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Combine two squares into a square with the sum of the two
Drink a Little Wine, Cut a Little RugPuzzle pieces, each in contact with 5 othersFair share of a square watermelon?The Erasmus dissection of a squareFairly Sharing a Frosted CakeSquaring a crossSquaring a cross one more timeOverlapping TilesCutting a cross made of 5 equal squares by 2 straight cut into 4 figure to together form a squareGeometry haberdasher problem - square to equilateral triangle variation
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
The red square is placed on top of a blue square
The goal is to cut the red square into 4 pieces and assemble them with the blue square to create a larger square.
The area of the resulting square is the sum of the area of the red and blue squares.
There are at least two solutions I am aware of.
The first one to show at least two solutions will be granted the 15 points:)
geometry
asked Jun 17 at 7:15
MotiMoti
9135 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
The red square is placed on top of a blue square
The goal is to cut the red square into 4 pieces and assemble them with the blue square to create a larger square.
The area of the resulting square is the sum of the area of the red and blue squares.
There are at least two solutions I am aware of.
The first one to show at least two solutions will be granted the 15 points:)
geometry
asked Jun 17 at 7:15
MotiMoti
9135 silver badges18 bronze badges
$endgroup$
$begingroup$
What is the significance of "The red square is placed on top of a blue square", as the first sentence? Or is this a red herring?
$endgroup$
– Stilez
Jun 17 at 13:39
add a comment |
$begingroup$
The red square is placed on top of a blue square
The goal is to cut the red square into 4 pieces and assemble them with the blue square to create a larger square.
The area of the resulting square is the sum of the area of the red and blue squares.
There are at least two solutions I am aware of.
The first one to show at least two solutions will be granted the 15 points:)
geometry
asked Jun 17 at 7:15
MotiMoti
9135 silver badges18 bronze badges
$endgroup$
The red square is placed on top of a blue square
The goal is to cut the red square into 4 pieces and assemble them with the blue square to create a larger square.
The area of the resulting square is the sum of the area of the red and blue squares.
There are at least two solutions I am aware of.
The first one to show at least two solutions will be granted the 15 points:)
geometry
geometry
asked Jun 17 at 7:15
MotiMoti
9135 silver badges18 bronze badges
asked Jun 17 at 7:15
MotiMoti
9135 silver badges18 bronze badges
asked Jun 17 at 7:15
MotiMoti
9135 silver badges18 bronze badges
asked Jun 17 at 7:15
MotiMoti
9135 silver badges18 bronze badges
asked Jun 17 at 7:15
MotiMoti
9135 silver badges18 bronze badges
9135 silver badges18 bronze badges
$begingroup$
What is the significance of "The red square is placed on top of a blue square", as the first sentence? Or is this a red herring?
$endgroup$
– Stilez
Jun 17 at 13:39
add a comment |
$begingroup$
What is the significance of "The red square is placed on top of a blue square", as the first sentence? Or is this a red herring?
$endgroup$
– Stilez
Jun 17 at 13:39
$begingroup$
What is the significance of "The red square is placed on top of a blue square", as the first sentence? Or is this a red herring?
$endgroup$
– Stilez
Jun 17 at 13:39
$begingroup$
What is the significance of "The red square is placed on top of a blue square", as the first sentence? Or is this a red herring?
$endgroup$
– Stilez
Jun 17 at 13:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think you can dissect the red square (on the left here) as follows
And then rearrange to form the larger square
This is always possible when the red square is larger than the blue.
Construction
If the length of the side of the larger square is $A$ and the smaller square $B$ then we find the point along each side of the large square which is a distance $fracA+B2$ from each vertex going in a clockwise direction. Then join opposing points in this construction.
The lines traversing the square intersect at right angles due to symmetry. Also, the length of each line $d$ can be determined with Pythagoras theorem by constructing a line from one of the vertices to the opposite side parallel to the side of the square. That is, $$ d^2 = A^2 + left(fracA+B2 - fracA-B2right)^2 = A^2 + B^2$$ It follows that the five pieces can be rearranged to form the combined square as shown.
answered Jun 17 at 9:12
hexominohexomino
54k5 gold badges158 silver badges250 bronze badges
$endgroup$
$begingroup$
This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares.
$endgroup$
– Moti
Jun 17 at 19:10
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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votes
$begingroup$
I think you can dissect the red square (on the left here) as follows
And then rearrange to form the larger square
This is always possible when the red square is larger than the blue.
Construction
If the length of the side of the larger square is $A$ and the smaller square $B$ then we find the point along each side of the large square which is a distance $fracA+B2$ from each vertex going in a clockwise direction. Then join opposing points in this construction.
The lines traversing the square intersect at right angles due to symmetry. Also, the length of each line $d$ can be determined with Pythagoras theorem by constructing a line from one of the vertices to the opposite side parallel to the side of the square. That is, $$ d^2 = A^2 + left(fracA+B2 - fracA-B2right)^2 = A^2 + B^2$$ It follows that the five pieces can be rearranged to form the combined square as shown.
answered Jun 17 at 9:12
hexominohexomino
54k5 gold badges158 silver badges250 bronze badges
$endgroup$
$begingroup$
This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares.
$endgroup$
– Moti
Jun 17 at 19:10
add a comment |
$begingroup$
I think you can dissect the red square (on the left here) as follows
And then rearrange to form the larger square
This is always possible when the red square is larger than the blue.
Construction
If the length of the side of the larger square is $A$ and the smaller square $B$ then we find the point along each side of the large square which is a distance $fracA+B2$ from each vertex going in a clockwise direction. Then join opposing points in this construction.
The lines traversing the square intersect at right angles due to symmetry. Also, the length of each line $d$ can be determined with Pythagoras theorem by constructing a line from one of the vertices to the opposite side parallel to the side of the square. That is, $$ d^2 = A^2 + left(fracA+B2 - fracA-B2right)^2 = A^2 + B^2$$ It follows that the five pieces can be rearranged to form the combined square as shown.
answered Jun 17 at 9:12
hexominohexomino
54k5 gold badges158 silver badges250 bronze badges
$endgroup$
$begingroup$
This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares.
$endgroup$
– Moti
Jun 17 at 19:10
add a comment |
$begingroup$
I think you can dissect the red square (on the left here) as follows
And then rearrange to form the larger square
This is always possible when the red square is larger than the blue.
Construction
If the length of the side of the larger square is $A$ and the smaller square $B$ then we find the point along each side of the large square which is a distance $fracA+B2$ from each vertex going in a clockwise direction. Then join opposing points in this construction.
The lines traversing the square intersect at right angles due to symmetry. Also, the length of each line $d$ can be determined with Pythagoras theorem by constructing a line from one of the vertices to the opposite side parallel to the side of the square. That is, $$ d^2 = A^2 + left(fracA+B2 - fracA-B2right)^2 = A^2 + B^2$$ It follows that the five pieces can be rearranged to form the combined square as shown.
answered Jun 17 at 9:12
hexominohexomino
54k5 gold badges158 silver badges250 bronze badges
$endgroup$
I think you can dissect the red square (on the left here) as follows
And then rearrange to form the larger square
This is always possible when the red square is larger than the blue.
Construction
If the length of the side of the larger square is $A$ and the smaller square $B$ then we find the point along each side of the large square which is a distance $fracA+B2$ from each vertex going in a clockwise direction. Then join opposing points in this construction.
The lines traversing the square intersect at right angles due to symmetry. Also, the length of each line $d$ can be determined with Pythagoras theorem by constructing a line from one of the vertices to the opposite side parallel to the side of the square. That is, $$ d^2 = A^2 + left(fracA+B2 - fracA-B2right)^2 = A^2 + B^2$$ It follows that the five pieces can be rearranged to form the combined square as shown.
answered Jun 17 at 9:12
hexominohexomino
54k5 gold badges158 silver badges250 bronze badges
edited Jun 17 at 9:24
answered Jun 17 at 9:12
hexominohexomino
54k5 gold badges158 silver badges250 bronze badges
answered Jun 17 at 9:12
hexominohexomino
54k5 gold badges158 silver badges250 bronze badges
answered Jun 17 at 9:12
hexominohexomino
54k5 gold badges158 silver badges250 bronze badges
54k5 gold badges158 silver badges250 bronze badges
$begingroup$
This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares.
$endgroup$
– Moti
Jun 17 at 19:10
add a comment |
$begingroup$
This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares.
$endgroup$
– Moti
Jun 17 at 19:10
$begingroup$
This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares.
$endgroup$
– Moti
Jun 17 at 19:10
$begingroup$
This is a great solution! Much better than what I had in mind. Though it seems like a "single" solution it holds for an infinite "blue" squares.
$endgroup$
– Moti
Jun 17 at 19:10
add a comment |
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$begingroup$
What is the significance of "The red square is placed on top of a blue square", as the first sentence? Or is this a red herring?
$endgroup$
– Stilez
Jun 17 at 13:39