Problem of the Month (August 2014)

Suppose we have n points in the interior of a square, and these points are rigid because they are constrained to be distance 1 from some subset of the sides of the square and each other. What are the possible sizes of the square, and the fewest number of points necessary to achieve them? What if the points are constrained to be distance 1 from some subset of the corners of the square and each other?

ANSWERS

We can almost always add points to a rigid configuration without changing the side of the square, so we only show configurations containing the fewest number of points for a given side.

Here are the known configurations where points may be 1 unit from a side:

Configurations with 1 Point

2

Configurations with 2 Points

2 - 1/√2 = 1.292+
2 + 1/√2 = 2.707+
3

Configurations with 3 Points

2-(√6+√2)/4 = 1.034+
2 - 2/√5 = 1.105+
2-(√6-√2)/4 = 1.741+
2+(√6-√2)/4 = 2.258+
5/2 = 2.5

2 + √3/2 = 2.866+
2 + 2/√5 = 2.894+
2+(√6+√2)/4 = 2.965+
4

Configurations with 4 Points

1.003+
2 - 3/√10 = 1.051+
7/5 = 1.4
2 - 1/√5 = 1.552+
(9 - √7)/4 = 1.588+

1.656+
1.874+
2.126+
3 - √3/2 = 2.133+
2.274+

2.278+
3 - 1/√2 = 2.292+
2.343+
(7 + √7)/4 = 2.411+
2 + 1/√5 = 2.447+

13/5 = 2.6
2.626+
2.721+
2.806+
(9 + √7)/4 = 2.911+

2 + 3/√10 = 2.948+
2.992+
2.996+
2 + √(3/2) = 3.224+
(5 + √3)/2 = 3.366+

2 + √2 = 3.414+
7/2 = 3.5
3 + 1/√2 = 3.707+
2 + √3 = 3.732+
3 + √3/2 = 3.866+

5

Here are the known configurations where points may be 1 unit from a corner:

Configurations with 1 Point

√2 = 1.414+

Configurations with 2 Points

√(3/2) = 1.224+
(1 + √7)/2 = 1.822+
(√2 + √6)/2 = 1.931+

Configurations with 3 Points

1.782+
√(2+√2) = 1.847+
1.95517+

Configurations with 4 Points

1.296+
1.347+
1.635+
1.680+
1.743+

1.816+
1.862+
√(7/2) = 1.870+
1.95500724+
1.95500785+

1.95521750+
1.95521754+
(√5 + √3)/2 = 1.984+
1.992+
1.999+

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 8/1/14.