Kimberling’s point X(20)

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.

vladimir.shelomovskii@gmail.com, vvsss

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Kimberling’s point X(20)

De Longchamps point X(20)