Let $ a,b$ be two complex numbers. Prove that roots of $ z^{4}+az^{2}+b$ form a rhombus with origin as center, if and only if $ \frac{a^{2}}{b}$ is a non-positive real number.

$ a,b,c$ are three different positive real numbers. Prove that:\[ \left|\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a}\right|>1\]

Find the largest real $ T$ such that for each non-negative real numbers $ a,b,c,d,e$ such that $ a+b=c+d+e$: \[ \sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}}\geq T(\sqrt a+\sqrt b+\sqrt c+\sqrt d+\sqrt e)^{2}\]

a) Let $ n_{1},n_{2},\dots$ be a sequence of natural number such that $ n_{i}\geq2$ and $ \epsilon_{1},\epsilon_{2},\dots$ be a sequence such that $ \epsilon_{i}\in\{1,2\}$. Prove that the sequence: \[ \sqrt[n_{1}]{\epsilon_{1}+\sqrt[n_{2}]{\epsilon_{2}+\dots+\sqrt[n_{k}]{\epsilon_{k}}}}\]is convergent and its limit is in $ (1,2]$. Define $ \sqrt[n_{1}]{\epsilon_{1}+\sqrt[n_{2}]{\epsilon_{2}+\dots}}$ to be this limit. b) Prove that for each $ x\in(1,2]$ there exist sequences $ n_{1},n_{2},\dots\in\mathbb N$ and $ n_{i}\geq2$ and $ \epsilon_{1},\epsilon_{2},\dots$, such that $ n_{i}\geq2$ and $ \epsilon_{i}\in\{1,2\}$, and $ x=\sqrt[n_{1}]{\epsilon_{1}+\sqrt[n_{2}]{\epsilon_{2}+\dots}}$

Prove that for two non-zero polynomials $ f(x,y),g(x,y)$ with real coefficients the system: \[ \left\{\begin{array}{c}f(x,y)=0\\ g(x,y)=0\end{array}\right.\] has finitely many solutions in $ \mathbb C^{2}$ if and only if $ f(x,y)$ and $ g(x,y)$ are coprime.

Let $ ABC$, $ l$ and $ P$ be arbitrary triangle, line and point. $ A',B',C'$ are reflections of $ A,B,C$ in point $ P$. $ A''$ is a point on $ B'C'$ such that $ AA''\parallel l$. $ B'',C''$ are defined similarly. Prove that $ A'',B'',C''$ are collinear.

a) Let $ ABC$ be a triangle, and $ O$ be its circumcenter. $ BO$ and $ CO$ intersect with $ AC,AB$ at $ B',C'$. $ B'C'$ intersects the circumcircle at two points $ P,Q$. Prove that $ AP=AQ$ if and only if $ ABC$ is isosceles. b) Prove the same statement if $ O$ is replaced by $ I$, the incenter.

Let $ I$ be incenter of triangle $ ABC$, $ M$ be midpoint of side $ BC$, and $ T$ be the intersection point of $ IM$ with incircle, in such a way that $ I$ is between $ M$ and $ T$. Prove that $ \angle BIM-\angle CIM=\frac{3}2(\angle B-\angle C)$, if and only if $ AT\perp BC$.

Let $ ABC$ be a triangle, and $ D$ be a point where incircle touches side $ BC$. $ M$ is midpoint of $ BC$, and $ K$ is a point on $ BC$ such that $ AK\perp BC$. Let $ D'$ be a point on $ BC$ such that $ \frac{D'M}{D'K}=\frac{DM}{DK}$. Define $ \omega_{a}$ to be circle with diameter $ DD'$. We define $ \omega_{B},\omega_{C}$ similarly. Prove that every two of these circles are tangent.

Let $ ABC$ be a triangle. Squares $ AB_{c}B_{a}C$, $ CA_{b}A_{c}B$ and $ BC_{a}C_{b}A$ are outside the triangle. Square $ B_{c}B_{c}'B_{a}'B_{a}$ with center $ P$ is outside square $ AB_{c}B_{a}C$. Prove that $ BP,C_{a}B_{a}$ and $ A_{c}B_{c}$ are concurrent.

Let $ n$ be a natural number, such that $ (n,2(2^{1386}-1))=1$. Let $ \{a_{1},a_{2},\dots,a_{\varphi(n)}\}$ be a reduced residue system for $ n$. Prove that:\[ n|a_{1}^{1386}+a_{2}^{1386}+\dots+a_{\varphi(n)}^{1386}\]

Let $ m,n$ be two integers such that $ \varphi(m) =\varphi(n) = c$. Prove that there exist natural numbers $ b_{1},b_{2},\dots,b_{c}$ such that $ \{b_{1},b_{2},\dots,b_{c}\}$ is a reduced residue system with both $ m$ and $ n$.

Let $ n$ be a natural number, and $ n = 2^{2007}k+1$, such that $ k$ is an odd number. Prove that \[ n\not|2^{n-1}+1\]

Find all integer solutions of \[ x^{4}+y^{2}=z^{4}\]

A hyper-primitive root is a k-tuple $ (a_{1},a_{2},\dots,a_{k})$ and $ (m_{1},m_{2},\dots,m_{k})$ with the following property: For each $ a\in\mathbb N$, that $ (a,m) = 1$, has a unique representation in the following form: \[ a\equiv a_{1}^{\alpha_{1}}a_{2}^{\alpha_{2}}\dots a_{k}^{\alpha_{k}}\pmod{m}\qquad 1\leq\alpha_{i}\leq m_{i}\] Prove that for each $ m$ we have a hyper-primitive root.

Something related to this problem: Prove that for a set $ S\subset\mathbb N$, there exists a sequence $ \{a_{i}\}_{i = 0}^{\infty}$ in $ S$ such that for each $ n$, $ \sum_{i = 0}^{n}a_{i}x^{i}$ is irreducible in $ \mathbb Z[x]$ if and only if $ |S|\geq2$. By Omid Hatami

Consider two polygons $ P$ and $ Q$. We want to cut $ P$ into some smaller polygons and put them together in such a way to obtain $ Q$. We can translate the pieces but we can not rotate them or reflect them. We call $ P,Q$ equivalent if and only if we can obtain $ Q$ from $ P$(which is obviously an equivalence relation). a) Let $ P,Q$ be two rectangles with the same area(their sides are not necessarily parallel). Prove that $ P$ and $ Q$ are equivalent. b) Prove that if two triangles are not translation of each other, they are not equivalent. c) Find a necessary and sufficient condition for polygons $ P,Q$ to be equivalent.

We call the mapping $ \Delta:\mathbb Z\backslash\{0\}\longrightarrow\mathbb N$, a degree mapping if and only if for each $ a,b\in\mathbb Z$ such that $ b\neq0$ and $ b\not|a$ there exist integers $ r,s$ such that $ a = br+s$, and $ \Delta(s) <\Delta(b)$. a) Prove that the following mapping is a degree mapping: \[ \delta(n)=\mbox{Number of digits in the binary representation of }n\] b) Prove that there exist a degree mapping $ \Delta_{0}$ such that for each degree mapping $ \Delta$ and for each $ n\neq0$, $ \Delta_{0}(n)\leq\Delta(n)$. c) Prove that $ \delta =\Delta_{0}$

Click for solution Official Solution: For (a), let $ a,b\in\mathbb{Z}$ be such that $ b\neq0$ and $ b\nmid a$. Now divide $ a$ by $ b$ and write $ a = br' + s'$ where $ 0 < s' < |b|$. If $ s' < \frac {|b|}2$, then put $ r = r',s = s'$; and if $ s'\geq \frac {|b|}2$ then put $ r = r'\pm1,s' = s - |b|$. (the sign $ \pm$ is the same as $ b$'s sign) One can easily see that $ a = br + s$ and $ |s|\leq\frac {|b|}2$ which shows that $ \delta(s)\leq\delta(b) - 1$. For (b), let $ \Delta_0(n)$ be the minimum of $ \Delta(n)$ for every degree map $ \Delta$. It suffices to show that $ \Delta_0$ is a degree map. Given $ a,b$, there exists a degree map $ \Delta$ for which $ \Delta_0(b) = \Delta(b)$. Now take the $ r,s$ given by $ \Delta$. So $ a = br + s$ and $ \Delta(s) < \Delta(b)$. But then $ \Delta_0(s)\leq\Delta(s) < \Delta(b) = \Delta_0(b)$ which shows that this $ r,s$ work for $ \Delta_0$ too. Now for (c): For $ n = \pm1$, $ \delta(n)\geq\Delta_0(n)\geq1 = \delta(n)$ which shows that $ \delta(n) = \Delta_0(n)$. Let $ n$ be a number such that $ \Delta_0(n) < \delta(n)$. If there are many, choose one with the minimum $ \Delta_0(n)$. $ n\neq\pm1$ because of the previous result. Now take $ a = \lfloor\frac n2\rfloor$ and $ b = n$. Then $ b\nmid a$ because $ a\neq\pm1$. Let $ r,s$ be such that $ \Delta_0(s) < \Delta_0(n)$ and $ a = br + s$. Since $ \Delta_0(s) < \Delta_0(n)$, we have $ \Delta_0(s) = \delta(s)$. But it can be easily shown that whatever the value of $ s$ is, $ |s|\geq\lfloor\frac {|n|}2\rfloor$ which is at least $ \delta(n) - 1$. So $ \delta(n)\geq\Delta_0(n) > \delta(n) - 1$ which shows that $ \Delta_0(n) = \delta(n)$. A contradiction!

We call a set $ A$ a good set if it has the following properties: 1. $ A$ consists circles in plane. 2. No two element of $ A$ intersect. Let $ A,B$ be two good sets. We say $ A,B$ are equivalent if we can reach from $ A$ to $ B$ by moving circles in $ A$, making them bigger or smaller in such a way that during these operations each circle does not intersect with other circles. Let $ a_{n}$ be the number of inequivalent good subsets with $ n$ elements. For example $ a_{1}= 1,a_{2}= 2,a_{3}= 4,a_{4}= 9$. If there exist $ a,b$ such that $ Aa^{n}\leq a_{n}\leq Bb^{n}$, we say growth ratio of $ a_{n}$ is larger than $ a$ and is smaller than $ b$. a) Prove that growth ratio of $ a_{n}$ is larger than 2 and is smaller than 4. b) Find better bounds for upper and lower growth ratio of $ a_{n}$.

Click for solution Official Solution: For the upper bound, to each arrangement assign a string of parentheses. First arrange the circles in such a way that each circle's center is on the x axis, and then cut the tops and bottoms of every circle. The remaining of each circle consists of two arcs shaped either like $ ($ or like $ )$. Assign this string to this arrangement. (Of course there may be several ways to assign a string to a single arrangement) This string is correctly parenthesized, because each $ ($ is matched to a corresponding $ )$, and the substring between these two, corresponds to a subarrangement with fewer circles (that is those circles contained in one of the circles in the original arrangement) It is straightforward to construct back the arrangement from the string of parentheses (Complete the circles by joining each $ ($ with its matching $ )$) Therefor $ a_n$ is smaller than or equal to the Catalan number of order $ n$ which is $ \frac 1{n+1}{2n\choose n}\leq 2^{2n}=4^n$. For the lower bound, select a string of $ 0$'s and $ 1$'s of length $ n$ and assign an arrangement to it. Reading the string from left to right, replace each $ 0$ by a circle containing no other circles, and each $ 1$ by a circle containing all the other previous circles. Of course some arrangements can not be achieved by this method. But if an arrangement can be achieved, it can be achieved from at most one string. This is because, we can easily reverse the procedure. first consider the outermost circles; at most one can contain other circles. Put as many $ 0$'s as there are empty circles at the end of the string. If no circle is left, then we are done. If there is one left, put a $ 1$ before these zeros, and then put the string representing the contents of this circle at the beginning of our string. This shows that $ a_n\geq 2^n$.

In the following triangular lattice distance of two vertices is length of the shortest path between them. Let $ A_{1},A_{2},\dots,A_{n}$ be constant vertices of the lattice. We want to find a vertex in the lattice whose sum of distances from vertices is minimum. We start from an arbitrary vertex. At each step we check all six neighbors and if sum of distances from vertices of one of the neighbors is less than sum of distances from vertices at the moment we go to that neighbor. If we have more than one choice we choose arbitrarily, as seen in the attached picture. Obviusly the algorithm finishes a) Prove that when we can not make any move we have reached to the problem's answer. b) Does this algorithm reach to answer for each connected graph?

Look at these fractions. At firs step we have $ \frac{0}{1}$ and $ \frac{1}{0}$, and at each step we write $ \frac{a+b}{c+d}$ between $ \frac{a}{b}$ and $ \frac{c}{d}$, and we do this forever \[ \begin{array}{ccccccccccccccccccccccccc}\frac{0}{1}&&&&&&&&\frac{1}{0}\\ \frac{0}{1}&&&&\frac{1}{1}&&&&\frac{1}{0}\\ \frac{0}{1}&&\frac{1}{2}&&\frac{1}{1}&&\frac{2}{1}&&\frac{1}{0}\\ \frac{0}{1}&\frac{1}{3}&\frac{1}{2}&\frac{2}{3}&\frac{1}{1}&\frac{3}{2}&\frac{2}{1}&\frac{3}{1}&\frac{1}{0}\\ &&&&\dots\end{array}\] a) Prove that each of these fractions is irreducible. b) In the plane we have put infinitely many circles of diameter 1, over each integer on the real line, one circle. The inductively we put circles that each circle is tangent to two adjacent circles and real line, and we do this forever. Prove that points of tangency of these circles are exactly all the numbers in part a(except $ \frac{1}{0}$). c) Prove that in these two parts all of positive rational numbers appear. If you don't understand the numbers, look at here.

Click for solution Official Solution: For part (a) we use induction. If $ bc-ad=1$ then $ b(a+c)-a(b+d)=ba+bc-ab-ad=bc-ad=1$ and $ (b+d)c-(a+d)c=bc+dc-ad-dc=bc-ad=1$ which proves the induction step. It is obvious that it is true at the beginning of the process. Now if $ bc-ad=1$ then $ (a,b)=1$. So every fraction appearing in this process is simplified. We will show that if $ x$ is a number generated in one of the processes, then $ \frac 1x,x+1,x-1$ are also produced. For the first operation: If the $ i^{th}$ number, counting from the beginning of the sequence, is $ \frac ab$, then the $ i^{th}$ number, counting from the end of the sequence, would be $ \frac ba$. This is true at the beginning. ($ \frac 10,\frac 01$) And using induction, it is no hard to prove it. Now note that the sequence generated by $ a+1,b+1$ as the first and the last fraction is exactly the sequence generated by $ a,b$ as the first and the last fraction, plus $ 1$. (that is a $ 1$ is added to each term of the sequence) Again this is easy to prove using induction. At the beginning it is true. If $ \frac ab,\frac cd$ are generated somewhere in the second sequence, then $ \frac {a+b}b,\frac {c+d}d$ are generated in the first sequence. The next fraction appearing between them is $ \frac{a+c+b+d}{c+d}$ which is one unit more than $ \frac{a+c}{b+d}$ (the one appearing in the second sequence) Now note that, in the $ 1^{st}$ step of the process, the number appearing just before $ \frac 10$ is $ \frac 11$. Therefor the numbers appearing between $ \frac 11,\frac 10$ are exactly one unit more than the numbers appearing between $ \frac 01,\frac 10$. (which are all the numbers appearing in the process) So if $ x$ appears in the process, $ x+1$ does, and if $ x+1$ does, then so does $ x$. Now for the second process: It is obvious that if $ x$ appears, then so does $ x+1$, because the first circles are one unit apart from each other. A unit inversion from the $ 0$ of the x axis, maps circles to circles, and tangent ones to tangent ones. The circle above $ 0$ is mapped to the line $ y=1$. So the circles tangent to it, are mapped to the circles tangent to $ y=0$ and $ y=1$. Each two consecutive circles among these ones are mapped to tangent circles. Therefor these circles are mapped to circles above natural numbers, with diameter $ 1$. (the first circles of the process) Since the inverse of inversion is itself, the circles above natural numbers are mapped to circles appearing in the process and tangent to the circle above $ 0$. This inversion maps the point $ x$ of the real line to $ \frac 1x$. Therefor if $ x$ appears in the process, $ \frac 1x$ appears in the image of the process under the inversion which also appears in the first sequence itself. So in both processes, if $ x$ appears, then so does $ \frac 1x$ and $ x+1$ and $ x-1$. (if it's not negative) Now consider the Euclidean algorithm for a pair of relatively prime numbers $ (a,b)$. In each step if our pair is $ (a,b)$ consider the fraction $ \frac ab$. Then a step consisting of swapping the numbers is the same as replacing our fraction $ x$ by $ \frac 1x$. A step consisting of replacing $ (a,b)$ by $ (a-b,b)$ is the same as replacing $ x$ by $ x-1$. Therefor after some number of moves of the form $ x\rightarrow\frac 1x$ and $ x\rightarrow x-1$, $ x$ becomes $ \frac 01$. (because $ (a,b)$ were relatively prime) Using the inverse operations, which are $ x\rightarrow\frac 1x$ and $ x\rightarrow x+1$, we see that we can reach each rational number from $ \frac 01$ which is present in both processes. So all the rational numbers appear in both processes. What remains to prove, is to show that only rational numbers appear in the second process. But it is a straightforward induction. Assume that two tangent circles $ C_1,C_2$ appeared in the process and we know that their tangency points with the x axis are rational numbers. Using some unit translations and inversion from $ 0$, we can map $ C_1$ to the circle above $ 0$. We have already shown that these operations and their inverse maps, map circles appearing in the process to circles appearing in the process. So now we have two circles appearing in the process, and one of them is the one above $ 0$. We know that the circles tangent to this one are the circles above $ \frac 1n$'s. So the circle between $ C_1,C_2$ is mapped to a circle above $ \frac 1n$ for some $ n$ which has a rational tangency point. Since our operations map rational numbers to rational numbers, the original tangency point should have been a rational number; which completes the proof.

Scientist have succeeded to find new numbers between real numbers with strong microscopes. Now real numbers are extended in a new larger system we have an order on it (which if induces normal order on $ \mathbb R$), and also 4 operations addition, multiplication,... and these operation have all properties the same as $ \mathbb R$. a) Prove that in this larger system there is a number which is smaller than each positive integer and is larger than zero. b) Prove that none of these numbers are root of a polynomial in $ \mathbb R[x]$.

Click for solution Let $ \alpha$ be a new number. If $ \alpha<0$ one can consider $ -\alpha$ which is greater than $ 0$. So assume that $ \alpha>0$. If for each positive real number $ r$, $ \alpha>r$ then, $ \frac 1{\alpha}$ is the number we are after, because for each positive number $ r$, $ \alpha>\frac 1{r}$ which gives $ \frac 1{\alpha}<r$. But $ \alpha>0$ which shows that $ \frac 1{\alpha}>0$. So assume that there is a positive real number $ R$ such that $ \alpha<R$. Let $ r$ be the superimum of all the real numbers less than $ \alpha$. (This exists because $ \alpha<R$ and $ 0$ is less than $ \alpha$) Let $ \epsilon=\alpha-r$. Then $ \epsilon\neq0$ because $ \alpha$ is not a real number. If $ \epsilon>0$, there is no positive real number $ a$, less than $ \epsilon$. Because if $ a$ is such a number, $ r+a<\alpha$ which contradicts $ r$ being the superimum. But if $ \epsilon<0$ then there is no negative real number $ a$ greater than $ \epsilon$. Because if $ a$ is such a number, then $ \alpha<r+a$ which shows that the superimum should have been less than or equal to $ r+a$. So in this case, $ -\epsilon$ is the required number. For the second part of the problem, assume $ P$ is a polynomial with real coefficients. $ P$ can be factored into linear and quadratic terms, each with real coefficients. So for $ \alpha$ to be a root of this polynomial, it has to be a root of one of the terms. (because the new system of numbers is a field) If $ ax+b$ is a linear term ($ a\neq 0$), then $ a\alpha+b=0$ gives $ \alpha=\frac{-b}{a}$ which shows that $ \alpha$ is a real number. If $ ax^2+bx+c$ is a quadratic term, ($ a\neq 0$ and $ b^2<4ac$ so that this term cannot be factored into linear ones) then $ a\alpha^2+b\alpha+c=a\left((\alpha+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}\right)$ But each term of the last expression is positive, because if $ c$ is a number in the new system, either $ c\leq 0$ which shows that $ c\times c\geq0\times 0=0$, or $ c\geq 0$ which shows that $ c\times c\geq0\times0=0$. So $ c^2\geq 0$. So $ (\alpha+\frac{b}{2a})^2\geq0$. And $ 4ac-b^2>0$ which gives $ \frac{4ac-b^2}{4a^2}>0$. Hence $ \alpha$ can not be a root of this term, which shows that $ \alpha$ is not a root of the original polynomial.

A ring is the area between two circles with the same center, and width of a ring is the difference between the radii of two circles. a) Can we put uncountable disjoint rings of width 1(not necessarily same) in the space such that each two of them can not be separated. b) What's the answer if 1 is replaced with 0?

Click for solution Official Solution: The answer for (a) is no. Assume that two rings $ A,A'$ with centers $ O,O'$ are tied together. We will show that $ |OO'|\geq 1$. Let $ R,R'$ denote the inner radii of the rings. Assume that $ R'\geq R$. The plane of the ring $ A'$ intersects the ring $ A$, in two intervals, one inside the ring, and the other outside. Let $ U$ be a point on the outer border of $ A'$ that is on the plane and inside the ring $ A$. Then $ R'+1=|O'U|\leq |O'O|+|OU|\leq |OO'|+R\leq |OO'|+R'$ which shows that $ |OO'|\geq 1$. Now for each ring with $ O$ as the center, consider the sphere with radius $ \frac 13$ around $ O$. This sphere contains a point with rational coordinates. So there is an injective mapping from the rings to the rational points. (It is injective because no two such spheres intersect) So the number of rings must be countable. The answer for (b) is yes. Consider a circle with center $ (0,0,0)$ lying in the xy plane and having radius $ 1$. For each $ 0\leq\theta\leq0.1$, move the circle by $ \theta$ along the positive direction of the x axis, and rotate it along this axis by $ \theta$ radians. It is obvious that the rings constructed in this way are pairwise tied to each other. They are in one-to-one correspondence to $ [0,0.1]$ which is uncountable.

In this question you must make all numbers of a clock, each with using 2, exactly 3 times and Mathematical symbols. You are not allowed to use English alphabets and words like $ \sin$ or $ \lim$ or $ a,b$ and no other digits.

Click for solution Official Solution: Some possible answers are shown in the following: \[ \begin{array}{ccc} 1=2^{2-2}&2=2+2-2&3=2+\frac 22\\ 4=\sqrt{2\times2}^2&5=2\times2+\lfloor\sqrt{2}\rfloor&6=2+2+2\\ 7=\lceil\sqrt{(2+2)!}\rceil+2&8=2\times2\times2&9=(2+\lfloor\sqrt{2}\rfloor)^2\\ 10=2\times\lceil\sqrt{(2+2)!}\rceil&11=\frac{22}{2}&12=\frac{(2+2)!}2\\ \end{array}\]