([b]4[/b]) Let $ a$, $ b$ be constants such that $ \lim_{x\rightarrow1}\frac {(\ln(2 \minus{} x))^2}{x^2 \plus{} ax \plus{} b} \equal{} 1$. Determine the pair $ (a,b)$.

Given two concentric circles and a pair of parallel lines. Find the locus of the fourth vertices of all rectangles with three vertices on the concentric circles, two vertices on one circle and the third on the other and with sides parallel to the given lines.

a) Each of the side of the convex hexagon is longer than $1$. Does it necessary have a diagonal longer than $2$? b) Each of the main diagonals of the convex hexagon is longer than $2$. Does it necessary have a side longer than $1$?

How many integers from $1$ to $1997$ have the sum of their digits divisible by $5$? (AI Galochkin)

Given positive integers $p$, $u$, and $v$ such that $u^2+2v^2=p$, determine, in terms of $u$ and $v$, integers $m$ and $n$ such that $3m^2-2mn+3n^2=24p$. (It is known that if $p$ is any prime number congruent to 1 or 3 modulo 8, then we can find integers $u$ and $v$ such that $u^2+2v^2=p$)

Consider the sequence $x_n>0$ defined with the following recurrence relation: \[x_1 = 0\] and for $n>1$ \[(n+1)^2x_{n+1}^2 + (2^n+4)(n+1)x_{n+1}+ 2^{n+1}+2^{2n-2} = 9n^2x_n^2+36nx_n+32.\] Show that if $n$ is a prime number larger or equal to $5$, then $x_n$ is an integer.

At a meeting of students of the 9th "G" class, it was decided to declare the 9th "G" a presidential republic. Four blocs nominated their candidates for the presidency: “Our Street”, “Our Yard”, “Our House” and “Our Entrance”. When discussing how to select a president four proposals were made. A. “What is there to think about! Have each student place a piece of paper with the name of the candidate they support in the box. Whoever gets the most votes is the president.” B. “No, you can’t do that. If no one gets more than half the votes, a repeat vote must be held, in which the top two from the first vote must participate.” C. “We must choose the one who is better than anyone else. How to do it? Let each person make a list: in the first place on his list he should put the best in his opinion, in the second place - the second, etc. If in most lists B is higher than A, then he is better than B. So, B is better than everyone if he is better than A, better than B and better than D.” D. “Let everyone really make their own list, as V said. For the first place on the list, the candidate receives three points, for the second - 2, for the third - 1 point, and for the fourth - 0. Whoever scores the most points will he’s the president.” As you can see, all four proposed methods are quite democratic. And yet, can it turn out that with method A one candidate wins, with method B another candidate wins, with C a third one, and in option D a fourth one? It is known that there are 29 people in the class, but the applicants (there are four of them) do not participate in the voting. Each student votes strictly according to his list (see speech B). After a heated discussion, option B was adopted. Interestingly, if elections had been held immediately, then after two rounds a representative of the Our Yard bloc would have become president. However, elections were scheduled for a week later. Students belonging to the “Our Yard” bloc, not knowing the true state of affairs, based on the principle “you can’t spoil porridge with butter,” launched a vigorous campaign in support of their candidate. As a result of this agitation, many students did not change their opinion. True, in some lists the position of the representative of “Our Yard” has improved. (All the changes boiled down to the fact that only this candidate’s position improved.) But as a result, another was elected president. How could this happen? [hide=might have typos, here is the original wording] На собрании учеников 9-го «Г» класса было принято решение — объявить 9-й «Г» президентской республикой. Своих кандидатов на пост президента выдвинули четыре блока: «Наша улица», «Наш двор», «Наш дом» и «Наш подъезд». При обсуждении способов выбора президента прозвучало четыре предложения. А. «Чего здесь думать! Пусть каждый ученик опустит в ящик бумажку с фамилией поддерживаемого им претендента. Кто наберет больше голосов, тот и президент.» Б. «Нет, так нельзя. Если никто не наберет больше половины голосов, надо устроить повторное голосование, в котором должны участвовать двое лучших по результатам первого голосования.» В. «Надо выбрать того, кто лучше любого другого. Как это сделать? Пусть каждый человек составит список: на первое место в своем списке он должен поставить самого лучшего по его мнению, на второе — второго и т.д. Если в большинстве списков В стоит выше А, то он лучше В. Значит, В лучше всех, если он лучше А, лучше Б и лучше Г.» Г. «Пусть и в самом деле каждый составит свой список, как сказал В. За первое место в списке кандидат получает три очка, за второе — 2, за третье — 1 очко, а за четвертое — 0. Кто наберет больше всех очков, тот и президент.» Как видим, все четыре предложенных способа вполне демократичны. И все же, может ли получиться так, что при способе А побеждает один кандидат, при способе Б — другой кандидат, при В — третий, ну, а в варианте Г — четвертый? Известно, что в классе 29 человек, но претенденты (их четверо) в голосовании не участвуют. Каждый ученик голосует строго в соответствии со своим списком (см. выступление В). После бурного обсуждения был принят вариант Б. Интересно, что если бы сразу же были проведены выборы, то после двух туров президентом стал бы представитель блока «Наш двор». Однако выборы были назначены на неделю позже. Ученики, входящие в блок «Наш двор», не зная ис тинного положения дел, исходя из принципа «кашу маслом не испортишь», развернули бурную агитацию в поддержку своего кандидата. В результате этой агитации многие ученики никак не изменили своего мнения. Правда, в некоторых списках улучшилось положение представителя «Наш двор». (Все изменения свелись к тому, что улучшилось положение только этого претендента.) Но в результате президентом был избран другой. Как такое могло случиться? [\hide]

In the adjoining plane figure, sides $AF$ and $CD$ are parallel, as are sides $AB$ and $EF$, and sides $BC$ and $ED$. Each side has length of 1. Also, $\measuredangle FAB = \measuredangle BCD = 60^\circ$. The area of the figure is [asy] size(200); defaultpen(linewidth(0.8)); pair A = dir(145), F = A + (0,-1), E = (0,-1), C = dir(35), D = C + (0,-1), B = origin; draw(A--B--C--D--E--F--cycle); label("$A$",A, dir(100)); label("$B$",B,2*N); label("$C$",C,dir(80)); label("$D$",D,dir(0)); label("$E$",E,S); label("$F$",F,W); label("$60^\circ$",A, 6*dir(295)); label("$60^\circ$",C, 6*dir(245)); [/asy] $\displaystyle \textbf{(A)} \ \frac{\sqrt 3}{2} \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ \frac{3}{2} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2$

Let $\mathcal{S}$ be the set of all positive integers which are both a multiple of $3$ and have at least one digit that is a $1$. For example, $123$ is in $\mathcal{S}$ and $450$ is not. The probability that a randomly chosen $3$-digit positive integer is in $\mathcal{S}$ can be written as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. [i]Proposed by GammaZero[/i]

Circles $S_1$ and $S_2$ meet at points $A$ and $B$. A line through $A$ is parallel to the line through the centers of $S_1$ and $S_2$ and meets $S_1$ and $S_2$ again $C$ and $D$ respectively. Circle $S_3$ having $CD$ as its diameter meets $S_1$ and $S_2$ again at $P$ and $Q$ respectively. Prove that lines $CP$, $DQ$, and $AB$ are concurent.

Let $x,y,z$ be real numbers such that, $x \geq y \geq z >0$. Prove that $$\frac{x^2-y^2}{z}+\frac{z^2-y^2}{x}+\frac{x^2-z^2}{y} \geq 3x-4y+z$$

Consider the parallelograms $ABCD$ and $AXYZ$, such that $X \in $[$BC$] and $D \in $[$YZ$]. Prove that the areas of the parallelograms are equal.

Let $P(z)$ be a polynomial with real coefficients whose roots are covered by a disk of radius R. Prove that for any real number $k$, the roots of the polynomial $nP(z)-kP'(z)$ can be covered by a disk of radius $R+|k|$, where $n$ is the degree of $P(z)$, and $P'(z)$ is the derivative of $P(z)$. can anyone help me? It would also be extremely helpful if anyone could tell me where they've seen this type of problems.............Has it appeared in any mathematics competitions? Or are there any similar questions for me to attempt? Thanks in advance!

Several students are competing in a series of three races. A student earns $5$ points for winning a race, $3$ points for finishing second and $1$ point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student? $\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$

Determine maximum real $ k$ such that there exist a set $ X$ and its subsets $ Y_{1}$, $ Y_{2}$, $ ...$, $ Y_{31}$ satisfying the following conditions: (1) for every two elements of $ X$ there is an index $ i$ such that $ Y_{i}$ contains neither of these elements; (2) if any non-negative numbers $ \alpha_{i}$ are assigned to the subsets $ Y_{i}$ and $ \alpha_{1}+\dots+\alpha_{31}=1$ then there is an element $ x\in X$ such that the sum of $ \alpha_{i}$ corresponding to all the subsets $ Y_{i}$ that contain $ x$ is at least $ k$.

Quadrilateral $APBQ$ is inscribed in circle $\omega$ with $\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$. Let $X$ be a variable point on segment $\overline{PQ}$. Line $AX$ meets $\omega$ again at $S$ (other than $A$). Point $T$ lies on arc $AQB$ of $\omega$ such that $\overline{XT}$ is perpendicular to $\overline{AX}$. Let $M$ denote the midpoint of chord $\overline{ST}$. As $X$ varies on segment $\overline{PQ}$, show that $M$ moves along a circle.