Nested Hexagrams and Circles

Geometry Level 4

For each positive integer $n$ , let $A_{p}(n)$ be the area of the hexagram inscribed in a circle of area $A_{c}(n)$ and $A_{c}(n + 1)$ be the area of the circle inscribed in the hexagram of area $A_{p}(n)$ . Let $A_{c} = \sum_{n = 1}^{\infty} A_{c}(n)$ and $A_{p} = \sum_{n = 1}^{\infty} A_{p}(n)$ .

If $\dfrac{A_{c} A_{p}}{A_{c}(1)^2} = \dfrac{1}{\pi}\left(\dfrac{a}{b}\right)^{b}\sqrt{a}$ , where $a$ and $b$ are coprime positive integers, find $a + b$ .

The answer is 5.

1 solution

Rocco Dalto
Jan 28, 2019

$\dfrac{2h_{1}}{r_{1}} = \tan(30^\circ) = \dfrac{1}{\sqrt{3}} \implies h_{1} = \dfrac{1}{2\sqrt{3}}r_{1}$ and $\dfrac{r_{1}}{2} = r_{2}\cos(30^\circ) = \dfrac{\sqrt{3}}{2}r_{2} \implies r_{2} = \dfrac{1}{\sqrt{3}}r_{1}$ .

$\implies h_{2} = \dfrac{1}{\sqrt{3}}r_{2} = \dfrac{1}{2\sqrt{3}}(\dfrac{1}{\sqrt{3}})r_{1}, \:\ r_{3} = \dfrac{1}{\sqrt{3}}r_{2} = (\dfrac{1}{\sqrt{3}})^2r_{1}, \:\ h_{3} = \dfrac{1}{2\sqrt{3}}r_{3} = \dfrac{1}{2\sqrt{3}}(\dfrac{1}{\sqrt{3}})^2r_{1}$

In General:

$r_{n} = (\dfrac{1}{\sqrt{3}})^{n - 1}r_{1}$ and $h_{n} = \dfrac{1}{2\sqrt{3}}(\dfrac{1}{\sqrt{3}})^{n - 1}r_{1}$

$\implies A_{c}(n) = (\dfrac{1}{3})^{n - 1}A_{c}(1)$ and $A_{p}(n) = \dfrac{\sqrt{3}}{\pi}(\dfrac{1}{3})^{n - 1}A_{c}(1)$

$\implies A_{c} = \dfrac{3}{2}A_{c}(1)$ and $A_{p} = \dfrac{3\sqrt{3}}{2\pi}A_{c}(1)$

$\implies \dfrac{A_{c} * A_{p}}{A_{c}(1)^2} = \dfrac{1}{\pi}(\dfrac{3}{2})^2\sqrt{3} = \dfrac{1}{\pi}(\dfrac{a}{b})^{b}\sqrt{a} \implies a + b = \boxed{5}$ .

Nested Hexagrams and Circles

The answer is 5.

1 solution

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