take all points within 1 unit of the perimeter, that would include points inside the nonagon that are within 1 unit of the perimeter. So, for example, if you have a point near the edge of the nonagon, it's within 1 unit of the perimeter. But points near the center might not be. Therefore, the union of the interior of the nonagon and the points within 1 unit of the perimeter would be the same as the set of points within 1 unit of the perimeter union the interior. Wait, but the interior is already included in the points within 1 unit of the perimeter? No, not necessarily. If the nonagon is large enough, then points near the center are more than 1 unit away from the perimeter. Therefore, the union would be the original nonagon plus a 1-unit buffer around it. Therefore, S is like a larger nonagon with rounded edges? Or is it?

Wait, actually, when you take all points within 1 unit of the perimeter of the nonagon, that's called an offset curve or a parallel curve. For a polygon, offsetting it outward by 1 unit would create a new polygon with the same number of sides, but each side is offset outward, and the corners are rounded with circular arcs. However, if you take the union of the original polygon and the offset region, then S would be the original polygon plus the offset region. But since the original polygon is already inside the offset region? Wait, no. If you take all points within 1 unit of the perimeter, that includes points inside the original polygon that are within 1 unit of the perimeter. So, actually, the union of the original polygon and the points within 1 unit of the perimeter is just the set of points within 1 unit of the perimeter. Wait, no. Because the original polygon is entirely within the set of points within 1 unit of its perimeter? No, that's not true. For example, take a point in the center of the polygon. If the polygon is large enough, that point is more than 1 unit away from the perimeter. Therefore, the union of the original polygon and the points within 1 unit of the perimeter is actually the original polygon expanded by 1 unit in all directions, but with the original polygon's interior included. Wait, but if you take the union, then it's the same as the set of points within 1 unit of the perimeter union the original polygon. But since the original polygon is entirely within the set of points within 1 unit of the perimeter? No, that's not true. Wait, if you take a point inside the polygon, is it within 1 unit of the perimeter? Only if it's within 1 unit of the perimeter. So, the union would be the original polygon plus the points within 1 unit of the perimeter. But actually, the points within 1 unit of the perimeter include the original perimeter and the area near it. However, the original polygon's interior is not entirely included. So, the union would be the original polygon plus the buffer zone. Therefore, S is like the original nonagon with a 1-unit wide strip around it. But how does that look?

Alternatively, maybe S is the Minkowski sum of the nonagon with a closed disk of radius 1. The Minkowski sum would be all points that are within 1 unit of the nonagon. Which would be the original nonagon expanded by 1 unit in all directions. However, the perimeter of this Minkowski sum would be the original perimeter plus twice the length of the perimeter? Wait, no. Wait, when you take the Minkowski sum of a polygon with a disk, the perimeter becomes the original perimeter plus twice the length of the perimeter? No, that doesn't make sense. Wait, actually, when you take the Minkowski sum of a polygon with a disk, the resulting shape is called an offset polygon. The perimeter of the offset polygon is equal to the original perimeter plus 2πr, where r is the radius of the disk. Wait, is that true? Let me think. If you offset a polygon outward by radius r, then each side is extended outward by r, but the corners become rounded with quarter-circles? Wait, no. For a polygon with n sides, when you offset it outward by radius r, each original side becomes a side offset outward, and each original vertex becomes a circular arc with radius r. The angle of each arc is equal to the exterior angle of the polygon. For a regular polygon, the exterior angles sum up to 2π. Therefore, the total length added by the arcs would be r times the sum of the exterior angles, which is 2πr. Therefore, the perimeter of the offset polygon would be the original perimeter plus 2πr. Wait, but the original sides are also extended. Wait, no. Wait, when you offset a polygon outward by radius r, the length of each side is increased by 2r * tan(π/n), or something like that? Wait, maybe not. Let me think again.

Imagine a polygon with side length L. When you offset it outward by radius r, each side is moved outward, but the distance between the original side and the offset side is r. However, the length of the offset side is not necessarily the same as the original side. Wait, actually, if you offset a straight line segment outward by distance r, keeping it parallel, then the length of the offset segment is the same as the original. However, when you offset a polygon, the sides are offset outward, but the corners are replaced by circular arcs. Therefore, the perimeter of the offset polygon would be equal to the original perimeter plus the circumference of a circle with radius r. Wait, because each corner is replaced by a circular arc. For a regular polygon, the sum of the exterior angles is 2π. Therefore, if each corner is replaced by an arc of angle equal to the exterior angle, then the total length added by the arcs is r * sum(exterior angles) = r * 2π. Therefore, the total perimeter of the offset polygon is original perimeter + 2πr. However, the original sides are still there? Wait, no. Wait, if you offset the polygon outward, the original sides are moved outward, but the sides themselves are still straight. However, the corners are replaced by circular arcs. Therefore, the perimeter of the offset polygon is equal to the original perimeter plus 2πr. Wait, but the original sides are not changed in length? Wait, no. If you offset a polygon outward, the length of each side remains the same? Wait, no. If you have a polygon and you offset each side outward by distance r, keeping them parallel, then the length of each side remains the same. However, the corners are replaced by circular arcs. Therefore, the total perimeter would be original perimeter + 2πr. Therefore, if S is the Minkowski sum of the nonagon with a disk of radius 1, then its perimeter would be original perimeter + 2π*1. But wait, the problem says S is the union of the interior of the nonagon and the points within 1 unit of the perimeter. Which is exactly the Minkowski sum of the nonagon with a disk of radius 1. Therefore, the perimeter of S would be original perimeter + 2π*1. Therefore, if I can compute the original perimeter of the nonagon, then add 2π, that would be the answer. But let me check if this is correct.

Wait, but the original nonagon has side length 2. Therefore, its perimeter is 9*2=18 units. Then, adding 2π*1, the perimeter of S would be 18 + 2π? But wait, that seems too straightforward. However, I need to make sure that the Minkowski sum is indeed the same as the union of the interior and the points within 1 unit of the perimeter. Let me think again. The Minkowski sum of the nonagon with a disk of radius 1 is all points that are within 1 unit of the nonagon. Which includes the interior of the nonagon and all points within 1 unit of the perimeter. Therefore, yes, S is exactly that. Therefore, the perimeter of S is original perimeter + 2π*1. Therefore, 18 + 2π. But wait, the problem says "the perimeter of S". However, I need to make sure that when you take the Minkowski sum, the perimeter is indeed original perimeter + 2πr. Let me think of a simple case. Take a square with side length 2. Its perimeter is 8. If we take the Minkowski sum with a disk of radius 1, the resulting shape is a square with rounded corners. The perimeter would be the original perimeter (8) plus the circumference of a circle with radius 1, but wait, no. Wait, each corner is replaced by a quarter-circle. There are four corners, so total circumference added is 4*(1/4)*2π*1 = 2π. Therefore, total perimeter is 8 + 2π. Which matches the formula original perimeter + 2πr. Therefore, same for a polygon. Therefore, for a regular nonagon, the perimeter of S would be original perimeter + 2π*1. Therefore, 9*2 + 2π = 18 + 2π. Therefore, the answer is 18 + 2π. But wait, the problem says "the perimeter of S". However, let me check again.

Wait, but maybe there's a mistake here. Because when you take the union of the interior of the nonagon and the points within 1 unit of the perimeter, is that the same as the Minkowski sum? Let me think. The Minkowski sum of the nonagon with a disk of radius 1 is all points that are within 1 unit of the nonagon. Which includes the interior of the nonagon and the points within 1 unit of the perimeter. Therefore, yes, S is exactly that. Therefore, the perimeter of S is the perimeter of the Minkowski sum, which is original perimeter + 2π*1. Therefore, 18 + 2π. However, let me check with another example. Suppose the nonagon is very small, such that the points within 1 unit of the perimeter include the entire original nonagon. Wait, but if the nonagon is small, then the points within 1 unit of the perimeter would include the original nonagon and a buffer around it. However, if the nonagon is so small that its radius is less than 1, then the Minkowski sum would be a larger shape. But regardless, the perimeter formula still holds. Therefore, maybe the answer is 18 + 2π. However, the problem says "the perimeter of S". Wait, but maybe I need to think about the shape again. When you take the Minkowski sum, the perimeter is the original perimeter plus 2πr. But in the case of the square, we saw that. However, in the case of a polygon, when you offset it, the sides are extended outward, but the corners are replaced by circular arcs. Therefore, the perimeter is original perimeter + 2πr. Therefore, in this case, 18 + 2π. However, let me check if the original perimeter is 18. The nonagon has 9 sides, each of length 2, so yes, 9*2=18. Therefore, the perimeter of S is 18 + 2π. But wait, 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 of these two sets. However, the set of all points inside the nonagon is already included in the set of points less than 1 unit away from the perimeter? No, not necessarily. For example, if the nonagon is large enough, then points near the center are more than 1 unit away from the perimeter. Therefore, the union is the set of points inside the nonagon plus the points within 1 unit of the perimeter. However, the points within 1 unit of the perimeter include the perimeter itself and the area near it. Therefore, the union is the set of points within 1 unit of the perimeter union the interior of the nonagon. But if the nonagon is large enough, then the interior of the nonagon is not entirely within the points within 1 unit of the perimeter. Therefore, the union is the Minkowski sum of the nonagon with a disk of radius 1. Therefore, the perimeter is original perimeter + 2π*1. Therefore, 18 + 2π. However, maybe I need to check if the original nonagon's perimeter is entirely included in the Minkowski sum. Yes, because the perimeter of the nonagon is the set of points at distance 0 from the nonagon, which is included in the set of points within 1 unit of the nonagon. Therefore, the union is indeed the Minkowski sum. Therefore, the perimeter is 18 + 2π. However, let me check again with the square example. Suppose the square has side length 2, perimeter 8. The Minkowski sum with radius 1 would have perimeter 8 + 2π. But if you take the union of the interior of the square and the points within 1 unit of the perimeter, that is the same as the Minkowski sum. Therefore, the perimeter is 8 + 2π. Therefore, same logic applies here. Therefore, answer is 18 + 2π. However, the problem says "the perimeter of S". Wait, but maybe there's a mistake here. Because when you take the Minkowski sum, the perimeter is not just original perimeter plus 2πr. Wait, let me think again. For the square, when you offset it by radius r, the perimeter becomes original perimeter + 2πr. But in reality, when you offset a square, each side is extended outward, but the corners are replaced by quarter-circles. Therefore, the total perimeter is original perimeter (which is 4*side length) plus the circumference of a circle with radius r. Wait, because each corner contributes a quarter-circle, so four corners contribute a full circle. Therefore, total perimeter is original perimeter + 2πr. Therefore, same as before. Therefore, for the nonagon, same thing. Therefore, perimeter of S is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, answer is 18 + 2π. However, the problem says "the perimeter of S". Wait, but maybe I need to check if the original nonagon's perimeter is entirely on the boundary of S. Because when you take the union of the interior and the points within 1 unit of the perimeter, the boundary of S would consist of the original perimeter offset outward by 1 unit, plus the circular arcs at the corners. Therefore, the perimeter of S is the same as the perimeter of the Minkowski sum, which is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, the answer is 18 + 2π. However, let me check if 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 of these two sets. However, the set of points less than 1 unit away from the perimeter includes the perimeter itself and the area within 1 unit of it. The set of points inside the nonagon includes the interior. Therefore, the union is the set of points that are either inside the nonagon or within 1 unit of the perimeter. However, the points within 1 unit of the perimeter include the perimeter and the area near it. However, the interior of the nonagon is not entirely included in the points within 1 unit of the perimeter. Therefore, the union is the set of points within 1 unit of the perimeter union the interior of the nonagon. Which is the same as the set of points within 1 unit of the nonagon. Therefore, S is the Minkowski sum. Therefore, perimeter is original perimeter + 2π*1. Therefore, 18 + 2π. However, maybe I need to think about the actual shape. If you have a nonagon, and you take all points within 1 unit of its perimeter, that would create a sort of "thickened" nonagon. The perimeter of this thickened nonagon would consist of the original sides offset outward by 1 unit, connected by circular arcs at each corner. Therefore, the length contributed by the sides would be the same as the original perimeter, but each side is offset outward. Wait, but if you offset each side outward by 1 unit, the length of each side remains the same? Wait, no. Wait, if you have a straight side of length L, and you offset it outward by distance r, keeping it parallel, then the length of the offset side is still L. However, the distance between the original side and the offset side is r. However, when you do this for all sides, the corners become rounded. Therefore, the total perimeter would be original perimeter + 2πr. Therefore, same as before. Therefore, the perimeter of S is 18 + 2π. Therefore, the answer is 18 + 2π. However, let me check if there's a different interpretation. Maybe the problem is not about the Minkowski sum, but something else. For example, maybe the set S is the union of the interior of the nonagon and the set of points within 1 unit of the perimeter. However, the set of points within 1 unit of the perimeter is a region that includes the perimeter and a buffer around it. However, the union with the interior would be the same as the set of points within 1 unit of the nonagon. Therefore, same as before. Therefore, perimeter is 18 + 2π. However, maybe the problem is in 3D? No, it's in 2D. Therefore, I think my reasoning is correct. Therefore, the perimeter of S is 18 + 2π. However, let me check if the original nonagon's side length is 2. Wait, the problem says "a regular nonagon with side length 2 units". Therefore, perimeter is 9*2=18. Therefore, adding 2π*1, the perimeter of S is 18 + 2π. Therefore, the answer is 18 + 2π. However, let me check if there's a different approach. For example, maybe the perimeter of S is just the perimeter of the original nonagon plus the perimeter of the buffer. But the buffer is a 1-unit wide strip around the nonagon. However, the perimeter of the buffer would be... Wait, if you have a buffer around a polygon, the perimeter is not straightforward. However, as we discussed earlier, it's original perimeter + 2πr. Therefore, same result. Therefore, I think the answer is 18 + 2π. However, let me check if the problem says "the set of all points inside... and the set of all points less than 1 unit away from a point on the perimeter". So, the union. Therefore, the boundary of S would be the boundary of the buffer region. Because the interior of the nonagon is already included in the buffer region? Wait, no. Wait, the buffer region is all points within 1 unit of the perimeter. The interior of the nonagon is not entirely within the buffer region. Therefore, the union is the buffer region plus the interior. Therefore, the boundary of S would be the boundary of the buffer region, but with the parts that are inside the original nonagon? Wait, no. Wait, the buffer region is all points within 1 unit of the perimeter. The interior of the nonagon is a separate set. Therefore, the union would have a boundary that is the boundary of the buffer region, but also the original perimeter? Wait, no. Wait, the buffer region includes the original perimeter. Therefore, the union of the interior of the nonagon and the buffer region is the same as the buffer region union the interior. But since the buffer region already includes the perimeter and the area near it, and the interior is the area inside the nonagon. However, the buffer region includes points inside the nonagon that are within 1 unit of the perimeter. Therefore, the union is the set of points within 1 unit of the nonagon. Therefore, the boundary of S is the same as the boundary of the buffer region, which is the original perimeter offset outward by 1 unit, with rounded corners. Therefore, the perimeter of S is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, the answer is 18 + 2π. However, let me check if there's a different way to compute it. For example, if I think about the perimeter of S as the original perimeter plus the perimeter contributed by the buffer. However, the buffer is a 1-unit wide region around the nonagon. However, the perimeter of the buffer would be... Well, if you imagine the buffer as a sort of "annular" region around the nonagon, but since the nonagon is a polygon, the buffer has rounded corners. Therefore, the perimeter of the buffer is equal to the original perimeter plus 2π*1. Therefore, same result. Therefore, I think the answer is 18 + 2π. Therefore, I think that is the answer. However, let me check if the problem says "the set of all points inside... and the set of all points less than 1 unit away from a point on the perimeter". Therefore, the union. So, if you imagine the original nonagon, and then you take all points within 1 unit of the perimeter, which is like a 1-unit wide strip around the nonagon. However, the union of the original nonagon and this strip would be the original nonagon plus the strip. However, the strip is already overlapping with the original nonagon near the edges. Therefore, the perimeter of S would be the perimeter of the strip, but with the parts that are inside the original nonagon removed? Wait, no. Wait, the perimeter of S is the boundary of the union. The union is the original nonagon plus the buffer. However, the buffer includes the original perimeter. Therefore, the boundary of the union would be the boundary of the buffer, because the original nonagon is entirely inside the buffer? No, because the buffer is all points within 1 unit of the perimeter. Therefore, the original nonagon is not entirely within the buffer. For example, the center of the nonagon is not within 1 unit of the perimeter if the nonagon is large enough. Therefore, the union is the original nonagon plus the buffer. Therefore, the boundary of the union would consist of two parts: the parts of the buffer that are outside the original nonagon, and the parts of the original nonagon that are not overlapped by the buffer. However, the original nonagon's perimeter is entirely overlapped by the buffer, because the buffer includes all points within 1 unit of the perimeter. Therefore, the boundary of the union would be the boundary of the buffer. Therefore, the perimeter of S is the same as the perimeter of the buffer, which is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, the answer is 18 + 2π. However, let me check if there's a different interpretation. For example, maybe the problem is considering the set S as the union of the interior of the nonagon and the set of points within 1 unit of the perimeter. Therefore, the boundary of S would be the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, because the buffer region is points within 1 unit of the perimeter. The interior of the nonagon is not entirely within the buffer region. Therefore, the union is the buffer region union the interior. Therefore, the boundary of the union is the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, that's not true. Wait, if you take the union of the interior of the nonagon and the buffer region, then the buffer region includes the perimeter and the area near it. The interior of the nonagon is the area inside the nonagon. Therefore, the union is the set of points that are either inside the nonagon or within 1 unit of the perimeter. Therefore, the boundary of this union would be the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, because the buffer region is points within 1 unit of the perimeter. The interior of the nonagon is points that are inside the nonagon. Therefore, the union is the set of points that are inside the nonagon or within 1 unit of the perimeter. Therefore, the boundary of this union would be the boundary of the buffer region, but also the parts of the original nonagon that are not overlapped by the buffer. Wait, but the buffer region includes the perimeter of the nonagon. Therefore, the union is the buffer region union the interior of the nonagon. However, the buffer region already includes the perimeter and the area near it. Therefore, the union is the buffer region union the interior. But since the buffer region includes the perimeter and the area near it, and the interior is the area inside the nonagon. Therefore, the union is the set of points within 1 unit of the perimeter union the interior. Therefore, the boundary of the union is the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, that's not true. For example, take a point in the center of the nonagon. If the nonagon is large enough, that point is more than 1 unit away from the perimeter, so it's not in the buffer region. Therefore, the union is the buffer region union the interior. Therefore, the boundary of the union is the boundary of the buffer region, plus the boundary of the interior? No, because the interior is entirely inside the buffer region? No, the interior is not entirely inside the buffer region. Therefore, the union is the buffer region union the interior. Therefore, the boundary of the union is the boundary of the buffer region, but with the parts that are inside the interior of the nonagon removed? Wait, this is getting confusing. Maybe I need to think of it as the union of two regions: the interior of the nonagon and the buffer region. The buffer region is all points within 1 unit of the perimeter. The interior of the nonagon is all points inside the nonagon. Therefore, the union is the set of points that are either inside the nonagon or within 1 unit of the perimeter. Therefore, the boundary of this union would be the boundary of the buffer region, because the interior of the nonagon is entirely inside the buffer region? No, because the buffer region is points within 1 unit of the perimeter. The interior of the nonagon is points inside the nonagon. Therefore, the union is the set of points that are inside the nonagon or within 1 unit of the perimeter. Therefore, the boundary of the union is the set of points that are either on the boundary of the buffer region or on the boundary of the interior of the nonagon. However, the boundary of the interior of the nonagon is the perimeter of the nonagon. But the buffer region includes the perimeter of the nonagon. Therefore, the union's boundary is the boundary of the buffer region. Because the perimeter of the nonagon is already included in the buffer region. Therefore, the boundary of the union is the same as the boundary of the buffer region. Therefore, the perimeter of S is the same as the perimeter of the buffer region, which is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, the answer is 18 + 2π. However, let me check if there's a different way to think about it. For example, if I imagine the perimeter of S as the original perimeter plus the perimeter contributed by the buffer. However, the buffer is a 1-unit wide region around the nonagon. However, the perimeter of the buffer is not just the original perimeter plus something. Wait, but earlier analysis with the square shows that it's original perimeter + 2πr. Therefore, same here. Therefore, answer is 18 + 2π. Therefore, I think this is correct. Therefore, I think the answer is 18 + 2π. However, let me check if the problem says "the set of all points inside... and the set of all points less than 1 unit away from a point on the perimeter". Therefore, the union. Therefore, the perimeter of S is the same as the perimeter of the Minkowski sum, which is original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, the answer is 18 + 2π. However, let me check if the problem is from an Olympiad or something, and maybe the answer is different. Wait, but I don't have that context. Alternatively, maybe I need to calculate the perimeter of S by considering the original nonagon's perimeter and adding the perimeter of the buffer. However, the buffer is a 1-unit wide region around the nonagon. However, the perimeter of the buffer is not straightforward. However, as we discussed, it's original perimeter + 2π*1. Therefore, 18 + 2π. Therefore, I think this is correct. Therefore, I think the answer is 18 + 2π. Therefore, I will go with that.

**Final Answer**
\boxed{18 + 2\pi}

To determine the perimeter of the set $ S $, we begin by understanding its definition:

> $ S $ is 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.

---

### Step 1: Understanding the Shape of $ S $

This union can be interpreted geometrically as the **Minkowski sum** of the nonagon with a closed disk of radius 1. In other words, $ S $ is the set of all points that are within 1 unit of the nonagon, including:
- The interior of the nonagon,
- The perimeter of the nonagon,
- A 1-unit wide "buffer" zone around the nonagon.

This shape is known as the **offset polygon** or **buffered polygon**. It has:
- Straight sides that are offset outward from the original nonagon,
- Rounded corners formed by circular arcs of radius 1.

---

### Step 2: Perimeter of the Offset Polygon

For a regular polygon with $ n $ sides, when you offset it outward by a distance $ r $, the resulting shape has:
- The same number of sides as the original polygon,
- Each side is offset outward, but the **length of each side remains unchanged** (since the offset is perpendicular to the side),
- The **corners are replaced by circular arcs** of radius $ r $, with each arc corresponding to the exterior angle of the polygon.

The total length of the arcs is:
$$
\text{Total arc length} = r \times \text{sum of exterior angles}
$$
For any convex polygon, the sum of the exterior angles is $ 2\pi $, so:
$$
\text{Total arc length} = r \times 2\pi
$$

Thus, the **perimeter of the offset polygon** is:
$$
\text{Perimeter of } S = \text{Original perimeter} + 2\pi r
$$

---

### Step 3: Apply to the Given Nonagon

- The nonagon is regular with side length 2.
- It has 9 sides, so its original perimeter is:
  $$
  9 \times 2 = 18
  $$
- The offset is 1 unit, so $ r = 1 $.

Therefore, the perimeter of $ S $ is:
$$
18 + 2\pi
$$

---

### 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.


Explanation:
Okay, so I need to figure out for how many real values of x the expression √(120 - √x) is an integer. Let me start by understanding the problem step by step.

First, let me note that the expression inside the square root, which is 120 - √x, must be non-negative because you can't take the square root of a negative number in the real number system. So, that gives me the inequality:

120 - √x ≥ 0
Which simplifies to:
√x ≤ 120
Then squaring both sides (since both sides are non-negative):
x ≤ 120²
So x ≤ 14400.
Also, since √x is defined, x must be non-negative. So x ≥ 0.
Therefore, x is in the interval [0, 14400].

Now, the main condition is that √(120 - √x) is an integer. Let me denote that integer as n. So:

√(120 - √x) = n, where n is an integer.
Then, squaring both sides:
120 - √x = n²
Which rearranges to:
√x = 120 - n²
Then, squaring both sides again:
x = (120 - n²)²

But since x must be real and non-negative, we need to ensure that (120 - n²)² is non-negative. However, since any real number squared is non-negative, this will always hold. However, we also have the earlier condition that √x must be real, which we already considered. But also, from the first step, we had that √x ≤ 120, so 120 - n² must be non-negative because √x is non-negative. Wait, let me check that.

Wait, √x = 120 - n². Since √x is non-negative, then 120 - n² must be ≥ 0. Therefore:

120 - n² ≥ 0
Which implies:
n² ≤ 120
Therefore, n² ≤ 120. Since n is an integer, n can be any integer such that n² ≤ 120. Also, since √(120 - √x) is a square root, it must be non-negative. Therefore, n must be a non-negative integer. So n ≥ 0.

So, n is a non-negative integer with n² ≤ 120. Let me find all such integers n.

First, let's find the maximum possible value of n. Since n² ≤ 120, then n ≤ √120. Let me compute √120. Well, 10² = 100, 11² = 121. So √120 is between 10 and 11. Therefore, the maximum integer n can be is 10, since 10² = 100 ≤ 120, and 11² = 121 > 120. Therefore, n can be 0, 1, 2, ..., 10. So that's 11 possible values of n.

But wait, hold on. Let me check if n can be zero. If n = 0, then √(120 - √x) = 0, which implies 120 - √x = 0, so √x = 120, so x = 120² = 14400. Which is valid. So n=0 is allowed. Similarly, n=10 gives √x = 120 - 10² = 120 - 100 = 20, so x = 20² = 400. Which is also valid. So n ranges from 0 to 10 inclusive. Therefore, there are 11 possible integer values for n. Each of these gives a unique x, right?

Wait, but the question is asking for how many real values of x. So if each n gives exactly one x, then there are 11 real values of x. But wait, let me check if there are any overlaps or if some n might lead to the same x. But since n is unique for each x, because if two different n's give the same x, then:

Suppose n1 ≠ n2, but (120 - n1²)² = (120 - n2²)². Then either 120 - n1² = 120 - n2² or 120 - n1² = -(120 - n2²). Let's check.

Case 1: 120 - n1² = 120 - n2² ⇒ n1² = n2² ⇒ since n1 and n2 are non-negative integers, n1 = n2.

Case 2: 120 - n1² = - (120 - n2²) ⇒ 120 - n1² = -120 + n2² ⇒ 240 = n1² + n2². So, if there are two different integers n1 and n2 such that n1² + n2² = 240, then they would produce the same x. Therefore, we need to check if such pairs exist.

But since n is between 0 and 10, let's check if there are any pairs (n1, n2) with n1 ≠ n2 such that n1² + n2² = 240.

Let me check possible squares:

n can be 0 to 10, so squares are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Looking for two squares in this list that add up to 240.

Let me check:

Start with the largest square less than 240, which is 100. Then 240 - 100 = 140. Is 140 a square? No. Next, 81. 240 - 81 = 159, not a square. 64: 240 - 64 = 176, not a square. 49: 240 - 49 = 191, nope. 36: 240 - 36 = 204, nope. 25: 240 - 25 = 215, nope. 16: 240 - 16 = 224, nope. 9: 240 - 9 = 231, nope. 4: 240 - 4 = 236, nope. 1: 240 - 1 = 239, nope. 0: 240 - 0 = 240, which is not a square. So there are no such pairs. Therefore, there are no two different integers n1 and n2 in 0 to 10 such that their squares add up to 240. Therefore, each n gives a unique x. Therefore, there are 11 real values of x.

Wait, but hold on. Let me check if there are any constraints I missed. For example, when we squared both sides of equations, could there be extraneous solutions? Let me check.

Starting with √(120 - √x) = n. Since square roots are non-negative, n must be non-negative integer. Then we derived that √x = 120 - n². Since √x is non-negative, 120 - n² must be non-negative, so n² ≤ 120. Which we already considered. Therefore, all the solutions we found are valid. Therefore, each n from 0 to 10 gives exactly one x, and there are no overlaps. Therefore, the answer is 11?

Wait, but let me check with specific examples. For example, n=0: x=(120 - 0)^2=14400. Then √x=120, so √(120 - √x)=√(120 - 120)=√0=0, which is integer. Good. For n=10: x=(120 - 100)^2=20²=400. Then √x=20, so √(120 - 20)=√100=10, which is integer. For n=1: x=(120 - 1)^2=119²=14161. Then √x=119, so √(120 - 119)=√1=1, which is integer. Similarly, n=2: x=(120 - 4)^2=116²=13456. √x=116, so √(120 - 116)=√4=2. All good. So seems like all these values work.

But wait, let me check if there are any other constraints. For example, when we have √x = 120 - n², then since √x must be real, 120 - n² must be non-negative, which we already considered. Also, x must be real, which is already satisfied. So, all these x's are valid. Therefore, 11 values. But wait, the problem says "real values of x". But each n gives exactly one x, so 11 different x's. Therefore, the answer is 11?

But wait, hold on. Wait, let me check if n can be negative. Wait, earlier I thought n must be non-negative because √(120 - √x) is a square root, which is non-negative. Therefore, n must be a non-negative integer. Therefore, n can't be negative. So n is 0,1,...,10. So 11 values. Therefore, answer is 11?

But wait, let me check again. Let me think if there's another way this could be approached. Maybe I made a mistake in assuming that n can be from 0 to 10. Let me check n² ≤ 120. So n can be integers from 0 up to floor(√120). Since √120 ≈ 10.954, so floor(√120) is 10. Therefore, n can be 0,1,...,10. So 11 integers. Therefore, 11 values. Therefore, answer is 11?

But wait, let me check if maybe for some n, 120 - n² is negative? But we already have n² ≤ 120, so 120 - n² is non-negative. Therefore, √x is non-negative, so x is non-negative. Therefore, all these x's are valid. Therefore, 11 values. Therefore, the answer is 11?

But wait, hold on. Wait, the problem says "real values of x". But maybe there are multiple x's for the same n? Wait, no. For each n, we have x = (120 - n²)^2. So each n gives exactly one x. Therefore, 11 different x's. Therefore, answer is 11. But let me check again with n=0 to n=10.

Wait, let me count them. From 0 to 10 inclusive, that's 11 numbers. So 0,1,2,...,10. So 11 numbers. Therefore, 11 real values of x. Therefore, answer is 11? But I have a feeling that maybe there's something wrong here. Let me check again.

Wait, let me think. Suppose n is an integer such that √(120 - √x) = n. Then 120 - √x must be a perfect square, right? Because n is integer. Wait, no. Wait, 120 - √x is equal to n², which is an integer. Therefore, √x = 120 - n². Therefore, √x must be an integer? Wait, no. Wait, √x is equal to 120 - n². But 120 - n² is an integer because n is integer. Therefore, √x is an integer. Therefore, x must be a perfect square. Therefore, x is a perfect square. Therefore, each x is a perfect square. Therefore, for each n, x is (120 - n²)^2. Therefore, each x is unique? Because if two different n's give same x, then (120 - n1²)^2 = (120 - n2²)^2. Which would imply that 120 - n1² = ±(120 - n2²). But since 120 - n² is non-negative (as we saw earlier), then 120 - n1² = 120 - n2² or 120 - n1² = - (120 - n2²). The second case would be 120 - n1² = -120 + n2² ⇒ n1² + n2² = 240. As before, but we saw that there are no such pairs. Therefore, each x is unique. Therefore, 11 values. Therefore, answer is 11. But let me check again if there are any mistakes here.

Alternatively, maybe I need to check if √x is an integer? Wait, √x = 120 - n². Since n is integer, 120 - n² is integer. Therefore, √x is integer. Therefore, x is a perfect square. Therefore, x must be a perfect square. Therefore, for each n, x is (120 - n²)^2. Therefore, each x is unique. Therefore, 11 values. Therefore, answer is 11. But maybe the problem is trickier?

Wait, let me check for n=10, x=(120 - 100)^2=20²=400. Then √x=20. Then √(120 - √x)=√(120 - 20)=√100=10, which is integer. For n=10, that works. For n=11, n²=121>120, which would make √x=120 - 121=-1, which is invalid because square root can't be negative. Therefore, n=11 is invalid. So n can't be 11. Therefore, n from 0 to 10 inclusive. So 11 values. Therefore, answer is 11. But maybe the answer is different?

Wait, but let me check if there are other possible values. For example, suppose that √(120 - √x) is integer, but maybe √x is not necessarily integer? Wait, no. Because if √(120 - √x) is integer, then 120 - √x must be a perfect square. Let me denote m = √(120 - √x), where m is integer. Then, m² = 120 - √x ⇒ √x = 120 - m². Therefore, √x must be equal to 120 - m², which is an integer because m is integer. Therefore, √x is integer, so x is a perfect square. Therefore, x must be a perfect square. Therefore, the logic holds. Therefore, there are 11 values. Therefore, answer is 11. But maybe I need to check if for some n, 120 - n² is negative? But we already have n² ≤ 120, so 120 - n² is non-negative. Therefore, √x is non-negative. Therefore, all good.

Wait, but let me check for n=0, x=(120 - 0)^2=14400. Then √x=120, so √(120 - √x)=√(120 - 120)=√0=0, which is integer. So that's valid. For n=1, x=(120 - 1)^2=119²=14161. Then √x=119, so √(120 - 119)=√1=1, which is integer. Similarly, n=2: x=(120 - 4)=116, so √x=116, then √(120 - 116)=√4=2. All good. So seems like all these work. Therefore, 11 values. Therefore, answer is 11. But maybe the answer is different? Wait, let me check if there are other possible integers m where m² ≤ 120. Wait, m is non-negative integer. So m can be 0,1,...,10. So 11 values. Therefore, answer is 11. Therefore, I think the answer is 11. But let me check again if there's a mistake in my logic.

Wait, but maybe there's a case where √x is not an integer, but √(120 - √x) is integer? Wait, but according to the equations, if √(120 - √x) is integer, then 120 - √x must be a perfect square. Let me denote m = √(120 - √x), so m is integer. Then, √x = 120 - m². Therefore, √x is equal to 120 - m², which is an integer. Therefore, √x is integer. Therefore, x is a perfect square. Therefore, there are no cases where √x is not integer. Therefore, my previous reasoning holds. Therefore, answer is 11. Therefore, I think the answer is 11. But maybe I need to check if there are more values?

Wait, let me check for m=10, which gives x=(120 - 10²)^2=(120 - 100)^2=20²=400. Then √x=20. Then √(120 - √x)=√(120 - 20)=√100=10. So that's valid. For m=10, that's okay. What about m= sqrt(120 - sqrt(x))? Wait, no, m is integer. So m can be from 0 to 10. Therefore, 11 values. Therefore, answer is 11. Therefore, I think the answer is 11. Therefore, I think the answer is \boxed{11}.

But wait, hold on. Wait, let me check if there are other possible values of m. For example, suppose m is a non-integer, but the problem says that √(120 - √x) is integer. Therefore, m must be integer. Therefore, no. Therefore, only m from 0 to 10. Therefore, 11 values. Therefore, answer is 11. Therefore, I think that's correct.

But let me check again. Let me think of another approach. Let me suppose that √(120 - √x) is integer. Let me denote that integer as k. Then:

k = √(120 - √x)
=> k² = 120 - √x
=> √x = 120 - k²
=> x = (120 - k²)²

Now, since x must be real and non-negative, as discussed earlier, 120 - k² must be ≥ 0. Therefore, k² ≤ 120. Since k is integer, k can be 0, 1, 2, ..., 10. Therefore, 11 values. Therefore, answer is 11. Therefore, I think that's correct.

But maybe the problem is in the interpretation of "real values of x". For example, if different k's lead to same x? But we saw that there are no such cases. Therefore, 11 different x's. Therefore, answer is 11. Therefore, I think the answer is 11. Therefore, I think the answer is \boxed{11}.

**Final Answer**
\boxed{11}

To determine how many **real values of $ x $** make the expression
$$
\sqrt{120 - \sqrt{x}}
$$
an integer, we proceed step by step.

---

### Step 1: Understand the Domain

The expression $ \sqrt{120 - \sqrt{x}} $ is defined only when the expression inside the square root is non-negative:

$$
120 - \sqrt{x} \geq 0 \Rightarrow \sqrt{x} \leq 120
$$

Since $ \sqrt{x} \geq 0 $, this implies:

$$
0 \leq \sqrt{x} \leq 120 \Rightarrow 0 \leq x \leq 120^2 = 14400
$$

So, $ x \in [0, 14400] $.

---

### Step 2: Let the Expression Be an Integer

Let:
$$
\sqrt{120 - \sqrt{x}} = n \quad \text{where } n \in \mathbb{Z} \text{ and } n \geq 0
$$

Squaring both sides:
$$
120 - \sqrt{x} = n^2 \Rightarrow \sqrt{x} = 120 - n^2
$$

Now square again:
$$
x = (120 - n^2)^2
$$

This gives a valid real value of $ x $ **if and only if** $ 120 - n^2 \geq 0 $, i.e.,

$$
n^2 \leq 120
$$

---

### Step 3: Determine Valid Integer Values of $ n $

We now find all non-negative integers $ n $ such that $ n^2 \leq 120 $.

Compute $ \sqrt{120} \approx 10.954 $, so the largest integer $ n $ satisfying $ n^2 \leq 120 $ is $ n = 10 $.

Thus, the possible values of $ n $ are:
$$
n = 0, 1, 2, \dots, 10
$$

That is, **11** values in total.

---

### Step 4: Verify Uniqueness of $ x $

Each value of $ n $ gives a unique value of $ x = (120 - n^2)^2 $. Since $ n $ is an integer and $ n^2 $ is unique for each $ n $, the expression $ (120 - n^2)^2 $ is also unique for each $ n $.

Therefore, each $ n $ corresponds to a **distinct** real value of $ x $.

---

### Final Answer

$$
\boxed{11}
$$

[Assistant]
Explanation:
Okay, so I need to figure out for how many real values of x the expression √(120 - √x) is an integer. Let me start by understanding the problem step by step.

First, let me note that the expression inside the square root, which is 120 - √x, must be non-negative because you can't take the square root of a negative number in the real number system. So, that gives me the inequality:

120 - √x ≥ 0
Which simplifies to:
√x ≤ 120
Then squaring both sides (since both sides are non-negative):
x ≤ 120²
So x ≤ 14400.
Also, since √x is defined, x must be non-negative. So x ≥ 0.
Therefore, x is in the interval [0, 14400].

Now, the main condition is that √(120 - √x) is an integer. Let me denote that integer as n. So:

√(120 - √x) = n, where n is an integer.
Then, squaring both sides:
120 - √x = n²
Which rearranges to:
√x = 120 - n²
Then, squaring both sides again:
x = (120 - n²)²

But since x must be real and non-negative, we need to ensure that (120 - n²)² is non-negative. However, since any real number squared is non-negative, this will always hold. However, we also have the earlier condition that √x must be real, which we already considered. But also, from the first step, we had that √x ≤ 120, so 120 - n² must be non-negative because √x is non-negative. Wait, let me check that.

Wait, √x = 120 - n². Since √x is non-negative, then 120 - n² must be ≥ 0. Therefore:

120 - n² ≥ 0
Which implies:
n² ≤ 120
Therefore, n² ≤ 120. Since n is an integer, n can be any integer such that n² ≤ 120. Also, since √(120 - √x) is a square root, it must be non-negative. Therefore, n must be a non-negative integer. So n ≥ 0.

So, n is a non-negative integer with n² ≤ 120. Let me find all such integers n.

First, let's find the maximum possible value of n. Since n² ≤ 120, then n ≤ √120. Let me compute √120. Well, 10² = 100, 11² = 121. So √120 is between 10 and 11. Therefore, the maximum integer n can be is 10, since 10² = 100 ≤ 120, and 11² = 121 > 120. Therefore, n can be 0, 1, 2, ..., 10. So that's 11 possible values of n.

But wait, hold on. Let me check if n can be zero. If n = 0, then √(120 - √x) = 0, which implies 120 - √x = 0, so √x = 120, so x = 120² = 14400. Which is valid. So n=0 is allowed. Similarly, n=10 gives √x = 120 - 10² = 120 - 100 = 20, so x = 20² = 400. Which is also valid. So n ranges from 0 to 10 inclusive. Therefore, there are 11 possible integer values for n. Each of these gives a unique x, right?

Wait, but the question is asking for how many real values of x. So if each n gives exactly one x, then there are 11 real values of x. But wait, let me check if there are any overlaps or if some n might lead to the same x. But since n is unique for each x, because if two different n's give the same x, then:

Suppose n1 ≠ n2, but (120 - n1²)² = (120 - n2²)². Then either 120 - n1² = 120 - n2² or 120 - n1² = -(120 - n2²). Let's check.

Case 1: 120 - n1² = 120 - n2² ⇒ n1² = n2² ⇒ since n1 and n2 are non-negative integers, n1 = n2.

Case 2: 120 - n1² = - (120 - n2²) ⇒ 120 - n1² = -120 + n2² ⇒ 240 = n1² + n2². So, if there are two different integers n1 and n2 such that n1² + n2² = 240, then they would produce the same x. Therefore, we need to check if such pairs exist.

But since n is between 0 and 10, let's check if there are any pairs (n1, n2) with n1 ≠ n2 such that n1² + n2² = 240.

Let me check possible squares:

n can be 0 to 10, so squares are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Looking for two squares in this list that add up to 240.

Let me check:

Start with the largest square less than 240, which is 100. Then 240 - 100 = 140. Is 140 a square? No. Next, 81. 240 - 81 = 159, not a square. 64: 240 - 64 = 176, not a square. 49: 240 - 49 = 191, nope. 36: 240 - 36 = 204, nope. 25: 240 - 25 = 215, nope. 16: 240 - 16 = 224, nope. 9: 240 - 9 = 231, nope. 4: 240 - 4 = 236, nope. 1: 240 - 1 = 239, nope. 0: 240 - 0 = 240, which is not a square. So there are no such pairs. Therefore, there are no two different integers n1 and n2 in 0 to 10 such that their squares add up to 240. Therefore, each n gives a unique x. Therefore, there are 11 real values of x.

Wait, but hold on. Let me check if there are any constraints I missed. For example, when we squared both sides of equations, could there be extraneous solutions? Let me check.

Starting with √(120 - √x) = n. Since square roots are non-negative, n must be non-negative integer. Then we derived that √x = 120 - n². Since √x is non-negative, 120 - n² must be non-negative, so n² ≤ 120. Which we already considered. Therefore, all the solutions we found are valid. Therefore, each n from 0 to 10 gives exactly one x, and there are no overlaps. Therefore, the answer is 11?

Wait, but let me check with specific examples. For example, n=0: x=(120 - 0)^2=14400. Then √x=120, so √(120 - √x)=√(120 - 120)=√0=0, which is integer. Good. For n=10: x=(120 - 100)^2=20²=400. Then √x=20, so √(120 - 20)=√100=10, which is integer. For n=1: x=(120 - 1)^2=119²=14161. Then √x=119, so √(120 - 119)=√1=1, which is integer. Similarly, n=2: x=(120 - 4)^2=116²=13456. √x=116, so √(120 - 116)=√4=2. All good. So seems like all these values work.

But wait, let me check if there are any other constraints. For example, when we have √x = 120 - n², then since √x must be real, 120 - n² must be non-negative, which we already considered. Also, x must be real, which is already satisfied. So, all these x's are valid. Therefore, 11 values. But wait, the problem says "real values of x". But each n gives exactly one x, so 11 different x's. Therefore, the answer is 11?

But wait, hold on. Wait, let me check if n can be negative. Wait, earlier I thought n must be non-negative because √(120 - √x) is a square root, which is non-negative. Therefore, n must be a non-negative integer. Therefore, n can't be negative. So n is 0,1,...,10. So 11 values. Therefore, answer is 11?

But wait, let me check again. Let me think if there's another way this could be approached. Maybe I made a mistake in assuming that n can be from 0 to 10. Let me check n² ≤ 120. So n can be integers from 0 up to floor(√120). Since √120 ≈ 10.954, so floor(√120) is 10. Therefore, n can be 0,1,...,10. So 11 integers. Therefore, 11 values. Therefore, answer is 11?

But wait, let me check if maybe for some n, 120 - n² is negative? But we already have n² ≤ 120, so 120 - n² is non-negative. Therefore, √x is non-negative, so x is non-negative. Therefore, all these x's are valid. Therefore, 11 values. Therefore, the answer is 11?

But wait, hold on. Wait, the problem says "real values of x". But maybe there are multiple x's for the same n? Wait, no. For each n, we have x = (120 - n²)^2. So each n gives exactly one x. Therefore, 11 different x's. Therefore, answer is 11. But let me check again with n=0 to n=10.

Wait, let me count them. From 0 to 10 inclusive, that's 11 numbers. So 0,1,2,...,10. So 11 numbers. Therefore, 11 real values of x. Therefore, answer is 11? But I have a feeling that maybe there's something wrong here. Let me check again.

Wait, let me think. Suppose n is an integer such that √(120 - √x) = n. Then 120 - √x must be a perfect square, right? Because n is integer. Wait, no. Wait, 120 - √x is equal to n², which is an integer. Therefore, √x = 120 - n². Therefore, √x must be an integer? Wait, no. Wait, √x is equal to 120 - n². But 120 - n² is an integer because n is integer. Therefore, √x is an integer. Therefore, x must be a perfect square. Therefore, x is a perfect square. Therefore, each x is a perfect square. Therefore, for each n, x is (120 - n²)^2. Therefore, each x is unique? Because if two different n's give same x, then (120 - n1²)^2 = (120 - n2²)^2. Which would imply that 120 - n1² = ±(120 - n2²). But since 120 - n² is non-negative (as we saw earlier), then 120 - n1² = 120 - n2² or 120 - n1² = - (120 - n2²). The second case would be 120 - n1² = -120 + n2² ⇒ n1² + n2² = 240. As before, but we saw that there are no such pairs. Therefore, each x is unique. Therefore, 11 values. Therefore, answer is 11. But let me check again if there are any mistakes here.

Alternatively, maybe I need to check if √x is an integer? Wait, √x = 120 - n². Since n is integer, 120 - n² is integer. Therefore, √x is integer. Therefore, x is a perfect square. Therefore, x must be a perfect square. Therefore, for each n, x is (120 - n²)^2. Therefore, each x is unique. Therefore, 11 values. Therefore, answer is 11. But maybe the problem is trickier?

Wait, let me check for n=10, x=(120 - 100)^2=20²=400. Then √x=20. Then √(120 - √x)=√(120 - 20)=√100=10, which is integer. For n=10, that works. For n=11, n²=121>120, which would make √x=120 - 121=-1, which is invalid because square root can't be negative. Therefore, n=11 is invalid. So n can't be 11. Therefore, n from 0 to 10 inclusive. So 11 values. Therefore, answer is 11. But maybe the answer is different?

Wait, but let me check if there are other possible values. For example, suppose that √(120 - √x) is integer, but maybe √x is not necessarily integer? Wait, no. Because if √(120 - √x) is integer, then 120 - √x must be a perfect square. Let me denote m = √(120 - √x), where m is integer. Then, m² = 120 - √x ⇒ √x = 120 - m². Therefore, √x must be equal to 120 - m², which is an integer because m is integer. Therefore, √x is integer, so x is a perfect square. Therefore, x must be a perfect square. Therefore, the logic holds. Therefore, there are 11 values. Therefore, answer is 11. But maybe I need to check if for some n, 120 - n² is negative? But we already have n² ≤ 120, so 120 - n² is non-negative. Therefore, √x is non-negative. Therefore, all good.

Wait, but let me check for n=0, x=(120 - 0)^2=14400. Then √x=120, so √(120 - √x)=√(120 - 120)=√0=0, which is integer. So that's valid. For n=1, x=(120 - 1)^2=119²=14161. Then √x=119, so √(120 - 119)=√1=1, which is integer. Similarly, n=2: x=(120 - 4)=116, so √x=116, then √(120 - 116)=√4=2. All good. So seems like all these work. Therefore, 11 values. Therefore, answer is 11. But maybe the answer is different? Wait, let me check if there are other possible integers m where m² ≤ 120. Wait, m is non-negative integer. So m can be 0,1,...,10. So 11 values. Therefore, answer is 11. Therefore, I think the answer is 11. But let me check again if there's a mistake in my logic.

Wait, but maybe there's a case where √x is not an integer, but √(120 - √x) is integer? Wait, but according to the equations, if √(120 - √x) is integer, then 120 - √x must be a perfect square. Let me denote m = √(120 - √x), so m is integer. Then, √x = 120 - m². Therefore, √x is equal to 120 - m², which is an integer. Therefore, √x is integer. Therefore, x is a perfect square. Therefore, there are no cases where √x is not integer. Therefore, my previous reasoning holds. Therefore, answer is 11. Therefore, I think the answer is 11. But maybe I need to check if there are more values?

Wait, let me check for m=10, which gives x=(120 - 10²)^2=(120 - 100)^2=20²=400. Then √x=20. Then √(120 - √x)=√(120 - 20)=√100=10. So that's valid. For m=10, that's okay. What about m= sqrt(120 - sqrt(x))? Wait, no, m is integer. So m can be from 0 to 10. Therefore, 11 values. Therefore, answer is 11. Therefore, I think the answer is 11. Therefore, I think the answer is \boxed{11}.

But wait, hold on. Wait, let me check if there are other possible values of m. For example, suppose m is a non-integer, but the problem says that √(120 - √x) is integer. Therefore, m must be integer. Therefore, no. Therefore, only m from 0 to 10. Therefore, 11 values. Therefore, answer is 11. Therefore, I think that's correct.

But let me check again. Let me think of another approach. Let me suppose that √(120 - √x) is integer. Let me denote that integer as k. Then:

k = √(120 - √x)
=> k² = 120 - √x
=> √x = 120 - k²
=> x = (120 - k²)²

Now, since x must be real and non-negative, as discussed earlier, 120 - k² must be ≥ 0. Therefore, k² ≤ 120. Since k is integer, k can be 0, 1, 2, ..., 10. Therefore, 11 values. Therefore, answer is 11. Therefore, I think that's correct.

But maybe the problem is in the interpretation of "real values of x". For example, if different k's lead to same x? But we saw that there are no such cases. Therefore, 11 different x's. Therefore, answer is 11. Therefore, I think the answer is 11. Therefore, I think the answer is \boxed{11}.

**Final Answer**
\boxed{11}

To determine how many **real values of $ x $** make the expression
$$
\sqrt{120 - \sqrt{x}}
$$
an integer, we proceed step by step.

---

### Step 1: Understand the Domain

The expression $ \sqrt{120 - \sqrt{x}} $ is defined only when the expression inside the square root is non-negative:

$$
120 - \sqrt{x} \geq 0 \Rightarrow \sqrt{x} \leq 120
$$

Since $ \sqrt{x} \geq 0 $, this implies:

$$
0 \leq \sqrt{x} \leq 120 \Rightarrow 0 \leq x \leq 120^2 = 14400
$$

So, $ x \in [0, 14400] $.

---

### Step 2: Let the Expression Be an Integer

Let:
$$
\sqrt{120 - \sqrt{x}} = n \quad \text{where } n \in \mathbb{Z} \text{ and } n \geq 0
$$

Squaring both sides:
$$
120 - \sqrt{x} = n^2 \Rightarrow \sqrt{x} = 120 - n^2
$$

Now square again:
$$
x = (120 - n^2)^2
$$

This gives a valid real value of $ x $ **if and only if** $ 120 - n^2 \geq 0 $, i.e.,

$$
n^2 \leq 120
$$

---

### Step 3: Determine Valid Integer Values of $ n $

We now find all non-negative integers $ n $ such that $ n^2 \leq 120 $.

Compute $ \sqrt{120} \approx 10.954 $, so the largest integer $ n $ satisfying $ n^2 \leq 120 $ is $ n = 10 $.

Thus, the possible values of $ n $ are:
$$
n = 0, 1, 2, \dots, 10
$$

That is, **11** values in total.

---

### Step 4: Verify Uniqueness of $ x $

Each value of $ n $ gives a unique value of $ x = (120 - n^2)^2 $. Since $ n $ is an integer and $ n^2 $ is unique for each $ n $, the expression $ (120 - n^2)^2 $ is also unique for each $ n $.

Therefore, each $ n $ corresponds to a **distinct** real value of $ x $.

---

### Final Answer

$$
\boxed{11}
$$

Solution:
Suppose that  $k = \sqrt{120 - \sqrt{x}}$ is an integer. Then $0\le k \le \sqrt{120}$, and because $k$ is an integer, we have $0\le k \le 10$. Thus there are 11 possible integer values of $k$. For each such $k$, the corresponding value of $x$ is $\left(120 - k^2\right)^2$. Because  $\left(120 -
k^2\right)^2$ is positive and decreasing for $0\le k \le 10$, the $\boxed{11}$ values of $x$ are distinct.