id stringlengths 12 12 | source stringlengths 76 76 | problem stringlengths 59 2.04k | solutions listlengths 0 11 |
|---|---|---|---|
USAMO-1972-1 | https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_1 | The symbols \((a,b,\ldots,g)\) and \([a,b,\ldots, g]\) denote the greatest common divisor and least common multiple, respectively, of the positive integers \(a,b,\ldots, g\). For example, \((3,6,18)=3\) and \([6,15]=30\). Prove that
\[
\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}
\]
. | [
"Consider an arbitrary prime \\(p\\). Let \\(p^\\alpha\\), \\(p^\\beta\\), and \\(p^\\gamma\\) be the greatest powers of \\(p\\) that divide \\(a\\), \\(b\\), and \\(c\\). WLOG let \\(\\alpha \\leq \\beta \\leq \\gamma\\).\n\nExamining each factor in the equation, we see that the largest power of \\(p\\) that divid... |
USAMO-1972-2 | https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_2 | A given tetrahedron \(ABCD\) is isosceles, that is, \(AB=CD, AC=BD, AD=BC\). Show that the faces of the tetrahedron are acute-angled triangles. | [
"Suppose \\(\\triangle ABD\\) is fixed. By the equality conditions, it follows that the maximal possible value of \\(BC\\) occurs when the four vertices are coplanar, with \\(C\\) on the opposite side of \\(\\overline{AD}\\) as \\(B\\). In this case, the tetrahedron is not actually a tetrahedron, so this maximum is... |
USAMO-1972-3 | https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_3 | A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after \(n\) selections (\(n>1\)), the product of the \(n\) numbers selected will be divisible by 10. | [
"For the product to be divisible by 10, there must be a factor of 2 and a factor of 5 in there.\n\nThe probability that there is no 5 is \\(\\left( \\frac{8}{9}\\right)^n\\).\n\nThe probability that there is no 2 is \\(\\left( \\frac{5}{9}\\right)^n\\).\n\nThe probability that there is neither a 2 nor 5 is \\(\\lef... |
USAMO-1972-4 | https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_4 | Let \(R\) denote a non-negative rational number. Determine a fixed set of integers \(a,b,c,d,e,f\), such that for every choice of \(R\),
\[
\left|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right|<|R-\sqrt[3]{2}|
\] | [
"Note that when \\(R\\) approaches \\(\\sqrt[3]{2}\\), \\(\\frac{aR^2+bR+c}{dR^2+eR+f}\\) must also approach \\(\\sqrt[3]{2}\\) for the given inequality to hold. Therefore\n\n\\[\n\\lim_{R\\rightarrow \\sqrt[3]{2}} \\frac{aR^2+bR+c}{dR^2+eR+f}=\\sqrt[3]{2}\n\\]\n\nwhich happens if and only if\n\n\\[\n\\frac{a\\sqrt... |
USAMO-1972-5 | https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_5 | A given convex pentagon \(ABCDE\) has the property that the area of each of the five triangles \(ABC\), \(BCD\), \(CDE\), \(DEA\), and \(EAB\) is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons ... | [
"Lemma: Convex pentagon \\(A_0A_1A_2A_3A_4\\) has the property that \\([A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]\\) if and only if \\(\\overline{A_{n - 1}A_{n + 1}}\\parallel\\overline{A_{n - 2}A_{n + 2}}\\) for \\(n = 0, 1, 2, 3, 4\\) (indices taken mod 5).\n\nProof: For the \"only if\" d... |
USAMO-1973-1 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_1 | Two points \(P\) and \(Q\) lie in the interior of a regular tetrahedron \(ABCD\). Prove that angle \(PAQ < 60^\circ\). | [
"Let the side length of the regular tetrahedron be \\(a\\). Link and extend \\(AP\\) to meet the plane containing triangle \\(BCD\\) at \\(E\\); link \\(AQ\\) and extend it to meet the same plane at \\(F\\). We know that \\(E\\) and \\(F\\) are inside triangle \\(BCD\\) and that \\(\\angle PAQ = \\angle EAF\\)\n\nN... |
USAMO-1973-2 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_2 | Let \(\{X_n\}\) and \(\{Y_n\}\) denote two sequences of integers defined as follows:
\[
X_0=1,
\]
\[
X_1=1,
\]
\[
X_{n+1}=X_n+2X_{n-1}
\]
\[
(n=1,2,3,\dots),
\]
\[
Y_0=1,
\]
\[
Y_1=7,
\]
\[
Y_{n+1}=2Y_n+3Y_{n-1}
\]
\[
(n=1,2,3,\dots).
\]
Thus, the first few terms of the sequences are:
\[
X:1, 1, 3, 5, 11, 21,... | [
"We can look at each sequence \\(\\bmod{8}\\):\n\n\\[\nX\n\\]\n\n:\n\n\\[\n1\n\\]\n\n,\n\n\\[\n1\n\\]\n\n,\n\n\\[\n3\n\\]\n\n,\n\n\\[\n5\n\\]\n\n,\n\n\\[\n3\n\\]\n\n,\n\n\\[\n5\n\\]\n\n,\n\n\\[\n\\dots\n\\]\n\n,\n\n\\[\nY\n\\]\n\n:\n\n\\[\n1\n\\]\n\n,\n\n\\[\n7\n\\]\n\n,\n\n\\[\n1\n\\]\n\n,\n\n\\[\n7\n\\]\n\n,\n\n\... |
USAMO-1973-3 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_3 | Three distinct vertices are chosen at random from the vertices of a given regular polygon of \((2n+1)\) sides. If all such choices are equally likely, what is the probability that the center of the given polygon lies in the interior of the triangle determined by the three chosen random points? | [
"There are \\(\\binom{2n+1}{3}\\) ways how to pick the three vertices. We will now count the ways where the interior does NOT contain the center. These are obviously exactly the ways where all three picked vertices lie among some \\(n+1\\) consecutive vertices of the polygon. We will count these as follows: We will... |
USAMO-1973-4 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_4 | Determine all the roots, real or complex, of the system of simultaneous equations
\[
x+y+z=3,
\]
\[
x^2+y^2+z^2=3,
\]
\[
x^3+y^3+z^3=3
\]
. | [
"Let \\(x\\), \\(y\\), and \\(z\\) be the roots of the cubic polynomial \\(t^3+at^2+bt+c\\). Let \\(S_1=x+y+z=3\\), \\(S_2=x^2+y^2+z^2=3\\), and \\(S_3=x^3+y^3+z^3=3\\). From this, \\(S_1+a=0\\), \\(S_2+aS_1+2b=0\\), and \\(S_3+aS_2+bS_1+3c=0\\). Solving each of these, \\(a=-3\\), \\(b=3\\), and \\(c=-1\\). Thus \\... |
USAMO-1973-5 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_5 | Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression. | [
"Assume that the cube roots of three distinct prime numbers can be three terms of an arithmetic progression. Let the three distinct prime numbers be \\(p\\), \\(q\\), and \\(r.\\) WLOG, let \\(p<q<r.\\)\n\nThen,\n\n\\[\nq^{\\dfrac{1}{3}}=p^{\\dfrac{1}{3}}+md\n\\]\n\n\\[\nr^{\\dfrac{1}{3}}=p^{\\dfrac{1}{3}}+nd\n\\]\... |
USAMO-1974-1 | https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_1 | Let \(a\), \(b\), and \(c\) denote three distinct integers, and let \(P\) denote a polynomial having all integral coefficients. Show that it is impossible that \(P(a)=b\), \(P(b)=c\), and \(P(c)=a\). | [
"It suffices to show that if \\(a,b,c\\) are integers such that \\(P(a) = b\\), \\(P(b)=c\\), and \\(P(c)= a\\), then \\(a=b=c\\).\n\nWe note that\n\n\\[\na-b \\mid P(a) - P(b) = b-c \\mid P(b)-P(c) = c-a \\mid P(c) - P(a) = a-b ,\n\\]\n\nso the quanitities \\((a-b), (b-c), (c-a)\\) must be equal in absolute value.... |
USAMO-1974-2 | https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_2 | Prove that if \(a\), \(b\), and \(c\) are positive real numbers, then
\[
a^ab^bc^c\ge (abc)^{(a+b+c)/3}
\] | [
"Consider the function \\(f(x)=x\\ln{x}\\). \\(f''(x)=\\frac{1}{x}>0\\) for \\(x>0\\); therefore, it is a convex function and we can apply Jensen's Inequality:\n\n\\[\n\\frac{a\\ln{a}+b\\ln{b}+c\\ln{c}}{3}\\ge \\left(\\frac{a+b+c}{3}\\right)\\ln\\left(\\frac{a+b+c}{3}\\right)\n\\]\n\nApply AM-GM to get\n\n\\[\n\\fr... |
USAMO-1974-3 | https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_3 | Two boundary points of a ball of radius 1 are joined by a curve contained in the ball and having length less than 2. Prove that the curve is contained entirely within some hemisphere of the given ball. | [
"Question: Why is the curve necessarily on a great circle? The curve is arbitrary–in space. Also, there is only one great circle through the two points, as three points determine a plane–a counterexample to this claim would then be readily found. Draw a Great Circle containing the two points and the curve. Then con... |
USAMO-1974-4 | https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_4 | A father, mother and son hold a family tournament, playing a two person board game with no ties. The tournament rules are:
(i) The weakest player chooses the first two contestants.
(ii) The winner of any game plays the next game against the person left out.
(iii) The first person to win two games wins the tournament... | [
"There are three possible strategies for the father:\n\n1) Sit out the first game\n\n2) Play the first game against the weaker of the other two players (the mother). This person is referred to throughout the solution as \"the weaker player\".\n\n3) Play the first game against the stronger of the other two players (... |
USAMO-1974-5 | https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_5 | Consider the two triangles \(\triangle ABC\) and \(\triangle PQR\) shown in Figure 1. In \(\triangle ABC\), \(\angle ADB = \angle BDC = \angle CDA = 120^\circ\). Prove that \(x=u+v+w\).
\[
[asy] size(400); defaultpen(1); pair C=(0,1), A=(1,6), B=(4,1), D=(1.5,2); draw(D--A--B--C--D--B); draw(A--C); label("$A$",A,N);... | [
"We rotate figure \\(PRQM\\) by a clockwise angle of \\(\\pi/3\\) about \\(Q\\) to obtain figure \\(RR'QM'\\):\n\n\\[\n[asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label(\"$P$\",P,... |
USAMO-1975-1 | https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_1 | (a) Prove that
\[
[5x]+[5y]\ge [3x+y]+[3y+x],
\]
where \(x,y\ge 0\) and \([u]\) denotes the greatest integer \(\le u\) (e.g., \([\sqrt{2}]=1\)).
(b) Using (a) or otherwise, prove that
\[
\frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}
\]
is integral for all positive integral \(m\) and \(n\). | [] |
USAMO-1975-2 | https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_2 | Let \(A,B,C,D\) denote four points in space and \(AB\) the distance between \(A\) and \(B\), and so on. Show that
\[
AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.
\] | [
"\\[\n[asy] defaultpen(fontsize(8)); pair A=(2,4), B=(0,0), C=(4,0), D=(4,2); label(\"A\",A,(0,1));label(\"D\",D,(1,0));label(\"B\",B,(-1,-1));label(\"C\",C,(1,-1)); axialshade(A--C--D--cycle, lightgray, A, gray, D); draw(A--B--C--A--D--C);draw(B--D, linetype(\"8 8\")); label(\"$m$\",(A+B)/2,(-1,1));label(\"$n$\",(... |
USAMO-1975-3 | https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_3 | If \(P(x)\) denotes a polynomial of degree \(n\) such that
\[
P(k)=\frac{k}{k+1}
\]
for \(k=0,1,2,\ldots,n\), determine \(P(n+1)\). | [
"Let \\(Q(x) = (x+1)P(x) - x\\), and clearly, \\(Q(x)\\) has a degree of \\(n+1\\).\n\nThen, for \\(k=0,1,2,\\ldots,n\\), \\(Q(k) = (k+1)P(k) - k = (k+1)\\cdot \\dfrac{k}{k+1} - k = 0\\).\n\nThus, \\(k=0,1,2,\\ldots,n\\) are the roots of \\(Q(x)\\).\n\nSince these are all \\(n+1\\) of the roots of the \\(n+1^{\\tex... |
USAMO-1975-4 | https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_4 | Two given circles intersect in two points \(P\) and \(Q\). Show how to construct a segment \(AB\) passing through \(P\) and terminating on the two circles such that \(AP\cdot PB\) is a maximum.
\[
[asy] size(150); defaultpen(fontsize(7)); pair A=(0,0), B=(10,0), P=(4,0), Q=(3.7,-2.5); draw(A--B); draw(circumcircle(A,P... | [
"A maximum \\(AP \\cdot PB\\) cannot be attained if \\(AB\\) intersects segment \\(O_1O_2\\) because a larger value can be attained by making one of \\(A\\) or \\(B\\) diametrically opposite \\(P\\), which (as is easily checked) increases the value of both \\(AP\\) and \\(PB\\). Thus, assume \\(AB\\) does not inter... |
USAMO-1975-5 | https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_5 | A deck of \(n\) playing cards, which contains three aces, is shuffled at random (it is assumed that all possible card distributions are equally likely). The cards are then turned up one by one from the top until the second ace appears. Prove that the expected (average) number of cards to be turned up is \((n+1)/2\). | [
"We begin by induction. Our base case is naturally when \\(n = 3\\), as there can be no less than \\(3\\) cards in the deck. The only way to turn up the second ace is to turn up the first, and then turn up the second, which requires \\(2\\) moves. This indeed is equal to \\((3 + 1)/2\\). Next, we assume that the ex... |
USAMO-1976-1 | https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_1 | \[
[asy] void fillsq(int x, int y){ fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray); } int i; fillsq(1,0);fillsq(4,0);fillsq(6,0); fillsq(0,1);fillsq(1,1);fillsq(2,1);fillsq(4,1);fillsq(5,1); fillsq(0,2);fillsq(2,2);fillsq(4,2); fillsq(0,3);fillsq(1,3);fillsq(4,3);fillsq(5,3); for(i=0; i<=7; ++i){draw((i... | [
"There are many ways to prove the first part, we will show one.\n\nConsider the first row. It contains at least four cells of the same color, say white. Pick any four such cells. Let's now consider these four columns only. If any of the remaining three rows contains two white cells in some of these columns, we are ... |
USAMO-1976-2 | https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_2 | If \(A\) and \(B\) are fixed points on a given circle and \(XY\) is a variable diameter of the same circle, determine the locus of the point of intersection of lines \(AX\) and \(BY\). You may assume that \(AB\) is not a diameter.
\[
[asy] size(300); defaultpen(fontsize(8)); real r=10; picture pica, picb; pair A=r*exp... | [
"WLOG, assume that the circle is the unit circle centered at the origin. Then the points \\(A\\) and \\(B\\) have coordinates \\((-a,b)\\) and \\((a,b)\\) respectively and \\(X\\) and \\(Y\\) have coordinates \\((r,s)\\) and \\((-r,-s)\\). Note that these coordinates satisfy \\(a^2 + b^2 = 1\\) and \\(r^2 + s^2 = 1... |
USAMO-1976-3 | https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_3 | Determine all integral solutions of \(a^2+b^2+c^2=a^2b^2\). | [
"Either \\(a^2=0\\) or \\(a^2>0\\). If \\(a^2=0\\), then \\(b^2=c^2=0\\). Symmetry applies for \\(b\\) as well. If \\(a^2,b^2\\neq 0\\), then \\(c^2\\neq 0\\). Now we look at \\(a^2\\bmod{4}\\):\n\n\\(a^2\\equiv 0\\bmod{4}\\): Since a square is either 1 or 0 mod 4, then all the other squares are 0 mod 4. Let \\(a=2... |
USAMO-1976-4 | https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_4 | If the sum of the lengths of the six edges of a trirectangular tetrahedron \(PABC\) (i.e., \(\angle APB=\angle BPC=\angle CPA=90^o\)) is \(S\), determine its maximum volume. | [
"Let the side lengths of \\(AP\\), \\(BP\\), and \\(CP\\) be \\(a\\), \\(b\\), and \\(c\\), respectively. Therefore \\(S=a+b+c+\\sqrt{a^2+b^2}+\\sqrt{b^2+c^2}+\\sqrt{c^2+a^2}\\). Let the volume of the tetrahedron be \\(V\\). Therefore \\(V=\\frac{abc}{6}\\).\n\nNote that \\((a-b)^2\\geq 0\\) implies \\(\\frac{a^2-2... |
USAMO-1976-5 | https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_5 | If \(P(x)\), \(Q(x)\), \(R(x)\), and \(S(x)\) are all polynomials such that
\[
P(x^5) + xQ(x^5) + x^2 R(x^5) = (x^4 + x^3 + x^2 + x +1) S(x),
\]
prove that \(x-1\) is a factor of \(P(x)\). | [
"In general we will show that if \\(m\\) is an integer less than \\(n\\) and \\(P_0, \\dotsc, P_{m-1}\\) and \\(S\\) are polynomials satisfying\n\n\\[\nP_0(x^n) + x P_1(x^n) + \\dotsb + x^{m-1} P_{m-1}(x^n) = \\sum_{k=0}^{n-1} x^k S(x),\n\\]\n\nthen \\(P_k(1) = 0\\), for all integers \\(0 \\le k \\le m-1\\). For th... |
USAMO-1977-1 | https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_1 | Determine all pairs of positive integers \((m,n)\) such that \((1+x^n+x^{2n}+\cdots+x^{mn})\) is divisible by \((1+x+x^2+\cdots+x^{m})\). | [
"Denote the first and larger polynomial to be \\(f(x)\\) and the second one to be \\(g(x)\\). In order for \\(f(x)\\) to be divisible by \\(g(x)\\) they must have the same roots. The roots of \\(g(x)\\) are the (m+1)th roots of unity, except for 1. When plugging into \\(f(x)\\), the root of unity is a root of \\(f(... |
USAMO-1977-2 | https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_2 | \(ABC\) and \(A'B'C'\) are two triangles in the same plane such that the lines \(AA',BB',CC'\) are mutually parallel. Let \([ABC]\) denote the area of triangle \(ABC\) with an appropriate \(\pm\) sign, etc.; prove that
\[
3([ABC]+ [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B']+ [A'BC]+[B'CA] + [C'AB].
\] | [] |
USAMO-1977-3 | https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_3 | If \(a\) and \(b\) are two of the roots of \(x^4+x^3-1=0\), prove that \(ab\) is a root of \(x^6+x^4+x^3-x^2-1=0\). | [
"Given the roots \\(a,b,c,d\\) of the equation \\(x^{4}+x^{3}-1=0\\).\n\nFirst, Vieta's relations give \\(a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abc+abd+acd+bcd=0, abcd = -1\\).\n\nThen \\(cd=-\\frac{1}{ab}\\) and \\(c+d=-1-(a+b)\\).\n\nThe other coefficients give \\(ab+(a+b)(c+d)+cd = 0\\) or \\(ab+(a+b)[-1-(a+b)]-\\... |
USAMO-1977-4 | https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_4 | Prove that if the opposite sides of a skew (non-planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the ... | [
"We first prove that if the opposite sides are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals.\n\nLet the vertices of the quadrilateral be A, B, C, D, in that order. Thus, the opposite sides congruent condition translates to (after squaring):\n\n\\[\nAB^2 = C... |
USAMO-1977-5 | https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_5 | If \(a,b,c,d,e\) are positive numbers bounded by \(p\) and \(q\), i.e, if they lie in \([p,q], 0 < p\), prove that
\[
(a+b +c +d +e)\left(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}\right) \le 25 + 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2
\]
and determine when there is equality. | [
"Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but \\(x\\). Then the expression on the LHS has the form \\((r + x)(s + \\frac{1}{x}) = (rs + 1) + sx + \\frac{r}{x}\\), where \\(r\\) and \\(s\\) are fixed. But this is convex. That is to say, as \\(x\\) increases if first dec... |
USAMO-1978-1 | https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_1 | Given that \(a,b,c,d,e\) are real numbers such that
\[
a+b+c+d+e=8,
\]
\[
a^2+b^2+c^2+d^2+e^2=16.
\]
Determine the maximum value of \(e\). | [
"By Cauchy Schwarz, we can see that \\((1+1+1+1)(a^2+b^2+c^2+d^2)\\geq (a+b+c+d)^2\\) thus \\(4(16-e^2)\\geq (8-e)^2\\) Finally, \\(e(5e-16) \\geq 0\\) which means \\(\\frac{16}{5} \\geq e \\geq 0\\) so the maximum value of \\(e\\) is \\(\\frac{16}{5}\\).\n\nfrom: Image from Gon Mathcenter.net",
"Seeing as we hav... |
USAMO-1978-2 | https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_2 | \(ABCD\) and \(A'B'C'D'\) are square maps of the same region, drawn to different scales and superimposed as shown in the figure. Prove that there is only one point \(O\) on the small map that lies directly over point \(O'\) of the large map such that \(O\) and \(O'\) each represent the same place of the country. Also, ... | [
"The point is obviously unique, because the two maps have different scales (but if P and Q where two fixed points the distance between them would be the same on both maps).\n\nLet the small map square be A'B'C'D' and the large be ABCD, where X and X' are corresponding points. We deal first with the special case whe... |
USAMO-1978-3 | https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_3 | An integer \(n\) will be called good if we can write
\(n=a_1+a_2+\cdots+a_k\),
where \(a_1,a_2, \ldots, a_k\) are positive integers (not necessarily distinct) satisfying
\(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1\).
Given the information that the integers 33 through 73 are good, prove that every integer \... | [
"Lemma: If \\(g\\) is a good integer, then so is \\(2g + 2\\) and \\(2g + 9\\).\n\nProof: Let \\(g = a_1 + a_2 + \\cdots + a_k\\) such that\n\n\\[\n\\frac{1}{a_1}+\\frac{1}{a_2}+\\cdots+\\frac{1}{a_k}=1,\n\\]\n\nwhere the \\(a_i\\) are positive integers. Then, notice that\n\n\\[\n1 = \\frac{1}{2} + \\frac{1}{2} = \... |
USAMO-1978-4 | https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_4 | (a) Prove that if the six dihedral (i.e. angles between pairs of faces) of a given tetrahedron are congruent, then the tetrahedron is regular.
(b) Is a tetrahedron necessarily regular if five dihedral angles are congruent? | [
"(a) Let \\(ABCD\\) be the said tetrahedron, and let the inscribed sphere of \\(ABCD\\) touch the faces at \\(W, X, Y, Z\\). Then, \\(OW, OX, OY, OZ\\) are normals to the respective faces. We know that the angle between any two normals is equal, so we have \\(|OW|=|OX|=|OY|=|OZ|\\) at equal angles. Now, since\n\n\\... |
USAMO-1978-5 | https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_5 | Nine mathematicians meet at an international conference and discover that among any three of them, at least two speak a common language. If each of the mathematicians speak at most three languages, prove that there are at least three of the mathematicians who can speak the same language. | [
"Suppose no three mathematicians speak the same language. Then person A can share some language with at most 3 other delegates, because if he shared some language with 4 delegates there would be 3 with the same language. So there are 5 delegates who do not share a language with A. Let one of them be B. Using the sa... |
USAMO-1979-1 | https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_1 | Determine all non-negative integral solutions \((n_1,n_2,\dots , n_{14})\) if any, apart from permutations, of the Diophantine Equation \(n_1^4+n_2^4+\cdots +n_{14}^4=1599\). | [
"Recall that \\(n_i^4\\equiv 0,1\\bmod{16}\\) for all integers \\(n_i\\). Thus the sum we have is anything from 0 to 14 modulo 16. But \\(1599\\equiv 15\\bmod{16}\\), and thus there are no integral solutions to the given Diophantine equation.",
"In base \\(16\\), this equation would look like:\n\n\\[\nn_1^4+n_2^4... |
USAMO-1979-2 | https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_2 | \(N\) is the north pole. \(A\) and \(B\) are points on a great circle through \(N\) equidistant from \(N\). \(C\) is a point on the equator. Show that the great circle through \(C\) and \(N\) bisects the angle \(ACB\) in the spherical triangle \(ABC\) (a spherical triangle has great circle arcs as sides). | [
"\\[\nUSAMO 1979 P2a.png\n\\]\n\nSince \\(N\\) is the north pole, we define the Earth with a sphere of radius one in space with \\(N=(0,0,1)\\) and sphere center \\(O=(0,0,0)\\) We then pick point \\(N\\) on the sphere and define the \\(xz\\)-plane as the plane that contains great circle points \\(A\\) , \\(B\\), a... |
USAMO-1979-3 | https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_3 | \(a_1, a_2, \ldots, a_n\) is an arbitrary sequence of positive integers. A member of the sequence is picked at random. Its value is \(a\). Another member is picked at random, independently of the first. Its value is \(b\). Then a third value, \(c\). Show that the probability that \(a + b +c\) is divisible by \(3\) is a... | [
"Let x equal the probability of picking an element of the sequence equivalent to 0 mod 3, y equal those 1 mod 3, and z equal those 2 mod 3. Then, considering that 0+0+0, 1+1+1, 2+2+2, and 0+1+2 are divisible by 3, we obtain the equivalent inequality in the First Hint. This simplifies to \\(x^3 + y^3 + z^3 + 6xyz \\... |
USAMO-1979-4 | https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_4 | \(P\) lies between the rays \(OA\) and \(OB\). Find \(Q\) on \(OA\) and \(R\) on \(OB\) collinear with \(P\) so that \(\frac{1}{PQ} + \frac{1}{PR}\) is as large as possible. | [
"Perform the inversion with center \\(P\\) and radius \\(\\overline{PO}.\\) Lines \\(OA,OB\\) go to the circles \\((O_1),(O_2)\\) passing through \\(P,O\\) and the line \\(QR\\) cuts \\((O_1),(O_2)\\) again at the inverses \\(Q',R'\\) of \\(Q,R.\\) Hence\n\n\\[\n\\frac{1}{PQ}+\\frac{1}{PR}=\\frac{PQ'+PR'}{PO^2}=\\f... |
USAMO-1979-5 | https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_5 | Let \(A_1,A_2,...,A_{n+1}\) be distinct subsets of \([n]\) with \(|A_1|=|A_2|=\cdots =|A_{n+1}|=3\). Prove that \(|A_i\cap A_j|=1\) for some pair \(\{i,j\}\). | [
"Suppose the problem statement does not hold. It is clear that \\(n \\ge 4\\). Choose the smallest \\(n\\) such that \\(|A_i \\cap A_j| \\neq 1\\) for each \\(\\{i, j\\}\\).\n\nFirst, the \\((n+1)\\) subsets have \\(3(n+1)\\) elements (some repeated) in conjunction. Because there are \\(n\\) elements of \\([n]\\) t... |
USAMO-1980-1 | https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_1 | A balance has unequal arms and pans of unequal weight. It is used to weigh three objects. The first object balances against a weight \(A\), when placed in the left pan and against a weight \(a\), when placed in the right pan. The corresponding weights for the second object are \(B\) and \(b\). The third object balances... | [
"The effect of the unequal arms and pans is that if an object of weight \\(x\\) in the left pan balances an object of weight \\(y\\) in the right pan, then \\(x = hy + k\\) for some constants \\(h\\) and \\(k\\). Thus if the first object has true weight x, then \\(x = hA + k, a = hx + k\\).\n\nSo \\(a = h^2A + (h+1... |
USAMO-1980-2 | https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_2 | Find the maximum possible number of three term arithmetic progressions in a monotone sequence of \(n\) distinct reals. | [
"Consider the first few cases for \\(n\\) with the entire \\(n\\) numbers forming an arithmetic sequence\n\n\\[\n(1, 2, 3, \\ldots, n)\n\\]\n\nIf \\(n = 3\\), there will be one ascending triplet (123). Let's only consider the ascending order for now. If \\(n = 4\\), the first 3 numbers give 1 triplet, the addition ... |
USAMO-1980-3 | https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_3 | \(A + B + C\) is an integral multiple of \(\pi\). \(x, y,\) and \(z\) are real numbers. If \(x\sin(A)+y\sin(B)+z\sin(C)=x^2\sin(2A)+y^2\sin(2B)+z^2\sin(2C)=0\), show that \(x^n\sin(nA)+y^n \sin(nB) +z^n \sin(nC)=0\) for any positive integer \(n\). | [
"Let \\(a=xe^{iA}\\), \\(b=ye^{iB}\\), \\(c=ze^{iC}\\) be numbers in the complex plane.\n\nNote that \\(A+B+C=k\\pi\\) implies \\(abc=xyz(e^{ik\\pi})=\\pm xyz\\) which is real. Also note that \\(x\\sin(A), y\\sin(B), z\\sin(C)\\) are the imaginary parts of \\(a, b, c\\) and that \\(x^2\\sin(2A), y^2\\sin(2B), z^2\\... |
USAMO-1980-4 | https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_4 | The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular. | [
"Let \\(ABCD\\) be the tetrahedron, and let \\(P\\) and \\(Q\\) be the points at which the insphere touches faces \\(ABC\\) and \\(ABD\\) respectively (and therefore the centroids of those faces). Looking at the plane containing \\(P\\), \\(A\\), and \\(Q\\), we see that the intersection of the sphere and the plane... |
USAMO-1980-5 | https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_5 | If \(x, y, z\) are reals such that \(0\le x, y, z \le 1\), show that \(\frac{x}{y + z + 1} + \frac{y}{z + x + 1} + \frac{z}{x + y + 1} \le 1 - (1 - x)(1 - y)(1 - z)\) | [
"Rewrite the given inequality so that \\(1\\) is isolated on the right side. Set the left side to be \\(f(x, y, z)\\). Now a routine computation shows\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2} = \\frac{2y}{(x + z + 1)^3} + \\frac{2z}{(x + y + 1)^3}\\geq 0\n\\]\n\nwhich shows that \\(f\\) is convex (concave up) i... |
USAMO-1981-1 | https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_1 | Prove that if \(n\) is not a multiple of \(3\), then the angle \(\frac{\pi}{n}\) can be trisected with ruler and compasses. | [
"Let \\(n=3k+1\\). Multiply throughout by \\(\\pi/3n\\). We get\n\n\\[\n\\frac{\\pi}{3} = \\frac{\\pi \\times k}{n} + \\frac{\\pi}{3n}\n\\]\n\nRe-arranging, we get\n\n\\[\n\\frac{\\pi}{3} - \\frac{\\pi \\times k}{n} = \\frac{\\pi}{3n}\n\\]\n\nA way to interpret it is that if we know the value \\(k\\), then the rema... |
USAMO-1981-2 | https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_2 | What is the largest number of towns that can meet the following criteria. Each pair is directly linked by just one of air, bus or train. At least one pair is linked by air, at least one pair by bus and at least one pair by train. No town has an air link, a bus link and a train link. No three towns, \(A, B, C\) are such... | [
"Assume \\(AB\\), \\(AC\\), and \\(AD\\) are all rail.\n\nNone of \\(BC\\), \\(CD\\), or \\(CD\\) can be rail, as those cities would form a rail triangle with \\(A\\).\n\nIf \\(BC\\) is bus, then \\(BD\\) is bus as well, as otherwise \\(B\\) has all three types.\n\nHowever, \\(CD\\) cannot be rail (as \\(\\triangle... |
USAMO-1981-3 | https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_3 | Show that for any triangle, \(\frac{3\sqrt{3}}{2}\ge \sin(3A) + \sin(3B) + \sin (3C) \ge -2\).
When does the equality hold? | [
"Given three angles that add to \\(180^\\circ\\), one can construct a triangle from them. However, its angles must all be nonnegative; thus the constraints on the angles of a triangle are \\(0^\\circ\\le A,B,C\\le180^\\circ\\) and \\(A+B+C=180^\\circ\\).\n\nIn fact, at this point, we only care about \\(3A, 3B,\\) a... |
USAMO-1981-4 | https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_4 | A convex polygon has \(n\) sides. Each vertex is joined to a point \(P\) not in the same plane. If \(A, B, C\) are adjacent vertices of the polygon take the angle between the planes \(PAB\) and \(PBC\). The sum of the \(n\) such angles equals the sum of the \(n\) angles in the polygon. Show that \(n=3\). | [] |
USAMO-1981-5 | https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_5 | Show that for any positive real \(x\), \([nx]\ge \sum_{1}^{n}\left(\frac{[kx]}{k}\right)\) | [
"First of all we write \\([kx]=kx-\\{kx\\}\\). So, we need to prove that \\(\\sum_{1}^{n}\\left(\\frac{\\{kx\\}}{k}\\right)\\geq \\{nx\\}.\\) Let's denote \\(a_k=\\{kx\\}\\). It is easy to see that \\(a_k+a_m \\geq a_{k+m}\\). We need to prove \\(\\sum_{1}^{n}\\left(\\frac{a_k}{k}\\right)\\geq a_n.\\)\n\nWe will pr... |
USAMO-1982-1 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_1 | In a party with \(1982\) people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else? | [
"We induct on \\(n\\) to prove that in a party with \\(n\\) people, there must be at least \\((n-3)\\) people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people \\(A, B, C\\), who do not know each other.)\n\nBase case: \\(n = 4\\) is obvious.\n\nInductive s... |
USAMO-1982-2 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_2 | Let \(S_r=x^r+y^r+z^r\) with \(x,y,z\) real. It is known that if \(S_1=0\),
\((*)\) \(\frac{S_{m+n}}{m+n}=\frac{S_m}{m}\frac{S_n}{n}\)
for \((m,n)=(2,3),(3,2),(2,5)\), or \((5,2)\). Determine all other pairs of integers \((m,n)\) if any, so that \((*)\) holds for all real numbers \(x,y,z\) such that \(x+y+z=0\). | [
"Claim Both \\(m,n\\) can not be even.\n\nProof \\(x+y+z=0\\) ,\\(\\implies x=-(y+z)\\).\n\nSince \\(\\frac{S_{m+n}}{m+n} = \\frac{S_m S_n}{mn}\\),\n\nby equating cofficient of \\(y^{m+n}\\) on LHS and RHS ,get\n\n\\(\\frac{2}{m+n}=\\frac{4}{mn}\\).\n\n\\(\\implies \\frac{m}{2} + \\frac {n}{2} = \\frac{m\\cdot n}{2... |
USAMO-1982-3 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_3 | If a point \(A_1\) is in the interior of an equilateral triangle \(ABC\) and point \(A_2\) is in the interior of \(\triangle{A_1BC}\), prove that
\(I.Q. (A_1BC) > I.Q.(A_2BC)\),
where the isoperimetric quotient of a figure \(F\) is defined by
\[
I.Q.(F) = \frac{\text{Area (F)}}{\text{[Perimeter (F)]}^2}
\] | [
"First, an arbitrary triangle \\(ABC\\) has isoperimetric quotient (using the notation \\([ABC]\\) for area and \\(s = \\frac{a + b + c}{2}\\)):\n\n\\[\n\\frac{[ABC]}{4s^2} = \\frac{[ABC]^3}{4s^2 [ABC]^2} = \\frac{r^3 s^3}{4s^2 \\cdot s(s-a)(s-b)(s-c)} = \\frac{r^3}{4(s-a)(s-b)(s-c)}\n\\]\n\n\\[\n= \\frac{1}{4} \\c... |
USAMO-1982-4 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_4 | Prove that there exists a positive integer \(k\) such that \(k\cdot2^n+1\) is composite for every integer \(n\). | [
"Indeed, \\(\\boxed{k=2935363331541925531}\\) has the requisite property.\n\nTo see why, consider the primes \\(3,\\ 5,\\ 17,\\ 257,\\ 65537,\\ 6700417,\\ 641\\), and observe that\n\n\\[\n\\begin{align*}k&\\equiv 1\\pmod{3}\\\\ k&\\equiv 1\\pmod{5}\\\\ k&\\equiv 1\\pmod{17}\\\\ k&\\equiv 1\\pmod{257}\\\\ k&\\equiv ... |
USAMO-1982-5 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_5 | \(A,B\), and \(C\) are three interior points of a sphere \(S\) such that \(AB\) and \(AC\) are perpendicular to the diameter of \(S\) through \(A\), and so that two spheres can be constructed through \(A\), \(B\), and \(C\) which are both tangent to \(S\). Prove that the sum of their radii is equal to the radius of \(S... | [
"Let the two tangent spheres be \\(S_1\\) and \\(S_2\\), and let \\(O, O_1, O_2\\) and \\(R, R_1, R_2\\) be the origins and radii of \\(S, S_1, S_2\\) respectively. Then \\(AO\\) stands normal to the plane \\(P\\) through \\(\\Delta ABC\\). Because both spheres go through \\(A\\), \\(B\\), and \\(C\\), the line \\(... |
USAMO-1983-1 | https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_1 | On a given circle, six points \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) are chosen at random, independently and uniformly with respect to arc length. Determine the probability that the two triangles \(ABC\) and \(DEF\) are disjoint, i.e., have no common points. | [
"First we give the circle an orientation (e.g., letting the circle be the unit circle in polar coordinates). Then, for any set of six points chosen on the circle, there are exactly \\(6!\\) ways to label them one through six. Also, this does not affect the probability we wish to calculate. This will, however, make ... |
USAMO-1983-2 | https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_2 | Prove that the zeros of
\[
x^5+ax^4+bx^3+cx^2+dx+e=0
\]
cannot all be real if \(2a^2<5b\). | [
"We prove the contrapositive: if the polynomial in question has the five real roots \\(x_1, x_2, x_3, x_4, x_5\\), then \\(5b \\le 2a^2\\).\n\nBecause \\(a = -(x_1 + x_2 + x_3 + x_4 + x_5)\\) and \\(b = x_1x_2 + x_1x_3 + ... + x_4x_5\\) by Vieta's Formulae, we have\n\n\\[\n2b = 2x_1x_2 + 2x_1x_3 + ... + 2x_4x_5 = (... |
USAMO-1983-3 | https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_3 | Each set of a finite family of subsets of a line is a union of two closed intervals. Moreover, any three of the sets of the family have a point in common. Prove that there is a point which is common to at least half the sets of the family. | [
"Let us first see that this works for anything less than three sets.\n\nObviously, due to the second condition given, they all share at least one point in common.\n\nNow, let us see that it works for three or more sets.\n\nFirst, take any three of the subsets in the family of sets.\n\nLet them be A, B, and C. We kn... |
USAMO-1983-4 | https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_4 | Six segments \(S_1, S_2, S_3, S_4, S_5,\) and \(S_6\) are given in a plane. These are congruent to the edges \(AB, AC, AD, BC, BD,\) and \(CD\), respectively, of a tetrahedron \(ABCD\). Show how to construct a segment congruent to the altitude of the tetrahedron from vertex \(A\) with straight-edge and compasses. | [
"Throughout this solution, we denote the length of a segment \\(S\\) by \\(|S|\\).\n\nIn this solution, we employ several lemmas. Two we shall take for granted: given any point \\(A\\) and a line \\(\\ell\\) not passing through \\(A\\), we can construct a line \\(\\ell'\\) through \\(A\\) parallel to \\(\\ell\\); a... |
USAMO-1983-5 | https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_5 | Consider an open interval of length \(1/n\) on the real number line, where \(n\) is a positive integer. Prove that the number of irreducible fractions \(p/q\), with \(1\le q\le n\), contained in the given interval is at most \((n+1)/2\). | [
"Let \\(I\\) be an open interval of length \\(1/n\\) and \\(F_n\\) the set of fractions \\(p/q\\in I\\) with \\(p,q\\in\\mathbb{Z}\\), \\(\\gcd(p,q)=1\\) and \\(1\\leq q\\leq n\\).\n\nAssume that \\(\\frac{p}{q}\\in F_n\\). If \\(k\\in\\mathbb{Z}\\) is such that \\(1\\leq kq\\leq n\\), and \\(p'\\in\\mathbb{Z}\\) i... |
USAMO-1984-1 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial \(x^4 - 18x^3 + kx^2 + 200x - 1984 = 0\), the product of \(2\) of its roots is \(- 32\). Find \(k\). | [
"Using Vieta's formulas, we have:\n\n\\[\n\\begin{align*}a+b+c+d &= 18,\\\\ ab+ac+ad+bc+bd+cd &= k,\\\\ abc+abd+acd+bcd &=-200,\\\\ abcd &=-1984.\\\\ \\end{align*}\n\\]\n\nFrom the last of these equations, we see that \\(cd = \\frac{abcd}{ab} = \\frac{-1984}{-32} = 62\\). Thus, the second equation becomes \\(-32+ac... |
USAMO-1984-2 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_2 | The geometric mean of any set of \(m\) non-negative numbers is the \(m\)-th root of their product.
\(\quad (\text{i})\quad\) For which positive integers \(n\) is there a finite set \(S_n\) of \(n\) distinct positive integers such that the geometric mean of any subset of \(S_n\) is an integer?
\(\quad (\text{ii})\quad... | [
"a) We claim that for any numbers \\(p_1\\), \\(p_2\\), ... \\(p_n\\), \\(p_1^{n!}, p_2^{n!}, ... p_n^{n!}\\) will satisfy the condition, which holds for any number \\(n\\).\n\nSince \\(\\sqrt[n] ab = \\sqrt[n] a * \\sqrt[n] b\\), we can separate each geometric mean into the product of parts, where each part is the... |
USAMO-1984-3 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_3 | \(P\), \(A\), \(B\), \(C\), and \(D\) are five distinct points in space such that \(\angle APB = \angle BPC = \angle CPD = \angle DPA = \theta\), where \(\theta\) is a given acute angle. Determine the greatest and least values of \(\angle APC + \angle BPD\). | [
"Greatest value is achieved when all the points are as close as possible to all being on a plane.\n\nSince \\(\\theta < \\frac{\\pi}{2}\\), then \\(\\angle APC + \\angle BPD < \\pi\\)\n\nSmallest value is achieved when point P is above and the remaining points are as close as possible to colinear when \\(\\theta > ... |
USAMO-1984-4 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_4 | A difficult mathematical competition consisted of a Part I and a Part II with a combined total of \(28\) problems. Each contestant solved \(7\) problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who, in Part I, solved either no p... | [] |
USAMO-1984-5 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_5 | \(P(x)\) is a polynomial of degree \(3n\) such that
\[
\begin{eqnarray*} P(0) = P(3) = \cdots &=& P(3n) = 2, \\ P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\ P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\ && P(3n+1) = 730.\end{eqnarray*}
\]
Determine \(n\). | [
"By Lagrange Interpolation Formula \\(f(x) = 2\\sum_{p=0}^{n}\\left ( \\prod_{0\\leq r\\neq3p\\leq 3n}^{{}}\\frac{x-r}{3p-r} \\right )+ \\sum_{p=1}^{n}\\left ( \\prod_{0\\leq r\\neq3p-2\\leq 3n}^{{}} \\frac{x-r}{3p-2-r}\\right )\\)\n\nand hence \\(f(3n+1) = 2\\sum_{p=0}^{n}\\left ( \\prod_{0\\leq r\\neq3p\\leq 3n}^... |
USAMO-1985-1 | https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_1 | Determine whether or not there are any positive integral solutions of the simultaneous equations
\[
\begin{align*} x_1^2 +x_2^2 +\cdots +x_{1985}^2 & = y^3,\\ x_1^3 +x_2^3 +\cdots +x_{1985}^3 & = z^2 \end{align*}
\]
with distinct integers \(x_1,x_2,\cdots,x_{1985}\). | [
"Lemma: For a positive integer \\(n\\), \\(1^3+2^3+\\cdots +n^3 = (1+2+\\cdots +n)^2\\) (Also known as Nicomachus's theorem)\n\nProof by induction: The identity holds for \\(1\\). Suppose the identity holds for a number \\(n\\). It is well known that the sum of first \\(n\\) positive integers is \\(\\frac{n(n+1)}{2... |
USAMO-1985-2 | https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_2 | Determine each real root of
\[
x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0
\]
correct to four decimal places. | [
"The equation can be re-written as\n\n\\[\n\\begin{align}\\label{eqn1} (x+10^5)^2(x-10^5)^2 -(x+10^5)(x-10^5) -x-1=0. \\end{align}\n\\]\n\nWe first prove that the equation has no negative roots. Let \\(x\\le 0.\\) The equation above can be further re-arranged as\n\n\\[\n\\begin{align*}[(x+10^5)(x-10^5)+1][(x+10^5)(... |
USAMO-1985-3 | https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_3 | Let \(A,B,C,D\) denote four points in space such that at most one of the distances \(AB,AC,AD,BC,BD,CD\) is greater than \(1\). Determine the maximum value of the sum of the six distances. | [
"Suppose that \\(AB\\) is the length that is more than \\(1\\). Let spheres with radius \\(1\\) around \\(A\\) and \\(B\\) be \\(S_A\\) and \\(S_B\\). \\(C\\) and \\(D\\) must be in the intersection of these spheres, and they must be on the circle created by the intersection to maximize the distance. We have \\(AC ... |
USAMO-1985-4 | https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_4 | There are \(n\) people at a party. Prove that there are two people such that, of the remaining \(n-2\) people, there are at least \(\lfloor n/2\rfloor -1\) of them, each of whom knows both or else knows neither of the two. Assume that "know" is a symmetrical relation; \(\lfloor x\rfloor\) denotes the greatest integer l... | [
"Consider the number of pairs (X, {Y, Z}), where X, Y, Z are distinct points such that X is joined to just one of Y, Z. If X is joined to just k points, then there are just k(n - 1 - k) ≤ (n - 1)2/4 such pairs (X, {Y, Z}). Hence in total there are at most \\(\\frac{n(n - 1)^2}{4}\\) such pairs. But there are \\(\\f... |
USAMO-1985-5 | https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_5 | Let \(a_1,a_2,a_3,\cdots\) be a non-decreasing sequence of positive integers. For \(m\ge1\), define \(b_m=\min\{n: a_n \ge m\}\), that is, \(b_m\) is the minimum value of \(n\) such that \(a_n\ge m\). If \(a_{19}=85\), determine the maximum value of \(a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}\). | [
"We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the \\(i\\)th column we fill the first \\(a_i\\) cells with one dot each. Then the \\(19\\)th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. ... |
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