Difference between revisions of "Isogonal conjugate"

Revision as of 02:03, 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 $P$ 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. $\blacksquare$

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. $\blacksquare$

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 2: / Line 2: @@
 == Definition of isogonal conjugate of a point ==
-[[File:Definitin 1.png|500px|right]]
+[[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>.
+<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 ==

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