In New York State, the standard high school physics course is Regents Physics, which is pretty straightforward but I thought definitely more difficult than the other regents science courses.

A few of my friends wanted to take AP Physics B next year and my school is offering it to students with having to have taken a basic first year course yet. I've heard that B is nowhere near as in-depth as C but is it difficult without the intro? I'm taking standard high school physics this year and will be taking C next year. I did terribly in the first semester but I have a tutor now and it's about a thousand times easier now.

Physics B is meant to be a "survery" course, so it covers far more than Physics C, but without the depth.  Physics C is closer to that of an freshman-year physics but depending on the university (like at most top tier ones), C is still nowhere close.  Also, most schools will not offer credit for Physics B whereas they will for C.

Your level of confidence with calculus should help with your decision.  If you don't know much calculus or aren't comfortable with it, don't take B.  Especially in the E&M portion, you'll need to know how to do some important integrals.

That is an important difference but as I pointed out, it's not the only one.

[quote="Yongyi781"]If I remember correctly, AP Physics B requires no calculus knowledge and is algebra-based...[/quote]Even as they say so, limits, derivatives and integrals may creep into course materials here and there. Calculus is not the only way to deduce the formulas, but calculus makes deduction of formulas easier to explain, so I think teachers and books slip into calculus mode without even realizing sometimes.
The breadth of AP Physics B also makes it hard. There is too much material to absorb in too short a time.
I would say that any physics course worth taking will be difficult with no previous course other than the bird's eye view from middle school. I also believe that a good teacher could help a great deal, and that with hard work it can be done.

Disclaimers/Notes: I made my son take AP Physics B (online) this year as a first high school physics course, and he complains a lot. He is behind mostly because he expects to get by with at most an hour of studying per day.

It's probably not a great idea to take an AP Physics as a first exposure (it certainly can be done, my own physics teacher asked me if I wanted to skip straight to AP despite never having finished an introductory course).  The problem is that doing that requires extra motivation and time commitments.

Calculus is certainly more difficult than pre-calculus, but I feel (and most physicists would probably agree) that calculus makes problems easier.  Limits, derivatives and integrals don't show up in the B curriculum which I think makes concepts a bit more difficult.  But it makes getting answers simpler since the formulas provided on the reference table are easy plug-and-chug formulas.  I also get the sense that B is far more disjointed since it covers so much more.  I doubt a high school physics teacher is going to carefully connect nuclear, thermo, E&M and everything else.  In C, there is a more natural flow.

Generally, physics is easier the second (or third, fourth, fifth,...) time around so it's wise to take an introductory course first.

I've been taking AP Physics B in my high school and I think it's pretty easy. However, you really do have to study for this class and re-read chapters just to get any concepts you may misunderstand when a teacher explains it. Although I'm one of the more brighter kids in the class, I'm sure that as long as you review any concepts/problems you don't know, you'll be okay. I just had my first semester final today and I think I got a perfect on it.  :wink:
And I'll agree with some others in this topic. The concepts you learn don't really connect well. First, I learned mechanical energy (springs, projectile motion, force, momenta, etc.), then jumped to fluids -> thermo -> waves/sound. Next is going to be electricity for my class. Most of the chapters re-cover over the previous ones usually so if you do bad on one of them, you're going to have to study it again or you'll continually do worse. I love this class though and it's very fun if you get the hang of it. I can't wait till I taking some of the competition tests.

AP Physics does not incorporate any integrals, that I can assure you. But there is a lot of materials, and a lot of concepts and problem solving.

AP Physics C divides into 2 parts, Mechanics and E&M. Mechanics is relatively simple, if you have a mastery of Calculus..then you should be fine.  E&M is a whole new ball games with the Maxwell's equations.

Personally, I made the choice (maybe a mistake?) to take all 3 AP exams last year (yeah, first time learning physics. ._.) which devastated all of the school grades (especially E&M).

So if you want to learn more in depth, then study AP Physics C Mechanics.  You will master mechanics. If you want a greater view of physics in general, go with AP Physics B (which covers all the physics up into Modern).  If you want a challenge, go with AP Physics C E&M (but be prepare for some multivar. calc symbols)

I agree with that assessment but I recommend that if you've never taken physics before, you don't skip straight to AP because the AP courses already assume you know the basics.

E&M is certainly harder than Mechanics, but unless you have a sadistic teacher, you won't need any multivariable calculus for the AP course.  In fact, at most universities, only the "physics for physics majors" E&M actually requires vector calculus.  As long as you know integrals and derivatives, you'll be fine with the math.

I'm a sophmore in AP physics B, and I think it's really easy, although a lot of people don't. It really depends on how well you can understand the concepts. I have 2 friends who took AP C their sophmore year, and they thought it was hard. For me, I don't think taking B was worth it, but I could see a lot of people benefiting from a low intensity course like that as an introduction to physics.
For B, you can pretty much get away with memorizing the basic formulas, as the course pretty much glides over all the topics. It does help to know the fundamentals of calculus, as knowing them will bring insight upon some of the formulas thrown at you.


Let $ D$ and $ E$ be two points on the side $ BC$ of triangle $ ABC$ such that $ BD/CD \equal{} 2CE/BE$. The circumcircle of $ ADE$ meets $ AB$ and $ AC$ at $ M$ and $ N$,respectively. Prove that regardless of the positions of the points $ D$ and $ E$ on $ BC$,the centroid of the triangle $ AMN$ lies on a fixed line.


[color=blue] [size=150]$18{th}$  Iberoamerican Olympiad[/size]
Buenos Aires, ARGENTINA. [2003]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 2.[/b] Sean $ C$ y $ D$ dos puntos en una semicircunferencia de diámetro $ AB$  tal que $ C$ y $ B$ están distintos lados de la recta $ AD.$ $ M,N$ y $ P$ son los puntos medios de $ CA,DB,CD,$ respectivamente y sean $ O_A$ y $ O_B$ los circuncentros de $ \triangle ACP$ y $ \triangle BDP.$ Probar que las rectas $ MN$ y $ O_AO_B$ son paralelas. [/quote]
Sea $ L$ el punto medio de $ AB$ y $ Q \equiv \odot(ACP) \cap \odot(BDP),$ diferente de $ P.$ Se tiene claramente  $ O_AO_B \perp PQ$ y  las rectas $ CA,DB$ y $ PQ$ concurren en el centro radical $ E$ de $ \odot(ACP) ,$ $ \odot(BDP)$ y la circunferencia de diámetro $ AB.$ Como $ CD$ es antiparalela a $ AB$ con respecto a los rayos $ EA,EB,$ existe el producto de la simetría axial con respecto a la bisectriz interior de $ \angle AEB$ por una homotecia positiva centrada en $ E$ que lleva $ P$ a $ L$ $ \Longrightarrow$ Los rayos $ EP$ y $ EL$ son isogonales con respecto a $ \angle AEB,$ i.e. $ EP$ es la E-simediana de $ \triangle EBC.$ Entonces como $ EMLN$ es cíclico en vista que $ \angle EML \equal{} \angle ENL \equal{} 90^{\circ},$ se tiene $ \angle MEL \equal{} \angle MNL \equal{} \angle PEN$ $ \Longrightarrow$ $ MN \perp PE$ y por tanto $ O_AO_B \parallel MN.$

[quote][b]Problema 5.[/b] En un cuadrado $ ABCD,$ sean $ P$ y $ Q$ dos puntos en $ BC$ y $ CD$ diferentes de sus puntos extremos. $ X$ e $ Y$ son dos puntos en  $ AP$ y $ AQ$ tales que $ X \not \equiv Y.$ Probar que por cada $ X,Y$ existe un triángulo cuyos lados miden $ BX,XY$ y $ DY.$[/quote]
Sean $ \omega_1$ y $ \omega_2$ las circunferencias centradas en $ X,Y$ con radios $ XB$ y $ YD,$ respectivamente. No hay más que probar que $ \omega_1$ y $ \omega_2$ son secantes. Considérese $ N \equiv BQ \cap AP$ y $ M \equiv DP \cap AQ.$  De $ \triangle ABP \cong \triangle BCQ$ y $ \triangle ADQ \cong \triangle DCP$ se desprende facilmente que $ AM \perp DP$ y $ AN \perp BQ$ $ \Longrightarrow$ $ T \equiv BQ \cap DP$ es el ortocentro de $ \triangle APQ.$ Ésto implica

$ TP^2 \minus{} TQ^2 \equal{} AP^2 \minus{} AQ^2 \equal{} BP^2 \minus{} DQ^2 \Longrightarrow TX^2 \minus{} XB^2 \equal{} TY^2 \minus{} YD^2$

$ \Longrightarrow  p(T,\omega_1) \equal{} p(T,\omega_2)$

Es decir que el eje radical $ \ell$ de $ \omega_1$ y $ \omega_2$ pasa por $ T.$  Por otro lado como $ BQ > BC \equal{} BA,$ resulta $ \angle BAQ > \angle AQB$ $ \Longrightarrow$ $ \angle DTQ > \angle TDQ$ $ \Longrightarrow$ $ QD > QT  ,$ $YD > YT.$ Entonces $ T$ es un punto interior a $ \omega_2.$ Así $ \ell$ corta a $ \omega_2,$ digamos en $ U,V$ $ \Longrightarrow$ $ \omega_1$ y $ \omega_2$ se cortan en $ U,V,$ como se deseaba probar.


inequalities
trigonometry
geometry
inequalities unsolved

[quote="msg_1989"]1. Let a, b, c be the sides of a triangle. Then show that
[(a/ (b + c)) + (b/ (c + a)) +(c/ (a + b))] lies between the limits 3/2 and 2.
Can equality hold at either limit???

2. A sheep is tied to the edge of a circular patch of grass of radius ‘r’. The length of the rope used to tie the sheep is ‘l’. Find the length of the rope if the sheep eats away exactly half of the grass.

please solve these problems as soon as possible

Thank u!!![/quote]
[hide="1"] We have
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b} \geq \frac{(a\plus{}b\plus{}c)^2}{2ab\plus{}2ac\plus{}2bc} \geq \frac{3}{2}$(becouse $ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}ac\plus{}bc$)(equality holds if a=b=c)[/hide]
[hide="2"]
We have that $ \frac{\pi \cdot r^2}{2}\equal{}\pi \cdot l^2$
so $ l\equal{}\frac{r}{\sqrt{2}}$[/hide]

[quote="msg_1989"]1. Let a, b, c be the sides of a triangle. Then show that
[(a/ (b + c)) + (b/ (c + a)) +(c/ (a + b))] lies between the limits 3/2 and 2.
Can equality hold at either limit???

please solve these problems as soon as possible
Thank u!!![/quote]

[hide="1"]
we should to prove that
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b} \leq 2$
which is equivolent to
$ \frac{b\plus{}c\minus{}a}{b\plus{}c}\plus{}\frac{a\plus{}c\minus{}b}{a\plus{}c}\plus{}\frac{a\plus{}b\minus{}c}{a\plus{}b} \geq 1$
Let's get $ b\plus{}c\minus{}a\equal{}x$
$ a\plus{}c\minus{}b\equal{}y$
$ a\plus{}b\minus{}c\equal{}z$
then we have to prove
$ \frac{2x}{2x\plus{}y\plus{}z}\plus{}\frac{2y}{x\plus{}2y\plus{}z}\plus{}\frac{2z}{x\plus{}y\plus{}2z} \geq 1$
we have
$ \frac{2x}{2x\plus{}y\plus{}z}\plus{}\frac{2y}{x\plus{}2y\plus{}z}\plus{}\frac{2z}{x\plus{}y\plus{}2z} \geq \frac{2(x\plus{}y\plus{}z)^2}{2(x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}xz\plus{}yz)}>1$
equality will not hold becouse the last inequality was strict.[/hide]

1. $ \sum \frac {a}{b \plus{} c} < \sum \frac {2a}{a \plus{} b \plus{} c} \equal{} 2$

You posted an INMO prob in no. 1.
[hide="solution to no 1"]
Well, here is my sol:::::::::
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}$
$ \equal{} \frac {a}{b \plus{} c} \plus{} 1 \plus{} \frac {b}{c \plus{} a} \plus{} 1 \plus{} \frac {c}{a \plus{} b} \plus{} 1 \minus{} 3$
$ \equal{} \frac {a \plus{} b \plus{} c}{b \plus{} c} \plus{} \frac {a \plus{} b \plus{} c}{c \plus{} a} \plus{} \frac {a \plus{} b \plus{} c}{a \plus{} b} \minus{} 3$
$ \equal{} (a \plus{} b \plus{} c)(\frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b}) \minus{} 3$
$ \equal{} \frac {1}{2}.(2a \plus{} 2b \plus{} 2c)(\frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b}) \minus{} 3$
$ \equal{} \frac {1}{2}.[(a \plus{} b) \plus{} (b \plus{} c) \plus{} (c \plus{} a)][\frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b}] \minus{} 3$
$ \geq\frac {1}{2}.9 \minus{} 3 \equal{} \frac {3}{2}$ :lol:
Hence  $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq\frac {3}{2}$

Also as $ a,b,c$ are the sides of a triangle, $ a \plus{} b > c , b \plus{} c > a, c \plus{} a > b$
Suppose $ a\le b\le c$
Hence $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\leq \frac {a}{a \plus{} c} \plus{} \frac {c}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \equal{} 1 \plus{} \frac {c}{a \plus{} b} < 1 \plus{} 1$ as a, b, c are the sides of a triangle.

Hence$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} < 2$
Thus   $ 2 > \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq \frac {3}{2}$

[/hide]
[hide="Click to reveal hidden content"]
For prob 2, SaYaT's sol is correct.....[/hide]

That INMO problem was surely not original, like many others.

[hide="The method for 2"]The triangle $ O_1O_2Y$ has sides $ r,r,l$. By cosine rule, we get the angles of it.

Now we have angle $ \angle YO_1X$ which gives the area of the portion of the grass-patch inside that angle. $ \cdots (i)$

Also we have angle $ \angle YO_2X$ which gives the area of the portion inside this angle and inside the 2nd circle. $ \cdots (ii)$

Also it is easy to measure the area of $ O_1O_2Y$ because its sides are $ r,r,l$. So we have the area of $ O_1YO_2X$ as well. $ \cdots (iii)$

We now have the equation: $ (i) \plus{} (ii) \minus{} (iii) \equal{} \frac {\pi r^2}{2}$[/hide]


geometry
function
perimeter
3D geometry
sphere
modular arithmetic
algebra

IMO 1978 ShortList Problems and Solutions.

[b][size=150]IMO 1978 ShortListed Problems: (Posted in Forums)[/size][/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1963307#p1963307]1.(BUL 1)[/url] The set $M = \{1, 2, \ldots , 2n \}$ is partitioned into $k$ non-intersecting subsets $M_1,M_2,\ldots , M_k$, where $n \geq k^3 + k$. Prove that there exist even numbers $2_{j_{1}}, 2_{j_{2}}, \ldots , 2_{j_{k+1}}$ in $M$ that are in one and the same subset $M_i \ (1 \leq i \leq k)$ such that the numbers $2_{j_{1}} - 1, 2_{j_{2}} - 1, \ldots, 2_{j_{k+1}} - 1$ are also in one and the same subset $M_j  \ (1 \leq j \leq k).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024440#p2024440]2.(BUL 4)[/url] Two identically oriented equilateral triangles, $ABC$ with center $S$ and $A'B'C$, are given in the plane. We also have $A'  \neq  S$ and $B' \neq S$. If $M$ is the midpoint of $A'B$ and $N$ the midpoint of $AB'$, prove that the triangles $SB'M$ and $SA'N$ are similar.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367368&#p367368]3.(CUB 3)[/url] [b]IMO1[/b] Let $ m$ and $ n$ be positive integers such that $ 1 \le m < n$.  In their decimal representations, the last three digits of $ 1978^m$ are equal, respectively, so the last three digits of $ 1978^n$.  Find $ m$ and $ n$ such that $ m \plus{} n$ has its least value.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024444#p2024444]4.(CZS 2)[/url] Let $T_1$ be a triangle having $a, b, c$ as lengths of its sides and let $T_2$ be another triangle having $u, v,w$ as lengths of its sides. If $P,Q$ are the areas of the two triangles, prove that
\[16PQ \leq a^2(-u^2 + v^2 + w^2) + b^2(u^2 - v^2 + w^2) + c^2(u^2 + v^2 - w^2).\]
When does equality hold?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024446#p2024446]5.(GDR 2)[/url] For every integer $d \geq 1$, let $M_d$ be the set of all positive integers that cannot be written as a sum of an arithmetic progression with difference $d$, having at least two terms and consisting of positive integers. Let $A = M_1$, $B = M_2  \setminus \{2 \}, C = M_3$. Prove that every $c \in C$ may be written in a unique way as $c = ab$ with $a \in A, b \in B.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367369&#p367369]6.(FRA 2)[/url] [b]IMO5[/b] Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024447#p2024447]7.(FRA 5)[/url] We consider three distinct half-lines $Ox, Oy, Oz$ in a plane. Prove the existence and uniqueness of three points $A \in Ox, B \in Oy, C \in Oz$ such that the perimeters of the triangles $OAB,OBC,OCA$ are all equal to a given number $2p > 0.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024448#p2024448]8.(GBR 4)[/url] Let $S$ be the set of all the odd positive integers that are not multiples of $5$ and that are less than $30m$, $m$ being an arbitrary positive integer. What is the smallest integer $k$ such that in any subset of $k$ integers from $S$ there must be two different integers, one of which divides the other?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367374&#p367374]9.(GBR 5)[/url] [b]IMO3[/b] Let $0<f(1)<f(2)<f(3)<\ldots$ a sequence with all its terms positive$.$ The $n-th$ positive integer which doesn't belong to the sequence is $f(f(n))+1.$ Find $f(240).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367388&#p367388]10.(NET 1)[/url] [b]IMO6[/b] An international society has its members from six different countries.  The list of members contain $1978$ names, numbered $1, 2, \dots, 1978$.  Prove that there is at least one member whose number is the sum of the numbers of two members from his own country, or twice as large as the number of one member from his own country.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024449#p2024449]11.(SWE 2)[/url] A function $f : I \to \mathbb R$, defined on an interval $I$, is called concave if $f(\theta x + (1 - \theta)y) \geq \theta f(x) + (1 - \theta)f(y)$ for all $x, y \in I$ and $0 \leq \theta \leq 1$. Assume that the functions $f_1, \ldots , f_n$, having all nonnegative values, are concave. Prove that the function $(f_1f_2 \cdots f_n)^{1/n}$ is concave.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367377&#p367377]12.(USA 1)[/url] [b]IMO4[/b] In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367379&#p367379]13.(USA 6)[/url] [b]IMO2[/b] We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024451#p2024451]14.(VIE 2)[/url] Prove that it is possible to place $2n(2n + 1)$ parallelepipedic (rectangular) pieces of soap of dimensions $1 \times  2 \times  (n + 1)$ in a cubic box with edge $2n + 1$ if and only if $n$ is even or $n = 1$.

[i]Remark[/i]. It is assumed that the edges of the pieces of soap are parallel to the edges of the box.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024453#p2024453]15. (YUG 1)[/url] Let $p$ be a prime and $A = \{a_1, \ldots , a_{p-1} \}$ an arbitrary subset of the set of natural numbers such that none of its elements is divisible by $p$. Let us define a mapping $f$ from $\mathcal P(A)$ (the set of all subsets of $A$) to the set $P = \{0, 1, \ldots, p - 1\}$ in the following way:

(i) if $B = \{a_{i_{1}}, \ldots , a_{i_{k}} \} \subset A$ and $\sum_{j=1}^k a_{i_{j}} \equiv n \pmod p$, then $f(B) = n,$

(ii) $f(\emptyset) = 0$, $\emptyset$ being the empty set.

Prove that for each $n \in P$ there exists $B \subset A$ such that $f(B) = n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024455#p2024455]16. (YUG 2)[/url] Determine all the triples $(a, b, c)$ of positive real numbers such that the system
\[ax + by -cz = 0,\]\[a \sqrt{1-x^2}+b \sqrt{1-y^2}-c \sqrt{1-z^2}=0,\]
is compatible in the set of real numbers, and then find all its real solutions.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024457#p2024457]17. (FRA 3)[/url] Prove that for any positive integers $x, y, z$ with $xy-z^2 = 1$ one can find nonnegative integers $a, b, c, d$ such that $x = a^2 + b^2, y = c^2 + d^2, z = ac + bd$.
Set $z = (2q)!$ to deduce that for any prime number $p = 4q + 1$, $p$ can be represented as the sum of squares of two integers.

hello, thank you for both links.
Sonnhard.

[size=150][b][u]IMO 1978 Longlist problems:[/u][/b][/size]
I have posted them in the forums and posted them in the contests section. It can be viewed [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=18&year=1978]here.[/url] Thanks to Amir for helping me with the pdf.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1963307&#p1963307]Problem 1:[/url] $(BUL 1)$ The set $M = \{1, 2, . . . , 2n\}$ is partitioned into $k$ nonintersecting subsets $M_1,M_2, \dots, M_k,$ where $n \ge k^3 + k.$ Prove that there exist even numbers $2j_1, 2j_2, \dots, 2j_{k+1}$ in $M$ that are in one and the same subset $M_i$ $(1 \le i \le k)$ such that the numbers $2j_1 - 1, 2j_2 - 1, \dots, 2j_{k+1} - 1$ are also in one and the same subset $M_j (1 \le j \le k).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2050068#p2050068]Problem 2:[/url] $(BUL 2)$ If
\[f(x) = (x + 2x^2 +\cdots+ nx^n)^2 = a_2x^2 + a_3x^3 + \cdots+ a_{2n}x^{2n},\]
prove that
\[a_{n+1} + a_{n+2} + \cdots + a_{2n} =\dbinom{n + 1}{2}\frac{5n^2 + 5n + 2}{12}\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2053018#p2053018]Problem 3:[/url] $(BUL 3)$ Find all numbers $\alpha$ for which the equation
\[x^2 - 2x[x] + x -\alpha = 0\]
has two nonnegative roots. ($[x]$ denotes the largest integer less than or equal to x.)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024440&#p2024440]Problem 4:[/url] $(BUL 4)$ Two identically oriented equilateral triangles, $ABC$ with center $S$ and $A'B'C$, are given in the plane. We also have $A'  \neq  S$ and $B' \neq S$. If $M$ is the midpoint of $A'B$ and $N$ the midpoint of $AB'$, prove that the triangles $SB'M$ and $SA'N$ are similar.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2050071#p2050071]Problem 5:[/url] $(CUB 1)$ Prove that for any triangle $ABC$ there exists a point P in the plane of the triangle and three points $A' , B'$ , and $C'$ on the lines $BC,
AC$, and $AB$ respectively such that
\[AB \cdot PC'= AC \cdot PB'= BC \cdot PA'= 0.3M^2,\]
where $M = max\{AB,AC,BC\}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2067995#p2067995]Problem 6:[/url] $(CUB 2)$ Prove that for all $X > 1$, there exists a triangle whose sides have lengths $P_1(X) = X^4+X^3+2X^2+X+1, P_2(X) = 2X^3+X^2+2X+1$, and $P_3(X) = X^4-1$. Prove that all these triangles have the same greatest angle and calculate it.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367368&#p367368]Problem 7:[/url] $(CUB 3)$ Let $ m$ and $ n$ be positive integers such that $ 1 \le m < n$.  In their decimal representations, the last three digits of $ 1978^m$ are equal, respectively, so the last three digits of $ 1978^n$.  Find $ m$ and $ n$ such that $ m \plus{} n$ has its least value.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2067997#p2067997]Problem 8:[/url] $(CZS 1)$ For two given triangles $A_1A_2A_3$ and $B_1B_2B_3$ with areas $\Delta_A$ and $\Delta_B$, respectively, $A_iA_k \ge B_iB_k, i, k = 1, 2, 3$. Prove that $\Delta_A \ge \Delta_B$ if the triangle $A_1A_2A_3$ is not obtuse-angled.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024444&#p2024444]Problem 9:[/url] $(CZS 2)$ Let $T_1$ be a triangle having $a, b, c$ as lengths of its sides and let $T_2$ be another triangle having $u, v,w$ as lengths of its sides. If $P,Q$ are the areas of the two triangles, prove that
\[16PQ \leq a^2(-u^2 + v^2 + w^2) + b^2(u^2 - v^2 + w^2) + c^2(u^2 + v^2 - w^2).\]
When does equality hold?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2050074#p2050074]Problem 10:[/url] $(CZS 3)$ Show that for any natural number $n$ there exist two prime numbers $p$ and $q, p \neq q$, such that $n$ divides their difference.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2065894#p2065894]Problem 11:[/url] $(CZS 4)$ Find all natural numbers $n < 1978$ with the following property: If $m$ is a natural number, $1 < m < n$, and $(m, n) = 1$ (i.e., $m$ and $n$ are relatively prime), then $m$ is a prime number.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2065896#p2065896]Problem 12:[/url] $(FIN 1)$ The equation $x^3 + ax^2 + bx + c = 0$ has three (not necessarily distinct) real roots $t, u, v$. For which $a, b, c$ do the numbers $t^3, u^3, v^3$ satisfy the equation $x^3 + a^3x^2 + b^3x + c^3 = 0$?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2065898#p2065898]Problem 13:[/url] $(FIN 2)$ The satellites $A$ and $B$ circle the Earth in the equatorial plane at altitude $h$. They are separated by distance $2r$, where $r$ is the radius of the Earth. For which $h$ can they be seen in mutually perpendicular directions from some point on the equator?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2058003#p2058003]Problem 14:[/url] $(FIN 3)$ Let $p(x, y)$ and $q(x, y)$ be polynomials in two variables such that for $x \ge 0, y \ge 0$ the following conditions hold:
$(i) p(x, y)$ and $q(x, y)$ are increasing functions of $x$ for every fixed $y$.
$(ii) p(x, y)$ is an increasing and $q(x)$ is a decreasing function of $y$ for every fixed $x$.
$(iii) p(x, 0) = q(x, 0)$ for every $x$ and $p(0, 0) = 0$.
Show that the simultaneous equations $p(x, y) = a, q(x, y) = b$ have a unique solution in the set $x \ge 0, y \ge 0$ for all $a, b$ satisfying $0 \le b \le a$ but lack a solution in the same set if $a < b$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2058004#p2058004]Problem 15:[/url] $(FRA 1)$ Prove that for every positive integer $n$ coprime to $10$ there exists a multiple of $n$ that does not contain the digit $1$ in its decimal representation.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367369#p367369]Problem 16:[/url] $(FRA 2)$ Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024457&#p2024457]Problem 17:[/url] $(FRA 3)$ Prove that for any positive integers $x, y, z$ with $xy-z^2 = 1$ one can find non-negative integers $a, b, c, d$ such that $x = a^2 + b^2, y = c^2 + d^2, z = ac + bd$.
Set $z = (2q)!$ to deduce that for any prime number $p = 4q + 1$, $p$ can be represented as the sum of squares of two integers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2056651#p2056651]Problem 18:[/url] $(FRA 4)$ Given a natural number $n$, prove that the number $M(n)$ of points with integer coordinates inside the circle $(O(0, 0),\sqrt{n})$ satisfies
\[\pi n - 5\sqrt{n} + 1<M(n) < \pi n+ 4\sqrt{n} + 1\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024447#p2024447]Problem 19:[/url] $(FRA 5)$ We consider three distinct half-lines $Ox, Oy, Oz$ in a plane. Prove the existence and uniqueness of three points $A \in Ox, B \in Oy, C \in Oz$ such that the perimeters of the triangles $OAB,OBC,OCA$ are all equal to a given number $2p > 0.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2058007#p2058007]Problem 20:[/url] $(GBR 1)$ Let $O$ be the center of a circle. Let $OU,OV$ be perpendicular radii of the circle. The chord $PQ$ passes through the midpoint $M$ of $UV$. Let $W$ be a point such that $PM = PW$, where $U, V,M,W$ are collinear. Let $R$ be a point such that $PR = MQ$, where $R$ lies on the line $PW$. Prove that $MR = UV$.

[u]Alternative version:[/u] A circle $S$ is given with center $O$ and radius $r$. Let $M$ be a point whose distance from $O$ is $\frac{r}{\sqrt{2}}$. Let $PMQ$ be a chord of $S$. The point $N$ is defined by $\overrightarrow{PN} =\overrightarrow{MQ}$. Let $R$ be the reflection of $N$ by the line through $P$ that is parallel to $OM$. Prove that $MR =\sqrt{2}r$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2065899#p2065899]Problem 21:[/url] $(GBR 2)$ A circle touches the sides $AB,BC, CD,DA$ of a square at points $K,L,M,N$ respectively, and $BU, KV$ are parallel lines such that $U$ is on $DM$ and $V$ on $DN$. Prove that $UV$ touches the circle.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=8993#p8993]Problem 22:[/url] $(GBR 3)$ Let $x$ and $y$ be two integers not equal to $0$ such that $x+y$ is a divisor of $x^2+y^2$. And let $\frac{x^2+y^2}{x+y}$ be a divisor of $1978$. Prove that $x = y$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024448&#p2024448]Problem 23:[/url] (GBR 4) Let $S$ be the set of all the odd positive integers that are not multiples of $5$ and that are less than $30m$, $m$ being an arbitrary positive integer. What is the smallest integer $k$ such that in any subset of $k$ integers from $S$ there must be two different integers, one of which divides the other?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367374&#p367374]Problem 24:[/url] (GBR 5) Let $0<f(1)<f(2)<f(3)<\ldots$ a sequence with all its terms positive$.$ The $n-th$ positive integer which doesn't belong to the sequence is $f(f(n))+1.$ Find $f(240).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2065903#p2065903]Problem 25:[/url] $(GDR 1)$ Consider a polynomial $P(x) = ax^2 + bx + c$ with $a > 0$ that has two real roots $x_1, x_2$. Prove that the absolute values of both roots are less than or equal to $1$ if and only if $a + b + c \ge 0, a -b + c \ge 0$, and $a - c \ge 0$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024446&#p2024446]Problem 26:[/url] $(GDR 2)$ For every integer $d \geq 1$, let $M_d$ be the set of all positive integers that cannot be written as a sum of an arithmetic progression with difference $d$, having at least two terms and consisting of positive integers. Let $A = M_1$, $B = M_2  \setminus \{2 \}, C = M_3$. Prove that every $c \in C$ may be written in a unique way as $c = ab$ with $a \in A, b \in B.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2056670#p2056670]Problem 27:[/url] $(GDR 3)$ Determine the sixth number after the decimal point in the number $(\sqrt{1978} +\lfloor\sqrt{1978}\rfloor)^{20}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2065907#p2065907]Problem 28:[/url] $(GDR 4)$ Let $c, s$ be real functions defined on $\mathbb{R}\setminus\{0\}$ that are nonconstant on any interval and satisfy
\[c\left(\frac{x}{y}\right)= c(x)c(y) - s(x)s(y)\text{  for any  }x \neq 0, y \neq 0\]
Prove that:
$(a) c\left(\frac{1}{x}\right) = c(x), s\left(\frac{1}{x}\right) = -s(x)$ for any $x = 0$, and also $c(1) = 1, s(1) = s(-1) = 0$;
$(b) c$ and $s$ are either both even or both odd functions (a function $f$ is even if $f(x) = f(-x)$ for all $x$, and odd if $f(x) = -f(-x)$ for all $x$).
Find functions $c, s$ that also satisfy $c(x) + s(x) = x^n$ for all $x$, where $n$ is a given positive integer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2065910#p2065910]Problem 29:[/url] $(GDR 5)$ Given a nonconstant function $f : \mathbb{R}^+ \longrightarrow\mathbb{R}$ such that $f(xy) = f(x)f(y)$ for any $x, y > 0$, find functions $c, s : \mathbb{R}^+ \longrightarrow \mathbb{R}$ that satisfy $c\left(\frac{x}{y}\right) = c(x)c(y)-s(x)s(y)$ for all $x, y > 0$ and $c(x)+s(x) = f(x)$ for all $x > 0$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367388&#p367388]Problem 30:[/url] $(NET 1)$ An international society has its members from six different countries.  The list of members contain $1978$ names, numbered $1, 2, \dots, 1978$.  Prove that there is at least one member whose number is the sum of the numbers of two members from his own country, or twice as large as the number of one member from his own country.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2067098#p2067098]Problem 31:[/url] $(NET 2)$ Let the polynomials
\[P(x) = x^n + a_{n-1}x^{n-1 }+ \cdots + a_1x + a_0,\]
\[Q(x) = x^m + b_{m-1}x^{m-1} + \cdots + b_1x + b_0,\]
be given satisfying the identity $P(x)^2 = (x^2 - 1)Q(x)^2 + 1$. Prove the identity
\[P'(x) = nQ(x).\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2069580#p2069580]Problem 32:[/url] $(NET 3)$ Let $\mathcal{C}$ be the circumcircle of the square with vertices $(0, 0), (0, 1978), (1978, 0), (1978, 1978)$ in the Cartesian plane. Prove that $\mathcal{C}$ contains no other point for which both coordinates are integers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068001#p2068001]Problem 33:[/url] $(SWE 1)$ A sequence $(a_n)^{\infty}_0$ of real numbers is called [i]convex[/i] if $2a_n\le a_{n-1}+a_{n+1}$ for all positive integers $n$. Let $(b_n)^{\infty}_0$ be a sequence of positive numbers and assume that the sequence $(\alpha^nb_n)^{\infty}_0$ is convex for any choice of $\alpha > 0$. Prove that the sequence $(\log b_n)^{\infty}_0$ is convex.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024449&#p2024449]Problem 34:[/url] $(SWE 2)$ A function $f : I \to \mathbb R$, defined on an interval $I$, is called concave if $f(\theta x + (1 - \theta)y) \geq \theta f(x) + (1 - \theta)f(y)$ for all $x, y \in I$ and $0 \leq \theta \leq 1$. Assume that the functions $f_1, \ldots , f_n$, having all nonnegative values, are concave. Prove that the function $(f_1f_2 \cdots f_n)^{1/n}$ is concave.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068091#p2068091]Problem 35:[/url] $(SWE 3)$ A sequence $(a_n)_0^N$ of real numbers is called concave if $2a_n\ge a_{n-1} + a_{n+1}$ for all integers $n, 1 \le n \le N - 1$.
$(a)$ Prove that there exists a constant $C >0$ such that
\[\left(\displaystyle\sum_{n=0}^{N}a_n\right)^2\ge C(N - 1)\displaystyle\sum_{n=0}^{N}a_n^2\:\:\:\:\:(1)\]
for all concave positive sequences $(a_n)^N_0$
$(b)$ Prove that $(1)$ holds with $C = \frac{3}{4}$ and that this constant is best
possible.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068092#p2068092]Problem 36:[/url] $(TUR 1)$ The integers $1$ through $1000$ are located on the circumference of a circle in natural order. Starting with $1$, every fifteenth number (i.e.,$1, 16, 31, \cdots$ ) is marked. The marking is continued until an already marked number is reached. How many of the numbers will be left unmarked?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068009#p2068009]Problem 37:[/url] $(TUR 2)$ Simplify
\[\frac{1}{\log_a(abc)}+\frac{1}{\log_b(abc)}+\frac{1}{\log_c(abc)},\]
where $a, b, c$ are positive real numbers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068094#p2068094]Problem 38:[/url] $(TUR 3)$ Given a circle, construct a chord that is trisected by two given noncollinear radii.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068096#p2068096]Problem 39:[/url] $(TUR 4)$ $A$ is a $2m$-digit positive integer each of whose digits is $1$. $B$ is an $m$-digit positive integer each of whose digits is $4$. Prove that $A+B +1$ is a perfect square.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068098#p2068098]Problem 40:[/url] $(TUR 5)$ If $C^p_n=\frac{n!}{p!(n-p)!} (p \ge 1)$, prove the identity
\[C^p_n=C^{p-1}_{n-1} + C^{p-1}_{n-2} + \cdots + C^{p-1}_{p} + C^{p-1}_{p-1}\]
and then evaluate the sum
\[S = 1\cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \cdots + 97 \cdot 98 \cdot 99.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367377&#p367377]Problem 41:[/url] $(USA 1)$ In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2067103#p2067103]Problem 42:[/url] $(USA 2)$ $A,B,C,D,E$ are points on a circle $O$ with radius equal to $r$. Chords $AB$ and $DE$ are parallel to each other and have length equal to $x$. Diagonals $AC,AD,BE, CE$ are drawn. If segment $XY$ on $O$ meets $AC$ at $X$ and $EC$ at $Y$ , prove that lines $BX$ and $DY$ meet at $Z$ on the circle.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068101#p2068101]Problem 43:[/url] $(USA 3)$ If $p$ is a prime greater than $3$, show that at least one of the numbers
\[\frac{3}{p^2} , \frac{4}{p^2} , \cdots, \frac{p-2}{p^2}\]
is expressible in the form $\frac{1}{x} + \frac{1}{y}$, where $x$ and $y$ are positive integers.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2069556#p2069556]Problem 44:[/url] $(USA 4)$ In  $ABC$ with $\angle C = 60^{\circ}$, prove that
\[\frac{c}{a} + \frac{c}{b} \ge2.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2069562#p2069562]Problem 45:[/url] $(USA 5)$ If $r > s >0$ and $a > b > c$, prove that
\[a^rb^s + b^rc^s + c^ra^s \ge a^sb^r + b^sc^r + c^sa^r.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367379&#p367379]Problem 46:[/url] $(USA 6)$ We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2069567#p2069567]Problem 47:[/url] $(VIE 1)$ Given the expression
\[P_n(x) =\frac{1}{2^n}\left[(x +\sqrt{x^2 - 1})^n+(x-\sqrt{x^2 - 1})^n\right],\]
prove:
$(a) P_n(x)$ satisfies the identity
\[P_n(x) - xP_{n-1}(x) + \frac{1}{4}P_{n-2}(x) \equiv 0.\]
$(b) P_n(x)$ is a polynomial in $x$ of degree $n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024451&#p2024451]Problem 48:[/url] $(VIE 2)$ Prove that it is possible to place $2n(2n + 1)$ parallelepipedic (rectangular) pieces of soap of dimensions $1 \times  2 \times  (n + 1)$ in a cubic box with edge $2n + 1$ if and only if $n$ is even or $n = 1$.

[i]Remark[/i]. It is assumed that the edges of the pieces of soap are parallel to the edges of the box.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2069571#p2069571]Problem 49:[/url] $(VIE 3)$ Let $A,B,C,D$ be four arbitrary distinct points in space.
$(a)$ Prove that using the segments $AB +CD, AC +BD$ and $AD +BC$, it is always possible to construct a triangle $T$ that is non-degenerate and has no obtuse angle.
$(b)$ What should these four points satisfy in order for the triangle $T$ to be right-angled?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2069572#p2069572]Problem 50:[/url] $(VIE 4)$ A variable tetrahedron $ABCD$ has the following properties:
Its edge lengths can change as well as its vertices, but the opposite edges remain equal $(BC = DA, CA = DB, AB = DC)$; and the vertices $A,B,C$ lie respectively on three fixed spheres with the same center $P$ and radii $3, 4, 12$. What is the maximal length of $PD$?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2069576#p2069576]Problem 51:[/url] $(VIE 5)$ Find the relations among the angles of the triangle $ABC$ whose altitude $AH$ and median $AM$ satisfy $\angle BAH =\angle CAM$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024453&#p2024453]Problem 52:[/url] $(YUG 1)$ Let $p$ be a prime and $A = \{a_1, \ldots , a_{p-1} \}$ an arbitrary subset of the set of natural numbers such that none of its elements is divisible by $p$. Let us define a mapping $f$ from $\mathcal P(A)$ (the set of all subsets of $A$) to the set $P = \{0, 1, \ldots, p - 1\}$ in the following way:

$(i)$ if $B = \{a_{i_{1}}, \ldots , a_{i_{k}} \} \subset A$ and $\sum_{j=1}^k a_{i_{j}} \equiv n \pmod p$, then $f(B) = n,$

$(ii)$ $f(\emptyset) = 0$, $\emptyset$ being the empty set.

Prove that for each $n \in P$ there exists $B \subset A$ such that $f(B) = n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2024455&#p2024455]Problem 53:[/url] $(YUG 2)$ Determine all the triples $(a, b, c)$ of positive real numbers such that the system
\[ax + by -cz = 0,\]\[a \sqrt{1-x^2}+b \sqrt{1-y^2}-c \sqrt{1-z^2}=0,\]
is compatible in the set of real numbers, and then find all its real solutions.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2068005#p2068005]Problem 54:[/url] $(YUG 3)$ Let $p, q$ and $r$ be three lines in space such that there is no plane that is parallel to all three of them. Prove that there exist three planes $\alpha, \beta$, and $\gamma$, containing $p, q$, and $r$ respectively, that are perpendicular to each other $(\alpha\perp\beta, \beta\perp\gamma, \gamma\perp \alpha).$


There are $ 5280$ feet in a mile.

$ \frac{100.9\times{5280}}{3600\times{60.5}}\approx\boxed{2.4}$

I got a different answer.

[hide]The distance/rate/time formula is d = rt, or r = d / t.

Distance = 60.5 / 5280.  (miles)

Time = .4 / (60 x 60).  (hours)

So . . . Rate = distance / time = 103.[/hide]

[quote="GameBot"]Nolan Ryan holds the record for the fastest pitch in baseball. He pitched a ball at a speed of 100.9 miles per hour. How many seconds, to the nearest tenth, did it take Ryan's pitch to travel the distance of 60 feet, 6 inches from the pitcher's mound to home plate?[/quote]
We first translate $100.9$ mph to ft/s. This is done by multiplying $\frac{5280}{3600} = \frac{22}{15}$. Now use the the distance formula; $d = rt$ to find the amount of time to get there:
$\frac{2219.8t}{15} = 60.5$
$t = \frac{907.5}{2219.8} \approx .4088$ seconds.
Rounded to the nearest tenth is therefore $\boxed{.4}$ seconds.
Birdmath did it correctly.


function
induction
floor function
algebra proposed
algebra

Find all functions $ f: Z \rightarrow R$ that verify the folowing two conditions:
(i) for each pair of integers $ (m,n)$ with $ m<n$ one has $ f(m)<f(n)$;
(ii) for each pair of integers $ (m,n)$ there exists an integer $ k$ such that $ f(m)\minus{}f(n)\equal{}f(k)$.

The first easy to prove that if f satisfy condition then $ \frac{f(x)}{a}$ also satisfy where a is an positive number . Therefore we can resume exist n such that $ f(n)\equal{}1$
Call $ M$ is  set of value of function f .
From (ii) condition we have  $ f(m)\minus{}1 \in M$
Other for every k there exist an integer $ k'$ such that $ f(k')\equal{}\minus{}f(k)$ ,it gives immediately when we change $ m\to n$.
Induction we have $ Z \in  M$
From condition (ii)  we have : $ t\in M \Leftrightarrow \{t\}\in M$  . (1)
Call $ k\plus{}m$ is the integer such that $ f(m\plus{}k)\equal{}1$
Consider set  values of $ \{f(m),...,f(m\plus{}k)\}$
We will prove that it must be an arithmetic sequence.
We have
$ f(m\plus{}k)\minus{}f(m)>f(m\plus{}k)\minus{}f(m\plus{}1)>...>f(m\plus{}k)\minus{}f(m\plus{}k\minus{}1)$
$ f(m\plus{}k)>f(m\plus{}k\minus{}1)>.....>f(m)$
Two sequence is a permutation so we have :
$ f(m\plus{}k)\minus{}f(m\plus{}k\minus{}1)\equal{}f(m\plus{}1)$
Similar $ f(m\plus{}k\minus{}1)\minus{}f(m\plus{}k\minus{}2)\equal{}f(m\plus{}1)$
....
It gives that this sequence is an arithmetic sequence. (2)
From (1),(2) we have $ f(n)\equal{}an\plus{}b$
Because exist k such that $ ak\plus{}b\equal{}0$ so $ \frac{b}{a}$ is an integer .
Therefore :
$ f(n)\equal{}a(n\plus{}t)$ where a is a positive real and t is a integer number .

[quote="TTsphn"]
From condition (ii)  we have : $ t\in M \Leftrightarrow \{t\}\in M$  . (1)
.[/quote]

What do you mean?


Prove that some power of $ 2$ has a decimal expansion that begins with $ 1999$; i.e. prove there exists n such that
$ 2^n \equal{} 1999...$

[hide]
Taking the log (base 10) of both sides,

$ n \log(2) \equal{} \log(1999\ldots)$.

Since $ 1999(10^k) \le RHS \le 2000(10^k)$,
we have $ \log(1999)\plus{}k \le n \log(2)\ < \log(2000)\plus{}k$.

But this is easily proven by the following lemma:

A man is walking around a circle with length $ k \in \mathbb{Z}$ (maybe even $ \mathbb{Q}$?) He has an irrational step length $ \alpha$. There is a hole on the circle with width > 0. Prove he will eventually fall into the hole.

[b]Proof.[/b] Consider the smallest $ n$ such that $ n\alpha > k$. Then, $ \{n\alpha\} < \alpha$ (where {x} is the fractional part of x) because if not, then $ (n\minus{}1)\alpha > k$, a contradiction. (here we use the fact that $ \alpha$ is irrational.)

Now make $ n\alpha$ your "new step length." (In reality, this is just $ n$ steps put together.) Every step, you "move" $ \{n\alpha\}$ (because you don't care how many times he moves around... just where he ends up.)

Since $ \{a\alpha\}$, for any $ a$, is always different (because $ \alpha$ is irrational), and we have just shown that there exists a sequence numbers $ S$ such that $ S_{n\plus{}1}/S_n \in \mathbb{Z}$ that is strictly decreasing, (rename the step length "$ \alpha$" in the above paragraph), the "step length" (which is actually lots of step lengths stuck together) will eventually be smaller than the width of the hole.

Then, the man must step into the hole. $ \blacksquare$

Using the lemma, we see that there must exist an $ n$ such that the man, with step length $ \log(2)$, must fall into the "hole" $ \log(1999)\plus{}k \le n\log(2) < \log(2000)\plus{}k$. So we're done. This generalizes to any irrational $ \log(n)$ and any sequence of numbers.

(Blah this is more like a proof outline, but filling in the details shouldn't be hard)
[/hide]


induction
number theory
modular arithmetic

[b]Prove[/b]: If $ a, b \in \mathbb{Z}$ and $ n \in \mathbb{N}$, prove that $ a \mod n \equal{} b \mod n \iff n|(a\minus{}b)$

I tried induction but didn't seem to get anywhere with it, if someone could show me how to do this it would be appreciated.  Also, I'm used to using congruency symbols in place of the equals, but my text has an equals symbol.  Is this significant?

The way my text defines modular arithmetic is $ a \equal{} qn\plus{}r$ where $ q$ is the quotient and $ r$ is the remainder upon dividing $ a$ by $ n$ gives us $ a \mod n \equal{} r$ or $ a \equal{} r \mod n$.  So by their definition, r is the smallest positive number or zero in the equivalence class for $ ^a/ _{\equiv n}$? I'm a bit confused on this.

So your text defines $ a \equiv b \bmod n$ if $ a, b$ leave the same remainder upon division by $ n$?

[hide="Then..."] Write $ a \equal{} qn \plus{} r, b \equal{} pn \plus{} r$.  Then $ n | a \minus{} b \equal{} (q \minus{} p)n$.  [/hide]


Let P, Q, R are the feets of the perpendiculars drawn from circumcentre to the sides of the triangle ABC. Circumcircles of the triangles APQ, BPR, CQR intersects circumcircle of ABC at the points X, Y, Z. Prove that the lines formed by XR, YQ, ZP intersects at a common point.

Is it a well known theorem?

It is easy to note that $ P, Q, R$ are midpoints of the sides of $ ABC$. (With respect to my labellings), the problem is equivalent to proving that the 3 medians of the triangle concur.

With respect to my labellings A=X, B=Y and C=Z... [b]borislav[/b] could you describe on which sides lie points P,Q,R?

OK thank you. Same as my labelling so A=X etc... Weird problem :P

i think mathangel is right
the cirumcircleof APQ intersects the circumcircle of triangle ABC at only one point (they are tangent to one another at A and so on )
hence the problem is equivalent to provingmedians are concurrent

Maybe he meant to say perpendiculars drawn from the circumference...don't know whether this can fix the problem.

I'm sorry - it was a problem from a math forum. "original" text was as is it in my first post. But it may be a good start of inventing a new problem(s).


inequalities
trigonometry
geometry
Putnam
triangle inequality

Determine the maximum value of

[tex]\displaystyle{\frac{(b^2+c^2-a^2)(c^2+a^2-b^2)(a^2+b^2-c^2)}{(abc)^2},}[/tex]

where [tex]a,b,c\in\mathbb{R}[/tex], and

[tex]\displaystyle{\frac{b^2+c^2-a^2}{bc}+\frac{c^2+a^2-b^2}{ca}+\frac{a^2+b^2-c^2}{ab}=2.}[/tex]

beta -- that only works if they're all positive.  If two of them were negative, something different might happen.

Thanks, this problem is tricky. Now it become much harder.

Just a little more work on this problem:

[hide]If, in fact, two terms must be negative, WLOG we can assume that they are the first two terms in ComplexZeta's second expression. Thus, multiplying by abc we obtain a^3 - ab^2 - ac^2 + b^3 - bc^2 - ba^2 + c^2 - ac^2 - ab^2 = 2abc, which factors into (a-b-c)(b-a-c)(c-a-b) - 2abc = 2abc. Can we prove that this equality cannot hold when two of the terms are negative?[/hide]

You forgot to multiply the RHS by abc.  Can you edit to fix?

JBL wrote:I really feel like the right route here is to use [hide]the fact that those expressions all come from the law of cosines.  In fact, does that give you that two of them can't be negative right away?[/hide]

[hide]Yes, the problem is equivalent to finding the max value of cos(x)cos(y)cos(z) given cos(x)+cos(y)+cos(z)=1.[/hide]

Where [hide]x, y and z are the angles of a triangle, most importantly -- it's not just any three angles.[/hide]

Later Edit:

Actually, I realized that this doesn't quite work, since a b and c are reals, not positive reals, so what I was saying before isn't really true.  However, if we deal with the two-of-them-are-negative case, we might be able to make it nicer by switching signs selectively.

Yeah...but this dealing with signs case by case seems ugly, tedious, boring, etc... I really don't think Simon would post that kind of a problem, would he? :)

Changing the sign of all 3 of a,b,c doesn't change either of the two values, so either all are positive or 2 are positive and 1 negative (WLOG). So the sign cases shouldn't be that tedious, since there are only 2 of them.

Then in the 2+, 1- case you can substitute A = -a and use cosines on the triangle with sides A, b, c ... does this lead anywhere?

the three positive numbers might not satisfy triangle inequality, so using cosines of angles wont work, unless we consider the forms-a-triangle and the doesnt-form-a-triangle cases separately.

In fact, it is impossible to have a,b,c satisfy triangle inequality according to masamune's equation

[hide](a-b-c)(b-a-c)(c-a-b) - 2abc = 2abc,

(a-b-c)(b-a-c)(c-a-b)=4abc

because if they are, a-b-c,b-a-c, c-a-b will all be negative and 4abc will be positive

therefore, LHS is negative, and RHS is positive.[/hide]

Yeah.  Well, back to the drawing board.

But, why would the Law of Cosines be there if you can't use it? :(

[quote="zabelman"]Are there any common properties of triangles that hold even if the triangle is degenerate or worse?[/quote]

I don't think so, since that's like asking "Are there any common properties of any 3 given real numbers?" ;)

Good point.

I just can't get over the fact that the question SCREAMS geometry and yet we can't do anything with it.

Multiplying out the equation we get
[tex]ab(a+b)+bc(b+c)+ac(a+c)=2abc+a^3+b^3+c^3[/tex].
But Schurs tells us
[tex]\begin{eqnarray*}
a^3+b^3+c^3+3abc & \ge & ab(a+b)+bc(b+c)+ac(a+c)
 \\ & = & 2abc+a^3+b^3+c^3
\end{eqnarray*}[/tex],
so
[tex]abc \ge 0 [/tex].

This tells us either none are negative or exactly two are negative.
(Hopefully this is not redundant!)

[quote="ComplexZeta"]Yeah, I was sure it had to be geometry, but I couldn't figure it out. (I still don't know how to do it, so I can't tell you if you're on the right track or not.)[/quote]

So where'd you get the problem from?

It was in a book. I think it's by Barbeau. Anyway, we got this problem in a photocopy of some book in Putnam practice, and I thought it would be fun to share it.

What if we reduce it to two variables by letting x = a/c and y = b/c?  I haven't tried it out, but it looks nice.  Also, we can impose basically any condition we want on a first- or second- (or higher-) degree homogeneous expression in terms of a, b and c because everything we have there is homogeneously 0th degree.  (My first suggestion amounts to the condition c = 1.)

I tried dividing them out, let a/c=x, b/c=y, a/b=x/y(same idea as Joel :) )

x/y+y/x-(1/xy)+x+1/x-y^2/x+y+1/y-x^2/y=2
multiply xy to both side
x^2+y^2-1+x^2(y)+y-y^3+y^2x+x-x^3-2xy=0
factor
(x-y)^2-(x-y)^2(x+y)+(x+y-1)=0
(x-y)^2(1-x-y)-(1-x-y)=0
((x-y)^2-1)(1-x-y)=0
either a+b=c, a+c=b, or b+c=a.
WLOG suppose a=b+c
sub into the original equation, we have
(a^2+b^2-c^2)/ab=2=(a^2+c^2-b^2)/ac
(b^2+c^2-a^2)/bc=-2
so the only possible product is 2(2)(-2)=-8
It is likely to be wrong.

Nice, beta. I got the same answer, but i did it a bit differently:

Like i mentioned earlier, I was trying to work in a geometric interpretation. I was reminded of the formula

[tex]\displaystyle \cos A + \cos B + \cos C = 1+\frac{r}{R}.[/tex]

After writing [tex]\displaystyle r=\frac{K}{s},[/tex] [tex]\displaystyle R=\frac{abc}{4K},[/tex] and [tex]K^2 = (a+b+c)(a+b-c)(b+c-a)(c+a-b)/16,[/tex] this identity becomes

[tex]\displaystyle \frac{a^2+b^2-c^2}{ab}+\frac{b^2+c^2-a^2}{bc}+\frac{c^2+a^2-bc^2}{ca}[/tex]

[tex]\displaystyle = 2+\frac{(a+b-c)(b+c-a)(c+a-b)}{abc}.[/tex]

Well, luckily it turns out that this geometric identity is actually an algebraic identity, so it is valid for any a, b, and c! With this knowledge, we conclude that [tex](a+b-c)(b+c-a)(c+a-b)=0[/tex], so WLOG, [tex]c=a+b[/tex]. Putting this c value into the expression for the product, the whole thing boils down to -8, so -8 is the only value for the product, and is thus the maximum value.


Hi!

I am looking for old (older than 2003) ICTM relays to pump up the math team practices at my school. Or other types of contest relays. Do you know where I could find them online?

Do you have any ideas how to organize math teams so that it would be fun and productive for everyone? (besides food)

Thanks!

I think the only reason that I come to math team practice is to eat food and hang out with friends. The NSML/ICTM stuff is just not mediocre compared to other math that I could be doing (on aops haha).


[quote="Rushil"]If $ [ \sqrt{n^2 +1} ] =\ [ \sqrt {n^2 + \lambda } ] $ , where $ n , \lambda \in \mathbb{N} $, find the no. of different values $ \lambda $ can have????[/quote]

lambda can have 2n possible values viz. 1,2,...2n

the answer is correct. Can you post your solution as  a hidden post plz!!!1

The next square number is $(n+1)^2=n^2+2n+1$, which is $2n+1$ greater than $n^2$, so you can have all natural numbers up $2n$, since if $\lambda=2n+1$, then the equation would be incorrect and show $n=n+1$


can someone solve for x?

(x^2)^2=27
apparently,
x belongs to set of NATURAL NUMBERS (ACCORDING TO A BOOK)
is this true or possible?

Sorry, is this a joke ?

EDIT. Then the book is a joke  :)

no...
its not
dont BE RUDE
:D jk
THIS IS IN AN ACTUAL BOOK...
SO YA...

We have $ \left( x^2 \right)^2\equal{}27$, or $ x^4\equal{}27$. Since $ \sqrt[4]{27}$ is not a natural number, no such $ x$ exists.

Sorry, is this a joke?
Solution:  x^4=27 therefore x=\sqrt[4]{27}


geometry
trapezoid
cyclic quadrilateral
projective geometry
geometry proposed

The opposite sides $ AB$ and $ CD$ of cyclic quadrilateral $ ABCD$ intersect at point $ M$. The tangent $ MN$ is parallel to the diagonal $ AC$. $ NB$ intersects $ AC$ at $ P$. Prove that the lines $ AN$, $ DB$, $ PM$ are concurrent.

I found a solution using projective geometry... i'd be interested in an eucledian one...

Let J be the intersection of BD and MN.
We have $ JN^2\equal{}JD.JB$
On the other side, $ \angle DMJ\equal{}\angle DCA\equal{}\angle DBA \Rightarrow JM^2\equal{}ID.JB$
So $ JM\equal{}JN$
Applying the lemma about trapezoid we are done!

So much simpler than mi solution... thanks

Dear Mathlinkers,
1. let Tn be the tangent to the circle at N  (Tn = MN).
2. According to the Pascal's theorem applied to Tn BDCAN, we are done...
Sincerely
Jean-Louis


inequalities
function
inequalities unsolved

While solving one of the latest Vasc's inequalities I came to it's equivavelent:

$ 3(a^{3}b^{2}c^{2} \plus{} b^{3}c^{2}a^{2} \plus{} c^{3}a^{2}b^{2}) \plus{} 2a^{5}b^{2} \plus{} 2b^{5}c^{2} \plus{} 2c^{5}a^{2}$ $ \plus{} a^{5}c^{2} \plus{} b^{5}a^{2} \plus{} c^{5}b^{2} \plus{} 2a^{4}b^{3} \plus{} 2b^{4}c^{3} \plus{} 2c^{4}a^{3}$ $ \geq4a^{4}b^{2}c \plus{} 4b^{4}c^{2}a \plus{} 4c^{4}a^{2}b \plus{} 2a^{4}c^{2}b \plus{} 2b^{4}a^{2}c \plus{} 2c^{4}b^{2}a \plus{} 2a^{4}c^{3} \plus{} 2b^{4}a^{3} \plus{} 2c^{4}b^{3}$

I found that this inequality is equivalent to (maybe this can be helpful):

$ (a \minus{} b)^{2}(2b^{3}c^{2} \plus{} a^{3}c^{2} \plus{} \frac {2a^{3}b^{3}}{a \minus{} b})$ $ \plus{} (b \minus{} c)^{2}(2c^{3}a^{2} \plus{} b^{3}a^{2} \plus{} \frac {2b^{3}c^{3}}{b \minus{} c}) \plus{} (c \minus{} a)^{2}(2a^{3}b^{2} \plus{} c^{3}b^{2} \plus{} \frac {2c^{3}a^{3}}{c \minus{} a})\geq 0$

I don’t know how to continue from this step so can anybody help me?

Is there any method for solving this kind of inequaliuties?

of course  :) There is a strong method to solve them  :D (but I don't think it is an nice method). Let $ a\equal{}min(a,b,c),b\equal{}a\plus{}p,c\equal{}a\plus{}1$. You will solve a function with coefficient. Consider the following functinon: $ g(a)\ge 0$

How can you assume that $ c\equal{}a\plus{}1$

Please can you post your complete solution [b]zaizai-hoang[/b]?

We can assume $ c\equal{}a\plus{}1$ because after we factored $ f(a,a\plus{}p,a\plus{}q)$ we will prove $ \sum a^n f(p,q)\ge 0$. We want to simplify the expression so we can put $ \frac{p}{q}\equal{}k$ then we only work with one variable.
I assume that $ c\equal{}a\plus{}1$ to make a simple calculation.  I talk about this method as a strong method to solve cyclic and also symetric inequalities but for your problem I don't think it will give us a nice solution although it still works very well.
Can you show me the latest Vasc's inequalities?

Actually this is not from new Vasc's book It is an equivalent form of one inequality that was proposed by Vasc in some magazine.


Dr. Worm crawls outside a square of side 1 cm, at all times remaining exactly 1 cm from the boundary of the square. In square cm, the area bounded by one complete circuit of the ant is what? Put answers in terms of pi.

[hide="hint"]
Think of a rounded square.
[/hide]

[hide="Solution                                                                                "]We can split the bounded area into several sections, the original unit square, four squares sharing a side with the original square, and four quarter-circles with radii one. The area of the five squares is 5*1=5. The four quarter-circles can be pieced together to make a unit circle, which has area pi. So the answer is 5+pi.[/hide]


algebra
function
domain
number theory open
number theory

Find all pairs of positive integers $ a,b$ such that $ q^2\minus{}r\equal{}2007$,
where $ q,r$ are respectively the quotient and the remainder obtained when dividing $ a^2\plus{}b^2$ by $ a\plus{}b$

In Gauss's ring $ a\plus{}ib,a,b\in Z$ any number can be factored.

Can you prove it with a simple method and don't use the complex number?

@Rust:
The possibility to factor doesn't suffice ;)
You need it to be a unique factorisation domain and that $ 2$ ramifies...

@thesunflower:
Yes one can do so (I did in an earlier thread), but it would still be better to learn the "complex" technique.

To Mods: Please lock this topic, this is a problem from M&Y, issue 10, and we can't discuss it now


group theory
abstract algebra
superior algebra
superior algebra solved

Let p>2 be a prime. Let C_p = {z in C | there exists n s.t. z p  n  = 1}

a) Show that C_p is a subgroup of the multiplicative group C*.
b) Let H be an infinite subgroup of C*, different from C*. Prove that every subgroup of H different from H is finite iff there exists a prime p s.t. H=C_p.

[i]Romanian NMO/final round/Vaslui 1998/problem 2/12'th grade[/i]

Seems related to quasicyclic group, isn't it ?

Errr... I don't know... I don't know what a quasicyclic group is... :D

1)If z\in C_p <=> z=exp(2i.pi.h/p^n)
1\in C_p
If z,z' \in C_p => z^(p^n)=z^(p^n')=1 => (zz')^(p^(n+n'))=1=> z.z' \in C_p
We also have z^(-1) \in C_p
C_p is a subrgroup of C*
2) Let z_0 \in C_p\H of order p^(k_0), z_0 is a root of X^(p^k)-1=0
and z\in C_p is of order p^k with k \geq k_0.Lets' prove that z is not in H
If z\in H => z'=z^(p^(k-k_0)) with order p^(k_0)
Let <z'> the group generated by z', <z'> its order is p^(k_0), and the
equation X^(p^ k_0)=1 has p^(k_0) solution =>z_0 \in <z'> contradiction.
The elements of order  \geq p^(k_0) is not in H, then the set of order
of all elements of H is bounded.Let k_1 its maximum.
If z has order k_1 => <z>={z, z^(p^(k_1))=1}=C_(p_1)
Then  {z , z^(p^(k_1))=1} is icluded in H and if z' \in H ,
(the order of z') \leq p^(k_1)=>z'^(p^(k_1))=1=> H is included C_(p_1)
H=C_(p_1)

 Z(p^oo)= \cup_{n\in N} {z\inC,  z^p^n=1} has no maximal subgroup
is called quasicyclic or Prfer p-group.
This showed there exist infinit group in wich  all elements are finite order.

I worked for quite some time on the problem:

If all the subgroups of G different from G are finite does it mean that G is finite?

Only after seeing the problem posted here I have understood that the answer for the above problem is "no".

Grobber does Romanian NMO change town each years ?

Well, I've never heard of the NMO being held in the same city in 2 consecutive years, but I think there are cities where it has been held more than once.


Let ($ W_B$) and($ W_C$) be excircles of triangle $ ABC$ opposite with $ B;C$ respectively .let ($ S_B$) ,($ S_C$) are symetric
with ($ W_B$) and($ W_C$) pass midpoint of $ AC$and $ AB$ respectively. $ m,N$ are intesection of ($ S_B$) and ($ S_C$)
 prove that $ MN$ divide triangle $ ABC$ become two part have same primeter.

This one is from ARMO 2005.

See here: http://www.mathlinks.ro/Forum/viewtopic.php?p=220226#220226


calculus
derivative
calculus computations

A train departures from the train station heading North with 50km/h. After 2h another departures from the train station with speed 60km/h heading East. Find the variation rate (dx/dy) between the distance of the 2 trains after 1.5h from the departure of the second train.

Let the +ve y-axis be the North direction and let the +ve x-axis be the East direction.

The position of the train, after 2hrs, going North is given by,

y = 100 + 50t

The position of the train going East is given by,

x = 60t

Let s be the distance between the two trains. Then s is given by,

s = √{x² + y²}

I suspect your question should have read,

[i]Find the variation rate of the distance between the 2 trains after 1.5h from the departure of the second train.[/i]

since otherwise it didn't make sense. dx/dy is independant of time, so specifying t = 1.5hrs was irrelevant.

i.e. what you want to find is ds/dt, not dx/dy.

s = √{x² + y²)
s = √{3600t² + (100+50t)²}
s = {3600t² + (100+50t)²}^(1/2)

using the chain rule,

ds/dt = (1/2){3600t² + (100+50t)²}^(-1/2)*{7200t + 2(100+50t)*50}
ds/dt = {7200t + 100(100+50t)} / 2√{3600t² + (100+50t)²}

at t = 1.5,

ds/dt = {10,800 + 17,500} / 2√{8,100 + 30,625}
ds/dt = 28,300 / 2√{38,725}
ds/dt = 71.9 km/hr
=============

I said earlier that dx/dy was independant of time. Here's why.

y = 100 + 50t
x = 60t

dy/dt = 50
dx/dt = 60

dx/dy = (dx/dt)/(dy/dt) = 60/50 = 1.2 = constant, hence independant of time.


Try drawing a line segment from C to E of length $\sqrt{5}$, now find area of two triangles. You have lots of right angles to work with.

[hide="solution"]OK. Draw E, BE, CE and an estimation of F and CF. Call CF z, BF y and FE x. We know that CE and BE are $\sqrt5$ and we use a little algebra to find that
$x^2+z^2=\sqrt5,y^2+z^2=5$
We subtract the first equation from the second
$y^2-x^2=4$
I'm to tired with this problem to go on. Sorry. Try to solve it from there.[/hide]

[hide="dont look at the solution yet, do the problem yourself"]Draw everything. Draw a segment $CE$ which is length $\sqrt5$
$BE$ also is $\sqrt5$ but $CF$ splits it into two parts. Label $BF$ as $x$ and $EF$ as $\sqrt{5}-x$.

Use pythagorean theorem on triangle $EFC$ and we get $CF^2 = 2\sqrt{5} - x^2$

For triangle $BCF$, we obtain $2^2 = x^2 + CF^2$

Plug stuff in and we get $x = \frac{2}{\sqrt{5}}$

Now you have $x$ so you can find the length of all segments. Plug stuff back into $CF$ and $EF$ to obtain $\frac{6}{5}$ but we need to add this to the area of $EDB$ which is $1+\frac{6}{5} = \boxed{\frac{11}{5}}$[/hide]

I think I got it, but it is a bit too long.
[hide]
I drew $\overline{CE}$.

The area of $\triangle CDE$ is:
$\frac{1\cdot 2}{2}=1$

Now, I find the area of $\triangle CEF$:
The area of $\triangle BEC$ is 2, so the height $\overline{CF}$ is:
$\frac{\sqrt{5}\cdot h}{2}=2$
$h=\frac{4\sqrt{5}}{5}$
Using the pythagorean theorem, I find $\overline{EF}$ to be:
$\frac{3\sqrt5}{5}$

So the area of $\triangle BEC$ is:
$\frac{6}{5}$

The Area of the quadrilateral $CDEF$ is then:
$\frac{6}{5}+1=\frac{11}{5}$
 So [b]C[/b][/hide]
Is that it?

[quote="jeez123"]Ok, next question:

Mientka Publishing Company prices its best seller Where's Walter as follows:

$12n$, if $1\le n \le 24,$    |
$11n$, if $25 \le n \le 48$  | }= C(n)
$10n$, if $49 \le n$,          |

where n is the number of books ordered, and C(n) is the cost indollars of n books. Notice that 25 books cost less than 24 books. For how many values of n is it cheaper to buy more than n books than to buy exactly n books?

the |||} is just the bracket

I'm not sure how to start.[/quote]

Well you know that when the price changes, some book amounts are less even when you buy more books. I would just try numbers near these amounts and when they become less you can just stop.
[hide]
23 and 24 books cost less than 25 books. 45, 46, 47, and 48 books are less than 49 books. The answer should be 6.. [/hide]

[quote="jeez123"]The whole question still confuses me, I have trouble sorting it out!

For how many vaues of n is it cheaper to buy more than n books than to buy exactly n books?
*a bit frustrated.[/quote]

Sorry.
It's basically asking how many book amounts cost more where there is another book amount where you get more books but you pay less total.

[hide="hint"]Try values close above the boundry and see if they are less than the highest one for the previous price[/hide]

[hide="dont look at the solution"]$12*24 = 288$
$25*11 = 275$ which is more than $11$ lower than $288$ so we can also include $26$, which is also lower than $288$

$11*48 = 528$
$528/10 = 52.8$ (approximately, but who cares)
So that means we can buy 49, 50, 51 and 52 books.

So thats 25, 26 and the previous 4 i just mentioned.
The answer is $\boxed{6}$
[/hide]

I think I might got it.

[hide]25 books cost less than 24 books.
49 books, 50 books, 51 books and 52 books cost less than 48 books

So that's [b]5[/b] values?
[/hide]

Did I missed any?

[quote="jeez123"]26 books also cost less!
[b]6[/b]
"Keeping track" lesson
Thanks!

-------------------------

How many two-digit positive integers N have the property that the sum of N and the number obtained by reversing the order of the digits of N is a perfect square?

So that's from 11-99, yeah that's where I am.[/quote]

Call the number 10a+b. Then the reversed digits is 10b+a, and the sum is 11(a+b). [hide]so the sum is a multiple of 11. This makes the perfect square 121. So 11(a+b)=121, a+b=11. Then...max number is 92, then 83, then...29 , so there are $\boxed{8}$ numbers. [/hide]

So we are looking at the sum,
$\sqrt{11(a+b)} \rightarrow \sqrt{11(11^2)}$
$11^2=121$
$a+b=121$
a can be from 99 to 22 and b can just follow.

$11^3$ is too much and $11^1$ is too little

I'm stuck here!

The number itself is (10a+b). The number with the reversed digits is (10b+a). a and b are just the digits of the two digit numbers. So if the sum of the numbers add up to 121, and 11(a+b)=121, then a+b=11...and then the digits are limited...

[quote="jeez123"]So we are looking at the sum,
$\sqrt{11(a+b)}\rightarrow \sqrt{11(11^{2})}$
[/quote]

That step doesn't make sense; I think you meant 11 instead of 121.

[quote="jeez123"]I need hints to these problems:

First of all:

$ABCD$ is a 2x2 square, $E$ is the midpoint of $\overline{AD}$, and $F$ is on $\overline{BE}$. If $\overline{CF}$ is perpendicular to $\overline{BE}$, then the area of quadrilateral $CDEF$ is....

A. $2$
B. $3-\frac{\sqrt{3}}{2}$
C. $\frac{11}{5}$
D. $\sqrt{5}$
E. $\frac{9}{4}$[/quote]
[hide]See the attached diagram for this question. (xfig is amazing!!) Since $AB$ is 2 and $AE = 1$, $BE = \sqrt{5}$. Note that $\triangle ABE$ and $\triangle BCF$ are similar. This means, \begin{eqnarray*}\frac{CF}{2}& = & \frac{2}{\sqrt{5}}\\ CF & = & \frac{4}{\sqrt{5}}\end{eqnarray*} So $BF = \frac{2}{\sqrt{5}}$. This tells us the area of $\triangle BCF = \frac{1}{2}\times \frac{2}{\sqrt{5}}\times \frac{4}{\sqrt{5}}= \frac{4}{5}$. The area of the trapezoid $BCDE = \frac{1}{2}\times (1+2) \times 2 = 3$.

So the area of the quadrilateral $CDEF = BCDE-BCF = 3-\frac{4}{5}= \frac{11}{5}$. So the answer is $\boxed{C}$.[/hide]


[quote="math92"]Anna wants to cover the outside of a rectangulare box bith paper.  The box has a square base with area of 16 square inches.  The volume of the box is 80 cubic inches.  How many square inches.  How many square inches of paper will Anna need to completly cover the box, with no paper left over?[/quote]

[hide]The dimensions of the box are $4x4xn$
The volume is 80 so n = 5
The dimensions of the box are now 4x4x5
The surface area of the box is 2(4x4 + 4x5 + 4x5)
The surface area is 112
$112 in^2$[/hide]


At the school of AAST, seniors have a tendency to dump menial tasks on underclassmen. Of course, this is AAST, so by "menial tasks", we mean "data input".

Ben has lots of data input due in eight hours - enough that it would take him seventeen hours to do himself. Although he is three times as efficient at data input than a freshman, he decides to relegate all of his work to eight unfortunate freshmen. Halfway into the job, he realizes that the freshmen will be able to complete the task with time to spare, so he allows $ x$ of them to leave. Assuming that all freshmen work at the same rate, what is the largest possible value of $ x$?

[hide="Where do you come up with these problems?"]

Okay, to get started lets set up an equation:

$ \frac{8\cdot(8-x)}{3\cdot17}= 1$

Hopefully, you can use logic to see where I got my equation.
Next, lets solve:

$ \frac{8\cdot(8-x)}{3\cdot17}= 1$
$ \frac{8\cdot(8-x)}= 3\cdot17$
$ 64-8x = 51$
$ 64-51 = 8x$
$ 13 = 8x$
$ x = 1.625$

Therefore, we can come to conclude that we can only take off 1 person from the group of freshman. (we have to round down).

[b][u]Answer to the "Manual Labor" Problem:[/u][/b] $ \boxed{1}$

[/hide]

I'm just wondering, did i get the right answer :?:

Um, for some reason I got [hide] 5, which seems like way to much, and I can't show how since the way I did it was rather long and messy, and probably wrong anyways, but still.  :?: [/hide]

No offense, but:

If we take away $ 5$ people, we are left with $ 3$ people...
As we can see from the problem, the freshman do $ \frac{1}{51}$ of the data in an hour.
Making it a total of $ \frac{3}{51}$ an hour with $ 3$ freshman...
Given only eight hours, they can only complete $ \frac{3\cdot8}{51}=\frac{24}{51}$...
As you can see, they will [b]not[/b] complete the data input on time.

 :(

[hide] oops, I didn't interpret my result carefully. OK, so I now realize that my solution for "x" is actually the number of people needed to remain behind, and 5.__ rounds up to 6, so you can dismiss 8-6 = 2 people ? [/hide]

Well... actually, we have to round down because if we round up we won't have enough time. You can try ($ 8-2=6$) 6 people if you want on your calculator, but you will get the answer that will tell you that the freshmen ran out of time.

[hide="Answer"]
3.

In four hours, eight freshmen complete $ \frac{32}{51}$ of the work, meaning there is $ \frac{19}{51}$ of the work left to do. This requires at least 4.75 freshmen to complete in four hours. $ 8-5=3$.
[/hide]

Oh!

I wasn't reading the problem right :blush: "[u][b]Halfway into the job[/b][/u]" I didn't seen that.
That was a great problem if you ask me.

When I first read it I thought "halfway into the job" meant when it was 1/2 done not when there was half the time left... :|


Two factors have a product of $3 \cdot 9^{2x} - 6 \cdot 27^x$.  One of the factors is $3^x - 2$.  What is the other factor?

[hide]
$\displaystyle 3 \cdot 9^{2x} - 6 \cdot 27^x$
$\displaystyle = 3^{4x+1} - 2 \cdot 3^{3x+1}$
$\displaystyle = 3^{3x+1}(3^{x} - 2)$
the other root is $3^{3x+1}$
[/hide]

I thought I'd bump this back so people can see it again.
[hide]3^(3x+1)[/hide]

Do $\LaTeX$ tests in the latex test forum.
http://www.artofproblemsolving.com/Forum/index.php?f=123
Not here.


[quote="GameBot"]Angie decided to use her class as a sample to predict the total number of students in her school wearing red on Valentine's Day. She counted 11 students wearing red in her class of 24 students. Using this ratio, how many of the 480 students in her school
will she estimate are wearing red?[/quote]

$ \frac{11}{24}\equal{}\frac{220}{480}$. 220.


inequalities
vector
inequalities proposed

Prove that $ k\equal{}\frac1{\sqrt{8}}$ is best possible such that the following holds for all $ a_i,b_i,c_i \ge0$ ($ i\equal{}1,2,3$):
\[ \boxed{\sqrt{3}\sum_{i\equal{}1}^3 a_ib_ic_i\plus{}\sqrt{\sum a_i^2\sum b_i^2\sum c_i^2}\ge k\sum a_i\sum b_i\sum c_i }\]
:D

Wow... nice inequality.
It took me some time to prove and i must say what i got was pretty ugly. I'm sure that it must have some nice geometric interpretation in $ \mathbb{R}^3$.

Here is what i did:
Let $ \alpha \equal{} (a_1,a_2,a_3)$ and $ \beta$ and $ \gamma$ be defined similarly. More over let's fix $ \alpha$ and $ \beta$ and denote $ m \equal{} (a_1b_1,a_2b_2,a_3b_3)$
the inequality can be rewritten as
$ \gamma \cdot \left(\sqrt {3}m \minus{} k(\alpha\cdot e)(\beta\cdot e) e\right) \plus{} \parallel{}\alpha\parallel{}\cdot\parallel{}\beta\parallel{}\cdot\parallel{}\gamma\parallel{}\geq 0$
($ e \equal{} (1,1,1)$)
It is clear that we can normalize to $ \parallel{}\gamma\parallel{} \equal{} 1$ and we see that when minimizing it $ \gamma$ must make the biggest possible angle with some fixed vector.
As $ \gamma$ is confined in the first quadrant it means that it has to be at the boundary. (the ugly vector in brackets can not have all negative coordinates. :wink: )
Thus we conclude that one of the $ a_i$'s one of the $ b_i$'s and one of the $ c_i$'s have to be $ 0$(by symmetry).
That $ k \equal{} \frac {1}{\sqrt {8}}$ works is obvious from AM-Gm.
To show that it can not be improved consider $ \alpha \equal{} (0,1,1)$ $ \beta \equal{} (1,0,1)$ and $ \gamma \equal{} (1,1,0)$
I wonder if you have a nice geometric argument. I'm sure it must involve $ \sqrt {3} \equal{} \parallel{}e\parallel{}$


Two circles touch internally at M.

A line touches the inner circle at P and cuts the outer circle at Q and R.

Prove that$ \angle QMP$ =$ \angle RMP$

I think it's well-known. And here is a simple proof for this nice, easy and useful problem.

Call the inner circle as $ (I)$ and the outer circle as $ (O)$. Let $ S$ be the intersection of $ MP$ with the circle $ (O)$. We have $ O$, $ I$, $ M$ are collinear, therefore $ \angle PMI\equal{}\angle SMO$. It follows that $ \angle MPI\equal{}\angle MSO$, i.e. $ IP\parallel OS$, which implies $ OS\perp QR$. Hence $ S$ is the middle point of arc $ QR$ of the circle $ (O)$, so $ \angle QMP\equal{}\angle RMP$. Done!


Find how many natural numbers $m,n\le 2001$ satisfy: $\sqrt{m+\sqrt{n}}$ and $\sqrt{n+\sqrt{m}}$ are natural numbers.

[quote="Rust"]Both expressions: $\sqrt{m+\sqrt{n}}$ and $\sqrt{n+\sqrt{m}}$ can not be naturals for naturals m,n.[/quote]

Actually either of them can be naturals, but not both at the same time.
Assume both are naturals for some $m,n$
$\sqrt{m+\sqrt{n}}$ is an integer
$\iff$ $m+\sqrt{n}$ is a perfect square
$\Rightarrow$ $n$ is a perfect square
Similarly for the other expression.
Let $m=a^{2},n=b^{2}$ where $a,b>0$
Then $a^{2}+b=x^{2}$ and $b^{2}+a=y^{2}$ for some integers $x,y>0$
$b=(x-a)(x+a)$
$a=(y-b)(y+b)$
$a<x+a\leq b<y+b\leq a$ $\Rightarrow\Leftarrow$
Therefore there are no solutions

[quote="Absolut"]There was a mistake. The right statement os m,n integers.
How to find them[/quote]

You find them by showing that no solutions exist for $m,n>0$ then the rest is trivial.


Do you think it is possible to understand your statement?  ;)

Let $S$ denote the LHS of the desired equality.

For all $x$, we have $\dsp (x+1)^{n+1} = \sum_{k=0}^{n+1} C_{n+1}^k x^k$.
Thus, differentiating, we obtain that :
$\dsp (n+1)\cdot (x+1)^n = \sum_{k=0}^{n+1} kC_{n+1}^k x^{k-1}$
Thus, for $x=1$ it leads to
$\dsp (n+1) \cdot 2^n = \sum_{k=0}^{n+1} kC_{n+1}^k = \sum_{k=0}^{n} (k+1)C_{n+1}^{k+1} = S + \sum_{k=0}^{n} C_{n+1}^{k+1} = S + 2^{n+1} - 1$.
The result follows.

Pierre.

Where is combinatoric solution, Pierre? ;)

here is a 'semi-combinatorial' solution:
  consider a group of n+1 people of different ages.we wish to form a non-empty committee and a leader for it without choosing the youngest member of the committee as its leader.clearly the summation in question is the number of ways of doing so.
 alternately choose a member of teh committee(in n+1 ways) and declare him the leader of the committee(for 2^n such committees).of course we form certain 'illegal' committees this way so we get rid of them-they are the ones inwhich our candidate for leader is the youngest;there are clearly (2^n+2^(n-1)+2^(n-2)+...+2+1) such bad committees,and this is simply 2^(n+1)-1.so the reqd number is
  (n+1)2^n-2^(n+1)+1 which is the desired number.

Aaargh...you posted it just before me.....

Pierre.

of course there is no necessity for the summation in what i wrote earlier(i just realized this after sending the post!).
 every single non-empty set of members counts one bad set with the youngest member as the leader(see the above post),so teh number of bad committees is indeed 2^(n+1)-1.
this makes this a combinatorial solution.


问一个问题，在哪个网站能找到《美国数学月刊》里的文章？能?能给个链接？我?胜感激。

有一个问题我一直想?通，我?常在国内的一些刊物或书?上看到《美国数学月刊》中的问题?解或其他文章，??知?这些作者怎么获得这些??料的。而且我还?现在国内?能办?邮购或订阅，请问这些作者们到底怎么?到《美国数学月刊》的？

我曾?在北京图书馆看过原版《美国数学月刊》?志，如果你在北京，?以到那儿看一看。

?外，下?的网?有近年的《美国数学月刊》问题?解?的问题

http://www.math.nwu.edu/~mlerma/problem_solving/

国内曾?出版过一本书，??字?《国际最佳数学?解问题分?》，汇集了几百?从《美国数学月刊》中精选出的题目，北京图书馆有此书，?外，超星电?图书馆有电?版书供下载。

[quote="zhaoli"]国内的一些大学图书馆?能也有《美国数学月刊》?志。

“中国数学在线?网站有2003.1-2004.5期《美国数学月刊》的问题供下载

http://www.chinamaths.org/Soft/jstd/gzjs/200406/287.html[/quote]

但是这个网站已?转到新地?上去了，注册?到新地?上去注册，但是我原?没有在旧论?上注册过，而现在旧论?以?能注册了，所以我没有??下载也没办法注册。
所以，如果您曾?下载过的?，能?能把它??到我的邮箱: alexanderlenin@126.com   我一定会??分感谢的. :)

Thanks, I have already recieved 2003-1 and 2003-2!  :)  :)  :D


On set $G=(2004,\infty)$ we define the composition law "$\circ$" like this: $x\circ y=2004+(x-2004)^{\ln{(y-2004)}}$ for all $x,y\in G$. If $B=\{x\in G | x\circ x\circ x =2004+e^{2004^{3}}\}$ where $e$ is the natural logarithm base compute $S=\sum_{x\in B}{x}$.

p.s. [i]I don't know if this is the right place to post this type of problems.[/i]

Yeah by substituting $\ln(x-2004)$ with $k$ it's done! I didn't saw that! Well done! :P


I'm doing a self-directed lab report about the kinetics of the oxidation of ascorbic acid in orange juice. I am wondering if anyone knows an accepted value for the activation energy of ascoribc acid which can serve as a secondary source for me to compare the value that I calculated from my raw data to. I tried searching on the internet but had no luck finding it.


Let a be a real positive number. Determine the maximum of the expression, if 0<=x<=z<=y<=a:

           x^3 + y^3 +z^3 -x^2*y -y^2*z -z^2*x - x*y*z.

$x^{3}+y^{3}+z^{3}-x^{2}\cdot y-y^{2}\cdot z-z^{2}\cdot x-x \cdot y \cdot z. \\ a \in \textbf{R}, a > 0; x, y, z \in [0; a].$
 Ok, now I have re-written the problem so that it could be intelligible  :wink:


Solve in integers the equation $(2x^2-5x+2)3^x=1-4x^2$

 :)

The equation factorises into $(1-2x)((2-x)3^x-(1+2x))=0$
Therefore, $(1-2x)=0, x=\frac{1}{2}$, or
$(2-x)3^x=2x+1$
We see that $x=1$ is a solution
For $x\geq2,(2-x)3^x\leq0$, but $2x+1>0$
For $x\leq-\frac{1}{2},2x+1\leq0$, but $(2-x)3^x>0$, and 0 is not solution

Hence, the only solution in integer is $x=1$

[color=red][modedit: corrected missing )][/color]


geometry articles probability AoPSwiki Support ARML MATHCOUNTS

[b]Q: What is it?[/b]

A: That's not quite clear.  It might be the name of a plot to take over the universe that was edited out of a Douglas Adams novel.  It might be a variant species of banana slug.  But probably not.  The truth is that nobody knows quite what will become of the AoPS wiki, but the goal is to create an online organized resource for students to which students (and teachers and anyone interested) can contribute.

[b]Q: When will the AoPS Wiki be available?[/b]

A: There's no telling when we'll get the next clearer glimpse of the Yeti or fuzzy photograph of Nessy, but we believe that the AoPS Wiki will come first!  Hopefully this week!

[b]Q: What can I do to improve my AMC score?[/b]

A: Eh, you're in the wrong forum.  However, if the Wiki does what it should do, there should be resources available to help students at many levels hone not only their mathematical abilities, but also their abilities in the sciences, computing, and scaling tall buildings with nothing but chewing gum and a ball or yarn (though we offer no guarantees of success).

[b]Q: What kind of information will the AoPS Wiki contain?[/b]

A: That depends on what we and you put in it.  We've already begun adding much of the information in our resources area such as the pages about math scholarships and competition lists.  Such information will be easier to update (since I won't have to update it all) and should be better organized so that students can find what they are looking for.  However, we expect much of the content of the Wikis to be articles about topics that students are interested in: modular arithmetic, DeMoivre's Theorem, Rolle's Theorem, Bayesian probability, etc.

Though there are broad resources on the net such as Wikipedia and Mathworld, much of the content is not particularly useful for many students.  Even the AoPS admins will admit to having a tough time reading many Mathworld articles!  By letting you, the users, contribute much of the material you feel was helpful to you, we hope that the Wiki fills that niche of Wiki math/sci/computers/etc. educational material online.

[b]Q: Who's idea was hiring Dennis Miller to announce football anyhow?[/b]

A: I don't know!  What I do know is that we're going to do our best to guide the development of the AoPS Wiki without Dennis Miller.  There will be conflicts and there will be locked entries that only AoPS staff can edit.  However, we'll do our best to resolve such problems and let students contribute as much of the content as seems reasonable.

So, the AoPS Wiki is to be one ginormous experiment.  Please help us figure out exactly where that experiment is going to go.  In fact, feel free to go ahead and theorize about what's going to happen below.  We certainly don't know and we'd feel relieved to be informed.

Why an AoPS wiki, and don't you think it'll get a lot of spam? What penalties will there be for doing ridiculous things, like moving "how to jump off buildings" to the category for olympiad stuff? (I am assuming it'll be categorized like the gentoo wiki @ gentoo-wiki.com , which is the only wiki I've had much experience with.) Finally, will it use AoPS tags or wiki tags?

Wikipedia has a LOT of staff members.

Also, when I tried to add my birthday on one of the day links, it got deleted very quickly (actually like a day).

Wait, is this available for all members?

Spammers will be dealt with more strictly than forum spammers, whom we have banned or deleted summarily at times.  As such, the forum seems to be under control to a reasonable extent.

Also, we have joml88 helping to manage the AoPSWiki, which is an advantage that we have over any other wiki that I know about!

[quote="robinhe"]Wikipedia has a LOT of staff members.

Wait, is this available for all members?[/quote]

Wikipedia has a large staff and a much broader audience/mission than ours.  Registration for the AoPSWiki will be the same as for the forums, which should help control the AoPSWiki to a great degree.

[quote="joml88"]wikipedia seems to do OK with respect to spam and people posting inappropriate stuff.  Why do you think an AoPS wiki would be more susceptible to these things and unable to protect against them?[/quote]

Joe, An AoPS wiki would be much more susceptible to these acts of "vandalism."  Wikipedia relies on a community of millions of "Recent Changes Patrollers" to watch an [url=http://en.wikipedia.org/wiki/Special:Recentchanges]auto-updating list of articles[/url] that were just updated.  These Patrollers then check to see if new information is actually true, or if edits make sense.  For example, edit any wikipedia article, no matter how obscure, and slightly edit a trivial fact.  It will be caught within the hour, usually within minutes.  Wikipedia's "Recent Changes Patrollers" are self-appointed, and there are a LOT of them.  That is how Wikipedia manages to stay accurate.  An AoPS wiki CAN be maintained accurately, but it would require a LOT more time and effort per individual, due to the fact that AoPS would have significantly less Patrollers than Wikipedia.  So, in conclusion, an AoPS wiki would not be impossible to moderate, but it would require more effort by admins, and it would be more susceptible to vandalism.  A nice way to stop vandalism is a temporary ban for flagrant or malicious vandalism.  Anyone who doesn't understand the logistics of a wiki, and how one manages to stay accurate, should visit Wikipedia and browse through their tutorials.  [url=http://en.wikipedia.org/wiki/Wikipedia:Community_Portal#Departments]This[/url] is a nice starting point.

Edit:  Between the time I responed, and the time I clicked submit, there were other responses.  Some of this is redundant, but I won't delete it...

By the way, I would be willing to help with the AoPS wiki, as I know a lot about how Wikipedia works, and I am a Wikipedia editor as well.

[url=http://researchweb.watson.ibm.com/history/results.htm]Here[/url] is an interesting study, and it may be useful for trend analysis.

Also, a random point that I thought was interesting:  Wikipedia already supports latex.

One thing that Wikipedia deals with that AoPSWiki won't have much of is politically sensitive content.  There will be much less motivation for auto-botting the AoPSWiki.  We'll also deal with these problems as best we can as they arise.

[quote="SplashD"]Wouldn't AoPS Wiki sort of be like mathworld?[/quote]

Hopefully it will be much more accessible to the average student than mathworld is.

[quote="MCrawford"]Though there are broad resources on the net such as Wikipedia and Mathworld, much of the content is not particularly useful for many students. Even the AoPS admins will admit to having a tough time reading many Mathworld articles! By letting you, the users, contribute much of the material you feel was helpful to you, we hope that the Wiki fills that niche of Wiki math/sci/computers/etc. educational material online.[/quote]

[quote="SplashD"]Wouldn't AoPS Wiki sort of be like mathworld?[/quote]

In addition to containing explanations of math concepts, an AoPSWiki can contain lessons, example problems, and links to AoPS posts and other resources.  The AoPSWiki can also serve science and computer students as well, whereas MathWorld serves mostly professional math users and extremely high level students for the most part.  Also, the AoPSWiki will contain organized information about academic competitions in every state and many nations.  Users could compile historical information about those competitions as well (like the top teams and individuals at ARML or MATHCOUNTS or IMO medalists).

[quote="robinhe"]I could help too I guess.

If you need any help... :(  :D[/quote]

We will need mountains of help.  AoPSWiki is a grand experiment of sorts, so we can't even tell yet exactly where we'll need help.  But we'll certainly need lots of help generating content!

[quote="mysmartmouth"]

By the way, I would be willing to help with the AoPS wiki, as I know a lot about how Wikipedia works, and I am a Wikipedia editor as well.[/quote]

Soni's theorem, anyone?

[quote="rrusczyk"][quote="Peter VDD"][quote="joml88"]As just one example, on the boards, to make something bold you put the [ b] [/ b] tag's around it.  For the wiki, you put apostrophes around it.  Taking a post from the message boards and putting it into an AoPSWiki article would take a LOT of work.[/quote]Indeed, and it is especially boring since apostrophes appear in those posts so would have to be removed too.

I'm serious, "supporting" it is not enough, it should really replace the standard wiki tags. And not to lose months of work, this should happen before the release. It's of course not me deciding, but from the experience I've had with comparable projects, I'd strongly suggest to do this [b]before[/b] the launch of the wiki, even if this delays it a week or more...[/quote]

We don't want to replace all the mediawiki stuff - there's a ton of things mediawiki can do that phpbb simply cannot do.   Probably what we're talking about here is having the Mediawiki be able to interpret phpbb codes as well.  So, not a replacement of, but rather accepting both markups.  (possibly by having some global phpbb tag within MW that says 'parse this as phpbb')[/quote]Actually we will. In fact there's a ton of things that phpbb [b]can do[/b] and mediawiki can't.

I'm really busy busy busy this time of year (as Peter said exams time :D) and the upcoming IMO 2006 but the initial plan was to use my own software to power this up. Since we are rushing it, we'll do first media wiki and then port the stuff over the summer, and then there will be no problem with connecting posts in the forum with the wiki or managing the spamming.

The best bet against spam is that guests will not be able to post. By doing this you eliminate from 50% up to 90% of spam.

In the end you should be able to write on media wiki just like you write posts on the forum (which is, how you will see, what has been already done with the blogs ;) ).

The resources pages some users mentioned before caused some confusion. On the MathLinks skin the resources pages is the pages with Contests problems on AoPS. The resources pages on AoPS skin is rather a collection of links and articles :)

The resources section will stay as is for the next month, but will be upgraded. Applications for being a rmanager can be sent to me (so whoever wants to help with this, and has at least a few months being on the board pm / email me about it).

[quote="Valentin Vornicu"]Actually we will. In fact there's a ton of things that phpbb [b]can do[/b] and mediawiki can't.[/quote]Exactly my thoughts. :)

Sure, there are pros and cons to both systems.  I think in the end, we'll probably have a hybrid approach, but it's definitely premature to say exactly what approach we'll use, or when we'll put it in place.  We have a lot of big projects on the horizon, and have an operational wiki now with MW versus waiting 1-2 years and having one with phpbb is an easy call.  (Yeah, I know some will say it won't take 1-2 years to build it in phpbb, but we have a lot of other higher priority tasks, and everything always takes longer than we think it will.)

[quote="rrusczyk"](Yeah, I know some will say it won't take 1-2 years to build it in phpbb, but we have a lot of other higher priority tasks, and everything always takes longer than we think it will.)[/quote]

To do it right, 1-2 years is probably a perfect estimate.  It's pretty ridiculous how much most web apps take to build and how little credit they are given.

Out of curiosity, who does all the programming around here?

There is nothing to write, since the thing is completely present in phpbb. ;)

Well, essentially, if all you are talking about is allowing all users to edit all posts.  But in doing that, we might as well call the forums as they are now a wiki.
(Or are you saying that there is a phpbb wiki mod?)

Plus the search function on phpbb just sucks.

I dunno, I just think the presentation and overall design of wikipedia is much better.  It's not a great leap to install Tex on a server with LatexRender and then add bbcode support.  I'm sure that making the phpbb login system work with the wiki software isn't a great leap either;  I've done some modding with phpbb and found it pretty nice to work with.

Let me worry about the programming, and you guys worry about the content :)

[quote="SnowStorm"](Or are you saying that there is a phpbb wiki mod?)[/quote]I am in fact writing one, so I do know something about their interference. ;)

The only nontrivial thing (at least for me it's nontrivial) is to allow http://www.site.com/wiki/ThisIsAWord redirect to the phpbb topic on ThisIsAWord, that part I'm currently trying to figure out how they did it in the MediaWiki software.

In fact, you're very right that forums and wikis are - from the structural-mathematical point of view - exactly the same structure, just in a slightly different skin.

[quote="Valentin Vornicu"]and you guys worry about the content :)[/quote]isn't the point of wikis exactly that people do not have to talk about the content, they can just add it? ;)

Ah,
In pondering your problem, I did a bit of googling. (I have always wondered how the php.net site and wiki achieve it).

I hope this helps: (pretty clever method)
http://nslog.com/archives/2003/02/26/404_search_function_code.php

Hmm... interesting link, I didn't know 404 pages were customizable to that degree. Thanks, I'll certainly check it out (but probably after my finals). :)

[quote="mysmartmouth"][quote="joml88"]
By the way, I would be willing to help with the AoPS wiki, as I know a lot about how Wikipedia works, and I am a Wikipedia editor as well.[/quote][/quote]

A disrupting one at that, I see you deleted 3 AfD votes, added a keep to someone elses and voted twice in an deletion discussion over a formula you decided to name after yourself.

First, please fix your quote tag as I did not say that.

Second, I don't think this is the right time or place for flaming someone else.

[quote="Peter VDD"]

[quote="Valentin Vornicu"]and you guys worry about the content :)[/quote]isn't the point of wikis exactly that people do not have to talk about the content, they can just add it? ;)[/quote]

Not at all - look at the massive number of collaborative projects going on in wikipedia.  Take a look at the Discussion tabs.  There's a great deal of talking about the content.  If people just added it, it would be, well, a lot more like the forum - a chaotic storm of information that is structured to be temporary.

[quote="PenguinIntegral"]So do we have an updated estimate on when the wiki will be up?[/quote]

Not long now, little smurflings...


Let $Ax = b$ be any consistent system of linear equations, and let $x_{0}$ be any of its solutions. Prove that for some generalized inverse $G$ of $A$, $x_{0}= Gb$.

Ok ok fred insists I should explicitly mention that $b\neq0$, instead of relying on your common sense.


Which of the following numbers is a perfect square?

$ \textbf{(A)}\ 98!\cdot 99!\qquad \textbf{(B)}\ 98!\cdot 100!\qquad \textbf{(C)}\ 99!\cdot 100!\qquad \textbf{(D)}\ 99!\cdot 101!\qquad \textbf{(E)}\ 100!\cdot 101!$

$ 98! \times 99! \equal{} (98!)^2 \times 99$

$ 98! \times 100! \equal{} (98!)^2 \times 99 \times 10^2$

$ 99! \times 100! \minus{} (99!)^2 \times 10^2$

Obviously, $ \boxed{\textbf{C}}$ is a perfect square.


Suppose you have a circle with any regular polygon innerally tangent to it. I tried to find a formula for it. Which happens to be \frac{s * sin(\frac{180}{s}) * cos(\frac{180}{s})}{\pi}. You can see it in my blog. [url=http://theblogofdoom.wordpress.com/2006/08/06/some-math/#more-20]linky[/url]

[quote="dr.silly"]poopy how do I get the math syntax work?[/quote]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5261[/url].


[quote="pavel kozlov"]Does anybody know about math. olympiad in Cyprus 7-12 March 2007?[/quote]

I do, I do ! :)

can anyone post here problems of Cyprus 2007 ? and How can take a plae in this competition? I mean registration? need it some money?

at http://www.seemous.com you can find the problems and the competition results.
BTW cangratulations to Pavel Kozlov for the gold medal. :first:

30 points   is the first , second ,third prize?

I can't open some page, please post here problem of Cyprus 2007

Hmmmm... maybe someone who participated can tell us what were the cutoffs for gold(I bet 40 :P ), silver and bronze medals?

I don't know exactly but:
GOLD - $29\pm1$ points;
SILVER - $>20$ points;
BRONZE - $5(!)$ points.
It was a lot of medals.

[quote="pavel kozlov"]As for Russian team:
1. Alexander Magazinov - 39
2. Vasily Astakhov - 38
3. Pavel Kozlov - 35(as on IMO 2005 :) )
5. Andrey Gavrilyuk - 32
10. Nikita Kalinin - 27[/quote]
Привет. Ты можеш скинуть задачи которые были предложены на этом конкурсе?

am I wrong or I think I saw the third problem around in the Forum... :)
Congrats to the Russians they really had the strongest team!

Does anybody know if such olympiad will be the next year? Where, when ...?

Next year such competition will be in Athens, Greece 5-10 March 2008

2nd South Eastern Europian Mathematical Olympiad
for (first and second year) University Students (SEEMOUS)
with international participation
http://www.hms.gr/seemous08/


$ a_1>a_2>a_3>a_4>a_5>a_6$ and $ p\equal{}a_1\plus{}a_2\plus{}a_3\plus{}a_4\plus{}a_5\plus{}a_6$, $ q\equal{}a_1a_3\plus{}a_3a_5\plus{}a_5a_1\plus{}a_2a_4\plus{}a_4a_6\plus{}a_6a_2$, $ r\equal{}a_1a_3a_5\plus{}a_2a_4a_6$. Are all the roots of the cubic equation $ 2x^3\minus{}px^2\plus{}qx\minus{}r\equal{}0$ real??

This is the cubic equation

$ f(x) \equal{} (x \minus{} a_1)(x \minus{} a_3)(x \minus{} a_5) \plus{} (x \minus{} a_2)(x \minus{} a_4)(x \minus{} a_6) \equal{} 0$

Now, some sign analysis.

- When $ x > a_1$ we have $ f(x)$ positive.
- When $ a_2 > x > a_3$ we have $ f(x)$ negative.
- When $ a_4 > x > a_5$ we have $ f(x)$ positive.
- When $ a_6 > x$ we have $ f(x)$ negative.

Hence $ f(x)$ has three distinct real roots.