Let $x,y,z,w$ be positive numbers.Find the necessary and sufficient condition
such that the following equation holds for arbitrary positive integers m,n.

\[\left(x^\frac{1}{m}+y^\frac{1}{m}\right)^n+\left(z^\frac{1}{m}+w^\frac{1}{m}\right)^n=\left[\left(x^\frac{n}{m}+z^\frac{n}{m}\right)^{\frac{1}{n}}+\left(y^{\frac{n}{m}}+w^{\frac{n}{m}}\right)^{\frac{1}{n}}\right]^n\]


If $ a * b \equal{} 2a \plus{} 3b$ for all $ a$ and $ b$, then what is the value of $ 4 * 3$?


An isosceles trapezoid has sides of length 4 cm, 10 cm, 16 cm and 10 cm. What is the area of this trapezoid, in square centimeters?

one of the bases is length 4 and the other is length 16
therefore, the difference between the two is 12
there are 12 cm total on both sides, so the bottom base is 6 cm longer on one side than the top base

knowing this, we can use the formula $ a^2\plus{}b^2\equal{}c^2$ to find the height

$ a^2\plus{}6^2\equal{}10^2$
$ a^2\plus{}36\equal{}100$
$ a^2\equal{}64$
$ a\equal{}8$

since the height is 8, we can plug that into our formula for finding the area of a trapezoid

$ A\equal{}\frac {(b1\plus{}b2)h} {2}$ where A is the area, h is the height, b1 is the length of the first base, and b2 is the length of the second base

$ A\equal{}\frac{(4\plus{}16)8} {2}$

$ A\equal{}\frac{20\times8} {2}$

$ A\equal{}\frac{160} {2}$

$ A\equal{}80$


FTW
Support
Harvard
college
Ross Mathematics Program

obama ftw,
but i supported clinton in the primaries

note: i don't know if this has already been set up, if yes, then just move it to there
note number 2!!!!: give reasons, not just "he's stupid" or anything like that

[color=green]Moderator says: I've corrected spelling and added additional options to the poll.[/color]

Obviously someone isn't too politically active, since 1) it's spelled "Nader", and 2) Nader isn't the candidate for the Green Party. :P

sry it was a typo

and Nader isn't Green Party, he's an independent
but then again Nader won't win either way, since 1) he never got close to the nomination in 00 or 04 and 2) voters are tired of the whole "independent thing" after bloomberg and the few who would have probably would vote for Obama or McCain, both of which attract independents.

Hmm...McCain will be hard on terror...

My personal view is that violence will only breed more violence. Violence is what terrorists thrive on.

Now at a practical level, I don't have an alternative to suggest. I just know that terrorism will not end by fighting it.

You make a valid point.  But if we do nothing about it, terror will grow out of control.  Perhaps a better idea is to find an Achilles' heel.  Either way, we have to be bold.

[quote="lingomaniac88"]...I saw [b]somewhere[/b] that McCain will be hard against terror... I also remember [b]seeing somewhere[/b] that Hamas would be happy to see Obama as president...[/quote]

So you just believe everything you read and hear (and from sources that you can't even recall)? That's not a good way to form opinions and make judgements. By doing so you're just amplifying the, possibly biased or misinformed, opinion of some (anonymous) other person.

If I could, probably Obama, though one or two of McCain's doings I agree with. I personally don't agree with the Independent platform.

[quote="lingomaniac88"]I think McCain would be better.  I saw somewhere that McCain will be hard against terror, which is what we need if we want to gain respect back.  I also remember seeing somewhere that Hamas would be happy to see Obama as president.  Now, why could that be...?[/quote]

Before I write down some points, I would like to say that even though I am from Iran, I do not support its government's actions and am not in favor of Islamic extremists or anything of that sort.

It is ridiculous to believe that America is currently engaged in war against terrorism.  The many coups that America has supported/planned throughout the 20th century show that America is (clearly) a country with a government that money and oil say the final word.

Also, to Hamas, Hezbolla, et cetera, it does not make much difference who becomes the next president.  McCain's victory will have obvious implications in this case.  Also, as far as I know, Obama's campaign is financially supported by the Crown (I think this is the correct name) family, which is a Jewish family that owns a company (in Chicago, I think).  Clearly, one who is supported by a Jewish family in America will not make Hamas, Hezbollah, et cetera "happy."  Also, remember that due to the fact that there are many wealthy and influential Jewish families that basically run America (business says the word).

Regarding the claim that America is in the vanguard on the war on terrorism, you have to observe that America was, is, and will be after complete control over Middle-East, which is equal to control over oil.  For example, America's desire for cheap oil led to a coup (CIA's director himself or his brother was (in 1953) was a shareholder of an oil company) in Iran in 1953 that tipped over Mohammad Mossadeq's (democratically elected) government and brought back the king (or Shah).  If America had not done this, Middle East would have been a much safer, better, and calm place (Iran would not have been under the control of Islamic extremists who support terrorism worldwide).

Therefore, if America was truly against terrorism worldwide, they would not have done such things throughout the 20th century.

Also, the actions of America in Guatemala  (coup that overthrew the democratically elected Arbenz (spelling?) for BANANA fields that were not used by the United fruit company (this led to a long civil war)) has led to the creation of the term "banana governments."

You and many others should not be credulous, especially in a place where media plays a great role on what the public knows.

I hope that I have not bored anyone and have not confused names and identities. :P

[url=http://en.wikipedia.org/wiki/Jacobo_Arbenz_Guzm%C3%A1n]Arbenz[/url] is the correct name. However, the Mossadeq and Arbenz incidents were over 50 years ago; there comes a time when you have to ask how you want to live your life now rather than dwelling that far in the past.

It would be a serious mistake to focus in a conspiratorial fashion on any particular person who is supporting Obama; his support, including his financial support, is broad and diffuse. But Obama is a mainstream American politician; he's not about to reverse such long-standing and popular positions as the U.S.'s basic support for Israel. I'll also point out that the moment you start using such phrases or ideas as "Jewish businessmen control America" you pretty much guarantee that a large fraction of those who read your words will immediately dismiss everything else you say - if you want to be taken seriously, try to avoid that kind of bigoted language.

The worst bit of misdirection the Bush administration ever did was encouraging and allowing the American public to believe that invading Iraq had anything to do with fighting terrorism and/or responding to 9/11.

The effects of that decision - the decision to invade Iraq - on Iran have been quite profound, and it's not at all clear where that will lead. On the one hand, having a large U.S. army engaging in combat right next to the Iranian border has to be publicly perceived in Iran as a provocation and a threat, and that perception has probably contributed to the popularity of politicians (like Ahmedinijad) who depend on anti-American rhetoric. On the other hand, the U.S. wound up doing some things that Iran would have wished for: removing the extremely hostile (to Iran) Baathist government of Saddam Hussein, ultimately replacing it with a government that is weak and much more likely to allow itself to be influenced by Iranian interests.

I guess I did not use the correct words to communicate my ideas.  I must admit that Jewish people do not control America.

1) The point that I was trying to make regarding Obama was that since he is being supported by a Jewish family, it is very unlikely, if not impossible, that he will act in such a way that Hamas will be pleased.

2) By claiming that wealthy Jewish families control America, I meant that they are very influential in American politics and business.  For example, a large portion of the media in New York, if I am not making a mistake, is under the control of Jewish individuals.

3) I believe that a group of individuals financially support a politician only if the group's general opinion agrees with that of the politician.  Generally, a Jewish person in the 21st century wants the well-being of Israel.

4) The point that I was trying to make was that Obama's actions will not be any different regarding the Arab-Israeli conflicts, especially since he is backed by a Jewish family.

5) Regarding the events, to which I referred, that occurred over half a century ago, I was trying to make the points that a) the American political system is highly influenced by business, especially energy, b) the coup in Iran greatly contributed to terrorism, and c) it is unlikely that the current wars are truly wars against terrorism.

6) There's a saying that says that history is the light for the future (or something along those lines).  It is, in my opinion of course, good to analyze politics with the aid of past events; dwelling half a century into the past is not much when analyzing political systems, especially those that do not change much.

7) Just a remark: President Ahmadiejad is certainly not liked among most Iranian people.

Mr. Kent Merryfield, I may be young and inexperienced, so my apologies if my words seem to be expressed in an llogical manner.  I hope that I have at least cleared some issues regarding my previous post.

[quote="Kent Merryfield"]But Obama is a mainstream American politician; he's not about to reverse such long-standing and popular positions as the U.S.'s basic support for Israel.[/quote]

it's actually even stronger than that -- in speaking to a pro-israel group recently, obama asserted that jerusalem should never be divided, which is a stronger stance than any recent president has held (i think he may have since backpedalled)

[quote="lingomaniac88"]I think McCain would be better.  I saw somewhere that McCain will be hard against terror, which is what we need if we want to gain respect back.  I also remember seeing somewhere that Hamas would be happy to see Obama as president.  Now, why could that be...?[/quote]

1) the 'war on terror' and the accompanying wars in afghanistan and iraq that sprung from it are a significant reason for the (worldwide) loss of respect towards the us.  waging war under false pretenses, without an exit strategy, torturing enemies, and killing innocent iraqis tends to have that effect.  if foreigners elected the next US president, obama would win in a landslide, so i don't think electing mccain will have the effects you think.

2) hamas is ostensibly a political group as well so it's reasonable for their officials to express political opinions.  of course, only an idiot would use said opinions as a significant factor in deciding whom to vote for.  hamas doesn't like john mccain because he talked about the US being in iraq another 100 years -- believe it or not, this is a very valid reason for americans to not like john mccain as well!

3) the role terrorism (and more generally, fear,) plays in the elections is absurd and ridiculous in the face of the many problems confronting americans at home.  poverty, healthcare, social security, recession -- these are real problems.

hah, I bet on here, half the people supporting mccain would just be trolling

[quote="MysticTerminator"]I'll be voting for Obama although I seem to self-identify as moderately to extremely conservative.[/quote]

lol why would you do that? so you could vote for who wins? (I do happen to know a few people who would do that so ya)

i would like u people to also vote in my poll at
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236063

I know I sound very cynical, but...

The U.S. is headed the wrong way. After a quarter of a century of continued growth and upwards movement, the second half of the Bush administration has undermined much of that work and is on its way to doing worse. Our jobs are being outsourced to Asia, the economy is headed downwards, global warming is rampant, thanks to Bush's love of the big oil companies, China is headed straight for the No. 1. GDP, education is lagging horribly, and foreign talents are undermining the U.S.'s dominance in intelligence, the European powers are becoming increasingly hostile to the Bush adminsitration, etc, etc...

A liberal is our best chance to turn this around. At this point, there is no risk in electing a leftist president. Even if he fails, it will not have changed anything. And Obama presents, true, just a chance, but a greater chance of saving America then John McCain.

Plus, Palin and Biden provide a stark contrast, which makes the GOP's statements about experience ridiculous. Never mind gaffes, Biden has been a Senate member for over a decade, while Palin, in her two years as governor of Alaska, has made the ecological situation much worse.

Finally, McCain's smear tactics show how desperate he is to win the election. He  is using politics as usual. Obama is not doing that as much. I trust Obama more.
And frankly, a socialist is exactly what this country needs when the free market fails!

I have noticed that intelligent people (120 IQs and up) tend to vote overwhelmingly for Obama. This tells something about the two...

[quote="evilhamster"]Plus, Palin and Biden provide a stark contrast, which makes the GOP's statements about experience ridiculous. [/quote]
One can also say their contrast makes the Dems' statements about Change ridiculous since Biden has been in Washington for decades.

[quote="evilhamster"]No, because Biden is widely known as a liberal. Saying that experience and the right are connected is saying that FDR was a conservative. FDR ran for president in 1921 as an independent, and nabbed 12% of the vote, which has probably never been equaled by anyone other than Ross Perot.[/quote]
Congress is currently liberal. Biden is known as a liberal. How is that change?

[quote="evilhamster"]Congress is currently [i]not[/i] liberal. If it was, the Bush administration would not have been able to pass many of its detrimental laws. There is a tendency to think that certain parties are related to certain positions and places. This is not true. If you look back to the 1880's, there was the "Solid South" for the Democrats, and the Northeast was Republican territory.[/quote]
The same argument could be used to say that not only is Congress liberal, but Bush [i]is[/i] also currently liberal, which further relates Biden to the current government.

Democrats hold the majority in Congress, and most Democrats are liberal. This doesn't mean that liberals control Congress, for two main reasons.
(1) Those majorities are thin, and the Republican minority does everything they can to slow down or kill liberal legislation. In the Senate, the procedures allow them to be successful nearly all of the time- a bill can be delayed indefinitely by filibuster, or by several days even if the supporters can put together 60 votes for cloture.
(2) There is an organized group of conservative "Blue Dog" Democrats who often vote with the president and other Republicans. If they defect, the liberal democrats don't have enough votes.

The practical result is that the current Congress has done very little of anything. Either side can block legislation, but neither has enough power to force much through.

Obama is a much more better candidate. For one thing, he has accomplished way more than McCain. I'd list but that could take a few days. Also Obama is a much more educated individual.

If you were to look at the scenario in which the president died who would you rather lead this country, Palin or Biden. I think the choice here would be very clear. The fact is that McCain made a terrible choice in picking Palin for the fact he ONLY chose her for picking up Hilary supporters and that plan obviously back fired.

[quote="ProtestanT"][quote="worthawholebean"]The past eight years have been mostly characterized by a united conservative government which was able to do pretty much whatever it wanted to. The president has been able to block pretty much anything done by the current congress; Biden would definitely present a change.[/quote]
If the president blocks anything from the current congress, how is that a "government which was able to do pretty much whatever it wanted to"? It's more like a government that can't make any progress.[/quote]

1. There is never a good reason to quote the entire post immediately preceding yours.

2. You have completely failed to parse worthawholebean's statement: the first sentence refers to the six years in which both houses of congress were Republican-controlled; the second refers to the two years since then ("the current congress").


number theory
relatively prime
Diophantine Equations

It follow from the Pytago solution.
The equation has no solution.

It follows from the statement $ 0\equal{}0$, too...
Seriously: the way you state works, but there still is a lot of ugly work to do. And the ugly (case)work is the main reason why I don't want to post a solution untill now...
The idea is the same as for $ x^4\plus{}y^4\equal{}z^2$, but this one has a nasty second case.

Can someone please outline a proof? If it is that ugly indeed, I'll just remove it from PEN, or replace with its nicer brother.

let d=gcd(x,y), then $ d^4$ divides $ z^2$, then $ (x/d)^4 \minus{} (y/d)^4 \equal{} (z/d^2)^2$, if p prime divides two of the terms, divides the third, then p divides x/d and y/d(contradiction). So we can assume that x,y,z are coprimes each two
$ (x^2 \plus{} z)(x^2 \minus{} z) \equal{} y^4$, then $ x^2 \plus{} z \equal{} m^4, x^2 \minus{} z \equal{} n^4$
$ 2x^2 \equal{} m^4 \plus{} n^4$ ;m>n. I will prove that this equation no has solution
we suppose otherwise, then analagously above we can assume that 2x,m,n are coprimes each two, then 16 divides $ m^4 \minus{} 1 \plus{} n^4 \minus{} 1$, then x is odd
we take the solution with x minimal
m=n+2k, since m+n is even.
then $ x^2 \equal{} (n^2 \plus{} 2nk \plus{} 4k^2)^2 \minus{} 8k^4$
$ 2k^4 \equal{} (A \plus{} x)/2 (A \minus{} x)/2$; since A and x are odd, if p prime divides (A+x)/2 and (A-x)/2, p divides A,x(p is odd), therefore p divides k
p divides A then p divides n, then p divides m(contradiction)

Case 1: $ (A \plus{} x)/2 \equal{} 2u^4, (A \minus{} x)/2 \equal{} v^4$
$ x \equal{} 2u^4 \minus{} v^4$. note that $ x^2 > (3k^2)^2 \minus{} 8k^4 \equal{} k^4$, then $ A^2 < 9x^2$,therefore A<3x, then u>v
note that $ m^2 \minus{} mn \plus{} n^2 \equal{} A \equal{} 2u^4 \plus{} v^4, (m \minus{} n)/2 \equal{} k \equal{} uv$; u and v coprimes
$ [(m \plus{} n)/2]^2 \equal{} m^2 \minus{} mn \plus{} n^2 \minus{} 3[(m \minus{} n)/2]^2 \equal{} 2u^4 \minus{} 3u^2v^2 \plus{} v^4 \equal{} (2u^2 \minus{} v^2)(u^2 \minus{} v^2)$; u>v
$ gcd(2u^2 \minus{} v^2,u^2 \minus{} v^2) \equal{} gcd(u^2,u^2 \minus{} v^2) \equal{} gcd(u^2,v^2) \equal{} 1$ then
$ 2u^2 \minus{} v^2 \equal{} r^2, w^2 \equal{} u^2 \minus{} v^2 \equal{} (u \plus{} v)(u \minus{} v)$
then $ u \plus{} v \equal{} s^2,u \minus{} v \equal{} t^2$ (s-t is even) or $ u \plus{} v \equal{} 2s^2,u \minus{} v \equal{} 2t^2$
$ r^2 \equal{} 2[(s^2 \plus{} t^2)/2]^2 \minus{} [(s^2 \minus{} t^2)]^2$ or $ r^2 \equal{} 2(s^2 \plus{} t^2)^2 \minus{} (s^2 \minus{} t^2)^2$
$ 2(r/2)^2 \equal{} [(s \plus{} t)/2]^4 \plus{} [(s \minus{} t)/2]^4 or 2r^2 \equal{} (s \plus{} t)^4 \plus{} (s \minus{} t)^4$
but $ x \equal{} 2u^4 \minus{} v^4 > 2u^4 \minus{} 2u^2v^2 \plus{} v^4 > (2u^2 \minus{} v^2)(u^2 \minus{} v^2) > 2u^2 \minus{} v^2 \equal{} r^2$ then x>r,r/2(contradiction)

Case 2: $ (A \plus{} x)/2 \equal{} u^4, (A \minus{} x)/2 \equal{} 2v^4$
$ u^4 > 2v^4 > v^4$, then u>v
$ m^2 \minus{} mn \plus{} n^2\equal{}A\equal{}u^4 \plus{} 2v^4, (m \minus{} n)/2 \equal{} k \equal{} uv$; u and v coprimes
$ [(m \plus{} n)/2]^2 \equal{} m^2 \minus{} mn \plus{} n^2 \minus{} 3[(m \minus{} n)/2]^2 \equal{} u^4 \minus{} 3u^2v^2 \plus{} 2v^4 \equal{} (u^2 \minus{} 2v^2)(u^2 \minus{} v^2)$; $ gcd(2v^2 \minus{} u^2,v^2 \minus{} u^2) \equal{} gcd(v^2,v^2 \minus{} u^2) \equal{} gcd(v^2,u^2) \equal{} 1$ then
$ u^2 \minus{} 2v^2 \equal{} r^2, w^2 \equal{} u^2 \minus{} v^2 \equal{} (u \plus{} v)(u \minus{} v)$
then $ 4u \plus{} v \equal{} s^2,u \minus{} v \equal{} t^2$ (s-t is even) or $ u \plus{} v \equal{} 2s^2,u \minus{} v \equal{} 2t^2$; s and t coprimes
$ 4r^2 \equal{} \minus{} s^4 \plus{} 6s^2t^2 \minus{} t^4$ or $ r^2 \equal{} \minus{} s^4 \plus{} 6s^2t^2 \minus{} t^4$
in both cases $ R^2 \equal{} \minus{} s^4 \plus{} 6s^2t^2 \minus{} t^4 \equal{} 4s^2t^2 \minus{} (s^2 \minus{} t^2)^2$, note that R,s,t are coprimes each two
$ (s^2 \minus{} t^2)^2 \equal{} (2st \minus{} R)(2st \plus{} R)$ ;gcd(2st-R,2st+R)=1,2,4
then $ 2st \minus{} R \equal{} a^2,2st \plus{} R \equal{} b^2$ or $ 2st \minus{} R \equal{} 2a^2,2st \plus{} R \equal{} 2b^2$
$ 4(s^2 \plus{} t^2)^2 \equal{} 4(8s^2t^2 \minus{} R^2) \equal{} 2(a^2 \plus{} b^2)^2 \minus{} (a^2 \minus{} b^2)^2$ or $ 4[2(a^2 \plus{} b^2)^2 \minus{} (a^2 \minus{} b^2)^2]$
then $ c^2 \equal{} 2(a^2 \plus{} b^2)^2 \minus{} (a^2 \minus{} b^2)^2$ then $ 2c^2 \equal{} (a \plus{} b)^4 \plus{} (a \minus{} b)^4$
$ c \le 2(s^2 \plus{} t^2) \le 4u$
$ 2x\equal{}2u^4 \minus{} 4v^4\equal{}2u^4 \minus{} (u^2 \minus{} r^2)^2 \ge 2u^4 \minus{} (u^2 \minus{} 1)^2 \equal{} u^4 \plus{} 2u^2 \minus{} 1 > u^4 \ge 8u \ge 2c$
; if $ u \ge 2$ this is true since $ u^2\equal{}2v^2\plus{}r^2>1$. then x>c(contradiction).

Here you've got a nicer approach, which makes your statement no ugly at all:

[b]Lemma[/b]. There are no positive integers $ a, b, c, d$ for which $ a^2 \minus{} b^2 \equal{} c^2$ and $ a^2 \plus{} b^2 \equal{} d^2$ both hold.
[i]Proof[/i]. Assume the contrary and let us consider one solution $ (a, b, c, d)$ for which the minimum of $ b$ is attained. Summing yields $ 2a^2 \equal{} c^2 \plus{} d^2$, or equivalently, since $ 2 | c \pm d$, $ \left( \frac {d \minus{} c}{2} \right)^2 \plus{} \left( \frac {c \plus{} d}{2} \right)^2 \equal{} a^2$. It is easy to check that if $ \frac {d \minus{} c}{2}$ and $ \frac {c \plus{} d}{2}$ have any common factor, this leads to a solution with smaller components, fact which contradicts the minimality of $ b$. Hence they are coprime, and by the Pythagorean equation there exist relatively prime $ m, n \in \mathbb{N}$ such that $ d \minus{} c \equal{} 2(m^2 \minus{} n^2)$, $ c \plus{} d \equal{} 4mn$ and $ a \equal{} m^2 \plus{} n^2$. Hence $ b^2 \equal{} 4mn(m \minus{} n)(m \plus{} n)$ with all $ m, n, m \minus{} n, m \plus{} n$ coprime. Thus there are some positive integers $ x, y, z, t$ for which $ m \equal{} x^2$, $ n \equal{} y^2$, $ m \minus{} n \equal{} x^2 \minus{} y^2 \equal{} z^2$ and $ m \plus{} n \equal{} x^2 \plus{} y^2 \equal{} t^2$ and $ y^2 \equal{} n < 4mn(m^2 \minus{} n^2) \equal{} b^2$, so that $ (x, y, z, t)$ is a solution with $ y < b$, a contradiction. The case $ d \minus{} c \equal{} 4mn$ and $ c \plus{} d \equal{} 2(m^2 \minus{} n^2)$ follows the same reasoning.

The proof of the proposed problem is now instantaneous: consider the solution with minimal $ z$ and notice that $ x, y$ need to be coprime. Hence either $ (x^2 \minus{} y^2, x^2 \plus{} y^2) \equal{} 1$, which in the view of $ (x^2 \minus{} y^2)(x^2 \plus{} y^2) \equal{} z^2$ shows that there exist some positive integers $ a, b$ such that $ x^2 \minus{} y^2 \equal{} a^2$ and $ x^2 \plus{} y^2 \equal{} b^2$, contradiction by the Lemma above, or $ (x^2 \minus{} y^2, x^2 \plus{} y^2) \equal{} 2$, which yields $ ((x^2 \minus{} y^2)/2)((x^2 \plus{} y^2)/2) \equal{} (z/2)^2$, and thus there are $ a, b \in \mathbb{N}$ for which $ x^2 \minus{} y^2 \equal{} 2a^2$ and $ x^2 \plus{} y^2 \equal{} 2b^2$, so $ a^2 \plus{} b^2 \equal{} x^2$ and $ b^2 \minus{} a^2 \equal{} y^2$, again a contradiction. Hence there are no solutions neither to this equation nor to the well-known $ x^4 \plus{} y^4 \equal{} z^4$.


trigonometry
inequalities
algebra unsolved
algebra

1. Find x,y
$ x^2\plus{}y^2\equal{}x\sqrt{1\minus{}y^2}\plus{}y\sqrt{1\minus{}x^2}$
$ 3x\plus{}4y\equal{}5$
2. Prove: (3+cosA)(3+cosB)(3+cosC)>32 with A,B,C<90

Because $ |x|,|y|\leq 1$
$ x = \cos{a},y = \cos{b}$ where $ a,b\in [0,  \pi]$
So we have
$ \cos^2{a} + \cos^2{b} = \sin(a + b)$
From $ 3\cos{a} + 4\cos{b} = 25$ we have $ \cos^2{a} + \cos^2{b}\geq 1$
So $ \sin(a + b)\geq 1$
Imply that ${ \sin(a + b) = 1,\frac {\cos{a}}{3} + \frac {\cos{b}}{4}}$
So $ a + b = \frac {\pi}{2}$
So $ \cos{a} = \sinh$
$ \Rightarrow 4\cos{a} = 3\sin{a}$
$ a = arctan(\frac {4}{3})$
For the b should be post in Inequalities and it is quite easy.


HOW Many different necklaces with 10 coral beads i can make if i have coral beads in 5 colours...I have a lof of coral beads so It is ppossible to make onecolour necklake.

You have 5^10  different rows. 5^10-5 of them have at least 2 colors. Each necklace is defined by 10 rows obtained by the rotations. Thus we have (5^10)/10 different necklaces with at least 2 colors and 5 necklaces wit one color only. Totally   (5^10)/10 +5 ;)

but that is not a whole number! i think you should look at each case seperately e.g. 1 colour used, 2 colours used ... and add the results for each case


salam bacheha . age moafegh bashin bad az marhale 2 hame biayn too in topic va nazarha va rahe halhatoono benevisin . ba arezooye moaffaghiat baraye hame :)

salam be hamegi. emtehana che tor dadin?
man  2 ta soala kamel hal kardam. soale 1 ba 4 zeleye mohati va ye hamneheshti. soale 3 ro ham ba esteghra va inke baraye ye noghte ,  parekhathay  beyne noghati ke beine nimsazhaye forjehaye do safheye amood bar hame  gozarande az an noghte gharar darand va an noghte  ba zamin zavieye kamtar az 45 misaze( ajab tataboe ezafati neveshtam :D)
faghat omidvaram farda ham jabr biad ham nazarieye adad.nazaretoon chie?

salam manam 2 tasho kamel hal kardam yani soale 1 va 2 soale 3 ham yechizaei neveshtam benazare man kheli rahat bood age intori bashe haddenesabe ghabooli az 4/5 bishtar mishe ;) movafagh bashid bye

Manam 1,2 ro kamel hal kardam 3 ro ham neveshtam (motmaen nistam kamelan) . soale 1 esh ba charzeli mohatio hamneheshti va soale 2 sh ro ba rabete bazgashti sabet mishe ke tedad masir haye O-O daghighan yeki bishtar az tedad masirhaye O-A ast . benazare man soala vaghean dar hade marhale 2 nabood eyne marhale 1 e emsal shod . omidvaram emtehane farda behtar bashe nazarie o jabram dashte bashe :)

baba yeki soala ro bezare maham bebinim! :oops:

man vaghean gand zadam,tanha soalio ke motmaennam dorost hal kardam soale 2 bood,ba hamoon rahe halle kashi, albate man ba esteghra sabet kardam, soale 3 va iek ro ham nesfe nime ie chizayi neveshtam:(

salam mellat,
man ke kheili bad dadam,
2 ta rooze 1 va yeki ham rooze 2 halidam,
omran ghabool nemisham dige!!
tasmim gereftam kollan bikhiale olampiad sham!

man faghat roo soale 6 motmaennam ke doros hal kardam. oonam chon age beshe adade 6n ro chap kard pas hatman mishe adade  n ro ham chap kard.va chon 13824=6^3* 64 pas kafie sabet konim adade 64 ro nemishe chap kard.  soale 4 ham ra motmaenam hame hal kardan shayad be joz oonaee ke formool balad nistan (mesle man) . soale 5 ham fek mikonam sakht bood .be har hal man motmaennam ghabool nemisham.omidvaram hamatoon ghabool shin.bye

be nazare man emtehan har dorooz vaghean soalaye badi dasht . aslan soalash khoob nabood . be nazaram bachehaye olampiyade computer behtar az ma mitoonestan in emtehano bedan , chon 3tash ke tarkibiat bood oonam az nazarie adadesh ke ... . hendeseye rooze dovvom nemidoonam chera injoori bood  vali 4,5 nafar az bacheha (az jomle khodam) $A$ ro roo dayereye $C_{1}$ gerefte boodan ke vaghean nemidoonam chera :D man soale 6 ro tamim dadam az khodam :) sabet kardam az beyne adadi ke be shekle $2^{a}3^{b}$ hastand adade $n=2^{x}3^{y}$ chap mishavad agar va tanha agar $3|x-y-1$ . inam ba esteghra rooye $a+b$ sabet kardam . dar har soorat manam omidi be emsal nadaram chon ba in emtehane asoon be nazar nemirese kaf kamtar az 5 bashe . baraye hame arezooye moafaghiat mikonam :)

man fekr konam yadeshoon rafte soale gashang vojood dare bara hamin in soala ro tarh kardan  :lol:  amma fekr konam kaf 5 menhaye epsilon bashe .

farz konid chap shodane 2^k*3^m ro ba (k,m) neshoon bedim va chap nashodanesh ro ba ~(k,m):

leme 1: (k,m)=>(k-1,m-1) borhan:(k,m)=>~(k-1,m) & ~(k,m-1)=>(k-1,m-1)

leme 2: (k,m) & k>0=>(k-1,m+2) borhan:
(k,m)=>~(k,m+1)* & ~(k+1,m)
(k,m)=>~(k-1,m) & ~(k-1,m+1)**
* & **=>(k-1,m+2)

13824=2^9*3*3
farze kholf: (9,3)=lem 2=>(8,5)=lem 2=>(7,7)=haft bar az leme 1 estefade mikonim=>(0,0) : ke be manaye chap shodane 1 boode va tanaghoz ast.

man dar kol 3 ta ro(1,46) kamel hal kardam soale 3 ro ham ta nesfe va shayad bishtar hal kardam . khak to saram .

agha age soal 4 ro ba rahe hal digei be gheir az oon formoole hal konim baz ham dorost mi girand?(manzooram esbate vojoodie)

[quote="Toxic_Frog"][quote="khashi70"] $A$ ro roo dayereye $C_{1}$ gerefte boodan ke vaghean nemidoonam chera :D[/quote]
mage $A$ jaye digeii bood ?[/quote]
khak to saram beshe mesle inke 3ta va nesfe soale 3 ro hal kardam gand zadam ganD ! :(

[quote="khashi70"]moalemaye ma migan ke kaf 4 ya 4+epsilon hast . dar zemn $A$ dakhele dayereye $C_{1}$ hast na rooye dayere :)[/quote] ade dakhele dayere bashe manzoor az kalameye taghatooe dovvom chie ?

Manzoor oon taghatoe doortar az dayereye $C_{2}$ hast . albatte fekr konam vase taghatoe avvalam hokm dorost bashe

manam soale 1 ,2,4,6 va soale 3 nesfesho hal kardam benazaram kafe ghabooli 29 mishe vali kollan emtehane kheli ashghali boood;) khashi movafagh bashi midoonam tala migiri :wink:

aya mosahheha be ide nomreye kamelo midan ?

[size=150][color=red]ye soal [/color]kashi moaletoon kie ?[/size]

salam

mishe yeki hendeseye rooze dovom ro bede?

[quote="khashi70"]$5)$do dayereye $C_{1}$ va $C_{2}$ dar noghteye $P$ bar ham momase khareji hastand va $A$ noghteyi dakhele dayereye $C_{1}$ ast . do momase $AM$ va $AR$ bar dayereye $C_{2}$ rasm mikonim . ($M$ va $R$ mahale tamase momas,ha hastand).noghate taghatoe dovvome $AM$ va $AR$ ba dayereye $C_{1}$ ra be tartib $N$ va $S$ minamim . neshan dahid : $\frac{PN}{PS}= \frac{MN}{RS}$ .[/quote]

test haye konkoor az in sakht tare.


MATHCOUNTS probability geometry AMC USA(J)MO USAMO ratio

Post all questions you have about the problems here, not in the other thread. That thread will be used for results only.

So questions like how many seats are there in #18 should go here?  Or discussion of the solutions after the rounds are over?

Both. If you think a problem is unclear, then post your question ehre.

*Whistle* *whistle*... I already pmed isabella2296 about this. Meanwhile, I gave the answer replacing the word "points" with "lines". (One reason I got a uber-phail 18... :()

[quote="5849206328x"]for #8, do you mean something like "what is the maximum number of intersection points inside the circle when 37 chords are drawn?"  that's what I assumed...[/quote]

Yes that is the meaning.

[quote="shentang"][quote="pythag011"][quote="shentang"]Are you sure that question 9 is right? Can someone rephrase it for me? Because the way I am interpreting it, the question requires some not mathcounts level math. The way I am interpreting it is that there is a 11-sided polygon inscribed in a circle and you draw all the diagonals in the polygon and you have to count the maximum number of non-overlapping regions.[/quote]

It can be solved using mathcounts math and some cleverness. (Or if you have seen the problem before :) )

Yes, there is an insribed polygon, how many regions is the circle split up into by the sides and diagonals?[/quote]

if you mean drawing and counting....... :|

also on number 11, I am assuming the words are just any group of different letters.[/quote]

I do not mean drawing and counting.

This thread is alike drug-taking in olympics...

Can't we just close this thread and pm our questions to isabella2296?

At the end of each round, I will post a document with the answers and solutions to each problem. In the meanwhile, it's better to post here rather than flood my inbox with previously answered questions. :)

I don't understand #20.

izzy told me to post it here.

Let the number of students be $ x$. Then the number of honor

sutdents is $ x\minus{}67$ Then the number of athletes are $ \frac{3}{100}(x\minus{}67)$, and the honor roll students are $ \frac{33}{100} \cdot \frac{3}{100}(x\minus{}67)\equal{}(x\minus{}67)$, and $ x\equal{}67$ so the answer
is $ 0$, which can't be right..........................

Can someone clarify? And on #9 i think that the official answer is

wrong unless i misinterpreted the question, which is very likely.[/hide]

Stop it with the complaints about question 9 please. The official answer is correct, I'm sure. And also, don't post your solution to a problem until everyone has finished (!!!).

*gasp* Yongyi!!!  :o

Tsk tsk, how could you tell me of all people? It's not a good idea to tell things like to this to the High-Mod-In-Training.  :wink:

Hmm, I may possibly take Number Theory problems from a source...  :oops:

That is a good idea. IMO Shorlist Number Theory problems are pretty interesting. Also get them fro the 2006 USAMO.

Sarcasm, anyone? (Points at pythag011)

No, seriously, get them from past Nationals tests (muahaha)!

OK, maybe that's not a good idea! Erm... seeing as I'm not *cough* beast *cough* enough to give up my position in the Olympics to give you problems... how about let izzy hone her writing skills and actually believe she can do it!

Cheer for izzy!! You can do it!

Edit: And pythag011, if you are planning to take them from the 2006 USAMO, also let her know that there are only 6 problems there! And the Olympics needs 20 to complete its set of problems.

Please don't take from another source because almost everyone will know it. (We have no life and do math 24/7)

I completely agree. Imagine this was the National MATHCOUNTS competition. We wouldn't want to give anyone an unfair advantage because they saw the problems before. :wink:

I'm warning you, you guys will all burst out laughing when you see the NT problems.

That's okay. Consider this "Problem writing practice"!

Hmm... is there a site called [url]http://www.artofproblemwriting.com[/url]?

Well, I mean they're going to suck because [i]I[/i] suck at Number Theory. :(

Well, can anybody rewrite the question 11 for me? I don't get what it meant. But as far as I see, it must relate to the formula in physics, so is it suitable for a Math contest?

#1 you mean greater than 2 right? Or else it doesn't work.
#7 any odd number

[quote="isabella2296"]Unfortunately, I'm really busy today, so I won't be able to post the questions for the next sport until Saturday - however, the answers were still due last night, so you can't send them in.

Here are the answers and solutions for Geometry - note that for #2, I also accepted 30 as an answer.[/quote]

How 270 can be a correct answer on #2?
It says 'what is the SMALLEST possible area of triangle if the length of each side is an integer?'.

Yes, since the ratio is in 39:15, this simplifies to 13:5, and indeed, the 5-12-13 triangle yields the smallest area. It asked for SMALLEST, so the SMALLEST should be the answer, not 2nd smallest.

What happened was that my source made a mistake, and by the time I realized this, I'd already counted everyone's answer of 270 correct, so I just decided to count it right.

a few questions for algebra round

5. does a slice of cake equal a cookie?
7. since 2 evens can't add up to be odd, isn't the answer sorta impossible to write?
14. are the sides at a scale of 1:17, or the area at a scale of 1:17?


floor function
number theory
Diophantine equation
number theory open

Hi,I have one question which seems simple but I don't know what to do :
suppose we have the diophantine equation $ \alpha x\plus{}\beta y \equal{}\gamma$,
.with $ \alpha,\beta,\gamma>0$.Can we find the number of the positive x that satisfy
this equation?(I couldn't find it on the internet.)

Below is the solution to related problem of counting non-negative solutions to $ \alpha x \plus{} \beta y \equal{} \gamma$ w.r.t. $ (x,y)$ where $ \alpha$ and $ \beta$ are any integers such that $ \gamma \equal{} \gcd(\alpha,\beta)$.

Let $ u,v$ be Bezout coefficients for $ \alpha,\beta$ so that
\[ \alpha u \plus{} \beta v \equal{} \gamma.
\]
Then the integer solutions to $ \alpha x \plus{} \beta y \equal{} \gamma$ must be of the form $ (x,y) \equal{} (u \plus{} k\beta,v \minus{} k\alpha)$ where $ k$ is integer. In particular, the number of non-negative integer solutions is
\[ \left\lfloor \frac {u}{\beta}\right\rfloor \plus{} \left\lfloor \vphantom{\frac {u}{\beta}}\frac {v}{\alpha}\right\rfloor.
\]
To get the number of positive solutions, simply count and subtract the number of solutions of the form $ (0,y)$ or $ (x,0)$ with $ x,y > 0.$


On each of 10 sheets of paper, some powers of 2 are written (possibly including duplicates on a single sheet as well as between sheets). The sum of the numbers on each sheet is the same. Show that some power of 2 is written at least six times.

Ughhhh I really suck at these random proofs. Can anyone help me, and maybe offer some advice on how to tackle tough pigeonhole proofs?

It's not pigeonhole.  It's a contradiction/estimate problem:
[hide="hint"]Let $ N$ be the common sum, and let $ n$ be the largest integer such that $ 2^n \le N$.  Suppose that each power appears at most 5 times.  What does that say about the sum of the numbers on all ten sheets?[/hide]

I think I got it....Thanks for the tip!

If $ N$ is the common sum of all the sheets, then all 10 sheets sum of $ 10N$. Let $ n$ be the biggest integer such that $ 2^n \le N$.
Now suppose each power of $ 2$ is used at most 5 times. Then the maximum sum possible would be $ 5(2^n) \plus{} 5(2^{n \minus{} 1}) \plus{} \dots \plus{} 5(2^1) \plus{} 5$
$ \equal{} 5(2^n \plus{} 2^{n \minus{} 1} \plus{} \dots \plus{} 2^1 \plus{} 1)$
$ \equal{} 5(2^{n \plus{} 1} \minus{} 1)$
$ \equal{} 5[2(2^n) \minus{} 1]$
And since $ 2^n$ is at most $ N$, the maximum of $ 5[2(2^n) \minus{} 1]$ is $ 5(2N \minus{} 1)$. However, this maximum is less than $ 10N$, so we have a contradiction. Hence, some power of 2 must be used at least 6 times.

And I guess we would need to show that $ 6$ is achievable.

[quote="Thunder365"]... so we have a contradiction. Hence, every power of 2 must be used at least 6 times.

[/quote]

This would not follow, because the problem stated that "some power of two is written at least
six times."

The opposite of this is that at least two powers of two are written at least six times.


modular arithmetic
function
algebra
binomial theorem
number theory proposed
number theory

[quote="Caelestor"]How do you find the last 3 digits of $ 60^{60^{60}}?$ I tried Euler's totient theorem, but it fails since I don't have coprime numbers.[/quote]

Euler's totient theorem? :rotfl:  :rotfl:  :rotfl:

$ 10|60 \Rightarrow 1000|60^{60^{60}} \Rightarrow$ last 3 digits are $ 000$

[quote="Caelestor"]Oh wait a minute. Wrong problem. (I should've chosen a number not with a multiple of 10.)

Like for instance $ 26^{26^{26}}$[/quote]

$ 2|26, 5|(26\minus{}1)$.

So the answer is $ 6$

[quote="Stephen"][quote="Caelestor"]Oh wait a minute. Wrong problem. (I should've chosen a number not with a multiple of 10.)

Like for instance $ 26^{26^{26}}$[/quote]

$ 2|26, 5|(26 \minus{} 1)$.

So the answer is $ 6$[/quote]

Sorry, I need the last 3 digits (It's obvious the last digit is 6, just like the first problem). Also, I don't understand the notation you gave. (Are you a different Stephen?)

[quote="Caelestor"]Oh wait a minute. Wrong problem. (I should've chosen a number not with a multiple of 10.)

Like for instance $ 26^{26^{26}}$[/quote]

$ (25\plus{}1)^{26^{26}} \equiv 26^{26}(25)\plus{}1 \equiv 25\plus{}1 \equiv 26 \pmod{125}$

Together with $ 8 | 26^{26^{26}}$ gives $ 26^{26^{26}} \equiv 776 \pmod{1000}$

So the last three digits are $ 776$.

[b]P.S.[/b] I am different from that Stephen .

The totient theorem doesn't fail; it just doesn't completely hold.  For this case, it holds in the sense that $ a^{400k\plus{}3} \equiv a^3 \bmod 1000$ for any positive integer $ a$.

The key is in the proof of a slightly stronger result, [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael's theorem[/url], which uses CRT.

[quote="stephencheng"]
$ (25 \plus{} 1)^{26^{26}} \equiv 26^{26}(25) \plus{} 1 \equiv 25 \plus{} 1 \equiv 26 \pmod{125}$

Together with $ 8 | 26^{26^{26}}$ gives $ 26^{26^{26}} \equiv 776 \pmod{1000}$

So the last three digits are $ 776$.

[b]P.S.[/b] I am different from that Stephen .[/quote]

For the first step, basically you split the 26 into two parts to get something like the Binomial theorem?
You lose me after the second equivalence.

Also what does the notation $ 8|26^{26^{26}}$ mean?

[quote="Caelestor"]
Also what does the notation $ 8|26^{26^{26}}$ mean?[/quote]

$ 8|26^{26^{26}}$ means $ 8$ divides $ 26^{26^{26}}$.  And this is a very common notation.

Oh, I see why 8 divides $ 26^{26^{26}}$ I think I'm understanding the extension of Euler's theorem as well.

One last question: How did we go from 26 to 776 again?

As $ 8$ and $ 125$ are relatively primes, one can use the [url=http://planetmath.org/encyclopedia/ChineseRemainderTheorem.html]Chinese remainder theorem[/url]


Define the sequence $\{a_{n}\}$ such that $a_{1}=-4,\ a_{n+1}=2a_{n}+2^{n+3}n-13\cdot 2^{n+1}\ (n=1,\ 2,\ 3,\ \cdots).$
Find the value of $n$ for which $a_{n}$ is minimized.


Find all triples of integers $ (x,y,z)$ such that $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$

looking at this mod2 the 2xyz term cancels so we get

$ x^2\plus{}y^2\plus{}z^2\equal{}1 (mod2)$

well, from this we know that either all three of x,y,z are odd, or exactly 1 or them is odd. maybe you can go from there

good luck! :D

[hide="Hint"]Solve for one variable[/hide]
[hide="More on that"][i]Solve for $ x$:[/i]

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$

$ x^2 \plus{} 2yzx \equal{} 1 \minus{} y^2 \minus{} z^2$

$ x^2 \plus{} 2(yz)x \plus{} (yz)^2 \equal{} (y^2)(z^2) \minus{} (y^2) \minus{} (z^2) \plus{} 1$

$ (x \plus{} yz)^2 \equal{} (y^2 \minus{} 1)(z^2 \minus{} 1) \; \; \iff \; \;(x \plus{} yz)^2 \equal{} (y \plus{} 1)(y \minus{} 1)(z \plus{} 1)(z \minus{} 1)$

Now we just need to find pairs $ (y,z)$ that will make the RHS a perfect square, and then solve for the two $ x$ values associated with each.[/hide]


Let $n$ be a positive odd integer. Consider sequences $a_0, a_1,\cdots, a_{n-1}$ such that $a_i\in{1,2,\cdots,n}$ for each $i$. Call a sequence $good$ if for all $i = 1,2,\cdots, n$ and $a_i-a_{i-1}$(not)$\equiv {i}$  $(mod n)$ where $a_n = a_0$.
A) Find the number of $good$ sequences
B) What if the condition is instead $a_i-a_{i-1}$ (not) $\equiv{i}$,$2i$  $(mod n)$ and $n$ is prime?

Great,how can you find the 2005 USA TST problems? :o

[quote="silouan"]Where did you find a problem of TST from USA ?Could you post all the problems ? Thanks[/quote]
That's the same i want to ask.

silouan asked that in Dec 04, 2005, when the TST was't public. But today you can find this tests in Resourses


function algebra domain calculus derivative inequalities conics

[quote="yagaron"]$a =-1$
$b = \frac{1}{k}$
$c = \frac{k-1}{k}$

$a^{2}+b^{2}+c^{2}= 2 \frac{1-k+k^{2}}{k^{2}}$

So. the maximum is Infinity :)[/quote]

$a, b, c \in [-1, 1]$.

Naturally, the maximum would seem to be 2, but I can't see how to prove it.

I think a convexity argument works for this one. $a^{2}+b^{2}+c^{2}$ is convex, so take $a,b,c$ at their bounds. You get one variable is $1$ or $-1$, then the second is the negative of that, and finally the last one is 0. So $2$ should be right.

[quote]Naturally, the maximum would seem to be 2, but I can't see how to prove it.[/quote]

If $a =-1 , b = \frac{1}{k}, c = \frac{k-1}{k}$  so $a, b, c \in [-1, 1].$ for larges $k > 1$

and the limit $a^{2}+b^{2}+c^{2}= 2 \frac{1-k+k^{2}}{k^{2}}$ such as $k$ tend to infinite is 2

$(-1 , \frac{1}{k}, \frac{k-1}{k}) \in [-1, 1]$

made a misscalculating before, sorry

That doesn't prove it though. What if $a$ isn't $-1$? You need to show you can't get any value over 2.
(obviously we do get $a = \pm 1$ by convexity, but that needs to be explicit)

$a+b+c = 0$ describes a plane in $\mathbb{R}^{3}$ on which the function $M(a, b, c)$, which is convex in all three variables, is defined.

The simplest definition of a convex function in one variable $f(x)$ is that a line from any two points on the graph of $f(x)$ is always above $f(x)$.  Accordingly, convex functions have a very useful feature:  in one variable, a convex function $f(x)$ on a domain $[a, b]$ takes on its maximum at either $a$ or $b$ (since we can draw the line from $(a, f(a))$ to $(b, f(b))$).

This property can be extended to an arbitrary number of variables; in particular, the restriction $a, b, c \in [-1, 1]$ defines some closed polygon on the plane $a+b+c = 0$ on which $M(a, b, c)$ is defined, and then the fact that $M$ is convex, implies that $M(a, b, c)$ takes on its maximum at the vertices of that polygon.

Thanks, but I'm not making all the connections yet. I see how a function of one variable can be convex on $[-1,1]$ like say $y=x^{2}$, but I don't see how $a+b+c=0$ or $a^{2}+b^{2}+c^{2}=M$ is convex. I see how the former defines a plane in space, but... how does convexity enter into it?

$a+b+c$ is not convex. The function $f(a,b,c) = a^{2}+b^{2}+c^{2}$ is. I know the formal definition for convexity on a single-variable function is that for all $x$ and $h \neq 0$, $f(x) < \frac{f(x-h)+f(x+h)}{2}$. For multivariable functions I'm not sure what the definition is. Probably you hold the other variables constant, so to say $f(x,y)$ is convex would mean $f(x,y) < \frac{f(x-h,y)+f(x+h,y)}{2}$ and $f(x,y) < \frac{f(x,y-h)+f(x,y+h)}{2}$

One way to confirm convexity is to take the second partial derivative with respect to each variable, and if you get a rseult that is never negative, it's convex. According to Mildorf's inequalities packet, this is the only legitimate way to use calculus on an olympiad. If you don't know about derivatives or partial derivatives, just forget about this paragraph.

While we're on this subject, concavity is when the inequality $f(x) > \frac{f(x-h)+f(x+h)}{2}$ (note the flipped sign) holds, equivalent to showing second partial derivative is always negative or 0. Then the bound rule applies if you're finding minima. If you're looking for a maximum on a concave function, similarly if you're looking for a minimum for a convex function, you're stuck solving the problem another way.

The definition of convexity goes something like, given two points $X$ and $Y$ in the domain and a real $0 \le \lambda \le 1$, we have

$\lambda f(X)+(1-\lambda)f(Y) \ge f(\lambda X+(1-\lambda)Y)$.

This can be applied in any dimension. In this case, we need to show that

$\lambda f(a, b, c)+(1-\lambda) f(a', b', c') \ge f(\lambda a+(1-\lambda)a', \lambda b+(1-\lambda)b', \lambda c+(1-\lambda)c')$.

Letting $f(a, b, c) = a^{2}+b^{2}+c^{2}$ and substituting accordingly, we want to show that

$\lambda \sum a^{2}+(1-\lambda) \sum (a')^{2}\ge \sum \left(\lambda a+(1-\lambda)a'\right)^{2}$

which is equivalent to

$\lambda(1-\lambda) \sum (a-a')^{2}\ge 0$.

So $f(a, b, c) = a^{2}+b^{2}+c^{2}$ is indeed convex.

if the second partial derivative wrt. a certain variable is positive for all numbers in a domain, then you have a convex function

[quote="chien than"]Let $a;b;c \in [-1;1]$ suh that:$a+b+c=0$.Find max of:
$M=a^{2}+b^{2}+c^{2}$[/quote]
[hide]
One of $ab,bc,ca$ is nonnegative since they are 3 numbers with nonnegative product. WLOG let $ab\geq 0$, then
\[a^{2}+b^{2}+c^{2}\leq a^{2}+2ab+b^{2}+c^{2}=2c^{2}\leq 2 \]
and we can set $a=1,b=0,c=-1$ for equality.
[/hide]

I am not sure if we need to use the second derivative or the definition of convexity on something like this.  We see that $a^{2}+b^{2}+c^{2}$ is a parabola in $a$, $b$, or $c$.  Since this parabola opens upwards, it has a minimum value at its vertex and it is increasing on either side of it.  And if we need to be rigorous about this we can complete the square like we all learned to do in algebra 1.

[hide]Since $a+b+c=0$, $-a-b=c$.  So:

$a^{2}+b^{2}+c^{2}=a^{2}+b^{2}+(-a-b)^{2}=2a^{2}+2ab+2b^{2}$.

Case 1:

If $a$ and $b$ are both greater than equal to 0, then we can set $a+b=1$.  We can see that $a+b$ cannot be greater than 1 because then $c$ would be less than 0.  If $a+b$ is less than 1, then we can increase $a$ so that $a+b=1$ which would increase $2a^{2}+2ab+2b^{2}$.  Then we get:

$2a^{2}+2ab+2b^{2}=2a^{2}+2a(1-a)+2(1-a)^{2}=2a^{2}-2a+2$.

We can see this parabola has a vertex when $a=\frac{1}{2}$ so its max value is when $a$ is 0 or 1.

The case where $a$ is less than or equal to 0 and $b$ is less than or equal to 0 is exactly the same because all the terms in $a^{2}+2ab+2b^{2}$ have a degree of 2 which means the two negatives will cancel out leaving the function exactly the same.

Case 2:

If $a$ is greater than or equal to 0 and $b$ is less than or equal to 0, then no matter the values of $a$ and $b$, $-a-b=c$ will be between -1 and 1 inclusive.  We can substitue $-b_{1}=b$.  So:

$2a^{2}+2ab+2b^{2}=2a^{2}-2ab_{1}+2b_{1}^{2}$

This is a parabola in $a$ or in $b$ (and it is symmetric) so we can just take the values at the endpoints which yields a maximum value of 2.

The case $a$ is less than or equal to 0 and $b$ is greater than or equal to 0 is exactly the same as the function $2a^{2}+2ab+2b^{2}$ is symmetrical in $a$ and $b$.[/hide]

My solution is much worse than that smart person's up there but I already typed this all out.

[quote="bthings_2000"]What is this "convexity" you use to say $a=-1$? That is, why can you logically assume $a=-1$?[/quote]

No!! the last "proof" is just was proofing that the maximum is infinity.

but i calculate the limite wrongly.

and in the previously post i just correct the limit. i did not give a new proof

dump

[hide="Idea"]
$a+b+c=0$ is a plane in 3-D space, so we could find the maximum distance of a point within the boundries specified on the plane.
[/hide]


Twin primes are pairs of prime numbers that have a difference of 2. For example, 11 and 13 are twin primes. What is the probability that the sum of a pair of twin is divisible by 12 if both primes in the pair are less 100? Express your answer as a common fraction.

We can actually list out the pairs of twin primes since we suspect there won't be that many. \[ (3,5)\newline(5,7)\newline(11,13)\newline(17,19)\newline(29,31)\newline(41,43)\newline(59,61)\newline(71,73).\] There are 8 of them. Testing all 8 we find only 1 pair does not have a sum divisible by 12, therefore the probability is $ \boxed{\frac{7}{8}}.$


You are right, allnames.
I got a big mistake in calculating, sorry. :(

I think that it involves squaring the equation a+b+c=3. Is that right?

[quote="Allnames"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \sqrt {\frac {a^2 \plus{} ab}{2b^2 \plus{} 1}} \plus{} \sqrt {\frac {b^2 \plus{} bc}{2c^2 \plus{} 1}} \plus{} \sqrt {\frac {c^2 \plus{} ca}{2a^2 \plus{} 1}} \ge \sqrt {6}
\]
[/quote]
But it is wrong,my brother,You can try with $ a \equal{} 10^{ \minus{} 5},b \equal{} 1,6,c \equal{} 3 \minus{} a \minus{} b$ :blush:
May be my check is wrong?????[/quote]

You could try it out. In the second square root, with b and c, you will have already turned c into terms of a and b, and then you will have everything in terms of a and b, which makes it a lot easier.


a,b,c are  reals ,  a<b<c , a+b+c = 6 , ab+bc+ca=9 .Show  that
0<a<1<b<3<c<4
thanks

Solved before here.
http://www.mathlinks.ro/viewtopic.php?t=4347&highlight=uk+95


AMC
AMC 12
AMC 12 A
email
AIME
geometry
AMC 10

I took AMC12 A this year at the local university.
The AMC manager there told me that he would send the score through the e-mail, but I haven't received it yet.
Has anyone received the official score yet? When will I get the official score report?

[quote="AIME15"]I got mine yesterday.  AMC 10A, if that's relevant.[/quote]
at school or by email?

Oh, really?
I thought we had longer to wait, I have to wait 1.5 weeks though, because we have break now :(

So does this mean all schools who took the AMC 10A got their score reports?

(crossing fingers)

I'm quoting this from the AMC 10A/12A Teacher's Manual for 2009. Most of questions like this can be found in this document.

[quote="AMC Manual"]The AMC office will send results by email ([b]if available[/b]) and first class mail [b]as soon as the answer forms are scored[/b]. If you have [b]not[/b] received your results from our office [b]within 30 days after the AMC 10/AMC 12 [/b] please contact us to verify that your answer forms were in fact received. If you would like to receive your results by e-Mail, and [b]have not previously sent us your email address [/b], send a message, including your name, school name, address, and CEEB # to:
hstran@amc.unl.edu
Results are not official until you receive the postal mailed paper copy of your report, and that should be no longer than 30 days after AMC 10/AMC 12. Please retain this report for future reference, either in your files or with the student counselor’s office. Many students cite these scores on college applications. [/quote]

I made bold fonts to point out the key parts. To put it simple, don't worry if you don't receive email right away. It might not have been graded or your teacher have made not registered his/her email with MAA and so forth. As you can see, there is [b]no[/b] exact date mentioned other than as soon as possible. So, no need to panic.  :)

Some of the score reports have been sent and received, so we can infer that the tests have been scored. The rest is just a matter of how fast the AMC can process everything.

I didn't see this anywhere else on the forum, so I thought I'd add it here:

"Students scoring at least 97.5 on AMC 12 A and students
   (in grades 10 or below) scoring at least 120.0 on AMC 10A
   qualify for the AMERICAN INVITATIONAL MATHEMATICS
   EXAMINATION (AIME) which will be held on Tuesday, March 17, 2009.  "

So it seems cut-off scores have been revealed

[quote="max221"]its from the score report email that was sent out yesterday evening.[/quote]

Score report email? Yesterday evening? What? Huh?

To whom were the emails sent?

[quote="blackbelt14253"]To whom were the emails sent?[/quote]
They were sent to the contest managers at the schools where the tests were taken. Or at least to some of them. (I'd be surprised if they're all out yet.) I heard this second hand from a teacher at a school in my area, and he can confirm the 97.5 AIME cutoff for the 12A.

[quote]When will I get my official AMC12A score report? [/quote]

Reports for the 2009 AMC 10 A and the 2009 AMC 12 A started going out by email on Friday, Feb 20.
We will put the the first paper mailed reports in the mail today, schools will start receiving them about the end of this week, if the postal mail works swiftly.

As of today, we have scored about 50% of the answer sheets received.  Collating the scores and processing them into school reports takes additional time.

We send the school reports in the order they are processed.  The processing order goes generally by states, with no particular order to the states, everything depends on what we receive from schools by mail.  Note that even today  Monday, Feb 23, we are still receiving packages of score sheets from schools.    Obviously those schools will receive their reports later than schools that mailed their answer forms immediately after Feb 10, 2009.

[quote] "Students scoring at least 97.5 on AMC 12 A and students
(in grades 10 or below) scoring at least 120.0 on AMC 10A
qualify for the AMERICAN INVITATIONAL MATHEMATICS
EXAMINATION (AIME) which will be held on Tuesday, March 17, 2009. " [/quote]

I confirm this information.

[quote]If I remember correctly, they generally come 2-3 weeks after the competition. It's only been 9 days.
[/quote]

See our FAQ: AMC 10/12 #4
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

I should note that the AMC staff has done an excellent job of getting the reports out earlier than promised. They should be acknowledged for their hard work.

[quote]	 Is it the same for the AMC 10? [/quote]

Yes.

[quote] Is the score report just the score or does it tell you which problems you got wrong as well? [/quote]

Included with the email is a CVS file that can be read into a spreadsheet.  That information is a complete transcript of what we read from the answer sheets from that school.  Your contest manager has all that information in the email.  Do not request your individual information from the AMC, we will not provide it.

[quote]  Score report email? Yesterday evening? What? Huh?

To whom were the emails sent? [/quote]

I suggest you take some time and read the AMC 10/AMC 12 Teachers' Manual.
I also suggest you read out FAQs referenced above.

Steve Dunbar
MAA Director, American Mathematics Competitions


algorithm
number theory
Euclidean algorithm

[color=olive]Show that if a and b are integers with (a,b)=1, then (a+b,a-b)=1 or 2[/color]

Should I use the fact that (a+cb,b)=(a,b)? I've also tried to start with (a,b)=1 => 1=xa+yb, but then I was stuck, don't know how to begin. Thanks for helping me out.

By the Euclidean algorithm, $gcd(a+b,a-b)=gcd(a+b,2b)$

Suppose that $gcd(a+b,2b)>2$. Then there exists a prime >2 that divides both $2b$ and $a+b$. But then this prime would have to divide $b$ as well (since it divides $2b$ and it is not 2) and therefore since it divides $a+b$, it divides $a$. But this contradicts $gcd(a,b)=1$. So $gcd(a+b,a-b) \leq 2 = 1$ or $2$.

Perhaps this should be in the high-school intermediate forum?

Here it is [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=56078[/url]

 :)


$a,b,c \ge 0,a^2+b^2+c^2=3$
then:
$a+b+c+\frac{bc}a + \frac{ca}b +\frac{ab}c \ge 6$?

It is not true when $a=b\rightarrow 0$, $c\rightarrow\sqrt{3}$

yes you are right,
but what about:
$a+b+c+\left(\sqrt 3+1\right)\left(\frac{bc}a + \frac{ca}b +\frac{ab}c \right) \ge 6+3\sqrt 3$?


induction
email
number theory solved
number theory

Please help me out :

For every natural number n, S(n) denotes the sum of the digits of n in base 10.
Prove that there are infinitely many integers k > 0 such that S(3^k) > S(3^(k + 1)).

Are you sure it is strict ineq? I have seen this problem many times: Prove that there are infinitely many k such that S(3^k) >= S(3^(k+1)). It was even given for training at last year?s TST camp in Romania. The proof goes like this: assume that there exists an n0 such that S(3^(n+1)) > S(3^n) for any n>=n0. Then S(3^(n+1))-S(3^n) >= 9 for any n>=n0, so S(3^(n0+k)) >= S(3^n0)+9k > 9k for any positive k. But it?s easy to see that, since 10^x/3^x-> infinity as x->infinity when k->infinity we will eventually get a k s.t. 3^(n0+k) has less than k digits, so S(3^(n0+k)) <= 9k, so we got a contradiction.

Here?s another problem where the ineq is strict:

Prove that there are infinitely many numbers n s.t. S(2^n) > S(2^(n+1)).

Anyway, where did you find this problem you posted (assuming it is different from the one I just solved).

this is a russian problem (dont rem the year,mayb 1995) but i remember having solved this one exactly the way grobber did

It was also used in the german mathematical olympiad 1996.

Induction ... how do you mean ?
I don't see how to apply it in this case  :?

So it's >=, not > (I mean in the problem you posted.

well basicly we have it like so:

If there exists such N so that S(3^n)<S(3^(n+1)), for any n>=N then we have that S(3^(n+1))>=S(3^n)+9 and now you can prove this using induction.... :D

[quote="ckellough"][quote="Arne"]I would also like to see other problems concerning digit sums (in base 10).
I'm sure everybody knows problems of this type !
So please post some other problems ...[/quote]

IF S(N) is the repeated sum of the digits of a number until a single digit sum is reached (ex.  S(1 048 576) = 1+0+4+8+5+7+6 = 31; 3+1=4 so S(1 048 576) = 4)

Then find S(2005^5002)

Email me if you need a result.

Clay[/quote]
Take modulo 9 and everything comes out directly.

[quote="Arne"]I would also like to see other problems concerning digit sums (in base 10).
I'm sure everybody knows problems of this type !
So please post some other problems ...[/quote]
For all natural numbers $n$, denote by $S(n)$ the sum of digits of $n$ in base $10$. I also define $\mathbb{N}$ to be the set containing all positive integers only (excluding $0$) (to avoid further disputes).

[u]Problem 1[/u]
It is known that for an $m\in\mathbb{N}$ there exists an $n\in\mathbb{N}$ such that $S(n)=mS(3n)$. Find all possible $m$.

[u]Problem 2 (a famous one)[/u]
Prove that for all $k\in\mathbb{N}$ there exists an $a\in\mathbb{N}$ such that $S(a)=k$ and $S(a^2)=k^2$.

leepakhin!
hi! can you solution of 2 problem?

We can change $3$ by $a$ and $S(a^{n+1})>S(a^{n})$
it isn't difficult,too


A clock is designed so that its face is divided into eight hours instead of twelve.  The minute hand goes around once per hour (just like a regular clock).  At what time do the hour hand and the minute hand coincide between 3 and 4 o'clock?

[hide]

Uhh... 3:25:43? Just an educated guess...

[/hide]


modular arithmetic
number theory unsolved
number theory

Let $0\le m,n,p,q\le p-1$ with $p$ is prime number.
Prove that:
If $m+n\equiv p+q(mod p);m^2+n^2\equiv p^2+q^2(mod p)$ then {$m,n$}={$p,q$}.

I don't understand.  $p \equiv 0 \bmod p$ and $p^2 \equiv 0 \bmod p$ so why write $p$ in the equations at all?  And $\le p - 1$ only applies to $q$?

I think condition must be such:
Let $0\leq m,n,p,q\leq r-1$ with $r$ is prime number.
Prove that:
If $m+n\equiv{p+q}\pmod{r}$ and $m^2+n^2\equiv{p^2+q^2}\pmod{r}$, then {$m,n$}={$p,q$}.
I think tkhtn simply use symbol $p$ for two variables.

Solution of this problem:
Our conditions are equivalent follows:
$m-p\equiv q-n\equiv k\pmod {r}$;  (1)
$k(m+p)\equiv (m-p)(m+p)\equiv m^2-p^2\equiv q^2-n^2\equiv (q-n)(q+n)\equiv k(q+n)\pmod {r}$.
If $k\equiv 0\pmod {r}$, then $m=p$ and $n=q$ - what we need.
Else since $k(m+p)\equiv k(q+n)\pmod {r}$, then
$m+p\equiv q+n\equiv k\pmod {r}$;  (2)
If $r=2$ there is contrexample: $m=n=1; p=q=0$, hence we shall proof the theorem if $r$ is odd prime number.
Then from (1) and (2) we get
(1)+(2): $2m\equiv 2q\pmod {r}$;
(2)-(1): $2p\equiv 2n\pmod {r}$.
 :arrow: $m=q$ and $n=p$, q.e.d.

Here is one interesting problem which is connected with this one: Let p be a prime number >2. Let Sp={0,1,2,...,p-1}. Consider all points (x,y) in the plane for which x,y are members of Sp. Show that we can choose p of these points such that there are no 3 which lies on a straight line and there are no 4 which are vertexs of parallelogram.


[quote="arqady"]Find the least value $k>0$ such that the following inequality
$\sum_{cyc}\frac{xy^{2}}{y+z}\leq\frac{k(x^{2}+y^{2}+z^{2})-(k-1)(xy+xz+yz)}{2}$
holds for all positive $x,y$ and $z.$[/quote]

Applying $x=y=1, z=0$, we obtain $k \geq 1$. Now I will prove that $k=1$ satisfies.

$\sum \frac{xy^{2}}{y+z}= \sum \frac{1}{2}\frac{xy^{2}+xz^{2}}{y+z}$
$\iff \sum \frac{1}{2}\frac{xy^{2}-xz^{2}}{y+z}=0$
$\iff \sum \frac{1}{2}(xy-xz)=0$ is true.
So $\sum \frac{xy^{2}}{y+z}\leq \frac{\sum x^{2}}{2}$
$\iff \sum (x^{2}-\frac{xy^{2}+xz^{2}}{y+z}) \geq 0$
$\iff \sum (x-y)^{2}\frac{xy}{(x+z)(y+z)}\geq 0$ is true.
(Equality holds when $(x,y,z)=(k,k,k), (k,k,0)$ or any cyclic forms.)


calculus
integration
trigonometry
limit
calculus computations

(1) Evaluate $ \int_0^{\plus{}\infty} \frac{\sin \alpha x}{x}\ dx.$

(2) For $ a>b>0,$ evaluate $ \int_0^{\plus{}\infty} \frac{\sin ax\cos bx}{x}\ dx.$

1) Via integration by parts, we can show that

$ \int_{0}^{\infty }e^{\minus{}\beta x}\cos \gamma xdx\equal{}\allowbreak \frac{\beta }{\gamma ^{2}\plus{}\beta ^{2}}\,\,\,\,\,(\beta >0)$

$ \int_{0}^{\infty }e^{\minus{}\beta x}\frac{\sin \alpha x}{x}dx$

$ \equal{}\int_{0}^{\infty }e^{\minus{}\beta x}\left( \int_{0}^{\alpha }\cos \gamma xd\gamma \right) dx$

$ \equal{}\int_{0}^{\infty }\left( \int_{0}^{\alpha }e^{\minus{}\beta x}\cos \gamma xdx\right) d\gamma$

$ \equal{}\int_{0}^{\infty }\frac{\beta }{\gamma ^{2}\plus{}\beta ^{2}}d\gamma$

$ \equal{}\lim_{\gamma \rightarrow \infty }\tan ^{\minus{}1}\frac{\gamma }{\beta }\equal{}\frac{\pi }{2}$

2) Using (1)

$ \int_{0}^{\infty }\frac{\sin ax\cos bx}{x}dx$

$ \equal{}\frac{1}{2}\int_{0}^{\infty }\frac{\sin (a\plus{}b)x\plus{}\sin (a\minus{}b)x}{x}dx$

$ \equal{}\frac{1}{2}\left( \int_{0}^{\infty }\frac{\sin (a\plus{}b)x}{x}dx\plus{}\int_{0}^{\infty }\frac{\sin (a\minus{}b)x}{x}dx\right) \equal{}\frac{\pi }{2}$

For (1), we don't need to take into consideration of sign of $ \alpha$?

$ \int_{0}^{\infty }\frac{\sin \alpha x}{x}dx\equal{}\allowbreak \frac{\pi \alpha }{2|\alpha |}$ for $ \alpha \neq 0$


geometry
rectangle
algebra
function
domain
combinatorics unsolved
combinatorics

Consider L-shaped tiles which can be formed by adjoining a unit tile to a 1 X 3 tile. Let R (m,n) be a rectangular domain with sides m and n where m and n are positive integers greater than 1. Prove that R (m,n) can be tiled with L-shaped pieces iff mn is a multiple of 8.

At first if $m=2k, n=4k$ we can make $2\times 4$ tile so we can divide the board into $2\times 4$ tiles so easily we can find that $4\mid mn$(necessary condition at the first sight).Then colouring the colunms with black and white alternatively .So we have two kind of the L-shape tile which one has three black and one white squares and other has one black and three white squares. Suppose that the number of first kind is $x$ and the number of second kind is $y$ then we have $3x+y$ black squares and $3y+x$ white squares .Symmetry says that $3x+y=3y+x \rightarrow x=y$ and we know $x+y=\frac{mn}{4}$ so $8\mid mn$.
When $8\mid mn$ ,we have two manner ,1) one of them ($m$ or $n$) divisible by $4$ and other divisible by $2$ ,2) one of them is odd and the other divisible by $8$ .We can easily prove the both manner by using the induction. (Sorry I haven't enough time to prove it by induction).


MATHCOUNTS probability geometry rectangle ratio inequalities quadratics

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

[hide="solutions"]1. The area of the entire dartboard is the area of a circle with radius 4 or 16pi. The area of the "ring" made up of the circle of radius 3 minus the circle of radius 2 is 9pi-4pi=5pi. 5pi/16pi=[b]5/16[/b].

2. First to find the volume of the can: If the radius of each ball is 4, their diameters are all 8. 8*3 makes up the height of the can since there are 3 balls; the height is 24. The radius of the base is 4 for obvious reasons; it is the same as the radius of the ball. So 16pi * 24 = 384pi. Now for the volume of one ball: Just plug the radius into the formula for the volume of a sphere. 4/3*pi*4 3  = 4/3* 64pi. Since there are three balls, we multiply this by three to get 4*64pi, or 236pi. 384pi-236pi=[b]148pi[/b].

3. A little confused as to what I'm supposed to find out.

4. We know that a*b = 8!, or 40320. Since we want a and b to be as close together as possible, they will be as close as possible to the square root of 8!, which is about 200. The closest I could get using guess and check was 210 (7*5*3*2) with the lower number being 192 (8*6*4*1). So, I'm hoping [b]210[/b].

5. Not terribly difficult. Roses are 20$/dozen; at 3/4 of one dozen (9 roses) that's 15$. Daisies are 5$/dozen and with half of a dozen daisies they cost $.50. Chrysanthemums (okay, I used copy and paste) are 10$/dozen at 1.5 dozens, that's also 15$. Add 'em up and you get [b]32.50[/b].

6. No!!!! I figured out the equation worked for 30 and 10 but 30 is a multiple of 10!!!! Augh!!!

7. Well, to find a, I just subtract 4 from 100 to get 96. Since 12, 14, 16...counts up by 2's and c counts up by 1's, I can look for a pattern that involves dividing 12, 14, 16...by 2. Sure enough, to get c, we just subtract 5 from half of the 12, 14, 16... term. So for 100 that gives 45. b is just 8 more than c, so it is 53. 96+45+53=[b]194[/b].

8. Tricky problem. Diagonal=d, width=w, length=l. Set up the proportion (d/l)=(l/w). Using means-extremes, l 2 =wd, or l 2 =2d since w=2. Because the length, width and diagonal make up a right triangle with the Pythagorean theorem with the diagonal as the hypotenuse, we have another equation to add: l 2 +w 2 =d 2 . Or, since w=2, l 2 +4=d 2 . Subtract l 2 =2d from this equation, and we get d 2 -2d-4=0. Quadratics! USing the Quadratic Formula, [b]d= 1 + sqrt(5).[/b]

9. Huh boy. I don't think this quadratic is factorable. So the formula yields 2+sqrt(5) or 2-sqrt(5). Let's try 2+sqrt(5). God, I can't think like this at 9:11. I'll try this one tomorrow.

10. Seems easy. Solve the inequalities and you get x is greater than or equal to 2.5 and it's less than or equal to 6. So there are your boundaries. The difference is [b]3   1/2[/b][/hide]


I'm not very gifted with computers...

I just updated my Adobe PDF Viewer, and now when I compile and view my Latex documents, I get a message that says, "Not all fonts could be loaded."  Does anybody here know how to fix it?

Thanks.

I fixed it!

With my minimal abilities with cimputers, I do not belong in the 21st century.  :wallbash_red:


calculus
integration
topology
analytic geometry
function
trigonometry
geometry

I have now to deal with surface integrals (the kind of creepy stuff you define on differential manifolds), and I read that $|det(J(u))|$, where $J$ is the Jacobian of $u$ (say $u$ is $C^{1}$), is often interpreted in physics as the volume of an elementary parallepipede. Alright, except that I am a physics dropout, so ... could you please explain to me ?

Consider a 2D-volume integral $\int \int f\ \text{d}V = \int \int f(x,y)\ \text{d}x\ \text{d}y.$ When changing coordinates to $u = u(x, y),\ v = v(x, y),$ the function f, the region of integration and $\text{d}V$ change according to this transformation. Let $x = x(u, v),\ y = y(u, v)$ be the inverse transformation. Then

$\text{d}x = \frac{\partial x}{\partial u}\text{d}u+\frac{\partial x}{\partial v}\text{d}v,\ \ \text{d}y = \frac{\partial y}{\partial u}\text{d}u+\frac{\partial y}{\partial v}\text{d}v$

Before the transformation, the elementary 2D-volume is

$\text{d}V = |\text{d}\vec{x}\times \text{d}\vec{y}| = |\text{d}x\ \vec{e}_{x}\times \text{d}y\ \vec{e}_{y}| = \text{d}x\ \text{d}y\ |\vec{e}_{z}| = \text{d}x\ \text{d}y$

Substituting $\text{d}\vec{x}= \frac{\partial x}{\partial u}\text{d}\vec{u}+\frac{\partial x}{\partial v}\text{d}\vec{v}= \frac{\partial x}{\partial u}\text{d}u\ \vec{e}_{u}+\frac{\partial x}{\partial v}\text{d}v\ \vec{e}_{v}$ and  $\text{d}\vec{y}= \frac{\partial y}{\partial u}\text{d}\vec{u}+\frac{\partial y}{\partial v}\text{d}\vec{v}= \frac{\partial y}{\partial u}\text{d}u\ \vec{e}_{u}+\frac{\partial y}{\partial v}\text{d}v\ \vec{e}_{v},$

we get the elementary 2D-volume in terms of $\text{d}u\ \text{d}v$ after the transformation:

$\text{d}V = |\text{d}\vec{x}\times \text{d}\vec{y}| = \left|\left(\frac{\partial x}{\partial u}\text{d}u\ \vec{e}_{u}+\frac{\partial x}{\partial v}\text{d}v\ \vec{e}_{v}\right) \times \left(\frac{\partial y}{\partial u}\text{d}u\ \vec{e}_{u}+\frac{\partial y}{\partial v}\text{d}v\ \vec{e}_{v}\right)\right| =$

$= \text{d}u\ \text{d}v\ \left|\left(\frac{\partial x}{\partial u}\cdot \frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\cdot \frac{\partial y}{\partial u}\right)\ \vec{e}_{w}\right| = \left|\frac{\partial(x, y)}{\partial(u, v)}\right| \text{d}u\ \text{d}v$

and the absolute value of jacobian $J(u, v) = \frac{\partial(x, y)}{\partial(u, v)}$ popped up. Similar thing could be done for a 3D-volume $\text{d}V = \text{d}x\ \text{d}y\ \text{d}z$ and the absolute value of jacobian would show up: $\text{d}V = \left|\frac{\partial(x, y, z)}{\partial(u, v, w)}\right| \text{d}u\ \text{d}v\ \text{d}w.$ Consider the most frequent transformations - from $(x,y)$ carthesian coordinates to $(r,\ \phi)$ polar coordinates and from $(x,y,z)$ carthesian coordinates to $(r,\ \phi,\ \theta)$ spherical coordinates. In polar coordinates $r \ge 0,$  $\phi \in [0, 2\pi),$

$x = r \cos \phi$
$y = r \sin \phi$

$J(r,\ \phi) = \frac{\partial x}{\partial r}\cdot \frac{\partial y}{\partial \phi}-\frac{\partial x}{\partial \phi}\cdot \frac{\partial y}{\partial r}=$

$= \cos \phi \cdot r \cos \phi-\sin \phi \cdot r (-\sin \phi) = r (\cos^{2}\phi+\sin^{2}\phi) = r$

In $(x,y)$ carthesian coordinates, the elementary 2D-volume is a little rectangle with sides $\text{d}x,\ \text{d}y$ and $\text{d}V = \text{d}x\ \text{d}y.$ In polar coordinates, the elementary 2D-volume is a little curved "rectangle" with sides $\text{d}r,\ \text{d}s,$ where the arc $ds = r\ \text{d}\phi.$ The 2D-volume is then $\text{d}V = r\ \text{d}r\ \text{d}\phi,$ where $r = |J(r, \phi)|$ is just the absolute value of the jacobian (the absolute value not necessary here). In spherical coordinates $r \ge 0,$  $\phi \in [0, 2\pi),$ $\theta \in [0, \pi],$

$x = r \cos \phi \sin \theta$
$y = r \sin \phi \sin \theta$
$z = r \cos \theta$

$J(r, \phi, \theta) =$

$= \frac{\partial x}{\partial r}\cdot \frac{\partial y}{\partial \phi}\cdot \frac{\partial z}{\partial \theta}+\frac{\partial x}{\partial \phi}\cdot \frac{\partial y}{\partial \theta}\cdot \frac{\partial z}{\partial r}+\frac{\partial x}{\partial \theta}\cdot \frac{\partial y}{\partial r}\cdot \frac{\partial z}{\partial \phi}-$

$-\frac{\partial x}{\partial \theta}\cdot \frac{\partial y}{\partial \phi}\cdot \frac{\partial z}{\partial r}-\frac{\partial x}{\partial \phi}\cdot \frac{\partial y}{\partial r}\cdot \frac{\partial z}{\partial \theta}-\frac{\partial x}{\partial r}\cdot \frac{\partial y}{\partial \theta}\cdot \frac{\partial z}{\partial \phi}=$

$= \cos \phi \sin \theta \cdot r \cos \phi \sin \theta \cdot r (-\sin \theta)+r (-\sin \phi) \sin \theta \cdot r \sin \phi \cos \theta \cdot \cos \theta+0-$

$-r \cos \phi \cos \theta \cdot r \cos \phi \sin \theta \cdot \cos \theta-r (-\sin \phi) \sin \theta \cdot \sin \phi \sin \theta \cdot r (-\sin \theta)-0 =$

$=-r^{2}\sin \theta (\cos^{2}\phi \sin^{2}\theta+\sin^{2}\phi \cos^{2}\theta+\cos^{2}\phi \cos^{2}\theta+\sin^{2}\phi \sin^{2}\theta) =-r^{2}\sin\theta$

The negative sign comes from the fact that when $\theta$ grows, $z = r \cos \theta$ drops, so that the integration limits $z_{1}< z_{2}$ result in $\theta_{1}> \theta_{2}$ and they have to be flipped, while changing the sign of the integral. In $(x,y,z)$ carthesian coordinates, the volume  is a little 3D-rectangle with sides $\text{d}x,\ \text{d}y,\ \text{d}z$ and $\text{d}V = \text{d}x\ \text{d}y\ \text{d}z.$  In spherical coordinates, the volume is a little 3D curved "rectangle" with sides $\text{d}r,\ \text{d}s,\ \text{d}t,$ where the arc $\text{d}s = r\ \text{d}\phi$ as in polar coordinates. When we change the angle $\theta$ by $\text{d}\theta$, length of the corresponding arc $\text{d}t$ is proportional to the distance from the origin r = 0, just like for the arc $\text{d}s$. An addition, it decreases as $\sin \theta,$ as we move from the equator to the poles of a sphere with radius r, and it is equal to $\text{d}t = r \sin \theta\ \text{d}\theta.$ The elementay 3D volume is then $\text{d}V = \text{d}r\ \text{d}s\ \text{d}t = r^{2}\sin \theta\ \text{d}r\ \text{d}\phi\ \text{d}\theta,$ where $r^{2}\sin \theta = |J(r, \phi, \theta)|$ is just the absolute value of the jacobian (since $\theta \in [0, \pi],$ $\sin \theta \ge 0$).


[hide]
The first day he uses $\frac 13$ of the paint.
The next day he uses $\frac 13 (1-\frac 13)=\frac 13 \frac 23 = \frac 29$.
So he has used $\frac 13 + \frac 29 = \frac 59$ of the paint.
He has $1-\frac 59=\frac 49$ left.
$D$[/hide]

[hide]On the first day, 1/3 of the paint is used. On the second day, 1-1/3/3=2/9 is used. 1-(1/3+2/9)=4/9 of the remaining paint left, or answer D.[/hide]

[quote="catcurio"]A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?

$\text{(A)}\ \frac1{10} \qquad\text{(B)}\ \frac19 \qquad\text{(C)}\ \frac13 \qquad\text{(D)}\ \frac49 \qquad\text{(E)}\ \frac59$[/quote]

[hide]1/3 is used. You have 2/3 left. 1/3 is used thus 2/3 is remaining. THus you have 4/9 to use the final day. The answer is D[/hide]

[hide]
Let the original amount be x. $\frac 13 x$ is used on the first day, leaving $\frac 23 x$. The second day, $\frac 23 \cdot \frac 13 x$ is used, leaving $\frac 23 \cdot \frac 23 x$ or $\frac 49 x$. The answer is D.
[/hide]

[quote="catcurio"]A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?

$\text{(A)}\ \frac1{10} \qquad\text{(B)}\ \frac19 \qquad\text{(C)}\ \frac13 \qquad\text{(D)}\ \frac49 \qquad\text{(E)}\ \frac59$[/quote]

[hide]$4/9$ because the first day they used up 1/3. The next day they had $2/3$ left so they used up $2/9$. Add them up and they used $5/9$. There is obviously $4/9$ left. :) That almost got me. I thought it was $5/9$ for a sec.[/hide]


calculus
integration
trigonometry
function
calculus computations

Is there a closed formula to the integral $ \int_{0}^{\infty} \frac{t^m}{(1\plus{}t)^n} dt$, when $ n,m$ are non-negative integers and $ n>m\plus{}2$?

hello Maple 12 gives $ \frac{\Gamma(\minus{}1\plus{}n\minus{}m)\cdot \Gamma(m\plus{}1)}{\Gamma(n)}$.
Sonnhard.

Let $ I(n,m) \equal{} \int_{0}^{\infty} \frac {t^m}{(1 \plus{} t)^n} \, dt$. For $ n \geq m \plus{} 2$, integration by parts shows that

$ I(n,m) \equal{} \begin{cases} \cfrac{m}{n \minus{} 1} \, I(n \minus{} 1, m \minus{} 1) & m \geq 1 \\
\cfrac{1}{n \minus{} 1} & m \equal{} 0 \end{cases}.$

Therefore we have $ I(n, m) \equal{} \frac {m}{n \minus{} 1} \cdots \frac {1}{n \minus{} m} I(n \minus{} m, 0) \equal{} \frac {m! (n \minus{} m \minus{} 2)!}{(n \minus{} 1)!}.$

If $ m$ and $ n$ are allowed to be real, just apply the substitution $ t \equal{} \tan^2 \theta$. Then you'll have a closed form involving beta function or gamma function, namely

$ \beta(m \plus{} 1, n \minus{} m \minus{} 1) \equal{} \frac {\Gamma(m \plus{} 1)\Gamma(n \minus{} m \minus{} 1)}{\Gamma(n)}$,

which is an analogy for the result above.

p.s. Dream Theater  d:lol: b

Cool... Also, is there a nice form to $ \int_{0}^{\infty} \frac {t^n \minus{} 1}{(t \minus{} 1)(1 \plus{} t)^{n \plus{} 3}} \, dt$ ? (this was to motivation to my original question...)

@sos440:
[hide] DT are great! They are coming to my country in a few months but it is very expensive... Still hoping to attend it[/hide]

Note that $ I(n, m) \equal{} \frac {1}{n \minus{} 1} \binom{n \minus{} 2}{m}$ and $ \frac {t^n \minus{} 1}{t \minus{} 1} \equal{} 1 \plus{} \cdots \plus{} t^{n \minus{} 1}$. Then we have

$ \int_{0}^{\infty} \frac {t^n \minus{} 1}{(t \minus{} 1)(1 \plus{} t)^{n \plus{} 3}} \, dt \equal{} \sum_{k \equal{} 0}^{n \minus{} 1} I(n \plus{} 3, k)$ $ \equal{} \frac {1}{n \plus{} 2} \sum_{k \equal{} 0}^{n \minus{} 1} \binom{n \plus{} 1}{k} \equal{} \frac {2^{n \plus{} 1} \minus{} (n \plus{} 1)}{n \plus{} 2}.$

@bambaman
[hide]Concert?! Wow, that's really enviable. [/hide]

Another way to arrive at the same result could be to set $ y \equal{} \frac{1}{1\plus{}t}$ , so that the integral becomes $ \int_{0}^{1} (1\minus{}y)^{m} y^{n\minus{}m\minus{}2} dy \equal{} \beta( m\plus{}1 , n \minus{} m \minus{} 1)$

Remark: using the above result we can easily evaluate integrals of the form
$ \int_{0}^{\frac{\pi}{2}}\frac{ \sin^{2m\minus{}1} \theta \cos^{2n\minus{}1} \theta}{(a^{2} \cos^2 \theta \plus{} b^{2} \sin^2 \theta )^{m\plus{}n}} d\theta$


[hide="solution"]
$ 3^{x\minus{}y} \equal{} 3^{4} \rightarrow x \minus{} y \equal{} 4$
$ 3^{x\plus{}y} \equal{} 3^{5} \rightarrow x \plus{} y \equal{} 5$
$ 2x \equal{} 9 \rightarrow x \equal{} \frac{9}{2}$
$ y \equal{} \frac{1}{2}$
$ \frac{1}{x} \plus{} \frac{1}{y} \equal{} \frac{20}{9}$
[/hide]


A triangle with side lengths in the ratio $ 3: 4: 5$ is inscribed in a circle of radius $ 3$. What is the area of the triangle?

$ \textbf{(A)}\ 8.64 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 5\pi \qquad \textbf{(D)}\ 17.28 \qquad \textbf{(E)}\ 18$

the answer is A.
it is obvious that the tringle is right and so hypotinuse is 6 as it equals twice thw radius of the circle.
therefore we can easily find the 2 other sides
and the area also...

[hide]
the other two sides would be 3.6 and 4.8
so the area is 8.64[/hide]

[hide]
Multiply the area of a 3-4-5 triangle by the square of the ratio of the sides (which is $\frac{6}{5}$), so the answer is $8.64, A$.[/hide]

How did you guess
definatly we know by Pythogeran Triples that the triangle is a right triangle
so all we see is that in a circle that the hypotenuse is the diameter
so we see that 6 is the diameter of the circle so 6/5 x 4
and 6/5 x 3
now mutiply these and devide by 2
 and there is the answer


Deternime $ p,\ q$ such that $ a_{n\plus{}1}\plus{}pb_{n\plus{}1}\equal{}q(a_n\plus{}pb_n).$

From the first equation, $ b_n \equal{} \ldots$ and so also $ b_{n \plus{} 1} \equal{} \ldots$.  Substitute these into the second equation to get an equation with $ b_i$s alone.  Then get a formula and use this to get $ a_n$.

Define $ c_n \equal{} a_n \plus{} b_n$.

Equation 1 = Equation 2:
$ a_{n \plus{} 1} \plus{} b_{n \plus{} 1} \equal{} 3 (a_n \plus{} b_n) \plus{} 2 b_n$
$ c_{n \plus{} 1} \minus{} 3 c_n \equal{} 2 b_n$

Equation 1 * 3 + Equation 2, shifted back one index:
$ 3 (a_n \plus{} b_n) \minus{} 2b_n \equal{} 7 (a_{n \minus{} 1} \plus{} b_{n \minus{} 1} )$
$ 3c_n \minus{} 7c_{n \minus{} 1} \equal{} 2b_n$

Combining the two results:
$ c_{n \plus{} 1} \minus{} 3c_n \equal{} 3c_n \minus{} 7c_{n \minus{} 1}$
$ c_{n \plus{} 1} \minus{} 6c_n \plus{} 7c_{n \minus{} 1} \equal{} 0$

And you know how to solve this.

That's what I got as well but I couldn't solve it. Could anyone point me in a good direction (in terms of e-books or something like that) how to solve recurrences like these?
Mathematica gives

$ c_n\equal{}\frac{7 (\sqrt{2}\minus{}2) (3\minus{}\sqrt{2} )^n\plus{}(\sqrt{2}\minus{}10) (3\plus{}\sqrt{2})^n}{14 (\sqrt{2}\minus{}3)}$

[quote="dgreenb801"]I came up with this in doing another problem, can anyone show me how to solve it?
$ a_1 \equal{} 1$, $ b_1 \equal{} 1$
$ a_{n \plus{} 1} \equal{} 2a_n \plus{} b_n$
$ b_{n \plus{} 1} \equal{} a_n \plus{} 4b_n$
Solve for $ a_{n } \plus{} b_{n}$ in terms of n.[/quote]
This can be brute-forced by solving one variable and substituting that into the other equation, like

$ a_n\equal{}b_{n\plus{}1}\minus{}4b_n$
$ b_{n\plus{}2}\minus{}4b_{n\plus{}1}\equal{}2(b_{n\plus{}1}\minus{}4b_n)\plus{}b_n$ and then solve the quadratic

Out of curiosity what problem were you doing? :D

I was doing [url]http://www.mathlinks.ro/viewtopic.php?p=508185#508185[/url].

[quote="Flame"]Could anyone point me in a good direction (in terms of e-books or something like that) how to solve recurrences like these? [/quote]  The first chapter or two of [url=http://www.math.upenn.edu/~wilf/DownldGF.html]generatingfunctionology[/url] are good.

Why does no one solve the problem with my hint?

[quote="kunny"]Deternime $ p,\ q$ such that $ a_{n + 1} + pb_{n + 1} = q(a_n + pb_n).$[/quote]

$ \Longleftrightarrow 2a_n+b_n+p(a_n+4b_n)=q(a\n+pb_n)$

$ \Longleftrightarrow p+2=q,\ 4p+1=q$, yielding $ (p,\ q)=(1+\sqrt{2},\ 3+\sqrt{2}),\ (1-\sqrt{2},\ 3-\sqrt{2}).$

Now from the first line we have $ a_n+pb_n=q^{n-1}(a_1+pb_1)=(1+p)q^{n-1}.$

Thus we get:

$ a_n+(1+\sqrt{2})b_n=(2+\sqrt{2})(3+\sqrt{2})^{n-1}$

$ a_n+(1-\sqrt{2})b_n=(2-\sqrt{2})(3-\sqrt{2})^{n-1}$

Adding the both sides gives $ \boxed{a_n+b_n=\frac{2+\sqrt{2}}{2}(3+\sqrt{2})^{n-1}+\frac{2-\sqrt{2}}{2}(3-\sqrt{2})^{n-1}\ (n\geq 1)}.$


Let $ x,y,z$ be the positive numbers satisfying $ x\plus{}y\plus{}z\equal{}1$.Let $ a\equal{}\sqrt{x^2\plus{}xy\plus{}y^2},b\equal{}\sqrt{y^2\plus{}yz\plus{}z^2},c\equal{}\sqrt{z^2\plus{}zx\plus{}x^2}$.Prove that $ ab\plus{}bc\plus{}ca$≥$ 1$

[quote="SUPERMAN2"]Let $ x,y,z$ be the positive numbers satisfying $ x \plus{} y \plus{} z \equal{} 1$.Let $ a \equal{} \sqrt {x^2 \plus{} xy \plus{} y^2},b \equal{} \sqrt {y^2 \plus{} yz \plus{} z^2},c \equal{} \sqrt {z^2 \plus{} zx \plus{} x^2}$.Prove that $ ab \plus{} bc \plus{} ca$≥$ 1$[/quote]

We can reduce the given inequality to homogeneous one and after expanding , we have :
$ \sum (x^2 \plus{} xy \plus{} y^2)(x^2 \plus{} xz \plus{} z^2) \plus{} 2\sum (x^2 \plus{} xy \plus{} y^2)\sqrt {(x^2 \plus{} xz \plus{} z^2)(y^2 \plus{} yz \plus{} z^2)}\geq (x \plus{} y \plus{} z)^4$
Easy to prove that
$ x^2 \plus{} xy \plus{} y^2 \geq \frac {3}{4}(x \plus{} y)^2$
The inequality is equivalent to :
$ 2\sum (x^2 \plus{} xy \plus{} y^2)(x^2 \plus{} xz \plus{} z^2) \plus{} \sum (x^2 \plus{} xy \plus{} y^2)(x \plus{} z)(y \plus{} z)\geq 2(x \plus{} y \plus{} z)^2$
Which is obviously true by
$ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2\geq xyz(x \plus{} y \plus{} x)$

[quote="SUPERMAN2"]Let $ x,y,z$ be the positive numbers satisfying $ x \plus{} y \plus{} z \equal{} 1$.Let $ a \equal{} \sqrt {x^2 \plus{} xy \plus{} y^2},b \equal{} \sqrt {y^2 \plus{} yz \plus{} z^2},c \equal{} \sqrt {z^2 \plus{} zx \plus{} x^2}$.Prove that $ ab \plus{} bc \plus{} ca$≥$ 1$[/quote]
$ \sum_{cyc}\sqrt{(x^2\plus{}xy\plus{}y^2)(x^2\plus{}xz\plus{}z^2)}\geq\sum_{cyc}\left(x^2\plus{}\frac{x(y\plus{}z)}{2}\plus{}yz\right)\equal{}$
$ \equal{}(x\plus{}y\plus{}z)^2\equal{}1.$


function
induction
limit
algebra unsolved
algebra

Find all monotoic functions $f: \mathbb R \to \mathbb R$ so that \[ f(x) \plus{} 2x \equal{} f(f(x))\] for all reals $x$.

There are infinetelymane solutions.
1)Let $ x$ is fixed point, then $ 2x \equal{} 0$. Because $ f(0)$ fixed point we have unique fixed point $ x \equal{} 0$.
2) If $ f(x) \equal{} 0$, then $ x \equal{} 0$. Therefore defined function $ g(x) \equal{} \frac {f(x)}{x}, x\not \equal{} 0$ and $ (g(xg(x)) \minus{} 1)g(x) \equal{} 2$.
If $ g(x)\equiv const$, then $ g(x) \equal{} \minus{} 1$ or $ g(x) \equal{} 2$.
Let $ g_2(x)\ge 2$ in interval $ (x_1,x_2), x_1 > 0, x_2 \equal{} g(x_1)x_1$ and $ g_2(x)x$ is monototonic. Then we get
$ g(y) \equal{} 1 \plus{} \frac {2}{g(x)},y \equal{} xg(x)$ and define $ g(x)$ in interval $ (x_2,x_3),x_3 \equal{} g(x_2)x_2 > x_2 \equal{} g(x_1)x_1$.
Then we can define function in interval $ (x_3,x_4)$ e.t.c.
For example, we can consider any decrease function $ f_0(x),f_0(0)\equal{}0,x>0$, suth that $ f_0(x)\plus{}2x$ increase
and define $ f(x)\equal{}f_0(x),x\ge 0, f(x)\equal{}x\plus{}2f_0^{\minus{}1}(x),x<0$.
Or consider any increase function $ f_0(x), 1<x<2$, suth that $ f_0(1)\equal{}2, f_0(2)\equal{}4$ and define
$ f_1(x)\equal{}f_0(x), 1\le x\le 2, f_1(x)\equal{}x\plus{}2f_0^{\minus{}1}(x), 2\le x\le 4$.
Define $ f(x)\equal{}sign(x) 4^{[ln_4|x|}f_1(4^{\{ln_4|x|\}})$. We can take distinct functions $ f_0(x),f_1(x)$ for distinct intervals $ (4^n,4^{n\plus{}1})$.

[quote="Rust"]There are infinetelymane solutions.
1)Let $ x$ is fixed point, then $ 2x \equal{} 0$. Because $ f(0)$ fixed point we have unique fixed point $ x \equal{} 0$.
2) If $ f(x) \equal{} 0$, then $ x \equal{} 0$. Therefore defined function $ g(x) \equal{} \frac {f(x)}{x}, x\not \equal{} 0$ and $ (g(xg(x)) \minus{} 1)g(x) \equal{} 2$.
If $ g(x)\equiv const$, then $ g(x) \equal{} \minus{} 1$ or $ g(x) \equal{} 2$.
Let $ g_2(x)\ge 2$ in interval $ (x_1,x_2), x_1 > 0, x_2 \equal{} g(x_1)x_1$ and $ g_2(x)x$ is monototonic. Then we get
$ g(y) \equal{} 1 \plus{} \frac {2}{g(x)},y \equal{} xg(x)$ and define $ g(x)$ in interval $ (x_2,x_3),x_3 \equal{} g(x_2)x_2 > x_2 \equal{} g(x_1)x_1$.
Then we can define function in interval $ (x_3,x_4)$ e.t.c.
For example, we can consider any decrease function $ f_0(x),f_0(0) \equal{} 0,x > 0$, suth that $ f_0(x) \plus{} 2x$ increase
and define $ f(x) \equal{} f_0(x),x\ge 0, f(x) \equal{} x \plus{} 2f_0^{ \minus{} 1}(x),x < 0$.
Or consider any increase function $ f_0(x), 1 < x < 2$, suth that $ f_0(1) \equal{} 2, f_0(2) \equal{} 4$ and define
$ f_1(x) \equal{} f_0(x), 1\le x\le 2, f_1(x) \equal{} x \plus{} 2f_0^{ \minus{} 1}(x), 2\le x\le 4$.
Define $ f(x) \equal{} sign(x) 4^{[ln_4|x|}f_1(4^{\{ln_4|x|\}})$. We can take distinct functions $ f_0(x),f_1(x)$ for distinct intervals $ (4^n,4^{n \plus{} 1})$.[/quote]

I'm sorry, Rust : I've not fully analyzed your demo but it seems to me that your conclusion is wrong :

I can easily show that the only decreasing solution is $ f(x)\equal{}\minus{}x$ and your conclusion seems to say that we can have a lot of others.
I think that the only increasing soluton is $ f(x)\equal{}2x$. I did not succeed showing this till now but I can show that if an increasing solution is different from $ f(x)\equal{}2x$, then it is continuous at $ 0$ and has discontinuity points in any interval owning $ 0$. And this constraint does not appear in your conclusion.

[quote="tdl"][quote="pco"]I can easily show that the only decreasing solution is $ f(x) \equal{} \minus{} x$[/quote]
Please give it more detail! Thanks![/quote]

If $ \exists u$ such that $ f(u) \equal{} u$, then $ u \plus{} 2u \equal{} u$ and so $ u \equal{} 0$
$ f(0) \plus{} 2\cdot 0 \equal{} f(f(0))$ and so $ f(f(0)) \equal{} f(0)$ and so (see above) $ f(0) \equal{} 0$

$ f(x)$ is injective and so, since monotonous, is either increasing, either decreasing. And so we know that the sign of $ \frac {f(x)}{x}$ is constant all over $ \mathbb R^*$ (since $ f(0) \equal{} 0$) : positive for any increasing solution and negative for any decreasing function.

Let then $ a\neq 0$ and $ b \equal{} f(a)$.
Let the sequence $ u_n$ defined as $ u_0 \equal{} a$, $ u_1 \equal{} b$ and $ u_{n \plus{} 2} \equal{} u_{n \plus{} 1} \plus{} 2u_n$
It's easy to show with induction that $ u_{n \plus{} 1} \equal{} f(u_n)$ and that $ u_n \equal{} \frac {2a \minus{} b}{3}( \minus{} 1)^n \plus{} \frac {a \plus{} b}{3}2^n$
So, if $ b\neq \minus{} a$, we have $ \lim_{n\to \plus{} \infty}\frac {u_{n \plus{} 1}}{u_n} \equal{} 2 > 0$ and so, since  $ \frac {u_{n \plus{} 1}}{u_n} \equal{} \frac {f(u_n)}{u_n}$, we get that $ f(x)$ is increasing.
So, if $ f(x)$ is decreasing, $ b \equal{} f(a) \equal{} \minus{} a$ $ \forall a$ and so $ f(x) \equal{} \minus{} x$ $ \forall x$.
Q.E.D.

Note : in order to use the same mechanism for showing that the only increasing solution is $ f(x)\equal{}2x$, it would be easy to have $ f(x)$ continuous or surjective. Since we dont have, it's more difficult.


function
abstract algebra
advanced fields
advanced fields unsolved

Let $X$ be an algebraic variety and let $\mathcal O$ be the sheaf of regular funtions on $X$. Let $\mathcal O^{*}$ be the sheaf of non-zero regular functions on $X$.
Prove that $H^{1}(X,\mathcal O^{*})$ is isomorphic to the set of isomorphism classes of invertible sheaves on $X$ or equivalently, to the set of isomorphism classes of line bundles over $X$.

Notation: an invertible $\mathcal{O}$-module $L$ of rank 1 on $X$ is given by the following data:

(a) an open covering $\mathcal{U}=(U_{i})$ of $X$, such that $L|U_{i}\cong\mathcal{O}|U_{i}$;

(b) for each $i,j$ an isomorphism of $\mathcal{O}|(U_{i}\cap U_{j})$-modules $\mathcal{O}|(U_{i}\cap U_{j})\cong\mathcal{O}|(U_{i}\cap U_{j})$.

The data in (b) is equivalent to giving, for each $i,j$, a unit $f_{ij}\in\Gamma(U_{i}\cap U_{j},\mathcal{O})$. The $f_{ij}$ give an element $f=(f_{ij})\in C^{1}(\mathcal{U},\mathcal{O}^{*})$. and since the isomorphism in (b) are compatible on the triple intersections $U_{i}\cap U_{j}\cap U_{k}$, $f$ is in fact a cocycle.
Conversely, given $f\in Z^{1}(\mathcal{U},\mathcal{O}^{*})$, you can construct a line bundle by glueing copies of $\mathcal{O}|(U_{i}\cap U_{j})$ as in (b).
Hence it is defined a map $\theta: Z^{1}(\mathcal{U},\mathcal{O}^{*})\to Pic X$, which is moreover a morphism of abelian groups (resp. multiplication in Z^1 and $\oplus$ in $PicX$).

Suppose now $f=(f_{ij})\in ker(\theta)$, that is, the line bundle $M: =\theta(f)$  is trivial. Let $g_{i}\in\Gamma(U_{i},M)$ the section corresponding to the global section $1\in\Gamma(X,\mathcal{O}^{*})$. Then by (b) $g_{i}=f_{ij}g_{j}$ for each $i,j$, that is, $f$ is a coboundary.
So $\theta$ induces an injection $H^{1}(\mathcal{U},\mathcal{O}^{*})\to PicX$.

Since every invertible sheaf is trivial over some covering, and the refinement maps are compatible with the maps $\theta$, the isomorphism $H^{1}(\mathcal{U},\mathcal{O}^{*})\cong PicX$ is obtained.

Corrections:
1) $\mathcal{O}|(U_{i}\cap U_{j})\cong\mathcal{O}|(U_{j}\cap U_{i})$ instead of $\mathcal{O}|(U_{i}\cap U_{j})\cong\mathcal{O}|(U_{i}\cap U_{j})$ in (b)
2) $\otimes$ instead of $\oplus$ in $PicX$
3) you can substitute $H^{1}(X,\mathcal{O}^{*})$ to the last $H^{1}(\mathcal{U},\mathcal{O}^{*})$


παιδια γινεται ενας διαγωνισμος στο middle school. θα γινει σε 2 γυρους και τα θεματα θα εχουν ενα καλο επιπεδο αυξουσας δυσκολιας για καθε γυρο.για περισσοτερα πηγαινεται στο  middle school another one. ;)  :)

ξεχασα να πω οτι αρχιζει στις 5 Ιουλιου.   :showoff:


Let $ ABC$ be an acute - angled triangle and let $ D, E, F$ be the feet of perpendiculars from $ A,B,C$ respectively to $ BC; CA; AB$ . Let the perpendiculars from $ F$ to $ CB, CA, AD, BE$ meet them in $ P,Q,M,N$ respectively. Prove that $ P,Q,M,N$ are collinear.

[hide]Please post full solutions and do not spam.[/hide]

See the attached file.
Cant post the whole solution as diagram is needed... :wink:

I have all the official solutions to the RMO & INMO problems available on the web ( :rotfl: ), but I wanted to add the RMO 2006 problems to the resources section; so I wanted some second opinions.  :)

Anyways, thanks for your reply.

let orthocentre be O

$ FMO \sim CDO$

and

$ FNO\sim CEO$

$ \Rightarrow\frac{NO}{OE}\equal{}\frac{MO}{OD}$

$ \Rightarrow  MON\sim DOE$

thus

$ MN$is parallel to$ ED$

join $ QM$

$ \widehat{AMF}\equal{}\widehat{AQF}$

thus $ AFMQ$ is cyclic

$ \widehat{MAF}\equal{}\widehat{FCB}\equal{}90\minus{}B$

also $ DMEC$ is cyclic

so

$ \widehat{FQM}\equal{}\widehat{MAF}\equal{}\widehat{OCD}\equal{}\widehat{MED}$

$ \widehat{FQM}\equal{}\widehat{MED}$

$ \Rightarrow$ $ QM$is parallel to $ ED$

similarly$ PN$ is parallel to $ ED$

$ \Rightarrow$they are collinear

Quadrilaeral $ BFMD$ is cyclic.
So, by the Pedal Theorem, $ P,N,M$ are collinear. Similarly, quadrilateral $ AEMF$is cyclic and by the pedal line theorem again, $ N,M,Q$ are collinear. Thus, we have that $ P,Q,M,N$ to be collinear.

Simson's Theorem: For a point P on the circumcircle of triangle ABC, the perpendiculars to AB,BC,CA at D,E,and F are collinear.

If H is the orthocenter of ABC then since $ \angle BFH\equal{}\angle BDH\equal{}90$, BFHD is cyclic so F lies on the circumcircle of BHD, so M,N,P are collinear by Simson's, similarly Q,M,N are collinear.


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superior algebra
superior algebra unsolved

The relation $ S$ contains the relation $ R$ if $ aSb$ whenever $ aRb$. Let $ R$ be the relation on $ Z$
$ aRb$ if $ b \equal{} a \plus{} 3$. How many equivalence relations on $ Z$ contain $ R$?

Discussion began, but did not end, over here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1306523687&t=308914
(I found that by searching for "equivalence relation contains")

The discussion's a little confusing.  I'd say start by figuring out how many equivalence classes $ R$ has.  Hint: the final answer depends only on this number.


There are $360$ people around an object. The first person pulls the object at $1^\circ$ with a force of $1$ newton, the second person pulls the object at $2^\circ$ with a force of $2$ newtons, the third person pulls the object at $3^\circ$ with a force of $3$ newtons, ... , and the 360th person pulls at $360^\circ=0^\circ$ with $360$ newtons. Find the net force acting on the object.
[hide="Hint"] Find the angle the net force will be acting on. [/hide]

Extension: There are $\frac{360}{n}$ people around an object, when $\frac{180}{n}\in\mathbb{Z}$. The first person pulls the object at $n^\circ$ with a force of $n$ newton, the second person pulls the object at $2n^\circ$ with a force of $2n$ newtons, the third person pulls the object at $3n^\circ$ with a force of $3n$ newtons, ... , and the $\frac{360}{n}$th person pulls at $360^\circ=0^\circ$ with $360$ newtons. Find the net force acting on the object, in terms of $n$.

Could a mod move this to Pre-Olympiad or an appropriate forum?

[hide="Answer"]Complex expression for the force is $\underline{F}=\frac{180}{\sin{n^\circ\over 2}}\,\mathrm{cis}\left(270^\circ+{n^\circ\over 2}\right)$ - that is, the net force is $F=\frac{180}{\sin{n^\circ\over 2}}$, acting under the angle $\theta=270^\circ+{n^\circ\over 2}$[/hide]