It's All Nested

Geometry Level 4

In the above diagram we have nested Pentagons, Pentagrams, and Circles.

For each positive integer n n , let r n r_{n} be the radius of each circle C n C_{n} and x n x_{n} be a side of each Pentagon P n P_{n} .

Let R = ∑ n = 1 ∞ r n R = \sum_{n = 1}^{\infty} r_{n} and X = ∑ n = 1 ∞ x n X = \sum_{n = 1}^{\infty} x_{n} .

If R r 1 = X x 1 = a + b c \dfrac{R}{r_{1}} = \dfrac{X}{x_{1}} = \dfrac{a + \sqrt{b}}{c} , where a , b a,b and c c are coprime positive integers, then find a + b + c a + b + c .

If R r 1 > X x 1 \dfrac{R}{r_{1}} > \dfrac{X}{x_{1}} , then find the sum R r 1 + X x 1 \dfrac{R}{r_{1}} + \dfrac{X}{x_{1}} to six decimal places.

If R r 1 < X x 1 \dfrac{R}{r_{1}} < \dfrac{X}{x_{1}} , then find the difference X x 1 − R r 1 \dfrac{X}{x_{1}} - \dfrac{R}{r_{1}} to six decimal places.

If R r 1 = X x 1 \dfrac{R}{r_{1}} = \dfrac{X}{x_{1}} and there does not exist coprime positive integers a , b a,b and c c such that R r 1 = X x 1 = a + b c \dfrac{R}{r_{1}} = \dfrac{X}{x_{1}} = \dfrac{a + \sqrt{b}}{c} , then enter 0 0 .


The answer is 8.

Rocco Dalto by Rocco Dalto , 67, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rocco Dalto
Jan 30, 2019

For Circles:

m ∠ A O P = π 5 m\angle{AOP} = \dfrac{\pi}{5} and by Inscribed Angle Theorem m ∠ E P F = 1 2 m ∠ E O F ⟹ m ∠ A P O = π 10 m\angle{EPF} = \dfrac{1}{2}m\angle{EOF} \implies m\angle{APO} = \dfrac{\pi}{10}

A C O C = tan ⁡ ( π 5 ) ⟹ O C = A C tan ⁡ ( π 5 ) \dfrac{AC}{OC} = \tan(\dfrac{\pi}{5}) \implies OC = \dfrac{AC}{\tan(\dfrac{\pi}{5})} and P C = A C tan ⁡ ( π 10 ) = r 1 − O C = r − A C tan ⁡ ( π 5 ) ⟹ PC = \dfrac{AC}{\tan(\dfrac{\pi}{10})} = r_{1} - OC = r - \dfrac{AC}{\tan(\dfrac{\pi}{5})} \implies

h = A C = tan ⁡ ( π 10 ) tan ⁡ ( π 5 ) tan ⁡ ( π 5 ) + tan ⁡ ( π 10 ) r 1 h = AC = \dfrac{\tan(\dfrac{\pi}{10})\tan(\dfrac{\pi}{5})}{\tan(\dfrac{\pi}{5}) + \tan(\dfrac{\pi}{10})}r_{1}

tan ⁡ ( π 5 ) = tan ⁡ ( 2 π 10 ) = 2 tan ⁡ ( π 10 ) 1 − tan ⁡ 2 ( π 10 ) ⟹ \tan(\dfrac{\pi}{5}) = \tan(\dfrac{2\pi}{10}) = \dfrac{2\tan(\dfrac{\pi}{10})}{1 - \tan^2(\dfrac{\pi}{10})} \implies h = 2 tan ⁡ ( π 10 ) 3 − tan ⁡ 2 ( π 10 ) r 1 h = \dfrac{2\tan(\dfrac{\pi}{10})}{3 - \tan^2(\dfrac{\pi}{10})}r_{1}

Let β = tan ⁡ ( π 10 ) 3 − tan ⁡ 2 ( π 10 ) ⟹ h 1 = 2 β r 1 = r 2 sin ⁡ ( π 5 ) \beta = \dfrac{\tan(\dfrac{\pi}{10})}{3 - \tan^2(\dfrac{\pi}{10})} \implies h_{1} = 2\beta r_{1} = r_{2}\sin(\dfrac{\pi}{5}) ⟹ r 2 = 2 β sin ⁡ ( π 5 ) r 1 \implies \boxed{r_{2} = \dfrac{2\beta}{\sin(\dfrac{\pi}{5})}r_{1}}

In General r n = ( 2 β sin ⁡ ( π 5 ) ) n − 1 r 1 \boxed{r_{n} = (\dfrac{2\beta}{\sin(\dfrac{\pi}{5})})^{n - 1}r_{1}} and R = ∑ n = 1 ∞ r n = sin ⁡ ( π 5 ) sin ⁡ ( π 5 ) − 2 β r 1 R = \displaystyle\sum_{n = 1}^{\infty} r_{n} = \dfrac{\sin(\dfrac{\pi}{5})}{\sin(\dfrac{\pi}{5}) - 2\beta}r_{1} .

Let u = tan ⁡ ( x 2 ) ⟹ u 2 = 1 − cos ⁡ ( x ) 1 + cos ⁡ ( x ) = sin ⁡ 2 ( x ) ( 1 + cos ⁡ ( x ) ) 2 ⟹ u = sin ⁡ ( x ) 1 + cos ⁡ ( x ) u = \tan(\dfrac{x}{2}) \implies u^2 = \dfrac{1 - \cos(x)}{1 + \cos(x)} = \dfrac{\sin^2(x)}{(1 + \cos(x))^2} \implies u = \dfrac{\sin(x)}{1 + \cos(x)} and u 2 = 1 − cos ⁡ ( x ) 1 + cos ⁡ ( x ) ⟹ cos ⁡ ( x ) = 1 − u 2 1 + u 2 ⟹ sin ⁡ ( x ) = 2 u 1 + u 2 u^2 = \dfrac{1 - \cos(x)}{1 + \cos(x)} \implies \cos(x) = \dfrac{1 - u^2}{1 + u^2} \implies \sin(x) = \dfrac{2u}{1 + u^2} and u = tan ⁡ ( x 2 ) u = \tan(\dfrac{x}{2})

Let x = π 5 ⟹ u = tan ⁡ ( π 10 ) ⟹ β = u 3 − u 2 x = \dfrac{\pi}{5} \implies u = \tan(\dfrac{\pi}{10}) \implies \beta = \dfrac{u}{3 - u^2} and sin ⁡ ( π 5 ) = 2 u 1 + u 2 \sin(\dfrac{\pi}{5}) = \dfrac{2u}{1 + u^2}

⟹ R = 3 − u 2 2 ( 1 − u 2 ) r 1 \implies R = \dfrac{3 - u^2}{2(1 - u^2)}r_{1} and u = tan ⁡ ( π 10 ) = 1 5 25 − 10 5 ⟹ R = ( 1 2 ) ( 5 + 5 5 ) r 1 = ( 1 + 5 2 ) r 1 ⟹ R r 1 = 1 + 5 2 u = \tan(\dfrac{\pi}{10}) = \dfrac{1}{5}\sqrt{25 - 10\sqrt{5}} \implies R = (\dfrac{1}{2})(\dfrac{5 + \sqrt{5}}{\sqrt{5}})r_{1} = (\dfrac{1 + \sqrt{5}}{2})r_{1} \implies \dfrac{R}{r_{1}} = \boxed{\dfrac{1 + \sqrt{5}}{2}} .

For Hexagons:

Let x 1 x_{1} be the length of a side of the initial Hexagon.

x 1 2 = r 1 sin ⁡ ( π 5 ) ⟹ r 1 = x 1 2 sin ⁡ ( π 5 ) \dfrac{x_{1}}{2} = r_{1}\sin(\dfrac{\pi}{5}) \implies r_{1} = \dfrac{x_{1}}2\sin(\dfrac{\pi}{5})

From above we have h 1 = 2 β r 1 = β sin ⁡ ( π 5 ) x 1 h_{1} = 2\beta r_{1} = \dfrac{\beta}{\sin(\dfrac{\pi}{5})}x_{1} and x 2 = 2 h 1 = 2 β sin ⁡ ( π 5 ) x 1 ⟹ x 3 = 2 h 2 = 2 β sin ⁡ ( π 5 ) x 2 = ( 2 β sin ⁡ ( π 5 ) ) 2 x 1 x_{2} = 2h_{1} = \dfrac{2\beta}{\sin(\dfrac{\pi}{5})}x_{1} \implies x_{3} = 2h_{2} =\dfrac{2\beta}{\sin(\dfrac{\pi}{5})}x_{2} = (\dfrac{2\beta}{\sin(\dfrac{\pi}{5})})^2 x_{1}

In General x n = ( 2 β sin ⁡ ( π 5 ) ) n − 1 x 1 \boxed{x_{n} = (\dfrac{2\beta}{\sin(\dfrac{\pi}{5})})^{n - 1}x_{1}} and from above ⟹ X x 1 = R r 1 = 1 + 5 2 = a + b c ⟹ a + b + c = 8 \implies \boxed{\dfrac{X}{x_{1}} = \dfrac{R}{r_{1}} = \dfrac{1 + \sqrt{5}}{2}} = \dfrac{a + \sqrt{b}}{c} \implies a + b + c = \boxed{8} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...