Explanation:
Okay, so I need to figure out the value of n for a 3×n grid of points where there are a total of 70 squares of three sizes. The Asymptote code shows a diagram with some dots and filled squares, but since I can't see the image, I need to rely on the description and the math.

First, let me recall how to count squares in a grid. In a grid of points, the number of squares depends on the size of the grid and the sizes of the squares. For a grid with m rows and n columns, the number of squares can be calculated by considering squares of different sizes.

But here, the grid is 3×n. Wait, does that mean there are 3 rows and n columns of points? Or is it a grid that is 3 units wide and n units tall? The Asymptote code seems to draw a 3x3 grid of points initially, but then extends it with more points. Wait, looking at the Asymptote code:

The first part has dots at (0,0) to (2,2), so that's a 3x3 grid of points (since coordinates go from 0 to 2 in both x and y). Then there are some filled squares, and then more dots at (4,0), (5,0), etc. So maybe the grid is extended? But in the problem statement, it says "a rectangular 3×n grid of points". So maybe the grid is 3 rows and n columns? Or 3 columns and n rows? Wait, in a grid, usually, the first number is rows and the second is columns. But let's think.

If it's a grid of points, then a 3×n grid would be 3 rows and n columns of points. So for example, the Asymptote code initially draws points from (0,0) to (2,2), which is 3 points horizontally (columns) and 3 points vertically (rows). But then it adds more points at (4,0), (5,0), etc. Maybe the diagram is showing an example with n=3, but the actual problem is for general n?

Wait, but the problem says there are squares of three sizes. So maybe there are squares of size 1x1, 2x2, and 3x3? Wait, but in a 3×n grid of points, how big can the squares be?

Wait, let me think again. The grid is 3 rows and n columns of points. So the number of units between points horizontally is (n-1) units, and vertically it's 2 units (since there are 3 rows). So squares can be of different sizes. But how?

For example, a square can have sides of length 1, √2, √5, etc. Wait, but the problem says "three sizes", so maybe they mean squares aligned with the grid, not rotated?

Wait, the Asymptote code shows filled squares. The initial figure has three small squares, then maybe some larger ones? Wait, maybe the diagram is showing a 3x3 grid (which would be 3 rows and 3 columns of points) and then extending it to n=?

But the problem states the total number of squares is 70, so maybe I need to find n such that the total number of squares in a 3×n grid of points is 70. But first, I need to understand how to count the squares in such a grid.

Let me recall that in a grid with m rows and n columns of points, the number of squares is the sum over all possible square sizes.

But first, clarify: if it's a grid of points, then the number of squares depends on the number of unit squares, 2x2 squares, etc. But in a grid with m rows and n columns of points, how many squares are there?

For example, in a grid with m rows and n columns, the number of squares of size k x k (where k is the number of units per side) is (m - k) * (n - k). Wait, but for squares aligned with the grid. So for example, in an m x n grid of points (so (m-1)x(n-1) grid of squares), the number of squares is sum_{k=1}^{min(m-1,n-1)} (m - k)(n - k).

Wait, but here the grid is 3×n points. So m=3, n columns. Therefore, the number of squares would be sum_{k=1}^{min(2, n-1)} (3 - k)(n - k). Wait, since m=3 rows of points, so vertical spacing is 2 units. So the number of squares with side length k units is (number of positions horizontally) times (number of positions vertically). For horizontal positions, since we have n columns of points, the number of positions for a square of width k is (n - k). For vertical positions, since there are 3 rows of points, the number of positions is (3 - k). Therefore, total number of squares of size k is (3 - k)(n - k). But k can be from 1 to 2, because if k=3, then (3 - 3) = 0. So the total number of squares would be:

For k=1: (3 - 1)(n - 1) = 2(n - 1)

For k=2: (3 - 2)(n - 2) = 1*(n - 2)

So total number of squares is 2(n - 1) + (n - 2) = 2n - 2 + n - 2 = 3n - 4.

But wait, but the problem mentions three sizes of squares. So if k=1, k=2? But in a grid of 3 points vertically, you can only have squares of size 1x1 and 2x2? Wait, but 2x2 squares would have side length 2 units. But in a 3-point grid vertically, can you fit a 2x2 square? Let me visualize:

Imagine points at (0,0), (1,0), (2,0), ..., (n-1,0) horizontally, and (0,1), (0,2) vertically. So a square of side length 1 unit would be between (0,0)-(1,0)-(1,1)-(0,1). Similarly, a square of side length 2 units would be between (0,0)-(2,0)-(2,2)-(0,2). Wait, but that square is 2 units wide and 2 units tall. However, in the vertical direction, there are only 3 points, so from y=0 to y=2, so a square of side length 2 would fit. But the grid is only 3 points high, so the number of positions vertically for a square of size 2x2 is 3 - 2 = 1. So that's why we have (3 - k)(n - k). So for k=2, it's (1)*(n - 2). So total squares would be 2(n - 1) + (n - 2) = 3n - 4. But the problem says there are 70 squares. So 3n - 4 = 70 => 3n = 74 => n = 74 / 3 ≈ 24.666... which is not an integer. That can't be.

Therefore, my previous reasoning must be wrong.

Wait, maybe there are more than two sizes? The problem mentions three sizes. So maybe squares of different orientations? Like squares that are not axis-aligned?

For example, in addition to the axis-aligned squares, there are squares that are tilted. For example, in a grid, you can have squares that are rotated by 45 degrees. But in that case, the number of squares would be more complicated. However, the problem says there are three sizes of squares. So perhaps the three sizes are 1x1, 2x2, and something else?

Wait, maybe the three sizes refer to different numbers of unit squares? For example, in a grid, there can be squares of different sizes. Wait, but in a 3×n grid of points, axis-aligned squares can be of sizes 1x1, 2x2. But maybe also squares of other sizes?

Wait, let me think. For example, if n is large enough, you can have squares that are not aligned with the grid. For example, in a 3×n grid of points, can you have squares with sides of √2, √5, etc.?

But the problem says there are three sizes. So maybe the three sizes are 1x1, 2x2, and something else? But how?

Alternatively, maybe the original problem is referring to the number of squares of three different sizes, which includes both axis-aligned and tilted squares?

But how would that work?

Wait, let me try to think again. If it's a 3×n grid of points, then the number of squares (axis-aligned) is 2(n - 1) + 1(n - 2) = 3n - 4, as above. However, that formula gives 3n - 4 squares. But if the problem says there are 70 squares of three sizes, then 3n - 4 = 70 would imply n ≈ 24.666, which is not possible. Hence, there must be other squares considered.

Therefore, maybe the problem is counting both axis-aligned squares and squares of other sizes? So total squares include axis-aligned and tilted squares. Then, total number of squares is more than 3n - 4.

Therefore, maybe we need to calculate all possible squares in a 3×n grid of points.

But how?

First, let me consider axis-aligned squares. As before, for a grid with 3 rows and n columns of points.

For axis-aligned squares, the number is:

For size 1x1: The number of 1x1 squares is (number of positions horizontally) times (number vertically). Since horizontally, between columns, there are (n - 1) positions, vertically between rows, there are (3 - 1) = 2 positions. Therefore, number of 1x1 squares is (n - 1) * 2.

For size 2x2: Similarly, horizontally, the number of positions is (n - 2), vertically (3 - 2) = 1. So number of 2x2 squares is (n - 2) * 1.

Therefore total axis-aligned squares: 2(n - 1) + (n - 2) = 3n - 4, same as before.

Now, if there are other squares, tilted ones, then we need to count those too. The problem says "three sizes of squares", so maybe there are three different sizes of squares, which could be combinations of axis-aligned and tilted.

Alternatively, maybe the problem is referring to three different sizes in terms of side lengths? For example, 1x1, sqrt(2)x sqrt(2), and 2x2? But then there's more possibilities.

Alternatively, maybe the grid is not 3 rows and n columns, but 3 units by n units? Wait, the problem says it's a "rectangular 3×n grid of points". So that would mean 3 rows and n columns of points. So the grid is 3 points high and n points wide.

But then, how can there be three sizes of squares?

Alternatively, maybe the original problem refers to a grid with 3 rows and n columns, and squares can be of sizes that depend on both dimensions. Wait, perhaps the Asymptote code shows a specific case where n is 3? Let me check.

Looking at the Asymptote code:

First, it draws points from (0,0) to (2,2), which is 3x3 grid of points. Then there's a few squares filled in, then more points added at (4,0), (5,0), etc. Maybe the diagram is showing n=3 and then extending it? But the problem says total squares of three sizes, so maybe in the diagram, there is a 3x3 grid with some squares, but the actual problem is for a larger grid?

Alternatively, maybe the grid is 3×n, but the squares are of three different sizes. Let me think about possible squares in a 3×n grid.

First, axis-aligned squares:

- 1x1: as above.

- 2x2: as above.

- Also, in some cases, maybe 1x1 squares in different orientations? No, they are same size.

But maybe squares of different sizes when considering that the grid has 3 rows. For example, if you have a square that is 1 unit tall and 1 unit wide, but also squares that are 1 unit wide and 2 units tall? No, those would be rectangles, not squares. So axis-aligned squares must have same width and height.

Therefore, only 1x1 and 2x2. Then, if there are tilted squares.

For example, in a grid, there can be squares that are diamonds, like with side length sqrt(2), formed by connecting points diagonally. For example, a square that goes from (0,0) to (1,1) to (0,2) to (-1,1) would be a square, but in our grid, we have only points from (0,0) up to some (n-1, 2). So depending on n, such squares may exist.

Similarly, there can be larger tilted squares. For example, with side length sqrt(5), etc.

So, perhaps the three sizes refer to:

1x1 axis-aligned squares,

2x2 axis-aligned squares,

and 1x1 tilted squares (with side length sqrt(2)).

But then, the problem says "three sizes", so maybe those are considered as different sizes? But 1x1 and sqrt(2)x sqrt(2) are different sizes.

But in that case, how many squares of each size?

So let me calculate the total number of squares:

First, axis-aligned:

- 1x1: 2(n - 1)

- 2x2: (n - 2)

Then, tilted squares:

Let me think about squares that have side length sqrt(2). These are squares whose sides are diagonals of 1x1 squares. So, for example, in a 3x3 grid, how many such squares are there?

In a grid, such squares can be formed by choosing two adjacent points horizontally and two adjacent points vertically, but shifted. Wait, actually, a square of side sqrt(2) would have vertices at (x,y), (x+1, y+1), (x, y+2), (x-1, y+1). But to have such a square, we need to have points at (x-1, y+1), so that would require x ≥ 1 and y+1 ≤ 2 (since there are only 3 rows). So for the grid of 3 rows and n columns, how many such squares?

Each such square requires a 2x2 block of points. Wait, no. Let me think again. If I want a square with side length sqrt(2), then it's determined by two adjacent points horizontally and two adjacent points vertically, connected by diagonals. So, for example, the square with vertices at (0,0), (1,1), (0,2), (-1,1) – but this would require a point at (-1,1), which is not in our grid. Therefore, in the grid of 3 rows and n columns of points, the squares of side length sqrt(2) would have to be placed such that all four vertices are within the grid.

For example, consider a square with vertices at (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). So, to have these points, x must be at least 1, and y+1 must be at most 2. Since y can be 0 or 1 (since there are 3 rows). If y = 0, then y+1 = 1, which is okay. If y = 1, then y+1 = 2, which is okay. Therefore, for each y in {0,1}, and x from 1 to n - 1, we can have such squares.

But wait, each such square is determined by the bottom-left corner (x, y) and then the square extends to (x+1, y+1), etc. However, for a square with side length sqrt(2), the number of such squares would be (number of positions horizontally) times (number vertically). Wait, for each y (rows), and for each x (columns) such that there's space to the left and right?

Wait, actually, the square spans horizontally from x-1 to x+1, and vertically from y to y+2. Therefore, in terms of the grid of points:

To have a square with vertices at (x-1, y), (x, y+1), (x+1, y), (x, y-1)? Wait, maybe I need to think differently.

Alternatively, the number of squares of side sqrt(2) is equal to the number of 2x2 blocks in the grid. Because each 2x2 block can form four such squares? Wait no. Wait, in a 2x2 block of points, how many squares of side sqrt(2) are there? Actually, each 2x2 block contains exactly one square of side sqrt(2). Wait, no. Wait, in a 2x2 grid of points, you can form a square by connecting the four points. Wait, no, that's a square of side length 1? Wait, connecting four points (0,0), (1,0), (1,1), (0,1) is a square of side length 1. But if you connect (0,0), (1,1), (0,2), (-1,1) – but that requires points outside of the 2x2 grid. Therefore, in a 2x2 grid of points, there are no squares of side sqrt(2). Wait, confusion here.

Wait, maybe I need to think of the squares that are rotated. Let me think in terms of coordinates. Suppose we have a square with vertices at (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). So to have all these points in the grid, we need:

For the point (x - 1, y + 1): x - 1 ≥ 0 => x ≥ 1; and y + 1 ≤ 2 => y ≤ 1. Since our grid has y from 0 to 2 (rows), so y can be 0 or 1. Therefore, for each y in {0,1}, and x in {1, 2, ..., n - 1}, we can have such squares. Therefore, the number of such squares is 2*(n - 1).

Wait, but each such square is unique? For example, if we take y = 0 and x = 1, then the square is (1,0), (2,1), (1,2), (0,1). Similarly, for y = 1 and x = 1, the square is (1,1), (2,2), (1,3)... Wait, but y = 1, y + 2 = 3, but there are only 3 rows (y=0,1,2). Therefore, y + 2 must be ≤ 2. Therefore, y + 2 ≤ 2 => y ≤ 0. Therefore, earlier reasoning is wrong.

Wait, if we have the square with vertices at (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). Then for the last point, (x - 1, y + 1), we need y + 1 ≤ 2, so y ≤ 1. Also, for the third vertex, (x, y + 2), we need y + 2 ≤ 2 => y ≤ 0. Therefore, only when y = 0. So y must be 0. Then, the square is determined by x ≥ 1 (so that (x - 1, 0 + 1) = (x - 1, 1) exists). So for y = 0, x can be from 1 to n - 1. Therefore, the number of such squares is (n - 1). Similarly, are there squares in other orientations?

For example, squares that go in the other direction? For example, square with vertices (x, y), (x + 1, y - 1), (x, y - 2), (x - 1, y - 1). But y - 2 needs to be ≥ 0 => y ≥ 2. Since y can be 0, 1, 2. So y = 2, then the square would have vertices (x, 2), (x + 1, 1), (x, 0), (x - 1, 1). Similarly, for y = 2, x can be 1 to n - 1, so another (n - 1) squares.

Therefore, in total, there are 2*(n - 1) squares of side length sqrt(2). Wait, but these are the squares that are "diamond" shaped, with side length sqrt(2). Therefore, that's another size of square. So now, we have three types of squares:

1. Axis-aligned 1x1: 2(n - 1)

2. Axis-aligned 2x2: (n - 2)

3. Tilted squares of side sqrt(2): 2(n - 1)

Wait, but then total squares would be 2(n - 1) + (n - 2) + 2(n - 1) = 2(n - 1) + 2(n - 1) + (n - 2) = 4(n - 1) + n - 2 = 4n - 4 + n - 2 = 5n - 6.

But if the total is 70, then 5n - 6 = 70 => 5n = 76 => n = 15.2. Not integer. Still not possible. So something is wrong here.

Alternatively, maybe there are other tilted squares. For example, squares with side length sqrt(5). Let me think.

A square with side length sqrt(5) would be formed by moving 1 unit in x and 2 units in y, or something like that. For example, the square with vertices at (0,0), (1,2), (3,1), (2,-1). But again, the last point is outside the grid. So, in our 3-row grid, can we have such squares?

Suppose we have a square with side length sqrt(5). For example, take a square that goes from (x, y) to (x + 1, y + 2). Wait, the distance between (x, y) and (x + 1, y + 2) is sqrt(1^2 + 2^2) = sqrt(5). Then, to form a square, we need another point. The next point would be (x + 2, y + 1), and then back to (x + 1, y). Let me check:

If we have points A(x, y), B(x + 1, y + 2), C(x + 2, y + 1), D(x + 1, y). Then AB vector is (1, 2), BC vector is (1, -1), CD vector is (-1, -2), DA vector is (-1, 1). Wait, is that a square? The lengths:

AB: sqrt(1 + 4) = sqrt(5)

BC: sqrt(1 + 1) = sqrt(2)

No, not equal. So that's not a square. Maybe another configuration.

Alternatively, take points A(x, y), B(x + 2, y + 1), C(x + 1, y + 3), D(x - 1, y + 2). But y + 3 would be out of bounds. Not helpful.

Alternatively, maybe squares of side length sqrt(5) can be formed with points in the grid? For example, consider a square with vertices at (0,0), (1,2), (3,1), (2,-1). But as before, (2,-1) is outside the grid. If our grid has 3 rows, then y is from 0 to 2, so if we try to have a square with y-coordinate up to 2, then maybe (0,0), (2,1), (1, 2), (-1, 1). Again, (-1,1) is outside. Alternatively, maybe (1,0), (2,2), (0,2), (0,0)? Wait, that's not a square. Let me check distances:

Between (1,0) and (2,2): sqrt(1 + 4) = sqrt(5)

Between (2,2) and (0,2): 2 units

Not equal. So not a square. Maybe a different square.

Alternatively, (0,0), (1,1), (2,0), (1,-1). Again, (1,-1) is outside.

Alternatively, perhaps no squares of side length sqrt(5) can be formed in a 3-row grid. Therefore, maybe only the two axis-aligned sizes and the tilted squares of side sqrt(2). But then total squares would be as before, which leads to n non-integer. So something is wrong here.

Alternatively, maybe I miscounted the number of tilted squares.

Let me double-check the tilted squares.

Earlier, I considered squares with vertices like (x,0), (x + 1,1), (x,2), (x - 1,1). For such a square to exist, we need x ≥ 1 (to have x - 1 ≥ 0), and y = 0. Then, for each x from 1 to n - 1, we have such a square. Similarly, for y = 2, we can have squares going downward. Wait, let me think again.

If we have a square with vertices (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). For this square, y + 2 must be ≤ 2, so y ≤ 0. Therefore, y = 0. Then, we need x - 1 ≥ 0 => x ≥ 1. And the rightmost point is (x + 1, y + 1) must be within the grid: since x + 1 ≤ n - 1 => x ≤ n - 2. Wait, no. Wait, in our grid, the columns go from 0 to n - 1. Therefore, (x + 1) must be ≤ n - 1 => x ≤ n - 2. Therefore, x ranges from 1 to n - 2. Therefore, number of such squares is (n - 2). Similarly, if we consider squares going downward, with y = 2, then we have:

Square with vertices (x, 2), (x + 1, 1), (x, 0), (x - 1, 1). Similarly, y = 2, so y - 2 = 0. So need x ≥ 1, and x + 1 ≤ n - 1 => x ≤ n - 2. Therefore, same number of squares: n - 2. Therefore total tilted squares of side sqrt(2) would be 2*(n - 2). Therefore total squares would be:

Axis-aligned:

1x1: 2(n - 1)

2x2: (n - 2)

Tilted sqrt(2): 2(n - 2)

Total: 2(n - 1) + (n - 2) + 2(n - 2) = 2n - 2 + n - 2 + 2n - 4 = 5n - 8

Set equal to 70:

5n - 8 = 70 => 5n = 78 => n = 15.6. Still not integer.

Hmm. Not helpful. Maybe there are more squares?

Alternatively, maybe squares with side length sqrt(5)? Let me try again.

Suppose we have a square that goes from (x, y) to (x + 1, y + 2) to (x + 2, y + 1) to (x + 1, y - 1). But again, y - 1 may be negative. Let's check coordinates in a 3x3 grid.

Take for example, in a 3x3 grid (n=3), points (0,0), (1,0), (2,0), (0,1), (1,1), (2,1), (0,2), (1,2), (2,2).

Can we form a square with side length sqrt(5)? Let's pick points (0,0), (1,2), (3,1), but (3,1) is outside. Wait, what about (0,0), (2,1), (1, 3), but again out of bounds. If we have points (1,0), (2,2), (0,2), ( -1, 0). No. Not helpful.

Alternatively, maybe a square with vertices at (0,1), (2,0), (1,-1), (-1,0)? No. Not within the grid. Seems impossible. Therefore, maybe no squares of side length sqrt(5) in a 3-row grid. Hence, only axis-aligned and side sqrt(2). But the total number is 5n - 8, which doesn't give integer.

Alternatively, perhaps I have miscalculated the number of tilted squares. Let me think again.

Suppose we think of all squares, not just the ones with side sqrt(2). For example, in addition to the ones I considered, are there other sizes?

For example, squares with side length 2*sqrt(2)? That would be larger squares, but in a 3-row grid, maybe not possible. For example, a square that spans from (0,0) to (2,2) to (0,4)... but y=4 is outside.

Alternatively, maybe squares with side length sqrt( (n - 1)^2 + 2^2 )? Not sure.

Alternatively, maybe there are other orientations for tilted squares? For example, squares that are not just diamond-shaped but in another way?

Wait, let me try to think differently. Maybe the problem is referring to squares with different side lengths in terms of grid units, not necessarily axis-aligned. For example, in a grid, the squares can have different sizes based on their side lengths.

But how many distinct sizes of squares are there in a 3×n grid?

Let me consider possible squares:

1. 1x1 axis-aligned.

2. 2x2 axis-aligned.

3. 1x1 tilted (sqrt(2)).

Are there any others?

Alternatively, perhaps squares with side length 2 units (axis-aligned) and tilted ones? Wait, no, 2 units is already considered.

Alternatively, squares with side length sqrt(1^2 + 2^2) = sqrt(5). As discussed before, can't be formed.

Alternatively, squares of side length 2x1 (which are rectangles, not squares). Not applicable.

Alternatively, maybe 2x2 squares are not the only axis-aligned ones. Wait, no. In a 3x n grid, axis-aligned squares can be 1x1 (side length 1) and 2x2 (side length 2). Because 3 units in height allows for a square of side length 2 (from y=0 to y=2). If you try to make a square of side length 3, it would require y going from 0 to 3, which isn't there. So only 1x1 and 2x2.

If that is the case, then the total number of squares is axis-aligned (1x1 and 2x2) and tilted squares (sqrt(2)).

Earlier calculation for tilted squares was 2(n - 2). So total squares:

1x1: 2(n - 1)

2x2: (n - 2)

sqrt(2): 2(n - 2)

Total: 2(n - 1) + (n - 2) + 2(n - 2) = 5n - 8

Which for total 70, 5n = 78, n=15.6. Not possible.

Alternatively, maybe my count for tilted squares is wrong. Let me think again.

How many squares of side sqrt(2) are there?

Each such square is determined by a 2x1 rectangle? Or a 1x2 rectangle?

Wait, if I have a square tilted by 45 degrees, with side length sqrt(2), it can be inscribed in a 2x2 grid of points. So in a 3x n grid, how many 2x2 blocks of points are there? The number of 2x2 blocks is (n - 1) * (3 - 1) = (n - 1)*2. Each 2x2 block can contain one square of side sqrt(2). Wait, no. In a 2x2 grid of points, you can have one square of side sqrt(2) which is the diamond formed by connecting the midpoints. Wait, but the midpoints are not points in the grid. Wait, no, confusion again.

Wait, in a 2x2 grid of points, which are four points forming a square. To form a square of side sqrt(2), you need to connect the corners, but that's a square of side length 1*sqrt(2). Wait, no. The distance between adjacent points is 1 unit. The distance between diagonal points is sqrt(2). So if you connect (0,0), (1,1), (0,2), (-1,1), but again, (-1,1) is outside. However, in a 2x2 grid of points (say, from (0,0) to (1,1)), the square of side sqrt(2) would require points outside. Therefore, actually, there are no squares of side sqrt(2) in a 2x2 grid of points. Therefore, perhaps my previous count is wrong.

Therefore, maybe there are no squares of side sqrt(2)? Which contradicts previous thought.

Alternatively, maybe squares that are not based on the 2x2 blocks?

Wait, if we look at the Asymptote code given in the problem. The Asymptote code seems to draw a 3x3 grid (points from (0,0) to (2,2)) and then adds some filled squares. Let me try to imagine that.

The Asymptote code draws:

First, some dots at (0,0) to (2,2). Then, it fills some small squares near the right side. For example:

filldraw((2.95,-0.05)--(3.05,-0.05)--(3.05,0.05)--(2.95,0.05)--cycle,black);

This seems to be a small square to the right of the grid. Similarly, other squares. Then, it adds more dots at (4,0), (5,0), etc. Maybe the diagram is showing a 3x3 grid with some squares, and the problem is generalized to a 3xn grid? But since the problem states that there are three sizes of squares, maybe in the Asymptote diagram, there are three sizes of squares.

If I can figure out the number of squares in the Asymptote diagram, maybe I can extrapolate.

But since I can't see the image, let me think: the Asymptote code draws a 3x3 grid of points, then adds more points to the right. The filled squares are small, maybe 1x1, 2x2, and something else?

Wait, in the first part, the filled squares are small, maybe 1x1 or something else. But if the grid is 3x3, then:

Number of axis-aligned squares:

- 1x1: In a 3x3 grid of points (which is 2x2 grid of squares), there are 2*2 = 4 squares of 1x1.

- 2x2: There is 1 square of 2x2.

Total axis-aligned squares: 4 + 1 = 5.

Then, if there are tilted squares, for example, the square that goes from (0,0), (1,1), (0,2), (-1,1). But since (-1,1) is not in the grid, it's not there. However, if we consider the square from (1,0), (2,1), (1,2), (0,1). This square is inside the grid. So that's one tilted square. Similarly, from (0,1), (1,2), (0,2)... wait no. If I take (0,1), (1,2), (2,1), (1,0). That's a tilted square of side sqrt(2), and it's inside the grid. So in the 3x3 grid, how many such squares are there?

Each of these tilted squares is formed by choosing a 2x2 block of points. Wait, no. The square (0,1), (1,2), (2,1), (1,0) is formed by the points in the middle of the grid. Similarly, there is another one on the top: (0,0), (1,1), (2,0), (1,-1)... no, (1,-1) is outside. But if we take (1,0), (2,1), (1,2), (0,1). That's one. Similarly, if we have (0,1), (1,0), (2,1), (1,2). Wait, same square. So in a 3x3 grid, how many tilted squares are there?

Actually, there are 2 such squares. One at the center and one... Wait, no. Wait, in the 3x3 grid, the number of tilted squares of side sqrt(2) would be 2. For example, the square mentioned above and another one if shifted? But since it's only 3 rows, shifting would go out of bounds. So maybe only 2 tilted squares?

Wait, no. Wait, if you consider all possible positions. For example, starting at (0,1), the square would require (1,2), (2,1), and (1,0). Similarly, starting at (1,0), you get the same square. So maybe only one unique square. But the Asymptote code shows three filled squares? Let me check the Asymptote code again.

The Asymptote code has:

filldraw((2.95,-0.05)--(3.05,-0.05)--(3.05,0.05)--(2.95,0.05)--cycle,black);

This seems to be a square to the right of the grid, maybe indicating the grid extends? Then:

filldraw((2.45,-0.05)--(2.55,-0.05)--(2.55,0.05)--(2.45,0.05)--cycle,black);

This is smaller square, maybe indicating a smaller square? And:

filldraw((3.45,-0.05)--(3.55,-0.05)--(3.55,0.05)--(3.45,0.05)--cycle,black);

Another small square. Then similarly above, at y=0.95 and y=1.95, same filled squares. So maybe there are 3 sizes of squares? The small ones (maybe 1x1), medium ones (maybe 2x2), and larger ones? But since the grid is only 3 rows, perhaps the squares are:

- Small squares (1x1),

- Medium squares (maybe 2x2),

- And another size? Maybe squares that are 1x1 tilted?

If the diagram shows three sizes of squares, then maybe in the 3x3 grid, there are 1x1, 2x2, and tilted squares of side sqrt(2). So total squares would be 5 axis-aligned (4 + 1) and 2 tilted squares, totaling 7? But the problem says there are 70 squares in total for some n. So maybe the Asymptote code is showing part of the problem, where with n=3, the number of squares is 7? But in reality, for n=3, total squares would be:

Axis-aligned: 2(n - 1) + (n - 2) = 2*2 + 1 = 5

Tilted: 2(n - 2) = 2(1) = 2

Total: 5 + 2 = 7. Which matches the hypothetical. If that's the case, then the problem says that for a 3xn grid, the total number of squares is 70. So using the formula:

Total squares = axis-aligned + tilted = [2(n - 1) + (n - 2)] + 2(n - 2) = 2n - 2 + n - 2 + 2n - 4 = 5n - 8

Set equal to 70: 5n - 8 = 70 => 5n = 78 => n = 15.6. Still not integer.

Hmm, contradiction. So there must be an error in my counting.

Wait, let me check for n=3, total squares is 5 + 2 = 7. But if I manually count, in a 3x3 grid of points:

Number of squares:

Axis-aligned:

1x1: 2*2 = 4 (since (3-1)*(3-1) ??? Wait no.

Wait earlier formula said for axis-aligned squares:

For k=1: (3 - 1)(n - 1) = 2*(3 - 1) = 4

For k=2: (3 - 2)(n - 2) = 1*(1) = 1

Total axis-aligned: 5.

Tilted squares: 2*(n - 2) = 2*(1) = 2.

Total: 7. Okay, so for n=3, total squares is 7.

If the problem says that there are 70 squares, then solving 5n - 8 = 70 => n=15.6, which is not integer. Therefore, my formula is wrong.

Alternatively, maybe the number of tilted squares is different? Let me think again.

For the tilted squares of side sqrt(2), in a 3xn grid, how many are there?

Each such square is determined by its topmost point. For example, in the 3x3 grid, the square is centered. But in general, for a grid with n columns, how many positions are there for such squares?

Let me consider that for each square of side sqrt(2), it spans horizontally from column x to column x + 1, and vertically from row y to row y + 1. Wait, no. Let me think of the square with vertices at (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). As before, to have all points in grid:

y + 2 ≤ 2 => y ≤ 0, so y = 0.

x - 1 ≥ 0 => x ≥ 1

x + 1 ≤ n - 1 => x ≤ n - 2

Therefore number of such squares is (n - 2). Similarly, squares going downward from y=2: vertices (x, 2), (x + 1, 1), (x, 0), (x - 1, 1). So similar constraints: x ≥ 1, x ≤ n - 2. So number of squares is (n - 2). Therefore total tilted squares: 2(n - 2)

So for n=3, it's 2(1)=2. Which matches earlier.

Therefore total squares is 5n - 8. For n=3, it's 5*3 - 8 = 15 - 8 = 7. Correct.

But if this formula is correct, then 5n - 8 = 70 => n = 78 / 5 = 15.6. Not possible. Therefore, there must be more squares.

Wait, maybe there are more tilted squares? Like squares of different sizes?

For example, squares with side length sqrt(5). Wait, let's try to find such squares.

Consider a square with vertices at (x, y), (x + 2, y + 1), (x + 1, y + 3), (x - 1, y + 2). But y + 3 would be outside the grid. What if we take y = 2? Then y + 3 = 5, still outside.

Alternatively, a square with vertices at (x, y), (x + 1, y + 2), (x + 2, y + 1), (x + 1, y - 1). But y - 1 needs to be ≥ 0, so y ≥ 1. For example, take y = 1. Then, points would be (x,1), (x + 1, 3) which is outside. No.

Alternatively, (x,0), (x + 2,1), (x + 1, 3), (x - 1, 2). Again, y=3 is outside.

Alternatively, (x,1), (x + 2, 2), (x + 1, 0), (x - 1, 1). Check if these points exist:

(x - 1, 1) must exist => x ≥ 1

(x + 1, 0) must exist => x + 1 ≤ n - 1 => x ≤ n - 2

(x + 2, 2) must exist => x + 2 ≤ n - 1 => x ≤ n - 3

Therefore, for such a square, x can range from 1 to n - 3. So number of such squares is (n - 3). Similarly, are there squares in the other direction?

For example, (x, 2), (x - 2, 1), (x - 1, 3), (x + 1, 2). But again, y=3 is outside. Not helpful.

So, if there are such squares, the number would be (n - 3). Then, total squares would be 5n - 8 + (n - 3) = 6n - 11. For n=3, it would be 6*3 - 11 = 7, which matches the previous total. But then if we set 6n - 11 = 70, 6n = 81, n = 13.5. Still not integer.

This is getting frustrating. Maybe there are more sizes? Or perhaps I'm missing something.

Wait, maybe the three sizes mentioned in the problem are not axis-aligned and tilted ones, but something else? For example, in the Asymptote code, maybe there are squares of sizes 1x1, 2x2, and 2x2 tilted? But that would be two sizes.

Alternatively, the problem may consider squares of different orientations but same side length? For example, 1x1 axis-aligned and 1x1 tilted are same size? But they are not. One is axis-aligned, the other is tilted.

Wait, going back to the problem statement:

"There is a total of 70 squares of three sizes whose vertices are points on this rectangular 3×n grid of points. What is the value of n?"

So there are three different sizes of squares, and total number is 70. So perhaps the three sizes refer to:

1. 1x1 axis-aligned

2. 2x2 axis-aligned

3. Some other size, like 1x1 tilted (sqrt(2))?

But according to previous calculation, total squares would be 5n - 8. But it doesn't fit 70.

Alternatively, maybe the three sizes are:

1. 1x1 axis-aligned

2. 2x2 axis-aligned

3. 3x3 axis-aligned? But in a 3x n grid, can we have 3x3 squares? No, because vertically there are only 3 rows, so a 3x3 square would need 3 units in both directions, but vertically, it would go from y=0 to y=3, but there are only 3 rows (y=0,1,2). Hence, no.

Alternatively, maybe the three sizes are 1x1, 2x2, and some larger tilted squares?

But as before, not sure.

Alternatively, maybe the three sizes are:

- Squares of size 1x1 (axis-aligned)

- Squares of size sqrt(2)x sqrt(2) (tilted, side length sqrt(2))

- Squares of size sqrt(5)x sqrt(5) (tilted, side length sqrt(5))

But then, need to count all these.

But how many squares of size sqrt(5)?

For example, in a 3x n grid, how many squares of side length sqrt(5) are there?

A square with side length sqrt(5) would have sides that are 1 unit in one direction and 2 units in another. So, for example, a square from (x, y) to (x + 1, y + 2) to (x, y + 4)... but y + 4 is outside the grid.

Alternatively, a square with vertices at (x, y), (x + 2, y + 1), (x + 1, y + 3), (x - 1, y + 2). Again, y + 3 is out.

Alternatively, in a 3-row grid, the maximum vertical extent is 2. So, to have a square with side length sqrt(5), we need a 1 unit horizontal and 2 units vertical move, but then the square would have to close back. For example, starting at (x, y), going to (x + 1, y + 2), then to (x + 2, y + 1), then to (x + 1, y). Let's check if this forms a square.

The distance between (x, y) and (x + 1, y + 2) is sqrt(1² + 2²) = sqrt(5).

The distance between (x + 1, y + 2) and (x + 2, y + 1) is sqrt(1² + (-1)²) = sqrt(2).

Not equal. Not a square.

Alternatively, starting at (x, y), going to (x + 2, y + 1), then to (x + 1, y + 3), but again, y + 3 is out.

Alternatively, starting at (x, y), going to (x + 2, y), then to (x + 1, y + 2), then to (x - 1, y + 2). Wait, this is getting too complex and likely outside the grid.

It seems impossible to have such squares in a 3-row grid. So maybe no squares of side sqrt(5). Hence, only two sizes? But the problem says three sizes.

Wait, perhaps the three sizes are axis-aligned 1x1, axis-aligned 2x2, and tilted 1x1 (side sqrt(2)), but I have to include them in the total.

If total squares are 5n - 8, and this equals 70, but it's not integer, maybe there is an error in the formula.

Wait, let's try with n=5. What is the total?

Axis-aligned: 2(5-1) + (5-2) = 8 + 3 = 11

Tilted: 2(5-2) = 6

Total: 17. If I use formula 5n - 8 = 5*5 - 8 = 17. Okay, matches. So 5n - 8 = total squares.

If n=10: 5*10 - 8 = 42

n=11: 5*11 - 8 = 47

n=12: 5*12 - 8 = 52

n=13: 5*13 - 8 = 57

n=14: 5*14 - 8 = 62

n=15: 5*15 - 8 = 67

n=16: 5*16 - 8 = 72

So 72 is for n=16. But we need 70. Hmm, between n=15 and n=16. But since n must be integer, this suggests that either the formula is incorrect, or there is another size of square.

Alternatively, maybe there is a miscalculation in how many tilted squares there are.

Wait, let me re-examine tilted squares. Perhaps I have undercounted or overcounted.

For each tilted square of side sqrt(2), it requires a specific position. In addition to the ones counted as 2(n - 2), maybe there are others?

For example, in the 3x3 grid, we had 2 tilted squares. But let me count manually:

In a 3x3 grid, how many squares of side sqrt(2) are there?

They are the squares that have vertices at (0,1), (1,2), (2,1), (1,0) and (1,0), (2,1), (1,2), (0,1). Wait, it's the same square. So only 1 unique square?

Wait, but in the Asymptote code, there are three filled squares, which might suggest more. Maybe the diagram shows multiple squares?

Alternatively, maybe there are more tilted squares. Let me think.

If we consider all possible placements:

For y = 0, the square is (x,0), (x+1,1), (x,2), (x-1,1). For x from 1 to n - 2.

For y = 2, the square is (x,2), (x+1,1), (x,0), (x-1,1). For x from 1 to n - 2.

But for y = 1, is there a square?

Let's try to form a square with y = 1. Let's say vertices at (x,1), (x+1,2), (x,3), (x-1,2). But y=3 is out of bounds. Hence, no. Similarly, downward, (x,1), (x+1,0), (x, -1), ... also out of bounds. Hence, no squares with y=1.

Hence, only 2(n - 2) tilted squares.

Therefore, the formula seems correct. So total squares = 5n - 8.

If the problem says total squares is 70, then 5n - 8 = 70 => n = 78/5 = 15.6. Not possible. Hence, something is wrong.

Wait, let me check the problem statement again:

"There is a total of 70 squares of three sizes whose vertices are points on this rectangular 3×n grid of points. What is the value of n?"

Wait, maybe the three sizes are not referring to different side lengths, but three different orientations? No, that doesn't make sense.

Alternatively, maybe the three sizes are 1x1, 2x2, and 3x3? But in a 3x n grid, can we have 3x3 squares? No, because vertically there are only 3 points, so a square of size 3x3 would need 3 units vertically, which would go from y=0 to y=3, but there are only y=0,1,2. Hence, not possible.

Alternatively, the three sizes are 1x1, 1x1 tilted, and 2x2 tilted? But that would be three sizes, but how many squares?

Wait, if we consider:

- 1x1 axis-aligned: 2(n - 1)

- 1x1 tilted: 2(n - 2)

- 2x2 axis-aligned: (n - 2)

Total: 2(n - 1) + 2(n - 2) + (n - 2) = 2n - 2 + 2n - 4 + n - 2 = 5n - 8. Same as before. So same result.

Still no good.

Wait, maybe the problem counts squares of different sizes, where sizes are defined by number of unit squares they cover? For example, a 1x1 square covers 1 unit square, a 2x2 covers 4, etc. But the problem says "three sizes", which might mean three distinct sizes, not three distinct counts.

Alternatively, maybe the three sizes are 1x1, 2x2, and 3x1? No, that's not a square.

Wait, this is very confusing. Let me try to search for similar problems.

Wait, I recall that in a grid, the number of squares is calculated by sum_{k=1}^{min(m,n)} (m - k + 1)(n - k + 1). But in this case, the grid is 3 rows and n columns of points, which corresponds to a 2x(n - 1) grid of squares? Wait, no.

Wait, the number of squares in a grid with m rows and n columns of points is:

For axis-aligned squares, it's sum_{k=1}^{min(m-1, n-1)} (m - k) * (n - k).

In our case, m = 3 rows, n columns. So sum_{k=1}^2 (3 - k)(n - k) = (3 - 1)(n - 1) + (3 - 2)(n - 2) = 2(n - 1) + 1(n - 2) = 2n - 2 + n - 2 = 3n - 4.

But if we include tilted squares, it's more.

But the problem says there are 70 squares of three sizes, which might mean 3 types, but how?

Maybe the three sizes are 1x1, 2x2, and 1x1 tilted? Then total squares would be 3n - 4 + 2(n - 2) = 3n - 4 + 2n - 4 = 5n - 8. Same as before.

But again, 5n - 8 = 70 => n = 15.6.

Wait, unless the problem is not counting tilted squares, and the three sizes are 1x1, 2x2, and 3x3? But 3x3 is not possible. Or maybe the three sizes are 1x1, 1x1, and 2x2? No.

Wait, perhaps the problem is in the Asymptote code, and the three sizes are 1x1, 2x2, and 3x3? But in a 3x n grid, 3x3 would require 3 rows and 3 columns of points? No, 3x3 grid of points would allow 2x2 squares, but not 3x3.

Wait, the Asymptote code draws a 3x3 grid of points, then adds more points at (4,0), (5,0), etc. So it's extending the grid to the right. The filled squares in the Asymptote code are small squares, which could be 1x1, and maybe larger squares. But since it's hard to see, maybe the original problem is different.

Maybe the grid is n×3, not 3×n? If it's a grid with n rows and 3 columns of points? Let's try that.

If it's a grid with n rows and 3 columns of points, then number of axis-aligned squares would be:

For k=1: (n - 1) * (3 - 1) = 2(n - 1)

For k=2: (n - 2) * (3 - 2) = (n - 2) * 1

Total axis-aligned: 2n - 2 + n - 2 = 3n - 4

Tilted squares of side sqrt(2): in this case, with 3 columns, how many?

For example, squares like (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). For this, x must be ≥ 1, and y + 2 ≤ n - 1 (assuming n rows). Wait, if it's n rows, then rows go from y=0 to y = n - 1. So for y + 2 ≤ n - 1 => y ≤ n - 3. For each such y, and x from 1 to 3 - 1 = 2 (since columns are 3). Wait, columns are 3, so x ranges from 0 to 2. To have x - 1 ≥ 0, x ≥ 1. So x can be 1 or 2. So for each y from 0 to n - 3, there are 2 squares. So total tilted squares: 2 * (n - 2)

Wait, if it's n rows and 3 columns, then number of tilted squares would be 2*(n - 2). Then total squares would be 3n - 4 + 2(n - 2) = 3n - 4 + 2n - 4 = 5n - 8. Same as before. So no change.

If the grid is n rows and 3 columns, and total squares is 70, then 5n - 8 = 70 => n = 15.6. Still no.

Wait, this suggests that no matter if it's 3×n or n×3, the formula is same.

But the problem says "rectangular 3×n grid of points", so it's 3 rows and n columns.

Wait, let me try to think outside the box. Maybe the three sizes are not axis-aligned and tilted, but something else. For example, the problem says "squares of three sizes", maybe it's referring to the number of squares of each size, not the number of distinct sizes? No, the wording says "three sizes".

Alternatively, maybe the grid is 3×n grid of squares, not points? If the grid is 3×n grid of squares, then it would have (3 + 1) × (n + 1) points, making it a 4×(n + 1) grid of points. But the problem says "3×n grid of points", so likely it's 3 rows and n columns of points.

Wait, maybe the translation is wrong? The original says "rectangular 3×n grid of points". Maybe it's a grid with 3 points in one direction and n in the other?

Wait, perhaps I made a mistake in counting axis-aligned squares. Let me re-derive it.

In a grid with m rows and n columns of points, the number of squares is:

For each possible square size k (in units), where k ranges from 1 to min(m - 1, n - 1), the number of such squares is (m - k) * (n - k).

In our case, m = 3 rows, n columns. So k can be 1 or 2.

Number of 1x1 squares: (3 - 1) * (n - 1) = 2(n - 1)

Number of 2x2 squares: (3 - 2) * (n - 2) = 1*(n - 2)

Total axis-aligned: 2n - 2 + n - 2 = 3n - 4

If the problem says there are 70 squares, and it's only counting axis-aligned, then 3n - 4 = 70 => 3n = 74 => n = 74/3 ≈ 24.666. Not integer.

But the problem says there are squares of three sizes. So maybe it's counting both axis-aligned and tilted squares, and the three sizes are 1x1, 2x2, and something else. But as before, leading to 5n - 8 = 70.

Wait, unless there are more tilted squares. For example, in addition to the 2(n - 2) squares of side sqrt(2), there are also squares of side 2*sqrt(2) or others.

Let me try to count squares of side 2*sqrt(2). For example, a square formed by moving 2 units in one direction and 2 units in another, but in a 3-row grid...

For example, a square with vertices at (x, y), (x + 2, y + 2), (x, y + 4), (x - 2, y + 2). But y + 4 is out of bounds. Not possible.

Alternatively, a square with vertices at (x, y), (x + 2, y + 1), (x + 1, y + 3), (x - 1, y + 2). Again, y + 3 is out.

No luck. So no such squares.

Wait, maybe the three sizes are 1x1, 2x2, and 1x1 tilted? Then, total squares would be 2(n - 1) + (n - 2) + 2(n - 2) = 5n - 8, as before.

But 5n - 8 = 70 => n = 15.6. Not integer. So, what if the problem has a mistake? Or maybe my counting is wrong.

Wait, let me try to think of n=16: 5*16 - 8 = 72. Close to 70. If there were 72 squares for n=16, but the problem says 70. Maybe there are 2 fewer squares, which could happen if there are some squares counted in the formula that don't actually exist, or if the three sizes are different.

Alternatively, maybe the three sizes are 1x1, 2x2, and 1x1 tilted, but in the formula, I should subtract some cases? For example, in the tilted squares, maybe some of them are counted twice or something?

Wait, in the case of n=3, total squares is 7: 5 axis-aligned and 2 tilted. Let me list them:

Axis-aligned:

1x1: 4 squares: (0,0)-(1,0)-(1,1)-(0,1), (1,0)-(2,0)-(2,1)-(1,1), (0,1)-(1,1)-(1,2)-(0,2), (1,1)-(2,1)-(2,2)-(1,2)

2x2: 1 square: (0,0)-(2,0)-(2,2)-(0,2)

Tilted:

2 squares: (0,1)-(1,2)-(2,1)-(1,0) and (1,0)-(2,1)-(1,2)-(0,1) - but these are the same square, so only 1 unique square. Wait, but earlier calculation said 2. What's the issue here?

Ah! Here is the mistake! In the count of tilted squares, I assumed 2(n - 2), but in reality, for each n, there are (n - 2) squares in one direction, and (n - 2) in the other, but some are overlapping or counted multiple times?

No, let's think again. In a 3x3 grid (n=3), for tilted squares of side sqrt(2), how many are there?

As above, only 1 unique square. But according to the previous count, it should be 2*(n - 2) = 2*(1) = 2. So there is overcounting.

Why? Because when n=3, for y=0, x ranges from 1 to n-2 = 1, so one square. For y=2, similarly, one square. But these two squares are the same as the one described above? Wait, no. Each is a different square?

Wait, if y=0, the square is (x,0), (x+1,1), (x,2), (x-1,1). For x=1, this is (1,0), (2,1), (1,2), (0,1). This is one square.

For y=2, the square is (x,2), (x+1,1), (x,0), (x-1,1). For x=1, this is (1,2), (2,1), (1,0), (0,1). This is the same square as before! So it's counted twice. Therefore, the total number of unique tilted squares is (n - 2), not 2(n - 2). Hence, my previous count was double-counting!

Ah! This is the error! The squares for y=0 and y=2 are the same square, just traversed in different direction. So, the actual number of unique tilted squares is (n - 2), not 2(n - 2). Hence, total squares is axis-aligned (3n - 4) + tilted (n - 2) = 4n - 6.

Let me check for n=3: 4*3 - 6 = 6. But manually, we have 5 axis-aligned + 1 tilted = 6. Correct! Previously, I had 5 + 2 = 7, which was wrong. So the correct total is 4n - 6.

Let's verify for n=3: 4*3 - 6 = 6. Correct.

For n=4: 4*4 - 6 = 10. Let's count manually.

Axis-aligned:

1x1: 2*(4 - 1) = 6

2x2: 1*(4 - 2) = 2

Total axis-aligned: 8

Tilted: n - 2 = 2

Total: 8 + 2 = 10. Correct.

So the correct formula is total squares = 3n - 4 (axis-aligned) + (n - 2) (tilted) = 4n - 6.

Now, set 4n - 6 = 70 => 4n = 76 => n = 19.

Ah! Finally, an integer!

So, the value of n is 19.

But wait, let me confirm this with another example.

For n=4, total squares is 4*4 - 6 = 10. Let's count:

Axis-aligned 1x1: 6

Axis-aligned 2x2: 2

Tilted: 2

Total: 10. Correct.

For n=5: 4*5 - 6 = 14

Axis-aligned: 3*5 - 4 = 11

Tilted: 5 - 2 = 3

Total: 14. Correct.

Yes, this makes sense. So the correct formula is total squares = 4n - 6.

Hence, solving 4n - 6 = 70 => 4n = 76 => n = 19.

Thus, the value of n is 19.

**Final Answer**
\boxed{19}

To solve the problem, we are given that there are **70 squares** in a **rectangular 3×n grid of points**, and we're told that **there are squares of three different sizes**, and we need to find the value of $ n $.

---

### Step 1: Understanding the Grid and Square Counting

The grid is defined as a **3×n grid of points**. This means:
- **3 rows** of points along the vertical axis (i.e., y-coordinates)
- **n columns** of points along the horizontal axis (i.e., x-coordinates)

This corresponds to a **2 × (n - 1)** grid of unit squares (since a unit square requires 2 points in width and 2 points in height).

---

### Step 2: Counting Axis-Aligned Squares

For axis-aligned squares, we need to consider the number of **1×1** and **2×2** squares that can fit within the 3×n grid:

- **1×1 squares**: These are the smallest unit squares that can fit in the grid. Since each such square is formed by 4 consecutive points (2 horizontal and 2 vertical), the number of 1×1 squares is:

  $$
  (n - 1) \times 2 = 2(n - 1)
  $$

- **2×2 squares**: These are larger squares, each formed by a 2×2 block of points. There are 2 ways to fit these in width (3 - 2 = 1) and $ n - 2 $ ways in height. So the number of 2×2 squares is:

  $$
  (3 - 2) \times (n - 2) = 1 \times (n - 2) = n - 2
  $$

- **Total axis-aligned squares**:
  $$
  2(n - 1) + (n - 2) = 2n - 2 + n - 2 = 3n - 4
  $$

---

### Step 3: Counting Tilted Squares

Now consider **tilted squares**, those that are diagonal in orientation. Specifically, the smallest such square has a side length of $ \sqrt{2} $, and is formed by the diagonals of 1×1 squares.

These have the following characteristics:
- They require 3 rows and at least 3 columns.
- For each column $ x $ where $ x \in [1, n - 2] $, we can form one such square by connecting:
  - $ (x, 0) $
  - $ (x + 1, 1) $
  - $ (x, 2) $
  - $ (x - 1, 1) $
- And similarly, each such square is unique, and the count is equal to $ n - 2 $.

So, the total number of tilted squares is:

$$
n - 2
$$

---

### Step 4: Total Number of Squares

Combining both types of squares:

- **Axis-aligned squares**: $ 3n - 4 $
- **Tilted squares**: $ n - 2 $

Total number of squares:

$$
(3n - 4) + (n - 2) = 4n - 6
$$

---

### Step 5: Solving for $ n $

We are told that the total number of squares is **70**. So we set up the equation:

$$
4n - 6 = 70
$$

Solve for $ n $:

$$
4n = 76 \Rightarrow n = \frac{76}{4} = 19
$$

---

### ✅ Final Answer

$$
\boxed{19}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of n for a 3×n grid of points where there are a total of 70 squares of three sizes. The Asymptote code shows a diagram with some dots and filled squares, but since I can't see the image, I need to rely on the description and the math.

First, let me recall how to count squares in a grid. In a grid of points, the number of squares depends on the size of the grid and the sizes of the squares. For a grid with m rows and n columns, the number of squares can be calculated by considering squares of different sizes.

But here, the grid is 3×n. Wait, does that mean there are 3 rows and n columns of points? Or is it a grid that is 3 units wide and n units tall? The Asymptote code seems to draw a 3x3 grid of points initially, but then extends it with more points. Wait, looking at the Asymptote code:

The first part has dots at (0,0) to (2,2), so that's a 3x3 grid of points (since coordinates go from 0 to 2 in both x and y). Then there are some filled squares, and then more dots at (4,0), (5,0), etc. So maybe the grid is extended? But in the problem statement, it says "a rectangular 3×n grid of points". So maybe the grid is 3 rows and n columns? Or 3 columns and n rows? Wait, in a grid, usually, the first number is rows and the second is columns. But let's think.

If it's a grid of points, then a 3×n grid would be 3 rows and n columns of points. So for example, the Asymptote code initially draws points from (0,0) to (2,2), which is 3 points horizontally (columns) and 3 points vertically (rows). But then it adds more points at (4,0), (5,0), etc. Maybe the diagram is showing an example with n=3, but the actual problem is for general n?

Wait, but the problem says there are squares of three sizes. So maybe there are squares of size 1x1, 2x2, and 3x3? Wait, but in a 3×n grid of points, how big can the squares be?

Wait, let me think again. The grid is 3 rows and n columns of points. So the number of units between points horizontally is (n-1) units, and vertically it's 2 units (since there are 3 rows). So squares can be of different sizes. But how?

For example, a square can have sides of length 1, √2, √5, etc. Wait, but the problem says "three sizes", so maybe they mean squares aligned with the grid, not rotated?

Wait, the Asymptote code shows filled squares. The initial figure has three small squares, then maybe some larger ones? Wait, maybe the diagram is showing a 3x3 grid (which would be 3 rows and 3 columns of points) and then extending it to n=?

But the problem states the total number of squares is 70, so maybe I need to find n such that the total number of squares in a 3×n grid of points is 70. But first, I need to understand how to count the squares in such a grid.

Let me recall that in a grid with m rows and n columns of points, the number of squares is the sum over all possible square sizes.

But first, clarify: if it's a grid of points, then the number of squares depends on the number of unit squares, 2x2 squares, etc. But in a grid with m rows and n columns of points, how many squares are there?

For example, in a grid with m rows and n columns, the number of squares of size k x k (where k is the number of units per side) is (m - k) * (n - k). Wait, but for squares aligned with the grid. So for example, in an m x n grid of points (so (m-1)x(n-1) grid of squares), the number of squares is sum_{k=1}^{min(m-1,n-1)} (m - k)(n - k).

Wait, but here the grid is 3×n points. So m=3, n columns. Therefore, the number of squares would be sum_{k=1}^{min(2, n-1)} (3 - k)(n - k). Wait, since m=3 rows of points, so vertical spacing is 2 units. So the number of squares with side length k units is (number of positions horizontally) times (number of positions vertically). For horizontal positions, since we have n columns of points, the number of positions for a square of width k is (n - k). For vertical positions, since there are 3 rows of points, the number of positions is (3 - k). Therefore, total number of squares of size k is (3 - k)(n - k). But k can be from 1 to 2, because if k=3, then (3 - 3) = 0. So the total number of squares would be:

For k=1: (3 - 1)(n - 1) = 2(n - 1)

For k=2: (3 - 2)(n - 2) = 1*(n - 2)

So total number of squares is 2(n - 1) + (n - 2) = 2n - 2 + n - 2 = 3n - 4.

But wait, but the problem mentions three sizes of squares. So if k=1, k=2? But in a grid of 3 points vertically, you can only have squares of size 1x1 and 2x2? Wait, but 2x2 squares would have side length 2 units. But in a 3-point grid vertically, can you fit a 2x2 square? Let me visualize:

Imagine points at (0,0), (1,0), (2,0), ..., (n-1,0) horizontally, and (0,1), (0,2) vertically. So a square of side length 1 unit would be between (0,0)-(1,0)-(1,1)-(0,1). Similarly, a square of side length 2 units would be between (0,0)-(2,0)-(2,2)-(0,2). Wait, but that square is 2 units wide and 2 units tall. However, in the vertical direction, there are only 3 points, so from y=0 to y=2, so a square of side length 2 would fit. But the grid is only 3 points high, so the number of positions vertically for a square of size 2x2 is 3 - 2 = 1. So that's why we have (3 - k)(n - k). So for k=2, it's (1)*(n - 2). So total squares would be 2(n - 1) + (n - 2) = 3n - 4. But the problem says there are 70 squares. So 3n - 4 = 70 => 3n = 74 => n = 74 / 3 ≈ 24.666... which is not an integer. That can't be.

Therefore, my previous reasoning must be wrong.

Wait, maybe there are more than two sizes? The problem mentions three sizes. So maybe squares of different orientations? Like squares that are not axis-aligned?

For example, in addition to the axis-aligned squares, there are squares that are tilted. For example, in a grid, you can have squares that are rotated by 45 degrees. But in that case, the number of squares would be more complicated. However, the problem says there are three sizes of squares. So perhaps the three sizes are 1x1, 2x2, and something else?

Wait, maybe the three sizes refer to different numbers of unit squares? For example, in a grid, there can be squares of different sizes. Wait, but in a 3×n grid of points, axis-aligned squares can be of sizes 1x1, 2x2. But maybe also squares of other sizes?

Wait, let me think. For example, if n is large enough, you can have squares that are not aligned with the grid. For example, in a 3×n grid of points, can you have squares with sides of √2, √5, etc.?

But the problem says there are three sizes. So maybe the three sizes are 1x1, 2x2, and something else? But how?

Alternatively, maybe the original problem is referring to the number of squares of three different sizes, which includes both axis-aligned and tilted squares?

But how would that work?

Wait, let me try to think again. If it's a 3×n grid of points, then the number of squares (axis-aligned) is 2(n - 1) + 1(n - 2) = 3n - 4, as above. However, that formula gives 3n - 4 squares. But if the problem says there are 70 squares of three sizes, then 3n - 4 = 70 would imply n ≈ 24.666, which is not possible. Hence, there must be other squares considered.

Therefore, maybe the problem is counting both axis-aligned squares and squares of other sizes? So total squares include axis-aligned and tilted squares. Then, total number of squares is more than 3n - 4.

Therefore, maybe we need to calculate all possible squares in a 3×n grid of points.

But how?

First, let me consider axis-aligned squares. As before, for a grid with 3 rows and n columns of points.

For axis-aligned squares, the number is:

For size 1x1: The number of 1x1 squares is (number of positions horizontally) times (number vertically). Since horizontally, between columns, there are (n - 1) positions, vertically between rows, there are (3 - 1) = 2 positions. Therefore, number of 1x1 squares is (n - 1) * 2.

For size 2x2: Similarly, horizontally, the number of positions is (n - 2), vertically (3 - 2) = 1. So number of 2x2 squares is (n - 2) * 1.

Therefore total axis-aligned squares: 2(n - 1) + (n - 2) = 3n - 4, same as before.

Now, if there are other squares, tilted ones, then we need to count those too. The problem says "three sizes of squares", so maybe there are three different sizes of squares, which could be combinations of axis-aligned and tilted.

Alternatively, maybe the problem is referring to three different sizes in terms of side lengths? For example, 1x1, sqrt(2)x sqrt(2), and 2x2? But then there's more possibilities.

Alternatively, maybe the grid is not 3 rows and n columns, but 3 units by n units? Wait, the problem says it's a "rectangular 3×n grid of points". So that would mean 3 rows and n columns of points. So the grid is 3 points high and n points wide.

But then, how can there be three sizes of squares?

Alternatively, maybe the original problem refers to a grid with 3 rows and n columns, and squares can be of sizes that depend on both dimensions. Wait, perhaps the Asymptote code shows a specific case where n is 3? Let me check.

Looking at the Asymptote code:

First, it draws points from (0,0) to (2,2), which is 3x3 grid of points. Then there's a few squares filled in, then more points added at (4,0), (5,0), etc. Maybe the diagram is showing n=3 and then extending it? But the problem says total squares of three sizes, so maybe in the diagram, there is a 3x3 grid with some squares, but the actual problem is for a larger grid?

Alternatively, maybe the grid is 3×n, but the squares are of three different sizes. Let me think about possible squares in a 3×n grid.

First, axis-aligned squares:

- 1x1: as above.

- 2x2: as above.

- Also, in some cases, maybe 1x1 squares in different orientations? No, they are same size.

But maybe squares of different sizes when considering that the grid has 3 rows. For example, if you have a square that is 1 unit tall and 1 unit wide, but also squares that are 1 unit wide and 2 units tall? No, those would be rectangles, not squares. So axis-aligned squares must have same width and height.

Therefore, only 1x1 and 2x2. Then, if there are tilted squares.

For example, in a grid, there can be squares that are diamonds, like with side length sqrt(2), formed by connecting points diagonally. For example, a square that goes from (0,0) to (1,1) to (0,2) to (-1,1) would be a square, but in our grid, we have only points from (0,0) up to some (n-1, 2). So depending on n, such squares may exist.

Similarly, there can be larger tilted squares. For example, with side length sqrt(5), etc.

So, perhaps the three sizes refer to:

1x1 axis-aligned squares,

2x2 axis-aligned squares,

and 1x1 tilted squares (with side length sqrt(2)).

But then, the problem says "three sizes", so maybe those are considered as different sizes? But 1x1 and sqrt(2)x sqrt(2) are different sizes.

But in that case, how many squares of each size?

So let me calculate the total number of squares:

First, axis-aligned:

- 1x1: 2(n - 1)

- 2x2: (n - 2)

Then, tilted squares:

Let me think about squares that have side length sqrt(2). These are squares whose sides are diagonals of 1x1 squares. So, for example, in a 3x3 grid, how many such squares are there?

In a grid, such squares can be formed by choosing two adjacent points horizontally and two adjacent points vertically, but shifted. Wait, actually, a square of side sqrt(2) would have vertices at (x,y), (x+1, y+1), (x, y+2), (x-1, y+1). But to have such a square, we need to have points at (x-1, y+1), so that would require x ≥ 1 and y+1 ≤ 2 (since there are only 3 rows). So for the grid of 3 rows and n columns, how many such squares?

Each such square requires a 2x2 block of points. Wait, no. Let me think again. If I want a square with side length sqrt(2), then it's determined by two adjacent points horizontally and two adjacent points vertically, connected by diagonals. So, for example, the square with vertices at (0,0), (1,1), (0,2), (-1,1) – but this would require a point at (-1,1), which is not in our grid. Therefore, in the grid of 3 rows and n columns of points, the squares of side length sqrt(2) would have to be placed such that all four vertices are within the grid.

For example, consider a square with vertices at (x, y), (x+1, y+1), (x, y+2), (x-1, y+1). So, to have these points, x must be at least 1, and y+1 must be at most 2. Since y can be 0 or 1 (since there are 3 rows). If y = 0, then y+1 = 1, which is okay. If y = 1, then y+1 = 2, which is okay. Therefore, for each y in {0,1}, and x from 1 to n - 1, we can have such squares.

But wait, each such square is determined by the bottom-left corner (x, y) and then the square extends to (x+1, y+1), etc. However, for a square with side length sqrt(2), the number of such squares would be (number of positions horizontally) times (number vertically). Wait, for each y (rows), and for each x (columns) such that there's space to the left and right?

Wait, actually, the square spans horizontally from x-1 to x+1, and vertically from y to y+2. Therefore, in terms of the grid of points:

To have a square with vertices at (x-1, y), (x, y+1), (x+1, y), (x, y-1)? Wait, maybe I need to think differently.

Alternatively, the number of squares of side sqrt(2) is equal to the number of 2x2 blocks in the grid. Because each 2x2 block can form four such squares? Wait no. Wait, in a 2x2 block of points, how many squares of side sqrt(2) are there? Actually, each 2x2 block contains exactly one square of side sqrt(2). Wait, no. Wait, in a 2x2 grid of points, you can form a square by connecting the four points. Wait, no, that's a square of side length 1? Wait, connecting four points (0,0), (1,0), (1,1), (0,1) is a square of side length 1. But if you connect (0,0), (1,1), (0,2), (-1,1) – but that requires points outside of the 2x2 grid. Therefore, in a 2x2 grid of points, there are no squares of side sqrt(2). Wait, confusion here.

Wait, maybe I need to think of the squares that are rotated. Let me think in terms of coordinates. Suppose we have a square with vertices at (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). So to have all these points in the grid, we need:

For the point (x - 1, y + 1): x - 1 ≥ 0 => x ≥ 1; and y + 1 ≤ 2 => y ≤ 1. Since our grid has y from 0 to 2 (rows), so y can be 0 or 1. Therefore, for each y in {0,1}, and x in {1, 2, ..., n - 1}, we can have such squares. Therefore, the number of such squares is 2*(n - 1).

Wait, but each such square is unique? For example, if we take y = 0 and x = 1, then the square is (1,0), (2,1), (1,2), (0,1). Similarly, for y = 1 and x = 1, the square is (1,1), (2,2), (1,3)... Wait, but y = 1, y + 2 = 3, but there are only 3 rows (y=0,1,2). Therefore, y + 2 must be ≤ 2. Therefore, y + 2 ≤ 2 => y ≤ 0. Therefore, earlier reasoning is wrong.

Wait, if we have the square with vertices at (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). Then for the last point, (x - 1, y + 1), we need y + 1 ≤ 2, so y ≤ 1. Also, for the third vertex, (x, y + 2), we need y + 2 ≤ 2 => y ≤ 0. Therefore, only when y = 0. So y must be 0. Then, the square is determined by x ≥ 1 (so that (x - 1, 0 + 1) = (x - 1, 1) exists). So for y = 0, x can be from 1 to n - 1. Therefore, the number of such squares is (n - 1). Similarly, are there squares in other orientations?

For example, squares that go in the other direction? For example, square with vertices (x, y), (x + 1, y - 1), (x, y - 2), (x - 1, y - 1). But y - 2 needs to be ≥ 0 => y ≥ 2. Since y can be 0, 1, 2. So y = 2, then the square would have vertices (x, 2), (x + 1, 1), (x, 0), (x - 1, 1). Similarly, for y = 2, x can be 1 to n - 1, so another (n - 1) squares.

Therefore, in total, there are 2*(n - 1) squares of side length sqrt(2). Wait, but these are the squares that are "diamond" shaped, with side length sqrt(2). Therefore, that's another size of square. So now, we have three types of squares:

1. Axis-aligned 1x1: 2(n - 1)

2. Axis-aligned 2x2: (n - 2)

3. Tilted squares of side sqrt(2): 2(n - 1)

Wait, but then total squares would be 2(n - 1) + (n - 2) + 2(n - 1) = 2(n - 1) + 2(n - 1) + (n - 2) = 4(n - 1) + n - 2 = 4n - 4 + n - 2 = 5n - 6.

But if the total is 70, then 5n - 6 = 70 => 5n = 76 => n = 15.2. Not integer. Still not possible. So something is wrong here.

Alternatively, maybe there are other tilted squares. For example, squares with side length sqrt(5). Let me think.

A square with side length sqrt(5) would be formed by moving 1 unit in x and 2 units in y, or something like that. For example, the square with vertices at (0,0), (1,2), (3,1), (2,-1). But again, the last point is outside the grid. So, in our 3-row grid, can we have such squares?

Suppose we have a square with side length sqrt(5). For example, take a square that goes from (x, y) to (x + 1, y + 2). Wait, the distance between (x, y) and (x + 1, y + 2) is sqrt(1^2 + 2^2) = sqrt(5). Then, to form a square, we need another point. The next point would be (x + 2, y + 1), and then back to (x + 1, y). Let me check:

If we have points A(x, y), B(x + 1, y + 2), C(x + 2, y + 1), D(x + 1, y). Then AB vector is (1, 2), BC vector is (1, -1), CD vector is (-1, -2), DA vector is (-1, 1). Wait, is that a square? The lengths:

AB: sqrt(1 + 4) = sqrt(5)

BC: sqrt(1 + 1) = sqrt(2)

No, not equal. So that's not a square. Maybe another configuration.

Alternatively, take points A(x, y), B(x + 2, y + 1), C(x + 1, y + 3), D(x - 1, y + 2). But y + 3 would be out of bounds. Not helpful.

Alternatively, maybe squares of side length sqrt(5) can be formed with points in the grid? For example, consider a square with vertices at (0,0), (1,2), (3,1), (2,-1). But as before, (2,-1) is outside the grid. If our grid has 3 rows, then y is from 0 to 2, so if we try to have a square with y-coordinate up to 2, then maybe (0,0), (2,1), (1, 2), (-1, 1). Again, (-1,1) is outside. Alternatively, maybe (1,0), (2,2), (0,2), (0,0)? Wait, that's not a square. Let me check distances:

Between (1,0) and (2,2): sqrt(1 + 4) = sqrt(5)

Between (2,2) and (0,2): 2 units

Not equal. So not a square. Maybe a different square.

Alternatively, (0,0), (1,1), (2,0), (1,-1). Again, (1,-1) is outside.

Alternatively, perhaps no squares of side length sqrt(5) can be formed in a 3-row grid. Therefore, maybe only the two axis-aligned sizes and the tilted squares of side sqrt(2). But then total squares would be as before, which leads to n non-integer. So something is wrong here.

Alternatively, maybe I miscounted the number of tilted squares.

Let me double-check the tilted squares.

Earlier, I considered squares with vertices like (x,0), (x + 1,1), (x,2), (x - 1,1). For such a square to exist, we need x ≥ 1 (to have x - 1 ≥ 0), and y = 0. Then, for each x from 1 to n - 1, we have such a square. Similarly, for y = 2, we can have squares going downward. Wait, let me think again.

If we have a square with vertices (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). For this square, y + 2 must be ≤ 2, so y ≤ 0. Therefore, y = 0. Then, we need x - 1 ≥ 0 => x ≥ 1. And the rightmost point is (x + 1, y + 1) must be within the grid: since x + 1 ≤ n - 1 => x ≤ n - 2. Wait, no. Wait, in our grid, the columns go from 0 to n - 1. Therefore, (x + 1) must be ≤ n - 1 => x ≤ n - 2. Therefore, x ranges from 1 to n - 2. Therefore, number of such squares is (n - 2). Similarly, if we consider squares going downward, with y = 2, then we have:

Square with vertices (x, 2), (x + 1, 1), (x, 0), (x - 1, 1). Similarly, y = 2, so y - 2 = 0. So need x ≥ 1, and x + 1 ≤ n - 1 => x ≤ n - 2. Therefore, same number of squares: n - 2. Therefore total tilted squares of side sqrt(2) would be 2*(n - 2). Therefore total squares would be:

Axis-aligned:

1x1: 2(n - 1)

2x2: (n - 2)

Tilted sqrt(2): 2(n - 2)

Total: 2(n - 1) + (n - 2) + 2(n - 2) = 2n - 2 + n - 2 + 2n - 4 = 5n - 8

Set equal to 70:

5n - 8 = 70 => 5n = 78 => n = 15.6. Still not integer.

Hmm. Not helpful. Maybe there are more squares?

Alternatively, maybe squares with side length sqrt(5)? Let me try again.

Suppose we have a square that goes from (x, y) to (x + 1, y + 2) to (x + 2, y + 1) to (x + 1, y - 1). But again, y - 1 may be negative. Let's check coordinates in a 3x3 grid.

Take for example, in a 3x3 grid (n=3), points (0,0), (1,0), (2,0), (0,1), (1,1), (2,1), (0,2), (1,2), (2,2).

Can we form a square with side length sqrt(5)? Let's pick points (0,0), (1,2), (3,1), but (3,1) is outside. Wait, what about (0,0), (2,1), (1, 3), but again out of bounds. If we have points (1,0), (2,2), (0,2), ( -1, 0). No. Not helpful.

Alternatively, maybe a square with vertices at (0,1), (2,0), (1,-1), (-1,0)? No. Not within the grid. Seems impossible. Therefore, maybe no squares of side length sqrt(5) in a 3-row grid. Hence, only axis-aligned and side sqrt(2). But the total number is 5n - 8, which doesn't give integer.

Alternatively, perhaps I have miscalculated the number of tilted squares. Let me think again.

Suppose we think of all squares, not just the ones with side sqrt(2). For example, in addition to the ones I considered, are there other sizes?

For example, squares with side length 2*sqrt(2)? That would be larger squares, but in a 3-row grid, maybe not possible. For example, a square that spans from (0,0) to (2,2) to (0,4)... but y=4 is outside.

Alternatively, maybe squares with side length sqrt( (n - 1)^2 + 2^2 )? Not sure.

Alternatively, maybe there are other orientations for tilted squares? For example, squares that are not just diamond-shaped but in another way?

Wait, let me try to think differently. Maybe the problem is referring to squares with different side lengths in terms of grid units, not necessarily axis-aligned. For example, in a grid, the squares can have different sizes based on their side lengths.

But how many distinct sizes of squares are there in a 3×n grid?

Let me consider possible squares:

1. 1x1 axis-aligned.

2. 2x2 axis-aligned.

3. 1x1 tilted (sqrt(2)).

Are there any others?

Alternatively, perhaps squares with side length 2 units (axis-aligned) and tilted ones? Wait, no, 2 units is already considered.

Alternatively, squares with side length sqrt(1^2 + 2^2) = sqrt(5). As discussed before, can't be formed.

Alternatively, squares of side length 2x1 (which are rectangles, not squares). Not applicable.

Alternatively, maybe 2x2 squares are not the only axis-aligned ones. Wait, no. In a 3x n grid, axis-aligned squares can be 1x1 (side length 1) and 2x2 (side length 2). Because 3 units in height allows for a square of side length 2 (from y=0 to y=2). If you try to make a square of side length 3, it would require y going from 0 to 3, which isn't there. So only 1x1 and 2x2.

If that is the case, then the total number of squares is axis-aligned (1x1 and 2x2) and tilted squares (sqrt(2)).

Earlier calculation for tilted squares was 2(n - 2). So total squares:

1x1: 2(n - 1)

2x2: (n - 2)

sqrt(2): 2(n - 2)

Total: 2(n - 1) + (n - 2) + 2(n - 2) = 5n - 8

Which for total 70, 5n = 78, n=15.6. Not possible.

Alternatively, maybe my count for tilted squares is wrong. Let me think again.

How many squares of side sqrt(2) are there?

Each such square is determined by a 2x1 rectangle? Or a 1x2 rectangle?

Wait, if I have a square tilted by 45 degrees, with side length sqrt(2), it can be inscribed in a 2x2 grid of points. So in a 3x n grid, how many 2x2 blocks of points are there? The number of 2x2 blocks is (n - 1) * (3 - 1) = (n - 1)*2. Each 2x2 block can contain one square of side sqrt(2). Wait, no. In a 2x2 grid of points, you can have one square of side sqrt(2) which is the diamond formed by connecting the midpoints. Wait, but the midpoints are not points in the grid. Wait, no, confusion again.

Wait, in a 2x2 grid of points, which are four points forming a square. To form a square of side sqrt(2), you need to connect the corners, but that's a square of side length 1*sqrt(2). Wait, no. The distance between adjacent points is 1 unit. The distance between diagonal points is sqrt(2). So if you connect (0,0), (1,1), (0,2), (-1,1), but again, (-1,1) is outside. However, in a 2x2 grid of points (say, from (0,0) to (1,1)), the square of side sqrt(2) would require points outside. Therefore, actually, there are no squares of side sqrt(2) in a 2x2 grid of points. Therefore, perhaps my previous count is wrong.

Therefore, maybe there are no squares of side sqrt(2)? Which contradicts previous thought.

Alternatively, maybe squares that are not based on the 2x2 blocks?

Wait, if we look at the Asymptote code given in the problem. The Asymptote code seems to draw a 3x3 grid (points from (0,0) to (2,2)) and then adds some filled squares. Let me try to imagine that.

The Asymptote code draws:

First, some dots at (0,0) to (2,2). Then, it fills some small squares near the right side. For example:

filldraw((2.95,-0.05)--(3.05,-0.05)--(3.05,0.05)--(2.95,0.05)--cycle,black);

This seems to be a small square to the right of the grid. Similarly, other squares. Then, it adds more dots at (4,0), (5,0), etc. Maybe the diagram is showing a 3x3 grid with some squares, and the problem is generalized to a 3xn grid? But since the problem states that there are three sizes of squares, maybe in the Asymptote diagram, there are three sizes of squares.

If I can figure out the number of squares in the Asymptote diagram, maybe I can extrapolate.

But since I can't see the image, let me think: the Asymptote code draws a 3x3 grid of points, then adds more points to the right. The filled squares are small, maybe 1x1, 2x2, and something else?

Wait, in the first part, the filled squares are small, maybe 1x1 or something else. But if the grid is 3x3, then:

Number of axis-aligned squares:

- 1x1: In a 3x3 grid of points (which is 2x2 grid of squares), there are 2*2 = 4 squares of 1x1.

- 2x2: There is 1 square of 2x2.

Total axis-aligned squares: 4 + 1 = 5.

Then, if there are tilted squares, for example, the square that goes from (0,0), (1,1), (0,2), (-1,1). But since (-1,1) is not in the grid, it's not there. However, if we consider the square from (1,0), (2,1), (1,2), (0,1). This square is inside the grid. So that's one tilted square. Similarly, from (0,1), (1,2), (0,2)... wait no. If I take (0,1), (1,2), (2,1), (1,0). That's a tilted square of side sqrt(2), and it's inside the grid. So in the 3x3 grid, how many such squares are there?

Each of these tilted squares is formed by choosing a 2x2 block of points. Wait, no. The square (0,1), (1,2), (2,1), (1,0) is formed by the points in the middle of the grid. Similarly, there is another one on the top: (0,0), (1,1), (2,0), (1,-1)... no, (1,-1) is outside. But if we take (1,0), (2,1), (1,2), (0,1). That's one. Similarly, if we have (0,1), (1,0), (2,1), (1,2). Wait, same square. So in a 3x3 grid, how many tilted squares are there?

Actually, there are 2 such squares. One at the center and one... Wait, no. Wait, in the 3x3 grid, the number of tilted squares of side sqrt(2) would be 2. For example, the square mentioned above and another one if shifted? But since it's only 3 rows, shifting would go out of bounds. So maybe only 2 tilted squares?

Wait, no. Wait, if you consider all possible positions. For example, starting at (0,1), the square would require (1,2), (2,1), and (1,0). Similarly, starting at (1,0), you get the same square. So maybe only one unique square. But the Asymptote code shows three filled squares? Let me check the Asymptote code again.

The Asymptote code has:

filldraw((2.95,-0.05)--(3.05,-0.05)--(3.05,0.05)--(2.95,0.05)--cycle,black);

This seems to be a square to the right of the grid, maybe indicating the grid extends? Then:

filldraw((2.45,-0.05)--(2.55,-0.05)--(2.55,0.05)--(2.45,0.05)--cycle,black);

This is smaller square, maybe indicating a smaller square? And:

filldraw((3.45,-0.05)--(3.55,-0.05)--(3.55,0.05)--(3.45,0.05)--cycle,black);

Another small square. Then similarly above, at y=0.95 and y=1.95, same filled squares. So maybe there are 3 sizes of squares? The small ones (maybe 1x1), medium ones (maybe 2x2), and larger ones? But since the grid is only 3 rows, perhaps the squares are:

- Small squares (1x1),

- Medium squares (maybe 2x2),

- And another size? Maybe squares that are 1x1 tilted?

If the diagram shows three sizes of squares, then maybe in the 3x3 grid, there are 1x1, 2x2, and tilted squares of side sqrt(2). So total squares would be 5 axis-aligned (4 + 1) and 2 tilted squares, totaling 7? But the problem says there are 70 squares in total for some n. So maybe the Asymptote code is showing part of the problem, where with n=3, the number of squares is 7? But in reality, for n=3, total squares would be:

Axis-aligned: 2(n - 1) + (n - 2) = 2*2 + 1 = 5

Tilted: 2(n - 2) = 2(1) = 2

Total: 5 + 2 = 7. Which matches the hypothetical. If that's the case, then the problem says that for a 3xn grid, the total number of squares is 70. So using the formula:

Total squares = axis-aligned + tilted = [2(n - 1) + (n - 2)] + 2(n - 2) = 2n - 2 + n - 2 + 2n - 4 = 5n - 8

Set equal to 70: 5n - 8 = 70 => 5n = 78 => n = 15.6. Still not integer.

Hmm, contradiction. So there must be an error in my counting.

Wait, let me check for n=3, total squares is 5 + 2 = 7. But if I manually count, in a 3x3 grid of points:

Number of squares:

Axis-aligned:

1x1: 2*2 = 4 (since (3-1)*(3-1) ??? Wait no.

Wait earlier formula said for axis-aligned squares:

For k=1: (3 - 1)(n - 1) = 2*(3 - 1) = 4

For k=2: (3 - 2)(n - 2) = 1*(1) = 1

Total axis-aligned: 5.

Tilted squares: 2*(n - 2) = 2*(1) = 2.

Total: 7. Okay, so for n=3, total squares is 7.

If the problem says that there are 70 squares, then solving 5n - 8 = 70 => n=15.6, which is not integer. Therefore, my formula is wrong.

Alternatively, maybe the number of tilted squares is different? Let me think again.

For the tilted squares of side sqrt(2), in a 3xn grid, how many are there?

Each such square is determined by its topmost point. For example, in the 3x3 grid, the square is centered. But in general, for a grid with n columns, how many positions are there for such squares?

Let me consider that for each square of side sqrt(2), it spans horizontally from column x to column x + 1, and vertically from row y to row y + 1. Wait, no. Let me think of the square with vertices at (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). As before, to have all points in grid:

y + 2 ≤ 2 => y ≤ 0, so y = 0.

x - 1 ≥ 0 => x ≥ 1

x + 1 ≤ n - 1 => x ≤ n - 2

Therefore number of such squares is (n - 2). Similarly, squares going downward from y=2: vertices (x, 2), (x + 1, 1), (x, 0), (x - 1, 1). So similar constraints: x ≥ 1, x ≤ n - 2. So number of squares is (n - 2). Therefore total tilted squares: 2(n - 2)

So for n=3, it's 2(1)=2. Which matches earlier.

Therefore total squares is 5n - 8. For n=3, it's 5*3 - 8 = 15 - 8 = 7. Correct.

But if this formula is correct, then 5n - 8 = 70 => n = 78 / 5 = 15.6. Not possible. Therefore, there must be more squares.

Wait, maybe there are more tilted squares? Like squares of different sizes?

For example, squares with side length sqrt(5). Wait, let's try to find such squares.

Consider a square with vertices at (x, y), (x + 2, y + 1), (x + 1, y + 3), (x - 1, y + 2). But y + 3 would be outside the grid. What if we take y = 2? Then y + 3 = 5, still outside.

Alternatively, a square with vertices at (x, y), (x + 1, y + 2), (x + 2, y + 1), (x + 1, y - 1). But y - 1 needs to be ≥ 0, so y ≥ 1. For example, take y = 1. Then, points would be (x,1), (x + 1, 3) which is outside. No.

Alternatively, (x,0), (x + 2,1), (x + 1, 3), (x - 1, 2). Again, y=3 is outside.

Alternatively, (x,1), (x + 2, 2), (x + 1, 0), (x - 1, 1). Check if these points exist:

(x - 1, 1) must exist => x ≥ 1

(x + 1, 0) must exist => x + 1 ≤ n - 1 => x ≤ n - 2

(x + 2, 2) must exist => x + 2 ≤ n - 1 => x ≤ n - 3

Therefore, for such a square, x can range from 1 to n - 3. So number of such squares is (n - 3). Similarly, are there squares in the other direction?

For example, (x, 2), (x - 2, 1), (x - 1, 3), (x + 1, 2). But again, y=3 is outside. Not helpful.

So, if there are such squares, the number would be (n - 3). Then, total squares would be 5n - 8 + (n - 3) = 6n - 11. For n=3, it would be 6*3 - 11 = 7, which matches the previous total. But then if we set 6n - 11 = 70, 6n = 81, n = 13.5. Still not integer.

This is getting frustrating. Maybe there are more sizes? Or perhaps I'm missing something.

Wait, maybe the three sizes mentioned in the problem are not axis-aligned and tilted ones, but something else? For example, in the Asymptote code, maybe there are squares of sizes 1x1, 2x2, and 2x2 tilted? But that would be two sizes.

Alternatively, the problem may consider squares of different orientations but same side length? For example, 1x1 axis-aligned and 1x1 tilted are same size? But they are not. One is axis-aligned, the other is tilted.

Wait, going back to the problem statement:

"There is a total of 70 squares of three sizes whose vertices are points on this rectangular 3×n grid of points. What is the value of n?"

So there are three different sizes of squares, and total number is 70. So perhaps the three sizes refer to:

1. 1x1 axis-aligned

2. 2x2 axis-aligned

3. Some other size, like 1x1 tilted (sqrt(2))?

But according to previous calculation, total squares would be 5n - 8. But it doesn't fit 70.

Alternatively, maybe the three sizes are:

1. 1x1 axis-aligned

2. 2x2 axis-aligned

3. 3x3 axis-aligned? But in a 3x n grid, can we have 3x3 squares? No, because vertically there are only 3 rows, so a 3x3 square would need 3 units in both directions, but vertically, it would go from y=0 to y=3, but there are only 3 rows (y=0,1,2). Hence, no.

Alternatively, maybe the three sizes are 1x1, 2x2, and some larger tilted squares?

But as before, not sure.

Alternatively, maybe the three sizes are:

- Squares of size 1x1 (axis-aligned)

- Squares of size sqrt(2)x sqrt(2) (tilted, side length sqrt(2))

- Squares of size sqrt(5)x sqrt(5) (tilted, side length sqrt(5))

But then, need to count all these.

But how many squares of size sqrt(5)?

For example, in a 3x n grid, how many squares of side length sqrt(5) are there?

A square with side length sqrt(5) would have sides that are 1 unit in one direction and 2 units in another. So, for example, a square from (x, y) to (x + 1, y + 2) to (x, y + 4)... but y + 4 is outside the grid.

Alternatively, a square with vertices at (x, y), (x + 2, y + 1), (x + 1, y + 3), (x - 1, y + 2). Again, y + 3 is out.

Alternatively, in a 3-row grid, the maximum vertical extent is 2. So, to have a square with side length sqrt(5), we need a 1 unit horizontal and 2 units vertical move, but then the square would have to close back. For example, starting at (x, y), going to (x + 1, y + 2), then to (x + 2, y + 1), then to (x + 1, y). Let's check if this forms a square.

The distance between (x, y) and (x + 1, y + 2) is sqrt(1² + 2²) = sqrt(5).

The distance between (x + 1, y + 2) and (x + 2, y + 1) is sqrt(1² + (-1)²) = sqrt(2).

Not equal. Not a square.

Alternatively, starting at (x, y), going to (x + 2, y + 1), then to (x + 1, y + 3), but again, y + 3 is out.

Alternatively, starting at (x, y), going to (x + 2, y), then to (x + 1, y + 2), then to (x - 1, y + 2). Wait, this is getting too complex and likely outside the grid.

It seems impossible to have such squares in a 3-row grid. So maybe no squares of side sqrt(5). Hence, only two sizes? But the problem says three sizes.

Wait, perhaps the three sizes are axis-aligned 1x1, axis-aligned 2x2, and tilted 1x1 (side sqrt(2)), but I have to include them in the total.

If total squares are 5n - 8, and this equals 70, but it's not integer, maybe there is an error in the formula.

Wait, let's try with n=5. What is the total?

Axis-aligned: 2(5-1) + (5-2) = 8 + 3 = 11

Tilted: 2(5-2) = 6

Total: 17. If I use formula 5n - 8 = 5*5 - 8 = 17. Okay, matches. So 5n - 8 = total squares.

If n=10: 5*10 - 8 = 42

n=11: 5*11 - 8 = 47

n=12: 5*12 - 8 = 52

n=13: 5*13 - 8 = 57

n=14: 5*14 - 8 = 62

n=15: 5*15 - 8 = 67

n=16: 5*16 - 8 = 72

So 72 is for n=16. But we need 70. Hmm, between n=15 and n=16. But since n must be integer, this suggests that either the formula is incorrect, or there is another size of square.

Alternatively, maybe there is a miscalculation in how many tilted squares there are.

Wait, let me re-examine tilted squares. Perhaps I have undercounted or overcounted.

For each tilted square of side sqrt(2), it requires a specific position. In addition to the ones counted as 2(n - 2), maybe there are others?

For example, in the 3x3 grid, we had 2 tilted squares. But let me count manually:

In a 3x3 grid, how many squares of side sqrt(2) are there?

They are the squares that have vertices at (0,1), (1,2), (2,1), (1,0) and (1,0), (2,1), (1,2), (0,1). Wait, it's the same square. So only 1 unique square?

Wait, but in the Asymptote code, there are three filled squares, which might suggest more. Maybe the diagram shows multiple squares?

Alternatively, maybe there are more tilted squares. Let me think.

If we consider all possible placements:

For y = 0, the square is (x,0), (x+1,1), (x,2), (x-1,1). For x from 1 to n - 2.

For y = 2, the square is (x,2), (x+1,1), (x,0), (x-1,1). For x from 1 to n - 2.

But for y = 1, is there a square?

Let's try to form a square with y = 1. Let's say vertices at (x,1), (x+1,2), (x,3), (x-1,2). But y=3 is out of bounds. Hence, no. Similarly, downward, (x,1), (x+1,0), (x, -1), ... also out of bounds. Hence, no squares with y=1.

Hence, only 2(n - 2) tilted squares.

Therefore, the formula seems correct. So total squares = 5n - 8.

If the problem says total squares is 70, then 5n - 8 = 70 => n = 78/5 = 15.6. Not possible. Hence, something is wrong.

Wait, let me check the problem statement again:

"There is a total of 70 squares of three sizes whose vertices are points on this rectangular 3×n grid of points. What is the value of n?"

Wait, maybe the three sizes are not referring to different side lengths, but three different orientations? No, that doesn't make sense.

Alternatively, maybe the three sizes are 1x1, 2x2, and 3x3? But in a 3x n grid, can we have 3x3 squares? No, because vertically there are only 3 points, so a square of size 3x3 would need 3 units vertically, which would go from y=0 to y=3, but there are only y=0,1,2. Hence, not possible.

Alternatively, the three sizes are 1x1, 1x1 tilted, and 2x2 tilted? But that would be three sizes, but how many squares?

Wait, if we consider:

- 1x1 axis-aligned: 2(n - 1)

- 1x1 tilted: 2(n - 2)

- 2x2 axis-aligned: (n - 2)

Total: 2(n - 1) + 2(n - 2) + (n - 2) = 2n - 2 + 2n - 4 + n - 2 = 5n - 8. Same as before. So same result.

Still no good.

Wait, maybe the problem counts squares of different sizes, where sizes are defined by number of unit squares they cover? For example, a 1x1 square covers 1 unit square, a 2x2 covers 4, etc. But the problem says "three sizes", which might mean three distinct sizes, not three distinct counts.

Alternatively, maybe the three sizes are 1x1, 2x2, and 3x1? No, that's not a square.

Wait, this is very confusing. Let me try to search for similar problems.

Wait, I recall that in a grid, the number of squares is calculated by sum_{k=1}^{min(m,n)} (m - k + 1)(n - k + 1). But in this case, the grid is 3 rows and n columns of points, which corresponds to a 2x(n - 1) grid of squares? Wait, no.

Wait, the number of squares in a grid with m rows and n columns of points is:

For axis-aligned squares, it's sum_{k=1}^{min(m-1, n-1)} (m - k) * (n - k).

In our case, m = 3 rows, n columns. So sum_{k=1}^2 (3 - k)(n - k) = (3 - 1)(n - 1) + (3 - 2)(n - 2) = 2(n - 1) + 1(n - 2) = 2n - 2 + n - 2 = 3n - 4.

But if we include tilted squares, it's more.

But the problem says there are 70 squares of three sizes, which might mean 3 types, but how?

Maybe the three sizes are 1x1, 2x2, and 1x1 tilted? Then total squares would be 3n - 4 + 2(n - 2) = 3n - 4 + 2n - 4 = 5n - 8. Same as before.

But again, 5n - 8 = 70 => n = 15.6.

Wait, unless the problem is not counting tilted squares, and the three sizes are 1x1, 2x2, and 3x3? But 3x3 is not possible. Or maybe the three sizes are 1x1, 1x1, and 2x2? No.

Wait, perhaps the problem is in the Asymptote code, and the three sizes are 1x1, 2x2, and 3x3? But in a 3x n grid, 3x3 would require 3 rows and 3 columns of points? No, 3x3 grid of points would allow 2x2 squares, but not 3x3.

Wait, the Asymptote code draws a 3x3 grid of points, then adds more points at (4,0), (5,0), etc. So it's extending the grid to the right. The filled squares in the Asymptote code are small squares, which could be 1x1, and maybe larger squares. But since it's hard to see, maybe the original problem is different.

Maybe the grid is n×3, not 3×n? If it's a grid with n rows and 3 columns of points? Let's try that.

If it's a grid with n rows and 3 columns of points, then number of axis-aligned squares would be:

For k=1: (n - 1) * (3 - 1) = 2(n - 1)

For k=2: (n - 2) * (3 - 2) = (n - 2) * 1

Total axis-aligned: 2n - 2 + n - 2 = 3n - 4

Tilted squares of side sqrt(2): in this case, with 3 columns, how many?

For example, squares like (x, y), (x + 1, y + 1), (x, y + 2), (x - 1, y + 1). For this, x must be ≥ 1, and y + 2 ≤ n - 1 (assuming n rows). Wait, if it's n rows, then rows go from y=0 to y = n - 1. So for y + 2 ≤ n - 1 => y ≤ n - 3. For each such y, and x from 1 to 3 - 1 = 2 (since columns are 3). Wait, columns are 3, so x ranges from 0 to 2. To have x - 1 ≥ 0, x ≥ 1. So x can be 1 or 2. So for each y from 0 to n - 3, there are 2 squares. So total tilted squares: 2 * (n - 2)

Wait, if it's n rows and 3 columns, then number of tilted squares would be 2*(n - 2). Then total squares would be 3n - 4 + 2(n - 2) = 3n - 4 + 2n - 4 = 5n - 8. Same as before. So no change.

If the grid is n rows and 3 columns, and total squares is 70, then 5n - 8 = 70 => n = 15.6. Still no.

Wait, this suggests that no matter if it's 3×n or n×3, the formula is same.

But the problem says "rectangular 3×n grid of points", so it's 3 rows and n columns.

Wait, let me try to think outside the box. Maybe the three sizes are not axis-aligned and tilted, but something else. For example, the problem says "squares of three sizes", maybe it's referring to the number of squares of each size, not the number of distinct sizes? No, the wording says "three sizes".

Alternatively, maybe the grid is 3×n grid of squares, not points? If the grid is 3×n grid of squares, then it would have (3 + 1) × (n + 1) points, making it a 4×(n + 1) grid of points. But the problem says "3×n grid of points", so likely it's 3 rows and n columns of points.

Wait, maybe the translation is wrong? The original says "rectangular 3×n grid of points". Maybe it's a grid with 3 points in one direction and n in the other?

Wait, perhaps I made a mistake in counting axis-aligned squares. Let me re-derive it.

In a grid with m rows and n columns of points, the number of squares is:

For each possible square size k (in units), where k ranges from 1 to min(m - 1, n - 1), the number of such squares is (m - k) * (n - k).

In our case, m = 3 rows, n columns. So k can be 1 or 2.

Number of 1x1 squares: (3 - 1) * (n - 1) = 2(n - 1)

Number of 2x2 squares: (3 - 2) * (n - 2) = 1*(n - 2)

Total axis-aligned: 2n - 2 + n - 2 = 3n - 4

If the problem says there are 70 squares, and it's only counting axis-aligned, then 3n - 4 = 70 => 3n = 74 => n = 74/3 ≈ 24.666. Not integer.

But the problem says there are squares of three sizes. So maybe it's counting both axis-aligned and tilted squares, and the three sizes are 1x1, 2x2, and something else. But as before, leading to 5n - 8 = 70.

Wait, unless there are more tilted squares. For example, in addition to the 2(n - 2) squares of side sqrt(2), there are also squares of side 2*sqrt(2) or others.

Let me try to count squares of side 2*sqrt(2). For example, a square formed by moving 2 units in one direction and 2 units in another, but in a 3-row grid...

For example, a square with vertices at (x, y), (x + 2, y + 2), (x, y + 4), (x - 2, y + 2). But y + 4 is out of bounds. Not possible.

Alternatively, a square with vertices at (x, y), (x + 2, y + 1), (x + 1, y + 3), (x - 1, y + 2). Again, y + 3 is out.

No luck. So no such squares.

Wait, maybe the three sizes are 1x1, 2x2, and 1x1 tilted? Then, total squares would be 2(n - 1) + (n - 2) + 2(n - 2) = 5n - 8, as before.

But 5n - 8 = 70 => n = 15.6. Not integer. So, what if the problem has a mistake? Or maybe my counting is wrong.

Wait, let me try to think of n=16: 5*16 - 8 = 72. Close to 70. If there were 72 squares for n=16, but the problem says 70. Maybe there are 2 fewer squares, which could happen if there are some squares counted in the formula that don't actually exist, or if the three sizes are different.

Alternatively, maybe the three sizes are 1x1, 2x2, and 1x1 tilted, but in the formula, I should subtract some cases? For example, in the tilted squares, maybe some of them are counted twice or something?

Wait, in the case of n=3, total squares is 7: 5 axis-aligned and 2 tilted. Let me list them:

Axis-aligned:

1x1: 4 squares: (0,0)-(1,0)-(1,1)-(0,1), (1,0)-(2,0)-(2,1)-(1,1), (0,1)-(1,1)-(1,2)-(0,2), (1,1)-(2,1)-(2,2)-(1,2)

2x2: 1 square: (0,0)-(2,0)-(2,2)-(0,2)

Tilted:

2 squares: (0,1)-(1,2)-(2,1)-(1,0) and (1,0)-(2,1)-(1,2)-(0,1) - but these are the same square, so only 1 unique square. Wait, but earlier calculation said 2. What's the issue here?

Ah! Here is the mistake! In the count of tilted squares, I assumed 2(n - 2), but in reality, for each n, there are (n - 2) squares in one direction, and (n - 2) in the other, but some are overlapping or counted multiple times?

No, let's think again. In a 3x3 grid (n=3), for tilted squares of side sqrt(2), how many are there?

As above, only 1 unique square. But according to the previous count, it should be 2*(n - 2) = 2*(1) = 2. So there is overcounting.

Why? Because when n=3, for y=0, x ranges from 1 to n-2 = 1, so one square. For y=2, similarly, one square. But these two squares are the same as the one described above? Wait, no. Each is a different square?

Wait, if y=0, the square is (x,0), (x+1,1), (x,2), (x-1,1). For x=1, this is (1,0), (2,1), (1,2), (0,1). This is one square.

For y=2, the square is (x,2), (x+1,1), (x,0), (x-1,1). For x=1, this is (1,2), (2,1), (1,0), (0,1). This is the same square as before! So it's counted twice. Therefore, the total number of unique tilted squares is (n - 2), not 2(n - 2). Hence, my previous count was double-counting!

Ah! This is the error! The squares for y=0 and y=2 are the same square, just traversed in different direction. So, the actual number of unique tilted squares is (n - 2), not 2(n - 2). Hence, total squares is axis-aligned (3n - 4) + tilted (n - 2) = 4n - 6.

Let me check for n=3: 4*3 - 6 = 6. But manually, we have 5 axis-aligned + 1 tilted = 6. Correct! Previously, I had 5 + 2 = 7, which was wrong. So the correct total is 4n - 6.

Let's verify for n=3: 4*3 - 6 = 6. Correct.

For n=4: 4*4 - 6 = 10. Let's count manually.

Axis-aligned:

1x1: 2*(4 - 1) = 6

2x2: 1*(4 - 2) = 2

Total axis-aligned: 8

Tilted: n - 2 = 2

Total: 8 + 2 = 10. Correct.

So the correct formula is total squares = 3n - 4 (axis-aligned) + (n - 2) (tilted) = 4n - 6.

Now, set 4n - 6 = 70 => 4n = 76 => n = 19.

Ah! Finally, an integer!

So, the value of n is 19.

But wait, let me confirm this with another example.

For n=4, total squares is 4*4 - 6 = 10. Let's count:

Axis-aligned 1x1: 6

Axis-aligned 2x2: 2

Tilted: 2

Total: 10. Correct.

For n=5: 4*5 - 6 = 14

Axis-aligned: 3*5 - 4 = 11

Tilted: 5 - 2 = 3

Total: 14. Correct.

Yes, this makes sense. So the correct formula is total squares = 4n - 6.

Hence, solving 4n - 6 = 70 => 4n = 76 => n = 19.

Thus, the value of n is 19.

**Final Answer**
\boxed{19}

To solve the problem, we are given that there are **70 squares** in a **rectangular 3×n grid of points**, and we're told that **there are squares of three different sizes**, and we need to find the value of $ n $.

---

### Step 1: Understanding the Grid and Square Counting

The grid is defined as a **3×n grid of points**. This means:
- **3 rows** of points along the vertical axis (i.e., y-coordinates)
- **n columns** of points along the horizontal axis (i.e., x-coordinates)

This corresponds to a **2 × (n - 1)** grid of unit squares (since a unit square requires 2 points in width and 2 points in height).

---

### Step 2: Counting Axis-Aligned Squares

For axis-aligned squares, we need to consider the number of **1×1** and **2×2** squares that can fit within the 3×n grid:

- **1×1 squares**: These are the smallest unit squares that can fit in the grid. Since each such square is formed by 4 consecutive points (2 horizontal and 2 vertical), the number of 1×1 squares is:

  $$
  (n - 1) \times 2 = 2(n - 1)
  $$

- **2×2 squares**: These are larger squares, each formed by a 2×2 block of points. There are 2 ways to fit these in width (3 - 2 = 1) and $ n - 2 $ ways in height. So the number of 2×2 squares is:

  $$
  (3 - 2) \times (n - 2) = 1 \times (n - 2) = n - 2
  $$

- **Total axis-aligned squares**:
  $$
  2(n - 1) + (n - 2) = 2n - 2 + n - 2 = 3n - 4
  $$

---

### Step 3: Counting Tilted Squares

Now consider **tilted squares**, those that are diagonal in orientation. Specifically, the smallest such square has a side length of $ \sqrt{2} $, and is formed by the diagonals of 1×1 squares.

These have the following characteristics:
- They require 3 rows and at least 3 columns.
- For each column $ x $ where $ x \in [1, n - 2] $, we can form one such square by connecting:
  - $ (x, 0) $
  - $ (x + 1, 1) $
  - $ (x, 2) $
  - $ (x - 1, 1) $
- And similarly, each such square is unique, and the count is equal to $ n - 2 $.

So, the total number of tilted squares is:

$$
n - 2
$$

---

### Step 4: Total Number of Squares

Combining both types of squares:

- **Axis-aligned squares**: $ 3n - 4 $
- **Tilted squares**: $ n - 2 $

Total number of squares:

$$
(3n - 4) + (n - 2) = 4n - 6
$$

---

### Step 5: Solving for $ n $

We are told that the total number of squares is **70**. So we set up the equation:

$$
4n - 6 = 70
$$

Solve for $ n $:

$$
4n = 76 \Rightarrow n = \frac{76}{4} = 19
$$

---

### ✅ Final Answer

$$
\boxed{19}
$$

Solution:
The number of $1\times1$ squares in the diagram is $2(n-1)$, the number of $2\times 2$ squares is $n-2$, and the number of $\sqrt{2} \times \sqrt{2}$ squares is also $n-2$ (see diagram).  Solving \[
2(n-1)+n-2+n-2=70
\] we find $n=\boxed{19}$.

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