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Reflections in a Square
Ray reflection inside the cubeDissecting a squareEnlarge the Square?Quadrilateral inside a squareThe Challenge SquareThe square and the compassNo ordinary magic squareSide length of square inside a triangleDissect a square into 3:1 rectanglesUgh! The extra SQUARE!Looking for Euler square solver
$begingroup$
An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.
What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?
Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.
Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).
logical-deduction geometry
asked Apr 22 at 10:49
Gilad MGilad M
1584
$endgroup$
add a comment |
$begingroup$
An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.
What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?
Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.
Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).
logical-deduction geometry
asked Apr 22 at 10:49
Gilad MGilad M
1584
$endgroup$
$begingroup$
Um, sorry, but ideal billiards table is shaped like a rectangle.
$endgroup$
– Zereges
Apr 22 at 20:46
1
$begingroup$
It has no pockets either - it's such an ideal billiards table that you can't play billiards with it. I'm a physicist. Making useful, non-ideal things is the engineer's job. :P
$endgroup$
– Gilad M
Apr 22 at 22:10
add a comment |
$begingroup$
An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.
What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?
Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.
Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).
logical-deduction geometry
asked Apr 22 at 10:49
Gilad MGilad M
1584
$endgroup$
An ideal billiards table (no friction, ideal reflections off of the walls, no pockets) is shaped like a square. From the bottom-left corner, shoot a point-sized cue ball at some angle.
What is the shortest path that hits all 4 edges of the table at least once each and ends in the top-right corner?
Note 1: hitting a corner ends the path without counting as hitting the walls on either side of it.
Note 2: I'm aware of this puzzle, which is higher-dimensional and more mathematically involved but has a similar answer to this one. I felt that this warranted its own puzzle, though, because I want more people to have the chance to find its elegant, visual solution without being intimidated by the apparent need to use math (and it's not asking for exactly the same thing).
logical-deduction geometry
logical-deduction geometry
asked Apr 22 at 10:49
Gilad MGilad M
1584
asked Apr 22 at 10:49
Gilad MGilad M
1584
asked Apr 22 at 10:49
Gilad MGilad M
1584
asked Apr 22 at 10:49
Gilad MGilad M
1584
asked Apr 22 at 10:49
Gilad MGilad M
1584
1584
$begingroup$
Um, sorry, but ideal billiards table is shaped like a rectangle.
$endgroup$
– Zereges
Apr 22 at 20:46
1
$begingroup$
It has no pockets either - it's such an ideal billiards table that you can't play billiards with it. I'm a physicist. Making useful, non-ideal things is the engineer's job. :P
$endgroup$
– Gilad M
Apr 22 at 22:10
add a comment |
$begingroup$
Um, sorry, but ideal billiards table is shaped like a rectangle.
$endgroup$
– Zereges
Apr 22 at 20:46
1
$begingroup$
It has no pockets either - it's such an ideal billiards table that you can't play billiards with it. I'm a physicist. Making useful, non-ideal things is the engineer's job. :P
$endgroup$
– Gilad M
Apr 22 at 22:10
$begingroup$
Um, sorry, but ideal billiards table is shaped like a rectangle.
$endgroup$
– Zereges
Apr 22 at 20:46
$begingroup$
Um, sorry, but ideal billiards table is shaped like a rectangle.
$endgroup$
– Zereges
Apr 22 at 20:46
1
1
$begingroup$
It has no pockets either - it's such an ideal billiards table that you can't play billiards with it. I'm a physicist. Making useful, non-ideal things is the engineer's job. :P
$endgroup$
– Gilad M
Apr 22 at 22:10
$begingroup$
It has no pockets either - it's such an ideal billiards table that you can't play billiards with it. I'm a physicist. Making useful, non-ideal things is the engineer's job. :P
$endgroup$
– Gilad M
Apr 22 at 22:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose we label the corner on the table like this:
Now we want to move from $A$ to $D$.
Now, imagine the table like this:
Here,
We can simulate the path as a single straight line.
As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.
Now, to hits all $4$ edges, that means
The path should pass $overlineAB$, $overlineCD$, $overlineAC$, and $overlineBD$.
In other words, the path must go beyond $2 times 2$ tables.
To get the shortest path,
We should choose the nearest $D$ (beyond $2 times 2$ tables).
One of the options is the $D$ located $3$ above and $3$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
We can choose the next nearest $D$ which is located $3$ above and $5$ right (or $5$ above and $3$ right). This is the shortest path.
The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
answered Apr 22 at 12:44
athinathin
9,04423181
$endgroup$
1
$begingroup$
Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
$endgroup$
– Ahmed Abdelhameed
Apr 22 at 12:50
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
Apr 22 at 12:57
$begingroup$
Nice! That's exactly right!
$endgroup$
– Gilad M
Apr 22 at 14:08
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose we label the corner on the table like this:
Now we want to move from $A$ to $D$.
Now, imagine the table like this:
Here,
We can simulate the path as a single straight line.
As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.
Now, to hits all $4$ edges, that means
The path should pass $overlineAB$, $overlineCD$, $overlineAC$, and $overlineBD$.
In other words, the path must go beyond $2 times 2$ tables.
To get the shortest path,
We should choose the nearest $D$ (beyond $2 times 2$ tables).
One of the options is the $D$ located $3$ above and $3$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
We can choose the next nearest $D$ which is located $3$ above and $5$ right (or $5$ above and $3$ right). This is the shortest path.
The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
answered Apr 22 at 12:44
athinathin
9,04423181
$endgroup$
1
$begingroup$
Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
$endgroup$
– Ahmed Abdelhameed
Apr 22 at 12:50
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
Apr 22 at 12:57
$begingroup$
Nice! That's exactly right!
$endgroup$
– Gilad M
Apr 22 at 14:08
add a comment |
$begingroup$
Suppose we label the corner on the table like this:
Now we want to move from $A$ to $D$.
Now, imagine the table like this:
Here,
We can simulate the path as a single straight line.
As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.
Now, to hits all $4$ edges, that means
The path should pass $overlineAB$, $overlineCD$, $overlineAC$, and $overlineBD$.
In other words, the path must go beyond $2 times 2$ tables.
To get the shortest path,
We should choose the nearest $D$ (beyond $2 times 2$ tables).
One of the options is the $D$ located $3$ above and $3$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
We can choose the next nearest $D$ which is located $3$ above and $5$ right (or $5$ above and $3$ right). This is the shortest path.
The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
answered Apr 22 at 12:44
athinathin
9,04423181
$endgroup$
1
$begingroup$
Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
$endgroup$
– Ahmed Abdelhameed
Apr 22 at 12:50
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
Apr 22 at 12:57
$begingroup$
Nice! That's exactly right!
$endgroup$
– Gilad M
Apr 22 at 14:08
add a comment |
$begingroup$
Suppose we label the corner on the table like this:
Now we want to move from $A$ to $D$.
Now, imagine the table like this:
Here,
We can simulate the path as a single straight line.
As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.
Now, to hits all $4$ edges, that means
The path should pass $overlineAB$, $overlineCD$, $overlineAC$, and $overlineBD$.
In other words, the path must go beyond $2 times 2$ tables.
To get the shortest path,
We should choose the nearest $D$ (beyond $2 times 2$ tables).
One of the options is the $D$ located $3$ above and $3$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
We can choose the next nearest $D$ which is located $3$ above and $5$ right (or $5$ above and $3$ right). This is the shortest path.
The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
answered Apr 22 at 12:44
athinathin
9,04423181
$endgroup$
Suppose we label the corner on the table like this:
Now we want to move from $A$ to $D$.
Now, imagine the table like this:
Here,
We can simulate the path as a single straight line.
As the line passes the border, it will simulate the result of the ball's reflection hence the table will be mirrored as above.
Now, to hits all $4$ edges, that means
The path should pass $overlineAB$, $overlineCD$, $overlineAC$, and $overlineBD$.
In other words, the path must go beyond $2 times 2$ tables.
To get the shortest path,
We should choose the nearest $D$ (beyond $2 times 2$ tables).
One of the options is the $D$ located $3$ above and $3$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
We can choose the next nearest $D$ which is located $3$ above and $5$ right (or $5$ above and $3$ right). This is the shortest path.
The overall length will be $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
answered Apr 22 at 12:44
athinathin
9,04423181
edited Apr 23 at 5:31
answered Apr 22 at 12:44
athinathin
9,04423181
answered Apr 22 at 12:44
athinathin
9,04423181
answered Apr 22 at 12:44
athinathin
9,04423181
9,04423181
1
$begingroup$
Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
$endgroup$
– Ahmed Abdelhameed
Apr 22 at 12:50
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
Apr 22 at 12:57
$begingroup$
Nice! That's exactly right!
$endgroup$
– Gilad M
Apr 22 at 14:08
add a comment |
1
$begingroup$
Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
$endgroup$
– Ahmed Abdelhameed
Apr 22 at 12:50
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
Apr 22 at 12:57
$begingroup$
Nice! That's exactly right!
$endgroup$
– Gilad M
Apr 22 at 14:08
1
1
$begingroup$
Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
$endgroup$
– Ahmed Abdelhameed
Apr 22 at 12:50
$begingroup$
Can you add another picture at the end with the line drawn over the first image? Just to save us from rotating our heads repeatedly :-D
$endgroup$
– Ahmed Abdelhameed
Apr 22 at 12:50
1
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
Apr 22 at 12:57
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
Apr 22 at 12:57
$begingroup$
Nice! That's exactly right!
$endgroup$
– Gilad M
Apr 22 at 14:08
$begingroup$
Nice! That's exactly right!
$endgroup$
– Gilad M
Apr 22 at 14:08
add a comment |
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$begingroup$
Um, sorry, but ideal billiards table is shaped like a rectangle.
$endgroup$
– Zereges
Apr 22 at 20:46
1
$begingroup$
It has no pockets either - it's such an ideal billiards table that you can't play billiards with it. I'm a physicist. Making useful, non-ideal things is the engineer's job. :P
$endgroup$
– Gilad M
Apr 22 at 22:10