A&S 153 4 the Icosahedron and Dodecahedron

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A&S 153 4 The Icosahedron and Do decahedron Here is a way to nd co ordinates for the vertices of an icosahedron. Lo ok at the diagram b elow. Begin with an o ctahedron. Let's assume the co ordinates of its vertices are 1; 0; 0, 0; 1; 0, 0; 0; 1. On each of its twelve edges we place a new p ointby dividing the edge into two unequal parts. These new p oints b ecome the twelvevertices of an icosahedron if they are placed appropriately. The problem is to determine the lo cation of the new p oints so that all of the resulting triangles pictured are equilateral. We can, for example, notice that the p oint P is closer to P than P , but the p oint P 61 6 1 36 is closer to P than P . Accordingly,we set P = P + tP P =0; 0; 1 + t1; 0; 1 = 3 6 61 6 1 6 t; 0;t 1 and P = P + tP P = 0; 1; 0 + t0; 1; 1 = 0; 1 t; t, and create 36 3 6 3 similar formulas for the ten other new p oints. Then by setting certain edge lengths of the icosahedron equal to each other, we can solvefort. Fortunately,oncewehave found the co ordinates for the icosahedron, those for the dodecahedron are easier: wecancho ose the centers of each of the triangles of the icosahedron to b e the vertices of the do decahedron. Let's break down the details a bit. 1. First, record the co ordinates of the six vertices of the o ctahedron. P 1; 0; 0 1 P 1; 0; 0 2 P 0; 1; 0 3 P 0; 1; 0 4 P 0; 0; 1 5 P 0; 0; 1 6 2. Now list the vertices of the icosahedron. There will b e one for each edge of the o ctahedron. P P + t[P P ]= 1; 0; 0 + t[0; 1; 0 1; 0; 0] = 1 t; t; 0 13 1 3 1 P P + t[P P ]= 1; 0; 0 + t[0; 1; 0 1; 0; 0] = 1 t; t; 0 14 1 4 1 P 23 P 24 P 35 P 36 P 45 P 46 P 51 P 52 P 61 P 62 3. Calculate the lengths of the following segments: P P : 35 36 P P : 35 13 4. Set these two lengths equal to each other and solvefor t. Simplify the expression for t so that there are no square ro ots in the denominator. 5. You nowknow the co ordinates for the vertices of the icosahedron. In order to display it, for example, with Maple, you need to know the twenty p olygons which are all triangles. List them b elow. P P P 13 35 51 P P P 13 35 36 6. To get the vertices of the do decahedron, for eachofthetwenty triangles ab oveyou need to know its center. You can get this by adding the co ordinates of its three corners, and dividing by3.You mightaswell continue using the symbol t even though bynow weknow its value. 1 1 1 1 Q P P P [1 t; t; 0 + 0; 1 t; t+ t; 0; 1 t] = ; ; 1 13 35 51 3 3 3 3 1t 2t 1 [1 t; t; 0 + 0; 1 t; t+ 0; 1 t; t] = ; ; 0 Q P P P 2 13 35 36 3 3 3 Q 3 Q 4 Q 5 Q 6 Q 7 Q 8 Q 9 Q 10 Q 11 Q 12 Q 13 Q 14 Q 15 Q 16 Q 17 Q 18 Q 19 Q 20 7. Using the ab ove list, nd eightvertices of the do decahedron that form the vertices of an inscrib ed cub e|a cub e that sits inside the do decahedron. 8. Finally, list the twelvepentagons for the do decahedron. Nowyou have enough infor- mation to display this Platonic solid as well..

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