1 Solution of Homework

Transcription

1 Math 3181 Dr. Franz Rothe February 2, 2012 All3181\4080_spr12h1.tex Name: Use the back pages for extra space due date: February 10 1 Solution of Homework Figure 1: A neutral construction of the tangents to a circle Construction 1 (A neutral construction of the tangent to a circle). Let point A be the intersection of the segment OP with the given circle C. At point A, we erect the perpendicular p onto segment OP. We need a second circle around O through point P. Let B be an intersection point of this second circle with the perpendicular p. Finally, we need an intersection point T of circle C with segment OB. This intersection point is the touching point of a tangent from point P to the given circle C. Validity of the Construction. The triangles OAB = OT P are congruent. This can be seen by SAS congruence, because of the common angle AOB = P OT = T OP and two pairs of congruent adjacent sides OA = OT and OB = OP. Hence we get congruent right angles OT P = OAB = R. Since point T lies on the circle C, too, the segment OT is a radius of that circle. Since T P is perpendicular to the radius OT, it is a tangent of circle C. 1

2 10 Problem 1.1 (Tangent to a circle in the Poincaré disk). Perform the above construction 1 in the Poincaré disk model of hyperbolic geometry. Use a circle with the same center O as the Poincaré disk. Actually do and describe your construction. Figure 2: Construction of a tangent to a circle in the Poincaré disk Answer. Draw segment OP and find the intersection point A with the given circle these two steps are absolute. The circle with diameter AA, around the center K, is the hyperbolic perpendicular to line OP at point A. Draw the circle around O through the given point P, this step is absolute. This circle intersects the perpendicular at points B and C. Draw the Euclidean ray OB and a third circle, around O through point K, which intersects the ray at point t. This point is the polar of the tangent t from point P to the original circle, to be constructed. 2

3 Figure 3: Tangents to a circle by Thales circle Construction 2 (Tangents to a circle via Thales Theorem). One begins by constructing a second circle T with diameter OP. (I call this circle the Thales circle over the segment OP ). The Thales circle intersects the given circle in two points T and S. The lines P T and P S are the two tangents from P to circle C. Validity of the Construction in Euclidean Geometry. By Thales theorem, the angle P T O is a right angle, because it is an angle in the semicircle over diameter OP. Since point T lies on the circle C, too, the segment OT is a radius of that circle. By Euclid III. 16, The line perpendicular to a diameter is tangent to a circle. Since T P is perpendicular to the radius OT, and hence to a diameter, it is a tangent of circle C. 10 Problem 1.2. In the Poincaré model are given: a circle C with center O, and a point P outside the circle. Literally do the construction 2. Check what kind of hyperbolic line you obtain. Why does it have two intersection points with the circle? Where lies the second intersection point? Answer. The circle with diameter OP turn out to be the same in the Euclidean and the hyperbolic sense. Let T 2 be an intersection point of the two circles. The Euclidean line 3

4 Figure 4: In hyperbolic geometry, Thales circle does not yield the tangent to a circle. P T 2 is the Euclidean tangent from point P to the given circle C. The hyperbolic line P T 2 is different from the Euclidean one. It intersects the circle C in a second point S between P and T Problem 1.3 (About the validity of the two constructions in hyperbolic geometry). We want to construct the tangents through a given point outside to a given circle. Which one of construction 2 and construction 1 explained in the section on neutral geometry of circles and continuity remains valid in hyperbolic geometry? Explain why. Answer. Construction 2 is no longer valid in hyperbolic geometry. On the other hand, construction 1 remains valid in hyperbolic geometry. Construction 2 is not valid in hyperbolic geometry, because the angle sum of a triangle is less than two right, and hence Thales theorem does not hold in hyperbolic geometry. On the other hand, construction 1 depends only on SAS-congruence, and the fact that tangent and radius being perpendicular. These statements hold in hyperbolic geometry, too. 10 Problem 1.4. Given is a point P inside the unit circle with Euclidean coordinates (1 p, 0) where 0 < p < 1. Find, in Euclidean analytic geometry, the equation of the polar P. Check that this line does not intersect the unit circle. 4

5 Answer. The reflected point P has coordinates ( 1, 0). Hence the midpoint M of the 1 p segment P P has coordinates (m, 0) with m = 1 2 The equation of the polar P is ( 1 p + 1 ) 1 p = (1 p) (1 p) = x = m, y = y free parameter The polar does not intersect the unit circle since 2 2p + p2 2(1 p) m 1 = 2 2p + p p 2(1 p) = p 2 2(1 p) > 0 Too, we see that in the limit p 0, the polar approaches the circle from outside in second order. 5

6 10 Problem 1.5. Given is a hyperbolic line δ = EF, with ideal endpoints E and F, and a point P on the line δ. Use the given drawing, with P, P, δ, E, F already available to erect the perpendicular σ on the line δ at point P. Use the ideal endpoints S and T of σ to check the accuracy of your drawing. In this first variant, make use of the ideal endpoints. Construction 3. The Euclidean lines EF and P intersect at point σ. Finally, to get σ is easy, one simply draws a circular arc with center σ through point P. Remark 1. One need not even construct the tangent from σ to D. Since δ σ, it is known that the ideal endpoints S and T of σ and δ lie on a Euclidean line. This fact can be used to check the accuracy of the construction. Figure 5: Erect a perpendicular on line δ at point P, using the right angle as explained in Remark 2. Finally, combining the two constructions yields better accuracy. Answer. Remark 2. As an alternative construction, one can also erect the perpendicular on the radial ray P δ at vertex P. Again, that perpendicular intersects line p at the polar point σ. Using both constructions gives another possibility for better accuracy. 6

7 10 Problem 1.6. Given is a hyperbolic line δ = EF, with ideal endpoints E and F, and a point P not on the line δ. Use the given drawing, with P, P, δ, E, F already available to erect the perpendicular σ on the line δ at point P. Use the ideal endpoints S and T of σ to check the accuracy of your drawing. Construction 4. The Euclidean lines EF and P intersect at point σ. Finally, to get σ is easy, one simply draws a circular arc with center σ through point P. Answer. Figure 6: Drop a perpendicular onto line δ from the given point P. 7