A conjecture about the intersections of three hyperboles related to any triangle

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Given any triangle $triangle ABC$, we build the hyperbole with foci in $A$ and $B$ and passing through $C$.



enter image description here



Similarly, we can build other two hyperboles, one with foci in $A$ and $C$ and passing through $B$ (red), and one with foci in $B$ and $C$ and passing through $A$ (green).



enter image description here




The first part of my conjecture is that the three hyperboles always intersect in two points $D$ and $E$.




enter image description here




Moreover, the ellipse with foci in these two points $D$ and $E$, and passing through one of the three vertices of the triangle $triangle ABC$, pass also through the other two vertices.




enter image description here



These are probably obvious results. However, is there an elementary proof for these conjectures?



Thanks for your help! Sorry in case this is too trivial.










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    up vote
    8
    down vote

    favorite
    1












    Given any triangle $triangle ABC$, we build the hyperbole with foci in $A$ and $B$ and passing through $C$.



    enter image description here



    Similarly, we can build other two hyperboles, one with foci in $A$ and $C$ and passing through $B$ (red), and one with foci in $B$ and $C$ and passing through $A$ (green).



    enter image description here




    The first part of my conjecture is that the three hyperboles always intersect in two points $D$ and $E$.




    enter image description here




    Moreover, the ellipse with foci in these two points $D$ and $E$, and passing through one of the three vertices of the triangle $triangle ABC$, pass also through the other two vertices.




    enter image description here



    These are probably obvious results. However, is there an elementary proof for these conjectures?



    Thanks for your help! Sorry in case this is too trivial.










    share|cite|improve this question

























      up vote
      8
      down vote

      favorite
      1









      up vote
      8
      down vote

      favorite
      1






      1





      Given any triangle $triangle ABC$, we build the hyperbole with foci in $A$ and $B$ and passing through $C$.



      enter image description here



      Similarly, we can build other two hyperboles, one with foci in $A$ and $C$ and passing through $B$ (red), and one with foci in $B$ and $C$ and passing through $A$ (green).



      enter image description here




      The first part of my conjecture is that the three hyperboles always intersect in two points $D$ and $E$.




      enter image description here




      Moreover, the ellipse with foci in these two points $D$ and $E$, and passing through one of the three vertices of the triangle $triangle ABC$, pass also through the other two vertices.




      enter image description here



      These are probably obvious results. However, is there an elementary proof for these conjectures?



      Thanks for your help! Sorry in case this is too trivial.










      share|cite|improve this question















      Given any triangle $triangle ABC$, we build the hyperbole with foci in $A$ and $B$ and passing through $C$.



      enter image description here



      Similarly, we can build other two hyperboles, one with foci in $A$ and $C$ and passing through $B$ (red), and one with foci in $B$ and $C$ and passing through $A$ (green).



      enter image description here




      The first part of my conjecture is that the three hyperboles always intersect in two points $D$ and $E$.




      enter image description here




      Moreover, the ellipse with foci in these two points $D$ and $E$, and passing through one of the three vertices of the triangle $triangle ABC$, pass also through the other two vertices.




      enter image description here



      These are probably obvious results. However, is there an elementary proof for these conjectures?



      Thanks for your help! Sorry in case this is too trivial.







      geometry triangle conic-sections geometric-construction






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      edited 3 hours ago

























      asked 4 hours ago









      Andrea Prunotto

      1,459726




      1,459726




















          2 Answers
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          Try this:



          Let E be a point of intersection of black and red hyperbole. Then



          black hyperbole: $EA-EB=AC-CB$



          red hyperbole: $EA-EC=AB-BC$



          Subtracting, you get



          $$EC-EB=AC-AB$$ therefore green hyperbole also passes through E.



          Do the same for point D, saying that D is a point of intersection of red and green hyperbole, and you will get that black hyperbole also passes through D.






          share|cite|improve this answer





























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            Second part: Let $a=BC$, $b = CA$ and $c =AB$.



            We have $$EB-EC = DC-DB = c-b$$
            $$colorredEB-EA = DA-DB = a-b$$ $$DA-DC = FC-FA =c-a$$



            If $$AD+AE = 2d$$ then since $$BD+BE = (DA+b-a)+(EA+a-b)=2d$$



            so $B$ also lies on this ellipse. The same is true for $C$.






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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              2
              down vote













              Try this:



              Let E be a point of intersection of black and red hyperbole. Then



              black hyperbole: $EA-EB=AC-CB$



              red hyperbole: $EA-EC=AB-BC$



              Subtracting, you get



              $$EC-EB=AC-AB$$ therefore green hyperbole also passes through E.



              Do the same for point D, saying that D is a point of intersection of red and green hyperbole, and you will get that black hyperbole also passes through D.






              share|cite|improve this answer


























                up vote
                2
                down vote













                Try this:



                Let E be a point of intersection of black and red hyperbole. Then



                black hyperbole: $EA-EB=AC-CB$



                red hyperbole: $EA-EC=AB-BC$



                Subtracting, you get



                $$EC-EB=AC-AB$$ therefore green hyperbole also passes through E.



                Do the same for point D, saying that D is a point of intersection of red and green hyperbole, and you will get that black hyperbole also passes through D.






                share|cite|improve this answer
























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Try this:



                  Let E be a point of intersection of black and red hyperbole. Then



                  black hyperbole: $EA-EB=AC-CB$



                  red hyperbole: $EA-EC=AB-BC$



                  Subtracting, you get



                  $$EC-EB=AC-AB$$ therefore green hyperbole also passes through E.



                  Do the same for point D, saying that D is a point of intersection of red and green hyperbole, and you will get that black hyperbole also passes through D.






                  share|cite|improve this answer














                  Try this:



                  Let E be a point of intersection of black and red hyperbole. Then



                  black hyperbole: $EA-EB=AC-CB$



                  red hyperbole: $EA-EC=AB-BC$



                  Subtracting, you get



                  $$EC-EB=AC-AB$$ therefore green hyperbole also passes through E.



                  Do the same for point D, saying that D is a point of intersection of red and green hyperbole, and you will get that black hyperbole also passes through D.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 3 hours ago

























                  answered 3 hours ago









                  MrDudulex

                  42519




                  42519




















                      up vote
                      2
                      down vote













                      Second part: Let $a=BC$, $b = CA$ and $c =AB$.



                      We have $$EB-EC = DC-DB = c-b$$
                      $$colorredEB-EA = DA-DB = a-b$$ $$DA-DC = FC-FA =c-a$$



                      If $$AD+AE = 2d$$ then since $$BD+BE = (DA+b-a)+(EA+a-b)=2d$$



                      so $B$ also lies on this ellipse. The same is true for $C$.






                      share|cite|improve this answer
























                        up vote
                        2
                        down vote













                        Second part: Let $a=BC$, $b = CA$ and $c =AB$.



                        We have $$EB-EC = DC-DB = c-b$$
                        $$colorredEB-EA = DA-DB = a-b$$ $$DA-DC = FC-FA =c-a$$



                        If $$AD+AE = 2d$$ then since $$BD+BE = (DA+b-a)+(EA+a-b)=2d$$



                        so $B$ also lies on this ellipse. The same is true for $C$.






                        share|cite|improve this answer






















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          Second part: Let $a=BC$, $b = CA$ and $c =AB$.



                          We have $$EB-EC = DC-DB = c-b$$
                          $$colorredEB-EA = DA-DB = a-b$$ $$DA-DC = FC-FA =c-a$$



                          If $$AD+AE = 2d$$ then since $$BD+BE = (DA+b-a)+(EA+a-b)=2d$$



                          so $B$ also lies on this ellipse. The same is true for $C$.






                          share|cite|improve this answer












                          Second part: Let $a=BC$, $b = CA$ and $c =AB$.



                          We have $$EB-EC = DC-DB = c-b$$
                          $$colorredEB-EA = DA-DB = a-b$$ $$DA-DC = FC-FA =c-a$$



                          If $$AD+AE = 2d$$ then since $$BD+BE = (DA+b-a)+(EA+a-b)=2d$$



                          so $B$ also lies on this ellipse. The same is true for $C$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          greedoid

                          31.6k114287




                          31.6k114287



























                               

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