Friday, April 2, 2010

Fox 271

6 comments:

  1. Consider a circle of radius r centred at (0,r) i.e. x^2+(y-r)^2=r^2. It will intersect with the parabola y=x^2 at y=2r-1 and if this is to be 0 then r=1/2 which means that the two will touch each other at (0,0). So one can infer that a ball of diameter=1 rolling iside the parabola will touch the latter at O. I'd, therefore, go for Option C.

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  2. Bob Ryden has submitted this yesterday:
    Let the radius of the circle be r, which puts center at (0, r). Equation is
    x^2 + (y - r)^2 = r^2
    x^2 + y^2 - 2ry = 0
    y = r + - sqrt (r^2 - x^2) by quad. formula. We are interested in the bottom half of the circle, so drop the plus sign before the square root. The circle must be above, or touching, the parabola for all values of x:
    r - sqrt (r^2 - x^2) >= x^2
    2r <= x^2 + 1
    For this to be true for all values of x,
    r <= 1/2
    or diameter < = 1
    -Bob Ryden

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  3. R=0;
    http://geometri-problemleri.blogspot.com/2010/04/problem-77-ve-cozumu.html

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  4. First, I, too, was thinking that point O is NOT reachable. But I am convinced that the answer is 1. I think you should think a little simpler than this. Newzad, do you see a problem in Joe's or Bob Ryden's solution?

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  5. sorry,
    R=1/2, which means diameter is 1,

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