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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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$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
Technically we can calculus the line segments into curves if we so please.
$endgroup$
– greenturtle3141
5 hours ago
$begingroup$
well, the line segments should be straight lines, if you want to clarify that.
$endgroup$
– ThomasL
5 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
5 hours ago
$begingroup$
Two related puzzles: Find a straight tunnel and Find a straight tunnel 2
$endgroup$
– Jaap Scherphuis
5 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
$endgroup$
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
geometry strategy
asked 8 hours ago
ThomasLThomasL
7332 silver badges19 bronze badges
7332 silver badges19 bronze badges
$begingroup$
Technically we can calculus the line segments into curves if we so please.
$endgroup$
– greenturtle3141
5 hours ago
$begingroup$
well, the line segments should be straight lines, if you want to clarify that.
$endgroup$
– ThomasL
5 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
5 hours ago
$begingroup$
Two related puzzles: Find a straight tunnel and Find a straight tunnel 2
$endgroup$
– Jaap Scherphuis
5 hours ago
add a comment
|
$begingroup$
Technically we can calculus the line segments into curves if we so please.
$endgroup$
– greenturtle3141
5 hours ago
$begingroup$
well, the line segments should be straight lines, if you want to clarify that.
$endgroup$
– ThomasL
5 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
5 hours ago
$begingroup$
Two related puzzles: Find a straight tunnel and Find a straight tunnel 2
$endgroup$
– Jaap Scherphuis
5 hours ago
$begingroup$
Technically we can calculus the line segments into curves if we so please.
$endgroup$
– greenturtle3141
5 hours ago
$begingroup$
Technically we can calculus the line segments into curves if we so please.
$endgroup$
– greenturtle3141
5 hours ago
$begingroup$
well, the line segments should be straight lines, if you want to clarify that.
$endgroup$
– ThomasL
5 hours ago
$begingroup$
well, the line segments should be straight lines, if you want to clarify that.
$endgroup$
– ThomasL
5 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
5 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
5 hours ago
$begingroup$
Two related puzzles: Find a straight tunnel and Find a straight tunnel 2
$endgroup$
– Jaap Scherphuis
5 hours ago
$begingroup$
Two related puzzles: Find a straight tunnel and Find a straight tunnel 2
$endgroup$
– Jaap Scherphuis
5 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
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
$endgroup$
$begingroup$
good finding! But I know that there is at least one more improvement...
$endgroup$
– ThomasL
5 hours ago
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 $2+sqrt2/2 approx 2.707107$.
$endgroup$
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$endgroup$
$begingroup$
good finding! But I know that there is at least one more improvement...
$endgroup$
– ThomasL
5 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
$endgroup$
$begingroup$
good finding! But I know that there is at least one more improvement...
$endgroup$
– ThomasL
5 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
$endgroup$
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
answered 8 hours ago
LOTGPLOTGP
1761 silver badge7 bronze badges
1761 silver badge7 bronze badges
$begingroup$
good finding! But I know that there is at least one more improvement...
$endgroup$
– ThomasL
5 hours ago
add a comment
|
$begingroup$
good finding! But I know that there is at least one more improvement...
$endgroup$
– ThomasL
5 hours ago
$begingroup$
good finding! But I know that there is at least one more improvement...
$endgroup$
– ThomasL
5 hours ago
$begingroup$
good finding! But I know that there is at least one more improvement...
$endgroup$
– ThomasL
5 hours ago
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 $2+sqrt2/2 approx 2.707107$.
$endgroup$
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 $2+sqrt2/2 approx 2.707107$.
$endgroup$
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 $2+sqrt2/2 approx 2.707107$.
$endgroup$
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 $2+sqrt2/2 approx 2.707107$.
edited 4 hours ago
answered 5 hours ago
Jaap ScherphuisJaap Scherphuis
19.2k1 gold badge34 silver badges84 bronze badges
19.2k1 gold badge34 silver badges84 bronze badges
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$begingroup$
Technically we can calculus the line segments into curves if we so please.
$endgroup$
– greenturtle3141
5 hours ago
$begingroup$
well, the line segments should be straight lines, if you want to clarify that.
$endgroup$
– ThomasL
5 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
5 hours ago
$begingroup$
Two related puzzles: Find a straight tunnel and Find a straight tunnel 2
$endgroup$
– Jaap Scherphuis
5 hours ago