Difference between revisions of "Isogonal conjugate"
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<cmath> \frac{\sin \angle q_a b}{\sin \angle c q_a} \cdot \frac{\sin \angle q_b c}{\sin \angle a q_b} \cdot \frac{\sin \angle q_c a}{\sin \angle b q_c} = \frac{\sin \angle p_a c}{\sin \angle b p_a} \cdot \frac{\sin \angle p_b a}{\sin \angle c p_b} \cdot \frac{\sin \angle p_c b}{\sin \angle a p_c} = 1, </cmath> | <cmath> \frac{\sin \angle q_a b}{\sin \angle c q_a} \cdot \frac{\sin \angle q_b c}{\sin \angle a q_b} \cdot \frac{\sin \angle q_c a}{\sin \angle b q_c} = \frac{\sin \angle p_a c}{\sin \angle b p_a} \cdot \frac{\sin \angle p_b a}{\sin \angle c p_b} \cdot \frac{\sin \angle p_c b}{\sin \angle a p_c} = 1, </cmath> | ||
so again by the trigonometric form of Ceva, the lines <math>q_a, q_b, q_c</math> concur, as was to be proven. <math>\blacksquare</math> | so again by the trigonometric form of Ceva, the lines <math>q_a, q_b, q_c</math> concur, as was to be proven. <math>\blacksquare</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Let points P and Q lie on the isogonals with respect angles <math>\angle B</math> and <math>\angle C</math> of triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | Then these points lie on isogonals with respect angle <math>\angle A.</math> | ||
== Second definition == | == Second definition == |
Revision as of 11:29, 28 February 2023
Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.
Contents
[hide]- 1 The isogonal theorem
- 2 Perpendicularity
- 3 Fixed point
- 4 Bisector
- 5 Isogonal of the diagonal of a quadrilateral
- 6 Isogonals in trapezium
- 7 Isogonal of the bisector of the triangle
- 8 Definition of isogonal conjugate of a point
- 9 Second definition
- 10 Distance to the sides of the triangle
- 11 Sign of isogonally conjugate points
- 12 Circumcircle of pedal triangles
- 13 Common circumcircle of the pedal triangles as the sign of isogonally conjugate points
- 14 Circles
- 15 Problems
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 reverse 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 be given. Let lines and intersect at point Let lines and intersect at point 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 reverse (also projective) transformation maps the points equidistant from onto isogonals.
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
Point abscissa
Equation of a straight line is
Equation of a straight line is
Point abscissa
Preimages of the points and lie on the isogonals.
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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 diameter of circumcircle of
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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.
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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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
is midpoint of and midpoint is parallelogramm
distances from and to are equal
Preimages and are isogonals with respect
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Definition of isogonal conjugate of a point
Let be a point in the plane, and let be a triangle. We will denote by the lines . Let denote the lines , , , respectively. Let , , be the reflections of , , over the angle bisectors of angles , , , respectively. Then lines , , concur at a point , 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
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 common 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 similar theorem in the case outside
Proof
Denote Similarly 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.
Then 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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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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Problems
Olympiad
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, i.e. these three lines intersect at a point. (Source)
Let be a given point inside quadrilateral . Points and are located within such that , , , . Prove that if and only if . (Source)