hw7-solutions

3 three hollowed triangular prisms are used to make a

• Notes
• 5
• 100% (2) 2 out of 2 people found this document helpful

This preview shows pages 3–5. Sign up to view the full content.

3. Three hollowed, triangular prisms are used to make a torus. Carefully count the num- ber of vertices, faces, and edges for this prismatic torus. Compute the Euler characteristic for this torus. (B+S 5.3.29) Solution. This torus has nine vertices, six on the “outside” of the hole and three on the “inside”. It has nine faces, three arising from each of the three original triangular prisms. It has 18 edges – these can be decomposed into three triangles going around the torus the “long” way and three triangles going around the short way. So the Euler characteristic is V - E + F = 9 - 18 + 9 = 0. 4. Carefully count the number of vertices, faces, and edges for a two-holed torus. One way to view this two-holed torus is as two copies of the torus in problem 3, with one side removed from each and then the open edges glued together. This operation is called the connected sum . Compute the Euler characteristic for this two-holed torus. (B+S 5.3.30) Solution. Glue two copies of the torus together along a four-sided face. The number of vertices of the resulting torus is twice the number of vertices of the original torus, minus the number which disappear in the gluing operation. In the gluing operation four vertices disappear – four pairs of corresponding vertices on the two copies become single vertices. So there are 2 · 9 - 4 = 14 vertices. Similarly, four edges disappear in the gluing operation, so there are 2 · 18 - 4 = 32 edges. Finally, two faces are removed in the course of the operation, so there are 2 · 9 - 2 = 16 faces. The Euler characteristic is thus 14 - 32 + 16 = - 2, as it is for any two-holed torus. 5. Recall that in class we constructed the Koch snowflake, a shape which had finite area but infinite perimeter, by starting with an equilateral triangle, replacing each edge with four edges of one-third the length in a certain way, and repeating this process infinitely many times. (See p. 435 of the text for a picture of what happens to one edge when we do this.) We can construct a similar “square Koch snowflake” by beginning with a square and replacing each edge with five edges of one-third the length, such that the three middle edges form a square. (Note that the boundary of this snowflake actually touches itself at various points.) The first few iterations of this appear as follows. Assume that the original square has side length 1. 3

This preview has intentionally blurred sections. Sign up to view the full version.

(a) What is the length of the n th curve in this series? (Note that the n th curve in this series is made up of a large number of short segments of the same length; you should find the length of each short segment and the number of them.) Solution. The first curve consists of four segments; in each iteration after that there are five times as many curves, so there are a total of 4 · 5 n - 1 segments in the n th curve in the series. The segments in each curve are one-third the length of those in the previous one, so the length of each segment in the n th curve is (1 / 3) n - 1 . Thus the total length is 4 · (5 / 3) n - 1 .
This is the end of the preview. Sign up to access the rest of the document.
• Summer '09
• Lugo
• Math, Angles, triangle, vertices, Cantor, Truncation, Archimedean solid

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern