Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.

Find the greatest positive integer $n$ for which there exist $n$ nonnegative integers $x_1, x_2,\ldots , x_n$, not all zero, such that for any $\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n$ from the set $\{-1, 0, 1\}$, not all zero, $\varepsilon_1 x_1 + \varepsilon_2 x_2 + \cdots + \varepsilon_n x_n$ is not divisible by $n^3$.

Let $ x,y\in \mathbb{R} $. Show that if the set $ A_{x,y}=\{ \cos {(n\pi x)}+\cos {(n\pi y)} \mid n\in \mathbb{N}\} $ is finite then $ x,y \in \mathbb{Q} $. Vasile Pop

Click for solution From the equality \[\left( \cos n\pi x -\cos n\pi y \right) ^2=2+\cos2n\pi x +\cos2n\pi y -\left( \cos n\pi x +\cos n\pi y \right)^2 \]we deduce that the set \[B_{x,y}=\{\cos n\pi x -\cos n\pi y |n\in \mathbb{N}\} \]is also finite. Since \[\cos n\pi x =\frac{(\cos n\pi x +\cos n\pi y )+(\cos n\pi x -\cos n\pi y )}{2} \]we obtain that the set \[A=\{\cos n\pi x |n \in \mathbb{N}\} \]is finite as well. It follows that there exist $n,m \in \mathbb{N},n \ne m$ such that $\cos n\pi x=\cos m\pi x$, hence $x=\frac {2k}{n\pm m}$, for some integer $k$. Similarly, $y \in \mathbb{Q}$.

Let $ ABCD $ be a cyclic quadrilateral and let $ M $ be the set of incenters and excenters of the triangles $ BCD $, $ CDA $, $ DAB $, $ ABC $ (so 16 points in total). Prove that there exist two sets $ \mathcal{K} $ and $ \mathcal{L} $ of four parallel lines each, such that every line in $ \mathcal{K} \cup \mathcal{L} $ contains exactly four points of $ M $.

Let $A$ and $B$ be points on a circle $\mathcal{C}$ with center $O$ such that $\angle AOB = \dfrac {\pi}2$. Circles $\mathcal{C}_1$ and $\mathcal{C}_2$ are internally tangent to $\mathcal{C}$ at $A$ and $B$ respectively and are also externally tangent to one another. The circle $\mathcal{C}_3$ lies in the interior of $\angle AOB$ and it is tangent externally to $\mathcal{C}_1$, $\mathcal{C}_2$ at $P$ and $R$ and internally tangent to $\mathcal{C}$ at $S$. Evaluate the value of $\angle PSR$.

Click for solution Let $ X = AC \cap BD $ and $ Y $ be the foot of the perpendicular from $ X $ to $ AB $. As $ X $ lies on the pole of $ M $, and $ XY $ is perpendicular to $ OM $, $ XY $ must be the pole of $ M $, and as $ Y $ is the foot of the perpendicular, this implies that $ M $ and $ Y $ are inverse points. Now, we know that the perpendicular to $ AB $ through $ Y $, $ AC $, and $ BD $, are all collinear at $ X $. Invert this statement with respect to $ O $. $ AC $ and $ BD $ turn into circles $ AOC $ and $ BOD $ while the perpendicular to $ AB $ through $ Y $ inverts into the circle with diameter $ OM $. These circles are all still concurrent at the inverse of $ X $ - call this point $ K $. Then we see that circles $ AOC $ and $ BOD $ intersect at $ K $, and $ OK $ is perpendicular to $ MK $ (as $ K $ lies on the circle with diameter $ OM $), as desired.

Let $ a\in \mathbb{R} $ and $ f_1(x),f_2(x),\ldots,f_n(x): \mathbb{R} \rightarrow \mathbb{R} $ are the additive functions such that for every $ x\in \mathbb{R} $ we have $ f_1(x)f_2(x) \cdots f_n(x) =ax^n $. Show that there exists $ b\in \mathbb {R} $ and $ i\in {\{1,2,\ldots,n}\} $ such that for every $ x\in \mathbb{R} $ we have $ f_i(x)=bx $.

Let $ p_1,p_2,\ldots,p_k $ be the distinct prime divisors of $ n $ and let $ a_n=\frac {1}{p_1}+\frac {1}{p_2}+\cdots+\frac {1}{p_k} $ for $ n\geq 2 $. Show that for every positive integer $ N\geq 2 $ the following inequality holds: $ \sum_{k=2}^{N} a_2a_3 \cdots a_k <1 $ Laurentiu Panaitopol

Let $ n\geq 3 $ be an integer and let $ x_1,x_2,\ldots,x_{n-1} $ be nonnegative integers such that \begin{eqnarray*} \ x_1 + x_2 + \cdots + x_{n-1} &=& n \\ x_1 + 2x_2 + \cdots + (n-1)x_{n-1} &=& 2n-2. \end{eqnarray*} Find the minimal value of $ F(x_1,x_2,\ldots,x_n) = \sum_{k=1}^{n-1} k(2n-k)x_k $.

Let $ n $ and $ r $ be positive integers and $ A $ be a set of lattice points in the plane such that any open disc of radius $ r $ contains a point of $ A $. Show that for any coloring of the points of $ A $ in $ n $ colors there exists four points of the same color which are the vertices of a rectangle.

Find all primes $ p,q $ such that $ \alpha^{3pq} -\alpha \equiv 0 \pmod {3pq} $ for all integers $ \alpha $.

Click for solution Obviously $3$, $p$ and $q$ have to be pairwise distinct. In fact, assuming the contrary and letting $p = q$ (for example), it would be $\alpha^{3p^2} \equiv \alpha \bmod p^2$, for any $\alpha \in \mathbb{Z}$. Anyway this is trivially checked to be false for $\alpha = p$. So suppose $p < q$ by simmetry. Then $p \neq 2$ and $3 < p < q$. In fact, letting $p = 2$ and $\gcd(\alpha, 3) = 1$, you'd find out $\alpha \equiv \alpha^{3pq} \equiv 1 \bmod 3$, which fails to be true for $\alpha \equiv 2 \bmod 3$. Now let $\alpha_1$ and $\alpha_2$ be primitive roots $\bmod\;\! p$ and $\bmod\;\! q$, respectively, and argue $(q-1) \mid (3p - 1)$ and $(p-1) \mid (3q - 1)$, by properties of generators and multiplicative orders. Then let i) $3p - 1 = (q-1)u$ and ii) $3q - 1 = (p-1)v$, for suitable $u, v \in \mathbb{Z}^+$. Because $q \geq 7$, you need $1 \leq u \leq 3$, since otherwise $(q-1)u \geq 3q - 1 > 3p - 1$. But $u = 1$ and $u = 3$ are actually impossible values (that's trivial!), so necessarily $u = 2$ and $q = \frac{3p+1}{2}$. Subsequently, it follows $p = \frac{2v+1}{2v-9}$ from ii). You need $p \geq 5$ as well as $v \geq 5$. Moreover you know the function $p(\cdot): [5, +\infty[ \; \mapsto \mathbb{R}: v \mapsto \frac{2v+1}{2v-9}$ is strictly decreasing for $v \geq 5$. You check $p(5) = 11$ and $p(4) < 5$ by hand, concluding that $v = 5$ only fits. Hence you get $p = 11$ and $q = 17$ (or viceversa, by symmetrization) and everything works.

Let $ n\geq 3 $ be an integer and let $ p\geq 2n-3 $ be a prime number. For a set $ M $ of $ n $ points in the plane, no 3 collinear, let $ f: M\to \{0,1,\ldots, p-1\} $ be a function such that (i) exactly one point of $ M $ maps to 0, (ii) if a circle $ \mathcal{C} $ passes through 3 distinct points of $ A,B,C\in M $ then $ \sum_{P\in M\cap \mathcal{C}} f(P) \equiv 0 \pmod p $. Prove that all the points in $ M $ lie on a circle.

Let $ x_1,x_2,\ldots,x_n $ be positive real numbers and $ x_{n+1} = x_1 + x_2 + \cdots + x_n $. Prove that \[ \sum_{k=1}^n \sqrt { x_k (x_{n+1} - x_k)} \leq \sqrt { \sum_{k=1}^n x_{n+1}(x_{n+1}-x_k)}. \] Mircea Becheanu

Click for solution By Jensen's applied to $ \sqrt{x} $, we get $ \sum_i \sqrt{x_i\left(\sum_{j\not=i} x_j\right)}\leq \sqrt{n\sum_{i\not=j} x_ix_j} $ But by AM-GM, we have that $ n\sum_{i\not=j} x_ix_j\leq (n-1)\left(\sum_i x_i\right)^2 $ so that $ \sum_i \sqrt{x_i\left(\sum_{j\not=i} x_j\right)}\leq \sqrt{(n-1)\left(\sum_i x_i\right)^2} $ which is exactly the inequality we want to prove.

Let $ x,y,z $ be real numbers. Prove that the following conditions are equivalent: (i) $ x,y,z $ are positive numbers and $ \dfrac 1x + \dfrac 1y + \dfrac 1z \leq 1 $; (ii) $ a^2x+b^2y+c^2z>d^2 $ holds for every quadrilateral with sides $ a,b,c,d $.

Let $ S $ be a set of $ n $ concentric circles in the plane. Prove that if a function $ f: S\to S $ satisfies the property \[ d( f(A),f(B)) \geq d(A,B) \] for all $ A,B \in S $, then $ d(f(A),f(B)) = d(A,B) $, where $ d $ is the euclidean distance function.

Click for solution Consider the two circles $ D $ and $ E $ such that $ d(D,E) $ is maximal. Then $ d(f(D),f(E))=d(D,E) $, and either $ f(D)=D, f(E)=E $ or $ f(D)=E, f(E)=D $. Now assume for two circles $ A,B $, $ d(f(A),f(B))>d(A,B) $. Then $ d(f(D),f(E)) = d(D,E) \leq d(D,A)+d(A,B)+d(B,E) < $ $ \min(d(f(D),f(A))+d(f(A),f(B))+d(f(B),f(E)), $ $ d(f(D),f(B))+d(f(B),f(A))+d(f(A),f(E))) = d(f(D),f(E)) $. Absurd.

Let $ n\geq 3 $ be an integer and let $ \mathcal{S} \subset \{1,2,\ldots, n^3\} $ be a set with $ 3n^2 $ elements. Prove that there exist nine distinct numbers $ a_1,a_2,\ldots,a_9 \in \mathcal{S} $ such that the following system has a solution in nonzero integers: \begin{eqnarray*} a_1x + a_2y +a_3 z &=& 0 \\ a_4x + a_5 y + a_6 z &=& 0 \\ a_7x + a_8y + a_9z &=& 0. \end{eqnarray*} Marius Cavachi