Thanks again!

Donna]]>

Sorry but I'm really not getting this. An octagon has 8 sides. A star of David has 6 points. So how to put these two together? What size is each?

A diagram would really help. Or a longer description such as (1) Draw a regular octagon ABCDEFGH. (2) Make a star of David by ???? please continue …

Bob

]]>If you enter a "star of David" in the very center of the octagon, and go from there to create the rest of the 195 (189) triangles to fit into the octagon (a regular octagon) , can you show me how to do this? ALL triangles!! No other shapes but triangles! I'm not "testing" you, rather I'm asking for your help. This is very important.

Thank you,

Donna]]>

Welcome to the forum.

If I'm understanding the problem correctly, then what difference does it make how big the octagon is? If you made an octagon of a certain size, counted the triangles, then shrunk it to half size, it would still have those many triangles.

I started with a regular octagon:

There are, if I've counted correctly, 65 points. I noticed that 195 = 65 x 3, so that's hopeful. If you picked a point (65 choices) then another different point (64 choices), then another (63 choices) then you could make a total of 65 x 64 x 63 triangles. Except you'd be counting every one several times (eg ABC, ACB, BAC, BCA, CAB, CBA) so divide by six. That's still way to many. You'd also be counting three points that actually don't make a triangle but rather a straight line. There's a lot of them.

But you didn't say it's a regular octagon. If not then that opens up more possibilities. I've arrowed a point where three diagonals cross at a single point. If you adjust the points at the edge of the octagon just slightly then that three way crossing becomes another triangle (see insert).

So how to proceed from here? When I get a problem that looks this hard I try simpler cases first to see if I can establish what to do.

Suppose the shape was just a single triangle. How many triangles? Easy. Just the one.

What about a quadrilateral and a pentagon?

I can easily count the triangles in the quadrilateral … 8, 4 small ones and another 4 that are made from two small triangles.

Is there a way to get this answer by reasoning it out? There are 5 points so if I try to make all the triangles I can; that's 5 choices for vertex 1, followed by 4 choices for vertex 2 followed by 3 for vertex 3, divide by the repeated triangle number explained above, that's 6 so I get (5x4x3)/6 = 10. Whoops, not 8. But there are two diagonals with 3 points on them. A 'triangle' made from those three points doesn't count as it's a straight line not a triangle. Two diagonals like that so subtract 2 … 10 -2 = 8 and I've got the right answer.

For the pentagon the number to subtract for 'straight line' triangles is much larger because each diagonal has four points, let's call them A,B,C and D. 'Triangles' ABC, ABD, ACD and BCD are all just straight lines so you'd need to subtract 4 for every diagonal, 5 so subtract 20 altogether.

General method. Draw the polygon. Adjust the points so that there are no 3 or more way crossings of lines. Count the number of points, N. Calculate Nx(N-1)x(N-2)/6. Count how many diagonals there are and how many ways of picking any 3 points on a diagonal to subtract the straight line cases.

At this point I'm going to pause and await your reply. I need to know if I'm interpreting the question correctly and whether we're meant to be considering a regular octagon.

Bob

]]>Sincerely, Donna]]>