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Difference between revisions of "Isogonal conjugate"

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== Second definition ==
 
== Second definition ==
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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>
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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 circumcenter of the <math>\triangle P_1P_2P_3</math> is the isogonal conjugate of <math>P.</math>
  
 
<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
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<cmath>PC = P_1C, PC = P_2C \implies P_1C = P_2C.</cmath>
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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]]
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<cmath>\angle ACQ = \angle BCP_1 \implies \angle QCP_1 = \angle ACB.</cmath>
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<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>
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<cmath>\angle BCQ = \angle ACP_2 \implies \angle QCP_2 = \angle ACB.</cmath>
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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>
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<math>\angle QCP_1 = \angle QCP_2, CP_1 = CP_2, QC</math> common <math>\implies</math>
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<cmath>\triangle QCP_1 = \triangle QCP_2 \implies QP_1 = QP_2.</cmath>
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Similarly  <math>QP_1 = QP_3 \implies Q</math> is  the circumcenter of the <math>\triangle P_1P_2P_3.</math> <math>\blacksquare</math>
  
 
== Problems ==
 
== Problems ==

Revision as of 06:34, 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

Definitin 1.png

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

For a given point $P$ in the plane of triangle $\triangle ABC,$ let the reflections of $P$ in the sidelines $BC, CA, AB$ be $P_1, P_2, P_3.$ Then the circumcenter 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 $\triangle P_1P_2P_3.$ $\blacksquare$

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)

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