linear algebra
matrix
probability
vector

Ok, so I have figured out most of the process of how to solve this problem, but I get to a certain point and then get stuck:

Paul travels back and forth in his work between New York and Boston. He never makes two trips on the same day. On a day when he wakes up in New York, the probability is 0.07 that he will fly to Boston during the day. On a day when he wakes up in Boston, the probability is 0.14 that he will fly to New York.

I have the matrix at Matrix formatting: $ A \equal{} \begin{bmatrix}.93 & .07 \\
.14 & .86\end{bmatrix}$ and have s (the initial probability distribution) at: x = 2/3 y = 1/3

However I do not know the next step/process:

If he wakes up in Boston on Monday, what is the probability he will wake up in New York on Wednesday?

Sorry about bad matrix formatting, still haven't figured it out.

Matrix formatting: $ A\equal{}\begin{bmatrix}.93&.07\\.14&.86\end{bmatrix}$
Quote me to see how it's done.

We can represent an initial state with a vector: $ \begin{bmatrix}0&1\end{bmatrix}$ if he starts in New York with probability $ 0$ and in Boston with probability $ 1$, for example. How can you use the matrix $ A$ to determine his probabilities for the next day? The day after that?

(Note: I would transpose everything, and you might want to as well)

[quote] How can you use the matrix  to determine his probabilities for the next day? The day after that? [/quote]

Thats exactly my problem. I have no idea.... :(

Write down the probabilities the next day in terms of the probabilities the previous day. Does this look like a matrix operation you know? It should, because there's a reason we write these things with matrices.

[quote="jmerry"](Note: I would transpose everything, and you might want to as well)[/quote]
The notational tradition for Markov chains is multiplication with row vectors to the left of square (stochastic) matrices. The analysis calls for looking at left eigenvectors of that matrix.

As I said, tradition.


Sally took a quiz. One of the question said "Points A,B,C, and D form a rectangle (it is longer horizontally). AD and BC are 1 unit long and AB and  CD are 2 units long. Take point E on the extention of segment BC such that ACED is a parallelogram. Find the area of parallelogram ACED"
Sally thought for a moment, then written down
"AD is one unit long.
AC is :sqrt:AD:^2:+CD:^2:. Substituting 1 for AD and 2 for CD, you get AC=:rt5:
AC*AD=1*:rt5:=:rt5:
The answer is :rt5:."
What did Sally do wrong and what is the correct answer?

Anyone can answer, but please use spoilers.

[hide]The area of a parallelogram is base * altitude to that base.

In this case, if you make the base (CE) 1, then the altitude to it is CD, which is still 2.  Therefore, the area is 1*2 = 2.

Edit: Alright, I thought saying what to do RIGHT would imply what she did wrong, but okay.  While it's true that DE = sqrt(5), it's not the altitude to the side of length one.  Since she chose the base to be the side with length 1, she needed to pick the altitude to it, which is CD, which has length 2.[/hide]

Right, but what did Sally do wrong? (in spoilers please)

I edited it to more explicitly say what she did wrong.

Correct. the moral here: Always look for what the question asks

I corrected your spelling error, I hope you don't mind...(you misspelled "alright")

[quote="Tare"]I corrected your spelling error, I hope you don't mind...(you misspelled "alright")[/quote]

Yes, I mind!  I mind VERY MUCH!  :twisted:  :twisted:

Naw, it's awwwwlright if you do that...  :wink:

all your doing is shifting the top.  area = b*h is unchanged, so its 2.

my question -- this owrks mathematically, but if u draw it out, it always looks like the slanted one is bigger and covers more area just because its,... slanted... do you guys get that too or do I just have a strange disposition towards seeing slanted figures as bigger..


calculus
integration
calculus computations

How can i compute the integral :
$\int \frac{sinx}{ln(2+x^2)} dx$


in moama ro az ketabe 1, 2, 3, binahayat naghle ghoul mikonam.
marde javani naghshaye ganji peyda kard. dastour peda kardan gang dar naghsheye ganj be sourate zir bood:
" ba ghayegh be 60 darajeye shomali va 70 darajeye sharghi boro. onja be jazirei matrook bar mikhory. dar shomale in jazire do darakh be name baloot va senoubar va yek choobeye dar mibini. be paye choobeye dar boro va be samte derakhte baloot harkat kon va ghadmat ra beshmar. sepas mokhalefe aghrabehaye saat be andazeye 90 daraje becharkh, va be haman andaze ke ghadmat ra shemordi harkat kon va anja ra alamat begozar.
be paye choobeye dar boro va be samte derakhte senouar harkat kon va ghadmat ra beshmar. sepas movafeghe aghrabehaye saat be andazeye 90 daraje becharkh, va be haman andaze ke ghadmat ra shemordi harkat kon va anja ra alamat begozar. nesfe faseleye do alamat ganj ast."
marde javan anja raft va hame chiz did bejoz choobeye dar.
chegoone ganj ra peyda konad.

khedmatetoon arezam ke:
ba tebgh shekl zir, ebteda mehvar haghighi va mohomi ra mikashim. "dur" dar har koja mikhat bashad masalai nis. dar inja gama farz shode. va bayesti besoorate "gamma=ai+b" bashad.
tebghe naghshe amal mikonim va javab "i" mishavad.{chera?}

fekr mikonam akhiran in masale erae shodeh. :!:

mitooni too  safheye aval e community peidash koni. :wink:

be har hal mitoonin inja ham saiyetoono bekonid
faghat ye tazakor bayad bedam ke dar adade mokhtalet har gah dar bordare kolle bordar ra dar 'i' zarb konim be sorate ghayeme nesbat be bordare avvaliye dar miayad.

azin ke tavvajohi be moamaye man nadashtid masrooreman kardid hala kodemoon javabe moamma ro midim  :lol:

nemoodary ke keshidam neshangar an ast ke:
agar faseleye do derakht ra vahed akhtiyar konim mitavan goft ke derakhte baloot dar noghteye $+1$ rooye mehvare haghighi gharar darad va derakhte senoobar dar noghteye $-1$ bar rooye an mehvar. ma namidanim ke choobeye dar dar kodam noghte boode ast pas mahalle farzi an ra dar noghtei ba harfe $\Gamma$ namayesh dade shode ast
moayan mikonim. choon choobeye dar elzaman bar rooye yeki az do mehvar naboode ast bayad an ra adade morakabi farz konim be soorate $\Gamma=a+b\i$.
aknoon ba ghaedeye zarbe adad haye mohoom chand hesabe sade mikonim. hargah choobeye dar dar $\Gamma$ va derakhte baloot dar $-1$ bashand faseleye anha az heise meghdar ba ebarate $(-1)-\Gamma=-(1+\Gamma)$ moayan mishavad. be vajhe moshabahe agar bekhahim in do fasele ra be andazeye yek zaviye ghayeme dar jahat harkate
aghrabe ha ya dar jahate mokhalefe harkate aghrabeha begardanim bayad bar tebgh ghavaed anha ra dar $-\i$ va $\i$ zarb konim va noghati ra ke bayad neshaneha va dar an gharar dahim ba in nahv taen konim:
neshaneye avval: $(-\i)[-(1+\Gamma)]+1=\i(\Gamma+1)-1$ neshaneye dovvom: $(+\i)(1-\Gamma)-1=\i(1-\Gamma)+1$ choon mahalle ganj dar vasate rah baine do neshane ast bayad nesfe majmooe do adad morakabi ra ke bedast avardim taen konim:
$\frac{1}{2}[\i(\Gamma+1)+1+\i(1-\Gamma)-1]=\frac{1}{2}[+\i\Gamma+\i+1+\i-\i\Gamma-1]=\frac{1}{2}(+2\i)=+\i$ har koja ke derakhtan bashandmibinim ke baz ham mahalle majhool choobeye dar noghteye $+\i$ ast.  :roll: movafagh bashid  :idea:


Formula for sum of sequenced numbers is

$ (Highest \plus{} Lowest)\left(\frac {Highest \minus{} Lowest \plus{} Interval}{2\times\text{Interval}}\right)$

So $ (N \plus{} 2)\left(\frac {N \minus{} 2 \plus{} 2}{4}\right) \equal{} (N \minus{} 2)\left(\frac {N}{4}\right) \equal{} \boxed{\frac {N^2 \plus{}2N}{4}}$

You need to use the arithmetic sequence.

2+4+6+...+n
= (n+2)/2 x ((n-2)/2+1)
= (n+2)(n-2+2)/2
= (n^2 + 2n)/2

@SuperNerd123
Where did you get -2 for the lowest and a 4 for your denominator

$ \begin{array}{crcrcrcccrcr}
& S & \equal{} & 2 & \plus{} & 4 & \plus{} & \cdots & \plus{} & (2m\minus{}2) & \plus{} & 2m \\
\plus{}) & S & \equal{} & 2m & \plus{} & (2m\minus{}2) & \plus{} & \cdots & \plus{} & 4 & \plus{} & 2 \\ \hline
& 2S & \equal{} & (2m\plus{}2) & \plus{} & (2m\plus{}2) & \plus{} & \cdots & \plus{} & (2m\plus{}2) & \plus{} & (2m\plus{}2)
\end{array}$

So, $ S\equal{}\frac{(2m\plus{}2) \times m}{2}$

even integers, so the easiest way is $ {N^2\plus{}N}$,which is what DANub's formula simplifies to.  :wink:

[quote="MCmaniac"]even integers, so the easiest way is $ {N^2 \plus{} N}$,which is what DANub's formula simplifies to.  :wink:[/quote]

Actually it was already at it's simplest form since N^2 doesn't have a factor of 2 as a coefficent.

then you have clearly factored wrong.

EDIT: what does N stand for? the number of integers or the last integer in the sequence?

EDIT (again): Ok, N=The last integer, in that case, SuperNerd's,TopOfMath's, and Kouichi Nakagawa's formulas are correct.

[quote="Cliu0301"]1. 2 + 4 + 6 + 8.........+ N.

Is there a formula rather than just adding them all up?[/quote]

Let $ S\equal{}2\plus{}4\plus{}6\plus{}8\plus{}\cdots\plus{}N$ Then write $ S$ backwards:

What do you get:

[hide="I got...(click here)"]$ S\equal{}2\plus{}4\plus{}6\plus{}8\plus{}\cdots\plus{}N, S\equal{}N\plus{}(N\minus{}2)\plus{}(N\minus{}4)\plus{}(N\minus{}6)\plus{}\cdots\plus{}2$ so add them together, and we have $ 2S\equal{}(N\plus{}2)\frac{N}{2}$ or $ S\equal{}\frac{N(N\plus{}2)}{4}$[/hide]Hope that helps:)

The formula for $ X$ even numbers in a row is $ X^2 \plus{} X$ (prove that with induction if you want)

Expanding from that, if there are $ X$ numbers in a row, then the last number $ N$ in terms of $ X$ is $ X \equal{} \frac {N}{2}$. Substituting in from the first formula $ X^2 \plus{} X$, we get $ (\frac {N}{2})^2 \plus{} \frac {N}{2}$, which simplifies to $ \frac {N^2 \plus{} 2n}{4}$

Okay...so 2 + 4 + 6 + 8 +......+ N = (N^2 + 2n)/4

And sum of the first N even integers is N^2 + N.

The sum of the first N odd integers is N^2.

1 + 3 + 5 + 7 + 9 + 11 + 13 +....+ N = [(N + 1)/2]^2?

Let me try this out...

First Formula: (N^2 + 2n)/4

What is the sum of 2 + 4 + 6 + 8 + . . . + 20?

[(20^2 + 2(20)]/4 = 110.

(Checks this using calculator :P It works! Thanks guys!)

Second Formula (I knew N^2 + N and N^2 :) )

What is the sum of 1 + 3 + 5 + . . . + 13?

[(13 + 1)/2] ^ 2 = 49

(Checks using calculator...It works!!)

Thanks so much!

[quote="SuperNerd123"]Formula for sum of sequenced numbers is

$ (Highest \plus{} Lowest)\left(\frac {Highest \minus{} Lowest \plus{} Interval}{2\times\text{Interval}}\right)$

So $ (N \plus{} 2)\left(\frac {N \minus{} 2 \plus{} 2}{4}\right) \equal{} (N \minus{} 2)\left(\frac {N}{4}\right) \equal{} \boxed{\frac {N^2 \plus{} 2N}{4}}$[/quote]
Just subsitute the new numbers in

[quote="Cliu0301"]1. 1 + 3 + 5 + 7 + 9 + 11 + 13 +....+ N[/quote]

Easy, first you find out how many number there are by $ P(N) \equal{} \frac {(N \plus{} 1)}{2}$. Now square $ P(N)$
Explaination: If you draw a dot, and then 3 more dots, you get a square of side length 2. Now adding 5 more dots, gives you a square of side length 3, and so on.

Edit: Uh SuperNerd, this is way easier, and I think your formula is wrong.

The question (or N) is in terms of the [u]last[/u] number in the sequence, so SuperNerd's equation is correct. (I thought the same in the beginning as well)[/u]


Given the domain {-2, -1, 0, 1},find the range of the following relation:
$g(x)=-x^2+2x-3$

the range would be -11 to -2, as g(-2)=-11 and g(1)=-2


There are 3 math clubs in the school district, with 5, 7, and 8 students respectively. Each club has two co-presidents. If I randomly select a club, and then randomly select three members of that club to give a copy of $ \emph{Introduction to Counting and Probability}$, what is the probability that two of the people who receive books are co-presidents?

Another solution: First, each school has $ 1/3$ chance of being selected. If a school with $ n$ members is chosen, the probability that each co-president gets a book is $ \Pr(n)\equal{}\frac{2}{n}\cdot\frac{1}{n\minus{}1}\cdot\frac{n\minus{}2}{n\minus{}2}\plus{}\frac{2}{n}\cdot\frac{n\minus{}2}{n\minus{}1}\cdot\frac{1}{n\minus{}2}\plus{}\frac{n\minus{}2}{n}\cdot\frac{2}{n\minus{}1}\cdot\frac{1}{n\minus{}2}$. Summing the three probabilities with weight gives $ \frac{1}{3}(\Pr(5)\plus{}\Pr(7)\plus{}\Pr(8))\equal{}\boxed{\frac{11}{60}}$.

Why did you just revive this 7 year old post...

[quote=maria.chrysafis][quote=math154]This is just casework: adding up the probabilities for each group, we get

$ \frac13\left(\frac{5\minus{}2}{\dbinom53} \plus{} \frac{7\minus{}2}{\dbinom73} \plus{} \frac{8\minus{}2}{\dbinom83}\right) \equal{} \boxed{\frac {11}{60}}$.[/quote]

NO COPYING[/quote]

what do you mean, no copying?

Edit: @1below, ok then!

here's another sol(its like the same as alcumus sol almost but with slightly different reasoning)

[hide=sol]First of all, there is a 1/3 probability that you select a given club. Also, there are nc3 ways to choose 3 people from each club. Now we proceed to find the number of ways you can choose the co-presidents. We let them be chosen by default, and so we have 5-2=3 ways for the first club, 7-2=5 for the second club, and 8-2=6 for the third. The desired answer is 1/3(3/10+5/21+6/28)=11/60[/hide]

[quote=wlm2]math154 probably won't see that post. :)[/quote]

They might


number theory
greatest common divisor
IMO
Euler
graph theory
combinatorics
imo 1991

Suppose $ \,G\,$ is a connected graph with $ \,k\,$ edges. Prove that it is possible to label the edges $ 1,2,\ldots ,k\,$ in such a way that at each vertex which belongs to two or more edges, the greatest common divisor of the integers labeling those edges is equal to 1.

[b]Note: Graph-Definition[/b]. A [b]graph[/b] consists of a set of points, called vertices, together with a set of edges joining certain pairs of distinct vertices. Each pair of vertices $ \,u,v\,$ belongs to at most one edge. The graph $ G$ is connected if for each pair of distinct vertices $ \,x,y\,$ there is some sequence of vertices $ \,x \equal{} v_{0},v_{1},v_{2},\cdots ,v_{m} \equal{} y\,$ such that each pair $ \,v_{i},v_{i \plus{} 1}\;(0\leq i < m)\,$ is joined by an edge of $ \,G$.

[hide]This is the problem that made our graph theory teacher study graph theory :D (mario pickavet, bronze '91)[/hide]

Start at any vertex A and number 1,2,3,... along any path. every vertex you pass will have edges k,k+1 so has gcd=1, until you reach a dead end. (which is either a vertex with degree 1 (no prob) or a vertex on which all edges are labeled yet (no prob). Now if there are unlabeled edges left, start at any vertex of such an edge, and repeat the procedure. Like that you can fill the graph such that all vertices have gcd(edges)=1.

[hide="remark"]gcd(2,3,6)=1...[/hide]

Can I make a small remark? It's kind of late now, maybe I'm missing something obvious.

[hide="Small (Redundant) Remark"]
I tried to solve the problem. But it didn't cross my mind such a simple trick. Maybe I should try solving problems at not so late hours.
So, about what we do after going through the first path:
We suppose that we're not done. Can we choose that "such an edge" just randomly, and then wander through the graph until we reach a dead-end? I don't think we can do it, maybe Peter thought that what should be done was obvious, but I'm trying to find out if I got it right.
Let $x_{1},\ldots,x_{t}$ be the vertices involved in the path. If $t=n$, just put random numbers on the other edges. Else let $v$ be another vertex. We can find a path $x_{1}, \ldots, v$, which we truncate to $x_{p},y_{1},\ldots,y_{s},v$, where $y_{1},\ldots,y_{s}$ are not among the $x_{i}$'s. None of the edges in between have been labeled, so we can safely repeat the procedure on that path, then upon reaching $v$ we "randomize". This increases the number of vertices involved by at least $1$, so, after finitely many steps, $t$ will be equal to $n$, in which case we do as above, i.e. label the remaining edges randomly as they don't matter.
[/hide]

I don't see your point... where would I use non-randomness? Note that we are labeling edges, not vertices.

Either a vertex has only one edge (in which case there's nothing to prove), or either there are at least 2, and by the algorithm we will ever run through it and make that vertex thus have at least 2 coprime edges.

I don't think this is the point that perfect_radio is making, but anyway... I don't see where your proof uses connectivity. When we have gone through the first set of labelling, you say that if there are unlabelled edges (Sorry about the typo that was here...) 'start at any vertex of such an edge'. How do we know that this vertex (the one we start on the second time we make a path) has gcd(edges)=1?
Can we get round this by saying 'start at a vertex that has some already labelled vertices', and hence use the fact that the graph is connected? Or am I just missing something altogether?

[quote="tim1234133"]I don't think this is the point that perfect_radio is making, but anyway... I don't see where your proof uses connectivity. When we have gone through the first set of labelling, you say that if there are unlabelled vertices 'start at any vertex of such an edge'. How do we know that this vertex (the one we start on the second time we make a path) has gcd(edges)=1?
Can we get round this by saying 'start at a vertex that has some already labelled vertices', and hence use the fact that the graph is connected? Or am I just missing something altogether?[/quote]

"Start at a vertex that has already labelled edges" is what he meant.  Such a vertex exists by connectivity.

[here was a wrong idea, thx tim1234133.]

For the rest of your remark, tim1234133, I will say it again: we are labeling edges, NOT vertices. If it was only a typo please reformulate your question in the right way as there are many interpretations possible.

How do we label the (unconnected) graph formed of two unconnected triangles (If we call the vertices A-F we have A connected to B and C with B and C also connected to each other, and D connected to E and F with E and F connected to each other) in the required manner?

Err... never mind, we really need connectedness indeed. :oops:

Yes, that was also my point. It's too bad I couldn't explain it in a more understandable manner  :(

I will induct on the number of edges and keep the number of verticies fixed.

For a base case, we need to prove that all trees can be labeled.  Notice that there is a path between all the verticies of degree >=2.  Thus, we should label these edges in order 1,2,3,4 ..., n.  (Include the two edges connected to a vertex of degree 1).
Call vertex with no edges labeled "unlabeled" since any permutation of the labels will work.
Note that there are only 2 labeled degree one verticies.

Now we progressively add edges to the graph until we get the desired number of edges.

Case 1: A vertex of degree >=2 connected to a vertex of degree >=2.
There is nothing needed to be done.
Case 2: An unlabeled vertex of degree 1 connected to a vertex of degree >=2.
Label the new edge and the unlabeled edge two consectutive numbers.
Case 3: An unlabeled vertex of degree 1 connected to an unlabeled vertex of degree 1.
Label the two unlabeled edges and the new edge three consecutive numbers.
Case 4: A labeled vertex of degree 1 connected to a vertex of degree >=2.
Label this new edge either n+1 if the label is n or keep it unlabeled if the label is 1.
Case 5: A labled vertex of degree 1 connected to a labeled vertex of degree 1.
Label the connection n+1.
Case 6: A labeled vertex of degree 1 connected to an unlabeled vertex of degree 1.
If the labled edge is n, label the two edges n+1 and n+2 (increase other edges if necessary).
If the labled edge is 1, label the two edges consecutive numbers.

Notice that there are at most two instances of connecting a labeled vertex to another vertex, so we are done.

This is easy probelm_hint_>we are doing the graph Euler Garlic by Connect  odd edges of the graph and we use  induaction.

[hide=wrong]Arbitrarily order the vertices. If a vertex has degree $1$ and $k$ vertices have been labelled, label the unique edge $k+1.$ Otherwise, label the edges $k+1,k+2,\cdots, k+\text{deg}(v).$ Since $\gcd(n,n+1)=1\ \forall\ n,$ we are done.

It's not so trivial. Suppose, we have 4 vertices: $a,b,c,d$ s.t. $\{b,c,d\}$ are connected with each other and $a$ is connected only with $b$. Following the above, we label $ab$ as $1$; $bc, bd$ as $2$ and $3$ respectively and $cd$ as 4. Look at the vertex $c$, its edges are labeled as $2,4$ !?

Not 100\% sure if this works, but here it is.

It suffices to have two adjacent labels at every vertex. Consider the path of maximum length. Label its edges consecutively, and then color all the edges red. Now, consider the maximum length path in non-red edges and repeat. Call path $k$ the maximum length path that we chose on the $k$th iteration. We claim that this works.

Note that all vertices that are not on the ends of any paths are automatically taken care of. The only potential issue is that there could be some vertex $v$ that is always only on the end of paths. If $v$ has degree $1$. then this isn't actually a problem. So suppose $v$ has degree at least $2$. It only appears on the end of paths, so there exist $i,j$ such that $v$ is on an end of path $i$ and path $j$ (WLOG $i<j$). But then, gluing together paths $i$ and $j$ results in a path of longer length at stage $i$ (all of path $j$ is still not red since it was not red at stage $j$). One possible issue is that if paths $i$ and $j$ also share their other respective ends, but this can easily solved by deleting one edge in the cycle that's formed when we glued them (still leads to a longer path on stage $i$). Therefore, we must have $\mathrm{deg}v=1$, so every vertex now has two adjacent labels adjacent to it.

EDIT: This is wrong, consider a cycle. I will try to fix soon...

Epic problem. Solution with [b]ewan[/b].

Say all degrees are even. Then, we have an Eulerian circuit, which we just label $1, 2, \ldots, k$ in succession. Then, each vertex is labeled either with 2 consecutive integers or with $1$ and $k$ (or both) so we are done.

Else, consider the graph $G'$ where we add edges to $G$ between pairs of odd-degree vertices. Then, $G'$ has an Eulerian circuit. Removing the added edges from the circuit turns it into a set of walks with union $E(G)$ such that each walk starts and ends at odd-degree vertices and each odd-degree is the endpoint of exactly one walk.

Now, if the walks have lengths $k_1, \ldots, k_m$, then label the first one with $[1, k_1]$ in order, the second one with $[k_1+1, k_1+k_2]$, etc., and the last one with $[k_1+\cdots+k_{m-1}+1, k_1+\cdots+k_m]$. We see that each non-leaf vertex must appear in the interior (i.e. non-endpoint) of some walk and thus have two edges labeled with consecutive numbers, and hence have $\gcd$ of edge labels equal to one. $\blacksquare$

[quote=MathStudent2002]
...Now, if the walks have lengths $k_1, \ldots, k_m$, then label the first one with $[1, k_1]$ in order, the second one with $[k_1+1, k_1+k_2]$, etc., and the last one with $[k_1+\cdots+k_{m-1}+1, k_1+\cdots+k_m]$. We see that each non-leaf vertex must appear in the interior (i.e. non-endpoint) of some walk and thus have two edges labeled with consecutive numbers, and hence have $\gcd$ of edge labels equal to one. $\blacksquare$[/quote]

What if the second walk, for example, begins and ends to one and the same vertex? So you assign $k_1+1$ and $k_1+k_2$ to those first and last edges  and of course the may not be coprime.
I know it can be patched somehow, but then that idea of Euler cycles will lost its originality and it would be dissolved  into something like in post #2.

I think what he/she is doing in the third paragraph is the following.

Let $v_1, v_2, \cdots, v_{2t}$ be the vertices of odd degree of $G$ (there's an even number of them by the Handshake Lemma, say). If $t = 0$, then simply take an Eulerian cycle and label the edges $1, 2, \cdots, k$ in order. Otherwise, we will add the edges $v_1v_2, v_3v_4, \cdots, v_{2t-1} v_{2t}.$ Let this modified graph be $G'.$ Note that this may result in the graph not being simple anymore, but that's fine since the problem still holds. Now, observe that $G'$ has all degree evens, and hence contains an Eulerian cycle. Then, it's clear that the removal of $v_1v_2, v_3v_4, \cdots, v_{2t-1} v_{2t}$ from the cycle would partition the cycle into walks from $v_2$ to $v_3$, from $v_4$ to $v_5$, $\cdots$, from $v_{2t}$ to $v_1.$ This makes it such that none of the walks can be cycles since the $v_i$'s are distinct, and so we finish as above.

Edit: Also, as far as I can tell, the idea in post #2 does not work, since it does not account for cycles (what if the walk ends at the same vertex it began with, and we label the edges $2, 3, 4$?). The solution in post #16, on the other hand, obviates the possibility of cycles using the odd degree trick.

[hide = Solution]
Call a labelling of the edges of $G$ [i]good[/i] if for every vertex with degree at least $2$, the greatest common divisor of its edges is $1$.

If $G$ is a cycle, then labelling the edges $1$, $2$, $\ldots$, $k$ in order works.

[color=#0B5394][b]Claim:[/b][/color] If $G$ is a tree with $k$ edges, then given any set $S$ of $k$ consecutive integers, there exists a good labelling of $G$ using the labels in $S$ (each used exactly once).

[i]Proof.[/i] Use strong induction on $k$, with base case $k = 0$. Take a path between two leaves of $G$, and label it with the $\ell$ largest integers of $S$, in order. Then delete all the edges (but not vertices) of this path, which splits $G$ into some number of trees. Split the remaining elements of $S$ (which are all consecutive) so that each tree is assigned a consecutive set of elements of $S$, and take a good labelling for each of these smaller trees. Then any lone vertex or leaf in a smaller tree was either a leaf in $G$, or was on the path in $G$, so this is a good labelling of $G$ as well. $\blacksquare$

Now if $G$ is a tree or a cycle, it has a good labelling. Otherwise take any cycle $C = (v_1v_2\cdots v_c)$ where $v_1$ has an edge not in $C$. Create a graph $G'$ by adding a new vertex $v_{c + 1}$ and replacing the edge $v_cv_1$ with $v_cv_{c + 1}$. This keeps the graph connected, and any good labelling of $G'$ is also a good labelling of $G$ (since all vertices in $G$ still have degree at least $2$ in $G'$). So by inducting on $e - v + 1$ (with base case $0$) we're done.
[/hide]

[hide=Solution]
We will prove this by strong induction on $k$.

If $k = 1$, then $G$ must be a single edge connecting two vertices, for which we can just place $1$ on that edge and all the conditions are satisfied.

Now, take some $r\geq 2$, and suppose that the problem is statement for $k = 1, 2, \cdots, r-1$. We will show that the statement is true for $k = r$. Consider the graph $G$ for $k = r$.

If there exists a vertex $v_1 \in G$ for which $\deg v_1 = 1$, then do the following: Let $v_2$ be the neighbor of $v_1$. If $\deg v_2 > 2$, then stop. Else, if $\deg v_2 = 2$, then let $v_3$ be the other neighbor of $v_2$. Repeat a similar process for $v_3$ and carry on.

Then, we have a path $v_1 \rightarrow v_2 \rightarrow \cdots \rightarrow v_\ell$ for which $\deg v_1 = 1$, and $\deg v_2 = \deg v_3 = \cdots = \deg v_{\ell-1} = 2$, $v_i$ is connected to $v_{i+1}$, and $\deg v_{\ell} \neq 2$. Then, label the edge $v_1v_2$ with $k$, $v_2v_3$ with $k-1$, $\cdots$, $v_{\ell-1}v_{\ell}$ with $k - \ell + 2$. Then, the labeled on each $v_k$ for $2\leq k\leq \ell-1$ differ by $1$, so their gcd is $1$. Thus, the condition is satisfied for all such $v_k$. Now, take the induced graph determined by removing $v_1, v_2, \cdots, v_{\ell-1}$. This graph would have $k - (\ell - 1) = k-\ell+1$ edges, and is obviously still connected, unless it is empty. If it is empty then $G$ is just the path $v_1\rightarrow v_2\rightarrow \cdots \rightarrow v_{\ell}$, and $\deg v_1 = \deg v_{\ell} = 1$, so the condition does not apply for those vertices, and we have shown that the condition holds for all other vertices, so we're done in that case. Else, we have that $\deg v_{\ell} \geq 3$, and the induced graph is nonempty and connected. Then, by the inductive hypothesis, we can label each edge of the induced graph $1, 2, \cdots, k-\ell+1$ so that the condition is satisfied. Now, consider these labelings in conjunction with our labelings of the other edges. We know that the labelings are fine for $v_1v_2, v_2v_3, \cdots, v_{\ell-2}v_{\ell-1}$ since they do not change based upon the labelings of the induced graph since they are only adjacent to other $v_iv_{i+1}$'s, and we know that the labelings are fine in the induced graph for every edge that is not connected to $v_{\ell}$. For $v_{\ell}$, note that the degree of $v_{\ell}$ in the induced graph is at least $2$, so the gcd of the edges connected to $v_{\ell}$ in the induced graph is $1$, and so the gcd of the labels in $v_{\ell}$ in $G$ is also $1$, since $\gcd(a_1, a_2, \cdots, a_{n+1}) \mid \gcd(a_1, a_2, \cdots, a_n)$, and the former must be $1$ if the latter is $1$. Thus, this edge labeling works if there exists a vertex $v_1 \in G$ for which $\deg v_1 = 1$.

If there does not exist a vertex $v\in G$ for which $\deg v = 1$, then for all $v\in G$, we have that $\deg v\geq 2$. Thus, there exists a cycle of length at least $2+1 = 3$ in $G$ (well-known fact that there exists a cycle of length $\text{mindeg } v+1$ in a graph). Let $c_1\rightarrow c_2\rightarrow \cdots \rightarrow c_{\ell}$ be a cycle in $G$, for some $\ell \geq 3$. If there exists more than one $c_i$ with $\deg c_i \geq 3$, then let $i_1$ be the smallest such index for which $\deg c_{i_1} \geq 3$, and let $i_2$ be the second-smallest such index for which $\deg c_{i_2} \geq 3$. Then, consider the path $c_{i_1} \rightarrow c_{i_1+1}\rightarrow \cdots \rightarrow c_{i_2}$. Note that for any $i_1 < j < i_2$, we have that $\deg c_j = 2$ by minimality of $i_1$, $i_2$. Now, label the edge $c_{i_1}c_{i_1+1}$ with $k$, $c_{i_1+1}c_{i_1+2}$ with $k-1$, $\cdots$, $c_{i_2-1}c_{i_2}$ with $k - (i_2-i_1-1)$, and consider the graph determined by removing all of those edges, and remove the vertices $c_{i_1+1}, c_{i_1+2}, \cdots, c_{i_2-1}$ (these vertices have degree $0$ after the removal of the edges since they had degree $2$ earlier). Then, the resulting graph is still connected, since you can get from one side to the other in the cycle by using the other direction, and it has $k - (i_2-i_1)$ edges. Thus, by the inductive hypothesis we can label the edges $1, 2, \cdots, k - (i_2-i_1)$ so that the condition is satisfied for the reduced graph. Now, consider the entire graph. Then, for each vertex $v\notin \{c_{i_1}, c_{i_1+1}, \cdots, c_{i_2}\}$, we have that the condition is satisfied for $v$ since it was satisfied in the reduced graph and no new edges are added to it. For $c_{i_1+1}, c_{i_1+2}, \cdots, c_{i_2-1}$, thy all have degree two and their labels are adjacent numbers (so they have gcd $1$). For $c_{i_1}$ and $c_{i_2}$, they have degree at least $3$, so they must have had degree at least $2$ in the reduced graph, and so the gcd of the labels in the reduced graph is $1$. Thus, since $\gcd(a_1, a_2, \cdots, a_{n+1}) \mid \gcd(a_1, a_2, \cdots, a_n)$, the former must be one if the latter is $1$, so the gcd is $1$ for $c_{i_1}$ and $c_{i_2}$ in $G$ as well. Thus, this edge labeling works if there are two vertices in the cycle with degree at least $3$.

If all vertices in the cycle have degree $2$, then $G$ must be just that cycle since $G$ is connected. Thus, we can just labeled the edges $1, 2, \cdots, k$ in that order and it clearly works.

Else, there is exactly one edge in the cycle with degree $3$, say $v_1$. Let $v_1, v_2, \cdots, v_r$ be the vertices in the cycle. Then, label $v_1v_2$ with $k$, $v_2v_3$ with $k-1$, $\cdots$, $v_{r-1}v_r$ with $k-(r-2)$, and $v_rv_1$ with $k - (r-1)$. Now, if $\deg v_1 > 3$, then we can just remove all the edges $v_iv_{i+1}$ and $v_rv_1$, and the vertices $v_2, \cdots, v_r$ to get a connected graph on $k-r$ edges, so we can label the edges $1, 2, \cdots, k-r$ so that it works on the reduced graph. Now, it is not hard to see that it works in the original graph -- for all vertices which are not a $v_i$, it works because no new edges are added. For all vertices which are $v_i$ for $i\neq 1$, it works because its two labels on the two edges emanating from it are adjacent, so their gcd is $1$. For $v_1$, note that in the reduced graph, the degree of $v_1$ is $>1$ (so it is $\geq 2$), so the gcds of the labels in the reduced graph is $1$, and so the gcds of the labels in $G$ is $1$ since $\gcd(a_1, a_2, \cdots, a_{n+1}) \mid \gcd(a_1, a_2, \cdots, a_n)$ as discussed before many many times. Else, $\deg v_1 = 3$. Do the same thing as we did in the degree $1$ case in the beginning to get a path $v_1 \rightarrow v_{1,2} \rightarrow v_{1,3} \rightarrow \cdots \rightarrow v_{1, \ell}$ for which $\deg v_1 = 3$, $\deg v_{1,2} = \deg v_{1,3} = \cdots = \deg v_{1, \ell-1} = 2$, and $\deg v_{1,\ell} \neq 2$. Then, label the edge $v_1v_{1,2}$ with $k-r$, the edge $v_{1,2}v_{1,3}$ with $k-(r+1)$, $\cdots$, the edge $v_{1, \ell-1}v_{1, \ell}$ with $k-(r+\ell-2)$. If $\deg v_{1,\ell} = 1$, then since $G$ is connected, then $G$ is just the path + the cycle, and it is not hard to see that the labeling works since the gcd of two adjacent numbers is $1$. Else, if $\deg v_{1, \ell} \geq 3$, then take the induced graph determined by removing $v_1, v_2, \cdots, v_r$, $v_{1, 2}, v_{1, 3}, \cdots, v_{1, \ell-1}$, and all edges that these vertices had, and then the remaining graph is a connected one on $k-(r+\ell - 1)$ edges, so by the inductive hypothesis we can label it and then this labeling works for $G$, since for every vertex not in the $v_i$ or the $v_{1, j}$'s, no new edges are added when we go from the induced graph to $G$, we already showed everything in the cycle works besides $v_1$, $v_1$ works since two of its edges are labeled $k-(r-1)$ and $k-r$, $v_{1,2}, v_{1, 3}, \cdots, v_{1, \ell-1}$ all work since their neighboring edges have labels that are adjacent numbers (and so they have gcd $1$), and $v_{\ell}$ works since it has degree at least $2$ in the inducted graph and we can use the same $\gcd(a_1, a_2, \cdots, a_{n+1}) \mid \gcd(a_1, a_2, \cdots, a_n)$ argument we have done several times before.

Thus, in all cases, our inductive step is complete and so we are done. $\blacksquare$

The idea here is that fixing two edges around a vertex essentially makes it work, so we can work greedily.

Call a vertex [i]valid[/i] if we have labeled two edges through it with relatively prime indices. It suffices to make all vertices valid.

Consider the longest walk in $G$, and label the edges along this walk in the order $1, 2, \dots, k$ consecutively. By maximality of the walk, every vertex in the walk (including the endpoints) is now valid; now, consider the subgraph $G'$ formed by removing all edges and any leaves in the walk, and continue the process until all edges are exhausted.

An inductive solution.
Let $G = (V,E)$. We induct on $|V|$, assuming $|V| \geq 3$. Assume the problem statement is true on all connected graph with the number of vertices less than $|V|$. Since $G$ is connected, let $T$ be the spanning tree of $G$. Let $v$ be a leaf of $T$. Then $G-v$ is still connected, so by the induction hypothesis, we can label the edges of $G-v$ satisfying the problem condition. Let $\deg v = s$. Then if $s \geq 2$, $\gcd(k, k-1, \dots, k-s+1) = 1$ so labeling the remaining edges in arbitrary order will be satisfying the condition. Thus we're done.

We induct with the base case being clearly true.

If $G$ has a cycle and the cycle is connected to some other component, then there is a vertex with $\ge 3$ degree.

Now remove an edge from this vertex thus breaking the cycle but the graph still remaining connected. Then by our induction hypothesis, we can get a number of the edges using $1,\ldots, k-1$.

Now, after removing the edge, we note taht the gcd of the vertex $=1$. So adding back the removed edge, the $\gcd$ would still be $=1$.

Now if the graph has only a cycle (i.e. $G$ has $k$ edges and $k$ vertices as shown below).

[center]
[img]https://i.imgur.com/bslnMTq.png[/img]
[/center]

Then we can just number the edges in an increasing order by $1$.

Otherwise, if $G$ is a tree, then also we can pick a path between the leaves of the tree and label the edges in a increasing order and also the subtrees in a similar increasing by $1$ order. It is easy to see that this works.

[center]
[img]https://i.imgur.com/QCzuiz8.png[/img]
[/center]
:yoda:

[Hide=Sketch of the solution]
Consider a path. Label all the edges on the path as consecutive numbers. The process will be over when there is a vertice with degree one or there is a vertice whose edges are already labeled .Therefore for all the vertices in our path our condition will be satisfied. Now consider a vertex who wasn't in this path. It will have no edges labelled. So we start the same process again. Then after repeating the process a few more times  we will be done.


inequalities
topology
triangle inequality
real analysis
real analysis unsolved

I have to prove that the distance $ p(x,E) \equal{} \inf(x,z)$ between the point $ x$ and the set $ E$ in the metric space $ X$ is uniformly continuous. I don't seem to be able to prove that $ p(x,E)\leq d(x,y) \plus{} p(y,E)$. Any hints?

Basic principle of analysis: when you're asked to prove that $ a\le b,$ back away from trying to get that at first and instead try proving that $ a\le b\plus{}\epsilon$ for any $ \epsilon>0.$

In this case, there exists $ z\in E$ such that $ d(y,z)\le p(y,E)\plus{}\epsilon.$

The triangle inequality applies: $ d(x,z)\le d(x,y)\plus{}d(y,z).$

Now expand that out:
\[ p(x,E)\le d(x,z)\le d(x,y)\plus{}d(y,z)\le d(x,y)\plus{}p(y,E)\plus{}\epsilon\]


geometry
incenter
perpendicular bisector
geometry proposed

Hello!

   Let $ ABC$ be a triangle, $ I$ is the incenter and $ \Omega$ is a circle through the points $ B,C$. Let $ \Gamma\left(J\right)$, $ J\in AI$, be a circle that tangent
to the lines $ AB,AC$ and to the circle $ \Omega$ and $ \left\{S\right\} \equal{} \Gamma\cap\Omega$. The perpendicular to the line $ BC$ through the point $ J$ intersects the
circle $ \Gamma$ at the point $ V$, $ AV < AJ$. Prove that $ I\in SV$.

P.S.
1.There are two circles $ \Gamma'\left(J'\right)$ and $ \Gamma"\left(J"\right)$ that tangent to the lines $ AB,AC$ and to the circle $ \Omega$.
2.The solution of this problem give us a easy way to construct the circles $ \Gamma'$, $ \Gamma"$ that tangent to the lines $ AB,AC$ and
to the circle $ \Omega$ ($ \Omega$ is a circle through the points $ B,C$).

Best regards, Petrisor Neagoe :)

Hello!

Please ignore my first message! :blush:
The name of this topic should be: Four circles that tangent to two lines and to a circle.

Proposed problem:

Let $ ABC$ be a triangle, $ I$ is the incenter, $ I_{a}$ is the $ A$-excenter and $ \Omega$ is a circle through the points $ B,C$.
Let $ \Gamma\left(J\right)$, $ J\in AI$, be a circle that tangent to the lines $ AB,AC$ and to the circle $ \Omega$
and $ \left\{S\right\} \equal{} \Gamma\cap\Omega$. The perpendicular to the line $ BC$ through the point $ J$ intersects the
circle $ \Gamma$ at the points $ V,W$ $ \left(AV < AJ,AW > AJ\right)$. Prove that $ I\in SV$ or $ I_{a}\in SW$.

P.S.
(a).There are four circles $ \Gamma\left(J\right)$, $ J\in AI$, that tangent to the lines $ AB,AC$ and to the circle $ \Omega$.
     (b).Let $ ABC$ be a triangle, $ I$ is the incenter, $ I_{a}$ is the $ A$-excenter and $ \Omega\left(O\right)$ is a circle through the points $ B,C$.
We can construct a circle $ \Gamma\left(J\right)$, $ J\in AI$, that tangent to the lines $ AB,AC$ and to the circle $ \Omega$ as follows:
     1. Let $ U$ be one of the intersection points of the perpendicular bisector of $ \left[BC\right]$ with the circle $ \Omega$
     2. $ \left\{U,S\right\} \equal{} UI\cap\Omega$ or $ \left\{U,S\right\} \equal{} UI_{a}\cap\Omega$
     3. $ \left\{J\right\} \equal{} OS\cap AI$
     4. Construct the circle $ \Gamma\left(J;JS\right)$.

My solution:

Problem1.

Let $ \Gamma$ be a circle with diameter $ \left[VW\right]$ and $ BC$ is a line such that $ BC\bot VW$. If $ S\in\Gamma$, $ \left\{S,X\right\} \equal{} SB\cap\Gamma$ and
$ \left\{S,Y\right\} \equal{} SC\cap\Gamma$ then prove that $ XY\parallel{}BC$ if and only if $ SV$ is the interior/exterior bisector of the angle $ \angle XSY$.

   Solution:
(1) $ XY\parallel{}BC$ and $ BC\bot VW$ $ \Rightarrow XY\bot VW$ $ \Rightarrow SV$ is the interior/exterior bisector of the angle $ \angle XSY$.

(2) $ SV$ is the interior/exterior bisector of the angle $ \angle XSY$ $ \Rightarrow XY\bot VW$
$ XY\bot VW$ and $ BC\bot VW$ $ \Rightarrow XY\parallel{}BC$.

Problem 2:

Let $ ABC$ be a triangle $ (AB<AC)$, $ I$ is the incenter, $ J\in AI$ and $ M\in AB$, $ N\in AC$ such that $ JM\bot AB$ , $ JN\bot AC$.
The perpendicular to the line $ BC$ through the point $ J$ intersects the circle $ \Gamma(J,JM)$ at the point $ V$ $ (AV<AJ)$ and
$ VI\cap BC=\{T\}$. Prove that $ \frac{TB}{TC}=\frac{BM}{CN}$.

     Solution:

$ BC=a$, $ CA=b$, $ AB=c$, $ p=\frac{a+b+c}{2}$
$ \{P\}=AI\cap BC$, $ PB=\frac{ac}{b+c}=\frac{ac}{2p-a}$, $ PC=\frac{ab}{b+c}=\frac{ab}{2p-a}$,
$ \frac{IP}{IA}=\frac{PB}{AB}=\frac{PC}{AC}=\frac{a}{b+c}=\frac{a}{2p-a}$

$ \{D\}=(I)\cap BC$, $ \{D,F\}=DI\cap(I)$, $ \{E\}=AV\cap BC$.
$ r$ is the radius of the circle $ (I)$ and $ \rho$ is the radius of the circle $ \Gamma(J)$.
(1) $ \frac{AI}{AJ}=\frac{r}{\rho}=\frac{IF}{JV}$ and $ IF\parallel{}JV\Rightarrow$ the points $ A,F,V$ are collinear and $ \frac{AF}{AV}=\frac{r}{\rho}$.
(2) $ \Delta DEF$, $ T-V-I$ transversal $ \Rightarrow\frac{TD}{TE}\cdot\frac{VE}{VF}\cdot\frac{IF}{ID}=1\Rightarrow\frac{TE}{VE}=\frac{TD}{VF}$.
(3) $ tan(\frac{\angle A}{2})=\frac{\rho}{AM}\Rightarrow AM=\frac{\rho}{tan(\frac{\angle A}{2})}\Rightarrow AM=\frac{\rho(p-a)}{r}=AN$
(4) $ \Delta APE$, $ I-T-V$ transversal $ \Rightarrow\frac{IA}{IP}\cdot\frac{TP}{TE}\cdot\frac{VE}{VA}=1\Rightarrow TP=\frac{IP}{IA}\cdot\frac{TE}{VE}\cdot VA\Rightarrow TP=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot TD$

$ A',B',C'\in AI$ such that $ A'$ lies on the perpendicular bisector of $ BC$, $ B'B\bot AB$ and $ C'C\bot AC$.
Let's consider the case $ J\in(IB')$.
$ \frac{VA}{VF}=\frac{\rho}{\rho-r}$, $ TD=DB-TB$ and $ TD=TC-DC$, $ TP=PB-TB$ and $ TP=TC-PC$.
$ TP=PB-TB$ and from (4) $ TP=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot TD=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot(DB-TB)\Rightarrow$
$ \Rightarrow PB-TB=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot(DB-TB)\Rightarrow$
$ \Rightarrow TB\cdot(\frac{IP}{IA}\cdot\frac{VA}{VF}-1)=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot DB-PB$   (5)
$ TP=TC-PC$ and from (4) $ TP=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot TD=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot(TC-DC)\Rightarrow$
$ \Rightarrow TC-PC=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot(TC-DC)\Rightarrow$
$ \Rightarrow TC\cdot(\frac{IP}{IA}\cdot\frac{VA}{VF}-1)=\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot DC-PC$   (6)
(5), (6) $ \Rightarrow\frac{TB}{TC}=\frac{\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot DB-PB}{\frac{IP}{IA}\cdot\frac{VA}{VF}\cdot DC-PC}=\frac{\frac{a\rho(p-b)}{(2p-a)(\rho-r)}-\frac{ac}{2p-a}}{\frac{a\rho(p-c)}{(2p-a)(\rho-r)}-\frac{ab}{2p-a}}=$
$ =\frac{a[\rho(p-b)-c(\rho-r)]}{a[\rho(p-c)-b(\rho-r)]}=\frac{cr-\rho(p-a)}{br-\rho(p-a)}\Rightarrow\frac{TB}{TC}=\frac{cr-\rho(p-a)}{br-\rho(p-a)}$   (*)

$ BM=AB-AM=\frac{cr-\rho(p-a)}{r}$ and $ CN=AC-AN=\frac{br-\rho(p-a)}{r}$
$ \Rightarrow\frac{BM}{CN}=\frac{cr-\rho(p-a)}{br-\rho(p-a)}$   (**)

(*), (**) $ \Rightarrow\frac{TB}{TC}=\frac{BM}{CN}$.

The proof for the cases $ J\in AI-[AI$, $ J\in(AI)$, $ J\in(B'A')$, $ J\in(A'C')$ and $ J\in AI-[C'A'$ is similar.

Problem 3:

Let $ ABC$ be a triangle $ (AB<AC)$, $ I_{a}$ is the $ A$-excenter, $ J\in AI_{a}$ and $ M\in AB$, $ N\in AC$ such that $ JM\bot AB$ , $ JN\bot AC$.
The perpendicular to the line $ BC$ through the point $ J$ intersects the circle $ \Gamma(J,JM)$ at the point $ W$ $ (AW>AJ)$ and
$ WI_{a}\cap BC=\{T\}$. Prove that $ \frac{TB}{TC}=\frac{BM}{CN}$.

The solution is similar to the solution of the problem 2.

The problems 2 and 3 $ \Rightarrow$ problem 4:

Let $ ABC$ be a triangle $ (AB<AC)$, $ I$ is the incenter, $ I_{a}$ is the $ A$-excenter, $ J\in AI$ and $ M\in AB$, $ N\in AC$ such that
$ JM\bot AB$ , $ JN\bot AC$. The perpendicular to the line $ BC$ through the point $ J$ intersects the circle $ \Gamma(J,JM)$ at the points
$ V,W$ $ (AV<AJ,AW>AJ)$. Prove that the lines $ VI$, $ WI_{a}$ and $ BC$ are concurrent at a point $ T$ and $ \frac{TB}{TC}=\frac{BM}{CN}$.

Problem 5:

Let $ ABC$ be a triangle $ \left(AB<AC\right)$, $ I$ is the incenter, $ I_{a}$ is the $ A$-excenter and $ J\in AI$. $ \Gamma\left(J\right)$ is a circle such that
$ AB$ and $ AC$ are tangents to the circle $ \Gamma$. The perpendicular to the line $ BC$ through the point $ J$ intersects the circle $ \Gamma$
at the points $ V,W$ $ \left(AV<AJ,AW>AJ\right)$. If $ S\in\Gamma$ such that $ SV$ is the interior/exterior bisector of the angle
$ \angle BSC$ then prove that $ I\in SV$ or $ I_{a}\in SW$.

   Solution:
$ \left\{M\right\}\equal{}AB\cap\Gamma$, $ \left\{N\right\}\equal{}AC\cap\Gamma$, $ \left\{S,X\right\}\equal{}SB\cap\Gamma$ and $ \left\{S,Y\right\}\equal{}SC\cap\Gamma$

problem 4 $ \Rightarrow VI\cap WI_{a}\cap BC\equal{}\left\{T\right\}$ and $ \frac{TB}{TC}\equal{}\frac{BM}{CN}$   (1)

$ \left\{T'\right\}\equal{}SV\cap BC$ or $ \left\{T'\right\}\equal{}SW\cap BC$ such that $ T',T\in\left(BC\right)$ or $ T',T\in BC\minus{}\left[BC\right]$.
So $ ST'$ is the interior/exterior bisector of the angle $ \angle BSC$ $ \Rightarrow\frac{T'B}{T'C}\equal{}\frac{BS}{CS}$   (2)

Problem1 $ \Rightarrow XY\parallel{}BC\Rightarrow\frac{BX}{BS}\equal{}\frac{CY}{CS}\Rightarrow\frac{BX}{CY}\equal{}\frac{BS}{CS}$   (3)
$ BM^{2}\equal{}BX\cdot BS$ and $ CN^{2}\equal{}CY\cdot CS$ $ \Rightarrow\frac{BM^{2}}{CN^{2}}\equal{}\frac{BX}{CY}\cdot\frac{BS}{CS}$   (4)
(3),(4)$ \Rightarrow \left(\frac{BM}{CN}\right)^{2}\equal{}\left(\frac{BS}{CS}\right)^{2}\Rightarrow\frac{BM}{CN}\equal{}\frac{BS}{CS}$   (5)

(2),(5)$ \Rightarrow\frac{T'B}{T'C}\equal{}\frac{BM}{CN}$   (6)
(1),(6)$ \Rightarrow\frac{T'B}{T'C}\equal{}\frac{TB}{TC}\Rightarrow T'\equal{}T$ $ \Rightarrow I\in SV$ or $ I_{a}\in SW$.

The solution of the proposed problem:

$ \left\{S,U'\right\} \equal{} SV\cap\Omega$, $ \left\{S,U"\right\} \equal{} SW\cap\Omega$.It's easy to prove that $ VW\parallel{}U'U"$.
$ VW\parallel{}U'U"$ and $ VW\bot BC\Rightarrow U'U"\bot BC$ $ \Rightarrow SU'$ is the interior/exterior bisector
of the angle $ \angle BSC$ $ \Rightarrow SV$ is the interior/exterior bisector of the angle $ \angle BSC$.
Problem 5 $ \Rightarrow I\in SV$ or $ I_{a}\in SW$.

Best regards, Petrisor Neagoe :)

Dear friends!
Finally, I found an easy solution for the proposed problem!

Let's consider the case $ J\in\left(AI\right)$ and let's prove that $ I\in SV$.
We have $ \Omega\left(O;R\right)$, $ \Gamma\left(J;r\right)$ and $ U$ is the intersection point of the perpendicular bisector of $ \left[BC\right]$ with $ \Omega$ ($ AU>AO$).
From [url=http://pagesperso-orange.fr/jl.ayme/Docs/Un%20remarquable%20resultat%20de%20Vladimir%20Protassov.pdf]here[/url] $ \Rightarrow SI$ is the interior bisector of the angle $ \angle BSC\Rightarrow I\in SU$   (1)
$ JV\parallel{}OU$, $ \frac{JV}{OU}\equal{}\frac{JS}{OS}\equal{}\frac{r}{R}$ and the points $ J,S,O$ are collinear $ \Rightarrow$ the points $ V,S,U$ are collinear.   (2)
(1),(2) $ \Rightarrow I\in SV$.

Best regards, Petrisor Neagoe :)


here is a simple problem ,hope you enjoy solving it.Solve it in 30 second or less or more :)
let $a,b,c \ge 0$

If $a+b+c\ge abc$ $\longrightarrow$ $a^2+b^2+ c^2\ge abc$

You can prove that $a^2 + b^2 + c^2 \geq \sqrt{3}abc$. Indeed, it is easy to see that \[a^2 + b^2 + c^2 \geq \sqrt{3abc(a + b + c)}\] (square this and expand, it follows from $a^4 + b^4 + c^4 \geq a^2b^2 + b^2c^2 + c^2a^2$ and $a^2b^2 + b^2c^2 + c^2a^2 \geq a^2bc + b^2ca + c^2ab$). Hence  \[a^2 + b^2 + c^2 \geq \sqrt{3abc(a + b + c)} \geq \sqrt{3abc \cdot abc} = \sqrt{3}abc\] and we're done.

([color=red]Edit: Oops, sorry, I didn't see this was posted in the pre-olympiad section[/color])

This inequality has been posted before. I gave a solution identical with Arne's. ;) Now,i'm searching the topic.Hope I'll find it. ;)

And Arne, we don't need to square and expand the inequality $ab+bc+ca\geq\sqrt{3abc(a+b+c)}$ since it follows immediately from $(x+y+z)^2\geq 3(xy+yz+zx) $ taking $x=ab,y=bc,z=ca$. ;)

Here is another solution for the inequality Arne was talking about. Let's assume by contradiction that, $a^2+b^2+c^2< abc\sqrt{3}$. By AM-GM we obtain $a^2+b^2+c^2\geq 3\cdot(a^2 b^2 c^2)^{\frac{1}{3}}$, hence $ abc> 3\sqrt{3}$.On the other hand, from Cauchy-Schwarz inequality we have $ abc\sqrt{3}> a^2 +b^2 +c^2\geq\frac{(a+b+c)^2}{3}\geq\frac{a^2 b^2 c^2}{3}$ and from here we deduce that $abc< 3\sqrt{3}$ so we get the contradiction desired. ;)

Nice way of thinking ,Arne.
Here is my solution, Smart and Primary solution.
Cezar ,here is $Pre$-olympiad ,not a place for showing that your are good at inequalities.
[color=blue]Solution.[/color]
Assume that problem is wrong.
$a^2+b^2+c^2 \le abc \longrightarrow \begin{cases} a^2 \le abc \longrightarrow a \le bc \\[.1in] b^2 \le abc \longrightarrow b \le ac \\[.1in] c^2 \le abc \longrightarrow c \le ab \end{cases}$ $\longrightarrow a+b+c \le ab+bc+ca \le a^2+b^2+c^2 \leftrightarrow \text{ABSURD}$

you see although it was nice but i think that it's level was less than pre-olympiad !!  :P

No, I think this problem is in the right forum. After all, it was given an a national olympiad, although it might seem easy for you, it might not be that easy for other people.

Unbelieveble, I found out that this problem (Arne's inequality) has been given at Balkan olympiad in 2001. :o

Yes, indeed. You see, often people say "this is trivial" etc, for example, to impress others, but it happens often that people start posting such "easy" problems in Pre-Olympiad, while the problem is actually way to hard for that forum. In my opinion, the Pre-Olympiad contains Olympiad problem which are not too hard, and the Olympiad problems contains (in general) very hard Olympiad problems - but both forums are basically about olympiad problems. :)


there is a random selection of 25 students to each present a project. five students are selected per day monday-friday. what is the probability that the shortest student is choosen on thursday?

My reasoning for this problem was that there is a 5/25 or 1/5 chance this student is choosen for that day. And there is a 1/5 chance that it is a Thursday. Therefore the answer would be 1/5 * 1/5 = 1/25... Unfortunately that is not the correct answer, any help?

I had this problem against you, and the answer is 1/5. The probability that the shortest student presents on Thursday is 1/5, because there are 5 Thursdays out of 25 days, so 5/25, or 1/5. Which Thursday he presents on does not matter.


Prove that for any positive integers $ m$ and $ n$ the equality \[ (5\plus{}3\sqrt2)^m\equal{}(3\plus{}5\sqrt2)^n\] is impossible.

[quote="jelly119"]Prove that for any positive integers $ m$ and $ n$ the equality
\[ (5 \plus{} 3\sqrt2)^m \equal{} (3 \plus{} 5\sqrt2)^n
\]
is impossible.[/quote]

[hide=" Hints"]
$ 1)$ show that $ a\plus{}b\sqrt{2}\equal{}c\plus{}d\sqrt{2}$ where $ a,b,c,d\in\mathbb{Z}$ implies $ a\equal{}c$ and $ b\equal{}d$

$ 2)$ Consider $ (m,n)$ minimum and then say m and n are not both even.

$ 3)$ consider the integer part of the equality modulus 3 and 5 and show there is a contradiction is m and n are not both even
[/hide]

[hide="The relevant lemma"] Let $ N(a \plus{} b \sqrt{2}) \equal{} a^2 \minus{} 2b^2$.  Then $ N(xy) \equal{} N(x) N(y)$.  In particular, $ N(z^n) \equal{} N(z)^n$. [/hide]
[hide="Solution"] Hence $ (5^2 \minus{} 2 \cdot 3^2)^m \equal{} 7^m \equal{} (3^2 \minus{} 2 \cdot 5^2)^n \equal{} (\minus{}41)^n$, absurd. [/hide]


limit
trigonometry
calculus
derivative
function
calculus computations

[hide="Number 2"]$ \lim_{x \to \infty} x \left(\frac{\pi}{4} \minus{} \arctan \frac{x}{x\plus{}1}\right) \equal{} \lim_{x \to \infty} x \cdot \arctan \frac{1}{2x\plus{}1}$

$ \equal{} \lim_{x \to \infty} x \cdot \left[\frac{1}{2x\plus{}1} \plus{} O(x^{\minus{}3})\right]$

$ \equal{} \lim_{x \to \infty} \left[\frac{1}{2 \plus{} \frac{1}{x}} \plus{} O(x^{\minus{}2})\right]$

$ \equal{} \boxed{\frac12}$[/hide]

[hide="Number 3"]$ L \equal{} \lim_{x \to 0} \frac{1 \minus{} \cos(6x) \minus{} \cos x \sin(2x) \sin(3x) \minus{} \sin x \cos(2x) \sin(3x) \minus{} \sin x \sin(2x) \cos(3x)}{x^2}$

$ \equal{} \lim_{x \to 0} \frac{1 \minus{} \cos(6x)}{x^2} \minus{} \lim_{x \to 0} \frac{\sin(2x) \sin(3x)}{x^2} \minus{} \lim_{x \to 0} \frac{\sin x \sin(3x)}{x^2} \minus{} \lim_{x \to 0} \frac{\sin x \sin(2x)}{x^2}$

$ \equal{} \lim_{x \to 0} \frac{[1 \minus{} \cos(6x)][1 \plus{} \cos(6x)]}{x^2 [1 \plus{} \cos(6x)]} \minus{} 6 \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot \frac{\sin(3x)}{3x} \minus{} 3 \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{\sin(3x)}{3x} \minus{}$

$ \minus{} 2 \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{\sin(2x)}{2x}$

$ \equal{} 18 \lim_{x \to 0} \left(\frac{\sin(6x)}{6x}\right)^2 \minus{} 6 \minus{} 3 \minus{} 2$

$ \equal{} \boxed{7}$[/hide]

[quote="AstroPhys"]Should be in Calculus forum.[/quote]
Agreed. (Specifically, the "Calculus Computations and Tutorials" forum.) A moderator should move it.

1 and 3 ones can be solved by power series expansions (the third can be solved simpler than whatever carcul is doing).

2nd one is just the derivative of -$ \tan ^{ \minus{} 1} \frac{1}{{1 \plus{} x}}$ at 0

[quote="L_Euler"]Find the limits:

$ 3.{\lim }\limits_{x \to 0} \frac {{1 - \cos xc{\rm{os}}2xc{\rm{os}}3x}}{{x^2 }}$[/quote]
The proof e\which be used Lopitan order is simple


hi
find all sets of positive integers (a,b,c) such that
(7^a)+1=(3^b) + (5^c)

$ a\equal{}b\equal{}c\equal{}1$ is the only solution in positive integers.
To see that there are no other solutions it is enough to consider the given equation modulo 3276.


I think we already have an unofficial one.
An official one might be out of question though. (Reference Mr. Rusczyk's comment [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=31&start=0&postdays=0&postorder=asc&highlight=chat+room]here[/url])

[quote="cats..."]how do we get into the unofficial one?  is it on AIM?[/quote]
No, it's a website. A link to it is in the forum somewhere, but it's like searching for a hay in a needlestack :)

oh, alright  thanks.

By the way, does your school do the MAML Olympiad?


modular arithmetic
number theory
greatest common divisor

Prove that for a prime $p$, $p|(a^2 + b^2)$ and $p \equiv 3 \pmod 4 \implies p|a$ and $p|b$.

[hide="something that might help"]Well, you know that a perfect square is either $0$ or $1$ mod 4, so $a^2 + b^2 \equiv 0, 1, 2 \mod {4}$[/hide]

[hide="Hint"]$a^2 \equiv -b^2 \pmod p$
When is -1 a square mod p?[/hide]

I read this problem in Burton's book.
First of all we assume that $ a \not \equiv 0 \bmod{p} $ The other cases are obviously trivial.
Now the above implies $ \exists \ c \  s.t. \ ac \equiv 1 \bmod{p} $
Now he says that use this to contadict the theorem that the congruence $ x^2 + 1 \equiv 0 \bmod{p} $ has a solution if and onli if $ p \equiv 1 \bmod{4} $.This step is what I am not able to deduce.

What to do now???

If $GCD(a, p) = 1, a^{\frac {p-1}{2}} \equiv 1 \pmod p \Leftrightarrow a$ is a quadradic residue.

Proof
Since $(a,p)=1$ there exists some $k$ such that $g^k \equiv a \pmod p$ where $g$ is a primitive root. Then $g^{\frac {k(p-1)}{2}} \equiv 1 \pmod p \Leftrightarrow (p-1)|\frac {k(p-1)}{2}$

$\frac {k(p-1)}{2} = q(p-1) \Rightarrow k = 2q$, $k$ is even and thus $a$ is a quadradic residue.

If $a$ is a quadradic residue there exists some unit $u$ such that $u^2 \equiv a \pmod p$. $u^{p-1} \equiv 1$ by Fermat's Little, $\Rightarrow u^{\frac {2(p-1)}{2}} \Rightarrow a^{\frac {p-1}{2}} \equiv 1 \pmod p$

Now if $-1$ is a quadradic residue $(-1)^{\frac {p-1}{2}} \equiv 1 \pmod p$.

Thus $\frac {p-1}{2}$ must be even and $p \equiv 1 \pmod 4$
If $p \equiv 1 \pmod 4$ then its a simple matter of reversing the steps to show that -1 is a quadradic residue.

In our problem, since $p \equiv 3 \pmod 4$ -1 isn't a square. What happens then?

[quote="tarquin"]If $GCD(a, p) = 1, a^{\frac {p-1}{2}} \equiv 1 \pmod p \Leftrightarrow a$ is a quadradic residue.

Proof
Since $(a,p)=1$ there exists some $k$ such that $g^k \equiv a \pmod p$ where $g$ is a primitive root... T[/quote]

Well if you  do not need the proof of existance of primitive root then you can easily:
 Assuming $(a,p)=1$ we have here:
$g^{2k_1}+g^{2k_2} \equiv  0 \pmod p$
or $g^{2k_1}(1+g^{2(k_2-k_1)}) \equiv 0 \pmod p$  and putting  $k=k_2-k_1$
or $1+g^{2k} \equiv 0 \pmod p$
but that's a contradiction.
 if p is of the form $4n - 1$ becuase  then  $g^{4n-2}  - 1 \equiv (g^{2n-1} -1)(g^{2n-1}+1) \equiv 0$
since $g^{2n-1} $ is not 1 is must be $-1$ or
only  $g^{odd numeber}$ can be   $-1$ (If p is of the form 4n-1)

Outline of  Another fun proof  (without advance number theory)

[hide]Let $a = pn + a_1$  and $b = pm+b_1$
(where $(n,m)$ are integers and $-(p-1)/2  \le  (a_1,b_1) \le  (p-1)/2 $

$a_1^2 + b_1^2 = kp $  with (k<p)
if k=0 both a_1 and a_2 must be zero (that is   (a,b) are multiples of p ..
k=1 case is not possible becuase  a_1^2   (or b_1^2) can only be (0 or 1) mod 4 so sum can not be 3 mod 4)
Now if k>1 then ...

Now  if  $a_1 = k n_1 + a_2 $ and $b_1 = k n_2 + b_2$
where $-(k-1)/2  \le (a_2,b_2) \le (k-1)/2 $
we get $0 \equiv a_1^2 + a_2^2 \equiv a_2^2+b_2^2 \bmod k$
or $a_2^2+b_2^2 = k q $    (with q strictly less than p)
or $(a_1a_2+b_1b_2)^2 + (a_1b_2 - a_2b_1)^2 = (a_1^2+b_1^2)(a_2^2+b_2^2) = k^2pq$
but $(a_1a_2+b_1b_2) \equiv 0 \bmod k $ can be easily checked  and
so is $(a_1b_2-b_1a_2) \equiv 0 \bmod k $
so let  $(a_1a_2+b_1b_2) = k a_3 $ and $(a_1b_2-b_1a_2) = kb_3$ we get

$a_3^2  + b_3^2 = q p $      with q strictly less than k  as long as k>1.

So  if k > 1 we can always reduce it to a smaller number..... (aka Fermat's infinte decent ...)
QED.

[/hide]

Can somebody post a solution which uses the result which i have stated ???

[quote]Now he says that use this to contadict the theorem that the congruence $x^2 + 1 \equiv 0 \bmod{p} $ has a solution if and onli if $p \equiv 1 \bmod{4}$  .This step is what I am not able to deduce. [/quote]

I  am sure you know that  if p is an odd  prime (and all mod p)   $x^2 \equiv k $ has a solution iff  $k ^{\frac {p-1} 2} \equiv  1$

Now if p is of  the form $4n+3$  than  ${\frac {p-1} 2}$ is odd and obviously$ (-1)^{(odd number)} $ is $-1$ so $x^2 \equiv  -1 $ has no solutions,

My proof goes something like this

$a^2 \equiv -b^2 \pmod p$

Suppose $a$ and $b$ are units mod $p$.  Since -1 is not a quadradic residue mod $p$ there $-b^2$ is not a quadradic residue mod $p$, a contradiction. Thus $a$ and $b$ can't be units, $p|a, p|b$.

$a$ is a unit mod $n$ iff $GCD(a,n)=1$. Thus if $n$ is a prime and $a$ is not a unit, $n|a$ since the GCD must be either 1 or n.

[quote="Rushil"]No waht I meant was that Burton says that since $ ac \equiv 1 \bmod{p} $ as I have shown in my post, use this (above) to contradict the thoerm. I know the proof of the theorm. I am just asking its connection to my result??[/quote]

May be I don't understand exactly what you  mean ...: The proof you quoted goes somethinng like:

All mods p and assuming both a and b are not  0 (mod p) etc... (with $a^2+b^2 \equiv 0 $ )

There exist $c$ such that $ac \equiv 1$
Now, of course,  we have $ c^2(a^2+b^2) \equiv 0 $
or $1 +(bc)^2 \equiv 0 $
But that's not  possible if   $p \equiv 3 \bmod 4$
Hope that helps

(I like this proof, because  all you need after this , is just to show the order of bc is 4 (that is $(bc)^4 \equiv  1$ and no power of bc , smaller than 4 is 1 IOW   if $(bc)^n  \equiv  1 $ then n is a multiple  of  $4$  but we do know $(bc)^{p-1} \equiv 1 $ so that is possible only if $(p-1)$ i is a multiple of  $4$ etc..)

Ya this is out. Doesn't the solution become simplest with my method!!!


Given the fraction:

${{{\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}}}}$

express the fraction as a simple fraction in lowest terms.

[hide]that equals something like 8/19, right?[/hide]


By any chance, does anyone know where I could find copies of past NYCIML contests as well as their solutions? If there are no copies online, does anyone know where I could order them from?

Thanks!

There is a book of the 1975-1984 contests, but it seems to be out of print now:
[url]http://www.amazon.com/exec/obidos/ASIN/0866513078[/url].
I don't know how to get a copy.

Yeah, that book is very hard to find since its out of print...in fact, I am not even sure there are any copies left to sell...

After searching through Google for a while, I found out that Daniel Jaye, a prinicpal or a teacher (one or the other) at Stuyvesant High School is the Executive Director for NYCIML, so I think you should contact him for *possible* ordering information.  Just explain who you are, and that you are looking for some past problems, and such. His e-mail address from the school's website is jayed@stuy.edu

Hope This Helps!

He's actually the assistant principal for mathematics.  (In other words, the head of the math department there.)  I strongly doubt that he will have the information you are looking for, but if you want to try, go ahead.  You might also contact a Mr. David Linker, who is also involved in running the contest.  As I say, though, I doubt you will be able to find a copy.

I e-mailed Mr. Jaye, and he said he would forward the request to the head coach, and they will contact me with more information, so I am still waiting on a reply. I still don't know why it would be so hard to find a copy? By the way JBL, do you know how I can contact the person you mentioned, as I can't seem to find his e-mail address :?

Because they stopped printing them a long time ago and there aren't just warehouses full of them sitting around.

perhaps this message is way out of date.. but if you still want this.. please contact me.. thanks.

Any one has idea where I could find copies of past NYCIML contests as well as their solutions? If there are no copies online, does anyone know where I could order them from?

They actually have a  copy of the nyciml contest problem book at the Library of Congress, but if you don't live near D.C. like me, you might be able to request it through an interlibrary loan at your local library. I don't know how that procedure works, but good luck.


vector
advanced fields
advanced fields unsolved

I wonder about this statement:

Given a vector space V. Then V* is separable implies V is separable.

Is is true?

http://www.mathlinks.ro/viewtopic.php?t=172571

topic has even (almost) the same name as this one  :wink:


How many units long is a segment whose endpoints are (2,3) and (7,15)?

Using distance formula, we see that the distance is $ \sqrt{5^2\plus{}12^2}\equal{}\boxed{13}$.


How many pairs of parallel faces does a right octagonal prism have?

$ 4$ pairs of sides and $ 1$ pair of bases:
Answer: $ \boxed{5}$

The sides can be paired up, giving us 4, and the bases make one more, so our answer is $ \boxed{5}$.

EDIT: Beaten to it.


How can i make two column proofs on latex?

Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=multicol]multicol package[/url] which allows you to mix single and multiple columns.

I don't think that's compatible with my TexShop (for mac)

I don't follow you. LaTeX is OS neutral so packages don't care what operating system you have. Also multicol is part of the tools bundle which is a [url=http://tug.ctan.org/pkg/required]required[/url] package of all LaTeX distros so I'd be surprised if you don't have it. It should be installed at .../tex/latex/tools and you can either check that directory or put \usepackage{multicol} in your tex document and see what the log says.

If you're trying to write a two-column proof (like the ones you do in a generic high school geometry class), wouldn't using tables be easier?

@AIME: When looking at package pages like the one stevem gave, you should consider the "Distributions with this package" section. The Mac OS and Linux use TeXLive, which is listed on the page. TeXShop is merely a frontend.


modular arithmetic
number theory proposed
number theory

i said integers, not nonnegative integers. :P but in some sense your solutions are all there are. this was also my solution. :D


Find the pair of integer $(a,b)$ such that $a^3-b^3=217$.

that seemed too easy for a pre-olympiad type problem.

You forgot $a-b=1 \Rightarrow a=9, b=8$. ;)

217=1x217=7x31
and i think both case is needed in order to get all the answer,
by setting up 2 trails,it turns out that (1,-6)(6,-1)(9,8)(-8,-9)is the answer
and this one is much below the olympic level....

[quote="TheJerk"]It is too easy for the pre-olympiad section, however judging by the problem source... you would probably need some justification for an answer.

[hide]$217=7 \cdot 31=a^3-b^3=(a-b)(a^2+ab+b^3)$

So we have two cases:

(1) $a-b=7$, and $a^2+ab+b^2=31$
(2) $a-b=31$, and $a^2+ab+b^2=7$

[u]Case 1[/u]: $a-b=7 \Rightarrow b=a-7 \Rightarrow a^2+a(a-7)+(a-7)^3=31 \Rightarrow 3a^2-21a+18=0 \Rightarrow (a-6)(a-1)=0 \Rightarrow (6,-1),(1,-6)$

[u]Case 2[/u]: $a-b=31 \Rightarrow b=a-31 \Rightarrow a^2+a(a-31)+(a-31)^2=7 \Rightarrow a^2-31a+318=0 \Rightarrow 31^2-4(318)<0 \Rightarrow (a,b) \not\epsilon \ \mathbb{Z}^2$

So we have two solutions: $(6,-1)$, and $(1,-6).  \ \ \clubsuit$[/hide][/quote]

$\epsilon$ and $\in$ are different commands.


polynomial
algebra
LaTeX
calculus
integration
number theory theorems
number theory

Could somebody give me proof of this theorem?  :P

it appears to be like this:

Theorem:Let A be a circle in a forum (x,y).Then it is a yellow one.

It says this:
 The polynomal P(x)=an*x^n+....+a1*x+a0 is given.Also there exists prime number p such that p is divisor of a0,a1,...,an-1,p is not divisor of an,and p^2 is not divisor of a0.Prove that P(x) cant be zero for any real number x.  :D
 I dont know Latex well but i hope you understand this.

The beginning of your statment sounds very like Eisensteins criterion, but the ending is curious (and wrong):

Eisenstein's Criterion: Let $p(x) = a_nx^n+ ... a_1 x + a_0$ be an integral polynomial and $p$ a prime with $p \nmid a_n , p| a_0,a_1,a_2 , ... , a_{n-1} , p^2\nmid a_0$. Then $p$ is irreducible (not a product of nontrivial rational polynomials).

So, he's asking to prove Eisenstein's criterion

So : http://planetmath.org/encyclopedia/EisensteinCriterion.html :)


Given $ a,b,c >0$. Prove that:
$ \frac{ab^2}{c^2(c\plus{}a)}\plus{}\frac{bc^2}{a^2(a\plus{}b)}\plus{}\frac{ca^2}{b^2(b\plus{}c)} \ge \frac{a}{c\plus{}a}\plus{}\frac{b}{a\plus{}b}\plus{}\frac{c}{b\plus{}c}$
 :)

substitution $ x\equal{}\frac {a}{b}...$$ xyz\equal{}1$
equivalent to $ \sum \frac{y^2}{z\plus{}1} \ge \sum \frac{1}{x\plus{}1}$
equivalent to $ (\sum xy^2 \minus{}3)\plus{}(\sum x^2\minus{}\sum xy)\plus{}(\sum xy^3 \minus{}xyz\sum x)\plus{}(\sum x^3\minus{}\sum x)\ge 0$
which is obviously true :)

[quote="hxy09"]substitution $ x \equal{} \frac {a}{b}...$$ xyz \equal{} 1$
equivalent to $ \sum \frac {y^2}{z \plus{} 1} \ge \sum \frac {1}{x \plus{} 1}$
equivalent to $ (\sum xy^2 \minus{} 3) \plus{} (\sum x^2 \minus{} \sum xy) \plus{} (\sum xy^3 \minus{} xyz\sum x) \plus{} (\sum x^3 \minus{} \sum x)\ge 0$
which is obviously true :)[/quote]
Nice solution :)

It's obviously trues by Cauchy Schwarts  :)
The hint to creat: $ LHS \equal{} \frac {(\frac {b}{c})^2}{\frac {c \plus{} a}{a}}$
it's not a new hint but is an interesting hint  :maybe:

[quote="nguoivn"]The second proof:
$ LHS \equal{} \sum\ \frac {(\frac {ab}{c})^2}{a(c \plus{} a)} \geq\ \frac {(\sum\ \frac {ab}{c})^2}{\sum\ a^2 \plus{} \sum\ ab} \geq\ RHS$
But we have: $ (\sum\ \frac {ab}{c})^2 \geq\ 3(a^2 \plus{} b^2 \plus{} c^2)$
And: $ \sum\ \frac {a}{a \plus{} b} \geq\ \frac {(a \plus{} b \plus{} c)^2}{\sum\ a^2 \plus{} \sum\ ab}$
We have q.e.d  :)[/quote]
Nice proof :lol:

The following inequality is also true
 $ a,b,c >0$. Prove that:
$ \frac{ab}{c(c\plus{}a)}\plus{}\frac{bc}{a(a\plus{}b)}\plus{}\frac{ca}{b(b\plus{}c)} \ge \frac{a}{c\plus{}a}\plus{}\frac{b}{a\plus{}b}\plus{}\frac{c}{b\plus{}c}$


search
geometry
geometric transformation
homothety
trigonometry
inequalities
blogs

So the INMO's the day after tomorrow...
[size=200][color=red]ALL THE BEST[/color][/size]
And remember, if in doubt say
[size=200][color=blue]ALL IZZ WELL!![/color][/size]

sahi bola Rijul,
ALL IZZ WELL!
All the best to you too and to everyone else! :lol:
 :D

I think it should be [i]AAL[/i] IZZ WELL  :rotfl:
anyways BEST OF LUCK to all appearing for INMO as well as KVPY(though this is not the right place)
BTW for those in Delhi region(though they must be knowing thru the camp)
u can get your interview date adjusted if its clashing with INMO
Contact R Ghosh immediately if u havn't

Best of luck to AAL apperaring in INMO tomorrow.
 :D  and ofc ALL IZZ WELL :thumbup:

[b][color=olive]All the best to all appearing for INMO 2010.[/color][/b]

[hide="What to expect in INMO 2010?"]
Expect the unexpected as usual! :ewpu:
[/hide]

though i amnt writing.. - ALL D BEST
to those who are writing..

ALL THE BEST TO EVERYONE..........

p.s. wow this is late.... :blush:  :blush:

how did you all do?
can someone post the problems.ive not written the exam

[quote="pardesi"]They set $ \infty$ questions this time is it?why is no one posting the problems?[/quote]
You missed it by a whisker....its $ \infty \minus{} 1$

Why is'nt anybody posting ???  :huh:  :huh:

POst here how have you done and so on...

 :)

Pirate, how people in West bengal done??? And how have Maharashtrians done????

[quote="pKa2010"]:rotfl:  :rotfl:[hide="Mr.POETS"]
Can you tell me why INMO_2010 results haven't been put up on the hbcse website?
[/hide][/quote]

Dr. Saini has been researching and investigating reasons for this mystery. Stay tuned for our next episode of $ \mathbb{Q.E.D.}$ to find out why exactly the results aren't put up on the NBSCE website.

This is the end of chapter 1, [b]Inductively Destructed[/b]. I request Dr. Saini to entitle the next chapter, and mention the introduction, from where I'll take over.

By the way, please upload your last year's, i.e., your INMO 2009's performance card, if possible. Thank you!

[quote="antiparticle"]
By the way, please upload your last year's, i.e., your INMO 2009's performance card, if possible. Thank you![/quote]

yepp
even i would like to see it.....

[quote="abhitayal"]even i would like to see it.....[/quote]
Really sorry dude :blush:
Actually I had scored [hide="..."]
11! last year and i had teared it off as soon i got it,although i remember that it was just a small slip containing the sign of RMO coordinator with my marks and cutoffs of X1 and X11.(26 and ~42 resp.)
Do you know this year cutoffs of X11?
i was gettin 2.5 correct ,and i am nowhere
[/hide]

no probs
and don't be sad about ur marks
and I think it was a good thing u teared it off if it was giving u pain
[hide="BTW"]I myself didn't get much ie 16
just wanted to see how an INMO performance card looks[/hide]

Due to the high standard of our posts, i have been unable to post very quickly. Nonetheless, the second chapter.... Brutally suppressed....shall be posted as soon as possible..

[b]Brutally Suppressed[/b]

Thank You all.... I shall now be releasing our next chapter....

Following tradition....I shall be hiding this also $ 10$ times...
[hide][hide][hide]

Ok... So Our research has led us to conclude that

$ 1$) Those who study geometry are fools.... :P  Just apply Trigonometry and Coordinate to every question having even a hint of geometry... and victims like Rijul Saini (with whom Dr. Archit had a private interview.. )are left sulking....which leads to even more happiness of people like Surbhi Goel....as Dr. Archit has observed.
For the second part, stay tuned.... :D
[/hide][/hide][/hide]

[hide="BTW"]I'm not so good at arithmetic....so I would please like others to make sure that I've not broken tradition... :oops: [/hide]

You do realize that
[hide="using the brutally powerful secret weapon"][hide][hide]pressing the quote button to see what's inside
[/hide][/hide][/hide]
ruins the entire purpose of your 'tradition', don't you?  :D

[quote="Rijul saini"]
$ 1$) Those who study geometry are fools.... :P  Just apply Trigonometry and Coordinate to every question having even a hint of geometry... and victims like Rijul Saini (with whom Dr. Archit had a private interview.. )are left sulking....which leads to even more happiness of people like Surbhi Goel....as Dr. Archit has observed.
[/quote]

Well, unfortunately, Rijul here has crossed a line, so I'm out of this.

[quote="pKa2010"]
Does somebody know the cut-offs for class X11?
[/quote]

Cut-off for 11th and below was 37 marks, and for 12th I guess it was around 48-50.

[quote="antiparticle"]
Cut-off for 11th and below was 37 marks, and for 12th I guess it was around 48-50.
[/quote]
Does getting 2.5~3 correct mean around 40~50 marks?

[quote="pKa2010"]
Does getting 2.5~3 correct mean around 40~50 marks?[/quote]

Well, I got 47 with about the same number of problems  :blush:, but it depends a lot on how you write your solution, I guess  :oops:.

A small request to Mathlinkers who will be attending  IMOTC this year.

Please add the problems of this year's TST's in the resources section.

Alright....now that
$ Professor\minus{}cum\minus{}Doctor\minus{}cum\minus{}Detective\minus{}cum\minus{}Student Archit$ (what to do....he had to be given a special mention... :) )
Has decided to leave this.....so I would like to take advantage :D and finish off out research work....
Not following tradition....I would hide it $ 10!$ times and not $ 10$ times.
[hide="1"][hide="Wow! 10!"]
1) Those who quoted this....(Like a person called Keerthan)....are likely to clear INOI...for their good information regarding internet and Mathlinks.... :rotfl:

2) Those who didn't quote this....(Like Myself)....and are in a habit of opening long hidden threads....can join me for my next book...."[b]How to teach stubborn fellows...![/b]"...Your attention would be appreciated... :lol:

3) Last ([b]AND[/b] and the least :P )....@pka2010:- The HBCSE people wanted to ward off a security threat.....you can see how the posting of results made the IIM site crash....and you can now continue by the tools we introduced earlier "BRUTE FORCE",and "INDUCTION"....finish off the proof quite easily.... :rotfl:
Thank You all!!!
x-x-x
[/hide][/hide]

the inoi results are out. http://www.iarcs.org.in/inoi/2010/inoi2010/results_inoi2010.php


An acute isosceles triangle, $ABC$ is inscribed in a circle. Through $B$ and $C$, tangents to the circle are drawn, meeting at point $D$. If $\angle ABC=\angle ACB=2\angle D$ and $x$ is the radian measure of $\angle A$, then $x=$

[asy]
defaultpen(linewidth(0.7)+fontsize(10));
real x=180/7;
pair D=origin, B=dir(3x), C=dir(4x), A=intersectionpoint(C--C+dir(2x), B--B+dir(5x)), O=circumcenter(A,B,C);
markscalefactor=0.015;
draw(B--C--D--B--A--C^^Circle(O, abs(O-C))^^anglemark(C,A,B));
dot(A^^B^^C^^D);
pair point=O;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$x$", A+0.1*dir(270), S);[/asy]

$\text{(A)} \ \frac37\pi \qquad \text{(B)} \ \frac49\pi \qquad \text{(C)} \ \frac5{11}\pi \qquad \text{(D)} \ \frac6{13}\pi \qquad \text{(E)} \ \frac7{15}\pi$

TYhere should be a better solutiuon than mine, as it's a little long for a number 14. :D

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=767&sid=42de29b6e7757397708d10367b9c3c1d[/img]

Let $O$ be the center of the circle. First, we see that the degree of arc $AC$ is twice that of $\angle ABC=2D$. Therefore, $\angle AOC=4D$. I realized I messed up my labelling, oops.  :blush:

Now we have an equation: $x+4\angle D=\pi$. ($\angle ACO=\angle CAO$ from isosceles triangle).

Now, we will show that $\angle DCB=x$. Because $CD$ is a tangent, $DCO=\pi /2$. If we draw an altitude from $A$ to base $BC$, we see that it bisects $\angle BOC$. a property of angles in circles is that the degree of arc BC is twice that of angle BAC, and equal to that of BOC. Therefore, $\angle COP=x$.  The rest follows. :P

Now in triangle $ADC$, we have that $x/2+2\angle D+x+\angle D/2=\pi$. Solving these two equations, $x=\frac37 \pi$.

[/hide]

There is indeed a much easier way.
[hide]Let angle $\ D = y$ such that angle $\ ABC=ACB=2y$  So arc $\ AC=4y$ and arc $\ AB=4y$.  Now we get the equation $\ y= 1/2 ( 8y - (2 \pi - 8y)$.  Just solve that for the value of $\ y$ and everything else is easy.[/hide]


Which of the following four values is largest?
a)  $\frac12\cdot\frac34\cdot\frac56\cdot\frac78\cdot\frac9{10}$
b)  $\frac1{2\sqrt{10}}$
c)  $\frac1{\sqrt{10}}$
d)  $\frac1{\sqrt{13}}$

[No calculators for the 1986 Target]

This is from a nationals? :?:

[hide]
d and b can be eliminated, because it's obvious they are smaller than c.
You might as well go ahead and calculate a, then cross multiply with c.
[/hide]

[quote="vamathletes"]This is from a nationals? :?: [/quote]Yes

It is C.  3 x 5 x 7 x 9 is 15 x 63 which is 630 + 315 or 945
The denominator is (2 x 4) 8 x 6 x 10 and 1/sqrt 10 is sqrt 10 over 10

so you have (945/(64 x 8 x 10)) is greater, less than, or equal to (sqrt 10)/10

Multiply each side by ten and you're comparing 945/(480 + 32) to something in between 3 and 4 (10 is between 9 and 16)

945/512 is about 1000/500, or 2.  So c is larger.

[hide]Just at first glance we can eliminate B and D because root 10 is between 3 and 4 so C is between 1/3 to 1/4. B is twice the denominator so it is between 1/6 to 1/8.  so b is out. obviously 1/root 10 is larger than 1/root 13. A is , just taking a few terms, is obviously smaller than c so the largest term is C.[/hide]


If $ x,y$ and $ z$ are consecutive even integers such that $ x < y < z$, then what is the value of $ z^2 \plus{} x^2 \minus{} 2y^2$?

We can also show this by letting $ x \equal{} y\minus{}2$ and $ z \equal{} y\plus{}2$

then $ z^2\plus{}x^2\minus{}2y^2 \equal{} y^2 \plus{}4y \plus{}4 \plus{} y^2 \minus{}4y \plus{}4 \minus{}2y^2 \equal{} \boxed{8}$

[quote="Mewto55555"]PLugging in -2,0,2 we get 4+4-0=8. The problem implies this alwasy works so our answer is 8.[/quote]

But the question said they are [u]consecutive[/u]

i got a different answer but it was wrong but i can't find anything wrong about it.
i plugged in x=1 y=2 z=3  so 3^2 + 1^2 - 2*2^2
So i got (9+1)-8 so I entered 2
Is there anything wrong?

They are consecutive [b]even[/b] integers, so 2 apart.


(1) Prove that the ineq

$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\le \frac {9}{2}$

holds for all positive $ a,b,c$ satisfying the condition

 $ \boxed{a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} k(ab \plus{} ac \plus{} bc) \equal{} 1 \plus{} k}$ where $ k\le \frac12$. (Vasc's one is the $ k\equal{}0$ case).

(2) Prove that the ineq

$ \frac1{1 \minus{} \bigl(\frac {a \plus{} b}2\bigl)^{2}} \plus{} \frac1{1 \minus{} \bigl(\frac {b \plus{} c}2\bigl)^{2}} \plus{} \frac1{1 \minus{} \bigl(\frac {c \plus{} a}2\bigl)^{2}}\le\frac92$

holds with the same condition where $ k_{max} \equal{} \frac {\sqrt {33} \minus{} 5}4\approx .18614.$ :D

For $ k\equal{}\minus{}2$, it does not hold as well......I think the range of $ k$ will be $ \minus{}1<k \leq \frac{1}{2}$
 :)

[quote="arqady"][quote="spanferkel"](1) Prove that the ineq

$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\le \frac {9}{2}$

holds for all positive $ a,b,c$ satisfying the condition

 $ \boxed{a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} k(ab \plus{} ac \plus{} bc) \equal{} 1 \plus{} k}$ where $ k\le \frac12$. (Vasc's one is the $ k \equal{} 0$ case).

[/quote]
But for a example, for $ k \equal{} \minus{} 1$ it's wrong. :wink:[/quote]
I think $ k \ge\ 0$ and with case $ k \equal{} 0$ It can be solve easily by PQR or Mixing avarible

[quote="onlylove_math"]
I think $ k \ge\ 0$ [/quote]yes, sorry.   :blush:

[quote="onlylove_math"]and with case $ k \equal{} 0$ It can be solve easily by PQR or Mixing avarible[/quote]
or with Cauchy  :)  :) see
http://www.mathlinks.ro/Forum/viewtopic.php?t=109029


Prove

$ (xy\plus{}yz\plus{}zx) \left(\frac 1{{(x\plus{}y)}^2}\plus{}\frac1{{(y\plus{}z)}^2}\plus{}\frac1{{(z\plus{}x)}^2}\right) \ge\frac{9}{4}$

for all positive reals x, y, z.

[quote="Dimitris X"][quote="MTM"]Prove

$ (xy \plus{} yz \plus{} zx) \left(\frac 1{{(x \plus{} y)}^2} \plus{} \frac1{{(y \plus{} z)}^2} \plus{} \frac1{{(z \plus{} x)}^2}\right) \ge\frac {9}{4}$

for all positive reals x, y, z.[/quote]
We have discuss in the forum one billion times for this!!!
It is the well-known Iran 1996 ineq...
Searchin the ineaqualities subforum and you will find the solution(the easiest i think is by expanding :wink: ).[/quote]

Sorry.


In a cube box with edge 4, can we put 65 balls of diameter 1?

Yes.  You can put 16 at the bottom, with 9 on top of those, then 16, 9, 16.  The height of the box used is $1+2\sqrt{2}$, which is less than 4.

Yes, because the minimum no. of balls you can put and fill it up is $64$ and if you arrange the balls to fit in then there is extra room created so $65$ is possible.

It makes me wonder if 66 is the most number or not...

hmm I think it might be possible but the calculations are very long! :P


how did everyone do??
my team got 84 :)  which is a pretty decent score...

Vestavia will probably win first place, in total....

[quote="Elemennop"]Vestavia won't win first -- we got 77.

Bummer. Congrats on beating us.[/quote]

yay (assuming no one else beats us :D )


A similar problem: $C_{2n}^n$ is divisible by $n+1$ (related to Catalan numbers).


I first found the idea of directed length, angle, and area in a MOP note on geometry. but I'm still confuse on its defintion and use (like, when will a segment\angle\area have negative value, and how to apply it on proofs). cansomeone explain it to me? thank you (because this is important in proving collinearity and concurrency)

This is a fairly old topic, but I am currently writing a note called "The Neuberg-Mineur circle" which will contain, on its first 5 pages, a brief introduction into some prerequirisites - including directed angles and lengths. A PDF of these pages is attached to this post. Feedback is greatly appreciated - feel free to tell me what can be improved, what is missing and what is plain wrong. Note that I am just itemizing the main properties, not proving them - in particular, I am not proving that the notions I define are indeed well-defined.

  Darij


Pythagorean Theorem
geometry
geometry proposed

Let [size=150]ABC[/size] be an acute-angled triangle. Let [size=150]D[/size] and [size=150]E[/size]
be points on [size=150]BC[/size] and [size=150]AC[/size], respectively, such that [size=150]AD[/size] is perpendicular to [size=150]BC[/size] and [size=150]BE[/size] is also perpendicular to [size=150]AC[/size]. Let [size=150]P[/size] be the point where [size=150]AD[/size] meets the semicircle constructed outwardly on [size=150]BC[/size], and [size=150]Q[/size] the point where [size=150]BE[/size] meets the semicircle constructed outwarldy on [size=150]AC[/size].

Prove that: [size=150]PC=QC[/size]

[color=white]..........[/color] :diablo:  :diablo:  :diablo: [color=white]..........[/color]

*************************************************************************************
[color=green]Any questions or suggestions?? feel free to leave me a message![/color]
*************************************************************************************

I'll probably post the answer the next day or some other day.. hmmm.. i'll just see whenever I'll be available! :lol:

[quote="maximos"]nice!

hmmm... it would be better though if you posted your solution! so that others may follow too...

nice work though... :thumbup:[/quote]
More detail $ \begin{array}{l}
 \Delta PDC \sim \Delta BPC \Rightarrow CD.BC \equal{} CP^2  \\
 \Delta BEC \sim \Delta ADC \Rightarrow CD.BC \equal{} EC.AC \\
 \Delta QEC \sim \Delta AQC \Rightarrow AC.EC \equal{} QC^2  \\
 \end{array}$ :)

So here's my proof too:  :lol:
[hide="Proof."]
By Pythagorean Theorem: $ QC^2 \equal{} EQ^2 \plus{} EC^2$ Since: $ EQ^2 \equal{} AE \cdot EC$ and $ PC^2 \equal{} DC\cdot BC$
Where:$ QC^2 \equal{} EQ^2 \plus{} EC^2 \equal{} AE \cdot EC \plus{} EC^2 \equal{} EC(AE \plus{} EC) \equal{} EC \cdot AC$
Notice that: $ \begin{array}{l} \Delta ADC \sim \Delta BEC \end{array}$,
So it follows that we'll obtain:
$ \frac {DC}{AC} \equal{} \frac {EC}{BC}$
which can also be $ DC \cdot BC \equal{} EC \cdot AC$ which is equivalent to $ PC^2 \equal{} QC^2$ and $ PC \equal{} QC$.

 :)

[/hide]


There's whole number $n \geq 1$. Designate all possible values of $km$, where $k$ and $m$ are positive whole numbers satisfying

\[n^{2}\leq k \leq m \leq (n+1)^{2}\]

i posted my solution here:

http://www.mathlinks.ro/Forum/viewtopic.php?p=821652#821652


$ \sum n^{4(x_i ^2\plus{}x_i)}\equal{}1$  for   $ 1<\equal{}i<\equal{}n$  solve this equation !

Maybe I misunderstood the problem or I made a mistake somewhere,but its solution seems to me very easy.Here is how I understand the problem:
Let $ n$ be a fixed natural number.Find all $ x_1,x_2,\dots,x_n\in\mathbb{R}$,such that
\[ \sum_{i \equal{} 1}^{n}n^{4(x_i^2 \plus{} x_i)} \equal{} 1
\]
[b]Solution:[/b]
Rewrite our equation as(by multiplying both sides to $ n$):
$ \sum_{i \equal{} 1}^{n}n^{(2x_i \plus{} 1)^2} \equal{} n$.
But as we know if $ x\geq 0$,then $ n^{x}\geq 1$.
Therefore,our equation can be true only if $ x_1 \equal{} x_2 \equal{} \dots \equal{} x_n \equal{} \minus{} \frac {1}{2}$,as there are exactly $ n$ terms in our sum.
Am I miss something?


modular arithmetic
combinatorics unsolved
combinatorics

Find $ S\equal{}\left( C_n^0 \right)^2\minus{}\left( C_n^1 \right)^2\plus{}\left( C_n^2 \right)^2\minus{}\left( C_n^3 \right)^2\plus{}...\plus{}(\minus{}1)^n\left( C_n^n \right)^2$

$ S$ is the coefficient of $ a^n\cdot b^n$ in the expansion of $ (a\plus{}b)^n\cdot (a\minus{}b)^n \equal{} (a^2\minus{}b^2)^n$, as can be seen by expanding the RHS. This is clearly just $ 0$ if $ n$ is odd, and $ (\minus{}1)^{n/2} \cdot \binom{n}{n/2}$ if $ n$ is even. We have:

$ S \equal{} \begin{cases}
(\minus{}1)^{n/2} \cdot \binom{n}{n/2} & n \equiv 0 \pmod{2} \\
0 & n \equiv 1 \pmod{2}\end{cases}$

[quote="Vworldv"]$ S$ is the coefficient of $ a^n\cdot b^n$ in the expansion of $ (a \plus{} b)^n\cdot (a \minus{} b)^n \equal{} (a^2 \minus{} b^2)^n$, as can be seen by expanding the RHS. This is clearly just $ 0$ if $ n$ is odd, and $ ( \minus{} 1)^{n/2} \cdot \binom{n}{n/2}$ if $ n$ is even. We have:

$ S \equal{} \begin{cases} ( \minus{} 1)^{n/2} \cdot \binom{n}{n/2} & n \equiv 0 \pmod{2} \\
0 & n \equiv 1 \pmod{2}\end{cases}$[/quote]

your solution is very nice,can you use same idea to find that:
$ S_1$= $ \sum (\minus{}1)^i.(C_{2n}^{2i})^2$

Bah humbug.  $ {n \choose k}^2$ is the number of pairs of subsets of $ \{ 1, 2, ... n \}$ with cardinality $ k$.  Given such a pair, find the smallest number that is in neither or both of the subsets and either add it or remove it from both.  When $ n$ is odd, this is always possible; when $ n$ is even, the only fixed points of this involution occur precisely when $ k \equal{} \frac {n}{2}$ and the subsets are disjoint.

[quote="math man"]your solution is very nice,can you use same idea to find that:
$ S_1$= $ \sum ( \minus{} 1)^i.(C_{2n}^{2i})^2$[/quote]

This doesn't even appear in Sloane, so I'd be rather skeptical. For odd $ n$, the sum is $ 0$ (logically), for $ n\equal{}2$ it's $ \minus{}34$, for $ n\equal{}4$ it's $ 3334$, for $ n\equal{}6$ it's $ \minus{}372436$.

The obligatory links:
http://www.mathlinks.ro/Forum/viewtopic.php?t=262752
http://www.mathlinks.ro/Forum/viewtopic.php?t=268056

  darij


trigonometry
geometry
circumcircle
trig identities
Law of Sines

In triangle ABC, a>=b>=c. If (a^3+b^3+c^3)/[(sinA)^3+(sinB)^3+(sinC)^3]=7, compute the maximum value of a. (AOPS vol. 2)

I don't have the faintest idea of what to do; any help would be very much appreciated. :)

So we have $ a\geq b\geq c$, and $ \frac{a^3\plus{}b^3\plus{}c^3}{\sin^3A\plus{}\sin^3B\plus{}\sin^3C}\equal{}7$.

By the Law of Sines, $ k\equal{}\frac a{\sin A}\equal{}\frac b{\sin B}\equal{}\frac c{\sin C}$, and cubing this chain, we get

$ k^3\equal{}\frac{a^3}{\sin^3A}\equal{}\frac{b^3}{\sin^3B}\equal{}\frac{c^3}{\sin^3C}\equal{}\frac{a^3\plus{}b^3\plus{}c^3}{\sin^3A\plus{}\sin^3B\plus{}\sin^3C}\equal{}7$

$ \implies k\equal{}\sqrt[3]7$.

Substituting this into the first chain (Law of Sines), we find that $ a\equal{}\sin A\sqrt[3]7$.

Since $ \sin A\leq1$, the maximum value of $ a$ is $ \boxed{\sqrt[3]7}$.

This is achieved when $ \angle A\equal{}90^\circ$.

For a more geometric interpretation, $ k \equal{} 2R$, so what the problem gives you is the size of the circumcircle and you want to maximize a side length, which clearly occurs when it's a diameter.


geometry
geometric transformation
rotation

Seven people arrive to dinner, but the table only seats six.  If two seatings such that one is a rotation of the other are considered the same, then in how many different ways can we choose six people and seat them at the table?

OK, there are $ \binom{7}{1}$ ways to choose the odd-man out.  Then, there are $ (6\minus{}1)!$ ways to arrange the remaining people, for the answer of $ \boxed{840}$.

[hide=my sol]
okay so basically this is in 2 parts: First we choose 6 people, and then we find the ways to arrange.

we do 7C6 to choose, which equals 7.
Then there are $6!$ ways to arrange

so it's $7*6!=840$.

Yes that is correct. Furthermore, the only reason why reflections don't matter here is because we assume the chairs are equally spaced apart, and there is an even number of chairs.

7*6*5*4*3*2*1/6= 840 because of rotational symmetry

[hide=sol]There are $\dbinom76$ ways to choose the people who get to sit at the table. Then we can fix the first person at the top of the table to account for symmetry so the number of possibilities is $5!$. Then we multiply and get $\boxed{840}$.[/hide]

[hide=My Solution]There are 7 ways to pick one person to exclude. Once we have 6 people left, there are 6! ways to seat them, but we must divide by 6 for rotational symmetry and we are left with 5!. So, the answer is 7 x 5! = 840.[/hide]


[url]http://www.tenthdimension.com/medialinks.php[/url]

What do you guys think of this guy, his book, and his theories? Genius, or BS. Discuss!

This is very confused. Could be quasi-science, which I'm very sick of.

No clue whether the theory is close to reality or not

but it's very interesting either way

I got lost beginning on the 3rd dimension. :rotfl:

The explanation of the third dimension reminds me of Madeleine L'Engle's books with the tesseract.

I think it's interesting, but scientifically has little value. But don't quote me on that; I'm no physicist.


linear algebra
matrix
linear algebra unsolved

Find all $ n\times n$ complex matrices that commute with all permutation matrices.

They are precisely the matrices which commute with all transposition matrices, and, unless I'm making some awfully stupid mistake :), it follows immediately from here that they are the matrices having equal entries on the main diagonal and also equal entries off the main diagonal. In other words, it's the subspace generated by $I_n$ and the matrix having $1$ everywhere.


calculus
integration
logarithms
calculus computations

$ \int \frac{e^{\minus{}x}}{1\plus{}e^{\minus{}x}} dx$

Should be pretty easy for you guys... :wink:

[quote="math92"]$ \int \frac {e^{ \minus{} x}}{1 \plus{} e^{ \minus{} x}} dx$[/quote]

$ \text{Also, }\frac {d}{dx}\left\{\ln\left(\frac {1}{1 \plus{} e^{ \minus{} x}}\right) \plus{} c\right\} \equal{} \frac {e^{ \minus{} x}}{1 \plus{} e^{ \minus{} x}} dx$

Right.

$ \int \frac{e^{\minus{}x}}{1\plus{}e^{\minus{}x}}\ dx\equal{}\int \frac{1}{e^x\plus{}1}\ dx\equal{}\int \left(1\minus{}\frac{e^x}{e^x\plus{}1}\right)\ dx$

$ \equal{}x\minus{}\ln(e^x\plus{}1)\plus{}C$.

$ \int \frac{e^{\minus{}x}}{1 \plus{} e^{\minus{}x}} \,dx \equal{} \minus{} \int \frac{(1 \plus{} e^{\minus{}x})'}{1 \plus{} e^{\minus{}x}}\,dx \equal{} \minus{} \ln(1 \plus{} e^{\minus{}x}) \plus{} C$.

[quote="I"]
$ \text{Also, }\frac {d}{dx}\left\{\ln\left(\frac {1}{1 \plus{} e^{ \minus{} x}}\right) \plus{} c\right\} \equal{} \frac {e^{ \minus{} x}}{1 \plus{} e^{ \minus{} x}} dx$

[/quote]

[quote="Carcul"]
$ \int \frac {e^{ \minus{} x}}{1 \plus{} e^{ \minus{} x}} \,dx \equal{} \minus{} \int \frac {(1 \plus{} e^{ \minus{} x})'}{1 \plus{} e^{ \minus{} x}}\,dx \equal{} \minus{} \ln(1 \plus{} e^{ \minus{} x}) \plus{} C$.

[/quote]

$ \minus{} \ln(1 \plus{} e^{ \minus{} x}) \equal{} \ln\left(\frac {1}{1 \plus{} e^{ \minus{} x}}\right) \equal{} \ln\left(\frac {e^x}{e^{x} \plus{} 1}\right) \equal{} x \minus{} \ln(1 \plus{} e^x)$

It's the same...

[quote="Carcul"]$ \int \frac {e^{ \minus{} x}}{1 \plus{} e^{ \minus{} x}} \,dx \equal{} \minus{} \int \frac {(1 \plus{} e^{ \minus{} x})'}{1 \plus{} e^{ \minus{} x}}\,dx \equal{} \minus{} \ln(1 \plus{} e^{ \minus{} x}) \plus{} C$.[/quote]

This is my first solution.

Basically we should check whether we can use $ \int \frac{f'(x)}{f(x)}\ dx\equal{}\ln |f(x)|\plus{}C$ or not.