Difference between revisions of "Isogonal conjugate"
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The circumcenter of <math>\triangle ABC</math> point <math>O,</math> and points <math>D</math> and <math>E</math> lies on the perpendicular bisector of <math>BC.</math> | The circumcenter of <math>\triangle ABC</math> point <math>O,</math> and points <math>D</math> and <math>E</math> lies on the perpendicular bisector of <math>BC.</math> | ||
− | < | + | <cmath>\angle BOD = \angle COE = \angle BAC.</cmath> |
− | < | + | <cmath>2 \angle BDO = \angle BDC = \overset{\Large\frown} {BC} = 360^\circ - \overset{\Large\frown} {CB} = 360^\circ - 2 \angle BPC.</cmath> |
− | < | + | <cmath>\angle BDO = 180^\circ - \angle BPC = \angle PBC + \angle PCB.</cmath> |
Similarly <math>\angle CEO = 180^\circ - \angle BQC = \angle QBC + \angle QCB.</math> | Similarly <math>\angle CEO = 180^\circ - \angle BQC = \angle QBC + \angle QCB.</math> | ||
<cmath>\angle PBC + \angle QBC = \angle PBC + \angle PBA = \angle ABC.</cmath> | <cmath>\angle PBC + \angle QBC = \angle PBC + \angle PBA = \angle ABC.</cmath> |
Revision as of 14:40, 6 February 2023
Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.
Contents
[hide]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.
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
Proof common Similarly is the circumcenter of the
Let point be the point with barycentric coordinates Then has barycentric coordinates
vladimir.shelomovskii@gmail.com, vvsss
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 vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss
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)