February 2005
Answer:
Let r[k] denote the length of the hypotenuse of triangle Tk, so that r[k-1] is the length of one leg of Tk, and r[0]=1 for starters. We have r[k-1]/r[k]=cos((
/2)/2^k). This recurrence relation, along with initial condition r[0]=1, is solved by r[k]=(2^k)sin((/2)/2^k). (Apply the formula for sin(2b)=2sin(b)cos(b) to b=((/2)/2^k).) Restate this as r[k]=(/2)(sin(c[k])/c[k]) where c[k]=(/2)/2^k, notice that c[k] tends to 0 as k increases, and use limit(sin(z)/z)=1 as z tends to 0, to deduce that r[k] tends to the limit p/2, which answers Part 1.
Notice that in the above expression, c[k] is also the angle between the x-axis and the hypotenuse of triangle Tk. So the point with polar coordinates (r,
)=(r[k],c[k]) is one of the vertices of triangle Tk. A curve passing through all such vertices is given by r=(/2)(sin()/). This answers Part 2.
Converting this to rectangular coordinates, we find y/x = tan [ y / 2(x^2+y^2) ] , to answer Part 3.
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