Given a real number $\lambda > 1$, let $P$ be a point on the arc $BAC$ of the circumcircle of $\bigtriangleup ABC$. Extend $BP$ and $CP$ to $U$ and $V$ respectively such that $BU = \lambda BA$, $CV = \lambda CA$. Then extend $UV$ to $Q$ such that $UQ = \lambda UV$. Find the locus of point $Q$.

Click for solution $\frac{BU}{AB}=\frac{CV}{AC}=\lambda$ and $\angle UBA=\angle ACV$ there for $\Delta ABU \cong \Delta ACV$ it means that $\Delta AUV \cong \Delta ABC$then $\frac{AV}{AC}=\frac{AU}{AB}=\frac{UV}{BC}$ we can get from this: $UV,UQ$ are constant now $AQ=\frac{\sin \angle B.QU}{\sin \angle QAY}$ and it's constant so the locus of Q is a circle by center A. Amir

There are $ n$ football teams in a round-robin competition where every 2 teams meet once. The winner of each match receives 3 points while the loser receives 0 points. In the case of a draw, both teams receive 1 point each. Let $ k$ be as follows: $ 2 \leq k \leq n - 1$. At least how many points must a certain team get in the competition so as to ensure that there are at most $ k - 1$ teams whose scores are not less than that particular team's score?

Prove that there exists $m \in \mathbb{N}$ such that there exists an integral sequence $\lbrace a_n \rbrace$ which satisfies: I. $a_0 = 1, a_1 = 337$; II. $(a_{n + 1} a_{n - 1} - a_n^2) + \frac{3}{4}(a_{n + 1} + a_{n - 1} - 2a_n) = m, \forall$ $n \geq 1$; III. $\frac{1}{6}(a_n + 1)(2a_n + 1)$ is a perfect square $\forall$ $n \geq 1$.

Find all real-coefficient polynomials $f(x)$ which satisfy the following conditions: i. $f(x) = a_0 x^{2n} + a_2 x^{2n - 2} + \cdots + a_{2n - 2} x^2 + a_{2n}, a_0 > 0$; ii. $\sum_{j=0}^n a_{2j} a_{2n - 2j} \leq \left( \begin{array}{c} 2n\\ n\end{array} \right) a_0 a_{2n}$; iii. All the roots of $f(x)$ are imaginary numbers with no real part.

Let $n$ be a natural number greater than 6. $X$ is a set such that $|X| = n$. $A_1, A_2, \ldots, A_m$ are distinct 5-element subsets of $X$. If $m > \frac{n(n - 1)(n - 2)(n - 3)(4n - 15)}{600}$, prove that there exists $A_{i_1}, A_{i_2}, \ldots, A_{i_6}$ $(1 \leq i_1 < i_2 < \cdots, i_6 \leq m)$, such that $\bigcup_{k = 1}^6 A_{i_k} = 6$.

Click for solution Ok here we go, Use the following double counting to stablish some lemmas. Call $m_k$ partial bound for the max number of subsets such that we cant find six $A_i$ with the desired property. First, count the number of sets of $6$ elements such that $5$ of them belong to one $A_i$, using this for $n=7$, we get that for $n=7$, $5{\binom{7}{6}}> 2m$, so we get $m_7=17$. Now counting the number of $7$ elements set, such that $5$ of them belong to an $A_i$, for $n=8$, we get $m_8=45$, and Finally for the other $n$, counting the number of $8$ element sets, $5$ of them in an $A_i$, we get a better bound that the desired one.

There are 1997 pieces of medicine. Three bottles $A, B, C$ can contain at most 1997, 97, 19 pieces of medicine respectively. At first, all 1997 pieces are placed in bottle $A$, and the three bottles are closed. Each piece of medicine can be split into 100 part. When a bottle is opened, all pieces of medicine in that bottle lose a part each. A man wishes to consume all the medicine. However, he can only open each of the bottles at most once each day, consume one piece of medicine, move some pieces between the bottles, and close them. At least how many parts will be lost by the time he finishes consuming all the medicine?