Steiner inellipse

The Steiner Inellipse

In geometry , the Steiner inellipse of a triangle is the uniquely determined ellipse that is inscribed in a triangle and touches the sides of this triangle in their centers. The Steiner inellipse is an example of an inellipse . The inscribed circle and the Mandart inellipse are also inellipses; however, they generally do not touch the sides of the triangle in the centers - unless it is an equilateral triangle . The Steiner Inellipse is attributed to Dörrie Jakob Steiner . The proof of uniqueness was carried out by Kalman.

The Steiner Inellipse is the counterpart to Steiner order ellipse (often, as in the following, Steiner ellipse called), which passes through the corners of the given triangle and has the triangle center of gravity as its center.

Definition and characteristics

definition

If the centers of the sides of a triangle are , then an ellipse that touches the sides of the triangle there is called a Steiner inellipse. ${\displaystyle M_{1},M_{2},M_{3}}$ ${\displaystyle ABC}$ ${\displaystyle M_{1},M_{2},M_{3}}$ ${\displaystyle ABC}$

Steiner ellipse (blue) with Steiner ellipse (red)

Steiner ellipse (blue) and Steiner ellipse (red) for an equilateral triangle

Properties:
The following applies to any triangle with the side centers : a) There is exactly one Steiner inellipse. b) The center of the Steiner inellipse is the common center of gravity of the triangles and . The Steiner ellipse of the triangle is thus the Steiner ellipse of the triangle . c) The Steiner ellipse of the triangle (blue in the picture) results from a centric expansion with the factor from the Steiner ellipse (red). Both ellipses have the same eccentricity (are similar). d) The area of ​​the Steiner inellipse results from multiplying the triangular area by . The area is a quarter of the content of the Steiner ellipse. e) The Steiner inellipse of a triangle has the largest area of ​​all inscribed ellipses of the triangle. ${\displaystyle ABC}$ ${\displaystyle M_{1},M_{2},M_{3}}$

${\displaystyle S}$ ${\displaystyle ABC}$ ${\displaystyle M_{1}M_{2}M_{3}}$ ${\displaystyle ABC}$ ${\displaystyle M_{1}M_{2}M_{3}}$
${\displaystyle ABC}$ ${\displaystyle {\tfrac {1}{2}}}$
${\displaystyle {\tfrac {\pi }{3{\sqrt {3}}}}}$

proof

Since any triangle results from an affine mapping from an equilateral triangle , an affine mapping maps the center points of lines and the center of gravity of a triangle to the center points of the image lines and the center of gravity of the image triangle and an ellipse and its center point onto an ellipse and its center point it is to prove the properties of an equilateral triangle.
To a): The inscribed circle of an equilateral triangle touches the sides in their centers. This follows either from symmetry considerations or through recalculation. There is no other conic section that touches the sides of the triangle at the side centers. Because a conic section is already clearly defined by 5 defining pieces (points, tangents).
To b): Recalculate
To c) The circumference changes into the inscribed circle due to the expansion with the factor at the common center point. The eccentricity is an invariant in the case of point expansion (similarity mapping). To d): Relationships between areas remain invariant in an affine mapping. To e): See inellipse or literature. ${\displaystyle {\tfrac {1}{2}}}$

Parametric representation and semi-axes

Parameter representation:

• Since the Steiner ellipse of a triangle results from a centric stretching with the factor from the Steiner ellipse, the parametric representation is obtained (see Steiner ellipse ): ${\displaystyle ABC}$ ${\displaystyle {\tfrac {1}{2}}}$

${\displaystyle {\vec {x}}={\vec {p}}(t)={\overrightarrow {OS}}\;+\;{\color {blue}{\frac {1}{2}}}{\overrightarrow {SC}}\;\cos t\;+\;{\frac {1}{{\color {blue}2}{\sqrt {3}}}}{\overrightarrow {AB}}\;\sin t\;,\quad 0\leq t<2\pi }$

• The 4 vertices of the ellipse are

${\displaystyle {\vec {p}}(t_{0}),\;{\vec {p}}(t_{0}\pm {\frac {\pi }{2}}),\;{\vec {p}}(t_{0}+\pi ),}$

wherein from ${\displaystyle t_{0}}$

${\displaystyle \cot(2t_{0})={\tfrac {{\vec {f}}_{1}^{\,2}-{\vec {f}}_{2}^{\,2}}{2{\vec {f}}_{1}\cdot {\vec {f}}_{2}}}\quad }$ With ${\displaystyle \quad {\vec {f}}_{1}={\frac {1}{2}}{\vec {SC}},\quad {\vec {f}}_{2}={\frac {1}{2{\sqrt {3}}}}{\vec {AB}}}$

results.

Semi-axes:

• With the abbreviations

${\displaystyle M:={\color {blue}{\frac {1}{4}}}({\vec {SC}}^{2}+{\frac {1}{3}}{\vec {AB}}^{2})}$

${\displaystyle N:={\frac {1}{{\color {blue}4}{\sqrt {3}}}}\left|\det({\vec {SC}},{\vec {AB}})\right|}$

results for the two semi-axes : ${\displaystyle a,b,\;a>b}$

${\displaystyle a={\frac {1}{2}}({\sqrt {M+2N}}+{\sqrt {M-2N}})}$

${\displaystyle b={\frac {1}{2}}({\sqrt {M+2N}}-{\sqrt {M-2N}})}$

• For the linear eccentricity of the Steiner inellipse we get: ${\displaystyle e}$

${\displaystyle e={\sqrt {a^{2}-b^{2}}}=\dotsb ={\sqrt {\sqrt {M^{2}-4N^{2}}}}}$

Trilinear equation

The equation of the Steiner inellipse in trilinear coordinates for a triangle with the side lengths is: ${\displaystyle u,v,w}$

${\displaystyle u^{2}x^{2}+v^{2}y^{2}+w^{2}z^{2}-2uvxy-2vwyz-2wuzx=0}$

Alternative calculation of the semi-axes

The lengths of the semi-major and minor axes for a triangle with side lengths are ${\displaystyle u,v,w}$

${\displaystyle {\frac {1}{6}}{\sqrt {u^{2}+v^{2}+w^{2}\pm 2Z}}}$

with the abbreviation

${\displaystyle Z:={\sqrt {u^{4}+v^{4}+w^{4}-u^{2}v^{2}-v^{2}w^{2}-w^{2}u^{2}}}.}$

An application

If you represent a triangle in the complex number plane , i.e. the coordinates of its corner points correspond to complex numbers, then for every third degree polynomial that has these corner points as zeros, the zeros of its derivative are the focal points of the Steiner inellipse ( Theorem von Marden ).

Constructions

Five generated points are sufficient for the representation of the Steiner inellipse. The arbitrarily chosen triangle can alternatively have three sides of unequal length or only two equal legs. In an equilateral triangle , which according to the modern definition is a special case of the isosceles triangle, the same five points result in the inscribed circle of the triangle.

The Steiner inellipse is a second degree algebraic curve . With the exception of the circle , such curves cannot be constructed with a compass and ruler . However, for each of the two constructive methods described below for determining the corresponding five elliptical points, there are tools with which the elliptical line can be approximated or drawn exactly.

Five elliptical points

Fig. 1: Steiner ellipse
with five constructively determined points ${\displaystyle D,F',E,D',F}$

In the selected triangle (Fig. 1) with three sides of unequal length, the three side bisectors with their intersections and are constructed. They meet in the center of gravity of the later ellipse. This is followed by the definition of the radius and by doubling the lines or on the bisector within the triangle. The five elliptical points sought are thus and . For the final drawing of the elliptical line, depending on whether the construction was done on the computer or created with a compass and ruler, the following tools are used: ${\displaystyle ABC}$ ${\displaystyle D,E}$ ${\displaystyle F}$ ${\displaystyle S}$ ${\displaystyle {\overline {SD'}}}$ ${\displaystyle {\overline {SF'}}}$ ${\displaystyle {\overline {SD}}}$ ${\displaystyle {\overline {SD}}}$ ${\displaystyle D,F',E,D'}$ ${\displaystyle F}$

• A dynamic geometry software (DGS) draws the ellipse as an (exact) parameterized curve .
• If, on the other hand, you simply connect the five ellipse points using a curved ruler, you get an approximate ellipse.

Major and minor axes as well as focal points

In order to enable the drawing of an exact elliptical line in a triangle with three sides of unequal length with the help of a mechanical aid, one or two of the following conditions are required:

• Major and minor axes of the ellipse with their vertices required when using an ellipsograph ${\displaystyle S_{1}\ldots S_{4},}$
• Focal points of the ellipse and additionally required when using an elliptical compass ${\displaystyle F_{1}}$ ${\displaystyle F_{2},}$

Both prerequisites can be constructed if at least two so-called conjugate radiuses of the inellipse, similar to the construction of conjugate radiuses for the Steiner ellipse, are first determined in the selected triangle .

method

In the selected triangle (Fig. 2) with three sides of unequal length, two side bisectors with their intersections and are constructed. They meet in the center of gravity of the later ellipse. Then the first relevant radius is determined by doubling the distance within the triangle. This is followed by the shear of the triangle into an isosceles and equal area triangle with (the same) height ${\displaystyle ABC}$ ${\displaystyle D_{1}}$ ${\displaystyle E}$ ${\displaystyle S}$ ${\displaystyle {\overline {SD_{2}}}}$ ${\displaystyle {\overline {SD_{1}}}}$ ${\displaystyle ABC}$ ${\displaystyle A'B'C'}$ ${\displaystyle {\overline {D_{1}'C'}}}$

Fig. 2: Steiner inellipse
Left: Triangle with three sides of unequal length, in the center the center of gravity, the main and minor axes with the vertices determined using the Rytz construction (points and ). Right: The isosceles triangle created by shear with the conjugate radius and according to the elliptical construction of de La Hire . ${\displaystyle S,}$ ${\displaystyle S_{1}\ldots S_{4}}$ ${\displaystyle J,K,L}$ ${\displaystyle M}$
${\displaystyle A',B',C'}$ ${\displaystyle S,D_{2}'}$ ${\displaystyle S,H',}$

It proceeds in an isosceles triangle with the medians of the center of gravity the route intersects, and the compound with Subsequently, according to the ellipses-engineering de La Hire , a circle with a radius in order to intersection to drawn and parallels of by drawn . The subsequent circle of radius to cuts in the closest parallel to starting to circle through runs, gives the intersection now with the connects. A parallel to through and through the line results in the point of intersection which, with the help of the circle with radius around projected onto the center of gravity axis, delivers with the second relevant radius . Thus the two conjugate radii and are determined. ${\displaystyle {\overline {A'E'}},}$ ${\displaystyle S'}$ ${\displaystyle {\overline {D_{1}'C'}}}$ ${\displaystyle S'}$ ${\displaystyle B'.}$ ${\displaystyle {\overline {S'E'}}}$ ${\displaystyle S'}$ ${\displaystyle F}$ ${\displaystyle {\overline {B'S'}}}$ ${\displaystyle {\overline {AB'}}}$ ${\displaystyle S}$ ${\displaystyle S'}$ ${\displaystyle {\overline {S'D_{1}'}}}$ ${\displaystyle S'}$ ${\displaystyle {\overline {D_{1}'C'}}}$ ${\displaystyle D_{2}'.}$ ${\displaystyle {\overline {A'B'}}}$ ${\displaystyle F}$ ${\displaystyle D_{1}'}$ ${\displaystyle G,}$ ${\displaystyle S'}$ ${\displaystyle {\overline {D_{1}'S'}}}$ ${\displaystyle F}$ ${\displaystyle {\overline {GS'}}}$ ${\displaystyle H,}$ ${\displaystyle {\overline {S'H}}}$ ${\displaystyle S'}$ ${\displaystyle {\overline {SS'}}}$ ${\displaystyle {\overline {S'H'}}}$ ${\displaystyle {\overline {SD_{2}}}}$ ${\displaystyle {\overline {S'H'}}}$

The construction is continued in the selected triangle. First, the radius just found is plotted on the center of gravity axis from with the point of intersection . The construction of the main and minor axes of the ellipse made possible by this is created using the six illustrated steps of the Rytz axis construction . Then the two focal points and are determined by taking the radius in the compass to pierce the vertex or as shown in and create the distances and . ${\displaystyle {\overline {S'H'}}}$ ${\displaystyle {\overline {SS'}}}$ ${\displaystyle S}$ ${\displaystyle H''}$ ${\displaystyle F_{1}}$ ${\displaystyle F_{2}}$ ${\displaystyle a}$ ${\displaystyle S_{3},}$ ${\displaystyle S_{4},}$ ${\displaystyle \left|S_{4}F_{1}\right|}$ ${\displaystyle \left|S_{4}F_{2}\right|}$

Finally, the ellipsoid line is drawn (exactly) with the aid of an ellipsograph or an elliptical compass.

Alternative construction of the second radius

Image 3: Steiner ellipse,
alternative construction of the radius with the help of the right triangle ${\displaystyle {\overline {SH}}}$ ${\displaystyle SFH}$

First the radius is calculated . The general formula for the height of the equilateral triangle with the side is used ${\displaystyle {\overline {SH}}.}$ ${\displaystyle h}$ ${\displaystyle a:}$

${\displaystyle h={\frac {\sqrt {3}}{2}}a}$

Half of this equilateral triangle is a right triangle with the (same) height:

${\displaystyle h={\frac {\sqrt {3}}{2}}\cdot 2\cdot {\frac {a}{2}}={\sqrt {3}}\cdot {\frac {a}{2}}}$

If you insert and   , this results in the right-angled triangle (see Fig. 3) with the height ${\displaystyle {\overline {AB}}=c,}$ ${\displaystyle {\frac {a}{2}}={\overline {SH}},}$ ${\displaystyle a=2\cdot {\overline {SH}}}$ ${\displaystyle h={\frac {c}{2}}}$ ${\displaystyle SFH}$

${\displaystyle {\frac {c}{2}}={\sqrt {3}}\cdot {\overline {SH}},}$

transformed applies

${\displaystyle {\overline {SH}}={\frac {1}{2}}\cdot {\frac {c}{\sqrt {3}}}.}$

It continues with the construction of the right triangle ${\displaystyle SFH:}$

It begins with drawing in a vertical line ( orthogonal ) from the center of gravity and transferring the route to the vertical line; it results in the line segment Now follows the construction of the angle width at the angle vertex by dividing the segment in half, drawing an arc with a radius around the point and another arc with the same compass opening around the point ; this results in the intersection point. By drawing a half-line , from through , the angle is generated at the apex of the angle . The final parallel to the line from the center of gravity creates the intersection on the half-straight line and thus provides the radius you are looking for ${\displaystyle {\overline {AB}},}$ ${\displaystyle S,}$ ${\displaystyle {\overline {AD_{1}}}={\frac {c}{2}}}$ ${\displaystyle {\overline {SF}}.}$ ${\displaystyle 30^{\circ }}$ ${\displaystyle F,}$ ${\displaystyle {\overline {SF}}}$ ${\displaystyle G}$ ${\displaystyle {\overline {GS}}}$ ${\displaystyle G}$ ${\displaystyle S}$ ${\displaystyle I.}$ ${\displaystyle F}$ ${\displaystyle I}$ ${\displaystyle F}$ ${\displaystyle 30^{\circ }}$ ${\displaystyle {\overline {AB}},}$ ${\displaystyle S,}$ ${\displaystyle H}$ ${\displaystyle {\overline {SH}}.}$

Since this right-angled triangle can be represented in a structurally simple manner, it is also possible to find the second conjugate radius in this way . ${\displaystyle {\overline {SH}}}$

Individual evidence

1. ↑ ^{a b c} Eric W. Weisstein : Steiner Inellipse . In: MathWorld (English).
2. ^ H. Dörrie: 100 Great Problems of Elementary Mathematics, Their History and Solution. (Translated by D. Antin), Dover, New York 1965, issue 98.
3. ^ ^{A b} Dan Kalman: An elementary proof of Marden's theorem . In: American Mathematical Monthly . tape 115 , no. 4 , 2008, p. 330–338 (English, An Elementary Proof of Marden's Theorem [PDF; 190 kB ; accessed on August 17, 2020]).
4. ↑ Eric W. Weisstein : Steiner Circumellipse . In: MathWorld (English).
5. ↑ GD Chakerian: Mathematical plums . In: Ross Honsberger (Ed.): A distorted view of geometry (=  The Dolciani Mathematical Expositions ). tape 4 . Mathematical Association of America, Washington, DC 1979, pp. 135-136 and 145-146 (English).
6. ^ Karl Strubecker : Lectures on descriptive geometry . Ed .: Mathematical Research Institute Oberwolfach. Vandenhoeck & Ruprecht, Göttingen 1967, 11. Ellipse constructions, 2. Construction of an ellipse with the help of its two vertex circles (Fig. 25), p. 25–27 ( harvard.edu [PDF; accessed August 17, 2020]).
7. ^ Karl Strubecker: Lectures on descriptive geometry . Ed .: Mathematical Research Institute Oberwolfach. Vandenhoeck & Ruprecht, Göttingen 1967, 11. Ellipse constructions, Rytz axis construction (Fig. 29), p. 29–30 ( harvard.edu [PDF; accessed August 17, 2020]).