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# February 2015 - Circles Inside Circles (Solution)
The answer is 225. Let the largest circle have radius R_{1}. The next largest circles have radius R_{1}/2. The second smallest circle's radius must be found using Pythagoras' theorem. Make a triangle with the centres of these circles: The largest circle, one of the next largest circles, and the second smallest circle. Let the second smallest circle have radius R_{2}. The horizontal side length is R_{1}- R_{2}. The vertical side length is R_{1}/2. The hypotenuse has length R_{1}/2+R_{2}. Thus (R_{1}- R_{2})^{2}+R_{1}^{2}/4=(R_{1}/2+R_{2})^{2}. Expanding, R_{1}^{2}- 2R_{1}R_{2}+ R_{2}^{2}+ R_{1}^{2}/4=R_{1}^{2}/4+ R_{1}R_{2}+ R_{2}^{2}. Subtracting R_{1}^{2}/4 - 2R_{1}R_{2}+ R_{2}^{2}, we have R_{1}^{2}=3R_{1}R_{2}. Thus R_{1}=3R_{2} and R_{2}=R_{1}/3. Consider another right-angled triangle with the centres of these circles: The largest circle, one of the second largest circles, and the smallest circle. Let the smallest circle have radius R_{3}. The vertical side is the same. The horizontal side has length R_{1}-2R_{2}- R_{3}. The hypotenuse has length R_{1}/2+R_{3}. Solving in a similar manner to the previous step, R_{3}=R_{1}/15. Now, the smallest circle has area πR_{1}^{2}/225=1. Thus πR_{1}^{2}=225. However, the largest circle has area πR_{1}^{2}. Thus the largest circle has area 225.
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