2
$\begingroup$

Assume we have a $\triangle ABC$ in which $b>a>c$. Then the Feuerbach point lies on arc $B'C'$ of incircle of $\triangle ABC$.
My question is why? Proofs that show the incircle and nine-point circle are tangent, do not specify the tangency point (Feuerbach point) location. enter image description here

  • 0
    What do you mean by "existence of Feuerbach point" ? A question should be self contained. – 2017-01-24
  • 0
    @JeanMarie I edited the question. – 2017-01-24
  • 0
    see (http://mathworld.wolfram.com/FeuerbachPoint.html) – 2017-01-24
  • 0
    @JeanMarie I have, but it didn't really help me with my question. – 2017-01-24
  • 0
    That just depends on the positions of the incenter $I$ and the centre $N$ of the nine-point circle, i.e. the midpoint of $OH$. $I$ has the same distance from the sides of $ABC$: what can be said about $N$? – 2017-01-24
  • 0
    @JackD'Aurizio I know where are you going with this. But $IN$ isn't really something you discuss it. Especially the intersection of $IN$ with incircle. – 2017-01-24
  • 0
    @SSepehr: who cares about the distance $IN$? The incenter splits $A'B'C'$ in three parts: we just need to understand which part $N$ lies in. Anyway, $IN$ is just $\frac{R}{2}-r$, so we may go that way, too. – 2017-01-24

1 Answers 1

2

enter image description here

Proving that $F$ lies on the wanted arc is the same as proving that $N$ (the midpoint of $OH$, center of the nine point circle) lies in the depicted grey region. The distance of $N$ from the $BC$-side is the average between the distance of $O$ from the $BC$-side and the distance of $H$ from the $BC$-side, hence $$ d(N,BC) = \frac{R}{2}\left(\cos A+2\cos B\cos C\right)=\frac{R}{2}\cos(B-C)$$ and the trilinear or barycentric coordinates of $N$ are easy to find.
Let $A'$ and $B'$ the vertices of the grey region on the $BC$ and $CA$ sides. We just need to check that the given constraints ensure $\det(A',N,I)>0$, $\det(I,N,B')>0$ and we are done.