# Euler line

In any triangle $\triangle ABC$, the Euler line is a line which passes through the orthocenter $H$, centroid $G$, circumcenter $O$, nine-point center $N$ and de Longchamps point $L$. It is named after Leonhard Euler. Its existence is a non-trivial fact of Euclidean geometry. Certain fixed orders and distance ratios hold among these points. In particular, $\overline{OGNH}$ and $OG:GN:NH = 2:1:3$

Euler line is the central line $L_{647}$.

Given the orthic triangle $\triangle H_AH_BH_C$ of $\triangle ABC$, the Euler lines of $\triangle AH_BH_C$,$\triangle BH_CH_A$, and $\triangle CH_AH_B$ concur at $N$, the nine-point circle of $\triangle ABC$.

## Proof Centroid Lies on Euler Line

This proof utilizes the concept of spiral similarity, which in this case is a rotation followed homothety. Consider the medial triangle $\triangle O_AO_BO_C$. It is similar to $\triangle ABC$. Specifically, a rotation of $180^\circ$ about the midpoint of $O_BO_C$ followed by a homothety with scale factor $2$ centered at $A$ brings $\triangle ABC \to \triangle O_AO_BO_C$. Let us examine what else this transformation, which we denote as $\mathcal{S}$, will do.

It turns out $O$ is the orthocenter, and $G$ is the centroid of $\triangle O_AO_BO_C$. Thus, $\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}$. As a homothety preserves angles, it follows that $\measuredangle O_AOG = \measuredangle AHG$. Finally, as $\overline{AH} || \overline{O_AO}$ it follows that $$\triangle AHG = \triangle O_AOG$$ Thus, $O, G, H$ are collinear, and $\frac{OG}{HG} = \frac{1}{2}$.

## Another Proof

Let $M$ be the midpoint of $BC$. Extend $CG$ past $G$ to point $H'$ such that $CG = \frac{1}{2} GH$. We will show $H'$ is the orthocenter. Consider triangles $MGO$ and $AGH'$. Since $\frac{MG}{GA}=\frac{H'G}{GC} = \frac{1}{2}$, and they both share a vertical angle, they are similar by SAS similarity. Thus, $AH' \parallel OM \perp BC$, so $H'$ lies on the $A$ altitude of $\triangle ABC$. We can analogously show that $H'$ also lies on the $B$ and $C$ altitudes, so $H'$ is the orthocenter.

## Proof Nine-Point Center Lies on Euler Line

Assuming that the nine point circle exists and that $N$ is the center, note that a homothety centered at $H$ with factor $2$ brings the Euler points $\{E_A, E_B, E_C\}$ onto the circumcircle of $\triangle ABC$. Thus, it brings the nine-point circle to the circumcircle. Additionally, $N$ should be sent to $O$, thus $N \in \overline{HO}$ and $\frac{HN}{ON} = 1$.

## Analytic Proof of Existence

Let the circumcenter be represented by the vector $O = (0, 0)$, and let vectors $A,B,C$ correspond to the vertices of the triangle. It is well known the that the orthocenter is $H = A+B+C$ and the centroid is $G = \frac{A+B+C}{3}$. Thus, $O, G, H$ are collinear and $\frac{OG}{HG} = \frac{1}{2}$

## The points of intersection of the Euler line with the sides of the triangle

Acute triangle

Let $\triangle ABC$ be the acute triangle where $AC > BC > AB.$ Denote $\angle A = \alpha, \angle B = \beta, \angle C = \gamma$ $$\implies \beta > \alpha > \gamma.$$

Let Euler line cross lines $AB, AC,$ and $BC$ in points $D, E,$ and $F,$ respectively.

Then point $D$ lyes on segment $AB, \frac {\vec BD}{\vec DA} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0.$

Point $E$ lyes on segment $AC, \frac {\vec AE}{\vec EC} = \frac {\tan \beta – \tan \gamma}{\tan \beta – \tan \alpha} > 0.$

Point $F$ lyes on ray $BC, \frac {\vec {BF}}{\vec {CF}} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} > 0.$

Proof

Denote $n = \frac {\vec BD}{\vec DA}, m = \frac {\vec CE}{\vec EA}, p = \frac {\vec CX}{\vec XB}, X \in BC, q = \frac {AY}{YX}, Y \in AX.$

We use the formulae $m + pn = \frac {p+1}{q}$ (see Claim “Segments crossing inside triangle” in “Schiffler point” in “Euler line”).

Centroid $G$ lyes on median $AA'' \implies X = A'' , Y = G, p = 1, q = 2 \implies m+n=1.$

Orthocenter $H$ lyes on altitude $AA' \implies X = A', Y = H, p = \frac {\vec {CX}}{\vec {XB}} = \frac {\tan \beta}{\tan \gamma}, q = \frac {AY}{YX}.$ $$q = \frac {\vec {AH}}{\vec {HA'}} = \frac {\cos \alpha}{\cos \beta \cdot \cos \gamma} = \tan \beta \cdot \tan \gamma – 1 \implies$$ $$m \tan \gamma + n \tan \beta = \frac {\tan \beta + \tan \gamma}{1 – \tan \beta \cdot \tan \gamma} = – \tan \alpha.$$ Therefore $$\frac {\vec {BD}}{\vec {DA}} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0,$$ $$\frac {\vec {CE}}{\vec {EA}} = m =\frac {\tan \beta – \tan \alpha}{\tan \beta – \tan \gamma} > 0.$$ We use the signed version of Menelaus's theorem and get $$\frac {\vec {BF}}{\vec {FC}} = \frac {\tan \alpha – \tan \gamma}{ \tan \alpha –\tan \beta} < 0.$$

Obtuse triangle

Let $\triangle ABC$ be the obtuse triangle where $BC > AC > AB \implies \alpha > \beta > \gamma.$

Let Euler line cross lines $AB, AC,$ and $BC$ in points $D, E,$ and $F,$ respectively.

Similarly we get $F \in BC, \frac {\vec {BF}}{\vec {FC}} = \frac {\tan \gamma – \tan \alpha}{ \tan \beta –\tan \alpha} > 0.$

$$E \in AC, \frac {\vec {CE}}{\vec {EA}} = \frac {\tan \beta – \tan \alpha}{\tan \beta – \tan \gamma} > 0.$$ $D \in$ ray $BA, \frac {\vec {BD}}{\vec {AD}} = \frac {\tan \gamma – \tan \alpha}{\tan \beta – \tan \gamma} > 0.$

Right triangle

Let $\triangle ABC$ be the right triangle where $\angle BAC = 90^\circ.$ Then Euler line contain median from vertex $A.$

Isosceles triangle

Let $\triangle ABC$ be the isosceles triangle where $AC = AB.$ Then Euler line contain median from vertex $A.$

Corollary: Euler line is parallel to side

Euler line $DE$ is parallel to side $BC$ iff $\tan \beta \cdot \tan \gamma = 3.$

Proof

$DE||BC \implies \frac {BD}{DA} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma}= \frac {CE}{EA} = \frac {\tan \beta – \tan \alpha}{\tan \beta – \tan \gamma}.$

After simplification in the case $\beta \ne \gamma$ we get $2 \tan \alpha = \tan \beta + \tan \gamma.$ $$180^\circ – \alpha = \beta + \gamma \implies \tan \alpha = \frac{\tan \beta + \tan \gamma}{\tan \beta \cdot \tan \gamma – 1} \implies \tan \beta \cdot \tan \gamma = 3.$$

## Angles between Euler line and the sides of the triangle

Let Euler line of the $\triangle ABC$ cross lines $AB, AC,$ and $BC$ in points $D, E,$ and $F,$ respectively. Denote $\angle A = \alpha, \angle B = \beta, \angle C = \gamma,$ smaller angles between the Euler line and lines $BC, AC,$ and $AB$ as $\theta_A, \theta_B,$ and $\theta_C,$ respectively.

Prove that $\tan \theta_A = \vert \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma} \vert,$ $$\tan \theta_B = |\frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}\vert, \tan \theta_C = |\frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}\vert.$$

Proof

WLOG, $AC > BC > AB \implies \frac {\vec {BF}}{\vec {CF}} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} > 0.$

Let $|BC| = 2a, M$ be the midpoint $BC, O$ be the circumcenter of $\triangle ABC \implies OM = \frac {a}{\tan \alpha}.$

$$MF = MC + CF, \frac {\vec {BF}}{\vec {CF}} = \frac {\vec {BC + CF}}{\vec {CF}} = \frac {2a}{|CF|}+ 1 =\frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} \implies$$ $$\frac {|MF|}{a} = \frac {\tan \beta – \tan \gamma}{2 \tan \alpha – \tan \beta – \tan \gamma} \implies$$ $$\tan \theta_A = \frac {|OM|}{|MF|} = |\frac {3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma}|.$$

Symilarly, for other angles.

## Distances along Euler line

Let $H, G, O,$ and $R$ be orthocenter, centroid, circumcenter, and circumradius of the $\triangle ABC,$ respectively. $$a = BC, b = AC, c = AB,$$ $$\alpha = \angle A,\beta = \angle B,\gamma = \angle C.$$

Prove that $HO^2 = R^2 (1 - 8 \cos A \cos B \cos C),$ $$GO^2 = R^2 - \frac {a^2 + b^2 + c^2}{9}.$$

Proof

WLOG, $ABC$ is an acute triangle, $\beta \ge \gamma.$

$$OA = R, AH = 2 R \cos \alpha, \angle BAD = \angle OAC = 90^\circ - \beta \implies$$ $$\angle OAH = \alpha - 2\cdot (90^\circ - \beta) = \alpha + \beta + \gamma - 180^\circ + \beta - \gamma = \beta - \gamma .$$ $$HO^2 = AO^2 + AH^2 - 2 AH \cdot AO \cos \angle OAC = R^2 + (2 R \cos \alpha)^2 - 2 R \cdot 2R \cos \alpha \cdot \cos (\beta - \gamma).$$ $$\frac {HO^2}{R^2} = 1 + 4 \cos \alpha (\cos \alpha - \cos (\beta - \gamma) = 1 - 4 \cos \alpha (\cos (\beta + \gamma) + \cos (\beta – \gamma)) = 1 - 8 \cos \alpha \cos \beta \cos \gamma.$$ $$\frac {HO^2}{R^2} = 1 + 4 \cos^2 \alpha - 4 \cos \alpha \cos (\beta - \gamma) = 5 - 4 \sin^2 \alpha + 4 \cos (\beta + \gamma) \cos (\beta - \gamma)$$ $$HO^2 = 5R^2 - 4R^2 \sin^2 \alpha + 2R^2 \cos 2\beta + 2R^2 \cos 2 \gamma = 9R^2 - 4R^2 \sin^2 \alpha - 4R^2 \sin^2 \beta - 4 R^2\sin^2 \gamma$$ $$HO^2 = 9R^2 - a^2 - b^2 - c^2, GO^2 = \frac {HO^2}{9} = R^2 - \frac {a^2 + b^2 + c^2}{9}.$$ vladimir.shelomovskii@gmail.com, vvsss

## Position of Kimberling centers on the Euler line

Let triangle ABC be given. Let $H = X(4), O = X(3), R$ and $r$ are orthocenter, circumcenter, circumradius and inradius, respectively.

We use point $\vec O = \vec X(3)$ as origin and $\vec {HO}$ as a unit vector.

We find Kimberling center X(I) on Euler line in the form of $$\vec X(I) = \vec O + k_i \cdot \vec {OH}.$$ For a lot of Kimberling centers the coefficient $k_i$ is a function of only two parameters $J = \frac {|OH|}{R}$ and $t = \frac {r}{R}.$

Centroid $X(2)$ $$X(2) = X(3) + \frac {1}{3} (X(4) - X(3)) \implies k_2 = \frac {1}{3}.$$ Nine-point center $X(5)$ $$X(5) = X(3) + \frac {1}{2} (X(4) - X(3)) \implies k_5 = \frac {1}{2}.$$ de Longchamps point $X(20)$ $$X(20) = X(3) - (X(4) - X(3)) \implies k_{20} = - 1.$$ Schiffler point $X(21)$ $$X(21) = X(3) + \frac {1}{3 + 2r/R} (X(4) + X(3)) \implies k_{21} = \frac {1}{3 + 2t}.$$ Exeter point $X(22)$ $$X(22) = X(3) + \frac {2}{J^2 - 3} (X(4) - X(3)) \implies k_{22} = \frac {2}{J^2 - 3}.$$ Far-out point $X(23)$ $$X(23) = X(3) + \frac {3}{J^2} (X(4) - X(3)) \implies k_{23} = \frac {3}{J^2}.$$ Perspector of ABC and orthic-of-orthic triangle $X(24)$ $$X(24) = X(3) + \frac {2}{J^2+1} (X(4) - X(3)) \implies k_{24} = \frac {2}{J^2 + 1}.$$ Homothetic center of orthic and tangential triangles $X(25)$ $$X(25) = X(3) + \frac {4}{J^2+3} (X(4) - X(3)) \implies k_{25} = \frac {4}{J^2 + 3}.$$ Circumcenter of the tangential triangle $X(26)$ $$X(26) = X(3) + \frac {2}{J^2 - 1}(X(4) - X(3)) \implies k_{26} = \frac {2}{J^2 - 1}.$$

Midpoint of X(3) and $X(5)$ $$X(140) = X(3) + \frac {1}{4} (X(4) - X(3)) \implies k_{140} = \frac {1}{4}.$$ vladimir.shelomovskii@gmail.com, vvsss

## Triangles with angles of $60^\circ$ or $120^\circ$

Claim 1

Let the $\angle C$ in triangle $ABC$ be $120^\circ.$ Then the Euler line of the $\triangle ABC$ is parallel to the bisector of $\angle C.$

Proof

Let $\omega$ be circumcircle of $\triangle ABC.$

Let $O$ be circumcenter of $\triangle ABC.$

Let $\omega'$ be the circle symmetric to $\omega$ with respect to $AB.$

Let $E$ be the point symmetric to $O$ with respect to $AB.$

The $\angle C = 120^\circ \implies O$ lies on $\omega', E$ lies on $\omega.$

$EO$ is the radius of $\omega$ and $\omega' \implies$ translation vector $\omega'$ to $\omega$ is $\vec {EO}.$

Let $H'$ be the point symmetric to $H$ with respect to $AB.$ Well known that $H'$ lies on $\omega.$ Therefore point $H$ lies on $\omega'.$

Point $C$ lies on $\omega, CH || OE \implies CH = OE.$

Let $CD$ be the bisector of $\angle C \implies E,O,D$ are concurrent. $OD = HC, OD||HC \implies CD || HO \implies$

Euler line $HO$ of the $\triangle ABC$ is parallel to the bisector $CD$ of $\angle C$ as desired.

Claim 2

Let the $\angle C$ in triangle $ABC$ be $60^\circ.$ Then the Euler line of the $\triangle ABC$ is perpendicular to the bisector of $\angle C.$

Proof

Let $\omega, O, H, I$ be circumcircle, circumcenter, orthocenter and incenter of the $\triangle ABC.$ $$\angle AHB = 180^\circ – \angle ACB = 180^\circ – 60^\circ = 120^\circ.$$ $$\angle AOB = 2 \angle ACB = 120^\circ.$$ $\angle AIB = 90^\circ + \frac {\angle ACB}{2} = 120^\circ \implies$ points $A, H, I, O, B$ are concyclic.

The circle $AIB$ centered at midpoint of small arc $AB \implies$

$EH = EO = CO, EO||CH \implies COEH$ is rhomb.

Therefore the Euler line $HO$ is perpendicular to $CI$ as desired.

Claim 3

Let $ABCD$ be a quadrilateral whose diagonals $AC$ and $BD$ intersect at $P$ and form an angle of $60^\circ.$ If the triangles PAB, PBC, PCD, PDA are all not equilateral, then their Euler lines are pairwise parallel or coincident.

Proof

Let $l$ and $l'$ be internal and external bisectors of the angle $\angle BPC, l \perp l'$.

Then Euler lines of $\triangle ABP$ and $\triangle CDP$ are parallel to $l'$ and Euler lines of $\triangle BCP$ and $\triangle ADP$ are perpendicular to $l$ as desired.

## Euler lines of cyclic quadrilateral (Vittas’s theorem)

Claim 1

Let $ABCD$ be a cyclic quadrilateral with diagonals intersecting at $P (\angle APB \ne 60^\circ).$ The Euler lines of triangles $\triangle APB, \triangle BPC, \triangle CPD, \triangle DPA$ are concurrent.

Proof

Let $O_1, O_2, O_3, O_4 (H_1, H_2, H_3, H_4)$ be the circumcenters (orthocenters) of triangles $\triangle APD, \triangle CPD, \triangle BPC, \triangle APB.$ Let $I_2I_4$ be the common bisector of $\angle APB$ and $\angle CPD.$ $$O_1O_4 \perp AC, H_3H_4 \perp AC, O_2O_3 \perp AC, H_1H_2 \perp AC \implies$$ $$O_1O_4 || H_3H_4 || O_2O_3 || H_1H_2.$$ $$O_1O_2 \perp BD, H_2H_3 \perp BD, O_3O_4 \perp BD, H_1H_4 \perp BD \implies$$ $$O_1O_2 || H_2H_3 || O_3O_4 || H_1H_4 .$$ Therefore $O_1O_2O_3O_4$ and $H_1H_2H_3H_4$ are parallelograms with parallel sides.

$\triangle APB \sim \triangle DPC \implies \angle O_2PH_2 = \angle O_4PH_4 \implies I_2I_4$ bisect these angles. So points $O_2, P, H_4$ are collinear and lies on one straight line which is side of the pare vertical angles $\angle O_2PH_2$ and $\angle O_4PH_4.$ Similarly, points $O_4, P, H_2$ are collinear and lies on another side of these angles. Similarly obtuse $\triangle APD \sim \triangle BPC$ so points $H_1, P$ and $O_3$ are collinear and lies on one side and points $H_3, P$ and $O_1$ are collinear and lies on another side of the same vertical angles.

We use Claim and get that lines $O_1H_1, O_2H_2, O_3H_3, O_4H_4$ are concurrent (or parallel if $\angle APD = 60^\circ$ or $\angle APD = 120^\circ$).

Claim 2 (Property of vertex of two parallelograms)

Let $ABCD$ and $EFGH$ be parallelograms, $AB||EF, AD||EH.$ Let lines $AE, BH,$ and $CG$ be concurrent at point $O.$ Then points $D, O,$ and $F$ are collinear and lines $AG, BF, CE,$ and $DH$ are concurrent.

Proof

We consider only the case $AB \perp AD.$ Shift transformation allows to generalize the obtained results.

We use the coordinate system with the origin at the point $O,$ and axes $Ox||AD, Oy||BA.$

We use $x_A, y_A, y_B, x_E, y_F$ and get $x_B = x_A, y_E = \frac {x_E \cdot y_A}{x_A}, x_F = x_E, y_H = y_E, x_H=\frac {x_E \cdot y_A}{y_B}, y_D = y_A,$ $$x_D=\frac {x_E \cdot y_A}{y_F}, x_C=x_D, y_C = y_B, x_G=\frac {x_E \cdot y_A}{y_B}, y_G = y_F \implies$$ $\frac {y_C}{x_C}=\frac {y_G}{x_G} \implies$ points $C, O,$ and $G$ are colinear.

We calculate point of crossing $AG$ and $BF, AG$ and $DH, AG$ and $CE$ and get the same result: $$x_I = \frac {y_A+ y_B – y_F – {\frac {x_E \cdot y_A}{x_A}}} {\frac{y_B}{x_E}– \frac {y_F}{x_A}}, y_I = \frac {\frac{y_A \cdot y_B}{x_A}+ \frac{y_B \cdot y_F}{x_E}– \frac{y_A \cdot y_F}{x_A} – \frac {y_B \cdot y_F}{x_A}} {\frac{y_B}{x_E} – \frac {y_F}{x_A}}$$ as desired (if $\frac{y_B}{x_E}= \frac {y_F}{x_A}$ then point $I$ moves to infinity and lines are parallel, angles $\angle APD = 60^\circ$ or $\angle APD = 120^\circ).$

vladimir.shelomovskii@gmail.com, vvsss ~minor edit by Yiyj1

## Concurrent Euler lines and Fermat points

Consider a triangle $ABC$ with Fermat–Torricelli points $F$ and $F'.$ The Euler lines of the $10$ triangles with vertices chosen from $A, B, C, F,$ and $F'$ are concurrent at the centroid $G$ of triangle $ABC.$ We denote centroids by $g$, circumcenters by $o.$ We use red color for points and lines of triangles $F**,$ green color for triangles $F'**,$ and blue color for triangles $FF'*.$

Case 1

Let $F$ be the first Fermat point of $\triangle ABC$ maximum angle of which smaller then $120^\circ.$ Then the centroid of triangle $ABC$ lies on Euler line of the $\triangle ABF.$ The pairwise angles between these Euler lines are equal $60^\circ.$

Proof

Let $G', O,$ and $\omega$ be centroid, circumcenter, and circumcircle of $\triangle ABF,$ respectevely.

Let $\triangle ABD$ be external for $\triangle ABC$ equilateral triangle $\implies F = CD \cap \omega.$ $\angle AFB = 120^\circ \implies AFBD$ is cyclic.

Point $O$ is centroid of $\triangle ABD \implies \vec O = \frac {\vec A + \vec B + \vec D}{3}.$ $$\vec G' = \frac {A + B + F}{3}, G = \frac {A + B + C}{3} \implies$$ $$\vec {OG} = \frac {\vec C – \vec D}{3} = \frac {\vec DC}{3}, \vec {G'G} = \frac {\vec C – \vec F}{3} = \frac {\vec FC}{3} \implies$$ $OG||G'G \implies$ Points $O, G',$ and $G$ are colinear, so point $G$ lies on Euler line $OG'$ of $\triangle ABF.$

$\vec {GG_0} = \frac {A – F}{3} \implies GG_0||AF, \vec {GG_1} = \frac {B – F}{3} \implies GG_1||BF.$

Case 2

Let $F$ be the first Fermat point of $\triangle ABC, \angle BAC > 120^\circ.$

Then the centroid $G$ of triangle $ABC$ lies on Euler lines of the triangles $\triangle ABF,\triangle ACF,$ and $\triangle BCF.$ The pairwise angles between these Euler lines are equal $60^\circ.$

Proof

Let $\triangle ABD$ be external for $\triangle ABC$ equilateral triangle, $\omega$ be circumcircle of $\triangle ABD \implies F = CD \cap \omega.$ $\angle ABD = 60^\circ, \angle AFD = 120^\circ \implies ABDF$ is cyclic.

Point $O$ is centroid of $\triangle ABD \implies$ $\vec {OG} = \frac {\vec DC}{3}, \vec {G'G} = \frac {\vec FC}{3} \implies OG||G'G \implies$

Points $O, G',$ and $G$ are colinear, so point $G$ lies on Euler line $OG'$ of $\triangle ABF$ as desired.

Case 3

Let $F'$ be the second Fermat point of $\triangle ABC.$ Then the centroid $G$ of triangle $ABC$ lies on Euler lines of the triangles $\triangle ABF',\triangle ACF',$ and $\triangle BCF'.$

The pairwise angles between these Euler lines are equal $60^\circ.$

Proof

Let $\triangle ABD$ be internal for $\triangle ABC$ equilateral triangle, $\omega$ be circumcircle of $\triangle ABD \implies F' = CD \cap \omega.$

Let $O_1, O_0,$ and $O'$ be circumcenters of the triangles $\triangle ABF',\triangle ACF',$ and $\triangle BCF'.$ Point $O_1$ is centroid of the $\triangle ABD \implies GO_1G_1$ is the Euler line of the $\triangle ABF'$ parallel to $CD.$

$O_1 O_0$ is bisector of $BF', O'O_1$ is bisector of $AF', O'O_0$ is bisector of $CF' \implies \triangle O'O_1O_0$ is regular triangle.

$\triangle O'O_0 O_1$ is the inner Napoleon triangle of the $\triangle ABC \implies G$ is centroid of this regular triangle. $$\angle GO_1O_0 = 30^\circ, O_1O_0 \perp F'B, \angle AF'B = 120^\circ \implies GO_0||F'A.$$

$\vec {GG_0} = \frac {\vec {F'A}}{3} \implies GG_0||F'A \implies$ points $O_0,G,$ and $G_0$ are collinear as desired.

Similarly, points $O',G,$ and $G'$ are collinear.

Case 4

Let $F$ and $F'$ be the Fermat points of $\triangle ABC.$ Then the centroid of $\triangle ABC$ point $G$ lies on Euler line $OG' (O$ is circumcenter, $G'$ is centroid) of the $\triangle AFF'.$

Proof

Step 1 We will find line $F'D$ which is parallel to $GG'.$

Let $M$ be midpoint of $BC.$ Let $M'$ be the midpoint of $FF'.$

Let $D$ be point symmetrical to $F$ with respect to $M.$

$M'M||F'D$ as midline of $\triangle FF'D.$ $$\vec {G'G} = \frac {\vec A + \vec B + \vec C}{3} – \frac {\vec A + \vec F + \vec F'}{3} = \frac {2}{3} \cdot (\frac {\vec B + \vec C}{2} – \frac {\vec F + \vec F'}{3})$$ $$\vec {G'G} = \frac {2}{3} (\vec M – \vec M') = \frac {2}{3} \vec {M'M} \implies M'M||F'D||G'G.$$

Step 2 We will prove that line $F'D$ is parallel to $OG.$

Let $\triangle xyz$ be the inner Napoleon triangle. Let $\triangle XYZ$ be the outer Napoleon triangle. These triangles are regular centered at $G.$

Points $O, z,$ and $x$ are collinear (they lies on bisector $AF').$

Points $O, Z,$ and $Y$ are collinear (they lies on bisector $AF).$

Points $M, X,$ and $y$ are collinear (they lies on bisector $BC).$ $$E = YZ \cap BF, E' = Zx \cap BC.$$ $$BF \perp XZ \implies \angle BEZ = 30^\circ.$$ $BC \perp Xy,$ angle between $Zx$ and $Xy$ is $60^\circ \implies \angle BE'Z = 30^\circ.$

$$\angle AF'B = \angle AZB = 120^\circ, AZ = BZ \implies \overset{\Large\frown} {BZ} = 60^\circ \implies$$

Points $A, Z, F', B, E',$ and $E$ are concyclic $\implies \angle OZx = \angle CBF.$ $$FM = MD, BM = MC \implies \angle CBF = \angle BCD.$$ Points $C, D, X, B,$ and $F'$ are concyclic $\implies \angle BCD = \angle BF'D.$

$\angle GZO = \angle GxO = 30^\circ \implies$ points $Z, O, G,$ and $x$ are concyclic

$$\implies \angle GOx = \angle OZx – 30^\circ = \angle DF'B – 30^\circ.$$ $$\angle AF'B = 120^\circ, Ox \perp AF' \implies OG||F'D.$$ Therefore $OG||G'G \implies O, G',$ and $G$ are collinear or point $G$ lies on Euler line $OG'.$

## Euler line of Gergonne triangle

Prove that the Euler line of Gergonne triangle of $\triangle ABC$ passes through the circumcenter of triangle $ABC.$

Gergonne triangle is also known as the contact triangle or intouch triangle. If the inscribed circle touches the sides of $\triangle ABC$ at points $D, E,$ and $F,$ then $\triangle DEF$ is Gergonne triangle of $\triangle ABC$.

Other wording: Tangents to circumcircle of $\triangle ABC$ are drawn at the vertices of the triangle. Prove that the circumcenter of the triangle formed by these three tangents lies on the Euler line of the original triangle.

Proof

Let $H$ and $I$ be orthocenter and circumcenter of $\triangle DEF,$ respectively. Let $A'B'C'$ be Orthic Triangle of $\triangle DEF.$

Then $IH$ is Euler line of $\triangle DEF,$ $I$ is the incenter of $\triangle ABC,$ $H$ is the incenter of $\triangle A'B'C'.$

$\angle DEF = \angle DB'C' = \angle BDF = \frac { \overset{\Large\frown} {DF}}{2} \implies B'C' || BC.$

Similarly, $A'C' || AC, A'B' || AB \implies A'B'C'\sim ABC \implies$

$AA' \cap BB' \cap CC' = P,$ where $P$ is the perspector of triangles $ABC$ and $A'B'C'.$

Under homothety with center P and coefficient $\frac {B'C'}{BC}$ the incenter $I$ of $\triangle ABC$ maps into incenter $H$ of $\triangle A'B'C'$, circumcenter $O$ of $\triangle ABC$ maps into circumcenter $I$ of $\triangle A'B'C' \implies P,H,I,O$ are collinear as desired.

## Thebault point

Let $AD, BE,$ and $CF$ be the altitudes of the $\triangle ABC,$ where $BC> AC > AB, \angle BAC \ne 90^\circ.$

a) Prove that the Euler lines of triangles $\triangle AEF, \triangle BFD, \triangle CDE$ are concurrent on the nine-point circle at a point T (Thebault point of $\triangle ABC.$)

b) Prove that if $\angle BAC < 90^\circ$ then $TE = TF + TD,$ else $TF = TE + TD.$

Proof

Case 1 Acute triangle

a) It is known, that Euler line of acute triangle $\triangle ABC$ cross AB and BC (shortest and longest sides) in inner points.

Let $O_0, O, O'$ be circumcenters of $\triangle AEF, \triangle BFD, \triangle CDE.$

Let $G_0, G,$ and $G'$ be centroids of $\triangle AEF, \triangle BFD, \triangle CDE.$

Denote $\angle ABC = \beta, K = DE \cap G'O', L = EF \cap G_0 O_0, M = EC \cap G'O', \omega$ is the circle $DEF$ (the nine-points circle).

$\angle CEH = \angle CDH = 90^\circ \implies O'$ is the midpoint $CH,$ where $H$ is the orthocenter of $\triangle ABC \implies O' \in \omega.$

Similarly $O_0 \in \omega, O \in \omega.$

$CO' = HO', BO = OH \implies OO'$ is the midline of $\triangle BHC \implies \triangle O_0OO' \sim \triangle ABC.$

Let $O'G'$ cross $\omega$ at point $T$ different from $O'.$

$\triangle ABC \sim \triangle AEF \sim \triangle DBF \sim \triangle DEC \implies$ spiral similarity centered at $E$ maps $\triangle AEF$ onto $\triangle DEC.$

This similarity has the rotation angle $180^\circ – \beta \implies$ acute angle between Euler lines of these triangles is $\beta.$

Let these lines crossed at point $T'.$ Therefore $\angle O_0T'O' = \angle O_0OO' \implies$ points $O, O_0, O',T$ and $T'$ are concyclic $\implies T = T'.$

Similarly, $OG \cap O'G' = T$ as desired.

b) $\triangle AEF \sim \triangle DEC \implies \frac {FL}{LE} = \frac {CM}{ME}.$ Point $G'$ lies on median of $\triangle DEC$ and divide it in ratio 2 : 1.

Point $G'$ lies on Euler line of $\triangle DEC.$

According the Claim, $\frac {DK}{KE}+ \frac {CM}{ME} = 1 \implies \frac {DK}{KE}+ \frac {FL}{LE} = 1.$ $FO_0 = EO_0 \implies \overset{\Large\frown} {FO_0} = \overset{\Large\frown} {EO_0} \implies \angle FTO_0 = \angle ETO_0 \implies \frac {TF}{TE} = \frac {FL}{LE}.$

Similarly $\frac {TD}{TE} = \frac {DK}{KE} \implies \frac {TF}{TE} + \frac {TD}{TE} = 1 \implies TE = TD + TF.$

Case 2 Obtuse triangle

a) It is known, that Euler line of obtuse $\triangle ABC$ cross AC and BC (middle and longest sides) in inner points.

Let $O_0, O, O'$ be circumcenters of $\triangle AEF, \triangle BFD, \triangle CDE.$

Let $G_0, G,$ and $G'$ be centroids of $\triangle AEF, \triangle BFD, \triangle CDE.$

Denote $\angle ABC = \beta, K = DF \cap GO, K' = DC \cap G'O',$ $L = EF \cap G_0 O_0, L' = EC \cap G'O', \omega$ is the circle $DEF$ (the nine-points circle).

$\angle AEH = \angle AFH = 90^\circ \implies O_0$ is the midpoint $AH,$ where $H$ is the orthocenter of $\triangle ABC \implies O_0 \in \omega.$

Similarly $O' \in \omega, O \in \omega.$

$CO' = HO', BO = OH \implies OO'$ is the midline of $\triangle BHC \implies \triangle O_0OO' \sim \triangle ABC.$

Let $O'G'$ cross $\omega$ at point $T$ different from $O'.$

$\triangle ABC \sim \triangle AEF \sim \triangle DBF \sim \triangle DEC \implies$ spiral similarity centered at $E$ maps $\triangle AEF$ onto $\triangle DEC.$

This similarity has the rotation angle $\beta \implies$ acute angle between Euler lines of these triangles is $\beta.$

Let these lines crossed at point $T'.$ Therefore $\angle O_0T'O' = \angle O_0OO' \implies$ points $O, O_0, O',T$ and $T'$ are concyclic $\implies T = T'.$

Similarly, $OG \cap O'G' = T$ as desired.

b) $\triangle AEF \sim \triangle DEC \implies \frac {EL}{LF} = \frac {EL'}{L'C}.$ $\triangle AEF \sim \triangle DBF \implies \frac {DK}{KF} = \frac {DK'}{K'C}.$

Point $G'$ lies on median of $\triangle DEC$ and divide it in ratio $2 : 1.$

Point $G'$ lies on Euler line of $\triangle DEC.$ According the Claim, $\frac {DK'}{K'C} + \frac {EL'}{L'C} = 1 \implies \frac {DK}{KF}+ \frac {EL}{LF} = 1.$ $FO_0 = EO_0 \implies \overset{\Large\frown} {FO_0} = \overset{\Large\frown} {EO_0} \implies \angle FTO_0 = \angle ETO_0 \implies \frac {TE}{TF} = \frac {EL}{LF}.$

Similarly $\frac {TD}{TF} = \frac {DK}{KF} \implies \frac {TE}{TF} + \frac {TD}{TF} = 1 \implies TF = TD + TE.$

Claim (Segment crossing the median)

Let $M$ be the midpoint of side $AB$ of the $\triangle ABC, D \in AC,$ $$E \in BC, G = DE \cap CM.$$ $$\frac {BE}{CE} = m, \frac {AD}{CD} = n.$$

Then $\frac {DG}{GE} = \frac{1+m}{1+n}, \frac {MG}{GC} = \frac{n+m}{2}.$

Proof

Let $[ABC]$ be $1$ (We use sign $[t]$ to denote the area of $t).$

Denote $[CDG] = x, [CEG] = y, [DGM] = z.$ $$[ACM] = [ BCM] = \frac {1}{2}.$$ $$x = \frac {CD \cdot CG}{2 \cdot CA \cdot CM}, y = \frac {CE \cdot CG}{2 \cdot CB \cdot CM} \implies$$ $$\frac {DG}{GE} = \frac {x}{y} = \frac {CD \cdot CB}{CA \cdot CE} = \frac {1+m}{1+n}.$$ $$x + y = \frac {CD \cdot CE}{AC \cdot BC} = \frac {1}{(1+m)(1+n)} \implies$$ $$x + y = x(1+\frac {y}{x}) = x (1 + \frac {1+n}{1+m} )= \frac {1}{(1+n)(1+m)} \implies x = \frac{1}{(1+n)(2+m+n)}.$$ $$z+x = \frac {[CDM]}{2[CAM]} = \frac {1}{2(1 + n)}.$$ $$\frac {MG}{GC} = \frac {z}{x} = \frac {z + x}{x} - 1 =\frac {1}{2(1 + n)} (1 + n)(2 + m + n) - 1 = \frac {n + m}{2}.$$ vladimir.shelomovskii@gmail.com, vvsss

## Schiffler point

Let $I, O, G, R, \alpha,$ and $r$ be the incenter, circumcenter, centroid, circumradius, $\angle A,$ and inradius of $\triangle ABC,$ respectively. Then the Euler lines of the four triangles $\triangle BCI, \triangle CAI, \triangle ABI,$ and $\triangle ABC$ are concurrent at Schiffler point $S = X(21), \frac {OS}{SG} = \frac {3R}{2r}$.

Proof

We will prove that the Euler line $O'G'$ of $\triangle BCI$ cross the Euler line $OG$ of $\triangle ABC$ at such point $S,$ that $\frac {OS}{SG} = \frac {3R}{2r}$.

Let $O'$ and $G'$ be the circumcenter and centroid of $\triangle IBC,$ respectively.

It is known that $O'$ lies on circumcircle of $\triangle ABC, \overset{\Large\frown} {BO'} = \overset{\Large\frown} {CO'}.$

Denote $E = OO' \cap BC, X = AE \cap G'O', Y = GG' \cap OO'.$

It is known that $E$ is midpoint $BC,$ point $G$ lies on median $AE,$ points $A, I, O'$ belong the bisector of $\angle A, \frac {AE}{GE} = \frac {IE}{IG'} = 3 \implies GY||AO', \frac {O'E}{YE} = 3, \frac {GG'}{G'Y} = \frac {AI}{IO'}.$

Easy to find that $AI = \frac {r} {\sin {\frac {\alpha}{2}}}$, $IO' = BO' = 2 R {\sin {\frac {\alpha}{2}}} \implies \frac {AI}{IO'} = \frac {r}{R \cdot (1 – \cos \alpha)}.$

We use sigh [t] for area of t. We get $$n = \frac {GG'}{G'Y} = \frac {[GXO']}{[YXO'} = \frac {[GXO']}{[EXO']} \cdot \frac {[EXO']}{[YXO'} = \frac {GX}{XE} \cdot \frac {3}{2} \implies$$

$$m = \frac {GX}{XE} = \frac {2n}{3}.$$ $$p = \frac {OE}{EY} = \frac {\cos \alpha}{(1 - \cos \alpha)/3} = \frac {3 \cos \alpha}{1 - \cos \alpha}$$ Using Claim we get $$\frac {OS}{SG} = \frac {p + 1}{m} - \frac {p}{n} = \frac {3(p + 1)}{2n} - \frac {p}{n} = \frac {p + 3}{2n} = \frac {3R}{2r}.$$ Therefore each Euler line of triangles $\triangle BCI, \triangle CAI, \triangle ABI,$ cross Euler line of $\triangle ABC$ in the same point, as desired.

Claim (Segments crossing inside triangle)

Given triangle GOY. Point $S$ lies on $GO, k = \frac {OS}{SG}.$

Point $E$ lies on $YO, p = \frac {OE}{EY}.$

Point $G'$ lies on $GY, n = \frac {GG'}{G'Y}.$

Point $X$ lies on $GE, m = \frac {GX}{XE}.$ Then $k = \frac {p + 1}{m} - \frac {p}{n}.$

Proof

Let $[OGY]$ be $1$ (We use sigh $[t]$ for area of $t).$ $$[GSG'] = \frac{n}{(n + 1)(k + 1)}, [YEG'] = \frac{1}{(n + 1)(p + 1)},$$ $$[SOE] = \frac{kp}{(k + 1)(p + 1)}, [ESG'] = \frac {[GSG']}{m},$$ $$[OGY] = [GSG'] + [YEG'] + [SOE] + [ESG'] = 1 \implies$$ $$\frac{n(p + 1)}{m}=nk + p \implies k = \frac {p + 1}{m} - \frac {p}{n}.$$ vladimir.shelomovskii@gmail.com, vvsss

## Euler line as radical axis

Let $\triangle ABC$ with altitudes $AA_1, BB_1,$ and $CC_1$ be given.

Let $\Omega, O, H$ and $R$ be circumcircle, circumcenter, orthocenter and circumradius of $\triangle ABC,$ respectively.

Circle $\omega_1$ centered at $Q_1$ passes through $A, A_1$ and is tangent to the radius AO. Similarly define circles $\omega_2$ and $\omega_3.$

Then Euler line of $\triangle ABC$ is the radical axis of these circles.

If $\triangle ABC$ is acute, then these three circles intersect at two points located on the Euler line of the $\triangle ABC.$

Proof

The power of point $O$ with respect to $\omega_1, \omega_2,$ and $\omega_3$ is $R^2.$

The power of point $H$ with respect to $\omega_1$ is $AH \cdot HA_1.$

The power of point $H$ with respect to $\omega_2$ is $BH \cdot HB_1.$

The power of point $H$ with respect to $\omega_3$ is $CH \cdot HC_1.$

It is known that $AH \cdot HA_1 = BH \cdot HB_1 = CH \cdot HC_1.$

Therefore points $H$ and $O$ lies on radical axis of these three circles as desired.

## De Longchamps point X(20)

Definition 1

The De Longchamps’ point of a triangle is the radical center of the power circles of the triangle. Prove that De Longchamps point lies on Euler line.

We call A-power circle of a $\triangle ABC$ the circle centered at the midpoint $BC$ point $A'$ with radius $R_A = AA'.$ The other two circles are defined symmetrically.

Proof

Let $H, O,$ and $L$ be orthocenter, circumcenter, and De Longchamps point, respectively.

Denote $B-$power circle by $\omega_B, C-$power circle by $\omega_C, D = \omega_B \cap \omega_C,$ $a = BC, b = AC, c = AB.$ WLOG, $a \ge b \ge c.$

Denote $X_t$ the projection of point $X$ on $B'C', E = D_t.$

We will prove that radical axes of $B-$power and $C-$power cicles is symmetric to altitude $AH$ with respect $O.$ Further, we will conclude that the point of intersection of the radical axes, symmetrical to the heights with respect to O, is symmetrical to the point of intersection of the heights $H$ with respect to $O.$

Point $E$ is the crosspoint of the center line of the $B-$power and $C-$power circles and there radical axis. $B'C' = \frac {a}{2}.$ We use claim and get:

$$C'E = \frac {a}{4} + \frac {R_C^2 - R_B^2}{a}.$$ $R_B$ and $R_C$ are the medians, so $$R_B^2 = \frac {a^2}{2}+ \frac {c^2}{2} - \frac {b^2}{4}, R_C^2 = \frac {a^2}{2}+ \frac {b^2}{2} - \frac {c^2}{4} \implies C'E = \frac {a}{4} + \frac {3(b^2 - c^2)}{4a}.$$

We use Claim some times and get: $$C'A_t = \frac {a}{4} - \frac {b^2 - c^2}{4a}, A_tO_t = \frac {a}{2} - 2 C'A_t = \frac {b^2 - c^2}{2a} \implies$$ $$O_t L_t = C'E - C'A_t - A_t O_t = \frac {b^2 - c^2}{2a} = A_t O_t = H_t O_t \implies$$ radical axes of $B-$power and $C-$power cicles is symmetric to altitude $AH$ with respect $O.$

Similarly radical axes of $A-$power and $B-$power cicles is symmetric to altitude $CH,$ radical axes of $A-$power and $C-$power cicles is symmetric to altitude $BH$ with respect $O.$ Therefore the point $L$ of intersection of the radical axes, symmetrical to the heights with respect to $O,$ is symmetrical to the point $H$ of intersection of the heights with respect to $O \implies \vec {HO} = \vec {OL} \implies L$ lies on Euler line of $\triangle ABC.$

Claim (Distance between projections)

$$x + y = a, c^2 - x^2 = h^2 = b^2 - y^2,$$ $$y^2 - x^2 = b^2 - c^2 \implies y - x = \frac {b^2 - c^2}{a},$$ $$x = \frac {a}{2} - \frac {b^2 - c^2}{2a}, y = \frac {a}{2} + \frac {b^2 - c^2}{2a}.$$

Definition 2

We call $\omega_A = A-$circle of a $\triangle ABC$ the circle centered at $A$ with radius $BC.$ The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of $A-$circle, $B-$circle, and $C-$circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under Definition 1.

Proof

Let $H, G,$ and $L_o$ be orthocenter, centroid, and De Longchamps point, respectively. Let $\omega_B$ cross $\omega_C$ at points $A'$ and $E.$ The other points $(D, F, B', C')$ are defined symmetrically. $$AB' = BC, B'C = AB \implies \triangle ABC = \triangle CB'A \implies$$ $$AB||B'C \implies CH \perp B'C.$$ Similarly $CH \perp A'C \implies A'B'$ is diameter $\omega_C \implies$ $$\angle A'EB' = 90^\circ, 2\vec {BG} = \vec {GB'}.$$

Therefore $\triangle A'B'C'$ is anticomplementary triangle of $\triangle ABC, \triangle DEF$ is orthic triangle of $\triangle A'B'C'.$ So $L_o$ is orthocenter of $\triangle A'B'C'.$

$2\vec {HG} = \vec GL_o, 2\vec {GO} = \vec HG \implies L_o = L$ as desired.

## De Longchamps line

The de Longchamps line $l$ of $\triangle ABC$ is defined as the radical axes of the de Longchamps circle $\omega$ and of the circumscribed circle $\Omega$ of $\triangle ABC.$

Let $\Omega'$ be the circumcircle of $\triangle DEF$ (the anticomplementary triangle of $\triangle ABC).$

Let $\omega'$ be the circle centered at $G$ (centroid of $\triangle ABC$) with radius $\rho = \frac {\sqrt{2}}{3} \sqrt {a^2 + b^2 + c^2},$ where $a = BC, b = AC, c = AB.$

Prove that the de Longchamps line is perpendicular to Euler line and is the radical axes of $\Omega, \Omega', \omega,$ and $\omega'.$

Proof

Center of $\Omega$ is $O$, center of $\omega$ is $L \implies OL \perp l,$ where $OL$ is Euler line. The homothety with center $G$ and ratio $-2$ maps $\triangle ABC$ into $\triangle DEF.$ This homothety maps $\Omega$ into $\Omega'.$ $R_{\Omega} \ne R_{\Omega'}$ and $\Omega \cap \Omega' = K \implies$ there is two inversion which swap $\Omega$ and $\Omega'.$

First inversion $I_{\omega'}$ centered at point $G = \frac {\vec O \cdot 2R + \vec H \cdot R}{2R + R} = \frac {2\vec O + \vec H}{3}.$ Let $K$ be the point of crossing $\Omega$ and $\Omega'.$

The radius of $\omega'$ we can find using $\triangle HKO:$

$$OK = R, HK = 2R, HG = 2GO \implies GK^2 = 2(R^2 – GO^2), GO^2 = \frac {HO^2}{9} \implies$$ $$R_G = GK = \frac {\sqrt {2(a^2 + b^2 + c^2)}}{3}.$$

Second inversion $I_{\omega}$ centered at point $L = \frac {\vec O \cdot 2R – \vec H \cdot R}{2R – R} = 2 \vec O – \vec H.$ We can make the same calculations and get $R_L = 4R \sqrt{– \cos A \cos B \cos C}$ as desired.

## CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)

Prove that the circumcenter of the tangential triangle $\triangle A'B'C'$ of $\triangle ABC$ (Kimberling’s point $X(26))$ lies on the Euler line of $\triangle ABC.$

Proof

Let $A_0, B_0,$ and $C_0$ be midpoints of $BC, AC,$ and $AB,$ respectively.

Let $\omega$ be circumcircle of $\triangle A_0B_0C_0.$ It is nine-points circle of the $\triangle ABC.$

Let $\Omega$ be circumcircle of $\triangle ABC.$ Let $\Omega'$ be circumcircle of $\triangle A'B'C'.$

$A'B$ and $A'C$ are tangents to $\Omega \implies$ inversion with respect $\Omega$ swap $B'$ and $B_0.$ Similarly, this inversion swap $A'$ and $A_0, C'$ and $C_0.$ Therefore this inversion swap $\omega$ and $\Omega'.$

The center $N$ of $\omega$ and the center $O$ of $\Omega$ lies on Euler line, so the center $O'$ of $\Omega'$ lies on this line, as desired.

After some calculations one can find position of point $X(26)$ on Euler line (see Kimberling's point $X(26)).$

## PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)

Let $\triangle A_1B_1C_1$ be the orthic triangle of $\triangle ABC.$ Let $N$ be the circumcenter of $\triangle A_1B_1C_1.$ Let $\triangle A'B'C'$ be the tangencial triangle of $\triangle ABC.$ Let $O'$ be the circumcenter of $\triangle A'B'C'.$

Prove that lines $A_1A', B_1B',$ and $C_1C'$ are concurrent at point, lies on Euler line of $\triangle ABC.$

Proof

$B'C'$ and $B_1C_1$ are antiparallel to BC with respect $\angle BAC \implies B'C' ||B_1C_1.$

Similarly, $A'C' ||A_1C_1, A'B' ||A_1B_1.$

Therefore $\triangle A_1B_1C_1 \sim \triangle A'B'C' \implies$ homothetic center of $\triangle A_1B_1C_1$ and $\triangle A'B'C'$ is the point of concurrence of lines $A_1A', B_1B',$ and $C_1C'.$ Denote this point as $K.$

The points $N$ and $O'$ are the corresponding points (circumcenters) of $\triangle A_1B_1C_1$ and $\triangle A'B'C',$ so point $K$ lies on line $NO'.$

Points $N$ and $O' = X(26)$ lies on Euler line, so $K$ lies on Euler line of $\triangle ABC.$

## Exeter point X(22)

Exeter point is the perspector of the circummedial triangle $A_0B_0C_0$ and the tangential triangle $A'B'C'.$ By another words, let $\triangle ABC$ be the reference triangle (other than a right triangle). Let the medians through the vertices $A, B, C$ meet the circumcircle $\Omega$ of triangle $ABC$ at $A_0, B_0,$ and $C_0$ respectively. Let $A'B'C'$ be the triangle formed by the tangents at $A, B,$ and $C$ to $\Omega.$ (Let $A'$ be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through $A_0A', B_0B',$ and $C_0C'$ are concurrent, the point of concurrence lies on Euler line of triangle $ABC,$ the point of concurrence $X_{22}$ lies on Euler line of triangle $ABC, \vec {X_{22}} = \vec O + \frac {2}{J^2 - 3} (\vec H - \vec O), J = \frac {|OH|}{R},$ where $O$ - circumcenter, $H$ - orthocenter, $R$ - circumradius.

Proof

At first we prove that lines $A_0A', B_0B',$ and $C_0C'$ are concurrent. This follows from the fact that lines $AA_0, BB_0,$ and $CC_0$ are concurrent at point $G$ and Mapping theorem.

Let $A_1, B_1,$ and $C_1$ be the midpoints of $BC, AC,$ and $AB,$ respectively. The points $A, G, A_1,$ and $A_0$ are collinear. Similarly the points $B, G, B_1,$ and $B_0$ are collinear.

Denote $I_{\Omega}$ the inversion with respect $\Omega.$ It is evident that $I_{\Omega}(A_0) = A_0, I_{\Omega}(A') = A_1, I_{\Omega}(B_0) = B_0, I_{\Omega}(B') = B_1.$

Denote $\omega_A = I_{\Omega}(A'A_0), \omega_B = I_{\Omega}(B'B_0) \implies$ $$A_0 \in \omega_A, A_1 \in \omega_A, O \in \omega_A, B_0 \in \omega_B, B_1 \in \omega_B, O \in \omega_B \implies O = \omega_A \cap \omega_B.$$

The power of point $G$ with respect $\omega_A$ is $GA_1 \cdot GA_0 = \frac {1}{2} AG \cdot GA_0.$

Similarly the power of point $G$ with respect $\omega_B$ is $GB_1 \cdot GB_0 = \frac {1}{2} BG \cdot GB_0.$

$G = BB_0 \cap AA_0 \implies AG \cdot GA_0 = BG \cdot GB_0 \implies G$ lies on radical axis of $\omega_A$ and $\omega_B.$

Therefore second crosspoint of $\omega_A$ and $\omega_B$ point $D$ lies on line $OG$ which is the Euler line of $\triangle ABC.$ Point $X_{22} = I_{\Omega}(D)$ lies on the same Euler line as desired.

Last we will find the length of $OX_{22}.$ $$A_1 = BC \cap AA_0 \implies AA_1 \cdot A_1A_0 = BA_1 \cdot CA_1 = \frac {BC^2}{4}.$$ $$GO \cdot GD =GO \cdot (GO + OD) = GA_1 \cdot GA_0$$ $$GA_1 \cdot GA_0 = \frac {AA_1}{3} \cdot ( \frac {AA_1}{3} + A_1A_0) = \frac {AA_1^2}{9} + \frac {BC^2}{3 \cdot 4} = \frac {AB^2 + BC^2 + AC^2}{18}= \frac {R^2 - GO^2} {2}.$$ $$2GO^2 + 2 GO \cdot OD = R^2 - GO^2 \implies 2 GO \cdot OD = R^2 - 3GO^2.$$ $$I_{\Omega}(D) = X_{22} \implies OX_{22} = \frac {R^2} {OD} = \frac {R^2 \cdot 2 GO}{R^2 - 3 GO^2} = \frac {2 HO}{3 - \frac {HO^2}{R^2}} = \frac {2}{3 - J^2} HO$$ as desired.

Mapping theorem

Let triangle $ABC$ and incircle $\omega$ be given. $$D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega.$$ Let $P$ be the point in the plane $ABC.$ Let lines $DP, EP,$ and $FP$ crossing $\omega$ second time at points $D_0, E_0,$ and $F_0,$ respectively.

Prove that lines $AD_0, BE_0,$ and $CF_0$ are concurrent.

Proof

$$k_A = \frac {\sin {D_0AE'}}{\sin {D_0AF'}} = \frac {D_0E'}{D_0A} \cdot \frac {D_0A}{D_0F'} = \frac {D_0E'}{D_0F'}.$$ We use Claim and get: $k_A = \frac {D_0E^2}{D_0F^2}.$ $$k_D = \frac {\sin {D_0DE}}{\sin {D_0DF}} = \frac {D_0E}{2R} \cdot \frac {2R}{D_0F} = \frac {D_0E}{D_0F} \implies k_A = k_D^2.$$ Similarly, $k_B = k_E^2, k_C = k_F^2.$

We use the trigonometric form of Ceva's Theorem for point $P$ and triangle $\triangle DEF$ and get $$k_D \cdot k_E \cdot k_F = 1 \implies k_A \cdot k_B \cdot k_C = 1^2 = 1.$$ We use the trigonometric form of Ceva's Theorem for triangle $\triangle ABC$ and finish proof that lines $AD_0, BE_0,$ and $CF_0$ are concurrent.

Claim (Point on incircle)

Let triangle $ABC$ and incircle $\omega$ be given. $$D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega, P \in \omega, F' \in AB,$$ $$PF' \perp AB, E' \in AC, PE' \perp AC, A' \in EF, PA' \perp EF.$$ Prove that $\frac {PF'}{PE'} = \frac {PF^2}{PE^2}, PA'^2 = PF' \cdot PE'.$

Proof

$$AF = AE \implies \angle AFE = \angle AEF = \angle A'EE'.$$ $$\angle EFP = \angle PEE' \implies \angle PFF' = \angle PEE' \implies$$ $$\triangle PFF' \sim \triangle PEA' \implies \frac {PF}{PF'} = \frac {PE}{PA'}.$$

Similarly $\triangle PEE' \sim \triangle PFA' \implies \frac {PE}{PE'} = \frac {PF}{PA'}.$

We multiply and divide these equations and get: $$PA'^2 = PF' \cdot PE', \frac {PF'}{PE'} = \frac {PF^2}{PE^2}.$$

## Far-out point X(23)

Let $\triangle A'B'C'$ be the tangential triangle of $\triangle ABC.$

Let $G, \Omega, O, R,$ and $H$ be the centroid, circumcircle, circumcenter, circumradius and orthocenter of $\triangle ABC.$

Prove that the second crosspoint of circumcircles of $\triangle AA'O, \triangle BB'O,$ and $\triangle CC'O$ is point $X_{23}.$ Point $X_{23}$ lies on Euler line of $\triangle ABC, X_{23} = O + \frac {3}{J^2} (H – O), J = \frac {OH}{R}.$

Proof

Denote $I_{\Omega}$ the inversion with respect $\Omega, A_1, B_1, C_1$ midpoints of $BC, AC, AB.$

It is evident that $I_{\Omega}(A') = A_1, I_{\Omega}(B') = B_1, I_{\Omega}(C') = C_1.$

The inversion of circles $AA'O, BB'O, CC'O$ are lines $AA_1, BB_1,CC_1$ which crosses at point $G \implies X_{23} = I_{\Omega}(G).$

Therefore point $X_{23}$ lies on Euler line $OG$ of $\triangle ABC, OG \cdot OX_{23} = R^2 \implies \frac {OX_{23}} {OH} = \frac {R^2}{OG \cdot OH} = \frac {3}{J^2},$ as desired.

## Symmetric lines

Let triangle $ABC$ having the circumcircle $\omega$ be given.

Prove that the lines symmetric to the Euler line with respect $BC, AC,$ and $AB$ are concurrent and the point of concurrence lies on $\omega.$

Proof

The orthocenter $H$ lies on the Euler line therefore the Euler line is $H-line.$ We use H-line Clime and finish the proof.

## H–line Claim

Let triangle $ABC$ having the orthocenter $H$ and circumcircle $\omega$ be given. Denote $H–line$ any line containing point $H.$

Let $l_A, l_B,$ and $l_C$ be the lines symmetric to $H-line$ with respect $BC, AC,$ and $AB,$ respectively.

Prove that $l_A, l_B,$ and $l_C$ are concurrent and the point of concurrence lies on $\omega.$

Proof

Let $D, E,$ and $F$ be the crosspoints of $H–line$ with $AB, AC,$ and $BC,$ respectively.

WLOG $D \in AB, E \in AC.$ Let $H_A, H_B,$ and $H_C$ be the points symmetric to $H$ with respect $BC, AC,$ and $AB,$ respectively.

Therefore $H_A \in l_A, H_B \in l_B, H_C \in l_C, AH = AH_B = AH_C, BH = BH_A = BH_C, CH = CH_A = CH_B \implies$ $$\angle HH_BE = \angle EHH_B = \angle BHD = \angle BH_CD.$$

Let $P$ be the crosspoint of $l_B$ and $l_C \implies BH_CH_BP$ is cyclic $\implies P \in \omega.$

Similarly $\angle CH_BE = \angle CHE = \angle CH_A \implies CH_BH_AP$ is cyclic $\implies P \in \omega \implies$ the crosspoint of $l_B$ and $l_A$ is point $P.$