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Line segments inside a square

Find a straight tunnelFind a straight tunnel 2Find shortest network connecting four pointsConnect four towers by roadsClash of arrowsQuadrilateral inside a squareInside or outside the square?IcosikaitrigonsA construction on an infinite 2d grid, part 1$verb|Eight Circles|$Form Common Geometric ShapesPentomino solution maximizing straight lines length in rectangle - wood cutter problem

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4

$begingroup$

A set of line segments inside or at the edge of a square with side length 1 should be positioned in such a way,
that any straight line going through the square must touch or intersect at least one of the line segment.
Find such a configuration where the total length of all such line segments is minimal?

Example: choose the 4 sides of the square as line segments. The length of those line segments is 4.
A better choice are the two diagonals of the square with a total length of $2timessqrt2$ ~ 2,828. Can you improve further?

geometry strategy

share|improve this question

asked 10 hours ago

ThomasLThomasL

73322 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Technically we can calculus the line segments into curves if we so please.
  $endgroup$
  – greenturtle3141
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  well, the line segments should be straight lines, if you want to clarify that.
  $endgroup$
  – ThomasL
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @greenturtle3141 While true, I can't think of any situations in this puzzle where we would prefer curves to straight lines.
  $endgroup$
  – LOTGP
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Two related puzzles: Find a straight tunnel and Find a straight tunnel 2
  $endgroup$
  – Jaap Scherphuis
  7 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

A set of line segments inside or at the edge of a square with side length 1 should be positioned in such a way,
that any straight line going through the square must touch or intersect at least one of the line segment.
Find such a configuration where the total length of all such line segments is minimal?

Example: choose the 4 sides of the square as line segments. The length of those line segments is 4.
A better choice are the two diagonals of the square with a total length of $2timessqrt2$ ~ 2,828. Can you improve further?

geometry strategy

share|improve this question

asked 10 hours ago

ThomasLThomasL

73322 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Technically we can calculus the line segments into curves if we so please.
  $endgroup$
  – greenturtle3141
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  well, the line segments should be straight lines, if you want to clarify that.
  $endgroup$
  – ThomasL
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @greenturtle3141 While true, I can't think of any situations in this puzzle where we would prefer curves to straight lines.
  $endgroup$
  – LOTGP
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Two related puzzles: Find a straight tunnel and Find a straight tunnel 2
  $endgroup$
  – Jaap Scherphuis
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

A set of line segments inside or at the edge of a square with side length 1 should be positioned in such a way,
that any straight line going through the square must touch or intersect at least one of the line segment.
Find such a configuration where the total length of all such line segments is minimal?

Example: choose the 4 sides of the square as line segments. The length of those line segments is 4.
A better choice are the two diagonals of the square with a total length of $2timessqrt2$ ~ 2,828. Can you improve further?

geometry strategy

share|improve this question

asked 10 hours ago

ThomasLThomasL

73322 silver badges1919 bronze badges

$endgroup$

share|improve this question

asked 10 hours ago

ThomasLThomasL

73322 silver badges1919 bronze badges

share|improve this question

asked 10 hours ago

ThomasLThomasL

73322 silver badges1919 bronze badges

asked 10 hours ago

ThomasLThomasL

73322 silver badges1919 bronze badges

asked 10 hours ago

ThomasLThomasL

73322 silver badges1919 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

Seems a slightly better solution would be to:

cover 2 of the sides that meet at one of the corners, then draw the half diagonal from the opposite corner to the middle.

Something like this:

The total length is then:

1 + 1 + sqrt(2)/2 = 2.707

share|improve this answer

answered 9 hours ago

LOTGPLOTGP

17611 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good finding! But I know that there is at least one more improvement...
  $endgroup$
  – ThomasL
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Building on LOTGP's answer, you could do this:

Assuming a unit square, the total length is:

The top left segment is $sqrt2/2$.
The three other segments are shortest when they meet at 120 degrees. This makes the triangle angles $(120, 45, 15)$. Using the sine rule, that gives
$sin45/sin120 approx 0.8164$ for the long sides
$sin15/sin120 approx 0.2988$ for the short sides

for a total of about $2.638958$.

This is a slight improvement over LOTGP's answer which is $2+sqrt2/2 approx 2.707107$.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Jaap ScherphuisJaap Scherphuis

19.2k11 gold badge3434 silver badges8484 bronze badges

$endgroup$

add a comment
 | 

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6

$begingroup$

Seems a slightly better solution would be to:

cover 2 of the sides that meet at one of the corners, then draw the half diagonal from the opposite corner to the middle.

Something like this:

The total length is then:

1 + 1 + sqrt(2)/2 = 2.707

share|improve this answer

answered 9 hours ago

LOTGPLOTGP

17611 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good finding! But I know that there is at least one more improvement...
  $endgroup$
  – ThomasL
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

6

$begingroup$

Seems a slightly better solution would be to:

cover 2 of the sides that meet at one of the corners, then draw the half diagonal from the opposite corner to the middle.

Something like this:

The total length is then:

1 + 1 + sqrt(2)/2 = 2.707

share|improve this answer

answered 9 hours ago

LOTGPLOTGP

17611 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good finding! But I know that there is at least one more improvement...
  $endgroup$
  – ThomasL
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

Seems a slightly better solution would be to:

cover 2 of the sides that meet at one of the corners, then draw the half diagonal from the opposite corner to the middle.

Something like this:

The total length is then:

1 + 1 + sqrt(2)/2 = 2.707

share|improve this answer

answered 9 hours ago

LOTGPLOTGP

17611 silver badge77 bronze badges

$endgroup$

share|improve this answer

answered 9 hours ago

LOTGPLOTGP

17611 silver badge77 bronze badges

answered 9 hours ago

LOTGPLOTGP

17611 silver badge77 bronze badges

answered 9 hours ago

LOTGPLOTGP

17611 silver badge77 bronze badges

4

$begingroup$

Building on LOTGP's answer, you could do this:

Assuming a unit square, the total length is:

The top left segment is $sqrt2/2$.
The three other segments are shortest when they meet at 120 degrees. This makes the triangle angles $(120, 45, 15)$. Using the sine rule, that gives
$sin45/sin120 approx 0.8164$ for the long sides
$sin15/sin120 approx 0.2988$ for the short sides

for a total of about $2.638958$.

This is a slight improvement over LOTGP's answer which is $2+sqrt2/2 approx 2.707107$.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Jaap ScherphuisJaap Scherphuis

19.2k11 gold badge3434 silver badges8484 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Building on LOTGP's answer, you could do this:

Assuming a unit square, the total length is:

The top left segment is $sqrt2/2$.
The three other segments are shortest when they meet at 120 degrees. This makes the triangle angles $(120, 45, 15)$. Using the sine rule, that gives
$sin45/sin120 approx 0.8164$ for the long sides
$sin15/sin120 approx 0.2988$ for the short sides

for a total of about $2.638958$.

This is a slight improvement over LOTGP's answer which is $2+sqrt2/2 approx 2.707107$.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Jaap ScherphuisJaap Scherphuis

19.2k11 gold badge3434 silver badges8484 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Building on LOTGP's answer, you could do this:

Assuming a unit square, the total length is:

The top left segment is $sqrt2/2$.
The three other segments are shortest when they meet at 120 degrees. This makes the triangle angles $(120, 45, 15)$. Using the sine rule, that gives
$sin45/sin120 approx 0.8164$ for the long sides
$sin15/sin120 approx 0.2988$ for the short sides

for a total of about $2.638958$.

This is a slight improvement over LOTGP's answer which is $2+sqrt2/2 approx 2.707107$.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Jaap ScherphuisJaap Scherphuis

19.2k11 gold badge3434 silver badges8484 bronze badges

$endgroup$

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Jaap ScherphuisJaap Scherphuis

19.2k11 gold badge3434 silver badges8484 bronze badges

answered 6 hours ago

Jaap ScherphuisJaap Scherphuis

19.2k11 gold badge3434 silver badges8484 bronze badges

answered 6 hours ago

Jaap ScherphuisJaap Scherphuis

19.2k11 gold badge3434 silver badges8484 bronze badges