id stringlengths 10 10 | source stringlengths 74 74 | problem stringlengths 57 1.29k | solutions listlengths 0 9 |
|---|---|---|---|
IMO-1975-3 | https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_3 | On the sides of an arbitrary triangle \(ABC\), triangles \(ABR, BCP, CAQ\) are constructed externally with \(\angle CBP = \angle CAQ = 45^\circ, \angle BCP = \angle ACQ = 30^\circ, \angle ABR = \angle BAR = 15^\circ\). Prove that \(\angle QRP = 90^\circ\) and \(QR = RP\). | [
"Consider \\(X\\) and \\(Y\\) so that \\(\\triangle CQA\\sim \\triangle CPX\\) and \\(\\triangle CPB\\sim \\triangle CQY\\). Furthermore, let \\(AX\\) and \\(BY\\) intersect at \\(F\\). Now, this means that \\(\\angle PBC = 45 = \\angle QAC = \\angle PXC\\) and \\(\\angle PCB = 15 = \\angle QCA = \\angle PCX\\), so... |
IMO-1975-4 | https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_4 | When \(4444^{4444}\) is written in decimal notation, the sum of its digits is \(A\). Let \(B\) be the sum of the digits of \(A\). Find the sum of the digits of \(B\). (\(A\) and \(B\) are written in decimal notation.) | [
"Note that\n\n\\[\n4444^{4444}<10000^{4444}=\\left(10^4\\right)^{4444}=10^{17776}\n\\]\n\nTherefore \\(4444^{4444}\\) has fewer than 17776 digits. This shows that \\(A<9\\cdot 17776=159984\\). The sum of the digits of \\(A\\) is then maximized when \\(A=99999\\), so \\(B\\leq 45\\). Note that out of all of the posi... |
IMO-1975-5 | https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_5 | Determine, with proof, whether or not one can find \(1975\) points on the circumference of a circle with unit radius such that the distance between any two of them is a rational number. | [
"Since there are infinitely many primitive Pythagorean triples, there are infinitely many angles \\(\\theta\\) s.t. \\(\\sin\\theta, \\cos\\theta\\) are both rational. Call such angles good. By angle-sum formulas, if \\(a,b\\) are good, then \\(a+b,a-b\\) are also good.\n\nFor points \\(A,B\\) on the circle \\(\\om... |
IMO-1975-6 | https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_6 | Find all polynomials \(P\), in two variables, with the following properties:
(i) for a positive integer \(n\) and all real \(t, x, y\)
\[
P(tx, ty) = t^nP(x, y)
\]
(that is, \(P\) is homogeneous of degree \(n\)),
(ii) for all real \(a, b, c\),
\[
P(b + c, a) + P(c + a, b) + P(a + b, c) = 0,
\]
(iii)
\[
P(1, 0) =... | [
"(i) If \\(n = 0\\) : Clearly no solution (ii) If \\(n = 1\\) : \\(P(x, y) = ax+by \\implies\\) the identification yields directly \\(P(x,y) = x-2y\\) (iii) If \\(n > 1\\), \\(a = 1, \\ b = 1, \\ c =-2 \\implies P(2,-2)+P(-1, 1)+P(-1, 1) = 0\\) \\(\\implies ((-2)^{n}+2) P(-1, 1) = 0 \\implies P(-1, 1) = 0 \\implies... |
IMO-1976-1 | https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_1 | In a convex quadrilateral (in the plane) with the area of \(32 \text{ cm}^{2}\) the sum of two opposite sides and a diagonal is \(16 \text{ cm}\). Determine all the possible values that the other diagonal can have. | [
"Label the vertices \\(A\\), \\(B\\), \\(C\\), and \\(D\\) in such a way that \\(AB + BD + DC = 16\\), and \\(\\overline{BD}\\) is a diagonal.\n\nThe area of the quadrilateral can be expressed as \\(BD \\cdot ( d_1 + d_2 ) / 2\\), where \\(d_1\\) and \\(d_2\\) are altitudes from points \\(A\\) and \\(C\\) onto \\(\... |
IMO-1976-2 | https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_2 | Let \(P_{1}(x) = x^{2} - 2\) and \(P_{j}(x) = P_{1}(P_{j - 1}(x))\) for \(j= 2,\ldots\) Prove that for any positive integer n the roots of the equation \(P_{n}(x) = x\) are all real and distinct. | [
"I shall prove by induction that \\(P_n(x)\\) has \\(2^n\\) distinct real solutions, where \\(2^{n-1}\\) are positive and \\(2^{n-1}\\) are negative. Also, for ever root \\(r\\), \\(|r|<2\\).\n\nClearly, \\(P_1(x)\\) has 2 real solutions, where 1 is positive and 1 is negative. The absolute values of these two solut... |
IMO-1976-3 | https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_3 | A box whose shape is a parallelepiped can be completely filled with cubes of side \(1.\) If we put in it the maximum possible number of cubes, each of volume \(2\), with the sides parallel to those of the box, then exactly \(40\) percent from the volume of the box is occupied. Determine the possible dimensions of the b... | [
"We name a,b,c the sides of the parallelepiped, which are positive integers. We also put\n\n\\[\n\\begin{align*} x &= \\left\\lfloor\\frac{a}{\\sqrt[3]{2}}\\right\\rfloor \\\\ y &= \\left\\lfloor\\frac{b}{\\sqrt[3]{2}}\\right\\rfloor \\\\ z &= \\left\\lfloor\\frac{c}{\\sqrt[3]{2}}\\right\\rfloor \\\\ \\end{align*}\... |
IMO-1976-4 | https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_4 | Determine the greatest number, who is the product of some positive integers, and the sum of these numbers is \(1976.\) | [
"Since \\(3*3=2*2*2+1\\), 3's are more efficient than 2's. We try to prove that 3's are more efficient than anything:\n\nLet there be a positive integer \\(x\\). If \\(3\\) is more efficient than \\(x\\), then \\(x^3<3^x\\). We try to prove that all integers greater than 3 are less efficient than 3:\n\nWhen \\(x\\)... |
IMO-1976-5 | https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_5 | We consider the following system with \(q = 2p\):
\[
\begin{matrix} a_{11}x_{1} + \ldots + a_{1q}x_{q} = 0, \\ a_{21}x_{1} + \ldots + a_{2q}x_{q} = 0, \\ \ldots , \\ a_{p1}x_{1} + \ldots + a_{pq}x_{q} = 0, \\ \end{matrix}
\]
in which every coefficient is an element from the set \(\{ - 1,0,1\}\)\(.\) Prove that there ... | [
"First of all note that we have \\((q + 1)^q - 1\\) possible nonzero vectors \\((x_1,\\cdots,x_q)\\) such that \\(0\\leq x_i\\leq q\\) are integers.\n\nBut \\(a_{j1}x_1 + \\cdots + a_{jq}x_q\\) can only assume \\(q^2 + 1\\) different values, because if it is maximized/minimized by \\((M_1,M_2,\\cdots,M_q)/(m_1,m_2,... |
IMO-1976-6 | https://artofproblemsolving.com/wiki/index.php/1976_IMO_Problems/Problem_6 | A sequence \((u_{n})\) is defined by
\[
u_{0} = 2 \quad u_{1} = \frac {5}{2}, u_{n + 1} = u_{n}(u_{n - 1}^{2} - 2) - u_{1} \ \text{ for } n = 1,\cdots
\]
Prove that for any positive integer \(n\) we have
\[
\lfloor u_{n} \rfloor = 2^{\frac {(2^{n} - ( - 1)^{n})}{3}}
\]
(where \(\lfloor x\rfloor\) denotes the smalle... | [
"Let the sequence \\((x_n)_{n \\geq 0}\\) be defined as\n\n\\[\nx_{0}=0,x_{1}=1, x_{n}=x_{n-1}+2x_{n-2}\n\\]\n\nWe notice\n\n\\[\nx_n=\\frac{2^n-(-1)^n}{3}\n\\]\n\nBecause the roots of the characteristic polynomial \\(x_{n}=x_{n-1}+2x_{n-2}\\) are \\(-1\\) and \\(2\\).\n\nWe also see \\(\\frac{2^1-(-1)^1}{3}=1=x_1\... |
IMO-1977-1 | https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_1 | In the interior of a square \(ABCD\) we construct the equilateral triangles \(ABK, BCL, CDM, DAN.\) Prove that the midpoints of the four segments \(KL, LM, MN, NK\) and the midpoints of the eight segments \(AK, BK, BL, CL, CM, DM, DN, AN\) are the 12 vertices of a regular dodecagon. | [
"Just use complex numbers, with \\(a = 1\\), \\(b = i\\), \\(c = - 1\\) and \\(d = - i\\). With some calculations, we have \\(k = \\frac {\\sqrt {3} - 1}{2}( - 1 - i)\\), \\(l = \\frac {\\sqrt {3} - 1}{2}(1 - i)\\), \\(m = \\frac {\\sqrt {3} - 1}{2}(1 + i)\\) and \\(n = \\frac {\\sqrt {3} - 1}{2}( - 1 + i)\\). Now ... |
IMO-1977-2 | https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_2 | In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence. | [
"Let \\(x_1,x_2,\\ldots\\) be the given sequence and let \\(s_n=x_1+x_2+\\ldots+x_n\\). The conditions from the hypothesis can be now written as \\(s_{n+7}<s_n\\) and \\(s_{n+11}>s_n\\) for all \\(n\\ge 1\\). We then have: \\(0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6<s_{17}<s_{10}<s_3<s_{14}<... |
IMO-1977-3 | https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_3 | Let \(n\) be a given number greater than 2. We consider the set \(V_n\) of all the integers of the form \(1 + kn\) with \(k = 1, 2, \ldots\) A number \(m\) from \(V_n\) is called indecomposable in \(V_n\) if there are not two numbers \(p\) and \(q\) from \(V_n\) so that \(m = pq.\) Prove that there exist a number \(r \... | [
"Lemma: there are \\(\\infty\\) many prime numbers \\(p \\not\\equiv 1 \\mod n\\).\n\nProof:\n\nAssume that there are only finitely many of them and let their product be \\(k\\). Then all prime factors of \\(nk - 1\\) must be \\(\\equiv 1 \\mod n\\) since this number is coprime with \\(k\\). But that would mean tha... |
IMO-1977-4 | https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_4 | Let \(a,b,A,B\) be given reals. We consider the function defined by
\[
f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x).
\]
Prove that if for any real number \(x\) we have \(f(x) \geq 0\) then \(a^2 + b^2 \leq 2\) and \(A^2 + B^2 \leq 1.\) | [
"\\(f(x) = 1-\\sqrt{a^2+b^2}\\sin (x+\\arctan\\frac{a}{b}) - \\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\geq 0\\). \\(f(x+\\pi) = 1+\\sqrt{a^2+b^2}\\sin (x+\\arctan\\frac{a}{b}) - \\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\geq 0\\)\n\nTherefore, \\(\\sqrt{A^2+B^2}\\sin (2x+\\arctan\\frac{A}{B}) \\leq 1\\... |
IMO-1977-5 | https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_5 | Let \(a,b\) be two natural numbers. When we divide \(a^2+b^2\) by \(a+b\), we the the remainder \(r\) and the quotient \(q.\) Determine all pairs \((a, b)\) for which \(q^2 + r = 1977.\) | [
"Using \\(r=1977-q^2\\), we have \\(a^2+b^2=(a+b)q+1977-q^2\\), or \\(q^2-(a+b)q+a^2+b^2-1977=0\\), which implies \\(\\Delta=7908+2ab-2(a^2+b^2)\\ge 0\\). Using AM-GM inequality and the fact a,b>0, we have 3(a+b)^2<=7908+8ab<=7098+8(a+b)^2/4, it follows \\(a+b\\le 88\\). If \\(q\\le 43\\), then \\(r=1977-q^2\\ge 12... |
IMO-1977-6 | https://artofproblemsolving.com/wiki/index.php/1977_IMO_Problems/Problem_6 | Let \(f(n)\) be a function \(f: \mathbb{N}^{+}\to\mathbb{N}^{+}\). Prove that if
\[
f(n+1) > f(f(n))
\]
for each positive integer \(n\), then \(f(n)=n\). | [
"We will prove this via induction. First we will prove there is a \\(t\\) such that \\(f(t)=1\\) and then that \\(t=1\\) is the only such solution.\n\nDefine the sequence \\(a_n\\) with \\(a_0>1\\) for \\(a_0\\in \\mathbb{N}\\) and \\(a_k=f(a_{k-1}-1)\\). By the given inequality we have that \\(f(a_n)>f(a_{n+1})\\)... |
IMO-1978-1 | https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_1 | Let \(m\) and \(n\) be positive integers such that \(1 \le m < n\). In their decimal representations, the last three digits of \(1978^m\) are equal, respectively, to the last three digits of \(1978^n\). Find \(m\) and \(n\) such that \(m + n\) has its least value. | [
"We have \\(1978^m\\equiv 1978^n\\pmod {1000}\\), or \\(978^m-978^n=1000k\\) for some positive integer \\(k\\) (if it is not positive just do \\(978^n-978^m=-1000k\\)). Hence \\(978^n\\mid 1000k\\). So dividing through by \\(978^n\\) we get \\(978^{m-n}-1=\\frac{1000k}{978^n}\\). Observe that \\(2\\nmid LHS\\), so ... |
IMO-1978-2 | https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_2 | We consider a fixed point \(P\) in the interior of a fixed sphere\(.\) We construct three segments \(PA, PB,PC\), perpendicular two by two\(,\) with the vertexes \(A, B, C\) on the sphere\(.\) We consider the vertex \(Q\) which is opposite to \(P\) in the parallelepiped (with right angles) with \(PA, PB, PC\) as edges\... | [
"\\[\nIMO 1978 P2a.png\n\\]\n\nLet \\(R\\) be the radius of the given fixed sphere.\n\nLet point \\(O\\) be the center of the sphere.\n\nLet point \\(D\\) be the 4th vertex of the face of the parallelepiped that contains points \\(P\\), \\(A\\), and \\(B\\).\n\nLet point \\(E\\) be the point where the line that pas... |
IMO-1978-3 | https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_3 | Let \(0<f(1)<f(2)<f(3)<\ldots\) a sequence with all its terms positive\(.\) The \(n-th\) positive integer which doesn't belong to the sequence is \(f(f(n))+1.\) Find \(f(240).\) | [
"Since the \\(n\\)-th number missing is \\(f(f(n))+1\\) and \\(f(f(n))\\) is a member of the sequence, it results that there are exactly \\(n-1\\) \"gaps\" less than \\(f(f(n))\\), which leads us to:\n\n\\(f(f(n))=f(n)+n-1\\) \\((\\star)\\)\n\nnow with a simple induction we can prove that \\(f(F_{n}+1) = F_{n+1}+f(... |
IMO-1978-4 | https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_4 | In a triangle \(ABC\) we have \(AB = AC.\) A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides \(AB, AC\) in the points \(P,\) respectively \(Q.\) Prove that the midpoint of \(PQ\) is the center of the inscribed circle of the triangle \(ABC.\) | [
"Denote \\(a = BC, b = AB = AC\\) the triangle sides, \\(r, R\\) the triangle inradius and circumradius and \\(s\\), \\(\\triangle\\) the triangle semiperimeter and area. Let \\(S, T\\) be the tangency points of the incircle (I) with the sides \\(AB, AC\\). Let \\(K, L\\) be the midpoints of of \\(BC, PQ\\) and \\(... |
IMO-1978-5 | https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_5 | Let \(f\) be an injective function from \({1,2,3,\ldots}\) in itself. Prove that for any \(n\) we have: \(\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.\) | [
"We know that all the unknowns are integers, so the smallest one must greater or equal to 1.\n\nLet me denote the permutations of \\((k_1,k_2,...,k_n)\\) with \\((y_1,y_2,...,y_n)=y_i (*)\\).\n\nFrom the rearrangement's inequality we know that \\(\\text{Random Sum} \\geq \\text{Reversed Sum}\\).\n\nWe will denote w... |
IMO-1978-6 | https://artofproblemsolving.com/wiki/index.php/1978_IMO_Problems/Problem_6 | An international society has its members from six different countries. The list of members has 1978 names, numbered \(1, 2, \ldots, 1978\). Prove that there is at least one member whose number is the sum of the numbers of two (not necessarily distinct) members from his own country. | [
"Suppose the contrary. Then there exists a partition of \\(\\{1,2,3,\\ldots, 1978\\}\\) into 6 difference-free subsets \\(A,B,C,D,E,F\\). A set S is difference-free if there are not \\(x,y,z \\in S\\) such that x = y - z.\n\nBy the Pigeonhole principle, one of these subsets, say \\(A\\), has a minimum of \\(\\left ... |
IMO-1979-1 | https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_1 | If \(p\) and \(q\) are natural numbers so that
\[
\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319},
\]
prove that \(p\) is divisible with \(1979\). | [
"We first write\n\n\\[\n\\begin{align*} \\frac{p}{q} &=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots-\\frac{1}{1318}+\\frac{1}{1319}\\\\ &=1+\\frac{1}{2}+\\cdots+\\frac{1}{1319}-2\\cdot\\left(\\frac{1}{2}+\\frac{1}{4}+\\cdots+\\frac{1}{1318}\\right)\\\\ &=1+\\frac{1}{2}+\\cdots+\\frac{1}{1319}-\\left(1+\\frac{1}... |
IMO-1979-2 | https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_2 | We consider a prism which has the upper and inferior basis the pentagons: \(A_{1}A_{2}A_{3}A_{4}A_{5}\) and \(B_{1}B_{2}B_{3}B_{4}B_{5}\). Each of the sides of the two pentagons and the segments \(A_{i}B_{j}\) with \(i,j=1,\ldots\),5 is colored in red or blue. In every triangle which has all sides colored there exists ... | [
"Let us prove first that the edges \\(A_1A_2,A_2A_3,\\cdots ,A_5A_1\\) are of the same color. Assume the contrary, and let w.l.o.g. \\(A_1A_2\\) be red and \\(A_2A_3\\) be green. Three of the segments \\(A_2B_l, (l = 1, 2, 3, 4, 5)\\), say \\(A_2B_i,A_2B_j,A_2B_k\\),have to be of the same color, let it w.l.o.g. be ... |
IMO-1979-3 | https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_3 | Two circles in a plane intersect. \(A\) is one of the points of intersection. Starting simultaneously from \(A\) two points move with constant speed, each travelling along its own circle in the same sense. The two points return to \(A\) simultaneously after one revolution. Prove that there is a fixed point \(P\) in the... | [
"Let \\(O_1, O_2\\) be the centers of the 2 circles and let the points B, C move on the circles \\((O_1), (O_2)\\), recpectively. Let A' be the other intersection of the 2 circles different from the point A. Since the angles \\(\\angle BO_1A = \\angle CO_2A\\) are equal, the isosceles triangles \\(\\triangle AO_1B ... |
IMO-1979-4 | https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_4 | We consider a point \(P\) in a plane \(p\) and a point \(Q \not\in p\). Determine all the points \(R\) from \(p\) for which
\[
\frac{QP+PR}{QR}
\]
is maximum. | [
"Let \\(T\\) be the orthogonal projection of the point \\(Q\\) on the plane \\(p\\). Then, the line \\(PT\\) is the orthogonal projection of the line \\(PQ\\) on the plane \\(p\\), and thus forms the least angle with the line \\(PQ\\) among all lines through the point \\(P\\) which lie in the plane \\(p\\); hence, ... |
IMO-1979-5 | https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_5 | Determine all real numbers a for which there exists non-negative reals \(x_{1}, \ldots, x_{5}\) which satisfy the relations \(\sum_{k=1}^{5} kx_{k}=a,\) \(\sum_{k=1}^{5} k^{3}x_{k}=a^{2},\) \(\sum_{k=1}^{5} k^{5}x_{k}=a^{3}.\) | [
"Let \\(\\Sigma_1= \\sum_{k=1}^{5} kx_{k}\\), \\(\\Sigma_2=\\sum_{k=1}^{5} k^{3}x_{k}\\) and \\(\\Sigma_3=\\sum_{k=1}^{5} k^{5}x_{k}\\). For all pairs \\(i,j\\in \\mathbb{Z}\\), let\n\n\\[\n\\Sigma(i,j)=i^2j^2\\Sigma_1-(i^2+j^2)\\Sigma_2+\\Sigma_3\n\\]\n\nThen we have on one hand\n\n\\[\n\\Sigma(i,j)=i^2j^2\\Sigma_... |
IMO-1979-6 | https://artofproblemsolving.com/wiki/index.php/1979_IMO_Problems/Problem_6 | Let \(A\) and \(E\) be opposite vertices of an octagon. A frog starts at vertex \(A.\) From any vertex except \(E\) it jumps to one of the two adjacent vertices. When it reaches \(E\) it stops. Let \(a_n\) be the number of distinct paths of exactly \(n\) jumps ending at \(E\). Prove that:
\[
a_{2n-1}=0, \quad a_{2n}={... | [
"First the part \\(a_{2n-1}=0\\) It's pretty obvious. Let's make a sequence \\(b\\) of 1 and -1, setting 1 when the frog jumps left and -1 when it jumps right. If the frog would want to come to vertex E from vertex A, then \\(\\sum b_{i}\\) from \\(i =1\\) to \\(i =2n-1\\) should be equal to either 4 or -4. But thi... |
IMO-1981-1 | https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_1 | \(P\) is a point inside a given triangle \(ABC\). \(D, E, F\) are the feet of the perpendiculars from \(P\) to the lines \(BC, CA, AB\), respectively. Find all \(P\) for which
\[
\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}
\]
is least. | [
"We note that \\(BC \\cdot PD + CA \\cdot PE + AB \\cdot PF\\) is twice the triangle's area, i.e., constant. By the Cauchy-Schwarz Inequality,\n\n\\[\n{(BC \\cdot PD + CA \\cdot PE + AB \\cdot PF) \\left(\\frac{BC}{PD} + \\frac{CA}{PE} + \\frac{AB}{PF} \\right) \\ge ( BC + CA + AB )^2}\n\\]\n\n,\n\nwith equality ex... |
IMO-1981-2 | https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_2 | Let \(1 \le r \le n\) and consider all subsets of \(r\) elements of the set \(\{ 1, 2, \ldots , n \}\). Each of these subsets has a smallest member. Let \(F(n,r)\) denote the arithmetic mean of these smallest numbers; prove that
\[
F(n,r) = \frac{n+1}{r+1}.
\] | [
"Clearly, the sum of the desired least elements is \\(\\sum_{k=1}^{n} k {n-k \\choose r-1} = \\sum_{k=1}^{n} {k \\choose 1}{n-k \\choose r-1}\\), which we claim to be equal to \\({n+1 \\choose r+1}\\) by virtue of the following argument.\n\nConsider a binary string of length \\(n+1\\) which contains \\(r+1\\) 1s. F... |
IMO-1981-3 | https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_3 | Determine the maximum value of \(m^2 + n^2\), where \(m\) and \(n\) are integers satisfying \(m, n \in \{ 1,2, \ldots , 1981 \}\) and \(( n^2 - mn - m^2 )^2 = 1\). | [
"We first observe that since \\(\\gcd(m,n)=1\\), \\(m\\) and \\(n\\) are relatively prime. In addition, we note that \\(n \\ge m\\), since if we had \\(n < m\\), then \\(n^2 -nm -m^2 = n(n-m) - m^2\\) would be the sum of two negative integers and therefore less than \\(-1\\). We now observe\n\n\\[\n(m+k)^2 -(m+k)m ... |
IMO-1981-4 | https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_4 | (a) For which values of \(n>2\) is there a set of \(n\) consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining \(n-1\) numbers?
(b) For which values of \(n>2\) is there exactly one set having the stated property? | [
"Let \\(k = \\prod p_i^{e_i}\\) be the greatest element of the set, written in its prime factorization. Then \\(k\\) divides the least common multiple of the other elements of the set if and only if the set has cardinality at least \\(\\max \\{ p_i^{e_i} \\}\\), since for any of the \\(p_i^{e_i}\\), we must go down... |
IMO-1981-5 | https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_5 | Three congruent circles have a common point \(O\) and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle and the point \(O\) are collinear. | [
"Let the triangle have vertices \\(A,B,C\\), and sides \\(a,b,c\\), respectively, and let the centers of the circles inscribed in the angles \\(A,B,C\\) be denoted \\(O_A, O_B, O_C\\), respectively.\n\nThe triangles \\(O_A O_B O_C\\) and \\(ABC\\) are homothetic, as their corresponding sides are parallel. Furthermo... |
IMO-1981-6 | https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_6 | The function \(f(x,y)\) satisfies
(1) \(f(0,y)=y+1,\)
(2) \(f(x+1,0)=f(x,1),\)
(3) \(f(x+1,y+1)=f(x,f(x+1,y)),\)
for all non-negative integers \(x,y\). Determine \(f(4,1981)\). | [
"We observe that \\(f(1,0) = f(0,1) = 2\\) and that \\(f(1, y+1) = f(0, f(1,y)) = f(1,y) + 1\\), so by induction, \\(f(1,y) = y+2\\). Similarly, \\(f(2,0) = f(1,1) = 3\\) and \\(f(2, y+1) = f(2,y) + 2\\), yielding \\(f(2,y) = 2y + 3\\).\n\nWe continue with \\(f(3,0) + 3 = 8\\); \\(f(3, y+1) + 3 = 2(f(3,y) + 3)\\); ... |
IMO-1982-1 | https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_1 | The function \(f(n)\) is defined on the positive integers and takes non-negative integer values. \(f(2)=0,f(3)>0,f(9999)=3333\) and for all \(m,n:\)
\[
f(m+n)-f(m)-f(n)=0 \text{ or } 1.
\]
Determine \(f(1982)\). | [
"Clearly \\(f(1) \\ge 1 \\Rightarrow f(m+1) \\ge f(m)+f(1) \\ge f(m)+1\\) so \\(f(9999) \\ge 9999\\).Contradiction!So \\(f(1)=0\\).This forces \\(f(3)=1\\).Hence \\(f(3k+3) \\ge f(3k)+f(3)>f(3k)\\) so the inequality \\(f(3)<f(6)<\\cdots<f(9999)=3333\\) forces \\(f(3k)=k \\forall k \\le 3333\\).Now \\(f(3k+2) \\ge k... |
IMO-1982-2 | https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_2 | A non-isosceles triangle \(A_{1}A_{2}A_{3}\) has sides \(a_{1}\), \(a_{2}\), \(a_{3}\) with the side \(a_{i}\) lying opposite to the vertex \(A_{i}\). Let \(M_{i}\) be the midpoint of the side \(a_{i}\), and let \(T_{i}\) be the point where the inscribed circle of triangle \(A_{1}A_{2}A_{3}\) touches the side \(a_{i}\)... | [
"Rename the vertices so that \\(A_1=A,A_2=B\\) and \\(A_3=C\\). Let \\(I\\) be the incenter of \\(\\triangle ABC\\).\n\nClaim 1: \\(S_1, S_2, S_3\\) all lie on the incircle. Proof: By construction, \\(IS_1\\) is the reflection of \\(IT_1\\) with respect to the \\(A\\) angle bisector since \\(I\\) lies on the angle ... |
IMO-1982-3 | https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_3 | Consider infinite sequences \(\{x_n\}\) of positive reals such that \(x_0=1\) and \(x_0\ge x_1\ge x_2\ge\ldots\).
a) Prove that for every such sequence there is an \(n\ge1\) such that:
\[
{x_0^2\over x_1}+{x_1^2\over x_2}+\ldots+{x_{n-1}^2\over x_n}\ge3.999.
\]
b) Find such a sequence such that for all \(n\):
\[
{x... | [
"By Cauchy, the LHS is at least: \\(\\frac{(x_{0}+...+x_{n-1})^{2}}{x_{1}+...+x_{n}}.\\)\n\nIt is clear that \\(x_{0}=1\\) and \\(x_{n}\\le \\frac{x_{1}+...+x_{n-1}}{n-1}\\)\n\nWe thus have that the LHS is at least: \\((\\frac{1+2(x_{1}+...+x_{n-1})+(x_{1}+...+x_{n-1})^{2}}{x_{1}+...+x_{n-1}})(\\frac{n-1}{n})\\)\n\... |
IMO-1982-4 | https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_4 | Prove that if \(n\) is a positive integer such that the equation \(x^3-3xy^2+y^3=n\) has a solution in integers \(x,y\), then it has at least three such solutions. Show that the equation has no solutions in integers for \(n=2891\). | [
"Suppose the equation \\(x^3-3xy^2+y^3=n\\) has solution in integers \\((x,y)\\) with \\(y=x+k\\). Then, completing the cube yields \\((y-x)^3-3x^2y+2x^3\\). Using the substitution \\(y=x+k\\) yields \\(k^3-3kx^2-x^3=n\\). Notice that equality directly implies that \\((k,-x)\\) is also a solution to the original eq... |
IMO-1982-5 | https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_5 | The diagonals \(AC\) and \(CE\) of the regular hexagon \(ABCDEF\) are divided by inner points \(M\) and \(N\) respectively, so that
\[
{AM\over AC}={CN\over CE}=r.
\]
Determine \(r\) if \(B,M\) and \(N\) are collinear. | [
"O is the center of the regular hexagon. Then we clearly have \\(ABC\\cong COA\\cong EOC\\). And therefore we have also obviously \\(ABM\\cong AOM\\cong CON\\), as \\(\\frac{AM}{AC} =\\frac{CN}{CE}\\). So we have \\(\\angle{BMA} =\\angle{AMO} =\\angle{CNO}\\) and \\(\\angle{NOC} =\\angle{ABM}\\). Because of \\(\\an... |
IMO-1982-6 | https://artofproblemsolving.com/wiki/index.php/1982_IMO_Problems/Problem_6 | Let \(S\) be a square with sides length \(100\). Let \(L\) be a path within \(S\) which does not meet itself and which is composed of line segments \(A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n\) with \(A_0=A_n\). Suppose that for every point \(P\) on the boundary of \(S\) there is a point of \(L\) at a distance from \(P\) ... | [
"Let the square be A'B'C'D'. The idea is to find points of L close to a particular point of A'D' but either side of an excursion to B'.\n\nWe say L approaches a point P' on the boundary of the square if there is a point P on L with PP' ≤ 1/2. We say L approaches P' before Q' if there is a point P on L which is near... |
IMO-1983-1 | https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_1 | Find all functions \(f\) defined on the set of positive reals which take positive real values and satisfy the conditions: (i) \(f(xf(y))=yf(x)\) for all \(x,y\); (ii) \(f(x)\to0\) as \(x\to \infty\). | [
"Let \\(x=y=1\\) and we have \\(f(f(1))=f(1)\\). Now, let \\(x=1,y=f(1)\\) and we have \\(f(f(f(1)))=f(1)f(1)\\Rightarrow f(1)=[f(1)]^2\\) since \\(f(1)>0\\) we have \\(f(1)=1\\).\n\nPlug in \\(y=x\\) and we have \\(f(xf(x))=xf(x)\\). If \\(a=1\\) is the only solution to \\(f(a)=a\\) then we have \\(xf(x)=1\\Righta... |
IMO-1983-2 | https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_2 | Let \(A\) be one of the two distinct points of intersection of two unequal coplanar circles \(C_1\) and \(C_2\) with centers \(O_1\) and \(O_2\) respectively. One of the common tangents to the circles touches \(C_1\) at \(P_1\) and \(C_2\) at \(P_2\), while the other touches \(C_1\) at \(Q_1\) and \(C_2\) at \(Q_2\). L... | [
"Let \\(A\\) be one of the two distinct points of intersection of two unequal coplanar circles \\(C_1\\) and \\(C_2\\) with centers \\(O_1\\) and \\(O_2\\) respectively. Let \\(S\\) be such point on line \\(O_1O_2\\) so that tangents on \\(C_1\\) touches it at \\(P_1\\) and \\(Q_1\\) and tangents on \\(C_2\\) touch... |
IMO-1983-3 | https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_3 | Let \(a\), \(b\) and \(c\) be positive integers, no two of which have a common divisor greater than \(1\). Show that \(2abc - ab - bc- ca\) is the largest integer which cannot be expressed in the form \(xbc + yca + zab\), where \(x\), \(y\) and \(z\) are non-negative integers. | [
"First off, I prove \\(2abc-ab-bc-ca\\) is un-achievable. Also, assume WLOG \\(a\\ge b\\ge c\\).\n\nAssume \\(2abc-bc-ca=xbc+yca+zab\\), then take the equation \\(\\pmod{ab}\\) to give us \\((x+1)bc+(y+1)ac\\equiv 0\\pmod{ab}\\). By CRT and \\(\\gcd(a,b)=1\\), we take this equation mod a and mod b to give us \\(y+1... |
IMO-1983-4 | https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_4 | Let \(ABC\) be an equilateral triangle and \(\mathcal{E}\) the set of all points contained in the three segments \(AB\), \(BC\) and \(CA\) (including \(A\), \(B\) and \(C\)). Determine whether, for every partition of \(\mathcal{E}\) into two disjoint subsets, at least one of the two subsets contains the vertices of a r... | [
"The answer is positive; there always a class of the partition will contain the three vertices of a right-angled triangle.\n\nFirst notice that if there are at least two points of the same color on a side, then all the points in \\(\\mathcal{E}\\) which project orthogonally onto points of that color on the respecti... |
IMO-1983-5 | https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_5 | Is it possible to choose \(1983\) distinct positive integers, all less than or equal to \(10^5\), no three of which are consecutive terms of an arithmetic progression? Justify your answer. | [
"The answer is yes. We will construct a sequence of integers that satisfies the requirements.\n\nTake the following definition of \\(a_n\\): \\(a_n\\) in base 3 has the same digits as \\(n\\) in base 2. For example, since \\(6=110_2\\), \\(a_6=110_3=12\\).\n\nFirst, we will prove that no three \\(a_n\\)'s are in an... |
IMO-1983-6 | https://artofproblemsolving.com/wiki/index.php/1983_IMO_Problems/Problem_6 | Let \(a\), \(b\) and \(c\) be the lengths of the sides of a triangle. Prove that
\[
a^{2}b(a - b) + b^{2}c(b - c) + c^{2}a(c - a)\ge 0.
\]
Determine when equality occurs. | [
"By Ravi substitution, let \\(a = y+z\\), \\(b = z+x\\), \\(c = x+y\\). Then, the triangle condition becomes \\(x, y, z > 0\\). After some manipulation, the inequality becomes:\n\n\\(xy^3 + yz^3 + zx^3 \\geq xyz(x+y+z)\\).\n\nBy Cauchy, we have:\n\n\\((xy^3 + yz^3 + zx^3)(z+x+y) \\geq xyz(y+z+x)^2\\) with equality ... |
IMO-1984-1 | https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_1 | Let \(x\), \(y\), \(z\) be nonnegative real numbers with \(x + y + z = 1\). Show that \(0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}\) | [
"Note that this inequality is symmetric with x,y and z.\n\nTo prove\n\n\\[\nxy+yz+zx-2xyz\\geq 0\n\\]\n\nnote that \\(x+y+z=1\\) implies that at most one of \\(x\\), \\(y\\), or \\(z\\) is greater than \\(\\frac{1}{2}\\). Suppose \\(x \\leq \\frac{1}{2}\\), WLOG. Then, \\(xy+yz+zx-2xyz=yz(1-2x)+xy+zx\\geq 0\\) sinc... |
IMO-1984-2 | https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_2 | Find one pair of positive integers \(a,b\) such that \(ab(a+b)\) is not divisible by \(7\), but \((a+b)^7-a^7-b^7\) is divisible by \(7^7\). | [
"So we want \\(7 \\nmid ab(a+b)\\) and \\(7^7 | (a+b)^7-a^7-b^7 = 7ab(a+b)(a^2+ab+b^2)^2\\), so we want \\(7^3 | a^2+ab+b^2\\). Now take e.g. \\(a=2,b=1\\) and get \\(7|a^2+ab+b^2\\). Now by some standard methods like Hensels Lemma (used to the polynomial \\(x^2+x+1\\), so \\(b\\) seen as constant from now) we get ... |
IMO-1984-3 | https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_3 | Given points \(O\) and \(A\) in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point \(X\) in the plane, the circle \(C(X)\) has center \(O\) and radius \(OX+{\angle AOX\over OX}\), where \(\angle AOX\) is measured in radians in the range \([0,2\pi)\). Prove that we can fi... | [
"Let \\(a_{n},\\ n\\ge 1\\) be a sequence of positive reals such that \\(\\left(\\sum_{n\\ge 1}a_{n}\\right)^{2}<2\\pi\\ (*)\\). For each \\(n\\ge 1\\), let \\(\\mathcal C_{n}\\) be the circle centered at \\(O\\) with radius \\(\\sum_{i=1}^{n}a_{i}\\).\n\nBecause of \\((*)\\), we can find points \\(x_{i,j}\\in\\mat... |
IMO-1984-4 | https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_4 | Let \(ABCD\) be a convex quadrilateral with the line \(CD\) being tangent to the circle on diameter \(AB\). Prove that the line \(AB\) is tangent to the circle on diameter \(CD\) if and only if the lines \(BC\) and \(AD\) are parallel. | [
"First, we prove that if \\(BC\\) and \\(AD\\) are parallel then the claim is true: Let \\(AB\\) and \\(CD\\) intersect at \\(E\\) (assume \\(E\\) is closer to \\(AD\\), the other case being analogous). Let \\(M,N\\) be the midpoints of \\(AB,CD\\) respectively. Let the length of the perpendicular from \\(N\\) to \... |
IMO-1984-5 | https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_5 | Let \(d\) be the sum of the lengths of all the diagonals of a plane convex polygon with \(n\) vertices (where \(n>3\)). Let \(p\) be its perimeter. Prove that:
\[
n-3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n+1\over 2}\Bigr]-2,
\]
where \([x]\) denotes the greatest integer not exceeding \(x\). | [
"Consider all the pairs of non-adjacent sides in the polygon. Each pair uniquely determines two diagonals which intersect in an interior point. The sum of these two diagonals is greater than the sum of the two sides (triangle inequality). Each side will appear in a pair \\(n-3\\) times. And each time the side appea... |
IMO-1984-6 | https://artofproblemsolving.com/wiki/index.php/1984_IMO_Problems/Problem_6 | Let \(a,b,c,d\) be odd integers such that \(0<a<b<c<d\) and \(ad=bc\). Prove that if \(a+d=2^k\) and \(b+c=2^m\) for some integers \(k\) and \(m\), then \(a=1\). | [
"Let \\(f:[1,b]\\rightarrow \\mathbb{R},\\ f(x)=x+\\dfrac{bc}{x}\\). As \\(f^\\prime (x)=1-\\dfrac{bc}{x^2}\\le 0\\), we infer that \\(f(x)\\ge f(b)=b+c,\\ \\forall x\\in [1,b]\\); in particular, \\(a+d=f(a)\\ge b+c\\Leftrightarrow k\\ge m\\).\n\nNow, \\(ad=bc\\Leftrightarrow a(2^k-a)=b(2^m-b)\\Leftrightarrow (b-a)... |
IMO-1985-1 | https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_1 | A circle has center on the side \(AB\) of the cyclic quadrilateral \(ABCD\). The other three sides are tangent to the circle. Prove that \(AD + BC = AB\). | [
"Let \\(O\\) be the center of the circle mentioned in the problem. Let \\(T\\) be the second intersection of the circumcircle of \\(CDO\\) with \\(AB\\). By measures of arcs, \\(\\angle DTA = \\angle DCO = \\frac{\\angle DCB}{2} = \\frac{\\pi}{2} - \\frac{\\angle DAB}{2}\\). It follows that \\(AT = AD\\). Likewise,... |
IMO-1985-2 | https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_2 | Let \(n\) and \(k\) be given relatively prime natural numbers, \(k < n\). Each number in the set \(M = \{ 1,2, \ldots , n-1 \}\) is colored either blue or white. It is given that
(i) for each \(i \in M\), both \(i\) and \(n-i\) have the same color;
(ii) for each \(i \in M, i \neq k\), both \(i\) and \(|i-k|\) have th... | [
"We may consider the elements of \\(M\\) as residues mod \\(n\\). To these we may add the residue 0, since (i) may only imply that 0 has the same color as itself, and (ii) may only imply that 0 has the same color as \\(k\\), which put no restrictions on the colors of the other residues.\n\nWe note that (i) is equiv... |
IMO-1985-3 | https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_3 | For any polynomial \(P(x) = a_0 + a_1 x + \cdots + a_k x^k\) with integer coefficients, the number of coefficients which are odd is denoted by \(w(P)\). For \(i = 0, 1, \ldots\), let \(Q_i (x) = (1+x)^i\). Prove that if \(i_1, i_2, \ldots , i_n\) are integers such that \(0 \leq i_1 < i_2 < \cdots < i_n\), then
\[
w(Q_... | [
"We first observe that \\((1+x)^{2^m} \\equiv 1 + x^{2^m} \\pmod{2}\\), so for any polynomial \\(P\\) of degree less than \\(2^m\\), \\(w(P\\cdot Q_{2^m}) = 2w (P)\\).\n\nLet \\(k\\) be the largest power of 2 that is less than or equal to \\(i_n\\). We proceed by induction on \\(k\\).\n\nFor \\(k = 0\\), the proble... |
IMO-1985-4 | https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_4 | Given a set \(M\) of \(1985\) distinct positive integers, none of which has a prime divisor greater than \(23\), prove that \(M\) contains a subset of \(4\) elements whose product is the \(4\)th power of an integer. | [
"We have that \\(x\\in M\\Rightarrow x=2^{e_1}3^{e_2}\\cdots 19^{e_8}23^{e_9}\\). We need only consider the exponents. First, we consider the number of subsets of two elements, such that their product is a perfect square. There are \\(2^9=512\\) different parity cases for the exponents \\(e_1,e_2,...,e_9\\). Thus, ... |
IMO-1985-5 | https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_5 | A circle with center \(O\) passes through the vertices \(A\) and \(C\) of the triangle \(ABC\) and intersects the segments \(AB\) and \(BC\) again at distinct points \(K\) and \(N\) respectively. Let \(M\) be the point of intersection of the circumcircles of triangles \(ABC\) and \(KBN\) (apart from \(B\)). Prove that ... | [
"\\(M\\) is the Miquel Point of quadrilateral \\(ACNK\\), so there is a spiral similarity centered at \\(M\\) that takes \\(KN\\) to \\(AC\\). Let \\(M_1\\) be the midpoint of \\(KA\\) and \\(M_2\\) be the midpoint of \\(NC\\). Thus the spiral similarity must also send \\(M_1\\) to \\(M_2\\) and so \\(BMM_1 M_2\\) ... |
IMO-1985-6 | https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_6 | For every real number \(x_1\), construct the sequence \(x_1,x_2,\ldots\) by setting \(x_{n+1}=x_n \left(x_n + \frac{1}{n}\right)\) for each \(n \geq 1\). Prove that there exists exactly one value of \(x_<x_n<x_{n+1}<1\) for every \(n\). | [
"By recursive substitution, one can write \\(x_n=P_n(x_1)\\) , where \\(P_n\\) is a polynomial with non-negative coefficients and zero constant term. Thus, \\(P_n(0)=0\\), \\(P_n\\) is strictly increasing in \\([0,+\\infty)\\) , and \\(\\displaystyle \\lim_{x_1 \\rightarrow + \\infty} P_n(x_1)=+\\infty\\). We can t... |
IMO-1986-1 | https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_1 | Let \(d\) be any positive integer not equal to \(2, 5\) or \(13\). Show that one can find distinct \(a,b\) in the set \(\{2,5,13,d\}\) such that \(ab-1\) is not a perfect square. | [
"We do casework with modular arithmetic.\n\n\\(d\\equiv 0,3 \\pmod{4}: 13d-1\\) is not a perfect square.\n\n\\(d\\equiv 2\\pmod{4}: 2d-1\\) is not a perfect square.\n\nTherefore, \\(d\\equiv 1, \\pmod{4}.\\) Now consider \\(d\\pmod{16}.\\)\n\n\\(d\\equiv 1,13 \\pmod{16}: 13d-1\\) is not a perfect square.\n\n\\(d\\e... |
IMO-1986-2 | https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_2 | Given a point \(P_0\) in the plane of the triangle \(A_1A_2A_3\). Define \(A_s=A_{s-3}\) for all \(s\ge4\). Construct a set of points \(P_1,P_2,P_3,...\) such that \(P_{k+1}\) is the image of \(P_k\) under a rotation center \(A_{k+1}\) through an angle \(120^\circ\) clockwise for \(k=0,1,2,...\). Prove that if \(P_{198... | [
"Consider the triangle and the points on the complex plane. Without loss of generality, let \\(A_1=0\\), \\(A_2=1\\), and \\(A_3=a\\) for some complex number \\(a\\). Then, a rotation about \\(A_1\\) of \\(120^\\circ\\) sends point \\(P\\) to point \\(e^{\\frac{4i\\pi}{3}}P\\). For \\(A_2\\), the rotation sends \\(... |
IMO-1986-3 | https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_3 | To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers \(x,y,z\) respectively, and \(y<0\), then the following operation is allowed: \(x,y,z\) are replaced by \(x+y,-y,z+y\) respectively. Such an operation is p... | [
"The algorithm always stops. Indeed, consider the five numbers on the vertices of the pentagon as forming a \\(5\\)-tuple \\(\\mathbf{x} = (x_1, x_2, x_3, x_4, x_5) \\in \\mathbb{Z}^5\\). The sum of the five entries of this \\(5\\)-tuple is positive (by assumption), and stays unchanged through the entire process (s... |
IMO-1986-4 | https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_4 | Let \(A,B\) be adjacent vertices of a regular \(n\)-gon (\(n\ge5\)) with center \(O\). A triangle \(XYZ\), which is congruent to and initially coincides with \(OAB\), moves in the plane in such a way that \(Y\) and \(Z\) each trace out the whole boundary of the polygon, with \(X\) remaining inside the polygon. Find the... | [
"Let \\(C \\not = A\\) the vertex which is adjacent to \\(B\\). While \\(XYZ\\) moves from \\(OAB\\) to \\(OBC\\), it is easy to see \\(XYBZ\\) is cyclic. Thus \\(X\\) lies on the bisector of \\(\\angle YBZ = \\angle ABC\\). Moreover, \\(X\\) is the intersection of a circle passing through \\(B\\) (the circumcircle... |
IMO-1986-5 | https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_5 | Find all (if any) functions \(f\) taking the non-negative reals onto the non-negative reals, such that
(a) \(f(xf(y))f(y) = f(x+y)\) for all non-negative \(x\), \(y\);
(b) \(f(2) = 0\);
(c) \(f(x) \neq 0\) for every \(0 \leq x < 2\). | [
"For all \\(x+y\\ge2\\), there exists a nonnegative real \\(t\\) such that \\(t+2=x+y\\) and \\(f(x+y)=f(t+2)=f(tf(2))f(2)=0\\). Hence, \\(f(x)=0\\) for all \\(x\\ge 2\\).\n\nFor \\(x<2\\), take \\(0=f(2)=f((2-x)+(x))=f((2-x)f(x))f(x)\\). Since \\(f(x)\\neq0\\) for \\(x<2\\), then \\(f((2-x)f(x))=0\\). Thus, \\((2-... |
IMO-1986-6 | https://artofproblemsolving.com/wiki/index.php/1986_IMO_Problems/Problem_6 | Given a finite set of points in the plane, each with integer coordinates, is it always possible to color the points red or white so that for any straight line \(L\) parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on \(L\) is not greater than \(1\)? | [
"I'll use a well-known result: if a connected graph has \\(2k>0\\) vertices of odd degree, then its edge set can be partitioned into \\(k\\) paths, and if all its vertices have even degree, then it has an eulerian circuit.\n\nWe have a bipartite graph with bipartition \\(A,B\\) here, constructed as follows: the hor... |
IMO-1987-1 | https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_1 | Let \(p_n (k)\) be the number of permutations of the set \(\{ 1, \ldots , n \} , \; n \ge 1\), which have exactly \(k\) fixed points. Prove that
\[
\sum_{k=0}^{n} k \cdot p_n (k) = n!.
\]
(Remark: A permutation \(f\) of a set \(S\) is a one-to-one mapping of \(S\) onto itself. An element \(i\) in \(S\) is called a fi... | [
"The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element \\(i\\) of the set, there are \\((n-1)!\\) permutations which have \\(i\\) as a fixed point. Therefore\n\n\\[\n\\sum_{k=0}^{n} k \\cdot p_n (k) = n!\n\\]\n\n,\n\nas desired.\n\nSlightly Clearer So... |
IMO-1987-2 | https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_2 | In an acute-angled triangle \(ABC\) the interior bisector of the angle \(A\) intersects \(BC\) at \(L\) and intersects the circumcircle of \(ABC\) again at \(N\). From point \(L\) perpendiculars are drawn to \(AB\) and \(AC\), the feet of these perpendiculars being \(K\) and \(M\) respectively. Prove that the quadrilat... | [
"We are to prove that \\([AKNM]=[ABC]\\) or equivilently, \\([ABC]+[BNC]-[KNC]-[BMN]=[ABC]\\). Thus, we are to prove that \\([BNC]=[KNC]+[BMN]\\). It is clear that since \\(\\angle BAN=\\angle NAC\\), the segments \\(BN\\) and \\(NC\\) are equal. Thus, we have \\([BNC]=\\frac{1}{2}BN^2\\sin BNC=\\frac{1}{2}BN^2\\si... |
IMO-1987-3 | https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_3 | Let \(x_1 , x_2 , \ldots , x_n\) be real numbers satisfying \(x_1^2 + x_2^2 + \cdots + x_n^2 = 1\). Prove that for every integer \(k \ge 2\) there are integers \(a_1, a_2, \ldots a_n\), not all 0, such that \(| a_i | \le k-1\) for all \(i\) and
\[
|a_1x_1 + a_2x_2 + \cdots + a_nx_n| \le \frac{ (k-1) \sqrt{n} }{ k^n - ... | [
"We first note that by the Power Mean Inequality, \\(\\sum_{i=1}^{n} x_i \\le \\sqrt{n}\\). Therefore all sums of the form \\(\\sum_{i=1}^{n} b_i x_i\\), where the \\(b_i\\) is a non-negative integer less than \\(k\\), fall in the interval \\([ 0 , (k-1)\\sqrt{n} ]\\). We may partition this interval into \\(k^n - 1... |
IMO-1987-4 | https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_4 | Prove that there is no function \(f\) from the set of non-negative integers into itself such that \(f(f(n)) = n + 1987\) for every \(n\). | [
"We prove that if \\(f(f(n)) = n + k\\) for all \\(n\\), where \\(k\\) is a fixed positive integer, then \\(k\\) must be even. If \\(k = 2h\\), then we may take \\(f(n) = n + h\\).\n\nSuppose \\(f(m) = n\\) with \\(m \\equiv n \\mod k\\). Then by an easy induction on \\(r\\) we find \\(f(m + kr) = n + kr\\), \\(f(n... |
IMO-1987-5 | https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_5 | Let \(n\) be an integer greater than or equal to 3. Prove that there is a set of \(n\) points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area. | [
"Consider the set of points \\(S = \\{ (x,x^2) \\mid 1 \\le x \\le n , x \\in \\mathbb{N} \\}\\) in the \\(xy\\)-plane.\n\nThe distance between any two distinct points \\((x_1,x^2_1)\\) and \\((x_2,x^2_2)\\) in \\(S\\) (with \\(x_1 \\neq x_2\\)) is:\n\n\\(d = \\sqrt{(x_1-x_2)^2+\\left(x^2_1-x^2_2\\right)^2} =\\) \\... |
IMO-1987-6 | https://artofproblemsolving.com/wiki/index.php/1987_IMO_Problems/Problem_6 | Let \(n\) be an integer greater than or equal to 2. Prove that if \(k^2 + k + n\) is prime for all integers \(k\) such that \(0 \leq k \leq \sqrt{n/3}\), then \(k^2 + k + n\) is prime for all integers \(k\) such that \(0 \leq k \leq n - 2\). | [
"First observe that if \\(m\\) is relatively prime to \\(b+1\\), \\(b+2\\), \\(\\cdots\\), \\(2b\\), then \\(m\\) is relatively prime to any number less than \\(2b\\). Since if \\(c\\leq b\\), then we can choose some \\(i\\) to make \\(2^ic\\) lies in range \\(b+1,b+2,\\cdots,2b\\), so \\(2^ic\\) is relatively prim... |
IMO-1988-1 | https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_1 | Consider 2 concentric circles with radii \(R\) and \(r\) (\(R>r\)) with center \(O\). Fix \(P\) on the small circle and consider the variable chord \(AP\) of the small circle. Points \(B\) and \(C\) lie on the large circle; \(B,P,C\) are collinear and \(BC\) is perpendicular to \(AP\).
1. For which values of \(\angle ... | [
"1. We claim that the value \\(BC^2+CA^2+AB^2\\) stays constant as \\(\\angle OPA\\) varies, and thus achieves its maximum at all value of \\(\\angle OPA\\). We have from the Pythagorean Theorem that \\(CA^2=AP^2+PC^2\\) and \\(AB^2=AP^2+PB^2\\) and so our expression becomes \\[ BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB... |
IMO-1988-2 | https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_2 | Let \(n\) be a positive integer and let \(A_1, A_2, \cdots, A_{2n+1}\) be subsets of a set \(B\).
Suppose that
(a) Each \(A_i\) has exactly \(2n\) elements,
(b) Each \(A_i\cap A_j\) \((1\le i<j\le 2n+1)\) contains exactly one element, and
(c) Every element of \(B\) belongs to at least two of the \(A_i\).
For which... | [
"Answer: All \\(n\\) such that \\(4|n\\)\n\nWe first make the following \\(2\\) claims:\n\nClaim \\(1\\): Each element of union belongs to exactly \\(2\\) subsets.\n\nProof:\n\nConsider a subset \\(A_i\\). Assume that some element \\(x \\in\\ A_i\\) also \\(\\in A_k, A_l\\). There are \\(n-1\\) elements remaining i... |
IMO-1988-3 | https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_3 | A function \(f\) defined on the positive integers (and taking positive integers values) is given by:
\[
\begin{matrix} f(1) = 1, f(3) = 3 \\ f(2 \cdot n) = f(n) \\ f(4 \cdot n + 1) = 2 \cdot f(2 \cdot n + 1) - f(n) \\ f(4 \cdot n + 3) = 3 \cdot f(2 \cdot n + 1) - 2 \cdot f(n), \end{matrix}
\]
for all positive integer... | [
"Considering that \\(f(n)=f(2n)\\), the two last equations give : \\(f(4n + 1)-f(4n) = 2(f(2n + 1) - f(2n))\\) \\(f(4n + 3)-f(4n+2)= 2(f(2n + 1) - f(2n))\\)\n\nAnd so, if \\(n\\) is even and \\(2^{p+1}>n\\geq 2^p>1\\), we have \\(f(n+1)-f(n)=2^p\\)\n\nSo, if we have an even \\(n=\\sum_{i=1}^{k} 2^{a_i}\\), where \\... |
IMO-1988-4 | https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_4 | Show that the solution set of the inequality
\[
\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}
\]
is a union of disjoint intervals, the sum of whose length is \(1988\). | [
"Consider the graph of \\(f(x)=\\sum_{k=1}^{70}\\frac{k}{x-k}\\ge\\frac{5}{4}\\). On the values of \\(x\\) between \\(n\\) and \\(n+1\\) for \\(n\\in\\mathbb{N}\\) \\(1\\le n\\le 69\\), the terms of the form \\(\\frac{k}{x-k}\\) for \\(k\\ne n,n+1\\) have a finite range. In contrast, the term \\(\\frac{n}{x-n}\\) h... |
IMO-1988-5 | https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_5 | In a right-angled triangle \(ABC\) let \(AD\) be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles \(ABD, ACD\) intersect the sides \(AB, AC\) at the points \(K,L\) respectively. If \(E\) and \(E_1\) dnote the areas of triangles \(ABC\) and \(AKL\) respectively, show ... | [
"Lemma: Through the incenter \\(I\\) of \\(\\triangle{ABC}\\) draw a line that meets the sides \\(AB\\) and \\(AC\\) at \\(P\\) and \\(Q\\), then:\n\n\\[\n\\frac{AB}{AP} \\cdot AC + \\frac{AC}{AQ} \\cdot AB = AB+BC+AC\n\\]\n\nProof of the lemma: Consider the general case: \\(M\\) is any point on side \\(BC\\) and \... |
IMO-1988-6 | https://artofproblemsolving.com/wiki/index.php/1988_IMO_Problems/Problem_6 | Let \(a\) and \(b\) be positive integers such that \(ab + 1\) divides \(a^{2} + b^{2}\). Show that \(\frac {a^{2} + b^{2}}{ab + 1}\) is the square of an integer. | [
"Choose integers \\(a,b,k\\) such that \\(a^2+b^2=k(ab+1)\\) Now, for fixed \\(k\\), out of all pairs \\((a,b)\\) choose the one with the lowest value of \\(\\min(a,b)\\). Label \\(b'=\\min(a,b), a'=\\max(a,b)\\). Thus, \\(a'^2-kb'a'+b'^2-k=0\\) is a quadratic in \\(a'\\). Should there be another root, \\(c'\\), th... |
IMO-1989-1 | https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_1 | Prove that in the set \(\{1,2, \ldots, 1989\}\) can be expressed as the disjoint union of subsets \(A_i, \{i = 1,2, \ldots, 117\}\) such that
i.) each \(A_i\) contains 17 elements
ii.) the sum of all the elements in each \(A_i\) is the same. | [
"Let us start pairing numbers in the following fashion, where each pair sums to \\(1990\\): \\((1, 1990-1), (2, 1990-2), \\ldots (936, 1054)\\)\n\nThere are a total of \\(117*8\\) pairs above. Let us start putting these pairs into each of the \\(117\\) subsets starting with the first pair \\((1, 1990-1)\\) going in... |
IMO-1989-2 | https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_2 | \(ABC\) is a triangle, the bisector of angle \(A\) meets the circumcircle of triangle \(ABC\) in \(A_1\), points \(B_1\) and \(C_1\) are defined similarly. Let \(AA_1\) meet the lines that bisect the two external angles at \(B\) and \(C\) in \(A_0\). Define \(B_0\) and \(C_0\) similarly. Prove that the area of triangle... | [
"Notice that since \\(A_1C\\) and \\(A_1B\\) substend the same angle in the circle, and so are equal. Thus \\(A_1BC\\) is iscoceles, and similarly for triangles \\(AB_1C\\) and \\(ABC_1\\). Also, since \\(AA_1BC\\) is a cyclic quadrilatera, \\(\\angle A_1BC=\\angle A_1AC=\\frac{A}{2}\\), and similarly for the other... |
IMO-1989-3 | https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_3 | Let \(n\) and \(k\) be positive integers and let \(S\) be a set of \(n\) points in the plane such that
i.) no three points of \(S\) are collinear, and
ii.) for every point \(P\) of \(S\) there are at least \(k\) points of \(S\) equidistant from \(P.\)
Prove that:
\[
k < \frac {1}{2} + \sqrt {2 \cdot n}
\] | [
"Let \\(\\{A_i\\}_{i=1}^n\\) be our set of points. We count the pairs \\((A_i,\\{A_j,A_\\ell\\}),\\ i\\ne j\\ne \\ell\\ne i\\) such that \\(A_iA_j=A_iA_\\ell\\). There are \\(n\\) choices for \\(A_i\\), and for each \\(A_i\\) there are \\(\\ge\\frac{k(k-1)}2\\) choices for \\(\\{A_j,A_\\ell\\}\\), so the number of ... |
IMO-1989-4 | https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_4 | Let \(ABCD\) be a convex quadrilateral such that the sides \(AB,AD,BC\) satisfy \(AB=AD+BC\). There exists a point \(P\) inside the quadrilateral at a distance \(h\) from the line \(CD\) such that \(AP=h+AD\) and \(BP=h+BC\). Show that:
\[
\frac{1}{\sqrt{h}}\ge\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}
\] | [
"Without loss of generality, assume \\(AD\\ge BC\\).\n\nDraw the circle with center \\(A\\) and radius \\(AD\\) as well as the circle with center \\(B\\) and radius \\(BC\\). Since \\(AB=AD+BC\\), the intersection of circle \\(A\\) with the line \\(AB\\) is the same as the intersection of circle \\(B\\) with the li... |
IMO-1989-5 | https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_5 | Prove that for each positive integer \(n\) there exist \(n\) consecutive positive integers none of which is an integral power of a prime number. | [
"There are at most \\(1+\\sqrt[2]{n}+\\sqrt[3]{n}+\\sqrt[4]{n}+...+\\sqrt[\\left\\lfloor \\log_2(n)\\right\\rfloor]{n} \\leq 1+ \\sqrt n log_2(n)\\) 'true' powers \\(m^k , k\\geq 2\\) in the set \\(\\{1,2,...,n\\}\\). So when \\(p(n)\\) gives the amount of 'true' powers \\(\\leq n\\) we get that \\(\\lim_{n \\to \\... |
IMO-1989-6 | https://artofproblemsolving.com/wiki/index.php/1989_IMO_Problems/Problem_6 | A permutation \(\{x_1, \ldots, x_{2n}\}\) of the set\(\{1,2, \ldots, 2n\}\) where \(\mbox{n}\) is a positive integer, is said to have property \(T\) if \(|x_i - x_{i + 1}| = n\) for at least one \(i \in \{1,2, \ldots, 2n - 1\}\). Show that, for each \(\mbox{n}\), there are more permutations with property \(T\) than wit... | [
"So for the specific case when \\(n = 2\\).\n\nWe have the set \\(\\{1,2,3,4\\}\\)\n\nTo satisfy the condition, the 2 numbers must be adjacent and we can have either \\((1,3), (3,1), (2,4), (4,2)\\) where \\((x,y)\\) represents an adjacent pair.\n\nTo find those that satisfy \\(T\\) we need to find: \\((1,3) \\cup ... |
IMO-1990-1 | https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_1 | Chords \(AB\) and \(CD\) of a circle intersect at a point \(E\) inside the circle. Let \(M\) be an interior point of the segment \(\overline{EB}\). The tangent line at \(E\) to the circle through \(D, E\), and \(M\) intersects the lines \(\overline{BC}\) and \({AC}\) at \(F\) and \(G\), respectively. If \(\frac{AM}{AB}... | [
"With simple angle chasing, we find that triangles \\(CEG\\) and \\(BMD\\) are similar.\n\nso, \\(\\frac{MB}{EC} = \\frac{MD}{EG}\\). .... \\((1)\\)\n\nAgain with simple angle chasing, we find that triangles \\(CEF\\) and \\(AMD\\) are similar.\n\nso, \\(\\frac{MA}{EC} = \\frac{MD}{EF}\\). .... \\((2)\\)\n\nso, by ... |
IMO-1990-2 | https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_2 | Let \(n\ge3\) and consider a set \(E\) of \(2n-1\) distinct points on a circle. Suppose that exactly \(k\) of these points are to be colored black. Such a coloring is “good” if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly \(n\) points from \(E\). Fin... | [
"We form a graph on the \\(2n-1\\) points by connecting two of these points iff one of the arcs they determine has exactly \\(n\\) points in its interior. If \\(3\\not|2n-1\\), then this graph is a cycle, and the question becomes: what's the minimal \\(k\\) s.t. whenever we color \\(k\\) vertices of a cycle of leng... |
IMO-1990-3 | https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_3 | Determine all integers \(n > 1\) such that \(\frac{2^n+1}{n^2}\) is an integer. | [
"Let \\(N = \\{ n\\in\\mathbb{N} : 2^n\\equiv - 1\\pmod{n^2} \\}\\) be the set of all solutions and \\(P = \\{ p\\text{ is prime} : \\exists n\\in N, p|n \\}\\) be the set of all prime factors of the solutions.\n\nIt is clear that the smallest element of \\(P\\) is 3. Assume that \\(P\\ne\\{3\\}\\) and let's try to... |
IMO-1990-4 | https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_4 | Let \(\mathbb{Q^+}\) be the set of positive rational numbers. Construct a function \(f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}\) such that \(f(xf(y)) = \frac{f(x)}{y}\) for all \(x, y\in{Q^+}\). | [
"If we let \\(y = x = 1\\) this implies \\(f(f(1)) = f(1)\\). Plugging \\(f(1) = y\\) in with this new fact. We get \\(f(f(f(1))) = \\frac{f(1)}{f(1)} = 1\\). Using \\(f(f(1)) = f(1)\\) again, we see that \\(f(f(f(1))) = f(f(1)) = f(1) = 1\\). Now plugging in \\(x = 1\\) we get \\(f(f(y)) = \\frac{1}{y}\\). Such a ... |
IMO-1990-5 | https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_5 | Given an initial integer \(n_{0}\textgreater1\), two players, \(\mathbb{A}\) and \(\mathbb{B}\), choose integers \(n_{1}, n_{2},n_{3}\), . . . alternately according to the following rules: Knowing \(n_{2k}\), \(\mathbb{A}\) chooses any integer \(n_{2k+1}\) such that
\[
n_{2k}\leq n_{2k+1}\leq n_{2k}^2
\]
. Knowing \(... | [
"If \\(n_0<6\\) it is clear that B wins, because A can only choose numbers with at most 2 prime factors (30 is the smallest with three and \\(30>5^2\\)), which B can either make smaller (choose the lesser of the two) or the number contains one prime factor in which case B wins. Once \\(n_{2k}=2\\), A can only choos... |
IMO-1990-6 | https://artofproblemsolving.com/wiki/index.php/1990_IMO_Problems/Problem_6 | Prove that there exists a convex \(1990\text{-gon}\) with the following two properties: (a) All angles are equal. (b) The lengths of the \(1990\) sides are the numbers \(1^2, 2^2,3^2,\dots, 1990^2\) in some order. | [
"Let \\(\\{a_1,a_2,\\cdots,a_{1990}\\}=\\{1,2,\\cdots,1990\\}\\) and \\(\\theta=\\frac{\\pi}{995}\\). Then the problem is equivalent to that there exists a way to assign \\(a_1,a_2,\\cdots,a_{1990}\\) such that \\(\\sum_{i=1}^{1990}a_i^2e^{i\\theta}=0\\). Note that \\(e^{i\\theta}+e^{i\\theta+\\pi}=0\\), then if we... |
IMO-1991-1 | https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_1 | Given a triangle \(ABC\) let \(I\) be the center of its inscribed circle. The internal bisectors of the angles \(A,B,C\) meet the opposite sides in \(A^\prime,B^\prime,C^\prime\) respectively. Prove that
\[
\frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}
\... | [
"We have \\(\\prod\\frac{AI}{AA^\\prime}=\\prod\\frac{1}{1+\\frac{IA^\\prime}{IA}}\\). From Van Aubel's Theorem, we have \\(\\frac{IA}{IA^\\prime}=\\frac{AB^\\prime}{B^\\prime C}+\\frac{AC^\\prime}{C^\\prime B}\\) which from the Angle Bisector Theorem reduces to \\(\\frac{b+c}{a}\\). We find similar expressions for... |
IMO-1991-2 | https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_2 | Let \(\,n > 6\,\) be an integer and \(\,a_{1},a_{2},\cdots ,a_{k}\,\) be all the natural numbers less than \(n\) and relatively prime to \(n\). If
\[
a_{2} - a_{1} = a_{3} - a_{2} = \cdots = a_{k} - a_{k - 1} > 0,
\]
prove that \(\,n\,\) must be either a prime number or a power of \(\,2\). | [
"We use Bertrand's Postulate: for \\(u\\ge 2\\) is a positive integer, there is a prime in the interval \\((u,2u)\\).\n\nClearly, \\(a_2\\) must be equal to the smallest prime \\(p\\) which does not divide \\(n\\). If \\(p=2\\), then \\(n\\) is a prime since the common difference \\(a_{i+1}-a_i\\) is equal to \\(1\... |
IMO-1991-3 | https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_3 | Let \(S = \{1,2,3,\cdots ,280\}\). Find the smallest integer \(n\) such that each \(n\)-element subset of \(S\) contains five numbers which are pairwise relatively prime. | [
"Let's look at this another way. We aim to find the least number of integers we can remove from S to leave a set which does not contain 5 pairwise coprime integers.\n\nLet \\(S_p = \\{p^k | p^k \\leq 280\\}\\) And more generally \\(S_{p_1,p_2,...,p_l} = \\{n = p_1^{q_1}p_2^{q_2}....p_l^{q_l}| n \\leq 280 \\}\\) And... |
IMO-1991-4 | https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_4 | Suppose \(\,G\,\) is a connected graph with \(\,k\,\) edges. Prove that it is possible to label the edges \(1,2,\ldots ,k\,\) in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1. | [
"We provide an algorithmic approach for labelling the edges: We start off with the set \\(V\\) of all vertices in \\(G\\). For each step in the process we remove a number of vertices from the set \\(V\\) if their exists an edge sorrounding it which is labelled. Additionally, we shall use up the numbers \\(1,2,...,k... |
IMO-1991-5 | https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_5 | Let \(\,ABC\,\) be a triangle and \(\,P\,\) an interior point of \(\,ABC\,\). Show that at least one of the angles \(\,\angle PAB,\;\angle PBC,\;\angle PCA\,\) is less than or equal to \(30^{\circ }\). | [
"Let \\(A_{1}\\) , \\(A_{2}\\), and \\(A_{3}\\) be \\(\\angle CAB\\), \\(\\angle ABC\\), \\(\\angle BCA\\), respectively.\n\nLet \\(\\alpha_{1}\\) , \\(\\alpha_{2}\\), and \\(\\alpha_{3}\\) be \\(\\angle PAB\\), \\(\\angle PBC\\), \\(\\angle PCA\\), respcetively.\n\nUsing law of sines on \\(\\Delta PAB\\) we get: \... |
IMO-1991-6 | https://artofproblemsolving.com/wiki/index.php/1991_IMO_Problems/Problem_6 | An infinite sequence \(x_0, x_1, x_2,\ldots\) of real numbers is said to be bounded if there is a constant \(C\) such that \(|x_i| \leq C\) for every \(i \geq 0\).
Given any real number \(a > 1\), construct a bounded infinite sequence \(x_0,x_1,x_2,\ldots\) such that \(|x_i-x_j|\cdot |i-j|^a\geq 1\) for every pair of ... | [
"Since \\(a>1\\), the series \\(\\sum_{k=1}^\\infty\\frac{1}{k^a}\\) is convergent; let \\(L\\) be the sum of this convergent series. Let \\(I\\subset \\mathbb{R}\\) be the interval \\([-L,L]\\) (or any bounded subset of measure \\(\\geq 2L\\)).\n\nSuppose that we have chosen points \\(x_0,x_1,\\ldots,x_{m-1}\\) sa... |
IMO-1992-1 | https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_1 | Find all integers \(a\), \(b\), \(c\) satisfying \(1 < a < b < c\) such that \((a - 1)(b -1)(c - 1)\) is a divisor of \(abc - 1\). | [
"\\[\n1<\\frac{abc-1}{(a-1)(b-1)(c-1)}<\\frac{abc}{(a-1)(b-1)(c-1)}\n\\]\n\n\\[\n1<\\frac{abc-1}{(a-1)(b-1)(c-1)}<\\left(\\frac{a}{a-1}\\right) \\left(\\frac{b}{b-1}\\right) \\left(\\frac{c}{c-1}\\right)\n\\]\n\nWith \\(1<a<b<c\\) it implies that \\(a \\ge 2\\), \\(b \\ge 3\\), \\(c \\ge 4\\)\n\nTherefore, \\(\\fra... |
IMO-1992-2 | https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_2 | Let \(\mathbb{R}\) denote the set of all real numbers. Find all functions \(f:\mathbb{R} \to \mathbb{R}\) such that
\[
f\left( x^{2}+f(y) \right)= y+(f(x))^{2} \hspace{0.5cm} \forall x,y \in \mathbb{R}
\] | [
"[quote=probability1.01]Set x = 0 to get \\(f(f(y)) = y+f(0)^{2}\\). We'll let \\(c = f(0)^{2}\\), so \\(f(f(y)) = y+c\\). Then\n\n\\(f(a^{2}+f(f(b))) = f(b)+f(a)^{2}\\) \\(f(a^{2}+b+c) = f(b)+f(a)^{2}\\) \\(f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}\\) \\(a^{2}+b+2c = b+(a+c)^{2}\\) \\(2c = c^{2}+2ac\\)\n\... |
IMO-1992-3 | https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_3 | Consider nine points in space, no four of which are coplanar. Each pair of points is joined by an edge (that is, a line segment) and each edge is either colored blue or red or left uncolored. Find the smallest value of \(n\) such that whenever exactly n edges are colored, the set of colored edges necessarily contains a... | [
"We show that for \\(n = 32\\) we can find a coloring without a monochrome triangle. Take two squares \\(R_1R_2R_3R_4\\) and \\(B_1B_2B_3B_4\\). Leave the diagonals of each square uncolored, color the remaining edges of \\(R\\) red and the remaining edges of \\(B\\) blue. Color blue all the edges from the ninth poi... |
IMO-1992-4 | https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_4 | In the plane let \(C\) be a circle, \(l\) a line tangent to the circle \(C\), and \(M\) a point on \(l\). Find the locus of all points \(P\) with the following property: there exists two points \(Q\), \(R\) on \(l\) such that \(M\) is the midpoint of \(QR\) and \(C\) is the inscribed circle of triangle \(PQR\). | [
"Note: This is an alternate method to what it is shown on the video. This alternate method is too long and too intensive in solving algebraic equations. A lot of steps have been shortened in this solution. The solution in the video provides a much faster solution,\n\nLet \\(r\\) be the radius of the circle \\(C\\).... |
IMO-1992-5 | https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_5 | Let \(S\) be a finite set of points in three-dimensional space. Let \(S_{x}\),\(S_{y}\),\(S_{z}\), be the sets consisting of the orthogonal projections of the points of \(S\) onto the \(yz\)-plane, \(zx\)-plane, \(xy\)-plane, respectively. Prove that
\[
|S|^{2} \le |S_{x}| \cdot |S_{y}| \cdot |S_{z}|,
\]
where \(|A|\... | [
"Let \\(Z_{i}\\) be planes with index \\(i\\) such that \\(1 \\le i \\le n\\) that are parallel to the \\(xy\\)-plane that contain multiple points of \\(S\\) on those planes such that all points of \\(S\\) are distributed throughout all planes \\(Z_{i}\\) according to their \\(z\\)-coordinates in common.\n\nLet \\(... |
IMO-1992-6 | https://artofproblemsolving.com/wiki/index.php/1992_IMO_Problems/Problem_6 | For each positive integer \(n\), \(S(n)\) is defined to be the greatest integer such that, for every positive integer \(k \le S(n)\), \(n^{2}\) can be written as the sum of \(k\) positive squares.
(a) Prove that \(S(n) \le n^{2}-14\) for each \(n \ge 4\).
(b) Find an integer \(n\) such that \(S(n)=n^{2}-14\).
(c) Pr... | [
"(a) Let \\(n \\geq 4\\) be a positive integer. We will prove that \\(S(n) \\leq n^2 - 14\\).\n\nAssume for the sake of contradiction that there exists a positive integer \\(n \\geq 4\\) such that \\(S(n) > n^2 - 14\\). Then, there exists a positive integer \\(m\\) such that \\(S(n) = n^2 - 14 + m\\).\n\nConsider t... |
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