Euler
inequalities unsolved
inequalities

[quote="pohoatza"]Prove the following geometric ineq:

               R/r >= a/b + b/a, where r,R are the classical notations ;)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%2A2A&t=4927

[quote="arqady"][quote="M4RI0"]This is equivalent to the Euler Inequality. :roll:[/quote]
If so then $\frac{a}{b}+\frac{b}{a}=2,$ but it is wrong. ;)[/quote]

But $\frac{a}{b}+\frac{b}{a}\geq 2,$

[quote="M4RI0"]This is equivalent to the Euler Inequality. :roll:[/quote]

[quote="M4RI0"]

But $\frac{a}{b}+\frac{b}{a}\geq 2,$[/quote]
OK. If $a=3,$ $b=2$ and if you are right then $\left(\frac{R}{r}\geq\frac{13}{6}\Leftrightarrow\frac{R}{r}\geq2\right),$ but this is wrong. ;)


geometry
inequalities unsolved
inequalities

a,b and c are the sides of a triangle.
$R=a+b+c $  and $ Area=S=(a+b+c) $
prove that $ \frac{1}{3}<=\frac{R}{S}<\frac{1}{2} $

The first is obviously Cauchy and the second is:
$a^2+b^2+c^2< 2(ab+bc+ca)$ wich is also obviously true


Let $ n\geq2$ be a positive integer. Consider the set of points with positive integers not greater than $ n$ as coordinates. Does there exist a subset of $ 2n$ points such that no three of these points lie on a line? For example if $ n\equal{}5$ the set of points $ \{(1,2),(1,4),(2,4),(2,5),(3,1),(3,2),(4,3),(4,5),(5,1),(5,3)\}$ has the desired property.

This is an open question.
As far as I know, the answer is "yes" for $ n \leq 52$ (by computer search, no general pattern is known). The best known construction, due to Hall gives at least $ (\frac 3 2 \minus{} e)n$ for any $ e > 0$ and $ n > n_0(e)$.

Pierre.


With the slew of recent and semi-recent threads all about Creationism/Evolution and such I think its also appropriate to bring to the surface some evidence that one and the other are the same, and there are some things that few people are taking into account.

6 Days of Creation = 16 billion years of creation using a little mix of evolution.

Ahem,

Most of us know about one of Einstein's theories and what it means (or atleast many of us) - it says that the rate of time is different depending on speed and gravity. Therefore, if one were standing, say, on a planet that was so massive time was slowed 350,000 times that of ours, you would live out 3 minutes in the time that us, on Earth, would live out 2 years. Time would feel the same for you, we would just seem to be going really, really fast. Vice Verca as well, you would seem to be moving very, very, slowly to us. Wave length of light is affected by time as well, we know this because sunlight's wavelength is slightly longer (stretched) than that of light produced on earth. Stretched light means it came from a slower time, which is the sun. Keep in mind that wavelength is the 'timepiece of the universe' - it records all.

The biblical calendar starts from after the creation of man. This can only be, since Earth was not present during the 6 days of creation (one can say that after day 3 there was Earth, but the view in the bible for all 6 days takes a universe-wide view, only afterwards does it focus on Earth).

To measure the age of the universe, we use light's waves, more specifically Cosmic Background Radiation (CBR) - which has been around since the Big Bang, and as such is the only accurate measurement for how time worked relative to Earth's currently. CBR estimates the universe at approcimately 16 billion years old.

Since we know our universe has expanded, at the Big Bang it was, in its entirety, packed into a tiny speck. Then it expanded, as we know, and still is expanding.  If we trace the CBR back to its original wavelength from the moment of quark confinement after the Big Bang, we get the state of time at that moment, approx. million million times that of now.

For our 16 billion years however, we are looking BACK in time. The bible however, starts looking FOWARD from creation. At the time of creation, the 24 hour period felt the same as it did now, yet time moved MUCH faster. Going back, and applying the rate it changed since our time, 16 billion years ago, [b]24 hours lasted 8 billion years of our time - day one of creation[/b]. By day two it had already slowed down to 4 bilion years in 24 hours, as per expansion, and by day 3, 2 billion years, and so forth. At the start of day 6 it is only 1/4 billion years, and at the end of day 6 it has already been 15 and 3/4 billion years. The rest, almost 6000 years, is ours, starting from Adam.

I could write out the formula for figuring this out, but i barely understand it, and to do so id have to completely plagiarize this book (see bottom for the book) - so I'll skip that.

It is interesting to note that the evidence for the start of ages matches perfectly with the biblical description. Day one- creation, 16 billion years ago - obvious. Day 2 (the bible is zooming in, slowly slowly towards earth)- creation of the firmament -  8 billion years ago - shaping of milky way. Day 3 - formation of earth - end of era which Earth was hit by meteors, immediately thereafter water and life appeared. Day 4 - heavenly bodies were visible from earth - atmosphere becomes transparent (due to plant life producing oxygen). Day 5 - creation of fish and flying things - Cambrian, sudden explosion of life in the ocean, and throughout the period of the 5th day in our time, winged insects appeared. Day 6 - animals (and man) - 250 million years ago, 90-95% of marine life dissapears, animal life starts on dry land.

Science, after all this time, has only come to discover what the bible hiddenly said 3300 years ago.

I got this from a really good book, "The Science of God" by Gerald L. Schroeder. I'm not done with it yet, but I will post more on the concurrence of the Bible and Science later as I understand more of it.

Tell me what you guys think. This only proves that it was, in fact, 6 days of creation.

Nobody has anything to say about that? Hehe, even a disproof?

I guess this discussion is going to get to the point where we have a big argument about whether God exists.  if you think about it, time is just an illusion.  The universe was never really created.  It's just an everlasting object if you look at it outside the universe.  So there must be something much bigger than the creation/creator of the universe.  Nothing can be created outside the universe, because time doesn't exist outside the universe...

Maybe I'm getting off topic, but my point is this: It really doesn't matter when the universe was created, since nothing can be created when you view things outside the universe...

i go to church, and our class saw this video with an aussie guy who explained this stuff.

he said that if there was a "Lucifer's flood" then, why did god say "everything was good"? Why didn't he say "Everyting BUT lucifer was good"?


A man went into the bank to cash a check. In handing over then money the cashier by mistake gave him dollars for cents and cents for dollars. He pocketed the money without examining it and spent a nickel on the way home. He then found that he possessed exactly twice the amount of the check. What was the exact amount of that check?

Good Job, that puzzle took me forever to figure out, even when I did know algebra.

Does anyone know of a way to solve this without making a table?

I think, other than that, it is pretty much guess and check, but your solution is the easiest.

[hide="Euclidean Algorithmic solution"]Assuming that the last equation was correct...
Rewrite as $ 98c \minus{} 199d \equal{} 5$
Reduce modulo 98:
$ \minus{} 3d\equiv{5}\mod98$
Then there is some integer $ e$ such that $ 98e \minus{} 3d \equal{} 5$
Reduce mod 3: $ 2e\equiv{2}\mod3$
Showing that $ e\equiv{1}\mod3$ and that there is some integer k such that $ e \equal{} 3k \plus{} 1$
Plug this in and get $ 98(3k \plus{} 1) \minus{} 3d \equal{} 5$
Solve for $ d$ to get $ \minus{} 3d \equal{} \minus{} 93 \minus{} 3(98)k$ or $ d \equal{} 31 \plus{} 98k$
Plug in for $ c$ and get $ 98c \minus{} 199(31 \plus{} 98k) \equal{} 5$
Solve for $ c$ to get $ c \equal{} 199k \plus{} 63$
The only value for $ k$ which this is in between 0 and 100 is 0, so $ k \equal{} 0$, $ c \equal{} 63$ and $ d \equal{} 31$
and the amount of the check was $ \$31.63$.[/hide]

Edit: Yeah, 111 posts!

Wow!!! Although that solves the problem, I would have rather done the table instead.


find beta such that the system be impossible

\[ \{ \begin{array}{ccccccc} x&+&2y&-&3z&=&4 \\ 3x&-&y&+&5z&=&2\\ 4x&+&y&+&(\beta^2-14)z&=&\beta+2 \\ \end{array} \]

[hide]one approach would be to consider cramer's rule, if the determinant that contains the coefficients of $x,y,z$ equals $0$, then when solving for the variables, you get undefined, so

$\begin{tabular}{|ccc|} 1&2&3\\ 3&-1&5\\ 4&1&(\beta^2-14)\\ \end{tabular}=0$

Then solving this system of equations, we get $\beta=\boxed{\pm\sqrt{22}}$[/hide]

You made the same mistake I almost made.... it's a -3 on that first $z$.

[hide]Anyways, just add the first two equations to get: \[ 4x + y + 2z = 6 \] Notice that this is the same as equation 3 excluding the $\beta$ terms. Find a $\beta$ such that $\beta^2 - 14 = 2$ and $\beta + 2$ is not equal to 6; the answer is clearly $-4$.[/hide]


algebra
polynomial
search
algebra proposed

Prove that the polinomial  $f(x)=x^4+26x^3+52x^2+78x+1989$  could not be written
in the form $f(x)=p(x)\cdot q(x)$  where  $p(x),q(x)$   are polinomials with  integer coefficients and  degree  $\leq 4$

Apply Eisenstein's criterion for the prime number $13$.

[quote="enescu"] Eisenstein's criterion for the prime number $13$.[/quote]

What is Eisenstein's criterion ?

Ok. I find it . Thank you very much nice solution.


Some facts: at the IMO 2009 there were about 560 contestants, among them about 50 girls (15 girls got medals).
Why :huh:

A [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214616]previous discussion[/url] of this and related issues. It's political and controversial, but I'd start with general social inequality and the messages it sends discouraging girls from pursuing math.

On the gender-gap thread mentioned by jmerry, I have posted a brief comment on the article "Cross-Cultural Analysis of Students with Exceptional Talent in Mathematical Problem Solving," by Titu Andreescu, Joseph A. Gallian, Jonathan M. Kane, and Janet E. Mertz, which appeared in Notices of the American Mathematical Society, Vol. 55, No.10, pages 1248-1260, November 2008.  I recommend this article highly, if you are interested in the topic of participation by girls in the IMO.

I know that some girls on the USA team were Melanie Wood and Sherry Gong... And they did very well. :|

Well I think the main reasons math competitions and forums like this one are mainly male is because in general (not to be stereotypical ) guys are mathematically and scientifically minded  while girls tend to be more creative.
I honestly think it may be a social thing too since most girls are not interested in math, so its weird to be a math loving girl so some girls maybe intimated by the lack of girls who are interested and enjoy doing math for fun.

Even harder when I chose an university on the basis of females vs. males...

University of Montreal (in mathematics) is about 1/3 girls for 2/3 guys total. And UQAM, with similar sexual proportions, had a much smaller mathematical pool to draw from so I would have found three years long as dating was a factor for my decision. And also I didn't want to make combinatorics out of my math life so UQAM was ruled out completely. (UQAM is best known in combinatorics, as far as math go)

If possible, I would raise the issue of women in Putnam... again, Melanie Wood was a non-trivial competitor.

I read an article which justify this effect by genetical theory:

So basically studies shown that the intelligence depends only in X-chromesome. Thus, if a guy is smart normally his mother is smart(because guy is XY while X comes from mom and Y from dad) and a girl's intelligence is usually the average of both father and mother.

Thus because girl's intelligence level depend on 2 entities, the general intelligence distribution of all girls tend to have normal distribution, ie lots of girls of average intelligence and few extremely smart ones and also few extremely stupid ones.

while as in the case of guys, because his intelligence depend only in his mom, the distribution of guy's intelligence level tendo to go to the extremes, i.e. lots of smart guys as well as lots of really stupid ones.

in effect in a competition while only extremely smart person is taken place, then it make sense to have more guys.

From what I can tell, the most significant scholarly article in favor of the theory described by chichi can be found here:

http://languagelog.ldc.upenn.edu/myl/llog/Turner1996.pdf

Imho, this is nothing more than an intriguing hypothesis.  The hypothesis seems to be based on that fact that there are many genes on the X-chromosome that are known to cause mental retardation.  The author essentially speculates that the X-chromosome must carry many genes that are important for high intelligence, and that this explains the observed intelligence differences in men and women.

My analysis is this:  The genetic basis of intelligence is undoubtedly extremely complicated, and it seems entirely reasonable that there are indeed sex-linked genes that contribute to "innate intelligence," whatever that is.  What upsets me is the huge logical leap that this somehow explains things like the under-representation of girls at the IMO.  Although there may be sex-linked genes for intelligence, we have zero science explaining how these genes express themselves in the overall picture of intelligence (or even what these genes are, short of the retardation genes mentioned above).

In my mind it seems obvious that nurture rather than nature is the dominant factor in the proportion of girls at the IMO, and my argument is simple:  If nature were the dominant factor in *most* countries, then the proportion of girls doing high level contest math in those countries should be roughly similar.  But it is not.

Moreover, consider that in the United States, around 1990 no female had ever gone to MOP.  Fast-forward to 2010, and although the boys still dominate MOP, it is now commonplace for girls to reach such heights.  And the fact is:  Back in 1990 you would have found MANY people who would have said that nature alone explained by there were never any girls at MOP.  I believe that history has shown them to be wrong, and those who believe in 2010 that nature alone explains why there are so few girls at MOP now will also be proven wrong.

[quote="yenlee"]In my mind it seems obvious that nurture rather than nature is the dominant factor in the proportion of girls at the IMO, and my argument is simple:  If nature were the dominant factor in *most* countries, then the proportion of girls doing high level contest math in those countries should be roughly similar.  But it is not.
[/quote]This is a fallacious argument. Of course there are significant social barriers between women and men, where sometimes a man might be persuaded to do math while a similarly-skilled woman is not. I'm pretty sure that the data show that there are more woman participating in the imo in gender-equal societies than in non-gender-equal societies.

Your argument is saying that because the proportion of girls doing high-level math is unequal across country lines, nurture is more important than nature in mathematics. This is a false assumption. All you've shown is that nurture has a big effect on math skills/participation. It's quite possible that nature could have an even bigger effect on this.

 What I'm trying to say is that we simply don't know how much of the difference is nature and how much is nurture. It's possible that in the future we will find out, but right now there isn't much proof either way.

If I had to make a guess, I would say that a large amount (note I did not say all) of the difference is because of nurture, but that's all it is. A guess.

I apologize for not being very precise, so let me be precise now about what it is I actually believe, and what I am arguing.  For the sake of argument, assume that math ability can be measured on a numerical scale.

Let P be the population of 16-year-olds whose *innate* potential math ability at time of birth was over 4 standard deviations above the mean (the 4 here is for the sake of being definite).  Let x be the fraction of people in P who are female.

Let Q be the population of 16-year-olds whose *actual* realized math ability is over 4 standard deviations above the mean.  Let y be the fraction of people in Q who are female.

Let z be some mysterious variable that measures how well girls' mathematical abilities are nurtured in comparison to boys.  It somehow includes all environmental factors.  Let's say z=0 corresponds to perfectly equal nurturing.

Roughly speaking, y is some function of x (nature) and z (nurture) that is monotone increasing in both x and z.  Other than that, we don't know what the function is, we don't know what x is, and we definitely don't know much about what z is.  However, we can measure y roughly.  And we know that x ought to be constant over all of modern history and across every human culture.

Here's my point.  y is observed to be small (say, less than 10%), which causes some people to believe that x is similarly small.  But this reasoning only has validity when z=0 or is close enough to zero as to have little significance.  My point about comparing different countries is that it proves that z is significantly nonzero enough that you cannot use a measurement of y to guess the value of x.  My point about the example of the United States is that some people contend that z might be close enough to zero in the US in 2010 (i.e. we live in a fairly gender-balanced society) in order to estimate x, but these people and those who said the same thing back in the 80s can't possibly both be right, because y keeps changing.

Moreover, there are those who might say, "Okay, fine, z is obviously nonzero.  But I'm not saying that x<10%.  I think it's probably somewhere in between 20% and 30%."  But those people are still just guessing.  There is still *no evidence* pointing to x being in any sort of range like that.  In fact, there isn't even any evidence that x is less than 50%.  Personally, I have no idea what x is.  The tempting gut feeling is that x is less than 50%.  (If I understand correctly, this is Ihatepie's best guess.)  But if you ask your gut why it feels that way, the only honest answer your gut will give you is, "Because y is so low!"  Therefore, even as a guess, the guess is illogical.

Final note:  I admit that I misspoke a bit when I wrote, "nurture rather than nature is the dominant factor."  It's sort a nonsensical statement.  If you have function of two variables, you can't really say that one of the two variables is the dominant factor or has a bigger effect, without being more precise.  (Wow, I can't believe I wasted so much time writing this all down.)

I know girls will go to math competitions, because at chapter, I saw that about (at least) 1/3 of them were girls. And there were about 300 total students. So a good 100 were girls. But only one girl got a prize, and she was second.

BTW, about half our math club are girls. So girls will do math.

This isn't IMO, but I was just giving it from my perspective.


Six straws of lengths 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, and 7 cm are placed in a box. Express as a common fraction the probability that three straws drawn at random without replacement form a triangle when connected at their endpoints.

After casework based on the length of the shortest straw, we find that the answer is $ \boxed{\frac{13}{20}}$.

could you be more specific? it's not very clear

A triangle can only be formed if the sum of the two shortest side lengths is greater than the longest length.

So we have

2,6,7
2,5,6
2,4,5
2,3,4

3,6,7
3,5,6
3,5,7
3,4,5
3,4,6

4,6,7
4,5,7
4,5,6

5,6,7

There are $\dbinom{6}{3}=20$ ways to choose three numbers.

Thus, the probability is $\boxed{\frac{13}{20}}$.


Hapland is a great game. it's easy to find using google. What do you guys think about this game and what do you like about it?

ps this can also be the place for ppl to ask questions about how to win the game :)

You mean hapland 1?
I don't want to tell you the whole solution. Tell me what you've discovered and where yu are stuck and I'll help you.

Well, If I tap the light in the cave a few times, the guy destroys the light. If I tap the spear, the spear moves. etc. Well, I have gotten to the part where the guy is shot out of a cannon onto the mountain and he pushes the bell tower, but I don't know where to go from there.

Sorry, this is a dead topic, but it should have more useful posts. There are only posts by me and rem. 2 each.

[quote="Iversonfan2007"][quote="bubka"]I'll play it when I have time. But have you noticed that there are exactly 21112 users now. It is quite a funny number I think.[/quote]

yes, it's divisible by $2$ and $4$ that's pretty crazy[/quote]
Yeah and it's a palindrome that's so flipping awesome.


super appears to be somewhat sarcastic here....but thank you, I appreciate your support. It's the little people who have gotten me this far! People like you, every one of you people I have crushed on my way to top....This is a touching moment, one I will treasure for the rest of today. Tomorrow: 1900!

Nice job mewto, but it would be more impressive if you would actually cd me.  :)

congrats mewmew!! only 200 more til the predicted 5th flag! u can do it! go super! cd mewmew! :D

actually, super wants to cd mewto.  it's the opposite thats not happening.  mewto doesn't want to cd super.

super thinks mewto is scared.

Mewto needs to cd me some more, because I want to leech away his rating....er, I mean, because it's good for him to cd a wide variety of people.

unfortunately, mewto doesn't cd anyone with a high rating
i got soooo close to beating him  :(
it was 4-3 *sob*

I've only ever beaten him 3 times in my life in a cd, and only twice in a game. :(

I appear to be stuck in the 1500s...I keep getting 1590 and then losing to vallon or kl in a cd.

*sigh* i was the only one who gave mewto 1337 few months ago (like 4 months ago)

I feel like getting stuck in 1400s...... I would barely make 1500 then lose to vallon, mewto, johnny, isabella, etc.

(kl doesn't cd me b/c i lost 4-3 and she got the last one by knowing a trivial fact)

[quote="dannyhamtx"]
(kl doesn't cd me b/c i lost 4-3 and she got the last one by knowing a trivial fact)[/quote]

lol trivial. i think the problem was like "The WNBA began in 1996, this was 49 years after the start of the NBA. What was the first year NBA began?" by using super knowledge, i got 1947 right away  :)

on the other hand, i think the rating system isnt really useful to compute whether or not someone is really good at ftw if he/she wont play someone with a high rating. so stay somone with a 1200 gets to 1700 because he/she cds 1200s or 1300s, then it wouldnt really be that impressive. but if someone actually plays 1500/1600s, and gets to a high rating, say 1700, then there would obviously be a big difference to the two people who got to the same rating.

ummm
just wondering but how did all of the ratings suddenly spike up by a couple hundred?
i saw the leaderboard and thought it was a joke for a second

Levans fixed a typo in the algorithm, i think.

And I beat mewto in a CD 4-2.  Yay!

Levans changed the rating system so it impacts ratings more. For example, if you beat someone 200 below you now you get more than you would have a month ago.

BTW How many posts do you have to have to get unmuted?? It's getting extremely annoying that I can't say anything back to people on AoPS and a couple of people even assume I'm a person other than myself!!

well, the number of posts is definitely not 3...
but anyway
nobody know how many posts you need
levans kept that information to himself
but number of posts isn't the only thing
you also need a number of games played and a number of [b]productive [/b]posts, judged by admins

if u have reached all that criteria, you can chat
unfortunately, until then, nobody can unmute you

best way to be able to chat: play a lot of games on FTW, and make a lot of [b]productive[/b] posts (and yes, spam does not count as productive, but u knew that.... :D )

hmm, i'm up for countdowning anybody

mewto you game? my rating is like 1400s. i'm kinda bad :\

Please refrain from spamming, Lazarus. Thank you.

Mewto reached 1817, last time I checked. What's your new high, Mewto?

link to leaderboard

[url]http://www.artofproblemsolving.com/Edutainment/g1/Pages/Leaderboard.php[/url]

mewtos now 1780.

Does anyone know at what point it will begin to get extremely hard to get rating? I'm at 1600 usually, and I find it slightly difficult but not much.


function
inequalities proposed
inequalities

X,Y,Z are positive real numbers

Find the maximum value of XYZ/(1+X)(1+Y)(Y+Z)(Z+16)

A beautiful problem!

Should the expression be
\[\frac{xyz}{(1 + x)(x + y)(y + z)(z + 16)}?\]

If so, the problem is equivalent to finding the minimum of
\[\frac{(1 + x)(x + y)(y + z)(z + 16)}{xyz} = (1 + x) \left( 1 + \frac{y}{x} \right) \left( 1 + \frac{z}{y} \right) \left( 1 + \frac{16}{z} \right).\]

It is easy to show that for $x_1$, $x_2$, $\ldots$, $x_n \ge 0$,
\[(1 + x_1)(1 + x_2) \dots (1 + x_n) \ge (1 + \sqrt[n]{x_1 x_2 \dots x_n})^n,\]
with equality iff $x_1 = x_2 = \dots = x_n$.

Applying this, we get
\[(1 + x) \left( 1 + \frac{y}{x} \right) \left( 1 + \frac{z}{y} \right) \left( 1 + \frac{16}{z} \right) \ge (1 + 2)^4 = 81,\]
with equality iff $x = 2$, $y = 4$, $z = 8$.

Michael, can you tell me which year this appeared?

Firstly,my apologies for stating the problem incorrectly.

You have guessed the correct form namely XYZ/(1+X)(X+Y)(Y+Z)(Z+16)

This problem appeared in 1987.It is not listed at the Kalva ,site.You can find it in"The Mathematical OLympiad Handbook-An introduction to problem solving" by A.Gardiner , on pages 61 and 148.DEspite the error I hope you found it an interesting problem to solve .

Michael


Ghm... It seems to me completely straight-forward: apply the determinant expansion by minors formula and after tons of boring computations you will get the result  :)  Or are you asking for some ingenious way to do it?


analytic geometry
graphing lines
slope
function
calculus

For the next two questions, express the slope of the secant line of each function in terms of x and h.  Be sure to simplify your answer.

(a) f(x) = -3x + 2

(b) f(x) = (1/x)

HINT GIVEN:

The slope of the secant line having two the points
(x, f(x)) and (x + h, f(x + h)) on the graph of the function may be given as f(x + h) - f(x)/h

NOTE:

I am not in a calculus course.  So, please do not reply thinking that you are teaching a calculus student.

wth???  if you know this is a calculus problem then y'd you post it here???

nexttime post it in the forum that is actually made for calculus problems.

locked

I just have to say this:  This is not a calculus question!  Stuff such as this is found in both Algebra II and PreCalculus, so this is the correct forum.

The approach the problem wants you to use is that of limits.

All you have to do is find the slope between the two given points, then let take the limit as $h\rightarrow0$.

[hide="(a)"]$\frac{-3(x+h)+2+3x-2}{h}=\frac{-3h}{h}=-3$.

This one did not require any limits at all, just simple definition.[/hide]

Actually, any Algebra I student should be able to solve this, given the hint.

[hide="(b)"]$\frac1{x+h}-\frac1{x}=\frac{x-(x+h)}{(x+h)(x)}=\frac{-h}{(x+h)(x)}$

As $h\rightarrow0$, this becomes 0.[/hide]

That should be enough (and I don't think I messed up...)


Given naturals $k<m<n$ such that $\gcd(k,n)=1$, i.e. $k$ and $n$ are coprime. For $a_{1},...a_{n}\ge0$ and indices cyclical $\mod n$, prove or disprove
$\sqrt[k]{\frac1n\sum_{i=1}^{n}a_{i+1}\cdots{a_{i+k}}}\geq\sqrt[m]{\frac1n\sum_{i=1}^{n}a_{i+1}\cdots{a_{i+m}}}$.  :roll:

as [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=617524#617524]zhaobin has pointed out[/url], taking $a_{1}=...=a_{m}=1, a_{m+1}=...=a_{n}=0$ shows that the ineq is false for $n^{m-k}>(m-k+1)^{m}$, i.e. putting $m-k=: \ell$, false for $n^\ell>(\ell+1)^{\ell+k}$. So maybe, it is better to state the following problem:

Given naturals $k<m$, find the biggest $n$ such that for all $a_{1},...,a_{n}\ge0$, the ineq $\sqrt[k]{\frac1n\sum_{i=1}^{n}a_{i+1}\cdots{a_{i+k}}}\geq\sqrt[m]{\frac1n\sum_{i=1}^{n}a_{i+1}\cdots{a_{i+m}}}$ holds.  :roll:  :roll:


Let $ g(x): \mathbb{R}\rightarrow \mathbb{R}$ a function such that $ g(x)\plus{}g(x\plus{}1)\equal{}x^2,\,\forall x.$ If $ g(101)\equal{}10001,$ find $ g(15).$

I don´t understand the last step (the one with the sum)
wouldn´t be more appropriate for it to be:
10001 + ∑(k^2, k, 15, 100) - 2*∑((2k)^2, k, 8, 50)?
Correct me if im wrong... :|

Sure its the same thing, I just made use of the fact that
$ 15^2\minus{}16^2 \equal{} (15\plus{}16)(15\minus{}16) \equal{} \minus{}(15\plus{}16)$
I think it makes it a little bit easier to evaluate, but your method is still correct.


trigonometry
quadratics
geometry
rectangle
Asymptote
inequalities
trig identities

A round table has radius 4. Six rectangular place mats are placed on the table. Each place mat has width 1 and length x as shown [in the attachment]. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length x. Further, the mats are positioned so that the inner corners touch an inner corner of an adjacent mat. What is x?

$ A. \; 2 \sqrt{5} \minus{} \sqrt{3} \qquad B. \; 3  \qquad C. \frac{3 \sqrt{7} \minus{} \sqrt{3}}{2} \qquad D. 2 \sqrt{3} \qquad{E. \frac{5 \plus{} 2 \sqrt{3}}{2}}$

[hide="Solution"]
- Note that the 6 little triangles are equilateral, and hence all sides are 1.

- If we connect the ccw circumference vertexes on two consecutive triangles, we have a single radius-length, equal to 4.
[i](consider doing this to all 6 pairs if you don't see why).[/i]

- It's not hard to see that the angle between the circumference edge of one of the small triangles and the circumference edge of the mat is 150 degrees.
[i](if you don't see why, consider what happens if we move the vertex to the other side of the circle -- this new angle and the original add to 180, and the new angle is half of the subtended 60 degree arc)[/i]

-  Hence, by law of cosines, we have $ 1^2 \plus{} x^2 \plus{} 2 (1)(x) \cos 150^o \equal{} 4^2$, which gives the answer to be $ \frac{3 \sqrt{7} \minus{} \sqrt{3}}{2} \implies \boxed{\text{C}}$.
[/hide]

This one was cool only because it had an optical illusion.

[hide="faster?"] Let one of the rectangles be ABCD, with A,B on the circle. Let the rectangle next to it be CEFG, with E,F on the circle. Note that BCE is equilateral, since BC=CE, <BCE=60. So BE=1.

Note that AE=4, since if we draw all 6 of these segments, we get a regular hexagon.

Note that <ABE=150. Law of cosines![/hide]

[hide="Potentially faster"]
Let ABCD be one mat and BEFG be a mat adjacent to it. O is the center of the circle. H is the midpoint of $ \overline{CG}$.

We quickly deduce that $ m<CBG\equal{}60^\circ$ due to the 90, 120, and 90 degree angles surrounding B. Since $ CB\equal{}BG$, we find that $ \triangle{CBG}$ is equilateral with side length 1 and so $ CH\equal{}\frac{1}{2}$ and $ BH\equal{}\frac{\sqrt3}{2}$.

Since the interior edges of length $ x$ form a regular hexagon, we find that $ BO\equal{}x\equal{}OH\minus{}BH$.

We also know that $ OC\equal{}OG\equal{}4$, since both segments are radii of the circle. Thus $ \triangle{COG}$ is isosceles, and H is the foot of the altitude from O. Thus $ OH\equal{}\sqrt{16\minus{}.25}\equal{}\sqrt{\frac{63}{4}}\equal{}\frac{3\sqrt7}{2}$.

We also know $ BH\equal{}\frac{\sqrt3}{2}$, so $ x\equal{}OH\minus{}BH\equal{}\frac{{3\sqrt7}\minus{}{\sqrt3}}{2}$.[/hide]

[quote="archimedes1"][hide="hint/walkthrough"][img]12545[/img]
Solve for the sides in terms of $ x$.  This can be done using 30-60-90 ratios.
Apply the Pythagorean Theorem!
Use the Quadratic Formula!
[/hide][/quote]

This was the way I did it. It took a fairly long time for me to think of it, but once I got on the hunch it went pretty quickly.

Was this seriously number 22 on the AMC 12? Wow. I think archimides' way is probably the fastest.  Most people probably don't think this, but the most important line in a circle is the diameter.

Yeah it was simple once you drew the line.

But I think lots of people got stuck looking at 1/6 of the table... (like me) :)

I used Power of a Point (a minute after the test ended), so that:

$ \displaystyle {\left( \frac{x}{2} \right)}^{2} \equal{}\left(4\plus{}\left(\frac{x\sqrt{3}}{2}\plus{}1\right)\right)\left(4\minus{}\left(\frac{x\sqrt{3}}{2}\plus{}1\right)\right)$

Multiply that through by 4 to get the same equation as drawing the diameter  :wink:

The Zuton Force, what did you use to draw your diagram? Asymptote?

[quote="archimedes1"][hide="hint/walkthrough"][img]12545[/img]
Solve for the sides in terms of $ x$.  This can be done using 30-60-90 ratios.
Apply the Pythagorean Theorem!
Use the Quadratic Formula!
[/hide][/quote]

Wait, how do you use your method? I get $ 2x\equal{}8$?

[hide="easiest one to see?"]from the center draw two lines to the verticies of one rectangle touching the circle. The height of this isoscles triangle is $ \sqrt {16 \minus{} \frac {x^2}{4}}$
Draw an equilateral triangle from the center to the same rectangle's verticies who dont touch the circle. The height of that is $ \frac {x\sqrt {3}}{2}$. The difference between these two heights is 1. So

$ \frac {x\sqrt {3}}{2} \plus{} 1 \equal{} \sqrt {16 \minus{} \frac {x^2}{4}}$
Solve.[/hide]

[quote="123456789"]The Zuton Force, what did you use to draw your diagram? Asymptote?[/quote]

Hahaha, I can't use asymptote for my life... that's a screen-shot from PowerPoint, which I do know how to use

To sonny: multiply that through by 2  :wink: Everything turns out the same. And drawing the diameter eliminates fractions

to 123456789: Note that it is a right triangle with legs of length $ x,2\plus{}x\sqrt{3}$ and a hypotenuse of length $ 8$.

The $ 2\plus{}x\sqrt{3}$ is found because the angle of the obtuse isosceles triangle is $ 120^\circ$. Solve from there.

[quote="mathcrazed"]To sonny: multiply that through by 2  :wink: Everything turns out the same. And drawing the diameter eliminates fractions

to 123456789: Note that it is a right triangle with legs of length $ x,2 \plus{} x\sqrt {3}$ and a hypotenuse of length $ 8$.

The $ 2 \plus{} x\sqrt {3}$ is found because the angle of the obtuse isosceles triangle is $ 120^\circ$. Solve from there.[/quote]

The equation will probably be the same, the way of doing it is different from achimedes1's....

[quote="mathcrazed"]To sonny: multiply that through by 2  :wink: Everything turns out the same. And drawing the diameter eliminates fractions

to 123456789: Note that it is a right triangle with legs of length $ x,2 \plus{} x\sqrt {3}$ and a hypotenuse of length $ 8$.

The $ 2 \plus{} x\sqrt {3}$ is found because the angle of the obtuse isosceles triangle is $ 120^\circ$. Solve from there.[/quote]

The equation will probably be the same, the way of doing it is different from achimedes1's way....

[quote="mathcrazed"]To sonny: multiply that through by 2  :wink: Everything turns out the same. And drawing the diameter eliminates fractions

to 123456789: Note that it is a right triangle with legs of length $ x,2 \plus{} x\sqrt {3}$ and a hypotenuse of length $ 8$.

The $ 2 \plus{} x\sqrt {3}$ is found because the angle of the obtuse isosceles triangle is $ 120^\circ$. Solve from there.[/quote]

Oh...stupid me. I thought the red triangle was a 30-60-90 triangle.

Hmm double post.  Yes, it is different, but drawing the diameter eliminates fractions, blah blah etc. =).  But every way has its merits.

Do you think that it would have been a good idea to redraw the diagram so that it was bigger or was the given one on the test fine?

If you don't see it, drawing the diagram bigger is never harmful. Once you see a method to reduce the problem to an equation, you can ignore the diagram. Personally, I think the 1,x,4 triangle with a 150 degree angle is the easiest solution to find, but the 30-60-90 is arguably more elegant.

This one is ridiculously easy to brute force, just because 4-(alt of eq triangle) is [b]extremely[/b] close to the right length.

That's how I did it, got the right answer, but then misbubbled :(

Brute force? Really?

What is (alt of eq triangle) and what is the right length?
The answers have radicals in them so how could you use an approximation without doing a bunch of arithmetic?

You [i]don't[/i] want to know how I did it. But I'll say it anyway :P

You know the tiny sector between every two placemats? Has straight lengths of 1...? I joined the ends of the placemats to get an "equilateral" triangle, and then approximated the altitude of this "triangle" to be $ \sqrt{3}/2$. Then,$ x\plus{}\sqrt{3}/2 \equal{} 4$, or $ x \approx 3.2$. Then I found what the answer choices approximately were, and I chose the best looking one, which was $ (C)$

Btw, idk if anyone mentioned this already, but this was #25 on the 10A...

I would rather have students solve the problem.  But I would agree that a numerical solution to this problem was quite feasible if you didn't immediately see the geometric / trigonometric solutions.

As a coach I advocate memorization of some simple constants to at least two decimals.  These include $ \pi\approx 3.14$, $ e\approx 2.718$, $ \sqrt {2}\approx 1.414$, $ \sqrt {3}\approx 1.732$, $ \sqrt {5}\approx 2.2$, $ \sqrt {10}\approx 3.16$, $ log_e{10}\approx 2.302$, $ log_{10}2\approx 0.301$, $ log_{10}3\approx 0.477$.  Also, you should remember perfect squares and primes, from your middle school days.

You can quickly derive *many* additional constants, if you know these.

Consider the answer choices to this problem:

A:  $ 2\sqrt {5} \minus{} \sqrt {3}\approx 2\times 2.2 \minus{} 1.732 \equal{} 4.4 \minus{} 1.732 \approx 2.7$
B:  $ 3$
D:  $ 2\sqrt {3}\approx 2\times 1.732 \equal{} 3.464$
E:  $ \frac {5 \plus{} 2\sqrt {3}}{2}\approx 2.5 \plus{} 1.732\approx 4.2$

As a previous poster observed, inspection limits the range of correct answers.  Consider a radius from the center of the circle passing through the point at which two placemats touch.  This yields two inequalities:
$ x \plus{} \sqrt {3}/2 < 4 < x \plus{} 1$.
Or $ 3 < x < 4 \minus{} \sqrt {3}/2$
Or $ 3 < x < 3.134$

All of the above answers are unquestionably out of range.  Choose answer "C" and move on.  Or, if time permits, we can try to calculate (C) to confirm it is within range:

$ 26^2 \equal{} 676$
$ 27^2 \equal{} 729$

So, $ \sqrt {700}\approx 26.5$

C:  $ \frac {3\sqrt {7} \minus{} \sqrt {3}}{2}\approx \frac {3\times 2.65 \minus{} 1.732}{2} \equal{} \frac {7.95}{2} \minus{} 0.866 \equal{} 3.975 \minus{} 0.866 \approx 3.11$


Find the electric field at (0,0,0) with a shell x^2 + y^2 + z^2 = R^2, z>0, and charge Q by method of Gaussian surfaces.

Since we don't have spherical or cylindrical symmetry that is going to be really messy. You're sure you have to use Gaussian surfaces?

sorry, but I didn't get what you were asking. :wink:


I read about Campbell on the USABO website and it seems that most of the questions are from that book. Why is it arranged like this?

I'm not sure that the USABO questions are [i]from[/i] Campbell, but rather that the material covered in the questions can all be found in the book.

yea, i mean that the material covered is in campbell.

campbell is cosidered bible of biology--esp 4 olympiads :rotfl:  :P


Prove that n+1 and n 2 -n+1 cannot both be perfect squares for all naturals n>0.

A square is 0 or 1 modulo 4.

Quick tries show that n+1 and n^2-n+1 are both 0 or 1 mod 4 iff n=0 mod 4.

If n+1=0 mod 4 and n^2-n+1 = 0 mod 4 then n^2+2=0 mod 4 but then n is even and here is the problem sinc we would have 2*(2n'^2+1)=0 mod 4  :D

even quicker: n^2-n+1 cannot be a perfect square at all, because it lies between (n-1)^2 and n^2 (except of course of the case n=0 in which both n+1 and n^2-n+1 are equal with 1^2 :D).

[quote="Valentin Vornicu"]even quicker: n^2-2n+1 cannot be a perfect square[/quote]

Sure but it is n^2-n+1 here  :D

But your trick can be adapted here : n^2-n+1=(n-1)^2+n

of course it's n^2-n+1 :) it was a typo, fixed :D


inequalities
rearrangement inequality
inequalities theorems

The same inequality appeared in a Kolmogorov cup (a Russian "math fight" competition) in the following form:

[color=blue][b]Problem.[/b] For any three positive numbers a, b, c, prove the inequality

$ \left( 1+abc\right) \left(\frac{1}{a\left( 1+b\right) }+\frac{1}{b\left( 1+c\right) }+\frac{1}{c\left( 1+a\right) }\right) \geq 3$.[/color]

___________________________________________________________________________

I think this is a beautiful inequality. But, as humble as I always am  :D , I think I also have a very beautiful proof of it. I really can't resist posting it here (you can try searching for other proofs, since I really don't think this was the intended solution!).

In fact, in general we have $ abc \neq 1$, we cannot apply Harazi's substitution $ a=\frac{v}{u}$, $ b=\frac{w}{v}$, $ c=\frac{u}{w}$ with positive u, v, w. But we can substitute $ a=k\cdot \frac{v}{u}$, $ b=k\cdot \frac{w}{v}$, $ c=k\cdot \frac{u}{w}$, where k, u, v, w are positive real numbers. Then $ abc=k^{3}$, and our inequality simplifies to

$ \left( 1+k^{3}\right) \left(\frac{u}{k\left( v+kw\right) }+\frac{v}{k\left( w+ku\right) }+\frac{w}{k\left( u+kv\right) }\right) \geq 3$.

This can be rewritten as

$ \frac{1+k^{3}}{k}\cdot \left(\frac{u}{v+kw}+\frac{v}{w+ku}+\frac{w}{u+kv}\right) \geq 3$.

Now this is a generalization of Nesbitt's inequality. For proving it, we define A = v + kw, B = w + ku, C = u + kv, and then we can readily find

$ u = \frac{C+k^{2}B-kA}{1+k^{3}}$;
$ v = \frac{A+k^{2}C-kB}{1+k^{3}}$;
$ w = \frac{B+k^{2}A-kC}{1+k^{3}}$.

Thus, the inequality we are intending to prove can be rewritten as

$ \frac{1+k^{3}}{k}\cdot \left( \frac{C+k^{2}B-kA}{A\left( 1+k^{3}\right) }+\frac{A+k^{2}C-kB}{B\left( 1+k^{3}\right) }+\frac{B+k^{2}A-kC}{C\left( 1+k^{3}\right) }\right) \geq 3$.

In other words,

$ \frac{1}{k}\cdot \left( \frac{C+k^{2}B-kA}{A}+\frac{A+k^{2}C-kB}{B}+\frac{B+k^{2}A-kC}{C}\right) \geq 3$.

Multiply with k and get

$ \frac{C+k^{2}B-kA}{A}+\frac{A+k^{2}C-kB}{B}+\frac{B+k^{2}A-kC}{C}\geq 3k$.

This can be rearranged (not in the meaning of using the rearrangement inequality) into

$ \left( \frac{B}{C}+k^{2}\frac{C}{B}\right)+\left( \frac{C}{A}+k^{2}\frac{A}{C}\right)+\left( \frac{A}{B}+k^{2}\frac{B}{A}\right)-3k \geq 3k$.

In other words,

$ \left( \frac{B}{C}+k^{2}\frac{C}{B}\right)+\left( \frac{C}{A}+k^{2}\frac{A}{C}\right)+\left( \frac{A}{B}+k^{2}\frac{B}{A}\right) \geq 6k$.

But this is trivial since AM-GM gives $ \frac{x}{y}+k^{2}\frac{y}{x}\geq 2k$.

___________________________________________________________________________

[b]EDIT:[/b] See also

http://www.mathlinks.ro/Forum/viewtopic.php?t=48261
http://www.mathlinks.ro/Forum/viewtopic.php?t=33550
http://www.mathlinks.ro/Forum/viewtopic.php?t=19469 problem 2

for this and related inequalities.

This inequality seems to have appeared in 1000 different places. You can discuss about its history at http://www.mathlinks.ro/Forum/viewtopic.php?t=85696 . As for now, the following sources of the above inequality are known:

- Balkan MO 2006, Problem 1, proposed by Sotiris Louridas from Greece;
- Crux Mathematicorum 1998, proposed by Mohammed Aassila;
- (maybe) the book "Old and New Inequalities" (at least the stronger version, http://www.mathlinks.ro/Forum/viewtopic.php?t=6044 , appeared in this book);
- UK Correspondence Course 2005;
- a Greek book on inequalities written by Skombris and Babis Stergiou (this is the MathLinks user Stergiu), where "Kömal 1993" is given as a source for this inequality (but this source may be imprecise);
- proposed by the Russian mathematician D. P. Mavlo in the Bulgarian journal Mathematica (number 4, 1987);
- Kvant problem M1136, number 11-12, 1988;
- USA IMO team training 1993;
- South African IMO team training;
- Titu Andreescu's book "Mathematical Miniatures";
- 6th Kolmogorov Cup, 1-8 December 2002, 1 round, 1 league;
- Indian IMO selection test 1997;
- Armenian National Olympiad 1996;
- Kiran Kedlaya's inequality notes (?);
- Po-Shen Loh's Inequality Notes (see http://www.mathlinks.ro/Forum/viewtopic.php?t=19469 ).

It is a very interesting inequality, but I hope we won't see it on a future olympiad anymore.

  Darij

[quote="darij grinberg"]$\displaystyle \frac{1+k^3}{k} \cdot \left(\frac{u}{v+kw}+\frac{v}{w+ku}+\frac{w}{u+kv}\right) \geq 3$.
[/quote]

Indeed, a beautiful problem and beautiful substitution. But I am surprised you didn't finish it up using your favorite Cauchy-Schwarz in Engel form:

$\displaystyle \frac{1+k^3}{k} \cdot \left(\frac{u}{v+kw}+\frac{v}{w+ku}+\frac{w}{u+kv}\right) \geq \frac{1+k^3}{k(k+1)} \cdot \frac{(u+v+w)^2}{uv+vw+wu}$.

The first factor is $\geq 1$ since $1+k^3\geq k^2+k$ by rearrangement.

The second factor is $\geq 3$ which is well-known.

I tried brute-force, it isn't very hard. Just grouping properly and then apply AM-GM.

But [b]harazi[/b] has a beautiful proof for a stronger version at:

http://www.mathlinks.ro/Forum/viewtopic.php?t=6044

He showed that:

$\displaystyle \left( 1+abc\right) \left(\frac{1}{a\left( 1+b\right) }+\frac{1}{b\left( 1+c\right) }+\frac{1}{c\left( 1+a\right) }\right) + 3$

$\displaystyle  = \frac{1+abc+a+ab}{a+ab}+\frac{1+abc+b+bc}{b+bc}+\frac{1+abc+c+ca}{c+ca} $

$\displaystyle = \frac{1+a}{ab+a}+\frac{b+1}{bc+b}+\frac{c+1}{ca+c} + \frac{b(c+1)}{b+1}+\frac{c(a+1)}{c+1}+\frac{a(b+1)}{a+1} $

$\displaystyle \geq\frac{3}{\sqrt[3]{abc}} + 3\sqrt[3]{abc}$ which is of course $\geq 6$.

Thanks indeed, both your and Harazi's proofs are much better than mine. Well, but to my defence I must say that what I did - with the last substitution - was essentially equivalent to proving Cauchy-Schwarz in the Engel form. And both your (Fuzzylogic)'s and my proofs can be modified to prove the stronger inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=6044 !

  Darij

you can just be courageous and try the traditional way it's very easy man!

(1+abc)(bc+ac+ab+3abc+bc+ca+ab)>=3(abc+abc(a+b+c)+abc(ab+bc+ac)+abc)
just try it it's easy !
bye

I cleared denominators and hacked it out using AM-GM.  I am horrible at fractional inequalities like that, but the whole thing comes down to:

$\displaystyle \left(\sum_{cyc} ab^2 +\sum_{cyc} a^3b^2c^2\right) +\left(\sum_{cyc} a^3bc^2 +\sum_{cyc} ab \right)-2\sum_{sym} a^2b^2c-2\sum_{sym} a^2bc\geq 0$, and the terms in brackets are easily seen to be bigger than the terms with negative signs by AM-GM

The inequality is equivalent to
 $ \sum_{cyc}\frac{abc+1}{a(1+b)}\geq 3$
 $ \sum_{cyc}(\frac{1+a}{a(1+b)}+\frac{abc+ab}{a(1+b)}) \geq 6$
  This ineq is right .( using AM_GM 3 times)

hi nttu, your solution is very nice!!
Anyway, may i know how can you think of converting (abc+1)/a(1+b) to (abc+ab)/a(1+b)+(1+a)/a(1+b) - 1 ??
Thanks!

Nice and easy. In fact, this is too easy! ;) We have
$ \frac{1+abc}{a+ab}=\frac{1+a+ab+abc}{a+ab}-1=\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}-1$.
Hence, rewrite the inequality in the form:

        $ \frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a}\geq 6$. This one follows immediately from AM-GM. ;)

Actually, this inequality can be proved using the following inequality of D.M.Pavlo which states that for any $ a,b,c>0$ the following inequality holds: $ 3+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\cdot\frac{(a+1)(b+1)(c+1)}{1+abc}$. So we only need to prove that $ \frac{1}{a+ab}+\frac{1}{b+bc}+\frac{1}{c+ca}\geq\frac{1}{{(a+1)(b+1)(c+1)}\cdot(3+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a})}$. Well, by making some calculaions here, we get $ \sum\frac{a+1)(b+1)}{c}\geq 3+\sum a+\sum\frac{1}{a}+\sum\frac{a}{b}$ which is equivalent with the following inequality:

       $ \sum\frac{ab}{c}+\sum\frac{a+b}{c}+\frac{1}{a}\geq 3+\sum a+\sum\frac{1}{a}+\sum\frac{a}{b}$ so the inequality reduces finally to this one: $ \sum\frac{ab}{c}+\sum\frac{a}{c}\geq 3+\sum a$ but from here it is easy, since we do know that $ \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq a+b+c$ and $ \frac{a}{c}+\frac{b}{a}+\frac{c}{b}\geq 3$.Both of then follow from AM-GM.  ;)

 Hope I haven't made any mistakes :?

Hello,

I'll post my attempt. It is based on computations, but it took me about 15 minutes to solve it this way.

I will also post a solution.
After clearing the denominators the ineq is equivalent to

$ ab(b+1)(ca-1)^{2}+bc(c+1)(ab-1)^{2}+ca(a+1)(bc-1)^{2}\geq 0$   which is true

[quote="X-man"]If $a, b, c \in \mathbb R$ are positive then prove \[ \frac{1}{a(b+1)} + \frac{1}{b(c+1)} + \frac{1}{c(a+1)} \ge \frac{3}{1 + abc}  \][/quote]
$\frac{1}{a(b+1)} + \frac{1}{b(c+1)} + \frac{1}{c(a+1)} \ge \frac{3}{1 + abc}\Leftrightarrow\sum_{cyc}(bc^2+bc)(ab-1)^2\geq0.$ :)

hi

i hv done the problem

clear the denominator and solve
in the end u wud be left with am gm inequality application

the process is quite long but not very tough to think

wud like to see a short proof :)

[quote=arqady][quote=Arne]Let $ a$, $ b$, $ c$ be positive real numbers. Prove the inequality
\[ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \][/quote]
It's КВАНТ 1988, problem М1136. ;)[/quote]
[b]1136. [/b]For any positive numbers A , M , S, the sum of the number 3, the numbers A , M , S and the numbers inverse to the numbers A , M , S , is not less than the tripled quotient of the product ( A + 1) ( B + 1) ( C + 1) for the sum of AMS + 1. Prove this.
http://www.kvant.info/zkm_1988.htm

[quote=Arne]Let $ a$, $ b$, $ c$ be positive real numbers. Prove the inequality
\[ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \][/quote]

Substituting ab=x^2  bc=y^2 ca=z^2 solves the problem easily

[quote=Arne]Let $ a$, $ b$, $ c$ be positive real numbers. Prove the inequality
\[ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \][/quote]
Let $a,b,c$ be positive real numbers . [url=https://artofproblemsolving.com/community/c6h1812768p12082679]Prove that[/url]$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$
[quote=sqing]Let $a_1,a_2,\cdots,a_n\geq 1$$(n\ge 3)$ . Prove that $$ \frac{1}{a_1\left(a_2+1\right)}+\frac{1}{a_2\left(a_3+1\right)}+\cdots+\frac{1}{a_{n-1}\left(a_n+1\right)}+\frac{1}{a_n\left(a_1+1\right)}\geq \frac{n}{1+a_1a_2\cdots a_n}.$$[/quote]
Let $a_1,a_2,\cdots,a_n\geq 1$$(n\ge 3)$ . Prove or disprove$$ \frac{1}{a_1a_2\left(a_2+1\right)\left(a_3+1\right)}+\frac{1}{a_2a_3\left(a_3+1\right)\left(a_4+1\right)}+\cdots+\frac{1}{a_{n-1}a_n\left(a_n+1\right)\left(a_1+1\right)}+\frac{1}{a_na_1\left(a_1+1\right)\left(a_2+1\right)}\geq \frac{n}{(1+a_1a_2\cdots a_n)^2}.$$

[quote=sqing][quote=Arne]Let $ a$, $ b$, $ c$ be positive real numbers. Prove the inequality
\[ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \][/quote]
Let $a,b,c$ be positive real numbers . [url=https://artofproblemsolving.com/community/c6h1812768p12082679]Prove that[/url]$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$
[/quote]
Let's notice that my inequality refines the Crux problem used in 2006.

[quote=sqing]Let $a_1,a_2,\cdots,a_n\geq 1$$(n\ge 3)$ . Prove that $$ \frac{1}{a_1\left(a_2+1\right)}+\frac{1}{a_2\left(a_3+1\right)}+\cdots+\frac{1}{a_{n-1}\left(a_n+1\right)}+\frac{1}{a_n\left(a_1+1\right)}\geq \frac{n}{1+a_1a_2\cdots a_n}.$$[/quote]

[b]Solution[/b]. Stipulate $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
\begin{align*}\displaystyle\sum_{i=1}^n\left(\frac{1+a_1a_2\cdots a_n}{a_i\left(a_{i+1}+1\right)}+1\right)=&\displaystyle\sum_{i=1}^n\frac{1+a_i+a_ia_{i+1}(1+a_1\cdots a_{i-1}a_{i+2}\cdots a_n)}{a_i\left(a_{i+1}+1\right)}\\
=&\displaystyle\sum_{i=1}^n\left(\frac{1+a_i}{a_i\left(a_{i+1}+1\right)}+\frac{a_{i+1}(1+a_1\cdots a_{i-1}a_{i+2}\cdots a_n)}{a_{i+1}+1}\right)\\
\left(\because a_i\ge1\,\forall i\right)\quad\ge&\displaystyle\sum_{i=1}^n\left(\frac{1+a_i}{a_i\left(a_{i+1}+1\right)}+\frac{a_{i+1}(1+a_{i+2})}{a_{i+1}+1}\right)\\
=&\displaystyle\sum_{i=1}^n\frac{1+a_i}{a_i\left(a_{i+1}+1\right)}+\sum_{i=1}^n\frac{a_{i+1}(1+a_{i+2})}{a_{i+1}+1}\\
\left(\text{by AM-GM}\right)\quad\ge&\frac{n}{\sqrt[n]{a_1a_2\cdots a_n}}+n\sqrt[n]{a_1a_2\cdots a_n}\ge2n.
\end{align*}Therefore we have $\displaystyle\sum_{i=1}^n\frac{1+a_1a_2\cdots a_n}{a_i\left(a_{i+1}+1\right)}\ge n$ thereby $\displaystyle\sum_{i=1}^n\frac{1}{a_i\left(a_{i+1}+1\right)}\ge\frac{n}{1+a_1a_2\cdots a_n}$. As required. $\blacksquare$

[quote=Arne]Let $ a$, $ b$, $ c$ be positive real numbers. Prove the inequality
\[ \frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\geq \frac{3}{1+abc}. \][/quote]

stronger is this:

Let $a,b,c>0$,prove that

\[\left( {\frac {1}{a \left( b+1 \right) }}+{\frac {1}{b \left( c+1
 \right) }}+{\frac {1}{c \left( 1+a \right) }} \right)  \left( 1+abc
 \right) \geq {\frac {27}{8}}\,{\frac { \left( b+c \right)  \left( c+a
 \right)  \left( a+b \right) }{ \left( a+b+c \right)  \left( ab+bc+ac
 \right) }}\]

[quote=sqing]Let $a_1,a_2,\cdots,a_n\geq 1$$(n\ge 3)$ . Prove that $$ \frac{1}{a_1\left(a_2+1\right)}+\frac{1}{a_2\left(a_3+1\right)}+\cdots+\frac{1}{a_{n-1}\left(a_n+1\right)}+\frac{1}{a_n\left(a_1+1\right)}\geq \frac{n}{1+a_1a_2\cdots a_n}.$$[/quote]
Let $a_1,a_2,\cdots,a_n\geq 1$$(n\ge 3)$ . Prove that $$ \frac{1}{a_1\left(a_2+n-2\right)}+\frac{1}{a_2\left(a_3+n-2\right)}+\cdots+\frac{1}{a_{n-1}\left(a_n+n-2\right)}+\frac{1}{a_n\left(a_1+n-2\right)}\geq \frac{n}{n-2+a_1a_2\cdots a_n}.$$

Let $a_1,a_2,\cdots,a_n>0$$(n\ge 2)$ . Prove that$$ \frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\geq \frac{n}{n-2+a_1a_2\cdots a_n}.$$
[size=50](Lijvzhi)[/size]

A pretty hard problem (idk why, but solutions like those in @nttu's post are almost impossible to find)

Suppose that $abc=k^3$. Then we can substitute $a=\frac x y k, b = \frac z x k, c = \frac y z k$. [color=#f00](This is a critical step. Notice how we did not apply the typical cyclic substitution.[/color] As in @dg's post, this simplifies down to showing that $\frac{k^3+1}{k}\sum_{cyc} \frac{x}{ky+z}\geq 3$. Of course, we forces a square into the numerator to get an equivalent inequality of $\frac{1+k^3}{k}\sum_{cyc}\frac{x^2}{kxy+xz}\geq 3$. Now we have from Engel that $\text{LHS}\geq \frac{1+k^3} k \cdot \frac{(x+y+z)^2}{(k+1)(xy+yz+zx)} = \frac{1+k^3}{k(k+1)}\frac{(x+y+z)^2}{xy+yz+zx}$. Notice that $k^3+1\geq k^2+k\leftrightarrow (k-1)^2(k+1)\geq  0,$ so we have that the first fraction $\geq 1,$ and it is well known that the second fraction $\geq 3,$ so we are done.

[color=#f00]In @fuzzylogic's sol, he says that[/color] $1+k^3\geq k^2+k$ [color=#f00]follows from rearrangement. How can this be possible? I thought rearrangements was only for products?[/color]

@above
Consider the tuples $(k^2,k)$ and $(k,1)$. They are equally sorted, so the result follows indeed by the rearrangement inequality.

Came here to generalize but Mr.Sqing had done the whole work  :D

Finally I have made a generalization to this problem.

The original problem is a special case of the following generalization where
$$n=1$$.

[b]Generalization 1[/b]
Let $a_{1},a_{2},\cdots,a_{3n}$ be positive reals ($n\geq 1$). Then prove that

$$\sum_{cyc}{\left(\dfrac{1}{a_{1}a_{2}\cdots a_{n}\left(a_{n+1}a_{n+2}\cdots a_{2n}+1\right)}\right)}\geq \dfrac{3n}{\prod{a_{1}}+1}$$


A candy-seller sells lollipops for $\$ 0.50$, Snickers for $\$ 3.00$, and Crunch bars for $\$ 10.00$.  You buy $100$ pieces of candy for $\$ 100$.  If you bought at least one Crunch bar, how many lollipops did you buy?

${10 C+\frac{L}{4}+3 S=100}$
${C+L+S=100}$

${L=-C-S+100}$

${\left(10 C+\frac{L}{4}+3 S=100\right)\Rightarrow (39 C+11 S=300)}$

The unique integer solution for the last equation is $(6,6)$.
We can solve this diofantine equation very very easy.

We can conclue $L = 88$

we don't need to know if he buy or not a crunch bar.

Please hide your solutions.

The reason the Crunch bar part doesn't matter is because I wrote the problem incorrectly.  Edited.

just because you wrote it wrongly isnt need to discard my solution

was your falt, not my  :D

[quote="The Zuton Force"][hide]
Let there be $C$ crunch bars, $S$ snickers, and $2L$ lollipops

$L+3S+10C=100$ and $2L+S+C=100$

If $C=6$, there are only 40 dollars left, with which a maximum of 80 pieces of candy can be bought. So, $C \in \{1,2,3,4,5 \}$

$L+3S=100-10C \implies 2L+6S=200-20C$ and $2L+S=100-C$, giving $5S=100-19C \implies 5|19C \implies 5|C \implies C=5$

$5S=100-19(5)=5 \implies S=1$

$2L+1=95 \implies \boxed{2L=94}$

[/hide][/quote][hide="Pretty much the same thing"]$L+S+C = 100$
$3S+10C+\frac{1}{2}(100-S-C) = 100$
$\frac{5}{2}S+\frac{19}{2}S = 50$
$5S+19C = 100$
$C \in \{0, 5\}$
$C \ne 0$
$C = 5$
$S = 1$
$L = 94$[/hide]


Cryptograms are puzzles in which each letter represents a certain letter in the alphabet.

For example, "abcdefd" could be a code for "physics" if a = p, b = h, c = y, d = s, e = i, and f = c.

Solve the following cryptogram:

"HZM GLAQ UOQ HG ZOWM O DYKMLB KE HG JM GLM."

Then you can try and post your own and see if people get that!

Good luck and have fun!

[quote="miyomiyo"]Maybe you could give us a hint?

I think it's
THE O--- -A- TO HAVE A ---EN- TO -E ONE

Am I on the right track?[/quote]

i think so... is the 2nd to last word be?

That was my theory as well.  However, that gives no extra information, so it could work or not.

[quote="miyomiyo"]Maybe you could give us a hint?

I think it's
THE O--- -A- TO HAVE A ---EN- -- TO -E ONE

Am I on the right track?[/quote]

You forgot one of the two letter words... that word may help you get other letters.  See if you can guess what the expression is gonna be from the letters you have here!

[quote="miyomiyo"]Maybe you could give us a hint?

I think it's
THE O--- -A- TO HAVE A ---EN- TO -E ONE

Am I on the right track?[/quote]

[hide]
HZM GLAQ UOQ HG ZOWM O DYKMLB KE HG JM GLM
THE ONLY WAY TO HAVE A FRIEND IS TO BE ONE
[/hide]

Done.

Note: I failed to see that HG = to :(
I just continue what you have done :wink:

Okay here's one:

P LAMB ALB OMEN.  OMG ALLY ALB OMEN.  OMG LAMB OMJ ALLY XAPZGL ALB OMEN.

Hint in white, size 1:  [size=9][color=white]You may have a harder time saying it that solving it![/color][/size]

P is easy to figure out. It has to be A or I. You only use it twice in the whole thing. the most common letter is o and m. E does not open a sentence much so O should be t


limit
calculus
integration
logarithms
real analysis
calculus computations

Find the limit:

$\lim_{n\to\infty}\frac{n!^\frac{1}{n}}{n}$

I think we can use integrals to evaluate it but I don't know how. Is there any other solution (without using integrals)?

Thanks a lot for help.
Nobody1111

You can use Stirling's formula:
$n! = \sqrt{ 2 \pi n} \left( \frac{n}{e} \right) ^n e^\frac{\theta (n)}{12n},0<\theta (n)<1$
which leads to $\lim_{n\to\infty}\frac{n!^\frac{1}{n}}{n} = \frac{1}{e}$

Stirling's formula gives correct solution (1/e) however I doubt if it could be used here because this formula is only an approximation of n!.  I'd prefer to evaluate it in different way. But thanks for "hint".

Nobody1111

Here's how I would approach it: start by dealing with the numerator.
$(n!)^{1/n}$ is the geometric mean of the first $n$ integers. Take its logarithm: $\ln((n!)^{1/n})=\frac1n\sum_{j=1}^n\ln j.$ Now, bring in the denominator:
$\ln\frac{(n!)^{1/n}}{n}=-\ln n+\frac1n\sum_{j=1}^n\ln j=\frac1n\sum_{j=1}^n(\ln j-\ln n)$
$=\frac1n\sum_{j=1}^n\ln \frac jn.$
This is a Riemann sum for the improper integral $\int_0^1 \ln x\, dx$

This and related methods are how you derive Stirling's formula. This problem requires only the dominant term, so we don't have to run error estimates on that integral.

We have that to find $l=\lim_{n\to\infty}\frac{n!^\frac{1}{n}}{n}$
So we have $l=\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt[n]{{n!}}}}{n}$

We know that $\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{a_n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{a_{n + 1} }}{{a_n }}$

And $\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt[n]{{n!}}}}{n}=\mathop {\lim }\limits_{n \to \infty } \frac{{(n + 1)!}}{{(n + 1)^{n + 1} }}\frac{{n^n }}{{n!}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n^n }}{{(n + 1)^n }} = \frac{1}{e}$

Ok, Thanks a lot for quick replay.
BTW I didn't know that Stirling's formula may give a correct value of n! (not an estimation)
$n! = \sqrt{ 2 \pi n} \left( \frac{n}{e} \right) ^n e^\frac{\theta (n)}{12n},0<\theta (n)<1$
Interesting.

Thanks a lot
Nobody1111

The "exact" Stirling formula can be obtained by the Mean-value-theorem for the Gamma-function.

Misha

Another take, very closely related to jmerry's post but with some inequalities.

Let $y_n=\frac{(n!)^{\frac1n}}n$. Then $\ln(y_n)=\frac1n\sum_{j=1}^n\ln j-\ln n.$

$\int_1^n\ln x\,dx < \sum_{j=1}^n\ln j < \int_1^{n+1}\ln x\,dx$

(Draw pictures; note that $\ln 1=0$ is useful.)

$n\ln n-n+1<\sum_{j=1}^n\ln j <(n+1)\ln(n+1)-(n+1)+1$

$\ln n - 1 + \frac1n < \frac1n\sum_{j=1}^n\ln j < \ln(n+1) + \frac1n\ln(n+1)-1$

$-1+\frac1n < \ln(y_n) < \ln\left(\frac{n+1}n\right) + \frac1n\ln(n+1)-1.$

The result follows.


function
induction
vector
algebra unsolved
algebra

Prove if there exist a function $f: C \rightarrow R$ such that
1. if $f(x)=f(y)$ then $x=y$
2. $f(x)+f(y) = f(x+y)$

Yes, I think such a function exists,  you need axiom of choice to construct it.

First we prove that f is linear over $\mathbb{Q}$:
$f(0)+f(0)=f(0) \Rightarrow f(0)=0$
$f(x)+f(-x) = 0 \Rightarrow f(-x)=f(x)$
$f((n+1)a) = f(na)+f(a) = nf(a)+f(a) = (n+1)f(a)$
by induction, $f(na) = nf(a) \forall n \in \mathbb{Z}$
$f(a) = f(n \frac{a}{n}) = nf(\frac{a}{n}) = \Rightarrow f(\frac{a}{n}) = \frac{f(a)}n$
$f(ka) = kf(a) \forall k \in \mathbb{Q}$
$f(q_{1}a_{1}+\ldots+q_{n}a_{n}) = q_{1}f(a_{1})+\ldots+q_{n}f(a_{n})$
for $q_{1},\ldots,q_{n}\in \mathbb{Q}$, by induction.

Let $B$ a basis of the vector space R over Q (a Hamel basis).
Then $B \cup iB$ is a basis of C over Q. In fact, every complex number a+ib can be written as a linear combination of the elements of $B$, it's obvius. If a linear combination of the elements of $B \cup iB$ is 0, then the linear combination of all real elements is 0, then all the coefficients are 0, and the same for the imaginary elements. Then this set is linearly indipendent.

Since $B$ is infinite, we have $|B|+|B| = |B|$, then exists a biiection $f: B \cup iB \rightarrow B$. Extend $f$ linearly over whole $\mathbb{C}$.

This function should work. If $f(a) = f(b)$ then $f(a-b)=0$. $a-b$ is a linear combination of the elements of $B \cup iB$. $f(a-b)$ is a linear combination of $\{f(b): b \in B \cup iB\}$, then a linear combination of $B$. Since B is a basis, $a-b$ must be 0. Then $f$ is injective.


Να αποδειχθεί ότι για όλους τους πραγματικούς αριθμούς $ x_1,x_2,y_1,y_2,z_1,z_2$  με  $ x_1,x_2 > 0$ και $ x_1y_1 > z_1^2 , x_2y_2 > z_2^2$  η ανισότητα

$ \dfrac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2}\leq{\dfrac{1}{x_1y_1 - z_1^2} + \dfrac{1}{x_2y_2 - z_2^2}}$  ικανοποίειται και να βρεθεί πότε ισχύει η ισότητα

file kritike sto RHS mipws ennoeis + anti gia -.
giati an einai opws exei to gia (x1,x2,y1,y2)=(1,2,2,1) kai z1=z2=1
pernume 8/5<=0...
an ennoeis + pantws exw vrei mia lysi. :)  perimenw apantisi!

File Athinaie to dior8wsa eixes dikio htan + anti gia - . Suggnwmh gia opoiodhpote problhma !


calculus
integration
trigonometry
function
calculus computations

Evaluate $I = \int^{\pi}_{0}x\sin x \cos nx\,dx,\;n = 0,1,2,3,...$.

[hide="My solution"]
Let's start by finding the primitives of the functions xsin(x) and xcos(x):

$\int x\sin x\,dx = \sin x-x\cos x$ and $\int x\cos x\,dx = x\sin x+\cos x$.

Let's now integrate the function xsin(x)cos(nx) by parts:

$P = \int x\sin x \cos(nx)\,dx = (\sin x-x\cos x)\cos(nx)+n\int (\sin x-x\cos x)\sin(nx)\,dx$
$= \sin x \cos(nx)-x\cos x \cos(nx)+n\int \sin x \sin(nx)\,dx-n\int x\cos x \sin(nx)\,dx$
$= \sin x \cos(nx)-x\cos x \cos(nx)-n\cos x \sin(nx)+n^{2}\int \cos x \cos(nx)\,dx-n[(x\sin x+\cos x)\sin(nx)-n\int (x\sin x+\cos x)\cos(nx)\,dx]$
$\Rightarrow P = \sin x \cos(nx)-x\cos x \cos(nx)-2n\cos x \sin(nx)-nx\sin x \sin(nx)+n^{2}P+2n^{2}\int \cos x \cos(nx)\,dx$

and so

$(1-n^{2})I = \left[\sin x \cos(nx)-x\cos x \cos(nx)-2n\cos x \sin(nx)-nx\sin x \sin(nx)\right]^{\pi}_{0}+2n^{2}\int^{\pi}_{0}\cos x \cos(nx)\,dx$.

This last integral is zero, because

$\int^{\pi}_{0}\cos x \cos(nx)\,dx = \frac{1}{2}\int^{\pi}_{0}[\cos[(n-1)x]+\cos[(n+1)x]]\,dx = \left[\frac{\sin[(n-1)x]}{2(n-1)}+\frac{\sin[(n+1)x]}{2(n+1)}\right]^{\pi}_{0}= 0$.

The final conclusion is that $I = \frac{\pi\cos(n\pi)}{1-n^{2}}= \frac{(-1)^{n}\pi}{1-n^{2}}$.[/hide]

Let $x=\pi-\theta , I=\cos n\pi \int_{0}^{\pi}(\pi-\theta)\sin \theta \cos \theta d\theta$. From this you can use integral by parts.


calculus
integration
function
inequalities
calculus computations

[u]So that you don't have to download:[/u]

Suppose $ f(x)$ is continuous for $ x \in [a,b]$. Let $ h$ be a positive parameter. Suppose that for all $ (x,h)$ such that $ [x\minus{}h,x\plus{}h] \in [a,b]$, the following inequality holds true:

$ f(x) \leq \frac{1}{2h} \int_{\minus{}h}^{h} f(x\plus{}t) dt$

Show that $ f(x)$ is a convex function over $ [a,b]$

[hide="Hint"]In words, it says that the average value f over an interval of x is greater than the value of f at the average value of x on this interval.

What does this sound like?[/hide]


[quote="snagg"]$\frac{2\displaystyle\sum^{i=0}_{n=i}10^i}{10^n}$[/quote]
Could you please tell us how do you get it and simplify your answers next time and every time?

Moreover, I don't think your answer is correct.

Would this converge to $\frac{200}{891}=.22446689\dots$?

[quote="hsiljak"][quote="surge"]$\displaystyle\sum_{n=1}^{\infty}\frac{2 \cdot 10^n + 2 \cdot 10^{n-1}...2 \cdot 10^0}{10 \cdot (100^{n-1})}$ ?[/quote]

This seems correct, but there is a small error, I think... I would write it this way

$\displaystyle\sum_{n=0}^{\infty}\frac{2 \cdot 10^n + 2 \cdot 10^{n-1}...2 \cdot 10^0}{10 \cdot (100^{n})}$[/quote]

yes, you're right.

[quote="hsiljak"][quote="snagg"]could you explain why my answer is incorrect?[/quote]

Well, can you write it in another way? This form in which you had put your solution is confusing: Are you using i or n as the "counting variable"?[/quote]
I'm sorry $\frac{2\displaystyle\sum^{+\infty}_{n=0}10^n}{100^n}$

[quote="amcavoy"]That converges to $\frac{20}{9}$, which isn't the original number that was posted.[/quote]

It should be correct...

I agree with amcavoy: 200/891 .

Your method?

[quote="amcavoy"]I just evaluated the series that was given.[/quote]

Can you explain it?

FWIW: Same as above, but perhaps a little easier to understand:
[hide]$S = \ \ \ \ 0.2 + 0.022+ 0.00222 +\dots$
$(.01)S= \ \ \ + 0.002 + 0.00022 + \dots$
so Subtracting
$.99 S = .2222\dots = 2/9$ which gives $S = 200/891$ [/hide]
Also Snagg
[hide]Yours:[quote]$\frac{2\displaystyle\sum^{+\infty}_{n=0}10^n}{100^n}$[/quote] Still makes no sense, as to where/what  you get  $100^n$  from?   (  $n$  out side sigma makes no sense.  [/hide]

[quote="amcavoy"]It just requires a small manipulation:

$\frac{1}{5}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{10^{k}}{100^{n}}=\frac{1}{5}\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{10^{-j}}{100^{i}}=\left(\frac{1}{5}\right)\left(\frac{10}{9}\right)\left(\frac{100}{99}\right)=\boxed{\frac{200}{891}}$

Notice the identity $\sum_{m=0}^{\infty}\sum_{n=0}^{m}a_{n,m-n}=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_{j,i}$.[/quote]

Thanks now

[quote="Yooper"]I agree with amcavoy: 200/891 .

[/quote]
I do too.

How do I know if a sequence is convergent or divergent????


Let $A, B, C, D, E$ be integers, $B \neq 0$ and $F=AD^{2}-BCD+B^{2}E \neq 0$. Prove that the number $N$ of pairs of integers $(x, y)$ such that \[Ax^{2}+Bxy+Cx+Dy+E=0,\] satisfies $N \le 2 d( \vert F \vert )$, where $d(n)$ denotes the number of positive divisors of positive integer $n$.

[quote="Peter"]Let $ A, B, C, D, E$ be integers, $ B \neq 0$ and $ F = AD^{2} - BCD + B^{2}E \neq 0$. Prove that the number $ N$ of pairs of integers $ (x, y)$ such that
\[ Ax^{2} + Bxy + Cx + Dy + E = 0,
\]
satisfies $ N \le 2 d( \vert F \vert )$, where $ d(n)$ denotes the number of positive divisors of positive integer $ n$.[/quote]
$ \Leftrightarrow Bx+D|Ax^2+Cx+E$
$ \Rightarrow Bx+D|ABx^2+BCx+BE$
$ \Rightarrow Bx+D|(Ax+C)(Bx+D)-(ADx+CD-BE)$
$ \Rightarrow Bx+D|ADx+CD-BE$
$ \Rightarrow Bx+D|ABDx+BCD-B^2E$
$ \Rightarrow Bx+D|AD(Bx+D)-AD^2+BCD-B^2E$
$ \Rightarrow Bx+D|AD^2-BCD+B^2E=F$
Imply that $ Bx+D|F$
So has at most $ 2d(|F|)$ satisfy the equation.

To be ahead of the questions: from his second last line, you have at most $ d(|F|)$ solutions for $ x$, and since the equation is quadratic, at most two $ y$ per $ x$, so at most $ 2d(|F|)$ pairs indeed.

but cant bx+d be a negative divisor as well for negative values of x? so wouldnt the answer be 4d(|F|)?


The ratio of boys to girls in Leland middle school is 8:9. How many girls are in the school if there are 731 students?

If the ratio of boys to girls is 8:9, then that means that their are 17 'parts'. If you divide 731 by 17, then you get 43, so that equals one of the 17 'parts'. So, if there are 9 'parts' for the girls you do $ 43*9$ to get the answer of 387. :)

Let $ g$ be the number of girls in the school. From the given info, $ g \plus{} \frac {8}{9}g \equal{} 731$ So $ \frac {17}{9}g \equal{} 731$ Multiplying both sides by $ \frac {9}{17}$ gives $ 387$

@origamiguy777: I don't think you are clear enough with the 'parts'

i THINK i know a faster way

9 if for the girls for the in the ratio

$\frac{8}{9}$ $=$ $\frac{boy}{girl}$

so the total is $\boxed{17}$ just for this ! (remember this is just the ratio !)
so the actual total is 731

so we get the equattion

$\frac{9}{17}$ $=$ $\frac{731}$

cross multiply we get the answer $\boxed{387}$


http://timelines.ws/countries/CHINA_2000_2030.HTML

2001  Apr 1, A US Navy EP-3 surveillance plane with 24 aboard collided with a Chinese fighter jet over the [color=red]South China Sea[/color] and was forced to land on China's Hainan island. The fighter jet crashed. Chinese pilot Wang Wei parachuted out of his F-8 jet but had not been found. Zhao Yu, a 2nd pilot, later blamed the US plane banked and hit Wei’s plane.

2001  Aug 9, It was reported that the US had decided to pay China [color=red]USD34,567  [/color]to cover the costs of the spy plane that was detained on Hainan island. China had asked for USD1 million and rejected the offer.

I want to listen to the opinion from the third side, so American and Chinese please don't reply.

Oy, not more of the US conspiracy to wipe out China.

Although I see that we seemed to have picked up some creationists and some religious fundamentalists, so I suppose you are in good company.  Enjoy!

Someone needs a little bit courage to admit after commit.


The product of two positive consecutive integers is 506. What is their sum?

we know 506 is greater than 20 squared, and less than 25 squared.
So you can deduce that 506=(22)(23)
22+23=[b][i][size=150]45[/size][/i][/b]

Instead of using guess and check to arrive at $ 22 \times 23$, you can use the units digit to determine it (I assume you use guess and check, as you did not explain your method).


probability
integration
probability and stats

An interval of length 1 is broken at a point uniformly distributed оver $ (0,1)$. Find the expected length of the subinterval that contains the point $ 0 < x < 1$. Show that it is maximized when $ x \equal{} 1/2$.

Let $ L$ be the length we're looking for. Let $ y$ be the point at which the interval is broken.

Then: $ L\equal{}\begin{cases}y&\text{if }y>x\\1\minus{}y&\text{if }y<x\end{cases}.$

(It's not entirely clear what $ L$ should be if $ x\equal{}y,$ but since that is an event of probability zero, it really doesn't matter what we say.)

So we can compute $ E(L\,|\,x)\equal{}\int_0^x(1\minus{}y)\,dy\plus{}\int_x^1y\,dy\equal{}\minus{}x^2\plus{}x\plus{}\frac12.$

Rewrite this as $ E(L\,|\,x)\equal{}\frac34\minus{}\left(x\minus{}\frac12\right)^2$ to make it clear that it has a maximum at $ x\equal{}\frac12$ and that maximum value is $ \frac34.$

You can then integrate that over $ x$ to get that

$ E(L)\equal{}E(E(L\,|\,x))\equal{}\int_0^1\left(\minus{}x^2\plus{}x\plus{}\frac12\right)\,dx\equal{}\frac23.$


function
linear algebra
linear algebra unsolved

suppose $V$ is a linear space and $f: V \mapsto V$
we know that $\dim V \geq 2 \dim f(V)$
prove that $f$ is a composition of two projections

is there anyone who knows how to do it??


induction
modular arithmetic
number theory proposed
number theory

Prove that for any positive integer k, there is a positive integer n such that

2k|(3m + 5).

It's kind of messy, but here it goes:

We see that the n's we're looking for will eventually have to be odd, because for k>=2 we have 4|3^n+5. It's enough to prove that any power of 2 | a number of the form (3^n+5)/4 = (3^n+1)/4+1 = 3^(n-1)-3^(n-2)+3^(n-3)-..+3^2-3+1+1. We group 3^(n-1) and 3^(n-2); 3^(n-3) and 3^(n-4) etc. We get 2(1+3+3^3+3^5+..+3^t), with t odd.

The new problem is proving that any power of 2 divides a number of the form a_t=1+3+3^3+3^5+..+3^t, with t odd. Let's assume 2^k | a_t but 2^(k+1) doesn't divide a_t. We should find n s.t. 2^k | a_n-a_t but 2^(k+1) doesn't divide a_n-a_t. But a_n-a_t=3^(t+2)(1+3^2+..+3^2x) for some positive integer x, so the new problem is this:

For each k there is an x s.t.  2^k | 3^x-1 but 2^(k+1) doesn't divide 3^x-1. We choose x to be 2^(k-2) and the property we want can be proved by induction after k. I hope you have a nicer one. Mine isn't hard, but it looks nasty.

No, I don't have a neater solution - that's why I posted it :)

I still believe that there must be a neat solution :)

that depends on what do you mean (understand) by neat :D

Neat means, according to me, elegant and short :)

Do you consider that proof REALLY ugly Arne? :D It's not that bad, is it? God knows I've seen much worse! :D

It's not too bad [it is my own solution also] :) but I would really like to see a 5 lines - solution :)

Note $2^{3}\mid (3+5)$ and admit there exists $m \in \mathbb{N}$ such that $2^{k}\mid (3^{m}+5)$ and $k \ge 3$. Then let $q = v_{2}(3^{m}+5)$, where $v_{2}(\cdot)$ is a 2-adic valuation. As $q \ge k$, this means $3^{m+n}+5 = 3^{m}({3^{n}}-1)+3^{m}+5$ is divisible at least by $2^{k+1}$, when choosing $n = 2^{q-2}$. The conclusion follows by induction.

Maybe this solution is neat:

$k=1$ and $k=2$ follow by inspection, so we will only discuss $k\geq 3$. The statement is also true for $k=3$ (we will need this for the sake of induction).
Since $ord_{2^{k}}(3)=\lambda(2^{k})=2^{k-2}$ (is this well known? If not, then you can easily prove this by induction), the numbers $3^{1},3^{2},\ldots,3^{2^{k-2}}$ all give different residues modulo $2^{k}$. However, since $ord_{2^{k+1}}(3)=\lambda(2^{k+1})=2^{k-1}>2^{k-2}$, we have $3^{i+2^{k-2}}\not\equiv 3^{i}\pmod{2^{k+1}}$ and thus $3^{i+2^{k-2}}\equiv 3^{i}+2^{k}\pmod{2^{k+1}}$. Hence, if $3^{i}\equiv-5\pmod{2^{k}}$, then either $3^{i}\equiv-5\pmod{2^{k+1}}$ or $3^{i+2^{k-2}}\equiv-5\pmod{2^{k+1}}$.


Five toppings for a pizza are available at Polina's Pizza. How
many combinations of two different toppings are possible?

There are 5C2=10 ways to choose 2 toppings


calculus
analytic geometry
integration
geometry
rotation
limit
geometric transformation

Let R be the region enclosed by the graphs of $ y\equal{}e^{\minus{}x}$, $ x\equal{}k {(k>0)}$, and the coordinate axes (first quadrant).

a. write an improper integral that represents the limit of the area of the region R as k increases without bound and find the value of the integral if it exists.
b. find the volume, in terms of k, of the solid generated if R is rotated about the y-axis.
c. find the volume, in terms of k, of the solid whose base is R and whose cross sections, perpendicular to the x-axis, are squares.

Thank you.

(Just a request: if you choose to answer this problem, please answer all the parts.)

$ a.)$

Take the limit as $ k$ approaches infinity of the integral from the $ y$-axis to $ x \equal{} k$ of $ e^{ \minus{} x} \minus{} 0$.

$ A \equal{} \lim_{k \to \infty} \int_0^k e^{ \minus{} x}dx \equal{} \lim_{k \to \infty} \minus{} e^{ \minus{} x}|_0^k \equal{} \lim_{k \to \infty} \minus{} e^{ \minus{} k} \plus{} 1 \equal{} 0 \plus{} 1 \equal{} 1$

$ b.)$

Using the shell method (the solid is formed through rotation about the $ y$-axis):

$ V \equal{} \lim_{k \to \infty} \int_0^k 2\pi x e^{ \minus{} x}dx \equal{} \lim_{k \to \infty} 2\pi ( \minus{} xe^{ \minus{} x} \minus{} e^{ \minus{} x}|_0^k)$
$ V \equal{} 2\pi \lim_{k \to \infty} ( \minus{} ke^{ \minus{} k} \minus{} e^{ \minus{} k} \minus{} (0 \minus{} e^{ \minus{} 0})) \equal{} 2\pi (0 \plus{} 0 \plus{} 1) \equal{} 2\pi$

$ c.)$

The area $ A(x)$ of each cross section is $ (e^{ \minus{} x} \minus{} 0)(e^{ \minus{} x} \minus{} 0)$.

$ V \equal{} \lim_{k \to \infty} \int_0^k A(x) dx \equal{} \lim_{k \to \infty} \int_0^k e^{ \minus{} 2x} dx \equal{} \lim_{k \to \infty} \minus{} \frac {1}{2}e^{ \minus{} 2x}|_0^k$
$ V \equal{} \lim_{k \to \infty} \minus{} \frac {1}{2}e^{ \minus{} 2k} \plus{} \frac {1}{2} \equal{} 0 \plus{} \frac {1}{2} \equal{} \frac {1}{2}$

[quote="Zakary"]$ \lim_{k\to\infty}\int_0^k e^{\minus{}x}\,dx$[/quote]
In the standard notation, the improper integral $ \int_0^{\infty} e^{\minus{}x}\,dx$ is defined to be this limit.

The other parts are similar.


Let $ABCD$ be a circumscribable quadrilateral and let $\omega$ be its incircle, which is tangent to $AD$ and $CD$ at $P$ and $Q$, respectively. Let $X$ and $Y$ be the points of intersection of $\omega$ and $BD$. If $M$ is the midpoint of $XY$, prove that $\angle AMP=\angle CMQ$.

See the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=113117]IbMO Day 2, Prob 5[/url].


[quote="math92"]Ther are six tags numbered 1, 2, 2, 3, 3, 4. using these tags, how many distinct three-digit numbers can be formed such that two of the digits are the same?[/quote]
[hide]
Since two of the digits must be same, we have that there must be $22?$. $?22$. $2?2$, and same for 3.

Here are 3 different possibilities for the $?$ so the total number is $3*6=18$[/hide]


geometry
analytic geometry
graphing lines
slope
trigonometry

$ H$ is the point of intersection of the altitudes of $ \triangle ABC$. If $ |AB|\equal{}8\sqrt 3$, $ |AC|\equal{}5\sqrt 3$ and $ |AH|\equal{}7$, then find all possible values of $ |BC|$.

If $ H \in [BC]$,
[asy]import geometry;
pair A=(0,7), B=(-sqrt(143), 0), C=(sqrt(26), 0), H=(0,0);
draw (A--B--C--cycle);
draw(A--H, dashed);
perpendicular(H, SW, A--H);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$H$", H, S);[/asy]

If $ H \not\in [BC]$,
[asy]import geometry;
pair A=(0,7), B=(-sqrt(143), 0), C=(-sqrt(26), 0), H=(0, 0);
draw(A--B--C--cycle);
draw(A--H^^C--H, dashed);
perpendicular(H, SW, A--H);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$H$", H, SE);[/asy]

$ H$ is not the altitude from $ A$ to $ BC$. $ H$ is the orthocenter.

[hide="Coordinate Bash..."]Let $ A\equal{}(0,0)$, $ B\equal{}(\sqrt{192\minus{}c^2},c)$, and $ C\equal{}(\pm\sqrt{75\minus{}c^2},c)$ so that $ AB\equal{}8\sqrt3$ and $ AC\equal{}5\sqrt3$, and this setup encompasses all possibilities.

Line $ AC$ has slope $ \pm\frac c{\sqrt{75\minus{}c^2}}$, so line $ BH$ has slope $ \pm\frac{\sqrt{75\minus{}c^2}}c$, and thus $ BH$ is $ y\minus{}c\equal{}\pm\frac{\sqrt{75\minus{}c^2}}c(x\minus{}\sqrt{192\minus{}c^2})$.

When $ x\equal{}0$, we get $ H\equal{}\left(0,c\pm\frac{\sqrt{(75\minus{}c^2)(192\minus{}c^2)}}c\right)$, so

$ AH\equal{}7\equal{}c\pm\frac{\sqrt{(75\minus{}c^2)(192\minus{}c^2)}}c\implies c^2(7\minus{}c)^2\equal{}(c^2\minus{}75)(c^2\minus{}192)$

$ \implies c^4\minus{}14c^3\plus{}49c^2\equal{}c^4\minus{}267c^2\plus{}14400$

$ \implies7c^3\minus{}158c^2\plus{}7200\equal{}(c\minus{}20)(c\plus{}6)(7c\minus{}60)\equal{}0$.

$ c\equal{}20$ makes the square roots nonreal.

Otherwise, we get the solutions $ B\equal{}(2\sqrt{39},\minus{}6),C\equal{}(\pm\sqrt{39},\minus{}6)$, and $ B\equal{}\left(\frac{44\sqrt3}7,\frac{60}7\right),C\equal{}\left(\pm\frac{5\sqrt3}7,\frac{60}7\right)$, which give

$ \boxed{BC\equal{}\left\{\sqrt{39},3\sqrt{39},\frac{39\sqrt3}7,7\sqrt3\right\}}$.[/hide]

[quote="math154"]$ H$ is not the altitude from $ A$ to $ BC$. $ H$ is the orthocenter.[/quote]

Sorry.
I seem to have misunderstood.
Thank you for pointing out my error.

[quote]$ \boxed{BC\equal{}\left\{\sqrt{39},3\sqrt{39},\frac{39\sqrt3}{7},7\sqrt3\right\}}$[/quote]

i tried to draw the triangle $ 8\sqrt 3, 5\sqrt 3, \sqrt {39}$, but i found $ AH \not \equal{} 7$.

Indeed, the orthocenter for $ A \equal{} (0,0)$, $ B \equal{} (2\sqrt {39}, \minus{} 6)$, $ C \equal{} (\sqrt {39}, \minus{} 6)$ is $ (0, \minus{} 19)$, which would make $ AH \equal{} 19$.

[b]Edit:[/b] OK, I checked on a program and $ A \equal{} (0,0)$, $ B \equal{} \left(\frac {44\sqrt3}7,\frac {60}7\right),C \equal{} \left(\frac {5\sqrt3}7,\frac {60}7\right)$ does not work either. The other two do, however, so the answers are $ 7\sqrt3$ and $ 3\sqrt {39}$.

acute angle
[geogebra]fdb5f90749aae6e1d37304c3586adfcc546e1168[/geogebra]

obtuse angle
[geogebra]df92a0980e69ae2c65ba814ad705c99d16580e65[/geogebra]

i) Asume $ \angle BAC \equal{} \alpha < 90^\circ$ and the triangle is acute. Draw parallel lines to $ AH$ and $ CH$ from $ C$ and $ A$, respectively. Let $ D$ be the point of intersection of these two lines. Obviously, $ \angle BAD \equal{} BCD \equal{} 90^\circ$. So $ BD$ is the diameter of  $ (ABC)$. $ \angle ABC \equal{} \angle BAC \equal{} \alpha \Rightarrow cos\alpha \equal{} \frac {7}{2R}$. And $ x^2 \plus{} 49 \equal{} 4R^2$

Use cosine theorem in $ \triangle ABC$ for $ \alpha$. $ AB^2 \plus{} AC^2 \minus{} 2\cdot AB \cdot AC \cdot cos\alpha \equal{} BC^2$.
$ 192 \plus{} 75 \minus{} 240cos\alpha \equal{} x^2 \Rightarrow 267 \minus{} 240\cdot \frac {7}{2R} \equal{} 4R^2 \minus{} 49$. Simplify the last equation as $ 79R \minus{} 210 \equal{} R^3$ where $ R\in \{7, \minus{} 10,3\}$. $ R \equal{} 3$ degenerates the right triangle, $ R \equal{} \minus{} 10$ is negative (but we will see it later), $ R \equal{} 7$ is the only solution. Thus $ \boxed {x \equal{} 7\sqrt 3}$. Also $ \alpha \equal{} 60^\circ$.

ii) Now assume $ \angle BAC \equal{} \alpha > 90^\circ$  (You can follow it from geogebras.). Similar steps as previous. But now from $ \triangle BDC$, we get $ cos(180 \minus{} \alpha) \equal{} \frac {7}{2R} \Rightarrow cos\alpha \equal{} \frac { \minus{} 7}{2R}$. Same pythagoras $ x^2 \plus{} 49 \equal{} 4R^2$. Thus the cosine formula equation is $ 79R \plus{} 210 \equal{} R^3$ where $ R\in \{10, \minus{} 7, \minus{} 3\}$. So we get $ R \equal{} 10$ and $ \boxed {x \equal{} 3\sqrt {39}}$.

iii) The third case which is $ \alpha < 90^\circ$ but $ \triangle ABC$ is obtuse is meaningless. Because $ \angle ACH > 90^\circ \Leftrightarrow AH > AC$ which is not.

So the solution set is $ \boxed  { \{7\sqrt 3 , 3\sqrt {39}\}}$


Is there any simple way to simplify this? I want this in a (wa + xb)(ya+zb) format.

[hide="Does this work for you?"] $a^{2}+kab+b^{2}= \left(\frac{k+\sqrt{k^{2}-4}}{2}a+b\right)\left(\frac{k-\sqrt{k^{2}-4}}{2}a+b\right)$[/hide]

[quote="pakagawa"]Thanks.

It does (doesn't that EXACTLY fit my format?)[/quote]

Well one problem is that it gives you a complex factorization for $k < 2$, but I don't know that that can be avoided.


When the two-digit integer $ MM$, with equal digits, is multiplied by the one-digit integer $ M$, the result is the three-digit integer $ NPM$. What is the greatest possible value of $ NPM$?

[asy]unitsize(0.4cm);
draw((-2.7,-3.3)--(3.2,-3.3),linewidth(0.7));

label("$M$",(0.5,0),S);
label("$M$",(2.5,0),S);

label("$M$",(2.5,-1.5),S);

label("$N$",(-1.5,-4),S);
label("$P$",(0.5,-4),S);
label("$M$",(2.5,-4),S);
label("$\ast$",(-2.3,-0.7),S);[/asy]

We want to find $ M$.

The units digit of $ M^2$ is $ M$, since the answer is $ NPM$.

$ (10M \plus{} M) * M \equal{} 10M^2 \plus{} M^2 \equal{} 11M^2.$

Which ($ M^2 * 11$) 's unit digit $ \equal{} M$?
After testing each integer starting from 10 and going down, we find that 6 is the [b]greatest[/b] digit with that property.

Testing, 6 * 66 = 396.

Thus, $ M \equal{} \boxed{6}$

They actually wanted the greatest value of NPM, so your test should be the correct answer: 396.

why does it have to have that property? it never says it does in the question


How would you know which organic compounds would have the highest boiling point such as the example below?

A    CH3CH2OH

B   CH3OH

C   CH4

D   H2C (double bonded to)  CH2

E  HC (triple bonded to)  CH

I think it's quite obvious that ethanol has the highest boling point from the substances above.Alchools in general have high boiling points because their molecules are associated in hydrogen bonds.And between two alchools the one with the higher molecular mass has the highest boiling point..... :roll:

Remember that a hydrogen-bond occurs (in general) when a hydrogen atom is bonded to a flourine, oxygen, or nitrogen atom.  Yes, there are other cases, but these are the most important.  The hydrogen atom's electrons are pulled away by the electronegative atom (N, O, or F), and so the hydrogen atom has a partial positive charge.  Likewise, the other atom has a partial negative charge.  Thus, the hydrogen of one molecule is attracted to the electronegative atom of another molecule, and thus the two molecules experience an attraction.  Higher attractions mean higher boiling points.

The quick answer is that I know:

A) is ethanol,
B) is methanol,
C) is methane (natural gas),
D) is ethylene (used to make polyethylene, a common plastic) and
E) is acetylene (used for gas welding)

Since C, D and E are gases at room tempeature and I know that the boiling point of ethanol is higher than methanol, the answer is ethanol.

However, someone who doesn't know that can figure this out with the general rules that inteluster are basically correct.  However, it is better to think of the electron probability distribution is moved.  This type of bond is called a hydrogen bond, since the (partially) naked proton has the ability to form such a bond that raises the boiling point (It is also what causes solid water (ice) to be less dense as the increased cross linking by hydrogen bonds as the thermal energy lessens increases the average molecular space.)

The highly electronegative atom mentioned by inteluster can also be provided by a benzene or pyridine ring.

I'm not sure I understand - are you saying benzene experiences hydrogen bonding?

Sorry, I wasn't very clear.  It's a complex and minor point.  I probably should have resisted the temptation to mention it.  I should have at least devoted sufficient space to fully lay out the concept.

If the highly electornegative atom (normally) oxygen is linked, through a single bond, I believe, to a benzene or pyridine ring, the resonance with the electrons in the ring give the highly electronegative atom a greater source of electons, generating a stronger hydrogen bond.  This may allow a borderline highly electronegative atom cross the threshold and form a bond with a [naked] proton strong enought to allow it to be considered a hydrogen bond, where it could not absent the ring source of electrons.

In chemical production this won't be an issue, it probably won't even in most labs, unless an experiement is set up to illustrate the point.

Oh, OK.  That makes sense now.  Don't apologize for mentioning it; it's good to know.  Who knows?  It might show up on this year's olympiad!


function
real analysis
real analysis unsolved

Do you know an example of a function $ f: ] \minus{} 1,1[\rightarrow \mathbb{R}$, such that $ f$ is increasing and has infinitely many discontinuity points  ?

edited

I don't understand your notation and I don't know what continuities of the "first kind" are, but if you consider the step-function (greatest integer value function), it increases constantly and has a discontinuity at every integer.

Here's the standard example.

Let $ r_k$ be an enumeration of the rationals in $ (\minus{}1,1).$ For each $ x,$ let $ A_x\equal{}\{k: r_k<x\}.$ Define $ f(x)\equal{}\sum_{k\in A_x}2^{\minus{}k}.$

This function is strictly increasing, has jump discontinuities at every rational, and is continuous at every irrational.


logarithms
ratio
calculus
calculus computations

Which of the following sequences converge as n tends to infinity (justify the answer)?

$ x_{n}\equal{}3n/(n\plus{}3)$

$ x_{n}\equal{}n^{100}/2^{n}$

$ x_{n}\equal{}\sqrt{n\plus{}1}\minus{}\sqrt{n}$

$ x_{n}\equal{}(n!)^{\frac{1}{n}}$

1) The idea that $ \frac{1}{n} \to 0$  will help.

2) Try using Squeeze theorem.

3) Take conjugate. This will make you clear.

4) I am not sure..

4) Take a natural logarithm : $ \frac {1}{n}\sum_{k \equal{} 1}^n \ln k$

[quote="DineshManohar"]
3) Take conjugate. This will make you clear.
[/quote]

What do you mean by saying "Take conjugate"?

I meant this...$ \frac{\sqrt{n\plus{}1}\minus{}\sqrt{n} ) (\sqrt{n\plus{}1}\plus{}\sqrt{n}) }{ \sqrt{n\plus{}1}\plus{}\sqrt{n} }$

For (iv) I considered the ratios $ \frac{x_{4n}}{x_{2n}} > \frac{x_{2n}}{x_{n}}$ and the fact that $ \left( a_n \right)_{n\equal{}1}^{\infty}$ is monotonic increasing. This shows that the sequence increases without limit.


calculus
integration
combinatorics unsolved
combinatorics

Let $ f$ be a map from the plane $ R^{2}$ into itself. For $ d \in R \plus{}$ we say that $ f$ preserves the distance $ d$ if
we have $ AB \equal{} d \Rightarrow f(A)f(B) \equal{} d$for every pair of points $ (A,B)$. We suppose that $ f$ preserves the
distance $ 1$ and is injective wich means that $ A \neq B \Rightarrow f(A) \neq f(B)$.
(a) Prove that $ f$ preserves the distance $ 2$ and then every integral distance.
(b) Prove that $ f$ preserves the distance $ \sqrt {3}$

details please.Thanks
 :P

You should probably look at some hexagons and their centers.


If I use geo_dev, I find this message: [b]error: could not load module 'geo_header_dev'.[/b]
Also if I use geometry_dev I find this message: [b]C:\Program Files\Asymptote/geometry_dev.asy: 1322.34: no matching variable 'filltype.fill'[/b]
What should I do ?

Hello,

1. You should NOT use geo_dev at all.
2. Use the latest version of geometry_dev with the latest version of Asymptote.