# De Longchamps point

The title of this article has been capitalized due to technical restrictions. The correct title should be de Longchamps point.
 $[asy] draw((0,0)--(44,60)--(44,-10)--cycle); draw((0,0)--(44,0),blue+dashed); draw((44,60)--(22,-5),blue+dashed); draw((44,-10)--(6.5,10),blue+dashed); label("H",(24,0),(1,1)); dot((24,0)); draw((22,30)--(44,14),red); draw((22,-5)--(34,46),red); draw((44,25)--(18,25),red); dot((29,25)); label("C",(29,25),(1,1)); draw(Circle((29,25),25),dashed); dot((34,50)); label("L",(34,50),(1,1)); [/asy]$ The de Longchamps point ($L$) is the the orthocenter ($H$) reflected through the circumcenter ($C$).

The de Longchamps point of a triangle is the reflection of the triangle's orthocenter through its circumcenter.

The point is collinear with the orthocenter and circumcenter.

## de Longchamps point

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 circle

De Longchamps circle $\Omega$ of the obtuse triangle $ABC$ is circle centered at de Longchamps point $L$ which is orthogonal to the $\omega_C.$ Prove that de Longchamps circle is orthogonal to the $\omega_A, \omega_B, A-$power, $B-$power and $C-$power cicles and the radius of de Longchamps circle is $R_{\Omega} = 4R \sqrt { – \cos A \cos B \cos C}.$

Proof

$\vec {A'B'} = 2\vec {BA}, A'B'$ – diameter $\omega_C.$

Let $E$ be the foot of perpendicular from $L$ to $AB'.$

Let $E_0$ be crosspoint of $\omega_C$ and $AB' \implies A'E_0 \perp AB' \implies E = E_0.$

Points $A'$ and $E$ are simmetric with respect to $BC, MA' = AM \implies$

Points $A'$ and $E$ lies on $A-$power circle and on $\omega_C$.

$\Omega \perp \omega_C,$ points $L, A',$ and $E$ are collinear $\implies \Omega \perp A-$ power circle.

$L$ is the radical center of all six circles, therefore $\Omega$ is perpendicular to each of these circles. $L$ is orthocenter of the anticomplementary triangle of $\triangle ABC$ so radius of $\Omega$ is twice radius of circle finded by Claim $$\implies R_{\Omega} = 4R \sqrt { – \cos A \cos B \cos C}.$$

Let $ABC$ be obtuse triangle $(\angle A > 90^\circ)$ with circumcircle $\Omega,$ circumradius $R,$ and orthocenter $H.$ Let $\omega'$ be the circle with diameter $AB.$ Let $\omega$ be the circle perpendicular to $\omega'$ centered at $H.$ Find $R_\omega,$ the radius of $\omega.$

Let altitude $AH$ cross $BC$ at $D$ and cross $\Omega$ second time at $H'.$

$AD \perp BD \implies D \in \omega',$ points $H,A, D$ are collinear $\implies$

Inversion with respect $\omega$ swap $A$ and $D \implies R_\omega^2 = HA \cdot HD.$

Well known that $HA = – 2R \cos A.$

$$BC \perp HD, AC \perp BH \implies \angle C = \angle BHD \implies HD = BH \cos C.$$ Points $H$ and $H'$ are symmetric with respect $BC \implies BH' = BH.$ $$\angle BAH' = 90^\circ – \angle B \implies BH' = 2R \sin \angle BAH' = 2R \cos B \implies$$ $$R_\omega^2 = – 2R \cos A \cdot 2R \cos B \cos C \implies R_\omega = 2R \sqrt{– \cos A \cos B \cos C}.$$

## 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.