Circle Area Geometric Series

Here is a question I was asked recently (click for zoomed-in version).

Let's look at a case where the first circle has a radius of 1 1 . It is a scaled version of the problem under consideration.

The center of the big circle is at (1,1) (1,1) , the center of the little circle is at (a,a) (a,a) , and the point of contact is (112,112) \Big(1 - \frac{1}{\sqrt{2}},1 - \frac{1}{\sqrt{2}} \Big) . We know that the radius of the small circle is equal to the distance from the center of the small circle to the point of contact.

a2=(a1+12)2+(a1+12)2a^2 = \Big(a - 1 + \frac{1}{\sqrt{2}} \Big)^2 + \Big(a - 1 + \frac{1}{\sqrt{2}} \Big)^2

Solving for a a yields a=322 a = 3 - 2 \sqrt{2} . The ratio of the area of the small circle to the big circle is a2=17122 a^2 = 17 - 12 \sqrt{2}.

Now back to the problem at hand. The area of the circle of radius 10 10 is 100π 100 \pi. The remaining circles follow in an infinite geometric series with common ratio 1712217 - 12 \sqrt{2}. The total area is therefore:

A=100π1(17122)=100π12216=25π324 A = \frac{100 \pi}{1 - (17 - 12 \sqrt{2})} = \frac{100 \pi}{12 \sqrt{2} - 16} = \frac{25 \pi}{3 \sqrt{2} - 4}

Note by Steven Chase
3 months, 3 weeks ago

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@Steven Chase Thanks you so much sir.

A Former Brilliant Member - 3 months, 3 weeks ago

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@Steven Chase BTW how do you predicted that they will follow geometric progression?
Thanks in advance.

A Former Brilliant Member - 3 months, 3 weeks ago

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Because the small circle just becomes the new big circle, and the whole thing repeats again

Steven Chase - 3 months, 3 weeks ago

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Clever deduction. Nice!

Mahdi Raza - 3 months, 3 weeks ago

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