erm...how can there be 1999 people on the island when err...1000+500+700=2200!!!can people choose more than one hobby!!!

Then I'm guessing there's more than 1 solution.

I know an answer(without demonstration) of $ 201$ liers.. :maybe:

[quote="sunnykim"]There are $ 1999$ persons on an island, some of which are liers(lie all the time) and the others always tell the truth. Each person has exactly one hobby among music, fotball and painting. $ 1000$ said they like music, $ 700$ fotball and $ 500$ painting. How many liers are there?[/quote]
Couldn't there be just $ 1$ liar? He might say he likes football and painting when he actually likes music. This would make for a total of $ 1999$ people (since the other $ 1998$ all told the truth).

They all could be liers. since the person whom liked football could say they like music instead.

how can there be 1999 people on the island when err...1000+500+700=2200!!!

[url=http://www.math123xyz.com]Interactive Math Resources[/url]

Since liars always lie and truthful people always tell the truth, truthful people always give one answer while liars always give two. Thus, the number of liars is equal to the number of extra answers. Since there are 201 more answers than people on the island, we know that there are 201 liars.

aren't liers spelled liars???

that came out of nowhere...

aw...I was gonna comment on that...

I notice (not naming any names) that about 50% of math geeks spell liar lier...

math154-bad stereotype.  I always spell it correctly: I AM A LIER

Please do not say something is spam, and keep bumping the topic up if it's an old thread. Thanks.

(And sorry for adding to the spam.)

[quote="isabella2296"][quote="sunnykim"]Yes, because there are liers who can say everything. So although we know that each has exactly one hobby some may argue that they have 2 ore even 3 :)[/quote]

And how could a liar give three answers, because one of those answers has to be true, and you said liars lie all the time.[/quote]

Who knows?  Maybe those liars like none of those three hobbies.

:spam:

[quote="isabella2296"]Aha! I have an answer!

First we add all the numbers: $ 1000 \plus{} 500 \plus{} 700 \equal{} 2200$

Now we subtract $ 1999$ from our total: $ 2200 \minus{} 1999 \equal{} 201$

And that is how we get to our answer of $ \boxed{201}$[/quote]

This is our answer. Let's allow this topic to die.

[quote="murder"]can't liars say that they like 3 answers in addition to some saying that they like 2 answers?[/quote]
They can!!!

This topic is pretty old - please let it die.

Izzy-why do you always tell me not to spam, and not ducky? she has WAY more spam posts than me...and thats saying something

[quote="isabella2296"]I'm saying it in general - as in, to EVERYONE.[/quote]

Including you, izzy.  No offense, but like 100 of your posts are somewhat similar to the following:

(1) Don't spam.

(2) This topic is pretty old.  Let this thread die.

Oooh.  And to wrap it up, I think the answer is $ \boxed{201 \text{ liars}}$

good point ernie
and about 200 of my posts are
:spam: this post is :spam: lol please dont :spam: actually its fun

I know I'm spamming too!!!!!!

I'm including myself in this!!!

I DON'T KNOW WHY WE SPAM SO MUCH. WHY AREN'T THE MODERATORS LOCKING THESE??


algebra
polynomial
number theory
algebra open

Find all $ P(x)$ such as
$ 1) P(x)$ is a polynomial with integer coeeficients and degree $ \geq 1$
$ 2) P(a \cdot b) \mid P(a^{2})^{2}\Rightarrow a=b$  $ \forall a,b \in \mathbb N$

PS: Should I put it in number theory forum?

This is a hard question. And , what do you have solution for it  :)


group theory
abstract algebra
topology
superior algebra
superior algebra unsolved

Suppose $G$ is a finitely generated group and $H\leq G$ that $[G: H]$ is finite. Prove that $H$ is also finitely generated.

Hey friends solve this problem. I think it is a challenging problem. I couldn't solve it until now.

Suppose we somehow manage to prove this for the case when $G$ is also a free group, i.e. $G=F_n$, the free group on $n$ generators. Then we can solve the general case as follows:

If $G$ is generated by $n$ elements, then there is a normal subgroup $D\lhd F_n$ such that $G=F_n/D$. All subgroups $H\le G,\ [G: H]<\infty$ are actually of the form $H=E/D$ for some $E\le F_n$ with $D\lhd E$ and $[F_n: E]=[G: H]<\infty$. Since we know the conclusion to be true for $F_n,E$ instead of $G,H$, we conclude that $E$ is generated by a fintie number of elements, say $m$. Then $H=E/D$ will be generated by the images of those $m$ elements through the natural projection $E\to H=E/D$, so we would be done.

Let's now prove the assertion for free groups $G$. In other words, we assume, just like I said in the first paragraph, that $G=F_n$ for some positive integer $n$. We can use covering space theory. Let $\mathcal G_G$ be a finite graph whose fundamental group is $G=F_n$ (for example, we can take $\mathcal G_G$ to be the [url=http://en.wikipedia.org/wiki/Wedge_sum]wedge sum[/url] of $n$ circles), and let $\mathcal G_H$ be a covering space (also a graph) whose fundamental group is $H$. Since $[G: H]$ is finite, $\mathcal G_G$ is finite, and $[G: H]$ equals the cardinal of the preimage of every point of $\mathcal G_G$ through the covering $\mathcal G_h\to\mathcal G_G$, it means that $\mathcal G_H$ must be finite as well, so, in particular, it's fundamental group (which is $H$) is a free group on finitely many generators (a moment's thought will even tell us the number of generators based on $n$ and $k=[G: H]$: it's $k(n-1)+1$).

LOL. I just thought for one minute about the problem and found EXACTLY the same solution as grobber :). Very nice, indeed.

we can solve it without using algebraic topology method.assume that $Hx_0,...,H{}_x{}_n$ are right cosets and suppose that $x_0=e$,also let $a_i$ be the set that generate $G$ and $r(x)=x_i$ such that $x\in Hx_i$.it not that hard $<x_i.a_i.r(x_i.a_i)^{-1}> = H$ ;)

[quote="sam-n"]it not that hard $<x_{i}.a_{i}.r(x_{i}.a_{i})^{-1}> = H$ ;)[/quote]

Could anybody give me a hint? I can't see this conclusion ...

[quote="sam-n"]it not that hard $<x_{i}.a_{i}.r(x_{i}.a_{i})^{-1}> = H$[/quote]

This is essentially the solution I reconstructed for the problem I posted, so I'll clarify.

Suppose $G$ is generated by $n$ elements $x_{1},\dots,x_{n}$ and $H$ is of index $m$, with left cosets $Hy_{1},Hy_{2},\dots,Hy_{m}$ ($y_{1}=e$ for convenience).
Consider all $m^{2}n$ elements of the form $y_{i}x_{j}y_{k}^{-1}$. For each choice of $i,j$, $y_{i}x_{j}\in Hy_{l}$ for some $l$, and $y_{i}x_{j}y_{k}^{-1}\in H$ iff $k=l$. For each choice of $j,k$, $y_{k}x_{j}^{-1}\in Hy_{l}$ for some $l$, and $y_{i}x_{j}y_{k}^{-1}=\left(y_{k}x_{j}^{-1}y_{i}^{-1}\right)^{-1}\in H$ iff $i=l$.
This means that exactly $mn$ of these elements are in $H$, and a choice of either $i$ or $k$ determines the other for a particular $j$. We will show that these $mn$ elements generate $H$ by algorithmic construction.

Let $h$ be an arbitrary element of $H$, and write it as a product of the $x_{i}$ and their inverses. For example, suppose $h=x_{5}x_{1}x_{4}^{-1}x_{3}$. Choose $i(1)$ so that $e x_{5}y_{i(1)}^{-1}\in H$. Choose $i(2)$ so that $y_{i(1)}x_{1}y_{i(2)}^{-1}\in H$. Choose $i(3)$ so that $y_{i(2)}x_{4}^{-1}y_{i(3)}^{-1}= \left(y_{i(3)}x_{4}y_{i(2)}^{-1}\right)^{-1}\in H$. Choose $i(4)$ so that $y_{i(3)}x_{3}y_{i(4)}^{-1}\in H$.
Now the product $e x_{5}y_{i(1)}^{-1}y_{i(1)}x_{1}y_{i(2)}^{-1}y_{i(2)}x_{4}^{-1}y_{i(3)}^{-1}y_{i(3)}x_{3}y_{i(4)}^{-1}= x_{5}x_{1}x_{4}^{-1}x_{3}y_{i(4)}=hy_{i(4)}\in H$. Since $h\in H$, $y_{i(4)}=y_{1}=e$, and we have written $h$ as a product of our alleged generators and their inverses. The procedure is general, and these elements generate $H$.


inequalities
combinatorics open
combinatorics

let N={1,2...n}. What is the maximum number f(n) s.t. for any partition of N into f(n) sets, there exists three elements in the same set (not necessarily distinct) s.t. a+b=c.

(typo corrected now)

i guess you wanted to say maximum number, since else we have f(n)=1(any partition of N into 1 set(which gives N clearly) has the desired property(expect for n=1) since 1+1=2).

Peter

The problem A4 from ISL 1999 is a bit similar (but there N={1,2..})
I consider that problem very hard

for N={1,2,...} i think it has been proven that given a partition into three sets, we can find a,b,c in the same set such that a+b=c. a counterexample with 4 sets is as follows: write every number n as $n=2^ab$, where b is odd. then distinct into 4 sets whether a is odd or even, and b is 1 or 3 mod 4.

Peter

I. Schur proved (1916) that for any given $k$ there exists an integer $n=n(k)$ such that for any partition of $ \{1,2,...,n \}$ into $k$ subsets, there is one of these sets which contains three numbers, say $a,b,c$ not necessarly distinct such that $a+b=c$.

The least integer $n=S(k)$ which has the above property (and which depends on $k$) is not known in general.
Schur proved that $ \frac {3^k+1} {2} < S(k) \leq [k!e] $, where $[.]$ denotes the integer part.
Now, it is known that $c(89)^{ \frac {k} {4} } < g(k) \leq [n!(e- \frac {1} {24})]$, for some constant $c > 0$
I'm not sure (ah, ah...sorry) but I think the lower bound has been sharpened to $315^ \frac {k-1} {5}$ (and no $c$).

Pierre.

ok...i recognized that my counterexample for four sets doesn't work(sorry)
i think i read a proof for schur's bounds... it is in: Arthur Engel, [i]Problem-solving strategies[/i], New York 1998: chapter 4, page 67 for the upper (i. e. right) bound and problem 45 for the lower one.

Peter

Note that Schur proved that result when working on Fermat's last theorem, since he deduced from the above that :
For all $m$, if $p$ is a prime sufficientely large then the equation $x^m+y^m=z^m$ has a nonzero solution in the integers modulo $p$.

Moreover, even he proved this result years ago before Ramsey famous theorem, it is often considered as the first nontrivial result in Ramsey Theory.

Pierre.

deducing that result is quite obvious. just suppose $p>N(m)$. we may define the m sets $A_0,...,A_{m-1}$ the following way: 0 is in $A_0$. since $(Z/pZ)*$ is isomorphic to $Z/(p-1)Z$, look at every number $a=1,...,N(m)$ as a member of $Z/pZ$ and send it to $Z/(p-1)Z$ where we look at it mod gcd(m,p-1), let that residue be a(k). then let k be in $A_{a(k)}$. now, if $a+b=c$ with $a,b,c$ in $A_k$, and g is a primitive root in $Z/pZ$, then $(ag^{-k})+(bg^{-k})=(cg^{-k})$ gives a desired triple, since sending every number into $Z/(p-1)Z$, and reducing mod gcd(m,p-1), we get 0, that means, it is an m-th power in $Z/pZ$.

Peter


vector
geometry
geometric transformation
homothety
projective geometry
geometry proposed

Um,I think the first one can be proved by Vector and Ceva.Let AC=mAC',AD=nAD',AB=pAB' and CB1/DB1=x/y;DC1/BC1=z/t;BD1/CD1=e/f then C'B1'/D'B1'=xn/ym; D'C1'/B'C1'=zp/tn;B'D1'/D1'C'=em/fp{If I don't have any mistake}.But
(x/y)(z/t)(e/f)=1 so (C'B1'/D'B1')(D'C1'/B'C1')(B'D1'/D1'C')=1.That means B'B'1, C'C'1, D'D'1 are concurent.
I think the second one can be done with the same way,but I have not checked it yet.So,let's try :)

[quote="treegoner"]Let $ BCD$ and $ B^{\prime}C^{\prime}D^{\prime}$ be triangles such that the lines $ BB^{\prime}$, $ CC^{\prime}$, $ DD^{\prime}$ are concurent at a point $ A$. Let $ B_1$, $ C_1$, $ D_1$ be points on the lines $ CD$, $ DB$, $ BC$ such that the lines $ BB_1$, $ CC_1$, $ DD_1$ are concurent at a point $ M$. Let $ B^{\prime}_1$ be the point of intersection of the lines $ AB_1$ and $ C^{\prime}D^{\prime}$. Let $ C^{\prime}_1$ be the point of intersection of the lines $ AC_1$ and $ D^{\prime}B^{\prime}$. Let $ D^{\prime}_1$ be the point of intersection of the lines $ AD_1$ and $ B^{\prime}C^{\prime}$.

[b](a)[/b] Prove that the lines $ B^{\prime}B^{\prime}_1$, $ C^{\prime}C^{\prime}_1$, $ D^{\prime}D^{\prime}_1$ are concurent at a point $ M^{\prime}$.
[b](b)[/b] Prove that the points $ A$, $ M$, $ M^{\prime}$ are collinear.[/quote]

[u]Two possible proofs of treegoner's result (proving (a) and (b) at once):[/u]

[b]Projective proof:[/b]

Using projective transformations, it is trivial: Undertake such a projective transformation which maps the axis of perspective of triangles BCD and B'C'D' to the line at infinity. Then, the (new) triangles BCD and B'C'D' are homothetic (since opposite sidelines are parallel). The center of homothety is A. Then, the homothety with center A mapping triangle BCD to triangle B'C'D' must, of course, map the points B1, C1, D1 to the points B1', C1', D1', respectively; thus, the point M lying on the lines BB1, CC1, DD1 is mapped to a point M1 lying on the lines B'B1', C'C1', D'D1' and on the line AM, of course (what follows from the properties of a homothety). Hence we are done.

[b]Elementary synthetic proof:[/b]

Now I have a purely synthetic proof:

We work in the projective plane, i. e. parallel lines are assumed to intersect each other at infinite points.

Since the triangles BCD and B'C'D' are perspective, i. e. the lines BB', CC' and DD' concur (at A), the Desargues theorem yields that the points X = CD /\ C'D', Y = DB /\ D'B' and Z = BC /\ B'C' are collinear, where the sign /\ means "intersection" (i. e. the intersection of two lines g and h is denoted by g /\ h).

Now consider the triangles BB'Y and B1B1'X. Since the points BB' /\ B1B1' = A, B'Y /\ B1'X = D' and YB /\ XB1 = D are collinear, the Desargues theorem shows that the lines BB1, B'B1' and YX concur. In other words, the point of intersection S of the lines BB1 and B'B1' lies on the line XYZ. Similarly, the point of intersection T of the lines DD1 and D'D1' lies on the line XYZ, too.

Finally we apply the Desargues theorem to the triangles DD'T and BB'S: Since the lines DB, D'B' and TS concur (at Y), the points DD' /\ BB', D'T /\ B'S and TD /\ SB are collinear. But DD' /\ BB' = A, D'T /\ B'S = D'D1' /\ B'B1' and TD /\ SB = DD1 /\ BB1 = M. Hence, the points A, D'D1' /\ B'B1' and M are collinear. In other words, the lines D'D1', B'B1' and AM concur. Similarly, the lines D'D1', C'C1' and AM concur. Hence, all four lines B'B1', C'C1', D'D1' and AM concur at one point. This proves both of your assertions a) and b) simultaneously.

  Darij

I think that it is a good idea to use projective transfomation (my friend said to me), but I don't know anything about it. :( I think I should learn it now to have a few ideas about it.
  Your elementary solution is not perfectly right when you apply Desgarses theorem since it can exist that the three lines can be parallel or concurs. And you can't have those points.
   My idea is to use space geometry to prove this one. I 'll give to you later.

Well... if the lines are parallel, then their point of intersection is at infinity; the Desargues theorem holds for such points, too!

However, there actually may be problems when some triangles the Desargues theorem is applied to are degenerate, but usually in such cases there is a simpler proof or we can apply continuity.

Treegoner, with the title of your post, do you want to say that the problem can be solved by Ceva? I can only think of solutions using projective geometry (as in my above post, by projective transformations, or by Desargues, maybe also by Pappos). Maybe there is also a solution using Menelaos (in fact, you can forget about D1 and D1', just prove that the meet of B'B'1 with C'C'1, the meet of BB1 and CC1, and the point A are collinear). How do you want to use Ceva in problem [b](b)[/b]?

  Darij

I Have just proved it by Ceva and Menelauyt

Using areas one can prove $ \frac{CD_1/D_1B}{C'D_1'/D_1'B'}\equal{}\frac{(ACD_1)/(AD_1B)}{(AC'D_1')/(AD_1'B')}\equal{}\frac{CA/AB}{C'A/AB'}$ doing this with every side one has:
$ \frac{CD_1}{D_1B}\frac{BC_1}{C_1D}\frac{DB_1}{B_1C} \equal{} \frac{C'D_1'}{D_1'B'}\frac{B'C_1'}{C_1'D'}\frac{D'B_1'}{B_1'C'}$


Hey, I know there is already a poll, but please put your 10*AIME+AMC index again so that we get a better idea of the distribution between 200 and 235.

PLEASE VOTE ONLY IF YOU ARE IN 11th OR 12th GRADE.

[quote="shbmsc90"]
PLEASE VOTE ONLY IF YOU ARE IN 11th OR 12th GRADE.[/quote]

Umm... 10th graders and below can still qualify on index, and would thus affect the index cutoff.

240.0 btw.

[quote="t0rajir0u"]AoPS isn't exactly the best place to get an accurate idea of the index  :P[/quote]

True, but the index is most likely to be lower.

When I was in 9th grade, the cutoff floor was 9 and the index cutoff was 233.5 or something. That was pre-expansion.

I think the index cutoff will be around 212/212.5.

Edit: Oops, I thought the poll was asking for what we thought the index cutoff will be. Umm, you can minus one from the 206.5-212 choice. Are we allowed to change our vote?

[quote="paladin8"][quote="xxazurewrathxx"]Edit: Oops, I thought the poll was asking for what we thought the index cutoff will be. Umm, you can minus one from the 206.5-212 choice. Are we allowed to change our vote?[/quote]

Heh, I did that too. Oh well.[/quote]
Wait doesn't it say, "What is your 2007 index?"  Doesn't that mean "What is your AMC 12 score plus 10 times your AIME score?", not "What do you think the 2007 USAMO index cutoff will be?"

I read the thread title and didn't bother to read the poll's question.


A large house contains a TV in each room that has an odd number of doors.  There is only one entrance to this house.  Show that it is always possible to enter this house and get to a room with a TV.

I don't get it, since there's TV in EACH room, why can't I get to it?

I understand.
Let $d$ be sum of doors (for each room you count number of doors in it and sum it all to $d$). We will show that $d$ is odd. There is one entrance=1 door. For each two roms A,B that are connected by door we count this door two times. So $d$ is odd. From this we know that there's at least one room with odd number of doors.
Is it what you wanted?

Hmm.. Not to cause trouble but  I can think of some un-usual topologies ... (If you see old movie Classics like Mughl-e-Azam you would understand the inspiration...:D)
(For more details see: http://www.rediff.com/movies/mughal.htm

Okay back to  the original problem:

House (Actually a palace)   Has One larger  room, and  inside it two small rooms .

Room A -  Is occupied by the Famous Anarkali -  King Mughl-e-azam did not like her.
Room B - Is occupied by her  Friend - King did not like him either.

One door to enter their Palace....
One door just hangs around for fun between the two chanbers...

NO there is no TV anywhere. .. And there are two  doors in one chamber and zero in other two...

Yes, it depends on how exactly you define "number of doors in a room" and whether a door can lead from one room to itself. If you rewrite this problem do graph terminology it would be much clearer :D.

Also you can imagine infinite set of rooms $R_1,R_2,\ldots$ such that $R_i$ contain $R_{i+1}$ and there's only one door in $R_{i+1}$ wall. You will never enter a room which has a TV :D

[quote], it depends on how exactly you define "number of doors in a room" and whether a door can lead[b] from one room to itself.[/b] [/quote]
..Also some times a door can  lead to a broom closet. only  :D  (So it depends also on how you define a room)
(While all will agree to calll that door a door, IMO no one, even the realestate person who is going to sell you that house, will  describe that broom closet as a room!... (besides even if it had a TV - flat screen type 'cause only that kind will fit  - it might  still be [b]impossible [/b] for you to get in,  if the  dimension of the closet is tiny...see the original problem)

P.S. I did like the what you mentioned about infinte  number of rooms. Beautiful!!!  (I did not think of that, but it was  beautiful.)
 ( Going away to beat some other dead horses,,,
  :D )


Show that any collection of pairwise disjoint discs in the plane is countable. What happens if we replace "discs" by "circles"?

since the rationals are dense in the reals, to every disc we may associate a point inside it with rational coordinates. since these points are pairwise distinct and $ \mathbb{Q}^2$ is countable, it follows that the set of discs is countable. on the other hand, for circles this isn't true: take the set of all circles centered at the origin with irrational radii.


Prove that $ \sum \left(a^{2}b^{2}+abc\right)\geq\left(a+b+c\right)\left(ab+bc+ac-1\right)$
for a,b,c positive reals and $ a^{2}+b^{2}+c^{2}=3$.

what does the summation notation? symmetric or cyclic

Can anybody posts a solution????I really need it.Thanks. :lol:


Claim: $A$ has independent columns if and only if $Ax=0$ implies that $x=0.$ (This is for column vectors $x$ of the appropriate size.

The reasoning behind this claim: $Ax$ is simply a linear combination of the columns of $A.$ $Ax$ equals $x_{1}$ times the first column of $A$ plus $x_{2}$ times the second column of $A,$ and so on. So $Ax=0$ is the statement that a certain linear combination of the columns is the zero vector. They are independent iff that can only be the trivial linear combination, which is the same as saying that $x$ is the zero vector.

So now, to the problem at hand:

$(AB)x=0\implies A(Bx)=0\implies Bx=0\implies x=0.$

Therefore $AB$ has independent columns.

q.e.d.

[quote="xirtamax"]suppose A and B are matrices with linearly independent columns. Prove that AB also has linearly independent columns
(assuming this matrix produc is defined.)[/quote]
We also can kill this problem by determinants: for any $\mathbf{M}$ fact that $\mathbf{M}$ has linear independent columns is equivalent to fact that $\det(\mathbf{M})\neq 0$, so $\det(\mathbf{A}),\det(\mathbf{B})\neq 0\implies \det(\mathbf{A})\cdot\det(\mathbf{B})\neq 0\iff \det(\mathbf{AB})\neq 0$ which ends the proof. But this solution is of course much worse than $\text{Kent Merryfield}$'s solution, because it requires proofs of some quite nontrivial facts about determinants.

Only square matrices have determinants; in your proof you are implicitly assuming that $A$ and $B$ are both square matrices. The theorem is true without that assumption; it is true for non-square matrices.

Oh, I see. My solution was very stupid.  :blush:


Here are my predictions:
AFC CONFERENCE CHAMPIONSHIP
Jets 20, Colts 17

NFC CONFERENCE CHAMPIONSHIP
Vikings 35, Saints 31

SUPER BOWL (with Favre taking on the Jets...)
[hide]Jets 14, Vikings 10[/hide]

[size=59]Okay, I'm from the New York area[/size]

you must be from NY lol.

My suberbowl pick:  Vikings: 31, Colts: 27

AFC Championship: Colts 24, Jets 10

NFC Championship: Saints 35, Vikings 31

Superbowl Game: Colts 31, Saints 24

Cough, I am also from the NY area.

[b]AFC Prediction and Analysis:[/b]

Peyton Manning enters the game iwth an astounding 4500 passing yards and a 99.0 Quarterback rating, compared to

Mark Sanchez's 63.0 rating and 2444 yards. Peyton Manning has long been regarded as one of the best in history, who

also has the experience factor, playing in his 12th season against a rookie quarterback.

However, the Jet defense plays very well, allowing only 14.8 points a game (the offense scores 21.8), compared to the

Colts' 19.2 PPG allowed.

If the defense can keep Peyton Manning in reasonable check, they have the oppertunity to break this game open with

their rushing game, that features Shonn Greene (Yes, that is how it is spelled), also a rookie, with an amazing 263

yards in the postseason thus far, compared to a disappointing Addai's 23 (Although the Colts have played only 1 game

to the Jets 2).

All in all, this should be an interesting game to watch, with twists all over the place, but Peyton Manning and his

offense will probably prevail, being the more experienced (and probably better) team

[b]Prediction: Colts 21, Jets 13.[/b]

[b]NFC prediction and analysis:[/b]

This one is much harder to call. Brett Farve certainly has the experience factor (now playing in his 19th(!) season),

an amazing 107.2 quarterback rating, and a +26 TD-to-interception rate.

These two offenses are monsters, being numbers 1 and 2 for points scored in the league, with the saints amassing an

amazing and intimidating [i]403.8[/i] passing yards a game, with Brees having a 109.6 quarterback rating, and a +23

TD-to-interception rate.

Reggie Bush has 390 yards on just 70 carries this season, while Peterson had 1383 on 314 carries.

This will be a heavy offensive battle, probably not unlike the Cardinals-Packers wildcard round game, with the Vikings

having a slightly more balanced offense, while the Saints tend to be a 1-dimensional team.

[b]Prediction: Vikings 34, Saints 28[/b] (Yes this is high-scoring)

DISCLAIMER:

These predictions and analysis are solely mine and may or may not agree with any, possibly more informed, other

predictions/analysis. I would also like the Jets and the Vikings to win, so my analysis may seem biased towards these

teams. This analysis may be reasonably accurate, or entirely wrong. If you use these predictions/analysis for any

independant use, do not hold me accountable for any problems. Or I will be mad.
[b]

GO JETS AND VIKINGS![/b]

EDIT:

[b]SUPER BOWL PREDICTION[/b]

No analysis, as that was pretty much covered above.

[hide="Prediction"][hide="C"][hide="O"]17

13[/hide][/hide][/hide]

My Preidiction:

Colts: 24
Vikings: 21

(Unless Peyton Manning is injured)

[b][color=blue]Colts: 31[/color][/b]
[color=olive]Saints: 27[/color]

[quote="ernie"]Colts 30, Jets 17
Saints 31, Vikings 28 (OT)[/quote]

I'm posting the real scores, lol.

Also, I already posted my prediction for the Super Bowl.

[quote="ernie"]Colts 31, Saints 24[/quote]

Guys the scores were:

COLTS 30 JETS 17

SAINTS 31 VIKINGS 28 (OT)

Ugh --> Vikings. They had like 12 easy wins.

WIN, bogtro, that's what I just said. :P

SUPER BOWL PREDICTIONS:
Colts 28
Saints 24
The number of predictions I get right this year: 0


geometry
circumcircle
parallelogram
power of a point
radical axis
perpendicular bisector
angle bisector

IN a quadrilateral ABCD,angle B=angle D,the bisector of A and it's supplementary angle intsection B and it's supplementary angle on E,F.
Prove EF bisector BD

$ (O_1) \cong (O_2)$ are circumcircles and of $ \triangle ABC, \triangle CDA.$ $ M$ is common midpoint of $ AC, O_1O_2.$ Internal angle bisectors  $ AF, CE$ cut $ O_1O_2$ at $ P_1, P_2.$ As $ \overline{MP_1} \equal{} \minus{}\overline {MP_2},$ $ AF \parallel CE$ and $ AFCE$ is a rectangle.

Let $ H_a, H_b, H_c$ be altitude feet of $ \triangle ABC$ and $ K_c, K_d, K_a$ altitude feet of $ \triangle CDA.$ Points $ E, F, H_a, H_c, K_a, K_c$ are all on a circle $ (M)$ with diameter $ AC.$ Let $ (N_1) \equiv \odot(MH_aH_bH_c)$ and $ (N_2) \equiv \odot(MK_cK_dK_a)$ be 9-point circles of $ \triangle ABC, \triangle CDA.$ Since $ (N_1) \sim (O_2)$ with similarity center $ B$ and coefficient $ \frac{_1}{^2},$ midpoint $ G$ of $ BD$ is on $ (N_1).$ Likewise, $ G \in (N_2)$ $ \Longrightarrow$ $ MG$ is radical axis of $ (N_1) \cong (N_2).$

$ H_aH_c$ is radical axis of $ (M), (N_1),$ while $ K_aK_c$ is radical axis of $ (M), (N_2),$ intersecting at the radical center $ R \in MG$ of $ (M), (N_1), (N_2).$ Since $ \overline{H_aH_c} \equal{} \overline{K_aK_c}$ and $ (N_1) \cong (N_2),$ it follows that $ \overline{RH_a} \equal{} \overline{RK_a},$ $ \overline{RH_c} \equal{} \overline{RK_c}$ $ \Longrightarrow$ $ MGR$ is common perpendicular bisector of $ H_aK_a \parallel H_cK_c.$

$ AH_a, AK_a$ are isogonals WRT $ \angle DAB$ $ \Longrightarrow$ $ AF$ also bisects $ \angle K_aAH_a$ and likewise, $ CE$ also bisects $ \angle H_cCK_c.$ Then $ F$ is midpoint of the arc $ H_aK_a$ of $ (M)$ opposite $ A$ and $ E$ is midpoint of the arc $ H_cK_c$ of $ (M)$ opposite $ C$ $ \Longrightarrow$ $ EF \equiv MGR$ is also the perpendicular bisector of $ H_aK_a \parallel H_cK_c$ $ \Longrightarrow$ $ G \in EF.$

Fisrtly easy to see $ AECF$ is a rectangle. Then let $ AE,AF$ intersect $ O(ABD)$ at $ J,K$; $ CE,CF$ intersect $ O(CBD)$ at $ T,S$ .
Apparently we have $ JKTS$⊥$ AB,G$ is on $ JK$, $ G$ is the midpoint of $ BD$. And because of $ JG*KG \equal{} TG*SG \equal{} BG*DG$.
So $ GJ/GT \equal{} GS/GK$, and △$ EJT$∽△$ FSK$, so △$ GET$∽△$ GFK$, then $ E,G,F$ is on a line .

For another way：$ G$ is on nine-point circles .

$ \bullet$ We suppose that $ \angle B \equal{} \angle D$ and $ \angle A > \angle C$ and let $ A',\ C'$ be, the points such that $ ABA'D$ and $ BCDC'$ to be parallelograms and so, we have that $ A'CAC'$ is also a parallelogram, because of its diagonals bisect each other, at midpoint $ G$ of the segment $ BD.$

Because of $ \angle B \equal{} \angle D,$ based, on the below [b]Lemma 1[/b], we have that the line segments $ AF,\ CE,$ as the angle bisectors of $ \angle A,\ \angle C,$ are parallels and so, $ AECF$ is a rectangle.

It is easy to show that $ \angle AC'D \equal{} \angle A'CB$ $ ,(1)$ and based on the [b]Lemma 2[/b], we have that $ \angle AC'D \equal{} \angle ACD$ $ ,(2)$

From $ (1),\ (2)$ $ \angle A'CB \equal{} \angle ACD$ $ ,(3)$

From $ (3)$ we conclude that the line segment $ CE,$ as the angle bisector of $ \angle C,$ is also the angle bisector of $ \angle A'CA.$

$ \bullet$ So, we have $ \angle A'CE \equal{} \angle ECS \equal{} 90^{o} \minus{} \angle SCF,$ where $ S\equiv AC\cap EF$ $ ,(4)$

From $ (3),\ (4)$ $ \Longrightarrow$ $ \angle A'CS \equal{} 180^{o} \minus{} 2\angle SCF \equal{} \angle CSF$ $ \Longrightarrow$ $ EF\parallel A'C\parallel AC$ $ ,(5)$

The line through the point $ A'$ and parallel to $ CF,$ intersects $ EF$ at the point so be it $ Z$ and then we have $ A'Z\parallel \equal{} CF\parallel \equal{} AE$ $ ,(6)$

From $ (6)$ and because of $ GA \equal{} GA',$ we conclude that the line segment $ EZ\equiv EF$ passes through the midpoint $ G$ of $ AA',\ BD$ and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA 1. - In every quadrilateral whose two opposite angles are equal, the angle bisectors of its other two angles, are parallels.[/color][/size][/b]

[b][size=100][color=DarkBlue]LEMMA 2. - A quadrilateral $ ABCD$ is given with $ \angle B \equal{} \angle D$ and $ \angle A > \angle C$ and let $ C'$ be, the point such that $ BCDC'$ to be a parallelogram. Prove that $ \angle AC'D \equal{} \angle ACD$.[/color][/size][/b]

Kostas Vittas.

PS. I will post here later, the easy proofs of the above [b]Lemmata[/b] I have in mind.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA 1. - In every quadrilateral whose two opposite angles are equal, the angle bisectors of its other two angles, are parallels.[/color][/size][/b][/quote]
[b]PROOF OF THE LEMMA 1[/b]. - I post only the schema because i think it is enough. This simple elementary solution has been presented  [url=http://www.mathematica.gr/forum/viewtopic.php?f=69&t=4763][b][color=red]Here[/color][/b][/url], by [b]Stavroula Tzimoulia[/b], a youngish girl student from [b]Thessaloniki - Hellas[/b].

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA 2. - A quadrilateral $ ABCD$ is given with $ \angle B \equal{} \angle D$ and $ \angle A > \angle C$ and let $ C'$ be, the point such that $ BCDC'$ to be a parallelogram. Prove that $ \angle AC'D \equal{} \angle ACD$.[/color][/size][/b][/quote]
[b]PROOF OF THE LEMMA 2.[/b] - Because of $ \angle B \equal{} \angle D,$ we have that the circucircles $ (O),\ (O'),$ of the triangles $ \bigtriangleup ABC,\ \bigtriangleup ADC,$ are congruent.

The line through $ A$ and parallel to $ CD,$ intersects $ (O),$ at point so be it $ E$ and it is easy to show that $ ADCE$ is a parallelogram.

( from $ \angle EAC \equal{} \angle ACD$ and $ \angle CEA \equal{}  \angle CDA$ $ \Longrightarrow$ $ \angle ECA \equal{} \angle CAD$ $ \Longrightarrow$ $ CE\parallel AD$ ).

So, we have that $ CE\parallel \equal{} AD$ and because of $ CB\parallel \equal{} DC',$ we conclude that the triangles $ \bigtriangleup EBC, \ \bigtriangleup AC'D$ are congruent and then, we have $ \angle AC'D \equal{} \angle EBC \equal{} \angle EAC \equal{} \angle ACD$.

Hence, we conclude that $ \angle AC'D \equal{} \angle ACD$ and the proof of the [b]Lemma 2[/b] is completed.

Kostas Vittas.


Given points $A, B, C, D, E$, $AB$ parallel to $CD$ construct a point $F$ different from $A, B, C, D, E$ such that points $A, B, C, D, E, F$ lie on a conic.

Of course point $F$ is different from points $A,B,C,D,E$!!!

Cvix knows this problem from me, because I invented it. I will wait for some other solutions before I post mine.

My solution:
Denote the midpoints of $AB$ and $CD$ by $M$ and $N$, respectively. Let $l$ be the line paralell to $AB$ through point $E$, $X$ be the intersection of $l$ and $MN$, $F$ the point on line $EX$, such that $XF=XE$ and $F$ is different from $E$.
Obviously, points $A, B, C, D, E, F$ lie on one conic and we are done  :D

I have just wrote that solution by cvix is good, but wait a moment...
what if $M,N,E$ are colinear? In the solution we have to remember about particular cases.

[quote="derezins"]In the solution we have to remember about particular cases.[/quote]

So what to do in your problem if $A=C$?  :P

In my problem $A,B,C,D,E$ are different points :!:


If 2 right lines  are parallell with each other and  another right line cut one of this parallell right line prove that It cut the other parallell right line to.  (right line I mean withaut ending)
Maybe this is more clear:
$ d_1//d_2$ and $ {d_1}\cap{d_3}$ prove that $ {d_2}\cap{d_3}$

Assume that $ d_1 \cap d_3 \equal{} \emptyset$, thus $ d_1 // d_3$. But we have $ d_1 // d_2$, so $ d_2 // d_3$ (it's a theorem and easy to prove). So our assumption is  wrong and $ d_1 \cap d_3 \ne \emptyset$.


[quote="USAMTS Solution Submission"]please email both the typeset document (meaning your PDF or DVI document) and the original source document (meaning all your LaTeX code).
[/quote]

I have .jpg images incorporated into my LaTeX document (using \includegraphics).  When I email it in, should I attach these .jpg files in addition to the typset document and the source?


CEMC
number theory
Pythagorean Theorem
geometry

On the website where it has the instructions for taking the CMOR, it says you can either handwrite solutions or type them. I am planning on hand-writing my solutions, and on the website it has the address to send it to. My question is do the solutions need to be [i]postmarked[/i] by the 11th or [i]received[/i] by the 11th. If its the latter, that would be kind of stupid.

If I could have an official CEMC person tell me the answer, or someone with previous experience, that would be great.

From experience, everything that is due on some day by mail has to be postmarked by that date. Basically they need to make sure that you don't get extra time, and if you get a postmark on the 11th, that means you finished and dropped them off on the 11th. Under normal circumstances.

Anyhow, I wish you luck on the CMOR!

:rotfl: I love your signature.

And thanks. Gl to you too!

I'm not doing CMOR, thankfully, though I did it the last two years. I'm off to the winter math camp soon (in fact, I'm about to leave right now).

And thanks for the signature complement. (:

What is Winter Math Camp? What do you have to do to be invited? Thanks.

Its a several day camp for IMO potentials. The best way to make it is to do well on the Olympiads. I think you can also make it by getting a really good COMC score but this year even 78 wasn't good enough to get invited. I think doing Olymon helps but it didn't help me enough this year because I didn't make it  :(

Hey for the CMOR, are we allowed to use well known theorems  , or do they count as an "outside source".

You can always use well known theorems. Otherwise problems would take too long (e.g. proving Fermat's Little Theorem in a solution would take an unnecessarily long time).

Thanks Bacteria. I really hope to make that camp next year. :)

No you may not use any known theorems. Everything you write must be derived from first principles. Lol.

Cheers,

Hunter

Thats completely pointless though. If one were to use an extremely well known theorem, whats the point of having to prove it? Theres a reason its called a [i]theorem[/i]...its already bee proven. Hmmm i use the Pythagorean theorem a bit...do I have to prove it? If i did, that would be stupid.

Are you sure about that, rofler? Because if thats true, i will be pretty pissed at the CME people...

hm well its pretty difficult to pick up on sarcasm when its written :P

Rofle, you werent serious?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let G be a finite group, N a normal subgroup of G and P a p-Sylow subgroup of G, where p is a prime dividing |G|.

Prove that:
1) The intersection of N with P is a p-Sylow subgroup of N.
2) $ PN/N$ is a p-Sylow subgroup of $ G/N$.
3) $ N_{G}(P)N/N \equal{} N_{G/N}(PN/N)$.

I proved 1 and 2 and I proved the containment $ \subseteq$ in 3,  but I'm having difficulties proving the converse containment. Can anyone help?

You may want to use the frattini's argument: $ G \equal{} NN_{G}(P)$.


algebra
polynomial
calculus
integration
number theory
greatest common divisor
Gauss

Let P(x) is polynomial with coeffient integers and Q(x) is polynomial with coeffient rations.and P(x)=Q(x)^2 .provethat Q(x) has coeffient integers.

Assume the contrary,hence: $Q(x)= \frac{  \sum^n_0{a_k}x^k} {M}$ where
 $a_o,...,a_n,M$ are natural numbers ,$gcd(a_0,...,a_n)=1$,$M>1$.
 Denote  $F(x)= \sum^n_0{a_k}x^k$ then $F(x)$ is elementary ( that is,integral polynomial with $gcd$ of its coefficients is 1)
 By [b]Gauss[/b] lemma we have $F(x)^2$ is also elemetary.
 But $Q(x)^2=\frac {F(x)^2} {M^2}$ and $M>1$ so we cannot have $Q(x)^2$ is integral polynomial.


$ a,b,c>0$ and $ abc\geq1$   prove:
$ \frac{a}{\sqrt{b\plus{}\sqrt{ac}}}\plus{}\frac{b}{\sqrt{c\plus{}\sqrt{ab}}}\plus{}\frac{c}{\sqrt{a\plus{}\sqrt{bc}}}\geq\frac{3}{\sqrt{2}}$

[quote="zaya_yc"]$ a,b,c > 0$ and $ abc\geq1$   prove:
$ \frac {a}{\sqrt {b \plus{} \sqrt {ac}}} \plus{} \frac {b}{\sqrt {c \plus{} \sqrt {ab}}} \plus{} \frac {c}{\sqrt {a \plus{} \sqrt {bc}}}\geq\frac {3}{\sqrt {2}}$[/quote]
$ \left(\sum_{cyc}\frac {a}{\sqrt {b \plus{} \sqrt {ac}}}\right)^2\sum_{cyc}a(b\plus{}\sqrt{ac})\geq(a\plus{}b\plus{}c)^3$ by Holder.
Id est, it remains to prove that
$ 2(a\plus{}b\plus{}c)^3\geq9\sum_{cyc}(ab\plus{}a\sqrt{ac})\sqrt[3]{abc}.$
But $ 2(a\plus{}b\plus{}c)^3\geq9\sum_{cyc}(ab\plus{}a\sqrt{ac})\sqrt[3]{abc}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^3\plus{}6a^2b\plus{}6a^2c\plus{}4abc\minus{}9a^{\frac{11}{6}}b^{\frac{1}{3}}c^{\frac{5}{6}}\minus{}9a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}})\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{3a^3}{2}\plus{}\frac{c^3}{2}\plus{}c^2a\plus{}2a^2b\plus{}3a^2c\plus{}abc\minus{}9a^{\frac{11}{6}}b^{\frac{1}{3}}c^{\frac{5}{6}}\right)\plus{}$
$ \plus{}\sum_{cyc}\left(3a^2b\plus{}3b^2a\plus{}3abc\minus{}9a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}\right)\geq0,$ which true by AM-GM.

[quote="zaya_yc"]$ a,b,c > 0$ and $ abc\geq1$   prove:
$ \frac {a}{\sqrt {b \plus{} \sqrt {ac}}} \plus{} \frac {b}{\sqrt {c \plus{} \sqrt {ab}}} \plus{} \frac {c}{\sqrt {a \plus{} \sqrt {bc}}}\geq\frac {3}{\sqrt {2}}$[/quote]
This is problem T4/352 in [b]Mathematics and Youth Magazine[/b].

[quote="arqady"][quote="zaya_yc"]$ a,b,c > 0$ and $ abc\geq1$   prove:
$ \frac {a}{\sqrt {b \plus{} \sqrt {ac}}} \plus{} \frac {b}{\sqrt {c \plus{} \sqrt {ab}}} \plus{} \frac {c}{\sqrt {a \plus{} \sqrt {bc}}}\geq\frac {3}{\sqrt {2}}$[/quote]
$ \left(\sum_{cyc}\frac {a}{\sqrt {b \plus{} \sqrt {ac}}}\right)^2\sum_{cyc}a(b \plus{} \sqrt {ac})\geq(a \plus{} b \plus{} c)^3$ by Holder.
Id est, it remains to prove that
$ 2(a \plus{} b \plus{} c)^3\geq9\sum_{cyc}(ab \plus{} a\sqrt {ac})\sqrt [3]{abc}.$
But $ 2(a \plus{} b \plus{} c)^3\geq9\sum_{cyc}(ab \plus{} a\sqrt {ac})\sqrt [3]{abc}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^3 \plus{} 6a^2b \plus{} 6a^2c \plus{} 4abc \minus{} 9a^{\frac {11}{6}}b^{\frac {1}{3}}c^{\frac {5}{6}} \minus{} 9a^{\frac {4}{3}}b^{\frac {4}{3}}c^{\frac {1}{3}})\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {3a^3}{2} \plus{} \frac {c^3}{2} \plus{} c^2a \plus{} 2a^2b \plus{} 3a^2c \plus{} abc \minus{} 9a^{\frac {11}{6}}b^{\frac {1}{3}}c^{\frac {5}{6}}\right) \plus{}$
$ \plus{} \sum_{cyc}\left(3a^2b \plus{} 3b^2a \plus{} 3abc \minus{} 9a^{\frac {4}{3}}b^{\frac {4}{3}}c^{\frac {1}{3}}\right)\geq0,$ which true by AM-GM.[/quote]

The right solution,but too complex!
Using CBS and AM-GM we have
$ LHS\ge \frac{(\sum\sqrt a)^2}{\sum\sqrt{b\plus{}\sqrt{ac}}}\ge \frac{2\sqrt2(\sum\sqrt a)^2}{\sum(2\plus{}b\plus{}\sqrt{ac})}$
We will prove
$ \frac{2\sqrt2(\sum\sqrt a)^2}{\sum(2\plus{}b\plus{}\sqrt{ac})} \ge \frac3{\sqrt2}$
<->$ a\plus{}b\plus{}c\plus{}5(\sqrt{ab}\plus{} \sqrt{bc}\plus{} \sqrt{ca})\ge 18$,true by AM-GM  :wink:


To anybody who took the AMC 8:
What do you think the hardest problem was?

Although I didn't take the AMC-8, statistically speaking problem 24 had the least percent correctness and problem 25 had the highest percent omissions.

I thought, having taken the test as a coach for experimentation after the official testing period, that some of the problems just after number 10 were harder than their placement in the test order would lead one to expect. Sure enough, problems pretty early on had some LOW percentages of correct answers in the general statistics.

The last geometry one was... interesting.

I personally thought the one with the two people taking turns walking a twelve calibrated wheel was hardest, i got that wrong..


4.) A hardware supplier manufactures 3 kinds of clamps; types A, B, and C. Production restrictions force it to make 20 more type C clamps than the total of the other types and twice as many as type B clamps as type A clamps. The shop must produce 260 clamps per day. How many of each type are made per day?

How many type A clamps are produced?


J.P vists his favorite local pub. While coming back from the restroom, he sees a stranger take his wallet from his jacket, which he left draped over his chair. J.P watches the thief spend the money. Can you explain this passive response?

[hide="Hint"]
The pub was known to have a shady reputation.
[/hide]

[hide]He is a counterfeiter, and he wants someone else to spend his fake money to see how much security there is in the shop.[/hide]

I have your book!!!! I saw that problem WORD FOR WORD!!!

it's a good book...barnes and nobles i presume?
anyone who attempts to solve this (without the book!) may verify with mathnerd's answer

Hmmm I thought of that but it seemed pretty weird.  How [hide]could he expect a thief to do that, and what about all his other ID junk in the wallet[/hide]?

[hide]
he left his wallet in plain view and expected it to be stolen, and as i specified in the hint, the pub had a shady reputation. i'm sure he took out anything of importance  :roll:
[/hide]

Another possible senario:

[hide] J.P is a police officer who is delibrately letting crooks steal and spend his money so he can arrest them.[/hide]

Hmm I don't think the police officer would have a pub with a shady reputation as his favourite. Unless he's corrupted.

Perhaps he likes the pub because he can arrest lots of people there?

but if JP is a police officer, why does he wait for the thief to spend his money before hypothetically arresting him? also, after making a number of arrests at the pub, the shady pub's manager will realize he is losing business...
:stretcher:

JP's safety and career lies in thievery


I am wondering how long will this continue. Man, you've understood the question, it is not necessary to post it 23 times!


geometry
symmetry
geometric transformation
reflection
trigonometry
angle bisector
trig identities

Given isosceles triangle ABC with $ AB \equal{} AC, \angle A \equal{} 80^{\circ},$ point O in the triangle such that $ \angle CBO \equal{} 10^{\circ}$ and $ \angle BCO \equal{} 30^{\circ},$ find the measure of $ \angle AOB$.

Edit: Oops. Sorry.

[quote="resurrection"]... such that $ \angle BOC \equal{} 10^{\circ}\angle$ and $ COB \equal{} 30^{\circ}$...[/quote]

this doesnt sound right?

[hide]Point O must lie on the angle bisector of $ \angle B$. The angle bisector to the base of an isoceles triangle creates a line of symmetry in the triangle. Thus $ \angle AOB \equal{} \angle COB \equal{} 180 \minus{} 10 \minus{} 30 \equal{} 140$[/hide]

$ A$ is the vertex? and $ O$ still doesn't lie on any angle bisector  :wink:

[hide]
[asy]size(250); pathpen = linewidth(0.7); real big = 5; pen dash = linetype("4 4") + linewidth(0.7), sm = fontsize(8);
pair extension(pair P, real r){return P+big*expi(r*pi/180);}
pair B = MP("B",D((0,0))), C = MP("C",D((2,0))), A = MP("A",D( expi(50 *pi/180)/cos(50 *pi/180) ),N), O = MP("O",D( IP(B--extension(B,10),C--extension(C,150)) ), NE), O_r = MP("O'", D( (C.x - O.x, O.y) ), NW); D(A--B--C--cycle); D(A--O--B--O--C); D(A--O_r--B,dash);D(C--O_r--O,dash);D(anglemark(C,B,O)); MP("10^{\circ}",B,(14,1.2),sm); D(anglemark(O,B,O_r,6)); MP("20^{\circ}",B,(10,3.5),sm);[/asy]
Let $ O'$ be the reflection of $ O$ across the perpendicular from $ A$ to $ \overline{BC}$. Then $ \overline{OO'} \parallel \overline{BC}$, so $ \angle BOO' \equal{} \angle OBC \equal{} 10^{\circ}$, and $ \angle BOC \equal{} 180 \minus{} 10 \minus{} 30 \equal{} 140^{\circ}$. Then, by the Law of Sines in $ \triangle BCO', \triangle ABC$, we have
\[ \frac {CO'}{BC} \equal{} \frac {\sin 30^{\circ}}{\sin 140^{\circ}}, \qquad \frac {\sin 50^{\circ}}{\sin 80^{\circ}} \equal{} \frac {AC}{BC}
\] But $ \sin 30^{\circ} \sin 80^{\circ} \equal{} \frac {1}{2} \sin 100^{\circ} \equal{} \sin 50^{\circ} \cos 50^{\circ} \equal{} \sin 50^{\circ} \sin 140^{\circ} \Longleftrightarrow AC \equal{} CO'$, and $ \triangle ACO \cong \triangle O'CO \Longrightarrow \angle AOB \equal{} \angle BAO \equal{} \angle BAC \minus{} \angle CO'O \equal{} \boxed{70^{\circ}}$.
[/hide]

Correct azjps but there is a purely geometric solution

[hide="Solution"]Extend $ CO$ to point $ P$ such that $ AP\equal{}AB\equal{}AC$. Notice then that $ \angle PAB\equal{}60^{\circ}$ and since $ AP\equal{}AB, \angle ABP\equal{}60^{\circ}$.

Now, since $ \angle ACO \equal{} 20^{\circ}, \angle APC\equal{}20^{\circ},$ and $ \angle OPB\equal{}40^{\circ}$. Furthermore, $ \angle PBO\equal{}60^{\circ}\plus{}40^{\circ}$ which gives us that $ \angle POB \equal{} 40^{\circ}$ and so, triangle $ POB$ isosceles, which means that $ PB\equal{}BO\equal{}AB\equal{}AC$. Since $ BO\equal{}AB$, triangle $ ABO$ is isosceles and the desired angle is $ \frac{1}{2}*(180^{\circ}\minus{}40^{\circ})\equal{}\boxed{70^{\circ}}$[/hide]


Let  a,b,c be a triangle. Prove that
$ (\frac{a}{\sqrt{a^2\plus{}8bc}})^k\plus{}(\frac{b}{\sqrt{b^2\plus{}8ca}})^k\plus{}(\frac{c}{\sqrt{c^2\plus{}8ab}})^k \geq \frac{3}{3^k}$
with $ k\equal{} \frac{54}{64\minus{}0,1^{100}}$

I think your inequality is not true for your $ k$ and $ a = 3; b = 4; c = 5$. Left side less then right  :wink:. But for $ k\geq 1$ it is true:
$ \sum_{cyc}{(\frac {a}{\sqrt {a^2 + 8bc}}})^k\geq \sum_{cyc}{(\frac {a^{\frac {4}{3}}}{a^{\frac {4}{3}} + b^{\frac {4}{3}} + c^{\frac {4}{3}}})^k} = \frac {a^{\frac {4k}{3}} + b^{\frac {4k}{3}} + c^{\frac {4k}{3}}}{(a^{\frac {4}{3}} + b^{\frac {4}{3}} + c^{\frac {4}{3}})^k}\geq \frac {3}{3^k}$
Last inequality could be written as:
$ \frac {a^{\frac {4k}{3}} + b^{\frac {4k}{3}} + c^{\frac {4k}{3}}}{3}\geq ( \frac {a^{\frac {4}{3}} + b^{\frac {4}{3}} + c^{\frac {4}{3}}}{3})^k$
Let $ a^{\frac {4}{3}} = x; b^{\frac {4}{3}} = y; c^{\frac {4}{3}} = z$ and we will have:
$ \frac {x^{k} + y^k + z^k}{3}\geq ( \frac {x + y + z}{3})^k$
This is Jensen's inequality for $ f(x) = x^k$, which is true for $ k\geq 1$.

Maybe you wanted to write: $ k = \frac {64}{64 - 0.1^{100}}$ instead of $ k = \frac {54}{64 - 0.1^{100}}$  :juggle:


Positive real numbers $x_{1},x_{2},\cdot\cdot x_{k}$ are such that $x_{1}^{2}+x_{2}^{2}+\cdot\cdot+x_{k}^{2}<\frac{x_{1}+x_{2}+\cdot\cdot+x_{k}}{2}$, and $x_{1}+x_{2}+\cdot\cdot+x_{k}<\frac{x_{1}^{3}+x_{2}^{3}+\cdot\cdot+x_{k}^{3}}{2}$.
a)Prove that $k>50$.
b)For some $k$ find such numbers $x_{1},x_{2}\cdot\cdot x_{k}$.
c)Find the minimal number $k$, for which exist such numbers.

* Just a remark: It is problem no.7 (10-11 grades) from the 28th Tournament of Towns.

i think year 2005? Do you have the solution silouan?


calculus
integration
LaTeX
algebra
difference of squares
special factorizations

Find three positive integral values for x that make $ 4^x \plus{} 4^{51} \plus{} 4^{54}$ a perfect square.

[hide="hint"]
divide the expression by $ 4^{51}$, a perfect square
the remaining expression, $ 65\plus{}4^{x\minus{}51}$ should be the square of a rational number

then use difference of squares to solve for the 3 x
[/hide]

[hide]$ 4^x \plus{} 4^{51} \plus{} 4^{54} \equal{} 2^{2x} \plus{} 2^{102} \plus{} 26^{108}$.
$ (2^a \plus{} 2^b)^2 \equal{} 2^{2a} \plus{} 2^{a \plus{} b \plus{} 1} \plus{} 2^{2b}$

Matching exponents,

$ 2a \equal{} 2x \Longrightarrow a \equal{} x$

$ 2b \equal{} 108 \Longrightarrow b \equal{} 54$

$ x \plus{} 54 \plus{} 1 \equal{} 102 \Longrightarrow x \equal{} 47$

$ 2a \equal{} 102 \Longrightarrow a \equal{} 51$

$ 2b \equal{} 108 \Longrightarrow b \equal{} 54$

$ 2x \equal{} 51 \plus{} 54 \plus{} 1 \Longrightarrow x \equal{} 53$

$ 2a \equal{} 102 \Longrightarrow a \equal{} 51$

$ 2b \equal{} 2x \Longrightarrow b \equal{} x$

$ x \plus{} 51 \plus{} 1 \equal{} 108\Longrightarrow x \equal{} 56$

So $ x \equal{} \{47,53,56\}$.[/hide]


Hey,

[b]1. The problem statement, all variables and given/known data[/b]
Solve the following for $ {r}$ where: $ {h}$ and $ {S}$ are constants,
$ {{{4}{{\pi}^{2}}{{r}^{4}}} \plus{} {{4}{\pi}{h}{S}{r}} \minus{} {{S}^{2}}} \equal{} {0}$

[b]2. Relevant equations[/b]
Knowledge of Solving Methods for Specific Quartics.

[b]3. The attempt at a solution[/b]
[hide="My Attempt at a Solution,"]
I recall that there was a general equation for solving a Quartic. However, I also remember that for specific quartics there were certain methods that could be applied to solve them efficiently. Unfortunately, I do not recall a method I know of to solve this quartic other than just using Maple. Any ideas?
[/hide]
Thanks,

-APSStudent

[quote="APSStudent"]Hey,

[b]1. The problem statement, all variables and given/known data[/b]
Solve the following for $ {r}$ where: $ {h}$ and $ {S}$ are constants,
$ {{{4}{{\pi}^{2}}{{r}^{4}}} \plus{} {{4}{\pi}{h}{S}{r}} \minus{} {{S}^{2}}} \equal{} {0}$

[b]2. Relevant equations[/b]
Knowledge of Solving Methods for Specific Quartics.

[b]3. The attempt at a solution[/b]
I recall that there was a general equation for solving a Quartic. However, I also remember that for specific quartics there were certain methods that could be applied to solve them efficiently. Unfortunately, I do not recall a method I know of to solve this quartic other than just using Maple. Any ideas?

APSStudent[/quote]

Dear [b]APSStudent[/b],

yes, you are right, roots of [i]some[/i] quartics can be expressed easier, then in general case. This happens when the [url=http://en.wikipedia.org/wiki/Galois_group]Galois group[/url] of the quartic is [b]not[/b] [url=http://en.wikipedia.org/wiki/Symmetric_group]S(4)[/url] (which happens in the general case, when the group contains 24 elements and is generated by permutations (1 2), (1 3) and (1 4)).

Let us substitute $ S\equal{}2\pi^2$ and $ h\equal{}\pi/2$ into the $ {{{4}{{\pi}^{2}}{{r}^{4}}} \plus{} {{4}{\pi}{h}{S}{r}} \minus{} {{S}^{2}}}\equal{}0$. In this case the equation will become:

\[ r^4\plus{}r\minus{}1\equal{}0\]

But the Galois group of the polynomial $ r^4\plus{}r\minus{}1$ is exactly $ S(4)$, so for $ S\equal{}2\pi^2$ and $ h\equal{}\pi/2$ nothing better then general quartic formula can be found for the roots.

Of course, we can find another $ S$ and $ h$, so that the polynomial will have another Galois group and will have better roots. For instance, let us substitute $ S\equal{}2\,\pi^2$ and $ h\equal{}2\,\pi$. Then we will have to solve the equation

\[ r^4\plus{}4\,r\minus{}1 \equal{} 0\]

Polynomial $ r^4\plus{}4\,r\minus{}1$ has Galois group $ D(4)$, e.g. the group contains 8 elements and can be generated using (1 3) and (1 2 3 4). This polynomial can be written as:

\[ r^4\plus{}4\,r\minus{}1 \equal{} \left(r^2\plus{}1\right)^2\minus{}2\,(r\minus{}1)^2 \equal{} (r^2\plus{}\sqrt{2}\,r\plus{}1\minus{}\sqrt{2})\,(r^2\minus{}\sqrt{2}\,r\plus{}1\plus{}\sqrt{2})\]

So it has 2 real and 2 compex roots:

\[ \frac{\pm\sqrt{4\sqrt{2}\minus{}2}\minus{}\sqrt{2}}{2} \qquad\mbox{and}\qquad \frac{\pm I \sqrt{4\sqrt{2}\plus{}2}\plus{}\sqrt{2}}{2}\]
Hope this will help.
Sergey Markelov
markelov@mccme.ru

Hey,

Thanks for the replies t0rajir0u and Markelov.

[quote="t0rajir0u"]
In this case there is no method especially more efficient than the [url=http://planetmath.org/encyclopedia/QuarticFormula.html]quartic formula[/url].  What's the motivation for this problem?  It looks like $ S$ is a surface area.
[/quote]

Hmm, I was afraid that I would have to resort to the quartic formula, in anycase the problem arises from my approach to a problem from the Putnam Exam, 1938-A2. Below is the link, the fifth post is where I explain how I arrived at this quartic equation. Thanks.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204960

[color=green]You don't really need to quote in its entirety a long post that comes before yours. -mod[/color]

Wow, thanks for the reply. I'll try that method myself and compare the results with the quartic formula. Hopefully, I make some progress on this problem.

Thanks,

-APSStudent


is there any prime number of the form
\[n+2^{n+2^{n}}\]
i presume all of them are composite but i couldn't prove it

$n = 0$ produces the trivial prime $2$.

Experimentally it seems $2n+1 | n+2^{n+2^{n}}$ for $n$ odd (clearly $n$ even cannot produce a prime), but I can't prove it.

[quote="t0rajir0u"]Experimentally it seems $2n+1 | n+2^{n+2^{n}}$ for $n$ odd.[/quote]

Il seems that, for $n=5$, $n+2^{n+2^{n}}=5+2^{37}=1$ (mod 11) and so $2n+1 | n+2^{n+2^{n}}$ is wrong for $n=5$

--
Patrick

n is a positive integer, $n\ge 1$

assume that the number is prime.
mod 3 gives $n\neq 6k+1$
i don't know what about $6k+3$ and $6k+5$

When n=23, the resulting 2.5 million digit number is not divisible by any of the first 10^8 primes.  So a different approach might be necessary.


[quote="GHAVR"][quote="The Master"]\[ \begin{array}{l} {P_{1} ,{\rm \; }P_{2} ,{\rm \; }P_{3} ,{\rm \; ...,\; }P_{12} ,{\rm { are\; consecutive\; vertices\; of\; a\; regular\; polygon\; with\; 12\; sides.\; If }}}\\{{\;}\overrightarrow{P_{1} P _{2}} =\overrightarrow{x} {\rm \; and}} {\overrightarrow{P_{1} P_{3} } \; =\; \overrightarrow{y} {\rm ,\; express\; the\; following\; vectors\; in\; terms\; of\; }\overrightarrow{x} {\rm \; and\; }\overrightarrow{y} {\rm : }} \\ {} \\ {{\rm a.\; \; }\overrightarrow{P_{1} P_{4} } } \\ {{\rm b.\; \; }\overrightarrow{P_{3} P_{7} } } \end{array}  \]
How do you do this? Do you have to plot points and then express the required vectors as a linear combination of vectors x and y? But if you generate points, you'd need to use trig ratios, right? Then you wouldn't get exact values......

There's gotta be an easy way of doin this. I don't think it's supposed to be a difficult question. I'm not thinkin straight today..........[/quote]

I'm sorry but I have no skills in Tex :blush:
So I'll just say the answer:

the first:  (1+sqrt(3))(x-y)
the second:  (3+2*sqrt(3))(y-2x)

(I'm very very sorry  :?  :oops: )[/quote]

$(1+\sqrt{3})(x-y)$ and $(3+2\sqrt{3})(y-2x)$?
Just put the thingy under the square root as {3} instead of (3) and dollar signs around the equation and..yeah.

well i'm no good at vectors, so can somebody explain these answers...?

Now I'll try...

\[ \begin{array}{{{{{c}\ \overrightarrow{P_{1} P_{4} }} {\rm \; } \parallel {\rm \; } \overrightarrow{P_{2} P _{3}} {\rm \; } \Rightarrow \overrightarrow{P_{1} P_{4} }} =k\overrightarrow{P_{2} P_{3} }} {\rm \;} {\rm { for\; some\; real\; k\;} {\rm ;\;} \ {\rm Let\; A\; be\; such\; point\; }{\rm : \;} {\rm \;} \overrightarrow{P_{2} A }} = \overrightarrow{P_{3} P_{4} }} {\rm \;} \Rightarrow {\rm \;} \overrightarrow{A P_{4} }} = \overrightarrow{P_{2} P_{3} }} {\rm ;\;} \ \angle P_{1}P_{2}A = 120^\circ {\rm \;} \Rightarrow {\rm \;} P_{1}A = \sqrt{3}P_{1}P_{2}= \sqrt{3}P_{2}P_{3} {\rm \;} {\rm {and\; \;} \overrigharrow{P_{1}A} \parallel \overrigharrow{P_{2}P_{3}} {\rm { \; so\; } \ \overrightarrow{P_{1}P_{4}} = \overrightarrow{P_{1}A} + \overrightarrow{AP_{4}} = \sqrt{3}\overrightarrow{P_{2}P_{3}} +\overrightarrow{P_{2}P_{3}} = (1+\sqrt{3}) \overrightarrow{P_{2}P_{3}} = (1+\sqrt{3})(\overrightarrow{y} - \overrightarrow{x}) \ \ P_{1}P_{7}{\rm \; is\; the\; diameter\; so\;} \angle P_{1}P_{3}P_{7} = \angle P_{1}P_{4}P_{7} = 90^\circ \ {\rm Let\; C\; be\; such \; point\; : \;} \overrightarrow{P_{3}C} = \overrightarrow{P_{5}P_{6}} {\rm \; then\; \;} \overrightarrow{P_{2}P_{3}} \parallel \overrightarrow{P_{1}P_{4}} \perp \overrightarrow{P_{4}P_{7}} \parallel \overrightarrow{P_{5}P_{6}} \parallel \overrightarrow{P_{3}C} \Rightarrow \angle P_{2}P_{3}C = 90^\circ \ \angle CP_{3}P_{4} = 60^\circ {\rm \; and\; \;} CP_{3} = P_{3}P_{4} {\rm \; so\; \;} CP_{3}P_{4} {\rm \; is\; a\; regular\; triangle\;} \ {\rm so\; we\; could\; see\;} \overrightarrow{P_{1}P_{2}} = \overrightarrow{CP_{4}} \Rightarrow \overrightarrow{P_{2}C} = (-\overrightarrow{x}+(1+\sqrt{3})( \overrightarrow{y}-\overrightarrow{x}) - \overrightarrow{x}) \ \overrightarrow{P_{1}P_{7}} = \overrightarrow{P_{1}P_{4}} + \overrightarrow{P_{4}P_{7}} = (1+\sqrt{3})(\overrightarrow{P_{2}P_{3}}+\overrightarrow{P_{5}P_{6}}) = (1+\sqrt{3})\overrightarrow{P_{2}C} \end{array}  \]
\[ \begin{array}{c}\ \overrightarrow{P_{3}P_{7}} =\overrightarrow{P_{1}P_{7}} - \overrightarrow{P_{1}P_{3}} = (3+2\sqrt{3})(\overrightarrow{y} - 2\overrightarrow{x}) \end{array}  \]

I think so...  :huh:   :)

lol i worked on it again this morning n i JUS got it......

it jus clicked all of a sudden.... but thanks a lot, now i kno my answer is right...
i jus did the 1st part tho, not the 2nd part.... a, not b...

ya, (1 + square root three)(y - x) is what i got as well.. good good......
it's a nice question, i like it.......


A printer  has to numberthe pages of a book from 1 to 150. Suppose the printer uses a seperate piece of type for each digit in each number. How many pieces of type will the printer have to use?

From 1-9, there are 9 one-digit numbers. From 10-99, there are 99-10+1=90 two-digit numbers. From 100-150, there are 150-100+1=51 three-digit numbers. So there are $9+2(90)+3(51)=\boxed{342}$ pieces of type.

I think it is
[hide] pages $1-9\rightarrow9$
$10-99\rightarrow180$
$100-150\rightarrow153$
So the total would be $\boxed{342}$[/hide] :|

I used the same logic as VenDiddy and Knuth's_check and got 342.

Nice,I also get 342. :D

Math is my life :rotfl:


conics parabola quadratics geometry perimeter inequalities

A farmer has 60 meters of fence with which to build a rectangular animal run (as shown) for her cows, horses, and pigs. She wants each type of animal to have the same area. What is the largest number of square meters which can be enclosed? Express your answer as a decimal.

[img]7204[/img]

I did something wrong because I didn't get an answer as a decimal, so I'm posting my solution so someone can tell me where I went wrong.

[hide]Let's call the side length $y$ and the top length $x$.

We're trying to maximize the value of $(6x)(4y) = 24xy$, so it's
Because the total fencing is $60$, $6x+4y = 60$.

Solving for $x$, we get \[x = \frac{60-4y}{6}\]
sufficient to find the maximum of $xy$. Plugging in the value for $x$ above, we get \[xy = \frac{60y-4y^{2}}{6}=-\frac{2y^{2}}{3}+10y\].

Because this is a parabola opening downward, we need to find the vertex. The vertex is given by $\frac{-b}{2a}$. Plugging in the values for our above quadratic, we get that \[y = \frac{-10}{-\frac{4}{3}}= \frac{15}{2}\].

Plugging this value back into $6x+4y = 60$, we get \[x = 5\].

Now plugging both of these into $24xy$, what we're looking for, we get \[24xy = 24 \times 5 \times \frac{15}{2}= 900\].
[\hide]

Isn't the sum of the side lengths $8x$ and not $6x$? I don't see how that will make a lot of a difference though...

@ Ubemaya - If $x$ is the length of $\frac13$ of the whole top and $y$ is the length of the side, the area enclosed is $3xy$. The perimeter is $6x+4y$.

Ubemaya, the area is $3xy$ so the answer would be $112.5$ with your method.

Yeah, sonny's right... I screwed up.

[hide="AM-GM solution"]
By the AM-GM Inequality, we have: \[\frac{6x+4y}{2}\geq  \sqrt{6x \times 4y}\]

Since $6x+4y = 60$, we have \[30 \geq  \sqrt{24xy}\]\[900 \geq  24xy\]\[112.5 \geq  3xy\]

Therefore, the maximum value the area of the farm can attain is when $6x = 4y$ and is equal to $\boxed{112.5}$.[/hide]

This has been posted before...
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111893

Its not the same question. They're different numbers


projective geometry
geometry
cyclic quadrilateral
geometry proposed

A circle $ (O)$ is given, with center $ O$ and diameter $ AB.$ We draw another circle $ (K),$ inwardly to $ (O),$ centered at an arbitrary point $ K,$ between $ A,$ $ O$ and let $ AC,$ $ BD$ be, the lines through $ A,$ $ B$ respectively, not to the same side with respect to $ AB$ and tanged to $ (K)$ at points $ C,$ $ D.$ We denote the points $ C'\equiv (O)\cap AC,$ $ D'\equiv (O)\cap BD,$ $ C''\equiv (O)\cap D'C$ and $ D''\equiv (O)\cap C'D.$
 Prove that the point $ M\equiv C'D'\cap C''D'',$ lies on the diameter $ AB$ of $ (O).$

Kostas Vittas.

$ M' \in AB \cap C'D'$. We have $ \frac{CA}{CC'}= \frac{KA}{KB}= \frac{sin \angle KBD}{sin \angle KAC}= \frac{sin \angle M'C'A}{sin \angle M'AC}= \frac{M'A}{M'C'}$ so $ M'C$ is the bisector of $ \angle AM'C'$. Similarly, $ M'D$ is the bisector of $ \angle BM'D'$, so $ M',C,D$ are collinear.
$ M'' \in AB \cap C''D''$, use Pascal's theorem with the hexagon $ AC'D''C''D'B$ we have $ M'',C,D$ are collinear.
So $ M' \equiv M'' \in AB \cap CD$, and $ AB,CD,C'D',C''D''$ are concurrent as we wanted.

Sorry for the late answer and thank you dear [b][size=100]PDatK40SP[/size][/b] for a sorter solution, ''half different'', the one I have in mid.

$ \bullet$ We denote the point $ M\equiv AB\cap C'D'$ and we will prove that the points $ C,$ $ M,$ $ D,$ are collinear.

Applying the [b][size=100]Menelaus’s theorem[/size][/b] in the triangle $ \bigtriangleup PAB,$ where $ P\equiv AC'\cap BD',$ it is enough to prove that

$ \frac{CA}{CP}\cdot\frac{DP}{DB}\cdot\frac{MB}{MA}\equal{} 1$ $ \Longrightarrow$ $ \frac{CA}{DB}\equal{}\frac{MA}{MB}$ $ ,(1)$ $ ($ because of $ CP \equal{} DP$ $ ).$

From $ KC\parallel BC'$ $ \Longrightarrow$ $ \frac{CA}{AC'}\equal{}\frac{KC}{BC'}$ $ ,(2)$

From $ KD\parallel AD'$ $ \Longrightarrow$ $ \frac{DB}{BD'}\equal{}\frac{KD}{AD'}$ $ ,(3)$

From $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac{CA}{DB}\cdot\frac{BD'}{AC'}\equal{}\frac{AD'}{BC'}$ $ \Longrightarrow$ $ \frac{CA}{DB}\equal{}\frac{(AC')\cdot (AD')}{(BC')\cdot (BD')}$ $ ,(4)$

From $ (1),$ $ (4)$ we conclude that it is enough to prove that $ \frac{MA}{MB}\equal{}\frac{(AC')\cdot (AD')}{(BC')\cdot (BD')}$ $ ,(5)$ which, however, is true in every cyclic quadrilateral as well known.

So, the relation $ (1)$ is also true and then the points $ C,$ $ M,$ $ D,$ are collinear.

Applying the [b][size=100]Pascal’s theorem[/size][/b] in the non convex cyclic hexagon $ ABD'C''D''C',$ we conclude that hte line segment $ C''D'',$ passes through the point $ M,$ as the intersection point of $ AB,$ $ CD.$

Hence, the line segments $ C'D',$ $ C''D'',$ intersect each other at the point $ M,$ lies on the diameter $ AB$ of $ (O)$ and the proof is completed.

Kostas Vittas.


Solve the equation: $ 6^x\plus{}2^x\equal{}5^x\plus{}3^x$

+) Equation $ \Longleftrightarrow 6^x - 5^x = 3^x - 2^x$

+) Let $ x_0$ is one root of this equation we have $ 6^{x_0} - 5^{x_0} = 3^{x_0} - 2^{x_0}$

+) Let $ y = f(t) = (t + 1)^{x_0} - t^{x_0}, t > 0$

+) Apply Largrange Theorem we get exit $ c \in (2;5)$ so that $ f'(c) = \frac {f(5) - f(2)}{5 - 2} = \frac {(6^{x_0} - 5^{x_0}) - (3^{x_0} - 2^{x_0})}{3} = 0$

$ \Longleftrightarrow x_0[(c + 1)^{x_0 - 1} - c^{x_0 - 1}] = 0 \Longleftrightarrow  [ \begin {array} {c} x_0 = 0 \\
x_0 = 1 \end {array}$

Hence this equation have roots are $ x=0, x=1$.


function
real analysis
real analysis solved

Prove that the rings (C,+,*) and (D,+,*) are not isomorphic, where C is the set of continuous functions f:R->R and D is the set of derivable functions f:R->R.

Let F be the morphism from C to D
Take f(x)=x and g s.t. F(g)=f. Then F(3 \sqrt g)3=f. Then F(3 \sqrt g)=3 \sqrt f, which is not derivable. Contradiction :D  :D


For any three positive reals a, b, c, we have

$a\sqrt{a^2+bc}+b\sqrt{b^2+ca}+c\sqrt{c^2+ab}\ge\sqrt{2}(ab+bc+ca)$

(I) Use Chebychev's Inequalities and Jensen's Inequalities.

We have to prove the inequality

$a\sqrt{a^2+bc}+b\sqrt{b^2+ca}+c\sqrt{c^2+ab}\geq\sqrt2\left(bc+ca+ab\right)$

for any three positive reals a, b, c.

Upon division by $\sqrt2$, this inequality becomes

$a\sqrt{\frac{a^2+bc}{2}}+b\sqrt{\frac{b^2+ca}{2}}+c\sqrt{\frac{c^2+ab}{2}}\geq bc+ca+ab$.

Now define $x=\sqrt a$, $y=\sqrt b$, $z=\sqrt c$. Then $x^2=\left(\sqrt a\right)^2=a$ and $\sqrt{bc}=\sqrt b\sqrt c=yz$, so that $bc=\left(\sqrt{bc}\right)^2=\left(yz\right)^2$. By the QM-AM inequality, $\sqrt{\frac{\left(x^2\right)^2+\left(yz\right)^2}{2}}\geq\frac{x^2+yz}{2}$. In other words, $\sqrt{\frac{a^2+bc}{2}}\geq\frac{x^2+yz}{2}$. Thus, $a\sqrt{\frac{a^2+bc}{2}}\geq a\frac{x^2+yz}{2}=x^2\frac{x^2+yz}{2}=\frac{x^4+x^2yz}{2}$. Similarly, $b\sqrt{\frac{b^2+ca}{2}}\geq\frac{y^4+y^2zx}{2}$ and $c\sqrt{\frac{c^2+ab}{2}}\geq\frac{z^4+z^2xy}{2}$. Thus, in order to prove the inequality

$a\sqrt{\frac{a^2+bc}{2}}+b\sqrt{\frac{b^2+ca}{2}}+c\sqrt{\frac{c^2+ab}{2}}\geq bc+ca+ab$,

it is enough to show

$\frac{x^4+x^2yz}{2}+\frac{y^4+y^2zx}{2}+\frac{z^4+z^2xy}{2}\geq bc+ca+ab$.

Since $bc=\left(yz\right)^2=y^2z^2$ and similarly, $ca=z^2x^2$ and $ab=x^2y^2$, this inequality is equivalent to

$\frac{x^4+x^2yz}{2}+\frac{y^4+y^2zx}{2}+\frac{z^4+z^2xy}{2}\geq y^2z^2+z^2x^2+x^2y^2$.

Multiplication with 2 transforms this into

$\left(x^4+x^2yz\right)+\left(y^4+y^2zx\right)+\left(z^4+z^2xy\right)\geq 2\left(y^2z^2+z^2x^2+x^2y^2\right)$.

But by the Schur inequality,

$x^2\left(x-y\right)\left(x-z\right)+y^2\left(y-z\right)\left(y-x\right)+z^2\left(z-x\right)\left(z-y\right)\geq 0$;

this immediately transforms into

$x^4-x^3y-x^3z+x^2yz+y^4-y^3z-y^3x+y^2zx+z^4-z^3x-z^3y+z^2xy\geq 0$,

or, equivalently,

$\left(x^4+x^2yz\right)+\left(y^4+y^2zx\right)+\left(z^4+z^2xy\right)\geq x^3y+x^3z+y^3z+y^3x+z^3x+z^3y$.

Hence, in order to establish the inequality

$\left(x^4+x^2yz\right)+\left(y^4+y^2zx\right)+\left(z^4+z^2xy\right)\geq 2\left(y^2z^2+z^2x^2+x^2y^2\right)$,

it is enough to prove that

$x^3y+x^3z+y^3z+y^3x+z^3x+z^3y\geq 2\left(y^2z^2+z^2x^2+x^2y^2\right)$.

But this inequality is clearly true, being the sum of the three inequalities

$y^3z+z^3y\geq 2y^2z^2$;
$z^3x+x^3z\geq 2z^2x^2$;
$x^3y+y^3x\geq 2x^2y^2$

(these three inequalities are trivial: for instance, $y^3z+z^3y\geq 2y^2z^2$ follows from $y^3z+z^3y-2y^2z^2=yz\left(y-z\right)^2\geq 0$). Proof complete.

  Darij


What is the smallest prime that is the sum of both 3 and 5 consecutive primes?

brute forcing:

[hide] we find that it is 83 [/hide]

not sure if there  is an easier or more methodical way (if there is, it would take me a lot longer to find it than brute forcing)


How much water must be evaporated from 160 litres of a 12% calium chloride solution to produce a 59% calcium chloride solution?

$12$ percent of $160$ liters would be $19.2$ liters. We want $19.2$ to be $59$ percent of the solution, so $.59x=19.2$. Therefore, $x$, or the amount of solution, would be about $32.54$ liters. So... the amount evapourated would be $160-32.54=$

$127.46$ liters evapourated.

I think that's right...

The math works out, I believe - I haven't checked, but the concept is incorrect.  You don't measure Calcium Carbonate in liters.  The correct solution would be to use the weight of the solution to determine the weight of Calcium Carbonate then convert to liters with the density of the water.  In a multiple choice exam there would be no penalty for getting the correct answer with the wrong concepts, but in a written exam, or in using the concepts further. . .

An alternative, more elegant method would be to determine how much (% or ratio) of the water would need to evaporate to convert a solution from the lower concentration to the higher then multiply this by the amount of the initial solution.


Και για αυτή την ενότητα ισχύουν οι γενικοί κανόνες που ισχύουν και για τη γεωμετρία. [b]Ας βάλουμε πρώτα 20 ασκήσεις [/b]και ... βλέπουμε !

 Προτείνω να μη μπλέξουμε με περιπτώσεις που θέλουν πολλαπλασιαστές Lagrange κλπ , διότι στις ΙΜΟ  δεν τις απαιτούν. Μπορούμε όμως αυτές να τις αναφέρουμε , όπου βρίσκουμε και άλλον τρόπο. Θυμηθείτε ότι και η περισυνή  ΙΜΟ - Ανισότητα , που ήταν από τις δυσκολότερες , έβγαινε με την  κλασσική θεωρία ! Να προσπαθήσουμε  να καλύψουμε όσο γίνεται περισσότερες περιπτώσεις. Το πεδίο είναι πολύ μεγάλο, αλλά ευτυχώς και η ελληνική βιβλιογραφία καλύπτει σε ικανοποιητικό βαθμό το σχετικό πεδίο. Καλύτερα να αφήσουμε λίγο χώρο για ακολουθίες , πολυώνυμα κλπ , όπου δεν έχουμε πολλέ δύσκολες ασκήσεις στη βιβλιογραφία μας.

 [color=red]Παρακαλώ τους φίλους της άλγεβρας να ...θριαμβεύσουν με τις ωραίες ασκήσεις τους !!![/color]

[b][size=150][color=darkblue]ΑΣΚΗΣΗ 1[/color][/size] (Carlson's inequality)[/b]

Αν $a,b,c>0$ να δειχθεί ότι

$\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}\geq  \sqrt{\frac{ab+bc+ca}{3}}$ .

[color=darkblue][u]Λύση[/u][/color]
Θα χρησιμοποιήσουμε το παρακάτω λήμμα.

[color=green]Λήμμα: [/color]Για $a,b,c>0$ ισχύει

$\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}\geq  \frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}$ .

[u]Απόδειξη λήμματος[/u]

$\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}=\frac{\left(a+b+c\right)\left(ab+bc+ca\right)-abc}{8}\geq \frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}$

η τελευταία ανισότητα προκύπτει λόγω της $\left(a+b+c\right)\left(ab+bc+ca\right)\geq 9abc \Leftrightarrow \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9$ (...ανισότητα αριθμητικού-αρμονικού μέσου).

Ας επιστρέψουμε τώρα στο πρόβλημα.
Λόγω του παραπάνω λήμματος προκύπτει ότι

$\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}\geq \sqrt[3]{\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}}$

συνεπώς αρκεί να αποδείξουμε ότι

$\sqrt[3]{\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}}\geq \sqrt{\frac{ab+bc+ca}{3}}\Leftrightarrow$

$\Leftrightarrow (a+b+c)^{2}\geq 3(ab+bc+ca)$

(...που ισχύει αφού $(a+b+c)^{2}-3(ab+bc+ca)=a^{2}+b^{2}+c^{2}-(ab+bc+ca)=$

$\frac{1}{2}\left(\left(a-b\right)^{2}+\left(b-c\right)^{2}+\left(c-a\right)^{2}\right)\geq 0$ ).

Η ισότητα ισχύει όταν $a=b=c$.

Στέλιος

[size=150][color=red]ASKHSH 2[/color][/size]

Na lithei to parakatw sistima gia tis pragmatikes times twn $x,y$

\[\left\{ \begin{array}{ll}2^{x^{2}+y}+2^{x+y^{2}}=8, \\ \sqrt{x}+\sqrt{y}=2 \end{array}\right.\]

Alexandros

[quote="silouan"]Σε αυτά τα πεδία  έχουμε ήδη πολλές ασκήσεις , γιατί γράφουμε και άλλες ? :?:  :huh:  . Νομίζω ότι πρέπει να βάλουμε λίγα πολυώνυμα αρκετές ακολουθίες και να γεμίσουμε τη συνδιαστική που είναι άδεια  :D  . Για όλα τα άλλα σήμερα που έριχνα μια γενική ματιά στο φόρουμ από τότε που έγινε έχουμε πολύ πράγμα.... Τι λέτε ?[/quote]

 [b]Ωραία λοιπόν !!! [/b]

 Να μαζέψουμε τότε ,  ένας - ένας , τα θέματα που ήδη έχουμε στείλει και είναι ήδη λυμένα. Απλά να φτιάξουμε καλά τις λύσεις τους.Θέλει και αυτό δουλειά.Θα πάμε στο άλλο πεδίο.

     Μπάμπης

[size=150][color=red][b]ΑΣΚΗΣΗ 4[/b][/color][/size]

Η άσκηση μαζί με τη λύση βρίσκεται ως συνήθως στο pdf   :) . Στοιχειοθετήθηκε με $\LaTeX$ .

Αφού συμφωνούμε κ. Μπάμπη ας ξεκινήσουμε από τις παλιές σελίδες και να μαζεύουμε ασκήσεις . Οκ ?
Παρακαλώ πολύ μην βάζετε καινούργιες.....

Ελέγχω όσες έχουν δημοσιευθεί προς το παρόν, και στο τέλος κάνουμε την επιλογή.

[color=red]ΔΙΟΡΘΩΣΕΙΣ:[/color]

[b]Άσκηση 1[/b]

Σωστή  :D

[b]Άσκηση 2[/b]

Σωστή  :D   (και οι 3 αποδείξεις)

[b]Άσκηση 3[/b]

Νομίζω πως η εκφώνηση και η τελευταία σχέση θα έπρεπε να είναι $(x^{2}+y^{2}+z^{2})( \frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ αντί για $(x^{2}+y^{2}+z^{2})( \frac{1}{x}+\frac{1}{x}+\frac{1}{x})$ .

[b]Άσκηση 4[/b]

Σωστή  :D
Αλλά μήπως στο σημείο που λες "με ακόμα μία μικρή τροποποίηση και προσοχή στις πράξεις, προκύπτει" να συμπληρωθούν οι τροποποιησούλες και οι πράξεις;  :roll:

[quote="darksaga"]$x,y,z$ θετικοι πραγματικοι να δειχθει
$(x^{2}+y^{2}+z^{2})(\frac{1}{x}+\frac{1}{x}+\frac{1}{x})\ge3(x+y+z)$

λυση
εχουμε οτι $x^{2}+y^{2}-xy \ge xy \Leftrightarrow (x+y)(x^{2}-xy+y^{2}) \ge (x+y)xy \Leftrightarrow x^{3}+y^{3}\ge (x+y)xy \Leftrightarrow \frac{x^{2}}{y}+\frac{y^{2}}{x}\ge x+y$
ομοια
$\frac{z^{2}}{y}+\frac{y^{2}}{z}\ge z+y$
$\frac{x^{2}}{z}+\frac{z^{2}}{x}\ge x+z$

προσθετουμε κατα μελη

$\frac{x^{2}}{y}+\frac{y^{2}}{x}+\frac{z^{2}}{y}+\frac{y^{2}}{z}+\frac{x^{2}}{z}+\frac{z^{2}}{x}\ge 2(x+y+z) \Leftrightarrow (x+y+z)+\frac{x^{2}}{y}+\frac{y^{2}}{x}+\frac{z^{2}}{y}+\frac{y^{2}}{z}+\frac{x^{2}}{z}+\frac{z^{2}}{x}\ge 3(x+y+z) \Leftrightarrow \frac{x^{2}}{x}+\frac{y^{2}}{y}+\frac{z^{2}}{z}+\frac{x^{2}}{y}+\frac{y^{2}}{x}+\frac{z^{2}}{y}+\frac{y^{2}}{z}+\frac{x^{2}}{z}+\frac{z^{2}}{x}\ge 3(x+y+z) \Leftrightarrow (x^{2}+y^{2}+z^{2})(\frac{1}{x}+\frac{1}{x}+\frac{1}{x})\ gneq3(x+y+z)$[/quote]
φίλε darksaga,
αν μπορείς διόρθωσε την αρχική ανισότητα προς απόδειξη...είναι φανερό το τι εννοείς στη δεύτερη παρένθεση :wink:
Επίσης η λύση μπορεί να γίνει και με χρήση της ανισότητας Tschebychev για τα ζεύγη x^2, y^2, z^2 kai 1/x, 1/y, 1/z τα αποία θα έχουν αντίθετη διάταξη.

Με συγψωρείτε που δεν την παραθέτω ολοκληρωμένη αλλά δεν έχω Latex :oops:

Παραθέτω και μία δική μου άσκηση. Δυστυχώς δεν έχω LATEX και μπορεί να δυσκολευτείτε στην ανάγνωση. Τη λύση θα τη στείλω αργότερα αν δεν είναι πρόβλημα. Όπου γράφω 3SQRT εννοώ ρίζα τρίτης τάξης.

Αν a, b, c πλευρές τριγώνου και R η ακτίνα του περιγεγραμμένου, να δειχθεί ότι:

3SQRT(a^2+b^2) * 3SQRT(b^2+c^2) * 3SQRT(c^2+a^2) <

2/3 * R * (a/cosA + b/cosB + c/cosC)

Θα το εκτιμούσα ιδιαίτερα αν κάποιος μπορούσε να την ξαναγράψει (με ή χωρίς τη λύση) σε LATEX και να μου πει αν είναι σωστή.

[quote="Protonios"]
Θα το εκτιμούσα ιδιαίτερα αν κάποιος μπορούσε να την ξαναγράψει (με ή χωρίς τη λύση) σε LATEX και να μου πει αν είναι σωστή.[/quote]
[size=150][color=red]
ASKHSH 5[/color][/size]

Αν $a, b, c$ πλευρές τριγώνου και $R$ η ακτίνα του περιγεγραμμένου, να δειχθεί ότι:

\[\sqrt[3]{a^{2}+b^{2}}\cdot \sqrt[3]{b^{2}+c^{2}}\cdot\sqrt[3]{c^{2}+a^{2}}< \frac{2}{3}R \left(\frac{a}{cosA}+\frac{b}{cosB}+\frac{c}{cosC}\right)\]

Oson afora tin lisi, perimenoume me ton diko sou tropo na tin parousiaseis etsi wste na min xathei (opws leei kai o kos Stergiou) o proswpikos tropos skepsis kai grafis. Apo ekei kai pera (meta apo tin lisi sou) mporei na akolouthisoun kai alloi tropoi lisis.

Alexandros

Βάζω μια άσκηση που είχε συζητηθεί παλιά εδώ και ήταν αρκετά δύσκολη .Εδώ δεν είχε συζητηθεί σε αυτή τη μορφή , εγώ την τροποποίησα ,την  έκανα πιο ισχυρή αλλά η λύση είναι όμοια με αυτή που είχε δώσει ο galeos τότε :

ΑΣΚΗΣΗ 6

Ακόμη μία από παλιά .Η λύση είχε δωθεί από τον Anto

ΑΣΚΗΣΗ 7
Άν$x>0$ και $y<1$ να αποδειχθεί οτι $x^{y}+y^{x}> 1$

ΛΥΣΗ
Θα γίνει χρήση διαφορικού λογισμού
Έστω ότι το $y$ είναι σταθερό. Θεωρώ την συνάρτηση $f(x) = x^{y}+y^{x}$, $x\in [0,1]$. Καθώς η συνάρτηση αυτή είναι συνεχής στο κλειστο διάστημα [0,1] παίρνει ελάχιστη τιμή στο διάστημα αυτό. Εάν η τιμή αύτή ταυτίζεται με ένα απο τα δύο άκρα του διαστήματος, τότε είναι προφανές ότι $f(x) > 1$ όταν $x\in (0,1)$. Έστω λοιπόν οτι η συνάρτηση παρουσιάζει το ελάχιστο της στο εσωτερικό του [0,1]. Έστω στο $a\in(0,1)$. Τότε απο το θ. Fermat, εφόσον η συνάρτηση μας είναι παραγωγίσιγη, έπετε ότι $f'(a)=0 \leftrightarrow ya^{y-1}+y^{a}ln(y) =0$ (1).
Εάν υποθέσουμε ότι $f(a) \leq 1$. Τότε $a^{y}+y^{a}\leq 1 < y^{a-1}$ άρα $\frac{1}{a}(y-1)y^{a-1}<-a^{y-1}\leftrightarrow_{(1)}\frac{1}{a}y<ln(y) \leftrightarrow y-1 < aln(y) \leftrightarrow ya^{y-2}[(y-1)-aln(y)]< 0 \leftrightarrow f''(a) < 0$
 Άρα το a δεν είναι δυνατόν να είναι τοπικό ελάχιστο. Αντίφαση. Άρα $f(a) > 1$

ΑΣΚΣΗΣΗ ....

[quote="it's me"]Αν
\[x,y,z>0\]  με
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}\]
να αποδειξετε οτι
\[\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\geq 1\]
[/quote] (It's me)

 ΛΥΣΗ ( Cetanman)

[quote="cretanman"]H dosmeni sxesi ginetai $2(xy+yz+xz)=3xyz \ \ (1)$

H pros apodeiksi anisotita grafetai $\frac{x^{2}}{x^{2}+xyz}+\frac{y^{2}}{y^{2}+xyz}+\frac{z^{2}}{z^{2}+xyz}\geq 1$

Omws apo tin anisotita Andreescu exoume oti

\begin{eqnarray}\frac{x^{2}}{x^{2}+xyz}+\frac{y^{2}}{y^{2}+xyz}+\frac{z^{2}}{z^{2}+xyz}&\geq&\frac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}+3xyz}\nonumber \\ &=&\frac{x^{2}+y^{2}+z^{2}+2(xy+yz+xz)}{x^{2}+y^{2}+z^{2}+3xyz}\nonumber \\ &\stackrel{(1)}{=}& 1 \nonumber \end{eqnarray}

Alexandros[/quote]

Είναι άμεσο πόρισμα της ανισότητας BCS και αποδεικνύεται επίσης επαγωγικά. Είναι η εξής βασική ανισότητα.

$x^{2}/a+y^{2}/b+z^{2}/c\geq (x+y+z)^{2}/(a+b+c)$

και γενικεύεται σε δυνάμεις και σε όρους με μια μικρή διαφορά στον παρονομαστή. Γενικά όμως για τρεις όρους είναι πιο χρήσιμη. Για περισσότερες πληροφορίες ανέτρεξε στο πρώτο βιβλίο του κύριου Στεργίου για τις ανισότητες.
Προσοχή όμως δεν μπορείς να το χρησιμοποιείς επίσημα χωρίς απόδειξη διότι δεν είναι καταχωρημένο στη διεθνή βιβλιογραφία με όνομα Andreescu. :wink:

Φίλε protonie ευχαριστώ πολύ για την πληροφορία.Την ανισότητα αυτή την γνωρίζω και τη χρησιμοποιώ δεν ήξερα όμως ότι είναι γνωστή με αυτή την ονομασία.

Είναι παρόμοιος με τον τρόπο λύσης που υπάρχει και στο βιβλίο "Κλασικές και Νέες Ανισότητες" , από εκδόσεις Σαββάλα, στη σελίδα 41.

Για να ξαναρχίσουμε να ζεσταίνουμε το κλίμα στο "ελληνικά" παραθέτω μια (σχετικά εύκολη...) ανισότητα από εθνική Ολυμπιάδα της Ρουμανίας το 2005.

Αν $ a,b,c,d$ θετικοί πραγματικοί αριθμοί ν.δ.ο:
$ \frac{a}{b\plus{}2c\plus{}d}\plus{}\frac{b}{c\plus{}2d\plus{}a}\plus{}\frac{c}{d\plus{}2a\plus{}b}\plus{}\frac{d}{a\plus{}2b\plus{}c}\geq 1$

Θα περιμένω να δω τη δική σας λύση γι' αυτό και δεν βάζω αμέσως τη λύση μου για να έχει λίγο ενδιαφέρον αν και όπως είπα επιφυλάσσομαι μήπως σας φανεί εύκολη :oops:

Απλά πολ/ζουμε τον κάθε όρο με τον αριθμητή πάνω και κάτω και εφαρμόζουμε το γνωστό λήμμα Andreescu που σκοτώνει όλες τις αδύναμες ανισότητες. Μετά χρειάζεται μόνο πράξεις. :)

[quote="gregg"]Μήπως θα μπορούσε κάποπιος να μου πει αν υπάρχει η γενικευμένη μορφή της ανισότητας Andreescu[/quote]

 Γρηγόρη ,
 τι εννοείς  γράφονατας '' αν υπάρχει '' . Αυτή την ανισότητα μπορείς να τη χρησιμποποιείς για να βγάζεις τις ασκήσεις , αλλά για να πέρεις όλες τις μονάδες θέλει να πεις και πώς βγαίνει.
  Στη βιβλιογραφία την έχουν έτσι οι Ρουμάνοι και εμείς , αλλά οι άλλοι δεν την καταλαβαίνουν. Από τους Ρουμάνους την πήρα και γω και μάλιστα από το μαθηματικό τους περιοδικό '' Gazeta Matematica''.Αν θέλεις και κάτι άλλο , ρώρησέ μας.
 Μπάμπης

Η απόδειξη δεν είναι τπτ. Πάρε την cauchy για δυάδες ${ \sqrt{a},\frac{x_{1}}{\sqrt{a}}}$ κλπ. Την απόδειξη θα τη βρεις λεπτομερώς στο βιβλίο -θησαυρός- των κυρίων Στεργίου, Σκομπρή, "Αλγεβρικές Ανισότητες" ! Αν δεν το έχεις και ασχολείσαι με διαγωνισμούς, στο συνιστώ ανεπιφύλαχτα!!  :)

A! Πρέπει να την αποδείξεις πρώτα !!!

Στέλιος

EDIT:

Σου βάζω κι ένα pdf με μια παλαιά άσκηση, για να δεις περίπου πώς λειτουργεί. (Υπόψη ότι η απόδειξη είναι λίγο σύντομη...)

[quote="gregg"]stergiu wrote:
[quote]Αυτή την ανισότητα μπορείς να τη χρησιμποποιείς για να βγάζεις τις ασκήσεις , αλλά για να πέρεις όλες τις μονάδες θέλει να πεις και πώς βγαίνει. [/quote]
Δεν το καταλαβαίνω αυτό :maybe: . Δηλαδή πρώτα θα λύνω την άσκηση και μετά τι πρέπει να γράψω;;
Ότι είναι η γενικευμένη μορφή της ανισότητας Andreescu ή μήπως πρέπει να την αποδείξω;;
Φέτος π.χ. το 2 (νομίζω) θέμα του Αρχιμήδη έβγαινε με αυτή την μορφή....[/quote]

 Mην ανησυχείς !  :)
 Βγάλε εσύ την άσκηση και αυτοί που διορθώνουν την ξέρουν καλά ! Ονόμασέ την καλύτερα ΄΄  ανισότητα  BCS '' .
 Στην Ελλάδα την ξέρουν και ως Ανδρεσκου.(Αν θέλεις , κάνε και μια αναφορά για την απόδειξή της , όπως σου είπε ο mostel.
 Προχώρα λοιπόν και μη σε ανησυχούν οι ονομασίες !

 Μπάμπης

[quote="gregg"]Παρκαλώ δείτε το συνημμένο γιατί για κάποιο λόγο δεν μπορώ να επικολλήσω κάτι τύπους

Όσο για το βιβλίο το έχω, μάλιστα αυτές τις μέρες ασχολούμαι αρκετά με αυτό.
Επίσης στο βιβλίο αυτό σε πολλές γενικεύσεις ανισοτήτων λέει ότι η απόδειξη είναι έξω από τα πλαίσια του βιβλίου ή κάπως έτσι.

Πάντως ευχαριστώ για τις υποδείξεις.[/quote]

 Γρηγόρη !

 Όλες τις ανισότητες που έχει θεωρία το βιβλίο αυτό μπορείς να τις χρησιμοποιείς χωρίς απόδειξη . Δεν πρόκειται κανένας να σου τις ζητήσει.
Προχώρα λοιπόν χωρίς φόβο και μάθε να λύνεις , χρησιμοποιώντας την κατάλληλη ανισότητα όπου χρειάζεται. Όλες αυτές οι ανισότητες είναι τόσο κλασικές για τις ολυμπιάδες , όσο η ταυτότητα $ (a\plus{}b)^{2}\equal{} ...$ για το λύκειο!.
    Αυτό σου το λέω υπεύθυνα και μην αμφιβάλλεις. Το μόνο που σου είπαν τα παιδιά είναι ότι άμα πας στη Βαλκανιάδα ή ΙΜΟ , την Andreescu  να τη λες Cauchy ή BCS . Με τα ονόματα υπάρχει διεθνώς μια μικρή σύγχιση και κανένας δε δίνει σημασία !!!
 [b]Καλή συνέχεια ![/b]

Μήπως μπορεί κάποιος να μου πει σε τι χρησιμεύει (δηλαδή αν την χρησιμοποιούμε καθόλου :huh: ) την ανισότητα Shapiro;;

Οτ ύπος της υπάρχει στο συννημένο για όποιον ενδιαφέρεται.

Τι να σου πω εγώ πρώτη φορά τη βλέπω. :oops:

Από τον τύπο της πάντως φαίνεται ότι δεν είναι ιδιαίτερα χρήσιμη, πόσο μάλλον για τις ανισότητες διαγωνισμών στυλ Ολυμπιάδας. :(

Φιλική συμβουλή. Μη χάνεις το χρόνο σου αναζητώντας σπάνιες ανισότητες. Ασχολήσου καλύτερα με τα τετριμμένα και ευρέως γνωστά γιατί οι ανισότητες είναι σίγουρο χαρτί σε διαγωνισμούς αν τις κατέχεις σε κάποιο ικανοποιητικό βαθμό, όπως και η γεωμετρία. Για τα άλλα δεν μπορείς να είσαι ποτέ σίγουρος. :wink:

Η συγκεκριμένη είχε πέσει σε μια ολυμπιάδα της ΕΣΣΔ για [b]LHS > n/4 [/b]νομίζω...


vector
quadratics
real analysis
real analysis unsolved

I have to show that on $ \mathbb{R}^2$ the following three norms are equivalent:
$ \|x\|_1 \equal{} |x_1| \plus{} |x_2|$
$ \|x\|_2 \equal{} \sqrt {|x_1|^2 \plus{} |x_2|^2}$
$ \|x\|_\infty \equal{} \max(|x_1|,|x_2|)$

(I know that all norms on finite dimensional vector spaces are equivalent, but obviously I'm not allowed to use that.)

It's obvious that we have $ \|x\|_\infty\leq \|x\|_2\leq \|x\|_1$. So I need to show that $ \|x\|_1\leq C_1\|x\|_2$ and $ \|x\|_2\leq C_2\|x\|_\infty$ for some $ C_1,C_2 > 0$.
Now, I made a drawing of the closed unit ball for the three norms. De 1-norm is a square and the 2-norm is a circle, so $ C_1 \equal{} \sqrt {2}$ should do the trick. My attempt to show that $ \|x\|_1\leq \sqrt {2}\|x\|_2$:
We have $ 2\|x\|_2^2 \minus{} \|x\|_1^2 \equal{} |x_1|^2 \plus{}|x_2|^2 \minus{} 2|x_1\parallel{}x_2|$. WLOG assume $ |x_1|\geq |x_2|$. Then $ |x_1|^2 \plus{} |x_2|^2\geq 2|x_2|^2$. But this estimate is too sharp, since I only know $ 2|x_1\parallel{}x_2|\leq 2|x_1|^2$...
Probably I am missing something very elementary. Thanks!

Basically, you want to show that $ a\plus{}b \leq \sqrt{2(a^2\plus{}b^2)}$ for positive $ a,b$. Square both sides, you get $ a^2\plus{}2ab\plus{}b^2 \leq 2(a^2\plus{}b^2)$. Cancelling $ a^2\plus{}b^2$ on both sides, $ 2ab \leq a^2\plus{}b^2$. Ring any bell?

Geometric mean $ \leq$ quadratic mean: $ \sqrt{ab}\leq \sqrt{\frac{a^2\plus{}b^2}{2}}$  :)

You don't need any computations.
[quote]Now, I made a drawing of the closed unit balls for the three norms. The $ 1$-norm is a square and the $ 2$-norm is a circle, and the $ \infty$-norm is a rhombus.[/quote]
(corrected and augmented)

By just geometrical simple reasoning, you can fit any into any other, by contractions and/or dilations.

Yep, geometrically it is clear. Since it was so obvious looking at the drawing, I thought the proof should be easy too. Thanks for the replies.


superior algebra
superior algebra unsolved

I know it sounds simple but actually i could not prove the following.Can you help me with this:

Let G be a group such that there is unique x in G with o(x) = 2,that is, the order of x is 2. Prove that xy = yx for all y in G.

Actually i could not figure out how to use uniqueness of x.

[quote="outsider"]Let G be a group such that there is unique x in G with o(x) = 2,that is, the order of x is 2. Prove that xy = yx for all y in G.[/quote]

Since the order of $x$ is $2$, we have $x\neq e$ but $x^{2}=e$, where $e$ is the unit of $G$. Thus, for any $y\in G$, we have $\left(y^{-1}xy\right)^{2}=y^{-1}xyy^{-1}xy=y^{-1}xxy=y^{-1}x^{2}y=y^{-1}ey=y^{-1}y=e$, but $y^{-1}xy\neq e$ (since $y^{-1}xy=e$ would lead to $xy=ye=y$, so that $x=e$, but we have $x\neq e$). Thus, the element $y^{-1}xy$ of $G$ has order $2$. Since the only element of $G$ with order $2$ is $x$, this yields $y^{-1}xy=x$, so that $xy=yx$, qed.

  darij

Thanks for your solution. Your help is appreciated.


function
integration
limit
logarithms
probability
calculus
probability and stats

I've encountered the following while studying a book on statistics. It stated without any proof that,
If $ x$ is random variable on $ [a,b]\subset (0,1)$ with distribution function $ F$. Then
\[ lim_{n\to\infty}\frac{1}{n} log \int exp (\frac{n}{2}x)dF(x)\equal{}\sup_{[a,b]}x.\]
I don't readily see how it is true. Even, I have no idea how to tackle this. Please enlighten me.

Suppose $ x$ is equal to $ b\in (0,1)$ almost surely. Then the left side is
$ \lim_{n\to\infty} \frac{1}{n}\log \exp(nb/2)\equal{}\frac{b}{2}$. It looks like $ 1/2$ in the exponential does not belong. I'll drop it

In the general case: if $ x$ takes values in $ [a,b]$ a.s., then $ \int \exp(nx) \le \exp(nb)$, which gives an upper bound. In the opposite direction, suppose $ x$ has a positive probability of being in $ [b\minus{}\epsilon,b]$ and estimate the integral from below.


calculus integration trigonometry LaTeX function geometry symmetry

I cannot figure this out:

$ \int_0^2{2\sqrt {4 - x^2}}$

The book says the answer is:

[hide]
(Lower bound 0, upper bound 2) $ {x\sqrt {4 - x^2} + 4\sin^{ - 1}{\frac {x}{2}}} = 2\pi$

Note: not sure how to latex the "evaluate integral from a to b" symbol.
[/hide]

I found this integral by an alternate method, reproduced below, but I do not know how to get it the book's way and would like to see. Here is how I did it:

[hide]

Given $ \int_0^2{2\sqrt {4 - x^2}}$

substitute $ x = 2\cos{u}$ and $ dx = - 2\sin{u}(du)$. New bounds:

${ \int_0^2{2\sqrt {4 - x^2}} = \int_{ - \frac {\pi}{2}}^0{ 4\sin{u}}( - 2\sin{u}(du))}$

${ = \int_{ - \frac {\pi}{2}}^0{ - 4 + 4\cos{2u}}du}$

$ = - 2\pi$

I'm also not sure why I'm getting a negative, since the answer is suppose to be positive (this is part of a volume integral), but I'm just not sure how to do it without the trig substitution.
[/hide]

It's due to your limits of integration.  The graph is similar to that of the upper half of a circle.  To get the positive answer, you need to integrate the positive region and to get the negative answer, you integrate the negative region.  You are integrating in a negative region.  Instead of $ \minus{}\frac{\pi}{2}$, you need $ \frac{\pi}{2}$.

What I do in such a situation is once I've made my substitution, I do the integral but I don't include the limits of integration-I just do the indefinite integral.  Then once I've evaluated the antiderivative, I plug back in the original functions and use the original limits of integration.  Works every time.

[quote="bthings_2000"]$ \left.x\sqrt {4 \minus{} x^2} \plus{} 4\sin^{ \minus{} 1}{\frac {x}{2}}\right|_0^2 \equal{} 2\pi$[/quote]
One version of the symbol- I usually use [ ] instead of |.

In your substitution, note that $ \sqrt {4 \minus{} 4\cos^2 u} \equal{} \minus{} 2\sin u$ for $ u\in [ \minus{} \frac {\pi}{2},0]$; your integrand is therefore wrong. Cosine substitutions run into this sort of sign problem a lot, and $ \sin$ substitutions usually don't. If you want the positive sign, use $ \frac\pi2$ and $ 0$ as the limits instead- this introduces a $ \minus{}$ sign since the limits are in the "wrong" order.

Of course, there's this alternate method: this is the area of a semicircle. We know the area of a semicircle, so we don't really need to compute any antiderivatives at all.

If someone in a calculus class of mine were to say that on a test, I would give the answer full credit.

$ A_n\equal{}2\int^n_0\sqrt{4\minus{}x^2}\,dx$

$ \sqrt{a^2\minus{}u^2}\Rightarrow u\equal{}x\equal{}2\sin\theta$
$ \Leftrightarrow A_n\equal{}4\int_0^n \cos{\theta}\,d\theta$

$ n\equal{}2\sin\theta\Rightarrow \theta(n)\equal{}\arcsin{\frac{n}{2}}$
$ \theta[2,0]\equal{}\{y\in\mathbb{Y}: y\equal{}f(x)\}\equal{}\mathbb{R}\equal{}[\minus{}\frac{\pi}{2},\frac{\pi}{2}]$

Thus, $ 4\int_0^{\theta(n)}\cos\theta\,d\theta \equal{}\boxed{A_n\equal{}4\sin{(\theta(n))}}$

You just wrote that $ 2\int_0^b \sqrt{4\minus{}x^2}\,dx\equal{}2b$, and the area of this semicircle is $ 4$. That's obviously ridiculous.

Wow that is, beyond doubt, the stupidest thing I have [b]ever[/b] done.  Not only is it not enough that $ f(f^{\minus{}1}(x))\equal{}x$ but $ \sin{x}$ doesn't even return increments of $ \pi$!

I should have at least seen that ahhhhhh.  :oops:

There's another shortcut available, and I strongly encourage my calculus students to use this one, as well.

First, go ahead and make the trigonometric substitution. Let $ x\equal{}2\sin\theta.$ (jmerry explained the practical motivation for using the sine substitution instead of the cosine substitution - not that the cosine is necessarily wrong, but that there are more mistakes that can be made that way.) Do pay attention to the limits of integration and change them appropriately. We get that the integral equals
\[ \int_0^{\frac{\pi}2}8\cos^2\theta\,d\theta.\]
Now I quote the result that I like to call the "yin-yang theorem":
\[ \int_0^{\frac{\pi}2}\cos^2\theta\,d\theta\equal{} \int_0^{\frac{\pi}2}\sin^2\theta\,d\theta\equal{}\frac{\pi}4.\]
More generally, for $ a\ne 0$ and $ j,k$ integers

$ \int_{\frac{j\pi}{2a}}^{\frac{k\pi}{2a}}\cos^2(ax)\,dx\equal{}
\int_{\frac{j\pi}{2a}}^{\frac{k\pi}{2a}}\sin^2(ax)\,dx\equal{}
\frac12\cdot\frac{(k\minus{}j)\pi}{2a}.$

Or to put it in a form in which it can be remembered: the definite integral of either $ \sin^2$ or $ \cos^2$ (of any linear expression) is "half the box", as long as the interval of integration starts and stops at ends of quadrants. I do find that this comes up often enough to be worth remembering.

If the interval of integration starts or stops somewhere other than the ends of quadrants, then it's not so simple, and you need the half-angle formula or something like that.

The "yin-yang" name comes from the picture for the simple symmetry argument used to prove that $ \int_0^{\frac{\pi}2}\cos^2\theta\,d\theta\equal{} \int_0^{\frac{\pi}2}\sin^2\theta\,d\theta$ and that the sum of those two integrals is the "whole box."

Thanks to all for the help, the benefit of using sin has been very instructive.

But, I still don't understand how we get the arcsin function. igiul explained but I'm not sure how he got $ n\equal{}2\sin{\theta}$

All help continues to be appreciated!

For integrals involving $ \sqrt{a^2\minus{}u^2}$ where $ u\equal{}f(x)$, it is customary to allow $ u\equal{}a\sin{\theta}$, which makes the expression equal to $ (a\sin{\theta})^{\prime} \equal{} a\cos{\theta}$.

\[ \therefore A_n \equal{}8\int^{\frac{\pi}{2}}_0 \cos^2{\theta}\,d\theta\]

By Yin-Yang, we find the integral to be $ 2\pi$.

As I stated in my earlier post, the domain for $ \theta(n)$ is $ [\minus{}\frac{\pi}{2},\frac{\pi}{2}]$. This is simply because inverse trigonometric functions are periodic too, but instead of outputting a single value for multiple arguments, it will output multiple values for a single argument (For instance, $ \arcsin{\frac{\sqrt{3}}{2}}$ has exactly one solution on $ [\minus{}\frac{\pi}{2},\frac{\pi}{2}]$).
This property directly affects us when converting the limits of integration: We equate $ x$ to $ \alpha$ and $ \beta$, then solve for $ \beta, \alpha$ in terms of $ x$. Because we redefined $ x$ as a trigonometric function,  an inverse trig function is both necessary and sufficient to the isolation of $ \theta$. Finally, to avoid the use of similar, but unequal values of theta, we restrict the domain of $ \theta$ to the optimal domain (closest to zero) for which $ \theta(x)$ passes the horizontal line test.

[i]***I think that method works, hope this post is better than last![/i]
 :)


real analysis
calculus
integration
function
real analysis solved

$f(x)$ is differentiable on $\mathbb{R}$. If $f'(x)$ is bounded on $[a,b]$, then $f'\in L([a,b])$ and
$\int_{[a,b]} f'(x) dx = f(b) - f(a)$

(please do not use absolute continuous etc)

The thing is I know next to nothing about measure theory, but I'm learning.. :) Doesn't Dominated Convergence work out just fine for this one? We can apply it for functions of the type $f_n=n(f(x+\frac 1n)-f(x))$, with some small corrections to account for the fact that $f_n$ is only defined for [$a,b-\frac 1n$].

hmm...it seems right...
the only thing left is to prove that $f'$ is measurable function, which is not difficult..

[quote="grobber"]The fact that the limit of a dominated sequence of $L^1$ functions is also $L^1$ is part of the Dominated Convergence Theorem.[/quote]
eh...yes...


[quote]
It really wouldnt be a tournament in my opinion, if its just seeing who can get the (lowest)EDIT:Highest rating...Since you aren't trying to win, I would say you don't need CTC approval.
[/quote]

Ok so yeah I'm making a tourney where you try to (lose) EDIT:gain the most points. Uhm just sign up and I'll give details l8r.

edit:

Yeah I decided to change it.

btw the quote is mewto55555

So are we trying to gain or lose points?

I changed it, now we are trying to GAIN points

WHAT???

I believe that this is far less nonsense than your post thank you very much!

lol

Unless you can make it CTC certified, its not going to work.

not a lot will join if it ain't ctc certified

Its not a tournament, more of a competition. I cant certify it unless I hear the rules.

Try to gain the most points.  Pi has a HUGE advantage.

ok pming the rules to mewto ...

OK GUYZ THIS IS HOW IT IS GOING TO WORK

[code]
Ok like people will start at like 8 in the morning and play and they must meet at 1 (so i can see the scores). The ones with the highest 'Today's Rating' gets to move on. Repeat that to find the champion.

Like if we start with 10

Day 1: Take Top 6
Day 2: Take Top 3
Day 3: THE CHAMPION IS ANNOUNCED

so your rating does not matter.
[/code]

No; Today rating is only your rating for that day.

[quote="superllamel"]Does today rating take into account your current rating?

Because if it does some people *cough* pi *cough* will have a huge advantage.[/quote]

I am not participating. I'm just hosting it.

[quote="PI-Dimension"]OK

MEWTO55555 SAYS ITS LEGIT!!!

OK I AM GOING TO START A NEW TOPIC JUST FOR SIGNING UP

IF YOU HAVE ANY QUESTIONS OR COMMENTS OR CONCERNS, SAY IT HERE OR PM ME[/quote]

Yeah, I have a concern.  You don't have to make a new topic just for sign-ups.


real analysis
integration
calculus
analytic geometry
real analysis unsolved

my book said solution to poisson eqaution: $ \Delta u\equal{}f$ in $ R^n$ (n>=3) is just the convolution of fundamental solution with $ f$, $ |x|^{2\minus{}n}*f\equal{}\int_{R^n}|y|^{2\minus{}n}f(x\minus{}y)dy$, and this solution is $ C^2(R^n)$.
Does this convolution even exist a.e?  because $ |y|^{2\minus{}n}f(x\minus{}y)$ is singular at y=0, it is not integrable w.r.t $ y$ !

The integral $ \int_{|y|\le 1}\frac{1}{|y|^p}dy$ converges if and only if $ p<n$.
The integral $ \int_{|y|\ge 1}\frac{1}{|y|^p}dy$ converges if and only if $ p>n$.
Use polar coordinates to reduce both integrals to one-dimensional ones: you'll pick up the factor of $ |y|^{n\minus{}1}$ along the way.