Olympiad #4
Two circles are drawn inside a square as shown in the picture. If the radius of the large circle is 3 r and the radius of the small circle is r , find the area of the square.
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1 solution
This is a solution given by the Olympiad can you think of better explanations??
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I did this the same way, but you could do essentially the same thing measuring horizontal projections of the lengths: if s is the side of the square, this gives s = 3 r + 2 3 r + 2 r + r = 2 r ( 1 + 2 )
and we get the same answer.
A different approach would be coordinate geometry; if we take the bottom left corner as an origin, the circles have equations ( x − 3 r ) 2 + ( y − 3 r ) 2 = 9 r 2 , ( x − s + r ) 2 + ( y − s + r ) 2 = r 2
and we want them to meet in exactly one point. Circle intersections involve solving a quadratic; having exactly one root means the discriminant of this quadratic will be zero.
This works, but seems more complicated than we need. Any other ideas?
Solution: Draw a diagonal passing through the centre of both circles. Label the top right corner of the square A, and the bottom left corner B, the centre of the smaller circle O1 and that of the larger circle O2. The diagonal passes through all these points. AO1 is √ 2r, BO2 is 3√ 2r, and O1O2 is 4r. So AB is 4(r + √ 2r), and so the area is 1/2 × AB2 = 8r 2 (3 + 2√ 2).