Difference between revisions of "Stereographic projection"
(→Moscow Math Olympiad, 1950) |
(→Moscow Math Olympiad, 1950) |
||
| Line 5: | Line 5: | ||
==Moscow Math Olympiad, 1950== | ==Moscow Math Olympiad, 1950== | ||
[[File:MMO 1950.png|300px|right]] | [[File:MMO 1950.png|300px|right]] | ||
| + | [[File:MMO 1950 a.png|300px|right]] | ||
A spatial quadrilateral is circumscribed about the sphere. | A spatial quadrilateral is circumscribed about the sphere. | ||
| Line 17: | Line 18: | ||
This circles lies in different planes and have the common points, so they are tangent in pares. | This circles lies in different planes and have the common points, so they are tangent in pares. | ||
| + | The stereographic projection from point <math>D</math> is shown on diagram. Points <math>A,B,C</math> maps into points <math>A', B', C',</math> tangent circles <math>\omega_b</math> and <math>\omega_c</math> maps into tangent circles <math>\omega'_b</math> and <math>\omega'_c,</math> circles <math>\omega_a</math> and <math>\omega</math> maps into parallel lines <math>\omega'_a</math> and <math>\omega',</math> tangent to circles <math>\omega'_b</math> and <math>\omega'_c,</math> respectively. | ||
| + | |||
| + | Condition that points <math>A,B,C,D</math> lie in one plane transforms into condition that points <math>A', B', C'</math> are collinear which is trivial. | ||
{{stub}}[[Category:Geometry]] | {{stub}}[[Category:Geometry]] | ||
Revision as of 07:05, 3 May 2025
A stereographic projection is a projection from a sphere to a tangent plane. Stereographic projections preserve angles.
To stereographically project a point on a sphere to a plane tangent to its south pole, draw the line from the north pole of the sphere to the point in question. The stereographic projection of this point is then the intersection of this line with the plane. As such, the stereographic projection of the north pole will be undefined.
Moscow Math Olympiad, 1950
A spatial quadrilateral is circumscribed about the sphere.
Prove that the four points of contact lie in one plane.
Proof
Let given quadrilateral be 
the points of contact be 
the north pole of the sphere be point 
Denote sphere as 
Let plane 
cross sphere at circle 
Points 
and 
lies on Similarly define circles 
with points 
and 
with points and 
with points 
and 
This circles lies in different planes and have the common points, so they are tangent in pares.
The stereographic projection from point is shown on diagram. Points 
maps into points 
tangent circles 
and 
maps into tangent circles 
and 
circles and 
maps into parallel lines 
and 
tangent to circles and respectively.
Condition that points 
lie in one plane transforms into condition that points 
are collinear which is trivial.
This article is a stub. Help us out by expanding it.
