geometry
geometric transformation
rotation

How many ways are there to form a triangle using the vertices of a regular 24-gon with all three angles greater than $ 30^{\circ}$?

[quote="aadil"]$ 24C3$[/quote]Wrong. Perhaps you should spend more time in the problem than doing random guessing

[quote="mathwizarddude"]Does anyone have a solution? I tried using the 24th roots of unity but failed... :( (keeping track all angles > 30 is not that easy)[/quote]

I am still trying to work it out...maybe halfway there. As the figure shows, we may view it another way; that is, finding the integer solutions of $ x\plus{}y\plus{}z\equal{}24, x,y,z\geq 5$. But this leads to an incorrect $ 55$. (The correct answer is $ 440$. )

[quote="JavaMan"][hide="Spoiler for jelly119"]
You have the right idea but you're not including rotations.

There's 1 equilateral triangle which has 8 rotations -> 8.
There are 4 isosceles (but not equilateral) triangles each with 24 rotations -> 4 * 24 = 96.
There are 14 scalene triangles each with 24 rotations -> 14 * 24 = 336 .

Which gives a total of 440.
[/hide][/quote]

I've seen someone who got the right answer this way:

$ 55\times \frac{24}{3}\equal{}440$

where the $ 55$ is just what I mentioned above. But I don't know what it means to multiply by $ \frac{24}{3}$.

For the one equilateral triangle which is part of the 55 you're just multiplying by 8. For the other 54 each unique triangle is represented 3 times as part of the 55, for example an isosceles triangle will have $ x\equal{}y\ne z$ which is essentially the same as if you swapped x and z, or y and z. Then once you have the number of unique triangles the multiplication by 24 is just taking into account the rotations.

Or in other words: a triangle is given by [i]specifying a first vertex[/i] and then finding the other vertices using your solution $ x \plus{} y \plus{} z \equal{} 24$.  This means each solution gives us actually 24 solutions.  However, we've counted each triangle three times now (once starting at each of its vertices), so we have to divide by 3.


calculus
algebra
polynomial
binomial theorem
real analysis
real analysis unsolved

I am working through Tom Apostol's calculus text, which, curiously, has absolutely no answers or guidance for any of the proofs. At this point I am proving some basic properties of polynomials. My questions involves this problem:

Let g(x) be a polynomial of degree m, such that

g(x)= sum:k=0...m: b(k) x^k

 and f(x) be a polynomial of degree n, such that

f(x) = sum: k=0...n: c(k) x^k

where m >= n. If g(x)=f(x) for m+1 distinct real values of x, prove that m=n, that each coefficient c(k)=b(k), and that f(x)=g(x) for all real x.

Now, I've already shown previously that given a polynomial h(x) of degree m, if h(x)=0 for m+1 distinct values of x, then every coefficient of h(x) is zero and h(x)=0 for all x. (Its interesting to note that this is really only a special case of what we are now trying to show).

Anyways, I set up a new m degree polynomial h(x) such that

h(x) = g(x) - f(x)

Now if g(x)=f(x) for m+1 distinct values, we have h(x)=0 for all x and then   of course g(x)=f(x) for all x.  Now, if we simply assume that m=n, it can easily be demonstrated by writing out the sigma nation that each c(k)=b(k). However, demonstrating directly that m=n  seems tricky. It seems that if we start out knowing that g(x)=f(x) independent of x, then we are saying that g(x) simply IS, f(x), implying that of course they have the same degree. Any suggestions?

Assume that $m > n$

The leading term of $h(x)$ is the leading term of $g(x)$. Since $h(x)=0$ for all $x$, its coefficients are all zero. Hence the leading term is zero, contradicting the hypothesis $\mathrm{deg}(g)=m$. Therefore $m \leq n$ and together with what we know we get that $m=n$.

The rest follows like you said.

Heres a polynomial problem from spivak.

Given a polynomial with the following properties:

1. Every coefficient is an integer.
2. The leading coefficient=1.

that if this polynomial is zero for some x, then the x is irrational unless it is an integer.

I simply extracted the last term x^n, and assumed that x was rational, and then subtracted the rest of the polynomial to the other side of the equation. From the last theorem we proved, if we can take the nth root of a number and get a rational number, this rational number is going to be a natural number, i.e., an integer. He however used what seemed to be a very different argument.....whats wrong with mine?

Theres also this:

Another problem asks to prove that if x=p + sqrt(q) where p and q are rational, then

x^m= a+b sqrt(q)  for some rational a and b if m is a natural number.

Spivak gives a very short inductive argument but I came up with something far more unecessarily complicated, but sufficiently different enough to warrant attention.

Starting with the binomial theorem, we can write

(p+sqrt(q))^m

In binomial form. We can rewrite the sum as two sums, one that sums through the even numbers and another that sums through the odds.

Now the sum with even numbers will be a rational number because these will cancel out the (1/2) power on q.

Now the sum with the odd numbers can be multiplied carefully by q/q, with q^1/2 going into the sum and leaving q^1/2   multiplying the sum from the outside. We see that the terms in the sum have now been shifted to be even and so it represents a rational number being multiplied by q^1/2, and I believe this demostrates the idea.

Nevermind about the first one. The theorem said that if we take the root of any NATURAL number it must be irrational unless that number was the square of another natural number.


This is from my instructor.
By relativity, energy and mass are equivalent, somebody said there are Weapon of MASS Destruction in Iraq, Q.E.D.
USA troops use/own weapon of minor/zero destruction only, maybe.  :rotfl:

Questions for Americans only:
1. What do you think the main reason for attacking Yugoslavia, Iraq etal?
2. Why each USA president started one or more wars against other contries?
3. Which country will be the next prey?

Taking without permition is called stealing; coming uninvited is called intrusion.
No matter how eloquent one is, he can't fool the world forever.

The American interventions in Yugoslavia had a clear and valid purpose.  Indeed, if all of our military actions were more like that, the world would be a better place -- thousands of Rawandans and Sudanese would not have been murdered by armed thugs.  The distinction between humanitarian and non-humanitarian missions is sometimes, but not always, vague -- in the case of the Serbian genocide, it was perfectly clear.

Every war has its own causes.  Sometimes they are good, sometimes not.  With respect to Iraq, it's worth noting that the majority of Americans no longer think it was a good idea; it's merely a shame it took them so long to realize it.  With respect to Serbia and Afghanistan and Iraq the first time around, the causes were all too clear.  I have no idea what your second question is meant to mean -- there have certainly been American presidents who did not start any wars.  (Technically, indeed, the United States hasn't gone to war since World War II and I believe we have only officially declared war in 7 instances since the founding of the nation.)

Finally, for your initial quip, it is wrong.  As many of us suspected to begin with, there were no weapons of mass destruction.  Physics can remain intact, George Walker Bush not withstanding.

3. Definitly not N. Korea....that is a big bomb that the US don't want to touch right now....not that N. Korea is strong but with one nuke at Japan's 30000 Americans stationed and the 3000 that are near the 38th parallel the anti-war feeling would be overwhelming....

I will say Syria....eventually Bush's plan is to secure the oil supply in middle east....as for starting a new front in Asian it is way too early....

[quote="JBL"]The American interventions in Yugoslavia had a clear and valid purpose.  Indeed, if all of our military actions were more like that, the world would be a better place -- thousands of Rawandans and Sudanese would not have been murdered by armed thugs.  The distinction between humanitarian and non-humanitarian missions is sometimes, but not always, vague -- in the case of the Serbian genocide, it was perfectly clear.[/quote]

I agree completely.  We need to draw the line between justified military actions by the U.S. such as those in Yugoslavia and completely unnecessary military actions such as the quagmire in Vietraq.

[quote="JBL"] (Technically, indeed, the United States hasn't gone to war since World War II and I believe we have only officially declared war in 7 instances since the founding of the nation.)[/quote]

Are you sure about this?  We declared war against Afghanistan and Iraq recently, and I'm pretty sure we declared war aginst Vietnam and North Korea during the cold war.  I might be wrong, though.

As for our next victim, who knows?  Maybe some other country will start producing "weapons of mass destruction" while North Korea makes more and more atomic bombs.

Gee, why did we have to invent the atomic bomb anyway?  It seems that many technological breakthroughs are accompanied by new, more powerful kinds of weapons.  One day, we humans will destroy ourselves.  I hope I never live to see that day.

Technically, we didn't declare war.  We just... sent some guys there and went hey they're just on vacation--wait you're attacking us hmmm we better start fighting but no, no war!

[quote="bubala"]One day, we humans will destroy ourselves.  I hope I never live to see that day.[/quote]

This thought has a very complicated structure, I think.  The kind of thing that banblink seems to like.

Probability is correct, by the way.  We've had a lot of things that get called wars, but not very many "official" wars (which require an act of Congress of a particular sort -- a declaration of war).  I had mistakenly thought that we went to war in Korea, but in fact we did not.  However, North and South Korea did officially declare war and have never revoked the declaration.  There is some quibbling as to whether Russia and Japan are at war -- the USSR declared war on Japan in WWII and never revoked it, so at issue is whether the disollution of the USSR left Russia with its war declaration valid or not.  (I tend to think not.)  The war in Afghanistan is an operation; the war in Iraq is (as a legalistic matter) as much a war as the war on drugs.  We use the word war far more loosely than its legalistic meaning applies.  Sometimes, those uses are literally accurate descriptions (Gulf Wars I and II), other times they are not (the War on Poverty).

[url=http://www.absoluteastronomy.com/encyclopedia/D/De/Declaration_of_war_by_the_United_States.htm]This website[/url] contains a brief history of U.S. wars.

Maybe you've forgot these, I won't.
1900, as a "gift" for the new century, Austria, British, France, German, Italy, Japan, Russia and USA(Belgium, Holand and Spain joined after the war to cut the melon) invaded China, they performed rape and massacre wherever they passed, when they entered the Forbidden City, they took everything they can carry, after the war,  'Xin Chou Treaty' or the 'Boxer Protocol of 1901' was signed, as a result, China payed 450,000,000 taels(=600,000,000 ounces)--plus the interest, totally 980,000,000 taels--of silver, foreign military forces can be stationed in the capital and coastal area.
Lucky enough, we survived, how many countries simply disappeared in the similar situations?
1999, USA bombed Chinese embassy in Yugoslavia as the perfect closing speech.

I have no idea what your point is.  I guess that's not true -- I know what your point is, but I don't see how it relates to what anyone else has said, and I certainly don't see anyone defending the actions you're talking about.

The bombing of an embassy was a stupid mistake, not an act of war.

[quote="probability1.01"]Well it's not fair to call these "attacks," as the US isn't really benefiting in the short term for either one. Not that I necessarily agree with them, but they're more like intervention, or attacks on very specific groups within these areas.[/quote]
[quote="JBL"]Technically, indeed, the United States hasn't gone to war since World War II and I believe we have only officially declared war in 7 instances since the founding of the nation.[/quote]
Now you know why.
[quote="JBL"]As many of us suspected to begin with, there were no weapons of mass destruction.[/quote]
To your surprise, I think Saddam did make lots of WMD, why USA can't find one? Because he hide them deeply--roughly 8000 miles--beneath the desert. How would the honorable American do such a drudgery to find them? They have more beneficial job to do, for example, SM. By the way, I 'm not surprised at the SM Gate, rape, robbery and massacre are the three pillars of any invasion.
[quote="JBL"]The bombing of an embassy was a stupid mistake, not an act of war.[/quote]
Five missiles hit the same taget at the same time by mistake? You can bet all your money on lotto now.

You obviously have a strange obsession with anti-US conspiracy theories.  8000 miles beneath the desert?  You are aware that 8000 miles is the diameter of the earth, right?

I also don't know why you seem to be arguing with me by claiming that the war in Iraq was a bad idea.  I was out on the streets of New York protesting the war -- I have no idea what you are trying to convince me of.

Yeah, if the United States really wanted to make war on China, we'd do it by blowing up an embassy in Serbia.  That makes lots of sense.  I suppose the next step of our invasion will be when we use global warming to drown Shanghai.  Give me a break.  5 bombs, [i]fired by the same attack group[/i] -- a single mistake.  More than 10 times as many Americans were killed by friendly fire in the first Gulf war than people were killed in the embassy bombing -- does that mean we're trying to invade ourself?

[quote="JBL"]I was out on the streets of New York protesting the war[/quote]
:thumbup: May I know why do you think you should do this? In what sense did you say the war in Iraq is a bad idea? For example, it hurt Iraqi, or it's not as beneficial as people once thought, or this task is too hard so that almost impossible to complete it, et al.
[quote="JBL"]More than 10 times as many Americans were killed by friendly fire in the first Gulf war than people were killed in the embassy bombing[/quote]
I believe much more than 10 times as many Americans were killed in the traffic accidents every year than that in 911, why do you make such a big fuss about 911?
What is "friendly fire"? I don't understand.

Friendly fire is when American/coalition troops shoot American/coalition soldiers (accidentally).  Accidents of war, except that in this case, Americans killing Americans.  The point is that mistakes happen in fighting and, because they involve guns and missles and bombs, people end up dead.  It shows that it is reasonable to think the embassy bombing could have been a mistake -- something that is not true of the World Trade Center attacks.

Why don't I support the war in Iraq?  Lots of reasons.  I think wars are bad ideas in general.  They are bloody and destructive and expensive and their outcomes are very difficult to control.  I don't trust the current American leadership.  And, it turns out, I was pretty much spot-on: we went to war on (many) false pretenses and at great cost in blood and treasure and the situation has turned out worse than it was when Saddam was in power.  And it's not clear how we're getting out or what state Iraq will be in when we do.

[quote="Buffalo"][quote="JBL"]As many of us suspected to begin with, there were no weapons of mass destruction.[/quote]
To your surprise, I think Saddam did make lots of WMD, why USA can't find one? Because he hide them deeply--roughly 8000 miles--beneath the desert. How would the honorable American do such a drudgery to find them? They have more beneficial job to do, for example, SM. By the way, I 'm not surprised at the SM Gate, rape, robbery and massacre are the three pillars of any invasion.  [/quote]

Yep, them WMDs are down with spongebob in his pineapple under the sea supporting homosexuality.

[quote="JBL"]You obviously have a strange obsession with anti-US conspiracy theories.  8000 miles beneath the desert?  You are aware that 8000 miles is the diameter of the earth, right?[/quote]

Heh. He's hiding them on the other side of the planet; we're just looking in the wrong place... of course! It all makes sense now. :rotfl:

Whom do you plan to vote for in your country's next presidential election?

Who was that addressed to, tokenadult?

I'm 5 months too young to vote next time around, but if I were old enough, I'd vote for the democratic candidate.  I don't know who either candidate's going to be yet, so I guess I'm partisan and close-minded, but at least I admit it :)  Not that I think Democrats are great candidates--it's just that the ultra-conservates that dominate the Republican party are 10000000000000000000000000 worse.

Anyhow, I'm really excited for the 2006 congressional elections.  I don't think a president's party has ever won seats in the sixth year.  In addition, if the Republicans keep up their abuse of power---from Terri Schiavo to filibusters, and Bush tries to do things that even his Republican friends don't like--i.e. Social security, and his Republican friends try to do things that conflict with his "culture of life" goals--i.e. stem cell research, the Republican party is done for in '06.  I'm just hoping that if the Dems do manage to retake Congress, they don't blow it like in 1994.  The RNC itself circulated a memo saying the Dubyas are going to lose 25 seats next election.

This is, of course, if there isn't another gigantic terrorist attack.  Then, they can say words that scare the populace like "terrorism" "September the 11th" "danger" "security is the essential roadblock to achieving the road map to peace" 100 times in a single speech and watch what happens.

Or they can try "life" "life" "life" "culture of life" "morality" "moral values" "our priorities is our faith."  That works, too.


If $ (a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^3$ the for any $ n$ positive integer prove that

$ (a\plus{}b\plus{}c)^{2n\plus{}1} \equal{} a^{2n\plus{}1} \plus{}b^{2n\plus{}1}\plus{}c^{2n\plus{}1}$

Good work :)

[quote="manlio"]If $ (a \plus{} b \plus{} c)^3 \equal{} a^3 \plus{} b^3 \plus{} c^3$ the for any $ n$ positive integer prove that

$ (a \plus{} b \plus{} c)^{2n \plus{} 1} \equal{} a^{2n \plus{} 1} \plus{} b^{2n \plus{} 1} \plus{} c^{2n \plus{} 1}$

Good work :)[/quote]

$ (a \plus{} b \plus{} c)^3 \equal{} a^3 \plus{} b^3 \plus{} c^3$ $ \iff$ $ (b \plus{} c)(a \plus{} b)(a \plus{} c) \equal{} 0$ Wlog say $ b \equal{} \minus{} a$ and so $ (a \plus{} b \plus{} c)^{2n \plus{} 1} \equal{} c^{2n \plus{} 1} \equal{} a^{2n \plus{} 1} \plus{} ( \minus{} a)^{2n \plus{} 1} \plus{} c^{2n \plus{} 1}$
Q.E.D.

[u]Edited [/u]: Sorry Sax for posting the same demo than yours 10 min later :)


calculus
integration
Euler
derivative
function
calculus computations

The first is the derivative of the Gamma function at $ k\plus{}1$.

I means if I can prove it by the knowledge which learned in grade one,if not,what should I learn in order to prove it?

[quote="zepanwucai"]Integral Problem:  ∫x^(k)e^(-x)lnxdx  (x from +Infinite to 0,k is a real number.)
I use vbs calculate some result,and guess the answer is k!*(c-∑1/i)  when k is + integer  (i fom 1 to k,c is Euler-Mascheroni constant),when k=0,it is c.
 Further,I guess when k is a real number,the answer is k!*(c+∑(ζ(i+3)*(-k)^i))     (i form 1 to +Infinite ，ζ(x) is Riemann function）
But I can't prove it...

Sum Problem:∑((1/m^2)*(∑1/i))    (m from 1 to +Infinite ，i from 1 to m)
I even can't guess out the answer...

I'm a fresh man,please help me,thanks for your answer![/quote]

Hello,for the integral problem you can learn some about Gamma Function!

$ \int_{0}^{\infty}e^{\minus{}r}$ln$ r$d$ r\equal{}\minus{}\gamma$,$ \gamma$is the Euler–Mascheroni constant .see.http://en.wikipedia.org/wiki/Euler_constant

$ \int_{0}^{\infty}re^{\minus{}r}$ln$ r$d$ r\equal{}1\minus{}\gamma$

$ \int_{0}^{\infty}r^{n}e^{\minus{}r}$ln$ r$d$ r\equal{}(n\minus{}1)!\plus{}n\int_{0}^{\infty}r^{n\minus{}1}e^{\minus{}r}$ln$ r$d$ r$,$ n\equal{}1,2,3,...$

Once you can prove out $ \int_{0}^{\infty}e^{\minus{}r}$ln$ r$d$ r\equal{}\minus{}\gamma$,the next will be very easy.

You can look for Arfken's Mathematical methods for physics


inequalities
logarithms
inequalities proposed

We have,
\[ \sum_{cyc}\sqrt{2\minus{}a^2}\le \sqrt{3(6\minus{}a^2\minus{}b^2\minus{}c^2)}.\]
Therefore it suffices to show that
\[ 6\minus{}a^2\minus{}b^2\minus{}c^2\le 3\Leftrightarrow a^2\plus{}b^2\plus{}c^2\ge 3,\]
which is obvious from AM-GM.

nayel, I'm verry sorry that I changed the problem.

kunny

This is the proof for my first post as follows.

Problem:

For $ 0<a\leq \sqrt{2},\ 0<b\leq \sqrt{2},\ 0<c\leq \sqrt{2}$ with $ abc\equal{}1$,

Prove that $ \sqrt{2\minus{}a^2}\plus{}\sqrt{2\minus{}b^2}\plus{}\sqrt{2\minus{}c^2}\leq 3$.

[quote="nayel"]We have,
\[ \sum_{cyc}\sqrt {2 \minus{} a^2}\le \sqrt {3(6 \minus{} a^2 \minus{} b^2 \minus{} c^2)}.\]
Therefore it suffices to show that
\[ 6 \minus{} a^2 \minus{} b^2 \minus{} c^2\le 3\Leftrightarrow a^2 \plus{} b^2 \plus{} c^2\ge 3,\]
which is obvious from AM-GM.[/quote]

The proof is correct.

kunny

[quote="kunny"]For positive numbers $ a,\ b, \ c$ with $ abc \equal{} 1$,
Prove that
\[ \sqrt [2009]{2 \minus{} a^{2009}} \plus{} \sqrt [2009] {2 \minus{} b^{2009}} \plus{} \sqrt [2009]{2 \minus{} c^{2009}}\leq 3.\]
[color=red]Last edited.[/color][/quote]

We have from Power-means inequality,
\[ \sum_{cyc}\sqrt[2009]{2 \minus{} a^{2009}}\le \sqrt[2009]{3^{2008}(6 \minus{} a^{2009} \minus{} b^{2009} \minus{} c^{2009})}.\]
Therefore it suffices to show that
\[ 6 \minus{} a^{2009} \minus{} b^{2009} \minus{} c^{2009}\le 3\Leftrightarrow a^{2009} \plus{} b^{2009} \plus{} c^{2009}\ge 3,\]
which is obvious from AM-GM.

Well, power-means is a corollary of Chebyshev's, Jensen's, Holder's and lots of other inequalities so if you're not comfortable with power-means you can think of it as any of the other ones.

Thanks. That's why I can't judge whether your proof is correct or not.
The judgement is left for inequaluty experts.

My solution is entirely different from you.

I would like to observe other's proof.

[quote="kunny"]Sorry, Power- means inequality is all Greek to me.[/quote]
It's just the following:
If $ \alpha \leq \beta$ are arbitrary real numbers, then for positiver real numbers $ x_i$, the following holds:
\[ \left(\frac{\sum_{i\equal{}1}^n x_i^{\alpha}}{n}\right)^\frac{1}{\alpha}\leq\left(\frac{\sum_{i\equal{}1}^n x_i^{\beta}}{n}\right)^\frac{1}{\beta}\mbox{.}\]
For $ \alpha\equal{}0$, we get (by definition) the geometric mean $ \sqrt[n]{\prod_{i\equal{}1}^n x_i}$.

Using this, you can easily generalize your inequality like this (for some $ n\geq1$):
\[ \sum_{cyc}\sqrt [n]{2\minus{}a^n}\leq 3\]

Thanks a lot.
Yes, the generalization is easy.

[quote="Potla"]We have (AM-GM) :
$ 2 \minus{} a^{2009} \plus{} \underbrace{1 \plus{} 1 \plus{} \cdots \plus{} 1}\geq 2009\cdot\sqrt [2009]{2 \minus{} a^{2009}}$
$ \implies 1 \plus{} \frac1{2009}\left(1 \minus{} a^{2009}\right)\geq \sqrt [2009]{2 \minus{} a^{2009}}$
So we reduce the problem to proving $ 3 \minus{} a^{2009} \minus{} b^{2009} \minus{} c^{2009}\leq 0$
Which is again, true from AM-GM since $ abc \equal{} 1$.
Equality occurs if and only if $ a \equal{} b \equal{} c \equal{} 1$. :)
I hope this is correct.
[hide="@kunny-san"] Thanks for your problems, sir. I think our solutions to this problem are the same.... :oops: [/hide][/quote]

Nice Solution, Potla! :lol:

I can't imagine such a solution.

[quote="Martin N."]you can easily generalize your inequality like this (for some $ n\geq1$):
\[ \sum_{cyc}\sqrt [n]{2 \minus{} a^n}\leq 3\]
[/quote]
\[ \boxed{\boxed{\sqrt [n]{2 \minus{} a^n}\leq 1 \minus{} \ln a}}\]


A is a set containing n elements. A subset P(A) is choosen at random . by returning the elements of P to A , another subset Q(A)  is taken. find the probablility that P,Q are mutually exclusive.

What exactly do you mean by a "random choice of a subset"?

I assumed the following problem was intended: two subsets are chosen uniformly at random from $ [n]$; compute the probability that they intersect.  In this case, it's some straightforward combinatorics: there are $ 3^n$ pairs of disjoint subsets and $ 4^n$ pairs of subsets, so the probability is $ 3^n/4^n$.


calculus
integration
trigonometry
algebra
partial fractions
calculus computations

Via partial fractions decomposition

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

Hence

$ \vspace{1pt}\int \frac {x \minus{} 1}{(x^{2} \plus{} 2x \plus{} 3)^{2}}dx$

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

$ \int \frac { \minus{} (2x \plus{} 2)}{\left( x^{2} \plus{} 2x \plus{} 3\right) ^{2}}\frac {x^{2} \plus{} 4x \plus{} 1}{ \minus{} 4\left( x \plus{} 1\right) }dx \minus{} \frac {1}{2\sqrt {2}}\tan ^{ \minus{} 1}\frac {1}{\sqrt {2}}\left( x \plus{} 1\right)$

$ \equal{} \frac {x^{2} \plus{} 4x \plus{} 1}{ \minus{} 4\left( x \plus{} 1\right) \left( x^{2} \plus{} 2x \plus{} 3\right) } \plus{} \int \frac {1}{4\left( x \plus{} 1\right) ^{2}}\,dx\allowbreak \minus{} \frac {1}{2\sqrt {2}}\tan ^{ \minus{} 1}\frac {1}{\sqrt {2}}\left( x \plus{} 1\right)$

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

$ \equal{} \minus{} \frac {x \plus{} 2}{2\left( x^{2} \plus{} 2x \plus{} 3\right) } \minus{} \frac {1}{2\sqrt {2}}\tan ^{ \minus{} 1}\frac {1}{\sqrt {2}}\left( x \plus{} 1\right)$ + C

$ \left(\minus{}\frac{1}{2}\cdot \frac{1}{x^2\plus{}2x\plus{}3}\right)'\equal{}\frac{x\minus{}1}{(x^2\plus{}2x\plus{}3)^2}\plus{}\frac{2}{(x^2\plus{}2x\plus{}3)^2}$.

The rest is left for you, because Japanese high school students don't learn $ \tan ^ {\minus{}1}x.$ :P


algebra
polynomial
function
geometry
circumcircle
national olympiad

[b][u]Ukrainian national team selection for IMO 2007[/u][/b]

[b][u]First test.[/u][/b](11.05.2007)

[b][u]1.[/u][/b] Numbers $ a$, $ b$, $ c$ are more then $ \frac{1}{\sqrt{6}}$  and $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1$ . Prove that
$ \frac{1\plus{}a^{2}}{\sqrt{2a^{2}\plus{}3ab\minus{}c^{2}}}\plus{}\frac{1\plus{}b^{2}}{\sqrt{2b^{2}\plus{}3bc\minus{}a^{2}}}\plus{}\frac{1\plus{}c^{2}}{\sqrt{2c^{2}\plus{}3ca\minus{}b^{2}}}\ge 2(a\plus{}b\plus{}c)$

[b][u]2.[/u][/b] $ ABCD$ is convex $ AD\parallel BC$ , $ AC\perp BD$ . $ M$ is interior point of  $ ABCD$ which is not a intersection of diagonals $ AC$  and  $ BD$ such that $ \angle AMB \equal{}\angle CMD \equal{}\frac{\pi}{2}$ .$ P$   is intersection of angel bisectors of $ \angle A$  and $ \angle C$ . $ Q$  is intersection of angel bisectors of $ \angle B$ and $ \angle D$ .Prove that $ \angle PMB \equal{}\angle QMC$ .

[b][u]3.[/u][/b] $ k$ and $ n$ are positive integers, $ k\plus{}1\le\sqrt{\frac{n\plus{}1}{ln(n\plus{}1)}}$.  Prove that there exist polynomial $ P(x)$ $ deg(P) \equal{} n$ with coefficients from $ {0,1,\minus{}1}$ such that $ (x\minus{}1)^{k}|P(x)$.

[b][u]Second test. (13.05.2007)[/u][/b]

[b][u]4.[/u][/b] Find all functions f $ Q\to Q$  such that $ f(x^{2}\plus{}y\plus{}f(xy)) \equal{} 3\plus{}(x\plus{}f(y)\minus{}2)f(x)$

[b][u]5.[/u][/b] $ AA_{3}$ and  $ BB_{3}$ are altitudes of  acute-angled  $ \triangle ABC$ . Points $ A_{1}$  and $ B_{1}$  are second points of intersection lines  $ AA_{3}$  and $ BB_{3}$  with circumcircle of  $ \triangle ABC$  respectively. $ A_{2}$ and $ B_{2}$  are points on $ BC$  and $ AC$ respectively. $ A_{1}A_{2}\parallel AC$ , $ B_{1}B_{2}\parallel BC$ . Point $ M$ is midpoint of $ A_{2}B_{2}$ . $ \angle DCA \equal{} x$ . Find $ \angle A_{3}MB_{3}$ .

[b][u]6.[/u][/b] Prime number $ p$  is called cubic if every integer could be represented as $ x^{3}\plus{}y^{3}\plus{}pz$ , where  $ x$,$ y$  ,$ z$   are integers. Find all cubic numbers.

[b][u]Third test. (15.05.2007)[/u][/b]

[b][u]7.[/u][/b] There are 25 people. Every two of them are use them some language to speak between. They use only one language even if they both know another one. Among every three of them there is one who speaking with two other on the same language. Prove that there exist one who speaking with 10 other on the same language.

[b][u]8.[/u][/b]  $ f(x)$ is polynomial with real coefficients. $ f(x) \equal{} x^{4}\plus{}a_{1}x^{3}\plus{}a_{2}x^{2}\plus{}a_{1}x^{1}\plus{}a_{0}$ . $ M$  is local maximum and $ m$  is absolute minimum. Prove that $ \frac{3}{10}(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}< M\minus{}m < 3(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}$

[b][u]9.[/u][/b] Points $ A_{1}$ ,$ B_{1}$ ,$ C_{1}$  are on the sides $ BC$, $ AC$ ,$ AB$ of $ \triangle ABC$ respectively. Circumcircles of $ \triangle AB_{1}C_{1}$ ,$ \triangle A_{1}BC_{1}$ and $ \triangle A_{1}B_{1}C$  intersects circumcircle of $ \triangle ABC$ in points  $ A_{2}$, $ B_{2}$ and $ C_{2}$   respectively ($ A_{2}\ne A$ ,$ B_{2}\ne B$ ,$ C_{2}\ne C$ ). Points  $ A_{3}$ ,$ B_{3}$ ,$ C_{3}$ are symmetric to  $ A_{1}$ ,$ B_{1}$ ,$ C_{1}$relatively to midpoints of $ BC$, $ AC$ ,$ AB$ respectively. Prove that $ \triangle A_{2}B_{2}C_{2}$  is similar to $ \triangle A_{3}B_{3}C_{3}$ .

[b][u]Fourth test. (17.05.2007)[/u][/b]

[b][u]10.[/u][/b] Find all positive integers $ n$  such that acute-angled $ \triangle ABC$ with  $ \angle BAC <\frac{\pi}{4}$  could be divided into $ n$ quadrilateral. Every quadrilateral is inscribed in circle and radiuses of circles are in geometric progression.

[b][u]11.[/u][/b] There are $ n\ge 2$ lamps in a row. Each of them can be in one of two conditions ‘on’ or ‘off’. Each second lamps together change their conditions according to next rules: if lamp $ L_{i}$ and her neighbours (or neighbour if $ i \equal{} 1$ or $ i \equal{} n$) are in the same condition then she becomes ‘off’ in other case she becomes ‘on’. In the started position  first left lamp is ‘on’ other lamps are ‘off’.
1) Prove that there are infinitely many positive integers $ n$ for which after some time all the lamps become ‘off’.
2) Prove that there are infinitely many positive integers $ n$ for which in any moment of time there exist a lamp in condition ‘on’.

[b][u]12.[/u][/b] Prove that there are infinitely many positive integers $ n$ for which all the prime divisors of $ n^{2}\plus{}n\plus{}1$ is not more then $ \sqrt{n}$.

Wow, thank you very much. Can you post links discuss?  :wink:

[b][u]1.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=155354
[b][u]2.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=167298
[b][u]3.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=167308
[b][u]4.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=167309
[b][u]5.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=167303
[b][u]6.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=167310
                       http://www.mathlinks.ro/Forum/viewtopic.php?t=90520
[b][u]7.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=167311
[b][u]8.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?p=930204&sid=971e4c9ab1f34ec1e4b8a0b45e159899#930204
[b][u]9.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?p=875036#p875036
[b][u]10.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=167313
[b][u]11.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?p=874978#p874978
[b][u]12.[/u][/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=167314


Let there be distinct integers X and Y
Let k also be an integer

Is it possible to get infinitely many solutions for X and Y for any value k when
X^2 = Y^2 + k

[hide]It seems to be going along the lines of implying that for each pair X and Y there can be only one value of k[/hide]

[hide]
$ x^2 \minus{} y^2 \equal{} k$
$ (x \minus{} y)(x \plus{} y) \equal{} k$

Clearly, $ k$ cannot have infinite factorizations, so the answer is "no".
[/hide]

Nope.

[hide]
Factor
(x-y)(x+y)=k
k has at most k factors (finite) trivial step
Therefore, (x-y)(x+y) combinations are finite
Therefore, the number of x and y is also finite.
[/hide]

[quote="max1546"]It seems to be going along the lines of implying that for each pair X and Y there can be only one value of k[/quote]  This is totally irrelevant, actually: for a given pair $ (x, y)$, there is a unique $ k$ such that $ x \equal{} y \plus{} k$, but this says nothing at all about how many pairs $ (x, y)$ there are for each $ k$.

I find it interesting that both preceding posters went immediately for factorizations; I personally would have noted that there are only finitely many pairs of [i]consecutive[/i] squares whose differences are less than or equal to $ k$, and concluded from there.  (This uses the rather important $ f(x) \equal{} x^2$ is monotone for nonnegative $ x$; if we replaced the squaring function with something oscillatory (say $ g(x) \equal{} x \cos (\pi x)$) this argument wouldn't work.)


Quadratic equation $ x^2\plus{}ax\plus{}b\plus{}1\equal{}0$ have 2 positive integer roots, for integers $ a,b$. Show that $ a^2\plus{}b^2$ is not a prime.

As I started writing it I got ashamed, so I hid it
[hide]For $ x,y$ the two integer (not necessarily positive, just not null) roots, Viete's relations yield $ x\plus{}y\equal{}\minus{}a$ and $ xy \equal{} b\plus{}1$, hence $ a^2 \plus{} b^2 \equal{} (x\plus{}y)^2 \plus{} (xy\minus{}1)^2 \equal{} (x^2\plus{}1)(y^2\plus{}1)$, where each factor is larger than $ 1$.[/hide]

I LOVE USING VIETE`S THEOREM.THAT IS VERY BEATIFULL

Here is my solution.

Let the two integer roots be $ p$ and $ q$. Then by Vieta's formulas,

$ a\equal{}\minus{}p\minus{}q$
$ b\plus{}1\equal{}pq$

Hence $ a^{2}\plus{}b^{2}\equal{}(p\plus{}q)^{2}\plus{}(pq\minus{}1)^{2}\equal{}p^{2}\plus{}q^{2}\plus{}p^{2}q^{2} \plus{} 1$.

Factoring yields that,

$ a^{2}\plus{}b^{2}\equal{}(p^{2}\plus{}1)(q^{2}\plus{}1)$

Since $ p$ and $ q$ are positive integers, $ a^{2}\plus{}b^{2}$ is divisible by two integers greater than $ 1$ and hence is not prime.


trigonometry
algebra
difference of squares
special factorizations

I'm not quite sure how to approach this problem,
but I have to prove the following equation true.

sin^4(x) - cos^4(x) = -cos(2x)

I haven't proven it yet, so I can't treat it like an equation.
I just have to somehow manipulate one side to look like the other.

[hide="Better..."]
Factoring using difference of squares, \[ \sin^4x-\cos^4x=(\sin^2x-\cos^2x)(\sin^2x+\cos^2x).\] Using the identity $ \sin^2x+\cos^2x=1,$ we have \begin{align*}(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)&=\sin^2x-\cos^2x\\&=-(\cos^2x-\sin^2x)..\end{align*} Here, we use the reverse double angle formula for cosine, to get \[ -(\cos^2x-\sin^2x)=-\cos2x.\] [/hide]

That's exactly Dr Sonnhard Graubner's solution, have you even read it?


ARML MATHCOUNTS AMC USA(J)MO USAMO trigonometry AIME

i think i'm going. when is it? the coordinator person said it was in early june. just wondering if anyone had any updates (i heard about at mathcounts state when that ARML guy asked me if i wanted to participate in ARML.). who else is going?

Whos the ARML guy?  Dr. Richert? (sp-sorry if wrong)

[quote="cognos599"]how can you get into ARML do you get invited or do you need to take a test?[/quote]
hehe i hope not, cuz i wasn't invited.. o_O

i don't mean to sound retarded, but how exactly does ARML work?

and BTW, i do not know the guy that asked me. if you remember from states, he was like the freakishly tall guy (like 7 foot) that was one of the judges in the countdown. he said he was the ARML coordinator.

and how do you make the teams? i thought there wasn't a qualifying exam?? do u think i could make it to any of the teams?

... lol you wanna know the coaches? Dr. Chi, that other guy... and isn't my dad just there to help or something? XD

uh..you can go too..they dont really care who goes right? (as long as you pay)

yeah, but there's really no point in going if i can't make the B team.(lol it makes it sound like tryouts for a sport)

it would be a good experience still. how do they pick the teams? do they pick them at the competition, cuz that guy basically told me to like drive to Terre Haute and hop on a bus that goes to ohio or something... i don't really remember all that much. i don't think he mentioned anything about practices, though

probalby all usamo qualifers on A team... (tahts 11)..they probalby take like 4 more seniors maybe? (to give them a shot)..B team..probalby high schools kids+fanatic, mathcrazed, and some other kid in 8th grade, sine people are so sensative aobut names..I dont know..if you just look at AMC scores and AIME, I probalby can make B team..but I dont think they would just do that..I can usually get 3-5 on ARML individuals..

Uh..why dont you go ask mathcrazed-his dad's a coach right?

okay. [color=red][size=150]MATHCRAZED[/size][/color]: do you think i could make it to any team (hopefully a better one)?

lol i could only do 7 on this year's AIME 1.

[quote="mathcrazed"]yuqing, if you can do 5 or 6, but you only get 2, you're not gonna make the white team... i think that's the higher right? Work on your accuracy, and just being able to do them doesn't say anything.  After all, we could all do like 9-11 on the AIME, yet we only got like 1-5's... so ya.[/quote]

I thought blue was higher (the way that he said it made white seem lower) but I bet you are going to make the higer one of the two.  But it depends on the peolple going, for example, I dont think Udayan can go.  He just called me today tellin gme this.

Ok and I could not even do 9-11 on the AIME...
I could only do 1,2,3,6,7,9 perfectly
4 was easy, just misjudging something-but thats wanst a careless mistake
5..now I know how to do it..
11..cant find square roots by hand..

So actually I could only do 6-7..

are the SAT II's the subject tests?

EDIT: 15^2 th post :)

[quote="stupidkid"][quote="Art of Owna"](for example some very good people fofrm carmel cannot go because SAT II is during that weekend. and its once a year)[/quote]
Nice, where did you get that from? You can take SAT II's many times a year.

Dude ARML teams aren't totally based on merit. They like seniors and people they know, e.g. Luke last year.[/quote]

[quote="Art of Owna"][quote="indianamath"]i'm prob not. Shawn you think i could make the second team?[/quote]

I probalby woundnt make the second team, and I dont think that they would really take any person less than 8th grade on the second team or above..because they arent reallky familar with you-even if you are better than the people in high school-like me (probalby all of you middle schoolers are better than me at math-and Im not trying to be modest) ..If I made USAMO I probalby still wouldnt be on the first team..because Indiana is somehow a good math state (as they probalby will have around 15 USAMO qualifers and the remaining spots they will probalby give out to seniors and stuff)[/quote]

I know its seniors and people they know-I said that earlier
Also, someone told me (aq 10th grader) they could only take the SAT II for like Euro History (or soemthign like that) that time in the summer.

[quote="stupidkid"][quote="Art of Owna"](for example some very good people fofrm carmel cannot go because SAT II is during that weekend. and its once a year)[/quote]
Nice, where did you get that from? You can take SAT II's many times a year.

Dude ARML teams aren't totally based on merit. They like seniors and people they know, e.g. Luke last year.[/quote]

Why woundt they be based on merit-do they wana lose?

ANd is luke some dumb kid that was put on the team just because he was a senior and killed the rest of the team?

guess what respect and generousity isnt worth losing.  if you win, you will get respect no matter what.

you cant go we have school it said on the math club email

[quote="stupidkid"][quote="mathcrazed"]Luke doesn't suck... he's better than alot of the people we have now.  Its only because he was a senior, so they put him on the higher team, more merit for college. lol[/quote]
I never said he sucked. But Lin Nan is definitely better than Luke.[/quote]

luke xie?  hmm my parents know that kids parents..went to MIT if thats who you are talking aobut-he was a senior last year-won some siemens award (like you) that helped him but I think they perceivced that he was good at math because he got like 5th or something at mathcounts state (according to arnold) but he was only 5th a school so that couldnve been a 1 time wonder....but that 5th place liked helped him get started..

[quote="mathcrazed"]huh... no? luke isn't stupid, but he was pulled up by tiffany and charlie... in the research time. lmao, tiffany ko and charlie tam ... that's so rigged :D...[/quote]
Haha, Lyndon, that's so true, I just didn't want to mention it since you might tell your dad or something...

huh, my dad already knew that.  He wanted luke to get some good stuff, so XD

hey sorry to bring this topic back...but speaking of research, what happened to the professor that mentored the two SC reg. finalist groups?

and stupidkid, if you are ever gonna be on, may i ask why you choose harvard math over mit math?  just curious

thanks


Let $f:\mathbb{R}\to\mathbb{R}$ be a convex function.

a) Prove that $f$ is continous;
b) Prove that there exists an unique function $g:[0,\infty)\to\mathbb{R}$ such that for all $x\geq 0$ we have \[ f(x+g(x)) = f(g(x))  - g(x) . \]

We see that $g(x)$ is a solution to the equation $f(x+y)=f(y)-y$. Therefore we are to prove it has unique solution. Indeed, for $x>0$ we have $\frac{f(x+y)-f(y)}{x}=-\frac{y}{x}$, where $LHS$ is incresing and $RHS$ is decresing. Moreover, $\lim_{y\to -\infty} RHS=+\infty$ and $\lim_{y\to +\infty} RHS=-\infty$.
That's all.


This is a generalization of French TST 2006 #6:
 If $ S \equal{} \{1,2,\ldots,4n\}$ where $ n \in \mathbb{N}$, and $ S$ is partitioned into four subsets $ A,B,C,D$ each having cardinality $ n$, prove that there exist 4 elements , one from each of the four parts such that the sum of some two of the elements is equal to the sum of the other two.
 This is like the Sidon Set problem as opposed to the French TST problem which was a sum-free problem.
 Ashwath

Is this that hard? I didn't think it was that hard - but apparently I was wrong :). Anyway, does anyone have any solutions?
The key idea seems to be the same as the 4n problem - you just need to worry about differences instead of sums. I'll work on this tomorrow and hopefully will get back with a solution. In the meantime - try it for what it's worth.


function
limit
calculus
calculus computations

[i]Firstoff, I must confess, I have no idea why I even care about this, or really even what exactly I'm trying to do with it. That being said:[/i]

I want to characterize the summation:
$F(x)=\sum_{k=1}^{\infty}\frac{x^{p_{k}}}{p_{k}!}$
where $p_{k}$ is the prime series.

In particular, I'm wondering about the strength of its growth. Strength of growth, I guess, is a sort of shifty way of putting it, so I'll make my own definition:

For two strictly increasing functions $f(x)$ and $g(x)$, $f(x)$ has stronger growth than $g(x)$ if for all positive numbers $n$, there exists another positive number $k$ such that for all $x>k$, $f(x)>n \cdot g(x)$.

So $e^{x}$ would be stronger than $x^{7}$, for example.
[b]
How would the growth strength of $F(x)$ compare to exponential functions? What type of function most closely matches $F(x)$ in growth strength?[/b]

All help, ideas, hints, insights, and beyond are appreciated.

~ tZf

Basically F(x) grows like $e^{x}$. More precisely, $F(x)=o(e^{x})$ (that would be the standard notation for your notion of $e^{x}$ growing stronger than $F(x)$; if you are unfamiliar with it, read Kent Merryfield's sticky about big O and little o in this subforum) but $e^{ax}=o(F(x))$ for all $a<1$ as $x\to+\infty$. Both facts have almost nothing to do with primes and are true for any sequence $\{p_{k}\}$ that is of uniform density $0$ (i.e., such that for every $\varepsilon>0$, there exists $N$ such that among any $N$ consecutive integers there are at most $\varepsilon N$ elements of the sequence) and doesn't have "big gaps" (more precisely, it should satisfy $\lim_{k\to\infty}\frac{p_{k+1}}{p_{k}}=1$). One can try to get tighter bounds for $F(x)$ but it would be equivalent to proving something about the error term in the prime number theorem, which is not easy at all.

As to the proofs, the idea is that the main part of the value of $e^{x}=\sum_{n\ge 0}\frac{x^{n}}{n!}$ comes from about $\sqrt x$ terms near $n=x$ that give almost equal contribution. Clearly, $F(x)$ has only a small portion of these terms in it, whence $F(x)=o(e^{x})$. On the other hand, just one term $x^{p}/p!$ is enough to beat $e^{ax}$ in $[p(1-\delta),p(1+\delta)]$ if $a<1$, $\delta=\delta(a)>0$ is small enough, and $p$ is large enough.


calculus
integration
geometry
3D geometry
tetrahedron
function
conics

i cant figgure why if im asked to find a triple integral of the region
bounded by $x+y+z=a;x=0;y=0;z=0$
i have to have the bounds of integration like this
$\int_{x=0}^{a}\int_{y=0}^{a-x}\int_{z=0}^{a-x-y}$ ?
i can understand only the last one.
:(

Imagine that you are trying to list all triples $(x,y,z)$ such that $x,y,z\ge 0$ and $x+y+z\le a$. What are the possible values of $x$ given these restrictions? It can be anywhere between $0$ and $a$. Once you have picked some value for $x$, your choice for $y$ is limited to the interval $[0,a-x]$. Finally, $z$ must be in the interval $[0,a-x-y]$.

In practice one usually sets the limits backwards, starting with $0\le z\le a-x-y$ (which you understand). Now that we are done with $z$, we must describe all pairs $(x,y)$ for which there is at least one $z$ such that $(x,y,z)$ lies within the region (a tetrahedron). In geometric terms, integration with respect to $x$ and $y$ occurs over the projection of the tetrahedron onto the $xy$-plane. This projection is a triangle bounded by the line $x+y=a$ in the first quadrant of the plane. Hence the limits for $y$ are $0\le y\le a-x$. Finally, we integrate w.r.t $x$ over the $x$-projection of this triangle, which is the segment $[0,a]$. At each step the number of variables is reduced by one, because a variable of integration has no meaning outside of its integral.

If you still don't understand why $\int_{x=0}^{a-y-z}\int_{y=0}^{a-x-z}\int_{z=0}^{a-x-y}f(x,y,z)\,dz\,dy\,dx$ is a wrong answer, try to evaluate this integral with some simple function, e.g. $f\equiv 1$.

ok great,ive understood , thank you!
you talk about a tetrahedron,from where i'm reading they also plot
the region and its a tetrahedron alright but...what if the equations
get more complicated,im getting blind :|
also can one figgure by pure intuition how's it going to look like ?
i mean x+y+z=a is eq of a plane its okay its a simple thing but
for more complicated ones ?

I would suggest just trying more complicated problems and post ones you get stuck on.

In general, if the region is not easy to picture like the aforementioned tetrahedron I make two sketches, one with three dimensions and one that is the projection onto a plane (usually xy).

For example: Find the bounds to evaluate a triple integral under the graph of $z=x^{2}$ above the region enclosed by the parabola $y=6-x^{2}$ and the line $y=x$ on the $xy$-plane.  In this case the 3D graph is quite boring while the majority of the work is done in the $xy$-plane.

Your book probably has better examples.

(And regarding your intuition question, do you mean can you figure out the bounds without visualizing the region?  Or can you figure out what the region will look like by looking at an equation?)

The general advice on figuring out limits for multiple integrals is to sketch the region you're integrating over. In more than two dimensions, you have to compromise with projections and the like; we can't draw three-dimensional objects on a two-dimensional piece of paper nicely.

This is from a quiz I gave last semester. In the following integral, change the order of integration to "$dx\,dy\,dz$".
$\int_{-1}^{1}\int_{-x}^{1}\int_{0}^{x+y}f(x,y,z)\,dz\,dy\,dx$
 :whistling:


linear algebra
matrix
algebra
polynomial
inequalities
linear algebra solved

$A\in M_n(C)$

$C(A)=\{B\in M_n(C); AB=BA\}$

Suppose that for any $B \in C(A)$ $|det(A+B)|\geq |det(B+I_n)|$

¨Prove that $A-I_n$ is nilpotent

We have to show that the characteristic polynomial of $A$ is $(x-1)^n$. If we take $B=-xI_n$ first, and then $B=-xA$, we get the inequalities $|P_A(x)|\le,\ge|(x-1)^n|$, so $|P_A(x)|=|(x-1)^n|$, and from here we easily get what we want.


analytic geometry
graphing lines
slope
Pythagorean Theorem
geometry

Find the equations of the tangents to the circle $ x^{2} \plus{} y^{2} \minus{} 2x \plus{} 4y \plus{} 1 \equal{} 0$ passing through point $ (1,2)$

[hide]$ x^2 \plus{} x^2 \minus{} 2x \plus{} 4y \plus{} 1 \equal{} 0 n\implies (x \minus{} 1)^2 \plus{} (y \plus{} 2)^2 \equal{} 4$
Now, we have the general form of a circle, so we know its center and radius.
The center is $ (1, \minus{} 2)$, and the radius is $ 2$.
We want the 2 tangents of this circle passing through $ (1,2)$, which is just 4 units above the center of our circle.
Since the radius is 2, we get, by the pythagorean theorem, that the length of the tangent segment is $ \sqrt {4^2 \minus{} 2^2} \equal{} 2\sqrt {3}$ and that both tangents make a 30 degree angle to the line segment connecting the center of the circle and the point $ (1,2)$: They both make a 30 degree to the vertical.
(we actually don't need that the angle is 30 degrees. If we did not know the angle, we could still find the cotangent using the right triangle formed by the tangent, radius, and thevertical line.)
In order to find the slope, we need rise over run, which is simply $ \pm \frac {\text{adjacent}}{\text{opposite}} \equal{} \pm \cot{30^{\circ}} \equal{} \pm\sqrt {3}$
Using point slope form, we get that
$ (y \minus{} 2) \equal{} \pm\sqrt {3}(x \minus{} 1) \implies y \minus{} 2 \equal{} \pm\sqrt {3}x\pm\sqrt {3} \implies \boxed{y \equal{} (\pm \sqrt {3})x \pm\sqrt {3} \plus{} 2}$[/hide]


function
ceiling function
number theory open
number theory

The generating function of this sequence is
\[ \sum_{n\equal{}0}^{\infty} a_n x^n \equal{} \frac{\sqrt{1\minus{}2x\plus{}3x^2}\plus{}x\minus{}1}{x^2}\]
implying that
\[ a_n \equal{} \sum_{k\equal{}\lceil (n\plus{}2)/2\rceil}^{n\plus{}2} {\frac12\choose k} {k\choose n\plus{}2\minus{}k} (\minus{}2)^{2k\minus{}n\minus{}2} 3^{n\plus{}2\minus{}k} \equal{} \frac{1}{2^{n\plus{}1}}\sum_{k\equal{}\lceil (n\plus{}2)/2\rceil}^{n\plus{}2} (\minus{}1)^{n\plus{}k\minus{}1} C_{k\minus{}1} {k\choose n\plus{}2\minus{}k}  3^{n\plus{}2\minus{}k},\]
where $ C_k$ are Catalan's numbers.


Let $A_{n}$ denote the set of all $2n$-digit numbers whose decimal representation conisists of exactly $n$ $1$'s and $n$ $0$'s (leading $0$'s NOT permitted). Show that there must be at least one number in the set $A_{n}$ that is evenly divisible by $n$.

Say, for $n = 3$, the set $A_{3}$ has $\frac{5!}{3! \times 2!}= 10$ numbers.
$A_{3}={100011, 100110, 101100, 111000, 110100, 110010, 110001, 101010, 101001, 100101}$ of which at least on be divisible by $3$. Here all of them are!!! That's a different story.

Prove it for any postitive number $n$ irrespective of how big or small $n$ is.

no attempts to this one?

May be the mods can move it to appropriate forum.


ratio
geometric sequence
arithmetic sequence
algebra proposed
algebra

[quote="moldovan"]The real numbers $ a_1,a_2,...,a_n (n \ge 3)$ form an arithmetic progression. There exists a permutation $ a_{i_1},a_{i_2},...,a_{i_n}$ of $ a_1,a_2,...,a_n$ which is a geometric progression. Find the numbers $ a_1,a_2,...,a_n$ if they are all different and the largest of them is equal to $ 1996$.[/quote]

Quite nice :)

[hide="My solution"]
Let $ q$ the common ratio of the geometric sequence ($ |q|\neq 1$)
Let $ r$ the common difference of the arithmetic sequence ($ r\neq 0$)

If $ q>0$, we can order the numbers : $ u, qu, q^2u, ...$ and this order is the same as the arithmetic one (changing $ r$ in $ \minus{}r$ if necessary) : $ u, u\plus{}r, u\plus{}2r, ...$
So $ qu\minus{}u\equal{}q^2u\minus{}qu$ and so $ (q\minus{}1)\equal{}q(q\minus{}1)$ ($ u\neq 0$ else all terms are equal) and so $ q\equal{}1$, which is impossible.

So $ q<0$ and the order of the geometric serie is $ (u, q^2u, q^4u, ..., q^5u, q^3u, qu)$
If $ n\geq 4$ we have again $ q^2u\minus{}u\equal{}qu\minus{}q^3u$ and so $ q\equal{}1$ or $ q\equal{}\minus{}1$ which is impossible (all terms different)

So $ n\equal{}3$ and the series are $ (u, q^2u, qu)$ and $ (u, u\plus{}r, u\plus{}2r)$

From there : $ q^2\minus{}1\equal{}q\minus{}q^2$ and so $ q\equal{}\minus{}\frac{1}{2}$ and then $ r\equal{}\minus{}\frac{3u}{4}$ and the sequence is $ (u,\frac{u}{4},\minus{}\frac{u}{2})$

So two solutions : $ u\equal{}1996$ and $ u\equal{}\minus{}3992$ and the two sequences :

$ \{a_k\}\equal{}(\minus{}998, 499, 1996)$ and $ \{a_{i_k}\}\equal{}(1996, \minus{}998, 499)$

$ \{a_k\}\equal{}(\minus{}3992, \minus{}998, 1996)$ and $ \{a_{i_k}\}\equal{}(\minus{}3992, 1996, \minus{}998)$

[/hide]


The techniques for solving linear equations and linear inequalities are similar, yet different. Explain and give an example of both a linear equation and a linear inequality that demonstrates this difference.

[quote="-oo-"]... = please say what I've got to write, I don't feel like thinkin'[/quote]

Thats exactly how the question presents itself...

I would say a different technique is multiplying by a negative constant. For the inequality the direction of the inequality changes whereas for an equation you keep equality.

[quote="Flash8o"]Thats exactly how the question presents itself...[/quote]

What -oo- meant was, do your own homework.


hi there everyone

im currently studying teaching, in my first year and my major is maths (which means i'll be able to teach maths from year 8 - 12 here in australia). Does anyone have any suggestions on how to make it more fun for students, because unfortunately they aren't all like us who do maths because we love it (or are good at it). Any help from teachers in what they have found useful or from students who have ideas about how they wish their maths classes were would be invaluable. Thanks!
-Klasey :)

Less exercise homework, but a few problems with a high level of difficulty.  Remember, less homework. :D :P

Masoud Zargar

ahh yes, but the only way you learn maths is by doing maths (well thats what was in my unit outline for my maths unit this semester at uni  :P ) I don't believe in mountains of homework, but I think doing a bit is well worth it and you learn more. I always found the  teacher writing notes on the board, then anotated examples, then working through problems worked really well for me, but I know some of my friends didn't like that. I guess it just depends on your teaching, stict, easygoing or firm but fair. Anyway thanks for your input. Anymore would be extremely helpful!
-Klasey

Don't teach from the textbook. They're REALLY boring and teach you almost nothing. If the school will pay for it, then you could try using the AoPS books, which are excellent and basically cover Algebra I through PreCalc, with added depth.

I suggest working with students one on one.  It's well worth the effort; always be available for students to talk to if they don't understand something.

I think if you enjoy maths yourself and find what you are doing interesting, then  that comes across when you teach it, and the students notice it.

I don't think there is a simple way to make maths interesting to everyone, but if you continue to learn about it, and find the beauty in different things (or atleast understand "the point" of doing some particular maths problem, which in some cases I've found doesn't happen until years after you first learn it), then you can communicate that.

A lot of maths is just taught in high school as a method, without trying to give the students any idea of why they are doing it that way, etc., and without communicating the meaning.  I've taught a pre-university course a couple of times and found the students often simply enjoy FINALLY understanding WHY they had to do something a particular way, or what a particular symbol actually MEANS, they are very happy.  You don't need to make it fun, just meaningful.

Sorry to go on for ever, but I forgot probably the most important thing:

I think a lot of the people who really hate doing maths, hate it because they have really low confidence in their ability, and so just feel stupid whenever they look at it. They end up in a cycle where instead of taking in what the teacher says they're too busy thinking "I'm so dumb. I can't possible learn this; everyone else understands; I'm never going to be able to do this", etc, etc. so then of course they can't understand and absolutely hate it.

There's a really big emotional factor to learning and teaching maths.

I think you have to be fairly sensitive to this, and try to make it so people don't feel bad asking "stupid" questions in the classroom, and also try to give people the confidence that they can understand whatever it is you are trying to teach. A lot of the time, this just requires that you are confident in them, and tell them so.  Then they have a chance of actually enjoying it - people enjoy being good at something.


Give four positive real number $a,b,c,d$ satisfying: $a+b+c+d=4$.
Find the minimum of the expression
$P = a^2+b^2+c^2+d^2+a^5b^5c^5d^5$

by the (n-1)EV principle the minimum is attained when three variables are equal sowe need to find the minimum of $3a^2+(4-a)^2+a^{15} (4-a)^5$

Oh, i just post it few minutes ago. Can you show me the way to apply (n-1)EV. Is it useful in this case:
$P = \sqrt{a^2+b^2+c^2+d^2}+a^5b^5c^5d^5$

yes !

$a+b+c+d=const a^2+b^2+c^2+d^2=const$ so the minimum of abcd happens when   three variables are equal!

[quote="Anh Cuong"]Oh, the power of $(n-1)EV$ really makes me admire, if i don't miss anything then this is Vasc's product. I want to check it power for some another inequalities, would you play this game with me:
Give $a,b,c,d > 0$. Prove that:
$\frac{3(a^4+b^4+c^4+d^4)}{4abcd} \geq 1 + \frac{3(a^2+b^2+c^2+d^2)}{ab+ac+ad+bc+bd+cd}$  :)[/quote]
$\frac{3(a^4+b^4+c^4+d^4)}{4abcd} \geq 1 + \frac{3(a^2+b^2+c^2+d^2)}{ab+ac+ad+bc+bd+cd}$ is wider than
$\frac{\sum (a^2-b^2)^2}{4abcd}\geq \frac{\sum (a-b)^2}{\sum ab}$ which is obvious  :)  :lol:


Find all functions$ f:R\to{R}$  such that
$f(f^{2}(x)+f(y))=xf(x)+y.$

$f$ is clearly bijective, so if we take $x$ s.t. $f(x)=0$ we get $f(f(x))=x,\ \forall x$. Since $f(f^2(0)+f(x))=x$, we also find $f(0)=0$. Now put $y=0$ to get $f(f^2(x))=xf(x),\ \forall x$.

If we replace $x$ with $f(x)$ and $y$ with $f(y)$, we find $f(x^2+y)=xf(x)+y$. If $y=0$ we obtain $f(x^2)=xf(x),\ \forall x\Rightarrow f(x^2+y)=f(x^2)+y,\ \forall x,y$, so $a\ge 0\Rightarrow f(a+x)=f(a)+x,\ \forall x\in\mathbb R$.

That's a pretty strong condition, and $f(x)=x,\ \forall x$ follows immediately: take $x=-a$ to get $f(a)=a,\ \forall a\ge 0$, and from $f(x^2)=xf(x)$ we get the other half: $f(x)=x,\ \forall x<0$, and we are done.

I think you missed something [b]grobber[/b], because $f(x)=-x$ is solution too.

Yes, it' a pretty big mistake and I don't know if it can be repaired in just a line. I'lkl think aboiut it. I don't have that much time at the moment..

The first paragraph is fine. We continue in a very similar manner: $f(x^2+y)=xf(x)+f(y)=f(x^2)+f(y)$. From here we get $f(a+x)=f(a)+f(x),\ \forall a\ge 0,\ \forall x$. It's easy to get the additivity of $f$ from here.

We now have the following properties: $f$ is additive, $f(f(x))=x$, and $f(x^2)=xf(x)$. Apply this last property to $x+y$ and use the additivity. You'll get $f(xy)=\frac{xf(y)+yf(x)}2$. Now, in this equality, make $y\to y+1$ and subtract them from one another. We get $f(x)=\frac{xf(1)+f(x)}2\Rightarrow f(x)=f(1)x$. From $f(f(x))=x$ we get the two solutions $f(x)=x$ and $f(x)=-x$.

[quote="Armo"][quote]I guess Armo has his own solution for this problem.
I think I saw this problem earlier. Do you have supposition about source, Armo?[/quote]
For the first part of your post:do you want me to submit my solution?
Actually this problem was given in some olympiad in  Armenia(but I am sure that it is not the original thread).Anyway, I'll try to find the original thread.[/quote]
Of course, if you have own original solution, please, send it! :)

Okay,here is it.
[b]Step 1[/b]Take $x=0$ and we get that f is sjurective.Take $f(x)=0$ and we obtain $f(f(x))=x,  \forall x.$
[b]Step 2[/b]Take $x=y=0$ then $f(f^{2}(0)+f(0))=0$.Now take $x=f(0), y=f^{2}(0)+f(0)$ and we get $f(0)=0$.
[b]Step 3[/b]Take $y=0$ and we have $f(f^{2}(x))=xf(x)$ and now put $x=f(t)$,so $f(t^{2})=tf(t), \forall t.$
[b]Step 4[/b]Now from step3 we have $ f(f^{2}(x))=xf(x)\Rightarrow   f^{2}(x)=f(xf(x))=f(f(x^{2}))=x^{2}.$
[b]Step 5[/b]We get that  $f^{2}(x)=x^{2}$, using this fact lets square our original equation and we get $f(x)f(y)=xy$, take $y=1$ and we get $f(x)=\frac{x}{f(1)}$.So the answer is:$f(x)=x$ if$ f(1)=1$ and $f(x)=-x$ if $f(1)=-1$.


Prove that the equation $m(m+1)(m+2)(m+3)=n(n+1)^{2}(n+2)^{3}(n+3)^{4}$ doesn't have any solution in positive integers.

[quote=Akzhol]the equation $m(m+1)(m+2)(m+3)=n(n+1)^{2}(n+2)^{3}(n+3)^{4}$ could be written in the form of $(n+3)^{4}=m(m+1)(m+2)(m+3)((n-1)!)((n)!)((n+1)!)((n+2)!)$ since
$(m+1)^{4}<m(m+1)(m+2)(m+3)<(m+3)^{4}$ and using same reason for $((n-1)!)((n)!)((n+1)!)((n+2)!)$  one  can findthat there is no solution :[/quote]
Can you write your solution obviously and  more fully. I cannot understand ending your solution.


vector
calculus
derivative
calculus computations

I decided to try to calculate the behavior of an ideal elastic material and reasoned that it could be modeled as the limit when a grid of point masses connected by ideal springs has the masses arbitrarily close together.

Anyway, I came up with the equation
$ \frac{\partial^2 F}{\partial^2 t}\equal{}\frac{\partial^2 F}{\partial^2 x}\plus{}\frac{\partial^2 F}{\partial^2 y}\plus{} ...$
where t is time, x, y, .. are components of the vector representing the intial positon of a point, and F is the current position of that point

Is this equation even right? If so, how do I solve it?
For example, I was trying to solve the initial value problem of a two dimensional sheet in three dimensional space

$ F(0,<x,y,0>) \equal{} <x,y,0>$

$ \frac{\partial F(0,<x,y,0>)}{\partial t} \equal{} <0,0,e^{\minus{}(x^2\plus{}y^2)}>$

but I couldn't figure out how to go about solving it

With that second derivative in time, this is the [i]wave equation[/i]. Or it would be if you had the notation correct. That should be $ \frac {\partial^2F}{\partial t^2},$ not $ \frac {\partial^2F}{\partial ^2t}$ as the notation for second derivative. And the object you're trying to compute on the right hand side is the [i]Laplacian[/i] in the space variables.

It looks like you arbitrarily set a bunch of physical constants equal to $ 1,$ and hence got a wave equation with $ 1$ as the speed of wave propagation. More generally, you'd get

$ \frac {\partial^2F}{\partial t^2} \equal{} c^2\nabla^2F,$

where $ \nabla^2F,$ also written as $ \Delta F,$ is the Laplacian of $ F$ and $ c$ is the speed of wave propagation.

For the start you have, Fourier transform methods would be in order. I've got a chapter or two in my course notes. I would also suggest using the key words "wave equation" to see what you can look up.

A further suggestion: try doing this in one space dimension before you work up to more dimensions than that. That is, the equation is $ \frac{\partial^2u}{\partial t^2}\equal{}c^2\frac{\partial^2u}{\partial x^2}.$ Here, $ u$ is the lateral displacement of an elastic string. You could put boundary conditions on this: $ u(0,t)\equal{}u(L,t)\equal{}0$ would be the condition of having a string of length $ L$ with both ends clamped, which is a description of a musical string instrument.

ok thanks. I figured there was probably a bunch of terminology I need to learn

By the way, couldn't you just remove the propagation speed coefficient by making an appropriate substitution for t?

You could. Mathematically, it's the same. That amounts to measuring time in some [i]ad hoc[/i] unit whose purpose is to make that speed 1. Whether you do that or not is a practical question and depends one the purposes to which you're putting this.

Ok, I think I figured out how to solve the one and two dimensional wave equations, corresponding physically to infinite strings or sheets.

But what do you do if you have a finite shape? I can't figure out how to solve this

[quote="Kent Merryfield"] ooo I've got a chapter or two in my course notes.[/quote]
Do you teach a class on differential equations?  I'm trying to find Authors and Professors with interest on solving ODEs and PDEs. Their is a Calculus-level Compiler that makes solving math problems as simple as possible. If interested, I'll start by solving any Implicit equation(s). Next, any Non-linear, then Constrained, Iniitial Value Problems, Boundary Value Problems, etc.

Please let me know if interested.

Thanks,
Phil Brubaker
Mathematician '66
http://www.digitalCalculus.com

Ok solve this then

$ \frac {\partial^2F}{\partial t^2} \equal{} \nabla^2F(x,y) for x^2 \plus{} y^2 < 1$

$ 2\frac {\partial^2F}{\partial t^2} \equal{} \nabla^2F(x,y) \plus{} c for x^2 \plus{} y^2 \equal{} 1$

Sorry, this problem looks like a homework problem. If not, where in industry does it appear?

Phil

Well it is intended to be a model for the motion of a perfectly elastic ball in two dimensions


function
induction
algebra unsolved
algebra

Determine all functions $f$  being defined on the set of positive integers with integer function values satisfying the following condition:

For all pairs $(x,y)$ of coprime positive integers we have:

\[f(x + y) = f(x + 1) + f(y + 1)\]

Did you mean:[b]Imf [/b]is subset of [b]Z[/b]-the set of  integers?

All $4m+r$ positive integer may be a sum of 2 coprime  positive integers :
$4m = 2m-1 + 2m+1 \ , \  4m+1 = 2m + 2m+1 \ , \  4m+2 = 2m-1 + 2m+3 $  , $  4m+3  = 2m+1 + 2m+2  $
Finally , by induction I found : $\forall m\geq 1 \ , \  f(2m)=(m-1) f(4) \ , \  f(2m+1)=(m-1) f(4) + f(3)$
 :cool:


vector
algebra
polynomial
linear algebra
linear algebra unsolved

Hi everyone,

I was reviewing solved problems of vector spaces, and even though this one seems very trivial I can't understand it:

[img]http://i50.photobucket.com/albums/f320/germoncho/problembasis.jpg[/img]

I have a specific question:

1) Why the dimension of P_n(t) = n + 1? I don't understand how that is so evident.

Many thanks for any help you could give me.

Kind regards,
Germán[/img]

1, t, ..., t^n forms a basis, thus it's of dimension n+1

Another way to see this is that if $ p(t)$ is in the span of those polynomials then $ p(\minus{}1) \equal{} 0$, but there are plenty of polynomials in $ {\bf P}_n(t)$ (like, say, any  nonzero constant polynomial) for which this is not true.

The idea behind Soarer's solution (that all bases of a finite-dimensional vector space have the same size) is much more important than the idea behind mine (which exploits a very peculiar property of the set of polynomials given).


22 points are equally spaced around a circle and numbered consecutively. A line is drawn through the center of the circle and passes between points 7 and 8. What is the sum of the other two consecutive numbers that the line passes between?

Those two numbers are correct. Can you explain how you arrived at that answer?

Well 7 and 18 are centrally symmetric and also 8 and 19 if we put the center of the circle as the center of symmetry. So the line between 7 and 8 will also pass through their symmetric points on the other sides.


function
algebra
functional equation
algebra unsolved

For a fixed $ n\in \mathbb{N}$, find all strictly increasing or strictly decreasing functions $ f : \mathbb{R} \rightarrow \mathbb{R}$ such that
$ f(x \plus{} f(y)) \equal{} f(x) \plus{} y^n$ for all reals $ x, y$.

Easy but nice!
My solution:

[hide]For $ y \equal{} 0$ we get: $ f(x \plus{} f(0)) \equal{} f(x) \plus{} 0 \equal{} f(x)$
And because f is monotonic we get: $ x \plus{} f(0) \equal{} x \Leftrightarrow f(0) \equal{} 0$

For $ x \equal{} 0$ we get: $ f(f(y)) \equal{} f(0) \plus{} y^n \equal{} y^n$ for all reals $ y$ (1)

For $ y \equal{} f(x)$ we get:
$ f(x \plus{} x^n) \equal{} f(x) \plus{} f^n(x) \Leftrightarrow f(f(x \plus{} x^n)) \equal{} f(f(x) \plus{} f^n(x)) \Leftrightarrow (x \plus{} x^n)^n \equal{} f(f(x)) \plus{} f^n(f(x)) \equal{} x \plus{} x^{n^2}$
for all reals $ x$

From the last equation for $ x \equal{} 1$ we get $ 2^n \equal{} 2 \Leftrightarrow n \equal{} 1$
And then (1) becomes: $ f(f(x)) \equal{} x$ for all reals $ x$ (2)

Now because f is monotonic, from (2) we get $ f(x) \equal{} f^{ \minus{} 1}(x)$ for all reals $ x$

For $ y \equal{} f^{ \minus{} 1}(y)$ (edit after 5 years: it should be $y \leftarrow {} f^{ \minus{} 1}(y)$) at the first equation we get: $ f(x \plus{} y) \equal{} f(x) \plus{} f^{ \minus{} 1}(y)$ and here for $ x \equal{} \minus{} y$ we get $ f( \minus{} y) \equal{} \minus{} f^{ \minus{} 1}(y) \equal{} \minus{} f(y)$ (because $ f(0) \equal{} 0$) and $ f$ is odd

Now we take 2 cases:

1) $ f$: strictly increasing
Suppose that there exists a $ k$ such that $ f(k) > k$, then $ f(f(k)) > f(k) > k$ which is a contradiction

Suppose that there exists a $ k$ such that $ f(k) < k$, then $ f(f(k)) < f(k) < k$, again a contradiction

So we have $ f(x) \equal{} x$ for all reals $ x$

2) $ f$: strictly decreasing
Suppose that there exists a $ k$ such that $ f(k) > \minus{} k$, then $ f(f(k)) < f( \minus{} k) \equal{} \minus{} f(k) \Leftrightarrow k < \minus{} f(k) \Leftrightarrow f(k) < \minus{} k$ which is a contradiction

Suppose that there exists a $ k$ such that $ f(k) < \minus{} k$, then $ f(f(k)) > f( \minus{} k) \equal{} \minus{} f(k) \Leftrightarrow k > \minus{} f(k) \Leftrightarrow f(k) > \minus{} k$ which is again a contradiction

So we have $ f(x) \equal{} \minus{} x$ for all reals $ x$

Hence the solutions of the functional equation are:
$ f(x) \equal{} x$ for all reals $ x$, and
$ f(x) \equal{} \minus{} x$ for all reals $ x$.[/hide]


ΑΝ $ a,b,c$ είναι γωνίες ώστε $ a \plus{} b \plus{} c \equal{}$π και $ x,y,z$ είναι τυχαίοι πραγματικοί αριθμοί ώστε $ xyz > 0$ να δείξετε ότι $ xcosa \plus{} ycosb \plus{} zcosc\leq\frac {yz}{2x} \plus{} \frac {zx}{2y} \plus{} \frac {xy}{2z}$ :)

Γιώργο δεν θα έλεγα το θέμα εύκολο για κάποιον που δεν έχει ξανα δεί παρόμοιο ...Τέλος πάντων, να μια λύση ( δεν ξέρω αν υπάρχει άλλη )

Γενικά ισχύει η ανισότητα :

$ x^2 \plus{} y^2 \plus{} z^2 \ge 2yzcosa \plus{} 2zxcosb \plus{} 2xycosc$  [color=darkred][b](1), [/b][/color]όταν είναι a+b+c = π.

Πραγματικά αν c = π - (a+b) έχουμε την ακόλουθη ταυτότητα

$ x^2 \plus{} y^2 \plus{} z^2 \minus{} 2(yzcosa \plus{} zxcosb \plus{} xycosc) \equal{}$

      $ (z \minus{} (xcosb \plus{} ycosa))^2 \plus{} (xsinb \minus{} ysina)^2 \ge0$

Αν θέσουμε στην ανισότητα [color=darkred][b](1)[/b][/color] στη θέση των x,y,z

$ \sqrt {\frac {yz}{x}}$,  $ \sqrt {\frac {xz}{y}}$,   $ \sqrt {\frac {yx}{z}}$

τότε προκύπτει η ζητούμενη... :)

Να γράψω και μια λύση που μπορούμε να κάνουμε αν x,y,z > 0 και a,b,c γωνίες τριγώνου.

Πολ/ζοντας με xyz > 0 η προς απόδειξη ανισότητα είναι ισοδύναμη με

$ (xy)^2\plus{}(yz)^2\plus{}(zx)^2\ge 2x^2yz(cosa)\plus{}2y^2xz(cosb)\plus{}2z^2xy(cosc)$

Θεωρούμε Τρίγωνο ΑΒΓ με μέτρα γωνιών a,b,c και έστω οι πλευρές που βρίσκονται απέναντι από τις κορυφές Α,Β,Γ ότι έχουν μήκη x,y,z αντίστοιχα.Από τον νόμο των συνημιτόνων παίρνουμε τις σχέσεις :

$ 2yz(cosa) \equal{} y^2\plus{}z^2\minus{}x^2$
$ 2yz(cosb) \equal{} x^2\plus{}z^2\minus{}y^2$
$ 2yz(cosc) \equal{} x^2\plus{}y^2\minus{}z^2$

όπου ανικαθιστώντας αυτές τις σχέσεις στο δεύτερο μέλος της (1) έχουμε :

$ (xy)^2\plus{}(yz)^2\plus{}(zx)^2\ge x^2(y^2\plus{}z^2\minus{}x^2)\plus{}y^2(x^2\plus{}z^2\minus{}y^2)\plus{}z^2(x^2\plus{}y^2\minus{}z^2)$

η οποία μετά απο πράξεις δίνει την $ x^4\plus{}y^4\plus{}z^4\ge (xy)^2\plus{}(yz)^2\plus{}(zx)^2$

που ισχύει  :)

Vangelismak, έχεις νομίζω λάθος. Δεν είναι δεδομένο ότι για τα x,y,z ισχύει η τριγωνική ανισότητα.

Nαι, η δεύτερη απόδειξη είναι στη περίπτωση που μπορεί να κατασκευαστεί τέτοιο τρίγωνο ...( το x,y,z > 0 από μόνο του δεν αρκεί ) Thanks για τη διευκρίνηση


Determine the smallest positive integer $ n$ such that there exists positive integers $ a_1,a_2,\cdots,a_n$, that smaller than or equal to $ 15$ and are not necessarily distinct, such that the last four digits of the sum,

\[ a_1!\plus{}a_2!\plus{}\cdots\plus{}a_n!\]

Is $ 2001$.

[hide="One value for one a_i"]notice that if $ a_i \ge 1 \Rightarrow a_i$ is even then at least one $ a_i\equal{}1$[/hide]
[hide="Hint 2"]There is no value of $ a_i$ with four last digits $ 2000$[/hide]
[hide="Solution"]$ n>2$ an example with $ \boxed{n\equal{}3}$ is $ a_1\equal{}1$, $ a_2\equal{}13$ and $ a_3\equal{}14$, with the following last four digits: $ 1\plus{}800\plus{}1200\equal{}2001$[/hide]


Let $a$ and $b$ be distinct real numbers. Prove that there exist integers $m$ and $n$ such that $am+bn<0$, $bm+an>0$.

[hide]Clearly m and n must be different as well.
Without loss of generality, a>b.  Then, (m,n)=(-x,x) will satisfy as long as x is a positive integer.  If the opposite was true, then (m,n)=(x,-x).
Proof:  -ax+bx<0 or x(b-a)<0.  Since x is a positive integer and b-a is negative, the product is indeed less than 0.
-bx+ax>0 or x(a-b)>0.  Since x is a positive integer and a-b is positive, the product is indeed greater than 0.  This same concept can be used when b>a.[/hide]
Am I underestimating this problem?  I think I am.

[hide="(Edited) Solution"]
$am < -bn$
$bm > -an$

Let $r = \frac{a}{b}, q = - \frac{n}{m}$, where $r \neq 1$.

We have two cases.

[b]Case:[/b]  $r > 1$.  Let $b$ be negative and $m$ positive.  Then the problem becomes

$r > q$
$1 < rq$

Simply choosing a $q$ such that $1 \le q < r$ is sufficient, and such a $q$ must exist.

[b]Case:[/b]  $r < 1$.  Let $b, m$ both be positive.  Then the problem becomes

$r < q$
$1 > rq$

Analogously, choosing a $q$ such that $r < q \le 1$ is sufficient, and such a $q$ must exist.  In the case that $r < 0$ we can choose any positive $q$.

[b]Remark:[/b]  Yes, you can just choose $q = 1$ for both cases, as dengmi did, but this fuller solution is nice because it confirms my initial problem-solver's instinct that the problem was either weaker than or equivalent to showing that the rationals are dense in the reals.

It's useful to examine the thought process that led to this solution.  The obvious observation to make of this problem is that the problem is homogeneous with respect to $a, b$ and to $m, n$, so that we only need to concern ourselves with ratios - one of which is rational, and one of which is real.  Once we rewrite suitably (being [b]very careful[/b] about signs!  :blush: ), the solution is immediate.
[/hide]


Round 6 Matchups

HashBrownie    vs.   white_horse_king88

kool_dudy    vs.    mlee06

And there is the winners bracket matchup (which is really round 5 for them)  going on:

Dune    vs.    mat_88

I believe that I sent the winners bracket problems out already... if you haven't got them pm me.  I am going to send out the loser's bracket problems as soon as possible which may be tomorrow or later.

kool_dudy is the only person to send answers in.  He got a 2.  Everyone else needs to get there answers in!

I need answers from:

white_horse_king88

mlee06

Dune

Please hurry up or else I will have to make you forfeit.

Also, I just got some gmail invites so the winner of this tournament gets a gmail invitation!

[quote="corpulus"]Remind me of my stupidity.  What is gmail?[/quote]

You can find out [url=http://www.gmail.com]here[/url].

:blush:  :P  :maybe:  ;)  :lol:  :cool:  :D

I think that only one more has submitted solutions for today.  I ll check this out later.  I have homework now. :(

what's happening in the tournament? no news has been posted for a while... just curious

Wait, wrong forum. Can I delete this post?

Just a question: what tournament is this, and is there going to be another one?

I doubt there will be another tournament like this until winter holidays or maybe when someone else has time to make one.

This tournament is similar to your regular sports playoff except you get two chances to lose before you're eliminated.  We each have five questions to answer within a time limit against an opponent.  Whoever solves the most correctly advances.

[quote="corpulus"]Remind me of my stupidity.  What is gmail?[/quote]

do you know Google?
I'm not asking you to search 'gmail' using Google, in fact gmail is 'A Google approach to email.'
http://gmail.google.com/

Sorry about the long delay.  We will now get back in action.

Results from this round:

Mat_88 def Dune  (4-forfeit)

Hashbrownie def white_horse_king88 (4-forfeit)

mlee06   def  kooldudy (3-2)


representation
number theory
Divisibility
IMO Shortlist

Let $a,b,n$ be positive integers, $b > 1$ and $b^n-1\mid a.$ Show that the representation of the number $a$ in the base $b$ contains at least $n$ digits different from zero.

Please check if my proof is correct...

We need to prove that for any integer $b>1$, $n$, and $c>0$, $c(b^n-1)$ has at least $n$ non-zero digits when written in base $b$. Let $c=\sum_{i=0}^{m} x_i b^i$, where $0\leq x_i<b$. We can use induction on $m$.

If $m = 0$, then $c < b$, and $a=c(b^n-1)=cb^n-c=\overline{(c-1)(b-1)...(b-1)(b-c)}$ in base $b$, giving $n$ non-zero digits for $c=1$ and $n+1$ non-zero digits for $c>1$.

Now assume $c(b^n-1)$ have at least $n$ non-zero digits for $m=k$, we want to prove the statement is true for $m=k+1$. Let $c=x_{k+1}b^{k+1}+c^'$, where $c^{'}=\sum_{i=0}^{k} x_i b^i$. Let $c^{'}(b^n-1)=\sum_{i=0}^{k+n} y_i b^i$, where $0\leq y_i<b$., then $a=x_{k+1}b^{k+n+1}+y_{k+n}b^{k+n}+...+y_{k+2}b^{k+2}+y_{k+1}b^{k+1}-x_{k+1}b^{k+1}+y_kb^k+...+y_0$.

If $x_{k+1}=y_{k+1}$, $a$ would have the same number of zeros in base $b$ as $c^{'}(b^n-1)$, because one digit would be changed to zero and a non-zero digit would be added, without changing anything else. If $x_{k+1}<y_{k+1}$, we will get one more. If $x_{k+1}>y_{k+1}$, then we can replace $y_{k+1}-x_{k+1}$ by $y_{k+1}-x_{k+1}+b$ (a non-zero number) and work on coefficients of larger exponents to get all coefficients in range. This would mean decrease the first non-zero coefficient met when going left from exponent $k+1$ by 1 and change all zero coefficients met on the way to $b-1$. Obviously at most 1 coefficient is changed from non-zero to zero, but again, the added non-zero coefficient for exponent $k+n+1$ would offset this decrease. In conclusion, $c(b^n-1)$ would have at least as much non-zero digits as $c^{'}(b^n-1)$.

We allow nubers in base b to have leading zeroes until their numer of digits in multiple of n. The value of a number with leading zeroes in the same as without zeroes. All operations will be in b base.
  Assume for the purpose of contradiction that there exist a multiple of $b^{n}-1$ with less than n nonzero digits.
   It is easy to prove that  if $g(t)$ is the number of nonzero digits of $t$ then $g(s+t)\le g(s)+g(t)$. And if $g(t),g(s)>0$ then $g(s+t)>0$

   Given $s=a_{1}a_{2}...a_{kn}$ a number multiple of $b^{n}-1$ we split his digits into blocks of n digits $c_{1}=a_{1}....a_{n}$,
$c_{2}=a_{n+1}...a_{2n},...,c_{k}=a_{nk-n+1}....a_{n}$ . It is easy to see that $s'=c_{1}+c_{2}+c_{3}+...+c_{k}\equiv s (\mod b^{n}-1)$ and $s'$ is strictly less than $s$ if s have more than n digits.

Cotinuing this transformation we will get smaler and smaler s's until they have less than or exactly n digits.  Now it is obvious that a number $s$ with $\le n$ digits and $<$ nonzero digits cannot be a multiple of $b^{n}-1$ because $0<s<b^{n}-1$. We got the desired contradction.

Is this a valid proof:
Let $a=k(b^n-1)$, where $k$ has $x$ nonzero digits and $y$ zeros in base $b$.  Also assume that $k$ has no trailing zeros because, if it did, we can simply consider the same number without the zeros and then multiply them in later for no net effect.  Then, if you consider the subtraction of $k(b^n)$ by $k$ in base $b$, we have that $k(b^n)$ is $k$ followed by $n$ zeros, and $k$ is simply $k$.  Thus, when we subtract, we carry over a $1$, and all the zeros but the last one become a $(b-1)$.  All the $(n-(x+y))$ $(b-1)$'s will stay that way and the $x$ nonzero digits on top will stay nonzero, and the $y$ zeros on bottom will become $(b-1)$'s. Thus, we have at least $(n-(x+y))+x+y=n$ nonzero digits, as desired.

Note: This is much easier to follow if you write out the two column subtraction, but I don't know how to align that in latex...


[quote="tetrahedr0n"]Just a message to you avid problem solvers in the Intermediate forum! Challenge yourselves with a mock USAMTS round available in the USAMTS forum. The deadline for the solutions will be next Sunday (as in about a week and a half.) It will be great practice for the actual USAMTS rounds.[/quote]

Hey, i'm kinda new here, where is the usamts forum? (and what does it stand for?)

thanks


Solve the system $ \{\begin{array}{c}x^{2}+xy+y^{2}=2\\\ y^{2}+yz+z^{2}=3\\\ z^{2}+zx+x^{2}=5\end{array}$, where $ \{x,y,z\}\subset (0,\infty )$  and show that  ${ xy+yz+zx=2\sqrt 2}$ without solving it !

Take a triangle $\Delta ABC$ of sides $AB=\sqrt{2}, AC=\sqrt{3}, BC=\sqrt{5}$ and a point $P$ inside it so that $PA=y, PC=z, PB=x$, and $\angle APB=\angle BPC=\angle APC=120^\circ$ ([b]Fermat[/b] point). See that $\Delta ABC$ is right-angled at $A$ and consequently has area $S=\sqrt{\frac{3}{2}}$. But its area $S=S_{ABP}+S_{APC}+S_{PBC}=\frac{sin\ 120^\circ}{2}\cdot(xy+yz+xz)$. Equalizing the two area relations we get the desired $xy+yz+xz=2\sqrt{2}$.

Best regards,
sunken rock


[i]A dog trainer wants to buy 8 dogs all of which are either cocker spaniels, Irish setters, or Russian wolfhounds.  In how many ways can she make the choice if she wants at least one from each category?[/i]
[hide="My Thoughts"]
For this problem, I would write: d d d d d d d d.  There are seven spaces, and I need to choose two dividers to separate the dogs into the three categories.  Thus, the answer is $ \binom{7}{2}\equal{}21$.[/hide]

[i]A dog trainer wants to buy 8 dogs all of which are either cocker spaniels, Irish setters, or Russian wolfhounds.  In how many ways can she make the choice?[/i]
[hide="My Thoughts"]
I don't think that I can do the same exact thing as above, since I could have no dogs from one category.  What should I do instead?[/hide]

Same thing, except that dividers can now be:
(1) at the beginning or end, and
(2) next to each other.

So now what you can do is treat the dividers as a "unit" along with the d's. Then you have 10 "units" and 2 of them have to be dividers. So the answer is now $ \binom{10}{2}\equal{}\boxed{45}$.

[quote="Yongyi781"]Same thing, except that dividers can now be:
(1) at the beginning or end, and
(2) next to each other.

So now what you can do is treat the dividers as a "unit" along with the d's. Then you have 10 "units" and 2 of them have to be dividers. So the answer is now $ \binom{10}{2} \equal{} \boxed{45}$.[/quote]

Ok, Thanks!


modular arithmetic
number theory unsolved
number theory

Prove that if a positive integer can be expressed as a sum of squares of two rational numbers, then it can be expressed as a sum of squares of two integers.

$ n\equal{}a^2b^{\minus{}2}\plus{}c^2d^{\minus{}2} \implies b^2d^2 n\equal{}a^2d^2\plus{}c^2b^2$ $ \implies b \mid d \mid b \implies |b|\equal{}|d| \implies b^2 n$ is sum of squares, so it is also $ n$.  :wink:

[quote="pco"][quote="jin thynj"]... $ b^2 n$ is sum of squares, so it is also $ n$.  :wink:[/quote]

Please, could you explain why ?[/quote]
consider the sufficient and needed condition for which an integer can be written as a sum of squares of two integers:
if $ p \equiv \minus{}1 \pmod {4}$ and $ p |n$ then $ v_p(n)$ is even.

[quote="shoki"]consider the sufficient and needed condition for which an integer can be written as a sum of squares of two integers:
if $ p \equiv \minus{} 1 \pmod {4}$ and $ p |n$ then $ v_p(n)$ is even.[/quote]

Ohh, I did not know. Thanks :)

Oh yes it was implicit the condition necessary and sufficient for being a sum of squares  :wink:
(Since I write yesterday (or two days ago) in a other topic of violetcraze http://www.mathlinks.ro/Forum/viewtopic.php?t=312093 )
Bye

1

if a positive integer can be expressed as a sum of squares of [color=red]three [/color]rational numbers, then it can be expressed as a sum of squares of [color=red]three[/color] integers???

2

if $ f(x) \in \mathbb{Z}[x]$ can be expressed as a sum of squares of $ g(x),h(x) \in \mathbb{Q}[x]$, that is , $ f(x) \equal{} g(x)^2 \plus{} h(x)^2$ ,  then it can be expressed as a sum of squares of  $ g_1(x), h_1(x)$, that is , $ f(x) \equal{} g_1(x)^2 \plus{} h_1(x)^2$ ,  where $ g_1(x), h_1(x) \in \mathbb{Z}[x]$  ?????????????


inequalities
integration
trigonometry
real analysis
real analysis solved

Supposed that $f''(x)$ is continuous at $a<x<b\ (a<-\pi,\pi<b)$.
Let \[a_n=\int^{\pi}_{-\pi} f(x)\sin nx dx\ ,b_n=\int^{\pi}_{-\pi} f''(x)\sin nx dx\ (n=1,2,3,\cdots)\]
If $\sum_{k=1}^{\infty} |b_k|$  converges and $f(-\pi)=f(\pi)$,then show the following inequality.
\[\sum_{k=1}^n \sqrt{|a_k|}<\frac{1}{2}\left(2+\sum_{k=1}^{\infty} |b_k|\right)\ (n=1,2,3,\cdots)\]

Integrate by parts twice to get that $b_n=-n^2a_n$.

The rest is just a Cauchy-Schwarz argument. No - not even Cauchy-Schwarz, but the lemma you can use to prove it, namely for positive $x$ and $y$,

$xy\le \frac12(x^2+y^2).$

$\sum_k|a_k|^{1/2}=\sum_k\frac1k\,|b_k|^{1/2}\le \frac12\sum_k \left( \frac1{k^2}+|b_k|\right)\le\frac12\left(\frac{\pi^2}6+\sum_k|b_k|\right)$

Where, of course, we can use the full estimate that $\sum_{k=1}^{\infty} \frac1{k^2}=\frac{\pi^2}6$ or we can use any of several easily obtained estimates that show that that sum is less than 2.

Thank you for your reply,kent Merryfield.

kunny


geometry probability AMC AIME USA(J)MO USAMO

I can do pretty good with Algebra and Combinatorics/NT to do Ireland, Canada and sometimes one or two of the Russian Olympiad problems. But, with Geometry, I start tapping my pencil even at the easiest of AMC 12 problems. What can I do to improve? I am rereading intro to geometry.

I have the same problem.  I am pretty good with algebra, probability, and NT, its just the GEOMETRY.  >_<

I used to be the same way with geometry, but I read through ACoPS (the geometry sections of course) and now I can consistently solve any geometry problem up to AIME difficulty.  USAMO geometry problems are still beyond me, but it's a start.  So yeah, that's what I'd recommend, and you should read ACoPS the whole way through regardless, its invaluable.

im not too great at combinatorics, but someone i know who is, said that aops's intermediate combinatorics book is phenomenal

idk what to do about algebra though

[quote="SnowEverywhere"]I am not very good at geometry but that is my best subject. I found Challenging Problems in Geometry to be very helpful. I strongly recommend getting it.

Also, my worst subjects are combinatorics and algebra -- the things you guys are great at. If you don't mind me asking, what books did you guys go through in these subjects?[/quote]
for combinatorics, get 101 problems in combinatorics. Its a GREAT book, has a lot of problems which is the most important thing in math.


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2002405#p2002405]Here[/url] you can the description of the German path to IMO.

Keywords: VAIMO, AIMO, Germany, Deutschland, Auswahlklausuren, Vorauswahl, Auswahl IMO, Team Selection Test, Auswahlwettbewerb zur Internationalen Mathematik-Olympiade, Auswahl der deutschen IMO-Mannschaft

I scanned the VAIMO (Vorauswahl zur Internationalen Mathematik-Olympiade) 1977-2003 and created .pdf documents which are in German. You can download from below this post. Alternatively, you can order hard copies:

[i]Post- und Hausanschrift:
Bundeswettbewerb Mathematik - IMO
Bildung und Begabung gemeinnützige GmbH
Kortrijker Str. 1
53177 Bonn

Telefon : +49 228-9 59 15-20
Fax : +49 228-9 59 15-29
E-Mail : info@bundeswettbewerb-mathematik.de[/i]

I translated the problems for VAIMO from 1978 to 2003 into English. However, some 2-3 images are missing from the scan and no solutions have been provided. I tried to compare the VAIMO problem sets with the ISL 1981 - 2002 to see which ISL problems have been used in the VAIMO. But I did make any ILL check yet. Please notify whether you found any mistakes or any not so proper translation.

The problems from the VAIMO 1977 are missing. However, those problems came all from previous IMOs

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849354&#p849354]Problem 1 = IMO 1975 problem 4[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357983&#p357983]Problem 2 = IMO 1974 problem 2[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366655&#p366655]Problem 3 = IMO 1972 problem 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366666&#p366666]Problem 4 = IMO 1972 problem 6[/url]

Was it possible 2004 TST have more tests than  2003 TST had problems?
Also I see 2003 TST were quite simple.
Weren't they too simple for a quadruple gold medalist like Christian Reiher ?

Hi Iura,

I think you got slightly confused. These problems I provided were German Pre-TSTs. These Pre-TSTs select a group of 16 students who attend IMO-Preparation units as further TSTs, seminars, lectures etc. These "real" TSTs written by the TOP16 students are taken into account to determine the German IMO team (all TSTs are equally weighted and points are added up to determine the final scores). Mostly seven TSTs are written and for those students. Unless there is a tie and further TSTs are required for the respective students.
I am not sure whether these pre-TSTs have always been pre-TSTs. Maybe some years ago they were even used for IMO TST.
Regarding Christian Reiher I think you are right. But as I said these pre-TSTs shall
determine the TOP16 not the actual team. Just a remark about the second pre-TST (second day in my files I posted): mostly they are the easy problems from the last ISL.

IMO pre-TSTs and TSTS 2004: http://www.mathlinks.ro/viewtopic.php?t=5783

[list][/list]Links to VAIMOs and AIMOs problems can be found from the resources section [url=http://www.artofproblemsolving.com/Forum/resources.php?c=64&cid=54]here.[/url]

I couldn't seem to find a TST in the zip file for 1981, so I translated it from the German pdf files you posted.


Pretty obvious...
[b]Solution:[/b]
Suppose that
\[ AD\cap BC\equal{}P
\]
\[ BA\cap CD\equal{}Q
\]
It is well-known fact that both points $ P,Q$ lies on the polar of $ I$ w.r.t to circumcircle of $ ABCD$.Therefore,$ PQ$ is the polar of $ I$.
It means that $ OI\perp PQ$,simultaneously $ OI\perp EF$,hence $ PQ\| EF$.
As we know division $ (CD,ZQ)$ is harmonic.Now consider a pencil $ P(CD,ZQ)$ and its intersection with $ EF$.
Finally,
\[ (EF,I\infty)\equal{}\minus{}1
\]
Hence $ EI\equal{}IF$ and we are done.

[quote="Erken"]Pretty obvious...
[b]Solution:[/b]
Suppose that
\[ AD\cap BC \equal{} P
\]

\[ BA\cap CD \equal{} Q
\]
It is well-known fact that both points $ P,Q$ lies on the polar of $ I$ w.r.t to circumcircle of $ ABCD$.Therefore,$ PQ$ is the polar of $ I$.
It means that $ OI\perp PQ$,simultaneously $ OI\perp EF$,hence $ PQ\| EF$.
As we know division $ (CD,ZQ)$ is harmonic.Now consider a pencil $ P(CD,ZQ)$ and its intersection with $ EF$.
Finally,
\[ (EF,I\infty) \equal{} \minus{} 1
\]
Hence $ EI \equal{} IF$ and we are done.[/quote]
Thank you very much!

[quote="stergiu"][quote="thanhnam2902"]Let $ ABCD$ is a quadrilateral and $ ABCD$ inscribed in a circle $ (O)$. Let $ AC$ meet $ BD$ at point $ I$. Let $ d$ is a line through point $ I$ and $ d \perp OI$. Let $ d$ meet $ AD$ and $ BC$ at $ E$ and $ F$ respectively. Prove that $ I$ is midpoint of $ EF$.[/quote]

 [b]NOTE(Nothing new to this solution !)[/b]

 If $ EF$ meets the circle at $ K,L$ , then I is the midpoint of $ KL$. By the butterfly theorem (with chord $ AC,BD$ through the midpoint $ I$ of $ KL$ we have then : $ IE \equal{} IF$

 Babis[/quote]
[color=darkblue]Thank stergiu very much.[/color]


logarithms
inequalities
inequalities unsolved

[quote="Hong Quy"]i think that you must choose ; $f_{1}(t) = t^{2}-\frac{9}{t}+10t$not $f_{2}(t) = t^{2}-\frac{9}{t}+10t+17lnt .$
    thus , $f_{2}(t) > f_{1}(t)$  iff $t > 1$  and ...
  can you explain for me ??[/quote]
 i'm stupid .
 sory,  i understood it ;  note that ;$0=17\ln(abc) =17 \lna+17\lnb+17\lnc$and $ab= \frac{1}{c}$ and ect...

[quote="arqady"]Sunjee, $f'(a)=\frac{(a-1)(2a^{2}-8a+9)}{a^{2}}.$
Hence, $a_{min}=1.$ Thus, $f(a)\geq f\left(a_{min}\right)=f(1)=0,$ where $a\in(0,+\infty).$[/quote]
Thank you! Dear Arqady, :blush:  :blush:  :blush:

[quote="arqady"]With same conditions the following inequality true too:
$a^{2}+b^{2}+c^{2}+10(ab+ac+bc)\geq11(a+b+c)$ :wink:[/quote]
can you post solutions

[quote="arqady"][quote="A. S. K."]$a, b, c > 0$
$abc = 1$
Prove that
\[a^{2}+b^{2}+c^{2}+9(ab+bc+ca) \geq 10(a+b+c). \]
[/quote]
$a^{2}+b^{2}+c^{2}+9(ab+bc+ca) \geq 10(a+b+c)\Leftrightarrow\sum_{cyc}\left(a^{2}+\frac{9}{a}-10a+17\ln a\right)\geq0$ is true since,
if $f(a)=a^{2}+\frac{9}{a}-10a+17\ln a$ then $f'(a)=\frac{(a-1)(2a^{2}-8a+9)}{a^{2}}$ and
$f(a)\geq f(1)=0.$:wink:[/quote]
Nice method :wink: Did you develop it on your own?

I have found the method arqady's one 3 years ago.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=149945]Here is my problem which I made it at that time.[/url]

No one have replied! The first problem was wrong.

[quote="me@home"]
Nice method :wink: Did you develop it on your own?[/quote]
Yes, this mine. I have found this method 30 years ago. :lol:
I think many have found this method independently. It's not hard..
But it not a strong method. Try to prove my inequality, and you will see it. :wink:

30 years ago?  arqady, Is my problem wrong?

kunny

Let solve the problem: Find the maximal  k such that

$a^{2}+b^{2}+c^{2}+k(ab+bc+ca) \ge (k+1)(a+b+c)$ for all a, b, c > 0, abc =1.

The answer is $k_{max}=6+4\sqrt{2}$. :lol:

$k_{min}=6-4\sqrt{2}$.


superior algebra
superior algebra unsolved

suppose K is the extension of the field F.and $ \bar{F_k}$is the closure F in K.proove every element of $ \dfrac{K}{\bar{F_k}}$ is not algebraic on $ \bar{F_k}$.in addition if K be algebraic closure then $ \bar{F_k}$ is algebraic closure too.

elements oif $ \overline{F}_k$ are clearly algebraic over $ \overline{F}_k$. but the other elements of $ K$ are not, which is immediate from the definition.

the second question might be: if $ K$ is algebraically closed, then also $ \overline{F}_k$. this can be seen from the fact that $ \overline{F}_k$ has no proper algebraic extensions. or directly from the transitivity of "algebraic".


Using Modular Arithmetic, prove that the sum of two odd square numbers can never be a square number. Can explain the answer please? I'm learning modular arithmetic right now and this sort of question puzzles me

every odd square is $ 1 \pmod{4}$, so a sum of two odd squares is $ 2\pmod{4}$, and thus cannot be a square because every square is either $ 0$ or $ 1 \pmod{4}$.

[hide="Proof of any square congruent to 0/1 mod 4"]It is obvious that an even number's square is congruent to $ 0$ modulo $ 4$. For an odd integer, denote that integer as $ 2n\minus{}1$. The square of this is equivalent to $ 4n^2\minus{}4n\plus{}1$, and note that this can be written as $ 4(n^2\minus{}n)\plus{}1$. If you take this in modulo $ 4$, you have $ 1$ left, so any perfect square is congruent to either $ 0$ or $ 1$ modulo $ 4$. [/hide]


$\boxed{1}$ When $(2x^{2}-3x+1)(x+1)$ is divided by $x^{2}+4$, find the quotient and residue.

$\boxed{2}$ Let $x=\frac{2-\sqrt{3}}{2+\sqrt{3}},\ y=\frac{2+\sqrt{3}}{2-\sqrt{3}}$, find the values of $\frac{y}{x^{2}}+\frac{x}{y^{2}}.$

$\boxed{3}$ Simplify the following equality.
\[\frac{2x^{2}-2x+1}{x-1}-\frac{2x^{2}+4x-1}{x+2}\]
$\boxed{4}$ Let $a+b=\sqrt{3\sqrt{3}-\sqrt{2}},\ a-b=\sqrt{3\sqrt{2}-\sqrt{3}}.$ Find the values of $ab=\boxed{A},\ a^{2}+b^{2}=\boxed{B}$ and $a^{4}+a^{2}b^{2}+b^{4}=\boxed{C}.$

4) Hint: A)(a+b)^2 + (a-b)^2= -------- (factorise the right hand side)
            B)(a+b)^2 - (a-b)^2=----------- (factorise the right hand side)
            C)(a^2 + b^2)^2 - (ab)^2
2) Rationalise the denominators of x and y and try solving

1) you may try polynomial division method. it wont be much long

3) {x^2 + (x-1)^2}/x-1 - {(x+2)^2+x^2-5}/x+2 and proceed....

Hum, no one can get correct solutions except $\boxed{2}.$ :(

[quote="Teki-Teki"](4)

$(a+b)^{2}-(a-b)^{2}= 4ab = 2\sqrt{3}-4\sqrt{2}\leftrightarrow ab = \frac{\sqrt{3}}{2}-\sqrt{2}$
[/quote]
You made a mistake here... It should be $(a+b)^{2}-(a-b)^{2}=4ab= (3\sqrt{3}-\sqrt{2})-(3\sqrt{2}-\sqrt{3}) = 4\sqrt{3}-4\sqrt{2}$, which gives $ab= \sqrt{3}-\sqrt{2}$.

Teki-Teki, you forgot to distribute a negative for number 3. Also, $-x^{2}/(x^{2}+4)=4/(x^{2}+4)-1$ for number 2.