I have a project in Accelerated Geometry and there is this one problem that is stumping me.

The question is: Can you create a network that has only one, three, or five odd vertices? Can you create a network that has an odd number of odd vertices? Explain why or why not.

My answer to this is no, that you cannot create a network with only one odd vertex, only three, or only five odd vertices. You also cannot create a network that has an odd number of odd vertices.

I just do not know my reasoning as to why you cannot. Some help would be great, please!

[quote="t0rajir0u"][hide="Hint"] Each edge in a network connects exactly two vertices. [/hide][/quote]

Alright, so my reasoning for the inability to construct an odd number of odd vertices is because since an edge in a network connects exactly two vertices, it it not possibly for there to be an odd number of vertices in general, and if there was it would not be a network?

And I'll delete that other post.

[quote="bryyy"]it it not possibly for there to be an odd number of vertices in general[/quote]

Close.  Of course, it is possible for there to be an odd number of vertices (consider, say, a triangle).  What I mean is this:

You can count the number of edges in a network by adding up the number of edges to which each vertex is connected and dividing by two.  This works because each edge connects exactly two vertices.

What happens when there are an odd number of odd vertices?

[quote="t0rajir0u"]What happens when there are an odd number of odd vertices?[/quote]

Then there are an odd number of edges? And you're right, I noticed that there can be an odd number of even edges, (i.e. the vertices of a triangle as you said).

For my answer to the question I put this:
  "No, you cannot create a network with only one, three, or five odd vertices. The reason is the paths (lines and curves) of the network. One path connects exactly two vertices, to have exactly one odd vertex you would have to have at least one path coming from that vertex, but it unfortunately connects to another point giving you a network with two odd vertices.  This applies to trying to make three or five odd vertices; it always seems to end up with an even amount of odd vertices. Thus, you cannot create a network that has an odd number of odd vertices. "

This is just about what I'm getting (my geometry book gives out very limited information on networks). I know that for a network conjecture you can say: a network can be traveled whenever there are zero or only two odd points (nodes, vertices, etc.). I just can't explain why it always ends up that there are even amounts of odd vertices and odd or even amounts of even vertices.  :(

[quote="t0rajir0u"]You can count the number of edges in a network by adding up the number of edges to which each vertex is connected and [b]dividing by two[/b].  This works because each edge connects exactly two vertices.[/quote]

[hide="So when there are an odd number of odd vertices..."] Adding up the number of edges to which each vertex is connected produces an odd number.  Dividing by two, we get a [i]non-integer number of edges.[/i] [/hide]


find the constant term in the binomial expansion of $ (x \plus{} 2/x)^12$

I know how to do this problem, but am curious to see how else it can be solved.

It's supposed to show as to the 12th power! What did I do wrong? odd...

To make an exponent, type:

[code]
$a^{bc}$
[/code]

You need the { and } signs to get a double digit or more exponent. :)

We want a term where the $ x$'s cancel out, leaving us with a constant. This occurs at the term $ \binom{12}{6}x^6 \cdot \left(\frac{2}{x}^6\right) \equal{} \binom{12}{6} \cdot 64 \equal{} \boxed{59136}$. (Hopefully, I didn't mess up.)

Her solution is correct; I'm just fixing a typo (she wrote the sixth power on the 2 instead of the whole fraction 2/x).

$ \binom{12}{6}x^{6}\cdot\left(\frac{2}{x}\right)^{6} \equal{}\binom{12}{6}\cdot 64 \equal{}\boxed{59136}$

That's correct! Your solution is a bit faster than mine... I used sequences...

EDIT: Now solve the question...

$ (x^2\plus{}2/x)^{10}$

$ \binom{10}{k}x^{2k}\left(\frac{2}{x}\right)^{10\minus{}x}$.

$ 2k\equal{}10\minus{}k\implies k\equal{}10/3\not\in\mathbb{Z}$

So there isn't a constant term.


[quote="mdk"]An urn has 900 chips, numbered 100 through 999. Chips are drawn at random and without
replacement from the urn, and the sum of their digits is noted. What is the smallest number of chips that must be drawn in order to guarantee that at least three of these digital sums be equal?[/quote]

Is this problem easy or am I missing something?

[hide]Just consider the worst case scenario - you get digital sums of
1, 2, 3, 4, 5, 6, 7, 8, 9, ..., 27

Then on the $ 28^{\text{th}}$ selection, you have to get two same. Then, suppose you get
2, 3, 4, 5, 6, 7, 8, 9, ..., 26 (* 1 and 27 are unique and already chosen)

So the answer is $ 27+25+1 = \boxed{53}$, since on the 53rd try, you are bound to get something between 3 and 25, which has already been counted 2 times.[/hide]


Okay. My math teacher gave me this problem, and I'm really at loss on how to solve it.

A merchant has a block that weighs 40 pounds, and he uses it to weigh the objects that he sells in his store. He accidentally drops the block on the floor, and it breaks into 4 pieces, but now, he can measure any weight from 1-40.

(Such as, if he has a 2 pound and a 20 pound, he can measure 18 and 22, because he can place them both on the same side of the scale, or on opposite sides, so as to know if the object is 18 pounds.)

All I know is that one block is 1 pound, aha.

Thank you!

[hide="Solution"]
Generally, if you can put blocks onto only same side of the scale, it's binary.

Also, if you can put blocks onto either same side or opposite side of the scale, it's ternary.

We can measure any weights from 1-40 by making four blocks' weight 1, 3, 9, 27.[/hide]


LaTeX
quadratics
algebra
quadratic formula

can somebody solve this for p?

10(p-0.4) = 1.28 times the square root of (p(1-p))

this was from a statistics test, go figure.
(i should probably learn latex)

[hide="The answer"]
p = approx .463832343683

I had to use a calculator, but here's the steps I did

First, open all the parentheses

$10p-4 = 1.28 \cdot \sqrt{p-p^2}$

Then, square both sides.

$100p^2 - 80p + 16 = 1.6384 (p-p^2)$

Expand the right side

$100p^2 - 80p + 16 = -1.6384p^2 + 1.6384p$

Bring everything to one side

$101.6384p^2 -81.6384p + 16 = 0$

Use a calculator to factor (or use the quadratic forumula if you are really dedicated)

$p = .463832343683$ and  $-.339391634656$

The second solution is extraneous.

Thus $p = .463832343683$

[/hide]

how did you get 10(p-0.4)=10p-40?
souldn't it be 10p-4 ?

My mistake, it was a typo. Thanks for tellin me, I fixed it.

That still doesn't affect the answer, cause when I squared it, I put 16.

thanks.
i feel stupid that i didn't use my calculator that was sitting right next to me.

[quote="ankur87"]

...Use a calculator to factor (or use the quadratic forumula if you are really dedicated)...

[/quote]

How did you use the calculator to factor?  I just wrote the quadratic formula in a program and factored that way.

It depends on your calculator.

I know that a TI 85, 86, and 89 all have the ability to factor polynomials. For the 85 and 86, there should be a "POLY" button on your calculator. For the 89, it is up top.

I don't own an 83 or 84, so I don't know what their capabilities are. I would guess that the 84 can factor.

Howewver, if you are smart enough to write a program (I cannot), then that works also. As long as the final answer is the same, it doesn't matter if you factor or use the quadratic formula.


This past week, Chinese reporters stumbled upon missile defense systems located on the perimeter of Athens, Greece. They were held in detention, but later released, and this piece of news evidently came out. Do you think that drastic measures such as missile defense systems are required to ensure safe competition, or is it too much? I mean, consider the fact that the ratio of security personel to athlete is 7:1.

There have been bombings in the past, and the Israeli Olympic Team (or part of it?) was once kidnapped by terrorists (I believe they were later killed). Heavy security seems like a mighty fine idea to me.

[quote="white_horse_king88"]This past week, Chinese reporters stumbled upon missile defense systems located on the perimeter of Athens, Greece.[/quote]

That sounds more like spies using a journalistic cover. Why would properly accredited journalists to the Olympic games be so far from the Olympic venues?

Yeah - that was my original thought. The reporters had no business around the city perimeters, but it's also quite interesting how easily something like this has been kept secret by the Greeks and by the United States who helped the Greeks set up the missile security system.

Care to post a link or several? I can't find what you've said anywhere.

[quote="Peter VDD"][quote="white_horse_king88"]I mean, consider the fact that the ratio of security personel to athlete is 7:1.[/quote]

That security personel won't stop a hijacked plane heading towards the olympic stadium :D[/quote]

Well, maybe if everybody in the stadium sneezed at the plane at the same time...
Back on topic, I think it's not a bad idea, since such a big gathering may be targetted, so if it's only being used for protecting the games, it should be fine.

[quote="Peter VDD"]I don't think Greece will do anything wrong with them :)[/quote]

Sneezes, or missiles?  ;)

It's quite hot in Greece. People don't sneeze there :)

We're still waiting for some links Whitehorse.

http://www.channelnewsasia.com/stories/afp_world/view/97914/1/.html

http://www.mindfully.org/Reform/2004/Olympics-Police-Outnumber12aug04.htm

Do you feel safe yet?
 70,000 Police
 Soldiers
 Private Armed Security
 1,320+ spy cameras
 Police helicopters
 US 6th Fleet
 Patriot missile launchers
 X-ray scanners
 Metal detectors

[quote="white_horse_king88"]What does heat have to do with sneezing??? I bet people there suffer from the demonic allergy monster there too.[/quote]

For the 100 people in the stadium that are allergic to something in greece :maybe:

that really won't blow a plane away :D


On the right, you see a silver and a golden pot. One of these pots contains a treasure and the other one is empty. Assume that you can determine from the text prints which pot contains the treasure.
The text prints on the pots are:
The silver pot: "This pot is empty."
The golden pot: "Exactly one of these texts is true."
 Which pot contains the treasure?

[hide]if it were in the silver pot, the silver text is true and in order for the golden text to be true, golden text itself has to be false, which causes a contradiction. However, if it were in the golden pot, silver text is false and golden text is true as long as itself is being true.[hide="therefore"]the golden pot[/hide][/hide]

I disagree with Gehoon.

[hide="My answer"]
There are two cases. The Golden pot text is true, or it is false.
If it is true, then the Silver pot has to be false. If it is false, then if the Silver pot were true, the golden pot would be true, so that's a contradiction. In both cases, the silver pot is false, and since it says it is EMPTY, the silver pot has the treasure.
[/hide]

I disagree with how you spell my name  :huuh:
but i like your answer better than mine.


a and b are positive coprime integers. A subset S of the non-negative integers is called admissible if 0 belongs to S and whenever k belongs to S, so do k + a and k + b. Find f(a, b), the number of admissible sets.


Consider those functions $ f$ that satisfy $ f(x \plus{} 4) \plus{} f(x \minus{} 4) \equal{} f(x)$ for all real $ x$. Any such function is periodic, and there is a least common positive period $ p$ for all of them. Find $ p$.

$ \textbf{(A)}\ 8\qquad
\textbf{(B)}\ 12\qquad
\textbf{(C)}\ 16\qquad
\textbf{(D)}\ 24\qquad
\textbf{(E)}\ 32$

[hide]
$ f(x\minus{}4)\minus{}f(x)\plus{}f(x\plus{}4) \equal{} 0 \qquad(1)$

$ f(x) \minus{} f(x\plus{}4) \plus{} f(x\plus{}8) \equal{} 0 \qquad (2)$

$ (1)\plus{}(2) \Rightarrow f(x\plus{}8) \equal{} \minus{}f(x\minus{}4)$

$ \Rightarrow f(x) \equal{} \minus{}f(x\plus{}12)$

Since this applies for all $ x$ we can assume that $ 12$ is half the period.
$ \Rightarrow p\equal{}12\cdot 2 \equal{} 24\longrightarrow \boxed{D}$
[/hide]


number theory
prime numbers
number theory unsolved

Hi!

Yesterday evening I discovered an interesting property with prime numbers:

If p (prime) is greater than or equal 3, then:

1) p divides (1 + 2 +  ... + (p-1)) (this holds for all odd numbers, and is therfore not very interesting) More interesting is that
2) p divides (1^2 + 2^2 + ... + (p-1)^2), p divides (1^3 + 2^3 + ... + (p-1)^3) up to
p divides (1^(p-2) + 2^(p-2) + ... (p-1)^(p-2))

For p = 3, we only have the trivial case: 3 divides (1+2)
For p = 5, we have: 5 divides (1+2+3+4), 5 divides (1^2+2^2+3^2+4^2) and 5 divides (1^3+2^3+3^3+4^3).

Why does this property exist??
Anybody know?

Because, since $p$ is a prime it has a primitive root, that is there exists $g$ such that $\{1,2, \cdots, p-1\} = \{g,g^2, \cdots, g^{p-1} \}.$
Note that from Fermat's little theorem, we have $g^{p-1} = 1$. Thus, for each $k\in \{1,2, \cdots, p-2 \}$, we have $g^k \neq 1.$
It follows that :
$1^k, 2 ^k , \cdots , (p-1)^k$ is a permutation of $g^k, g^{2k}, \cdots, g^{(p-1)k}.$
Therefore the sum is
$S_k = g^k + (g^k)^2 \cdots +( g^k)^{p-1} = g^k \frac{1 - (g^k)^{p-1} } {1 - g^k } = 0$ from Fermat's littlme theorem.

Pierre.


inequalities
number theory proposed
number theory

how many solutions can you give me for this problem:
consider x,y:natural number and 3<=x<=y then we have this inequality:
      x^y>=y^x.

Let me refer to a post by Gyan in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22393&start=15]Which is greater[/url], post #16

Can anybody give me at least 3 solutions that you don't use logarithm.I think it's more interesting and harder if you give another solution.


So I've decided to start a thread about the upcoming North Fulton Varsity (and now also JV) Invitational Tourney.  Who's coming? :D

Yes, I know this is a bit too early, but that just shows my enthusiasm.  :|

Boru and I will come, maybe (hopefully) with some other folks.

Hmm, only one reply to this thread makes this a very bad thread.  But anyways...

Congrats to Monica and Chamblee and Allen and Walton for winning!  :)


Ok, so here's the thread. This is where you can discuss the book Harry Potter and the Half-Blood Prince, and it's contents. Spoilers etc. can be put in here, but no where else.

Well......I'm only about a third of the way through the book, so I'll be keeping out of here until I'm done, which shouldn't be too much longer. :)

1/3 done here as well. Doesn't seem like too much has really happened so far, at least nothing really huge.

well, i am only on chapter 5 and nothing has happened.

I'm done. Kept me up a bit last night, but eh. I won't spill the juicy details until someone else finishes too.

took me like 10 hours to finish it (yeah I'm a slow reader) went to bed at 4:55 am *yawn*
I'll reveal some spoilers later

i'm on chapter 10 and i'm trying to find out who the half blood prince is...
i'm guessing its
[hide]Lucious Malfoy[/hide]
or
[hide]Voldemort[/hide]
but i'm not sure

your second one is obviously wrong, jkr's been going on that its NOT harry and NOT voldy since the title was released.

[hide]BUT OMG, WHO IS R.A.B.? everyone whose finished, please help me think. aaaah! its driving me crazy.

and the last chapter was a total rip. no way to end a book. NO CLOSURE! very jkr tho. *sigh*[/hide]

Agreed, last chapter was basically cut in half. So, predictions for book 7?

[hide]I think R.A.B. is Sirius's brother, Regulus Black. Remember he was a Death Eater but switched sides so Voldemort killed him.[/hide]

WORST Harry Potter Book EVER!!!!!!!!!!!!!!!!
 :mad:  :mad:  :mad:
There was no plot!

And Rowling made it way too melodramatic.

Finished the book.....it wasn't that great....got it at midnight, but kinda surprised at the end...

There wasn't enough action!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[quote="bleumoose"][hide]SNAP. i was thinking "black" or (maybe?) "bones" but... i forgot about that. i bet you are right....

ok, what is UP with snape? i really want to side with dumbledore on this one.  especially with the argument in the woods- i think they were arguing about whether or not snape should go through with the Vow or not. also, i think that the "severus... please..." was more "please do what i asked you to beforehand" than "please spare me."

and why the body-bind on harry? i guess that was because he knew harry would try to save him, and THAT wouldn't help him (harry) at all.

and... oh gosh. so much to talk about. [/hide][/quote]

probably the body bind was to protect harry...you know, so he wouldn't do anything stupid...dumbledore mentioned harry was more important than himself...

[quote="h_s_potter2002"]Yes, even less plot than the fifth book.

I'm not sure what it is, but I thought this book was missing something.  I think it spent less time on all the things before.  Like classes (besides potions), the feasts and holidays, or something.  Instead, it spent more time on boyfriend girlfriend stuff with people, and potions.  I dunno though.[/quote]

I think the books are getting worse and worse... the best was book 3.

neither have i

and klebian, i suggest u edit ure signature

and... dumbledores dead, snapes evil, harry's not a horcrux, regulus black is not R.A.B.

[quote="Klebian"]Posting 6 times in a row?
Congrats, never seen that  before.[/quote]

if people post more.. it wouldnt happen

not really. all 6 were in one day, and u made 5 of them in a row in a matter of 11 minutes.

harry didn't shake hand at the leaky cauldron

[quote="jhollenbeck"]Harry Potter is a bad book! I dont read it![/quote] :rotfl:

[quote="HoaXeD"]would you like to cry some more?

fine, ill lower my posts and put it all in one from now on...

after reading the 1st book again.. a few Q's popped up..

1. How come Harry and Quirrel shook hands in the Leaky Cauldron, but when they meet again Harry burns him? i know why it burns him, but i wana kno why it DIDNT...

2. Why is Snape always wanting to know how far Quirrel's gotten to getting to the stone? If he was working for Voldemort, then why wasnt Quirrel keen to help Snape, if they both wanted the Stone for Voldemort. If he wasnt working for Voldemort, why didn't he turn Quirrel into Dumbledore?

Book 6:
Is it just coincidence that the chapter ''The lightning struck tower'' Is from trelawney's card? I dont think so. Could it have anything to do with harry's scar?

And is it possible that Dumbledore is an Animagus... As a pheonix?
Dumbledore's wand was never found, was it? Because he still needed it... His body wasnt seen at the funeral. His resting place set fire around the edges. He turned into a pheonix, harry swears he saw a pheonix joyfully fly into the blue. By the time the ring of fire was exhausted, a tomb was over the 'body'... There wasnt a body, and it was dumbledore singing the lament, soothing and calming everyone.

Book 4: Dumbledore learns that Harry's blood is in Voldemort. His eyes gleam when Harry tells him this. Could it be because he realised that the connection between Voldemort and Harry would be even stronger? Their wands, both from Fawkes. Harry's blood rushing through both of them. Seeing eachother at times and controlling one another... Bonds strong enough to destroy one if the other dies, just like the prophecy says?
[url]http://www.the-leaky-cauldron.org/#book:7[/url]
Theres some good stuff there... Check it out.[/quote]

these r some good questions.

for 1. i m not exactly sure Harry and Quirrel shook hands or not,
i no for sure that he shook hands with that dedelasus diggle guy or something like that

2. in book 6, when talking to draco's mom and bellatrix lestrange, he says that all he saw was quirrel going for the stone. he had no clue that voldemort was involved.

book 6. 1. i do not think it was a coincidence, something magical happened.

book 6. 2. i dont think dumbledore could be a pheonix, but he could certainly be an animagus

book 4. 1. nah i dont think so, the prophecy says neither can live while the other survives.

book 1. he did shake hands, it says so.
book 6. if snape said he saw quirrel going for the stone, why didnt he tell dumbledore? and why was he always asking how far he'd gotten in book 1 so fiercely? why pursue the matter if he was on dumbledores side instead of just turning him in?

the prophecy does say neither can live while the other survives, so shouldnt that mean that they both are doomed? if one survives, neither can live. read the prophecy backwards.
While the other survives, neither can live.

Uh...I'm not sure i follow your logic. While is a conditional statement. If the other doesn't survive then the statement "neither can live" is no longer necessarily true. I'm not sure I understand the statement either. Based on it, there is a distinction between surviving and living.

OK
I seriously like the idea that Dumbledore could have turned into Fawkes or something

It was a HUGE disappointment when Dumbldore died...like it's not the end

...... and it isn't

spooky!
tspa

I think that possibly Dumbledore and Snape turned into eachother with the polyjuice potion. Voldemort had probably told Snape about the Horcrux. When 'Dumbledore' was drinking the potion and saying "Please dont hurt them, its all my fault, kill me, kill me instead" could be talking about the night at the Potters when Lily and James died. And did we ever see the the old ring during this time?

I think that when Snape was saying about how he knew about ''the plan'' at spinners end, he was just bluffing or reding narcissa's mind. he never got into details about the plan, and he hesitated at the end when they were talking about how he must do the job draco was assigned to if draco should fail.

half the book is about Harry falling in love with Ginny!!

[quote="kheg_k"]hoaxed, what the prophecy means is [color=red]neither[/color] can live while the other survives. not neither as in both of them, neither as in it wouldnt be that only harry would die if voldemort lived, voldemort would [b]also[/b] die if harry lived.

am i being clear enough? i dont no how to explain it.[/quote]

And the prophecy doesnt suggest that potter shall kill voldemort or voldemort harry; Draco could kill Voldemort :)

In book 7, I'm betting on some internal dissent in the Death Eaters myself...no evil organization is complete without internal dissent. :D

And I don't think R.A.B. is Regulus; I don't get the feeling that he'd destroy a horcrux (based on what we know about the character).

And dumbledore isn't dead; considering half the book was about nonverbal spells, I personally subscribe to the theory that Snape used some other spell on Dumbledore nonverbally while saying "Avada Kedavra"


There is a square plot of land, and there is a sidewalk around
the plot in a square shape that has a uniform width to it.
If the outer perimeter of the sidewalk is 32 feet longer than
the perimeter around the square plot, then how wide is the
sidewalk?

[hide]If the plot of land has side $ s$ and the sidewalk has width $ x$, we can write

$ 4(s\plus{}2x)\equal{}32\plus{}4s\\
4s\plus{}8x\equal{}32\plus{}4s\\
\boxed{x\equal{}4}$.[/hide]


Looks like you have the smiley thing solved; click profile after logging in to set your flag.

[quote="VIDIANI"]-1) how put a smiley in my messages ?  :(

(2) how have a frenc flag under my name ? :D[/quote] I have manually set up your flag, so don't worry about it anymore. :)

[quote=samrocksnature]Wait just looking at old posts but what did VIDIANI mean by frenc flag? It sounds fun...or is this some removed feature[/quote]

I'm pretty sure it just meant he wanted a French flag under his name (maybe back then you had flags to say your nationality or stuff)

[quote=X_star12321]if you want to know how to do a normal :) without the emoji, do this:
:$ $)
[code]:$ $)[/code][/quote]

I don't get how that relates to the topic, and you could just disable BBCode.

[quote=awin][quote=TheFlamingoHacker]Please do not bump old threads. I’m pretty sure all of these posts will be deleted.[/quote]

(S)he had a valid question. There's nothing wrong with bumping a thread if you're adding something.[/quote]

Although it probably won't get a response from rrusczyk this time ;)

I don't think he's that active anymore :(

[quote=OliverA][quote=pith0n]Bump. I was also wondering about the flag. I feel like that would be a really cool feature.[/quote]

2004-era AoPS was very different. In 2011 or 2015 (or both!) there were major changes to the site that have brought it much closer to its current look. While the old website design had some really neat features, most will agree that its design is dated, and I don't image AoPS re-implementing features that they took out when they did a major overhaul on the sites appearance due to the amount of effort put in to remove such features (removing wouldn't be bad, but making new code is).

Another thing to consider is that the old AoPS community was very different compared to now in many ways, the flag just doesn't fit this community now.[/quote]

We aren't exactly asking for the feature to be in AoPS now, but we are wondering what the flag was exactly.

I quoted "I feel like that would be a really cool feature" specifically, because it implies a want for that specific feature. I wasn't responding to the rest of the posts.


I know that this topic has been discussed many times on this forum.  I've spent hours reading through this forum trying to find formulas. :blush:   Does anyone have a link to a document with formulas or a website with formulas?  I could really use them because i have state on Saturday.

Well, maybe you could look in the formulas topic  ;) .

It has about 14 pages of stuff thats just waiting to be read.  :arrow:  :arrow:

Just browse around, there are good links, lock.


geometry
geometry proposed
U sviti matematyky

The incircle of triangle $ ABC$ with center $ I$ touches the sides $ AB$ and $ BC$ at points $ K$ and $ P$ respectively. The bissector of angle $ C$ intersects the segment $ KP$ at point $ Q$ and the straight line $ AQ$ intersects the side $ BC$ at point $ N.$ Prove that points $ A,I,N$ and $ B$ lie at a common circle.

notice A,P,Q,I   B,K,I,P are concyclic respectively
the result follows immediately

Here is my solution:
Using Inversion and simple angle chasing.
Let M be the tagency point of (I) with AC. Let A' and B' respectively be the midpoints of KM, KP.
Invert around I, with radius r, that maps B into B', A into A', N into N'. We need to prove that A', N', B' are collinear so that A, I, B, N are concyclic.
It is easy to see that through the inversion of pole I, radius r, we have:
Q maps in to Q', Q'A'IN' is concyclic. BC maps into the circle with IP its diameter, IN'B'P  is concyclic.
It needs to prove that (N'A',N'I)=(PB',PI).
We have (N'A',N'I)=(Q'A',Q'I)=(AA',AQ)=(AI,AQ). So we need to prove that (IQ,IA)=(IP,IB), which is true because (IP,IB)=(KP,KB)=(KQ,KA)=(IQ,IA) (IAQK is concyclic- well known result :lol: ).
So triangle IAQ are similar to triangle IBP. Hence (AI,AQ)=(BI,BP)=(PB',PI). Therefore (N'A',N'I)=(PB',PI)=(N'B',N'I). This leads to the fact that A', N', B' are collinear. :lol:
Our proof is completed. :blush:

Another approach.

$ \angle CIA\equal{}\frac{B}{2}\plus{}90^{o}, \angle CKQ\equal{}\angle IKC\plus{}\angle QKI\equal{}90^{o}\plus{}\frac{B}{2}$
and, $ \angle BCI\equal{}\angle ICA$
so, $ \triangle CKQ and \triangle CIA$ are spiral simiral triangle.
so, $ \angle IBK\equal{}\angle IAN$
Then, we are done.

Another solution :
We have $ \angle PKC\equal{}\angle BPK\minus{}\angle BCQ\equal{}\frac{\angle BAC}{2}\equal{}\angle KAI$
$ \Rightarrow$ quarilateral $ AIKQ$ is cyclic.
In addition , $ \angle PBI\equal{}\angle PKI\equal{}\angle IAN$
We have done

Other approach is to see that $ AQ \perp CQ$, and to prove this we shall work backwards, it seems to be very easy: take $ AN \perp CI$, {Q} = AN∩CI, $ N \in BC$. It’s easy to see that CN=AC, CP=p-c, PN=p-a=AK, PB=BK and, by reciprocal of Menelaos, P, K and Q are collinear. Hence, in our problem, PK intersects CI at a point Q | $ CQ \perp AQ$. Further from DACQ we get $ \angle QAI \equal{} \frac {\angle B} {2} \equal{} \angle CBI$, hence A, N, B and I are concyclic.

Best regards,
sunken rock


inequalities
algebra
polynomial
inequalities unsolved

[u]Problem 1[/u]
[i] If a,b,c[/i] $ \in[\frac {1}{\sqrt {2}}, \sqrt {2}]$
           Prove that
  $ \frac {3}{a \plus{} 2b} \plus{} \frac {3}{b \plus{} 2c} \plus{} \frac {3}{c \plus{} 2a} \geq \frac {2}{a \plus{} b} \plus{} \frac {2}{b \plus{} c} \plus{} \frac {2}{c \plus{} a}$

[u]Problem 2  [/u][b] [i] If a,b,c[/i][/b] $ \in$[i] [b]R[/b][/i]
 Prove that
  $ ( a^2 \plus{} b^2 \plus{} c^2 )( a^4 \plus{} b^4 \plus{} c^4 ) \geq ( (a \minus{} b)(b \minus{} c)(c \minus{} a))^2$
[i][b] Argady , Ji Chen , Virgil Nicula, Darij .....  Can you help me ? :blush:[/b][/i]

the first one is very nice where did you find it? i didn't find any solution yet. :(

The first solution is easy! But nice :D
Write that inequality as follows
$ \sum_{cyc}(\frac{3}{a+2b}-\frac{2}{a+b})\geq 0$
 $ \sum_{cyc}(\frac{a-b}{(a+2b)(a+b)})\geq 0$
 $ \sum_{cyc}(\frac{a-b}{(a+2b)(a+b)}+\frac {1}{6}(\frac {1}{a}-\frac{1}{b}))\geq 0$
 $ \sum_{cyc}(\frac{(a-b)^2(2b-a)}{6ab(a+2b)(a+b)})\geq 0$
Since $ 2b-a\geq \frac{2}{\sqrt{2}}-\sqrt{2}=0$
The inequality is obviously true. :D

[quote="nkht-tk14"]

[u]Problem 2  [/u][b] [i] If a,b,c[/i][/b] $ \in$[i] [b]R[/b][/i]
 Prove that
  $ ( a^2 \plus{} b^2 \plus{} c^2 )( a^4 \plus{} b^4 \plus{} c^4 ) \geq ( (a \minus{} b)(b \minus{} c)(c \minus{} a))^2$
[/quote]
$ ( a^2 \plus{} b^2 \plus{} c^2 )( a^4 \plus{} b^4 \plus{} c^4 ) \geq ( (a \minus{} b)(b \minus{} c)(c \minus{} a))^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6\plus{}2a^3b^3\plus{}2a^4bc\minus{}2a^3b^2c\minus{}2a^3c^2b\plus{}2a^2b^2c^2)\geq0,$ which true by Miurhead.

We can assume that c=min{a,b,c}.
It is easy to show that $ (a^2\plus{}b^2\plus{}c^2)(a^4\plus{}b^4\plus{}c^4)\geq(x^2\plus{}y^2)(x^4\plus{}y^4)$
where $ x\equal{}a\minus{}c, y\equal{}b\minus{}c\geq 0$
Thus, we still have to show that $ (x^2\plus{}y^2)(x^4\plus{}y^4)\geq x^2y^2(x\minus{}y)^2(\equal{}((a\minus{}b)(b\minus{}c)(c\minus{}a))^2)$
or $ x^2y^2(x\plus{}y)^2\geq 0.$
The inequality is proved. :D
nkht_tk14: Are you Nhat Linh, or Nguyen???

[quote="nkht-tk14"]If $ a,b,c\in\left[\frac {1}{\sqrt {2}}, \sqrt {2}\right]$, prove that

$ \frac {3}{b \plus{} 2c} \plus{} \frac {3}{c \plus{} 2a} \plus{} \frac {3}{a \plus{} 2b} \geq \frac {2}{b \plus{} c} \plus{} \frac {2}{c \plus{} a} \plus{} \frac {2}{a \plus{} b}$.[/quote]$ \frac {3}{b \plus{} 2c} \plus{} \frac {3}{c \plus{} 2a} \plus{} \frac {3}{a \plus{} 2b} \minus{} \frac {2}{b \plus{} c} \minus{} \frac {2}{c \plus{} a} \minus{} \frac {2}{a \plus{} b}$

$ \equiv\frac {F(a,b,c)}{(b \plus{} c)(c \plus{} a)(a \plus{} b)(b \plus{} 2c)(c \plus{} 2a)(a \plus{} 2b)}$.

$ a,b,c\in\left[\frac {1}{\sqrt {2}}, \sqrt {2}\right]\Longrightarrow b \plus{} c > a, c \plus{} a > b,a \plus{} b > c$. Without losing generality, we assume $ a \equal{} \max\{a,b,c\}$.

$ F(a,b,c) \equal{} F(y \plus{} z,z \plus{} x,x \plus{} y) \equiv G(x,y,z) \equal{} G(x,x \plus{} s,x \plus{} t)$

$ \equal{} 96\left(s^2 \minus{} st \plus{} t^2\right)x^3 \plus{} 12\left(10s^3 \minus{} 11s^2t \plus{} 5st^2 \plus{} 10t^3\right)x^2$

$ \plus{} 8(2s \plus{} t)(s \plus{} t)\left(3s^2 \minus{} 7st \plus{} 6t^2\right)x \plus{} (2s \plus{} t)\left(3s^4 \minus{} 2s^3t \minus{} 8s^2t^2 \plus{} 7st^3 \plus{} 6t^4\right)$

$ \geq 0$,

which is clearly true for

$ 3s^4 \minus{} 2s^3t \minus{} 8s^2t^2 \plus{} 7st^3 \plus{} 6t^4 \equal{} \frac {(3s \plus{} 4t)(4s \plus{} 3t)(2s \minus{} 3t)^2 \plus{} 6s^4 \plus{} 8s^3t \plus{} 45st^3}{18}$.

By the way, if $ a,b,c\in\left[\frac {1}{d}, d\right]$, then

$ \frac {3}{b \plus{} 2c} \plus{} \frac {3}{c \plus{} 2a} \plus{} \frac {3}{a \plus{} 2b} \geq \frac {2}{b \plus{} c} \plus{} \frac {2}{c \plus{} a} \plus{} \frac {2}{a \plus{} b}$,

where $ d \equal{} 1.9676\cdots$ be a root of the following irreducible polynomial

$ 11492d^{32} \plus{} 59736d^{30} \plus{} 200208d^{28} \minus{} 733396d^{26} \minus{} 4894679d^{24} \minus{} 6662736d^{22}$

$ \plus{} 2856552d^{20} \plus{} 12029308d^{18} \plus{} 4316382d^{16} \minus{} 8027300d^{14} \minus{} 6103776d^{12}$

$ \plus{} 2970528d^{10} \plus{} 3045985d^8 \minus{} 959764d^6 \minus{} 498600d^4 \plus{} 139080d^2 \plus{} 11492$;

with equality if $ a \equal{} d, b \equal{} \frac {1}{d}, c \equal{} 0.90689\cdots$ be a root of the following irreducible polynomial

$ 132066064c^{32} \minus{} 130181376c^{30} \plus{} 1979783240c^{28} \plus{} 5956780144c^{26}$

$ \minus{} 35317512999c^{24} \minus{} 25133584540c^{22} \plus{} 167907591990c^{20} \minus{} 83364196068c^{18}$

$ \minus{} 112132713231c^{16} \plus{} 71179712840c^{14} \plus{} 72310474344c^{12} \minus{} 10073962016c^{10}$

$ \minus{} 25506666376c^8 \minus{} 8427896928c^6 \minus{} 862368320c^4 \plus{} 438336c^2 \plus{} 1296$.

Ji chen , your proof is very ugly and hard
  please try another .... :maybe:


induction
combinatorics unsolved
combinatorics

Find all subsets A, B satisfying three conditions :
 a) $ A $  :cup:  $ B $ = $ Z $
 b) If $ x \in A $ then $ x-1 \in B $
 c) If $ x, y \in B $ then $ x+y \in A $

- If $0 \in B$ then $0 = 0+0 \in A$, so that $-1 \in B$. Thus if $x \in B$ then $x-1 \in A$. (1)
By an easy induction, we deduce that, since $0 \in A \cap B$ then any $n \leq 0$ belongs to $A\cap B$.

If there is no $a \geq 3$ in $A$ then $3 \in B$ and $6 \in A$, a contradiction. Thus, we may find $a \geq 3$ which belongs to $A$.
Note that if $x \in A$ then $x-1 \in B$ and $2(x-1) \in B.$
Let $x_0 = a$ and $x_{n+1} = 2x_n - 2$. From above we deduce that all $x_n$'s are in $A$. Since $a \geq 3$, this sequence has limit $+ \infty$ so that $A$ contains even integers as large as we want. It follows that $B$ contains odd integers as large as we want, and from (1) and the analogous for $B$, we easily deduce that all even non-negative integers are in $A$, and then all the odd non-negative integers are in $B$.
But, since $-2 \in B$, we deduce that if $x \in B$ then $x-2 \in A$. Therefore, all the odd integers are in $A$ too, and the all the even integers are in $B$ too.
Thus, $A = B = \mathbb {Z}$, which clearly form a solution.

- If $0 \notin B$ then $0 \in A$ and $-1 \in B$. Thus we still have (1).
An easy induction shows that $-2 \mathbb {N} + 1 \subset B$ and $-2 \mathbb {N} \subset A$. (2)
In another hand, using (1)and the similar relation for $B$, we deduce that if $x \in A$ then $x-2 \in A$ and the same for $B$. Since $0 \in A$ and $0 \notin B$ it easily follows by induction that no non-negative even integers can belong to $B$ and no non-negative odd integer can belong to $B$. Thus $2 \mathbb {N} \subset A$ and $2 \mathbb {N} \cap B = \emptyset$. And $2 \mathbb {N} + 1 \subset B$ and $2 \mathbb {N} +1 \cap A = \emptyset.$ Thus, working backward, we deduce that the inclusions in (2) are strict.
Therefore, in that case, $A = 2 \mathbb {Z}$ and $B = 2 \mathbb {Z} + 1$, which form a solution of the problem.

Pierre.


analytic geometry
function
vector
trigonometry
calculus
derivative
geometry

Let P be an interior point of a regular n-gon $ A_1 A_2 ...A_n$, the lines $ A_i P$ meet the regular n-gon at another point $ B_i$, where $ i\equal{}1,2,...,n$. Prove that sums of all $ PA_i\geq$ sum of all $ PB_i$.

The most [i]crucial idea [/i]is the following lemma:
[b]
Lemma:[/b]
Let $ n$ be some positive integer, greater than $ 3$.For an arbitrary set of points, $ A_1,A_2,\dots, A_n$, define point $ P$, which lies in the plane of points $ A_1,A_2,\dots, A_n$, as a point, sum of distances from which to the points $ A_i$ reaches minimum. Then the following equation holds:
\[ \sum_{i \equal{} 1}^{n}\frac {\overline{PA_i}}{|PA_i|} \equal{} 0
\]
[b]
Proof:[/b]

Let me just to sketch it out:

Suppose that $ (x,y)$ are the coordinates of the point $ P$, and $ (x_i,y_i)$, where $ i \equal{} 0,1,2,\dots,n$, are the coordinates of points $ Q,A_1,A_2,\dots, A_n$, respectively.

Denote by $ f_Q(x,y) \equal{} |\overline{PQ}|$. If positive reals $ \alpha, \beta$ satisfy to the following conditions:

[u]1.[/u] $ \alpha \plus{} \beta \equal{} 1$

[u]2.[/u] There exist two points $ P'(x',y')$ and $ P''(x'',y'')$, so that:
\[ x \equal{} \alpha\cdot x' \plus{} \beta\cdot x'', y \equal{} \alpha\cdot y' \plus{} \beta\cdot y''
\]
Therefore,

$ f_Q(\alpha\cdot x' \plus{} \beta\cdot x'', \alpha\cdot y' \plus{} \beta\cdot y'') \equal{} |\overline{PQ}| < \alpha|\overline{QP'}| \plus{} \beta|\overline{QP''}| \equal{}$

$ \equal{} \alpha f_Q\left(x',y'\right) \plus{} \beta f_Q\left(x'',y''\right)$.

It means that $ f_Q$ is a strictly convex function, hence, it has only one minimum point.

Afterwards, we might use the fact that, in order to reach minimum at some point, for every unit vector $ \overline{e} \equal{} (\cos{\alpha},\sin{\alpha})$, derivative of the function in the line of $ e$ must be nonnegative.

Using this property and recalling that sum of derivatives is derivative of the sum, we get the desired equation. $ \square$.

[i]Obviously if these points, $ A_1,A_2,\dots A_n$ form regular $ n$-gon, then the point $ X$ coincides with center of $ n$-gon. [/i]

I am going to prove a slightly stronger result, namely, define points $ B_i$ as intersection with circumcircle of $ n$-gon, instead of intersection with its boundary.

If $ O$ is the center and $ R$ is the circumradius, then

$ \sum_{i \equal{} 1}^{n}PA_i \plus{} \sum_{i \equal{} 1}^{n}PB_i\leq 2 \sum_{i \equal{} 1}^{n} OA_i\leq 2\sum_{i \equal{} 1}^{n} PA_i$.

Therefore,

$ \sum_{i \equal{} 1}^{n} PA_i\ge \sum_{i \equal{} 1}^{n} PB_i$ $ \blacksquare$

Is there any easier, non-calculus proof of Erken's lemma?

We look $A_i=A_j$ iff $n \mid i-j$
Note that if $k=\lfloor \frac{n}{2} \rfloor +1$, $A_1A_k$ is longest diagonal of that polygons
We have $A_iP+PB_i =A_iB_i \leq d$ and $A_iP+PA_{i+k}\geq A_iA_{i+k}=d$ for each $i=1,2,...,n$
Summing up the inequality give us $2\sum_{i=1}^{n}{PA_i} \geq nd \geq \sum_{i=1}^{n}{PA_i +PB_i}$
So $\sum_{i=1}^n{PA_i} \geq \sum_{i=1}^{n}{PB_i}$

[quote=ThE-dArK-lOrD]
Note that if $k=\lfloor \frac{n}{2} \rfloor +1$, $A_1A_k$ is longest diagonal of that polygons
We have $A_iP+PB_i =A_iB_i \leq d$ and $A_iP+PA_{i+k}\geq A_iA_{i+k}=d$ for each $i=1,2,...,n$
[/quote]
when $n$ is even, $A_iA_{i+k}$ is not the longest diagonal of that polygon, but $A_iA_{i+k-1} is$.


Let $ ABCD$ be a trapezoid inscribed in a circle $ \mathcal{C}$ with $ AB\parallel CD$, $ AB\equal{}2CD$. Let $ \{Q\}\equal{}AD\cap BC$ and let $ P$ be the intersection of tangents to $ \mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.

Denote by $ 2a=AB$ and $ b=AD=CB$. We deduce that $ DC=a$. Denote by $ x=S_{\triangle PQD}$ and by $ h$ the altitude the triangle $ QAD$.

$ \angle CBP=\angle QDP$, $ CB=DQ$ and $ PD=PB$ $ \Rightarrow S_{\triangle QDP}=S_{\triangle PCB}$ $ \Rightarrow S_{\triangle QPA}=S_{\triangle QPB} \Rightarrow ABPQ$ is a trapezoid.

Denote by $ R=DP \cap AB \Rightarrow \triangle RAD \sim \triangle ADC \Rightarrow AD^2=RA \cdot DC \Rightarrow RA=\frac {b^2} {a}$.But $ PQ=RA \Rightarrow PQ=\frac {b^2} {a}$

$ S_{\triangle PQA}=2x=\frac {h \cdot PQ} {2} \Rightarrow h=\frac {4x} {PQ}=\frac {4ax} {b^2}$

$ S_{ABPQ}=S_{ABQ}+S_{BPQ}=\frac {h \cdot 2a} {2}+2x=\frac {4ax \cdot 2a} {2 \cdot b^2}+2x=\frac {4a^2x} {b^2}+2x$

I think there is a mistake, you can't take altitude of ADQ because these three points are colinear.


algebra
polynomial
function
algebra unsolved

Let p(x) be the polynomial [tex]x^3 + 14x^2 - 2x + 1[/tex]. Let [tex]p_{n}(x)[/tex] denote [tex]p(p_{n-1}(x))[/tex]. Show that there is an integer N such that [tex]p_{N}(x) - x[/tex] is divisible by 101 for all integers x

well It is a nice problem:

It is easy to see $P(x)=(x+72)^3-72^3+1=Q(x)$ modulo 101, thats it, both polynomials are identical, so we only work with $Q(x)$.

Lemma $a^3=b^3$ mod101 only has the solution $a=b$. Proof: $ab^{-1}$ must have order $1$ or $3$, but it cannot be $3$ since it doesnt divides $100$, otherwise we also have $a=b=0$

For each $a$, consider the sequence $0,Q(a)-a, Q_2(a)-a....$ And see it mod101, then it eventually becomes periodic, from Lemma 1 we deduce it is a pure period, that means $Q_{mT}(a)=0$ for some $T$ and all $m$.

So from here, just take Icm of all periods obtained and this number works for $N$

oh,i forgot this topic.But i don't think your solution is true,Pascual 2005.Remember that e must prove there is an integer N such that [tex] p_{N}(x)-x [/tex] is divisible by 101 for all integers x,not for some known value

Already on this site : http://www.mathlinks.ro/Forum/viewtopic.php?p=14148#14148

Anyway, ... Let $Q(x)$ be the function  $Q:\{0,...,p-1\}\longrightarrow \{0,...,p-1\}$. Let  $Q_n(t)$ denote $Q_n(t)=Q(Q_{n-1}(t))$.
 If $Q$ is injective then there is $N$ such that $Q_N(t)=t$ (try to prove it :D )

But Pascual2005 proved that $Q(t)$ defined by $Q(t)=P(t)(mod\  p)$ is injective so .. that's all  ..

  :cool:


It's almost here!
I'm going to the HCSSiM 2009, and if you are going or planning to go in the future, then you should definitely post here, along with something about yourself.
My name is Yushi wang, and I'm going to be a junior. This is also my first time going to the east coast, so I'll see what it's like!

Hi, Yushi!
I'm going to HCSSiM this year too! I can't believe that it's only three days away.
I will be a junior and I live in Pennsylvania.
Where do you live?

Hey, I guess it's kinda late to be posting on the forum, but I'm Rachel, and I'm from Georgia. I'll be a junior next year. I hope to meet all of you tomorrow!

I'm so excited! :lol:


function
real analysis
real analysis unsolved

let f,g be functions from R to R s.y.
 f(x)<g(x) for all x in R.show there is infinite subset P  of R s.t.
 f(x)<g(y) for all x,y in P.


For any n > 1, find n points on the plane such that no three are collinear and the distance between any two is an integer.

define positive integers a_1,a_2,...,a_n and b_1,b_2,...,b_n as follows (Pell's equation):
(i) a_i^2+1=2b_i^2, i=1,2,...,n
(ii) a_1<a_2<...<a_n

define points P_i=(2a_i/(a_i^2+1),(a_i^2-1)/(a_i^2+1)), i=1,2,...,n

the distance between any two distinct points is rational

after multiplying by a suitable large integer, the distance between any two distinct points is integer

PS: This solution is based on the solution of IMO 1975/5. To prove that no three points are collinear, apply the substitution P_i=(cos(x_i),sin(x_i)).

In another hand, a well-known result of Anning and Erds states that if an infinite plane set E of points is such that for any two points A and B of E the distance AB is an integer then E is contained in a line.

Another exercise in the same taste :

Let E be an infinite plane set of points such that for any three points A,BC of E, with B =/= C, the distance from Ato the line (BC) is an integer.
Prove that E is contained in a line.

Pierre.

On last problem.

Suppose E isn't contained in one line.
Let A, B, C are three points from E, ABC is nondegenerative triangle.
Since E is infinite set we can choose sequence X1, X2, ... from it. Let en be such unit vestor that AXn=en*|AXn|. Now we can select converging subsequence enk. Suppose WLOG en converges ==> d(B,AXn) converges too. Since d(B,AXn) is an integer we have d(B,AXn)=m \in N for all n>N, i.e. all Xn (n>N) and A are contained in one line L.
For each point X from E triple (d(X,AB),d(X,BC),d(X,CA)) (where d(X,AB) is oriented distance) is unique. Hence Xn is unbounded sequence. Suppose WLOG |AXn| is infinitely increasing sequence. We know (B is not in L) or (C is not in L). Suppose B is not in L. Consider d(A,BXn). We can state d(A,BXn)->D for some D and d(A,BXn)<>D for sufficiently large n (geometricaly obvious statement). But d(A,BXn) is an integer, so it is necessary d(A,BXn)=D for sufficiently large n. Contradiction.


$ P \in Z[x]$. $ P$ is irreducible over $ Z$. Show that $ \forall N \in R$, $ \exists n \in Z \cap (N,\plus{}\infty)$, $ p$ is a prime, $ p | P(x), p^2\nmid P(x)$.

[quote="1001001"]$ P \in Z[x]$. $ P$ is irreducible over $ Z$. Show that $ \forall N \in R$, $ \exists n \in Z \cap (N, \plus{} \infty)$, $ p$ is a prime, $ p | P(x), p^2\nmid P(x)$.[/quote]
Did you mean n instead of x.

are you asking that every irreducible polynomial takes on infinitely many squarefree values?
(this is not true if $ gcd(f(n))\not{\equal{}}1$ in general...) or you just want the existence of such a prime?

(as far as i know this has been proven only for deg f<4, and the general conjecture is open, in the cas eof squarefree numbers)

I am sorry, $ n$ instead of $ x$ and I just want the existence of such a prime $ p$.


[quote="arqady"][quote="skywalker"]Given $x_{1},...,x_{n}\in [-1,1],x_{1}+...+x_{n}=0$ prove that
$\sqrt{1+x_{1}+x_{2}^{2}}+\sqrt{1+x_{2}+x_{3}^{2}}+...+\sqrt{1+x_{n}+x_{1}^{2}}\ge n$
 :D[/quote]
It's true for all $x_{i}\in\mathbb R,$ such that $\sum_{i=1}^{n}x_{i}=0.$:wink:[/quote]
Sorry you are wrong
Try when $n=3, x_{1}=2,x_{2}=-2,x_{3}=0$ then it doesn't exist the root of the express $\sqrt{1+x_{2}+x_{3}^{2}}$ :D

I don't understand
I meant $x_{i}\in [-1,1]$ need for this inequality while you said it didn't need, did you? That's all

In "Sang Tao BDT" of Hung, a particular case of this problem (for n=3 and similarly for n=4) is in Chapter II (problem 2.3x, I don't remember exactly). Phan Thanh Nam proposed this problem (for n=3,4). But I don't now it is still true forn $n$. Nice result.

But after all of them you should know more this problem is of me Luke Skywalker :D

It can be pro by minkowki ' ineq
[url]http://en.wikipedia.org/wiki/Minkowski_inequality[/url]

If you only know to write a name of a famous inequality then the best you shouldn't post any thing  :D , now if you know more, post all solution.

OK... This problem is of yours (as general problem) but of Phanthanhnam (as particular case)  :P  Are you agree?  :)

Solution in Hung's book in Vietnamese:

If $xy\ge 0$ then
\[\sqrt{1+x}+\sqrt{1+y}\ge 1+\sqrt{1+x+y}.\]

Suppose that $(x+y^{2})(y+z^{2}) \ge 0$  (we can always suppose that) then
\[\sqrt{1+x+y^{2}}+\sqrt{1+y+z^{2}}+\sqrt{1+z+x^{2}}\ge\]
\[1+\sqrt{x+y+y^{2}+z^{2}}+\sqrt{1+z+x^{2}}\]
\[=1+\sqrt{(1-z+z^{2})+y^{2}}+\sqrt{(1+z)+x^{2}}\]
\[\ge 1+\sqrt{(\sqrt{1-z+z^{2}}+\sqrt{1+z})^{2}+(x+y)^{2}}\]
It remains to prove that
\[\sqrt{({1-z+z^{2}}+\sqrt{1+z})^{2}+(1-z)^{2}\ge 4.}\]
This last one is so easy.

Very nice proof? Isn't t?

ok it is nice for three variables  :D , but does anyone have solution for mine in n-variable case ?


(x^-1)+(y^-1)+(z^-1)=1
x,y,z in R+      then prove 8+(z-1)((x-1)(y-1))^2   >=   4xy/z^2 + 4xy

sorry my handwritten is very bad!! :oops:  :blush:

do you mean:

$ \frac 1x \plus{} \frac 1y \plus{} \frac 1z \equal{} 1\Rightarrow 8 \plus{} (z \minus{} 1)\left((x \minus{} 1)(y \minus{} 1)\right)^2\geq \frac {4xy}{z^2} \plus{} 4xy$

 :?:  :maybe:


AMC
AIME
AMC 10
geometry
search
function
number theory

Hi,

I'm going to be a 10th grader next year. I took the AMC10 this year and I think i cam pretty close 113.5/120. I have tried AIMEs from previous years and got like a 2 or 3. Any tips on how to prepare for these exams?

THanks

Anh Nguyen

Well, first of all, there is the obvious: take previous AMC/AIMEs to practice.  Identify the kinds of problems you tend to miss (geometry, algebra, number theory, whatever you think is your weakest area) and work on that area.  Another thing you can do is find problems of the same style from other sources; you may very well see a similar problem on an AMC or AIME, and if you do, you'll know how to solve it this time around.  The AoPS books are a big help too; if you have them, study them!

Look for stuff in the forums. Try solving problems everyone puts up or joining a contest. Also, check out the AMC Contests FAQ that's a sticky.

Use this forum's search function to look for earlier threads with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12172]advice on how to prepare[/url] for the AMC contests.

Good luck.


group theory
abstract algebra
superior algebra
superior algebra unsolved

$ K\equal{}\mathbb{Q}(\sqrt{2\plus{}\sqrt{2}})$ is a Galois extension of $ \mathbb{Q}$ [I showed this]. Determine $ \text{Gal}(K/\mathbb{Q})$ and describe the lattice of subfields $ \mathbb{Q} \subset F \subset K$.

I found that $ \text{Gal}(K/\mathbb{Q})\equal{}\mathbb{Z}_4$. I do not know how to draw the lattice of subfields, so far I have this: [see image], but I do not think it is right. I need help with the lattice of subfields.

[img]http://lh4.ggpht.com/_miqRA1rlzTU/SUB8n1XQUJI/AAAAAAAAADw/KYIJYYCyUXc/1.PNG[/img]

How many subgroups of $ \mathbb{Z}_4$ are there, besides the identity and $ \mathbb{Z}_4$ itself?  The answer is the number of fields you are missing.

Also, try to think of some simpler field that must be included in $ K$.


function
number theory proposed
number theory

Find f:Q->Q if for any q  ,  f^-1(|q|) is infinite.

I don't understand the problem.  Take $f(x) = x \forall x \in \mathbb{Q}$ and the problem statement is clearly false, as $f^{-1} (x) = x \neq \infty$!

Maybe he asked if there is such a function ?


geometry
trigonometry
inequalities unsolved
inequalities

Let ABCD have AB=BC=AD=1.Prove that the area of ABCD is at least [3*sqrt(3)]/4

i think you mean *at most* $\frac{3 \sqrt{3}}{4}$.... because we can take an equilateral triangle of side length 1, whose area is only $\frac{\sqrt{3}}{4}$, and move two of its sides apart slightly to get a quadrilateral of sides 1, 1, 1, and $\epsilon$ (some small value) whose area is very close to $\frac{\sqrt{3}}{4}$

let $y$ denote the fourth side of the quadrilateral, and let $S$ denote the area. then $S^2 = (\frac{y+1}2)^3(\frac{3-y}2) - y\cos ^2 (\frac{A+B}{2})\leq (\frac{y+1}2)^3(\frac{3-y}2)$. by AM-GM,

$(\frac{y+1}2)^3(\frac{3-y}2) = \frac13 (\frac{y+1}2)^3(\frac{9-3y}2) \leq \frac13 (\frac{3(y+1)+9-3y}{8})^4 = \frac13 (\frac32)^4 = \frac{27}{16}$

so $S\leq \sqrt{\frac{27}{16}} = \frac{3\sqrt{3}}{4}$

equality occurs when $A+B = \pi$ and $9-3y = y+1$, i.e. when the sides are 1, 1, 1, 2 (and the angles are $\pi/3, \pi/3, 2\pi/3, 2\pi/3$)


vector
invariant
superior algebra
superior algebra solved

Does anyone know how to prove (i)? If it is proved, (ii) and (iii) follow easily.

But isn't i) obvious? I mean, it's clear that we have $\sigma(x-\sigma(x))=\sigma(x)-\sigma(\sigma(x))=0$ and thus one inclusion is proved. For the other, suppose that $ \sigma(x)=0 $. Then $ x=x-\sigma(x) $ and the other inclusion is proved.

Oh, dear, I didn't think of $x=x-\sigma(x)$ if $\sigma(x) = 0$. :blush:

For ii)
$V=\ker(\sigma)\oplus Im(\sigma) \Leftrightarrow \{ \begin{array}{cc} V=\ker(\sigma)+Im(\sigma) \\ \ker(\sigma)\cap Im(\sigma)=0\end{array}$

For $x\in V$ we have $x=(x-\sigma(x))+\sigma(x)$. By i) we have $(x-\sigma(x))\in \ker(\sigma)\ \forall x\in V$. Obviously $\sigma(x)\in Im(\sigma)\ \forall x\in  V$. So $V=\ker(\sigma)+Im(\sigma)$.

Let $x\in \ker(\sigma)\cap Im(\sigma)$. Then $\exists z\in V$ such as $\sigma(z)=x$ and $\sigma(x)=0$. This means $\sigma(\sigma(z))=0 \Rightarrow \sigma(z)=0 \Rightarrow x=0$.

Done with ii)

[quote="liyi"][b]V[/b] is a vector space on a number field $F$. $\sigma$ is a linear transformation on [b]V[/b] and satisfy $\sigma^2 = \sigma$. Show that
(i) $Ker(\sigma) = \{ \xi - \sigma(\xi) | \xi \in V\}$
(ii) $V = Ker(\sigma) \oplus Im(\sigma)$
(iii) if $\tau$ is a linear transformation on $V$, then the sufficient and necessary condition such that $Ker(\sigma)$ and $Im(\sigma)$ are invariant for $\tau$ is $\sigma\tau = \tau\sigma$.[/quote]

For iii) The converse.
If $\sigma\tau=\tau\sigma$ then obviously $\tau(Im(\sigma))\subset Im(\sigma)$, so $Im(\sigma)$ is $\tau$ invariant.
If $x\in\ker(\sigma)$ then $\sigma(\tau(x))=\tau(\sigma(x))=\tau(0)=0$, so $\tau(\ker(\sigma))\subset \ker(\sigma)$, so $\ker(\sigma)$ is $\tau$ invariant.

Now assume $\ker(\sigma)$ and $Im(\sigma)$ are invariant for $\tau$.

We have $\sigma(\sigma\tau(x)-\tau\sigma(x))=\sigma\tau(x)-\sigma\tau\sigma(x)=\sigma\tau(x-\sigma(x))$. By (i), $x-\sigma(x)\in\ker\sigma$ and $\ker\sigma$ is $\tau$ invariant, so $\sigma\tau(x-\sigma(x))=0$, so $\sigma\tau(x)-\sigma\tau(x)\in\ker\sigma\ \forall x\in V$.

Because $Im\sigma$ is $\tau$ invariant, $\forall x\in V\ \exists y\in V$ such that $\tau\sigma(x)=\sigma(y)$.
So $\sigma\tau(x)-\tau\sigma(x)=\sigma\tau(x)-\sigma(y)=\sigma(\tau(x)-y)\in Im\sigma$.

This means that $\sigma\tau(x)-\tau\sigma(x)\in \ker\sigma \cap Im\sigma=0$, so $\sigma\tau=\tau\sigma$.


[quote="Peter"]Yes, IUB is an international university in Bremen. They are also one of the sponsors of the IMC. So they clearly like international people. :)[/quote]

My university is also a sponsor of the IMC. Jacobs University Bremen was previously known as International University Bremen (IUB) but had serious financial issues and thus was looking for a sponsor. They managed to get funding by the Swiss Jacobs Foundation, i.e. 200 million in total with 15 million each year to support research and teaching. [url=http://www.jacobs-university.de/]Jacobs University Bremen[/url] has graduates from 64 countries including 207 bachelors, 80 masters, and 46 doctoral graduates.

Yes, now it's Jacobs University Bremen :). It's very young university with good professors, at least in Math Faculty.

[quote="Valentin Vornicu"][quote="Peter"]Yes, IUB is an international university in Bremen. They are also one of the sponsors of the IMC. So they clearly like international people. :)[/quote]I got admitted in IUB when I applied 4 years ago. In the end I didn't go (mostly because of the language problem - I don't know German).[/quote]But the only language spoken in IUB is English!  :huh: That's why they call it an international university.


logarithms
Euler
real analysis
real analysis unsolved

Use Raabe Duhamel test and you should get that the series converges for $a>e$ and diverges for $a\leq e$.

That would be converges for $a<e^{-1}$ and diverges for $a\ge e^{-1}.$

For a more elementary argument, show that this series is limit comparable to

$\sum_{n=1}^{\infty}n^{\ln a}.$

yes, you are right Kent. Sory for the mistake.

[quote="kouboy"]find all $a>0$ such that the series $\sum_{n=1}^{\infty}a^{(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{n})}$ converges[/quote]

$\sum_{n=1}^{\infty}a^{(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{n})}$=
$\sum_{n=1}^{\infty}a^{lnn+\gamma+o(1)}$ where $\gamma$=0,5772... Euler const


geometry
circumcircle
trigonometry
trig identities
Law of Sines
Law of Cosines
Pythagorean Theorem

Here are some geometry problems. Enjoy!

1. Given a square $ABCD$, a line is drawn from point $A$ intersecting $BC$ at $E$. Line segment $DE$ is drawn.  Find the measure of angle $AED$.

2. The circumcircle of triangle $ABC$ is drawn as well as diameter $BK$.  $BK$ cuts side $AC$ of triangle $ABC$ at $D$.
Given that $AB = 60$ , $BC = 52$ and $AK = 25$, find the length of side $AC$.

3. The circumcircle of equilateral triangle $XYZ$ of side length $10$ is drawn.  Three tangents are drawn meeting the circle at points $X$, $Y$, and $Z$. The tangents are extended until they meet so that they form a new triangle. Find the area of this new triangle in terms of $[XYZ.]$

#1

[hide]
To Boride:  Angle AED will not always be an inscribed angle.  Only if E falls on B or C will <AED be an inscribed angle.

Actually, the measure of <AED would change depending on where E is.

To come up with a general formula for the measure of <AED, lets suppose E divides side BC into two segments:  a and b.  The measure of <AED is:

arctan(a/(a+b)) + arctan(b/(a+b))
[/hide]

[quote="Boridi"][hide="2"]
In short:

BAK/BCK are right triangles, used the pyth theorem to find BK and CK.
Then used Law of Sines to find <ABK and <CBK, and added them together to get <ABC. Used segment AB, segment BC, <ABC in Law of Cosines to find AC

 I got 56 units?
[/hide][/quote]
Yeah, that's what I got.

After using Pythagorean theorem to find $BK$ and $CK$, you can simply use Ptolemy's theorem ;)

[quote="weiyan"]#3

[hide]
I'm not sure what "Find the area of this new triangle in terms of [XYZ]" means but I'll just find the area of the new triangle.

Let's name the vertex of the new triangle opposite X to be A, the vertex opposite Y to be B, and the vertex opposite Z to be C.

On the circle, arc YZ is 120 degrees because it is exactly 1/3 of the circle.  Angle A is 60 degrees because it is the angle formed by two tangents cutting off an 120 degree arc, therefore:  (240-120)/2=60.  The same can be done for angles B and C.  Because angles A, B, and C are all 60 degrees, angle ABC is also equilateral.

Angle AYZ is 60 degrees because it is an angle formed by a tangent and a secant cutting off an 120 degree arc, therefore 120/2=60.  The same of angle AZY.  Because all of its verteces measure 60 degrees, triangle AYZ is also equilateral (doing the same would prove triangles BXZ and CYX equilateral as well), which would mean YZ=AZ.  Since YZ=AZ and YX=YZ, YX=AZ as a result of the transitive property.  Since BXZ is also equilateral, BZ=XZ.  Since BZ=XZ and XZ=YX, BZ=YX as a result of the transitive property.  Since YX=AZ and BZ=YX, AZ=BZ as a result of the transitive property.  Because AB=AZ+BZ, and AZ=BZ, we get AB=2AZ.  And since YX=AZ, AB=2YX.

Therefore:  Triangle ABC is an equilateral triangle whose sides are twice as long as those of triangle XYZ.

Since each side of triangle XYZ is 10, each side of triangle triangle ABC is 20, which would make the area of triangle ABC [b]100sqrt3[/b].

[/hide][/quote]

[hide]

That's exactly correct.  What he meant by find it in terms of $[XYZ]$ is to find the area of the new triangle in terms of the area of $XYZ$.  As you correctly stated, $ABC$ is an equilateral triangle with twice the side length of $XYZ$.  If you draw a picture of this, it's pretty clear that
$[ABC] = 4[XYZ]$, however you can also find the same result by applying area formulas.

[/hide]

[quote="weiyan"]#1

Angle AED will not always be an inscribed angle.  Only if E falls on B or C will <AED be an inscribed angle.

Actually, the measure of <AED would change depending on where E is.

[/quote]

Yes, you are absolutely right. I didn't realize that while making up the question :blush:. I thought it would be 45 degrees . For number 2, I used the same strategy as frt and got 56 as well. And for number 3, I found a side length of the new equilateral triangle through trig and got 100 sqrt 3.  :)


inequalities
trigonometry
inequalities proposed

Solve the following inequality.

\[\left (\frac{\sin x+\sin 3x+\sin 5x}{\cos x+\cos 3x+\cos 5x}\right )^2  \leqq\left(\frac{\sin 2x+\sin 4x+\sin 6x}{\cos 2x+\cos 4x+\cos 6x}\right)^2  \ \left(0\leqq x\leqq \frac{\pi}{2}\right)\]

Noting that $\sin x + \sin 5x = 2{\sin 3x}{\cos 2x}$, substituing similarly in the other terms and cancelling, this becomes $\tan^2 3x \leq \tan^2 4x$. It's fairly clear that this holds for $0 \leq x \leq \frac {\pi}{7}, \frac{2\pi}{7} \leq x \leq \frac {3\pi}{7}$.

Thank you for your reply, Feachra!

You are right.

kunny


vector
linear algebra
linear algebra unsolved

Let $T:V\rightarrow V$ be a linear transformation in a finite dimensional vector space $V$. Prove that $\mbox{rank }T = \mbox{rank } T^2$ iff $\mbox{range } T \cap \ker T=\{0\}$.

as far as i understand by $rangeT$ yo meant $ImT$:
Since $dim ImA=dim ImA^2+dim({ImA} \bigcap{KerA})$ you got the result. There is also more general relation between $dim ImA$ and $dim ImA^{n+1}$ for any $n\in\mathbb{N}$: actually
$dim ImA=dim ImA^{n+1}+\displaystyle\sum_{k=1}^{n}dim ({ImA^k} \bigcap{KerA})$


The length of the longer side of rectangle R is 10 percent more than the length of a side of square S. The length of the shorter side of rectangle R is 10 percent less than the length of a side of square S. What is the ratio of the area of rectangle R to the area of square S? Express your answer as a common fraction.


geometry
3D geometry
search
Putnam
combinatorics unsolved
combinatorics

1. An n-cube is an undirected graph with 2^n vertices labeled with 2^n-digit binary numbers. There is an edge between two vertices if their binary labels differ exactly at one digit. Show that an n-cube has a Hamiltonian circuit for any n greater than or equal to 1.

2.Towns T0, T1,…. , Tn−1 are situated clockwise on a circular route of length 1. Assume that the town Ti can supply you with the amount of fuel that is sufficient to drive your car for distance pi, and that (sigma i=0 to n-1)Pi .That is, the total amount of fuel is exactly enough to take you once around the circle. Prove that you can start a car with an empty tank in one of the towns, drive clockwise around the entire circuit (picking up the fuel at each town on the route) and finish at the same spot (with an empty tank, of course).

3. S is the set of all (x, y) with x, y non-negative integers such that x + y < n. Each element of S is colored yellow or black, so that if (x, y) is yellow and x1 ≤ x, y1 ≤ y, then (x1, y1) is also yellow. A BLACK subset of S has n black elements with different first member and a YELLOW subset of S has n black elements with different second member. Show that there are the same numbers of BLACK and YELLOW subsets.

HEY GUYS.....plz solve these question and other unsolved question by msg_1989...as soon as possible...!!!!

me struggling to solve these questions....!!!

 :(

I am wondering,
1. why these questions are that urgent, and
2. why are so many people interested in them.
But then again, if a contest chooses 3 problems and 2 of them are just well-known, they have nobody else to blame for people copying the solutions from the net.

Try a search for "Hamilton* hypercube".
Try a search for "fuel" in the Combinatorics section.

Wait, both 1 and 2 seem to be Putnam problems. Do you know where to find Putnam solutions on the net? Otherwise, google.

  darij


Find all solutions in the range $ 0$$ \leq$ $ x$ $ \leq$ $ \pi$ , of the equation $ \sqrt2$.$ Sin2x$ +$ \sqrt{Sin4x}$ =$ 0$

$ 2\sin^{2}2x \equal{}\sin4x$

$ 2\sin^{2}2x \equal{} 2\sin2xcos2x$

$ 2\sin2x(\sin2x\minus{}\cos2x) \equal{} 0$

so $ \boxed{\sin2x \equal{} 0}$ or $ \boxed{\sin2x \equal{}\cos2x}$

let $ u \equal{} 2x$

so $ sinu \equal{} 0$ domain: $ 0\leq{u}\leq2\pi$
$ u \equal{} 0,180,360$

\\\ly $ sinu \equal{} cosu$ $ 0\leq{u}\leq2\pi$
$ u \equal{} 45^{o}$

therefore $ x \equal{} 0,90,180,22.5$

change to radians


calculus
integration
trigonometry
parameterization
function
limit
complex analysis

(i) Easy: use the residue theorem to show $I:=\oint_\Gamma \frac{dz}{1+z^n}\ (n=2,3,...)\ =\frac{2\pi}{n}(\sin\frac{\pi}{n}-i\cos\frac{\pi}{n})$, $\Gamma$ being the contour of 2 circle radii with an arc of $\frac{2\pi}{n}$ between them. (so closed contour but with a pole inside)

(ii) Use the previous to calculate $J:=\int^{\infty}_0\frac{dx}{1+x^n}$.
[hide="Answer"]$\frac{\pi/n}{\sin(\pi/n)}$[/hide]

No one? Is the problem statement clear?

[late edit: thanks fedja and nene, I got the concept. This also solves most other exercices in the paper. :)]

Seems like you have no problem with part 1 , which I'll prefer to write as $I=-\frac{2\pi i}ne^{\frac{\pi i}n}$. For part 2, notice that $I$ consists of three integrals: almost $J$, an integral over a faraway arc, and an integral over $e^{\frac{2\pi i}n}\mathbb R_+$ travelled from almost $\infty$ to $0$. Now, the integral over an arc is very small when the arc is far away if $n>1$ (it decays like $R^{-(n-1)}$ where $R$ is the radius of the arc). On the other hand, the integral over $e^{\frac{2\pi i}n}\mathbb R_+$ is almost $J$ times $-e^{\frac{2\pi i}n}$: just use the parametrization $z=e^{\frac{2\pi i}n}x$; then the function is the same $\frac 1{1+x^n}$ and $dz=-e^{\frac{2\pi i}n}dx$ ($-$ because we go from $\infty$ to $0$). Therefore, we get $J(1-e^{\frac{2\pi i}n})=I=-\frac{2\pi i}ne^{\frac{\pi i}n}$ from where the result follows immediately.

I think $n\ge 2\;\;n\in\mathbb{N}$. Then:
\[{f(z)=\dfrac 1{1+z^n}\Rightarrow \int_{\Gamma}f(z)=2\pi i Res(f,e^{\frac \pi ni})=\int_0^R\dfrac{dx}{1+x^n}+\int_0^{\frac{2\pi }n}iRe^{i\theta}f(Re^{i\theta})\,d\theta\;+\int_R^0\dfrac{e^\frac{2\pi i}n}{1+x^n}}\,dx=(1-
e^\frac{2\pi i}n)\int_0^R\dfrac{dx}{1+x^n}+\int_0^{\frac{2\pi }n}iRe^{i\theta}f(Re^{i\theta})\,d\theta.\]
because f(z) has one simple pole $ z=e^{\frac\pi ni}$ inside $\Gamma$.
Since:
\[\lim_{R\to +\infty}\int_0^{\frac{2\pi }n}iRe^{i\theta}f(Re^{i\theta})\,d\theta=0\]
and
\[{Res(f,e^{\frac{\pi i}n})=\dfrac 1{(1+z^n)^'}\Big\vert_{z=e^{\frac{\pi i}n}}\;=\dfrac 1{ne^\frac{(n-1)\pi i}n}}\]
Then letting ${R\to +\infty}$ we get \[J=\dfrac {\pi/n}{\sin{\frac{\pi}n}}\]

Have you thaught of it that z^n = a. Were a = a complex number has n solutions? That means there is more then one pole for n > 1. Bij defining the length of the radius, it is possible to have less poles.

[quote="Peter VDD"]It is the normal case. That's why our contour has arc between 0..2Pi/n[/quote]

jep, sorry. I read the post incorrectly. I thought it wasn't said that the contour should exist of y1 = t, y2 = e^(2 pi / n) and y3 = e^(2 pi / n)(R - t). But as I can see I read the post to fast.


A solid wheel of radius $R$ and mass $M$ can move without sliding on a horizontal table. We fix a rod of mass $M$ and length $l$ to the middle of the wheel. The rod can swing in the vertical plane. Find the period of small oscillations of the system.

I got $\omega^{2}= \frac{10g}{3l}$. Is that ok?

Btw, as a bonus: solve this problem with the rod fixed to a point on the cylinder a distance $r < R$ from its center  :lol:

no thats wrong unfortunately. :(  Yes I will try to solve the bonus problem :D

Ok, let's start slowly once again. The kinetic energy of the cylinder is $M\dot x^{2}/2+I\omega^{2}/2 = 3M\dot x^{2}/4,$ and the kinetic energy of the rod is $\frac{M}{2}[(\dot x+l\dot\phi\cos\phi/2)^{2}+l^{2}\dot\phi^{2}\sin^{2}\phi/4] = \frac{M}{2}[\dot x^{2}+l\dot x\dot\phi\cos\phi+l^{2}\dot\phi^{2}/4].$ The potential energy of the rod is $V =-Mgl\cos\phi/2$ (with $V = 0$ at the center of the cylinder). Am I right so far?

Firstly let's write the total energy at the position $\phi$.

$E=\frac{I\omega^{2}}{2}-Mg\frac{L}{2}cos\phi+MgR$

Here,$I$ is the total moment of inertia due to touching point to the ground.

$I=\frac{2}{5}MR^{2}+MR^{2}+\frac{ML^{2}}{12}+Mx^{2}$

$x$ is the distance between the center of the rod and the touching point.

After using Lagrangian,I got the following:

$T=2\pi \sqrt{\frac{2}{15gL}(36R^{2}+5L^{2}-15RL)}$


Predict the major product from each of the following reactions:

1) n-Propyl bromide with methoxide ion, in methanol, at 50ºC.
2) 3-Bromopentane with $HS^{-}$ in methanol, at 50ºC.
3) 1-Chloropentane with potassium tert-butoxide in tert-butanol, at 50ºC.
4) (S)-2-Chloropentane with $CH_{3}O^{-}$ in $CH_{3}OH$, 50ºC.
5) (S)-2-Chloropentane with $CH_{3}CO_{2}^{-}$ in acetic acid, 50ºC.
6) (S)-2-Chloropentane with $AgCN$.
7) (S)-2-Chloropentane with $CH_{3}C \equiv C^{-}$.
8) (S)-2-Chloropentane with triphenylphosphine.
9) 2-Bromo-2-methylhexane with potassium tert-butoxide in tert-butanol, at 50ºC.

1. first the solvent is poalr protic and further it's methanol so methoxide ion will be properly solvated so there is very less chance of anything bimolecular happeneing so it has to be $E1$ or $SN_{1}$
in some sense bot are as deifficult as other to proceed so basically though the temp is huge and the solvent is weakly polar but still since the substrate is a primary one so in some sense it will gradually proceed via $S_{N}1$ nearly equal to $E_{1}$ $>>S_{N}2$ so ithink the substituefd product should be just greater than the elimnation product

No. How can the reaction occur via SN1 or E1 if this is a primary halide, and primary carbocations are not particularly stable? So the reaction must occur via SN2 or E2.

yes that's what i have mentioned as you read on .....the reaction basically is blocked from all the sides ...no one can deny a reaction can't proceed via a particular path just that we calculate teh probability of it's passing.here the second order reactions are hugely restricted because the base methoxide is properly solvaated so it is less free to move  so in some sense no  path can be called as a valid path but since substitution involves less energy change so it may just win over

Yes, for the first one we get more substitution product because the base/nucleophile is small. How about the other 8?


[quote="livetolove212"]Given triangle $ ABC$ and its circumcircle $ (O)$, its orthocenter $ H$. Let $ P$ be an arbitrary point inside triangle $ ABC$. $ AP,BP,CP$ intersect $ (O)$ again at $ A_1,B_1,C_1$,respectively. Let $ A_2B_2C_2$ be the pedal triangle of $ P$ wrt $ \Delta ABC, A_3,B_3,C_3$ be the reflections of $ A_1,B_1,C_1$ wrt $ A_2,B_2,C_2$, respectively. Prove that $ A_3,B_3,C_3,H$ are concyclic.[/quote]

could you plz post your solution?

[quote="jayme"]Dear Linh and Mathlinkers,
this is a tentavite of a synthetic proof
1. Let A'2, B'2, C'2 the symmetrics of A2, B2, C2 wrt H.
2. let (1), (2), (3) the respective circles with diameter HA'2, HB'2, HC'2.
3. If we can prove that the second points of intersection of these circles are collinear then H is on the circle A'2B'2C'2.
4. By symmetry wrt H we are done…
What do you think?
Sincerely
Jean-Louis[/quote]

Sorry dear [b]Jean-Louis[/b] but I think that the second points of intersection of 3 circles (1), (2), (3) are not collinear.

Dear Livetolove,
I verify your observation on my figure but the collinearity held when I am moving some points.
Can You send us your figure?
Sincerely
Jean-Louis

Dear [b]Jean-Louis[/b],
 Your notation is different from mine. I'm sorry for misunderstand you. I think that we can prove the same for circles $(HA_2), (HB_2), (HC_2)$.

Let $ Q $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ and $ V $ be the Poncelet point of $ ABCP. $ It's well-known that the reflection $ U $ of $ H $ in $ V $ is the pole of the simson line of $ \triangle ABC $ with direction $ \perp $ $ OQ $ and $ V $ is the pole of the simson line of $ \triangle A_2B_2C_2 $ with direction $ \parallel $ $ OQ $ (see [url=http://www.artofproblemsolving.com/community/c6h290991]Simson line wrt pedal triangle[/url]), so notice $ \triangle A_1B_1C_1 $ $ \stackrel{+}{\sim} $ $ \triangle A_2B_2C_2 $ we get $$ \measuredangle UB_1A_1 = \measuredangle (AU, AP) = \measuredangle (\parallel OQ, AQ) = \measuredangle (\perp OQ, B_2C_2) = \measuredangle VB_2A_2 \Longrightarrow A_1B_1C_1U \stackrel{+}{\sim} A_2B_2C_2V, $$ hence we conclude that $ A_1B_1C_1U $ $ \stackrel{+}{\sim} $ $ A_2B_2C_2V $ $ \stackrel{+}{\sim} $ $ A_3B_3C_3H $ $ \Longrightarrow $ $ A_3, $ $ B_3, $ $ C_3, $ $ H $ lie on a circle.


geometry search probability AMC AIME USA(J)MO USAMO \/closed

On the "How to Write a Solution" page it says that the excerpts are taken from the "Olympiad Geometry" class.  When I clck "Online Classes" there is no Olympiad Geometry!  Has that class been discontinued?

Also, is Intermediate Number THeory and Intermediate Geometry coming out soon?

The class has been subsumed into WOOT. If I remember correctly, it wasn't really financially or otherwise viable for AoPS to run it.

Mr. Rusczyk, would it be possible for you to sell the olympiad geometry class transcripts or sell something similar to that?

Dang...that stinks.

Anyway:

[quote="DPatrick"][quote="horselvr101"]Speaking of books and classes and the like, what are the next books that are going to come out?[/quote]Introduction to Algebra and Intermediate Counting & Probability should come out in early 2007.  Intermediate Number Theory should come out in Spring 2007.[/quote]

This is from the thread that xpmath directed me to:  There's still no Intermediate Number Theory, and it's already 1.5 years since spring 2007...

Well, is there a projected release date for Intermediate Number THeory and Intermediate Geometry now?  The Intro to geometry book says inter. geometry is coming in fall 2008...

On the classes page, Olympiad Geometry is listed as a class for Spring 2009, however, when I click on it, it says you've reached this page in error...

[quote="xpmath"]That's basically happened with every single book they've published. Many things happen and the books got delayed.[/quote]

You can say this for basically anything. These books arent the only ones that are delayed. People always delay the date that movies come out. Theres this road near where i live. Its Interstate 79. Theyve been working on it for over 2 years now. They spent the first year fixing a bridge.

Sorry for getting off topic.

I realize that, but as aime15 was talking about the books, I only talked about the books.

@Aime15: It's probably just a mistake. Don't worry about it.

will all of the stuff in olympiad geometry be covered in WOOT?

[quote="rrusczyk"][url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=7]Back by popular demand, the Olympiad Geometry course[/url].[/quote]

Whoa, cool!!!!!  (not for me though, i'm not gonna take it until later possibly)

Is an Intermediate Geometry coming soon also, then?

No.  I'll probably finally write that book in a few years, and will combine that with the Olympiad Geometry class to produce an 18-week advanced geometry of some sort.  But that's a while off yet.

cognos: no.  A little of it will, but not all of it.  WOOT is covering a lot more than just geometry, so can't hit all the geometry that's covered in the Olympiad Geometry class.  Of the 24 hours of instruction in the Olympiad Geometry class, maybe around 1-2 hours of the same content (i.e., same discussion) are in WOOT, and 4-5 hours of Olympiad geometry are on topics discussed in WOOT but with problems that are different than those discussed in WOOT.

will it be possible for the class to start and end earlier?
it seems like it will end during the aime, and the usamo...

What type of graded assesments will this class have?

Two Challenge Sets.  They won't have A, B, C, D grades -- there will be thorough feedback on how you write solutions, with suggestions on areas for future study, etc.  There will also be message board problems, but these will not be formally evaluated like the Challenge Sets will be.

[quote="rrusczyk"]No - we're pushing it starting as early as we are.  When we start classes in January, no one signs up.[/quote]
Can you sell the transcripts then? I think some people would pay decent money for it.

We won't be selling the transcripts.  Eventually we'll build a book out of the class.

Quick question will we do any problems like number 4 on 2008 USAMO (the one about the triangulation of the n-o-gon into isosceles triangles)? ie Somewhat discrete geometry. :starwars:

No; the key insights in that problem are not really geometric.

[quote="hurdler"]Roughly when will (most likely) the next time this class is offered again (excluding this coming february) be?[/quote]

We haven't decided.  Certainly not before February, 2010, but it's possible we might not offer it again until Feb, 2011, or even later if enrollment is low this time around.


hello cool doods.  :D

can you find

$ \sum_{j\equal{}1}^\infty\sum_{i\equal{}1}^j\frac{1}{i(i\plus{}2)j(j\plus{}2)}$

[quote="misan"][quote="Extremal"]hint:
$ \sum_{i \equal{} 1}^{j}\frac {1}{i(i \plus{} 2)} \equal{} \frac {1}{2}(1 \plus{} 1/2 \minus{} \frac {1}{i \plus{} 1} \minus{} \frac {1}{i \plus{} 2})$[/quote]

 :?: where did $ j$ go ?  :([/quote]
I mean :
$ \sum_{i \equal{} 1}^{j}\frac {1}{i(i \plus{} 2)} \equal{} \frac {1}{2}(1 \plus{} 1/2 \minus{} \frac {1}{j \plus{} 1} \minus{} \frac {1}{j \plus{} 2})$   :blush:

Just a numerical calculation...
$ \textbf S_n\equal{}\sum_{j\equal{}1}^{n}\sum_{i\equal{}1}^j\frac{1}{i(i\plus{}2)j(j\plus{}2)}\equal{}\frac{1}{2}\cdot\left(\frac{1}{2}\minus{}\frac{1}{n\plus{}2}\right)\plus{}\frac{1}{4}\cdot\left(\sum_{j\equal{}1}^{n\plus{}2}{\frac{1}{j^2}}\minus{}1\minus{}\frac{1}{4}\right)$
$ \textbf S\equal{}\lim_{n\to\infty}\textbf S_n\equal{}\frac{2\pi ^2\minus{}3}{48}$


algebra
polynomial
quadratics
algebra unsolved

Find all values of $ b$ for which the equations $ 1988x^2\plus{}bx\plus{}8891\equal{}0$ and $ 8891x^2\plus{}bx\plus{}1988\equal{}0$ have a common root.

[hide]If two polynomials $ P(x)$ and $ Q(x)$ have a common root $ x\equal{}r$, then $ P(r) \minus{} Q(r) \equal{} 0 \minus{} 0 \equal{} 0$. Thus, the polynomial $ P(x)\minus{}Q(x)$ also has $ x\equal{}r$ as a root.

Thus, any common root shared by the two quadratics must be a root of:
$ (8891x^2\plus{}bx\plus{}1988) \minus{} (1988x^2\plus{}bx\plus{}8891) \equal{} (8891\minus{}1988)(x^2\minus{}1) \equal{} 0$

Thus, the only possible values for the common root are $ x \equal{} \pm1$

$ x \equal{} 1$ yields $ 1988(1)^2 \plus{} b(1) \plus{} 8891 \equal{} 0 \Rightarrow b \equal{} \minus{}10879$

$ x \equal{} \minus{}1$ yields $ 1988(\minus{}1)^2 \plus{} b(\minus{}1) \plus{} 8891 \equal{} 0 \Rightarrow b \equal{} 10879$

So, the only possible values of $ b$ are $ \pm 10879$. [/hide]


The other day, I started thinking about the following problem:  Given a group G, under what circumstances can one find two subgroups A and B, such that if each element of A is operated on the corresponding element of B, the result will always be the identity element?

That got me thinking about another question.  A subgroup H of a group G is called a normal subgroup iff, for all x in G, xHx^-1 = H.  What, if any, subgroups exist s.t., for all x in G and all h in H, xhx^-1 = h^-1 ?  What changes if we are allowed to restrict the x's to various subgroups of G?

Group theory  :D

[quote="Anduin"]The other day, I started thinking about the following problem:  Given a group G, under what circumstances can one find two subgroups A and B, such that if each element of A is operated on the corresponding element of B, the result will always be the identity element?
[/quote]

Do you mean : for all x in A there exists y in B such as x*y = e ??

But then x^-1 is in B and then x is in B and so A is included in B.

[quote="Anduin"]That got me thinking about another question.  A subgroup H of a group G is called a normal subgroup iff, for all x in G, xHx^-1 = H.  What, if any, subgroups exist s.t., for all x in G and all h in H, xhx^-1 = h^-1 ?  What changes if we are allowed to restrict the x's to various subgroups of G?[/quote]

If we take x=e (which commutes with everybody) then h = h^-1 or h=e (a (sub-)group with a=e for all a is commutative : classical exercice) and we have xh=hx for all h and all x. So the condition is H is a sub-group of Z(G) (the subgroup of the elements which are kind enough to communte with everybody).

Is it correct ?


1. Let $ n$ is positive number such that $ n \ge 2$. Prove that:
$ {C_{n}}^0+{C_{n}}^1+{C_{n}}^2+..................+{C_{n-1}}^{n-1}+{C_{n}}^{n} \le {(\frac{2^n-2}{n-1})}^{n-1}$

2. Let $ a,b,c$ are positive numbers, choose the right feature:
a/ $ \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} < \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ with all $ a,b,c \in N*$.
b/  $ {\frac{a}{b+c}}+{\frac{b}{c+a}}+{\frac{c}{a+b}} < \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{{\frac{c}{c+a}}}$ with all $ a,b,c \in N*$.

[quote="L_Euler"]1. Let $ n$ is positive number such that $ n \ge 2$. Prove that:
$ {C_{n}}^0 + {C_{n}}^1 + {C_{n}}^2 + .................. + {C_{n - 1}}^{n - 1} + {C_{n}}^{n} \le {(\frac {2^n - 2}{n - 1})}^{n - 1}$

2. Let $ a,b,c$ are positive numbers, choose the right feature:
a/ $ \frac {a}{a + b} + \frac {b}{b + c} + \frac {c}{c + a} < \sqrt {\frac {a}{b + c}} + \sqrt {\frac {b}{c + a}} + \sqrt {\frac {c}{a + b}}$ with all $ a,b,c \in N*$.
b/  $ {\frac {a}{b + c}} + {\frac {b}{c + a}} + {\frac {c}{a + b}} < \sqrt {\frac {a}{a + b}} + \sqrt {\frac {b}{b + c}} + \sqrt {{\frac {c}{c + a}}}$ with all $ a,b,c \in N*$.[/quote]

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


I'm not sure if the concpets of my solution are appropriate for getting started, so I'll post it briefly.

[hide]Because the opposite angles are supplementary, the quadrilateral is cyclic. Inscribe it in a circle, add an auxilliary rectangle, then use power of a point to get 4.2[/hide]

Does anyone have a diagram for this, because in my diagram I don't see why $ AE \equal{} CF$ (such that $ BF \equal{} 3*7/5$)?

$ \triangle AED \cong \triangle CFB$, therefore $ BF \equal{} \boxed{\textbf{(E)} \ 5}$

Is this right?  $ ABCD$ should form a rectangle, I think.

How do you know those two triangles are congruent?

[hide]
we have
$ \triangle{EDC} \sim \triangle{FCB}$ and $ \triangle{AFB} \sim \triangle{DFA}$

it follows that

$ \frac {3}{5} \equal{} \frac {BF}{AF}$ and $ \frac {5}{7} \equal{} \frac {10 \minus{} AF}{BF}$
thus
$ BF \equal{} \frac {3AF}{5} \equal{} \frac {70 \minus{} 7AF}{5} \rightarrow AF \equal{} 7$

$ \boxed{BF \equal{} \frac {21}{5} \equal{} 4.2}$

the triangles are not congruent. they are however similar proof of similarity lies in the fact that angle C and A are both 90 degrees
this means
$ \angle{BCF} \equal{} \angle{EDC}$ and $ \angle{BAF} \equal{} \angle{ADE}$
since each triangle created is right a right triangle it means that they are similar if they share one same angle
[/hide]


geometry
geometric transformation
rotation
geometry proposed

Given a convex quadrilateral $ABCD$. Suppose there is a point $M$ such that $\triangle AMB$ and $\triangle CMD$ are isosceles with $\angle AMB=\angle CMD=120^\circ$. Prove that there exists a point N, such that $\triangle BNC$ and $\triangle DNA$ are equilateral.

Given two congruent segments $XY,ZT$ in the plane, if $\overrightarrow{XY}$ can be taken onto $\overrightarrow{ZT}$ through a rotation of angle $\alpha$, then it can be taken onto $\overleftarrow{TZ}$ through a rotation of angle $\pi-\alpha$ (translations are considered rotations of angle $0$, centered at the infinity point). We apply this to $XY=AC,ZT=BD$.


Let ABC be any triangle, $r$ be radius of incircle of ABC, $R$ be circumradius of ABC  and  $d$ be the shortest distance  between  circles. Find a algebraic  relatives  between $r, R ,d$

Can you show your solution,please?Thank you.

The relation $IO^2=R(R-2r)$ is wellknown. If you don't know it, then I shall prove it here to-morrow.

[u]I note:[/u]
the circumcircle $e=C(O,R)$;
the incircle $i=C(I,r)$;
the power $p_e(X)=XO^2-R^2$ of the point $X$ w.r.t. the circle $e$;
the intersections $\{A,E\}\subset AI\cap e$ and $\{E,F\}\subset EO\cap e$;
the point $D\in (AC)$, $ID\perp AC$.

Prove you that $EB=EC=EI$. Therefore,

$\triangle AID\sim \triangle FEC\Longrightarrow\frac{IA}{EF}=\frac{ID}{EC}\Longrightarrow$

$\frac{IA}{2R}=\frac{r}{EC}\Longrightarrow IA\cdot IE=2Rr\Longrightarrow$

$-p_e(I)=2Rr\Longrightarrow R^2-OI^2=2Rr\Longrightarrow OI^2=R^2-2Rr.$


Jennifer’s Bakery sold half as many marble cakes as strawberry
shortcakes. The price of a marble cake is 6 dollars and the price of a strawberry
shortcake is 8.50. If the total amount of sales for these cakes was
391 dollars, what was the total number of each kind of cake that was sold?

Is one equation x6 + 8.5y = 391?

If so, what is the other equation?

y=2x, so then the equation would be:
6x+ 8.5(2x)=391
6x+17x=391
23x=391
x=17 and y=34

Yes, they (subtly) gave you the other equation with the words [b]sold half as many marble cakes as strawberry[/b]. That means you're going to have, as the poster above said, the equation y = 2x in there as well.


[quote="xxreddevilzxx"]1. Evaluate: $\displaystyle\sum_{k=1}^{\infty}\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}$
[/quote]

[hide]

We find the partial fraction decomposition:
$\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})} = \frac{6^k}{3^k-2^k}-\frac{6^{k+1}}{3^{k+1}-2^{k+1}}$.

From here its clear that all the terms cancel out except the first one, $\frac{6}{3-2} = 6.$[/hide]


geometry
area of a triangle
Heron\u0027s formula

1.  A tribble moves 2 miles per hour on land. A Kilogian moves 20 miles per hour on land. The tribble never
stops during a race; the Kilogian always takes a nap for 1 hour every 2 miles during a race. If these two
different animals race on a 95040 ft. land course, by how many miles will the winner beat the loser?

2.  Point P is the center of a circle with radius 30 cm.  Square ABCD has vertices A and B on the circle and point P on side CD.  In square centimeters, what is the area of the square?

3.  Two lifeguard stations on a long, straight beach are 4 km apart.  Using radar, a boat’s captain determines that the boat is 7 km from one station and 9 km from the other.  To the nearest tenth of a kilometer, how far is the boat from the beach?

4.  The numbers 1 through 16 are entered into a square grid with four rows
and four columns.  The sum of the numbers in each of the columns is the same.
What is that sum?

5.  The new census taker has been told that Ms. Pimath always answers truthfully, but not always clearly.  With great fear, the census taker knocks on Ms. Pimath’s door.  When she opens the door, he asks, “How many children do you have?”  She responds, “Three.”  He continues, “What are their ages?”  Ms. Pimath answers, “The product of their ages is 90 and the sum of their ages is the address of my house.”  The census taker looks at the house number, and complains, “That’s not enough information.  Do you have a one-year old child?”  When Ms. Pimath responds truthfully, he says with relief, “Thank you, now I know their ages.”
	What is Ms. Pimath’s house number?  (All of the ages are whole numbers.)

1. [hide]1056[/hide]
2. [hide]15sqrt 2[/hide]
3. I'm not sure about this one.[hide]8.8[/hide]
4. I'm not sure about this one either.[hide]30[/hide]
5. Does she have a one-year old child or not?

[quote="Chinaboy"]Umm, anyone want to give these a try?[/quote]
I will.

Mine just got erased...  :(  Have to do it over again.

Can I just post the answers with brief explanation?

I didn't get #2.
Do you mean that point A and B lies on circumference or inside the circle?
1.[hide]$ \boxed{8640}$ miles

[u]Explanation: Plug and Chug[/u]

[u]Info.[/u]:

$ t\equal{}2x$

$ k\equal{}\frac{20}{11}x$

$ x\equal{}47520$[/hide]
2.$ ?$

3.[hide]$ \approx\boxed{6.7}$

[u]Explanation: Heron's Formula[/u][/hide]
4.[hide]$ \boxed{34}$

[u]Explanation: Make pairs like (1,2,15,16)[/u][/hide]

5.[hide]Either can be $ (9,2,5),(6,3,5),(1,15,6),(1,10,9),(15,3,2)...$ There might be more.

[u]Explanation: Factors[/u][/hide]

what was the heron's formula?
is that the square root of [code]s[/code] times [code](s-a)[/code] and yada yada?

A circle is technically a one-dimesional curve, so it has to be the circumference.

@spiders_hate_chestnut - Herons formula gives the area of a triangle given the side lengths $ a$, $ b$, $ c$, and the semiperimeter $ s$, which is $ \frac{a\plus{}b\plus{}c}{2}$.  The formula is $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$.

AZN:  you almost got #1 correct...sorry about the others

DK:  You correctly answered #3 and #4.  Try again on the others

[hide="Hint on #5"]

"Do you have a one-year old child?” indicates something about which numbers to use.[/hide]

[hide="Extended Hint #5"]
"Do you have..." indicates that there are two that give the same sum...[/hide]

hehe I didnt read the question on #1
1. [hide].2[/hide]
3. [hide]6.7[/hide]
4. [hide]34[/hide]

AZN:  Good Job!  Now you just have two more to go!


$ABCD$ is a quadrilateral with $AD=BC$. If $\angle ADC$ is greater than $\angle BCD$, prove that $AC>BD$.

Try  to solve by using sine and cosine functions

${|AC|}^{2}=|AD|^{2}+|CD|^{2}-2|AD\parallel CD|\cos{\angle{ADC}}$

${|BD|}^{2}=|AD|^{2}+|CD|^{2}-2|AD\parallel CD|\cos{\angle{BCD}}$

Cosine is strictly decreasing in the interval $[0; \pi]$ thus $AC>BD$

Please hide your answers & solutions and refrain from reviving old topics.

[hide="Solution"]$AD=BC$ and $CD$ is common to both triangles $\triangle ADC,\ \triangle BCD$.  Since $\angle ADC>\angle BCD$,clearly, $AC>BD$.[/hide]


[quote="matt276eagles"]For letters $a$ through $e$, it would help to see a few more terms of the sequence.[/quote]

I added an extra term for b and d (I think that's definitely enough). a is pretty easy anyway and the next term in e is so large, that it wouldn't help. If you need more terms in c, you're really making it more complicated than it is. These are supposed to be a bit challenging but not too hard (maybe except for a in part 2 if you've never seen it before) :)

You got all of them except a in part 2. Actually, your next term was correct (by chance) but your explanation wasn't what I was looking for.


modular arithmetic
number theory unsolved
number theory

find all solutions if any

$ q_1,q_2,p$ be primes

$ (q_1\minus{}1)(q_2\minus{}1) \equiv 2 \mod (p)$
$ q_1q_2 \equiv 2 \mod (p)$

by doing what PhilG did in your last post which was like this one , we get
 $ (q_1\minus{}q_2)^2 \equiv (q_1\plus{}q_2)^2\minus{}4q_1q_2 \equiv 1\minus{}4.2 \equiv \minus{}7 \pmod {p} \implies \left(\frac{\minus{}7}{p}\right)\equal{}1$
but we have
$ 1\equal{}\left(\frac{\minus{}7}{p}\right)\equal{}\left(\frac{\minus{}1}{p}\right) \left(\frac{7}{p}\right)\equal{} (\minus{}1)^{\frac{p\minus{}1}{2}}. \left(\frac{p}{7}\right) .(\minus{}1)^{\frac{p\minus{}1}{2}}\equal{} \left(\frac{p}{7}\right) \implies p \equiv 1,4,2 \pmod {7}$
so if now $ p \equiv 1,4,2 \pmod {7}$then there exists an odd number $ x$ having this property $ x^2 \equiv \minus{}7 \pmod {p} \implies   (q_1\minus{}q_2)^2 \equiv x^2 \pmod {p}$ so we have these two cases:

1)$ q_1\minus{}q_2 \equiv \minus{}x \pmod {p},q_1\plus{}q_2 \equiv 1 \pmod {p} \implies q_1 \equiv \frac{1\minus{}x}{2} \pmod {p}$
,$ q_2 \equiv \frac{1\plus{}x}{2} \pmod {p}$.
one can easily check that if $ q_1,q_2,x$ satisfy these conditions then the modular equations mentioned hold,note that there exists infinitely many primes $ q_1,q_2$ satisfying the above modular equations.

2)this case is similarly done just see that this time we have $ q_1\minus{}q_2 \equiv x \pmod {p},q_1\plus{}q_2 \equiv 1 \pmod {p}$.

it was an error in the original equations.... Sorry


the problem is to solve for integral x,y,z,w

 x^4 + y^4 +z^4 -w^4=1995

ok how does one know to concider $\mod 16$?

You could expand (2x)^4 and see that it is 16x^4 and expand (2x+1)^4 and see it is 16x^2+32x^3+24x^2+8x+1.  We can see that x^4 is always 1 or 0 mod 8.

Dividing 24x^2+8x by 16 we get

(3x^2+x)/2
(x(3x+1))/2

and we see that x and 3x+1 are of opposite parity and it works...but I just remember mods when people like Singular show it to me.

[u]Another problem[/u]:

Solve $x^{3}+y^{3}+z^{3}=2012$ for $x,y,z \in \mathbb{Z}$.

thanks a lot for the help guys.appreciated


articles
geometry
AMC
AIME
USA(J)MO
USAMO
AMC 10

Seeing as I can answer just about every problem on the AMC 10's that I've seen, I'm thinking I'd like to take the 10 and try to get a perfect score.

So, I am wondering what people have for advice.

Which books or articles should I study? Where should I look for practice? Should I do old AHSME when I run out of AMC 10? If so, which problems on those should I focus on?

I would like to brush up so that I have a better chance of being able to solve every problem and not make any stupid mistakes.

Practice alot of AMC 10's.

Also, If one type of problem is stopping you from a perfect score, do alot of those.

I got 2 150's and a 146.5 at the 3 AMC 10's I did for practice. The one problem I left blank was a #20-25 Geometry problem. I didn't think of the smart way to do that problem.

Then on the real test, I don't see the smart way to approach 2006AMC10B#23. I spend 15 minutes on it and never figure it out.

I am probably going to do every #20-25 AMC 10 Geometry problem that I can do before the real thing next year.

You shouldn't bother trying too hard for a perfect score on AMC 10. Sure, do a few practices, but there's no point in squeezing out that last AMC 10 problem by doing a ton of problems that aren't any challenge for you.

Why not just practice amc-12's?  They're more challenging and a lot of the amc 10 questions are borrowed from them anyway.  The harder problems will be good practice for easier ones.  If you're doing so well on the amc's to worry about a perfect score, just practice with harder aime problems.

[quote="calc rulz"][quote="Xantos C. Guin"]Practice alot of AMC 10's. [/quote]

Yeah, but...there are only a few AMC 10's out there to practice  :wink:

@probability1.01 I doubt I'll be doing any practice problems that are no challenge at all - yes, I can usually get pretty much all the problems on the AMC 10, but doing some end-AMC 10 or slightly harder AMC-type problems will probably not be completely trivial for me.[/quote]

"Not completely trivial" and "challenging" are two very different things, though. It's easy to get into the trap of doing "instant gratification" problems -- just hard enough to make you feel like you worked to solve them, but not hard enough to actually pose any serious challenge to you; for some reason our subconsciouses like solving problems or something, go figure (that's a guess btw, I'm not a psychologist).

Btw do you go to ABRHS? If so, and if you're a rising sophomore, I think I know who you are, which means you qualified for the USAMO last year...I would be worrying about that (the USAMO and perhaps AIME) much more than the AMC 10.

[quote="JSteinhardt"]I would be worrying about that (the USAMO and perhaps AIME) much more than the AMC 10.[/quote]

True enough.  They're much more of a challenge and, I think, more intellectually invigorating than the AMC10/12.  :)

You know the math.  Practice thinking, logic, and trick-type questions, though I don't know where you can find those...

[quote="JSteinhardt"]
Btw do you go to ABRHS? If so, and if you're a rising sophomore, I think I know who you are, which means you qualified for the USAMO last year...I would be worrying about that (the USAMO and perhaps AIME) much more than the AMC 10.[/quote]

Yep, that's me. Yes, I'm mainly focusing on USAMO and AIME (doing AIME PS and WOOT), but I just noticed that I can do pretty much all of the AMC 10 problems, so I thought it would be kind of fun to try to get a perfect score  :D

Okay, it seems pretty straightforward to just do a few practice AMC 10/12's. I was just wondering if anyone else happened to have some other thoughts.

I agree with JSteinhardt and probability1.01.  The AMC 10 is a really trivial affair, compared to contests that have more substance.  In truth, if you prepare for harder contests, it pretty much takes care of the easier ones, which you might not care that much about any more.  I practically never prepare for stuff like the AMC 12, apart from what my math teacher might give us, because I think that more difficult problems are more worth my time.

To conclude, the AMCs are more random than others, and anyhow, perparing for harder contests is actually probably a better way to improve your score at those anyhow, becuase it will actually make you better at math, whereas doing a million AMC 10-type problems will make you kind of light on your feet, but not much better at math, so why bother, if you can assuredly qualify for USAMO?

I agree that you shouldn't be worrying about getting a perfect on the AMC 10. If you know you can do that well on the 10, don't worry about getting a perfect score. Instead, spend your time working on things you might not be as good as, like the AMC 12 or the AIME. Of course, it's nice to get a perfect on the AMC 10 but there are better things to be working towards this early in the year.

PRactice AMC 12.  They certainly could help you in AMC 10.  And yes, make sure you know the important stuff up to Algebra II for AMC 10, I believe.

All of the said above is true. Since you can obviously pass AMC 10, practice harder stuff and go for the next level: the AIME. An extra problem or two on the AMC 10 isn't going to make a difference in the big picture. If you're really intent on acing it, just be careful and do lots of practices (as said before, harder stuff)

[quote="mathgeniuse^ln(x)"]this is random

But where on MathLinks do you get some Past Contest Problems?[/quote]

Click the button that says "Contests" at the top.

AMC 10 is pointless .. highest score in each state is almost always an 8th grader since good high school students should be taking 12. I almost think they should change the name of AMC 10 to AMC 8 version 2.0, and only let junior high kids take it.

If you are in high school it would be to your benefit to take the 12. If you can qualify for AIME through AMC 10 then you will qualify through the 12, see some better problems, and it won't affect your USAMO chances. If you are shooting for a perfect on the AMC 10 then you can definitely make AIME through the 12. Sure it's nice to get a perfect on something but nobody out there is too fooled by "perfect score AMC 10" on a college application or something (well, that's probably not 100% true but it should be and close enough..). If you look at the big picture, will you benefit more as a mathematician by taking the 12 or the 10? I think the answer would be to take the 12. Getting 150 on AMC 12 is never that easy, no matter how good you are because of the time limit. Getting 15 on AIME is never easy no matter how good you are because you make stupid mistakes and occasionally they throw a strange problem out there that eats up time. Getting 42 on USAMO is never easy because there are jerks like me trying to submit really hard problems to the problem selection committee.

But perfect AMC 10..? Just take the 12.