0
$\begingroup$

Let $ABC$ any triangle. Consider $C_1$ the incircle and $C_2$ one of the three excircles.

The question is, what is the radical axis? That is the only information we have.

Working a lot with this problem, if we consider $A', B', C'$ the middle points of $BC, AC, AB$, respectively, we have that the radical axis of $C_1$ and $C_2$ is one external bisector of triangle $A'B'C'$.

How to prove this? Remembering that any point $P$ in the radical axis of $C_1$ and $C_2$ verify that $PO_1^2-r_1^2=PO_2^2-r_2^2$, where $O_1, O_2$ and $r_1, r_2$ are centers and radios of $C_1, C_2$, respectively.

This is abstract geometry, so we are not able to use results of analytic geometry.

Help! :(

$\endgroup$
1
  • $\begingroup$ btw Gauss, how may the radical triangle be related to ABC? $\endgroup$
    – Narasimham
    Oct 4, 2020 at 10:36

1 Answer 1

Reset to default
2
$\begingroup$

The radical axis of two circles possesses the following properties.

(1) It is perpendicular to the line of centers; and (2) It bisects their direct common tangents.

Therefore, to find the radical axis of the in-circle and the corresponding ex-circle (wrt A), we do:-

(1) Locate P, the point of contact of the in-circle to the line BC.

(2) Locate Q, the point of contact of the ex-circle to the line BC.

enter image description here

(3) Locate M, the midpoint of PQ.

(4) Through M, drop…..

$\endgroup$
2
  • $\begingroup$ One can show that $\overline{BP}\cong\overline{CQ}$, making $M$ even easier to construct (and the target radical axis even easier to describe). $\endgroup$
    – Blue
    Oct 4, 2020 at 12:59
  • $\begingroup$ @Blue A good point. $\endgroup$
    – Mick
    Oct 4, 2020 at 13:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .