Difference between revisions of "Isogonal conjugate"
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Distance formulas for isogonal points== | ||
+ | Let triangle <math>\triangle ABC</math> and point <math>P</math> be given. Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | ||
+ | Let lines <math>AP</math> and <math>AP'</math> cross sideline <math>BC</math> at <math>D</math> and <math>E</math> and circumcircle of <math>\triangle ABC</math> at <math>F</math> and <math>G,</math> respectively. | ||
+ | We apply the Isogonal’s property and get <math>\frac {BD}{DC} \cdot \frac{BE}{EC} = \frac {AB^2}{AC^2}.</math> | ||
+ | <math>EF || BC.</math> We apply the Ptolemy's theorem to <math>BCGF</math> and get <math>BC \cdot FG = BG^2 – BF^2.</math> | ||
+ | We apply the barycentric coordinates and get | ||
+ | <math>\left| \frac{BE}{EC} - \frac {BD}{DC} \right| = \frac {AB \cdot BC \cdot FG}{AC \cdot PF \cdot P'G}.</math> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
[[Category:Geometry]] | [[Category:Geometry]] |
Revision as of 05:54, 13 August 2023
Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.
Contents
- 1 The isogonal theorem
- 2 Parallel segments
- 3 Perpendicularity
- 4 Fixed point
- 5 Bisector
- 6 Isogonal of the diagonal of a quadrilateral
- 7 Isogonals in trapezium
- 8 Isogonals in complete quadrilateral
- 9 Isogonal of the bisector of the triangle
- 10 Points on isogonals
- 11 Trapezoid
- 12 IMO 2007 Short list/G3
- 13 Definition of isogonal conjugate of a point
- 14 Three points
- 15 Second definition
- 16 Distance to the sides of the triangle
- 17 Sign of isogonally conjugate points
- 18 Circumcircle of pedal triangles
- 19 Common circumcircle of the pedal triangles as the sign of isogonally conjugate points
- 20 Two pares of isogonally conjugate points
- 21 Circles
- 22 1995 USAMO Problems/Problem 3
- 23 2011 USAMO Problems/Problem 5
- 24 Distance formulas for isogonal points
The isogonal theorem
Isogonal lines definition
Let a line and a point
lying on
be given. A pair of lines symmetric with respect to
and containing the point
be called isogonals with respect to the pair
Sometimes it is convenient to take one pair of isogonals as the base one, for example, and
are the base pair. Then we call the remaining pairs as isogonals with respect to the angle
Projective transformation
It is known that the transformation that maps a point with coordinates into a point with coordinates
is projective.
If the abscissa axis coincides with the line and the origin coincides with the point
then the isogonals define the equations
and the lines
symmetrical with respect to the line
become their images.
It is clear that, under the converse transformation (also projective), such pairs of lines become isogonals, and the points equidistant from lie on the isogonals.
The isogonal theorem
Let two pairs of isogonals and
with respect to the pair
be given. Denote
Prove that and
are the isogonals with respect to the pair
Proof
Let us perform a projective transformation of the plane that maps the point into a point at infinity and the line
maps to itself. In this case, the isogonals turn into a pair of straight lines parallel to
and equidistant from
The converse (also projective) transformation maps the points equidistant from onto isogonals. We denote the image and the preimage with the same symbols.
Let the images of isogonals are vertical lines. Let coordinates of images of points be
Equation of a straight line
is
Equation of a straight line is
The abscissa of the point
is
Equation of a straight line is
Equation of a straight line is
The abscissa of the point
is
Preimages of the points and
lie on the isogonals.
The isogonal theorem in the case of parallel lines
Let and
are isogonals with respect
Let lines and
intersect at point
Prove that and line
through
parallel to
are the isogonals with respect
Proof
The preimage of is located at infinity on the line
The equality implies the equality the slopes modulo of
and
to the bisector of
Converse theorem
Let lines and
intersect at point
Let and
be the isogonals with respect
Prove that and
are isogonals with respect
Proof
The preimage of is located at infinity on the line
so the slope of
is known.
Suppose that
The segment and the lines
are fixed
intersects
at
but there is the only point where line intersect
Сontradiction.
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Parallel segments
Let triangle be given. Let
and
be the isogonals with respect
Let
Prove that lies on bisector of
and
Proof
Both assertions follow from The isogonal theorem in the case of parallel lines
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Perpendicularity
Let triangle be given. Right triangles
and
with hypotenuses
and
are constructed on sides
and
to the outer (inner) side of
Let
Prove that
Proof
Let be the bisector of
and
are isogonals with respect to the pair
and
are isogonals with respect to the pair
and
are isogonals with respect to the pair
in accordance with The isogonal theorem.
is the diameter of circumcircle of
Circumradius and altitude are isogonals with respect bisector and vertex of triangle, so
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Fixed point
Let fixed triangle be given. Let points
and
on sidelines
and
respectively be the arbitrary points.
Let be the point on sideline
such that
Prove that line
pass through the fixed point.
Proof
We will prove that point symmetric
with respect
lies on
.
and
are isogonals with respect to
points
and
lie on isogonals with respect to
in accordance with The isogonal theorem.
Point symmetric
with respect
lies on isogonal
with respect to
that is
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Bisector
Let a convex quadrilateral be given. Let
and
be the incenters of triangles
and
respectively.
Let and
be the A-excenters of triangles
and
respectively.
Prove that is the bisector of
Proof
and
are isogonals with respect to the angle
and
are isogonals with respect to the angle
in accordance with The isogonal theorem.
Denote
WLOG,
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Isogonal of the diagonal of a quadrilateral
Given a quadrilateral and a point
on its diagonal such that
Let
Prove that
Proof
Let us perform a projective transformation of the plane that maps the point to a point at infinity and the line
into itself.
In this case, the images of points and
are equidistant from the image of
the point (midpoint of
lies on
contains the midpoints of
and
is the Gauss line of the complete quadrilateral
bisects
the preimages of the points and
lie on the isogonals
and
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Isogonals in trapezium
Let the trapezoid be given. Denote
The point on the smaller base
is such that
Prove that
Proof
Therefore
and
are isogonals with respect
Let us perform a projective transformation of the plane that maps the point to a point at infinity and the line
into itself.
In this case, the images of points and
are equidistant from the image of
contains the midpoints of
and
, that is,
is the Gauss line of the complete quadrilateral
bisects
The preimages of the points and
lie on the isogonals
and
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Isogonals in complete quadrilateral
Let complete quadrilateral be given. Let
be the Miquel point of
Prove that is isogonal to
and
is isogonal to
with respect
Proof
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Isogonal of the bisector of the triangle
The triangle be given. The point
chosen on the bisector
Denote
Prove that
Proof
Let us perform a projective transformation of the plane that maps the point to a point at infinity and the line
into itself.
In this case, the images of segments and
are equidistant from the image of
Image of point is midpoint of image
and midpoint image
Image is parallelogramm
distances from
and
to
are equal
Preimages and
are isogonals with respect
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Points on isogonals
The triangle be given. The point
chosen on
The point
chosen on
such that
and
are isogonals with respect
Prove that
Proof
Denote
We use the Law of Sines and get:
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Trapezoid
The lateral side of the trapezoid
is perpendicular to the bases, point
is the intersection point of the diagonals
.
Point is taken on the circumcircle
of triangle
diametrically opposite to point
Prove that
Proof
WLOG, is not the diameter of
Let sidelines
and
intersect
at points
and
respectively.
is rectangle
is isogonal to
with respect
is isogonal to
with respect
In accordance with The isogonal theorem in case parallel lines
is isogonal to
with respect
in accordance with Converse theorem for The isogonal theorem in case parallel lines.
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IMO 2007 Short list/G3
The diagonals of a trapezoid intersect at point
Point lies between the parallel lines
and
such that
and line
separates points
and
Prove that
Proof
and
are isogonals with respect
is isogonal to
with respect
From the converse of The isogonal theorem we get
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Definition of isogonal conjugate of a point
Let triangle be given. Let
be the circumcircle of
Let point
be in the plane of
Denote by
the lines
respectively. Denote by
the lines
,
,
, respectively.
Denote by
,
,
the reflections of
,
,
over the angle bisectors of angles
,
,
, respectively.
Prove that lines ,
,
concur at a point
This point is called the isogonal conjugate of
with respect to triangle
.
Proof
By our constructions of the lines ,
, and this statement remains true after permuting
. Therefore by the trigonometric form of Ceva's Theorem
so again by the trigonometric form of Ceva, the lines
concur, as was to be proven.
Corollary
Let points P and Q lie on the isogonals with respect angles and
of triangle
Then these points lie on isogonals with respect angle
Corollary 2
Let point be in the sideline
of
Then the isogonal conjugate of a point is a point
Points and
do not have an isogonally conjugate point.
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Three points
Let fixed triangle be given. Let the arbitrary point
not be on sidelines of
Let
be the point on isogonal of
with respect angle
Let
be the crosspoint of isogonal of
with respect angle
and isogonal of
with respect angle
Prove that lines and
are concurrent.
Proof
Denote
and
are isogonals with respect
and S lie on isogonals of
is isogonal conjugated of
with respect
and
lie on isogonals of
Therefore points and
lie on the same line which is isogonal to
with respect
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Second definition
Let triangle be given. Let point
lies in the plane of
Let the reflections of
in the sidelines
be
Then the circumcenter of the
is the isogonal conjugate of
Points and
have not isogonal conjugate points.
Another points of sidelines have points
respectively as isogonal conjugate points.
Proof
is common therefore
Similarly
is the circumcenter of the
From definition 1 we get that is the isogonal conjugate of
It is clear that each point has the unique isogonal conjugate point.
Let point be the point with barycentric coordinates
Then
has barycentric coordinates
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Distance to the sides of the triangle
Let be the isogonal conjugate of a point
with respect to a triangle
Let and
be the projection
on sides
and
respectively.
Let and
be the projection
on sides
and
respectively.
Then
Proof
Let
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Sign of isogonally conjugate points
Let triangle and points
and
inside it be given.
Let be the projections
on sides
respectively.
Let be the projections
on sides
respectively.
Let Prove that point
is the isogonal conjugate of a point
with respect to a triangle
One can prove a similar theorem in the case outside
Proof
Denote
Similarly
Hence point
is the isogonal conjugate of a point
with respect to a triangle
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Circumcircle of pedal triangles
Let be the isogonal conjugate of a point
with respect to a triangle
Let be the projection
on sides
respectively.
Let be the projection
on sides
respectively.
Prove that points are concyclic.
The midpoint is circumcenter of
Proof
Let
Hence points are concyclic.
is trapezoid,
the midpoint is circumcenter of
Similarly points are concyclic and points
are concyclic.
Therefore points are concyclic, so the midpoint
is circumcenter of
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Common circumcircle of the pedal triangles as the sign of isogonally conjugate points
Let triangle and points
and
inside it be given. Let
be the projections
on sides
respectively.
Let
be the projections
on sides
respectively.
Let points be concyclic and none of them lies on the sidelines of
Then point is the isogonal conjugate of a point
with respect to a triangle
This follows from the uniqueness of the conjugate point and the fact that the line intersects the circle in at most two points.
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Two pares of isogonally conjugate points
Let triangle and points
and
be given. Let points
and
be the isogonal conjugate of a points
and
with respect to a triangle
respectively.
Let cross
at
and
cross
at
Prove that point is the isogonal conjugate of a point
with respect to
Proof
There are two pairs of isogonals and
with respect to the angle
are isogonals with respect to the
in accordance with The isogonal theorem.
Similarly are the isogonals with respect to the
Therefore the point is the isogonal conjugate of a point
with respect to
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Circles
Let be the isogonal conjugate of a point
with respect to a triangle
Let be the circumcenter of
Let be the circumcenter of
Prove that points and
are inverses with respect to the circumcircle of
Proof
The circumcenter of point
and points
and
lies on the perpendicular bisector of
Similarly
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1995 USAMO Problems/Problem 3
Given a nonisosceles, nonright triangle let
denote the center of its circumscribed circle, and let
and
be the midpoints of sides
and
respectively. Point
is located on the ray
so that
is similar to
. Points
and
on rays
and
respectively, are defined similarly. Prove that lines
and
are concurrent.
Solution
Let be the altitude of
Hence
and
are isogonals with respect to the angle
and
are isogonals with respect to the angle
Similarly and
are isogonals with respect to
Similarly and
are isogonals with respect to
Let be the centroid of
is the isogonal conjugate of a point
with respect to a triangle
Corollary
If median and symmedian start from any vertex of the triangle, then the angle formed by the symmedian and the angle side has the same measure as the angle between the median and the other side of the angle.
are medians, therefore
are symmedians, so the three symmedians meet at a point which is triangle center called the Lemoine point.
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2011 USAMO Problems/Problem 5
Let be a given point inside quadrilateral
. Points
and
are located within
such that
,
,
,
.
Prove that if and only if
.
Solution
Case 1 The lines and
are not parallel. Denote
Point
isogonal conjugate of a point
with respect to a triangle
and
are isogonals with respect to
Similarly point isogonal conjugate of a point
with respect to a triangle
and
are isogonals with respect to
Therefore points lies on the isogonal
with respect to
is not parallel to
or
Case 2 We use The isogonal theorem in case parallel lines and get
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Distance formulas for isogonal points
Let triangle and point
be given. Let point
be the isogonal conjugate of a point
with respect to a triangle
Let lines
and
cross sideline
at
and
and circumcircle of
at
and
respectively.
We apply the Isogonal’s property and get
We apply the Ptolemy's theorem to
and get
We apply the barycentric coordinates and get
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