You may use pencil, pens, protractor, ruler, compass and scissors on this test. You may not use a calculator, book or notes. Please take your time on each problem and show your work in the space provided. If you get stuck on a problem, move on to the next and return when you have time.
In hyperbolic geometry, the circumference of a circle of radius r is
a) greater than 2πr.
b) less than 2πr.
d) a function of the location of the center of the circle.
The answer is (a). Because of the curvature of the hyperbolic plane, there is "more stuff" at distance r from the center of a circle in a hyperbolic plane than there is in the Euclidean plane. That's why our doily won't lie flat but instead wrinkles when you try to flatten it.
The list of steps given on pages 71-72 of Chapter 3 also works in the hyperbolic plane. The
following Euclidean constructions and theorems were used on page 72 and also work in the
a) (10 pts) Prove or explain why you cannot: every point P on line MN has the property |PA| = |PB|.
Most proofs of this fact use the side-angle-side theorem. If you are unable to prove congruence of angles in hyperbolic space, you are probably unable to prove this statement.
Let point T be the intersection of line MN and line AB. Because the proof of the pons assinorum and of side-side-side work in hyperbolic as well as Euclidean geometry (p. 118), the proof that MN intersects AB at right angles at the midpoint of segment AB (p. 72) is still valid (congruent supplementary angles are right angles in hyperbolic plane as well as the Euclidean one.)
By side-angle-side, triangle ATP is congruent to triangle BTP. Therefore |PA| = |PB|.
b) (10 pts) Prove or explain why you cannot: every point P that has the property |PA| = |PB| is on line MN.
Our Euclidean proof of this relied on the fact that there could be only one line through point T that was perpendicular to line AB. This remains true in the hyperbolic plane, but it would be understandable if you were unable to convince yourself of this.
By side-side-side, triangle ATP is congruent to triangle BTP. Since angles ATP and BTP are congruent supplementary angles they must be right angles. Therefore P lies on a line through T perpendicular to line AB; line MN is the only line through T perpendicular to AB, so P must lie on line MN.
a) (10 pts) Draw E6(AB) for two points A and B that are three blocks apart in the x-direction and one block apart in the y-direction.
The solution looks like a rectangle of points around A and B; the four corners of the rectangle are not included in the answer.
b) (10 pts) Read the definition of an ellipse posted at http://www.mathopenref.com/ellipse.html. Would you say that Er(AB) is an ellipse in taxicab geometry? Write a paragraph supporting your answer.
There is more than one right answer to this question.
Er(AB) is not an ellipse in taxicab geometry because an ellipse must be "A curved line forming a closed loop". Er(AB) will be either empty or a collection of discrete points for any choice of r, A and B. There are no lines, curves, or closed loops in the taxicab plane, so this definition does not apply.
Er(AB) could be defined to be an ellipse in taxicab geometry. The definition of an Euclidean ellipse states that "he sum of the distances from two points (foci) to every point on the line is constant." In Er(AB), the sum of the distances from two points A and B to every point of Er(AB) equals the constant r. Therefore Er(AB) should be considered a taxicab ellipse.