Isogonal conjugate
Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.
Contents
[hide]- 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
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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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