prelim1solns - Math 3110 Prelim 1 Solutions Problem 1 Upon...

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Unformatted text preview: Math 3110 Prelim 1 Solutions Problem 1 Upon rotation, the 2 n-sided polygon is a subset of the 2 n +1-sided polygon, and thus the 2 n +1-sided polygon has larger area than the 2 n-sided polygon. That is, { A n } is monotone increasing. Since the polygons are inscribed in the unit circle, they are subsets of the unit circle, and thus have smaller areas than the unit circle. That is, { A n } is bounded above by π , the area of the unit circle. As the number of sides of the polygon increases, the difference in area between the polygon and the unit circle shrinks to zero. Therefore, as n → ∞ , A n → π , the area of the unit circle. The area of the 2 n-sided polygon is obtained by dividing it into 2 n congruent triangular wedges. Two of the sides of each triangle are radiuses of the unit circle, and thus have length 1. The angle between these sides is 2 π 2 n . The area of each triangle is thus 1 2 sin( 2 π 2 n ). Since the area of the polygon is the sum of 2 n such triangles, the area of the polygon is A n = 2 n- 1 sin 2 π 2 n . We use a double-angle formula for sin to obtain A n = 2 n- 1 2 sin 2 π 2 n +1 cos 2 π 2 n +1 = A n +1 cos 2 π 2 n +1 . Since 2 n +1 ≥ 8, we have 0 < 2 π 2 n +1 ≤ π 4 . Therefore, 1 √ 2 ≤ cos 2 π 2 n +1 < 1, so A n ≤ A n +1 . That is, { A n } is monotone increasing. Since 2 n ≥ 4, we have 0 < 2 π 2 n ≤ π 2 . Using the fact that 0 < sin x < x for 0 < x ≤ π 2 , we see that A n = π 2 n 2 π sin 2 π 2 n < π....
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).

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prelim1solns - Math 3110 Prelim 1 Solutions Problem 1 Upon...

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