It is hard to find a good game to play, for me at least. My (hopefully not too demanding) standards are:

1. Determined, Complete Information, Nothing to be attributed to chance
Theoretically, there should be a computable winner (or tie, which is acceptable but not preferred).

2. Fair (doesn't have to be equal, although preferable), Does not tie often, if at all
(May contradicts #1, but...) That is, so far as current knowledge has it. May have a rule to make it fair.

Hex does well here.

3. Simplicity, Universality, Generality, Uniqueness in Rules
not Arbitrary, Complicated
--Rules should be nice. There shouldn't be a whole bunch of similar games that can branch off.

Chess and variants, unfortunately, utterly fail at this.

4. Complexity in Play
An amount of complexity and uncertainty is required for a game to gain interest.

5. A Good Amount, but not too much of, Predictability
Players should be able to have comebacks after poor play, but the game shouldn't sway too much in terms of which player is in the lead.

6. Decisive
You should be able to tell in a few seconds from the ending position who has won.

Go fails here.

7. Balance of Strategy (long-term) and Tactics (short-term)
Often games get boring because they are too heavily one way or another. There shouldn't be a huge amount of micromanagement going on, but one should still be able to calculate sequences of play in one's head.

Hex seems to do the best here... but somehow it gets boring...
hmm....

[quote="leoxnlin"]6. Decisive
You should be able to tell in a few seconds from the ending position who has won.

Go fails here.[/quote]
This shortcoming is removed if you play with an online client, which computes the final score for you.  I'd say that Go does the best job of satisfying your other requirements, too.

[quote="leoxnlin"]Nothing to be attributed to chance [/quote]

Unless I'm misinterpreting you, I disagree with this requirement.  I believe a certain amount of luck is necessary for a good game (I search for games with a nice blend of skill and luck).

For me, an element of chance and luck is good for a short, light, brief break from heavy, slow thought. But chance, to me at least, ruins the concept of fairness.

EDIT: Wow, my writing sounds horrible.

[quote="mewto55555"]Connect 4?[/quote]

Solved, I believe.

Although this may sound odd on a math forum, how about Scrabble?  It has an element of luck, but good players (i.e. better than me  :| ) can generally overcome bad luck.  It seems (at least to me) to meet most of your other requirements as well.

No offense if you like the game, but even if luck were somehow removed, Scrabble rather fails the simplicity game - it incorporates the entire English lexicon in it - not unique (why not Turkish?), general, or universal.

I don't know, but I like the game [url=http://en.wikipedia.org/wiki/Arimaa]Arimaa.[/url]

Wow, no one seems to have the same idea of a good game as me...

Arimaa is the type of game I may play 2-3 games against someone online or something, and then forget about. It, in the same way as chess, is far too complex for me.

[quote=t0rajir0u]I'd say that Go does the best job of satisfying your other requirements, too.[/quote]

I'd agree that Go does the best job for the criteria I listed among popular games. But counting squares? That seems a tad bit inelegant to me... and the komi compensation? Why not 7? Why not 5? Why not 6.36510253603? There's no wonderfully unique number that clearly solves the problem. And superkos? Anything that requires players to memorize or record the entire position of the board from all previous moves is... bleh.

All rather frustrating difficulties, for me at least, to an otherwise beautiful game. It's like little specks of impurities in a clear gem - so annoying...

Anyone heard of Hex? Can someone suggest more games?

Go isn't really about winning or losing.  A close game is satisfying to both players - the goal isn't to defeat the other player so much as to have a "hand conversation" where you explore the world of the game as intellectual partners.  As for the ko rule, again, that's what online clients are for.

I'm not sure what you have against counting squares.  The fact that $ 361$ is a large number relative to the numbers that are easy to deal with doesn't affect, to me at least, the beauty of the game.

t0rajir0u: The last sentence of yours has changed my views about the counting large territory issue.  :) (I didn't realize that I could change my mind that easily!)

I now conclude that Go is most likely the most beautiful game I have found that satisfies the criteria of the first post, but it is unplayable by stupid people like me  :( - if I can't count squares, I don't know where I am in this intellectual conversation, as t0rajir0u put it.

Hi everybody,

[url=http://joujoujou.over-blog.com/article-25658471.html]URL Morpion au carré[/url]

It's a free game. You can play it using paper and pen.
It's a two-players game. A variant of Tic Tac Toe.

For connect 4 and Hex don't they utterly fail in "fairness" - first to move has an obvious win.


Is there an AMC-8 class. If not, can someone create one?

...
no
...
:huuh: You obviously don't understand how much effort the instructors have to put into each class. :roll: I would recommend the Intro courses and MC Problem Series, which is at a more difficult level than AMC 8, but will help anyway.

Currently we have no plans to create an AMC 8 problem series, but chess64's suggestion is a good one -- the MATHCOUNTS Problem Series covers much of the same material that would be covered in an AMC 8 problem series.  The Intro subject classes are also good, though they go a good bit beyond AMC 8.


Let [b]a,b,c>0[/b] be real numbers with sum 1. Prove that

\frac{a^2}b + \frac{b^2}c + \frac{c^2} a \ geq 3(a^2 + b^2 + c^2)

The ineq is
[quote="Fitim"]Let [b]a,b,c>0[/b] be real numbers with sum 1. Prove that

$\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \geq 3(a^{2}+b^{2}+c^{2})$[/quote]
and it has allready been discused here:
http://www.mathlinks.ro/Forum/viewtopic.php?p=515092#p515092


25 students are to join a school which will take place in three different classrooms consisting of 3, 4 and 5 students. The simultaneous classes will be held once every day and the school director will determine the classrooms for each student to join.

At least how many days are necessary if each student is to participate in a meeting at least once with all the other students?

( The question means:
Each day, there will be exactly 3 classes - one of 3 student, one of 4 people and one of 5 people. Nobody can be in more than 1 class on any day. By the end, each of the 25 student has to have been in at least 1 class with each other student.)


can any one give a proof to this theorem ? is it so difficulte as Brouwer's theorem  :?  !!

Some info can be found on http://planetmath.org/encyclopedia/HairyBallTheorem.html . Very advanced stuff. (for me at least)

think you , it's seem's to be very interessting  :D


calculus
integration
trigonometry
calculus computations

Calculate
\[ \int_{0}^{\pi}\frac{\cos (rx)}{1\minus{}2a\cos x\plus{}a^2}\;dx\]

hello, what do we know about $ a$ and $ r$?
Sonnhard.

But it diverges for $ |a|\ge 1.$ And it's a nicer problem for $ r$ an integer than it is for $ r$ not an integer.

If we assume that $ 0\le a<1,$ the following may be a useful formula:

$ \sum_{k\equal{}\minus{}\infty}^{\infty}a^{|k|}e^{ikx}\equal{}1\plus{}2\sum_{k\equal{}1}^{\infty}a^k\cos kx\equal{}\frac{1\minus{}a^2}{1\minus{}2a\cos x\plus{}a^2}.$

Could you please elaborate why the integral is diverges when $ |a|\geq1$ ?

Oops, sorry about that. It certainly diverges for $ a\equal{}1,$ since in that case the denominator would be $ 2\minus{}2\cos x\equal{}4\sin^2\left(\frac x2\right),$ so the integral would be comparable to $ \frac1{x^2}$ near zero.

But it turns out that $ a>1$ is OK. We can write $ 1\minus{}2a\cos x\plus{}a^2\equal{}|1\minus{}ae^{ix}|^2,$ which is never zero for either $ a>1$ or $ a<1.$

Can you find a different geometric sum, similar to the one I posted, but valid for $ a>1?$

[quote="Kent Merryfield"]
$ \sum_{k \equal{} \minus{} \infty}^{\infty}a^{|k|}e^{ikx} \equal{} 1 \plus{} 2\sum_{k \equal{} 1}^{\infty}a^k\cos kx \equal{} \frac {1 \minus{} a^2}{1 \minus{} 2a\cos x \plus{} a^2}.$[/quote]

At first, I confused with the second equality but now I understand. :lol:

For $ a>1$, $ \sum_{k\equal{}\minus{}\infty}^{\infty}a^{\minus{}|k|}e^{ikx}\equal{}1\plus{}2\sum_{k\equal{}1}^{\infty}a^{\minus{}k}\cos kx$

$ \equal{}\frac{1\minus{}\left(\frac{1}{a}\right)^{2}}{1\minus{}2\left(\frac{1}{a}\right)\cos x\plus{}\left(\frac{1}{a}\right)^{2}}\equal{}\frac{a^{2}\minus{}1}{1\minus{}2a\cos x\plus{}a^{2}}$.


Are  you going to be at awesome math next year? Hopefully I will be able to.

It's a little early to say...since I might have a summer job or other camps or something...

I sure hope that I can go though, since it was soooooooo much fun.

I really think i am not going.

I already went last year (2007) and it was fun and it was fun also this year.

But, i better make MOP from what i learned :D

I might go... I definitely want to. But like Carbon... I have other school-related camps to do...

AMSP was cool, but it's time for me to move on  :(  . MOP if everything goes well, or maybe Mathcamp or something if I fail.

It was an awesome experience and I hope I do go back. Just a random comment: I will probably never eat as many different brands of cereal in one weekend as I did in Awesomemath.

hmmm
either i make rsi and i go to rsi
or
i dont make rsi (likely) and i go to mathcamp and mop, if i make it again
i may return the year after that as a counselor!
maybe
xD

I'd love to come back and all, but like Zephyr said, maybe it's time to move on. In all honesty, it depends on who else that I knew would be going back.

No one I know will go, so no go for me. :(
@infinity4ever: Are you kimmystar94???

AMSP was a lot of fun!!  I really want to go back next year!!

[quote="dannyhamtx"][quote="zephyredx"][quote="kl2836"]i must make pink mop next year![/quote]

Purple is mine![/quote]

(^_^) already went to purple...... too bad.

Open ones: yellow, green, orange, and rainbow (personally rainbow FTW :D )[/quote]

Is white MOP open?

I hope to come back next year; I certainly had a great time this year! But I still wonder how college will effect my attitudes towards problem solving.

i am going for a very beautiful colored mop

hmmm
cgmo moppers should be called pink mop...!

[quote="dannyhamtx"][quote="zephyredx"][quote="kl2836"]i must make pink mop next year![/quote]

Purple is mine![/quote]

(^_^) already went to purple...... too bad.

Open ones: yellow, green, orange, and rainbow (personally rainbow FTW :D )[/quote]

But they don't have to be visible colors do they?...

"I made X-Ray MOP!"

This MOP discussion is getting off topic sorry.

[quote="yaj1729"]I really miss AMSP  :(
I doubt I'll come back next year, but maybe, just maybe

First Post!![/quote]

Welcome!
I really miss AMPS too.  I really want to go back next year!

[quote="yaj1729"]I really miss AMSP  :(
I doubt I'll come back next year, but maybe, just maybe

First Post!![/quote]

Is this Yajit?


The largest integer $n$ for which $n^{200} < 5^{300}$ is

$\textbf{(A) }8\qquad
\textbf{(B) }9\qquad
\textbf{(C) }10\qquad
\textbf{(D) }11\qquad
\textbf{(E) }12$

[hide="Solution"]
$n^{200}<5^{300}$

$\Longleftrightarrow n^{\frac{200}{100}}<5^{\frac{300}{100}}$

$\Longleftrightarrow n^2<5^3$

$\Longleftrightarrow n^2<125$

$\Longleftrightarrow n<11$

$\therefore\boxed{(D)}$[/hide]

[hide]$n^{200}<5^{300} \Longrightarrow n^2<5^3 \Longrightarrow n^2<125$, so the biggest integer possible is $\boxed{11}$.[/hide]

[hide]
Since all the numbers involved are positive reals, we can take the hundredth root of both sides to arrive at
$n^2<5^3$ or $n^2<125$
The smallest perfect square less than 125 is 121, whose square root is 11. Thus the answer is D
[/hide]


[quote="reidster"]There are 7 digits in a telephone number.
If the first number cannot be 0 or 1, and the first 3 numbers cannot be 0, 0, 0 (they could however be 2, 0, 0.. or 3, 0, 1, etc) and they last four numbers cannot be 0, 0, 0, 0 (same goes here), How many combinations are there?[/quote]

8 * 100 * 9999 = [b]7999200[/b].

8 possibilities for the first digit, second and third can be anything, and the rest can be anything except 0000.

[quote="reidster"]There are 7 digits in a telephone number.
If the first number cannot be 0 or 1, and the first 3 numbers cannot be 0, 0, 0 (they could however be 2, 0, 0.. or 3, 0, 1, etc) and they last four numbers cannot be 0, 0, 0, 0 (same goes here), How many combinations are there?[/quote]
There are (8*10*10)(10*10*10*10-1) possible combinations.  I'll leave it to you to figure out why.

Edit: Apparently someone posted while I was typing this.


[quote="orl"]I chose option 5. Thanks for the great effort but I think it is more important to have more problems in the resources section than to have your posted AoPS/ML AIME problems/solutions piped into a .pdf-file as well! And an advice on top. Not sure right now whether you are a resources manager but it is a good idea to have your .pdf-files referenced in the resources sections, e.g. here for 1983: http://www.mathlinks.ro/Forum/resources.php?c=182&cid=45&year=1983 Keep up the good work![/quote]

orl, correct me if I'm wrong but Valentin once said that there won't be PDF documents for solutions of the AoPS/ML Resources problems because it will make things messy (or something close to it). So, maybe we need to put this for the "special list" in "our" forum.  :D

[quote="4everwise"]Resource managers can't do that. Only an administrator can change the layout to include links.
[/quote]

Well, of course you can. Check this URL: http://www.mathlinks.ro/Forum/resources.php?c=182&cid=45&year=1983

Just go here: "Math Forum » Olympiad Resources » Add Resources » Edit/Delete Problems' Data" and enter USA -> AIME -> 1983. Press the button "Submit Problems" yielding that site which is partly shown as screen shot below. There you can enter the data as I did.

[quote="Silverfalcon"] orl, correct me if I'm wrong but Valentin once said that there won't be PDF documents for solutions of the AoPS/ML Resources problems because it will make things messy (or something close to it). So, maybe we need to put this for the "special list" in "our" forum.  :D[/quote]

Well, the issue of being messy is only factor. Extracting all these solutions each time heavily inreceases the work load for the server. People don't just download free solutions but come back to ML/AoPS checking the available solutions. This way we also increase the probabilty that people post those solutions they found when they were working on the downloaded problem sets. Or in other words. People don't only gain but also contribute to our ML/AoPS community.

[quote="4everwise"][quote="JGeneson"]These are great. You might want to take a look at the equality case on 1983.9 though.[/quote]I'm not sure what's wrong with it. Can you be more specific?  :)[/quote]
[hide]You needed to divide by 9 instead of multiply by it. This gives $y=\frac{2}{3}$ (which can actually be attained!) other than $y=6$ which was in the document.[/hide]

Reply: Thanks.

[quote="samath"]On AIME 1983.10, with numbers that have two 1's and two other numbers that do not repeat, there would be 3 ways to arrange the numbers.  Wouldn't there be six? (For example, if the number were 1123, since the number must start with 1, you could rearrange the digits to get: 1123, 1132, 1213, 1231. 1312. 1321.)

Thanks for your help and nice job with the problems.[/quote]

Yeah.  I agree with both statements.  These are a huge help!

[quote="b-flat"]Umm, now, the link says URL not available on the server, so I can't get the solutions.[/quote]

Neither can I

It probably had something to do with the server being changed.

[quote="4everwise"]Hmm.....  :|

I'll repost them after I finish this chapter in CPIG. :D[/quote]

Thanx, I appreciate it.

Sorry for reviving a topic, but I can't access them. :wink:

that happened to those solutions and questions for AMC 10 and 12 on Kalva or whatever it's called. There were old links, but none work anymore.  :(  :(

Okay. The AIME's are now on my website, at :

theani.isgreat.org/tmp/contests.php

I'm just having them there temporarily though, as the format will be changed (but the files willl stay)

[quote="4everwise"]Here are the Solution Documents. Enjoy!  :)

[url=http://www.artofproblemsolving.com/Forum/download.php?id=3956]Aime 1983[/url]
[url=http://www.artofproblemsolving.com/Forum/download.php?id=3960]Aime 1984[/url]
[url=http://www.artofproblemsolving.com/Forum/download.php?id=4017]Aime 1986[/url]
[url=http://www.artofproblemsolving.com/Forum/download.php?id=4334]Aime 1992[/url][/quote]

When I click the links they don't work?

[quote="anirudh"]Exactly.

As I said, the Solution for the 1984 AIME is on my website.[/quote]
No need to lie...

Sorry. I meant the 1983 one. It comes with the document.

That or you could just download them from here...

[quote="anirudh"]They're the same ones. :roll:

http://theani.isgreat.org/tmp/files/AIME1983.pdf[/quote]
Yes, I know but I figured it would save people the trouble of having to go to your site when they could just get it instantly from here...

[quote="4everwise"]Here are two more old ones.

Hmm, I don't think I'll have time to type up solutions to all these AIME problems. I mean, I barely have time to do math when school starts, so um... I might type one up every once in a while, but there probably won't be a complete solution set.

I'll definately have all the problems up in a few days though. Missing that one test... argh... (AIME 2000 I)[/quote]

Oh. Thanks.

thanks for all the problems and solutions 4everwise.

Hi! I don't know if this has been posted before, or if anyone really cares, but in the 1986 AIME solution to pr. 8, I think it should be 23 instead of 26. Also, solution to problem 7 was missing in the Wiki.


best of 5 because it is enuff that luck cannot win and not as long as 7

please abandon this thread. it is decided that both the Countdown Cup and Ernie's Tounrey will be best-of-5.  thank you.


Two players $ A$, $ B$ and another 2001 people form a circle, such that $ A$ and $ B$ are not in consecutive  positions. $ A$ and $ B$ play in alternating turns, starting with $ A$. A play consists of touching one of the people neighboring you, which such person once touched leaves the circle. The winner is the last man standing.

Show that one of the two players has a winning strategy, and give such strategy.

Note: A player has a winning strategy if he/she is able to win no matter what the opponent does.

[hide]$ A$ wins. The strategy is to keep the number of people between them odd.[/hide]


probability
geometry
rectangle
Pythagorean Theorem

these r from 07-08 school handbook, but idk how to do em:
1)There is a method traditionally used in some Russian villages to see which of the young women in the village are to be married the next year. Three blades of grass are folded in half and held in such a way that the six ends of the blades are visible, but the rest of the blades are hidden. A young women ties the ends together in pairs at random such that there are three knots and each end is tied to exactly one other end. If, on release, a three-blade loop is formed, the woman will be married the next year. What is the probability of getting a three-blade loop? Express your answer as a common fraction.

2)Points M and N are the midpoints of sides AB and BC, respectively, of rectangle ABCD. If segments AN and CM intersect at point X, what fraction of the area of the rectangle ABCD lies in the quadrilateral BNXM? Express your answer as a common fraction.

3)A square is inscribed in a right triangle with legs of 8 units and 15 units. If two of the vertices of the square lie on the hypotenuse and the other two vertices of the square lie on the legs of the triangle, what is the length of a side of the square? Express your answer as a common fraction.

4)The symbol * represents a sequence of mathematical operations. If 12 * 6=35, 4 * 2=3, 11 * 14=8, 5 * 7=3, and 10 * 7=8, what is the value of 8 * 4?

5)Point A is located at (5,5) on the Cartesian plane. Point B is also in the plane and has integer coordinates. If 0 < AB (less than or equal to) 4, at how many points could B be located?

I think #3 supposed to do with harmonic mean and stuff, but there is another way to do it besides memorizing all that geo stuff? (hate geo mem)
I did triangle similarity and got this funny solution.

for #5, there is a way to do it besides constructing a radius 4 circle about (5,5) and counting one by one the points (i draw horrible and draw circles bad too, so yeah...)  :)

[hide="For #4,"] we can see that all of the outputs are $ 1$ less than a perfect square.[/hide]

[hide="the asterisk means..."]By closely examining the first two numbers of each, we see that the square root of the square previously mentioned is always equal to the difference between the two numbers.[/hide]

[hide="using the hypothesis...."]Using our hypothesis that the output is equal to the difference between the first and second numbers, squared, and then minus $ 1$, we can guess that $ 8*4 \equal{} 4^2 \minus{} 1 \equal{} 16 \minus{} 1 \equal{} \boxed{15}$[/hide]

[quote="CA Math"]For #5:  You don't necessarily have to draw out the whole circle, as you can:
count the number of points in quadrant 1, multiply by 4.
count the number of points on the positive x-axis, multiply by 4.
and then add one for the origin.

In fact, you do not need to draw out the circle at all (all though it helps) because imagining the points above are not really that hard (for many people).  Just be careful when counting.

Here's how I would count the number of points in quadrant 1:
[hide]
There are four possible $ x$ values; $ 1, 2, 3, \text{or} 4$.
For $ x$-value of $ 1$, we know that the $ y$-value can be any number from $ 1$ to $ 4$.  We know this because the Pythagorean theorem shows that $ 1^2 \plus{} 4^2 \le 5^2$.

For $ x$-value of $ 2$, we know that $ y$-value can also be any number from $ 1$ to $ 4$ for the same reason as above (in this case, with $ 2^2$ rather than $ 1^2$).

For $ x$-value of $ 3$, we know that $ y$-value can also be any number from $ 1$ to $ 4$, using same reason.

And finally, for $ x$ value of $ 4$, the $ y$-values can be anywhere from $ 1$ to $ 3$.  It can not be $ 4$ because $ 4^2 \plus{} 4^2 > 5^2$

Summing these values, we get $ 4 \plus{} 4 \plus{} 4 \plus{} 3 \equal{} 15$.[/hide]

While this process may seem long and tedious on paper (screen), in your mind it will probably be very quick and easy.[/quote]
edit:woops, misread ur solution.  yeah, but note that there exists four points that have the same distance from (5,5). i got 64, but the answer wuz 48... :|

#3 is pretty funny
draw a diagram, im too lazy to. make the hypotenuse the base.
The area of the triangle is 8*15/2=60. The area of the triangle is also the sum of the area of the square and the areas of tthe smaller triangles the square creates (let this equal A). The height of the triangle is 60*2/17=120/17. Let the side length of the square equal s. Combine the two lower smaller triangles to make a triangle with base 17-s and height s. Then
$ A\equal{}\dfrac{s(17\minus{}s)}{2}\plus{}\dfrac{s\left(\dfrac{120}{17}\minus{}s\right)}2\plus{}s^2$
Fortunately the s^2 terms cancel out, and we are left with
$ A\equal{}\dfrac{409}{34}s$
$ 60\equal{}\dfrac{409}{34}s$
$ \boxed{s\equal{}\dfrac{2040}{409}}$
hmm...whered i mess up... i recall the answer was like 120/37...o well


Show that if f(x) is coersive (lim f(x) x->infinity =infinity) then it has a global minima

Maybe you forgot to say that $f$ is continuous?  :?

Thanks for your reply.yes it is but I would need to a proof

Take $r$ so big that $f(x)>f(0)$ for all $x$ with $|x|>r$.
Now $f$ has a minimum on $[-r,r]$

It can be unattainable, or it can be $-\infty$ :?

I thought in the second post he meant that we shall use continuety¿

How big it shold be and how can you proof there exist such a r.

You can't give a term saying how big it will be, and it exists since the function grows to infinity, which is more or less by definition what I used


I have not recived my checkmark yet for round 4.  Just wonding if any one else had not yet got their checkmark

I've had my check mark since like 2 days after the deadline (and I turned it in just about at the deadline!) :maybe:

[quote="timwu"]It's Mar.30th now, and I haven't received my check mark yet (I submitted it on Mar.7th, four days before the deadline)...is this normal?[/quote]No...everything that we have received (on time) should be checked in by now.  Please contact us (usamts@usamts.org) and provide your USAMTS ID# so we can research the matter.  Please also let us know when and how you submitted your solutions (email, mail, or fax, and approximately what day and time).

I tried to resubmit the solutions, but it gave me an error message:

[hide]
This is an automatically generated Delivery Status Notification

Delivery to the following recipient failed permanently:

    solutions@usamts.org

Technical details of permanent failure:
PERM_FAILURE: SMTP Error (state 17): 552 5.3.4 Error: message file too big

  ----- Original message -----

Received: by 10.142.158.17 with SMTP id g17mr4368750wfe.157.1207007385313;
       Mon, 31 Mar 2008 16:49:45 -0700 (PDT)
Received: by 10.142.99.17 with HTTP; Mon, 31 Mar 2008 16:49:38 -0700 (PDT)
Message-ID: <8676dc250803311649j397494c9h1a91287f784ac515@mail.gmail.com>
Date: Mon, 31 Mar 2008 19:49:38 -0400
From: "Tim Wu" <wtq0000@gmail.com>

To: tinac@artofproblemsolving.com, solutions@usamts.org
Subject: Fwd: Tianqi Wu, 9731, Round 4
In-Reply-To: <8676dc250803071304j1a4e004dpf9b17ec47adbccc1@mail.gmail.com>
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Resubmission of my solutions to Round 4, thanks.

  ----- Message truncated -----
[/hide]

EDIT: never mind, I figured out it. I'm waiting for further notice of reception of my solutions.

One more thing: when I resent my solutions, should I resend to solutions@usamts.org or something else? This might seem redundant, but I haven't got any reply yet, so I kind of panicked...

Have anyone received my resent solutions for Round 4? I still haven't seen my checkmark... :maybe:


(1) Factorize $p^4-2ap^2+a^2-p-a$

(2) Solve $x=a+\sqrt{a+\sqrt{x}}\ (a>0)$

Source: 1972 Tokyo Medical and Dental University

Hmmm interesting source.  They have math competitions in medical schools in Japan?

[hide="hint on 2"]When you plug in $x$... see what happens.  You get a continued radical.[/hide]

cool method and the $1$st one is the same as the $2$one :D

another method for the 2nd one when you solve the $2$nd one you see it comes to be the same as the $1$st factorisation, and hence it is easy... but the factorisation isnt :P

[quote="eryaman"]Hmmm interesting source.  They have math competitions in medical schools in Japan?[/quote]
No, it's an entrance exam for the university. A lot of Japanese universities have their own extrance exams :)


Given the function f(x)=ax^2+bx+c
and |f(0)|<=1 ; |f(2)|<=1
Find all functions f(x) such that D=b^2-4ac is min

Solution for this question was posted on your question at
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=887975#887975


Find the $n$ th term of the sequence $\{a_{n}\}$ such that $\sum_{k=1}^{n}a_{k}=n^{3}+3n^{2}+2n.$ and Calculate $\sum_{k=1}^{n}\frac{1}{a_{k}}.$

I must say I don't see a mistake. The general term for $a_{n}$ works in the given equation, and the rest is plainly correct :)

In the first line, you should write $n\geq 2.$ and conclude that $a_{n}=....\ (n\geq 1).$

You're right, though in the given case it happens to lead to the correct result anyway, because $\sum_{k=1}^{0}a_{k}=0=a_{0}$

Certainly in this case You are right, in this connection, in Japanese university entrance exam almostly 95%, we deal the first term as $a_{1}.$ :)

Aw...Farenhajt, did you get to all of these?


Let $a$ be a postive integer.
How many not-negative integer-solutions for $x$ has the equation $\left\lfloor \frac{x}{a}\right\rfloor = \left\lfloor \frac{x}{a+1}\right\rfloor$ ?
Annotation: For each real number $z$, you describe with $\left\lfloor z \right\rfloor$ the biggest integer, that isn't bigger than $z$.

If $x=ka+r$, where $k$ and $r$ are integer quotient and residue respectively, then we get

$[{x\over a}]=[{x\over a+1}]\iff [k+{r\over a}]=[k+{r-k\over a+1}]\iff k=[k+{r-k\over a+1}]$

and that is equivalent to $0\leqslant{r-k\over a+1}<1\iff r-a-1<k\leqslant r$.

Since the left part is always negative ($r<a$), that reduces to $0\leqslant k\leqslant r$. Therefore for each $r$ we have $r+1$ values for $k$, which means that the total number of solutions is $N=1+2+\dots+a={a(a+1)\over 2}$


inequalities
geometry
circumcircle
trigonometry
area of a triangle
Heron\u0027s formula
trig identities

[hide]I love algebra so I will post the Algebra approaching way, despite it's kind of off topic :P :
Using Cauchy-Schwartz's inequality we have:
$ \sum_{\text{cyc}} \frac {a^3}{a^2 \plus{} ab \plus{} b^2} \ge \frac {(a^3 \plus{} b^3 \plus{} c^3)^2}{a^5 \plus{} b^5 \plus{} c^5 \plus{} a^4b \plus{} b^4c \plus{} c^4a \plus{} a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2}$
Now we just need to prove: $ 3(a^3 \plus{} b^3 \plus{} c^3)^2 \geq (a^5 \plus{} b^4 \plus{} c^5 \plus{} a^4b \plus{} b^4c \plus{} c^4a \plus{} a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2)(a \plus{} b \plus{} c)$
Algebraic manipulation, we're done(we also need to use Schur 6th-degree:$ a^3(a \minus{} b)(a \minus{} c) \plus{} b^3(b \minus{} c)(b \minus{} a) \plus{} c^3(c \minus{} a)(c \minus{} b)\geq 0$)

Alternative way: Using $ (a^3 \minus{} b^3)/(a^2 \plus{} ab \plus{} b^2) \equal{} a \minus{} b$ and summing all these three identities, you get the $ LHS \equal{} \sum_{\text{cyc}} \frac {b^3}{a^2 \plus{} ab \plus{} b^2}$[/hide]

how can you assume that a,b,c are sides of a triangle?it is not specified

[quote="Anaxerzia"]For $ a,b,c > 0$, show that
\[ \sum_{\text{cyc}} \frac {a^3}{a^2 \plus{} ab \plus{} b^2} \ge\frac {a \plus{} b \plus{} c}{3}
\]
What I have so far is to
[hide]Consider triangle XYZ, with angles $ X,Y,Z < \frac {2\pi}{3}$. Construct on each side of the triangle an equilateral triangle. Let the intersections of the circumcircles of these equilateral triangles be point $ P$. Let $ PY \equal{} a$, $ PZ \equal{} b$, and $ PX \equal{} c$. It is simple to show that $ a,b,c$ each can take any positive value, independent of each other. By the law of cosines, $ x^2 \equal{} a^2 \plus{} ab \plus{} c^2$, $ y^2 \equal{} b^2 \plus{} bc \plus{} c^2$, and $ z^2 \equal{} c^2 \plus{} ac \plus{} a^2$. Thus, we must prove that:
\[ \sum_{\text{cyc}} \frac {a^3}{x^2} \ge\frac {a \plus{} b \plus{} c}{3}
\]
and... from here I don't know what to do. I think the idea of making it into a triangle is a good one, but I'm not sure what to do with it. You can use Law of Sines to find the area of the triangle, to be $ \frac {\sqrt {3}}{4}(ab \plus{} bc \plus{} ac)$, and I think possibly relating that to Heron's formula might be useful, but I'm not sure.[/hide][/quote]
by AM-GM ineq,we get:
$ \sum {\frac {{{a^3}}}{{{a^2} \plus{} ab \plus{} {b^2}}}} \equal{} \sum {\left( {a \minus{} \frac {{{a^2}b \plus{} a{b^2}}}{{{a^2} \plus{} ab \plus{} {b^2}}}} \right)} \ge \sum {\left( {a \minus{} \frac {{a \plus{} b}}{3}} \right) \equal{} a \plus{} b \plus{} c}$
and another proof:
because $ \sum {\frac{{{a^3}}}{{{a^2} \plus{} ab \plus{} {b^2}}}}  \equal{} \sum {\frac{{{b^3}}}{{{a^2} \plus{} ab \plus{} {b^2}}}}$
so,$ 2LS\equal{}\sum {\frac{{{a^3} \plus{} {b^3}}}{{{a^2} \plus{} ab \plus{} {b^2}}}}  \ge \frac{2}{3}\sum {\frac{{{a^3} \plus{} {b^3}}}{{{a^2} \plus{} {b^2}}}}  \ge a \plus{} b \plus{} c$(by chebyshev ineq)

I didn't assume that a,b,c are sides of a triangle.

Thanks for your responses =)

[quote="Anaxerzia"]For $ a,b,c > 0$, show that
\[ \sum_{\text{cyc}} \frac {a^3}{a^2 \plus{} ab \plus{} b^2} \ge\frac {a \plus{} b \plus{} c}{3}
\]
What I have so far is to
[hide]Consider triangle XYZ, with angles $ X,Y,Z < \frac {2\pi}{3}$. Construct on each side of the triangle an equilateral triangle. Let the intersections of the circumcircles of these equilateral triangles be point $ P$. Let $ PY \equal{} a$, $ PZ \equal{} b$, and $ PX \equal{} c$. It is simple to show that $ a,b,c$ each can take any positive value, independent of each other. By the law of cosines, $ x^2 \equal{} a^2 \plus{} ab \plus{} c^2$, $ y^2 \equal{} b^2 \plus{} bc \plus{} c^2$, and $ z^2 \equal{} c^2 \plus{} ac \plus{} a^2$. Thus, we must prove that:
\[ \sum_{\text{cyc}} \frac {a^3}{x^2} \ge\frac {a \plus{} b \plus{} c}{3}
\]
and... from here I don't know what to do. I think the idea of making it into a triangle is a good one, but I'm not sure what to do with it. You can use Law of Sines to find the area of the triangle, to be $ \frac {\sqrt {3}}{4}(ab \plus{} bc \plus{} ac)$, and I think possibly relating that to Heron's formula might be useful, but I'm not sure.[/hide][/quote]
We can prove that :$ \sum\ \frac{a^3}{a^2\plus{}ab\plus{}b^2} \ge \frac{\sum{a^2}}{\sum{a}}$
USE CAUCHY-SCHWARZ INEQUALITY :)

@AAdil:
It is not assumed that $ a,b,c$ are sides of a triangle.
It is because of the expressions $ a^2\plus{}ab\plus{}b^2$ and the like that the auhtor used  the Fermat Point properties using Cosine RUle.

@quykhtn-qa1 Yeah I thought the same too.
Great. ANy improved form of the ineequality u got other than the obvious ones?

[quote="Agr_94_Math"]
@quykhtn-qa1 Yeah I thought the same too.
Great. ANy improved form of the ineequality u got other than the obvious ones?[/quote]
What do you mean?
We have:
$ LHS \ge \frac{(\sum \ a^2)^2}{\sum \ a(a^2+ab+b^2)}$
But ${ \sum \ a(a^2+ab+b^2}=(\sum \ a^2)(\sum \ a)$
Then ,we have Q.E.D
What about the following,arqady?
$ \sum \ \frac{a^3}{a^2+ab+b^2} \ge \frac{\sum \ a^3}{\sum \ a^2}$ :)


a) Let a, b be two vectors in 3D and for every integer n consider the vector Vn = a X (a X (a X (... X (a X b))) ... ), where “aX” appears n times. What are the conditions on a and b such that the limit of Vn exists as n approaches infinity, and what is the limit then?

b) Four vectors a, b, c and x satisfy the relation: (a . x)b = c + x. Show that (b . a - 1)a . x = c . a, and if
b . a does not equal to 1,  then express the vector x in terms of the vectors a, b and c.

[hide="(a)"]  $ V_{n\plus{}1}\equal{}a \times V_n \to a, V_n \perp V_{n\plus{}1}$, $ V_0\equal{}b$

$ a\equal{}(a_1,a_2,a_3), V\equal{}(c_1,c_2,c_3)$

$ V\equal{}a\times V \equal{} (c_3a_2\minus{}c_2a_3, c_1a_3\minus{}c_3a_1,c_2a_1\minus{}c_1a_2)$

$ \to c_3a_2\minus{}c_2a_3\equal{}c_1, c_1a_3\minus{}c_3a_1\equal{}c_2,c_2a_1\minus{}c_1a_2\equal{}c_3$

which is an invertible system for c, which will have a unique solution, namely the trivial solution.

aka V must be the zero vector.

so how do we ensure that $ |V_n| \to 0$ with a and b?

well if a and b are parrellel, then $ V_1\equal{}0$ and we win,

elsewise,  $ |a\times b| \le |a\parallel{}b| \to |a|<1$ will also work[/hide]

[hide="(b)"]  let $ a\cdot b \ne 1,$

$ (a \cdot x)b\equal{}c\plus{}x \to a\cdot (a \cdot x)b\equal{}a\cdot(c\plus{}x)$

 $ (a \cdot x)(a\cdot b)\equal{}(a\cdot c)\plus{}(a\cdot x)\to (a\cdot c) \equal{}(a \cdot x) ((a \cdot b) \minus{} 1)$

$ \alpha \equal{} \frac{1}{(a \cdot b) \minus{} 1}\to 0\equal{} (a\cdot \alpha c)\minus{}(a\cdot x)\equal{}a\cdot(\alpha c \minus{} x)$

then $ x\in \{ \alpha c\plus{} v: a \perp v \}$
[/hide]


It says ive mastered "Counting Independent Events", but ive only gotten 33% right.

Alcumus doesn't just look at a subject to determine your proficiency in it---it also looks at other subjects that depend on that subject.  So, since you've done well on many harder problems that depend on Counting Independent Events, Alcumus isn't going to make you slog through 20 "Steve has 4 shirts and 3 pants, how many outfits can Steve make" problems.  (Don't you wish your regular math classes were like that :) )

Alcumus will go back and test you on passed subjects from time to time, just to make sure...


Just a reminder that this is the [b]American Mathematics Competitions[/b] forum.  Items posted in this forum should be questions about, or problems from, the US AMC/AIME/USAMO contests.

In particular, please don't post problems from other contests and/or "general" problems in this forum.  Instead use the appropriate high school or regional/national forum.

Thanks! :)

I guess the IMO should be considered as final stage of the AMC process.  :wink:

Bumping this to remind people that this is the [b]AMC[/b] forum.  Only topics [b]directly[/b] related to the AMC series of contests (AMC 8/10/12, AIME, USA(J)MO) should be in this forum.

I have recently moved several topics that are unrelated (or only incidentally related) to the AMC.


analytic geometry
geometry
complex numbers

[quote="ben12345_5"]For problem 1, I was exploring the case where the second roll is 5.  I wrote that n^2 is congruent to 0, 1, or 4 (mod5).  So I concluded that the first roll can be 1, 4, and 5, BUT I stupidly forgot to mention that the first roll could also be 6.

I did everything else correctly.  Will points be deducted?

For problem 4, the problem asks us to "find a and b, and prove your answer can result in the conversation above."  In my solution, I started out by stating that a=4 and b=13 without showing how I derived this answer, and then for the proof, I showed that IF a=4 and b=13, THEN the conversation is true.  Is this OK, or will points be deducted?

for problem 5, the math jam proved that AB = 19 and AC = 17.

If I found that AB = 17 and AC = 19, is there any difference?[/quote]
For your last question - probably you made a mistake setting up your diagram, but if that's the only error they probably won't deduct points.

For number 1, a few points will be deducted.

You just labeled your triangle backwards. Depends on the leniency of the grader.  :)

I did the same for number 4. I think it's fine.

Edits: Can't spell

[quote="ben12345_5"]For problem 1, I was exploring the case where the second roll is 5.  I wrote that n^2 is congruent to 0, 1, or 4 (mod5).  So I concluded that the first roll can be 1, 4, and 5, BUT I stupidly forgot to mention that the first roll could also be 6.

I did everything else correctly.  Will points be deducted?

For problem 4, the problem asks us to "find a and b, and prove your answer can result in the conversation above."  In my solution, I started out by stating that a=4 and b=13 without showing how I derived this answer, and then for the proof, I showed that IF a=4 and b=13, THEN the conversation is true.  Is this OK, or will points be deducted?

for problem 5, the math jam proved that AB = 19 and AC = 17.

If I found that AB = 17 and AC = 19, is there any difference?[/quote]

I'm assuming points will be deducted if you missed a roll on 1.

On number 4 i did what you did... I think it should be ok.. that's what it says to prove... just make sure you didnt set the limits that a,b<20 when proving it

For 5, of course there is a difference between AB=19, AC=17 and AB=17, AC=19. That messes the midline up (everything will be reversed. The problem specifically stated that Y was between X and N.

If I were a grader, I would probably deduct 1 point for missing the last congruence, but no more.  Number four seems to be worded specifically to imply writing the proof in the form that you did, so I think you're fine there.  For number five, I personally wouldn't care about the flip, but since you had a month to go over the solution, a grader might take a point off.  In any case, unless we get an actual grader here, this is all speculation.

[quote]
For number 1, a few points will be deducted. [/quote]

Do you think more than 1 point will be deducted for question 1?  :o

[quote="Shogia"]For #3 I forgot to have the y coordinates contain $i$ soI ended up with an area of $8$ instead of $8i$, any ideas on how much of a deduction that would be?[/quote]
For a trivial error like that, I'd say one point, but none if the grader is nice and doesn't fault you for one small slip-up.

Er, the actual answer was 8. You can't have complex numbers for the area. The coordinates of the vertices are real numbers, and so is the area of the quadrilateral.

Oh, well thats nice to know, I thought something was amiss there but i missed the math jam and the transcript hasnt shown up yet so I thought 8i was the right answer thanks for clearing that up.

Yeah, it's $8$. For some reason, a lot of people got $8i$, $10$, or $16$, etc... An easy way to check was simply substitute 1 for $r$! Duh! :roll:

well, in our defense, $8i$ is a different case than the others.  If one got $8i$ he/she didn't mess up the roots, so telling someone to "substitute 1" would've been useless.  The problem was that he/she did not see/know that area must be real.

My turn.  :)

For number 2, I just compared my solution to the one provided on the math jam. My answer was about .08 loops more than his. Will I get a point deducted or will the grader not notice? (The solution is nice)

For number 3,I didn't arrive at the answer, but provided my thoughts and method I might use to solve the problem. If the method was okay, how many points will I probably get?

[quote="chess64"][quote="johnsmith"]For number 3, I simply found the area of the solution by substituting 1 for r, since r can be any nonzero real number. I got that the final area was 8. Do you think I'll be fine with that as my primary method?[/quote]

You didn't prove that it is independent of $r$![/quote]

Yeah I figured that out soon enough, but do you think I'll still get some credit for getting the answer?

Number 5, I don't think points will be deducted from your score for that.


linear algebra
matrix
linear algebra unsolved

Let $ T$ be an arbitrary $ n\times n$ matrix. Are there any linear operators $ f$ satisfying that $ T$ is its standard matrix?

What's your definition of standard matrix, and why isn't $ x \mapsto Tx$ a valid example?

Hi t0rajir0u,

I got the definition of "standard matrix" and "linear operator" from the set of notes [url=http://www.maths.mq.edu.au/~wchen/lnlafolder/lnla.html]Linear Algebra by WWL Chen[/url] (Chapter 8, p1-2).
[quote]A transformation $ T : \mathbb{R}^n \mapsto \mathbb{R}^m$ is called a linear transformation if there exists a real matrix
\[ A \equal{} \begin{pmatrix} a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn}\end{pmatrix}
\]
such that for every $ x \equal{} (x_1, \ldots, x_n) \in\mathbb{R}_n$, we have $ T(x_1, \ldots, x_n) \equal{} (y_1, \ldots, y_m)$, where
\[ \begin{pmatrix}y_1 \\
\vdots \\
y_m\end{pmatrix} \equal{} \begin{pmatrix} a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn}\end{pmatrix}\begin{pmatrix}x_1 \\
\vdots \\
x_n\end{pmatrix}
\]
The matrix $ A$ is called the standard matrix for the linear transformation $ T$.[/quote]
[quote]A linear transformation $ T : \mathbb{R}^n \mapsto \mathbb{R}^m$ is said to be a linear operator if $ n \equal{} m$. In this case, we say that $ T$ is a linear operator on $ \mathbb{R}^n$.[/quote]

In other words, $ x\mapsto Tx$ is the valid example.


Lower Slobovia is a developing country. Its GDP last year was 75 million zugats. Upper Slobovia is a developed country. Its GDP last year was 300 billion zugats. It enjoys an average annual growth of 3%. If Lower Slobovia wants to have the same annual GDP as Upper Slobovia in 50 years, what does its annual rate of growth have to be?

[hide]Lower Slobovia's annual rate of growth must be 5.9%. in 50 years, Upper Slobovia's GDP will be about 1,315 billion zugats. that is about 17.5 times Lower Slobovia's current GDP. so, over the course of 50 years, Lower Slobovia needs an annual rate of increase of about 5.9% to match Upper Slobovia's GDP.[/hide]

no, because it's 75 [b]million[/b] v.s. 300 [b]billion [/b]  ;)

whoa. okay. :wallbash_red:  :wallbash_red:  :wallbash_red: i thought it was 75 billion. yea, then you are correct.


Prove that for each positive integer $ n$, the sum of the numbers of digits of $ 4^n$ and $ 25^n$ (in the decimal system) is odd.

[hide]Using Scientific Notation:
Let $ 4^n \equal{} a_n \times 10^{p_n}$ and let $ 25^n \equal{} b_n \times 10^{q_n}$
where $ 1 < a_n,b_n < 10$ (Since neither $ 4^n$ nor $ 25^n$ is an integer power of $ 10$).

Note that $ 4^n$ has $ p_n\plus{}1$ digits and $ 25^n$ has $ q_n\plus{}1$ digits.

Thus:
$ (4^n)(25^n) \equal{} (a_n \times 10^{p_n})(b_n \times 10^{q_n})$

$ 100^{n} \equal{} a_n b_n \times 10^{p_n\plus{}q_n}$

$ 10^{2n} \equal{} a_n b_n \times 10^{p_n\plus{}q_n}$

So $ a_n b_n$ must be a power of $ 10$.

Since $ 1 < a_n,b_n < 10$, we have $ 1 < a_n b_m < 100$ and thus, $ a_n b_n \equal{} 10$.

Therefore, $ p_n\plus{}q_n \equal{} 2n\minus{}1$, and thus, $ (p_n\plus{}1)\plus{}(q_n\plus{}1) \equal{} 2n\plus{}1$ which is odd as desired. [/hide]


There are no solution ,isn't there ? :(
I think it not easy but not very hard  :)

@Do Tien Dang, please wait for a few days, maybe anybody are solving your problems. :(


For any three positive reals a, b, c, we have

${{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge\frac{a+b+c}{a^2+b^2+c^2}(\frac{a^2+b^2}{\sqrt{a+b}}+\frac{b^2+c^2}{\sqrt{b+c}}}+\frac{c^2+a^2}{\sqrt{c+a}}})$

Oh my dear, this is so canonical... We have to prove the inequality

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

for any three positive reals a, b, c.

Rewrite this inequality in the form

$\sum\sqrt{b+c}-\frac{a+b+c}{a^2+b^2+c^2}\sum\frac{b^2+c^2}{\sqrt{b+c}}\geq 0$.

But

$\sum\sqrt{b+c}-\frac{a+b+c}{a^2+b^2+c^2}\sum\frac{b^2+c^2}{\sqrt{b+c}}=\sum\sqrt{b+c}-\sum\left(\frac{a+b+c}{a^2+b^2+c^2}\cdot\frac{b^2+c^2}{\sqrt{b+c}}\right)=\sum\left(\sqrt{b+c}-\frac{a+b+c}{a^2+b^2+c^2}\cdot\frac{b^2+c^2}{\sqrt{b+c}}\right)$
$=\sum\frac{\sqrt{b+c}\cdot\sqrt{b+c}\cdot\left(a^2+b^2+c^2\right)-\left(a+b+c\right)\cdot\left(b^2+c^2\right)}{\left(a^2+b^2+c^2\right)\cdot\sqrt{b+c}}=\sum\frac{\left(b+c\right)\cdot\left(a^2+b^2+c^2\right)-\left(a+b+c\right)\cdot\left(b^2+c^2\right)}{\left(a^2+b^2+c^2\right)\cdot\sqrt{b+c}}$
$=\sum\frac{ab\left(a-b\right)-ca\left(c-a\right)}{\left(a^2+b^2+c^2\right)\cdot\sqrt{b+c}}$.

Thus, it remains to prove the inequality

$\sum\frac{ab\left(a-b\right)-ca\left(c-a\right)}{\left(a^2+b^2+c^2\right)\cdot\sqrt{b+c}}\geq 0$.

Upon multiplication with $a^2+b^2+c^2$, this inequality becomes

$\sum\frac{ab\left(a-b\right)-ca\left(c-a\right)}{\sqrt{b+c}}\geq 0$.

But

$\sum\frac{ab\left(a-b\right)-ca\left(c-a\right)}{\sqrt{b+c}}=\sum\left(\frac{ab\left(a-b\right)}{\sqrt{b+c}}-\frac{ca\left(c-a\right)}{\sqrt{b+c}}\right)$
$=\sum\frac{ab\left(a-b\right)}{\sqrt{b+c}}-\sum\frac{ca\left(c-a\right)}{\sqrt{b+c}}=\sum\frac{ab\left(a-b\right)}{\sqrt{b+c}}-\sum\frac{ab\left(a-b\right)}{\sqrt{c+a}}$
$=\sum\left(\frac{ab\left(a-b\right)}{\sqrt{b+c}}-\frac{ab\left(a-b\right)}{\sqrt{c+a}}\right)=\sum\left(ab\left(a-b\right)\left(\frac{1}{\sqrt{b+c}}-\frac{1}{\sqrt{c+a}}\right)\right)$.

Thus, the inequality in question is equivalent to

$\sum\left(ab\left(a-b\right)\left(\frac{1}{\sqrt{b+c}}-\frac{1}{\sqrt{c+a}}\right)\right)\geq 0$.

But this is trivial: If we have $a\geq b$, then $b\leq a$, so that $b+c\leq c+a$, and thus $\sqrt{b+c}\leq\sqrt{c+a}$, so that $\frac{1}{\sqrt{b+c}}\geq\frac{1}{\sqrt{c+a}}$, and $\frac{1}{\sqrt{b+c}}-\frac{1}{\sqrt{c+a}}\geq 0$, what, together with $a-b\geq 0$ (this is since $a\geq b$), yields $\left(a-b\right)\left(\frac{1}{\sqrt{b+c}}-\frac{1}{\sqrt{c+a}}\right)\geq 0$. Similarly, in the case when $a\leq b$, we also have $\left(a-b\right)\left(\frac{1}{\sqrt{b+c}}-\frac{1}{\sqrt{c+a}}\right)\geq 0$. Thus, $\left(a-b\right)\left(\frac{1}{\sqrt{b+c}}-\frac{1}{\sqrt{c+a}}\right)\geq 0$ holds in every case, and therefore, $\sum\left(ab\left(a-b\right)\left(\frac{1}{\sqrt{b+c}}-\frac{1}{\sqrt{c+a}}\right)\right)\geq 0$. This completes the proof of our inequality.

  Darij

If (a,b,c)satisfy the inequality then (ka,kb,kc) too for any k>0.

So we can suppose whithout lost of generality : a+b+c=1.

And the inequality becomes:

 :sqrt: (a+b)+ :sqrt: (b+c)+ :sqrt: (a+c) >= [(a 2 +b 2)/ :sqrt: (a+b)+(b 2 +c 2 )/ :sqrt: (b+c)+(a 2 +c 2 )/ :sqrt: (a+c)]/(a 2 +b 2 +c 2 ).

Or  :sqrt: (a+b)-(a 2 +b 2 )/[(a 2 +b 2 +c 2 ) :sqrt: (a+b)]
=[(a+b)(a 2 +b 2 +c 2 )-(a 2 +b 2 )]/(a 2 +b 2 +c 2 ) :sqrt: (a+b)]
=[(1-c)(a 2+b 2 +c 2 )-(a 2 +b 2 ]/(a 2 +b 2 +c 2 ) :sqrt: (a+b)
=[c 2 -c(a 2 +b 2+c 2 )]/( a 2+b 2 +c 2) :sqrt: (a+b);.

Doing he same wth others and suppose that : :beta: =inf(1/ :sqrt: (a+b),1/ :sqrt: (b+c),1/ :sqrt: (a+c))

Then  >= :beta:  [(a 2 +b 2 +c 2 )-(a+b+c)(a 2 +b 2 +c 2)]/(a 2 +b 2 +c 2)

Or a+b+c =1 .

And we done

AYMANE! But what if $c^2-c(a^2+b^2+c^2)<0$, then your final estimate is untrue.

BTW, maybe it is canonical, but performing all these transformations is not an obvious task, Darij!

[quote="AYMANE"]Really i don't undrestand where i did a mistake. I just used :

A  >= a , B >= b, C >= c  then A+B+C >= a+b+c

So is there any conditions to use this inequality :?[/quote]

yes.
Even when $ A \ge B$, $AC \ge BC$ is true only when $C \ge 0$.

No, you use that if $a>b$ then $Aa>Ab$, and it is untrue for $A<0$, $a,b>0$.

THANK YOU I SEE NOW !
 :blush:

 BUT I AM SURE WE CAN FIND SMART PROOF  :D

I WILL TRY


http://www.randomdragoon.com/other/thegame.swf

You'll know when you've figured it out.

Please hide your solutions please.

BTW: The answer is not "right click and press play". I'm too much of a flash noob to know how to disable that without looking through a helpfile which I didn't do.

omg thats so cool
the solution:
[hide]
figure it out yourself.
[hide]
you really should try the puzzle
[hide]
before looking at the solution

if you insist

exercise, man!

do you exercise by watching other people?

didn't think so.

youll hate yourself once you look

i am not responsible for any suicides.

each number is equal to teh previous number.
[/hide][/hide][/hide]


Susan's March birthday is on a Saturday this year, but she doesn't celebrate it until 100 days later. On what day of the year will she celebrate?

$ 100\equiv 2\text{ (mod 7)}$, so 2 days after Saturday is $ \boxed{\text{Monday}}$


geometry
geometric transformation
rotation
Ring Theory
combinatorics unsolved
combinatorics

Let is say that a chess piece is a subset $ A\subset \mathbb{G}$. A chess piece is reduced if $ A$ is symmetric with respect to right angle rotations and $ A\equal{}\{ z,\minus{}z,iz,\minus{}iz\}$ for some $ z\in \mathbb{G}\setminus \{0\}$.

Prove

Let $ A$ be a chess piece which is symmetric with respect to right angle rotations and $ X\subset\mathbb{G}$ be the set of those squares that you can go from origin by finitely many moves. Then there is a reduced chess piece $ B$ which can go from origin to all the squares of $ X$ by finitely many moves and no other squares.

Sorry, I copied the wrong definition: a chess piece is a subset $ A\subset \mathbb{G}\setminus\{0\}$. This follows from the fact that Gaussian integers forms a principal ideal domain.


algebra
polynomial
complex analysis
complex analysis unsolved

Prove that for any points $P_{1},...,P_{n}$ in the complex plane and any $d>0$ one can find some disks with the sum of their radii $2d$ and such that for any point outside any of these disks, the product of distances of this point to all $P_{i}$ is at least $(d/e)^{n}$? Is the last bound optimal?

I remember reading a proof of this result in a book by Borwein and Erdelyi, a while ago. We want to cover the zeros of a monic polynomial $p(z)=\prod_{k=1}^{n}(z-z_{k})$ by disks so that $|p(z)|$ is not too small outside of them. The idea to cover each zero with a disk of radius $2d/n$ is quickly dismissed: it gives no better bound than $(2d/n)^{n}$, as the polynomial $p(z)=\left(\frac{zn}{2d}\right)^{n}-1$ shows. Considering this example, we realize that it's better to use one large disk to cover several zeros when possible. However, with this approach we can no longer be certain that a zero is at the center of the disk that covers it. This forces us to use smaller disks (with total radius $d$) so that we can double them later.

Let $m_{1}$ be the largest integer for which there is a closed disk of radius $m_{1}d/n$ that covers exactly $m_{1}$ zeros of $p$. Let $D_{1}$ denote this disk. Next, let $m_{2}$ be the largest integer for which there is a closed disk of radius $m_{2}d/n$ that covers exactly $m_{2}$ of the zeros not already covered... The process stops when all zeros are covered by disks $D_{1},\dots,D_{k}$. It's clear that the sum of radii of these disks is $d$. It remains to prove that $2D_{1},\dots, 2D_{k}$ work as advertised. Let $z$ be a point outside of all $2D_{k}$. A somewhat tedious case-by-case analysis shows that for $1\le m\le n$ the closed disk of radius $md/n$ centered at $z$ contains at most $m$ zeros of $p$. Hence $|p(z)|\ge n!(d/n)^{n}>(d/e)^{n}$.

I don't believe that $e$ is optimal, but have no idea how to improve the estimate.


If $ \log_{10} (x^2\minus{}3x\plus{}6)\equal{}1$, the value of $ x$ is:
$ \textbf{(A)}\ 10\text{ or }2 \qquad\textbf{(B)}\ 4\text{ or }\minus{}2 \qquad\textbf{(C)}\ 3\text{ or }\minus{}1 \qquad\textbf{(D)}\ 4\text{ or }\minus{}1\\
\textbf{(E)}\ \text{none of these}$

[hide]
Rewrite in exponential form.
$ x^2 \minus{} 3x \plus{} 6 \equal{} 10$
Subtract 10 from both sides.
$ x^2 \minus{} 3x \minus{} 4 \equal{} 0$
Factor.
$ (x \minus{} 4)(x \plus{} 1) \equal{} 0$
We see x = 4, -1. Plugging back in to look for extraneous solutions, we find all the roots are valid.
$ \boxed{\text{D}}$[/hide]


6.A throws three dice in a special game of ludo. If it is known that he needs 15 or more to win then what are his chances of winning?
a.	5/54
b.	17/216
c.	13/216
d.	15/216

i get a solution of 14/216.. there are 14 fvourable events
[5,5,6] -> 3 Arrangements
[6,6,5] -> 3 Arrangements
[5,5,5] -> 1 Arrangement
[6,6,6] -> 1 Arrangement
[4,5,6] -> 6 Arrangements
Total   -> 14 Arrangements

What am i missing here???

7.In a horse race there are 18 horses numbered 1- 18. the probability that horse 1 would win is 1/6, horse 2 is 1/10 and 3 is 1/8. Assuming a tie is impossible find the chance that one of the three would win.
a.	47/120
b.	119/120
c.	11/129
d.	1/5

i worked it out...
[1/6 x 9/10 x 7/8] + [5/6 x 1/10 x 7/8] + [5/6 x 9/10 x 1/8] = 143/420
agaian missing something here.....

6) (4, 6, 6)

7) For this problem, I don't think you multiply one person's probability of winning by the probability of the other two people losing. The probability of that one person winning already requires that the other people lose since there can only be one winner. The probability of one of those three people winning is just the sum of the individual probabilities of each one of those people winning, so 47/120.

for the first i still get 17/216  while the given solution is 5/54...
anything else missing????

more help with the following case...

The probability that a student will pass in maths is 3/5 and that he will pass in English is 1/3. If the probability that he will pass in both is 1/8. what is the probability that he will pass in at least 1 subject?
a.	97/120
b.	87/120
c.	53/120
d.	120/297

here wont 14/15 be the probability that he passes in both the subject??? if not how to solve this one?


On the AMC 10A answer sheet I forgot to write my grade, but I put my age, less than or equal to 12.  Will I still recieve recognition for my performance at the 8th grade and below level?

Thanks,
lyra

I think you'd fall under "Unknown Grade" in this case and not 8th grade.

Is there any way to correct this?

~lyra

I would suggest contacting the AMC directly.


How many ways can a committee consisting of two Republicans and three Democrats be chosen from eight Republicans and six Democrats?

This problem deals with combinatorics. There are eight Republicans to choose two from, and six Democrats to choose three from; thus, we solve an expression with two separate combination cases:
$ \binom{8}{2}\times\binom{6}{3}$

$ \equal{} \frac {8!}{2!(8 \minus{} 2)!}\times\frac {6!}{3!(6 \minus{} 3)!}$

$ \equal{} \frac {8!}{2!6!}\times\frac {6!}{3!3!}$

$ \equal{} \frac {8\times7}{2\times1}\times\frac {6\times5\times4}{3\times2\times1}$

$ \equal{} \frac {56}{2}\times\frac {120}{6}$

$ \equal{} 28\times20$
$ \equal{} \fbox{560}$


AwesomeMath
summer program
email
ratio
blogs

What is the all USA Girls Math Olympiad team ? Who are the members?

so I think sherry is one but like wtf why is this in awesomemath forum

"All USA Girls [snippet] in track 3"
or something like that was in the email we got.

WHAT IS THIS WORLD COMING TO??? Google doesn't return anything...

Does this mean that all '' will go to this camp? I know one of them...That would be a suprise...

I think that the deal was like she just got some random letter: you are on '' team. Congratulations.

At least the boy to girl ratio will not be like 100 to 1...

Also, AMSP can get on the news again.

p.s. Arnav, are you going to amsp or what? or do you just like spamming on aops :P p.p.s. arnav, solve that problem in my blog

[quote="MysticTerminator"]which one
[/quote]

Number 14


[hide="Click to reveal hidden text"]7, and its obvious why if you do it algebraically.

$\frac{3(x+9)-13} {2}-\frac{3x} {2}$

That's all you're basically saying.

Solving, you get $\frac{3x+14} {2}-\frac{3x} {2}$

Subtract them now to get $\frac{14} {2}$ or $\boxed {7}$[/hide]

[quote="easyas3.14159..."]Pick a number. Add 9 to it. Multiply the total by 3. Subtract 13. Divide the total by 2. Subtract 3/2 the original number. What is the answer and why?[/quote]

[hide]first represent the number as x

x+9
3(x+9)
3x+27-13
3/2x+14/2
3/2x+14/2-3/2x
Always equals 14/2, which is 7[/hide]

+edits+
Whoops didn't simplify

[hide]
You always get 7, do the algebra if you really want :sleeping: [/hide]

[hide]$\frac{3(x+9)-13}{2}$ $-\frac{3}{2}$ And this will give you seven.[/hide]

Enclosing the following text in dollar signs or "\ ["  "\ ]" with the spaces and quotes removed

 \frac {a}{b}

gives
\[ \frac{a}{b}  \]


It is relatively simple to prove the squeeze theorem on the reals, using the usual metric. My question is, can you prove the squeeze theorem on R for an arbitrary metric (on R)? Does this even hold for an arbitrary metric on R? It seems to me that part way through the proof, you would need to show that x<=y<=z implies both that d(x,y)<=d(x,z) and d(y,z)<=d(x,z), where d is the metric. I'm not sure wether or not this is true for an arbitrary metric because I've had little experience with metrics on R other than the usual one. Is there a way to prove this/ is it even true?

Should this be moved to the calculus section?


geometry
3D geometry
sphere
vector
combinatorics proposed
combinatorics

prove that for any broken and closed curve ( which is composed of just lines)in the space if we add all changing direction we get atleast $2\pi$ note that if our curve could be placed in a plane then the result is $2\pi$because the sum of external angel is $2\pi$ (i hope u understand what I mean! :maybe: )

maybe I can't imply what I mean otherwise I'm sure that there were so many problem-solver who devote their time to this interesting problem :D

No sam, I think we're not getting what you mean.

You mean the absolute values of the change everytime? If not, that is a lemniscate-figurine a figure where it's 0?

If you do, it seems quite logical to me. Consider the plane perpendicular to the starting direction. To get back to that plane you need to make an angle of at least $\frac\pi2 + x$ so then you go through it at an angle of $\frac\pi2 - x$ with the axis parallel to your starting direction.

So to get back to the place where you started, you need to make an angle of $\frac\pi2 + (\pi - x)$ which makes a total of $2^\pi$ minimally? :?

Another approach could be to project it on all possible planes. If it's not solvable as 1 planar figure, the given statement is true for all possible planes to project it on :)

If this is all not it you'll have to explain us this interesting problem :P

when u want to calculate the sum of the external angel of poligon in the plane what do we do? :D but in this problem our polygon doesn't lie on the certain plane now we evaluate the sum of the  external angel of this polygon( in othe word :the least dimention that this polygon can lie on it is 3).

This is an adaptation of a problem I found in Klingenberg's "A Course in Differential Geometry". I will use [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=curve&t=5060]this problem[/url], which states that a curve on a unit sphere which is not contained in any open hemisphere must have length $\ge 2\pi$.

Orient the polygonal curve $\mathcal C$ in space, and consider the vectors representing its sides, $v_1,\ldots,v_n$ (in this order). For each of these vectors $v_i$ consider the ray $r_i$ starting from $O$ along it, and let $X_i$ be the intersection between $r_i$ and the unit sphere $\mathbb S^2$ centered at the origin. Now let $c$ be the curve obtained by joining $X_i,X_{i+1}$ with a geodesic (the smallest of the two great arcs which connect them) on the unit sphere for every $i$ ($X_n$ is joined to $X_1$).

Notice that the length of $c$ is precisely the sum of the external angles of $\mathcal C$, so if we prove that  $c$ has length $\ge 2\pi$ we're done. In turn, this will follow from the result cited in the first paragraph if we show that $c$ cannot be contained in an open hemisphere of $\mathbb S^2$. We can say more, actually, namely that the points $X_i$ cannot be contained in an open hemisphere. Otherwise the vectors $v_i$ would all be contained in the same open half-space, but this is impossible, since their sum is zero (they represent the sides of an oriented closed polygonal curve).

Grobber, I admire your wide-known about........ our forum. As far as I know, it would be very surprise if there existed a problem in our forum you could solve without others, without some words like "[b]according to this, because of this etc.[/b]" :D :)


Find last two digits of the number
\[ 1!\plus{}2!\plus{}3!\plus{}...\plus{}2007!\plus{}2008!\]

Yeah, here's an explanation of this - in order to find 2 last digits of the whole number we may take into account only 9 first members of the sum instead of 2008 since the other 1999 members don't affect the 2 last digits(they have zeros on these positions). Hence, 1!+2!+3!+4!+5!+6!+7!+8!+9!, the units digit is equal to  (1!+2!+3!+4!)mod10=3(also we should carry 1 to the tens digit). Now, the tens position can be obtained by calculating 5!=120, x=2 -> x+(x=(x*6))mod10+(x=(x*7))mod10+(x=(x*8))mod10+
+(x=(x*9))mod10+1=1. So, the answer is just as the same as apollo's. :D


Let $ x, y, z$ be nonnegative reals such that $ x \plus{} y \plus{} z\geq 5$. Prove that at least two of the following three inequalities

$ 2x \plus{} 3y \plus{} 6z\geq 14 , 2y \plus{} 3z \plus{} 6x\geq 14,2z \plus{} 3x \plus{} 6y\geq 14$

are true.

Suppose at most one is true, so at least two are false. By cyclically permuting variables, WLOG have
\[ 2x\plus{}3y\plus{}6z<14,2y\plus{}3z\plus{}6x<14\]
Now $ x\plus{}y\plus{}z\geq 5$ and the first inequality implies
\[ 5(2x\plus{}3y\plus{}6z)<14(x\plus{}y\plus{}z)\Rightarrow y\plus{}16z<4x\]
Similarly, $ z\plus{}16x<4y$. So we get
\[ 4y\leq 4y\plus{}64z<16x\leq z\plus{}16x<4y\]
a contradiction.

Nice solution, scorpius119. My proof is based on

$ 4(2x \plus{} 3y \plus{} 6z) \plus{} (2y \plus{} 3z \plus{} 6x) \minus{} 14(x \plus{} y \plus{} z) \equal{} 13z\geq0$.

Similar problem see http://www.mathlinks.ro/Forum/viewtopic.php?t=139

Another problem is

Let $ x, y, z$ be nonnegative reals such that $ yz \plus{} zx \plus{} xy\geq 2xyz \plus{} 1$. Prove that at least two of the following three inequalities

$ 2x \plus{} 3y \plus{} 6z\geq 4\sqrt {3} , 2y \plus{} 3z \plus{} 6x\geq 4\sqrt {3},2z \plus{} 3x \plus{} 6y\geq 4\sqrt {3}$

are true.


search
number theory proposed
number theory

Prove that a necessary and sufficient for the existence of a set $ S \subset \{1,2,...,n \}$ with the property  that the integers $ 0,1,...,n\minus{}1$ all have an odd number of representations in the form $ x\minus{}y, x,y \in S$, is that $ (2n\minus{}1)$ has a multiple of the form $ 2.4^k\minus{}1$

[i]L. Lovasz, J. Pelikan[/i]

harazi posted it [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=15382730&t=111860]here[/url] in a different form.


trigonometry
LaTeX
geometry unsolved
geometry

we have x,y<60° and x+y<90°
prove that:

1/sin(60°-x)+1/sin(60°-y)>=(4sinx+4siny)/[sin(60°+x)*sin(60°+y)]

i would appreciate if someone switched it to LaTeX

$\frac 1{sin(60 ^{\circ}-x)}+\frac 1{sin(60 ^{\circ}-y)}= \frac{4 sin (x)+4 sin (y)}{sin (60 ^{\circ}+x)*sin (60 ^{\circ}+y)}$

I can't help you with the demonstration...


geometry
geometric transformation
superior algebra
superior algebra theorems

In one of my college books, I found this paragraph, which keeps confusing me... Maybe someone could help me to understand it in the right way:

[quote]Let $ (X,\leq )$ be an ordered (partially ordered) set and $ A$ nonempty subset of $ X$. Element $ m$ from $ X$ is a minorant of $ A$ in $ (X,\geq )$ if for each $ a$ form $ A$ inequality $ m \leq a$ holds. At most one of the minorants is an element of $ A$. When such minorant exists, it is called the smallest element of A.
[u]The smallest element defined in such a manner should not be in general confused with the minimal (or first) element $ a_{0}$ which is defined with the formula:

*for each $ a$ from $ A$, $ a\leq a_{0}$ implies $ a \equal{} a_{0}$. *

A set can have one or more minimal elements without having a smallest element.
If the set has the smallest element, than it is, obviously, the only minimal element of that set.[/u][/quote]

Now, what confuses me here? The underlined part. Can someone give me an example for the case of a set having more minimal elements and no smallest element?

P. S. I hope my translation of this paragraph isn't too bad, and that someone will help me... thanks.

EDIT: After some more reading, I realized that the example I ask for is some kind of partially ordered set (which can obviously have more minimal elemnets, and neither one of them can be treated as the smallest because of the partial ordering)... I think I finally understand it, but it would be useful to hear an opinion from someone more experieced in the sets theory...

Sure. Consider the empty partial order on a set $ X$; $ a\le b$ only if $ a\equal{}b$. Every element is minimal, and there is no least element.


In the fraction (29  28  27  ...  16)/(15  14  ...  2), parenthese may be repeatedly placed anywhere on the numerator, granted they are also placed on the identical locations in the denominator.

a) Find the least possible integral value of the resulting expression.
b) Find all possible integral values of the resulting expression.

This is from somebody's 1999 olympiad -- was it the Czech-Slovak one?

BUMP.

I don't like this question at all.  The minimum is at least 1292646 = 2*3*17*19*23*29.

Rebump.  Do you have a solution for this, Stephen?

Yah.

[hide]

This fraction can be written as a ratio A/B of two integers A,B where AB = 29!.

Now we do the big factor into primes thing, 29! = 2^25 * 3^13 * 5^6 * 7^4 * 11^2 * 13^2 * 17 * 19 * 23 * 29.

Any primes with an odd exponent can't be cancelled from A/B. So no integer value can be less than 2*3*17*19*23*29 = 1292646.

On the other hand..

29/((...((28/27)/26)/.../17)/16)

all over

15/((...((14/13)/12)/.../3)/2)

= (29 * 14)/(15.28) * (27 * 26 * ... * 16)/(13 * 12 * .. * 2)

= ((29 * 14^2) / 28) * 27!/(15!^2)

= 29 * 7 * blah, where blah we work out all the exponents again and cancel, and we get 1292646. hm. but anyway, to continue:

Second part is easier. In each of the 14 pairs of numbers {29,15},{28,14],...{16,2}, one is a factor in A and the other is a factor in B. So the resulting value can be written as (29/15)^e1 * (28/14)^e2... *(16/2)^e14

where each ei equals :pm: 1 and e1 = 1 and e2 = -1.

Because 27/13, 26/12, ..., 16/2 are greater than 1, the resulting value must be <= (29/15) * (14/28) * (27/13)... (16/2), which is 1292646. So in fact thats the only integer value possible.

[/hide]


Is there any more or less "canonical" way in which two uniform r.v. may be correlated?

For example: For two normal r.v. we have jointly normal distribution and other "non canonical" ways for correlation.

there is a huge literature on [url=http://en.wikipedia.org/wiki/Copula_(statistics)]copulae[/url], but this is nothing else than specifying the joint law. But I guess that there are some nice and well-studied parametric families of joint distribution.


real analysis
vector
geometry
real analysis unsolved

Prove that if $B$ is a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$, then $B$ is not a Borel set.
Can $B$ be a Lebesgue measurable set ?

I'll think about the first one. I remember reading somewhere that it appeared in a paper of Sierpinski. What's the book you mentioned in the source, alekk?

As for the second one, the answer is 'yes'. Take any set $A\subset\mathbb R$ of null measure and such that $A$ generates $\mathbb R$ as a vector space over $\mathbb Q$. For instance, $A$ could be a union of translates of Cantor's middle-third set by integers (i.e. take a copy of Cantor's set in each of the intervals $[n,n+1],\ n\in\mathbb Z$; their union is $A$). It's well-known that Cantor's set $C$ satisfies $C+C=[0,2]$, so $A$ satisfies $A+A=\mathbb R$, and it will do for our purposes. Now, from the set of generators $A$ extract a basis of $\mathbb R$ over $\mathbb Q$. This basis has null measure, and is thus measurable.

my book is a french one about measure where it is said that it can be proved that blablabla ..

Assume $B$ is a Hamel basis that is a Borel set. Then

$E=\mathbb Q\cdot B=\bigcup_{q\in\mathbb Q}q\cdot B$
is Borel as well.

Let $E_n=\{a_1+\dots+a_n| a_1, \dots, a_n\in E\; (\text{distinct})\}\Longrightarrow \mathbb R=\bigcup_{n=1}^{\infty}E_n$.

But each $E_n$ is Lebesgue measurable. If $\mu(E_k)>0$ for some $k$, then $E_k-E_k$ contains some open interval and that leads to a vanishing linear combination of elemens in $E$.

Therefore $\mu(E_k)=0$ for all $k\geq 1$ and hence $\mu(\mathbb R)=0$, contradiction.

Are you positive that if $E$ is Borel, the $E_n$'s have to be measurable? I'm not convinced. I remember fedja posting a problem where he asked for an example of a pair of measurable sets $A,B\subset\mathbb R$ such that $A+B$ is not measurable (please don't give away the example, and don't give any hints, fedja, if you're reading this :)).

The continuous image of a Borel (or, more generally, Suslin) set is a Suslin set again and, therefore, it is Lebesgue measurable. So [b]cimabue[/b]'s argument is fine.

I realize that now. In fact, cimabue said the exact same thing earlier, but he deleted his message, for some reason :?. Anyway, thank you (both of you :)) for the explanation. My stupid question was due to my lack of knowledge in the area :).


the p.m.f. for a pair of discrete random variables X and Y are given in the following table:

Y\X 0... 1.... 2... 3... 4
1.. .050 .00 .050 .10 .15
2.. .100 .05 .000 .05 .10
3.. .150 .10 .050 .00 .05

1. Determine the marginal p.m.f. for each of the random variables X and Y. Determine the mean and standard deviation of each of them.

2 Determine the conditional p.m.f. for the random variable Y given that the r.v. X has the value 4. Determine mean[sub] YlX=4 and variance [sub] YlX=4.

My answers:
1.p[sub]X (0)= P[X=1] + P[X=2] + P[X=3]
=.05 + .10 + .15 = .30

etc. for other values of X and Y

b)for the mean I took the sum of X={0,1,2,3,4} multiplied by its decimal value.
ex. p[sub]X (0) = 1x.05 + 2x.10 + 3x.15 = 0.70
I'm not sure that this is what I'm supposed to do for the mean???

c) as for the variance I have no clue... we have so many different formulas and I'm not sure if any of them even apply to this question?

2. I have been doing the nearly the same thing again struggling with the variance an mean portion.

For the same question above I am asked the covariance between two random variables:

Cov[X,Y] = E[(X-mean[sub]X) (Y- mean[sub]Y)
or = E[XY] - mean[sub]X * mean[sub]Y

and the correlation

= Cov[X,Y] / (stand. dev. X * stand. dev. Y)

for the covariance I determined that the mean for both X and Y is 2 by finding the sum of the values of X or Y times their proportion in relation to other values of X or Y. However, I don't know where to go with the E[XY]???


you shouldn't be posting this in the MC forum

There are no logs in mathcounts :(

All you really need to know is that if $\log_b{x}=y$ then x=b^y$.  You can derive all the properties from that.

\[\begin{eqnarray*}
\log_b{x}+\log_b{y} &=& \log_b{xy}\\
\log_b{x}-\log_b{y}  &=&\log_b{\frac xy}\\
\log_b{x^c} &=& c\log_b{x}\\
\log_{b^c}{x} &=& \frac 1c \log_b{x}\\
\log_b{x}\log_c{y} &=& \log_b{y}\log_c{x}\\
\log_{b^c}{x^c} &=& \log_b{x}\\
\end{eqnarray*}\]

And finally, the one that I think is most important (and which makes a lot of the above identites obvious) is the change of base property:

\[\log_b{x}=\frac{\log_c{x}}{\log_c{b}}.\]

You only need to memorize the first 3 and the change of base.  But become familiar with deriving the others.

[b]

If f(x*y)=f(x)+f(y), f(x) is a log function.[/b]

Thats assuming you know f is continuous, etc.

Hmm, there were discussions about this in the intermediate forums.

I'm not too sure this is appropriate for getting started.

You see log properties not being used in competitions like AMC 8, or AMC 10, which getting started is more geared towards.

Moved.


I am at my wits end. I cannot seem to explain properly how to subtract a negative number from a positive number to my wife. I understand very well how to do this, yet my wife insists that what I am telling her makes no sense so I'm hoping someone can point me in another direction that will help me explain it to my wife. give the following problem: 43-60 we know that the answer would be -17. My wife sees the answer as -23. She insists that you subtract 0 from 3 and then 6 from 4. I have tried to explain to her that you just cannot do it that way, yet I cannot explain it so that she believes me. My wife is not stupid, she is stubborn, and she is very frustrated that she cannot understand this. She's not saying that I am wrong, but she needs to understand why I am right. Is there anyone out here that can help me get this across to her? Some type of tutorial or web program etc...? This will be the last time I allow her to take a math class online instead of face to face. I believe that would help elevate most, if not all, of her problems.

I understand how to do it... I don't understand how to explain it so that she may understand. Looking for the "why" not "how" please.

A useful method would be to reverse the numbers (Carcul's way). $ 43\minus{}60 \equal{} 60\minus{}43$. Now, she can perform subtraction. However, this answer is positive, and does not make sense in the original problem. Therefore, you would take the negative answer.

$ 60\minus{}43 \equal{} 17 \equal{} \minus{}17$

When you do the subtraction method, you subtract 0 from 3 and get 3. Then you subtract 6 from 4 and get -2. The thing is that the negative sign is only on the 2, not the 3, so the answer is $ \minus{}20\plus{}3$ rather than $ \minus{}(20\plus{}3)$.

There we go... that's helped A LOT! Thank you very much for saving me hours more of headache explaining.


I understand the first part of the solution given on kalva... but I don't get this part:
The 12 numbers which are multiples of 81 reduce to multiples of 3. There are no numbers which are multiples of 372, so we cannot get numerators which are multiples of 37. Thus 648 + 12 = 660 in total.

http://www.kalva.demon.co.uk/aime/aime92.html

$999=3^3\cdot 37$.  Take, for example, $\frac{81}{999}$.  This reduces to $\frac{3}{37}$.  Therefore, this number is in lowest terms.  However, this number has not been counted in the $648$ because we counted those numerators that were not multiples of $3$ or $37$.  Yet this numerator is required and begins with a multiple of $3$.  Therefore, we must count all of these multiples of $81$, because they will reduce to a multiple of $3$ (since $\frac{81}{3^3}=3$.)  There are $12$ of these numbers so we must add them to our total.

For something like this to happen to $37$ it would have to be like $\frac{37^2}{999}$.  However, $37^2>999$, so we don't have to consider this.

I remember this problem well because I fell for the multiples of $81$ trick while doing it.  :D


Let $ a,b,c,d$ be positive real numbers. Prove that

\[ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+d)}+\frac{1}{d(1+a)}\ge \frac{16}{1+8\sqrt{abcd}}.\]

This nice inequality also has a very very nice solution.

The equality holds for (e.g)  $ a=b=c=d=\frac{1}{2}$

Put $ a: =\frac{kx}{y}$ and $ b: =\frac{kz}{x}$ we need to show that :
$ \sum \frac{a}{d+kb}\geq\ \frac{16k}{8k^{2}+1}$
Apply Cauchy-Schwarz we have :
$ \sum \frac{a}{d+kb}\geq\ \frac{(a+b+c+d)^{2}}{\sum a(d+kb)}\geq\ \frac{16k}{8k^{2}+1}$
The last inequality equivalent to :
$ 8k^{2}[(a+c)^{2}+(b+d)^{2}]-16k(a+c)(b+d)+(a+b+c+d)^{2}\geq\ 0$
Put : $ x=a+c$ and $ y=b+d$ , apply AM-GM we have :
$ (8k^{2}+1)(x^{2}+y^{2}) \geq\ 2(8k^{2}+1)xy \geq\ (16k-2)xy$
Done !

Very nice inequality, Hungkhtn! :lol:
And very nice solution, Tanpham90! :)
But I think the substitution is $ k=\sqrt[4]{abcd}$ and $ a=k\frac y{x}$, $ a=k\frac z{y}$, $ a=k\frac t{z}$, $ a=k\frac x{t}$.
Another solution is by rearrangement method.

Yes. My solution is quite the same as VASC idea and tanpham's solution, but I made it more clearly

[quote="Vasc"]It is not necessary to consider two cases. We have to show that
$ \frac{(A+B)^{2}}{AB+\frac k{2}(A^{2}+B^{2})}\ge \frac{16k}{1+8k^{2}}$,
$ A^{2}+B^{2}+2(8k^{2}-8k+1)AB \ge 0$,
which holds if
$ 2AB+2(8k^{2}-8k+1)AB \ge 0$.[/quote]

Yes, thanks your comment, VASC. Now I realize that this solution can be reduced as you say.


inequalities
integration
function
calculus
probability and stats

Assume that $ S_t$ follows the next Ito's process
$ dS_t\equal{}\theta(\mu\minus{}S_t)dt\plus{}\sigma dW_t$ - well-knwon Ornstein-Uhlenbeck model.
$ \alpha_t\equal{}(\minus{}(r\plus{}\theta)S_t\plus{}\theta \mu)/\sigma$
How can  I check if the following inequality holds: $ \mathbb{E}[\exp[0.5 \int \limits_0^T \alpha_t^2 dt]]<\infty$
I hope that someone has ever seen the enequalities like that. :wink:

Tarasenya

Well I don't want to make these tedious calculations but you can try to get the result following these steps :

First : Integrate the SDE of $ S_t$ this is classical using Itô with $ f(S_t)\equal{}e^{\theta t}S_t$

Second : Using Itô again express $ \alpha_t^2$ without term looking like $ (\int_0^{t}something.dW_s)^2$

Third : Use stochastic Fubini to get something of the form $ \exp[0.5 \int \limits_0^T \alpha_t^2 dt]\equal{}e^{\int_0^T f(s)ds\plus{}\int_0^T g(s)dW_s }$

Then try to finish the job

It might happen than even before you get to step 3 you could be able to get your bound

Of course there is no garantee and this method could fail

PS: With this kind of change in measure what do you want to do ?

see ya

ok, I'll try.
first step was done long time ago(our function $ f(t,x) \equal{} e^{\theta t}x$) and using Ito'  lemma it's easy to get the result we need.
(second step looks like Ito -isometry) , but I stopped before the 3 step.
Thanks a lot.
 PS This transformation appears in one problem of stochastic finance. :ninja:
 About "tedious calculations"-  there are only a bit students that can avoid problems like this and go in for sophisticated math with magnificent ideas.

Well if you go to the end of the steps

I think you may gain a view about the law of $ e^{1/2\int_0^T\alpha_s^2ds}$ (if all steps work of course)

Nevertheless I'm glad I could be of any help.

The hardest step is calculating the following thing( I've been trying to do it for a week, but I didn't managed to)$ \mathbb{E} exp( \int \limits^{t}_0 \int \limits^{s}_0 e^{u \minus{} s} W_u dudW_s)$
Maybe I don't know some standard theorems that can help here.

The form of your stochastic integral make me think of iterated stochasic integrals
Maybe you should look at a decomposition of this integral in terms of  Itô-Wiener Chaoses

But I don't know if this works it's only an idea


trigonometry
geometry
rectangle
geometric transformation
geometry open

Prove  papus theorem logicaly not  intuiton using pascal theorem

[quote="K81o7"]It also follows from five applications of Menelaus.[/quote]

recantly i proved that:

EACH PROBLEM WHICH CAN BE SOLVED WITH MENELAUS,CAN ALSO BE SOLVED WITH SPIRAL SIMILARITY

then i claim that papus theorem will be solved with spiral similarity too :wink:

as you all know it has a solution using harmonic division too

We have:
$ (A,R,D,B')\equal{}A'(A,R,D,B')\equal{}(A,B,C,O)\equal{}C'(A,B,C,O)\equal{}(E,P,C,B')$
And $ AE,DC$ pass through $ Q$ then $ RP$ pass through $ Q$.
It mean $ P,Q,R$ are collinear!

[img]http://i225.photobucket.com/albums/dd18/nktp/pappus.png[/img]

Projection  :D
Just see nktp's figure as a 3D rectangle and the vanishing point is O.


Find all positive integers $ n$ having exactly $ 16$ divisors $ 1\equal{}d_1<d_2<...<d_{16}\equal{}n$ such that $ d_6\equal{}18$ and $ d_9\minus{}d_8\equal{}17.$

[quote="moldovan"]Find all positive integers $ n$ having exactly $ 16$ divisors $ 1 \equal{} d_1 < d_2 < ... < d_{16} \equal{} n$ such that $ d_6 \equal{} 18$ and $ d_9 \minus{} d_8 \equal{} 17.$[/quote]

[hide="My solution"]
If $ n \equal{} \prod p_i^{n_i}$, the number of divisors of $ n$ is $ \prod(n_i \plus{} 1) \equal{} 16$, so $ 5$ possibilities :

$ 16 \equal{} 2\times 2\times 2\times 2$ and so $ n \equal{} p_1p_2p_3p_4$
$ 16 \equal{} 2\times 2\times 4$ and so $ n \equal{} p_1p_2p_3^3$
$ 16 \equal{} 4\times 4$ and so $ n \equal{} p_1^3p_2^3$
$ 16 \equal{} 2\times 8$ and so $ n \equal{} p_1p_2^7$
$ 16 \equal{} 16$ and so $ n \equal{} p_1^{15}$

Since $ d_6 \equal{} 18 \equal{} 2\times 3^2$, we can eliminate the first and the last and it remains three possibilities :

1) $ n \equal{} 2\times 3^3\times p$ with prime $ p > 3$
Then the 16 divisors are :
$ 1,2,3,6,9,18,27,54$ and $ p,2p,3p,6p,9p,18p,27p,54p$

Since $ 18$ is the sixth in the first list, we get $ p > 18$ and 3 possibilities :

1.1) $ 18 < p < 27$. In this case, $ d_8 \equal{} 27$ and $ d_9 \equal{} 2p$ and so $ 2p \minus{} 27 \equal{} 17$ and $ p \equal{} 22$, wrong since not prime

1.2) $ 27 < p < 54$. In this case, $ d_8 \equal{} p$ and $ d_9 \equal{} 54$ and so $ 54 \minus{} p \equal{} 17$ and $ p \equal{} 37$, OK and so $ n \equal{} 2\times 3^3\times 37 \equal{} 1998$
The divisor list is then : $ 1,2,3,6,9,18,27,37,54$ $ ,74,111,222,333,666,999,1998$

1.3) $ 54 < p$. In this case, $ d_8 \equal{} 54$ and $ d_9 \equal{} p$ and so $ p \minus{} 54 \equal{} 17$ and $ p \equal{} 71$, OK and so $ n \equal{} 2\times 3^3\times 71 \equal{} 3834$
The divisor list is then : $ 1,2,3,6,9,18,27,54$ $ ,71,142,213,426,639,1278,1917,3834$

2) $ n \equal{} 2^33^3$
The divisor list is then $ 1,2,3,4,6,8,9,12,$ $ 18,24,27,36,54,72,108,216$ and then $ d_6 \equal{} 8$, so wrong

3) $ n \equal{} 2\times 3^7$.
The divisor list is then $ 1,2,3,6,9,18,27,54,$ $ 81,162,243,486,729,1558,2187,4374$
Then $ d_6 \equal{} 18$ but $ d_9 \minus{} d_8 \equal{} 81 \minus{} 54 \equal{} 27$ so wrong too

So the only solutions are $ \boxed{n\in\{1998,3834\}}$

[/hide]

Nice problem :)

Firstly, the condition that $d_6=18$ gives that $d_2=2$ and $d_3=3$. While considering the fact that $n$ has 16 divisors and $d_6=18$, we must have $n=2.3^7$ or $n=2.3^3.p$ where $p$ is some prime.Its easy to note that the first condition fails to satisfy $d_9-d_8 = 17$. Hence we deal only with the $n=2.3^3.p$.

The first $6$ divisors of $n=2.3^3.p$ are $1,2,3,6,9,18$. It's obvious that all primes $p$ less than 18 fails. Indeed, checking primes greater than $19$ and less than $30$ fails, this forces $d_7 = 27$. Now, due to size reasons, $d_8$ or $d_9$ could be $54$.

Case 1 : $d_8 = 54$. Thus, $d_9 = 17+d_8 = 71$, which is a prime. Thus, $p = 71$ and $n=2.3^3.71=3834$
Case 2 : $d_9 = 54$. Thus, $d_8 = d_9-17 = 37$, which is a prime too. Thus, $p = 37$ and $n=2.3^3.37=1998$

Hence the required numbers are $n = 3834,1998$

@above,how did you get $d_2=2$ and $d_3=3??$
Please enlighten me.Thank You!

Clearly, $d$ must be of the form $p^{15}$, $pq^7$, $p^3q^3$, $pqr^3$ or $pqrs$. However, as $d_6=18=2\cdot 3^2$ is a divisor of $d$, $p^{15}$ and $pqrs$ are impossible. Also, it is quick to check that $d_6=2\cdot 3^2$ readily determines $d_1=1,d_2=2,d_3=3,d_4=6,d_5=8,d_6=18$, since if $2$ appears with a power bigger than one, $4$ is a divisor smaller than $18$, and thus $d_6<18$. Hence $p^3q^3$ is impossible as well. Now if $d=2\cdot 3^7$, we quickly get that $d_9=81$ and $d_8=54$, but $d_9-d_8=27\neq 17$.

Hence the only remaining possibility is $pqr^3=2\cdot 3^3\cdot q$, where $q>18$.

   - If $d_7=q$, then we must have $18<q<27$, so $q=19,23$. If $q=19$, then $d_8=27$ and $q_9=38$, so $q_9-q_8=11\neq 17$. If $q=23$, then
     $d_8=27$ and $d_9=46$, so $d_9-d_8=19\neq 17$. Hence $d_7=27$.
   - If $d_8=q$, then $27<q<54$. However, since $2q>54$, we must have $d_9=54$. Hence $d_8=d_9-17=37$, which is indeed prime, so $q=37$
     gives a valid solution.
   - If $d_9=q$, then $d_8=54$, so $q=54+17=71$, which is also prime and gives a valid solution.

Note that $d_i=q$ for $i>9$ cannot happen, as $d_8=54$ already contains $2\cdot 3^3$, so there must be another prime not equal to $q$ dividing $d$, impossible. Hence all the answers are $2\cdot 3^3\cdot 37$ and $2\cdot 3^3\cdot 71$, which clearly work.


Hi,
I'm now ninth grade in Spain.I'm learning German and I want to study math in Germany.So can someone from Germany or someone who is studing in Germay tell which are the best universities to study math.And how many foreigner they take(in %).
Thans you.

uni bonn  $\Rightarrow$ hausdorff institut for pure math

heidelberg,berlin,lmu for applied

Ask Darij Grinberg, or another German AoPSer?

Also I have no clue because I don't live/have never lived in Germany, but isn't Gottingen really good?

[quote="yugrey"]Ask Darij Grinberg?

 isn't Gottingen really good?[/quote]

yes but not as good as the ones mentioned


I was wondering if you use one of them first in the subtracting formula to get m? Because I need to know...
*finds answer* but NOT in textbook..

Given two solutions, $(x_{1},y_{1})$ and $(x_{2},y_{2})$, first find the slope. \[m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\] Then, plug m in for one of the points (yes, I guess you're right) to find b.

EX: Find the line going through (2,3) and (4,7). \[m=\frac{7-3}{4-2}=2\]  \[y=2x+b \Rightarrow 3=2(2)+b \Rightarrow b=-1\] so the equation of the line is 2x-1.


Can anyone give me a specific link to download"MATHEMATICAL CIRCLE - A RUSSIAN EXPERIENCE" pls pls pls ? :huh:  :blush: (OBVIOUSLY FOR FREE!! :D )

[quote="sayak38kolay"]I have a copy of the e-book (in djvu format)....
i can send it to you...
Just give me your e-mail address....   :)[/quote]
 Can you send me this book on adress

krenkovr@yahoo.com


[quote="Yuriy Solovyov"]For positive $ a, b, c, d$ prove the inequality
$ \sqrt {\frac {ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd}{6}}\geq \sqrt [3]{\frac {abc \plus{} abd \plus{} acd \plus{} bcd}{4}}$[/quote]
It's a private case of the Maclaren's inequality. See  my proof ( M565 ) here:
http://kvant.mccme.ru/1980/04/resheniya_zadachnika_kvanta_ma.htm

[quote=Yuriy Solovyov]For positive $ a, b, c, d$ prove the inequality
$ \sqrt {\frac {ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd}{6}}\geq \sqrt [3]{\frac {abc \plus{} abd \plus{} acd \plus{} bcd}{4}}$[/quote]
[url=https://artofproblemsolving.com/community/c6h1062166p9659137][b]Germany 1970[/b][/url]

[quote=Takeya.O][quote=sqing]For positive $ a, b, c, d$ prove the inequality
$$\frac{2}{3}(ab+ac+ad+bc+bd+cd)\geq \sqrt{(abc+bcd+cda+dab)(a+b+c+d)}$$
(Vietnam， Math Forum)[/quote]

Squaring both sides, we get

$\frac{4}{9}(ab+ac+ad+bc+bd+cd)^2\geq (abc+bcd+cda+dab)(a+b+c+d)$

This holds by AM-GM inequality :D[/quote]
Thanks.

Find the range of value of real number $k$ such that the inequality $$ abc+bcd+cda+dab-kabcd\leq\frac{4-k}{6}(ab+ac+ad+bc+bd+cd)$$ holds for any nonnegative number $a,b,c,d$ satisfying $a+b+c+d=4.$
Old.
Where?


A square and an equilateral triangle have the same perimeter. Let $ A$ be the area of the circle circumscribed about the square and $ B$ be the area of the circle circumscribed about the triangle. Find $ A/B$.

$ \textbf{(A)}\ \frac{9}{16} \qquad
\textbf{(B)}\ \frac{3}{4} \qquad
\textbf{(C)}\ \frac{27}{32} \qquad
\textbf{(D)}\ \frac{3\sqrt{6}}{8} \qquad
\textbf{(E)}\ 1$

For the square, it's easy.  Say it has some side length $ a$.  Then the radius of the circle circumscribed around it is $ \frac{a\sqrt{2}}{2}$ so you can now find $ A$.  For the triangle, I think there is some formula for the area of such a circle but I don't know it off the top of my head.

[hide]Ok the easiest way i think to solve this is too assign a value for the perimeter of both, well say 12 to keep the number pretty.  first we know that the the diameter of the circle that is circumscribe about the square is just $ 3\sqrt {2}$ so the radius of this is $ \frac {3\sqrt {2}}{2}$ squaring and multiplying by $ \pi$ yeilds $ \frac {18\pi}{4}$ and the area of a triangle we know to be $ \frac {AB \times AC \times BC}{4R}$ where A,B, and C are verticies of the triangle and R is the radius.  in our equalateral triangle we have all sides equal to 4 so our are is $ \frac {16}{R}$ and because it is equalateral we know also that the area is ((s^2)\sqrt{2})/(4) = $ 4\sqrt {3}$ this is equal to $ \frac {16}{R}$ so cross multiplying yeilds $ R(4\sqrt {3}) \equal{} 16$ dividing by $ 4\sqrt {3}$ we get $ R \equal{} \frac {4\sqrt {3}}{3}$ squaring and mulitplying by $ \pi$ yeilds that the area of the circle is $ \frac{16\pi}{3}$ so we have $ \frac {A}{B}$ = ((18pi)/(4))\((16pi)/(3)) whcih gives us $ \frac {18\pi}{4} \times \frac {3}{16\pi}$ this gives us $ \frac {9}{4} \times \frac {3}{8}$ = $ \frac {27}{32} \equal{} \boxed{C}$[/hide]

Let x be a side of the triangle and y a side of the square. Then y=3/4x (both figures are equilateral and have the same perimeters). Let Rt be radious of the circle circumscribed on the triangle, Rs - on the square.

Since the triangle is equilateral Rt=x/sqrt(3); Rs=3/4x*(sqrt(2)/2). A/B=(Rs/Rt)^2=27/32.


Let $ a$, $ b$, and $ c$ be real numbers such that $ a\minus{}7b\plus{}8c\equal{}4$ and $ 8c\plus{}4b\minus{}c\equal{}7$.  Then $ a^2\minus{}b^2\plus{}c^2$ is

A-0
B-1
C-4
D-7
E-8

[quote="AIME15"]and $ 8c \plus{} 4b \minus{} c \equal{} 7$.  [/quote]

is that supposed to be $ 8a\plus{}4b\minus{}c\equal{}7$?

The coefficient of the first term of the second equation, he/she means.  Isn't is supposed to be $ a$?

A related problem:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207016

Nice Problem! What year?

a-7b+8c=4, 8a+4b-c=7
Square both eqq's:
a(a-7b+8c) -7b(a-7b+8c) +8c(a-7b+8c)=16,     8a(8a +4b -c) +4b(8a +4b -c) -c(8a +4b -c)=49
Simplifying the first eqq:
a^2 -7ab +8ac -7ab +49b^2 -56bc +8ac -56bc +64c^2=16
a^2 + 49b^2 +64 c^2 -14ab +16ac -112bc=16
Simplifying 2nd eqq:
64a^2 +32ab -8ac +32ab +16b^2 -4bc -8ac -4bc +c^2 =49
64a^2 +16b^2 +c^2 +64ab -16ac -8bc=49
Adding both eqq's:
65a^2 +65b^2 +65c^2 +50ab -120bc = 65
65a^2 -65b^2 +65c^2 = 65-130b^2-50ab+120bc
65(a^2 -b^2 +c^2)= 65 -130b^2 -50ab +120bc
a^2-b^2+c^2= 1-2b^2 -(10ab+24bc)/13
Since our answer is an integer, (10ab+24bc)/13 is also an integer
Thus 10ab +24bc is a multiple of 13
Thus 5ab +12bc is a multiple of 13
We can write this as:
5ab +12bc = 13n
We can recognize this as the Pythagorean triplet 5-12-13
Thus ab=bc=n
a=c
Going back to the first 2 eqq's:
a-7b +8c=4 and 8a +4b-c=7
Multiply the first eqq by 4 and the second by 7:
4a-28b+32c=16 and 56a+28b-7c=49
Add the eqq's together:
60a +25c=65
substitute c in for a:
85c=65
c=(65/85)=(13/17)=a
Substituting this in the first eqq:
(13/17) -7b +(8*13/17)=4
-7b=(68-13-104)/17
-7b=-49/17
b=7/17
Thus a^2 -b^2 +c^2=
(13/17)^2 -(7/17)^2 +(13/17)^2
[size=150]=B: 1[/size]

[quote="AIME15"]Isn't that what I put...[/quote]

no you put $ 8c \minus{} 4b \plus{}c \equal{} 7$ instead of $ 8a \minus{} 4b \plus{}c \equal{} 7$

Stupid me....

I forget what year this is, but it's a #20...

yea i did this on a symmetry handout similar problem was the one that was like 3cosA - 4sinB = .....
4sina-3sinb =....
and you had to square and find a + b

Notice that these two equations represented in a 3 dimensional graph represent planes and they intersect at a line. Thus, we don't have much work to do but find a suitable solution that works to our favor, so we may not need to number crunch and do heavy algebra.

Okay, label the equations 1 and 2.
multiply equation 2 by 8  and add it to equation 1 to get :

13a + 5b = 12

(sorry for not being able to latex)

now, we can just generate any (a,b), which will in turn satisfy the conditions for c for the equation.
I used a = 4 and b = -8.

Plug these back in to the original to get c = -7.
Now, just

16-64 + 49 = 1

Hence (B)
could someone hide the solution for me as well?

if anyone has a cool solution just based on the variables alone and manipulating them due to the the fact that the coefficients and constants are permuted across the equations, could you show it? I spent a while trying to find it, but nothing came out of it....

hence my cheap solution.

Thanks guys!

BTW, to hide:

[code][hide] 12345 [/hide][/code]
comes out as
[hide]
12345
[/hide]


Prove that exist only one sequence of positive integers $(a_{n}) (n=1,2,3,...)$ satisfying:$a_{1}=1,a_{2}=2,a_{4}=12$ and $a_{n+1}a_{n-1}=a_{n}^n \pm 1 \forall n \ge 2$
(I don't remember $a_{n}^n \pm 1$or$a_{n}^2 \pm 1$.May be this problem was in a competition)

[quote] (I don't remember $a_{n}^n \pm 1$or$a_{n}^2 \pm 1$.May be this problem was in a competition)[/quote]

OK! $a_{n+1}a_{n-1}=a_n^2\pm 1$.

Edit:  Sorry, didn't realize the $\pm$ could change for each term.

Why do you know that you can only use the $+$ sign again¿
Note that the sign is not determined by the way it was choosen for a special $n$.


analytic geometry
graphing lines
slope
limit

So what if the slopes are equal..
What exactly does that prove...??

To me at x = 1, slopes do not appear to be equal...(by definition...)
LHS we have slope = 1 and RHS slope = infinity....

Where is the prob...

ok soumya ur not supposed to post ur solutions here but rather in a different thread with the problme written on the top...also take this example if u r uncomfartable with that
$ f(x) \equal{} 2x \plus{} 1,x\leq 0$
$ f(x) \equal{} x^{2} \plus{} 2x ,x > 0$

neither to it appears there is any conceptual problem...
we have for $ x \leq 1$ $ f(x)\equal{}x$ so
$ \lim_{x \to 0^{\minus{}}} \frac{f(x)\minus{}f(0)}{x\minus{}0}\equal{}1$ while the RHS diverges
Same explaination holds true in my case also

Its the same thing here too....

At x=0, LHS slope = 2 and RHS slope = -infinity...
where is the prob...??