<Question>
Let $S$ be the union of the set of all points inside a regular nonagon with side length $2$ units and the set of all points less than $1$ unit away from a point on the perimeter of the nonagon.  What, in units, is the perimeter of $S$?
</Question>

<think>
<Plan_1>
Okay, so I need to find the perimeter of set S, which is the union of two things: the interior of a regular nonagon with side length 2 units, and all points within 1 unit of the perimeter of the nonagon. Hmm, let me visualize this.
First, a regular nonagon has nine sides, all equal length, and all internal angles equal. The side length is 2 units, so each side is 2 units long. The perimeter of the nonagon itself would be 9 times 2, which is 18 units. But here, S is the union of the interior of the nonagon and all points within 1 unit of its perimeter. So essentially, S is like the original nonagon with a 1-unit buffer zone around it, but only if that buffer zone extends outside the original nonagon.

Wait, but if the nonagon is regular, the distance from the center to each side (the apothem) is important. Maybe I need to calculate that first? Let me recall the formula for the apothem of a regular polygon.
The apothem a is given by a = (s)/(2 tan(π/n)), where s is the side length and n is the number of sides. For a nonagon, n=9, s=2. So a = 2/(2 tan(π/9)) = 1/tan(π/9). Let me compute that value.

But maybe I don't need the exact numerical value yet. Let's see. The buffer zone of 1 unit around the nonagon would create a sort of expanded shape. However, since the nonagon is already a closed figure, the union of its interior and the 1-unit buffer around its perimeter would result in a shape that is the original nonagon expanded outward by 1 unit in all directions, but only where the buffer doesn't overlap with the original interior.

Wait, actually, the union would be the original nonagon plus a 1-unit wide border around it. But if the original nonagon has an apothem (distance from center to a side) greater than 1 unit, then the buffer zone would extend outside, creating a sort of "rounded" perimeter. However, if the apothem is less than 1 unit, then the buffer would overlap with the original nonagon, and the union would just be a larger polygon.

So first, I need to check if the apothem of the nonagon is greater than 1. If it is, then the perimeter of S will be the perimeter of the original nonagon plus the perimeter added by the buffer zone. If not, then the buffer zone is entirely within the original nonagon, and the perimeter remains the same as the original.

Therefore, first step: calculate the apothem of the regular nonagon with side length 2. Let's compute that.

<prompt_1.1> Calculate the apothem of the regular nonagon using the formula and determine if it is greater than 1. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The formula for the apothem a is:

a = (s) / (2 tan(π/n))

Plugging in s=2, n=9:

a = 2 / (2 tan(π/9)) = 1 / tan(π/9)

Now, π/9 radians is 20 degrees (since π radians is 180 degrees, so π/9 is 20 degrees). So tan(20°) is approximately 0.3640. Therefore, a ≈ 1 / 0.3640 ≈ 2.747 units.

So the apothem is approximately 2.747 units, which is greater than 1. Therefore, a 1-unit buffer around the nonagon will extend outside of it. Therefore, the union S will be the original nonagon with a 1-unit wide border around it, but since the buffer is only 1 unit, the parts of the buffer that extend outward will form a sort of "rounded" perimeter. However, the original sides of the nonagon will still contribute to the perimeter, but the buffer may add some rounded edges or extensions.

Wait, but actually, when you take the union of the interior of the nonagon and all points within 1 unit of its perimeter, you are effectively creating a Minkowski sum of the nonagon with a disk of radius 1. The perimeter of the Minkowski sum is equal to the perimeter of the original polygon plus 2πr, where r is the radius, but only if the original polygon is convex. Since a nonagon is convex, this should apply.

Wait, is that correct? Let me recall. The Minkowski sum of a convex polygon with a disk results in a shape that has the original edges "rounded" with circular arcs. However, the perimeter of the Minkowski sum is equal to the perimeter of the original polygon plus the circumference of the disk. Wait, but that can't be exactly right because when you offset a polygon by a radius r, the perimeter increases by 2πr. Let me check with an example.

Take a square with side length 2. The perimeter is 8. If we take the Minkowski sum with a disk of radius r, the resulting shape is a square with rounded corners, each corner being a quarter-circle of radius r. The total perimeter would be the original perimeter plus the circumference of a full circle (since each corner's arc is a quarter-circle, and there are four corners, so 4*(πr/2) = 2πr). Therefore, the perimeter becomes 8 + 2πr. So in general, for a convex polygon, the perimeter of the Minkowski sum with a disk of radius r is equal to the original perimeter plus 2πr.

But in our case, the original polygon is a nonagon with perimeter 18. If we take the Minkowski sum with a disk of radius 1, then the perimeter of S would be 18 + 2π*1 = 18 + 2π. However, is this applicable here?

Wait, but in our problem, S is the union of the interior of the nonagon and all points within 1 unit of the perimeter. That is exactly the Minkowski sum of the nonagon with a disk of radius 1. Therefore, the perimeter of S would be the perimeter of the nonagon plus the circumference of the disk. So 18 + 2π*1 = 18 + 2π units.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But wait, hold on, the problem states "the set of all points inside a regular nonagon... ".

<prompt_2.1> Verify if the Minkowski sum explanation applies to the problem and confirm the perimeter calculation. </prompt_2.1>

But let me confirm this. Let me think again. If you have a convex polygon and you take all points within distance r from it, the resulting shape is indeed the Minkowski sum of the polygon with a disk of radius r. This operation rounds all the corners with radius r, and the perimeter increases by 2πr. However, in the case where the original polygon has very sharp corners, but since a nonagon is a 9-sided polygon, each internal angle is 140 degrees (since the formula for internal angle is (n-2)*180/n, which for n=9 is 140 degrees). So each corner is 140 degrees. When you offset the polygon by 1 unit, each corner is replaced by a circular arc of radius 1 and angle equal to the external angle of the polygon.

<prompt_2.2> Analyze the effect of offsetting the nonagon on its perimeter and verify the perimeter calculation. </prompt_2.2>

But let's verify this.

<prompt_2.3> Verify the perimeter calculation using the Minkowski sum method. </prompt_2.3>

But wait, wait a minute. Let me think again. The original nonagon has an apothem of approximately 2.747, which is greater than 1. Therefore, when we take the union of the nonagon and the 1-unit buffer, the buffer does not overlap with the interior of the nonagon. Therefore, the perimeter of the union is equal to the perimeter of the original nonagon plus the perimeter contributed by the buffer.

<prompt_2.4> Reconsider the effect of the apothem on the buffer and verify the perimeter calculation. </prompt_2.4>

But let me check with another approach.

<prompt_2.5> Confirm the perimeter calculation using an alternative method. </prompt_2.5>

Therefore, the answer should be 18 + 2π. But let me check again.

<prompt_2.6> Finalize the perimeter calculation and verify it thoroughly. </prompt_2.6>

But let me check if this is correct.

Wait, but when you offset a polygon, the perimeter does increase by 2πr. For example, take a regular polygon with a large number of sides; it approximates a circle. If you offset it by r, the perimeter becomes 2π(R + r), where R is the original radius. But the original circumference is 2πR. The difference is 2πr, which matches the increase. So in the limit, for a circle, the Minkowski sum with a disk of radius r increases the perimeter by 2πr.

<prompt_2.7> Verify the perimeter calculation using an example and finalize the answer. </prompt_2.7>
</Plan_2>

<Execution_2>
<execution_2.1>
Because the set is the union of the interior and the points within 1 unit of the perimeter. So yes, that's equivalent to the closure of the nonagon expanded by 1 unit. So in that case, the perimeter is the original perimeter plus 2π*1. So 18 + 2π. Therefore, the perimeter of S is 18 + 2π units. Is that correct?

Wait, but let me think again. If the nonagon is already a closed figure, then taking all points inside it (the interior) and all points within 1 unit of its perimeter (which includes the boundary and 1 unit outside) would result in a shape that is the original nonagon expanded by 1 unit. So yes, that should be the Minkowski sum, which adds 2πr to the perimeter. So 18 + 2π*1 = 18 + 2π.
</execution_2.1>

<execution_2.2>
Wait, the external angle is 180 - internal angle = 40 degrees. Wait, no. Wait, the internal angle is 140 degrees, so the external angle is 40 degrees. But when you offset the polygon, each corner is replaced by a circular arc whose angle is equal to the external angle. So each arc is 40 degrees. But 40 degrees is π/4.5 radians. Wait, but 40 degrees is 2π/9 radians (since 180 degrees is π radians, so 40 degrees is (40/180)*π = 2π/9). So each arc is 2π/9 radians. The length of each arc is r * θ, where θ is in radians. So each arc length is 1 * 2π/9 = 2π/9. Since there are 9 corners, the total added perimeter from all arcs is 9*(2π/9) = 2π. Therefore, the total perimeter is the original perimeter (18) plus 2π. So the perimeter of S is 18 + 2π.
</execution_2.2>

<execution_2.3>
The original polygon's edges are straight, and when offset outward by radius r, each edge is replaced by a circular arc. The total length added per original edge is 2r divided by the sine of half the angle at the vertex. Wait, maybe my initial thought was wrong.

Alternatively, the formula for the perimeter of the offset polygon (Minkowski sum) is P' = P + 2πr, where P is the original perimeter. But I need to verify that.

Wait, actually, when you offset a polygon with radius r, each vertex is replaced by a circular arc of radius r and angle equal to the exterior angle of the polygon. The exterior angle of a regular polygon is 2π/n. For a nonagon, n=9, so exterior angle is 2π/9. Therefore, each arc length is r * angle = r*(2π/9). Since there are 9 vertices, total added arc length is 9*(2πr/9) = 2πr. Additionally, the original edges are extended outward by r, but since the edges are straight, their contribution to the perimeter is the same as before. Wait, but actually, when you offset a polygon, the original edges are translated outward, but their lengths remain the same. However, the corners are replaced by circular arcs. Therefore, the total perimeter is the original perimeter plus the circumference of the circle with radius r. Because each corner's arc contributes to a full circle's circumference. Since each exterior angle is 2π/n, and there are n corners, the total arc length is n*(r*2π/n) = 2πr. Therefore, the perimeter of the Minkowski sum is indeed P + 2πr.

Therefore, in our case, the original perimeter is 18, and the radius is 1, so the new perimeter is 18 + 2π*1 = 18 + 2π.
</execution_2.3>

<execution_2.4>
But how exactly does the perimeter change?

When you offset a polygon outward by radius r, each vertex, which is a sharp corner, is replaced by a circular arc. The length added per vertex is the length of the arc. The amount of angle "cut off" by the original vertex is equal to the external angle of the polygon. For a regular polygon, the external angle is 2π/n radians. For a nonagon, n=9, so external angle is 2π/9 ≈ 0.698 radians (about 40 degrees). Wait, but each original vertex is a convex angle of 140 degrees (internal angle). The external angle is 180 - 140 = 40 degrees, which is 2π/9 radians. So when you offset the polygon, each vertex is replaced by a circular arc of radius r (1 unit here) and angle equal to the external angle, which is 2π/9. Therefore, the length added per vertex is r * θ = 1 * 2π/9 = 2π/9. Since there are 9 vertices, total added length is 9*(2π/9) = 2π. Therefore, the total perimeter of the Minkowski sum is the original perimeter plus 2π. Therefore, 18 + 2π.
</execution_2.4>

<execution_2.5>
Alternatively, the perimeter of S is the boundary of the union. Since the nonagon is already a closed figure, and the buffer around it is 1 unit away from the perimeter, the boundary of S will consist of the original perimeter of the nonagon and the 1-unit buffer around it. But wait, no. The union's perimeter is actually the outer boundary of the expanded shape. So if you have a nonagon and you expand it by 1 unit, the new perimeter is the original perimeter plus the circumference of a circle with radius 1. But that seems too simplistic.

Wait, no. When you offset a polygon outward by radius r, the new perimeter is the original perimeter plus 2πr. This is because each vertex of the polygon, which is a sharp angle, is replaced by a circular arc. The total angle turned around the polygon is 2π radians, so the total length added by all the arcs is equal to the circumference of a circle with radius r, which is 2πr. Therefore, the perimeter of the offset polygon is the original perimeter plus 2πr. Therefore, in this case, since the nonagon is convex, offsetting it by 1 unit would indeed result in a perimeter of 18 + 2π*1 = 18 + 2π.
</execution_2.5>

<execution_2.6>
Alternatively, the Minkowski sum's perimeter is the original perimeter plus 2πr. So 18 + 2π*1 = 18 + 2π. Therefore, the perimeter of S is 18 + 2π units.

But wait, but let me confirm this with another approach. Let's imagine constructing the set S. It's all points that are either inside the nonagon or within 1 unit of its perimeter. So the boundary of S consists of the original perimeter of the nonagon, but each side is now "expanded" outward by 1 unit. However, since the original nonagon's apothem is about 2.747, which is more than 1, the expansion by 1 unit doesn't reach the original center. Wait, actually, if you offset each side outward by 1 unit, the new sides would form a parallel side at a distance of 1 unit from the original. However, the corners would become rounded with radius 1. But since the original nonagon's internal angles are 140 degrees, the external angle is 40 degrees (since external angle = 180 - internal angle). Wait, no. Wait, external angles of a regular polygon sum up to 360 degrees, so each external angle is 360/n = 360/9 = 40 degrees. Therefore, when you offset each side outward, the corners between the offset sides will be rounded with a radius equal to the offset distance (1 unit) and angle equal to the external angle (40 degrees). Therefore, each corner contributes a circular arc of radius 1 and angle 40 degrees (which is π/4.5 radians). The length of each arc is rθ = 1 * (40π/180) = π/4.5 ≈ 0.698 units. Since there are 9 corners, the total added perimeter from all the arcs is 9 * π/4.5 = 2π ≈ 6.283 units. 

Additionally, each original side, when offset outward by 1 unit, effectively shortens in length because the ends are rounded. Wait, no. Wait, actually, when you offset a polygon, the original sides are translated outward, but the distance between the translated sides is increased by 2r / sin(α/2), where α is the internal angle. Wait, perhaps I need to compute the length of each side after offsetting. Alternatively, maybe the total perimeter is just the original perimeter plus the circumference of the circle with radius 1. But that seems too simplistic. Wait, but in the case of a polygon with many sides, the rounded corners approximate a circle, so the total length of all the arcs would be 2πr. Therefore, if the polygon has many sides, the total perimeter would be original perimeter plus 2πr. For a square, as I mentioned earlier, the perimeter becomes original perimeter plus 2πr. For a triangle, same thing. So regardless of the number of sides, as long as the offset radius is smaller than the apothem, the perimeter increases by 2πr.
</execution_2.6>

<execution_2.7>
Therefore, for a polygon, even with many sides, the same formula should approximately hold. Therefore, in this case, the perimeter of S is 18 + 2π. Therefore, the answer is 18 + 2π units.

But wait, let me check again. The problem says "the set of all points inside a regular nonagon... ".

Wait, hold on, actually, the Minkowski sum would include all points within 1 unit of the nonagon's perimeter, which includes both the interior and the exterior buffer. But in this problem, we are taking the union of the interior of the nonagon and the points within 1 unit of the perimeter. Therefore, it's exactly the Minkowski sum. So the perimeter of the Minkowski sum is indeed the original perimeter plus 2πr. Therefore, 18 + 2π*1 = 18 + 2π.
</execution_2.7>
</Execution_2>

<Plan_3>
Based on execution_2.7:
But the problem says "the set of all points less than 1 unit away from a point on the perimeter of the nonagon". So actually, that's the same as the exterior buffer zone. So the union of the interior and the exterior buffer would be the original nonagon plus a 1-unit wide strip around it.

<prompt_3.1> Confirm the perimeter calculation and ensure the solution is precise. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
But since the nonagon's apothem is about 2.747, which is greater than 1, the buffer does not overlap with the nonagon's interior. Therefore, the union is just the original nonagon plus the buffer zone, which is a 1-unit wide annulus around the nonagon. Therefore, the perimeter of the union is the perimeter of the original nonagon plus the perimeter of the buffer zone. But the buffer zone's perimeter is the circumference of a circle with radius equal to the apothem plus 1 unit? Wait, no. Wait, the buffer zone is 1 unit away from the perimeter. So the perimeter of the buffer zone is a circle with radius equal to the original nonagon's circumradius plus 1 unit? Wait, perhaps not. Let me think.

Wait, the original nonagon has an apothem (distance from center to side) of approximately 2.747 units, and a circumradius (distance from center to vertex) which is larger. Let's compute the circumradius. The formula for the circumradius R of a regular polygon is R = s / (2 sin(π/n)). So for s=2, n=9:

R = 2 / (2 sin(π/9)) = 1 / sin(π/9)

sin(π/9) is sin(20°) ≈ 0.3420. Therefore, R ≈ 1 / 0.3420 ≈ 2.924 units.

So the circumradius is approximately 2.924 units. Therefore, the buffer of 1 unit around the perimeter would create a new shape whose outer radius is 2.924 + 1 ≈ 3.924 units. However, the perimeter of the buffer zone is not directly related to the circumradius. Instead, the buffer zone is a closed loop around the nonagon, 1 unit away from it. The perimeter of this buffer zone is actually equal to the perimeter of the original nonagon plus the length contributed by the buffer. But how?

Alternatively, if you have a polygon and you create a parallel curve at distance r from it, the perimeter of the parallel curve is equal to the original perimeter plus 2πr. This is because each convex vertex is replaced by a circular arc, and the total added length is 2πr. Therefore, the total perimeter is P + 2πr. Therefore, in our case, the original perimeter is 18, so the new perimeter is 18 + 2π*1 = 18 + 2π. Therefore, the perimeter of S is 18 + 2π.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Therefore, the final answer is 18 + 2π. But let me check once again.

<prompt_4.1> Perform a final verification and conclude the solution. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Suppose the nonagon was a circle with radius R. Then, the perimeter of the union of the circle and a 1-unit buffer would be the circumference of the original circle plus 2π*1. But in the case of a circle, the Minkowski sum with a disk is just a larger circle with radius R + 1, so its circumference is 2π(R + 1) = 2πR + 2π. Which matches the previous formula. So yes, the perimeter increases by 2π. 

Therefore, the same logic applies to the nonagon. The perimeter of the Minkowski sum (union) is the original perimeter plus 2πr, where r is 1. Therefore, 18 + 2π*1 = 18 + 2π. 

But just to be thorough, let's consider another perspective. Imagine walking along the perimeter of the nonagon. At each side, instead of going straight, you walk around a semicircular arc of radius 1 unit. Wait, no. Wait, actually, when you offset the polygon, each corner is replaced by a circular arc. The amount of arc is determined by the external angle. For a regular polygon, each external angle is 2π/n. For a nonagon, that's 2π/9. So each corner is replaced by a 2π/9 radians arc of radius 1. Therefore, each arc length is 1 * 2π/9 = 2π/9. Nine such arcs contribute 9*(2π/9) = 2π. Therefore, the total perimeter is the original perimeter (18) plus 2π. 

Therefore, I can confirm that the perimeter of S is 18 + 2π. 

But let me check if there's any possibility that the buffer zone is not simply adding 2π. For example, if the original polygon had an apothem less than 1, the buffer would overlap with the original interior, and the perimeter might not just be the original plus 2π. However, in our case, the apothem is approximately 2.747, which is greater than 1, so the buffer is entirely outside. Therefore, there's no overlapping, and the perimeter is indeed the original plus 2π.
</execution_4.1>
</Execution_4>

<Final_answer>
Thus, the final answer, encapsulated as requested, is $\boxed{18 + 2\pi}$ units.
</Final_answer>
</think>