## Thursday, December 8, 2005

### Connect the Dots. Topic: Space Geometry. Level: AMC/AIME.

Problem: (2004 AIME1 - #3) P is a convex polyhedron with $26$ vertices, $60$ edges and $36$ faces. $24$ of the faces are triangular and $12$ are quadrilaterals. A space diagonal is a line segment connecting two vertices which do not belong to the same face. How many space diagonals does P have? [Source: http://www.kalva.demon.co.uk/aime/aime04a.html]

Solution: What defines a space diagonal? The question tells us that "a space diagonal is a line segment connecting two vertices which do not belong to the same face." So we're basically choosing two points and connecting them, with several exceptions. Let's get the basics down first.

We have $26$ vertices. If we take any two at a time, there are $26C2 = 325$ ways of doing that.

But do we count anything we don't want? How about edges and the diagonals on the faces of the squares? We don't want those, but all of them are counted anyway. So we subtract away the number of edges and twice the number of squares ($2$ diagonals per square) to get $325 - 60 - 2(12) = 241$ space diagonals. QED.

--------------------

Comment: A relatively easy AIME problem, requiring only basic counting and knowledge of a little space geometry. The AIME usually has a few space geometry problems so it's good to brush up on your 3-dimensional problem solving.

--------------------

Practice Problem #1: Suppose in the above problem that the number of edges was not given. How would you solve the problem then?

Practice Problem #2: Suppose in the above problem that neither the number of edges nor the number of vertices were given. Is the problem still solvable?

1. For 1, just use Euler's Formula. For 2, I say it's still solvable, you just have to visualize it, which is what I did on the actual test. Unfortunately, I think I was off by one or something stupid like that when I tried counting them all up.

2. 1. V-E+F=2 and solve for E.

3. 2. It is solvable, actually, as the number of edges can be calculated by half of 4(# of quadrilaterals)+3(#triangles) = 1/2 (4*12+3*24) = 60 edges as given. Then it's just #1 again.