Day 1
Suppose that $ a_1$, $ a_2$, ..., $ a_n$ are positive real numbers such that $ a_1 \leq a_2 \leq \dots \leq a_n$. Let \[ {{a_1 + a_2 + \dots + a_n} \over n} = m; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{a_1^2 + a_2^2 + \dots + a_n^2} \over n} = 1. \] Suppose that, for some $ i$, we know $ a_i \leq m$. Prove that: \[ n - i \geq n \left(m - a_i\right)^2 \]
Assume $ABC$ is an isosceles triangle that $AB=AC$ Suppose $P$ is a point on extension of side $BC$. $X$ and $Y$ are points on $AB$ and $AC$ that: \[PX || AC \ , \ PY ||AB \] Also $T$ is midpoint of arc $BC$. Prove that $PT \perp XY$
Click for solution There is also a different solution: Problem (slightly extended). Let ABC be an isosceles triangle such that AB = AC. Let P be a point on the line BC, and let X and Y be two points on the lines AB and AC such that PX || AC and PY || AB. Finally, let T be the midpoint of the arc BC on the circumcircle of triangle ABC. Prove that $PT\perp XY$. Solution. Since PX || AC, Thales yields BX : XA = BP : PC. Since PY || AB, Thales yields AY : YC = BP : PC. Thus, BX : XA = AY : YC. Let O be the circumcenter of triangle ABC. Then, since the triangle ABC is isosceles with AB = AC, the triangles AOB and AOC are congruent. But since the triangle AOC is isosceles (in fact, OA = OC, since the point O is the circumcenter of triangle ABC), the triangle AOC is congruent to the triangle COA (please do look at the order of the vertices), and thus, the triangles AOB and COA are congruent. Since the points X and Y lie on the respective sides BA and AC of these two triangles and divide they in the same ratio BX : XA = AY : YC, they are corresponding points of these two triangles; hence, since corresponding points of congruent triangles have equal distances, we have OX = OY. Thus, the point O lies on the perpendicular bisector of the segment XY. In other words, if R is the midpoint of the segment XY, then $RO\perp XY$. We have PX || AC and PY || AB; in other words, PX || AY and PY || AX. Thus, the quadrilateral AXPY is a parallelogram. Since the diagonals of a parallelogram bisect each other, therefore, the segments AP and XY must bisect each other. In other words, the midpoint R of the segment XY is also the midpoint of the segment AP. Since the triangle ABC is isosceles with AB = AC, symmetry observations show that its apex A, its circumcenter O and the midpoint T of the arc BC on the circumcircle of the triangle ABC all lie on the perpendicular bisector of the segment BC. Hence, the segment AT, being a chord of the circumcircle of triangle ABC and passing through the center O of this circumcircle, is a diameter of this circumcircle. Hence, the center O of the circumcircle is the midpoint of the segment AT. On the other hand, we know that the point R is the midpoint of the segment AP. Hence, the line RO is a centers line in triangle APT, and thus it is parallel to the side PT of this triangle. Thus, $RO\perp XY$ yields $PT\perp XY$. Problem solved. By the way, this problem is equivalent to Argentina 5 / geo problem 2 from the op02 project. Darij
Suppose there are 18 lighthouses on the Persian Gulf. Each of the lighthouses lightens an angle with size 20 degrees. Prove that we can choose the directions of the lighthouses such that whole of the blue Persian (always Persian) Gulf is lightened.
Day 2
Find all $f : N \longmapsto N$ that there exist $k \in N$ and a prime $p$ that: $\forall n \geq k \ f(n+p)=f(n)$ and also if $m \mid n$ then $f(m+1) \mid f(n)+1$
Click for solution Suppose $n \geq k$ and $p$ doesn't divide $n-1$. You see there is $k$ that $n-1\ |\ n+kp$. So $f(n)\ |\ f(n+kp)+1$ but we know that $f(n)=f(n+kp)$, Therefore $f(n)\ |\ 1$ and $f(n)=1$. Consider an arbitary $n \neq 1$, We know that $n-1\ |\ (n-1)kp$, So $f(n)\ |\ f\big((n-1)kp\big)+1=2$. So for every $n\neq 1,\ f(n) \in \{1,2\}$. Now we have 2 cases: a ) $f(n)=2, \forall n\geq k$ and $p\ |\ n-1$, Consider $n$ that $p$ doesn't divide $n-1$. There is $m$ that $n-1|m$ and $p|m-1$. So $f(n)\ |\ f(m)+1=3$ and $f(n)=1$. So $f(n)=1$ for all $n \geq k$ or $m$ doesn't divide $n$; And is arbitarily defined for every $n<k$ and $p\ |\ n-1$ b ) $f(n)=1, \forall n\geq k$ and $p\ |\ n-1$, In this case $f(n)=1, \forall n \geq k$, and if we suppose: $S=\{a\ |\ f(a)=2 \}$ then it doesn't exist $m,n \in S $ that $m-1\ |\ n$. So all of functions with the problem properties are defined like this: Suppose $S$ is a finite subset of natural numbers that there doesn't exist $m,n\in S$ that $m-1\ |\ n$ and for $n>1,\ f(n)=2$ iff $n\in S$. $f(1)$ is arbitarily defined with this condition that $f(2)\ |\ f(1)+1$
Suppose there are $n$ distinct points on plane. There is circle with radius $r$ and center $O$ on the plane. At least one of the points are in the circle. We do the following instructions. At each step we move $O$ to the baricenter of the point in the circle. Prove that location of $O$ is constant after some steps.
Click for solution assume that the sequence of circles is $C_i$ then we want to say that $C_i$ is constant eventually,so suppose that $A_{1,C_i},\dots,A_{k_i,C_i}$ are in $C_i$ then define such a function $F(C_i)=\sum_{j=1}^{i_k} (R^2-|O_iA_{j,C_i}|^2)$ then this function if $O_i\neq O_{i+1}$ increase .(this part is cmputational) defining such afunction is not that hard i think that if you define $F(C_i)=\sum (|O_iA_{j,C_i}|^2)+(n-i_k)R^2$ this decrease so it will stop.any way no one could'nt solve it in exam.
Suppose $S= \{1,2,\dots,n\}$ and $n \geq 3$. There is $f:S^k \longmapsto S$ that if $a,b \in S^k$ and $a$ and $b$ differ in all of elements then $f(a) \neq f(b)$. Prove that $f$ is a function of one of its elements.
Click for solution I have another solution. May be it is wrong, or it is similar to Myth's solution, I don't know. We will prove it by induction on $ k $. The case $ k = 1 $ is trivial. Suppose now that $ k \geqslant 2 $. Suppose that we have a function $ f : S^{k} \rightarrow S $, satisfying the conditions of the problem. Then there are two cases: 1) There exist $ a_{2},a_{3}, \ldots , a_{k} \in S $, such that the function $ \varphi : S \rightarrow S $, given as $ \varphi (a):= f(a,a_{2},a_{3}, \ldots ,a_{k}) $, is one-to-one, and hence bijective. Take now some $ b_{2},b_{3}, \ldots ,b_{k} \in S $, such that $ b_{2} \neq a_{2}, \ b_{3} \neq a_{3}, \ \ldots , \ b_{k} \neq a_{k} $. Then if we take some $ b \in S $, then for every $ a \in S, a \neq b $ we have that the sequences $ (a,a_{2}, \ldots , a_{k}) $ and $ (b,b_{2}, \ldots , b_{k}) $ differ in all of elements, therefore $ x:= f(b,b_{2},b_{3}, \ldots ,b_{k}) \neq f(a,a_{2}, \ldots , a_{k}) = \varphi (a) $, so $ x \neq \varphi (a) $ for every $ a \neq b $, therefore $ x = \varphi (b) $. Therefore $ (b_{2},\ldots ,b_{k}) $ satisfies also that $ f(b,b_{2}, \ldots , b_{k}) = \varphi (b) $. Doing this step one more time, we can reach every sequence $ c_{2}, c_{3}, \ldots , c_{k} $ ( I mean , for arbitrary sequence $ c_{2}, \ldots , c_{k} \in S $, we can find , because of $ S $ has at least $ 3 $ elements, a sequence $ b_{2}, \ldots , b_{k} $, such that $ a_{i} \neq b_{i} \neq c_{i} $ for every $ 2 \leqslant i \leqslant k $). So we get that for every $ (c_{1},c_{2}, \ldots, c_{k}) \in S^{k} $, that $ f(c_{1},c_{2}, \ldots , c_{k}) = \varphi(c_{1}) $. 2) For every $ a_{2},a_{3}, \ldots , a_{k} \in S $, there exist distinct $ \alpha(a_{2},a_{3}, \ldots , a_{k}) $, $ \beta(a_{2},a_{3}, \ldots, a_{k}) \in S $, such that $ f(\alpha,a_{2},a_{3}, \ldots ,a_{k}) = f(\beta,a_{2},a_{3}, \ldots ,a_{k}) $. Then for every such $ a_{2},a_{3}, \ldots , a_{k} \in S $ make an arbitrary choice of such $ \alpha , \beta $, and define $ g: S^{k-1} \rightarrow S $ by $ g(a_{2},a_{3}, \ldots , a_{k}) := f(\alpha,a_{2},a_{3}, \ldots ,a_{k}) = f(\beta,a_{2},a_{3}, \ldots ,a_{k}) $. Then I claim that $ g: S^{k-1} \rightarrow S $ satisfies the conditions of the problem. Indeed, if $ (a_{2},a_{3}, \ldots , a_{k}) , (b_{2},b_{3}, \ldots , b_{k}) \in S^{k-1} $, such that they differ in all of elements, then if $ a',a'' = \alpha,\beta(a_{2}, \ldots, a_{k}) $ and $ b',b''= \alpha,\beta(b_{2}, \ldots, b_{k}) $, then or $ a' \neq b' $ , or $ a' \neq b'' $. Without loss of generality, assume that $ a' \neq b' $. Then $ (a',a_{2},\ldots, a_{k}), (b',b_{2}, \ldots, b_{k}) \in S^{k} $ differ in all of elements, therefore $ g(a_{2},\ldots,a_{k}) = f(a',a_{2},\ldots, a_{k}) \neq f(b',b_{2}, \ldots, b_{k}) = g(b_{2},\ldots, b_{k}) $. Therefore $ g: S^{k-1} \rightarrow S $ satisfies the conditions of the problem, so by induction, without loss of generality, we can assume that $ g(a_{2}, \ldots, a_{k}) = \varphi(a_{2}) $, for some $ \varphi : S \rightarrow S $. Of course then the map $ \varphi $ must be bijective (the argument is similar as before for $ g $ satisfying the conditions of the problem). So we get that for every $ a_{2},a_{3}, \ldots, a_{k} \in S $ that taking $ \alpha,\beta(a_{2},a_{3},\ldots,a_{k}) $ we have that $ f(\alpha,a_{2}, \ldots,a_{k}) = f(\beta,a_{2}, \ldots,a_{k}) = \varphi(a_{2}) $. Now I claim that for every $ (b_{1},b_{2}, \ldots, b_{k}) \in S^{k} $ we have that $ f(b_{1},b_{2}, \ldots, b_{k}) = \varphi(b_{2}) $. Indeed, if it is not so, then assume that there exists $ (b_{1},b_{2}, \ldots, b_{k}) \in S^{k} $, such that $ f(b_{1},b_{2}, \ldots, b_{k}) = c \neq \varphi(b_{2}) $, then take $ b_{2}' \in S $ such that $ \varphi(b_{2}') = c $, and then of course $ b_{2}' \neq b_{2} $. Take now arbitrary $ b_{3}', \ldots, b_{k}' \in S $, such that $ b_{3}' \neq b_{3}, \ldots , b_{k}' \neq b_{k} $. Take also $ \alpha,\beta = \alpha,\beta(b_{2}',b_{3}', \ldots, b_{k}') $. Then if $ b_{1} = \alpha $, take $ b_{1}' = \beta $, otherwise take $ b_{1}'=\alpha $. At we end we have found a sequence $ (b_{1}',b_{2}', \ldots,b_{k}') \in S^{k} $, which differs from the sequence $ (b_{1},b_{2}, \ldots ,b_{k}) $ at each element and $ f(b_{1},b_{2}, \ldots ,b_{k}) = c = \varphi(b_{2}') = f(b_{1}',b_{2}', \ldots, b_{k}') $ and we come to contradiction. I hope it is correct .