By Gordon, Basil; Âglom, Isaak Moiseevič; McCawley, James; Âglom, Akiva Moiseevič
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Additional resources for Challenging mathematical problems with elementary solutions Volume I : Combinatorial Analysis and Probability Theory
Suppose we could arrange seven points AI> A z, A a, A" As, Ae, A7 and seven lines Pb P2, Pa. p" Ps, Pe, P7 in a configuration satisfying the conditions of the problem. We show first that in this case any line joining two of the points AI> As, Aa, A 4, As. Ae. A7 is one of the seven lines PI> pz, Ps. P4. Ps, Pe. P7. and that any point of intersection of two of the lines Pl. pz. Ps. P4. Ps. Pe. P7 is one of the points AI> A z• Aa. A" As. Ae. A 7. Suppose. for example. that PI' Pz. Pa are the three lines which pass through the point AI' By hypothesis two of our points (apart from AI) lie on each of these lines.
Next, if p is any integer, and q is a positive integer, we define a 11/ q to be (a 1/ q )11. Thus we have defined a" for all rational numbers r. The laws of exponents, namely ar+s = ar . a", arB = (ary, where rand s are rational, are not hard to prove using the above definitions. It is also easy to show that ar is an increasing ° ° 29 Xl. Areas of regions bounded by curves y = xt y y -t----------tJ---y = c x q>i a. q
41 Thus the only case remaining is the one in which A, B, C, D are at the vertices of the trapezoid illustrated in fig. 37f. In this case A, B, C, E must form a congruent trapezoid, and it is easy to see that the five points A, B, C, D, E must lie at the vertices of a regular pentagon (fig. /5)/2. • Note that if we circumscribe a circle about the trapezoid ABCE (fig. 37f), then the points A, B, C, D will be four of the vertices of a regular pentagon inscribed in this circle. Toprovethis,putLDAB= LADB=oc.
Challenging mathematical problems with elementary solutions Volume I : Combinatorial Analysis and Probability Theory by Gordon, Basil; Âglom, Isaak Moiseevič; McCawley, James; Âglom, Akiva Moiseevič