geometry rectangle perimeter function probability ratio trigonometry

Here is the IMO shortlist problems 1985, I got it from internet.

Here is a new version of the above International Mathematial Olympiad (IMO) Longlist 1985 attachment which is a bit more comprehensive. For Windows there is a [url=http://www.stdutility.com/download/stduviewer.exe]STDU viewer.[/url]

[quote="TALENTO"]Hello Orl not get to see the file longlisted 1985 djvu (Appears as php) Please I wish I sent him back or if someone has Talves in pdf would be great thank you very much in advance.[/quote]

In the link above you have a link to a DJVU-reader for Windows. [url=http://www.4shared.com/dir/19895350/70e80fee/International_Mathematical_Olympiad__IMO__Shortlisted_Problems.html]Here[/url] you can download all the documents, i.e. ILL 1985 in djvu.

I appreciate that you are very kind

Greetings  :D

[b][size=150]IMO 1985 LongListed Problems: (Posted in Forums)[/size][/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000885&#p2000885]1. (AUS 1)[/url][b](SL85-4)[/b]Each of the numbers in the set $N = \{1, 2, 3, \cdots, n - 1\}$, where $n \geq 3$, is colored with one of two colors, say red or black, so that:

[i](i)[/i] $i$ and $n - i$ always receive the same color, and

[i](ii)[/i] for some $j \in N$, relatively prime to $n$, $i$ and $|j - i|$ receive the same color for all $i \in N, i \neq j.$

Prove that all numbers in N must receive the same color.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014809#p2014809]2. (AUS 2)[/url] We are given a triangle $ABC$ and three rectangles $R_1,R_2,R_3$ with sides parallel to two fixed perpendicular directions and such that their union covers the sides $AB,BC$, and $CA$; i.e., each point on the perimeter of $ABC$ is contained in or on at least one of the rectangles. Prove that all points inside the triangle are also covered by the union of $R_1,R_2,R_3.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014812#p2014812]3. (AUS 3)[/url] A function f has the following property: If $k > 1, j > 1$, and $\gcd(k, j) = m$, then $f(kj) = f(m) (f(k/m) + f(j/m))$. What values can $f(1984)$ and $f(1985)$ take?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014821#p2014821]4. (BEL 1)[/url]Let x, y, and z be real numbers satisfying $x + y + z = xyz.$ Prove that
\[x(1 - y^2)(1 - z^2) + y(1 -z^2)(1 - x^2) + z(1 - x^2)(1 - y^2) = 4xyz.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000962&#p2000962]5. (BEL 2)[/url] [b](SL85-16)[/b] If possible, construct an equilateral triangle whose three vertices are on three given circles.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014826#p2014826]6. (BEL 3)[/url] On a one-way street, an unending sequence of cars of width $a$, length $b$ passes with velocity $v$. The cars are separated by the distance $c$. A pedestrian crosses the street perpendicularly with velocity $w$, without paying attention to the cars.

[b](a)[/b] What is the probability that the pedestrian crosses the street uninjured?

[b](b)[/b] Can he improve this probability by crossing the road in a direction other than perpendicular?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014828#p2014828]7. (BRA 1)[/url] A convex quadrilateral is inscribed in a circle of radius $1$. Prove that the difference between its perimeter and the sum of the lengths of its diagonals is greater than zero and less than $2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014832#p2014832]8. (BRA 2)[/url] Let $K $ be a convex set in the $xy$-plane, symmetric with respect to the origin and having area greater than $4 $. Prove that there exists a point $(m, n) \neq (0, 0)$ in $K$ such that $m$ and $n$ are integers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2001038&#p2001038]9. (BRA 3)[/url] [b](SL85-2)[/b] A polyhedron has $12$ faces and is such that:

[b][i](i)[/i][/b] all faces are isosceles triangles,
[b][i](ii)[/i][/b] all edges have length either $x$ or $y$,
[b][i](iii)[/i][/b] at each vertex either $3$ or $6$ edges meet, and
[b][i](iv)[/i][/b] all dihedral angles are equal.

Find the ratio $x/y.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000901&#p2000901]10. (BUL 1)[/url] [b](SL85-13)[/b] Let $m$ boxes be given, with some balls in each box. Let $n < m$ be a given integer. The following operation is performed: choose $n$ of the boxes and put $1$ ball in each of them. Prove:

[i](a) [/i]If $m$ and $n$ are relatively prime, then it is possible, by performing the operation a finite number of times, to arrive at the situation that all the boxes contain an equal number of balls.

[i](b)[/i] If $m$ and $n$ are not relatively prime, there exist initial distributions of balls in the boxes such that an equal distribution is not possible to achieve.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014840#p2014840]11. (BUL 2)[/url] Let $a$ and $ b$ be integers and $n$ a positive integer. Prove that
\[\frac{b^{n-1}a(a + b)(a + 2b) \cdots (a + (n - 1)b)}{n!}\]
is an integer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014849#p2014849]12. (CAN 1)[/url]Find the maximum value of
\[\sin^2 \theta_1+\sin^2 \theta_2+\cdots+\sin^2 \theta_n\]
subject to the restrictions $0 \leq \theta_i , \theta_1+\theta_2+\cdots+\theta_n=\pi.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014850#p2014850]13. (CAN 2)[/url] Find the average of the quantity
\[(a_1 - a_2)^2 + (a_2 - a_3)^2 +\cdots + (a_{n-1} -a_n)^2\]
taken over all permutations $(a_1, a_2, \dots , a_n)$ of $(1, 2, \dots , n).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014913#p2014913]14. (CAN 3)[/url] Let $k$ be a positive integer. Define $u_0 = 0, u_1 = 1$, and $u_n=ku_{n-1}-u_{n-2} , n \geq 2.$ Show that for each integer $n$, the number $u_1^3 + u_2^3 +\cdots+ u_n^3 $is a multiple of $u_1 + u_2 +\cdots+ u_n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014925#p2014925]15. (CAN 4)[/url] [i]Superchess[/i] is played on a $12 \times 12$ board, and it uses [i]superknights[/i], which move between opposite corner cells of any $3\times4$ subboard. Is it possible for a [i]superknight[/i] to visit every other cell of a superchessboard exactly once and return to its starting cell ?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2001045&#p2001045]16. (CAN 5)[/url] [b](SL85-18)[/b] Let $x_1, x_2, \cdots , x_n$ be positive numbers. Prove that
\[\frac{x_1^2}{x_1^2+x{}_2x{}_3} + \frac{x_2^2}{x_2^2+x{}_3x{}_4} + \cdots +\frac{x_{n-1}^2}{x_{n-1}^2+x{}_nx{}_1} +\frac{x_n^2}{x_n^2+x{}_1x{}_2} \leq n-1\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014933#p2014933]17. (CUB 1)[/url] Set
\[A_n=\sum_{k=1}^n \frac{k^6}{2^k}.\]
Find $\lim_{n\to\infty} A_n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014937#p2014937]18. (CYP 1)[/url] The circles $(R, r)$ and $(P, \rho)$, where $r > \rho$, touch externally at $A$. Their direct common tangent touches $(R, r)$ at B and $(P, \rho)$ at $C$. The line $RP$ meets the circle $(P, \rho)$ again at $D$ and the line $BC$ at $E$. If $|BC| = 6|DE|$, prove that:
[b](a)[/b] the lengths of the sides of the triangle $RBE$ are in an arithmetic progression, and

[b](b)[/b] $|AB| = 2|AC|.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014940#p2014940]19. (CYP 2)[/url] Solve the system of simultaneous equations
\[\sqrt x - \frac 1y - 2w + 3z = 1,\]\[x + \frac{1}{y^2} - 4w^2 - 9z^2 = 3,\]\[x \sqrt x - \frac{1}{y^3} - 8w^3 + 27z^3 = -5,\]\[x^2 + \frac{1}{y^4} - 16w^4 - 81z^4 = 15.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014942#p2014942]20. (CZS 1)[/url] Let $T$ be the set of all lattice points (i.e., all points with integer coordinates) in three-dimensional space. Two such points $(x, y, z)$ and $(u, v,w)$ are called [i]neighbors[/i] if $|x - u| + |y - v| + |z - w| = 1$. Show that there exists a subset $S$ of $T$ such that for each $p \in T$ , there is exactly one point of $S$ among $p$ and its [i]neighbors[/i].

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014944#p2014944]21. (CZS 2)[/url] Let $A$ be a set of positive integers such that for any two elements $x, y$ of A, $|x-y| \geq \frac{xy}{25}.$ Prove that $A$ contains at most nine elements. Give an example of such a set of nine elements.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000897&#p2000897]22. (CZS 3)[/url] [b](SL85-7)[/b] The positive integers $x_1, \cdots , x_n$, $n \geq 3$, satisfy $x_1 < x_2 <\cdots< x_n < 2x_1$. Set $P = x{}_1x{}_2 \cdots x_n.$ Prove that if $p$ is a prime number, $k$ a positive integer, and $P$ is divisible by $pk$, then $\frac{P}{p^k} \geq n!.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014948#p2014948]23. (CZS 4)[/url] Let $\mathbb N = {1, 2, 3, . . .}$. For real $x, y$, set $S(x, y) = \{s | s = [nx+y], n \in \mathbb N\}$. Prove that if $r > 1$ is a rational number, there exist real numbers $u$ and $v$ such that
\[S(r, 0) \cap S(u, v) = \emptyset, S(r, 0) \cup S(u, v) = \mathbb N.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017727#p2017727]24. (FRA 1)[/url]Let $d \geq 1$ be an integer that is not the square of an integer. Prove that for every integer $n \geq 1,$
\[(n \sqrt d +1) \cdot | \sin(n \pi \sqrt d )| \geq 1\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017728#p2017728]25. (FRA 2)[/url] Find eight positive integers $n_1, n_2, \dots , n_8$ with the following property: For every integer $k$, $-1985 \leq k \leq 1985$, there are eight integers $a_1, a_2, \dots, a_8$, each belonging to the set $\{?1, 0, 1\}$, such that $k=\sum_{i=1}^{8} a_i n_i .$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2001021&#p2001021]26. (FRA 3)[/url] [b](SL85-15)[/b] Let $K$ and $K'$ be two squares in the same plane, their sides of equal length. Is it possible to decompose $K$ into a finite number of triangles $T_1, T_2, \cdots , T_p$ with mutually disjoint interiors and find translations $t_1, t_2,\cdots , t_p$ such that
\[K'=\bigcup_{i=1}^{p} t_i(T_i) \ ? \]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017730#p2017730]27. (FRA 4)[/url] Let $O$ be a point on the oriented Euclidean plane and $(\mathbf i, \mathbf j)$ a directly oriented orthonormal basis. Let $C$ be the circle of radius $1$, centered at $O$. For every real number $t$ and non-negative integer$ n$ let $M_n$ be the point on $C$ for which $< \mathbf i , \overrightarrow{OM_n} > = \cos 2^n t.$ (or $\overrightarrow{OM_n} =\cos 2^n t \mathbf i +\sin 2^n t \mathbf j$).
Let $k \geq 2$ be an integer. Find all real numbers $t \in [0, 2\pi)$ that satisfy

[b](i)[/b] $M_0 = M_k$, and

[b](ii)[/b] if one starts from $M0$ and goes once around $C$ in the positive direction, one meets successively the points $M_0,M_1, \dots,M_{k-2},M_{k-1}$, in this order.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017732#p2017732]28. (FRG 1)[/url] [i]a)[/i] Let $M$ be the set of the lengths of an octahedron whose sides are congruent quadrangles. Prove that $M$ has at most three elements.

[i]b)[/i]  Let an octahedron whose sides are congruent quadrangles be given. Prove that each of these quadrangles has two equal sides meeting at a common vertex.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017733#p2017733]29. (FRG 2)[/url] [i]a)[/i] Call a four-digit number $(xyzt)_B$ in the number system with base $B$ stable if $(xyzt)_B = (dcba)_B - (abcd)_B$, where $a \leq b \leq c \leq d$ are the digits of $(xyzt)_B$ in ascending order. Determine all stable numbers in the number system with base $B.$

[i]b)[/i] With assumptions as in [i]a[/i] , determine the number of bases $B \leq 1985$ such that there is a stable number with base $B.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017735#p2017735]30. (GBR 1)[/url] A plane rectangular grid is given and a “rational point” is defined as a point $(x, y)$ where $x$ and $y$ are both rational numbers. Let $A,B,A',B'$ be four distinct rational points. Let $P$ be a point such that $\frac{A'B'}{AB}=\frac{B'P}{BP} = \frac{PA'}{PA}.$ In other words, the triangles $ABP, A'B'P$ are directly or oppositely similar. Prove that $P$ is in general a rational point and find the exceptional positions of $A'$ and $B'$ relative to $A$ and $B$ such that there exists a $P$ that is not a rational point.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017738#p2017738]31. (GBR 2)[/url] Let $E_1, E_2$, and $E_3$ be three mutually intersecting ellipses, all in the same plane. Their foci are respectively $F_2, F_3; F_3, F_1$; and $F_1, F_2$. The three foci are not on a straight line. Prove that the common chords of each pair of ellipses are concurrent.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017741#p2017741]32. (GBR 3)[/url] A collection of $2n$ letters contains $2$ each of $n$ different letters. The collection is partitioned into $n$ pairs, each pair containing $2$ letters, which may be the same or different. Denote the number of distinct partitions by $u_n$. (Partitions differing in the order of the pairs in the partition or in the order of the two letters in the pairs are not considered distinct.) Prove that $u_{n+1}=(n+1)u_n - \frac{n(n-1)}{2} u_{n-2}.$

(GBR 3) A pack of $n$ cards contains $n$ pairs of $2$ identical cards. It is shuffled and $2$ cards are dealt to each of $n$ different players. Let $p_n$ be the probability that every one of the $n$ players is dealt two identical cards. Prove that $\frac{1}{p_{n+1}}=\frac{n+1}{p_n} + \frac{n(n-1)}{2p_{n-2}}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000931&#p2000931]33. (GBR 4)[/url] [b](SL85-12)[/b] A sequence of polynomials $P_m(x, y, z), m = 0, 1, 2, \cdots$, in $x, y$, and $z$ is defined by $P_0(x, y, z) = 1$ and by
\[P_m(x, y, z) = (x + z)(y + z)P_{m-1}(x, y, z + 1) - z^2P_{m-1}(x, y, z)\]
for $m > 0$. Prove that each $P_m(x, y, z)$ is symmetric, in other words, is unaltered by any permutation of $x, y, z.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000186&#p2000186]34. (GBR 5)[/url] [b](SL85-20)[/b] A circle whose center is on the side $ED$ of the cyclic quadrilateral $BCDE$ touches the other three sides. Prove that $EB+CD = ED.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017756#p2017756]35. (GDR 1)[/url] We call a coloring $f$ of the elements in the set $M = \{(x, y) | x = 0, 1, \dots , kn - 1; y = 0, 1, \dots , ln - 1\}$ with $n$ colors allowable if every color appears exactly $k$ and $ l$ times in each row and column and there are no rectangles with sides parallel to the coordinate axes such that all the vertices in $M$ have the same color. Prove that every allowable coloring $f$ satisfies $kl \leq n(n + 1).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017761#p2017761]36. (GDR 2)[/url] Determine whether there exist $100$ distinct lines in the plane having exactly $1985$ distinct points of intersection.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017762#p2017762]37. (GDR 3)[/url] Prove that a triangle with angles $\alpha, \beta, \gamma$, circumradius $R$, and area $A$ satisfies
\[\tan \frac{ \alpha}{2}+\tan \frac{ \beta}{2}+\tan \frac{ \gamma}{2} \leq{9R^2}{4A}.\]

Remark. Can we determine all of equality cases ?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2001046&#p2001046]38. (IRE 1)[/url][b](SL85-21)[/b] The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $ABC$ meet at $X$. Let $M$ be the midpoint of $BC$. Prove that

[i](a)[/i] $\angle BAM = \angle CAX$, and

[i](b)[/i] $\frac{AM}{AX} = \cos\angle BAC.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017774#p2017774]39. (IRE 2)[/url] Given a triangle $ABC$ and external points $X, Y$ , and $Z$ such that $\angle BAZ = \angle CAY , \angle CBX = \angle ABZ$, and $\angle ACY = \angle BCX$, prove that $AX,BY$ , and $CZ$ are concurrent.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017785#p2017785]40. (IRE 3)[/url] Each of the numbers $x_1, x_2, \dots, x_n$ equals $1$ or $-1$ and
\[\sum_{i=1}^n x_i x_{i+1} x_{i+2} x_{i+3} =0\]
where $x_{n+i}=x_i. $ Prove that $4|n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000940&#p2000940]41. (IRE 4)[/url] [b](SL85-14)[/b] A set of $1985$ points is distributed around the circumference of a circle and each of the points is marked with $1$ or $-1$. A point is called “good” if the partial sums that can be formed by starting at that point and proceeding around the circle for any distance in either direction are all strictly positive. Show that if the number of points marked with $-1$ is less than $662$, there must be at least one good point.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017793#p2017793]42. (ISR 1)[/url]Prove that the product of two sides of a triangle is always greater than the product of the diameters of the inscribed circle and the circumscribed circle.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017815#p2017815]43. (ISR 2)[/url] Suppose that $1985$ points are given inside a unit cube. Show that one can always choose $32$ of them in such a way that every (possibly degenerate) closed polygon with these points as vertices has a total length of less than $8 \sqrt 3.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000178&#p2000178]44. (ISR 3)[/url] [b](SL85-19)[/b] For which integers $n \geq 3$ does there exist a regular $n$-gon in the plane such that all its vertices have integer coordinates in a rectangular coordinate system?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017817#p2017817]45. (ITA 1)[/url] Two persons, $X$ and $Y$ , play with a die. $X$ wins a game if the outcome is $1$ or $2$; $Y$ wins in the other cases. A player wins a match if he wins two consecutive games. For each player determine the probability of winning a match within $5$ games. Determine the probabilities of winning in an unlimited number of games. If $X$ bets $1$, how much must $Y$ bet for the game to be fair ?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017821#p2017821]46. (ITA 2)[/url] Let $C$ be the curve determined by the equation $y = x^3$ in the rectangular coordinate system. Let $t$ be the tangent to $C$ at a point $P$ of $C$; t intersects $C$ at another point $Q$. Find the equation of the set $L$ of the midpoints $M$ of $PQ$ as $P$ describes $C$. Is the correspondence associating $P$ and $M$ a bijection of $C$ on $L$ ? Find a similarity that transforms $C$ into $L.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017827#p2017827]47. (ITA 3)[/url] Let $F$ be the correspondence associating with every point $P = (x, y)$ the point $P' = (x', y')$ such that
\begin{eqnarray} x'= ax + b,\qquad  y'= ay + 2b. \end{eqnarray}
Show that if $a \neq 1$, all lines $PP'$ are concurrent. Find the equation of the set of points corresponding to $P = (1, 1)$ for $b = a^2$. Show that the composition of two mappings of type $(1)$ is of the same type.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017830#p2017830]48. (ITA 4)[/url] In a given country, all inhabitants are knights or knaves. A knight never lies; a knave always lies. We meet three persons, $A, B$, and $C$. Person $A$ says, “If $C$ is a knight, $B$ is a knave.” Person $C$ says, “$A$ and I are different; one is a knight and the other is a knave.” Who are the knights, and who are the knaves ?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000920&#p2000920]49. (MON 1)[/url] [b](SL85-1)[/b] Given a set $M$ of $1985$ positive integers, none of which has a prime divisor larger than $26$, prove that the set has four distinct elements whose geometric mean is an integer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017835#p2017835]50. (MON 2)[/url] From each of the vertices of a regular $n$-gon a car starts to move with constant speed along the perimeter of the $n$-gon in the same direction. Prove that if all the cars end up at a vertex $A$ at the same time, then they never again meet at any other vertex of the $n$-gon. Can they meet again at $A \ ?$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017840#p2017840]51. (MON 3)[/url] Let $f_1 = (a_1, a_2, \dots , a_n) , n > 2$, be a sequence of integers. From $f_1$ one constructs a sequence $f_k$ of sequences as follows: if $f_k = (c_1, c_2, \dots, cn)$, then $f_{k+1} = (c{}_{i{}_{1}}, c{}_{i{}_{2}}, c{}_{i{}_{3}} + 1, c{}_{i{}_{4}} + 1, . . . , c{}_{i{}_{n}} + 1)$, where $(c{}_{i{}_{1}}, c{}_{i{}_{2}},\dots , c{}_{i{}_{n}})$ is a permutation of $(c_1, c_2, \dots, c_n)$. Give a necessary and sufficient condition for $f_1$ under which it is possible for $f_k$ to be a constant sequence $(b_1, b_2,\dots , b_n), b_1 = b_2 =\cdots = b_n$, for some $k.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017843#p2017843]52. (MON 4)[/url] In the triangle $ABC$, let $B_1$ be on $AC, E$ on $AB, G$ on $BC$, and let $EG$ be parallel to $AC$. Furthermore, let $EG$ be tangent to the inscribed circle of the triangle $ABB_1$ and intersect $BB_1$ at $F$. Let $r, r_1$, and $r_2$ be the inradii of the triangles $ABC, ABB_1$, and $BFG$, respectively. Prove that $r = r_1 + r_2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017846#p2017846]53. (MON 5)[/url] For each $P$ inside the triangle $ABC$, let $A(P), B(P)$, and $C(P)$ be the points of intersection of the lines $AP, BP$, and $CP$ with the sides opposite to $A, B$, and $C$, respectively. Determine $P$ in such a way that the area of the triangle $A(P)B(P)C(P)$ is as large as possible.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017849#p2017849]54. (MOR 1)[/url] Set $S_n = \sum_{p=1}^n (p^5+p^7)$. Determine the greatest common divisor of $S_n$ and $S_{3n}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017852#p2017852]55. (MOR 2)[/url] The points $A,B,C$ are in this order on line $D$, and $AB = 4BC$. Let $M$ be a variable point on the perpendicular to $D$ through $C$. Let $MT_1$ and $MT_2$ be tangents to the circle with center $A$ and radius $AB$. Determine the locus of the orthocenter of the triangle $MT{}_1T{}_2.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018284#p2018284]56. (MOR 3)[/url] Let $ABCD$ be a rhombus with angle $\angle A = 60^\circ$. Let $E$ be a point, different from $D$, on the line $AD$. The lines $CE$ and $AB$ intersect at $F$. The lines $DF$ and $BE$ intersect at $M$. Determine the angle $\angle BMD$ as a function of the position of $E$ on $AD.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018314#p2018314]57. (NET 1)[/url] [i]a)[/i] The solid $S$ is defined as the intersection of the six spheres with the six edges of a regular tetrahedron $T$, with edge length $1$, as diameters. Prove that $S$ contains two points at a distance $\frac{1}{\sqrt 6}.$

[i]b)[/i] Using the same assumptions in [i]a)[/i], prove that no pair of points in $S$ has a distance larger than $\frac{1}{\sqrt 6}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017854#p2017854]58. (NET 2)[/url] Prove that there are infinitely many pairs $(k,N)$ of positive integers such that $1 + 2 + \cdots + k = (k + 1) + (k + 2)+\cdots + N.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2001069&#p2001069]59. (NET 3)[/url] [b](SL85-3)[/b].

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2017857#p2017857]60. (NOR 1)[/url] The sequence $(s_n)$, where $s_n= \sum_{k=1}^n \sin k$ for $n = 1, 2,\dots$ is bounded. Find an upper and lower bound.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018311#p2018311]61. (NOR 2)[/url] Consider the set $A = \{0, 1, 2, \dots , 9 \}$ and let $(B_1,B_2, \dots , B_k)$ be a collection of nonempty subsets of $A$ such that $B_i \cap B_j$ has at most two elements for $i \neq j$. What is the maximal value of $k \ ?$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018307#p2018307]62. (NOR 3)[/url] A “large” circular disk is attached to a vertical wall. It rotates clockwise with one revolution per minute. An insect lands on the disk and immediately starts to climb vertically upward with constant speed $\frac{\pi}{3}$ cm per second (relative to the disk). Describe the path of the insect

[i](a)[/i] relative to the disk;

[i](b)[/i] relative to the wall.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2001019&#p2001019]63. (POL 1)[/url] [b](SL85-6)[/b]. Let $x_n = \sqrt[2]{2+\sqrt[3]{3+\cdots+\sqrt[n]{n}}}.$ Prove that
\[x_{n+1}-x_n <\frac{1}{n!} \quad n=2,3,\cdots\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018315#p2018315]64. (POL 2)[/url] Let $p$ be a prime. For which k can the set $\{1, 2, \dots , k\}$ be partitioned into $p$ subsets with equal sums of elements ?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018320#p2018320]65. (POL 3)[/url] Define the functions $f, F : \mathbb N \to \mathbb N$, by
\[f(n)=\left[ \frac{3-\sqrt 5}{2} n \right] , F(k) =\min \{n \in  \mathbb N|f^k(n) > 0 \},\]
where $f^k = f \circ \cdots  \circ f$ is $f$ iterated $n$ times. Prove that $F(k + 2) = 3F(k + 1) - F(k)$ for all $k \in  \mathbb N.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000899&#p2000899]66. (ROM 1)[/url] [b](SL85-5)[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018322#p2018322]67. (ROM 2)[/url] Let $k \geq 2$ and $n_1, n_2, . . . , n_k \geq  1$ natural numbers having the property $n_2 | 2^{n_1} - 1, n_3 | 2^{n_2} -1 , \cdots, n_k | 2^{n_k-1}-1$, and $n_1 | 2^{n_k} - 1$. Show that $n_1 = n_2 = \cdots = n_k = 1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=631934#p631934]68. (ROM 3)[/url] Show that the sequence $\{a_n\}_{n\geq1}$ defined by $a_n = [n \sqrt 2]$ contains an infinite number of integer powers of $2$. ($[x]$ is the integer part of $x$.)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018324#p2018324]69. (ROM 4)[/url] Let $A$ and $B$ be two finite disjoint sets of points in the plane such that any three distinct points in $A \cup B$ are not collinear. Assume that at least one of the sets $A, B$ contains at least five points. Show that there exists a triangle all of whose vertices are contained in $A$ or in $B$ that does not contain in its interior any point from the other set.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018326#p2018326]70. (ROM 5)[/url] Let $C$ be a class of functions $f : \mathbb N \to \mathbb N$ that contains the functions $S(x) = x + 1$ and $E(x) = x - [\sqrt x]^2$ for every $x \in  \mathbb N$. ($[x]$ is the integer part of $x$.) If $C$ has the property that for every $f, g \in C, f + g, fg, f \circ g \in C$, show that the function $\max(f(x) - g(x), 0)$ is in $C.$ For every $f,g \in C.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018330#p2018330]71. (ROM 6)[/url] For every integer $r > 1$ find the smallest integer $h(r) > 1$ having the following property: For any partition of the set $\{1, 2, . . ., h(r)\}$ into $r$ classes, there exist integers $a \geq 0, 1 \leq x \leq y$ such that the numbers $a + x, a + y, a + x + y$ are contained in the same class of the partition.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018331#p2018331]72. (SPA 1)[/url] Construct a triangle $ABC$ given the side $AB$ and the distance $OH$ from the circumcenter $O$ to the orthocenter $H$, assuming that $OH$ and $AB$ are parallel.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018332#p2018332]73. (SPA 2)[/url] Let $A{}_1A{}_2,B{}_1B{}_2, C{}_1C{}_2$ be three equal segments on the three sides of an equilateral triangle. Prove that in the triangle formed by the lines $B{}_2C{}_1, C{}_2A{}_1,A{}_2B{}_1$, the segments $B{}_2C{}_1, C{}_2A{}_1,A{}_2B{}_1$ are proportional to the sides in which they are contained.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018333#p2018333]74. (SPA 3)[/url] Find all triples of positive integers $x, y, z$ satisfying
\[\frac 1x +\frac 1y + \frac 1z = \frac 45 .\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018335#p2018335]75. (SPA 4)[/url] Let $ABCD$ be a rectangle, $AB = a, BC = b$. Consider the family of parallel and equidistant straight lines (the distance between two consecutive lines being $d$) that are at an the angle $\phi, 0 \leq \phi \leq 90^{\circ},$ with respect to $AB$. Let $L$ be the sum of the lengths of all the segments intersecting the rectangle. Find:

[i](a)[/i] how $L $ varies,

[i](b)[/i] a necessary and sufficient condition for $L$ to be a constant, and

[i](c)[/i] the value of this constant.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018336#p2018336]76. (SWE 1)[/url] Are there integers $m$ and $n$ such that
\[5m^2 - 6mn + 7n^2 = 1985 \ ?\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018337#p2018337]77. (SWE 2)[/url] Two equilateral triangles are inscribed in a circle with radius $r$. Let $A$ be the area of the set consisting of all points interior to both triangles. Prove that $2A \geq r^2 \sqrt 3.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000207#p2000207]78. (SWE 3)[/url] [b](SL85-17)[/b] The sequence $f_1_, f_2, \cdots, f_n, \cdots $ of functions is defined for $x > 0$ recursively by
\[f_1(x)=x , \quad f_{n+1}(x) = f_n(x) \left(f_n(x) + \frac 1n \right)\]
Prove that there exists one and only one positive number $a$ such that $0 < f_n(a) < f_{n+1}(a) < 1$ for all integers $n \geq 1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018338#p2018338]79. (SWE 4)[/url] Let $a, b$, and $c$ be real numbers such that
\[\frac{1}{bc-a^2} + \frac{1}{ca-b^2}+\frac{1}{ab-c^2} = 0.\]
Prove that
\[\frac{a}{(bc-a{}^2){}^2} + \frac{b}{(ca-b{}^2){}^2}+\frac{c}{(ab-c{}^2){}^2} = 0.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018339#p2018339]80. (TUR 1)[/url] Let $E = \{1, 2, \dots , 16\}$ and let $M$ be the collection of all $4 \times 4$ matrices whose entries are distinct members of $E$. If a matrix $A = (a_{ij} )_{4\times4}$ is chosen randomly from $M$, compute the probability $p(k)$ of $\max_i \min_j a_{ij} = k$ for $k \in E$. Furthermore, determine $l \in E$ such that $p(l) = \max \{p(k) | k \in E \}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018353#p2018353]81. (TUR 2)[/url] Given the side $a$ and the corresponding altitude $h_a$ of a triangle $ABC$, find a relation between $a$ and $h_a$ such that it is possible to construct, with straightedge and compass, triangle $ABC$ such that the altitudes of $ABC$ form a right triangle admitting $h_a$ as hypotenuse.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849673#p849673]82. (TUR 3)[/url] Find all cubic polynomials $x^3 + ax^2 + bx + c$ admitting the rational numbers $a, b$, and $c$ as roots.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018354#p2018354]83. (TUR 4)[/url] Let $\Gamma_i, i = 0, 1, 2, \dots$ , be a circle of radius $r_i$ inscribed in an angle of measure $2\alpha$ such that each $\Gamma_i$ is externally tangent to $\Gamma_{i+1}$ and $r_{i+1} < r_i$. Show that the sum of the areas of the circles $\Gamma_i$ is equal to the area of a circle of radius $r =\frac 12 r_0 (\sqrt{ \sin \alpha} + \sqrt{\text{csc} \alpha}).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2001062&#p2001062]84. (TUR 5)[/url] [b](SL85-8)[/b] Let $A$ be a set of $n$ points in space. From the family of all segments with endpoints in $A$, $q$ segments have been selected and colored yellow. Suppose that all yellow segments are of different length. Prove that there exists a polygonal line composed of $m$ yellow segments, where $m \geq \frac{2q}{n}$ , arranged in order of increasing length.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018513#p2018513]85. (USA 1)[/url] Let $CD$ be a diameter of circle $K$. Let $AB$ be a chord that is parallel to $CD$. The line segment $AE$, with $E$ on $K$, is parallel to $CB$; $F$ is the point of intersection of line segments $AB$ and $DE$. The line segment $FG$, with $G$ on $DC$, extended is parallel to $CB$. Is $GA$ tangent to $K$ at point $A \?$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018516#p2018516]86. (USA 2)[/url] Let $l$ denote the length of the smallest diagonal of all rectangles inscribed in a triangle $T$ . (By inscribed, we mean that all four vertices of the rectangle lie on the boundary of $T$ .) Determine the maximum value of $\frac{l^2}{S(T)}$ taken over all triangles ($S(T )$ denotes the area of triangle $T$ ).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000927&#p2000927]87. (USA 3)[/url] [b](SL85-9)[/b] Determine the radius of a sphere $S$ that passes through the centroids of each face of a given tetrahedron $T$ inscribed in a unit sphere with center $O$. Also, determine the distance from $O$ to the center of $S$ as a function of the edges of $T.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018369#p2018369]88. (USA 4)[/url] Determine the range of $w(w + x)(w + y)(w + z)$, where $x, y, z$, and $w$ are real numbers such that
\[x + y + z + w = x^7 + y^7 + z^7 + w^7 = 0.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018372#p2018372]89. (USA 5)[/url] Given that $n$ elements $a_1, a_2,\dots, a_n$ are organized into $n$ pairs $P_1, P_2, \dots, P_n$ in such a way that two pairs $P_i, P_j$ share exactly one element when $(a_i, a_j)$ is one of the pairs, prove that every element is in exactly two of the pairs.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1826878#p1826878]90. (USS 1)[/url] Decompose the number $5^{1985}-1$ into a product of three integers, each of which is larger than $5^{100}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018524#p2018524]91. (USS 2)[/url] Thirty-four countries participated in a jury session of the IMO, each represented by the leader and the deputy leader of the team. Before the meeting, some participants exchanged handshakes, but no team leader shook hands with his deputy. After the meeting, the leader of the Illyrian team asked every other participant the number of people they had shaken hands with, and all the answers she got were different. How many people did the deputy leader of the Illyrian team greet?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000176&#p2000176]92. (USS 3)[/url] [b](SL85-11)[/b] Find a method by which one can compute the coefficients of $P(x) = x^6 + a_1x^5 + \cdots+ + a_6$ from the roots of $P(x) = 0$ by performing not more than $15$ additions and $15$ multiplications.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018528#p2018528]93. (USS 4)[/url] The sphere inscribed in tetrahedron $ABCD$ touches the sides $ABD$ and $DBC$ at points $K$ and $M$, respectively. Prove that $\angle AKB = \angle DMC$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366594&#p366594]94. (USS 5)[/url] [b](SL85-22)[/b] 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 $\angle OMB=90^{\circ}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000939&#p2000939]95. (VIE 1)[/url] [b](SL85-10)[/b] Prove that for every point $M$ on the surface of a regular tetrahedron there exists a point $M'$ such that there are at least three different curves on the surface joining $M$ to $M'$ with the smallest possible length among all curves on the surface joining $M$ to $M'$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018374#p2018374]96. (VIE 2)[/url] Determine all functions $f : \mathbb R \to \mathbb R$ satisfying the following two conditions:

[i](a)[/i] $f(x + y) + f(x - y) = 2f(x)f(y)$ for all $x, y \in \mathbb R$,

[i](b)[/i] $\lim_{x\to \infty} f(x) = 0.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2018530#p2018530]97. (VIE 3)[/url] In a plane a circle with radius $R$ and center $w$ and a line $\Lambda$ are given. The distance between $w$ and $\Lambda$ is $d, d > R$. The points $M$ and $N$ are chosen on $\Lambda$ in such a way that the circle with diameter $MN$ is externally tangent to the given circle. Show that there exists a point $A$ in the plane such that all the segments $MN$ are seen in a constant angle from $A.$


AMC
AMC 12
AIME
USA(J)MO
USAMO
geometry
AMC 12 B

I'll give the scores for my whole team and their usernames just cuz i feel like it:
me: 116
joe (joml88):124
david (furious):106.5
brian: 107.5
will: 104.5
Joey:119.5

I think we did pretty well overall, considering the fact that brian and will are both freshmen, and will is in an ordinary 9th grade math class.  None of us had really good scores, but at least all of us qualified for aime, and we had a team score of 359.5.  I think we are gonna take the amc 12 b just so that we can get a little better scores and have a better usamo index.

[quote="goirulz"]11/11 answered
14/14 blank

101 for me. I'm just glad I made the cut for AIME. Last year, I took the AMC 10 and earned a 112.5, thinking that there was no way I could possibly advance. But then remarkable things just sort of happened....lol.[/quote]

Congratulations of making your first AIME.

I got a 116.  Except for making a careless error on the "pentagonal arithmetic sequence of the sum of the vertices" problem, I am very happy with my score.  :)

I got a 109.5. After finishing the test, I had convinced myself I was getting a 139.5. But then I started comparing answers. I mixed up Josh and Mike(you know which problem i'm talking about). I messed up the dice problem(I knew it was a trick question). I remembered not to cancel out all the fours in 444 on the odometer one(patted myself on the back for not falling for the trap), only realizing afterwards that I shouldn't have remembered that cuz that's wrong. Forgot one is not a composite number and I think I counted one thousand twice. And I subtracted wrong on the graph one, getting nine instead of eight points. All in all, I can't complain(<--sarcasm, just thought I'd point it out)

127.5, same issues - switch of Josh and Mike, and mistake on the odometer.

Rather dissappointing, will go for better on B.

[quote="origamimasterjared"]#1-11=101 :)

I got 12, 13, 15, 17, 19, 22, 23 afterward...

I seriously can't believe how easy the odometer one was...[/quote]

Hey...it's somewhat tricky. Like, very very hard. Incredably hard. Unbelievably hard.

the top five at jefferson were all 139.5 or above.

i scored a 139.5 and qualified for the aime, unlike last year

1-19, 21-23, rest blank.

I took a second look at it yesterday while waiting for the bus.  I realized [hide] "wait, 2005 is just base 9!"...[/hide]...

[quote="Elemennop"][quote="paladin8"]Err... you can't solve trig problems without... trig.[/quote]

Yeah, but that's not calculus/pre-cal.[/quote]

I know, which is exactly why it can show up on the AMC12. I was replying to a statement that mentioned that every problem was solvable with Algebra II methods. And besides, most precalculus curriculums cover some trig (at least mine did).

I only got a 96........  :(

but luckily for me, I am only in 8th grade,

there was another 8th grader at my school that got a 112.5

[quote="mcalderbank"]who got the other 150?[/quote]

Zabelman, or Zack.

[quote="nr1337"]How did you get your scores already?[/quote]

The answers come with the test booklet, so we were all making extremely highly educated guesses. Why we weren't 100% sure is because the machine might mis-read our answers.

I don't know if anyone has pointed it out, but for those of you freaked out that even though you barely qualified, you are only in the middle of this poll, realize that this is a voluntary response poll, thus people with higher scores are more amenable to replying, thus you are more likely in the top 5% at least; on the contrary to what it seems here.

Score: 102
12/13 answered
12 blank

Not quite what I was hoping for, but at least I made AIME. Even though I freaked out under test conditions and should have done better, I really can't complain. Last year I probably would have only gotten 1 or 2 problems on the test, so this does reflect progress. I do have one more year to try to make USAMO.

Someone might have said this before, but is there any way to estimate your AIME score based on your AMC score?  I'm a senior but this is my first time qualifying for AIME.

Score=119.5

[quote="joml88"]AIME is much harder than AMC.  Predicting your AIME score from AMC is tricky and unreliable.  A person who is better at speed would be probably overestimate their AIME score given their AMC.

But, if I had to put a 119.5 in an AIME range, I would say that would probably correspond to a score of around 4-10 (i know, wide range).[/quote]

Four right? I don't know... Last year I only got two right and this year I scored 116 on 12A and 120.5 on 12B....

Your right though, its very hard to tell. The AIME is almost like Three Hours of Hard Core math. Actually, it is three hours of Hardcore math.

Honestly, its only fifteen questions, you'd think that you wouldn't need THAT much time... but then you do... and then you solve the other half of the test in like ten minutes... but whatever...

Seriously, its good stuff, see both of you guys there.

[quote="joml88"]AIME is much harder than AMC.  Predicting your AIME score from AMC is tricky and unreliable.  A person who is better at speed would be probably overestimate their AIME score given their AMC.

But, if I had to put a 119.5 in an AIME range, I would say that would probably correspond to a score of around 4-10 (i know, wide range).[/quote]

Eh, I got 119.5.

Hopefully I can do better than a 4.

When I was a freshman, I got a 105.5 on the AMC-12 and I got a 6 on the AIME, so clearly it varies.

answered: 23 (all but #24,25)
right: 21
score: 131

and im really pissed at my own reading skills -
for #3 i read "diagonal" as "length" somehow, which made it a rectangle area problem given length and width, which would be absurdly easy but since it was only #3 i didnt give it a second thought.
for another problem w/ 3 digit numbers i read the "average" of the first and last digits as their "sum"... dont ask how...
[the really tricky thing was that the wrong answers i got were both amongst the answer choices given]

the same kind of thing happened to me on the UMD competition, for anyone who lives in MD

moral of the story: check work thoroughly

Yep, no stupid mistakes for me, so 122.

Hope I get a better than 3 score on the AIME this year. I'm probably going to need about a 10 or 11 this year to make the USAMO -- possible, but I'm going to have to hope for a good set of questions.

In 10th grade, I got 116 and got an 8 on the AIME, so we'll see.

131.  21 right and 2 wrong, and I'm only a freshman.

Made a stupid mistake on the point on a dice problem, and answered #25 with the choice that probably the most people chose, (A).

I left #21/#24 blank.

didnt take the AMC12A, but took it as a practice (with the real timing and stuff)....
i got 22 right, 1wrong, 2 blank (24, 25) giving me 137. Which was better than my AMC12B, 124. (oops)


invariant
inequalities
inequalities proposed

Let $ x,\ y,\ z$ be real numbers such that $ x^{2}+y^{2}+z^{2}-xy-yz-zx=3$.
For real constants $ p,\ q,\ r$, find the maximum value of $ (q-r)x+(r-p)y+(p-q)z$.

By cauchy-schwarz, the answer is $ 2 \sqrt{ p^{2}+q^{2}+r^{2}-pq-qr-rs }$ :D
with equality iff $ (x-y,y-z,z-x) \sim (q+p-2r, r+q-2p, p+r-2q)$ (that is always possible).

First solution, the "magical" one... I rewrote the text as follow:
\[ 3(q-r)x+3(r-p)y+3(p-q)z = \]

\[ = 3(q-r)x+3(r-p)y+3(p-q)z-(x+y+z) ( (q-r)+(r-p)+(p-q) ) = \]

\[ = (q-r)(2x-y-z)+(r-p)(2y-x-z)+(p-q)(2z-x-y) \leq^{CS}\]

\[ \leq \sqrt{ (q-r)^{2}+(r-p)^{2}+(p-q)^{2}}\sqrt{ (2x-y-z)^{2}+(2y-x-z)^{2}+(2z-x-y)^{2}}= \]

\[ = \sqrt{ 2(p^{2}+q^{2}+r^{2}-pq-pr-rq ) \cdot 6( x^{2}+y^{2}+z^{2}-xy-yz-zx ) }= \]

\[ = 6 \sqrt{ p^{2}+q^{2}+r^{2}-pq-qr-rs }\]
Second solution, the "brute force" one... :D

Let's toggle one parameter... in fact both the text and the hypotesis are invariant if the add a costant c to x,y,z.
Call $ (x,y,z)=(x,x-a,x+b)$ and now we have $ (q-r)x+(r-p)y+(p-q)z = (p-r)a+(p-q)b$ (now we can suppose WLOG p is the max, in reality we had to do this at the beginning...).

Now rewriting the hypotesis we have $ a^{2}+b^{2}+ab=3$ ==> $ (2a+b)^{2}+(\sqrt{3}b)^{2}=12$ Now we try to manipule again the text in order of finding $ 2a+b$ and $ \sqrt{3}b$ to use Schwarz. We find that:
\[ (p-r)a+(p-q)b = \frac{ p-r}2 (2a+b)+\frac 1{\sqrt{3}}\left(p-q-\frac{ p-r}2 \right )( \sqrt{3}b) \]
Now using Cauchy Schwarz we'll get
\[ text^{2}=((p-r)a+(p-q)b )^{2}\leq \left( \frac 1{12}( 3(p-r)^{2}+(p+r-2q)^{2}) \right) \cdot ( 12 ) \]
and then the thesis.

note: by a simpler Cauchy-Schwarz, we could find the weaker bound $ (q-r)x+(r-p)y+(p-q)z\leq\sqrt{6\left(p^{2}+q^{2}+r^{2}\right)}$, but equality can only hold here if $ p+q+r=0$.
In fact, if you try to see the case of equality, it's not always possible 'cause we ought have $ (x-y,y-z,z-x) \sim (r,p,q)$ and we are able to find such a triple iff $ p+q+r=0$.

[Moderator edit: References to incorrect post removed.]

Oh... Very nice and meaningful inequality  :) At the first time seeing it, I just think it is trivial, but the way of applying Cauchy is really special.

Thank you, kunny and Simo.


[quote=Twins]$a,b,c,d>0$ $\frac{1}{a+c}+\frac{1}{b+d}=1$
Let $S=\frac{ab}{b+c+d}+\frac{bc}{c+d+a}+\frac{cd}{d+a+b}+\frac{da}{a+b+c}$
(1)Find the maximal $k$ that $S\geq k$.
(2)Let $a\geq b\geq c\geq d>0$. Prove that $S\geq \frac{4}{3}$[/quote]
[url=https://artofproblemsolving.com/community/c6h135376p765688]Any one?[/url]


I have a wonderful problem,I am sure you wll enjoy it.
Suppose the positive integers have been expressed as a disjoint union of arithmetic progressions [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/99196e618d5a5e8acd0468dd30b5d23b.gif[/img],i=1,2,...k.Show that
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/8d63408e3b94a19f888388e25a89868a.gif[/img]

Beautiful!!!! Observe that for all $x$ sufficiently small we have $\sum_{i=1}^{k}{\frac{x^{a_i}}{1+x+...+x^{d_i-1}}}=1$  (since $\sum_{i=1}^{k}{\frac{x^{a_i}}{1-x^{d_i}}}=\frac{1}{1-x}$). Take $x$ close to 1 in this and you will find that $\sum_{i=1}^{k}{\frac{1}{d_i}}=1$. Next, derive that relation, take again x close to 1 an \d use the previous result. You will find exactly the conclusion.

Good job,your solution is better than mine :) .But I think you have forgotten to multiple x with R.H.S. :D

Please someone give the original statement...

i continue pro. show atleast 2 common diff. of some 2 a.p. are same.
 :)


AMC
AIME
algebra
polynomial
absolute value

The number $ \sqrt{X\sqrt{6}\plus{}Y\sqrt{10}\plus{}Z\sqrt{15}\plus{}2006}$ can be written as $ a\sqrt{2}\plus{}b\sqrt{3}\plus{}c\sqrt{5},$ where $ a, b,$ and $ c$ are positive integers. Find $ abc$.

[quote="1=2"]The number $ \sqrt {X\sqrt {6} \plus{} Y\sqrt {10} \plus{} Z\sqrt {15} \plus{} 2006}$ can be written as $ a\sqrt {2} \plus{} b\sqrt {3} \plus{} c\sqrt {5},$ where $ a, b,$ and $ c$ are positive integers. Find $ abc$.[/quote]
[hide="Solution"]
Set them equal to each other and square, so $ X\sqrt {6} \plus{} Y\sqrt {10} \plus{} Z\sqrt {15} \plus{} 2006 \equal{} 2a \plus{} 3b \plus{} 5c \plus{} 2ab\sqrt {6} \plus{} 2bc\sqrt {15} \plus{} 2ac\sqrt {10}$.  Matching squareroots, this becomes $ X \equal{} 2ab$, $ Y \equal{} 2ac$, $ Z \equal{} 2bc$, so multiplying everything, we get that $ XYZ \equal{} 8a^2b^2c^2\implies abc \equal{} \sqrt {\frac {XYZ}{8}} \equal{} \boxed{\frac {\sqrt {2XYZ}}{4}}$  [/hide]
EDIT: Yes, [b]the future[/b], it does; I am not sure how [b]1=2[/b] wanted us to express our answer.  I made the (probably false) assumption that he wanted it in terms of $ X$, $ Y$, and $ Z$.
EDIT again: Oh, wait, I made a calculation mistake... it should be $ 2a^2 \plus{} 3b^2 \plus{} 5c^2 \equal{} 2006$, which would have only one solution, as the hint says below...

[quote="The QuattoMaster 6000"][quote="1=2"]The number $ \sqrt {X\sqrt {6} \plus{} Y\sqrt {10} \plus{} Z\sqrt {15} \plus{} 2006}$ can be written as $ a\sqrt {2} \plus{} b\sqrt {3} \plus{} c\sqrt {5},$ where $ a, b,$ and $ c$ are positive integers. Find $ abc$.[/quote]

Set them equal to each other and square, so $ X\sqrt {6} \plus{} Y\sqrt {10} \plus{} Z\sqrt {15} \plus{} 2006 \equal{} 2a \plus{} 3b \plus{} 5c \plus{} 2ab\sqrt {6} \plus{} 2bc\sqrt {15} \plus{} 2ac\sqrt {10}$.  Matching squareroots, this becomes $ X \equal{} 2ab$, $ Y \equal{} 2ac$, $ Z \equal{} 2bc$, so multiplying everything, we get that $ XYZ \equal{} 8a^2b^2c^2\implies abc \equal{} \sqrt {\frac {XYZ}{8}} \equal{} \boxed{\frac {\sqrt {2XYZ}}{4}}$  [/quote]

but doesnt 2a+3b+5c have to still equal 2006?

[hide="Hint"]$ (a\sqrt {2} \plus{} b\sqrt {3} \plus{} c\sqrt {5})^2 \equal{} 2ab \sqrt {6} \plus{} 2 bc \sqrt {15} \plus{} 2 ac \sqrt {10} \plus{} (2a^2 \plus{} 3b^2 \plus{} 5c^2)$

Hence, $ 2a^2 \plus{} 3b^2 \plus{} 5c^2 \equal{} 2006$

It now suffices to prove that $ (a,b,c) \equal{} (13,4,18)$ is the only positive integer solution to the above equation, which implies the answer, $ \boxed{abc \equal{} 936}$
[/hide]

Now I have a 2008 version:

There is a number of the form $ a\sqrt{2}\plus{}b\sqrt{3}\plus{}c\sqrt{5}$ such that it equals $ \sqrt{100\sqrt{6}\plus{}20\sqrt{10}\plus{}125\sqrt{15}\plus{}2008}$. Find the absolute value of the sum of the coefficients of the polynomial with roots a, b, and c.

I decided to make it a little harder so it would be suitable for the intermediate forum. :)

Ah yes, but noone's solved your first problem yet  :wink:

The answer is independent of X Y and Z because there's only one possible triple (a,b,c)... see my previous post  :)

[b]Edit:[/b] The second question is actually much easier than the first question because X Y and Z are given.

[hide="Outline"]
Expand the left. The coefficients of the squareroots in the second expression can be used to prove that b and c are once divisible by 5, and we can easily see following that that a=2. Then, we look for solutions of the form (a,b,c)=(2,5m,5n), and using $ 2a^2\plus{}3b^2 \plus{} 5c^2 \equal{} 2008$, quickly find two solutions: (2,0,20) and (2,25,5). It is not hard to see that the second is the correct answer. We then expand (1-2)(1-5)(1-25)= -96, and the answer is 96.[/hide]


it was given in a book by Atkins that during the reaction
$2H_{2}+O_{2}\rightarrow2H_{2}O$
the difference $\Delta H-\Delta U$ $=$ $\Delta n_{g}RT$
where $\Delta n_{g}$ is the change in  number of moles when the substances are taken according to their stoichiometric coefficients .

[b]My Doubt[/b]
now $\Delta H-\Delta U$ $=$ $\Delta(PV)$ $=$ $P_{water}V_{water}-P_{gas}V_{gas}$
now iff one assumes that the $V_{water}<<V_{gas}$ one can arrive at this conclusion which is logical .Am i right or is there some other logic to it

You're pretty much right, except about one thing.  Work done is not $\Delta{PV}$; rather, it is $P\delta{V}$.  If pressure is also changing, you need to integrate it.  This is all really just an expression of the formula $w = F\cdot d$ anyway, and it is certainly NOT path independent.

Now to answer your question: one assumes that the volume of water changes very little throughout the reaction.  That way, its formation or disappearance does negligible work on the atmosphere.  So yes, you are correct.  It is an interesting exercise to calculate the ratio of molar volumes of gas to water is at--it is a pretty good approximation:).


conics
ellipse
ratio
geometry
trigonometry
trig identities
Law of Cosines

I thought I would help Rep123max out by starting a thread and posting some solutions to the problems.  If other people would contribute it would really be helpful, since it's important to look at the problems you got wrong so you learn from your mistakes and can answer that type of question correctly next time.  Please post the question, then post your solution, so it's easy to read.

Here's a few:

[quote][b]*[/b]8. Find the sum of [tex]\displaystyle\frac{7}{12}+\frac{25}{144}+\ldots+\frac{3^n+4^n}{12^n}+\ldots[/tex].[/quote]

(8)

W/the ellipses you are implying the infinite sum:

[tex]\displaystyle{\sum_{n=1}^{\infty}{\frac{3^n+4^n}{12^n}}}=\displaystyle{\sum_{n=1}^{\infty}{(\frac{1}{4})^{n}}+\sum_{n=1}^{\infty}{(\frac{1}{3})^{n}}}[/tex]

Evaluate each of these sums separately, using the formula [tex]S=\frac{a_1}{1-r}[/tex]

[tex]\sum_{n=1}^{\infty}{(\frac{1}{4})^{n}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}[/tex]

[tex]\sum_{n=1}^{\infty}{(\frac{1}{3})^{n}}}=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}[/tex]

Add the two sums together, 1/2 + 1/3 = 5/6

[quote] [b]*[/b]9. What is the sum of the infinite series [tex]\displaystyle\frac{3}{4}+\frac{3}{10}+\ldots+\frac{3}{k(k+3)}[/tex]?
[/quote]

(9) 11/6

Okay, so ya want:
[tex]\displaystyle{\sum_{n=1}^{\infty}{\frac{3}{k(k+3)}}[/tex]

Using partial fraction decomposition:
3 / k(k+3) = A/k + B/(k+3)
3=A(k+3) + B(k)
let k=0
3=3A
1=A
let k=-3
3 = -3B
-1 = B

Therefore you can split this up into the sum of sums:
[tex]\displaystyle{\sum_{n=1}^{\infty}{\frac{3}{k(k+3)}}}=\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{k}}} - \displaystyle{\sum_{n=1}^{\infty}{\frac{1}{k+3}} [/tex]

When you write out a bunch of terms of both of these, all of them cancel out (telescope) except for:

1/1 + 1/2 + 1/3 = 11/6

[quote][b]*[/b]13. The total area of all the faces of a rectangular solid is 22 centimeters squared, and the total length of all of its edges is 24 centimeters. Then the length in cm of any one of its internal diagonals is what?[/quote]

(13) [tex]\sqrt{14}[/tex]

Draw a picture of the box (label l=length, w=width, h=height).  Given are the equations:
2lh+2wh+2wl=22
4h+4w+4l=24 ---> h+w+l=6

Now you want to express the inner diagonal in terms of l, w, and h:

Draw the box, it is a right triangle w/a diagonal of one of the sides equal to sqrt(w^2+l^2) and h.  So its hypotenuse is equal to sqrt(l^2+w^2+h^2) b/c of the Pythagorean theorem.

Now manipulate the given equations:
*square both sides of l+w+h=6 to get:
(h^2+w^2+l^2) + (2hw+2lh+2lw)=36
*sub in the other equation:
(h^2+w^2+l^2)+22=36
*solve for what you want:
sqrt(h^2+w^2+l^2)=sqrt(36-22)=sqrt(14)

I'll do most of the easy ones for you guys.

[quote] 1. In a group of cows and geese, the number of legs was 84 more than twice the number of heads. How many cows were there? [/quote]

We obtain the equation 4c+2g = 84 + 2(c+g)  *Since there is four legs for a cow and two legs for a goose, and one head for each*

The gs cancel each other out, and we're left with equation:

2c = 84
c= 42.

Plugging c back into the original equation
42(4) = 84 + 2(42)
168=168
Correct.

[quote] 2. A farmer divides his heard of n cows between his four sons so that one son gets one-half the herd; a second son, one-fourth; a third son, one-fifth and the fourth son gets 7 cows. What is n? [/quote]

First problem how much the fourth son got.   To find this we subtract how much the other sons got from 1.  Therefore,
1 - 1/2 - 1/4 - 1/5 = 1/20.

Since the 1/20n = 7, we just multiply both sides by 20 to received the answer of 140.

[quote] 5. A student has four science books, three literature books and two math books. He wishes to keep all books of the same kind together on his shelf. How many ways can the books be arranged? [/quote]

We first need to find how many ways three things can be organized, which is 3! or 6.   Next, we have to see how many ways the same books can be organized WITHIN each other.  For science, we have 4!, literature, 3!, and math (Haha, um, hello, isn't math MORE important?  :-D)  2!.  Multiplying all the possibilities together yields the answer of 1,728.

With some MAJOR guidance from Julie (AFTER the contest of course) I obtained the answer for 11, which states:  [quote] 11. In an arithmetic progression of distinct terms, the second, seventh and twenty-seventh terms form successive terms of a geometric progression. What is the ratio of this geometric progression if the arithmetic progression is written in increasing order? [/quote]

We let:
a_2 = a + d
a_7 = a + 6d
a_27 =  a + 26d

We know to obtain the common ratio, we take nth term and divide it by n-1 term.   Ergo we obtain these two equations:

(a+6d)/(a+d)
(a+26d)(a+6d)

Since both equation equal each other we can cross-multiply.  Expanding and reducing, we obtain:

10d :^2: /15ad

Solve for a, and we get a = 2/3d

Plugging in 2/3d to our first equation [(a+6d)/(a+d)] we obtain the answer of (6/2/3)/(1/2/3).  Simplied, we get the answer 4.   Therefore, the common ratio is 4.

Well.  Um, it's 4:23am, which means I'm sleeping.

I just learned about complex stuff very recently, so I will try this one
#15)multiply both side of the equation by [tex](x^2-2)[/tex]
we have
 [tex]x^{12}-2^6=0[/tex]
[tex]x^{12}=(\sqrt2)^{12}[/tex]
By the roots of unity stuff, the roots are
[tex](\sqrt2, 0), [/tex] [tex](\sqrt2, 30)[/tex] ......... [tex](\sqrt2, 330)[/tex]

The roots forms a dodecagon on the complex plane with radius r.
Area of one of the little triangle
[tex]= \frac{(\sqrt2)(\sqrt2)\sin30}{2}[/tex]
[tex]=sin30=\frac{1}{2}[/tex]
[tex]12(\frac{1}{2})=6[/tex]=area of dodecagon
notice that[tex] (\sqrt2, 0), (\sqrt2, 180)[/tex] are not the roots.
we need subtract 2 triangles from the area.
By Law of Cosines, a side [tex]s^2=2+2-\sqrt{3}=4-\sqrt3[/tex].
Area of a triangle formed by 3 adjacent vertices of the dodec. is
[tex]\frac{(4-\sqrt3)(sin150)}{2}[/tex]
[tex]=\frac{(4-\sqrt3)}{4}[/tex]
area of dodec-area of 2 triangle[tex]=\frac{8+\sqrt3}{2}[/tex]

I think these the ones that are left

4) in the form a,b,c possibilities are:
9,8,0<= c<8
8 possibilities
9,7,0<= c<7
7 possibilities
continues to 9,1,0
with one possibility
with a=8, we have 7,6,5,4,3,2,1 possibilities of when b=7,6,5,4,3,2,1 respectively
this pattern continues until a=2 where there is only one possibility (2,1,0)
when we organize the number of posibilities we get
8*1+7*2+6*3+5*4+4*5+3*6+2*7+1*8
=8+14+18+20+20+18+14+8
=120

6)total area = Pir Squared (r=d/2=3/2)
=Pi 3/2 Squared
=9Pi /4
divided into 4 areas, each will be 9Pi /16
first circle -> Pi r Squared -> 9Pi/16= Pi r Squared -> sqrt 9/16= r -> r= 3/4
second circle -> area -> 2*(9Pi/16) 9 Pi /8 -> radius = 3/2sqrt2
third cirlce -> area -> (3*9Pi/16) 27:pi: /16 -> radius - 3sqrt 3/2
fourth cirlce -> 3/2

7) possible
0+0+10, 0+1+9, 0+2+8, ..., 0+10+0 (11)
1+0+9, ..., 1+9+0 (10)
....
10+0+0(1)
sum 1+2+...+11
=n*(a+t at n)/2
=11*(1+11)/2
=11*12/2
=66
1+1+8...1+8+1 (8)
2+1+7....2+7+1 (7
....
8+1+1 (1)
sum 1+2+...+8
=n*(a+t at n)/2
=8*(1+8)/2
=8*9/2
=36
n=1005-2=1003, a=1, t at n = 10003
s at n = n(a+t at n)/2
=1003(1+1003)/2
=1003(1004)/2
=1003(502)
=503506

10) plot the triangle on an x-y grid
since BC is 14 point B will be (0.0) and C will be (0,14)
make equation for the other 2 circles to see where the point will be 13 and 15 away from B and C respectively
xSquared+ySquared=169
xSquared+(y-14)Squared=225
using subsitution we get the the point is (12,5), wont bore you with the math
equation of AC is y=5/12 x
perpendicular line with y intercept of 14 is y=-12/5 x +14
intercept point with line y=5
is (15/4,5)
since the altitude is on a horizontal line we can ingnore the value of y for each
so DH is 15/4
AH = 12-15/4
=33/4
15x/4=5
15x=20
x=4/3
33/4*4/3=11
therefore k=11

12) factor ab+bc=44 to b(a+c)=44
factor ac+bc=23 to c(a+b)=23
since 23 is a prime number c can only equal 1 or 23 and since both a and b must be intergers c cant be 23 because the sum of two intergers can't equal 1, so c=1
b can only equal 1, 4, 11, 2, and 22
a can equal 43(b=1), 10(b=4), 3 (b=11), 21(b=2), and 1(b=22)
since a+b=23 only two ordered triplets will satisfy the equations

14) I got a bit of help with this one
1/1+2/10+3/100+
=1/1+1/10+1/10+1/100+1/100+1/100+
first we sum 1/1+1/10+1/100
proved before that if r\leq|1|, then sum of infinite geometric series is a/(1-r)
so 1/(1-1/10)=1/(9/10)=10/9
then we sum 1/10+1/100+1/1000
(1/10)/(1-1/10)=(1/10)/(9/10)=1/9
then sum 1/100+1/1000+1/10000
(1/100)/(1-1/10)=(1/100)/(9/10)=1/90
so we now have to sum 10/9+1/9+1/90
a=10/9, r=1/10
a/(r-1)=(10/9)/(1-1/10)=(10/9)/(9/10)=100/81
since 100 over 81, the sum of the infinite geometric series, is rational, the sum of the geometric series is rational

What about three?  And for number 14, I just said that it was a repeating decimal...and repeating decimals can be reduced to fractions.

[quote="beta"]
By Law of Cosines, a side [tex]s^2=2+2-\sqrt{3}=4-\sqrt3[/tex].
[/quote]
I see, that's the mistake.
it's [tex]s^2=(\sqrt2)^2+(\sqrt2)^2-2(\sqrt2)(\sqrt2)(\cos30)[/tex]
simplify it to get:
[tex]4-2\sqrt3[/tex], not [tex]4-\sqrt3[/tex].

How did you figure out the area, rep?


a,b,c πραγματικοί να δείξετε ότι

$ \sqrt{a^2\plus{}b^2\minus{}ab}\plus{}\sqrt{b^2\plus{}c^2\minus{}bc}\geq{\sqrt{a^2\plus{}c^2\minus{}ac}}$

Δινω μια λυση περιορισμενη στους θετικους αριθμους.

Οι υποριζες ποσοτητες θυμιζουν νομο συνημιτονων.  Οποτε παιρνουμε τρεις ημιευθειες Οχ,Οψ,Οζ που σχηματιζουν δυο γωνιες 60 μοιρων και σε αυτες θεωρουμε τμηματα ΟΑ=a,ΟΒ=b,ΟΓ=c αντιστοιχα.

Τοτε
            ΑΒ^2=a^2+b^2-2abσυν60=a^2+b^2-ab
            ΒΓ^2=b^2+c^2-2bcσυν60=b^2+c^2-bc
            ΑΓ^2=a^2+c^2-2acσυν120=a^2+c^2+ac

Αρκει λοιπον να δειξουμε οτι      ΑΒ+ΒΓ>=sqrt(ΑΓ^2-2ac)
Ομως απο  τριγωνικη ανισοτητα στο τριγωνο ΑΒΓ παιρνουμε ΑΒ+ΒΓ>=ΑΓ
Αρκει λοιπον ΑΓ^2>=ΑΓ^2-2ac<=>2ac>=0 που ισχυει αφου a,c θετικοι.

Ισως να δουλευει και για αρνητικους αν αλλαξουμε σε γωνιες 120 μοιρων και θεσουμε a=-x.

The following proof is dedicated to vanglelismak . :)

Me διαδοχικές υψώσεις εις το τετράγωνο μας μένει να δείξουμε ότι

$ 3(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \plus{} 3b^4\geq 2ab^3 \plus{} 2bc^3 \plus{} 2a^bc \plus{} 2abc^2$

που είναι ισοδύναμη με $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} (ac \minus{} ab \minus{} bc)^2 \plus{} (ab \minus{} b^2)^2 \plus{} (bc \minus{} b^2)^2 \plus{} (ca \minus{} b^2)^2\geq 0$ που ισχύει προφανώς.  :wink:


Prove that if $x,y,z>0$ and $x^2+y^2+z^2=1$, then
\[ \frac1{1-\bigl(\frac{x+y}2\bigl)^2}+\frac1{1-\bigl(\frac{y+z}2\bigl)^2}+\frac1{1-\bigl(\frac{z+x}2\bigl)^2}\geq\frac92.  \]

[quote="e.lopes"]use that $\frac{1}{1-(\frac{x+y}{2})^2} \geq \frac{1}{1-xy}$

and the new problem was posted many times. (the new problem is from crux if i'm not wrong)[/quote]

e.lopes, you're wrong. The sign reversed.

The inequality is not true. See $x=1$,  $y=z=0$. :o


Let $ABC$ be acute-angled with $AC>BC.$ Let $M$ be the midpoint of $AB$ and let $CD,\,AP$ and $BQ$ be the altitudes.
Denote circumcircle of $\triangle PQC$ by $k_{1}$ and the circumcircle of $\triangle DRP$ by $k_{2}$ where $R=AB\cap PQ.$
 Prove that:
$(a)\;\, MP$ is tangent to both $k_{1}$ and $k_{2}$
$(b)\;\, RH\perp CM$ where $H$ is the orthocenter of $\triangle ABC$

a) It is obvious that $(ABDR) =-1$. Since $M$ is the midpoint of $AB$, we know that $MD \cdot MR = MB^{2}=MP^{2}$, then $MP$ touches $k_{1}$.
The centre of $k_{1}$ is the midpoint $O_{1}$ of $CH$. We have $\angle MPO_{1}= \angle MPH+\angle HPO_{1}= \angle MAP+\angle PHC = \angle DAH+\angle AHD = 90$. Then $MP$ touches $k_{1}$.

b) I assume you meant $RH \perp CM$. $AQPB$ is cyclic with centre $M$. The polar of $C$ is the line $HR$, which is perpendicular to $CM$  :)


one can easily modify addition and subtraction for binary(or any other base).
how would one modify multiplication for binary, and how would one modify division?

for example:in base 8 the baseten number 33 would be 41, and baseten 3 would be 3
so  in base 10 , 33/3=11,
and in base 8-->41/3=13

For multiplication, I would just do it thinking the same way as for addition (so if you had 13*3: 3*3=9...9=1(mod 8) so put a one on the bottom...carries 1 over...1*3=3...3+1=4...41 ans). For division, I personally would convert everything to base 10 and then convert back.

Division:

41/3 in base 8: 3 goes into 4 1 time, so we have 10+11/3. 3 goes into 11 3 times (multiplication table in base 8), so we have 10+3=13. Just like long division.

Random polls: Please don't put completely irrelevant polls in your posts.

Post count arguments: Who cares? Useless posts might get deleted, and it's always possible to reduce your postcount.

Multiplication and division are essentially the same except that you need to convert your times tables to the appropriate base.

Yeah, basically for addition and subtraction, when you borrow, instead of carrying a 10, you carry one of whatever base you're working with.
And yeah, for multiplication and division, just use that base's multiplication table instead of the base ten one you guys all memorized.


The probability for an event to success is 10%. You do the event N times in a row. If you failed the first N-1 times, what is the porbability that you will success at the Nth time.

Yeah.  All of the cases are independent trials.

There was a really silly question like this on the AP Stats examination.


number theory
Diophantine equation
number theory unsolved

Find $ m,n\in N*$
such that
$ m(m\plus{}1)(m\plus{}2)(m\plus{}3)(m\plus{}4)(m\plus{}5)\equal{}n^2\minus{}1$

We have given the Diophantine equation

$(1) \; m(m+1)(m+2)(m+3)(m+4)(m+5) = n^2-1$,

where $m$ and $n$ are positive integers.

By multiplying both sides of (1) with 64 and setting $x=2m+5$ (so $x$ is an odd integer $\geq$ 7), we end up with the equation

$(x - 5)(x - 3)(x - 1)(x + 1)(x + 3)(x + 5) = 64n^2 - 64$,

i.e.

$(x^2 - 25)(x^2 - 9)(x^2 - 1) = 64n^2 - 64$,

or equivalently

$(2) \; x^6 - 35x^4 + 259x^2 - 161 = (8n)^2$.

Hence by (2)

$(3) \; (16n)^2 = 4(x^6 - 35x^4 + 259x^2 - 161) = (2x^3 - 35x)^2 - 189x^2 - 644$

and

$(4) \; (16n)^2 = 4(x^6 - 35x^4 + 259x^2 - 161) = (2x^3 - 35x - {\textstyle \frac{48}{x})^2 + 3x^2 - 4004 - {\textstyle (\frac{48}{x}})^2}$.

Clearly $-189x^2 - 644 < 0$ and $3x^2 - 4004 - {\textstyle (\frac{48}{x}})^2 > 0$ for $x \geq 37$, implying

$(2x^3 - 35x - {\textstyle \frac{48}{x})^2 < (16n)^2 <  (2x^3 - 35x)^2}$,

i.e.

$(5) \; 2x^3 - 35x - {\textstyle \frac{48}{x}} < 16n <  2x^3 - 35x$

when $x \geq 37$.
If $x \geq 48$, then ${\textstyle \frac{48}{x} \leq 1}$, so $2x^3 - 35x - 1 < 16n <  2x^3 - 35x$ by (5), which obviously is impossible.  Hence (2) has no solution when $x \geq 48$.
Assume $37 \leq x \leq 47$.  Then, due to the fact that ${\textstyle \frac{48}{x}< 2}$, the only possibility according to (5) is $16n = 2x^3 - 35x - 1$, which inserted in (2) gives

$(6) \; 4(x^6 - 35x^4 + 259x^2 - 161) = (2x^3 - 35x - 1)^2$,

implying $x$ is a divisor of $4 \cdot 161 + 1 = 645 = 3 \cdot 5 \cdot 43$.  Consequently, since $37 \leq x \leq 47$, the only possible solution in this case is $x=43$.  By inspection $x=43$ does not satisfy (6).  Hence there is no solution of (2) when $37 \leq x \leq 47$.

Summa summarum we have proved that (2) has no solution when $x \geq 37$.  Using (2) to examine the  remaining 15 cases $x \in \{7,9,11, \ldots ,35\}$, we find that there is only in the case where $x=9$ that (2) has a  solution.

[b]Conclusion:[/b] The only solution of (1) is $(m,n)=(2,71)$.


probability
factorial
Alcumus
function
number theory
prime factorization

ok. Does anyone know what the exclamation point stands for? :blush:

It is used in counting and probability a lot. Factorials are only defined for non-negative integers:

$ 0!\equal{}1$
$ 1!\equal{}1$
$ 2!\equal{}2 \cdot 1\equal{}2$
$ 3!\equal{} 3 \cdot 2 \cdot 1\equal{}6$
$ 4!\equal{}4 \cdot 3 \cdot 2 \cdot 1\equal{}24$
$ 5!\equal{}5\cdot 4 \cdot 3 \cdot 2 \cdot 1\equal{}120$
$ 6!\equal{}720$
$ 7!\equal{}5040$
$ 8!\equal{}40320$
$ 9!\equal{}362880$
$ 10!\equal{}3628800$

From there on, they get pretty big. For more information, you can try [url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]Alcumus[/url].

Or...defined recursively as
\[ x!\equal{}\begin{cases}1&\qquad\text{if }x\equal{}0\\x\times(x\minus{}1)!&\qquad\text{if }x>0\end{cases}\forall x\in\mathbb{N}.\]

um... powerofpi? I have another question.

in my homework, it says what is the largest power of 7 that divides 500!? That is hard to calculate. Are there any tricks?

Thx!

Yes. Imagine taking the prime factorization of every number up to 500. Every multiple of 7 will have one 7 in its prime factorization, every multiple of 49 will have two 7's in its prime factorization, and every multiple 343 will have three 7's in its prime factorization. Now, we have to count these multiples the specified amount of times.

We can start with 7's. We have $ 500 \div 7\equal{}71$ with a remainder of 3, but those three don't have a 7 so we can forget about it.

Now, to count the 49's. We already counted them once with the multiples of 7, but we want to count them again. Doing a similar thing as with the 7's, we find $ 500 \div 49\equal{}10$ with a remainder of 10, but again, that doesn't matter to us.

Now, we want to count the 343's 3 times. We already counted them once with the multiples of 7, once with the multiples of 49, and once now. $ 500 \div 343\equal{}1$ and we forget about the remainder.

Now, in our prime factorization, we have a bunch of primes being multiplied together. If we look at the 7's, we have:

$ 7^{71} \cdot 7^{10} \cdot 7^1\equal{}7^{82}$.

So our final answer would be $ \boxed{82}$.

You can do a similar thing with powers of other numbers too.

but I don't get why we have to divide by seven. Why seven? why not eight or nine, or eleven?

How come the remainders don't matter?

How did we get 82?

It is pretty confusing, but it works. To avoid this confusion, I will start with a smaller number, 25!. We can write 25! as:

$ 25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$

Now, we can take the prime factorization of each of them (I won't actually do all of them):

$ 25 \cdot 24 \cdot ... \cdot 21 \cdot ... \cdot 14 \cdot ... \cdot 7 \cdot ... \cdot 2 \cdot 1\equal{}(5^2) \cdot (2^3 \cdot 3) \cdot ... \cdot (7\cdot 3) \cdot ... \cdot (7 \cdot 2) \cdot ... \cdot (7) \cdot ... \cdot 2 \cdot 1$

Now we can combine these and take out the parentheses because we are multiplying. We see that every multiple of $ 7$ has at least one $ 7$ in its prime factorization, and we want to count these. We can see that in the example of $ 25!$, the numbers with a prime factorization with a 7 are 7, 14 and 21. We can write these as $ 7 \cdot 1$, $ 7\cdot 2$, and $ 7 \cdot 3$. So we find the least multiple of 7 less than 25 and stop there. Since we want to find $ n$, where $ n \cdot 7$ is the greatest multiple of 7 less than 25, we can just find $ 25 \div 7 \equal{} 3$.

To answer your specific questions:
We divide by $ 7$ because we want to find the number of multiples of seven less than 500.
Remainders don't matter because none of those numbers are multiples of seven.
We get 82 because we combine the sevens and add their exponents.

Do you understand why we count the 49's and 343's?

Since $ 500! \equal{} 500 * 499 * 498 * ... * 1$, we have to count how many times $ 7$ is used as a factor of all the factors of $ 500!$. First, we count all the $ 7$s in $ 500$, so we divide $ \frac{500}{7} \equal{} 71 R 3$, but the three remainders are  500, 499, and 498, which have no factor of 7, so we don't count them. Next we find all the second power factors ($ 49$) of $ 500!$. This is $ \frac{500}{49} \equal{} 10 R 10$. Again, these extra ten contain no multiples of 49, so we don't count them. So far we have $ 7^{71\plus{}10} \equal{} 7^{81}$. Next we find all the third power multiples ($ 343$) in $ 500!$.  This is $ \frac{500}{343} \equal{} 1 R 157$, but we disregard the remainders because they don't contain $ 343$. We have the highest factor as $ 7^{71 \plus{} 10 \plus{} 1} \equal{} 7^{82}$.

Just to be a bit more direct, we first count the sevens as PowerOfPi and goldenboy have done. Those are counting the numbers with *at least* one seven in their prime factorization. Then we count with by 49's. But why? Aren't we overcounting them as we have already counted them once with 7? The answer is no. When diving 500 by 7, we have counted only *one* 7 from their prime factorization. And some of those numbers have more than just one seven in their prime factorization. Thus, us dividing 500 by 49, counts just the ones with sevens(it recounts some of the ones we have already counted again, but we look at it from the perspective that we are only counting the number of sevens that occur in their prime factor. So when we recount a number n, we are counting its second seven in its prime factor.) The same exact thing occurs with 343. Thus we add up the exponents of $ 7^{71} \cdot 7^{10} \cdot 7^1 \implies 7^{71 \plus{} 10 \plus{} 1} \implies \boxed{7^{82}}.$

This is kinda straying away from the other question, but one common type of question that use factorials (!) are questions like the following: how many ways are there to arrange the letters ABCDE? The answer is 5!,but why? There are 5 choices of letters for the first spot,4 choices for the next,3 choices,then 2,then finally there is only one letter you can use for the last spot. 5*4*3*2*1=5!=120. Just be aware that factorial questions can get a lot trickier than this. say the letters were AAABC instead. The answer would now be 20,not120. Do you see why? We need to divide 5! By 3! To account for the 3 A's.

Um... now I KNOW how to solve a question like "What is the largest power of 7 that divides 500!?"

But, does the process of FULLY comprehended the process come with a matter of time?

Thunder's problem appears a lot in Alcumus, so you should really try it.

What do you mean giraffe?

I think what giraffe is triying to say is that will it take some time for one to become a master at factorials and knowing when to use them

[quote="giraffe1"]is alcumus hard?

What I mean is:
Is the process of FULLY understanding how exactly to answer that kind of problem, come with a matter of time?[/quote]

Well, depending on what you consider hard. To me, i think its relatively easy, but maybe for you, it might be hard. To answer your second question, i say no, but you need to practice these kinds of problems to do them fast, but the overall concept can be understood within like 3 practice problems.

Yes, I would believe so. Some concepts like that just need to sink in and you need to realize it. Just sit down with a paper and pencil with a clear mind and just think about it. The majority of the realization, however, I think will just come with time, as you have said, and this is true for I think for many concepts in math.

Read intro(and intermediate) to C&P, work on ALCUMUS, ask smart questions, make smart problems, and wh-la! You are a master!(of !)

Factorial is too easy to be in Interm C+P. It's in the Intro book though.

[hide="Comment on PoPs first post in this thread"]
There is a factorial defined for complex values of $ x.$ It's called the [url=http://en.wikipedia.org/wiki/Gamma_function]Gamma Function[/url] and denoted by $ \Gamma(x).$

However, this is far beyond the scope of classroom math, so I'll just leave it at that.[/hide]

I think his question has already been answered, so we can just let this post die out.


This is $ \binom{4}{2} \equal{} 6$ lines.

You can just think of this as how many ways can I choose 2 points to form a line out of 4 which is the same as $\binom{4}{2}$ which is $6$, so the answer is $\boxed{6}$.

[quote=Pi_]You can just list them all out:
AB, AC, AD, BC, BD, CD
$\boxed{6}$[/quote]

You will probably not get the answer first if you do that

6 lines--- 4 to connect each horizantally/vertically and 2 to connect diagonally. 4+2 = 6 yayayay

since there are 2 lines, we have to choose 2 of them from 4. We can choose them by doing choose. This is simply 4c2, 4*3/2, or 6.


An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $\, AB \,$ intersects altitude $\, CC' \,$ and its extension at points $\, M \,$ and $\, N \,$, and the circle with diameter $\, AC \,$ intersects altitude $\, BB' \,$ and its extensions at $\, P \,$ and $\, Q \,$. Prove that the points $\, M, N, P, Q \,$ lie on a common circle.

[hide]

Let H be the orthocenter, the intersection AA' and BB'. Call the midpoint of AB $ O_{C}$ and the midpoint of AC $ O_{B}$. These are the centers of the two circles with AB and AC as diameter.

B' is the intersection of AC and circle $ O_{B}$ since $ \angle AB'B$ is a right angle and AB is a diameter. Similarily, C' is the intersection of AB and circle $ O_{C}$. We also see that for altitude AA', A' is the intersection of BC and circles $ O_{C}$ and $ O_{B}$.

By Power of a Point,
$ AH\bullet A'H=NH\bullet MH$
$ AH\bullet A'H=PH\bullet QH$

Then $ NH\bullet MH=PH\bullet QH$ which implies quadrilateral NPMQ is cyclic by the converse of power of a point.

QED
[/hide]


You might want to think about this title...

Joe and his sister both bought pumpkins.  The total weight of the two pumpkins was 20 pounds.  Joe bought a pumpkin that was 10 cents more per pound than her sisters, and his pumpkin weighed more than his sister's.
Joe paid $\$$3.52 for his pumpkin and his sister paid $\$$ 0.48 for her pumpkin.  How many pounds did Joe's pumpkin weigh?

This problem seems harder than it really is with a little logic.
[size=75][color=orange]
[solafidefarms edit: Be careful with dollars! $\$$ \  $\$$ $\$$][/color][/size]

Oh yuck, thanks!

[quote="kyyuanmathcount"]You might want to think about this title...

Joe and his sister both bought pumpkins.  The total weight of the two pumpkins was 20 pounds.  Joe bought a pumpkin that was 10 cents more per pound than her sisters, and his pumpkin weighed more than his sister's.
Joe paid $\$$3.52 for his pumpkin and his sister paid $\$$ 0.48 for her pumpkin.  How many pounds did Joe's pumpkin weigh?

This problem seems harder than it really is with a little logic.
[size=75][color=orange]
[solafidefarms edit: Be careful with dollars! $\$$ \  $\$$ $\$$][/color][/size][/quote]

[hide]Since the pounds of each pumpkin is the cost $\div$cost per pound, we have
$\frac{3.52}{x+.1}+\frac{.48}{x}=20$. Solving, $x=.12$. Cost per pumpkin for Joe is $.22$, so $\frac{3.52}{.22}=\boxed{16 \text{ pounds}}$. [/hide]

Checkin My answer (It was lame when I did it a long time ago) I had to use some guess and check in the end... but reasonable guess and check... (only like 4 different possibilities)
[hide]16 lbs.[/hide]


Each vertex of a convex hexagon is the center of a circle whose radius is equal to the shorter side of the hexagon that contains the vertex. Show that if the intersection of all six circles (including their boundaries) is not empty, then the hexagon is regular.

[hide="My Solution"] [geogebra]0ebe18aef3d872aa38c316a20c3f35c517f39e8a[/geogebra]

Consider a point lying in the nonempty intersection of the six circles. Connect this point to the vertices of the hexagon. Since $ h,g \le a$ (since the point lies on the circles from the two vertices), we have $ \angle 1 \ge \angle 2, \angle 3$, so $ \angle 1 \ge 60$. But we similarly find that all the other central angles are greater than or equal to $ 60$ as well, and, since they sum to $ 360$, they all equal $ 60$. But then $ \angle 2 \equal{} \angle 3 \equal{} 60$, and since all the six triangles are now congruent, we're done.

Is it correct?  :maybe: [/hide]

Let ABCDEF be a hexagon and let the point X belong to all circles.Connect X with A, B, C, D, E, F.
AX≤min(AB,AF)≤AB
BX≤min(AB,BC)≤AB
So the longest side of the triangle ABX is AB. So <AXB ≥60.
the same <BXC,<CXD,<DXE,<EXF,<FXA ≥60.
So everything is equal to 60 and point X lies at the same distance from the ends. So the hexagon is regular. Proved.


What is the largest number which, when divided into each of 1905, 2587, 3951, 7020 and 8725, leaves the same remainder each time?

Did you read "divided [b]into[/b]" correctly?

[hide="Hint"] Say that $ m$ leaves the same remainder when divided into each of those numbers.  Then $ m$ divides any of their pairwise differences.  ;) [/hide]


[quote="life1"]Hi all -- I was surprised to find out that my son (I'm serving as mom and coach) is making it to the states this year. We're in NJ. But I was curious to know approx. how many kids go from each of the 6 regions to the state (including team and individuals) and I can't find these stats anywhere.  I know each state is different, but I've read anywhere from ONLY "2 individuals and one team of 4" -- to 8 teams of 4 and 30 individuals (62 kids) from a single region, and surely, both can't be true -- can it??  What is the Average from your state? And how does it work from there? I mean -- do the top 4 INDIVIDUAL SCORES make it - including those scores of kids on a team -- or are teams counted separately? Do Teams compete in team rounds against other teams at nationals or do only individuals compete at that final level? I DID check mc website and am more confused than ever. PLEASE HELP me understand.

Thanks[/quote]
PM treething.

Here are the official instructions given to coordinators.

Yeah like in Maryland they take the top 29 people in my chapter (wow a lot huh) but then at states only the top 4 people qualify for nationals.

[quote="indianamath"]the number of kids who make states from each chapter varies a lot because some chapters are a lot stronger than others.

Some chapters take top 2 individuals and the top team, but others like mine take top 12 individual and top 6 teams.[/quote]It just said that that's the minimum


[quote="payman_pm"]let $x_1,x_2,...,x_n \in Z$ such that
$\sum _{i=1}^{n} x_i^2 +n^3 \le (\sum_{i=1}^{n}x_i)(2n-1)+n^2$
prove that :
I)$x_i\ge0$ for $1\le i \le n$
II)$\sum_{i=1}^{n} x_i +n+1$ is NOT a perfect square.[/quote]

Let $f(x)= x^2+n^2-x(2n-1)-n = (x- \frac{2n-1}{2})^2-\frac{1}{4}$ for integer x.
The minimum occurs when x=n or n-1, and f(n)=f(n-1)=0.
Hence for all integer x, $f(x)= x^2+n^2-x(2n-1)-n \geq 0$ or $ x^2+n^2 \geq x(2n-1)+n$. Thus,  $\sum _{i=1}^{n} x_i^2 +n^3 \geq (\sum_{i=1}^{n}x_i)(2n-1)+n^2$
The given condition implies that equality holds and hence we have $x_i=n$ or $n-1$ for each i. Thus, $x_i \geq n-1 \geq 0$

On the other hand,
$n^2<n^2+1 = n(n-1)+n+1 \leq \sum_{i=1}^{n} x_i +n+1 \leq n(n)+n+1=n^2+n+1<n^2+2n+1=(n+1)^2$
Thus $\sum_{i=1}^{n} x_i +n+1$ cannot be  a perfect square


Can anybody help me please in this problem i have? When i place an image in the tex file, at the pdf file it allways appears at the end of the file. Thank you :)

It depends on what you have or haven't told LaTeX what to do with your image. Rather than speculating could you give a minimal example of your document? This would include the preamble (the stuff before \begin{docment}) and example text and image which goes to the end instead of where you want it. Then it should be easy to see where the problem lies.

Hello again. This is the way i include my image to the tex file. All the rest are the document which is not nececary to place it. But where ever i put the comand \includegraphics, my image appears at the end of the document at the pdf file...Thanks again

\documentclass[11pt]{article}
\usepackage{amsfonts,amsmath,greektex}
\usepackage{graphicx}
\begin{document}

\begin{figure}
  % Requires \usepackage{graphicx}
  \includegraphics[width=350pt length=200pt]{tra}\\
  \caption{3}\label{7}
\end{figure}

\end{document}

Firstly there are a few errors in the line
[quote]\includegraphics[width=350pt length=200pt]{tra}\\[/quote]
[i]length[/i] should be [i]height[/i], there needs to be a comma before it, and you don't need \\ so you get
[code]\includegraphics[width=350pt, height=200pt]{tra}[/code]

Secondly, if you want the graphics to appear as close to where you've typed it as possilbe then change the line
[code]\begin{figure} [/code]to [code]\begin{figure}[h][/code]h meaning [i]here[/i]

If that still doesn't work then try
[code]\begin{figure}[!h][/code]where ! means it [i]must[/i] be here and ignores other spacing.

The following document works putting the figure between the 2 lines of text:
[code]\documentclass[11pt]{article}
\usepackage{amsfonts,amsmath}
\usepackage{graphicx}
\begin{document}
This is text before the image
\begin{figure}[h]
% Requires \usepackage{graphicx}
\includegraphics[width=350pt, height=200pt]{tra}
\caption{3}\label{7}
\end{figure}
\\This is text after the image
\end{document}[/code]Note the \\ is used before the second text line to separate it from the first line so that there is room for the image.

(I don't have fonts used in greektex.sty but I wouldn't have thought it would make a difference.)

Thank you, it works fine now! Can i ask please another silly question? How can i place my image in the left side of my document and keep writing on the left side? Thanks again :lol:

I don't understand what you mean. Normally you have

text here
image here
.
.
.
text here

so everything stays left aligned. Presumably you want something else?

Just thought - do you mean that you want the caption to be left-aligned? If so, use the caption package with the line[code]\usepackage[justification=raggedright,singlelinecheck=false]{caption}[/code]

Sorry i was not understood. I mean how can i have this style:

text here

text here + image on right side

text here
thanks again :lol:

You can use minipages to put things side by side. However, to use \caption you need a float (\begin{figure}...\end{figure}) and that can't be used inside a minipage. The solution is to put these outside the minipages. Here's an example:
[code]\documentclass[11pt]{article}
\usepackage{amsfonts,amsmath}
\usepackage{graphicx}
\begin{document}
This is text before the image
\begin{figure}[h]
\begin{minipage}{4cm}
This is text to the left of the image
This is text to the left of the image
This is text to the left of the image
This is text to the left of the image
This is text to the left of the image
This is text to the left of the image
This is text to the left of the image
\end{minipage}
\begin{minipage}{2cm}
\
\end{minipage}
\begin{minipage}{6cm}
% Requires \usepackage{graphicx}
\includegraphics[width=5cm, height=2cm]{image}
\caption{3}\label{7}
\end{minipage}
\end{figure}
\\This is text after the image
\end{document} [/code]
Here I've used 3 minipages with the middle one only having a space \ in it, which gives a gap between the text and image. Obviously you'll need to adjust the width of all the minipages to suit your image and the width of the document.

You'll also find a couple of suggested packages at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134343]Getting text to the right of a picture?[/url]


algebra
polynomial
Vieta
Divisibility Theory
pen

[list=a][*] Find infinitely many pairs of integers $a$ and $b$ with $1<a<b$, so that $ab$ exactly divides $a^{2}+b^{2}-1$. [*] With $a$ and $b$ as above, what are the possible values of \[\frac{a^{2}+b^{2}-1}{ab}?\] [/list]

a) the equation is symmetric in $a$ and $b$. $(2,3)$ is a solution with $k=2$. by vieta-jumping we get for the 2 zeroes $a_{1}$ and $a_{2}$ of $a^{2}+b^{2}-1=kab$  the equations $a_{1}+a_{2}=kb$ and $a_{1}a_{2}=b^{2}-1$. so $a_{2}>b$ for $k>1$. now we change $a$ and $b$ and get infinitely many pairs $(a,b)$. if we want to we could name them exactly...

b) we want to show that all positive numbers greater than 1 are possible.
so we set $a=k$ and $b=k^{2}-1$ where $k\in\mathbb{N}>1$:
\[\frac{a^{2}+b^{2}-1}{ab}=\frac{k^{2}+k^{4}-2k^{2}+1-1}{k(k^{2}-1)}=\frac{k^{2}(k^{2}-1)}{k(k^{2}-1}=k\]
$a$ and $b$ are greater than $1$ so $a^{2}+b^{2}-1>ab$ so we get $k>1$.

Naphthalin

[quote="Naphthalin"]
a) the equation is symmetric in $ a$ and $ b$. $ (2,3)$ is a solution with $ k \equal{} 2$. by vieta-jumping we get for the 2 zeroes $ a_{1}$ and $ a_{2}$ of $ a^{2} \plus{} b^{2} \minus{} 1 \equal{} kab$ the equations $ a_{1} \plus{} a_{2} \equal{} kb$ and $ a_{1}a_{2} \equal{} b^{2} \minus{} 1$. so $ a_{2} > b$ for $ k > 1$. now we change $ a$ and $ b$ and get infinitely many pairs $ (a,b)$. if we want to we could name them exactly...
[/quote]

how does the above logic imply infinitely many solutions

How exactly do we name them??

could anyone give a more analytical method. i.e without having to guess for the proper forms. something on the lines of solving the following equations

\[ \frac{{\left( {a \plus{} b \minus{} 1} \right)\left( {a \plus{} b \plus{} 1} \right)}}
{{ab}} \equal{} n \plus{} 2\]

(1) When $b=a+1$, $a^2 + b^2 - 1 = a^2 + (a+1)^2 - 1 = 2a(a+1)$, which is divided by $ab=a(a+1)$.

[quote="FelixD"]Actually your solution to b) implies a)... so why did you write a) too?^^[/quote]

(2) This problem can be solved by using Vieta-jumping method conversely.

$\frac{a^2 + b^2 - 1}{ab} = m$ is equal to $a^2 - mab + b^2 - 1 = 0$.
WLOG $a>b$, we can make another positive integer solution $(a', b)$ by setting $a'=mb-a$.
By vieta theorem, $a \times a' = b^2 - 1 < a^2$; thus $a'<a$.
This procedure lasts until $b=1$. Therefore the integer solution which gives the smallest sum is $(m,1)$.
Then, the solution of the previous step would be $(a',b')$ which satisfies $m=mb'-a'$ and $1=b'$, or $a'=m^2 - 1$ and $b'=m$.

[quote="arkanm"]The first equation implies that $v_2$ is an integer, the second that $v_2>0$ since $A>B>1$. Thus, $(B,v_2)\in S$. Furthermore, $a+b$ is minimal at $(A,B)$...[/quote] I assume that your definition of $S$ also requires $a,b > 1$ since you're using the fact that $A > B > 1$.  In that case, this is wrong: you need $v_2 > 1$ to get the pair in $S$.

In fact, $(a,b) = (8,3)$ yields $k=3$ if I'm not mistaken, so the claim you're trying to prove was wrong to begin with (and I just debugged it by plugging in $(a,b) = (8,3)$ and seeing what it Vieta jumped to).  See the second post in the thread.

[quote=mxgo][quote=Naphthalin]
b) we want to show that all positive numbers greater than 1 are possible.
so we set $a=k$ and $b=k^{2}-1$ where $k\in\mathbb{N}>1$:
\[\frac{a^{2}+b^{2}-1}{ab}=\frac{k^{2}+k^{4}-2k^{2}+1-1}{k(k^{2}-1)}=\frac{k^{2}(k^{2}-1)}{k(k^{2}-1}=k\]
$a$ and $b$ are greater than $1$ so $a^{2}+b^{2}-1>ab$ so we get $k>1$.

Naphthalin[/quote]

What is the motivation for this construction?[/quote]

The vieta jumping naphthalin did in part a) can be reversed until the equation produces the pair (a,1). This is clearly invalid, but then you note that roots $a_1, a_2$ satisfy $a_1 + a_2 = kb$ and $a_1 \cdot a_2 = b^2-1$. Letting $a_1$ = 1 implies that k = b. Moreover, we must be able to do this due to the "finite descent" the vieta jumping invoked. Since b can be varied over all integers, this implies that all k > 1 work.

Tracing back our steps provides the solution a = k and b = $k^2-1$.

[quote=mxgo]
What is the motivation for this construction?[/quote]
I got the same construction, so here is my thought process:
With no idea on how to start with (b), I started with (a).
Naturally, I first set $a=2$ to get $3b|b^2-8$. It is clear that $b=3$ works.
Here is when it struck me- does every solution of the form $(a,a+1)$ work? Substitute it, and we get
$$E=\frac{a^2+b^2-1}{ab}=\frac{a^2+a^2+2a+(1-1)}{a(a+1)}=\frac{2a(a+1)}{a(a+1)}=2$$
That's how some wishful thinking can come in handy!

For the part (b), I guessed that every natural number is possible, for which all that was needed was to make a (clever) construction. Naturally, I tried to extend the above construction to $(a,b)=(a,a+k)$. I tried this, but it did not work. What is it that is not working here, but worked for $k=1$?
After some thought, I realized it is that the $1$ gets canceled by the (annoying) $-1$! So I rather tried $(a,b)=(a,ka-1)$. We then get
$$E=\frac{a^2+(ka-1)^2-1}{a(ka-1)}=\frac{(k^2+1)a-2k}{ka-1}=k+\frac{a-k}{ka-1}$$
So we want $ka-1|a-k$. The simplest thing that we can do is to make $a-k=0$ by setting $a=k$! Boom, we are done!
We have the construction $(a,b)=(k,k^2-1)$.