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Shapes of Constant Width and Beyond Shapes of Constant Width and Beyond Sept. 18, 2019 — Math Club, CU Boulder Yuhao Hu University of Colorado Boulder A. 50 Pence A. 50 Pence A. 50 Pence A. 50 Pence B. Manhole Cover B. Manhole Cover C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) C. Ruler-Compass Construction (Starting with a regular (2n + 1)-gon . ) D. Ruler-Compass Construction (Removing corners . ) D. Ruler-Compass Construction (Removing corners . ) D. Ruler-Compass Construction (Removing corners . ) D. Ruler-Compass Construction (Removing corners . ) D. Ruler-Compass Construction (Removing corners . ) D. Ruler-Compass Construction (Removing corners . ) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) E. Ruler-Compass Construction (From any triangle with sides a > b > c ...) 2 F. Strictly Convex Regions in R Definition. Given a region K ⊂ R2, it is said to be convex if the line segment connecting any two points p; q 2 K remains entirely in K, and it is said to be strictly convex if, for any two points p; q 2 K the interior of the line segment connecting p; q lives entirely in the interior of K. 2 F. Strictly Convex Regions in R Definition. Given a region K ⊂ R2, it is said to be convex if the line segment connecting any two points p; q 2 K remains entirely in K, and it is said to be strictly convex if, for any two points p; q 2 K the interior of the line segment connecting p; q lives entirely in the interior of K. 2 F. Strictly Convex Regions in R Definition. Given a region K ⊂ R2, it is said to be convex if the line segment connecting any two points p; q 2 K remains entirely in K, and it is said to be strictly convex if, for any two points p; q 2 K the interior of the line segment connecting p; q lives entirely in the interior of K. 2 F. Strictly Convex Regions in R Definition. Given a region K ⊂ R2, it is said to be convex if the line segment connecting any two points p; q 2 K remains entirely in K, and it is said to be strictly convex if, for any two points p; q 2 K the interior of the line segment connecting p; q lives entirely in the interior of K. G. The Support Function p(θ) p(θ) = (x(θ); y(θ)) · (sin θ; − cos θ) = x(θ) sin θ − y(θ) cos θ: G. The Support Function p(θ) p(θ) = (x(θ); y(θ)) · (sin θ; − cos θ) = x(θ) sin θ − y(θ) cos θ: G. The Support Function p(θ) p(θ) = (x(θ); y(θ)) · (sin θ; − cos θ) = x(θ) sin θ − y(θ) cos θ: G. The Support Function p(θ) p(θ) = (x(θ); y(θ)) · (sin θ; − cos θ) = x(θ) sin θ − y(θ) cos θ: G. The Support Function p(θ) p(θ) = (x(θ); y(θ)) · (sin θ; − cos θ) = x(θ) sin θ − y(θ) cos θ: H. From p(θ) to (x(θ); y(θ)) Theorem. Given a (closed, bounded) strictly convex region K, the support function p(θ) is C1 (i.e. continuously differentiable). Moreover, we have x(θ) sin θ cos θ p(θ) = : y(θ) − cos θ sin θ p0(θ) Idea: Let x(θ) sin θ x(θ) := ; u := : y(θ) − cos θ We have p = uT x: If we can show that p is differentiable, then T p0 = uT x0 + u0 x: Since x0 is perpendicular to u, uT x0 = 0, and we obtain p uT = x: p0 u0T Now solve this linear system for x. H. From p(θ) to (x(θ); y(θ)) Theorem. Given a (closed, bounded) strictly convex region K, the support function p(θ) is C1 (i.e. continuously differentiable). Moreover, we have x(θ) sin θ cos θ p(θ) = : y(θ) − cos θ sin θ p0(θ) Idea: Let x(θ) sin θ x(θ) := ; u := : y(θ) − cos θ We have p = uT x: If we can show that p is differentiable, then T p0 = uT x0 + u0 x: Since x0 is perpendicular to u, uT x0 = 0, and we obtain p uT sin θ − cos θ = x = x: p0 u0T cos θ sin θ Now solve this linear system for x. H. From p(θ) to (x(θ); y(θ)) Theorem. Given a (closed, bounded) strictly convex region K, the support function p(θ) is C1 (i.e. continuously differentiable). Moreover, we have x(θ) sin θ cos θ p(θ) = : y(θ) − cos θ sin θ p0(θ) Idea: Let x(θ) sin θ x(θ) := ; u := : y(θ) − cos θ We have p = uT x: If we can show that p is differentiable, then T p0 = uT x0 + u0 x: Since x0 is perpendicular to u, uT x0 = 0, and we obtain p uT sin θ − cos θ = x = x: p0 u0T cos θ sin θ Now solve this linear system for x. H. From p(θ) to (x(θ); y(θ)) Theorem. Given a (closed, bounded) strictly convex region K, the support function p(θ) is C1 (i.e. continuously differentiable). Moreover, we have x(θ) sin θ cos θ p(θ) = : y(θ) − cos θ sin θ p0(θ) Idea: Let x(θ) sin θ x(θ) := ; u := : y(θ) − cos θ We have p = uT x: If we can show that p is differentiable, then T p0 = uT x0 + u0 x: Since x0 is perpendicular to u, uT x0 = 0, and we obtain p uT sin θ − cos θ = x = x: p0 u0T cos θ sin θ Now solve this linear system for x. H. From p(θ) to (x(θ); y(θ)) Theorem. Given a (closed, bounded) strictly convex region K, the support function p(θ) is C1 (i.e. continuously differentiable). Moreover, we have x(θ) sin θ cos θ p(θ) = : y(θ) − cos θ sin θ p0(θ) Idea: Let x(θ) sin θ x(θ) := ; u := : y(θ) − cos θ We have p = uT x: If we can show that p is differentiable, then T p0 = uT x0 + u0 x: Since x0 is perpendicular to u, uT x0 = 0, and we obtain p uT sin θ − cos θ = x = x: p0 u0T cos θ sin θ Now solve this linear system for x. H. From p(θ) to (x(θ); y(θ)) Theorem. Given a (closed, bounded) strictly convex region K, the support function p(θ) is C1 (i.e. continuously differentiable). Moreover, we have x(θ) sin θ cos θ p(θ) = : y(θ) − cos θ sin θ p0(θ) Idea: Let x(θ) sin θ x(θ) := ; u := : y(θ) − cos θ We have p = uT x: If we can show that p is differentiable, then T p0 = uT x0 + u0 x: Since x0 is perpendicular to u, uT x0 = 0, and we obtain p uT sin θ − cos θ = x = x: p0 u0T cos θ sin θ Now solve this linear system for x. I. The Width Function W (θ) p(θ) = x(θ) sin θ − y(θ) cos θ;W (θ) = p(θ) + p(θ + π): J. Using Fourier Series Idea: To obtain curves of constant width, first find a C1, 2π-periodic function p(θ) that satisfies p(θ) + p(θ + π) = D for some constant (diameter) D > 0, then use the formula x(θ) sin θ cos θ p(θ) = y(θ) − cos θ sin θ p0(θ) to construct a curve. Q: How to find a desired function p(θ)? Fourier Series. J. Using Fourier Series Idea: To obtain curves of constant width, first find a C1, 2π-periodic function p(θ) that satisfies p(θ) + p(θ + π) = D for some constant (diameter) D > 0, then use the formula x(θ) sin θ cos θ p(θ) = y(θ) − cos θ sin θ p0(θ) to construct a curve. Q: How to find a desired function p(θ)? Fourier Series. J. Using Fourier Series Idea: To obtain curves of constant width, first find a C1, 2π-periodic function p(θ) that satisfies p(θ) + p(θ + π) = D for some constant (diameter) D > 0, then use the formula x(θ) sin θ cos θ p(θ) = y(θ) − cos θ sin θ p0(θ) to construct a curve.
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