Gauss
combinatorics proposed
combinatorics

I think this can be proved by following way;;
Consider the tournament result as a directed graph.
Note that every complete directed graph either has a full length cycle or can be divided into two sets $ A, B$ such that every edges between them are directed to $ B$. This can be proved by considering maximum cycle of the graph...
If we apply this lemma repeatedly, we can divide a complete directed graph $ G$ into some cycles $ A_{1}, A_{2}, \cdots , A_{n}$ such that if $ i < j$, every edges among $ A_{i}, A_{j}$ are directed to $ A_{j}$, so it is enough to prove that each cycles have length less than $ M \plus{} 1$. If it can, we can get a desired arrangement of $ G$ by aligning points like $ A_{1}, A_{2}, A_{3}, \cdots$ and each points in the same cycle by the reverse order of original cycle.
We'll prove that if there are no $ n$-cycle in a complete directed graph, any $ n \plus{} 1$-cycles also doesn't exist. if not so, let $ C$ be that cycle. Remove one point $ \{x\}$ in the $ C$. Applying the lemma to $ C/\{x\}$, we can divide it to $ A$, $ B$, which satisfies the fact mentioned above... because there are no full cycle. and by its definition, initial $ n \plus{} 1$-cycle should be in the form of $ x \rightarrow A \rightarrow B \rightarrow x$. Let $ a$ be the lastest point of $ A$ in the cycle. than the cycle would proceed as $ a \rightarrow b_{1} \rightarrow b_{2} \rightarrow \cdots$. However, there is also $ a \rightarrow b_{2}$, and this reduces the length of the cycle by 1. This contradicts the assumption...

[quote="Little Gauss"]I think this can be proved by following way;;
Consider the tournament result as a directed graph.
We'll prove that if there are no $ n$-cycle in a complete directed graph, any $ n \plus{} 1$-cycles also doesn't exist. ...[/quote]
Dear Little Gauss, nice solution! In fact there is always at least one chain passing all the points in a complete graph.Then by the lemma above we get the proof immediately. :)

We prove the result by induction on $M$ and $N$. Notationally, we say $p_1>p_2$ if player $p_1$ beats player $p_2$.

Suppose that $M=2$. Then the relation between players is transitive, which easily implies that they can be uniquely ordered. Rigorously, take the player with the most wins, and note that by transitivity if any player beat the first, the second player would have strictly more wins. So there is an undefeated player, and the result follows by induction on the rest of the tournament.

Now suppose $M>2$. The condition is equivalent to there being no $M+1$ cycle. We now split up the proof into two cases.

[b]Case 1:[/b] There is an $M$-cycle $v_1>v_2>\cdots > v_M>v_1$, denoted $C$, in the tournament. Consider some $w$ not in $C$. If $w>v_i$ for some $i$, then
\[
w>v_i>v_{i+1}>\cdots >v_{i-1}
\]
This is a cycle of length $M+1$, where the indices are taken modulo $M$, so $w>v_{i-1}$. But from $w>v_i\implies w>v_{i-1}$ it easily follows that if $w>v_1$, then $w>v_i$ for all $i$. Similarly, if $w<v_1$, then $w<v_i$ for all $i$. Thus we can partition the tournament into sets $A,B,C$, where $A$ beats all members of $C$, $C$ beats all members of $B$, and $a+b+M=N$, where $a$ and $b$ are the sizes of $A$ and $B$. Now for any member $x$ of $A$ and any member $y$ of $B$, $x>v_1>v_2>\cdots >v_{m-1}>y$, so $x>y$.

We now assign the labels $\{1,\cdots b\}$ to $B$, $\{b+1,\cdots N-a\}$ to $C$, and $\{N-a+1,\cdots N\}$ to $A$. By the inductive hypothesis, we can assign some relative ordering of the labels in each set that is internally consistent with the desired property of the problem in both $A$ and $B$, and we can label $C$ arbitrarily in an internally consistent way, as long as the player labelled $M$ beats the player labelled $1$. Also, as all members of $A$ beat all members of $C$ which beat all members of $B$, the property $a\ge b+M-1\implies$ player $a$ beats $b$ is vacuously true between two members of any two of $A,B,C$. Thus we can construct a valid labelling.

[b]Case 2:[/b] There is no $M$-cycle. Then by the inductive hypothesis on $M$, we can label the players in such a way that $a\ge b+M-2$ implies $a$ beats $b$, a strictly stronger condition than the one needed.

Dear Little Gauss,your soluntion is very nice.But actually we can't always divide a complete directed graph without a full length cycle into two sets [b]A,B[/b] such that every edges between them are directed to [b]B[/b]

[quote=Tanyingjie]But actually we can't always divide a complete directed graph without a full length cycle into two sets [b]A,B[/b] such that every edges between them are directed to [b]B[/b][/quote]

Yes you can. :)

Consider any maximal cycle $C$, say  \[ v_1 \to v_2 \to \dots \to v_k \to v_1. \] By maximality, this means we cannot have $v_i \to w \to v_{i+1}$ for any $i$ and any vertex $w \notin C$.  It follows that for any $w \notin C$, either [list] 	[*]$v_i \to w$ for every $i$, 	in which case we say $w$ is [i]weak[/i], or 	[*]$w \to v_i$ for every $i$, 	in which case we say $w$ is [i]strong[/i]. [/list] It then follows that every strong vertex beats every weak vertex too (if $w$ is strong and $x$ is weak, then $w \to v_1 \to \dots \to v_k \to x$ is a path). Since $C$ wasn't the entire graph, there is either at least one strong or at least one weak vertex. Done.

[asy] size(6cm); dotfactor *= 1.5;

pair A = dir(90); pair B = dir(210); pair C = dir(330);

dot("$v_1$", A, A, red); dot("$v_2$", B, B, red); dot("$v_3$", C, C, red); draw(A--B, red, EndArrow, Margin(2,2)); draw(B--C, red, EndArrow, Margin(2,2)); draw(C--A, red, EndArrow, Margin(2,2));

draw(ellipse( (2.6,0.3), 0.6, 1.6), blue); draw(ellipse( (-2.6,0.3), 0.6, 1.6), blue); draw((1,0.3)--(2,0.3), blue+1, EndArrow, Margin(2,2)); draw((-2,0.3)--(-1,0.3), blue+1, EndArrow, Margin(2,2)); draw((-2.3,-1.2)--(2.3,-1.2), blue+dashed, EndArrow, Margin(2,2));

dot((2.8,0.7), blue); dot((2.7,1.2), blue); dot((2.5,0.1), blue); dot((-2.8,0.7), blue); dot((-2.7,-1.0), blue); dot((-2.3,0.1), blue); [/asy]

Incidentally, this is also the main idea underlying [b]pi37[/b]'s solution, which is the same as my own. Rather than taking $C$ to be a maximal cycle we take $C$ to be an $M$-cycle which also solves the problem.

[quote=MellowMelon]Let $ N > M > 1$ be fixed integers. There are $ N$ people playing in a chess tournament; each pair of players plays each other once, with no draws. It turns out that for each sequence of $ M \plus{} 1$ distinct players $ P_0, P_1, \ldots P_M$ such that $ P_{i \minus{} 1}$ beat $ P_i$ for each $ i \equal{} 1, \ldots, M$, player $ P_0$ also beat $ P_M$. Prove that the players can be numbered $ 1,2, \ldots, N$ in such a way that, whenever $ a \geq b \plus{} M \minus{} 1$, player $ a$ beat player $ b$.

[i]Gabriel Carroll.[/i][/quote]

We induct on $N>M>1$. For $M=2$, pick a player $v$, then partition the remaining players in two groups $A, B$ so that $\overrightarrow{va}, \overrightarrow{bv}$ are edges for all $a, b \in A, B$ respectively. The condition ensures that $\overrightarrow{ba}$ is an edge for all $a \in A, b \in B$. Hence, we can label $B$ with $1, 2, \dots, |B|$ by induction on $N$ and $A \cup \{v\}$ by $|B|+1, \dots, N$, so that both are internally satisfied. This provides a valid labelling.

Now for the induction step. For fixed $N$, suppose $1<M<N$ is the smallest integer with this property.

[b]Claim.[/b] The maximal cycle has size at least $M$.

(Proof) Pick two points $P_0, P_d$ with the maximum graph theoretic distance. Then $P_0 \rightarrow P_1 \rightarrow  \dots \rightarrow P_d$; so $d>1 \implies P_d \rightarrow P_0$ then we have a $(d+1)$ cycle. If $d<M-1$ then for any chain $P_0 \rightarrow P_1 \rightarrow \dots \rightarrow P_{M-1}$ we have $P_0 \rightarrow P_{M-1}$ contradicting the minimality of $M$. Hence, $(d+1) \ge M$, as desired.

Now we pick this maximal cycle $\mathcal{C}$. For any vertex $u$ if we can find $u, w \in \mathcal{C}$ with $\overrightarrow{vu}, \overrightarrow{wv}$ as edges then it is possible to append $v$ in $\mathcal{C}$. So for all $v \not \in \mathcal{C}$ all edges between $v$ and vertices of $\mathcal{C}$ have the same direction.

Suppose $A$ is the set of those vertices where these edges lead inwards and $B$ where they lead outwards. For any $a \in A, b \in B$, if $\overrightarrow{ab}$ is an edge, then we have a directed path $a \rightarrow b \rightarrow v_1 \rightarrow \dots \rightarrow v_{M-1}$ for $v_1 \rightarrow \dots$ as a chain of $\mathcal{C}$, so $a \rightarrow v_{M-1}$, a contradiction! Thus, all players outside $A$ defeat everyone in $A$ (if $A$ is empty, $\mathcal{C}$ fits this role). Labelling $1, \dots, |A|$ to vertices in $A$ in a suitable way, we can translate a natural labelling of the remaining vertices by $|A|$.

The following claim solves the problem easily:
In each tournament, either there is a hamiltonian cycle or the participants can be partitioned into two nonempty sets $A,B$ auch that each player in $A$ beats each player in $B$.
Edit: I see it is Little Gauss’s solution.

Did this in November and posting this now... I will use two lemmas.

Lemma 1(Very standard)
There exists a path of length $N-1$.

We proceed by strong induction. Suppose $v\rightarrow v_1,\cdots,v_k$. Then by inductive hypothesis, there is a path of length $k-1$ in $v_1,\cdots,v_k$ and a path of length $N-k-2$ in the other $N-k$ vertices. Hence the path (length $N-k-2\rightarrow v\rightarrow $ path of length $k-1$) works.

Hence we can take this path and label $v_N\rightarrow v_{N-1}\rightarrow \cdots \rightarrow v_1$. To finish the problem, we use the following lemma.

Lemma 2.
There exists no cycle of length $\ge M$.

We instead show if there exist a cycle of length $x+1$ then there is a cycle of length $x$ for $x\ge 3.$

Consider the paths $v_{i}\rightarrow v_{i+3}\rightarrow v_{i+4}\rightarrow \cdots \rightarrow v_{i-1}\rightarrow v_{i+2}\rightarrow v_{i}$. For these paths to not exist, $v_{2i}\rightarrow v_{2i\pm 3}$ for all $i$ or $v_{2i\pm 3}\rightarrow v_{2i} $. This gives a contradiction if $l$ is odd. We will deal with the case $l$ is even and $l\ge 8$.

For $l=4$, we can't even deal with $v_2\rightarrow v_0\rightarrow v_2$, so it is okay.

For $l\ge 8$,
the cycle $v_0\rightarrow v_3\rightarrow v_1\rightarrow v_2\rightarrow v_5\rightarrow v_6\rightarrow v_7\rightarrow \cdots \rightarrow v_{l-1}\rightarrow v_0$ works. Notice how $v_4$ is not in the cycle.

Therefore, taking the longest path $v_n\rightarrow v_{n-1}\rightarrow \cdots \rightarrow v_1$, and labelling $v_i$ with $i$ indeed works.

Twitch lemma unit

Note that this implies no cycles of length $M+1$ or larger exist.
Let $C$ be the cycle of maximal length. Then the rest of vertices can be partition into $A \sqcup B$ where $A$ points into $C$ and $C$ points into $B$.
Since $C$ has length at most $M$, we can solve this inductively for $A, B, C$ and sort them in that order.


quadratics
function
conics
parabola
special factorizations

Suppose the revenues R(X) of producing x items is modled by R(x)=-(x-40)(x-210)

Find and interpret R(75)=

Graphically where does the maximum occur?

What is the maximum revenue?

What number of items should be sold to produce maximum revenue?

When is the revenue greater than 5000  i.e.  R(x)greater than equal to 5000?

I just dont know how to start.  :(

All you need to start on is simple quadratics, first multiply the quardratc equation out in form of (ax^2+bx+c), use -b/2a to find the number of items, and then plug -b/2a into the equation to get the maximum.

jimhu may be getting a bit ahead there. If you're not even sure where to start, let's take it a bit slower. This sounds like a homework question to me, but I hope you'll get something out of it anyway.

Also, if you don't mind, I'll rewrite your function as $\displaystyle R(x) = (-1)(x-40)(x-210)$, since just having a - sign in front of a bracket with nothing to subtract it from is ugly.

$a)$ This is simple. You just have to find $R(75)$. So, therefore: \[\displaystyle R(75) = (-1)(75-40)(75-210)\]
\[R(75) = (-1)(35)(-135)\]
\[R(75) = 4725\]
Therefore, the revenue is 4725. Simple, eh?

$b)$ This one is explainable in words. This function is second degree, if we multiply it out. It'll take the form of $Y = Ax^2+Bx+C$ so it must be a parabola. Therefore the maximum will occur, if we draw out this parabola, at it's [i]vertex[/i], which is the highest point, the peak, of a parabola.

$c)$ Since we know the maximum value is at the vertex, we should write the function in it's "vertex form" (aka $\displaystyle y = a(x - h)^2 + k$, in which $(h,k)$ is the vertex). So, first let's multiple the factored form out.\[\displaystyle R(x) = (-1)(x-40)(x-210)\]
\[\displaystyle R(x) = (-1)\left[(x)^2-250x+8400)\right]\]
And, completing the square, we get:
\[\displaystyle R(x) = (-1)\left[(x-125)^2-7225\right]\]
\[\displaystyle R(x) = (x-125)^2+7225\]
Therefore, the maximum value is $7225$, and it happens when $x=125$.

$d)$ Looking at the formula above, $R(125) = 7225$, which is the maximum value. Therefore, $125$ units should be produced.

$e)$ This is back to the repetitiveness. Let's just set up: \[\displaystyle R(X) = 5000\]
\[\displaystyle (-1)(x-40)(x-210) = 5000\]
\[\displaystyle (-1)(x^2-250x+8400) = 5000\]
\[\displaystyle (-x)^2+250x-13400 = 0\]
\[\displaystyle x^2-250x+13400 = 0\]
I'm too lazy to factor that so I'll let you do it ^^;;

Sorry if I've been overly verbose, but this sounded like a homework question so I felt it needed a more thorough explanation...


linear algebra
matrix
function
induction

How about it? It seems like the last one failed - perhaps there are more people on this forum now(or maybe not).

Here's a problem:
A Cell Lacking in it's ability to secret glycoproteins would most likely be deficient in its:
a) nuclear DNA
b) extracellular matrix
c) golgi apparatus
d) b and c
e) a,b,c

explain your answer too.

choice d)" b and c" although the answer should propably be "b [b]or[/b] c"
i'm not 100% sure and i don't have another question.

Yeah D is the correct answer because almost all cell surface proteins involve the Golgi to get there. Of course, a cell that lacking in its ability to secrete the glycoproteins will not have a very strong extracellular matrix due to the fact that glyoproteins are a very important component of the ECM.

Here is my biology trivia question (from the USABO Open Exam last year):

What type of epithelium is found lining the inside of the urethra?

That's correct. Here's another one:

What is the name of the enzyme that catalyzes the $\text{ATP}\rightarrow \text{cAMP}$ reaction?

Adenylyl Cyclase A

how about this...
What are the flattened discs that make up the Golgi Bodies called?

I think they're called cisternae.

A trisomy of chromosome 21 qould result in what disorder?

down's syndrome

what is the most common neurotransmitter in the body?

If I had to guess I'd say acetylcholine...

What is the most important type of bond that makes the secondary structure of a protein different from the primary structure of a protein?

hydrogen

what's the difference between nitrifying, nitrogen-fixing, and denitrifying bacteria?

The actual differences? As in species differences? I'm not sure about that in so much depth. Of course, nitrogen-fixing bacteria catalyze $\text{N}_{2}\rightarrow\text{NH}_{3}$, nitrifying bacteria catalyze $\text{NH}_{3}\rightarrow\text{NO}_{2}^{-}$ and $\text{NO}_{2}^{-}\rightarrow\text{NO}_{3}^{-}$, and then, of course, denitrifying bacteria catalyze $\text{NO}_{3}^{-}\rightarrow\text{N}_{2}$.

sorry, I should have been more specific.  Your answer was what I was looking for, anyway.  Post a question?

In in when [i]lac[/i] proteins are transcribed, we can infer that the repressor is ________.

Sorry if it's too easy.

I was looking for [i]not able to bind to the promoter[/i] or [i]inactive[/i]

Yeah cause lactose acts as the allosteric inhibitor. But I was thinking "inducable operon."

oi... we should keep this going.

since the answer's out and no-one is posting one, I guess I will go...

what enzyme stimulates uterus contraction during childbirth?

I believe it's my good brotha secreted from the posterior pituitary, oxytocin!

What is the difference between autopolyploidy and allopolyploidy?

Autopolyploidy is when an organism (commonly plants) produces gametes which are polyploidy, and this results in polyploidy offspring. Allopolypoidy is when 2 gametes of different species join during hybridization. This can be 2 normal gametes or aneuploidy gametes of 2 species. The two processes (allo/autoploidy) both produce polyploidy offspring which are usually unable to mate with other species, and thus this results in sympatric speciation.

Why do steriods usually trigger intracellular signal transduction pathways while protein hormones usually trigger intercellular signal transduction pathways?

Steroids are hydrophobic and can thus pass through the plasma membrane to trigger the receptors located inside the cell, while protein hormones are unable to pass through the membrane (because they are hydrophilic and too large), so they must bind to receptors on the outside of the cell. I'm not completely sure if this is the right answer (I'm just learning bio and this is one of the few questions in this thread I think I can get). If it's right then someone else can post a new question.

I believe that is pretty much correct... I'll post another question

What is the name of the "blisters" on tree trunks that allow gas diffusion to occur in those places?

This is an obscure guess, but would it happen to be lenticles? I think lenticles are the cracks formed on the cork of the trees that allow air flow.

What is a the function of a morphogen in the development of organisms?

I invented morphogens. They're cool substances that are important in development and differentiation. They form gradients in cells or regions of cells and will usually serve as transcription factors to activate transcription of other genes. They are very important in axis formation in embryos.

My Question:

Sticking with developmental biology, what is cellular induction?

I believe that cellular (or embryonic) induction is when one or a group of cells influences neighboring cells by secreting chemicals that act through diffusion. By the way, these influential cells are called organizers.

On a relatively similar note: describe epistasis.  :D

Epistasis is where the function of one gene can block the function of another gene. I'm not sure if that was a good way to describe it...

If that's right, then someone else can post a question.

Describe supercoiling and what enzyme/s are involved in regulating it.

Its just the coiling of DNA into really really compact coils... and, as for the proteins... I only remember the histones(what were they, h1-h5?)

Yeah daveed I think they're probably still histones. Post a question.

Supercoiling is not the coiling of DNA around histones.

[hide="hint"]It occurs during DNA replication.[/hide]

Then could it be....

[hide="this?"]The condensing of DNA (i.e. increased visibility) in DNA replication phases like prophase to convert chromatin into chromosomes?[/hide]

EDIT: I think I'm simplifying things too much.

[hide="hint"]Topoisomerase is involved; it regulates this process during DNA replication.[/hide]

I think it's oxytosin, but I don't have any good question right now.


Guess I should stop proposing problems at 2:00 AM, as this can lead to ones like this here:

Let $a$, $b$, $c$ be three positive reals. Prove the inequality
$\frac{a^2+2b^2}{b+c}+\frac{b^2+2c^2}{c+a}+\frac{c^2+2a^2}{a+b}\geq\frac32\left(a+b+c\right)$.

[quote="darij grinberg"]
Let $a$, $b$, $c$ be three positive reals. Prove the inequality
$\frac{a^2+2b^2}{b+c}+\frac{b^2+2c^2}{c+a}+\frac{c^2+2a^2}{a+b}\geq\frac32\left(a+b+c\right)$.[/quote]
I think, $\sum_{cyc}\frac{a^8+2b^8}{b^7+c^7}\geq\frac{3(a+b+c)}{2}.$ ;)

I will write my solution to Darij's one and then I will think about the one posted by Arqady

.We have by Cauchy $\sum\frac{a^2+2b^2}{b+c}=\sum\frac{a^2}{b+c}+2\sum\frac{b^2}{b+c}\geq\frac{a+b+c}{2}+2\frac{a+b+c}{2}=\frac{3}{2}(a+b+c)$


2. A chessboard is an 8*8 grid. If a chessman is onthe bottom left square, how many paths may he take to the top right square if he moves only up and to the right.
Given the 4*6 grid shown, in how many way can one go from the bottom left corner to the upper right corner along the lines if one can only go up and right?
Are these two question same

[quote="linj"]A chessboard is an 8*8 grid. If a chessman is onthe bottom left square, how many paths may he take to the top right square if he moves only up and to the right.[/quote]
Can you specify what kinds of moves are allowed? For example, if the piece is like a rook, then it could move as many spaces up/right as it wants, and could, for example, get to the top-right in two moves.
If only unit moves are allowed, it's fairly trivial, and is given by $ \binom{16}{8}$.

[quote="linj"]Given the 4*6 grid shown, in how many way can one go from the bottom left corner to the upper right corner along the lines if one can only go up and right?
Are these two question same[/quote]
I don't see a grid shown, but I don't see how any such grid could make the question the same as the first one.


Let be $ x,y,z>0$ such that $ xyz\equal{}1$ . Show that :

$ \frac{x^3}{x\plus{}y}\plus{}\frac{y^3}{y\plus{}z}\plus{}\frac{z^3}{z\plus{}x}\ge \frac{1}{2}$

I know it's easy , but i hope you'll like it  :lol: .

[quote="alex2008"]Let be $ x,y,z > 0$ such that $ xyz \equal{} 1$ . Show that :

$ \frac {x^3}{x \plus{} y} \plus{} \frac {y^3}{y \plus{} z} \plus{} \frac {z^3}{z \plus{} x}\ge \frac {1}{2}$

I know it's easy , but i hope you'll like it  :lol: .[/quote]
Thanks,alex2008, :lol: but this inequality very weak
we can easily prove that:
$ \frac {x^3}{x \plus{} y} \plus{} \frac {y^3}{y \plus{} z} \plus{} \frac {z^3}{z \plus{} x} \ge \frac {3}{2}$ with condition$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$ or stronger $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$
I hope you will like it :)

[quote="quykhtn-qa1"]
Thanks,alex2008, :lol: but this inequality very weak
we can easily prove that:
$ \frac {x^3}{x \plus{} y} \plus{} \frac {y^3}{y \plus{} z} \plus{} \frac {z^3}{z \plus{} x} \ge \frac {1}{2}$ with condition$ x^2\plus{}y^2\plus{}z^2\equal{}3$ or stronger $ x^3\plus{}y^3\plus{}z^3\equal{}3$
I hope you will like it :)[/quote]

Well with your conditions the inequalities are stronger , but i think they are easy to prove (only by CS)

[quote="alex2008"][quote="quykhtn-qa1"]
Thanks,alex2008, :lol: but this inequality very weak
we can easily prove that:
$ \frac {x^3}{x \plus{} y} \plus{} \frac {y^3}{y \plus{} z} \plus{} \frac {z^3}{z \plus{} x} \ge \frac {1}{2}$ with condition$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$ or stronger $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$
I hope you will like it :)[/quote]

Well with your conditions the inequalities are stronger , but i think they easy to prove (only by CS)[/quote]
Alex,you can solve it with $ x^3\plus{}y^3\plus{}z^3 \equal{} 3$ ?:maybe:

[quote="FelixD"]Why should there be any mistake?[/quote]
I dont know...
When equality occurs???
Is it a strict inequality????
Because in my way it seems to be strict...

I guess the inequality is strict, just note that it is similar to Nesbitt's inequality.
Indeed a proof confirmes that $ \sum{\dfrac{x^3}{x \plus{} y}}\ge\dfrac{3}{2}$ if $ x \plus{} y \plus{} z\ge 3$. or $ x^2\plus{}y^2\plus{}z^2\ge 3$...
I'll see what happens with $ x^3...$

:)

[quote="alex2008"]$ LHS \equal{} \sum_{cyc}\frac {x^6}{x^4 \plus{} x^3y}\ge \frac {9}{2(x^4 \plus{} y^4 \plus{} z^4)}\ge \frac {1}{2}$[/quote]
Are you sure? $ x^4\plus{}y^4\plus{}z^4 \ge x^3\plus{}y^3\plus{}z^3 \equal{}3$ :wink:

[quote="quykhtn-qa1"]
Are you sure? $ x^4 \plus{} y^4 \plus{} z^4 \ge x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$ :wink:[/quote]

$ \frac{9}{2(x^4\plus{}y^4\plus{}z^4)}\ge \frac{1}{2}\Leftrightarrow 9\ge x^4\plus{}y^4\plus{}z^4\Leftrightarrow 3(x^3\plus{}y^3\plus{}z^3)\ge x^4\plus{}y^4\plus{}z^4$

[quote="alex2008"][quote="quykhtn-qa1"]
Are you sure? $ x^4 \plus{} y^4 \plus{} z^4 \ge x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$ :wink:[/quote]

$ \frac {9}{2(x^4 \plus{} y^4 \plus{} z^4)}\ge \frac {1}{2}\Leftrightarrow 9\ge x^4 \plus{} y^4 \plus{} z^4\Leftrightarrow 3(x^3 \plus{} y^3 \plus{} z^3)\ge x^4 \plus{} y^4 \plus{} z^4$[/quote]
Sorry,it must be:
$ \frac{x^3}{x\plus{}y}\plus{}\frac{y^3}{y\plus{}z}\plus{}\frac{z^3}{z\plus{}x} \ge \frac{3}{2}$,not$ \frac{1}{2}$.
Then your solution 's false :wink: '

Ok , then i think everybody must take again their solution  :)

[quote="Dimitris X"][quote="quykhtn-qa1"]with $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$ ?:maybe:[/quote]
I think there is a mistake in my profe :blush:
We consider the $ f(x) \equal{} \frac {1}{x}$
$ f''(x) \geq 0$
So we have that:
$ x^3f(x \plus{} y) \plus{} y^3f(y \plus{} z) \plus{} z^3f(z \plus{} x)\geq 3f(\frac {x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \plus{} y^3z \plus{} z^3x}{3})$
So the $ LHS \geq \frac {9}{x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \plus{} y^3z \plus{} z^3x}$
This can be also proved by B.C
So we have to prove that:
$ 18 \geq x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \plus{} y^3z \plus{} z^3x$
But $ (x \plus{} y \plus{} z)(x^3 \plus{} y^3 \plus{} z^3) \equal{} (x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \plus{} y^3z \plus{} z^3x) \plus{} x^3z \plus{} y^3x \plus{} z^3y$
So the $ RHS \equal{} 3(x \plus{} y \plus{} z) \minus{} (x^3z \plus{} y^3x \plus{} z^3y)$
So we have to prove that:
$ 18 \plus{} x^3z \plus{} y^3x \plus{} z^3y\geq 3(x \plus{} y \plus{} z)$
But $ \sqrt [3]{\frac {x^3 \plus{} y^3 \plus{} z^3}{3}}\geq \frac {x \plus{} y \plus{} z}{3}$
So $ 9 \geq 3(x \plus{} y \plus{} z)$
So our ineq is $ 9 \plus{} x^3z \plus{} y^3x \plus{} z^3y\geq 0$ Which is true...Where is my mistake????[/quote]
Sorry,must be $ LHS \ge \frac{3}{2}$ :) ,then your solution is false

[quote="quykhtn-qa1"][quote="Dimitris X"][quote="quykhtn-qa1"]with $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$ ?:maybe:[/quote]
I think there is a mistake in my profe :blush:
We consider the $ f(x) \equal{} \frac {1}{x}$
$ f''(x) \geq 0$
So we have that:
$ x^3f(x \plus{} y) \plus{} y^3f(y \plus{} z) \plus{} z^3f(z \plus{} x)\geq 3f(\frac {x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \plus{} y^3z \plus{} z^3x}{3})$
So the $ LHS \geq \frac {9}{x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \plus{} y^3z \plus{} z^3x}$
This can be also proved by B.C
So we have to prove that:
$ 18 \geq x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \plus{} y^3z \plus{} z^3x$
But $ (x \plus{} y \plus{} z)(x^3 \plus{} y^3 \plus{} z^3) \equal{} (x^4 \plus{} y^4 \plus{} z^4 \plus{} x^3y \plus{} y^3z \plus{} z^3x) \plus{} x^3z \plus{} y^3x \plus{} z^3y$
So the $ RHS \equal{} 3(x \plus{} y \plus{} z) \minus{} (x^3z \plus{} y^3x \plus{} z^3y)$
So we have to prove that:
$ 18 \plus{} x^3z \plus{} y^3x \plus{} z^3y\geq 3(x \plus{} y \plus{} z)$
But $ \sqrt [3]{\frac {x^3 \plus{} y^3 \plus{} z^3}{3}}\geq \frac {x \plus{} y \plus{} z}{3}$
So $ 9 \geq 3(x \plus{} y \plus{} z)$
So our ineq is $ 9 \plus{} x^3z \plus{} y^3x \plus{} z^3y\geq 0$ Which is true...Where is my mistake????[/quote]
Sorry,must be $ LHS \ge \frac {3}{2}$ :) ,then your solution is false[/quote]
Hmmm.
My way works for $ LHS\geq\frac{1}{2}$ and $ LHS \geq1$...But for  $ LHS \geq \frac{3}{2}$ i don't know...


geometry
analytic geometry
geometry solved

A quadrilateral has sides $a$, $b$, $c$, $d$ (in that order) and area $F$. Prove that \[ 2a^2+5b^2+8c^2-d^2 \geq 4F  \]. When does equality hold ?

The solution in Crux [1997: 55] uses analytic geometry and it's long... someone of you has another solution  :?

$Tipe$

[quote="tipe"]A quadrilateral has sides $a$, $b$, $c$, $d$ (in that order) and area $F$. Prove that \[ 2a^2+5b^2+8c^2-d^2 \geq 4F  \]. When does equality hold ?[/quote]

Denote the quadrilateral by ABCD such that a = AB, b = BC, c = CD, d = DA. Let $F_1$ be the area of triangle ABC, and $F_2$ the area of triangle CDA. Then, the area F of the quadrilateral ABCD equals $F=F_1+F_2$.

Apply the Corollary from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#p204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url] to the triangle ABC with the area $F_1$ and the reals u = 2, v = 5, $w=-\frac97$; then, since the numbers $v+w=\frac{26}{7}$, $w+u=\frac57$, u + v = 7 and vw + wu + uv = 1 are nonnegative, we get

$u\cdot AB^2+v\cdot BC^2+w\cdot AC^2\geq 4\sqrt{vw+wu+uv}\cdot F_1$, i. e.
$2a^2+5b^2+\left(-\frac97\right)\cdot AC^2\geq 4F_1$.

Apply the Corollary from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#p204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url] to the triangle CDA with the area $F_2$ and the reals u = 8, v = -1, $w=\frac97$; then, since the numbers $v+w=\frac27$, $w+u=\frac{65}{7}$, u + v = 7 and vw + wu + uv = 1 are nonnegative, we get

$u\cdot CD^2+v\cdot DA^2+w\cdot AC^2\geq 4\sqrt{vw+wu+uv}\cdot F_2$, i. e.
$8c^2+\left(-1\right)d^2+\frac97\cdot AC^2\geq 4F_2$.

Thus,

$2a^2+5b^2+8c^2-d^2=\left(2a^2+5b^2+\left(-\frac97\right)\cdot AC^2\right)+\left(8c^2+\left(-1\right)d^2+\frac97\cdot AC^2\right)$
$\geq 4F_1+4F_2=4\left(F_1+F_2\right)=4F$,

qed.

  Darij


Stanford
college
combinatorics open
combinatorics

A subset M of {1, 2, 3, . . . , 15} does not contain three elements whose product
is a perfect square. Determine the maximum number of elements of M.

"Math Olympiad Problem Solving
Stanford University EPGY Summer Institutes 2008" :!:  :!:

Posted [url=http://www.mathlinks.ro/viewtopic.php?t=219935]before[/url]  :wink:


For any pythagorean triple a,b,c, show that the greatest common factor of (c-b)/2 and (c+b)/2 is a perfect square [of course, this is with (c-b)/2 and (c+b)/2 as integers, which we can always find]

I'm almost certain this is true for primitive pythagorean triples, but I haven't found a pattern for the non-primitive ones yet...

I am not sure if I am allowed to answer this ( :D ) but I just can't help myself.

It is known that if (a,b,c) is a primitive pyt. triplet, then  $a=2mn$, $b=m^2-n^2$ and $c=m^2+n^2$, for some integers $\gcd (m,n)=1$. Thus, a is even, while b and c are both odd. So $ \frac { b+c} 2 = m^2 $ and $\frac {c-b} 2 = n^2 $, thus obviously $ \gcd (m^2,n^2 ) = 1$ which is a perfect square :P

If the triplets are not primitive, then we can just multiply them by a non-perfect square constant, and obtain that the gcd we are seeking is exactly that constant.

LoL. Actually, I was trying to prove this lemma I posted in order to show that all pythagorean triples could be expressed as what you proposed. However, I realized that I was wrong...

But I was really trying to show that (c-b)/2 and (c+b)/2 are always squares....however, now I'm trying to find the generalized exceptions...

If it's a primitive triplet it's clear. If it's not, let d be the gcd of a,b,c. If d is a perfect square, then the numbers $ \frac {c+b}2$ and $\frac {c-b}2$ are indeed perfect squares. Otherwise they are not.


If $ x,y,z > 1$ and $ x\plus{}y\plus{}z \equal{} 6$, then
$ \frac{1}{14\minus{}x\minus{}yz}\plus{}\frac{1}{14\minus{}y\minus{}zx}\plus{}\frac{1}{14\minus{}z\minus{}xy}\leq\frac{3}{8}$

Let $ a\equal{}x\minus{}1,b\equal{}y\minus{}1,c\equal{}z\minus{}1$,then $ a\plus{}b\plus{}c\equal{}3$,then inequality becomes
\[ \frac{1}{9\minus{}ab}\plus{}\frac{1}{9\minus{}bc}\plus{}\frac{1}{9\minus{}ca}\leq \frac{3}{8}\]
it's an old problem....I think arqady can help you find the ever posts...

[quote="filomen"]If $ x,y,z > 1$ and $ x + y + z = 6$, then
$ \frac {1}{14 - x - yz} + \frac {1}{14 - y - zx} + \frac {1}{14 - z - xy}\leq\frac {3}{8}$[/quote]

Let $ x=1+\frac{3a}{a+b+c},y=1+\frac{3b}{a+b+c},z=1+\frac{3c}{a+b+c},a,b,c>0$, ineq becomes

$ \sum{\frac{1}{(a+b+c)^2-bc} \le \frac{27}{8(a+b+c)^2}.\qquad (1)}$

actually,

$ (1)\Longleftrightarrow 5abc\sum {\left( {a + b} \right)\left( {a - b} \right)^2 }  + 3\sum {ab\left( {a^2  - b^2 } \right)^2 }  +$

$ 4\sum {ab\left( {a^4  + b^4 } \right)}  + \sum {a^2 b^2 \left( {a^2  + b^2 } \right)}  + 3\sum {a^6 }  - 39a^2 b^2 c^2 \ge 0,$

which is true by AM-GM.


LaTeX
induction
combinatorics open
combinatorics

Direct if you notice that
$\frac{n!}{a_{1}!a_{2}!....a_{m}!(n-a_{1}-a_{2}-\dots-a_{m})!}$
is a multinomial coefficient (always in $\mathbb{N}$).

i m not understand what abou you talking

Maybe this link will help you:
[url=http://en.wikipedia.org/wiki/Multinomial_theorem]Wikipedia[/url]

Or you could also consider this combinatorial proof
if a1+a2+.............am=n
Consider n objects arranged in m rows of a1,a2.....am
No. of ways = n!/a1!a2!....am! which is an integer as no. of ways can only be an integer
Similarly if a1+a2+.............am=n-1
(n-1)!/a1!a2!....am! is an integer which implies n!/a1!a2!.....am! is an integer
Similarly it can be proved for all other cases

1. That it has a combinatorial representation is exactly the same thing as saying it is a multinomial coefficient.

2. You have raised up a topic that was successfully finished a year ago.

3. You are not using [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX[/url].

A different proof is by induction: show $ a_{1}!\cdot a_{2}! | (a_{1}+a_{2})!$, then apply repeatedly.


[quote="236factorial"]What is the number of units in the length of AB wtih endpoints at A (0,0) and B (5,9)? Express your answer as a decimal to the nearest tenth.[/quote]

[hide]The length is the hypotenuse of a triangle with legs of length 5 and 9.

$\sqrt{25+81}$

$\sqrt{106}\approx10.3$[/hide]

[quote="Drunken_Math"][quote="236factorial"]What is the number of units in the length of AB wtih endpoints at A (0,0) and B (5,9)? Express your answer as a decimal to the nearest tenth.[/quote]

[hide]The length is the hypotenuse of a triangle with legs of length 5 and 9.

$\sqrt{25+81}$

$\sqrt{106}$[/hide][/quote]

What happened to the decimal?  :(

[hide]
using distance formula
i get
$\sqrt{106}=\boxed{10.3}$[/hide]

[quote="math92"][hide]
using distance formula
i get
$\sqrt{106}=\boxed{10.295630140987000315797369464198}$[/hide][/quote]

Why are you guys betraying the question? IT says the nearest tenth...


linear algebra
matrix
function
linear algebra unsolved

A1,..., An are finite sets and Aij = card(Ai ∩ Aj). If A is the matrix formed by Aij show that det(A) ≥ 0

The sets $A_1,A_2,\dots,A_n$ are all subsets of some finite set $E.$ (If nothing else, you could just take $E$ to be the union.) Assume that $|E|=m.$

Construct an $m\times n$ matrix $M$ as follows: Label, in order the elements of $E.$ Let $M_{jk}=1$ if the $j$th element of $E$ is in $A_k$ and $M_{jk}=0$ otherwise.

Another way to say this is that the $k$th column of $M$ is the characteristic function of the set $A_k.$

Claim: $M^TM=A,$ the matrix of this problem. The $(i,j)$ element of this matrix is the inner product of of the $i$th and $j$th column, and this is the same as $|A_i\cap A_j|.$

But any real matrix of the form $M^TM$ is always positive semidefinite. Proof: Let $x$ be any real column vector.

$x^T(M^TM)x=(x^TM^T)(Mx)=(Mx)^T(Mx)=\|Mx\|^2\ge0.$

All of the eigenvalues of a positive semidefinite matrix are nonnegative, and the determinant, which is the product of those eigenvalues, must also be nonnegative.

Here is an elementary proof, for students of 11 grade.

Matrix A has the following properties:
1. A is symmetric.
2. $\mid a_{ij} \mid \leq a_{ii} , \forall i \neq j$
3. $a_{ii} \geq 0$

We can assume that $a_{ii} \neq 0 \forall i$.
[If $a_{ii} = 0$ for one i then $A_i = \emptyset$ and one row of matrix A is 0 so det(A) = 0]

We apply to matrix A the following transformation which will not change the value of its determinant, and it will bring A to a matrix A' which has on the first column elements $a_{11}, 0, ..., 0$.

For this we multiplicate :

first row with $-a_{21}/a_{11}$ and add this to second row,
first row with $-a_{31}/a_{11}$ and add this to third row,
...

We get:

$A' = \left(\begin{array}{cccccccc}a_{11}&a_{12}&...&a_{1i}&...&a_{1k}&...&a_{1n} \\ 0&a_{22} - a_{12}^2/a_{11}&...&a_{2i} - a_{12}*a_{1i}/a_{11}&...&...&...&... \\ \mid&&&&&&& \\ 0&...&...&a_{ii} - a_{1i}^2/a_{11}&...&a_{ik} - a_{1k}*a_{1i}/a_{11}&...&... \\ \mid&&&&&&& \\ 0&...&...&a_{ki} - a_{1i}*a_{1k}/a_{11}&...&...&...&... \\ \mid&&&&&&& \\ 0&&&&&&& \end{array}\right)$

So, det(A) = det(A') = $a_{11}$ * det(B), and we can see that B is a symmetric matrix [A is symmetric].
It remains to proof the points 2 and 3.
It will result that B has the 3 properties from above and apply the transformation recursive down, we get:

det (A) = $a_{11}*\alpha_2*...*\alpha_{n-2}*\begin{array}{|cc|}a&b\\b&d\\\end{array}$

where $a_{11}, \alpha_2, ..., \alpha_{n-2} \geq 0$ and $\mid b \mid \leq a, \mid b \mid \leq d$, so $det(A) \geq 0$.

It remains to proof the points 2 and 3.
For 2: $b_{ii} \geq 0$ means $a_{ii}*a_{11} \geq a_{1i}^2$ which is true.
For 3: $\mid b_{ij} \mid \leq b_{ii} \Leftrightarrow -b_{ii} \leq b_{ij} and b_{ij} \leq b_{ii}$

$b_{ij} \leq b_{ii} \Leftrightarrow a_{ij} - a_{1i}*a_{1j}/a_{11} \leq a_{ii} - a_{1i}^2/a_{11}$,
$a_{1i}(a_{1i} - a_{1j}) \leq a_{11}(a_{ii} - a_{ij})$ [1].
Denote $\mid A_i \cap A_j \cap A_1 \mid = t, \mid A_i \cap A_j \mid = x + t, \mid A_j \cap A_1 \mid = t + z, \mid A_1 \cap A_i \mid = y + t, \mid A_i \mid = \alpha + x + y + t, \mid A_1 \mid = u + z + y + t$,
[1] can be written as:
$(y + t)(y - z) \leq (\alpha + y)(y + t + z + u)$, which is true.
$-b_{ii} \leq b_{ij}$ is similar.

[i hope i did not miss something :roll: ]


quadratics
number theory
prime factorization
number theory unsolved

For all integers number $x,y,z$; prove:
$x^{2}+y^{2}=6z^{2}$

 :wink:

I'll asume you meant "Solve".

It is clear that the exponent of $3$ in the prime factorization of the right side is odd (unless both sides are $0$).  Suppose it is $2k+1$.
$3^{2k+1}|x^{2}+y^{2}\implies x^{2}\equiv-y^{2}\mod{3^{2k+1}}$
But unless $x \equiv y \equiv 0$, we have $-1$ is a quadratic residue mod $3^{2k+1}$, which is impossible because then $-1 \equiv (-1)^{\frac{3^{2k+1}-1}{2}}\equiv 1$ by Euler's c riterion (-1 is not even a quadratic residue mod 3, which is a requirement).
Thus, $x \equiv y \equiv 0\mod{3^{2k+1}}$
But the exponents of $3$ in $x^{2}$ and $y^{2}$ are even, of course, and are at least $2k+1$.  Thus, both $x^{2}$ and $y^{2}$ are divisible by $2k+2$.  Thus,

$\frac{x^{2}}{3^{2k+2}}+\frac{y^{2}}{3^{2k+2}}=\frac{6z^{2}}{3^{2k+2}}$

But the exponent of $3$ on the right was $2k+1$, which means the the right side is not an integer.  Contradiction.

Then the only solutions is $(0, 0, 0)$.

Well, this gets totally trivial when using the following theorem:

$x^{2}+y^{2}=az^{2}$ has a solution with $z \neq 0$ iff $x^{2}+y^{2}=a$ has a solution.

Proof: try yourself.

More general:

Let $a,b,c$ be pairwise coprime and all squarefree.
$ax^{2}+by^{2}+cz^{2}=0$ has a nontrivial (meaning: not all numbers $=0$) solution iff:
- not all numbers $a,b,c$ have the same sign.
- $-ab$, $-bc$ and $-ca$ are quadratic residues $\mod c$, $a$ and $b$ respectively.

Proof: usage of the theorem of Chevalley-Warning, Hensel's Lemma and the theorem of Hasse-Minkowski for three variables (probably doesn't sound so, but this theorems have all elementary solutions); or even better: just mimic the proof of Hasse-Minkowski for the form $ax^{2}+by^{2}+cz^{2}$, then neither of the mentioned theorems is needed.

Or you could use the fact a number is the sum of two squares iff it has the form \[(\prod_{i=1}^{m}{p_{i}^{a_{i}}})(\prod_{j=1}^{n}{q_{j}^{2b_{j}}})2^{g}\] where each $p_{i}$ is a 1mod4 prime and each $q_{j}$ is a 3mod4 prime...


AMC
USA(J)MO
USAJMO
USAMO
AIME
AMC 10
calculus

Here's how it is:

To make the USAJMO: Take the AMC 10 and AIME and get a high index: AMC 10 score+AIME score*10. It's generally around 215, but it varies year to year. Also, you must get below an 11 on the AIME.

To make the USAMO: Either get a 11+ on the AIME through the AMC 10 route or take the AMC 12/AIME with a high index. This is also generally around 215.

Also, remember that if you make both, it is automatically assumed that you would prefer to take the USAMO.

In addition, it is impossible (well possible, but highly improbable) to make blue or black MOP through the USAJMO. Two of the problems (#1 and #4 on USAMO/#3 and #6 on the USAJMO) coincide on both tests and only your scores on those two on the USAJMO put you up for consideration for the MOP. So there is a high chance of a few red MOPpers coming from the USAJMO.

Oh, so I probably want to try to get a perfect AMC 10 or near perfect so then in AIME I can just focus on the first couple of problems.

[quote="modularmarc101"]I'm confused with the new rules, I've read em about 2 or 3 times.

I'm in 9th grade. Correct me if Im wrong please:

I must:

-Take AMC 10 NOT AMC 12 because it will get me to USAMO not USAJMO

-Then take AIME and must get 10*score + amc10 > 210(depends on each year?)[/quote]

The USAMO qualifying index varies greatly and since this is the first USAJMO, we can't really estimate accurately this year's index.

I guess. But most people that want to take USAJMO are not capable of easily getting 11+ on AIME.

One other thing: it is extremely common, even for the very best students, to think that you've solved an AIME problem only to find out later that you overlooked something or fell into some trap. So if you ever reach the point of thinking, "gee, I already have 10 and if I solve this one I'll have 11," them maybe where you really are is that you have 8, and it would be awfully nice to get that 9th problem.

[quote="maybach"][quote="remy1140"]Wait, so if I get 11+ AIME (as well as high AMC10) USAMO is the only choice? If I'm not mistaking, I should "purposely" get less than 11 in AIME to get in USAJMO...[/quote]

Of course I'll never make USAMO or USAJMO in a million years, if I could, I'd definitely take USAMO, since you have more opportunities for high level MOP.[/quote]

Unfortunately, it's not that simple. Do you have calculus yet? Or real olympiad training? Then for right now, it's probably USAJMO.

[quote="basketball9"]To awesome toad

     You think its possible to do usajmo without olympiad training[/quote]

Possibly. But if you're seriously trying to do well on USAJMO, then you might as well start early, but don't go to far.

Do the Intermediate Series, do more AIME Problems.
To push yourself more, do USAMO problems too, but don't get discouraged if right now you can't solve many. These are very hard.

Good luck:)

So I have another question, to get to USAJMO, we can't get 11 on AIME, [b]regardless[/b] of the AMC 10 score?

Yes, if you get a 11 or 11+ on the AIME, you will already be among the top indices. But since you got 11+ on the AIME, you will be selected for both USAMO, and USAJMO. But since you got selected for both, you will be selected to take the USAMO.

Correct me if I am wrong anywhere...  :|

You don't [b]need[/b] W00T or any other formal olympiad training. Some hard work, determination, and experience is a pretty good substitute. My goal is USAMO(even though I didn't even make state MC this year, I think I have improved a crazy lot =P). Good luck to all!

No one who scores an 11+ on the AIME [b]wants[/b] to take USAJMO, btw.
Also, you don't really get much out of USAJMO. Other than the valuable experience, of course.

[quote="james4l"][quote="AwesomeToad"]Do you have calculus yet?[/quote]

BTW, calculus is never a requisite for solving Olympiad problems (please correct me on here if there are). Putnam, yes. Olympiad problems, no.[/quote]

However... higher math... is often useful on olympiads... i.e. with linear algebra 2008 #6 became a LOT easier...

[quote="AwesomeToad"]Don't need Calculus for Olympiad, but it makes it a lot easier.

@pythag: How hard is Olympiad Counting/Probability?[/quote]

"a lot easier" is actually typically very inaccurate... i'd estimate that it'll improve your score by 1 problem on 1/10 usamos.

Olympiad counting/probability... most of olympiad stuff is combinatorics, not counting/probability...

Also, olympiad counting/probability is olympiad level :P

[quote="pythag011"]
Also, olympiad counting/probability is olympiad level :P[/quote]

Eh, ok...

So harder than Intermediate right? Or... very hard?

Pythag is right about higher-level math - it's never necessary but occasionally helps motivate problems, and if you know the background you can get a better grasp of what is going on in some situations. If your only focus is olympiad math, don't try to just get a superficial understanding of, say, linear algebra - that won't help. Either really learn it or ignore it.


Inside a convex quadrilateral $ABCD$ there is a point $P$ such that the triangles $PAB, PBC, PCD, PDA$ have equal areas. Prove that the area of $ABCD$ is bisected by one of the diagonals.

$B,D$ are equidistant from $AP$, so $AP$ must pass through the midpoint of $BD$. In the same way we conclude that $CP$ must pass through the midpoint of $BD$, while $BP,DP$ pass through the midpoint of $AC$. If $AP=CP$ or $BP=DP$ we're done: $AC$ or $BD$ is the diagonal we're looking for; if, on the other hand, we assume that $AP\ne CP$ and $BP\ne DP$, from the observation above it follows that $P$ is the common midpoint of the diagonals $AC,BD$, contradicting $AP\ne CP$.

My solution uses vectors. Here it is.

Note that if we have shown that $P \in AC$ or $P \in BD$ then the conclusion follows. Then assume the contrary and note that the equality of the areas of $PAB$ and $PBC$ implies that $\vec{PA}\times\vec{PB}=\vec{PB}\times\vec{PC}$ or equivalently $(\vec{PA}+\vec{PC})\times \vec{PB}=0$. Thus $\vec{PA}+\vec{PC}=0$ which means that $P\in AC$ or $\vec{PA}+\vec{PC}$ and $\vec{PB}$ are collinear. In the same fashion we get that $\vec{PA}+\vec{PC}$ and $\vec{PD}$ are collinear and hence $\vec{PB}$ and $\vec{PD}$ are collinear which means $P\in BD$. In such a way we get a contradiction.


vector
number theory unsolved
number theory

solve this equation

a^2+b^2+c^2+d^2 = 2005

for a b c d integers

The sharp version of the Four-Squares-Theorem gives that there are $8\cdot (1+5+401+2005) = 19296$ solutions, so if you want all of them: good luck  :P
But at least one solution is $2005=39^2+22^2+0^2+0^2$  ;)

$(a-b)^2+(c-d)^2+2ab+2cd=2005$

must be

$2ab=p^2$
$2cd=q^2$

so if $(a,b,c,d)$ is solution also $(a-b,c-d,2ab,2cd)$ is too
for $a,b,c,d$ of the form
$a=2^{2f-1},b=m^2$
$b=2^{2g-1},d=n^2$

now you can use a precalculated vector with powers of 2
bounded upto 2005 and also squares up to that number.
than youll have to try all combinations and see what it gets you
to,if you study further this stuff probably you'll get all the
solutions

I don't see how this works and why this is easier than subtracting all different perfect squares lower than $2005$ and trying to write the result as sum of $3$ squares and so on...  :?


Hey, where can I find the answers to the class problems and homework? All I can see are these posts...

What's hodge conjecture?

hello hello

[quote="npark"]What's hodge conjecture?[/quote]
See [url]http://www.claymath.org/millennium/Hodge_Conjecture/[/url] or [url]http://en.wikipedia.org/wiki/Hodge_conjecture[/url]. Google is sometimes useful.
Anyway, the seven millenium problems are use to denote ranks of MathLinks users. Let $p$ be the number of posts of some MathLinker. Then:
$0\leqslant p\leqslant 19$ gives [b]New Member[/b] (0 stars),
$20\leqslant p\leqslant 49$ gives [b]P versus NP[/b] (0.5 star),
$50\leqslant p\leqslant 99$ gives [b]Hodge Conjecture[/b] (1 star),
$100\leqslant p\leqslant 249$ gives [b]Poincare Conjecture[/b] (2 stars),
$250\leqslant p\leqslant 499$  gives [b]Riemann Hypothesis[/b] (2.5 stars),
$500\leqslant p\leqslant 999$ gives [b]Yang-Mills Theory[/b] (3 stars),
$1000\leqslant p\leqslant 2499$ gives [b]Navier-Stokes Equations[/b] (4 stars),
$2500\leqslant p<+\infty$ gives [b]Birch & Swinnerton-Dyer[/b] (5 stars).
If you become an administrator, you will have 6 stars.

u must mean the jls thread go to jls mathcounts page

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112276&start=120

its being spammed though

[color=blue]Learn to not double post ~ccy[/color]


Does anyone know how to prove that if V is finite dimensional, then its dual space V* is finite dimensional too, and dimV=dimV*??
I think that one way to prove it is to take a basis of V and show that there exists a bijection between V and V*. But then I have some problems proving the bijection...

Let $\{e_{i}\}_{i=1}^{n}$ be a basis. Once you know the values of a linear functional on $\{e_{i}\}$, you know the functional, because any element of $V$ is a linear combination of $e_{i}$. Conversely, given any numbers $c_{1},\dots,c_{n}$ (i.e. elements of the field under consideration), you can define a linear functional $f$ as follows: let $f(e_{i})=c_{i}$, and extend $f$ to $V$ by linearity.

let $e_{1},...........,e-{n}$ be basis of $E$.then take $f_{i}$ such for every $j\not=i,f_{i}(e_{j})=0 and f_{i}(e_{i})=1$.

so its easy to prove that $(f_{1},..........,f_{n})$ is a basis for $E*$. :wink:

alright, what I understand is that for every element in the basis of V there exists only one, unique element in V* that assigns it a 1 in R. The set of such elements of V* is the basis of V*. If dim(V*)<dim(V) then there exists an element in V* that assigns a 1 to more than one element of the basis of V => impossible. So dim(V)=dim(V*). Is what I understand correct or am I missing any point? Thanks for helping!

[quote="terebide"]then there exists an element in V* that assigns a 1 to more than one element of the basis of V[/quote]
Such elements exist- but they're linear combinations of what we already have. fermat3 gave an alleged basis for $V^{*}$; you need to prove that it is linearly independent and that it spans $V^{*}$. The first is easy. For the second, suppose $f$ is an arbitrary functional. Can you construct the linear combination $\sum a_{i}f_{i}$ which should be $f$?

Yes, I think I got it now :) Thanks for the help!


The sequence $\{a_1,a_2,\ldots\}$ is recursively defined by  $a_1 = 1$, $a_2 = 1$, $a_3 = 2$, and \[ a_{n+3} = \frac 1{a_n}\cdot (a_{n+1}a_{n+2}+7), \ \forall \ n > 0. \] Prove that all elements of the sequence are integers.

Lets rewrite the expression like so:

a_(n+3)an=a_(n+1)a_(n+2)+7 (1)
and also we write the expression for n-->n+1 (2)

we then do (1)-(2) and get after some computations that:
(a_(n+4)+a_(n+2))/a_(n+3)=(a_(n+2)+a_n)/a_(n+1)=5 for n even and 3 for n odd.
so we get two reccurence relationships ,
a_(2n+1)-3a_(2n)+a_(2n-1)=0 and a_(2n+2)-5a_(2n+1)+a_(2n)=0

I would have proposed to make the problem more interesting, (though very simple questions) :D
a)show that all the terms are integers;
b)show that the sum of any two consecutive odd positioned terms are divisible by 3;
b)show that the sum of any two consecutive even positioned terms are divisible by 5;

cheers! :D
hope it's okay!


I have a forum with 50 members.
http://www.atfreeforum.com/math92
I was wondering if some more AoPSers could join and give me some feedback about the site.  :D

WOW I CAN'T GET THE CONFIRMATION CODE RIGHT THE FONT IS SO GODDAMMM MESSED UP AND ITS LIKE "YOU EXCEEDED NUMBER OF ATTEMPTS TO REGISTER MORON" AHHHH

I just joined now,  and it looks like a good site.  All you need are some more topics and people and you'll be fine!! :thumbup:

"The page isn't redirecting properly

Firefox has detected that the server is redirecting the request for this address in a way that will never complete."

Site should be made functional first.


Let A be a countable ordered set. whose order relation satisfies
1) for any x,y in A , there exist z such that x<z<y
2)for any z belonging to A, there exist x,y belonging to A such that x<z<y.
Show that A is order isomorphic to Q.


inequalities
geometry
circumcircle
geometry unsolved

[quote="111222333444555"]Let $P$ be a point in a triangle $ABC$ with circumradius $R$. Prove that

$\frac{PA}{AB.AC}+\frac{PB}{BC.BA}+\frac{PC}{CA.CB}\geq\frac{1}{R}$[/quote]
PA.BC>=2(S-PAB+S-PAC)    S-PAB mean triangle PAB's S
PB.AC>=2(S-PAB+S-PBC)
PC.AB>=2(S-PAC+S-PBC)
So PA.BC+PB.AC+PC.AB>=4S-ABC
And AB.BC.CA=4R.S-ABC
So left=(PA.BC+PB.AC+PC.AB)/(AB.BC.CA)>=1/R
Sorry for my poor English  :lol:


topology
real analysis
real analysis unsolved

Hey folks I am sure this is an easy question but I and many other people working on it have no idea where to start.

Suppose that f:R -> R is continuous. Let S={x in R : f(x) = 0} be nonempty. Prove that S is a closed set.

I am trying to prove that the set is compact and therefore closed but I am not sure if that is where I should be headed, Help?

You can't prove compactness, because it isn't true. The set could be all of $\mathbb{R}$ if we choose the zero function.
This is trivial from the topological definition of continuity, so I'll assume you're working with the usual metric definition.

A nonempty set $S$ in $\mathbb{R}$ (or any metric space) is closed if and only if every convergent sequence of elements of $S$ converges to a member of $S$. What can you say about $f(c)$, if we have $f(x_{n})=0$ for some sequence $x_{n}$ converging to $c$?

what can we say about f(c) provided that x_n converges to c, that f(x_n) converges to f(c), right?

I think this idea might work:

$A=f^{-1}((-\infty, 0])$ $B=f^{-1}([0,\infty))$.
Then $S=A \cap B$, which is a closed set being the intersection of two closed sets.


algebra
polynomial
arithmetic sequence
Divisor Functions

Let $m$, $n$ be positive integers. Prove that, for some positive integer $a$, each of $\phi(a)$, $\phi(a+1)$, $\cdots$, $\phi(a+n)$ is a multiple of $m$.

By the chinese remainder theorem, simply take an $a \equiv -i \mod p_i$ for $i=0,1,...,m$ with some distinct primes $p_i \equiv 1 \mod m$ (existing by Dirichlet's theorem).

To avoid the full Dirichlet theorem:
Let $m=r_1 \cdot r_2 \cdot ... \cdot r_k$ be the prime factorisation of $m$.
Take distinct primes $q_{i,j}$ such that $q_j \equiv 1 \mod r_j$ (the existence of these primes was shown in http://www.problem-solving.be/pen/viewtopic.php?t=176 ).
Now it suffices to take each $p_i = q_{i,1} \cdot q_{i,2} \cdot... \cdot q_{i,k}$ to get the result again.

But even choosing primes $p_i \equiv 1 \mod m$ avoids the full Dirichlet theorem, since there exists an elementary proof for this, given by Rotkiewicz (using cyclotomic polynomials) for the case when the first term of the arithmetic progression is $1$.

That elementary proof goes back to Gauß. It is still easier to prove it for prime moduli only as it avoids a couple of more subtle arguments, but if anyone wants to take a look at it: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=722&t=393347 .


is there an AMC site for each state where we can look at who got what?

because if i can remember correctly, there were sites for every states which had scores of all the individuals who took AMC.

For Washington State (That's where you are from, right Young?) the site is http://www.math.wsu.edu/amc

Stats for 2005 not up though. That's the only state I could find since I am also from Washington.

thanks. Derek, are you in high school? then where do u go to?
i go to Thomas Jefferson in Auburn.

Here's a boy with Asperger's Syndrome, obsessed with standardized test scores. THus link INSTANTLY GOT ME ADDICTED.

Ok - so where did you get that site? I wanna check other state stats too. Ah - Thomas Jefferson - is math actually cool at that school? It sends so many students to the Northwest Math CHampionships.

A lot of people had their scores drop from 10th grade to 12th grade, for some reason. Ashley Keilman got 111 in 10th grade, then only 97.5 in 12th. Tammy Guo got 96 and then 85 two years later. Tammy Guo, BTW, is a Siemens Westinghouse Semifinalist. But in Bio so she needn't be good at math.

Meru Sadhu went to Caltech, only got 105.5 in his senior year. Hah I have a chance.

Jie Tang apparently didn't even participate in his sophomore year for some reason. Either that or he was owned. But he was damn good when he was at Odle. AH, he got 116.5 in junior year. He went to RSI. Maybe I could get my score to catch up to someone at RSI.

Damn...Lakeside...

I think all the state stats are up on the AMC webpage now:  http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05AMC1012dir-pict.html#States

Overall, 1221 people who took the 12 got a 99.5. That's a lot of pissed off people right there.

lol, it's quite unfair for someone with a 98.5 usamo index to make usamo...

It would be really cool to live in a state with only one AIME qualifier.

From Pennsylvania, only 2 ninth graders qualified by AMC 12...

LOL.

And I think Shaunak is going to make USAMO.

[quote="paladin8"][quote="Simfish"]LMAO.

Wyoming has absolutely no AIME qualifiers. Since there must be a USAMO qualifier for every state, I wonder how the USAMO qualifier for Wyoming feels. ;)

Still, Alaska, Mississippi, Idaho, and North Dakota all have AIME qualifiers.[/quote]

I believe if there are no AIME qualifiers then nobody progresses to the USAMO either. That wouldn't make sense for someone who didn't make AIME to go to USAMO... they wouldn't last very long, either.[/quote]

That's right...in past years (and this year too, it seems), Wyoming has not had any USAMO qualifiers because no one made AIME.  Last year was an exception to the rule, as there was at least one AIME qualifier from Wyoming.

Don't they choose the highest AMC score from the state if no one makes the cut in that state?

I heard that they have at least on person from each state taking AIME.

JB

But very few people in Wyoming even took the AMC, probably the schools at Wyoming don't care about AMC at all.

Hey Yifan; you're the only perfect in CT out of the AMC 10 and 12! see the state stats here [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05hsStateStats/CT.HTML]www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2005-12a/05hsStateStats/CT.HTML[/url].

Btw how many at CHOATE made the AIMEs? I think six made it at the High School...

23. 21 from AMC12, 2 from AMC10.

Strangely enough, the two from the AMC10 were the highest in the state.

Wow... ya the high school has only one for the 10s (me) and six for the 12s, but only five took the test (the number one rank didn't want to take the AIMEs, Barry Chang)


All numbers greater than 4 can be written in the form $2(n+2) - n = n + 4$.  So, no.


Find all integer pairs $ (a,b)$ such that $ a|(9b\minus{}1)$ and $ b|(9a\minus{}1)$

$ ab|(9a\minus{}1)(9b\minus{}1)$
$ ab|9a\plus{}9b\minus{}1$
Then use inequalities.


So I'm sure some of you have probably looked into national olympiads other than USAMO.  In your opinion which country has the hardest national olympiad?  How does USAMO compare to it?

You may also want to check this thread [url=http://www.mathlinks.ro/viewtopic.php?t=247799]here.[/url]  :)

I think Russia and Iran are supposed to be pretty hard also, but I may be mis-remembering my conversation with a certain pythag011...

[quote="not_trig"]Yes - China, Russia, and Iran are the harder ones. East Asian countries (China, Vietnam, Korea) like geometry. There are quite a few more inequalities on other national olympiads than USAMO (Iran, Vietnam come to mind?) For some reason I don't really like Chinese olympiads because they look messy/computational (though they are probably not >.>)

USAMO generally has 2 combo/2 geo or sometimes 2 NT/alg (combined together?)

Ireland is notoriously easy though :O[/quote]

China MO is easy, believe or not. China TST test and quiz are hard. For MO, I think Russian is the hardest.

By the way, Ireland is not that easy.

Really? I had the impression that Ireland was fairly easy (see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=650331#650331[/url] (discussion about difficulty), and [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204258[/url] (21st IrMO)). The problems are rather easy (1 is AIME-level, 2 is almost automatic, 3 and 4 I think I've been able to do before in practice in at most half an hour each)

edit: oops thanks carl

How about those? Some are not that easy, correct?

http://web.archive.org/web/20040602045803/www.kalva.demon.co.uk/irish/irish91.html
http://web.archive.org/web/20040602051710/www.kalva.demon.co.uk/irish/irish94.html


quadratics
LaTeX
search
number theory unsolved
number theory

This is exercise 10.14.6 in Uspensky and Heaslett's Elementary Number Theory.  Chapter 10 introduces Legendre and Jacobi symbols and proves the quadratic reciprocity law.  I validated the solution at the link below (some coincidence) by scrounging grobber's (first) response together with lhvietbao's response.

Link:  http://www.mathlinks.ro/Forum/topic-6197.html

Although the solution is valid, it seems unfair to require the reader to conjure this.  Is this the intended solution, or is there a simpler one?

Subject should be:

Intended solution for impossibility of $ t^2 \equal{} 4abc\minus{}b\minus{}c$ (pos integers)

New to forum: didn't realize that Latex not allowed in subject or that subject would truncate.

Found simpler solution at http://www.mathlinks.ro/viewtopic.php?search_id=1686185741&t=291225

Still seems unfair of Uspensky and Heaslett to not give hint, but hard to imagine a simpler solution.


algebra
polynomial
calculus
derivative
algebra proposed

Let P(x) be a polynomial satisfied P<sup>(m)</sup>(1)=0. (1<=m).
$P(x)=\sum a_{i}x^{i}$ i=0 to n.
Prove that:
For any Q(x) whose degree is smaller then m then we have:
$\sum a{i}Q(i)$$=0$.

I think there's something missing here. Maybe you meant $P^{(k)}(1)=0,\ \forall\  0\le k\le m$ (by $P^{(0)}$ I mean $P$)? Just the fact that the $m$'th derivative is $0$ at $1$ can't be enough..

Why that condition can't be enough...? :?
$Pm(1)=0 with m is a positive integer.$
Try it again! :D

Assume $m=1$ and $Q$ is the constant polynomial taking value $1$ everywhere. Then all we know is that $P'(1)=0$, so $na_n+(n-1)a_{n+1}+\ldots+a_1=0$, but we need to show that $a_n+a_{n-1}+\ldots+a_1+a_0=0$, and this isn't always true.


Hey, how was the UM math competition this year?
I didn't do as well as I hoped :( --it was harder than I expected.
I eventually got all 5 problems, unfortunately not all during the test.
I zipped through 1-3, 4a pretty easily ~40 minutes. Spent all the rest of the time trying to figure out 4b, 5, with little success. Eventually got 4b (used 4a by [hide]dividing quadrilateral into isosceles triangles [/hide] but unfortunately there was a slight flaw that I did not patch in time. Drew diagrams furiously :mad: for #5, thought I should have gotten it, having a lot of practice with box principle and having seen coloring problems before, but I didn't. Had a lot of false starts, ended up writing a lot of stuff making no sense. Prob. 5 [hide]My friend got the idea that it can be proved with 3 pts of the same color in a row, but wasn't able to prove the lemma. Realized that the problem reduced to a classic box-principle problem--given 9 numbers (pts) prove 3 are in arithmetic sequence(equally spaced) , or van der Waal/waarden(?) theorem[/hide]

I got 1, 3, and 4a completely.  On 2, I made an absurdly stupid mistake in the last line, which hopefully won't cost me much since everything else was right.  On 4b I wrote a bunch of random stuff that might lead to a solution after a long time.  On 5, I created a huge, complicated diagram that might work - I didn't have time to make the proof rigorous.

So overall pretty bad - I was tied for third after Part I, but now I'm just hoping for Baltimore County winner.  :(

I think I got 1-4a and 5. So I'm very happy that I was lucky enough to find a solution for five. If you inscribe a midpoint triangle in the original triangle, and then make the original triangle the midpoint triangle of a larger triangle, you can show that there must be three points of the same color, one of which is the midpoint of the other two.

I'm sure i got 1-2 completely, maybe partials on 3-5. :(
for #2b though i got like 144 or something which is a really high age...

haha yes! i was a little bit worried about that.
for #1 i just made a bunch of equations and from then got this:
$ h_1 \equal{} h_4 \equal{} h_7 \equal{} ... \equal{} h_{100} \equal{} h_3 \equal{} ...$
$ h_2 \equal{} h_5 \equal{} h_8 \equal{} ... \equal{} h_{98} \equal{} h_1 \equal{} ...$
$ h_3 \equal{} h_6 \equal{} h_9 \equal{} ... \equal{} h_{99} \equal{} h_2 \equal{} ...$

and from there its easy to see that they are all equal.

i thought #1,2 were pretty simple. im very bad with questions like #5, so i didnt spend too much time on it. i pretty much faked through my "solution".  :blush:

Could you post the questions?

It would be cool to see these challenging problems.

The problems, as well as problems from previous years, can be found at [url]http://www.math.umd.edu/highschool/mathcomp/2007.shtml[/url] (You may have to wait a few days before the problems are posted)


inequalities
function
logarithms
inequalities proposed

Dear Vasc , you should check it again  .
 $ a=1,b=1/2,c=2 $

Why 7 ? You put k>0 and seek the values of k for which the ineq. is true !

Solve first for k=7, then we will see....

This ineq is homogeneous , so we can suppose $ abc=1 $ .
  Consider $ f(x) = \sqrt[3]{x^3+7} $ .
  $ f''(x) <0 $ . So f is a concave function .
 Apply Jensen's , we have :
 $ f(a) + f(b) + f(c) \leq 3.f(\frac{a+b+c}{3}) = 3.\sqrt[3]{(\frac{a+b+c}{3})^3 + 7} \leq 2(a+b+c) $

i don't know exactly but this problem , ok

i change it into:

$\frac{a^3}{k}+\frac{b^3}{n}+\frac{c^3}{m}\geq abc(\frac{1}{k}+\frac{1}{n}+\frac{1}{m})$

long time ago , i remember posting some prob like this , and vasc gave  the nicest solution :huh:

[quote="Vasc"]Let $a,b,c$ be positive numbers. Prove that

$\sqrt[3]{7a^3+abc}+\sqrt[3]{7b^3+abc}+\sqrt[3]{7c^3+abc} \le 2(a+b+c)$.

This one is wrong, sorry. The correct inequality is the following:

$\sqrt[3]{a^3+7abc}+\sqrt[3]{b^3+7abc}+\sqrt[3]{c^3+7abc} \le 2(a+b+c)$.[/quote]

By Holder,
\[(\sqrt[3]{a^3+7abc}+\sqrt[3]{b^3+7abc}+\sqrt[3]{c^3+7abc})^3\]
 \[\le (1+1+1)(a+b+c)(a^2+7bc+b^2+7ca+c^2+7ab)\]
\[=3(a+b+c)(a^2+7bc+b^2+7ca+c^2+7ab).\]
\[3(a+b+c)(a^2+7bc+b^2+7ca+c^2+7ab) \le 8(a+b+c)^3 \Leftrightarrow\]
\[ab+bc+ca \le a^2+b^2+c^2.\]
So the original inequality is true.

[quote="nttu"]You are so fast , yptsoi  :)
  Your method can also be used to solve
$ \sqrt[4]{\frac{a^4+15}{16}} + \sqrt[4]{\frac{b^4+15}{16}} + \sqrt[4]{\frac{c^4+15}{16}} \leq a+b+c $
  but impossible for this one :
 $ \sqrt{\frac{a^2+1}{2}} + \sqrt{\frac{b^2+1}{2}} + \sqrt{\frac{c^2+1}{2}} \leq a+b+c $[/quote]

For second one, we have

$\sqrt{\frac{1+a^2}{2}} \le \frac{3a^2+2a+3}{4a+4}$

Hence it suffices to show

$\sum_{cyc} a+1 - \frac{4}{a+1} \ge 0$

which is evident upon clearing denominators.

[quote="harazi"]I bet that would be a little hard for $n$ variables, since the inequality remains true for $n$ variables. I proposed it on the forum long long time ago. It has a solution in 1 line.[/quote]
wt's the general problem?

Actually, it seems only minimal alterations are necessary. Apply the same lemma. Now, let $f(x) = e^x+1-\frac{4}{e^x+1}$. Then $f''(x) = \frac{e^x\left(5  -e^x+3e^{2x}+e^{3x}\right)}{\left(1 + e^x\right)^3} \ge 0$. Upon substituting $a_i = e^{b_i}$, we have $b_1 + \cdots + b_0 = 0$, and the desired is evident by Jensen.

Here is another solution: $\sqrt{\frac{x^2+1}{2}}+\sqrt{x}\leq x+1$ and it's done by AM-GM.

[quote="harazi"]If $a_1...a_n=1$ then $\sqrt{\frac{a_1^2+1}{2}}+...+\sqrt{\frac{a_n^2+1}{2}}\leq a_1+...+a_n$.[/quote]
The following more general inequality holds:
If  $0<p\le 3+2\sqrt{2}$ and $a_1...a_n=1$ then
 $\sqrt{\frac{pa_1^2+1}{p+1}}+...+\sqrt{\frac{pa_n^2+1}{p+1}}\leq a_1+...+a_n$.
And this inequality holds for some larger $p$.

[quote="Vasc"]Using Holder (as shown yptsoi), we can prove that for $p \ge 2$, the inequality holds
$\sqrt[3]{\frac{a^3+pabc}{1+p}}+\sqrt[3]{\frac{b^3+pabc}{1+p}}+\sqrt[3]{\frac{c^3+pabc}{1+p}} \le a+b+c$
But this inequality is true for larger $p$. :!:[/quote]
The following inequality is also true.
For non-negatives $a$, $b$ and $c$ prove that:
\[3(a+b+c)\geq\sqrt[3]{13a^3+14abc}+\sqrt[3]{13b^3+14abc}+\sqrt[3]{13c^3+14abc}\]
[hide="remark"]
It seems that $5(a+b+c)\geq\sqrt[3]{62a^3+63abc}+\sqrt[3]{62b^3+63abc}+\sqrt[3]{62c^3+63abc}$ is true, but
$\sqrt[3]2(a+b+c)\geq\sqrt[3]{a^3+abc}+\sqrt[3]{b^3+abc}+\sqrt[3]{c^3+abc}$ is wrong already. :(
[/hide]

an old and effective method prove it ,you used it for many times in this forum.

[hide]$f(a)=3a-\sqrt[3]{13a^2+14}-\frac{55}{27}In(a)\ge 0$[/hide]

[quote="red3"]an old and effective method prove it ,you used it for many times in this forum.

$f(a)=3a-\sqrt[3]{13a^2+14}-\frac{55}{27}In(a)\ge 0$[/quote]
I think you are meaning $f(a)=3a-\sqrt[3]{13a^3+14}-\frac{14}{9}\ln{a}$. :wink:
No, in my proof I used Holder.
For $5(a+b+c)\geq\sqrt[3]{62a^3+63abc}+\sqrt[3]{62b^3+63abc}+\sqrt[3]{62c^3+63abc}$ these methods fail.

could you please post your solution? :oops:


[quote="cubedude"]anybody go to runescape.... the massive online game??? if you do i have some great cheats.if you dont....... your a loser!

                                                    ps. no affence to any of you losers.    pss. answer my poll.... i only have 4  people to do it.  pssss. answer my posts.... :lol:  :D  :)[/quote]

Sry, but this post should either be mass edited by a moderator or locked.

[quote="cubedude"]anybody go to runescape.... the massive online game??? if you do i have some great cheats.if you dont....... your a loser!

                                                    ps. no affence to any of you losers.    pss. answer my poll.... i only have 4  people to do it.  pssss. answer my posts.... :lol:  :D  :)[/quote]

I used to play runescape but then i realized life was more important than some silly game.

[quote="cubedude"]anybody go to runescape.... the massive online game??? if you do i have some great cheats.if you dont....... your a loser!

                                                    ps. no affence to any of you losers.    pss. answer my poll.... i only have 4  people to do it.  pssss. answer my posts.... :lol:  :D  :)[/quote]
could you not spam my inbox? I don't care about this gay game so pming me wont do anything :mad:

[quote="ragnarok23"]runescape is a plot to take over kids' minds.[/quote]

Ha, runescape entertained me for a month, it was sad.

Then, I realized how stupid, "Wait while your guy chops wood" and doing that for hours straight just to improve your useless woodcutting skill was pointless, as well as the rest of the game.  So, yeah...

HAHA. Runescape is indeed a quite useless game. But quite time eating. It made me waste endless hours for 4 months since my whole school always played in computer lab,, etc.

The next school year, everybody quit.

It's addictive, stupid, and ridiculous.

offence? i am sorry but there is a point where i stop.

[quote="kyyuanmathcount"]Ha, runescape entertained me for a month, it was sad.

Then, I realized how stupid, "Wait while your guy chops wood" and doing that for hours straight just to improve your useless woodcutting skill was pointless, as well as the rest of the game.  So, yeah...[/quote]
Same here. Except I played for a year and got level 93. :P

This topic should be locked and Runescape should be destroyed.

[quote="i_like_pie"]Runescape should be destroyed.[/quote]
maple story first

You messed up with the quote...

I think all RPG's should be discontinued as they serve absolutely no purpose.

I spent some hours in the tutorial, but it takes only minutes for me to be killed.

What a waste of time! :mad:

[quote="cubedude"]anybody go to runescape.... the massive online game??? if you do i have some great cheats.if you dont....... your a loser!

                                                    ps. no affence to any of you losers.    pss. answer my poll.... i only have 4  people to do it.  pssss. answer my posts.... :lol:  :D  :)[/quote]

I'm sorry, but this is really irking me:

it's [i]offense[/i], not affence.
and I despise runescape for the terrible, horrible game it is.
(not to mention boring and tedious).

*gasp*  I say that this topic should be deleted. Why all this anamocity? Just because this stupid game was brought up by someone, a few people disagree, and then people use (put very vulgar language here) As I said in cubedud's topic "no taste", I told him to ont share opinions at the expense of others. Do I reall y have to say it again in this topic?

If I see any more anamocity in this topic below this very post, I'm going to ask MithsApprentice to delete this topic!! (echo) delete this topic!!! (echo) delete this topic!!!

Just to get everyone's notice, I'm echoing that statement.

My opinion: Have you ever wet to a disney massive multiplayer game called "Toontown"? It is USELESS!!!!

Replace Toontown with Runescape and edit out the disney.

P.S. A word of caution: Do not go there. If you value your brain, don't go there.

[quote="junggi"][quote="moogra"][quote="junggi"]lol edited.
wait no not all rpgs. there are some pwning RPGs out there, that are um useless but really good(not some bs like maplestory or runescape(that means not just super repetitive thing) but actually has story/other good stuff).[/quote]

shut up taylor. runescape is ok. its just long and people still miss at lvl 60s. and maplestory is fun, even though i don't play anymore, and people ask me for hacks[/quote]
david, go die[/quote]

hahaha junggi when you told me to die! ha guess what? the same day some guy told me not to die on him! did i look dead?

@nutz fore: you are right we should delete the topic. cubedude made it.

guess what? I already pmed mithsaprentice to delete this topic!!

I wasted a few years of my life on runescape and I still sucked at it.  only got to lvl 57

[quote="Treething"][quote="kstan013"][quote="cubedude"]anybody go to runescape.... the massive online game??? if you do i have some great cheats.if you dont....... your a loser!

                                                    ps. no affence to any of you losers.    pss. answer my poll.... i only have 4  people to do it.  pssss. answer my posts.... :lol:  :D  :)[/quote]

P.S. Oh yeah, just smile away after calling everyone a [b]loser[/b] for not playing the dumbest game on the face of the Earth. Oh ya, I'm cool, I'm chopping wood in runescape :sleeping: Whoa! Oh yeah. Must defeat spider. :| You're wrong with a capital w that's Wrong in case you didn't know.

P.S.S. offence O-F-F-E-N-C-E offence :wink:
P.S.S.S. Have a nice day in world 265 while i'm doing cool stuff like not playing stupid massive midevil online chat games.
P.S.S.S.S. that is all. for now. at least. maybe later. or not.[/quote]

P.S.S.S.S.S. Medieval ;) .

I really don't have that much against Runescape, it was fun for a while, and while it's repetitive, it is a nice time waster for a while. I don't play it anymore, but honestly, all games get boring after a while.

Except Connect Four. That never gets old.[/quote]

P.P.P.P.P.P.S. You don't stack the S's, you stack the P's. It's not Post Script Script Script Script Script... it's Post Post Post Post Post Script.

why not just lock this.
im sure cubedudes already banned

[quote="junggi"]god i feel so offended that someone called me a loser for not playing such a gay game.[/quote]
Seeing as the game is both genderless and incapable of feeling attraction, I don't see how it can possibly be gay.

[quote="Inspired By Nature"][quote="nutz_for2.718281828"]Now we all have VERY strong arguements against him![/quote]
I have heard about Runescape, and sounds like a pure waste of time. I have also heard from countless people that the graphics are horrible. :D[/quote]
Graphics aren't necessarily important. For example, roguelikes don't have any graphics and they can still be fun.

[quote="cubedude"]anybody go to runescape.... the massive online game??? if you do i have some great cheats.if you dont....... your a loser!

                                                    ps. no affence to any of you losers.    pss. answer my poll.... i only have 4  people to do it.  pssss. answer my posts.... :lol:  :D  :)[/quote]
Oh no, some random person I don't know on the internet just called me a loser! I'm going to go sit in the corner and cry now. Why must life be so painful? I just want to be loved...

[quote="fireforce"][quote="cubedude"]anybody go to runescape.... the massive online game??? if you do i have some great cheats.if you dont....... your a loser!

                                                    ps. no affence to any of you losers.    pss. answer my poll.... i only have 4  people to do it.  pssss. answer my posts.... :lol:  :D  :)[/quote]

There is the great doubling your value trick. Simply get your most valuable items, all your money, get on runescape at 8 during a weekday on world 1, goto the Varrok town square and wait for it....Press Alt and F4 together. REmember it only works when you have ALL of your most valuable items. If you don't then god will smite you and take all your stuff, you are carrying minus 3 things.[/quote]you forgot , you have to drop your stuff before pressing alt f4!!([b][i]snicker, snicker[/i][/b])

[quote="Klebian"][/quote][quote="Treething"][quote="kstan013"][quote="cubedude"]anybody go to runescape.... the massive online game??? if you do i have some great cheats.if you dont....... your a loser!

                                                    ps. no affence to any of you losers.    pss. answer my poll.... i only have 4  people to do it.  pssss. answer my posts.... :lol:  :D  :)[/quote]

P.S. Oh yeah, just smile away after calling everyone a [b]loser[/b] for not playing the dumbest game on the face of the Earth. Oh ya, I'm cool, I'm chopping wood in runescape :sleeping: Whoa! Oh yeah. Must defeat spider. :| You're wrong with a capital w that's Wrong in case you didn't know.

P.S.S. offence O-F-F-E-N-C-E offence :wink:
P.S.S.S. Have a nice day in world 265 while i'm doing cool stuff like not playing stupid massive midevil online chat games.
P.S.S.S.S. that is all. for now. at least. maybe later. or not.[/quote]

P.S.S.S.S.S. Medieval ;) .

I really don't have that much against Runescape, it was fun for a while, and while it's repetitive, it is a nice time waster for a while. I don't play it anymore, but honestly, all games get boring after a while.

Except Connect Four. That never gets old.[/quote][quote="Klebian"]

P.P.P.P.P.P.S. You don't stack the S's, you stack the P's. It's not Post Script Script Script Script Script... it's Post Post Post Post Post Script.[/quote]
P.P.P.P.P.P.P.S.I know. I was just mocking that lame cubedude character. :D

[quote="jmadsen"]Graphics aren't necessarily important. For example, roguelikes don't have any graphics and they can still be fun.[/quote]

NetHack is win.

runescape? i quit when i died. that was when i was lvl 10


On side $BC$ of a triangle $ABC$ point $A'$ is taken.The midperpendicular to segment $A'B$ intersects side $AB$ at point $M$ and the midperpedincular to segment $A'C$ intersects side $AC$ at point $N$.Prove that the symmetric to point $A'$ through line $MN$ lies on the circumscribed circle of triangle $ABC$.

[quote="fotis"][color=darkred]On side $BC$ of a triangle $ABC$ point $A'$ is taken. The midperpendicular to segment $A'B$ intersects side $AB$ at point $M$ and the midperpedincular to segment $A'C$ intersects side $AC$ at point $N$. Prove that the symmetric $S$ to point $A'$ through line $MN$ lies on the circumscribed circle of triangle $ABC$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Prove easily that the points $M$, $N$ are the circumcenters of the triangles $BA'S$, $CA'S$ respectively and $\left\{\begin{array}{c}m(\widehat{BSA'})=90^{\circ}-B\\\ m(\widehat{CSA'})=90^{\circ}-C\end{array}\right\|$ $\Longrightarrow$ $m(\widehat{BSC})=m(\widehat{BSA'})+m(\widehat{CSA'})$ $\Longrightarrow$ $m(\widehat{BSC})=A$ $\Longrightarrow$ the quadrilateral $ASBC$ is cyclically, i.e. the point $S$ belongs to the circumcircle of the triangle $ABC$.

[b]An equivalent enunciation.[/b] Let $ABC$ be a triangle and let $M,N$ be two points of the side $[BC]$ so that $M\in (BN)$ and $2\cdot MN=BC$. Denote the points $O_{1}\in AB$, $O_{2}\in AC$ for which $O_{1}M\perp BC$, $O_{2}N\perp BC$ and $O_{1}B=R_{1}$, $O_{2}N=R_{2}$. Prove that the circles $C(O_{1},R_{1})$. $C(O_{2},R_{2})$ and the circumcircle $C(O,R)$ of the triangle $ABC$ have a common point. [/color]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=22639 .

  darij


Let $ f(x)\equal{}2009x^5\minus{}x^4\minus{}x^3\minus{}x^2\minus{}2006x\plus{}1$. Prove that $ f(n);f(f(n));f(f(f(n)))...$ are pairwise coprime for all integer $ n$.

General problem: Let $ f\in\mathbb{Z}[x]$ such that $ f(0)\equal{}f(1)\equal{}1$. Prove that $ f(n),f(f(n)),...$ are pairwise coprime for all integer $ n$.  :)

[quote="N.T.TUAN"]General problem: Let $ f\in\mathbb{Z}[x]$ such that $ f(0) \equal{} f(1) \equal{} 1$. Prove that $ f(n),f(f(n)),...$ are pairwise coprime for all integer $ n$.  :)[/quote]
we have $ \forall x\in\mathbb{Z}: \ f(x)\equal{}x(x\minus{}1)g(x)\plus{}1$ for some $ g\in\mathbb{Z}[X]$
we take $ m\in \mathbb{N}^*$ and $ f^0(m)\equal{}m$ $ f^{j}(m)\equal{}fofo..j\ times...of(m))$
$ \forall i\in\mathbb{N}\minus{}\{0\}: \ f^{i}(m)\equal{}1\plus{}(m\minus{}1)mg(m)\prod_{i\minus{}1}^{k\equal{}0}g(f^{k}(m))f^{k}(m)$ gives $ \forall i\in\mathbb{N}\minus{}\{0\}: \ gcd(f^{i}(m),m)\equal{}1$
and $ m\equal{}f^{j\minus{}i}(n)$ gives  $ gcd(f^{j}(n),f^i(n))\equal{}1$

[quote="ricardokaka"]It's so difficult to understand![/quote]

[color=green]Which part of this solution?[/color]

Here
[quote="aviateurpilot"]
$ \forall i\in\mathbb{N} \minus{} \{0\}: \ f^{i}(m) \equal{} 1 \plus{} (m \minus{} 1)mg(m)\prod_{i \minus{} 1}^{k \equal{} 0}g(f^{k}(m))f^{k}(m)$
[/quote]

[color=green]uhm, you can see the submition $ \forall x\in\mathbb{Z}: \ f(x)\equal{}x(x\minus{}1)g(x)\plus{}1$ to the equation

\[ f^i(m)\equal{}f(f^{i\minus{}1}(m))\equal{}1\plus{}f^{i\minus{}1}(m)(f^{i\minus{}1}(m)\minus{}1)g(f^{i\minus{}1}(m))\plus{}1
\\
.....
\][/color]

$ x\mid f(x)\minus{}1$ in $ \mathbb{Z}[x]\Rightarrow f^k(x)\mid f(f^k(x))\minus{}1\equal{}f^{k\plus{}1}(x)\minus{}1$ in $ \mathbb{Z}[x]$ for all $ k\in\mathbb{N}_0$
and
$ x\minus{}1\mid f(x)\minus{}1$ in $ \mathbb{Z}[x]\Rightarrow f^{k\plus{}1}(x)\minus{}1\mid f^{k\plus{}2}(x)\minus{}1\mid\cdots \mid f^{l}(x)\minus{}1$ in $ \mathbb{Z}[x]$ for all $ k,l\in\mathbb{N}_0, k<l$.

So $ f^k(m)\mid f^l(m)\minus{}1$ for all $ m\in\mathbb{Z}$ and $ k,l\in\mathbb{N}_0, k<l$.

Nice problem.


Let ABC be a triangle. A circle passing through A and B intersects segments AC and BC at D and E, respectively. Lines AB and DE intersect at F, while lines BD and CF intersect at M. Prove that $ MF\equal{}MC$ if and only if $ (MB)(MD)\equal{}MC^2$.

[geogebra]2e39fac68b212f0b7c4075961241aba95473ee25[/geogebra]
Replace that $ G$ with an $ M$...

We first prove the if direction ($ \Rightarrow$): If  $ MB\cdot MD\equal{}\displaystyle MC^2$ then $ \triangle MBC\sim\triangle MCD$.

We wish to prove that $ \frac{MC}{MF}\equal{}1$. Note that by Ceva's, $ \frac{FA}{AB}\cdot\frac{BE}{EC}\cdot\frac{MC}{MF}\equal{}1$.

Thus it suffices to show that $ \frac{FA}{AB}\cdot\frac{BE}{EC}\equal{}1$ or $ \frac{FA}{AB}\equal{}\frac{EC}{BE}\iff \frac{BF}{BA}\equal{}\frac{BC}{BE}\iff \triangle BAE \sim \triangle BFC\iff AE\parallel{}FC$.

This is clearly true since $ ADEB$ is cyclic and thus $ \angle MCA\equal{}\angle DBE \equal{}\angle DAE$ so $ \angle MCA\equal{}\angle DAE$. Since they are alternate interior angles, the claim follows.

For the only if direction ($ \Leftarrow$): We use Ceva's to deduce that $ \frac{FA}{AB}\equal{}\frac{EC}{BE}\Rightarrow \triangle BAE\sim \triangle BFC$ since they share $ \angle ABE$.

This directly implies that $ AE\parallel{}FC$ and hence $ \angle DBE\equal{}\angle DAE\equal{}\angle MCD \Rightarrow \triangle MDC \sim \triangle MCB$ and the claim follows.

Hey nice solution!!  Lol its funny, Indiana had like 20 active users like a year ago, but now, we only have 2.

I worked on the other problems, and some of them, I manage to get pretty far (maybe even a complete USAMO problem for me!).  However, on some, I did not know how to prove the last couple of steps (usually non-trivial ones--I did write out everything I proved) and for 3 of the problems (2,4,6), I dont even know where to begin.

You will probably be busy solving more difficult problems this summer (btw how are you going to do both research + MOP? <--- cause thats like amazing if you make Red MOP and Siemens Reg. Finalist in summer of 10th grade/begining of 10th grade), but maybe you can solve them like later.  Im interested in seeing how they are proved.

Oh yea and lol, btw, the Indiana forum will officially be a ghost forum again because you will probably be busy at camp/research and I am also doing research this summer (maybe Ill get lucky and get something good at Siemens =), but Ill be okay if I don't because its more about the experience than anything else) so that will take away most of my free time.


The students in Borden Junior High School sold $ \$13,000$ worth of candy bars last year. If their profit is $ \$1,000$ on the first $ \$10,000$ in sales and 20% on all sales over $ \$10,000$, which of the following represents their total profit selling candy bars?
[color=darkblue]
Moderator's Note: Try to learn some $ \text{\LaTeX}$ before using it and experimenting in examples.[/color]

A. [code]1000[/code] + (0.20 x [code]13 000[/code])

B. [code]1000[/code] + (0.20 x [code]10 000[/code])

C. [code]1000[/code] + 0.20 ([code]10 000[/code]-[code]1000[/code])

D. [code]1000[/code] + 0.20 ([code]13 000[/code]-[code]1000[/code])

E. [code]1000[/code] + 0.20 ([code]13 000[/code]-[code]10 000[/code])

I suggest you clean up your syntax.  I can't really tell what you mean.  When you say 13000 worth of candy bars, is that profit or number of bars?

For basic stuff, the syntax is rather simple.  Just type what you want, but put $ \$$ around it.  For example,
$ ax^{2}\plus{}bx\plus{}c\equal{}0$
is just
[code]$ax^{2}+bx+c=0$[/code]

and something like
$ a\times b\equal{}b\times a$
is
[code]$a\times b=b\times a$[/code]

I think it is because you put dollar signs after the numbers. Either type [code]$\$$[/code] which gives $ \$$, or replace $ \$$ with USD. Other than that, just take out those code boxes, which will really help. Also, if you don't LaTeX $ \times$, use the asterick (*) instead of x.

i have no idea what the 'following' says

so ill just solve it

problem: The students in Borden Junior High School sold 13000 dollar worth of candy bars last year. If their profit is 1000 dollars on the first 10 000 dollars in sales and 20% on all sales over 10000,which of the following represents their total profit selling candy bars?

think thats it

over 10000=3000 dollars
$ 3000*\frac{2}{10}\equal{}600$

1000+600=1600

answer : 1600 dollars


[quote="zhaobin"]let $a_k=\frac{1}{2^k}$ for $k=1,2,\cdots,n$ for n is large enough,I find it is wrong.[/quote]

I agree zhaobin,so we need consider at first when $n=3$,that is:

Let $a_i>0(i=1,2,3)$, prove or disprove that:
\[ \frac{a_1}{a_2} ln{\frac{a_1+a_2}{a_1}}+\frac{a_2}{a_3} ln{\frac{a_2+a_3}{a_2}}+\frac{a_3}{a_1} ln{\frac{a_3+a_1}{a_3}} \le 3\ln2.  \]

[quote="jjchen611"]A rare and difficult Log-inequality:

Let $n>1$ be a positive integer,$a_i>0(i=1,2,...)$, prove:
\[ \frac{a_1}{a_2} ln{\frac{a_1+a_2}{a_1}}+\frac{a_2}{a_3} ln{\frac{a_2+a_3}{a_2}}+...+\frac{a_{n-1}}{a_n} ln{\frac{a_{n-1}+a_n}{a_{n-1}}}+\frac{a_n}{a_1} ln{\frac{a_n+a_1}{a_n}} \le n\ln2.  \]

when $n=2$,the inequality has been proved by Virqil Nicula recently.[/quote]

Sorry,it was Virgil Nicula (not Virqil Nicula ) first gave a  proof for the case $n=2$,

See http://www.mathlinks.ro/Forum/viewtopic.php?t=72818


complex analysis
advanced fields
advanced fields unsolved

Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_{1},z_{2},\ldots,z_{n},\ldots$ are its zeros $(|z_{k}| < 1)$, then

$\sum_{n} (1 - |z_{n}|) < \infty$.

Hint: Consider the Blaschke product

$B(z)=\prod_n\frac{z-z_n}{1-\overline{z_n}z}.$

$\frac{f(z)}{B(z)}$ should be bounded on the disk, and this should also be true for any partial products. What would the convergence of the product tell you about the zeros?


Please help me with this question. ABC is a triangle. D, E and F are points on BC, AC and AB such that AD, BE and CF are concurrent. Show that lines parallel to AD, BE and CF from the midpoints BC, AC and AB are also concurrent

[quote="yetti"]The lines parallel to the concurrent cevians AD, BE, CF through the midpoints A', B', C' of the sides BC, CA, AB are cevians of the medial triangle $\triangle A'B'C'$, which has the sides $A'B' \parallel AB,\ B'C' \parallel BC,\ \C'A' \parallel CA$ parallel to the sides of the original triangle $\triangle ABC$.[/quote]
How can they be parrallel to the sides. they are already given to be parallel to AD, BE and CF. Moreover these cevians are fromthe vertex of the triangle to opposite sides. I think you didn't understand the question.
  I have anyway made a negligible mistake in typing. That is lines are drawn from the midpoint of the opposite sides parallel to the cevians.

Well yetti's idea should still work. Basically, the three lines in question are to A'B'C' the same way that AD, BE, CF are to ABC.


calculus
integration
logarithms
Putnam
calculus computations

Does anyone happen to know how to evaluate this $ \displaystyle\int_0^{\infty}\frac{x^{p\minus{}1}}{1\plus{}x}(\ln x) dx$ where $ 0<p<1$.

$ \int_0^{\infty}\frac{x^{p\minus{}1}}{1\plus{}x}\,dx \equal{} \pi \csc(\pi p)$

$ \equal{}> \frac{d}{dp}\int_0^{\infty}\frac{x^{p\minus{}1}}{1\plus{}x}\,dx \equal{} \int_0^{\infty}\frac{d}{dp}\frac{x^{p\minus{}1}}{1\plus{}x}\,dx \equal{} \int_0^{\infty} \frac{(\ln x) x^{p\minus{}1}}{1\plus{}x}\,dx \equal{} \frac{d}{dp} [\pi \csc(\pi p)]$

$ \therefore \int_0^{\infty} \frac{(\ln x) x^{p\minus{}1}}{1\plus{}x}\,dx \equal{} \minus{} \pi^2 \csc(\pi p) \cot(\pi p)$

Hi Carcul, thank you very much for showing me how to do this!

After seeing the solution, I was wondering that if this integral appeared in some contest, say the Putnam, then does one have to justify why he/she can differentiate under the integral sign.  I mean does one have to show/say anything about uniform convergence or what is in Carcul's post enough?


How strong do you think the correlation between interest in math and achievement in math is?

I suspect that there is a strong correlation between long-term achievement and interest. It's hard to be bored by a subject and end up making huge progress in the field. With math, I think you really need to be interested in it to go really far, especially when your work does not appear to have practical applications.

My intuition agrees with paladin8, but I had an experience several years ago which shakes that intuition.  I was visiting graduate schools (to which I'd been accepted) in order to decide where I wanted to go.  My visit to one institution coincided with that of three other prospective graduate students, so the four of us sat around chatting a bunch.  At one point I asked them all why they were choosing to go to grad school in mathematics.  One was clearly excited about studying math, but the other two had no real answers or any demonstrated passion.  Of those two, one was returning to school from a position in industry, on a lark.  Now, it wouldn't surprise me to encounter that sort of thing at a non-prestigious/not-very-selective/not-high-quality institution, but the place we were visiting was (at the time) in the top 10 graduate schools in the country.  True, I have no idea what those guys have done since then (I don't even remember their names); they may no longer be in mathematics.  But it does seem to be possible to be a mathematical high achiever in the 5-10 year term and not be all that interested in mathematics.


$ P_{AGHD}\equal{}2x\plus{}2y\equal{}23$
$ P_{ABCD}\equal{}2x\plus{}8y\equal{}47$

Then $ 6y\equal{}24\Rightarrow y\equal{}4$ and $ x\equal{}7.5$.  Then the area is $ 4\times 7.5\equal{}30$.

The correct [hide=answer is.]
$\boxed{120}$.

I believe you forgot to multiply the height by 4, because it asks for the area of the larger rectangle


geometry
circumcircle
trigonometry
geometry unsolved

Let n be an odd positive integer. Consider a regular polygon with n sides inscribed in a circle. If the vertices are numbered from 1 to n and P is a point on the circle on the arc between the first and the n-th vertex, show that the sum of the distances of P  from the even numbered vertices equals the sum of its distances from the odd numbered distances.

Let $A_1A_2\ldots A_{2n+1}$ be the regular polygon, $O$ its circumcenter and consider $P$ on the arc $A_1A_{2n+1}$. Denote $\angle{POA_1}=2x$ and $\frac{2\pi}{2n+1}=2a$.
We have $PA_1=2R\sin x,PA_2=2R\sin(x+a)$, etc. (here $R$ is the circumradius). We have to prove that
\[\sin x-\sin(x+a)+\sin(x+2a)-\ldots +\sin(x+2na)=0.
\]
Muliplying by $2\cos\frac a2$ the sum telescopes to
\[\sin\left(x-\frac a2 \right)+\sin\left(x+2na+\frac a2 \right)=2\sin(x+na)\cos\frac{(2n+1)a}{2}.
\]
But $\cos \frac{(2n+1)a}{2}=\cos \frac{\pi}{2}=0,$
as needed.

[quote="Arne"]Or you can just use Ptolemy (like n times or something).[/quote]
That's interesting. I'd like to see that.

[quote="Arne"]Or you can just use Ptolemy (like n times or something).[/quote]
Really, Arne, I am very interested in your solution using Ptolemy theorem!

If you heard for geometry book written by Prasolov...
 well you have that problem there with elementary solution and it is via Ptolomy theorem (I think, I am not 100% sure that it is in that book, but I am sure that you can solve this by Ptolomy, because some of my friends had this on their written test paper only with $7$-gon not $n$-gon :D )

Yes, I found the solution in Prasolov's book, thank you! It is indeed short and nice: one applies Ptolemy in every quadrilateral $PA_iA_{i+1}A_{i+2}$. Carefully adding up the equalities yields the result.


Columbia
email
Math Prize for Girls
AMC 10

Just before presenting awards at the Math Prize for Girls (sponsored and organized by the Advantage Testing Foundation), we will be having a lecture by a distinguished mathematical scientist.  I am delighted to announce that this year's speaker will be [url=http://en.wikipedia.org/wiki/Janna_Levin]Janna Levin[/url], Professor of Physics at Barnard College of Columbia University.  Professor Levin’s research is in the Early Universe, Chaos, and Black Holes.  She is the author of the popular science book [i]How the Universe Got its Spots: Diary of a Finite Time in a Finite Space[/i] and the novel [i]A Madman Dreams of Turing Machines[/i].  She received her PhD in Physics from MIT.  In 2006, she was [url=http://www.comedycentral.com/search/gsa/index.jhtml?searchterm=janna+levin]interviewed[/url] on The Colbert Report.

P.S.  Our application deadline is coming up -- this Friday, May 22nd.  We will send out invitations by email a few days later.

A direct link to the interview is [url=http://www.colbertnation.com/the-colbert-report-videos/73484/august-24-2006/janna-levin]here[/url].  It's a very good interview for only 6 minutes: he lets her talk quite a bit, and she's very good at getting to the point quickly enough for the medium.  I bet she'll make a great speaker!

I am hijacking my younger daughter’s user account (actually she took my account, so it is hers now).
How’s the application process for Math Price for Girls? Did you get lots of applicants? Is there any preliminary cut off for invitation? My younger daughter (a current 7th grader) has applied and is hoping for a trip to New York. I myself will be accompanying her to New York if she gets invited. I cannot wait to listen to Janna Levin’s lecture.

The application process is coming along fine.  We received more than 700 applications from all over the country.  I expect the cutoff for the AMC 10 to be in the high 110's and the cutoff for the AMC 12 to be in the high 90's.  I will announce the exact cutoffs here next week as soon as we set them.  I will also send an email message to all the contestants next week to let them know their status.

I hope your daughter makes it!  If you come to New York, feel free to stop me and say hello.

Thank you for your quick response. I assume my younger daughter made it.

An addendum: While the contestants take our exam in the morning, another speaker will be speaking to the parents and other guests.  Our morning speaker will be Richard Rusczyk, founder of the Art of Problem Solving.  I believe Richard will be speaking on resources and opportunities for bright math students.  At our awards ceremony, Richard will also give a few brief remarks.