How Many Triangles in a Triangle?

Fig. 27 Checkered Triangles

          This diagram appeared in the, afore mentioned, puzzle book (see The Art Of Mathematical Origami).  The problem?  Prove that in this series of triangles, there appear as many triangles as the cube of the order.  In other words, if we examined triangle n, we would find that it contained n³ unique triangles within in it (eg. 2³ = 8, 3³ = 27 are easily verified).  I especially like this proof as it shows how unifying mathematics can reap benefits.  This particular proof uses Combinations, used in probability theory and binomial expansions, to generate a result in geometry.  Although not specifically shown, I assume the triangle that corresponds to n = 1, would look like this.

Fig. 28 Triangle of the First Order

          To prove the relation given, firstly, we recognize that any triangle created must contain either of the points A, or C, since all other points are either not connected or lie on the same line as each other.  Now, we tackle each case separately.

          For those triangles that contain A, there are n lines produced from C that are intersected by the n lines produced from A, and also inherently by C, giving a total of n + 1 intersection points.  Three points must be chosen from this selection of points in order to make up a triangle.  The number of arrangements possible is given by the number of lines (n) multiplied by the number of possibilities on each line.  Assuming A is already chosen, this leaves nⁿ⁺¹C₂.

          The situation is similar for those triangles containing C, however those triangles containing A have already been accounted for and thus there is one less point available to be chosen on each line (as A lies on all of these lines).  Using the same logic as that followed for A, we can see that the number of possibilities for triangles containing B, and not A, is nⁿC₂.  Since we have covered all possible triangles, the total number of possibilities is the sum of those expressions already derived.

        Total N^o ∆s = n(ⁿ⁺¹C₂ + ⁿC₂)

                            = n[(n + 1)!/2!(n – 1)! + n!/2!(n – 2)!]

                            = n[n(n + 1) + n(n – 1)]/2

                            = n(2n²/2)

                            = n³

          Isn’t that beautiful?