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 : $$(a+b)(x^2+y^2)+2h(x\beta + \alpha y) + (a-b)(x\alpha - y\beta)=0$$ 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, $$(a+b)+2h\sec \psi \sin {(\theta + \phi + \psi)}=0$$ Yet this simplification is useless, unless I have a relation between the point through the base and the vertex. Any suggestions are welcome.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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).

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.