Simson Line

In geometry, given a triangle and a point on its circumcircle, the three closest points to on lines , , and are collinear. The line through these points is the Simson line of , named for Robert Simson in 1799.

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

Equation

Placing the triangle in the complex plane, let the triangle with unit circumcircle have vertices whose locations have complex coordinates , , , and let P with complex coordinates be a point on the circumcircle. The Simson line is the set of points satisfyingTodor Zaharinov, "The Simson triangle and its properties", ''Forum Geometricorum'' 17 (2017), 373--381. http://forumgeom.fau.edu/FG2017volume17/FG201736.pdf

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

where an overbar indicates complex conjugation.

Properties

*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 and are points on the circumcircle, then the angle between the Simson lines of and is half the angle of the arc . 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 denote the orthocenter of the triangle , the Simson line of bisects the segment 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 on the circumcircle for both triangles does not depend of .
*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 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. The Simson point of a trapezoid is the point of intersection of the two nonparallel sides.
* No convex polygon with at least 5 sides has a Simson line.

Proof of existence

The method of proof is to show that $\angle NMP + \angle PML = 180^\circ$. $PCAB$ is a cyclic quadrilateral, so $\angle PBA + \angle ACP = \angle PBN + \angle ACP = 180^\circ$. $PMNB$ is a cyclic quadrilateral (Thales' theorem), so $\angle PBN + \angle NMP = 180^\circ$. Hence $\angle NMP = \angle ACP$. Now $PLCM$ is cyclic, so $\angle PML = \angle PCL = 180^\circ - \angle ACP$. Therefore $\angle NMP + \angle PML = \angle ACP + (180^\circ - \angle ACP) = 180^\circ$.

Generalizations

Generalization 1

* 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 ''A_p, B_p, C_p'' respectively. Let ''A''₀, ''B''₀, ''C''₀ be the projections of ''A_p, B_p, C_p'' onto ''BC, CA, AB'', respectively. Then ''A''₀, ''B''₀, ''C''₀ are collinear. Moreover, the new line passes through the midpoint of ''PH'', where ''H'' is the orthocenter of Δ''ABC''. If ℓ passes through ''P'', the line coincides with the Simson line.Nguyen Le Phuoc and Nguyen Chuong Chi (2016). 100.24 A synthetic proof of Dao's generalisation of the Simson line theorem. The Mathematical Gazette, 100, pp 341-345. doi:10.1017/mag.2016.77.
The Mathematical Gazette

Generalization 2

* 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 ''A''₁, ''B''₁, ''C''₁ respectively. ''QA''₁ intersects ''BC'' at ''A''₂, ''QB''₁ intersects ''AC'' at ''B''₂, and ''QC''₁ intersects ''AB'' at ''C''₂. Then the four points ''A''₂, ''B''₂, ''C''₂, and ''P'' are collinear if only if ''Q'' lies on the conic Γ.

Generalization 3

* R. F. Cyster generalized the theorem to cyclic quadrilaterals i
The Simson Lines of a Cyclic Quadrilateral

See also

*Pedal triangle
*Robert Simson

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