linear algebra
matrix
combinatorics unsolved
combinatorics

Let $a_{1},a_{2}, \ldots, a_{2n+1} \in \mathbb R$ and $n \in \mathbb N^{*}$. For every $1 \leq i \leq 2n+1$, assume that there exist two sets $G_{1}$ and $G_{2}$ so that $\left\{1,2, \ldots, i-1, i+1, \ldots, 2n+1\right\} = G_{1} \cup G_{2}$, $G_{1} \cap G_{2} = \varnothing$ and $|G_{1}| = |G_{2}|$ and $\displaystyle \sum_{j \in G_{1}} a_{j} = \sum_{j \in G_{2}} a_{j}$. Prove that $a_{1} = a_{2} = a_{3} = \ldots = a_{2n+1}$.

It is easy to see that all $a_i$ can simultaneously multiplied or added by a constant.
Without loss of generality we can assume that all $a_1,\dots,a_{2n+1}$ are integer such that $a_1=0$.

Let $S=\sum_{j=1}^{2n+1} a_j$.
Then for every $i=1,\dots,2n+1$ there exist two sets $G_1$ and $G_2$ such that
\[ S - a_i = \sum_{j=1\atop j\ne i}^{2n+1} a_j = \sum_{j\in G_1\cup G_2} a_j = 2\sum_{j\in G_1} a_j \]
meanning that $a_i$ has the same oddness as $S$.

Since $a_1=0$, $S$ as well as any other $a_i$ must be even. Dividing all $a_i$ by 2 we still will have numbers that satisfy the condition. Note that such transformation decreases the sum $|a_1|+\dots+|a_{2n+1}|$. So after a finite number of such dividings we will end up with all zeros. But all applied transformations are invertible implying that all initial numbers $a_1,\dots,a_{2n+1}$ are equal.

Another approach (they may be equivalent; I didn't read what maxal wrote):

Consider the system of $2n+1$ equations with $2n+1$ unknowns that you get. If we prove that its matrix has rank $2n$, we're done. If its matrix had rank $<2n$, then so would its matrix reduced modulo $2$, but (after rearranging the equations, if necessary) this is simply the matrix having $0$ on the main diagonal and $1$ everywhere else, and its eigenvalues (over the field with two elements) are $2n\ 1$'s and one $0$. It's clear by a simple parity argument that the only nonempty subset of its rows which sums up to $0$ is the entire set, i.e. the rank of the matrix is indeed $2n$, and not less. This is exactly what we wanted to prove.

[quote="georg"][quote="maxal"]
Without loss of generality we can assume that all $a_1,\dots,a_{2n+1}$ are integer such that $a_1=0$.

[/quote]

why?  :blush:[/quote]
First multiply them all by a large number to make all non-zero pairwise distances greater than 1. And then apply Kronecker's approximation theorem for $t=2n$ (here and below $n$ stands for $2n+1$ in the original problem):

[quote]If $\gamma_1,\dots,\gamma_n$ are arbitrary reals, $n\geq 1$, and $t\in\mathbb{N}$, then there exist $s_1,\dots,s_n\in\mathbb{Z}$ and $q\in\mathbb{N}$ such that
\[ |q\gamma_k - s_k|<\frac{1}{t},\qquad k=1,\dots,n,\qquad 1\leq q\leq t^n.  \][/quote]
Integers $s_k$ can be used instead of $a_k$ since:
1) $s_k$ are distinct as soon as $a_k$ distinct;
2) $\sum_{j \in G_{1}} s_{j} = \sum_{j \in G_{2}} s_{j}$ as soon as $\sum_{j \in G_{1}} a_{j} = \sum_{j \in G_{2}} a_{j}$.


Here's a problem I just dug up and haven't solved yet:

Define a sequence by x(1) = 1,  and for n > 1,
x(n) = 2(x(n - 1) + x(n - 1)^(-2))/3.

a)  Compute x(2):^3:, x(3):^3:, and x(4):^3: (use a caclulator).  Make a conjecture about the limit of x(n) as n goes to infinity.

b) prove your conjecture.

If it approaches a limit (it does, but that's a separate proof), then

[hide]x(n) = x(n-1) (almost), which we can call z.  Solving the resulting equation when we plug in z for x(n) and x(n-1) gives z=2^(1/3)[/hide]


search
algebra
polynomial
number theory unsolved
number theory

prove that the sequence $a_{n}=(n!-1)n!+1$ contains infinitely many composite numbers... :?:

One of my favourite problems: search the ISL 2005 (it must be in the ressources, with a solution), for a more general version with $P(n!)$ for a polynomial $P$, not too trivial...

Let $f(x)$ is polinom with integer coefficients degree k.
If $f(n!)=p$ is prime, then $(p-n-1)!=\frac{(p-1)!}{(p-1)(p-2)...(p-n)}\equiv \frac{(-1)^{n+1}}{n!}(mod \ p)$.
Therefore if $f(x)$ is symmetric ($x^{k}f(\frac{1}{x})\equiv f(x)$, were $k=deg(f(x))$), then $p|f((p-n-1)!)>p$ if n is odd.
It prove these statment for symmetric polinoms, and $f(x)=x^{2}-x+1\equiv x^{2}(\frac{1}{x^{2}}-\frac{1}{x}+1)$.

well done guys...
thanx rust,its a great solution...  :P
and thank you harazi for the resource of the problem...

[quote="Rust"]Let $ f(x)$ is polinom with integer coefficients degree k.
If $ f(n!)=p$ is prime, then $ (p-n-1)!=\frac{(p-1)!}{(p-1)(p-2)...(p-n)}\equiv \frac{(-1)^{n+1}}{n!}(mod \ p)$.
Therefore if $ f(x)$ is symmetric ($ x^{k}f(\frac{1}{x})\equiv f(x)$, were $ k=deg(f(x))$), then $ p|f((p-n-1)!)>p$ if n is odd.
It prove these statment for symmetric polinoms, and $ f(x)=x^{2}-x+1\equiv x^{2}(\frac{1}{x^{2}}-\frac{1}{x}+1)$.[/quote]

Can anyone give explanation for this?
thx..

assume that $ f(x)=x^{2}-x+1$ and for a odd natural number $ n$ we have $ f(n!)=p$ where $ p$ is prime we claim that $ f(p-n-1)$ is composite...
we have:

$ (p-n-1)!=\frac{(p-1)!}{(p-1)(p-2)...(p-n)}\equiv \frac{(-1)^{n+1}}{n!}=\frac{1}{n!}(\bmod p)$

$ \Rightarrow f((p-n-1)!)=(p-n-1)!^{2}-(p-n-1)!+1$

$ =(p-n-1)!((\frac{1}{(p-n-1)!})^{2}-\frac{1}{(p-n-1)!}+1)$

in the other hand we have:

$ (\frac{1}{(p-n-1)!})^{2}-\frac{1}{(p-n-1)!}+1 \equiv n!^{2}-n!+1 \equiv f(n!)=p \equiv 0 (\bmod p)$

$ \Rightarrow p\mid (\frac{1}{(p-n-1)!})^{2}-\frac{1}{(p-n-1)!}+1$

$ \Rightarrow p\mid f((p-n-1)!)>p$

so $ f(p-n-1)$ is composite...

note that as RUST said the above solution works for every symmetric polynomials (i.e. $ x^{k}f(\frac{1}{x}) \equiv f(x)$ where $ \deg f=k$)...
and as harazi said the shortlist 2005,N7 is somehow a generalization of this problem:
[b]$ N7:$let $ P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}$ s.t $ a_{0},...,a_{n}\in \mathbb{Z}$ and $ a_{n}>0$ and $ n\geq 2$,prove that there exists a positive integer $ m$ s.t $ P(m!)$ is a composite number...[/b]

[quote="harazi"]One of my favourite problems: search the ISL 2005 (it must be in the ressources, with a solution), for a more general version with $ P(n!)$ for a polynomial $ P$, not too trivial...[/quote]
sorry for the spam but what is ISL and where is it in the ressources? ty

@srulikbd:IMO short list...
see my last post


floor function
inequalities
graph theory
combinatorics

Let $S$ be a set consisting of $m$ pairs $(a,b)$ of positive integers with the property that $1 \leq a < b \leq n$.  Show that there are at least
\[ 4m \cdot \dfrac{(m - \dfrac{n^2}{4})}{3n} \]
triples $(a,b,c)$ such that $(a,b)$, $(a,c)$, and $(b,c)$ belong to $S$.

I am not sure if this works. But I will try anyway.

Draw a graph and its vertices are the integers $1,2,.......,n$. An edge is drawn between $x$ and $y$ if and only if $(x,y)$ or $(y,x)$ belongs to $S$. For $3$-point subsets of $S$, call it good if any two points are connected by an edge. Denote $F(x)$ as the number of edges the vertex $x$ belongs to.

Let $x$ be joined to $y$. We have that $F(x)+F(y)-2$ edges are attached to the other $n-2$ vertices and thus we also have at least $F(x)+F(y)-2-(n-2)$ vertices that are attached to both $x$ and $y$. Therefore we have at least $F(x)+F(y)-n$ good triples that contains $x$ as well as $y$. Hence the total number of good triples is at least:

$\sum_{(x,y) \in S}\frac{F(x)+F(y)-n}{3}$. Now we have that $\sum_{(x,y) \in S}(F(x)+F(y)) = \sum^{n}_{x=1}d(x)^{2}$.

Therefore $\sum_{(x,y) \in S}\frac{F(x)+F(y)-n}{3}= \frac{\sum^{n}_{x=1}F(x)^{2}-nm}{3}\geq \frac{(\sum^{n}_{x=1}F(x))^{2}}{3n}-\frac{nm}{3}= \frac{4m(m-\frac{n^{2}}{4})}{3n}$
as $\sum_{x=1}^{n}F(x) = 2m$.

[quote="shobber"]Let $S$ be a set consisting of $m$ pairs $(a,b)$ of positive integers with the property that $1 \leq a < b \leq n$.  Show that there are at least
\[4m \cdot \frac{(m-\frac{n^{2}}{4})}{3n}\]
triples $(a,b,c)$ such that $(a,b)$, $(a,c)$, and $(b,c)$ belong to $S$.[/quote]

Don't you mean S is a set of sets of two elements $\{a,b\}$  (and not pairs, as $(a,b)$ and $(b,a)$ are distinct?) and we need to count the number of sets of three elements $\{a,b,c\}$ instead of triplets $(a,b,c)$?

In fact we can prove the stronger statement, if $m > \frac{n^{2}}{4}$ (there is nothing to show in above problem, if $m \leq \frac{n^{2}}{4}$), then there are atleast $\lfloor\frac{n}{2}\rfloor$ such sets $\{a,b,c\}$ such that $S$ contains $\{a,b\}, \{b,c\}, \{a,c\}$

Yeah, missed the $a < b$ condition (if not, we can find a counter example)

Anyway, does not matter. Why do these olympiad question try to disguise well known graph theory problems in such silly ways? Cities with airplanes, pairs with $a < b$ etc.

It is a rhetorical question. You don't have to answer.

[quote="TaiPan~SP!"]I am not sure if this works. But I will try anyway.

Draw a graph and its vertices are the integers $1,2,.......,n$. An edge is drawn between $x$ and $y$ if and only if $(x,y)$ or $(y,x)$ belongs to $S$. For $3$-point subsets of $S$, call it good if any two points are connected by an edge. Denote $F(x)$ as the number of edges the vertex $x$ belongs to.

Let $x$ be joined to $y$. We have that $F(x)+F(y)-2$ edges are attached to the other $n-2$ vertices and thus we also have at least $F(x)+F(y)-2-(n-2)$ vertices that are attached to both $x$ and $y$. Therefore we have at least $F(x)+F(y)-n$ good triples that contains $x$ as well as $y$. Hence the total number of good triples is at least:

$\sum_{(x,y) \in S}\frac{F(x)+F(y)-n}{3}$. Now we have that $\sum_{(x,y) \in S}(F(x)+F(y)) = \sum^{n}_{x=1}d(x)^{2}$. [size=150]I don't understand!![/size] What is $d(x)$?

Therefore $\sum_{(x,y) \in S}\frac{F(x)+F(y)-n}{3}= \frac{\sum^{n}_{x=1}F(x)^{2}-nm}{3}\geq \frac{(\sum^{n}_{x=1}F(x))^{2}}{3n}-\frac{nm}{3}= \frac{4m(m-\frac{n^{2}}{4})}{3n}$
as $\sum_{x=1}^{n}F(x) = 2m$.[/quote]

Interpret as a graph in the obvious way as TaiPan did; we need to prove that a graph with $n$ vertices and $m$ edges has at least $4m\cdot\frac{(m-\frac{n^{2}}{4})}{3n}$ 3-cliques. Let $f(G(n,m))$ be the number of such cliques for such a graph $G(n,m)$ and $g(G(n,m))$ be the number of triples of vertices s.t. exactly two pairs of them are connected.

[b]Lemma:[/b] $g(G(n,m))\le m \left(n-1-\frac{2m}{n}\right)$.
[b]Proof:[/b] Let $a(V)$ be the set of vertices adjacent to vertex $V$, $d(V)$ its degree; it is apparent that $\sum d(V) = 2m$. Then we find that
\[g(G(n,m))=\sum_{X,Y\in G,X\not \in a(Y)}|a(X)\cap a(Y)|\le \sum_{X,Y\in G,X\not \in a(Y)} \min\{d(X),d(Y)\}\]
The last term, were it not to take the minimum of each pair, counts a vertex $V$ with degree $d(V)$ exactly  $n-1-d(V)$ times. But the larger degree isn't counted in each pair, so it's bounded above by
\[\frac{1}{2}\sum d(V)(n-1-d(V))\le \frac{n}{2} \cdot\frac{2m}{n}\left(n-1-\frac{2m}{n}\right)\]
the last inequality following from Jensen's.

Now notice that $3f(G(n,m))+g(G(n,m))=\sum \binom{d(X)}{2}\ge m\left(\frac{2m}{n}-1\right)$ again from Jensen's. Thus \[f(G(n,m))\ge \frac{m}{3}\left(\frac{2m}{n}-1\right) - \frac{g(G(n,m))}{3}\]
Using our lemma, this is bounded below by
\[\frac{m}{3}\left(\frac{2m}{n}-1\right)-\frac{m}{3}\left(n-1-\frac{2m}{n}\right)=4m\cdot\frac{(m-\frac{n^{2}}{4})}{3n}\]
as desired.

Oh, I see. Mantel's theorem is just Turan's theorem for $k=3$. :) I don't believe that this problem is solvable through Mantel's, since if you look at the $m-\frac{n^2}{4}$ term, it's actually a stronger result. If you try to go the "naive" route of removing $\frac{4m}{3n}\left(m-\frac{n^2}{4}\right)$ edges to create a $3$-free graph and bounding the edges using Turan's, it leads nowhere, and sharper bounding on the fewest edges one would need to remove seems to be as difficult as the problem itself.

[quote="robinson123"]Can we use Mantel Theorem to solve this problem?

Mantel Theorem states that: " Given a graph $ G  $ with $ n $ points and $ e $ edges, if $G $ doesn't contain any triangle then the maximum value of $ e $ is $ \left[\frac{n^2}{4}\right] $[/quote]
do you can point out the case of equal? And prove its?

[quote=shivangjindal]We basically need to show that the graph with $k$ edges and $n$ vertices has at least  $ \frac{k(4k-n^2)}{3n} $ triangles

We label the points $ 1 , 2 , 3  ..... , n $  and let point $i$ have degree $ d_i $(number of edges) . then if $i$ and $j$ are joined they have at least $d_i + d_j -2 $ other edges between them and these edges join then to  $n-2$ other points . So there must be at least $d_i + d_j - n $ triangles  which have $ i $ and $j$ as two vertices . Hence the total number of triangles must be at least $ \sum (d_i + d_j - n )/3 $ . But $ \sum ( d_i + d_j) = \sum (d_i)^2 $ because each point in $i$ occurs in just $ d_i $ terms .
Thus the total number of triangle is at least $ \frac{\sum (d_i)^2}{3} - \frac{nk}{3} $ . But $ \sum (d_i)^2  >= \frac{ (\sum d_i )^2}{n}$ ( can be proved using chebyshev's  inequality ) $= \frac{4k^2}{n}$

HENCE PROVED
$ QED $  :Idea[/quote]
I am not getting which part implementing $a<b$ case here  :maybe:
Can anyone explain how he is saying this -
[quote]So there must be at least $d_i + d_j - n $ triangles  which have $ i $ and $j$ as two vertices . [/quote]

@above: Take two adjacent vertices, with degrees $d_i$ and $d_j$. There are $n-2$ vertices excluding the two vertices then there are at least $(d_i -1) + (d_j-1) -(n-2)$ triangles

I do not understand how this follows from Jensen's: \[\frac{1}{2}\sum d(V)(n-1-d(V))\le \frac{n}{2} \cdot\frac{2m}{n}\left(n-1-\frac{2m}{n}\right)\]

[quote=shivangjindal]We label the points $ 1 , 2 , 3  ..... , n $ and let point $i$ have degree $ d_i $(number of edges) . then if $i$ and $j$ are joined they have at least $d_i + d_j -2 $ other edges between them and these edges join then to  $n-2$ other points . So there must be at least $d_i + d_j - n $ triangles which have $ i $ and $j$ as two vertices . Hence the total number of triangles must be at least $ \sum (d_i + d_j - n )/3 $ . But $ \sum ( d_i + d_j) = \sum (d_i)^2 $ because each point in $i$ occurs in just $ d_i $ terms .
Thus the total number of triangle is at least $ \frac{\sum (d_i)^2}{3} - \frac{nk}{3} $ . But $ \sum (d_i)^2  >= \frac{ (\sum d_i )^2}{n}$ ( can be proved using chebyshev's inequality ) $= \frac{4k^2}{n}$[/quote]

I still don't get how there are at least $d_i+d_j-n$ triangles  as well. IF someone can explain that part of shivangjindal's proof, that would also be great!

We use the obvious graph theoretic interpretation. The key idea is that the number of triangles involving the edge $vw$ is at least $\mathrm{deg}(v)+\mathrm{deg}(w)-2n$. We then see that thrice the number of triangles is at least
\begin{align*}
    \sum_{vw\in E}(\mathrm{deg}(v)+\mathrm{deg}(w)-n)&=\sum_{v\in V}\mathrm{deg}(v)^2-mn \\
    &\ge n(2m/n)^2-mn,
\end{align*}
which gives the desired result.

Interpret the problem in terms of graph theory.

Consider a vertex $v$ with neighbors $N(v)=\{w_1,w_2,\dots,w_{\text{deg}(v)}\}.$ Since the graph has $n$ vertices, each element $w_i$ of $N(v)$ is connected to at least $\text{deg}(w_i)-(n-\text{deg}(v))$ other elements of $N(v).$ Therefore, since summing over all vertices counts each edge twice, the number of triangles passing through $v$ is at least $$\frac{1}{2}\sum_{w\in N(v)}(\text{deg}(w)-(n-\text{deg}(v)))=\frac{1}{2}\left(\sum_{w\in N(v)}\text{deg}(w)-n\cdot\text{deg}(v)+\text{deg}(v)^2\right).$$ Summing over all $v,$ we find that the number of pairs $(T,v)$ where $T$ is a triangle passing through a vertex $v,$ is at least $$\frac{1}{2}\left(2\sum_{v\in G}\text{deg}(w)^2-2mn\right)\ge\frac{(\sum_{v}\text{deg}(v))^2}{n}-mn\ge\frac{4m^2}{n}-mn.$$ But since triangles have three vertices, the number of pairs $(T,v)$ is simply three times the number of triangles. Thus, the number of triangles in $G$ is at least $$\frac{1}{3}\left(\frac{4m^2}{n}-mn\right)=\frac{m}{3n}(4m-n^2),$$ as desired.

Let $e=\{u,v\}$ be an arbitrary edge. There are at least $\deg u+\deg v-n$ triangles that $e$ is part of by pigeonhole principle. Hence, there are at least
\[\frac{\displaystyle\sum_{u\in G}(\deg u)^2-mn}{3} \ge \frac{4m^2}{3n}-mn/3=\frac{m}{3n}(4m-n^2)\]
triangles as desired.

Very standard result; I'm surprised I didn't know of this earlier.

Obviously, count triples of edges that form a triangle. For each edge $e = uv$, there are $\deg u + \deg v - n$ vertices that are connected to both $u$ and $v$. Thus the number of triangles is $$T = \frac{\sum_{uv \in E} (\deg u + \deg v - n)}3 = \frac{\sum_v (\deg v)^2 - mn}3 \geq \frac m{3n}(4m-n^2)$$ by C-S as needed.

For any edge $uv$, there are at least $\text{deg}(u)-1+\text{deg}(v)-1-(n-2)=\text{deg}(u)+\text{deg}(v)-n$ triangles with $uv$ as an edge. Hence there are at least

$$\frac{1}{3}\sum_{\text{edge uv}} (\text{deg}(u)+\text{deg}(v)-n) = \frac{1}{3}\left(\sum_{\text{vertex v}} \text{deg}(v)^2 - mn\right) \ge \frac{1}{3}\left(n\left(\frac{2m}{n}\right)^2-mn\right) = \frac{1}{3}\left(\frac{4m^2}{n} - mn\right)$$

as desired, where the main inequality step follows from Jensen/QM-AM.


Yale college Harvard Stanford Duke Princeton MIT

By JBL's request, here is a thread for discussing the Budapest Semester in Mathematics.

I'll start with an obvious question: which colleges in the United States sent students to the Budapest program this year?

Does anyone know if there is an age requirement for this program?  Must be 18 or not over 24 type of thing?

can someone provide a link to this program?

There was also another small thread about BSM [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=403577#p403577]here[/url].

We still want to hear about JBL's experience though (and anyone else that attended) ;)

I'm interested in knowing what Budapest is like. Both in terms of academics, math, etc... and more generally life there.

So, colleges that sent students this year (with possible misspellings and certain omissions):

St. Olaf
Grinnell
Carelton
Wartburg
Trinity College (CT)
Trinity University (TX)
St. Lawrence
Kalamazoo
Swarthmore
Oberlin
Furman
Pomona
Weslyan
Virginia Tech
University of Colorado Boulder
UNC Chapel Hill
University of South Carolina
UMass Amherst
Dartmouth
Yale
Harvard
Stanford
Cornell
Duke

Somewhere on their website there is a list of all schools that have ever participated, I believe.  Also, the program really has two websites -- blah$^3$ linked to the one run by the American organizer (based at St. Olaf's College), but there is also a Hungary-based website.  I believe there is a link to it in the thread joml pointed out.  I think it may be http://www.bsmath.hu but I'm not 100% sure.

There is no age requirement -- they've had students as young as 15 before, I believe, as well as relatively older students.  (I think there was someone this semester who was older than 25.)  You probably need to be a college student (or just finished being a college student, anyhow -- there were a few people who had finished their senior year already.)

So, BSM is a very interesting program.   The student body, as you can see from the list of colleges, is coming from all over the country.  Although large state and private universities are represented, the (substantial) majority of the class is composed of students from small liberal arts schools, largely in the midwest.  This leads to a substantially different student body than those of us from the coasts are used to.  There are also a handful of international students who come -- this semester, there were two Romanians, two Chinese, one Japanese and one Vietnamese national among the student body.  (Also, a Canadian and Australian.)  The ability and experience level of the students is mixed, but so are the class offerings, so there will typically be both classes and students at approximately your level.  (To be continued -- I'm writing a required evaluation of the program for my school, and I'll post relevant parts of that along with other comments by tonight, I hope, unless I fall asleep (still recovering from jetlag).)

If we are interested in applying, which year of undergrad studies is it recomended to go?

Simon:  I believe (but am not 100% sure) that there was a student who was planning on going for the entire year who got to Budapest, decided he didn't like the city, and went home.  I have associated that person with the name Branden, but I'm not entirely sure why.  So, I guess the answer is "maybe, but if so, he didn't stay very long."

Manuel -- it depends.  If you attend a larger university and are serious about math, odds are that you want to go in your junior year or else you may be too advanced.  A number of people do go in their senior (or post-senior) year, but it really sucks to apply to grad school from a foreign country.  If you go to a smaller school, there are concerns about what classes are being taught which semester by whom that should also play a role.  It is possible to go earlier in your college career, as well, but you are supposed to have taken a reasonable number of introdutory classes first (analysis, algebra, for example), and I only took algebra in my sophomore year.  You can also just e-mail them and ask any sorts of questions like this.  As far as we (I and the other students I talked to about it) could tell (they don't make this information publicly available, so I am just guessing) they don't reject any applicants, because their applicant body is highly self-selecting.

So, I really will keep posting on this, but not right immediately -- I promise I'll get back to it within the next few days, tetrahedr0n.

Joel does it matter if our college is not on the list? i mean does the college has to approve or something like that?
...

it seemed kind of strange that princeton and mit are not in the list.. when theres so many people interested in math in those two...

No, it doesn't.  However, to study abroad, you need your own college's permission and willingness to grant credit (at least, if you want credit).  If you go to a school that hasn't sent students before, you will need to convince some administrator to grant you credit, most likely.  The program is open to any North American university student.

Here are some excerpts from my general evaluation of the program.  If you have more questions or want more specifics, feel free to ask.

[quote]Comment on your academic experience, including teaching mode, specific information on assessment, and whether the courses and program met your expectations, indicating any strengths and weaknesses.[/quote]
My academic experience was wonderful – the classes are exceedingly well taught (in my experience, better than Harvard classes) in small lecture classes by fully English-fluent professors.  (Very small classes – those with 4 or fewer registered students – were taught as reading classes, with less-frequent class meetings.)  Courses could be taken for letter grade or for “audit.”  Mathematically (academically), the program was excellent.  Strengths include a good, problem-solving based approach to mathematics, strength in areas (esp. graph theory and combinatorics) in which Harvard is weak and the general wonderfulness of the staff.  I found no significant academic weaknesses.

[quote]What were your living arrangements?  How well did they serve your needs/meet your expectations?[/quote]
The program arranged for you to rent a privately owned, furnished apartment along with one or two other members of the program.  I roomed with a friend who I previously knew.  I was pleased with my apartment and general living circumstances – people should be aware that they are traveling into a big city in a poor country in Europe, and so there are general differences to be expected (no clothes dryers, for instance).

[quote]Would you recommend this program to other students?  Why or why not?  Please feel free to make any additional comments that might be helpful to other students.[/quote]
Absolutely, with no reservations.  This is a really wonderful program.  Unlike some study abroad programs, it is academically quite serious.  You’re likely to be one of the better students there, but you’ll take the hard classes and that will be fine.  The math is really, really good, and living abroad is incredibly fun and exciting.

[quote]On a scale of 1 to 5 (where 5 is ‘That was wonderful and life-changing,’ and 1 is ‘I’ve had it, I’m never leaving the country again!’), please rate your overall satisfaction with your study abroad experience.[/quote]

5

[quote]In comparison with Harvard, how would you evaluate overall:[/quote]
Overall educational experience: Excellent
Quality of instruction: Excellent
Course content: Excellent
Intellectual challenge: Good
Intellectual caliber of fellow students: Fair
Library Facilities: Poor

wow, nice ratings :) maybe i'll show them to my parents! they dont' seem to see the need to go abroad during college (my family and i moved to the US so that i could go into high school here and have opportunities for a "good solid education")

as of now, i'm tentatively double majoring in math and CS, i'm not quite sure about exact details though!

Hi, JBL. Good to see you back.
Was anything else (outside of math) offered? I know I would get bored of doing math all day halfway through the program. I didn't take any humanities last term and got exhausted from writing proofs each night.
Did you have to go to every class listed in the syllabus? What was the workload like?

Hi bictor!

There are other courses offered.  This semester, there were 3 Hungarian courses (1 beginner, 2 intermediate), a course on central European and Hungarian history from WWI onwards, a class in Hungarian art, culture and film, a philosophy course, and a class on the history of physics and mathematics.  The Budapest website for the course (http://www.bsmath.hu) should have a full list.

There are typically around 25 or so classes offered, of which you may take as many as you like (additional fees apply for more than 5 or 6 math classes), with most people taking your average 3-5 courses (I took 3 math, 2 non-math).

I took fewer math classes than many, but they were the more advanced classes (Combinatorics 2, Advanced Graph Theory, Topics in Number Theory (which may have been the hardest class in the program, although obviously that's a little hard to judge)) and I found the workload to be very manageable, even light.  (By way of bragging disclosure, I was the strongest student in CO2 and GT -- I'm not entirely sure about NT, but there were only 4 of us in the class anyhow.)  You obviously have a fair amount of control over the quantity of work you will do by having a lot of flexibility in the classes you take.  Homework assignments are typically weekly problem sets -- for GT, it was 5 questions every week, for CO2 it was about 10 questions counting extra credit, and for NT it was 3 questions plus extra credit.  I found GT to be lots of fun but didn't think the problem sets were difficult at all, while the NT was very challenging.  You can basically choose for yourself how much of a burden you want to take on.

[quote="JBL"]As far as we (I and the other students I talked to about it) could tell (they don't make this information publicly available, so I am just guessing) they don't reject any applicants, because their applicant body is highly self-selecting.
[/quote]

That's interesting, because the scuttlebutt on the faculty end (at least those I know) is that only the strongest students get in.  Of course, that's really nothing more than scuttlebutt---it's not like I've asked the organizers or anything.  It could be that this impression among some faculty leads to further self-selection among applicants.

Well, I did say I was just guessing.  Let me rephrase: no one I spoke to in BSM has spoken to anyone who had applied to BSM and been rejected.

One person asked the US office how many applications they received and they refused to answer, thus my not actually having any idea.  I also don't know if the people in the program were the strongest students from their schools -- in some cases, I have no doubt that they were, but I just don't have any idea how strong the student body at most colleges is in mathematics to know if what you say is true.  In fact, even if we all were top in the department, I still wouldn't know (perhaps no weaker students had applied).

I just noticed that Bictor717 is posting in this forum (about Brown), and I know that he just spent a semester in Budapest (on my recommendation!), so I'm bumping this thread in the hopes that he would like to add his opinion of the program (and maybe put on the record that I didn't horribly mislead him with my positive reviews).

I agree with about everything JBL said. It was a very good experience for me, both academically and culturally. The teachers were top-notch, all better than a few of the professors I've had at Brown.
My main motivation for attending were the combinatorics and graph theory classes, which are lacking at Brown. I took the following classes. I'm a junior, by the way.
Complex Functions
Conjecture and Proof
Combinatorics 2
Topics in Number Theory
Geometric Graph Theory
Beginning Hungarian
(Basically, I avoided all of the real analysis and algebra courses, which I planned to take at Brown.) Of these, number theory was the most challenging for me, but based on how others talked about it, the toughest course was probably Real Functions and Measures. The workload was still very manageable; it was a lucky coincidence that I never had more than 2 sets due on any given day.
During my semester, BSM started a collaboration with the Renyi Institute, allowing students to take classes there with some of the best mathematicians in Hungary. For example, my graph theory course was taught by Laszlo Lovasz and Katalin Vesztergombi, although they were very busy. Fortunately, the grad student they left in charge was very capable. I don't know whether this collaboration will be continued.
I got the impression that the classes were varied enough so that anyone could get a lot out of the program, regardless of prior experience. Even the introductory courses were very rigorous and taught in a way different from what I'd expected. I felt that several students had not done much problem solving, but there were many more who simply thought about math on a higher level than I did.

Nathan Pflueger from Stanford recently came back to Washington to give a guest lecture, and while he did he exhorted us all to try this out.  It seems interesting :)

Quick question:  if I've already taken some of the introductory courses at my college, would it be worth while to take them over again for a different perspective, or should I just move on?

The introductory number theory class seemed comparable to the one I took at Columbia, so I think that one isn't very different from the number theory classes usually offered, in terms of content.

I think the algebra course is very different from most beginning algebra classes. For some reason, they started with rings instead of groups and covered much more than I did in my first algebra course.

I can't comment on the other introductory courses (graph theory and combinatorics) because I haven't taken them before and can't compare them to anything. I didn't take them at BSM because their course descriptions didn't sound very interesting to me.

In retrospect, I would have liked to sit in on those courses simply because they did things differently enough. Some students did that and I don't think the professors mind. I probably would not have officially enrolled because of the complications arising from transferring the credit.

Nathan is an excellent speaker. His presentations were so articulate.

Number theory was among the hardest classes while I was there, too -- was Professor Balog teaching?  I'm somewhat jealous that you got Professor Lovasz (although I absolutely loved my graph theory professor; when I was there, everyone was very excited because Lovasz was about to or just had returned to Hungary).

I don't think there's any particular reason to repeat for credit a course you've already taken.  If you're just curious, then I'm quite confident that no professor would object to your sitting in on lectures (unless there aren't enough chairs ;)).


search
inequalities unsolved
inequalities

let a,b,c be positive reals whose product does not exceed their sum.
prove that
$ a^2\plus{}b^2\plus{}c^2>\equal{} sqrroot{3}abc$

I believe you meant to write:
if $ a,b,c\in\mathbb{R}_{\plus{}}$, such that $ a\plus{}b\plus{}c\geq abc$,then
\[ a^2\plus{}b^2\plus{}c^2\geq\sqrt{3}abc
\]
I assure you,that whenever the problem was at IMO or BMO, it's highly probable that the problem was posted before.
Take a look at the Resource section or try the search function.

http://www.mathlinks.ro/viewtopic.php?p=495326#495326


[quote="keyree10"]Prove that there is no function $ f : Z$ → $ Z$ such that for all $ m, n$ ∈ $ Z$
$ f(m) \equal{} n$ implies $ f(n) \equal{} m$ and $ f(n \plus{} 3) \equal{} 3m.$[/quote]

If $ f(n \plus{} 3) \equal{} 3m.$ we have $ f(3m) \equal{} n\plus{}3$

We can then rewrite to get $ f(3m) \equal{} f(m) \plus{} 3$

We have then that 2m divides 3

which means m = .5 or 1.5

But this contradicts the fact that m is an integer.

[quote="befuddlers"]

We have then that 2m divides 3

which means m = .5 or 1.5

But this contradicts the fact that m is an integer.[/quote]

Hey couln't grasp this point........please elaborate..... :maybe:


I'll give you some hints. Read them one by one, and think about one before proceeding.

[hide="Hint 1"]What does it mean if a number ends in a 0?[/hide]

[hide="Hint 2"]What kind of factor does a number that ends in 0 have?[/hide]

[hide="Hint 3"]If a number is a multiple of 10, it is a multiple of ...[/hide]

[hide="Hint 4"]Are there more factors of 2 or 5?[/hide]

[hide="Hint 5"]We need thus to count the factors of ...?[/hide]

EDIT - Don't read G-UNIT's solution until you try the hints. They'll help you understand better.

[quote="erik-the-red"]One of the problems on the Mock AMC #1 is [i]how many zeroes are at the end of (2003!) / (1003!) ?[/i]

How do you answer this question? Namely, what is the strategy for solving this type of question?[/i][/quote]

[hide="Answer"]  Finding the number of 0's in an expression, is the same thing as finding out the number of 10's it contains.  Since the composition of 10 is: $(2 \cdot5)$.  We must find the number of 2's and 5's the expression contains.  However, the number of 2's in any expression always exceeds the number of 5's, meaning we only to find out how many 5's the expression contains.

  2003! has (rounding the fractions):

  $\frac{2003}{5}  +  \frac{2003}{25}  +  \frac{2003}{125}  +  \frac{2003}{625} =499-5's$

  Since each 5 can be paired with a 2, $(5\cdot2)^{499}=10^{499}$ meaning it has 499 zeros.

  1003! has (rounding the fractions):

  $\frac{1003}{5}  +  \frac{1003}{25}  +  \frac{1003}{125}  +  \frac{1003}{625} = 249-5's$

  As we saw before each 5 can be paired with a 2, meaning it has 249 zeros.

  $\frac{2003!}{1003!}= 1004 \cdot 1005 \cdot 1006..................2003$

  Which is the samething as subtracting the zeros from before that point.

  So we have the 499 zeros from 2003! minus the 249 zeros from 1003!.

  Thus, we have $499-249=\boxed{250}$zeros in the expression $\frac{2003!}{1003!}$ [/hide]

Actually, it's $\lfloor \rfloor$ (floor function), not rounding. ;)

[quote="erik-the-red"]One of the problems on the Mock AMC #1 is [i]how many zeroes are at the end of (2003!) / (1003!) ?[/i]

How do you answer this question? Namely, what is the strategy for solving this type of question?[/i][/quote]

[hide]$2003!$ ends in this many number of zeroes:

$\lfloor\frac{2003}{5}\rfloor=400$

$\frac{400}{5}=80$

$\frac{80}{5}=16$

$\lfloor\frac{16}{5}\rfloor=3$

$3+16+80+400=499$

$1003!$ ends in this many zeroes:

$\lfloor\frac{1003}{5}\rfloor=200$

$\frac{200}{5}=40$

$\frac{40}{5}=8$

$\lfloor\frac{8}{5}\rfloor=1$

$1+8+40+200=249.$

$499-249=\boxed{250}$[/hide]

Right?  :(

Is this question type a special case of another question type?

That is, is there an even more general form of this question?

The most famous type of question like this is something like,
[i]How many zeros end 236! when it is expressed as an integer? [/i]

Or this is different, but sometimes they ask,
[i]How many digits are in the decimal expansion of [/i]$2^{24}\cdot5^{22}$?

[quote="chess64"][hide="2nd question"]
$2^{24}\cdot5^{22}=2^2 \cdot 2^{22} \cdot 5^{22}=4 \cdot 10^{\boxed {22}}$ :D
[/hide][/quote]

So, then, the principle is the same?

Express it in "scientific notation" and add one to whatever the power of ten is.

[quote="bob123"][quote="chess64"][hide="2nd question"]
$2^{24}\cdot5^{22}=2^2 \cdot 2^{22} \cdot 5^{22}=4 \cdot 10^{\boxed {22}}$ :D
[/hide][/quote]

It said how many digits in the decimal expansion, not how many zeroes there are at the end...[/quote]

W/e, :roll: okay, $23$.

[quote="erik-the-red"]Is the answer 57 zeroes to the first question?[/quote]

Yes, 47+9+1=57.

[quote="erik-the-red"]One of the problems on the Mock AMC #1 is [i]how many zeroes are at the end of (2003!) / (1003!) ?[/i]

How do you answer this question? Namely, what is the strategy for solving this type of question?[/i][/quote]

There are two ways to solve these kind of problems:

1st Way:

When you have $N!$, keep dividing $N$ by $5$, so:

$2003!--->\frac{2003}{5}=400--->\frac{400}{5}=80--->\frac{80}{5}=16--->\frac{16}{5}=3$

Thus $400+80+16+3=499$

2nd Way:

Divide by the powers of $5$

$1003!--->\frac{1003}{5}=200--->\frac{1003}{25}=40--->\frac{1003}{125}=8--->\frac{1003}{625}=1$

Thus $200+40+8+1=249$.

Now you should subtract the number of zeroes because you are dividing, so $499-249=\boxed{250}$

[quote="joml88"][quote="Iversonfan2005"]
There are two ways to solve these kind of problems:
[/quote]

Two ways which are essentially the same...[/quote]

Agreed :P It's essentially just

\[\lfloor n/5 \rfloor+\lfloor n/5^2 \rfloor+\lfloor n/5^3 \rfloor+\cdots\]

[hide="Quick Solution"]
2003! has a power of ten of 499. (400+80+16+3). 1003! has a power of ten of 249. (200+40+8+1)

2003!/1003! therefore has a power of ten of 250, meaning it has 250 zeroes at the end. I think.
[/hide]


No because then you could have g=0.0000001
EDIT: Whoops, my mistake, I was thinking tan.

Sorry, I've put g incorrectly.There's actually no g. :blush: .

[quote="TRAN THAI HUNG"]Let M is a point in $ \Delta ABC$
$ \angle MAB \equal{} x$, $ \angle MBC \equal{} y$,$ \angle MCA \equal{} z$
Show that $ \cot gx \plus{} \cot gy \plus{} \cot gz \ge 3\sqrt 3$ :lol:[/quote]

So we need to show that cot x + cot y + cot z >= root(3), eh?

cot A + cot B = ?

We do know that tan(A+B) = (tan(A)+tan(B))/(1-tan(A)tan(B))...

tan A + tan B = tan(A+B) (1-tan(A)tan(B))...

tan A tan B = 1-(tan A + tan B)/tan(A+B)

And since 1/A+1/B = (A+B)/AB...

cot A + cot B = (tan A + tan B)/(1-(tan A + tan B)/tan(A+B))...

cot A + cot B = ((tan A + tan B) (tan (A + B))) / (1-(tan A + tan B))...

Eww...

I just derived a random formula for cot A + cot B and I don't think it's right...

OK, never mind. Back to the drawing board, everyone.

I get as far as:
cot(x)+cot(y)+cot(z)= [tan(x)tan(y)+tan(x)tan(z)+tan(y)tan(z)]/ [tan(x)tan(y)tan(z)]

knowing that x+y+z=2pie
and tan(x+y+z)=0
I need help

I think the problem simplifies to cot(x) > 3^.5
with y or z
but I don't know where to go from there


What is the total surface area, in square centimeters, of a right rectangular prism with edges measuring three centimeters, four centimeters, and five centimeters?


How many ten-figure number(made from 5 figures:1,2,3,4,5) satisfy both follow conditions?:
- Each of 5 figures appears 2 times
-The same figure doesn't stay beside the other.

[hide="Solution"]There are $ \frac {10!}{2!2!2!2!2!}$ 10-figure numbers which satisfy the first condition. Now we have to subtract number of 10-figure numbers which satisfy the first condition and do not satisfy the second condition.

1st case) All pairs of same figures stay beside each other.
We can assume that pairs 11, 22, 33, 44 and 55 are different figures, then there are $ 5!$ of ways to arrange them.

2nd case) Exactly 4 pairs of same figures stay beside each other.
We can assume that we have 6 figures (aa,bb,cc,dd,e,e) , then there are $ \frac {6!}{2} \minus{} 5!$ ways to arrange them (we subtracted 5! for the subcase then e and e stays beside each other). Also there are 5 ways to choose e, so the number of 10-figure numbers where exactly 4 pairs of same figures stay beside each other is $ 5(\frac {6!}{2} \minus{} 5!)$

3rd case) Exactly 3 pairs of same figures stay beside each other.
We can assume that we have 7 figures (aa,bb,cc,d,d,e,e) , then there are $ \frac {7!}{2!2!} \minus{} 2(\frac {6!}{2} \minus{} 5!) \minus{} 5! \equal{} \frac {7!}{4} \minus{} 6! \plus{} 5!$ ways to arrange them (subtracting for similar reasons as in the 2nd case). Also there are $ \binom{5}{2} \equal{} 10$ ways to choose d and e, so the number of 10-figure numbers where exactly 3 pairs of same figures stay beside each other is $ 10(\frac {7!}{4} \minus{} 6! \plus{} 5!)$

4th case) Exactly 2 pairs of same figures stay beside each other.
We can assume that we have 8 figures (aa,bb,c,c,d,d,e,e), then there are $ \frac {8!}{2!2!2!} \minus{} 3(\frac {7!}{4} \minus{} 6! \plus{} 5!) \minus{} 3(\frac {6!}{2} \minus{} 5!) \minus{} 5! \equal{} \frac {7!}{4} \plus{} \frac {3\cdot6!}{2} \minus{} 5!$ ways to arrange them. Also there are $ \binom{5}{3} \equal{} 10$ ways to choose c,d and e, so the number of 10-figure numbers where exactly 2 pairs of same figures stay beside each other is $ 10(\frac {7!}{4} \minus{} \frac {3\cdot6!}{2} \minus{} 5!)$

5th case) Exactly 1 pair of same figures stay beside each other.
We can assume that we have 9 figures (aa,b,b,c,c,d,d,e,e), then there are $ \frac {9!}{2!2!2!2!} \minus{} 4(\frac {7!}{4} \minus{} \frac {3\cdot6!}{2} \minus{} 5!) \minus{} 6(\frac {7!}{4} \minus{} 6! \plus{} 5!) \minus{} 4(\frac {6!}{2} \minus{} 5!) \minus{} 5! \equal{} \frac {9!}{16} \minus{} \frac {5\cdot7!}{2} \plus{} 9\cdot6! \plus{} 7\cdot5!$ ways to arrange them. Also there are $ \binom{5}{4}\equal{}5$ ways to choose b,c,d and e, so the number of 10-figure numbers where exactly 1 pair of same figures stays beside each other is $ 5(\frac {9!}{16} \minus{} \frac {5\cdot7!}{2} \plus{} 9\cdot6! \plus{} 7\cdot5!)$

So the answer is $ \frac {10!}{2!2!2!2!2!} \minus{} 5! \minus{} 5(\frac {6!}{2} \minus{} 5!) \minus{} 10(\frac {7!}{4} \minus{} 6! \plus{} 5!) \minus{} 10(\frac {7!}{4} \minus{} \frac {3\cdot6!}{2} \minus{} 5!) \minus{}$
$ \minus{} 5(\frac {9!}{16} \minus{} \frac {5\cdot7!}{2} \plus{} 9\cdot6! \plus{} 7\cdot5!) \equal{} \boxed{\boxed{\boxed{17880}}}$[/hide]

 :stretcher:

Some part or other of your inclusion-exclusion is wrong (but I don't know what part).  This is sequence [url=http://www.research.att.com/~njas/sequences/A114938]A114938[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

Some past discussions:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155637
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262290
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65961

(I found these by searching for "A114938.")


A person is standing on a bathroom scale. Which of these is [i]not[/i] a force exerted on the scale?

(A) a contact force due to the floor
(B) a contact force due to the person's feet
(C) the weight of the scale
(D) the weight of the person

[hide]
$ C$ since the scale's weight is a forced exerted by the scale not on the scale.
[/hide]

The weight of the scale acts on the scale (it's the force exerted by the earth on the scale).
The answer is (D) since the weight of the person acts on the person, not on the scale.


[quote="ertanrock"]ABCD is a trapezoid. DB is one of the diagonals. E and F are points on DB such that AE and CF are perpendicular to DB.
If AE = 10, CF = 5 and DB = 17 then EF = ?[/quote]

[size=150][color=red]SORRY FOR THE MISSING PART.
AD = BC[/color][/size]

[hide="here it is"] $ AC\cap BD\equal{}M$
$ \frac {5}{10}\equal{}\frac {FM}{ME}\equal{} \frac {CM}{MA}$ $ \implies$ $ |CM|\equal{}\frac {17}{3}$ (because of $ |DB|\equal{}|AC|\equal{}17$)
so $ |FM|\equal{}\frac {8}{3}$ with pyth in $ \Delta FMC$  and finally $ |EM|\equal{}2|MF|\equal{}\frac {16}{3}\implies |EF|\equal{}8$[/hide]

[quote="sinankaral53"][hide="here it is"] $ AC\cap BD \equal{} M$
$ \frac {5}{10} \equal{} \frac {FM}{ME} \equal{} \frac {CM}{MA}$ $ \implies$ $ |CM| \equal{} \frac {17}{3}$ (because of $ |DB| \equal{} |AC| \equal{} 17$)
so $ |FM| \equal{} \frac {8}{3}$ with pyth in $ \Delta FMC$  and finally $ |EM| \equal{} 2|MF| \equal{} \frac {16}{3}\implies |EF| \equal{} 8$[/hide][/quote]
aferin kocum.


Let ABC be an acute-angled triangle and let D, E, F be the feet of perpendiculars from A, B, C respectively to BC, CA, AB. Let the perpendiculars from F to CB, CA, AD, BE meet them in P, Q, M, N respectively. Prove that P, Q, M, N are collinear.

$ \angle{FNP} \equal{} \pi \minus{} \angle{B}$
$ \angle{FNM} \equal{} \angle{FHA} \equal{} \angle{B}$ ($ H$ is the orthocenter of the triangle)

Then $ \angle{PNM} \equal{} \pi$ or $ M, N, P$ are collinear. Similar argument yields the result.

My question is: Is there any method that only using Simson's line or something like that ?

[color=darkblue]Denote $ F_{AC}$, $ F_{AD}$, $ F_{BE}$ and $ F_{BC}$ proiections of the point $ F$ onto the respectiv lines.

Let $ H$ be the orthocenter. Then $ F$ lies on the circumcircle of $ AEH$, so $ F_{AC},F_{BE}$ and $ F_{AD}$ lie on the Simson line of $ F$. Simultaneously $ F_{AD}$, $ F_{BC}$, $ F_{BE}$ are collinear. In conclusion, all 4 points are collinear.[/color]

Dear Mathlinkers,
PQMN is the Simson's line of F wrt the delta determined by the triangle ADC and the line BHE.
Sincerely
Jean-Louis

I've always been poor at Geometry, but could (feasibly) an argument work which demonstrate that QAM is the enlargement of NFP centre the intersection of NP and FB?

Dear Mathlinkers,
PQMN is also known as the Miquel's line of the delta quoted in my last message.
Sincerely
Jean-Louis

[quote="SimonM"][b]I've always been poor at Geometry[/b], but could (feasibly) and argument work which demonstrate that QAM is the enlargement of NFP centre the intersection of NP and FB?[/quote]

I seriously doubt that after reading the rest of the sentence.  :rotfl:

[b]mathangel [/b]wrote:
[quote] [i] My question is: Is there any method that only using Simson's line or something like that ?[/i][/quote]

Solution: BCEF is cyclic, hence P, N, Q are collinear, PNQ being Simson's line of F w.r.t. triangle BEC; similarly PMQ being Simson's line of F w.r.t. triangle ACD.

Best regards,
sunken rock


geometry
3D geometry
sphere
trigonometry
calculus
calculus computations

if there is sphere is given with radius, and 2 points on the sphere. How do we find path between this two points(assume traveling along the sphere).

Do you mean the shortest path between the two points?  If so, it's just an arc of the great circle on which the two points lie; this [url=http://mathworld.wolfram.com/GreatCircle.html]Mathworld[/url] page has a good discussion.

if you mean how do we parametrize, we can take the idea of
$ (x_1,y_1,z_1)$ to $ (x_2,y_2,z_2)$

$ (x_i,y_i,z_i) \to (R,\theta_i,\phi_i )$

so $ f(t)\equal{}(R,t\theta_2  \plus{} (1\minus{}t)\theta_1,t\phi_2 \plus{} (1\minus{}t)\phi_1), t\in [0,1]$

and so we have a path

$ x (t)\equal{}(R\cos (t\theta_2  \plus{} (1\minus{}t)\theta_1)\sin (t\phi_2 \plus{} (1\minus{}t)\phi_1),$
$ y(t)\equal{}R\sin (t\theta_2  \plus{} (1\minus{}t)\theta_1)\sin (t\phi_2 \plus{} (1\minus{}t)\phi_1),$
$ z(t)\equal{}R\cos (t\phi_2 \plus{} (1\minus{}t)\phi_1)$

so $ f(t) \equal{} (R,t\theta_2 \plus{} (1 \minus{} t)\theta_1,t\phi_2 \plus{} (1 \minus{} t)\phi_1), t\in [0,1]$

can u explain me how u got it


Morgan walks $x$ yards north, then walks $(2x-1)$ yards west. She is now $(2x+1)$ yards from her starting point. What is the distance, in yards, she travels?

[hide]Just use Pythagorean:
$x^{2}+(2x-1)^{2}= (2x+1)^{2}$
$x^{2}+4x^{2}-4x+1 = 4x^{2}+4x+1$
$x^{2}-8x = 0$
$x = 0$ or $x = 8$
Depending on whether you consider moving -1 yards west to not count, $x = 0$ might be considered extranneous.  If not, she has traveled 1 yard.  Otherwise, she's traveled 23 yards.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hello,
in my algebra notes the action of a group A of automorphisms of the group G, on the group G is said to be semiregular if $ \forall a \in A,a \neq 1$ it is $ C_G(a)\equal{}1$.

How is defined the centralizer of an [i]automorphism?[/i], $ C_G(a)$??

thanks

The set of automorphisms forms a group, and this just means the centralizer, in the usual sense of that group: the set of all automorphisms $ b$ such that $ ab\equal{}ba$.

ab and ba as automorphisms so the equality has to be intended applied to every g in G right?

There is another possible interpretation. Since $ A$ operates on $ G$ by groupautomorphisms one can form the semidirect product $ S \equal{} G\ltimes A$ of $ A$ and $ G$. Now $ A$ and $ G$ are subgroups of $ S$ and $ C_G(a)$ can be interpretated as the usual centralizer: $ C_G(a) \equal{} \{x\in G | a\cdot x \equal{} x\cdot a\}$ where the product is taken in $ S$. Using the definition of this multiplication one can rewrite the definition as $ C_G(a) \equal{} \{x\in G | ^ax \equal{} x\}$, i.e. $ C_G(a)$ the set of all fixpoints of $ a$.

This interpretation of centralizers (and analogue normalizers) is used by Aschbacher. And, more important, this interpretation makes the definition of "semiregular" equal to the usual one.

Yes, I found it in Aschbacher's "Finite Group Theory". So it means that every non trivial automorphism has no fixed points?

thanks

[quote="dphexen"]Yes, I found it in Aschbacher's "Finite Group Theory". So it means that every non trivial automorphism has no fixed points?[/quote]

Correct. No non-identity automorphism in A fixes any non-identity element in G, so that the permutation group A acts semiregularly on the set of non-identity elements of G.

Examples are the multiplicative group A of a field acting on the additive group G of the field, or more generally a Frobenius complement on the Frobenius kernel. A very specific example is the multiplicative group {+-1} acting on the additive group {-1,0,1} of the field of three elements.


geometry trigonometry circumcircle vector algorithm inequalities incenter

I made this thread to see how many triangle formulas people can come up with.

Please post as many as possible =)  :roll:

[b]Heron's formula:[/b] The area of a triangle is $\sqrt{s(s-a)(s-b)(s-c)}.$

[b]Pythagoras' Theorem:[/b] In a right triangle, $a^{2}+b^{2}=c^{2}.$

In all triangles, $m\angle A+m\angle B+m\angle C=180.$

[b]SSS Postulate:[/b] If all three sides of a triangle are congruent to all three sides of another triangle, the triangles are congruent.

[b]SAS Postulate (angle touches both sides):[/b] If two sides and an angle of a triangle are congurent to two sides and an angle of another tirangle, the triangles are congruent.

[b]ASA Postulate (both angles touch side):[/b] If two angles and a side of a triangle are congruent to two angles and a side of another triangle, the triangles are congruent.

[b]AAS Theorem (angle inbetween angle and side):[/b] If two angles and a side of a triangle are congruent to two angles and a side of another triangle, the triangles are congruent.

Law of Cosines: $a^{2}=b^{2}+c^{2}-2bc\cos A$

Triangle Area: $\frac{1}{2}absinC$

Sine Rule: $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R$

Circumcircle: $Area=\frac{abc}{4R}$

Incircle: $Area=sr$  s: semiperimetre

among some I have found today.

Triangle in circle

A= abc/4r

Circle in Triangle

A=RS

and to find the length of angle bisector

2/a+b root abs(s-c) = length of angle bisector of C

Area of Triangle

$\frac{1}{2}bh$ where b is the length of the base and h is the height.

Menalaus and Ceva's Theorems are good too.

[color=darkred][b]Exercise 1.[/b] Prove that in an isosceles triangle $ABC\ (AB=AC)$ for any point $X\in BC$ there is the relation $AX^{2}=AB^{2}+\overline{XB}\cdot\overline{XC}\ .$
Let $ABC$ be a triangle for which exists at least a point $X$ which belongs to the sideline $BC$ so that exists the relation $AX^{2}=AB\cdot AC+\overline{XB}\cdot\overline{XC}\ .$
Prove that the triangle $ABC$ is isosceles or the ray $[AX$ is the bisector of the angle $\widehat{BAC}\ .$

[b]Exercise 2.[/b] Let $ABC$ be a triangle for which $b\ne c\ .$ Denote the interior bisector $[AD$ and the exterior bisector $[AD_{1}$ of the angle $\widehat{BAC}\ ,$ where the points $D$ and $D_{1}$ belong to the sideline $BC\ .$ Prove that there are the following remarkable relations : $AD^{2}=bc-DB\cdot DC$ and $AD1^{2}=D_{1}B\cdot D_{1}C-bc\ .$
Infer that $AD^{2}=\frac{2}{b+c}\cdot\sqrt{bcp(p-a)}=\frac{2bc}{b+c}\cdot\cos\frac{A}{2}$ and $AD_{1}^{2}=\frac{2}{|b-c|}\cdot\sqrt{bc(p-b)(p-c)}=\frac{2bc}{|b-c|}\cdot\sin\frac{A}{2}\ .$

[b][u]I present lower down some remarkable relations in any triangle[/u][/b] $ABC$ [u][b](which became afterwards formulas) and a multiple characterization for [/u][/b] $A=90^{\circ}\ .$

$\{\begin{array}{c}p(p-a)+(p-b)(p-c)=bc\\\\ p(p-a)-(p-b)(p-c)=bc\cos A\end{array}$ $\Longrightarrow$ $\{\begin{array}{c}\sin \frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}\\\\ \cos\frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}\\\\ \tan\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{p(p-a)}}\end{array}$

$\{\begin{array}{c}a=b\cos C+c\cos B\\\\ b=c\cos A+a\cos C\\\\ c=a\cos B+b\cos A\end{array}$ $\|\begin{array}{c}\ \ \bullet\ (-a)\ |\ \ \bullet\ (+a)\ |\ \ \bullet\ (+a)\\\\ \ \ \bullet\ (+b)\ |\ \ \bullet\ (-b)\ |\ \ \bullet\ (+b)\\\\ \ \ \bullet\ (+c)\ |\ \ \bullet\ (+c)\ |\ \ \bullet\ (-c)\end{array}|\ \ \bigoplus$ $\Longrightarrow$ $\{\begin{array}{c}b^{2}+c^{2}-a^{2}=2bc\cos A\\\\ c^{2}+a^{2}-b^{2}=2ca\cos B\\\\ a^{2}+b^{2}-c^{2}=2ab\cos C\end{array}$

$4m_{a}^{2}=b^{2}+c^{2}+2bc\cos A\ ;\ S=r_{a}(p-a)=$ $r_{b}(p-b)=r_{c}(p-c)=p(p-a)\tan\frac{A}{2}=$ $2R^{2}\sin A\sin B\sin C=\frac{1}{4}(b^{2}+c^{2}-a^{2})\tan A\ .$

$=====================$ $===========================$ $============================$

$A=90^{\circ}\Longleftrightarrow$ $\{\begin{array}{c}b^{2}+c^{2}=a^{2}\\\\ 2p(p-a)=bc\\\\ 2(p-b)(p-c)=bc\\\\ a=2m_{a}\\\\ a=2R \end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}r=p-a\\\\ r_{a}=p\\\\ r_{b}=p-c\\\\ r_{c}=p-b\\\\ 2R+r=p\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}p(p-a)=(p-b)(p-c)\\\\ 2S=bc\ \ ;\ \ bc=ah_{a}\\\\ S=p(p-a)\\\\ S=(p-b)(p-c)\\\\ S=aR\cos B\cos C\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}\underline{\mathrm{Exercises\ : }}\\\\ \cos B+\cos C=\sin B+\sin C\\\\ a^{2}+b^{2}+c^{2}=8R^{2}\\\\ 4S=a^{2}\sin 2B\\\\ b\sin B=(a-c)(1+\cos B)\mathrm{\ ,\ if\ }C\ne 90^{\circ}\\\\ \cos 2A+\cos 2B+\cos 2C=-1\end{array}$

[b]Question.[/b] What become these last equivalencies if $A<90^{\circ}$ or $A>90^{\circ}\ ?$ For example, $A>90^{\circ}\Longleftrightarrow$ $b^{2}+c^{2}<a^{2}$ $\Longleftrightarrow$ $a>2m_{a}$ $\Longleftrightarrow$ $r>p-a$ a.s.o.

[b]Exercise (bonus !).[/b] Let $ABC$ be a triangle for which $\underline{b\ne c}\ .$ Prove that $\boxed{\ b\cdot\tan\frac{B}{2}-c\cdot\tan\frac{C}{2}=(b-c)\cdot\tan\frac{A}{2}\Longleftrightarrow A=90^{\circ}\ }\ .$[/color]

[color=darkblue]I presented lower up (in the my last reply) some simple exercises which use the mentioned relations and formulas. Here is a "block" of exercises :[/color]

$1.\blacktriangleright\{\begin{array}{c}r_{a}+r_{b}+r_{c}=4R+r\\\\ r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}=p^{2}\\\\ r_{a}r_{b}r_{c}=p^{2}r\\\\ A=90^{\circ}\Longleftrightarrow r+r_{a}+r_{b}+r_{c}=2p\end{array}|$ $\mathrm{\ \ \ and\ \ \ }$ $|\begin{array}{c}(b^{2}-c^{2})^{2}+16S^{2}=4a^{2}m_{a}^{2}\\\\ ab+bc+ca=p^{2}+r^{2}+4Rr\\\\ \boxed{\ 3r\sqrt 3\le p\le \frac{3R\sqrt 3}{2}\ }\end{array}\}$

$2.\blacktriangleright\{\begin{array}{c}\sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=\frac{p}{R}\\\\ \cos A+\cos B+\cos C=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=1+\frac{r}{R}\\\\ \sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C=\frac{2pr}{R^{2}}\\\\ \cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C\end{array}$ $\Longrightarrow$  $\{\begin{array}{c}\sin A+\sin B+\sin C\le \frac{3\sqrt 3}{2}\\\\ \cos A+\cos B+\cos C\le \frac{3}{2}\\\\ \sin A\sin B\sin C\le \frac{3\sqrt 3}{8}\\\\ \cos A\cos B\cos C\le \frac{1}{8}\end{array}|$ $\mathrm{\ \ \ and\ \ \ }$  $|\begin{array}{c}\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\le \frac{1}{8}\\\\ \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\le \frac{3\sqrt 3}{8}\\\\ \sin 2A+\sin 2B+\sin 2C\le\frac{3\sqrt 3}{2}\\\\ \cos 2A+\cos 2B+\cos 2C\ge-\frac{3}{2}\end{array}\}$

$3.\blacktriangleright\{\begin{array}{c}h_{a}=\frac{2p(p-a)x}{x^{2}+p(p-a)}\Longleftrightarrow x\in \{r_{b}\ ,\ r_{c}\}\\\\ h_{a}=\frac{2(p-b)(p-c)x}{|x^{2}-(p-b)(p-c)|}\Longleftrightarrow x\in \{r\ ,\ r_{a}\}\end{array}$ \[\blacktriangleleft\boxed{\ \mathrm{\ \ SUCCESS\ !\ }\ }\blacktriangleright\]

Nice, Virgil Nicula.

Another one:
Area of a triangle is equal to half of the vector product of two defining vectors originating from the same vertex. (I'm not sure how to write that out nicely...).

And of course:
[url=http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf]Shoelace Algorithm[/url]

The exterior angles sum to 360 in a triangle.

Area of an equilateral triangle
$A = \frac{\sqrt{3}}{4}s^{2}$
where s is the side length.

Law of tangents:

$\frac{a-b}{a+b}= \frac{tan\frac{1}{2}(A-B)}{tan\frac{1}{2}(A+B)}$

where a and b are side lengths and A and B are angles.

$Area = \frac{bc*sinA}{2}$

$Area = \frac{2a*sinB*sinC}{2sinA}$

$cot A = \frac{2a+2b+2c}{4*Area}$

[hide="A nice ... trifle."]$\{\begin{array}{ccccc}a+(p-a)=p & \Longleftrightarrow & \frac{2S}{h_{a}}+\frac{S}{r_{a}}=\frac{S}{r}& \Longleftrightarrow & \boxed{\ \frac{2}{h_{a}}+\frac{1}{r_{a}}=\frac{1}{r}\ }\\\\ (p-b)+(p-c)=a & \Longleftrightarrow & \frac{S}{r_{b}}+\frac{S}{r_{c}}=\frac{2S}{h_{a}}& \Longleftrightarrow & \boxed{\ \frac{1}{r_{b}}+\frac{1}{r_{c}}=\frac{2}{h_{a}}\ }\end{array}$ $\Longrightarrow$ $\boxed{\ h_{a}=\frac{2rr_{a}}{r_{a}-r}=\frac{2r_{b}r_{c}}{r_{b}+r_{c}}\ }\ .$[/hide]

The notations used are as in O.Bottema,R.Z.Djordjevic,D.S.Mitrinovic and P.M.Vasic  [i] Geometric Inequalities[/i],Wolters-Noordhoff Publ.Groningen 1969.

[b](I)[/b]In the following $X_{1},X_{2},X_{3}$ are the roots of $\begin{array}{|c|}\hline P(X)=X^{3}-AX^{2}+Bx-C \\ \hline\end{array}.$ \[\begin{array}{|c|c|c|c|c|}\hline\hline{\bf A}&{\bf B}&{\bf C}&{\bf X_{j}}&{\bf \Delta}: ={\bf 2A^{2}-6B}\\ \hline\hline &&&&\\ 2s & s^{2}+r(4R+r)& 4srR & a,b,c & 2\left(s^{2}-3r(4R+r)\right)\\ &&&&\\ \hline &&&& 2(4R+r-s\sqrt{3})\times \\ 4R+r& s^{2}& s^{2}r& r_{a},r_{b},r_{c}& \times (4R+r+s\sqrt{3})\\ &&&&\\ \hline &&&&\\ &&&&E_{1}: = \frac{(r^{2}+s^{2}-2rR)^{2}}{2r^{2}R^{2}}\\ \frac{r^{2}+s^{2}-2rR}{2rR}& \frac{2\left(s^{2}-rR-r^{2}\right)}{2rR}& \frac{2rR+r^{2}+s^{2}}{2rR}& \frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}& \Delta=E_{1}-E_{2}=... \\ &&&& E_{2}=\frac{12rR(s^{2}-rR-r^{2})}{2r^{2}R^{2}}\\ &&&&\\ \hline &&& b+c-a & \\ \frac{2F}{r}& 4r(4R+r)& 8rF & c+a-b&8\left(s^{2}-3r(4R+r)\right) \\ &&&a+b-c& \\ \hline &&&&E_{1}: =\frac{\left(s^{2}+r(4R+r)\right)^{2}}{2R^{2}}\\ \frac{s^{2}+r(4R+r)}{2R}& \frac{2F^{2}}{rR}& \frac{2F^{2}}{R}& h_{a},h_{b},h_{c}&\Delta=E_{1}-E_{2}\\ &&&& E_{2}=\frac{24rRs^{2}}{2R^{2}}\\ \hline\hline \end{array}\] [b](II)[/b] If $\Delta ABC$ has the (length) sides $a, b, c$ then there exists a triangle $\Delta A_{1}B_{1}C_{1}$ having the sides   $\sqrt{a},\sqrt{b},\sqrt{c}$.
  Let $F_{1}$ be area of triangle $\Delta A_{1}B_{1}C_{1}$. Then \[F_{1}=\frac{1}{2}\sqrt{r(r+4R)}\; \; \; \; ,\; \; \; \; \; \; \cos{A_{1}}=\frac{s-a}{\sqrt{bc}}\; ,\] and observe that $\Delta A_{1}B_{1}C_{1}$ is an [b]acute triangle[/b].

[b](III) Triangle $OIH$ .[/b] $O: =$ circumcentre , $I: =$incentre . $H: =$orthocentre . \[\begin{array}{ccl}\parallel OI\parallel &=&\sqrt{R(R-2r)}\\ \parallel OH \parallel &=&\sqrt{9R^{2}+2r(r+4R)-2s^{2}}\\ \parallel IH\parallel &=&\sqrt{3r^{2}+4rR+4R^{2}-s^{2}}\\ \end{array}\; .\] If $F_{*}$ is area of $\Delta OIH$ , then \[\begin{array}{|c|}\hline \\ F_{*}^{2}=\;-\; \frac{1}{16}\left[s^{4}-2(2R^{2}+10rR-r^{2})s^{2}+r(r+4R)^{3}\right]=\\ \\ =  \frac{(a-b)^{2}(b-c)^{2}(c-a)^{2}}{64r^{2}}\\ \\ \hline \end{array}\] [b]Questions:[/b][color=green]
i) It's true that $\cos{OIH}<0$ ?
ii) $\Delta OIH$ is equilateral ?
iii)In which case, i.e. which are the triangles having $\Delta OIH$ isoscel ?
iV) Write $F_{*}^{2}\ge 0$ in order to obtain interesting inequalities between $s,r , R .$
(See R.Sondat,Nouv.Ann.Math.(3)10(1891) 43-47 and W.J.Blundon, Canad.Math.Bull. 8(1965)615-626.)

[/color]

[color=darkblue][b]Thanks[/b], Dl. Alex. Lupas, for the your table in ... LaTeX ! Never I knew how achieve it. I hope soon to present a similar thing. To-morrow, here is 1:31 AM. Numai bine ![/color]

45-45-90 Triangle

Sides of the triangle are x, x, and $x\sqrt{2}$.

Well, only if you can determine the base and height.  If you don't know that, you'll have to use other formulas.

About the $\frac{s^{2}\sqrt{3}}{4}$ formula:

My teacher tought me to memorize it as the 1-2-3-4 formula:

1 s to the 2 power times the sqrt of 3 over 4.

I just thought that was a good mnemonic-type thing.

Very nice and interesting the Flip2004's table and the adition w.r.t. the triangle $A_{1}B_{1}C_{1}\ (\sqrt a,\sqrt b\sqrt c)\ ,$ in which, I add, $\boxed{\ R_{1}=\sqrt{\frac{pR}{4R+r}}\ }\ .$
[hide="Hint."]In any triangle $ABC$ exists the relation $\sum a^{2}=8R^{2}\left(1+\prod\cos A\right)\ .$[/hide]

[quote="Virgil Nicula"]... in triangle $A_{1}B_{1}C_{1}\ (\sqrt a,\sqrt b\sqrt c)\ ,$ in which, I add, $\boxed{\ R_{1}=\sqrt{\frac{pR}{4R+r}}\ }\ .$
[/quote]
Indeed, and using an inequality proved by O.Kooi, Simon Stevin 32(1958)97-101, it may be shown that \[k_{1}\cdot R(R-2r)\le \frac{R^{2}}{3}-R_{1}^{4}\le k_{2}\cdot R(R-2r)\] with $k_{1}=\frac{1}{12}$  and $k_{2}=\frac{4}{9}$ .

[b]Some rarer identities with[/b] $\cos A$ a.s.o. :

$\begin{array}{c}\sum a\cos B\cos C=\frac{S}{R}\ .\\\\ \{\begin{array}{c}\sum\cos A=1+\frac{r}{R}\le\frac{3}{2}\ .\\\\ \sum\cos B\cos C=\frac{p^{2}+r^{2}}{4R^{2}}-1\ .\\\\ \prod\cos A=\frac{\sum a^{2}}{8R^{2}}-1=\frac{p^{2}-(2R+r)^{2}}{4R^{2}}\ .\end{array}\end{array}$ $\ldots\ \maltese\ \ldots$ Examples : $\{\begin{array}{c}A=90^{\circ}\Longleftrightarrow\{\begin{array}{c}a\cos B\cos C=\frac{S}{R}\ .\\\\ p^{2}+r^{2}=4R^{2}(1+\cos B\cos C)\ .\\\\ p=2R+r\ .\end{array}\\\\ a^{2}+b^{2}+c^{2}\le 9R^{2}\Longrightarrow\cos A\cos B\cos C\le \frac{1}{8}\ .\end{array}$

[quote="mysmartmouth"]
And of course:
[url=http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf]Shoelace Algorithm[/url][/quote]

For a triangle, you could also do:

If the triangle has vertices (a,b), (c,d), (e,f)

Determinant of this 3x3 matrix:
{
 {a b 1}
 {c d 1}
 {e f 1}}
Multiplied by [u]+[/u] $\frac12$

[b]Exercise :[/b] $a^{2}+b^{2}+c^{2}=8R^{2}\Longleftrightarrow 90^{\circ}\in\{A,B,C\}\ .$ See the last my reply.

[color=darkred]$A=90^{\circ}\Longleftrightarrow$ $\{\begin{array}{c}b^{2}+c^{2}=a^{2}\\\\ 2p(p-a)=bc\\\\ 2(p-b)(p-c)=bc\\\\ a=2m_{a}\\\\ a=2R \end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}r=p-a\\\\ r_{a}=p\\\\ r_{b}=p-c\\\\ r_{c}=p-b\\\\ 2R+r=p\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}p(p-a)=(p-b)(p-c)\\\\ 2S=bc\ \ ;\ \ bc=ah_{a}\\\\ S=p(p-a)\\\\ S=(p-b)(p-c)\\\\ S=aR\cos B\cos C\end{array}$ $\Longleftrightarrow$ $\{\begin{array}{c}\underline{\mathrm{Exercises\ : }}\\\\ p^{2}+r^{2}=a^{2}+bc\\\\ \cos B+\cos C=\sin B+\sin C\\\\ a^{2}+b^{2}+c^{2}=8R^{2}\\\\ b\sin B=(a-c)(1+\cos B)\mathrm{\ ,\ if\ }C\ne 90^{\circ}\\\\ b\tan\frac{B}{2}-c\tan\frac{C}{2}=(b-c)\tan\frac{A}{2}\mathrm{\ ,\ if\ }b\ne c\\\\ \frac{S}{aR}+(\frac{a}{b+c})^{2}=1\end{array}$

[b]Question.[/b] What become these last equivalencies if $A<90^{\circ}$ or $A>90^{\circ}\ ?$ For example, $A>90^{\circ}\Longleftrightarrow$ $b^{2}+c^{2}<a^{2}$ $\Longleftrightarrow$ $a>2m_{a}$ $\Longleftrightarrow$ $r>p-a$ a.s.o.[/color]

[quote="now a ranger"]tetrahedron

s cubed root 2 over 12[/quote]
you realize that that says nothing whatsoever.
It should be, the volume of a regular tetrahedron is
$\frac{s^{3}\sqrt{2}}{12}$. If the tetrahedron isn't regular, the formula is a lot more complicated.
Another is area of a triangle is $\frac{ab\sin(C)}{2}$ where the sides are a, b, and c, and angle C is opposite c.
Also the triangle inequality says that if a, b, and c are the sides of a triangle then
$a+b>c$
$a+c>b$
$b+c>a$

For $O$ the circumcentre and $H$ the orthocentre:

$OH^{2}=R^{2}(1-8 \cos A \cos B \cos C)$

could anyone explain theorem of medians for me?

You mean $m_{c}=\sqrt{\frac{a^{2}}{2}+\frac{b^{2}}{2}-\frac{c^{2}}{4}}$?

This means that if the sides of a triangle are a, b, and c, the median to c has length shown above.

[color=darkblue]Consider the parallelogram $ABDC\ .$ Apply the [u]generalized Pythagoras' theorem[/u] to the side $AD=2m_{a}$ in the triangle $ABD\ :$

$AD^{2}=AB^{2}+BD^{2}-2\cdot AB\cdot BD\cdot\cos \widehat{ABD}$ $\Longrightarrow$ $4m^{2}_{a}=b^{2}+c^{2}-2bc\cos\widehat{ABD}$ $\Longrightarrow$

$4m^{2}_{a}=b^{2}+c^{2}+2bc\cos A$ $\Longrightarrow$ $4m^{2}_{a}=b^{2}+c^{2}+(b^{2}+c^{2}-a^{2})$ $\Longrightarrow$ $\boxed{\ 4m^{2}_{a}=2(b^{2}+c^{2})-a^{2}\ }\ .$

[b]Remark.[/b]Is easily remembered : $\|\begin{array}{c}b^{2}+c^{2}-2bc\cos A=a^{2}\\\\ b^{2}+c^{2}+2bc\cos A=4m^{2}_{a}\end{array}\ .$[/color]

[quote="13375P34K43V312"][quote="bpms"]
Another is area of a triangle is $\frac{ab\sin(C)}{2}$ where the sides are a, b, and c, and angle C is opposite c.
[/quote]
Also $\frac{bc\sin(A)}$ and $\frac{ac\sin(B)}$.[/quote]
YOu messed up the formulas. Also, that can be said WLOG.


geometry
inequalities
number theory
IMO
IMO Shortlist

IMO 1959-1966 - partial collection of LongList problems:
(in the second post, below, you will find links to where you should discuss the solutions of the problems)

[b][size=150]Problems on the board: [/size][/b]

[u][size=150]By category:[/size][/u]

 [b]ALGEBRA[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16216]Question number  18  ( hungary 1 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16207]Question number  9  ( Romania 1 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16260]Question number  48  ( Sovjet Union 5 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16247]Question number  35  ( Poland 5 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16258]Question number  46  ( Yugoslavia 3 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16252]Question number  40  ( Socialist Republic of Czechoslovakia 6 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16241]Question number  31  ( Romania 5 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16234]Question number  25  ( Democratic Republic of Germany 3 )[/url]
[b] COMBINATORICS [/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16456]Question number  4  ( Poland 1 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16264]Question number  52  ( Sovjet Union 9 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16257]Question number  45  ( Socialist Republic of Czechoslovakia 10 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16262]Question number  51  ( Sovjet Union 8 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16254]Question number  43  ( Socialist Republic of Czechoslovakia 9 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16232]Question number  24  ( Poland 4 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16212]Question number  14  ( Poland 2 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16205]Question number  8  ( Sovjet Union 4 )[/url]
[b] GEOMETRY[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16199]Question number  6  ( Sovjet Union 2 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16237]Question number  28  ( Socialist Republic of Czechoslovakia 5 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16243]Question number  32  ( Bulgaria 6 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16215]Question number  17  ( Romania 2 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16251]Question number  39  ( Romania 7 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16255]Question number  44  ( Yugoslavia 2 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16253]Question number  41  ( Socialist Republic of Czechoslovakia 7 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16248]Question number  36  ( Poland 6 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16269]Question number  56  ( Sovjet Union 13 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16263]Question number  50  ( Sovjet Union 7 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16249]Question number  37  ( Poland 7 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16267]Question number  55  ( Sovjet Union 12 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16265]Question number  53  ( Sovjet Union 10 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16261]Question number  49  ( Sovjet Union 6 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16259]Question number  47  ( Romania 8 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16250]Question number  38  ( Romania 6 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16246]Question number  33  ( Bulgaria 7 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16236]Question number  27  ( Democratic Republic of Germany 4 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16213]Question number  15  ( Poland 3 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16228]Question number  23  ( Bulgaria 5 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16217]Question number  19  ( Hungary 2 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16220]Question number  22  ( Bulgaria 4 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16219]Question number  21  ( Bulgaria 3 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16218]Question number  20  ( Hungary 3 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16214]Question number  16  ( Socialist Republic of Czechoslovakia 3 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16206]Question number  7  ( Sovjet Union 3 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16193]Question number  1  ( Socialist Republic of Czechoslovakia 1 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16195]Question number  3  ( Bulgaria 1 )[/url]
[b] INEQUALITIES[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16235]Question number  26  ( Socialist Republic of Czechoslovakia 4 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16242]Question number  30  ( Romania 4 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16266]Question number  54  ( Sovjet Union 11 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16256]Question number  42  ( Socialist Republic of Czechoslovakia 8 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16211]Question number  13  ( Yugoslavia 1 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16194]Question number  2  ( Democratic Republic of Germany 1 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16196]Question number  5  ( Sovjet Union 1 )[/url]
[b] NUMBERTHEORY[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16240]Question number  29  ( Romania 3 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16245]Question number  34  ( Bulgaria 8 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16208]Question number  11  ( Socialist Republic of Czechoslovakia 2 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16210]Question number  12  ( Bulgaria 2 )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16209]Question number  10  ( Democratic Republic of Germany 2 )[/url]

[u][size=150]By Country:[/size][/u]

[b]Bulgaria[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16195]1.) Question 3 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16210]2.) Question 12 ( number theory )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16219]3.) Question 21 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16220]4.) Question 22 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16228]5.) Question 23 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16243]6.) Question 32 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16246]7.) Question 33 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16245]8.) Question 34 ( number theory )[/url]

[b]Democratic Republic of Germany[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16194]1.) Question 2 ( inequalities )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16209]2.) Question 10 ( number theory )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16234]3.) Question 25 ( algebra )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16236]4.) Question 27 ( geometry )[/url]

[b]Hungary[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16216]1.) Question 18 ( algebra )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16217]2.) Question 19 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16218]3.) Question 20 ( geometry )[/url]

[b]Poland[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16456]1.) Question 4 ( combinatorics/geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16212]2.) Question 14 ( combinatorics )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16213]3.) Question 15 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16232]4.) Question 24 ( combinatorics )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16247]5.) Question 35 ( algebra )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16248]6.) Question 36 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16249]7.) Question 37 ( geometry )[/url]

[b]Romania[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16207]1.) Question 9 ( algebra )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16215]2.) Question 17 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16240]3.) Question 29 ( number theory )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16242]4.) Question 30 ( inequalities )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16241]5.) Question 31 ( algebra )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16250]6.) Question 38 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16251]7.) Question 39 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16259]8.) Question 47 ( geometry )[/url]

[b]Socialist Republic of Czechoslovakia[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16193]1.) Question 1 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16208]2.) Question 11 ( number theory )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16214]3.) Question 16 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16235]4.) Question 26 ( inequalities )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16237]5.) Question 28 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16252]6.) Question 40 ( algebra )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16253]7.) Question 41 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16256]8.) Question 42 ( inequalities )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16254]9.) Question 43 ( combinatorics )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16257]10.) Question 45 ( combinatorics )[/url]

[b]Sovjet Union[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16196]1.) Question 5 ( inequalities )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16199]2.) Question 6 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16206]3.) Question 7 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16205]4.) Question 8 ( combinatorics )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16260]5.) Question 48 ( algebra )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16261]6.) Question 49 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16263]7.) Question 50 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16262]8.) Question 51 ( combinatorics )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16264]9.) Question 52 ( combinatorics )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16265]10.) Question 53 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16266]11.) Question 54 ( inequalities )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16267]12.) Question 55 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16269]13.) Question 56 ( geometry )[/url]

[b]Yugoslavia[/b]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16211]1.) Question 13 ( inequalities )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16255]2.) Question 44 ( geometry )[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16258]3.) Question 46 ( algebra )[/url]

[u][size=150]By official ranking[/size][/u] (mainly for reference to the PDF document)

1.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16193]Socialist Republic of Czechoslovakia 1 ( geometry )[/url]
2.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16194]Democratic Republic of Germany 1 ( inequalities )[/url]
3.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16195]Bulgaria 1 ( geometry )[/url]
4.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16456]Poland 1 ( combinatorics/geometry )[/url]
5.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16196]Sovjet Union 1 ( inequalities )[/url]
6.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16199]Sovjet Union 2 ( geometry )[/url]
7.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16206]Sovjet Union 3 ( geometry )[/url]
8.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16205]Sovjet Union 4 ( combinatorics )[/url]
9.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16207]Romania 1 ( algebra )[/url]
10.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16209]Democratic Republic of Germany 2 ( number theory )[/url]
11.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16208]Socialist Republic of Czechoslovakia 2 ( number theory )[/url]
12.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16210]Bulgaria 2 ( number theory )[/url]
13.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16211]Yugoslavia 1 ( inequalities )[/url]
14.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16212]Poland 2 ( combinatorics )[/url]
15.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16213]Poland 3 ( geometry )[/url]
16.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16214]Socialist Republic of Czechoslovakia 3 ( geometry )[/url]
17.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16215]Romania 2 ( geometry )[/url]
18.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16216]hungary 1 ( algebra )[/url]
19.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16217]Hungary 2 ( geometry )[/url]
20.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16218]Hungary 3 ( geometry )[/url]
21.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16219]Bulgaria 3 ( geometry )[/url]
22.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16220]Bulgaria 4 ( geometry )[/url]
23.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16228]Bulgaria 5 ( geometry )[/url]
24.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16232]Poland 4 ( combinatorics )[/url]
25.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16234]Democratic Republic of Germany 3 ( algebra )[/url]
26.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16235]Socialist Republic of Czechoslovakia 4 ( inequalities )[/url]
27.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16236]Democratic Republic of Germany 4 ( geometry )[/url]
28.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16237]Socialist Republic of Czechoslovakia 5 ( geometry )[/url]
29.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16240]Romania 3 ( number theory )[/url]
30.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16242]Romania 4 ( inequalities )[/url]
31.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16241]Romania 5 ( algebra )[/url]
32.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16243]Bulgaria 6 ( geometry )[/url]
33.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16246]Bulgaria 7 ( geometry )[/url]
34.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16245]Bulgaria 8 ( number theory )[/url]
35.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16247]Poland 5 ( algebra )[/url]
36.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16248]Poland 6 ( geometry )[/url]
37.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16249]Poland 7 ( geometry )[/url]
38.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16250]Romania 6 ( geometry )[/url]
39.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16251]Romania 7 ( geometry )[/url]
40.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16252]Socialist Republic of Czechoslovakia 6 ( algebra )[/url]
41.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16253]Socialist Republic of Czechoslovakia 7 ( geometry )[/url]
42.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16256]Socialist Republic of Czechoslovakia 8 ( inequalities )[/url]
43.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16254]Socialist Republic of Czechoslovakia 9 ( combinatorics )[/url]
44.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16255]Yugoslavia 2 ( geometry )[/url]
45.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16257]Socialist Republic of Czechoslovakia 10 ( combinatorics )[/url]
46.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16258]Yugoslavia 3 ( algebra )[/url]
47.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16259]Romania 8 ( geometry )[/url]
48.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16260]Sovjet Union 5 ( algebra )[/url]
49.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16261]Sovjet Union 6 ( geometry )[/url]
50.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16263]Sovjet Union 7 ( geometry )[/url]
51.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16262]Sovjet Union 8 ( combinatorics )[/url]
52.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16264]Sovjet Union 9 ( combinatorics )[/url]
53.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16265]Sovjet Union 10 ( geometry )[/url]
54.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16266]Sovjet Union 11 ( inequalities )[/url]
55.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16267]Sovjet Union 12 ( geometry )[/url]
56.) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16269]Sovjet Union 13 ( geometry )[/url]

ok I broke Pierre's record for longest post on the board :D

Wow. Thank you very much, Peter. Soon we will introduce an "solved-symbol-emoticon". Then we can indicate which problems were solved up to now though we always would like to see new solutions. This means a problem is never closed.

Indeed Orl. I just wondered, how will this happen exactly?
Perhaps a field in posting "Do you think this is a full solution?" (Y/N)
could be a good idea... It will require no work from mods or from anyone, but you don't get "Solved" just because someone points out an error or asks a question on it. :)

[quote="orl"]Wow. Thank you very much, Peter.[/quote]
I think you're the one to be thanked on this ;)

Update: the last missing questions have been updated the numbering etc. should be ok now :)


farz konid $x,y,z > 0$ va dashte bashim $xy+yz+zx=1$ . sabet konid :
$\frac{x}{x^{2}+1}+\frac{y}{y^{2}+1}+\frac{z}{z^{2}+1}\leq \frac{3\sqrt 3}{4}$

[quote="N.T.TUAN"][hide="A hint"]Put $x=\tan{\frac{A}{2}},$ e.t.c. Here $A,B,C$ are three angles of a triangle.[/hide][/quote]
nice hint [b]N.T.TUAN[/b]...

in a triangle we have:

$\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{B}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{A}{2}}=1$

so here we can assume that $x=\tan{\frac{A}{2}},y=\tan{\frac{B}{2}},z=\tan{\frac{C}{2}}$

now multiply the both sides of the inequality in $2$,so we have to prove that:

$\frac{2x}{x^{2}+1}+\frac{2y}{y^{2}+1}+\frac{2z}{z^{2}+1}\leq \frac{3\sqrt{3}}{2}$

or if we put $\tan{\frac{A}{2}},...$ instead of $x,...$ we have to prove that:

$\frac{2\tan{\frac{A}{2}}}{\tan^{2}{\frac{A}{2}}+1}+\frac{2\tan{\frac{B}{2}}}{\tan^{2}{\frac{B}{2}}+1}+\frac{2 \tan{\frac{C}{2}}}{\tan^{2}{\frac{C}{2}}+1}\leq \frac{3\sqrt{3}}{2}$

but we know that $\frac{2\tan{\alpha}}{1+\tan^{2}{\alpha}}=\sin{2\alpha}$

so we only need to show that $\sin A+\sin B+\sin C\leq \frac{3\sqrt{3}}{2}$

which can be easily proved by [b]jensen[/b]...

Are oon rahe hal kheili ba Jensen ghashange ;) . hala yeki dige ke inam khoobe
$a+b+c=3$ sabet konid :
$a^{2}b^{2}c^{2}\geq(3-2a)(3-2b)(3-2c)$

avalan tavajoh dashte bashid ke $3-2a=b+c-a$ khob hala mishe farz kard ke $a,b,c$ azlae mosalasan chera ke dar gheire in soorat samte raste namosavi manfi va samte chapesh mosbat mishe ke badihatan bargharare...

pas farz konid $a,b,c$ azlae mosalase $\triangle ABC$ ba mohite $2p$,masahate $S$,shoae dayere mohiti va mohatie $R,r$ bashe...

hokmemoon be in soorat sade mishe:

$8(p-a)(p-b)(p-c)\leq a^{2}b^{2}c^{2}$

khob hala tooye mosalase $ABC$ midoonim $S=\frac{abc}{4R}$ hamchenin midoonim $r=\frac{S}{p}$ pas azin dota natije
mishe ke $abc=4Rrp \Rightarrow a^{2}b^{2}c^{2}=16R^{2}r^{2}p^{2}$ hamchenin tebghe rabete [b]heron[/b] midoonim $S=\sqrt{p (p-a)(p-b)(p-c)}$ pas natije mishe ke $(p-a)(p-b)(p-c)=pr^{2}$

khob hala in dota meghdaro tooye hokm jagozari mikonim,pas hokmemoon mishe:

$8pr^{2}\leq 16R^{2}r^{2}p^{2}\Rightarrow 1\leq 2R^{2}p$ amma dashtim $a+b+c=3$ pas $2p=3$ pas kafie sabet konim $1\leq 3R^{2}$

amma midoonim $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R \Rightarrow a=2R\sin A,...$

hala too rabete $a+b+c=3$ bejaye $a,b,c$ motanazeraye oonaro bezarid natije mishe ke $2R(\sin A+\sin B+\sin C)=3$

pas $\sin A+\sin B+\sin C=\frac{3}{2R}\leq \frac{3\sqrt{3}}{2}\Rightarrow \frac{1}{\sqrt{3}}\leq R$ yeni hamoon chizi ke
mikhastim...

[quote="khashi70"]Vase In namosavie ye rahnamayi mikonam :
age $a,b,c$ azlae yek mosalas bashand (yani $A+B+C=180$) angah $\sin{A}+\sin{B}+\sin{C}$ $\leq$ $\frac{3\sqrt3}{2}$ . albatte shayad in badihi bashe vali tooye in masale aslan be cheshm nemiad ![/quote]
are too in yekiam azin nokte estefade mishe,vali na be shedate ghablie...

[color=darkblue]We have:
\[\sum\frac{x}{x^{2}+1}=\sum\frac{x}{x^{2}+xy+yz+zx}=\sum\frac{x}{(x+y)(x+z)}=\frac{2\sum yz}{\prod (y+z)}\quad (1) \]
On the otherhand, $\prod (y+z)\ge\frac{8}{9}(x+y+z)(xy+yz+zx)\ge\frac{8}{9}\sqrt 3\quad (2)$

From $(1)$ and $(2)$ we have: $\sum\frac{x}{x^{2}+1}\le\frac{3\sqrt 3}{4}$[/color]

afarin babak , doroste . hala yeki dige mizaram harki doost dasht roosh fekr kone
agar $ab+bc+ca+2abc=1$ sabet konid
$\sqrt ab$+$\sqrt bc$+$\sqrt ca$ $\leq$ $\frac{3}{2}$

gharar bedid $\sqrt{ab}=x,\sqrt{bc}=y,\sqrt{ca}=z$ pas farzemoon be shekle $x^{2}+y^{2}+z^{2}+2xyz=1$ dar miad,hokmemoonam be shekle $x+y+z\leq \frac{3}{2}$...
khob az noe taghire moteghaieremoon natije mishe ke avalan $x,y,z\geq 0$ va badesham az farze masale natije mishe ke $x,y,z\leq 1$ pas $x,y,z \in [0,1]$...
khob hala biaid farze masalaro bishtar tahlil konim:
$x^{2}+x(2yz)+(y^{2}+z^{2}-1)=0$ in rabete ye moadele daraje dovom bar hasbe $x$ hastesh,pas rishe mosbatesh barabar mishe ba:
$x=-yz+\sqrt{y^{2}z^{2}-y^{2}-z^{2}+1}$
hala biaiad ino too hokm jagozari konid,pas kafie sabet konim:
$(-yz+\sqrt{y^{2}z^{2}-y^{2}-z^{2}+1})+y+z\leq \frac{3}{2}$
hala biaid hokmo sade konid pas kafie sabet konim:
$-(y-1)(z-1)+\sqrt{(y^{2}-1)(z^{2}-1)}\leq \frac{1}{2}$
baghiasho motmaen nistam doroste ya na chon kheili kasif karie...
[hide]radikalo ye taraf negah darido baghiaro bebarid ye tarafo dar $2$ zarb konido be tavane do beresoonid mishe:

$4(y^{2}-1)(z^{2}-1)\leq 4(y-1)^{2}(z-1)^{2}+4(y-1)(z-1)+!$

$\Rightarrow 0\leq y^{2}(8-8z)+y(20z-8z^{2}-12)+(5-12z+8z^{2})$

in ye moadele daraje dovom bar hasbe $y$e,ke zaribe $y^{2}$ mosbate,pas in tabe noghte minimum dare,hala age in noghte too baze $[0,1]$ bashe kheili khoobe...

yeni halati ke $y=\frac{8z^{2}-20z+12}{16-16z}$ pas chon $y\leq 1$ kheili khoobe age dashte bashim:

$8z^{2}-20z+12\leq 16-16z \Rightarrow 8z^{2}-4z-4\leq 0 \Rightarrow 2z^{2}-z-1\leq 0$

$\Rightarrow (z-1)(2z+1)\leq 0$

ke badihatan bargharare...

pas faghat kafie meghdare tabaro too noghte minimumesh check konim

amma midoonim hasele tabe too noghte minimomesh barabare ba $\frac{\Delta}{4a^{2}}$ pas kafie sabet konim $0\leq \Delta$

ke inam vazehe chon age intori nabashe oonvagh $\Delta<0$ va chon zaribe $y^{2}$ mosbate pas maghadire tabe hamishe mosbateo masale hal mishe,pas $\Delta\geq 0$...
[/hide]

[quote="khashi70"]Karanesho ke behtar nemishe kard masalan age $x=y=z=\frac{1}{2}$ tooye dovvomie mishe : $\sqrt ab+\sqrt bc+\sqrt ca$ = $\frac{3}{2}$
va albatte man yejaye rahe halleto nemifahmam . bebin oon jayi ke miangin zadi ghalate chon $y,z<1$ pas age roo oon radikale miangin bezani
azash in biroon miad $\frac{(1-y)+(1-z)}{2}$ :) . fekr konam ye khorde ziadi zayifesh kardi chon oonjayi ke gofti
$\sqrt{(1-y)(1-z)}\sqrt{2(y+z)}\leq \frac{1}{2}$ ghalate masalan gharar bede $y=\frac{1}{2}$ va $z=\frac{1}{3}$ oon moghe natije bayad beshe ke
$\frac{\sqrt5}{3}\leq \frac{1}{2}$ke ghalate  :)[/quote]
editesh kardam,fek konam in dge doros bashe,azoon jai ke hidden shode be bado tashih kardam

[quote="khashi70"][color=darkred]Let $a,\,b,\,c$ be positive real numbers such that: $ab+bc+ca+2abc=1.$Prove that:
\[\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq\frac{3}{2}\]
[/color]
(Is it true? Because I don't understand Iranian  :) )[/quote]
[hide="Hint."]
[color=darkblue][b]Proof 1.[/b] Put $\sqrt{bc}=\sin\frac{A}{2}.$ e.t.c. with $A,\,B,\,C$ are angles of atriangle.
[b]Proof 2.[/b] Put $x=\sqrt{bc}.$ e.t.c. and denote $s=x+y+z$
Note that: $(s-1)^{2}=2\prod (1-x)\le 2\left(\frac{\sum 1-x}{3}\right)^{3}$[/color][/hide]

[quote="khashi70"]Dar morede rahe hallet fekr konam age $\Delta$ ro hesab konim hal she :).
alave bar oon ye rabete tooye mosalas hast ke mitoonin khodetoon sabet konid : $\cos^{2}+\cos^{2}{B}+\cos^{2}{C}+2\cos{A}\cos{B}\cos{C}=1$
hala age gharar bedin $ab=\cos^{2}{C}$ , $bc=\cos^{2}{A}$ va $ca=\cos^{2}{B}$ oon moghe rabetaro sade konin be :
$\cos{A}$+$\cos{B}$+$\cos{C}\leq \frac{3}{2}$ ke inam ba Jensen vazehe :)[/quote]
are age $\Delta$ ro hesab konimam dar miad,vali kheili kasif kari dare,man yeki ke tarafesh nemiram... :!:
in rahe hal ham mese rahaye ghabli (taghire moteghayere mosalasati too mosalas),rahe kheili ghashangie... :)

[quote="khashi70"]Inja Chera Enghadr bad shode baba yekhorde fa'aliat dashte bashin :(( . ye soale namosavie dige mizaram ke sakhte ! Bebinam ki hallesh mikone :
farz konid $a,b,c>0$ sabet konid :
$a^{2}+b^{2}+c^{2}+2abc+3 \geq (1+a)(1+b)(1+c)$[/quote]
[hide]
aval az hame ye namosavie por karbord ro be dast miarim:

tebghe namosavie shur midoonim $a^{3}+b^{3}+c^{3}+3abc\geq a^{2}b+ab^{2}+a^{2}c+ac^{2}+b^{2}c+bc^{2}$

pas age be do tarafesh ebarate $6abc$ ro ezafi konim va $a^{3}+b^{3}+c^{3}$ ro be tarafe dge namosavi bebarim natije mishe:

$2(a^{2}b+ab^{2}+b^{2}c+bc^{2}+ca^{2}+c^{2}a+3abc)-(a^{3}+b^{3}+c^{3}+a^{2}b+ab^{2}+a^{2}c+ac^{2}+bc^{2}+b^{2}c)\leq 9abc$

pas age do paranteze samte chapo etehad benevisim mishe: $2(ab+bc+ca)(a+b+c)-(a^{2}+b^{2}+c^{2})(a+b+c)\leq 9abc$

pas $2(ab+bc+ca)-(a^{2}+b^{2}+c^{2})\leq \frac{9abc}{a+b+c}$...

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

khob hala mirim sorghe hokmemoon:

bayad sabet konim $a^{2}+b^{2}+c^{2}+2+abc\geq ab+bc+ca+a+b+c$ hala do tarafe ino zarbdare do konid mishe:

$2a^{2}+2b^{2}+2c^{2}+4+2abc\geq 2(ab+bc+ca)+2(a+b+c)$ hala midoonim $a^{2}+1\geq 2a$

pas kafie sabet konim $a^{2}+b^{2}+c^{2}+1+2abc\geq 2(ab+bc+ca)$

amma tebghe oon chizi ke avale esbat goftim darim $2(ab+bc+ca)\leq \frac{9abc}{a+b+c}+a^{2}+b^{2}+c^{2}$

pas kafie sabet konim $1+2abc\geq \frac{9abc}{a+b+c}$

pas kafie sabet konim $a+b+c+2a^{2}bc+2ab^{2}c+2abc^{2}\geq 9abc$

amma ba miangin midoonim ke $a+b+c+a^{2}bc+a^{2}bc+ab^{2}c+ab^{2}c+abc^{2}+abc^{2}\geq 9\sqrt[9]{a^{9}b^{9}c^{9}}=9abc$
yeni hamoon chizi ke mikhastim... :!: [/hide]

ye rahe asoone dige:
ba esteghra rooye a+b+c dar bazeye (i-1,i] ke i=1,2,... mitoonim injoori esbat konim:
baraye i=1 ke (0,1] ke hokm badihie chon
a^2+b^2+c^2+abc+2>=2>=2(a+b+c)>=ab+ac+bc+a+b+c

 & hala age khaste bashim baraye(n,n+1] sabet konim
kafie majmooe ro besoorate a+b+c+x (c>=a>=b)benevisim ke x anghadr bozorg ast ke a+b+c dar bazeye (0,n] gharar begire.hal ba in forooze esbate
 a^2+b^2+(c+x)^2+ab(c+x)+2>=ab+a(x+c)+b(c+x)+a+b+c+x
badihie...chon faghat bayad sabet konim ke x^2+2cx+abx>=ax+bx+x ke chon 2xc>=x(a+b)
& x+ab>=1(bana be nahveye entekhabe x) hokm badihie...

[quote="sunset"]ye rahe asoone dige:
ba esteghra rooye a+b+c dar bazeye (i-1,i] ke i=1,2,... mitoonim injoori esbat konim:
baraye i=1 ke (0,1] ke hokm badihie chon
a^2+b^2+c^2+abc+2>=2>=2(a+b+c)>=ab+ac+bc+a+b+c

 & hala age khaste bashim baraye(n,n+1] sabet konim
kafie majmooe ro besoorate a+b+c+x (c>=a>=b)benevisim ke x anghadr bozorg ast ke a+b+c dar bazeye [0,n] gharar begire.hal ba in forooze esbate
 a^2+b^2+(c+x)^2+ab(c+x)+2>=ab+a(x+c)+b(c+x)+a+b+c+x
badihie...chon faghat bayad sabet konim ke x^2+2cx+abx>=ax+bx+x ke chon 2xc>=x(a+b)
& x+ab>=1(bana be nahveye entekhabe x) hokm badihie...[/quote]
ye moshkeli dare rahe hal oonam ine ke vase farze esteghra bayad 3 ta bazeye $(0,1]$ , $(1,2]$ va $(2,3]$ ro esbat koni masalan rahe to baraye esbate
$a=b=c=\frac{2}{3}$ moshkel dare chon to tooye rahe hallet az in estefade kardi ke $x\geq 1$ . vali benazaram inam esbatesh rahate chon in rabete ye rabeteye daraje dovvom bar hasbe $a$ hast va fekr konam mishe esbatesh kard :)

thanx khashi,havasam be in nabood ke momkene maximuimeshoon az 1 kamtar bashe.vali esbatesh baraye (2,3],(1,2] ham asoone.

Afarin oon namosavi karbordie kheili bedard mikhore baraye mesal 2 ta dige az karbordhash :
1)agar $a,b,c>0$ sabet konid : $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$
2)agar $a,b,c>0$ sabet konid : $a^{2}+b^{2}+c^{2}+2abc+1 \geq 2(ab+bc+ca)$
hala ye namosavie jadid ke inam khoobe :)
3)agar $a,b,c>0$ sabet konid
$\frac{(2a+b+c)^{2}}{2a^{2}+(b+c)^{2}}+\frac{(2b+c+a)^{2}}{2b^{2}+(a+c)^{2}}+\frac{(2c+a+b)^{2}}{2c^{2}+(a+b)^{2}}\leq 8$

Doroste , oonam editesh kardam :) . ye rahe halle dige
chon namosavi hamgene mishe farz kard $a+b+c=1$ ke dar oon soorat kasre avval mishe : $\frac{a^{2}+2a+1}{3a^{2}-2a+1}$
hala be sadegi sabet mishe ke $\frac{a^{2}+2a+1}{3a^{2}-2a+1}\leq \frac{12a+4}{3}$ . hala in namosaviaro jam bezanin va az $a+b+c=1$ estefade konin hokm esbat mishe :)

bahse hamgeni shod...

ba farze hamgenie $a+b+c=1$ natije mishe $a,b,c \in (0,1)$ in ye rahe kasif ba jensen dare...

vali ye rahe asoontar ba hamgeni ine ke farz konim $a+b+c=3$ oonvagh mishe:

$(2a+b+c)^{2}=(a+3)^{2}$ pas samte chape namosavimoon be shekle zir dar miad:

$\frac{(a+3)^{2}}{2a^{2}+(3-a)^{2}}+\frac{(b+3)^{2}}{2b^{2}+(3-b)^{2}}+\frac{(c+3)^{2}}{2c^{2}+(3-c)^{2}}\leq 8$

yekami ke sade konim mishe:

$\frac{a^{2}+6a+9}{3a^{2}-6a+9}+\frac{b^{2}+6b+9}{3b^{2}-6b+9}+\frac{c^{2}+6c+9}{3c^{2}-6c+9}\leq 8$

amma $a^{2}+6a+9=(a^{2}-2a+3)+(8a+6)$ pas $\frac{a^{2}+6a+9}{3a^{2}-6a+9}=\frac{8a+6}{3a^{2}-6a+9}+\frac{1}{3}$

pas age too hokmemoon jagozari konim mishe:

$\frac{8a+6}{3a^{2}-6a+9}+\frac{1}{3}+\frac{8b+6}{3b^{2}-6b+9}+\frac{1}{3}+\frac{8c+6}{3c^{2}-6c+9}+\frac{1}{3}\leq 8$

hamchenin vaseye makhraje har kasr darim:

$3a^{2}-6a+9=3(a^{2}-2a+3)=3[(a-1)^{2}+2]=3(a-1)^{2}+6\geq 6$ (chon $3(a-1)^{2}$ ye ebarate mosbate...)

pas age azin namosavi too makhraje kasramoon estefade konim kafie sabet konim:

$\frac{8a+6}{6}+\frac{8b+6}{6}+\frac{8c+6}{6}+1\leq 8$

pas kafie sabet konim $\frac{8(a+b+c)+18}{6}\leq 7$

amma $a+b+c=3$ pas kafie sabet konim:

$\frac{24+18}{6}\leq 7$...

yeni $7\leq 7$... :!:


calculus
integration
function
real analysis
topology
real analysis unsolved

Assume that f(x) is a finite Lebesgue measure fuction on [a,b],can we modify the value of f(x) on a zero measure set in [a,b] such that f(x) is reimann integrable(Being equal to the discontinue set of f(x) has zero meaure )? NOTE:See that we can modify the value of f(x) on a set with arbitrary small postive measure such that f(x) is continue on [a,b]

Consider the characteristic function of a Cantor-type set $ E$ with positive measure. Not only being of positive measure, but it also has an additional property that $ \mathbb{R} \backslash E$ is an open dense subset of $ \mathbb{R}$. This will enable us to conclude that $ \chi_{E}$ serves as a counter-example.

Let $ f$ be a function obtained by modifying the value of $ \chi_{E}$ on a set of measure zero. If $ x \notin E$, then there is an open neighborhood contained in $ \mathbb{R} \backslash E$ since $ E$ is closed. If $ x \in E$, then any $ \epsilon$-neighborhood of $ x$ has intersection with $ \mathbb{R} \backslash E$. In any cases, any open neighborhood of $ x$ contains a subset of positive measure where $ f$ vanishes. (Note that every nonempty open set has a positive measure.) This shows that any lower sum of $ f$ cannot be positive.

But if $ f$ is Riemann integrable, the limit of lower sum of $ f$ must coincide its Lebesgue integral. This contradicts the fact that the Lebesgue integral of $ f$ is positive. Therefore no such $ f$ is Riemann integrable.

assume mE=1/2,can you help me find the method:to modify the value of f(x) on a set with measure c<1/2 such that f(x) becomes continue.


trigonometry
inequalities
AMC
USA(J)MO
USAMO
inequalities unsolved

Does any one know how to solve this problem?

Let $ a_0, a_1, ... a_n$ be numbers from the interval (0, $ \pi/2$) such that
\[ \tan (a_0\minus{}\frac{\pi}{4})\plus{} \tan (a_1\minus{}\frac{\pi}{4})\plus{}\cdots \plus{}\tan (a_n\minus{}\frac{\pi}{4})\geq n\minus{}1.\]
Prove that
\[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n\plus{}1}.\]

This is USAMO problem.

Here is a link related the problem written in French, although you could understand it.

Take a look Exercise #5.

[url]http://www.animath.fr/old/tutorat/dossier_01024sol.html[/url]


[quote="Poincare"]Problem: How many base 6 integers less than $ 1000_6$ include five as a digit?

[hide="Solution(the question is in here)"]So, we use complementary counting so we get $ (5*6*6) \minus{} (4*5*5) \equal{} 80$ but intro to number theory says its 91 and uses very weird method at first but then it just does$ 6^3 \minus{} 5^3$. I don't understand why. :([/hide]
EDIT: never mind...[/quote]

Well, the $ 6^3\minus{}5^3$ comes from counting the total number of base 6 integers less than $ 1000_6$ (3 digits, 6 choices each).  Now, subtract out the ones that don't have a 5 in any digit to leave you with the ones with at least one 5.

The way you are doing it, you are forgetting that the first digit can be 0.