Provide a synthesis of:

1) 3-ethylaniline from benzene;

2) PABA (p-aminobenzoic acid) from benzene;

3) 3-(p-bromophenyl)propanoic acid from toluene;

4) 1,3,5-tribromobenzene from benzene;

5) 2,6-dichloroaniline from benzene;

6) 2-methyl-4-nitroaniline from benzene;

7) 2-methyl-4-phenylbutan-2-ol from benzene;

8) 3-bromo-5-(2-bromoethyl)benzoic acid from toluene;

9) Isophthalic acid from toluene.

Hi
Here's my solution to a few of those:
1) 3-ethylaniline from benzene;
Ans:Steps:
* nitration
*NaNO2 + HCl
*CH3COOH
*Friedel Craft’s alkylation with C2H5Cl (only para-product)
*Hydrolysis (acidic)

2) PABA (p-aminobenzoic acid) from benzene;
Ans:
Use the above synthesis-method and at last oxidize using alk. KMnO4
5) 2,6-dichloroaniline from benzene;
*Sulphonation (para-product would dominate because of steric hindrance )
*Free radical chlorination (in excess of chlorine)

Hope the above methods are right...

[quote="Shyamkumar"]* nitration
*NaNO2 + HCl [/quote]

Nitro groups do not react with sodium nitrite in this context.

[quote="Shyamkumar"]*NaNO2 + HCl
*CH3COOH[/quote]

Assuming we have a diazonium group, what is the outcome by treating it with acetic acid?

[quote="Shyamkumar"]Friedel Craft’s alkylation with C2H5Cl (only para-product)[/quote]

Note 1: Friedel Crafts reactions cannot be made when amino groups are present in the ring.
Note 2: we want a meta product, not a para one.

[quote="Shyamkumar"]Use the above synthesis-method and at last oxidize using alk. KMnO4[/quote]

Oxidations with KMnO4 cannot be performed when an amino is in the ring.

[quote="Shyamkumar"]*Sulphonation (para-product would dominate because of steric hindrance )[/quote]

What steric hindrance?

[quote="Shyamkumar"]*Free radical chlorination (in excess of chlorine)[/quote]

In general, free radicals do not attack benzene rings.

I strongly advive you to review your Organic Chemistry knowledge, if that is relevant to you.

4.a)mono-nitration
    b)Sn/HCl
    c)Bromine/Water
    d)$ \ce{NaNO2/HCl}$
    e)$ \ce{H3PO3}$ $ \Delta$

[quote="Carcul"]
2) PABA (p-aminobenzoic acid) from benzene;

[/quote]

What about:

1. Friedel-Craft Alkylation ($ CH_3Cl/AlCl_3$) to get toluene
2. $ Br_2/FeBr_3$ and separate to get the para substituted product
3. $ KMnO_4/H_2SO_4/\Delta$
4. $ NH_2^{ \minus{} }/NH_3$

[quote="BanishedTraitor"]Just a question. Because for some questions I may be tempted to use Friedel-Craft Alkylation but we know that the activated benzene ring will give a range os substituted products. What should we do about that?
[/quote]

Just use a Friedel-Crafts Acylation.

[quote="Pardesi"]4.a)mono-nitration
b)Sn/HCl
c)Bromine/Water
d)$ NaNO_2/HCl$
e) $ H_3PO_3$ $ \Delta$
[/quote]

The correct acid should be hypophosphorous acid, $ H_3PO_2$, and don't forget the diazotization must be done at a temperature close to 0ºC, otherwise the diazonium salt decomposes as it forms. Also, we can make this synthesis more elegant by writing "$ d)\,\,NaNO_2/H_3PO_2/0^oC$" - this way, the diazonium salt is reduced as soon as it forms and we don't need a separate step.

[quote="BanishedTraitor"]1. Friedel-Craft Alkylation ($ CH_3Cl/AlCl_3$) to get toluene
2. $ Br_2/FeBr_3$ and separate to get the para substituted product
3. $ KMnO_4/H_2SO_4/ \Delta$
4. $ NH_2^\minus{}/NH_3$
[/quote]

First, the oxidations of alkyl groups to CO2H is done with hot [i]alkaline[/i] potassium permanganate: then in the end we acidify the mixture to liberate the acid. Second, in your last step we are treating a benzoic acid with a strong base (amide anion), which will result in an acid-base reaction.

thanks carcul .
when u check our answers it's almost like we are writing our school exams and we have a strict teacher...very strict  :)

for paba, benzene -> toluene
toluene -> p nitro toluene foll by free chlorination with excess of cl2 to p nitro benzo tri chloride foll by hydrolyis

VARUNARASU, why make a chlorination and then an hydrolysis, if you can do the same with a simple permanganate oxidation? While in the paper one extra step in a synthesis may not seem to much, in the laboratory that could mean one extra day of work.

sorry forgot to reduce no2 to nh2.
use sn/hcl b4 chlorination. i know simple permanganate oxidation should oxidize ch3 but wont it also oxidize nh2 group. hydrolysis will lead to formation of acid only without affecting nh2 group

hi
i am new to this site so help me out i just thought i could try this one
Isophthalic acid from toluene.
convert toluene to benzoicacid by adding kmno4 and then brominate it to get in the meta position and then add mg to form grignard and then add co2 and hydrolyse to get a -cooh instead of br and there it is isopthalic acid am i right???

6.2-methyl-4-nitroaniline from benzene
step 1 : nitrate
step 2 : Fe/HCl OH-
step 3 : Convert NH2 into NHCOCH3
step 4 : methylate by F & C
step 5 : nitrate
step 6 : OH-

4) 1,3,5-tribromobenzene from benzene;

hi back again i thought i could try this one too to benzene alkylate it and then oxidise using kmno4 and get benzoic acid now brominate it to get 3,5 dibromo benzoic acid and then add excess nh3 to convert the -cooh to conh2 and then do a hofmann bromamide reaction and the add nano2/hcl at 0c and do a sandmayer reaction to get a -br there thats all am i right???

[hey rohit perfect da but one small problem dont you think if you methylate you will also get methyl group in para position then how can you attach a NO2 group at para position ??? :D

It's not very wise to try to oxidize a phenol... (why?).

No but i think Chemrock wants to get the Benzoquinone product  :maybe:

[quote="Carcul"]Question: what is the general method to obtain unsaturated cycloketones from phenols?[/quote]

Hint: what reduction methods do you know for aromatic rings?

[quote="Carcul"]Not with potassium dichromate.[/quote]

Can we use CrO3 and O2 instead ?

[quote="Madness"]Can we use CrO3 and O2 instead?[/quote]

Of course not.

[quote="VARUNARASU"]birch reduction should do the trick[/quote]

Yes, that's right. Now, can you propose a synthesis of 4-n-Propylcyclohex-3-en-1-one?

can you tell us the answer Carcul????????

P.S Carcul can you tell me in which step my synthesis would not hold good??

No. Think by yourself, if you want.

P.S. I have already told you that, in my post immediatly after from yours.

Why not? My school guide tells me that to convert phenol into benzoquinone, CrO3 and O2 can be used  :maybe:

[quote="carcul"]
VARUNARASU wrote:
birch reduction should do the trick

Yes, that's right. Now, can you propose a synthesis of 4-n-Propylcyclohex-3-en-1-one?[/quote]
from what?

Sorry, I saw it rite now.
Benzene to phenol(the usual methods)
Phenol +bromination in CS2(para product)
Use si(me)3cl to protect oh group and react with mg in ether to give organometallic
+propylchloride to give para product
Now treat this with na,liq.nh3 foll by acidification (removes si(me)3cl and converts enol to keto)

Grignard reagents do not react with alkyl halides like that.

then how do they react. do you get an elimination reaction

No. Although that might look surprising, Grignard reagents do not react with alkyl halides in normal conditions. However, the reaction occurs when metal salt like $ Li_2CuCl_4$ is present (which convertes the Grignard into a Gilman-type reagent, which is more reactive towards alkyl halides). With this modification, your synthesis is correct. Here's another possibility:

[hide="Click here"]1) Benzene + CH3CH2COCl/AlCl3 -----> C6H5-CO-CH2CH3

2) C6H5-CO-CH2CH3 + CH3CO3H -----> C6H5-O-CO-CH2CH3

3) C6H5-O-CO-CH2CH3 + AlCl3 -----> p-CH3CH2CO-C6H4-OH

4) p-CH3CH2CO-C6H4-OH + CH_3I/NaHO -----> p-CH3CH2CO-C6H4-OCH3

5) p-CH3CH2CO-C6H4-OCH3 + NH2NH2/DMSO -----> p-CH3CH2CH2-C6H4-OCH3

6) p-CH3CH2CH2-C6H4-OCH3 + Na/NH3 -----> 1-Metoxy-4-n-propylcyclohexa-1,4-diene

7) 1-Metoxy-4-n-propylcyclohexa-1,4-diene + H3O+ -----> 4-n-propylcyclohex-3-en-1-one[/hide]


calculus
integration
logarithms
limit
floor function
real analysis
real analysis unsolved

[b](i)[/b] Catalan's integral

$ I: \equal{}\int_0^1 \frac{1}{1\plus{}x}\sum_{n\equal{}1}^\infty x^{2^n\minus{}1}\, dx\equal{}\sum_{n\equal{}1}^\infty \sum_{k\equal{}0}^\infty \int_0^1 (\minus{}1)^k\, x^k\, x^{2^n\minus{}1}\, dx$

$ \equal{}\sum_{n\equal{}1}^\infty \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{2^n\plus{}k}\equal{}\sum_{n\equal{}1}^\infty \sum_{k\equal{}2^n}^\infty \frac{(\minus{}1)^k}{k}$ .

Now define $ A_m': \equal{}\sum_{2^m\le k<2^{m\plus{}1}} \frac{(\minus{}1)^k}{k}$ and $ A_m: \equal{}\sum_{2^m\le k<2^{m\plus{}1}} \frac{1}{k}$ .

$ A_m'\plus{}A_m\equal{}\sum_{2^m\le k<2^{m\plus{}1}} \frac{(\minus{}1)^k\plus{}1}{k}\equal{}\sum_{2^m\le 2k<2^{m\plus{}1}} \frac{2}{2k}\equal{}\sum_{2^{m\minus{}1}\le k<2^m} \frac{1}{k}\equal{}A_{m\minus{}1}$

$ \Rightarrow A_m'\equal{}A_{m\minus{}1}\minus{}A_m$ .

And because of $ \sum_{m\equal{}n}^\infty A_m'\equal{}\sum_{k\equal{}2^n}^\infty \frac{(\minus{}1)^k}{k}$ we obtain $ I\equal{}\sum_{n\equal{}1}^\infty \sum_{m\equal{}n}^\infty A_m'$ .

And this is $ \sum_{m\equal{}1}^\infty \sum_{n\equal{}1}^m A_m'\equal{}\sum_{m\equal{}1}^\infty m A_m'$.

$ S_n: \equal{}\sum_{m\equal{}1}^n m A_m'\equal{}\sum_{m\equal{}1}^n m(A_{m\minus{}1}\minus{}A_m)$

$ \equal{}\sum_{m\equal{}1}^n A_{m\minus{}1}\plus{}\sum_{m\equal{}1}^n \Big((m\minus{}1) A_{m\minus{}1}\minus{}m A_m\Big)\equal{}\sum_{1\le k<2^n} \frac{1}{k}\minus{}nA_n$

$ \equal{}H_{2^n}\minus{}\frac{1}{2^n}\minus{}n\left[\left(H_{2^{n\plus{}1}}\minus{}\frac{1}{2^{n\plus{}1}}\right)\minus{}\left(H_{2^n}\minus{}\frac{1}{2^n}\right)\right]$

$ \equal{}H_{2^n}\minus{}n\left(H_{2^{n\plus{}1}}\minus{}H_{2^n}\right)\plus{}o(1)$ . And because of $ H_{2^n}\equal{}\gamma\plus{}\log 2^n\plus{}o(1)$

$ S_n\equal{}\gamma\plus{}n\log 2\minus{}n\Big((n\plus{}1)\log 2\minus{}n\log 2\Big)\plus{}o(1)\equal{}\gamma\plus{}o(1)$ . Hence $ I\equal{}\lim_{n\to\infty} S_n\equal{}\gamma$ .

Since $ \lfloor \log_2 k\rfloor \equal{}m$ for $ 2^m\le k<2^{m\plus{}1}$

Vacca's series $ \sum_{k\equal{}1}^\infty \frac{(\minus{}1)^k\, \lfloor \log_2 k\rfloor}{k}\equal{}\sum_{m\equal{}1}^\infty m \sum_{2^m\le k<2^{m\plus{}1}} \frac{(\minus{}1)^k}{k}$

is also $ \sum_{m\equal{}1}^\infty m A_m'\equal{}\lim_{n\to\infty} S_n\equal{}\gamma$ .


Four boys and three girls will be seated in a row.  When a boy is next to a girl, we will call this a "meeting point."  When the seven kids are seated, there may be only one one meeting point, as in BBBBGGG, or there may be as many as six meeting points, as in BGBGBG.  Given all of the possible seating arrangements for these seven kids, what is the average number of meeting points per seating arrangement?  Express your answer as a common fraction.

Answer [hide]24/7[/hide]

i assume you are asking for expected value
[hide]multiply the probablity of each happening by the number of meeting points. then divide by the total
$ \frac {1*2 \plus{} 2*5 \plus{} 3*12 \plus{} 4*9 \plus{} 5*6 \plus{} 6*2} {\binom{7} {4}}$

$ \frac {2 \plus{} 10 \plus{} 36 \plus{} 36 \plus{} 30 \plus{} 12} {35}$

$ \frac {126} {35}$

$ \frac {18} {5}$
[/hide]

what did i do wrong. probably counting.

I still haven't gotten this problem.  Can anyone help me??  :|

well...i think you can do quick counting

[hide]
0 meetings - 0
1 meeting - 2
2 meetings...

etc.

then just add and divide (there is a faster way but I want *you* to find it)[/hide]

mamathbug-

Pls. explain the fast solution.  I tried counting, but it took more than a page of casework!  :huh:

Thanks.

Listing out all combinations isn't as tedious as it seems.  If this was a target problem, as it probably is,  then the other problem in the pair should take around half a minute.  4 minutes is enough to list all of these out.

notice that you don't have to find the number of ways to rearrange the boys and girls (4!*3!) because it would cancel out in the numerator and denominator.

[quote]notice that you don't have to find the number of ways to rearrange the boys and girls (4!*3!) because it would cancel out in the numerator and denominator.[/quote]

Why did you do this, unimpossible?

[hide]For this problem, don't even bother considering $ 3!\cdot4!$. The number of permutations for every single arrangement (arrangements are BBBBGGG, GGBBBBG, etc.) will equal this. Rather, let us set up the equation: $ \frac{n_{1}\plus{}2n_{2}\plus{}3n_{3}\plus{}4n_{4}\plus{}5n_{5}\plus{}6n_{6}}{35}$.
We put 35 as our denominator because we know that there are 35 possibilities from calculating $ \frac{7!}{4!3!}$. Knowing this will also serve as a good way to check that we correctly count possible arrangements later on in this problem. Clearly $ n_{1}\equal{}2, n_{2}\equal{}5$. Now, we can count the possible combinations for the other meeting points. For $ n_{3}$, we see that the possible arrangements are GBGGBBB, GBBGGBB, GBBBGGB, GGBGBBB, GGBBGBB, GGBBBGB, and these arrangements backwards. Therefore, $ n_{3}\equal{}12$. For $ n_{4}$, we see that the possible arrangements are  BGBGGBB, BGBBGGB, BBGBGGB, BBGBGBG, these arrangements backwards, and GBBGBBG, which is symmetrical, so we don't count this backwards. Therefore, $ n_{4}\equal{}9$. For $ n_{5}$, we see that the possible arrangements are BGBGBBG, BGBBGBG, BBGBGBG, and these arrangements backwards. Therefore $ n_{5}\equal{}6$. It's pretty clear that $ n_{6}\equal{}1$. Now we check to see if we counted everything, and indeed we did. Now plug these values into the equation we previously created, and we get, $ \frac{2\plus{}2(5)\plus{}3(12)\plus{}4(9)\plus{}5(6)\plus{}6}{35}\equal{}\frac{120}{35}\equal{}\frac{24}{7}$.[/hide]

[quote="lokesh"][hide]For this problem, don't even bother considering $ 3!\cdot4!$. The number of permutations for every single arrangement (arrangements are BBBBGGG, GGBBBBG, etc.) will equal this. Rather, let us set up the equation: $ \frac {n_{1} \plus{} 2n_{2} \plus{} 3n_{3} \plus{} 4n_{4} \plus{} 5n_{5} \plus{} 6n_{6}}{35}$.
We put 35 as our denominator because we know that there are 35 possibilities from calculating $ \frac {7!}{4!3!}$. Knowing this will also serve as a good way to check that we correctly count possible arrangements later on in this problem. Clearly $ n_{1} \equal{} 2, n_{2} \equal{} 5$. Now, we can count the possible combinations for the other meeting points. For $ n_{3}$, we see that the possible arrangements are GBGGBBB, GBBGGBB, GBBBGGB, GGBGBBB, GGBBGBB, GGBBBGB, and these arrangements backwards. Therefore, $ n_{3} \equal{} 12$. For $ n_{4}$, we see that the possible arrangements are  BGBGGBB, BGBBGGB, BBGBGGB, BBGBGBG, these arrangements backwards, and GBBGBBG, which is symmetrical, so we don't count this backwards. Therefore, $ n_{4} \equal{} 9$. For $ n_{5}$, we see that the possible arrangements are BGBGBBG, BGBBGBG, BBGBGBG, and these arrangements backwards. Therefore $ n_{5} \equal{} 6$. It's pretty clear that $ n_{6} \equal{} 1$. Now we check to see if we counted everything, and indeed we did. Now plug these values into the equation we previously created, and we get, $ \frac {2 \plus{} 2(5) \plus{} 3(12) \plus{} 4(9) \plus{} 5(6) \plus{} 6}{35} \equal{} \frac {120}{35} \equal{} \frac {24}{7}$.[/hide][/quote]

lokesh - What does $ n_(subscript)$ stand for?


Putnam
number theory unsolved
number theory

prove that if the sum of divisors of N is 2N+1, then N is the square of an odd integer..

Let $n=2^{k}m$, were m is odd, then $\sigma(n)=(2^{k+1}-1)\sigma(m)$ odd if and only if m is square (because the sum of divisors d and m/d give even number).
Let $N=2^{k}n^{2}, \sigma(N)=2N+(2^{k+1}-1)\sigma(n^{2})-2^{k+1}n^{2}=2N+1$. It give $(2^{k+1}-1)\sigma(n^{2})=2^{k+1}n^{2}+1$.
It is easy to check, that n=p don't solution.

Hi I was just working on this problem and found this in the search.

I do not understand the solution (specifically how the last equation has no solutions)

Can anyone clarify?

PS: I'm not sure if I'm supposed to post here or make a new topic sorry.

From the last equation, you can get $2^{k+1}-1 | n^2+1$. Then $n^2+1$ has a prime factor $p=4m+3$ when $k>0$.
Or $n^2=-1 \mod p$. Then $n^{p-1}=(n^2)^{2m+1}=-1\mod p$, contradiction with Fermat!

Ahah, that's an easy one. The one I can solve...


A bug crawls from vertex to vertex on a cube along the edges. If the bug makes 7 moves, what is the probability that he visits every vertex exactly once?

[hide] There are 2 possibilities for the first 3 moves: They are all on the same plane, or they aren't.

Case 1: on the same plane. There is a probability of $1$ for the first move, $\frac{2}{3}$ for the second move, and $\frac{1}{3}$ for the 3rd move to keep it on the same plane.
Then, the path must go out of the plane for $\frac{1}{3}$ probability. Then, on the new plane, it can choose from 2 paths, clockwise or counterclockwise (kind of  :huh: ) for $\frac{2}{3}$. Then all the paths are set for $\frac{1}{3}$ and $\frac{1}{3}$.
The product is $\frac{4}{3^{6}}$

Case 2: $1$ and $\frac{2}{3}$. Then, after 3 of the points in the plane have been visited, the path must turn to another plane $\frac{1}{3}$. The path may only continue in one direction, clockwise or counterclockwise, or it will not be able to reach the 4th point of the first plane. $\frac{1}{3}$. Thereafter, all the paths are set $\frac{1}{3}$ and $\frac{1}{3}$.
The product is $\frac{2}{3^{6}}$

$\frac{4+2}{3^{6}}=\frac{3\cdot2}{3^{6}}=\frac{2}{3^{5}}=\boxed{\frac{2}{243}}$ [/hide]


geometry
trigonometry
circumcircle
ratio
inradius
geometry proposed

Three congruent circles have a common point O and the circles (taken in pairs) meet at A, B and C as well.
Let A'B'C' be the triangle containing the three given circles, such that every side of the triangle is tangent to exactly two of the circles.
Prove that $[A'B'C'] \geq 9[ABC]$. (Here [P] is the area of polygon P.)

I have a solution using trigonometry but is there a solution which does not use trig?

Sure there is :).

If $O_1,O_2,O_3$ are the centers of the three circles, then it's easy to see that $ABC$ and $O_1O_2O_3$ are congruent and $O_1O_2O_3$ is homothetic with $A'B'C'$ s.t. $B'C'\|O_2O_3$ and the distance between $O_2O_3$ and $B'C'$ is $R$, the common radius of the three circles, which is also the circumradius of $O_1O_2O_3$ (the same holds for the pairs of lines $(O_3O_1,C'A'),(O_1O_2,A'B')$, of course).

We need to show that the similarity ratio between $A'B'C'$ and $O_1O_2O_3$ is at least $3$. However, the inradius of $O_1O_2O_3$ is a number $r$, while the inradius of $A'B'C'$ is, according to the observations above, $r+R\ge 3r$, and we are done.


if $p$ be a prime number prove that $2^p+3^p$ can't be perfect power.

I think this one appeared many times here...
The key is to prove that $ 2^p+3^p $ is divisible by $ 5 $ but not by $ 25 $ (for odd $ p $).

Pierre.

except for $ p = 5 $ since $ 2^5+3^5=11*5^2 $ :) (but then $ 2^5+3^5 $ is not a perfect power as easily seen...)
I was just solving this stuff ... ;)


Here are some questions from someone who knows nothing about quants, the mathematics of finance, or the world of finance.

1)  Do quants deserve any blame for the current (Sept. 2008) U.S. financial crisis?

2)  How will this financial crisis affect the job market for quants?

3)  Does this financial crisis reveal that the mathematical models quants have been using are really a bunch of garbage?

4)  How have firms like Jane Street Capital been affected by this situation?

Although I also know little about quants, I will attempt to answer your question, because no one else has.  (Or you might try re-posting in the career forum.)

1.  Undoubtedly, some of the quants must shoulder some of the blame, but most are probably blameless.  From what I understand (which is very little), quants perform a wide variety of tasks.  The basic problem was that the risk various mortgage-backed securities (MBS) was seriously underestimated.  I would guess that at some banks, there were quants who were entrusted with the task of estimating risk, and maybe they blew it.  But the biggest problem was that the banks were buying up tons of MBS that was graded AAA (super-duper safe), and they may have made no independent assessments of the risks.  If that's the case, then it was some non-quant management person's fault for not doing due diligence, and also the fault of the agencies that rated this crap AAA.  Presumably the ratings agencies had quanty people on board, because you need people like that to run the numbers and assess the risk in order to come up with these AAA grades.  I would say that more than anyone else, these are the people who dropped the ball in the worst possible way.

I believe that most quants work for hedge funds, which are not to be confused with the banks.  I don't think that the hedge funds really had a hand in creating this mess, but I could be wrong.  The effect of hedge funds on the rest of the world seems mysterious and subtle to me.

2.  I would assume that it will have an effect in the short-term, but the long-term is anyone's guess.  Like I mentioned before, most quants work in the hedge funds, each of which may or may not have been burned by recent events.  But since everyone wants to get their hands on their capital right now, they might remove a lot of money from the hedge funds.

3.  Some people somewhere were definitely using garbage models.  The ratings agencies for sure.  This isn't so surprising since the ratings agencies get paid by the very customers who they are rating, and they apparently have no accountability to anyone.

4.  I believe they're a hedge fund, so I have no idea.

I invite better posts from people who are more knowledgeable.


algebra
function
domain
integration
real analysis
real analysis theorems

I want to show that $ \mu(\liminf E_n)\le \liminf \mu (E_n)$ using Fatou's lemma. Here $ \mu$ is a measure with domain $ \mathcal{A}$ and $ E_n$ is a sequence of sets in $ \mathcal{A}$.

Here is my approach: set $ \mu(E_n) \equal{} \int \chi_n d\mu$. Then we have $ \mu(\liminf \chi_n d\mu) \leq \liminf \int \chi_n d\mu \Rightarrow$ $ \mu (\liminf E_n) \leq \liminf \mu(E_n)$. I think it is basically correct, but something seems wrong here.  :mad: Can it be fixed?

Los

:maybe:

When you say $ \mu(\liminf A)\leq \liminf\mu(A)$ do you mean $ \mu(\liminf E_n)\leq \liminf\mu(E_n)$?

ähm, yes actually :-) I have made the correction

Los

Hmm...I would probably say:

$ \mu(E_n) \equal{} \int\chi_{E_n}d\mu$, so

$ \mu(\liminf E_n) \equal{} \int\liminf\chi_{E_n}d\mu\leq \liminf\int\chi_{E_n}d\mu \equal{} \liminf\mu(E_n)$

(I can't seem to make the $ E_n$ look like they are subscripted on the chi's, but they are.)

But I'm only in an undergrad measure theory class right now, so am probably missing some detail where you have to write everything as intersections so they are decreasing or something.

You'd still have to prove that $ \liminf \chi_{E_n}\equal{}\chi_{\liminf E_n}$ though !

but this is trivial given what the identity function is.  :D


This is a problem proposed by me.

Let $x,y,z>0$ satisfies$xy+yz+zx=1$ Prove that \[\sum \frac{1}{(2x(x+y+z)+7)^{\frac32}}\leq \frac19\]

.I find a strong inequality and I change it to this weeker inequality which looks harder :P. Theoretically,it can be solved by SOS because it is not strong enough like some inequalities given by Vasc.But SOS brings a lot of calculating.I'll post the orginal inequality some times.

I am very sorry that no one are interested in my post....... :huh:

Come on Hawk Tiger, what is your original inequality? I guess it is something without the $\tfrac32$.

Yes ,you are right.And I have found that my orginal inequality has been posted in this forum. :P.
I think it is not a strong inequality,and it may can be proved derectly.

I'll post my orignial inequality day after tomorrow evening if no one solve it.

My computer doesn't work last night.Let me give you some hint:[hide]Try to get rid of the $\frac32$ by some inequality :P [/hide]

Using Bernoulli's Inequality, we stil have to show that
$\sum\frac 1{x(x+y+z)+2}\leq 1$.
This is equivalent to
$\sum\frac x{x(x+y+z)+2}\geq \frac 1{x+y+z}$.
Cauchy-Schwarz Inequality ends the proof. :)

Totally same as my proof!!I really admire you capacity for solving inequality,Vasc.

And I have found the inequality in Vasc post is a problem from China TST.


geometry
circumcircle
parallelogram
geometry proposed

Let $ ABCD$ be an inscribed quadrilateral and let $ E$ be the midpoint of the diagonal $ BD$. Let $ \Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4$ be the circumcircles of triangles $ AEB$, $ BEC$, $ CED$ and $ DEA$ respectively. Prove that if $ \Gamma_4$ is tangent to the line $ CD$, then $ \Gamma_1,\Gamma_2,\Gamma_3$ are tangent to the lines $ BC,AB,AD$ respectively.

Using an inversion of pole D, the figure becomes a paralelogram $ DA'E'C'$, with $ B'$ the midpoint of $ DE'$. The conclusion is trivial from here.
If someone wants further explanations, please ask.


Could anybody tell me of  real-world applications of diophantine equations?
(What about higher degree equations?)

Thanks a lot.

Probably nothing right now. ;)  But people are just discovering some of the uses of number theory now, especially in cryptography.


Given $ a, b, c > 0$. Prove that:
$ \frac{1}{a\plus{}b} \plus{} \frac{1}{b\plus{}c} \plus{} \frac{1}{c\plus{}a} \geq\ \frac{a\plus{}b\plus{}c}{2(ab\plus{}bc\plus{}ca)} \plus{} \frac{3}{a\plus{}b\plus{}c}$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq\ \frac {a \plus{} b \plus{} c}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {3}{a \plus{} b \plus{} c}$
 :)[/quote]
Multiplying by $ (a \plus{} b \plus{} c)$ the inequality becomes:

$ \sum \frac {a}{b \plus{} c} \geq \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}$

which is obvious by C.S as follow:

$ ( a.(b \plus{} c) \plus{} b.(a \plus{} c) \plus{} c.(a \plus{} b) ).( \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} ) \geq (a \plus{} b \plus{} c)^2$


Circle $O$ is circumscribed about the equilateral triangle $ABC$. $P$ is a point on the arc $AB$ of the circle $O$.

Prove that $PA+PB=PC$

(Is this hard for Getting Started?)

Apply Ptolemy's Theorem on cyclic quadrilateral $APBC$.
$(PC)(AB)=(PA)(BC)+(PB)(AC)$
Since $ABC$ is equilateral, $AB=BC=AC$, so we have our desired $PC=PA+PB$.

Yup, that's it :)

There was another way to prove this (without using Ptolemy), but I don't remember...

Can anyone do it?


When several thieves tried to divide a sum of money by giving $ \$4$ to each thief, one theif received nothing. When each thief took $ \$3$, they had $ \$1$ left over. What is the sum of the number of dollars and the number of thieves?

Ok but how did you get the number of dollars and thieves?

[quote="dragon96"]You can also use algebra:

Let t be number of thieves.
We have
4(t-1)=3t+1
4t-4=3t+1
t=5
Sum of money=16
16+5=21[/quote]
That makes sense now

[quote="AIME15"]i think brute force would actulaly be faster in this situation.[/quote]

Is brute force just guessing and checking?

A lot of the countig problems on ftw can be brute forced quickly :P

But it's important to know the "real" way to solve problems. Remember, the process is more important then the solution!

But on FTW, if there's like 12 seconds to solve it, then you brute force it?!?

It's a good idea if you can, just make sure you know how to solve it for real.

Slightly off topic: Is it possible to solve a problem in 5 seconds, like this one??

Yes. If you randomly guess that there are 5 thieves and get 16 dollars, you might get it.


Hey!

So my situation is quite similar to that of Alumen's, who wanted me to join this forum, since his friend, Andrew Qian, wanted more representation from the San Mateo-Foster City area in this forum.  I'm a freshmen at Aragon HS, a school pretty much out of the entire mathematics loop (although, one person did earn a gold a few years ago at the imos, that's nothing compared to gunn and paly  :) ).  Nonetheless, I am currently working with the math department to develop a better math program/ team and we will probably begin to participate in the local competitions next year.  Now, there is probably no chance that any of you know me, as I placed a relatively low 11th place at states mathcounts last year and have not participated in many other things than school-based amcs.

So...to make a long story short- Hi!

[quote]although, one person did earn a gold a few years ago at the imos[/quote]
That's quite an achievement! Few students in the Bay Area got that honor.
Congratulations!

[quote="CircleSquared"]Welcome to the bay area forums!
Here's your room key, the elevator is to the right.  Enjoy your stay.

11th place at states isn't bad.[/quote]
After talking to him on gtalk, he seems to have a Ubeyish attitude ;) don't be surprised if more of that stuff comes.

[quote]although, one person did earn a gold a few years ago at the imos[/quote]
Lol, you call 8 years ago a few? That's too long ago to attribute to our school... lol.
Anyways hi Omelette XD

[quote="DataBox"][quote="CircleSquared"]Welcome to the bay area forums!
Here's your room key, the elevator is to the right.  Enjoy your stay.[/quote]

that's a rather lively welcome, hopefully i will enjoy my stay  :lol:
anyways, if i may inquire, what is "a Ubeyish attitude?"
fine, alumen, "although, one person did earn a gold at the 2000 imos"[/quote]
Oh, you'll find out :)

ohhh not at all, mr. databox.  :evilgrin:

[quote="DataBox"]:huh: that sounds ominous[/quote]

Yeah, they get weird.

I think its the pressure, but don't mind me - I fail at math.

 :lol:

[quote="happyme"][quote="xscapezaer"]@databox: If you [i]really[/i] want to know, seek out the one they call Ubemaya...  :ninja:

...or don't. Personally, I'd much recommend the latter, and so would a lot of ppl. Including Ubemaya himself, strangely enough :huh: (There's a hint there, that is) Eventually, you'll run into him, and then...you'll wish you hadn't (Ubemaya, this is you cue to burst in laughing evilly)[/quote]

Since 'Bert appears to not be present (or hes sulkily crying about his misfortunes with Meggy-Meg).

BWHAHAHAHAHAHAHHAHAHAH! TEAPOTSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS.

[/bad impression][/quote]
Yes horrible impression indeed...
I think he'll need another cue tomorrow :D

Ah, there we go. DataBox, meet Ubemaya...

EDIT: huh, I just noticed this. Ubemaya, ur accomplish-...er, I mean, "failures" are out of date...

EDIT again: I mean, in your signature...

Not to mention, those MathCounts failures are soon gonna be changed...

Well, I'll see you getting your pointy trophy there!

[quote="DataBox"]....Hello Ubemaya?[/quote]
Mr. SAXENA (tis what you get for using my name), you don't wait and expect other people to see the thread and necessarily answer...on a forum that's how things work.

sorry Mr. serialk11r, I am not well versed in the decorum of this forum....you may want to chill out  :|

[quote="CircleSquared"][quote="Ubemaya"]Okay, I updated my sig.

Of course, my second goal is WAY more important than my first. I'M BOUND TO GET LUCKY SOMEDAY!![/quote]

And I updated my avatar.  I'd like comments. (I would update my sig but the equation takes up the first 220 characters or so)[/quote]
Go thread hijacking!
Its decent, I like the older one better simply because it was the one that was there when I joined.

[quote="CircleSquared"][quote="Ubemaya"]Okay, I updated my sig.

Of course, my second goal is WAY more important than my first. I'M BOUND TO GET LUCKY SOMEDAY!![/quote]

And I updated my avatar.  I'd like comments. (I would update my sig but the equation takes up the first 220 characters or so)[/quote]

What does your sig's equation even mean? I'm lost  :| .

[quote="bowei"]i dont get the new one
but i did get the old one
and it was a cute smiley
so i liked that one better
maybe if you explain the new one
it will be funny or smth?[/quote]

The small red thing is a panic button.

:D

The equation is just some complicated stuff....

I should get myself a new avatar!

[quote="archimedes1"]Off topic, but I actually own one of those plastic "PANIC" keys...  Unfotunately, it doesn't fit when I try to stick it on my keyboard. :)[/quote]

Aside from serial's computer, that is the 2nd coolest thing I have ever heard of.

Well, simply put.. Ubemaya has a nice long history about complaining how poorly he performs on all math related things, when in reality, he scores quite highly and makes the rest of us poor schmucks feel bad.
 :lol:

He's a really cool guy in person xD

[quote="serialk11r"][quote="junggi"][quote="DiscreetFourierTransform"]
He's a really cool guy in person xD[/quote]
eh i dont get how he acts different off line

dude i swear there was someone who looks exactly like albert at SMT[/quote]
From homestead right? he was on my team (you people placed me on the leftover team ARGH and I think I did better than some of the people on the team). I think Albert acted just like he does on forums in chinese school last year, but on gtalk he's pretty serious.[/quote]

Dude ya, I was like wait is that Albert?? I think he was like a junior or senior, I'm  not sure...

[quote="sonny"][/quote][quote="serialk11r"][quote="junggi"][quote="DiscreetFourierTransform"]
He's a really cool guy in person xD[/quote]
eh i dont get how he acts different off line

dude i swear there was someone who looks exactly like albert at SMT[/quote]
From homestead right? he was on my team (you people placed me on the leftover team ARGH and I think I did better than some of the people on the team). I think Albert acted just like he does on forums in chinese school last year, but on gtalk he's pretty serious.[/quote][quote="sonny"]

Dude ya, I was like wait is that Albert?? I think he was like a junior or senior, I'm  not sure...[/quote]
He was a junior. Apparently steven (krustyteklown) shouted "albert albert!" according to junggi, and the guy who looks like albert, his friend apparently was named albert or something and turned around. lol

[quote="serialk11r"][/quote][quote="sonny"][/quote][quote="serialk11r"][quote="junggi"][quote="DiscreetFourierTransform"]
He's a really cool guy in person xD[/quote]
eh i dont get how he acts different off line

dude i swear there was someone who looks exactly like albert at SMT[/quote]
From homestead right? he was on my team (you people placed me on the leftover team ARGH and I think I did better than some of the people on the team). I think Albert acted just like he does on forums in chinese school last year, but on gtalk he's pretty serious.[/quote][quote="sonny"]

Dude ya, I was like wait is that Albert?? I think he was like a junior or senior, I'm  not sure...[/quote][quote="serialk11r"]
He was a junior. Apparently steven (krustyteklown) shouted "albert albert!" according to junggi, and the guy who looks like albert, his friend apparently was named albert or something and turned around. lol[/quote]

Whoa, whoa. Wait. We only had one junior on our team (who left before award ceremony anyway). He did CERTAINLY NOT look like Albert.

But we had two seniors, one of whom has a slight resemblance to Albert. (In case you know who he is, I'm talking about A Wong.)

[quote="vishalarul"][/quote][quote="serialk11r"][/quote][quote="sonny"][/quote][quote="serialk11r"][quote="junggi"][quote="DiscreetFourierTransform"]
He's a really cool guy in person xD[/quote]
eh i dont get how he acts different off line

dude i swear there was someone who looks exactly like albert at SMT[/quote]
From homestead right? he was on my team (you people placed me on the leftover team ARGH and I think I did better than some of the people on the team). I think Albert acted just like he does on forums in chinese school last year, but on gtalk he's pretty serious.[/quote][quote="sonny"]

Dude ya, I was like wait is that Albert?? I think he was like a junior or senior, I'm  not sure...[/quote][quote="serialk11r"]
He was a junior. Apparently steven (krustyteklown) shouted "albert albert!" according to junggi, and the guy who looks like albert, his friend apparently was named albert or something and turned around. lol[/quote][quote="vishalarul"]

Whoa, whoa. Wait. We only had one junior on our team (who left before award ceremony anyway). He did CERTAINLY NOT look like Albert.

But we had two seniors, one of whom has a slight resemblance to Albert. (In case you know who he is, I'm talking about A Wong.)[/quote]
Oh okay, then we must've gotten them mixed :P
Nice, you have 1337 posts :D


Okay this game is very simple. Every time you post, you can raise one administrator/staff person's point by one and decrease one administrator/staff person's point by one. No double posting, no raising and decreasing the same admin's points:

(Note to all staff: Don't take this too seriously, it's only a game.)

Original List
-------------
  DPatrick 30
  levans 30
  markan 30
  nsato 30
  rrusczyk 30
  Valentin Vornicu 30
  Rusczyk 30

My turn
----------

  DPatrick 30
  levans 30
  [color=red]markan 29  Sorry.[/color]
  [color=green]nsato 31[/color]
  rrusczyk 30
  Valentin Vornicu 30
  Rusczyk 30

My turn
----------

  DPatrick 30
  [color=green]levans 31[/color]
  markan 29
  nsato 31
  rrusczyk 30
  Valentin Vornicu 30
  [color=red]vRusczyk(???) 29 Sorry, you are all cool![/color]

My turn
----------

  DPatrick 30
  levans 31
  markan 29
  nsato 31
  rrusczyk 30
  [color=green]Valentin Vornicu 31[/color]
  [color=red]vRusczyk 28 well, I haven't seen you post at all....[/color]

My turn
----------

DPatrick 30
levans 31
markan [color=red]28[/color]
nsato 31
rrusczyk [color=green]31[/color]
Valentin Vornicu 31
vRusczyk 28

DPatrick 30
levans 31
markan 28
nsato 31
rrusczyk [color=green]32[/color]
Valentin Vornicu 31
vRusczyk [color=red]27[/color]

DPatrick 30
levans [color=green]32[/color]
markan 28
nsato 31
rrusczyk 32
Valentin Vornicu 31
vRusczyk [color=red]26[/color]

Levans is a BEAST (even though he has me muted until I stop spamming in the forums, which is equivalent to a permanent mute :D )

DPatrick 30
[color=green]levans 33[/color]
markan 28
nsato 31
rrusczyk 32
[color=red]Valentin Vornicu 30 [/color]
vRusczyk 26

[color=orange]DPatrick 30 [/color]
[color=green]levans 33[/color]
[color=red]markan 27[/color]
[color=orange]nsato 31
rrusczyk 32
Valentin Vornicu 30
vRusczyk 26[/color]

i did orange cuz yellow is hard to see

DPatrick 30
levans 33
markan 27
nsato 31
[color=green]rrusczyk 33[/color]
Valentin Vornicu 30
[color=red]vRusczyk24[/color]

DPatrick 30
[color=green]levans 34 [/color]
markan 27
nsato 31
rrusczyk 33
Valentin Vornicu 30
[color=red]vRusczyk23[/color]

im sorry, youre all awesome, but levans is too cool

DPatrick 30
[color=green]levans 35 [/color]
[color=red]markan 26[/color]
nsato 31
rrusczyk 33
Valentin Vornicu 30
vRusczyk 23

DPatrick 30
[color=green]levans 36[/color]
markan 26
nsato 31
rrusczyk 33
[color=red]Valentin Vornicu 29[/color]
vRusczyk 23

DPatrick 30
[color=green]levans 37[/color]
[color=red]markan 25 [/color]
nsato 31
rrusczyk 33
Valentin Vornicu 29
vRusczyk 23

DPatrick 30
[color=green]levans 38[/color]
markan 25
nsato 31
rrusczyk 33
Valentin Vornicu 29
[color=red]vRusczyk 22[/color]

lol this will be funny...
DPatrick 30
[color=green]levans 39 [/color]
[color=red]markan 24[/color]
nsato 31
rrusczyk 33
Valentin Vornicu 29
vRusczyk 22

omg levans will beast this whole game
every1 idolizes levans
he has the ownagenest sounding name out of all of them
(i mean im not sure how to pronounce rrusczyk)

I want it to end...
rrusczyk - 59
nsato - 20
However, I agree that TZF is being dictator-ish.

rrusczyk - 59
nsato - 18

this is going to end soon :D

rrusczyk - 60
nsato - 16

I don't know either of the people in the AoPS staff, but I know that rrusczyk wrote the AoPS books, and I got 2 of them.

Sorry nsato...
[color=green]rrusczyk - 61 [/color]
[color=red]nsato - 15[/color]

If all ends well, 7 more posts.
If someone up there hates us, 74 more...

rrusczyk - 62
nsato - 13

rrusczyk - 66
nsato - 5

We're almost done. I'm so excited;  :)

Valentin Vornicu- 100
rrusczyk - 67
nsato - 3

rrusczyk - 68
nsato - 1

Finish it off ernie.

[color=green][b]rrusczyk 69[/b][/color]
[color=red][b]nsato 0 (eliminated)[/b][/color]
=================================

[b] :winner_first:  Valentin Vornicu
 :winner_second:  rrusczyk
 :winner_third:  nsato[/b]
[color=blue][b]HM1.[/b][/color] levans
[color=blue][b]HM2.[/b][/color] DPatrick
[color=blue][b]HM3.[/b][/color] vRusczyk
[color=blue][b]HM4.[/b][/color] markan

See post 341, as well as ernie's post before your last move.

Oh, that is why people kept on adding Valentin Vornicu to the list.


[quote="caubetoanhoc94"]Let a;b;c are a;b;c be the sides leight of a triangle;$ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ \sum \sqrt [3]{\frac {2a^3}{b^3 \plus{} 5c}}\ge \sqrt [3]{9}$
 :)[/quote]

It's easy by holder  :)

Holder :
$ LHS^3( \sum a(b^3 \plus{} 5c) ) \ge 2.81$
Next, we prove:
$ 18 \ge \sum ab^3 \plus{} 5ab$
Which is trivial if using Vasc's lemma : $ (\sum a^2)^2 \ge 3 \sum ab^3$ . ( $ a,b,c$ are sides length of a triangle , therefore :$ 2\sum  ab \ge \sum a^2$ )


I noticed that nobody has been doing anything in this forum for a while, so I decided to make my own problem.  Here it is:

Find a way to evaluate the remainder of 593402 when divided by 101 without doing the actual arithmetic.

Define "actual arithmetic".  Just about all problems make you do at least a little.

That doesn't account for what I'm pretty sure you mean, which is $ 593402\div 101$.

You can express $ 593402$ as $ 505000\plus{}80800\plus{}7070\plus{}505\plus{}27$, so the answer would be $ \boxed{27}$, assuming I did that correctly.

That's an excellent solution.  I was thinking more along the lines of using the fact that 101 is 1 more than 100, and hence using the fact that any power of 10 is congruent to -1 (mod 101), but I'm impressed by that.

Well remember that mods are built around different bases so by going into a different base (like 101 although don't ask me how), it could help.

When you defined "actual arithmetic" as prime factorization, that made me think that you could just use long division. Was this intended or not?

Your definition was a little vague.

to do it without actual arithmetic, i'd rather just pull out my handy desk calculator.  :P


geometry
geometric transformation
homothety
circumcircle
incenter
ratio
cyclic quadrilateral

prove the lemmaRL22022006

at

http://www.mathlinks.ro/Forum/viewtopic.php?t=37236

Let $I_a,I_b,I_c$ be the excenters and $AD,BE,CF$ the internal bisectors of $ABC$ triangle.
We known $EF,BC,I_cI_b$ concur at $P$. Now $PEF$ cut the circumcircle of $II_bI_c$ at $K', L'$.($I$ is the incenter of $ABC$).If $M$ (midpoint of $I_bI_c$) is the center of cyclic quadrilateral $BCI_bI_c$. Now $M$ is the orthocenter of $I_aPI$, (by an old theorem: In a cyclic quadrilateral the circumcenter is the orthocenter of its autopolar triangle) $PI$ cut $MI_a$ at $Q$, now $PQ$ is orthogonal to $MI_a$. Then $PIQ$ is the radical axis of the two circumcircles of the quadrilaterals $I_aCIB$, $IAMQ$ and $II_bI_c$ triangle. So: $PC.PB=PI.PQ=PI_b.PI_c=PA.PM$. Further $EF$ meet the circumcircle of $II_bI_c$ at $K', L'$  we have: $PC.PB=PK'.PL'=PA.PM$, i.e. $C,K',A,M, L',B$ are on the circumcircle of $ABC$.Therefore $K'$ and $L'$ concides with $K,L$.
It is well known fact the circumcircles of $II_bI_c, II_aI_b, II_cI_a$ and $I_aI_bI_c$ are equals with radii $2R$, where $R$ is the circumradius of $ABC$. The rest is easy.

if P is the A-excenter of ABC and K and L the ends of a chord of circumcircle of ABC through the feet of B- and C-bisector on AC and AB respectively, then c(IKL), where I is the incenter of ABC, is homothetic with c(ABC) with center P and ratio 2.

My Proof
Let Q and R be the B-excenter and C-excenter of ABC. It is easy to see that ABC is the orthic triangle of  PQR and I is the orthocenter of ABC. The circum circle of ABC is the nine-point circle of PQR. Is easy to verify that c(IQR) is homothetic with c(ABC) with center P and ratio 2.
Since AICQ is cyclic, by radical axes theorem it holds  that KL, AC and QI concur at BI/\AC. This means that L lies on c(IQR). Similarly we can deduce that K lies on c(IQR). That’s all it was claimed.

[quote="juancarlos"]by an old theorem: In a cyclic quadrilateral the circumcenter is the orthocenter of its autopolar triangle[/quote]

Whould you please formulate this theorem? What is an autopolar triangle of a cyclic quadrilateral?

You can download this theorem in the pdf file:
http://www.mathlinks.ro/Forum/viewtopic.php?t=15133
(PQS is the autopolar triangle of ABCD).


Hi, this is my first post. I'm stumped on this question. Sorry there's no pic.

Let Triangle ABC be an isosceles triangle(AB=AC) with Angle BAC=20 degrees. Point D is on side AC such that Angle DBC=60 degrees. Point E is on side AB such that Angle ECB=50 degrees. Find, with proof, the measure of Angle EDB.

Can you please try to find a solution not involving trigonometry? Thanks. :P

It'll be very helpful to draw it out.

THANKS, ANDREW TOM!

I did this some time ago and recently came across a number of solutions as well as similar problems and their solutions here: http://www.cut-the-knot.org/triangle/80-80-20/index.shtml#Original

Hope this helps you.


group theory
abstract algebra
advanced fields
advanced fields unsolved

Hello, below is a problem I was not successful at solving.
Since we have not yet covered Langrange's Theorem on the order of a group and the order of its subgroup, I am not suppose to use it. Here is the problem.

Let $\phi : G \rightarrow G'$ be a group homomorphism. Show that if $|G|$ is finite, then $|\phi[G]|$ is finite and is a divisor of $|G|$.

You probably don't have first isomorphism theorem either then. That would make this nice. Suppose $\varphi : G \to G'$ is a homormorphism and that $|G|<\infty$, then by the F.I.T. $\text{im}\varphi \cong G/\ker\varphi$, so $|\text{im}\varphi |=\big| G/\ker\varphi \big| = \frac{|G|}{|\ker\varphi |}$ .

I am not sure if we are allowed to use F.I. T either. Your proof looks like a nice one. Also the proof that uses Langrange's is very short. But too bad, the poor lecturer thinks there is a simpler proof, which I am not seeing (perhaps I am not trying harder).

The most elementary way I can think of is to show that all fibres over $\phi(G)$ have the same number of elements, i.e. $\#\phi^{-1}(\phi(g))$ does not depend on $g$. Then, $\#G$ is the product of $\#\phi(G)$ and this fixed number. owk

How about showing that the $card \phi^{-1}(x) = card ker(\phi)$ ?

Pick $y\in\phi^{-1}(x)$. Define a map $\ker (\phi)\to \phi^{-1}(x)$ by $z\mapsto yz$. It's a bijection, with the inverse given by $z\mapsto y^{-1}z$.


parameterization
algebra unsolved
algebra

[color=blue][b]Let the system of equation:

$ \{ \begin{array}{l} x(2x + y + 1) = 4 + m \\  {(1 + 2x)^3} = (m + 3\sqrt 3 )y \\  \end{array} $

Find all the values of $ m$ parameter such that the system of equation has the roots $ x;y>0$ with $ y\le x^2$[/b][/color]

[color=blue][b]May this have no solution?????????????????[/b][/color]


[u][b]Final Participants for Who Wants to be a Mathematician:[/b][/u]

[list]- George Xing: [b]Winner of $ \$$3000[/b]. Rebecca Jacobs: [b]2nd Place[/b].
- Raku Watari, Stephen Yoo, Kevin Yin, Ricky Demer, David Ke, Oleksandr Stubailo, Hansen Han, Matthew Alexander[/list]

[u][b]Final Participants for Southern California Mathematical Olympiad:[/b][/u]

[list]- [b]Green Team: The Winning Team: 12 Points, won Tie Breaker.[/b] Benjamin Cosman, David Yang, Jocelyn Tang, Jason Jong, Tuan Le, Wendy Hsu, Michael Murata, Jan Wimmer
[b]- Yellow Team: 2nd Place: 12 Points.[/b] Richard Zhang, Devin Corr-Robinett, Stephanie Tai, Bryan Bocong Chen, Ben Barard, Zongwei Li, Christopher Underhill, Ryan Yoo
[b]- Blue Team: 8 Points.[/b] Johnny Li, Brandon Sim, Chelsey Qiaoshu Lin, Derek Ouyang, Katharine Daly, Gregory Simonian, Hugh Bahe, Sean Wang
[b]- Red Team: 5 Points.[/b] Dai Yang, Edward Woo, Jong Wook Kim, Su-Hwa Yu, Richard Cho, Eba Kim, Tiffany Chen, Sonam Dhiman, Michael V. Loginov[/list]

Photos are viewable on Math Zoom [url]websitehttp://www.mathzoom.org[/url].

Recap of the day's events:

[list]9:00am - Attendees arrive.  Contestants for "Who Wants to be a Mathematician" went on stage to practice.
9:45am - Kevin Wang (DrKevin): welcome the audience, introduce the hosts for WWTBAM: Mike Breen and Bill Butterworth.  Explanation of rules for SCMO Team Contest.
9:55am - DrKevin announces that keynote speaker of the day - Dr. Paul Dreyer - is stuck in traffic.  If he doesn't arrive in five minutes, SCMO shall start ahead of time.
9:58am - The cell phone rings.  Paul Dreyer on the phone (speakerphone audible to the audience) saying he will be right here.
9:59am - Paul Dreyer arrives, right on time for his talk.
10:00am - Keynote Speech: Six Degrees of Graph Theory: Kevin Bacon, Paul Erd$ \"o$s, William McKinley and me by Dr. Paul Dreyer
10:45am - Southern California Mathematical Olympiad - Team Contest begins.  Contestants are divided into Red, Blue, Green, and Yellow Teams.  Score Chart was displayed on screen to show the current scores of each time, as well as how many chances are still available for each time to answer each question.
11:45am - The Team Contest finishes: Yellow Team and Green Team ties at 12 points.  Each team selects a representative for the tie-breaker: Richard Zhang for Yellow Team, and David Yang for Green Team.
11:50am - Green Team (David Yang) wins tie-breaker.
12:00pm - Who Wants to be a Mathematician first round.
12:25pm - Kevin Yin and Rebecca Jacobs ties for the first round.  Rebecca Jacobs wins the tie-breaker.
12:30pm - Who Wants to be a Mathematician second round.
12:50pm George Xing wins the second round.
1:00pm - George Xing and Rebecca Jacobs square-off round.  George Xing wins.
1:00pm-1:30pm - Award Ceremony.

[/list]

Photos and report from Mike Breen are posted on the AMS website:

[url]http://www.ams.org/wwtbam/archive/la08.html[/url]


On the refrigerator, MATHCOUNTS is spelled out with 10 magnets, one letter per magnet.  Two vowels and three consonants fall off and are put away in a bag.  If the Ts are indistinguishable, how many distinct possible collections of letters could be in the bag?

[hide]
Vowels: AOU
Consonants: TTMHCNS

There are clearly 3 combinations of vowels that can come off. For the consonants, we have two cases: two T's fall off and one T falls off. If two T's fall off, we just choose one of the other 5, if one falls off, we choose to of the other 5, giving 15 total. Hence, there are $3\cdot15=45$ such collections.[/hide]

[quote="Anaxerzia"]I think he forgot the possibility that no T's fall off - but I think even that is off...[/quote]

Oh right, but that does give the correct answer, since we would add $3{5\choose2}=30$. I believe in my solution I intended that the second case was "either zero or one T falls off," which would result in ${6\choose3}=20$.

oops - yeah i had my combinatorics off heh :blush:

[hide]
There are 3 vowels: A, O, and U
There are 7 consonants: C, H, M, N, S, T, and T

So, there are 3 ways for 2 vowels to fall off.
There are 5 consonants left after the two T's are taken off, 5
There are 5*4 ways to have 2 other consonants fall off after one T falls off, 20.
3*25=75[/hide]


Zan has created this rule for generating sequences of whole
numbers.

If a number is 25 or less, double the number.
 If a number is more than 25, subtract 12 from it.

For example, if Zan starts with 10, she gets the sequence 10, 20, 40, 28, 16, $ \ldots$. If the third number in Zan's sequence is 36, what is the sum of the four distinct numbers that could have been the first number in her sequence?

I believe the only way is to try out the different possibilities.

[b]Case 1:[/b] The second number was less than 25.
The second number would be 36/2=18.

If the first number was greater than 25, the first number is 18+12=30.
If the first number was less than 25, the first numbers is 18/2=9.

[b]Case 2:[/b] The second number was greater than 25.
The second number would be 36+12=48.

If the first number was greater than 25, the first number is 48+12=60.
If the first number was less than 25, the first number would be 48/2=24.

The four numbers are 30, 9, 60, and 24.

The sum would be $ 30 \plus{} 9 \plus{} 60 \plus{} 24 \equal{} 123$, like AIME said (how did you get it anyways?)

I tried all the possibilites. I just wondered whether there was some quick way.


Hey guys here goes the question:

How many hours of math do you do a day?
And with that number of hours how far have you gone in olympiads, research, or how much math do you think you know?

I probably end up with at most 1 hour of math a day.
And I probably haven't gone as far in competitions as I could have, had I started seriously doing math earlier.  Heck, I'm not all that serious about it now anyways.

hmm on average like 1 to 2 hours on weekdays(unless a lot of homework), and around like.... 3 to 4 hours on weekends(unless a lot of homework).

about 3 on weekdays, 4 on weekends (pulled 5 or 6 once).

5 hours..i try to end up my homework in an hour then i can do math :lol:

Hmm I need to practice more. Right now I do about 0 hours a day on weekdays and 1 hour a day on weekends.

i do about 2-3-4 hours when i must go to school and between 4 and 8 when i don't(weekends and holidays)

Sooner or later, someone's bound to say $4!$

2 on weekdays, up to 14 on weekends (pulled off 16 hours over winterbreak)


In this figure $ XO \equal{} 8$, $ ZO \equal{} 6$, and the vertex $ Y$ of the rectangle $ XYZO$ lies on the circle $ O$. Find the area of the shaded region. Express your answer in terms of $ \pi$.

[asy]draw(circle((0,0),10));
filldraw(arc((0,0),10,90,180)--(-10,0)--(0,0)--cycle,black);
filldraw((0,0)--(-8,0)--(-8,6)--(0,6)--cycle,white);
label("O",(0,0),SE);
label("X",(-8,0),S);
label("Y",(-8,6),NW);
label("Z",(0,6),NE);[/asy]

This is the area of the quarter-circle minus the area of the rectangle.

By the pythagorean theorem, XZ=10. Because diagonals of rectangles are congruent, YO=the radius of the circle=10, so a quarter circle has area $ \pi 10^2 /4\equal{} 25\pi$

Our answer is $ 25\pi\minus{}48$


I have noticed that there is no "Post Report" button to be clicked on improper topic...

I wanted to report advertisement topic which should be removed but I found no button for reporting posts... So, can that be implemented, like on other forums?

Also that button could serve not only for reporting posts/topics that are advertisement, but can be used to sort out misplaced problems and misplaced topics... double posts and such... It has very wide specter of use :)

I hope you will consider this idea...

Note: a moderator [i]in the part of the forum that the message is in[/i].  PMing me is not helpful for spam messages not in the Round Table, for example.

Question for the OP: is the post you wanted to complain about still there? Or is it already gone? The various people associated with this site are pretty aggressive about cleaning up certain kinds of things.


How many pairs $(a,b)$ of non-zero real numbers satisfy the equation
\[ \frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}?  \]
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0$
$\text{(E)} \ \text{two pairs for each} ~b \neq 0$

[hide]
$\frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}$ is equivalent with $(a+b)^2 = ab$.
Now, $a^2+ab+b^2 = \frac{a^3-b^3}{a-b}$ must be equal to $0$, which is only possible when $a^3-b^3=0$, but then $a^3=b^3$, so that $a=b$, which would make the fraction not to exist. Therefore, there are no solutions.
[/hide]

[quote="Scrambled"][quote="4everwise"][hide="Solution"]$\text{Answer: (A)}$

We hope to simplify this expression into a quadratic in order to find the solutions. To do this, we find a common denominator to the LHS by multiplying by $ab$. $a+b=\frac{ab}{a+b}$.

Rewriting the equation,
$a^2+2ab+b^2=ab$
$a^2+ab+b^2=0$

By the quadratic formula, this has no real solutions.[/hide][/quote]
um, maybe im being stupid, but what if a=-b? dont both sides equal to 0?[/quote]

$a^2 + ab + b^2 = a^2 - a^2 + a^2 = a^2$

 :)

[hide="Moreover..."]
We can divide by $b^2$ and our equation becomes $( \frac{a}{b} )^2 + \frac{a}{b} + 1 = 0$, so we substitute $x = \frac{a}{b}$ to look for the solutions to $x^2 + x + 1 = 0$, which has no real solutions if we simply complete the square: $(x + \frac{1}{2})^2 = - \frac{3}{4}$.
[/hide]

[quote="t0rajir0u"][quote="catcurio"][hide]Multiply the equation by $ab$ to get $a+b=\frac{ab}{a+b}$.
$a^2+2ab+b^2=ab$
$a^2+b^2=-ab$.
This cannot be possible unless a and b are both 0, because the sum of 2 squares cannot be a negative number. $\boxed{A}$.[/hide][/quote]

The problem asks for nonzero reals - consider if $a$ is negative and $b$ is positive.  Try again.   :D[/quote]

The square of a negative number is still positive, so the sum is still positive? Unless I'm missing something... :?


What do you do in your spare summer vacation time aside from awesome math?

Work, watch TV, and play video games. :lol:

Does any do another academic summer activity after or before awesome math?

Take the time available to learn as much math as possible.

read books, some sort of math that I'll self-study, exercise (running, frisbee, etc), spend time with friends, video games, i worked at a hospital one year

tennis, ap psych @ WCATY-ALP ( WI only), violin, piano, math, sleeping (a lot),  volunteering, shopping, hanging, swimming,

butttt...

AMSP=BEST THING OF THE SUMMERRRR. (or life) by farr.ahaha.  XP.

Plan to do something productive.

Or, rather, plan to plan to do something productive.

Actually, plan to plan to plan to do something productive. Literally. Consciously. Like, what runs through my mind: "I tried planning for the summer yesterday, and that didn't work. Probably because I didn't give myself enough time to plan. I need a good few consecutive hours to get concentrated. So, since is too late today, and I have piano lessons tomorrow, tomorrow I'll plan out a schedule for the day after tomorrow for planning for the rest of the summer."

Never happens.

ahahaha.
leoxnlin.
ur thoughts are HILARIOUS  :rotfl:

three things:
teach math to incoming freshmen (often, this makes me  :mad: )
golf!!! (lol now that i think of it, this makes me much more annoyed than incoming frosh)
soaringgggg. so awesome

haha michelle you have so muchhh going on girl!
hmm what do i do..
sleep eat...go to the park and hang out..
occasionally beach..
no shopping-_-
but tons of pool
and studying
and violin-ing[ewwww]
usually im stuck at one of those really asian prep academies to take classes in this freaking shitty city called IRVINE.
yeah our mayor's korean...that shows how asian this freakin city is..
anywayyyy
as for next year..
no tengo idea xD


let $f(x)=x\cdot[x\cdot[x\cdot[x]]]$ for all positive real numbers x, and [x] denotes the greatest integer less than or equal to x.
1) find x such that f(x)=2001;
2) prove that f(x)=2002 has no solution.

Induction is: step 1.  Find, that $6<x<7$, therefore $[x[x]]=[6x]$.
step 2 find [6x]=41, therefore $f(x)=x[41x], \ \frac{41}{6}\le x<7$ and find $[41x]=286$.

[quote="Rust"]Induction is: step 1.  Find, that $6<x<9$, therefore $[x[x]=[6x]$.
step 2 find [6x]=41, therefore $f(x)=x[41x]$ and find $[41x]=286$.[/quote]

THANKS!


algebra
polynomial
function
number theory unsolved
number theory

Show that for each positive integer $ n$, $ \sum_{a=1, (a,n)=1}^n {e^{2\pi i a/n} = \mu(n)}$

$ \mu$ is an arithmetic function, so the LHS should be too.  Prove it; then it suffices to show that this conclusion holds for prime powers.


Let:$ a;b;c>0;a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
$ (a^2.\sqrt{b^4\plus{}c}\plus{}b^2.\sqrt{c^4\plus{}a}\plus{}c^2\sqrt{a^4\plus{}b}) \le 3\sqrt{2}$

[quote="caubetoanhoc94"][quote="2424"][quote="caubetoanhoc94"]Let:$ a;b;c > 0;a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ (a^2.\sqrt {b^4 \plus{} c} \plus{} b^2.\sqrt {c^4 \plus{} a} \plus{} c^2\sqrt {a^4 \plus{} b}) \le 3\sqrt {2}$[/quote]
Yes,It's true and very easy! :P[/quote]
Prove it  :lol:
Thanks :lol:[/quote]
With $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}3$. We have $ ab\plus{}bc\plus{}ca\plus{}2ab^2\plus{}2bc^2\plus{}2ca^2 \leq 9$

[quote="2424"]
With $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. We have $ ab \plus{} bc \plus{} ca \plus{} 2ab^2 \plus{} 2bc^2 \plus{} 2ca^2 \leq 9$[/quote]
Try $ a\equal{}1\minus{}\epsilon,$ $ b \equal{} 2,$  $ c\equal{}\epsilon$ and $ \epsilon\rightarrow0^\plus{}.$ :wink:

[quote="2424"][/quote][quote="caubetoanhoc94"][quote="2424"][quote="caubetoanhoc94"]Let:$ a;b;c > 0;a^2 + b^2 + c^2 = 3$.Prove that:
$ (a^2.\sqrt {b^4 + c} + b^2.\sqrt {c^4 + a} + c^2\sqrt {a^4 + b}) \le 3\sqrt {2}$[/quote]
Yes,It's true and very easy! :P[/quote]
Prove it  :lol:
Thanks :lol:[/quote][quote="2424"]
With $ a,b,c > 0$ such that $ a + b + c = 3$. We have $ ab + bc + ca + 2ab^2 + 2bc^2 + 2ca^2 \leq 9$[/quote]

Prove it  :)


Define a sequence $ a_n$ as $ \sum_{d|n} a_d \equal{} 2^n$ where $ n \in \mathbb{N}$. Prove that $ n|a_n$ for all $ n$.[hide]

My approach: Define a repeating binary sequence of length $ n$ if there exists a partition into $ d$ identical blocks of binary sequences for some $ d|n$.
So, we can generate any sequence of length $ n$ by repeating its unique longest non repeating initial block. Probably a recursion can help here. This idea is very combinatorial. Maybe some can help me in this. Also, please post other solutions. A solution by Mobius inversion is easy. So I would like to see osme other solutions. Thank you.[/hide]

That approach is fine.  This is, in my opinion, a deep property of words.


$ a_{0}\equal{} 4,a_{1}\equal{} 22$
$ a_{n\plus{}1}\equal{} 6a_{n}\minus{}a_{n\minus{}1}$
Prove that there exist sequence $ (x_{n}),(y_{n})$ such that
$ a_{n}\equal{}\frac{y_{n}^{2}\plus{}7}{y_{n}\minus{}x_{n}}$

Do you mean for $ x_{n}, y_{n}$ to also be integer sequences?  Otherwise the problem is trivial.

Yes $ (x_{n}),(y_{n})$ are natural sequence.

[quote="aviateurpilot"]it's equivalent to prove that $ \forall n\in\mathbb{N},\exists y_{n}\in\mathbb{N}:\ a_{n}|y_{n}^{2}\plus{}7$[/quote]
can you prove it
This problem is a collary of other problem .We can find a sequence of $ (y_{n})$ from $ a_{n}$ :lol:


Hey Illinoisians!

When is your chapter contest, and what are your goals for Mathcounts this year?

If you're in high school, do you have someone to cheer for?

Post your thoughts here! Also, feel free to introduce yourself.

I'm gauss1181 (a.k.a. Gaussy or Guass), and I'm in 8th grade. I'm in the northern suburbs chapter, so yeah, I live in the Chicago area. My chapter contest is this coming Saturday (YES, IN JUST FOUR DAYS!) and my goal is to at least make Nationals. If I do make Nationals, I'm gonna strive for top 30.

[quote="guineapiggies"]hi gaussy!!
i'm too old for math counts =[
i uber phailed it last year. so did our team b/c we were thrown together 2 days before the chapter contest.
i cheer for everyone in Illinois especially gaussy, stargirl, oasiscake (if she's going), and...and... dang i feel like i'm forgetting an Illinoisan!! arg I'M SORRY IF I FORGOT YOU!!! but anyway go Illinois!!!

ok i'm to introduce myself

i'm guineapiggies (aka pig, piggies, piggy, guinea, g-pig, etc...) i'm a freshman in high school. i'm also in the chicago area. i'm too old for mathcounts (duh i'm a freshie) and my goal is to not epic fail at math competitions  :D[/quote]
I guess oasisxyz is probably in HS...(I think)? i'm not actually sure though

BTW:

Check out the mathcounts problem in the other thread that I just posted. See if anyone can find out a faster way besides guess and check!

Oh and there was one time where our MC team got thrown together 2 days before the competition. We failed...7th place team. And I had to do every problem alone on the team round..stupid team. And we got a 6/10 on team round cuz of me. That happened in 6th grade....0.o

Hi I'm suma_milli and I'm in 8th grade!  This year is my first year doing any sort of competitive math.  I'm going to strive to make mathcounts nationals but I probably won't make it,  I've got to worry about chapter first. One step at a time :P  Also I'm going to try to make AIME this year.  Oh yeah my chapter is February 14th (Valentines Day lol  :rotfl:).  I live in the Champaign area.  Call me suma or milli or smillie :D

omg...smilie truce... :rotfl:  :D  :P

And...unlucky me. I have chapter earlier than all of y'all do.

:rotfl: Strangely some people at my school started calling me smillie.

The latest anyone has it is the 21st, darn lol.  But at least I don't have it the 7th (sorry guassy :P).

GOOD LUCK TO EVERYONE!!!! :D

:D  :D  :D

lol smillie sounds like smiley

Suma is your chapter strong?


probability
calculus
ratio
geometry
integration
analytic geometry
percent

Let 0 :le: a :le: 10 and 0 :le: b :le: 2. If values for a and b are randomly chosen, what is the probability that ab :le: 10?

If you mean a and b to be integers, then this problem is pretty easy. If you don't mean a and b have to be integers, is there a way to do it without using calculus?

You could do it geometrically i suppose,

If we do this continuously (a, b any real numbers), The probability is the ratio of the area of the region in which it works to the area of the total region.  The problem is that this particular region is not nicely defined -- you probably need an integral to calculate its area.

I do in fact believe calculus is involved in the problem... I was more posting this as a way of learning how to solve it, because indeed I do not know how... a and b can be all real numbers... Joel, could you please explain a bit more as to what you are talking about with the areas.

The graph of the region we are examining is in the black rectangle.  The curve ab = 10 divides that region into two parts: where ab < 10 and where ab > 10.  The probability of a randomly selected point being in each part is the area of that portion (blue or red in my diagram) of the graph.

this might work.

[hide]0  x  10 and 0  y  2

so its on the coordinate system. and then then we  that its bounded by y<10/x and y<2 and x<10

y=2 is an horizontal line which cross at x=5 so take the integral with the sum of the two parts

int(2,x,0,5)+int(10/x,x,5,10)

so 10+ 10(ln2)

over 10*2=20

so (1+ln(2))/2

it probably wrong i did it in a rush. [/hide]

church, that answer is correct, very good.

Joel, thanks for the explanation, it makes perfect sense now... what program did you use to make that diagram?

I used Geometer's Sketchpad version 4.00 and Paint.

I was sitting in my physics class a little earlier tonight and magnetism didn't thrill me too much, so I decided to make this problem a bit more abstract.

Let 0 :le: a :le: R_1 and 0 :le: b :le: R_2. If values for a and b are randomly chosen, what is the probability that ab :le: C where C :le: R_1*R_2?

For those of you do not know calculus and would care to memorize a solution to such a problem, the solution will be:

C/(R_1*R_2) * [ 1 + ln(R_1*R_2/C) ]

Derive your own solution, i'm not typing it all out here, sorry.

I did it like this.

2 :le: x :le: 10
0 :le: y :le: 2

xy=10
y=10/x

let S be the integral symbol

S (10/x)= 10 ln|x|
this is the portion of the 5 by 2 square whose product is less than 10. So 10-10 ln|x| is the area of the greater than ten region.
so 10-10ln|2| is the area of the satisfying region which is  :cong: 3.06853.
divide by total area (20) and get 15.3426 percent or
(10-10 ln|2|)/20

hmm i feel slightly silly. i did it perfectly for greater than or equal to, also didn't see the post which solved for less than in the same manner. seems though i did it right though. interesting question.


Let $ a,b,c > 0$.Prove that:
$ \sum_{cyc}{a^4c} \plus{} \sum_{cyc}{a^3c^2} \plus{} \sum{a^2b^2(a \plus{} b)} \plus{} 2\sum{abc(ab \plus{} bc)}\geq 8abc(a^2 \plus{} b^2 \plus{} c^2)$

[quote="Inequalities Master"]Let $ a,b,c > 0$.Prove that:
$ \sum_{cyc}{a^4c} + \sum_{cyc}{a^3c^2} + \sum{a^2b^2(a + b)} + 2\sum{abc(a + b)}\geq 8abc(a^2 + b^2 + c^2)$[/quote]
$ \sum{abc(a + b)}$But this one isn't 5-th degree.Not very easy. :?: "Easy 5-th degree cyclic inequality"
It's still right?

[quote="Cai Ni"][quote="Inequalities Master"]Let $ a,b,c > 0$.Prove that:
$ \sum_{cyc}{a^4c} + \sum_{cyc}{a^3c^2} + \sum{a^2b^2(a + b)} + 2\sum{abc(a + b)}\geq 8abc(a^2 + b^2 + c^2)$[/quote]
$ \sum{abc(a + b)}$But this one isn't 5-th degree. :?: "Easy 5-th degree cyclic inequality"
It's still right?[/quote]

I have edited.Thanks for appointing the mistake.

The examination is coming……I have no time calculating.I think it seems that if we use SOS is not very hard.Just maybe……
At all,I know you have something which you want to tell us.So I guess it has some nice solution.

$ \sum_{cyc}{bc^4} \plus{} \sum_{cyc}{b^2c^3} \plus{} \sum{a^3(b^2 \plus{} c^2)} \plus{} 4abc\sum{bc} \minus{} 8abc\sum{a^2}$

$ \equal{} \sum_{cyc}{a(a\minus{}b)^2(b\minus{}2c)^2}\geq 0.$


trigonometry
Gauss
modular arithmetic
algebra proposed
algebra

Prove or disprove that there exist a positive integer $n$ and two systems $\left\{c_{1},c_{2},...,c_{n}\right\},\; \left\{b_{1},b_{2},\dots ,b_{n}\right\},\; \; b_{k}\ne \frac{3}{10}, 1\le k\le n,$ of rational numbers ,   such that
\[\frac{1+\sqrt{5}}{2}= \sum_{k=1}^{n}c_{k}\sin{(\pi b_{k})}. \]

[quote="pco"][quote="harazi"]There exist: this follows from the expression of the Gauss sum $\sum_{k=0}^{p-1}{e(k^{2})}=\sqrt{p}$ if $p=1\pmod 4$ is a prime and $e(x)=exp(2i\pi/p)$.[/quote]
Very nice, but :
1) I think you mean $e(x)=exp(2i\pi x/p)$
2) You should verify that none of the $b_{k}$ is $\frac{3}{10}$ (actually, it's OK)
In fact, this formula gives :$\sin(\frac{\pi}{10})=\frac{\sqrt{5}-1}{2}$ and so $\frac{1+\sqrt{5}}{2}=\sin(\frac{\pi}{10})+\sin(\frac{\pi}{2})$[/quote]
Indeed, thanks !


[quote="math92"]this is not the right place for that problem. ;)[/quote]
plz tellme where is it?i mean where should i post it because i think it is good for here and very easy. :D

this is too hard for classroom math. mods, move it please.

[quote="jensen"]for all $x,n$ find $\frac{sinx}{n}$. :D[/quote]

[hide="Very interesting problem"]
both of the n's cancel out which will make it six.
[/hide]

I moved this from Classroom Math.

There will be no more warnings for Jensen about posts in the Middle School forums.  Follow the forum guidelines when they are posted at the top of a forum and don't try to trick the students.  The whole point is to keep scary looking problems out of those forums so that students don't become disinterested.  I'll simply remove your view of the middle school forums if the problem continues.


Given triangle $ABC$ prove that locus of $P$ (all $P$ on plane) such that $PA+PB+PC \le 3R$ is a convex set.

[hide]Use : If $P\in [AB]$ then exist $k\in [0,1]$ such that $MP\leq kMA+(1-k)MB\forall M$.[/hide]

The best you should post all solution instead of a trivial hint  :D

Ok! This is my solution.
[b]lemma.[/b] If $P\in [AB]$ then exist $k\in [0,1]$ such that $MP\leq kMA+(1-k)MB\forall M$.

[i]proof.[/i] Suppose $\vec{BP}=k\vec{BA}$(*) for some $k\in [0,1]$, from (*) we have $\vec{MP}=k\vec{MA}+(1-k)\vec{MB}\forall M$. Now, by $|\vec{a}+\vec{b}|\leq |\vec{a}|+|\vec{b}|\forall \vec{a},\vec{b}$ lemma is proved.

I will use lemma above to slove your problem.

Denote $\mathcal{H}=\{P|PA+PB+PC\leq R\}$. We need prove $\mathcal{H}$ is convex set. For any $P_{1},P_{2}\in\mathcal{H}$, we will prove $[P_{1}P_{2}]\subset\mathcal{H}$, each $P\in [P_{1}P_{2}]$ by lemma exist $0\leq k\leq 1$ such that $MP\leq kMP_{1}+(1-k)MP_{2}\forall M$. In particular $AP\leq kAP_{1}+(1-k)AP_{2}$,$BP\leq kBP_{1}+(1-k)BP_{2}$ and $CP\leq kCP_{1}+(1-k)CP_{2}$,

 therefore $PA+PB+PC\leq k(P_{1}A+P_{1}B+P_{1}C)+(1-k)(P_{2}A+P_{2}B+P_{2}C)\leq k3R+(1-k)3R=3R$
(note: $P_{1},P_{2}\in\mathcal{H}$), hence $P\in\mathcal{H}$.

 QED

ok that's right goo solution!
My solution is base on this nice inequality
Given segment $MN$, forall triangle $ABC$ and $P\in NM$ then
$f(P)\le \max\{f(M),f(N)\}$ with $f(X)=XA+XB+XC$
Thus we can prove the general problem. :)

[quote="gemath"]
Given segment $MN$, forall triangle $ABC$ and $P\in NM$ then
$f(P)\le \max\{f(M),f(N)\}$[/quote]

Post your solution, please!  :D

My solution is from convex Analysis !

My solution use that above lemma  :D  :D !
We can prove general for polygon and with weight.
Because forall $P\in MN$ then $XP\le kXM+(1-k)XN$
With polygon (not necessary convex) $A_{1}...A_{n}$ forall $\alpha_{i}\ge 0,i=\overline{1,n}$ let $f(X)=\sum_{i=1}^{n}\alpha_{i}XA_{i}$
give $X\equiv A_{1},...,A_{n}$
$f(P)=\sum_{i=1}^{n}\alpha_{i}PA_{i}\le \sum_{i=1}^{n}\alpha_{i}(kA_{i}M+(k-1)A_{i}N)=k\sum_{i=1}^{n}\alpha_{i}A_{i}M+(1-k)\sum_{i=1}^{n}\alpha_{i}A_{i}N\le (k+1-k)\max\{f(M),f(N)\}=\max\{f(M),f(N)\}$
Because here $k,1-k\ge 0$
 :D


The equation $x^2+y^2+z^2=x^3+y^3+z^3$ has infinitely many integer solutions.

Davron


conics
analytic geometry
trigonometry
geometry
3D geometry
calculus
calculus computations

Find the maximum and minimum of $z$ given that
$x^{2}+y^{2}=z^{2}$ and $x+y+2z=1$.

Thank you for everyone's help.

By using polar coordinates with $x = r \cos\theta$ and $y = r \sin\theta$, the two given equations is equivalent to $z = \pm r$ and

$(1) \;\; r(\cos\theta+\sin\theta \pm 2) \;=\; 1.$

Since $\cos\theta+\sin\theta \,=\, \sqrt{2}\, \cos(\theta \:+\: \frac{\pi}{4})$ and $r = \pm z$, equation (1) are equivalent to

$(2) \;\; z \;=\; \pm \: \frac{1}{\sqrt{2}\, \cos(\theta \:+\: \frac{\pi}{4}) \: \pm\: 2}\;=\; \frac{1}{2 \: \mp \: \sqrt{2}\, \cos(\theta \:+\: \frac{\pi}{4})}$

From (2) we see that

$z_{min}\: =\: \frac{1}{2+\sqrt{2}}\;=\; 1 \:-\:{\textstyle \frac{1}{\sqrt{2}}}$

and

$z_{max}\: =\: \frac{1}{2-\sqrt{2}}\;=\; 1 \:+\:{\textstyle \frac{1}{\sqrt{2}}}.$

this is just a general question, but how does one go about knowing when to convert an equation to polar coordinates?

the rule would be to look for $x^{2}+y^{2}$, eg. in this case $r^{2}=x^{2}+y^{2}= z^{2}$ is a regular cone, easily parameterized by polar coordinates..

$(x+y)^{2}+(x-y)^{2}=2(x^{2}+y^{2})\Longleftrightarrow (1-2z)^{2}+(x-y)^{2}=2z^{2}.$

Since $x,\ y$ real numbers, we have $(1-2z)^{2}\leq 2z^{2}$, yielding $\frac{2-\sqrt{2}}{2}\leq z\leq \frac{2+\sqrt{2}}{2}$. :lol:

Consequently, this problem can be solved just by using Cachy-Schwarz's inequality. :P

alternately we can do it in this way

$(1-y-2z)^{2}+y^{2}=z^{2}$
$\leftrightarrow 2y^{2}+2y(2z-1)+3z^{2}-4z+1=0$

since $y$ is real , the discriminant must be $\geq 0$

so we have $(2z-1)^{2}-2(3z^{2}-4z+1)\geq 0$ $\leftrightarrow 2z^{2}-4z+1\leq 0$

Sure! probably Japanese high school students will solve such as your solution, shyong. :)

Alternative solition.

Consider on the $xy$ plane, we are to find the condition for which the circle $x^{2}+y^{2}=z^{2}$ with radius $|z|\geq 0$ and the line $x+y=1-2z$.

Thus use the formula of the distance of a point and a line we have $\frac{|1-2z|}{\sqrt{2}}\leq |z|$.

i haven't tried it yet and no one has posted a solution with legrange multipliers so i was wondering if it would be worth trying.

Go ahead. Lagrange multipliers aren't quite as nice with multiple constraints, but they still work well for problems with some symmetry.

One can consider $x^{2}+y^{2}\leq (x+y)^{2}$ and then use the givens.


In equilateral $ \Delta ABC$, points $ M$ and $ N$ are the midpoints of $ AB$ and $ AC$, respectively. Given that the area of $ \Delta BPC$ is 18 square units, how many square units are in the area of $ \Delta ABC$?

[asy]pair A,B,C,P,M,N;
A = (0,2); B = (-1,0); C = (1,0); M = (A+B)/2; N = (A+C)/2;
P = IntersectionPoint(B--N,M--C); draw(N--B--A--C--M); draw(B--C);
label("$A$",A,N);label("$B$",B,W);label("$C$",C,E);label("$M$",M,W);label("$N$",N,E);label("$P$",P,N);[/asy]

$ BP\equal{}PC\equal{}\frac{a\sqrt3}3$
$ S_{BPC}\equal{}\frac{BP\cdot PC\cdot\sin120^\circ}{2}\equal{}\frac{a^2\sqrt3}{12}$
$ S_{ABC}\equal{}\frac{a^2\sqrt3}4\equal{}3S_{BPC}\equal{}54$

[quote="vjnmath"]Since it is an equilateral triangle, BPC = BPA = CPA.
Hence area of triangle ABC = 18 * 3 = 54 sq units[/quote]

$ BPC\equal{}BPA\equal{}CPA$ even if $ ABC$ isn't an equilateral triangle, since $ BN$ and $ CM$ are medians

Extend AP to point D on BC, centroid P divides median AP into a 2:1 ratio, so our answer is 3*18=54.


algebra
function
domain
quadratics
quadratic formula

If   $x , y$ are real numbers such that   $x^2 + xy + y^2 = 1$ , find the least and the greatest value( minimum and maximum) of the expression   $K = x^3y + xy^3$

 [u]Babis[/u]

[b] Sorry !!! I forgot to write that these 4 problems( 4 topics) were [u]JUNIOR LEVEL[/u][/b]

Nice yes. but why Easy??

What trick did you use??
 :?

Let (x^2 + y^2) be m, (xy) be n.
Then m + n  = 1, i. e. m = 1 - n
Now K = m * n = n - n^2 = - (n - 1/2)^2 + 1/4.
Note that m >= 0, so n =< 1.
So, minimum value of K is 0, for n = 1;
      maximum value of K is 1/4, for n=1/2.

That's simple but nice.  :)

For the maximum...

[hide]
Let $a = x^2+y^2$, $b = xy$.

We have $a+b = 1$ and also $a \ge 2b$ (trivial), so $b \le \frac{1}{3}$.

By AM-GM, we know

$\sqrt{2ab} \le \frac{a+2b}{2} \Rightarrow ab \le \frac{1}{2}\left(\frac{a+2b}{2}\right)^2 = \frac{(1+b)^2}{8} \le \frac{\left(\frac{4}{3}\right)^2}{8} = \frac{2}{9}$,

with equality at $x = y = \frac{1}{\sqrt{3}}$.[/hide]

BTW, rem your solution doesn't work because $xy \neq \frac{1}{2}$.

Yours answer is incorrect, except paradin8  :(

I don't think this problem is easy.

Few people can solve this problem completely for junior level.

You should find the domain of $xy.$ accurately.   :)

The answer is $-2\leq K\leq \frac{2}{9}.$

kunny

[quote="Altheman"]rem and paladin8, how do you know that your maximum and minimum give real solutions for x and y? i think that is generally the key that you have to make sure that x and y are real (i think quadratic formula would help)[/quote]

I think the best way is to find the range of $xy$ (like Kunny already said):

[hide]By AM-GM,  $x^2+y^2=1-xy \ge 2xy \implies xy \le \frac{1}{3}$ and,

$x^2+y^2+xy=(x+y)^2-xy=1 \implies (x+y)^2=xy+1$
$\Longleftrightarrow xy+1 \ge 0 \implies xy \ge -1$

$K=xy(x^2+y^2)=xy(1-xy)$ and since $-1 \le xy \le \frac{1}{3} \implies -2 \le K \le \frac{2}{9}$[/hide]

$\sqrt{(xy)^2}=xy?$ Please correct the last line, Gabriel(fr). :)

kunny

[quote="Altheman"]rem and paladin8, how do you know that your maximum and minimum give real solutions for x and y? i think that is generally the key that you have to make sure that x and y are real (i think quadratic formula would help)[/quote]

I gave the equality case, didn't I? I think that's enough to show that they are real...

[u] To socrates.[/u]

 Yes , I wanted to write << Nice rather simple >>. Of course the problem is not simple for the greek standards. I was in Athens ,  when Silouan gave me the problem. We solved it , eating in a restaurant using  the method of the last posts ( $-1 \leq xy \leq \frac{1}{3}$ . But in olympiads saying << simple >> we mean << a l;ittle difficult , but not hard >>. I hope you understand me.If you are a Junior Student , I must agree with you: << not so simple ! >>.

        [u]babis[/u]

[quote="paladin8"][quote="kunny"]paladin8, you can't use A.M-G.M. in this case because $-1\leq b\leq \frac{1}{3}.$ :D[/quote]

Well it works for the maximum case, right? Because I'm just applying it on $xy$ and $x^2+y^2$ and if $xy < 0$ then $K < 0$ so it's not a maximum; therefore we may assume $xy > 0$ and apply AM-GM. I didn't specify this in my previous post, but it does give the correct answer, I believe...[/quote]

This is the explanation which I want. :)

kunny


Let x,y be a real number amd x,y different from 0 satisfying that
$(x+y)xy=x^{2}+y^{2}-xy$.find max of $\frac{1}{x^{3}}+\frac{1}{y^{3}}$ :wink:

$x^{2}+y^{2}=r^{2}$  Then exists $\alpha \in [\frac{-\pi}{2},\frac{\pi}{2}]$  such that  $x=rcos(\alpha)$ and $y=rsin(\alpha)$

$(x+y)xy=x^{2}+y^{2}-xy \Rightarrow$
$r^{3}(cos(\alpha)+sin(\alpha))cos(\alpha)sin(\alpha)=r^{2}(cos^{2}(\alpha)+sin^{2}(\alpha)-cos(\alpha)sin(\alpha))$

$r(cos(\alpha)+sin(\alpha))sin(2\alpha)=2(1-cos(\alpha)sin(\alpha))$

$\frac{2}{r}=\frac{sin(2\alpha)(cos(\alpha)+sin(\alpha))}{(1-cos(\alpha)sin(\alpha))}$  ;     we see $cos(\alpha)sin(\alpha) \neq 1$ for all $\alpha$

Now:
$\frac{1}{x^{3}}+\frac{1}{y^{3}}=\frac{(cos(\alpha)+sin(\alpha))(1-cos(\alpha)sin(\alpha))}{r^{3}cos^{3}(\alpha)sin^{3}(\alpha)}$

$=\frac{(cos(\alpha)+sin(\alpha))^{2}}{r^{2}cos^{2}(\alpha)sin^{2}(\alpha)}$

$=\left(\frac{2}{r}\right)^{2}\left(\frac{1+sin(2\alpha)}{sin^{2}(2\alpha)}\right)$

$=\left(\frac{sin(2\alpha)(cos(\alpha)+sin(\alpha))}{(1-cos(\alpha)sin(\alpha))}\right)^{2}\left(\frac{1+sin(2\alpha)}{sin^{2}(2\alpha)}\right)$

$=\frac{(1+sin(2\alpha))^{2}}{1-sin(2\alpha)+cos^{2}(\alpha)sin^{2}(\alpha)}$

$=\left(\frac{1+sin(2\alpha)}{1-\frac{sin(2\alpha)}{2}}\right)^{2}$

we see that max of $\frac{1}{x^{3}}+\frac{1}{y^{3}}$  in $\alpha=\frac{\pi}{4}$

$\frac{1}{x^{3}}+\frac{1}{y^{3}}=16$

$x=\frac{1}{2}$
$y=\frac{1}{2}$

[quote="Virgil Nicula"]$x>0$, $y>0$, $(x+y)xy=x^{2}-xy+y^{2}\Longrightarrow$ $\Lambda\equiv \frac{1}{x^{3}}+\frac{1}{y^{3}}=\frac{x^{3}+y^{3}}{x^{3}y^{3}}=\frac{(x+y)(x^{2}-xy+y^{2})}{x^{3}y^{3}}=\left(\frac{x+y}{xy}\right)^{2}\ ;$ $\lambda\equiv\frac{x+y}{xy}=\frac{(x+y)^{2}}{x^{2}-xy+y^{2}}\ ,\ \Lambda=\lambda^{2}$.
$\Lambda\mathrm{\ \--\ max.}\Longleftrightarrow \lambda\mathrm{\ \--\ max.}$ $\Longleftrightarrow$ $\frac{x^{2}+2xy+y^{2}}{x^{2}-xy+y^{2}}\mathrm{\ \--\ max.}$ $\Longleftrightarrow$ $1+\frac{3xy}{x^{2}-xy+y^{2}}\mathrm{\ \--\ max.}$ $\Longleftrightarrow$ $\frac{xy}{x^{2}-xy+y^{2}}\mathrm{\ \--\ max.}$ $\Longleftrightarrow$ $\frac{x^{2}-xy+y^{2}}{xy}\mathrm{\ \--\ min.}$ $\Longleftrightarrow$ $\left(-1+\frac{x}{y}+\frac{y}{x}\right)\mathrm{\ \--\ min.}$ $\Longleftrightarrow$ $\left(\frac{x}{y}+\frac{y}{x}\right)\mathrm{\ \--\ min.}\ \ \wedge\ \ \frac{x}{y}\cdot\frac{y}{x}=1$ $\mathrm{(constant)}$ $\Longleftrightarrow$ $x=y\ \ \wedge\ \ (x+y)xy=x^{2}-xy+y^{2}$ $\Longleftrightarrow$ $x=y=\frac{1}{2}\ .$[/quote]
x,y can be negative number,too. :wink:

Let's see that $\frac{1}{x^{3}}+\frac{1}{y^{3}}=\frac{x^{3}+y^{3}}{x^{3}y^{3}}=\frac{(x+y)(x^{2}-xy+y^{2})}{x^{3}y^{3}}= \left(\frac{x+y}{xy}\right)^{2}$
Let's make $a=\frac{1}{x}$ and $b=\frac{1}{y}$ then: $(x+y)xy = x^{2}+y^{2}-xy \Rightarrow a+b = a^{2}+b^{2}-ab \mbox{\; (1)}$
While $\left(\frac{x+y}{xy}\right)^{2}= (a+b)^{2}$
Working with (1):
$a+b-ab= (a-b)^{2}\Rightarrow a+b \geq ab \mbox{\; also \;}a+b+3ab = (a+b)^{2}$
From these last relations$(a+b)^{2}= a+b+3ab \leq 4(a+b)$
It's easy to note that $(a+b) \geq 0$, then  $(a+b)\leq 4$. And the max is $(a+b)^{2}= 16$, when $x=y=\tfrac{1}{2}$.


geometry
trigonometry
calculus
derivative
geometry proposed

Given a triangle $ ABC$ with area 1 and $ AB \equal{} 2.$ Find the measure of $ \angle{BAC}$ for which the value of $ BC^2 \plus{} (2\sqrt {3} \minus{} 1)AC^2$ is minimized.

$ a \equal{} BC \ge 1,\ b \equal{} CA \ge 1,\ c \equal{} AB \equal{} 2.$

$ S(\triangle ABC) \equal{} \frac {bc}{2} \sin A \equal{} 1,\ c \equal{} 2 \Longrightarrow \sin A \equal{} \frac {1}{b},\ \cos A \equal{} \pm \frac {\sqrt {b^2 \minus{} 1}}{b}$

$ a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc \cos \angle A \Longrightarrow a^2 \equal{} b^2 \plus{} 4 \mp 4 \sqrt {b^2 \minus{} 1}$

$ a^2 \plus{} (2 \sqrt {3} \minus{} 1) b^2 \equal{} 4 \mp 4 \sqrt {b^2 \minus{} 1} \plus{} 2 \sqrt {3} b^2 \equal{} f(b)$

Since science course applicants were not allowed to use derivatives, put $ x^2 \equal{} b^2 \minus{} 1 \ge 0,$ $ f(b) \equal{} q(x) \equal{} 4 \mp 4x \plus{} 2\sqrt {3} (x^2 \plus{} 1),$ which has minimum for $ x \equal{} \pm \frac {4}{4\sqrt {3}},$ $ x^2 \equal{} \frac {1}{3},$ $ b^2 \equal{} x^2 \plus{} 1 \equal{} \frac {4}{3},$ $ b \equal{} \frac {2}{\sqrt {3}} > 0,$ $ \sin A_{12} \equal{} \frac {1}{b} \equal{} \frac {\sqrt {3}}{2},$ $ \angle A_{12} \equal{} 60^\circ$ or $ 120^\circ.$ But $ \cos A_{12} \equal{} \pm \frac {1}{2},$ $ a_{12}^2 \equal{} \frac {4(4 \mp \sqrt {3})}{3}.$ $ a_1^2 < a_2^2 \Longrightarrow$ minimum is for $ \angle A_1 \equal{} 60^\circ.$

This problem is a cute control question
  on  understanding what Stewart theorem
  is all about.

  M.T.


[quote="nirvanatear"]do u mean a British Community?
well i think the local communities are just for ppl who speak english as the second language or don't understand english.[/quote]
This is not correct. The national community is not only for the people who speak the same language, but also for the people with same culture, same history and same topics. Language is only a tool when they discuss.

[quote="riddler"]I think as England being very popular, and many english on this site, I do think it is neccesary that an English forum is present. Do you agree? :D[/quote]A good idea! :)

Therefore, we introduce the [url=http://www.mathlinks.ro/Forum/index.php?f=407]British Forum[/url]  :yup:


What are the last two digits of $ 8^{25} \plus{} 12^{25}$?

[quote=carlos8]Can someone please explain the intuition for doing 8 = 10-2 and 12 = 10 +2?[/quote]

first, you want to find digits of a number. second, the average of 8 and 12 is 10

[quote=carlos8][quote]first, you want to find digits of a number. second, the average of 8 and 12 is 10[/quote]

But how did you know that getting the average of 8 and 12 will help you?[/quote]

because of binomial theorem(stuff cancels out)

This can be quickly solved with lifting the exponent. Note that 8 and 12 are both divisible by 4, so the last two digits are divisible by 4. Applying LTE with base 5 gives that the exponent of 5 is 3. So it is divisible by 25, and the number is divisible by 100, and the sum of the last two digits is $\boxed{0}$.

[quote=ilovepizza2020][quote=mathleticguyyy][quote=BooBooTM]This one was my last problem before becoming an alcumus grand master :)[/quote]

I am so jealous of you guys... [b]Now after Intermediate Algebra no one can get Alcumus grandmaster anymore...[/b]
[hide=sol]$8^{25}=37,778,931,862,957,161,709,568$
$12^{25}=953,962,166,440,690,129,601,298,432$
Add them together, we get that $8^{25}+12^{25}=37,778,931,862,957,161,709,568+953,962,166,440,690,129,601,298,432=953,999,945,372,553,086,763,008,000$ so the sum of last two digits is $0+0=0$[/hide][/quote]

Very funny. :rotfl:

[quote=BooBooTM]This one was my last problem before becoming an alcumus grand master :)[/quote]

Let's see your profile: 212 passed, 199 mastered. Very grand when there are 295 total topics.[/quote]

That was before IA.

[quote=heheman][quote=samuel][quote=math154]Insignificant Note: The answer is $ \boxed{00}$.[/quote]

i put 0, it was right
(I thought it was asking for the last digit :P)[/quote]

same,lol[/quote]

same,lol

[hide=sol][hide=if the sum of two numbers x and y is a multiple of 10, when they are raised to any odd power, then the result is also a multiple of 10]which means $x\equiv -y\pmod{10}$ and let $k$ be any odd number, then $x^k+y^k\equiv x^k+(-1)^k x^k\equiv x^k-x^k\equiv 0 \pmod{10}$[/hide], so the answer is $\boxed{0}$[/hide]

[hide]00[/hide] there a troll about this...

[quote=heheman][quote=samuel][quote=math154]Insignificant Note: The answer is $ \boxed{00}$.[/quote]

i put 0, it was right
(I thought it was asking for the last digit :P)[/quote]

same,lol[/quote]

omg same

[quote=ProbablyNot][hide=Solution]
Every fifth power of an integer ends in the units digit of the integer. So $(8^5)^5$ ends in a zero and $(12^5)^5$ ends in a zero. It is important to note that they both end with two zeros. The answer is $\boxed{0}$  :)
[/hide][/quote]
Your first sentence is correct, but, following its logic, $(8^5)^5$ ends in an 8 and $(12^5)^5$ ends in a 2.  (Neither is divisible by 10.)


I was wondering if anyone else here is taking AP CS in java.  I've been doing independent study for most of the year.  Does anyone know what the scoring standards are for the AB test?  Like what percent do you need to get a 5?  Also to people who received 5s, did you study a lot before hand?  Was the test very easy for you?
Right now, most of the free response is fairly easy, but the multiple choice tends to get me.  I have trouble with the recursion on the multiple choice and also the MBS case study questions.  I missed 4 out of 5 of them on a practice test I finished today and 3 or 4 on the rest of the test.

i'm not very sure, but it could be here  :blush:

http://www.collegeboard.com/prod_downloads/ap/students/compsci/APCompSciCD040915.pdf

I'm not sure about this, but I remember in a book that it said 70%.

for most APs you need around 65-70% to get a 5...

I took AB last year and thought that I didn't do very well on it (I'm not a programming person) but I still pulled a 5. *shrug* Didn't study outside of the class.

Last time I checked, 37.5% of people who take the test get 5s, which is one of the higher percentages around (Environmental Science is 8.8% and Gov is about 5% if I remember correctly).

All the best to you for the AP.

[quote="rippledance"]
Last time I checked, 37.5% of people who take the test get 5s, which is one of the higher percentages around (Environmental Science is 8.8% and Gov is about 5% if I remember correctly).
[/quote]

A little high. It was 27% last year. All the grade distributions can be found [url=http://www.collegeboard.com/student/testing/ap/about.html]here[/url] on the left panel.

Anyway, I took APCS AB last year too and got a 5; I studied quite a bit outside of class (because I was in an A class). Missing 10 on the MC will probably be ok as long as you do very well on the FR -- just make sure you don't bomb one of the FRs if you're going to do this. My best advice would be to brush up on the MBCS because it is a fairly significant portion of the test.

BTW, definitely get a review book too, but I would expect that you already have one.

oops. I could have sworn it was higher than that the year before though.

Yea, I do have a review book.  I've read the whole thing and done 2 of 4 of the tests.  I tend to make lots of stupid mistakes though heh.

What i didnt realize though until today is that we get a reference packet.  That ought to make it quite a bit easier.

CS AB has a very generous distribution thingy.  I bombed at least two of the free response problems, and I didn't do that well on the multiple choice, but I still got a 5.

I had issues with the free response because I didn't really know how to use treemaps and whatnot because I had to teach them to myself about a week before the test and didn't bother practicing writing programs with them because I was in an A class.  Then I freaked out a lot because the guy next to me finished about an hour early and laughed at me a lot.

Yea,
I really don't think writing programs on paper is the best way to really test someone in java.  When I program, it usually takes a lot of experimentation and testing to get it exactly the way I want it.


[quote="tjhance"]hm $ R^{+}$ refers only to positive reals, right? So you shouldn't be able to use 0...[/quote]

Does it matter? $ \lim_{x\rightarrow 0^{+}}x>0$ but it's as good as $ 0$ in this case.


function
complex numbers
imaginary numbers

[quote="binomial_4eva"]i googled it but i couldnt exactly undrestand what a root of unity is. can someone explain what exactly a root of unity is and what is the use of it plz?
thnx[/quote]
They are complex numbers that are solutions to the equation $ x^n \equal{} 1$.
I googled "root of unity" and came up with a lot of explanations.
In the real numbers, we know that $ ( \minus{} 1)^2 \equal{} 1$, but when we talk about square root of $ 1$, we refer to the positive solution to $ x^2 \equal{} 1$.
Complex numbers were "imagined" to have solutions to certain problems, like $ x^2 \equal{} \minus{} 1$. One solution to that equation is the imaginary number $ i$. Complex numbers look like $ a \plus{} ib$. They have a very practical use when calculating with alternating electrical currents. Other uses are not so close to everyday life.

the nth roots of unity are the roots to the equation x^n=1 generally abbeviated as w(omega in greek),w^2,w^3,...,w^n.
Any book which explains Generating functions should have numerous problems which needs the use of roots of unity.You can also view IMO 1995#6. :D


calculus
integration
function
real analysis
real analysis unsolved

let f a function from R+ to R wich is C^2  and

integral of absolutevalue(f) over R+ is convergent
and

integral of (f " )^2 over R+ is convergent

prove or disproven that f '(x) tend to 0 when x tend to +oo ?


For $(a),$ calculate $\sum_{k=1}^{n}\sin \theta \sin (2k-1)\theta.$

by the formula $\sin \alpha \sin \beta =\frac{1}{2}\{\cos (\alpha-\beta)-\cos (\alpha+\beta)\},$ then use telescoping method.

For $(b),$ you can use the half-angle formula for $\sin^{2}x.$

[quote="13375P34K43V312"]Isn't it also possible to use complex numbers here?[/quote]

and also $n^{th}$ roots of unity

[quote="13375P34K43V312"]Isn't it also possible to use complex numbers here?[/quote]
ty kunny adn how complex numbers?


algebra polynomial calculus integration function domain LaTeX

Lately I've been studying problems related to:

Given:
   an integral domain "R",
   the field of quotients "F" of R.

What is known about polynomials "f" such that that:
   f has coefficients in F,  and,
   f maps elements of R to elements of R.

It should be pointed out that if  "f" is a function from R to R,  and "f" can be expressed as a polynomial,  then the polynomial has coeffients in F.   This can be seen by taking n + 1 points,  where "n" is the degree of "f",   and finding an interpolating polynomial using the Vandermonde matrix.

In particular,   I'm interested in bases of these polynomials.   For example,  if R is the integers,  then F is the rationals,  and it is reasonably well known that the binomial polynomials form a basis.   I.E.  if "f" is a polynomial under consideration,  then:

   f (x) = c_0 + c_1 * (x choose 1) + c_2 * (x choose 2) + ... c_n * (x choose n)

where the coefficients are integers and "n" is the degree of  "f".

For starters,  I've been looking at the polynomials that map gaussian integers to gaussian integers.   I've had no trouble finding a number of these polynomials,  but I still haven't had any real insight.   For example:

x * (x + 1) / (1+i)
x * (x + 1) * (x + 2)  * (x+3) / 2
x * (x + 1) * (x + i) * (x + 1 + i) / [2 * (1 + i)]
x * (x + 1) * (x + i) * (x + 1 + i) * (x + 1 + 2 i) / [ (2 + i) * 2 * (1 + i) ]
x * (x + 1) * (x + i) * (x + 1 + i) * (x + 2 + i) / [ (1 + 2 i) * 2 * (1 + i) ]
x * (x + 1) * (x + 2) * (x + 3) * (x + 4)  /  10

The last three are particularly interesting... they are all of degree 5,   but they show how different choices of a "pattern"  in the complex plane lead to differing guarantees of divisibility.    Also,  if we were only concerned with mapping the integers to the integers,  then normally we could divide the last polynomial by 120.   However,  since we are concerned with mapping the gaussian integers to the gaussian integers,  we can only divide by 10.   (and maybe a few more,   I dunno).

Finally,  it appears that if we use an interpolating polynomial "g"  for a function f : Z[i] -> Z[i]  at three points as follows:

f(x) = y_0
f(x + 1) = y_1
f(x + 1 + i) = y_2

Then "g" appears to map Z[i] to Z[i] as well.   However,  if we pick the interpolating polynomial using the three points

f(x) = y_0
f(x+1) = y_1
f(x+2) = y_2

Then "g" most certainly can map a gaussian integer to a gaussian rational.  On the other hand,  if this case was restricted to the integers,  then "g" would certainly map integers to integers.

If anybody has any insight and/or references to this problem,  I'm most interested.

There were some papers by Manjul Bhargava on "p-orderings" as a way to describe and (for a finite domain) enumerate polynomials with this property, for any Dedekind domain.  Either his paper or its references will pertain to what you are looking for.

For nice enough $R$, you should be able to find a basis for these polynomials as an $R$-module. This can even be done using one polynomial of each degree.

On the other hand, for anything other than $\mathbb{Z}$, it's not going to work out with a universal nice formula.

The fifth-degree basis element in the Gaussian integers should have denominator $(1+i)^3(2+i)(2-i)$, up to multiplication by units. The tenth-degree basis element should have denominator $(1+i)^8\cdot(2+i)^2\cdot(2-i)^2\cdot3=1200$.

Bhargava's results were that it does generalize in a nice way if you define the correct analogue of $n!$.

Thanks.   I'll definitely track down those papers.

Ahh yes... I see now that

x (x + 1) (x + i) (x + 1 + i) (x + 3 + 3i) / 10 (1 + i)

is a gaussian-integer-valued polynomial.   JMerry,  how are you coming up with those numbers for the denominator?    I suppose you could optimize factors in the numerator for each individual prime and then apply the Chinese Remainder Theorem...

My intuition is that there are no polynomials as described  when  R = Q[x].   It seems that if F is the quotient field of Q[x],  namely the rational functions of a single variable "x",  then any polynomial F[y]  will map some element Q[x] to a rational function.  (Well, any F[y] with properly rational function coefficients,   that is....)  Is this true,  or can somebody give me a polynomial in F[y] that maps Q[x] to Q[x]?

and, sorry for the newbie question here,  but how do you conveniently use LaTeX in these forums?

For $\LaTeX$ math mode, enclose your code in dollar signs.

I was thinking independently in each prime. For each Gaussian integer $r=a+bi$, define $N(r)=a^2+b^2$. For each prime $p$, $\mathbb{Z}(i)/p\mathbb{Z}(i)$ is a field with $N(p)$ elements. You clearly need degree $N(p)$ to get even one factor if $p$ in the denominator, so the numerator can be a zero-valued polynomial in that quotient field. The total exponent of $p$ you can get with degree $n$ is $\sum_k\left\lfloor \frac n{(N(p))^k}\right\rfloor$. You presumably understand the relations of primes in the Gaussian integers to primes in the integers, so this is sufficient to find the denominator for all $n$, which always works out to an integer or an integer times $1+i$ (up to multiplication by powers of $i$). The denominator is fixed, but there is still a very wide choice for the exact polynomial, with nothing as obvious as the binomial coefficients..

In the case $R=\mathbb{Q}[x]$, $R$ has no ideals of finite index, and therefore only polynomials with "integer" coefficients are "integer" valued. For each irreducible polynomial $p$, the only zero-valued polynomial with coefficients in $R/pR$ is the zero polynomial.

By this reasoning, setting $R=\mathbb{Z}[x]$ would give identical results to $R=\mathbb{Z}$.