## Tuesday, January 19, 2010

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

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

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.

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

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

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