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MA 125: Introduction to Geometry: Quickie 8. (1) What’s the maximal angle sum for a triangle on the sphere? (2) (a) What’s the angle sum for an n-gon in the plane? (b) What’s the interior angle for a regular n-gon? (c) With which regular n-gons can you tile the plane? All this is of course in the lecture notes, but try to answer the questions without looking at the lecture notes. (3) What can you say about the angle sum of a 4-gon (or n-gon) on the sphere? (4) Show that in the diagram below, AÔB = 2 · AĈB, where O is the centre of the circle (Euclid III.20). C O A (a) (b) (c) (d) B Draw the segment OC. Think of ∠ACB as ∠ACO + ∠BCO. Where else can you find ∠ACO and ∠BCO? Use that the angle sum in triangles equals π and that the angle sum at a point equals 2π. 1 2 Answers to Quickie 7. (1) Which of the following statements are true? (a) Any two great circles on the sphere are congruent. (b) Triangles on the sphere have angle sum π. (c) Two segments of two great circles are congruent if and only if the segments have the same length. (d) Isosceles triangles on the sphere have two equal angles. Answer. (1), (2) and (4) are true. You can prove (4) using SAS the same way we proved the statement for planar triangles. Statement (2) is incorrect, the angle sum is always more than π. (2) Show that any two great circles meet in two points. Answer. Let C1 and C2 be two distinct great circles. They are defined by two planes P1 and P2 in R3 passing through the origin. Any two planes in R3 intersect in a line L. Now denote by Q1 and Q2 the two intersection points of the line L with the sphere. Then Q1 and Q2 are precisely the intersection points of C1 and C2 . (5) In the lecture notes, Proposition 5.8, a proof is given that any triangle in the plane is isosceles. Find the mistake! Answer. If you do a careful sketch of the problem you will notice that either R or Q lies in the triangle ABC. This implies that Step (5) in the ‘proof’ does not work. (Note that unfortunately the ‘proof’ also has two typos: In line two it should say ‘bisector DE of CB’ and in (6) it should say CR = BQ.)