Math104: Non-Euclidean Geometry

Demonstrations:

1. Problem 7 (p. 34)

2. Morley's Theorem



3. Elliptic Geometry (Upper Hemisphere)

Consider the real projective plane in which the "points" [x,y,z] are the equivalence classes of all real-valued ordered triples (x,y,z) under the equivalence relation (x,y,z) ~ (kx,ky,kz) for all real numbers k ≠ 0. A "line" is a real-valued ordered triple (u,v,w) defined as the set of all points [x,y,z] that satisfy the equation ux + vy + wz = 0
"Incidence" is defined as set membership. Show that this interpretation satisfies all of the axioms for a projective plane.

Greenberg does not actually state in the problem that the points are defined as equivalence classes on R³\{0}, so we can start by showing that it is.

Reflexivity:
(x,y,z) ~ (x,y,z). This is true for k = 1.
Symmetry:
Suppose (x,y,z) ~ (a,b,c). Then, a=kx, b=ky, and c=kz for some nonzero real number k. Since it is nonzero, it has an inverse k' = 1/k which is also real and nonzero. So, (a,b,c) ~ (k'a,k'b,k'c) = (x,y,z).
Transitivity:
Let (x,y,z) ~ (a,b,c), and let (a,b,c) ~ (d,e,f). Then, (a,b,c) = (k₁x, k₁y, k₁z) for some nonzero k₁ and (d,e,f) = (k₂a, k₂b, k₂c) for some nonzero k₂. So, (d,e,f) = (k₁k₂x, k₁k₂y, k₁k₂z). And because k₁k₂ is nonzero, (d,e,f) ~ (x,y,z)

So, (x,y,z) ~ (kx,ky,kz) defines an equivalence relation.

In the real projective plane, we write the "points" as equivalence classes [x,y,z]. In R³, these classes correspond to lines passing through the origin. We can, therefore, represent the real projective plane in 3-space by choosing representative points from each of the lines that pass through the origin. One way of doing this is to choose points on the lines that are unit distance from the origin. This leads us to consider all of the points on a sphere of radius 1. But, since each line passing through the origin hits the sphere twice, by using the whole sphere we are actually using two representatives from each line. Consequently, we need to throw away half of the unit sphere. By convention, it is convenient to throw away the southern hemisphere along with half of the equator.

Now that we have defined what points are in this representation of the real projective plane, we can ask about what the lines look like. According to the definition, the line (u,v,w) consists of the set of points [x,y,z] that satisfy the equation ux + vy + wz = 0. Having already described [x,y,z] as a unit vector confined to the northern hemisphere, we can similarly think of (u,v,w) in the same sense. With this is mind the defining equation of a line can be written as a dot product:

(u,v,w)·(x,y,z) = 0
So, in this real projective plane, the line (u,v,w) consists of all points that are perpendicular to the vector (u,v,w). This is commonly known as the definition of a plane with normal vector (u,v,w). Taking the intersection of this plane with all of the points in the northern hemisphere gives us a representation of a line.

From this picture it is easy to see that all three incidence axioms are satisfied. Incidence axiom 1 holds because two unique points along with the origin determine a unique plane, which in turn, corresponds to a unique line. Incidence Axiom 2 is satisfied because the intersection of any plane through the origin with the northern hemisphere will consist of an infinite number of distinct points. Incidence Axiom 3 is satisfied because there is no plane that can intersect the entire northern hemisphere, so for every line there are infinitely many distinct points that do not lie on it. This implies the third axiom.

Furthermore, we can deduce the elliptic parallel property from the fact that the intersection of any two planes going through the origin will be a line passing through the origin. This line will meet the northern hemisphere at a single point, which is the the point of intersection of the two lines corresponding to the two planes.



In this demonstration, two points P and Q can be chosen on the northern hemisphere. These two points determine a plane whose intersection with the northen hemisphere produces a red line through the two points.

4. Desargues' Theorem





This demonstration draws two different pictures: a two-dimensional one and a three-dimensional one. The 2D graph represents the problem as it is stated for triangles lying in the plane with their vertices located on a set of three parallel lines. When the sides of the three triangles are extended, the corresponding edges meet at three points P, Q, and R. These points are collinear.

To show that this is true for any pair of triangles we choose, we move up to 3-space. In the 3D graph, three different lines with the same direction are drawn. Along these lines we plot six points A, B, C and A', B', C' which serve as the vertices for △ABC and △A'B'C'. Assuming the plane containing △ABC is not parallel to the plane containing △A'B'C', the two planes will intersect each other at a line. In the control panel, the variable labeled "max" determines how far you want to extend the planes so that they intersect. Their intersection appears in the 3D graph as a white line. Because each line that passes through a side of a triangle must lie in the same plane as the triangle, the intersections of the extended sides of the two triangles must lie in the intersection of the two planes. In the control panel, you can use the "ShowAB", "ShowBC", and "ShowAC" variables to display the extended sides of the triangles and verify that they do intersect at the white line.

The demonstration, however, actually shows equivalent pictures, because even though one is a 2D graph and the other a 3D graph, they are both drawn on a flat surface. If we rotate the 3D graph in a certain way, it is possible to create a picture that is identical to the 2D graph. There are two cases in which the white line does not show up in the 3D graph. In the first case, if the white line is directed out of the screen, it will appear as a point. This can only happen if the planes of the two triangles are both perpendicular to the plane of the screen. But if this were true, then the triangles would have not visible area, in fact, they would project onto the screen as line segments. In the second case, if the planes of the two triangles were parallel, then they would not intersect and there will be no white line.

Now, consider the case where AB and A'B' are segments lying on parallel lines. The other two pairs of sides will intersect at points Q and R, and they will uniquely determine a line. Since the line passing through AB and the line passing through A'B' do not meet, they must both be parallel to the line passing through Q and R. If we were to think about this problem in terms of a projective space, then these parallel lines would all meet at a point at infinity, which we would call P. Using this interpretation, P, Q, and R would be collinear.

We can also look at the extreme case where AB || A'B', BC || B'C', and AC || A'C'. In this case, we have three pairs of parallel lines and, therefore, no intersection points P, Q, or R. But, again, thinking about the problem in terms of a projective space means that AB and A'B' meet at a point P at infinity, BC and B'C' meet at a point Q at infinity, and AC and A'C' meet at a point R at infinity. P, Q, and R are collinear in this interpretation because in a projective space there is an infinity line passing through all of the infinity points. So, P, Q, and R would lie on this infinity line.

5. Existence of a Perpendicular in UHP


Given a point P not on an H-line L in the UHP, show that it is possible to find an H-line through P that is perpendicular to L.

If L is the vertical H-line x = x₀ and P is a point not on L, then the H-line through P and perpendicular to L is the semicircle with center (x₀,0) and radius ||P - (x₀,0)||.

Now, consider a semicircle H-line L centered at (0,0) with radius R₁, and a point P not on L. Then, let (x₀,0) and R₂ be the center and radius of the H-line through P = (x₁,y₁) and perpendicular to L. If we connect the point of intersection of the two H-lines and draw the two segments from this point to the centers of the two H-lines, we form a right triangle. Applying the pythagorean theorem to this triangle gives the following equation:

R₁² + R₂² = x₀²

But, by the distance formula, R₂ = √((x₁-x₀)² + y₁²). So,

(x₁-x₀)² + y₁² + R₁² = x₀²
x₀² - 2x₀x₁ + x₁² + y₁² + R₁² = x₀²
2x₁x₀ = x₁² + y₁² + R₁²
x₀ = (x₁² + y₁² + R₁²)/(2x₁)

So, if we know the radius of L and the location of point P with respect to the center of L, we can calculate the location of the center of the H-line through P that is perpendicular to L. With the center point and P, we can draw the unique H-line through P that is perpendicular to L.

Demonstration



This demonstration draws a green semicircle H-line with radius R₁. In the 2D window, you can choose a point P anywhere in the upper half-plane. The red curve is the unique H-line passing through P that is perpendicular to the red H-line.

Given two H-lines that do not meet, under what circumstances will there be an H-line perpendicular to both of them?

(1)If the two H-lines that do not meet are semicircles with different radii but the same center point (x₀,0), then the H-line perpendicular to both of them is the vertical line x = x₀.

(2)If the two H-lines that do not meet are both vertical lines, then there is no H-line that is perpendicular to both of them, since the perpendicular line would have to be horizontal at two different values of x.

(3) Suppose the two H-lines are semicircles: one with radius R₁ and center at (x₁,0), and the other with radius R₂ and center at (x₂,0), and x₁≠x₂. There can be no vertical H-line that passes through both of these semicircles, because the only case in which that can happen was treated in (1). So, the perpendicular H-line must also be a semicircle. Let its radius be R₀ and let it be centered at (x₀,0). If you connect the point of intersection between the first H-line and the perpendicular H-line with the centers of the two H-lines, you get a right triangle, and the pythagorean theorem gives the equation:

R₀² + R₁² = (x₁-x₀)²
Similarly, for the intersection with the second H-line, the pythagorean theorem yeilds:

R₀² + R₂² = (x₂-x₀)²
These are two equations that allow us to solve for the two unknowns (R₀ and x₀). So, this gives us both the radius and location of the center of the perpendicular H-line.

(4) Suppose one of the H-lines is a vertical line x = x₁ and the other H-line is a semicircle with radius R₁ and centered at (x₂,0). To meet the vertical H-line at a right angle, the perpendicular H-line must be a semicircle with center at (x₂,0). Let the radius of the perpendicular H-line be R₂. If you connect the point of intersection between the semicircle H-line and the perpendicular H-line with the centers of the two H-lines (x₂,0) and (x₁,0), you get a right triangle. From this observation, the pythagorean theorem gives us:

R₁² + R₂² = (x₂-x₁)²
Since R₁, x₁, and x₂ are all known values, we can solve for R₂ and take the single positive solution. We can, therefore, calculate both the radius R₂ and the center (x₂, 0) of the perpendicular H-line.


6. H-line perpendicular to two given H-lines in UHP model



7. Congruences that preserve i.


We can write down the LFTs that preserve i as a matrix R_θ with 1 as its determinant :

[cosθ -sinθ]
[sinθ cosθ]

Since the image of the imaginary axis z = it has its center at x₀ = (cos²θ - sin²θ)/(2sinθcosθ) = cos(2θ)/sin(2θ) = cot(2θ), x₀ can have any value between -∞ and +∞. So the images of the imaginary line under R_θ consist of all semicircles in the UHP that pass through i.

Demonstration



This demonstration draws the point i and the imaginary axis. The part of the axis above i is colored red while the part of the axis below i is colored green. In the control panel, you can use the tapedeck to change the angle θ of the rotation of the axis in the UHP (here I use the letter "q" instead of "θ"). As θ goes from 0 to π/2, the axis gets mapped onto semicircle H-lines. It looks as if 2θ is the angle between the imaginary axis and the line tangent to the image curve at i. At θ = π/2, the red part has been mapped to the lower part of the axis and the green part has been mapped to the upper part of the axis. Note that the point (0,i) is preserved and observe that at θ = π/4, the image curve is the unit circle centered at the origin. In this case, a = -b = c = d = 1/√2.

Now, choose a point P on the imaginary axis. As you rotate the axis, P also gets rotated. Suppose that at θ = 0 P = (0,t). Then, at θ = π/2, P = (0,1/t). Using the variable called "ShowPath", you can display the path traced out by P. Notice how the path of P always seems to be perpendicular to the rotated images of the imaginary axis. Since rotations are supposed to preserve length, the path traced out by P should be a circle with radius ln(t) and center i. Remarkably, circles in the UHP resemble Euclidean circles.



8. Midpoint Polygon Problem



This demonstration draws a triangle, a quadrilateral and a pentagon as well as the corresponding midpoint triangle, midpoint quadrilateral, and midpoint pentagon. In the control window, the ratio of the area of the polygon to its midpoint polygon is calculated. This ratio is 1/4 for triangles and 1/2 for quadrilaterals. Notice that the ratio for quadrilaterals is 1/2 even if the quadrilateral is not convex. For pentagons the ratio varies.

9. Isometry of Poincare Disc, Beltrami-Klein, and UHP Models of Hyperbolic Geometry





10. Perpendicularity in the Beltrami-Klein Model



The demosntration allows you to choose a an open chord m in the Klein model as well as a point C. Then, there is a unique perpendicular chord k to m that passes through C. It is constructed by drawing the Euclidean line through C and the pole of m, P(m), and then taking its intersection with the interior of the circle γ. When we map these lines straight up onto the hemisphere we can see that the lines intersect at right angles.

11. Equidistant Curve in UHP



12. Laguurre Geometry (Tangent Circles)



13. Tangent Circles



14. Orthogonal Circles: (P-15 p. 284)



15. Laguerre Geometry (Dilation)



16. 9-Point Circle



The 9-point circle in 2D:






The 9-point circle in 3D:




The three-dimensional model begins with the two dimensional perspective. By rotating the figure in three dimensions, we can actually see how the two prisms are formed out of the different points. Furthermore, the 9-point circle in 3D is either an ellipsoid, if H lies inside the red circle, or a hyperboloid, if H lies outside the red circle. As the points of the triangle ΔABC are moved around, we can see the ellipsoid turn into a cylinder and then into a hyperboloid. All 8 vertices of the prism lie on the ellipsoid/hyperboloid. So, in three dimensions, we get an 11-point ellipsoid/hyperboloid, since it passes through the original 9 points of the 9-point circle as well as the points O and H. The point U = (O + H)/2 becomes the center of the ellipsoid/hyperboloid.

17. Spherical Trigonometry

As in class, consider a spherical triangle with vertices A, B, and C on the unit sphere centered at O with a right angle at C. Let the sides opposite vertices A, B, and C be labelled a, b, and c. Let D be on OA, E on OB, and F on OC such that FD and ED are perpendicular to OA, and EF is perpendicular to OC. Prove the following rules for Spherical Trigonometry:

1. sin(B)sin(c) = sin(b)
2. tan(c)cos(A) = tan(b)
3. tan(A)sin(b) = tan(a)
4. cos(c) = cot(A)cot(B), and
5. cos(A) = cos(a)sin(B).

Helpful diagram:


18. Arbelos



19. Ptolemny's Theorem

Here is a demo that illustrates the relationship between angle sum and area for spherical geometry.

The demo draws a geodesic triangle on the unit sphere whose three vertices are controlled by moveable hotspots. In the control panel, the three double lunes of angles α, β, and γ can be toggled on and off. Notice that when all three double lunes are toggled on, the sphere is completely covered once, except for two triangular regions, which are covered three times.
With this observation we can find an expression for the area of the geodesic triangle in terms of the areas of the double lunes.

20. Pythagorean Theorem



21. Equivalence of Klein and Poincare Disc Models via Hyperboloid

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