Difference between revisions of "Feuerbach point"

Revision as of 15:02, 27 December 2023

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 prove

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

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

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

Let $H$ be the base of height $AH.$

Prove that points $F, E, D,$ and $H$ are concyclic.

Proof

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

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. So $BH = \frac {a}{2} - \frac {b^2 - c^2}{2a} \implies HM = \frac {b^2 - c^2}{2a}.$

Therefore $MH \cdot MD = MT^2 = ME \cdot MF \implies$ points $F, E, D,$ and $H$ are concyclic.

@@ Line 47: / Line 47: @@
 <math>MT</math> tangent to <math>\omega \implies MT^2 = ME \cdot MF.</math>
 Denote <math>a = BC, b = AC, c = AB.</math>
 <cmath>BD = \frac {ac}{b+c}, BM = \frac {a}{2} \implies MD = \frac {a(b-c)}{2(b+c)}.</cmath>
 <cmath>BT = \frac {a+c-b}{2} \implies MT = \frac {b-c}{2}.</cmath>
 Point <math>H</math> lies on radical axis of circles centered at <math>B</math> and <math>C</math> with the radii <math>c</math> and <math>b,</math> respectively. So <math> BH = \frac {a}{2} - \frac {b^2 - c^2}{2a} \implies HM =  \frac {b^2 - c^2}{2a}.</math>
 Therefore <math>MH \cdot MD = MT^2 = ME \cdot MF \implies</math>   points <math>F, E, D,</math> and <math>H</math> are concyclic.

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Difference between revisions of "Feuerbach point"

Revision as of 15:02, 27 December 2023

Sharygin’s prove