This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 3110 Prelim 1 Solutions Problem 1 Upon rotation, the 2 nsided polygon is a subset of the 2 n +1sided polygon, and thus the 2 n +1sided polygon has larger area than the 2 nsided 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 nsided 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 doubleangle 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....
View
Full
Document
This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).
 '08
 RAMAKRISHNA
 Polygons

Click to edit the document details