Three black circles of unit radius have centers at points (-1,0) (0,0) and (1,0)?

Question

Infinitely many smaller red circles are tightly packed between black circles, as shown in the picture:
http://i18.tinypic.com/6xion4p.gif

(All circles which can tauch are tangent)
What is combined area of red circles?

(a) 1 / 12
(b) π / 12
(c) π (π+3) / 12
(d) π (π+3) (π-3) / 12

Answer 1

The answer is (d). Proof:
Let the radius of "nth" red circle is Rn /n=1,2,3,..., sorry, no subscripts in the font used by Yahoo Answers!/,
R1 > R2 > R3 > . . .
Applying Pythagorean Theorem for the triangles with vertexes (0,0), (1,0) and the center of "nth" red circle we obtain:
(1 + R1)² = 1² + (1 - R1)²;
(1 + R2)² = 1² + (1 - 2R1 - R2)²;
(1 + R3)² = 1² + (1 - 2R1 - 2R2 - R3)²;
(1 + R4)² = 1² + (1 - 2R1 - 2R2 - 2R3 - R4)², etc.
what leads consecutively to:
R1 = 1/4, R2 = 1/12, R3 = 1/24, R4 = 1/40, etc.,
the general relationship being:
(1 + Rn)² = 1 + (1/n - Rn)², or
1 + 2Rn + Rn² + 1 + 1/n² - (2/n)Rn + Rn², or
2(1 + 1/n)Rn = 1/n² and we obtain the first important result:

Rn = 1/(2n(n+1)) /n=1,2,3,.../

All this can be easily proven inductively.
The sum A of areas of all circles is therefore:

A = π ∑[n=1..∞] Rn² = (π/4)∑[n=1..∞] 1/(n²(n+1)²)

so we'll need to find the sum of the infinite series
∑[n=1..∞] 1/(n²(n+1)²), whose convergence is easily proven by comparison criterion, comparing it with the famous Euler's series
(E) ∑[n=1..∞] 1/n² = π²/6
/please follow the link below to see my answer to a related question some time ago/
We'll need also the well-known series
∑[n=1..∞] 1/(n(n+1)) = ∑[n=1..∞] ((n +1) - 1)/(n(n+1)) =
= ∑[n=1..∞] 1/n - 1/(n+1) =
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - . . . = 1
Now we are ready for the main thing:
∑[n=1..∞] 1/(n²(n+1)²) =
= ∑[n=1..∞] ((n+1) - n)/(n²(n+1)²) =
= ∑[n=1..∞] 1/(n²(n+1)) - ∑[n=1..∞] 1/(n(n+1)²) =
= ∑[n=1..∞] ((n+1) - n)/(n²(n+1)) -
- ∑[n=1..∞] ((n+1) - n)/(n(n+1)²) =
=∑[n=1..∞] 1/n² - ∑[n=1..∞] 1/(n(n+1)) -
- ∑[n=1..∞] 1/(n(n+1)) + ∑[n=1..∞] 1/(n+1)² =
= π²/6 - 1 - 1 + (π²/6 -1) = π²/3 - 3 = (π² - 9)/3
/the last series is (E) without 1st term:
∑[n=1..∞] 1/(n+1)² = ∑[n=2..∞] 1/n² = π²/6 -1/
So, finally:
A = (π/4)*(π² - 9)/3 = π(π + 3)(π - 3)/12

Very entertaining problem, thank You! I've seen a lot with the infinitely many circles, inscribed in an angle between 2 straight lines, their radiuses usually form a geometric progression in such cases, but this was much more interesting and challenging!

Answer 2

IF I understand properly, you're inquiring for the component on the circle x²+y²=121 it is 10 gadgets removed from {11,0} ok. The set of all factors that are 10 gadgets removed from {11,0} variety a circle whose equation is (x-11)²+y²=one hundred remedy the two quadratics concurrently (for advantageous y values, there's a 'replicate image' component 'under' the x axis) and you have your answer. Doug