# math

Let S be the locus of all points

(x, y) in the first quadrant such that

(x/t)+(y/(1− t))= 1 for some t with 0<t<1. Find the area of S.

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1. S is a collection of line segments with x- and y-intercepts at

(0,1-t) and (t,0)

Divide the interval (0,1) into n intervals of width 1/n. Then the two lines at t=k/n and t=(k+1)/n intersect at

y = (n-k)(n-k-1)/n^2

So, adding the (k+1)st line adds a thin triangle of base 1/n and height (n-k)(n-k-1)/n^2. It has area

(n-k)(n-k-1)/(2n^3)

Thus, the sum of the n triangles is

n-1
∑(n-k)(n-k-1)/(2n^3) = (1/6)(1 - 1/n^2)
k=0

as n->∞ the sum -> 1/6

So, the area of S is 1/6
*whew*
any other ideas?

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