Difference between revisions of "Isogonal conjugate"

Revision as of 11:47, 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 $P$ be a point in the plane, and let $ABC$ be a triangle. We will denote by $a,b,c$ the lines $BC, CA, AB$ . Let $p_a, p_b, p_c$ denote the lines $PA$ , $PB$ , $PC$ , respectively. Let $q_a$ , $q_b$ , $q_c$ be the reflections of $p_a$ , $p_b$ , $p_c$ over the angle bisectors of angles $A$ , $B$ , $C$ , respectively. Then lines $q_a$ , $q_b$ , $q_c$ concur at a point $Q$ , called the isogonal conjugate of $P$ with respect to triangle $ABC$ .

Proof

By our constructions of the lines $q$ , $\angle p_a b \equiv \angle q_a c$ , and this statement remains true after permuting $a,b,c$ . Therefore by the trigonometric form of Ceva's Theorem $\frac{\sin \angle q_a b}{\sin \angle c q_a} \cdot \frac{\sin \angle q_b c}{\sin \angle a q_b} \cdot \frac{\sin \angle q_c a}{\sin \angle b q_c} = \frac{\sin \angle p_a c}{\sin \angle b p_a} \cdot \frac{\sin \angle p_b a}{\sin \angle c p_b} \cdot \frac{\sin \angle p_c b}{\sin \angle a p_c} = 1,$ so again by the trigonometric form of Ceva, the lines $q_a, q_b, q_c$ concur, as was to be proven. $\blacksquare$

Second definition

Let triangle $\triangle ABC$ be given. Let point $P$ lies in the plane of $\triangle ABC,$ $P \notin AB, P \notin BC, P \notin AC.$ Let the reflections of $P$ in the sidelines $BC, CA, AB$ be $P_1, P_2, P_3.$

Then the circumcenter $Q$ of the $\triangle P_1P_2P_3$ is the isogonal conjugate of $P.$

Proof $PC = P_1C, PC = P_2C \implies P_1C = P_2C.$ $\angle ACQ = \angle BCP_1 \implies \angle QCP_1 = \angle ACB.$ $\angle BCQ = \angle ACP_2 \implies \angle QCP_2 = \angle ACB.$ $\angle QCP_1 = \angle QCP_2, CP_1 = CP_2, QC$ common $\implies$ $\triangle QCP_1 = \triangle QCP_2 \implies QP_1 = QP_2.$ Similarly $QP_1 = QP_3 \implies Q$ is the circumcenter of the $\triangle P_1P_2P_3.$ $\blacksquare$

Let point $P$ be the point with barycentric coordinates $(p : q : r),$ $p = [(P-B),(P-C)], q = [(P-C),(P-A)], r = [(P-A),(P-B)].$ Then $Q$ has barycentric coordinates $(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.$

vladimir.shelomovskii@gmail.com, vvsss

Distance to the sides of the triangle

Let $Q$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$ Let $E$ and $D$ are the projection $P$ on sides $AC$ and $BC,$ respectively. Let $E'$ and $D'$ are the projection $Q$ on sides $AC$ and $BC,$ respectively. Then $\frac {PE}{PD} = \frac{QD'}{QE'}.$ Proof Let $\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.$ $\frac {PE}{PD} = \frac {PC \sin \theta}{PC \sin \Theta} = \frac {QC \sin \theta}{QC \sin \Theta} = \frac {QD'}{QE'}.$ vladimir.shelomovskii@gmail.com, vvsss

Problems

Olympiad

Given a nonisosceles, nonright triangle $ABC,$ let $O$ denote the center of its circumscribed circle, and let $A_1, \, B_1,$ and $C_1$ be the midpoints of sides $BC, \, CA,$ and $AB,$ respectively. Point $A_2$ is located on the ray $OA_1$ so that $\triangle OAA_1$ is similar to $\triangle OA_2A$ . Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1,$ respectively, are defined similarly. Prove that lines $AA_2, \, BB_2,$ and $CC_2$ are concurrent, i.e. these three lines intersect at a point. (Source)

Let $P$ be a given point inside quadrilateral $ABCD$ . Points $Q_1$ and $Q_2$ are located within $ABCD$ such that $\angle Q_1 BC = \angle ABP$ , $\angle Q_1 CB = \angle DCP$ , $\angle Q_2 AD = \angle BAP$ , $\angle Q_2 DA = \angle CDP$ . Prove that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if $\overline{Q_1 Q_2} \parallel \overline{CD}$ . (Source)

@@ Line 31: / Line 31: @@
 Then <math>Q</math> has barycentric coordinates <cmath>(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.</cmath>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Distance to the sides of the triangle==
+Let <math>Q</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> Let <math>E</math> and <math>D</math> are the projection <math>P</math> on sides <math>AC</math> and <math>BC,</math> respectively. Let <math>E'</math> and <math>D'</math> are the projection <math>Q</math> on sides <math>AC</math> and <math>BC,</math> respectively. Then <math>\frac {PE}{PD} = \frac{QD'}{QE'}.</math>
+<i><b>Proof</b></i>
+Let <math>\theta = \angle ACP = \angle BCQ, \Theta =  \angle ACQ = \angle BCP.</math>
+<math>\frac {PE}{PD} = \frac {PC \sin \theta}{PC \sin \Theta} = \frac {QC \sin \theta}{QC \sin \Theta}  = \frac {QD'}{QE'}.</math>
 '''vladimir.shelomovskii@gmail.com, vvsss'''

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