I need to find the minimum sum S for the vertices of each square in this bar network. The intersections are labeled from 1 to 10, and all three squares have the same sum for their four corners.

First, I should understand the diagram. It shows three diamonds or squares connected horizontally. Each square shares a side with the next one. Let me sketch this out.

There are points: let's say the top-left, bottom-left, top-right, bottom-right, and the middle points where they connect. Since it's a network of bars, I think it's like a grid but with diamonds.

Looking at the description, it's three squares side by side, sharing common vertices. So for three squares, how many unique vertices do we have?

Each square has 4 vertices, but they share vertices with adjacent squares. So for three squares in a row, sharing sides, the total number of unique vertices should be: for the first square, 4 vertices; second square adds 2 new vertices (since two are shared); third square adds 2 new vertices. So 4 + 2 + 2 = 8 vertices? But the problem says points 1 to 10, which is 10 points, so I must be missing something.

Perhaps the squares are not aligned in a way that they share only vertices but not full sides? Or maybe it's a different configuration.

The problem says "the network of bars shown," but since I can't see it, I need to infer from the text. It says "the points of intersection of the network of bars," and it's shown with points 1 to 10.

Perhaps it's a set of bars forming a grid or something. Let me think differently.

Another idea: it might be like a graph where horizontal and vertical bars connect points, but here it says "bars," so perhaps it's like a grid with cells.

The problem mentions "squares," so there are square cells formed by the bars.

Let me assume it's a 3x3 grid or something, but a 3x3 grid has 4x4 = 16 points, which is too many, but here only 10 points, so not a full grid.

Perhaps it's a different shape. Let me count the points from the description.

The diagram has three diamonds: each diamond is a square rotated 45 degrees, I think. So like a row of three such squares.

For example, the left diamond has points: top, left, right, bottom. But when you place three in a row, they share the left-right points with neighbors.

Let me label the points.

Say, for the left square: let's call the top vertex A, bottom vertex B, left vertex C, right vertex D.

Then the middle square shares the right vertex D with the left square and has its own top E, bottom F, and left is D, right is G.

But then the right square shares G with the middle, and has top H, bottom I, left G, right J.

But then we have points: A, B, C, D, E, F, G, H, I, J. That's 10 points: A to J.

A, B, C, D for left; E, F for middle (since D and G shared); G, H, I for right? Let's list:

• Left square: vertices A (top), C (left), B (bottom), D (right)

• Middle square: shares D (left-top?), no, in terms of the squares, let's think of the positions.

Perhaps it's better to think of the points in a coordinate system.

Set a coordinate: let’s say the network has points at (0,1), (1,0), (1,2), (2,1), etc., but for diamonds.

A standard way: for a diamond shape, points at (0,1), (1,0), (1,2), (2,1) for one unit, but that's only 4 points.

For three in a row, the first diamond: points at (0,1), (1,0), (1,2), (2,1)

Second diamond: shares (2,1) and (1,2)? No, to have a second diamond to the right, it might be at (2,1), (3,0), (3,2), (4,1)

But then points: (0,1), (1,0), (1,2), (2,1), (3,0), (3,2), (4,1) — that's 7 points, but we need 10.

Perhaps there are additional points. For the bars, between (1,0) and (2,1), etc.

I think I'm confusing myself.

Another idea: perhaps the "network of bars" means that there are horizontal and vertical bars connecting points, forming cells.

For example, it could be a 2x2 grid of cells, but with only 10 points instead of 9 or 12.

A 2x2 grid has 3x3=9 points, but 10 points, so not that.

Perhaps it's a row of three squares with an additional point or something.

Let's look for the minimum S. Since the sums are all equal, and numbers 1 to 10, the total sum of all numbers is 55, since 1+2+...+10 = 55.

Each point is a vertex of the squares, but some points are shared.

In this configuration, each square has four vertices, but some vertices are shared.

Let me denote the three squares as S1, S2, S3 from left to right.

For S1, vertices: let's say A, B, C, D

S2 shares two vertices with S1, say D and E, but S2 has its own vertices.

From the diamond shape: for S1, points: top T1, bottom B1, left L1, right R1.

S2 shares R1 with S1, and has its own top T2, bottom B2, and left is R1, right R2.

Then S3 shares R2 with S2, and has T3, B3, left R2, right R3.

But then points: L1, T1, R1, B1 for S1; R1, T2, R2, B2 for S2; R2, T3, R3, B3 for S3.

So unique points: L1, T1, R1, B1, T2, B2, R2, T3, B3, R3. That's 10 points: let's call them P1=L1, P2=T1, P3=R1, P4=B1, P5=T2, P6=B2, P7=R2, P8=T3, P9=B3, P10=R3.

In terms of the squares:

• S1 vertices: P1 (left), P2 (top), P3 (right), P4 (bottom)

• S2 vertices: P3 (left), P5 (top), P7 (right), P6 (bottom) — P3 is left for S2, P7 is right for S2

• S3 vertices: P7 (left), P8 (top), P10 (right), P9 (bottom)

P7 is shared between S2 and S3.

The sums:

For S1: P1 + P2 + P3 + P4 = S

For S2: P3 + P5 + P7 + P6 = S

For S3: P7 + P8 + P10 + P9 = S

P6 is bottom for S2, P9 for S3.

All points are distinct except P3 and P7 are shared but not repeated in the sum for each square.

The total sum of all points is P1 to P10, sum to 55.

If I add the sums of the three squares: (P1+P2+P3+P4) + (P3+P5+P6+P7) + (P7+P8+P9+P10) = 3S

But P3 and P7 are counted twice, so this equals total sum + P3 + P7

Because total sum has P1 to P10 once each, but in the sum of squares, P3 and P7 are counted twice, so:

Sum of three square sums = (P1+P2+P3+P4) + (P3+P5+P6+P7) + (P7+P8+P9+P10) = (P1+P2+P3+P4+P5+P6+P7+P8+P9+P10) + P3 + P7 = 55 + P3 + P7

And this equals 3S.

So 55 + P3 + P7 = 3S

Therefore, 3S = 55 + P3 + P7

Since S must be integer, and P3 and P7 are between 1 and 10, so P3 + P7 between 2 and 20, so 3S between 57 and 75, so S between 19 and 25.

But we need min S, so S min when P3 + P7 min, but P3 and P7 are specific points.

S = (55 + P3 + P7)/3

For S to be integer, 55 + P3 + P7 must be divisible by 3.

55 div 3 is 18*3=54, remainder 1, so 55 ≡ 1 mod 3, so P3 + P7 ≡ 2 mod 3, so that 1 + 2 = 3 ≡ 0 mod 3.

So P3 + P7 must be ≡ 2 mod 3.

Now, to minimize S, we need to minimize P3 + P7, since S = (55 + P3 + P7)/3.

Min P3 + P7 with P3 + P7 ≡ 2 mod 3.

Min possible P3 + P7: smallest numbers, 1 and 2, sum 3 ≡ 0 mod 3, not 2.

Next, 1 and 3, sum 4 ≡ 1 mod 3, not 2.

1 and 4, sum 5 ≡ 2 mod 3, yes.

Similarly, 1 and 5, sum 6≡0, no; 1 and 6,7≡1, no; 1 and 7,8≡2, yes.

But min is 1 and 4, sum 5.

Also, 2 and 3, sum 5≡2, yes.

2 and 4, sum 6≡0, no.

So possible min P3+P7=5, with {1,4} or {2,3}.

Then S = (55+5)/3 = 60/3 = 20.

Is that possible? S=20.

But we need to check if we can assign the numbers so that all sums are equal with S=20.

Moreover, we have the sums for each square.

S1: P1 + P2 + P3 + P4 = 20

S2: P3 + P5 + P6 + P7 = 20

S3: P7 + P8 + P9 + P10 = 20

And P3 + P7 =5, for example.

Now, total points sum 55.

If S=20, then from above, 3S = 60, and 55 + P3 + P7 =60, so P3 + P7=5, as we have.

Now, can we assign numbers 1 to 10 to P1 to P10 with P3 and P7 being 1 and 4 or 2 and 3.

Let me assume P3 and P7 are 1 and 4, say P3=1, P7=4.

Then S1: P1 + P2 +1 + P4 =20, so P1 + P2 + P4 =19

Similarly, S2: 1 + P5 + P6 +4 =20, so P5 + P6 =15

S3: 4 + P8 + P9 + P10 =20, so P8 + P9 + P10=16

Now, all numbers 1 to 10 except P3=1 and P7=4, so available: 2,3,5,6,7,8,9,10 and P1,P2,P4,P5,P6,P8,P9,P10.

P1,P2,P4,P5,P6,P8,P9,P10 to be assigned from {2,3,5,6,7,8,9,10}

S1: P1 + P2 + P4 =19

S2: P5 + P6 =15

S3: P8 + P9 + P10=16

Now, P5 and P6 sum to 15, possible pairs from available: 5 and 10, 6 and 9, 7 and 8.

Similarly, P8,P9,P10 sum to 16, with three numbers.

And P1,P2,P4 sum to 19.

Available numbers: 2,3,5,6,7,8,9,10

But P5 and P6 take two of these, sum 15.

Case 1: P5 and P6 are 5 and 10 or 10 and 5.

Then available for P1,P2,P4 and P8,P9,P10: 2,3,6,7,8,9

P1+P2+P4=19, P8+P9+P10=16

Possible for P1,P2,P4: 19 with three numbers from {2,3,6,7,8,9}, min sum 2+3+6=11, max 7+8+9=24, 19 possible, e.g., 9,8,2 or 9,7,3 etc.

Similarly P8+P9+P10=16, e.g., 2,3,11 not, 2,3,11 not available, min 2+3+6=11, max 8+9+7=24, 16 possible, e.g., 3,6,7=16 or 2,6,8=16, etc.

But we need to use all numbers.

If P5=5, P6=10, then available: 2,3,6,7,8,9 for P1,P2,P4,P8,P9,P10.

P1+P2+P4=19

P8+P9+P10=16

Suppose P1,P2,P4: 9,8,2 sum 19

Then P8,P9,P10: 3,6,7 sum 16? 3+6+7=16, yes.

But numbers: P1=9, P2=8, P4=2, P5=5, P6=10, P8=3, P9=6, P10=7? P10=7, but P7=4, P3=1.

Points: P1=9, P2=8, P3=1, P4=2, P5=5, P6=10, P7=4, P8=3, P9=6, P10=7.

Check sums:

S1: P1+P2+P3+P4=9+8+1+2=20, yes.

S2: P3+P5+P6+P7=1+5+10+4=20, yes.

S3: P7+P8+P9+P10=4+3+6+7=20, yes.

And all numbers 1 to 10: 1,2,3,4,5,6,7,8,9,10 all present.

P4=2, P8=3, P10=7, etc.

So S=20 is possible.

But the problem is to find minimum S, and we have S=20.

But is it minimum? Earlier I thought P3+P7 min 5, giving S=20.

But P3+P7 could be smaller? Min sum with ≡2 mod 3 is 5, as 1+4 or 2+3.

1 and 4 is 5, 2 and 3 is 5, next is 1 and 7=8, larger, so 5 is minimum possible for P3+P7.

Thus S=20 is achievable, as above.

But in the assignment, we have S=20 with the numbers.

But the problem says "the sums S of the four numbers on the vertices of each square are all the same", and we have it.

But is there a smaller S? S=19 or less?

From the equation, S = (55 + P3 + P7)/3

P3 and P7 at least 1 and 1, but 1+1=2 ≡2 mod 3? 2≡2 mod 3, yes.

55 + 2 =57, 57/3=19, so S=19 possible if P3+P7=2.

But P3 and P7 are distinct points, so min P3+P7=1+2=3, but 1+2=3≡0 mod 3, not 2.

P3 and P7 must be different since all points distinct.

So possible pairs: min sum with ≡2 mod 3.

1 and 4:5≡2

1 and 7:8≡2

1 and 10:11≡2

2 and 3:5≡2

2 and 6:8≡2

2 and 9:11≡2

3 and 4:7≡1, not 2

3 and 5:8≡2

etc.

So min sum is 5, with {1,4} or {2,3}.

Thus S min 20.

But in our assignment, we have S=20 with P3=1, P7=4, for example.

In the assignment above, it works.

But the problem is to find the minimum S, and we have it.

But let me confirm the point labels.

In the beginning, I assumed the labeling, but the points are labeled 1 to 10, but we assign the values.

In the diagram, the points are fixed, but we assign numbers to them.

In our case, with the values, it works.

Perhaps for S=20, it is minimum.

But the problem says "what is the minimum value of S", and we have S=20 possible.

But let me see if S=19 is possible.

If S=19, then from 3S=57, so 55 + P3 + P7 =57, so P3 + P7=2.

But P3 and P7 are at least 1 and 2, sum 3>2, impossible. Or 1 and 1, but points distinct, min sum 1+2=3>2, so P3+P7=2 impossible. Thus S=19 impossible.

Similarly, S=18 would require 3S=54, 55 + P3+P7=54, P3+P7=-1, impossible.

So S=20 is indeed the minimum.

But in our assignment, we have it working.

But in the diagram, are there any constraints I missed?

The problem says "the network of bars shown", and "points of intersection", but I think we covered it.

Perhaps the squares are not the only cells, but the problem says "each square", and we have three squares.

In this configuration, are there only three squares? In the diamond chain, with three diamonds, there might be additional cells, but in this case, with the points given, only three squares are formed, I think.

For example, with points at (0,1), (1,0), (1,2), (2,1), (3,0), (3,2), (4,1), but that's 7 points, not 10.

I think I have a different configuration.

Perhaps it's a different grid.

Another common one: perhaps it's like a plus shape or something.

I think I should stick with the initial approach.

Since we have an assignment for S=20, and it works, and S cannot be smaller, so min S=20.

But the problem is from a context where maybe there is a trick.

Perhaps the squares are not all the same size or something, but in the diagram, it should be.

Another thought: in the bar network, there might be more than three squares, but the problem says "each square", and "the sums are all the same", implying all squares in the network.

But in this case, with three diamonds in a row, are there only three squares?

For example, with points: let's list the points as per my earlier labeling.

Points: P1, P2, P3, P4 for left diamond.

P3, P5, P7, P6 for middle? Earlier I had S2 with P3, P5, P7, P6.

In terms of connections: P1 to P2, P2 to P3, P3 to P4? No.

In a diamond: for S1, vertices P1, P2, P3, P4, with edges P1-P2, P2-P3, P3-P4, P4-P1? That would be a square, but in diamond shape, it might be P1 left, P2 top, P3 right, P4 bottom, so edges P1-P2, P2-P3, P3-P4, P4-P1, but P1 to P2 is vertical? I think I have coordinate confusion.

Assume P1 at (0,0), P2 at (1,1), P3 at (2,0), P4 at (1,-1) or something. Better not.

Since it's a bar network, the points are connected by horizontal and vertical bars, but the squares are formed.

Perhaps it is a grid.

For example, a 2x2 grid of cells would have 3x3=9 points, but here 10 points, so not.

A 3x2 grid or something.

3x2 cells would have 4x3=12 points, too many.

Perhaps it's a different configuration.

Another idea: perhaps it's a row of three squares with an additional row or column.

I think I should accept the initial solution.

Perhaps the diagram has the points labeled, but since not shown, we have to assume.

In many such problems, it's the diamond chain.

With 10 points for three diamonds: each diamond has 4 points, but sharing, so 4 + 2 + 2 = 8, but 10 points, so must have two extra points or something.

In my earlier, I had P1, P2, P3, P4 for left; then for middle, P3 and P4? No.

Let's define properly.

Consider the points: the left diamond has points: top, bottom, left, right. Call them T1, B1, L1, R1.

Then the middle diamond: it shares the right point of left, which is R1, and has its own top T2, bottom B2, and left is R1, but for the middle diamond, its left vertex is R1, right vertex is, say, R2.

Then the right diamond shares R2, has T3, B3, left R2, right R3.

But then points: L1, T1, R1, B1, T2, B2, R2, T3, B3, R3. That's 10 points.

Now, the squares are:

• Square 1: vertices L1, T1, R1, B1

• Square 2: vertices R1, T2, R2, B2 (since left is R1, top T2, right R2, bottom B2)

• Square 3: vertices R2, T3, R3, B3

In terms of the values, same as before.

And we have the sums equal.

In my assignment, it worked with S=20.

Perhaps for the minimum, we need to consider that all numbers must be used, and in our case, it is.

Perhaps S=20 is not minimum because of the bar network, but I think it's fine.

Another thought: in the bar network, there might be additional constraints from the bars, but the problem is only about the sums of the vertices for each square.

So I think S=20 is the minimum.

But let me see the answer.

Perhaps I can think without coordinates.

Let the shared points be A and B, where A is shared between S1 and S2, B between S2 and S3.

In my notation, P3 is A, P7 is B.

S1: P1, P2, A, P4

S2: A, P5, B, P6

S3: B, P8, P9, P10

Sums: (P1+P2+A+P4) + (A+P5+B+P6) + (B+P8+P9+P10) = 3S

Total sum: P1+P2+A+P4 + P5+P6 + P8+P9+P10 + B, but A and B are included, so total sum is P1+P2+A+P4 + P5+P6 + P8+P9+P10 + B, which is all points, but P3 and P7 are A and B, so sum is 55.

In the sum of squares: S1 has A, S2 has A and B, S3 has B, so A and B counted twice, so sum of square sums = total sum + A + B = 55 + A + B = 3S

Same as before.

A and B are P3 and P7.

Min S when A+B min with 55+A+B divisible by 3, A+B ≡2 mod 3, min 5, S=20.

And we have an example.

Perhaps the minimum is higher due to other constraints, but in our assignment, it works.

In the assignment, we have P1=9, P2=8, P4=2, A=P3=1, P5=5, P6=10, B=P7=4, P8=3, P9=6, P10=7.

All numbers used, sums 20.

If we use A and B as 2 and 3.

Say P3=2, P7=3.

Then S1: P1+P2+2+P4=20, so P1+P2+P4=18

S2: 2 + P5 + P6 +3 =20, so P5+P6=15

S3: 3 + P8 + P9 + P10=20, so P8+P9+P10=17

Available numbers: 1,4,5,6,7,8,9,10

P5+P6=15, e.g., 5+10, 6+9, 7+8.

Say P5=5, P6=10.

Then available: 1,4,6,7,8,9 for P1,P2,P4,P8,P9,P10.

P1+P2+P4=18

P8+P9+P10=17

P1,P2,P4 sum 18: possible 9,8,1 or 9,7,2 not, 2 not available, 9,8,1=18, or 9,7,2 no, 8,7,3 no, 8,7,3=18 but 3 not, available 1,4,6,7,8,9.

So 9,8,1=18 or 9,7,2 no, 8,7,3 no, 9,6,3 no, so only 9,8,1 or permutations.

9,8,1 sum 18.

Then P8,P9,P10: 4,6,7 sum 17? 4+6+7=17, yes.

Then P1=9, P2=8, P4=1, P5=5, P6=10, P7=3, P8=4, P9=6, P10=7.

S1: P1+P2+P3+P4=9+8+2+1=20? P3=2, yes 9+8+2+1=20.

S2: P3+P5+P6+P7=2+5+10+3=20.

S3: P7+P8+P9+P10=3+4+6+7=20.

P8=4, P9=6, P10=7, sum 3+4+6+7=20, yes.

Also works.

So S=20 is achievable.

Therefore, the minimum S is 20.

But the problem says "the minimum value of S", and in the context, perhaps it is.

I think I should go with that.