Must inscribed rectangles share a center point with each other?
2
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If a rectangle is inscribed inside another rectangle, will they always share a center point? My intuition says yes, but I am not sure how to prove it to myself.
geometry euclidean-geometry
asked Dec 18 '18 at 22:32
MarkMark
132
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add a comment |
2
$begingroup$
If a rectangle is inscribed inside another rectangle, will they always share a center point? My intuition says yes, but I am not sure how to prove it to myself.
geometry euclidean-geometry
asked Dec 18 '18 at 22:32
MarkMark
132
$endgroup$
1
$begingroup$
Prove it by symmetry. Any rectangle rotated to any angular position about it's center point will have its diagonally opposite corners the same horizontal and vertical distances from the center.
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– Phil H
Dec 18 '18 at 22:54
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Have you checked that it is possible to inscribe a rectangle in a nonsquare rectangle? I'm not sure you can. If that's the case, then for a square the answer is clearly "yes". for a nonsquare it's "yes" but a vacuous assertion.
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– Ethan Bolker
Dec 18 '18 at 23:26
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@EthanBolker Why not? Imagine you already have the inner rectangle, through the opposite vertices, draw two pairs of parallel lines which are mutually perpendicular. I highly doubt you would get a square.
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– Quang Hoang
Dec 19 '18 at 3:01
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@QuangHoang Right you are.
$endgroup$
– Ethan Bolker
Dec 19 '18 at 14:33
add a comment |
2
2
2
0
$begingroup$
If a rectangle is inscribed inside another rectangle, will they always share a center point? My intuition says yes, but I am not sure how to prove it to myself.
geometry euclidean-geometry
asked Dec 18 '18 at 22:32
MarkMark
132
$endgroup$
If a rectangle is inscribed inside another rectangle, will they always share a center point? My intuition says yes, but I am not sure how to prove it to myself.
geometry euclidean-geometry
geometry euclidean-geometry
asked Dec 18 '18 at 22:32
MarkMark
132
asked Dec 18 '18 at 22:32
MarkMark
132
edited Dec 18 '18 at 22:40
Mark
asked Dec 18 '18 at 22:32
MarkMark
132
asked Dec 18 '18 at 22:32
MarkMark
132
asked Dec 18 '18 at 22:32
MarkMark
132
132
1
$begingroup$
Prove it by symmetry. Any rectangle rotated to any angular position about it's center point will have its diagonally opposite corners the same horizontal and vertical distances from the center.
$endgroup$
– Phil H
Dec 18 '18 at 22:54
$begingroup$
Have you checked that it is possible to inscribe a rectangle in a nonsquare rectangle? I'm not sure you can. If that's the case, then for a square the answer is clearly "yes". for a nonsquare it's "yes" but a vacuous assertion.
$endgroup$
– Ethan Bolker
Dec 18 '18 at 23:26
$begingroup$
@EthanBolker Why not? Imagine you already have the inner rectangle, through the opposite vertices, draw two pairs of parallel lines which are mutually perpendicular. I highly doubt you would get a square.
$endgroup$
– Quang Hoang
Dec 19 '18 at 3:01
$begingroup$
@QuangHoang Right you are.
$endgroup$
– Ethan Bolker
Dec 19 '18 at 14:33
add a comment |
1
$begingroup$
Prove it by symmetry. Any rectangle rotated to any angular position about it's center point will have its diagonally opposite corners the same horizontal and vertical distances from the center.
$endgroup$
– Phil H
Dec 18 '18 at 22:54
$begingroup$
Have you checked that it is possible to inscribe a rectangle in a nonsquare rectangle? I'm not sure you can. If that's the case, then for a square the answer is clearly "yes". for a nonsquare it's "yes" but a vacuous assertion.
$endgroup$
– Ethan Bolker
Dec 18 '18 at 23:26
$begingroup$
@EthanBolker Why not? Imagine you already have the inner rectangle, through the opposite vertices, draw two pairs of parallel lines which are mutually perpendicular. I highly doubt you would get a square.
$endgroup$
– Quang Hoang
Dec 19 '18 at 3:01
$begingroup$
@QuangHoang Right you are.
$endgroup$
– Ethan Bolker
Dec 19 '18 at 14:33
1
1
$begingroup$
Prove it by symmetry. Any rectangle rotated to any angular position about it's center point will have its diagonally opposite corners the same horizontal and vertical distances from the center.
$endgroup$
– Phil H
Dec 18 '18 at 22:54
$begingroup$
Prove it by symmetry. Any rectangle rotated to any angular position about it's center point will have its diagonally opposite corners the same horizontal and vertical distances from the center.
$endgroup$
– Phil H
Dec 18 '18 at 22:54
$begingroup$
Have you checked that it is possible to inscribe a rectangle in a nonsquare rectangle? I'm not sure you can. If that's the case, then for a square the answer is clearly "yes". for a nonsquare it's "yes" but a vacuous assertion.
$endgroup$
– Ethan Bolker
Dec 18 '18 at 23:26
$begingroup$
Have you checked that it is possible to inscribe a rectangle in a nonsquare rectangle? I'm not sure you can. If that's the case, then for a square the answer is clearly "yes". for a nonsquare it's "yes" but a vacuous assertion.
$endgroup$
– Ethan Bolker
Dec 18 '18 at 23:26
$begingroup$
@EthanBolker Why not? Imagine you already have the inner rectangle, through the opposite vertices, draw two pairs of parallel lines which are mutually perpendicular. I highly doubt you would get a square.
$endgroup$
– Quang Hoang
Dec 19 '18 at 3:01
$begingroup$
@EthanBolker Why not? Imagine you already have the inner rectangle, through the opposite vertices, draw two pairs of parallel lines which are mutually perpendicular. I highly doubt you would get a square.
$endgroup$
– Quang Hoang
Dec 19 '18 at 3:01
$begingroup$
@QuangHoang Right you are.
$endgroup$
– Ethan Bolker
Dec 19 '18 at 14:33
$begingroup$
@QuangHoang Right you are.
$endgroup$
– Ethan Bolker
Dec 19 '18 at 14:33
add a comment |
1 Answer
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Hint: $angle FAB = angle HCD$ as they have (pairwise) parallel sides. So $triangle FAB = triangle HCD$. And we conclude that $FH$ intersects $BD$ at their common midpoints.
answered Dec 19 '18 at 3:08
Quang HoangQuang Hoang
13.2k1233
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$begingroup$
Hint: $angle FAB = angle HCD$ as they have (pairwise) parallel sides. So $triangle FAB = triangle HCD$. And we conclude that $FH$ intersects $BD$ at their common midpoints.
answered Dec 19 '18 at 3:08
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
1
$begingroup$
Hint: $angle FAB = angle HCD$ as they have (pairwise) parallel sides. So $triangle FAB = triangle HCD$. And we conclude that $FH$ intersects $BD$ at their common midpoints.
answered Dec 19 '18 at 3:08
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
1
1
1
$begingroup$
Hint: $angle FAB = angle HCD$ as they have (pairwise) parallel sides. So $triangle FAB = triangle HCD$. And we conclude that $FH$ intersects $BD$ at their common midpoints.
answered Dec 19 '18 at 3:08
Quang HoangQuang Hoang
13.2k1233
$endgroup$
Hint: $angle FAB = angle HCD$ as they have (pairwise) parallel sides. So $triangle FAB = triangle HCD$. And we conclude that $FH$ intersects $BD$ at their common midpoints.
answered Dec 19 '18 at 3:08
Quang HoangQuang Hoang
13.2k1233
answered Dec 19 '18 at 3:08
Quang HoangQuang Hoang
13.2k1233
answered Dec 19 '18 at 3:08
Quang HoangQuang Hoang
13.2k1233
answered Dec 19 '18 at 3:08
Quang HoangQuang Hoang
13.2k1233
13.2k1233
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add a comment |
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1
$begingroup$
Prove it by symmetry. Any rectangle rotated to any angular position about it's center point will have its diagonally opposite corners the same horizontal and vertical distances from the center.
$endgroup$
– Phil H
Dec 18 '18 at 22:54
$begingroup$
Have you checked that it is possible to inscribe a rectangle in a nonsquare rectangle? I'm not sure you can. If that's the case, then for a square the answer is clearly "yes". for a nonsquare it's "yes" but a vacuous assertion.
$endgroup$
– Ethan Bolker
Dec 18 '18 at 23:26
$begingroup$
@EthanBolker Why not? Imagine you already have the inner rectangle, through the opposite vertices, draw two pairs of parallel lines which are mutually perpendicular. I highly doubt you would get a square.
$endgroup$
– Quang Hoang
Dec 19 '18 at 3:01
$begingroup$
@QuangHoang Right you are.
$endgroup$
– Ethan Bolker
Dec 19 '18 at 14:33