16 students took part in a competition. All problems were multiple choice style. Each problem had four choices. It was said that any two students had at most one answer in common, find the maximum number of problems.

Click for solution Suppose there are $m$ problems $P_1,P_2...,P_m$ for problem $P_i$, suppose $a_i$ students answer first choice, $b_i$ secound choice, $c_i$ third choice and $d_i$ fourth choice. Since $a_i+b_i+c_i+d_i=16$ we have at least $4\binom{4}{2}=24$ pairs with an answer in common for each problem. Since there are $m$ problems there are at least $24m$ such pairs, but on the other hand there are at most $\binom{16}{2}=120$ such pairs. So we have that $m$ is at most $5$. Now can equality hold? since for $m=5$ we need stronger conditions (namely for every problem every choice is selected by 4 students, and every pair has a common choice for some problem) I think such construction is possible since i dont fin any other constraint.

Let $n \geq 2, n \in \mathbb{N},$ find the least positive real number $\lambda$ such that for arbitrary $a_i \in \mathbb{R}$ with $i = 1, 2, \ldots, n$ and $b_i \in \left[0, \frac{1}{2}\right]$ with $i = 1, 2, \ldots, n$, the following holds: \[\sum^n_{i=1} a_i = \sum^n_{i=1} b_i = 1 \Rightarrow \prod^n_{i=1} a_i \leq \lambda \sum^n_{i=1} a_i b_i.\]

For any prime $p$, prove that there exists integer $x_0$ such that $p | (x^2_0 - x_0 + 3)$ $\Leftrightarrow$ there exists integer $y_0$ such that $p | (y^2_0 - y_0 + 25).$

Click for solution The first one holds iff there is an $x_0$ s.t. $p|(2x_0+1)^2+11\ (*)$, while the second one holds iff there is an $y_0$ s.t. $p|(2y_0+1)^2+99\ (**)$. $(*)$ is equivalent to $-11$ being a quadratic residue $\pmod p$, while $(**)$ is equivalent to $-99=9\cdot(-11)$ being a quadratic residue, which is equivalent to $-11$ being a quadratic residue, and we're done.

A triangle $ABC$ is given in the plane with $AB = \sqrt{7},$ $BC = \sqrt{13}$ and $CA = \sqrt{19},$ circles are drawn with centers at $A,B$ and $C$ and radii $\frac{1}{3},$ $\frac{2}{3}$ and $1,$ respectively. Prove that there are points $A',B',C'$ on these three circles respectively such that triangle $ABC$ is congruent to triangle $A'B'C'.$

Click for solution Let $ a = BC = \sqrt {13},\ b = CA = \sqrt {19},\ c = AB = \sqrt 7$ be the triangle sides opposite to the vertices A, B, C. Let F be the 1st Fermat point of the triangle $ \triangle ABC$ and x = AF, y = BF, z = CF. The angles $ \angle AFB = \angle BFC = \angle CFA = 120^o$ are all equal. Using the cosine theorem for the triangles $ \triangle AFB, \triangle BFC, \triangle CFA$ with the sides (x, y, c); (y, z, a); (z, x, b) and the angles $ \phi = \widehat{(x, y)} = \widehat{(y, z)} = \widehat{(z, x)} = 120^o$, $ \cos \phi = - \frac 1 2$, $ \triangle AFB: \ \ x^2 + y^2 + xy = c^2 = 7$ $ \triangle BFC: \ \ y^2 + z^2 + yz = a^2 = 13$ $ \triangle CFA: \ \ z^2 + x^2 + zx = b^2 = 19$ Subtracting the 1st equation from the 2nd one, 2nd from 3rd and 3rd from 1st, $ z^2 - x^2 + yz - xy = (z - x)(x + y + z) = a^2 - c^2 = 6$ $ x^2 - y^2 + zx - yz = (x - y)(x + y + z) = c^2 - b^2 = - 12$ $ y^2 - z^2 + xy - zx = (y - z)(x + y + z) = b^2 - a^2 = 6$ Dividing any pair of equations always yields nothing else but z = 2x - y. Substistuting z into the original set of equations, $ x^2 + y^2 + xy = 7$ $ 4x^2 + y^2 - 2xy = 13$ $ 7x^2 + y^2 - 5xy = 19$ The 3rd equation is a linear combination of the 1st and 2nd, hence, we can disregard it: $ (3) = 2 \times (2) - (1)$. Substracting the 1st equation from the 2nd one, in order to eliminate at least one square, we get $ 3x^2 - 3xy = 6,\ \ y = x - \frac 2 x$ Substituting y into the 1st equation, $ x^2 + x^2 - 4 + \frac {4}{x^2} + x^2 - 2 = 7$ $ 3x^4 - 13x^2 + 4 = 0$ $ x^2 = \frac {13 \pm \sqrt {169 - 48}}{6} = \frac {13 \pm 11}{6} = 4$ . or . $ \frac 1 3$ Since we are not interested in negative roots, x = 2 or $ \frac {\sqrt 3}{3}$. Calculating y, z from the previous equations $ y = x - \frac 2 x = 1$ . or . $ - \frac {5 \sqrt 3}{3}$ $ z = 2x - y = 3$ . or . $ \frac {7 \sqrt 3}{3}$ Since we are interested only in the solution with x > 0, y > 0, z > 0, the distances of the 1st Fermat point from the vertices A, B, C are x = AF = 2, y = BF = 1, z = CF = 3 and x : y : z = 2 : 1 : 3 On the other hand, radii of the circles $ (A), (B), (C)$ are in the ratios $ r_A : r_B : r_C = \frac 1 3 : \frac 2 3 : 1 = 1 : 2 : 3$. I have solved the problem before and it came with exactly the same error - it should be $ r_A = \frac 2 3, r_B = \frac 1 3,\ r_C = 1$, which implies $ r_A : r_B : r_C = 2 : 1 : 3$. If this was the case, then $ \frac {r_A}{x} = \frac {r_B}{y} = \frac {r_C}{z} = \frac 1 3$ i.e., the circles $ (A), (B), (C)$ would be seen from the 1st Fermat point F under the same angle $ 2 \theta = 4 \arcsin{\frac {r_A}{2x}} = 4 \arcsin{\frac 1 6}$. If the triangle $ \triangle ABC$ was then rotated by the angle $ \pm \theta = \pm 2 \arcsin{\frac 1 6}$ around the point F into a congruent riangle $ \triangle A'B'C'$, the vertices $ A', B', C'$ would simultaneously fall on the circles $ (A), (B), (C)$. Alternately, the problem could be cured by keeping the radii of the circles $ (A), (B), (C)$ as $ r_A = \frac 1 3,\ r_B = \frac 2 3,\ r_c = 1$, but specifying the triangle side lengths as $ a = BC = \sqrt {19},\ b = CA = \sqrt {13},\ c = AB = \sqrt 7$. Then we would get x = AF = 1, y = BF = 2, z = CF = 3 and $ x : y : z = 1 : 2 : 3 = r_A : r_B : r_C$. EDIT: See http://www.mathlinks.ro/viewtopic.php?t=194847 for the correct solution.

A $(3n + 1) \times (3n + 1)$ table $(n \in \mathbb{N})$ is given. Prove that deleting any one of its squares yields a shape cuttable into pieces of the following form and its rotations: ''L" shape formed by cutting one square from a $2 \times 2$ squares.

Click for solution A partial solution: If $3n+1\ge 13$, then the board is completely covered by the $4\ (3n+1-6)\times(3n+1-6)$ squares in the corners, so we can assume the missing piece comes from one of those, we can cover that with $L$'s, according to the induction hypothesis (we assume we've shown it's true for smaller boards), and we can cover the rest with $2\times 3$ rectangles (this is easy to see). If $3n+1=10$, the board is covered by the $9\ 4\times 4$ squares which have even distances from their sides to the sides of the board. We can cover any of these $9$ squares with one small square missing according to teh case $3n+1=4$, and for each one of the $9$ squares, the rest can, again, be tiled with $2\times 3$ rectangles. If $3n+1=4$, it's easy to see that a $4\times 4$ minus a $2\times 2$ in a corner is tileable with $L$'s, so we can first place an $L$ near the missing small square so as to complete a $4\times 4$ minus a $2\times 2$ in a corner, and then cover the rest with $L$'s. I guess this leaves $3n+1=7$. The case $3n+1=7$ is not hard at all, but it's very hard to explain it here without some pictures. The idea is to complete the hole created by taking out a small square to a $2\times 2$ square, since there are a lot fewer cases to check. In fact, there are three: we can form a $2\times 2$ square in the corner, or one adjacent to a $2\times 2$ in a corner, or one which contains the center of the $7\times 7$ board (there are other possible positions for a $2\times 2$ square, but we can get one of these, because we also have a certain freedom in choosing the $2\times 2$ square we construct by putting an $L$ near the hole). In these three cases, it's easy to get coverings of the rest of the board.

For any $n,T \geq 2, n, T \in \mathbb{N}$, find all $a \in \mathbb{N}$ such that $\forall a_i > 0, i = 1, 2, \ldots, n$, we have \[\sum^n_{k=1} \frac{a \cdot k + \frac{a^2}{4}}{S_k} < T^2 \cdot \sum^n_{k=1} \frac{1}{a_k},\] where $S_k = \sum^k_{i=1} a_i.$