Kimberling’s point X(21)

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 $[s]$ for area of $s.$ 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. $X(21) = \frac {3R X(2) + 2r X(3)}{3R + 2r} = \frac {R (X(4) + 2X(3)) + 2r X(3)}{3R + 2r} \implies$ $X(21) = X(3) + \frac {1}{3+ \frac {2r}{R}} (X(4) – X(3)) \implies X(21) = X(3) + k_{21}(X(4) – X(3)), k_{21} = \frac {1}{3+\frac {2r}{R}}.$

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

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Kimberling’s point X(21)

Schiffler point