Elem. Math. 57 (2002) 26 – 31
0013-6018/02/010026-6 $ 1.50+0.20/0 Elemente der Mathematik
The notable configuration of inscribed equilateral triangles in a triangle
Blas Herrera Go´mez
Blas Herrera Go´mez obtained his Ph.D. in mathematics at the University Auto`noma of Barcelona in 1994. Presently, he is a professor at the University Rovira i Virgili of Tarragona. His main fields of interest are: the geometry of foliations, the dynamics of galaxies, classical geometry, and the mechanics of fluids and turbulence.
1 Introduction
Many notable configurations for a given triangle have been described by different ge- ometers in the past. Among the most famous classic configurations are the Euler and Simson lines, the Gergonne, Lemoine, Brocard and Nagel concurrences, the Feuerbach, Brocard, Lemoine, Tuker and Taylor circumferences, etc. More modern configurations can be found in the Soddy circles [9], the Euler-Gergonne-Soddy triangle [8], the Tor- ricelli configuration [10], etc. In fact, one can see in [3] four hundred configurations (triangle centers), in [4] even eight hundred configurations are picked up. Recently, new configurations continue to appear, for example Morley triangles related to the Feuerbach circumferences [5], associated rectangular hyperbolas [1], triangle centers associated with a rhombus [6], Euler lines concurrent on the nine-point circle [7], etc.
Among modern configurations we focus our attention on the Lucas circles. Edouard Lucas studied the configuration given by three inscribed squares in a triangle with a side parallel to each one of the sides of the triangle. With them appear three circles. Recently [2], Yiu and Hatzipolakis have been continuing the study of this configuration and, in particular, have shown recently that the Lucas circles are tangent to each other.
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Bekanntlich gibt es eine Vielzahl bereits untersuchter Konfigurationen zu einem ge- gebenen ebenen Dreieck. Der Autor des nachfolgenden Beitrags knu¨pft an eine von E. Lucas studierte Konfiguration an, die ein Dreieck mit drei einbeschriebenen Qua- draten zum Ausgangspunkt hat und zu den sogenannten Lucas-Kreisen fu¨hrt. Hier wird nun ein Dreieck mit drei einbeschriebenen gleichseitigen Dreiecken, von denen jeweils genau eine Seite zu einer der Seiten des Ausgangsdreiecks parallel ist, untersucht. Bei dieser Konfiguration zeigt sich unter anderem, dass jeweils vier der insgesamt neun Eckpunkte der einbeschriebenen Dreiecke auf drei Kreisen liegen, die sich in einem Punkt schneiden.
In this paper we take on Lucas’ idea and consider not the three inscribed squares, but the three inscribed equilateral triangles with one side parallel to each one of the sides of the triangle (see Fig. 1).
The notable configuration which appears is outlined below and will be summarized in the theorem of the following section.
We will show that the centers of these equilateral triangles are aligned (see Fig. 2). We will see also that the mid-points of the sides of the equilateral triangles are aligned three by three (see Fig. 2). We will find that the simple ratios of these mid-points and the simple ratios of the vertices are equal to the simple ratio of the centers. But the most remarkable feature will be the appearance of three circumferences which link the vertices four by four, and these three circles are concurrent in a point (see Figs. 3, 4).
A
A
A
C
C C
B
A1
A1
A2
A2
A3
A3
C1
C1
C2
C2
C3
C3
B
B1
B1
B2
B2
B3
B3
Fig. 1 Construction of the three inscribed equilateral triangles.
2 Configuration
First of all, in order to establish our notations we recall that the “simple ratio”(A,B,C) of three aligned points is(A,B,C) =±ACBC with a positive sign ifC is not between A andB, and a negative sign otherwise.
We recall that in order to construct an inscribed equilateral triangle inABC with a side parallel to sideAB, we consider the exterior equilateral triangle with sideAB,ABC. LetC3be the intersection of the lineCC with the lineAB. The lines through the point C3 parallel toAC andBC intersect with the linesAC andBC at the pointsC1 andC2
respectively. The triangleC1C2C3 is the inscribed equilateral triangle (see Fig. 1).
The other two triangles A1A2A3 andB1B2B3, with sides parallel to BC and AC respectively, can be constructed in a similar way.
The description of the configuration can be summarized in the following theorem.
A
C B
Fig. 2 The centersOA,OB,OCare aligned. The linesAA3,BB3,CC3are concurrent (outward Fermat point F=X13). The mid-pointsA12,B13,C13are aligned; the mid-pointsA13,B12,C23are aligned; the mid- pointsA23,C12,B23are aligned. All the simple ratios coincide with the simple ratio(OA,OB,OC).
Theorem 1 LetABCbe a triangle. LetA1A2A3,B1B2B3,C1C2C3be the three inscribed equilateral triangles with A1A2 parallel toBC, B1B2 parallel toAC, C1C2
parallel toAB and A1,B1,C3on the lineAB.
LetOA,OB,OC be the respective centers of the three equilateral triangles, and letAi j, Bi j,Ci j be the respective mid-points of the sides AiAj,BiBj,CiCj.
Then we have:
1. The center pointsOA,OB,OC are aligned.
2. The linesAA3,BB3,CC3 are concurrent(in the outward Fermat point).
3. The mid-pointsA12,B13,C13are aligned, the mid-pointsA13,B12,C23are aligned and the mid-pointsA23,C12,B23are also aligned.
4. The following simple ratios are equal:
(OA,OB,OC) = (A2,B3,C1) = (A1,B1,C3) = (A3,B2,C2)
= (A12,B13,C13) = (A13,B12,C23) = (A23,B23,C12).
5. The points A3,C2,A1,C3 are concyclic, the pointsA3,B2, A2,B3 are concyclic and the pointsC3,B1,C1,B3 are also concyclic.
6. The above circumferences A3C2A1C3, A3B2A2B3,C3B1C1B3 are concurrent in a pointη.
Proof.Take a system of Cartesian coordinates in such a way that A= (a,b) with 0≤a,b≤1, B = (0,0), C= (1,0).
A straightforward computation proves the parts 1 and 3 of the theorem (see Fig. 2).
Part 2 of the theorem is well-known. The concurrence of the linesAA3,BB3,CC3 is the outward Fermat pointF (X13in [4]).
A
C B
A1 A2
A3
C1
C2
C3
B1
B2
B3
η
Fig. 3 The pointsA3,C2,A1andC3are concyclic, the pointsA3,B2,A2andB3are concyclic and the pointsC3,B1,C1andB3are concyclic. Also, the three circumferences are concurrent at the pointη.
Part 4 can be found with a straightforward computation. The value of the simple ratio obtaining in every case is the same expression:
(OA,OB,OC) = −1
√3
3a−√
3b 3a2−6a+3b2+2√ 3b+3
2b+√
3 3a2−3a+3b2−√ 3b .
Part 5 (see Figs. 3, 4) can be found with a long computation and we have that the points A3,C2,A1,C3 belong to the circumference of centerOAC given by
OACx=1 2
√33a3+3√
3a2b+3ab2+3a2+2√
3ab+3√
3b3+5b2
3a2+3b2+2√
3b 2b+√ 3
,
OACy=−1 2
3a3−5√
3a2b+3ab2−3a2+2√
3ab−5√
3b3−9b2
3a2+3b2+2√
3b 2b+√ 3
,
and radiusrAC:
3(a2+b2)(3a4−4√
3a3b+10a2b2−4√
3ab3−6a3+8√
3a2b−14ab2+3a2−4√
3ab+7b4+8√ 3b3+7b2) (3a2+3b2+2√3b)2(2b+√3)2 . Analogously, the pointsA3,B2,A2,B3belong to the circumference of centerOAB given by:
OABx=1 2
√33a3+√
3a2b+3ab2−6a2+3a+√
3b3+6b2+3√ 3b
3a2−6a+3b2+2√
3b+3 2b+√ 3
,
OABy=1 2
3a3+5√
3a2b+3ab2−6a2−8√
3ab+3a+5√
3b3+6b2+3√ 3b
3a2−6a+3b2+2√
3b+3 2b+√ 3
,
A
C
A1
A2
A3
C1
C2
C3
B
B1
B2
B3 η
Fig. 4 Concurrence of the three circumferences.
and radiusrAB:
3(a2−2a+b2+1)(3a4+4√3a3b+10a2b2+4√3ab3−6a3−4√
3a2b−6ab2+3a2+7b4+4√ 3b3+3b2) (3a2−6a+3b2+2√
3b+3)2(2b+√3)2 .
Analogously the pointsB1,C3,B3,C1 belong to the circumference of centerOBC given by:
OBCx=3 2
3a4+4√
3a3b+6a2b2+4√
3ab3−6a3−2√
3a2b−2ab2+3a2+2√
3ab+3b4+2√ 3b3+5b2
(3a2+2√
3b+3b2)(3a2−6a+3+2√
3b+3b2) ,
OBCy=−1 2
√3
3a2−3a+3b2+√
3b a−√
3b−1 √ 3b+a
3a2+3b2+2√
3b 3a2−6a+3b2+2√
3b+3 ,
and radiusrBC:
9(a2+b2)(a2−2a+b2+1)(3a4+6a2b2−6a3−4√
3a2b−6ab2+3a2+4√
3ab+3b4+4√ 3b3+7b2) (3a2+3b2+2√3b)2(3a2−6a+3b2+2√
3b+3)2 .
Finally, part 6 corresponds to the above circumferences
A3C2A1C3, A3B2A2B3, C3B1C1B3
are concurrent in a point η (see Figs. 3, 4). To prove this statement letη be the image of pointA3 under the axial symmetry with respect to the lineOABOAC. We obtain
η=
1 2 ς τ,−1
6
√3υ φ
with
ς=3a6+8√
3a5b+9a4b2+16√
3a3b3+9a2b4+8√
3ab5−3a5
−11√
3a4b−6a3b2−14√
3a2b3−3ab4−3a4−2√ 3a3b
−10√
3ab3+3a3+5√
3a2b+3ab2+3b6−3√
3b5−5b4+5√ 3b3 , τ=3a6+9a4b2+9a2b4+9√
3a4b−9a5−18a3b2+10√ 3a2b3
−9ab4+12a4−18√
3a3b+18a2b2−10√
3ab3−9a3+9√ 3a2b +3a2−9ab2+3b6+√
3b5−2b4+√
3b3+3b2 ,
υ=
3a2−3a+3b2−√
3b 3a4−2a2b2−6a3+2ab2+3a2−5b4+3b2 , φ=3a6+9a4b2+9a2b4−9a5+9√
3a4b−18a3b2+10√
3a2b3−9ab4 +12a4−18√
3a3b+18a2b2−10√
3ab3−9a3−9ab2+9√
3a2b+3a2 +3b6+√
3b5−2b4+√
3b3+3b2.
This pointη, by construction, belongs to the two circumferencesA3C2A1C3,A3B2A2B3. It only remains to prove thatη belongs to the circumferenceC3B1C1B3. But when we compute the distance betweenηandOBCwe obtain exactly the same expression obtained above forrBC.
This completes the proof. 䊐
Acknowledgement
I would like to thank Agustı´ Revento´s Tarrida, Departament de Matema`tiques, Universitat Auto`noma de Barcelona for his constructive criticism of this paper.
References
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Blas Herrera Go´mez
Departament d’Enginyeria Informa`tica i Matema`tiques Universitat Rovira i Virgili
43007 Tarragona, Spain
e-mail:bherrera@etse.urv.es
