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→Hilbert's construction: distinct |
m →Proof in the Poincaré half-plane model: LaTeX spacing clean up, replaced: \,</math> → </math> (5) using AWB |
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Compose the following two [[hyperbolic motion]]s:
:<math>x \to x-a
:<math>\mbox{inversion in the unit semicircle.}
Then <math>a \to \infty</math>, <math>b \to (b-a)^{-1},\quad c \to (c-a)^{-1},\quad d \to (d-a)^{-1}.</math>
Now continue with these two hyperbolic motions:
:<math>x \to x-(b-a)^{-1}
:<math>x \to \left [ (c-a)^{-1} - (b-a)^{-1} \right ]^{-1} x
Then <math>a</math> stays at <math>\infty</math>, <math>b \to 0</math>, <math>c \to 1</math>, <math>d \to z</math> (say). The unique semicircle, with center at the origin, perpendicular to the one on <math>1z</math> must have a radius tangent to the radius of the other. The right triangle formed by the abscissa and the perpendicular radii has hypotenuse of length <math>\begin{matrix} \frac{1}{2} \end{matrix} (z+1)</math>. Since <math>\begin{matrix} \frac{1}{2} \end{matrix} (z-1)</math> is the radius of the semicircle on <math>1z</math>, the common perpendicular sought has radius-square
:<math>\frac{1}{4} \left [ (z+1)^2 - (z-1)^2 \right ] = z.
The four hyperbolic motions that produced <math>z</math> above can each be inverted and applied in reverse order to the semicircle centered at the origin and of radius <math>\sqrt{z}</math> to yield the unique hyperbolic line perpendicular to both ultraparallels <math>p</math> and <math>q</math>.
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