0
$\begingroup$

Info: A $P$-Appolonius circle of two points $A,B$ is the locus of point P such that $\frac{AP}{BP}$ is a constant $AP\neq PB$. This locus is a circle and is not difficult to prove.

Question: Given a scalene triangle $ABC$, show that the $A-$,$B-$,$C-$appolonius circles concur at a point inside $ABC$.

I was trying radical axis theorem using circumcircle and the appolonius circles, but cant proceed, please help. Thank You.

$\endgroup$
2
$\begingroup$

Let $X$ be the intersection point of the $B$- and $C$-Apollonius circles of $\triangle ABC$ that lies inside $\triangle ABC$. Then $$ \frac{AX}{CX} = \frac{AB}{CB} \quad \mbox{and} \quad \frac{AX}{BX} = \frac{AC}{BC}, $$ so ...

$\endgroup$
  • $\begingroup$ I have a small quibble - the 1st isodynamic point is not always inside the triangle. It is the isogonal conjugate of the first Fermat point, and therefore is inside the triangle iff the triangle has no angle greater than 120 degrees. With that disclaimer user133281's answer works. $\endgroup$ – JamesDixon Sep 1 '17 at 15:59

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.