Recent content by melese

I agreed with you up to መለሰ

(BGR,1989) Prove that for any integer $n>1$ the equation $\displaystyle \frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}\cdots+\frac{x^2}{2!}+\frac{x^1}{1!}+1=0$ has no rational roots.

(HUN,1979) Prove the following statement: If a polynomial $f(x)$ with real coefficients takes only nonnegative values, then there exists a positive integer $n$ and polynomials $g_1(x),g_2(x),...,g_n(x)$ such that $f(x)=g_1(x)^2+g_2(x)^2+\cdots+g_n(x)^2$. A related question of my own, but I...

Hint: A known formula, involving a trigonometric function, for the area of triangles. solution:

I tried to find a formula depending only on $k$, namely, given a positive integer $k$, to find its location $(C_k,R_k)$ in the table. For brevity of reference, we say that $k=(C_k,R_k)$ (abuse of notation) Considering the obvious diagonals, it's clear that '$k$ is in the $n$'th diagonal' is...

3. Prove that for any triangle with sides $\displaystyle a,b,c$ and area $P$ the following inequality holds: $\displaystyle P\leq\frac{\sqrt3}{4}(abc)^{2/3}$ Find all triangles for which equality holds.

Re: IMO 1981,(ROU). shortlisted One should be carful of thinking that the answer is somehow $\displaystyle P(n+1)=\binom{n+1}{n+1}^{-1}$; since $P$ is of degree $n$ has already been determined at $n+1$ points ($k=0,1,...,n$). We try to build a polynomial identical to $P$. My idea was to find...

IMO 1981,(ROU). shortlisted This problem is probably not as difficult, but was misleading at first to me. 2.Let $P$ be a polynomial of degree $n$ satisfying $P(k)=\binom{n+1}{k}^{-1}$ for $k=0,1,...,n$. Determine $P(n+1)$. መለሰ

Re: ressurecting a previous post. Coprime mod $n$ implies coprime-ish mod $n$. Here's what I think, From $ap + bq+n \gamma =1$, it's evident that $(a,b)$ is coprime to $\gamma$. Then I look for solutions in the form $p'=p+nx$,$q'=q+ny$, so if we want $ap'+bq'=1$ we are led to $ax+by=\gamma$...

I was curious to see what the answer would be if the same question were for equilateral triangle. Looking at the triangle centered at the origin with $(0,1)$ as one of its vertices, I found for each side independently when a point $(x,y)$ is equidistant from it and the origin (Ackbach's method)...

a nice combinatorial proof To each yellow circle we can associate a pair of green circles in the way suggested by the drawing. This shows a bijection between (and hence the same number of) all the yellow circles and all the pairs of green circles. Here, there are $1+2+3+4+5+6+7+8$ yellow...

If $a\equiv b\pmod n$, then $(a,n)=(b,n)$. This can be seen by writing $a=b+nk$, for some integer $k$. Here, $(a^{n-1},n)=(1,n)=1$.

$\displaystyle\sum_{n=2}^{\infty }(1-\zeta(n))=\sum_{n=2}^{\infty }\sum_{k=2}^{\infty }\frac{1}{k^n}=\sum_{k=2}^{\infty }\sum_{n=2}^{\infty }\frac{1}{k^n}=\sum_{k=2}^{\infty }\frac{1}{k^2}(1+\frac{1}{k}+\frac{1}{k^2}+\cdots)=\sum_{k=2}^{\infty }\frac{1}{k^2}\cdot\frac{k}{k-1}=\sum_{k=2}^{\infty...

I think there's a simpler reason. When multiplying out, we find that $\displaystyle\prod_{0\leq k\leq n}(1+x^{2^k})$ is the sum of terms of the form $x^m$ (except for $1$), where $m$ is a sum of powers of two. Then we may notice that any $1\leq m\leq2^{n+1}-1$ is obtained exactly once because...

Here's what I tried: From $\displaystyle(2n+1-(2k-1))a=(2n+1-(2k-1))\frac{2\pi}{2n+1}=2\pi-(2k-1)\frac{2\pi}{2n+1}=2\pi-(2k-1)a$, it follows that $\sin{(2n+1-(2k-1))a}=\sin(-(2k-1)a)=-\sin{(2k-1)a}$. So we simply have, $\displaystyle\frac{\sin{(2n+1-(2k-1))a}}{{\sin(2k-1)a}}=-1$(*).The identity...