Is the Torricelli/Fermat point unique?












2












$begingroup$


In a triangle with all angles less than 120 degrees can there be two such points?



I am wondering because in the geometry game Euclidea it gives two "v-stars" for this problem (theta 8.6). This usually means that there are two solutions possible.



I found one solution, basically by constructing equilateral triangles on the sides and connecting their external vertices.



However I am confused what the second solution is supposed to be.






geometry







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    2












    $begingroup$


    In a triangle with all angles less than 120 degrees can there be two such points?



    I am wondering because in the geometry game Euclidea it gives two "v-stars" for this problem (theta 8.6). This usually means that there are two solutions possible.



    I found one solution, basically by constructing equilateral triangles on the sides and connecting their external vertices.



    However I am confused what the second solution is supposed to be.






    geometry







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      In a triangle with all angles less than 120 degrees can there be two such points?



      I am wondering because in the geometry game Euclidea it gives two "v-stars" for this problem (theta 8.6). This usually means that there are two solutions possible.



      I found one solution, basically by constructing equilateral triangles on the sides and connecting their external vertices.



      However I am confused what the second solution is supposed to be.






      geometry







      share|cite|improve this question









      $endgroup$




      In a triangle with all angles less than 120 degrees can there be two such points?



      I am wondering because in the geometry game Euclidea it gives two "v-stars" for this problem (theta 8.6). This usually means that there are two solutions possible.



      I found one solution, basically by constructing equilateral triangles on the sides and connecting their external vertices.



      However I am confused what the second solution is supposed to be.






      geometry




      geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked May 4 '18 at 22:11









      user1583209user1583209

      1133




      1133






















          1 Answer
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          $begingroup$

          There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible



          There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle



          When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle



          So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Indeed they wanted the second isogonic centre.
            $endgroup$
            – user1583209
            May 5 '18 at 9:55










          • $begingroup$
            There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
            $endgroup$
            – Henry
            May 5 '18 at 10:33











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible



          There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle



          When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle



          So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Indeed they wanted the second isogonic centre.
            $endgroup$
            – user1583209
            May 5 '18 at 9:55










          • $begingroup$
            There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
            $endgroup$
            – Henry
            May 5 '18 at 10:33
















          4












          $begingroup$

          There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible



          There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle



          When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle



          So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Indeed they wanted the second isogonic centre.
            $endgroup$
            – user1583209
            May 5 '18 at 9:55










          • $begingroup$
            There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
            $endgroup$
            – Henry
            May 5 '18 at 10:33














          4












          4








          4





          $begingroup$

          There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible



          There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle



          When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle



          So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points






          share|cite|improve this answer











          $endgroup$



          There is one Fermat–Torricelli point, i.e. the point such that the total distance from the three vertices of the triangle to the point is the minimum possible



          There are two isogonic centres of a triangle (except for special cases where one of the original angles is $120^circ$ or $60^circ$ such as an equilateral triangle) which form angles of $120^circ$ or $60^circ$ with each of the three pairs of vertices of the original triangle



          When all the original angles are less than $120^circ$ then the Fermat–Torricelli point is one of the isogonic centres (the one inside the original triangle). When one of original angles is greater $120^circ$ then that vertex is the Fermat–Torricelli point while both isogonic centres are outside the original triangle



          So perhaps if something suggests there are two solutions to this type of question, it is referring to isogonic centres rather than Fermat–Torricelli points







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited May 4 '18 at 22:31

























          answered May 4 '18 at 22:26









          HenryHenry

          100k481167




          100k481167












          • $begingroup$
            Indeed they wanted the second isogonic centre.
            $endgroup$
            – user1583209
            May 5 '18 at 9:55










          • $begingroup$
            There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
            $endgroup$
            – Henry
            May 5 '18 at 10:33


















          • $begingroup$
            Indeed they wanted the second isogonic centre.
            $endgroup$
            – user1583209
            May 5 '18 at 9:55










          • $begingroup$
            There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
            $endgroup$
            – Henry
            May 5 '18 at 10:33
















          $begingroup$
          Indeed they wanted the second isogonic centre.
          $endgroup$
          – user1583209
          May 5 '18 at 9:55




          $begingroup$
          Indeed they wanted the second isogonic centre.
          $endgroup$
          – user1583209
          May 5 '18 at 9:55












          $begingroup$
          There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
          $endgroup$
          – Henry
          May 5 '18 at 10:33




          $begingroup$
          There is a picture at an earlier question I asked. $A,B,C$ are the vertices, $F$ is the Fermat–Torricelli point and first isogonic centre, and $S$ is the second isogonic centre
          $endgroup$
          – Henry
          May 5 '18 at 10:33


















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