Prove that locus of vertex is $(a+b)(x^2+y^2)+2h(xbeta + alpha y) + (a-b)(xalpha - ybeta)=0$











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The base of a triangle passes through a fixed point $(alpha ,beta )$. Let the perpendicular bisectors of the sides be the lines $ax^2+2hxy+by^2=0$. It is to prove that the locus of the vertex is :
$$(a+b)(x^2+y^2)+2h(xbeta + alpha y) + (a-b)(xalpha - ybeta)=0$$
Clearly, the origin is the circumcentre of the triangle. So , it is easy to take polar coordinates and define,
$$x:=cos theta , y:=sin theta$$
$$alpha :=cos phi , beta :=sin phi$$
$$tan psi = frac {a-b}{2h}$$
This greatly simplifies the desired expression to,
$$(a+b)+2hsec psi sin {(theta + phi + psi)}=0$$
Yet this simplification is useless, unless I have a relation between the point through the base and the vertex. Any suggestions are welcome.










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  • You seem to assume circumradius $=1$, is that so?
    – Aretino
    Nov 27 at 15:32










  • @Aretino, yes , just for the sake of simplicity.
    – Awe Kumar Jha
    Nov 27 at 15:33















up vote
0
down vote

favorite
1












The base of a triangle passes through a fixed point $(alpha ,beta )$. Let the perpendicular bisectors of the sides be the lines $ax^2+2hxy+by^2=0$. It is to prove that the locus of the vertex is :
$$(a+b)(x^2+y^2)+2h(xbeta + alpha y) + (a-b)(xalpha - ybeta)=0$$
Clearly, the origin is the circumcentre of the triangle. So , it is easy to take polar coordinates and define,
$$x:=cos theta , y:=sin theta$$
$$alpha :=cos phi , beta :=sin phi$$
$$tan psi = frac {a-b}{2h}$$
This greatly simplifies the desired expression to,
$$(a+b)+2hsec psi sin {(theta + phi + psi)}=0$$
Yet this simplification is useless, unless I have a relation between the point through the base and the vertex. Any suggestions are welcome.










share|cite|improve this question
























  • You seem to assume circumradius $=1$, is that so?
    – Aretino
    Nov 27 at 15:32










  • @Aretino, yes , just for the sake of simplicity.
    – Awe Kumar Jha
    Nov 27 at 15:33













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





The base of a triangle passes through a fixed point $(alpha ,beta )$. Let the perpendicular bisectors of the sides be the lines $ax^2+2hxy+by^2=0$. It is to prove that the locus of the vertex is :
$$(a+b)(x^2+y^2)+2h(xbeta + alpha y) + (a-b)(xalpha - ybeta)=0$$
Clearly, the origin is the circumcentre of the triangle. So , it is easy to take polar coordinates and define,
$$x:=cos theta , y:=sin theta$$
$$alpha :=cos phi , beta :=sin phi$$
$$tan psi = frac {a-b}{2h}$$
This greatly simplifies the desired expression to,
$$(a+b)+2hsec psi sin {(theta + phi + psi)}=0$$
Yet this simplification is useless, unless I have a relation between the point through the base and the vertex. Any suggestions are welcome.










share|cite|improve this question















The base of a triangle passes through a fixed point $(alpha ,beta )$. Let the perpendicular bisectors of the sides be the lines $ax^2+2hxy+by^2=0$. It is to prove that the locus of the vertex is :
$$(a+b)(x^2+y^2)+2h(xbeta + alpha y) + (a-b)(xalpha - ybeta)=0$$
Clearly, the origin is the circumcentre of the triangle. So , it is easy to take polar coordinates and define,
$$x:=cos theta , y:=sin theta$$
$$alpha :=cos phi , beta :=sin phi$$
$$tan psi = frac {a-b}{2h}$$
This greatly simplifies the desired expression to,
$$(a+b)+2hsec psi sin {(theta + phi + psi)}=0$$
Yet this simplification is useless, unless I have a relation between the point through the base and the vertex. Any suggestions are welcome.







analytic-geometry polar-coordinates






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edited Nov 27 at 15:13

























asked Nov 27 at 13:59









Awe Kumar Jha

30511




30511












  • You seem to assume circumradius $=1$, is that so?
    – Aretino
    Nov 27 at 15:32










  • @Aretino, yes , just for the sake of simplicity.
    – Awe Kumar Jha
    Nov 27 at 15:33


















  • You seem to assume circumradius $=1$, is that so?
    – Aretino
    Nov 27 at 15:32










  • @Aretino, yes , just for the sake of simplicity.
    – Awe Kumar Jha
    Nov 27 at 15:33
















You seem to assume circumradius $=1$, is that so?
– Aretino
Nov 27 at 15:32




You seem to assume circumradius $=1$, is that so?
– Aretino
Nov 27 at 15:32












@Aretino, yes , just for the sake of simplicity.
– Awe Kumar Jha
Nov 27 at 15:33




@Aretino, yes , just for the sake of simplicity.
– Awe Kumar Jha
Nov 27 at 15:33










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I'll describe here a construction of the triangle: you could possibly use it to obtain the equation of the locus.



Let $AB$ be the base of the triangle (containing point $P=(alpha,beta)$) and $C$ its third vertex. Notice that the angle $psi$ between the perpendicular bisectors of $AC$ and $BC$ (red and blue dashed lines in the diagram) is the same as $angle ACB$ and also the same as $angle AOQ$, where $OQ$ is the perpendicular bisector of base $AB$ and $O$ the circumcenter of triangle $ABC$.



To construct the triangle, choose then line $OQ$ at will and draw $OQ'$ such that $angle QOQ'=psi$. Drop from $P$ the perpendicular to line $OQ$, which will meet line $OQ'$ at $A$. Reflect then $A$ about $OQ$ to get $B$, and reflect either $A$ or $B$ about one of the perpendicular bisectors to get $C$.



As line $OQ$ varies, point $C$ varies too and its locus is indeed a circle (purple circle below).



enter image description here






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    I'll describe here a construction of the triangle: you could possibly use it to obtain the equation of the locus.



    Let $AB$ be the base of the triangle (containing point $P=(alpha,beta)$) and $C$ its third vertex. Notice that the angle $psi$ between the perpendicular bisectors of $AC$ and $BC$ (red and blue dashed lines in the diagram) is the same as $angle ACB$ and also the same as $angle AOQ$, where $OQ$ is the perpendicular bisector of base $AB$ and $O$ the circumcenter of triangle $ABC$.



    To construct the triangle, choose then line $OQ$ at will and draw $OQ'$ such that $angle QOQ'=psi$. Drop from $P$ the perpendicular to line $OQ$, which will meet line $OQ'$ at $A$. Reflect then $A$ about $OQ$ to get $B$, and reflect either $A$ or $B$ about one of the perpendicular bisectors to get $C$.



    As line $OQ$ varies, point $C$ varies too and its locus is indeed a circle (purple circle below).



    enter image description here






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      I'll describe here a construction of the triangle: you could possibly use it to obtain the equation of the locus.



      Let $AB$ be the base of the triangle (containing point $P=(alpha,beta)$) and $C$ its third vertex. Notice that the angle $psi$ between the perpendicular bisectors of $AC$ and $BC$ (red and blue dashed lines in the diagram) is the same as $angle ACB$ and also the same as $angle AOQ$, where $OQ$ is the perpendicular bisector of base $AB$ and $O$ the circumcenter of triangle $ABC$.



      To construct the triangle, choose then line $OQ$ at will and draw $OQ'$ such that $angle QOQ'=psi$. Drop from $P$ the perpendicular to line $OQ$, which will meet line $OQ'$ at $A$. Reflect then $A$ about $OQ$ to get $B$, and reflect either $A$ or $B$ about one of the perpendicular bisectors to get $C$.



      As line $OQ$ varies, point $C$ varies too and its locus is indeed a circle (purple circle below).



      enter image description here






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        I'll describe here a construction of the triangle: you could possibly use it to obtain the equation of the locus.



        Let $AB$ be the base of the triangle (containing point $P=(alpha,beta)$) and $C$ its third vertex. Notice that the angle $psi$ between the perpendicular bisectors of $AC$ and $BC$ (red and blue dashed lines in the diagram) is the same as $angle ACB$ and also the same as $angle AOQ$, where $OQ$ is the perpendicular bisector of base $AB$ and $O$ the circumcenter of triangle $ABC$.



        To construct the triangle, choose then line $OQ$ at will and draw $OQ'$ such that $angle QOQ'=psi$. Drop from $P$ the perpendicular to line $OQ$, which will meet line $OQ'$ at $A$. Reflect then $A$ about $OQ$ to get $B$, and reflect either $A$ or $B$ about one of the perpendicular bisectors to get $C$.



        As line $OQ$ varies, point $C$ varies too and its locus is indeed a circle (purple circle below).



        enter image description here






        share|cite|improve this answer














        I'll describe here a construction of the triangle: you could possibly use it to obtain the equation of the locus.



        Let $AB$ be the base of the triangle (containing point $P=(alpha,beta)$) and $C$ its third vertex. Notice that the angle $psi$ between the perpendicular bisectors of $AC$ and $BC$ (red and blue dashed lines in the diagram) is the same as $angle ACB$ and also the same as $angle AOQ$, where $OQ$ is the perpendicular bisector of base $AB$ and $O$ the circumcenter of triangle $ABC$.



        To construct the triangle, choose then line $OQ$ at will and draw $OQ'$ such that $angle QOQ'=psi$. Drop from $P$ the perpendicular to line $OQ$, which will meet line $OQ'$ at $A$. Reflect then $A$ about $OQ$ to get $B$, and reflect either $A$ or $B$ about one of the perpendicular bisectors to get $C$.



        As line $OQ$ varies, point $C$ varies too and its locus is indeed a circle (purple circle below).



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 at 14:24

























        answered Nov 28 at 17:01









        Aretino

        22.5k21442




        22.5k21442






























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