In a scalene triangle $ABC$ with $\angle A = 90^\circ,$ the tangent line at $A$ to its circumcircle meets line $BC$ at $M$ and the incircle touches $AC$ at $S$ and $AB$ at $R.$ The lines $RS$ and $BC$ intersect at $N,$ while the lines $AM$ and $SR$ intersect at $U.$ Prove that the triangle $UMN$ is isosceles.

Click for solution In a triangle $ ABC$, where $ I$ is the incenter, and $ D,E,F$ the tangenty points of the incircle with $ BC,CA,AB$, and $ L \in BI \cap EF$, we have that $ BL \perp CL$. (A proof of this you can found in the begging of my proof of this problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=132338) Now we just consider $ V,W$ as the intersection points of $ RS$ with the circumcircle of $ ABC$. Then $ \widehat{AV} = \widehat{BV}$ and $ \widehat{AW} = \widehat{CW}$, and $ \angle{MNU} = \frac {\widehat{CW} - \widehat{BV}}{2}$ , and $ \angle{MUN} = \angle{AUS} = \frac {\widehat{AW} - \widehat{AV}}{2}$, thus $ MUN$ is isosceles.

For n real numbers $a_{1},\, a_{2},\, \ldots\, , a_{n},$ let $d$ denote the difference between the greatest and smallest of them and $S = \sum_{i<j}\left |a_i-a_j \right|.$ Prove that \[(n-1)d\le S\le\frac{n^{2}}{4}d\] and find when each equality holds.

The numbers $1,\, 2,\, \ldots\, , n^{2}$ are written in the squares of an $n \times n$ board in some order. Initially there is a token on the square labelled with $n^{2}.$ In each step, the token can be moved to any adjacent square (by side). At the beginning, the token is moved to the square labelled with the number $1$ along a path with the minimum number of steps. Then it is moved to the square labelled with $2,$ then to square $3,$ etc, always taking the shortest path, until it returns to the initial square. If the total trip takes $N$ steps, find the smallest and greatest possible values of $N.$

Find all pairs $(a,\, b)$ of positive integers such that $2a-1$ and $2b+1$ are coprime and $a+b$ divides $4ab+1.$

The sides $AD$ and $CD$ of a tangent quadrilateral $ABCD$ touch the incircle $\varphi$ at $P$ and $Q,$ respectively. If $M$ is the midpoint of the chord $XY$ determined by $\varphi$ on the diagonal $BD,$ prove that $\angle AMP = \angle CMQ.$

Consider a regular $n$-gon with $n$ odd. Given two adjacent vertices $A_{1}$ and $A_{2},$ define the sequence $(A_{k})$ of vertices of the $n$-gon as follows: For $k\ge 3,\, A_{k}$ is the vertex lying on the perpendicular bisector of $A_{k-2}A_{k-1}.$ Find all $n$ for which each vertex of the $n$-gon occurs in this sequence.