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Feuerbach point

Revision as of 13:36, 12 December 2024 by Vvsss (talk | contribs) (→‎Scalene triangle with angle 60^\circ)

The incircle and nine-point circle of a triangle are tangent to each other at the Feuerbach point of the triangle. The Feuerbach point is listed as X(11) in Clark Kimberling's Encyclopedia of Triangle Centers and is named after Karl Wilhelm Feuerbach.

Sharygin’s proof

$1998, 24^{th}$ Russian math olympiad

Feuerbach 1.png

Claim 1

Let $D$ be the base of the bisector of angle A of scalene triangle $\triangle ABC.$

Let $DE$ be a tangent different from side $BC$ to the incircle of $\triangle ABC (E$ is the point of tangency). Similarly, we denote $D', E', D'',$ and $E''.$

Prove that $EE'||AB, \triangle ABC \sim \triangle EE'E'', AE, BE', CE''$ are concurrent.

Proof

Let $T, T',$ and $T''$ be the point of tangency of the incircle $\omega$ and $BC, AC,$ and $AB.$

Let $\angle A = 2 \alpha, \angle B = 2 \beta, \angle C = 2 \gamma, \alpha + \beta + \gamma = 90^\circ.$ WLOG, $\beta > \gamma.$ \[\angle TIT'' = 180^\circ - 2 \beta, \angle ADB = 180^\circ - \alpha - 2 \beta,\] \[\angle DIT = 90^\circ - \angle ADB = \alpha + 2 \beta - 90^\circ = \beta -\gamma, \angle EID = \angle TID \implies\] \[\angle T''IE = \angle T''IT + 2 \angle TID = 180^\circ - 2 \beta + 2(\beta - \gamma) = 180^\circ - 2 \gamma.\] Similarly, $\angle T''IE' = 180^\circ – 2 \gamma \implies$ points $E$ and $E'$ are symmetric with respect $T''I \perp AB \implies AB || EE'.$

Similarly, $BC || E''E', AC || E''E \implies \triangle ABC \sim \triangle EE'E''.$

$AE, BE', CE''$ are concurrent at the homothetic center of $\triangle ABC$ and $\triangle EE'E''.$

Claim 2

Feuerbach 2.png

Let $M, M',$ and $M''$ be the midpoints $BC, AC,$ and $AB,$ respectively. Points $E, E',$ and $E''$ was defined at Claim 1.

Prove that $ME, M'E',$ and $M''E''$ are concurrent.

Proof

\[\triangle ABC \sim \triangle MM'M'' \implies\] \[\triangle MM'M'' \sim \triangle EE'E'' \implies\] $ME, M'E', M''E''$ are concurrent at the homothetic center of $\triangle MM'M''$ and $\triangle EE'E''.$

Claim 3

Feuerbach 3a.png

Let $H$ be the base of height $AH.$ Let $F_0 = ME \cap \omega \ne E.$ Prove that points $F_0, E, D,$ and $H$ are concyclic.

Proof

$MT$ tangent to $\omega \implies MT^2 = ME \cdot MF_0.$

Denote $a = BC, b = AC, c = AB.$ \[BD = \frac {ac}{b+c}, BM = \frac {a}{2} \implies MD = \frac {a(b-c)}{2(b+c)}.\] \[BT = \frac {a+c-b}{2} \implies MT = \frac {b-c}{2}.\] Point $H$ lies on radical axis of circles centered at $B$ and $C$ with the radii $c$ and $b,$ respectively. \[BH = \frac {a}{2} - \frac {b^2 - c^2}{2a} \implies HM = \frac {b^2 - c^2}{2a}.\] Therefore $MH \cdot MD = MT^2 = ME \cdot MF_0 \implies$ points $F_0, E, D,$ and $H$ are concyclic.

Claim 4

Feuerbach 4.png

Prove that points $F_0, M, M',$ and $H$ are concyclic.

Proof

\[\angle EDM = \angle TIE = 2 \angle TID = 2(\beta - \gamma).\] $F_0, E, D,$ and $H$ are concyclic $\implies$ \[\angle EF_0H = \angle EDM = 2(\beta - \gamma) = \angle MF_0H.\] \[\angle M'HM = \angle ACB = 2 \gamma.\] \[MM'||AB \implies \angle M'MC = 2 \beta.\] \[\angle HM'M = \angle CMM' - \angle MHM' = 2\beta - 2 \gamma = \angle MF_0H \implies\] points $F_0, M, M',$ and $H$ are concyclic.

Sharygin’s proof

The incircle $\omega$ and the nine-point circle $\Omega$ of a triangle are tangent to each other.

Proof

Let $F_0 = ME \cap \omega \ne E, F' = M'E' \cap \omega \ne E', F'' = M''E'' \cap \omega \ne E''.$

According claim 4, each of this point lyes on $\Omega.$

$\omega$ and $\Omega$ have not more then two common point, so two of points $F_0, F',$ and $F''$ are coincide.

Therefore these two points coincide with point $F$ witch means that $F = \omega \cap \Omega.$

$F$ is the center of similarity of $\omega$ and $\Omega,$ therefore there is no second point of intersection of $\omega$ and $\Omega.$

We conclude that these circles are tangent to each other at point $F.$

vladimir.shelomovskii@gmail.com, vvsss

Scalene triangle with angle 60^\circ

16 2025 I.png

The Feuerbach point of a scalene triangle lies on one of its bisectors. Prove that the angle corresponding to the bisector is $60^\circ.$

Proof

Denote $ABC -$ given triangle,$\angle ABC = 2 \beta, s -$ semiperimeter of $\triangle ABC, \omega, \omega', \theta$ are incircle, B-excircle, and nine points circle centered at $I, I',$ and $Q, F \in BI -$ Feuerbach point, $r = ID, D \in BC -$ the inradius, $r' = I'E, E \in BC - B-$ exradius.

It is known that $\theta$ has diameter $R$ (half of diameter of circumcircle).

$\omega'$ is tangent to $\theta,$ points $Q$ and $I'$ lies on $B-$ bisector, so \[II' = FI' - r = FK + KI' - r = R + r' - r \implies \sin \beta = \frac {r' - r} {R + r' - r}.\] \[BD = s - b, \frac {r}{s-b} = \tan \beta, r'(s-b) = rs, \frac {b}{R} = 2 \sin 2 \beta \implies\] \[\sin \beta = \frac { 1}{1+ 4\sin^2 \beta} \implies \sin \beta = \frac {1}{2} \implies 2 \beta = 60^\circ.\]

vladimir.shelomovskii@gmail.com, vvsss

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