# Inscribed equilateral triangle - More...

We have already seen in the parent problem how to generate all equilateral triangles inscribed in a given ABC.
An other way of getting the whole family of these triangles is from the "isodynamic centers".
This will give us a different way to construct the minimal equilateral inscribed triangle.
First of all a few theorems.

#### (a) Miquel theorem

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 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) - All the triangles UVW for a given Q are similar. - Their common center of similitude is Q - The oriented angles of lines (QA,QB) = (CA,CB) + (UW,VW)
Existence of Q is prooved by showing that the intersection point Q of two circles lies on the 3rd. Writing (a,b) the oriented angle of lines a and b.
QUCV concyclic : (QU,QV) = (CU,CV), also QVAW concyclic : (QV,QW) = (AV,AW), hence (QU,QW) = (CU,CV) + (AV,AW) = (CB,CA) + (AC,AB) = (BC,BA), hence QUBW concyclic.
Conversely, if (QU',QV') = (CB,CA) etc...
The first property immediately results from the above angles.
(V'Q,V'U') = (CQ,CU') = (CQ,CB) hence triangle U'V'Q stays similar to UVQ, in a similitude with center Q (fixed point)
And the same for the others, therefore also for the whole triangle U'V'W', similar to UVW.

(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)

#### Isodynamic centers - definition

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 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 Apollonius circles are then the locus of points P with PB/PC = AB/AC etc...
(of course these circles go through the opposite vertex !)
B is inside circle (XX') and C outside this circle
C is inside circle (YY') and B is on this circle.
Hence circles (XX') and (YY') intersect.
Let R an intersection point of these two circles :
RA/RB × RB/RC = RA/RC is then CA/CB × AB/AC = BA/BC, and R lies on the 3rd circle.

Theorem :

 The pedal triangle of an isodynamic center is equilateral. The isodynamic centers are the only point whose pedal triangle is equilateral.
Recall the pedal triangle of a point P is the triangle made by the perpendicular projections of P on the sides of ABC.
The Apollonius circles are the locii of the points whose pedal triangle is isosceles : (Draggable point P on circle (XX') in the applet)
U' and V' on circle with diameter PC, hence U'V' = PC.sin(C)
U' and W' on circle with diameter PB, hence U'W' = PB.sin(B)
U'V'/U'W' = PC/PB × sin(C)/sin(B) = PC/PB × AB/AC = 1

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.

#### Isogonic center - constructions

Sorry, your browser is not Java compliant.. The isogonic center S is the point from which we see the sides of ABC under equal angles (that is 120° mod 180°).
We already met that point : It is the common intersection point of the three circumcircles to the equilateral triangles constructed on the sides of ABC !

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.
A proof here

Theorem :

 The isogonic center and the isodynamic center are isogonal conjugates.
That is the angles SCA = RCB etc...

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")

#### Conclusion

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'.

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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