Difference between revisions of "Kimberling’s point X(21)"
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<math>IO' = BO' = 2 R {\sin {\frac {\alpha}{2}}} \implies \frac {AI}{IO'} = \frac {r}{R \cdot (1 – \cos \alpha)}.</math> | <math>IO' = BO' = 2 R {\sin {\frac {\alpha}{2}}} \implies \frac {AI}{IO'} = \frac {r}{R \cdot (1 – \cos \alpha)}.</math> | ||
− | We use sigh [ | + | We use sigh <math>[s]</math> for area of <math>s.</math> We get |
<cmath>n = \frac {GG'}{G'Y} = \frac {[GXO']}{[YXO'} = \frac {[GXO']}{[EXO']} \cdot \frac {[EXO']}{[YXO'} = \frac {GX}{XE} \cdot \frac {3}{2} \implies</cmath> | <cmath>n = \frac {GG'}{G'Y} = \frac {[GXO']}{[YXO'} = \frac {[GXO']}{[EXO']} \cdot \frac {[EXO']}{[YXO'} = \frac {GX}{XE} \cdot \frac {3}{2} \implies</cmath> | ||
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<cmath>\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}.</cmath> | <cmath>\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}.</cmath> | ||
Therefore each Euler line of triangles <math>\triangle BCI, \triangle CAI, \triangle ABI,</math> cross Euler line of <math>\triangle ABC</math> in the same point, as desired. | Therefore each Euler line of triangles <math>\triangle BCI, \triangle CAI, \triangle ABI,</math> cross Euler line of <math>\triangle ABC</math> in the same point, as desired. | ||
− | <cmath>X(21) = 3R X(2) + 2r X(3) = R (X(4) + 2X(3)) + 2r X(3) \implies</cmath> | + | <cmath>X(21) = \frac {3R X(2) + 2r X(3)}{3R + 2r} = \frac {R (X(4) + 2X(3)) + 2r X(3)}{3R + 2r} \implies</cmath> |
− | <cmath>X(21) = X(3) + \frac {1}{3+ \frac {2r}{R}} (X(4) – X(3)) \implies k_{21} = \frac {1}{3+ | + | <cmath>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}}.</cmath> |
<i><b>Claim (Segments crossing inside triangle)</b></i> | <i><b>Claim (Segments crossing inside triangle)</b></i> |
Latest revision as of 16:41, 23 November 2022
Schiffler point
Let and
be the incenter, circumcenter, centroid, circumradius,
and inradius of
respectively. Then the Euler lines of the four triangles
and
are concurrent at Schiffler point
.
Proof
We will prove that the Euler line of
cross the Euler line
of
at such point
that
.
Let and
be the circumcenter and centroid of
respectively.
It is known that lies on circumcircle of
Denote
It is known that is midpoint
point
lies on median
points
belong the bisector of
Easy to find that ,
We use sigh for area of
We get
Using Claim we get
Therefore each Euler line of triangles
cross Euler line of
in the same point, as desired.
Claim (Segments crossing inside triangle)
Given triangle GOY. Point lies on
Point lies on
Point lies on
Point lies on
Then
Proof
Let be
(We use sigh
for area of
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