2015 Countdown Problem #18: A Vicious Circle

First, we draw a circle.

Then draw the vertical and horizontal lines.

We now draw the shifted lines where the dashed lines are their original places.

We need to know where they will eventually end.

Since $\lim_{m \to \infty} \sum_{n = 0}^{m} \frac{1}{x^n} = \frac{x}{x - 1}$

then $\lim_{m \to \infty} \sum_{n = 0}^{m} \frac{1}{3^n} = \frac{3}{2}$

The horizontal line will end $30$ cm above of the center while the vertical line will end $22.5$ cm to the right of the center.

We now label the sections of the previous figure.

We can see that (from the figure) the following relationships $E = I + F$ $H = I + G$ $I = (30 cm)(22.5 cm) = 675 {cm}^{2}$

The area of the first quadrant and third quadrant (of the original axes) is $\frac{\pi}{2} r^2 = A + I + F + G + C$

while the area of the second quadrant and fourth quadrant (of the original axes) is $\frac{\pi}{2} r^2 = B + E + D +H$

Now, the shaded areas(SA) and the unshaded areas(UA) are $SA = A + E + I + H + C$ $UA = B + G + F + D$

Substituting $E$ and $H$ in terms of $I$ for $SA$ while substituting $F$ and $G$ in terms of $I$ for $UA$ $SA = A + F + 3I + G + C$ $UA = B + H + E + D - 2I$

Now substituting these equations using the previous relationships with $\pi$ , then $SA = \frac{\pi}{2} r^2 + 2I$ $UA = \frac{\pi}{2} r^2 - 2I$

Now take the difference between the shaded and unshaded areas. $|SA - UA| = 4I$ $|SA - UA| = 4(675 {cm}^{2})$ $\boxed{|SA - UA| = 2700 {cm}^{2}}$