If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$, $BCD$, $CDA$, $DAB$ are the vertices of a rectangle.

Let $ a_1$, $ a_2$, ..., $ a_n$ and $ b_1$, $ b_2$, ..., $ b_n$ be $ 2 \cdot n$ real numbers. Prove that the following two statements are equivalent: i) For any $ n$ real numbers $ x_1$, $ x_2$, ..., $ x_n$ satisfying $ x_1 \leq x_2 \leq \ldots \leq x_ n$, we have $ \sum^{n}_{k = 1} a_k \cdot x_k \leq \sum^{n}_{k = 1} b_k \cdot x_k,$ ii) We have $ \sum^{s}_{k = 1} a_k \leq \sum^{s}_{k = 1} b_k$ for every $ s\in\left\{1,2,...,n-1\right\}$ and $ \sum^{n}_{k = 1} a_k = \sum^{n}_{k = 1} b_k$.

Click for solution The intention must be that (i) is true for all possible increasing n-tuples $ x$, otherwise the problem is false. Also, the inequality in (ii) must go the opposite way, as follows from the proof below. Here is the corrected version: [Moderator edit: I have corrected Orl's original post above according to this. $ -$ Darij] This problem is very interesting, because (ii) is the condition for $ b$ to majorize $ a$, but without the usual condition that $ a$, $ b$ are ordered. So it looks like a majorization condition for the class of increasing functions in place of convex ones. Therefore, there should be several more equivalent conditions such as (iii) Sequence $ b$ can be obtained from $ a$ by a series of transformations of the following type: given indices $ i < j$ and a positive real number $ p \geq 0$, change $ a_i \to a_i - p$ and $ a_j \to a_j + p$. Question : what is the analogue of the convex majorization criterion using double stochastic matrices (row and column sums = 1, Birkhoff polytope)? Anyway, the proof for the problem: Inequality (i) holds for all n-tuples $ x$ if and only if it holds for constant n-tuples $ (c,c,c,...,c)$ and all n-tuples of the form $ (0,0,...,1,1,1,...)$. This is because any increasing sequence $ x$ can be written as a sum (with non-negative coefficients) of such vectors. (i) for constant vectors is the condition $ \Sigma a_i = \Sigma b_i$. (i) for the vector with $ x_i = 0$ for $ i < k$ and $ x_i = 1$ for $ i \geq k$ is the condition $ \sum^{n}_{i = k} a_i \leq \sum^{n}_{i = k} b_i$. That's all.

Given a positive integer $A$ written in decimal expansion: $(a_{n},a_{n-1}, \ldots, a_{0})$ and let $f(A)$ denote $\sum^{n}_{k=0} 2^{n-k}\cdot a_k$. Define $A_1=f(A), A_2=f(A_1)$. Prove that: I. There exists positive integer $k$ for which $A_{k+1}=A_k$. II. Find such $A_k$ for $19^{86}.$

Given a triangle $ABC$ for which $C=90$ degrees, prove that given $n$ points inside it, we can name them $P_1, P_2 , \ldots , P_n$ in some way such that: $\sum^{n-1}_{k=1} \left( P_K P_{k+1} \right)^2 \leq AB^2$ (the sum is over the consecutive square of the segments from $1$ up to $n-1$). Edited by orl.

Given a square $ABCD$ whose side length is $1$, $P$ and $Q$ are points on the sides $AB$ and $AD$. If the perimeter of $APQ$ is $2$ find the angle $PCQ$.

Click for solution Let the excircle of triangle APQ opposite to the vertex A have center C' and touch the sideline AP of the triangle at the point B'. Then, $C^{\prime}B^{\prime}\perp AP$, and the distance AB' equals the semiperimeter of triangle APQ. Since the perimeter of triangle APQ equals 2, its semiperimeter must be 1; thus, we have AB' = 1. On the other hand, we know that AB = 1 (in fact, the square ABCD has sidelength 1). Since the points B and B' both lie on the ray AP, we thus can conclude that the points B and B' coincide. Hence, $C^{\prime}B^{\prime}\perp AP$ becomes $C^{\prime}B\perp AP$; in other words, $C^{\prime}B\perp AB$. Hence, the point C' lies on the perpendicular to the line AB at the point B. But since ABCD is a square, the perpendicular to the line AB at the point B is the line BC. Hence, the point C' lies on the line BC. Similarly, the point C' lies on the line DC. As the lines BC and DC have only one common point, namely the point C, it follows that the point C' coincides with the point C. Since we have defined the point C' as the center of the excircle of triangle APQ opposite to the vertex A, it follows that the point C is the center of the excircle of triangle APQ opposite to the vertex A. Thus, this point C lies on the external angle bisector of the angle APQ. Hence, $\measuredangle CPQ=90^{\circ}-\frac{\measuredangle APQ}{2}$. Similarly, $\measuredangle CQP=90^{\circ}-\frac{\measuredangle AQP}{2}$. Hence, by the sum of the angles in triangle PCQ, we have $\measuredangle PCQ=180^{\circ}-\measuredangle CPQ-\measuredangle CQP=180^{\circ}-\left(90^{\circ}-\frac{\measuredangle APQ}{2}\right)-\left(90^{\circ}-\frac{\measuredangle AQP}{2}\right)$ $=\frac{\measuredangle APQ+\measuredangle AQP}{2}$. But since the triangle PAQ is right-angled at A (this is because the quadrilateral ABCD is a square), we have < APQ + < AQP = 90°, and thus $\measuredangle PCQ=\frac{90^{\circ}}{2}=45^{\circ}$. darij

Given a tetrahedron $ABCD$, $E$, $F$, $G$, are on the respectively on the segments $AB$, $AC$ and $AD$. Prove that: i) area $EFG \leq$ max{area $ABC$,area $ABD$,area $ACD$,area $BCD$}. ii) The same as above replacing "area" for "perimeter".

Click for solution Fix $E,F$, and move $G$ on $AD$. If we regard the area/perimeter of $EFG$ as a function of $x=AG$, then it's easy to see that this function is convex, so it reaches its maximum in one of the endpoints of the segment $AD$. We can thus move $G$ to a vertex of the tetrahedron and increase the area/perimeter. We repeat the procedure until all three vertices $E,F,G$ are placed in some vertices of the tetrahedron, and the conclusion follows.

Let $x_i,$ $1 \leq i \leq n$ be real numbers with $n \geq 3.$ Let $p$ and $q$ be their symmetric sum of degree $1$ and $2$ respectively. Prove that: i) $p^2 \cdot \frac{n-1}{n}-2q \geq 0$ ii) $\left|x_i - \frac{p}{n}\right| \leq \sqrt{p^2 - \frac{2nq}{n-1}} \cdot \frac{n-1}{n}$ for every meaningful $i$.

Mark $4 \cdot k$ points in a circle and number them arbitrarily with numbers from $1$ to $4 \cdot k$. The chords cannot share common endpoints, also, the endpoints of these chords should be among the $4 \cdot k$ points. i. Prove that $2 \cdot k$ pairwisely non-intersecting chords can be drawn for each of whom its endpoints differ in at most $3 \cdot k - 1$. ii. Prove that the $3 \cdot k - 1$ cannot be improved.