Consider a sphere and a plane $\pi$. For a variable point $M \in \pi$, exterior to the sphere, one considers the circular cone with vertex in $M$ and tangent to the sphere. Find the locus of the centers of all circles which appear as tangent points between the sphere and the cone. Octavian Stanasila

Let $OABC$ be a trihedral angle such that \[ \angle BOC = \alpha, \quad \angle COA = \beta, \quad \angle AOB = \gamma , \quad \alpha + \beta + \gamma = \pi . \] For any interior point $P$ of the trihedral angle let $P_1$, $P_2$ and $P_3$ be the projections of $P$ on the three faces. Prove that $OP \geq PP_1+PP_2+PP_3$. Constantin Cocea

Consider all regular convex and star polygons inscribed in a given circle and having $n$ sides. We call two such polygons to be equivalent if it is possible to obtain one from the other using a rotation about the center of the circle. How many classes of such polygons exist? Mircea Becheanu

Prove that for all positive integers $0<a_1<a_2<\cdots <a_n$ the following inequality holds: \[ (a_1+a_2+\cdots + a_n)^2 \leq a_1^3+a_2^3 + \cdots + a_n^3 . \] Viorel Vajaitu

Click for solution I will prove it by induction. It is true for $n=1$, so suppose it is true for $n$. We have: \[ \sum\limits_{i=1}^n a_i ^3 \geq \sum\limits_{i=1}^n a_i ^2 + 2\sum\limits_{i<j}^n a_ia_j \] Adding $a_{n+1}^3$ to both sides we have to prove that: \[ \sum\limits_{i=1}^n a_i ^2 + 2\sum\limits_{i<j}^{n} a_ia_j +a_{n+1}^3 \geq \sum\limits_{i=1}^{n+1} a_i ^2 + 2\sum\limits_{i<j}^{n+1} a_ia_j \] equivalently: \[ a_{n+1}^2+2a_{n+1}\sum\limits_{i=1}^n a_i \leq a_{n+1}^3 \] \[ a_{n+1}+2\sum\limits_{i=1}^n a_i \leq a_{n+1}^2 \] I will prove this by induction as well - suppose it is true for $n+1$, adding to both sides $a_{n+1}+a_{n+2}$ we are left to prove: \[ a_{n+1}^2 +a_{n+1}+a_{n+2} \leq a_{n+2}^2 \] \[ a_{n+1}(a_{n+1}+1)\leq a_{n+2}(a_{n+2}-1) \] which is true due to condition $a_{n+1}<a_{n+2}$ Hence the result

The cells of a $11\times 11$ chess-board are colored in 3 colors. Prove that there exists on the board a $m\times n$ rectangle such that the four cells interior to the rectangle and containing the four vertices of the rectangle have the same color. Ioan Tomescu

Find all vectors of $n$ real numbers $(x_1,x_2,\ldots,x_n)$ such that \[ \left\{ \begin{array}{ccc} x_1 & = & \dfrac 1{x_2} + \dfrac 1{x_3} + \cdots + \dfrac 1{x_n } \\ x_2 & = & \dfrac 1{x_1} + \dfrac 1{x_3} + \cdots + \dfrac 1{x_n} \\ \ & \cdots & \ \\ x_n & = & \dfrac 1{x_1} + \dfrac 1{x_2} + \cdots + \dfrac 1{x_{n-1}} \end{array} \right. \] Mircea Becheanu

In the plane there are given the lines $\ell_1$, $\ell_2$, the circle $\mathcal{C}$ with its center on the line $\ell_1$ and a second circle $\mathcal{C}_1$ which is tangent to $\ell_1$, $\ell_2$ and $\mathcal{C}$. Find the locus of the tangent point between $\mathcal{C}$ and $\mathcal{C}_1$ while the center of $\mathcal{C}$ is variable on $\ell_1$. Mircea Becheanu

The positive integer $n$ is given and for all positive integers $k$, $1\leq k\leq n$, denote by $a_{kn}$ the number of all ordered sequences $(i_1,i_2,\ldots,i_k)$ of positive integers which verify the following two conditions: a) $1\leq i_1<i_2< \cdots i_k \leq n$; b) $i_{r+1}-i_r \equiv 1 \pmod 2$, for all $r \in\{1,2,\ldots,k-1\}$. Compute the number $a(n) = \sum\limits_{k=1}^n a_{kn}$. Ioan Tomescu

Click for solution In other words, we are asked to find the number of increasing sequences of elements in $[n]$ which are alternatively even and odd. Now, I just guessed the answer, and this makes things a lot easier . For each $n$, let $o_n$ be the number of such sequences which begin with an odd number, and $e_n$ the number of sequences which begin with an even number (so the value we're trying to find is $a(n)=o_n+e_n$). We have $o_n=(o_{n-1}+1)+(o_{n-3}+1)+\ldots$ (we define $o_0=0$, and the sum goes on for as long as the indices are non-negative), because if your sequence begins with $1$, then you can either leave it like this, or extend it in $o_{n-1}$ ways, if it begins with $3$, you can leave it like this or extend it in $o_{n-3}$ ways, and so on. It is also obvious that $e_n=o_{n-1}$. We can observe now that $x_n=F_{n+2}-1$ has the same initial values as $o_n$ and satisfies the same recurrence, where $F_0=0,F_1=1,F_n=F_{n-1}+F_{n-2},\ \forall n\ge 2$ is the Fibonacci sequence. This means that we have $o_n=F_{n+2}-1$, and we thus have $a(n)=o_n+e_n=F_{n+3}-2$.

Prove that for all positive integers $n\geq 1$ the number $\prod^n_{k=1} k^{2k-n-1}$ is also an integer number. Laurentiu Panaitopol.

Let $ p > 2$ be a prime number. Find the least positive number $ a$ which can be represented as \[ a = (X - 1)f(X) + (X^{p - 1} + X^{p - 2} + \cdots + X + 1)g(X), \] where $ f(X)$ and $ g(X)$ are integer polynomials. Mircea Becheanu.

Click for solution The remainder of $x^{p-1}+\ldots+x+1$ when divided by $x-1$ is $p$, so $a=p$ works. It's also the smallest one, since plugging $1$ into $(x-1)f(x)+(x^{p-1}+\ldots+x+1)g(x)=a$ gives $p|a$. Why do we need $p$ to be prime? The remainder when we divide $\frac{x^n-1}{x-1}$ to $x-1$ is always $n$ (since $x^{n-1}+\ldots+x+1=(x-1)(x^{n-2}+2x^{n-3}+\ldots+(n-2)x+(n-1))+n$), and the exact same argument as above shows that $n|a$ in the general case.. Sorry if I'm missing something obvious.

Let $x,y,z$ be real numbers with $x+y+z=0$. Prove that \[ |\cos x |+ |\cos y| +| \cos z | \geq 1 . \] Viorel Vajaitu, Bogdan Enescu

The four vertices of a square are the centers of four circles such that the sum of theirs areas equals the square's area. Take an arbitrary point in the interior of each circle. Prove that the four arbitrary points are the vertices of a convex quadrilateral. Laurentiu Panaitopol

Click for solution It suffices to show that if $A,B,C$ are three consecutive vertices of the square and $\ell$ is a line, with $X_A,X_B,X_C$ being the projections of $A,B,C$ on $\ell$, then the sum of the areas of the circles with radii $AX_A,BX_B,CX_C$ is greater than the area of $ABCD$. First of all, in order for the sum of the areas of the circles to be $<S[ABCD]$, we need all the radii $AX_A,BX_B,CX_C$ to be smaller than $\frac{\sqrt 2}2$ (assume the side of the square has length $1$). Now fix $BX_B$, and rotate $\ell$. It's easy to see that in order for the sum of the areas of the three circles to be minimal, we need $\ell$ to intersect the segments $BA,BC$, so let's assume it does, and let $T_A,T_C$ be the points of intersection. Also, let $a=BT_A,c=BT_C$. It's clear that $\sqrt{\frac 1{a^2}+\frac 1{c^2}}$ is constant, and we want to minimize $\left(\frac{AT_A}{T_AB}\right)^2+\left(\frac{CT_C}{T_CB}\right)^2=\left(\frac 1a-1\right)^2+\left(\frac 1c-1\right)^2$. This happens when $a=c$, i.e. when $\ell$ makes angles of $45^{\circ}$ with $CA,CB$, and the desired inequality is easy to get now.

Let $a$ be a positive integer. The sequence $\{x_n\}_{n\geq 1}$ is defined by $x_1=1$, $x_2=a$ and $x_{n+2} = ax_{n+1} + x_n$ for all $n\geq 1$. Prove that $(y,x)$ is a solution of the equation \[ |y^2 - axy - x^2 | = 1 \] if and only if there exists a rank $k$ such that $(y,x)=(x_{k+1},x_k)$. Serban Buzeteanu

Click for solution These are the only solutions if we impose the condition that $y,x>0$, for example. Otherwise, we can clearly find others (like $(-y,-x)$ or $(-x,y)$ for instance). Notice that if $(y,x)$ is a solution, so is $(x,y-ax)$. Since $y,x>0$, the sum of the components decreases when we perform this procedure, and all the solutions are obtained from minimal solutions (i.e. positive solutions which cannot be “decreased” through this procedure without making a component $\le 0$) by applying the inverse transformation. It thus suffices to prove that the only minimal solution is $(y,x)=(a,1)$. This, however follows immediately: since $(y,x)$ is minimal, it must be that $y-ax\le 0$. If we were to have $y-ax<0$, then we would also have $|y^2-ayx-x^2|=|y(y-ax)-x^2|>1$, contradiction. This shows that $y-ax=0\Rightarrow x=1\Rightarrow y=a$, and we’re done.

Let $\Delta$ denote the set of all triangles in a plane. Consider the function $f: \Delta\to(0,\infty)$ defined by $f(ABC) = \min \left( \dfrac ba, \dfrac cb \right)$, for any triangle $ABC$ with $BC=a\leq CA=b\leq AB = c$. Find the set of values of $f$.

Let $[a,b]$ be a given interval of real numbers not containing integers. Prove that there exists $N>0$ such that $[Na,Nb]$ does not contain integer numbers and the length of the interval $[Na,Nb]$ exceedes $\dfrac 16$.

The finite sets $A_1$, $A_2$, $\ldots$, $A_n$ are given and we denote by $d(n)$ the number of elements which appear exactly in an odd number of sets chosen from $A_1$, $A_2$, $\ldots$, $A_n$. Prove that for any $k$, $1\leq k\leq n$ the number \[{ d(n) - \sum\limits^n_{i=1} |A_i| + 2\sum\limits_{ i<j} |A_i \cap A_j | - \cdots + (-1)^k2^{k-1} \sum\limits_{i_1 <i_2 <\cdots < i_k} | A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_k}}| \] is divisible by $2^k$. Ioan Tomescu, Dragos Popescu

Click for solution Firstly note that the $r^{th}$ summand (without the $2^r$) is precisely #$\{(x,A{}_i{}_1,A{}_i{}_2,\ldots,A{}_i{}_r)|x \in A{}_{i{}_1} \cap A{}_{i{}_2} \cap \cdots \cap A{}_{i{}_k}\}$. Counting from the perspective of the points instead, we get that this sum is the same as $\sum_{x\in S} \binom{|A_x|}{r}$ where $A_x=\{i|x\in A_i\}$ and $S$ is the superset of the $A_i's$. what we want to show is that for each $1\leq k\leq n$ we have that $d(n)-\sum_x \binom{|A_x|}{1} +2\sum_x \binom{|A_x|}{2}-\cdots+(-1){}^k2{}^{k-1} \sum_x \binom{|A_x|}{k}$ = \[ d(n)-\sum_x\big{(} \binom{|A_x|}{1} +2\binom{|A_x|}{2}-\cdots+(-1){}^k2{}^{k-1}\binom{|A_x|}{k}\big{)} \] is divisible by $2^k$. Note that to prove this it suffices to show that $d(n)-\sum_x\big{(} \binom{|A_x|}{1} +2\binom{|A_x|}{2}-\cdots+(-1){}^n2{}^{n-1}\binom{|A_x|}{n}\big{)}$ is divisible by $2^n$. but this summand (for each $x$ ) is simply $\dfrac{(1-\theta)^{|A_x|}-1}{\theta}|_{\theta=2}$ which is $-1$ if $|A_x|$ is odd and zero if it is even. Hence the result.