# Solution to the triangle problem

**
Let h_i and l_i be the height and length of the i-th rectangle, respectively.
Since we are always adding a rectangle of area 1 to the previous rectangle,
h_i x l_i = i. We use this to write the equation (i+1) x h_i x l_i =
i x h_(i+1) x l_(i+1). Suppose that i is odd: this means that we are
adding the new rectangle to the right hand side, so h_i = h_(i+1). We
substitute these into the equation to get (i+1) x h_(i+1) x l_i =
i x h_i x l_(i+1). This yields the recursive relationship l_(i+1)/h_(i+1)
= i/(i+1) x l_i/h_i. We get the same relationship when i is even. Solving
this relationship involves the infinite product 2 x 2/3 x 4/3 x 4/5 x 6/5 x
6/7 x ...... This is Wallis' product, which can be shown to be Pi/2, which
is the limiting ration of length to height.
**