Isogonal conjugate

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

1 The isogonal theorem
2 Perpendicularity
3 Fixed point
4 Bisector
5 Isogonal of the diagonal of a quadrilateral
6 Isogonals in trapezium
7 Isogonal of the bisector of the triangle
8 Parallel segments
9 Definition of isogonal conjugate of a point
10 Three points
11 Second definition
12 Distance to the sides of the triangle
13 Sign of isogonally conjugate points
14 Circumcircle of pedal triangles
15 Common circumcircle of the pedal triangles as the sign of isogonally conjugate points
16 Circles
17 Problems
- 17.1 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.

Corollary

If $XY || X'Y'$ then $OZ' || XY.$

If $XY' || X'Y$ then $OZ || X'Y.$

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Perpendicularity

Let triangle $ABC$ be given. Right triangles $ABD$ and $ACE$ with hypotenuses $AD$ and $AE$ are constructed on sides $AB$ and $AC$ to the outer (inner) side of $\triangle ABC.$ Let $\angle BAD = \angle CAE, H = CD \cap BE.$ Prove that $AH \perp BC.$

Proof

Let $\ell$ be the bisector of $\angle BAC, F = BD \cap CE.$

$AB$ and $AC$ are isogonals with respect to the pair $(\ell,A).$

$AD$ and $AE$ are isogonals with respect to the pair $(\ell,A) \implies$

$AH$ and $AF$ are isogonals with respect to the pair $(\ell,A)$ in accordance with The isogonal theorem.

$\angle ABD = \angle ACE = 90^\circ \implies AD$ is diameter of circumcircle of $\triangle ABC \implies AH \perp BC.$

vladimir.shelomovskii@gmail.com, vvsss

Fixed point

Let fixed triangle $ABC$ be given. Let points $D$ and $E$ on sidelines $BC$ and $AB$ respectively be the arbitrary points.

Let $F$ be the point on sideline $AC$ such that $\angle BDE = \angle FDC. G = BF \cap CE.$

Prove that line $DG$ pass through the fixed point.

Proof

We will prove that point $A',$ symmetric $A$ with respect $\ell = BC,$ lies on $DG$ .

$\angle BDE = \angle FDC \implies DE$ and $DF$ are isogonals with respect to $(\ell, D).$

$A = BE \cap CF \implies$ points $A$ and $G$ lie on isogonals with respect to $(\ell, D)$ in accordance with The isogonal theorem.

Point $A'$ symmetric $A$ with respect $\ell$ lies on isogonal $AD$ with respect to $(\ell, D),$ that is $DG.$

vladimir.shelomovskii@gmail.com, vvsss

Bisector

Let a convex quadrilateral $ABCD$ be given. Let $I$ and $J$ be the incenters of triangles $\triangle ABC$ and $\triangle ADC,$ respectively. Let $I'$ and $J'$ be the A-excenters of triangles $\triangle ABC$ and $\triangle ADC,$ respectively. $E = IJ' \cap I'J.$

Prove that $CE$ is the bisector of $\angle BCD.$

Proof

$\angle ICI' = \angle JCJ' = 90^\circ \implies CI'$ and $CJ'$ are isogonals with respect to the angle $\angle ICJ.$

$A = II' \cap JJ' \implies AC$ and $EC$ are isogonals with respect to the angle $\angle ICJ$ in accordance with The isogonal theorem.

Denote $\angle ACI = \angle BCI = \alpha, \angle ACJ = \angle DCJ = \beta.$

WLOG, $\beta \ge \alpha.$ $\angle ACJ = \angle ACE + \alpha = \beta,$ $\angle BCE = 2 \alpha + \beta - \alpha = \alpha + \beta = \angle DCE.$

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 $\ell \implies$

$AC$ contains the midpoints of $AC$ and $BD \implies$

$\ell$ 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

Isogonals in trapezium

Let the trapezoid $AEFC, AC||EF,$ be given. Denote $B = AE \cap CF, D = AF \cap CE.$

The point $M$ on the smaller base $AC$ is such that $EM = MF.$

Prove that $\angle AMB = \angle AMD.$

Proof

$EM = MF \implies \angle AME = \angle MEF = \angle MFE = \angle CMF.$ Therefore $EM$ and $FM$ are isogonals.

Let us perform a projective transformation of the plane that maps the point $M$ to a point at infinity and the line $\ell = AC$ into itself.

In this case, the images of points $E$ and $F$ are equidistant from the image of $\ell \implies AC$ contains the midpoints of $AC$ and $EF$ , that is, $\ell$ is the Gauss line of the complete quadrilateral $ABCDEF \implies$

$\ell$ bisects $BD \implies BB_0 = DD_0 \implies$

The preimages of the points $B$ and $D$ lie on the isogonals $MB$ and $MD.$

vladimir.shelomovskii@gmail.com, vvsss

Isogonal of the bisector of the triangle

The triangle $ABC$ be given. The point $D$ chosen on the bisector $AA'.$

Denote $B' = BD \cap AC, C' = CD \cap AB, E = BB' \cap A'C', F = CC' \cap A'B'.$

Prove that $\angle BAE = \angle CAF.$

Proof

Let us perform a projective transformation of the plane that maps the point $A$ to a point at infinity and the line $\ell = AA'$ into itself.

In this case, the images of segments $BC'$ and $B'C$ are equidistant from the image of $\ell \implies BC' || B'C || \ell.$

$D$ is midpoint of $BB'$ and midpoint $CC' \implies BCB'C'$ is parallelogramm $\implies$

$BC = B'C' \implies \frac {DE}{BD} = \frac {DF}{CD} \implies$ distances from $E$ and $F$ to $\ell$ are equal

$\implies$ Preimages $AE$ and $AF$ are isogonals with respect $(\ell,A).$

vladimir.shelomovskii@gmail.com, vvsss

Parallel segments

Let triangle $ABC$ be given.

Let $AD$ and $AE$ be the isogonals with respect $\angle BAC.$

Let $BD ||CE, P= BE \cap CD.$

Prove that $P$ lies on bisector of $\angle BAC$ and $BD||AP.$

Proof

Both assertions follow from the corollary of The isogonal theorem.

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. $\blacksquare$

Corollary

Let points P and Q lie on the isogonals with respect angles $\angle B$ and $\angle C$ of triangle $\triangle ABC.$

Then these points lie on isogonals with respect angle $\angle A.$

Three points

Let fixed triangle $ABC$ be given. Let the arbitrary point $D$ not be on sidelines of $\triangle ABC.$ Let $E$ be the point on isogonal of $CD$ with respect angle $\angle ACB.$ Let $F$ be the crosspoint of isogonal of $BD$ with respect angle $\angle ABC$ and isogonal of $AE$ with respect angle $\angle BAC.$

Prove that lines $AD, BE,$ and $CF$ are concurrent.

Proof

Denote $D' = BF \cap CE, S = CF \cap BE.$

$AE$ and $AF$ are isogonals with respect $\angle BAC \implies$

$D'$ and S lie on isogonals of $\angle BAC.$

$\angle DBC = \angle D'BA, \angle DCB = \angle D'CA \implies$

$D'$ is isogonal conjugated of $D$ with respect $\triangle ABC \implies$

$D'$ and $D$ lie on isogonals of $\angle BAC.$

Therefore points $A, S$ and $D$ lie on the same line which is isogonal to $AD'$ with respect $\angle BAC.$ $\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

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)

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Isogonal conjugate

Contents

The isogonal theorem

Perpendicularity

Fixed point

Bisector

Isogonal of the diagonal of a quadrilateral

Isogonals in trapezium

Isogonal of the bisector of the triangle

Parallel segments

Definition of isogonal conjugate of a point

Three points

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