Too many circles in a triangle

Calculus Level 3

Suppose that circles of equal diameter are packed tightly in n n rows inside an equlateral triangle such that there are n circles in n n th row. If A A is the area of the triangle and A n A_n is the total area occupied by the circles in n n rows, find lim n A n / A \displaystyle\lim_{n\to\infty}{A_n}/{A} (the diagram shows an example of n = 4 n=4 rows).

π 3 \pi\sqrt{3} π / 3 \pi/ \sqrt{3} π / 6 \pi/6 π / ( 2 3 ) \pi/ ({2}\sqrt{3})

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1 solution

Chew-Seong Cheong
May 13, 2017

Let the radius of the circle be r r and the distance between the point where the end circle touches side of the triangle to the vertex of the triangle be a a . Then, a = r tan 3 0 = 3 r a = \dfrac r{\tan 30^\circ} = \sqrt 3 r . The side length the triangle with n n rows is given by b = ( n 1 ) ( 2 r ) + 2 a = 2 r ( n 1 + 3 ) b = (n-1)(2r) + 2a = 2r(n-1+\sqrt 3) . Then, we have:

lim n A n A = lim n 1 2 n ( n + 1 ) π r 2 1 2 3 2 b 2 = lim n 1 2 n ( n + 1 ) π r 2 3 4 4 r 2 ( n 1 + 3 ) 2 = π 2 3 \begin{aligned} \lim_{n \to \infty} \frac {A_n}A & = \lim_{n \to \infty} \frac {\frac 12 n(n+1) \pi r^2}{\frac 12 \cdot \frac {\sqrt 3}2b^2} \\ & = \lim_{n \to \infty} \frac {\frac 12 n(n+1) \pi r^2}{\frac {\sqrt 3}4 \cdot 4r^2(n-1+\sqrt 3)^2} \\ & = \boxed{\dfrac \pi {2\sqrt 3}} \end{aligned}

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