MATH Non-Euclidean Geometry Exercise Set #9 Solutions

Transcription

1 MATH Non-Euclidean Geometry Exercise Set #9 Solutions 1. Consider the doubly asymptotic triangle AMN in H where What is the image of AMN under the isometry γ 1? Use this to find the hyperbolic area of AMN. We need to find the images of A, M, and N under γ. 8 + i 1+ i γ A i 8 + i + i γ M 1 + ( ) γ N + This is the following doubly asymptotic triangle: 8 + i A, M, and N. i Exercise Set #9 Page 1 1 Spring 004

2 Now this doubly asymptotic triangle clearly has a right angle at γ A, so the area of this doubly asymptotic triangle as well as that of AMN is given by π π K π. Draw the asymptotic triangle ABM in H with A i, B i, and M 1. What is the Poincaré length d( A, B )? What is the area of ABM? Since B lies above A, we have that the length of AB is d( A, B ) ln( ) 1 ln To find the area we need both angles. We know that the angle at A is π. What is the angle at B? Let's try a little different technique to find the center of the circle that represents the Poincaré line through 1 and B. It will be at a number on the x-axis, and since > 1, this number must be to the left of 0. Let the point be at ( x,0). Then, the distance from ( x,0) ( 0, ). Setting this up we get: to( ) i i 1, 0 must be the same as the distance from ( x,0) ( 1+ x) ( ) x + 1+ x+ x x + x 1 to Thus, the center of the circle lies at the point ( 1, 0). The radius to B then has slope, 0 1 meaning the tangent line to the radius has slope. Thus, the tangent line makes an angle of π π with the x-axis, making the angle at the y-axis to be. 6 1 Exercise Set #9 Page Spring 004

3 Thus, A π and B π. Thus, the area of the triangle is π K π ( A+ B) 6. In H we have a triangle with angles A 10, B 0, and C 40. Find the sides of this triangle. We have to use the Hyperbolic Law of Cosines for Angles cosc cos Acos B+ sin Asin Bcosh c. This means that cos B cosc+ cos A cos 0 cos 40 + cos10 cosh a sin Bsin C sin 0sin 40 a cos AcosC+ cos B cos10 cos 40 + cos 0 cosh b sin Asin C sin10sin 40 b cos Acos B+ cosc cos10 cos 0 + cos 40 cosh c sin Asin B sin10sin 0 c In H we have a triangle with sides a, b 4, and c. Find the angles in this triangle. Is it a right triangle? We have to use the Hyperbolic Law of Cosines for Sides cosh c cosh acosh b sinh asinh bcosc. This means that cosh b cosh c cosh a cosh 4 cosh cosh cos A sinh bsinh c sinh 4sinh A radians.916 cosh acosh c cosh b cosh cosh cosh 4 cos B sinh asinh c sinh sinh B radians cosh a cosh b cosh c cosh cosh 4 cosh cosc sinh asinh b sinh sinh 4 C radians What is the area of the triangle with sides, 4, and? The area is K π ( A+ B+ C). This makes the area of this triangle to be: K π ( A+ B+ C) π ( ) What happens if you use the Hyperbolic Heron s Formula: s 6 and 4sinh ssinh( s a)sinh( s b)sinh( s c) 1 cosk (1+ cosh a)(1+ cosh b)(1+ cosh c) Exercise Set #9 Page Spring 004

4 This gives us: 4sinh 6sinh sinh sinh1 1 cos K (1+ cosh )(1+ cosh 4)(1+ cosh ) cos K K Given a triangle in H with sides and 4 and included angle of 0º, what is the third side and what are the other two angles? The Law of Cosines gives us the third side. cosh c cosh acosh b sinh asinh bcosc π cosh c cosh cosh 4 sinh sinh 4cos c.6890 Now, using the Law of Sines sinh a sinh c sin A sin C sinh a sinh π sin A sin C sin sinh c sinh c 6 A radians Similarly, we find that B radians Given a triangle in H with angles º and 4º and included side of 7, what is the third angle, the other two sides, and the area of this triangle? From the Hyperbolic Law of Cosines for Angles we have cosc cos Acos B+ sin Asin Bcosh c cos cos 4 + sin sin 4 cosh C radians The other sides are given by the Law of Sines: sinh a sinh c sin A sin C sin A sin sinh a sinh c sinh sin C sin 70. a sin B sinh b sinh c sin C b The area is given by K π A + B + C + + radians ( ) 180 ( 4 70.) Exercise Set #9 Page 4 Spring 004

5 8. Let ABC be a right triangle with the right angle at C. Prove that cos A cosh asin B. We will use the Hyperbolic Pythagorean Theorem and the results about the sine and cosine of the angle A. tanh b sinh b cosh c cos A tanh c cosh b sinh c sinh b cosh acosh b Hyperbolic Pythagorean Theorem cosh b sinh c sinh b cosh a sinh c Formula 11.4 cosh asin B 9. Let ABC be a right triangle with the right angle at C. Prove that cot Acot B cosh c. cos A cos B cot Acot B sin A sin B tanh b tanh a tanh c tanh c sinh a sinh b sinh c sinh c tanh a tanh b sinh c sinh a sinh b tanh c 1 1 ( cosh c) cosh a cosh b cosh c 10. Find all sides of the isosceles right triangle with angles A B 0. From above, cosh c cot Acot B ( cot 0) c Likewise, cosh acosh b cosh c ( a) cosh cosh a a Exercise Set #9 Page Spring 004