Reflections in a Square
$begingroup$
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 2 hours ago
Gilad MGilad M
161
New contributor
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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 2 hours ago
Gilad MGilad M
161
New contributor
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 2 hours ago
Gilad MGilad M
161
New contributor
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 2 hours ago
Gilad MGilad M
161
New contributor
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Gilad MGilad M
161
New contributor
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Gilad MGilad M
161
New contributor
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Gilad MGilad M
161
asked 2 hours ago
Gilad MGilad M
161
161
New contributor
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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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 $overline{AB}$, $overline{CD}$, $overline{AC}$, $overline{BD}$.
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 $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
This is the shortest path.
The overall length will be $sqrt{5^2 + 3^2} = sqrt{34}$ times the table length.
Which is like this:
answered 1 hour ago
athinathin
8,74422877
$endgroup$
$begingroup$
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
57 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
50 mins ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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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 $overline{AB}$, $overline{CD}$, $overline{AC}$, $overline{BD}$.
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 $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
This is the shortest path.
The overall length will be $sqrt{5^2 + 3^2} = sqrt{34}$ times the table length.
Which is like this:
answered 1 hour ago
athinathin
8,74422877
$endgroup$
$begingroup$
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
57 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
50 mins ago
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 $overline{AB}$, $overline{CD}$, $overline{AC}$, $overline{BD}$.
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 $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
This is the shortest path.
The overall length will be $sqrt{5^2 + 3^2} = sqrt{34}$ times the table length.
Which is like this:
answered 1 hour ago
athinathin
8,74422877
$endgroup$
$begingroup$
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
57 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
50 mins ago
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 $overline{AB}$, $overline{CD}$, $overline{AC}$, $overline{BD}$.
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 $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
This is the shortest path.
The overall length will be $sqrt{5^2 + 3^2} = sqrt{34}$ times the table length.
Which is like this:
answered 1 hour ago
athinathin
8,74422877
$endgroup$
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 $overline{AB}$, $overline{CD}$, $overline{AC}$, $overline{BD}$.
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 $4$ above and $4$ right. But we can't choose this because the line will hit only both corners $A$ and $D$.
Thus,
This is the shortest path.
The overall length will be $sqrt{5^2 + 3^2} = sqrt{34}$ times the table length.
Which is like this:
answered 1 hour ago
athinathin
8,74422877
edited 51 mins ago
answered 1 hour ago
athinathin
8,74422877
answered 1 hour ago
athinathin
8,74422877
answered 1 hour ago
athinathin
8,74422877
8,74422877
$begingroup$
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
57 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
50 mins ago
add a comment |
$begingroup$
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
57 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
50 mins ago
$begingroup$
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
57 mins ago
$begingroup$
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
57 mins ago
1
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
50 mins ago
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
50 mins ago
add a comment |
Gilad M is a new contributor. Be nice, and check out our Code of Conduct.
Gilad M is a new contributor. Be nice, and check out our Code of Conduct.
Gilad M is a new contributor. Be nice, and check out our Code of Conduct.
Gilad M is a new contributor. Be nice, and check out our Code of Conduct.
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