User:Arthur Lenskold

My Mathematical Association of America ID is 59715. https://www.github.com/nskold 327-4 Evans St, Williamsville NY 14221 American Mathematical Monthly PROBLEMS 12216. Proposed by Zachary Franco, Houston, TX. A regular icosahedron with volume 1 is rotated about an axis connecting opposite vertices. What is the volume of the resulting solid? SOLUTION Arbitrarily choose one pair of opposite vertices, (P1, P12) , from the 6 pairs. These are the 2 poles. The resulting solid is a cylinder capped, from each base, with a right circular cone that extends to a pole. The volume of the solid, T, is a real number between 20 percent and 30 percent larger than the volume of the icosahedron. We propose that T = a3k , where a is the edge of the regular icosahedron of unit volume, and k = a constant. Let the point O denote the centroid of the icosahedron. Consider two circumradii, R and S. R is the circumradius of the icosahedron and is well-known. S is the circumradius of the regular pentagon that is inscribed in the circular base of a polar conical cap. We get 10 angles at the center, Q, of this base, and each angle measures � 5 radians. Since S is a leg of an isosceles triangle, we have sin � 5 = a 2 S = q 5􀀀 p 5 2 2 . Now we choose another pair of opposite vertices, (P6, P11) , where P11 and P12 are adjacent vertices in the southern hemisphere. 1 Triangle OP11P12 has equal legs R and has an isosceles triangle base a . S = P11Q , a line segment that is perpendicular to OQ. Let angle P12OP11 = 2�. Then sin 2� = S R, and sin � = a 2 R . Remark: The angle with measure 2� is interesting because it is paired with a vertical angle in the opposite hemisphere. So there are two diagonally opposite half-wedges of cheese, that, when matched with the other two half-wedges, form two stacked cylinders in the resulting solid. End of Remark. Because the sphere has equal diameters, both through the poles and across the equator, there is a disk, with area G, that is shared between the top of the cylinder and the bottom of the cone. The top half-cylinder has height (R-j ) , where j is the altitude of the cone. Since S = a 2 sin � 5 , we have: S2 = 8 a2 4 (5􀀀 p 5) , and G = �S2 = 8� a2 4 (5􀀀 p 5) . Each member of the pair of stacked cylinders has volume G(R-j ). Each member of the pair of cones has volume Gj 3 . The volume of the resulting solid is: T = 2(G(R-j ) + Gj 3 ) = 2( G(R - 2j 3 ) ), where j= p a2 􀀀 S2= r a2 􀀀 8 a2 4 (5􀀀 p 5) =a r 1 􀀀 8 1 4 (5􀀀 p 5) . Two results, from MathWorld.wolfram.com/RegularIcosahedron.html, are that: R = a 4 p 10 + 2 p 5 , and that the volume of the regular icosahedron, that is given as 1, is equal to: 5 12 (3 + p 5)a3 . Therefore, a3 = 1 5 12 (3+ p 5) , and we have 2 T = a3k = 1 5 12 (3+ p 5) 2(8� 1 4 5􀀀 p 5 ( 1 4 p 10 + 2 p 5 􀀀 2 3 q 1 􀀀 8 1 4 5􀀀 p 5 )) , and we are done.