geometry Support inequalities Harvard college Stanford

[quote="Idiot_without_a_village"]Tiankai Liu is the shiznit, but I mean, why bother taking the B if you already have a perfect on the A. [/quote]

I'm not sure, but I think Stuy signs up for both early enough that you're signed up for the B before the A.  Either way, I took the B after perfecting the A, and I also think that I hold the individual record for greatest decrease in scores between the A and B exams in the same year :)

[quote="Idiot_without_a_village"]*shrug* only two indians got perfects this year. Oh well, the indians will rise to power eventually. A special holla to mah indian homies.

And only two perfects in CA? Man, musta been some biznitch of a test.[/quote]

uh chinese have already risen to power and will continue to hold it. and before any indians will rise Mystic Terminator needs to gets very souped up math engine.

Hey JBL, did you get the pm's that I sent you?  Don't respond to them here but you can tell me if you've received them.  Thanks.

[i]In the future, use some other way of communicating to ask your question.  Forums are for topics of interest to many people.

On another note, you could see if JBL received your message by looking at the 'outbox' and 'sendbox' folders of your own private-messaging system.  If a message is in 'outbox' it hasn't been read by the recipient yet.

- moderator mathfanatic[/i]

[quote="Idiot_without_a_village"]Besides, you know all those math homies up in MOP and IMO and shiyt from the USA are all Taiwanese. NOT to be confused with the Chinese, after all. . . they ARE an independent nation.

Ooh, let's see what responses this will incite.[/quote]

Kudos to you!

Well, not to turn this into a political discussion board, I'll keep it short.

Go Chen!  Yes Taiwan!

(and Lien, please accept the fact that you lost.  Don't lose Taiwan's face in the international community.)

I don't think this is a good place to talk about which race is better than which in the AMCs.  Afterall, why does it matter?  Anybody can succeed if they work hard enough.

-interesting_move

[quote="Idiot_without_a_village"][quote="churchilljrhigh"]
uh chinese have already risen to power and will continue to hold it. and before any indians will rise Mystic Terminator needs to gets very souped up math engine.[/quote]

Enjoy your power while you can, as far as I'm concerned, we are fortunate to be born now, right at the brink of when India's population shall spin out of control, and you'll see Indian's all over the place, even in Cosmo!

Besides, you know all those math homies up in MOP and IMO and shiyt from the USA are all Taiwanese. NOT to be confused with the Chinese, after all. . . they ARE an independent nation.

Ooh, let's see what responses this will incite.[/quote]

Uh what country has been making the top 3 in IMO for practicallly all but one year? CHINA! and theres only one CHINA! and we never see India or Taiwan in the top 3 for IMO! HA! Last of all, you should recheck your statement about MOP!.

Taiwan is a rebellious province who is 98% Ethnic Han Chinese (Taiwan's Governmnet Factbook).

Well, when our population DOES spin out of control, there won't be really anyone left to defeat so we'll automatically be first. -------------

Hey mathfanatic, I'll say what I want. FIGHT THE OPRESSIVE ESTABLISHMENT!

I thank the moderator of the AMC Forum for moving these posts to a more on-topic home in the Round Table Forum.

[quote="Idiot_without_a_village"]----------------------------------------------------[/quote]

That is a random comment and should be deleted!

[i]Wish granted.

- mathfanatic[/i]

It is easy to guess though, because I am a  Mormon.

Okay, now that this split-off thread is in the Round Table Forum, where it is on-topic (thank you, Mathfanatic), let's apply some mathematical thinking to the social policy issues here.

The first question to ask is, what would we expect, using our math knowledge, would be the rank ordering of different countries in the world in performance on math tests? Why? What other, hidden assumptions do we have to check before we can decide what countries are overperforming, and what countries are underperforming?

Whoa guys, didn't mean to piss anyone off. I was joking anyways. I mean look at my avatar, please don't take any offense to what I say, I'm a nice guy. =)

Yeah, he got a perfect AMC 12A, give the man some slack.

Really, can you ever take a person seriously with The Cheat on his avatar?

I've landed on some state math competition Web sites from time to time, and I have the impression that Mystic Terminator has a very souped-up math engine.

[quote="MCrawford"]One thing about research on the Ashkenazi Jews is that there is no accounting for the fact that in modern times, decendents are in fact the single most affluent ethnic population in the world -- and by a large margin.  It was largely this Jewish population that established the Western banking world in a time when Christians had objections to interest on money exchanges.  The result was a plural monopoly on financial services that, even as it has been diluted, has resulted in immense wealth and resources.

Additionally, all this is during a time where (from about the 1400s in Europe starting in the Netherlands in particular) we see the first steady state economic growth due to the exponential growth in technology.

Interest on capital takes on several forms including a premium that might as well be called "I want to spend it today, not tomorrow" (which essentially makes up what's usually called the [i]risk free interest rate[/i]).  However, the real wealth-producing part of interest (that isn't just fighting the tide of inflation) is what economists call the [i]Technological Factor of the Economy[/i].

In essence, the Jewish population became wealthy for precise the same reason that the United States did -- capitalization on technological development.

If you look at capitalists as a whole, it's not clear whether or not Jews still outperform intellectually -- nobody's ever done the study, so nobody can claim to have stripped away all the variables and we're left still slicing the genetic argument thinner (even if it can never be cut away entirely).

Additionally, the extreme circumstances of any small population affect its culture values, which is to say that it affects the standard environment of the children of that culture -- particularly during formative years.[/quote]

Have you looked at the testable hypothesis of the Cochran-Hardy-Harpending paper?  It cuts through all the uncertainty you describe.  There is no reasonable way in which Cochran et al. could be fundamentally wrong about Jewish genetics yet be right about their so far untested prediction that Jewish Tay Sachs heterozygotes (or heterozygotes for any of the other genetic diseases in that cluster) have noticeably higher average cognitive aptitudes than their non-carrier siblings.

The comments section of [url=http://www.gnxp.com/blog/2006/01/tay-sachs-carriers.php]this recent GNXP post[/url] has some discussion of the matter.

As biotech continues to advance, we will be forced to confront more and more facts like this which we previously could address with plausible deniability.

That said, there is nothing complicated about this from an educator's perspective.  You emphasize to students what they are capable of changing, rather than what they cannot.  I found the following concrete result along these lines particularly fascinating:

[i]But one impediment to growing grit may be -- surprisingly -- the seemingly innocent act of parents praising a child's intelligence. In one fascinating series of studies, Stanford University psychologist Carol Dweck and her colleagues showed that children who were praised for their intelligence cared more about their grades than about learning during subsequent tasks. And after experiencing a failure, these children were less persistent than their peers who had been praised for their effort. "When you praise kids' intelligence and then they fail, they think they're not smart anymore, and they lose interest in their work," Dweck explains. "In contrast, kids praised for effort show no impairment and often are energized in the face of difficulty."

In another study, seventh graders were categorized according to whether they felt that intelligence is fixed or malleable. Students who viewed intelligence as something that can be cultivated earned progressively higher grades during the next two years, whereas those who had a fixed view of intelligence did not improve. "When you have a malleable view of intelligence, you believe in learning and you believe in effort," Dweck says. "When you have the fixed view, you don't."

Based on these findings, Dweck and colleagues designed a program (the Brainology Workshop) that explains to kids that learning creates new neuronal connections in the brain and that they themselves can foster this process. When seventh graders whose grades had been declining took the workshop, their marks shot right back up. Among a control group that received only the study skills, grades continued to fall. The program is currently being tested in New York City schools.[/i]

(from [url]http://www.psychologytoday.com/articles/pto-20051017-000003.xml[/url])

There is still a way in which this theory can be flawed.

I'm sure you're aware of intelligence correlations with Briggs-Meyers personality type.  There are correlations with other psychological markers as well.  It could be that the genetic links are personality links and that environment encourages procreation of people with certain personality types and withdrawn from those conditions, the balanced of personality types reverts to new cultural norms.  We could call these factors genetic, but only the in the sense of being short-term, in which case they are a matter of game theoretic equilibriums of the social structure (environment) and not a necessarity of the robust genetic structure.

The Gene Expression people are smart (I know many of them), but they're fooling themselves if they think we've come close to testing away game theoretic circumstances of environmental equilibriums -- something that would take almost as much time as it took to happen and would be quite resources consuming, not to mention brutally unethical.

I'm not saying I disbelieve in some purely answers purely based on long term genetic robustness, but I feel that they are negligible in relation to the necessary amount of gaming going on in the genetic pool on behalf of environmental conditions.

Beyond that, I think intelligence is best defined in relation to its environment, in which case, arguments can be made for each of the races within their environment on the basis of the idea that the genes would have simply come to different equilibrium points were they not [very nearly] optimal.

I added a fair amount to my last post after you replied, so you might want to take a brief look.

[quote="MCrawford"]Beyond that, I think intelligence is best defined in relation to its environment, in which case, arguments can be made for each of the races within their environment on the basis of the idea that the genes would have simply come to different equilibrium points were they not [very nearly] optimal.[/quote]
I do buy this principle.  Most of us are now living in a literate urban or suburban environment that more closely resembles Ashkenazi Jewish living conditions in the past millennium than that of any other large group.  [i]g[/i] matters more in this environment than in most others our ancestors have lived in.  (Though, paradoxically, it may have already ceased to matter in a strict evolutionary sense.)

[quote="K81o7"]To quote Russell Peters...

[i]One day, everybody in the world will be some mix of Chinese and Indian[/i]

 :D Perhaps not the most reliable source, but whatever..[/quote]

   Rofl, all of humanity will be "beige" one day. :lol:

[quote="jimhu"][quote="G-UNIT"][quote="K81o7"]To quote Russell Peters...

[i]One day, everybody in the world will be some mix of Chinese and Indian[/i]

 :D Perhaps not the most reliable source, but whatever..[/quote]

   Rofl, all of humanity will be "badge" one day. :lol:[/quote]

YAY! There is hope for chinese curry  :)[/quote]

  There's a restaurant specializing in " Indian style Chinese Food " just two block further than my place of residence.  Apparently, the dishes are Chinese, and the spices Indian; but what I find most humorous is that the owner speaks both languages and his transition from one language to the other is quite plausible indeed.

[quote="JBL"][quote="G-UNIT"]...the owner speaks both languages ...[/quote]
Chinese and Indian?  (At least one can say "Chinese" and be correctly understood to mean Mandarin most of the time).[/quote]

  Mandarin and Hindi.  When a specific language is not mentioned, given a multilingual country, the dominant languages are implied.  :D

[quote="JBL"]Hindi is spoken by only 30% of the Indian population.  While it enjoys official status, there are many Indians who would be annoyed by the substitution of "Indian" for "Hindi."  (Most of my Indian friends are from the south and speak Dravidian languages, so I have developed an awareness of the significance of the issue.)[/quote]

I'd say that saying a "country" 's language would mean the country's official language.

[quote="Mariag"]so is Hindi [i]the[/i] official language of India, or is it possible to have more than one official language?[/quote]

I'm pretty sure that there are countries with multiple official languages, but now that you mention it, that may be the case with India... :blush:

India has many official languages.  Hindi appears to have some special place among them (as a "national language").  I quote from the [url=http://www.cia.gov/cia/publications/factbook/]CIA World Factbook[/url]:
[quote="The CIA"]Languages:
English enjoys associate status but is the most important language for national, political, and commercial communication; Hindi is the national language and primary tongue of 30% of the people; there are 14 other official languages: Bengali, Telugu, Marathi, Tamil, Urdu, Gujarati, Malayalam, Kannada, Oriya, Punjabi, Assamese, Kashmiri, Sindhi, and Sanskrit; Hindustani is a popular variant of Hindi/Urdu spoken widely throughout northern India but is not an official language.[/quote]


modular arithmetic
number theory proposed
number theory

Find a positive integer $n$ with 1000 digits, all distinct from zero, with the following property: it's possible to group the digits of $n$ into 500 pairs in such a way that if the two digits of each pair are multiplied and then add the 500 products, it results a number $m$ that is a divisor of $n$.

We will look for a number with all digits equal 1 except one equal 4. Then the sum of their pairwise products is 503 (which is a prime number). We need to find a multiple of 503 of the form $\frac{10^{10000}-1}{9}+3\cdot 10^{k}$ where $k<10000.$
The smallest non-negative solution to
$\frac{10^{10000}-1}{9}+3\cdot 10^{k}\equiv 0\pmod{503}$
is $k=423$.

Therefore, $n=\frac{10^{10000}-1}{9}+3\cdot 10^{423}$ satisfies the required property.

Nice solution. This is a creative problem with many many answers. Here is a solution presented in kalva.demon.co.uk using $m=528$:

Answer

11...1 2112 2112 ... 2112 (960 1s followed by 10 2112s)

Solution

Suppose we take 980 digits to be 1 and 20 digits to be 2. Then we can take 8 pairs (2,2), 4 pairs (2,1) and 488 pairs (1,1) giving a total of 528 = 16·3·11. The sum of the digits is 1020 which is divisible by 3, so n is certainly divisible by 3. We can arrange that half the 1s and half the 2s are in odd positions, which will ensure that n is divisible by 11. Finally, n will be divisible by 16 if the number formed by its last 4 digits is divisible by 16, so we take the last 4 digits to be 2112 (=16·132). So, for example, we can take n to be 11...1 2112 2112 ... 2112, where we have 960 1s followed by 10 2112s.

-----------------------------

I have another of my own with $m=625$. Let's see other values of $m$ that people can find and hopefully post here for fun :)

well, i have another one, with m=1024.

We proceed similiary as with $m=528$   :wink:

we look for a number in wich 1024 divides the number formed by the last 10 digits
then we filled up to 1000 digits with 1's and 2's, in such way that we can took the 500 products in order to sum 1024.

in fact (i guess), m can be: 2048, 4096,...

with a limit, of course.  :)


linear algebra
matrix
linear algebra unsolved

I would like to know the statement is always true or sometimes false, and what is the reason:
A is a square matrix
P/S: I denote transpose A as A^T
1)If AA^T is singular, then so is A;
2)If A^2 is symmetric, then so is A.

1) is true, because $\det (AA^{T})=\det(A)\det(A^{T})=\det(A)^{2}$;
2) is false: $\begin{pmatrix}0& 1\\ 0& 0\end{pmatrix}\begin{pmatrix}0& 1\\ 0& 0\end{pmatrix}=\begin{pmatrix}0& 0\\ 0& 0\end{pmatrix}$


The sum of two numbers is 25 and their product is 126. What is the absolute value of the difference of the two numbers?

By trial and error, you can find that the numbers are 18 and 7, so the difference is 11.

Let the numbers be $ x$ and $ y$ respectively. You get two equations
\begin{align*} x + y & = 25 \\
xy & = 126 \end{align*}
Now square the first equation to get $ x^2 + 2xy + y^2 = 625$. Then substitute $ xy = 126$ to get $ x^2 + y^2 + 252 = 625\rightarrow x^2 + y^2 = 373$. Then by guess and check you get that $ x = 7$ and $ y = 18$. (both answers are symmetric so it really doesn't matter) so the difference is $ x - y = 18 - 7 = \boxed{11}$

edit: Gotta type faster.

Using the equations above, you square the first equation to get $ x^2\plus{}2xy\plus{}y^2\equal{}625$.  Subtracting $ 4xy$ from both sides gives you $ x^2\minus{}2xy\plus{}y^2\equal{}625\minus{}4(126)\equal{}121$.  Taking the square root, noting that $ (x\minus{}y)^2\equal{}x^2\minus{}2xy\plus{}y^2$, you get $ x\minus{}y\equal{}11$.

For this problem, it's easier with trial and error, but what if you had
\[ x\plus{}y\equal{}3\]
\[ xy\equal{}\minus{}14\]

As you've seen two different equations already, I don't want to repeat them. Just a tip-it's much faster to just use your mathematical instincts on problems like these. In other words, guess and check all the way.


In how many ways can one arrange the numbers 21, 31, 41, 51, 61, 71, and 81 such that the sum of every four consecutive numbers is divisible by 3?

[hide]
$ 21\equiv 0 \mod{3}$
$ 31\equiv 1 \mod{3}$
$ 41\equiv-1 \mod{3}$
$ 51\equiv 0 \mod{3}$
$ 61\equiv 1 \mod{3}$
$ 71\equiv-1 \mod{3}$
$ 81\equiv 0 \mod{3}$

So, we need to first find the number of arrangements of 3 0's, 2 1's, and 2 -1's such that each set of 4 adds to 0 (it can't add to 3 because that would imply 3 1's).

Each arrangement is just dependent on the first 3 numbers, because after that each number is determined by the previous 3.

So, the only possibilities for the first 3 are:

0,0,0
0,0,1
0,0,-1
0,1,1
0,1,-1
0,-1,-1

We can immediately discount the possibilities of 0,-1,-1 (because the next number would have to be 2), 0,1,1 (because the next number would have to be -2) and 0,0,0 (because the next number would have to be 0, and there are only 3).

This leaves us with 0,0,1, 0,0,-1, and 0,1,-1. The first and the second are impossible (there have to be 2 more 0's), leaving us with 0,1,-1 as the only possibility we need to count.

The counting is trivial: $ 6 \cdot 3 \cdot 2 \cdot 2 = \boxed{72}$

[/hide]


A locksmith charges 40 dollars plus 28 dollars for each lock installed. How many locks can be replaced for 236 dollars?

After being discounted 20 percent, a radio sells for 63.96 dollars. Find the original price.

Juan scored 5 points higher on his midterm and 13 points higher on his final than he did on his first exam. If his mean (average) score was 90, what was his score on the first exam?

Please help and show work. I'm not really quite sure what the formulas are and it's giving me some trouble. Thanks!

1) every lock costs 40+28=68 dollars
    236 dollars divided by 68=3R32
so you can buy 3 locks

2)100%-20%=63.96
   80%=63.96
   20%=15.99
   100%=79.95
   The radio costs $9.95
3) Let x represent the score on the first exam
    (x+x+5+x+13)/3=90
     (3x+18)/3=90
      x+6=90
      x=90-6
      x=84
 His score on the first exam was 84.

Thank you for the second two. But for some reason my book says the answer to number one is 7. I just don't see how the number can be that big.

The percentage one still doesn't make sense...is there another way you can show me?

[hide="Number 1"]The cost for the guy coming is $ \$40$.  So we have $ \$196$ left.  With this, we can buy $ \boxed{7}$ locks.[/hide]

Wow, that was so easy it's scary.
Any thoughts on the second one please?
I really appreciate the help...

[hide="Number 2"]OK, since the radio is $ 20\%$ off, the price is $ (100 \minus{} 20)\%$ or $ 80\%$ the original price.  Call this price $ p$.  Then we have:

$ 0.8p \equal{} 63.96$

Divide both sides by $ 0.8$.

$ p \equal{} \boxed{79.95}$[/hide]

cost of lock = $ 40 fixed charge and$28 for each lock.
total cost of lock = $ 236
Therefore  ($236 - $ 40)/$28 = 7 locks can be done

Telling someone not to revive old topics:  :spam:
See: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23215[/url] :wink:


integration
calculus
geometry
calculus computations

Find the volume of the region in the first octant that lies beneath the surface
(x/a)^(1/2) + (y/b)^(1/2) + (z/c)^(1/2) = 1
where a,b,c are constants.

$ V\equal{}\iiint_E\,dx\,dy\,dz$ where $ E$ is the region in question.

Let's make the change of variable $ x\equal{}au^2,y\equal{}bv^2,z\equal{}cw^2.$ We get $ dx\,dy\,dz\equal{}8abcuvw\,du\,dv\,dw$ and the new description of the region is those $ (u,v,w)$ in the first octant such that $ u\plus{}v\plus{}w\le 1.$

$ V\equal{}8abc\int_0^1\int_0^{1\minus{}u}\int_0^{1\minus{}u\minus{}v}uvw\,dw\,dv\,du$

$ \equal{}4abc\int_0^1\int_0^{1\minus{}u}uv(1\minus{}u\minus{}v)^2\,dv\,du$

From here, I'm going to do a sequence of integrations by parts.

$ \equal{}4abc\int_0^1\left[\left.\minus{}\frac13uv(1\minus{}u\minus{}v)^3\right|_{v\equal{}0}^{v\equal{}1\minus{}u}\plus{}\frac13\int_0^{1\minus{}u}u(1\minus{}u\minus{}v)^3\,dv\right]\,du$

$ \equal{}\frac{4abc}{3}\int_0^1\left[\left.\minus{}\frac14u(1\minus{}u\minus{}v)^4\right|_{v\equal{}0}^{v\equal{}1\minus{}u}\right]\,du$

$ \equal{}\frac{abc}3\int_0^1u(1\minus{}u)^4\,du$

$ \equal{}\frac{abc}3\left[\left.\minus{}\frac15u(1\minus{}u)^5\right|_0^1\plus{}\frac15\int_0^1(1\minus{}u)^5\,du\right]$

$ \equal{}\frac{abc}{90}$

That may seem very small, but this is a fairly thin and hollowed out region. If $ a\equal{}b\equal{}c\equal{}1,$ note that the point $ \left(\frac19,\frac19,\frac19\right)$ is on the surface.

Here is my solution for Japanese highschool students, because they don't study triple integral in their lesson.

Cut the solid by the plane $ z=t\ (0\leq t\leq c)$, we have the cross section in the $ xy$ plane:$ \sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=k$, where $ k=1-\sqrt{\frac{t}{c}}$. Thus the area $ S(t)=b\int_0^{ak^2}(k-\sqrt{\frac{x}{a}})^2dx=\int_0^{ak^2} ydx=\frac{b}{a}\int_0^{ak^2} (\sqrt{a}k-\sqrt{x})^2dx$

$ =\frac{b}{a}\int_0^{ak^2} (ak^2-2\sqrt{a}k\sqrt{x}+x)dx$

${ =\frac{b}{a}[ak^2x-2\sqrt{a}k\cdot \frac{2}{3}x^{\frac{3}{2}}+\frac{x^2}{2}}]_0^{ak^2}$

$ =\frac{b}{a}\cdot a^2k^4(1-\frac{4}{3}+\frac{1}{2})=\frac{1}{6}abk^4$.

The desired volume $ V=\int_0^c S(t)dt=\frac{1}{6}ab\int_0^c k^4dt$

$ =\frac{1}{6}ab\int_1^0 k^4\cdot c\cdot 2(1-k)(-1)dk$

$ =\frac{1}{3}abc\int_0^1 k^4(1-k)dk$

$ =\boxed{\frac{1}{90}abc}$.

Hey,
[quote="kunny"]
Here is my solution for Japanese highschool students, because they don't study triple integral in their lesson.
[/quote]
Nice solution, do you have a solution without using calculus?

Thanks,

-[u][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/u]

Thanks, regrettably I don't have solution without using Integration.

I belive there is no such a solution.


AMC
AIME
algebra
polynomial
trigonometry
complex numbers

Let $H$ be a heptagon in the plane centered at the origin such that $(1, 0)$ is a vertex.  Calculate the product of the distances from $(1, 1)$ to each vertex of $H$.

[hide="Massive hint"] Complex numbers ;) [/hide]

this was also on mildorf's first mock aime

Since the heptagon is centred at the origin and a vertex is on $(1,0)$, you could easily consider the seventh root of unity, in other words, the polynomial $f(z)=z^{7}-1=(z-1)(z-\zeta)(z-\zeta^{2})\dots(z-\zeta^{6})$, where $\zeta=e^{i\frac{2\pi}{7}}$.  We are to find the product of the distances between $(1,1)=1+i$ and $1,\zeta, \zeta^{2},...,\zeta^{6}$, i.e. $|f(1+i)|=|(1+i)^{7}-1|$.
\[\\ |(1+i)^{7}-1|=\left|\left(\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\right)^{7}-1\right|\\ =|7-i8|\\ =\sqrt{113}\]
Masoud Zargar

It's not hard to generalize to an n-gon also.


modular arithmetic
number theory unsolved
number theory

$k$ is an integer number greater than $1$.
Integer numbers $a_1,a_2, \cdots, a_n$ satisfy:
$a_1+2^ia_2+3^ia_3+\cdots+n^ia_n=0$ for $i=1,2,\cdots,k-1$

Prove, that:

$a_1+2^ka_2+3^ka_3+\cdots+n^ka_n\equiv{0}\pmod{k!}$


Find all natural numbers $n$ such that the number $n(n+1)(n+2)(n+3)$ has exactly three different prime divisors.

Among the integers $ n,n\plus{}1,n\plus{}2,n\plus{}3$ at least one is divisible by $ 2$ and at least one is divisible by $ 3$, so these are two of the three prime divisors. The third prime divisor $ p$ is at least $ 5$ so only one of the numbers $ n,n\plus{}1,n\plus{}2,n\plus{}3$ is divisible by $ p$, and the other $ 3$ must be divisible by $ 2$ and/or $ 3$.
Among the numbers $ n,n\plus{}1,n\plus{}2,n\plus{}3$ at most two are divisible by $ 3$ so among the three numbers divisible by $ 2$ and/or $ 3$ one is not divisible by $ 3$  and only by $ 2$ so it is a power of $ 2$. By the same reasoning we get one of the numbers $ n,n\plus{}1,n\plus{}2,n\plus{}3$ is a power of $ 3$. Then the number among $ n,n\plus{}1,n\plus{}2,n\plus{}3$ that is not a power of $ 2$ or $ 3$ and not divisible by $ p$ can be either a power of $ 2$, a power of $ 3$, or divisible by $ 2$ and $ 3$.
1. If it is a power of $ 2$ then we get two powers of $ 2$ differing by $ 2$ so they must be $ 2$ and $ 4$. So we either get $ n\plus{}3\equal{}4$ giving $ n(n\plus{}1)(n\plus{}2)(n\plus{}3)\equal{}24$ -- divisible only by $ 2$ and $ 3$ -- so this case fails, or $ n\plus{}2\equal{}4$ giving $ n\equal{}2$ and $ n(n\plus{}1)(n\plus{}2)(n\plus{}3)\equal{}120$ -- divisible by $ 2,3,5$ and everything is fine.
2. If it is a power of $ 3$ get two powers of $ 3$ differing by $ 3$ which is impossible.
3. If it is divisible by $ 2$ and $ 3$ the only case when two of the numbers $ n,n\plus{}1,n\plus{}2,n\plus{}3$ are divisible by $ 3$ is when they are $ n$ and $ n\plus{}3$.
i. $ n$ is a power of $ 3$ ie $ n\equal{}3^{k}$, then $ n\plus{}3\equal{}3^{k}\plus{}3$ while $ n\plus{}1, n\plus{}3$ are even but $ n\plus{}3$ is not a power of $ 2$ as it is divisible by $ 3$ so $ n\plus{}1$ is a power of $ 2$ ie $ n\plus{}1\equal{}3^{k}\plus{}1\equal{}2^{l}$. Also, $ 3^{k}\plus{}3\equal{}3(3^{k\minus{}1}\plus{}1)$ and the expression in brackets is not divisible by $ 3$ so it is a power of $ 2$, so get $ 3^{k\minus{}1}\plus{}1\equal{}2^{m}$. Then, $ 3^{k}\equal{}(3^{k\minus{}1})*3\equal{}(2^{m}\minus{}1)*3$ but $ 3^{k}\equal{}2^{l}\minus{}1$ so get $ 3*2^{m}\minus{}3\equal{}2^{l}\minus{}1$ so $ 2^{l\minus{}1}\plus{}1\equal{}3*2^{m\minus{}1}$ then $ 2^{m\minus{}1}\equal{}1$ and $ 2^{l\minus{}1}\equal{}2$ so $ 2^{l}\equal{}4$ so $ 2^{l}\minus{}1\equal{}3\equal{}n$ so $ n\equal{}3$. Then $ n(n\plus{}1)(n\plus{}2)(n\plus{}3)\equal{}360$ divisible only by $ 2,3,5$ -- everything is fine.
ii. $ n\plus{}3$ is a power of $ 3$ then $ n\plus{}2\equal{}2^{l}$, $ n\plus{}3\equal{}3^{k}$, $ n\equal{}3^{k}\minus{}3\equal{}3(3^{k\minus{}1}\minus{}1)$ so $ 3^{k\minus{}1}\minus{}1\equal{}2^{m}$ so $ 3*2^{m}\plus{}2\equal{}2^{l}$ so $ 3*2^{m\minus{}1}\plus{}1\equal{}2^{l\minus{}1}$ so $ 2^{m\minus{}1}\equal{}1$ as it must be odd so $ 2^{l}\equal{}4$ so $ n\plus{}2\equal{}4$ so $ n\equal{}2$ and this case works from 1.

Therefore, $ n\equal{}2$ and $ n\equal{}3$ are the only solutions.

For $n \leqslant 3$, $n = 2,3$ are solutions, now we suppose $n \geqslant 4$.
Among the integers $n,n + 1,n + 2,n + 3$， there are two odd numbers $2k - 1,2k + 1,k \geqslant 3$, and $2k - 1,2k,2k + 1$ are pairwise coprimes. So there are exactly three different prime divisors of $2k - 1,2k,2k + 1$,this force $2k$ is power of $2$, and ${2^3}\left| {2k} \right.$.
Among the integers $2k - 1,2k,2k + 1$, there is a multiple of $3$, so one of $2k - 1,2k + 1$ must be a power of $3$,hence it is a multiple of $9$.
Now we consider the fourth number, it has not any new prime divisor, so it can only divisible by $2,3$. Among the integers $n,n + 1,n + 2,n + 3$， there are not two numbers can divisible by a number great than $3$, so the only possible is $6$, and $n = 6$.
Therefore, $n = 2,3,6$ are the only solutions.


search
combinatorics unsolved
combinatorics

1. There are n>1 different points on the plane. The center of every segment from one point to another is painted pink. What is the minimal number of pink points?

2. For any set of 2n-1 integers prove that there exists a subset containing n of them such that the sum of its elements is divisible by n.

3. How many ways are there of dividing the set {1,2,...,1995} into 3 non-empty subsets such that none of them contains two consequent natural numbers?

1) http://www.mathlinks.ro/Forum/viewtopic.php?p=164304&highlight=#164304
2) it's famous, i think it's Erdos' theorem, you should search for that here, but if you will not find, i remember it was somewhere at http://www.staszic.waw.pl
3) first position we can choose on $3$ ways other on $2$, so there are $3 \cdot 2^{1994}$ ways

[quote="Zorro"]
2) it's famous, i think it's Erdos' theorem, you should search for that here, but if you will not find, i remember it was somewhere at http://www.staszic.waw.pl
[/quote]
Yes, this is Erdos Ziv Ginzburg theorem ;)

Erdös-Ginzburg-Ziv is here: http://www.mathlinks.ro/Forum/viewtopic.php?t=57546


limit
logarithms
calculus
calculus computations

hello, you must write $ \lim_{x \to 0\plus{}0} (e^x \minus{} 1)^{1\minus{}\ln x}$. You can use the rule of L'Hospital the searched limit is $ 0$.
Sonnhard.

Perhaps the limit should be $ \lim_{x \to 0^\plus{}} (e^x\minus{}1)^{1/\ln x}$.

hello,in this case we have $ \lim_{x \to 0^\plus{}} (e^x\minus{}1)^{1/\ln x}\equal{}e$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, you must write $ \lim_{x \to 0 \plus{} 0} (e^x \minus{} 1)^{1 \minus{} \ln x}$. You can use the rule of L'Hospital the searched limit is $ 0$.
Sonnhard.[/quote]

Or:

$ \lim_{x \to \plus{} 0} (e^x \minus{} 1)^{1 \minus{} \ln x} \equal{} \lim_{x \to \plus{} 0} e^{(1 \minus{} \ln x)ln(e^x \minus{} 1)} \equal{} 0$


algebra
function
domain
Gauss
AMC
AMC 12
AMC 12 A

found this on the AMC website, need some help:

Let f(x) = sqrt (ax^2 + bx), for how many real values of [i]a[/i] is there at least one positive value of [i]b[/i] for which the domain of [i]f[/i] and the range of [i]f[/i] are the same set???

i dont really even get what the question is asking...can someone explain?

oh yea, the choices are 0, 1, 2, 3, or infinite

thanks...

Here's some definitions that may help you out:

A [b]function[/b] is like a machine: if you put in a number for x, the function will change that number into another number (denoted f(x)) according to a set of rules that define the function.

For instance, if f(x) = 2x + 3, then if you plug in 3 for x, the "function machine" does 2*(3) + 3 = 6 + 3 = 9.

The [b]domain[/b] is the set of values that x can be.
For instance, if you had a function f(x) = 1/(x-2), then x can be all real numbers except for 2, since plugging in 2 for x would produce 1/0 which is undefined.

The [b]range[/b] is the set of values that f(x) can be.  For instance, in the function f(x) = 1/(x-2), f(x) can be all real numbers except 0 because no matter what you put in for x, you can't make 0.

In this problem, no matter what a or b is, x can be any number.  Thus, in order to have the domain set equal the range set, t is necessary to have f(x) be any number also.

Case #1: ax^2 + bx is positive.  Then sqrt(ax^2 + bx) MUST BE positive.  (Do you see why?)  Therefore, f(x) must be positive.

So in this case, x can be any number but f(x) must be > 0.  Thus the domain and the range are two different sets.  So this is no good.

Case #2: ax^2 + bx is negative.  Then sqrt(ax^2 + bx) is not real.  (Do you see why?)  Therefore, f(x) cannot be any number.

Case #3: ax^2 + bx is 0.  Then no matter what x is, sqrt(ax^2 + bx) is 0.  Thus the range for f(x) is {0}, which differs from the domain of {all real numbers}.

These three cases sum up all possible cases, and for each one, the domain and range differ.  Thus, the answer is [b]0[/b] since no cases exist for which this works.

I think this is right.  ComplexZeta?  Gauss?  anyone? Check, please.

Hope this helps.
mathfanatic

P.S. Keep coming to this site!  If you've got any more questions, be sure to post them here.  We'd be happy to help and we're trying to increase the size of our community.

What about the case where a = 0? Then a < 0 and a > 0. Draw some pictures, I think that will help.

#25 from AMC12A 2003 I presume? I solved this problem by considering the cases a>0, a=0, and a<0. a=0 clearly works. You can prove that a>0 doesn't have any solutions because we have lots of negative values of x that will be in the domain and thus not in the range. The toughest case is a<0. Here's what I did:

Then we square it because
square roots are messy, so we have ax^2+bx. So then taking a derivative and
setting equal to zero gives x=b/(2a), if we use -a instead of a because a is
negative, and positive numbers are slightly nicer to work with. Also, the
maximum of the range is b/a. So the square of the maximum value will need to
be b^2/a^2. Putting x=-b/(2a) into ax^2+bx
gives -b^2/(4a)+b^2/(2a)=b^2/(4a). We need b^2/(4a) to be equal to b^2/a^2.
Then a=4, which is really a=-4.

This is copied from an email, and the notation is not that great. Also you don't have to take a derivative to find the vertex of a parabola, but that's just the way I do it.

Therefore the solutions are a=0 and a=-4. What dismayed me about this problem is that some people missed it by forgetting the trivial case a=0, while some people missed it by missing the nontrivial a=-4. But all those people got the same score for the problem. I don't know what can be done about it, but it seems a shame to penalize the two groups of people in the same way.

ok, i know why a=0 works, but i dont get that stuff you did to get the -4...

why did u get x=b/(2a) instead of -b/(2a)?

how did you know b/a was the maximum of the range? and why did u square it?

and
"Putting x=-b/(2a) into ax^2+bx gives -b^2/(4a)+b^2/(2a)=b^2/(4a)."
wouldnt you get b^2/(4a)-b^2/(2a) = -b^2/(4a) ??  the negative of what you got??

and lastly, why does that thing you get have to equal b^2/a^2 ??

this +/- thing is really messing me up, i think im doing the same thing you're doing...if thats the case...ignore the 3rd question...

thanks alot!


calculus
derivative
limit
calculus computations

Prove that the derivative of $e^{x}$ with respect to $x$ is $e^{x}$ by using only the limit definition.

what i am having trouble with is showing that $\lim_{h\to\ 0}\frac{e^{h}-1}{h}=1$


Let m be the number of rows of dots in a rectangular lattice array
Let n be the number of columns of dots in the same rectangular lattice array

For every minimum size set of three noncollinear points (dots), that determines
a circle.

Let m > or = 2 and n > or = 2.

Please come up with a formula in terms of m and n for the total number
of circles in this m by n rectangular lattice array.

If you're not already considering this as part of the
problem/solving of the problem, you may want to
see this Partial *SPOILER*/clarification next . . .

[hide]While each set of 3 noncollinear points (dots) has
one circle passing through it, that doesn't mean that
every set of 3 noncollinear points (dots) will give rise
to circles all different from the ones before them (the ones
you just looked at or just counted).[/hide]


symmetry
geometry
geometric transformation

请问liyi，?图形A关于x轴和y轴对称；
?图形C关于原点对称；
(1)，?(2)
英文怎么说啊？谢谢

以?还有很多如此的笨问题 :blush:
希望?教
 :P

(1)，?(2)
?太明白你的??，是?1??2?？
?图形A关于x轴和y轴对称；
A is symmetric about x-axis and y-axis
?图形C关于原点对称
C is symmetric about the origin.

对啊，是1??2?的??
还有，对称性怎么说？
谢谢你了 :lol:  :D

谢谢啦，问题有点笨，麻烦??了 :blush:

?过好?英文的?明当中??常出现(1)?(2)这样的...  :?
起?那些书上?常看到.
?以用些其他表述.

真的?？
我是想表达：
从1得到2，?3
这个样?。
你觉得怎么说比较好？

你有QQ么，?以加你?，我的是81869008[/i]

那么还是?以的.
看起?, or?常出现在排版上居中的公?之间
[code]
If follows that
                           <formula>,              (2)
or
                           <formula>              (3)
[/code]
这样的.

[quote="Alexander Lenin"]我觉得，有许多英文数学??或?组的缩写我都?了解，比如说WLOG。
能?能有哪?热心人帮忙上传一些相关的资料或帖?供俺?考？[/quote]
WLOG我想是?妨设的???

WLOG的??是without lose of general，??失一般性，对?？

A more common translation is "Without loss of generality", which has the same meaning essentially as zhaoli stated.

有个问题:
"不动点"应该怎么翻译??

[quote="shobber"]有个问题:
"不动点"应该怎么翻译??[/quote]
fixed point, I suppose :?


The U-shaped body of mass $ M$ shown in the figure is at rest and lies on a smooth horizontal surface (e.g. on ice). Its centre of mass is the midpoint of the semicircle of radius $ R$. The length of the straight arms of the U is $ 2R$. A small disc of mass m slides tangentially to point $ A$ on one of the arms, with a velocity of $ v$.

a) Where will point $ A$ be when the disc is at point $ B$ on the other arm?

b) Draw the path of the two bodies.

c) Find the velocity versus time and acceleration versus time functions for both bodies.

(Friction and air drag are negligible everywhere. Data: $ M\equal{}m\equal{}1$ kg, $ R\equal{}0.1$ m, $ v\equal{}0.5$ m/s.)

I didn't get the problem.
Is this side view or top view? :)
Is Body-Disk interface smooth?

[b]@cancer[/b] why not continue the discussion [url=http://www.mathlinks.ro/viewtopic.php?search_id=1911540163&t=152580]here[/url]  :wink:


What is Moems? Can I find it in Puerto Rico?


geometry
circumcircle
incenter
Euler
geometry unsolved

please prove that the center of the circle passing through the excenters  , the circumcenter and the incenter of a triangle are collinear.

That's easy. Let our triangle be $ABC$ and triangle formed by the excenters $DEF$. Then, it's easy to prove that $ABC$ is an orthic triangle of triangle $DEF$ and the incenter of $ABC$ is orthocenter of $DEF$. Also, because the circumcenter of $ABC$ is the nine-point circle of $DEF$, so the circumcenter of $ABC$ is an nine-point circle, circumcenter, incenter of $ABC$ and circumcenter of $DEF$ lie on Euler Line of triangle $DEF$.


I noticed that when doing linear algebra over finite fields, weird things can happen. For example there are nonzero symmetric matrices with all eigenvalues zero.

I would like to know more about all these things. Our linear algebra course only covered fields of characteristic zero, and I feel I'm missing something. Can anyone refer me to some good books/websites/... specificly on linear algebra over finite fields (eigenvalues, nullspaces, ... just the regular stuff)?

im not positive but i think hoffman and kunze might. if not, try greub's linear algebra. that is i think the king of linear algebra books for undergrads. difficult, though.


limit
trigonometry
logarithms
calculus
calculus computations

Consider what happens to the logarithm of that expression as $x\to 0$:
$\tan x\ln\frac{1}{x}=-\tan x\ln x$ tends to $0$ (say, by L'Hopital's rule).

[hide="Answer"]$\lim_{x\to0^{+}}x^{-\tan x}= A \Rightarrow  ln A = \lim_{x\to0^{+}}-\tan x \ln x (0\cdot\infty)$
$= \lim_{x\to0^{+}}-\frac{\ln x}{\cot x}(\frac{\infty}{\infty}) = \lim_{x\to0^{+}}\frac{\sin^{2}x}{x}= \lim_{x\to0^{+}}\frac{\sin x}{x}\sin x = 1\cdot0 = 0$
$\Rightarrow  A = e^{0}= 1$[/hide]

[quote="misan"]$\lim_{x\to0^{+}}\;\tan{x}\rightarrow\;x$[/quote]

This doesn't make any sense.

What misan is doing is abusing language and abusing notation. My mental processes in looking at this limit would go something like this:

[quote]$\tan x$ looks like $x$ as $x\to 0$ so $x^{-\tan x}$ looks like $\frac1{x^{x}}$ and I know $x^{x}\to1,$ so this limit is 1.[/quote]
That takes me just a few seconds. But now comes the second-guessing, starting with those ultra-informal words "looks like." Of course, I mean they are asymptotically equivalent; that $\frac{\tan x}x\to1$ as $x\to0.$ Now is that good enough to complete the rest of the argument? For formality, I would probably take the step mlok took above, where he considered the logarithm. And there the asymptotic equivalence is clearly good enough to say that $\tan x\ln x$ might as well be $x\ln x.$ Here, I then rely on a collection of limits long ago computed and long ago committed to memory with the phrase "as $x\to0^{+},\ \ln x$ goes to infinity slower than any (negative) power of $x.$" So I know that limit is zero, which means the original limit is $1.$

I don't dispute that misan has the correct answer for essentially the correct reason, but I would reserve the symbol $\to$ for actual limits, and find other ways to express asymptotic equivalence.

[quote="xscapezaer"]$ \lim_{x\rightarrow0^{+}}\frac{1}{x}^{\tan{x}}=$[/quote]
Or we can do like below:
[hide="solution"]
$ =\exp\lim_{x\to 0^{+}}\tan x\ln\frac{1}{x}=\exp\lim_{x\to 0^{+}}-\frac{\tan x}{x}\cdot \ln x^{x}=\exp-1\cdot \ln 1=\exp 0=1$
And I should show $ \lim_{x\to 0^{+}}\ln x^{x}=0$, but it's easy :P
We can take $ t=\frac{1}{x}$, then compute: $ \lim_{t\to\infty}\ln\sqrt[t]{t}$ and limit $ \lim_{t\to\infty}\sqrt[t]{t}$ is well-known, and it isn't so hard to compute it :)[/hide]


Seems like all of us have begun posting unparseable or potentially dangerous latex formulas now... (Okay, now it works again.)

  darij

the tests are over and the problems solved (hopefully all), there should be no more problems with latex. the images are png8 with binary transparency (no more gif). if from now on , anyboday has problems seeing the formulas should say that to valentin and it will be fixed (hopefully). we have tested it on mozilla and ie and it should work just fine.

sorry for the temporary problems, have fun on the new server...

Test: $ABC$ $D$ $E$ $F$ $GHIJKLM$ $NOPQRST$ $UVWXYZ$

Do some of the letters look funny to you?  For example, C, O and Z look a bit different from what I remember.

By the way, "DEF" is apparently either unparsable or dangerous. :?

[quote="Valiowk"]By the way, "DEF" is apparently either unparsable or dangerous. :?[/quote]

Yeah, it's clear now!
Thanks!

  Darij

l got this  "dangerous...." after having typed \mathbbN
$\mathbb N$

$\displaystyle \left(\frac{\partial ^2}{\partial x^2}+ \frac{\partial^2}{\partial ^2} \right) |\phi (x+iy)|^2=0$


Να επιλυθει στους ακεραιους η εξισωση $x^{3}+x^{2}+x+1=y^{2}$(ζητω διαφορες μεθοδους αν μπορειτε...good luck) :wink:

Την έχω προχωρήσει αρκετά αλλά για να μην κάνω τσάμπα κόπο και τη βάλλω και έχει ελλείψεις μπορείς να μου πεις ποιες είναι οι λύσεις τελικά ?
Ευχαριστώ .

Συγνώμη Σιλουαν που αργησα τοσο πολυ να σου απαντησω (πραγματικα σου λεω οτι ο ελευθερος μου χρονος ειναι ελαχιστος.....)αλλα εστω και τοσο αργα δες εδω: http://www.mathlinks.ro/viewtopic.php?t=149349

ξαναπροτάθηκε και συζητήθηκαν κι άλλες ιδέες [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=371&t=184784]εδώ[/url]


inequalities
analytic geometry
advanced fields
advanced fields unsolved

[quote]... an oriented graph ... that has no oriented cycles ...[/quote]
Question: is that the same as a partial order?

Yes, of course: you can introduce a partial order by saying that $x>y$ if there is an oriented path from $x$ to $y$ and any partial order on a countable set will surely give you such a graph. Does it make the problem any easier? :?

How about the following idea?  Because the vertex set is countable, we can enumerate the vertices.  Let's take the vertices one at a time in that enumeration.  Suppose we are currently at vertex $v$.  We assign $v$ any number bigger than all its already-assigned ancestors, and smaller than all its already-assigned descendants.

We won't run into any problems, because if $u$ is an ancestor of $v$, and $w$ is a descendant of $v$, then $u$ is an ancestor of $w$.  So when we enumerated the later of $u$ and $w$, we already made sure that $u$'s number was smaller than $w$'s number.

What if we change the question to assigning integers to the vertices?

The answer becomes (quite obviously) negative.

The question can be restated thus: can every countable poset be embedded in the integers (with the usual ordering)? If the poset is a totally and densely ordered set (i.e. between any two elements there is another element), then it can't be embedded in $(\mathbb Z,\le)$.

[quote="fedja"]Consider an oriented graph with countably many vertices that has no oriented cycles. Is it always true that then one can assign a real number to each vertex in such a way that the numbers strictly decrease in the directions given by the edges?[/quote]

Also, for the case of an uncountable (and of cardinality equal to  $R$) set of vertices, is there an example where no such order-embedding exists?

There even exists a total order of cardinality $\aleph_{1}$ (without the assumption of the Continuum Hypothesis) which cannot be embedded in $\mathbb R$ with the usual order: just consider the first uncountable ordinal number. It's a well-ordered uncountable set with the property that every element has only countably many elements preceding it. Such a set cannot live inside $\mathbb R$ (one reason being, for example, that with only countably many exceptions, every point of an uncountable subset $S$ of $\mathbb R$ is an accumulation point of $S$).

A more explicit, but still somehow artificial, construction is possible.

Consider the space $S$ of sequences of positive integers with the ordering "is asymptotically larger than", that is, $x > y$ if  $x_{n}-y_{n}> 0$ for large $n$.   Assume without loss of generality that the range of the order-embedding is unbounded (if not, change coordinates in $R$ to make it unbounded).
Then, choosing an increasing and unbounded sequence of real numbers that are images of elements $s_{i}\in S$, and constructing a diagonal sequence $t = (s_{i})_{i}+1$,  the image of $t$ cannot be finite.


The Blog Details section of the blog is a little weird...it says that I have no entries and under "blog started:" it always says the time that I made my last entry under...

Yeah that's a bug. Probably related to the fact that you have deleted your previous blog :|

I have a blog, created after the initiation of the new website, and it does that too. Also, there is no link from aops to my blog (besides the button "my blog") it doesn't show on "aops blogs" Also, another thing, I can't find an "edit" or "delete" option for my entries...

This is the AoPS MathLinks blogs page: http://www.mathlinks.ro/Forum/apsjour.php

Click on the picture below for visualisation.


Prove the following identity:
$\sum_{k=0}^nC_{m+k}^k2^{n-k}+\sum_{k=0}^mC_{n+k}^k2^{m-k}=2^{n+m+1}$.
Try to find a combinatorics using solution.

Easy! :D

Consider all binary sequences of length $m+n+1$. There are $2^{m+n+1}$ such sequences. We can partion them into two disjoint sets $A_0$ and $A_1$, where $A_0$ is a set of sequences, whcih contain at least $m+1$ zeros, and $A_1$ is a set of sequences, which contain at least $n+1$ "1".
It is easy to see that first summand is $|A_0|$ and second summand is $|A_1|$. Indeed, suppose $x=(x_1,...,x_{m+n+1})\in A_0$, ad fix position $s$ of $m+1$-th zero in $x$. It can be number from $m+1$, $m+2$, ..., $m+n+1$. So number of sequences $x\in A_0$ with $m+1$-th zero on the $(m+1+k)$-th place is $C_{m+k}^{m}2^{n-k}$. Therefore
\[|A_0|=\sum_{k=0}^{n}C_{m+k}^{m}2^{n-k}=\sum_{k=0}^{n}C_{m+k}^{k}2^{n-k}.\]
Analoguously, for $A_1$.

That's all!  :)

[quote="Myth"]
Consider all binary sequences of length $m+n+1$. There are $2^{m+n+1}$ such sequences. We can partion them into two disjoint sets $A_0$ and $A_1$, where $A_0$ is a set of sequences, whcih contain at least $m+1$ zeros, and $A_1$ is a set of sequences, which contain at least $n+1$ zeros.
[/quote]

Can't there be sequences common to $A_0$ and $A_1$? Or maybe you meant to say that $A_1$ is the set of those sequences with at least $n+1$ $1$'s (not $0$'s)?


Va rog sa ma ajutati; nu stiu sa calculez limita sirului
[1/sqr(n^2+1) +1/sqr(n^2+2) +1/(sqr(n^2+3)+ ......1/sqr(n^2+n)]^n

se aplica metoda de la sirul lui Euler si trebuie calculata limita expresiei n(S-1), unde S este suma din paranteza enuntului. se obtine nS-n , se scrie n=1+1+1+....+1, unde 1 apare de n ori si se grupeaza fiecare fractie cu cate un -1.se aduce la acelasi numitor in fiecare grupare, se rationalizeaza si apoi se aplica criteriul clestelui( facandu-se majorari si minorari pentru numitori). se obtine limita e la puterea -1/4.

[quote="spix"]$\frac{n}{\sqrt{n^{2}+n}}\leq\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+\frac{1}{\sqrt{n^{2}+3}}+...+\frac{1}{\sqrt{n^{2}+n}}\leq\frac{n}{\sqrt{n^{2}+1}}$
 de unde iese cu teorema clestelui, teorema celor 2 jandarmi, teorema celor 2 paznici, si alte denumiri....[/quote]
Cred că dacă mai citeşti o dată enunţul, o să fii bucuros pentru avatar-ul ales :)

off... :blush: ... text non-latex.... $\left( \frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+\frac{1}{\sqrt{n^{2}+3}}+...+\frac{1}{\sqrt{n^{2}+n}}\right)^{n}$ .... solutia postata de sweetonect e cea mai potrivita...


LaTeX
search
number theory proposed
number theory

Let i,j >= 1. Show :
gcd(i,j) = 1  => there is a k0 € N , a,b € N U{0}, such that for every k >= k0,  ai + bj = k.

Proof. gcd(i,j) = 1  => there are some a0,b0 € Z with a0*i + b0*j = 1 .
Let k = rij + c,   r>=1,  0 <= c < ij.
Let q € Z with 0 <= c*a0 - qj =: a1 < j.
Let b1 € Z with a1*i + b1*j = k.
Show that b1 € N U{0}.
We have c*a0*i + c*b0*j = c
<=> (c*a0 - qj) i  + (c*b0 + qi) j = c
Because of 0 <= (c*a0 - qj) i < j, and 0 <= c < ij, we have |(c*b0 + qi)*j| < ij,
so b1 = (c*b0 + qi) + ri >= 0 , cause |c*b0 + qi| < i and ri >= i.
So k0 = i*j

I just came up with it. I dont know if its correct. Even if so, there have to be far better ways to get to it. Any ideas?
PS: sorry for non-latexing

In fact, the minimal value is $ k_0 \equal{} ab \minus{} a \minus{} b \plus{} 1$.

Please use [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX[/url] in the future.  It's not hard to learn and makes your math much more readable.  From a cursory examination, it looks like you have the right idea, but I'm not sure.

Wouldn´t k0 depend on i and j? We are given i and j, and we search for a and b.
For r=1, c=0, my smallest k0 = i*j!

Sorry; $ k_0 \equal{} ij \minus{} i \minus{} j \plus{} 1$.  For example, when $ i \equal{} 3, j \equal{} 5$ we have $ k_0 \equal{} 15 \minus{} 3 \minus{} 5 \plus{} 1 \equal{} 8$.


Let $a,b,c$ be the side lenghts of a triangle, and let $r \ge 3$. Prove that

$\frac{a^rb+b^rc+c^ra}{3} \ge (\frac{a+b+c}{3})^{r+1}$.

Quite Nice.
First,We Prove that If $k$ is a posive real satisfy:

$\frac{a^kb+b^kc+c^ka}{3}\ge (\frac{a+b+c}{3})^{k+1}$

Then For all real $r \ge k$ we have:

$\frac{a^rb+b^rc+c^ra}{3}\ge (\frac{a+b+c}{3})^{r+1}$

The reaminder Is Prove when $k=3$,
Because
$3(a^3b+b^3c+c^3a) \ge (ab+bc+ca)(a^2+b^2+c^2)$

And we need Prove:

$9(ab+bc+ca)(a^2+b^2+c^2) \ge (a+b+c)^4$

Which this one Is Easy If $a,b,c$ are lengths Of a triangle!


Each of two interior angles of a quadrilateral has measure 100 degrees. Another angle is a right angle. How large is the last remaining angle?
[hide]sum of angles in quadrilateral is 360, 100+100+90 = 290 360-290 = 70degrees[/hide]

[quote="shinwoo"][quote]360-(100+90)
360-290
70[/quote]

100 + 90 can't be 290

since it's 2 interior 100 degrees angles, it's supposed to be 200 degrees and when you add 200 and 90, it's 290
and subtract 360 - 290 = 70

I think you just forgot to add 100 ;)[/quote]

yea i did....thx for pointing that out.

k
the total degrees is 360 and the other two are 100 and the other one is 90
so 100+100+90=290
360-290=70
so the answer is 70 :D

[hide]degrees in a quadrilateral is 360 degrees.
100+100+90=290
360-290=70 degrees.[/hide]


trigonometry
geometry
geometric transformation
reflection
geometry proposed

[quote="gemath"]It is quite strange because
$\frac{MB}{MA}+\frac{NC}{ND}\ge 0$ we have equality when $M\equiv B,N\equiv C$
 :D[/quote]Yes, sorry ! I forgot the condition $O\in MN$, where $O\in AC\cap CD$. I corrected.
Thanks a lot. In your case $O\not \in MN$ !
In my opinion, this problem is very nicely. Soon I"ll offer my solution.

If I don't have mistake then this a solution!
Let $\frac{MB}{MA}=x,\frac{NC}{ND}=y$ we have
because $M\in AB$ we have
$x\vec{MA}+\vec{MB}=\vec{0}\Leftrightarrow x\vec{OA}+\vec{OB}=(x+1)\vec{OM}\Leftrightarrow x\frac{OA}{OC}\vec{OC}+\frac{OB}{OD}\vec{OD}=(x+1)\frac{OM}{ON}\vec{ON}$
Because $N\in CD$ we get
$x\frac{OA}{OC}+\frac{OB}{OD}=(x+1)\frac{OM}{ON}(1)$
similar for $N$ we get
$\frac{OC}{OA}+y\frac{OD}{OB}=(y+1)\frac{ON}{OM}(2)$
Multiply $(1),(2)$ we get
$(x\frac{OA}{OC}+\frac{OB}{OD})(\frac{OC}{OA}+y\frac{OD}{OB})=(x+1)(y+1) \Leftrightarrow x+y+\frac{OBOC}{ODOA}+xy\frac{OAOD}{OCOB}=x+y+1+xy \Leftrightarrow xy=\frac{1-\frac{OBOC}{ODOA}}{\frac{OAOD}{OCOB}-1}$
Thus $xy=const$ we easily see $x+y$ min when $x=y=\sqrt{\frac{1-\frac{OBOC}{ODOA}}{\frac{OAOD}{OCOB}-1}}$
 :D

[quote="Virgil Nicula"][color=darkred]Let $ABCD$ be a convex quadrilateral. Denote $O\in AC\cap BD$. Ascertain and [b][u]construct[/u][/b] the positions of the points $M\in (AB)$ and $N\in (CD)$, $O\in MN$ so that the sum $\frac{MB}{MA}+\frac{NC}{ND}$ is minimum.[/color][/quote]
[size=84][color=darkblue][b][u]Proof.[/u][/b] Denote the minimum position $M_{0}$ of the mobil point $M$ and $N_{0}\in CD\cap OM_{0}$.

$\left\{\begin{array}{c}m(\widehat{MOA})=x\\\ m(\widehat{MOB}=y\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}\frac{MB}{MA}=\frac{OB}{OA}\cdot\frac{\sin y}{\sin x}\\\\ \frac{NC}{ND}=\frac{OC}{OD}\cdot \frac{\sin x}{\sin y}\end{array}\right\|$ $\Longrightarrow$ $\frac{MB}{MA}\cdot\frac{NC}{ND}=\frac{OB\cdot OC}{OA\cdot OD}$.

([b][u]constant[/u][/b]). Then the minimum of the given sum is achieved when

$\frac{M_{0}B}{M_{0}A}=\frac{N_{0}C}{N_{0}D}=\sqrt{\frac{OB\cdot OC}{OA\cdot OD}}$. Denote the points $\left\{\begin{array}{c}\boxed{\ T\in AB\ ,\ TO\parallel CD\ }\\\\ S\in TO\cap AD\end{array}\right\|\Longrightarrow$

$\frac{SA}{SD}=\frac{OA}{OC}$. Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{SOT}$ of the triangle $ABD$:

$\frac{TB}{TA}\cdot \frac{SA}{SD}\cdot\frac{OD}{OB}=1$ $\Longrightarrow$ $\frac{TB}{TA}=\frac{OB\cdot OC}{OA\cdot OD}$, i.e. $\frac{TB}{TA}=\left(\frac{M_{0}B}{M_{0}A}\right)^{2}$.

Denote the reflection $M'_{0}$ of the point $M_{0}$ w.r.t. the point $T$. Then the points $M_{0}$, $M'_{0}$

are harmonical conjugate w.r.t. the points $A$, $B$. [b][u]Construct with the line and compasses[/u][/b]

the position of the minimum point $M_{0}$ for which $\boxed{\ \frac{M_{0}B}{M_{0}A}=\sqrt{\frac{TB}{TA}}\ }$. It is easily ![/color][/size]


geometry MATHCOUNTS counting distinguishability

Going to grade 8
I join
How many rounds are there? Are solutions needed?

OK i will join:

going to 7th grade
i join
i like number theory :).

What do you mean by "small prize"?

And, what level is it? (How hard?)

EDIT: So I have no chance of winning.

blah. i'll blackmail/beat up athunder until he gives me answers :P. going to ninth (i may or may not have time for this). join!

Going into 8th.
My school prevents me from skipping geometry in 8th grade. Hope to do Multi variable calc. in 9th grade!

Ya I'm pretty sure I'm already a helper already, right? Well I'm going to be in California till Sunday night, so I wont be able to help you during round 1.

Going to $ 8^{th}$ grade
I join...
My strongest are is probably geometry?

Going into 5th
I join
I'm good at algebra.

EDIT: SRY

I don't know but i was bored so i organized who joined:
[hide="participants"]
cgyao15 (joins 3 times lol)
SuperNerd123
abcak
varunrocks
PerpetualMotion
nikeballa96
yaofan
arb95
Fermatsmymom
PowerOfPi

I don't know if Thunder365 joined or just asked to be a helper
[/hide]

going to 9th
i join
er geo? at least mathcounts geo :P otherwise idk

Going into eight grade
I Join

Strongest in Algebra

[quote="varunrocks"]RoFlLoLcOpT is lying. He is going into 8th grade to join Algebra 2.[/quote]

And how would you know that?

I don't really see any motivation for me to lie...Why would you care anyway?

Sorry, I was thinking of someone else mabe?

Maybe you should think before you post something, especially an accusation.

no resources meaning no calculators?

EDIT: for #6, are all the donuts that are the same type indistinguishable?

Do we have to send solutions to every problem, or just answers for the first 9 and a proof for the last?

I think you mixed up the points too. It says 3 points for round 1.

Really? I expected fermatsmymom to get a perfect!

Also, since you gave an unlimited time limit, lots of people might turn it in later, like on sunday night or something.

Maybe you shouldn't give so much time...since it's only 10 problems.

Can I send you the answers to the first 9 problems in one PM and then the proof in another?


algebra
polynomial
greatest common divisor
number theory unsolved
number theory

Suppose$ P(x)\in Z[x]$,and $ P(a)P(b) \equal{} \minus{} (a \minus{} b)^2$ for some distinct $ a,b\in Z$.Prove that $ P(a) \plus{} P(b) \equal{} 0$

Edit: Kirill beat me to it

[hide]It is well known that for $ Q(x)\in\mathbb{Z}[x]$, $ a \minus{} b|Q(a) \minus{} Q(b)$.  So, we have $ a \minus{} b|P(a) \minus{} P(b)$.   Thus, we have $ (a \minus{} b)^2|(P(a) \minus{} P(b))^2 \equal{} P(a)^2 \plus{} P(b)^2 \minus{} 2P(a)P(b)$.  Moreover, since $ 2P(a)P(b)$ is a multiple of $ (a \minus{} b)^2$, we have

$ (a \minus{} b)^2|P(a)^2 \plus{} P(b)^2$

Let $ P(a)^2 \plus{} P(b)^2 \equal{} k(a \minus{} b)^2$.  Then, we have from $ P(a)P(b) \equal{} \minus{} (a \minus{} b)^2$ that

$ (P(a) \plus{} P(b))^2 \equal{} (k \minus{} 2)(a \minus{} b)^2$

and

$ (P(a) \minus{} P(b))^2 \equal{} (k \plus{} 2)(a \minus{} b)^2$

Thus, we must have $ k \minus{} 2,k \plus{} 2$ be perfect squares.  The only integer value for $ k$ is $ k \equal{} 2$.  But then $ (P(a) \plus{} P(b))^2 \equal{} 0$ and so $ P(a) \plus{} P(b) \equal{} 0$ as desired.
[/hide]

[quote="Kirill"]We have $ (P(a) \minus{} P(b))^2$ is divisible by $ (a \minus{} b)^2$, so $ P(a)^2 \plus{} P(b)^2$ is divisible by $ \minus{} (a \minus{} b)^2 \equal{} P(a)P(b)$.

If $ P(a) \equal{} P(b)$, then we have $ P(a) \equal{} P(b) \equal{} 0$. Let $ P(a) \neq P(b)$.

Lemma. If $ m^2 \plus{} n^2$ is divisible by $ mn \neq 0$, then $ |m| \equal{} |n|$.

Proof: suppose $ (m,n)$ - a pair of numbers.
We can think that $ gcd(m,n) \equal{} 1$ (in other side, if $ d$ - the greatest common divisor of $ m,n$, then $ (m/d, n/d)$ - "good" pair, too).
If $ m$ is divisible by prime number $ p$, then $ m^2 \plus{} n^2$ isn't divisible by $ p$, but $ mn$ is. So, $ |m| \equal{} |n| \equal{} 1$.

So, using lemma, we see that $ |P(a)| \equal{} |P(b)|$, but $ P(a) \neq P(b)$. This means that $ P(a) \plus{} P(b) \equal{} 0$.[/quote]
I have just found that proof,thank a lot
With cosinator's thank for other nice one :lol:


ARML
trigonometry
geometry
3D geometry
analytic geometry
ratio
induction

Alright, the test is finished!  Many, many thanks to Altheman for his very helpful feedback on the questions.  :)

You all know the rules: 10 minutes per pair, no calculators, etc. I'll set a preliminary deadline of midnight (Eastern) on Thursday, but if people want it extended through ARML, that's fine too.

While it is not required, a comment on how you solved the problem would be much appreciated.

I hope you enjoy the test!

[hide="Set 1"]
1) Compute the sum of all prime numbers which divide $123369369123$.

2) Compute $x$ in degrees if $0\leq x\leq 180$ and

$\cos x+\csc x+\tan x=\sec x+\sin x+\cot x$.
[/hide]

[hide="Set 2"]
3) A cone with height $6$ and base radius $r$ points upward with its apex at $(0,4)$. A second cone, identical to the first and also pointing upward, has its apex at $(0,0)$. Let $U$ be the volume of the union of the two cones and $I$ be the volume of the intersection of the two cones. If $U-I=\frac{13\pi}{6}$, compute $r$.

4) Let $A_{n}$ be the number of $n$-digit numbers with no two consecutive digits the same such that the units digit is $1$, and $B_{n}$ be the number of $n$-digit numbers with no two consecutive digits the same such that the units digit is $0$. Compute $A_{2007}-B_{2007}$.
[/hide]

[hide="Set 3"]
5) Compute the number of paths of length $20$ from $(-5,0)$ to $(5,0)$ in the coordinate plane that travel only along the lattice grid and do not pass through any lattice point $(x,y)$ with $|x|+|y|\leq 4$.

6) $a$, $b$, and $c$ are distinct positive integers, with $a<b<c$. The triangle with sides $a$, $b$, and $c$ has area $\frac{\sqrt{455}}{4}$. Compute the ordered triple $(a,b,c)$.
[/hide]

[hide="Set 4"]
7) A square is divided into four triangles by its diagonals. A circle is inscribed in each triangle, and then a square is inscribed in each circle. Compute the ratio of the total area of the smaller squares to that of the large square.

8) Set $f^{0}(x)=\frac{2x}{x+1}$ and $f^{n+1}(x)=f^{0}(f^{n}(x))$ for $n=0,1,\ldots$. Let $a_{1}, a_{2}, \ldots$ be all real numbers such that $f^{k}\left(\frac{1}{a_{i}+1}\right)$ is undefined for some positive integer $k$. Compute $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots$.
[/hide]

[hide="E^(pi*i)=-1"]In question #8, $f$ is not a defined function. Are we to assume $f$ and $f^{0}$ are the same?[/hide]

I've attached a printable version of the mock ARML, although it is not quite as nicely typeset as Altheman's usually are (maybe I should ask him for a template...).

Note that I tacitly assumed that by $f$, you meant $f^{0}$.

Edit: I've retooled the printable version slightly, including changing all the $f^{0}$'s to $f^{1}$'s, to make it more consistant with standard definitions.

[hide="About #8"]
Yes, I was using $f$ and $f^{0}$ to mean the same thing.  I have edited my post to conform to roadnottaken's PDF.
[/hide]

Eh that is kind of weird then as $f^{1}(x) = f(f(x))$.

Yeah, I know, it isn't the best notation.  But at least now it is consistent.

Come to think of it, it would have made more sence to start with $f^{1}(x)$, which is typically $f(x)$, where $f^{0}(x)$ is typically $x$.  But it still gets the point across as it stands.

Sorry to revive, but I couldn't find an answers/solutions thread for this. Does anyone have solutions they'd like to share?

[hide="Problem #1 Solutions"]
[b]Bashing/Stupid Solution/Solution 1[/b]
Notice that it is divisble by $3$ $7$ and $11$. After dividing we get
$534066533$. This is still divisible by $11$ so
$48551503$, still divisible by $11$,
$4413773$, divisible by $7$,
$630539$, divisible by $7$
$90077$, divisible by $13$,
$6929$, divisible by $13$,
$533$, divisible by $13$,
$41$, PRIME!!
So the prime factorization is $3\cdot 7^3\cdot 11^3\cdot 13^3\cdot 41$ and the sum of the distinct primes is $\boxed{75}$.
There's probably an easier way compared to this bash.

EDIT: Found an easier way  :wallbash:

[b]Clever Solution/Solution 2[/b]
Notice that the number has to be divisible by $123=3\cdot 41$, after dividing we get
$1003003001=(1000+1)^3=1001^3=7^3\cdot 11^3\cdot 13^3$
So the sum of the prime factors is $\boxed{75}$.
[/hide]

[hide="Problem #2 Solution"]
$ \cos x+\csc x+\tan x=\sec x+\sin x+\cot x$
 $\cos x+\frac{1}{\sin x}+\frac{\sin x}{\cos x}=\frac{1}{\cos x}+\sin x+\frac{\cos x}{\sin x}$
$\sin x\cos ^2 x+\cos x+\sin ^2 x=\sin ^2 x\cos x+\sin x+\cos ^2 x$
$\sin x\cos x(\cos x-\sin x)+\cos x-\sin x+\sin ^2 x-\cos ^2 x=0$
$(\cos x-\sin x)(\sin x\cos x-\sin x-\cos x+1)=0$
$(\cos x-\sin x)(\sin x-1)(\cos x-1)=0$
$\cos x=\sin x$
$x=45$
$\sin x=1$
$x=90$
However $\sec x$ and $\tan x$ would be undefined so no.
$\cos x=1$
$x=0$
However $\csc x$ and $\cot x$ would be undefined so no.
$\boxed{x= 45}$
[/hide]

[hide="Problem #3 Solution"]
Let $V$ be the volume of one of the cones. Then $U=2V-I$ so
$2V-2I=\frac{13\pi}{6}$
$V-I=\frac{13\pi}{12}$
Now $V=2\pi r^2$.
The intersection of the two cones is a smaller cone with radius $\frac{r}{3}$ and height $2$. The height is obtained by the difference between $y=0$ and $y=-2$, the bases of the two cones and the radius is obtained by similar triangles. Note that $I=\frac{V}{27}$ because they are similar.
So we desire
$V-\frac{V}{27}=\frac{26}{27}V=\frac{52\pi r^2}{27}=\frac{13\pi}{12}$
$r^2=\frac{9}{16}$
$r=\boxed{\frac{3}{4}}$
[/hide]

If there's something wrong with my solutions, please correct.

More answers :D

[hide="#6"]
The area resembles Heron's.  We see that the numerator will be equal to $\sqrt{455}$, so now we have $\sqrt{(a + b + c)(a + b - c)(a - b + c)(-a + b + c)} = \sqrt{455}$ and $(a + b + c)(a + b - c)(a - b + c)(-a + b + c) = 455$.  Note that $455 = 13 \cdot 7 \cdot 5$, and the four expressions are distinct, so we must have $a + b + c = 13$, $a + b - c = 7$, $a - b + c = 5$, and $-a + b + c = 1$ (here we just randomly assign variables; we can order them later).  The corresponding values are $3$, $6$, and $4$, so the answer is $\boxed{(3,4,6)}$.
[/hide]
[hide="#7"]
Let the side of the larger square be $2$.  Then, the semiperimeter of a small triangle is $1 + \sqrt{2}$ and the area is $1$.  By the $A = rs$ formula, the radius of the small circles is $\sqrt{2} - 1$.  Now note that the radius is half the diagonal of the smaller square, so half the diagonal is $\frac{_{\sqrt{2} - 1}}{^{\sqrt{2}}}$, or $\frac{_{2 - \sqrt{2}}}{^{2}}$.  So the side of the smaller square is $2 - \sqrt{2}$, and the area of all four is $4(6 - 4\sqrt{2})$.

Since the area of the larger square is $4$, the ratio is $\boxed{6 - 4\sqrt{2}}$.
[/hide]

[hide="#4"]
Let $S_n=A_{2007}-B_{2007}$. Notice that we can almost create a bijection between $A_n$ and $B_n$ by simply changing the units digit from $1$ to $0$ or vice versa. The only time that this does not work is when the tens digit is $0$ or $1$. There are $B_{n-1}$ exceptions in the first case, and $A_{n-1}$ exceptions in the other case. Thus, $A_n-B_n=B_{n-1}-A_{n-1}$, or $S_n=-S_{n-1}$. Since $S_2$ can easily be verified to be $-1$, $S_{2007}=\boxed{1}$.
[/hide]

[hide="#5"]
Note that we must go to the right $10$ times, we must go up $5$ times, and we must go down $5$ times. Putting this together, we find that we must go through either $(0,5)$ or $(0,-5)$. These two cases are the same by symmetry.
There are $C_5=\frac{1}{5+1}{{10}\choose{5}}=42$ ways to get from $(-5,0)$ to $(0,5)$ without going over the segment connecting $(-4,0)$ to $(0,4)$ (where $C_n$ is the $n$th Catalan number). Thus, the answer is $2\times 42^2=\boxed{3528}$.
[/hide]

[hide="#8?"]
Let $f^n(x)$'s inverse be $g^n(x)$.  Then, we have $g^0(f^{n + 1}(x)) = f^n{x}$.  So basically, we apply $g$ over and over again until we reach something undefined.  But what is $g^0{x}$?  This happens when $x = \frac{2g^0(x)}{g^0(x) + 1}$, so $g^0(x) = \frac{x}{2 - x}$.  Plugging this into $g^0(x)$ again we have $\frac{x}{4 - 3x}$.  Plugging into $g^0(x)$ again we have $\frac{x}{8 - 7x}$.

By now we realize that the $n$th term is $\frac{x}{2^n - (2^n - 1)x}$.  Just to prove it, here:
[hide="using induction, of course"]
Base case is $\frac{x}{2 - x}$, true, blah blah blah...
Now for the inductive step, we assume that this is true.  Plugging it back into $g^0(x)$ we have $\frac{\frac{x}{2^n - (2^n - 1)x}}{2 - \frac{x}{2^n - (2^n - 1)x}}$, or $\frac{x}{2^{n + 1} - 2^{n + 1}x + x}$ which is $\frac{x}{2^{n + 1} - (2^{n + 1} - 1)x}$ which is the desired result.
[/hide]

Okay good, now that we've proved the above.  Plugging in $\frac{1}{a_i + 1}$ in we have $\frac{\frac{1}{a_i + 1}}{2^n - \frac{2^n}{a_i + 1} + \frac{1}{a_i + 1}}$, or $\frac{1}{(a_i + 1)2^n - 2^n + 1}$, which is $\frac{1}{a_i \cdot 2^n + 1}$.  So this works if $a_i \cdot 2^n = -1$, or $a_i = -\frac{1}{2^n}$.  So the terms are $-\frac{1}{2}$, $-\frac{1}{4}$, and so on.  By the $\frac{a}{1 - r}$ formula the answer is $\boxed{-1}$.
[/hide]

Oops!  Beaten by a lot!

[hide="#1, how I did it"]

Clearly, the number will be divisible by 123. Dividing by 123 gives $1003003001$ (you can consider 123 369 369 123 as seperate blocks). You should recognize 1, 3, 3, 1, but if you don't, write it as $10^9+3 \cdot 10^6+3 \cdot 10^3+1$ to make it clearer. Anyway, this is $(10^3+1)^3=1001^3=7^3 \cdot 11^3 \cdot 13^3$, so the answer is $3+41+7+11+13=75$. [/hide]

[hide="#3"]

Write the volume of the cone as x. By similarity, we have $I=\frac{x}{27}$, and $U=2x-I$. Thus, $U-I=2x-2I=2x-\frac{2x}{27}=\frac{52x}{27}$. Writing $x=2\pi r^2$ and solving gives $x=\frac{3}{4}$.[/hide]

[hide="#6"]
By herons, we have $\sqrt{s(s-a)(s-b)(s-c)}=\frac{\sqrt{455}{4}} \implies s(s-a)(s-b)(s-c)=\frac{455}{16} \implies p(p-2a)(p-2b)(p-2c)=455$. Since $a<b<c, we have p-2a>p-2b>p-2c$, and thus $p=13, p-2a=7, p-2b=5, p-2c=1$. Solving this gives $(a,b,c)=(3,4,6)$.[/hide]

all problems have been posted in separate threads in HSM  , I just created the related post collections

[url=https://artofproblemsolving.com/community/c3734441_2007_mock_arml_ii_individual]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


Find all triples (a,b,c) of positive integers such that \[(1 + \frac{1}{a} ) (1 + \frac{1}{b} ) (1 + \frac{1}{c} )=3\]

Note that:

3 = (1+1/a)(1+1/b)(1+1/c)
3^(1/3) = [(1+1/a)(1+1/b)(1+1/c)]^(1/3) <= 1/3*(3+1/a+1/b+1/c)
3*(3^(1/3)-1) <= 1/a+1/b+1/c

It then becomes clear that one of a,b,c are equal to 1 or 2. Examining all the few related cases gives the only solutions (1,4,5) and (2,2,3) (along with all the corresponding permutations of course).

Close. But you do not have the complete solutions.
[hide]We start with the following equation-
\[\left(1+ \frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = 3\]
\[\left(1+\frac{1}{b}+\frac{1}{a}+\frac{1}{ab}\right)\left(1+\frac{1}{c}\right)= 3\]
\[\frac{abc+ab+ac+bc+a+b+c+1}{abc} = 3\]
\[(c+1)(a+b+1) = ab(2c-1)\]
If $c>2$, $2c-1>c+1$. If $a\geq2$, $b>3$, then $ab>a+b+1$. Under these conditions the equality above cannot be true since $L.H.S<R.H.S.$.
Thus, at least one of the the three conditions must be violated so that the equality is true.
Since the variables, $a$, $b$ and $c$ are interchangeable, if the equation is solved for $a=k$, where $k$ is an integer, it does not need to be solved for $b=k$ and $c=k$.
Thus, $a$ can be 1, 2, 3.

If $a = 1$, we get

\[(c+1)(b+2) = b(2c-1)\]
\[c = \frac{2b+2}{b-2}\]

Thus,

\[2b+2 \cong 0 (mod \mbox{ }b-2)\]
\[6 \cong 0 (mod \mbox{ } b-2)\]

Thus, $b-2$ can be 1, 2, 3, 6. Therefore, $b$ is 3, 4, 5, 8, giving the unique triplets $(1,3,8)(1,5,4)$.
If $a=2$, we get

\[(c+1)(b+3) = 2b(2c-1)\]
\[b = \frac{c+1}{c-1}\]

Thus,

\[c+1 \cong 0 (mod \mbox{ }c-1)\]
\[2 \cong 0 (mod \mbox{ }c-1)\]

Thus, $c-1$ can be $1$ or $2$. Thus, $c$ is $2$ or $3$, giving the only unique triplet $(2,3,2)$. Finally, if $a = 3$, we get

\[(c+1)(b+4) = 3b(2c-1)\]
\[c = \frac{4b+4}{5b-4}\]

Thus,

\[4b+4 \cong 0 (mod \mbox{ }5b-4)\]
\[36 \cong 0 (mod \mbox{ }5b-4)\]

Here, $5b-4$ can be 1, 2, 3, 4, 6, 9, 12, 18, 36. The only integer solutions for $b$ are 1, 2, 8, giving the triplets $(3, 1, 8)(3, 2, 2)(3,8,1)$, which were obtained earlier.
Therefore the only unique triplets that satisfy the equality are $(1,3,8)(1,4,5)(2,2,3)$. [/hide]


if we have $x^2+y^2+z^2=1$ then find the maximum value of $2x+3y+6z$.

Yes jensen it is really easy. And probably posted in the some thread of Pre-Olympiad.

[hide=""Solution""]
Using Cauchy Schwarz Inequality[/hide]

Well, if it was easy then why did you post it. Maybe your reasoning will inspire me too to post problems which are easy. :D

[quote="jensen"]i think we should solve easy problems too,to make our selves strong.[/quote]

That's a good reason and a very important advice for me. Thanksss. :)
You may like to see my signature.

[quote="kunny"]You should show the values $x,y,z$ which give the maximum value. ;)[/quote]

Good call.

[hide]
Cauchy equality condition is reached when $\frac{4}{x^2} = \frac{9}{y^2} = \frac{36}{z^2} \Rightarrow x = \frac{2}{7}, y = \frac{3}{7}, z = \frac{6}{7}$.[/hide]

I don't think you forgot the condition.  :)

[quote="shobber"]$\frac{4}{x^2} = \frac{9}{y^2} = \frac{36}{z^2} \Rightarrow x = -\frac{2}{7}, y = -\frac{3}{7}, z = -\frac{6}{7}$ should also be an answer.[/quote]

Well Cauchy only applies to positive reals, so yeah... and you basically implicitly assume positives since the maximum is obviously going to occur with positive $x,y,z$.

Yep. I was wrong. I was talking about the minimum.


(A.Akopyan, V.Dolnikov) Given a set of points inn the plane. It is known that among any three of its points there are two such that the distance between them doesn't exceed 1. Prove that this set can be divided into three parts such that the diameter of each part does not exceed 1.

Nice problem!

We will first dispatch of the case where there are two points which are at least $\sqrt3$ apart. Let $S$ denote the set of points in question.

[b] Case 1. [/b] The diameter of $S$ is $\ge \sqrt 3.$

Let $A, B$ be points in $S$ so that $AB \ge \sqrt3.$ Let $\omega_A, \omega_B$ denote the unit circles centered at $A, B,$ respectively. Let $\omega$ denote the region which is the (closed) intersection of $\omega_A$ and $\omega_B.$ Note that since $AB \ge \sqrt 3$, $\omega$ is either empty or has diameter at most $1.$

We claim that all of the points of $S$ in $\omega_A / \omega$ comprise a set of diameter at most $1.$ Indeed, if not, then consider two points $P_1, P_2 \in (\omega_A / \omega) \cap S$ so that $P_1 P_2 > 1.$ Then $P_1, P_2, B$ contradict the condition of the problem.

A similar thing holds for $S \cap (\omega_B / \omega),$ and so from the previous results we know that $S \cap (\omega_A / \omega), S \cap (\omega_B / \omega), S \cap \omega$ is a partition of $S$ into three parts, each of diameter at most $1.$

[b] Case 2. [/b] The diameter of $S$ is at most $\sqrt 3.$

Notice that for any three points $A, B, C$ so that $AB, BC, CA \le \sqrt 3$, there is a unit circle covering $\triangle ABC.$ Indeed, simply consider cases on whether $\triangle ABC$ is acute or obtuse (if acute, take the circumcircle; if obtuse take the circle with the longest side of $\triangle ABC$ as diameter).

Hence, by Helly's Theorem, the set of unit circles centered at the points in $S$ all have a common point, say $P$. This means that the entirety of $S$ is contained in the unit circle centered at $P.$

Let $Q \in S$ be the point furthest from $P$, and let $PQ = r \le 1.$ Let $\omega_P$ denote the circle centered at $P$ of radius $r$; note that $S \subset \omega_P.$ Let $\omega_Q$ denote the unit circle centered at $Q.$ Note that the set of points of $S$ contained in $\omega_P / \omega_Q$ has diameter at most one, as else we can obtain a contradiction with two points of distance $>1$ and $Q.$

It only remains to notice that $\omega_P \cap \omega_Q$ can be split into two sets of diameter $1$ rather easily, by simply bisecting it along line $PQ.$ From here, the partition of $S$ is clear.

As we've exhausted all cases, we're done.

$\square$


function
trigonometry
complex analysis
complex analysis unsolved

Let
\[ u(x,y) \equal{} \frac{\sin x}{\cos x \plus{} \cosh y}, \qquad |x| < \pi, \qquad y \in \mathbb{R}.
\]
Find the analytic function $ f(z),$ defined in the same band $ |\Re z| < \pi$,
and satisfying $ f(0) \equal{} 0.$

Are $ f$ and $ u$ assumed to be related in any way?


Hope i don't make stupid mistakes.
We need to prove:
            $\frac{1}{5-2ab}+\frac{1}{5-2bc}+\frac{1}{5-2ca} \leq 1$
            $\Longleftrightarrow \sum \frac{1-ab}{5-2ab} \geq 0$
            $\Longleftrightarrow \sum \frac{1-a^3b^3}{(5-2ab)(1+ab+a^2b^2)} \geq 0$
Put:$f(x)=(5-2x)(1+x+x^2)$ ($x \in R$)
    $\Longrightarrow f(x)=-2x^3+3x^2+3x+5$ for all $x \in R$
we will prove that $f(ab) \geq f(ac)$ when $ab \geq ac$.Put $x=ab,y=ac$
In fact;$f(x)-f(y)=(x-y)(3+3(x+y)-2(x^2+xy+y^2))$
We have $(ab+ac)^2=a^2(b+c)^2 \leq 2a^2(b^2+c^2) \leq \frac{9}{2}$
            $\Longrightarrow (x+y) \leq \frac{3}{\sqrt{2}}$
Now:$3+3(x+y)-2(x^2+xy+y^2) >3+3(x+y)-2(x+y)^2 >0$ (because $x+y <\frac{3}{\sqrt{2}}$)
So that:$f(ab) \geq f(ac)$ when $ab \geq ac$
Wlog $ab \geq bc \geq ac$.Hence we have $f(ab) \geq f(bc) \geq f(ac)$
Apply Trebusep's inequality we have:
            ${{\sum \frac{1-a^3b^3}{(5-2ab)(1+ab+a^2b^2)}=\sum \frac{1-a^3b^3}{f(ab}} \geq 3(3-\sum a^3b^3)(\sum \frac{1}{f(ab}})$
Use your lemma:when $x,y,z \geq 0$ and satifying $x+y+z=3$ then we have $(xy)^k+(yz)^k+(zx)^k \leq Max\{3,(\frac{3}{2})^k\}$ ($k \geq 0$)
So that $a^3b^3+b^3c^3+c^3a^3 \leq 3$ from the condition $a^2+b^2+c^2 =3$
Therefore $\sum \frac{1-a^3b^3}{(5-2ab)(1+ab+a^2b^2)} \geq 0$ (QED)
The equality hold when $a=b=c=1$ :D  :D

There is a problem:
 $a^2+b^2+c^2=3$ doesn't imply $a^3b^3+b^3c^3+c^3a^3 \leq 3$.
 For example, a=0 and b=c.
 You mistook it for
 $a^3+b^3+c^3=3$ implies $a^4b^4+b^4c^4+c^4a^4 \leq 3$.  :blush:

but then the solution would need just a small change?!

$f(x)=(5-2x)(1+x)(1+x^2)$

[quote="Vasc"]Prove that if $a,b,c$ are non-negative reals such that $a^2+b^2+c^2=3$, then:

$\frac{1}{5-2ab}+\frac{1}{5-2bc}+\frac{1}{5-2ca} \leq 1$.[/quote]
I think this can be solved in the same manner as my two solutions to Vasc's inequality
\[ \frac1{1-ab}+\frac1{1-bc}+\frac1{1-ca}\leq\frac92, \]
with $a,b,c\geq0$ and $a^2+b^2+c^2=1$.

All three problems

\[ \frac{1}{5-2ab}+\frac{1}{5-2bc}+\frac{1}{5-2ca} \le 1 \]
\[ \frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-1ca} \le \frac{9}{2} \]
\[ \frac{1}{3-abc}+\frac{1}{3-bcd}+\frac{1}{3-cda}+\frac{1}{3-abc} \le 2 \]

Can be solved by the same method. And I think It's still vaild for

\[ \frac{1}{7-3abc}+\frac{1}{7-3bcd}+\frac{1}{7-3cda}+\frac{1}{7-3abc} \le 1 \]

Indeed it can be generalized into n-var ;)


Let P(x) be a polynomial; $\large Q(x) \in Z[x]$. Suppose: $\large T(x)=P(x^3)+xQ(x^3)$; T(x) is divisible by $\large (x^2+x+1)$. Prove that: $\large gcd(P(2000); Q(2000))$ is divisible by 1999.

I assume you mean that $P, Q$ have integer coefficients, right? Then it is easy: since $T$ is divisible by $x^2+x+1$, it has the third roots of unity $\omega, \omega^2$ as roots. Since $\omega^3=1$, plugging into the original equation this leads to $0=P(1)+\omega Q(1)=P(1)+\omega^2 Q(1)$. Then $(\omega^2-\omega)Q(1)=0$, so $Q(1)=0$. Then $P(1)=0$ too. This means that $P, Q$ are divisible by $x-1$, so in particular $P(2000), Q(2000)$ are divisible by $1999$.


Let $ S(n)$ is the sum of digits of number $ n$. Find all positive integers $ n$ such, that:

$ \textbf{(1)}\; S(n)(S(n) \plus{} 1) \equal{} n$
$ \textbf{(2)}\; S(n)(S(n) \plus{} k) \equal{} n$ for fixed $ k\in \textbf{Z}$
$ \textbf{(3)}\; S(n)^{3} \equal{} 3n \minus{} 56$

$ (1)$ is easy. Here's a solution:
Let $ n$ have $ t$ digits.
Then, $ n\geq 10^{t \minus{} 1}$ and $ S(n)\leq 9t$. We have $ S(n)(S(n) \plus{} 1) \leq 9t(9t \plus{} 1)$ and, therefore $ 10^{t \minus{} 1}\leq 9t(9t \plus{} 1)$. We can then see, since the left size grows much faster than the right, that $ t\leq 4$.
If $ t \equal{} 1$, $ S(n) \equal{} n$, we have $ n^2 \plus{} n \equal{} n$, which leads us to a contradiction.
If $ t \equal{} 2$, let $ n \equal{} 10a \plus{} b$ ($ a > 0$, $ 0\leq b\leq 9$). We have $ (a \plus{} b)^2 \equal{} 9a$. Then, $ a \equal{} 1$, $ a \equal{} 4$ or $ a \equal{} 9$. If $ a \equal{} 1$, we have $ b \equal{} 2$ ($ n \equal{} 12$). If $ a \equal{} 4$, we again have $ b \equal{} 2$ ($ n \equal{} 42$). If $ a \equal{} 9$, we have $ b \equal{} 0$ ($ n \equal{} 90$).
If $ t \equal{} 4$, let $ n \equal{} 1000a \plus{} 100b \plus{} 10c \plus{} d$ (same as before, only with $ 4$ digits). Then we have $ (a \plus{} b \plus{} c \plus{} d)^2 \equal{} 9(111a \plus{} 11b \plus{} c)$. However $ (a \plus{} b \plus{} c \plus{} d)^2\leq {36}^2$ so $ 111a \plus{} 11b \plus{} c\leq \frac {{36}^2}{9} \equal{} 144$. Hence, $ a \equal{} 1$. However, if $ a \equal{} 1$ then $ (a \plus{} b \plus{} c \plus{} d)^2\leq {27}^2$ ($ a \plus{} b \plus{} c \plus{} d$ must be divisible by $ 3$), so $ 111a \plus{} 11b \plus{} c\leq 81$, and that gives us a contradiction.
If $ t \equal{} 3$, let $ n \equal{} 100a \plus{} 10b \plus{} c$ (same as before, only with $ 3$ digits). Then we have $ (a \plus{} b \plus{} c)^2 \equal{} 9(11a \plus{} b)$. Again, we try to limit $ a$. $ (a \plus{} b \plus{} c)^2\leq {27}^2$, so $ 11a \plus{} b\leq \frac {{27}^2}{9} \equal{} 81$. Hence, $ a\leq 7$. But then $ (a \plus{} b \plus{} c)^2$, which must be divisible by $ 9$ is less or equal to $ {24}^2$. From this we gather $ 11a \plus{} b\leq 64$, which leads to $ a\leq 5$. We do the similar argument once more $ (a \plus{} b \plus{} c)^2\leq {21}^2$, so $ 11a \plus{} b\leq 49$, so $ a\leq 4$.
If $ a \equal{} 1$, we have $ 180\geq 9(11a \plus{} b)\geq 99$, which means $ 13\geq a \plus{} b \plus{} c\geq 10$. However, $ a \plus{} b \plus{} c$ must be divisible by $ 3$, so $ a \plus{} b \plus{} c \equal{} 12$. From here we get $ n \equal{} 12(12 \plus{} 1)$, $ n \equal{} 156$. We check and see this is one of the solutions.
If $ a \equal{} 2$, we have $ 279\geq 9(11a \plus{} b)\geq 198$, which means $ 16\geq a \plus{} b \plus{} c\geq 15$. However, $ a \plus{} b \plus{} c$ must be divisible by $ 3$, so $ a \plus{} b \plus{} c \equal{} 15$. From here we get $ n \equal{} 15(15 \plus{} 1)$, $ n \equal{} 240$. However, this is obviously not a good solution.
If $ a \equal{} 3$, we have $ 378\geq 9(11a \plus{} b)\geq 297$, which means $ 19\geq a \plus{} b \plus{} c\geq 18$. However, $ a \plus{} b \plus{} c$ must be divisible by $ 3$, so $ a \plus{} b \plus{} c \equal{} 18$. From here we get $ n \equal{} 18(18 \plus{} 1)$, $ n \equal{} 342$. However, this is obviously not a good solution.
If $ a \equal{} 4$, we have $ 477\geq 9(11a \plus{} b)\geq 396$, which means $ 21\geq a \plus{} b \plus{} c\geq 20$. However, $ a \plus{} b \plus{} c$ must be divisible by $ 3$, so $ a \plus{} b \plus{} c \equal{} 21$. From here we get $ n \equal{} 21(21 \plus{} 1)$, $ n \equal{} 462$. However, this is obviously not a good solution.

So, we reach the conclusion: The only solutions are $ 12$, $ 42$, $ 90$ and $ 156$.

I might have missed something somewhere (wrongly limiting or something), but I think this is generally OK. I will try to see about $ (2)$ and $ (3)$, but I don't guarantee anything.

EDIT: You can also limit the number of digits of $ n$ in $ (3)$ to less than or equal to $ 5$. A similar strategy using further limiting might solve the task, however, I'm not sure and it does seem somewhat more complex. I'll keep trying.
EDIT2: If nothing, you can always try all possible $ S(n)$ in $ (3)$. Here's the solution:
$ n \equal{} \frac {S(n) \plus{} 56}{3}$. Now, since $ n$ has less than or equal to $ 5$ digits (Use similar argument as in $ (1)$: Let $ n$ have $ t$ digits. Then, $ n\geq 10^{t \minus{} 1}$ and $ S(n)^3\leq (9t)^3$. We have $ 3\cdot 10^{t \minus{} 1} \minus{} 56\leq 9^3\cdot t^3$. We can then see, since the left size grows much faster than the right, that $ t\leq 5$.) So, $ S(n)\leq 45$. However, since $ 3n \minus{} 56$ gives $ 1 (mod 3)$, then $ S(n)$ gives $ 1 (mod 3)$. You can just try the following possibilities for $ S(n)$: $ 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43$. As you can see there's not too much of them, but that's really not a nice solution.


Posted many times.
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=2002+taiwan&t=41543

This is problem is great than above problem:
Find all $n$ such that for all $0<x_{1},x_{2},...,x_{n}\leq\frac{1}{2}$ we have
$\frac{\prod_{i=1}^{n}x_{i}}{\prod_{i=1}^{n}(1-x_{i})}\leq\frac{\sum_{i=1}^{n}x_{i}^{n}}{\sum_{i=1}^{n}(1-x_{i})^{n}}$.


Thank you in advance to whoever answers this for me. Appreciate it so much.

If the temperature  at 8:00 a.m. was 10 degrees C. and at 4:00 p.m. the same day it was -6 degrees C., what was the average decrease per hour during this period?

A. 0.5 degrees C

B. 1.5 degrees C

C. 2 degrees C

D. 4 degrees C

E. 8 degrees C

[quote="nick42"]temperature went down 16 degrees in 8 hours. so the average is 16/[b]2[/b]= $ \boxed{2}$ degrees C.

answer

[b]C[/b][/quote]

I believe you mean 8.

The average decrease is $ \frac{\text{total decrease}}{\text{time}}$. For total decrease, we have $ 10\minus{}(\minus{}6)\equal{}16$. For time, 4:00 p.m. - 8:00 a.m. = 8 hours.

So, the average decrease is $ \frac{16}{8}$, or 2 degrees Celsius every hour.

[quote="fishythefish"]dgemor, generally speaking, average changes will be found using

change/time.

Or if a problem asks "An [i]x[/i] foot tall tree grew [i]y[/i] feet. What is the percent change?"

Then the answer will be the change over the previous height. (y/x) And then convert to a percent: 100y/x %

And if you have trouble with finding how much time has elapsed, try using 24-hour timing (1 pm = 13, 2 pm = 14, etc.).[/quote]
More simply
If $ y$ is the bigger number and $ x$ is the smaller number...

$ y\minus{}x$/the original number

Multiply the answer by 100 to convert to a percent.

2 degrees, because if you do the formula

D=RT

The time is 8 hours, and the distance is 16 degrees, so you have;

16 = 8R

Then if you divide by eight you get 2 degrees as your answer.


ratio
geometry
circumcircle
geometry unsolved

ABC is an isosceles triangle (AB=BC) and <B= $ \beta$ .the line that pass from the middles of AB and BC sides ,intersect the circumcircle of ABC at D,E.(DE ll AC). find the ratio of areas of ABC and DBE.

We have $ \angle BDE \equal{} \angle BED \equal{} \angle DAB$. Let $ M$ the midpoint of $ AB$. Since the angle $ \angle ABD$ is common, triangles $ BDM$ and $ BAD$ are similar. Thus $ BD^2 \equal{} BM.BA \Rightarrow BD \equal{} \frac {AB}{\sqrt {2}}$. Let $ N$ the midpoint of $ DE$. Clearly $ BN \equal{} \frac {1}{2}ABcos\frac {\beta}{2}$. Therefore $ DN^2 \equal{} \frac {AB^2}{2} \minus{} \frac {AB^2}{4}cos^2\frac {\beta}{2} \Rightarrow DN \equal{} \frac {AB}{2}\sqrt {2 \minus{} cos^2\frac {\beta}{2}}$. If $ P$ is the midpoint of $ AC$, the ratio is

$ \frac {BP.AP}{BN.DN} \equal{} \frac {BN.AC}{BN.DN} \equal{} \frac {AC}{DN} \equal{} \frac {4sin\frac {\beta}{2}}{\sqrt {2 \minus{} cos^2\frac {\beta}{2}}}$

Beautiful, isn't it?


find all integral values n for which the polynomial
$f(n)=n^{4}-4n^{3}+10n^{2}-12n+86$
is aperfect square

[hide="A hint"]Try subtracting 77 from that expression.[/hide]

Another problem in a similar vein: compute all integer values of $n$ such that $n^{4}-4n^{3}-n^{2}+10n+7$ is a perfect square.


quadratics
modular arithmetic
number theory
relatively prime
absolute value

Hi,

In Chapter 23, of AoPS Vol. 2, it says: If there is a number $ a$ such that $ a|bc,$ we can find two numbers $ a_1,a_2,$ such that $ a_1a_2\equal{}a,$ and $ a_1|b,$ and $ a_2|c.$

That I am clear with. But after that it says: If $ p$ is prime, the only way to split it into two parts is as $ p \cdot 1.$ Thus, if $ p|bc,$ then we have $ p|b,$ or $ p|c.$

Does this assume that $ b$ and $ c$ are relatively prime, because I don't think it is mentioned in the book?

And also, I don't really understand the proof of why a prime number greater than 2 must always have the maximum number of residues for odd numbers, which is $ (p\plus{}1)/2.$ More precisely, I am not sure why the first $ (p\minus{}1)/2$ terms cannot have any duplications.

Thank you :)

1. No, $ b$ and $ c$ do not need to be relatively prime for the property that $ p \mid bc$ implies $ p \mid b$ and/or $ p \mid c$.

2. If $ x^2 \equiv y^2 \pmod{p}$ with $ 0 \leq x, y \leq (p\minus{}1)/2$, then $ (x\minus{}y)(x\plus{}y) \equiv 0 \pmod{p}$, but both factors are less than $ p$ in absolute value, so this implies $ x\equal{}y$, therefore there are no duplications.

While on a similar topic, I am just learning about fermat's little theorem, and about primitive roots, so the things kind of seem all over the place.

So, just to clarify up a few things:

1) Why is the period of a number at most p-1?

2) So, a primitive root, by definition, is a number with the property that the period of it is exactly p-1? Or is there more to it?

Thanks :)

EDIT: Thank you very much for the reply mavropnevma :)

1. Not only the period is at most $ p\minus{}1$, but it divides $ p\minus{}1$. This is because as you calculate $ a^k$, $ k \equal{} 1,\ldots,p\minus{}1,p$ (where $ a$ not divisible by $ p$), some remainder must repeat (there are only $ p\minus{}1$ not-null). Further analysis proves Fermat's.
2. Correct!

I agree that the period must divide $ p \minus{} 1.$ And also that some of them must repeat because there are only a number of residues in modulo m, for any m. But I am still somewhat not "convinced" that the most the period can be is 1.

Thank you. :)

[quote]But I am still somewhat not "convinced" that the most the period can be is 1. [/quote]

You most likely mean ... the most the period can be is $ p\minus{}1$. But if you agree that period of length $ \ell \mid p\minus{}1$, then $ \ell \leq p\minus{}1$, no?

mavropnevma, yes, that what I meant. I haven't exactly though about it that way. I was thinking in terms of the residues. But that reasoning works a charm.

Now that is clear, thank you. :)

But looking back at my initial question about the duplications, I still don't exactly understand it.
Here's how Vol. 2 has it:

     Let $ p$ be a prime other than 2. If there are two numbers $ u$ and $ v$ such that $ u^2 \equiv v^2 \pmod{p},$ so that $ u^2\minus{}v^2 \equal{} (u\plus{}v)(u\minus{}v)$ is a multiple of $ p.$ Since $ p$ is prime, this implies that either $ u\minus{}v$ is a multiple of $ p,$ so that $ u \equiv v \pmod{p},$ or $ u\plus{}v$ is a multiple of $ p,$ so that $ u \equiv \minus{}v \pmod{p}.$
     Thus, when we write down the quadratic residues $ 0^2, 1^2, \dots, (p\minus{}1)^2,$ there can be no duplications, except that the last $ (p\minus{}1)/2$ are copies of the first $ (p\minus{}1)/2$ nonzero squares, much like descried in Example 23-6. In other words, an odd prime $ p$ always has the maximum $ (p\plus{}1)/2$ residues.

1) But here it says that: since $ p|(u\plus{}v)(u\minus{}v),$ we have either $ p|(u\minus{}v),$ or $ p|(u\plus{}v).$ So, does this mean that it can be $ p|(u\minus{}v)$ and/or $ p|(u\plus{}v)?$

2) What am I missing here that implies that there can be no duplications in the first $ (p\minus{}1)/2$ nonzero terms?

Ok, now we're talking ... The  textbook assumes $ u^2 \equiv v^2 \pmod{p}$ and reaches $ u \equiv v \pmod{p}$ or $ u \equiv \minus{}v \pmod{p}$. So, except the case $ u \equal{} v \equal{} 0$ (when in effect, [b]both[/b] $ u\equal{}v$ [b]and[/b] $ u\equal{}\minus{}v$), all other cases of equality are grouped in pairs $ (u, \minus{}u)$, since $ u \equiv \minus{}u \pmod{p}$ would be equivalent to $ 2u \equiv 0 \pmod{p}$, i.e. $ u \equiv 0 \pmod{p}$ ($ p$ is odd).
Therefore there are $ 1 \plus{} \frac {p\minus{}1} {2} \equal{} \frac {p\plus{}1} {2}$ distinct quadratic residues.
Now, modulo $ p$, $ \minus{}u$ and $ p\minus{}u$ are the same, so the $ \frac {p\minus{}1} {2}$ residues in the second half of the list repeat the $ \frac {p\minus{}1} {2}$ non-zero residues from the first half, which are distinct, according to the above.

I don't quite understand this part:
[quote]all other cases of equality are grouped in pairs (u, -u), since u \equiv -u \pmod{p} would be equivalent to 2u \equiv 0 \pmod{p}, i.e. u \equiv 0 \pmod{p} (p is odd). [/quote]

I don't see how they are grouped as $ (u,\minus{}u),$ and how that implies $ u \equiv \minus{}u \pmod{p}.$ The rest is fine.

Thanks :)


inequalities
ratio
geometry
similar triangles
perpendicular bisector

Because I am quite sick of all of the inequalities that are being posted (really, people, enough is enough, eh?), I am going to start posting random non-inequality problems that I enjoy.

First up,

Construct a right triangle with a givne hypotenuse, if it is given that the median to the hypotenuse is equal to the geom mean of the 2 legs.

[quote="jelyman"]Because I am quite sick of all of the inequalities that are being posted (really, people, enough is enough, eh?), I am going to start posting random non-inequality problems that I enjoy.[/quote]

You go girl.

But inequalities are cool...

One VERY VERY CONVOLUTED solution:

1. Construct a right angle with one side parallel to c. Mark this side x.
2. Trisect this right angle. (Trisection of a right angle is doable with a straightedge and compass.)
3. Trisect one of the 30o angles to obtain a 15o angle, with one of its sides x.
4. Mark the midpoint of c as O. construct circle O with radius c/2.
5. Translate the 15o angle to c such that the angle is contained in the circle, and its vertex falls on the circle. Mark the translated segment's other intersection with the circle O as P.

The required triangle is the triangle formed with segment c and point P.

Proof:

Suppose c = 2. Then we're required to find a and b such that a2 + b2 = 4 and :sqrt:(ab) = 1.

From the second equation obtain b = 1 / a. Sub this into the first equation:

a2 + 1 / a2 = 4

a2 + 2 + 1 / a2 = 6

(a + 1 / a)2 = 6

a > 0, thus a + 1 / a > 0, thus a + 1 / a = :sqrt:6.

a = (:sqrt:6 :pm: :rt2:) / 2.

Suppose a represents the shorter side, then a = (:sqrt:6 - :rt2:) / 2. Then,

sin (angle between b and c) = a / c = (:sqrt:6 - :rt2:) / 4 = sin 15o.

Thus the angle opposite the shorter side MUST have measure 15o!

So we're done for c = 2. Then by similar triangles, all other cases of c hold, so the construction is valid.

ruler and compass?

Analysis: let h be the altitude on the hypotenuse c, then hc=ab where a and b are the two legs. So h = ab/c = c/4 since sqrt(ab) = c/2.

Construct a circle with hypotenuse c as the diameter.

Constrcut a radius (c/2) perpendicular to the diameter(the hypotenuse).

Construct a  perpendicular bisector of the radius.

Any one of intersection points of the perpendicular bisector with the circle together with two end points of the diameter forms a right triangle satisfying the requirement.

Used algebra to find ratio of legs are [tex]2+\sqrt{3}:1[/tex].  Then constructed triangle to match.  See attached illustration.  Similar to previous constructions, so I won't re-explain.

Also shows a construction of the geometric mean of the two sides of the right triangle.  Gray arc demonstrates (non-rigorously) that geometric mean is equal in length to the median of the right triangle.

Is there a solution that is algebra-free?


the perimeter of a right triangle is 60
the height to the hypotenuse is 12
what is the area?

how can you do this w/o messy equations? and without knowing right triangles that will work, not guess/check?

Oh right... eh.  :blush:

That makes things different. Then there's lots of algebra.

ummm
how do you get $\frac{a+b}{c}$?  is it ab/c=12?

the answer, from guessing, is 15-20-25 triangle

[quote="themonster"][hide]
Sides a, b, c=12
$a+b+c=60 \Rightarrow a+b=48\\
a^2+b^2=c^2=144\\
(a+b)^2=a^2+2ab+c^2=2304\\
144+2ab=2304\\
2ab=2160\\
\frac{1}{2}ab=540$
which is the area

[/hide][/quote]

well i think the question says the [b]Height[/b] to the hypotenuse is 12, not the hypotenuse itself......

[quote="themonster"][quote="shobber"]
well i think the question says the [b]Height[/b] to the hypotenuse is 12, not the hypotenuse itself......[/quote]
Sorry, i didn't read the question well enough  :blush:[/quote]

themonster, here's a hint. If the height of the hypotenuse is 12, then either a or b have to be 12. Now use the same method you used before, and solve. :lol:

[quote="d343seven"]no, a or b both have to be greater than 12 because of the right triangles formed by the altitude to c, ie, the hypotenuse has to be longer than the legs.

does anyone have a solution that would lead to the answer of 15,20,25 right triangle?

this was a mathcounts national sprint round..so do you think they expected people to know the answer by knowlegde of right triangles or actually solve it some way?[/quote]

I thought that the height of the hyptunese is the same thing as the height of the triangle?

The fact that it's a right triangle means that $a^2 + b^2 = c^2$, and it's perimeter being 60 means that $a + b + c = 60$.

The last condition, that the height to the hypotenuse is 12, means that $a*b = 12*c$. Do you see why? (think about the area of the triangle from two different perspectives).

[hide="So we have that"]
$a + b = 60 - c$
$(a + b)^2 = (60 - c)^2$
$a^2 + 2ab + b^2 = c^2 - 120c + 60^2$

Rearranging and substituting the other conditions gives:
$(a^2 + b^2) + 2ab = c^2 - 120c + 60^2$
$c^2 + 2(12c) = c^2 - 120c + 60^2$
$24c = -120c + 60^2$
$144c = 60^2$
So $c = 25$.

The area of the triangle is then $1/2*12*c = 150$.[/hide]

Marcus, you're back!!!   :)  :cool: Where have you been?  (you can PM me so that we don't go off track in this thread).


induction
advanced fields
advanced fields unsolved

If the torus with $ k$ holes is cut along $ k\plus{}1$ disjoint
circles, prove that it is not connected anymore.

It took me quite some time to understand the problem, but I am still not sure:  Do you mean "circles " in the punctured torus by just circles in the unpuctured torus, i.e. it is possible that the circle contains the puctured point and it is still called a "circle"?  And more, two circles intersecting only at puncture points are considered disjoint, is that what you mean?  Otherwise I don't see why you need $ k\plus{}2$ circles not just $ 2$.  By the way if it is $ k\plus{}1$ then the statement already fails when $ k\equal{}0$

I believe he means that the torus has genus k?

yes, i think he's thinking about the genus $ k$ surface: in this case, it should suffice working on induction: case $ k\equal{}0$ is just jordan theorem (which is not trivial ;) ); now cutting a surface of genus $ g\plus{}1$ along a non-separating circle you just obtain a $ g$ surface minus two discs.

by inductive hypothesis $ g\plus{}1$ circles on the genus $ g$ surface disconnect: but they disconnect also a twice punctured genus $ g$ surface, because removing discs doesn't affect connectedness.

so our surface turns to be disconnected by $ g\plus{}2$ circles, too.