Tuesday, January 19, 2010

Fox 230

6 comments:

  1. Let the smaller and larger semicircles have radii of a & b resply. with the centre of the latter as the origin and let (p,q) be the point of tangency. Hence we've: p^2+q^2=b^2 ----(1) and px+qy=b^2-------(2). The tangent passes thru'[-(a+b),a]. So we have -(a+b)p+aq=b^2 -(3). Now the A(Square)=[p+(a+b)]^2+[q-a]^2=p^2+2(a+b)p+(a+b)^2+q^2-2aq+a^2. Use equations (1) and (3) to obtain A(Square)=b^2+(a+b)^2-2b^2=2(a+b)a=A(Rectangle).
    Ajit

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  2. A typo in the final line above: A(Square)=b^2+(a+b)^2+a^2-2b^2 giving A(Square)=2a^2+2ab=(a)*2(a+b)=A(Rectangle)
    Ajit

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  3. Very good! But there is a little simpler way for the proof - as simple as writing two Pythagoras'. Also, I couldn't find one but if there is a geometric way of showing the proof (by drawing simple lines), that would be wonderful.
    Thank you Ajit.

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  4. Yet another slightly easier way is to consider the vertex of the square on the smaller circle. Let's call this point E and let EC meet the larger circle in F1 and F2 resply. Now since ET is a tangent, ET^2=EF1*EF2=(a+b-√(b^2-a^2)*(a+b+√(b^2-a^2))=(a+b)^2-(b^2-a^2)=2a^2+2ab=A(Rect.)
    Ajit

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  5. both good Ajit. here is a soln with pythagoras only:
    O1 : center of circle 1 (radius: a)
    O2 : center of circle 2 (radius: b)
    E: point square touches the smaller circle
    x: side of square

    EO1O2 and ETO2 are right triangles sharing the same hypo:
    a^2 + (a+b)^2 = x^2 + b^2
    2a^2 + 2ab = x^2
    a(2a+2b) = x^2
    A(rectangle) = A(square)
    Very nice question!
    -binary

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  6. Good solutions! Thanks.
    Janos Pataki: do you see a geometric solution for this one?

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