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All five problems below are about models of hyperbolic geometry (HG). In the first two tasks, we use Poincar´e’s upper half plane (PÖH). All points have the y-coordinate> 0. The geodesy between two points A and B is the unique circular arc that passes through the two points and where the center of the circle lies on the x-axis. the x-axis should be perceived as ∞ (infinity).
The three concluding tasks are about the model of HG called Poincar´es circle (PC). See the link: Geodeters in P oincar´es circle model on Moodle.
http://homepage.lnu.se/staff/hfrmsi/1ma113/lines_p…
I refer to the article in the text below. All points lie inside a circle, c, with radius r and center O. The geodesy between two points A and B is the unique arc of a circle that passes through the two points and intersects c at right angles. The stripe of c should be perceived as ∞ (infinity). Learn to invert a point A in a circle (construction 1.2). The relationship between a point A and its inverted point A ^ −1 is that OA · OA^ − 1 = r ^ 2 where r denotes the radius of the circle. Note that all geodets through O are diameters.
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Geometry Questions:
1. Construct two geodeters, l1 and l2, in Poincar´e’s upper half plane (PÖH) that do not intersect. Then draw a geodet m that intersects both l1 and l2 and where the alternate angles are different. As before, the alternate angles are on each side of m, but this time it is angles between tangents at the intersection points. You have thus shown that Euclid’s 29th theorem does not apply in HG. This is where he got use of his fifth axiom that does not apply in HG. Theorem 27, on the other hand, also applies in HG. It says that if the alternate angles are equal, then the two geodeters cannot intersect each other.
2. In this problem you need the equation of the circle. Given the surveyor, m, which passes through (- √ 1 2, √ 1 2) and (√ 1 2, √ 1 2) in PÖH and a point A = (0, 2). Find the geode (circular arc), n, through A that meets (tangents) m on the positive x-axis (i.e. the infinity in the model). The angle between the y-axis and the key to n at point A is called the parallel angle. For smaller angles, cut geodeters through the A geodesy m. Calculate the parallel angle.
3. Show that a circle c1 passing through point A and its inverted point B is orthogonal to circle c with the radius OP. See figure. The fact that two circles are orthogonal to each other means that the angle between the keys at the point of intersection is 90 degrees. We need this result in the next task.
Point B is inverse to point A in the circle C. Are the triangles OBP and OPA uniform? Are the angles OBP and APO equal? If the circle C1 is orthogonal to C, it means that the angle O1PO is right.
4. Given two points A and B in PC, find the geodetic through the two points (construction 2.1). Explain your design!
5. Construct a triangle ABC in PC with the angles π / 4, π / 4 and π / 4. Select the center of the circle c as one of the corners (A) so that two surveyors go along diameters. Thus, the third geodetic remains, which must be an arc of a circle. Let the corner B lie on the x-axis with the coordinate b. We set r = 1 and then the inverted point has x = 1 / b. The arc of the circle must therefore intersect the two diameters at an angle √ π / 4. Find an equation for b and solve it. That tan π / 8 = 2 – 1 can be used. Tip: Focus on the center of the desired arc. you get the x-coordinate from the requirement that the circle should be orthogonal to c. The y-coordinate you get from the requirement that the circle should go through B and the angle between the tangent and the x-axis should be π / 4 there. Since the angle should be π / 4 even at C, we have a line of symmetry. The desired triangle is an isosceles hyperbolic triangle with an angle sum of 3π / 4.
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