ON PROPELLERS FROM TRIANGLES

ZVONKO ČERIN

Abstract. We prove several improvements and analogs of Leon Bankoff’s theorem on asymmetric propellers from directly similar triangles.

1. Propellers from directly similar triangles

Figure 1. Asymmetric propeller from directly similar triangles determines in every position of blades directly similar triangle UV W.

The propeller from triangles has a triangle ABC as a shaft and three triangles AP X, Y BQ, and RZC as blades. There are some interesting results on propellers from triangles which show that regardless of the way in which the blades are glued to the shaft the triangle UV W on the midpoints of segments Y R, ZP, and XQ shares certain properties with the shaft and the blades. For example, if the shaft and the blades are equilateral triangles, then UV W is also an equilateral triangle. This equilateral propeller theorem and its special cases and extensions have long history and numerous contributions. Many of them are collected in section 2 of the excellent survey [8]. From it we recall papers by A. Hofmann [6] referring to the historical origin of the propeller theorem, by W. Götz [4] showing a subcase of Bankoff’s theorem which corresponds with a joint figure from kinematics already known in the seventies of the 19th century and connected with names like W. Clifford and A. Cayley, and (regarding the interesting variety of proof methods) the papers [12], [7], [9], and [11].(The latter paper shows the connection between the famous three-circles theorem and propellers, see also Math. Reviews 81 c:51010.)

Another example is the following improvement of the equilateral propeller theorem by Leon Bankoff [3] where the shaft and the blades are directly similar (see the above Figure 1). Its proof in [3] is geometric while our proof below uses complex numbers and is algebraic.

Theorem 1. Let ABC, P QR, and XY Z be triangles such that ABC, AP X, Y BQ, and RZC are directly similar. Then the triangle UV W on midpoints of segments Y R, ZP, and XQ is also directly similar to ABC.

Proof. Let us use the convention that a point with a capital Latin letter as a label is represented with the complex number (its affix) which is denoted with the same small Latin letter. The complex conjugate of p is p̄ and P̄ (the reflection of P in the real axis) is a point whose affix is p̄.

It is well-known (see [5, p.57]) that triangles ABC and P QR are directly similar if and only if the determinant D(ABC, P QR) of the matrix with the first row a, b, c, the second row p, q, r, and the third row 1, 1, 1 is zero.

Hence, if ABC and AP X are directly similar, then D(ABC, AP X) = 0 so that we can solve for x to get x = −(d_{b} p + a d_{a})/d_{c}, where d_{a} = b − c, d_{b} = c − a, and d_{c} = a − b. Similarly, y = −(d_{c} q + b d_{b})/d_{a} and z = −(d_{a} r + c d_{c})/d_{b}.

The determinant D(ABC, UV W) is (D_{1} − D_{2} − D_{3})/2, where

D_{1} = |\begin{matrix}a & b & c \\ r & p & q \\ 1 & 1 & 1 \end{matrix}| ,  D_{2} = |\begin{matrix}a & b & c \\ \frac{q d_{c}}{d_{a}} & \frac{r d_{a}}{d_{b}} & \frac{p d_{b}}{d_{c}} \\ 1 & 1 & 1 \end{matrix} | ,  D_{3} = |\begin{matrix}a & b & c \\ \frac{b d_{b}}{d_{a}} & \frac{c d_{c}}{d_{b}} & \frac{a d_{a}}{d_{c}} \\ 1 & 1 & 1 \end{matrix} | .

One can easily check that D_{1} = D_{2} and D_{3} = 0. Hence, D(ABC, UV W) = 0 so that triangles ABC and UV W are directly similar. □

From the above theorem we can get many related results if we observe that for directly similar triangles ABC and P QR and for any real number λ the triangle λ_{A}^{P}λ_{B}^{Q}λ_{C}^{R} is also directly similar to ABC, where λ_{A}^{P} is A for A = P and for A ≠ P and λ = −1 it is P while for A ≠ P and λ ≠ −1 it is the unique point X on the line AP such that the quotient |AX|/|XP | of oriented distances is λ. Since the affix of λ_{A}^{P} is (a + λ p)/(λ + 1) (for λ ≠ −1), the above claim follows from the fact that the determinant is a linear functional (of rows). In particular, from this we immediately get the following result.

Corollary 1. Let ABC, P QR, and XY Z be triangles such that ABC, AP X, Y BQ, and RZC are directly similar. Then the triangle DEF on centroids of triangles ARY, BP Z, and CQX is directly similar to ABC.

Proof. Since the centroids D, E, F divide segments AU, BV, CW in ratio 2, we can use Theorem 1 and the above observation for λ = 2 and triangles ABC and UV W. □

Another supply of directly similar to ABC triangles in Bankoff’s propellers is offered by the following lemma.

Lemma 1. Let ABC, P QR, and XY Z be triangles such that P X, QY, RZ have equal oriented length and are parallel to segments joining A, B, and C with the circumcentre O of ABC. Then the triangles ABC and XY Z are directly similar if and only if the triangles ABC and P QR are directly similar.

Proof. Let the common oriented length of segments P X, QY, and RZ be a real number h. We can assume that the unit circle of all unimodular complex numbers is the circumcircle of ABC. Then x = p + h a, y = q + h b, and z = r + h c. The linearity of the determinant with respect to rows implies that D(ABC, P QR) = 0 if and only if D(ABC, XY Z) = 0. □

Corollary 2. With the notation and the assumptions of Theorem 1, the triangle DEF on endpoints of segments at midpoints U, V, and W of equal oriented length and parallel to segments joining A, B, and C with the circumcentre O of ABC is directly similar to ABC.

2. Rotating blades in Bankoff’s propellers

In this section we explore what happens with the triangle UV W while points P, Q, and R rotate with constant angular speed on circles of equal radius with centres at A, B, and C, respectively.

Theorem 2. Let points P, Q, and R rotate with the constant angular speed φ around the vertices of ABC on circles of radius k. Let X, Y, and Z be points such that triangles AP X, Y BQ, and RZC are directly similar to ABC. The midpoints U, V, and W of segments Y R, ZP, and XQ rotate with the angular speed φ on circles with centres at midpoints A_{m}, B_{m}, and C_{m} of BC, CA, and AB and with radii k times \frac{|BB_{m}|}{|BC|}, \frac{|CC_{m}|}{|CA|}, and \frac{|AA_{m}|}{|AB|}, respectively.

Proof. We again assume that the unit circle of all unimodular complex numbers is the circumcircle of ABC. Then p = a + k e^{i φ}, q = b + k e^{i φ}, and r = c + k e^{i φ}. It follows that x = a − k e^{i φ} d_{b}/d_{c}, y = b − k e^{i φ} d_{c}/d_{a}, and z = c − k e^{i φ} d_{a}/d_{b}, so that u = (b + c)/2 + k e^{i φ} (d_{a} − d_{c})/(2 d_{a}). From this the claim about the rotation of the point U follows easily. The points V and W are treated similarly. □

We shall use S to denote the area of ABC and ℓ_{a}, ℓ_{b}, and ℓ_{c} for lengths of its sides. The symmetric functions of ℓ_{a}, ℓ_{b}, and ℓ_{c} we denote as follows.

s = ℓ_{a} + ℓ_{b} + ℓ_{c}, m = ℓ_{a} ℓ_{b} ℓ_{c}, t = ℓ_{b} ℓ_{c} + ℓ_{c} ℓ_{a} + ℓ_{a} ℓ_{b}, s_{a} = −ℓ_{a} + ℓ_{b} + ℓ_{c},

s_{b} = ℓ_{a} − ℓ_{b} + ℓ_{c}, s_{c} = ℓ_{a} + ℓ_{b} − ℓ_{c}, m_{a} = ℓ_{b} ℓ_{c}, m_{b} = ℓ_{c} ℓ_{a}, m_{c} = ℓ_{a} ℓ_{b},

r_{a} = ℓ_{b}−ℓ_{c}, r_{b} = ℓ_{c}−ℓ_{a}, r_{c} = ℓ_{a}−ℓ_{b}, z_{a} = ℓ_{b}+ℓ_{c}, z_{b} = ℓ_{c}+ℓ_{a}, z_{c} = ℓ_{a}+ℓ_{b}.

For each k ≥ 2, s_{k}, s_{ka}, s_{kb}, and s_{kc} are derived from s, s_{a}, s_{b}, and s_{c} with the substitution ℓ_{a} = ℓ_{a}^{k}, ℓ_{b} = ℓ_{b}^{k}, ℓ_{c} = ℓ_{c}^{k}. In a similar fashion we can define analogous expressions using letters m, r, t, and z.

Theorem 3. Under the assumptions of Theorem 2,

• The circumcentre of UV W rotates with the angular speed φ on the circle with the centre at the centre of the nine-point circle of ABC and with the radius k times \frac{\sqrt{2 s_{4}−5 t_{2}}}{8 S}.

• The centroid of UV W rotates with the angular speed φ on the circle with the centre at the centroid of ABC and with the radius k times \frac{\sqrt{21 m_{2}+5 g_{1}−s_{6}−3 g_{2}}}{6 m}, where g_{1} = g_{4,2}^{a,c} = ℓ_{a}^{4} ℓ_{c}^{2} + ℓ_{c}^{4} ℓ_{b}^{2} + ℓ_{b}^{4} ℓ_{a}^{2} and g_{2} = g_{4,2}^{a,b}.

• The orthocentre of UV W rotates with the angular speed φ on the circle with the centre at the circumcentre of ABC and with the radius k times \frac{\sqrt{g}}{8 S m}, where g = g_{10}^{a} − g_{8,2}^{a,b} − 4 g_{8,2}^{a,c} − 3 g_{6,4}^{a,b} + 6 g_{6,4}^{a,c} + g_{6,2,2}^{a,b,c} + 3 g_{4,4,2}^{a,b,c}.

Proof. While it is possible to prove this theorem using complex numbers, with the help of computers, it is simpler to apply Cartesian coordinates because in this approach it is easier to identify the radii of revolving circles.

Here is an outline of the key steps in the proof for the circumcentre of UV W. Let A(0, 0), B(0, h (f + g)), and C(\frac{(f^{2}−1) g h}{f g−1}, \frac{2 f g h}{f g−1}), where h is the inradius of ABC and f and g are cotangents of angles A/2 and B/2. The coordinates of P are \frac{k (1−ψ^{2})}{1+ψ^{2}} and \frac{2 k ψ}{1+ψ^{2}} , where ψ = tan \frac{φ}{2}. In coordinates of Q and R we must add coordinates of B and C to these values.

In the next step we determine the coordinates of points X, Y, and Z and then U, V, and W. Let T be the circumcentre of UV W with coordinates T_{x} and T_{y}. When we eliminate the parameter ψ from equalities x = T_{x} and y = T_{y} the equation of the required locus will emerge. This equation has the form (x − F_{x})^{2} + (y − F_{y})^{2} − r_{O}^{2} = 0, where F_{x} and F_{y} are coordinates of the centre F of the nine-point circle of ABC and r_{O} is an expression in terms of f, g, h, and k which has the above form when we apply equalities

f = \frac{z_{a}^{2} − ℓ_{a}^{2}}{4 S} , g = \frac{z_{b}^{2} − ℓ_{b}^{2}}{4 S} , h = \frac{2 S}{s} .

□

3. Propellers from reversely similar and orthologic triangles

Recall that triangles ABC and P QR are reversely similar provided triangles ABC and P̄Q̄R̄ are directly similar. In other words, D(ABC, P̄Q̄R̄) = 0 is a necessary and sufficient condition for ABC and P QR to be reversely similar. In the next theorem we shall show that reverse similarity is equivalent with two interesting geometric properties of triangles known as paralogy and orthology whose definitions are as follows.

Triangles ABC and XY Z are orthologic provided the perpendiculars from the vertices of ABC on the sides Y Z, ZX, and XY of XY Z are concurrent. The point of concurrence of these perpendiculars is denoted by [ABC, XY Z]. It is well-known (see [2] or [10]) that the relation of orthology for triangles is reflexive and symmetric. Hence, the perpendiculars from the vertices of XY Z on the sides BC, CA, and AB of ABC are concurrent at the point [XY Z, ABC].

Lemma 2. Triangles XY Z and P QR are orthologic if and only if

O(XY Z, P QR) = x(q̄−r̄)+x̄(q −r)+y(r̄−p̄)+ȳ(r−p)+z(p̄−q̄)+z̄(p−q) = 0.

Proof. The line QR is [q̄ − r̄, r − q, q r̄ − q̄r] so that the perpendicular per_{QR}^{X} through X onto QR is the line [q̄ − r̄, q − r, x (r̄ − q̄) + x̄ (r − q)]. The perpendiculars per_{RP}^{Y} and per_{P Q}^{Z} through Y and Z onto RP and P Q are relatives of per_{QR}^{X}. These three perpendiculars are concurrent if and only if Θ = 0, where Θ denotes the determinant

| \begin{matrix}q̄ − r̄ & q − r & x (r̄ − q̄) + x̄ (r − q) \\ r̄ − p̄ & r − p & y (p̄ − r̄) + ȳ(p − r) \\ p̄ − q̄ & p − q & z (q̄ − p̄) + z̄(q − p) \end{matrix} | .

But, Θ = O(XY Z, P QR) Ω, where Ω = p (q̄ − r̄) + q (r̄ − p̄) + r (p̄ − q̄). Since Ω = 0 if and only if points P, Q, and R are collinear (and our assumptions exclude this possibility), we conclude that the triangles XY Z and P QR are orthologic if and only if O(XY Z, P QR) = 0. □

Triangles ABC and XY Z are paralogic provided the parallels through the vertices of ABC with the sides Y Z, ZX, and XY of XY Z are concurrent. The point of concurrence of these parallels is denoted by ∥ABC, XY Z∥. The relation of paralogy for triangles is symmetric so that the parallels through the vertices of XY Z with the sides BC, CA, and AB of ABC are concurrent at the point ∥XY Z, ABC∥.

Lemma 3. Triangles XY Z and P QR are paralogic if and only if

P (XY Z, P QR) = x(q̄−r̄)−x̄(q −r)+y(r̄−p̄)−ȳ(r−p)+z(p̄−q̄)−z̄(p−q) = 0.

Proof. The line QR is [q̄ − r̄, r − q, q r̄ − q̄r] so that the parallel par_{QR}^{X} through X with QR is the line [q̄ − r̄, r − q, x (r̄ − q̄) − x̄ (r − q)]. The parallels par_{RP}^{Y} and par_{P Q}^{Z} through Y and Z with RP and P Q are relatives of par_{QR}^{X}. The rest of the proof is the same as the proof of Lemma 2. □

Theorem 4. Triangles ABC and XY Z are reversely similar if and only if they are both paralogic and orthologic.

Proof. Let Δ = D(XY Z, P̄Q̄R̄), Ω = O(XY Z, P QR), and Π = P (XY Z, P QR). Since Π + Ω = 2 Δ and Π − Ω = −2 Δ̄, we see that Δ = 0 if and only if Π = 0 and Ω = 0. □

Theorem 5. Let ABC, P QR, and XY Z be triangles such that AP X, Y BQ, and RZC are reversely similar to ABC. Then the triangle UV W on midpoints of segments Y R, ZP, and XQ is orthologic to ABC and never paralogic to it.

Proof. Since ABC and AP X are reversely similar, D(ABC, ĀP̄X̄) = 0 so that we can solve for x to get x = −(d̄_{b} p + a d̄_{a})/d̄_{c}. Similarly, y = −(d̄_{c} q + b d̄_{b})/d̄_{a} and z = −(d̄_{a} r + c d̄_{c})/d̄_{b}.

We can now easily compute affixes of midpoints U, V, and W and verify that O(ABC, UV W) = 0. On the other hand, P (ABC, UV W) is equal (up to a sign) to Λ = ā d_{a} + b̄ d_{b} + c̄d_{c}, where Λ = 0 is the condition for points A, B, and C to be collinear. □

With a similar argument, the above theorem could be slightly improved to the following statement: If ABC, P QR, and XY Z are triangles such that AP X, Y BQ, and RZC are orthologic to ABC, then the triangle UV W on midpoints of segments Y R, ZP, and XQ is also orthologic to ABC. On the other hand, we have: If ABC, P QR, and XY Z are triangles such that AP X, Y BQ, and RZC are paralogic to ABC, then the triangle UV W on midpoints of segments Y R, ZP, and XQ is never paralogic to ABC.

The method employed in the proof of Corollary 1 implies the following.

Corollary 3. Let ABC, DEF, P QR, and XY Z be triangles such that AP X, Y BQ, and RZC are orthologic to ABC. Then the triangle UV W on centroids of triangles DY R, P EZ, and XQF is orthologic to ABC if and only if DEF is orthologic to ABC.

Our last theorem on propellers with the blades orthologic to the shaft is similar to Corollary 2 and gives a large supply of triangles orthologic to the shaft ABC. For an efficient formulation of this result we need the following notation and definitions.

For an expression ε in terms of ℓ_{a}, ℓ_{b}, and ℓ_{c} and a real number h, let ε[h] denote the triple (h ε, h φ(ε), h ψ(ε)). More precisely, the coordinates ε[h]_{1}, ε[h]_{2}, ε[h]_{3} of ε[h] are products with h of ε, the first cyclic permutation φ(ε) of ε, and the second cyclic permutation ψ(ε) of ε, respectively. For example, a[h] = (h a, h b, h c) and s_{a}[h] = (h s_{a}, h s_{b}, h s_{c}).

Let T_{i}, for i = 1, …, 13, denote the following expressions. 1, ℓ_{a}^{2}, s_{2a}, ℓ_{a}/z_{a}, 1/z_{a}, 3 s_{2a} + 4 \sqrt{3} S, 3 s_{2a} − 4 \sqrt{3} S, s_{2a} + 4 \sqrt{3} S, s_{2a} − 4 \sqrt{3} S, z_{2a}, ℓ_{a}^{2} (z_{2a}^{2} − m_{a}^{2}), (r_{2a}^{2} − ℓ_{a}^{2} z_{2a})(s_{2}^{2} − 2 m_{a}^{2}), z_{4a} − ℓ_{a}^{2} z_{2a}.

For a triple h = (s_{1}, s_{2}, s_{3}) of real numbers and for triangles ABC and XY Z, let [ABC, XY Z, h] denote the triangle UV W such that the segments XU, Y V, ZW are parallel to lines AO, BO, CO joining the circumcentre O of ABC with its vertices and the directed distances |XU|, |Y V |, |ZW | are equal to s_{1}, s_{2}, s_{3}, respectively. When s_{1} = 0, we put U = X, and we do similar assignments when s_{2} and s_{3} are zero. For s_{1} > 0 the vector X⃗U points towards outside of ABC while for s_{1} < 0 it points towards inside.

Theorem 6. Let ABC, P QR, and XY Z be triangles such that triangles AP X, Y BQ, and RZC are orthologic to ABC. Let U, V, and W be midpoints of segments Y R, ZP, and XQ. Then for every real number h and every integer i from 1 to 13 the triangle [ABC, UV W, T_{i}[h]] is also orthologic to ABC.

Proof. We shall give proof only for i = 2 because the procedure for other values of i is similar.

Let us assume again that the circumcircle of ABC is the set of all unimodular complex numbers. Since ABC and AP X are orthologic, O(ABC, AP X) = 0 so that we can solve for x to get that x is any complex number whose complex conjugate is

\frac{x}{a b} + \frac{(c − a)(p − c a p̄)}{c a (a − b)} + \frac{(b − c)(a^{2} − b c)}{a b c (a − b)}.

Similarly, y and z are any complex numbers whose complex conjugates are related to themselves by relations that are cyclic relatives of the above relation between x̄ and x.

We can now easily find that the affixes of the vertices of [ABC, UV W, T_{2}[h]] are

\frac{y + r}{2} + \frac{h a (b − c)^{2}}{b c} ,  \frac{z + p}{2} + \frac{h b (c − a)^{2}}{c a} ,  \frac{x + q}{2} + \frac{h c (a − b)^{2}}{a b} ,

and verify that O(ABC, [ABC, UV W, T_{2}[h]]) = 0. □

4. Propellers from isocentric triangles

In the rest of this paper we consider propellers in which the blades share centroids with the shaft. Once again this property is retained by the triangle UV W.

Recall that triangles ABC and P QR are isocentric provided they have the same centroid. For example, ABC and its Brocard triangle A_{b}B_{b}C_{b} are isocentric.

Theorem 7. Let ABC, P QR, and XY Z be triangles such that triangles ABC, AP X, Y BQ, and RZC are isocentric. Then the triangle UV W on midpoints of segments Y R, ZP, and XQ is isocentric with ABC.

Figure 2. For a propeller from isocentric triangles the triangle UV W is also isocentric with ABC.

Proof. The affixes of the centroids of ABC and AP X are (a + b + c)/3 and (a + p + x)/3. They will coincide if and only if x = b + c − p. In the same way we get that the affixes of Y and Z are y = c + a − q and z = a + b − r. Then u = (c + a − q + r)/2, v = (a + b − r + p)/2, and w = (b + c − p + q)/2. Hence, ABC and UV W are isocentric. □

The following theorem is analogous to Corollaries 1 and 3.

Theorem 8. Let ABC, DEF, P QR, and XY Z be triangles such that ABC, AP X, Y BQ, and RZC are isocentric. Then the triangle UV W on centroids of triangles DY R, P EZ, and XQF is isocentric with ABC if and only if DEF is isocentric with ABC.

Proof. Let U, V, and W be the centroids of triangles DY R, P EZ, and XQF. Since x = b + c − p, y = c + a − q, and z = a + b − r it follows that

u = \frac{d + c + a − q + r}{3}, v = \frac{e + a + b − r + p}{3}, w = \frac{f + b + c − p + q}{3} .

The centroid G of UV W has affix (d + e + f + 2(a + b + c))/9. This complex number will be (a + b + c)/3 (the affix of the centroid of ABC) if and only if d + e + f = a + b + c which holds if and only if ABC and DEF are isocentric.

□

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