Difference between revisions of "Feuerbach point"
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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. | 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 | + | ==Sharygin’s proof== |
− | 1998, | + | <math>1998, 24^{th}</math> Russian math olympiad |
− | + | [[File:Feuerbach 1.png|500px|right]] | |
− | + | ===Claim 1=== | |
− | |||
Let <math>D</math> be the base of the bisector of angle A of scalene triangle <math>\triangle ABC.</math> | Let <math>D</math> be the base of the bisector of angle A of scalene triangle <math>\triangle ABC.</math> | ||
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WLOG, <math>\beta > \gamma.</math> | WLOG, <math>\beta > \gamma.</math> | ||
<cmath>\angle TIT'' = 180^\circ - 2 \beta, \angle ADB = 180^\circ - \alpha - 2 \beta,</cmath> | <cmath>\angle TIT'' = 180^\circ - 2 \beta, \angle ADB = 180^\circ - \alpha - 2 \beta,</cmath> | ||
− | <cmath>\angle DIT = 90^\circ - \angle ADB = \alpha + 2 \beta | + | <cmath>\angle DIT = 90^\circ - \angle ADB = \alpha + 2 \beta - 90^\circ = \beta -\gamma, \angle EID = \angle TID \implies</cmath> |
<cmath>\angle T''IE = \angle T''IT + 2 \angle TID = 180^\circ - 2 \beta + 2(\beta - \gamma) = 180^\circ - 2 \gamma.</cmath> | <cmath>\angle T''IE = \angle T''IT + 2 \angle TID = 180^\circ - 2 \beta + 2(\beta - \gamma) = 180^\circ - 2 \gamma.</cmath> | ||
Similarly, <math>\angle T''IE' = 180^\circ – 2 \gamma \implies</math> points <math>E</math> and <math>E'</math> are symmetric with respect <math>T''I \perp AB \implies AB || EE'.</math> | Similarly, <math>\angle T''IE' = 180^\circ – 2 \gamma \implies</math> points <math>E</math> and <math>E'</math> are symmetric with respect <math>T''I \perp AB \implies AB || EE'.</math> | ||
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<math>AE, BE', CE''</math> are concurrent at the homothetic center of <math>\triangle ABC</math> and <math>\triangle EE'E''.</math> | <math>AE, BE', CE''</math> are concurrent at the homothetic center of <math>\triangle ABC</math> and <math>\triangle EE'E''.</math> | ||
+ | |||
+ | ===Claim 2=== | ||
+ | [[File:Feuerbach 2.png|500px|right]] | ||
+ | Let <math>M, M',</math> and <math>M''</math> be the midpoints <math>BC, AC,</math> and <math>AB,</math> respectively. Points <math>E, E',</math> and <math>E''</math> was defined at Claim 1. | ||
+ | |||
+ | Prove that <math>ME, M'E',</math> and <math>M''E''</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>\triangle ABC \sim \triangle MM'M'' \implies </cmath> | ||
+ | <cmath>\triangle MM'M'' \sim \triangle EE'E'' \implies</cmath> | ||
+ | <math>ME, M'E', M''E''</math> are concurrent at the homothetic center of <math>\triangle MM'M''</math> and <math>\triangle EE'E''.</math> | ||
+ | |||
+ | ===Claim 3=== | ||
+ | [[File:Feuerbach 3a.png|500px|right]] | ||
+ | Let <math>H</math> be the base of height <math>AH.</math> Let <math>F_0 = ME \cap \omega \ne E.</math> | ||
+ | Prove that points <math>F_0, E, D,</math> and <math>H</math> are concyclic. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>MT</math> tangent to <math>\omega \implies MT^2 = ME \cdot MF_0.</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. | ||
+ | <cmath>BH = \frac {a}{2} - \frac {b^2 - c^2}{2a} \implies HM = \frac {b^2 - c^2}{2a}.</cmath> | ||
+ | Therefore <math>MH \cdot MD = MT^2 = ME \cdot MF_0 \implies</math> points <math>F_0, E, D,</math> and <math>H</math> are concyclic. | ||
+ | |||
+ | ===Claim 4=== | ||
+ | [[File:Feuerbach 4.png|450px|right]] | ||
+ | Prove that points <math>F_0, M, M',</math> and <math>H</math> are concyclic. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>\angle EDM = \angle TIE = 2 \angle TID = 2(\beta - \gamma).</cmath> | ||
+ | <math>F_0, E, D,</math> and <math>H</math> are concyclic <math>\implies</math> | ||
+ | <cmath>\angle EF_0H = \angle EDM = 2(\beta - \gamma) = \angle MF_0H.</cmath> | ||
+ | <cmath>\angle M'HM = \angle ACB = 2 \gamma.</cmath> | ||
+ | <cmath>MM'||AB \implies \angle M'MC = 2 \beta.</cmath> | ||
+ | <cmath>\angle HM'M = \angle CMM' - \angle MHM' = 2\beta - 2 \gamma = \angle MF_0H \implies</cmath> | ||
+ | points <math>F_0, M, M',</math> and <math>H</math> are concyclic. | ||
+ | |||
+ | ===Sharygin’s proof=== | ||
+ | The incircle <math>\omega</math> and the nine-point circle <math>\Omega</math> of a triangle are tangent to each other. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>F_0 = ME \cap \omega \ne E, F' = M'E' \cap \omega \ne E', F'' = M''E'' \cap \omega \ne E''.</math> | ||
+ | |||
+ | According claim 4, each of this point lyes on <math>\Omega.</math> | ||
+ | |||
+ | <math>\omega</math> and <math>\Omega</math> have not more then two common point, so two of points <math>F_0, F',</math> and <math>F''</math> are coincide. | ||
+ | |||
+ | Therefore these two points coincide with point <math>F</math> witch means that <math>F = \omega \cap \Omega.</math> | ||
+ | |||
+ | <math>F</math> is the center of similarity of <math>\omega</math> and <math>\Omega,</math> therefore there is no second point of intersection of <math>\omega</math> and <math>\Omega.</math> | ||
+ | |||
+ | We conclude that these circles are tangent to each other at point <math>F.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 10:23, 29 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 proof
Russian math olympiad
Claim 1
Let be the base of the bisector of angle A of scalene triangle
Let be a tangent different from side
to the incircle of
is the point of tangency). Similarly, we denote
and
Prove that are concurrent.
Proof
Let and
be the point of tangency of the incircle
and
and
Let
WLOG,
Similarly,
points
and
are symmetric with respect
Similarly,
are concurrent at the homothetic center of
and
Claim 2
Let and
be the midpoints
and
respectively. Points
and
was defined at Claim 1.
Prove that and
are concurrent.
Proof
are concurrent at the homothetic center of
and
Claim 3
Let be the base of height
Let
Prove that points
and
are concyclic.
Proof
tangent to
Denote
Point
lies on radical axis of circles centered at
and
with the radii
and
respectively.
Therefore
points
and
are concyclic.
Claim 4
Prove that points and
are concyclic.
Proof
and
are concyclic
points
and
are concyclic.
Sharygin’s proof
The incircle and the nine-point circle
of a triangle are tangent to each other.
Proof
Let
According claim 4, each of this point lyes on
and
have not more then two common point, so two of points
and
are coincide.
Therefore these two points coincide with point witch means that
is the center of similarity of
and
therefore there is no second point of intersection of
and
We conclude that these circles are tangent to each other at point
vladimir.shelomovskii@gmail.com, vvsss