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Simson line
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Simson line
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The Simson line LN (red) of the triangle ABC with respect to point P on the circumcircle

In geometry, given a triangle ABC and a point P on its circumcircle, the three closest points to P on lines AB, AC, and BC are collinear.[1] The line through these points is the Simson line of P, named for Robert Simson.[2] The concept was first published, however, by William Wallace in 1799,[3] and is sometimes called the Wallace line.[4]

The converse is also true; if the three closest points to P on three lines are collinear, and no two of the lines are parallel, then P lies on the circumcircle of the triangle formed by the three lines. Or in other words, the Simson line of a triangle ABC and a point P is just the pedal triangle of ABC and P that has degenerated into a straight line and this condition constrains the locus of P to trace the circumcircle of triangle ABC.

Equation

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Placing the triangle in the complex plane, let the triangle ABC with unit circumcircle have vertices whose locations have complex coordinates a, b, c, and let P with complex coordinates p be a point on the circumcircle. The Simson line is the set of points z satisfying[5]: Proposition 4 

where an overbar indicates complex conjugation.

Properties

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Simson lines (in red) are tangents to the Steiner deltoid (in blue).
  • The Simson line of a vertex of the triangle is the altitude of the triangle dropped from that vertex, and the Simson line of the point diametrically opposite to the vertex is the side of the triangle opposite to that vertex.
  • If P and Q are points on the circumcircle, then the angle between the Simson lines of P and Q is half the angle of the arc PQ. In particular, if the points are diametrically opposite, their Simson lines are perpendicular and in this case the intersection of the lines lies on the nine-point circle.
  • Letting H denote the orthocenter of the triangle ABC, the Simson line of P bisects the segment PH in a point that lies on the nine-point circle.
  • Given two triangles with the same circumcircle, the angle between the Simson lines of a point P on the circumcircle for both triangles does not depend of P.
  • The set of all Simson lines, when drawn, form an envelope in the shape of a deltoid known as the Steiner deltoid of the reference triangle. The Steiner deltoid circumscribes the nine-point circle and its center is the nine-point center.
  • The construction of the Simson line that coincides with a side of the reference triangle (see first property above) yields a nontrivial point on this side line. This point is the reflection of the foot of the altitude (dropped onto the side line) about the midpoint of the side line being constructed. Furthermore, this point is a tangent point between the side of the reference triangle and its Steiner deltoid.
  • A quadrilateral that is not a parallelogram has one and only one pedal point, called the Simson point, with respect to which the feet on the quadrilateral are collinear.[6] The Simson point of a trapezoid is the point of intersection of the two nonparallel sides.[7]: p. 186 
  • No convex polygon with at least 5 sides has a Simson line.[8]

Proof of existence

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It suffices to show that .

is a cyclic quadrilateral, so . is a cyclic quadrilateral (since ), so . Hence . Now is cyclic, so .

Therefore .

Generalizations

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Generalization 1

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The projections of Ap, Bp, Cp onto BC, CA, AB are three collinear points
  • Let ABC be a triangle, let a line ℓ go through circumcenter O, and let a point P lie on the circumcircle. Let AP, BP, CP meet ℓ at Ap, Bp, Cp respectively. Let A0, B0, C0 be the projections of Ap, Bp, Cp onto BC, CA, AB, respectively. Then A0, B0, C0 are collinear. Moreover, the new line passes through the midpoint of PH, where H is the orthocenter of ΔABC. If ℓ passes through P, the line coincides with the Simson line.[9][10][11]
A projective version of a Simson line

Generalization 2

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  • Let the vertices of the triangle ABC lie on the conic Γ, and let Q, P be two points in the plane. Let PA, PB, PC intersect the conic at A1, B1, C1 respectively. QA1 intersects BC at A2, QB1 intersects AC at B2, and QC1 intersects AB at C2. Then the four points A2, B2, C2, and P are collinear if only if Q lies on the conic Γ.[12]

Generalization 3

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See also

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References

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Revisions and contributorsEdit on WikipediaRead on Wikipedia

Simson line

View on Grokipedia
from Grokipedia
In , the Simson line (also known as the Wallace-Simson line) of a point PP on the of a ABCABC is the straight line containing the feet of the perpendiculars dropped from PP to the lines containing the sides BCBC, CACA, and ABAB. This of the feet, known as Simson's , holds if and only if PP lies on the of ABCABC. 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 in 1797 or 1799. The line arises as a degeneration of the pedal triangle of PP with respect to ABCABC, which collapses into a line precisely when PP is on the . Key properties of the Simson line include its role in bisecting the segment from PP to the orthocenter HH of ABCABC, with the midpoint of PHPH lying on the . The angle between the Simson lines of two points on the equals half the angular measure of the arc between them, and the Simson lines of diametrically opposite points are perpendicular and intersect on the . These features connect the Simson line to broader , including the and isogonal conjugates, and it generalizes to oblique versions via Carnot's theorem for points not on the . The envelope of all Simson lines for a fixed forms a with area half that of the .

Definition and History

Definition

In , the circumcircle of a triangle ABC is the unique circle that passes through all three vertices A, B, and C. Given a triangle ABC and its , 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. 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 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. As P traverses the circumcircle, the Simson line rotates around the triangle.

History

The Simson line was first proved by the Scottish mathematician 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 of the feet of the perpendiculars from a point on a triangle's to its sides. The line is named after Robert Simson (1687–1768), another Scottish mathematician renowned for restoring geometric texts and exploring porisms and loci. Simson conducted investigations into properties around 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. The attribution to Simson gained prominence through in the early 19th century, leading to ongoing debates and the alternative designation Wallace-Simson line to acknowledge Wallace's priority. This development occurred amid the golden age of triangle geometry in the 18th and 19th centuries, paralleling advances in properties and the orthocenter by figures like Leonhard Euler and .

Properties

Basic Properties

One fundamental property of the Simson line arises in special cases when the point PP coincides with a vertex of the triangle △ABC\triangle ABC. If PP is at vertex AA, the feet of the perpendiculars from AA to the sides BCBC, ABAB, and ACAC degenerate such that the Simson line coincides with the altitude from AA to side BCBC. Similar degenerations occur for points PP at vertices BB or CC, yielding the respective altitudes from those vertices. The Simson line also exhibits a notable relation to the orthocenter HH of △ABC\triangle ABC. Specifically, the Simson line of PP bisects the line segment PHPH, with the midpoint MM of PHPH lying on the Simson line itself. This midpoint MM additionally resides on the of △ABC\triangle ABC. Another distinctive case occurs when PP is the point on the diametrically opposite to a vertex, say opposite to AA. In this configuration, the Simson line of PP coincides with the side BCBC opposite to AA. Analogous results hold for the points diametrically opposite to BB or CC, yielding sides ACAC or ABAB, respectively. The Simson line is unique for each point PP on the of △ABC\triangle ABC, as the of the feet of the perpendiculars from PP to the sides holds precisely when PP lies on this . If PP is not on the , the three feet are generally not , and the pedal triangle remains non-degenerate.

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 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 of the Simson lines as P and Q vary on the , reflecting the inscribed angle theorem in the . 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 , with M also residing on the of ABC. As the point P varies around the , the family of Simson lines envelopes a three-cusped known as the Steiner deltoid. This curve circumscribes the 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 and serves as the boundary tangent to all Simson lines. 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 of the tangential triangle formed by the lines through the vertices perpendicular to the sides. This identification links the Simson line to , where the trilinear polar represents the polar line of P under the trilinear polarity defined by ABC. 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.

Mathematical Formulation

Equation

In coordinate geometry, the Simson line of a point PP with respect to a triangle ABCABC can be expressed algebraically using complex numbers to represent the points in the plane. The vertices AA, BB, CC and the point PP on the circumcircle are denoted by complex numbers aa, bb, cc, and pp, respectively. For simplicity, assume the circumcircle is the unit circle centered at the origin, so ∣a∣=∣b∣=∣c∣=∣p∣=1|a| = |b| = |c| = |p| = 1; this setup leverages the symmetry in aa, bb, cc and can be generalized to arbitrary positioning via affine transformations. The explicit equation of the Simson line is 2abczˉ−2pz+p2+(a+b+c)p−(ab+bc+ca)−abcp=0,2abc \bar{z} - 2pz + p^2 + (a + b + c)p - (ab + bc + ca) - \frac{abc}{p} = 0, where zz is the complex coordinate of a point on the line. Alternative representations include the parametric form in terms of the feet DD, EE, FF of the perpendiculars from PP to the lines BCBC, CACA, ABAB, respectively: since DD, EE, FF lie on the Simson line, it can be parametrized as r(t)=D+t(E−D)\mathbf{r}(t) = \mathbf{D} + t (\mathbf{E} - \mathbf{D}) for real tt, where boldface denotes position vectors. In vector form, the Simson line is the affine span of these projection points onto the sides of the triangle.

Derivation

To derive the equation of the Simson line in the complex plane, assume without loss of generality that the circumcircle of triangle ABCABC is the unit circle centered at the origin, with vertices represented by complex numbers aa, bb, cc and point PP by pp, all satisfying ∣a∣=∣b∣=∣c∣=∣p∣=1|a| = |b| = |c| = |p| = 1 and thus aˉ=1/a\bar{a} = 1/a, bˉ=1/b\bar{b} = 1/b, etc.. The foot of the perpendicular from pp to side BCBC (the line through bb and cc) is the point dd on that line minimizing the distance to pp. For points vv and ww on the unit circle, this projection is given by the formula d=12(b+c+p−bcpˉ).d = \frac{1}{2} \left( b + c + p - bc \bar{p} \right). Since ∣p∣=1|p| = 1, pˉ=1/p\bar{p} = 1/p, simplifying to d=12(b+c+p−bcp).d = \frac{1}{2} \left( b + c + p - \frac{bc}{p} \right). Analogously, the foot ee from pp to side CACA (through cc and aa) is e=12(c+a+p−cap),e = \frac{1}{2} \left( c + a + p - \frac{ca}{p} \right), and the foot ff to side ABAB (through aa and bb) is f=12(a+b+p−abp).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 tt in the line parametrization b+t(c−b)b + t(c - b) (with tt real) via dot products involving conjugates, Re((p−b)(c−b)ˉ)/∣c−b∣2((p - b) \bar{(c - b)})/|c - b|^2, and simplifying under pˉ=1/p\bar{p} = 1/p.. The points dd, ee, and ff are collinear on the Simson line since pp lies on the . To obtain the equation of this line, represent it in the general form for a line in the : αz+βzˉ+γ=0\alpha z + \beta \bar{z} + \gamma = 0, where α\alpha, β\beta, γ∈C\gamma \in \mathbb{C} are coefficients (not all zero) determined up to scalar multiple. Substitute the known positions of dd, ee, and ff into this equation, yielding a homogeneous of three equations in α\alpha, β\beta, γ\gamma. Let s1=a+b+cs_1 = a + b + c, s2=ab+bc+cas_2 = ab + bc + ca, and s3=abcs_3 = abc. Solving the system involves clearing denominators (multiplying by 2p2p) and using the relations zˉ=1/z\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 ppˉ=1p \bar{p} = 1 and expanding products such as (b+c+p−bc/p)dˉ(b + c + p - bc/p) \bar{d}, reducing cross terms involving s1s_1, s2s_2, and s3s_3. After algebraic manipulation—collecting coefficients of zz, zˉ\bar{z}, and constants—the solution (up to scale) yields the Simson line equation in : 2s3zˉ−2pz+p2+s1p−s2−s3p=0.2 s_3 \bar{z} - 2 p z + p^2 + s_1 p - s_2 - \frac{s_3}{p} = 0. This form is linear in zz and zˉ\bar{z}, confirming it represents a straight line, and the coefficients depend symmetrically on the triangle's elementary symmetric polynomials.. To verify, substitute a vertex, say p=ap = a, into the : 2s3zˉ−2az+a2+s1a−s2−s3a=0.2 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 of the altitude from AA to BCBC. Specifically, when p=ap = a, the feet become d=12(b+c+a−bc/a)d =\frac{1}{2} (b + c + a - bc/a) (the orthocenter foot HaH_a on BCBC) and e=f=ae = f = a, so the line passes through aa and HaH_a. The derived satisfies this, as the coefficients reduce to those defining the from aa to the line BCBC (whose involves bˉ=1/b\bar{b} = 1/b, etc.), confirming degeneration to the altitude..

Proof of Collinearity

Geometric Proof

Consider ABC with point P lying on its . Let D, E, and F be the feet of the perpendiculars from P to sides BC, CA, and AB, respectively. The 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 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 PFBD, angle chasing yields ∠FPD = ∠BPC, as both angles are related through inscribed angles in the . Similarly, in the 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 directions are aligned such that the feet D, E, F are collinear. The follows from these angle equalities establishing that the feet subtend supplementary angles at P, consistent with lying on a common straight line.

Analytic Proof

To provide an analytic proof of the of the feet of the perpendiculars from a point PP on the circumcircle of triangle ABCABC to its sides, place the circumcircle as the unit circle centered at the origin, so that A=(cos⁡α,sin⁡α)A = (\cos \alpha, \sin \alpha), B=(cos⁡β,sin⁡β)B = (\cos \beta, \sin \beta), and C=(cos⁡γ,sin⁡γ)C = (\cos \gamma, \sin \gamma) for distinct angles α,β,γ\alpha, \beta, \gamma. Let P=(x,y)P = (x, y) be a point in the plane. The feet of the perpendiculars are computed using the orthogonal projection formula. For the foot XX from PP to line BCBC, parameterize the line as Z=B+t(C−B)Z = B + t(C - B), where t=(C1−B1)(x−B1)+(C2−B2)(y−B2)(C1−B1)2+(C2−B2)2t = \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=(B1+t(C1−B1),B2+t(C2−B2))X = (B_1 + t(C_1 - B_1), B_2 + t(C_2 - B_2)). The feet YY on CACA and ZZ on ABAB are obtained analogously by cycling the vertices. Collinearity of XX, YY, ZZ is tested by the vanishing of the of the matrix formed by their : ∣X1X21Y1Y21Z1Z21∣=0.\begin{vmatrix} X_1 & X_2 & 1 \\ Y_1 & Y_2 & 1 \\ Z_1 & Z_2 & 1 \end{vmatrix} = 0.
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