Difference between revisions of "Isogonal conjugate"
(→Construction and Theorem) |
(→Definition of isogonal conjugate of a point) |
||
Line 2: | Line 2: | ||
== Definition of isogonal conjugate of a point == | == Definition of isogonal conjugate of a point == | ||
− | [[File:Definitin 1.png| | + | [[File:Definitin 1.png|400px|right]] |
Let <math>P</math> be a point in the plane, and let <math>ABC</math> be a triangle. We will denote by <math>a,b,c</math> the lines <math>BC, CA, AB</math>. Let <math>p_a, p_b, p_c</math> denote the lines <math>PA</math>, <math>PB</math>, <math>PC</math>, respectively. Let <math>q_a</math>, <math>q_b</math>, <math>q_c</math> be the reflections of <math>p_a</math>, <math>p_b</math>, <math>p_c</math> over the angle bisectors of angles <math>A</math>, <math>B</math>, <math>C</math>, respectively. Then lines <math>q_a</math>, <math>q_b</math>, <math>q_c</math> [[concurrence | concur]] at a point <math>Q</math>, called the isogonal conjugate of <math>P</math> with respect to triangle <math>ABC</math>. | Let <math>P</math> be a point in the plane, and let <math>ABC</math> be a triangle. We will denote by <math>a,b,c</math> the lines <math>BC, CA, AB</math>. Let <math>p_a, p_b, p_c</math> denote the lines <math>PA</math>, <math>PB</math>, <math>PC</math>, respectively. Let <math>q_a</math>, <math>q_b</math>, <math>q_c</math> be the reflections of <math>p_a</math>, <math>p_b</math>, <math>p_c</math> over the angle bisectors of angles <math>A</math>, <math>B</math>, <math>C</math>, respectively. Then lines <math>q_a</math>, <math>q_b</math>, <math>q_c</math> [[concurrence | concur]] at a point <math>Q</math>, called the isogonal conjugate of <math>P</math> with respect to triangle <math>ABC</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]] | ||
+ | <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> | ||
== Proof == | == Proof == |
Revision as of 01:03, 6 February 2023
Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.
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.
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)