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# 6 Squares and a Common Point

## Question

As a follow up to my circles problem:

Given six squares where any group of five have two common points.  Does there exist a common point for all 6?

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If the squares A and B share a common part of a side, which we'll call AB, then we know that any three of the remaining squares must share exactly two common points on this line segment. By the same argument and assumptions as before, each of these squares must necessarily intersect with eight distinct points on AB, which is impossible for a square unless the overlapping area is another line segment. However, this leads to the entire line segment being shared among all six squares, which is another clear contradiction. A similar argument can be made if the squares A and B share two common line segments.

I made a mistake. 'eight distinct' should be 'six distinct'.

Edited by gavinksong
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Assume the squares are congruent

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If all six squares lie in the same plane and exactly two common points are shared among any of the five squares, then each of three squares of the six will share one common point with one each of the other three squares.

Labelling these squares, one vertex (corner point) of square-A shares a point on the perimeter with with square-B, one vertex of square-C shares a point on the perimeter with square-D, and one vertex of square-E shares a point on the perimeter with square-F. When selecting any five of these squares only two points will be shared. Labelling and permuting the three shared points one finds that no common point will be shared for all six combinations of five squares for the six squares.

If the squares are not required to lie in the same plane, the shared points may be anywhere on the perimeter of the squares. And a six square may share one or both points with the other five.
Edited by DejMar
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If all six squares lie in the same plane and exactly two common points are shared among any of the five squares, then each of three squares of the six will share one common point with one each of the other three squares.

Labelling these squares, one vertex (corner point) of square-A shares a point on the perimeter with with square-B, one vertex of square-C shares a point on the perimeter with square-D, and one vertex of square-E shares a point on the perimeter with square-F. When selecting any five of these squares only two points will be shared. Labelling and permuting the three shared points one finds that no common point will be shared for all six combinations of five squares for the six squares.

If the squares are not required to lie in the same plane, the both shared points may be anywhere on the perimeter of the squares. And a sixth square may share one or both points with the other five.

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I need some clarifications:

1) Does every group of five squares share exactly two common points, or can they have more?

2) Are the squares solid?

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I need some clarifications:

1) Does every group of five squares share exactly two common points, or can they have more?

2) Are the squares solid?

Label the squares A, B, C, D, E, and F. Let A' refer to the pair of points that is shared by all the squares other than A, and so on.

Suppose that the six squares do not share a common point. Then, A' cannot be in A, B' cannot be in B, and so on. Since if A' is not in A, but B' is in A, they cannot refer to the same pair of points. Extending that logic suggests that A', B', C', D', E', and F' all refer to distinct pairs of points.

Let's look at the squares A and B. They must both contain C', D', E', and F' - which is a total of eight distinct points.

Wait a second. I just realized this won't work.

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I need some clarifications:

1) Does every group of five squares share exactly two common points, or can they have more?

2) Are the squares solid?

Label the squares A, B, C, D, E, and F. Let A' refer to the pair of points that is shared by all the squares other than A, and so on.

Suppose that the six squares do not share a common point. Then, A' cannot be in A, B' cannot be in B, and so on. Since if A' is not in A, but B' is in A, they cannot refer to the same pair of points. Extending that logic suggests that A', B', C', D', E', and F' all refer to distinct pairs of points.

Let's look at the squares A and B. They must both contain C', D', E', and F' - which is a total of eight distinct points.

Wait a second. I just realized this won't work.

and the squares are not solid

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BMAD, It seems you left point 1 of gavinksong's request for clarifications unclarified. Your answer is ambiguous in that regard. And, to add a question of my own, are the squares to be considered residing in the same Euclidean plane?

Edited by DejMar
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BMAD, It seems you left point 1 of gavinksong's request for clarifications unclarified. Your answer is ambiguous in that regard. And, to add a question of my own, are the squares to be considered residing in the same Euclidean plane?

my apologies.  Yes. every five squares share exactly two points.

and to your question: they are all in the same Euclidean plane

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Another question, as the problem sounds like a trick. Though the five squares share exactly two points, will it require some of the squares to intersect more than two points in total with all squares? Or are the shared two points only referring to the points shared between any two squares?

Edited by DejMar
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Another question, as the problem sounds like a trick. Though the five squares share exactly two points, will it require some of the squares to intersect more than two points in total with all squares? Or are the shared two points only referring to the points shared between any two squares?

any five squares you pick, all of them will share precisely two points.  This does not preclude the selection of a fewer amount having more than two points in common.

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Let two squares share a side. (These two squares will share an infinite number of points on the line segment that represents the side of the two squares.)

Let a third square be placed such that one of its corners will form an isoceles triangle with its base about 1/3 of the length of the line segment in the middle of the line segment.
(At this juncture, the three squares share only two points.)

Let a fourth square mirror the third along this line segment.
(The four squares now share the same two points.

Without loss of generality, let the distance between the two points be an interger distance. It is not nessecary that these points be integral distance, but only to assist in demonstrating an infinite number of squares can share these two points.

Place the fifth square such that its corner forms a Pythagorean triple with the two points.
A sixth square can mirror this square along the line segment, as well.

Repeating the former two steps, using a different pythagorean triple (scaled as necessary), an infinite number of pythagorean triples, you can be placed, and each will only share the two points. Thus, the answer is that there does exist a common point for 6 (and 7, and 8, and 9, etc.).

*Remove any one of the first two squares, as these two shared more than one point, and were placed only for personal construction.

Edited by DejMar
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I need some clarifications:

1) Does every group of five squares share exactly two common points, or can they have more?

2) Are the squares solid?

Label the squares A, B, C, D, E, and F. Let A' refer to the pair of points that is shared by all the squares other than A, and so on.

Suppose that the six squares do not share a common point. Then, A' cannot be in A, B' cannot be in B, and so on. Since if A' is not in A, but B' is in A, they cannot refer to the same pair of points. Extending that logic suggests that A', B', C', D', E', and F' all refer to distinct pairs of points.

Let's look at the squares A and B. They must both contain C', D', E', and F' - which is a total of eight distinct points.

Wait a second. I just realized this won't work.

Imagine any two squares intersecting at exactly two points. This is not hard to do.

Now suppose that each square is actually three squares lying exactly on top of each other. If we take any five of the squares, they will obviously have exactly two common points. All six of the squares will also have exactly two common points. Thus, there does exist a common point for all six.

If a general answer exists, it's yes.

Edit: fixed a typo

Edited by gavinksong
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Both of these proofs presented appear to argue specific cases for the op but do they extend to prove that no matter what orientation of six squares that meet the op requirements have to share 1 point?

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I need some clarifications:

1) Does every group of five squares share exactly two common points, or can they have more?

2) Are the squares solid?

Label the squares A, B, C, D, E, and F. Let A' refer to the pair of points that is shared by all the squares other than A, and so on.

Suppose that the six squares do not share a common point. Then, A' cannot be in A, B' cannot be in B, and so on. Since if A' is not in A, but B' is in A, they cannot refer to the same pair of points. Extending that logic suggests that A', B', C', D', E', and F' all refer to distinct pairs of points.

Let's look at the squares A and B. They must both contain C', D', E', and F' - which is a total of eight distinct points.

Wait a second. I just realized this won't work.

I decided this post needs some love.

Assuming that the six squares do not share a common point, we know that any two squares must share at least eight distinct points. If they are positioned directly on top of each other then it is obvious that all six squares must share a common point, so we note that this cannot be the case with any two squares. Then, we may note that two congruent squares can only intersect at eight points at most. Thus, we can say that the squares A and B overlap at exactly eight distinct points: C', D', E', and F'. The remaining four squares must go through at least six of these eight points, but this forces them directly on top of either of A or B, leading to a clear contradiction. Therefore, the six squares must share a common point.

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I need some clarifications:

1) Does every group of five squares share exactly two common points, or can they have more?

2) Are the squares solid?

Label the squares A, B, C, D, E, and F. Let A' refer to the pair of points that is shared by all the squares other than A, and so on.

Suppose that the six squares do not share a common point. Then, A' cannot be in A, B' cannot be in B, and so on. Since if A' is not in A, but B' is in A, they cannot refer to the same pair of points. Extending that logic suggests that A', B', C', D', E', and F' all refer to distinct pairs of points.

Let's look at the squares A and B. They must both contain C', D', E', and F' - which is a total of eight distinct points.

Wait a second. I just realized this won't work.

I decided this post needs some love.

Assuming that the six squares do not share a common point, we know that any two squares must share at least eight distinct points. If they are positioned directly on top of each other then it is obvious that all six squares must share a common point, so we note that this cannot be the case with any two squares. Then, we may note that two congruent squares can only intersect at eight points at most. Thus, we can say that the squares A and B overlap at exactly eight distinct points: C', D', E', and F'. The remaining four squares must go through at least six of these eight points, but this forces them directly on top of either of A or B, leading to a clear contradiction. Therefore, the six squares must share a common point.

I failed to account for the case where two squares share part of a side or sides.

If the squares A and B share a common part of a side, which we'll call AB, then we know that any three of the remaining squares must share exactly two common points on this line segment. By the same argument and assumptions as before, each of these squares must necessarily intersect with eight distinct points on AB, which is impossible for a square unless the overlapping area is another line segment. However, this leads to the entire line segment being shared among all six squares, which is another clear contradiction. A similar argument can be made if the squares A and B share two common line segments.

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