Difference between revisions of "Euler line"
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<i><b>Case 1</b></i> | <i><b>Case 1</b></i> | ||
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Let <math>F</math> be Fermat point <math>F_1</math> of <math>\triangle ABC</math> maximum angle of which smaller then <math>120^\circ.</math> Then the centroid of triangle <math>ABC</math> lies on Euler line of the <math>\triangle ABF.</math> | Let <math>F</math> be Fermat point <math>F_1</math> of <math>\triangle ABC</math> maximum angle of which smaller then <math>120^\circ.</math> Then the centroid of triangle <math>ABC</math> lies on Euler line of the <math>\triangle ABF.</math> | ||
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+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\angle AFB = 120^\circ, CF</math> is bisector <math>\angle AFB.</math> | ||
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+ | As shown above, the Euler line <math> O_1H_1</math> of the <math>\triangle ABF</math> is parallel to <math>CF.</math> | ||
+ | |||
+ | Let <math>H, M,</math> and <math>O</math> be orthocenter, centroid and circumcenter of <math>\triangle ABC,</math> respectively. | ||
+ | Let <math>M_1</math> be centroid of <math>\triangle ABF.</math> | ||
+ | <cmath>\vec M = \frac {\vec A + \vec B + \vec C}{3}, \vec M_1 = \frac {\vec A + \vec B + \vec F}{3} \implies \vec {M_1M} = \frac {\vec {FC}}{3}.</cmath> | ||
+ | Point <math>M_1 </math> lies on Euler line of <math>\triangle ABF</math>, this line is parallel to <math>FC \implies M \in O_1 H_1.</math> | ||
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+ | Symilarly, <math>M</math> lies on Euler lines of <math>\triangle BCF</math> and <math>\triangle ACF</math> as desired. | ||
==See also== | ==See also== |
Revision as of 03:01, 20 October 2022
In any triangle , the Euler line is a line which passes through the orthocenter , centroid , circumcenter , nine-point center and de Longchamps point . It is named after Leonhard Euler. Its existence is a non-trivial fact of Euclidean geometry. Certain fixed orders and distance ratios hold among these points. In particular, and
Euler line is the central line .
Given the orthic triangle of , the Euler lines of ,, and concur at , the nine-point circle of .
Contents
[hide]Proof Centroid Lies on Euler Line
This proof utilizes the concept of spiral similarity, which in this case is a rotation followed homothety. Consider the medial triangle . It is similar to . Specifically, a rotation of about the midpoint of followed by a homothety with scale factor centered at brings . Let us examine what else this transformation, which we denote as , will do.
It turns out is the orthocenter, and is the centroid of . Thus, . As a homothety preserves angles, it follows that . Finally, as it follows that Thus, are collinear, and .
Another Proof
Let be the midpoint of . Extend past to point such that . We will show is the orthocenter. Consider triangles and . Since , and they both share a vertical angle, they are similar by SAS similarity. Thus, , so lies on the altitude of . We can analogously show that also lies on the and altitudes, so is the orthocenter.
Proof Nine-Point Center Lies on Euler Line
Assuming that the nine point circle exists and that is the center, note that a homothety centered at with factor brings the Euler points onto the circumcircle of . Thus, it brings the nine-point circle to the circumcircle. Additionally, should be sent to , thus and .
Analytic Proof of Existence
Let the circumcenter be represented by the vector , and let vectors correspond to the vertices of the triangle. It is well known the that the orthocenter is and the centroid is . Thus, are collinear and
Euler line for a triangle with an angle of 120
Let the in triangle be Then the Euler line of the is parallel to the bisector of
Proof
Let be circumcircle of
Let be circumcenter of
Let be the circle symmetric to with respect to
Let be the point symmetric to with respect to
The lies on lies on
is the radius of and translation vector to is
Let be the point symmetric to with respect to Well known that lies on Therefore point lies on
Point lies on
Let be the bisector of are concurrent.
Euler line of the is parallel to the bisector of as desired.
vladimir.shelomovskii@gmail.com, vvsss
Concurrent Euler lines and Fermat points
Consider a triangle with Fermat–Torricelli points and The Euler lines of the triangles with vertices chosen from and are concurrent at the centroid of triangle
Case 1
Let be Fermat point of maximum angle of which smaller then Then the centroid of triangle lies on Euler line of the
Proof
is bisector
As shown above, the Euler line of the is parallel to
Let and be orthocenter, centroid and circumcenter of respectively. Let be centroid of Point lies on Euler line of , this line is parallel to
Symilarly, lies on Euler lines of and as desired.
See also
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