Triangles Rotating

Geometry Level 4

Δ 1 { \Delta }_{ 1 } is an equilateral triangle with unit length and we have a function f ( x ) f(x) which is used to create other triangles.

We create further triangles with the following rules

We take Δ n { \Delta }_{ n } 's side and produce it's sides clockwise till it's new length is f ( n ) f(n) times the original length,and the final end points are joined to make Δ n + 1 { \Delta }_{ n+1 }

For example, if f ( x ) = 2 f(x)=2 Here if the B l u e \color{#3D99F6}{Blue} triangle is Δ n { \Delta }_{ n } then the R e d \color{#D61F06}{Red} triangle is Δ n + 1 { \Delta }_{ n+1 } ,and the ratio of lengths B l u e + O r a n g e B l u e \frac { \color{#3D99F6}{Blue}+ \color{#EC7300}{Orange} }{ \color{#3D99F6}{Blue} } = = f ( n ) f(n) ,and the triangle rotates by 0.3334 0.3334 radians

If f ( n ) = 2 3 n 2 − 1 2 3 n 2 − 3 f(n)=\frac { 2\sqrt { 3 } { n }^{ 2 }-1 }{ 2\sqrt { 3 } { n }^{ 2 }-3 } , find the total rotation of Δ n { \Delta }_{ n } in radians when compared to Δ 1 { \Delta }_{ 1 } as n → ∞ n\rightarrow \infty .


The answer is 0.785398163397.

Shauryam Akhoury by Shauryam Akhoury , 18, India

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1 solution

Let x n denote the side length of the triangle Δ n In the above figure, applying sine rule for the GREEN triangle,we get \begin{aligned} &\text{Let } x_{n} \text{ denote the side length of the triangle } \Delta_{n}\\ &\text{In the above figure, applying sine rule for the } \color{#20A900}\text{GREEN}\color{#333333}\text{ triangle,we get} \end{aligned}

( f ( n ) − 1 ) x n sin ⁡ θ n = f ( n ) ⋅ x n sin ⁡ ( 60 − θ n ) = x n + 1 sin ⁡ ( 12 0 ∘ ) ⟹ f ( n ) f ( n ) − 1 = sin ⁡ ( 60 − θ n ) sin ⁡ θ n ( 1 ) We have, f ( n ) f ( n ) − 1 = 1 1 − 1 f ( n ) = 1 1 − 2 3 n 2 − 3 2 3 n 2 − 1 f ( n ) = 2 3 n 2 − 1 2 3 n 2 − 3 (Given) = 2 3 n 2 − 1 2 ( 2 ) Also, sin ⁡ ( 60 − θ n ) sin ⁡ θ n = sin ⁡ ( 60 − θ n ) ⋅ sin ⁡ ( 6 0 ∘ ) sin ⁡ θ n ⋅ sin ⁡ ( 6 0 ∘ ) = sin ⁡ ( 6 0 ∘ ) × ( cot ⁡ θ n − cot ⁡ ( 6 0 ∘ ) ) sin ⁡ ( A − B ) sin ⁡ A ⋅ sin ⁡ B = cot ⁡ B − cot ⁡ A = 3 2 cot ⁡ θ n − 1 2 ( 3 ) From ( 1 ) , ( 2 ) and ( 3 ) , we get 3 2 cot ⁡ θ n − 1 2 = 2 3 n 2 − 1 2 cot ⁡ θ n = 2 n 2 θ n = cot ⁡ − 1 ( 2 n 2 ) We are required to find the total angle which is given by, ∑ n = 1 ∞ θ n = ∑ n = 1 ∞ cot ⁡ − 1 ( 2 n 2 ) = ∑ n = 1 ∞ tan ⁡ − 1 ( 1 2 n 2 ) = ∑ n = 1 ∞ tan ⁡ − 1 ( 2 4 n 2 ) = ∑ n = 1 ∞ tan ⁡ − 1 ( 2 1 + ( 4 n 2 − 1 ) ) = ∑ n = 1 ∞ tan ⁡ − 1 ( ( 2 n + 1 ) − ( 2 n − 1 ) 1 + ( 2 n + 1 ) ⋅ ( 2 n − 1 ) ) = ∑ n = 1 ∞ tan ⁡ − 1 ( 2 n + 1 ) − tan ⁡ − 1 ( 2 n − 1 ) We can see that the above sum telescopes ⟹ ∑ n = 1 ∞ θ n = tan ⁡ − 1 ( ∞ ) − tan ⁡ − 1 ( 1 ) = π 2 − π 4 = π 4 ≈ 0.7853 \begin{aligned} \dfrac{(f(n)-1)x_{n}}{\sin\theta_{n}}&=\dfrac{f(n)\cdot x_{n}}{\sin(60-\theta_{n})}=\dfrac{x_{n+1}}{\sin(120^{\circ})}\\ \implies\dfrac{f(n)}{f(n)-1}&=\dfrac{\sin(60-\theta_{n})}{\sin\theta_{n}}\hspace{4mm}\color{#3D99F6}\small (1)\\\\ \text{We have,}\\ \dfrac{f(n)}{f(n)-1}&=\dfrac{1}{1-\dfrac{1}{f(n)}}\\ &=\dfrac{1}{1-\dfrac{2\sqrt3 n^2-3}{2\sqrt3 n^2-1}}\hspace{4mm}\color{#3D99F6}\small f(n)=\dfrac{2\sqrt3 n^2-1}{2\sqrt3 n^2-3}\hspace{2mm}\text{(Given)}\\ &=\dfrac{2\sqrt3 n^2-1}{2}\hspace{4mm}\color{#3D99F6}\small (2)\\\\ \text{Also,}\hspace{7mm}&\\ \dfrac{\sin(60-\theta_{n})}{\sin\theta_{n}}&=\dfrac{\sin(60-\theta_{n})\cdot \sin(60^{\circ})}{\sin\theta_{n} \cdot\sin(60^{\circ})}\\ &=\sin(60^{\circ})\times(\cot\theta_{n}-\cot(60^{\circ}))\hspace{6mm}\color{#3D99F6}\small\dfrac{\sin(A-B)}{\sin A \cdot\sin B}=\cot B-\cot A\\ &=\dfrac{\sqrt3}{2} \cot\theta_{n}-\dfrac12\hspace{4mm}\color{#3D99F6}\small (3)\\\\ \text{From } \color{#3D99F6}(1), (2) \color{#333333}& \text{ and }\color{#3D99F6} (3), \color{#333333}\text{ we get}\\ \dfrac{\sqrt3}{2} \cot\theta_{n}-\dfrac12&=\dfrac{2\sqrt3 n^2-1}{2}\\ \cot\theta_{n}&=2n^2\\ \theta_{n}&=\cot^{-1} (2n^2)\\\\ \text{We are required to } &\text{find the total angle which is given by,}\\ \sum_{n=1}^{\infty}\theta_{n}&=\sum_{n=1}^{\infty}\cot^{-1} (2n^2)\\ &=\sum_{n=1}^{\infty}\tan^{-1} \left(\dfrac{1}{2n^2}\right)\\ &=\sum_{n=1}^{\infty}\tan^{-1} \left(\dfrac{2}{4n^2}\right)\\ &=\sum_{n=1}^{\infty}\tan^{-1} \left(\dfrac{2}{1+(4n^2-1)}\right)\\ &=\sum_{n=1}^{\infty}\tan^{-1} \left(\dfrac{(2n+1)-(2n-1)}{1+(2n+1)\cdot (2n-1)}\right)\\ &=\sum_{n=1}^{\infty}\tan^{-1}(2n+1)-\tan^{-1}(2n-1)\\ \text{We can see that}&\text{ the above sum telescopes}\\ \implies\sum_{n=1}^{\infty}\theta_{n}&=\tan^{-1}(\infty)-\tan^{-1}(1)\\ &=\dfrac{\pi}{2}-\dfrac{\pi}{4}\\ &=\dfrac{\pi}{4}\approx\color{#EC7300}\boxed{\color{#333333}0.7853}\end{aligned}

2 Helpful 2 Interesting 3 Brilliant 0 Confused

The third angle is 60-theta instead of 120-theta.

A Former Brilliant Member - 2 years, 2 months ago

Thanks for noticing ,i will fix it!

Anirudh Sreekumar - 2 years, 2 months ago

Could someone point out my mistake i can't seem to fix this

Anirudh Sreekumar - 2 years, 2 months ago

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It might be i made the same mistake as you,i will edit this f(n) shortly

Shauryam Akhoury - 2 years, 2 months ago

Does f ( n ) = 2 3 n 2 − 1 2 3 n 2 − 3 f(n)=\frac { 2\sqrt { 3 } { n }^{ 2 }-1 }{ 2\sqrt { 3 } { n }^{ 2 }-3 } work?

Shauryam Akhoury - 2 years, 2 months ago

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yp, i think so

Anirudh Sreekumar - 2 years, 2 months ago

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@Anirudh Sreekumar – Thanks, i edited it,but funny thing,i got correct angels for 3 different f(n)'s even with my wrong result

Shauryam Akhoury - 2 years, 2 months ago

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