Product of diagonals of a parallelogram












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I came across a property which I am not sure if true in general. Suppose you have a parallelogram whose vertices are $v_1, v_2, v_3, v_4inmathbb R^2$. Let's say that the side $[v_1,v_4]$ is parallel to $[v_2,v_3]$. Is it true that



$||v_1-v_4||^2< ||v_1-v_3||cdot ||v_2-v_4||,$



that is,
the square of one side is less than the product of the diagonals?
I could also be just missing an obvious counterexample, but so far I can't prove it either.










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    1












    $begingroup$


    I came across a property which I am not sure if true in general. Suppose you have a parallelogram whose vertices are $v_1, v_2, v_3, v_4inmathbb R^2$. Let's say that the side $[v_1,v_4]$ is parallel to $[v_2,v_3]$. Is it true that



    $||v_1-v_4||^2< ||v_1-v_3||cdot ||v_2-v_4||,$



    that is,
    the square of one side is less than the product of the diagonals?
    I could also be just missing an obvious counterexample, but so far I can't prove it either.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I came across a property which I am not sure if true in general. Suppose you have a parallelogram whose vertices are $v_1, v_2, v_3, v_4inmathbb R^2$. Let's say that the side $[v_1,v_4]$ is parallel to $[v_2,v_3]$. Is it true that



      $||v_1-v_4||^2< ||v_1-v_3||cdot ||v_2-v_4||,$



      that is,
      the square of one side is less than the product of the diagonals?
      I could also be just missing an obvious counterexample, but so far I can't prove it either.










      share|cite|improve this question









      $endgroup$




      I came across a property which I am not sure if true in general. Suppose you have a parallelogram whose vertices are $v_1, v_2, v_3, v_4inmathbb R^2$. Let's say that the side $[v_1,v_4]$ is parallel to $[v_2,v_3]$. Is it true that



      $||v_1-v_4||^2< ||v_1-v_3||cdot ||v_2-v_4||,$



      that is,
      the square of one side is less than the product of the diagonals?
      I could also be just missing an obvious counterexample, but so far I can't prove it either.







      geometry






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      asked Dec 10 '18 at 19:56









      chhrochhro

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          $begingroup$

          Consider a rhombus with unit sides which is very long and thin. As it gets
          pointier, one diagonal tends to zero and the other to length $2$, so the product
          of the diagonals gets very small.






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            active

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            1












            $begingroup$

            Consider a rhombus with unit sides which is very long and thin. As it gets
            pointier, one diagonal tends to zero and the other to length $2$, so the product
            of the diagonals gets very small.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Consider a rhombus with unit sides which is very long and thin. As it gets
              pointier, one diagonal tends to zero and the other to length $2$, so the product
              of the diagonals gets very small.






              share|cite|improve this answer









              $endgroup$
















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                1





                $begingroup$

                Consider a rhombus with unit sides which is very long and thin. As it gets
                pointier, one diagonal tends to zero and the other to length $2$, so the product
                of the diagonals gets very small.






                share|cite|improve this answer









                $endgroup$



                Consider a rhombus with unit sides which is very long and thin. As it gets
                pointier, one diagonal tends to zero and the other to length $2$, so the product
                of the diagonals gets very small.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 '18 at 20:00









                Lord Shark the UnknownLord Shark the Unknown

                104k1160132




                104k1160132






























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