[alogo] 1. Wallace

I'll narrate the story of the Deltoid starting with the Wallace line with respect to a triangle (more commonly known as Simson line, albeit Simson did nothing about that).


[1] The projections of a point P of the circumcircle of triangle ABC on its sides are aligned on a line W(P) (see Simson.html ).
[2] The orthopole PO of the tangent tP of P to the circumcircle of ABC is on the Wallace line W(P) (see Orthopole2.html ).
[3] As P varies on the circumcircle lines W(P) envelope an algebraic curve of degree four called Deltoid.
[4] The orthopole PO of tP is the contact point of W(P) with the deltoid.

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[alogo] 2. Deltoid

The fact that point PO generates the envelope of Wallace lines was discussed in Orthopole2.html .The discussion that follows will clarify assertion [3]. First commes the usual definition of the deltoid as an Hypocycloid (see Hypocycloid.html ).
In the most general case hypocycloids are curves generated by rolling a circle of radius b inside a circle of radius a > b.
Taking the center E of the fixed circle as origin, the x-axis in the direction of EA2 and the polar angle u = angle(A2EQ), where Q the center of the rolling circle, the parametric equations of a rolling point PO(x,y) (which for u=0 is at A2) are


The angle(P''QPO) = -(a/b)u.
In the case of the deltoid the radius of the rolling circle is 1/3 of the radius of the circle on which it rolls (a=3b), angle(P''QPO) = -3u = -3angle(A2EQ), and the equation becomes

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[alogo] 3. The mean or derivative equilateral

To start establishing the connexion of the deltoid with the triangle consider the equilateral mean or derivative triangle which is an equilateral triangle A1B1C1 with the same circumcircle as ABC and whose symmetry-axes are parallel to those of the deltoid.


This equilateral is defined as follows: (i) Let A0 be the intersection point of the circumcircle of ABC with the medial line of BC lying on the same side as A. Define {B0, C0} analogously. (ii) On the arc A0A take a point A1 such that the ratio of arcs A0A/A0A1 = 3. Define {B1, C1} analogously.
It is an easy exercise to show that triangle A1B1C1 is equilateral (see Derivative.html ).

[alogo] 4. Euler's intervention

Then commes Euler's circle and line (see Euler.html ) which are very closely related to the subject. Euler's circle has half the radius of the circumcircle and the line joining its center E with the circumcenter O is Euler's line of the triangle. Draw from E parallels to the symmetry-axes {OA1,OB1,OC1} of the equilateral derivative.

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[1] These parallels {EA2,EB2,EC2} are axes of symmetry of the deltoid. They coincide respectively with the Wallace lines {W(A1), W(B1), W(C1)} of the vertices of the equilateral derivative.
[2] The deltoid is tangent to the Euler circle.
[3] The cusps of the deltoid are on a circle of radius three times that of the Euler circle.

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[4] The triangle formed by the cusps of the deltoid and the equilateral derivative are two homothetic triangles. The homothety-center is a point H' lying on the Euler line OE of ABC and being symmetric to the orthocenter H of ABC with respect to the circumcenter O (H' is X(20) or the De Longchamps point of triangle ABC). The homothety-ratio is the homothety ratio of the circumscribing circles i.e. 3/2.

[alogo] 5. Rolling and enveloping

Next figure displays the connexion of the rolling circle to the enveloping property of the Wallace line of a certain point P of the circumcircle of triangle ABC. P is one intersection point with the circumcircle of a line OP parallel to line EQ which joins the centers of the fixed and the rolling circle.

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The figure is best analysed by starting with point P on the circumcircle of ABC. Its Wallace line W(P) intersects the Euler circle at a point P' which is the middle of the segment PH joining P to the orthocenter of ABC. This basic fact (see SteinerLine.html ) implies that taking the symmetric of the Euler circle with respect to P' we produce a second circle (the rolling one) with radius equal to the radius of the Euler circle and tangent to it at P'.
[1] The quadrilateral EHQP is a parallelogram since P' is common middle of its diagonals PH and EQ.
[2] Segment QP is always parallel to the Euler line OE and has fixed length equal to OH.
[3] When the Wallace line is W(P) = EA2, then line OP is parallel to EA2.
[4] Since two Wallace lines are inclined to each other by half the corresponding central angle of their defining points (see SimsonProperty2.html ) we have angle(A2EQ) = 2angle(W(P),A2E).
Here w = angle(W(P),A2E) denotes the angle of the two lines. It follows that angle(QE,W(P)) = 3w and consequently the central angle(P''QPO) = 6w i.e. angle(P''QPO) = -3angle(A2EQ). This shows that the circle constructed through the symmetry on P' coincides with the rolling one.

[alogo] 6. Dependence on angles

Next figure proves the initial claim that the orthopole PO of the tangent tP is the point describing the deltoid as the moving circle rolls into the fixed one. This is simply done by calculating the distance of the Wallace line from E which is r*sin(3u), r being the radius of the Euler circle and u being the angle between the fixed line EA2 and W(P).

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When W(P) obtains the position orthogonal to tP i.e.parallel to EQ, then u has been augmented to 2u+pi, hence the distance of the corresponding Wallace line has become r*sin(3(2u+pi)). This is equal to r*(6u-pi) which gives the distance of PO from EQ. This shows that PO is on the Wallace line which is orthogonal to tP. But the orthopole of tP lies on W(P) (see Orthopole2.html ) and also on the Wallace line which is orthogonal to tP (see Orthopole.html ). Thus it coincides with PO.

[alogo] 7. Other symmetries

There are lots of relations hidden in the figure studied so far. Below I record some of them resulting by most simple arguments (which I ommit since they are trivial or discussed in Orthopole2.html ).

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[1] The symmetric P2 of the orthopole PO of tP with respect to P' is on the Euler circle.
[2] P2P3 is the Wallace line of the antipodal P7 of P, intersecting W(P) orthogonally at P2.
[3] P1 is the orthopole of W(P) and lies on the Steiner line S(P) of P. The distance P1P2 is equal to the distance of P from W(P).
[4] P3 is the symmetric of P' with respect to E. The symmetric of P2 with respect to P3 is point P4 on the Wallace line orthogonal to W(P) and also the orthopole of the tangent at the antipodal P7 of P.
[5] The orthopoles PO and P4 of tangents at antipodal points of the circumcircle are on the Simson line POP4 which is parallel to the diameter defined by the antipodal points PP7 and also to the diameter of the Euler circle P'P3.
[6] P6 is the point on the circumcircle whose Wallace line coincides with line POP4. It is the other intersection point of the circumcircle with the orthogonal from P on W(P).
[7] Thus P on the circumcircle determines a unique right-angled triangle P0P2P4 whose sides are tangent to the deltoid.
[8] The right-angled triangles PP6P7 and P'P2P3 are antihomothetic with respect to the centroid G and ratio 2:1.
[9] Since, by appropriate selection of P, POP4 can obtain the position of any tangent to the deltoid it follows that the tangents to the deltoid at the intersection points of a tangent with the deltoid intersect at a right angle on a point of the Euler circle and define a diameter of this circle parallel to the tangent.
[10] Point P2 coincides with the orthopole of diameter PP7 consequently is on the Wallace line which is orthogonal to this diameter (and also on the Euler circle).
[11] Thus POP4 has the length of a diameter of the circumcircle. Since this is an arbitrary tangent to the deltoid we deduce that the segments cut off by the deltoid on its tangents have all the length of a diameter of the circumcircle (see Orthopole2.html ).
[12] The middles of segments POP4 are on the Euler circle. Thus all right-angled triangles P0P2P4 inscribed in the deltoid have the same Euler circle.
[13] The third Wallace line through P2 is the altitude of the right-angled triangle P0P2P4. This is the Wallace line W(H0) of a point H0 which is symmetric to the orthocenter with respect to P2.
[14] PH0P7 is the STriangle corresponding to point P2 and the three tangents to the deltoid through that point (see S_Triangles.html ). It is the parallel translate of P0P2P4 by the vector HP2.

[alogo] 8. Particular positions

[1] P is antipodal of C. Then W(P) coincides with side AB, point P7 is C, P2 obtains the position of the foot of the altitude from C, P' takes the position of the middle of AB and PO, the contact point of side AB with the deltoid is symmetric to P2 with respect to the middle of side AB.
[2] P coincides with C. Then W(P) coincides with the altitude to AB, points {P1,P2} coincide with the foot of the altitude to AB and the contact point PO of the altitude with the deltoid is twice the distance of this foot from the other intersection of this altitude with the Euler circle.

[alogo] 9. References

[1] Butchart, J. H. The Deltoid Regarded as the Envelope of Simson LinesThe American Mathematical Monthly, Vol. 46, No. 2. (Feb., 1939), pp. 85-86.
[2] Gallatly William The modern geometry of the triangle London, Francis Hodgson 1913, p.49.
[3] Honsberger, R. Episodes in Nineteenth and Twentieth Century Euclidean Geometry. Washington DC, Math. Assoc. Ammer., 1995, pp. 106-110.
[4] Goormaghtigh, R. On Some Loci Connected with the Orthopole-GeometryThe American Mathematical Monthly, Vol. 37, No. 7. (Aug. - Sep., 1930), pp. 370-371.
[5] Karl Cordia Mary The Projective Theory of Orthopoles The American Mathematical Monthly, Vol. 39, No. 6, (Jun. - Jul., 1932), pp. 327-338
[6] Ramler, J. O. The Orthopole Loci of Some One-Parameter Systems of Lines Referred to a Fixed Triangle The American Mathematical Monthly, Vol. 37, No. 3, (Mar., 1930), pp. 130-136
[7] Van Horn, C. E. The Simson Quartic of a Triangle The American Mathematical Monthly, Vol. 45, No. 7. (Aug. - Sep., 1938), pp. 434-437.

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