This problem is so easy, @microsoft_office_word solved it looooool

Haha lmao ikr

Here's how to crack it: Let $ABC$ be a triangle with $\angle A = 90, AB=n-1, BC=n$. Let $D$ be a point on $AB$ such that $BD=1$ and $B$ is in between $A$ and $D$. The rest is obvious, we get $E_n=2$

XxProblemDestroyer1337xX wrote: Prove that $E_n=\frac{\arccos {\frac{n-1}{n}} } {\text{arccot}{\sqrt{2n-1} }}$ is a natural number for any natural number $n$. (A natural number is a positive integer) Solution. Put $\alpha_n=\text{arccot}\sqrt{2n-1}$, thus $\cot\alpha_n=\sqrt{2n-1}$. Noticing $\cos^2x=\displaystyle\frac{1}{1+\tan^2x}$, we have \begin{align*}\cos(2\alpha_n)=2\cos^2\alpha_n-1=\frac{2}{1+\tan^2\alpha_n}-1=\frac{2}{1+\frac{1}{\cot^2\alpha_n}}-1=\frac{n-1}{n}, \end{align*}thereby $2\alpha_n=\arccos\frac{n-1}{n}$, namely $\frac{\arccos\frac{n-1}{n}}{\text{arccot}\sqrt{2n-1}}=2$. The result follows. $\blacksquare$

ytChen wrote: thereby $2\alpha_n=\arccos\frac{n-1}{n}$ Why?

rmtf1111 wrote: ytChen wrote: thereby $2\alpha_n=\arccos\frac{n-1}{n}$ Why? $\cos x=r\iff x=\arccos r$ for $x\in\left[0,\frac{\pi}{2}\right]$.

This problem was proposed by me and published in " Revista de Matematica a Elevilor din Timisoara"- Mathematical Magazine of Students fromTimisoara.

I'm probably being dumb, but I a little confused by the use of $E_n$ in the original problem. Is this necessary? Or is it just stating that the fraction simplifies to a natural natural number for all natural numbers $n$? It was the "E" that just tripped me up.

naenaendr wrote: I'm probably being dumb, but I a little confused by the use of $E_n$ in the original problem. Is this necessary? Or is it just stating that the fraction simplifies to a natural natural number for all natural numbers $n$? It was the "E" that just tripped me up. https://artofproblemsolving.com/community/c6h1848892p12461318