problems stringlengths 81 1.57k | solutions stringlengths 0 5.65k |
|---|---|
For every integer n prove that the fraction $\frac{21n+4}{14n+3}$ cannot be reduced any further. | The desired result $(14n + 3, 21n + 4) = 1$ follows from $3(14n + 3) β 2(21n + 4) = 1.$ |
Let x be an angle and let the real numbers $a$, $b$, $c$, $\cos{x}$ satisfy the following equation: $a \cos^2{x} + b \cos{x} + c = 0$. Write the analogous quadratic equation for $a$, $b$, $c$, $\cos{2x}$. Compare the given and the obtained equality for a = 4, b = 2, c = β1. | Multiplying the equality by $4(a \cos^{2x} β b \cos{x}+c)$, we obtain $4a^2 \cos^4{x} + 2(4ac β 2b^2)\cos^2{x} + 4c^2 = 0$. Plugging in $2 \cos{2x} = 1 + \cos{2x}$ we obtain (after quite a bit of manipulation): $^2 \cos^2{2x} + (2a^2 + 4ac β 2b^2)\cos{2x} + (a2 + 4ac β 2b^2 + 4c^2)=0$. For a = 4, b = 2, and c = β1 we ... |
Construct a right-angled triangle whose hypotenuse c is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangle. | Analysis. Let a and b be the other two sides of the triangle. From the conditions of the problem we have $c^2 = a^2 +b^2$ and $c/2 = \sqrt{ab} \iff 3/2c^2 = a^2+b^2+2ab = (a+b)^2 \iff \sqrt{3/2}c = a+b$. Given a desired \triangle ABC let D be a point on (AC such that CD = CB. In that case, $AD = a + b = \sqrt{3/2}c$, a... |
Determine all prime numbers p and all positive integers x and y satisfying $x^3+y^3 = p(xy + p)$. | Up to a swap of the first two entries, the only solutions are (x, y, p) = (1, 8, 19), (x, y, p) = (2, 7, 13) and (x, y, p) = (4, 5, 7). The verification is routine. Set $s = x+y$. Rewrite the equation in the form $s(s^2 β 3xy) = p(p + xy)$, and express $xy = \frac{s^3 β p^2}{3s + p}$ . (β) In particular, $s^2 \ge 4xy =... |
Determine all prime numbers p and all positive integers x and y satisfying $x^3+y^3 = p(xy + p)$. | Set again $s = x + y$. It is readily checked that $s \le 8$ provides no solutions, so assume $s \ge 9$. Notice that $x^3+y^3 = s(x^2βxy+y^2 ) \ge \frac{1}{4}s^3 and xy β€ \frac{1}{4} s^2$. The condition in the statement then implies $s^2 (s β p) \le 4p^2$, so $s < p + 4$. Notice that p divides one of s and $x^2 β xy + y... |
Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$. Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$. | Solution 1. Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q=\operatorname{deg} Q \mid \operatorname{deg} S=s$. We will show that $S(x)=T(Q(x))$ for some polynomial $T$. Then $P(Q(x))=R(S(x))=R(T(Q(x)))$, so the polynomial $P(t)-R(T(t))$ vanishes upon substitution $t=S(x)$; it therefore vanishes identically... |
Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$. Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$. | Solution 2. All polynomials in the solution have real coefficients. As usual, the degree of a polynomial $f(x)$ is denoted $\operatorname{deg} f(x)$. Of all pairs of polynomials $P(x), R(x)$, satisfying the conditions in the statement, choose one, say, $P_{0}(x), R_{0}(x)$, so that $P_{0}(Q(x))=R_{0}(S(x))$ has a minim... |
Problem 6. Let $r, g, b$ be non-negative integers. Let $\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\Gamma$ are each coloured red, green or blue. It turns out that $\Gamma$ has - a spanning tree in which exactly $r$ of the edges are red, - a spanning tree in which exactly $g$ of the edges are gree... | Solution 1. Induct on $n=r+g+b$. The base case, $n=1$, is clear. Let now $n>1$. Let $V$ denote the vertex set of $\Gamma$, and let $T_{r}, T_{g}$, and $T_{b}$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases. Case 1: There exists a partition $V=A \sqcup B$ o... |
Problem 6. Let $r, g, b$ be non-negative integers. Let $\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\Gamma$ are each coloured red, green or blue. It turns out that $\Gamma$ has - a spanning tree in which exactly $r$ of the edges are red, - a spanning tree in which exactly $g$ of the edges are gree... | Solution 2. For a spanning tree $T$ in $\Gamma$, denote by $r(T), g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively. Assume that $\mathcal{C}$ is some collection of spanning trees in $\Gamma$. Write $$\begin{array}{rlrl} r(\mathcal{C}) & =\min _{T \in \mathcal{C}} r(T), & g(\mathcal{C})=\m... |
Find all pairs of positive integers $(a,b)$ such that, \[ab^2+b+7|a^2b+a+b\]\n \n | We have the following divisibility relations, \[ab^2+b+7|a^2b+a+b|b(a^2b+a+b)=a^2b^2+ab+b^2\]\n\[ab^2+b+7|a(ab^2+b+7)=a^2b^2+ab+7a\]\nSubtracting, \[ab^2+b+7| \;|b^2-7a|\]\n If $b=1$, we may have \[a+8|7a-1\] \nOtherwise, we would have \[|b^2-7a|<ab^2+b+7\]\nIn the first case, \[a+8|7a+56\]\nThis gives, \[a+8|57\]\nS... |
(Mongolia)\nLet $a_1, a_2, \dotsc, a_n$ be positive real numbers such that $a_1 + a_2 + \dotsb + a_n < 1$. Prove that\n\[\frac{a_1 a_2 \dotsm a_n \bigl[ 1- (a_1 + a_2 + \dotsb + a_n) \bigr]}{ (a_1 + a_2 + \dotsb + a_n)(1-a_1)(1-a_2) \dotsm (1-a_n)} \le \frac{1}{n^{n+1}} .\]\n | Let $a_{n+1}$ be the positive real number such that $a_{n+1} = 1 - \sum_{k=1}^n a_k$. Then\n\[\frac{a_1 a_2 \dotsm a_n \bigl[ 1- (a_1 + a_2 + \dotsb + a_n) \bigr]}{ (a_1 + a_2 + \dotsb + a_n)(1-a_1)(1-a_2) \dotsm (1-a_n)} = \frac{\prod_{k=1}^{n+1} a_k} {\prod_{k=1}^{n+1} \sum_{j\neq k} a_j} .\]\nBut by AM-GM, $\sum_{j... |
(Australia)\nLet $r_1, r_2, \dotsc, r_n$ be real numbers greater than or equal to 1. Prove that\n\[\frac{1}{r_1 + 1} + \frac{1}{r_2 + 1} + \dotsb + \frac{1}{r_n + 1} \ge \frac{n}{\sqrt[n]{r_1 r_2 \dotsm r_n} + 1} .\]\n | Let $f$ denote the function $x \mapsto 1/(e^x + 1)$.\n Lemma 1. For $x \ge 1$, the function $gΒ : x \mapsto x/(x +1)^2$ is decreasing.\n Proof. Note that $g(x) = \frac{1}{x + 2 + 1/x}$. Since $x + 1/x$ is increasing for $x\ge 1$, the lemma follows. $\blacksquare$\n Lemma 2. For positive $x$, $f(x)$ is convex.\n Proof... |
An $m \times n$ array of real numbers has the sum of each row and column integral. Show that each non-integral element $x$ can be changed to either $\left\lfloor x \right\rfloor$ or $\left\lfloor x \right\rfloor + 1$ so that the row and column sums are unchanged.\n \n | Coming soon...\n |
Let $T$ denote the set of all ordered triples $(p,q,r)$ of nonnegative integers. Find all functions $f:T \rightarrow \mathbb{R}$ such that\n $f(p,q,r) = \begin{cases} 0 & \text{if} \; pqr = 0, \\ 1 + \tfrac{1}{6}\{f(p + 1,q - 1,r) + f(p - 1,q + 1,r) & \\ + f(p - 1,q,r + 1) + f(p + 1,q,r - 1) & \\ + f(p,q + 1,r - 1) + f... | We can see that $h(p,q,r)=\frac{3pqr}{p+q+r}$ for $pqr\neq0$ and $h(p,q,r)=0$ for $pqr=0$ satisfies the equation. Suppose there exists another solution $f(p,q,r)$. Let $g(p,q,r)=f(p,q,r)-h(p,q,r)$. Plugging in $f=g+h,$ we see that $g$ satisfies the relationship $g(p,q,r)=\begin{cases} \tfrac{1}{6}\{g(p + 1,q - 1,r) + g... |
Let $a_0, a_1, a_2, \ldots$ be an arbitrary infinite sequence of positive numbers. Show that the inequality $1 + a_n > a_{n - 1} \sqrt [n]{2}$ holds for infinitely many positive integers $n$.\n | We proceed with a proof by contradiction. Suppose the statement were false. Then, there exists a sequence $a_0, a_1, \ldots$ of positive integers for which there are only finitely many $a_n$ with $1+a_n> a_{n-1}\sqrt[n]{2}$. Let the largest such $n$ be $N-1$, so that $1+a_n\le a_{n-1}\sqrt[n]{2}$ whenever $n\le N-1$. T... |
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$, satisfying\n $f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$ for all $x,y$.\n | Assume $f(1) = 0$. Take $y = 1$. We get $f^2(x) = 0,\ \forall x$, so $f(x) = 0,\ \forall x$. This is a solution, so we can take it out of the way: assume $f(1)\ne 0$. \n $y = 1\Rightarrow f(x)[f(x) - f(1)] = (x - 1)f(x)f(1)$. We either have $f(x) = 0$ or $f(x) = f(1)x$, so for every $x$, $f(x)\in\{0,f(1)x\}$. In partic... |
Find all positive integers $a_1, a_2, \ldots, a_n$ such that\n $\frac {99}{100} = \frac {a_0}{a_1} + \frac {a_1}{a_2} + \cdots + \frac {a_{n - 1}}{a_n},$ where $a_0 = 1$ and $(a_{k + 1} - 1)a_{k - 1} \geq a_k^2(a_k - 1)$ for $k = 1,2,\ldots,n - 1$.\n | We claim that there is only one such sequence: $a_1=2, a_2=5, a_3=56, a_4=56\times 1400$. This works because\n\t\t\t\[\frac{1}{2}+\frac{2}{5}+\frac{5}{56}+\frac{56}{56\times 1400}\]\n\[=\frac{700}{1400}+\frac{560}{1400}+\frac{125}{1400}+\frac{1}{1400}=\frac{1386}{1400}=\frac{99}{100}.\]\n \t\t\tIt is also easy to check... |
Prove that for all positive real numbers $a,b,c$,\n $\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.$ Generalization The leader of the Bulgarian team had come up with the following generalization to the inequality:\n $\frac {a}{\sqrt {a^2 + kbc}} + \frac {b}{\sqrt {b^2... | We will use the Jenson's inequality. \n Now, normalize the inequality by assuming $a+b+c=1$\n Consider the function $f(x)=\frac{1}{\sqrt{x}}$. Note that this function is convex and monotonically decreasing which implies that if $a > b$, then $f(a) < f(b)$.\n Thus, we have\n $\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sq... |
Let $A = (a_1, a_2, \ldots, a_{2001})$ be a sequence of positive integers. Let $m$ be the number of 3-element subsequences $(a_i, a_j, a_k)$ with $1 \le i < j < k \le 2001$ such that $a_j = a_i + 1$ and $a_k = a_j + 1$. Considering all such sequences $A$ find the greatest value of $m$.\n | Solution 1 Consider what happens if $A$ is ordered from least to greatest. Then, all the original subsequences will still be subsequences because, since $a_k > a_j > a_i$, the order they appear in is $a_i, a_j, a_k$, so it is still a subsequence. So, if A attains the maximal value, so does it unstrictly increasing vers... |
Let $n$ be an odd integer greater than 1 and let $c_1, c_2, \ldots, c_n$ be integers. For each permutation $a = (a_1, a_2, \ldots, a_n)$ of $\{1,2,\ldots,n\}$, define $S(a) = \sum_{i = 1}^n c_i a_i$. Prove that there exist permutations $a \neq b$ of $\{1,2,\ldots,n\}$ such that $n!$ is a divisor of $S(a) - S(b)$.\n | We shall prove this by contradiction. Assume that for some $n$-tuple of $c_i$ there does not exist two permutations $a$ and $b$ of $s_n=\{ 1, 2, \ldots,n\}$ such that $n!|S(a)-S(b)$. Note that there are $n!$ distinct permutations of $s_n$, which means there are $n!$ distinct sums $S(a)$. Because no two of them are cong... |
Define a $k$-clique to be a set of $k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.\n | Solution 1 If there exists only one 3-clique, remove anyone in that clique. (If there are no 3-cliques, we are done!) Otherwise, consider the following cases:\n Case 1: There exist a pair of 3-cliques that share 2 people.\n Let these 3-cliques be $\{A,C,D\}$ and $\{B,C,D\}$. If every other 3-clique contained either $C$... |
A set of three nonnegative integers $\{x,y,z\}$ with $x < y < z$ is called historic if $\{z - y,y - x\} = \{1776,2001\}$. Show that the set of all nonnegative integers can be written as the union of pairwise disjoint historic sets.\n | We describe a greedy algorithm to cover all integers. If an integer has been included in a set, we call it 'colored'. We also say that a number x is 'in column A' (A being x, y, or z) if x was in the x, y, or z position in its historic set (e.g. in $\{1,2,4\}$, 4 is in column z, 2 in column y, and 1 in column x). \n S... |
Find all finite sequences $(x_0, x_1, \ldots,x_n)$ such that for every $j$, $0 \leq j \leq n$, $x_j$ equals the number of times $j$ appears in the sequence.\n | See Problem 23, Chapter 4, in 102 Combinatorial Problems by Andreescu and Feng.\n |
For a positive integer $n$ define a sequence of zeros and ones to be balanced if it contains $n$ zeros and $n$ ones. Two balanced sequences $a$ and $b$ are neighbors if you can move one of the $2n$ symbols of $a$ to another position to form $b$. For instance, when $n = 4$, the balanced sequences $01101001$ and $0011010... | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
A pile of $n$ pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column... | Note that if a column moves a pebble to the right, either the first column is still bigger or they have the same size. This means that at every point in the process, the amount of pebbles in every column decreases from left to right.\n At no point during this process can there be three or more columns with the same num... |
Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three... | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $A_1$ be the center of the square inscribed in acute triangle $ABC$ with two vertices of the square on side $BC$. Thus one of the two remaining vertices of the square is on side $AB$ and the other is on $AC$. Points $B_1,\ C_1$ are defined in a similar way for inscribed squares with two vertices on sides $AC$ and ... | Let $BC=a$, $CA=b$, $AB=c$, $\angle BAC=\alpha$, $\angle CBA=\beta$, and $\angle ACB=\gamma$. Let $A_2$ be the point on the other side of $BC$ than $A$ such that $BA_2C$ is an isosceles right triangle. Define $B_2$ and $C_2$ similarly. Let $E$ and $F$ be the points on $AB$ and $CA$ that are the vertices of the square c... |
Consider an acute-angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $ABC$. Assume that $\angle C \geq \angle B + 30^{\circ}$. Prove that $\angle A + \angle COP < 90^{\circ}$.\n | See 2001 IMO 1 page.\nhttps://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_1\n |
Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG + BP{\cdot}BG + CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$.\n | We claim that the expression is minimized at $P=G$, resulting it having a value of $(a^2+b^2+c^2)/3$ ($a,b,c$ being the side lengths of $ABC$).\n \t\t\tWe will use vectors, with $G=\vec{0}$ (meaning that $\vec A+\vec B+\vec C=\vec 0$). Note that by Cauchy-Schwarz, \n\t\t\t\[\|\vec A\| \|\vec A-\vec P \|+\|\vec B\|\|\ve... |
Let $M$ be a point in the interior of triangle $ABC$. Let $A'$ lie on $BC$ with $MA'$ perpendicular to $BC$. Define $B'$ on $CA$ and $C'$ on $AB$ similarly. Define\n $p(M) = \frac {MA' \cdot MB' \cdot MC'}{MA \cdot MB \cdot MC}.$ Determine, with proof, the location of $M$ such that $p(M)$ is maximal. Let $\mu(ABC)$ d... | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $ABC$ be an acute triangle. Let $DAC,EAB$, and $FBC$ be isosceles triangles exterior to $ABC$, with $DA = DC, EA = EB$, and $FB = FC$, such that\n $\angle ADC = 2\angle BAC, \quad \angle BEA = 2 \angle ABC, \quad \angle CFB = 2 \angle ACB.$ Let $D'$ be the intersection of lines $DB$ and $EF$, let $E'$ be the inte... | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $ABC$ be a triangle and $P$ an exterior point in the plane of the triangle. Suppose the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ (or extensions thereof) in $D$, $E$, $F$, respectively. Suppose further that the areas of triangles $PBD$, $PCE$, $PAF$ are all equal. Prove that each of these areas is equa... | Solution 1 by Mewto55555:\n We use barycentric coordinates.\n So $A$ is $(1,0,0)$, $B$ is $(0,1,0)$, $C$ is $(0,0,1)$, and $P$ is $(p,q,r)$, with $p+q+r=1$.\n Now, the equation of line $AP$ is just the line $qz=ry$, $BP$ is just $pz=rx$, and $CP$ is $qx=py$.\n Also, $AB$ is just $z=0$, $BC$ is $x=0$, and $AC$ is $y=0$.... |
Let $O$ be an interior point of acute triangle $ABC$. Let $A_1$ lie on $BC$ with $OA_1$ perpendicular to $BC$. Define $B_1$ on $CA$ and $C_1$ on $AB$ similarly. Prove that $O$ is the circumcenter of $ABC$ if and only if the perimeter of $A_1B_1C_1$ is not less than any one of the perimeters of $AB_1C_1, BC_1A_1$, and $... | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?\n | |
Prove that there is no positive integer $n$ such that, for $k = 1,2,\ldots,9$, the leftmost digit (in decimal notation) of $(n + k)!$ equals $k$.\n | Suppose that there is such a number $n$. Let $a$ be the number of digits of $(n + 8)!$, and let $b$ be the number of digits of $n + 9$. Then we have:\n $8 \cdot 10^{a-1} \leqslant (n + 8)! < 9 \cdot 10^{a-1}$\n and\n $10^{b-1} \leqslant n + 9 < 10^b$\n Combining these inequalities via multiplication we get:\n $8 \cdot ... |
Consider the system $x + y = z + u,$ $2xy = zu.$ Find the greatest value of the real constant $m$ such that $m \leq x/y$ for any positive integer solution $(x,y,z,u)$ of the system, with $x \geq y$.\n | First consider the real solutions to the system. We have by AM-GM that $\frac{z+u}{2}\ge\sqrt{zu}$ and substituting we get $\frac{x+y}{2}\ge\sqrt{2xy}$. Squaring and simplifying and dividing by $y^2$, we get the inequality $r^2-6r+1\ge0$, where $r=\frac{x}{y}$. Then $r^2-6r+9\ge8$, so $r\ge3+2\sqrt2$ or $r\le3-2\sqrt2$... |
Let $a_1 = 11^{11}, \, a_2 = 12^{12}, \, a_3 = 13^{13}$, and $a_n = |a_{n - 1} - a_{n - 2}| + |a_{n - 2} - a_{n - 3}|, n \geq 4.$ Determine $a_{14^{14}}$.\n | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $a > b > c > d$ be positive integers and suppose that\n $ac + bd = (b + d + a - c)(b + d - a + c).$ Prove that $ab + cd$ is not prime.\n | Equality is equivalent to \n$a^2 - ac + c^2 = b^2 + bd + d^2 (1)$.\n Let $ABCD$ be the quadrilateral with $AB = a$, $BC = d$, $CD = b$, $AD = c$, $\angle BAD = 60^\circ$, and $\angle BCD = 120^\circ$. Such a quadrilateral exists by $(1)$ and the Law of Cosines.\n By Strong Form of Ptolemy's Theorem, we find that;\n $B... |
Is it possible to find 100 positive integers not exceeding 25,000, such that all pairwise sums of them are different?\n | The answer is yes.\n First Solution For example, let's say that the integers are, $k, k+a_1, k+a_2, ..., k+a_{99}$. Now this turns into a problem of solving for the $99$ integers $a_i$. This then each ai takes on the form, $j+b_1, j+b_2,..., j+b_{98}$. Then we must find the $98$ $b$ integers. By doing this process over... |
\nWhat is the smallest positive integer $t$ such that there exist integers $x_1,x_2,\ldots,x_t$ with $x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}$? | Observe that $2002^{2002}\equiv 4^{2002}\equiv 64^{667}\cdot 4\equiv 4\pmod{9}$. On the other hand, each cube is congruent to 0, 1, or -1 modulo 9. So a sum of at most three cubes modulo 9 must among $0,\pm 1,\pm 2,\pm 3$ none of which are congruent to 4. Therefore $t\geq 4$.\n \nTo show that 4 is the minimum value of ... |
Circles $\displaystyle S_1$ and $\displaystyle S_2$ intersect at points $\displaystyle P$ and $\displaystyle Q$. Distinct points $\displaystyle A_1$ and $\displaystyle B_1$ (not at $\displaystyle P$ or $\displaystyle Q$) are selected on $\displaystyle S_1$. The lines $\displaystyle A_1P$ and $\displaystyle B_1P$ mee... | We will use directed angles mod $\displaystyle \pi$.\n Since $\displaystyle A_1, B_1, C$ are collinear, $\angle CA_1Q = \angle B_1A_1Q$. Since $\displaystyle A_1, B_1, P, Q$ all lie on $\displaystyle S_1$, $\angle B_1A_1Q = \angle B_1PQ$. Hence, $\angle CA_1Q = \angle B_1PQ$. Similarly, $\angle CA_2Q = \angle B_2PQ$... |
Let $\displaystyle n$ be a positive integer. A sequence of $\displaystyle n$ positive integers (not necessarily distinct) is called full if it satisfies the following conditions: for each positive integers $\displaystyle k \ge 2$, if the number $\displaystyle k$ appears in the sequence then so does the number $\displa... | We claim that there is a bijection between the permutations of the numbers $\displaystyle 1, \ldots , n$ and the full sequences of length $\displaystyle n$.\n To obtain a full sequence $\displaystyle a_1, \ldots a_n$ from a permutation $\displaystyle \sigma$ of the first $\displaystyle n$ positive integers, begin by le... |
(Bulgaria)\nLet $\displaystyle T$ be the set of ordered triples $\displaystyle (x,y,z)$, where $\displaystyle x$, $\displaystyle y$, $\displaystyle z$ are integers with $0 \le x,y,z \le 9$. Players $\displaystyle A$ and $\displaystyle B$ play the following game. Player $\displaystyle A$ chooses a triple $\displaystyl... | In mod 2, we see that\n \n$\displaystyle |x+y -a-b| + |y+z -b-c| + |z+x -c-a| \equiv 2(x+y+z -a-b-c)$,\n\n so the outcome of $\displaystyle B$'s move must always be even. Furthermore, the outcome must be no greater than 54 and no less than 0, so there are at most 28 different possible outcomes per move. Since there a... |
(Hojoo Lee) Let $n \geq 3$ be an integer. Let $t_1, t_2, \dots , t_n$ be positive real numbers such that \n \[n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).\]\n Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with ... | For $n=3$, suppose (for sake of contradiction) that $t_3 = t_2 + t_1 + k$ for $k \ge 0$; then (by Cauchy-Schwarz Inequality)\n \begin{align*}10 &> [2(t_1 + t_2) + k]\left(\frac {1}{t_1} + \frac {1}{t_2} + \frac 1{t_1 + t_2 + k}\right) = 2(t_1+t_2)\left(\frac 1{t_1} + \frac{1}{t_2}\right) + \left(\frac{k}{t_1} + \frac k... |
(Mihai BΔlunΔ, Romania)\nAn infinite sequence $a_0, a_1, a_2, \ldots$ of real numbers satisfies the condition\n \n$a_n = | a_{n+1} - a_{n+2} | \qquad$ for every $n \ge 0$,\n\n with $a_0$ and $a_1$ positive and distinct. Can this sequence be bounded?\n This was also Problem 4 of the 2005 German Pre-TST and Problem 1 of... | Solution 1 We note that each of the $a_i$ must be nonnegative.\n Lemma 1. If the two initial terms of the sequence are nonzero and distinct, then every term of the sequence is nonzero and no two consecutive terms are equal.\n Proof. We proceed by induction; we are given a base case. If ${}a_k \neq a_{k+1}$, then $| a... |
(Canada)\nDoes there exist a function $sΒ : \mathbf{Q} \rightarrow \{ -1,1 \}$ such that if $\displaystyle x$ and $\displaystyle y$ are distinct rational numbers satisfying $\displaystyle {} xy=1$ or $x+y \in \{ 0,1 \}$, then $\displaystyle s(x)s(y) = -1$? Justify your answer.\n | For a number $\displaystyle x$, we define the function $\displaystyle t(x)$ to be 0, if $\displaystyle x$ is an integer, or $\frac{1}{x-\lfloor x \rfloor}$ if $\displaystyle x$ is not an integer. We use the notation $\displaystyle t^n$ to denote $\displaystyle t$ composed $\displaystyle n$ times. We note that if $\di... |
Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab + bc + ca = 0$ we have the following relations\n \[f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c).\]\n Contents\n\n1 Problem\n2 Solution\n\n2.1 Solution 1\n2.2 Solution 2\n\n\n3 See also\n\n | Solution 1 From $b=c=0$, we have $f(a) + f(-a) = 2f(a) \Longrightarrow f(a) = f(-a)$, so $f$ is even, and all the degrees all of its terms are even. Let $\text{deg}\, f(x) = n$\n Let $(a,b,c) = (6x, 3x, -2x)$*; then we have $f(3x) + f(5x) + f(8x) = 2f(7x)$. Comparing lead coefficients, we have $3^n + 5^n + 8^n = 2 \cdo... |
(Puerto Rico)\nThere are 10001 students at an university (sic). Somer students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form societies (a club may belong to different societies. There are a total of $\displaystyle k$ societies. Suppose that the follo... | Solution 1 Let us replace 10001 with an arbitrary odd integer $\displaystyle n$. Suppose that some special student is a member of clubs $C_1, \ldots, C_j$, each with $2m_1 + 1, \ldots, 2m_j + 1$ members (including our special student). We note that for $1 \le i \le j$, no two $\displaystyle C_i$ may contain a common s... |
(Germany)\nLet $\displaystyle n$ and $\displaystyle k$ be positive integers. There are given $\displaystyle n$ circles in the plane. Every two of them intersect at two distinct points, and all points of intersection they determine are pairwise distinct. Each intersection point must be colored with one of $\displayst... | The answer is $\displaystyle k=2$ and $n \le 3$, or $3 \le k \le n$. We must have $\displaystyle k \le n$, since there are $\displaystyle n$ colors.\n If $\displaystyle k=1$, then all points on any circle must be the same color, so all points must be the same color and $\displaystyle n=1$, a contradiction.\n For $\dis... |
(Australia)\nThe following operation is allowed on a finite graph: Choose an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge in that cycle, and delete it from the graph. For a fixed integer $n \ge 4$, find the least number of edges of a graph that can be obtained by repeated applications of th... | Surprisingly, the answer is $\displaystyle n$.\n We first claim that our graph must always stay connected. Indeed, if an edge $\displaystyle AB$ is removed, then it must have been part of a 4-cycle, so there must remain some other way of travelling from $\displaystyle A$ to $\displaystyle B$. Hence the graph must alw... |
(Kazakhstan)\nThe circle $\displaystyle \Gamma$ and the line $\ell$ do not intersect. Let $\displaystyle AB$ be the diameter of $\displaystyle\Gamma$ perpendicular to $\ell$, with $\displaystyle B$ closer to $\ell$ than $\displaystyle A$. An arbitrary point $C \neq A,B$ is chosen on $\displaystyle \Gamma$. The line ... | Solution 1 We use directed angles mod $\displaystyle \pi$.\n Let $\displaystyle CF$ meet $\displaystyle \Gamma$ at $\displaystyle H$. The problem is equivalent to showing that lines $\displaystyle GH$ and $\ell$ are parallel, which happens if and only if $\angle AGH \equiv \angle AFD$. But by cyclic quadrilaterals an... |
(South Korea)\nLet $\displaystyle O$ be the circumcenter of an acute-angled triangle $\displaystyle ABC$ with $\angle B < \angle C$. The line $\displaystyle AO$ meets the side $\displaystyle BC$ at $\displaystyle D$. The circumcenters of the triangles $\displaystyle ABD$ and $\displaystyle ACD$ are $\displaystyle E$ ... | Lemma. In any triangle $\displaystyle ABC$ with circumcenter $\displaystyle O$, the altitude from $\displaystyle A$ is the reflection of $\displaystyle AO$ over the angle bisector of $\displaystyle A$.\n Proof. This is well-known, but we prove it anyway. Let $\displaystyle AO, BO, CO$ meet sides $\displaystyle a,b,c... |
A cyclic quadrilateral $ABCD$ is given. The lines $AD$ and $BC$ intersect at $E$, with $C$ between $B$ and $E$; the diagonals $AC$ and $BD$ intersect at $F$. Let $M$ be the midpoint of the side $CD$, and let $N \neq M$ be a point on the circumcircle of $\triangle ABM$ such that $\frac{AN}{BN} = \frac{AM}{BM}$. Prove th... | Let $P = CD \cap EF$. Let $Q = AB \cap CD$. Let $R = AB \cap EF$. Let$(ABM)$ denote the circumcircle of $\triangle ABM$. Let $N' = EF \cap (ABM)$. Note that $N' = PR \cap (ABM)$\n \nClaim: $P$ is on $(ABM)$. Proof: $(C, D; P, Q) = -1$ as complete quadrilaterals induce harmonic bundles. $QP \cdot QM = QC \cdot QD$ by Le... |
(Russia)\nThe function $\displaystyle \psi$ from the set $\mathbf{N}$ of positive integers to itself is defined by the equality\n \n$\psi(n) = \sum_{k=1}^{n}(k,n), \qquad n \in \mathbf{N}$,\n\n where $\displaystyle (k,n)$ denotes the greatest common divisor of $\displaystyle k$ and $\displaystyle n$.\n a) Prove that $\... | Let $d$ be a divisor of $n$. We note that for $k \leq n/d$, $(dk,n) = d$ if and only if $k$ is relatively prime to $n/d$. It follows that each divisor $d$ of $n$ is found in the sum $\sum_{i=1}^{n}(k,n)$ exactly $\phi(n/d )$ times, where $\phi(m)$ is defined as the number of natural numbers less than or equal to $m$ ... |
Find all monic polynomials $\displaystyle p(x)$ of degree two for which there exists an integer polynomial $\displaystyle q(x)$ such that $\displaystyle p(x)q(x)$ is a polynomial having all coefficients $\pm 1$.\n \nThis was also the last problem of the final round of the 2006 Polish Mathematics Olympiad.\n | Since the constant term of $\displaystyle p(x)q(x)$ is $\pm 1$, and $\displaystyle p(x)$ and $\displaystyle q(x)$ both have integral constant terms, the constant term of $\displaystyle p(x)$ must be $\pm 1$.\n We note that for $\displaystyle |z| \ge 2$, $\displaystyle n \ge 2$ ($n \in \mathbb{N}$), we have\n \n$|z|^n >... |
(Australia)\nA house has an even number of lamps distributed among its rooms in such a way that there are at least three lamps in every room. Each lamp shares a switch with exactly one other lamp, not necessarily from the same room. Each change in the switch shared by two lamps changes their states simultaneously. P... | It is sufficient to prove that we can always reduce the number of monochromatic rooms by one after finitely many steps.\n Suppose there is a monochromatic room. We start in that room, and perform a sequence of moves according to the following rules:\n If we are in a monochromatic room, we toggle a switch connected to ... |
(Iran)\nLet $\displaystyle k$ be a fixed positive integer. A company has a special method to sell sombreros. Each customer can convince two persons to buy a sombrero after he/she buys one; convincing someone already convinced does not count. Each of these new customers can convince two others and so on. If each one... | Suppose $m$ persons receive videos. We wish to prove $n \ge m(k+2)$. Since this is clear for $m=0$, let us WLOG assume that $m$ is positive. Under this assumption, we will now prove the stronger bound\n $n \ge (m+1)(k+1) + m$ by induction on $m$.\n We say a person $A$ is a direct successor of $B$ if $B$ directly c... |
(Iran)\nIn an $m \times n$ rectangular board of $\displaystyle mn$ unit squares, adjacent squares are ones with a common edge, and a path is a sequence of squares in which any two consecutive squares are adjacent. Each square of the board can be coloured black or white. Let $\displaystyle N$ denote the number of colo... | We will call black paths which connect the left edge of the board to the right edge good paths.\n Lemma. Consider a board as described in the problem, and let columns run from top to bottom, and rows run from left to right, with $\displaystyle m$ columns and $\displaystyle n$ rows. We say that one path $\displaystyle... |
(Ukraine)\nLet $\displaystyle ABCD$ be a parallelogram. A variable line $\ell$ passing through the point $\displaystyle A$ intersects the rays $\displaystyle BC$ and $\displaystyle DC$ at points $\displaystyle X$ and $\displaystyle Y$, respectively. Let $\displaystyle K$ and $\displaystyle L$ be the centres of the ex... | Let $\ell_1, \ell_2$ be the interior angle bisectors of $\displaystyle ABX, YAD$. Let $\displaystyle m_1, m_2$ be the exterior angle bisectors of $\displaystyle ABC, CDA$. Then $\displaystyle K$ is the intersection of $\ell_1, m_1$ and $\displaystyle L$ is the intersection of $\ell_2, m_2$.\n ISL2005G3.png Let us d... |
Let $a_1,a_2,\ldots$ be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer $n$ the numbers $a_1,a_2,\ldots,a_n$ leave $n$ different remainders upon division by $n$.\n Prove that every integer occurs exactly once in the sequence $a_1,a_2,\ldots$.\n | It is clear that $a_i=a_j$ if and only if $i=j$, or the sequence would not satisfy the specified property.\n If $|a_i-a_j|\geq \max(i,j)$, then $a_i$ and $a_j$ leave the same remainder when divided by $|a_i-a_j|$, which violates the given condition for the sequence when $n=|a_i-a_j|$. It then follows that $|a_i-a_j|<\m... |
(Mongolia)\nLet $a$, $b$, $c$, $d$, $e$, and $f$ be positive integers. Suppose that the sum $S = a+b+c+d+e+f$ divides both $abc + def$ and $ab+bc+ca - de-ef-fd$. Prove that $S$ is composite.\n This was also Problem 1 of the 2nd 2006 German TST, and a problem at the 2006 Indian IMO Training Camp.\n | For all integers $\displaystyle x$ we have\n \n$(x+a)(x+b)(x+c) \equiv x^3 + (a+b+c)x^2 + (ab+bc+ca)x + abc \equiv x^3 - (d+e+f)x^2 + (de+ef+fd)x - def$ $\equiv (x-d)(x-e)(x-f) \pmod{S}$,\n\n since each coefficient of the first two polynomials is congruent to the corresponding coefficient of the second two polynomials,... |
(Estonia)\nA sequence of real numbers $a_0, a_1, a_2, \dots$ is defined by the formula\n \n$a_{i+1} = \lfloor a_i \rfloor \cdot \langle a_i \rangle$\nfor $i \ge 0$;\n\n here $\displaystyle a_0$ is an arbitrary real number, $\lfloor a_i \rfloor$ denotes the greatest integer not exceeding $\displaystyle a_i$, and ${} \la... | We first note that for all nonnegative integers $\displaystyle i$,\n \n$| a_{i+1} | = | \lfloor a_i \rfloor \cdot \langle a_i \rangle | < | \lfloor a_i \rfloor |$,\n\n so ${} \left| \lfloor a_i \rfloor \right|$ is a non-increasing function of $\displaystyle i$. We also note that if $\displaystyle a_i$ is not positive ... |
(Poland)\nThe sequence of real number $a_0, a_1, a_2, \dots$ is defined recursively by\n \n$\displaystyle a_0 = -1$,\n$\sum_{k=0}^n \frac{a_{n-k}}{k+1} = 0$\nfor $n \ge 1$.\n\n Show that $\displaystyle {} a_n > 0$ for $n \ge 1$.\n This was also Problem 6 of the 2007 Poland Math Olympiad.\n | We proceed by induction on $\displaystyle n$. For the base case, we note that $\displaystyle a_1 = 1/2$. Suppose that $a_1, \dots, a_{n-1}$ are positive. We note that $\displaystyle a_n$ is positive if and only if $\sum_{k=1}^n \frac{a_{n-k}}{k+1}$ is negative. Now, since $a_1, \dots, a_{n-1}$ are all positive, we ... |
Prove the inequality\n\[\sum_{i<j} \frac{a_ia_j}{a_i+a_j} \le \frac{n}{2(a_1 + a_2 + \dotsb a_n)} \sum_{i<j} a_i a_j\]\nfor positive real numbers $a_1, \dotsc, a_n$.\n | Note that\n\[\sum_{i<j} \frac{a_ia_j}{a_i+a_j} = 1/2 \sum_{i\neq j} \frac{a_ia_j}{a_i + a_j} = 1/2 \sum_{j=1}^n \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} .\]\nSuppose that $1 \le k \neq \ell \le n$. Note that $1/(1/a_i + 1/a_j)$ is an increasing function of both $a_i$ and $a_j$. It follows that if $a_k \le a_\ell$, then... |
(Hojoo Lee, South Korea) Let $a,b,c$ be the sides of a triangle. Prove that\n\[\frac{\sqrt{b+c-a}}{\sqrt{b}+ \sqrt{c} - \sqrt{a}} + \frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a} - \sqrt{b}} + \frac{\sqrt{a+b-c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} \le 3.\]\n This problem also appeared on the 2007 IMO TSTs of Italy and Bangladesh.... | Lemma. For any positive reals $x,y,z$,\n\[\sum_{\rm cyc} \left( \frac{y}{x} - 1 \right) \left( \frac{z}{x} -1 \right) \ge 0 .\]\n Proof 1. This is the Vornicu-Schur Inequality on the function $x \mapsto 1/x$. $\blacksquare$\n Proof 2. Without loss of generality, suppose that $x\ge y \ge z$. Evidently,\n\[\left( \fra... |
(France)\nWe have $n \ge 2$ lamps $L_1, \dots, L_n$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamps as follows:\n if the lamp $\displaystyle {} L_i$ and its neighbours (only one neighbour for $\displaystyle i=1$ or $\displaystyle {i=n}$, two neighbours for o... | We suppose that $\displaystyle L_1$ is the leftmost lamp. We note that for $k \le n$, in the $\displaystyle k$th second, $\displaystyle L_k$ is on, and is the rightmost lamp which is on. This follows from induction.\n Lemma. If $2^k \le n$, then in the $\displaystyle 2^k$th second, ${} L_1, \dots, L_{2^k}$ are exactl... |
(Argentina) An $(n,k)$-tournament is a contest with $n$ players held in $k$ rounds such that:\n (i) Each player plays in each round, and every two players meet at most once.\n(ii) If player $A$ meets player $B$ in round $i$, player $C$ meets player $D$ on round $i$, and player $A$ meets player $C$ in round $j$, then pl... | Let $t$ be the greatest integer such that $2^t$ divides $n$. Then there exists an $(n,k)$-tournament if and only if $k \le 2^t - 1$.\n We first prove that if $k \le 2^t - 1$, then there exists an $(n,k)$-tournament. Since we may partition our $n$ players into $n/2^t$ different groups of size $2^t$, it suffices to pro... |
(Ukraine)\nLet $\displaystyle ABCD$ be a trapezoid with parallel sides ${} \displaystyle AB >CD$. Points $\displaystyle K$ and $\displaystyle L$ lie on the line segments $\displaystyle AB$ and $\displaystyle CD$, respectively, so that ${} \displaystyle AK/KB = DL/LC$. Suppose that there are points $\displaystyle P$ a... | Since ${} \displaystyle A,B,K$ and $\displaystyle D,C,L$ are collinear, the condition ${} \displaystyle AK/KB = DL/LC$ is equivalent to the condition that lines $\displaystyle AD$, $\displaystyle KL$, and $\displaystyle BC$ are concurrent. Let $\displaystyle X$ be the point of concurrence.\n Let $\displaystyle \omega_... |
(Canada)\nFor $x \in (0,1)$ let $y \in (0,1)$ be the number whose $\displaystyle n$th digit after the decimal point is the $\displaystyle (2^n)$th digit after the decimal point of $\displaystyle x$. Show that if $\displaystyle x$ is rational then so is $\displaystyle y$.\n | For any real ${ a \in (0,1) }$ and any natural number $\displaystyle n$, let $\displaystyle f_a(n)$ the $\displaystyle n$th digit after the decimal point of $\displaystyle a$. We note that $\displaystyle a$ is rational if and only if $\displaystyle f_a(n)$ is periodic for sufficiently large $\displaystyle n$, i.e., if... |
(New Zealand)\nYou are given a sequence $a_1,a_2,\dots ,a_n$ of numbers. For each $i$ ($1\leq i\leq n$) define\n $d_i=\max\{a_j:1\leq j\leq i\}-\min\{a_j:i\leq j\leq n\}$ and let\n $d=\max\{d_i:1\leq i\leq n\}$. (a) Prove that for arbitrary real numbers $x_1\leq x_2\leq \dots \leq x_n$,\n $\max\{|x_i-a_i|:1\leq i\leq n... | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n Resources |
(Bulgaria)\nConsider those functions $f:\mathbb{N}\to\mathbb{N}$ which satisfy the condition\n $f(m+n)\ge f(m)+f(f(n))-1$ for all $m, n\in\mathbb{N}$. Find all possible values of $f(2007).$\n ($\mathbb{N}$ denotes the set of all integers.)\n | |
(Thailand)\nFind all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that\n\[f(x+f(y)) = f(x+y) + f(y)\]\nfor all $x,y \in \mathbb{R}^+$. (Symbol $\mathbb{R}^+$ denotes the set of all positive real numbers [sic].)\n | We will show that $f(x) = 2x$ is the unique solution to this equation. To this end, let $g(x) = f(x) - x$. The given condition then translates to\n\[g(x+y+ g(y)) + x+y+g(y) = g(x+y) + x+y + g(y) + y ,\]\nor\n\[g(x+y+g(y)) = g(x+y) + y .\]\n Lemma 1. The function $g$ is injective.\n Proof. Suppose $g(a) = g(b)$. The... |
Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves first. On any move of his, Bob replaces the number $n$ on the blackboard with a number of the form $n-a^{2}$, where $a$ is a positive integer. On any move of hers, Amy replaces th... | The answer is in the negative. For a positive integer $n$, we define its square-free part $S(n)$ to be the smallest positive integer $a$ such that $n / a$ is a square of an integer. In other words, $S(n)$ is the product of all primes having odd exponents in the prime expansion of $n$. We also agree that $S(0)=0$. Now w... |
Given any positive real number $\varepsilon$, prove that, for all but finitely many positive integers $v$, any graph on $v$ vertices with at least $(1+\varepsilon) v$ edges has two distinct simple cycles of equal lengths. (Recall that the notion of a simple cycle does not allow repetition of vertices in a cycle.) | Fix a positive real number $\varepsilon$, and let $G$ be a graph on $v$ vertices with at least $(1+\varepsilon) v$ edges, all of whose simple cycles have pairwise distinct lengths.\n\nAssuming $\varepsilon^{2} v \geq 1$, we exhibit an upper bound linear in $v$ and a lower bound quadratic in $v$ for the total number of ... |
Let $a, b, c, d$ be positive integers such that $a d \neq b c$ and $\operatorname{gcd}(a, b, c, d)=1$. Prove that, as $n$ runs through the positive integers, the values $\operatorname{gcd}(a n+b, c n+d)$ may achieve form the set of all positive divisors of some integer. | We extend the problem statement by allowing $a$ and $c$ take non-negative integer values, and allowing $b$ and $d$ to take arbitrary integer values. (As usual, the greatest common divisor of two integers is non-negative.) Without loss of generality, we assume $0 \leq a \leq c$. Let $S(a, b, c, d)=\left\{\operatorname{g... |
Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow... | The required number is $\left(\begin{array}{c}2 n \ n\end{array}\right)$. To prove this, trace the circumference counterclockwise to label the points $a_{1}, a_{2}, \ldots, a_{2 n}$.\n\nLet $\mathcal{C}$ be any good configuration and let $O(\mathcal{C})$ be the set of all points from which arrows emerge. We claim that ... |
Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow... | \nUse the counterclockwise labelling $a_{1}, a_{2}, \ldots, a_{2 n}$ in the solution above.\n\nLetting $D_{n}$ be the number of good configurations on $2 n$ points, we establish a recurrence relation for the $D_{n}$. To this end, let $C_{n}=\frac{(2 n) !}{n !(n+1) !}$ the $n$th Catalan number; it is well-known that $C_... |
Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow... | Let $C_{n}=\frac{1}{n+1}\left(\begin{array}{c}2 n \ n\end{array}\right)$ denote the $n$th Catalan number and recall that there are exactly $C_{n}$ ways to join $2 n$ distinct points on a circumference by $n$ pairwise disjoint chords. Such a configuration of chords will be referred to as a Catalan n-configuration. An or... |
Problem 2. Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying\n\n$$\nP(x)^{10}+P(x)^{9}=Q(x)^{21}+Q(x)^{20} .\n$$ | \nThe answer is in the negative. Comparing the degrees of both sides in $(*)$ we get $\operatorname{deg} P=21 n$ and $\operatorname{deg} Q=10 n$ for some positive integer $n$. Take the derivative of $(*)$ to obtain\n\n$$\nP^{\prime} P^{8}(10 P+9)=Q^{\prime} Q^{19}(21 Q+20) \text {. }\n$$\n\nSince $\operatorname{gcd}(10... |
Problem 2. Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying\n\n$$\nP(x)^{10}+P(x)^{9}=Q(x)^{21}+Q(x)^{20} .\n$$ | \nSolution 2. Letting $r$ and $s$ be integers such that $r \geq 2$ and $s \geq 2 r$, we show that if $P^{r}+P^{r-1}=$ $Q^{s}+Q^{s-1}$, then $Q$ is constant.\n\nLet $m=\operatorname{deg} P$ and $n=\operatorname{deg} Q$. A degree inspection in the given relation shows that $m \geq 2 n$.\n\nWe will prove that $P(P+1)$ has... |
Ann and Bob play a game on an infinite checkered plane making moves in turn; Ann makes the first move. A move consists in orienting any unit grid-segment that has not been oriented before. If at some stage some oriented segments form an oriented cycle, Bob wins. Does Bob have a strategy that guarantees him to win? | \nThe answer is in the negative: Ann has a strategy allowing her to prevent Bob's victory.\n\nWe say that two unit grid-segments form a low-left corner (or LL-corner) if they share an endpoint which is the lowest point of one and the leftmost point of the other. An up-right corner (or UR-corner) is defined similarly. T... |
For a positive integer $a$, define a sequence of integers $x_{1}, x_{2}, \ldots$ by letting $x_{1}=a$ and $x_{n+1}=2 x_{n}+1$ for $n \geq 1$. Let $y_{n}=2^{x_{n}}-1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_{1}, \ldots, y_{k}$ are all prime. | \nThe largest such is $k=2$. Notice first that if $y_{i}$ is prime, then $x_{i}$ is prime as well. Actually, if $x_{i}=1$ then $y_{i}=1$ which is not prime, and if $x_{i}=m n$ for integer $m, n>1$ then $2^{m}-1 \mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \ldots, y_{k}$ are primes ... |
Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? | \nSuch a tester pair exists. We may biject $\mathbb{R}$ with the closed unit interval, so it suffices to find a tester pair for that instead. We give an explicit example: take some positive real numbers $\alpha, \beta$ (which we will specify further later). Take\n\n$$\ng(x)=\max (x-\alpha, 0) \quad \text { and } \quad ... |
Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? | \nLet us agree that a sequence $\mathbf{x}=\left(x_{n}\right)_{n=1,2, \ldots}$ is cofinally non-constant if for every index $m$ there exists an index $n>m$ such that $x_{m} \neq x_{n}$.\n\nBiject $\mathbb{R}$ with the set of cofinally non-constant sequences of 0's and 1's, and define $g$ and $h$ by\n\n$$\ng(\epsilon, \... |
Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? | \nSuppose that $f(\mathbf{x})$ and $\mathbf{x}$ agree for the first $n$ terms, whatever $\mathbf{x}$. Consider any sequence, and write it as $\mathbf{x}=\epsilon, \mathbf{y}$. Without loss of generality, we may (and will) assume that $\epsilon=0$, so $f(\mathbf{x})=0, \mathbf{y}^{\prime}$ by the base case. Yet then $f(... |
Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? | \nWe will show that there exists a tester pair of bijective functions $g$ and $h$.\n\nFirst of all, let us find out when a pair of functions is a tester pair. Let $g, h: \mathbb{R} \rightarrow \mathbb{R}$ be arbitrary functions. We construct a directed graph $G_{g, h}$ with $\mathbb{R}$ as the set of vertices, its edge... |
Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$. The lines $A B$ and $C D$ meet at $P$, the lines $A D$ and $B C$ meet at $Q$, and the diagonals $A C$ and $B D$ meet at $R$. Let $M$ be the midpoint of the segment $P Q$, and let $K$ be the common point of the segment $M R$ and the circle $\omega$. Prove ... | \nLet $O$ be the centre of $\omega$. Notice that the points $P, Q$, and $R$ are the poles (with respect to $\omega$ ) of the lines $Q R, R P$, and $P Q$, respectively. Hence we have $O P \perp Q R, O Q \perp R P$, and $O R \perp P Q$, thus $R$ is the orthocentre of the triangle $O P Q$. Now, if $M R \perp P Q$, then th... |
Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:\n\n$$\nx=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor\n$$\n\nProve that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$ | \nWe prove that the $a_{n}$ are precisely the $k$ th-power-free positive integers, that is, those divisible by the $k$ th power of no prime. The conclusion then follows.\n\nLet $B$ denote the set of all $k$ th-power-free positive integers. We first show that, given a positive integer $c$,\n\n$$\n\sum_{b \in B, b \leq c... |
Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:\n\n$$\nx=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor\n$$\n\nProve that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$ | \nFor every $n=1,2,3, \ldots$, introduce the function\n\n$$\nf_{n}(x)=x-1-\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor\n$$\n\nDenote also by $g_{n}(x)$ the number of the indices $i \leq n$ such that $x / a_{i}$ is the $k$ th power of an integer. Then $f_{n}(x+1)-f_{n}(x)=1-g_{n}(x)$ for every inte... |
\n$2 n$ distinct tokens are placed at the vertices of a regular $2 n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2 n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged e... | \nStep 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2 n$-gon accordingly. Consider any three tokens $i<j<k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactly when two of these... |
\nDoes there exist an infinite sequence of positive integers $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{m}$ and $a_{n}$ are coprime if and only if $|m-n|=1$ ?\n | \nThe answer is in the affirmative.\n\nThe idea is to consider a sequence of pairwise distinct primes $p_{1}, p_{2}, p_{3}, \ldots$, cover the positive integers by a sequence of finite non-empty sets $I_{n}$ such that $I_{m}$ and $I_{n}$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_{n}=\prod_{... |
\nFor an integer $n \geq 5$, two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another counter. A... | \nWe shall prove that the first player wins if and only the exponent of 2 in the prime decomposition of $n-3$ is odd.\n\nSince the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins od Losses according as whether a player faced with that po... |
\nA finite list of rational numbers is written on a blackboard. In an operation, we choose any two numbers $a, b$, erase them, and write down one of the numbers\n\n$$\na+b, a-b, b-a, a \times b, a / b(\text { if } b \neq 0), b / a(\text { if } a \neq 0)\n$$\n\nProve that, for every integer $n>100$, there are only finit... | \nWe prove the problem statement even for all positive integer $n$.\n\nThere are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_{1}, \ldots, x_{n}$ only using the four elementary arithmetic operations, and each $x_{k}$ exactly once. Each such formula for $k>1$ is obtained by an e... |
\nLet $A B C$ be a triangle, let $D$ be the touchpoint of the side $B C$ and the incircle of the triangle $A B C$, and let $J_{b}$ and $J_{c}$ be the incentres of the triangles $A B D$ and $A C D$, respectively. Prove that the circumcentre of the triangle $A J_{b} J_{c}$ lies on the bisectrix of the angle $B A C$.\n | \nLet the incircle of the triangle $A B C$ meet $C A$ and $A B$ at points $E$ and $F$, respectively. Let the incircles of the triangles $A B D$ and $A C D$ meet $A D$ at points $X$ and $Y$, respectively. Then $2 D X=D A+D B-A B=D A+D B-B F-A F=D A-A F$; similarly, $2 D Y=D A-A E=2 D X$. Hence the points $X$ and $Y$ coi... |
\nLet $p \geq 5$ be a prime number. For a positive integer $k$ we denote by $R(k)$ the remainder of $k$ when divided by $p$. Determine all positive integers $a<p$ such that\n\n$$\nm+R(m a)>a\n$$\n\nfor every $m=1,2, \ldots, p-1$.\n | \nThe required integers are $p-1$ along with all the numbers of the form $\lfloor p / q\rfloor, q=$ $2, \ldots, p-1$. In other words, these are $p-1$, along with the numbers $1,2, \ldots,\lfloor\sqrt{p}\rfloor$, and also the (distinct) numbers $\lfloor p / q\rfloor, q=2, \ldots,\left\lfloor\sqrt{p}-\frac{1}{2}\right\rf... |
\nGiven a positive integer $n$, determine the largest real number $\mu$ satisfying the following condition: for every $4 n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greate... | \nThe required maximum is $\frac{1}{2 n+2}$. To show that the condition in the statement is not met if $\mu>\frac{1}{2 n+2}$, let $U=(0,1) \times(0,1)$, choose a small enough positive $\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\left(\frac{i}{n+1} \pm \epsilon\r... |
\nLet $A B C$ be a triangle and let $D$ be a point on the segment $B C, D \neq B$ and $D \neq C$. The circle $A B D$ meets the segment $A C$ again at an interior point $E$. The circle $A C D$ meets the segment $A B$ again at an interior point $F$. Let $A^{\prime}$ be the reflection of $A$ in the line $B C$. The lines $... | \nLet $\sigma$ denote reflection in the line $B C$. Since $\angle B D F=\angle B A C=$ $\angle C D E$, by concyclicity, the lines $D E$ and $D F$ are images of one another under $\sigma$, so the lines $A C$ and $D F$ meet at $P^{\prime}=\sigma(P)$, and the lines $A B$ and $D E$ meet at $Q^{\prime}=\sigma(Q)$. Consequen... |
\nLet $A B C$ be a triangle and let $D$ be a point on the segment $B C, D \neq B$ and $D \neq C$. The circle $A B D$ meets the segment $A C$ again at an interior point $E$. The circle $A C D$ meets the segment $A B$ again at an interior point $F$. Let $A^{\prime}$ be the reflection of $A$ in the line $B C$. The lines $... | \nAs in the first solution, $\sigma$ denotes reflection in the line $B C$, the lines $D E$ and $D F$ are images of one another under $\sigma$, the lines $A C$ and $D F$ meet at $P^{\prime}=\sigma(P)$, and the lines $A B$ and $D E$ meet at $Q^{\prime}=\sigma(Q)$.\n\nLet the line $A D$ meet the circle $A B C$ again at $M... |
\nLet $A B C$ be a triangle and let $D$ be a point on the segment $B C, D \neq B$ and $D \neq C$. The circle $A B D$ meets the segment $A C$ again at an interior point $E$. The circle $A C D$ meets the segment $A B$ again at an interior point $F$. Let $A^{\prime}$ be the reflection of $A$ in the line $B C$. The lines $... | \nAs in the previous solutions, $\sigma$ denotes reflection in the line $B C$. Let the lines $B E$ and $C F$ meet at $X$. Due to the circles $B D E A$ and $C D F A$, we have $\angle X B D=$ $\angle E A D=\angle X F D$, so the quadrilateral $B F X D$ is cyclic; similarly, the quadrilateral $C E X D$ is cyclic. Hence $\a... |
\nA cubic sequence is a sequence of integers given by $a_{n}=n^{3}+b n^{2}+c n+d$, where $b, c$ and $d$ are integer constants and $n$ ranges over all integers, including negative integers.\n\n(a) Show that there exists a cubic sequence such that the only terms of the sequence which are squares of integers are $a_{2015}... | \nThe only possible value of $a_{2015} \cdot a_{2016}$ is 0 . For simplicity, by performing a translation of the sequence (which may change the defining constants $b, c$ and $d$ ), we may instead concern ourselves with the values $a_{0}$ and $a_{1}$, rather than $a_{2015}$ and $a_{2016}$.\n\nSuppose now that we have a ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.