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Simson line

In geometry, the Simson line (also known as the Wallace-Simson line) of a point P on the circumcircle of a triangle ABC is the straight line containing the feet of the perpendiculars dropped from P to the lines containing the sides BC, CA, and AB.^[1] This collinearity of the feet, known as Simson's theorem, holds if and only if P lies on the circumcircle of ABC.^[2] The concept is attributed to the Scottish mathematician Robert Simson (1687–1768), though it does not appear in his published works and was likely first proven by William Wallace in 1797 or 1799.^[3]^[2] The line arises as a degeneration of the pedal triangle of P with respect to ABC, which collapses into a line precisely when P is on the circumcircle.^[3] Key properties of the Simson line include its role in bisecting the segment from P to the orthocenter H of ABC, with the midpoint of PH lying on the nine-point circle.^[1] The angle between the Simson lines of two points on the circumcircle equals half the angular measure of the arc between them, and the Simson lines of diametrically opposite points are perpendicular and intersect on the nine-point circle.^[3] These features connect the Simson line to broader Euclidean geometry, including the Euler line and isogonal conjugates, and it generalizes to oblique versions via Carnot's theorem for points not on the circumcircle.^[3] The envelope of all Simson lines for a fixed triangle forms a deltoid curve with area half that of the circumcircle.^[1]

Definition and History

Definition

In geometry, the circumcircle of a triangle ABC is the unique circle that passes through all three vertices A, B, and C.^[4] Given a triangle ABC and its circumcircle, consider a point P lying on this circumcircle. From P, drop perpendiculars to the lines containing the sides BC, CA, and AB (or their extensions if necessary), letting the feet of these perpendiculars be D, E, and F respectively.^[1]^[3] The points D, E, and F are collinear, and the line passing through them is known as the Simson line of the point P with respect to triangle ABC. This collinearity holds precisely because P lies on the circumcircle; for points P not on the circumcircle, the feet generally form a triangle rather than degenerating to a line.^[1]^[3]^[5] As P traverses the circumcircle, the Simson line rotates around the triangle.^[5]^[2]

History

The Simson line was first proved by the Scottish mathematician William Wallace in 1799, marking a key contribution to triangle geometry during a period of renewed interest in synthetic methods. Wallace, born in 1768, published his result in the Mathematical Repository, demonstrating the collinearity of the feet of the perpendiculars from a point on a triangle's circumcircle to its sides.^[6]^[7] The line is named after Robert Simson (1687–1768), another Scottish mathematician renowned for restoring ancient Greek geometric texts and exploring porisms and triangle loci. Simson conducted investigations into triangle properties around 1750 but did not publish a proof of the theorem, and it does not appear in his known works, including his posthumous collection Opera Quaedam Reliqua, edited and published in 1776.^[8]^[9] The attribution to Simson gained prominence through Jean-Victor Poncelet in the early 19th century, leading to ongoing debates and the alternative designation Wallace-Simson line to acknowledge Wallace's priority.^[1]^[7] This development occurred amid the golden age of triangle geometry in the 18th and 19th centuries, paralleling advances in circumcircle properties and the orthocenter by figures like Leonhard Euler and Joseph-Louis Lagrange.^[8]

Properties

Basic Properties

One fundamental property of the Simson line arises in special cases when the point P coincides with a vertex of the triangle \triangle ABC. If P is at vertex A, the feet of the perpendiculars from A to the sides BC, AB, and AC degenerate such that the Simson line coincides with the altitude from A to side BC.^[2]^[10] Similar degenerations occur for points P at vertices B or C, yielding the respective altitudes from those vertices.^[2] The Simson line also exhibits a notable relation to the orthocenter H of \triangle ABC. Specifically, the Simson line of P bisects the line segment PH, with the midpoint M of PH lying on the Simson line itself.^[11]^[12] This midpoint M additionally resides on the nine-point circle of \triangle ABC.^[11] Another distinctive case occurs when P is the point on the circumcircle diametrically opposite to a vertex, say opposite to A. In this configuration, the Simson line of P coincides with the side BC opposite to A.^[13]^[10] Analogous results hold for the points diametrically opposite to B or C, yielding sides AC or AB, respectively.^[13] The Simson line is unique for each point P on the circumcircle of \triangle ABC, as the collinearity of the feet of the perpendiculars from P to the sides holds precisely when P lies on this circumcircle.^[14] If P is not on the circumcircle, the three feet are generally not collinear, and the pedal triangle remains non-degenerate.^[14]

Advanced Properties

One advanced property of the Simson line concerns the angle between the Simson lines of two distinct points P and Q on the circumcircle of triangle ABC. The angle between these two Simson lines is equal to half the measure of the arc PQ. This relation arises from the rotational symmetry of the Simson lines as P and Q vary on the circumcircle, reflecting the inscribed angle theorem in the circumcircle.^[15] Another key metric property involves the position of the Simson line relative to the orthocenter H of triangle ABC. The Simson line of P intersects the line segment PH at its midpoint M, which lies at a distance of half the length of PH from H. This bisection property highlights the Simson line's role in connecting the orthocenter and points on the circumcircle, with M also residing on the nine-point circle of ABC.^[2] As the point P varies around the circumcircle, the family of Simson lines envelopes a three-cusped hypocycloid known as the Steiner deltoid. This curve circumscribes the nine-point circle of triangle ABC as its incircle, with the cusps of the deltoid located at the midpoints of the sides of ABC. The Steiner deltoid has an area equal to half that of the circumcircle and serves as the boundary tangent to all Simson lines.^[16] The Simson line of P is also the trilinear polar of P with respect to triangle ABC, meaning it is the line joining the points of contact of the tangents from P to the circumcircle of the tangential triangle formed by the lines through the vertices perpendicular to the sides.^[17] This identification links the Simson line to projective geometry, where the trilinear polar represents the polar line of P under the trilinear polarity defined by ABC.^[18] Finally, the Simson line of P is tangent to the nine-point circle of ABC at the midpoint M of PH. This tangency ensures that the Simson line intersects the nine-point circle precisely at M, underscoring its geometric affinity with the Euler points and midpoints defining the circle.^[19]

Mathematical Formulation

Equation

In coordinate geometry, the Simson line of a point P with respect to a triangle ABC can be expressed algebraically using complex numbers to represent the points in the plane. The vertices A, B, C and the point P on the circumcircle are denoted by complex numbers a, b, c, and p, respectively.^[20] For simplicity, assume the circumcircle is the unit circle centered at the origin, so |a| = |b| = |c| = |p| = 1; this setup leverages the symmetry in a, b, c and can be generalized to arbitrary positioning via affine transformations. The explicit equation of the Simson line is

$2abc \bar{z} - 2pz + p^2 + (a + b + c)p - (ab + bc + ca) - \frac{abc}{p} = 0,

where z is the complex coordinate of a point on the line.^[20] Alternative representations include the parametric form in terms of the feet D, E, F of the perpendiculars from P to the lines BC, CA, AB, respectively: since D, E, F lie on the Simson line, it can be parametrized as \mathbf{r}(t) = \mathbf{D} + t (\mathbf{E} - \mathbf{D}) for real t, where boldface denotes position vectors.^[1] In vector form, the Simson line is the affine span of these projection points onto the sides of the triangle.^[1]

Derivation

To derive the equation of the Simson line in the complex plane, assume without loss of generality that the circumcircle of triangle ABC is the unit circle centered at the origin, with vertices represented by complex numbers a, b, c and point P by p, all satisfying |a| = |b| = |c| = |p| = 1 and thus \bar{a} = 1/a, \bar{b} = 1/b, etc..^[21] The foot of the perpendicular from p to side BC (the line through b and c) is the point d on that line minimizing the distance to p. For points v and w on the unit circle, this projection is given by the formula

d = \frac{1}{2} \left( b + c + p - bc \bar{p} \right).

Since |p| = 1, \bar{p} = 1/p, simplifying to

d = \frac{1}{2} \left( b + c + p - \frac{bc}{p} \right).

Analogously, the foot e from p to side CA (through c and a) is

e = \frac{1}{2} \left( c + a + p - \frac{ca}{p} \right),

and the foot f to side AB (through a and b) is

f = \frac{1}{2} \left( a + b + p - \frac{ab}{p} \right).

These formulas arise from the vector projection in the complex plane, leveraging the unit circle condition to express the real parameter t in the line parametrization b + t(c - b) (with t real) via dot products involving conjugates, Re((p - b) \bar{(c - b)})/|c - b|^2, and simplifying under \bar{p} = 1/p..^[21] The points d, e, and f are collinear on the Simson line since p lies on the circumcircle. To obtain the equation of this line, represent it in the general form for a line in the complex plane: \alpha z + \beta \bar{z} + \gamma = 0, where \alpha, \beta, \gamma \in \mathbb{C} are coefficients (not all zero) determined up to scalar multiple. Substitute the known positions of d, e, and f into this equation, yielding a homogeneous system of three equations in \alpha, \beta, \gamma. Let s_1 = a + b + c, s_2 = ab + bc + ca, and s_3 = abc. Solving the system involves clearing denominators (multiplying by $2p) and using the unit circle relations \bar{z} = 1/z for each point, along with properties of complex conjugates to eliminate imaginary parts and enforce the real projection condition. The conjugates facilitate simplification by pairing terms like p \bar{p} = 1 and expanding products such as (b + c + p - bc/p) \bar{d}, reducing cross terms involving s_1, s_2, and s_3. After algebraic manipulation—collecting coefficients of z, \bar{z}, and constants—the solution (up to scale) yields the Simson line equation in quadratic form:

$2 s_3 \bar{z} - 2 p z + p^2 + s_1 p - s_2 - \frac{s_3}{p} = 0.

This form is linear in z and \bar{z}, confirming it represents a straight line, and the coefficients depend symmetrically on the triangle's elementary symmetric polynomials..^[21] To verify, substitute a vertex, say p = a, into the equation:

$2 s_3 \bar{z} - 2 a z + a^2 + s_1 a - s_2 - \frac{s_3}{a} = 0.

Under the unit circle, this simplifies to the equation of the altitude from A to BC. Specifically, when p = a, the feet become d =\frac{1}{2} (b + c + a - bc/a) (the orthocenter foot H_a on BC) and e = f = a, so the line passes through a and H_a. The derived equation satisfies this, as the coefficients reduce to those defining the perpendicular from a to the line BC (whose equation involves \bar{b} = 1/b, etc.), confirming degeneration to the altitude..^[21]

Proof of Collinearity

Geometric Proof

Consider triangle ABC with point P lying on its circumcircle. Let D, E, and F be the feet of the perpendiculars from P to sides BC, CA, and AB, respectively. The quadrilaterals formed by P, the feet, and the adjacent vertices—such as PFBD for vertex B, PDCE for vertex C, and PEAF for vertex A—are each cyclic. This follows because each such quadrilateral contains two opposite right angles at the feet of the perpendiculars (e.g., ∠PFB = 90° and ∠PDB = 90° in PFBD), and the sum of those opposite angles is 180°. In the cyclic quadrilateral PFBD, angle chasing yields ∠FPD = ∠BPC, as both angles are related through inscribed angles in the circumcircle. Similarly, in the cyclic quadrilateral PEAF, the supplementary property gives ∠EPF = 180° - ∠BPC. Therefore, ∠FPD + ∠EPF = ∠BPC + (180° - ∠BPC) = 180°. This supplementary sum of adjacent angles at P implies that the rays from P through F, E, and D lie in a straight line configuration, meaning the perpendicular directions are aligned such that the feet D, E, F are collinear.^[22] The collinearity follows from these angle equalities establishing that the feet subtend supplementary angles at P, consistent with lying on a common straight line.^[22]

Analytic Proof

To provide an analytic proof of the collinearity of the feet of the perpendiculars from a point P on the circumcircle of triangle ABC to its sides, place the circumcircle as the unit circle centered at the origin, so that A = (\cos \alpha, \sin \alpha), B = (\cos \beta, \sin \beta), and C = (\cos \gamma, \sin \gamma) for distinct angles \alpha, \beta, \gamma. Let P = (x, y) be a point in the plane.^[14] The feet of the perpendiculars are computed using the orthogonal projection formula. For the foot X from P to line BC, parameterize the line as Z = B + t(C - B), where t = \frac{(C_1 - B_1)(x - B_1) + (C_2 - B_2)(y - B_2)}{(C_1 - B_1)^2 + (C_2 - B_2)^2}, with subscripts denoting coordinates; thus, X = (B_1 + t(C_1 - B_1), B_2 + t(C_2 - B_2)). The feet Y on CA and Z on AB are obtained analogously by cycling the vertices.^[14] Collinearity of X, Y, Z is tested by the vanishing of the determinant of the matrix formed by their homogeneous coordinates:

\begin{vmatrix} X_1 & X_2 & 1 \\ Y_1 & Y_2 & 1 \\ Z_1 & Z_2 & 1 \end{vmatrix} = 0.

Denote this determinant by W(x, y). Direct computation yields W(x, y) = m (x^2 + y^2 - 1), where m = \frac{\sin(\alpha - \beta) + \sin(\beta - \gamma) + \sin(\gamma - \alpha)}{4} is a nonzero constant depending only on the triangle (since the angles are fixed and distinct). Thus, W = 0 precisely when x^2 + y^2 = 1, meaning P lies on the unit circumcircle, which implies X, Y, Z are collinear.^[14]

Generalizations

Projections from Lines through Circumcenter

A significant generalization of the Simson line theorem, known as Dao's generalization, extends the collinearity property to projections involving a line through the circumcenter. Consider a triangle ABC with circumcenter O and a point P on its circumcircle. Let L be a line passing through O. The lines AP, BP, and CP intersect L at points A', B', and C', respectively. The feet of the perpendiculars from A' to side BC, from B' to side CA, and from C' to side AB—denoted A₀, B₀, and C₀—are collinear on a line that passes through the midpoint of the segment PH, where H is the orthocenter of triangle ABC.^[23] This collinearity holds specifically because L passes through the circumcenter O; for lines not through O, the feet A₀, B₀, C₀ generally do not lie on a straight line. The construction replaces the direct projection from a single point P with projections from the derived points A', B', C' on L, maintaining the essence of pedal projections while incorporating the directional structure of L. Synthetic proofs of this theorem typically rely on properties of cyclic quadrilaterals, angle chasing in the circle, and the Euler line relations between O, H, and P, demonstrating the collinearity without coordinate geometry.^[24] The theorem reduces to the classical Simson line when L passes through P, in which case A' = B' = C' = P, and the feet A₀, B₀, C₀ coincide with the standard feet of the perpendiculars from P to the sides, lying on the Simson line of P. This connection highlights how the generalization embeds the original theorem as a special case, where the line L "degenerates" in the sense that the intersection points collapse to P itself. Furthermore, the collinear line in the general case, often termed the "generalized Simson line" or Dao's line, bisects the segment from the orthocenter H to P, providing an additional locus property linking the configuration to key triangle centers.^[23]

Extensions to Conics

The Simson line extends to general conics through projective geometry, where orthogonal projections are replaced by projective analogs defined via a conic. Consider a triangle ABC. Using a fixed center of projection Z and a fixed axis f, the projective feet—defined as the intersections of the lines ZA₁, ZB₁, ZC₁ with the sides BC, CA, AB, where A₁, B₁, C₁ are the intersections of ZA, ZB, ZC with f—lie on a straight line called the projective Wallace line if the point X (defining the configuration) lies on a certain conic k. This conic k belongs to a two-parameter family of circumconics, with its equation in homogeneous coordinates ξ = (ξ₀ : ξ₁ : ξ₂) given by (a + b)ξ₀ξ₁ + (b + c)ξ₁ξ₂ + (c + a)ξ₂ξ₀ = 0, where the ratios a : b : c parameterize the axis f.^[25] In this setup, pole-polar relations with respect to the conic play a key role; for example, the polar of Z is given by (w + v)x₀ + (u + w)x₁ + (v + u)x₂ = 0. This framework unifies affine and projective properties, where the Euclidean metric emerges as a special case when the conic is the circumcircle, Z is the orthocenter H, and f is the line at infinity, reducing to the standard Simson line with orthogonal projections.^[25] These extensions link directly to pole-polar relations in projective geometry, where the collinear line serves as the polar of a point with respect to the conic, facilitating applications in studying incidences, dualities, and conic envelopes in triangle configurations. Seminal work emphasizes the role of such generalizations in unifying Euclidean and projective theorems, with high-impact contributions from algebraic methods in coordinate geometry.^[25]

References

[1]
Simson Line -- from Wolfram MathWorld
The Simson line is the line containing the feet P_1, P_2, and P_3 of the perpendiculars from an arbitrary point P on the circumcircle of a triangle to the ...
[2]
[PDF] Simson Theorem - Whitman College
Abstract. This paper is a presentation and discussion of several proofs of. Simson's Theorem. Simson's Theorem is a statement about a specific type of line ...
[3]
Simson line - Interactive Mathematics Miscellany and Puzzles
Given ΔABC and a point P, the feet of the perpendiculars from P to the sides of ΔABC form a triangle which is known as the pedal triangle of P. The definition ...
[4]
Circumcircle -- from Wolfram MathWorld
The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the triangle's three vertices.
[5]
Simson Line Theorem | Brilliant Math & Science Wiki
In Euclidean geometry, the Simson line theorem states that the three closest points from a point P P P on the circumcircle on △ A B C \triangle ABC △ABC to ...<|control11|><|separator|>
[6]
William Wallace (1768 - 1843) - Biography - University of St Andrews
Wallace's work was on geometry and Simson's line (which is definitely not due to Simson!) appears first in a paper of Wallace in 1799. See THIS LINK. One of ...Missing: details | Show results with:details
[7]
A curious misattribution: the early history of 'Simson's line' - jstor
This line is commonly called the 'Simson line' ... although it was first mentioned by W. Wallace, eleven years after Simson's death.'.
[8]
Robert Simson (1687 - 1768) - Biography - MacTutor
Simson also made many discoveries of his own in geometry and the Simson line is named after him. However the Simson line does not appear in his work but ...
[9]
Mathematical Treasure: Robert Simson's Opera Quaedam Reliqua
A complete digital scan of the first edition of Simson's Opera Quaedam Reliqua, call number QA3.S5 1776, can be found in the Linda Hall Library Digital ...
[10]
9 The Simson Line
The equality ϕ1 = ϕ4 implies that lT and FH are parallel, whence we also have TS = SH, or, in words: the line segment joining T and the orthocenter H is ...Missing: bisects | Show results with:bisects
[11]
[PDF] NINE POINT CIRCLE
Aug 16, 2015 · Recall that the Simson Line of a point P bisects the segment. PH where H is the orthocenter of 4ABC. 10. Angle chasing. 11. Introduce some ...
[12]
[PDF] Bashing Geometry with Complex Numbers - Evan Chen
Aug 29, 2015 · (Simson Line) Let ABC be a ... (b) Moreover, prove that the line through these points bisects PH, where H is the orthocenter of ABC.
[13]
[PDF] College Geometry
By definition, every Simson line contains at least one point on line BC. So, if a. Simson line is parallel to BC, this Simson line must actually be BC. Is it ...
[14]
[PDF] The Simson Line
The points X, Y , Z are collinear if and only if P lies on the circumcircle of ABC. Proof. The point P lies on the circumcircle of ABC if and only if. ZAP C ...
[15]
[PDF] arXiv:2311.10772v1 [math.HO] 15 Nov 2023
Nov 15, 2023 · The circle Γ transforms into the circumcircle of △ABC, and O becomes the circumcenter. ... The Simson line of Ta is perpendicular to OR.
[16]
[PDF] The envelope of the Wallace-Simson lines of a triangle. A simple ...
Wallace-Simson lines are formed by orthogonally projecting a point on a triangle's sides. Their envelope is the Steiner deltoid, a curve tangent to the ...
[17]
Analytic Geometry of the Triangle - jstor
In this case the line is called the Simson's line or pedal line of the point ... trilinear polar of P with respect to A1A2A3. The trilinear polar of ...
[18]
(PDF) An Introduction to the Modern Geometry of the Triangle and ...
The Simson line bisects the line joining its pole to the orthocenter of the triangle. ... The trilinear polar of a point for a triangle is also its trilinear ...
[19]
[PDF] Parabolas Escribed to a Triangle
Simson lines, and the Simson line rotates half as fast in the opposite sense to the corresponding radius of the ninepoint circle. If the triangle is ...
[20]
http://forumgeom.fau.edu/FG2017volume17/FG201736.pdf
[21]
None
Nothing is retrieved...<|control11|><|separator|>
[22]
[PDF] GEOMETRY REVISITED H. S. M. Coxeter S. L. Greitzer - Aproged
parallel to the Simson line of P. Finally, we observe that, in APHV, the Simson line AIBl is parallel to the side HV and bisects the side PV (at A1 ). Hence ...<|control11|><|separator|>
[23]
A Generalization of Simson Line
Let R be the Poncelet point of \{ A, B, C, P \} (It's well-known that R is the midway between P and the orthocenter H of \Delta ABC). From a lemma at post ...Missing: attribution | Show results with:attribution
[24]
[PDF] another synthetic proof of dao's generalization of the simson line ...
Let E, M, K be the midpoint of AP, PH, PC, respectively. As ZC1YM = 90. Page 3. ANOTHER SYNTHETIC PROOF OF DAO'S GENERALIZATION OF THE SIMSON LINE THEOREM AND ...<|control11|><|separator|>
[25]
[PDF] Affine and Projective Generalization of Wallace Lines
This construction allows the following affine and projective generalization. 1. Affine generalization. 1.1. Affine Wallace lines. In the real affine plane ...
[26]
Kiepert Hyperbola -- from Wolfram MathWorld
The Kiepert hyperbola is a hyperbola and triangle conic that is related to the solution of Lemoine's problem and its generalization to isosceles triangles.