Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 39 (1998), No. 2, 263-267.

Regular Hexagons Associated to Centroid Sharing Triangles

Zvonko Čerin

Kopernikova 7, 10020 Zagreb, Croatia e-mail: ZCERIN@X400.SRCE.HR

1. Introduction

In the Problem 1493 from the February 1996 issue of Mathematics Magazine and in the Problem 10514 from the March 1996 issue of American Mathematical Monthly, Jiro Fukuta describes two methods of generating regular hexagons associated to any triangle ABC. More precisely, Fukuta observed that for certain real numbers s, after performing certain constructions we always get regular hexagons F_{1}^{s}F_{2}^{s} ⋯ F_{6}^{s} and G_{1}^{s}G_{2}^{s} ⋯ G_{6}^{s}.

The purpose of this note is to give an analogous procedure which will provide families of regular hexagons that depend on two parameters s and t associated to three triangles with the same centroid.

In order to describe our result we shall use the following notation. For different points A and B in the plane, let AB denote the line passing through A and B. For lines m and n that are not parallel, we shall use m ∧ n to denote their intersection.

Let s denote a real number different from −1. For points A and B, let s_{A}^{B} be the point A when A = B and the unique point P on the line AB such that |AP |/|P B| = s when A ≠ B.

Theorem . Let s and t be real numbers different from −1 such that neither st + s + 1 = 0 nor st + t + 1 = 0. Let ABC, P QR, and XY Z be triangles whose centroids coincide such that neither A, P, X, nor B, Q, Y, nor C, R, Z are collinear. Let

P_{1} = As_{P}^{X} ∧ P t_{X}^{A}, P_{2} = As_{X}^{P} ∧ P t_{A}^{X}, P_{3} = Bs_{Q}^{Y} ∧ Qt_{Y}^{B},
P_{4} = Bs_{Y}^{Q} ∧ Qt_{B}^{Y},  P_{5} = Cs_{R}^{Z} ∧ Rt_{Z}^{C},  P_{6} = Cs_{Z}^{R} ∧ Rt_{C}^{Z}.

Let P_{7} = P_{1}. For i = 1, …, 6, let Q_{i}P_{i}P_{i+1}, be the equilateral triangles built on the sides of the hexagon P_{1}P_{2} ⋯ P_{6} all with the same orientation. Let Q_{0} = Q_{6} and Q_{7} = Q_{1}. Let F_{i}

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 1998 Heldermann Verlag

be the centroid of the triangle Q_{i−1}Q_{i}Q_{i+1}, i = 1, …, 6. Let R_{i}Q_{i−1}Q_{i+1}, i = 1, …, 6, be the equilateral triangles built on the small diagonals of the hexagon Q_{1}Q_{2} ⋯ Q_{6} all with the same orientation as equilateral triangles used in the construction of points Q_{1}, Q_{2}, …, Q_{6}. Let R_{0} = R_{6} and R_{7} = R_{1}. Let G_{i} be the centroid of the triangle R_{i−1}R_{i}R_{i+1}, i = 1, …, 6. Then F_{1}F_{2} ⋯ F_{6} and G_{1}G_{2} ⋯ G_{6} are regular hexagons whose centers coincide with the centroid of the triangle ABC. Moreover, the segments Q_{1}Q_{4}, Q_{2}Q_{5}, and Q_{3}Q_{6} have equal length ℓ, the segments R_{1}R_{4}, R_{2}R_{5}, and R_{3}R_{6} have equal length 2ℓ, |F_{1}F_{2}| = ℓ/3, and the hexagon G_{1}G_{2} ⋯ G_{6} is the image of the hexagon F_{1}F_{2} ⋯ F_{6} under the homothety with the center at the centroid of ABC and the stretching factor 2.

Observe that the above theorem includes two groups of related hexagons since there are two orientation classes. Recall that a triangle ABC is positively oriented provided we move counterclockwise when going from A to B and then to C and that it is negatively oriented when this journey is in the clockwise direction.

There are many triangles associated to the triangle ABC which share with it the centroid. For example, the complementary triangle A_{m}B_{m}C_{m} whose vertices are midpoints of sides, the anticomplementary triangle A_{a}B_{a}C_{a} whose vertices are intersections of parallels through vertices to opposite sides, the first Brocard triangle A_{b}B_{b}C_{b} of ABC, the triangles whose vertices are vertices (centers) of equilateral triangles built on sides either outwards or inwards, and the degenerate triangle GGG, where G is the centroid of ABC.

2. Proof of the Theorem

In the proof we shall use complex numbers. The point and its affix will be denoted by the same capital roman letter. The corresponding small roman letter denotes the complex conjugate of its affix.

For an expression Ex that depends on the set of variables S and a permutation b of S, let b(Ex) denote the result of substitution of variables given by b. For example, if Ex = 3uv^{5}w^{3} and b(u) = v, b(v) = w, and b(w) = u, then b(Ex) = 3u^{3}vw^{5}.

Let j, σ, and τ denote substitutions

{s → t, t → s, A → P, B → Q, C → R, P → A, Q → B, R → C},

{A → B, B → C, C → A, P → Q, Q → R, R → P, X → Y, Y → Z, Z → X},

and

{A → C, B → A, C → B, P → R, Q → P, R → Q, X → Z, Y → X, Z → Y }.

Then P_{1} is (Ast + Xs + P)m, where m = 1/(st + s + 1). Moreover, we have

P_{2} = j(P_{1}), P_{3} = σ(P_{1}), P_{4} = σ(P_{2}), P_{5} = τ(P_{1}), and P_{6} = τ(P_{2}).

Let η denote the complex number (−1 + i\sqrt{3})/2 (the cubic root of unit), where i denotes the imaginary unit (square root from −1).

Let n = 1/(st + t + 1). Let us assume that we are building negatively oriented equilateral triangles. The expressions for the positively oriented case are similar. The affixes of Q_{1} and Q_{2} are

[(A − P)s^{2}t^{2}η + As^{2}t^{2} + (X − P)s^{2}tη + (A − X)st^{2}η + Xs^{2}t + Ast^{2} + (A + P + X)st + (X − A)sη + (P − X)tη + Xs + P t + (P − A)η + P]mn,

and

[(P − B)s^{2}t^{2}η + P s^{2}t^{2} + (P − Y)s^{2}tη + (X − B)st^{2}η + P s^{2}t +

Xst^{2} + (A − B + P − Q + X − Y)stη + (A + P + X)st + (A − Y)sη + (X − Q)tη + As + Xt + (A − Q)η + A]mn,

while Q_{3} = σ(Q_{1}), Q_{4} = σ(Q_{2}), Q_{5} = τ(Q_{1}), and Q_{6} = τ(Q_{2}). Notice that Q_{1} and Q_{2} are in fact (1 + η)P_{1} − ηP_{2} and (1 + η)P_{2} − ηP_{3}.

Similarly, the affixes of R_{1} and R_{2} are

[(R − B)s^{2}t^{2}η + (A − B + P)s^{2}t^{2} + (R − Y)s^{2}tη + (Z − B)st^{2}η +

(P + X − Y)s^{2}t + (A − B + X)st^{2} + (C − B + R − Q + Z − Y)stη +

(2A − B + 2P − Q + 2X − Y)st + (C − Y)sη + (Z − Q)tη +
(A + X − Y)s + (P − Q + X)t + (C − Q)η + A + P − Q]mn,

and

[(A − Q)s^{2}t^{2}η + (B + P − Q)s^{2}t^{2} + (X − Q)s^{2}tη + (A − Y)st^{2}η +

(P − Q + Y)s^{2}t + (B + X − Y)st^{2} + (A − B + P − Q + X − Y)stη +

(A + P + X)st + (X − B)sη + (P − Y)tη + (A − B + Y)s +

(Q + X − Y)t + (P − B)η + A − B + Q]mn,

while R_{3} = σ(R_{1}), R_{4} = σ(R_{2}), R_{5} = τ(R_{1}), and R_{6} = τ(R_{2}). Notice that R_{1} and R_{2} are in fact (1 + η)Q_{6} − ηQ_{2} and (1 + η)Q_{1} − ηQ_{3}.

The affixes of F_{1} and F_{2} are

[(R − B)s^{2}t^{2}η + (A + P + R)s^{2}t^{2} + (R − Y)s^{2}tη + (Z − B)st^{2}η +

(P + R + X)s^{2}t + (A + X + Z)st^{2} + (C − B + R − Q + Z − Y)stη +

(2A + C + 2P + R + 2X + Z)st + (C − Y)sη + (Z − Q)tη +
(A + C + X)s + (P + X + Z)t + (C − Q)η + A + C + P]mn/3,

and

[(A − Q)s^{2}t^{2}η + (A + B + P)s^{2}t^{2} + (X − Q)s^{2}tη + (A − Y)st^{2}η +

(P + X + Y)s^{2}t + (A + B + X)st^{2} + (A − B + P − Q + X − Y)stη +

(2A + B + 2P + Q + 2X + Y)st + (X − B)sη + (P − Y)tη + (A + X + Y)s

+ (P + Q + X)t + (P − B)η + A + P + Q]mn/3, while F_{3} = σ(F_{1}), F_{4} = σ(F_{2}), F_{5} = τ(F_{1}), and F_{6} = τ(F_{2}). Notice that F_{1} and F_{2} are in fact equal to (Q_{6} + Q_{1} + Q_{2})/3 and (Q_{1} + Q_{2} + Q_{3})/3.

Finally, the affixes of G_{1} and G_{2} are

[(A − B + C + R − P − Q)s^{2}t^{2}η + (2A + P − Q + R)s^{2}t^{2} +

(R − P − Q + X − Y + Z)s^{2}tη + (A − B + C − X − Y + Z)st^{2}η +

(P − Q + R + 2X)s^{2}t + (2A + X − Y + Z)st^{2} +
2(C − B + R − Q + Z − Y)stη + (3A − B + C + 3P − Q + R + 3X − Y + Z)st
+ (C − A − B + X − Y + Z)sη + (P − Q + R − X − Y + Z)tη +
(A − B + C + 2X)s + (2P + X − Y + Z)t +
(C − A − B + P − Q + R)η + A − B + C + 2P]mn/3,

and

[(A − B − C + P − Q + R)s^{2}t^{2}η + (A + B − C + 2P)s^{2}t^{2} +

(P − Q + R + X − Y − Z)s^{2}tη + (A − B − C + X − Y + Z)st^{2}η +

(2P + X + Y − Z)s^{2}t + (A + B − C + 2X)st^{2} +
2(A − B + P − Q + X − Y)stη + (3A + B − C + 3P + Q − R + 3X + Y − Z)st
+ (A − B + C + X − Y − Z)sη + (P − Q − R + X − Y + Z)tη +
(2A + X + Y − Z)s + (P + Q − R + 2X)t +

(A − B + C + P − Q − R)η + 2A + P + Q − R]mn/3,

while G_{3} = σ(G_{1}), G_{4} = σ(G_{2}), G_{5} = τ(G_{1}), and G_{6} = τ(G_{2}). Again, observe that G_{1} and G_{2} are in fact equal to (R_{6} + R_{1} + R_{2})/3 and (R_{1} + R_{2} + R_{3})/3.

Since the affix of the centroid G of ABC is (A + B + C)/3, the affix of the centroid G_{p} of P QR is (P + Q + R)/3, and the affix of the centroid G_{x} of XY Z is (X + Y + Z)/3, we see that the assumption that G, G_{p}, and G_{x} coincide is equivalent to the assumption that P = A + B + C − Q − R and X = A + B + C − Y − Z. Let π denote the substitution

{P → A + B + C − Q − R, p → a + b + c − q − r, X → A + B + C − Y − Z, x → a + b + c − y − z}.

One can show easily by hand or by programs in symbolic computation (like Mathematica and Maple) that π(|Q_{1}Q_{4}|), π(|Q_{2}Q_{5}|), and π(|Q_{3}Q_{6}|) have equal value ℓ, that π(|GF_{i}|) and π(|F_{i}F_{i+1}|) all have the same value ℓ/3 and that G is the midpoint of segments π(F_{1}F_{4}), π(F_{2}F_{5}), and π(F_{3}F_{6}). Moreover, π(|R_{1}R_{4}|), π(|R_{2}R_{5}|), and π(|R_{3}R_{6}|) have equal value 2ℓ, π(|GG_{i}|) and π(|G_{i}G_{i+1}|) (i = 1, …, 6) all have the same value 2ℓ/3 and G is the midpoint of segments π(G_{1}G_{4}), π(G_{2}G_{5}), and π(G_{3}G_{6}), the points G, π(F_{i}), and π(G_{i}) (i = 1, …, 6) are collinear, and the lines π(F_{i}F_{i+1}) and π(G_{i}G_{i+1}) are parallel.

3. Appendix

In this appendix we shall provide details for the proof of the statements:

(a) π(|Q_{2}Q_{5}|^{2}) = ℓ^{2}, where ℓ = π(|Q_{1}Q_{4}|),
(b) π(|R_{1}R_{4}|^{2}) = 4ℓ^{2},
(c) π(|GG_{1}|^{2}) = 4ℓ^{2}/9, and
(d) π(|G_{1}G_{2}|^{2}) = 4ℓ^{2}/9.

In order to prove (a), note that |AX|^{2} = (A − X)(a − x). It follows that |Q_{1}Q_{4}|^{2} = kK, where K is

[(A + C − P − Q)s^{2}t^{2}η + (A − Q)s^{2}t^{2} + (X + Z − P − Q)s^{2}tη + (A + C − X − Y)st^{2}η + (X − Q)s^{2}t + (A − Y)st^{2} + (C − B + R − Q + Z − Y)stη

+ (A − B + P − Q + X − Y)st + (X + Z − A − B)sη + (P + R − X − Y)tη +

(X − B)s + (P − Y)t + (P + R − A − B)η + P − B]mn,

Similarly, |Q_{1}Q_{4}|^{2} = uU, where U is

[(P + R − B − C)s^{2}t^{2}η + (P − C)s^{2}t^{2} + (P + R − Y − Z)s^{2}tη +

(X + Z − B − C)st^{2}η + (P − Z)s^{2}t + (X − C)st^{2} + (A − B + P − Q + X − Y)stη + (A − C + P − R + X − Z)st + (A + C − Y − Z)sη

+ (X + Z − Q − R)tη + (A − Z)s + (X − R)t + (A + C − Q − R)η + A − R]mn,

If we now expand the products kK and uU, use the fact that η^{2} = −1 − η, and apply the subsitution π, we get that π(|Q_{1}Q_{4}|^{2} − |Q_{2}Q_{5}|^{2}) = 0.

The proofs for statements (b), (c), and (d) are similar. For example, we first compute |R_{1}R_{4}|^{2} = vV, where V is

[2(B − R)s^{2}t^{2}η + (B − A + C − P + Q − R)s^{2}t^{2} + 2(Y − R)s^{2}tη +

2(B − Z)st^{2}η + (Q − P − R − X + Y + Z)s^{2}t + (B − A + C − X + Y − Z)st^{2} + 2(B − C + Q − R + Y − Z)stη + 2(B − A + Q − P + Y − X)st + 2(Y − C)sη + 2(Q − Z)tη + (B−A−C −X +Y +Z)s+(Q−P +R−X +Y −Z)t+2(Q−C)η+B−A−C −P +Q+R]mn,

If we now expand the products kK and vV, use the fact that η^{2} = −1 − η, and apply the subsitution π, we get that π(4|Q_{1}Q_{4}|^{2} − |R_{1}R_{4}|^{2}) = 0.

References

[1] Deaux, R.: Introduction to the geometry of complex numbers. Ungar Publishing Co., New York 1956.

[2] Hahn, L.: Complex numbers and geometry. Mathematical Association of America, Washington 1994.

[3] Morley, F.; Morley, F. V.: Inversive Geometry. Chelsea Publ. Co., New York 1954.

[4] Schwerdtfeger, H.: Geometry of complex numbers. Oliver and Boyd, Toronto 1962.

[5] Yaglom, I. M.: Complex numbers in geometry. Academic Press, New York 1968.

Received August 30, 1996