Thales Configuration · Live Viewer

ETC Loci Sandbox Discovery Sandbox — Thales Configuration

$A=(1,0)$ and $B=(-1,0)$ are fixed while $C$ moves on the unit circle. The loci use exact rational parametric forms $x(t),y(t)$ with $t=\tan(\theta/4)$, obtained by normalising barycentric weights $\bar{a}=1-t^2$, $\bar{b}=2t$, $\bar{c}=1+t^2$ and applying the Weierstrass half-angle identities.

6 ETC centers Rational parametric loci JSXGraph live board ∠C = 90° always

Compact derivation

With $A=(1,0)$, $B=(-1,0)$, and $C=(\cos\theta,\sin\theta)$, each center starts from a barycentric triple $(u:v:w)$. The Thales normalization turns the Cartesian conversion into \[ x=\frac{u-v+w\cos\theta}{u+v+w},\qquad y=\frac{w\sin\theta}{u+v+w}. \] Using the substitution $t = \tan \frac{\theta}{4}$, we replace trigonometric terms by rational ones: \[ \cos\theta = \frac{1-t^2}{1+t^2},\qquad \sin\theta = \frac{2t}{1+t^2}. \]

Why this matters The locus formulas become rational in $t$, which is why the viewer can plot exact-looking curves so smoothly.

How to use the sandbox Move the angle slider, compare the radian and degree readouts, and inspect how the plotted center responds to the same parameter $t$.

Incenter · Centroid · Nagel Point

X(1) : X(2) : X(8)

$C$

$X(1)$

$X(2)$

$X(8)$

$\theta$ = —

Angle slider

Radians: — Degrees: —

Parametric curves — $t = \tan\frac{\theta}{4}$, $t \in (0,1)$

X(1) Incenter — degree 2

$x = -\dfrac{t^2 + 2t - 1}{t^2 + 1}$

$y = \dfrac{2t(1-t)}{t^2 + 1}$

X(2) Centroid — degree 4, locus is circle r = 1/3

$x = \dfrac{(t^2-2t-1)(t^2+2t-1)}{3(t^2+1)^2}$

$y = \dfrac{4t(1-t^2)}{3(t^2+1)^2}$

X(8) Nagel — degree 4, y ≥ 0 for t ∈ (0,1)

$x = \dfrac{(t^2+2t-1)(3t^2-2t+1)}{(t^2+1)^2}$

$y = \dfrac{4t^2(1-t)^2}{(t^2+1)^2}$

Nine-Point · Symmedian · Feuerbach

X(5) : X(6) : X(11)

$C$

$X(5)$

$X(6)$

$X(11)$

$\theta$ = —

Angle slider

Radians: — Degrees: —

Parametric curves — $t = \tan\frac{\theta}{4}$, $t \in (0,1)$

X(5) Nine-Point — degree 4, locus is circle r = 1/2

$x = \dfrac{(t^2-2t-1)(t^2+2t-1)}{2(t^2+1)^2}$

$y = \dfrac{2t(1-t^2)}{(t^2+1)^2}$

X(6) Symmedian — degree 4, locus is unit circle (= C)

$x = \dfrac{(t^2-2t-1)(t^2+2t-1)}{(t^2+1)^2}$

$y = \dfrac{2t(1-t^2)}{(t^2+1)^2}$

$X(2), X(5), X(6)$ share the same numerator factor, which visually echoes Euler-line collinearity.

X(11) Feuerbach — degree 6, smooth on t ∈ (0,1)

$x = \dfrac{(t^2+2t-1)(t^4-6t^3+2t-1)}{(t^2+1)^2(5t^2-4t+1)}$

$y = \dfrac{2t(1-t)(t^2+2t-1)^2}{(t^2+1)^2(5t^2-4t+1)}$