The size of the biggest square that can be inscribed within a circle
$begingroup$
(a) Is the size the biggest square that can be fit inside a circle always $2r^2$?
(b) How do you do when you show that is indeed the case?
I want to compare my boyfriend's proofs with that of other people, since he thinks his proof is unique.
euclidean-geometry
edited Jan 7 at 19:46
Eevee Trainer
5,3981936
asked Jan 7 at 19:13
simulacrasimulacra
1
New contributor
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
(a) Is the size the biggest square that can be fit inside a circle always $2r^2$?
(b) How do you do when you show that is indeed the case?
I want to compare my boyfriend's proofs with that of other people, since he thinks his proof is unique.
euclidean-geometry
edited Jan 7 at 19:46
Eevee Trainer
5,3981936
asked Jan 7 at 19:13
simulacrasimulacra
1
New contributor
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
There's a subtle difference between "fits inside" and "inscribes." It can indeed be shown that a square which fits inside a circle of radius $r$ has an area $A leq 2r^2$, with $A = 2r^2$ occurring when the square is inscribed. That said, there is rarely ever a "unique" proof of some fact; take the Pythagorean theorem as an example: books full just of hundreds of proofs of it have been published.
$endgroup$
– Eevee Trainer
Jan 7 at 19:41
1
$begingroup$
As for one proof I'd make. I'd make the assumption the inscribed square has diagonal $2r$, show said square is indeed inscribed, then show a square can have no other diagonal (w.r.t. to its length), and then calculate the area of said square. (This is the barest of bones of the proof, going into depth on said proof probably isn't going to be very useful.)
$endgroup$
– Eevee Trainer
Jan 7 at 19:44
$begingroup$
As Eevee Trainer stated, there are many proofs of what you are asking. I have just given a different one here, plus I can also think of $1$ or $2$ more with some effort. However, like Eevee also stated, I'm not sure how useful that will be for you.
$endgroup$
– John Omielan
Jan 7 at 19:49
add a comment |
$begingroup$
(a) Is the size the biggest square that can be fit inside a circle always $2r^2$?
(b) How do you do when you show that is indeed the case?
I want to compare my boyfriend's proofs with that of other people, since he thinks his proof is unique.
euclidean-geometry
edited Jan 7 at 19:46
Eevee Trainer
5,3981936
asked Jan 7 at 19:13
simulacrasimulacra
1
New contributor
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
(a) Is the size the biggest square that can be fit inside a circle always $2r^2$?
(b) How do you do when you show that is indeed the case?
I want to compare my boyfriend's proofs with that of other people, since he thinks his proof is unique.
euclidean-geometry
euclidean-geometry
edited Jan 7 at 19:46
Eevee Trainer
5,3981936
asked Jan 7 at 19:13
simulacrasimulacra
1
New contributor
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 7 at 19:46
Eevee Trainer
5,3981936
asked Jan 7 at 19:13
simulacrasimulacra
1
New contributor
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 7 at 19:46
Eevee Trainer
5,3981936
edited Jan 7 at 19:46
Eevee Trainer
5,3981936
edited Jan 7 at 19:46
Eevee Trainer
5,3981936
5,3981936
asked Jan 7 at 19:13
simulacrasimulacra
1
New contributor
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 19:13
simulacrasimulacra
1
asked Jan 7 at 19:13
simulacrasimulacra
1
1
New contributor
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
simulacra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
There's a subtle difference between "fits inside" and "inscribes." It can indeed be shown that a square which fits inside a circle of radius $r$ has an area $A leq 2r^2$, with $A = 2r^2$ occurring when the square is inscribed. That said, there is rarely ever a "unique" proof of some fact; take the Pythagorean theorem as an example: books full just of hundreds of proofs of it have been published.
$endgroup$
– Eevee Trainer
Jan 7 at 19:41
1
$begingroup$
As for one proof I'd make. I'd make the assumption the inscribed square has diagonal $2r$, show said square is indeed inscribed, then show a square can have no other diagonal (w.r.t. to its length), and then calculate the area of said square. (This is the barest of bones of the proof, going into depth on said proof probably isn't going to be very useful.)
$endgroup$
– Eevee Trainer
Jan 7 at 19:44
$begingroup$
As Eevee Trainer stated, there are many proofs of what you are asking. I have just given a different one here, plus I can also think of $1$ or $2$ more with some effort. However, like Eevee also stated, I'm not sure how useful that will be for you.
$endgroup$
– John Omielan
Jan 7 at 19:49
add a comment |
1
$begingroup$
There's a subtle difference between "fits inside" and "inscribes." It can indeed be shown that a square which fits inside a circle of radius $r$ has an area $A leq 2r^2$, with $A = 2r^2$ occurring when the square is inscribed. That said, there is rarely ever a "unique" proof of some fact; take the Pythagorean theorem as an example: books full just of hundreds of proofs of it have been published.
$endgroup$
– Eevee Trainer
Jan 7 at 19:41
1
$begingroup$
As for one proof I'd make. I'd make the assumption the inscribed square has diagonal $2r$, show said square is indeed inscribed, then show a square can have no other diagonal (w.r.t. to its length), and then calculate the area of said square. (This is the barest of bones of the proof, going into depth on said proof probably isn't going to be very useful.)
$endgroup$
– Eevee Trainer
Jan 7 at 19:44
$begingroup$
As Eevee Trainer stated, there are many proofs of what you are asking. I have just given a different one here, plus I can also think of $1$ or $2$ more with some effort. However, like Eevee also stated, I'm not sure how useful that will be for you.
$endgroup$
– John Omielan
Jan 7 at 19:49
1
1
$begingroup$
There's a subtle difference between "fits inside" and "inscribes." It can indeed be shown that a square which fits inside a circle of radius $r$ has an area $A leq 2r^2$, with $A = 2r^2$ occurring when the square is inscribed. That said, there is rarely ever a "unique" proof of some fact; take the Pythagorean theorem as an example: books full just of hundreds of proofs of it have been published.
$endgroup$
– Eevee Trainer
Jan 7 at 19:41
$begingroup$
There's a subtle difference between "fits inside" and "inscribes." It can indeed be shown that a square which fits inside a circle of radius $r$ has an area $A leq 2r^2$, with $A = 2r^2$ occurring when the square is inscribed. That said, there is rarely ever a "unique" proof of some fact; take the Pythagorean theorem as an example: books full just of hundreds of proofs of it have been published.
$endgroup$
– Eevee Trainer
Jan 7 at 19:41
1
1
$begingroup$
As for one proof I'd make. I'd make the assumption the inscribed square has diagonal $2r$, show said square is indeed inscribed, then show a square can have no other diagonal (w.r.t. to its length), and then calculate the area of said square. (This is the barest of bones of the proof, going into depth on said proof probably isn't going to be very useful.)
$endgroup$
– Eevee Trainer
Jan 7 at 19:44
$begingroup$
As for one proof I'd make. I'd make the assumption the inscribed square has diagonal $2r$, show said square is indeed inscribed, then show a square can have no other diagonal (w.r.t. to its length), and then calculate the area of said square. (This is the barest of bones of the proof, going into depth on said proof probably isn't going to be very useful.)
$endgroup$
– Eevee Trainer
Jan 7 at 19:44
$begingroup$
As Eevee Trainer stated, there are many proofs of what you are asking. I have just given a different one here, plus I can also think of $1$ or $2$ more with some effort. However, like Eevee also stated, I'm not sure how useful that will be for you.
$endgroup$
– John Omielan
Jan 7 at 19:49
$begingroup$
As Eevee Trainer stated, there are many proofs of what you are asking. I have just given a different one here, plus I can also think of $1$ or $2$ more with some effort. However, like Eevee also stated, I'm not sure how useful that will be for you.
$endgroup$
– John Omielan
Jan 7 at 19:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All squares inscribed in a given circle have equal size. There is no 'biggest' one. You can easily prove that their area is $2r^2$ by noting that the area of a rhombus is given by $frac12d_1d_2$, where $d_1,d_2$ are the lengths of its diagonals. In your case, $d_1=d_2=2r$, as the diagonals are diameters of the circle.
answered Jan 7 at 19:18
Shubham JohriShubham Johri
4,676717
$endgroup$
1
$begingroup$
Please ignore my previous comment as I just saw the title say "biggest square that can be inscribed within a circle". However, the text in the question differs slightly from this by saying "fit inside" which implied to me that it's the set of all squares contained in a circle.
$endgroup$
– John Omielan
Jan 7 at 19:29
$begingroup$
Yes, there is a dichotomy in the title and first line. Although it can be resolved by noting that the area $frac12d_1d_2$ is maximized when the vertices lie on the circle, for the largest line segment that fits in a circle is its diameter
$endgroup$
– Shubham Johri
Jan 7 at 19:40
add a comment |
$begingroup$
In the diagram above, $ABCD$ are the points of the inscribed square on the circle, $O$ is the center and let $r$ be the radius. Note that $AC$ is the diameter as $angle ABC$ is a right angle (since if the angle subtending a chord is a right angle, the chord must be the diameter), so $AC$ must be of length $2r$. Also, $OB$ is perpendicular to $AC$ (since the diagonals of a square are perpendicular to each other) and of length $r$. Thus, it's the height of $triangle ABC$ using $AC$ as the base, so the triangle area is $frac{2r times r}{2} = r^2$. As the area of $triangle ADC$ is the same, the total area of $ABCD$ is $r^2 + r^2 = 2r^2$.
answered Jan 7 at 19:41
John OmielanJohn Omielan
1,26529
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
All squares inscribed in a given circle have equal size. There is no 'biggest' one. You can easily prove that their area is $2r^2$ by noting that the area of a rhombus is given by $frac12d_1d_2$, where $d_1,d_2$ are the lengths of its diagonals. In your case, $d_1=d_2=2r$, as the diagonals are diameters of the circle.
answered Jan 7 at 19:18
Shubham JohriShubham Johri
4,676717
$endgroup$
1
$begingroup$
Please ignore my previous comment as I just saw the title say "biggest square that can be inscribed within a circle". However, the text in the question differs slightly from this by saying "fit inside" which implied to me that it's the set of all squares contained in a circle.
$endgroup$
– John Omielan
Jan 7 at 19:29
$begingroup$
Yes, there is a dichotomy in the title and first line. Although it can be resolved by noting that the area $frac12d_1d_2$ is maximized when the vertices lie on the circle, for the largest line segment that fits in a circle is its diameter
$endgroup$
– Shubham Johri
Jan 7 at 19:40
add a comment |
$begingroup$
All squares inscribed in a given circle have equal size. There is no 'biggest' one. You can easily prove that their area is $2r^2$ by noting that the area of a rhombus is given by $frac12d_1d_2$, where $d_1,d_2$ are the lengths of its diagonals. In your case, $d_1=d_2=2r$, as the diagonals are diameters of the circle.
answered Jan 7 at 19:18
Shubham JohriShubham Johri
4,676717
$endgroup$
1
$begingroup$
Please ignore my previous comment as I just saw the title say "biggest square that can be inscribed within a circle". However, the text in the question differs slightly from this by saying "fit inside" which implied to me that it's the set of all squares contained in a circle.
$endgroup$
– John Omielan
Jan 7 at 19:29
$begingroup$
Yes, there is a dichotomy in the title and first line. Although it can be resolved by noting that the area $frac12d_1d_2$ is maximized when the vertices lie on the circle, for the largest line segment that fits in a circle is its diameter
$endgroup$
– Shubham Johri
Jan 7 at 19:40
add a comment |
$begingroup$
All squares inscribed in a given circle have equal size. There is no 'biggest' one. You can easily prove that their area is $2r^2$ by noting that the area of a rhombus is given by $frac12d_1d_2$, where $d_1,d_2$ are the lengths of its diagonals. In your case, $d_1=d_2=2r$, as the diagonals are diameters of the circle.
answered Jan 7 at 19:18
Shubham JohriShubham Johri
4,676717
$endgroup$
All squares inscribed in a given circle have equal size. There is no 'biggest' one. You can easily prove that their area is $2r^2$ by noting that the area of a rhombus is given by $frac12d_1d_2$, where $d_1,d_2$ are the lengths of its diagonals. In your case, $d_1=d_2=2r$, as the diagonals are diameters of the circle.
answered Jan 7 at 19:18
Shubham JohriShubham Johri
4,676717
answered Jan 7 at 19:18
Shubham JohriShubham Johri
4,676717
answered Jan 7 at 19:18
Shubham JohriShubham Johri
4,676717
answered Jan 7 at 19:18
Shubham JohriShubham Johri
4,676717
4,676717
1
$begingroup$
Please ignore my previous comment as I just saw the title say "biggest square that can be inscribed within a circle". However, the text in the question differs slightly from this by saying "fit inside" which implied to me that it's the set of all squares contained in a circle.
$endgroup$
– John Omielan
Jan 7 at 19:29
$begingroup$
Yes, there is a dichotomy in the title and first line. Although it can be resolved by noting that the area $frac12d_1d_2$ is maximized when the vertices lie on the circle, for the largest line segment that fits in a circle is its diameter
$endgroup$
– Shubham Johri
Jan 7 at 19:40
add a comment |
1
$begingroup$
Please ignore my previous comment as I just saw the title say "biggest square that can be inscribed within a circle". However, the text in the question differs slightly from this by saying "fit inside" which implied to me that it's the set of all squares contained in a circle.
$endgroup$
– John Omielan
Jan 7 at 19:29
$begingroup$
Yes, there is a dichotomy in the title and first line. Although it can be resolved by noting that the area $frac12d_1d_2$ is maximized when the vertices lie on the circle, for the largest line segment that fits in a circle is its diameter
$endgroup$
– Shubham Johri
Jan 7 at 19:40
1
1
$begingroup$
Please ignore my previous comment as I just saw the title say "biggest square that can be inscribed within a circle". However, the text in the question differs slightly from this by saying "fit inside" which implied to me that it's the set of all squares contained in a circle.
$endgroup$
– John Omielan
Jan 7 at 19:29
$begingroup$
Please ignore my previous comment as I just saw the title say "biggest square that can be inscribed within a circle". However, the text in the question differs slightly from this by saying "fit inside" which implied to me that it's the set of all squares contained in a circle.
$endgroup$
– John Omielan
Jan 7 at 19:29
$begingroup$
Yes, there is a dichotomy in the title and first line. Although it can be resolved by noting that the area $frac12d_1d_2$ is maximized when the vertices lie on the circle, for the largest line segment that fits in a circle is its diameter
$endgroup$
– Shubham Johri
Jan 7 at 19:40
$begingroup$
Yes, there is a dichotomy in the title and first line. Although it can be resolved by noting that the area $frac12d_1d_2$ is maximized when the vertices lie on the circle, for the largest line segment that fits in a circle is its diameter
$endgroup$
– Shubham Johri
Jan 7 at 19:40
add a comment |
$begingroup$
In the diagram above, $ABCD$ are the points of the inscribed square on the circle, $O$ is the center and let $r$ be the radius. Note that $AC$ is the diameter as $angle ABC$ is a right angle (since if the angle subtending a chord is a right angle, the chord must be the diameter), so $AC$ must be of length $2r$. Also, $OB$ is perpendicular to $AC$ (since the diagonals of a square are perpendicular to each other) and of length $r$. Thus, it's the height of $triangle ABC$ using $AC$ as the base, so the triangle area is $frac{2r times r}{2} = r^2$. As the area of $triangle ADC$ is the same, the total area of $ABCD$ is $r^2 + r^2 = 2r^2$.
answered Jan 7 at 19:41
John OmielanJohn Omielan
1,26529
$endgroup$
add a comment |
$begingroup$
In the diagram above, $ABCD$ are the points of the inscribed square on the circle, $O$ is the center and let $r$ be the radius. Note that $AC$ is the diameter as $angle ABC$ is a right angle (since if the angle subtending a chord is a right angle, the chord must be the diameter), so $AC$ must be of length $2r$. Also, $OB$ is perpendicular to $AC$ (since the diagonals of a square are perpendicular to each other) and of length $r$. Thus, it's the height of $triangle ABC$ using $AC$ as the base, so the triangle area is $frac{2r times r}{2} = r^2$. As the area of $triangle ADC$ is the same, the total area of $ABCD$ is $r^2 + r^2 = 2r^2$.
answered Jan 7 at 19:41
John OmielanJohn Omielan
1,26529
$endgroup$
add a comment |
$begingroup$
In the diagram above, $ABCD$ are the points of the inscribed square on the circle, $O$ is the center and let $r$ be the radius. Note that $AC$ is the diameter as $angle ABC$ is a right angle (since if the angle subtending a chord is a right angle, the chord must be the diameter), so $AC$ must be of length $2r$. Also, $OB$ is perpendicular to $AC$ (since the diagonals of a square are perpendicular to each other) and of length $r$. Thus, it's the height of $triangle ABC$ using $AC$ as the base, so the triangle area is $frac{2r times r}{2} = r^2$. As the area of $triangle ADC$ is the same, the total area of $ABCD$ is $r^2 + r^2 = 2r^2$.
answered Jan 7 at 19:41
John OmielanJohn Omielan
1,26529
$endgroup$
In the diagram above, $ABCD$ are the points of the inscribed square on the circle, $O$ is the center and let $r$ be the radius. Note that $AC$ is the diameter as $angle ABC$ is a right angle (since if the angle subtending a chord is a right angle, the chord must be the diameter), so $AC$ must be of length $2r$. Also, $OB$ is perpendicular to $AC$ (since the diagonals of a square are perpendicular to each other) and of length $r$. Thus, it's the height of $triangle ABC$ using $AC$ as the base, so the triangle area is $frac{2r times r}{2} = r^2$. As the area of $triangle ADC$ is the same, the total area of $ABCD$ is $r^2 + r^2 = 2r^2$.
answered Jan 7 at 19:41
John OmielanJohn Omielan
1,26529
edited Jan 7 at 20:30
answered Jan 7 at 19:41
John OmielanJohn Omielan
1,26529
answered Jan 7 at 19:41
John OmielanJohn Omielan
1,26529
answered Jan 7 at 19:41
John OmielanJohn Omielan
1,26529
1,26529
add a comment |
add a comment |
simulacra is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
There's a subtle difference between "fits inside" and "inscribes." It can indeed be shown that a square which fits inside a circle of radius $r$ has an area $A leq 2r^2$, with $A = 2r^2$ occurring when the square is inscribed. That said, there is rarely ever a "unique" proof of some fact; take the Pythagorean theorem as an example: books full just of hundreds of proofs of it have been published.
$endgroup$
– Eevee Trainer
Jan 7 at 19:41
1
$begingroup$
As for one proof I'd make. I'd make the assumption the inscribed square has diagonal $2r$, show said square is indeed inscribed, then show a square can have no other diagonal (w.r.t. to its length), and then calculate the area of said square. (This is the barest of bones of the proof, going into depth on said proof probably isn't going to be very useful.)
$endgroup$
– Eevee Trainer
Jan 7 at 19:44
$begingroup$
As Eevee Trainer stated, there are many proofs of what you are asking. I have just given a different one here, plus I can also think of $1$ or $2$ more with some effort. However, like Eevee also stated, I'm not sure how useful that will be for you.
$endgroup$
– John Omielan
Jan 7 at 19:49