id stringlengths 10 10 | source stringlengths 74 74 | problem stringlengths 57 1.29k | solutions listlengths 0 9 |
|---|---|---|---|
IMO-1993-1 | https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_1 | Let \(f\left(x\right)=x^n+5x^{n-1}+3\), where \(n>1\) is an integer. Prove that \(f\left(x\right)\) cannot be expressed as the product of two non-constant polynomials with integer coefficients. | [
"For the sake of contradiction, assume that \\(f\\left(x\\right)=g\\left(x\\right)h\\left(x\\right)\\) for polynomials \\(g\\left(x\\right)\\) and \\(h\\left(x\\right)\\) in \\(\\mathbb{R}\\). Furthermore, let \\(g\\left(x\\right)=b_mx^m+b_{m-1}x^{m-1}+\\ldots+b_1x+b_0\\) with \\(b_i=0\\) if \\(i>m\\) and \\(h\\lef... |
IMO-1993-2 | https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_2 | Let \(D\) be a point inside acute triangle \(ABC\) such that \(\angle ADB = \angle ACB+\frac{\pi}{2}\) and \(AC\cdot BD=AD\cdot BC\).
(a) Calculate the ratio \(\frac{AB\cdot CD}{AC\cdot BD}\).
(b) Prove that the tangents at \(C\) to the circumcircles of \(\Delta ACD\) and \(\Delta BCD\) are perpendicular. | [
"Let us construct a point \\(B'\\) satisfying the following conditions: \\(B', B\\) are on the same side of AC, \\(BC = B'C\\) and \\(\\angle BCB' = 90^{\\circ}\\).\n\nHence \\(\\triangle ADB \\sim \\triangle ACB'\\).\n\n\\[\n\\implies \\frac{AB}{BD} = \\frac{AB'}{B'C}\n\\]\n\nAlso considering directed angles mod \... |
IMO-1993-3 | https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_3 | On an infinite chessboard, a game is played as follows. At the start, \(n^2\) pieces are arranged on the chessboard in an \(n\) by \(n\) block of adjoining squares, one piece in each square. A move in the game is a jump in a horizontal or vertical direction over an adjacent occupied square to an unoccupied square immed... | [
"The solution as described in the video is that all values of \\(n\\) are valid except when \\(n\\) is divisible by 3. He describes why those values are not valid using modular math and why the other ones are by creating an algorithmic pattern of reducing the matrices by taking out pieces in subarrays of 3 by m.\n\... |
IMO-1993-4 | https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_4 | For three points \(A,B,C\) in the plane, we define \(m(ABC)\) to be the smallest length of the three heights of the triangle \(ABC\), where in the case \(A\), \(B\), \(C\) are collinear, we set \(m(ABC) = 0\). Let \(A\), \(B\), \(C\) be given points in the plane. Prove that for any point \(X\) in the plane,
\(m(ABC) \... | [
"First we prove the claim for all points \\(X\\) within or on the triangle \\(ABC\\): In this specific case, suppose without loss of generality that \\(a\\ge b\\ge c\\). Then the length of the smallest height of \\(ABC\\) is that of the height from vertex \\(A\\). This has a value of \\(\\frac{2[ABC]}{a}\\). Simila... |
IMO-1993-5 | https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_5 | Let \(\mathbb{N} = \{1,2,3, \ldots\}\). Determine if there exists a strictly increasing function \(f: \mathbb{N} \mapsto \mathbb{N}\) with the following properties:
(i) \(f(1) = 2\);
(ii) \(f(f(n)) = f(n) + n, (n \in \mathbb{N})\). | [
"Here is my Solution https://artofproblemsolving.com/community/q2h62193p16226748\n\nFind as ≈ Ftheftics"
] |
IMO-1993-6 | https://artofproblemsolving.com/wiki/index.php/1993_IMO_Problems/Problem_6 | There are \(n\) lamps \(L_0, \ldots , L_{n-1}\) in a circle (\(n > 1\)), where we denote \(L_{n+k} = L_k\). (A lamp at all times is either on or off.) Perform steps \(s_0, s_1, \ldots\) as follows: at step \(s_i\), if \(L_{i-1}\) is lit, switch \(L_i\) from on to off or vice versa, otherwise do nothing. Initially all l... | [] |
IMO-1994-1 | https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_1 | Let \(m\) and \(n\) be two positive integers. Let \(a_1\), \(a_2\), \(\ldots\), \(a_m\) be \(m\) different numbers from the set \(\{1, 2,\ldots, n\}\) such that for any two indices \(i\) and \(j\) with \(1\leq i \leq j \leq m\) and \(a_i + a_j \leq n\), there exists an index \(k\) such that \(a_i + a_j = a_k\). Show th... | [
"Let \\(a_1, a_2, \\dots a_m\\) satisfy the given conditions. We will prove that for all \\(j, 1 \\le j \\le m,\\)\n\n\\[\na_j+a_{m-j+1} \\ge n+1\n\\]\n\nWLOG, let \\(a_1 < a_2 < \\dots < a_m\\). Assume that for some \\(j, 1 \\le j \\le m,\\)\n\n\\[\na_j + a_{m-j+1} \\le n\n\\]\n\nThis implies, for each \\(i, 1 \\l... |
IMO-1994-2 | https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_2 | Let \(ABC\) be an isosceles triangle with \(AB = AC\). \(M\) is the midpoint of \(BC\) and \(O\) is the point on the line \(AM\) such that \(OB\) is perpendicular to \(AB\). \(Q\) is an arbitrary point on \(BC\) different from \(B\) and \(C\). \(E\) lies on the line \(AB\) and \(F\) lies on the line \(AC\) such that \(... | [
"Let \\(E'\\) and \\(F'\\) be on \\(AB\\) and \\(AC\\) respectively such that \\(E'F'\\perp OQ\\). Then, by the first part of the problem, \\(QE'=QF'\\). Hence, \\(Q\\) is the midpoint of \\(EF\\) and \\(E'F'\\), which means that \\(EE'FF'\\) is a parallelogram. Unless \\(E=E'\\) and \\(F=F'\\), this is a contradic... |
IMO-1994-3 | https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_3 | For any positive integer \(k\), let \(f(k)\) be the number of elements in the set \(\{k + 1, k + 2,\dots, 2k\}\) whose base 2 representation has precisely three \(1\)s.
- (a) Prove that, for each positive integer \(m\), there exists at least one positive integer \(k\) such that \(f(k) = m\).
- (b) Determine all positi... | [
"a) Surjectivity of f\n\nFor space-time saving we say that a positive integer is a T-number or simply is T if it has exactly three 1s in his base two representation.\n\nIt's easy to see that (one has to place two 1s in the less n significative bit of a (n+1)-bit number)\n\n- \\(f(2^n) = \\tbinom n2 = \\sum_{i=1}^{n... |
IMO-1994-4 | https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_4 | Find all ordered pairs \((m,n)\) where \(m\) and \(n\) are positive integers such that \(\frac {n^3 + 1}{mn - 1}\) is an integer. | [
"Suppose \\(\\frac{n^3+1}{mn-1}=k\\) where \\(k\\) is a positive integer. Then \\(n^3+1=(mn-1)k\\) and so it is clear that \\(k\\equiv -1\\pmod{n}\\). So, let \\(k=jn-1\\) where \\(j\\) is a positive integer. Then we have \\(n^3+1=(mn-1)(jn-1)=mjn^2-(m+j)n+1\\) which by cancelling out the \\(1\\)s and dividing by \... |
IMO-1994-5 | https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_5 | Let \(S\) be the set of real numbers strictly greater than \(-1\). Find all functions \(f:S \to S\) satisfying the two conditions:
1. \(f(x+f(y)+xf(y)) = y+f(x)+yf(x)\) for all \(x\) and \(y\) in \(S\);
2. \(\frac{f(x)}{x}\) is strictly increasing on each of the intervals \(-1<x<0\) and \(0<x\). | [
"The only solution is \\(f(x) = \\frac{-x}{x+1}.\\)\n\nSetting \\(x=y,\\) we get\n\n\\[\nf(x+f(x)+xf(x)) = x+f(x)+xf(x).\n\\]\n\nTherefore, \\(f(s) = s\\) for \\(s = x+f(x)+xf(x).\\)\n\nNote: If we can show that \\(s\\) is always \\(0,\\) we will get that \\(x+f(x)+xf(x) = 0\\) for all \\(x\\) in \\(S\\) and theref... |
IMO-1994-6 | https://artofproblemsolving.com/wiki/index.php/1994_IMO_Problems/Problem_6 | Show that there exists a set \(A\) of positive integers with the following property: For any infinite set \(S\) of primes there exist two positive integers \(m \in A\) and \(n \not\in A\) each of which is a product of \(k\) distinct elements of \(S\) for some \(k \ge 2\). | [] |
IMO-1995-1 | https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_1 | Let \(A,B,C,D\) be four distinct points on a line, in that order. The circles with diameters \(AC\) and \(BD\) intersect at \(X\) and \(Y\). The line \(XY\) meets \(BC\) at \(Z\). Let \(P\) be a point on the line \(XY\) other than \(Z\). The line \(CP\) intersects the circle with diameter \(AC\) at \(C\) and \(M\), and... | [
"Since \\(M\\) is on the circle with diameter \\(AC\\), we have \\(\\angle AMC=90\\) and so \\(\\angle MCA=90-A\\). We similarly find that \\(\\angle BND=90\\). Also, notice that the line \\(XY\\) is the radical axis of the two circles with diameters \\(AC\\) and \\(BD\\). Thus, since \\(P\\) is on \\(XY\\), we hav... |
IMO-1995-2 | https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_2 | Let \(a, b, c\) be positive real numbers such that \(abc = 1\). Prove that
\[
\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}.
\] | [
"We make the substitution \\(x= 1/a\\), \\(y=1/b\\), \\(z=1/c\\). Then\n\n\\[\n\\begin{align*} \\frac{1}{a^3(b+c)} + \\frac{1}{b^3(c+a)} + \\frac{1}{c^3(a+b)} &= \\frac{x^3}{xyz(1/y+1/z)} + \\frac{y^3}{xyz(1/z+1/x)} + \\frac{z^3}{xyz(1/x+1/z)} \\\\ &= \\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} . \\end{a... |
IMO-1995-3 | https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_3 | Determine all integers \(n>3\) for which there exist \(n\) points \(A_1,\ldots,A_n\) in the plane, no three collinear, and real numbers \(r_1,\ldots,r_n\) such that for \(1\le i<j<k\le n\), the area of \(\triangle A_iA_jA_k\) is \(r_i+r_j+r_k\). | [] |
IMO-1995-4 | https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_4 | The positive real numbers \(x_0, x_1, x_2,.....x_{1994}, x_{1995}\) satisfy the relations
\(x_0=x_{1995}\) and \(x_{i-1}+\frac{2}{x_{i-1}}=2{x_i}+\frac{1}{x_i}\)
for \(i=1,2,3,....1995\)
Find the maximum value that \(x_0\) can have. | [
"First we start by solving for \\(x_{i}\\) in the recursive relation\n\n\\[\nx_{i-1}+\\frac{2}{x_{i-1}}=2x_{i}+\\frac{1}{x_{i}}\n\\]\n\n\\[\n\\frac{x_{i-1}^{2}+2}{x_{i-1}}=\\frac{2x_{i}^{2}+1}{x_{i}}\n\\]\n\n\\[\n\\left( x_{i} \\right)\\left( x_{i-1}^{2}+2 \\right)=\\left( x_{i-1} \\right)\\left( 2x_{i}^{2}+1 \\rig... |
IMO-1995-5 | https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_5 | Let \(ABCDEF\) be a convex hexagon with \(AB=BC=CD\) and \(DE=EF=FA\), such that \(\angle BCD=\angle EFA=\frac{\pi}{3}\). Suppose \(G\) and \(H\) are points in the interior of the hexagon such that \(\angle AGB=\angle DHE=\frac{2\pi}{3}\). Prove that \(AG+GB+GH+DH+HE\ge CF\). | [
"Draw \\(AE\\) and \\(BD\\) to make equilateral \\(\\triangle EFA\\) and \\(\\triangle BCD\\), and draw points \\(I\\) and \\(J\\) such that \\(IA=IB\\), \\(JD=JE\\), directed angle \\(\\measuredangle IAB=-\\measuredangle CDB\\), and directed angle \\(\\measuredangle JDE=-\\measuredangle FAE\\) to make equilateral ... |
IMO-1995-6 | https://artofproblemsolving.com/wiki/index.php/1995_IMO_Problems/Problem_6 | Let \(p\) be an odd prime number. How many \(p\)-element subsets \(A\) of \({1,2,\ldots,2p}\) are there, the sum of whose elements is divisible by \(p\)? | [
"Let \\(A(x,y)\\) be the generating function\n\n\\[\nA(x,y) = (1+yx)(1+yx^2)\\cdots(1+yx^{2p})\n\\]\n\nWe apply the roots of unity filter on \\(x\\) to get\n\n\\[\n\\frac{A(1,y)+A(w,y)+\\cdots+A(w^{p-1},y)}{p} = \\frac{(1+y)^{2p}+(p-1)(1+yw)\\cdots(1+yw^{2p})}{p}\n\\]\n\nWe call this function on \\(y\\), \\(B(y)\\)... |
IMO-1996-1 | https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_1 | We are given a positive integer \(r\) and a rectangular board \(ABCD\) with dimensions \(|AB|=20\), \(|BC|=12\). The rectangle is divided into a grid of \(20 \times 12\) unit squares. The following moves are permitted on the board: one can move from one square to another only if the distance between the centers of the ... | [
"First we define the rectangular board in the cartesian plane with centers of the unit squares as integer coordinates and the following coordinates for the squares at the corners of \\(A\\), \\(B\\), \\(C\\), \\(D\\), as follows: \\(A=(1,1)\\), \\(B=(20,1)\\), \\(C=(20,12)\\), \\(D=(1,12)\\)\n\nLet \\((x_i,y_i)\\) ... |
IMO-1996-2 | https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_2 | Let \(P\) be a point inside triangle \(ABC\) such that
\[
\angle APB-\angle ACB = \angle APC-\angle ABC
\]
Let \(D\), \(E\) be the incenters of triangles \(APB\), \(APC\), respectively. Show that \(AP\), \(BD\), \(CE\) meet at a point. | [
"\\[\n[asy] import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.124923887131423, xmax = 11.886638474419073, ymin = -8.067061524000941, y... |
IMO-1996-3 | https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_3 | Let \(S\) denote the set of nonnegative integers. Find all functions \(f\) from \(S\) to itself such that
\(f(m+f(n))=f(f(m))+f(n)\) \(\forall m,n \in S\) | [
"Plugging in m = 0, we get f(f(n)) = f(n) \\(\\forall n \\in S\\). With m = n = 0, we get f(0) = 0.\n\nIf there are no fixed points of this function greater than \\(0\\), then \\(f(x) = 0 \\forall x \\in \\mathbb{N}\\), which is a valid solution.\n\nLet \\(n_{0}\\) be the smallest fixed point of \\(f(x)\\) such tha... |
IMO-1996-4 | https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_4 | The positive integers \(a\) and \(b\) are such that the numbers \(15a+16b\) and \(16a-15b\) are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares? | [
"Let us first set up two equations given the information in the problem: Let \\(15a + 16b = m^2\\) Let \\(16a - 15b = n^2\\)\n\nIn general, when we are given alternating sums and differences like this, it is a good idea to square both equations and add them up to cancel out any cross terms. Doing this results in th... |
IMO-1996-5 | https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_5 | Let \(ABCDEF\) be a convex hexagon such that \(AB\) is parallel to \(DE\), \(BD\) is parallel to \(EF\), and \(CD\) is parallel to \(FA\). Let \(R_{A}\), \(R_{C}\), \(R_{E}\) denote the circumradii of triangles \(FAB\), \(BCD\), \(DEF\), respectively, and let \(P\) denote the perimeter of the hexagon. Prove that
\[
R_... | [
"Let \\(s_{1}=\\left| AB \\right|,\\;s_{2}=\\left| BC \\right|,\\;s_{3}=\\left| CD \\right|,\\;s_{4}=\\left| DE \\right|,\\;s_{5}=\\left| EF \\right|,\\;s_{6}=\\left| FA \\right|\\)\n\nLet \\(d_{1}=\\left| FB \\right|,\\;d_{2}=\\left| BD \\right|,\\;d_{1}=\\left| DF \\right|\\)\n\nLet \\(\\alpha_{1}=\\angle FAB,\\;... |
IMO-1996-6 | https://artofproblemsolving.com/wiki/index.php/1996_IMO_Problems/Problem_6 | Let \(p, q, n\) be three positive integers with \(p+q<n\). Let \((x_0,x_1,\cdots ,x_n)\) be an \((n+1)\)-tuple of integers satisfying the following conditions:
(i) \(x_0=x_n=0\);
(ii) For each \(i\) with \(1 \le i \le n\), either \(x_i-x_{i-1}=p\) or \(x_i-x_{i-1}=-q\).
Show that there exists indices \(i<j\) with \(... | [] |
IMO-1997-1 | https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_1 | In the plane the points with integer coordinates are the vertices of unit squares. The squares are colored alternatively black and white (as on a chessboard).
For any pair of positive integers \(m\) and \(n\), consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths \(m\) and... | [
"For any pair of positive integers \\(m\\) and \\(n\\), consider a rectangle \\(ABCD\\) whose vertices have integer coordinates and whose legs, of lengths \\(m\\) and \\(n\\), lie along edges of the squares.\n\nLet \\(A\\), \\(B\\), \\(C\\), and \\(D\\), be the lower left vertex, lower right vertex, upper right ver... |
IMO-1997-2 | https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_2 | The angle at \(A\) is the smallest angle of triangle \(ABC\). The points \(B\) and \(C\) divide the circumcircle of the triangle into two arcs. Let \(U\) be an interior point of the arc between \(B\) and \(C\) which does not contain \(A\). The perpendicular bisectors of \(AB\) and \(AC\) meet the line \(AU\) and \(V\) ... | [] |
IMO-1997-3 | https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_3 | Let \(x_{1}\), \(x_{2}\),...,\(x_{n}\) be real numbers satisfying the conditions
\[
|x_{1}+x_{2}+...+x_{n}|=1
\]
and
\(|x_{i}| \le \frac{n+1}{2}\), for \(i=1,2,...,n\)
Show that there exists a permutation \(y_{1}\), \(y_{2}\),...,\(y_{n}\) of \(x_{1}\), \(x_{2}\),...,\(x_{n}\) such that
\[
|y_{1}+2y_{2}+...+ny_{n}... | [] |
IMO-1997-4 | https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_4 | An \(n \times n\) matrix whose entries come from the set \(S={1,2,...,2n-1}\) is called a \(\textit{silver}\) matrix if, for each \(i=1,2,...,n\), the \(i\)th row and the \(i\)th column together contain all elements of \(S\). Show that
(a) there is no \(\textit{silver}\) matrix for \(n=1997\);
(b) \(\textit{silver}\)... | [] |
IMO-1997-5 | https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_5 | Find all pairs \((a,b)\) of integers \(a,b \ge 1\) that satisfy the equation
\[
a^{b^{2}}=b^{a}
\] | [
"Case 1: \\((1 \\le a \\le b)\\)\n\n\\[\n(a^{b})^{b}=b^{a}\n\\]\n\nLooking at this expression since \\(b \\ge a\\) then \\(a^{b} \\le b\\).\n\nHere we look at subcase \\(a>1\\) which gives \\(a^{b}>b\\) for all \\((1 < a \\le b)\\). This contradicts condition \\(a^{b} \\le b\\), and thus \\(a\\) can't be more than ... |
IMO-1997-6 | https://artofproblemsolving.com/wiki/index.php/1997_IMO_Problems/Problem_6 | For each positive integer \(n\), let \(f(n)\) denote the number of ways of representing \(n\) as a sum of powers of \(2\) with nonnegative integer exponents. Representations which differ only in the ordering of their summands are considered to be the same. For instance, \(f(4)=4\), because the number 4 can be represent... | [] |
IMO-1998-1 | https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_1 | In the convex quadrilateral \(ABCD\), the diagonals \(AC\) and \(BD\) are perpendicular and the opposite sides \(AB\) and \(DC\) are not parallel. Suppose that the point \(P\), where the perpendicular bisectors of \(AB\) and \(DC\) meet, is inside \(ABCD\). Prove that \(ABCD\) is a cyclic quadrilateral if and only if t... | [
"First, let's prove that if ABCD is cyclic, then the triangles ABP and CDP have equal areas. The circumcenter of ABCD is the intersection of perpendicular bisectors of the sides of the quadrilateral ABCD, so P is the circumcenter. From this, we derive that PC=PB=PA=PD. Since the diagonals AC and BD are perpendicula... |
IMO-1998-2 | https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_2 | In a competition, there are \(a\) contestants and \(b\) judges, where \(b\ge3\) is an odd integer. Each judge rates each contestant as either “pass” or “fail”. Suppose \(k\) is a number such that, for any two judges, their ratings coincide for at most \(k\) contestants. Prove that \(\frac{k}{a}\ge\frac{b-1}{2b}\). | [
"Let \\(c_i\\) stand for the number of judges who pass the \\(i\\)th candidate. The number of pairs of judges who agree on the \\(i\\)th contestant is then given by\n\n\\begin{align*} {c_i \\choose 2} + {{b - c_i} \\choose 2} &= \\frac{1}{2}\\left(c_i(c_i - 1) + (b - c_i)(b - c_i - 1) \\right) \\\\ &= \\frac{1}{2}\... |
IMO-1998-3 | https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_3 | For any positive integer \(n\), let \(d(n)\) denote the number of positive divisors of \(n\) (including 1 and \(n\) itself). Determine all positive integers \(k\) such that \(d(n^2)/d(n) = k\) for some \(n\). | [
"First we must \\(d\\)etermine gener\\(a\\)l values for \\(d(n)\\): Let \\(n=p1 ^ a1 * p2 ^ a2 * .. * pc ^ ac\\), if \\(d\\) is an ar\\(b\\)itr\\(a\\)ry divisor of \\(n\\) then \\(d\\) must have the same prime factors of \\(n\\), each with an exponent \\(b_i\\) being: \\(0\\leq b_i\\leq a_i\\). Hence there are \\(A... |
IMO-1998-4 | https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_4 | Determine all pairs \((a, b)\) of positive integers such that \(ab^{2} + b + 7\) divides \(a^{2}b + a + b\). | [
"We use the division algorithm to obtain \\(ab^2+b+7 \\mid 7a-b^2\\) Here \\(7a-b^2=0\\) is a solution of the original statement, possible when \\(a=7k^2\\) and \\(b=7k\\) where \\(k\\) is any natural number. This is easily verified.\n\nOtherwise we obtain the inequality (by basic properties of divisiblity): \\(7a-... |
IMO-1998-5 | https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_5 | Let \(I\) be the incenter of triangle \(ABC\). Let the incircle of \(ABC\) touch the sides \(BC\), \(CA\), and \(AB\) at \(K\), \(L\), and \(M\), respectively. The line through \(B\) parallel to \(MK\) meets the lines \(LM\) and \(LK\) at \(R\) and \(S\), respectively. Prove that angle \(RIS\) is acute. | [
"Denote the length of the side MB with x. Denote the angle BKM by a and angle BAC by 2b. Using angle chasing and trigonometry, we can derive: RB = x*cos(b)/cos(a-b); BS = x*cos(a-b)/cos(b); RM = x*sin(a)/cos(a-b); SK = x*sin(a)/cos(b). Let's prove that RI^2 + IS^2 > RS^2 because this will imply that angle RIS is ac... |
IMO-1998-6 | https://artofproblemsolving.com/wiki/index.php/1998_IMO_Problems/Problem_6 | Determine the least possible value of \(f(1998),\) where \(f:\Bbb{N}\to \Bbb{N}\) is a function such that for all \(m,n\in {\Bbb N}\),
\[
f\left( n^{2}f(m)\right) =m\left( f(n)\right) ^{2}.
\] | [] |
IMO-1999-1 | https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_1 | Determine all finite sets \(S\) of at least three points in the plane which satisfy the following condition:
For any two distinct points \(A\) and \(B\) in \(S\), the perpendicular bisector of the line segment \(AB\) is an axis of symmetry of \(S\). | [
"Upon reading this problem and drawing some points, one quickly realizes that the set \\(S\\) consists of all the vertices of any regular polygon.\n\nNow to prove it with some numbers:\n\nLet \\(S=\\left\\{ P_{0},P_{1},P_{2},...,P_{n-1} \\right\\}\\), with \\(n\\ge 3\\), where \\(P_{i}\\) is a vertex of a polygon w... |
IMO-1999-2 | https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_2 | Let \(n \geq 2\) be a fixed integer.
- (a) Find the least constant \(C\) such that for all nonnegative real numbers \(x_1, \dots, x_n\),
\[
\sum_{1\leq i<j \leq n} x_ix_j (x_i^2 + x_j^2) \leq C \left( \sum_{i=1}^n x_i \right)^4.
\]
- (b) Determine when equality occurs for this value of \(C\). | [
"The answer is \\(C=1/8\\), and equality holds exactly when two of the \\(x_i\\) are equal to each other and all the other \\(x_i\\) are zero. We prove this by induction on the number of nonzero \\(x_i\\).\n\nFirst, suppose that at most two of the \\(x_i\\), say \\(x_a\\) and \\(x_b\\), are nonzero. Then the left-h... |
IMO-1999-3 | https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_3 | Consider an \(n \times n\) square board, where \(n\) is a fixed even positive integer. The board is divided into \(n^{2}\) units squares. We say that two different squares on the board are adjacent if they have a common side.
\(N\) unit squares on the board are marked in such a way that every square (marked or unmarke... | [] |
IMO-1999-4 | https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_4 | Determine all pairs \((n,p)\) of positive integers such that
\(p\) is a prime, \(n\) not exceeded \(2p\), and \((p-1)^{n}+1\) is divisible by \(n^{p-1}\) | [
"Clearly we have the solutions \\((1,p)\\) and \\((2,2)\\), and for every other solution \\(p \\geq 3\\). It remains to find the solutions \\((x,p)\\) with \\(x \\geq 2\\) and \\(p \\geq 3\\). We claim that in this case \\(x\\) is divisible by p and \\(x y 2p\\), whence \\(x = p\\). This will lead to\n\n$\n\n\\[\np... |
IMO-1999-5 | https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_5 | Two circles \(G_{1}\) and \(G_{2}\) are contained inside the circle \(G\), and are tangent to \(G\) at the distinct points \(M\) and \(N\), respectively. \(G_{1}\) passes through the center of \(G_{2}\). The line passing through the two points of intersection of \(G_{1}\) and \(G_{2}\) meets \(G\) at \(A\) and \(B\). T... | [] |
IMO-1999-6 | https://artofproblemsolving.com/wiki/index.php/1999_IMO_Problems/Problem_6 | Determine all functions \(f:\Bbb{R}\to \Bbb{R}\) such that
\[
f(x-f(y))=f(f(y))+xf(y)+f(x)-1
\]
for all real numbers \(x,y\). | [
"Let \\(f(0) = c\\). Substituting \\(x = y = 0\\), we get:\n\n\\[\nf(-c) = f(c) + c - 1. \\hspace{1cm} ... (1)\n\\]\n\nNow if c = 0, then:\n\n\\[\nf(0) = f(0) - 1\n\\]\n\nwhich is not possible.\n\n\\(\\implies c \\neq 0\\).\n\nNow substituting \\(x = f(y)\\), we get\n\n\\[\nc = f(x) + x^{2} + f(x) - 1\n\\]\n\n.\... |
IMO-2000-1 | https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_1 | Two circles \(G_1\) and \(G_2\) intersect at two points \(M\) and \(N\). Let \(AB\) be the line tangent to these circles at \(A\) and \(B\), respectively, so that \(M\) lies closer to \(AB\) than \(N\). Let \(CD\) be the line parallel to \(AB\) and passing through the point \(M\), with \(C\) on \(G_1\) and \(D\) on \(G... | [
"\\(\\textbf{Proof of problem:}\\) Let ray \\(NM\\) intersect \\(AB\\) at \\(X\\). By our lemma, \\(\\textit{(the two circles are tangent to AB)}\\), \\(X\\) bisects \\(AB\\). Since \\(\\triangle{NAX}\\) and \\(\\triangle{NPM}\\) are similar, and \\(\\triangle{NBX}\\) and \\(\\triangle{NQM}\\) are similar implies \... |
IMO-2000-2 | https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_2 | Let \(a, b, c\) be positive real numbers with \(abc=1\). Show that
\[
\left( a-1+\frac{1}{b} \right)\left( b-1+\frac{1}{c} \right)\left( c-1+\frac{1}{a} \right) \le 1
\] | [
"There exist positive reals \\(x\\), \\(y\\), \\(z\\) such that \\(a = \\frac{x}{y}\\), \\(b = \\frac{y}{z}\\), \\(c = \\frac{z}{x}\\). The inequality then rewrites as\n\n\\[\n\\left(\\frac{x-y+z}{y}\\right)\\left(\\frac{y-z+x}{z}\\right)\\left(\\frac{z-x+y}{x}\\right)\\leq 1\n\\]\n\nor\n\n\\[\n(x-y+z)(y-z+x)(z-x+y... |
IMO-2000-3 | https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_3 | Let \(n \ge 2\) be a positive integer and \(\lambda\) a positive real number. Initially there are \(n\) fleas on a horizontal line, not all at the same point. We define a move as choosing two fleas at some points \(A\) and \(B\) to the left of \(B\), and letting the flea from \(A\) jump over the flea from \(B\) to the ... | [] |
IMO-2000-4 | https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_4 | A magician has one hundred cards numbered \(1\) to \(100\). He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card.
A member of the audience selects two of the three boxes, chooses one card from each and announces the sum of the numbers on the chosen cards. Gi... | [
"Consider \\(1\\), \\(2\\) and \\(3\\). If they are in different boxes, then \\(4\\) must be in the same box as \\(1\\), \\(5\\) in the same box as \\(2\\) and so on. This leads to the solution where all numbers congruent to each other mod \\(3\\) are in the same box.\n\nSuppose \\(1\\) and \\(2\\) are in box \\(A\... |
IMO-2000-5 | https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_5 | Does there exist a positive integer \(n\) such that \(n\) has exactly 2000 prime divisors and \(n\) divides \(2^n + 1\)? | [
"THIS SOLUTION IS WRONG.. ANOTHER SOLUTION IS NEEDED..\n\nLet \\(N=2^n+1\\). We will assume for the sake of contradiction that \\(n|N\\).\n\n\\(2^n+1 \\equiv 0 \\pmod{n} \\Rightarrow 2^n \\equiv -1 \\pmod{n}\\). So 2 does not divide \\(n\\), and so \\(n\\) is odd.\n\nSelect an arbitrary prime factor of \\(n\\) and ... |
IMO-2000-6 | https://artofproblemsolving.com/wiki/index.php/2000_IMO_Problems/Problem_6 | Let \(\overline{AH_1}\), \(\overline{BH_2}\), and \(\overline{CH_3}\) be the altitudes of an acute triangle \(ABC\). The incircle \(\omega\) of triangle \(ABC\) touches the sides \(BC\), \(CA\), and \(AB\) at \(T_1\), \(T_2\), and \(T_3\), respectively. Consider the reflections of the lines \(H_1H_2\), \(H_2H_3\), and ... | [] |
IMO-2001-1 | https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_1 | Consider an acute triangle \(\triangle ABC\). Let \(P\) be the foot of the altitude of triangle \(\triangle ABC\) issuing from the vertex \(A\), and let \(O\) be the circumcenter of triangle \(\triangle ABC\). Assume that \(\angle C \geq \angle B+30^{\circ}\). Prove that \(\angle A+\angle COP < 90^{\circ}\). | [
"Take \\(D\\) on the circumcircle with \\(AD \\parallel BC\\). Notice that \\(\\angle CBD = \\angle BCA\\), so \\(\\angle ABD \\ge 30^\\circ\\). Hence \\(\\angle AOD \\ge 60^\\circ\\). Let \\(Z\\) be the midpoint of \\(AD\\) and \\(Y\\) the midpoint of \\(BC\\). Then \\(AZ \\ge R/2\\), where \\(R\\) is the radius o... |
IMO-2001-2 | https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_2 | Let \(a,b,c\) be positive real numbers. Prove that \(\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1\). | [
"Firstly, \\(a^{2}+8bc=(a^{2}+2bc)+6bc \\leq^{AG} (a^{2}+b^{2}+c^{2})+6bc=S+6bc\\) (where \\(S=a^{2}+b^{2}+c^{2}\\)) and its cyclic variations. Next note that \\((a,b,c)\\) and \\(\\left( \\frac{1}{\\sqrt{S+6bc}}, \\frac{1}{\\sqrt{S+6ca}}, \\frac{1}{\\sqrt{S+6ab}} \\right)\\) are similarly oriented sequences. Thus\... |
IMO-2001-3 | https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_3 | 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... | [
"For each girl, we know that there is a boy who solved a problem in common with her. Since a girl solves at most six problems, for a girl and her set of six problems there are at least 11 boys who solved a problem in common with the girl such that at least three boys solved that common problem. (This is a simple ap... |
IMO-2001-4 | https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_4 | Let \(n_1, n_2, \dots , n_m\) be integers where \(m>1\) is odd. Let \(x = (x_1, \dots , x_m)\) denote a permutation of the integers \(1, 2, \cdots , m\). Let \(f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m\). Show that for some distinct permutations \(a\), \(b\) the difference \(f(a) - f(b)\) is a multiple of \(m!\). | [
"Notice that if \\(\\{0,1,2,...,m!-1\\}\\not=\\{f(1),f(2),f(3),...,f(m!)\\}\\pmod{m!}\\) then by the pigeon hole princible, there must be some \\(a,b\\in\\{1,2,...,m!\\}\\) such that \\(f(a)\\equiv f(b)\\pmod{m!}\\) as desired. Thus, we must prove that \\(\\{0,1,2,...,m!-1\\}\\not=\\{f(1),f(2),f(3),...,f(m!)\\}\\pm... |
IMO-2001-5 | https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_5 | \(ABC\) is a triangle. \(X\) lies on \(BC\) and \(AX\) bisects angle \(A\). \(Y\) lies on \(CA\) and \(BY\) bisects angle \(B\). Angle \(A\) is \(60^{\circ}\). \(AB + BX = AY + YB\). Find all possible values for angle \(B\). | [
"\\[\n[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); pair A = (0,sqrt(3)), D= (-1, 0), E=(1,0); pair Bb = rotate(40,E)*A; pair B = extension(A,D,E,Bb); pair H = foot(A,D,E); pair X = extension(A,H,B,E); pair Yy = bisectorpoint(A,B,E); pair Y =extension(A,E,B,Yy); pair C = E - (... |
IMO-2001-6 | https://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_6 | \(K > L > M > N\) are positive integers such that \(KM + LN = (K + L - M + N)(-K + L + M + N)\). Prove that \(KL + MN\) is not prime. | [
"First, \\((KL+MN)-(KM+LN)=(K-N)(L-M)>0\\) as \\(K>N\\) and \\(L>M\\). Thus, \\(KL+MN>KM+LN\\).\n\nSimilarly, \\((KM+LN)-(KN+LM)=(K-L)(M-N)>0\\) since \\(K>L\\) and \\(M>N\\). Thus, \\(KM+LN>KN+LM\\).\n\nPutting the two together, we have\n\n\\[\nKL+MN>KM+LN>KN+LM\n\\]\n\nNow, we have:\n\n\\[\n(K+L-M+N)(-K+L+M+N)=KM... |
IMO-2002-1 | https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_1 | \(S\) is the set of all \((h,k)\) with \(h,k\) non-negative integers such that \(h + k < n\). Each element of \(S\) is colored red or blue, so that if \((h,k)\) is red and \(h' \le h,k' \le k\), then \((h',k')\) is also red. A type \(1\) subset of \(S\) has \(n\) blue elements with different first member and a type \(2... | [
"Consider the points \\(x+y < n\\) where \\(x,y \\geq 0\\) in the cartesian plane. Let \\((j_1,k_1),...,(j_l,k_l)\\) be the maximal red points. That is, neither \\((j_i+1,k_i)\\) nor \\((j_i,k_i+1)\\) is red for all \\(i\\). We may assume WLOG \\(j_1 < j_2 < \\cdots < j_l\\). Note then \\(k_1 > k_2 > \\cdots > k_l\... |
IMO-2002-2 | https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_2 | \(BC\) is a diameter of a circle center \(O\). \(A\) is any point on the circle with \(\angle AOC \not\le 60^\circ\). \(EF\) is the chord which is the perpendicular bisector of \(AO\). \(D\) is the midpoint of the minor arc \(AB\). The line through \(O\) parallel to \(AD\) meets \(AC\) at \(J\). Show that \(J\) is the ... | [] |
IMO-2002-3 | https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_3 | Find all pairs of positive integers \(m,n \ge 3\) for which here exist infinitely many positive integers \(a\) such that
\[
\frac{a^m+a-1}{a^n+a^2-1}
\]
is itself an integer | [] |
IMO-2002-4 | https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_4 | Problem: Let \(n>1\) be an integer and let \(1=d_{1}<d_{2}<d_{3} \cdots <d_{r}=n\) be all of its positive divisors in increasing order. Show that
\[
d=d_1d_2+d_2d_3+ \cdots +d_{r-1}d_r <n^2
\] | [
"We proceed with two parts:\n\n## Part 1: Proof of Inequality\n\nWe have:\n\n\\[\n\\sum_{i=1}^{r-1} d_i d_{i+1} = d_1d_2 + d_2d_3 + \\cdots + d_{r-1}d_r\n\\]\n\nFor each term \\( d_i d_{i+1} \\), note that \\( d_{i+1} \\leq \\frac{n}{d_i} \\) (since divisors come in pairs). Thus:\n\n\\[\nd_i d_{i+1} \\leq d_i \\cdo... |
IMO-2002-5 | https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_5 | Find all functions \(f:\Bbb{R}\to \Bbb{R}\) such that
\[
(f(x)+f(z))(f(y)+f(t))=f(xy-zt)+f(xt+yz)
\]
for all real numbers \(x,y,z,t\). | [
"Given the problem \\( (f(x) + f(y))(f(u) + f(v)) = f(xu - yv) + f(xv - yu) \\), we aim to find a function that satisfies it.\n\nWe start by considering the case when \\( x = y = u = v = 0 \\). This leads us to \\( 4f(0)^2 = 2f(0) \\), implying \\( f(0) = 0 \\) or \\( f(0) = 1/2 \\).\n\nIf \\( f(0) = 1 \\), then pu... |
IMO-2002-6 | https://artofproblemsolving.com/wiki/index.php/2002_IMO_Problems/Problem_6 | Let \(n \ge 3\) be a positive integer. Let \(C_1,C_2,...,C_n\) be unit circles in the plane, with centers \(O_1,O_2,...,O_n\) respectively. If no line meets more than two of the circles, prove that
\[
\sum_{1\le i< j \le n}^{}\frac{1}{O_iO_j}\le\frac{(n-1)\pi}{4}
\] | [] |
IMO-2003-1 | https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_1 | \(S\) is the set \(\{1, 2, 3, \dots ,1000000\}\). Show that for any subset \(A\) of \(S\) with \(101\) elements we can find \(100\) distinct elements \(x_i\) of \(S\), such that the sets \(\{a + x_i \mid a \in A\}\) are all pairwise disjoint. | [
"Consider the set \\(D=\\{x-y \\mid x,y \\in A\\}\\). There are at most \\(101 \\times 100 + 1 = 10101\\) elements in \\(D\\). Two sets \\(A + t_i\\) and \\(A + t_j\\) have nonempty intersection if and only if \\(t_i - t_j\\) is in \\(D\\). So we need to choose the \\(100\\) elements in such a way that we do not us... |
IMO-2003-2 | https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_2 | Determine all pairs of positive integers \((a,b)\) such that
\[
\frac{a^2}{2ab^2-b^3+1}
\]
is a positive integer. | [
"The only solutions are of the form \\((a,b) = (2n,1)\\), \\((a,b) = (n,2n)\\), and \\((8n^4-n,2n)\\) for any positive integer \\(n\\).\n\nFirst, we note that when \\(b=1\\), the given expression is equivalent to \\(a/2\\), which is an integer if and only if \\(a\\) is even.\n\nNow, suppose that \\((a,b)\\) is a so... |
IMO-2003-3 | https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_3 | Each pair of opposite sides of convex hexagon has the property that the distance between their midpoints is \(\frac{\sqrt{3}}{2}\) times the sum of their lengths. Prove that the hexagon is equiangular. | [] |
IMO-2003-4 | https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_4 | Let \(ABCD\) be a cyclic quadrilateral. Let \(P\), \(Q\), and \(R\) be the feet of perpendiculars from \(D\) to lines \(\overline{BC}\), \(\overline{CA}\), and \(\overline{AB}\), respectively. Show that \(PQ=QR\) if and only if the bisectors of angles \(ABC\) and \(ADC\) meet on segment \(\overline{AC}\). | [
"Clearly \\(PQR\\) is the Simson Line and \\(APDQ\\), \\(BPDR\\), \\(CQDR\\) is cyclic. By angle chasing we have \\(\\triangle DPQ\\sim\\triangle DBC\\), \\(\\triangle DQR\\sim\\triangle DAB\\). Then by \\(PQ=QR\\) we have \\(\\frac{DC}{CB}=\\frac{DQ}{QP}=\\frac{DQ}{QR}=\\frac{DA}{AB}\\). Rearranging and using the ... |
IMO-2003-5 | https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_5 | Let \(n\) be a positive integer and let \(x_1 \le x_2 \le \cdots \le x_n\) be real numbers. Prove that
\[
\left( \sum_{i=1}^{n}\sum_{j=i}^{n} |x_i-x_j|\right)^2 \le \frac{2(n^2-1)}{3}\sum_{i=1}^{n}\sum_{j=i}^{n}(x_i-x_j)^2
\]
with equality if and only if \(x_1, x_2, ..., x_n\) form an arithmetic sequence. | [
"We have \\begin{align*}\\left(\\sum_{i,j=1}^{n}|x_i-x_j|\\right)^2 &=\\left(2\\sum_{1\\le i\\le j\\le n}(x_j-x_i)\\right)^2 \\\\ &= \\left((2n-2)x_n+(2n-6)x_{n-1}+\\dots +(2-2n)x_1\\right)^2 \\\\ &\\le ((2n-2)^2+(2n-6)^2+(2n-10)^2+\\dots + (2-2n)^2)(x_1^2+x_2^2+\\dots + x_n^2) \\\\ &= \\frac{4(n-1)(n)(n+1)}{3}(x_1... |
IMO-2003-6 | https://artofproblemsolving.com/wiki/index.php/2003_IMO_Problems/Problem_6 | Let \(p\) be a prime number. Prove that there exists a prime number \(q\) such that for every integer \(n\), the number \(n^p-p\) is not divisible by \(q\). | [
"Let N be \\(1 + p + p^2 + ... + p^{p-1}\\) which equals \\(\\frac{p^p-1}{p-1}\\) \\(N\\equiv{p+1}\\pmod{p^2}\\) Which means there exists q which is a prime factor of n that doesn't satisfy \\(q\\equiv{1}\\pmod{p^2}\\). \\\\unfinished",
"For \\(p\\) prime and \\(gcd(n, p) = 1\\), \\(n^{p}\\equiv{n}\\pmod{p}\\) si... |
IMO-2004-1 | https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_1 | Let \(ABC\) be an acute-angled triangle with \(AB\neq AC\). The circle with diameter \(BC\) intersects the sides \(AB\) and \(AC\) at \(M\) and \(N\) respectively. Denote by \(O\) the midpoint of the side \(BC\). The bisectors of the angles \(\angle BAC\) and \(\angle MON\) intersect at \(R\). Prove that the circumcirc... | [
"\\[\n[asy] unitsize(2.5cm); size(60); pair A,B,C,O,K,M,N,R,D; C=(0,0); B=(3.2,0); A=(1.1,3.35); O=(1.6,0); K=(1.51,0); R=(1.42,0.73); M=(2.3,1.44); N=(0.31,0.95); D=(1.1, 2.02); draw(arc((1.6,0),1.6,0,180)); draw(A--B--C--cycle); draw(M--N); draw(A--K); draw(O--D); draw(R--M); draw(C--R); draw(B--R); draw(N--R); d... |
IMO-2004-2 | https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_2 | 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
\[
f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c).
\] | [
"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\nLet \\((a,b,c) = (6x, 3x, -2x)\\)*; then we have \\(f(3x) + f(5x) + f(8x) = 2f(7x)\\). Comparing lead coefficients, we have \\(... |
IMO-2004-3 | https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_3 | Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.
\[
[asy] unitsize(0.5 cm); draw((0,0)--(1,0)); draw((0,1)--(1,1)); draw((2,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3... | [] |
IMO-2004-4 | https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_4 | Let \(n \geq 3\) be an integer. Let \(t_1, t_2, \dots , t_n\) be positive real numbers such that
\[
n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).
\]
Show that \(t_i\), \(t_j\), \(t_k\) are side lengths of a triangle for all \(i\), \(j\), \(k\) wi... | [
"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\n\\[\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}\\righ... |
IMO-2004-5 | https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_5 | In a convex quadrilateral \(ABCD\), the diagonal \(BD\) bisects neither the angle \(ABC\) nor the angle \(CDA\). The point \(P\) lies inside \(ABCD\) and satisfies
\[
\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.
\]
Prove that \(ABCD\) is a cyclic quadrilateral if and only if \(AP = CP.\) | [
"Assume \\(ABCD\\) is cyclic, let \\(K\\) be the intersection of \\(AC\\) and \\(BE\\), let \\(L\\) be the intersection of \\(AC\\) and \\(DF\\),\n\n\\[\n[asy] size(6cm); draw(circle((0,0),7.07)); draw((-3.7,-6)-- (3.7,-6)); draw((-6.8,-2)-- (6.8,-2)); draw((-5,5)-- (5,5)); draw((-5,5)-- (-3.7,-6)); draw((-5,5)-- (... |
IMO-2004-6 | https://artofproblemsolving.com/wiki/index.php/2004_IMO_Problems/Problem_6 | We call a positive integer alternating if every two consecutive digits in its decimal representation are of different parity. Find all positive integers \(n\) which have an alternating multiple. | [
"We claim that all positive integers \\(n\\) except multiples of \\(20\\) have a multiple that is alternating. If \\(n\\) is a multiple of \\(20\\), then the units digit is \\(0\\) and the tens digit is a multiple of \\(2\\), so both digits are even. Now, we will prove that if \\(20\\nmid n\\), then there is a mult... |
IMO-2005-1 | https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_1 | Six points are chosen on the sides of an equilateral triangle \(ABC\): \(A_1, A_2\) on \(BC\), \(B_1\), \(B_2\) on \(CA\) and \(C_1\), \(C_2\) on \(AB\), such that they are the vertices of a convex hexagon \(A_1A_2B_1B_2C_1C_2\) with equal side lengths. Prove that the lines \(A_1B_2, B_1C_2\) and \(C_1A_2\) are concurr... | [
"Let \\(D = C_2A_1 \\cap A_2B_1\\), and similarly define \\(E\\) and \\(F\\). We claim that \\(DEF\\) is equilateral.\n\nProof: Note that the vectors \\(\\overrightarrow{A_1A_2}\\), \\(\\overrightarrow{A_2B_1}\\), \\(\\overrightarrow{B_1B_2}\\), \\(\\overrightarrow{B_2C_1}\\), \\(\\overrightarrow{C_1C_2}\\), \\(\\o... |
IMO-2005-2 | https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_2 | Let \(a_1, a_2, \dots\) be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer \(n\) the numbers \(a_1, a_2, \dots, a_n\) leave \(n\) different remainders upon division by \(n\). Prove that every integer occurs exactly once in the sequence. | [
"\\({a_n}\\) satisfies the conditions if and only if \\({a_n-a_1}\\) does. Therefore we can assume that \\(a_1=0\\)\n\nFirst of all, \\(|a_n|<n\\) Otherwise, \\(a_n = 0\\) mod \\(a_{a_n}\\)\n\nClaim: \\(a_{k+1}\\) is either the smallest positive number not in \\({a_1, a_2, ..., a_k}\\) or the largest negative numbe... |
IMO-2005-3 | https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_3 | Let \(x, y, z > 0\) satisfy \(xyz\ge 1\). Prove that
\[
\frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{x^2+y^5+z^2} + \frac{z^5-z^2}{x^2+y^2+z^5} \ge 0.
\] | [] |
IMO-2005-4 | https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_4 | Determine all positive integers relatively prime to all the terms of the infinite sequence
\[
a_n=2^n+3^n+6^n -1,\ n\geq 1.
\] | [
"Let \\(k\\) be a positive integer that satisfies the given condition.\n\nFor all primes \\(p>3\\), by Fermat's Little Theorem, \\(n^{p-1} \\equiv 1\\pmod p\\) if \\(n\\) and \\(p\\) are relatively prime. This means that \\(n^{p-3} \\equiv \\frac{1}{n^2} \\pmod p\\). Plugging \\(n = p-3\\) back into the equation, w... |
IMO-2005-5 | https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_5 | Let \(ABCD\) be a fixed convex quadrilateral with \(BC = DA\) and \(BC \nparallel DA\). Let two variable points \(E\) and \(F\) lie of the sides \(BC\) and \(DA\), respectively, and satisfy \(BE = DF\). The lines \(AC\) and \(BD\) meet at \(P\), the lines \(BD\) and \(EF\) meet at \(Q\), the lines \(EF\) and \(AC\) mee... | [] |
IMO-2005-6 | https://artofproblemsolving.com/wiki/index.php/2005_IMO_Problems/Problem_6 | In a mathematical competition, in which 6 problems were posed to the participants, every two of these problems were solved by more than 2/5 of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 problems each. | [] |
IMO-2006-1 | https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_1 | Let \(ABC\) be triangle with incenter \(I\). A point \(P\) in the interior of the triangle satisfies \(\angle PBA+\angle PCA = \angle PBC+\angle PCB\). Show that \(AP \geq AI\), and that equality holds if and only if \(P=I.\) | [
"We have\n\n\\[\n\\angle IBP = \\angle IBC - \\angle PBC = \\frac{1}{2} \\angle ABC - \\angle PBC = \\frac{1}{2}(\\angle PCB - \\angle PCA).\n\\]\n\nand similarly\n\n\\[\n\\angle ICP = \\angle PCB - \\angle ICB = \\angle PCB - \\frac{1}{2} \\angle ACB = \\frac{1}{2}(\\angle PBA - \\angle PBC).\n\\]\n\nSince \\(\\an... |
IMO-2006-2 | https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_2 | Let \(P\) be a regular 2006-gon. A diagonal of \(P\) is called good if its endpoints divide the boundary of \(P\) into two parts, each composed of an odd number of sides of \(P\). The sides of \(P\) are also called good. Suppose \(P\) has been dissected into triangles by 2003 diagonals, no two of which have a common po... | [
"Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity. Let ABC be an iso-odd triangle, with AB and BC odd sides. This means that there are an odd number of sid... |
IMO-2006-3 | https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_3 | Determine the least real number \(M\) such that the inequality
\[
\left| ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})\right|\leq M(a^{2}+b^{2}+c^{2})^{2}
\]
holds for all real numbers \(a,b\) and \(c\). | [] |
IMO-2006-4 | https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_4 | Determine all pairs \((x, y)\) of integers such that
\[
1+2^{x}+2^{2x+1}= y^{2}.
\] | [
"If \\((x,y)\\) is a solution then obviously \\(x\\geq 0\\) and \\((x,-y)\\) is a solution too. For \\(x=0\\) we get the two solutions \\((0,2)\\) and \\((0,-2)\\).\n\nNow let \\((x,y)\\) be a solution with \\(x > 0\\); without loss of generality confine attention to \\(y > 0\\). The equation rewritten as\n\n\\[\n2... |
IMO-2006-5 | https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_5 | Let \(P(x)\) be a polynomial of degree \(n>1\) with integer coefficients, and let \(k\) be a positive integer. Consider the polynomial \(Q(x) = P( P ( \ldots P(P(x)) \ldots ))\), where \(P\) occurs \(k\) times. Prove that there are at most \(n\) integers \(t\) such that \(Q(t)=t\). | [
"We use the notation \\(P^k(x)\\) for \\(Q(x)\\).\n\nLemma 1. The problem statement holds for \\(k=2\\).\n\nProof. Suppose that \\(a_1, \\dotsc, a_k\\), \\(b_1, \\dotsc, b_k\\) are integers such that \\(P(a_j) = b_j\\) and \\(P(b_j) = a_j\\) for all indices \\(j\\). Let the set \\(\\{ a_1, \\dotsc, a_k, b_1, \\dots... |
IMO-2006-6 | https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_6 | Assign to each side \(b\) of a convex polygon \(P\) the maximum area of a triangle that has \(b\) as a side and is contained in \(P\). Show that the sum of the areas assigned to the sides of \(P\) is at least twice the area of \(P\). | [] |
IMO-2007-1 | https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_1 | Real numbers \(a_1, a_2, \dots , a_n\) are given. For each \(i\) (\(1\le i\le n\)) define
\[
d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}
\]
and let
\[
d=\max\{d_i:1\le i\le n\}.
\]
(a) Prove that, for any real numbers \(x_1\le x_2\le \cdots\le x_n\),
\[
\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} (*)
\... | [
"Since \\(d_i=\\max\\{a_j:1\\le j\\le i\\}-\\min\\{a_j:i\\le j\\le n\\}\\), all \\(d_i\\) can be expressed as \\(a_p-a_q\\) where \\(1\\le p\\le i\\le q \\le n\\). Thus \\(d\\) can be expressed as \\(a_p-a_q\\) for some \\(p\\) and \\(q\\) with \\(1\\le p\\le q\\le n\\)\n\nLemma: \\(d\\ge 0\\).\n\nSince \\(a_p \\ge... |
IMO-2007-2 | https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_2 | Consider five points \(A,B,C,D\), and \(E\) such that \(ABCD\) is a parallelogram and \(BCED\) is a cyclic quadrilateral. Let \(\ell\) be a line passing through \(A\). Suppose that \(\ell\) intersects the interior of the segment \(DC\) at \(F\) and intersects line \(BC\) at \(G\). Suppose also that \(EF=EG=EC\). Prove ... | [
"\\[\n[asy] import cse5; import graph; import olympiad; dotfactor = 3; unitsize(1.5inch); path circle = Circle(origin, 1); draw(circle); pair D = (-sqrt(3)/2, -0.5), C = (sqrt(3)/2, -0.5); //G = bisectorpoint(C, B, D); pair Ee = rotate(38,C)*D; pair E = IP(C--Ee, circle,1); pair Gg = rotate(76,C)*D; path circle2 ... |
IMO-2007-3 | https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_3 | In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size. Given that, in this competition, the ... | [] |
IMO-2007-4 | https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_4 | In \(\triangle ABC\) the bisector of \(\angle{BCA}\) intersects the circumcircle again at \(R\), the perpendicular bisector of \(BC\) at \(P\), and the perpendicular bisector of \(AC\) at \(Q\). The midpoint of \(BC\) is \(K\) and the midpoint of \(AC\) is \(L\). Prove that the triangles \(RPK\) and \(RQL\) have the sa... | [
"\\(\\angle{RQL} = 90+\\angle{QCL} = 90+\\dfrac{C}{2}\\), and similarly \\(\\angle{RPK} = 90+\\angle{PCK} = 90+\\dfrac{C}{2}\\). Therefore, \\(\\angle{RQL} = \\angle{RPK}\\). Using the triangle area formula \\(A = \\dfrac{1}{2}bc\\sin{\\angle{A}}\\) yields \\(RQ \\cdot QL = RP \\cdot PK = \\dfrac{PK}{QL} = \\dfrac{... |
IMO-2007-5 | https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_5 | Let \(a\) and \(b\) be positive integers. Show that if \(4ab-1\) divides \((4a^2-1)^2\), then \(a=b\). | [
"Lemma. If there is a counterexample for some value of \\(a\\), then there is a counterexample \\((a,b)\\) for this value of \\(a\\) such that \\(b<a\\).\n\nProof. Suppose that \\(b >a\\). Note that \\(4ab -1 \\equiv -1 \\pmod{4a}\\), but \\((4a^2-1)^2 \\equiv 1 \\pmod{4a}\\). It follows that \\((4a^2-1)^2/(4ab-1) ... |
IMO-2007-6 | https://artofproblemsolving.com/wiki/index.php/2007_IMO_Problems/Problem_6 | Let \(n\) be a positive integer. Consider
\[
S=\{(x,y,z)~:~x,y,z\in \{0,1,\ldots,n \},~x+y+z>0\}
\]
as a set of \((n+1)^3-1\) points in three-dimensional space. Determine the smallest possible number of planes, the union of which contain \(S\) but does not include \((0,0,0)\). | [
"We will prove the result using the following Lemma, which has an easy proof by induction.\n\nLemma Let \\(S_1 = \\{0, 1, \\ldots, n_1\\}\\), \\(S_2 = \\{0, 1, \\ldots, n_2\\}\\) and \\(S_3 = \\{0, 1, \\ldots, n_3\\}\\). If \\(P\\) is a polynomial in \\(\\mathbb{R}[x, y, z]\\) that vanishes on all points of the gri... |
IMO-2008-1 | https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_1 | An acute-angled triangle \(ABC\) has orthocentre \(H\). The circle passing through \(H\) with centre the midpoint of \(BC\) intersects the line \(BC\) at \(A_1\) and \(A_2\). Similarly, the circle passing through \(H\) with centre the midpoint of \(CA\) intersects the line \(CA\) at \(B_1\) and \(B_2\), and the circle ... | [
"Let \\(M_A\\), \\(M_B\\), and \\(M_C\\) be the midpoints of sides \\(BC\\), \\(CA\\), and \\(AB\\), respectively. It's not hard to see that \\(M_BM_C\\parallel BC\\). We also have that \\(AH\\perp BC\\), so \\(AH \\perp M_BM_C\\). Now note that the radical axis of two circles is perpendicular to the line connectin... |
IMO-2008-2 | https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_2 | (i) If \(x\), \(y\) and \(z\) are three real numbers, all different from \(1\), such that \(xyz = 1\), then prove that \(\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\). (With the \(\sum\) sign for cyclic summation, this inequality could be ... | [
"Consider the transormation \\(f:\\mathbb{R}/\\{1\\} \\rightarrow \\mathbb{R}/\\{-1\\}\\) defined by \\(f(u) = \\frac{u}{1-u}\\) and put \\(\\alpha = f(x), \\beta = f(y), \\gamma = f(z)\\). Since \\(f\\) is also one-to one from \\(\\mathbb{Q}/\\{1\\}\\) to \\(\\mathbb{Q}/\\{-1\\}\\), the problem is equivalent to sh... |
IMO-2008-3 | https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_3 | Prove that there are infinitely many positive integers \(n\) such that \(n^{2} + 1\) has a prime divisor greater than \(2n + \sqrt {2n}\). | [
"Solution 1\n\nThe main idea is to take a gaussian prime \\(a+bi\\) and multiply it by a \"twice as small\" \\(c+di\\) to get \\(n+i\\). The rest is just making up the little details.\n\nFor each sufficiently large prime \\(p\\) of the form \\(4k+1\\), we shall find a corresponding \\(n\\) such that \\(p\\) divides... |
IMO-2008-4 | https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_4 | Find all functions \(f: (0, \infty) \mapsto (0, \infty)\) (so \(f\) is a function from the positive real numbers) such that
\[
\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}
\]
for all positive real numbers \(w,x,y,z,\) satisfying \(wx = yz.\) | [
"Considering \\(w=1\\) and \\(z=y=\\sqrt{x}\\) which satisfy the constraint \\(wx=yz\\) we get the following equation:\n\n\\[\n\\frac{(f(1))^2 + (f(x))^2}{f(x) + f(x)} = \\frac{1+x^2}{x+x} \\Leftrightarrow x((f(1))^2 + (f(x))^2) = (1+x^2)f(x)\n\\]\n\nAt once considering \\(x=1\\) we get \\((f(1))^2 = f(1)\\) and kn... |
IMO-2008-5 | https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_5 | Let \(n\) and \(k\) be positive integers with \(k \geq n\) and \(k - n\) an even number. Let \(2n\) lamps labelled \(1\), \(2\), ..., \(2n\) be given, each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from... | [
"For convenience, let \\(A\\) denote the set \\((1,2,\\ldots n)\\) and \\(B\\) the set \\((n+1,n+2,\\ldots,2n)\\).\n\nWe can describe each sequences of switching the lamps as a \\(k\\)-dimensional vector \\((a_1, a_2, \\ldots, a_k)\\), where \\(a_i \\in A \\cup B\\) signifies which lamp was switched on the \\(i\\)-... |
IMO-2008-6 | https://artofproblemsolving.com/wiki/index.php/2008_IMO_Problems/Problem_6 | Let \(ABCD\) be a convex quadrilateral with \(BA \ne BC\). Denote the incircles of triangles \(ABC\) and \(ADC\) by \(\omega _1\) and \(\omega _2\) respectively. Suppose that there exists a circle \(\omega\) tangent to ray \(BA\) beyond \(A\) and to the ray \(BC\) beyond \(C\), which is also tangent to the lines \(AD\)... | [
"Here are some hints:\n\nLet B be the top vertex of triangle ABC, O and K are the centers of the incircles of triangles ABC and ADC with radii R and r, respectively. S is the center of the circumcircle tangential to the extensions of AB, AC and DC. And let\n\nE be the foot of the projection of O to AB. U be the foo... |
IMO-2009-1 | https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_1 | Let \(n\) be a positive integer and let \(a_1,\ldots,a_k (k\ge2)\) be distinct integers in the set \(\{1,\ldots,n\}\) such that \(n\) divides \(a_i(a_{i+1}-1)\) for \(i=1,\ldots,k-1\). Prove that \(n\) doesn't divide \(a_k(a_1-1)\). | [
"Let \\(n=pq\\) such that \\(p\\mid a_1\\) and \\(q\\mid a_2-1\\). Suppose \\(n\\) divides \\(a_k(a_1-1)\\). Note \\(q\\mid a_2-1\\) implies \\((q,a_2)=1\\) and hence \\(q\\mid a_3-1\\). Similarly one has \\(q\\mid a_i-1\\) for all \\(i\\)'s, in particular, \\(p\\mid a_1\\) and \\(q\\mid a_1-1\\) force \\((p,q)=1\\... |
IMO-2009-2 | https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_2 | Let \(ABC\) be a triangle with circumcentre \(O\). The points \(P\) and \(Q\) are interior points of the sides \(CA\) and \(AB\) respectively. Let \(K,L\) and \(M\) be the midpoints of the segments \(BP,CQ\) and \(PQ\), respectively, and let \(\Gamma\) be the circle passing through \(K,L\) and \(M\). Suppose that the l... | [
"## Diagram\n\n\\[\n[asy] dot(\"O\", (50, 38), N); dot(\"A\", (40, 100), N); dot(\"B\", (0, 0), S); dot(\"C\", (100, 0), S); dot(\"Q\", (24, 60), W); dot(\"P\", (52, 80), E); dot(\"L\", (62, 30), SE); dot(\"M\", (38, 70), N); dot(\"K\", (27, 42), W); draw((100, 0)--(24, 60), dotted); draw((0, 0)--(52, 80), dashed);... |
IMO-2009-3 | https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_3 | Suppose that \(s_1,s_2,s_3,\ldots\) is a strictly increasing sequence of positive integers such that the subsequences
\[
s_{s_1},s_{s_2},s_{s_3},\ldots
\]
and
\[
s_{s_1+1},s_{s_2+1},s_{s_3+1},\ldots
\]
are both arithmetic progressions. Prove that the sequence \(s_1,s_2,s_3,\ldots\) is itself an arithmetic progressi... | [
"then\n\ni.s.w.\n\nput S(3) = b, S(2) = a, S(1) = k\n\ntherefore, every S(n) is an arithmetic sequence. Q.E.D."
] |
IMO-2009-4 | https://artofproblemsolving.com/wiki/index.php/2009_IMO_Problems/Problem_4 | Let \(ABC\) be a triangle with \(AB=AC\). The angle bisectors of \(\angle CAB\) and \(\angle ABC\) meet the sides \(BC\) and \(CA\) at \(D\) and \(E\), respectively. Let \(K\) be the incenter of triangle \(ADC\). Suppose that \(\angle BEK=45^\circ\). Find all possible values of \(\angle CAB\). | [
"Extend \\(CK\\) to meet \\(BE\\) at \\(I\\). Then, we can see that \\(I\\) is the incenter of \\(\\triangle ABC\\), so \\(IM=ID\\), where \\(M\\) is the intersection of the incircle with \\(\\overline{AC}\\).\n\nSince \\(CI\\) bisects \\(\\angle ACB\\), we have \\(\\triangle IDC \\cong \\triangle IMC\\), so \\(\\a... |
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