Solutions 1 MT1100 From Euclid to Mandelbrot 1. (i) Two triangles are congruent if one can be transformed to the the other using a combination of translations, rotations and reflections. Another definition (easier to check, but only works for triangles): Two triangles are congruent if their edges can be paired up so that the corresponding sides are equal in length and the corresponding angles are equal. (ii) Two triangles are congruent if their edges can be paired up so that one of the following holds: (SAS) two corresponding edges are equal in length, and the included angles are also equal; (SSS) all three of the corresponding edges are equal in length; (ASA) two of the three corresponding angles are equal, and the included sides are equal in length; (AAS) two of the three corresponding angles are equal, and one correspond- ing pair of non-included sides are equal in length. (iii) Here is one example. Let A = (0,2), B = (-1,0), C = (1,0) and D = (2,0). The triangles BAD and CAD satisfy SSA, but are not congruent. See the diagram below. 1
A 1 -2 B -1 c 1 O D 2 2. See the diagram below. Draw a line L through A and B. Draw a circle c1 with centre B through A. Let D be the second point of intersection of c1 with L. Draw a circle c2 with centre D through B. Then C is the second point of intersection of c2 with L. A B D C c1 c2 Since AB and DB are radii of the same circle c1, |DB| = |AB| = 1. Since DB and DC are radii of the same circle c2, |DC| = |DB| = 1. Hence |AC|=|AB|+|BD|+ | DC| = 1 + 1 + 1 = 3. 3. (i) By Pythagoras' Theorem applied to the triangle AB1B2, |AB1|2 + | B1B2|2 = |AB2|2. 2
Since the left hand side is 12 + 12 = 2, we see that |AB2|2 = 2 and so |AB2| = v2. Similarly, |AB3| = v3, since Pythagoras' Theorem applied to the triangle AB2B3 shows that 3 = (v2)2 + 12 = |AB2|2 + |B2B3|2 = | AB3|2. Pythagoras' Theorem applied to AB3B4 shows that | AB4| = V4 = 2, and then Pythagoras' Theorem applied to AB4B5 shows that |AB5| = v5. (ii) This is a very repetitive construction! Draw the line AB1. Erect a perpendicular to that line at B1. (Construc- tion 1.5). Draw a circle through A, centre B1. Define B2 to be at the intersection of the perpendicular and this circle. Draw the line AB2. Erect a perpendicular to that line at B2. (Construc- tion 1.5). Draw a circle through B1, centre B2. Define B3 to be at the intersection of the perpendicular and this circle. Draw the line AB3. Erect a perpendicular to that line at B3. (Construc- tion 1.5). Draw a circle through B2, centre B3. Define B4 to be at the intersection of the perpendicular and this