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Chapter 3
Solutions to Selected Chapter 3 Problems Problem 5 In Imamura’s Inki Sanka there are no solutions, so we provide a modern confirmation of the results. In any first-year geometry text it is shown that the area of a regular polygon is A = n 21 sa, where n is the number of sides, s is the length of the side, and a is the apothem, the perpendicular distance (altitude) from the center of the polygon to one of the sides. You can easily prove that s /2 a= , tan(180/n ) and so the area of the polygon is n A= s 2. 4 tan(180/n ) Plugging in n = 1, . . . , 9 gives the values for A near—but not quite the same as—those listed by Imamura. For part 2, the reason the square is omitted is that it is trivial, A = a 2. The hexagon is merely six times the area of an equilateral triangle. For part 3, the altitude of a regular tetrahedron of side s is h = ( 6 /3)s and the volume is (1/3) × (area base) × (altitude) = ( 2 /12)s 3 = 0.11785113s 3. The Inki Sanka’s numerical answer is of course approximate. Problem 6 (2) Considering, say, the lower triangle, then from figure 3.1 we see that 15 = 2r. From the half-angle formula 15 = (1 − cos 30)/(sin 30) and so 2 r = 2 − 3 = 0.26794. This value was written in the book. (3) In this case we have 2r = tan 27 = (1 − cos 54)/ sin 54 ≅ 0.50952. Again, this approximation was written in the book. (In chapter 4, problem 32, we show how to work out the trigonometric functions of such strange angles.) Problem 8 If rn , and Rn are the radii of the inscribed and circumscribed circles of an n- sided regular polygon of side 1, then it is easy to show that sin(180/n) = 1/(2Rn) and R n2 = 1/4 + r n2 . (1) For the triangle, we use the triple-angle formula sin 3θ = − 4 sin3 θ + 3 sin θ, where θ = 180/3. Plugging in sin θ = 1/2R 3 from above gives the 2 2 formula 3R 3 − 1 = 0, and with the Pythagorean theorem 12r 3 − 1 = 0.