[alogo] 1. Turning the pedal

Consider the pedal triangle A'B'C' of the point P with respect to the triangle ABC. Its vertices are the feet of the perpendiculars from P to the sides of the triangle of reference ABC. By drawing lines {PA'', PB'', PC''} from P equal inclined to these perpendiculars and taking their traces on the sides we create triangle A''B''C'' similar to the pedal A'B'C'.

[0_0] [0_1]

In fact, the right-angled triangles {PA'A'', PB'B'', PC'C''} are all similar. This implies the similarity of triangles {PA'B', PA''B''} and also of triangles {PB'C', PB''C''}. From these similarities follows also the claimed similarity.

Corollary-1 For all triangles A''B''C'', resulting by the previous procedure from A'B'C', the angles {A''PB'', B''PC'', C''PA''} remain constant and are equal to respective angles of ABC (to account for points P outside the area of the triangle) or their complementaries.
Corollary-2 The position of P relative to A'B'C' is uniquely determined by the angles {A''PB'', B''PC'', C''PA''}, which also determine the ratios PA':PB':PC', hence the trilinears of P with respect to ABC.

[alogo] 2. The twelve positions of P

Fixing triangle ABC and the angles of A'B'C' one can ask how many P's there are, such that the resulting pedal triangle is similar to A'B'C'. The possibilities depend on two factors (i) which vertex of A'B'C' lies on which side of ABC and (ii) orientation of A'B'C' relative to the one of ABC.

[0_0] [0_1]

The first possibility gives 6 cases and the second doubles them to 12. Thus, there are at most twelve different locations for P relative to ABC and consequently also relative to A'B'C'. In the previous figure we have a case in which the orientation of A'B'C' is the inverse of the orientation of ABC (see Inverse_Pedals.html ).
Consequently (Corollary-2 above) also at most twelve possible positions of P relative to ABC.
Next figure shows the positions of these twelve points relative to A'B'C' (the inscribed one) in a case of two triangles for which the twelve positions are all different. The positions relative to triangle ABC present a different and more interesting structure, which is studied in the file TwelvePivots.html .

[0_0] [0_1]

[alogo] 3. Varying the circumscribed triangle

Instead of turning the triangle A'B'C' so that its vertices glide on the sides of ABC one can vary inversely triangle ABC, so that A'B'C' remains invariant, ABC remains similar to a fixed triangle, while the sides of ABC pass respectively through vertices of A'B'C'.


In fact, given the position of P relative to A'B'C', draw the circumcircles of triangles {PA'B', PB'C', PC'A'}. Take then some point A on the first circle and join it to some vertex of A'B'C' lying on the same circle. Extend the previous line to intersect a second time the adjacent circle at B and repeat the procedure from B to find C. An angle chasing argument shows that {A, C, B'} are collinear and the triangle thus constructed has the correct relations of angles to the angles {PA'B', PB'C', PC'A'}.
Remark-1 In this procedure the position of ABC for which A'B'C' becomes the pedal of P is the one maximizing the perimeter of ABC for which also the sides of this triangle are parallel to the lines joining the centers of the three circles.
Remark-2 P is a so-called pivot of inscription of A'B'C' into ABC or circumscription of ABC to A'B'C'. It is also the Miquel point of the three points {A', B', C'} on the sides of triangle ABC.

[alogo] 4. Construction of a particular case

There is an elementary construction of an inscribed triangle similar to A'B'C' and its corresponding pivot P for each one of the twelve possible configurations discussed in section 2. In this section I construct the case in which the vertices {A', B', C'} are respectively on the sides {BC, BA, AC} and the orientation is that of ABC.


The construction of a triangle with this prescriptions (similar to A'B'C' and inscribed in this way in ABC) can be carried out in the following way.
Select an arbitrary point A0 on BC and an arbitrary point C1 on CA. Construct a triangle A0B1C1 similar to A'B'C' and oriented according to the prescription. Fix A0 and repeat the construction with a second auxiliary point C2 on CA delivering a second triangle A0B2C2 similar to A'B'C'. By well known theorem on similar triangles pivoting around a fixed point and having one vertex (Ci) gliding on a line (see reference (1) below), the other vertex will glide also on a fixed line. Thus line B1B2 is the fixed line on which vertex Bi will glide for all triangles A0BiCi similar to A'B'C and having Ci gliding on CA.

[0_0] [0_1]
[1_0] [1_1]

Let C0 be the particular point on AC for which triangle A0B0C0 is similar to A'B'C' and B0 is the intersection of line B1B2 with AB. Triangle A0B0C0 satisfies the prescriptions. Its pivot P can be found by intersecting two of the circumcircles of {AB0C0, BB0A0, CC0A0} since all three pass through it.

[alogo] 5. The set of pivots relative to ABC

The set of pivots of A'B'C' relative to ABC is the set of all points P whose pedals are similar to A'B'C'. Fixing ABC this set contains at most 12 elements which are permuted by a group of transformations naturally connected with triangle ABC. This is studied in the file TwelvePivots.html .

See Also


Return to Gallery

Produced with EucliDraw©