Question 637939: The possible triangulations T(n) of an n-gon, for n = 3, 4, and 5
are
T(3) = 1 (A triangle is its own triangulation)
T(4) = 2 (A convex quadrilateral can be triangulated diagonally on each of two diagonals)
T(5)= 5 ((A pentagon can be triangulated with two segments joining each vertex to its two opposite vertices)
Determine T(n) for n = 6, 7, and 8
Need Formula please! Thank you in advance.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! I'm not sure how it works or how it's derived entirely, but the formula for triangulating any n-gon (n is an integer and n >= 3) is
 = \frac{2\cdot6\cdot10\cdots(4n-10)}{(n-1)!} "x")
Source:
Pickover, Clifford A. "Euler's Polygon Division Problem." The Math Book: From
Pythagoras to the 57th Dimension, 250 Milestones in the History of
Mathematics. New York, NY: Sterling Pub., 2009. 184. Print.
More Source Info:
Formula found in page 184 of this book.
See this page (page 184) in google books here so you can see the formula for yourself.
In the case of a triangle, T(3) = (2)/(3-1)! = 2/2 = 1, which means that there is only one way to triangulate a triangle.
In the case of a quadrilateral, T(4) = (2*6)/(4-1)! = 12/6 = 2, which means that there are two ways to triangulate a quadrilateral.
In the case of a pentagon, T(5) = (2*6*10)/(5-1)! = 120/24 = 5, which means that there are 5 ways to triangulate a pentagon.
In the case of a hexagon, T(6) = (2*6*10*14)/(6-1)! = 1680/120 = 14, which means that there are 14 ways to triangulate a hexagon.
You continue this pattern to find the triangulation for any convex n-gon.
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