Circle!

Geometry Level 3

Let C 0 C_0 be a unit circle. For n ≥ 1 n\geq 1 , let C n C_n be a circle whose area equals the area of a square inscribed in C n − 1 C_{n-1} , where n = 1 , 2 , 3 , … n=1,2,3,\ldots .

Compute ∑ i = 0 ∞ Area ( C i ) . \large \sum_{i=0}^{\infty}\text{Area}(C_i) \; .

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π 2 π − 2 \frac{{\pi}^2}{\pi-2} 3 4 \frac{3}{4} 1 π 2 \frac{1}{{\pi}^2} π 2 {\pi}^2

Akshay Sharma by Akshay Sharma , 21, India

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1 solution

Hassan Abdulla
Dec 27, 2017

the Area of the square inscribed in C n C_{ n } is 2 ( r n ) 2 2{ \left( r_{ n } \right) }^{ 2 }

so the area of C n + 1 = 2 ( r n ) 2 = π ( r n + 1 ) 2 C_{ n+1 }=2{ \left( r_{ n } \right) }^{ 2 }=\pi { \left( r_{ n+1 } \right) }^{ 2 }

⇒ r n + 1 = 2 π r n \Rightarrow r_{ n+1 }=\sqrt { \frac { 2 }{ \pi } } r_{ n }

⇒ r n = ( 2 π ) n n = 0 , 1 , 2 , 3 , . . . . . . \Rightarrow r_{ n }=\left( \sqrt { \frac { 2 }{ \pi } } \right) ^{ n }n=0,1,2,3,......

∑ A r e a ( C n ) = ∑ n = 0 ∞ π ( r n ) 2 = π ∑ n = 0 ∞ ( 2 π ) n \sum { Area(C_{ n }) } =\sum _{ n=0 }^{ \infty }{ \pi { \left( r_{ n } \right) }^{ 2 } } =\pi \sum _{ n=0 }^{ \infty }{ { \left( \frac { 2 }{ \pi } \right) }^{ n } }

= π 1 − 2 π = π 2 π − 2 =\frac { \pi }{ 1-\frac { 2 }{ \pi } } =\frac { { \pi }^{ 2 } }{ \pi -2 }

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