Difference between revisions of "Isogonal conjugate"

Revision as of 15:32, 26 February 2023

Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.

1 The isogonal theorem
2 Isogonal of the diagonal of a quadrilateral
3 Definition of isogonal conjugate of a point
4 Second definition
5 Distance to the sides of the triangle
6 Sign of isogonally conjugate points
7 Circumcircle of pedal triangles
8 Common circumcircle of the pedal triangles as the sign of isogonally conjugate points
9 Circles
10 Problems
- 10.1 Olympiad

The isogonal theorem

Isogonal lines definition

Let a line $\ell$ and a point $O$ lying on $\ell$ be given. A pair of lines symmetric with respect to $\ell$ and containing the point $O$ be called isogonals with respect to the pair $(\ell,O).$

Sometimes it is convenient to take one pair of isogonals as the base one, for example, $OA$ and $OB$ are the base pair. Then we call the remaining pairs as isogonals with respect to the angle $\angle AOB.$

Projective transformation

It is known that the transformation that maps a point with coordinates $(x,y)$ into a point with coordinates $(\frac{1}{x}, \frac {y}{x}),$ is projective.

If the abscissa axis coincides with the line $\ell$ and the origin coincides with the point $O,$ then the isogonals define the equations $y = \pm kx,$ and the lines $(\frac{1}{x}, \pm k)$ symmetrical with respect to the line $\ell$ become their images.

It is clear that, under the reverse transformation (also projective), such pairs of lines become isogonals, and the points equidistant from $\ell$ lie on the isogonals.

The isogonal theorem

Let two pairs of isogonals $OX – OX'$ and $OY – OY'$ be given. Let lines $XY$ and $X'Y'$ intersect at point $Z.$ Let lines $X'Y$ and $XY'$ intersect at point $Z'.$ Prove that $OZ$ and $OZ'$ are the isogonals with respect to the pair $(\ell,O).$

Proof

Let us perform a projective transformation of the plane that maps the point $O$ into a point at infinity and the line $\ell$ maps to itself. In this case, the isogonals turn into a pair of straight lines parallel to $\ell$ and equidistant from $\ell.$

The reverse (also projective) transformation maps the points equidistant from $\ell$ onto isogonals.

Let the images of isogonals are vertical lines. Let coordinates of images of points be $X(-a, 0), X'(a,u), Y(-b,v), Y'(b,w).$ Equation of a straight line $XY$ is $\frac{x+a}{a-b} = \frac {y}{v},$

Equation of a straight line $X'Y'$ is $\frac{x-a}{b-a} = \frac {y-u}{w-u},$

Point $Z$ abscissa $Z_x = \frac {v a - a w + u b}{u-v-w}.$

Equation of a straight line $XY'$ is $\frac{x+a}{b+a} = \frac {y}{w},$

Equation of a straight line $X'Y$ is $\frac{x-a}{-b-a} = \frac {y-u}{v-u},$

Point $Z'$ abscissa $Z'_x = \frac {v a - a w + u b}{-u+v+w} = - Z_x \implies$

Preimages of the points $Z$ and $Z'$ lie on the isogonals.

vladimir.shelomovskii@gmail.com, vvsss

Isogonal of the diagonal of a quadrilateral

Given a quadrilateral $ABCD$ and a point $P$ on its diagonal such that $\angle APB = \angle APD.$ Let $E = AB \cap CD, F = AD \cap BC.$ Prove that $\angle EPB = \angle FPD.$

Proof

Let us perform a projective transformation of the plane that maps the point $P$ to a point at infinity and the line $\ell = AC$ into itself. In this case, the images of points $B$ and $D$ are equidistant from the image of $AC \implies$

the point $M$ (midpoint of BD) lies on $L \implies AC$ contains the midpoints of $AC$ and $BD,$ that is, $\ell$ is the Gauss line of the complete quadrilateral $ABCDEF \implies \ell$ bisects $EF \implies EE_0 = FF_0 \implies$ the preimages of the points $E$ and $F$ lie on the isogonals $PE$ and $PF.$

vladimir.shelomovskii@gmail.com, vvsss

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 $P.$

Points $A, B,$ and $C$ have not isogonal conjugate points.

Another points of sidelines $BC, AC, AB$ have points $A, B, C,$ respectively as isogonal conjugate points.

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 $\triangle P_1P_2P_3.$ $\blacksquare$

From definition 1 we get that $P$ is the isogonal conjugate of $Q.$

It is clear that each point $P$ has the unique isogonal conjugate point.

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 $D$ be the projection $P$ on sides $AC$ and $BC,$ respectively.

Let $E'$ and $D'$ be 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

Sign of isogonally conjugate points

Let triangle $\triangle ABC$ and points $P$ and $Q$ inside it be given.

Let $D, E, F$ be the projections $P$ on sides $BC, AC, AB,$ respectively.

Let $D', E', F'$ be the projections $Q$ on sides $BC, AC, AB,$ respectively.

Let $\frac {PE}{PD} = \frac{QD'}{QE'}, \frac {PF}{PD} = \frac{QD'}{QF'}.$ Prove that point $Q$ is the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

One can prove similar theorem in the case $P$ outside $\triangle ABC.$

Proof

$\frac {PE}{PD} = \frac {PE}{PC} : \frac {PD}{PC} = \frac {\sin \angle ACP}{\sin \angle BCP},$ $\frac {QD'}{QE'} = \frac {QD'}{QC} : \frac {QE'}{QC} = \frac {\sin \angle BCQ}{\sin \angle ACQ}.$

Denote $\angle ACP = \varphi, \angle BCQ = \psi, \angle ACB = \gamma.$ $\sin \varphi \cdot \sin (\gamma - \psi) = \sin \psi \cdot \sin (\gamma - \varphi) \implies$ $\cos (\varphi - \gamma + \psi) - \cos(\varphi + \gamma - \psi) = \cos (\psi - \gamma + \varphi) - \cos(\psi + \gamma - \varphi)$ $\cos (\gamma + \varphi -\psi) = \cos(\gamma - \psi + \varphi) \implies$ $\cos \gamma \cos (\varphi - \psi) - \sin \gamma \sin (\varphi - \psi) = \cos \gamma \cos (\varphi - \psi) + \sin \gamma \sin (\varphi - \psi)$ $2 \sin \gamma \cdot \sin (\varphi - \psi) = 0, \varphi + \psi < 180^\circ \implies \varphi = \psi.$ Similarly $\angle ABP = \angle CBQ \implies$ point $Q$ is the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Circumcircle of pedal triangles

Let $Q$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$ Let $E, D, F$ be the projection $P$ on sides $AC, BC, AB,$ respectively.

Let $E', D', F'$ be the projection $Q$ on sides $AC, BC, AB,$ respectively.

Then points $D, D', E, E', F, F'$ are concyclic.

The midpoint $PQ$ is circumcenter of $DD'EE'FF'.$

Proof

Let $\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.$ $CE \cdot CE' = PC \cos \theta \cdot QC \cos \Theta = PC \cos \Theta \cdot QC \cos \theta = CD \cdot CD'.$ Hence points $D, D', E, E'$ are concyclic.

$PQE'E$ is trapezoid, $E'E \perp PE \implies OE = OE' \implies$

the midpoint $PQ$ is circumcenter of $DD'EE'.$

Similarly points $D, D', F, F'$ are concyclic and points $F, F', E, E'$ are concyclic.

Therefore points $D, D', E, E', F, F'$ are concyclic, so the midpoint $PQ$ is circumcenter of $DD'EE'FF'.$

vladimir.shelomovskii@gmail.com, vvsss

Common circumcircle of the pedal triangles as the sign of isogonally conjugate points

Let triangle $\triangle ABC$ and points $P$ and $Q$ inside it be given. Let $D, E, F$ be the projections $P$ on sides $BC, AC, AB,$ respectively. Let $D', E', F'$ be the projections $Q$ on sides $BC, AC, AB,$ respectively.

Let points $D, E, F, D', E', F'$ be concyclic and none of them lies on the sidelines of $\triangle ABC.$

Then point $Q$ is the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

This follows from the uniqueness of the conjugate point and the fact that the line intersects the circle in at most two points.

vladimir.shelomovskii@gmail.com, vvsss

Circles

Let $Q$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$ Let $D$ be the circumcenter of $\triangle BCP.$ Let $E$ be the circumcenter of $\triangle BCQ.$ Prove that points $D$ and $E$ are inverses with respect to the circumcircle of $\triangle ABC.$

Proof

The circumcenter of $\triangle ABC$ point $O,$ and points $D$ and $E$ lies on the perpendicular bisector of $BC.$ $\angle BOD = \angle COE = \angle BAC.$ $2 \angle BDO = \angle BDC = \overset{\Large\frown} {BC} = 360^\circ - \overset{\Large\frown} {CB} = 360^\circ - 2 \angle BPC.$ $\angle BDO = 180^\circ - \angle BPC = \angle PBC + \angle PCB.$ Similarly $\angle CEO = 180^\circ - \angle BQC = \angle QBC + \angle QCB.$ $\angle PBC + \angle QBC = \angle PBC + \angle PBA = \angle ABC.$ $\angle QCB + \angle PCB = \angle QCB + \angle QCA = \angle ACB.$

$\angle CEO +\angle BDO = \angle ABC + \angle ACB = 180^\circ - \angle BAC \implies$ $\angle OBD = 180^\circ - \angle BOD - \angle BDO = \angle OEC \implies$ $\triangle OBD \sim \triangle OEC \implies \frac {OB}{OE} = \frac {OD}{OC} \implies OD \cdot OE = OB^2. \blacksquare$

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 41: / Line 41: @@
 Preimages of the points <math>Z</math> and <math>Z'</math> lie on the isogonals.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Isogonal of the diagonal of a quadrilateral==
+[[File:Quadrungle isogonals.png|350px|right]]
+Given a quadrilateral <math>ABCD</math> and a point <math>P</math> on its diagonal such that <math>\angle APB = \angle APD.</math> Let <math>E = AB \cap CD, F = AD \cap BC.</math> Prove that <math>\angle EPB = \angle FPD.</math>
+<i><b>Proof</b></i>
+Let us perform a projective transformation of the plane that maps the point <math>P</math> to a point at infinity and the line <math>\ell = AC</math> into itself. In this case, the images of points <math>B</math> and <math>D</math> are equidistant from the image of <math>AC \implies</math>
+the point <math>M</math> (midpoint of BD) lies on <math>L \implies AC</math> contains the midpoints of <math>AC</math> and <math>BD,</math> that is, <math>\ell</math> is the Gauss line of the complete quadrilateral <math>ABCDEF \implies \ell</math> bisects <math>EF \implies EE_0 = FF_0 \implies</math> the preimages of the points <math>E</math> and <math>F</math> lie on the isogonals <math>PE</math> and <math>PF.</math>
 '''vladimir.shelomovskii@gmail.com, vvsss'''

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Difference between revisions of "Isogonal conjugate"

Revision as of 15:32, 26 February 2023

Contents

The isogonal theorem

Isogonal of the diagonal of a quadrilateral

Definition of isogonal conjugate of a point

Second definition

Distance to the sides of the triangle

Sign of isogonally conjugate points

Circumcircle of pedal triangles

Common circumcircle of the pedal triangles as the sign of isogonally conjugate points

Circles

Problems

Olympiad