ratio
geometry
angle bisector
geometry solved

[quote="marko avila"]Let $ABC$ be a triangle and let $D$, $E$ and $F$ be the midpoints of its sides $AB$, $BC$ and $CA$ respectively. Suppose that the angle bisector of $\angle{BDC}$ meets $BC$ at the point $M$, and the angle bisector of $\angle{ADC}$ meets $AC$ at the point $N$.  Let $MN$ and $CD$ intersect at $O$. Let the line $EO$ meet $AC$ at $P$, and let the line $FO$ meet $BC$ at $Q$.
Prove that $CD=PQ$.[/quote]

Easy but very neat exercise.

The angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides. Applying this to the triangle BDC with its angle bisector DM, we obtain $\frac{CM}{MB}=\frac{CD}{DB}$. Since D is the midpoint of the segment AB, we have $DB=\frac12\cdot AB$, so this becomes $\frac{CM}{MB}=\frac{CD}{\frac12\cdot AB}=\frac{2\cdot CD}{AB}$. Similarly, $\frac{CN}{NA}=\frac{2\cdot CD}{AB}$. Thus, $\frac{CM}{MB}=\frac{CN}{NA}$, so by Thales we have MN || AB. Therefore, again by Thales, $\frac{CO}{OD}=\frac{CM}{MB}$, so that $\frac{CO}{OD}=\frac{2\cdot CD}{AB}$. Now, since D and E are the midpoints of the sides AB and BC of triangle ABC, we have DE || CA, and thus by Thales $\frac{PO}{OE}=\frac{CO}{OD}$; thus, $\frac{PO}{OE}=\frac{2\cdot CD}{AB}$. Similarly, $\frac{QO}{OF}=\frac{2\cdot CD}{AB}$, and therefore $\frac{PO}{OE}=\frac{QO}{OF}$, so that Thales again yields PQ || EF, and thus, applying Thales for the last time, we get $\frac{PQ}{EF}=\frac{PO}{OE}$, so that $\frac{PQ}{EF}=\frac{2\cdot CD}{AB}$. Now, since E and F are the midpoints of the sides BC and CA of triangle ABC, we have $EF=\frac12\cdot AB$ (please don't count this as an application of Thales, there are easier ways to prove it ;) ), and thus this becomes $\frac{PQ}{\frac12\cdot AB}=\frac{2\cdot CD}{AB}$. This simplifies to PQ = CD. And it's done.

Now you can count how often I applied Thales above. :D

  Darij

very nice darij¡¡ thank you for solutions to all the problems i will read them carefully to understand them and get some new ideas. :P

dear marco here is a solution .but i dont wether its like darijs solution. :) .
FD andCQ are paralell.then OD/OC=OF/OQ.
DE andPC are parallel. then OD/OC=OE/OP.
it means that the triangles OFE andOPQ are similar.then PQshould be paralell to EF.

we have DC/AD=CN/NA, DC/AD=CM/MB (AD=AB)and it means that CN/NA=CM/MB.
then: CN/CN+NA =CM/CM+MB. and again it means that MN is parallel to EF.
and we have AD=DB=DO(because angle NDM=90).
look at the triangles FON andFQP. they are similar to each other.then we have OF/FQ=NO/PQ.
again we have OF/FQ-OF=NO/PQ-NO.it means OF/OQ=NO/PQ-NO.as i said at first we have :OF/OQ=OD/OC.then we have:NO/PQ-NO=OD/OC.we know NO=OD.then we should have PQ-NO=CO.
and:PQ=NO+CO=OD+CO=CD and the problwm is solved.  ;)


LaTeX
abstract algebra
linear algebra
matrix
linear algebra unsolved

the second one is in lenear algebra:

http://img215.imageshack.us/my.php?image=img8127wp2.jpg

they  solved a question about span equalisation
first they prooved that the first one is a part of the second one
then they prooved the second one is a part of the other

and then they did another step
which i dont know why???

the third question is in lenear algebra:

http://img512.imageshack.us/my.php?image=img8126fw0.jpg

in the link i wrote the question and the way i tried to solve it

but i got a realy weird answer in the first part

and in the second part i solved it too
but i am not sure if my theoretical knowledge about finding ker and Im are right

Hi! Welcome to the forum :welcomeani: There are a few "unwritten rules" here that you should know about.

1) Everyone is expected to learn basic $ \text{\LaTeX}$ and use it for posting formulae. The posts with jpg images of handwritten text occupy unnecessary server space, require special effort to read and for those and other reasons are usually ignored :)

2) When posting routine homework problems, state the problem clearly, show what you've done and tell exactly where you are stuck. Then you may expect a hint on what to do next.

3) Don't double post and be patient when waiting for an answer. This is not a help hotline, just a place where people hang out in their free time to discuss some mathematics.

There are a few more rules, but these three are good for a start. :)

Addendum to #1: If you don't know $ \text{\LaTeX}$ yet, write out the problem in plain text with a bit of markup. We can understand things like x^2+4.

as you saw in the link i
showed my question and my solution in a very good hand writing

its a lot of text and symbols
matrices
it will be a total nightmare to translate it into chars

Well, as I said, you are expected to learn and use $ \text{\LaTeX}$ eventually :) It is true that it takes some time to type matrices in $ \text{\LaTeX}$ but it is much less of a headache than you think...
[quote="transgalactic"]
they  solved a question about span equalisation
first they prooved that the first one is a part of the second one
then they prooved the second one is a part of the other
and then they did another step
which i dont know why???
[/quote]
What is written there can hardly be called a proof of anything. Step 1 is unfinished, Step 2 is rather useless, and Step 3 is just wrong. Whoever you copied those notes from certainly doesn't know what he or she writes as long as linear algebra is concerned.

As to your third question, your idea to use the row echelon form is correct but you seem to be confused about what to do with it. Let me just remind you that the dimension of the kernel of an $ m\times n$ matrix ($ m$ rows and $ n$ columns) is just $ n$ minus the number of non-zero rows in the row echelon form, so to maximize the dimension of the kernel, you want to make as many rows as possible zero. You cannot do much about the first 2 rows in this respect, but you can make the third row zero if you choose $ a,b,c$ appropriately. The dimension of the image (or range, if you prefer) is the number of the not identically zero rows in the row echelon form, so, to maximize it, you want as many non-zero rows as possible. Again, you cannot do much about the first two rows but the third one can be made both zero and non-zero and now you want it to be non-zero.

what about the Im

how to find the dim(Im(t))???

Have you read my post :? Let me repeat: [color=red]The dimension of the image is the number of the not identically zero rows in the row echelon form.[/color] This is not the only possible way to find it, of course, but once you have the row echelon form ready, this is the most direct way.


calculus
derivative
calculus computations

Any help would be appreciated!
A police helicopter is flying due north at 100mi/h and at a constant altitude of .5mi. Below, a car is traveling west on a highway at 75 mi/h. At the moment the helicopter crosses over the highway the car is 2 miles east of the helicopter.
$a)$ How fast is the distance between the car and helicopter changing at the moment the helicopter crosses the highway?
$b)$ Is the distance between the car and helicopter increasing or decreasing at that moment?

One of the key things in a related rates problem is determining which facts or relationships are always true (remain true even as things change) and which ones are "snapshots": true only for an instant, false immediately afterward.

The helicopter is flying north at 100 mi/h? Always true, but this is telling me the derivative of something. The derivative of what? We need to name a variable that this can be the derivative of, and that would be something reflecting the horizontal position of the helicopter.

Idea: let $y$ be the net horizontal distance the helicopter is north of the vertical plane that extends upwards from the road. Then (in miles per hour), $\frac{dy}{dt}=100.$

At a constant altitude of .5 mi? That's constant; that remains true all of the time. We'll just have to remember to build it into distance formulas.

The car is traveling west at 75 mi/hr? Also always true; again we need a variable for this to be the derivative of. We've got a choice to make about which way to make positive - this choice is arbitrary. Neither choice is wrong, but we must visualize this properly and maintain consistency with whatever choice we make. I'll choose to make $x$ be the horizontal distance that the car is east of the point under where the helicopter crosses. Then $\frac{dx}{dt}=-75$ miles/hr. (Do you see why this is negative?)

"At the moment the helicopter crosses over the highway the car is 2 miles east": THAT'S what I mean by a "snapshot." There is an instant at which $x=2$ and $y=0,$ and that's the instant we'll use to answer the questions. Don't ever differentiate "snapshot" information; it does not remain true over time.

What are we asked about? The distance between the helicopter and the car, and how it's changing. We need a variable for that.  With most of the good letters taken (I don't want to use $d$ because I'm also using Leibniz notation for derivatives.) So I'll call it $R.$

Quick check: are our units of measurement compatible or do we need to do any conversions? Everything is in miles and hours, so we're OK.

First important task: Find an equation that relates $R, x, y$ and $.5.$ This equation must remain true even as things change. (In particular, you must not use any snapshot information to set up this equation.) Some variant of a distance formula is what you need. (Don't bother solving for any one variable. In other words, leave it as squares, don't bother taking any square roots.)

Back to you kevyy: can you do this much?


why dosent the iit-jee papers contain eucledian geometry problems? and after qualifying is the exam is the study all free?

No, qualifying in the IIT entrance exam does not lead to free education. it is just a means to get admission.
Though it is cheaper than other institutes.

About euclidean geometry only the paper setters can answer..

@communist
forget IIT. u shud try for [hide="MIT"]Maoist Institute of Terrorism  :ninja: [/hide]

[hide]
bcoz mostly ppl go to IIT to secure a future in the [hide="home of capitalism"]america[/hide]
commies not allowed[/hide]

What are you talking about? Have you seen this year's JEE paper. Paper 2 had an euclid geometry problems in single digit answer type problem. PLus most of the solutions of triangle can be killed off by euclid. Physics questions also demand a lot of euclid geometry.

[quote="Conjurer"]Physics questions also demand a lot of euclid geometry.[/quote]

did they??
i dont even kno wat is euclidean ..but didnt hav problems in physics.... :?:

The geometry we did in classes 6th to 10th.

'euclidean geometry' aka 'pure geometry' as against 'coordinate geometry'
like those axioms and stuff  :)

i see there is lot of co-ordinate geometry problems. i am talking about real euclidean geometry problems of high school level. as we have in IMO. engineers and no geometry, is worth questioning. and about the fees for engineering can it be afforded by a middle class family? and just giving some rigorous problems wont make the exam harder than the MIT. i guess we need IMO type problems to really test the future engineer;s skill.
@Sagar_Vaze
i and mit,  i am not rich enough coz i am a commie. we people dont have the opportunity for such stuffs

I personally think IIT-JEE syllabus is similar to 12th and geometry is not there in 12th.

but in some of prev year ques,i think geometry was applicable and wasn't that easy(as said already by conjurer here)

and if they set IMO type prob,there is simple answer:
Don't attempt :rotfl: (time matters  :rotfl: )

@commie
IIT doesn't require one to be rich. I can say this on behalf of myself and ppl around me  :(
If getting admission into IIT was abt money alone, IIT wudn't be IIT, IIT wud be something like Amity,LPU,VIT :rotfl:

this said, i dont know if ppl other than indians are allowed in iit

Of course foreign nationals are allowed in IIT.They even had a JEE center in Dubai few students qualified.
[hide]Those dumb foreign nationals are no match to the intelligent Indian students :clap: [/hide]

@aashay
itz indian expats and not foreigners who giv jee in dubai
but still most expats dont qualify jee and instead go to some notworthmention university in canada/nz by payin a hefty sum  :wink:

@Sagar_Vaze
But its mentioned in the JEE brochure(page 2) that IITs conduct JEE for admissions of Indian and Foreign nationals

[quote="communist"]From which level is co-ordinate geometry and trigonometry is taught in India?[/quote]
We don't reveal the sekritz of our educational system yo communists!  :rotfl:


search
number theory
prime numbers
prime factorization
number theory unsolved

Show that the set of all prime numbers is infinite

You should use the search button located in the upper right. This is an old problem and its proof(s) are easily found on this site and anywhere in general.

Suppose that prime numbers are finite. Let P be the product of all prime numbers. Let's add one to P. For P+1 we have to cases:
1- P+1 is prime ---- We found another prime number so we get a contradiction.
2- P+2 is not prime. If we make the prime factorization of P+1, we will have at least to prime factors of P+1  and this to factors will be different with all the factors of P because we added one. --- Also in the second case we found another prime number, so we get a contradiction. Prime numbers are infinite.

To prove this you can use also Euclidian Algorithm.


1. Paul ist im Keller.

2. Gretchen ist auf dem Speicher.

3. Hans und Susi sind im Garten.

4. Ich möchte ins Wirtshaus gehen.

Do you need any further help with German: I notice that your post is over a year old?


Prove that if $\mathbb N$ is partitioned into arithmatic progressions $\{na_{1}+b_{1}\}_{n=1}^\infty,\{na_{2}+b_{1}\}_{n=1}^\infty\dots, \{na_{k}+b_{1}\}_{n=1}^\infty$ then two of $a_{i}$ are equal.

Doctor jamali told me about this one  :)

http://www.mathlinks.ro/Forum/viewtopic.php?t=56334

Mr.harazi ,Can you please say how that formal series Solution works?

[quote="lomos_lupin"]Mr.harazi ,Can you please say how that formal series Solution works?[/quote]

It is very famous.  Define $f_{i}(x) = \Sigma_{n \in P_{i}}x^{n}$.  Consider the equation $\Sigma f_{i}= 1/(1-x)$ near the singularities of the $f_{i}$.


if a+b =8
and a ,b positive

prove that
$(a+\frac{1}{b})^{2}+(b+\frac{1}{a})^{2}\geq \frac{289}{ 8}$


Can anyone post the problems from Turkish MO 2001 and 2007 - I have almost all remaining years.

2002,2004,2005,2006: http://imocompendium.com/index.php?options=oth|Tur|Ne|Ne&ttn=Turkey&p=0

2003: http://www.mathlinks.ro/resources.php?c=174&cid=95&year=2003

I have all problems from Turkish MO but I don't have only Turkish MO 2001 and Turkish MO 2007 it will be good if someone post them. Thanks in advance.
I and a friend of mine translated lots of the missing years of Bulgarian Math Olympiad so I can ask such a question it is the link. I think there are lots of excellent problems:
http://imomath.com/index.php?options=oth|Bul|Ne|Ne&ttn=Bulgaria&p=0

[quote="borislav_mirchev"]I found 2007 problems. Can anyone post Turkey MO 2001 problems?[/quote]

Can you please give us 2007 problems?


Happy Chinese New Year! I'm sure theres a topic about this somewhere else, but I want to wish you one with my very own. We're supposed to have a lunch from General Tso's Restaurant tomorrow, but we have a Black Saga pizza party during our lunch time. How Chinese, huh? That's one way to celebrate! We're also having a special dinner. My uncle went to Taiwan.

What are you doing?

it's tomorrow...
well.. in china/asian time, it's right now...

I know it's tomorrow, I wanted to get an early start.

Yeah. It's either they all die or prosper and grow double. I hope it's not the second option. Go Mickey, go Minnie..

I only like the year of the rat because of a character in a manga series I read called Fruits Basket. Go Yuki!

Happy year of rat!

Let's go to Disneyland and celebrate lol

[quote="math_explorer"]Yesterday I stayed up till 3:30 am. I was born in the year of the Rat.[/quote]

Me too...

'me too' for what?

Staying up till 3:30 am or born in the year of the rat? or both...? :wink:

I know it's over, but I went to bed at my regular curfew...But I got twenty dollars in a red envelope.  :D


parameterization integration factorial Gauss MATHCOUNTS calculus derivative

You didnt partition the set of all ages into disjoint sets.  :rotfl:

My conjecuture: the distribution of all ages will form a Gaussian curve.

where is the option for if you're 2 years old?

I'm one of the oldest students in 7th grade. I just turned 13 on September 12.

Only 10, now  :D That's why when I first joined, I didn't post much in High School Basics.

[quote="hunter34"]I am $ 13!$ Yes that is a factorial.[/quote]

WOW YOU'RE OLD

There's a lot of 12/13 year olds on AoPS...

16. There's a lot of younger people on here.

YO I AM SOMEWHERE BETWEEN 17 AND 29

THIS WILL PROBABLY CHANGE IN A COUPLE DECADES

just a random comment: YAY BIRTHDAY'S IN 2 WEEKS!!!!!!!! (turning 12) holy crap i gotta get those invites out FAST!

im 10.

(am i the youngest person on here? i think so)

splat, demon shadow is also 10.

im 11, almost 12

I turn 15 next february 24th. wow my bday is errily close to 7h3.demon.113

Whazzat??? 10-13 and 13-16? I seem to fit in both ranges...

wahh i'm 14. meh or 24(according to aops family)  :D


geometry
AMC
AIME
ratio
3D geometry
probability
rectangle

Let's see
Messed up #4

Took the area of the 4 quarter circles rather than the enclosed area thinking the 1/4 circles were the enclosed area... so cheap...
my brain tends to find a convex area more of an area than a concave area :-P 4-Pi instead of Pi argh

Messed up #7
added wrong

Not cool...

WHAT???!!!!

x^36-x^19-x^17-1=0 can be FACTORED by GROUPING?
i got there, decided that i would divide through by x^18 then forgot how to solve it

This was a rough AIME for me with careless errors... I guess the fates were making up for my making no mistakes last year and getting a 13. Basically, I correctly solved 1-14, but made the following errors:

#1- Didn't account for 3210 (positive INTEGER, not positive digits)
#6- Used the value of 9 choose 2 instead of 9 choos 3
#11- Found ratio of the cone to the whole, not the cone to the other portion, which I calculated first for convenience and forgot to convert back
#12- Used 1/5 instead of 4/5

Blah. Of course I can't complain about my score, but it disturbs me that I made that many careless errors...

My test strategy this year was 80 minutes solving problems and 100 min. checking, so the only mistake that I didn't catch before the test was over was using 1/5 instead of 4/5 in #12.
I made some interesting mistakes though... 2^5 does not equal 34, for example (I think that was problem 9?) and on #5 I marked the answer as 349, forgetting to add the denominator... a bunch of adding mistakes for problems 1,6,7

clarifiaction... that log problem? he summed the infinite series correctly.  He multiplied correctly.  He simplified correctly.  End result? 5/9.  What did he do? put down 13 (for 5+9).

i think it's funny; we all like doing math contests so much, but afterwards we almost always feel awful, thinking of the mistakes we made and what we should have gotten right.

#2, I added 198+3=202. Aren't I just smart

Ouch...I know how it feels like to mess up :(

#9 -- misbubbled: 135, not 035.
#10 -- generally confused about the shape of this thingie
#11 -- what, the slant height isn't 4?
#12 -- somehow I got the probability that the two logs are equal and even
#15 -- subtle mistake in combinatorial bookkeeping -- my value of m was only off by 2^8

I made numerous minor bookkeeping errors along the way -- amazingly, I managed to catch all my arithmetic mistakes.

On the whole, much better than my freshman year, when I answered 12 and got a 6.  I'm annoyed at the misbubble -- three 11's in a row would be cool.

Yes, I know the question.  ;-)  What confused me was the shape of the possible positions of the center.  I think I was looking at it the wrong way -- concentrating on the region of centers for which the circle intersected the diagonal, instead of answered the question they asked directly.

[quote="tAsDoOm"]#10 is a 15 by 36 rectangle with one diagonal, and you are trying to find the probability of landing a circle of radius 1 completely within the two triangles formed by the diagonal.[/quote]

#1 - forgot 3210
#3 - did 26*21/2, figuring that each vertex was directly connected to 5 vertices (instead of 26C2 - ..., etc.)
#5 - added wrong
#6 - "
#7 - "
it was not pretty

Whoo, I feel a ton better reading this thread.

#3 - Did something like subtracted the edges twice.
#4 - Put 100 :pi:  instead of 100(4- :pi: )
#9 - I think I didn't reduce correctly or something.

:(.

3. just plain stupid
6. forgot to count zero
11. somehow 72 magically changed to 36 on my paper
12. forgot to skip 1/4, 1/16, etc. and 4/25, 4/625, etc. when counting

8. Counted a regular 1000-gon as a star
12. 10/18 is in reduced form

#5 - misbubbled, had right answer
#6 -  addition error during brute force calculations
#8 - logic mistake
#10 - misread (switched sheets of paper, so i put a wrong number down)
#12 - summed series wrong

Impressive... 11 -> 6.

#3-stupid mistake
#6-stupid mistake
#8-stupid mistake
#9-stupid mistake

Those 4 brought my score from a 10 to a 6.  Maybe I should have done a few extras.  Is it just me, or are 11, 12 and 14 pretty easy?

-interesting_move

I did the same as some others seem to have done - for number four, I took the non-enclosed area (100 :pi: ).  Nice to know other people were thinking along the same wave length ;)


inequalities
induction
probability
combinatorics unsolved
combinatorics

Let $A_1,A_2,.....,A_n$ be $n$ finite sets such that for every $i=1,2,...,n$ we have
$\left|A_{i}\cap A_{i+1}\right|>\frac{n-2}{n-1}\left|A_{i+1}\right|$.
Prove that $\left|\bigcap_{i=1}^{n}A_{i}\right|>0$.

i made a post yesterday but later discovered a mistake in it( the solution came out rather neat so i was pretty sure that it had to be right!) and so deleted that post.later i realised that the problem is easier than what i initially figured.
  suppose $x_S=\vert \bigcap_{i\in S} A_i\cap\bigcap_{i\notin S} \overline{A_i}\vert$. we are given the following information:
 \[\sum_{S|i,i+1\in S} x_S >(1-\frac{1}{n-1})\sum_{S|i+1\in S} x_S\forall i\].
rearranging terms we get
\[\sum_{S|i+1\in S} x_S > (n-1) \sum_{S|i+1\in S,i\notin S} x_S \forall i\].summing this over all $i$ and interchanging sums we get
\[ \sum_{S\neq \emptyset} \vert S\vert x_S >(n-1) \sum_{S\neq\emptyset} (\sum_{i: i+1\in S,i\notin S} 1)\cdot x_S\].
Now come the crucial part:if $S\neq \{1,2,\ldots,n\}$ then $\sum_{i: i+1\in S,i\notin S} 1\geq 1$ so that the RHS $\geq \sum_{S\neq\emptyset,S\neq [n]} \vert S\vert x_S$.
hence if $x_{[n]}=0$ we arrive at a contradiction.
this time i'm sure everything is fine.

[quote="K09"]Thanks for you.You have tried to solve my problem but I donnot understand .I have several question for yo.Can you explain? May be I am stupid?
1,What is $\overline{A_{i}}$
2,Why $ \sum_{S|i,i+1\in S}x_{S}>(1-\frac{1}{n-1})\sum_{S|i+1\in S}x_{S}\forall i $
3,If $S=(1,2,...,n-1)$ then $\sum_{i|{i+1}\in S,i\notin S}1=0$[/quote]

ok. first $\overline{A}$ means complement of $A$.
next,the next condition comes as a simplification of the given condition.let me explain a little more now:
the given condition says $\vert A_i\cap A_{i+1}\vert >\frac{n-2}{n-1} \vert A_{i+1}\vert$. in terms of $x_S$'s if we write out these expressions we get
  $\vert A_i\cap A_{i+1}\vert=\sum_{i,i+1\in S} x_S$.likewise $\vert A_{i+1}\vert = \sum_{i+1\in S} x_S$.
think of a venn diagram with the numbers $x_S$ written in all the distinct regions created.so if you want to get the size of $A_i\cap A_{i+1}$ you add all the terms $x_S$ contained in $A_i\cap A_{i+1}$ and that is precisely the sum here.and $\frac{n-2}{n-1} =1-\frac{1}{n-1}$ so that should answer your second question.
finally, the problem needs a 'circular condition'(so your example is not actually appropriate),i.e. we need to have $A_{n+1}=A_1$ and the corresponding inequality in this case also.otherwise the statement of the problem is not true:
$A_1=\{1,2,3,4,5\},A_2=\{3,4,5,6,7\},A_3=\{6,7,8\}$.this will be a counterexample.
in fact the conditon(i.e. the constant $\frac{n-2}{n-1}$) cannot be bettered in the general sense. also we need strict inequality or else we could alter $A_3$ in the above example to be $\{6,7,1,2\}$ and that again gives a counterexample.
hopefully your doubts have been answered.

Complement with ? What is complement of $A_i$.I didnot say about circle,I think   I said  last two inequallity of your solution?

oh iam sorry.I think iwas very tired.I understood .Thank you .But I think you should explain clearly for  reader in forum.THANKS

Another solution, in case someone is interested:

We will prove a stronger (but easier to establish...) assertion:

[color=blue][b]Problem.[/b] Let n finite sets $A_1$, $A_2$, ..., $A_n$ satisfy

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\right|>\frac{n-2}{n-1}\sum_{i=1}^n\left|A_i\right|$,

where indices are cyclic modulo n (so that, in particular, $A_{n+1}=A_1$).

Then, prove that $\left|A_1\cap A_2\cap ...\cap A_n\right|>0$.[/color]

[i]Solution.[/i] We will prove:

[color=blue][b]Lemma 1.[/b] For every integer r with $1\leq r<n$, we have

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r}\right|>\frac{n-1-r}{n-r}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$.[/color]

[i]Proof of Lemma 1.[/i] We will prove Lemma 1 by induction over r:

[i]Induction base:[/i] For r = 1, the assertion of Lemma 1 becomes $\sum_{i=1}^n\left|A_i\cap A_{i+1}\right|>\frac{n-2}{n-1}\sum_{i=1}^n\left|A_i\right|$, what is true since it is a condition of the problem.

[i]Induction step:[/i] Now, assume that for some integer r with $2\leq r<n$, we know that Lemma 1 is true for r - 1, i. e. that we have

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|>\frac{n-r}{n-r+1}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-2}\right|$,

and we have to prove that Lemma 1 also holds for r, what means that we have to prove that

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r}\right|>\frac{n-1-r}{n-r}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$.

Now, rewrite the inequality

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|>\frac{n-r}{n-r+1}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-2}\right|$

as

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-2}\right|<\frac{n-r+1}{n-r}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$.

On the other hand,

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r}\right|=\sum_{i=1}^n\left|\left(A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right)\cap\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right)\right|$
$=\sum_{i=1}^n\left(\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|+\left|A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right|-\left|\left(A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right)\cup\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right)\right|\right)$
$=\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|+\sum_{i=1}^n\left|A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right|-\sum_{i=1}^n\left|\left(A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right)\cup\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right)\right|$.

Now, we have $\sum_{i=1}^n\left|A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right|=\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$ (just a cyclic permutation of the index). Similarly, $\sum_{i=1}^n\left|A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r-1}\right| = \sum_{i=1}^n\left|A_{i}\cap A_{i+1}\cap ...\cap A_{i+r-2}\right|$.

Also, every $i\in\left\{1,2,...,n\right\}$ satisfies

$\left|\left(A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right)\cup\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right)\right|\leq\left|A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r-1}\right|$

(since $\left(A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right)\cup\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right)=\left(A_i\cup A_{i+r}\right)\cap\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r-1}\right)$ is a subset of $A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r-1}$). Hence,

$\sum_{i=1}^n\left|\left(A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right)\cup\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right)\right|$
$\leq\sum_{i=1}^n\left|A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r-1}\right|$
$=\sum_{i=1}^n\left|A_{i}\cap A_{i+1}\cap ...\cap A_{i+r-2}\right|<\frac{n-r+1}{n-r}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$.

Thus,

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r}\right|$
$=\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|+\sum_{i=1}^n\left|A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right|-\sum_{i=1}^n\left|\left(A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right)\cup\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right)\right|$
$>\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|+\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|-\frac{n-r+1}{n-r}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$

(because of $\sum_{i=1}^n\left|\left(A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right)\cup\left(A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right)\right| < \frac{n-r+1}{n-r}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$ and $\sum_{i=1}^n\left|A_{i+1}\cap A_{i+2}\cap ...\cap A_{i+r}\right|=\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$). This gives

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r}\right|$
$>\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|+\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|-\frac{n-r+1}{n-r}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$
$= \left(2-\frac{n-r+1}{n-r}\right)\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|=\frac{n-1-r}{n-r}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+r-1}\right|$.

Hence, we have proven Lemma 1 for r, and thus, the induction proof of Lemma 1 is complete.

Now, applying Lemma 1 to the integer r = n - 1 (with $1\leq n-1<n$), we get

$\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+\left(n-1\right)}\right|>\frac{n-1-\left(n-1\right)}{n-\left(n-1\right)}\sum_{i=1}^n\left|A_i\cap A_{i+1}\cap ...\cap A_{i+\left(n-1\right)-1}\right|$.

Now, the left hand side of this inequality is a sum of n equal terms, all equal to $\left|A_1\cap A_2\cap ...\cap A_n\right|$ (this is because $A_i\cap A_{i+1}\cap ...\cap A_{i+\left(n-1\right)}=A_1\cap A_2\cap ...\cap A_n$ for every index i), while the right hand side is 0 (since n - 1 - (n - 1) = 0). Hence, we get $n\left|A_1\cap A_2\cap ...\cap A_n\right|>0$, so that $\left|A_1\cap A_2\cap ...\cap A_n\right|>0$. And the problem is solved.

  darij

Thinks the sets on a circle, then the condition is equivalent to  "the probability of choosing a number from some set which it is in the preceding one is greater than n-2/n-1 ".
 So there must be some element on the union of some set with probability greatyer than n-2/n-1". of choosing it and being on the preceding subset, this one must be on all subset.

 Ok, i wasnt very clear, but thats the main point.


[quote="math92"]An auditorium that holds 2,500 people is 30% empty. How many people are in the
auditorium?[/quote]

[hide]  If an auditorium if 30% empty it must be 70% full.  Meaning there are 2500 * .7 = [b]1750 people inside the auditorium.[/b][/hide]


Construct complete quad given 3 diagonal lengths.

  T.Y.
  M.T.

It's just according to three circle have the same axis，and the Orthocenter' proportion of Orthocenter Group .
But the time is too long，I have forget something，Next I will solve it.

I just have made a picture and found this problem actually is easy .


[quote="Potla"]For $ a,b,c > 0$, show that
\[ \frac a{5a \plus{} 4b \plus{} 3c} \plus{} \frac b{5b \plus{} 4c \plus{} 3a} \plus{} \frac c{5c \plus{} 4a \plus{} 3b} < \frac 19 \left[\frac a{a \plus{} b} \plus{} \frac b{b \plus{} c} \plus{} \frac c{c \plus{} a}\right] \plus{} \frac {10}{27}\]
It is really easy , and I hope that
[hide]You will not laugh at me... .:( [/hide][/quote]

Thank you Sayan.

\[ \sum{\frac{a}{5a \plus{}4b \plus{}3c}} <\sum{\frac{a}{3a \plus{}3b \plus{}3c}}\equal{}\frac{1}{3} <\frac{10}{27} <\frac{1}{9}\left(\frac{a}{a\plus{}b} \plus{}\frac{b}{b\plus{}c} \plus{}\frac{c}{c\plus{}a} \right) \plus{}\frac{10}{27}\]

I'd like to see your proof !

[quote="enndb0x"][quote="Potla"]For $ a,b,c > 0$, show that
\[ \frac a{5a \plus{} 4b \plus{} 3c} \plus{} \frac b{5b \plus{} 4c \plus{} 3a} \plus{} \frac c{5c \plus{} 4a \plus{} 3b} < \frac 19 \left[\frac a{a \plus{} b} \plus{} \frac b{b \plus{} c} \plus{} \frac c{c \plus{} a}\right] \plus{} \frac {10}{27}\]
It is really easy , and I hope that
[hide]You will not laugh at me... .:( [/hide][/quote]

Thank you Sayan.
\[ \sum{\frac {a}{5a \plus{} 4b \plus{} 3c}} < \sum{\frac {a}{3a \plus{} 3b \plus{} 3c}} \equal{} \frac {1}{3} < \frac {10}{27} < \frac {1}{9}\left(\frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a} \right) \plus{} \frac {10}{27}\]
I'd like to see your proof ![/quote]
Thanks for your solution, but my inequality seems to be unexpectedly weak..... :(

However, I proved it using Cauchy schwarz:
\[ \sum{\frac {a}{5a \plus{} 4b \plus{} 3c}} \equal{} \sum \frac a{(a \plus{} b) \plus{} 3(a \plus{} b \plus{} c) \plus{} a}\]

\[ < \sum \left[\frac 19 \left(\frac a{a \plus{} b} \plus{} \frac a{3(a \plus{} b \plus{} c)} \plus{} \frac aa\right)\right]\]

\[ \equal{} \frac 19 \sum \frac a{a \plus{} b} \plus{} \frac 19 \left(3 \plus{} \frac 13\right) \equal{} \frac 19 \sum \frac a{a\plus{}b}\plus{}\frac {10}{27}\]
Also, Can_hang 2007 proved two stronger results by cauchy schwartz (I do not know the proof):
\[ (a) \sum{\frac {a}{5a \plus{} 4b \plus{} 3c}} < \frac 19 \sum \frac a{a \plus{} b} \plus{} \frac 16\]

\[ (b) \sum{\frac {a}{5a \plus{} 4b \plus{} 3c}} < \frac 19 \sum \frac a{a \plus{} b} \plus{} \frac {24 \minus{} 7\sqrt {5}}{85}\]
Where $ \frac {24 \minus{} 7\sqrt {5}}{85} \approx 0.0982061\cdots$

Can anyone please give a hint on finding the best positive constant that satisfies
\[ \frac a{5a \plus{} 4b \plus{} 3c} \plus{} \frac b{5b \plus{} 4c \plus{} 3a} \plus{} \frac c{5c \plus{} 4a \plus{} 3b} < \frac 19 \left[\frac a{a \plus{} b} \plus{} \frac b{b \plus{} c} \plus{} \frac c{c \plus{} a}\right] \plus{} k\]
Thanks a lot for your interest, enndb0x and can_hang :) .

[quote="Potla"]
\[ (a) \sum{\frac {a}{5a \plus{} 4b \plus{} 3c}} < \frac 19 \sum \frac a{a \plus{} b} \plus{} \frac 16\]
[/quote]
The proof for this problems is as follows:

Since $ 5a \plus{} 4b \plus{} 3c \equal{} 2(a \plus{} b) \plus{} 2(a \plus{} c) \plus{} (a \plus{} 2b \plus{} c),$ by Cauchy-Schwarz Inequality, we have
$ \frac {1}{5a \plus{} 4b \plus{} 3c} \le \frac {1}{9}\left[ \frac {1}{2(a \plus{} b)} \plus{} \frac {1}{2(a \plus{} c)} \plus{} \frac {1}{a \plus{} 2b \plus{} c}\right] < \frac {1}{9} \left[ \frac {1}{2(a \plus{} b)} \plus{} \frac {1}{2(a \plus{} c)} \plus{} \frac {1}{a \plus{} b}\right].$
Thus,
$ \sum \frac {a}{5a \plus{} 4b \plus{} 3c} \le \frac {1}{9}\left[ \frac {1}{2} \sum \left( \frac {a}{a \plus{} b} \plus{} \frac {a}{a \plus{} c}\right) \plus{} \sum \frac {a}{a \plus{} b}\right] \equal{} \frac {1}{6} \plus{} \frac {1}{9}\sum \frac {a}{a \plus{} b}.$

I will post the proof for (b) later.


induction
combinatorics unsolved
combinatorics

Let $n$ be a fixed positive integer. The points $A_1$, $A_2$, $\ldots$, $A_{2n}$ are on a straight line. Color each point blue or red according to the following procedure: draw $n$ pairwise disjoint circumferences, each with diameter $A_iA_j$ for some $i \neq j$ and such that every point $A_k$ belongs to exactly one circumference. Points in the same circumference must be of the same color.

Determine the number of ways of coloring these $2n$ points when we vary the $n$ circumferences and the distribution of the colors.

This was an easy number six.

First, every circle goes from an odd point to an even point, so there is aan equal number of odd blue points and even blue points and the same for red.

There are $\binom{2n}{n}$ permutations satisfying that.

Now we prove all of them can be obtained by the coloring process.

We make induction on $n$

Clearly there are two consecutive points of the same color, putting a circle trought them, and removing them leaves a nice scenario for $n-1$ case

So we are done

Can you post other problems of test this?Thank you!

They are allready posted here:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=339857#339857]HERE[/url]

[quote="Pascual2005"]This was an easy number six.

[/quote]

It becomes easy only when I see the solution..  ;)


synthesize dibenzcoronene from benzene in minimal steps.

can u please give  the structure or IUPAC name of dibenzcoronene

it is a huge molecule (9) benzene rings .do you have a software to draw benzene rings?I think its iupac name is the same(eg.azulene,anthracene etc)

how about trying to approach by the benzyne mechanism to combine the two benzene rings (by having small concentration of nucleophile in the mediium)

Repeated Diels Alder in the presence of Pd/C

consider two benzene molecules. after diels alder reaction i think they form something like biphenylene{as in the benzyne mechanism with little nucleophile) and not naphthalene.
 the diels alder reaction probably gives alternate six membered and 4 membered rings and not continuous six membered rings as u say.

i did not ask to use two benzene molecules.  Benzenes
 do not give Diels Alder since the loss od aromaticity makes
 enthalpy positive. for a Diles - Alder synthesis , the
 entropy change is -ve. Under such conditions, if
 you use two benzene rings, the reactin will never occur.
 Use branched diene and branched dienophile.


calculus
integration
function
search
derivative
Galois Theory
real analysis

Compute:

$\displaystyle \int e^{x^{2}} dx.$

Yes, please! Why do you ask? Is it very hard?

Something tells me that it is impossible.

And is there a way I can prove that it is impossible? :D

I know that there is criterion which allow to determine is it possible to compute indefinite integral via elementary (basic) functions (cos, x^n, e^x, etc.).
I think you should ask anyone who knows this criterion.

${\int e^{x^{2}}\, dx=\frac{1}{2}\sqrt{\pi }\text{erfi}(x)=\frac{1}{2}(-i) \sqrt{\pi }\text{erf}(x i)}$

http://en.wikipedia.org/wiki/Error_function

search something about differential galois theory and this will prove you what you want
http://en.wikipedia.org/wiki/Differential_galois_theory

And that would perhaps be the easiest way to check weather the anti derivative is elementary, compare its taylor series with a library of the series of elementary functions. If it is not there, then it is not elementary.


symmetry
geometry
3D geometry
number theory
relatively prime
number theory proposed

Let integers $ a,x,y,z,t$ satisfying $ a^3x \plus{} a^2y \plus{} az \plus{} t$ is divisible by 5. Prove that there exists integer $ b$ satisfying $ x\plus{} yb \plus{} zb^2 \plus{} tb^3$ is divisible by 5.

There is a missing condition : if $ a,y,z,t$ are divisible by $ 5$ but not $ x$ the result cannot hold.

I prove the result with the extra condition that $ a \neq 0 \mod[5]$.
Then $ a^2 \equal{} 1 \mod[5]$ or $ a^2 \equal{} \minus{} 1 \mod[5]$.

- In the first case, we have
$ x\plus{}ay\plus{}a^2z\plus{}a^3t \equal{} a^6x\plus{}a^5y\plus{}a^4z\plus{}a^3t \mod[5]$
$ \equal{}a^3(a^3x\plus{}a^2y\plus{}az\plus{}t)$
$ \equal{}0 \mod[5]$.
Thus, we may choose $ b\equal{}a.$

- In the second case, we have
$ x\plus{}(\minus{}a)y\plus{}a^2z\plus{}(\minus{}a)^3t \equal{}\minus{}( a^6x\plus{}a^5y\plus{}a^4z\plus{}a^3t) \mod[5]$
$ \equal{}\minus{}a^3(a^3x\plus{}a^2y\plus{}az\plus{}t)$
$ \equal{}0 \mod[5]$.
Thus, we may choose $ b\equal{}\minus{}a.$

Pierre.

[quote="mr.danh"]Let integers $ a,x,y,z,t$ satisfying $ a^3x \plus{} a^2y \plus{} az \plus{} t$ is divisible by 5. Prove that there exists integer $ b$ satisfying $ x \plus{} yb \plus{} zb^2 \plus{} tb^3$ is divisible by 5.[/quote]
I believe that we also need for $ y^2\minus{}4xz\equiv 0, \pm 1\bmod 5$ to hold true.
[hide="Solution"]
We have that if $ a$ is not a multiple of $ 5$, then $ a$ is relatively prime to $ 5$, so we consider $ b\equiv \frac{1}{a}\bmod 5$, meaning that $ x\plus{}yb\plus{}zb^2\plus{}tb^3\equiv x\plus{}\frac{y}{a}\plus{}\frac{z}{a^2}\plus{}\frac{t}{a^3}\bmod 5$.  Yet, $ 5$ is not a factor of $ b$ or $ a$, so the expression is equivalent to $ a^3x\plus{}a^2y\plus{}za\plus{}t\equiv 0\bmod 5$.  Now, if $ a$ is a mutliple of $ 5$, then we can see that $ a^3x\plus{}a^2y\plus{}az\plus{}t\equiv t\bmod 5$, so $ 5|t$.  It follows that $ x\plus{}yb\plus{}zb^2\plus{}tb^3\equiv zb^2\plus{}yb\plus{}x\bmod 5$.  If $ z\equiv 0\bmod 5$, then we could make $ b\equiv \frac{\minus{}x}{y}\bmod 5$ as long as $ 5$ is not a factor of $ y$.  If it is, then we would see that $ 5$ must be a factor of $ x$ by [b]pbornsztein[/b]'s condition (otherwise, the problem would be false), so $ 5$ would be a factor of the expression for all $ b$.

Now, consider if $ z$ were not a multiple of $ 5$.  We want to find $ b$ so that $ 4z^2b^2\plus{}4yzb\plus{}4xz\equiv 0\bmod 5$, so $ (2zb\plus{}y)^2\minus{}(y^2\minus{}4xz)\equiv 0\bmod 5$.  It follows that we wish for $ (2zb\plus{}y)^2\equiv y^2\minus{}4xz\bmod 5$.  Now, we need $ y^2\minus{}4xz\equiv 0, \pm 1\bmod 5$ (this is not always true, for example: $ y\equiv 1\bmod 5$, $ z\equiv 3\bmod 5$, and $ x\equiv 2\bmod 5$; I believe that we also need this condition to be added to the problem for the problem statement to hold).  Once we can establish this, we can let $ y^2\minus{}4xz\equiv k\bmod 5$, so $ 2zb\plus{}y\equiv k^2\bmod 5$, so $ b\equiv \frac{k\minus{}y}{2z}\bmod 5$, which is possible since $ 5$ is not a factor of $ z$ and $ 2$ and $ 5$ are relatively prime.  We have found a soution for $ b$, so we are done. [/hide]

with the condition that $ a$ is not divisible by 5 it is easy because there exist $ a^{\minus{}1}$(mod5).the condition 5|$ a^3x\plus{}a^2y\plus{}az\plus{}t$ implies directly the answer by multiplying it with the cube of $ a^{\minus{}1}$


An $ 18^\prime \times 18^\prime$ carpet is partitioned into $ 3^\prime\times 3^\prime$ squares. How many squares will there be?

$ \frac{18}{3}$=6
So 6 squares on each side
now $ 6^2\equal{}36$ so $ 36$ total.


Find:
$ \displaystyle\sum_{i\equal{}1}^{160} \frac{(9i\minus{}4)}{(3i\minus{}2)(3i\minus{}1)(3i)}$

hello, rewrite $ \frac{9i\minus{}4}{(3i\minus{}2)(3i\minus{}1)3i}$ as $ \frac{1}{3i\minus{}1}\minus{}\frac{2}{3i}\plus{}\frac{1}{3i\minus{}1}$.
Sonnhard.

Dr. Graubner, there is a typo in your post. I think you meant to say $ \frac{1}{3i \minus{} 1} \plus{} \frac{1}{3i \minus{} 2} \minus{} \frac{2}{3i}$.

In any case, I'm going to guess that there's a typo in the problem. According to [url=http://www34.wolframalpha.com/input/?i=sum+(9a+-+4)+%2F+((3a+-+2)(3a+-+1)(3a))%2C+a+%3D+1+to+160]Wolfram|Alpha[/url], this doesn't have a very nice final result.


inequalities
geometry
circumcircle
incenter
inequalities open

$\ a\geq b$,
$\ b\geq c$,
$\ a\geq c$
so $\ 2a\geq b+c$ $\rightarrow$ $\frac{a}{b+c}\geq \frac{1}{2}$ $\rightarrow$ $\sqrt{\frac{a}{b+c}}\geq \frac{1}{\sqrt{2}}$

So it follows that:$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}\geq \frac{3}{\sqrt{2}}$

i don't quite understand your solution. Just because $\sqrt{\frac{a}{b+c}}\geq \frac{1}{\sqrt{2}}$ does not mean the other two terms are greater than or equal to $\frac{1}{\sqrt{2}}$. In fact, we will have $\frac{c}{a+b}\le \frac{1}{2}$ and so this term is not greater than or equal to $\frac{1}{\sqrt{2}}$ :wink:

[quote="Alisher"] \[a,b,c-\] is sides of triangle

prove: \[\sqrt{\frac{a}{{b+c}}}+\sqrt{\frac{b}{{a+c}}}+\sqrt{\frac{c}{{a+b}}}\ge \frac{3}{{\sqrt 2 }}\] [/quote]
It is wrong. Try $a=b=14$ and $c=1.$ :wink:

Yes, it's wrong. We just can prove that

$\sum \sqrt{\frac{a}{b+c}}> 2$

by using $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$.

The inequality cant be hold, but we cant replace 2 by bigger constant (take a = b and c --> 0)

If I'm not wrong, it was Romanian MO problem.

If $a,b,c$ are the sides of a  triangle, $R$ is the radius of circumcircle, $r_{a},r_{b},r_{c}$ are the radii of  excircles, then

\[\sqrt{\frac{a}{{b+c}}}+\sqrt{\frac{b}{{a+c}}}+\sqrt{\frac{c}{{a+b}}}\ge \frac{\sqrt 2}{3R}(r_{a}+r_{b}+r_{c}).\]

x=(b+c-a)/2,y=(a+c-b)/2,z=(a+b-c)/2,  as you see x>0,y>0,z>0
so we have we have to prove
sqrt((y+z)/(2x+y+z))+sqrt((x+z)/(x+2y+z))+sqrt((x+y)/(x+y+2z))>=
3/sqrt(2)
if s=x+y+z
sqrt((s-x)(s+x))+sqrt((s-y)(s+y))+sqrt((s-z)(s+z))>=3/sqrt(2)
let x>=y>=z
we see that  (sqrt((s-x)(s+x)))''=s(s-2x)((s+x)^(-5/2))((s-x)^(-3/2))
as s-2y>0 and s-2z>0 we can use inequality Iensena
so we can say y=z=(s-x)/2
after that we only have to prove
sqrt((s-x)/(s+x))+2sqrt((s+x)(3s-x))>=3/sqrt(2)  [/hide]

If $a,b,c$ are the sides of a  triangle, $r$ is the radius of incircle, $IA,IB,IC$ are the distances of  incentre $I$ to vertices $A,B,C$, then
\[\sqrt{\frac{a}{{b+c}}}+\sqrt{\frac{b}{{a+c}}}+\sqrt{\frac{c}{{a+b}}}\geq \sqrt{2}r\left(\frac 1{IA}+\frac 1{IB}+\frac 1{IC}\right).\]

[quote="Shishkin"]$x=\frac{b+c-a}{2}$,$y=\frac{a+c-b}{2}$, $z=\frac{a+b-c}{2}$, as you see $x>0,y>0,z>0$
so we have to prove
$\sqrt{\frac{y+z}{2x+y+z}}+\sqrt{\frac{x+z}{x+2y+z}}+\sqrt{\frac{x+y}{x+y+2z}}\ge \frac{3}{\sqrt{2}}$
if $s=x+y+z$
$\sqrt{\frac{s-x}{s+x}}+\sqrt{\frac{s-y}{s+y}}+\sqrt{\frac{s-z}{s+z}}\ge \frac{3}{\sqrt{2}}$
we see that $(\sqrt{\frac{s-x}{s+x}})''=(-s(s-x)^{\frac{-1}{2}}(s+x)^{\frac{-3}{2}})'= (s-x)^{\frac{-3}{2}}*(s+x)^{\frac{-5}{2}}*(s^{2}-xs+x^{2})>0$
so using inequality Yensena we get what we need.[/quote]
I think
$\left(\sqrt{\frac{s-x}{s+x}}\right)''=\cdots$
is wrong.
Actually, we have
$\left(\sqrt{\frac{s-x}{s+x}}\right)''=(s-x)^{-3/2}(s+x)^{-5/2}(s^{2}-2sx)$
And actually we have
$S=\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2$
and let $a,b\rightarrow 1, c\rightarrow 0$ then $S\rightarrow 2$
So this inequality is also wrong!

I'm sory.I have made a mistake second time and the first was true.I'll continue it.

let $x \ge y \ge z$
we see that $(\sqrt{\frac{s-x}{s+x}})''=s(s-2x)(s+x)^{-5/2}(s-x)^{-3/2}$
as $s-2y \ge 0$ and $s-2z \ge 0$ we can use inequality Iensena
so we can say $y=z=\frac{s-x}{2}$
after that we only have to prove
$\sqrt{\frac{s-x}{s+x}}+2\sqrt{\frac{s+x}{3s-x}}\ge K$
$(\sqrt{\frac{s-x}{s+x}}+2\sqrt{\frac{s+x}{3s-x}})'=-s(s-x)^{-\frac{1}{2}}(s+x)^{-\frac{3}{2}}+4s(s+x)^{-\frac{1}{2}}(3s-x)^{-\frac{3}{2}}=s(s+x)^{-\frac{3}{2}}(s-x)^{-\frac{1}{2}}(3s-x)^{-\frac{3}{2}}(4(s+x)(s-x)^{\frac{1}{2}}-(3s-x)^{\frac{3}{2}})=0$
$16(s+x)^{2}(s-x)=(3s-x)^{3}$
as $2s \ge 2x \ge s$
$x=\frac{14+-\sqrt{86}}{10}s$ so it is monotonic on (s/2;s)
and min will be when s=x
so K=2 is the best.


1. How many possible poker hands are there where
a) You have 1 and only 1 pair.
b) There are no pairs (so the 5 cards are all different)

2. How many distinct four letter words (doesn't have to be real words) can you form from the word parallelogram?

EDIT: I realized that question one might be a bit unclear. Poker hands are 5-card hands and part a says you only have one pair (not that you have the number 1 and one more pair).

#2
[hide]

P A R A L L E L O G R A M
1 3 2    3   1    1 1       1

Total = 8 distinct words

Now we can either form 4 letter words with all letters being distinct, 2 letters being the same and rest 2 being different, 2 sets of 2 same letters, and finally 3 same letters and 1 different... the ways of doing them are respectively

8x7x6x5 = 1680
4 x 7c2 x 4!/2! = 1008
4 x 3 x 4!/2!2!  = 72
2 x 7c1 x 4!/3! = 56

total = 2806 ways[/hide]

[quote="computer_nuke"]#1

[hide]
a) (13C1)(4C2)(12C3)(4C1)(4C1)(4C1)

b) (13C5)(4C1)(4C1)(4C1)(4C1)(4C1)
[/hide][/quote]

Could you please explain your results?  :)

[hide]#1
There are 13 possible values the pair could be, and you can pick any 2 out of the 4 possible cards of a certain value to get a pair. Then, since all the other cards have to be of different values but can be any 1 of the four cards of that different value, the total answer is:
$(13)\binom{4}{2}(12)(4)(11)(4)(10)(4)$
$= 6589440$
[/hide]


What subsets of the unit circle are possible as the set of points at which a power series $ f(z) \equal{} \sum c_n z^n$ with radius of convergence $ 1$ converges?

Harazi has already asked this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68078]here[/url]. Alas, I haven't heard of any progress since then. :(

I'm not an expert of this topic, but if we can't answer this question maybe we can answer the following:
Do we know subsets of the unit circle that cannot be exactly the set of points at which any power series with radius of convergence $ 1$ converges?

Oh, plenty of them. The set of convergence is always $ F_{\sigma\delta}$ (just write the Cauchy sequence condition) and, as far as I remember, it is not too hard to show that every $ F_\sigma$ set is a set of convergence. Unfortunately, this doesn't help much :(.


algebra
polynomial
calculus
integration
real analysis
real analysis unsolved

if for $ c_0,c_1,\dots,c_n\in \mathbb{R}$ we have :$ c_0\plus{}\frac{c_1}{2}\plus{}\dots\plus{}\frac{c_{n\minus{}1}}{n}\plus{}\frac{c_n}{n\plus{}1}\equal{}0$ prove that :
$ c_0\plus{}c_1x\plus{}\dots\plus{}c_{n\minus{}1}x^{n\minus{}1}\plus{}c_nx^n\equal{}0$ has at least one root in $ [0,1]$

Integrate you polynom from $ 0$ to $ 1$, if it hasn't roots the value of integral will be positive or negative(strictly), so we have contradaction.

you can also do this by using the mean value theorem $ f'(c) = \frac{f(b)-f(a)}{b-a}$ for $ c$ in $ (a,b)$  with $ f(x) =c_0x+\frac{c_1x^2}{2}+\dots+\frac{c_{n-1}\&x^n}{n}+\frac{c_nx^{n+1}}{n+1}=0$ and letting a = 0, b = 1


Starting $n=1$?  Isn't this just plug and chug?  I don't think this belongs in the Olympiad section (and I know that these are DEFINITELY not olympiad problems.)

[quote="i_like_pie"][quote="suedenation"][color=olive]Find the smallest prime in the arithmetic progression an + b , where
a) a = 5, b = 1
b) a = 7, b = 2
c) a = 23, b =13[/color]

Thanks very much guys!!![/quote]
[hide]$n=2$

$n=3$

$n=2$[/hide][/quote]

Edit:
ummm... terrible misread of the question,  :oops:  the primes are $11, 23, 59$ agreeing with i_like_pie

Wow, very hard problem ;)

Moved to Getting Started now.

My $n$ values are correct, but the question asked for the prime numbers. They are, as tim1234133 said, $11,\ 23,\ 59.$

[quote="i_like_pie"][quote="suedenation"][color=olive]Find the smallest prime in the arithmetic progression an + b , where
a) a = 5, b = 1
b) a = 7, b = 2
c) a = 23, b =13[/color]

Thanks very much guys!!![/quote]
[hide]$n=2$

$n=3$

$n=2$[/hide][/quote]

I agree with i_like_pie. i got the same


1. Imagine holding a really light object on top of a really heavy object, and then dropping both of them at the same time onto the ground. If all the collisions are elastic, and the objects are dropped from height of h, how high will the little object bounce?

2. In the previous question, what should be the ratio of the masses if the bottom mass had no velocity after all the collisions? How high would the little mass bounce?

Could someone please show me the answers and how you got them??

thx a bunch!

Well if one is on top of the other, I don't see how they can be at the same height.  But otherwise, use your kinematic equations to determine the velocity of each, and conservation of momentum and so on.


There is a theorem that states "If two chords intersect in a circle, then the products of the measures of the segments of the chords are equal", my geometry teacher told me 2 prove it 4 hw & i am stumped
[img]http://intermath.coe.uga.edu/newInterMath/algebra/functns/resrces/intchord.gif[/img]

wut would i do after i prove the 2 triangles similar?

this type of problem does not belong in this forum...please post these types of questions in getting started or intermediate

That theorem says:

$a \cdot n = b \cdot m$

To prove it, just unite $C$ with $E$, $D$ with $F$, then you gotta find a couple of similar triangles.

Let $A$ be the intersection of both chords.


geometry
circumcircle
incenter
angle bisector
geometry proposed

Given a triangle $ ABC$ and its circumcircle$ O$,Let $ D$ be a point on the plane,and a circle$ O'$ touches $ DB$ $ DC$ circle$ O$ at $ M,N,A$ repectively,
prove the angle bisector of $ <BAC$ passes through the incenter of triangle $ DBC$

[quote="jayme"]Dear Mathlinkers,
see for example,
http://perso.orange.fr/jl.ayme    vol. 4  A new mixtilinear incircle adventure p.3-6 the result is on page 6.
Sincerely[/quote]
Oh,Thank you,your article included so many interesting result!But can you translate your solution into English and post it here,I am eager to see it :)

p.s.Did tthese articles have English version?

My article haven't any traduction in English, but with a little effort they have so many figures with the most possible standart notations that you are able to follow the proof...
Perhaps in the future...
Sorry
Sincerely
Jean-Louis

But can you just post the solution for this topics in English?Because I think your solution seems very synthetic which is much different from which I knew.
Tell you the truth I have knew three solutions for these problems but all of them used complicated trigonometric  :(
Thank you so much!

anybody any ideas?
I am waiting for your beautiful proof :)

Last week I finally found a solution using [u]Sawayama Theorem[/u]
After applying it,alll we need to do is an angle chasing :)
a link for Sawayama's: forumgeom.fau.edu/FG2003volume3/FG200325.pdf


geometry
3D geometry
combinatorics proposed
combinatorics

A $ 20\times20\times20$ block is cut up into 8000 non-overlapping unit cubes and a number is assigned to each. It is known that in each column of 20 cubes parallel to any edge of the block, the sum of their numbers is equal to 1. The number assigned to one of the unit cubes is 10. Three $ 1\times20\times20$ slices parallel to the faces of the block contain this unit cube. Find the sume of all numbers of the cubes outside these slices.

The sum of all numbers is 400. The sum of numbers in each slice is 20. The intersection of any two slices is 1. Hence, by the inclusion-exclusion principle, we get the answer: $ 400\minus{}3\cdot 20\plus{}3\cdot1 \minus{}10\equal{}333$.


are the products correct ? or will the product B be a lactone?

My I ask by what mechanism does your A forms?


So what were the top 6 scores at the contest?

No, not Miller, I'm talking about the mathleague competition at UC Berkeley yesterday.

i know the top 3 scores were 185, 186, and 192.
i forget the rest.

[quote="Nerd_of_the_Ages"]Are you talking about the one that was in Miller MS?[/quote]

First place 7th grade, 2nd place CD :D

Berkeley was too far away, so I didn't go  :(

anyone have the problems?

I left early, but I know I got 6 on Target, for 60 points, and hopefully 22-24 on Sprint, and about 4 wrong, for 60+88=148.  This is for the mathleague.org one.

And to james, I have the problems :D but IDK if im allowed to post them...

Oh god, you're telling me they're copying Mathcounts now?? I'm glad I didn't go.

Well it's 10 points for each Target question, 4 pairs of questions, 10 minutes per pair, 30 questions on sprint, 4 points apiece, -1 if wrong answer, and it's multiple choice, and a relay round, and team so it's not like a TOTAL copy xD

[quote] The fact that they stole the names from Mathcounts of all places implies crappiness.
[/quote]

....maybe, but the problems are much harder, and our arml a team is awesome  :P

Well, the problems are harder than MATHCOUNTS, but at an ARML standpoint, these problems are almost ridiculously easy.

Hmmm....well idk, last year the qualifying rounds were ridiculously easy compared to ARML, but I feel that this year's contests are not all that much easier than ARML...But that's just my opinion...

[quote="DataBox"]but I feel that this year's contests are not all that much easier than ARML..[/quote]

Yeah, but I'm never gonna get a 6 on an ARML individual round... :\


conics
ellipse
analytic geometry
geometry
geometric transformation

What are the coordinates of the foci of the ellipse with equation?
$ 3x^{2} \plus{} 4y^{2} \minus{} 6x \plus{} 16y \plus{} 7 \equal{}0$

I can't figure it out  :huh:

$ 3x^2 \plus{} 4y^2 \minus{} 6x \plus{} 16y \plus{} 7 \equal{} 0$

$ \Longleftrightarrow 3(x \minus{} 1)^2 \plus{} 4(y \plus{} 2)^2 \equal{} 12$

$ \Longleftrightarrow \frac {(x \minus{} 1)^2}{4} \plus{} \frac {(y \plus{} 2)^2}{3} \equal{} 1$

This ellipse can be obtained by paralell translation of the ellipse $ \frac{x^2}{4}\plus{}\frac{y^2}{3}\equal{}1$

As it turns out, the foci can be computed by taking the center of the ellipse, and going $ \sqrt{a^2\minus{}b^2}$ away from the center towards the major axes. The distance is $ 1$, and the major axis is the $ x$-axis, so the foci are at $ (0, \minus{}2)$ and $ (2, \minus{}2)$ unless I screwed up somehow.

First you must factor it.

$ 3x^2 \minus{} 6x \plus{} 3 \plus{}  4y^2 \plus{} 16y \plus{} 16 \equal{} \minus{}7 \plus{} 16 \plus{} 3$

$ 3(x^2 \minus{} 2x \plus{} 1) \plus{} 4(y^2 \plus{} 4y \plus{} 4) \equal{} 12$

$ 3(x\minus{}1)^2 \plus{} 4(y\plus{}2)^2 \equal{} 12$

$ \frac{(x\minus{}1)^2}{4} \plus{} \frac{(y\plus{}2)^2}{3} \equal{} 1$

So $ a \equal{} 2$ and $ b \equal{} \sqrt{3}$. We know that the foci are $ (1 \pm c, \minus{}2)$, and since $ c \equal{} \sqrt{a^2 \minus{} b^2} \equal{} \sqrt{4 \minus{}3} \equal{} 1$, we have that the foci are

$ \boxed{(2,\minus{}2) \ and \ (0,\minus{}2)}$


Do men have the same number of sisters as women?

On average, no

In a given family, each woman has one fewer sister than each man does.

Heh, nice try, but you're actually wrong.  After all, try the first two cases:  if we have 1-child families, there are 0 sisters.  If we have 2-child families, there are 4 possible families (MM, MW, WM, WW).  We have two men with one sister and two women with one sister -- thus, two sisters of men and two sisters of women, for equal probability.  If you don't believe me, try it with three sibling families.

The total number of sisters of men in all families with $n$ siblings is $\sum_{i=0}^n {n \choose i}i(n-i) = \sum_{i=1}^{n-1} \frac{n!\cdot (n - i)i}{(n-i)! i!} = n(n-1)\sum_{i=1}^{n-1} \frac{(n-2)!}{(n-i-1)!(i-1)!} =$ $n(n-1)\sum_{i=1}^{n-1}{n-2\choose i-1} = n(n-1)2^{n-2}$

The total number of sisters of women in all families with $n$ siblings is $\sum_{i = 0}^n {n\choose i} i(i-1) = \sum_{i = 2}^n \frac{n!\cdot i(i-1)}{i!(n-i)!} = n(n-1)\sum_{i=2}^n \frac{(n-2)!}{(i-2)!(n-i)!} =$ $n(n-1)\sum_{i=2}^n {n-2\choose i-2} = n(n-1)2^{n-2}$

The most basic reason k81's reasoning is wrong is that a pair of sisters counts as two sisters of women, while and sister and a brother count as only one sister of a man.

Another way to present the flaw: in a family with both sons and daughters, each son will have more sisters than each daughter will.  This on its own is not enough to ensure that men in general have more sisters than women in general will (I think an instance of Simpson's Paradox?), and it is further exacerbated by the fact that there are families with only daughters and families with only sons.

In fact, men and women will have on average the same number of sisters no matter what the relative birthrates are, so long as births are independent.  That is, the result still holds even if 52% (or 60%, or 99%) of births are of women.  But this should be clear anyhow: my gender clearly has no bearing on the number of sisters that I have.


[quote="gemath"]Given triangle $ ABC$, let $ A',B',C'$ be in side $ BC,CA,AB$, resp such that $ AA'=BB'=CC'$ and $ \angle BAA'=\angle CBB'=\angle ACC'$. Prove that triangle $ ABC$ is equilateral.[/quote]
Draw $ A'D \perp AB,B'E \perp BC,C'F \perp AC$
We have: $ \Delta ADA'=\Delta BEB'=\Delta CFC'$
$ \Rightarrow A'D=B'E=C'F;AD=BE=CF$
 Suppose that $ \hat{A}\geq \hat{B}$
$ \Rightarrow AF \leq BD \Rightarrow AC\eq AB$
$ \Rightarrow \hat{B}\leq \hat{C}\Rightarrow BD \geq CE$
$ \Rightarrow AB \geq BC \Rightarrow \hat{C}\geq \hat{A}$
$ \Rightarrow CE \leq AF \Rightarrow BC\leq AC$
$ \Rightarrow$$ \hat{A}\leq \hat{B}$=>$ \hat{A}=\hat{B}$
Similar,we have $ \hat{A}=\hat{B}=\hat{C}$
So triangle $ ABC$ is equilateral


After collecting for the candy drive, Sally had eight $ \$1$-bills, five $ \$5$-bills and three $ \$10$-bills. She randomly selected three bills without replacement. What is the probability that she chose one of each type of bill? Express your answer as a common fraction.

There are $ 8 \plus{} 5 \plus{} 3 \equal{} 16$ bills in total. The probability of picking each kind of bill is as follows:

$ \frac{8}{16} \cdot \frac{5}{16} \cdot \frac{3}{16} \equal{} \frac{15}{512}$

But wait - we're not done yet! There are $ 3! \equal{} 6$ ways to order these bills, so we multiply by $ 6$, to get $ \boxed{\frac{45}{256}}$.

Bah but once u pick one there are 15 left to choose from

$ \frac{8\cdot5\cdot3\cdot6}{16\cdot15\cdot14}\equal{}\frac{3}{14}$

3/14 is right, but where is the 6 coming from in the numerator?

That's the number of ways we can order the choice of bills.

[quote="isabella2296"] There are $ 3!\equal{}6$ ways to order these bills[/quote]

we know that ther are 16 bills

so we can easily get the equation

8*5*3/16

hmm. this looks like our equation BUT wait theres more!

there are 3! ways to arrange these why ? because there are

3-#1
2-#2
1-#3

so the real equation is 8*5*3*6/16=3/14

so our answer is 3/14 or $\boxed{3/14}$

[hide]There are 16bills 8x5x3/16
and there are 3ways
8x5x3x6/16=3\14 answer 3/14[/hide]
Hello, Problem and solution above!


$S$ majmoeiy na motenahi az nogat dar safhe hast, s.t. agar $A,B,C$ se nogteye motemayez az $S$ bashand, faseleye $A$ ta$BC$ adadi sahih ast. Sabet konid hameye nogat roye khati rast hastand.

bache ha nazaretoon chie rahe halaye hamdigaro tashih konim??
yeni age doroste begim doroste va age ghalate joofesho begim
ghaboole??:huh:

[quote="amir2"]Bacheha in rahe hale mano check mikonid?

$16-36=25-45$
$4^2-4\times 9=5^2-9\times 5$
$4^2-4\times 9+\frac{81}{4}=5^2-9\times 5+\frac{81}{4}$
$(4-\frac{9}{2})^2=(5-\frac{9}{2})^2$
$4-\frac{9}{2}=5-\frac{9}{2}$
$4=5$
$2\times 2=5$

bellakhare fahmidim chera $2\times 2$ mishe $5$[/quote]
wow  :o
ta hala no nadideh boodam
akhe badish inike rabetash bargasht paziram hast :(

ye rigi too kafshesh hast.manam midoonam chie.vali nemigam tablo nashe.

bikhodi pichoondish amir joon.
ye taraf $-\frac12$ye taraf$+\frac12$

[quote="amir2"]Bacheha in rahe hale mano check mikonid?

$16-36=25-45$
$4^2-4\times 9=5^2-9\times 5$
$4^2-4\times 9+\frac{81}{4}=5^2-9\times 5+\frac{81}{4}$
$(4-\frac{9}{2})^2=(5-\frac{9}{2})^2$
$4-\frac{9}{2}=5-\frac{9}{2}$
$4=5$
$2\times 2=5$

bellakhare fahmidim chera $2\times 2$ mishe $5$[/quote]
are mohammad rast mige
ye taraf $\frac{-1}{2}$ ye taraf $\frac{+1}{2}$
vali ghashang bood tahala ino nadide boodam :D

[quote="Omid Hatami"]In masaleh ra ma parasal shenidim (tooye doreye 12 nafar) va hich kas natoonest hallesh koneh age rahe hallesh ra midooni befrest[/quote]
na rahe hallesho nemidoonam faghatn midoonam ke too ketabe doktor mosaab oomade ;)
bebinam too oon dore nima ahmadi pouram bood dige?? :rotfl:

[quote="Omid Hatami"]Are chetor mage?[/quote]
akhe in soale male oone
ye jaye dige porside bood baghie ham behesh gofte boodan opene ;)

in soal Opene
kasi ta hala halesh nakarde.

[quote="BaBaK Ghalebi"]bebinam in dota soal openan??:
$1)$agar $2^\alpha,3^\alpha,5^\alpha \in \mathbb N$ anagah $\alpha\in \mathbb N$
$2)$agar $n^\alpha \in \mathbb N$ baraye tamame $n$ haye tabii angah $\alpha\in \mathbb N$[/quote]

 :huh:  :huh:

[quote="BaBaK Ghalebi"]khob farz konid intor nabashad va noghtei manade $A_1$ vojood dashte bashad ke rooye khate morede nazar nabashad va noghte $C$ paye amoode noghte $A_1$ be khate matloob bashad
az noghte $A_1$ be baghie noghat vasl konid
hal baraye noghte $A_2$ midanim ke faseleye an ta khotoote vasel beyne $A_1$ va baghie noghat adadi sahih ast
amma az tarafi chon dar har marhale paye amoode vasel az noghte $A_2$ az $A_2$ door mishavad pas hamvare toole ertefa(fasele noghte $A_2$)afzayesh miabad amma in meghdar $\leq AC$pas in meghdar meghdari mahdood ast pas tedad noghate rooye khate matloob ke dar samte chape $A_2$vaghe hastand mahdood ast
hamchenin ba estedlali moshabeh natije mishavad ke tedade noghat dar samte raste $A_2$ niz mahdood ast vali in amr ba namotanahi boodane majmooe $S$ tanaghoz darad
doroste aghaye yazdan bod?? :?[/quote]

Babak jan bebakhshid ke dir shod amma fekr konam ke rahe halet droste ,albate kamel motmaen nistam ;)

[quote="ehsan2004"]rasti inchiye ke ziresh neveshti [quote]doroste aghaye yazdan bod?? [/quote]???????[/quote]
valla ma javoon ke boodim ba ye babai be name sadra yazdanbod kalkal mikardim vali az zamani ke ishoon mashghalehaie digari peida kardan dige aslan be site sar nemizanan
manzooram hamoon syboy khodemoone :(  ;)
ehsan merc az inike rahe hallamo khoondi
age baghie ham nazareshoono bedan khoshhal mishim

ei baba...chand ta maseleye khoshgele dg ham bezarin ta ma ham ye kam hal konim...age mishe ye zare maselahey computery bezarin?
 :D

[quote="BaBaK Ghalebi"]bebinam in dota soal openan??:
$1)$agar $2^\alpha,3^\alpha,5^\alpha \in \mathbb N$ anagah $\alpha\in \mathbb N$
$2)$agar $n^\alpha \in \mathbb N$ baraye tamame $n$ haye tabii angah $\alpha\in \mathbb N$[/quote]

Babak

Soal 1 ro Nima ghablan post karde , Harazi gofte masale hanoz Opene :
http://www.mathlinks.ro/Forum/viewtopic.php?t=42786
(Oh nadidam khodet ino be omid hatami gofti, pas hichi)

Soal 2 ham ye masale ghadimiye Putnam:
http://www.kalva.demon.co.uk/putnam/psoln/psol716.html

saman joon felan bia komake man ke hanooz too kafe in soalam gerafaro migam!!! :(


[quote="pohoatza"]Fie $ a,b,c$ pozitive cu $ abc \equal{} 1$. Sa se arate ca:

$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \plus{} 7\geq 5(a \plus{} b \plus{} c)$[/quote]

[b]Solutia mea:[/b]

Folosim identitatea: $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1$, si obtinem ca:
\[ \left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right) \plus{} 7 \geqslant 5\left( {a \plus{} b \plus{} c} \right) \Leftrightarrow \left( {a \plus{} b \plus{} c} \right)\left( {ab \plus{} bc \plus{} ca} \right) \plus{} 6 \geqslant 5\left( {a \plus{} b \plus{} c} \right)
\]
, sau $ ab \plus{} bc \plus{} ca \plus{} \frac {6} {{a \plus{} b \plus{} c}} \geqslant 5$, aceasta fiind acum inegalitatea pe care trebuie sa o demonstram pentru a rezolva problema.

Pentru inceput $ ab \plus{} bc \plus{} ca \geqslant 3\root 3 \of {a^2 b^2 c^2 } \equal{} 3$. Mai avem de demonstrat doar ca $ \frac {6} {{a \plus{} b \plus{} c}} \geqslant 2$
Este binecunoscuta inegalitatea: $ \left( {x \plus{} y \plus{} z} \right)^2 \geqslant 3\left( {xy \plus{} yz \plus{} zx} \right)$, valabila oricare ar fi $ x,y,z$ pozitive. Egalitatea are loc atunci cand $ x \equal{} y \equal{} z$

In cazul nostru obtinem ca $ \left( {ab \plus{} bc \plus{} ca} \right)^2 \geqslant 3\left( {a^2 bc \plus{} b^2 ac \plus{} c^2 ab} \right) \equal{} 3\left( {a \plus{} b \plus{} c} \right)$, si tinand cont de inegalitatea demonstrata anterior, deducem ca
\[ 9 \geqslant 3\left( {a \plus{} b \plus{} c} \right) \Leftrightarrow a \plus{} b \plus{} c \leqslant 3 \Rightarrow \frac {1} {{a \plus{} b \plus{} c}} \geqslant \frac {1} {3} \Rightarrow \frac {6} {{a \plus{} b \plus{} c}} \geqslant 6 \cdot \frac {1} {3} \equal{} 2.
\]
Solutia se incheie.

[b]Observatie:
[/b] O alta inegalitate asemanatoare acesteia este urmatoarea:

[i]Prove that if $ a,b,c > 0$ have product 1, then $ (a \plus{} b)(b \plus{} c)(c \plus{} a)\geqslant 4(a \plus{} b \plus{} c) \minus{} 1$[/i].
[b]MOSP, 2001[/b].

[b]Bibliografie [/b]

[1]T. Andreescu, V. Cartoaje, G. Dospinescu si M. Lascu - Old and New Inequalities, GIL Publishing House, 2004 .

Deci daca $ \ds ab+bc+ca\geq 3$ si $ a+b+c\leq ab+bc+ca$, atunci $ a+b+c\leq 3?$  :)
In plus, din medii avem $ a+b+c\geq 3\sqrt[3]{abc}=3$..

Da, imi cer scuze nu am fost atent un moment...si m-am incurcat.


function
number theory solved
number theory

Find all functions $f$ $\mathbb{N} \mapsto \mathbb{N}$ such that $(f(m))^2 + f(n)$ divides $(m^2 +n )^2$ $\forall m,n \ \in \ \mathbb{N}$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=life&t=27188

Pierre.

Thanks!!! Sorry I didn't check earlier!!! Another copy by the Indians!!! When will they post original problems???


trigonometry
analytic geometry
calculus
geometry
trig identities
Law of Sines

How does one find sin of $\sqrt2$ degrees by hand?

[quote="uglyman621"]I don't really get what you mean by hands, but I can attempt to explain the sine of root2 degrees.

If you graph it on a coordinate plane, you have the terminal side and the initial side making a root 2 degree angle from the origin. Call this angle theta. You know the measures of all the angles. You can't really do much more from there but you can try law of sines[/quote]

by hand means find the answer utilizing algebra or somthing like it

My calculus teacher did says it is impossible, is it?

[quote="Bruno Bonagura"]I think it's not possible for $\sqrt 2$ degrees, but for radians it is possible.
Use MacLaurin series...[/quote]

Can you explain why is impossible for $\sqrt2$ degrees?

[quote="fourierseries"]Convert sqrt(2) degrees into radians, then try McClaurin series.  It might not be a closed form, however.[/quote]

Can you show how i can use Mac Laurin series for $\sqrt2$degrees or radians ?

[quote="Ravi B"]Even the first-order approximation $\sin x \approx x$ (for $x$ a small radian value) is pretty good.  We get
\[
  \sin {\sqrt{2}\,}^\circ
    = \sin \frac{\pi\sqrt{2}}{180}
    \approx \frac{\pi\sqrt{2}}{180}
    \approx .024683.
\]
The first 5 digits after the decimal point are correct.[/quote]

can you calculate by hand without approximations ( fraction):?:

[quote="Bruno Bonagura"]felipesa,

It's impossible to calculate exactly.

(1) You don't know the exact value of $\sqrt{2}$.
(2) MacLaurin serie for sine is: $\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1} = \frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$
(3) You can't calculate infinite terms of MacLaurin series by hand.[/quote]

(1)right
(2)How Can I calculate sin() use MacLaurin Series?(Give an example)
(3)Only by calculator?

The McClaurin series of sine is

$\displaystyle \sin x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1} = \frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+... $

Just plug in the value (in radians) as your x and you get an infinite series for sine.


Let a,b,c be three sides of a triangle.Prove that
$\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}+\frac{ab+bc+ca}{a^2+b^2+c^2}<\frac{17}{6}$

[quote="minhkhoa"]Let a,b,c be three sides of a triangle.Prove that
$\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}+\frac{ab+bc+ca}{a^2+b^2+c^2}>\frac{17}{6}$[/quote]
I don't know my calculation is correct or not,but it seem that the inequality is wrong,at least when $a=b=c$ :?

Indeed, it doesn't work, please post the correct version.

[quote="ychjae"]I think the correct one should be
$\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}+\frac{ab+bc+ca}{a^2+b^2+c^2}<\frac{17}{6}$
while the left-hand-side of the inequality "equal" to $\frac{17}{6}$ when $a=2b=2c$.
(I don't know any inequality which is useful here :| )[/quote]
Yes!The sign is reversed.Corrected


Three standard dice are tossed. What is the probability that the sum of the numbers on the tops of the three dice is $ 17$ or greater? Express your answer as a common fraction.

Either a 17 is rolled, or a 18 is rolled. There are 4 ways total: 5-6-6, 6-5-6, 6-6-5, and 6-6-6. So the total probability is $ \frac{4}{216}\equal{}\boxed{\frac{1}{54}}$.

There are $ 2$ basic ways to do it: $ 5\minus{}6\minus{}6$ or $ 6\minus{}6\minus{}6$. There are $ \dbinom{3}{2}$ ways to order the first one, and only $ 1$ way to order the second one. Also, since each roll is independent, there is a $ \dfrac{1}{6}$ chance for a single number.

We have $ \dbinom{3}{2} \cdot \dfrac{1}{6^3} \plus{}\dfrac{1}{6^3}\equal{} \dfrac{3\plus{}1}{216}\equal{}\boxed{\dfrac{1}{54}}$


geometry
3D geometry
probability
LaTeX
number theory
prime factorization

1. (1999 Team #6) Given positive integers x and y such that x doesnt equal y, and 1/x + 1/y = 1/12, what is the smallest possible value of x + y ?

2. (2000 Team #6) Four number cubes, with numbers 1 to 6, one per face, are rolled. What is the probability that two of the numbers are the same even number and that the other two numbers are the same odd number? Express answer as a common fraction.

3. (2001 Team #3) An Internet service provider allows a certain number of free hours each month and then charges for each additional hour used. Wells, Ted, and Vino each have  separate accounts. This month the total hours used by Wells and Ted was 105, and each used all of their free hours. Their total cost was 10 dollars . Vino used 105 hours by himself and had to pay 26 dollars. What is the numbe of cents charged for each extra hour? (edit: forgot dollar sign is latex)

4. (2001 Team #8) A turn consists of rolling a standard die and tossing a fair coin. The game is won when the die shows a 1 or a 6 and the coin shows heads. What is the probability the game will be won before the fourth turn? Express answer as common fraction.

5. (2003 Team #3) Chris and Ashley are playing a game. First, Chris removes one number from the set {1,2,3,4,5}. Then Ashley picks two of the remaining numbers at random replacement. Ashley wins if the sum of the two number she picks is not prime. otherwise Chris wins. What number should Chris remove to maximize his chances of winning?

6. (2003 Team #8) N^2 is a divisor of 8! . What is the greatest possible integer value of N ?

[hide]Answers:
1. 49
2. 1/24
3. 40
4. 91/216
5. 5
6. 24[/hide]

for number one, 1/x+1/y=1/12, so (x+y)/xy=1/12, and 12x+12y=xy(crossmultiply) solving for x, xy-12x=12y, so x(y-12)=12y, and x=12y/(y-12) here is the tricky part. add and subtract 144 to the numerator, without changing the value, so x=(12y-144+144)/(y-12)=12+144/(y-12)

since x is an integer, y-12 must be a factor of 144. go from there

For number two:

[hide]There are 4C2 ways there can be two even numbers. There are 3 even numbers possible and 3 odd numbers possible so the total number of ways there can be 2 of the same even numbers and two of the same odd numbers is 6*3*3=54. Then are $ 6^4$ total outcomes. So the probablity is $ \frac{54}{6^4}\equal{}\frac{1}{432}$[/hide]

For Number 3:

[hide]$ W\plus{}T\equal{}105$
$ X\plus{}N(W\minus{}X)\plus{}X\plus{}N(T\minus{}X)\equal{}\$10$
$ V\equal{}105$
$ X\plus{}N(V\minus{}X)\equal{}\$26$
Where W,T,V are the number of hours; X=the number of free hours; N=number of cents charged for each extra hour. Solve for N and multiply by 100.[/hide]

For #4:

[hide]I solved this by calculating the probability the game doesnt end after 4 turns and then subtracted from 1. The proability that the game is lost is $ 1 \minus{} \frac {1}{2}\cdot\frac {1}{3} \equal{} \frac {5}{6}$ There for the proability that the game is not won before 4 games is $ \left(\frac {5}{6}\right)^{3} \equal{} \frac {125}{216}$. Subtracting that from $ 1$, our answer is $ \frac {91}{216}$
[/hide]

Hint for #6:
Find prime factorization of 8!. The product of two perfect squares is also a perfect square

5.
[hide]For this, Chris wants the sum to be prime.
The first 3 primes numbers are 3,5, and 7.
5 is the only number than can be paired with another number to get a prime number..
so it's 5[/hide]

Find all pairs from prime factorization.
[hide]8!=8*7*6*5*4*3*2*1=2^3*7*3*2*5*2^2*3*2=2^7*3^2*5*7

so you can find pairs of 2^3*3

so 24 is the highest n.[/hide]


linear algebra
matrix
inequalities
combinatorial geometry
inequalities proposed

Let $a_{ij}, m_{ij}$ be two matrices of real numbers such that the sum of the elements of any line and column in $a_{ij}$ is 1 and $a_{ij}\geq 0$. Prove that we can find a permutation $\sigma$ of the numbers $1,2,..,n$ such that $\sum_{i,j=1}^{n}{a_{ij}m_{ji}}\leq\sum_{i=1}^{n}{m_{i\sigma(i)}}$.

The last time I posted in this forum was like half a year ago :P.

Any doubly-stochastic matrix (i.e. a matrix with the property $(a_{ij})$ enjoys) is a convex combination of permutation matrices. This means that the LHS in the inequality is a convex combination of the sums $S_{\sigma}=\sum_{i=1}^n m_{i\sigma(i)}$ (for $\sigma\in S_n$), and the conclusion follows.

It is nice I stimulate you with nice problems, grobber!! :P  :P   Yet, this is not the solution I was expecting, since the fact that you use is not at all easy to prove, I think. Anyway, more difficult than the problem itself. I want a simple solution and what't nice is that a simple solution for this problem implies a simple solution for Birkoff theorem (what you actually used).  :)

I think this is wrong, but anyway, does the multiplication of a bisctochastic matrix of $a_{ij}$ means that {$m_{ij}$} majorizes {$a_{ij} m_{ij}$}?

I'll try to prove the theorem I used (the proof I thought I had when I posted the above was wrong, so I had to modify it :)). I will, however, use the Krein-Milman Theorem for finite-dimensional spaces:

A convex compact subset of $\mathbb R^d$ is the closed convex hull of its set of extremal points.

In view of the theorem stated above, it suffices to show that a doubly-stochastic matrix which is not a permutation matrix is not an extremal point of the convex compact set $\mathcal K\subset\mathbb R^{n^2}$ of doubly-stochastic matrices.

Given a matrix $A=(a_{ij})\in\mathcal K$ which is not a permutation matrix, we can find at least one real $x_1=a_{i_1j_1}\in (0,1)$. Then, we can find at least one $x_2$ in the same column as $x_1$, but in a different position, which also belongs to $(0,1)$. Afterwards, we can find at least one real $x_3$ in the same row as $x_2$ but in a different position, also belonging to $(0,1)$, and so on, until, at some point, we will encounter either some $x_{2j}$ which has the same column as some $x_{2i+1},\ 2i+1<2j$, or an $x_{2j+1}$ sharing the same line with an $x_{2i},\ 2i<2j+1$. We may assume WLOG that we started the procedure from $x_{2j}$ or $x_{2j+1}$, depending on which case among the two mentioned above we are in, so we assume WLOG that $x_1=x_{2j}$ or $x_{2j+1}$.

Take $\varepsilon>0$ s.t. $\varepsilon<|x_i|,|1-x_i|,\ \forall i$. Increase $x_1$ with $\varepsilon$, decrease $x_2$ with $\varepsilon$, then increase $x_3$ with $\varepsilon$, and so on, until we reach $x_1$ again. We have obtained a different matrix, $A'$. Now reverse the process: starting with the matrix $A$ again, decrease $x_1$ with $\varepsilon$, then increase $x_2$ with $\varepsilon$, then decrease $x_3$ with $\varepsilon$, and so on, until we reach $x_1$. This is a third matrix, $A''$. It's easy to see now that $A=\frac 12(A'+A'')$ and $A',A''\in\mathcal K$, so we have shown that any matrix $A\in\mathcal K$ which is not a permutation matrix lies in some open interval having matrices in $\mathcal K$ as endpoints, so it's not an extremal point. On the other hand, it's obvious that the permutation matrices are extremal points of $\mathcal K$, and the conclusion follows.

I hope it's correct. For that procedure with the $i$'s and $j$'s which was so difficult to explain, drawing a picture might be helpful.

P.S.

I've found my last messages in the "Inequalities" forum:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35418[/url]

Please check out that thread, harazi, and post (if you have the time, of course) the solution you promissed back then :).


function
invariant
abstract algebra
number theory
prime factorization
superior algebra
superior algebra solved

Show that ( Z[x] , + ) and ( Q* , . ) are isomorphic, where Q* is the set of all nonzero rational numbers.

(Please give a detail solution....)

Thank you :)

$ \phi( x^{k}) \equal{} p_{k}$, where $ p_{k}$ denotes the sequence of primes (and $ p_{0}\equal{} 2$).  The result follows by prime factorization.  ;)

sir......kindly explain elaborately.

thank you sir.

the long version  :blush: (and note, that instead of the multiplicative group of rationals it is an isomorphism to the multiplicative group of strictly positive rationals)

For any prime $ p$ let $ a_{p}:\mathbb{Z}\rightarrow\mathbb{Z}$ denote an integer function such that $ a_{p}(m)$ denotes the greatest power of $ p$, which divides $ m$. Note, that $ a_{p}$ is a "multiplicative" and symmetric function, i.e. $ a_{p}(nm)=a_{p}(n)+a_{p}(m)\,,\,a_{p}(m)=a_{p}(-m)$. If $ \mathbb{Q}_{>0}^{*}$ denotes the multiplicative group of strictly positive rationals, we define a function $ \Psi :\mathbb{Q}_{>0}^{*}\rightarrow\mathbb{Z}[x]$ as follows. Let the complete set of primes be indexed as $ p_{0}<p_{1}<\ldots$ and note, the representation $ m=sign(m)\,\prod_{k=0}^{\infty}p_{k}^{a_{p_{k}}(m)}$.

For $ \frac{A}{B}\in\mathbb{Q}_{>0}^{*}$ let $ \Psi\left(\frac{A}{B}\right)=\sum_{k=0}^{\infty}[a_{p_{k}}(A)-a_{p_{k}}(B)]x^{k}$. The function $ \Psi$ is well-defined, since $ \forall p_{k}>max\{|A|,|B|\}: a_{p_{k}}(A)=a_{p_{k}}(B)=0$ holds and since it is invariant under general representation switches $ \frac{A}{B}\rightarrow\frac{tA}{tB}$ with $ t\in\mathbb{Z}_{>0}$ or $ t$ being the inverse of a common divisor of $ A,B$.

For $ \frac{A_{1}}{B_{1}},\frac{A_{2}}{B_{2}}\in\mathbb{Q}_{>0}^{*}$ we calculate $ \Psi\left(\frac{A_{1}}{B_{1}}\frac{A_{2}}{B_{2}}\right)=\sum_{k=0}^{\infty}[\underbrace{a_{p_{k}}(A_{1}A_{2})}_{a_{p_{k}}(A_{1})+a_{p_{k}}(A_{2})}-\underbrace{a_{p_{k}}(B_{1}B_{2})}_{a_{p_{k}}(B_{1})+a_{p_{k}}(B_{2})}]x^{k}=\Psi\left(\frac{A_{1}}{B_{1}}\right)+\Psi\left(\frac{A_{2}}{B_{2}}\right)$ and conclude that $ \Psi$ is a group homomorphism.

As for injectivity of $ \Psi$, the equation $ \Psi\left(\frac{A_{1}}{B_{1}}\right)=\Psi\left(\frac{A_{2}}{B_{2}}\right)$ implies for all $ k\geq 0:\,a_{p_{k}}(A_{1})-a_{p_{k}}(B_{1})=a_{p_{k}}(A_{2})-a_{p_{k}}(B_{2})$, i.e. $ \frac{A_{1}}{B_{1}}=\prod_{k=0}^{\infty}p_{k}^{[a_{p_{k}}(A_{1})-a_{p_{k}}(B_{1})]}=\prod_{k=0}^{\infty}p_{k}^{[a_{p_{k}}(A_{2})-a_{p_{k}}(B_{2})]}=\frac{A_{2}}{B_{2}}$.

As for surjectivity of $ \Psi$ let $ f(x)=\sum_{k=0}^{n}c_{k}x^{k}\in\mathbb{Z}[x]$ be an arbitrary element. Define $ A=\prod_{k=0,c_{k}>0}^{n}p_{k}^{c_{k}},B=\prod_{k=0,c_{k}<0}^{n}p_{k}^{-c_{k}}$, then $ \Psi\left(\frac{A}{B}\right)=f(x)$ is valid.

omg ... it's just that t0rajir0u said, so many formulas are out of place and hide the (trivial!) reason for the isomorphism.

$ \mathbb{Z}[x]$ is by definition free abelian with basis $ 1,x,x^{2},...$ and $ \mathbb{Q}^{\plus{}}$ is free abelian with basis $ p_{1},p_{2},...$, the letter because of prime factorization. thus $ x^{k}\mapsto p_{k}$ extends uniquely to a isomorphism $ \mathbb{Z}[x]\to\mathbb{Q}^{\plus{}}$.

please explain, how Q* is generated by the primes

Any positive rational $ q$ can be written uniquely as a product of primes

$ q \equal{}\prod_{k \equal{} 1}^{n}p_{k}^{e_{k}}$

where $ e_{k}\in\mathbb{Z}$.  This follows trivially from prime factorization over $ \mathbb{N}$.  :)


Prove that with each $ n>3$ there exist $ x_1,x_2,x_3,...,x_n$ are nonnegative integers numbers satisfying
                             $ x_1x_2...x_n\equal{}(x_2\plus{}x_3\plus{}...\plus{}x_n)(x_1\plus{}x_3\plus{}...\plus{}x_n)....(x_1\plus{}x_2\plus{}..x_{n\minus{}1})$

the only solution is:$ x_1\equal{}x_2\equal{}...\equal{}x_n\equal{}0$
am I wrong?

Apply AM GM ineq on each term of RHS which results in an equality with LHS
this implies x1=x2=...=xn
=>x1*x2*..*xn=(n-1)^n*x1*x2*..*xn
=>{(n-1)^n-1}*x1*x2*...*xn=0
For n>3 (n-1)^n-1 > 0
=>x1=x2=x3=...=xn=0

Is AM-GM really necessary?  If the $ x_i$ are non-negative then clearly $ (x_2 \plus{} ... \plus{} x_n) \ge x_2, (x_1 \plus{} x_3 \plus{} ...) \ge x_3$ and so forth, with equality if they are all equal to zero.

I'm sorry the numbers $ x_1,x_2..x_n$ is only the integers numbers.

Here is my solution
   If we put $ S\equal{}x_1\plus{}x_2\plus{}x_3\plus{}...\plus{}x_n$ then the equality is(I will post continous on my free time...)


Find the best positive real constant $ \delta$ such that
$ \frac{ab^2\plus{}bc^2\plus{}ca^2\minus{}3abc}{a^2b\plus{}b^2c\plus{}c^2a\minus{}3abc}\geq\delta$
Holds for all $ a,b,c$ which are side-lenghts of a trangle.

[quote="Ibrogimov"]Find the best positive real constant $ \delta$ such that
$ \frac {ab^2 \plus{} bc^2 \plus{} ca^2 \minus{} 3abc}{a^2b \plus{} b^2c \plus{} c^2a \minus{} 3abc}\geq\delta$
Holds for all $ a,b,c$ which are side-lenghts of a trangle.[/quote]
Let $ a\equal{}y\plus{}z,$ $ b\equal{}x\plus{}z$ and $ c\equal{}x\plus{}y.$
$ \sum_{cyc}(ab^2\minus{}abc)\geq\delta\cdot\sum_{cyc}(a^2b\minus{}abc)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3\minus{}xy^2)\geq\delta\cdot\sum_{cyc}(x^3\minus{}x^2y)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left((x\minus{}y)(x^2\plus{}xy)\minus{}\frac{2(x^3\minus{}y^3)}{3}\right)\geq\delta\cdot\sum_{cyc}\left((x\minus{}y)x^2\minus{}\frac{x^3\minus{}y^3}{3}\right)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x\minus{}y)^2\left((1\minus{}2\delta)x\plus{}(2\minus{}\delta)y\right)\geq0.$
Let $ x\equal{}\min\{x,y,z\},$ $ y\equal{}x\plus{}u$ and $ z\equal{}x\plus{}v.$
Hence, $ \sum_{cyc}(x\minus{}y)^2\left((1\minus{}2\delta)x\plus{}(2\minus{}\delta)y\right)\geq0\Leftrightarrow$
$ \Leftrightarrow2(1\minus{}\delta)(u^2\minus{}uv\plus{}v^2)x\plus{}(1\minus{}\delta)u^3\plus{}\delta u^2v\minus{}uv^2\plus{}(1\minus{}\delta)v^3\geq0.$
We obtain $ \delta\leq1.$
Let $ u\equal{}vt.$ Hence, we need to find $ \delta_{best}$ for which the following inequality
\[ (1\minus{}\delta)t^3\plus{}\delta t^2\minus{}t\plus{}1\minus{}\delta\geq0\]
holds for all $ t\geq0.$
We obtain $ (1\minus{}\delta)t^3\plus{}\delta t^2\minus{}t\plus{}1\minus{}\delta\geq0\geq\Leftrightarrow\delta\leq\frac{t^3\minus{}t\plus{}1}{t^3\minus{}t^2\plus{}1}.$
Let $ f(t)\equal{}\frac{t^3\minus{}t\plus{}1}{t^3\minus{}t^2\plus{}1}.$ Thus, $ \delta_{best}\equal{}\min_{t\geq0}f.$
$ f'(t)\equal{}\minus{}\frac{t^4\minus{}2t^3\plus{}t^2\minus{}2t\plus{}1}{(t^3\minus{}t^2\plus{}1)^2}.$
Hence, $ t_{min}\equal{}\frac{1\plus{}\sqrt2\minus{}\sqrt{2\sqrt2\minus{}1}}{2}$ and $ \delta_{best}\equal{}f\left(t_{min}\right)\equal{}0.713009453...$

Similar problem:
Find the best positive real constant $ k$ such that
$ \frac{a^5\plus{}b^5\plus{}c^5\minus{}3}{a^3\plus{}b^3\plus{}c^3\minus{}3} \geq k$
where $ a;b;c>0$ and satisfying $ a\plus{}b\plus{}c\equal{}3$

[quote="chien than"]Similar problem:
Find the best positive real constant $ k$ such that
$ \frac {a^5 \plus{} b^5 \plus{} c^5 \minus{} 3}{a^3 \plus{} b^3 \plus{} c^3 \minus{} 3} \geq k$
where $ a;b;c > 0$ and satisfying $ a \plus{} b \plus{} c \equal{} 3$[/quote]

Posted before by zazai-hoang


Is this problem on the forum?
How close to 1 one can get adding n different Egyptian fractions? (i mean <1) :)
It is well known but very hard.

A fraction $\frac{1}{n},\ n\in\mathbb{N}$.

No one?  :( This problem is really worth thinking ;) Not to say that the solutioon was firt found in 1922 by Curtiss :P Also a solution was given by Erdos. But i am expecting your solutions. Till now i just know one which of course is not mine :blush:

I see that no one has solve the problem :( If anyone is interested to see a solution from "the Book" let me know :)

This problem is mentioned [url=http://www.mathlinks.ro/Forum/post-168246.html#p168246]here[/url], together with [url=http://arxiv.org/PS_cache/math/pdf/0502/0502247.pdf]a link to the proof[/url].

The Lemma at the end was also on the forum, i.e. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29109]here[/url].


true or false
if $k\in N$ and positive numbers $a,b,c$ such that $abc=1$ then
\[a^{2}+b^{2}+c^{2}+k(ab+bc+ca)\geq (k+1)(a+b+c)\]

[quote="Sunjee"]true or false
if $k\in N$ and positive numbers $a,b,c$ such that $abc=1$ then
\[a^{2}+b^{2}+c^{2}+k(ab+bc+ca)\geq (k+1)(a+b+c) \]
[/quote]
It's wrong for all $k>\min_{0<x<1}\frac{2x^{4}+2x^{3}+2x^{2}+2x+1}{x^{2}-x^{4}}.$ :wink:

[quote="Sunjee"]Do you mean if $k<\min_{0<x<1}\frac{2x^{4}+2x^{3}+2x^{2}+2x+1}{x^{2}-x^{4}}.$
then
\[a^{2}+b^{2}+c^{2}+k(ab+bc+ca)\geq (k+1)(a+b+c) \]
?
here $abc=1$ positive numbers.[/quote]
No! I meant that for all $k>\min_{0<x<1}\frac{2x^{4}+2x^{3}+2x^{2}+2x+1}{x^{2}-x^{4}}=15.08...$ your inequality is wrong.
By the way, your inequality is true for all $k\leq10.$ :wink:


$ a$ $ b$ and $ c$ are reals numbers such that : $ a\neq0$ and  $ a$ and $ 4a\plus{}3b\plus{}2c$ have the same sign.
prove that there isn't any polynomials  $ ax^2\plus{}bx\plus{}c$ has two roots $ \in$ $ [1;2]$
from moroccan  TST 2002

[hide]Assume the contrary.

let the two roots be $ r$ and $ s$

$ b\equal{}\minus{}a(r\plus{}s)$
$ c\equal{}ars$

Therefore, $ a$ having the same sign as $ 4a\plus{}3b\plus{}2c\equal{}a(4\minus{}3(r\plus{}s)\plus{}2rs)$ is equivalent to $ 1$ having the same sign as $ 4\minus{}3(r\plus{}s)\plus{}2rs$. That is, $ 4\minus{}3(r\plus{}s)\plus{}2rs>0\Leftrightarrow (r\minus{}\frac{3}{2})(s\minus{}\frac{3}{2})>\frac{1}{4}$. However, $ |(r\minus{}\frac{3}{2})(s\minus{}\frac{3}{2})|\le |\frac{1}{2}\parallel{}\frac{1}{2}|\equal{}\frac{1}{4}$, contradiction[/hide]


[quote="Interval"]My Dad and I are building a round pen for our horse.  We have 16
16ft. panels and a 10 ft. gate and a 4ft. gate (270 ft. total).  We
want to use a radius and mark the places to dig holes for each post
that will support the panels, but we don't know how long the radius
should be.[/quote]
[hide]That means that the circumference is 270 ft. to find the radius, $\pi r^{2}=270 \rightarrow \sqrt{r^{2}}=\sqrt{\frac{270}{\pi}}=9.27...$ so the radius should be about 9.3 feet or so.[/hide]


here is a type of question that I see come up alot... AMC and MATHCOUNTS love it.

George and Walter are playing a game, which consists of removing 1,2,3,4,5 or 6 pennies from a pile of 100. If George goes first, how many pennies should he remove to ensure that he wins?

Or how about a general case:
They are playing a game that consists of removing, 1,2,3... m pennies from a pile of n. If George goes first, how many pennies should he remove to ensure that he wins the game.
hint: [hide]
there are multiple scenarios in the general case[/hide]

I'm going to assume that winner has to take the last penny.

[hide]You want to make the other player end up with $n$ pennies, where $n \equiv 0 \pmod{m+1}$.  That way, no matter how many pennies the other player gets, you can always pick a number that will make $n \equiv 0 \pmod{m+1}$.  That means that the number of pennies to pick in the first turn is $n \pmod{m+1}$, which will make $n \equiv 0 \pmod{m+1}$.

So the answer to 1 is 2, and 2 is $n \pmod{m+1}$[/hide]

you assumed correctly
good job... now lets change the question:
Suppose that n cong. to -1 (mod m+1)... how many pennies should a person remove to win
[hide]hint: trick question[/hide]

[hide]I solved this by imagining a simplier case.  I thought if there was 10, what would I do to force a win a for me.  I realized I had to leave my opponent with the maximium amount of coins you are allowed to take in one turn plus one.  So in the case of 10, I needed to take 3.  In the case of 100, I had to take enough to the nearest multiple of (6+1) so I should take 2.  In the case of n and m, I should take enough from n to make the amount left a multiple of m+1.[/hide]


How many integers between $ 100$ and $ 150$ have three different digits in increasing order? One such integer is $ 147$.

You could work with distinct sets of numbers and one to one correspondence, but casework is rather easy for this problem. The first digit is 1. If the tens digit is 2, then 3-9, or 7 numbers, are possibilities for the ones digit. The rest follows:

Tens Digit.........Number of Possible Numbers
2                       .....................7
3                       .....................6
4                       .....................5

We cannot have 5 as the units digit, since our numbers must be between 100 and 150.

The total is $ 7 \plus{} 6 \plus{} 5 \equal{} \boxed{18}$


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arithmetic sequence
number theory unsolved
number theory

show that

the product of 5 consecutive positive integer doesn't equal to a perfect square

P.s: I know the  Erdös et Selfridge theorem

I think this has been posted many times before..
You can use Bertrand's postulate [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=660441045&t=112923]here[/url]

i don't see the relation btw bertrand's postulat and my exercice

I happen to know the theorem that 4 positive integers in an  arithmetic progression cant be all squares. Is that what that Erdos theorem is?

according to Erdös-Selfridge theorem

the product of four or more consecutive terme of a co-prime arithmetique progression cannot be a power.

Any prime 5 or greater must divide only one of the consecutive integers, and it must divide with even multiplicity. BTW we're going to assume n is pretty big. So at most one number is a square. Also at most one number is twice a square (otherwise we get $2a^{2}$ and $2b^{2}$ being within 5 of each other). Also at most one number is thrice a square. So this is a contradiction as there are still two numbers left that can't really be anything.

Edit--oops, it's not quite that simple. It is conceivable that some number is 6 times a square. But only one number could be that. You can do casework to get rid of the cases where n is small; it should be fairly easy.

thanks mr probability1.01 but i want a proof to my exercice wich has only 5 consecutive number


calculus
integration
algebra
polynomial
quadratics
number theory unsolved
number theory

Prove that for every $k\in N$ we can find $n\in N$ s.t. $n\cdot 2^k +17$ is a perfect square.
Can this be generalized somehow?

I'm sure this occured before, but wasn't able to find it.

The result says nothing else than that there is one solution for $x^2 \equiv 17 \mod 2^k$; indeed there are $\infty$ many solutions.

Lets assume that $k \geq 3$ because the other cases are boring.
Let $a$ be odd, then $a^2 \equiv 1 \mod 8$. Thus all odd (the term "odd" is also defined $\mod$ all even numbers) squares $\mod 2^k$ are $1 \mod 8$, so only $\frac{1}{4}$ of the odd residues $\mod 2^k$ can be squares.
In the other direction for odd $a,b$, we can only have $a^2 \equiv b^2 \mod 2^k$ iff $2^k | a^2-b^2 = (a+b)(a-b)$. Clearly $a+b$ and $a-b$ are even, but only one of them can be divisible by $4$ since we would need $a \equiv b \equiv -a \mod 4$, so especially that $a$ is even, which is excluded.
Thus either $2^{k-1}|a+b$ or $2^{k-1}|a-b$, so $a \equiv \pm b \mod 2^{k-1}$.
So for any odd $a$ there are exactly four different $b$ with $a^2 \equiv b^2 \mod 2^k$. Since one out of four odd residues is possible to be a square , all of them must be a square (four times each),especially $17 \equiv 1 \mod 8$.

Conclusion: every residue $\equiv 1 \mod 8$ is a square $\mod$ every power of two; in other words: $x^2 = a$ (and $v_2(a)=0$) has a $2$-adic solution iff $a \equiv 1 \mod 8$.

Generalization:
For $p$ an odd prime and $a$ not divisble by $p$ we have that $x^2 \equiv a \mod p^k$ has a solution iff it has one $\mod p$ (note the difference to the case $p=2$ where it doesn't suffice to look $\mod p$).

Both of them can more directly and constructively shown by (the integral version of) Hensel's Lemma:
Let $f$ be a polynomial with integer coefficients, $p$ a prime and $x$ an integer (or a $p$-adic number). When $n: = v_p(f(x)) > 2 v_p(f'(x))=: k$ ($f'$ the formal derivation of $f$), then there is a $\xi \in \mathbb{Z}_p$ with $\xi \equiv x \mod p^{n-k}$ and $f(\xi)=0$.
In other words: we have a sequence $(x_m)_m$ of integers with $x_0=x$, $x_{m+1} \equiv x_m \mod p^{n+m-k}$ (especially $x_m \equiv x \mod p^{n-k}$) and $f(x_m) \equiv 0 \mod p^{n+m}$ (and $v_p(f'(x_m)=k)$).

ZetaX does this resemble somehow with Proth's theorem ?

[quote="ZetaX"]
So for any odd $a$ there are exactly four different $b$ with $a^2 \equiv b^2 \mod 2^k$. Since one out of four odd residues is possible to be a square , all of them must be a square (four times each),especially $17 \equiv 1 \mod 8$.
[/quote]
Sorry, somehow I don't understand this part, could you please explain?

Another possible solution (by induction).
[hide]
[b]Let $p$ be an odd prime and $a$ be an integer with $p\nmid a$. Then $a$ is quadratic residue $\mod {p^\alpha}$ iff $a$ is quadratic residue $\mod p$[/b]
Well, if $a$ is quadratic residue $\mod {p^\alpha}$ then it is obviously $\mod p$ so lets only prove the other direction.
Suppose that $\exists x: x^2\equiv a \mod {p^\alpha}$.
$\Rightarrow \exists b: x^2=a+bp^\alpha$.
Consider $(x+rp^\alpha)^2$ for some integer $r$. We have
$(x+rp^\alpha)^2= x^2+2xrp^\alpha + r^2p^{2\alpha} = a + p^\alpha(b+2xr) + r^2p^{2\alpha}$.
Now since we have $2\alpha\geq \alpha + 1$ we obtain
$(x+rp^\alpha)^2 \equiv a + p^\alpha(b+2xr) \mod{p^{\alpha+1}}$.
Now obviously $b+2xr\equiv 0 \mod{p}$ has solution in $r$, hence there exists an $r$ so that $(x+rp^\alpha)^2\equiv a \mod{p^{\alpha+1}}$.

An alike solution for $p=2$:
[b]Let $a$ be odd and $\alpha \ge 3$. Then $a$ is quadratic residue $\mod 2^\alpha$ iff $a$ is quadratic residue $\mod 8$[/b]
If $a$ is quadratic residue $\mod 2^\alpha$ then it is obviously $\mod 8$.
Suppose that $\exists x: x^2\equiv a \mod {2^\alpha}$. Obviously $x$ is odd.
$\Rightarrow \exists b: x^2=a+2^\alpha b$. If $b$ is even, it is trivial so let's suppose that $b$ is odd.
We have
$(x+2^{\alpha-1})^2=a+b2^\alpha + 2^\alpha x + 2^{2\alpha - 2}$
Since we have $\alpha \ge 3 \leftrightarrow 2\alpha-2 \ge \alpha + 1$ we have
$(x+2^{\alpha-1})^2 \equiv a + 2^\alpha(b+x) \mod{2^{\alpha+1}}$
Since both $b$ and $x$ are odd, we have
$(x+2^{\alpha-1})^2\equiv a \mod{2^{\alpha+1}}$

$\Box$
[/hide]

You gave the Hensel-like proof  :)

I proved that any value that is archieved once is at most archieved four times (since there are only four different $b$ by the previous). Now just assume that at least one residue $\equiv 1 \mod 8$ is not archieved, then there are $2^{k-3}-1$ left. Every residue is the square of at most $4$ odd residues, thus there could have been only ${4\cdot (2^{k-3}-1})= 2^{k-1}-4 < 2^{k-1}$ odd numbers $\mod 2^k$, clearly impossible.

We have $m_0=\sqrt{1+4x}=1+2\sum_{n=0}^{\infty}Ca_nx^{n+1},$ were $Ca_n$ is Catalan's number.
x=4 in 2-adic numbers give exactly form for n and m: $m=m_0(mod 2^k),n=(m^2-17)/2^k.$

[quote="Yimin Ge"]

We have
$(x^2+2^{\alpha-1})^2=a+b2^\alpha + 2^\alpha x + 2^{2\alpha - 2}$
Since we have $\alpha \ge 3 \leftrightarrow 2\alpha-2 \ge \alpha + 1$ we have
$(x^2+2^{\alpha-1})^2 \equiv a + 2^\alpha(b+x) \mod{2^{\alpha+1}}$
Since both $b$ and $x$ are odd, we have
$(x^2+2^{\alpha-1})^2\equiv a \mod{2^{\alpha+1}}$

$\Box$
[/quote]

You just have one typo, you wrote $(x^2+2^{\alpha -1})$ but it should be $(x+2^{\alpha -1})$, but never mind, beside that, this is the only solution from here that I understand completly :)

bye

Well I suppose that everything is OK, but there are lot of stuff
that I don't understand. First of all, theory of quadratic
residues-I don't know anything about that... If I start to ask
questions, there would be a hole bunch of them, but if you don't mind,
 I can look at your proof more carefully and ask
everything (and again EVERYTHING) that I don't understand :) Are you
in or not? :)