group theory
abstract algebra
superior algebra
superior algebra solved

Hello again.

I have another Sylow type question that I'm trying to figure out.

The Problem:

If [tex]|G|=6[/tex], then either [tex]G[/tex] is cyclic of order 6 or [tex]G=S_3[/tex].

What (I think) I know:

By the third Sylow theorem, G has a 2-Sylow subgroup and a 3-Sylow subgroup.  Because the order of the two subgroups are primes, then the two such subgroups are cyclic. From there, my brain comes to a screaching halt.

Thanks for your help and for setting me straight.

Regards,

dogma

There is (2. or 3. Sylowian theorem, don't know which one) exactly one 3-Sylow subgroup $H$ cyclical generated by $x$.
Now choose some $y \in G-H$.
1. case: $ord(y)=6$ so $G$ is cyclic.
2. case: $ord(y)=2$ and $xy=yx$, then $x$ and $y$ commutate and $ord(xy)=6$, so $G$ is cyclic.
3. case: $ord(y)=2$ and $xy \neq yx$, thus $xy=yx^{-1}$ and $G= \{ e,x,x^{-1},y,xy,x^{-1}y \}$ is isomorphic to $S_3$ as can be seen by the isomorphism $T,T(y)=(12),T(x)=(123)$ in cycle notation.

Another (more general) solution: we call the only 3-Sylowsubgroup $H$. By the Sylow theorems it must be normal. Now take a 2-Sylowsubgroup $K$. Then we have $G=HK$. Since $H$ is normal $G$ is a semi-direct product of $H$ and $K$ with a homomorphism $\phi: K\rightarrow Aut(H)$. In our case each of $Aut(H)$ and $K$ have two elements. So, $\phi$ is the trivial homomorphism or the only isomorphism. In the first case $G$ is abelian, in the second isomorphic to $S_3$. The only abelian group of order $6$ is $\Bbb Z_6$.

Misha

Thanks Misha and ZetaX for your help.  I apologize for not getting back sooner...a long weekend  :)

Thanks and take care.

dogma


The planning committee at school has 10 members. Exactly four
of these members are teachers. A four-person subcommittee
with at least one member who is a teacher must be formed from
the members of the planning committee. How many distinct
subcommittees are possible?

Without restrictions, there are $ \binom{10}{4}$ ways to choose the committees.  However, we must subtract $ \binom{6}{4}$ (committees without a teacher), so we have $ 210 \minus{} 15$, or $ \boxed{195}$.

why couldn't we do (4 choose 1) * (6 choose 3)?

There could be two, three, or four teachers on the subcommittee.

Sorry for the revive, but I have a question.

Why can't you do $4*9*8*7$?

First, you choose one of the $4$ teachers. Then there is $9$ people left. Choose $1$ there is $8$ people left. Choose $1$ more there is $7$!

$4*9*8*7$ is $2016$.

Why does this NOT work?

There's a lot of things wrong here.

1) People aren't unique; all the teachers are the same, and all of the non-teachers are the same.  You need to take combinations, not permutations.

2) There isn't just 1 teacher; there can be 2, 3, or 4 teachers.  Also, when you multiply by 9, 8, and 7, you aren't making a distinction between teacher and non-teacher.

Does that make sense?

[quote=Mathcounts_ALT]Surely, this should work: You have 4 options for the one teacher. THEN THE REST OF THE 3 PEOPLE DO NOT MATTER AT ALL!!! So why can't you do 4*(9C3)? That = 336. To rephrase, those 9 people could be anyone, teacher or student, so long as there is already 1 of 4 teachers. Are the members distinguishable? They must be.[/quote]
[quote=RamanConjecture]Bumping this, because this was what I thought as well.[/quote]

The issue with this argument: what if you have $2$ teachers, say, $A$ and $B$? Then, you'll count the committee $A, B, c, d$ (where $c$ and $d$ are arbitrary) twice: once with $A$ and once with $B$. Does this make sense?

why can't be C(10, 1) * C(10, 3) + C(10, 2) * C(10, 2) + C(10, 3) * C(10, 1) + 1? is just taken 1 student and 3 teachers, 2 students and 2 teachers, 3 teachers and 1 student, and just 4 teachers and a committe of 4 teachers respectively


Let $ x,y,z>0$ such that $ x^3\plus{}y^3\plus{}z^3\equal{}3$. Prove that $ \frac{xy}{z}\plus{}\frac{yz}{x}\plus{}\frac{zx}{y}\geq3$.

[quote="magicgeometry"]Let $ x,y,z > 0$ such that $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$. Prove that $ \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y}\geq3$.[/quote]
This inequality is not always true but it is true for one of the following conditions
1. $ x,y,z$ are the side lengths of a  triangle.
2. $ x^3y^3\plus{}y^3z^3\plus{}z^3x^3 \ge xyz(x^3\plus{}y^3\plus{}z^3)$
:)


1. Let P equal the sum of the measures of the interiors angles of a convex polygon of n + 4 sides. Let Q equal the sum of the measures of the interior angles of a convex polygon of n - 4 sides. If n is an integer greater than 6, what is the value of P - Q. in degrees?

2. In how many different ways can 10 be written as a sum using only the numbers 1, 3, and 5? Note that 3+3+3+1 and 1+3+3+3 are two different ways that should be included.

3. How many integers from 1 to 1000, inclusive, are divisible by neither 2 nor 5?

wait, isn't this from the 09-10 MATHCOUNTS handbook?

1. Since the sum of the interior angles of an n-gon is 180(n-2),
P=180(n+2),
Q=180(n-6)

P-Q=180*8=1440

3) Since by making a ven-diagram we get that there are 500 numbers divisible by 2 and 200 divisible by 5 and there are also 100 divisible by the LCM(2,5)=10. so we get that the total number of numbers divisible by 2 and 5 are 200+500-100 which is 600 and since we need to find the numbers not divisible by 2 and 5 we get 1000-600=400

2)Well we first use case work......
sum 10: 5,5= 1 way
             5,3,1,1= 4!/(2!) the 2! is from the two 1's that are the same 4! from the 4 total numbers listed
             5,1,1,1,1,1= 6!/(5!) the 5! is also from the 5 consecutive 1's 6! from the 6 total numbers listed
             3,3,3,1= 4!/(3!) 3! from the 3 consecutive 3's
             3,3,1,1,1,1= 6!/(4!*2!)  4! from the 4 consecutive 1's and 2! from the 2 consecutive 3's
             3,1,1,1,1,1,1,1 = 8!/(7!)
             1,1,1,1,1,1,1,1,1,1= 1
             From the cases above we get a total of 1+6+6+4+15+8+1=41 total ways


Of the 210 juniors at XYZ High School, 120 are enrolled in history and 30 are enrolled in art.  If the total of 100 juniors are enrolled in ONLY one of the two courses, find the number of juniors who are NOT enrolled in EITHER course.

NOTE:  I worked around this question through and through but totally in vain.

The book's answer is 85.

How do I get 85?

[hide]
PIE
Say x is the number of people enrolled in both classes.
Then $120+30-2x=100$.  It's 2x because we have to take out the people who take both classes.
So $x=25$.
The number of people who take at least one of those 2 classes, then, is $100+x=125$.  There are 210 students, so $210-125=85$ don't take either class.[/hide]


You can now access Round 1 problems [url=http://www.artofproblemsolving.com/USAMTS/AoPS_U_Problems.php]here[/url].

Also, the solution forms and LaTeX templates for those who choose to write solutions in LaTeX can be found by [url=http://www.artofproblemsolving.com/USAMTS/AoPS_U_MyForms.php]clicking here[/url].  Note that it is not mandatory to write your solutions in LaTeX.

As always, please do not discuss the problems until well after the round is over.  The solutions submission deadline is October 4.  Please do not discuss the problems until after October 11.

That was fast.  You caught me off guard :-)

Yay  :)   I didn't learn about USAMTS until about half-way through Year 15 - Round 3 so I've been waiting for months for this year to start up.

I believe that LaTeX is very much preferred.  It says on one page that a PDF is acceptable.  Just don't do it in word... they wont take it if you do that.

[quote="SingTheSorrow"]I don't know what is wrong with my computer....but I want to use LaTeX for my solutions...and that file that tells you how to use LaTeX for the USAMTS won't work on my computer...Uh, help?  Can I just make a .pdf file?[/quote]

You can just send in a PDF file if you like.  What do you mean by 'the file that tells you how to use LaTeX for the USAMTS won't work'?

[quote="Demon"]If we use [tex]\LaTeX[/tex] do we have to use the template?[/quote]

It's pretty strongly preferred, but you don't have to, as long as all the information that is on the solution form is on each page of your submission.

[quote]Also, the solution forms and LaTeX templates for those who choose to write solutions in LaTeX can be [b]found by clicking here[/b]. Note that it is not mandatory to write your solutions in LaTeX. [/quote]

Somehow, my computer messes up when I try to download that stuff so I can use Latex...I think I'm just going to send a .pdf file.  Seems the most easiest.

I can't seem to find my answer on this forum!  Anyway, what score or what percentile do you have to get/be in, so that you are invited to take the AIME?

[quote="SingTheSorrow"]Let me explain this problem a little clearer.   :P   Whenever I download the "temples" that I need to have so that if I do a really good solution, it will be posted, I try to open them, and it says that "an error occurred on Internet Explorer" and it needs to close.[/quote]

The answer to that question, my friend, is to use Mozilla.

As for AIME qualification, the past few years the benchmark has been 68 after three rounds.  It might change, though.

Mr. Rusczyk - do you mind if we remove the comments (e.g. "%% If you have any problems, contact usamts@usamts.org. %%") from the solutions file?

[quote="SingTheSorrow"]Wow, Mozilla rocks!  It's so fast - even with my so dialup.  Wow, thanks so much.  Let me see if the downloads work now!  (Whoa, 68 is 7 points from being perfect!  Hopefully it's lowered....a little.)

EDIT:  Okay, I d/led Mozilla, and um, it says the temples are IE files...now when I try to open the up with Mozilla they say that it's an invalid WIN 32 file or something..then I try to open up the file with Word, Wordperfect, Adobe Reader, and the like, and  they all don't like the file.  Uh, is something wrong with my computer or something?  Um, help?   :?[/quote]

They are [tex]\TeX[/tex] files. They shouldn't be opened with internet browsers. try using TexnicCenter or some other [tex]\TeX[/tex] programm.

[quote="mathfanatic"]

Mr. Rusczyk - do you mind if we remove the comments (e.g. "%% If you have any problems, contact usamts@usamts.org. %%") from the solutions file?[/quote]

No problem.

[quote="ComplexZeta"][quote="DPopov"]What does everyone think of the difficulty of the 1st round problems for USAMTS? (Which ones do u find hard/easy)

P.S. Please don't put any hints/answers in your replies[/quote]

I think people should not answer this question until the round is over.[/quote]

Correct; please do not even discuss the test in such general terms until after this round is over.

Danny, I assume that when you said 'temples', you mean the templates from the USAMTS --> Forms section of the website.  In that case, Demon is right - you can't open them with any browser, even Mozilla.  However, you can read them with Notepad; they're plain text.  I recommend TeXnicCenter personally, but Notepad should be fine.

... and you'll need MiKTeX as well.  Come to think of it, just read the whole LaTeX section of this site.  That's what it's here for.

Are we allowed to include additional packages, mainly the ones to draw pictures right in $\text{\TeX}$.

I'm thinking about using the pgf package and maybe some of the packages that go with it, (pgfarrows, pgfnodes and pgfshade). These seem to be the simplest pdf drawing packages.

Ideally you'd find a way to do it without any other packages, but we won't penalize you if you find you have to use other packages.

The xy package is very good for drawing commutative diagrams and stuff like that.

If we want to send a TeX file, must we attach the images along with the TeX file and include them as if they were in the correct folder?

Can you not publish pdf submissions or convert pdf to LaTeX so that you can publish (assuming we used LaTeX to make the pdf)?

Finally, not to press the issue, but since there are very few published solutions, why can't someone convert only those that get published?

Thanks

[quote="rrusczyk"]It would be much easier for us if we already have the images rather than having to reproduce them ourselves - this is why we ask that you send the external images ourselves.  It's not impossible for us to do the conversion, just much simpler if we have you send an email with 6 attachments (or whatever) which we save in a folder and can easily retrieve if needed.  It shouldn't be too taxing on your end to attach the extra files with most mailers, I hope.[/quote]

As the person who typeset the selected solutions for rounds 3 and 4 of year 13, all rounds of year 14, and round 1 of year 15, I know that transcribing the students' solutions and diagrams from paper into software is thirty hours work per round. It will be much faster with you providing the LaTeX files, but if the Art of Problem Solving people had to reproduce your diagrams from scratch, that would be many hours of work again. You know how long it took you to create your diagrams: it would take them the same amount of time. We do not want to overwork these nice people who volunteered to host and manage the USAMTS.

Erin Schram
USAMTS grader


1. How do you become a moderator, or even moderate something?
2. Where can you find a earch place for other questions such as these?

Please answer (I'm new here)


let numbers a,b,c,X,Y,Z,k are all positive satisfying a+X=b+Y=c+Z=k,
show a*Y+b*Z+c*X < k^2.
this inequality is very untight ... (hint: dont use inequalities ... )

Solution 1:[tex]k^3=(a+x)(b+y)(c+z)\geq xbc+ayc+abz+ayz+xbz+cyz \\ =ay(c+z)+bz(a+x)+cx(b+y)=k(ay+bz+cx)[/tex]

Solution 2:WLOG Let [tex]a=\max (a,b,c,x,y,z)[/tex], we have [tex]ay+bz+cx\leq ay+ba+kx=ak+kx=k^2[/tex]


$ (\forall)n\ge 2\ ,\ n\in \mathbb{N}$ , prove the identity :

$ [\sqrt{1}]\plus{}[\sqrt{2}]\plus{}...\plus{}[\sqrt{n^2\minus{}1}]\plus{}1^2\plus{}2^2\plus{}...\plus{}n^2\equal{}n^3$

[hide]
In $ xy$ plane,consider the function $ y \equal{} x^2$,note that whose inverse function is $ y \equal{} \sqrt {x}$ for $ x>0$.

Therefore

$ 1^2 \plus{} 2^2 \plus{} \cdots \plus{} n^2$ is equal to the number of  lattice points $ (x,y)$ such that $ y\le x^2$ and $ 0 < x\le n,y > 0$ and

,$ [\sqrt {1}] \plus{} [\sqrt {2}] \plus{} ... \plus{} [\sqrt {n^2 \minus{} 1}] \plus{} [\sqrt {n^2}]$ is equal to the number of  lattice points $ (x,y)$ such that $ x\le \sqrt {y}$ and $ 0 < y\le n^2,x > 0$.

notice lattice points on the curve $ y \equal{} x^2$ are double counted,thus

$ [\sqrt {1}] \plus{} [\sqrt {2}] \plus{} ... \plus{} [\sqrt {n^2 \minus{} 1}] \plus{} [\sqrt {n^2}] \plus{} 1^2 \plus{} 2^2 \plus{} ... \plus{} n^2 \minus{} n \equal{} [\sqrt {1}] \plus{} [\sqrt {2}] \plus{} ... \plus{} [\sqrt {n^2 \minus{} 1}] \plus{} 1^2 \plus{} 2^2 \plus{} ... \plus{} n^2$
is equal to the number of lattice points $ (x,y)$ such that $ 0 < x\le n$ and $ 0 < y\le n^2$,which is $ n^3$.[/hide]

Wow ... , your solution is very nice , Anavel . I never thought to use lattice points .

In $ [\sqrt{1}]\plus{}[\sqrt{2}]\plus{}...\plus{}[\sqrt{n^2\minus{}1}]$,

$ [\sqrt{i^2}],[\sqrt{i^2 \plus{}1}],[\sqrt{i^2 \plus{}2}],...,[\sqrt{(i\plus{}1)^2\minus{}1}]$ all yield i.

The number of such terms are $ (i\plus{}1)^2 \minus{} i^2 \equal{} 2i \plus{}1$
Therfore,$ [\sqrt{1}]\plus{}[\sqrt{2}]\plus{}...\plus{}[\sqrt{n^2\minus{}1}] \equal{} \sum_{i\equal{}1}^{n\minus{}1}i(2i\plus{}1)$

$ 1^2\plus{}2^2\plus{}...\plus{}n^2 \equal{} \frac{n(n\plus{}1)}{2}$

Adding both,we get $ n^3$

Now try a difficult one :

Let be $ k\in \mathbb{N}\ ,\ k\ge 2$ . $ (\forall)n\in \mathbb{N}\ ,\ n\ge 2$ , show that :

$ [\sqrt[k]{1}]\plus{}[\sqrt[k]{2}]\plus{}[\sqrt[k]{3}]\plus{}...\plus{}[\sqrt[k]{n^k\minus{}1}]\equal{}n^{k\plus{}1}\minus{}(1^k\plus{}2^k\plus{}...\plus{}n^k)$

This is like the above sum.

$ LHS\equal{}\sum_{i\equal{}1}^{n\minus{}1}i[ (i\plus{}1)^k \minus{} i^k ]$       (Reason same as above problem's)

      $ \equal{}\sum_{i\equal{}1}^{n\minus{}1}i [ (i\plus{}1)^{k\minus{}1} \plus{} (i\plus{}1)^{k\minus{}2}*i \plus{} ... \plus{} (i\plus{}1)*i^{k\minus{}2} \plus{} i^{k\minus{}1} ]$

      $ \equal{}\sum_{i\equal{}1}^{n\minus{}1} (i\plus{}1)^{k\minus{}1}*i \plus{} (i\plus{}1)^{k\minus{}2}*i^2 \plus{} ... \plus{} (i\plus{}1)*i^{k\minus{}1} \plus{} i^{k}$

      $ \equal{}\sum_{i\equal{}1}^{n\minus{}1} (i\plus{}1)^{k\plus{}1} \minus{} i^{k\plus{}1} \minus{} (i\plus{}1)^k$

     Telescopically cancelling,

      $ \equal{}n^{k\plus{}1} \minus{} 1 \minus{} \sum_{i\equal{}1}^{n\minus{}1}(i\plus{}1)^k$

      $ \equal{}n^{k\plus{}1} \minus{} (1^k \plus{} 2^k \plus{}....\plus{}n^k)$

      $ \equal{}RHS$


vector
linear algebra
linear algebra unsolved

Suppose that $\omega,\omega_{1},\omega_{2},\dots,\omega_{n}\in\mathbb R^{n}$ that for each $v\in\mathbb R^{n}$ that $v.\omega>0$ then there exist $i$ that $v.\omega_{i}>0$. Prove that there exist $\lambda_{1},\lambda_{2},\dots,\lambda_{n}\geq0$ that $\omega=\lambda_{1}\omega_{1}+\lambda_{2}\omega_{2}+\dots+\lambda_{n}\omega_{n}$

Prove first of all that the set of linear combinations with nonnegative coefficients of these vectors is closed, then use projection on closed convex sets.


It's very innocent question; but...

        Are there infinitely many primes of the form n^2+1?

It's an unsolved problem, and this is definitely not the right section for the question. It should be somewhere in Number Theory.

Alex! "if you wonna drawn then you must dive into a deep sea" and i guess that i dived into a deep sea...  Thanx for your warning!
                                                                fatih sahin


inequalities
trigonometry
inequalities solved

I don't know if this is a right direction.
  Let x = max{x,y,z}.
  If x>= y+z then -x^2 +y^2 +z^2 <= -2yz <0
      => the result.
  If x< y+z then x,y,z are sides of a triangle ABC.
  And we will have a trigonometry inequality.
  Unfortunately, I have no paper now so I can't prove it. I 'll post my solution (if I can solve it) later.  :D

well you guys might wanna check [url=http://www.mathlinks.ro/viewtopic.php?p=3045]THIS[/url] out first :)

[quote="harazi"]Prove that for all positive reals x,y,z we have:
    (x+y+z) 2 (x+y-z) 2 (y+z-x) 2 (z+x-y) 2 >=3( \sum x 2 )(x 2 +y 2 -z 2 )(y 2 +z 2 -x 2 )(x 2 +z 2 -y 2 ).[/quote]

Dear friends!

I just found that this inequality is exactly Schur for x^2, y^2, z^2 and r = 2, i.e
    Sum x^8 + Sum x^4y^2z^2 >= 2Sum x^6y^2
where the sum are symmetric sum (6 terms each)

Namdung

Exactly! It can also be reduced easily to the famous Hadwiger-Finsler inequality 2 \sum ab- \sum a 2  \geq 4 \sqrt 3S


A solid right circular cone of altitude h = 304.6 mm and radius of base r = 76.2 mm has its centre of gravity C on its geometric axis at the distance h/4 = 76.2 mm  :?: above the base. This cone rests on an inclined plane AB , which makes an angle of 30 degrees with the horizontal and for which the coefficent of friction is [img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/3/a/4/3a4e56595df1d02f21e5c3ff7b0e9d80921858ff.gif[/img] = 0.5 . A horizontal force P is applied on the vertex O of the cone and acts in the vertical plane of the figure as shown. Find the maximum and minimum values of P consistent with equilibrium of the cone if weight W = 44.5 N .

Ans : P max = 20.52 N and P min  = 2.67 N

For translational purposes, I think the best components to consider are the normal and parallel to the inclined plan.

For minimum P, the friction force points upwards an its modulus is the normal component (sum of the weight's and P's components) multiplied by the friction coefficient.

When P increases, it's not hard to see that the limit of rotational equilibrium occurs sooner than the translational one., so you have to balance the torques and the best pole is the upper edge. As long as the radius and the gravity center's height are taken rigorously as h/4 it doesn't matter it's numerical value.


$a,b,c>0$.Find the minimun of:

$(\frac{a}{b+c})^k+(\frac{b}{a+c})^k+(\frac{c}{a+b})^k$

If  $k$ is apositive real and $k \ge \frac{1}{3}$

I belive it have the solution if $k\le \frac{1}{3}$,But now I can't solve it.

I think that :$\sum(\frac{a}{b+c})^k\ge $min{3/(2^k);2}

Please,Explain you answer.My result isn't that.

$\sum{(\frac{a}{b+c})^k}>=3/(2^k) $ is rignt with k>=log_2 3-1 and >=2 with other case!

the minimum is $ \frac{3}{2^k} $ is just true for $ k \geq 2/3 $
 (the solution for $ k \geq 2/3 $ was posted on M&Y )

What should we do if $\frac{1}{3} \le k \le \frac{2}{3}$
I want to show the complete Solution.

It was discussed also here

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%28%5C%5Cfrac%7Ba%7D%7Bb%2Bc%7D%29%5Ek&t=12763

I Wait other people post their solution by some days.

[quote="hungkhtn"]Ok,I've just complete this problem:

$\sum(\frac{a}{b+c})^k \ge max(2,\frac{3}{ 2^k})$

For all $k\ge 0$

I think it's better.[/quote]

do you mean $\sum(\frac{a}{b+c})^k \ge min(2,\frac{3}{ 2^k})$

I couldn't prove it.If you know the solution please post it.
My idea was to take $a+b+c=1$ and to show that $\left(\frac{a}{1-a}\right)^k\ge 2a$.The equality case holds for $a=1/2$ or $a=1/3$.(Of course here $k=log_23-1$).But it's not always true.
Betwean I was very busy in these days. :(

[quote="hungkhtn"]My solution use  Missing varible.
Try to prove that if $k \ge 1/3$ and $a\ge b \ge c$

$f(a,b,c)=\sum(\frac{a}{b+c})^k$ then

$f(a,b,c) \ge f(\frac{a+b}{2},\frac{a+b}{2},c)$[/quote]

can you show me the proof of this?  :blush:

Ok,It's not easy but it's not so difficult.Let $t=\frac{a+b}{2},u=\frac{a-b}{2}$ we have $f'(u) \ge 0$

-------
But I think better I don't post the solution by a near day.

I'd like to post my solution (though I don't know if there are still some mistakes in it), but it's not short.
When I worked it out I didn't know "Nesbitt" so I just called it "an inequality about $a,b,c,\alpha$".


I am struggling to explain why 2,6-dimethylbenzoic acid can't be esterified in methanol with acid but CAN be esterified by treatment with diazomethane or by dissolving it in sulphuric acid and then pouring into anhydrous methanol.  Could it be due to neighbouring group participation?

Any help or pointers gratefully received - thank you

Snowdrop

First, welcome to this forum Snowdrop. Regarding your question, start by noting that the carbonyl carbon of 2,6-dimethylbenzoic acid is strongly hindered due to the proximity of the two methyl groups. This inhibits the common esterification with alcohol and acid, because its mechanism involves the nucleophilic attack at the carbonyl carbon. On the other hand, the mechanism with diazomethane is completely different since it does not involve carbonyl carbon attack. Can you propose a mechanism Snowdrop?

I suspect you can start off by drawing the resonance structure of diazomethane.

[hide="Possible mechanism"]

First consider the resonance structure of diazomethane. Note one of structure has the negative charge residing on the carbon atom.

The carboxylic acid is deprotonated to give $ \ce{CH3}\minus{}\ce{N\plus{}} \equiv N$ and a carboxylate ion. Now we can have a $ S_N2$ (or quasi-$ S_N2$) mechanism to get the desired ester.

Notice this way no species attack the carbonyl carbon hence steric hindrance plays less of a role.[/hide]

Yes, that's right, but let Snowdrop try to solve this by himself.

btw, 2,6 dimethyl benzoic acid can be esterified under more drastic conditions but the mechanism wud b different. it wouldnt be the ususal nucleophilic addition elimination type. it wud be nucleophilic substitution by formation of acylium ion to reduce the steric hindrance which wud otherwise be present in the tetrahedral intermediate of esterification

Another thing I read just now is that the preparation of Methyl Esters using diazomethane can yield esters in such purity that subsequent purification is unnecessary.

Thanks guys!  I am a lot further foward now!


I don't know where to post this question but i am a high school student and i can't find the answer for this question so anyone can help me
(1*2)+(2*3)+(3*4)+...+(99*100)=?
And can i get the general formula for this question? Thanks

$ \sum_{k\equal{}1}^n k(k\plus{}1)\equal{} \sum_{k\equal{}1}^n k^2\plus{}\sum_{k\equal{}1}^n k\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}6\plus{}\frac{n(n\plus{}1)}2.$

See [url=http://mathworld.wolfram.com/ChristmasStockingTheorem.html]Hockey Stick / Christmas Stocking[/url].  :)


Determine all pairs $ (x, y)$ of positive integers such that $ x^2y \plus{} x \plus{} y$ is
divisible by $ xy^2 \plus{} y \plus{} 7.$

ειναι,
$ xy^2 \plus{} y \plus{} 7 | y(x^2y \plus{} x \plus{} y) \minus{} x(xy^2 \plus{} y \plus{} 7) \equal{} y^2 \minus{} 7x$

εαν $ y^2 > 7x$ τοτε $ xy^2 \plus{} y \plus{} 7 > 7x^2 \plus{} y \plus{} 7 \equal{} u \equal{} xy^2 \plus{} y \plus{} 7 \minus{} x(y^2 \minus{} 7x) > 0$

αρα ο αριθμος u ειναι θετικος  αριθμος ,μικροτερος του $ xy^2 \plus{} y \plus{} 7$ ο οποιος ομως τον διαρει. Ατοπο.

εαν $ y^2 \equal{} 7x$ τοτε την λυνουμε και βρισκουμε τις εξης λυσεις, $ (x,y) \equal{} (7k^2,7k)$ οπου k στο Ζ*+. H επαληθευση μας πιστοποιει οτι ειναι δεκτες.

εαν $ y^2 < 7x$ τοτε $ xy^2 \plus{} y \plus{} 7 \leq 7x \minus{} y^2 \leftrightarrow (7 \minus{} y^2)x > y^2 \plus{} y \plus{} 7 > 0$
σχεση που ευκολα μας περιοριζει τις τιμες του y σε 1 ή 2.

$ y \equal{} 1, \frac {x^2y \plus{} x \plus{} y}{xy^2 \plus{} y \plus{} 7} \equal{} \frac { x^2 \plus{} x \plus{} 1}{x \plus{} 8} \equal{} \frac { 57}{x \plus{} 8} \plus{} x \minus{} 7$ και για να ειναι ακεραιος x=49 ή x=11

$ y \equal{} 2$,$ \frac {x^2y \plus{} x \plus{} y}{xy^2 \plus{} y \plus{} 7} \equal{} x \minus{} \frac {7x \minus{} 4}{4x \plus{} 9}$

αρα$ 7x \minus{} 4 \geq k(4x \plus{} 9) \leftrightarrow k \equal{} 1$ , για x>0 ακεραιο που ομως ως ισοτητα η σχεση μας οδηγει παλι σε ατοπο.

Συνεπως ολες οι λυσεις,
$ (x,y) \equal{} (7k^2,7k), ( 49,1) , (11,1)$    :)


If sinx = -sqrt{2}/2 and cosx = sqrt{2}/2, find the measure of angle x.

Symmetry, almost all of the problems you post are much easier than what you normally find in "Getting Started." Is this your homework or what?

Too many topics from the same person... Looks like homework  :P


If the plane is partitioned into two disjoint subsets, show that one of the subsets contains three points forming the vertices of an equilateral triangle.

I would like to know what others make of this problem.

[hide="Hint"]
Call the subsets "black" and "white". Look at a regular hexagon, along with its center. Assume the center is black. What can you say about the colors of the vertices? Can you add another small triangle to finish this off?
[/hide]

I had a solution, I just didn't know whether there were any other ways of doing it.

[hide]Call the subsets S and T.  If |S|<2 then it's clear that T has the triangle, otherwise choose two points in S, say A and F.  Some nearby points in the plane (which happen to make a regular hexagon) are B,C,D,E,F,G (The midpoint of AF) and center O, as in the image.  If C is in S then the triangle ACF is made, otherwise C is in T.

Case:  O is in S.  In that case, if either one of G or E is in S then the triangle is GOF or OEF.  Otherwise G,C,E are all in T and GCE is the triangle.

Case:  O is in T.  In that case, if either one of B or D is in T then the triangle is BCO or CDO.  Otherwise B,D,F are all in S, leaving the triangle BDF.[/hide]

If $ G$ and $ E$ are not in $ \textbf{S}$, that does not mean that $ C$ is in $ \textbf{T}$. ... Right?  :huh:

[quote="CornFlakes"]If C is in S then the triangle ACF is made, otherwise C is in T.[/quote]

From this point, we are working under the assumption that C is in T.


Slove the system of equations
$ x^3\plus{}y^3\plus{}x^2(y\plus{}z)\equal{}xyz\plus{}14$
$ y^3\plus{}z^3\plus{}y^2(x\plus{}z)\equal{}xyz\minus{}21$
$ z^3\plus{}x^3\plus{}z^2(x\plus{}y)\equal{}xyz\plus{}7$

[quote="chien than"]Slove the system of equations
$ x^3 \plus{} y^3 \plus{} x^2(y \plus{} z) \equal{} xyz \plus{} 14$ (1)
$ y^3 \plus{} z^3 \plus{} y^2(x \plus{} z) \equal{} xyz \minus{} 21$   (2)
$ z^3 \plus{} x^3 \plus{} z^2(x \plus{} y) \equal{} xyz \plus{} 7$    (3)[/quote]
$ (1)\plus{}(2)\plus{}(3) <\equal{}> p^3 \minus{} 3pq \plus{} 3r \plus{} p(p^2\minus{}q) \equal{} 3r <\equal{}> p(p^2\minus{}2q) \equal{} 0 <\equal{}> x\plus{}y\plus{}z \equal{} 0$ or  x=y=z
(p = x+y+z; q = xy+yz+zx ; r =xyz)


What is the area of a regular hexagon with sides 1 in. long?

[quote="E^(pi*i)=-1"][hide]
We can dissect the hexagon into 6 equilateral triangles with side length 1 in.  So $6(\frac{\sqrt{3}}{4})=\frac{3\sqrt{3}}{2}$ square inches.
[/hide][/quote]
This is where the formula [hide]$\frac{s^{2}\sqrt{3}}{4}$ [/hide]


quadratics algebra polynomial Gauss Euler logarithms trigonometry

imaginary numbers:
the square root of -1 is i
so the square root of -4 is 2i
real numbers:
any negative or positive number line numbers

complex numbers are imaginary numbers added with real numbers
for example
(i+1)
(6-3i)
( :sqrt: 2-5i)

comp. numbers can be multiplied
(i+1)(i-1)=-2
ii+1i-1i-1
-1-1=-2

It's in the AoPS book :D. Only reason I remember now.

There is also another term, pure imagenary. Pure imaginary is when the $a$ part of $a+bi$ is 0. Real numbers are numbers where $b$ is 0. Complex numbers are when neither are 0. When both are 0 its a real number.

I believe complex numbers include real numbers and pure imaginaries.

[quote="Dr. No"][quote="math92"]can someone explain the square root of a negative number.  Can you also explain complex and imaginary numbers?[/quote]

I believe I know a thing or two about complex numbers so I'll give it a shot.

To begin with, every positive real number has a square root. (Euclid proved this way way way back when.)  Second, square roots themselves are always positive. In Algebra 2 one is accustomed to getting answers to quadratic equations (polynomials whose highest exponent is "2") like: $x=\pm\sqrt{a}$.

Notice that in that formula $\sqrt{a}$ is not negative. What the "$\pm$" tells you is that $x$ could be negative (or it could be positive, either one).

But what about when $a$ is negative? such as in $a = -37$ then $\sqrt{a} =\sqrt{-37}$. The problem comes because no real number when multiplied by itself results in a negative number. If you multiply $(+)\cdot(+) =(+)$; likewise, $(-)\cdot(-)=(+)$.

Carl Friedrich Gauss noticed the discrepency and coined the term "imaginary number" to describe such a case. For several years, imaginary numbers remained nothing but a curiosity. I don't know who first used the notation $\imath$ to denote $\sqrt{-1}$, but Lenhard Euler discovered that $e$, where $e$ is the base of the Napierian logarithm ($e =2.7182818285...$), raised to an imaginary power, $e^{ix}=\cos x +i\sin x$.

From then on "imaginary numbers" became useful because Euler's result meant that $e^{ix}$ has essentially the same form as $A\cos x + B\sin x$. This same essential form means that one can use "complex exponentials" to describe trigonometry functions. Trigonometry functions are useful in solutions of all kinds of engineering problems, but complex exponentials are often easier to work with.

I've used the term complex exponential without explaining it, and complex numbers were one of the things you asked about. The function $f(x)= e^x$, where $e$ is the same number described above is called the exponential function. A complex number is the sum of a real number and an imaginary number. Now Euler's formula is $e^{ix}=\cos x + i\sin x$. It has a real part, $\cos x$ and an imaginary part $\sin x$. (Notice that the imaginary part is not $i\sin x$. The $\imath$ is an operator that indicates the thing it is multiplied by is the imaginary part.)  All of that said, $e^{ix}$ is a complex exponential, because it is the exponential function of an imaginary number which gives a complex number, a real number plus an imaginary number, as a result.

Now one other fact about complex numbers: The real numbers are a "field" of numbers. (A field of numbers is just a set of numbers that you can add, subtract,  and multiply and you'll never get an answer that is not itself part of the field you started with.) It turns out the complex numbers are also a field of numbers, but the field of real numbers is well ordered, meaning that one can say -3 < +14 < +25.4 for example. The field of complex numbers is [u]not[/u] well ordered. For example, the statement $5+6\imath < 67 -32\imath$ makes no sense. This is because there is no "greater than/less than" relationship for imaginary numbers.[/quote]
thanks Dr. No.  that really helped

[quote="math92"]thanks Dr. No.  that really helped[/quote]

You're welcome.

[quote="eryaman"]Dr. No, I'd just like to remind you that this is for middle schoolers, so don't get into too advanced material.  ;)[/quote]

Thanks for the reminder, eryaman.  Honestly, I know where I am "The Basics" and such, and I try to keep that in mind. However, I reason that anyone bright enough to ask the question has got to have been exposed to the material at some point already, and with regard to this topic in particular I tried to answer math92's question keeping in mind things that puzzled me when I first discovered that my teacher had "lied." One can indeed take the square root of a negative number.

[quote="mathgeek2006"]Why do we have to have "imaginary numbers" anyway?  I think that they're a pure waste of time.[/quote]

Why do we have negative numbers?  We have imaginary numbers because they are very useful when you come across an otherwise impossible quadratic.

[quote="mathgeek2006"]Why do we have to have "imaginary numbers" anyway?  I think that they're a pure waste of time.[/quote]

They are NOT a waste of time. If you get into more advanced stuff you'll see why.

[quote="4everwise"]I'm not sure if Algebra 2 material belongs in this section.[/quote]

Probably doesn't, but I don't believe in being cryptic, evasive, or deceptive when asked a direct question. The answer to "What is an imaginary number?" is the same whether that question is asked by a grade-schooler or a college sophomore.

Complex numbers are used for calculating electric currents in physics. Or something along that line.

[quote="JesusFreak197"]Hmm... I don't remember those two relating, but then, I didn't learn that part of Physics very well, so it's possible.

As others have stated, [b]complex numbers[/b] are in the form $a+bi$ and [b]imaginary numbers[/b] are simply $bi,$ where $i^2=-1.$ I've never really thought about their specific uses, since the only times I've come across them are when I'm asked a question such as "What are the roots of the equation $x^2+9=0$?" In this case, the roots come out to be $x=\pm 3i,$ since $(\pm 3i)^2=9(-1)=-9.$[/quote]

JesusFreak,

Complex numbers occur as solutions to vibration or oscillaton problems in physics.
As such they are particularly useful for solving alternating current (AC) circuit problems in electrical theory.


Let {$ x_n$} be a sequence. Suppose that there is an a > 1 such that$ \mid\(x_{k+1}-x_k\mid\le a^{-k}$
for all k$ \in$ N. Prove that $ x_n\longrightarrow x$ for some x $ \in$ R.

[quote="Golden Rule"]Let {$ x_n$} be a sequence. Suppose that there is an a > 1 such that$ \mid\(x_{k + 1} - x_k\mid\le a^{ - k}$
for all k$ \in$ N. Prove that $ x_n\longrightarrow x$ for some x $ \in$ R.[/quote]
$ |x_{m + n} - x_n|\leq \frac {1}{a^n}({\frac {1 - \frac {1}{a^m}}{1 - \frac {1}{a}})\to 0}$
So apply Cauchy law easy to check $ x_n$ is converge.

For $ m>n,$ $ |x_m\minus{}x_n|\le|x_m\minus{}x_{m\minus{}1}|\plus{}|x_{m\minus{}1}\minus{}x_{m\minus{}2}|\plus{}\cdots\plus{}|x_{n\plus{}1}\minus{}x_n|$

$ \le\sum_{k\equal{}n}^{m\minus{}1}a^{\minus{}k}\equal{}\frac{a^{\minus{}n}\minus{}a^{\minus{}m}}{1\minus{}a^{\minus{}1}}<
\frac{a^{1\minus{}n}}{a\minus{}1}.$

This quantity can be made as small as we please by choosing $ n$ large enough. Hence this is a Cauchy sequence. We now quote the theorem that every Cauchy sequence in $ \mathbb{R}$ converges.

Application: suppose that $ f: [a,b]\mapsto[a,b]$ and there is some $ K<1$ such that $ |f(x)\minus{}f(y)|\le K|x\minus{}y|$ for all $ x,y\in[a,b].$ Then there exists a unique $ c\in[a,b]$ such that $ f(c)\equal{}c.$

(Hint: choose an arbitrary $ x_0\in[a,b].$ Let $ x_{k\plus{}1}\equal{}f(x_k).$ Show that $ x_k$ satisfies an estimate very similar to the one in this problem.)


I just saw the amc test. I need to seriously cram  for it. Can someone please teach me [color=red]mod edit[/color] real quick?
Thanks!1

I guess it was a mistake deleting it.

1. You got an unauthorized view of the recent AMC test.

2. (Which is illegal)

3. (Which is also cheating)

4. In which the only punishment is to [i]not[/i] help you.

You saw the AMC test and you found out you forgot probability. Pretend it never happened. Take the test legit.

Discussion of current AMC tests are prohibited.


[quote="DuongChinh_K41SPHN"]let a,b,c be non number, no two of which are zero ,prove that
\sum \frac{1}{b+c}\ge \frac {2a}{3a^2+bc}[/quote]

1, Don't use Vietnamese to creat Subject'name

2, Use $ \text{\LaTeX}$

$ \sum \frac{1}{b+c}\ge \sum\frac {2a}{3a^2+bc}$

3, Try SOS or SS  :)


In ponicare's blog, when I roll my mouse over the Spam? button, the blog starts flashing. What happened?


what does the graph of $y^2>x^2+1$ look like?

You already asked this question, and got an answer :
http://www.mathlinks.ro/Forum/viewtopic.php?t=73045


modular arithmetic
algebra
polynomial
number theory unsolved
number theory

Find all integer numbers $ A$ . $ A$ have $ 2009$ digits , $ 3| A$ and all digits of $ A\in\{2;3;7;9\}$

let $ a,b,c,d$ be the number of digits $ 2,3,7,9$ respectively in number $ A$, so $ 0 \le a,b,c,d \le 2009$.
then $ 3|2a\plus{}3b\plus{}7c\plus{}9d$, so $ 3|2(a\minus{}b)$ so $ a \equiv b \pmod{3}$.
the last condition is $ a\plus{}b\plus{}c\plus{}d\equal{}2009$.

here are the only conditions needed: it has $ a$ digits $ 2$, $ b$ digits $ 3$, $ c$ digits $ 7$, $ d$ digits $ 9$ where $ a,b,c,d$ are non-negative integers with $ a\plus{}b\plus{}c\plus{}d\equal{}2009$ and also $ a \equiv b \pmod{3}$.

is there something wrong with the problem?

Two things:
- That should be $ a\equiv c$, not $ a\equiv b$.
- "Find all" is definitely the wrong question. "How many" makes more sense.

Define the polynomial $ p(x): \equal{}(x^2\plus{}x^3\plus{}x^7\plus{}x^9)^{2009}\equal{}\displaystyle \sum_{0 \le i \le 2009 \cdot 9}{a_ix^i}$ for some fixed integers $ a_0,a_1,\ldots,a_{2009 \cdot 9}$. Let $ S$ the required number and $ \omega$ a primitive cubit root of unit. Then $ S\equal{}\displaystyle \sum_{0 \le i \le 2009 \cdot 3}{a_{3i}}\equal{}\frac{1}{3}\left(p(1)\plus{}p(\omega)\plus{}p(\omega^3)\right)\equal{}\frac{2}{3}\left(2^{4017}\plus{}1\right)$. []


inequalities
search
inequalities unsolved

[quote="bokinhvan3"][quote="thanhnam2902"]Let $ a,b,c > 0$. Prove that: $ \sqrt [3]{\frac {a}{b \plus{} c}} \plus{} \sqrt [3]{\frac {b}{c \plus{} a}} \plus{} \sqrt [3]{\frac {c}{a \plus{} b}} > 2$[/quote]

In my idea, If a=b=c >0, we have:
$ \frac {3}{\sqrt [3]{2}} < 2$  wrong.[/quote]

Oh, no! $ \frac {3}{\sqrt [3]{2}} > 2 \Longleftrightarrow \frac{27}{2}>8 \Longleftrightarrow 27 > 16$, then it's true.

This inequality is still wrong. For instance, let $ c \rightarrow 0^{\plus{}}$. Then, it suffices to prove that

\[ \sqrt[3]{\frac{a}{b}} \plus{} \sqrt[3]{\frac{b}{a}} <2\]

The above claim is false due to AM-GM inequality.

[quote="thanhnam2902"]Who can give me a solution for this problem please! Thank you so much  :P .[/quote]

Search for it...

http://www.mathlinks.ro/viewtopic.php?t=301484
http://www.mathlinks.ro/viewtopic.php?t=301485


function
real analysis
real analysis unsolved

[quote="suadnovic"]Find all $ f\in{C(R)}$  which maps every open set to closed one.[/quote]

First find an example of such a function.  Then find a non-constant example.
Then you can probably guess the answer, and prove it.


Using exactly the following 17 characters and no others##:
6 zeroes, 2 pairs of parentheses, 2 factorial signs (exclamation points), 2 "^" symbols for exponentiation, and 3 "-" characters to be used for any combination of negation or subtraction within the same overall expression.

The expression that you are trying to minimize in size first, and then another to maximize in size later, by a different rearrangement (expected), is to be valid on a typical
TI-Graphics calculator.

WITH MY ADDED RESTRICTIONS, even if they are redundant with the calculators' systems or EVEN if the calculator recognizes the expression I am banning, are the following:

1. No zeroes are to be adjacent to each other.
2. No factorial signs are to be adjacent to each other.
3. Reminder, when you're checking the expression with mixed
    negative and subtraction symbols, different buttons are in
    use on the TI-Graphics keyboards.
4. Your expressions must have a total of the 17 characters
    mentioned at the top.
5. No subforms of the type a^b^c are allowed for the
    solutions, as shown here, because they do not have
    parentheses in them.  At least "(a^b)" or "(b^c)"
    must be present in your overall answer, if you try this route.
    (I am not stating there is an opening for that approach to
     work.)

## Amounts to no fewer, no more, and/or no substitute
symbols that otherwise would accomplish the same arithmetic operations

====----- Questions -----====-----====-----====-----====

A.  What is the minimum integer that the 17 characters
     above, and them only, can be made to equal using the
     above restrictions?

B.  What is the maximum integer that the 17 characters
     above, and them only, can be made to equal using the
     above restrictions?


point set
geometry
euclidean distance
maximization
IMO Shortlist
IMO Longlist

The $n$ points $P_1,P_2, \ldots, P_n$ are placed inside or on the boundary of a disk of radius 1 in such a way that the minimum distance $D_n$ between any two of these points has its largest possible value $D_n.$ Calculate $D_n$ for $n = 2$ to 7. and justify your answer.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


AMC USA(J)MO USAMO quadratics calculus polynomial algebra

[quote="fourierseries"]To what extent is higher math needed in the USAMO and IMO?

For example, are topics such as analytic number theory needed?

On one side, I've seen Eisenstein's Criterion used in a problem (albeit a Romanian one).  On the other, higher concepts usually do not have much applicability in the USAMO, and traps (such as having to use quadratic reciprocity to brute force problems) happen.[/quote]
Not much, I think.  There isn't [i]supposed[/i] to be any "higher math" (in the sense of anything with calculus necessary) at all.  And how is Eisenstein's Criterion analytic number theory? :huh:   The proof is fairly elementary.  (Or at least, there exists a fairly elementary proof.  Maybe there's a version for higher variables of which I'm not aware.)

[hide="The Criterion of Eisenstein"](It's usually known as Eisenstein's Criterion, but if the hide tags allow apostrophes, I don't know how to make them do it.)

The theorem states that if $P$ is a polynomial of $x$ of degree $n$ with relatively prime integer coefficients, and there exists a prime $p$ such that $p$ divides all the terms of $P$ except the leading coefficient and $p^{2}$ does not divide the constant term, then $P$ is irreducible over the integers.

We denote the coefficient of the $x^{k}$ term of the polynomial $A(x)$ as $a_{k}$ througout.

Suppose that $Q$ and $R$ are polynomials of $x$ with integer coefficients such that $P(x) = Q(x)R(x)$.

Clearly, exactly one of the constant terms of $Q$ and $R$ must be divisible by $p$.  Without loss of generality, let $R$'s leading constant term be divisible by $p$.  We now note that in general, $p_{k}= \sum_{i=0}^{k}q_{i}r_{k-i}$ (some of these terms are allowed to be zero), so, proceeding in an inductive manner, we obtain the result that all of $R$'s coefficients are divisible by $p$ (since $p \nmid q_{0}$, or we would have $p^{2}| p_{0}$), except for it's $x^{n}$ coefficient, which cannot be divisible by $p$ and hence is [b]not zero[/b].  Hence $R$ has degree $n$, and $Q$ has degree 0.  But since the coefficients of $P$ are relatively prime, we must have $Q(x) = 1$, and therefore $P$ is irreducible over the integers, Q.E.D.

This can obviously be really useful in problems of the form, "Prove that $P(x)$ is irreducible over the integers."  The extended version gives a similar condition for a polynomial having an irreducible factor (over the integers, of course) of at least a certain degree, which should not be too difficult to derive once you've understood this proof.[/hide]

[quote="Phelpedo"]Another way to think about this: Just work in the field $Z[p]$. Then $f$ becomes $kx^{n}$. If this is factorable, it must be by two polynomials, eachs whose constant terms is $0 \pmod{p}$.[/quote]
Hmm, that works, too :lol: .  I guess that the idea of "field" could be considered higher math, though, which I shall falsely cite as the reason that I gave my rather inferior proof.

only aime and above if it is needed at all. you barely need trig in the amc 10s and 12


geometry
circumcircle
perpendicular bisector
angle bisector
geometry unsolved

In convex quadrilateral $ ABCD$, $ AB \equal{} BC$ and $ AD \equal{} DC$. Point $ E$ lies on segment $ AB$ and point $ F$ lies on segment $ AD$ such that $ B$, $ E$, $ F$, $ D$ lie on a circle. Point $ P$ is such that triangles $ DPE$ and $ ADC$ are similar and the corresponding vertices are in the same orientation (clockwise or counterclockwise). Point $ Q$ is such that triangles $ BQF$ and $ ABC$ are similar and the corresponding vertices are in the same orientation. Prove that points $ A$, $ P$, $ Q$ are collinear.

The triangles AED and AFB are similar; if we take E' on AD and D' on AB such as AE' = AE and AD' = AD, then D'E' || BF and, by an appropriate dilating they may coincide, therefore the problem can be simplified :

Let ABC be a triangle, BPC and BQC isosceles triangles (such that PQ is the perpendicular bisector of BC, with P and Q opposite A w.r.t. BC) having angle PBC + angle QBC = angle BAC; then AP and AQ are simmetrical w.r.t. angle bisector of the angle BAC.

Proof of the simplified problem:
Let M be the middle of the arc BC not containing A in the circumcircle of the triangle ABC, and T a variable point on this arc. Then BT and CT intersect the perpendicular bisector of BC at P and Q, with the property of our problem. Obviously BM is the bisector of the angle PBQ and the circle (ABC) is the Appolonius circle of the triangle PQB, hence AM is the bisector of the angle PAQ.

Important notes:
1) In the given problem, building two triangles on each BF and DE similar to ABC and ADC, i.e. DPE and DP'E, then BQF and BQ'F, then the lines APQ and AP'Q' are simmetrical w.r.t. the bisector of the angle BAD.
2) The solution was given to me by my son, too busy to post it himself.

Best regards,
sunken rock

anyone mind helping me explain clearly how APQ and AP'Q' being symmetrical can lead to collinearity?

Solution:
Let $O$ be the circumcenter of $BEFC$ then since $\angle FBA=\angle ADE$,$\angle FAB=\angle EAD$, $\angle BFO=\angle FBO=90-\angle FDB=\angle ACD=\angle PDE=\angle PED$ then $BAF\sim DAE$ $DPE\sim BOF$ and they have the same orientation so $ABOF\sim ADPE$ and $\angle DAP=\angle BAO$ Similarly $\angle DAQ=\angle EAO$ Since $\angle EAO=\angle BAO$ then $\angle DAP=\angle DAQ$ so $A,P,Q$ are collinear.


On the chessboard 8x8 place 21 bars 3x1. Which fields could be not covered?

$4$ spaces, but not necessarily a $2\times2$ square will be left uncovered.


Ring Theory
superior algebra
superior algebra unsolved

i m splitting the questions in the post below into separate topics:

Let R be a commutative ring with M a maximal ideal. Showt aht R/(M^k) is a local ring for k>0 an integer.

Hi,
Consider any maximal ideal $N$ in $R \setminus M^n$.It corresponds to a maximal ideal $\tilde{N}$ of $R$ containing $M^n$,now as $\tilde{N}$ is also prime, we get
$M \subseteq \tilde{N}$ and hence $M = \tilde{N}$.The statement now follows clearly.


Let be K≥1 an integrer. In a group of 2K+1 peoples some are honest (always say the truth) and the rest are unpredictable (sometimes say the truth, and sometimes they lie). It is known that the unpredictable are less or equal that K. An outsider of the group wants to determinate who is honest and who is unpredictable by a sequence of steps. En each step he chooses two people of the group, A and B, and he asks A: ¿is B honest?. Prove that with 3K steps (questions) the outsider will clasify 100% surely, and without doubt, who in the group of 2K+1 people is honest and who is unpredictable.

(Before each step the outsider knows the answers of the questions made before).
(Each of the 2K+1 people in the group know who is honest and who unpredictable).

No one??? I was told that Darij was very good solving problems. With all my respect, can you try to solve it Darij??? (Don´t missunderstand me please, I´m not trying to say that you can´t).

[quote="spix"]You should determine in $2k$ steps a person who always tells the truth.[/quote]

How??? I don´t think that there finishes the problem no???


geometry
3D geometry
probability
factorial
puzzles

Consider a 3x3x3 cube. It is easy to cut it into 27 1x1x1 cubes using a total of 6 straight cuts.

Now suppose you're allowed to pick up the pieces between cuts and restack them any way you wish. Is it possible to get 27 small cubes in less than 6 cuts?

There is a way for 5.  Not that easy to explain though.

Could you elaborate probability?

By the way, I just figured out your signature!! I'm so happy.  I didn't understand what H*ll meant until know.  I thought it was like some geometry thing with H being parallel to something... :rotfl:

6 is the minimum.
pf) Consider the centre cube!

Oops, I missed something.  Yes 6 is the minimum.

[quote="GSPgeek"]it's possible to get up to 64 cubes with six cuts by restacking sliced pieces, but i'm not sure more than 16 in five cuts is possible[/quote]
$64=2^6$!(this exclaimation mark is not factorial :rotfl: )  But this is not we are talking about. We are talking about cutting unit cubes.

Where does it say unit cubes?

The problem involves stacking sliced pieces of a given cube. yes?

To get 64 smaller cubes, cut a cube in half (no one said you can't make your slices between units). Now stack the halved pieces vertically parallel to the slice, without changing their orientation, and slice them in half again.  You should now have four 4x4x1 pieces.  Arrange them to form a cube again and slice the remaining pieces analogously along the other two axes (four more cuts) for 64 cubes

Yes, 2^6 is 64, which is irrelevant, as you so eloquently pointed out (although, it could lead to the conjecture that for any 3n number of slices of a cube, 2^3n is the maximum number of smaller cubes that can be formed under the given stacking rules)

Regardless, my point remains, I'm reasonably convinced the greatest number of congruent small cubes in 5 cuts is 16, and those only add up to a quarter of the original cube's volume.  Of course, the problem doesn't even specify the cubes are the same size, so it's still possible that there may be a way to get 27 smaller cubes of different side lengths, as long as you aren't limited to unnecessary assumptions that oversimplify the problem.


I was doing the problems in "All-Time Greatest Mathcounts Problems" and realized that the 1993 State Sprint #6 (Question 23 for those who have the book) is the exact same question as #5 from MOEMS round one. Did anyone else notice this and has this happened before?

Round 1 from this year? Can't you not talk about it in case some other people have not taken it??

it says (read the stickies) not to discuss any problem until [b]one month after.[/b] we can now discuss.

I don't have MOEMS in our school, so I don't know anything about it.

Is the problem you are talking about a very common problem?

It has been at least two month...

It is the one with the rectangle with the star in it and they want you to find how many triangles are in it. I don't have a picture. I just thought it was strange that it was the same question. I doubt it was on purpose and I am not accusing MOEMS of anything. I just thought it was strange that they used the same question.


function
inequalities
algebra unsolved
algebra

[quote="kihe_freety5"]determine all function f such that
f is a continnuous function on R
and
6f(f(x))=2f(x)+x[/quote]

From the given function $ 6f\left( {f\left( x \right)} \right) \equal{} 2f\left( x \right) \plus{} x$, we have
\[ f\left( {f\left( {f\left( x \right)} \right)} \right) \equal{} f\left( {\frac{{2f\left( x \right) \plus{} x}}{6}} \right)
\]
On the other hand we also have
\[ f\left( {f\left( {f\left( x \right)} \right)} \right) \equal{} \frac{{2f\left( {f\left( x \right)} \right) \plus{} f\left( x \right)}}{6} \equal{} \frac{{\frac{{2f\left( x \right) \plus{} x}}{3} \plus{} f\left( x \right)}}{6} \equal{} \frac{5}{{18}}f\left( x \right) \plus{} \frac{x}{{18}}
\]
Hence we have
\[ 18f\left( {\frac{{2f\left( x \right) \plus{} x}}{6}} \right) \equal{} 5f\left( x \right) \plus{} x
\]
Now replacing $ x \equal{} 4f\left( x \right)$ then we have
\[ 18f\left( {f\left( x \right)} \right) \equal{} 9f\left( x \right)
\]
But we also have
\[ 6f\left( {f\left( x \right)} \right) \equal{} 2f\left( x \right) \plus{} x
\]
\[ \Rightarrow 6f\left( x \right) \plus{} 3x \equal{} 9f\left( x \right)
\]
\[ \Leftrightarrow f\left( x \right) \equal{} x
\]
Now check the function $ f\left( x \right) \equal{} x$ for the condition $ 6f\left( {f\left( x \right)} \right) \equal{} 2f\left( x \right) \plus{} x$ we have
\[ \left\{ \begin{array}{l}
 6f\left( {f\left( x \right)} \right) \equal{} 6f\left( x \right) \equal{} 6x \\
 2f\left( x \right) \plus{} x \equal{} 2x \plus{} x \equal{} 3x \\
 \end{array} \right.
\]
Therefor there is no fuction that satisfy the given condition. :)

[quote="Honey_S"]Now replacing $ x \equal{} 4f\left( x \right)$ ... [/quote]
 :?:
[quote="Honey_S"]Therefor there is no fuction that satisfy the given condition. :)[/quote]
I think $ f(x) \equal{} cx$, where $ c$ is a real root of $ 6a^2 \equal{} 2a \plus{} 1$ works. :maybe:

[quote="Akashnil"][quote="Honey_S"]Now replacing $ x \equal{} 4f\left( x \right)$ ... [/quote]
[quote="Honey_S"]Therefor there is no fuction that satisfy the given condition.[/quote]
I think $ f(x) \equal{} cx$, where $ c$ is a real root of $ 6a^2 \equal{} 2a \plus{} 1$ works. :maybe:[/quote]

Oh yes Akashnil I make a mistake.  :P

Sorry for my wrong solution. :(

who can solve it? it is like a problem in my exam in HUS( hanoi university of science)?

Uhm,so you are a Vietnamese
You can see it in MYM (I dont remember accurate month)
Any way,You can let $ a_n \equal{} f(f(..(n))..))$ (there  are n words "f" in total)
so we have the equation $ 6a_{n \plus{} 1} \equal{} 2a_n \plus{} a_{n \minus{} 1}$
ANd now it is very easy for a HUS student :wink:

oh, yes I thought that, but I get difficult in that!! could you write full solution!!ok?
I don't think it is easy!!

Obviously $ f(x)$ is injective. Then continuity implies strict monotonicity.
If $ b$ is in the range of $ f(x)$, say $ f(a)\equal{}b$. We have $ 6f(b)\equal{}6f(f(a))\equal{}2f(a)\plus{}a\equal{}2b\plus{}a$. So $ a\equal{}6f(b)\minus{}2b$.We know that is the only root to equation$ f(x)\equal{}b$.

Now we prove $ f(x)$ is also surjective.
Suppose it is not true, we have $ \inf_xf(x)>\minus{}\infty$ and/or $ \sup_xf(x)<\infty$. We will disprove the former and it will be similar for the latter.
Let $ M\equal{}\inf_xf(x)>\minus{}\infty$. We can find series $ b_n$ decreasingly converging to $ M$ and all $ b_n$ are in the range of $ f(x)$.
For each $ b_n$, we have $ a_n\equal{}6f(b_n)\minus{}2b_n$ such that $ f(a_n)\equal{}b_n$.
Since $ f(x)$ is continuous, we have $ a_n\rightarrow a\equal{}6f(M)\minus{}2M$ such that $ f(a)\equal{}M$.
Hence $ M$ is in the range of $ f(x)$.
But $ f(x)$ is strictly monotone, we can always find $ a'\neq 6f(M)\minus{}2M$ such that $ f(a')<M$.
Contradiction with $ M\equal{}\inf_xf(x)$.

Let us fix $ x_0$ and assume $ x_1\equal{}f(x_0)$. We can easily define a series $ x_n$ such that $ x_n\equal{}f(x_{n\minus{}1})\equal{}\frac{2x_{n\minus{}1}\plus{}x_{n\minus{}2}}6$ for $ n>0$.
Since $ f$ is surjective, we can also expand the series to the other direction. In other words, $ x_{\minus{}n}\equal{}f^{\minus{}1}(x_{\minus{}n\plus{}1})\equal{}6x_{\minus{}n\plus{}2}\minus{}2x_{\minus{}n\plus{}1}$.
Now we have series $ x_n$ all all integer $ n$.
It is not hard to prove $ x_n\equal{}c_1\alpha^n\plus{}c_2\beta^n$ for all integer $ n$, where $ \alpha>0>\beta$ are two roots of equation $ 6x^2\minus{}2x\minus{}1\equal{}0$, $ c_1\equal{}\frac{\beta x_0\minus{}x_1}{\beta\minus{}\alpha}$ and $ c_2\equal{}\frac{\alpha x_0\minus{}x_1}{\alpha\minus{}\beta}$.

Case 1) $ f(x)$ is strictly increasing, i.e., $ (x\minus{}y)(f(x)\minus{}f(y))>0$ for all $ x\neq y$.
$ (x_n\minus{}x_{n\plus{}1})(x_{n\plus{}1}\minus{}x_{n\plus{}2})>0$, or
$ ((c_1\minus{}c_1\alpha)\alpha^n\plus{}(c_2\minus{}c_2\beta)\beta^n)((c_1\minus{}c_1\alpha)\alpha^{n\plus{}1}\plus{}(c_2\minus{}c_2\beta)\beta^{n\plus{}1})>0$
Since $ \alpha>0>\beta$, the inequality will not hold if $ c_1\minus{}c_1\alpha\equal{}0$. Divide the inequality above by $ (c_1\minus{}c_1\alpha)^2\alpha^{2n\plus{}1}$ and denote $ z\equal{}\frac{\beta}{\alpha}$ and $ d\equal{}\frac{c_2\minus{}c_2\beta}{c_1\minus{}c_1\alpha}$, we get
$ (1\plus{}dz^n)(1\plus{}dz^{n\plus{}1})>0$.
To solve the inequality, $ d$ has to fall in between $ \minus{}z^{\minus{}n}$ and $ \minus{}z^{\minus{}n\minus{}1}$.
These two bounds all converge to $ 0$ when $ n\rightarrow\minus{}\infty$ because $ |z|<1$.
Hence, $ d\equal{}0$, or $ c_2\equal{}0$, or $ \alpha x_0\equal{}x_1\equal{}f(x_0)$.
Since $ x_0$ is arbitrary, we have $ f(x)\equal{}\alpha x$ for all $ x$.

Case 2) $ f(x)$ is strictly decreasing, i.e., $ (x\minus{}y)(f(x)\minus{}f(y))<0$ for all $ x\neq y$.
$ (x_n\minus{}x_{n\plus{}1})(x_{n\plus{}1}\minus{}x_{n\plus{}2})<0$, or
$ ((c_1\minus{}c_1\alpha)\alpha^n\plus{}(c_2\minus{}c_2\beta)\beta^n)((c_1\minus{}c_1\alpha)\alpha^{n\plus{}1}\plus{}(c_2\minus{}c_2\beta)\beta^{n\plus{}1})<0$
Since $ \alpha>0>\beta$, the inequality will not hold if $ c_2\minus{}c_2\beta\equal{}0$. Divide the inequality above by $ (c_2\minus{}c_2\beta)^2\alpha^{2n\plus{}1}$ and denote $ z\equal{}\frac{\beta}{\alpha}$ and $ d\equal{}\frac{c_1\minus{}c_1\alpha}{c_2\minus{}c_2\beta}$, we get
$ (d\plus{}z^n)(d\plus{}z^{n\plus{}1})<0$.
To solve the inequality, $ d$ has to fall in between $ \minus{}z^n$ and $ \minus{}z^{n\plus{}1}$.
These two bounds all converge to $ 0$ when $ n\rightarrow\infty$ because $ |z|<1$.
So we have $ d\equal{}0$ or $ c_1\equal{}0$, or $ \beta x_0\equal{}x_1\equal{}f(x_0)$.
Since $ x_0$ is arbitrary, we have $ f(x)\equal{}\beta x$ for all $ x$.


I was looking at a post by PiGuy and noticed that his website was displayed like this:
[b]WWW[/b]
It wasn't a link.
So I went to his profile and saw his websites, it looked like this
http://ilikepi.com
http://piguy.com
Is this a glitch?
PiGuy's profile is [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=38343]here[/url].

It's not a glitch. You can put there whatever you want (including two websites). The downside of that is that you can't click on the WWW button anymore if you do that.


a, b, c, and d are all positive real numbers.

Given that

a + b + c + d = 12
abcd = 27 + ab +ac +ad + bc + bd + cd

Find all positive real solutions for a, b, c, d

Consider AMGM on the sum $\sum ab$; we have

\[ \frac{ab+ac+ad+bc+bd+cd}{6}\geq \sqrt[6]{a^3b^3c^3d^3} \]

Substituting $abcd-27$ for the numerator on the LHS, and simplifying to the left, we have

\[ abcd - 6\sqrt{abcd} - 27 \geq 0\Longleftrightarrow (\sqrt{abcd} - 9 )(\sqrt{abcd} + 3 ) \geq 0 \]

Since $a,b,c,d>0$, then we have $\sqrt{abcd}\geq 9$ (*).

Now, consider AMGM on the four variables $a,b,c,d$; we have

\[ \frac{a+b+c+d}{4}\geq \sqrt[4]{abcd} \]

Substituting $a+b+c+d=12$ and squaring, we have $9\geq\sqrt{abcd}$ (**).

Combining (*) and (**), we have that $\sqrt{abcd} = 9$; taking the equality case on $AMGM$, we must have $a=b=c=d$, yielding the only solution $(3,3,3,3)$.

This is from the British Olympiad somewhere in the ninties.

Masoud Zargar


Let a,b, c,d be distinct integers such that the equation (x−a)(x−b)(x−c)(x−d)−9 = 0
has an integer root r. Show that 4r = a+b+c+d.

Solution:

[hide]
Since both r and a,b,c,d are integers, we are talking about four distinct factors multiplying to 9, which can only be +/- 3 and +/- 1. Thus we can write:

(x-a)(x-b)(x-c)(x-d)=9

$ [r-(r-3)][r-(r+3)][r-(r-1)][r-(r+1)]=9$

Then clearly a+b+c+d=4r

[/hide]

[hide="Fast"]$ x-a,x-b,x-c,x-d \in \mathbb{Z}$ and $ (x-a)(x-b)(x-c)(x-d)=9=3^{2}$. Note that $ 9$ only has $ 6$ factors; $ \pm 1, \pm 3, \pm 9$.

The only way to express 9 as the product of $ 4$ distinct integers is as $ (-3)(-1)(1)(3)$

Thus, wlog $ r-a=-3$, $ r-b=-1$, $ r-c=1$, and $ r-d=3$. Adding the three give $ 4r-a-b-c-d=0 \implies 4r=a+b+c+d$.[/hide]

[b]Edit 1:[/b] Mmkay, same as above.


calculus
integration
logarithms
calculus computations

Let $a,b$ be costant numbers such that $a^2\geq b.$ Find the following indefinite integrals.

(1) $I=\int \frac{dx}{x^2+2ax+b}$

(2) $J=\int \frac{dx}{(x^2+2ax+b)^2}$

(1) let the roots be $\alpha$ , $\beta$

if $\alpha>\beta$ then $I=\int\frac{1}{\alpha-\beta}\left(\frac{1}{x-\alpha}-\frac{1}{x-\beta}\right)dx=\frac{1}{\alpha-\beta}\ln\frac{x-\alpha}{x-\beta}+C$

or $I=\frac{1}{2\sqrt{a^2-b}}\left(\ln\frac{x+a-\sqrt{a^2-b}}{x+a+\sqrt{a^2-b}}\right)+C$

if $\alpha =\beta$ then $a^2=b$ so $I=\int\frac{dx}{(x+a)^2}=-\frac{1}{x+a}+C$

You need to write $|\ |$ for the integral of $log$ and $a^2>b.$
The rest is correct.


algebra
polynomial
search
algebra unsolved

[quote="mikeynot"]The $ n$ roots of a complex coefficient polynomial $ f(z)=z^{n}+a_{n-1}z^{n-1}+\cdots+a_{n-1}z+a_{n}$ are $ z_{1}, z_{2}, \cdots, z_{n}$. If $ \sum_{k=1}^{n}|a_{k}|^{2}\leq 1$, then prove that $ \sum_{k=1}^{n}|z_{k}|^{2}\leq n$.[/quote]

Please check the problem statement for typos...

  Darij

I think it should be f(z) = z^n + (a1)z^(n-1) + ... + an.

Here is Myth's famous proof: http://www.mathlinks.ro/viewtopic.php?search_id=84403368&t=27903


Hello!
i'm looking for A and B  (2*2 matrices) such that
AB-BA=I (identity Matrix)

any idea how to start ? Do they exist?!

They don't exist. Look at the trace: $ \text{tr}(AB)=\text{tr}(BA) $ so $ AB-BA $ can only be a matrix with trace zero.

In any field of characteristic 2, $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&0\\1&0\end{bmatrix}$ work.


Hi, this is my first time participating in the AMC (I just got my school to register this school year). I noticed that on past tests it says "No aids are permitted other than... calculators that are accepted for use on the SAT" (as well as a lot of other stuff! like protractors) However this year's brochure says "Students may not use calculators on the contests." I know calculators aren't [i]necessary[/i]  but could someone please clarify? Thank you![/i]

This year is a change: no calculators are allowed anymore. (also applies to AMC 10 and 8 I think)

i was allowed a calculator on AMC 8. 4 function

this is like the 20th time someone is asking this question, go to the amc website and look under frequently asked questions.


$lim\frac{tan3x-sin3x}{tan2x-sin2x}=?$
            $x\rightarrow 0$

As before, L'Hopital's rule gives the answer as

$\displaystyle\lim_{x\to 0} \displaystyle\frac{\frac{d}{dx} \left(\tan 3x - \sin 3x\right)} {\frac{d}{dx} \left(\tan 2x - \sin 2x\right)} = \displaystyle\lim_{x\to 0} \displaystyle\frac{3 \sec^2 3x - 3 \cos 3x} {2 \sec^2 2x - 2 \cos 2x}$

$= \displaystyle\lim_{x\to 0}\ \displaystyle\frac{3 \left(1 - \cos^3 3x \right)} {2 \left(1 - \cos^3 2x \right)} \times \displaystyle\frac{\sec^2 3x} {\sec^2 2x}$

$= \displaystyle\lim_{x\to 0} \displaystyle\frac{3 \left(0 - 9 \cos^2 3x \sin 3x \right)} {2 \left(0 - 6 \cos^2 2x \sin 2x \right)} \times \displaystyle\frac{1} {1}$

$= \displaystyle\lim_{x\to 0} \displaystyle\frac{9 \left(\cos^2 3x \right)} {4 \left(\cos^2 2x \right)} \times \displaystyle\frac{\sin 3x} {\sin 2x} = \displaystyle\frac{27}{8}$

Comments:

1.  If you're just using this forum to get your homework done, stop.  Or at least go to the appropriate forum and ask for help on concepts instead of posting individual homework questions (which is what I suspect these to be).
2.  Please have the foresight to place your topics in the right forum.

This really is getting intolerable.  Don't you know which section certain problems belong in by now?

you mean thios question is not belongs here?sorry! :oops:

[quote="jensen"]but see he has solved it. :?:[/quote]

That's completely immaterial.  Enough people browse a variety of forums that I could post a calculus problem in the combinatorics forum and get a solution within a day.  But that still doesn't make it the correct place to do it.

i see ,i am so sorry.
but you mean i shouldn't post calculus problems here?

Indeed.  There is a section called 'Calculus Computations and Tutorials' which is the correct place for posts related to, but not limited to, limits, derivatives, and integrals.


[hide="Solution"]Sum to first 2005 numbers $=\frac{2005 \cdot 2006}{2}$

Since $2005= 5 (mod 100)$ and $2006 = 6 (mod 100),$
Sum to first 2005 numbers $=\frac{5 (mod 100)\cdot 6 (mod 100)}{2}=15$[/hide]

[quote="karthik_k"]Find the last two digits of
$1 + 2 + 3 + 4 + \cdots + 2003 + 2004 +2005$[/quote]

Hmm... that formula kind of makes the problem way too easy.

The way I did it was that the sum $1+2+\cdots+99=4950$, and occurs twice. It ends in $00$. $1+2+3+4+5=15$.

i know a way but it's sort of the same.  Disregard all the numbers below that number/100 without remainders.  Then add.  So add up.

for example the triangle number of 1006 is the same last two digits as 1+2+3...6 and 1+2+3...2005 shares the same last two digits as 1+2+3...5

sry a bit late here, but i wanna share the way i did it too.

1+2+3+4.... +100+.... 2005 = xxxxx15

how?

converging number bonds. i.e. 1+99 = 100. 2+98 = 100. 3+97 = 100 ......... til the last bond is 49+51. it all adds up to 100, leaving a 50 behind.
the same goes to 101+199 except that it has an increment of 100, but that is something we can overlook, because we are only concern about the LAST 2 DIGITS.

from 1 to 2000, there are 20 sets of hundreds. that means, the 50 being left behind as the last 2 digits are being multiplied (20x50) and can be forgone, as the last 2 digits of them are 00.

so therefore, we can conclude that sum of 1 to 2000 has last 2 digits of 00.

simply add up 1 to 5 to find the last 2 digits.

sounds kinda long.. but when this process goes thru the mind, it takes no more than 10 seconds.

seems that everyone is giving the answer fifteen over and over again. If you don't contribute something, it is considered spamming.


$ \alpha, \beta, \gamma$, are real number, distinct 0. and a,b,c,d,p,q are arbitrary real number, prove that matrix
$ B\equal{}\left(\begin{array}{ccc}a&b\frac{\alpha }{\beta }&c\frac{\alpha }{\gamma }\\b\frac{\beta }{\alpha }&d&p\frac{\beta }{\gamma }\\c\frac{\gamma }{\alpha }&p\frac{\gamma }{\beta }&q\end{array}\right)$ has all eigenvalues are real numbers

We have $ H_3\left(\frac{1}{\gamma}\right) H_2 \left(\frac{1}{\beta}\right)H_1 \left(\frac{1}{\alpha}\right) B H_1(\alpha) H_2(\beta) H_3(\gamma) \equal{} \left[\begin{array}{ccc}a&b&c \\ b&d&p\\c&p&q\end{array}\right]$. But if $ H \equal{} H_3\left(\frac{1}{\gamma}\right) H_2 \left(\frac{1}{\beta}\right)H_1 \left(\frac{1}{\alpha}\right)$ then $ H^{\minus{}1} \equal{} H_1(\alpha) H_2(\beta) H_3(\gamma)$. Therefore,

$ HBH^{\minus{}1} \equal{} \left[\begin{array}{ccc}a&b&c \\ b&d&p\\c&p&q\end{array}\right]$

$ \equal{}> B \equal{} H^{\minus{}1} \left[\begin{array}{ccc}a&b&c \\ b&d&p\\c&p&q\end{array}\right] H$.

But this means that B is similar to a symmetric matrix, and similar matrices have the same eigenvalues. On the other hand, all eigenvalues of a symmetric matrix are real. Therefore, all eigenvalues of B are real.


$a,b,c,x,y,z>0.$ Prove $(ax+by+cz)(\frac1{abc}+\frac1{xyz})\ge9(\frac1{a+b+c}+\frac1{x+y+z})$
and show that equality holds iff $a=b=c=x=y=z$. :)

let us denote by $p$ and $q$ the products $\sqrt[3]{abc}$ and$\sqrt[3]{xyz}$ then we have
$(ax+by+cz)\left(\frac{1}{abc}+\frac{1}{xyz}\right)\geq 3pq(\frac{1}{p^3}+\frac{1}{q^3})=3\frac{p^3+q^3}{p^2 q^2}\geq 3 (\frac{1}{p}+\frac{1}{q})\geq 9(\frac{1}{a+b+c}+\frac{1}{x+y+z})$
we used here the fact that $\frac{p^3+q^3}{p^2 q^2}=\frac{(p+q)(p^2+q^2-pq)}{p^2 q^2}\geq \frac{pq(p+q)}{p^2 q^2}=\frac{1}{p}+\frac{1}{q}$


Euler
function
number theory
prime factorization
totient function
number theory proposed

$ a,b$ are natural numbers.prove that there are infinitely many positive integer pairs $ (x,y)$ such that $ (x,y) \equal{} 1$ and $ \frac {\phi(x)}{\phi(y)} \equal{} \frac {a}{b}$

We may assume that $ (a,b) \equal{} 1$ (for if $ (a,b) \equal{} g$ we may consider $ \frac {a}{g}$ and $ \frac {b}{g}$ instead of $ a$ and $ b$).

Let $ a \equal{} p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k}$ and $ b \equal{} q_1^{f_1} q_2^{f_2} \ldots q_t^{f_t}$ be the prime factorization of $ a$ and $ b$.

Let $ p$ be a prime not equal to any of the $ p_i$ and $ q_i$

We can then easily observe that
$ (x,y) \equal{} (pp_1^{e_1 \plus{} 1} p_2^{e_2 \plus{} 1} \ldots p_k^{e_k \plus{} 1} q_1 q_2 \ldots q_t , p p_1 p_2 \ldots p_k q_1^{f_1 \plus{} 1} q_2^{f_2 \plus{} 1} \ldots q_t^{f_t \plus{} 1})$ is a solution. (Just plug it into the Euler totient function formula, and it's obvious!)

The infinitude of $ p$ is visible, and also is the infinitude of $ (x,y)$

You forget, that $ (x,y)\equal{}1$. It is hard problem.


Prove that there exists a choice of signs such that :
$(\sin 1 \pm \sin 2 \pm ... \pm \sin 10)^{2}+(\cos 1 \pm \cos 2 \pm ... \pm \cos 10)^{2}\leq 10$
Generalization.

err....degrees? I did degrees

$\sum \sin x\approx 0.957$ and $\sum (-1)^{x}\cos x=-0.008$...so yes.

Perhaps there is a tricky pidgeonhole argument...

Can this work for the generalization? I mean, can you still count degrees?  :ninja:


function
limit
logarithms
integration
calculus
derivative
calculus computations

Determine the limit of the sequence:

$ x_n \equal{} \sqrt[n\plus{}1]{2008 \plus{} \frac{1}{n\plus{}1}} \plus{}  \sqrt[n\plus{}2]{2008 \plus{} \frac{1}{n\plus{}2}} \plus{} \ldots \plus{} \sqrt[2n]{2008 \plus{} \frac{1}{2n}} \minus{} n$.

Let $ f(x) \equal{} (2008 \plus{} x)^x$. We find that $ f'$ is increasing on $ [0, \infty)$.

In fact, $ f'(x) \equal{} f(x)\left[ \ln(2008 \plus{} x) \plus{} x/(2008 \plus{} x) \right]$, and since both $ f(x)$ and $ \ln(2008 \plus{} x) \plus{} x/(2008 \plus{} x)$ are positive increasing functions on $ [0, \infty)$, $ f'(x)$ iteslf is increasing on $ [0, \infty)$.

Then by mean value theorem, $ f(1/(n \plus{} k)) \minus{} 1 \equal{} f'(c_k)/(n \plus{} k)$ for some $ 0 < c_k < 1/(n \plus{} k)$. Then

$ x_n \equal{} \sum_{k \equal{} 1}^{n} \left[f\left(\frac {1}{n \plus{} k}\right) \minus{} 1\right] \equal{} \sum_{k \equal{} 1}^{n} \frac {f'(c_k)}{n \plus{} k}$

and

$ \sum_{k \equal{} 1}^{n} \frac {f'(0^ \plus{} )}{n \plus{} k} \leq x_n \leq \sum_{k \equal{} 1}^{n} \frac {f'(1/n)}{n \plus{} k},$

where $ f'(0^ \plus{} ) \equal{} \lim_{x\to0^ \plus{} } f'(x) \equal{} \ln 2008$. Therefore, by squeezing theorem,

$ \lim_{n\to\infty} x_n \equal{} f'(0^ \plus{} )\int_{0}^{1}\frac {dx}{1 \plus{} x} \equal{} \ln 2008 \cdot \ln 2$.

Thanks, sos440! But I don't know derivatives and integrals. Can you give a more elementary solution, please?

You can get away with "two remarhkable limits": $ \lim_{h\to 0}\frac{e^h\minus{}1}h\equal{}1$ and $ \lim_{h\to 0}\frac{\log(1\plus{}h)}h\equal{}1$ (which, in fact, are just two sides of the same coin) but do you know them?

[quote="fedja"]You can get away with "two remarhkable limits": $ \lim_{h\to 0}\frac {e^h \minus{} 1}h \equal{} 1$ and $ \lim_{h\to 0}\frac {\log(1 \plus{} h)}h \equal{} 1$ (which, in fact, are just two sides of the same coin) but do you know them?[/quote]

Yes, I know them...

$ x_n\equal{}\sum_{k\equal{}1}^n \left(\sqrt[n\plus{}k]{2008\plus{}\frac{1}{n\plus{}k}}\minus{}1\right)$

$ x^p\minus{}1\equal{}(x\minus{}1)(x^{p\minus{}1}\plus{}x^{p\minus{}2}\plus{}\cdots \plus{}x\plus{}1)$


Let $n\geq 4$, and let $P_0=A_0A_1\ldots A_{n-1}$ be a polygon, such that $A_{2i}A_{2i+1}=A_{2i+1}A_{2i+2}=2^i$ for all sides of $P_0$ (obviously, taking the proper indices modulo $n$, and stopping at $A_{n-1}A_0$). Now, let $P_1=A'_0A'_1\ldots A'_{n-1}$ be the symmetric of $P_0$ with respect to $A_0$, let $P_2=A''_0A''_1\ldots A''_{n-1}$ be the symmetric of $P_1$ wrt $A'_1$, $P_3$ the symmetric of $P_2$ wrt $A''_2$ and so on. For what values of $n$ do infinitely many $P_i$'s coincide with $P_0$?

does somebody ahs a solution for thuis one?

it's not such an impossible problem... and i can say it's nice :)

sorry for this stupid question ... but what do you mean by 'symmetric of'?  :?

The polygon obtained by reflecting the original polygon in that vertex.

you are right its not impossible,, after Imo i can post my solution


linear algebra
matrix
search
real analysis
real analysis unsolved

Let $ n \in \mathbb{N}$ Prove that :
1) $ 1 \plus{} cosx \plus{} \frac {1}{2}cos(2x) \plus{} \frac {1}{3}cos(3x) \plus{} .... \plus{} \frac {1}{n}cos(nx) \geq 0$ , $ \forall x \in \mathbb{R}$
2) $ sinx \plus{} \frac {1}{2}sin(2x) \plus{} \frac {1}{3}sin(3x) \plus{} .... \plus{} \frac {1}{n}sin(nx) \geq 0$ , $ \forall x \in \begin{bmatrix} 0 & \pi \end{bmatrix}$
 [color=blue]Thanks much .[/color]

[quote="Myth"]Look at
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Jackson&t=20812[/quote]
 A proof very nice . Thanks very much  :)

http://www.mathlinks.ro/viewtopic.php?t=251484
I think there is no big differnece between sin serie  is or cos serie..


algebra
polynomial
calculus
integration
least common multiple
binomial theorem
Rational Root Theorem

Find all polynomials $p(x) \in \mathbb{R}[x]$ such that the equation \[f(x) = n\] has at least one rational solution, for each positive integer $n$.

We can first prove, by using the Lagrange interpolation formula, that the polynomial has rational coefficients. After that, post number $2$ in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=47243]this topic[/url] takes care of it (actually, the coefficients are integral in that post, but it doesn't matter: we multiply everything by the least common multiple $m$ of the denominators of the coefficients and it works just the same, except that instead of having "every $n$" we have "every multiple of $m$"; the solution still works as outlined there).

Isnt the answer just ax+b for a and b rational? From the prevoius post and in general we can show that the polynomial must have rational coefficients. By the rational root theorem, the denominator of the input must divie the numerator of the leading coeficients so we can bound the denominator of the input with some integer W. Now use the binomial theorem on the polnomial to expand p(x+1/W) and it is easy to notice that for large enough x, the we have p(x+1/w)-p(x) > c for any constant c we choose so eventually the polynomial skips over a large number of integers so we cant equate them all but also note this binomial expansion is only increasing if the original polynomial was degree 2 or higher.


Let $n$ be the smallest positive integer that is a multiple of $75$ and has exactly $75$ positive integral divisors, including $1$ and itself. Find $n/75$.

[hide]Because $75=3\cdot 5^2$, The smallest $n$ with $75$ divisors will be in the form $n=a^2b^4c^4$. If $n$ is a multiple of $75$, we know that it must contain $3$ and $5^2$, so $n=5^23^4c^4$. The $c$ that will minimize $n$ is obviously $2$, so we have $n=2^43^45^2$. Our answer is $n/75=\boxed{432}$.[/hide]

[hide]The number is of the form $5^2*3*k$. With some trial and error, we obtain $k$ as $3^3*2^4$ and the number will be $5^2*3^4*2^4$ with $75$ distinct factors. Therefore, $n/75 = 3^3*2^4=432$.[/hide]

edit: I didn't realize you had posted the solution already 4everwise  :roll:

[hide="Answer"]$75=3\cdot 5^2$
For $n$ to have exactly $75$ positive integral divisors, obviously the smallest possibility is $p_1^2p_2^4p_3^4$. Say $p_2=2$, $p_3=3$, $p_1=5$. Coincidentally, this is divisible by $75$, so our answer is $2^4\cdot 3^3=\boxed{432}$.[/hide]

[quote=Ultroid999OCPN][hide=sol]Since $n$ has an odd number of divisors, it must be a perfect square. We write $n=2^4\cdot 3^4\cdot 5^2,$ where we have minimized its value by placing higher exponents on lower prime factors and choosing exponents that will give a total of 75 divisors. The requested answer is then $\frac{2^4\cdot 3^4\cdot 5^2}{3\cdot 5^2}=\boxed{432}.$[/hide]
[/quote]

so 75=5^2 *3
so n has to have 5^2 *3
exponent +1 on top have to x up to 75
we need smallest so make the 3 exponent +1 be 5
so 5^2 * 3^4
and then we need one more make it the smallest so 2 so
5^2 *3^4*2^4/75=432

[hide=Solution]We find the smallest possible number having $75$ positive divisors. Since $75 = 5 \cdot 5 \cdot 3$, this would just be $2^4 \cdot 3^4 \cdot 5^2$. This is divisible by $75$, and dividing gives $\boxed{432}$.[/hide]

we must have k*3*5^2 have 3*5*5 divisors so minimizing this is 2^(5-1)*3^(5-1)*5^(3-1) = 2^4*3^4*5^2 and dividing by 3*5^2 gives 16*27 or 432


Let P be the product of all primes from 1-100. What is the remainder when P is divided by 16?

there is no real shortcut
but there is a minor shortcut
obviously the answer is even since there is a '2'
from there you just have to find all the other primes and take them mod 8 (since you know the answer is even)
and multiply them a few at a time and keep taking mod 8 when necessary
then multiply that by 2
if you didnt get that i will show you:
these are the first few primes mod 8
3*5*7*3*5*3*7...

suppose the number gets too big after like 10 primes
divide it by 8 and find the remainder
then multiply that by 8 and multiply the other primes
suppose you end up with 7 (you wont) then you multiply by 2 and the answer would be 14


I looked around to find something fun to play and found thatquiz.org.  I wanted to see how I did, so I tried their "Algebra" section to see how long it would take me to solve "10" Level 3 problems(38 seconds, their problems are easy as cake).  But higher level problems aren't.  POst results here! :D

Score	100%
Completed	10
Unanswered	0
Right	10
Wrong	0
Time	0:19
Seconds Per Problem	1.9

Hmmmm :)...
I tried the exponents.... level.... 2-3?


Solve the equation $7^x -3^y =4$ in positive integers.

[quote="Peter"]Solve the equation $ 7^x \minus{} 3^y \equal{} 4$ in positive integers.[/quote]
If $ x\equal{}1\equal{}>y\equal{}1$ or  if $ y\equal{}1$ =>$ x\equal{}1$
If $ x \geq 2;y \geq 2$
[b]*Case 1:[/b]$ x\equal{}2k$ with $ k \in Z\plus{}$
We have $ 3^{2k}\plus{}4\equal{}7^x$
=>$ 9^k\plus{}4\equal{}7^x$=>$ 7^x \equiv 1(mod 4)$
=>$ x \equal{}2m$ with $ m \in Z^\plus{}$
Then we have $ (7^m\minus{}3^k)(7^m\plus{}3^k)\equal{}4$=>$ 7^m\minus{}3^k\equal{}1 and 7^m\plus{}3^k\equal{}4$
=>$ 2.7^m\equal{}5$=>$ m$ isn't integer number
[b]*Case 2:[/b]$ x\equal{}2k\plus{}1$
Then $ 3^{k\plus{}1}\plus{}4\equal{}7^x$=>$ 3.9^k\plus{}4\equal{}7^x$
=>$ 7^x \equiv 3(mod 4)$ wrong because $ 7^x \equiv (\minus{}1)^x (mod 4)$
So $ (x;y)\equal{}(1;1)$

[quote="chien than"][quote="Peter"]Solve the equation $ 7^x \minus{} 3^y \equal{} 4$ in positive integers.[/quote]
If $ x \equal{} 1 \equal{} > y \equal{} 1$ or  if $ y \equal{} 1$ =>$ x \equal{} 1$
If $ x \geq 2;y \geq 2$
[b]*Case 1:[/b]$ x \equal{} 2k$ with $ k \in Z \plus{}$
We have $ 3^{2k} \plus{} 4 \equal{} 7^x$
=>$ 9^k \plus{} 4 \equal{} 7^x$=>$ 7^x \equiv 1(mod 4)$
=>$ x \equal{} 2m$ with $ m \in Z^ \plus{}$
Then we have $ (7^m \minus{} 3^k)(7^m \plus{} 3^k) \equal{} 4$=>$ 7^m \minus{} 3^k \equal{} 1 and 7^m \plus{} 3^k \equal{} 4$
=>$ 2.7^m \equal{} 5$=>$ m$ isn't integer number
[b]*Case 2:[/b]$ x \equal{} 2k \plus{} 1$
Then $ 3^{k \plus{} 1} \plus{} 4 \equal{} 7^x$=>$ 3.9^k \plus{} 4 \equal{} 7^x$
=>$ 7^x \equiv 3(mod 4)$ wrong because $ 7^x \equiv ( \minus{} 1)^x (mod 4)$
So $ (x;y) \equal{} (1;1)$[/quote]
Wrong!

Maybe you can give the correct solution? :)

[quote=mhdr_haamiii]There is an easier way to prove that!
Lemma :
If ab =cd,  and  (a, b) = (c, d) = (a, d) =1 , then :
   a = c  and b = d.  (a,b,c,d ≠ 0)
That's really easy!

[ a|cd , (a,d)=1 => a|c. (1)
  c|ab , (a,b)=1 => c|a or c|b
  if c|a then the lemma is proved.
  if c|b , from (1) we know a|c so a|b,
  but (a,b) =1 and it's wrong. ]

Now let's get back to our problem!

Let x=b+1 and y=a+1.
We know :  3(3^b - 1) = 7(7^a - 1)  (1)
Besides,  we know :
( 3 , 3^b - 1 ) = ( 7 , 7^a - 1) = (3,7) = 1.

So:
3 = 7^a -1 and that's impossible ! :D
So the equation (1) is 0=0 , which means
a = b = 0.  So x = y = 1!  :)[/quote]

$c|ab , (a,b)=1 => c|a or c|b$

This is wrong.

consider $c=6 ,a=2 ,b=3$

Clearly $x = y = 1$ provides a solution. We claim that there are no others. When $y \ge 2$, we get $3^y \pmod{234} \in \{9, 27, 81\} = A$, while $7^x \pmod{234} \in \{1, 7, 49, 109, 61, 193, 181, 97, 211, 73, 43, 67\} = B$ for all $x \in \mathbb{N}$. Since $(A + 4) \cap B = \emptyset$, there is no solution for $y \ge 2$.


ratio
geometry
projective geometry
geometry proposed

Dear mathlinkers

For a circle w with its center O and a point P outside the circle w,
a line L through P intersects the circle w at Q,R and the polar of P at S.
P,Q,S,R are harmonic conjugate. :arrow: I have seen only the projective geometric proof for this.

Are there some elementary proofs for it? If you have,may I ask you a favor for them?
 :help: please~ thank you.

If the line is the diameter of the circle the problem is quite trivial. Let's denote A and B the points where the tangents from P touch the circle, so AB is the polar of P w.r.t. circle O and RQP a secant. The quadrilateral ARBQ is harmonic, as the tangents from A and B concur on RQ and we have AR*BQ = AQ*BR ( * )

[i][b]Proof[/b]:
AP being tangent to the circle (ARQ), we have PQ/PR=AQ^2/AR^2 ( 1 ) and, similarly, from PB and circle (BRQ) we get PQ/PR=BQ^2/BR^2 ( 2 ). From (1) and (2) the relation (*) is proved.[/i]

From relation (*) we can easily see that the tangents at R and Q concur on AB, hence AS is symedian in triangle ARQ and, consequently,
SQ/RS=(AQ/AR)^2 ( 3 ). From (1) and (3) we got the proof.

Best regards,
sunken rock

Thanks.
Sorry that your proof is a little different from elementary proofs which I referred to.
I am surprised at sudden formulas like ◇/◇'=◎²/☆² (I guess this formula stems from projective geometry,right?).
In addition,I don't know the formulas  :blush:

I have got that the formula is called "Steiner" theorem,just right now. :blush: I'm ashamed of myself.ToT

We denote as $ (O),$ the given circle with center $ O$ and let be the point $ Q,$ between the points $ P$ and $ R.$

We define the point $ S,$ between the points $ Q,\ R,$ as the harmonic conjugate of $ P,$ with respect to the poits $ Q,\ R.$

It is enough to prove that the point $ S,$ lies on a constant line, as the polar of $ P,$ with respect to the circle $ (O).$

Let be the points $ B\equiv (O)\cap PO$ and $ C\equiv (O)\cap PO$ $ ($ $ C,$ between $ B,\ P$ $ ).$

Let also be the points $ A\equiv BR\cap CQ$ and $ H\equiv BQ\cap CR.$

Because of $ BQ\perp AC$ and $ CR\perp AB,$ we conclude that the point $ H,$ is the orthocenter of the triangle $ \bigtriangleup ABC.$

So, we have $ AH\perp BC$ and let be the point $ T\equiv BC\cap AH.$

From the complete quadrilateral $ AQHRBC,$ we have that the point $ S'\equiv QR\cap AH,$ is the harmonic conjugate of $ P\equiv BC\cap QR,$ with respect to the points $ Q,\ R$ and so, we conclude that $ S'\equiv S.$

The point $ T\equiv BP\cap AH$ is also, the harmonic conjugate of $ P,$ with respect to the points $ B,\ C,$ as well.

That is, the point $ S$ lies on the line through the constant point $ T$ and perpendicular to $ BP,$ which is the polar of $ P,$ with respect to the circle $ (O)$ and the proof is completed.

Kostas Vittas.

[quote][color=darkred][b][u]Lemma.[/u][/b] Let $ w$ be the circumcircle of the triangle $ ABC$ . Denote $ \{\begin{array}{c} T\in BB\cap CC \\
\ D\in AT\cap BC \\
\ \{A,E\} = w\cap AT\end{array}$ .

Then exist the relations : $ \frac {DB}{DC} = (\frac {AB}{AC})^2\ \ ,\ \ \frac {BA}{BE} = \frac {CA}{CE}\ \ , \ \ \frac {DA}{DE} = \frac {TA}{TE}$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Show easily that $ \frac {DA}{AB\cdot AC} = \frac {DB}{BE\cdot BA} = \frac {DE}{EC\cdot EB} = \frac {DC}{CA\cdot CE}$ . Thus, $ \frac {DA}{DE} = \frac {BA}{BE}\cdot \frac {CA}{CE}\ \ (1)$

and $ \frac {DB}{DC} = \frac {BA\cdot BE}{CA\cdot CE}$ . From the well-known property, $ \{\begin{array}{cc} \triangle ABE\ : & \frac {TA}{TE} = (\frac {BA}{BE})^2 \\
 \\
\triangle ACE\ : & \frac {TA}{TE} = (\frac {CA}{CE})^2\end{array}\|$ $ \implies$ $ \boxed {\ \frac {BA}{BE} = \frac {CA}{CE}\ }\ \ (3)$

([b]the cyclical quadrilateral $ ABEC$ is harminically[/b]), $ \boxed {\ \frac {DB}{DC} = (\frac {AB}{AC})^2\ }$ , ([b]the point $ D$ is the $ A$ - symmedian

in the triangle[/b] $ ABC$ ) and $ \frac {TA}{TE} = \frac {BA}{BE}\cdot\frac {CA}{CE}\ \ (2)$ . The relations $ (1)$ , $ (2)$ , $ (3)$ $ \implies$ $ \boxed {\ \frac {DA}{DE} = \frac {TA}{TE} = (\frac {BA}{BE})^2 = (\frac {CA}{CE})^2\ }$

([b]the points $ D$ , $ T$ are harmonical conjugate w.r.t. the pair of points $ \{A,E\}$ , i.e. the division $ (A,D,E,T)$ is harmonically).[/b]

[b][u]Remark.[/u][/b] The point $ T$ is the [b]pole[/b] of the line $ BC$ and the line $ BC$ is the [b]polar[/b] of the point $ T$ w.r.t. the circle $ w$ .[/color]

I appreciate your kind explanation,Virgil Nicula. :)


[quote="Moubinool"]Prove that
 $\frac 1{\pi} = \sum_{n \geq 0} \binom {2n}{n}^3 \cdot \frac{42n+5}{2^{12n+4}}$
[/quote]

 :ninja:


Calculate: $ \frac{1\plus{}\frac12}{\frac12 \plus{} \frac13}$. Express your answer as a common fraction.


keler marku is conveniently #69

he stole it from zuming

dude these topics suck

GASP AM I FINALLY MATURING

hah probably not


Solve in the set of real numbers:
\[ 3\left(x^2 \plus{} y^2 \plus{} z^2\right) \equal{} 1,
\]

\[ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \equal{} xyz\left(x \plus{} y \plus{} z\right)^3.
\]

If one of them , say $ x\equal{}0$, then we have at least one of the others, say $ y\equal{}0$ also, so $ (0,0,\frac{1}{\sqrt{3}})$ and its premutations are solutions.

Now for $ x,y,z \neq 0$, we have $ \frac{x^2\plus{}y^2\plus{}z^2}{3} \ge \frac{(|x|\plus{}|y|\plus{}|z|)^2}{3^2}$
$ \Leftrightarrow |x|\plus{}|y|\plus{}|z| \le 1$

So now we have $ x^2y^2\plus{}y^2z^2\plus{}z^2x^2\equal{}xyz(x\plus{}y\plus{}z)^3 \le |x\parallel{}y\parallel{}z|(|x|\plus{}|y|\plus{}|z|)$
But, $ \sum (x^2y^2z^0) \ge \sum (|x|^2|y\parallel{}z|)$ by muirhead.

So equality holds $ \Rightarrow x\equal{}y\equal{}z$ is positive , which gives $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$ as solution after checking.

So $ (x,y,z)\equal{}(\frac{1}{3},\frac{1}{3},\frac{1}{3}), (0,0,\frac{1}{\sqrt{3}}),(0,\frac{1}{\sqrt{3}},0),(\frac{1}{\sqrt{3}},0,0)$ are the only solutions.

What about $ \Big( \frac{\minus{}1}{3}, \frac{\minus{}1}{3}, \frac{\minus{}1}{3} \Big)$ and $ \Big(0, 0, \frac{\minus{}1}{\sqrt 3}\Big)$?  :wink:

[quote="djuro"]What about $ \Big( \frac { \minus{} 1}{3}, \frac { \minus{} 1}{3}, \frac { \minus{} 1}{3} \Big)$ and $ \Big(0, 0, \frac { \minus{} 1}{\sqrt 3}\Big)$?  :wink:[/quote]

Sorry , correction to my solution:

The case $ x\equal{}0$ gives $ y\equal{}0, z^2\equal{}\frac{1}{3}$, so $ (0, 0, \frac {\minus{}1}{\sqrt 3})$ and its premutations are solutions also.

On the second line from the bottom, it should be equality holds $ \Rightarrow |x|\plus{}|y|\plus{}|z|\equal{}|x\plus{}y\plus{}z| \Rightarrow x,y,z$ are all positive or all negative, and $ |x|\equal{}|y|\equal{}|z|$,  these gives $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$ or $ x\equal{}y\equal{}z\equal{}\minus{}\frac{1}{3}$ after checking.

So $ (x,y,z)\equal{}(\frac{1}{3},\frac{1}{3},\frac{1}{3}), (\minus{}\frac{1}{3},\minus{}\frac{1}{3},\minus{}\frac{1}{3}),(0, 0, \frac {1}{\sqrt 3})(0, 0, \frac {\minus{}1}{\sqrt 3})$ and their premutations are the solutions.

This is Indonesia IMO TST 2004..
$ (x\plus{}y\plus{}z)^2 \le 3(x^2\plus{}y^2\plus{}z^2)\equal{}1$
$ xyz(x\plus{}y\plus{}z)\ge xyz(x\plus{}y\plus{}z)^3\equal{}x^2y^2\plus{}y^2z^2\plus{}z^2x^2 \ge xyz(x\plus{}y\plus{}z)$

I think this is the shortest path..

$3(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)=xyz(x+y+z)^3)$. In the left hand we have only positive numbers, so we will work with positive $x,y,z$(only have to take $|x|,|y|,z$and permutations. The main ideea( i won't study particular cases) is $3(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)\ge xyz(x+y+z)^3)$. The inequality is not hard, after expending we get:$3 \displaystyle\sum\limits_{\text{cyc}}(x^4y^2+x^2y^4)+3x^2y^2z^2\ge3 \displaystyle\sum\limits_{\text{cyc}}(x^3y^2z+x^3yz^2)+x^4yz+xy^4z+xyz^4$. We can aply AM-GM for the numbers $x^4y^2+x^2y^2z^2\gex^3y^2z$. The rest of the inequality is easy to prove using Muirhead. ($4\ge3, 4+2\ge3+2, 4+2+0=3+2+1$, and $4\ge4, 4+2\ge4+1, 4+2+4+1+1$,etc)

We claim the solutions are $(0,0,\pm1/\sqrt3)$, $\pm1(1/3,1/3,1/3)$, and permutations. These are easily seen to work. Now, using AM-GM and the triangle inequality,
\begin{align*}
|xyz(x+y+z)^3|&=|x^2y^2+y^2z^2+z^2x^2|\\
|xyz||(x+y+z)^3|&=x^2y^2+y^2z^2+z^2x^2\\
|xyz||x+y+z|^3&\ge |xyz|(|x|+|y|+|z|)\\
|xyz||x+y+z|^3&\ge|xyz||x+y+z|\\
|xyz||x+y+z|((x+y+z)^2-1)&\ge0.
\end{align*}
Now, if $xyz(x+y+z)=0$, then it follows two of $x,y,z$ are $0$, from which we get the solutions $(0,0,\pm1/\sqrt3)$. Otherwise, $(x+y+z)^2\ge1$. But then, $xy+yz+zx\ge\frac13=x^2+y^2+z^2\ge xy+yz+zx$, so from the equality case of the trivial inequality, $x=y=z$. This gives $\pm1(1/3,1/3,1/3)$. $\blacksquare$


Let [tex] a,b,c [/tex] be three positive numbers and [tex] n [/tex] is a positive integer. Prove that:
[tex] \frac{a^n+b^n}{(a+b)^n}+\frac{b^n+c^n}{(b+c)^n}+\frac{c^n+a^n}{(c+a)^n} \leq \frac{3}{2^{n-1}}+\frac{(a-b)^2}{4ab}+\frac{(b-c)^2}{4bc}+\frac{(c-a)^2}{4ca} [/tex]  :ewpu:

This inequality is only true for n=1,2,3
for n>3,it is false for (a,b,c)=(1,1,2) :?

Umm... for [tex] a=b=1, c=2 [/tex], the inequality  is equal to:
[tex] \frac{2^n+1}{3^n} \leq \frac{1}{2^{n-1}}+\frac{1}{8} [/tex]
[tex] \Leftrightarrow \frac{3^n}{2^{n-1}}+\frac{3^n}{8} \geq 2^n+1 [/tex]
For [tex] n \geq 6, \frac{3^n}{8} > 2^n, \frac{3^n}{2^{n-1}} >1 [/tex] so it is true.
For [tex] n=4,5 [/tex] we can easily check it, right...?  :roll:


Yea, so the problem is find all f s.t. \[f(x+t)=f(x)+f(t)+2\sqrt{f(x)f(t)}\] and they say $f(x)=0$ is only function but doesn't $f(x)=cx^{2}$ work ??

[quote="Elemennop"]Doesn't it only matter that $f(x)$ and $f(t)$ are positive? Which they would be since they are squares...[/quote]

But $\sqrt{(-x)^{2}t^{2}}= xt$, not $-xt$.

Ooooooooh

the left side

RIGHT

I get it now

thanks


Hi,
When I was doing the first round, I realized right before I turned it in that I could do #3.  (I left #4 blank).  However, I just realized that by doing #3, I misaligned my #5 solution so it doesn't start on a new sheet of paper.  Is there a way I can ensure my grader will read mine correctly?

Thanks in advance.

EDIT: I forgot to mention my ID is 10653.

I think you may have to send an email to the graders and explain.

Remember to put your ID, name, and description of your problem.


Hello,

I think it would be nice to have some small section, may be not a forum but a small "bibliographic reference" for the cathegory that the users have. I mean, like the "P versus NP", "Hodge Conjecture", "Yang-Mills Theory", "Navier-Stokes Equations" and "Birch & Swinnerton Dyer", and the rest. Some are very well known, but others don't. Also, it would be nice to know why the person who decided them (Probably Valentin) decided for those! (I find kind of funny those cathegories)

Best regards,

Actually I came up with the stars, and Richard & Mathew had the nice idea of the 7 millenium problems :)


This figure has a total area of 108 square units and is a rectangle composed of three congruent, smaller rectangles. Find the sum of the length and width of one of the smaller rectangles. Express your answer in simplest radical form.
[asy]draw((0,0)--(3,0)--(3,2)--(0,2)--cycle);
draw((0,1)--(2,1));
draw((2,0)--(2,2));[/asy]

Well each rectangle has area 108/3=36. It's clear that the length is 2xwidth (looking at diagram) so 2w^2=36 so w=sqrt18=3sqrt2. 3sqrt2+2x3sqrt2=9sqrt2


Let $ ABCD$ be a square. Let $ M$ be the midpoint of $ DC$, $ N$ the midpoint of $ AC$, and $ P$ the intersection of $ BM$ and $ AC$. What is the ratio of the area of  $ \begin{array}{l} \Delta MNP \end{array}$ to that of the square $ ABCD$??

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[img]http://img231.imageshack.us/img231/8100/squarehx3.png[/img]

Let the length of a side of the square=1
Since N is the midpoint of AC, MN=$ \dfrac{1}{2}$.
To find the altitude of $ \bigtriangleup MNP$, you can do several things. One way is by positioning the square such that CD and AD are on the x and y axis, respectively.
So, A is at (0,1), B is at (1,1), C is at (1,0), and D is at (0,0).
The equation of the line containing AC is $ y = - x + 1$
The equation of the line containing BM is $ y = 2x - 1$

So, the point of intersection, P, is $ (\dfrac{2}{3} , \dfrac{1}{3} )$.

Thus, the altitude of $ \bigtriangleup MNP$ is $ \dfrac{2}{3} - \dfrac{1}{2} = \dfrac{1}{6}$.

The area of $ \bigtriangleup MNP$ is then $ \dfrac{\dfrac{1}{6} \cdot \dfrac{1}{2}}{2}$, or $ \dfrac{1}{24}$
The area of square $ ABCD$ is just 1, so the ratio of the area of triangle MNP to the area of square ABCD is $ \boxed{\dfrac{1}{24}}$.
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[hide="Non-Analytical"]
Refer to 27r's diagram and let a side of the square be $ s$. Now drop the altitudes from $ P$ to sides $ \overline{MN}$ and $ \overline{BC}$ and call the feet $ P_1$ and $ P_2$, respectively. Then since $ \triangle MNP\sim \triangle BCP$ (Why?) and $ PP_1\plus{}PP_2\equal{}\frac s2\implies PP_2\equal{}\frac s2\minus{}PP_1$, $ \frac{MN}{BC}\equal{}\frac{PP_1}{PP_2}\implies\frac{\frac s2}s\equal{}\frac{PP_1}{\frac s2\minus{}PP_1}\implies\frac12\equal{}\frac{PP_1}{\frac s2\minus{}PP_1}$
$ \implies2PP_1\equal{}\frac s2\minus{}PP_1\implies3PP_1\equal{}\frac s2\implies PP_1\equal{}\frac s6$. Thus, $ [MNP]\equal{}\frac12\cdot\frac s2\cdot\frac s6\equal{}\frac{s^2}{24}$, and the desired ratio is $ \frac{\frac{s^2}{24}}{s^2}\equal{}\boxed{\frac1{24}}$.
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1) $ (X_n)$ i.i.d. r. v. non negative, with $ EX_1 \in (0, \plus{} \infty)$.
$ S_n \equal{} X_1 \plus{} ... \plus{} X_n$,
$ M(t) \equal{} Sup(n: S_n \le t )$ for $ t \ge 0$.
Show $ P(M(t) < \infty) \equal{} 1$.

2) Let $ (X_n)$ and $ (Y_n)$ r. v. such that
$ Y_n$ are independent and $ Var(Y_n) \equal{} 1/n^2$,
$ \sum_1^{ \plus{} \infty}P(X_n \plus{} EY_n \ne Y_n) < \infty$.
Study the convergence a. s. of $ \sum_1^nX_i$.

1. $ A \equal{} \{M_t \equal{} \infty\} \equal{} \{S_n < t \ \forall n \}$, but $ \frac {S_n}{n}\to EX_1$, so $ S_n\to \infty$ a.s., hence $ P(A) \equal{} 0$.
2. The given condition implies that series $ \sum X_n$ converges if and only if $ \sum(Y_n \minus{} EY_n)$ converges, but since $ \sum E(Y_n \minus{} EY_n)^2 \equal{} \sum\frac {1}{n^2} < \infty$,  $ \sum(Y_n \minus{} EY_n) < \infty$ a.s.

Others problems:
1) $ (X_n)$ iid with $ E(X_n) \equal{} 0$, $ E(X_n^2) \equal{} 1$. Set $ S_n \equal{} \sum_1^nX_i$. Show
$ \frac {S_n}{\sqrt {n}\log{n}} \minus{} > 0$ a.s.
2) Let $ X_0$ have a uniform distribution on $ [0,1]$ and for $ n \ge 1$, $ X_{n \plus{} 1}$ has a uniform distribution on $ [0,X_n]$. Show that $ \frac {\log{X_n}}{n}$ converges a.s. and find the limit.
3) If $ F_n \minus{} > F$ in distribution, and if $ F_n$ and $ F$ have an unique median, then $ median(F_n) \minus{} > median(F)$.
4) Let $ (X_n)$ be independent r.v. such that
$ P(X_n \equal{} \minus{} 1) \equal{} P(X_n \equal{} 1) \equal{} \frac {1}{2n}$,
$ P(X_n \equal{} 0) \equal{} \frac {n \minus{} 1}{n}$.
Study the convergence in distribution of
$ \frac {(\sum_1^nX_i)^2}{\sum_1^n\frac {1}{i}}$.

It seems that I have to remind everyone of the [b]one problem per post[/b] policy.


The longest daisy chain documented was made in 7 hours by 16 people in Good Easter, Great Britain, and measured 6980 ft. 7 in. Given that each of the 16 people worked at the same rate and for the full 7 hours, how many feet of daisy chain per hour did each person make? Round your answer to the nearest whole foot.

im practicing for cd cus i suck lol

can anyone tell me of a good page to go when i need help on MC stuff?