geometry
3D geometry
tetrahedron
function
geometry unsolved

Let $S_{i}$, $i=\{1,2,3,4\}$, be the areas of the faces of a tetrahedron.
$S=\max\{S_{i}\}$.

Prove that the area of any section of the tetrahedron is $\leq S$.

If $X,Y$ are fixed points in space and $Z$ moves on a line in space, then it can be shown that the area of $XYZ$ is a convex function, in the sense that if we restrict our attention to the movement of $Z$ on a segment of the line, then the area reaches its maximum in one of the ends of the segment.

By using this, if the section is a triangle, we can move it so that it becomes a face of the tetrahedron, so that through this movement its area can only increase. Similarly, if $ABCD$ is the tetrahedron, and $MNPQ$ is the section, where $M\in AB,N\in BC,P\in CD,Q\in DA$, then we can move the plane $MNPQ$ continuously so that $MP$ is constant. Again, we can show that the area of the section is convex in the sense mentioned above.

for triangles it is quite clear.

for quadrilaterals i don't understand why the area of $MNPQ$ is "convex"

A related question (which would immediately imply this problem): for any plane $P$, let $h(P)$ be the maximum distance between two planes, both parallel to $P$, and both intersecting the tetrahedron.   Does $h(P)$ always lie between the shortest and longest altitude of the tetrahedron?

Also, does a section that cuts edges $OA, OB, OC$ always have area less than triangle $ABC$?

I still haven't figured out what to do in case the section is a quadrilateral... Could you help me?


:rotfl:  :juggle:  :love:
  plz  please prove it

This inequality is equivolent to:
$ \sum_{sym}{x^3y}\plus{}\sum_{sym}{x^2y^2}\plus{}2\sum_{sym}{x^2yz} \geq 2\sum_{sym}{x^2y^2}\plus{}2\sum_{sym}{x^2yz}$
or
$ \sum_{sym}{x^3y} \geq \sum_{sym}{x^2y^2}$
which is right by Muirhead inequality.


[quote="indianamath"]is this supposed to be in the MC forum?   :o[/quote]

yeah maybe because of the years 2003 and 2005 and 2007 but I don't remember in MC you had to prove anything...

[hide="nope"]I tried to rationalize denominator, then look for telescoping, but it didn't telescope and I got it down to
$4011\sqrt{2007}-6(\sqrt3+\sqrt5+\cdots+\sqrt{2005})-5<328/669$
but I'm stuck at this point......[/hide]

inom .. how did you open the brackets ? :) it it's obvious that

$\frac{1}{(\sqrt 1+\sqrt 3)^{3}}+\frac{1}{(\sqrt 3+\sqrt 5)^{3}}+\cdots+\frac{1}{(\sqrt{2003}+\sqrt{2005})^{3}}> 0$


Hi all!

i have a nice question, which i'll thank if someone will solve:

3^(x/2)+1=2^x.

sure, you can see x=2.

now, you need to prove it...

i tried to exchange paramteres, to make ln. meanwhile didn't work.
anyone?

[quote="jmerry"]This is not the kind of equation you solve algebraically. Since you already know that $x=2$ is a solution, how about proving that $(\sqrt{3})^x+1>2^x$ for $x<2$ and  $(\sqrt{3})^x+1<2^x$ for $x>2$? That shouldn't be too hard (and would justify placing the post here).[/quote]

thanks.... but now can you prove that? i am sorry, i don't remember how to prove it...... :-(
any more suggestion? or can you solve this jmerry's hint?

[quote="jmerry"]This is not the kind of equation you solve algebraically. Since you already know that $x=2$ is a solution, how about proving that $(\sqrt{3})^x+1>2^x$ for $x<2$ and  $(\sqrt{3})^x+1<2^x$ for $x>2$? That shouldn't be too hard (and would justify placing the post here).[/quote]

i thought about it again:

you need to answer the question as if you DON'T KNOW THAT X=2 is the answer. you need to solve in analiticly, not guess the answer and then prove you are right (so it's unlike like finding limits of functions...)

but, besides that, i'll be glad to see the other solution as well JMERRY..... :-)
Thanks...

Unfortunately, there are no analytic techniques for "solving" this type of equations. There are only techniques for showing the existence and/or uniqueness of the solution. In this particular case, to show uniqueness, it is enough to divide by $2^x$ and rewrite the equation in the form $\left(\frac{\sqrt 3}{2}\right)^x+\left(\frac{1}{2}\right)^x=1$. Now it is clear that the left hand side is a strictly decreasing function of $x$, so it can attain any particular value only once.

Again, I'm deleting your identical posts in Calculus Unsolved Problems. Please, stop the multiple posting practice: it only confuses and irritates people.  :)


Demonstraţi că dacă $a,b,c>0, \, abc = 1$, atunci \[\frac1{a+b^{2}+1}+\frac1{b+c^{2}+1}+\frac1{c+a^{2}+1}\leqslant 1 .\] Eu cred că e adevărată, iar [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=12908]arqady[/url] susţine că ar avea o demonstraţie urâtă.

Poate găsiţi voi o demonstraţie (elegantă) sau poate chiar un contraexemplu (desi sper să nu fie cazul).

Pai o demonstratie "straightforward" ar fi notarea numerelor cu $\frac{x}{y},\frac{y}{z},\frac{z}{x}$,numitor comun etc + Miurhead[dar trebuie sa aiba si o demonstratzie mai frumoasa]

nu stiu sigur daca e buna solutia ca am facut tot asa in 20 minute ... (sunt pe fuga)
aduc la acelasi numitor :)
iesea asa :
$\sum_{cyc}a^{3}b+\sum_{cyc}a^{2}b^{3}\geq \sum a+\sum a^{2}$.
acum sper sa demonstrez ca prima suma din membru stang e mai mare decat prima suma din membrul drept si a doua la fel...
omogenizez si dupaia imi da
$\sum_{cyc}x^{5}y^{4}\geq \sum_{cyc}x^{4}y^{2}z^{3}$.
$8x^{5}y^{4}+2y^{5}z^{4}+11z^{5}x^{4}\geq 21\sqrt[21]{x^{84}y^{42}z^{63}}=21x^{4}y^{2}z^{3}$.
insumez ciclic astea si a iesit prima parte.
sper ca iese la fel si a doua parte..

ps : nu iese cu muirhead din cauza ca e doar ciclica si nu simetrica

Se pare că doar inegalităţile \[\sum_\textrm{cyc}a^{3}b \geqslant \sum_\textrm{cyc}a, \, \sum_\textrm{cyc}a^{2}b^{3}\geq \sum_\textrm{cyc}a, \, \sum_\textrm{cyc}a^{2}\geqslant \sum_\textrm{cyc}a\] sunt adevărate.

Cu $\sum a^{2}$ nu prea iese nimic interesant. (am făcut cu un program $\textrm{C}++$, sper să nu fi făcut vreo greşeală)

PS: Nici eu nu cred că e Muirhead-abilă, din acelaşi motiv.

atunci hai sa rezolvam un sistem de trei ecuatii cu sase necunoscute, astfel incat vectorul solutie sa aiba toate componentele pozitive. eu nu am rabdare pentru asa ceva :)
ps : sigur nu e gresita ? (si mie mi se pare "la ochi" asa ca e corecta, dar am intrebat :) nu am dat cu parul ... repet, si mie mi se pare coretca, dar acum nu ma simt capabil absolut deloc sa rezolv nici macar cea mai banala problema)

E bine :D

Se pare că sistemul care se obţine are trei ecuaţii şi şase necunoscute:

$\left\{ \begin{array}[c]{c}0x-5y-1z+1u-3v-4v=0 ; \\ 3x+4y-z+0u+5v+1w=0 ; \\-3x+1y+2z-1u-2v+3w=0 . \end{array}\right.$

Spatiul $\textrm{ker}\, A$, unde $A = \begin{pmatrix}0 &-5 &-1 & 1 &-3 &-4 \\ 3 & 4 &-1 & 0 & 5 & 1 \\-3 & 1 & 2 &-1 &-2 & 3 \end{pmatrix}$, este "spawned" de \[\left\{ \begin{pmatrix}11 \\-12 \\ 0 \\ 0 \\ 0 \\ 15 \end{pmatrix}, \begin{pmatrix}-13 \\-9 \\ 0 \\ 0 \\ 15 \\ 0 \end{pmatrix}, \begin{pmatrix}-4 \\ 3 \\ 0 \\ 15 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix}3 \\-1 \\ 5 \\ 0 \\ 0 \\ 0 \end{pmatrix}\right\}.\] $\begin{pmatrix}1 \\ 10 \\ 103 \\ 223 \\ 10 \\ 10 \end{pmatrix}$ aparţine $\textrm{ker}\, A$ deoarece \[\begin{pmatrix}1 \\ 10 \\ 103 \\ 223 \\ 10 \\ 10 \end{pmatrix}= \frac23 \cdot \begin{pmatrix}11 \\-12 \\ 0 \\ 0 \\ 0 \\ 15 \end{pmatrix}+\frac23 \cdot \begin{pmatrix}-13 \\-9 \\ 0 \\ 0 \\ 15 \\ 0 \end{pmatrix}+\frac{223}{15}\cdot \begin{pmatrix}-4 \\ 3 \\ 0 \\ 15 \\ 0 \\ 0 \end{pmatrix}+\frac{103}5 \cdot \begin{pmatrix}3 \\-1 \\ 5 \\ 0 \\ 0 \\ 0 \end{pmatrix}.\] De asemenea, programul cu care am făcut calculele de mai sus mi-a dat următoarea soluţie: \[\begin{pmatrix}1 \\ 1 \\ 13 \\ 25 \\ 1 \\ 1 \end{pmatrix}= \frac1{15}\cdot \begin{pmatrix}11 \\-12 \\ 0 \\ 0 \\ 0 \\ 15 \end{pmatrix}+\frac1{15}\cdot \begin{pmatrix}-13 \\-9 \\ 0 \\ 0 \\ 15 \\ 0 \end{pmatrix}+\frac53 \cdot \begin{pmatrix}-4 \\ 3 \\ 0 \\ 15 \\ 0 \\ 0 \end{pmatrix}+\frac{13}5 \cdot \begin{pmatrix}3 \\-1 \\ 5 \\ 0 \\ 0 \\ 0 \end{pmatrix},\] adică $15 s_{1}+27 s_{2}\geq 42 t_{2}$.

defapt sunt doua ecuatii, pentru ca a treia e prima minus a doua :)


This is quite a tricky problem.. but what you have to prove is quite amazing really..

Let b(n) be the number of ways of expressing n as the sum of powers of 2 using each exponent at most twice. (we define b(0) = 1 also).
For example, b(13) = 3, because
13 = 8 + 4 + 1 = 8 + 2 + 2 + 1 = 4 + 4 + 2 + 2 + 1.

Prove:
All the ratios b(n)/b(n+1) are in lowest terms, and every positive rational number occurs exactly once in this list.

I'm working hard on the amazing problem

I managed to prove a few things :

[hide]

b(2n) = b(2n+1) + b(2n-1)

b(2^n) = n+1

b(2^n-1) = 1

b(2^n+1) = n

gcd(b(n),b(n+1)) = 1

and a few other relations

[/hide]

Now I'm stuck  :cry:

Is it the right way (= finding some relations) or does the solution use an other tool (like generating functions) ?? Thanks !

Yep you're on the right track (and proved the first part about being in lowest terms! Well at least you say you have   )

You should be able to find a few more specific relations about b(2s-1), b(2s) in terms of previous bs. Then [hide]if you take, say, the "smallest" rational that doesn't appear (for some definition of smallest ) you can assume all "smaller" rationals do appear and you may be able to get a contradiction.l[/hide]

[color=cyan]Without having done any math, can you say anything about the relationship between your function and the number of 1's in the binomial expansion?  I might be off the mark here entirely, but I think it would let you prove the every-rational-number bit, if a reasonable relationship exists.[/color]


How to do this kind of question?
1let S={(1,0,i),(1,2,1)} in C^3 ( with the usual inner product)
compute S perp(orthogonal complement of S)
2Let W=span({i,0,1)} in C^3 .
find orthonormal bases for W and W perp (orthogonal complement of W)

$1)$

Take   $w=(x,y,z) \in S ^{\perp}$ . Then if $a=(1,0,i) \, , b=(1,2,1)$ we have $<w,a> \, = \, <w,b> \, =0$ or

$x+iz =0$ and $x+2y+z=0$ . From these equations we take $x= -iz$ and $y = \frac{-z+iz}{2}$ therefore the solution

to the system is the set $A= \{ (it,\frac{-t+it}{2},t) \, ,t \in \mathbb{C} \}$ . If you take $t=1$ then $S ^{\perp} = <(i,\frac{-1+i}{2},1)> \, =A$.

Or simpler $S ^{\perp}$ is the solution space of the system $\left(\begin{array}{ccc} 1&0&i\\\ 1&2&1\end{array}\right) \cdot \left(\begin{array}{ccc} x\\\ y\\\ z\end{array}\right) = 0$

$2)$

First you find $W ^{\perp}$ as above and then if $W ^{\perp} =<a,b>$ and $W =<c>$ you find an orthonormal

basis using Gram - Schmidt orthogonalization for $W ^{\perp}$ and you take $\frac{c}{ \|c \| }$ as a basis for $W$ .

[quote="TeeNKiT"]i have an other question:
let A ,B be subset of an finite dimensional inner product space V
prove:
1 A is a subset of B implies B perp is a subset of A
2 A= ((Aperp) perp) where A perp is orthogonal complement of A[/quote]

This is theory.

$B^{\perp} =\{ x \in V / <x,b> \, =0\, , \, \forall \, b \in B \}$ but since $A \subseteq B$  ,$\{ <x,b> \, =0\, , \, \forall \, b \in B \} \Rightarrow \{ <x,a> \, =0\, , \, \forall \, a \in A \}$ therefore $x \in B^{\perp} \Rightarrow x \in A^{\perp}$ .

For the second question $V=A \oplus A^{\perp} \, \Rightarrow dim A^{\perp} = dimV -dim A$ and $dim( A^{\perp })^{\perp} = dimV -dim A^{\perp}$ therefore $dimA = dim( A^{\perp })^{\perp}$. Also from the definition of orthogonal complement $A \subseteq (A^{\perp })^{\perp}$ hence $A=(A^{\perp })^{\perp}$ .


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1. Human intelligence is largely hereditary.

2. Civilization depends totally upon innate intelligence. Without innate intelligence, civilization would never have been created. When intelligence declines, so does civilization.

3. The higher the level of civilization, the better off the population. Civilization is not an either-or proposition. Rather, it's a matter of degree, and each degree, up or down, affects the well-being of every citizen.

4. At the present time, we are evolving to become less intelligent with each new generation. Why is this happening? Simple: the least-intelligent people are having the most children.

5. Unless we halt or reverse this trend, our civilization will invariably decline. Any decline in civilization produces a commensurate increase in the collective "misery quotient."

Logic and scientific evidence stand behind each statement listed above, see http://www.neoeugenics.com/

So, what are your thoughts?

[quote="Dog of Justice"]It takes centuries for dysgenics to do significant damage.  While extrapolation of technological trends is highly error-prone, I think that it is reasonable to believe that we will have some human genetic engineering abilities in 30-60 years.  That can be expected to "solve" the problem you discuss (though other problems will be created in the process).[/quote]

So you believe we should ignore the problem based on [b]speculation[/b] that genetic engineering to alter human intelligence and personality will become available, that all the unintelligent people will be able to afford to pay for it, that the public will even generally support the use of this technology, even though they already want a ban on human cloning and other reproductive technologies, and that not only will the public agree that such technologies should be legal, but also they will actively use the technology on a mass scale to ensure that dysgenics does not continue, but reverses.

[quote="Press alt f4"]Oro? According to the topic, you state that the human race is gradually becoming less intelligent, when your link and the rest of your post mention no evidence stating that mankind is indeed becoming less intelligent. Your point's are correct, but I don't see how this has any relavance seeing as there isn't anything supporting that we are gradually growing stupider.

If so, it probably is a evolutionary period of little change or backwards change before a shorter period of large forward change -- punctuated equilibrium.[/quote]

Here is some more direct research that does not require me to hold your hand:

http://www.eugenics.net/papers/evidence.html
http://www.eugenics.net/papers/Intelfert.html
http://www.eugenics.net/papers/lynnrev.html

Regards.

[quote="Asian-American"]So you believe we should ignore the problem based on [b]speculation[/b] that genetic engineering to alter human intelligence and personality will become available, that all the unintelligent people will be able to afford to pay for it, that the public will even generally support the use of this technology, even though they already want a ban on human cloning and other reproductive technologies, and that not only will the public agree that such technologies should be legal, but also they will actively use the technology on a mass scale to ensure that dysgenics does not continue, but reverses.[/quote]

Yes, with one exception -- I am concerned about affordability.  I see the creation of a permanent genetic underclass due to affordability problems as a definite possibility which we have to actively prevent.

My other optimistic predictions do not arise from a belief that public opposition will magically vanish.  Instead, I believe the competitive advantages provided by genetic engineering can be expected to become so great that, if anything remotely resembling our current globalized economy remains, in the long run I don't think the public has a choice in the matter.  Genetic enhancements may very well be banned in the West for decades, given how democracy works.  But what makes you think the East will cooperate?

One fun things would be to find a mathematical system in order to maximize intelligence.

1:  People have been saying this for several hundred years now.  Before people had any real idea about genetics, there was speculation that the world would be overrun with the less capable and intelligent, since those with better blood (i.e., the upper classes) were having less children.  The same story has been repeated periodically, with slightly different ideas of what good versus bad blood is.  And what has happened?  Aside from various atrocities on the part of some of the eugenicists -- absolutely nothing.

2:  The "studies" you linked to (they read more like propaganda for eugenics than science) insist that there is a (very slight) negative correlation between number of siblings and IQ test scores.  But to conclude from this that our population will thus end up decreasing in intelligence in general is absurd.  There is a very simple explanation for this slight correlation, and it has absolutely nothing to do with innate intelligence:  people with more siblings tend to have come from lower-class families, and people from lower-class families tend to get less of the preparation that enables them to do well on IQ tests.  End of story.  No "dysgenics"; no doomsday panic necessary.

[quote="Setofallsets"] End of story.  No "dysgenics"; no doomsday panic necessary.[/quote]

What a completely non-scientific declaration: "End of story": no one should bother to do any further research because you have [b]dogmatically[/b] stated that it is the "end of [the] story," that your statement is absolute, unfallable, as if set in truth by supernatural forces.

Science involves not simply stating an anecdotal story as proving an absolute truth.  Rather, research from all different angles, the more studies the better, are used to show where the body of evidence stands.  And once a theory has been accepted, the theory is always supceptible to debunking by further research.

Now, maybe if you had made your statement at a place like Christianforums.com, it would have more merit.

I find aprt of that true, except i dont belive that those people you are refrring to as unintelligent are necesarily stupid, maybe you should call them lazy or ignorant. It does not necesarily that theyre children are that way, but i guess it does have an impact on the way they were raised up in soceity...

[quote="Dog of Justice"]Yes, with one exception -- I am concerned about affordability.  I see the creation of a permanent genetic underclass due to affordability problems as a definite possibility which we have to actively prevent.

My other optimistic predictions do not arise from a belief that public opposition will magically vanish.  Instead, I believe the competitive advantages provided by genetic engineering can be expected to become so great that, if anything remotely resembling our current globalized economy remains, in the long run I don't think the public has a choice in the matter.  Genetic enhancements may very well be banned in the West for decades, given how democracy works.  But what makes you think the East will cooperate?[/quote]

In the East, only China has eugenics laws.  The only other country that has eugenics policies is Israel which has more fertility clinics per capita than any other country in the world and use these clinics for agressive screening of embroyos for genetic diseases and so forth.

The West though is more or less doomed.  It would take a cataclysmic event to radically alter the Western ethos, such as a nuclear World World III, or a massive economic collapse.  But small gradual changes for the worse are not enough to change the minds of the masses, esp. when the elites who control the media and government promote radical environmentalism (the belief that genes play no part in human intelligence and personality).

From http://www.jewishpress.com/page.do/3151/_History,_Eugenics,_And_The_Jews.html

[quote]In Israel itself many eugenic measures have become widely accepted.
There are now more fertility clinics per capita there than in any other country in
the world. Surrogacy was legalized in 1996. In vitro fertilization and embryo
transfer are preferred by some rabbis as a form of fertility treatment that does
not violate the literal halachic precepts against adultery. And, although human
reproductive cloning is currently not permitted because the technology is not
yet considered safe, the Chief Rabbinate of Israel sees no inherent religious
interdiction in reproductive cloning as a form of treatment for infertility.
[/quote]
[ . . . ]

Complete text at http://www.jewishpress.com/page.do/3151/_History,_Eugenics,_And_The_Jews.html

Interesting article on why China will become the World Power and the West will become Third World:  http://www.amren.com/0111issue/0111issue.htm#article1

I strongly recommend reading the whole article, it is something you definitely don't want to miss, but here is an interesting quote:

[quote]Prof. Lynn is convinced, however, that an Asian country—most likely China—will soon institute a mandatory ES program for its population, and that the resulting improvement in its gene pool will tip the international balance of power decisively in its favor. Attitudes in China radically differ from those in the West. Chinese law already requires sterilization of mental retardates and those with genetic illnesses. Prenatal testing of fetuses is mandatory, and defectives must be aborted. No one with mental illness, venereal disease, or hepatitis may marry.

In the mid-1990s, a poll-taker asked Chinese and Western doctors the following question: Should there be mandatory sterilization for a single, blind woman on public welfare who has already had three children by three different men, all of whom are absent from the household? Only five percent of Western doctors but 82 percent of Chinese doctors said “yes.”

Now that socialism is discredited, Prof. Lynn thinks the Chinese will fill the ideological void with eugenics. He predicts it will become the first, full-fledged eugenic state: all 12-year-old girls will be fitted with contraceptives, only approved couples will be permitted to have children, and ES will be used for all births. Psychopathy and genetic diseases will be eliminated, and IQ will stabilize at the theoretical maximum of about 200 in six or seven generations. Licensing parents will seem just as reasonable as licensing drivers.

Prof. Lynn predicts that in the short run, China’s rulers will clone themselves. In most cases this will mean talent and ability are passed on to the next generation, and it will make it easier for the oligarchs to pass on power to people they can trust—their own twins.

Prof. Lynn’s best guess at a timetable is that ES will be perfected and in obligatory use in China within ten years. Twenty years later there will be the first generation of ES adults, and 20 years after that, half the working population will have come from selected embryos. In 50 years, therefore, China will be the world’s most formidable power.[/quote]

[quote="Asian-American"]In the East, only China has eugenics laws.  The only other country that has eugenics policies is Israel which has more fertility clinics per capita than any other country in the world and use these clinics for agressive screening of embroyos for genetic diseases and so forth.

The West though is more or less doomed.  It would take a cataclysmic event to radically alter the Western ethos, such as a nuclear World World III, or a massive economic collapse.  But small gradual changes for the worse are not enough to change the minds of the masses, esp. when the elites who control the media and government promote radical environmentalism (the belief that genes play no part in human intelligence and personality).[/quote]

Singapore has implemented policies as tacky as [url=http://www.csmonitor.com/2002/0716/p01s02-woap.html]government-sponsored singles bars[/url].  South Korea was comfortable enough with genetic engineering to assign celebrity status to a researcher in the field... who turned out to have falsified data.  China's government is not alone; there is support for genetic enhancement among the broad East Asian populace.

On the other hand, it's unwise to write off the West.  A lot of biology-literate folk are conscious of these issues.  They are just very cautious -- because of some painful historical lessons.  When the East Asian challenge comes, they may rise to the occasion.

[quote="Dog of Justice"]Singapore has implemented policies as tacky as [url=http://www.csmonitor.com/2002/0716/p01s02-woap.html]government-sponsored singles bars[/url][/quote]

Notice how the Christian Science Monitor put a negative spin on Singapore's eugenics policies.  I think you have too much faith that the West is going to become pro-eugenics.  They had their chance: I speculate that they made a fatal error by rejecting the early hereditarianism and supplanting it for cultural Marxism or Boasianism.  A population group has only so much adaptive flexability, and I believe the current environment has changed to a point outside their range of adaptability: natural selection will work against them, choosing a different population group over them.  But who knows, I can be wrong, you never know. . .

[quote]On the other hand, it's unwise to write off the West.  A lot of biology-literate folk are conscious of these issues.  They are just very cautious -- because of some painful historical lessons.  When the East Asian challenge comes, they may rise to the occasion.[/quote]

How will the biologically literate Westerners save the west when they have no access to the mass media to reindoctrinate the West?

How eugenics made the Jews so smart:

[b]The evolution of intelligence[/b]

[color=red][b]Natural genius?[/b][/color]

Jun 2nd 2005
From The Economist print edition

[b]The high intelligence of Ashkenazi Jews may be a result of their persecuted past[/b]

THE idea that some ethnic groups may, on average, be more intelligent than others is one of those hypotheses that dare not speak its name. But Gregory Cochran, a noted scientific iconoclast, is prepared to say it anyway. He is that rare bird, a scientist who works independently of any institution. He helped popularise the idea that some diseases not previously thought to have a bacterial cause were actually infections, which ruffled many scientific feathers when it was first suggested. And more controversially still, he has suggested that homosexuality is caused by an infection.

Even he, however, might tremble at the thought of what he is about to do. Together with Jason Hardy and Henry Harpending, of the University of Utah, he is publishing, in a forthcoming edition of the Journal of Biosocial Science, a paper which not only suggests that one group of humanity is more intelligent than the others, but explains the process that has brought this about. The group in question are Ashkenazi Jews. The process is natural selection.

[ . . . ]

Complete article at http://www.economist.com/science/displaystory.cfm?story_id=4032638

blahblahblah, you should be careful in what you're saying:  Asians outscore Blacks [i]on average[/i], which is not at all the same thing that you just said.

[quote="blahblahblah"]Wow, an amusing take on my name.  I am dazzled by your wit and originality.

Your statement is essentially unprovable.  What is the proper socioeconomic environment for Asians?  Is it the same proper socioeconomic environment for whites?  For blacks?  For Hispanics?  Why do Asians from any socioeconomic background outscore the richest blacks?  Why do Ashkenazi Jews have such a high IQ as a group?  Why are there more males than females in the upper echelons of academia?  Why are there more males than females scoring at the bottom end of intelligence tests?  Nature (same mean, higher variance) or nurture?

I doubt you have any 'provable' answers to any of those questions.[/quote]

Dear blahblahblah, my statement is no less provable than yours...you make such blanket statements as 'Why do Asians from any socioeconomic background outscore the richest blacks?"  wow, I can't understand where is your head...what Asians are you talking about and where does this take place...how many are you using as your test group...by your comments I have to draw the conclussion that you have no idea of what you are talking about...do research your topic before making such blanket statements in reference to blacks...who happen to have a very old and enriching history in the world...it did not start with Asians or whites for that matter...peace to the world, Carlos Enrique

[quote="wisdomseeker"]
Dear blahblahblah, ...[/quote]

Sorry, this just sounded so funny  :lol:
I apologize for making a useless post...

*mouth drops open*

I simply don't know where to start.
Let's see. How about we do bullets, or something?

- Just because a website LOOKS important doesn't mean it's solid. I am very leery of .coms in general.

-The article talked about economic and education differences. It did not discuss color-blindness, geography, culture, or test-making
When you see a person, you make certain inferences about them. You aren't color-blind, even if you'd like to pretend like you are. If you see an Asian person, you tend to assume "I bet they are good at math," just like if you see a tall black person, you think "wow, they are probably a good basketball player." In both cases, you might be wrong. I have a fabulous Asian friend who is incredibly gifted at music, but is not a mathematician. I have a tall black friend who has been studying advanced math and science for four years. The thing is, when we make inferences about people, we treat them differently. An example I can cite is about genders. There was this study done (nope, I can't cite it, it was discussed in a philosophy class a WHILE back) on the way we react to different genders. It was noticed that when a person greeted a male, they had a certain set of reactions, and when they greeted a female, they had another set. Like an invisible rulebook. But then, when faced with an androgenous person, the subject always paused, like they had lost their little rulebook and had NO idea how to respond or treat this person. Just because they looked different from their male/female guidelines.
Alabama has a large proportion of African Americans, and a fast-growing population of Hispanics. Alabama also ranks dead-last or second-to-last in education. Coincidence? Yes. Alabama has a history of not raising taxes, because Alabamians do not like the word "taxes." I think if you look in an Alabama dictionary, it's actually not there (that was a joke). Anyways, since that state does not have any amount of taxes, they do not have a lot of federal funds to spend on education. Some places are silly that way.
The Asian-American culture pushes their children. I feel confident in saying this, even though I am not, myself, Asian-American, as I have quite a few very close Asian-American friends [notice I am continuing to say "Asian-American" and not simply "Asian," because I have a theory that the average Asian is not as pushy as the first-generation Asian-American parents tend to be towards their children. And by "average," I mean: considering that there are a billion people each in China and India]. Anyways, if you've ever listened to any of Bill Cosby's speeches, you might come to the conjecture that the African-American community has a very different take on things. But there are a lot of things regarding the African-American culture, including ridiculously long drug-possession sentences that keep fathers away from their children, that I just don't want to get into right now. Suffice it to say, there are cultural differences within a race that cross all socio-economic lines.
Who makes the SAT and other standardized tests? I think they are old white men. I'm almost sure they are white males. Guess what kind of person tends to do the best on standardized tests? WHITE. MALES. Big shocker, right?

-I'm getting kind of sick of all this. I want to cry every time I read another post in this thread. So I'm just going to give up. Go on thinking the world is simple and calculatable and explainable with just a few simple rules. Just remember that once, someone tried to tell you that the world wasn't that easy, and that there was no set formula for how someone was going to turn out.

[quote="wisdomseeker"][quote="blahblahblah"]The standardized test in question is the SAT 1, and I'm not sure how big the test group is exactly (obviously it was dependent on the specific year) but more than enough to be statistically conclusive.  These are hardly facts that I am pulling out of my rear end (see [url=http://www.lagriffedulion.f2s.com/testing.htm]here[/url])

Stop wasting your time and my time by accusing me of ignorance - try substantiating your arguments instead.

And, while you're at it, stop using ellipses - a single period and two spaces following that, or occasional usage of the 'Enter' key, would be infinitely preferable.[/quote]

Look blah….so all this time…you have been basing your asinine logic on the SAT1 test…you’ve got to be kidding…this is all a joke in bad taste…insulting several groups in the human race in order to elevate one (which one do you belong to…let me guess).  This is most definitely a waste of time…I thought that this was a group that discussed issues with facts that carry substance…adiós, arrevoir. sayonara, shalom, and good riddance.[/quote]

How ridiculous.  "blahblahblah" is the one providing a reference; [b]you're[/b] the one ignoring substantive facts, instead engaging in an [i]ad hominem[/i] attack on his motives.  How 'bout you answer this: if he really were an "Asian supremacist", why would he reference the fact that Ashkenazi Jews outperform Asians by a significantly larger margin, on average, than Asians outperform Caucasians on these g-loaded standardized tests?

I don't consider it insulting when someone tells me I'm not six feet tall, or even that Asian average height is lower than Caucasian average height.  It's just a fact.  If I don't like it, then the proper response is to focus on how to change it, e.g. better nutrition, and maybe a dash of genetic engineering in the future.  I don't argue the fact doesn't exist, saying things like "Look at Yao Ming!", unless I want to make a fool of myself.  Intelligence is a lot more complicated than height; in particular, it's not immediately obvious how to measure it.  But when folks that are apparently trying their best to measure it keep reproducing these racial performance differences, we shouldn't dismiss them just because the conclusions pointed to by the data are "insulting", we should try to figure out what's going on.

There is a part of me that is not very patient with racists (and no...not being a racist does not mean to be wrong)  Regardless of what race is blahblahblah, his comments are blatant racism.  The SAT 1 is not the measure stick with which the world measures  intelligence, the world is bigger then the SAT 1...humans are so resilient, that everyone no matter their race, religion, socioeconomic background, or any other difference, can be very intelligent or an a simple ass...this is a sad discussion...as far as I am concerned I apologize for not been very patient with racists.

Wow...this place has turned awfully silly. My opinion, whether you care for it or not: I do not think it's racism to suggest that some races, *on average*, have higher intellectual ability than others--just like it isn't racism to suggest that some races, on average, have higher physical ability than others. We don't know right now if that's the case or not, so it is possible. (I personally doubt it, but.) It is racism to suggest action (like eugenics) based on this, since an individual's performance has little to do with his/her race's average performance. No one except A-A has done that, and they seem to have left. Therefore, no one currently participating in this thread is a racist, and you should all calm down and quit with the strawmen.

To bleumoose:

It seems that you are trying to get across the point that there are individuals within a group that may have characteristics far different than the mean of the group.  This is obviously true.  Moreover, if individuals have normally distributed traits, this is to be expected.

You are also saying that you believe that Asians score well on standardized tests because they are driven to (e.g, nurture).  You are very possibly correct.  Note that I originally asked a question, and did not offer a theory.  Apparently to you and others the mere asking of the question is tantamount to being a racist.

Do I believe that people of all ethnicities are given a fair go in the states?  No.  They are not.  But the idea that you cannot study the intelligence of individuals without perfectly controlled circumstances is absurd.  People are complex, and statistical can only be effective if it is kept reasonably simple.  Economic status is a 'good' measure of the opportunities afforded an individual, and it makes sense to study the effects of it on intelligence tests.

To 'wisdomseeker'

The SAT is not the universal measuring stick of intelligence, no.  But it does, in some crude way, measure a sort of intelligence, and a huge body of data from across the US is available on it.  Do you believe that that data is fundamentally flawed, to the point that the obvious statistical conclusions which are drawn are completely wrong?  Note that you are immediately willing to reject the racial aspect of SAT scores (which is borne out by the data) but just as quickly accept the economic aspect of SAT scores (scores increase with household income, also borne out by the data).

I am no racist.  My opinion is very similar to 'bookaholic's' (another asian! horrors) - I do not think that it is inherently racist to explore the differences in intellectual ability between races, and I do not know if an intrinsic difference in intellectual ability, independent of socioeconomic status, even exists.  I believe that this thread reflects that, and if you have any other reason for calling me a racist, which is a rather serious allegation, I would like to hear it.

Let us pause for a moment to de-escalate here, please.  blahblahblah, I'd like to suggest to you a possible reason you're coming under attack: your most recent set of comments have stepped in to fill a void that was left by someone who lists among his favorite websites at least 5 which actively advocate eugenics programs.  You also made an error that I pointed out by talking about "Asians of all socioeconomic groups outperforming the richest Blacks."  This error was arguably factual but even more clearly strategic: by not putting in a simple disclaimer ("on average" would have been fine) you inadvertantly associated yourself more strongly with a radical set of views.

The second problem here is that people are talking around each other.  There are serious issues of references to groups as opposed to references to individual members of groups.  I've dealt with blah^3's contribution to this problem above.  It's mirrored by wisdomseeker's claim that "all humans have the same capacity to learn if placed in the proper socioeconomic enviroment," a statement which is clearly false on its face as a statement about individuals (my deceased uncle who suffered from Downs Syndrome simply had lower innate capacity than the people on this board) and which doesn't quite make sense as a statement about groups.  bleumoose has a personal strong preference to discuss issues of intelligence as individual issues.  Dog of Justice continues the problem of group versus individual commentary by conflating his personal height with the innate intelligence of a group of people.
(There is also an entirely different set of arguments going on, of course, pertaining to the extent to which "intelligence" is measurable.  Reliance on SAT scores as a proxy for "intellectual ability" is the issue here.)

If you want to have this discussion, you should all back up and take a deep breath.  Those of you who have stepped into the "pro-eugenics" role: decide if this is really the context in which you want to present your views (since at this point it doesn't seem like any of you are actually pro-eugenics, even at a low level).  If it is, then be careful in how you present yourselves, because you [i]have[/i] stepped into the pro-eugenics side.
The rest of you: relax.  These people [i]aren't[/i] actually pro-eugenics.  Try to discern what they really are arguing, because it doesn't appear to me that any of them continues in the belief that there is a significant inherent/genetic component to racial test-score differences.  Try to meet them at the level they actually seem to believe, which at this moment is significantly less pernicious.

Another thing that gets people so offended and riled up about this subject is that it has a huge historical context to it.

Intelligence is one of the most prized and precious qualities that we as humans have. Its not the same thing as someone's height or physical abilities. To point out differences in people's "intelligence" is, in a lot of ways, an implicit question of their degree of humanity. And historically, once you deny a group of people any degree of humanity it becomes more likely and more justifiable to abuse them and deny them their human rights. This has been shown through history time and time again with truly terrible results.

So when we ask, what's wrong with just asking the question? I can see lots of things wrong with it, especially if its not accompanied by lots of caveats and cautions. Any study that purports to show differences in intelligence based on "race" is inherently dangerous. (Especially since race isn't even a well defined categorical variable). Even if the authors of a such a study themselves have no (conscious) malicious intent, they have to be smart enough to realize that a lot of people in this world do.

That's one of the things about the ethics of science - scientists don't just have a responsibility to ask questions, they have a responsibility to think about the impact that their discoveries could make on the world, and to put them into historical perspective.

As for the studies themselves, such as the SAT study, I'm not particularly convinced that you can use that data to say anything about innate racial differences. There are lots of possible confounding factors which have nothing to do with genetic racial differences. Especially studies done using an American population, you have to be aware of confounding factors.

[quote="tokenadult"][quote="gauss202"]Any study that purports to show differences in intelligence based on "race" is inherently dangerous. (Especially since race isn't even a well defined categorical variable).[/quote]

That's one of the big problems here. There is a HUGE degree of fallacy of equivocation in most studies of "race," because people who care about race can't agree on the racial categorization of the various individuals who consitute humankind.

The problems in the early posts in looking at the literature on mental ability tests

http://learninfreedom.org/iqbooks.html

also vitiate many of the conclusions some would like to leap to here. The best scholars on human intelligence do NOT take IQ tests as anything at all like a "measure" of "intelligence."[/quote]
To add to the IQ thing, not all of your intelligence is biologically determined. Just look at the conjecture of László Polgar, and the results of his experiment.


Find all functions $f: \mathbb N^* \to \mathbb N^*$ such that
 \[f(f(m)f(n))=f(n)+f(m),\quad \forall m,n \in\mathbb N^{*}.\]

[quote="duongqua"]find $ f: N^{*}\to\ N^{*}$ such that:
 $ f(f(m)f(n)) \equal{} f(n) \plus{} f(m) \forall m,n \in N^{*}$[/quote]

[hide="My solution"]
Let $ \mathbb A \equal{} f(\mathbb N)$. The equation is $ f(xy) \equal{} x \plus{} y$ $ \forall x,y\in\mathbb A$

So, if $ x,y,z,t\in\mathbb A$ are such that $ xy \equal{} zt$, then $ f(xy) \equal{} f(zt)$ and so $ x \plus{} y \equal{} z \plus{} t$ and it is easy to conclude that either $ (x,y) \equal{} (z,t)$, either $ (x,y) \equal{} (t,z)$

$ u,v\in\mathbb A$ $ \implies$ $ u \plus{} v\in\mathbb A$ (since $ f(uv) \equal{} u \plus{} v$) and so $ pu \plus{} qv\in\mathbb A$ $ \forall p,q\in\mathbb N_0$

Let then $ u,v\in\mathbb A$ and :
let $ x \equal{} uu \plus{} uv$
let $ y \equal{} u \plus{} 2v$
let $ z \equal{} uu \plus{} 2uv$
let $ t \equal{} u \plus{} v$

All these four numbers are in the form $ pu \plus{} qv$ and so belong to $ \mathbb A$ and we have $ xy \equal{} zt \equal{} u(u \plus{} v)(u \plus{} 2v)$ and so :
Either $ (x,y) \equal{} (z,t)$ and so $ z \minus{} x \equal{} uv \equal{} 0$, which is impossible
Either $ (x,y) \equal{} (t,z)$ and so $ x \minus{} t \equal{} (u \minus{} 1)(u \plus{} v) \equal{} 0$ and so $ u \equal{} 1$

So $ u,v\in\mathbb A$ $ \implies$ $ u \equal{} 1$ and so $ \mathbb A \equal{} \{1\}$, which is obviously wrong ($ f(x) \equal{} 1$ is not a solution).

So no solution.
[/hide]


Is given a function $ f$ such that $ f(a,b) \equal{} max(a^{2},b^{2})$   for a,b,c reals . If we know that $ a \plus{} b \plus{} c \equal{} 0$ and $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$ then find the smallest value of $ f$ .

[hide]
Assume wlog $ a\le b \le c$
Then the question asks for the minimum value of $ b$.
Sub $ c\equal{}\minus{}(a\plus{}b)$
$ \Rightarrow a^2\plus{}b^2 \plus{} (a\plus{}b)^2 \equal{} 1$ where $ a\le b$

Maximising $ a$ will also minimise $ b$, so let $ a\equal{}b$
$ \Rightarrow 6b^2 \equal{} 1$

To check that this minimises $ b$, let $ b\equal{}b\plus{}h$ and $ a\equal{}b\minus{}h$, and $ h >0$ is very small
$ \Rightarrow 2(b\plus{}h)^2 \plus{}2(b\plus{}h)(b\minus{}h) \plus{}2(b\minus{}h)^2 \equal{} 6b^2 \plus{} 2h^2 \ge 6b^2$

So finally $ Min: f(a,b)\equal{}b^2 \equal{} \frac{1}{6}$

[/hide]

Hi ,

Did you use any standart solution to do this ?

I need some methodes to do this kind of exercices with (max and min)

I only know that $ max(a,b)\ge \sqrt{\frac{a^{2}\plus{}b^{2}}{2}}$ but this can not be used because a,b,c can not be positive .


We will show that $a+b=ae^b+be^a$ implies $a=0$ or $b=0$.
Indeed, if $a=0$ or $b=0$ then we have equality. Suppose $a\neq 0$ and $b\neq 0$. Then it is necessary $ab<0$, otherwise, if $a,b>0$ then $a+b<ae^b+be^a$ and if $a,b<0$ then again $a+b<ae^b+be^b$. Suppose WLOG $a<0$ and $b>0$. We have $(ae^b+be^a-a-b)'_b=ae^b+(e^a-1)<0$. Since for $b=0$ we have equality $ae^b+be^a-a-b=0$, we conclude $ae^b+be^a-a-b<0$ for $b>0$.

For initial problem. We just need to solve $x^2-1=0$ and $x=0$. So answer is -1, 0, 1.


I don't think so.Who are having a nice solution $ ?$

[quote="every"]Let $ a_1,a_2,a_3,a_4,a_5$ be non-negative reals numbers whose sum equal 1.Prove the inequality $ a_2a_3a_4a_5 \plus{} a_1a_3a_4a_5 \plus{} a_1a_2a_3a_5 \plus{} a_1a_2a_3a_4\leq \frac {1}{256} \plus{} \frac {3275}{256}a_1a_2a_3a_4a_5$ When does equality occur?[/quote]
AC (UMV) method kills it easily.
$ f(a_1,a_2,a_3,a_4,a_5) \le \max [f((a_1\plus{}a_2)/2,(a_1\plus{}a_2)/2,a_3,a_4,a_5),f(a_1\plus{}a_2,0,a_3,a_4,a_5)]$
By the way, we can prove the general problem :
If $ a_1,a_2,...,a_n$ are nonnegative reals with sum 1 and $ k$ is a real number, then
\[ F \le \frac{n^2\plus{}k}{n^n}
\]
for all $ k \ge \minus{}n^2\plus{}\frac{n^n}{(n\minus{}1)^{n\minus{}1}}$
and
\[ F \le \frac{1}{(n\minus{}1)^{n\minus{}1}}
\]
for all $ k \le \minus{}n^2\plus{}\frac{n^n}{(n\minus{}1)^{n\minus{}1}}$
where $ F\equal{}\sum_{i\equal{}1}^n \prod_{j \neq i}a_j\plus{}k\prod_{i\equal{}1}^na_i$.

I think the solution won't use AC method.Do you have a different solution?


"What is the sum of the 13 smallest positive palindromes that have a tens digit of 3, and a ones digit of 7?"

I don't keep a calculator with me that much ever since I missed a problem because of a pressing the buttons error

And most of the large number calculations are ones that can be done in your head if you think a little

It actually makes him sound hypocritical, considering how often he "forgets" to bring his calculator to math team stuff.

But if anyone disagrees with the answer, feel free to post

[hide="Sum"]
The sum is 1472981.
[/hide]

So, either one of you guys made an error, or both of you did, I suppose

And Peijin, there's a thing called "I don't feel like it"

Uh..I added up the numbers alanchou gave me, and I got 1,472,981. I don't get who is wrong or right.  My answer goes with hunter34, but I don't want to blame anyone else for getting the wrong answer! I guess I will let that for you people to solve. :)  I don't have a calculator with me at the moment,( I calculated it manually) so anyone with a calculator could calculate the answer, and post it here.

Guys, dont fight!  :mad:  I know Alanchou could have calculated it, but he doesn't have to. He helped me anyway. Its up to him. And the rest of you also helped, because you took it a step further, and gave me the answer, so all of you helped.  :)   So now, back to business. :wink:  The final answer for this problem is now:1,472,981.

[hide=" :( "]
[hide=" Boo."]
This is what we do every day at school.  :lol:
[/hide]
[/hide]

Please refrain from making disparaging remarks on the forum, through private message, or life in general.


induction
number theory
number theory proposed

Given positive integers a, b such that gcd(a,b)=1, and let p be a odd prime show that

gcd(a+b, (a^p+b^p)/a+b)= 1 or p

It's enough to show it for $b=-1$ because if $q|a+b$ then $(q,b)=1$, so $\frac {a^p+b^p}{a+b}=0$ in $\mathbb Z_q$ is equivalent to $\frac {(a(-b^{-1}))^p-1}{a(-b^{-1})-1}=0$.

On the other hand, it's not hard to show by induction that $(\frac {a^n-1}{a-1},a-1)=(a-1,n)$ for all integers $a>1$ and all positive integers $n$, and we're done (we apply this for $n=p$).

Leo try to use the following lemma, I wrote in this link:

http://www.mathlinks.ro/viewtopic.php?t=5123

Leo, this is the most powerful lemma i know!, it solves almost every number theory problem or makes it easier.

Hey, that lemma it's really nice...where did you read it? :D

I dont know, i guess in last year imo training :)


geometry
circumcircle
inequalities
inequalities proposed

In a triangle ABC , let :   $ a,b,c $= the (lengths) sides,   $ r$=radius of incircle, $R$=radius of circumcircle, $ s$=the   semiperimeter. Let us denote   $ {\bf Q}=\sqrt{s^2-3r(4R+r)} \;   . $
[b]Question:[/b]
 [i]If    $ a\ge b\ge c \; ,\; $ prove the inequalities [/i] :
  \[(1)\; \; \;  {\bf Q}\le a-c \le \frac{2\sqrt{3}}{3}{\bf Q}\; ,\; \; \;
\; \; (2)\; \; \; \;  a-\frac{b+c}{2}\le {\bf Q}\; \; , \; (3)\; \; \; \;  \frac{a+b}{2}-c\le {\bf Q} \; .
\]
Remarks:
[A] If ABC is an isosceles triangle with ($ a=b $), then equality case occurs in LHS of (1) and in (3).
[B] The equality is attained in RHS of (1) when  $ b=\frac{a+c}{2} \; , \; (a<3c<9a )\; .$
[C] If ABC is isosceles  ($ c=b \; ,\; ( a<2b)\; $ ), then  in (2) holds equality.

These inequalities come from this equality
\[ (a-b)^2 + (b-c)^2 + (c-a)^2 = 2Q \]

 We denote \[x=a-b, y=b-c , z=a-c \]

So we have \[x+y-z=0 \] and

\[x^2 + y^2 +z^2 = 2Q \] (*)

1) equivalent to 2/sqrt(3)Q >= z >= Q
   We have    2(x^2 + y^2) >= z^2 and z^2 >=x+y)^2

2) equivalent to x+z <=2Q

    Apply BCS inequality we can get (2)

3) equivalent to x+2y <= 2Q

We denote  T= x+2y

 We can easily prove  T^2 <= 4Q^2

This complete the proof.


let a,b,c,d is reall number such that $ a\plus{}c \ge b\plus{}d$and$ a\plus{}b\equal{}c\plus{}d$
prove that:
$ a^2\plus{}\frac{1}{2}(a\plus{}b)^2 \ge d(a\plus{}b)\plus{}ab$

[quote="great math"]I think we should consider the cases since a, b,c d are real numbers  :wink:[/quote]

ohhh... I misread the question  :lol:

$ a\plus{}b\equal{}c\plus{}d$
$ c\equal{}a\plus{}b\minus{}d$
$ a\plus{}c\geq b\plus{}d$
$ a\geq d$

$ 2(a^2\plus{}\frac{(a\plus{}b)^2}{2})\equal{}3a^2\plus{}b^2\plus{}2ab \geq 2a^2\plus{}4ab\geq 2(d(a\plus{}b)\plus{}ab)$

$ (a\minus{}b)^2\geq 0$
$ a^2\plus{}b^2\geq 2ab$
For any reals a , b.

[quote="Matematika"]

$ 2a^2 \plus{} 4ab\geq 2(d(a \plus{} b) \plus{} ab)$

$ (a \minus{} b)^2\geq 0$
$ a^2 \plus{} b^2\geq 2ab$
For any reals a , b.[/quote]

how does this follow?

[quote="Matematika"]$ 2a^2 \plus{} 4ab \equal{} 2a(a \plus{} b) \plus{} 2ab\geq 2d(a \plus{} b) \plus{} 2ab \equal{} 2(d(a \plus{} b) \plus{} ab)$[/quote]

what if $ a\plus{}b\le 0$?  :(

sorry my mistake(I'm fool)
for a=-3, b=-3, c=0, d=-6
9+18>=36+6
27>=42
this statement is wrong


Euler
analytic geometry
integration
function
linear algebra
matrix
calculus

Here is a something I fail to understand from our analysis III course. Any help during the next 24 hours would be appreciated, since this is the time remaining before the first exam.

We know that if $ Q\subset\mathbb{R}^n$ is an open subset, and $ L: Q\times\mathbb{R}^n\to\mathbb{R}$ is a $ \mathcal{C}^2$ mapping written in the form $ L\left(q,v\right)$ (id est, the first $ n$ coordinates of the input are called $ q$, and the last $ n$ coordinates are called $ v$), then the $ \mathcal{C}^2$ stationary solutions $ \gamma$ of the variation problem $ \delta\int_{t_0}^{t_1}L\left(\gamma,\dot\gamma\right)dt \equal{} 0$ satisfy the Euler-Lagrange equations

$ \frac {d}{dt}\left(\frac {\delta L}{\delta v}\left(\gamma\left(t\right),\dot\gamma\left(t\right)\right)\right) \equal{} \frac {\delta L}{\delta q}\left(\gamma\left(t\right),\dot\gamma\left(t\right)\right)$.

Now, our analysis text tries to apply these equations to find geodesics on manifolds. Namely, let $ M$ be an $ n$-dimensional $ \mathcal{C}^3$-submanifold of $ \mathbb{R}^{\text{something}}$. Let $ \phi: U\to Q\subset\mathbb{R}^n$ be a $ \mathcal{C}^3$ map with $ Q$ being open, and let $ \psi \equal{} \phi^{ \minus{} 1}$ be the inverse function of this map. Denote $ g_{ij}\left(q\right) \equal{} \left < \delta_i\psi\left(q\right),\delta_j\psi\left(q\right)\right >$. For any curve $ \gamma: \left[t_0,t_1\right]\to U\subset M$ we can define a function $ q \equal{} \phi\circ\gamma$. Now, minimizing the length of $ \gamma$ is easily seen to be equivalent to minimizing $ \int_{t_0}^{t_1}\sqrt {\sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n g_{ij}\left(q\left(t\right)\right)\cdot\dot q^i\left(t\right)\cdot\dot q^j\left(t\right)}dt$.

Up to here the whole matter is clear to me. Now, the text claims that if we set $ L\left(q,v\right) \equal{} \sqrt {\sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n g_{ij}\left(q\right)\cdot v^i\cdot v^j}$ and add in the condition $ \left|\left|\dot\gamma\right|\right| \equal{} 1$, the Euler-Lagrange equations transform (after some computations [b](*)[/b]) into

$ 0\equal{}\dot{\dot\gamma}^m \plus{} \frac12 \sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n\sum_{k \equal{} 1}^ng^{mk}\left(g_{ij,k} \minus{} 2g_{ik,j}\right)\dot\gamma^i\dot\gamma^j$ for all $ m\in\left\{1,2,...,m\right\}$.

Here, $ \left(g^{mk}\right)$ means the inverse matrix to the matrix $ \left(g_{ij}\right)$, and $ g_{ij,k}$ is a shorthand for $ \delta_kg_{ij}$.

The text then proceeds to rewrite the above equation in a more symmetric way, with the Christoffel symbols appearing on the right hand side.

My questions are:

- Why can we assume that $ \left|\left|\dot\gamma\right|\right| \equal{} 1$ ? Doesn't then our optimization problem change into an optimization problem with boundary conditions, so that the Euler-Lagrange equations cannot be used anymore?
- Can anyone refer me to a place where the above computational step [b](*)[/b] is explicitely performed? I have tried and got something extremely long and obscure, but I didn't use $ \left|\left|\dot\gamma\right|\right| \equal{} 1$ in my computations, since, as I said, I don't understand what allows us to just assume this condition.

Feel free to answer in hints, after all I should be able to compute some derivatives on my own, but [b]please[/b] do not use dirty notations (Einstein conventions, "hidden" derivations as in $ L_k$ for the $ k$-th derivative of $ L$), because these are driving me crazy on a regular basis.

  Darij

Okay, don't bother reading the above. I just got it explained by a friend.

Apparently, everyone who makes this computation uses the fact that $ \int_{t_0}^{t_1}\sqrt {\sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n g_{ij}\left(q\left(t\right)\right)\cdot\dot q^i\left(t\right)\cdot\dot q^j\left(t\right)}dt$ is minimal under the side condition $ \left|\left|\dot\gamma\right|\right| \equal{} \text{const}$ if and only if $ \int_{t_0}^{t_1}\sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n g_{ij}\left(q\left(t\right)\right)\cdot\dot q^i\left(t\right)\cdot\dot q^j\left(t\right)dt$ is minimal (where $ t_0$ and $ t_1$ are fixed), because the Cauchy-Schwarz inequality yields

$ \int_{t_0}^{t_1}\sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n g_{ij}\left(q\left(t\right)\right)\cdot\dot q^i\left(t\right)\cdot\dot q^j\left(t\right)dt\cdot \int_{t_0}^{t_1}1dt$
$ \geq\left(\int_{t_0}^{t_1}\sqrt {\sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n g_{ij}\left(q\left(t\right)\right)\cdot\dot q^i\left(t\right)\cdot\dot q^j\left(t\right)}dt\right)^2$

with equality if and only if $ \sum_{i \equal{} 1}^n\sum_{j \equal{} 1}^n g_{ij}\left(q\left(t\right)\right)\cdot\dot q^i\left(t\right)\cdot\dot q^j\left(t\right)dt$ is proportional to $ 1$, e. g. if and only if $ \left|\left|\dot\gamma\right|\right| \equal{} \text{const}$.

Of course, minimizing the latter integral seems a much nicer work to do.

  darij


A horrible horrible surd, I might add. This problem is from an exam yesterday (yes, it is okay for me to post it). It is from a beginning course in mathematics at university, so I'm not quite sure whether this is at olympiad level or whether it is at pre-olympiad high school. Anyway, here is the problem:

Show for example with induction that
\[ {2n \choose n} \ge \frac{4^n}{2\sqrt{n}} \]
for all integers $n \ge 1$.

It is quite easy to use the induction step, in this way:
[hide]
\[ LHS_{p+1} = {2p+2 \choose p+1} = \frac{(2p+2)!}{((p+1)!)^2} = \frac{(2p)!(2p+1)(2p+2)}{(p!)^2(p+1)^2} = LHS_p\frac{2(2p+1)}{p+1} \ge \frac{4^p(2p+1)}{\sqrt{p}(p+1)} \]
[/hide]

However, I can't figure out how to use this in order to show that $LHS_{p+1}\ge RHS_{p+1}$. I sat with this problem for a long time, but I just can't figure it out. And it's not supposed to be that hard. :(

edit: the multiline parsing kind of sucks, but I don't know LaTeX well enough to fix it.

just compare the thing you've obtained with R.H.S. (p+1). It is an easily proved strict inequality. This should be around intermediate topics stuff.

$\frac{4^p ( 2p+1)}{\sqrt{p} (p+1)} \ge \frac{4^{p+1}}{2 \sqrt{p+1}}$
is equivalent to $(2p+1)\sqrt{p+1} \ge 2(p+1)\sqrt{p}$ or $(4p^2+4p+1)(p+1) \ge 4p(p^2+2p+1)$ or $4p^3+8p^2+5p+1 \ge 4p^2+8p+4p$

Yeah, I guess this was actually the easiest part of the problem. I don't know why I got so stuck on it. I did it pretty much exactly like this, but I kept getting that it was greater than something negative. Maybe I did some calculation error or something.

[quote="Soarer"]This should be around intermediate topics stuff.[/quote]

I decided to move it to Pre-Olympiad. In case you are interested - we also have ${2n \choose n} \leq 4^n$ (that's a weak but also famous inequality...)


In any two triangles with sides $ a,b,c$ and $ x,y,z$ with area $ S$ and $ S'$ respectively.

(dar har do mosallas ba tool haye $ a,b,c$ va $ x,y,z$ be tartib ba masahathaye $ S$ va $ S'$)

this inequality holds and has been proven by Daniel Pedoe(American mathematician):

(in namosavi ke tavassote Daniel Pedoe riazidan(hendese dan)e amricayi saabet shode bargharar ast):

$ z^{2}(a^{2}+b^{2}-c^{2})+y^{2}(a^{2}+c^{2}-b^{2})+x^{2}(b^{2}+c^{2}-a^{2})\geq16SS'$

and equality holds if and only if the triangles are similar in these pairs$ (x,a),(y,b),(z,c)$
(tasaavi bargharaar ast agar va faghat agar do mosallas be in tartib moshabeh bashand: $ (x,a),(y,b),(z,c)$)

khob midoonim $ S=\frac{1}{2}ab\sin C,S'=\frac{1}{2}xy\sin Z$ hamchenin tebghe ghazie $ \cos$ha dar mosalas darim:

$ c^{2}=a^{2}+b^{2}-2ab\cos C$

pas khahim dasht:

$ a^{2}+b^{2}-c^{2}=2ab\cos C$

$ b^{2}+c^{2}-a^{2}=2b^{2}-2ab\cos C$

$ c^{2}+a^{2}-b^{2}=2a^{2}-2ab\cos C$

pas soorate hokemoono mishe be shekle zir nevesht:

$ z^{2}(2ab\cos C)+y^{2}(2a^{2}-2ab\cos C)+x^{2}(2b^{2}-2ab\cos C)\geq 4abxy\sin C\sin Z$

hala tebghe ghazie $ \cos$ ha dar $ \triangle XYZ$ darim:

$ z^{2}=x^{2}+y^{2}-2xy\cos Z$

$ \Rightarrow z^{2}(2ab\cos C)=(x^{2}+y^{2}-2xy\cos Z)(2ab\cos C)$

pas namosavie akharo mishe intori neveshtesh:

$ (x^{2}+y^{2}-2xy\cos Z)2ab\cos C+y^{2}(2a^{2}-2ab\cos C)+x^{2}(2b^{2}-2ab\cos C)\geq 4abxy\sin C\sin Z$

khob hala biaid ino bazesh konido jomlehaye moshabeh ro sade konid mishe:

$ -4abxy\cos C\cos Z+2a^{2}y^{2}+2x^{2}b^{2}\geq 4abxy\sin C\sin Z$

pas kafie sabet konim:

$ 2(a^{2}y^{2}+x^{2}+b^{2})-4abxy(\cos C\cos Z+\sin C\sin Z)\geq 0$

hala biaid be samte chapesh,ebarate $ 4abxy$ ro ezafe va kam konid mishe:

$ 2(a{}^{y}{}^{2}+x^{2}b^{2}-2abxy)-4abxy(\cos C\cos Z+\sin C\sin Z-1)\geq 0$

amma midoonim $ \cos (C-Z)=\cos C\cos Z+\sin C\sin C$ pas namosavie akhar be shekle zir dar miad:

$ 2(ay-bx)^{2}-4abxy(\cos (C-Z)-1)\geq 0$

$ \Rightarrow 2(ay-bx)^{2}+4abxy[1-\cos (C-Z)]\geq 0$

ke inam vazehe chon darim $ cos \alpha\leq 1$...

hala vaseye halate tasavi ham chon $ (ay-bx)^{2}\geq 0$ va $ 1-\cos (C-Z)\geq 0$,halate tasavi zamani mishe ke $ ay=bx\rightarrow \boxed{\frac{a}{b}=\frac{x}{y}}(I)$ va hamchenin $ \cos (C-Z)=1$ yeni $ \angle (C-Z)=0$ yeni $ \boxed{\angle C=Z}(II)$

hala az ravabete $ (I),(II)$ natije mishe ke halate tasavi zamanie ke $ \triangle ABC \sim \triangle XYZ$...  :)


geometry
incenter
perimeter
trigonometry
geometry solved

Help me do this problem:

Given any triangle [i][b]ABC[/b][/i], a circle is inscribed and its diameter drawn perpendicular to [i][b]AC[/b][/i] with endpoints [i][b]D[/b][/i] and [i][b]E[/b][/i] on the circle, with [i][b]D[/b][/i] on [i][b]AC[/b][/i]. Line [i][b]BE[/b][/i] is extended to meet [i][b]AC[/b][/i] at [i][b]F[/b][/i]. Prove that [i][b]AF=DC[/b][/i].

Let A', B', C' the midpoints of BC, CA, AB, respectively, I the incenter and s the semiperimeter of the triagle $\triangle ABC$. The incenter I of the triangle $\triangle ABC$ is identical with the Nagel point of the medial triangle $\triangle A'B'C'$, which means that the line B'I splits in half the perimeter of the medial triangle. Let the line through B parallel to B'I meet the opposite side CA at $F'$. From central similarity of the triangle $\triangle ABC$ and its medial triangle $\triangle A'B'C'$, the line BF' is the perimeter splitter of the triangle $\triangle ABC$, i.e.,

$BC + CF' = s,\ CF' = s - BC$

But D is the incircle tangency point with CA, $CD = s - BC$ as well, so that CF' = AD. It follows that B' (the midpoint of CA) is the midpoint of DF' as well. Since $BF' \parallel B'I$, $DF' = 2DB'$ and DI is a radius of the incircle, the line BF' passes through the other enpoint of the diameter DE. It follows the the points $F' \equiv F$ are identical and $CF \equiv CF' = AD$. WLOG, let $\angle A > \angle C$. Then $CD = CF' + F'D = AD + DF = AF$.

The fact that the perimeter splitter through B is parallel to B'I can be shown without using the Nagel point in the following way: Denote a = BC, b = CA, c = AB and r the inradius. Let the perimeter splitter through B meet CA at F', i.e., BC + CF'  = s, CF' = s - a. D is the incircle tangency point with CA, hence AD = s - a as well. Let Q be the foot of the B-altitude. WLOG, assume $\angle A > \angle C$. It is sufficient to show that the right angle triangle $\triangle BQF' \sim \triangle IDB'$ are similar.

$\frac{DB'}{DI} = \frac{\frac b 2 - (s - a)}{r} = \frac{a - c}{2r}$

$\frac{QF'}{QB} = \frac{b - (s - a) - c \cos A}{c \sin A} =$

(substituting the cosine theorem $c \cos A = \frac{b^2 + c^2 - a^2}{2b}$)

$= \frac{2b^2 - (ab + b^2 + bc - 2ab) - (b^2 + c^2 - a^2)}{2bc \sin A} =$

(substituting the triangle area $\triangle = \frac 1 2 bc \sin A$)

$= \frac{ab - bc + a^2 - c^2}{4\triangle} = \frac{b(a - c) + (a + c)(a - c)}{4\triangle} =$

$= \frac{2s(a - c)}{4\triangle} = \frac{a - c}{2r}$

Hence, $\frac{DB'}{DI} = \frac{QF'}{QB},\ \triangle BQF' \sim \triangle IDB'$ and $BF' \parallel B'I$.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=19035 .

  Darij

draw a perpendicular line to $DE$ at $E$ and draw ex-circle of$\triangle ABC$ wrt $B$($X_a$ is the tangency point of this circle with $AC$) .and we see two similair triangle.
hence, $B,E,X_a$ are collinear.
from the wellknown relation we have :$AX_a=CD=p-c$


In any triangle prove that :

$ \frac{\cos\ A}{a(b\plus{}c)}\plus{}\frac{\cos\ B}{b(c\plus{}a)}\plus{}\frac{\cos\ C}{c(a\plus{}b)}\le \frac{1}{8Rr}$

[quote="alex2008"]In any triangle prove that :

$ \frac {\cos\ A}{a(b \plus{} c)} \plus{} \frac {\cos\ B}{b(c \plus{} a)} \plus{} \frac {\cos\ C}{c(a \plus{} b)}\le \frac {1}{8Rr}$[/quote]
[hide="My Solution"]$ \sum\frac{8Rr\cos A}{a(b \plus{} c)}\equal{}\sum\frac{4r\cos A}{\sin A(b\plus{}c)}\equal{}\sum\frac{4r\cos A}{\frac{2pr}{bc}(b\plus{}c)}\equal{}\frac{1}{2p}\sum\frac{4bc\cos A}{b\plus{}c}$
We need to prove that $ \sum\frac{4bc\cos A}{b\plus{}c}\le a\plus{}b\plus{}c$
W.L.O.G, assume that $ \cos A\ge \cos B\ge\cos C\implies \frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}\implies \frac{bc}{b\plus{}c}\le\frac{ca}{c\plus{}a}\le\frac{ab}{a\plus{}b}$
By Chebyshev inequality, we have:
$ \sum\frac{4bc}{b\plus{}c}\cos A\le \frac{1}{3}\left(\sum\frac{4bc}{b\plus{}c}\right)(\sum\cos A)\le a\plus{}b\plus{}c$[/hide]


Ethanolamine, $ \ce{HOCH2CH2NH2}$, and diethanolamine, $ \ce{(HOCH2CH2)2NH}$, are used commercially to form emulsifying agents and to absorb acidic gases. Propose syntheses of these two compounds.

for the first one:
ethene + MMPP(or something like dat. dont remember the name rite now) to ethylene oxide.
ethylene oxide + nh3

That's correct. How about the other compound?

MMPP = Magnesium MonoPeroxi Phthalate

oh ok tx.
treat ethanolamine with ethylene oxide again.

Yes, that's right. However, this problem was too easy. How about the others I posted? Too hard?

well some of them. am trying them currently. how come you posted so many problems all of a sudden? no complaints tho. :D


search
inequalities unsolved
inequalities

Let $ x,y,z >0$ such that $ xyz\equal{}1$. Prove that
$ \sqrt{(4\plus{}9x^2)} \plus{}\sqrt{(4\plus{}9y^2)} \plus{} \sqrt{(4\plus{}9z^2)} \leq \sqrt{13}(x\plus{}y\plus{}z)$

Posted before. Find in a post of arqady.


AMC
AIME
probability
number theory
relatively prime

In a five-team tournament, each team plays one game with every other team. Each team has a 50% chance of winning any game it plays. (There are no ties.) Let m/n be the probability that the tournament will produce neither an undefeated team nor a winless team, where m and n are relatively prime positive integers. Find m+n.

[hide="another way"]
There are 10 games played total, so there are $2^{10}$ possible outcomes.  There are $5*2^{6}$ ways for a team to win all their games or loose all their games.  There are $2*{5\choose2}*2^{3}= 5*2^{5}$ ways for a team to win all and loose all.  So the probability that at least one team wins all or looses all is

$\frac{5*2^{6}+5*2^{6}-5*2^{5}}{2^{10}}= \frac{5(2^{7}-2^{5})}{2^{10}}= \frac{5(3)}{32}$

Then $1-\frac{15}{32}= \frac{17}{32}$.

17 + 32 = 49.
[/hide]

[quote="lotrgreengrapes7926"][hide][quote="towersfreak2006"]Here was my method:

One team has probability $\left(\frac{1}{2}\right)^{4}$ of winning all of its games if no other conditions are given.
By the same reasoning, one team has probability $\frac{1}{16}$ of losing all its games.

Since there are five teams, then the probability of any one team winnning or losing all of its games is $5*\left(\frac{1}{16}+\frac{1}{16}\right)=\frac{5}{8}$. But we have overcounted the situations in which one team wins all its games and one team loses all its games.

The probability that a certain two teams do this is $\left(\frac{1}{2}\right)^{7}=\frac{1}{128}$. There are $5*4$ ways to choose two teams to do this.

Thus, the probability that at least one team wins all its game or one team loses all its games is $\frac{5}{8}-\frac{20}{128}=\frac{60}{128}=\frac{15}{32}$.

Hence, the desired answer is $1-\frac{15}{32}=\boxed{\frac{17}{32}}$ since we want the probability that no team wins or loses all its games.[/quote][/hide][/quote]
Why is the probability that there is a team that won 4 games and another that loss all 4 games = 1/2?


modular arithmetic
number theory proposed
number theory

Find all integers $ x,y$ such that $ 3x^2\plus{}8xy\plus{}3y^2$ is a perfect square.

[quote="spanferkel"]Find all integers $ x,y$ such that $ 3x^2 \plus{} 8xy \plus{} 3y^2$ is a perfect square.[/quote]

 $ 3x^2 \plus{} 8xy \plus{} 3y^2 \equal{} c^2$ $ \implies$ $ (3x \plus{} 4y)^2 \minus{} 7y^2 \equal{} 3c^2$ and so we need to study solutions of $ u^2 \minus{} 7v^2 \equal{} 3w^2$

But $ u^2\in\{0,1,2,4\}\pmod 7$ while $ 3w^2\in\{0,3,5,6\}\pmod 7$ and so $ u \equal{} w \equal{} 0\pmod 7$ and so $ 49u_1^2 \minus{} 7v^2 \equal{} 3\cdot 49w_1^2$, which implies $ v \equal{} 0\pmod 7$ and $ u_1^2 \minus{} 7v_1^2 \equal{} 3w_1^2$

And so $ u \equal{} v \equal{} w \equal{} 0$ is the only solution to $ u^2 \minus{} 7v^2 \equal{} 3w^2$

And so $ x \equal{} y \equal{} c \equal{} 0$ is the only solution of $ 3x^2 \plus{} 8xy \plus{} 3y^2 \equal{} c^2$

3x²+8xy+3y²=(2x+2y)²-x²-y²=c²
(2x+2y)²=x²+y²+c²
4 divide x²+y²+c² so 2 divide x, 2 divide y, 2 divide c.
x=2k y=2l c=2m
(2k+2l)²=k²+l²+m²
An infinit descent gives us x=y=c=0.

So far so good. Now the same question for $ 3x^2\plus{}10xy\plus{}3y^2$.  :)

$ 3x^2\plus{}10xy\plus{}3y^2\equal{}(x\plus{}3y)(3x\plus{}y)\equal{}c^2$
$ \gcd(x,y)\equal{}d$
$ x\equal{}da$ and $ y\equal{}db$
We get $ (a\plus{}3b)(b\plus{}3a)\equal{}k^2$
and $ \gcd(a\plus{}3b,b\plus{}3a)\equal{}1,2,4,8$
$ a\plus{}3b\equal{}u^2$
$ b\plus{}3a\equal{}v^2$
So, $ 4(a\plus{}b)\equal{}u^2\plus{}v^2$ but $ 2$ divide $ u$ and $ 2$ divide $ v$ wich is a contradition!
$ a\plus{}3b\equal{}2u^2$
$ b\plus{}3a\equal{}2v^2$
$ 6v^2\minus{}2u^2\equal{}8a\Longleftrightarrow 3v^2\minus{}u^2\equal{}4a$ but $ 3v^2\minus{}u^2\equiv 3\minus{}1\equal{} 2 \mod 4$ cont!
$ a\plus{}3b\equal{}4u^2$
$ b\plus{}3a\equal{}4v^2$
$ 12v^2\minus{}4u^2\equal{}8a$ and $ 3v^2\minus{}u^2\equal{}2a$    wich is true and so you deduce b;
$ a\plus{}3b\equal{}8u^2$
$ b\plus{}3a\equal{}8v^2$
$ 24v^2\minus{}8u^2\equal{}8a$ so $ a\equal{}3v^2\minus{}u^2$ and so you deduce b!
After you can deduce $ x$ and $ y$ !


when P(z) is divided by x-1 the remainder is 1 when P(z) is divided by (x-2)(x-3) the remainder is 5 wat is the remainder when P(z) is divided by (x-1)(x-2)(x-3)

plz help

using remainder thereom p(1)=1 p(2)=5 P(3)=5 since its (x-2)(x-3)qx^2+5 so when p=2 its p(2)=0+5 and same for p(3) so a^2+b+c=1 4a^2+2b+c=5 and 9a^2+6b+c=5 so -2a=4 a=-2 b=-5a=10 and c=1-a-b=-7 so remainder is -2x^2+10x-7

[hide="yes, you are correct  :yup: "]

you can express the polynomial P(x) as

$ P(x) \equal{} (x\minus{}1)(x\minus{}2)(x\minus{}3)Q(x) \plus{} ax^2 \plus{} bx \plus{} c$

$ P(1) \equal{} 1 \ \Rightarrow \ a\plus{}b\plus{}c\equal{}1 \\ \\ P(2) \equal{} 5 \ \Rightarrow \ 4a\plus{}2b\plus{}c\equal{}5 \\ \\ P(3) \equal{} 5 \ \Rightarrow \ 9a\plus{}3b\plus{}c\equal{}5$

Solving for $ a,b,c$ gives $ a\equal{}\minus{}2,b\equal{}10,c\equal{}\minus{}7$

Thus, the required remainder is $ \minus{}2x^2\plus{}10x\minus{}7$.

[/hide]

[quote="Nora.91"]
you can express the polynomial P(x) as

$ P(x) \equal{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)Q(x) \plus{} ax^2 \plus{} bx \plus{} c$
[/hide][/quote]

Wait how did you get Q(x)?

[quote="generalmotors"]
Wait how did you get Q(x)?[/quote]

you don't need to find it for this question ...

i used the fact that if a polynomial $ P(x)$ leaves remainder $ k$ on dividing by $ x\minus{}a$, then

$ P(x) \equal{} (x\minus{}a)Q(x) \plus{} k$ where $ Q(x)$ is another polynomial with lower degree than $ P(x)$.

 :)


Umm..why do you have a word document just for:

\[\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}=?\]


When something is uniquely determined, what does that mean?

It means that something is the only solution for a certain problem, with the given restrictions.

... frequently, up to some concept of "isomorphism."


what is the probability to get a full house from 13 cards?
what is the probability that no husband sits next to his wife 5 If 4 married couples are arranged in a row?

Which 13 cards are u talking about in #1?  Do u mean a chosen suit?  The answer is different depending on if you're talking about a specific set of 13 cards or a random set of 13 cards.

Should #2 read:  What is the probability that no husband sits next to his wife If 4 married couples are arranged in a row?

For #1 he means that 3 of the five cards are one particular card (Ace, King, Queen, ... , 2) and 2 are another particular card (not the same card, of course, as there are only 4 of each card in a deck).
Oh, and suit doesn't matter.

#1 choose the 13 card randomly from a deck
#2 Yes.

And Hokkage explained what full house is :)


If $ p$ is a prime greater than 2, define $ p\equal{}2q\plus{}1$. Prove that $ (q!)^2\plus{}(\minus{}1)^q$ is divisible by $ p$.

[hide]By Wilson's Theorem, $ (p \minus{} 1)!\equiv \minus{} 1 \pmod{p} \implies (2q)!\equiv \minus{} 1\pmod{p}$. Thus,
\[ q! (q \plus{} 1)(q \plus{} 2)\cdots (2q) \equiv q! ( \minus{} q)( \minus{} q \plus{} 1)\cdots ( \minus{} 1)\equiv (q!)^{2}( \minus{} 1)^{q}\equiv \minus{} 1\pmod{p}
\]
Note that $ p|( \minus{} 1)^{q}((q!)^{2} \plus{} ( \minus{} 1)^{q})$ as seen by
\[ ( \minus{} 1)^{q}((q!)^{2} \plus{} ( \minus{} 1)^{q})\equiv \minus{} 1 \plus{} ( \minus{} 1)^{2q}\equiv 0 \pmod{p}
\]
Hence, since $ p\nmid ( \minus{} 1)^{q}$, $ p$ must divide $ (q!)^{2} \plus{} ( \minus{} 1)^{q}$. [/hide]

Why is $ (q\plus{}1) \equiv \minus{}q \pmod p$ ?  :blush:


The triangle ABC has BC=1 and $ \angle BAC \equal{} a$. Find the shortest distance between its incenter and its centroid. Denote this shortest distance by $ f(a)$. When a varies in the interval $ (\frac {\pi}{3},\pi)$, find the maximum value of $ f(a)$.


Is there a list of latex commands on FTW, i often miss questions because I don't know the coding.

Thanks

you dont hafta use LATEX on FTW. i use

pi=$ \pi$
sqrtx=$ \sqrt{x}$
x^2=$ x^2$

Take a look at this.  (LaTeX codes for FTW are more to the bottom of the page)
http://www.artofproblemsolving.com/Edutainment/g1/Pages/Documentation.php
:)

lol, on one game i used sqrt.(2) as an answer, won't screw that up again


Let $ ABCD$ convex quadrilateral. Let $ O_1$ circle intersects $ A,B$ and tangent with $ CD$ at $ P$. Let $ O_2$ circle intersects $ C,D$ and tangent with $ AB$ at $ Q$. Let $ O_1$ and $ O_2$ meet at $ E$ and $ F$.
Prove that $ BC$ parallel with $ DA$ if and only if $ EF$ intersects the midpoint of $ PQ$

This the sketch, but BC not parallel with DA


arithmetic sequence
number theory unsolved
number theory

Find infinitely many triples (a,b,c) of positive integers such that a,b,c are in arithmetic progression and such that ab+1, bc+1, and ca+1 are perfect squares.

If $ab+1$ be a perfect square ,thus suppose that $c=a+b+2\sqrt{ab+1}$  :D

Hi guys,this problem is easy.
Looking this[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=378715#p378715]www.mathlinks.ro/Forum/viewtopic.php?p=378715#p378715[/url]But this problem arn't hard too :(  :D

but a,b,c are in arithmetic progression  ???

Easy ????


Hi folks,

I have what may be a stupid question. How do you inscribe (construct) a semi-circle in a right triangle?

Thanks,
Tom

Draw an angle bisector, and intersect it with the opposite side. That's the center. For the radius, drop a perpendicular from that point to another side.
The nicest one is the bisector of the right angle.

I knew it was something stupid. But I had a brain freeze.

Thanks!


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $G$ be a finite abelian  group and $x,y\in G$ .suppose that the factor group of $G$ with respect to the subgroup generated by $x$ and factor group of $G$ respect to the subgroup generated by $y$ are isomorphism .prove that $G$ has an automorphim that maps $x$ to $y$.

We note $p_{x}: G \rightarrow G/<x>$ and $p_{y}: G \rightarrow G/<y>$ the canonical projections, and $\phi: G/<x> \rightarrow G/<y>$ a group isomorphism. For each element $C$ of $G/<x>$ (resp. $G/<y>$), we pick a representant $\alpha(C)$ of $C$ in $G$.
Consider the mapping $g = \alpha(p_{x}(g))+kx \mapsto \alpha(\phi(p_{x}(g)))+ky$. It is easy to show that this mapping is well defined and is an automorphism of $G$ sending $x$ to $y$. Note that in order to show the surjectivity we need the fact that $x$ and $y$ have same order in $G$ (which can be derived from the fact that $|G/<x>|=|G/<y>|$ since $\phi$ is an isomorphism).


The length of the circumradius of regular hexagon ABCDEK equals [i]R[/i]. Find the length of inradius of triangle ABD

Note that in a regular hexagon the lenght of the circumradius is the same as the side of the hexagon.

 :)

[hide="hint"]The sides of the trianglse in question are R, 2R and $ \sqrt{3} R$.
$ \Delta \equal{} rs \equal{} \frac {abc}{4R}$[/hide]


LaTeX
inequalities proposed
inequalities

[quote="Fermat -Euler"]i'm sorry i have latex but i dont know how to use it in this forum[/quote]
you don't need latex at all to use it in the forum you only need to put your latex code in between two dollar signs.


geometry
circumcircle
trapezoid
perpendicular bisector
geometry proposed

Let $w=C(O,R)$ be the circumcircle of the triangle $ABC$ and the points $E$, $F$ so that $BEO\sim COF$, $[AE]\cap OB=\emptyset$, $[AF]\cap CO=\emptyset$. I note the intersection $D\in BC\cap EF$. Prove that the line $OD$ is tangent to the circumcircle of the triangle $EOF$.

[u]Remark.[/u] If $BEO\sim ABC$ then we obtain the problem from
http://www.mathlinks.ro/Forum/viewtopic.php?t=67235

[color=brown][b]Lemma[/b]
Let $k=(I,r)$ and $B$ an external point of $k$. Let $O$ be a point such that $BI^2-r^2=BO^2$ and the line $d$ such that $B\in d, d\perp OI$.
Then $\forall P \in d$ we have $PI^2-r^2=PO^2$

[hide="Proof of the Lemma"]
$BI^2-BO^2 = r^2$
Let $M \in d \cap OI$

$PI^2 = PM^2+MI^2$
$PO^2 = PM^2+MO^2$
$\Rightarrow$
$PI^2-PO^2 = MI^2-MO^2 = \text{ fixed } = r^2$
(Actually, the point $M$ is the midpoint of $OO'$, where $O'$ is the inverse of $O$)[/color][/hide]

In the problem we have $D\in BC \cap EF,$ so $BC \not \parallel EF$

$\boxed{[AE] \cap OB = \emptyset, [AF]\cap CO = \emptyset} \ \ (1)$

The conditions $(1)$ are not sufficient.
We can choose $A$ on $(O,R)$ such that these conditions both hold but $OD$ is not a tangent of the circumcircle of $\triangle EOF$.

[hide="A sufficient condition - an assumption"]
A necessary and sufficient condition is that the angles $OBE,OCF$ are "on the same side" of the angles $OBC,OCB$.
I know that it sounds too complicated, so I'll try to explain it.

Let $E',F'$ be the points where the rays $BE,CF$ intersect the circle $(O,R)$.
If $BB',CC'$ are diameters of $(O,R)$ then we have:

$\angle OBE = \angle OCF \Rightarrow \angle B'BE' = \angle C'CF' \Rightarrow$

$(B'E') = (C'F') \ \ (2),$ where $(XY)$ is the arc $XY$

once we know the place of $E'$, we can find two points for $F'$ such  that $(2)$ holds.
One of these points (the convinient) is when $E'F' \parallel BC$, in other words the points $E',F'$ are symmetric w.r.t. the perpedicular bisector of $BC$.
So we make this assumption:

[u][color=green]We assume that the points E',F' are symmetric w.r.t. the perpedicular bisector of BC.[/color][/u]
[/hide]

$\triangle BEO \sim \triangle COF \Rightarrow  \frac{BE}{CO} = \frac{BO}{CF} \Rightarrow  BO^2 = BE \cdot CF \ \ (3)$

Let $H$ be the symmetric point of $F$ w.r.t. the perpendicular bisector of $BC$.
According to our assumption, the point $H$ belongs to the ray $BE$ (this is important because ensures that $B \not \in [HE]$)
Also, $H$ is different than $E$ because $EF \not \parallel BC$.

The segment $BC,FH$ have common perpendicular, hence $BCFH$ is an isosceles trapezoid and $BH=CF$

$(3) \Rightarrow  BO^2 = BE \cdot BH$

[b]Apply the Lemma[/b]
Let $k=(I,r)$ be the circumcircle of the triangle $EFH$
From $(3)$, the power of $B$ w.r.t. $k$ is $BO^2$
The perpendicular bisector of $BCFH$ pass through $I$, so $OI \perp BC$
$B \not \in [HE] \Rightarrow  B \not \in k$

According to the Lemma, every point $P$ on the line $BC$ has power equal to $PO^2$.

So for the point $D$ we have $\boxed{DF\cdot DE=DO^2} \ \ (4)$

[color=brown][b]Comment[/b]
We made the assumption that $E'F' \parallel BC$. The points $E,F$ are on the rays $BE',CF'$.
So $E,F,E',F'$ are all on the same side of $BC$.
This implies that $[EF] \cap BC = \emptyset \Rightarrow  D \not \in EF \Rightarrow  D$ is out of the circle $k$

So the relation $(4)$ says that $DO$ is tangent to $k$[/color]


Vieta algebra polynomial MATHCOUNTS calculus derivative

problem:

((1/3)*t^3)+t=s

find t in terms of s?

answer should just be t in terms of just s.

[hide] Since $\frac{t^{3}}{3}+t=s$, we can say that $\frac{t^{3}+3t}{3}=s$, so $t^{3}+3t-3s=0$.  Now, we must solve for t, but this is a depressed cubic equation, which I don't know how to solve.  [/hide]

[hide="partial solution"]

$\frac{t^{3}}{3}+t=s$

$t(\frac{t^{2}}{3}+1)=s$

$t=\frac{s}{\frac{t^{2}}{3}+1}=\frac{s}{\frac{t^{2}+3}{3}}=\frac{3s}{t^{2}+3}$

$t^{3}+3t=3s$

$t^{3}+3t-3s=0$

We use Vieta's formula, and see that the sum of the roots is 0.[/hide]

[hide="an attempt"]
$\frac{t^{3}}{3}+t = s$
$t^{3}+3t=3s$
$t^{3}+3t-3s=0$
since solving it algebraicly gets it into a cubic equation that i cannot solve, i have to solve it this way:
Answer: $t=f(s)$ where $f(s)$ is $t$ in terms of $s$

[/hide]

you know, there's no use for posting the same thing someone else did
  :|

[url]http://en.wikipedia.org/wiki/Cubic_equation[/url]

Have a lot of fun !  :oops:

Cubic polynomials aren't quite Mathcounts problems, so I'm moving this topic to Intermediate.

I'm wondering if this can even be solved in intermediate.

ok. an attempt to solve the equation,

[hide="in here"]The coeficciant of the t cubed term is 1, and there is no t squared term. so that eliminates 2 steps.

So let's say that $t=v-u$, where $u^{3}-v^{3}=-3s$ and $uv=1$

Therefore, $v=\frac{1}{u}$ and $u=\sqrt[3]{\frac{-3s}{2}\pm\sqrt{\frac{-3s}{4}+1}}$

So

$t=\frac{1}{\sqrt[3]{\frac{-3s}{2}\pm\sqrt{\frac{-3s+4}{4}}}}-\sqrt[3]{\frac{-3s}{2}\pm\sqrt{\frac{-3s+4}{4}}}=\frac{1}{\sqrt[3]{\frac{-3s}{2}\pm\frac{\sqrt{-3s+4}}{2}}}-\sqrt[3]{\frac{-3s}{2}\pm\frac{\sqrt{-3s+4}}{2}}$

$=\frac{1}{\sqrt[3]{\frac{-3s\pm\sqrt{-3s+4}}{2}}}-\sqrt[3]{\frac{-3s\pm\sqrt{-3s+4}}{2}}$[/hide]

Someone please simplify that. :rotfl:

yea

it's cool to solve cubic equations, but in reality, i think that the sum, product, symmetric sums, etc. of the polynomials in general are better and eventually lead to more elegant solutions

This is the [b]third[/b] topic you've posted about this problem, at least.  Cut it out.

t0rajir0u is right, links to the other spamming topics are below:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=736933#736933[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=736935#736935[/url]

ok. I'm refreshed enough so that I might just be able to solve it.

[hide="in here"]We left off with

$t=\frac{1}{\sqrt[3]{\frac{-3s\pm\sqrt{-3s+4}}{2}}}-\sqrt[3]{\frac{-3s\pm\sqrt{-3s+4}}{2}}$

We can simplify that to

$\sqrt[3]{\frac{2}{-3s\pm\sqrt{-3s+4}}}-\sqrt[3]{\frac{-3s\pm\sqrt{-3s+4}}{2}}$

We find a common denominator and add:

$\sqrt[3]{\frac{4}{-6s\pm2\sqrt{-3s+4}}}-\sqrt[3]{\frac{9s^{2}-3s+4\mp3s\sqrt{-3s+4}}{-6s\pm2\sqrt{-3s+4}}}=\frac{\sqrt[3]{4}-\sqrt[3]{9s^{2}-3s+4\mp3s\sqrt{-3s+4}}}{2\sqrt[3]{-6s\pm2\sqrt{-3s+4}}}$[/hide]

uhh.... I hope that satisfies mli7... :rotfl:

Yuck... I didn't know the cubic formula was so ugly.  :wacko: Is there more of a method to solving a cubic equation?

[quote="white_horse_king88"]What's interesting is that on my Calculus test last week this exact problem came up and we had to find the derivative of the inverse of that function. It was funny because I actually was *smart* :rotfl:  enough to try and do it algebraically before realizing how *smart* I was. So if this is part of a larger problem, use a calculator to find whatever it is that needs to be found.[/quote]

Instead of taking the reciprocal of the derivative...? :D

But that's getting into calculus.

Yep, the cubic formula is ugly. I believe 1=2's method was first used by Tartaglia and Cardano.


Find all prime numbers $ a,b,c$ that fulfill the equality:
$ (a\minus{}2)!\plus{}2b!\equal{}22c\minus{}1$

In right side number is odd, in left it odd only if $ a\equal{}2,3$. It give $ 11c\equal{}1\plus{}\frac{2b!}{2}$.
If you mean $ 2b!\equal{} (2b)!$, then no solution. Else $ 11c\equal{}1\plus{}b!\to b\equal{}5,c\equal{}11.$


algorithm
number theory
Euclidean algorithm

A] For positive integer $ q$ prove that the fraction:

$ \frac{15q^3\plus{}4q^2\plus{}5q\plus{}1}{15q^3\plus{}19q^2\plus{}9q\plus{}1}$ is irreducible.

B] For prime $ p$ prove that the fraction:

$ \frac{35p^3\plus{}5p!\plus{}p}{105p^4\plus{}10p^2\plus{}15p \cdot p! \plus{}(p\minus{}1)!}$ is irreducible.

 :)

[hide="hint for A"]

Euclidean Algorithm

[/hide]

[hide="solution for A"]

Using the Euclidean Algorithm, we get:

$ (15q^3\plus{}19q^2\plus{}9q\plus{}1, 15q^3\plus{}4q^2\plus{}5q\plus{}1)$
$ (15q^3\plus{}4q^2\plus{}5q\plus{}1, 15q^2\plus{}4q)$
$ (15q^2\plus{}4q, 5q\plus{}1)$
$ (5q\plus{}1,q)$
$ (q,1)$
$ (1,0) \rightarrow \boxed{1}$.  Thus, the numerator and denominator are relatively prime.[/hide]

[hide="B"]Let the numerator be $ a_1$ and the denominator be $ a_2$.

Notice that $ a_3 \equal{} a_2 \minus{} 3pa_1 \equal{} 7p^2 \plus{} (p\minus{}1)!$.

We also have $ a_4 \equal{} a_1 \minus{} 5p^2a_3 \equal{} p$.

Observe that $ a_2$ mod $ p$ is $ (p\minus{}1)!$. This is nonzero since $ p$ is prime. By the Euclidean Algorithm, $ (a_1,a_2) \equal{} (a_4,a_2)$. But $ a_4 \equal{} p$ and $ p \nmid a_2$. So $ (a_4,a_2) \equal{} 1$. Thus $ (a_1,a_2) \equal{} 1$ and the fraction is irreducible.[/hide]


is there any relation between the roots of a polynomial?
by this i mean take for example $ x^2\minus{}3\plus{}2$ the roots are $ x\equal{}1,2$
is it possible to get other roots if u knew one ? not only in case of quadratics but in general i mean.

You could use division (or synthetic division) to simplify the polynomial into one of a lesser degree...
Oh, I almost forgot some other important stuff:
In a polynomial of degree n:
Assign a as the coefficient of x^n, and b as the coefficient of x^(n-1)
The sum of the roots is -b/a
Also, the product of the roots is the constant term (if the polynomial is of an odd degree) and the opposite of the constant term (if the polynomial is of an even degree)

Your question's not very clear, so here's my attempts at understanding it.

Consider the polynomial $ P(x) \equal{} x^3 \minus{} 6x^2 \plus{} 11x \minus{} 6$. By inspection, maybe by observing the sum of coefficients being $ 0$, you can find that $ x \equal{} 1$ is a root. This means that $ P(x)$ is divisible by $ x \minus{} 1$, so it can be written as $ P(x) \equal{} (x \minus{} 1)Q(x)$ for another polynomial $ Q(x)$, and $ Q$ can be easily determined through synthetic division or otherwise. The result is $ Q(x) \equal{} x^2 \minus{} 5x \plus{} 6$. All roots of $ Q(x)$ are also roots of $ P(x)$, so by finding a root of $ P(x)$ we have it much simpler to find other roots of $ P(x)$.

Or maybe you were looking for something like [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas]Vieta's[/url].

And the above poster just made the same guesses I did. :roll:

EDIT: Yar, I did forget something, but notice that you did in your first post as well.

@mellowmellon:-i think u forgot to put a power of 2
noooooooo i dont mean those, thanks but i knew all of those you two said. i am lookin for another thing.
$ p(x)\equal{}x^3\minus{}6x^2\plus{}11x\minus{}6$ if we has the root you said the $ 1$. would it be possible to substitue that one into another formula that could get the other two roots, stright away ? not by synthetic division or long devision.

No.  If that was the case, you could solve a general quintic just by multiplying it by $ (x\minus{}1)$.  :wink:

if the u take a quintic that has a root $ 1$ then u can do that. i think its possible

He's implicitly invoking that the general quintic is not solvable in radicals.

http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

I don't understand exactly what you're trying to do, but it looks like this theorem might be a bit of a problem to it.


function
integration
calculus
real analysis
geometry
irrational number
analytic geometry

How is this theorem proved?

$\exists \int_{a}^{b}f(x)dx\Longleftrightarrow f(x) \mbox{ is piecewise continuous on [a,b]}$

That's not an if and only if. There are integrable functions which are not piecewise continuous.

You didn't mention what theory of integration you are using, but I'll assume that you are using the Riemann theory.

Example (famous):

Let $f(x)=\begin{cases}\frac1q&\text{if }x=\frac{p}{q}\text{ in lowest terms}\\0&\text{if }x\text{ is irrational}\end{cases}$

There is no interval on which this function is continuous, so it cannot be piecewise continuous. It is continuous at every irrational number and discontinuous at every rational number. Nonetheless it it Riemann integrable on any interval, and has integral zero.

--

It is true that a piecewise continuous function is Riemann integrable. This would follow from the following:

(1) A continous function on a bounded interval is Riemann integrable. The proof uses the notion of uniform continuity.

(2) A piecewise integrable function is integrable, just by partitioning appropriately.

--

There is a theorem which characterizes the Riemann integrable functions: A bounded function on a bounded interval is Riemann integrable if and only if the set of its discontinuities has Lebesgue measure zero.

[quote="Kent Merryfield"]That's not an if and only if. There are integrable functions which are not piecewise continuous.
[/quote]
I read this in the following book:
Calculus and Analytic Geometry, Al Shenk, Page 217
Goodyear Publishing Company, Inc.

It said that this theorem is proved in advanced calculus.

*The integral I'm using is the Riemann Integral.

Are you sure Shenk wasn't just claiming the correct one-way implication? That piecewise continuity implies integrability? I have my doubts that he would have claimed the false converse (or that his editors or reviewers would let him get away with that.) What was his exact wording?

[quote="Kent Merryfield"] What was his exact wording?[/quote]

This is his exact wording:

[i]Note.[/i] In advanced calculus it is proved that the integral

(1) $\int_{a}^{b}f(x)dx$

exists if $f$ is PIECEWISE CONTINUOUS for $a\leq x\leq b$.

As I expected. The key word is that "if" on the last line. The author's claim is correct. He did not say, nor did he intend to say, "iff." The claim is also nowhere near the strongest possible result, and its proof is deferred beyond the end of the first-year calculus course for which the textbook was written.

He's willing to accept less than the strongest possible result, becuase piecewise continuous functions are a broad enough class to contain all of the examples he intends to use in his book. He's unwilling to face the proof because it would interrupt the flow and, at that point, simply confuse the average student. Indeed, the proof requires the development of the concept of uniform continuity and the (compactness-related) proof that a continuous function on a closed bounded interval is uniformly continuous. These are concepts I would hesitate to mention at the level of that course.

One could proof a different sufficient condition for Riemann integrability using more elementary techniques: a function is Riemann integrable if it is piecewise monotone.

This theorem neither contains nor is contained in the other theorem. Its proof is somewhat easier, requiring only the notion of upper and lower sums and a telescoping sum argument. It neither contains nor is contained in the other result becuase it is possible to find functions that are piecewise contininuous but not piecewise monotone. (Example: $f(x)=x\sin(1/x)$ and $f(0)=0,$) and it is possible to find functions that are piecewise monotone but not piecewise continuous. For an example of that, let $r_{n}$ be an enumeration of all of the rationals in $[0,1]$ and let

$g(x)=\sum_{r_{n}<x}2^{-n}.$

This function is monotone increasing on $[0,1]$ but discontinuous at every rational.

Yes, you are right. I can't believe I made this mistake... :blush:


Find all the positive integers x,y,z satisfiing : x^{2}+y^{2}+z^{2}=2xyz

Do you think it is algebra?  :?
The problem is from Larson and I think it's well known.  :)

sure it's not algebra but what i think is that no body gave me a good solution
i will put it in number theory
and i will wait for your solution :D


Simplify $ \displaystyle\frac{3x^2 \plus{} 9x}{4x^3 \plus{} 12x^2}$.

You can pull out a $ 3x$ up top and $ 4x^{2}$ below:
$ \frac{3x(x\plus{}3)}{4x^{2}(x\plus{}3)}\equal{}\frac{3}{4x}$

FTW only accepts "3/(4x)" not "3/4x".  I got it wrong with the latter.  :(

It doesn't accept "3/4x" as the answer because from the order of operations, 3/4x is interpreted as $\frac{3}{4}x$.


Calculate:
$I_{n}=\int\limits_{0}^{n}\frac{sinnx}{sinx}dx, n\in Z$