[quote="Iversonfan2005"]sorry if this is a dumb question, but what do the paranthesis stand for? :|[/quote]

Number of equation

"haha this is confusing me.. if you put $x=1$ into the first equatio and the second, the sum of the three is equal to 2 different values... so there are no constant solutions. maybe theres one in terms of $x$?"

^^*edit -- that was my little brother's attempt at doing pre-olympiad math... :).

[quote="Barnacle"]haha this is confusing me.. if you put $x=1$ into the first equatio and the second, the sum of the three is equal to 2 different values... so there are no constant solutions. maybe theres one in terms of $x$?[/quote]

Adding the three equations, we obtain $(a+b+c)x^{2}+(a+b+c)x+a+b+c=0$.  Dividing by $a+b+c$, we have $x^{2}+x+1=0$.  So you can't just substitute $x=1$. ;)

EDIT:

From $(a+b+c)x^{2}+(a+b+c)x+a+b+c=0$, we can factor to obtain $(a+b+c)(x^{2}+x+1)=0$.  Therefore, $a+b+c=0$ or $x^{2}+x+1=0$.  :)

[quote="TRAN THAI HUNG"]Find a,b,c(a,b,c belong toZ) satisfying
$\ ax^2+bx+c=0 (1)$
$\ bx^2+cx+a=26 (2)$
$\ cx^2+ax+b= -26 (3)$ :P

Math is my life  :!:[/quote]

Adding (2) and (3), we get $(b+c)x^2+(a+c)x+(a+b)=ax^2+bx+c=0 \ \ (4)$
Subtracting (2) from (1), we get $(a-b)x^2+(b-c)x+(c-a)=cx^2+ax+b=-26 \ \ (5)$.

Equating coefficients from (4) we get $b+c=a \ , \ \ a+c=b \ , \ \ a+b=c$, which has only one solution at $a=b=c=0$. But this doesnt work in (5) so there are $\boxed{ \text{No Solutions}}$. (I think).

[quote="Altheman"][quote="TRAN THAI HUNG"]Find a,b,c(a,b,c belong toZ) satisfying
$\ ax^2+bx+c=0 (1)$
$\ bx^2+cx+a=26 (2)$
$\ cx^2+ax+b= -26 (3)$ :P

Math is my life  :!:[/quote]

this question is ambigious; what is the point of x?

p.s. sam if you were just going to equate coefficients, you would just get rid of the RHS, and immediately get $a=b=c$, $b=c=a$, and $c=a-26=b+26$, which obviously has no solution

but since the question presumes that there is an answer...i think we are misinterpreting it...[/quote]

ok.. yea, im not sure what the point of the $x$ is either... maybe the $a,b,c$ are in terms of $x$, but i dont think that would work. Ummm... yea...

Well, i don't see how we could be misinterpreting it.

An interpretation you haven't tried yet is the following: find $x \in \bf R$ and $a, b, c \in \bf Z$ such that those three equations hold.

[quote="10000th User"]matt276eagles was heading in the right direction but his solution is not finished. Now that you got $a+b+c=0$ or $x^2+x+1=0$, you have two cases to check on the original equations. Also, you have to think $x$ is a real number (a constant) and not a variable. That's why Altheman and others committed the same mistake. Look at Barnacle's solution: it disproves the case $a+b+c=0$. Now, check when $x^2+x+1=0$ and see if you can find integers $a,b,c$. :)[/quote]
Yes, a+b+c=0.And x belong to Z. And it has three solutions :lol:

Math is my life :!:

Thanks to TRAN THAI HUNG's reply, I went back to check and Barnable's brother's (or himself's) solution does not disprove the $a+b+c=0$. Hehe my bad. The case $x^2+x+1=0$ is ruled out then. $x=0$ is one solution :D
[EDIT1]$x=3$ is another. Trying to get the third one.
[EDIT2]$x=-1$ :D

The key equations (in the order of my solution):

$a+b+c=0$

$x\ne 1$

$a(x+1)+b=0$

$a(x^3-1)+26=0$

:D

If  (1),(2),(3) sum side by side we get
x^2+x+1=0
Then
x_1=-1+ i[/i]


John flips a fair coin 3 times. Find the probability that exactly 2 of the flips come up the same.

Just the probability that three of the same don't come up. So 1-1*1/2*1/2=3/4

[hide]Your possibilities are:
HTT
HHH<--
HTH
HHT
THH
TTT<--
THT
TTH

Of these, the arrowed ones don't work. nat mc's solution also works.[/hide]

[quote="biffanddoc"]Its definetely 1/4[/quote]

wth!

how?

[quote="chess64"][quote="G-UNIT"]a book i found says the answer is 1/2, and i can see why[/quote]

why?[/quote]

  It said that you can either get 3, 2, 1,  or 0 tails (heads vice versa) on your flip.  Meaning their are 4 combinations.  Out of these you can get either "HHT" or "TTH" which gives you a total of 2 combinations.  Thus, the probability is 1/2.  My first approach to the problem was the same as the books, but you guys seem to differ.

[quote="sirorange"]That would be because the probability of getting say two heads, is twice as great as getting say three heads.[/quote]
3x, not 2.  remember Pascal, 1 3 3 1

Yes, Pascal is made from combinations and the sum of each row is a power of 2.  If you flip a coin n times and you want to find the probability that there are k heads, then the probability is
${n\choose{k}}/2^n$.  That is the same thing as the k+1 th term of the nth row over the sum of the terms of the nth row in PAscal

yea its 3/4 the book puposely gave the wrong answer and then explained the correct one

They show you why it is not the common misconception


induction
combinatorics proposed
combinatorics

Let $G$ be a tournoment such that it's edges are colored either red or blue.
Prove that there exists a vertex of $G$ like $v$ with the property that, for every other vertex $u$ there is a mono-color directed path from $v$ to $u$.

is there some information missing? See the attached $K_4$

No, there is no information missing.
In your example the buttom-right vertex has the requested property, it has a red path to any other vertex.

We'll prove it by induction. Show that it is true for $n=3$ vertices (trivial) and assume it is true for $n-1$. We'll say that a point that satisfies the given condition is a 'powerful point'.

Take $G$ a graph with $n$ vertices. Suppose it has no powerful point. For any vertex $v$, the subgraph $G-v$ has $n-1$ vertices and thus has a powerful point. Then there's a powerful point for $G-v_1$, for $G-v_2$, ..., for $G-v_n$. If two of these were the same, this vertex would be a powerful point for $G$. So they must all be distinct. Thus each vertex has monochromatic paths to exactly $n-2$ other vertices; for each vertex $v$, let $f(v)$ be the vertex it doesn't reach. Consider the sequence $v, f(v), f(f(v)), \ldots$. Eventually some vertex must appear twice. Consider the first time this happens. If this vertex is not $v$ itself, then it must have two distinct 'ancestors' (points that don't reach it), which is absurd (since each vertex has exactly one other vertex that doesn't reach it). Then the first vertex to appear again must be $v$, and we must have a cycle. However, if the cycle contains fewer than $n$ vertices, then there must be a powerful point in this subgraph, which contradicts the fact that each vertex can't reach the next one in the cycle. Thus the cycle has $n$ vertices, that is, it is the whole graph.

So order the vertices according to their position in the cycle as $v_1, v_2, \ldots, v_n$, where $f(v_i)=v_{i+1}$. Now we will draw more edges than there originally were. If there's a red path from $v$ to $u$, draw the red edge $vu$; the same goes for blue paths (of course, if the path from $v$ to $u$ is just the edge $vu$, we don't draw it again). Is it possible for two vertices $v, u$ to have both a blue edge $vu$ and a red edge $vu$? No, because then $v$ would reach all the points that $u$ reaches, and in particular $v$ would reach $f(v)$, absurd. It is easy to see, then, that each two vertices $v, u$ are joined by two edges $vu, uv$, each of which can be red or blue, unless $u=f(v)$ or vice-versa.

Now, for each vertex $v$, consider the number of red edges $vx$. Clearly it is at most $n-2$ and at least $0$. So there must be two vertices with the same number of red edges $v, u$. Now, if $vu$ is red, then $v$ has at least as many red edges as $u$ does; in particular, for $u$ to have the same number of red edges, $uv$ must be red too. It follows that $vu, uv$ are either both red or both blue. Assume that they're both red (the other case is analogous). Now take $f(v), f(u)$. If $vf(u)$ was red, then there would be a red path $uvf(u)$ from $u$ to $f(u)$, absurd. Then $vf(u)$ is blue. Analogously $uf(v)$ is blue too. Now, if $f(v)v$ were blue, then there would be a blue path $uf(v)v$, from $u$ to $v$; then there would be a blue $uv$ edge too, absurd. So $f(v)v, f(u)u$ are red. Finally, if $f(v)f(u)$ were blue,  then there would be a blue path $uf(v)f(u)$ from $u$ to $f(u)$, absurd. So $f(v)f(u), f(u)f(v)$ are red. By induction, $f(f(u)), f(f(v))$ are joined by red edges too, and so on. It follows, also, that $f(f(v))f(v)$ is red, and so on; by induction, $v_{i+1}v_i$ is red for all $i$. But then we can go from any vertex to any other vertex through a red path on the cycle!

[quote="Nima Ahmadi Pour"]A tournoment is a directed graph where between each two vertices there is exactly one directed edge.
That is exactly one of these two edges are in the graph : $uv$ and $vu$[/quote]

Thank you. :D

[quote="Severius"] If this vertex is not $v$ itself, then it must have two distinct 'ancestors' (points that don't reach it)[/quote]
Why is this so ?

[quote="Severius"]So order the vertices according to their position in the cycle as $v_1, v_2, \ldots, v_n$, where $f(v_i)=v_{i+1}$. Now we will draw more edges than there originally were. If there's a red path from $v$ to $u$, draw the red edge $vu$; the same goes for blue paths (of course, if the path from $v$ to $u$ is just the edge $vu$, we don't draw it again). Is it possible for two vertices $v, u$ to have both a blue edge $vu$ and a red edge $vu$? No, because then $v$ would reach all the points that $u$ reaches, and in particular $v$ would reach $f(v)$, absurd. It is easy to see, then, that each two vertices $v, u$ are joined by two edges $vu, uv$, each of which can be red or blue, unless $u=f(v)$ or vice-versa.

Now, for each vertex $v$, consider the number of red edges $vx$. Clearly it is at most $n-2$ and at least $0$. So there must be two vertices with the same number of red edges $v, u$. Now, if $vu$ is red, then $v$ has at least as many red edges as $u$ does; in particular, for $u$ to have the same number of red edges, $uv$ must be red too. It follows that $vu, uv$ are either both red or both blue. Assume that they're both red (the other case is analogous). Now take $f(v), f(u)$. If $vf(u)$ was red, then there would be a red path $uvf(u)$ from $u$ to $f(u)$, absurd. Then $vf(u)$ is blue. Analogously $uf(v)$ is blue too. Now, if $f(v)v$ were blue, then there would be a blue path $uf(v)v$, from $u$ to $v$; then there would be a blue $uv$ edge too, absurd. So $f(v)v, f(u)u$ are red. Finally, if $f(v)f(u)$ were blue,  then there would be a blue path $uf(v)f(u)$ from $u$ to $f(u)$, absurd. So $f(v)f(u), f(u)f(v)$ are red. By induction, $f(f(u)), f(f(v))$ are joined by red edges too, and so on. It follows, also, that $f(f(v))f(v)$ is red, and so on; by induction, $v_{i+1}v_i$ is red for all $i$. But then we can go from any vertex to any other vertex through a red path on the cycle![/quote]
Can't we simplify this second half a bit? If $v_n\to v_{n-1}\to\cdots\to v_1\to v_n$ is a monochromatic cycle, we're obviously done. Otherwise, there exists $i$ such that $v_{i+1}v_i$ and $v_iv_{i-1}$ are colored differently, say red and blue respectively. Now consider the monochromatic path from $v_{i-1}$ to $v_{i+1}\ne v_i=f(v_{i-1})$. If it's red, then we get a red path from $v_{i-1}$ to $v_i=f(v_{i-1})$, and if it's blue, then we get a blue path from $v_i$ to $v_{i+1}=f(v_i)$; either way yields a contradiction.

I had a slightly different proof than the ones I've seen so far.

We will prove it by induction on the number of vertices $n$. Clearly for $n=2,3$ it is true, now suppose it is true for $n-1$. First label the graph $V=\{v_1,v_2,...v_n\}$. We say that a vertex $u$ "reaches" $v$ if there is a monochromatic directed path from $u$ to $v$ For each vertex $v_i$ we can apply the induction hypothesis on $V\setminus v_i$  and get that there is some vertex $v_j$ such that $v_j$ cannot reach $v_i$ but can reach every other vertex. Call this vertex the "king" for that set. Also for each set of $n-1$ a different king must be chosen otherwise if some vertex is chosen as the king twice it can reach every vertex so we are done. Hence each vertex in the graph can reach every vertex but one, and this choice is unique for different vertices.

Although the start is the same I did not reach the same cycle idea as in the solution above. Instead I looked at $n=4$ and tried to generalize the contradiction I got there to all $n$, which worked well. Once you reach this point the idea is mostly about getting two paths of different colors between two points.
Now there are two cases to consider:

1. There is some vertex with both a red and blue directed path coming from it
Let's call that vertex $A$ and suppose that $A$ cannot reach $C$. However $A$ can reach every other vertex. Let $R=\{r_1,r_2,...,r_s\}$ be the vertices which $A$ can reach with a red path and $B=\{b_1,b_2,...,b_t\}$ the vertices $A$ can reach with a blue path. Now some vertex must not be able to reach $A$, suppose it is blue and call it $b_j$. Then $b_j$ can reach all vertices in $B$. Also by the induction hypothesis some vertex $r_i$ in $R$ can reach every vertex in $R$, so it can't reach some vertex $b_k$ in $B$. Now $r_iA$ must be red otherwise $r_iAb_k$ would be a blue path from $r_i$ to $b_k$. This means that $b_jr_i$ must be blue otherwise $b_jr_iA$ would be a red path from $b_j$ to $A$. But now if $r_iC$ is red, $Ar_iC$ is a red path from $A$ to $C$, and if it is blue, $Ab_jr_iC$ is a blue path from $A$ to $C$, contradiction.

2. Every vertex has only one color path from it to the vertices it can reach
Suppose some vertex $A$ has red paths from it to $R=\{r_1,r_2,...,r_m\}$, and cannot reach some vertex $C$. Then $r_1C,r_2C,...,r_mC$ are all blue. Hence all paths coming from $r_1,r_2,...,r_m$ are blue. Suppose $r_i$ cannot reach $A$. Since $n\ge 4$, we have $m\ge 2$ so there is some other vertex $r_j$ which can reach $A$, but then $r_ir_jA$ is a blue path from $r_i$ to $A$, contradiction. So we are done.

Hopefully this works...

[hide]
[i]Lemma 1: [/i]Atleast one of the red and blue edges form a weakly connected graph (a directed graph is weakly connected if its undirected version is connected).
[i]Proof: [/i]Suppose the red edges are not weakly connected, so there is a partition of the vertices $V= S \cup T$ such that there exists no red edge between $S$ and $T$. But then for any $u \in S, v \in T$ there must be a blue edge between $u,v$. Thus the undirected version of the blue edges contains the complete bipartite graph $K_{S,T}$ as a subgraph, which is connected.

WLOG assume the red edges are weakly connected.

[i]Lemma 2: [/i]There exists a vertex $v$ with the following property: for any $w \neq v$ if there exists a red path from $v$ to $w$, then there also exists a red path from $w$ to $v$.
[i]Proof: [/i]Decompose the red graph into strongly connected components and consider the compressed directed acyclic graph. Since there are no cycles in the compressed graph, there must exist a vertex with outdegree 0. Suppose it corresponds to a strongly connected component $S$. Pick a $v \in S$. This $v$ satisfies the property: if there exists a red path from $v$ to $w$, then $w$ must also belong to $S$ (as the vertex corresponding to $S$ in the compressed graph has outdegree 0) and thus there must also exist a red path from $w$ to $v$ (as $S$ is a strongly connected component).

Call a vertex $u$ central for a set $S$ if there exists a monochromatic path from $u$ to $w$ for all $w \in S$. Choose a vertex $v$ satisfying the condition in lemma 2. By induction there is a central vertex $c$ for the remaining graph $G \setminus \{v\}$. Let $S$ be the set of vertices reachable from $c$ via red edges only. If $v \in S$ then $c$ is central for  $G$ and we are done so assume that is not the case. Note that the vertices in $G \setminus (S \cup \{v\})$ are reachable from $c$ via blue edges only (since $c$ is central for $G \setminus \{v\}$). Also note that if the edge between $c$ and $v$ is directed from $c$ to $v$ then $c$ is central for $G$ so assume it is directed the other way.

[i]Claim 1: [/i]For all $w \in S$ the edge between $v$ and $w$ has to be blue.
[i]Proof: [/i]Suppose the edge between $v$ and some $w \in S$ is red. If the edge is directed from $w$ to $v$ then this implies $v \in S$, contradiction. Otherwise there is a red path from $v$ to $w$. By the condition in lemma 2 this implies there is a red path from $w$ to $v$, again implying $v \in S$.

By claim 1 the edge from $v$ to $c$ must be blue, so there is a blue path from $v$ to all vertices in $G \setminus (S \cup \{v\})$.

[i]Claim 2: [/i] $|S \cup \{v\}| < |G|$
[i]Proof: [/i] Since the red edges form a weakly connected graph there must be some vertex $w$ that is connected to $v$ by a red edge. By claim 1 $w$ cannot belong to $S$, so $w \in G \setminus (S \cup \{v\}) \implies |S \cup \{v\}| < |G|$.

By claim 2 and induction there exists a vertex $c'$ central for $S \cup \{v\}$. We do some casework.

[b]Case 1 - $c'=c$: [/b] Then $c$ is central for $G$ and we are done.
[b]Case 2 - $c'=v$:[/b] Then $v$ is central for $G$ and we are done (we showed before that there is a blue path from $v$ to all vertices not in $S$).
[b]Case 3 - $c' \in S, c' \neq c$:[/b] By claim 1 the monochromatic path from $c'$ to $v$ must be blue. Thus there is a blue path from $c'$ to $c$ and thus to all vertices not in $S$. By assumption there is a monochromatic path from $c'$ to all vertices in $S$, so $c'$ is central for $G$.

We are done.

[/hide]

what a great problem
for a directed graph $G$ let such $v$ vertex be the president of $G$
let $n$ be smallest integer such that there exists a graph $G$ doesn't satisfy the problem condition with $n$ vertices
if we hide $v$ the let $f(v)$ be the president of $G - v$
[b][u][color=#f00]the main claim:[/color][/u][/b] $v_1 \longrightarrow v_2\longrightarrow \dots v_n$ is a directed cycle
[b][u][color=#00f]proof:
[/color][/u][/b]suppose that there are $u,v$ such that $w=f(v)=f(u)$ then there exist a colored path from $w$ to every vertex in $G-v \cup G-u =G$ contradiction
so WLOG let $v_i=f^{i-1}(v_1)$
$\blacksquare$
now it's obvious that there isn't any colored path  from $f(v)$ in $v$ so the edges $v_1v_2,v_2v_3 \dots v_nv_1$ aren't monochromatic
WLOG let $v_1v_2$ be red and $v_2v_3$ be white
consider the colored path  $\vec{v_3v_1}$
 if it is red then there's a red path $\vec{v_3v_2}$ contradiction
 if it's white then there's a white path $\vec{v_2v_1}$ contradiction
 and we win :D

[url=https://core.ac.uk/download/pdf/82547341.pdf]Sands, B., Sauer, N. and Woodrow, R., 1982. On monochromatic paths in edge-coloured digraphs. Journal of Combinatorial Theory, Series B, 33(3), pp.271-275.[/url] Corollary 2.

We induct on $|G|$, where the base cases $|G| = 1, 2, 3$ are trivial.

For the inductive step, assume for contradiction that $G$ does not satisfy the desired property. Note that for every vertex $v \in G$, there exists some $f(v) \in G$ which has a directed edge pointing to each other vertex $G \setminus v$ per the inductive hypothesis. If $f(v) \rightarrow v$ then we are done, so assume henceforth that $v \rightarrow f(v)$ for all $v \in G$. Repeating $f$ iteratively, we have \[ v \to f(v) \to \dots \to f^{k-1}(v) \to f^k(v)  \] is a cycle $\mathcal{C}$ for some $k$.

Assume $\mathcal{C}$ that it is not monochromatic, or else we are done. We now direct our attention to the vertices $v$, $f(v)$, and $f^2(v)$, as looking at them is sufficient to finish. WLOG $v \to f(v)$ is colored red and $f(v) \to f^2(v)$ is colored blue; consider the following casework.
[list]
[*] If $v \sim f^2(v)$ is colored red, then $f^2(v)$ works as the desired vertex, and
[*] if $v \sim f^2(v)$ is colored blue, then $f(v)$ works.
[/list]
The inductive step is complete, finishing.

The result for one colored tournaments seems like folklore. Anyways, nice problem.

We show this inductively. The base case of $n = 1, 2$ is immediate.
FTSOC suppose not for $n \ge 3$. Let $V_i$ be the such vertex of the tournament when removing the vertex $i$, which follows inductively. If $V_i = V_j$, then contradiction follows. Thus the map $i \mapsto V_i$ is injective and thus bijective.
Else, all $V_i$ are distinct.

[b][color=pink]Claim:[/color][/b] There exists no monochromatic directed path from $V_i$ to $i$.
[i]Proof.[/i] If there did, then this leads to contradiction because then $V_i$ would also lead to $i$. $\blacksquare$
As such, it follows that the edge from $i$ to $V_i$ is directed towards $V_i$.

[b][color=pink]Claim:[/color][/b] The path from $V_i$ to $V_{V_i}$ and the path from $i$ to $V_i$ are the same color.
[i]Proof.[/i] We have a monochromatic path $V_{V_{i}} \to i$, as well as monochromatic direct paths from $i \to V_i$ and $V_i \to V_{V_{i}}$.
Then neither of the paths $V_{V_i} \to i \to V_i$ and $V_i \to V_{V_i} \to i$ can be monochromatic. It thus follows that $i \to V_i$ and $V_i \to V_{V_{i}}$ both have the opposite color of $V_{V_{i}} \to i$ and thus have the same color. $\blacksquare$
However, now the chain $i \to V_i \to V_{V_{i}} \to \dots$ is eventually cyclic and monochromatic, contradicting the existence of a monochromatic path from $V_i$ to $i$.


I know they come from something called MATHCOUNTS, but do they also come from AMC 8/10/12?

There are some AMC 8 and AMC 10 questions, I believe (not so sure about the latter). The predominant source of the question is the MATHCOUNTS competition, though. There are also some questions courteously provided by the people of AoPS.  :)

Some of the easier AMC 10 questions are included.

I think they are mostly from AMC and MATHCOUNTS promlems. :blush:


Given an integer $ n\geq 3$. Find the maximum of

$ X_1^2\cdot X_2\plus{}X_2^2\cdot X_3\plus{}...\plus{}X_n^2\cdot X_1$,

where $ X_1$, $ X_2$, ..., $ X_n$ are $ n$ nonnegative reals such that $ X_1\plus{}X_2\plus{}...\plus{}X_n\equal{}1$.

I have taken the freedom to replace the original post (which contained an old problem that was posted before) by this problem, which was posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=24113 by Michael Niland, and generalizes http://www.mathlinks.ro/Forum/viewtopic.php?t=216 .

  darij


graph theory
combinatorics unsolved
combinatorics

How to prove that there is no hamiltonian cycle in Petersen's graph? Apart from typing all possibilities and cross out them one by one.


I am currently working on another Mock test. I hope to finish it this weekend. If you want to take it/want to help me check over the test, post/ send me a pm.

--Alex

Yes.
I will definitely take it.
You're a beast man.

I would like to participate. Thank you so much for your efforts!

Plain beastly.  :D
I might actually take it, this time.

[quote="mustafa"]Will it be ready before the real AIME on tuesday?[/quote]

that is my goal; i hope to finish by early sunday, i need somebody to check though...

this test will probably be a bit easier than the previous tests...

i might take it but i am not fully sure yet.

um sure send me the hard (good) problems and I might check them if I am not utterly lazy and/or swamped by work? question mark?

if i have time and/or if arnav doesn't doesn't screw everything up on purpose, i'll do it (both are likely)

I'll take it (if I have time - the state math league tournament is Monday).

i think i'll try it, too, although i'm not positive. (keyword: [i]try[/i]).  :D Thanks!

this is my first time doing this, so could someone possibly tell me how this works or direct me to somewhere that can?

cool another one of those brillant mock AIME.

this one and one of Mildorf's should keep me sharp at math :rotfl:

[quote="MysticTerminator"][quote="Altheman"][quote="solafidefarms"]Uh aren't you supposed to just chill the day before the test, i.e. Monday? So we'd have to do it tomorrow?[/quote]

i don't particularly agree with the whole "chill" thing...there is no reason not to take a practice test the night before to prepare yourself for the contest conditions, however, i see where that comment is coming from, i believe that the teachers here don't want people to try and cram formulas, that is a waste of your time, but as i said, a practice is helpful because it prepares your for the test[/quote]

considering I don't practice the rest of the year I figure I need to do at least one the day before[/quote]
Eh, skip school the day before, it works.


Let $ x_1,...,x_n$ are positive numbers such that $ x_1\plus{},...,x_n\equal{}1$.Let be S the biggest number from the serie:

$ \frac{x_1}{1\plus{}x_1},\frac{x_2}{1\plus{}x_1\plus{}x_2},...,\frac{x_n}{1\plus{}x_1\plus{}x_2\plus{}...\plus{}x_n}$.

Find the value of $ x_1,x_2,...,x_n$ in way to get the minimum value of $ S$ ?

May go to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=238910[/url] for reference.

I think for numbers $ x_1,x_2 ,...,x_n$ should hold $ x_1 <x_2 <...<x_n$


calculus
integration
trigonometry
calculus computations

I've tried to do this problem but I'm not getting the right answer. Can anyone help me out? Here's the problem with my latest attempt:

Find the arc length of the polar curve described by:
r = 9 + 9cos(t)

I used the definition:
[Int] (r^(2) + (dr/dt)^(2)) ^(1/2)

so:
18(2)^(1/2) * (1-cos(t))^(1/2) from 0-->2pi
which is:
(18(2)^(1/2) * (1-cos(2pi))^(1/2)) - (18(2)^(1/2) * (1-cos(0))^(1/2))
= 0

which is wrong. Anyone see where I've gone wrong?
Thanks =)

The graph is of a cartoid:
http://www.teachers.ash.org.au/mikemath/calculators.html

I don't have the graph in front of me.  However, if the curve is not one-to-one on the entire interval, then the integral will be wrong.  That is, does the curve overlap itself in the interval in question?  Check that.

Curve overlaps are irrelevant to the problem of finding arc length.  The integrand is non-negative; you should be getting a positive answer unless you just calculated the integral wrong.  We have $ \frac {dr}{dt} \equal{} \minus{} 9 \sin t$ and the integral

$ \int_{0}^{2\pi} \sqrt { (9 \plus{} 9 \cos t)^2 \plus{} ( \minus{} 9 \sin t)^2 } \, dt \equal{} \int_{0}^{2\pi} \sqrt { 162 \plus{} 162 \cos t} \, dt$.  You don't seem to have actually integrated.

$ \int_0^{2\pi}9\sqrt {(1 \plus{} \cos t)^2 \plus{} ( \minus{} \sin t)^2}\,dt \equal{} \int_0^{2\pi}9\sqrt {2 \plus{} 2\cos t}\,dt$

$ \equal{} 9\int_0^{2\pi}\sqrt {4\cos^2\left(\frac t2\right)}\,dt \equal{} 18\int_0^{2\pi}\left|\cos\left(\frac t2\right)\right|\,dt$

$ \equal{} 36\int_0^{\pi}\cos\left(\frac t2\right)\,dt \equal{} 72.$

This is a little bit longer than the $ 18\pi$ arclength of the circle $ r \equal{} 9.$


function
induction
algebra unsolved
algebra

Find all function $f : Q+ \Rightarrow Q+$
$f(x+\frac{y}{x}) = f(x)+ \frac{f(y)}{f(x)} + 2y$ $for x,y \in Q+$
I can solve that for all $x \in N$ $f(x) = x^2$
How to continue that ,sir.

Taking $y=x$ we have $f(x+1)=f(x)+2x+1$   $(1)$
By easy induction we can prove that for all natural numbers $k$ ,$f(x+k)=f(x)+2xk+k^2$ $(2)$
Taking $x=1$ we have $f(y+1)=f(1)+\frac{f(y)}{f(1)}+2y$ $(3)$
From $(1)$ and $(3)$ we have $f(1)=1$ and from $(2)$ we get $f(n)=n^2$ $(n\in N)$
Replasing $y$ with $yx$ we get $f(x+y)=f(x)+\frac{f(xy)}{f(x)}+2xy$
Taking $y=\frac{p}{q}$ and $x=q$ $(q,p\in N)$ here we get
$f(q+\frac{p}{q})=f(q)+\frac{f(p)}{f(q)}+2p$ and from $(2)$
$f(q+\frac{p}{q})=f(\frac{p}{q})+2p+q^2$
And we get $f(\frac{p}{q})=\frac{f(p)}{f(q)}=\frac{p^2}{q^2}$
So $f(x)=x^2$ for all $x \in Q^+$
      That's all... :lol:

Another solution
WE have:$f(x)=x^2 \forall x\in N*$
Otherwise,we have:
$f(x+\frac{y}{x}=f(x)+\frac{f(y)}{f(x)}+2y$
$\Rightarrow f(x+\frac{y}{x})=x^2+\frac{y^2}{x^2}+2y$(1)
$f(x+\frac{y}{x})=f(\frac{y}{x}+\frac{y}{\frac{y}{x}})=f(\frac{y}{x})+\frac{y^2}{f(\frac{y}{x})}+2y$(2)
$(1)(2)\Rightarrow x^2+\frac{y^2}{x^2}=f(\frac{y}{x})+\frac{y^2}{f(\frac{y}{x}}$
$\Rightarrow f(\frac{y}{x})=\frac{y^2}{x^2}\Rightarrow f(x)=x^2 \forall x\in Q^+$

[quote=mps]$f : \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}:$
$f\left(x+\frac{y}{x}\right) = f(x)+ \frac{f(y)}{f(x)} + 2y.$[/quote]
My solution without using $f(n)=n^2:$
Let $f(x)=x^2+g(x), \left(g: \mathbb{Q}^{+}\rightarrow \mathbb{Q}\right).$
Take $x=y$ to receive $f(x+1)=f(x)+2x+1$, so $g(x+1)=g(x). \ (1)$
Now take $y=p\cdot x$ to receive $f(x+p)=f(x)+2px+\frac{f(p\cdot x)}{f(x)} \ \left( where \ x=\frac{q}{p-1}\right).$
On the other hand, by taking $(1) \ p$ times $f(x+p)=f(x)+2px+p^2$, so $f(px)=p^2\cdot f(x).$
$(px)^2+g(px)=f(px)=p^2\cdot f(x)=p^2(x^2+g(x))$, so $g(px)=p^2g(x)$, but from $(1)$ we also have that $g(px)=g\left(q+\frac{q}{p-1}\right)=g\left(\frac{q}{p-1}\right)=g(x)$, so $g\equiv 0.$


modular arithmetic
number theory proposed
number theory

given $ k\ge 2$, find all $ m\in\mathbb{N}$ s.t. $ a^k\equiv 1\pmod{m}$  for all $ a\in\{1,\dots,m\minus{}1\}$

[quote="knoppix"]given $ k\ge 2$, find all $ m\in\mathbb{N}$ s.t. $ a^k\equiv 1\pmod{m}$  for all $ a\in\{1,\dots,m \minus{} 1\}$[/quote]
The first we prove that $ m$ must be a prime number.
Suppose $ m \equal{} pq$ where $ p,q > 1$
So $ p\leq m \minus{} 1$
Take $ a \equal{} p$ then $ p\not |a^k \minus{} 1$
It gives contradiction .
So $ m$ must be a prime .
Take $ a$ is a primitive root of m .
Then $ m|a^k \minus{} 1 \Leftrightarrow m \minus{} 1|k$
Easy to check that if $ m \minus{} 1|k$ then $ m|a^k \minus{} 1,\forall a \equal{} 1,..,m \minus{} 1$
So all m satisfy condition are :
$ A \equal{} \{m\in P|m \minus{} 1|k\}$


Find a number that fits all the following clues. It is greater than 200 but less than 400. It is divisible by 5 and 9, not a square number and not even.

Since the number is divisible by both $ 5$ and $ 9$, it must be divisible by $ 45$. The smallest number that is divisible by $ 45$ that is greater than $ 200$ is $ 225$. This is a square number. The next one is $ 270$, and this is even. The next one is $ \boxed{315}$, which is neither even nor a square number.


Let $ABC$ be a triangle with angle bisectors $AA',BB',CC'$($A'$ on the side $BC$, $B'$ on the side $CA$, $C'$ on the side $AB$).Prove that:$B'C'+C'A'+A'B' \leq \frac{1}{2} (BC+CA+AB)$

[quote="leepakhin"]If $\triangle ABC$ is acute, your inequality works for all $A'$ on $BC$, $B'$ on $CA$ and $C'$ on $AB$. By Franano's problem, the triangle of minimal perimeter incribed in a given acute-angled triangle is its orthic triangle, whose perimeter is half of the original one.[/quote]

This should be: whose perimeter is less or equal to half of the original one. The problem is that Fagnano's theorem doesn't help here, since the perimeter of triangle A'B'C' is generally greater or equal and not less or equal to the perimeter of the orthic triangle.

The actual problem was solved on http://www.mathlinks.ro/Forum/viewtopic.php?t=29083 .

  darij


inequalities
search
inequalities proposed

I'm proving problem from IMO 1996. But who know an easy solution?

I'm sure it has been discussed many many times before,try to search it:

anyway here is another quick solution:
we want to prove that:
[b]if $ a,b,c > 0$ and $ abc = 1$ then $ \sum \frac {ab}{a^5 + b^5 + ab}\leq 1$[/b]

note that:

\begin{eqnarray*}a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4) = (a + b)\left((a - b)^2(a^2 + b^2 + ab) + a^2b^2\right)\geq a^2b^2(a + b)\end{eqnarray*}

so we have:

\begin{eqnarray*}\frac{ab}{a^5 + b^5 + ab}\leq\frac {ab}{a^2b^2(a + b) + ab} = \frac1{ab(a + b) + 1} = \frac {c}{abc(a + b) + c} = \frac {c}{a + b + c}\end{eqnarray*}

thus:

$ \sum \frac {ab}{a^5 + b^5 + ab}\leq \frac {c}{a + b + c} + \frac {a}{a + b + c} + \frac {b}{a + b + c} = 1$

and the equality occurs when $ a = b = c = 1$

To BaBaK Ghalebi:
You can aslo write that inequality
$ a^5 \plus{} b^5\geq a^2b^2(a \plus{} b)$, follows from the Muirherd for two numbers.
To Zhas:
This problem is ISL 1996,not IMO 1996. :wink:
And before posting it(by the way,twice),you should look for it here,using search.

Thanks(Inequality from IMO 1996) It's very easy :rotfl: . Who help me prove the inequality from Costa-Rica? I used a Ravi substitution, but it's no solve a problem.

Zhas post each problem in a new topic...
by the way your inequality is wrong just take $ a \equal{} b \equal{} c$,it should be:

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

which is true for every positive real numbers $ a,b,c$...

[quote="Erken"]
You can aslo write that inequality
$ a^5 \plus{} b^5\geq a^2b^2(a \plus{} b)$, follows from the Muirherd for two numbers.
[/quote]

Or rearangement twice.


sin a+sin b =sin c
find c in terms of a,b

Wouldn't it be $c=\frac{(4k+1)}{2}\pi \pm \left( \sin^{-1}(\sin a+\sin b)-\frac{\pi}{2}\right)$ or something like that


Andrew's school has 1024 lockers and all of them are closed.

He starts from locker number one, and opens every locker.

Then he goes and opens (or closes if the locker is open) every 2nd locker, then every 3rd locker, and so on and so forth.

By the time he finishes opening/closing every 1024th locker, how many lockers are closed?

Yes towersfreak is correct - the problem is phrased wrongly.

I think you mean he opens them if they're closed and closes them if they're open or else the lockers will just stay open. ^_^

Nice problem! :)

[hide="Hint"]A locker is open if it is switched an odd number of times or even number of times? What does this say about the number of divisors it has?
[/hide]

[hide="Hint"]Think about the number of factors every number has.[/hide]
[hide="Solution"]For a locker to remain open at the end, it must have an odd number of factors.  Only perfect square meet this requirement.  Because $32^{2}=1024$, there are $32$ lockers that will be open.  So there are $1024-32=992$ lockers that are closed at the end.[/hide]

[hide]for a locker to be open, there has to be an odd number of switches which means it has to be a perfect square.

the largest perfect square is 1024.

it is 32 square

1024-32=[b]990[/b]

the locker 1024 gives the problem away[/hide]

I think it was an AIME question (with 1000 lockers) once.


[quote="shyong"][quote="jaquz"]Given: Triangle ABD, Angle BAD = 90* altitude AC (C lying on segment BD)
AD = 6, BC = 5..

Find AB and CD[/quote]

Is AC perpendicular to BD ? If so then by letting AB=p , AC=q , CD=r and form three equation from pythagoras theorem and you are done .[/quote]
There's an easier way to find $r$ ($CD$).

Since $\triangle BCD$ and $\triangle ABD$ are similar,  $\frac{r}{6}=\frac{6}{r+5}\Longleftrightarrow r^2+5r-36=0\Longleftrightarrow (r-4)(r+9)=0$

So, $CD=4$ and we can find $AB$ easily (either by using Pythagorian Theorem or by similar triangles).

Let $CD=x$.  Then $36=x(x+5) \implies x=4$.  Thus, $CD=4$ and $AB=3\sqrt{5}$.


1. Let $K$ be a [b]proper[/b] nonempty compact subset of the unit circle $|z| = 1$. Show that there exists a polynomial $p(z) \in \mathbb{C}[z]$ such that $p(0) = 1$ and $|p(z)| < 1$ for every $z \in K$.

2. Does there exist a polynomial $p(z) \in \mathbb{C}[z]$ such that $p(0) = 1$ and $|p(z)| < 1$ for all $z$ on the unit circle $|z| = 1$?

I'd be very surprised to see an elementary (olympiad level) proof of the first assertion, which is a trivial consequence of Runge's theorem... The second question is of course a joke.

[quote="harazi"]The second question is of course a joke.[/quote]

Yes  :P  .

But the first question was known before Runge proved his theorem, I believe...


calculus
integration
geometry
rectangle
calculus computations

hey is any1 online please help me!

Estimate the area under the graph of f(x) = e^-x^2
from x = -2 to x = 2 using four approximating rectangles and midpoints.

Just use the midpoint formula, it's an explicit calculation.
Place your points at $ x\equal{}\minus{}1.5$, $ x\equal{}\minus{}.5$, $ x\equal{}.5$, and $ x\equal{}1.5$.  So now we evaluate at these points: $ f(\minus{}1.5)\equal{}e^{\minus{}2.25}$, $ f(\minus{}.5)\equal{}e^{\minus{}.25}$, $ f(.5)\equal{}e^{\minus{}.25}$, and $ e^{\minus{}2.25}$.
These are the heights of the 4 rectangles.  Each rectangle has a width of 1 (make sure you understand why).  So we find the area of each rectangle the usual way and add: $ A\approx 2(e^{\minus{}2.25}\plus{}e^{\minus{}.25})$.


[hide]
There are 16 men and they dug 5 holes in 9 days
then
how many days will it take for a single man to dig .5 ( or half) hole?

work = rate*time
5/16 = rate*9

rate=5/144 holes/day.

.5 holes / (5/144 holes/day) = 14.4 days

[/hide]

[quote="shinwoo"]good problem

There are 16 men and they dug 5 holes in 9 days
then
how many days will it take for a single man to dig .5 ( or half) hole?[/quote]

[hide="click text"]
5(work)=16(power)*9(time)

5/9=16 (work/time for 16 people), for a single only divide by 16
5/144 = 1 then we have time and holes for a single
5 = 144
.5 = x
((.5)(144))/(5)=x
time = 14.4
[/hide]

[hide]
It takes one man $\frac{144}{5}\cdot \frac{1}{2}$, or $14.4$ days to dig half a hole.[/hide]

[hide]
16 people can dig $\frac{5}{9}$ of a hole in one day.
1 person can dig $\frac{5}{144}$ of a hole in a day.
there fore, it takes 14.4 days to dig 0.5 of a hole.

[/hide]


algebra
polynomial
calculus
integration
inequalities
algebra proposed

[b]The equation $x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_n=0\; ,\; n\ge 3,$ has $n$ real roots $x_j$ (not all equal).  Show that $(n-1)a_1^2-2na_2>0 .$
 If  $D: =\max\limits_{j}x_j -\min\limits_{j}x_j , \;$  prove that
\[ (*)\; \; \; \; \; \; \; \; \; \frac{n}{2}\le\frac{(n-1)a_1^2-2na_2}{D^2}\le \left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]\; ,  \]
where $[\cdot]$ denotes integral part. Prove that (*) inequalities cannot be improved.
[/b]
[i]In connection with Problem 2 proposed at Seventeenth annual U.S.A. Mathematical Olympiad (case n=3 in the right-side of (*)).[/i]

[quote="flip2004"][b]The equation $x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_n=0\; ,\; n\ge 3,$ has $n$ real roots (not all equal).  Show that $(n-1)a_1^2-2na_2>0 .$
 [/quote]
Solution to this nice problem.[quote="darij grinberg"]Since the polynomial P(x) has the leading coefficient 1 and is the product of first degree factors with real coefficients, it can be written in the form

$P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)...\left(x-x_n\right)$

with real $x_1$, $x_2$, ..., $x_n$. Opening the brackets, we obtain

$P\left(x\right)=x^n-\sum\limits_{i=1}^n x_i\cdot x^{n-1}+\sum\limits_{1\leq i<j\leq n} x_ix_j\cdot x^{n-2}+\underbrace{...}_{\text{lower powers of }x}$.

On the other hand, we know that

$P\left(x\right)=x^n+a_1x^{n-1}+a_2x^{n-2}+\underbrace{...}_{\text{lower powers of }x}$.

Hence, comparing coefficients, we get

$a_1=-\sum\limits_{i=1}^n x_i$; $a_2=\sum\limits_{1\leq i<j\leq n} x_ix_j$

Now,

$\left(n-1\right)a_1^2-2na_2=\left(n-1\right)\left(-\sum\limits_{i=1}^n x_i\right)^2-2n\sum\limits_{1\leq i<j\leq n} x_ix_j$
$=\left(n-1\right)\left(\sum\limits_{i=1}^n x_i\right)^2-2n\sum\limits_{1\leq i<j\leq n} x_ix_j$
$=\left(n-1\right)\left(\sum\limits_{i=1}^n x_i^2+2\sum\limits_{1\leq i<j\leq n} x_ix_j\right)-2n\sum\limits_{1\leq i<j\leq n} x_ix_j$
$=\left(n-1\right)\sum\limits_{i=1}^n x_i^2-2\sum\limits_{1\leq i<j\leq n} x_ix_j$.

Now, $\left(n-1\right)\sum\limits_{i=1}^n x_i^2=\sum\limits_{1\leq i<j\leq n}\left(x_i^2+x_j^2\right)$ (in fact, both of these sums count every $x_i^2$ exactly n - 1 times); hence, this becomes

$\left(n-1\right)a_1^2-2na_2=\sum\limits_{1\leq i<j\leq n}\left(x_i^2+x_j^2\right)-2\sum\limits_{1\leq i<j\leq n} x_ix_j=\sum\limits_{1\leq i<j\leq n}\left(x_i^2+x_j^2-2x_ix_j\right)$
$=\sum\limits_{1\leq i<j\leq n}\left(x_i-x_j\right)^2\geq 0$,

since squares are always $\geq 0$. Thus, $\left(n-1\right)a_1^2\geq 2na_2$, and we are done.

  darij[/quote]
Now could someone prove the second?


Gary performed the following three operations: 15 ÷ x, 21 ÷ x and 6 ÷ x. If he never got a remainder and x is an integer greater than 1, what is the value of x?

[hide]$ x$ is the gcd of 15,21,6, which is simply $ \boxed{3}$.[/hide]


a section of niagra falls has water flowing down a distance of 50 m at a rate of 1.2 x 10^6 kg/s (yes it's kilograms per sec). How much power is being produced by the water?

I know you need to find some way to convert the units into meters/second but i'm not sure how. I've managed to find the time (i think), however. This is what i did.

assuming the water is being accelerated by gravity (9.81 m/s^2), then g = v/t

but v = d/t so g = (d/t)/t = d/t^2 = 9.81 m/s

so g(t^2) = d and solving for t we get that t = (d/g)^1/2 and substituting the values we know i got t = (50m/9.81 m/s^2)^1/2 = 2.6 sec

by the way if you are able to help me on this problem, the book says that the answer is 590 MW

It's actually really simple. You don't need the velocity of the water or any such thing. The power that is produced is

$P=\dfrac{E}{t}$

where E is the energy the water gains from falling down 50 m:

$P=\dfrac{E}{t}=\dfrac{mgh}{t}=\dfrac{(Qt)gh}{t}$.

Q = 1.2 x 10^6 kg/s so the mass of water if Q x t. Now we get:

$P=Qgh$

or 588.6 MW.

well... the work of gravity all goes into giving the water energy...

[quote="ROLEX4life"]thanks for the help but i am still a little bit confused.

P = w/t but you used P = E/t....

did you do that because of the work energy theorem? (W = final potential energy - initial potential energy)[/quote]Work = the change in kinetic energy. By the conservation of energy, the change in kinetic energy = minus the change in potential energy. In this case, minus the change in potential energy = mgh. (The water starts at height h and ends at height 0.)

From there, $P = \dfrac{\partial W}{\partial t} = \dfrac{\partial \left(mgh\right)}{\partial t} = gh\dfrac{\partial m}{\partial t}$.

[quote="ROLEX4life"]thanks for the help but i am still a little bit confused.

P = w/t but you used P = E/t....

did you do that because of the work energy theorem? (W = final potential energy - initial potential energy)[/quote]Yes, I'm sorry, I use E instead of W all the time. And I also left out all the deltas, Dr. No has done it properly here ------^


[quote="Tiger3"]I was wondering why the sum of the exterior angles on ANY polygon is ALWAYS 360 degrees.  Please explain...[/quote]

[hide]The sum of the angles in an $ n$ sided polygon is $ 180(n\minus{}2)$.  To find the exterior angle, subtract the interior from $ 180$.  Thus $ 180\cdot n\minus{}180(n\minus{}2)\equal{}360$.[/hide]

Imagine you are walking around the polygon.  Eventually, you must end up where you started.  That means you must have turned in a full circle, $ 360^{\circ}$.

The sum of the interior angles in a regular [i]n-gon[/i] is given by $ \pi(n\minus{}2)$.
So, the value of one angle internal angle is given by $ \frac{\pi(n\minus{}2)}{n}$.
Think carefully. Lets refer to this angle as $ \alpha$

[hide="WHAT WE KNOW"]
$ \alpha <\pi$ because $ \frac{n\minus{}2}{n}< 1$.
But why is that relevant? If you extend any one of the sides, it is clear that;
$ \pi\minus{}\alpha\equal{}\theta$ and that if everything i've said thus far is true;

$ n\theta \equal{}2\pi$

Lets see if this holds.
$ n\theta \equal{} 2\pi$
$ \Longrightarrow n\left(\pi\minus{}\frac{\pi(n\minus{}2)}{n}\right)\equal{}2\pi$
$ \Longrightarrow n\pi\minus{}\pi(n\minus{}2)\equal{}2\pi$
$ \Longrightarrow n\minus{}n\plus{}2\equal{}2$
And, at last; $ 2\equal{}2$ verifying that all convex [i]n-gon[/i]'s must have an external angle sum of $ 2\pi$.[/hide]

I think it base on the sum of the exterior angles in a triangle...

I feel as though some of you have used the fact that the sum of the interior angles in any convex n-gon is $ 180(n\minus{}2)$ (In Degrees).  However, I see this almost as circular reasoning, since this obviously implies that the sum of the exterior angles is 360.  For this reason, I will prove why the said fact is true.

[hide="proof"]Choose any point in the interior of the polygon.  Draw from this point n lines that connect this point to the vertices.  This creates n triangles, whose angles add up to 180 each.  However, subtracting the 360 degrees formed from the angles at the point, we get that the sum of the interior angles is $ 180n\minus{}360\equal{}180(n\minus{}2)$[/hide]

I don't mean to sound rude, but just make sure that you can fully prove your argument...

ok..here it goes
[hide="reason why"]assume u have an n-sided polygon with angles A, B, C,...N.
so, by the formula that we know for the sum of the interior angles $ A\plus{}B\plus{}C\plus{}...\plus{}N\equal{}180(N\minus{}2)$. now the exterior angle at each interior angle can be represented by $ 180\minus{}A, 180\minus{}B..$ ectr. so the sum of the exterior angles is $ 180\minus{}A\plus{}180\minus{}B\plus{}180\minus{}C\plus{}...\plus{}180\minus{}N\equal{}180(N)\minus{}(180(N\minus{}2))\equal{}360$[/hide]


Hi

Where can I get more LaTeX fonts?

I just now \usepackage[T2A]{fontenc}

Thanks.

See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=psfchoice]Choice of scalable outline fonts[/url] and [url=http://tug.ctan.org/tex-archive/macros/latex/doc/fntguide.pdf]LaTeX2e font selection[/url]

Usually I take it from [url=http://www.tug.dk/FontCatalogue]LaTeX Font Catalogue.[/url]


algebra
function
domain
calculus
integration
abstract algebra
Ring Theory

I recall that $R$ is called Dedekind domain if it is an integral domain, integrally closed, $1$-dimensional (any nonzero prime ideal is maximal) and finally noetherian. Find examples of rings which verify every set of three conditions among these, but not the fourth one!

not integrally closed: Any non-maximal order in a number field or $\mathbb{C}[X,Y]/(Y^{2}-X^{3})$
not an integral domain: non-trivial direct sum of two Dedekind domains
not $1$-dimensional: $k[X,Y]$

And last but not least :), not Noetherian: the ring of all algebraic integers. It is, of course, an integrally closed integral domain, it's $1$-dimensional because it's integral over the $1$-dimensional ring $\mathbb Z$ and integral extensions preserve dimension, and it should be easy to prove that the ideal generated by $\sqrt[n]2,\ n\ge 2$ is not finitely generated.


Hi,
Does the infinite series $ \sum_{n\equal{}1}^{\infty}(e\minus{}(1\plus{}\frac{1}{n})^n)$ diverge?

It is divergent, refer to this thread for details.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1545439#1545439


All you do is solve $f(x)=\frac{ax+b}{cx+d}$ for $x$ to get $x=\frac{b-df(x)}{cf(x)-a}$.  And thus $f^{-1}(x)=\frac{b-dx}{cx-a}$.

It has the same method as the inverse of a matrix.

That is, if $f(x)=\frac{ax+b}{cx+d}$, then $f^{-1}(x)=\frac{dx-b}{-cx+a}$.

Right.
Basically, you can treat $f^{-1}(x)$ as just another variable in this case, so that $f(f^{-1}(x))=x=\frac{af^{-1}(x)+b}{cf^{-1}(x)+d}$ and just solve for $f^{-1}(x)$:
$(cx)f^{-1}(x)+dx=(a)f^{-1}(x)+b$
$(cx-a)f^{-1}(x)=(b-dx)$
$f^{-1}(x)=\frac{b-dx}{cx-a}$

you have post an extremely important property of the inverse of a function:
\[f[f^{-1}(x)]=x\]
and there is also:
\[f^{-1}[f(x)]=x\]

Note that $f$ has no inverse at $x=\frac{c}{a}$, even though $f$ itself is defined there.

[quote="shobber"][quote="jli"][quote="towersfreak2006"]If $\displaystyle f(x)=\frac{5x-6}{2x+3}$, and $\displaystyle f^{-1}(x)=\frac{rx+s}{tx+u}$, what is $r+s+t+u$?

In general, for $\displaystyle f(x)=\frac{ax+b}{cx+d}$, what is $\displaystyle f^{-1}(x)$?[/quote]

[hide]switch the x and the y

$x=\frac{5y-6}{2y+3}$

$2xy+3x=5y-6$

$2xy-5y=-3x-6$

$y=\frac{-3x-6}{2x-5}$

$-3-6+2-5=-12$  is there more than 1 solution?[/hide][/quote]
i don't think there is.
the function has only one inverse.[/quote]

However....

move the negative ffrom the numerator to the denominator and the answer is 12 ;)


algebra
polynomial
search
algebra unsolved

Let $ P(x)$ be a given polynomial with integer coefficients. Prove that there exist two polynomials $ Q(x)$ and $ R(x)$, again with integer coefficients, such that (i) $ P(x)Q(x)$ is a polynomial in $ x^{2}$; and (ii) $ P(x)R(x)$ is a polynomial in $ x^{3}$.

[quote="makar"]Let $ P(x)$ be a given polynomial with integer coefficients. Prove that there exist two polynomials $ Q(x)$ and $ R(x)$, again with integer coefficients, such that [b](i)[/b] $ P(x)Q(x)$ is a polynomial in $ x^{2}$; and [b](ii)[/b] $ P(x)R(x)$ is a polynomial in $ x^{3}$.[/quote]

(i) : $ P(x)P(\minus{}x)$ is an even polynomial and so may be written as $ H(x^2)$ and so $ \boxed{Q(x)\equal{}P(\minus{}x)}$

(ii) : If degree of $ P(x)$ is $ n$, if coefficient of $ x^n$ is $ a_n$ and if $ z_i$, for $ i\in[1,n]$ are all roots of $ P(x)$, then :
$ P(x)P(\frac{\minus{}1\minus{}i\sqrt 3}2x)P(\frac{\minus{}1\plus{}i\sqrt 3}2x)$ $ \equal{}a_n^3\prod(x^3\minus{}z_i^3)$

And, obviously, in $ P(\frac{\minus{}1\minus{}i\sqrt 3}2x)P(\frac{\minus{}1\plus{}i\sqrt 3}2x)$ :
All $ i$ terms disappear (since the two factors are conjugate terms)
All $ \sqrt 3$ terms disappear (since attached to $ i$ : all their odd powers disappear)
We can make all $ \frac 12$ terms disappear too when multiplying by $ 2^{2n}$ and we get a polynomial with integer coefficients : $ \boxed{R(x)\equal{}2^{2n}P(\frac{\minus{}1\minus{}i\sqrt 3}2x)P(\frac{\minus{}1\plus{}i\sqrt 3}2x)}$

pco, you have a nice solution, with your method, this problem can prove with x^k.


calculus
derivative
absolute value
inequalities unsolved
inequalities

[quote="Xevarion"]Max is 0 since we cannot have $a > b, b > c, c > a$.

I believe the min is $\frac{-1}{27}$ but I'm not confident that my reasoning is perfectly rigorous.[/quote]

you can have two negative and one positive.

[quote="snake_pissken"]Let $a,b,c$ satisfy $a,b,c \geq 0$ and $a+b+c=1$.

Find the $maximum$ and $minimum$ value of

$P(a,b,c)= (a-b)(b-c)(c-a)$[/quote]

Let $a$ be the minimum one. In case of $a\leq b\leq c$, P is positive, and in other case, namely $a \leq c \leq b$, P is negative. Moreover, $P(a,b,c)=-P(a,c,b)$. So the minimum value of P has the same absolute value with the maximum value of P.

Let $b=a+x, c=a+x+y$. Then $1=3a+2x+y \geq 2x+y$, so $P=xy(x+y) \leq x(1-2x)(1-x)$. The derivative of this expression wrt $x$ is $6x^{2}-6x+1$, so the maximum takes place at $x=\frac{3-\sqrt{3}}{6}$, and the value is $\frac{\sqrt{3}}{18}$. (Minimum is $-\frac{\sqrt{3}}{18}$)

[quote="Rzeszut"][quote="A. S. K."]if we set A-G inquality, we get
P<= 0
so max is 0.[/quote]
$\mathrm{AM-GM}$ works only for positive numbers.[/quote]

oh yes...

[quote="snake_pissken"]Let $a,b,c$ satisfy $a,b,c \geq 0$ and $a+b+c=1$.

Find the $maximum$ and $minimum$ value of

$P(a,b,c)= (a-b)(b-c)(c-a)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3533254#p3533254


[url=http://ca.today.reuters.com/news/NewsArticle.aspx?type=topNews&storyID=2007-11-27T191207Z_01_N26366544_RTRIDST_0_NEWS-PALESTINIANS-ISRAEL-COL.XML]This URL was too long and a moderator fixed it.[/url]

Israel and the Palestinians have started talks of creating a peace treaty.  Do you think it will last?

Honestly, I don't think it will work too well.  Even if there is a peace treaty, some Palestinians will still hold a resentment toward Israel, and an attack might slip through.  If it does work at all, it will probably be only short-term.

I would think you should wait until it's been completed and signed before you ask whether it will last.  Also, please use the URL tags to make your urls shorter so they don't stretch my screen.

Because, if anything, the Arab lesson from sixty years of Israeli-Palestinian conflict should be that bloody-mindedness, violence, warmongering, hatred and unrealistic goals are [i]extremely effective[/i] at bringing peace, safety, and comfort to Israel, Arab states and the Palestinians themselves.

Grow up.

[quote]... if Israel is removed from the Map.[/quote]
Israel's neighbors tried this, more than once. Israel won those wars, and still has a very strong military to defend itself.

The country has been around for 60 years. Stop talking about eliminating it- that sort of rhetoric doesn't get used anywhere else.

There are plenty of extremists with ugly positions on both sides (notably "We have a divine right to all of the land"), and the political process in Israel and the occupied areas isn't helping. For peace to have a chance, things need to change first, with the extremists marginalized and unpopular. Until then, there won't be much change.

Consider the resistance of Hezbollah and Hamas and their achievements , how can we say fighting with Israel is in Israel's benefits  ?

Who is that question addressed to and what is it supposed to mean?  The behavior of Hamas and Hezbollah hasn't been good for the Israeli state or people, and it hasn't been good for the Palestinians or Lebanese, either.

[quote="Toxic_Frog"]It's only a photo opportunity .The true governing party of Palestine is Hamas and it hasn't attended the conference so with all western Medias's boasting and propaganda the conference is NOTHING indeed.Palestine issue would be solved completely if and only if Israel is removed from the Map.[/quote]

I'm gonna run contrary to what you guys deem politically correct and agree with this statement.

However... if "Israel is removed from the Map", this is going to spawn so many problems. What would the perpetrators do about allegations of total disregard for human life and rights? Retaliation by other communities? Total racist/ideologist/religious backlash? It's a totally unrealistic policy that sounds more like attention whoring than actual diplomacy.

[quote="Lazarus"]I'm gonna run contrary to what you guys deem politically correct and agree with this statement.[/quote]

This sentence is stupid and content-free: calling your own statement politically incorrect does not substitute for providing an argument to support it.  Calling someone else's statement politically correct does not substitute for providing an argument to counter it.  More generally, the phrase "politically correct" as used by you in this context does not, as far as I can tell, have any meaning.  Finally, it's not clear that you're actually disagreeing with anyone on this thread: jmerry and I both responded to the last of the three sentences, while you appear to agree only with the first two.

I'm afraid this is just some show by president Bush (Clinton did the same thing at the end of his second term), and perhaps intended to please Arab allies of the USA.

I find it hard to judge from here.... on both sides I see questionable things.

Well, what does Bush have to show for his presidency?  During the Bush administration, the United States went from a surplus to a deficit.


AMC USA(J)MO USAMO AIME geometry calculus integration

We will be hosting a Math Jam on Friday April 23 at 7:30 EST (4:30 PST) to discuss strategies for approaching the USAMO.  We will discuss topics such as how to allocate your time, how to efficiently brainstorm, how to organize your work, general problem solving strategies, and tips for how to write a clean proof.

It would be hard (impossible?) to determine whether a person under a username qualified.  Anyone should be allowed to go, anyway.

[quote="beta"]Yeah, didn't make it this year doesn't mean you won't make it next year.(although I will probably never make it in my life)[/quote]

You seemed to understand how to solve a good number of AIME problems.  I'm not sure why you would decide that the USAMO is out of reach.  I know many students who have improve their AIME scores significantly over the course of a year.  Between my first two years taking the AIME my score improved by 8 problems.  I did work very hard, but that is an improvement that is attainable.  A student from Texas made a 15 this year after making a 6 last year.  Some students just hit a stride as their enthusiasm for the subject grows.  It appears that you enjoy problem solving. :)

[quote="MCrawford"] A student from Texas made a 15 this year after making a 6 last year. [/quote]

YES!!...that means there's a hope of 15 next year for every USAMO qualifier out there (and the hundreds of 6's that barely missed it)! :-D

gj fergie :-)

Thanks, Mr.Crawford. I just got pretty hopeless after missed AIME on both AMCs this year. I really like problem solving.

I'm really excited for this Math Jam! Thank you so much for doing it... im sure it will help alot! (and I need alot of help :) )

[quote="zabelman"] (and I need alot of help :) )[/quote]

haha...yea...me too. I have like almost no clue about how the USAMO works at all :-( The only things I've been able to pick up are from the AMC site and from threads here on AoPS...too bad I won't be able to make the MathJam, but I'll pick up a copy of the transcript.

For those who feel like they don't know what to expect on the USAMO, I have a few book recommendations.

First, there is a book published by the MAA that includes USAMO problems from 1972-1986.  These problems are mostly easier than current USAMO problems (with a couple of exceptions of the kinds of outrageously lengthy problems that are typically excluded from more recent USAMOs), but they give a good indication of what you can expect (as well as some practice).

The Paul Zeitz book [u]The Art and Craft of Problem Solving[/u] is a good primer for Olympiad level problem solving.

There is a great book called [u]Problem Solving Strategies[/u] by Arthur Engel that I consider a particularly good read for students who can already solve some Olympiad level problems and are trying to progress further.

These books seem to do a lot, but they don't explain the basics of proof-writing.  We'll try to fill in that gap at the Math Jam.  Once you master the basics of proof-writing you will quickly begin to see the difference between the process that you walk through to solve a problem and the way you present those results as a proof.  These processes are typically reversed.  When solving a problem, you work from any information you have to work with until you get to the very basic calculations/ideas that are the solution.  When writing the proof, you start with the very basic calculations/ideas (those things that are trivially valid) and work the opposite direction until the last you demonstrate is the original problem statement. (at which point you say "voila!", "qed", "awd", or some other goofiness).

[quote="MCrawford"](at which point you say "voila!", "qed", "awd", or some other goofiness).[/quote]

hmm...how bout that lil black square?...although I don't know if CAMC likes looking at black squares on answer sheets...

Also, what if our solution is longer than the one page they give?

[quote="MithsApprentice"]
Also, what if our solution is longer than the one page they give?[/quote]

You can turn in other sheets of paper.  I can remember problems at MOP on which I wrote 6-12 page solutions (not many, but several).  Usually solutions will only require 1 to 4 pages depending on handwriting, organization, and diagrams (expect geometry solutions to take more pages due to diagrams).

[quote="MCrawford"] (expect geometry solutions to take more pages due to diagrams).[/quote]

In my experience, (classical) geometry problems usually have the shortest solutions using synthetic methods. It's hard to find, but usually quick to write. (Diagrams are often helpful in a write-up, except in situations where there are distinct configurations, then you need to be careful on what you draw)

The longest solutions are those where you have to explain things (e.g. combinatorics) and where there is a lot of cases. In general, number theory and combinatorical problems have long solutions. When typed, functional equations may have rather long solutions, because of the number of display equations, but if you count the number of words, it really isn't much.

In conclusion, the length of the solution have NOTHING to do with the difficulty of the problem. Short solutions don't mean that they are easy to find. And more straight-forward solutions may need long write-ups (e.g. if you were to write a olympiad-style solution to some of the AIME problems, it may take a while to explain your ideas, despite the fact that they are very straightforward).

I agree with that.  Many of the harder problems have short solutions and synthetic geometry is shorter than other geometry.

Some hard problems do however have very long solutions, but that is usually for reasons you mentioned (casework or lengthy, but integral examples).

what does "awd" mean? and then there's the halmos square: [tex]\blacksquare[/tex] is cool too. And there's W^5 (which was what we wanted) and ... heh they're not goofy they're cute and huggable


quadratics
number theory proposed
number theory

I indeed misunderstand the problem :blush: .Please delete my post....

It is equivalent to $\frac{a}{b}=b+\frac{a}{10}$ or $a(10-b)=10b^{2}$ and since $(a,b)=1$ we have that $a|10$ which gives $4$ quadratics to solve.

[quote="Hawk Tiger"]Is it means that $\frac{a}{b}=b+\frac{a}{10},(0\leq a\leq 9,1\leq b<9$?[/quote]

Nothing says that $a$ has just one digit. So it really means (according to me) :
$\frac{a}{b}=b+\frac{a}{10^{n}},(10^{n-1}\leq a<10^{n},1\leq b$

Then $10^{n}a=b(10^{n}b+a)$ and $a|10^{n}$ and $b|10^{n}$ $\Rightarrow$ either $a=2^{u}$ and $b={5^{v}}$, either $a=5^{u}$ and $b={2^{v}}$

1) case $a=2^{u}$ and $b=5^{v}$
$2^{u+n}5^{n}=5^{v}(5^{v+n}2^{n}+2^{u})$ with $10^{n-1}\leq 2^{u}<10^{n}$, and $u\geq n$ (the case $n=1$, $u=0$ gives no solution)
So $2^{u}5^{n}=5^{v}(5^{v+n}+2^{u-n})$, and $u=v=n$ and no solution.

2) case $a=5^{u}$ and $b=2^{v}$
$2^{n}5^{u+n}=2^{v}(2^{v+n}5^{n}+5^{u})$ with $10^{n-1}\leq 5^{u}<10^{n}$, and $u\geq n$ (the case $n=1$, $u=0$ gives no solution)
So $2^{n}5^{u}=2^{v}(2^{v+n}+5^{u-n})$, and $u=v=n=1$

And the solution found by Hawk Tiger (but without showing $n=1$):
$\frac{5}{2}=2.5$

[b][u]Edited [/u][/b]:
The same problem in any base gives a unique solution :
The base must be $base=n(n^{2}+1)$ with $n>1$ and the unique solution is $\frac{n^{2}+1}{n}=n_{base}.(n^{2}+1)_{base}$
With $n=2$, we have the current problem.

--
Patrick

Why $(a,b)=1$?
If we found all solutions, then:
$a=5^{2n-u}(5^{u}-2^{v})^{2}2^{2n-v}, b=2^{n}(5^{u}-2^{v})5^{n-u}$
But we must chose $5^{u}>2^{v}, \ (5^{u}-2^{v})^{2}<2^{v-n}5^{u-n}, u\le n,v\le 2n$.
Obviosly solution $u=n=1,v=2$.
But I don't no, had it solutions u>1 or not.
From abc conjecture it is easy to show, that number of solutions are finite.

Statement of the problem says that $(a, b)=1$...

[quote="N.T.TUAN"]Find all pairs of coprime natural numbers $a$ and $b$ such that the fraction $\frac{a}{b}$ is written in the decimal system as $b.a.$[/quote]

Assume that $10^{n-1}\leq a < 10^{n}$ where $n$ is a positive integer. We have:

$\frac{a}{b}= b . a \iff \frac{a}{b}= \frac{10^{n}b+a}{10^{n}}\iff \frac{ab}{a-b^{2}}= 10^{n}$

We shall prove that $ab$ and $a-b^{2}$ are relatively prime. Assume they have a common prime factor $p$.

Notice $a^{3}= a^{2}(a-b^{2})-ab \times ab$ and $b^{3}= ab-b(a-b^{2})$ which immediately shows a contradiction to $(a,b)=1$.

Therefore it follows that $a-b^{2}=1$ and that $ab = 10^{n}$. Since $(a,b)=1$ and $a= 1+b^{2}>b$ thus $a=5^{n}$, $b=2^{n}$.

Plugging the above into $a= 1+b^{2}$ we have $4^{n}+1 = (4+1)^{n}$. So the only possibility is $n=1$.

Therefore $a=5$, $b=2$ is the only solution.

[quote="Hawk Tiger"]I indeed misunderstand the problem :blush: .Please [b]delete my post[/b]....[/quote]
Hawk Tiger,
Not necessary and you so not necessary must edit your post if you misunderstand the problem. We are discussing and we can misunderstand, wrong, false, e.t.c . Don't worry about that.  :wink: And if who quote your post then please don't edit. Final, this is a good topic!  :)


Kind of rough but hey, it's getting quite late.
[hide]The Inequality is equivalent to $(a+b)^{2}(a-b)^{2}+4ab(a+b)(a-b)+4a^{2}b^{2}\leq 2$
By AM-GM we see that $\frac{a^{2}+b^{2}}{2}\leq \sqrt (a^{2}b^{2})$ and thus we deduce the fact that $\rightarrow$

1. $1 \leq a^{2}b^{2}\leq 2ab$

2. $-2ab \leq a^{2}b^{2}$

And after a bit of rearraniging and manipulation we find that the orginal Inequality is true.[/hide]

[quote="shobber"]Let $a=m\cos{x}$ and $b=m\sin{x}$, then $|a^{2}+2ab-b^{2}|=m^{2}|\sin{2x}+\cos{2x}| \leq \sqrt{2}m^{2}$. Since $a^{2}+b^{2}=m^{2}\leq 1$, so $\sqrt{2}m^{2}\leq \sqrt{2}$.[/quote]

Okay, you beat me. I like this solution better than my one. :D