# Isogonal conjugate

**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 Simplified distance formula for isogonal points
- 25 Point on circumcircle
- 26 Fixed point on circumcircle
- 27 Distance formula for isogonal points
- 28 Miquel point for isogonal conjugate points
- 29 Point on circumcircle
- 30 Isogonal of line BC with respect to angle BAC
- 31 Isogonal bijection lines and points
- 32 Miquel point for two pare 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

*and get*

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## Simplified distance formula for isogonal points

Let triangle points and and be given. Let point be the isogonal conjugate of a point with respect to a triangle Prove that

**Proof**

and are both subtended by arc Similarly Product of isogonal segments

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## Point on circumcircle

Let triangle points and be given.

Denote Prove that

**Proof**

WLOG, the order of the points is as shown on diagram.

The spiral symilarity centered at maps to and point to point

is the external angle of

**Corollary**

is the isogonal conjugate to with respect

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## Fixed point on circumcircle

Let triangle point on circumcircle and point be given.

Point lies on point be the isogonal conjugate of a point with respect to a triangle

Prove that is fixed point and not depends from position of

**Proof**

WLOG, the order of points on sideline is point is closer to than to

Denote

Spiral similarity centered at which maps into transform point into point Points and are collinear.

It is known ( Ratio of isogonal segments) that

We use the ratio of the areas and get: Denote Therefore which means ( Problems | Simple) that is the radical axes of and

and not depends from position of

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## Distance formula 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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## Miquel point for isogonal conjugate points

Let triangle points and be given. Let point be the isogonal conjugate of a point with respect to a triangle Let be the Miquel point of a complete quadrilateral

Prove that lies on the circumcircle of

**Proof**

Point is the isogonal conjugate of a point with respect to a triangle so point is the isogonal conjugate of a point with respect to a triangle

Points and lies on the same line, therefore

Point lies on circles and spiral similarity centered at transform triangle to

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## Point on circumcircle

Let triangle and points and be given.

Let

Let lines and be the isogonals with respect to the angle

Let be an arbitrary point on

Prove that lies on

### Simplified problem

Let and points and be given,

Let lines and be the isogonals with respect to

Prove that

**Proof, Simplified problem**

points are concyclic on

**Proof**

Let points and be the isogonal conjugate of a points and with respect to a triangle

It is known that

points are concyclic on

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## Isogonal of line BC with respect to angle BAC

Let triangle be given,

Let lines and be the isogonals with respect to

Prove that is tangent to

**Proof**

Let and be the circumcenter and the orthocenter of respectively. is isogonal to with respect to is tangent to

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## Isogonal bijection lines and points

Let triangle and line be given,

Define the point with property

Prove that is equal the angle between and

**Proof**

WLOG, the configuration is the same as shown on diagram, is the tangent to

is isogonal to is isogonal to with respect to A bijection has been established between the set of lines parallel to a given one and the set of points of the circumcircle.

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## Miquel point for two pare isogonal points

Let triangle and points and be given.

Let points and be the isogonal conjugate of the points and with respect to is the Miquel point of quadrilateral

Prove that

**Proof**

Denote

Then is the Miquel point of quadrilateral

Denote

Let be the point with property

WLOG, configuration is similar as shown in diagram.

( Isogonal_bijection_lines_and_points).

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