Poincare plane, sum of triangle angles < \pi

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Poincare plane, sum of triangle angles < \pi

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Source: Brazilian Mathematical Olympiad 1986 P3

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#1 h Jul 27, 2018, 6:54 AM • 2 Y

Y by Adventure10, Mango247

The Poincare plane is a half-plane bounded by a line $R$ . The lines are taken to be
(1) the half-lines perpendicular to $R$ , and
(2) the semicircles with center on $R$ .
Show that given any line $L$ and any point $P$ not on $L$ , there are infinitely many lines through $P$ which do not intersect $L$ . Show that if $ABC$ is a triangle, then the sum of its angles lies in the interval $(0, \pi)$ .

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#2 h Jul 27, 2018, 7:10 AM • 2 Y

Y by Adventure10, Mango247

parmenides51 wrote:

The Poincare plane is a half-plane bounded by a line $R$ . The lines are taken to be
(1) the half-lines perpendicular to $R$ , and
(2) the semicircles with center on $R$ .
Show that given any line $L$ and any point $P$ not on $L$ , there are infinitely many lines through $P$ which do not intersect $L$ . Show that if $ABC$ is a triangle, then the sum of its angles lies in the interval $(0, \pi)$ .

Poincare model of Lobachevskii geometry is well-known...
Statement (1) is obvious.
Statement (2) can be proved by the following construction. Consider Lobachevskii triangle $ABC$ and let $A'$ be the point symmetric to $A$ with respect to the absolute $R$ . Consider The inversion centered in point $A'$ . Then we get the Poincare model in circle (this circle is the image of $R$ ), where the traingle $ABC$ has vertex $A$ centered in the center of this circle. For such triangle statement (2) is obvious. As inversion is conformal transformation, statement (2) is also true for the first triangle $ABC$ .

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