Difference between revisions of "Isogonal conjugate"
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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> | ||
− | == | + | == Second definition == |
+ | For a given point <math>P</math> in the plane of triangle <math>\triangle ABC,</math> let the reflections of <math>P</math> in the sidelines <math>BC, CA, AB</math> be <math>P_1, P_2, P_3.</math> Then the center of the circle <math>P_1P_2P_3</math> is the isogonal conjugate of <math>P.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
By our constructions of the lines <math>q</math>, <math>\angle p_a b \equiv \angle q_a c</math>, and this statement remains true after permuting <math>a,b,c</math>. Therefore by [[Ceva's Theorem | the trigonometric form of Ceva's Theorem]] | By our constructions of the lines <math>q</math>, <math>\angle p_a b \equiv \angle q_a c</math>, and this statement remains true after permuting <math>a,b,c</math>. Therefore by [[Ceva's Theorem | the trigonometric form of Ceva's Theorem]] |
Revision as of 01:12, 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
For a given point in the plane of triangle
let the reflections of
in the sidelines
be
Then the center of the circle
is the isogonal conjugate of
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.
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)