Pentagonal Tiling Type 9

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Michael Tritle MATH 401 3D printing George Mason University 09/18/2019 Type 9 pentagons must satisfy the following criteria: If A, B, C, D and E are the interior angles and a, b, c, d, e are the lengths of edges of the pentagon, then 2A + C = 360°, D + 2E = 360°, and b = c = d = e I. FINDING THE INTERIOR ANGLES Since we are working with a pentagon, we know that the sum of the interior angles is 540°, so A + B + C + D + E = 540°. We would like to find values for A, B, C, D and E that simultaneously the system of equations: A + B + C + D + E = 540° 2A + C = 360° D + 2E = 360°. This system is equivalent to this system A - B + E = 180° 2B + C - 2E = 0° D + 2E = 360° This system has an infinite number of solutions. Choose a B and E such that 0° < B < 180° and 0° < E < 180°, then we get: A = 180° + B - E C = 2E - 2B D = 360° - 2E. II. FINDING THE MISSING EDGE For now, we will assume b = c = d = e = 1, This will simplify the calculation for finding a. We will generalize this later on. We now will make use a coordinate plane to represent our pentagon. Using vectors, we will construct our pentagon in such a way that the vector of length e will have a direction parallel to the x axis. **For later, It would be helpful to draw out a pentagon described above on a coordinate axis, and label its vertices and edges.** Since we are assuming that a vector of length e, v_1 is parallel to the x-axis, we can see that the angle v_1 makes with the x-axis is A. Let this angle be θ_1. To find all of the θ_n for 1 < n <= 5, think of a line parallel to the x-axis that intersects where each of the edges meet on the pentagon. We examine the alternate angle that forms with adjacent angle to θ_(n-1) and we can see that θ_(n) = nth interior angle - the adjacent angle to the θ_(n-1). So, more formally we get that: θ_(n) = nth interior angle - (180° - θ_(n-1)). So, θ_(1) = A θ_(2) = B - (180° - θ_ (1)). θ_(3) = C - (180° - θ_ (2)). θ_(4) = C - (180° - θ_ (3)). θ_(5) = C - (180° - θ_ (4)). To find a, we sum up all of the x-components of our v_n for 1 <= n <= 4, since the pentagon was constructed so that v_5 (that has length a) would be parallel to the x-axis, and thus would have a zero y-component. So, a = cos(θ_ (1)) + cos(θ_ (2)) + cos(θ_ (3)) + cos(θ_ (4)). III. FINDING A REPRESENTATION From earlier, we have that <0,0> = v_1 + v_2 + v_3 + v_4 + v_5 = <cos(θ_ (1)), sin(θ_ (1))> + <cos(θ_ (2)), sin(θ_ (2))> + <cos(θ_ (3)), sin(θ_ (3))> + <cos(θ_ (4)), sin(θ_ (4))> + a*<cos(θ_ (5)), sin(θ_ (5))>. To scale this, we can multiply both sides be some scalar k, giving us <0,0> = k<cos(θ_ (1)), sin(θ_ (1))> + k<cos(θ_ (2)), sin(θ_ (2))> + k<cos(θ_ (3)), sin(θ_ (3))> + k<cos(θ_ (4)), sin(θ_ (4))> + k*a*<cos(θ_ (5)), sin(θ_ (5))>. To finally generate all of the points P_ n on the plane the represent the pentagon, we will sum the v_ n's up for all natural numbers less than n and view them as ordered pairs. By default we have, p_ 1 = (0,0). p_ 2 = (k*cos(θ_ (1)), k*sin(θ_ (1))) p_ 3 = (k*cos(θ_ (2)), k*sin(θ_ (2))) + p_ 2 p_ 4 = (k*cos(θ_ (3)), k*sin(θ_ (3))) + p_ 3 p_ 5 = (k*cos(θ_ (4)), k*sin(θ_ (4))) + p_ 4