Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle…
Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle of dimensions $1times2$. If the largest square has area 64, and the other three squares have side lengths that are whole numbers no larger than 7, what are their areas?
Attempt
Possible areas are 1,4,9,16,25,36 and 49.
Found this 2,3,5,8 as the answer (@pic). But I want to know that if there exist any other solutions or not. If not, what's the reason?
geometry inequality self-learning rectangles
asked Nov 24 at 23:07
jayant98
472115
|
show 8 more comments
Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle of dimensions $1times2$. If the largest square has area 64, and the other three squares have side lengths that are whole numbers no larger than 7, what are their areas?
Attempt
Possible areas are 1,4,9,16,25,36 and 49.
Found this 2,3,5,8 as the answer (@pic). But I want to know that if there exist any other solutions or not. If not, what's the reason?
geometry inequality self-learning rectangles
asked Nov 24 at 23:07
jayant98
472115
1
Have you tried to draw a picture?
– saulspatz
Nov 24 at 23:11
1
Hint: Start with the 2x1 piece and keep adding squares to it. You wanr to keep it a rectangular shape as you do this, because with so few square pieces you cannot afford creating irregular shapes and hope to make that back into a nice rectangle
– Bram28
Nov 24 at 23:15
1
Another hint: Fibonacci.
– David K
Nov 24 at 23:21
1
You are correct in that this is a solution. Note that the sides of the squares you have used, $2,3,5,8$ are Fibonacci numbers as David K suggested. What is the problem now? Are you trying to prove that this solution is the only one, or just looking for confirmation?
– saulspatz
Nov 25 at 0:00
1
If you have the $1times2$ rectangle in the corners, as pictured, the larges square you can put on top of it is $3times3.$ Otherwise, you have a rectangular area of height $!$ that you couldn't cover with squares of different sizes. If you place the $3times3$ square, then you have to place a $1times1$ square to fill the hole. No matter how you place the $8times8$ square, there's no way to complete a rectangle with only one more square. Now what if the $1times2$ is placed along an edge, of if it's completely surrounded by squares?
– saulspatz
Nov 25 at 0:17
|
show 8 more comments
Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle of dimensions $1times2$. If the largest square has area 64, and the other three squares have side lengths that are whole numbers no larger than 7, what are their areas?
Attempt
Possible areas are 1,4,9,16,25,36 and 49.
Found this 2,3,5,8 as the answer (@pic). But I want to know that if there exist any other solutions or not. If not, what's the reason?
geometry inequality self-learning rectangles
asked Nov 24 at 23:07
jayant98
472115
Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle of dimensions $1times2$. If the largest square has area 64, and the other three squares have side lengths that are whole numbers no larger than 7, what are their areas?
Attempt
Possible areas are 1,4,9,16,25,36 and 49.
Found this 2,3,5,8 as the answer (@pic). But I want to know that if there exist any other solutions or not. If not, what's the reason?
geometry inequality self-learning rectangles
geometry inequality self-learning rectangles
asked Nov 24 at 23:07
jayant98
472115
asked Nov 24 at 23:07
jayant98
472115
edited Nov 25 at 0:43
asked Nov 24 at 23:07
jayant98
472115
asked Nov 24 at 23:07
jayant98
472115
asked Nov 24 at 23:07
jayant98
472115
472115
1
Have you tried to draw a picture?
– saulspatz
Nov 24 at 23:11
1
Hint: Start with the 2x1 piece and keep adding squares to it. You wanr to keep it a rectangular shape as you do this, because with so few square pieces you cannot afford creating irregular shapes and hope to make that back into a nice rectangle
– Bram28
Nov 24 at 23:15
1
Another hint: Fibonacci.
– David K
Nov 24 at 23:21
1
You are correct in that this is a solution. Note that the sides of the squares you have used, $2,3,5,8$ are Fibonacci numbers as David K suggested. What is the problem now? Are you trying to prove that this solution is the only one, or just looking for confirmation?
– saulspatz
Nov 25 at 0:00
1
If you have the $1times2$ rectangle in the corners, as pictured, the larges square you can put on top of it is $3times3.$ Otherwise, you have a rectangular area of height $!$ that you couldn't cover with squares of different sizes. If you place the $3times3$ square, then you have to place a $1times1$ square to fill the hole. No matter how you place the $8times8$ square, there's no way to complete a rectangle with only one more square. Now what if the $1times2$ is placed along an edge, of if it's completely surrounded by squares?
– saulspatz
Nov 25 at 0:17
|
show 8 more comments
1
Have you tried to draw a picture?
– saulspatz
Nov 24 at 23:11
1
Hint: Start with the 2x1 piece and keep adding squares to it. You wanr to keep it a rectangular shape as you do this, because with so few square pieces you cannot afford creating irregular shapes and hope to make that back into a nice rectangle
– Bram28
Nov 24 at 23:15
1
Another hint: Fibonacci.
– David K
Nov 24 at 23:21
1
You are correct in that this is a solution. Note that the sides of the squares you have used, $2,3,5,8$ are Fibonacci numbers as David K suggested. What is the problem now? Are you trying to prove that this solution is the only one, or just looking for confirmation?
– saulspatz
Nov 25 at 0:00
1
If you have the $1times2$ rectangle in the corners, as pictured, the larges square you can put on top of it is $3times3.$ Otherwise, you have a rectangular area of height $!$ that you couldn't cover with squares of different sizes. If you place the $3times3$ square, then you have to place a $1times1$ square to fill the hole. No matter how you place the $8times8$ square, there's no way to complete a rectangle with only one more square. Now what if the $1times2$ is placed along an edge, of if it's completely surrounded by squares?
– saulspatz
Nov 25 at 0:17
1
1
Have you tried to draw a picture?
– saulspatz
Nov 24 at 23:11
Have you tried to draw a picture?
– saulspatz
Nov 24 at 23:11
1
1
Hint: Start with the 2x1 piece and keep adding squares to it. You wanr to keep it a rectangular shape as you do this, because with so few square pieces you cannot afford creating irregular shapes and hope to make that back into a nice rectangle
– Bram28
Nov 24 at 23:15
Hint: Start with the 2x1 piece and keep adding squares to it. You wanr to keep it a rectangular shape as you do this, because with so few square pieces you cannot afford creating irregular shapes and hope to make that back into a nice rectangle
– Bram28
Nov 24 at 23:15
1
1
Another hint: Fibonacci.
– David K
Nov 24 at 23:21
Another hint: Fibonacci.
– David K
Nov 24 at 23:21
1
1
You are correct in that this is a solution. Note that the sides of the squares you have used, $2,3,5,8$ are Fibonacci numbers as David K suggested. What is the problem now? Are you trying to prove that this solution is the only one, or just looking for confirmation?
– saulspatz
Nov 25 at 0:00
You are correct in that this is a solution. Note that the sides of the squares you have used, $2,3,5,8$ are Fibonacci numbers as David K suggested. What is the problem now? Are you trying to prove that this solution is the only one, or just looking for confirmation?
– saulspatz
Nov 25 at 0:00
1
1
If you have the $1times2$ rectangle in the corners, as pictured, the larges square you can put on top of it is $3times3.$ Otherwise, you have a rectangular area of height $!$ that you couldn't cover with squares of different sizes. If you place the $3times3$ square, then you have to place a $1times1$ square to fill the hole. No matter how you place the $8times8$ square, there's no way to complete a rectangle with only one more square. Now what if the $1times2$ is placed along an edge, of if it's completely surrounded by squares?
– saulspatz
Nov 25 at 0:17
If you have the $1times2$ rectangle in the corners, as pictured, the larges square you can put on top of it is $3times3.$ Otherwise, you have a rectangular area of height $!$ that you couldn't cover with squares of different sizes. If you place the $3times3$ square, then you have to place a $1times1$ square to fill the hole. No matter how you place the $8times8$ square, there's no way to complete a rectangle with only one more square. Now what if the $1times2$ is placed along an edge, of if it's completely surrounded by squares?
– saulspatz
Nov 25 at 0:17
|
show 8 more comments
1 Answer
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Unless one side of the big rectangle is $8$, the $8times 8$ must touch smaller parts on two edges, which already accounts for all parts. So one of the edges must touch two smaller squares - which leaves a gap at the smaller square that cannot be filled. We conclude that one side of the rectangle is $8$.
After removing the $8times 8$, we are left with three smaller squares of side-lengths $a<b<c<8$ and the $2times 1$, forming a rectangle. Again, if the large $ctimes c$ square has neighbours on two edges, we run into problems. We conclude that one edge of the rectangle is $c$. Hence we have
$$2+a^2+b^2+c^2=8c. $$
Numerically, we could have $c=7$, then $a^2+b^2=5$, i.e., $a=1$, $b=2$. Or $c=6$, then $a^2+b^2=10$, so $a=1$, $b=3$. Or $c=5$, then $a^2+b^2=13$, so $a=2$, $b=3$. $c=4$ is not possible, nor is $cle 3$.
One readily sees that it is impossible to fill a $1times 7$ when one part os $2times 2$, or fill a $2times 6$, when one part is $3times 3$. The remaining case $c=5$ leads to the well-known solution.
answered Nov 25 at 1:39
Hagen von Eitzen
276k21269496
add a comment |
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Unless one side of the big rectangle is $8$, the $8times 8$ must touch smaller parts on two edges, which already accounts for all parts. So one of the edges must touch two smaller squares - which leaves a gap at the smaller square that cannot be filled. We conclude that one side of the rectangle is $8$.
After removing the $8times 8$, we are left with three smaller squares of side-lengths $a<b<c<8$ and the $2times 1$, forming a rectangle. Again, if the large $ctimes c$ square has neighbours on two edges, we run into problems. We conclude that one edge of the rectangle is $c$. Hence we have
$$2+a^2+b^2+c^2=8c. $$
Numerically, we could have $c=7$, then $a^2+b^2=5$, i.e., $a=1$, $b=2$. Or $c=6$, then $a^2+b^2=10$, so $a=1$, $b=3$. Or $c=5$, then $a^2+b^2=13$, so $a=2$, $b=3$. $c=4$ is not possible, nor is $cle 3$.
One readily sees that it is impossible to fill a $1times 7$ when one part os $2times 2$, or fill a $2times 6$, when one part is $3times 3$. The remaining case $c=5$ leads to the well-known solution.
answered Nov 25 at 1:39
Hagen von Eitzen
276k21269496
add a comment |
Unless one side of the big rectangle is $8$, the $8times 8$ must touch smaller parts on two edges, which already accounts for all parts. So one of the edges must touch two smaller squares - which leaves a gap at the smaller square that cannot be filled. We conclude that one side of the rectangle is $8$.
After removing the $8times 8$, we are left with three smaller squares of side-lengths $a<b<c<8$ and the $2times 1$, forming a rectangle. Again, if the large $ctimes c$ square has neighbours on two edges, we run into problems. We conclude that one edge of the rectangle is $c$. Hence we have
$$2+a^2+b^2+c^2=8c. $$
Numerically, we could have $c=7$, then $a^2+b^2=5$, i.e., $a=1$, $b=2$. Or $c=6$, then $a^2+b^2=10$, so $a=1$, $b=3$. Or $c=5$, then $a^2+b^2=13$, so $a=2$, $b=3$. $c=4$ is not possible, nor is $cle 3$.
One readily sees that it is impossible to fill a $1times 7$ when one part os $2times 2$, or fill a $2times 6$, when one part is $3times 3$. The remaining case $c=5$ leads to the well-known solution.
answered Nov 25 at 1:39
Hagen von Eitzen
276k21269496
add a comment |
Unless one side of the big rectangle is $8$, the $8times 8$ must touch smaller parts on two edges, which already accounts for all parts. So one of the edges must touch two smaller squares - which leaves a gap at the smaller square that cannot be filled. We conclude that one side of the rectangle is $8$.
After removing the $8times 8$, we are left with three smaller squares of side-lengths $a<b<c<8$ and the $2times 1$, forming a rectangle. Again, if the large $ctimes c$ square has neighbours on two edges, we run into problems. We conclude that one edge of the rectangle is $c$. Hence we have
$$2+a^2+b^2+c^2=8c. $$
Numerically, we could have $c=7$, then $a^2+b^2=5$, i.e., $a=1$, $b=2$. Or $c=6$, then $a^2+b^2=10$, so $a=1$, $b=3$. Or $c=5$, then $a^2+b^2=13$, so $a=2$, $b=3$. $c=4$ is not possible, nor is $cle 3$.
One readily sees that it is impossible to fill a $1times 7$ when one part os $2times 2$, or fill a $2times 6$, when one part is $3times 3$. The remaining case $c=5$ leads to the well-known solution.
answered Nov 25 at 1:39
Hagen von Eitzen
276k21269496
Unless one side of the big rectangle is $8$, the $8times 8$ must touch smaller parts on two edges, which already accounts for all parts. So one of the edges must touch two smaller squares - which leaves a gap at the smaller square that cannot be filled. We conclude that one side of the rectangle is $8$.
After removing the $8times 8$, we are left with three smaller squares of side-lengths $a<b<c<8$ and the $2times 1$, forming a rectangle. Again, if the large $ctimes c$ square has neighbours on two edges, we run into problems. We conclude that one edge of the rectangle is $c$. Hence we have
$$2+a^2+b^2+c^2=8c. $$
Numerically, we could have $c=7$, then $a^2+b^2=5$, i.e., $a=1$, $b=2$. Or $c=6$, then $a^2+b^2=10$, so $a=1$, $b=3$. Or $c=5$, then $a^2+b^2=13$, so $a=2$, $b=3$. $c=4$ is not possible, nor is $cle 3$.
One readily sees that it is impossible to fill a $1times 7$ when one part os $2times 2$, or fill a $2times 6$, when one part is $3times 3$. The remaining case $c=5$ leads to the well-known solution.
answered Nov 25 at 1:39
Hagen von Eitzen
276k21269496
answered Nov 25 at 1:39
Hagen von Eitzen
276k21269496
answered Nov 25 at 1:39
Hagen von Eitzen
276k21269496
answered Nov 25 at 1:39
Hagen von Eitzen
276k21269496
276k21269496
add a comment |
add a comment |
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1
Have you tried to draw a picture?
– saulspatz
Nov 24 at 23:11
1
Hint: Start with the 2x1 piece and keep adding squares to it. You wanr to keep it a rectangular shape as you do this, because with so few square pieces you cannot afford creating irregular shapes and hope to make that back into a nice rectangle
– Bram28
Nov 24 at 23:15
1
Another hint: Fibonacci.
– David K
Nov 24 at 23:21
1
You are correct in that this is a solution. Note that the sides of the squares you have used, $2,3,5,8$ are Fibonacci numbers as David K suggested. What is the problem now? Are you trying to prove that this solution is the only one, or just looking for confirmation?
– saulspatz
Nov 25 at 0:00
1
If you have the $1times2$ rectangle in the corners, as pictured, the larges square you can put on top of it is $3times3.$ Otherwise, you have a rectangular area of height $!$ that you couldn't cover with squares of different sizes. If you place the $3times3$ square, then you have to place a $1times1$ square to fill the hole. No matter how you place the $8times8$ square, there's no way to complete a rectangle with only one more square. Now what if the $1times2$ is placed along an edge, of if it's completely surrounded by squares?
– saulspatz
Nov 25 at 0:17