geometry
geometric transformation
rotation

Generally, what effects do double and triple bonds have on typical physical properties such as boiling points (e.g. compared with single bonds)?

Double and triple bonds differ in two important ways from single bonds: they are shorter, and they do not allow rotation.  That said, generally the shorter bonds will allow less space for van der Waals dispersion forces, and boiling points decrease, vapor pressure increases, etc.

About the rotation, generally this either has no real effect (although it is good to keep in mind we haven't really defined relative to what) or this lowers boiling point due to poor shapes for packing.

Correct me if I am wrong but doesn't Molecular Shape, Polarity and Bond types in general (Ionic Bonds, Covalent Bonds and Metallic Bonds) have more to do with Phase Changes?

Thanks for the responses.
[quote="gregx007"]generally the shorter bonds will allow less space for van der Waals dispersion forces, and boiling points decrease, [b]vapor pressure increases,[/b] etc.[/quote]

Why is it that the vapor pressure increases?

Franz -- you are correct, if I have interpreted your post right.  Note that multiple bonds affect the molecular shape, and hence also affect phase changes.  However, molecular shape, polarity, and bond type also affect other physical properties (arguably, they affect all of the species's physical properties).

jen7-- Boiling point is directly related to the vapor pressure (actually, boiling point is the temperature at which the vapor pressure is equal to atmospheric pressure).  If that doesn't make sense, don't worry about it.  Keep in mind the origin of vapor pressure--molecules at the surface of a liquid (or solid I suppose) are in constant motion, and some have enough energy to move off the surface.  The weaker the intermolecular forces are, the easier it is to move off, and the more molecules are in gas phase, which contributes to higher pressure.

side note about the rotation,
keep in mind that presence of a single bond does not mean rotation, conjucated double bonds will affect the single bonds in between due to resonance and will not allow rotation. (1,3 butadien for example)

Good point.  Other moieties also exhibit this phenomemon... amides for example.  The question becomes, can such bonds really be considered single?

[quote="wizard2378"]An increased order of bonds tends to cause the molecules to favor the more dense state, solid, then liquid, then gas.  This lowers freezing and boiling points.  This is because of the interaction of the electron dense bonds with the electropostive hydrogen or carbon bonds.   (Think of water's hydrogen bonds done light.)[/quote]

I think you mean "This raises freezing and boiling points"...


Find all functions $f: \mathbb R \to \mathbb R$ satisfying
\[ f(x\plus{}y)\equal{}f(x)f(y)f(xy)\] for all $ x,y\in \mathbb R$.

[quote="apratimgtr"]Find all $ f: R\rightarrow R$ satisfying,
$ f(x \plus{} y) \equal{} f(x)f(y)f(xy)$ for all $ x,y\in R$

I can't guess the function,please give me a hint.[/quote]

[hide="Some hint"]
Compute $ f(x \plus{} y \plus{} z)$
[/hide]

[hide="Solution"]
If $ \exists u$ such that $ f(u) \equal{} 0$, then $ P(x \minus{} u,u)$ $ \implies$ $ f(x) \equal{} f(x \minus{} u)f(u)f((x \minus{} u)u) \equal{} 0$ and we get the solution $ f(x) \equal{} 0$ $ \forall x$

Suppose now $ f(x)\neq 0$ $ \forall x$

$ f(x \plus{} y \plus{} z) \equal{} f(x \plus{} y)f(z)f(xz \plus{} yz)$ $ \equal{} f(x)f(y)f(xy)f(z)$ $ f(xz)f(yz)f(xyz^2)$

So $ f(x \plus{} y \plus{} z) \equal{} f(x)f(y)f(z)f(xy)f(xz)f(yz)f(xyz^2)$

So (since no factor is zero) $ f(xyz^2) \equal{} f(x^2yz) \equal{} f(xy^2z)$ $ \forall x,y,z$ and so $ f(x) \equal{} a$ $ \forall x\neq 0$ ,with $ a \equal{} a^3$
The case $ f(0)$ is easy to solve and gives $ f(0)^2 \equal{} 1$. Considering then $ P(x,\minus{}x)$, we get that $ f(0)\equal{}a$ too.

Hence the solutions :
$ f(x) \equal{} 0$ $ \forall x$
$ f(x) \equal{} \minus{} 1$ $ \forall x$
$ f(x) \equal{} \plus{} 1$ $ \forall x$
[/hide]


Harvard college MIT MATHCOUNTS videos email \/closed

The math competitions during this spring give community members a chance to bring more students from other schools to the forum.  Please invite your friends and competitors to join us.

We will also be spreading the word about the community at several prominent competitions such as the Harvard-MIT Tournament, the UNCC competition, the Great Plains Math League, and others.  We will be offering free classs enrollments as prizes at several contests.

If you know of a contest that we should contact regarding publicizing the site, please let us know about the contest, and who to contact (if you know).

Thanks!

You could contact [url=http://www.imlem.org]IMLEM[/url] for middle schoolers and offer them like MATHCOUNTS class enrollment and such, or offer [url][url=http://www.delta.edu/math/mmpc/]MMPC[/url][/url] or [url=http://www.maml.org]MAML Olympiad[/url]high scorers prizes too on the state level (and I'm sure there's at least one for each state too). Good luck!

The Math competition that I compete in is TMSCA but I don't know if you contacted them yet but serveral from that competition on here.

[quote="rrusczyk"]If you know of a contest that we should contact regarding publicizing the site, please let us know about the contest, and who to contact (if you know).[/quote]

The Minnesota Junior High Mathematics League site

http://members.aol.com/mathleague/

is where official results of that contest will be posted soon. There are only division contests in that league, not a statewide contest, but there are statewide standings. The director of the contest, Marlys Henke, can be reached from the site, and might very well be persuaded if you asked her to put a link on her site to this site.

Hope this helps!

Maybe we could dig up that idea about a T-shirt and we can each wear one to school or something, all it has to say is "www.artofproblemsolving.com" so it shouldn't be that hard to make...I could write that with a pen on a white T-shirt :)

Richard, does AoPS have a flyer that students or coaches could distribute at Mathcounts or other competitions?

Because Mathcounts Nats is on TV, and because the list of all people who made Nats is ascertainable before Nats starts, it seems to me that there might be some marketing value in sending an AoPS T-shirt to every registered AoPS member who makes Nats this year. Whether that would be ECONOMICALLY SOUND marketing value is another story [grin], as I know from my own sad experience that one can have a very good Web site without necessarily making much (or any) money off of it.

Or do Mathcounts participants have to wear official Mathcounts T-shirts, as participants in the national spelling bee have to wear spelling bee T-shirts? In that case, other brands of T-shirts wouldn't have any visibility on ESPN. (The new-to-video movie Spellbound is a must-see for all national contest competitors; it's very interesting.)

i think all bees are very biased competitions although (i made 3rd and 1st in state in geography). especially speeling bee. everybody should get the same 100 words compile the paper and seee who wins right? just like how math comptetition work
or at least include some questions that everybody answers like in the geography bee
and what is even more ridiculous is the spelling bee's one AMM and you are out (geography bee gives you two and in preliminary its the total points added up)
whenver they do the one question and you're out is ridiculous
i got kicked out in the class in 5th, 6th, and 7th grade. In elementary the teacher included three more sutdents for the school bee through a testing paper. in 7th grade the teacher felt pity for me and since alot dind't show up i was in.

[quote]We will also be spreading the word about the community at several prominent competitions such as the Harvard-MIT Tournament[/quote]  does this mean you will actually be there?

Sorry it's taken me so long to respond to the questions/comments in this thread - I'm on vacation.

Yes, we have a flyer people can distribute.  We'd be happy to send a stack of them to whomever would like to distribute them.  Just send me an email with the details.

Unfortunately we won't be at the MIT/Harvard contest in person, but we hope to some day.

We'll be in touch with the other contests suggested here; thanks much!


geometry
3D geometry
sphere
trigonometry
analytic geometry

/*This topic also posted on the Intermediate forum, please delete if illegal, apologize if illegal*/

Consider two points on a perfect sphere. Let the radius of the sphere be r, and the latitudes and longitudes of the two points be (a, b) and (x, y) (Or whatever you prefer to call it). It has been discovered that the points are relatively close together (Say, the are in the same octant). What is the distance between the points, 1) along the surface of the sphere, and 2) through Euclidean space?

For 1), I'm thinking you apply $\theta * r = s$, but I'm nt sure how to get the $\theta$.

For 2), its an SAS trig problem if you can find $\theta$, but again, I'm having trouble with $\theta$.

Any help would be appreciated. Thanks!

Your $\theta $ is given by:
$\cos \theta =  \cos a  \  \cos x  \ \cos (b-y) + \sin a \  \sin x $

One easy way to think  [hide="why it is so is"]  to find x,y,z coordinates of the two points and find the distance between these two points. Now think of a triangle with the sides  (r,r, the_computed_distance).  Rest is simple.[/hide]


geometry
3D geometry
combinatorics proposed
combinatorics

At the vertices $ A,B,C,D,E,F,G,H$ of a cube (see picture), $ 1,2,3,4,5,8,7,6$ stones respectively are placed. It is allowed to move three stones from a vertex to each of its three neighbors, or to move three stones to a vertex from each of its three neighbors. Which of the following arrangements of stones at $ A,B,...,H$ can be obtained:

(a) $ 1,2,3,4,6,7,8,5$
(b) $ 2,3,4,1,6,5,8,7$
(c) $ 4,2,3,1,5,8,7,6$.

(c) cannot be obtained since A+B+G+H must preserve its oddness. For the original placement of stones, it's 1+2+7+6=16 (even), while for the placement (c), it's 4+2+7+6=19 (odd).


function
limit
algebra proposed
algebra
functional equation

Prove there is no function $f:R+ \rightarrow R+$ that for $x,y>0$
\[ f(x+y)\geq f(x)+yf(f(x))\]

Here's the outline:

First of all, it's obvious that $f$ should be strictly increasing. Then, it's also obvious that $\displaystyle \lim_{x\to \infty}f(x)=\infty$. Let $g(x)=f(x)-x$. Then $g(x+y)-g(x)\ge y(f(f(x))-1)\to \infty$ when $x\to \infty$ for fixed $y$. This shows that $g(x)\to \infty$ when $x\to \infty$ (we also use the fact that $g$ becomes strictly increasing from a certain point on).

Now let $y=1$. We have $1>\frac {f(f(x))}{f(x+1)}\ (*)$, but for large enough $x$, $f(x)>x+1\Rightarrow f(f(x))>f(x+1)\Rightarrow$ we have a contradiction with $(*)$.

it's also from IMC2001 in Czech Republic.


modular arithmetic
number theory unsolved
number theory

[quote="shirshendu"]all odd prime factors of numbers which are 1+ an odd power of 3
are -1 mod 3[/quote]
Did you mean:
[quote="shirshendu"]Let $n=1+3^{2k+1}$ for some $k\in\mathbb{N}$. Then for any odd prime $p\mid n$ we have $p\equiv-1\pmod{3}$. [/quote]
 :?:
If you meant this, something is wrong here. Take $k=1$. Then $n=1+3^{3}= 28$ and $7\mid n$, but $7\equiv 1\pmod{3}$. :maybe:

If odd prime $p|1+3^{2k+1}$, then $(\frac{-3}{p})=1\Longrightarrow p=1(mod \ 6)$.


Show that there exists an infinite sequence $ (a_n)_{n\equal{}0,1,2,\dots}$ of positive integers such that for all non-negative integers $ N$ the number $ \sum_{k\equal{}0}^{N}a_k^2$ is the square of an integer. Find a reccurence formula for such a sequence.

Take $ a_0\equal{}3$ and suppose that $ a_0,a_1,...,a_{n\minus{}1}$ are integers with the property stated. We want to construct $ a_n$ so that $ a^2\plus{}a_n^2\equal{}c^2$ where $ a$ is an integer such that $ a^2\equal{}\sum_{i\equal{}0}^{n\minus{}1}a_i^2$ and $ c$ is some integer. If $ a$ is even, take $ a_n\equal{}\frac{a^2}{4}\minus{}1$. If $ a$ is odd, take $ a_n\equal{}\frac{a^2\minus{}1}{2}.$


counting
distinguishability
email
AMC
AMC 8
MATHCOUNTS
search

[quote="mliang"]Ann has 3 roses and 3 carnations. She will put two flowers in each of three pots. How many different arrangements of potted flowers are there?[/quote]
[hide]$\frac{6!}{3!3!2!}=10$ I think[/hide]

I can't explain my method really well, so I'll try the best I can do:
I agree with bpms:
[hide="I think I have it"] Without any restriction (type of flowers, how many flowers per pot), we have $6!$ ways. We divide this by $3!$ and by another $3!$ to account for the types of flowers. We must also divide by $2!$ to account for the number of flowers per pot. Doing this gets $10$ as the answer. [/hide]

[hide]
first solution: if the flowers are being arranged only by the type.

let C = carnation and R = rose

We have three flower pots X, Y and Z.

The possible arrangements are:

 X______Y______Z

CR_____RR_____CC

CR_____CC_____RR

CC_____CR_____RR

CC_____RR_____CR

RR_____CR_____CC

RR_____CC_____CR

CR_____CR_____CR

That would be 7 different arrangements.

second solution: if the carnations were all different and if all the roses were all different. (for example, their color; something that gives them individuallity)

for pot X we would have 2 flowers to pick out of 6. The number of possible combinations for pot X would be:

$_{6}C_{2}= \frac{_{6}P_{2}}{2!}= 15$

for pot Y we would have 2 flowers to pick out of 4. The number of possible combinations for pot Y would be:

$_{4}C_{2}= \frac{_{4}P_{2}}{2!}= 6$

for pot Z we would have only 2 flowers left, which would leave only one possible combination.

so that would be $15 \times 6 \times 1$ or $90$ different combinations

for pot Y if we would have 2 flowers to pick out of 6, the number of ADDITIONAL possible combinations for pot Y would be $9$

for pot Z if we would have 2 flowers to pick out of 4, the number of ADDITIONAL possible combinations for pot Z would be $5$

so that would be $9 \times 5 \times 1$ or $45$ ADDITIONAL combinations.

for pot Z if we would have 2 flowers to pick out of 6, the number of ADDITIONAL possible combinations for pot Z would be $9$

so that would be $9$ ADDITIONAL combinations.

the total number of arrangements would be $90+45+9$ or $144$

[/hide]

[quote="frost13"]mliang, do you have the answer?  I don't think 10 is correct...[/quote]
Can you please post your reasoning, sir? I did this the same way as vishalarul, but I may be wrong...

[quote="mliang"]Ann has 3 roses and 3 carnations. She will put two flowers in each of three pots. How many different arrangements of potted flowers are there?[/quote]
[hide]If the pots and flowers are both indistinguishable, there are $2$ arrangements.

If the pots and flowers are distinguishable, then there are $252$ arrangments.

If the pots are indistiguishable but the flowers are, there are $72$ arrangements.

If the pots are distiguishable but the flowers aren't, there are $7$ arrangements.[/hide]

no need for the formality bmps :lol:

Here's how I see it.
[hide]
rc    rc     rc
cc   cr      rr

Any other arrangement is the same as one of these two the way I see it.  I admit the question is vague and 13375P34K43V312 is right to ask if the pots are distinguishable.  I believe that since we are not told if the pots are labeled or fixed in position, we have to assume we can not tell the difference from one pot to the next, just as I would assume I can't tell the difference from one carnation to another.

Do you see what I mean.  If the source was posted I could check the answer, but I am in no mood to sort through every question ever written:-)
[/hide]

That only helps if you believe me....and of course if I am right.  I guess it would be embarrasing if I am not :lol:

[quote="frost13"]no need for the formality bmps :lol: [/quote]
It's a habit, I grew up learning to be respectful to elders.
Anyways, I thought that the flowers were distinct, but the pots aren't. It definitely matters...
If pots and flowers are indistinguishable than this is similar to this old nationals sprint:
Four indistinguishable red chips and two indistinguishable white chips are arranged in a circle. How many different arrangements are there?

[quote="bpms"][quote="frost13"]no need for the formality bmps :lol: [/quote]
It's a habit, I grew up learning to be respectful to elders.
[/quote]

Yeah, like I said the question is pretty vague.

Being respectful is great and your parents taught you well, but at 33 I'm not ready to be considered an "elder" just yet :lol:

The pots could be distinguishable just by how they are situated in relation to each other.

I have already posted the answer for each case, so all we need is a clarification from mliang.

[quote="frost13"][quote="bpms"][quote="frost13"]no need for the formality bmps :lol: [/quote]
It's a habit, I grew up learning to be respectful to elders.
quote]

Yeah, like I said the question is pretty vague.

Being respectful is great and your parents taught you well, but at 33 I'm not ready to be considered an "elder" just yet :lol:[/quote][/quote]
By elder I mean any adult that is out of college pretty much...

hey bmps, didn't want to keep going back and forth on the forum.  You didn't offend me or anything with the "elder" comment, I was just kidding with you.

Did you get my email with the AMC8 solution scan?

Edited : Seriously I am losing my mind, I meant that to be a PM :lol:

[quote="frost13"]hey bmps, didn't want to keep going back and forth on the forum.  You didn't offend me or anything with the "elder" comment, I was just kidding with you.

Did you get my email with the AMC8 solution scan?[/quote]
Yeah, thanks a lot for that.
Anyways, does anyone get I like pie's solution, some of it doesn't seem to be right...

[quote="i_like_pie"][hide="Solutions"]If the pots and flowers are both indistinguishable, there are $\boxed{2}$ arrangements: $\text{cc rr cr}$ and $\text{cr cr cr.}$

If the pots are distiguishable but the flowers aren't, there are $\boxed{7}$ arrangements. The pots in the first arrangement (from above) can be switched to make six arrangements, plus the other one.

If the pots are indistiguishable but the flowers are, there are $\boxed{72}$ arrangements. Take the original two arrangements. In each arrangement, all roses can be switched for six different arrangements, and the same can be done for the carnations. Therefore, there are $2\cdot6^{2}=72$ arrangements.

If the pots and flowers are distinguishable, there are $\boxed{252}$ arrangments. Here, we take the number of arrangements if only the pots are distinguishable, $7,$ and again multiply this by $6^{2}.$[/hide]
Still waiting for a clarification...[/quote]
Okay, never mind your method makes more sense then mine.

I agree with Ilikepie

[hide] 6! divided by 3! divided by 3! divided by 2!. 6!=720, divided by 3! equals 6= 120. Divided by 3! = 20. Divided by 2! = 10.  :P [/hide]

I found the actual mathcounts competition problem stated as :

"Amy had three Powerful Pink geraniums and three Roaring Red geraniums.  If she plants two flowers in each of three pots, how many distinct 3-pot arrangements could she make"

Idon't know if mliang just rephrased it for typing ease, but this was in an actual competition.

[hide=" answer"]
2 arrangements
[/hide]

[quote="mliang"]Sorry for the clarification delay people. I just got the answer yesterday. It was one of the practice problems from my coach. Frost has the exact problem posted. I was confused about the distinguishment as well. But yes, it is only 2 ways. I guess pots and flowers are both undistinguishable.
@frost, how did you find the problem? it was from a really old sprint.[/quote]
Which one was it from?

well frost may/may not (don't want to say anything.....) have a lot of tests

-jorian

i didn't find the test actually.  I have all problems from like 1983 - 1995 sorted by topic.  I just opened the document and did a search for "flower"

And since I keep getting requests for a document I made my 2nd or 3rd coaching called "Tips and Tricks" that somehow ended up posted on Goodlymaths and is now not working, here it is.  I haven't had a chance to update it with a lot of the "tips or tricks" I have learned since, but here's how it stands.

[quote="bpms"][quote="frost13"]i didn't find the test actually.  I have all problems from like 1983 - 1995 sorted by topic.  I just opened the document and did a search for "flower"[/quote]
Educaide?[/quote]

yup

thanks for the document

i will have time to look at it later

-jorian


geometry
circumcircle
algebra
function
domain
combinatorial geometry
combinatorics unsolved

Does there exist a set of $n > 2$ points in the plane such that no three are collinear and the circumcenter of any three points of the set is also in the set?

[i]Brasil 1996[/i]

No, there is no such set.

Suppose, for a contradiction, that there exists such a finite set, say $S$.
Let $P,Q$ be two consecutive vertices on the convex hull of $S$. Then, all the other points from $S$ are on the same side of the line $(PQ)$, and none is on this line.
Let $E$ be the domain delimited by the line $(PQ)$ and the two lines perpendicular to $(PQ)$ from $P$ and $Q$, and on the same side of $(PQ)$ than the points of $S$.

Since $|S| \geq 3$ there is at least one point in $S$ which is not $P$ or $Q$, and then the center of the circumcircle of $PQM$, say $\Omega$, belongs to $E$ (since it is on the perpendicular bissector of the segment $PQ$).
It follows that the set $S'$ of the points from $S$ which belongs to $E$ is non-empty, and finite.
Let $M \in S'$ such that the distance of $M$ to the line $(PQ)$ is minimal.
Since $M \in S'$ we deduce that both angles $MPQ$ and $MQP$ are not greater than $\frac{\pi} 2.$
- If the angle $PMQ$ is not greater than $\frac{\pi} 2$ then the circumcenter of $MPQ$, say $\Omega$ again, is an interior point of $PMQ$, which is impossible from the minimality of $M$, or is the midpoint of the segment $PQ$ which is impossible from above.
- If the angle $PMQ$ is greater than $\frac{\pi} 2$ then $\Omega$ is on the other side of $(PQ)$ than $M$ which is a contradiction too.

Pierre.

Yes.
It is combinatorial geometry so it is combinatorics.

Pierre.

Here is my solution.
There is no such set.

Suppose, for a contradiction, that there exists such a finite set, say S.

Let $d$ be a smallest distance between any two points of $S$. Suppose $AB=d$, where $A, B \in S$.

Call $l$ is the perpencular bisector of the segment $AB$.

Because $n\ge 3$, there is at least one point $C$ in $S$ which is not $A$ and $B$
and then the center of the circumcircle of $ABC$ , say $D$ , belongs to $l$.
.Furthere more, the center of the circumcircle of $ABD$, say $E$, belongs to $l$,too.
Besides, no three point of S are collinear, hence the center of there circumcircle of $ABE$ is $D$.
Therefore, we have $DA=DB=DC=CA=CB$, it follows that $\hat{ADB}=\hat{ACB}=120^{o}$.
thus, $CD=\frac{d}{\sqrt{3}}<d$. A contraction. $(C, D\in S)$

Another problem.

Prove that there exists n>2 such that there exist n points in plane such that the circumcenter of any three points of the set is also in the set.

Choose any $n$ collinear points  ;)

Pierre.


[quote="Beat"]Let ABC be an acute triangle. Circle k with diameter AB intersects AC and BC again at M and N respective ly. The tangents to k at M and N meet at point P. Given that CP=MN, determine angle ACB.[/quote]
Try to prove that,if $H$ is orthocentre of $ABC$,then $P$ is midelpoint of $CH$,and we have that $CP=\frac{CH}{2}=Rcos\angle{C}$,also $MN=ABcos\angle{C}$=>
$CP=MN$<=>$cos\angle{C}=1/2$=>$\angle{C}=60^0$

For the middlepoint $D$ of the side $AB$ we have $PD=\frac{c}{2\cos (90^{\circ}-C)}=$ $\frac{c}{2\sin C}=$ $R=2\cdot \frac R2$, i.e. the point $P$ is the middlepoint of the segment $[CH].$

[b]Remark.[/b] The middlepoint $U$ of the segment $[CH]$ and the middlepoint $D$ of the side $[AB]$ belong to the Euler's circle $C\left(E,\frac R2\right)$ and the segment $[DU]$ is its diameter, i.e. $DU=R.$

[quote="Beat"]I`m sick of not clear explanations. Can you write from where did you get those equalities!

Does only darij can explain in this forum?[/quote]
What do you can't understand? :?

I cannot understand why when i`m asking for help you are answering me:
Try to prove that,..
Do you think that i did`n tried?

Ok,ok,ok!!!!!!! :roll:
Her is explaining to my solution.
We have that $H\in{AN}\cap{BM}$,and $\angle{CMH}=\angle{CNH}=90^0$=>
$CNHM$ inscribed in circle with deometre $CH$.=>If$O$ is midelpoint of $CH$=>
$OM=ON$ and $\angle{OMN}=\angle{ONM}=90^0-\angle{C}$,but also$\angle{MAN}= \angle{MBN}=90^0-\angle{C}$=>$P=O$
Now I hope you can understand evrithing in my solution :) .

[quote="Beat"]Thanks Tiks     :yup:[/quote]
You are welcome [b]Beat [/b] ;)


Try representing the vectors as their components i+j+k.  Then, the cross product is essentially the determinant of the matrix below.
$ 3 \times 3$ matrix
\[ \begin{pmatrix}
i & j & k \\
a & b & c \\
d & e & f
\end{pmatrix} \]

Where the first component vector is $ (a,b,c)$ and the second component vector is $ (d,e,f)$.

[quote="cdymdcool"]if v=(0 1 0) and w=(1 0 0) what's v*w?
the 3d vectors are so hard to grasp so does the right hand rule, anyone can explain it to me?[/quote]
Using that rule, the cross product is (0, 0, -1) (sorry no Latex) :(

Wow I just studied this yesterday in review of physics =o

Regarding hte right hand rule, hold your hand in a fist like manner but like keep your fingers straight after it curls back. The quickest path is then the way your finger points

Its kind of hard to describe in words; does your book have a picture?

the way i learned to picture the right hand rule was a guy lying on his back just after waking up with his hands perpendicular to his body.

it's a stretch, but i assure you that you will not forget the concept

Oh...that rule.

Put your right hand/palm on a wheel.  When you push your hand out to make the wheel spin, if your thumb is up, it's positive.

Why can't you just remember that counterclockwise is positive? :mad: It's the same for vector bearings in physics.


AMC
USA(J)MO
USAMO
algebra proposed
algebra

[quote="CDP100"]Let $ a_{i} \in N, i \equal{} 1,2,3...k, k \in N$. Find the values of $ k$ for which the equation

$ a_{1} \plus{} a_{2} \plus{} a_{3} ... \plus{} a_{k} \equal{} a_{1}a_{2}a_{3} ... a_{k}$

has integer solutions.[/quote]

For any k.

$ k\equal{}1$ : $ 1\equal{}1$
$ k\equal{}2$ : $ 2\plus{}2\equal{}2\times 2$
$ k>2$ : $ 1\plus{}1\plus{}1\plus{}\ldots\plus{}1\plus{}2\plus{}k\equal{}2k$


If $ a^x \equal{} c^q \equal{} b$ and $ c^y \equal{} a^z \equal{} d$, then

$ \textbf{(A)}\ xy \equal{} qz \qquad\textbf{(B)}\ \frac {x}{y} \equal{} \frac {q}{z} \qquad\textbf{(C)}\ x \plus{} y \equal{} q \plus{} z \qquad\textbf{(D)}\ x \minus{} y \equal{} q \minus{} z$
$ \textbf{(E)}\ x^y \equal{} q^z$

[hide="Solution"]
Suppose we solve for $ a$ and $ c$ in terms of $ b$, we have:

$ a^x \equal{} b \implies a \equal{} b^{\frac{1}{x}}\\
c^q \equal{} b \implies c \equal{} b^{\frac{1}{q}}$

Now we plug that information into the equation $ c^y \equal{} a^z$.

$ (b^{\frac{1}{q}})^y \equal{} (b^{\frac{1}{x}})^z\\
b^{\frac{y}{q}} \equal{} b^{\frac{z}{x}}\\
\frac{y}{q} \equal{} \frac{z}{x}\\
\boxed{(A) xy \equal{} qz}$
[/hide]

I dont think your solution is correct. What if $q=x=0$ so $b=1$? You cannot divide by 0.

I think if $q = x = 1$, $xy = qz = 0$ (right)
Then we do as TachyonPulse

You could also do this problem using logarithms.
$a^x=b$
$\therefore x=log_ab$
$c^q=b$
$\therefore q=log_cb$
$c^y=d$
$\therefore y=log_cd$
$a^z=d$
$\therefore z=log_ad$

$x \cdot y = log_ab \cdot log_cd = log_ad\cdot log_cb = z \cdot q = q \cdot z$


ABC is an equalirateral triangle .and (O)is its circumcircle.

for each point like P on (O) inwich PB>PAandPC,prove:

PA+PC=PB

excuse me kunny,what do u mean by :the point is rotating 60.?

Hint

[hide]This is a special case of the Ptolemy theorem.[/hide]

thanks arne,
 ;)
im waiting for theother solutions.

Special case can have some special solutions.

Let $AP \cap BC = X$. Then $\angle{BXP}=\angle{BAP}+60^o=\angle{BCP}+60^o=\angle{ACP}$. So:

$\frac{\triangle{BXP}}{\triangle{ACP}}=\frac{BX\cdot XP}{AC \cdot CP}$ and $\frac{\triangle{PXC}}{\triangle{ACP}}=\frac{PX \cdot XC}{AC \cdot CP}$

Hence: $\frac{\triangle{BXP}+\triangle{CXP}}{\triangle{ACP}}=\frac{\triangle{BPC}}{\triangle{ACP}} =\frac{XP \cdot BC}{AC \cdot CP}$

Since $\angle{BXA}=180^o-\angle{ACP}$, so we can use the same trick on $\triangle{BXA}, \triangle{AXB}$ and $\triangle{ACP}$. We will get:
\[\frac{\triangle{ABC}}{\triangle{ACP}}=\frac{AX \cdot BC}{AC \cdot CP}\]
Add it to the previous equation we got:
\[\frac{S_{ABPC}}{\triangle{ACP}}=\frac{BC \cdot AP}{AC \cdot CP}\]
Similarly, we have $\frac{S_{ABPC}}{\triangle{ABP}}=\frac{AP \cdot BC}{BP \cdot AB}$

Since $\triangle{ABP}+\triangle{ACP}=S_{ABPC}$, so we will end with $\frac{AC \cdot CP+ BA \cdot BP}{AP \cdot BC}=1$.
Because $AB=AC=BC$, so we finally get $PA=BP+CP$.

[quote="kunny"]The point is rotating $60^\circ$. :lol:[/quote]
Can you please show your proof, kunny?

I suppose that the line BC separes the points A,P. Let $R\in PC$ be the point for which the line AC separes the points P,R and $m(\angle BAR)=m(\angle BAP)$. Easily shows that the triangle APR is equilateral and CR=PB, i.e. PA=PB+PC.


Is there a number K such that:

                                      (k)(k+1)(k+2)···(k+2006)/2007!

ends in exactly 2007 zeros?

A condition was missing: all of the numbers k,k+1,k+2,k+3.....k+2006 end in, at most three zeros.
 :)


Aflati inversa functiei f:(0, infinit) cu valori in (0, infinit), f(x) = radical din x + radical de ordinul 3 din x.

Scue, nu stiu cum sa bat aici radicalul.

Multumesc

Tu trebuie sa rezolvi ecuatia $y=\sqrt{x} + \sqrt[3]{x}$. Fie $x=z^6$. Atunci $y=z^3+z^2$. Apoi vezi http://mathworld.wolfram.com/CubicFormula.html ...

Problema e foarte urata  :D

Intr-adevăr, problema este urâtă. Ideea de rezolvare se mai poate găsi şi în "Tabele şi formule matematice" de E. Rogai.

Va multumesc foarte mult.  :)

Intrebarea este ce cauta intr-un manual de clasa a zecea? Teoria necesara rezolvarii tine de anul intai facultate.
si mai este o intrebare. Autorii de manuale ar trebui obligati prin lege sa anexeze si solutiile complet rezolvate la probleme. Profesorii oricum nu-ti spun solutiile la problemele nerezolvate in clasa.

Numai bine

[quote="jeanvaljean"]Va multumesc foarte mult.  :)

Intrebarea este ce cauta intr-un manual de clasa a zecea? Teoria necesara rezolvarii tine de anul intai facultate.
si mai este o intrebare. Autorii de manuale ar trebui obligati prin lege sa anexeze si solutiile complet rezolvate la probleme. Profesorii oricum nu-ti spun solutiile la problemele nerezolvate in clasa. [/quote]
M-am uitat si eu in manual si nu ai dreptate. Problema nu cere determinarea inversei, ci doar studiul injectivitatii si surjectivitatii. E adevarat ca surjectivitatea e mai greu de argumentat, dar in nici un caz nu ai nevoie de cunostinte de facultate. Toate cele bune!

Ai dreptate draga Bogdan. Uitasem enuntul initial deoarece aveam problema in cap de mult timp.
totusi, cum se poate justifica surjectivitatea utilizand numai cunostinte de clasele noua si zece? Fara analiza sau ecuatii de gradul trei. Eu nu vad momentan decat metoda grafica; schitand graficul si paralela la Ox etc. Daca ai timp sugereaza-mi ideea.

Multumesc


It looks challenging and smart!!!
I tried it , unfortunately not finding the complete solution. :(
I thought that since $0\leq(x)<1$ we get $[x]<0$ but then.......?
Do you have any hint?

I am not so acquainted with such problems
although I like them! :)

[quote="Rushil"]Solve the equation

$ [x] \times (x) = x $ where $ [x] $ is the greatest integer fn and $(x) $ is the fractional part of x.

Is there any general method for such problems....[/quote]

The soln. set should be  {  b2/(b-1) where b is a [b]-ve[/b] integer  or 0 }

Please verify the ans.

[quote="L_Li"]try considering different groups of values for $x$:

1) $x=0$ is a trivial solution

2) $x$ cannot be positive because either:
$0<x<1$, which gives $[x]=0$ and makes the LHS 0;
$x$ is an integer, which gives $(x)=0$ and also makes LHS 0;
all other positive $x$, for which $(x)<x$ and $[x]<x$ so their product cannot possibly equal $x$

3) negative values of $x$ are trickier:
negative integers don't work for the same reason positive integers don't;
$-1<x<0$ are all solutions because $[x](x)=(-1)(-x)=x$;
all other negative $x$, let $a$ represent its integer part (a negative integer) and $b$ represent its decimal part (a positive number <1), $[x](x)=x\Leftrightarrow (a-1)(b)=a-b\Leftrightarrow ab=a\Leftrightarrow b=1$, which is a contradiction because b is a decimal portion so there are no solutions.

conclusion: $-1<x\le 0$[/quote]

Well, according to my soluton,
x=b2/(b-1) where b is some -ve integer or zero.

Since , zero is the trivial solution, let's test for say b= -3

we have x= (-3)2/(-3-1) = -9/4

  Also [x] = [-9/4]=[-2.25] = -3

 and (x) = (-2.25) = (-3 + 0.75 )= 0.75

So [x](x)= -3 * 0.75 = -3*(3/4) = -9/4 = x

Similarly we can prove that any number of the form b2/(b-1) where b is some -ve integer or zero, is a solution to the given equation.

Also we note that x=-1/2 is a solution .

[quote="socrates"]How did you reach that?? :)[/quote]

A small correction in my solution..note the square on b..

shall xplain the solution later to u ...'m succumbing to my workload (at office  :D ) now

[quote="L_Li"]oh ok yours is right. i misundertood (x) for negative numbers to be just what's after the decimal point, i.e. i made (-2.25) to be .25 instead of .75[/quote]

Actually, by definition (x) or {x},i.e. the fractional part of any real no x,is defined as

(x) = x- [x].

So it turns out to be something which always lies in the range [ 0, 1) for any real no. x.

Well you could let the fractional part be a and the integer part b, then you know
$ab=a+b$ where $a\in (0, 1)$ and b is a negative integer...
Then for each b, there is an a which works.  You find it's something like
$-\frac{b^2}{b+1}$ which gives us $a=\frac{b}{b+1}$
This gives us 0, -1/2, -4/3, -9/4, -16/5, ....

[quote="nat mc"]If $x$ is negative, say $-2.25$, is the fractional part $\frac14$ or $-\frac14$?[/quote]
The fractional part is $\frac{3}{4}$, because $-2.25$ rounded down gives $-3$.

[quote="nat mc"]So did we agree the answer is $x=-\frac{b^2}{b+1}$ with $b\in\mathbb{N}\cup\{0\}$?[/quote]

Yes,

actually the solution -b2/(b+1) where b is a natural no. or zero

is equivalent to b2/(b-1)where b is a NEGATIVE integer or zero

we can prove that easily

indeed the answer IS $ x = \frac{b^2}{b-1} $ where $ b \leq 0 $

[color=green]EDIT : Sorry there was a typo!!![/color]

[quote="Rushil"]indeed the answer IS $ x = \frac{b^2}{b+1} $ where $ b \leq 0 $[/quote]

Shouldn't it be  b2/(b-1) for b <= 0

and NOT b2/(b+1) for b <= 0 ,for if b=-1, the solution  b2/(b+1) is NOT defined.

Actually there are two equivalent solutions

 x = -b2/(b+1) where b is some Natural no. or zero

and x = b2/(b-1) where b is some Negative integer or zero


Solve the equation with $ x,y,z\in \mathbb Z^{\plus{}}$: $ x^x\plus{}2y^y\equal{}3z^z$

[hide="hint"]
$ x\equal{}y\equal{}z$
use the fact that $ 2 \le (1\plus{}\frac{1}{z})^z<3$.
[/hide]

$ x^{x}\plus{}2y^{y}\equal{}3z^{z}$
If $ x\equal{}y\implies x\equal{}y\equal{}z$
Supposex,y,z different .Hence $ max (x,y,z)\ge 3$ .If  $ z\equal{}max(x,y,z)$
$ 3z^{z}\equal{}2z^{z}\plus{}z^{z}>2y^{y}\plus{}x^{x}$
If $ x\equal{}max(x,y,z)$ we have
$ x^{x}\equal{}x\cdot x^{x\minus{}1}>3z^{z}$ ,$ (x\ge 3,x\minus{}1\ge z)$
Similary if $ x\equal{}max(x,y,z)$  we have $ x^{x}>3z^{z}$
Answers :(n,n,n) :P


Teepees are made such that the length of a supporting pole is at least four feet longer than the diameter of the circular base.  If a teepee has a base circumference of 40 feet, at least how long must a supporting pole be?  Express your answer to the nearest whole number.

The diameter is 40/pi. So the supporting pole must be at least 40/pi + 4, which is approximately 17


anyone who knows how to solve for "n" in permutations
given:

P(n,5)=2520

You are to find $n\geq 5$ such that $n(n-1)(n-2)(n-3)(n-4)=2520.$
That is to say, express $2520$ as the product of 5 consecutive numbers.

Yeah, unless you just want to, you shouldn't actually solve this. I would recomment just plugging in values of n>5 to check, it shouldn't take you long.


integration
trigonometry
logarithms
calculus
calculus computations

try substituting $ x \equal{} \tan \theta$

@ NORA and KUNNY

actually i have got the answer but i don know if its correct.
plese give ur answers so that i can check with mine


Demonstrati ca exista o infinitate de patrulatere convexe pentru care daca stim lungimile a trei laturi, putem afla si lungimea ultimei laturi. :)

Exista o infinitate de patrulatere convexe circumscriptibile.

Da, asa e, folosim Lema lui Pitot cu patrulaterele circumscriptibile.


Is nth a word?

Apparently it doesn't have any vowels, but it is defined in many dictionaries, used as an adjective. Vote, post, support.

nth as in the nth side of a figure.

I guess we math people use it a lot. However, I don't think it is a real word.... :|

I say yes, but only in mathematical terminology (whatever that is). It is similar to "an n-gon".

It's used in real life when you want to say a number is large, as in "for the nth time, could you please stop?".

Also, a word doesn't need to have vowels, take [url=http://en.wiktionary.org/wiki/cwm]cwm[/url], for instance (although you could say that w is being used as a vowel, I guess).

The letters A, B, C, D, E, F, G, H, I, ... are in the dictionary as "The [i]nth[/i]( :rotfl: ) word in the English alphabet." So there are entries in the dictionary without vowels.


Can I format nicely an illustration of the long division involved with calculating the first few terms of a Laurent series, say for $ \frac {z}{Sin^2(z)}$.  I tried using the \polynom package but that seems to be inapplicable (didn't work).

Consider for example, the first three sets of calculations for
\[ \frac {z}{z^2 \minus{} \frac {z^4}{3} \plus{} \frac {2z^6}{45} \plus{} \cdots}
\]
I saw the tabular example given in another post here and I can start the code:
\[ \setlength{\tabcolsep}{30pt} z^2 \minus{} \frac {z^4}{3} \plus{} \frac {2z^6}{45} \plus{} \cdots\,\begin{tabular}{|lllll} \hline z \\
\end{tabular}
\]
So the first line under the z (from multiplication by $ 1/z$ would be $ \minus{} (z \minus{} \frac {z^3}{3} \plus{} \frac {2z^5}{45})$  that would give the first term of $ 1/z$.  Next would be the multiplication by $ z/3$, and so on.  Guess I'd be interested in perfectly formatting it (everything aligned nicely)  but maybe that's a tough request to ask LaTex.  How do I  place the result over the division symbol: $ 1/z \plus{} z/3 \plus{} z^3/15 \plus{} \cdots$?

The code here does the first line under z but does not seem to work in the forum.  Also when I attempt to use \frac it does not compile in TeXnicCenter also:

[code]
\setlength{\tabcolsep}{30pt}
z^3-\frac{z^4}{3}+\frac{2z^6}{45}+\cdots \,
\begin{tabular}{|lllll}
\hline
\begin{align*}
&z\\
-&\!\!\!(z-z^3/3+2z^5/45)\\
\hline
\end{align*}
\end{tabular}
[/code]

Any ideas?
Shaw

The problem is not \frac but the fact that you are trying to use the align environment with its associated new lines inside a cell. A cell cannot contain more than one line so solve the problem by putting align inside a parbox.
[code]\setlength{\tabcolsep}{30pt}
$z^3-\frac{z^4}{3}+\frac{2z^6}{45}+\cdots \,$
\begin{tabular}{|lllll}
\hline
\parbox{1cm}{\begin{align*}
-&(z-z^3/3+2z^5/45)\\
\hline
\end{align*}}
\end{tabular}[/code]

Also have a look at the code at http://www.physicsforums.com/showthread.php?p=373931 which you may want to adapt.

Thanks Steve!  I modified the PF code for my problem.  It's just about perfect.  May need to tweak it a bit.  I'm unable to post it here though:

[code]\begin{equation*}
\begin{array}{rc@{}c}
& \multicolumn{2}{l}{\, \, \, \frac{1}{z}+\frac{z}{3}+\frac{z^3}{15}-\dotsb} \vspace*{0.12cm} \\ \cline{2-3}
\multicolumn{1}{r}{z^2-\frac{z^4}{3}+\frac{2z^6}{45}-\dotsb \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} z} \\
& \multicolumn{2}{l}{\, \, \, z-\frac{z^3}{3}+\frac{2z^5}{45}-\dotsb} \vspace*{0.12cm} \\ \cline{2-3}

& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} \frac{z^3}{3}-\frac{2z^5}{45}+\dotsb} \\
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} \frac{z^3}{3}-\frac{z^5}{9}+\frac{2z^7}{135}+\dotsb} \vspace*{0.12cm} \\ \cline{2-3}
& \multicolumn{2}{l}{\, \, \, \phantom{z{}-{}\frac{z^3}{3}{}-{}} \frac{z^5}{15}-\dotsb}
\end{array}
\end{equation*}
[/code]

Works fine on AoPS as well if you use dollars instead of \begin/\end{equation}: I would spread the lines a bit though...
\[ \begin{array}{rc@{}c}
& \multicolumn{2}{l}{\, \, \, \frac{1}{z}\plus{}\frac{z}{3}\plus{}\frac{z^3}{15}\minus{}\dotsb} \vspace*{0.12cm} \\ \cline{2\minus{}3}
\multicolumn{1}{r}{z^2\minus{}\frac{z^4}{3}\plus{}\frac{2z^6}{45}\minus{}\dotsb \hspace*{\minus{}4.8pt}} & \multicolumn{1}{l}{ \hspace*{\minus{}5.6pt} \Big) \hspace*{4.6pt} z} \\
& \multicolumn{2}{l}{\, \, \, z\minus{}\frac{z^3}{3}\plus{}\frac{2z^5}{45}\minus{}\dotsb} \vspace*{0.12cm} \\ \cline{2\minus{}3}

& \multicolumn{2}{l}{\, \, \, \phantom{1{}\minus{}{}} \frac{z^3}{3}\minus{}\frac{2z^5}{45}\plus{}\dotsb} \\
& \multicolumn{2}{l}{\, \, \, \phantom{1{}\minus{}{}} \frac{z^3}{3}\minus{}\frac{z^5}{9}\plus{}\frac{2z^7}{135}\plus{}\dotsb} \vspace*{0.12cm} \\ \cline{2\minus{}3}
& \multicolumn{2}{l}{\, \, \, \phantom{z{}\minus{}{}\frac{z^3}{3}{}\minus{}{}} \frac{z^5}{15}\minus{}\dotsb}
\end{array}\]

Can't seem to get the spacing between the horizontal lines just right but that's ok for now.  Thanks guys!

\[ \begin{array}{rc@{}c}
& \multicolumn{2}{l}{\, \, \, \frac{1}{z}+\frac{z}{3}+\frac{z^3}{15}-\dotsb} \vspace*{0.12cm} \\ \cline{2-3}
\multicolumn{1}{r}{z^2-\frac{z^4}{3}+\frac{2z^6}{45}-\dotsb \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} z} \\
& \multicolumn{2}{l}{\, \, \, z-\frac{z^3}{3}+\frac{2z^5}{45}-\dotsb}
\\
\vspace{1mm}
\cline{2-3}
\vspace{1mm}
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}}
 \frac{z^3}{3}-\frac{2z^5}{45}+\dotsb} \\
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} \frac{z^3}{3}-\frac{z^5}{9}+\frac{2z^7}{135}+\dotsb} \vspace*{0.12cm}
\\
\vspace{1mm}
\cline{2-3}
\vspace{1mm}
& \multicolumn{2}{l}{\, \, \, \phantom{z{}-{}\frac{z^3}{3}{}-{}} \frac{z^5}{15}-\dotsb}
\end{array}\]

I'm working on another residue problem and created another version of long division which works well.  Some notes:

(1) I eliminated the multicolumn syntax above.  The code below is simpler.  Each term has a separate column.  So the figure below has six columns.

(2) The six columns give better control over the length of the horizontal lines.

(3) In order to create a nicer spacing above and below the horizontal lines, I used the phantom operator with a \Big) command.  This forces the succeeding line to be larger thus giving a nicer spacing.

(4) In the particular case below, I needed to enlarge the column containing the \cdots so that the other elements in that column would fit.

(5) The case below is for even $ n$ just for the record.
\[ \begin{array}{llllll} & w^{n - 2} + & \pi w^{n - 3} + & \phantom{tk}\cdots\phantom{tk} + & \pi^{n - 2} + &{ \frac {\pi^{n - 1}}{w - \pi}} \\
\cline{2 - 6} w - \pi & \hspace{ - 6pt}\Big)w^{n - 1} \\
& w^{n - 1} - & \pi w^{n - 2} \\
\cline{2 - 6} \phantom{t\Big)} & & \pi w^{n - 2} \\
& & \pi w^{n - 2} + & \pi^2w^{n - 3} \\
\cline{3 - 6} \phantom{t\Big)} & & & \pi^2w^{n - 3} \\
\phantom{t\Big)} & & & \vdots \\
\cline{4 - 6} \phantom{t\Big)} & & & \cdots & & \pi^{n - 1} \end{array}
\]


[quote="arqady"][quote="fjwxcsl"]Positive numbers $ x,y,z$   are such that $ xy + yz + zx = 3$ . Prove the inequality

$ A.\frac {x + y + z}{3}\ge{\sqrt [21]{\frac {x^3 + y^3 + z^3}{3}} \ge{\sqrt [14]{\frac {x^2 + y^2 + z^2}{3}}.}}$

$ B.\frac {x + y + z}{3}\ge{\sqrt [6]{\frac {x^2 + y^2 + z^2}{3}} \ge{\sqrt [18]{\frac {x^3 + y^3 + z^3}{3}}.}}$[/quote]
$ \frac {x + y + z}{3}\ge{\sqrt [16]{\frac {x^3 + y^3 + z^3}{3}}}$ stronger. :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78337[/quote]

Thank you,arqady!

 We have   the strongest  and  best

$ AB.\frac {x + y + z}{3}\ge{\sqrt [6]{\frac {x^2 + y^2 + z^2}{3}} \ge{\sqrt [16]{\frac {x^3 + y^3 + z^3}{3}}\ge{\sqrt [11]{\frac {x^2 + y^2 + z^2}{3}}.}}}$

This is the most perfect!

[quote="fjwxcsl"]
 We have   the strongest  and  best

$ AB.\frac {x + y + z}{3}\ge{\sqrt [6]{\frac {x^2 + y^2 + z^2}{3}} \ge{\sqrt [16]{\frac {x^3 + y^3 + z^3}{3}}\ge{\sqrt [11]{\frac {x^2 + y^2 + z^2}{3}}.}}}$

This is the most perfect![/quote]
The right inequality follows from $ \sqrt[3]{\frac{x^3+y^3+z^3}{3}}\geq\sqrt{\frac{x^2+y^2+z^2}{3}}$ and $ x^3+y^3+z^3\geq3.$

[quote="arqady"][quote="fjwxcsl"]
 We have   the strongest  and  best

$ AB.\frac {x + y + z}{3}\ge{\sqrt [6]{\frac {x^2 + y^2 + z^2}{3}} \ge{\sqrt [16]{\frac {x^3 + y^3 + z^3}{3}}\ge{\sqrt [11]{\frac {x^2 + y^2 + z^2}{3}}.}}}$

This is the most perfect![/quote]
The right inequality follows from $ \sqrt [3]{\frac {x^3 + y^3 + z^3}{3}}\geq\sqrt {\frac {x^2 + y^2 + z^2}{3}}$ and $ x^3 + y^3 + z^3\geq3.$[/quote]

You are right,arqady!

$ \sqrt [3]{\frac {x^3 + y^3 + z^3}{3}}\geq\sqrt {\frac {x^2 + y^2 + z^2}{3}}$

is stronger than

\[ {\sqrt [16]{\frac {x^3 + y^3 + z^3}{3}}\ge{\sqrt [11]{\frac {x^2 + y^2 + z^2}{3}}.}}\]

[quote="arqady"][quote="fjwxcsl"]Positive numbers $ x,y,z$   are such that $ xy + yz + zx = 3$ . Prove the inequality

$ A.\frac {x + y + z}{3}\ge{\sqrt [21]{\frac {x^3 + y^3 + z^3}{3}} \ge{\sqrt [14]{\frac {x^2 + y^2 + z^2}{3}}.}}$

$ B.\frac {x + y + z}{3}\ge{\sqrt [6]{\frac {x^2 + y^2 + z^2}{3}} \ge{\sqrt [18]{\frac {x^3 + y^3 + z^3}{3}}.}}$[/quote]
$ \frac {x + y + z}{3}\ge{\sqrt [16]{\frac {x^3 + y^3 + z^3}{3}}}$ stronger. :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78337[/quote]

${ \frac {x + y + z}{3}\ge (\frac {x^3 + y^3 + z^3}{3}})^{\frac{1}{k}}$ stronger. :wink:

[quote="fjwxcsl"]

 We have   the strongest  and  best

$ AB.\frac {x + y + z}{3}\ge{\sqrt [6]{\frac {x^2 + y^2 + z^2}{3}} \ge{\sqrt [16]{\frac {x^3 + y^3 + z^3}{3}}\ge{\sqrt [11]{\frac {x^2 + y^2 + z^2}{3}}.}}}$

This is the most perfect![/quote]
The minimal $ k$ for which the inequality
\[ \left(\frac {x^2 + y^2 + z^2}{3}\right)^k \geq\frac {x^3 + y^3 + z^3}{3}\]
is true for all non-negative numbers $ x,$ $ y$ and $ z$ such that $ xy+xz+yz=3$
is $ k=\max_{t\geq0}\frac{3(t^2+2t+2)(t^2+2)}{2(t+2)(t^3+2)}=2.526828496...$


Putnam
conics
ellipse
geometry
perimeter
college contests

Putnam 1950/B2)  An ellipse with semi-axes a and b has perimeter length $p(a, b)$. For $\frac{b}{a}$ near 1, is $\pi (a+b)$ or $2\pi \sqrt{ab}$ the better approximation to $p(a, b)$?

[hide="My weak (without any hint of rigor) physics like solution to this using differentials"]For clarity, I shall do [color=blue]the calculation for a circle[/color] of radius $r$ first:

[color=blue]The area of such an circle is given by $A(r)=\pi r^{2}$. Now one might approximate the circumference $C$ of said circle by the method of differentials according to the following geometric observation: that the $C$ is approximately given by the difference of areas of concentric circles with radii $r$ and $r+\Delta r$ divided by the "width" of the circular annulus, namely $\Delta r$, that is put
\[C\approx \frac{A(r+\Delta r)-A(r)}{\Delta r}= \frac{\Delta A}{\Delta r}\]
at this point we may use differentials (wherein we put $\Delta r = dr$) to approximate the difference of areas to get
\[C\approx \frac{dA}{dr}= \frac{d}{dr}\left(\pi r^{2}\right) = 2\pi r \]
[/color]

Now, as for that same [color=red]calculation for said ellipse[/color]:
[color=red]
Recall that the area of such an ellipse is given by $A(a,b)=\pi ab$ so that reasoning similar to the method of approximating $C$ by geometric observation used the above, we boldly state that the perimeter $p(a,b)$ is approximately given by
\[p(a,b)\approx \frac{A(a+\Delta a, b+\Delta b)-A(a,b)}{\Delta r}= \frac{\Delta A}{\Delta r}\]
where the quantity $\Delta r$ is something like, say the mean width of the elliptical annulus, but since we know that $\frac{b}{a}$ is near 1, we have
\[a\approx b\Rightarrow \Delta a\approx \Delta b\Leftrightarrow da\approx db\Rightarrow \frac{da}{db}\approx 1 \]
so set $\Delta r= dr \approx da\approx db$ and hence by the method of differentials we have
\[p(a,b)\approx \frac{dA}{\Delta r}= \frac{\pi b\da+\pi a\db}{dr}\approx \frac{\pi b\dr+\pi a\dr}{dr}= \pi (a+b) \]
and for $a=b: =r$ this is the same as the above approximation for $C$.[/color] [/hide]

Edit: fixed dimensionality mix-up (that is I had volumes in there, but now I have areas).

[quote="Kent Merryfield"]Look at what I did [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=55166]here[/url] and tell me whether it helps.[/quote]

Maybe [url=http://www.kalva.demon.co.uk/putnam/psoln/psol508.html]this[/url] would be interesting to you.

BTW, what do you think my solution to this would have been scored as?


inequalities
function
induction
inequalities unsolved

In http://www.mathlinks.ro/Forum/viewtopic.php?t=22923 , a more general result was shown:

[color=blue]If $f:\mathbb{R}\to\mathbb{R}$ is a convex function, and $a_1$, $a_2$, ..., $a_n$ are real numbers such that $a_1\leq a_2\leq ...\leq a_n$, then

$\sum_{i=1}^n a_{i+1}f\left(a_i\right)\leq\sum_{i=1}^n a_if\left(a_{i+1}\right)$;

hereby, we set $a_{n+1}=a_1$.[/color]

Your problem follows from this general result if we take the convex function $f\left(x\right)=x^4$.

  Darij


quadratics
ARML
number theory proposed
number theory

Let a,b ,n be positive integers and p be prime number.Determine all pair of (a,b) which satisfies the following equation.

\[a^3+b^3=p^n\]

Let $d = \gcd(a,b)$. Then $d = p^t$ for some $t$, and we may divide each term of the equation by $p^{3t}$.
Thus, we assume that $\gcd(a,b) = 1$. (1)
Thus $p^n = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2-3ab)$
Since $\gcd(a,b) = 1$ we have $\gcd(a+b,(a+b)^2-3ab)=1$ or $3$.

- If $\gcd(a+b,(a+b)^2-3ab)=1$ then since $a+b > 1$, we have $a+b = p^n$ and $a^2+b^2-ab = 1$.
Wlog, we may assume that $a \geq b$.
If $a > 1$ then $a> b$ (they are coprime thus not equal) so that $a^2 + b^2 = 1 +ab \leq 1 + a(a-1)$ and then $a+b^2 \leq 1$ which is impossible. Thus $a=b=1$ and $p=2$ and $n=1$.
Returning to (1) it leads to the solutions $a=2^t=b$ and $p=2$, $n = 3t+1$.

- If $\gcd(a+b,(a+b)^2-3ab)=3$ then $p=3$ and $a+b = 3^{\alpha}$ and $(a+b)^2 -3ab = a^2 + b^2 - ab = 3^{\beta}$, for some integers $\alpha \geq 1$ and $\beta \geq 0$.
Moreover neither $a$ nor $b$ is divisible by $3$ otherwise both are divisible by $3$ which contradicts that they are coprime.
But $3^{2 \alpha} - 3ab = 3^{\beta}$ so that $\beta  = 1$.
It follows that $ab = 3^{2 \alpha -1} - 1$ and $a+b = 3^{\alpha}$.
Thus $a,b$ ae the solutions of the quadratic $X^2 - 3^{\alpha} X + 3^{2\alpha-1} - 1 = 0$.
Its discriminant is $\Delta = 4 - 3^{2\alpha -1}$ which has to be a non-negative integer. This forces $\alpha = 1$ thus $n=2$ and $a=1,b=2$ (or $a=2,b=1$).
Back to (1) this leads to the solutions $a=3^t,b=2\cdot 3^t,p=3,n=3t+2$ and $b=3^t,a=2\cdot 3^t,p=3,n=3t+2$.

Pierre.

Bonjour,Pierre!

Merci beaucoup de votre reponse.

There is no doubt about your solution.You are solving rapidly.
May be this problem is EASY for you.Is this problem famous?

kunny

I do not know if it is famous but it is quite straightforward (even if I did some usless things  :D  ).

Pierre.

Hmm...this problem looks very familiar, actually, a problem from ARML Power Question during 1990's. I forgot the year though  :(


calculus
integration
rotation
algebra unsolved
algebra

[quote="ZHANGWENZHONGKK"]On the table before you is a cylindrical ice cream cake with chocolate icing on top. From it you cut successive wedges of angle $ x$, where $ x$ is [b]arbitrary[/b]. Each time a wedge is cut, it is turned upside-down and reinserted into the cake.
 :arrow: Prove that after a [b]finite[/b] number of such operations, all the icing is back on top of the cake.

It's really a surprising conclusion... But can you give a proof? $ x$ is really arbitrary,and it can be irregular.[/quote]
It's wrong, IMHO.

Let the sequence $ x_1 \equal{} x\in(0,2\pi)$ and $ x_{k \plus{} 1} \equal{} 2\pi\{\frac {x_k \plus{} x}{2\pi}\}$ (where $ \{u\}$ is the fractional part of $ u$).
$ x_k$ is the part of the top which is either iced either non-iced (depending on the number of full circles we did) after $ k$ operations.

If we want the top full iced or full non_iced after $ n$ operations, we need $ x_n \equal{} 0$ and, since we clearly have $ x_k \equal{} 2\pi\{\frac {kx}{2\pi}\}$, we need $ \frac {x}{2\pi}\in\mathbb Q$

The scary thing is, the problem is correct as stated (think about what happens to the icing when it is flipped, it is also reversed!!!!)

But seriously, it is correct.

[quote="rofler"] But seriously, it is correct.[/quote]

If the problem is correct, I'm seriously interested in the proof, or in any flaw in my demo above.

 :ewpu:

[quote="rofler"]http://demonstrations.wolfram.com/TheCakeIcingPuzzle/

:)[/quote]

Humm, there is no demonstration in this link. Just an assertion and an animation which demonstrate nothing.
Could you just give us a demo or point the flaw in my own one ?

Thanks.

The flaw I believe is that if a wedge that is partially iced and partially not iced is flipped, the positions of the iced and non-iced sides reverse. So

Plain
Iced
Iced

flips over to become

Plain
Plain
Iced

as opposed to

Iced
Plain
Plain

[quote="MellowMelon"]The flaw I believe is that if a wedge that is partially iced and partially not iced is flipped, the positions of the iced and non-iced sides reverse.[/quote]

Wow ! you're right! I dont know if the problem is true, but my own counter-demo is surely false !
Thanks a lot. I used a bad model.  :blush:

:)  Mr Zuming Feng picked it into his lecture at classes in China TST Activity. If you believe him, then, please believe me. But he forgot at last to give the solution.
:!:  I still remember that his last hint was that, never prove that at a certain moment all the chocolate icing is on the bottom. Although this makes the problem more and more peculiar, I wish my hint would save your time.

So, dear MellowMelon, you'd better change your idea.

I have solved it - the result is true. I cannot fathom how Peter Winkler managed to come up with this problem.

[hide="solution"]If $ x$ is an integral divisor of $ 2\pi$ the result is obvious. Assume $ x$ is not such a divisor.

We will define the operation $ cut$ on a cake to mean cutting and flipping slices $ k$ times, going counterclockwise from the positive x-axis such that $ kx < 2\pi$ and $ (k \plus{} 1)x > 2\pi$, and then rotating the cake by $ p \equal{} 2\pi \minus{} kx$ degrees counterclockwise. This is equivalent to cutting and flipping $ k$ slices and rotating the cake to be ready for the next $ cut$.

We will define -1 as white and 1 as chocolate. So if we have a wedge of color 1, it's a chocolate wedge.

We will define an $ awesome_n$ cake to consist of $ n$ alternating $ p$ wedges of color $ ( \minus{} 1)^{n \plus{} 1}$ and $ x \minus{} p$ wedges of color $ ( \minus{} 1)^n$ and ending in a $ p$ wedge, going counterclockwise.

The complete chocolate cake, which we start out with, is a $ awesome_0$ cake. That is our base case.

Taking any $ awesome_n$ cake, where $ n \leq k$, and performing $ cut$ on it will flip each $ p$, $ x \minus{} p$ group to become a $ x \minus{} p$ of color $ ( \minus{} 1)^{n \plus{} 1}$, $ p$ of color $ ( \minus{} 1)^n$ group. It will leave a $ p$ slice of the same color as the other $ p$ slices below the positive x-axis. The cake will then be rotated $ p$ radians to become an $ awesome_{n \plus{} 1}$ cake.

Performing this operation a finite number of times will finally lead us to the $ awesome_{k \plus{} 1}$ cake.

We then define a $ spectacular_n$ cake to be an $ awesome_n$ cake, except with the wedges going clockwise instead of counterclockwise.

The $ awesome_{k \plus{} 1}$ cake is also the $ spectacular_{k \plus{} 1}$ cake. That is our base case.

Taking any $ spectacular_n$ cake, where $ n \geq 1$, and performing $ cut$ on it will flip each $ p$, $ x \minus{} p$ group to become a $ x \minus{} p$ of color $ ( \minus{} 1)^{n \plus{} 1}$, $ p$ of color $ ( \minus{} 1)^n$ group. It will leave a $ x \minus{} p$ slice of color $ ( \minus{} 1)^{n \plus{} 1}$ in the group reached by going counterclockwise from the positive x-axis, and will leave the $ p$ slice of color $ ( \minus{} 1)^{n \plus{} 1}$ below the positive x-axis. It will then rotate the figure to become the $ spectacular_{n \minus{} 1}$ cake.

Performing this operation a finite number of times will lead us to the $ specactular_0$ cake, which is the chocolate cake.

Notes: keeping track of how many total cuts have occurred will lead to the formula $ 2k(k \plus{} 1)$ total $ x$ cuts. I left it out of the proof because it's not really related.
Notesnotes: again, that's assuming non-integral divisors of $ 2\pi$.
[/hide]

[quote="Differ"]I have solved it - the result is true. [/quote]

Wow, great congrats !  :clap:

And very nice demo, indeed. I had some difficulties to figure out each case but finally it's OK.

Great!

:clap:  :clap:
And the demonstration is also excellent. Sometimes I forget to find ideas there.


Find $ a,b$ such that $ Max|\sqrt {x} \plus{} ax \plus{} b|$ is smallest possible, where $ a,b$ are 2 real numbers and $ x \in [0;16]$

Another ricardokaka or lalala clone... Lock this up mods, please.

(M&Y, already posted by some of above mentioned:))


Do you assume that each $ a_i$ is an integer? If so, the conclusion is clear. Otherwise it is not right. For example, let $ n\equal{}2$ and $ \det A\equal{}a_2\minus{}a_1$.

The exact proof depends on your definition of determinant, but in any case $ \det A$ will be equal to some sum of products of integers, hence an integer itself.


calculus integration Gauss inequalities derivative LaTeX geometry

let f(n)=n^(1/n)-1. for example, if n=2, then its sqrt2-1. if n=3, then its cbrt3-1.
prove that f(1)+f(2)+f(3)+f(4)+... is infinite
[hide="hint"] prove that this series never goes below one and f(n) where n is sufficiently large is very close to one[/hide]
[hide="big hint"] show that f(k) where k>3 is more than 1/k[/hide]
let g(a)=f(a)+1
prove that g(1)*g(2)*g(3)*g(4)*... is divergent also

Very nice problem Scrambled. It could perhaps more easily be stated by saying:

Prove that $\sum_{n=1}^{\infty} (\sqrt[n]{n} - 1)$ diverges.

This may properly belong to the Calculus forum (though there are non-calculus ways of attacking it).

Why would you want to use calculus?

Applying the integral test gets ugly, unless I'm missing something.

But for $n\geq 3$, we have $\left(1+\frac {1}{n}\right)^n<n$, which is a strong enough result.

This is similar to a related problem: prove that $n^{n+1}>(n+1)^n$ for $n>2$

true, but the fact you mentioned which makes it diverge (if you didnt look at the hint) was supposedly hard to see (seeing that gauss didnt catch that fact).
the next result you stated (n+1)^n<n^(n+1)is trivially proven using this fact

[quote=blahblahblah]Why would you want to use calculus?[/quote]

Other than the fact that it has to do with infinite series... The assertions that you pointed out are naturally easy to prove with calculus. But I guess its just an aesthetic distinction.

And I agree, the integral test is probably not a good idea for this question.

[quote="blahblahblah"]Why would you want to use calculus?[/quote]

Other than the fact that it has to do with an infinite series... The assertions that you pointed out seem more natural consequences of calculus to me than algebra. But I think that's just an aesthetic distinction.

I agree with you though, the integral test is probably not a good idea. But that's not what I had in mind...

Hmm.  I've known that result for such a long time that I take it for granted, but when I first proved it back in the day, I remember noting that a) monoticity + boundedness (by a geometric series) makes the proof of the above inequality quite quick.  Conversely, differentiation will require logarithmic derivatives and a whole mess of algebra, which tends to make me unhappy quite quickly.

for the calculus solution:

[hide]just use the limit comparison principle w/ the harmonic series[/hide]

I would hide the full solution, but for me, LaTeX is still a mystery.

more detail:

[hide] the limit is infinity, so cuz the harmonic series diverges, so does this one[/hide]

w-hat?

There's no calculus there.  First of all, you need to show that its dominated by the harmonic series.  I showed this without using calculus.  You can do it with calculus, but it's ugly.  The 'limit comparison test' has nothing to do with calculus, unfortunately, as its just pigeonholed into that area of math due to the fact that its usually taught concurrently with integration.

EDIT: the above response was to bubala, and not gauss

Gauss, anyways, it's kind of funny, because even in, say, analysis texts, there's no consensus as to which part of the book series belong.  Certainly there's a great temptation to stick it near the end of the text so that you can use the integral test, but in other text, like baby Rudin, series are right smack up front.  I tend to agree with the latter approach more, as it motivates things like Cauchy sequences better (what good is a Cauchy sequence if there aren't any applications for it) but that's just me.

wait, so does my solution not work?  Oh...

Your solution works.  But you need some way to establish the inequality.  I stated two ways in which it can be done.  But you need to establish at least one of those  :)

[quote="gauss202"]Right. Which is a result that I first learned in calculus. So I guess that's where we differ first and foremost.

But there are still some metamathematical arguments to be made about dealing with infinite series without a proper regard for calculus. I realize this problem isn't [i]that[/i] complicated, and that there are purely algebraic ways to show all the steps of this problem (even that the harmonic series diverges). But I think of divergence and things dealing with infinity as results belonging more naturally to calculus.

Its really like splitting hairs on this problem though.[/quote]
um so proofs of any kind belong in calculus?

[quote="blahblahblah"]I could tell you, but it's more educational if you do it yourself

1) Prove that it's monotonic increasing, that is, that $\left(1+\frac {1}{n}\right)^n<\left(1+\frac {1}{n+1}\right)^{n+1}$

2) Prove that its bounded above by three (split off a chunk of terms and show that they are dominated by a geometric series which sums to two).[/quote]
i can already do it, but i wanted to see your solution, since it appeared that my solution was radically different from almost anyone else whom i asked

[quote="Scrambled"]
um so proofs of any kind belong in calculus?[/quote]

No. Not at all. But proofs that deal with infinite series and divergence probably do.

[quote="Scrambled"][quote="blahblahblah"]well, for example, $\left(1+\frac {1}{n}\right)^n<3$ for all natural $n$.[/quote]
and how do you go about doing that?[/quote]

Actually, I find it interesting that as n approaches infinity, the limit of that expression is e. Try it out.

Yes, that is well known.  But it increases monotonically to $e$ - it doesn't 'jump' around.

Isn't it one of the definitions of $e$ as well? I know it is used in interest compounded continuously, too, but that's about it.

The definition of $e^x$ that I use, and that every analysis text will use, is just

$e^x=\sum^{\infty}_{i=0}\frac{x^i}{i!}$

I see. So its just defined by the taylor polynomial usually.

Well, this method has the advantage of very fast convergence.


function
vector
calculus
derivative
linear algebra
matrix

is {$ e^x,e^{2x}$} ,linear dependent or not?

For the above to be linearly independent we must prove:

for all a,b,x $ ae^x \plus{} be^{2x}\Longrightarrow a\equal{}b\equal{}0$

AND for linear dependency we must prove:

there exist a,b,x such that $ ae^x \plus{} be^{2x} \equal{}0$ AND $ a\neq 0,b\neq 0$

CORRECT??

This should be more or less obvious by taking $ x$ large enough.  You can also compute the [url=http://en.wikipedia.org/wiki/Wronskian]Wronskian[/url] of the two functions.

[quote="τρικλινο"]AND for linear dependency we must prove:

there exist a,b,x such that $ ae^x \plus{} be^{2x} \equal{} 0$ AND $ a\neq 0,b\neq 0$

CORRECT??[/quote]

no, you need to prove that there exist $ a$ and $ b$, not both $ 0$, such that for all $ x$, $ ae^x \plus{} be^{2x} \equal{} 0$.

The first condition is also wrong; linear independence means that if $ ae^x \plus{} be^{2x} \equal{} 0$ for all $ x$, then $ a \equal{} b \equal{} 0$.  (To see why your wording is incorrect, take $ a \equal{} \minus{}b, x \equal{} 0$.)

[b]WHY when we negate a definition that has the expression "for all x' ,that expression will not become : there exists an ,x[/b]

[quote="t0rajir0u"]Because you seem to be confused about the definition of equality of functions.  $ x$ always has the same quantifier in that definition, and equality is not what is being negated in the definitions of dependence and independence.[/quote]

What is your definition for the linear Independence of the {$ e^x, e^{2x}$} functions in quantifier form??

[quote="t0rajir0u"]$ \not \exists a \neq 0, b \neq 0 : \forall x, ae^x \plus{} be^{2x} \equal{} 0$.[/quote]

You mean $ \neg\exists a\exists b[ a\neq 0\wedge b\neq 0\wedge\forall x( ae^x \plus{} be^{2x} \equal{}0)]$

Actually it should be $ \not\exists (a, b) \neq (0, 0) \colon \ldots$ (in t0r's formulation) or $ \neg\exists a\exists b[ (a\neq 0\vee b\neq 0)\wedge \ldots$ (in τρικλινο's formulation).

In any case, understandable, usable, correct definitions were given in the second and third responses (third and fourth posts).

[quote="JBL"]Actually it should be $ \not\exists (a, b) \neq (0, 0) \colon \ldots$ (in t0r's formulation) or $ \neg\exists a\exists b[ (a\neq 0\vee b\neq 0)\wedge \ldots$ (in τρικλινο's formulation).[/quote]

But the formula:

$ \neg\exists a\exists b[ a\neq 0\wedge b\neq 0\wedge\forall x( ae^x \plus{} be^{2x} \equal{} 0)]$ is [b] equivalent [/b] to the formula:

$ \forall a\forall b[ (a\neq 0\wedge b\neq 0)\Longrightarrow\exists x( ae^x \plus{} be^{2x}\neq 0)]$ ,which means ,given :

$ a\neq 0$ and $ b\neq 0$ , then there exists an x such that :  $ ae^x \plus{} be^{2x}\neq 0$

But that is possible [b] for all functions,except the null functions[/b], meaning that all functions are independent.

[b] Is that possible??[/b]

 [quote="JBL"]  In any case, understandable, usable, correct definitions were given in the second and third responses (third and fourth posts).[/quote]

What is the meaning of the words ( understandable ,usable ) in a definition and consequent mathematical proofs??

[quote]
What is the meaning of the words ( understandable ,usable ) in a definition and consequent mathematical proofs??
[/quote]

It means that reasonable people with a background suitable for studying linear
algebra can read, understand, and apply these definitions. When every thread
degenerates into an argument about symbolic logic rather than the higher-order
mathematical concepts, maybe the formalism is getting in the way of the
mathematics.

I have no idea what your question is trying to communicate or what it has to do with what I said, in part because the natural-language portion of it lacks the quantifiers necessary for me to decipher it and in part because it's far from clear whether you understood what I wrote.  (Certainly, I have no idea in what sense your comment is a response to mine.)

Here's a simple analysis to go through: if your definition implies that $ f(x) \equal{} x$ and $ g(x) \equal{} 2x$ are linearly independent, you're using the wrong definition and/or using the definition incorrectly.  Similarly, if your definition implies that $ f(x) \equal{} 0$ and $ g(x) \equal{} e^x$ are linearly independent.  (This latter case is what my comment was addressing: in the linear combination, we need not have all the constants non-zero, but rather there must exist a nonzero constant.)

The terms you've quoted refer to the processes of human cognition; they are not formal mathematical terms.  If you don't understand the definitions (in natural language) provided by t0r and blahblahblah, you aren't going to think usefully about their formal counterparts.

Edit: Nukular posted while I was typing.  I strongly second what he wrote.

The definition for linear Independence which i wrote in my very 1st post can be proved in the following way:

 [b] Let a,b.x be any real Nos ,and $ ae^x \plus{} be^{2x} \equal{}0$[/b].

But ,  $ ae^x \plus{} be^{2x}\equal{}0\Longrightarrow a \plus{} be^x \equal{} 0\Longrightarrow\frac{d( a \plus{}be^x)}{dx} \equal{} 0\Longrightarrow be^x \equal{} 0\Longrightarrow a\equal{}b\equal{}0$

[b] And if that proof is correct[/b],then we have proved:

for all a,b, x: if a,b,x are real Nos and $ ae^x \plus{} be^{2x} \equal{} 0$,then a=b=0.

CORRECT??

[quote="τρικλινο"]And if that proof is correct[/quote]  It is.
[quote="τρικλινο"]then we have proved:

for all a,b, x: if a,b,x are real Nos and $ ae^x \plus{} be^{2x} \equal{} 0$,then a=b=0.

CORRECT??[/quote] No.  As t0r pointed out, the statement you've written here is false: for example, $ x \equal{} 0, a \equal{} 1$ and $ b \equal{} \minus{}1$ are real numbers such that $ a e^x \plus{} be^{2x} \equal{} 0$ and we do not have $ a \equal{} b \equal{} 0$.  The statement that you have proven (which is the statement of the linear independence of $ e^x$ and $ e^{2x}$ as functions over $ \bf R$) is that for all real numbers $ a, b$, if $ a e^x \plus{} be^{2x} \equal{} 0$ for all real $ x$ then $ a \equal{} b \equal{} 0$.

Let $ V$ be a vector space over the field $ K$, and let $ u_1,..u_n$ be elements of $ V$. We shall say that $ u_1,...u_n$ are[b] linearly dependent  [/b] if there exist elements $ a_1,...,a_n$ in K not all equal to 0 such that
                                                         $ a_1 u_1 \plus{} ...a_n u_n \equal{} 0$
For our functions. We suppose that there are numbers $ a,b$ such that
                                                              $ ae^t \plus{} be^{2t} \equal{} 0$
(for all values of t). Differentiate this relation. We have
                                                              $ ae^t \plus{} 2be^{2t} \equal{} 0$.
Subtract the first from the second relation. We obtain $ 2be^{2t} \equal{} 0$ and hence $ b \equal{} 0$. From this relation,it follows that $ ae^t \equal{} 0$ and hence $ a \equal{} 0$. Hence our functions are linear independent  :)

Note: you need { } for exponents of more than one character. I've edited this correction into your post.

Using derivatives that way leads to the [url=http://en.wikipedia.org/wiki/Wronskian]Wronskian[/url], a very useful tool for higher-order differential equations. We have the matrix $ \begin{bmatrix}f(x)&g(x)\\f'(x)&g'(x)\end{bmatrix}\equal{}\begin{bmatrix}e^x&e^{2x}\\e^x&2e^{2x}\end{bmatrix},$ which has determinant $ e^{3x}\neq 0.$ That determinant being nonzero proves linear independence.

Here was the first time where I show the Wronskian... Thank you jmerry. :lol:


On sides $ AB$ and $ AC$ of triangle $ ABC$ there are given points $ D,E$ such that  $ DE$ is tangent of circle inscribed in triangle $ ABC$ and $ DE \parallel BC$. Prove
$ AB\plus{}BC\plus{}CA\geq 8DE$

[quote="Matematika"]On sides $ AB$ and $ AC$ of triangle $ ABC$ there are given points $ D,E$ such that  $ DE$ is tangent of circle inscribed in triangle $ ABC$ and $ DE \parallel BC$. Prove
$ AB \plus{} BC \plus{} CA\geq 8DE$[/quote]
Quite easy as a TST. :)
\[ \frac{DE}{a}\equal{}\frac{h_a\minus{}2r}{h_a}\iff DE\equal{}a\minus{}\frac{2ar}{h_a}\equal{}a\minus{}\frac{2a^2sr}{ash_a}\equal{}a\minus{}\frac{2a^2\Delta}{2\Delta s} \equal{}a\minus{}\frac{a^2}{s}\] We need to prove that, \[ 2s \geq 8a\minus{}\frac{8a^2}{s} \iff 4a^2\minus{}4as\plus{}s^2 \geq 0 \iff (2a\minus{}s)^2 \geq 0\]
Done! :)

Wow! Now I find that it is Italy 1999. ;)

Don't make mistake this isn't IMO TST this is MEMO tst.

Hmm...I know.
BTW how did Adrian do?  :)

Let $r$ and $r_a$ denote the inradius and $A-$exradius, respectively. It is standard to see that $rs=[ABC]=r_a(s-a)$, where $s$ is the semiperimeter. Then, since $\triangle ADE\sim\triangle ABC$ and the $A-$excircle of $\triangle ADE$ is the incircle of $\triangle ABC$, we find that $\frac{DE}{BC}=\frac{r}{r_a}=\frac{s-a}{s}=\frac{b+c-a}{a+b+c}$. Thus, $DE=\frac{a(b+c-a)}{a+b+c}$. Finally,
\begin{align*}
(b+c-3a)^2&\ge0\\
9a^2+b^2+c^2-6ab-6ca+2bc&\ge0\\
a^2+b^2+c^2+2ab+2ca+2bc&\ge8ab+8ca-8a^2\\
(a+b+c)^2&\ge8a(b+c-a)\\
a+b+c&\ge\frac{8a(b+c-a)}{a+b+c}\\
AB+BC+CA\ge8DE.\,\blacksquare
\end{align*}


Hello....
I have this problem which i am trying to solve
{The currency of a country i lets say a euro....for one euro there are coins of 50, 20, 10, 5, 2 ,1 cents...Now for any price less than 100 cents (99 cents) i need to find the minimum number of coins that will be returned as change...The solution must be found in only 6 steps....The 6 steps must give the solution for any number between 1 and 99  cents.....

[hide]Let $x$ be the amount of change. Subtract 50 from $x$ as many times as possible, then 20, then 10, then 5, then 2, then 1. For example, if you need 0.88 change, you get one of each coin.[/hide]


It's not true , try $ a=\frac{\sqrt{14}}{\sqrt{5}}, b=c=\frac{1}{\sqrt{10}}$

[quote="nguoigo"]For$ a,b,c>0$   and $ \sqrt{a}+\sqrt{b}+\sqrt{c}=3$.Pro that     $ \frac{1}{a+b}$+$ \frac{1}{b+c}$+$ \frac{1}{c+a}$ <=$ \frac{3}{2}$ :blush:
   I'm sorry!!![/quote]
Try $ a=7.84$ and $ b=c=0.01.$ :wink:

It is still wrong.Take $ a\rightarrow 0,b\rightarrow 0$ :D


linear algebra
matrix
absolute value
combinatorics proposed
combinatorics

In a chess tournament , each of 2n players plays every other player once in each of two rounds. A win is worth 1 point ,a draw is worth 0.5 point and a loss is worth nothing. Prove that if for every player , the total score in the first found differs from that in the second round by at least n points ,then this difference is exactly n points for every player.

We show that the sum of differences does not exceed 2n^2  if the difference for each player is over n, then the difference has to be n.

Construct a matrix that describes round 1. Make the entry for (i,j) the points i obtained for playing j, with possible entries being 0, 0.5, 1. Ignore all (i,i) entries. Do the same for round 2.

Subtract the matrix for round 1 to the matrix for round 2. notice that in the result matrix the possible entries are 0, -0.5, 0.5, 1, - 1 with (i,j) = -(j,i) given i != j.

The absolute value for the sum of each column must be greater than or equal to n. However, the sign may be either negative or positive. Suppose there are m players for whom the difference is positve. Rearrage the matrix such that these m players have columns 1 to m. So sum for the positve differences is |(1,2) + (1,3) + (1,4)... (1,n)| + |(2,1) + (2,3)...| +...+ |(m,1) + (m,2)...+ (m,n)|. Since they are all of the same sign (i.e. positive) we may remove the absolute value sign. Now notice that for any (i,j), j < m, there exist (j,i) with (j,i) = -(i,j), so they cancel out. Exactly m^2 - m numbers cancel out that way and there are (2n-1)m components to the sum. So with the numbers that cancel out there are exactly 2nm-m^2 components. Since the maximal entry is 1, the maximal sum is 2nm-m^2. The same can be done for all negative differrences.

Thus the maximal for m positve differences and k negative differences with m + k = 2n is 2n(m+k) - m^2 - k^2 = 2n^2 - m^2 - k^2. But m^2 - k^2 is minimal if m = k = n. So the maximal is 2n(2n) - n^2 - n^2 = 2n^2, as discribed in the initial statement. So the only possible differences are n for each player.


calculus
integration
number theory unsolved
number theory

Can anyone please send a solution for this:-

Prove that on the average the number of ways of expressing a positive integer n as a sum of two integral squares
                          $ x^{2}$+$ y^{2}$=n, is $ \pi$

hi anup welcome to the forum i am moving this to numnber theory section


search
number theory unsolved
number theory

Let $k$ be a given positive integer. Deﬁne $x_{1}= 1$ and, for each $n > 1$, set $x_{n+1}$ to be the smallest positive integer not belonging to the set $\{x_{i}, x_{i}+ik | i = 1, . . . , n\}$. Prove that there is a real number $a$ such that $x_{n}= [an]$ for all $n \in\mathbb{ N}$.

lemma: two irretional numbers a,b such that $\frac{1}{a}+\frac{1}{b}=1$ then $[an]$ and $[bn]$ is partition of N
proof:exercise for you
 consider a the positive root of equation $\frac{1}{a}+\frac{1}{a+k}=1$
try to prove it's the required number.

Solution above same as solution I know :P
lemma is Beatty Theorem , here is a nice apply http://www.mathlinks.ro/Forum/viewtopic.php?t=136049

Sorry, but I haven't had sufficient understanding about the Beatty Theorem and the lemma written by dhthstn. Can you explain more clearly and post the proof ? I really appreciate that, thank you..  :)

If you search on forum for keywords : beatty theorem then you will have http://www.mathlinks.ro/Forum/viewtopic.php?highlight=theorem+beatty&t=26352


calculus
integration
geometry
trigonometry
calculus computations

Find the area of the surface

S=Int(Int(sqrt(1+(dz/dx)^2+(dz/dy)^2))

on the plane x+3y-z=2 inside the cylinder x^2+y^2=9

[Note that it does not tell you the limits nor what you are supposed to be integrating with respect to]

How are you suppose to do this? Find the partials and sub them into the S equation? What does S stand for? Isn't S the double integral of the magnitude of velocity? Argh!!!

$S$ is the surface area. You are given a formula for it: $S=\iint_{D}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}}\,dA$. This formula is valid for the part of the graph $z=z(x,y)$ above (or below -- which is irrelevant) a region $D$ in the $xy$-plane. Another way to put it: $D$ is the projection of the surface onto the $xy$-plane. In your case $z=x+3y-2$ and the region $D$ is a disk of radius $3$.

this can be done in an elementary way...consider the cosine of the angle formed with the given plane and a plane perpendicular to the cylinder

Working off mlok's post:

\[S=\iint_{D}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}}\,dA=\iint_{D}\sqrt{1+1^{2}+3^{2}}\,dA \]
\[= \sqrt{11}\iint_{D}\,dA = \sqrt{11}\mbox{Area}(D) = \sqrt{11}\pi 3^{2}=9\sqrt{11}\pi\]


Lisa Sauermann (Germany):	                7	2	7	7	5	7	35	12	Gold medal
Zhuo Chen (People's Republic of China):	7	7	7	7	7	0	35	12	Gold medal

Check: http://www.imo-official.org/year_individual_r.aspx?year=2008&column=total&order=desc&gender=filter&nameform=western&language=en

But actually Germany wins as the German contestant solved problem 3 and 6!  :D  :first:

[quote="orl"][quote="Yuriy Solovyov"]Why Farhod Haydarov ♀ from Uzbekistan is in that list?  :D[/quote]

I guess Farhod likes girls. But I must admit somebody must have scanned the linked list thoroughly to find out the odd one!  :D[/quote]

there is another typo : Redi Haderi ♀ from Albania  :D

Also
Juan Andrés Dodyk ♀. So there are probably also some girls listed as boys.

Actually it's the other way round: Juan is a boy who is listed as a girl at the IMO webpage.

Anyway, isn't it a bit funny?  :P


Give a,b,c are positive real and min{a,b,c} \geq 1/4max{a,b,c}.
Prove that:
a 2 b(a-b)+b 2 c(b-c)+c 2 a(c-a) \geq 0

The Kalva website gives the intial condition as a, b, and c are the sides of a triangle, not min{a,b,c}>=1/4max{a,b,c}.  I'll assume that you intend the former.

Let a=y+z, b=z+x, c=x+y.  The triangle conditions implies that x,y,z>0.  Then the inequality to be proven becomes, after some trivial transformations,

y^3 x + z^3 y + x^3 z >= xyz(x+y+z).

Observe that 2/7 y^3 x + 1/7 z^3 y + 4/7 x^3 z >= x^2yz, by the weighted AM-GM inequality.  Adding this inequality to its cyclic counterparts gives the required result.

No, I think it is a new inequality made with an inspiration from the IMO one.

Pierre.

You are right ,pbornsztein?. :) I know the IMO problem Michael Lipnowski tell us have a very short solution.May be my inequality is created from that proof.

can you tell me why we put  min{a,b,c}>=1/4max{a,b,c}

I still don't have solution yet,but I think min{a,b,c}>=1/4max{a,b,c} to let we have  \sqrt a, \sqrt b, \sqrt c are three sides of a triangles.So we can let a=(x+y) 2 ,b=(y+z) 2 ,c=(x+z) 2 .....

Finally,I found the greater condition for this one.That is  \sqrt ab, \sqrt ac, \sqrt bc are three sides of a triangle which can be deduced from min{a,b,c}  1/4max{a,b,c}.
So do the new or old problem,your choice...{However, i still wait the answer for both}. :)

i have the solution for your inequality, anhcuong
we have
a 2 b(a-b)+b 2 c(b-c)+c 2 a(c-a) \geq 0
<=> (a-c) 2 (ac+ab-bc)+(b-c) 2 (bc+ac-ab)+(b-a) 2 (ab+bc-ac) \geq 0
let a be the max{a,b,c}
case 1: a \geq b \geq c \geq a/4 (it's easy to check)
case 2: a \geq c \geq b \geq a/4
from b \geq a/4 => (b-a) 2 (2bc-ac) \geq -1/2(b-a) 2 ac (1)
from cauchy: [(a-c) 2 +(c-b) 2] \geq 1/2(b-a) 2 ac (2)
b(a-c) 2 [(b-a) 2 -(c-b) 2] \geq 0 (3)
b(c-a) 2 (a-c) \geq 0 (4)
(1)+(2)+(3)+(4) and we have what we want  :D
for this problem i think you can make it tigher, right? anhcuong
please show me your result

So,the first problem is proved by a nice solution:) .Please tell me how can you change the first one to (a-c) 2  (ac+ab-bc)+(b-c)  2 (bc+ac-ab)+(b-a)  2  (ab+bc-ac)  \geq  0
And what about the second condition,  \sqrtab,  \sqrtac,  \sqrtbc are three sides of a triangle.I wait for your ideas...

Anh Cuong
i have another solution
we have the solution with teacher said
i don't remember that carefully
  c>b>a
so if we have a>b/2 this is end
if a<b/2
we put c=x+a so a>1/4c and 3a>x
put  b=y+a so a<y
then we prove A1A2+A2A3+A3A1>0
sorry but i only prove that with min{a,b,c} \geq 1/4max{a,b,c}
can you help me with this way

Sorry,but I don't think this one helps .Let try it and replyed to me when you find something.But what about the second condition of mine.Think about it too,ok??? :)

hi anhcuong. I have the solution for your ineq with the second condition
and here's my solution
a 2 b(a-b)+b 2 c(b-c)+c 2 a(c-a) \geq 0
<=> (a-c) 2 (ac+ab-bc)+(b-c) 2 (bc+ac-ab)+(b-a) 2 (ab+bc-ac) \geq 0 (*)
we can assume a be the max of {a,c,b}
case1: a \geq b \geq c (easy)
case2: a \geq c \geq b
because  \sqrt ab, \sqrt bc, \sqrt ca are the sides of a triangle so we can assume that
ab=(x+y) 2
bc=(y+z) 2
ca=(z+x) 2
with x,y,z > 0
then we have
(*) <=> (a-c) 2 (x 2 +xy+xz-yz) + (b-c) 2 (z 2 +zy+zx-xy) + (a-b) 2 (y 2 +yx+yz-xz)  \geq 0
from a \geq c \geq b we have x \geq z \geq y
from cauchy: (a-c) 2 x 2 + (b-c) 2 z 2  \geq 2abs((a-c)(b-c))xz  (1)
xz[(a-c) 2 +(b-c) 2 ]=xz[(a-b) 2 - 2abs((a-c)(b-c))xz] (2)
(a-b) 2 (y 2 +yx+yz) +(a-c) 2 (xy-yz) + (b-c) 2 (zy-xy) \geq 0 (3)
(1)+(2)+(3) we have your inequality  :D
is this the same as your solution, anhcuong
if not, can you show me how you prove it, ok?

hi  anhcuong's problem is very beautyful but arosis i's prove

Very nice solution for it,arosisi. :) I have two another solution and it don't need to examine two case.However it can be proved by BCS.... :)


Could someone provide me with some information regarding the quality of Harvard's physics department? I am mostly interested in quantum physics and its other sub divisions.

It's one of the best in the country.  Was there something more specific you wanted to know?


Find the constants x,y,z such that in any acute triangle we have
  a^2+b^2+c^2<=xR^2+yRr+zr^2 and such that this is the strongest inequality of this type. I don't have a solution.

In  any  triangle we  have  a^2+b^2+c^2=2(s^2-r^2-4Rr)
and  4R+r>=s*sqrt3.   s=semiperimeter
Maybe this will  help  you...

So  you  must  find the  best  inequality  of the  form s<=f(R,r).
I will  take  a  look  at  my  notes..and  i will  keep  in  touch.

I  would  like  to  see  your  solution  Langragia....

well... the solution is actually based on the following fact:

a^2+b^2+c^2<=4(R^2+3Rr-r^2-sqrt(R)(R-2r)^(2/3)); an inequality which is well-known; this is the smoothest inequality!! (which involves R,r); then it's easy to show that a^2+b^2+c^2<=xR^2+yRr+zr^2  for
x=9; y=-6; z=12;

cheers! :D :D


[quote="Amazigh"]You would be a small genius if you wouldn't now the "trick" for 111...11. :p[/quote]
 :D
lol dik chi li kisla7 lhad 97ab d lihoud  :rotfl:    khso y kon genie zamel lakhor ma3arfchi bi ana aklakh nass homa sohyoun  biba lilk a khatabi


abstract algebra
topology
superior algebra
superior algebra theorems

Let $\mathcal F,\ \mathcal F',\ \mathcal G$ be sheaves of abelian groups on a topological space $X$.
Assume that \[0\to\mathcal F'\to\mathcal F\to\mathcal G\to 0\] is an exact sequence of sheaves and that $\mathcal F'$ is flasque i.e. for every pair of open subsets $U\subseteq V$, the restriction map $\mathcal F'(V)\to\mathcal F'(U)$ is surjective.

Prove that there is an exact sequence of abelian groups:
$0\to\mathcal F'(X)\to\mathcal F(X)\to\mathcal G(X)\to 0$

It is well-known that an exact sequence of sheaves $0 \rightarrow \mathcal{F'}\xrightarrow{f}\mathcal{F}\xrightarrow{g}\mathcal{F''}\rightarrow 0$ induces an exact sequence of abelian groups $0 \rightarrow \mathcal{F'}(U) \xrightarrow{f(U)}\mathcal{F}(U) \xrightarrow{g(U)}\mathcal{F''}(U)$, for any open set $U\subset X$. So we only need to prove that the map $\mathcal{F}(U) \xrightarrow{g(U)}\mathcal{F''}(U)$ is surjective.

Let $P\in U$. Since the map of sheaves $\mathcal{F}\xrightarrow{g}\mathcal{F''}$ is surjective, it follows that the induced map on stalks $\mathcal{F}_{P}\xrightarrow{g_{P}}\mathcal{F''}_{P}$ is also surjective. Pick $s\in \mathcal{F''}(U)$. There is a neighborhood $W$ of $P$, $W\subset U$ and a section $r\in \mathcal{F}(W)$ such that $g(W)(r)=s$. Pick a maximum such preimage, as shown below.

 Let $M=\{(V,t) \mid P\in V\subset U \ \mbox{and}\ s\restriction_{V}=g(V)(t)\}.$ By the surjectivity of $g_{P}$, $M$ is nonempty. An easy argument shows that $M$, equipped with the order relation $(V_{1},t_{1})\leq(V_{2},t_{2})$ iff $V_{1}\subseteq V_{2}\ \mbox{and}\ t_{1}=t_{2}\restriction_{V_{1}}$, is a totally ordered set. Hence, by Zorn's Lemma there is a maximal element in $M$. Suppose the maximal element is $(W,r)$ and suppose $W\neq U$. Let $Q\in U\setminus W$. By the same argument as before, there is an open set $W'\subset U$, $W' \not\subset W$ and a section $r' \in \mathcal{F}(W')$ such that $r'=s\restriction_{W'}$.

We would like to glue $r$ and $r'$ together, but this may not be possible. However, $(r-r')\restriction_{W\cap W'}=0$, so $(r-r')\restriction_{W\cap W'}=f(W\cap W')(\phi)$ for some $\phi \in \mathcal{F'}(W\cap W')$. Since $\mathcal{F}$ is flasque, $\phi$ extends to a section in $\mathcal{F'}(U)$. Now glue together $r$ and $r''\in \mathcal{F}(W')$, where $r''=f(W')(\phi)-r\restriction_{W\cap W'}$ (this is possible since they restrict to the same map on $W\cap W'$). We obtain a section on $W\cup W'$  which is taken by $f(W\cap W')$ to $s$. This contradicts the maximality assumption for $(W,r)$.


10.Through the point $P$ inside of the given circle are driven two perpendicular lines which meets the circle at $A$ and $B$. Find the locus of the projections of $P$ onto $AB$. $\clubsuit$

[color=darkred][b][u]Lermma. The properties of the cyclic and orthodiagonal quadrilateral.[/u][/b]

Let $ABCD$ be a cyclic, orthodiagonal $(AC\perp BD)$ and convex quadrilateral  inscribed in the circle $w=C(O,R)\ .$ We will use the following standard notation : $AB=a$, $BC=b$, $CD=c$, $DA=d$,$AC=e$, $BD=f\ .$ Denote the intersection $E\in AC\cap BD$ and the middlepoints $M$, $N$, $P$, $Q$ of the sides $[AB]$, $[BC]$, $[CD]$, $[DA]$ respectively. Then there are the following properties :

$1.\blacktriangleright ME\perp CD\ ;\ \delta_{AB}(O)=\frac 12\cdot CD\ ,$ where $\delta_{AB}(O)$ is the distance of the point $O$ to the line $AB\ .$

$2.\blacktriangleright \delta_{AB}(O)+\delta_{BC}(O)+\delta_{CD}(O)+\delta_{DA}(O)=p\ .$

$3.\blacktriangleright \sigma [ABCD]=\frac 12(ac+bd)=\frac 12ef\ .$

$4.\blacktriangleright a^2+c^2=b^2+d^2=4R^2\ ;\ MP=NQ=\frac 12\sqrt{e^2+f^2}\ .$

$5\blacktriangleright e^2+f^2=4\left[R^2+p_w (E)\right]\ ,$ where $p_w(E)=R^2-OE^2$ is the power of the point $E$ w.r.t. the circle $w\ .$

$6.\blacktriangleright$ The points $M,N,P,Q$ and the projections of the point $P$ on the sides of the quadrilateral $ABCD$ belong to the circle with the centre in the middlepoint of the segment $OE$ and the length of the diameter equal to $R^2+p_w(E)=2R^2-OE^2$ ([u]the circle of the eight points[/u]) .[/color]

[color=blue][b]Remark.[/b] The proposed problem is a immediate consequence of the  above mentioned $6^{th}$- property of a cyclic and orthodiagonal quadrilateral.[/color]


A man travels $ m$ feet due north at $ 2$ minutes per mile. He returns due south to his starting point at $ 2$ miles per minute. The average rate in miles per hour for the entire trip is:

$ \textbf{(A)}\ 75\qquad
\textbf{(B)}\ 48\qquad
\textbf{(C)}\ 45\qquad
\textbf{(D)}\ 24\qquad\\
\textbf{(E)}\ \text{impossible to determine without knowing the value of }{m}$

The rate for going north is $ .5mpm$ (miles per minute). The rate for going south is $ 2mpm$.

The total minutes traveled north is $ 2m$. The total minutes traveled south is $ .5m$. That is a total of $ 2.5m$ minutes.

The total miles traveled north is m. The total miles traveled south is m.

To get the average, we divide the total miles by the total time (in minutes). $ \frac{miles}{minutes}\equal{}\frac{m\plus{}m}{2.5m}\equal{}\frac{2m}{2.5m}\equal{}\frac{4}{5}$

We have a rate of $ \frac{4}{5}mpm$. We multiply that by 60 (because there is 60 minutes in an hour). Our final answer is $ \frac{4}{5}\times60\equal{}48$. That is answer choice $ \mathbf{(B)}$.


Triangle $ ABC$ is inscript in a circumference $ \Gamma$. A chord $ MN\equal{}1$ of $ \Gamma$ intersects the sides $ AB$ and $ AC$ at $ X$ and $ Y$ respectively, with $ M$, $ X$, $ Y$, $ N$ in that order in $ MN$. Let $ UV$ be the diameter of $ \Gamma$ perpendicular to $ MN$ with $ U$ and $ A$ in the same semiplane respect to $ MN$. Lines $ AV$, $ BU$ and $ CU$ cut $ MN$ in the ratios $ \frac{3}{2}$, $ \frac{4}{5}$ and $ \frac{7}{6}$ respectively (starting counting from $ M$). Find $ XY$

[quote="castigioni"]With analitycal geometry the value is 1/11[/quote]
you did get the answer,but it can be solved by just some ratio substitutions,not analytical.

shall I give some hint here?
the key step(at least it seems to me) is to notice if chord $AB,MN$ intersect at $X$,then
$\frac{MX}{XN}=\frac{MA*MB}{AN*BN}$.

1/11 is the correct answer. Sorry I plugged in a wrong number and came up with a different numerical result. Please delete my post above. Thanks.


logarithms
limit
function
Euler
calculus
calculus computations

Having: $f:R \to (0, + \infty )$ satisfating:
\[
f(x) \cdot \ln f(x) = e^x ,\forall x \in R
\]\[
\mathop {\lim }\limits_{x \to \infty } (1 + \frac{{\ln x}}{{f(x)}})^{\frac{{f(x)}}{x}}  = ?,\forall x \in R.
\]

We have $f(x)\ln{f(x)}=e^{x}$ thus $f(x)>e^{\frac{x}{2}}$ $\forall x\in{\mathbb{R}}$.It is clear that we have to count $\lim({\frac{f(x)}{x}}\cdot\ln({1+\frac{\ln{x}}{f(x)}}))$.Since $f(x)>e^{\frac{x}{2}}$ then $\frac{\ln{x}}{f(x)}{\rightarrow}{0}$ so ${\frac{f(x)}{x}}\cdot\ln({1+\frac{\ln{x}}{f(x)}){\sim}\frac{f(x)}{x}}{\cdot}\frac{\ln{x}}{f(x)}=\frac{\ln{x}}{x}   {\rightarrow}{0}$ so the initial limit is $1$.

[quote="silviu"]
$f(x) \cdot \ln f(x) = e^x ,\forall x \in R$
[/quote]

$f(x)$ is related to $W$ Lambert's function proposed by Lambert 1758 studied by Euler 1779

$f(x)=exp(W(e^x))$


Just wondering, who here goes to the Waterloo Math Circles? I'm the Indian guy in the senior class looking lost the whole time :blush: .

me! :D

aww.. it says my message is too small.. hopefully it's long enough now..

I *may* be making an appearance as an instructor.

yeah I used to go regulary, didn't go for a year, now going again

It's a good place to waste time in :D
No, actually one can learn some formulas :)


group theory
abstract algebra
Ring Theory
superior algebra
regular element

Prove that there does not exist a ring with exactly 5 regular elements.
($ a$ is called a regular element if $ ax \equal{} 0$ or $ xa \equal{} 0$ implies $ x \equal{} 0$.)

A ring is not necessarily commutative, does not necessarily contain unity element, or is not necessarily finite.

You should however specify what you understand by ring: does it have a unity, for instance? A very similar problem was discussed very long time ago: if $ A$ is a unitary ring, then it cannot have $ 5$ invertible elements. If I remember correctly, the solution used actually that they were regular, so maybe in this case one can adapt the proof. When the ring has no unity, I guess it's more tricky.

OK, I will.

P.S. [i]Here the ring does not necessarily contains unity element, but one part of my proof was that it has the unity element.[/i]

hm, if the ring has a unit, then it's easy, right? it seems like the regular elements form a cyclic group of order 5 and then nobody can be $ \minus{}1$ (since it would then have even order)

The set $ G$ of regular elements of the ring $ R$ is obviously closed under multiplication.
The left-multiplication in $ G$ with any $ a\in G$ is injective, and since $ G$ is finite, it is also surjective.
Analog for the right-multiplication, so the equations $ ax \equal{} b$ and $ ya \equal{} b$ have solutions $ x,y\in G$ for any $ a,b\in G$.
This already gives $ G$ a group, so $ G\cong C_5$ cyclic of order $ 5$ and has a neutral element $ 1$.
($ 1$ is not necessarily neutral in $ R$)

But now as MysticTerminator pointed out we get $ \{\pm 1\}$ a subgroup of $ G$, and by Lagrange we get $ 1 \equal{} \minus{} 1$.
Then for $ 1\not \equal{} c\in G$ we get the canonical ring homo $ \Phi: \mathbb{F}_2[X]\to R, 1\mapsto 1,X\mapsto c$.

The image $ S$ of $ \Phi$ is a subring of $ R$ with neutral element $ 1$, and since $ G \equal{} \langle c\rangle \subseteq S$,
we get in fact every regular element of $ R$ a unit in $ S$.

Conversely any unit $ a$ of $ S$ must be a regular element of $ R$,
because otherwise assume WLOG $ xa \equal{} 0$ with some $ 0\not \equal{} x\in R$.
But then for $ b\in S$ with $ ab \equal{} 1$ we get $ x1 \equal{} xab \equal{} 0$, which contradicts $ 1\in G$ regular in $ R$.
So $ S$ has $ 5$ units.

$ c^5 \equal{} 1$ by Lagrange, so the kernel of $ \Phi$ contains $ X^5 \minus{} 1 \equal{} (X \minus{} 1)(X^4 \plus{} X^3 \plus{} X^2 \plus{} X \plus{} 1)$.
Let $ \zeta$ be root of $ X^4 \plus{} X^3 \plus{} X^2 \plus{} X \plus{} 1$ over $ \mathbb{F}_2$.
Then $ \mathbb{F}_2(\zeta) \equal{} \mathbb{F}_{2^n}$ for some $ n\le 4$,
and by Lagrange $ 5 \equal{} |\langle \zeta\rangle|$ divides $ |\mathbb{F}_{2^n}^\times| \equal{} 2^n \minus{} 1$, which gives $ n \equal{} 4$
and $ X^4 \plus{} X^3 \plus{} X^2 \plus{} X \plus{} 1$ the irreducible minimalpolynomial of $ \zeta$ over $ \mathbb{F}_2$.

And now the [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese remainder theorem[/url] gives $ S$ isomorphic to a factor-ring of
$ \mathbb{F}_2[X]/(X^5 \minus{} 1)\cong \mathbb{F}_2[X]/(X \minus{} 1)\times \mathbb{F}_2[X]/(X^4 \plus{} X^3 \plus{} X^2 \plus{} X \plus{} 1)\cong \mathbb{F}_2\times \mathbb{F}_{16}$,
so $ S$ is isomorphic to $ 0$ or $ \mathbb{F}_2$ or $ \mathbb{F}_{16}$ or  $ \mathbb{F}_2\times \mathbb{F}_{16}$, and has either $ 1$ or $ 15$ units.
Contradiction!

Generalization:

Characterize those finite semigroup $ G$ with the property that there is a ring whose semigroup of regular elements is isomorphic to $ G$.

(Just as in the initial problem, the ring need not have a unit, be commutative, finite, etc.)

I pretty much proved above, that $ G$ must be the unit-group of some finite unitary ring.
(take $ S \equal{} \sum_{g\in G}\mathbb{Z}g$ the finite unitary subring with $ G \equal{} S^\times$)
But beyond that? Take the unit group $ G$ of any group algebra $ RH$ for any finite unitary ring $ R\not \equal{} 0$
and any finite group $ H$. Then always $ H\le G$.

EDIT
Ok, here is something better using [url=http://en.wikipedia.org/wiki/Artin-Wedderburn_theorem]Artin–Wedderburn[/url] and [url=http://en.wikipedia.org/wiki/Wedderburn_theorem]Wedderburn[/url].
In fact every finite unitary ring $ R$ has a nilpotent ideal $ J(R)$,
with $ R/J(R)\cong \mathcal{M}_{n_1}(\mathbb{F}_{q_1})\times \cdots\times\mathcal{M}_{n_r}(\mathbb{F}_{q_r})$ for some finite fields $ \mathbb{F}_{q_1},\ldots,\mathbb{F}_{q_r}$ and some $ n_1,\ldots,n_r\in\mathbb{N}$.

Then the unit-group $ G \equal{} R^\times$ has the nilpotent normal subgroup $ N \equal{} 1 \plus{} J(R)\unlhd G$ with
$ G/N\cong (R/J(R))^\times\cong GL_{n_1}(\mathbb{F}_{q_1})\times \cdots\times GL_{n_r}(\mathbb{F}_{q_r})$.

For that show, that the group homo $ R^\times\to (R/J(R))^\times$ is an epi with kernel $ N \equal{} 1 \plus{} J(R)$.
And for $ 1 \plus{} J(R)$ nilpotent we need $ J(R)$ nilpotent and $ [1 \plus{} J(R),1 \plus{} J(R)^n]\le 1 \plus{} J(R)^{n \plus{} 1}$ for all $ n\in\mathbb{N}$.
For that take $ x\in J(R),y\in J(R)^n$, then
$ [1 \plus{} x,1 \plus{} y] \equal{} (1 \plus{} x)^{ \minus{} 1}(1 \plus{} y)^{ \minus{} 1}(1 \plus{} x)(1 \plus{} y) \\
\hspace*{0.77in} \equal{} (1 \plus{} x)^{ \minus{} 1}(1 \plus{} y)^{ \minus{} 1}((1 \plus{} y)(1 \plus{} x) \plus{} xy \minus{} yx) \\
\hspace*{0.77in} \equal{} 1 \plus{} (1 \plus{} x)^{ \minus{} 1}(1 \plus{} y)^{ \minus{} 1}(\underbrace{xy \minus{} yx}_{\in J(R)^{n \plus{} 1}})\in 1 \plus{} J(R)^{n \plus{} 1}$


calculus
integration
real analysis
real analysis solved

Compute lim(n->00)  \int (n to n+1) {x 2 *sin(1/x)}dx
{x}=decimal part of x

I think it is, indeed, 1/2. We write it as \int{n->n+1}f(x)dx-\int{n->n+1}[f(x)]dx, where f(x)=x2sin(1/x). It's easy to see that \int{n->n+1}[f(x)]dx-\int{n->n+1}ndx->0, so \int{n->n+1}[f(x)]dx-n->0, so we have to compute the limit of \int{n->n+1}f(x)dx-n=\int{n->n+1}f(x)dx-\int{n->n+1}xdx+1/2=\int{n->n+1}(f(x)-x)dx+1/2, and since f(x)-x->0 as x->00, it means that the limit is 1/2.


calculus
integration
algebra unsolved
algebra

Sorry this problem should be in Algebra Unsolved Problems Forum; not in the proposed and own problems!

So, how do you know this is an easy one? :wink:

Pierre.

I've solved it :lol:  and anyway it is a regional olympiad problem for 9th and 10th graders; so it should be easy.

Still no takers for this problem?...
[hide="Hint"]Try using Number Theory and modulo properties[/hide]

i actually don't get what you mean by product of powers of [b]alternate[/b] primes

Oh, I am sorry for the typo. q is the product of arbitrary powers of alternate primes i.e., q = $2^{x}.5^{y}.11^{z}...$.Here alternate primes mean 2,5,11,17,23,...


What does the rank mean? Is a higher or lower rank better?

[quote="levans"]It's just your rank compared to other users. Higher is better, generally.[/quote]

so a rank of 9 is better than a rank of 10?

Isn't 1 of 32 better than 15 of 32...

Lower is better, I think... :o

So like as of now am 29th in Karma out of 1433. That means there are 28 people better than me, and 1404 below me.