Dividing Regular ShapesDate: 10/22/96 at 19:56:54 From: Victoria Moss Subject: Dividing Regular Shapes Hi, My geometry class was given some problems and I was wondering if you could help me with some of them. I got most of them, but some problems I just don't understand: 1. Every vertex of a triangle is joined by straight lines to 6 points on the opposite side of the triangle. No three of the joining lines pass through the same point. Into how many regions do these 18 lines divide the interior of the triangle? To get this one, I looked for a pattern by first trying it with 1 line per vertex, then two, and so on, but I didn't find any pattern. I came up with 128 as my answer, but I don't think that is right. 2. If every vertex of a regular pentagon is connected to every other vertex, how many triangles are formed? I counted 40, but I really have no idea. I may have missed some somewhere. Date: 10/23/96 at 21:58:41 From: Doctor Charles Subject: Re: Dividing Regular Shapes (1) Draw the lines from each vertex in turn. The first six lines from the first vertex cut the triangle into seven regions. Now drawing one of the lines from the second vertex, note that we cut each of the first six lines once. Every time we cross a line we cut one region into two. In other words, we have created an extra region. This also happens when we reach the other side of the triangle. Each line we draw from the second vertex gives us 6+1 = 7 extra regions. Finally from the last vertex each of the 6 lines cuts the twelve other lines once and hits the far side. This gives 12 + 1 = 13 extra regions for each line. Adding these up : 7 + 6*7 + 6*13 = 127 regions In general for n lines from each vertex: number of regions = 1 + n + n*(n+1) + n*(2n+1) Can you see why? I have spaced the terms meaningfully so that they correspond with my argument. -Doctor Charles, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 01/25/97 at 13:48:04 From: Doctor Donald Subject: Re: Dividing Regular Shapes (2) There are 35 triangles. Count them this way: All triangles whose vertices are vertices of the pentagon. There are exactly 10 of these, one for every different set of three vertices. Next count the triangles using exactly two adjacent vertices of the pentagon, plus one vertex inside the figure. Each pair of adjacent vertices has three such triangles, so we get 15 more. Now count the triangles which use two NON-adjacent vertices of the pentagon plus an inside vertex. There is just one triangle for each such pair, and there are five pairs, so 5 more triangles. Now count the triangles using one pentagon vertex and two inside vertices. There is only one for each vertex, so 5 more. There are NO triangles using only inside vertices, so we are done, and it adds up to 35. -Doctor Donald, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/