[quote="StefanS"]Solve the function $ (x \plus{} y)f(yf(x)) \equal{} x^2f(f(x) \plus{} f(y))$ $ f: R^ \plus{} \to R^ \plus{}$ for all x,y>0[/quote]
Let $ P(x,y)$ be the assertion $ (x\plus{}y)f(yf(x))\equal{}x^2f(f(x)\plus{}f(y))$

If $ f(a)\equal{}f(b)$ : dividing $ P(a,y)$ by $ P(b,y)$, we get $ \frac{a\plus{}y}{b\plus{}y}\equal{}\frac{a^2}{b^2}$ and so $ (a\minus{}b)(ay\plus{}by\plus{}ab)\equal{}0$ $ \forall y$ and so $ a\equal{}b$ and so $ f(x)$ is injective.

Then $ P(\frac{1\plus{}\sqrt 5}2,1)$ $ \implies$ $ \frac{3\plus{}\sqrt 5}2f(f(\frac{1\plus{}\sqrt 5}2))$ $ \equal{}\frac{3\plus{}\sqrt 5}2f(f(\frac{1\plus{}\sqrt 5}2)\plus{}f(1))$ and so, since $ f(x)$ is injective :

$ f(\frac{1\plus{}\sqrt 5}2)\equal{}f(\frac{1\plus{}\sqrt 5}2)\plus{}f(1)$ and so $ f(1)\equal{}0$, which is impossible since $ f: \mathbb R^\plus{}\to\mathbb R^\plus{}$

So no solution.

(where is this silly problem coming from ?)


Let $A,B \in M_n(\mathbb C)$ such that $AB=A+B$
Prove that $rank(A)=rank(B)$
by Constantin Cocea si Mihai Piticari
------------------------------------------------------------------------------------------
I would have another solution here mine with mathematic in  MPSI/MP*

The relation $AB=A+B$ => $A, B$ commutes, so there are simultaneous
trigonalisable in $M_n(C)$.They have the same rank.

The condition $AB=A+B$ can be rewritten as $(A-I)(B-I)=I.$
Since matrices commute with their inverses, this also means that $(B-I)(A-I)=I.$

Suppose $Av=0$. Then,
$(A-I)v=-v$
Multiply both sides of this by $B-I$:
$v=-(B-I)v$
$(B-I)v=-v$
$Bv=0$.
So the null space of $A$ is the same as the null space of $B$; of course that means they have the same rank.

Another quick solution (although it uses the well-known fact that if $X$ is invertible, then $rank(AX)=rank(A)$, which I won't prove here).

Let $X=I-A,\ X^{-1}=I-B$. We have $rank(A)=rank(I-X)=rank(X^{-1}(I-X))$, which is equal to $rank(X^{-1}-I)=rank(I-X^{-1})=rank(B)$.

I have another methode.

Scince $AB=A+B$, we have $B=A(B-I)$, and then :
\[{\text{rank}}(B)={\text{rank}}\left(A(B-I)\right)\leq\min\left({\text{rank}}A,{\text{rank}}(B-I)\right)\leq{\text{rank}}(A).\]
Now we have $A=(A-I)B$, and thus :
\[{\text{rank}}(A)={\text{rank}}\left((A-I)B\right)\leq\min\left({\text{rank}}(A-I),{\text{rank}}B\right)\leq{\text{rank}}(B).\]
Therefore ${\text{rank}}(A)={\text{rank}}(B)$.

Sorry for bad English.


abstract algebra
linear algebra
matrix
Ring Theory
superior algebra
superior algebra unsolved

Denote by $K(R)$ the [url=http://planetmath.org/encyclopedia/TotalRingOfFractions.html]total ring of fractions[/url] of a reduced ring $R$. Prove that every prime ideal of $K(R)$ is maximal.
EDIT: If it helps, you may suppose, that $R$ is noetherian

I had posted a proof for the weaker problem (when $R$ is Noetherian), but it was flawed. I hope this one does work.

Instead of working with the total ring of fractions, let's just assume the extra condition that all elements of $R$ are either invertible or zero divisors. This characterizes total rings of fractions. We can show that with these stronger hypotheses, $R$ must actually be a finite direct product of fields.

Consider a primary decomposition $(0)=\bigcap_{i=1}^n\frak q_i\ (*)$ of the zero ideal of $R$, and let $\frak p_i$ be the radical of $\frak q_i$. Because the ring is reduced, the radical of $(0)$ is $(0)$, and passing to radicals in $(*)$ we get another primary decomposition $(0)=\bigcap_{i=1}^r\frak p_i$, where $\frak p_i,\ i\in\overline{1,r}$ are minimal prime ideals. The set of zero divisors in $R$ is exactly $\bigcup_{i=1}^r\frak p_i$, and since any maximal ideal $\frak m$ contains only non-invertible elements and is thus a subset of the set of zero divisors, each such ideal $\frak m$ coincides with one of the minimal prime ideals $\frak p_i$. In particular, every maximal ideal is a minimal prime ideal, and we've obtained what we initially wanted: every prime ideal is maximal.

It follows from the above that every localization $R_\frak m$ at a maximal ideal $\frak m$ is a field (for a commutative ring this is equivalent to the ring being reduced and of dimension zero). Now take $a\in R$, and let $\overline R=R/\mbox{Ann}(a)$. $\overline R$ is still reduced and of dimension zero, so again, each localization $\overline R_\frak m$ is a field. If $\bar a$ (the image of $a$ in $\overline R$) is not invertible, then we can find $\frak m$ which contains it. Then on the one hand $\frac{\bar a}1$ is not zero in $\overline R_\frak m$, and on the other hand it's not invertible, and we have a contradiction (because $\overline R_\frak m$ is a field). This means that $\bar a$ must be invertible, i.e. there is an $x\in R$ such that $a=xa^2$. In other words, $R$ is a von Neumann regular ring.

Being von Neumann regular and Noetherian, it's semisimple, so by the Wedderburn-Artin Theorem it's a finite direct product of matrix rings $M_{n_i}(D_i)$ over division rings $D_i$. From the fact that our ring is commutative it follows that all $n_i$ are $1$ and all $D_i$ are in fact fields, so we're done.

I think I have a counterexample for the initial question (in which the ring was not assumed to be Noetherian).

Let $k$ be any field, and consider the ring $S$ of formal power series over $k$ in indeterminates $x_{i},\ i\ge 1$ subject to $x_{i}x_{j}=0,\ \forall i\ne j$. The ring $R$ will be the subring of $S$ consisting of those power series in which only finitely many of the $x_{i}$ appear.

There are now a couple of things one can easily check: (a) $R$ is reduced; (b) $R$ is local, and its maximal ideal consists of the series with no free term. By (b), a non-invertible element $f$ must be a finite sum of series of the form $a_{1}x_{i}+a_{2}x_{i}^{2}+\ldots$ for various $i$ (remember that $x_{i}x_{j}=0,\ \forall i\ne j$). If we multiply $f$ by some $x_{j}$ different from all $x_{i}$ appearing in $f$, we get $0$, so every non-invertible element is a zero divisor. Equivalently, $R$ is its own total ring of fractions. However, it cannot be true that every prime ideal is maximal, since for a local reduced ring this is equivalent to being a field. Of course, this is not the case here.


Hi, I have a random question.

If you could meet with any person in history, who would it be, what would you say, and what is it about them that you relate to?

I'm interested in your responses.

mmm putting this in the college forum
must be teasing me...
i just had a practice SAT CR section about who someone would want to meet from history, it was saying how they wouldn't want to meet a famous person, but rather someone who was there to observe lots of them

i don't really have any of my own ideas about who i'd like to meet heh :D

probably should move this to round table

[quote="Koda"]It will surely be either Einstein or Feynman, two of the most interesting people in history. Can't decide on Erdos though ...[/quote]

Umm interesting, of all people in history you chose all scientists? I agree it must have been very interesting to be acquainted with both of them but I certainly don't think they are the ones who have most to teach about history. About physics sure, in the case of Erdos about math but about history...

It is an interesting question. I find myself unable to decide. So I've made a chronological list

Before Greece, Siddharta Gautama then, I'd like to know Socrates, Alexander the Great, Hadrian and Marcus Aurelius, Jesus, and Mary Magdalene if she did exist, Muhammad, da Vinci and that's nearly nine hundred years without one person that really interests me, no wait, Marco Polo is in between, there are a lot of great people in the Renaissance so I'll leave it, Newton, Fouché, Christopher Colombus, this is becoming too lengthy I know, Lincoln, Bolivar, Lenin, Goebbels, Mandela...

[color=red]Edit[/color] I forgot to say why. Gautama, Muhammad and Jesus, because they are icons of some of the greatest religions of our time and I'm interested about their true personality, time and history surely distort somehow their humanity

In the case of Mary Magdalene I'm intrigued about the truth in the Bible. This has been recently put forward in a dubious way by things like the Da Vinci Code phenomena but it is a subject that indeed has a lot of misterious.

Hadrian and Marcus Aurelius were good roman emperors, and I'm impressed by their achievements when Rome got to know its greatest time. And history puts them as surprisingly advanced minds for their time which makes me interested.

Marco Polo and Christopher Colombus are the kind of person I would like to be in some other life. Curious explorer of strange lands.

Newton and da Vinci because it must be an overwhelming experience to meet a mind of such calibre.

Fouché is a guy who has to tell us about French history at its apex, at the time it was strongly influencing world history, when there was monarchy, revolution, empire and all in thirty years or so. Fouche lived and played a role in all epochs so it must be interesting. Not a good guy it appears but wait til Goebbels.

Lincoln and Bolivar for their major role in American history, north and south respectively. And because they seem to have been great honourable men.

Goebbels to understand how nazism could have occured to humanity. I want to hear first hand how you convince many millions of such a horrible 'truth'. What is there behind it?

Lenin because he marks the beginning of a world with two poles, I wan't to know what was the communism he planned, how he would have done things if he wouldn't have died.

En fin, Mandela, because he represents the modern sacrifice for a just cause, because he decided not to be a ruler and fall in the webs of power and because I find him a particularly nice old man

Sorry about the length :)

Ursula

I can only pick da Vinci from that least.
What about Puskas?

I would pick Hannibal, because he was such a great leader. I wonder what he REALLY lookd like, and how he spoke. I've only seen videos and stories in books about the Punic Wars and how he almost helped Carthage win the 2nd Punic war.

it would be great to have a chance to meet someone who lived in a completely different era.

History seems to convey war heroes to most people.
Well, let me tell you, war is no adventure.

nostradamus, cause he knew the msot about what was going to go down in history at the time

[quote="Koda"]It will surely be either Einstein or Feynman, two of the most interesting people in history. Can't decide on Erdos though ...[/quote]

feynman would be my choice too  :)

Yeah it should be the natural choice.
By the way, Ignite, Nostradamus knew nothing about the future. :roll:

[quote="koda"]It will surely be either Einstein or Feynman, two of the most interesting people in history. Can't decide on Erdos though ...
[/quote]

[quote="bubka"]I will select two different kinds of people:
Group I: Carroll, Wodehouse, Douglas Adams
Group II: Marx, Lenin, Stalin, Mao, Ho, Che[/quote]

why i find this strange? :huh:

Yeah, Even I would go for groups-
Group I-Wodehouse, Shakespeare, Asimov, and Christie
Group II-Einstein and Ramanujan
Group III-Only Indians may understand this one- Akbar, and if u would allow me to go to  really really ancient history, then each of the Dashavatharams(10 incarnations of God :D ).
 :lol:
Who knows, maybe someone will come up with some kind of weird time machine and we may get to do this!! :)

Reimann, Gauss, or Yang Mills in Mathematics.

In political issues, Ronald Reagan.

In philosophy, Socrates, Plato, or Aristotle.

In writing, Edgar Allen Poe.

a lot of people.....

[quote="ursula"]About Feynman...
[quote="Koda"]Yeah it should be the natural choice.
[/quote]

Why? What is it about Feynman that is natural to pick him?

[quote="bubka"]
I will select two different kinds of people:
Group I: Carroll, Wodehouse, Douglas Adams
Group II: Marx, Lenin, Stalin, Mao, Ho, Che
[/quote]

Boy, I'm not sure the people in your first group will approve people in your second, ja! But I'm interested, why do you want to meet this guys?

Ursula[/quote]

They are respectively the greatest in the fields of nonsense/farce and communism/revolution, my two true passions.

[quote="deimos"][quote="ursula"]About Feynman...
Why? What is it about Feynman that is natural to pick him?
[/quote]

great man. and did you read his famous: feynman lectures on physics?[/quote]

Yeah, as a matter of fact, I recommend them to every non physicist person I know but still with an interest for physics. But Feynman's great insight and physical intuition and even greater capacity for teaching does not make him the "natural" person to meet among all the other great people in history, at least not for me. I also read the book that apparently has made him so widely known to the general public, "Surely, you're joking Mr Feynman". And I also feel admiration for him. But I happen to know also that Feynman and others celebrated the Hiroshima bombing, he must have seen the thing as the logical conclusion to their scientific efforts in Los Alamos. Nevertheless other scientists involved, particularly Oppenheimer, were already regretting to have been contributors to project, in  Op's case the scientific leader, and warning about the catastrophic consequences. Indeed Einstein never agreed to participate in the project. So what is Feynman to me? Certainly an outstanding physicist with a good bunch of charisma and a whole lot of other amazing qualities unrelated to physics nonetheless but among other geniuses, or other outstanding people he does not occupies, for me an special place. Sure, if I list a hundred people Feynman will make it but since I studied theoretical physics [i]his[/i] contribution to mankind is familiar to me. That's why I didn't put Einstein in the list either, another modern great mind.

[quote="nutz"]
Reimann, Gauss, or Yang Mills in Mathematics.
[/quote]
Yang and Mills are two different guys, you know this, do you? And I don't know about their work in mathematics but I think they are more widely known for their work in physics. They proposed for the first time one model of the kind of theory we now call Yang-Mills.

[quote="bubka"]
They are respectively the greatest in the fields of nonsense/farce and communism/revolution, my two true passions.
[/quote]
Yes, I see you have a passion for revolution and communism. I myself do, but I don't think Stalin or Mao have been the greatest in this field. Probably they have been the greatest in destroying revolutions and certainly the greatest in killing innocent people. You remind me of Ignite  except for you are in the opposite corner of the ring. But I'm again interested, although this probably deserves another thread, have you ever tried to work out a 'revolutionary' world that works? Oh make another thread if you are going to answer seriously.

Ursula

[quote="ursula"]Yes, I see you have a passion for revolution and communism. I myself do, but I don't think Stalin or Mao have been the greatest in this field. Probably they have been the greatest in destroying revolutions and certainly the greatest in killing innocent people. You remind me of Ignite  except for you are in the opposite corner of the ring. But I'm again interested, although this probably deserves another thread, have you ever tried to work out a 'revolutionary' world that works? Oh make another thread if you are going to answer seriously.

Ursula[/quote]

Does this mean that I am the creator of revolutions and the greatest in saving the innocent?

she was referring to me in that sentence and stalin/mao in the previous one, so your joke is annulled. better luck next time.

having said that, i must add that i hope to change her ideas about these two stalwarts of communism.

Uh-hem.  I'd meet a relation of Nancy Reagan, or at least someone who'd say for once and for all whether she was born in 1923, like she says in her autobiography, or in 1921, which Wikipedia says.

Wikipedia?  Or herself?  Really there's no contest, but still.

Hypatia because there was very small number of woman in science/philosophy and she was one of them.

Tolkien because he created new world and much more than that.

Galileo and Newton because I admire them.

[quote="Nienna"]Hypatia because there was very small number of woman in science/philosophy and she was one of them.
[/quote]

Beautiful, I inmediately thought after my post that there was actually only one woman in my list and that it was very sad. I didn't editted it nonetheless because I thought that if they didn't came to my mind at first, it would be because objectively the most of the contributions were done before by men. Luckily, we're coming out of that epoch. Some other women that produce me admiration are Emmy Noether, Marie Curie, Anais Nin, Plotina(wife of Trajan, romam emperor).

Ursula


trigonometry
analytic geometry
combinatorics unsolved
combinatorics

In the Cartesian plane, let $C$ be a unit circle with center at origin $O$. For any point $Q$ in the plane distinct from $O$, define $Q'$ to be the intersection of the ray $OQ$ and the circle $C$. Prove that for any $P\in C$ and any $k\in\mathbb{N}$ there exists a lattice point $Q(x,y)$ with $|x|=k$ or $|y|=k$ such that $PQ'<\frac{1}{2k}$.

Consider the case when $OP$ makes an angle $\theta$ such that $0\leq\theta\leq\frac{\pi}{4}$.  The argument then extends to all of $C$ by symmetry.  Let $k$ be given.  Consider all the lattice points of the form $Q(k,y)$ with $0\leq y\leq k$.  Then by drawing all the rays $OQ$, we separate the eighth-circle into $k$ arcs.  The length of the $j^{th}$ arc is $\tan^{-1}\left(\frac{j}{k}\right)-\tan^{-1}\left(\frac{j-1}{k}\right)$ by the formula $s=r\theta$, with $r=1$ in this case.  Since $\tan x$ is concave down, i.e. $\frac{d^{2}}{dx^{2}}\tan x \leq 0$, when $x\geq 0$, the longest arc will be the first one, between the first and second rays, formed by the origin and $(k,0)$ and $(k,1)$, respectively.  This arc has length $\tan^{-1}\left(\frac{1}{k}\right)-\tan^{-1}\left(\frac{1-1}{k}\right)=\tan^{-1}\left(\frac{1}{k}\right)$.  Since $\tan^{-1}x<x$ for all $x>0$, $\tan^{-1}\left(\frac{1}{k}\right)<\frac{1}{k}$.  So all the arcs have length less than $\frac{1}{k}$.  This means that no matter what arc $P$ lies in, the length of the arc between $P$and the closest $Q'$ must be less than $\frac{1}{2k}$, which implies $PQ'<\frac{1}{2k}$, as required.


MATHCOUNTS
probability
geometry
3D geometry
tetrahedron

[b]Another Mock MATHCOUNTS competition [/b]
Will consist of Sprint (40 min) and Target (24 min)

Competition is over, but If you want you can still submit answers as long as you dont cheat.

[b][u]Please reply if you are going to do it[/b][/u]

Umm, I think i can make it though i won't do very good. (Didn't participate in MATHCOUNTS but hoping to next year)...

Everyone remember to turn their clocks tomorrow!

And if you do have countdown round, how does it work?

[quote="tarquin"]He'll probably have everybody invited to some AIM chatroom, tell the competitors to mute everybody but himself and the other competitor, and then ask the question. I don't know if the person will have to "buzz in" or just type the answer... but that's what I'm guessing...[/quote]Good explanation. It will be an AIM chatroom, and you will have to buzz. Same rules as normal, but I might do It national style (bracket vs. ladder).

[quote="DarkKnight"]Ahh... In that case,  i won't participate in countdown. No AIM but i have an account. I might be able to go on AIM Express.[/quote]Its free to download. I can send a link.

Umm... I'm not allowed to have AIM but I think my parents may let me on AIM Express to do the Countdown round.

[quote="tarquin"]If there are more than 10, then the top 10 would make countdown round. Of course if you have less, the cutoff can be somewhat higher I guess.[/quote]The cutoff will probably be 12 if we have that many. Else everybody

[quote="biffanddoc"]What is AIM express?? If you can join a chatroom its fine[/quote]

I'm not sure if you can join a chatroom. I'll see.

Yes you can. I'll ask my parents if i can use AIM Express tomorrow once we're sure we have enough people.

Haha. Yeah, we're going to need a whole lot more. ;)

TURN YOUR CLOCKS. We don't want to have anyone an hour late.

1
You can tell that the intercepted arcs of AC and AB are all 60 degrees, so there's an equilateral triangle. $\frac 16$ of the circle minus the equilateral triangle with a side of 7 times four plus two times the area of the equilateral triangle with side 7 is the answer.

6
Only way that the sum can be odd is if 2 is in it because 2 is the only even prime and an even plus an odd is the only way to get an odd sum. The number of ways to pick 6 numbers is $\frac {18!}{12!6!}$. Since you know there has to be a 2 in there, you find the number of ways to pick the other 5 numbers, $\frac {17!}{5!12!}$ and the final answer is $\frac {\frac {17!}{5!12!}}{\frac {18!}{12!6!}} = \frac 13$.

How to do #3
Is the intersection of the perpendicular bisectors the center of the circle?

Yes that is the center of the circle.
Also note that the perpendicular bisectors of the sides of a triangle are concurrent at the center of its circumscribed circle.

[quote="tarquin"]8
$\eqnarray 4\pi r^2 = 256 \\
& = & \pi r^2 = 64 \\
& = & r = \frac {8}{\sqrt \pi} \\
& = & r^3 = \frac {512}{\pi * \sqrt \pi} \\
& = & \frac 43 \pi r^3 = \frac {2048}{3\pi \sqrt \pi} * \pi$
Rationalize..
$\frac 43 \pi r^3 = \frac {2048 \sqrt \pi}{3 \pi}$

12
Five vowels, twenty-one consanants.
VCVC
$5 * 21 * 4 * 20$
Times four again because you can switch vowels with eachother and consanants with eachother... $8400 * 4 = 33600$

13
$(5, 2) ( -3 , 4) ( 8 , -2)$
Scalefactor of $\frac 25$ ...
$2 + \frac 45 - \frac 65 + \frac 85 - \frac {16}{5} - \frac 45 = \frac {28}{5}$

18
Common fraction... $\frac {3}{25}$

21
Probability for prime $\frac 35$. Has to get in this order PRIME - NOPRIME or NOPRIME - PRIME, so $\frac 35 * \frac 25 * 2 = \frac {12}{25}$

22
Possible points are $(3,3)(3,2)(3,1)(2,2)(2,1)(1,1)$.
Possible midpoints are $(3,2)(2,1)(2,2)$.
Must choose two points for every mid point backwards or forwards so $\frac 13 * \frac 15 * 2 * 3 = \frac 25$

28
Volume of tetrahedron is $528 * \frac 12 * \frac 13 = 88$

I messed the test up horribly...[/quote]
#28, Let x be the side of the orginal cube, the the volume of the tetrahedron is:
(1/2) * (x/2) * (x/2) * (x/2) * (1/3)= x^3/48=528/48=11.

#22, {{{1, 1}, {1, 3}}, {{1, 1}, {3, 1}}, {{1, 1}, {3, 3}}, {{1, 2}, {3, 2}}, {{1, 3}, {1, 1}}, {{1, 3}, {3, 1}}, {{1, 3}, {3, 3}}, {{2, 1}, {2, 3}}.
There are 8 of them.
There are (9 choose 2)=36 total possiblities.
the answer is: 8/36 = 2/9.

#21. It can be prime-prime or noprime-noprime.

13 and 18 are messed up. Ill change scores. :blush:  :blush:

Scores are updated.

Tarquin-30
Xantos C. Guin-29
Chess64-10
Solfidefarms-4

i only did the target round - i got 10  :blush:

[quote="tarquin"]#21
Says [b]exactly one[/b] prime.

Rest are right, sorry about that :) I'm an idiot.[/quote]
Ooh youre right about exactly one prime. But, The probability of Prime-noprime is $ \frac 25 * \frac {12}{19} = \frac {24}{95}$ and the probability of noprime-prime is  $ \frac 35 * \frac {8}{19} = \frac {24}{95}$. Add together and get $ \frac {48}{95}$ Am I not right?

[quote="tarquin"]I think it said with replacement.
And how did you get 19's and 20's? I see only 5 numbers there.[/quote]
O you're right about replacement. When I did the problems, I wrote the wrong thing. Ill give you credit.

Lol for sprint theres 31 downloads and 2 submissions?? As long as you dont cheat, You can still submit


[quote="ckck"]A man was born in the early part of the 19th century was x years old in the year $x^{2}$, while a woman born in the later part of the 19th century was y years old in the year $y^{2}$. How many years apart were they born?[/quote]
[hide]Well, the only perfect square in the 1800s is 1849 (43) so the man was born in 1806. The next perfect square is 1936 so the lady was born in 1892. Our answer is 86.[/hide]


algebra
polynomial
trigonometry
inequalities unsolved
inequalities

Let x,y be real number such that $x^{2}+xy+y^{2}=3$. Find the greatest value of $M=x^{2}y+x-y$

$x^{2}+xy+y^{2}-3=0\Leftrightarrow x=$ $-\frac{1}{2}y\pm \frac{1}{2}\sqrt{\left(-3y^{2}+12\right) }$

$M=\left(-\frac{1}{2}y\pm \frac{1}{2}\sqrt{\left(-3y^{2}+12\right) }\right)^{2}y+\left(-\frac{1}{2}y\pm \frac{1}{2}\sqrt{\left(-3y^{2}+12\right) }\right)-y=-\frac{1}{2}y^{3}-\frac{1}{2}y^{2}\sqrt{\left(-3y^{2}+12\right) }+\frac{3}{2}y+\frac{1}{2}\sqrt{\left(-3y^{2}+12\right) }$

And it's become polynomial in $y$

Sorry about my English.I will improve it :rotfl:

We can set $x=\sqrt{3}\cos \theta-\sin \theta ,\ y=2\sin \theta$.

For these kinds of problems, I think Lagrangian Multipliers are best.

Kunny, can you explain how did you get to that substitution?

This one stumped me for a while but I managed to attack it with the substitution Kunny proposed.

$x^{2}+xy+y^{2}=3$
$\left(x+\frac{y}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}y\right)^{2}=3$
$\left(x+\frac{y}{2}\right)^{2}=3\cos^{2}\phi$, $\left(\frac{\sqrt{3}}{2}y\right)^{2}=3\sin^{2}\phi$
$x=\sqrt{3}\cos\phi-\sin\phi$, $y=2\sin\phi$
$x^{2}y+x-y=\sqrt{3}\cos\phi-\sin\phi+4\sin\phi\cos\phi(\cos\phi-\sqrt{3}\sin\phi)$

Now we have to 'factor' somehow those summands. Notice that $\sqrt{3}=\tan\frac{\pi}{3}$.

\begin{eqnarray*}\sqrt{3}\cos\phi-\sin\phi &=& \tan\frac{\pi}{3}\cos\phi-\sin\phi \\ &=& \frac{1}{\cos\frac{\pi}{3}}\left(\sin\frac{\pi}{3}\cos\phi-\cos\frac{\pi}{3}\sin\phi\right) \\ &=& 2\sin\left(\frac{\pi}{3}-\phi\right) \end{eqnarray*}

\begin{eqnarray*}\cos\phi-\sqrt{3}\sin\phi &=& \cos\phi-\tan\frac{\pi}{3}\sin\phi \\ &=& \frac{1}{\cos\frac{\pi}{3}}\left(\cos\phi\cos\frac{\pi}{3}-\sin\phi\sin\frac{\pi}{3}\right) \\ &=& 2\cos\left(\phi+\frac{\pi}{3}\right) \end{eqnarray*}

\begin{eqnarray*}x^{2}y+x-y &=& 2\sin\left(\frac{\pi}{3}-\phi\right)+8\sin\phi\cos\phi\cos\left(\phi+\frac{\pi}{3}\right) \\ &=& 2\sin\left(\frac{\pi}{3}-\phi\right)+4\sin2\phi\cos\left(\phi+\frac{\pi}{3}\right) \\ &=&-2\sin\left(\phi-\frac{\pi}{3}\right)+2\sin\left(3\phi+\frac{\pi}{3}\right)+2\sin\left(\phi-\frac{\pi}{3}\right) \\ &=& 2\sin\left(3\phi+\frac{\pi}{3}\right)\leq\boxed{2}\end{eqnarray*}

So, maximum of M is 2. Correct?


When Helen jogs, every $ 20$ minutes she checks the total distance she traversed. One day, the first time she checked, Helen had jogged a distance of $ 2.5$ miles, the second time she checked, she had jogged a total distance of $ 4.8$ miles, and the third and final time that she checked, Helen found out that she jogged a total distance of $ 6.8$ miles. What is the difference between her average speeds, in miles per hour, during the first $ 20$ minutes of jogging and the last $ 20$ minutes of jogging?

The first $ 20$ minutes, Helen jogged $ 2.5$ miles, so her mph is $ 7.5$. The last $ 20$ minutes, she jogged $ 6.8\minus{}4.8\equal{}2$ miles, so her mph is $ 6$. So the difference in mph is $ 1.5$.


quadratics
logarithms
algebra
calculus
calculus computations

Let $ \alpha ,\ \beta$　be the real roots of the quadratic equation with respect to $ x$ : $ x^2 \plus{} x\log_8 a \plus{} (\log_8 a)^2 \plus{} \log_8 a \minus{} 1 \equal{} 0$. Find the range for which $ \alpha ^3 \plus{} \beta ^3$　can be valued.

I still don't understand what the "question" is..
for which $ \alpha^3\plus{}\beta^3$ can be valued,, so $ \alpha$ and $ \beta$ must be valued too..(is that "real" according to the problem?)

let $ ^8\log{a}\equal{}p$ sorry it's not commonly viewed,, but in my country we write like that..

$ x^2\plus{}px\plus{}(p^2\plus{}p\minus{}1)\equal{}0$
if the roots are real,, so $ D\geq0 \rightarrow p^2\minus{}4(p^2\plus{}p\minus{}1)\equal{}\minus{}3p^2\minus{}4p\plus{}4\geq0$
so that's why $ 3p^2\plus{}4p\minus{}4\leq0 \rightarrow (3p\minus{}2)(p\plus{}2)\leq0 \rightarrow \minus{}2\leq p\leq \frac{2}{3}$

so $ \minus{}2\leq ^8\log{a}\leq \frac{2}{3} \rightarrow \frac{1}{64} \leq a \leq4$

sorry if I'm mistaken..

ah,,yeah..
I was mistaken in understanding the problem..

[continue from above..]
$ \alpha\plus{}\beta\equal{}\minus{}p$
$ \alpha\beta\equal{}p^2\plus{}p\minus{}1$
$ \alpha^3\plus{}\beta^3\equal{}(\alpha\plus{}\beta)(\alpha^2\minus{}\alpha\beta\plus{}\beta^2)\equal{}(\alpha\plus{}\beta)[(\alpha\plus{}\beta)^2\minus{}3\alpha\beta]\equal{}(\minus{}p)(\minus{}2p^2\minus{}3p\plus{}3)\equal{}(2p^3\plus{}3p^2\minus{}3p)$

okay,,subtitude the vallue of the previous $ p$ to $ \alpha^3\plus{}\beta^3$
if I'm not mistaken.. $ 2\leq \alpha^3\plus{}\beta^3\leq\frac{14}{3}$

sorry if my solution is wrong.. :blush:  :blush:  :blush:

when we get $ \minus{}2 \leq p \leq \frac23$,

we need to plot the graph of $ y \equal{} 3p^3 \plus{} 3p^2 \minus{}3p$ to get the range of y when $ \minus{}2 \leq p \leq \frac23$

I used the graph program to help me out...

then i get $ \minus{}0.5981 \leq \alpha^3 \plus{} \beta^3 \leq 4.5981$

...my two cents.

[quote="kunny"][quote="mhar_teens"]

let $ ^8\log{a} \equal{} p$ sorry it's not commonly viewed,, but in my country we write like that..

$ x^2 \plus{} px \plus{} (p^2 \plus{} p \minus{} 1) \equal{} 0$
if the roots are real,, so $ D\geq0 \rightarrow p^2 \minus{} 4(p^2 \plus{} p \minus{} 1) \equal{} \minus{} 3p^2 \minus{} 4p \plus{} 4\geq0$
so that's why $ 3p^2 \plus{} 4p \minus{} 4\leq0 \rightarrow (3p \minus{} 2)(p \plus{} 2)\leq0 \rightarrow \minus{} 2\leq p\leq \frac {2}{3}$

so $ \minus{} 2\leq ^8\log{a}\leq \frac {2}{3} \rightarrow \frac {1}{64} \leq a \leq4$

$ \alpha \plus{} \beta \equal{} \minus{} p$
$ \alpha\beta \equal{} p^2 \plus{} p \minus{} 1$
$ \alpha^3 \plus{} \beta^3 \equal{} (\alpha \plus{} \beta)(\alpha^2 \minus{} \alpha\beta \plus{} \beta^2) \equal{} (\alpha \plus{} \beta)[(\alpha \plus{} \beta)^2 \minus{} 3\alpha\beta] \equal{} ( \minus{} p)( \minus{} 2p^2 \minus{} 3p \plus{} 3) \equal{} (2p^3 \plus{} 3p^2 \minus{} 3p)$
[/quote]

That's correct so far.[/quote]

Well I found the local maxima and minima of $ (2p^3 \plus{} 3p^2 \minus{} 3p)$ and obtained that it happens for $ p \equal{} \frac{\pm\sqrt{3} \minus{} 1}{2}$

Substituting these gives:

$ \minus{}0.59807 < \alpha^3 \plus{} \beta^3 < 4.59807$

Is there any alternative clever way kunny?  :ninja:


calculus
analytic geometry
graphing lines
slope
real analysis
real analysis unsolved

Hello,

I am seeking for different solutions to a simple problem.  Now, I am seeking for a calculus solution to the problem below:

The length of the legs of a certain right traingle are 18 and 63.  What is the radius of the circle which is tangent to both legs of the traingle and has its centre on the hypotenuse?

Let $r$ be the desired radius, we have $\frac{r}{63-r}=\frac{18}{63}$, yielding $r=14.$

Yes, but I am trying to solve it using calculus, just to see some different solutions. :)

yo kunny where does the $\frac{r}{63-r}=\frac{18}{63}$ come from?  Not sure of the identity used here.  For the calulus solution it will be a differential equation or so i think.

i may have misunderstood the question though, but for both legs of the triangle to be tangents of the circle, they have to be at the half way points of the circle, where the slope is zero and where it approaches infinity.  so the triangle's legs have to be equal to go through the center.

if the corners of the triangle have to touch the circle, and the slopes of the leg and circle didnt releate i wouldnt know how to apporach it with calculus.

if the triangle wasnt a right triangle, i would make a system, one equation for the point where one leg touches, one where the other leg touches, an equation for the slope each intersection and possibly another one to put the hyponaneous through the center.  but that'd be a mess and i may have misunderstood the question


Solve the following system of equation:

//edit (sorry i had a typo): \[x^{2}-5x-y^{2}-5y = 0\]  \[x^{2}+xy+y^{2}= 7\]

[quote="Aralc"]First we take $x=-y$. When we substitute this in the second equation, we get $x^{2}= 7$. [/quote]
I think you mistook the constant of the $x^{2}$ term of the second equation. It is $0$, not $1$:
[quote="akimatsu"]$\left\{ \matrix{ 0 \hfill \cr x^{2}+xy+y^{2}= 7 \hfill \cr}\right\}$
[/quote]

No, there was intended to be a break between $0$ and $x^{2}$.

I think you mistook the constant of the  term of the second equation. It is 0, not 1:

rewrite $x^{2}+xy+y^{2}=7$ as $(x+y)^{2}-xy=7$ Now if x=-y then

$x^{2}=7$
$x=\pm 7$

[quote="Aralc"]The first equation can be written as $(x+y)(x-y-5)=0$.
First we take $x=-y$. When we substitute this in the second equation, we get $x^{2}= 7$. So we have already 2 solutions : $x=\sqrt 7$, $y=-\sqrt7$ and $x=-\sqrt7$, $y=\sqrt 7$.
Then we have $x-y=5$. We write the second equation like this : $(x-y)^{2}+3xy=7$. So we have $x-y=5$ and $x(-y)=6$. We concider $x$ and $-y$ as the solutions of the equation $a^{2}-5a+6=0$ and we find $a=2$ or $a=3$. So the other solutions are $x=3$, $y=-2$ and $x=2$, $y=-3$[/quote]
How do you know $x(-y)=6$?


geometry
AMC
AMC 10
MATHCOUNTS
inequalities
induction
trigonometry

[quote="rnwang2"]What sort of geometry and algebra do I need to know for the AMC10?[/quote]
This is the geometry you need for Mathcounts and AMC 8/ 10:
Pythagorean
30-60-90
Identifying similar and congruent triangles
Measures of angles inscribed in circles and power of a point
Triangle Inequality
Shoestring (maybe)
This is absolutely everything I can think of. Almost any problem can crushed by cleverly combining these methods/formulas.

[quote="rnwang2"]What is shoestring?[/quote]
It's a really complicated formula that is one of the few that I have no idea how to prove. It gives area of a polygen when you know the coordinates.

[quote="Elemennop"]It's not that complicated at all

and I think it's induction off of Pick's theorem or something
or maybe cross product induction[/quote]
It's complicated to explain. I always thought it was from matrices or something.

what exactly does the theorem state?

I tried looking it up but didn't find anything.

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf

You should get the Introduction to Geometry book from the AoPS bookstore.

[quote="Phelpedo"]You don't need "shoestring" at all.[/quote]
Is there a shoestring? I always call the shoelace the shoestring...

Knowing

quadratic formula and synethetic division helps too

although both are easily proven.

[quote="Karth"][quote="Phelpedo"]You don't need "shoestring" at all.[/quote]

Well, you don't NEED shoestring, but that's like saying you don't NEED Heron's formula... after all, Heron's formula doesn't necessarily have to be used, right? You can just use the Law of Cosines to get an angle and find the area of the triangle.  :wink:

Likewise, shoestring is particularly useful when trying to blast out a geometry problem on the AMC (and I believe I have found 2 or 3 AIME problems...) with analytic geometry. Shoestring is quite helpful if you know how to apply it properly.  :)[/quote]
Actually you don't even need trig for heron's. Heron's is derived by dropping an altitude of a triangle, and setting up equations. For example, drop the altitude to 14 in a 13,14,15 triangle. Call everything to its left x and the stuff to the right 14-x. Then call the altitude h and use pythagorean twice. It is only slightly more work than heron's itself. For shoestring, unless the numbers are pretty, it takes way too long by any other method.

It's easier for me just to take the absolute value of \[\frac{1}{2}\left|\begin{matrix}x_{1}& y_{1}& 1\\ x_{2}& y_{2}& 1\\ x_{3}& y_{3}& 1\end{matrix}\right|.\]

[quote="chess64"]It's easier for me just to take the absolute value of
\[\frac{1}{2}\left|\begin{matrix}x_{1}& y_{1}& 1\\ x_{2}& y_{2}& 1\\ x_{3}& y_{3}& 1\end{matrix}\right|. \]
[/quote]
But that only works for triangles. Shoestring is for all polygons.

I never actually knew the name for "shoestring" until this topic.

It's pretty easy to prove using Green's theorem for line integrals though...
(although it may suck up too much of your time on AMC).


Four pens and three pencils cost $ \$2.24$. Two pens and five pencils cost $ \$1.54$. No prices include tax. In cents, what is the cost of a pencil?


[hide="Answer"]$ x \equal{} 0$ or $ x\equal{}1$[/hide]

It is too easy to give such a answer like that, isn't it?  :wink:


A high school's football stadium has 11 rows of 199 seats each.  One night, several groups, each not exceeding 39 people, attend a game.  If people from the same group sit in the same row, what is the maximum number of tickets that can be sold so that everyone is guaranteed a seat?

$\frac{199}{39}=5.1...$, so $(5)(39)(11)=2145$ tickets can be sold to guarantee that each group sits together.


[quote="junggi"]I'm about $\frac{2e^{\pi}}{\pi}$[/quote]
Is that supposed to be 14?

[quote="Klebian"]14.73182248... usually rounds to 15...[/quote]
well convert the decimal to months

rofl I'm 19... not many active posters here older than me besides [b]fredbel6[/b] (in this subforum I mean -- I'm a youngster in College Playground :D)

I am 73 [color=white]...ok, i'm 14[/color]

[quote="pianoforte"]I am 73 [color=red]...ok, i'm 14[/color][/quote]

[quote="junggi"][quote="i_like_pie"][quote="Tiks"]I will be 19 in 14.12.2006 :P[/quote]
December 14? Why don't people in Europe and USA decide on common things? (Metric vs. Imperial, mm/dd/yy vs. dd/mm/yy, etc.)[/quote]
because the US thinks that theyre THE BEST and dont want to change theirs to europeans.

ps i just noticed, why is there a category for 65-120(120?wtf?)[/quote]
People live to be that old... However that is very rare and even if someone is 120+, I don't think they're a member of AoPS. But you should always consider all possibilities. :wink:

[quote="i_like_pie"][quote="Tiks"]I will be 19 in 14.12.2006 :P[/quote]
December 14? Why don't people in Europe and USA decide on common things? (Metric vs. Imperial, mm/dd/yy vs. dd/mm/yy, etc.)[/quote]
Yes, you are right,December 14.Or maybe you think that I was born in 12 of 14th month :D .

[quote="K81o7"][quote="not_trig"]lol

I'm a certain prime < 20 (guess which one)[/quote]

I'm gonna take a stab at [hide]13[/hide][/quote]
I'll guess 17, which would be my age :)

[quote="goldendomer"]I guess it's 19.[/quote]

ditto.

[quote="not_trig"]One right answer, 4 wrong. NOT 2.[/quote]
I say you're 13

[quote="i_like_pie"][/quote][quote="junggi"][quote="i_like_pie"][quote="Tiks"]I will be 19 in 14.12.2006 :P[/quote]
December 14? Why don't people in Europe and USA decide on common things? (Metric vs. Imperial, mm/dd/yy vs. dd/mm/yy, etc.)[/quote]
because the US thinks that theyre THE BEST and dont want to change theirs to europeans.

ps i just noticed, why is there a category for 65-120(120?wtf?)[/quote][quote="i_like_pie"]
People live to be that old... However that is very rare and even if someone is 120+, I don't think they're a member of AoPS. But you should always consider all possibilities. :wink:[/quote]

maybe u should make it til 124, a french lady once lived til 123 (record)
-jorian

as of october 2006 my age is  :rotfl: -2 :rotfl:

the coffin dance emoji:  :rotfl:  :stretcher:  :rotfl:

in 2006, my age was -2... wow ancient threads

[quote=nkouevda]I'm 13, too.[/quote]

Wow!! Right now you're 27!!!! :wow:

[quote=13375P34K43V312]How old are you? (AS OF OCTOBER 2006)

I'm 13, probably of the younger people on the forum.[/quote]

You're not young! You're 27!!


calculus
derivative
function
limit
real analysis
real analysis unsolved

Maybe I'm missing something, but what's wrong with $ f(x)\equal{}0$?

$ e^{\alpha x}$ works for any $ |\alpha| < 1$; the limit is $ 0$ everywhere.  Actually, I think any function $ \sum f_n \frac {x^n}{n!}$ where $ f_n \equal{} O(\alpha^n)$ for some $ |\alpha| < 1$ should also work.

[quote="taxofonas"]Find such function $ f(x)$ for which the n-th derivative exists for $ \forall n \in N$ and the limit $ \lim_{n\to\infty} f^n(x) \equal{} g(x)$ converges.

($ e^x$ does not count)

Lecturer asked me this question after the lecture about power series, but not power series function will do as well.[/quote]You can get all nontrivial analytical limits by considering power series $ f(x) \equal{} \sum_{k\equal{}0}^\infty a_k \frac{x^k}{k!}$ for which $ \lim_k a_k \equal{} \alpha$ exists. Then $ f^{(n)}(x) \equal{}  \sum_{k\equal{}0}^\infty a_{k\plus{}n} \frac{x^k}{k!}$. The limiting function will be $ g(x) \equal{} \alpha e^x$.

In particular, the radius of convergence must be infinite.

Thanks.

So far looks like, regardless of constant functions, some sort of exponential ones (but expressed as power series) dominates.  Funny thing those exponentials in the world of derivatives...


advanced fields
advanced fields unsolved

A and B are open sets. Denote by frA the boudary of A , and by * the intersection and by + th union of sets;
Show that :
( A *frB) +( B*frA) is in fr(A*B) which is in  ( A *frB) +( B*frA)+(frA*frB)

In standard notation:
$ (A\cap\partial B)\cup (B\cap\partial A) \subseteq \partial (A\cap B) \subseteq (A\cap\partial B) \cup (B\cap\partial A)\cup (\partial A\cap \partial B)$
Just do it case by case: if $ x\in A$ and $ x \in \partial B$, then any sufficiently small neighborhood of $ x$ is contained in $ A$ and includes the points from $ B$ and from its complement. Hence $ x\in\partial (A\cap B)$. And so on... the whole problem is about understanding the definitions of boundary points and of open sets.


geometry
incenter
circumcircle
parallelogram
trigonometry
function
geometry unsolved

Consider a triangle $ ABC$ and its circumcircle. Let $ I$ be the incenter of this triangle, let $ B'$ be a point on the circumcircle, s.t. $ BB'$ is a diameter in this circle. Let $ M$ be the tangency point of the incircle with the side $ CA$. Consider $ K\in (AB)$ and $ L\in (BC)$, s.t. $ AM\equal{}BL$ and $ MC\equal{}BK$. Prove that $ B'I\perp KL$.

[quote="plane geometry"]KD/KE=(c-a)/2(p-c)sin(B/2)  where p=1/2(a+b+c)
RS=B’S=tan(B/2)   IR=2Rsin(B/2)-2Rsin(C-A)/2*tan(B/2)
B’R=B’S/cos(B/2)=B’S/sin(A+C/2)
So KD/KE= B’R/ IR=[2sin(C-A)/2]/[sinA+sinB-sinC]
 △KDE∽△B’RI
DONE![/quote]

I'm sorry, but could you give more explanations in your solution? Starting with your notations, what are the points $ D,E,R$ and so on in your draw? Thank you very much.

DE=DK=MC  DKLE is a parallelogram
R is the intersection of B'C and DI

The problem is rather easy, we get it by direct calculations without trig functions: all we have to prove is $ B^'K^2 - B^'L^2 = IK^2 - IL^2$.

Best regards,
sunken rock


function
real analysis
real analysis solved

Let $f:\mathbb R\to\mathbb R$  be a decreasing function and $F$ a primitive of $f$. Prove that $F$ is concave.

If $F$ is not concave there exist $a<c<b$ such that $F(c)<xF(a)+(1-x)F(b)$ where $c=xa+(1-x)b$.So $x(F(c)-F(a))<(1-x)(F(b)-F(c))$.There exist also $e1$ in $(a,c)$ and $e2$ in $(c,b)$ such that $f(e1)=\frac {F(c)-F(a)}{c-a}$ and $f(e2)=\frac{F(b)-F(c)}{b-c}$.So $xf(e1)(c-a)<(1-x)f(e2)(b-c)$ $=>$ $f(e1)<f(e2)$ for $e1<e2$.This is a contradiction since f is decreasing.

[quote="Killer"] $xf(e1)(c-a)<(1-x)f(e2)(b-c)$ $=>$ $f(e1)<f(e2)$ for $e1<e2$[/quote]

Can someone give the details of this implication ?

Nice, killer has just shown to me his solution to this very easy problem, so I am able to justify the implication. It follows from the fact that x is pozitive, c>a and $ x(c-a)=(1-x)(b-c)$.


We have the surface defined by the equation \[ x^/a^2 \plus{} y^2/b^2 \minus{} z^2/c^2\equal{}1\].

Prove that given a point X on this surface there exist two lines that pass through X and are completely contained in the surface.

  i hope it is appropiate to post this problem here.

If such lines exist, they will be tangent to the surface and therefore they will be contained in the tangent plane. Find the equation of tangent plane at $ (x_0,y_0,z_0)$ and then find its intersection with the surface.

$ \frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \minus{} \frac {z^2}{c^2} \equal{} 1 \Longleftrightarrow \left( \frac {x}{a} \minus{} \frac {z}{c} \right) \left( \frac {x}{a} \plus{} \frac {z}{c} \right) \equal{} \left( 1 \minus{} \frac {y}{b} \right) \left( 1 \plus{} \frac {y}{b} \right)$. So you can find two different sheafs of lines completely contained in the surface by chosing

$ \left\{\begin{array}{l} \frac {x}{a} \minus{} \frac {z}{c} \equal{} \alpha \left( 1 \minus{} \frac {y}{b} \right) \\
\frac {x}{a} \plus{} \frac {z}{c} \equal{} \frac {1}{\alpha} \left( 1 \plus{} \frac {y}{b} \right) \end{array} \right.$

or

$ \left\{\begin{array}{l} \frac {x}{a} \minus{} \frac {z}{c} \equal{} \alpha \left( 1 \plus{} \frac {y}{b} \right) \\
\frac {x}{a} \plus{} \frac {z}{c} \equal{} \frac {1}{\alpha} \left( 1 \minus{} \frac {y}{b} \right) \end{array} \right.$

with $ \alpha \in \mathbb{R} \setminus \{0\}$
Now I think it's not difficult to proof that for each point X on the surface you can find two different lines passing through X, one from the first sheaf and one from the second

hello

 sorry ,but what do you mean by sheafs? would you mind explaining the proof with al ittle more detail. thanks

maybe my english is bad, with sheaf I mean a family of lines, for example $ y \equal{} mx$, $ m \in \mathbb{R}$ is a sheaf of lines in the plane passing through $ (0,0)$. The system

$ \left\{\begin{array}{l} \frac {x}{a} \minus{} \frac {z}{c} \equal{} \alpha \left( 1 \minus{} \frac {y}{b} \right) \\
\frac {x}{a} \plus{} \frac {z}{c} \equal{} \frac {1}{\alpha} \left( 1 \plus{} \frac {y}{b} \right) \end{array} \right.$

represent, for any fixed $ \alpha$, the intersection of 2 planes in the $ \mathbb{R}^3$ space, that is a line. Those lines are completely contained in your surface, because each point $ (x_0,y_0,z_0)$ that satisfies

$ \left\{\begin{array}{l} \frac {x_0}{a} \minus{} \frac {z_0}{c} \equal{} \alpha \left( 1 \minus{} \frac {y_0}{b} \right) \\
\frac {x_0}{a} \plus{} \frac {z_0}{c} \equal{} \frac {1}{\alpha} \left( 1 \plus{} \frac {y_0}{b} \right) \end{array} \right.$

must satisfy their product, too and so $ \frac {x_0^2}{a^2} \plus{} \frac {y_0^2}{b^2} \minus{} \frac {z_0^2}{c^2} \equal{} 1$

Conversely, each point $ (x_0,y_0,z_0)$ that is on the surface satisfies the equation $ \left( \frac {x_0}{a} \minus{} \frac {z_0}{c} \right) \left( \frac {x_0}{a} \plus{} \frac {z_0}{c} \right) \equal{} \left( 1 \minus{} \frac {y_0}{b} \right) \left( 1 \plus{} \frac {y_0}{b} \right)$, and so you can choose an $ \alpha$ such that this point satisfies one of the 2 systems I wrote.


In $\triangle{ABC},A+C=2B, \frac{1}{\cos{A}}+\frac{1}{\cos{C}}=-\frac{\sqrt{2}}{\cos{B}}.$ Find the value of $\cos{\frac{A-C}{2}}.$

[hide]
Since $A+C=2B$ and $A+C+B=180$, we get $B=60$ and $A+C=120$

after multiplying through by $\cos A \cos B \cos C$, you get

$\cos B (\cos A + \cos C) = \sqrt{2} \cos A \cos C$
since $\cos B=\frac1{2}$, we have $\cos A + \cos C = -2\sqrt2 \cos A \cos C$

using product to sum and sum to product formulas, we get

$2 \cos {60} \cos {\frac{A-C}{2}}=-2\sqrt{2} \cdot \frac1{2} (\cos {120}+\cos {A-C})$
simplifying and letting $x=\frac{A-B}{2}$ and also using the double angle formula for cos,

we get $\cos x = \frac{-\sqrt2}2 - \sqrt2 (2 \cos^2 x -1)$

solving for cos x, we get, $\cos{\frac{A-C}{2}} = \frac{\sqrt{2}}2$  (used quad equation)

so A=105, C=15

[/hide]


algebra
polynomial
Galois Theory
algebra unsolved

The beginning is as mszew did:
$ (-3x^{3}+6x^{2}\sqrt{3}-8\sqrt{3})(-3x+2\sqrt{3})=0$. We obtain the obvious solution $ x=\frac{2}{\sqrt{3}}$; now let $ x=\frac{2t}{\sqrt{3}}$; we obtain $ p(t)=-t^{3}+3t^{2}-3=0$. The $ p$-Galois group is $ \mathbb{Z}_{3}$; thus we may think that if $ a$ is a $ p$-root then $ \exists{n}$ s.t. ${ \mathbb{Q}(a)=\mathbb{Q}(cos(\frac{2\pi}{n}}))$. Necessarily $ \phi(n)=2.3=6$, thus $ n\in\{7,9,14,18\}$. That works with $ n=18$: the ${ cos(\frac{\pi}{9}})$-minimal polynomial,  $ q(y)=8y^{3}-6y-1$, has the roots ${{{ cos(\frac{\pi}{9}}),cos(\frac{7\pi}{9}}),cos(\frac{5\pi}{9}})$. Moreover $ p(1-2y)=q(y)$.
Conclusion: the $ p$-roots are: ${{{ 1-2cos(\frac{\pi}{9}}),1-2cos(\frac{7\pi}{9}}),1-2cos(\frac{5\pi}{9}})$.


number theory
greatest common divisor
modular arithmetic

Ok! this is a game ~If not proper to the rules of which i cant find i think then just lock please

Rules:
Do not spam!
Check if the same guess you have made is posted or not!
Be nice!
[i]Also, If you guess if it is true then please explain how you came to that conclusion[/i]

Object:
the object of this game is to figure out if a large number (really really large) is a prime number or not, and what it is divisable by if it is not
[i]You must find the GCD if it is not a prime number[/i]
[b]If you get it right post a new number! and if you do you get to tell someone if they are right =)[/b]

First number, is 1111151141
Guess away~ =)

$1111151141=19^{3}\cdot161999$

Do I post another number now?

Yep, And nice one. But you came to it quite quick =[ lol

it can be devided by 2

new one: 9999978347

[quote="The Zuton Force"]Eugh... $461|9999978347$

3657500101

This is cool
[hide=" :) "]
If $p(N)$ represents the Nth prime, then, this is $p^{13}(1)$

[/hide][/quote]

correct, whatever the hidden part means,
also can people please make sure to check back here to say if something is correct =)

[quote="hunter34"]

Next:8787132021[/quote]

Divisible by 3

Got it in like 10 sec.

1234567897

I think you've stumped us all, if not... WOW @ AoPS'ers

Wow                                              ?

See http://math.about.com/library/bldivide.htm , snee.

$ 0 \equiv 37492875034937 \pmod {7}$

2069384750923453

(987654321!*9984234.6!^3958203!)^573038!

Incase this is impossible (my scientific calc gave me some number ending in 'e'. What does that mean?) try:
781633287*2531273^48

Um, both of those are obviously not prime, since they're products of factorials.

And $ a \text{ \tiny{E} } b \equal{} a\cdot 10^b$

Try 1732743457.

[quote="Temperal"]$ 3|1234567897$[/quote]

You are wrong.

 :diablo:

[quote="pingala"][quote="i_like_pie"]Next: $ 123454321$[/quote]
divisible by 3.[/quote]
Umm...

$ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 4 \plus{} 3 \plus{} 2 \plus{} 1 \equal{} 25\equiv1\mod3$

[quote="snee"]e = ^ in calculators.[/quote]

....no. That's totally different.

$ a \text{\tiny{E} } b = a\cdot 10^b$, like I said. $ a\^\, b = a^b$

as for the other number pingala gave an incorrect answer to...
$ 11111|123454321$

Remember that old trick?

164327


the first two inequalities 1) and 2) are not valid as we take the [i][u]equilateral triangle with all sides have length 1.[/u][/i]____________________________________________________________

Hi!!
$1)a^{3}b+b^{3}c+c^{3}a+2a^{2}b^{2}c^{2}\geq ab^{3}+bc^{3}+ca^{3}+2abc(a+b+c)$
$\Leftrightarrow \sum b(a+b-c)(a-c)^{2}\geq 0$;

$2)a^{3}b+b^{3}c+c^{3}a+a^{2}b^{2}c^{2}\geq ab^{3}+bc^{3}+ca^{3}+abc(a+b+c)$
  In $\Delta ABC$  $a^{2}=b^{2}+c^{2}-2bccosA$;
  $\Rightarrow a^{3}b+b^{3}c+c^{3}a=ab(b^{2}+c^{2}-2bccosA)+bc(c^{2}+a^{2}-2cacosA)+ca(a^{2}+b^{2}-2abcosA=ab^{3}+bc^{3}+ca^{3}+abc(a+b+c)-2abc(bcosA+cCosB+acosC)$
I will prove that:
  $2abc(bcosA+ccosB+acosC \leq (a^{2}+b^{2}+c^{2})$;
   $x=ab;t=bc;z=ca$;
  $\Rightarrow 2abc(bcosA+cCosB+acosC)=2(abbccosA+bccacosB+caabcosC)=2(xycosa+yzcosB+zxcosC) \leq x^{2}+y^{2}+z^{2}$;
  It's true.     :P
And problem 3) k min =1; it's hard :lol:


Hello!!!

Let S(m,n) be the set of all integers k such that m mod k + n mod k >= k. What is S(7,9)?

I tried to make it more general solving as following:

A mod B = A - B[A/B], where [x] is the floor function.

m mod k + n mod k >= k
m - k[m/k] + n - k[n/k] >= k
m/k - [m/k] + n/k - [n/k] >= 1 (1)

A/B - [A/B] results in a value in the range [0,1) as can be seen in the example below:

2.3 - [2.3] = 2.3 - 2 = 0.3
-2.3 - [-2.3] = -2.3 + 3 = 0.7

So, the left side of (1) is in the range [0,2*0.999...)

But I don't know how to solve the inequality in order to find S(m,n).

Thanks!!!

[hide="Answer"]If $ S (m, n) = \{k \in \mathbb  Z^{+}\ |\ k \leq m\;(mod\ k) + n\;(mod\ k)\}$, then

$ S (7, 9) = \boxed{\left\{2, 4, 5, 8, 10, 11, 12, 13, 14, 15, 16\right\}}$.

It's not asking you to generalize; it's asking you to find the elements of the set for a given $ m$ and $ n$ (in this case, $ 7$ and $ 9$)[/hide]


ratio geometry inradius Vieta probability algebra polynomial

Ok. math marathon! just continue. Like, just post a solution and a new problem. Btw, this is sorta to help you in mathcounts.
K.
I made this one up:
The angles of a triangle are in the ratio of 2:4:6. What is the measure of the smallest angle in the triangle?

[/hide]

The numbers add to twelve, so $ 12x \equal{} 180$, and $ x \equal{} 15$. Thus, the smallest angle is 30.

2. How many factors of $ 2^{105}$ are bigger than 1,000,000?

$ 2^{20} \equal{} (2^{10})^2 \equal{} 1048576$, and every integer through $ 105$ can be made after that. Thus, the number of factors is $ 105 \minus{} 20 \plus{} 1 \equal{} \boxed{86}$.

3. Find the area of a triangle with side lengths $ 5$, $ 6$ and $ 7$.

A 5,6,7 is not a right triangle so that doesn't work.

Restarting from 3.
We are given the sides of a triangle, and what better to use than Heron's right now?
We have

$ s = \frac{a+b+c}{2} = \frac{5+6+7}{2} = 9$

$ = \sqrt{s*(s-a)(s-b)(s-c)} = {\sqrt{9*(9-5)(9-6)(9-7)}}$

$ \boxed{6\sqrt{6}}$

4. Find the inradius of a triangle with sides $ 10$, $ 12$, and $ 13$.

We have that $ rs\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$

$ \implies r(\frac{35}{2})\equal{}\sqrt{\frac{35}{2}(\frac{35}{2}\minus{}10)(\frac{35}{2}\minus{}12)(\frac{35}{2}\minus{}13)}$.

$ \implies r\approx3.257$.

NP: Find the sum of the roots of $ x^4\plus{}8x^3\minus{}16x^2\plus{}39x\minus{}5\equal{}0$.

5. We have this cubic equation in the form $ ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e \equal{} 0$, and by Vieta's Formula, we have the sum of the roots equal to $ \frac { \minus{} b}{a}$, so therefore, the sum of the roots is $ \frac{\minus{} 8}{1} \equal{} \minus{}8$.

Problem 6: What is the probability that when I flip $ 10$ coins, $ 3$ or less coins come up as heads?

$ \frac {\binom{10}{0} \plus{} \binom{10}{1} \plus{} \binom{10}{2} \plus{} \binom{10}{3}}{2^{10}} \implies \boxed{\frac {11}{64}}$.

New Problem (7): If a polynomial $ 20x^3 \minus{} 9x^2 \plus{} ax \plus{} b$ is divisible by $ x^2 \plus{} 4$, find the sum of $ c$ and $ d$ such that $ ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ is divisible by $ 40x^2 \minus{} 58x \plus{} 40$.

EDIT: Oops, didn't see the MathCounts part...

New Problem (7A): What is a general ratio of surface area of a cube with side $ r$ to the volume of a cube with side $ r$?  Express your answer in $ a: b$ form, where $ a$ and $ b$ are in simplest terms and reduced.

The surface area is given by $ 6r^2$, and the volume by $ r^3$, so $ \frac{6r^2}{r^3}\equal{}\boxed{\frac{6}{r}}$.

8. Find the largest prime factor of $ 2^{15}\minus{}1$.

8. We see that $ 2^{15} \minus{} 1$ is not a Mersenne Prime, so therefore, we start off using the long way. $ 2^{15} \minus{} 1$, in numerical form, is 32767. We check using our divisibility rules, and see that it is divisible by $ 7$, leaving $ 4681$. Now, through tedious work, we check by dividing the prime numbers up to $ \sqrt {4681}$, and find that it's divisible by 31. Now, we are left with $ 151$. $ 151$ is not divisible by anything as we see by testing the primes up to $ \sqrt{151}$, so therefore, $ 151$ is a prime number. Thus, the largest prime factor is $ \boxed{151}$.
There seems to be an easier way to factor and find the prime by factoring, but I'm not sure how.

9. I dip a cube in red, such that all the exposed surfaces are all colored red. Then, I separate this cube into $ 125$ different cubes. Then, I roll a cube. The top face is red. What is the probability that the bottom face is also red?

8. difference of cubes gives us 2^15-1=(2^5-1)(2^10+2^5+1) so 2^15-1 is divisible by 31.  using difference of 5th powers, we find that 2^15-1 is also divisble by 7. so 2^15-1 is divisible by lcm(31,7)=31*7.  divide.

[quote="xxrxxhxx"]9. I dip a cube in red, such that all the exposed surfaces are all colored red. Then, I separate this cube into $ 125$ different cubes. Then, I roll a cube. The top face is red. What is the probability that the bottom face is also red?[/quote]

In math terms...0

In science terms...not enough information.

1) If you don't cut the cube at all, and take 124 nanosized particles, who knows?
2) The cube itself is made of atoms. Not a cube, because it has air spaces (which causes friction) between them.
3) Cutting the cube will cause some extremely small pieces to chip off. Therefore, you cannot ensure that they are cubes.

10. Find the sum of the coefficients of $ (x \plus{} y)^{96}$.

Expanding by the binomial theorem, the coefficients are $ \binom{96}{0}\plus{}\binom{96}{1}\plus{}\cdots\plus{}\binom{96}{95}\plus{}\binom{96}{96}\equal{}\boxed{2^{96}}$

11.
Find the distance from the incenter and circumcenter of a triangle with side lengths 3,4,5.


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

AMC, AIME, USAMO target different skills. You can read more in my blog:
http://blog.tanyakhovanova.com/?p=116

I suggest the following: USAMO participants are allowed to go to next year's AIME no matter what their AMC scores are. USAMO winners are allowed to go to the next year's USAMO no matter what their AIME results are.

This way kids who have proved that they are great swimmers do not need to compete in running again. By swimmers I mean kids who are good at proofs, not accurate speed calculation.

That is wrong:
If someone was a USAMO qualifier, then he/she has demonstrated that they are capable of the speed and accuracy not only of making AIME, but having a high enough AMC score to make the USAMO.
If someone was a USAMO winner one year, they have demonstrated that they have the necessary skills to make USAMO and do well on it.
This proposition would be superfluous and reward slackers.

Even Olympic Gold Medalists have to go through the Olympic Trials again (atleast for the U.S.)

While I do agree that somebody who is capable of acheiving a high score on the USAMO should concentrate only on the USAMO, the proposed idea applies to too few people.

Somebody who has made the USAMO should make the AIME. And if s/he doesn't make the AIME and is still allowed to compete, the hit on the index because of the low AMC score would almost certainly eliminate that person from USAMO qualification.

Somebody who has been a USAMO winner probably won't screw up AIME bad enough to drop the USAMO index low enough to not qualify. It does happen, but at most one or two students (probably just none) who have a bad day on the AIME do not require a change of rules.

We are more concerned with those on the borderline instead of those who have proven themselves more than capable for making the USAMO.

What if only about 80 12th graders make it one year? That would create 420 already made spots, and 80 ones for the new 9th-graders.

That the AMC/AIME/USAMO target different skills is definitely something I agree with. Not sure about the proposal itself. For one thing, if a USAMO qualifier doesn't qualify for the AIME the year after, their qualification for the USAMO was more likely some fluke then them actually deserving to take AIME. If a USAMO winner doesn't qualify for the USAMO, something is [i]really[/i] messed up. Of all of MOP (a much bigger group), I think it's only like one or two reds, if anyone, that miss USAMO the next year. So I'm not sure how this would change things.

I see why you might want to change things, based on your original point of them testing different things. However I think you might underestimate how much improving proof-writing skills and olympiad prowess makes you better at the AMCs/AIMEs too. There is still a speed element in olympiads, as only the best can do three problems that difficult in only(!) 4.5 hours, and there are still a few olympiad-level problems that involve computation.

Although they target different skills, isn't that the point? So that people are being tested not just in one area but all?
Anyways, like simo14 said, if they've already made the usamo, haven't they demonstrated that they're capable of these tests and the different skills required for each? plus, if you really deserve to be there, you have to be good enough too & i think these tests do a good job at assessing that.

It's not the point. Someone who can do really well on the IMO might not even be able to pass the AIME (he can pass AMC10 through USAMTS). The goal of the AMC series is to select an IMO team, not test how quickly you can solve easy problems (as far as I know).

There are people that always don't "barely" make the AIME, USAMO, AIME I mean all you have to do is look on this site and see everyone going "epic phaillll .5 away from making aime >< " or something around the lines of that and no matter how it changes, there's always going to be people out there who don't make it, barely. So, keeping the current qualifications and system is just fine.

[quote="worthawholebean"]If you can succeed at IMO, you can pass AIME. Let's not be ridiculous.[/quote]

Yes.  And if you can't pass AIME, you probably don't have the skills to do IMO or USAMO.  Although AIME does have a lot of computation, someone would have to really suck at arithmetic, and I mean [b]really[/b] suck in order to not be able to finish the AIME.  Arithmetic is easy.  If you can't do arithmetic, you probably can't do olympiad math.

people are overexaggerating computation time. the most time-consuming part of the aime is finding solutions and experimenting. a person who succeeds at the olympiad level should be able to find quick solutions to aime problems which would give him/her time to do the "heavy arithmetic" that people point to (although I find it extremely unprobable that a usamo winner can't do computation...).

[quote="jb05"]I don't think that's a reasonable assumption at all. The AIME is EXTREMELY heavy on computation. I would say about half of the time content of the AIME is spent on computation. It is entirely feasible to assume that someone who would do quite well at IMO would be unable to do this (especially without making lots of careless mistakes). In fact, I know people who have gone to IMO who consistently just BARELY made USAMO. Doesn't that imply that there exist people who should go to IMO who consistently just barely did NOT make USAMO?[/quote]

I would say computation takes me 10 minutes or less, and I attempt at least 12 of the problems (which means I'm not a newb who just does the first 5 problems then gives up).  Though I am somewhat fast at computation (I did well in Countdown Rounds back in the days of Mathcounts for me), I think "half of the time content" is not a "reasonable assumption at all" (I'm just quoting your words).
Also, I don't think worthawholebean's statement is an assumption (but a fact based on the history of these contests), and I think you're lying.  Name the people who have gone to IMO and just barely made USAMO.  The scores of previous participants do exist on the AMC website, so I can check (assuming that it was in, say, the 2000's, because the difficulty level of AMC/AIME/USAMO fluctuates if you look before the 21st century and so an argument using such information is not valid), or perhaps I can ask the AMC people for records if needed.

[quote="the future"]Actually, now that I think about it, dont they have a similar procedure for the USNCO.  Its something like if you make the camp, you make it next year too?  Correct me if I am wrong.[/quote]

I believe it's the other way around ... if you get a silver or gold at IChO, you can't go to the study camp again.

I agree that someone can be able to solve USAMO- and IMO-type problems without being good at the "fast and furious" style of the AMC (and the AIME, to a slightly lesser extent). However, don't forget that "the main purpose of the AMC 10/12 is to spur interest in mathematics and to develop talent through solving challenging problems in a timed multiple-choice format" (as stated on the AMC website). Selection of the IMO team is secondary. As a result, the contest must be accessible to the average test-taker. Around 200,000 students (a rough estimate) take the AMC 10 or AMC  12 each year. Of those, only around 500 qualify for the USAMO. There are certainly many more who would like to take the USAMO, but those interested in a proof-style contest still constitute a small portion of the 200,000 taking the AMC. Making the lower levels of the AMC ladder more like the USAMO would be certain to scare off the vast majority of participants. Such a change is also infeasible; every USAMO solution must be read by an actual person (or actual people), not just a machine. There are indeed tests with many thousands of participants that have essay-type questions, like AP tests. Unlike questions in AP tests, however, USAMO questions are very in-depth and require graders to closely scrutinize solutions. They can't just say, "okay, they got A: plus 1. They got B: plus 2..." Also, I imagine few would pay $80+ to take the USAMO, so not as many scorers could be hired as are used for AP tests to grade thousands more papers (although I'm not sure exactly how the USAMO grading is carried out).

Your proposal, I'm afraid, would be of little help in ensuring that those who are good at proof-type questions and not as good at AMC-type tests would have a better chance at qualifying for the USAMO. As others in the topic have mentioned, someone who qualifies for the USAMO is unlikely to fail to qualify for the AIME the next year if they attempt to (I don't have any evidence that supports this, and there are certainly people who might qualify for the USAMO one year and not qualify for the AIME the next, but I don't imagine there are many). Someone who is in the top 12 of all USAMO participants is probably not likely to miss the USAMO cutoff the next year (again, I'm only speculating, and there may be exceptions).

being in the top 12 of the usamo demands a level of creativity and intelligence very few of us have. a person like that would certainly not miss the usamo index.

If you qualify for AIME by way of USAMTS, what is your AMC score as part of your index? By the way, I have to lean towards the view that aime is computation based, as I solved 10 problems in the first hour and a half, spending some of that time cheking my work, then spent anouther hour checking, found 5 mistakes, then solved two more problems. Although you could make the case that if you know the material well enough you should have time to check. Although I've heard many MOPers just barely qualify for USAMO.

personally i think it's much harder to make aime by usamts than amc, so it's kind of redundant.
and you get a 100 amc 12.

[quote="Phelpedo"]OK, people who are good at Olympiad level problems might not be as good at the AIME - but surely in 3 hours they can be expected to solve 6 problems correctly, which has been the floor the past couple of years?[/quote]
Getting the floor score does not guarantee AIME qualification. Last year, 3070 people scored 6 or higher on the AIME. 505 people qualified for the USAMO. That's less than a sixth of the people who had the floor score or higher.

[quote="simo14"]personally i think it's much harder to make aime by usamts than amc, so it's kind of redundant.[/quote]
To borrow the TC's analogy, if the AIME is to the USAMO as running is to swimming, then the AMC is to the USAMTS as a 100m dash is to a marathon. The two styles are so different that I'm certain there are many who consider getting 100+ on the AMC 12 to be harder than qualifying for the AIME through the USAMTS (which typically requires around 70 points after the first three rounds IIRC). I happen to agree that qualification through the AMC is easier, but our personal evaluations of the two contests' relative difficulties are of no consequence.

I am not saying that I agree with the the original post but I think it is very possible for someone who is capable of being a USAMO winner or IMO level participant to not do well on the AIME.

I attended the AwesomeMath Summer Program the first year that it was held and there was a student there from Columbia who had been 7th on the Columbian TST, so basically one spot away from going to IMO. I believe he was in the Columbian equivalent of 9th or 10th grade. We had a test at the beginning of AMSP to place us into groups based on our mathematical abilities. The test was composed of an AIME format test and a USAMO format test. This particular student did very well on the USAMO format test, but did horribly on the AIME format test for whatever reason. He was placed in the 2nd lowest group (out of 5) because the AIME test was weighted heavily enough for his cumulative score to suffer severely. I also was placed in the same group that he was in and after the first day of classes, when he immediately solved every proof problem that Zuming wrote on the board in a matter of minutes, he was told to go to the highest group from then on.

So yeah, it can happen.


In a class of 20 students, 8 girls were absent due to a basketball team trip. Fifty percent of the students present are girls. Given that no boys are absent, what percent of the total class is boys?

12 students are in class, and we are told that half of them are girls, so there are 6 girls currently in class and 14 total.  So 6 students are boys, giving $ 6/20\equal{}.3$, or $ 30\%$.


number theory
least common multiple
greatest common divisor
prime factorization

Find the smallest positive integer $n$ for which there exist $n$ different positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying [list] [*] $\text{lcm}(a_1,a_2,\cdots,a_n)=1985$,[*] for each $i, j \in \{1, 2, \cdots, n \}$, $gcd(a_i,a_j)\not=1$, [*] the product $a_{1}a_{2} \cdots a_{n}$ is a perfect square and is divisible by $243$, [/list] and find all such $n$-tuples $(a_{1}, \cdots, a_{n})$.

The problem doesn't make sense with $ 1985$, since the first and third conditions can't be satisfied simultaneously. Judging from the topic description, $ 1995$ was meant, and here's the solution assuming that.

$ n \equal{} 7$: $ 15,21,57,105,285,399,665$ is the only sequence that satisfies every condition up to permutation.

First, since the least common multiple of all the numbers is $ 1995 \equal{} 3 \cdot 5 \cdot 7 \cdot 19$, every $ a_i$ must be the product of some subset of $ \{3,5,7,19\}$. By the second condition, the subset must be nonempty. The third condition requires that the product of all elements of the sequence is divisible by $ 3^5$, but since it also must be a perfect square it must actually be divisible by $ 3^6$. Since each element of the sequence can have at most one $ 3$ in the prime factorization, this gives $ n \equal{} 6$ as a lower bound (for now).

Suppose $ n \equal{} 6$ works. All elements must then be divisible by $ 3$, so the second condition is trivially satisfied. There are now $ 8$ possible numbers to choose for the $ 6$ elements of the sequence: The product of any subset of $ \{5,7,19\}$ times $ 3$. We notice that the product of all $ 8$ of these elements is a perfect square, so in order for the product of the $ 6$ elements of the sequence to be a square we must omit two distinct elements whose product is a perfect square. But since every prime has an exponent of $ 0$ or $ 1$ in these $ 8$ numbers, the product of two of them is a perfect square if and only if they are the same number. Contradiction. $ n \equal{} 6$ is not possible.

$ n \equal{} 7$ is, however. In this case we have $ 6$ elements divisible by $ 3$. The final element is not divisible by $ 3$, since if it were the product would have $ 3^7$ in the prime factorization and not be a perfect square. Let $ a_7$ be the number not divisible by $ 3$. If $ a_7$ is prime, then the second condition requires that all numbers be divisible by $ a_7$, and then the product of the sequence has $ a_7^7$ in the prime factorization, so it can't be a square. So $ a_7$ has to be a product of at least two elements of $ \{5,7,19\}$.

Suppose $ a_7$ is the product of two of these, which we will call $ q$ and $ r$, and is not divisible by one of them, which we will call $ p$. That is, $ a_7 \equal{} qr$. Of the other $ 6$ elements divisible by $ 3$, we can choose four that are divisible by $ p$: $ 3p, 3pq, 3pr, 3pqr$. But including one of them, $ 3p$, contradicts the second condition. So we only have three choices of numbers divisible by $ p$. Because the product of all sequence elements must be a perfect square, we can only select two of the numbers. Now we need $ 4$ more numbers divisible by $ 3$ in the sequence. We can only choose from $ 3, 3q, 3r, 3qr$, so we must have them all. But including $ 3$ and $ qr$ in the sequence contradicts the GCD condition. Thus, $ a_7$ must be $ 5 \cdot 7 \cdot 19$.

So $ a_7 \equal{} 5 \cdot 7 \cdot 19$. For the other $ 6$ elements of the sequence, we can only select the product of any subset of $ \{5,7,19\}$ times $ 3$ as before. But notice that one of these, $ 3$, contradicts the GCD condition since $ a_7$ is not divisible by $ 3$. So we only can select from seven elements. The product of these seven elements and $ a_7$ is $ 3^7 5^5 7^5 19^5$. So for this to be a perfect square we have to divide out $ 1995 \equal{} 3 \cdot 5 \cdot 7 \cdot 19$, and this is the element we omit. So the other $ 6$ elements must be $ 3 \cdot 5,\ 3 \cdot 7,\ 3 \cdot 19,\ 3 \cdot 5 \cdot 7,\ 3 \cdot 5 \cdot 19,\ 3 \cdot 7 \cdot 19$. This gives the unique sequence described in the first paragraph, and we're done.

(edit: typo)


\[ \ a,b,c>0;a\plus{}b\plus{}c\equal{}3.\]Prove that:\[ \frac{a^2}{b\plus{}c^3}\plus{}\frac{b^2}{c\plus{}a^3}\plus{}\frac{c^2}{a\plus{}b^3} \geq \frac{3}{2}.\] :blush:  :blush:  :blush:

[quote="mai quoc thang"]
\[ \ a,b,c > 0;a \plus{} b \plus{} c \equal{} 3.
\]
Prove that:
\[ \frac {a^2}{b \plus{} c^3} \plus{} \frac {b^2}{c \plus{} a^3} \plus{} \frac {c^2}{a \plus{} b^3} \geq \frac {3}{2}.
\]
[/quote]
$ \sum_{cyc}\frac {a^2}{b \plus{} c^3}\equal{}\sum_{cyc}\frac{a^4}{a^2b\plus{}c^3a^2}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^2b\plus{}a^3b^2)}.$
Thus, it remains to prove that $ 2(a^2\plus{}b^2\plus{}c^2)^2\geq3\sum_{cyc}(a^2b\plus{}a^3b^2).$
But $ 2(a^2\plus{}b^2\plus{}c^2)^2\geq3\sum_{cyc}(a^2b\plus{}a^3b^2)\Leftrightarrow$
$ \Leftrightarrow2(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)^2\geq(a\plus{}b\plus{}c)^2(a^2b\plus{}b^2c\plus{}c^2a)\plus{}9(a^3b^2\plus{}b^3c^2\plus{}c^3a^2)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^5\plus{}a^4b\plus{}2a^4c\minus{}7a^3b^2\plus{}3a^3c^2\minus{}2a^3bc\plus{}a^2b^2c)\geq0.$
But by AM-GM $ \sum_{cyc}(a^4c\plus{}a^2b^2c\minus{}2a^3bc)\geq0$ is true and
$ \sum_{cyc}(2a^5\plus{}a^4b\plus{}a^4c\minus{}7a^3b^2\plus{}3a^3c^2)\equal{}\sum_{cyc}(a^5\plus{}a^4b\minus{}7a^3b^2\plus{}3a^2b^3\plus{}ab^4\plus{}b^5)\equal{}$
$ \equal{}\sum_{cyc}(a\minus{}b)(a^4\plus{}2a^3b\minus{}5a^2b^2\minus{}2ab^3\minus{}b^4)\equal{}$
$ \equal{}\sum_{cyc}\left((a\minus{}b)(a^4\plus{}2a^3b\minus{}5a^2b^2\minus{}2ab^3\minus{}b^4)\plus{}a^5\minus{}b^5\right)\equal{}$
$ \equal{}\sum_{cyc}a(a\minus{}b)^2(2a^2\plus{}5ab\plus{}b^2)\geq0.$ :)

[quote="arqady"][quote="mai quoc thang"]
\[ \ a,b,c > 0;a \plus{} b \plus{} c \equal{} 3.
\]
Prove that:
\[ \frac {a^2}{b \plus{} c^3} \plus{} \frac {b^2}{c \plus{} a^3} \plus{} \frac {c^2}{a \plus{} b^3} \geq \frac {3}{2}.
\]
[/quote]
$ \sum_{cyc}\frac {a^2}{b \plus{} c^3} \equal{} \sum_{cyc}\frac {a^4}{a^2b \plus{} c^3a^2}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^2b \plus{} a^3b^2)}.$
Thus, it remains to prove that $ 2(a^2 \plus{} b^2 \plus{} c^2)^2\geq3\sum_{cyc}(a^2b \plus{} a^3b^2).$
But $ 2(a^2 \plus{} b^2 \plus{} c^2)^2\geq3\sum_{cyc}(a^2b \plus{} a^3b^2)\Leftrightarrow$
$ \Leftrightarrow2(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)^2\geq(a \plus{} b \plus{} c)^2(a^2b \plus{} b^2c \plus{} c^2a) \plus{} 9(a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^5 \plus{} a^4b \plus{} 2a^4c \minus{} 7a^3b^2 \plus{} 3a^3c^2 \minus{} 2a^3bc \plus{} a^2b^2c)\geq0.$
But by AM-GM $ \sum_{cyc}(a^4c \plus{} a^2b^2c \minus{} 2a^3bc)\geq0$ is true and
$ \sum_{cyc}(2a^5 \plus{} a^4b \plus{} a^4c \minus{} 7a^3b^2 \plus{} 3a^3c^2) \equal{} \sum_{cyc}(a^5 \plus{} a^4b \minus{} 7a^3b^2 \plus{} 3a^2b^3 \plus{} ab^4 \plus{} b^5) \equal{}$
$ \equal{} \sum_{cyc}(a \minus{} b)(a^4 \plus{} 2a^3b \minus{} 5a^2b^2 \minus{} 2ab^3 \minus{} b^4) \equal{}$
$ \equal{} \sum_{cyc}\left((a \minus{} b)(a^4 \plus{} 2a^3b \minus{} 5a^2b^2 \minus{} 2ab^3 \minus{} b^4) \plus{} a^5 \minus{} b^5\right) \equal{}$
$ \equal{} \sum_{cyc}a(a \minus{} b)^2(2a^2 \plus{} 5ab \plus{} b^2)\geq0.$ :)[/quote]
We have a harder is
\[ \sum \frac{a^2}{b\plus{}3c^3} \ge \frac{3}{4}\]
:)


The denominator 3-2^2=3-4= -1.
The numerator reduces to 3-5-1=-3.
-3/-1=3.[/b]


function
algebra
polynomial
quadratics
complex numbers

In these exercises use the given zero to find all the zeros of the function

56. g(x) = 4x^3 + 23x^2 + 34x - 10
-3+[i]i[/i]

57. h(x) = 3x^3 - 4x^2 + 14x + 20
1-radical (3) [i]i[/i]

please... i need it as soon as possible

hello, $g(x)=0$ has the solutions $-3-i,-3+i$ and $\frac{1}{4}$.
Are you sure, that $h(x)=3x^{3}-4x^{2}+14x+20$ is right?
Sonnhard.

o dang

um
for the second one

it is h(x) = 3x^3 - 4x^2 + 8x + 8
my bad

with one solution being 1 - radical 3 i

hello, if you have the solution $1-\sqrt{3}i$ then must be
$1+\sqrt{3}i$ the other one, additionally $-\frac{2}{3}$ ist a solution.
Sonnhard.

[quote="cincodemayo5590"]The point of these exercises is usually to show that for a polynomial with [i]real[/i] coefficients, if a complex number is a root then its complex conjugate is a root. (i.e. $P(n \text{ cis}[\theta])=0 \Longrightarrow P(n \text{ cis}[-\theta])=0$, for $P(x)$ with real coefficients. )

Usually, after you factor out the two complex roots (by synthetic or long division, or any other method you want) there will remain something easily factorable ,usually by inspection or by the quadratic formula.[/quote]

all you do is take the conjugate of P(z)=0, and using the conjugate rules, you get P(z conjugate)=0, so it is also a root


Hints:

[hide="1"]quarter circle[/hide]

[hide="2"]How would the players be needed to be placed so that #2 player does not make it to the finals?[/hide]

[hide="3"]Find the minimum and maximum lengths.  Note symmetry.[/hide]

[quote="mihail911"][b]1. A point (x,y) is randomly selected from a rectangular grid in the coordinate plane with vertices at (0,0), (8,0), (0,4), and (8,4). What is the probability that $ x^2 \plus{} y^2\leq 4$?
2. In a single elimination tournament that consisted of three rounds, the better player always won. The winner of Round 3 was the champion, and the loser was runner-up. 8 players were randomly assigned slots in Round 1. What is the probability that the runner-up was not second best player in the tournament?

3. Point P is 9 units from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths?

[/b][/quote]

[hide]

1.find y and x intercepts, they are both 2. So it is a quarter-circle with radius 2.

pi*2^2/4 so 4pi/4 or pi total possible is 32 so 32/pi

2. The ways that the 2nd best team didn't get 2nd would be if they were on the same bracket side as the 1st place one. so 3/7.

3. longest chord possible is 30 (diameter) shortest is 24. So 24, 25,26,27,28,29,30. But there can be two of each not including biggest and smallest. So 24,25,25,26,26,27,27,28,28,29,29,30, or 12.[/hide]

[quote="mihail911"]i understand where i went wrong on the geometric probability one but im still not sure about the team probability. 3/7? i know that's the right answer, but not sure how you guys got it.[/quote]

ok, think about it this way. Since there are 3 single elimination rounds, there are 8 different players in the tournament. The top (best player) is in one side of the bracket. So that means that the 2nd best player (not necessarily 2nd place) can be anywhere in those remaining 7 spots. So, the spots that the 2nd best player can be to make it so he faces off against the best player before the finals are when he is in the same side of the bracket as him. i.e. either playing him in first round or in 2nd round. There are 3 available spots on that side, so 3 out of 7.


Let $ a,b,c\in\mathbb R^\plus{}$ then under what condition does the following inequality $ a\plus{}b\plus{}c\ge ab\plus{}bc\plus{}ac$ hold?

[quote="Null"]Let $ a,b,c\in\mathbb R^ \plus{}$ then under what condition does the following inequality $ a \plus{} b \plus{} c\ge ab \plus{} bc \plus{} ac$ hold?[/quote]

Suppose $ {a,b,c}\in{(0,k]},$ then

  $ a \plus{} b \plus{} c\ge{ab \plus{} bc \plus{} ca}$

holds iff $ k\in{(0,1]}.$


A salesperson can choose one of these two monthly salary plans.  With $ \$25,\!000$ of  sales this month, what is the positive difference in earnings under the two plans?\\		A basic salary of $ \$500$ plus a $ 10\%$ commission on sales.	\\
A basic salary of $ \$1000$ plus a $ 5\%$ commission on sales.

well, let's find out which one gets you more money with a $ \$25000$ of sales

using the first plan, we find that $ \$500$ plus $ 10\%$ would be $ 500 \plus{} 0.1 \times 25000 \equal{} 3000$
using the second plan, we find that $ \$1000$ plus $ 5\%$ would be $ 1000 \plus{} 0.05 \times 25000 \equal{} 2250$

therefore, we find that the positive difference in earnings between the two plans is $ 3000 \minus{} 2250 \equal{} 750$


Nobody?

BTW, why my latex code does NOT work ?   :( +:mad:

Plus, an Idea: It will be nice if somebody create a simple program which will be called "Muirhead-Schur". I mean, if we input a homogeneous polynomial inequality, then, the output is the result of the proof by muirhead's theorem and Schur's inequality. Someone can do it ?

[quote=ideahitme]Let $ a$, $ b$, and $ c$ be positive real numbers. Show that
$ (abc)^{2}(a \plus{} b \plus{} c)^{3}( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c) \geq (a^{2} \plus{} b^{2} \plus{} c^{2})^{3}( \minus{} a^{2} \plus{} b^{2} \plus{} c^{2})(a^{2} \minus{} b^{2} \plus{} c^{2})(a^{2} \plus{} b^{2} \minus{} c^{2})$.[/quote]
We can prove it by hand without computer, even without calculator.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where
$f(w^3)=(3u^2-2v^2)^3(729u^6-1458u^4v^2+648u^2v^4+216u^3w^3-144uv^2w^3+8w^6)-u^3(27u^3-36uv^2+8w^3)w^6$
But $f'(w^3)=(3u^2-2v^2)^3(216u^3-144uv^2+16w^3)-54u^6w^3+72u^4v^2w^3-24u^3w^6\geq$
$\geq u^6\cdot88u^3-54u^6w^3>0$,
which says that $f$ is an increasing function and it's enough to prove our inequality for a minimal value of $w^3$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or $w^3=x^3-3ux^2+3v^2x$.
Thus, a line $y=w^3$ and graph of $y=x^3-3ux^2+3v^2x$ have three common points and $w^3$ gets a minimal value in the following cases.
1. $w^3\rightarrow0^+$.
Let $c\rightarrow0^+$ and $b=1$. We obtain $(a^2+1)^4(a^2-1)^2\geq0$;
2. $b=c=1$, which gives $(a-1)^2(a^5+2a^4+7a^3+12a^2+16a+16)a^5\geq0$.
Done!

[quote=luofangxiang][img]http://s9.sinaimg.cn/middle/006ptkjAzy75NN5V0N208&690[/img][/quote]

done , 挡什么？
z>0 ?????????????????? 哈哈

[quote]By computer,we have[/quote]
If not a secret, what program can do that ?

Dear luofanxiang,

very nice your proof but I cannot prove

$9m \geq 2(mn)^2 +2mn +3n$

Can you please tell me if your prove is by full expansion?

Thank you very much

[quote=szl6208][quote=ideahitme]Let $ a$, $ b$, and $ c$ be positive real numbers. Show that
$ (abc)^{2}(a \plus{} b \plus{} c)^{3}( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c) \geq (a^{2} \plus{} b^{2} \plus{} c^{2})^{3}( \minus{} a^{2} \plus{} b^{2} \plus{} c^{2})(a^{2} \minus{} b^{2} \plus{} c^{2})(a^{2} \plus{} b^{2} \minus{} c^{2})$.[/quote]

By computer,we have

\[LHS-RHS=2(a^2+b^2+c^2)\sum_{cyc}{a^3b^3(a-b)^2(a+b-c)^2}\]
\[+\sum_{cyc}{ab(2a^4+a^3b+6a^2b^2+ab^3+9abc^2+2b^4)(a-b)^2(a^2+b^2-c^2)^2}\]
\[+\frac{1}{2}\sum_{cyc}{ab(3ab+ac+bc+3c^2)(a-b)^4(a+b-c)^4}\]
\[+\frac{1}{2}\sum_{cyc}{(a-b)^4(a^2+b^2-c^2)^4}\]
\[+\frac{1}{2}\prod{(a-b)^2(a+b-c)^2}\]
\[\ge{0}\][/quote]

评注：
1.这个不等式有一个特点，你们觉得它是什么？
[hide]两个符号不确定的量构成了不等式。这类不等式目前看到的好象是“非常稀少”[/hide]

2.从式差的表达式可以看出，这个不等式仍含有较多的泥巴。
如何让它更强呢？

amount of symbol uncertainty:
=================================
$x,y,z>0$,if $f(x,y,z)\geq 0$ is not true,and $f(x,y,z)\le 0$ is not true,then call $f(x,y,z)$ is amount of symbol uncertainty.

with $M$ and $N$ are amount of symbol uncertainty,then call inequality
\[M\geq N\]

is amount of symbol uncertainty inequality.


inequalities
trigonometry
geometry
inradius
circumcircle
triangle inequality
area of a triangle

In the triangle $ABC$ $AM$ is a median. Prove that : $\sin{\angle{MAB}}\ge{\frac{2r}{R}}$.
Hereby r is inradius, R is circumradius.
Please post your solution very quickly!!

Very nice problem, but it has a typo, and moreover, please don't write things like "Please post your solution very quickly!!" since people tend to wonder, when seeing such requests, whether the problem is from some current mathematical olympiad.

Here is the corrected version of the problem:

[color=blue][b]Problem.[/b] Let M be the midpoint of the side BC of a triangle ABC. Prove that $\sin\measuredangle MAB>\frac{r}{2R}$, where r is the inradius and R is the circumradius of the triangle ABC.[/color]

[i]Solution.[/i] The Sine Law in triangle BAM yields

$\sin\measuredangle MAB=\sin\measuredangle ABM\cdot\frac{BM}{AM}$.

But < ABM = B and $BM=\frac{a}{2}$ (since the point M is the midpoint of the side BC of triangle ABC). Thus, we get

$\sin\measuredangle MAB=\sin B\cdot\frac{\left(\frac{a}{2}\right)}{AM}=\frac{a\sin B}{2\cdot AM}$.

Now, the area $\Delta$ of triangle ABC can be expressed in the form $\Delta=\frac12 ca\sin B$ on the one hand, while, on the other hand, it can be found from the formula $\Delta=\frac12\left(a+b+c\right)r$ (this is just the well-known fact that the area of a triangle equals the semiperimeter times the inradius). Thus, $\frac12 ca\sin B=\frac12\left(a+b+c\right)r$, what simplifies to $ca\sin B=\left(a+b+c\right)r$, and thus $a\sin B=\frac{\left(a+b+c\right)r}{c}$. Since a + b > c by the triangle inequality, we have $a\sin B=\frac{\left(a+b+c\right)r}{c}>\frac{\left(c+c\right)r}{c}=2r$. Thus,

$\sin\measuredangle MAB=\frac{a\sin B}{2\cdot AM}>\frac{2r}{2\cdot AM}=\frac{r}{AM}$.

Finally, we have AM < 2R; in fact, since the point A lies on the circumcircle of triangle ABC, and the point M, being the midpoint of the side BC, lies inside this circumcircle, the segment AM is contained in the circumcircle of triangle ABC, and thus it is shorter than the diameter 2R of this circumcircle. From AM < 2R, it follows that $\frac{r}{AM}>\frac{r}{2R}$; altogether,

$\sin\measuredangle MAB>\frac{r}{AM}>\frac{r}{2R}$,

and the problem is solved.

  darij

Thank you, darij.I'll do after as you asked.

[quote="Armo"]And what about this?
$\sin{AMB}>\frac{2r}{R}$.[/quote]

I think it suffices to prove this inequality:
In triangle $ABC$, $AM$ is the median, then
\[2R \ge \frac{bc}{AM} \ge 4r.\]
However I am not sure if it is really true.  :?

[quote="Armo"]And what about this?
$\sin{AMB}>\frac{2r}{R}$.[/quote]

I guess you mean $\sin\measuredangle AMB\geq\frac{2r}{R}$. And this can be proven as follows:

If H is the foot of the altitude of triangle ABC from the vertex A, then the triangle AHM is right-angled, so that $\sin\measuredangle AMH = \frac{AH}{AM}$. But we have either < AMH = < AMB, or < AMH = 180° - < AMB; in both of these cases, sin < AMH = sin < AMB, and thus we get $\sin\measuredangle AMB = \frac{AH}{AM}$. Hence, the inequality in question, $\sin\measuredangle AMB\geq\frac{2r}{R}$, rewrites as $\frac{AH}{AM}\geq\frac{2r}{R}$. This is clearly equivalent to $\frac{R}{2r}\geq\frac{AM}{AH}$. And this can be written in the form $\frac{R}{2r}\geq\frac{m_a}{h_a}$, where $m_a=AM$ is the median and $h_a=AH$ is the altitude of triangle ABC from the vertex A. In this form, the inequality was (ingeniously!) proven by Grobber in http://www.mathlinks.ro/Forum/viewtopic.php?t=2558 post #2.

[quote="yptsoi"]I think it suffices to prove this inequality:
In triangle $ABC$, $AM$ is the median, then
\[2R \ge \frac{bc}{AM} \ge 4r.\][/quote]

Yes, this is true, although I don't see why you need the left part of this inequality (actually, only the right part, $\frac{bc}{AM}\geq 4r$, is equivalent to $\sin\measuredangle AMB\geq\frac{2r}{R}$). The left part is almost trivial, by the way (we have $2R=\frac{bc}{h_a}$ and $h_a\leq AM$).

  Darij


function
real analysis
Ramsey Theory
real analysis unsolved

Does there exist  a function $ f$ such that $ f$ is not measurable and
$ f(\frac{x\plus{}y}{2})\leq \frac{1}{2}(f(x)\plus{}f(y))$.

Yes. Take one of the monsters that satisfies the equality $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ and add any convex function.

A lot of thanks!
Please give any  of the monsters that satisfies the equality:$ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$

You can't give them explicitly, as constructions rely on a Hamel basis for $ \bf R$ over $ \bf Q$.

Any time you see "not measurable" (in the Lebesgue sense), you should immediately accept that you're not going to see an explicit construction or example. Constructing stuff that's not Lebesgue-measurable requires the Axiom of Choice.

[quote="jmerry"]I'm pretty sure there's a model of analysis without the AC that has all sets measurable. It isn't really what we want mathematics to be; we just give up way too much.[/quote]
This doesn't contradict my statement. :) I can say anything I want about non-measurable sets in a model where they don't exist!

I used to be interested in that question of how much you can do without AC, but a lot of people whom I respect think that not having AC is such a loss that there's no point in even considering it, so I've thought about it less of late. An interesting thing I noticed recently was that the ergodic proof of Szemeredi's theorem as given by Furstenberg requires AC, but a couple of years ago Terence Tao published a paper which uses ergodic theory but not AC to prove that result.


function
calculus
derivative
logarithms
limit
topology
real analysis

Let $f: [a,b] \to R$ be a continuous function , $\exists f'(x)$ $\forall x \in (a,b)$ and $\exists c \in (a,b)$ such that $f'(c)=0$.
Prove that $\forall k \in R^{+}, \exists d \in (a,b)$ such that $f'(d)=k(f(d)-f(a))$

Assume additionally that $f$ is not constant; the statement is trivially true in that case.

WLOG, $f(a)=0$. The quotient $\frac{f'(x)}{f(x)}$ is the derivative of $g(x)=\log |f(x)|$ (on the open set where $f$ is nonzero), so it has the intermediate value property on each of its intervals of definition.
Let $P$ be a point in $(a,b)$ with $f'(P)=0$, $f(P)\neq 0$; if there is no such point, we have some $x$ with $f(x)=f'(x)=0$, and the statement is trivially true. Let $Q$ be the greatest $x$ less than $P$ with $f(Q)=0$

Now, consider $g$ on $(Q,P]$. The limit $\lim_{x\to Q^{+}}g(x)$ is $-\infty$, so $g'$ must be unbounded above. Since $g'(P)=0$ as well, $g'$ takes all positive values by its intermediate value property.


The set of all numbers x for which \[x+\sqrt{x^{2}+1}-\frac{1}{x+\sqrt{x^{2}+1}}\] is a rational number is the set of all:
$\textbf{(A)}\ \text{ integers } x \qquad
\textbf{(B)}\ \text{ rational } x \qquad
\textbf{(C)}\ \text{ real } x\qquad
\textbf{(D)}\ x \text{ for which } \sqrt{x^2+1} \text{ is rational} \qquad
\textbf{(E)}\ x \text{ for which } x+\sqrt{x^2+1} \text{ is rational }$

Put dollar signs ($\$$) before and after your LaTeX codes to make it look nicer ;)

[quote="Zootieroaz"]The set of all numbers $x$ for which $x+\sqrt{x^{2}+1}-\frac{1}{x+\sqrt{x^{2}+1}}$ is a rational number is the set of all:
a, integers $x$
b, rational $x$
c, $x$ for which $x+\sqrt{x^{2}+1}$ is rational
d, real $x$
e, $x$ for which $x+\sqrt{x^{2}+1}$ is irrational[/quote]LaTeXed :D

[quote="Zootieroaz"]The set of all numbers x for which $x+\sqrt{x^{2}+1}-\frac{1}{x+\sqrt{x^{2}+1}}$ is a rational number is the set of all:
a, integers x
b, rationa x
c, x for which $x+\sqrt{x^{2}+1}$ is rational
d, real x
e, x for which $x+\sqrt{x^{2}+1}$ is irrational[/quote]

[hide][size=200][b]C?[/b][/size][/hide]

[hide="hint"]rationalize the denominator of the second term and see what happens...

[hide="answer"]it simplfies to $2x$, so $B$[/hide]

[/hide]

Sorry, al, I think you made a tiny mistake when doing that. ;)

[hide="Answer"]$x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}$
Rationalizing the denominator of the fraction, we have:
$\frac{1}{x+\sqrt{x^2+1}}\left(\frac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}\right)=x-\sqrt{x^2+1}$
Therefore, we have $x+\sqrt{x^2+1}-(x-\sqrt{x^2+1})=2\sqrt{x^2+1}$.
So, our answer is $\boxed{D}$.[/hide]

[quote="JesusFreak197"]
$\frac{1}{x+\sqrt{x^2+1}}\left(\frac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}\right)=x-\sqrt{x^2+1}$
[/quote]

Sorry, but the denominator would be $(x + \sqrt{x^2 + 1}) (x - \sqrt{x^2 + 1}) = x^2 - (\sqrt{x^2 + 1})^2 = (x^2) - (x^2 +1) = -1$, not $+1$.

So that expression is equal to $\sqrt{x^2+1} - x$, and simplifies the way Altheman said it does.  The answer is, indeed, $\boxed{B}$.  (A calculator verifies this quickly.)

:oops: I guess that's what happens when you don't bother to fully calculate it out. :roll:

i know...i should write out the problems more...

Sorry - I'm a bit confused.  Doesn't the expression simplify to
\[ x\plus{}\sqrt{x^2\plus{}1}\minus{}((x \plus{} \sqrt{x^2 \plus{} 1}) (x \minus{} \sqrt{x^2 \plus{} 1}) \equal{} x^2 \minus{} (\sqrt{x^2 \plus{} 1})^2 \equal{} (x^2) \minus{} (x^2 \plus{}1) \equal{} \minus{}1)\equal{}2x
\]

If so, then wouldn't whether or not $ x$ is real depend on $ \sqrt{x^2\plus{}1}$ being real?  Did I do something wrong?

Thanks in advance!

The problem contains the implicit assumption that $ x$ is real; otherwise, the square root function is multi-valued and the problem is not well-defined.


geometry
incenter
circumcircle
IMO 1981
similar triangles
IMO

Three circles of equal radius have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle are collinear with the point $O$.

[quote="yetti"]Let A, B, C be the centers of the 3 congruent circles (A), (B), (C) intersecting at the point O. Since the point O is equidistant from the centers A, B, C, it is the circumcenter of the triangle $\triangle ABC$. Let a, b, c be the common external tangents of the circle pairs (B), (C); (C), (A); (A), (B), neither of them intersecting the remaining circle and let the lines a, b, c intersect at points $A' \equiv b \cap c,\ B' \equiv c \cap a,\ C' \equiv a \cap b$, forming a triangle $\triangle A'B'C'$. Let O' be the circumcenter of this new triangle. Since the common external tangent of 2 congruent circles is parallel to their center line, $a \equiv B'C' \parallel BC,\ b \equiv C'A' \parallel CA,\ c \equiv A'B' \parallel AB$. Thus the triangles $\triangle A'B'C' \sim \triangle ABC$ are centrally similar, having the corresponding sides parallel. The lines A'A, B'B, C'C connecting the corresponding vertices of the 2 triangles meet at their homothety center. But the lines A'A, B'B, C'C are the bisectors of the angles $\angle A', \angle B', \angle C'$ (and also of the angles $\angle A, \angle B, \angle C$), hence, the homothety center is the common incenter I of the triangles $\triangle A'B'C',\ \triangle ABC$. The circumcenters O', O are the corresponding points of these 2 centrally similar triangles, hence, the line O'O also passes through the homothety center I.[/quote]

Why do A'A, B'B, C'C meet at the incentre?

Because A', B', C' are equidistant from the sides so they all lie on the respective angle bisectors.


[hide="A Start"]
$ \theta = \frac{180^{\circ}}{7}$
$ \frac{1}{\sin \theta}= \frac{1}{\sin 2 \theta}+\frac{1}{\sin 3 \theta}$
$ \frac{1}{\sin \theta}= \frac{1}{2 \sin \theta \cos \theta}+\frac{1}{3 \sin \theta-4 \sin^{3}\theta}$
$ \sin \theta = \frac{2 \sin \theta \cos \theta+3 \sin \theta-4 \sin^{3}\theta }{(2 \sin \theta \cos \theta) (3 \sin \theta-4 \sin^{3}\theta )}$.
[/hide]

I think up to your second equation it was good, then it got a little crazy.

Instead, try writing it as $ \sin 2\theta\sin 3\theta=\sin\theta(\sin 2\theta+\sin 3\theta)$ and notice that $ \sin 3\theta = \sin 4\theta$.


algebra function domain calculus derivative conics parabola

how would the graph of square root(X^4-16X^2) look like?
where's the rel. extrema and intercepts?

$ f(x) \equal{} \sqrt{x^4\minus{}16x^2}$
intercepts:

$ \sqrt{x^4\minus{}16x^2} \equal{} 0$
$ x^4\minus{}16x^2 \equal{} 0$
$ x^2(x\minus{}4)(x\plus{}4)\equal{}0$

$ x\equal{} \pm 4, 0$

These x-intercepts are also going to be extrema because $ f(x)$'s domain is $ (\minus{}\infty, \minus{}4] \cup [0] \cup [4,\infty)$.  The other extrema can be found by differentiation:

$ y^2 \equal{} x^4\minus{}16x^2$
$ 2yy' \equal{} 4x^3\minus{}32x \equal{} 4x(x^2\minus{}8)$
So $ y'\equal{}0$ when $ x\equal{}\pm 2 \sqrt{2}$ (both outside our domain) and $ x\equal{}0$

Well, notice that $ \sqrt {x^4 \minus{} 16 x^2} \approx \sqrt {x^4 \minus{} 16 x^2 \plus{} 64} \equal{} x^2 \minus{} 8$ for $ |x| > > 0$

So you can draw that parabola in with a dotted line or whatever. Now, notice that the smaller $ |x|$ is, the lower $ f(x)$ is compared to $ x^2 \minus{} 8$, because the difference of 64 under the squareroot makes a bigger and bigger difference. Finally, when it gets to $ |x| \equal{} 4$, the graph, of $ f(x)$ hits $ 0$, and is undefined for $ |x| < 4$ (except at the origin).

The one additional piece of information that you might want is the points of inflection. These are the point at which the graph transitions from being locally hyperbola-like (shooting straight up near the intercept and flattening out) to being parabola like (being convex).

To do this, we find the second derivative and equate it to $ 0$, and solve:

$ (\sqrt {x^4 \minus{} 16x^2})' \equal{} \minus{} \frac {\left( \minus{} 32 x \plus{} 4 x^3\right)^2}{4 \left( \minus{} 16 x^2 \plus{} x^4\right)^{3/2}} \plus{} \frac { \minus{} 32 \plus{} 12 x^2}{2 \sqrt { \minus{} 16 x^2 \plus{} x^4}} \equal{} 0 \implies x \equal{} \pm 2 \sqrt {6}$

So, $ f$ is concave for $ 4 < |x| < 2 \sqrt {6}$, and convex for $ |x| > 2 \sqrt {6}$

The final graph is attached! I've circled the intercept and the point of inflection on the right half of the graph, and also included the parabola "guide", to show how the two are related, since you would probably plot the parabola first if you were drawing it by hand.


Suppose that $ A$ is a compact subset of $ \mathbb{R}$ and $ f: A \rightarrow \mathbb{R}$ is continuous. Prove that the set $ \{x \in A | 0 \leq f(x) \leq 1 \}$ is a compact subset of $ \mathbb{R}$.

I know that the set $ \{x \in A | 0 \leq f(x) \leq 1 \}$ is bounded because $ A$ is bounded, so I just need to prove that $ \{x \in A | 0 \leq f(x) \leq 1 \}$ is closed. I was thinking to take a sequence and show that it converges in the set. This is confusing me. Thanks for any help in advance.

[b]Solved it.  :lol: [/b]

$ (x_n\rightarrow L) \ \Longrightarrow \ (f(x_n)\rightarrow f(L))\ \Longrightarrow \ (0\leq f(L) \leq 1)\  \Longrightarrow \ (L\in A)$
 :cool:

It is closed because $ \{x\in A\}|0\leq f(x)\leq 1\}\equal{}f^{\minus{}1}([0,1])$ is the preimage of a closed set. Because $ f$ is continuous this preimage is closed.

Mark