Consider f(x)=a/(x-1)^2-1. For which a>1 is the triangle formed by (0,0) and the intersections of f(x) with the positive x- and y-axis isosceles?

We'll first compute these intersections by setting x=0 and y=0 consecutively. This gives y=a-1 and a/(x-1)^2-1=0. Hence we find (x-1)^2=a, so x=1+-sqrt(a). As we have a>1 and we want the intersection with the positive x-axis, we need x=1+sqrt(a). So our two intersections are (0,a-1) and (1+sqrt(a),0).

Now, as the angle at (0,0) is 90 degrees, we need the sides meeting at (0,0) to have the same length, as the third side will always be longer than the other two by the Pythagorean theorem. So, we need d((0,a-1),(0,0))=d((1+sqrt(a)),(0,0)). This gives a-1=1+sqrt(a). Now, as a>1, we can say a=b^2 for some b>0. This gives us b^2-1=1+b and hence b^2-b-2=0. This factorises to (b+1)(b-2)=0. Then, as b>0, we need b=2. This gives us a=4, the required a.