Golden ratio problems

Geometry Level 4

If half the inverse of the golden ratio equals sin ( π n ) \sin \left(\dfrac \pi n\right) , what is n n ?


The answer is 10.

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.

2 solutions

Tapas Mazumdar
May 11, 2017

The golden ratio is given by

ϕ = 1 + 5 2 \phi = \dfrac{1+\sqrt{5}}{2}

Our problem wants us to find n n such that

sin ( π n ) = 1 1 + 5 = 5 1 4 = 1 + 5 4 \sin \left( \dfrac{\pi}{n} \right) = \dfrac{1}{1 + \sqrt{5}} = \dfrac{\sqrt{5}-1}{4} = \dfrac{-1 + \sqrt{5}}{4}

The result is standard and is equal to the value of sin 1 8 = sin ( π 10 ) \sin 18^{\circ} = \sin \left( \dfrac{\pi}{10} \right) . Thus n = 10 \boxed{ n = 10} .

Are you a moderator or can you modify answers to have Latex script ?

Vijay Simha - 4 years, 1 month ago

Log in to reply

No, I'm not a moderator here. Anyone can type in Latex. If you want to learn the basics of Latex you can click here .

Tapas Mazumdar - 4 years, 1 month ago

Log in to reply

I am trying to clean up the presentation of this problem Convergence of Triangular number sums. Let us mop it up !

Vijay Simha - 4 years, 1 month ago

Log in to reply

@Vijay Simha I'll comment a proper write up here itself, you can copy-paste it there.

Tapas Mazumdar - 4 years, 1 month ago

@Vijay Simha

Original:

Suppose the sum of reciprocals of cubes of all the triangular numbers converges to a number A

1 + 1/3^3 + 1/6^3 + 1/10^3 + 1/15^3 + 1/21^3 + 1/28^3 + 1/36^3 + 1/45^3 + .... = A

and that the sum of reciprocals of squares of all the triangular numbers converges to B

1 + 1/3^2 + 1/6^2 + 1/10^2 + 1/15^2 + 1/21^2 + 1/28^2 + 1/36^2 + 1/45^2 + ....= B

and A/B can be expressed as M*(P - pi^2)/(pi^2 - O) where M, O and P are all integers.

What is the value of M + O + P ?


Latexed and edited:

If the sum of reciprocals of cubes of all the triangular numbers converges to the number A A ,

A = 1 + 1 3 3 + 1 6 3 + 1 10 3 + 1 15 3 + 1 21 3 + 1 28 3 + 1 36 3 + 1 45 3 + A = 1 + \dfrac{1}{3^3} + \dfrac{1}{6^3} + \dfrac{1}{{10}^3} + \dfrac{1}{{15}^3} + \dfrac{1}{{21}^3} + \dfrac{1}{{28}^3} + \dfrac{1}{{36}^3} + \dfrac{1}{{45}^3} + \cdots

and the sum of reciprocals of squares of all the triangular numbers converges to the number B B ,

B = 1 + 1 3 2 + 1 6 2 + 1 10 2 + 1 15 2 + 1 21 2 + 1 28 2 + 1 36 2 + 1 45 2 + B = 1 + \dfrac{1}{3^2} + \dfrac{1}{6^2} + \dfrac{1}{{10}^2} + \dfrac{1}{{15}^2} + \dfrac{1}{{21}^2} + \dfrac{1}{{28}^2} + \dfrac{1}{{36}^2} + \dfrac{1}{{45}^2} + \cdots

then A B \dfrac AB can be expressed as M ( P π 2 ) ( π 2 O ) M \dfrac{\left( P - \pi^2 \right)}{\left( \pi^2 - O \right)} where M , O M, O and P P are positive integers. Find M + O + P M+O+P .


Raw format for the Latexes:

If the sum of reciprocals of cubes of all the triangular numbers converges to the number *(A*),

*[A = 1 + \dfrac{1}{3^3} + \dfrac{1}{6^3} + \dfrac{1}{{10}^3} + \dfrac{1}{{15}^3} + \dfrac{1}{{21}^3} + \dfrac{1}{{28}^3} + \dfrac{1}{{36}^3} + \dfrac{1}{{45}^3} + \cdots *]

and the sum of reciprocals of squares of all the triangular numbers converges to the number *(B*),

*[B = 1 + \dfrac{1}{3^2} + \dfrac{1}{6^2} + \dfrac{1}{{10}^2} + \dfrac{1}{{15}^2} + \dfrac{1}{{21}^2} + \dfrac{1}{{28}^2} + \dfrac{1}{{36}^2} + \dfrac{1}{{45}^2} + \cdots *]

then *(\dfrac AB*) can be expressed as *(M \dfrac{\left( P - \pi^2 \right)}{\left( \pi^2 - O \right)}*) where *(M, O*) and *(P*) are positive integers. Find *(M+O+P*).

Note: Before copy-pasting the above Latex formatted write-up remove all the asterix symbols (*) and replace them with a backslash ( \ ).

Tapas Mazumdar - 4 years, 1 month ago

Log in to reply

@Tapas Mazumdar Thanks much

Vijay Simha - 4 years, 1 month ago

Same as Mr. Tapas Mazumdar , but slight differently.
n = π A r c S i n ( 1 2 2 1 + 5 ) = 10 n=\pi *{ArcSin(\frac 1 2 *\frac 2 { 1+\sqrt5} )}=10

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...