Let U,V,W any points on the sides of ABC.
Then the three circumcircles to AVW, BUW and CUV intersect in a same point Q.
Conversely, given a point Q, we can construct infinitely many such circles through Q, defining a family of inscribed triangles UVW in ABC. (by choosing any U and constructing V,W as other intersection of AB and AC with the circles BQU and CQU for instance).
- Angles UPV, VPW and UPW are constant (= π-C, π-A, π-B)
(QA,QB) = (QA,AB) + (AB,QB) = (QV,VW) + (UW,UQ) = (QV,QU) + (UW,VW) = (CV,CU) + (UW,VW) = (CA,CB) + (UW,VW)
In the above applet, A,B,C,U,V,W are draggable, and the green point defines the angle of the similitude UVW → U'V'W' (Click on U'V'W' button to display)
Apollonius circles :
Let X,Y,Z,X',Y',Z' the feet of angle bissectors in ABC.
The circles with diameters XX', YY', ZZ' are called the Apollonius circles of ABC.
They have two common points, named the isodynamic centers of ABC.
The pedal triangle of an isodynamic center is equilateral.
The isodynamic centers are the only point whose pedal triangle is equilateral.
In the applet, the initial shape of ABC is choosen so flat as to display entirely the Apollonius circles.
Of course A,B,C are fully draggable, but then the circles (as well as point P on it) may get out of the picture.
Move ABC to better see the pedal triangles.
Points R and R' are interchanged if ABC is in the other direction.
If ABC is (about) isosceles, one of the Apollonius circles degenerates into the perpendicular bissector of the base.
This circles becomes then much inaccurate on the applet.
Theorem (said "Napoléon theorem") :
Let ABC', BCA' and ACB' the equilateral triangles constructed on the sides of ABC, outside.
AA', BB', CC' intersect in the isogonic center S.
|The isogonic center and the isodynamic center are isogonal conjugates.|
Let UVW the pedal triangle of R, hence equilatéral. CURV concyclic etc...
Therefore [Miquel] : (RA,RB) = (CA,CB) + (UW,VW) = (CA,CB) + 60°
Let T the isogonal conjugate of R.
(TA,TB) = (TA,AB) + (AB,TB)
(RA,RB) = (RA,AB) + (AB,RB) = (AC,TA) + (TB,BC) (R and T isogonal conjugates)
(TA,TB) + (RA,RB) = (AC,TA) + (TA,AB) + (AB,TB) + (TB,BC) = (AC,BC)
Therefore (TA,TB) = -60° = 120° etc. and T is really the isogonic center S.
This gives a rather "simple" construction of the isodynamic center(s), without the need to construct the Apollonius circles.
Lets construct the equilateral triangles ABC', BCA' and ACB' outside ABC.
Then the lines Aa", Bb" Cc" with angles BAa" = CAA' etc...
These lines intersect at the isodynamic point R.
(of course we just need to construct A', Aa", B' and Bb")
Let construct in that way the isodynamic center R of ABC and its pedal triangle UVW.
AVRW are concylic (right angles) and the same for BURW and CURV
R is then a Miquel point for UVW.
Considering the lines RU, RV, RW as a rigid system, free to rotate around R.
The intersections of these lines with the sides of ABC build a family of
triangles U'V'W' similar to UVW [Miquel] hence equilateral.
We generate in that way the two families of inscribed equilateral triangles and "escribed" triangles. The "escribed" triangles are those generated from the previous method, but with equilateral triangles inside ABC, resulting into triangles with "120° angles" !
(That is from the other isodynamic center R').
Note : In the applet R and R' are interchanged when ABC is in the reverse direction.
The green draggable point defines the rotation for the current triangle U'V'W'.
Of course the size of the equilateral inscribed triangle
is minimal when distance RU' is minimal !
(for they are all similar, with similitude center R, hence UV/RU = const)
|The minimal inscribed equilateral triangle is the pedal triangle of the isodynamic center|