Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.

Find all natural numbers $n$ the product of whose decimal digits is $n^2-10n-22$.

Let $a,b,c$ be real numbers with $a$ non-zero. It is known that the real numbers $x_1,x_2,\ldots,x_n$ satisfy the $n$ equations: \[ ax_1^2+bx_1+c = x_{2} \]\[ ax_2^2+bx_2 +c = x_3\]\[ \ldots \quad \ldots \quad \ldots \quad \ldots\]\[ ax_n^2+bx_n+c = x_1 \] Prove that the system has zero, one or more than one real solutions if $(b-1)^2-4ac$ is negative, equal to zero or positive respectively.

Click for solution Consider the function \[ f(x)=\left(\frac{2ax+b-1+\sqrt{\Delta}}{2ax+b-1-\sqrt{\Delta}}\right)^{\frac{1}{\sqrt{\Delta}}} \] where $\Delta=(b-1)^2-4ac$. If $x_i$ is an approximation to a root of this function, then $x_{i+1}$ is next approximation to the root of this function obtained by using the Newton-Raphson method. So a solution of the system of equations corresponds to a failure of the Newton-Raphson method to find a root of this function, producing instead a cycle of length $n$.

Prove that every tetrahedron has a vertex whose three edges have the right lengths to form a triangle.

Let $f$ be a real-valued function defined for all real numbers, such that for some $a>0$ we have \[ f(x+a)={1\over2}+\sqrt{f(x)-f(x)^2} \] for all $x$. Prove that $f$ is periodic, and give an example of such a non-constant $f$ for $a=1$.

Let $n$ be a natural number. Prove that \[ \left\lfloor \frac{n+2^0}{2^1} \right\rfloor + \left\lfloor \frac{n+2^1}{2^2} \right\rfloor +\cdots +\left\lfloor \frac{n+2^{n-1}}{2^n}\right\rfloor =n. \] RemarkFor any real number $x$, the number $\lfloor x \rfloor$ represents the largest integer smaller or equal with $x$.

Click for solution It immediately comes from base 2 representation. Indeed, suppose $n=\overline{a_{n}...a_2a_1a_0}$ (actually, we know that it containes at most $n-1$ digit for $n\geq 1$). And we have $\left[\frac{n+2^{k-1}}{2^{k}}\right]=\left[\frac{n}{2^{k}}+\frac{1}{2}\right]=\overline{a_{n-1}...a_{k}}+a_{k-1}$ . Therefore the whole sum is $\sum a_k(2^k-1)+\sum a_{k-1}= \sum a_k2^k=n$.