This is spiralling out of control!
What is the value of the smallest positive integer N such that k = 1 ∑ N arctan ( k 1 ) > 2 π ?
The answer is 17.
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
Could you please explain how you proved that it ends at 1 7 ?
In my case I used Riemann sums to determine that the answer was either 15,16 or 17. Then, I just guessed the answer. :D
Log in to reply
Draw out another right triangle with sides 1 , 1 7 , 1 8 touching the last triangle, you will see that its side passes the very first triangle drawn, which means the total angle exceeds 3 6 0 ∘ or 2 π
Log in to reply
Yes, I had understood from the title of your question that if we continue drawing triangles this way we may get the answer. But, I'm asking if you had a proof in mind (because this answer uses only construction)?
Actually, even after understanding the title of your question I had not bothered to do the construction, fearing that it would take too long and assuming that there was an easy solution which I was missing. So, I used Riemann sums to narrow down the possibilities and guessed.
Brilliant presentation of your solution!
Spiral of Theodorus