Solutions 4 MT1100 From Euclid to Mandelbrot 1. (i) Since the triangle QPX is right-angled, Pythagoras' Theorem gives |QX|2 = |QP|2 + |PX|2 = 11 + 12 = 2. So |QX| = v/2. Since S trisects the line segment QX, we see that |QS| = |QX|/3 = v2/3. The angle 8 at PQS is equal to the angle PQX. But the angles at PQX and PXQ are equal (as the triangle QPX is isosceles), and add up to T/2 as they are the non-right angles in the right-angled triangle QPX. So ß = T/4. (ii) The cosine rule says that [PS|2 = |QP|2 + |QS|2 - 2|QP|QS| cos(T/4) = 12+ (v/2/3)2 - 2(v/2/3)(1/v/2) = 1+2/9 - 2/3 = 5/9. Hence |PS| = v5/3. The sine rule says that PS| sin(T/4) = sin a |QS| . Hence sin Q = |PS| sin(T/4)|QS| (v5/3) = (1/1/2)(v2/3) = 1/15. So Q = sin-1(1/5) ~ 0.4636 radians. We have (T/2)/3 ~ 0.5236. Hence a (T/2)/3, and so the method does not trisect in this case. 2. (i) There are e places to start our arrow, since there are e edges. Once we have picked out starting point, there are two possibilities for the arrow's 1
end point since an edge is adjacent to two faces. So there are 2e arrows. Counting again, there are f faces and each face contains n arrows (since it has n sides, and there is an arrow from each side to its centre). So there are fn arrows. Since we have counted the same collection of objects in two different ways, fn = 2e. (ii) There are v places to start our arrow, since there are v vertices. Once we have picked our starting point, there are q possibilities for the arrow's end point since this is in an adjacent face and since q faces meet at each vertex. Counting again, there are f faces and each face contains n arrows (since it has n sides, so has n corners, and there is an arrow from each corner to its centre). Since we have counted the same collection of objects in two different ways, fn = qv. 3. We count the number of arrows from the mid-point of an edge to the centre of an adjacent face. There are e edges to start from. Once we have picked the starting point, there are two possibilities for the adjacent face to go to, so there are 2e arrows. Counting again, there are N faces for an arrow to point to. Every face has 3 sides, so there are three arrows (one per side) once we have chosen a face. So there are 3N arrows. Hence we get the equation 2e = 3N, and so a deltahedron with N faces must have 3N/2 edges. The number v of vertices is then given by Euler's formula: v = e + 2 - f = = > . 2+2-N= >(N+4). 1 Clearly v must be an integer (you cannot have half a vertex!).