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Reflections in a Square
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
New contributor
asked 2 hours ago
Gilad MGilad M
111
111
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add a comment |
add a comment |
1 Answer
1
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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 $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.
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 $4$ 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 $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
$endgroup$
$begingroup$
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
32 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
25 mins ago
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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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 $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.
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 $4$ 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 $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
$endgroup$
$begingroup$
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
32 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
25 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 $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.
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 $4$ 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 $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
$endgroup$
$begingroup$
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
32 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
25 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 $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.
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 $4$ 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 $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
$endgroup$
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 $overlineAB$, $overlineCD$, $overlineAC$, $overlineBD$.
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 $4$ 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 $sqrt5^2 + 3^2 = sqrt34$ times the table length.
Which is like this:
edited 25 mins ago
answered 38 mins ago
athinathin
8,73422877
8,73422877
$begingroup$
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
32 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
25 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
32 mins ago
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
25 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
32 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
32 mins ago
1
1
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
25 mins ago
$begingroup$
@AhmedAbdelhameed added :)
$endgroup$
– athin
25 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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