Prove that locus of vertex is $(a+b)(x^2+y^2)+2h(xbeta + alpha y) + (a-b)(xalpha - ybeta)=0$
$begingroup$
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
asked Nov 27 '18 at 13:59
Awe Kumar JhaAwe Kumar Jha
41313
$endgroup$
add a comment |
$begingroup$
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
asked Nov 27 '18 at 13:59
Awe Kumar JhaAwe Kumar Jha
41313
$endgroup$
$begingroup$
You seem to assume circumradius $=1$, is that so?
$endgroup$
– Aretino
Nov 27 '18 at 15:32
$begingroup$
@Aretino, yes , just for the sake of simplicity.
$endgroup$
– Awe Kumar Jha
Nov 27 '18 at 15:33
add a comment |
$begingroup$
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
asked Nov 27 '18 at 13:59
Awe Kumar JhaAwe Kumar Jha
41313
$endgroup$
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
analytic-geometry polar-coordinates
asked Nov 27 '18 at 13:59
Awe Kumar JhaAwe Kumar Jha
41313
asked Nov 27 '18 at 13:59
Awe Kumar JhaAwe Kumar Jha
41313
edited Nov 27 '18 at 15:13
Awe Kumar Jha
asked Nov 27 '18 at 13:59
Awe Kumar JhaAwe Kumar Jha
41313
asked Nov 27 '18 at 13:59
Awe Kumar JhaAwe Kumar Jha
41313
asked Nov 27 '18 at 13:59
Awe Kumar JhaAwe Kumar Jha
41313
41313
$begingroup$
You seem to assume circumradius $=1$, is that so?
$endgroup$
– Aretino
Nov 27 '18 at 15:32
$begingroup$
@Aretino, yes , just for the sake of simplicity.
$endgroup$
– Awe Kumar Jha
Nov 27 '18 at 15:33
add a comment |
$begingroup$
You seem to assume circumradius $=1$, is that so?
$endgroup$
– Aretino
Nov 27 '18 at 15:32
$begingroup$
@Aretino, yes , just for the sake of simplicity.
$endgroup$
– Awe Kumar Jha
Nov 27 '18 at 15:33
$begingroup$
You seem to assume circumradius $=1$, is that so?
$endgroup$
– Aretino
Nov 27 '18 at 15:32
$begingroup$
You seem to assume circumradius $=1$, is that so?
$endgroup$
– Aretino
Nov 27 '18 at 15:32
$begingroup$
@Aretino, yes , just for the sake of simplicity.
$endgroup$
– Awe Kumar Jha
Nov 27 '18 at 15:33
$begingroup$
@Aretino, yes , just for the sake of simplicity.
$endgroup$
– Awe Kumar Jha
Nov 27 '18 at 15:33
add a comment |
1 Answer
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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).
answered Nov 28 '18 at 17:01
AretinoAretino
23.1k21443
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add a comment |
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1 Answer
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$begingroup$
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).
answered Nov 28 '18 at 17:01
AretinoAretino
23.1k21443
$endgroup$
add a comment |
$begingroup$
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).
answered Nov 28 '18 at 17:01
AretinoAretino
23.1k21443
$endgroup$
add a comment |
$begingroup$
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).
answered Nov 28 '18 at 17:01
AretinoAretino
23.1k21443
$endgroup$
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).
answered Nov 28 '18 at 17:01
AretinoAretino
23.1k21443
edited Nov 29 '18 at 14:24
answered Nov 28 '18 at 17:01
AretinoAretino
23.1k21443
answered Nov 28 '18 at 17:01
AretinoAretino
23.1k21443
answered Nov 28 '18 at 17:01
AretinoAretino
23.1k21443
23.1k21443
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$begingroup$
You seem to assume circumradius $=1$, is that so?
$endgroup$
– Aretino
Nov 27 '18 at 15:32
$begingroup$
@Aretino, yes , just for the sake of simplicity.
$endgroup$
– Awe Kumar Jha
Nov 27 '18 at 15:33