Must inscribed rectangles share a center point with each other?












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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







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    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
















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







share|cite|improve this question











$endgroup$








  • 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.
    $endgroup$
    – 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














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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 22:40







Mark

















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














  • 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










1 Answer
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enter image description here



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.






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    $begingroup$

    enter image description here



    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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      enter image description here



      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        enter image description here



        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.






        share|cite|improve this answer









        $endgroup$



        enter image description here



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 3:08









        Quang HoangQuang Hoang

        13.1k1233




        13.1k1233






























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