Fermat-Punkt
[color=#000000][b] Gibt es in jedem Dreieck einen Punkt F so, daß die Summe der Entfernungen von F zu den drei Eckpunkten minimal ist?[br] [/b]Diese Problemstellung taucht zum ersten Mal um die Mitte des 17. Jahrhunderts auf. Urheber ist der französischer Mathematiker, Jurist und Parlamentsrat [url=http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Fermat.html]Pierre de Fermat [/url] (1601-1665). [br][color=#444444][color=#000000][u]Konstruktion des Fermat-Punkts:[/u][/color][/color][br]● Über den Seiten eines beliebigen Dreiecks ([i]Innenwinkel ≤ 120°[/i]) werden gleichseitige Dreiecke errichtet und deren freie Eckpunkte mit den gegenüberliegenden Eckpunkten des Ausgangsdreiecks verbunden.[/color][br][color=#000000]● Wenn ein Dreieck einen [i]Winkel[/i] von [i]mehr als 120 °[/i] hat, befindet sich der Fermat-Punkt am stumpfen Winkel.[/color][br]
Bernat Link:[url=https://www.geogebra.org/material/show/id/wksuzsqc] https://www.geogebra.org/material/show/id/wksuzsqc[/url][br]zu einem sehr schönen Applet mit einer Illustration auf einer zweidimensionalen Oberfläche
Euler-Gerade und Feuerbach-Kreis
eine dynamische Konstruktion
2D Dynamic construction for studying the properties of the geometric center.
This applet presents a dynamic construction for studying the properties of the [color=#ff00ff]centroid[/color] or [color=#ff00ff]geometric center[/color] [url=https://en.wikipedia.org/wiki/Centroid]https://en.wikipedia.org/wiki/Centroid[/url] .[br][color=#333333]The centroid or Geometric Center [/color]of figure is the arithmetic mean position [math]\vec{P_i}[/math] of all n- points in the figure:[br] [math]\vec{C_m}=\frac{\sum\vec{P_i}}{n},i=1,...,n[/math]. Two special expressions are associated with centroid.[br] - From its definition: [math]\sum\left(\vec{C_m}-\vec{P_i}\right)=0[/math] : The Addition of radius vectors of all points relative to the centroid is zero.[br] - Difference of the two sums: over the squared distances for all points from B and from the centroid is equal to the n times squared distance between centroid and B. It follows, that the sum of the squared distances for all points from the centroid [math]C_m[/math] is the smallest. You can compare results with the Steiner's theorem in the case of unit point masses. [url=https://en.wikipedia.org/wiki/Parallel_axis_theorem]https://en.wikipedia.org/wiki/Parallel_axis_theorem[/url] [br] Creation of this applet was inspired by alfinio [url=https://www.geogebra.org/material/show/id/DZbG9HMZ— February 26, 2015 - 11:36 PM]https://www.geogebra.org/material/show/id/DZbG9HMZ— February 26, 2015 - 11:36 PM[/url] to prove and implement it for more general case.[br] Change the number of particles n in a system, the position of points P[math]_i[/math], B. Make sure that the formula is correct and try again.[br]
2D. Explanation of Invariance of the sum of squares of distances by using the Steiner theorem.
n/2 points of [color=#0000ff]A[sub]1[/sub],...,A[sub]n[/sub][/color] may be moving in a circle of radius r. Each [color=#0000ff]point[/color] has [color=#ff7700]antipodal point[/color] on the circle. Thus, the center of gravity [color=#ff00ff]Cm[/color] is the centre of the circle. [br] It is shown, that the sum of squares of distances from any point [color=#ff0000]D[/color] of a circle to → {[color=#0000ff]A[sub]i[/sub][/color] } , divided by radius squared, is always equal of twice the number of points. [br][url=https://en.wikipedia.org/wiki/Parallel_axis_theorem]https://en.wikipedia.org/wiki/Parallel_axis_theorem[/url][br][url=https://en.wikipedia.org/wiki/Antipodal_point]https://en.wikipedia.org/wiki/Antipodal_point[/url]
X(85) Isogonal conjugate of X(41)
isogonal conjugate of X(41)
Triangle center X(41) is the X(6)-ceva conjugate of X(31).[br]X(6) is the symmedian point[br]X(31) is the 2nd power point of the triangle ABC.[br]A power point is a point whose coordinates are defined by a power function.[br]For the 2nd power point the [url=http://mathworld.wolfram.com/TrilinearCoordinates.html]trilinear coordinates[/url] are a² : b² : c².[br]P, triangle center X(41) is the X(6)-[url=http://mathworld.wolfram.com/CevaConjugate.html]Ceva conjugate[/url] of X(31). This means that P, the X(6)-Ceva conjugate of X(31) is given by the perspector of the Cevian triangle of X(6) and the anticevian triangle of X(31).[br]The isogonal conjugate of X[sub]41[/sub], triangle center X(41) can be constructed as follows:[br][list][*]Reflect the lines AX[sub]41[/sub], BX[sub]41[/sub], CX[sub]41[/sub] about the bisectors of the triangle ABC (=blue lines)[/*][*]These blue lines cross at the triangle center X(85).[/*][/list]The barycentric coordinates of this point depend on the lenghts of the triangle.
isogonale toegevoegde van X(41)
Driehoekscentrum X(41) is de X(6)-ceva toegevoegde van X(31).[br]X(6) is het punt van Lemoine.[br]X(31) is het tweedemachtspunt van de driehoek ABC.[br]Een machtspunt is een punt waarvan de coördintaten bepaald worden door een machtsfunctie.[br]Voor het tweedemachtspunt worden de [url=http://mathworld.wolfram.com/TrilinearCoordinates.html]trilineaire coördinaten[/url] a² : b² : c².[br]P, driehoekscentrum X(41) is de X(6) [url=http://mathworld.wolfram.com/CevaConjugate.html]Ceva toegevoegde[/url] van X(31). Dit betekent dat P, de X(6)-Ceva toegevoegde X(31) het perspectiefcentrum is va de Ceva driehoek van X(6) en de anticeva driehoek van X(31).[br]Het isogonale toegevoegde punt van X[sub]41[/sub], het driehoekscentrum X(41) construeer je als volgt:[br][list][*]Spiegel de rechten AX[sub]41[/sub], BX[sub]41[/sub], CX[sub]41[/sub] t.o.v. de bissectrices van ABC (=blauwe lijnen).[/*][*]Deze blauwe lijnen snijden elkaar in het driehoekscentrum X(85).[/*][/list]De barycentrische coördinaten van dit punt worden bepaald door de lengtes van de zijden van de driehoek.