Solutions 3 MT1100 From Euclid to Mandelbrot 1. We know that | AP| = |BP|, |AQ| = |BQ| and (clearly) | PQ| = |PQ|. So, by SSS, the triangles APQ and BPQ are congruent. In particular, since a and ß are corresponding angles in congruent triangles, a = B. Since | AP| = |BP|, a = B and (clearly) |PR| = |PR|, the triangles APR and BPR are congruent by SAS. So the third sides AR and BR of these triangles have equal length, which is what we wanted to show. 2. A solution is given in the diagram below. Given points A and B, we can construct the points C, D, E, F and G in turn by intersecting constructible circles. / D C E F A / = G B 1
3. See the diagram below. C1 X1 Q C2 X2 -P A O B P -Q X3 X4 Draw the line AB through A and B (this is the x-axis). Using Construc- tion 1.2, draw a circle centre A and radius p. This is possible since p is a constructible length. Let P= (p, 0) and -P= (-p, 0) be the points where this circle intersects AB. Erect a perpendicular L to AB at A (this is the y-axis). Using Construction 1.2, draw a circle centre A and radius q. This is possible since q is a constructible length. Let Q= (0, q) and -Q= (0, -q) be the points where this circle intersects L. Erect perpendiculars M1 and M2 to AB at P and -P respectively. Erect perpendiculars N1 and N2 to L at Q and -Q respectively. Construct the points X1, X2, X3 and X4 where Mi and Nj meet. 4. Construct the mid-point X of the line segment AB (this is the intersection of the perpendicular bisector of A and B with the line AB). Construct the circle C1 with centre X, through A. Construct the line L through the point C, parallel to the line AB (using one of the constructions in the lectures). Let P be one of the points of intersection of L with C1. Then PAB has the properties required. 2
C B X A To prove correctness, first note that C1 is a circle with diameter AB, since X is the mid-point of the line segment AB. Any point Q on this circle will have the property that AQB is a right-angle, by Thales' Theorem. In particular, since P lies on this circle, PAB is a right-angled triangle with the right-angle at P. Any point Q on the line L is the same height above the line AB as C, so the formula '}bh' for the area of a triangle shows that the areas of the triangles ABQ and ABC are equal. Since P lies on L, the triangles PAB and ABC have equal area. So our construction is correct. 3