Suppose real numbers $A,B,C$ such that for all real numbers $x,y,z$ the following inequality holds: \[A(x-y)(x-z) + B(y-z)(y-x) + C(z-x)(z-y) \geq 0.\] Find the necessary and sufficient condition $A,B,C$ must satisfy (expressed by means of an equality or an inequality).

Click for solution very not sure about my "proof" but thought i would post anyway $\displaystyle A(x-y)(x-z) + B(y-z)(y-x) + C(z-x)(z-y) \geq 0.$ (*) Suppose (*). Without loss of generality, $x \geq y \geq z$. [We can repeat the same argument if $x \geq z \geq y$, etc.] Let $a = x-y$ and $b = y-z$ and understand that $a,b$ are positive reals. Then: $(*) \Leftrightarrow A(x-y)(x-y + y-z) + C(x-y + y-z)(y-z) \geq B(x-y)(y-z)$ $\Leftrightarrow Aa(a+b) + C(a+b)b \geq Bab$ $\Leftrightarrow A(\frac{a}{b})^2 + (A+C-B)(\frac{a}{b}) + C \geq 0$ (If $b = 0$, the condition would be just $A \geq 0$). Notice that $\frac{a}{b} \geq 0$. If $A \leq 0$, for extreme values of $a/b$, we have it being negative. So certainly $A > 0$. If there is a positive root, then some segment of the graph of the quadratic is in the 4th quadrant. So we want $- (A+C-B) + \sqrt{A^2 + B^2 + C^2 - 2AB - 2BC - 2CA} < 0$, or $A+C-B > \sqrt{(A+C-B)^2 - 4AC}$, which is true if $AC > 0$. From here, using AMGM it is clear that the condition is $4\sqrt{AC} \geq B$. Because our steps are reversible, it is clear it is sufficent.

Find all functions $f: \mathbb{Q} \mapsto \mathbb{C}$ satisfying (i) For any $x_1, x_2, \ldots, x_{1988} \in \mathbb{Q}$, $f(x_{1} + x_{2} + \ldots + x_{1988}) = f(x_1)f(x_2) \ldots f(x_{1988})$. (ii) $\overline{f(1988)}f(x) = f(1988)\overline{f(x)}$ for all $x \in \mathbb{Q}$.

In triangle $ABC$, $\angle C = 30^{\circ}$, $O$ and $I$ are the circumcenter and incenter respectively, Points $D \in AC$ and $E \in BC$, such that $AD = BE = AB$. Prove that $OI = DE$ and $OI \bot DE$.

Click for solution I think this was a Crux problem some years ago. I will prove a more general result: Let ABC be an arbitrary triangle, and D and E two points on the halflines AB and CB such that AD = CE = CA. Prove that $OI\perp DE$ and $DE=2\sin B\cdot OI$, where O is the circumcenter and I is the incenter of triangle ABC. Here is a quite smart proof of this (not from me; Peter surely will remember my loooong proof using vectors, but this one is much simpler): We will assume that b = CA is the shortest side of the triangle ABC, but in all other cases the proof will work as well, with some slight modifications. The circumcenter O of the triangle ABC lies on the perpendicular bisectors of the segments BC, CA, AB. Let D' and E' be the orthogonal projections of the point I on the perpendicular bisectors of the segments BC and AB. Also, let A' and C' be the midpoints of the segments BC and AB; and let X and Z be the orthogonal projections of the point I on the lines BC and AB, i. e. the points where the incircle of triangle ABC touches the sides BC and AB. It is well-known that BX = s - b, where $s=\frac{1}{2}\left( a+b+c\right)$ is the semiperimeter of the triangle ABC. Thus, $XA^{\prime }=BX-BA^{\prime }=\left( s-b\right) -\frac{1}{2}a=\left( \frac{1}{2}\left( a+b+c\right) -b\right) -\frac{1}{2}a=\frac{1}{2}\left( c-b\right) $. Since the quadrilateral IXA'D' is a rectangle (it has three right angles: at X, at A' and at D'), we have ID' = XA', so $ID^{\prime }=\frac{1}{2}\left( c-b\right) $. Now we have BD = AB - AD = AB - CA = c - b; therefore, $ID^{\prime }=\frac{1}{2}\left( c-b\right) =\frac{1}{2}\cdot BD$. Similarly, $IE^{\prime }=\frac{1}{2}\cdot BE$. Finally, we have $D^{\prime }I\perp A^{\prime }D^{\prime }$ and $A^{\prime }D^{\prime }\perp BC$, i. e. $A^{\prime }D^{\prime }\perp EB$, so that D'I || EB. Similarly, E'I || DB. Hence, the angle < D'IE' is equal to < DBE, since the sides of the two angles are respectively parallel. Together with $ID^{\prime }=\frac{1}{2}\cdot BD$ and $IE^{\prime }=\frac{1}{2}\cdot BE$, this shows that the triangles DBE and D'IE' are similar, with the ratio of similitude $\frac{1}{2}$. Hence, $D^{\prime }E^{\prime }=\frac{1}{2}\cdot DE$. Now, since < OD'I = 90° and < OE'I = 90°, the points D' and E' lie on the circle with diameter OI. In other words, the circumcircle of triangle D'IE' has the segment OI as diameter. Therefore, after the extended law of sines, $D^{\prime }E^{\prime }=OI\cdot \sin \measuredangle D^{\prime }IE^{\prime }$. Hence we have $\frac{1}{2}\cdot DE=D^{\prime }E^{\prime }=OI\cdot \sin \measuredangle D^{\prime }IE^{\prime }=OI\cdot \sin \measuredangle DBE=OI\cdot \sin B$, and thus $DE=2\cdot OI\cdot \sin B=2\sin B\cdot OI$. This already proves one part of our assertion. Remains to show $OI\perp DE$. Now, we will work with directed angles modulo 180°. Actually, the triangles DBE and D'IE' are not just similar, they are inversely similar (since the angles < DBE and < D'IE' are oppositely equal, if we consider them as directed angles). Hence, < ID'E' = - < BDE. But since the points D' and E' lie on the circle with diameter OI, we have < IOE' = < ID'E', so < IOE' = - < BDE. Hence, < (OI; OE') = < IOE' = - < BDE = < EDB = < (DE; AB), and thus < (OI; DE) = < (OI; OE') + < (OE'; AB) + < (AB; DE) = < (OI; OE') + < (OE'; AB) - < (DE; AB) = < (OE'; AB) = 90° (since $OE^{\prime }\perp AB$). Therefore, $OI\perp DE$, qed.. Darij

Let $k \in \mathbb{N},$ $S_k = \{(a, b) | a, b = 1, 2, \ldots, k \}.$ Any two elements $(a, b)$, $(c, d)$ $\in S_k$ are called "undistinguishing" in $S_k$ if $a - c \equiv 0$ or $\pm 1 \pmod{k}$ and $b - d \equiv 0$ or $\pm 1 \pmod{k}$; otherwise, we call them "distinguishing". For example, $(1, 1)$ and $(2, 5)$ are undistinguishing in $S_5$. Considering the subset $A$ of $S_k$ such that the elements of $A$ are pairwise distinguishing. Let $r_k$ be the maximum possible number of elements of $A$. (i) Find $r_5$. (ii) Find $r_7$. (iii) Find $r_k$ for $k \in \mathbb{N}$.

Let $f(x) = 3x + 2.$ Prove that there exists $m \in \mathbb{N}$ such that $f^{100}(m)$ is divisible by $1988$.

Click for solution I used the obvious lemma: $a, b, c$ are positive integers, if $(a, b) = 1$, then $am + c$ (mod $b$) are all distinct for $m = 1, 2, \ldots, b$. Thus, let $m = 1, 2, \ldots, 1988$, which are all distinct (mod $1988$), since 3 is prime, $(3, 1988) = 1$, then using the lemma. We have $f(1), f(2), \ldots, f(1988)$ are all distinct (mod $1988$), that is a permutation of $1, 2, 3, ..., 1988$. Repeat above disscusion 100 times, then the conclusion follows easily: Among $f^{100}(1), f^{100}(2), \ldots, f^{100}(1988)$, just one value is divisible by 1988. By the way, from the above solution, we can say the minimum value of $m$ satisfying the condition is no larger than 1988.

Let $ABCD$ be a trapezium $AB // CD,$ $M$ and $N$ are fixed points on $AB,$ $P$ is a variable point on $CD$. $E = DN \cap AP$, $F = DN \cap MC$, $G = MC \cap PB$, $DP = \lambda \cdot CD$. Find the value of $\lambda$ for which the area of quadrilateral $PEFG$ is maximum.

A polygon $\prod$ is given in the $OXY$ plane and its area exceeds $n.$ Prove that there exist $n+1$ points $P_{1}(x_1, y_1), P_{2}(x_2, y_2), \ldots, P_{n+1}(x_{n+1}, y_{n+1})$ in $\prod$ such that $\forall i,j \in \{1, 2, \ldots, n+1\}$, $x_j - x_i$ and $y_j - y_i$ are all integers.

Click for solution Divide the plane into unit squares. Let $O$ be the origin. Let $U$ be one of the unit squares, with vertices $XYZT$ where $X$ is the lower-left one. Then consider the translation $t_U$ with vector $XO$. It maps $U$ onto the unit square $S$ with lower-left vertex $O$. Note that for each point $M$ in $U$, the vector $Mt(M)$ has integer coordinates (since it is $XO$). The common part of the given polygon and $U$, say $P_U$, which is a polygon, is map to a congruent polygon $t(P_U)$, so that it has the same area. Now, except for a part with area $0$ (the intersection of the given polygon with the lines with equation $x=k$ of $y=k$, where $k$ is an integer), the area of the given polygon is the sum of the areas of the polygons $P_U$, where the sum is over all the unit squares $U$. But all the $t(P_U)$ are in $S$, which has area $1$. Since the sum of the areas is greater than $n$, the pigeon-hole principle ensures that a point, say $M$, of $S$, is covered at least $n+1$ times by the $t(P_U)$. It means that there exists $M_1, \cdots, M_{n+1}$ pairwise distinct points in the given polygon such that each of them has image $M$ by the appropriate translation. Since all the translations have vectors with integer coordinates and that a composition of translation is a translation obtained by summing the vectors, it follows that the vector $M_iM_j$ has integer coordinates for all $i<j$. Pierre.

There is a broken computer such that only three primitive data $c$, $1$ and $-1$ are reserved. Only allowed operation may take $u$ and $v$ and output $u \cdot v + v.$ At the beginning, $u,v \in \{c, 1, -1\}.$ After then, it can also take the value of the previous step (only one step back) besides $\{c, 1, -1\}$. Prove that for any polynomial $P_{n}(x) = a_0 \cdot x^n + a_1 \cdot x^{n-1} + \ldots + a_n$ with integer coefficients, the value of $P_n(c)$ can be computed using this computer after only finite operation.