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Dividing Regular Shapes

Date: 10/22/96 at 19:56:54
From: Victoria Moss
Subject: Dividing Regular Shapes


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

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
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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!   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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