LaTeX search Harvard college Putnam geometry function

For this special event let us reminisce about the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=154375#p154375]history of ML.[/url] Let me quote a former description of ML here:

"[b]Hi folks! [/b]

Let me say Welcome (once again) and tell you that if you need to ask more questions about things below of about the things that are not answered below you can also check out the About Mathlinks forum.

[b]What is MathLinks.ro ?[/b]

MathLinks EveryOne is a mathematical website, featuring a large database of problems, ranging from pre-olympiad levels to hard olympiad problems, with a lively populated, LaTeX based, forum, download site with tons of problems, a strong (and now traditional) weekly contest with international participation, and also last but not least a large collection of mathematical links.

That's how a tehnical description (maybe for a search engine) would sound. But MathLinks is more than that, it's a large problem-solving community that interacts in many ways, not just in solving difficult (and beautifull) math problems. Most of the people on MathLinks know other people from MathLinks from previous enounters at different mathematical competitions, such as IMO, BMO, national olympiads and other regional contests.

[b]What can you do on MathLinks? [/b]

You can solve problems and post their solutions so that others can see your solution, also you can search for a specific problem using the search after source option in the search window, or you can ask your fellow MathLinkers to guide you in the steps of solving a problem, you can ask or give a reference, or even a friendly advice, and if we are talking about friendly, you can also make a lot of friends.

[b]How can you contribute? [/b]

It's simple: by spreading the word about this site in your mathematical community, and also by participating actively in the discussions on the forum. Also a special request for uploading whole papers from your national competitions (at least the ones you took part in one way or another).

[b]In the end, a little bit of history ...[/b]

It all started back in 2001, after I qualified to the IMO 2001. As I was lacking materials to prepare (at that time in Romania internet was not as developed as it is today, in fact it was quite unused in general) I had to put in a lot of effort to gain access to some new problems (other than the ones presented in the Romanian books at that time). After the IMO, I had already made a few connections, and also gain access home to internet, so I decided to put up a (personal) web-page in the spirit of the ones that [url=http://www.ajorza.org/math/]Andrei Jorza[/url] (fellow IMO team member in 2001, now Harvard student), [url=http://my.netian.com/~ideahitme/eng.html]Hojoo Lee[/url] (Korrean graduate student), [url=http://de.geocities.com/darij_grinberg/]Darij Grinberg[/url] (German high-school student), [url=http://www.kalva.demon.co.uk/]John Scholes[/url] (his site was a lot smaller back then, containing only IMO and Putnam problems) and others had at the moment, so I made one myself, including some Romanian and Balkan problems, some with solutions. The idea was/is/will be that people are linked (or connected) by Mathematics.

Later then, I changed my (free) provider, and the site went to http://www.mathlinks.go.ro where it growed, passing through different stages (I even tried working with Flash), untill I discovered phpBB2 forums. Intially I found a provider that hosted the forum and the site remained at the go.ro address, but then I moved it completely to the new provider. The forum started on 3 February 2003, with just a couple or so of users. Weeks passed and not much was happening, except for Grobber which started posting (and never stopped :) ). Then REEA came with a sponsorship for hosting the site on their servers, and that's when http://www.mathlinks.ro became alive. With the new hosting, all comercials adds had been eliminated, and new people started joining in the discussions on the board.

Soon, the IMO came. I was along with the Romanian team as an Observer to the leader, and this (along with the fact that the Japanese official site was not so well organized) gave Mathlinks.ro the chance of being the first site in the world with the IMO 2003 questions online. As a result, more than 6,000 new visitators checked Mathlinks, and the forum started growing in numbers. First Moubi, Pierre Bornstein, Namdung, well known teachers in problem-solving communities started posting, and they were followed by top students as Fedor Petrov (gold medalist), Le Viet Bao (full score in IMO 2003), Andrei Negut (two times IMO silver medalist, future Princetonian) just to say a couple of names that come first to my head, because there are many others as good as them (if not better).

Next, again the providers changed in need for more bandwidth and server power, going to Roedu.net hosting, and from then, in April 2004, it is hosted on one of the servers of the SpareTimeGroup, a similar community in informatics and computer science. With the help of Alexandru Armean and Steme Mayer the movement went fine, and you can now post in LaTeX on MathLinks.

So, word by word, from person to person, with the very much appreciated help of Orlando Doehring, and also many others, which you can find even right now online, MathLinks has grown into a full size, large and diversificated mathematical community, with a strong accent on Olympiad Style problems and tehniques. Hopefully the community will soon include many other people working in other areas but Olympiads of this wonderfull discipline that Mathematics is.

Have fun on MathLinks,
Valentin Vornicu.

Last updated at April 22, 2004"

I have been waiting for this thread for a long time :D .

//brings Blackforest cake in
//brings prezzies in :P

Happy Birthday MathLinks! :) :) :)

EDIT: I just saw the post that I made last year and realise that this year's post looks remarkably similar to last year's one :oops: .

Well,

I al so say HAPPY BIRTHDAY MATHLINKS.

best regards,

[quote="orl"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3472]HAPPY BIRTHDAY TO MATHLINKS AND VALENTIN.[/url] Congrats to the second anniversary of ML and the 22nd of Valentin. A happy and long life to both.  :)[/quote]While I am feeling  :blush:  :blush:  about this, my birthday is not on 14th of February as some may think, but it's rather a nameday, or a coincindence. When I was born, in the communist Romania era, there wasn't any refferece of a Saint Valentine :D

This being said, I am happy to report that I am still 21 years old :D :D

Happy birhtday Mathlinks

Open bottles of Champagne, red wine Pommard 1er cru

[quote="djimenez"]Well,

I al so say HAPPY BIRTHDAY MATHLINKS.

best [b]retards[/b],[/quote]

Haha. I got a kick out of that typo.

HAPPY BIRTHDAY MATHLINKS!

Even though I've really been on Aops all my math life...

[quote="Valentin Vornicu"]This being said, I am happy to report that I am still 21 years old :D :D[/quote]

It really doesn't matter what your age is, Valentin, we all know that you'll never grow old :P .

[quote="bénabar"]Happy birthday Mathlinks (and Valentin? I didn't understand well)[/quote]Just recently [url=http://www.pictureframes.co.uk/pages/saint_valentine.htm]Saint Valentine's Day[/url] took place (each 14th of February) which is a nameday for me :P (that is the reason I avoided putting up a holiday message in the first place for this :P  :blush: ).

yup... happy mathlinksday all :)
<let's call it mathlinksday, ok? :P>

well, 2 years isn't bad, still I feel like many more years are awaiting us! :D

[quote="djimenez"]Just a little of stadistic:

The date the site were created, there were only two members registered: Valentin and Math1 (I don't know if the second is a real user or a test user), after that

By the end of the first month there were 8 users.
By the end of the first 3 months there were 32 users.
By the end of the first 6 months there were 409 users.
By the end of the first year there were 1265 users.
By the end of the second year there were 6389 users.  (some function that approximate the number of users?)

Also, something courius is that the days from May 25th to 30th, 2003 there were a pick in the registration (almost a 25% of the people registered by the end of the first 6 months registered in those five days!!)

Well, just something that people could find interesting.

And finally, Congratulations to the site and to Valentin, the truth is that the site is great, and even if it is not your birthday too, you have done a great job, and so many people enjoy it right now.

(I know that may be I am a little unfair with other people who have helped a lot in the site mantainace, but I don't really know about you. I know a lot of people helps moderating, and some even administrating, but I don't know exactly who, and what else you do besides moderate, that is, itself, very helpfull)

Well, now for real, finally, THREE HURRAS FOR VALENTIN AND HIS TEAM... AND THE GREAT SITE THEY HAVE CREATED!!!!

Best,[/quote]

wow...thats really cool, where did you get all that info, or did actually record it

[quote="RC-7th"]Go to the member list, then click "sort by date joined".[/quote]

Very smart!!!, that is exactly what I did. And after posting my last message on this thread, I found that if you go to usergroups you can actually find who are exactly the adminstrators, and substract them accondingly. Of course, Valentin already gave us the list!

Best regards,


analytic geometry
graphing lines
slope
trigonometry
geometry
angle bisector
trig identities

No calculator.

An acute angle is formed by two lines of slope 1 and 7.  What is the slope of the line that bisects the angle?

I'm pretty sure there's a formula that would make this really easy, but I don't remember it.

Depends on what you would call really easy. There's certainly a formula for the answer, but it's not worth remembering and defeats the purpose of the problem. There is a useful formula, of course:
[hide="this"]
the angle bisector theorem. Now just relate slopes to lengths/angles in a nice way.
[/hide]

Well, this page describes the angle between two lines...
http://www.tpub.com/math2/5.htm

Let the two slopes be $ m$ and $ n$, where $ m > n$. Then, the angle formed by the line with slope $ m$ and the x-axis would be $ \tan^{ \minus{} 1} m$, and the angle formed by the line with slope $ n$ and the x-axis would be $ \tan^{ \minus{} 1} n$.

Then, the angle formed by the angle bisector and the x-axis would be:
\[ \frac {\tan^{ \minus{} 1} m \minus{} \tan^{ \minus{} 1} n}{2} \plus{} \tan^{ \minus{} 1} n \equal{} \frac {\tan^{ \minus{} 1} m \plus{} \tan^{ \minus{} 1} n}{2}
\]
The slope would therefore be:
\[ \tan{\left(\frac {\tan^{ \minus{} 1} m \plus{} \tan^{ \minus{} 1} n}{2}\right)}
\]
I don't know how to simplify from there...
[hide]By the way, how come $ \tan{\left(\frac {\tan^{ \minus{} 1} 7 \plus{} \tan^{ \minus{} 1} 1}{2}\right)}\equal{}2$?[/hide]

Another way, using the Angle Bisector Theorem, is:
\[ n\plus{}\frac{\left(m\minus{}n\right)\sqrt{n^2\plus{}1}}{\sqrt{m^2\plus{}1}\plus{}\sqrt{n^2\plus{}1}}\]However, I do not know how to simplify this any further.

[quote="123456789"]Another way, using the Angle Bisector Theorem, is:
\[ n \plus{} \frac {\left(m \minus{} n\right)\sqrt {n^2 \plus{} 1}}{\sqrt {m^2 \plus{} 1} \plus{} \sqrt {n^2 \plus{} 1}}
\]
However, I do not know how to simplify this any further.[/quote]

$ \frac{ \frac{n}{\sqrt{n^2\plus{}1}}\plus{} \frac{m}{\sqrt{m^2\plus{}1}}} {\frac{1}{\sqrt{n^2\plus{}1}}\plus{}\frac{1}{\sqrt{m^2\plus{}1}}}$

[quote="123456789"]Let the two slopes be $ m$ and $ n$, where $ m > n$. Then, the angle formed by the line with slope $ m$ and the x-axis would be $ \tan^{ \minus{} 1} m$, and the angle formed by the line with slope $ n$ and the x-axis would be $ \tan^{ \minus{} 1} n$.

Then, the angle formed by the angle bisector and the x-axis would be:
\[ \frac {\tan^{ \minus{} 1} m \minus{} \tan^{ \minus{} 1} n}{2} \plus{} \tan^{ \minus{} 1} n \equal{} \frac {\tan^{ \minus{} 1} m \plus{} \tan^{ \minus{} 1} n}{2}
\]
The slope would therefore be:
\[ \tan{\left(\frac {\tan^{ \minus{} 1} m \plus{} \tan^{ \minus{} 1} n}{2}\right)}
\]
I don't know how to simplify from there...
[hide]By the way, how come $ \tan{\left(\frac {\tan^{ \minus{} 1} 7 \plus{} \tan^{ \minus{} 1} 1}{2}\right)} \equal{} 2$?[/hide][/quote]

Consider...[hide]$ \tan (\alpha \plus{} \beta) \equal{} \frac{\tan \alpha \plus{} \tan \beta}{1\minus{}\tan \alpha \tan \beta}$  and..the tangent double angle formula.  [/hide]

[hide="answer"]\
Let $ \tan{\left(\frac {\tan^{ \minus{} 1} 7 \plus{} \tan^{ \minus{} 1} 1}{2}\right)}\equal{}x$

$ \tan{\left(\\tan^{ \minus{} 1} 7 \plus{} \tan^{ \minus{} 1} 1\right)} \equal{} \frac{1\plus{}7}{1\minus{}7}$.

Using the double angle formula, we have $ \frac{\minus{}4}{3} \equal{} \frac{2x}{1\minus{}x^2}$

Solving for $ x$ we get $ x \equal{} \frac{\minus{}1}{2} , 2$.  Uh, it's obvious that the angle bisector must have positive slope..and between the two numbers...so $ 2$.
[/hide]


arithmetic sequence
number theory unsolved
number theory

Prove that there are infinitely many positive integers $n$ with the following property: For any $n$ integers $a_{1},a_{2},...,a_{n}$ which form in arithmetic progression, both the mean and the standard deviation of the set $\{a_{1},a_{2},...,a_{n}\}$ are integers.
[i]Remark[/i]. The mean and standard deviation of the set $\{x_{1},x_{2},...,x_{n}\}$ are defined by $\overline{x}=\frac{x_{1}+x_{2}+...+x_{n}}{n}$ and $\sqrt{\frac{\sum (x_{i}-\overline{x})^{2}}{n}}$, respectively.


calculus
integration
geometry
trapezoid
real analysis
real analysis unsolved

$ \frac{1}{n^2}\equal{}s_n\minus{}s_{n\plus{}1}, s_n\equal{}\sum_{k\equal{}n}^\infty \frac{1}{k^2}$
How to prove:
\[ s_n\equal{} 1/n \plus{}O(1/n^2)\]
thanks

[quote="zhoraster"][quote="adroit."]what is meant by O ?[/quote]
[url]http://www.mathlinks.ro/viewtopic.php?t=31517[/url]

[/quote]

ya thanks for the link.  but its way over my curren knowledge.  anwyays would try to understand something out of it.

[quote="zhoraster"][quote="adroit."]what is meant by O ?[/quote]
[url]http://www.mathlinks.ro/viewtopic.php?t=31517[/url]

As for the original question, write
\[ \sum_{k\ge n} \Big(\frac {1}{k(k \plus{} 1)} \plus{} \frac {1}{k^2} \minus{} \frac1{k(k \plus{} 1)}\Big) \equal{} \frac1n \plus{} \sum_{k\ge n} \frac {1}{k^2(k \plus{} 1)} \equal{} \frac 1n \plus{} O\Big(\frac1{n^2}\Big)\]
where the last e.g. by Schwartz (discrete L'Hospital).[/quote]

You mean the last is by stolz ?  what do you mean by Schwartz?

Thanks

Perhaps another method... compare the series to an integral.

[quote="water2011"]
You mean the last is by stolz ?  what do you mean by Schwartz?

Thanks[/quote]
Ouch, Stolz, of course  :blush:

Or use the integral test.

$ s_n\equal{}\sum_{k\equal{}1}^{\infty}\frac1{k^2}$

$ s_{n\plus{}1}<\int_n^{\infty}\frac1{x^2}\,dx<s_n$ (Draw a picture!)

$ s_n\minus{}\frac1{n^2}<\frac1n<s_n,$ so

$ \frac1n<s_n<\frac1n\plus{}\frac1{n^2}.$

Which is already more that what was asked for.

By using trapezoids rather than left or right sums, we can get

$ \frac12(s_n\plus{}s_{n\plus{}1})>\int_n^{\infty}\frac1{x^2}\,dx,$ which rearranges to

$ s_n>\frac1n\plus{}\frac1{2n^2}$

But then there's the midpoint rule approach, from which we get that

$ s_n<\int_{n\minus{}\frac12}^{\infty}\frac1{x^2}\,dx\equal{}\frac1{n\minus{}\frac12}<\frac1n\plus{}\frac1{2n^2}\plus{}\frac1{4n^3}.$

So now I have that $ s_n\equal{}\frac1n\plus{}\frac1{2n^2}\plus{}O(n^{\minus{}3}).$


[i]Find all pairs of positive integers $ \left( m \ ; \ n \right)$  such that :[/i]

 $ m^{2} \ \minus{} \ 1 \ | \ 3^{m} \ \plus{} \ (n! \ \minus{} \ 1 )^{m}$

[i]  Zetax , I am waiting for your solution [/i] :D

If $ n\equal{}1$, then $ m\equal{}2$.
If $ n\equal{}2$, then $ (m\minus{}1)(m\plus{}1)|3^m\plus{}1$. If m odd, then $ 3^m\plus{}1\equal{}4\mod 8$, but $ 8|(m^2\minus{}1$. Therefore m is even. All odd prime divisors p of $ 3^m\plus{}1$ are $ p\equal{}1\mod 4$. But $ m\minus{}1\equal{}3\mod 4$ or $ m\plus{}1\equal{}3\mod 4$ and had prime divisor form $ p\equal{}3\mod 4$. Therefore $ n\equal{}2$ had not solution.
If $ n>2$ then $ n!\minus{}1\equal{}2\mod 3$, therefore $ 3|m$ (else $ 3|(m^2\minus{}1)$, but $ 3\not |3^m\plus{}(n!\minus{}1)^m$.
If $ n>2,m$ -even, then all odd prime divisors of $ 3^m\plus{}(n!\minus{}1)^m$ are $ p\equal{}1\mod 4$. But $ (m\minus{}1)(m\plus{}1)$ had prime divisor form $ p\equal{}3\mod 4$.
Therefore $ n>2,m$- odd. If $ n>3$, then $ 3^m\plus{}(n!\minus{}1)^m\equal{}2\mod 8$, but $ 8|m^2\minus{}1$.
Therefore $ n\equal{}3, m$ -odd and $ m^2\minus{}1|3^m\plus{}5^m\to 3|m$.
Obviosly $ (n,m)\equal{}(3,3)$ solution. If $ m>3$, then from $ 3|m$ we get all odd prime divisors $ 3^m\plus{}5^m$ are $ p\equal{}1\mod 6,p>7.$
Easy to chek. that if we had another solution, then $ n\equal{}3, 3|m, m$ odd and $ m\ge 63$.
I found only $ (n,m)\equal{}(1,2),(3,3)$.


Find the units digit of $ 19^{19}\plus{}99^{99}$.

[hide="clickie!"]I know i would find the the digits of both the $ 19^{19}$ and the $ 99^{99}$ with moding it with two (since the units digit repeats after a power of two) but, when I get 1 as the residue then does that mean that 9 would be the digit or would that mean that 1 is the digit? [/hide]


function
integration
real analysis
real analysis unsolved

1. Prove that if $ n=1\$$ and $ u\in W^{1,p}(0,1)\$$ for some $ 1\leq p<\infty\$$, then $ u\$$ is equal a.e to an absolutely continuous function, and $ u'\$$ (which exists a.e) belongs to $ L^p(0,1)\$$.\\

2. Prove directly that if $ u\in W^{1,p}(0,1)\$$ for some $ 1<p<\infty\$$, then $ |u(x)-u(y)|\leq |x-y|^{1-\frac{1}{p}}(\int_{0}^1|u'|^pdt)^{\frac{1}{p}}\$$ for a.e $ x,y\in [0,1]\$$. (I have a solution, but I think it is not good, beacuase it can not directly show the expression $ |x-y|^{1-\frac{1}{p}\$}$.)

1) [hide="hint"]Just try to prove $ u(x)\equal{}\int_0^xu_1(t)dt\plus{}C$ where $ C$ is a constant and $ u_1$ is u's generilized derivative.[/hide]

2) [hide="hint"]Use 1) and $ \text{Hold}\ddot{e}\text{r}$ inequality.[/hide]

[quote="pluricomplex"]$ \text{Hold}\ddot{e}\text{r}$[/quote]

see my type $ \rm Hold\ddot{e}r$, you know how?  :rotfl:


From the Introduction to Counting and Probability, we have the Grunters and the Screamers, two teams in a tournament. So let's call this a 7-match tournament. In order for the Grunters to win in exactly 7 matches, we must have the Grunters and the Screamers each win 3 of the first 6 games. Thus the number of distinct outcomes the tournament could have is $ \binom{6}{3}$ and the last one is reserved for the Grunters. But what is wrong with saying that out of the seven games that have to be played, the Grunters must choose 4 of them to win or the Screamers must choose 3 of them to loose, thus turning out with $ \binom{7}{4} \equal{} \binom{7}{3}$?

I am not quite sure what is wrong with that.

Thank you very much for any help. :)

Both of those combinations are the same thing!

You will later learn that $ \dbinom{n}{r}\equal{}\dbinom{n}{n\minus{}r}$

\[ \binom{n}{r} \equiv {}_nC_r \equiv \frac{n!}{(n\minus{}r)!r!}\]
So,\[ \binom{n}{r} \equal{} \frac{n!}{(n\minus{}r)!r!} \equal{} \frac{n!}{r!(n\minus{}r)!} \equal{} \binom{n}{n\minus{}r}\]

[quote="Ihatepie"][quote="limac"]Uhhh...I see where I could go wrong with $ \binom{7}{3},$ for example. In that example, we know that the Grunters *must* win the last game, but with $ \binom{7}{3},$ or $ \binom{7}{4},$ we are also to let the Screamers win the last game.

I am not sure how to delete this...I click on the 'edit/delete' button, and then what?[/quote]

I'm pretty sure that's not what is wrong with your answer. What's wrong with 7C3/4 is that you are not taking into account that the Grunters can win in less than 7 matches.[/quote]

Ihatepie, The problem asked for the the number of combination of the Grunters' victory in exactly 7 games, I think you might have misread that part ;)

While on a related topic, I have another question:

This is the problem: A penny is tossed 3 times and a nickel is tossed 2 times. What is the probability that more heads are tossed using the penny than the nickels?

I know this is not the best approach, but what is wrong with this?

We know that $ P(\text{more head with pennies}) \equal{} 1 \minus{} P(\text{more or equal heads with nickels}).$
Thus, we have a sequence like: $ T_p,T_p,T_p,H_n,T_n$ or $ T_p,T_p,T_p,H_n,H_n.$ Two heads is the maximum we can have with nickels. So if we have 1 heads with a nickel, we can have it for either nickel. And we could have no heads with the pennies for a total of $ \binom{3}{0}\binom{2}{1},$ or we could have one heads with the pennies each for a total of $ \binom{3}{1}\binom{2}{1}.$ We could have 2 heads now. So, we have 0 heads with the pennies $ \binom{3}{0}\binom{2}{2},$ or have 1 heads with the pennies $ \binom{3}{1}\binom{2}{2},$ or finally, $ \binom{3}{2}\binom{2}{2},$ for a total of 8 for 1 heads with the nickel and 10 for 2 heads with a nickel and we could have all tails for another one. Thus, we can have $ 2^5 \minus{} 19 \equal{} 13.$ So, we have $ 13/32.$ The correct answer however was $ 1/2.$ I am not sure where I am going wrong.

[hide="solution"]
Each ordering has an equal probability of $ \frac{1}{32},$ so at the end, just add up the number of successful orderings and put it over $ 32.$
\[ (3,2)\Rightarrow\binom{3}{3}\binom{2}{2}\equal{}1\newline(3,1)\Rightarrow\binom{3}{3}\binom{2}{1}\equal{}2\newline(3,0)\Rightarrow\binom{3}{3}\binom{2}{0}\equal{}1\newline (2,1)\Rightarrow\binom{3}{2}\binom{2}{1}\equal{}6\newline(2,0)\Rightarrow\binom{3}{2}\binom{2}{0}\equal{}3\newline(1,0)\Rightarrow\binom{3}{1}\binom{2}{0}\equal{}3\] Add everything up and put it over $ 32$ to get $ \frac{16}{32}\equal{}\boxed{\frac{1}{2}}.$

Brute forcing this isn't that hard, but I think there is an easier solution.[/hide]

[quote="limac"][quote="Ihatepie"][quote="limac"]Uhhh...I see where I could go wrong with $ \binom{7}{3},$ for example. In that example, we know that the Grunters *must* win the last game, but with $ \binom{7}{3},$ or $ \binom{7}{4},$ we are also to let the Screamers win the last game.

I am not sure how to delete this...I click on the 'edit/delete' button, and then what?[/quote]

I'm pretty sure that's not what is wrong with your answer. What's wrong with 7C3/4 is that you are not taking into account that the Grunters can win in less than 7 matches.[/quote]

Ihatepie, The problem asked for the the number of combination of the Grunters' victory in exactly 7 games, I think you might have misread that part ;)
[/quote]Nope, I didn't. What I was saying is that 7C3/4 includes cases where the Grunters win in under 7 matches. That's not allowed in the problem, so that's why it didn't work.

[quote="limac"]While on a related topic, I have another question:

This is the problem: A penny is tossed 3 times and a nickel is tossed 2 times. What is the probability that more heads are tossed using the penny than the nickels?

I know this is not the best approach, but what is wrong with this?

We know that $ P(\text{more head with pennies}) = 1 - P(\text{more or equal heads with nickels}).$
Thus, we have a sequence like: $ T_p,T_p,T_p,H_n,T_n$ or $ T_p,T_p,T_p,H_n,H_n.$ Two heads is the maximum we can have with nickels. So if we have 1 heads with a nickel, we can have it for either nickel. And we could have no heads with the pennies for a total of $ \binom{3}{0}\binom{2}{1},$ or we could have one heads with the pennies each for a total of $ \binom{3}{1}\binom{2}{1}.$ We could have 2 heads now. So, we have 0 heads with the pennies $ \binom{3}{0}\binom{2}{2},$ or have 1 heads with the pennies $ \binom{3}{1}\binom{2}{2},$ or finally, $ \binom{3}{2}\binom{2}{2},$ for a total of 8 for 1 heads with the nickel and 10 for 2 heads with a nickel and we could have all tails for another one. Thus, we can have $ 2^5 - 19 = 13.$ So, we have $ 13/32.$ The correct answer however was $ 1/2.$ I am not sure where I am going wrong.[/quote]

Direct counting:

$ \begin{gathered}\binom{3}{1}\binom{2}{0} + \binom{3}{2}\binom{2}{0} + \binom{3}{3}\binom{2}{0} + \binom{3}{2}\binom{2}{1} + \binom{3}{3}\binom{2}{1} + \binom{3}{3}\binom{2}{2} = & 3 + 3 + 1 + 6 + 2 + 1 = 16\end{gathered}$

You are missing $ \binom{3}{0}\binom{2}{0}$ (no heads) and the counting should be

$ \begin{gathered}\binom{3}{0}\binom{2}{0} + \binom{3}{0}\binom{2}{1} + \binom{3}{1}\binom{2}{1} + \binom{3}{0}\binom{2}{2} + \binom{3}{1}\binom{2}{2} + \binom{3}{2}\binom{2}{2} = & 1 + 2 + 6 + 1 + 3 + 3 = 16\end{gathered}$

EDIT: OK I see you have "we could have all tails for another one". You've miscounted, it's not 10, it should be 7.

Thank you guys very much for the responses. 10000th User, can you please specify where/how I overcounted. I tried checking it but couldn't figure out how I was overcounting? Thank you again :)

[quote="limac"]... and 10 for 2 heads with a nickel ...[/quote]Like I said before, it's not 10, it's 7.

Got it! Thank you very much for the replies!


function
calculus
logarithms
derivative
calculus computations

Hi, can you show me the way to solve for $ i$ for the following equation?

$ i \equal{} 45e^{\minus{}100i} \minus{} 20$

thanks.

Numerically.

We can use the Intermediate Value Theorem to guarantee that it has a solution (for some small positive $ i$) and since $ i$ is increasing while $ 45e^{ \minus{} 100i} \minus{} 20$ is decreasing, it has only one solution. But you're not going to get that solution in closed form using the usual functions of Calculus I-II.

Here's a quick take. First I rewrite the equation as $ i \equal{} \minus{} 0.01\ln\left(\frac {20 \plus{} i}{45}\right).$

The function on the right appears to have a rather small derivative, so we iterate it to solve the fixed point problem.

That is, let $ x_0 \equal{} 0$ and then let $ x_{k \plus{} 1} \equal{} \minus{} 0.01\ln\left(\frac {20 \plus{} x_k}{45}\right).$

Upon displaying 14 digits in my spreadsheet, I see that the sequence stops showing any changes after $ x_4,$ settling at $ 0.00810525035795.$ So there's your approximate answer.

You could also do this pretty quickly with a calculator - but when I'm seated at a computer, I prefer the spreadsheet.

hello, you can solve this equation using the LambertW function, the answer is
$ \minus{} 20 \plus{} {\frac {1}{100}}\,{\it LambertW} \left( 4500\,{{\rm e}^{2000}} \right)$
or
$ x \approx 0.0081052503579510569634565323133755821968915199$
Sonnhard.

The $ W$ function seems to me to come up often in math and for sure it's not in Calculus I or II, still though I think the technique to arrive at the solution posted above is good to be familiar with when you have implicit functions and you're interested in knowing what an explicit expression would look like.  So if we're given the expression $ f \equal{} g(x)e^{g(x)}$, then the $ W$ function is it's inverse, that is: $ g(x) \equal{} W(f)$.  So that if we have the expression $ y \equal{} 45e^{ \minus{} 100y} \minus{} 20$, then what do we have to do to get that expression into a form so that we could take the $ W$ function of both sides and arrive at:
\[ y \equal{} \frac {1}{100}W(45e^{2000}) \minus{} 20
\]
Edit:  I see I got 45 instead of 4500 above.  Maybe someone can show me what I did wrong.


algorithm
probability
group theory
number theory
superior algebra
superior algebra unsolved

Find the number of primitive 18th roots of unity in $ GF(19)$.

Since $ GF(19)$ is a finite field of order 19, and so is $ \mathbb{Z}_{18}$, they both are isomorphic. Thus we can look at $ \mathbb{Z}_{18}$.

A number x is a root of unity if gcd(18,x) = 1. So the number of such elements is $ \phi(18) \equal{} 2*3 \equal{} 6$.

Is there another way of doing this?

[quote="daboa"]Since $ GF(19)$ is a finite field of order 19, and so is $ \mathbb{Z}_{18}$, they both are isomorphic.[/quote]

This is both semantically and syntactically nonsense. What you want to say is that $ \left(\text{GF}\left(19\right)\right)^{\times}$ is isomorphic to $ \mathbb{Z}_{18}$, but you need a different argument for this than just the equality of order.

[quote="daboa"]A number x is a root of unity if gcd(18,x) = 1.[/quote]

I hope you want to say the following: A number $ x$ is a [i]primitive[/i] $ 18$-th root of unity if and only if $ \gcd\left(18,x\right) \equal{} 1$. And this is still informal; a 100% correct way to state this would be: An element $ x\in\left(\text{GF}\left(19\right)\right)^{\times}$ is a primitive $ 18$-th root of unity if and only if $ \gcd\left(18,i\left(x\right)\right) \equal{} 1$, where $ i$ is a given isomorphism $ \left(\text{GF}\left(19\right)\right)^{\times}\to\mathbb{Z}_{18}$.

  darij

1) $ 19$ is a prime then $ GF(19)^*$ is a cyclic group for the multiplication and thus is isomorphic to $ \mathbb{Z}_{18}$ with the addition.
2) A primitive element is (definition!) a generator of this group; for example $ 2$.
There exists $ 6$ such generators.
3) According to Fermat, each element of $ GF(19)^*$ is a 18th root of the unity (that is to say $ x^{18}\equal{}1$).

1)Just out of curiosity since primitive roots are important in cryptograhpy, am I correct to say there is no general method in determining a primitive root for a prime?
2)If #1 is no, then do their certain algorithms in computational number theory that give the primitive roots.

Hi sylow_theory,
often in cryptography the "owner" of the system must know a primitive element. Here $ p$ is a very big prime ($ >10^{150}$). The test is simple:
$ a$ is a generator iff $ \forall{q}$ prime and divisor of $ p\minus{}1$,  $ a^{(p\minus{}1)/q}\not\equal{}1 \;mod\,p$.
1) The owner calculates the decomposition in prime of $ p\minus{}1$; in general if $ p$ is a random prime then he can do that (in few hours). Else he picks another $ p$.
2) The owner picks a random $ a$-value and he tests this $ a$ (the complexity is polynomial). The probability that $ a$ is a generator is $ \phi(p\minus{}1)/(p\minus{}1)\equal{}\pi{(1\minus{}1/q)}$ where the product $ \pi$ is taken over the primes $ q$ which are divisor of $ p\minus{}1$. Thus this probability is near $ 1/2$ if the prime factors $ q\not\equal{}2$ are big. (that is contradictory with 1)).
Conclusion: after few hundred tests we are done.


logarithms
calculus
integration
calculus computations

this came to my mind by a related topic posted here...

Show that $ \sum_2^{\infty}\frac{1}{n(\log{n})^p}$ converges if $ p>1$ and diverges if $ p\le 1$.

Integral test.  Alternately, [url=http://en.wikipedia.org/wiki/Cauchy_condensation_test]Cauchy condensation[/url].

Discuss the convergence of the integral:

$ \int\limits_{a}^{\infty}\frac {dx}{x(\ln x)^{p}}$,

which can be studied by using the substitution $ \ln x \equal{} y$.

using $ u \equal{} \ln{x}$

$ \int_a^\infty \frac{1}{u^p}du$
This is a p-series integral which only converges when p>1. This can be easily shown:

$ \int_a^\infty u^{\minus{}p} du$
If p=1, the antiderivative is $ \ln{u}$ which clearly diverges

Otherwise, the antiderivative is
$ \frac{u^{1\minus{}p}}{1\minus{}p}$
The numerator diverges when p<1. Converges when p>1.


algebra
polynomial
induction
algebra proposed

F(x) is a polynomial whose degree is greater than zero.

Find all F(x) satisfying the equation F(F(x)) =$[F(x)]^{n}$ where n is a positive integer

Suppose that $P(x)=\sum_{k=0}^{m}{a_{k}x^{k}},a_{m}\not=0$
it is obvious that $a_{m}=1$and $m=n$ so we can prove by induction the $a_{n-k}=0$for all $k=1,2,\dots,n$


analytic geometry
geometry
geometry theorems

$ P(x_1: y_1: z_1)$ and $ Q(x_2: y_2: z_2)$ with $x_1+y_1+z_1=x_2+y_2+z_2$ represent the barycentric coordinates of $P,Q$ WRT $\triangle ABC.$ The distance betwen $P,Q$ is given by

$PQ^2 \equal{} S_A(x_2 \minus{} x_1)^2 \plus{} S_B(y_2 \minus{} y_1)^2 \plus{} S_C(z_2 \minus{} z_1)^2$

With the usual Conway notation

$ S_A \equal{} \frac { b^2 \plus{} c^2 \minus{} a^2}{2} \ , \ S_B=\frac{c^2+a^2-b^2}{2} \ , \ S_C=\frac{a^2+b^2-c^2}{2}$

[b]Leibniz formula:[/b] Let $ Q$ be a point with homogeneus barycentric coordinates $ (u: v: w)$ with respect to $ \triangle ABC.$ For any point $ P$ on the plane $ ABC,$ the following relation holds:

$uPA^2 \plus{} vPB^2 \plus{} wPC^2 \equal{} (u \plus{} v \plus{} w)PQ^2 \plus{} uQA^2 \plus{} vQB^2 \plus{} wQC^2$

Thank you very much for your message, dear Luis! I am not familiar with algebraic geometry (especially about barycentric coordinates) so I found your post is really helpful. Thank you!

Due to a request via PM, I will write the proof of Leibniz theorem

By definition $(u: v: w)$ are three scalars that satisfy $ u\overrightarrow{QA} \plus{} v\overrightarrow{QB} \plus{} w\overrightarrow{QC} \equal{} \overrightarrow{0} \ (\star)$

For  any point $ P$ in the plane $ ABC,$ we'll have

$ u\overrightarrow{PA}^2 \equal{} uPQ^2 \plus{} uQA^2 \plus{} 2u \cdot \overrightarrow{PQ} \cdot \overrightarrow{QA}$

$ v\overrightarrow{PB}^2 \equal{} vPQ^2 \plus{} vQB^2 \plus{} 2v \cdot \overrightarrow{PQ} \cdot \overrightarrow{QB}$

$ w\overrightarrow{PC}^2 \equal{} wPQ^2 \plus{} wQC^2 \plus{} 2w \cdot \overrightarrow{PQ} \cdot \overrightarrow{QC}$

Now, summing up the 3 latter equations keeping in mind $(\star),$ we get then

$uPA^2 \plus{} vPB^2 \plus{} wPC^2 \equal{} (u \plus{} v \plus{} w)PQ^2 \plus{} uQA^2 \plus{} vQB^2 \plus{} wQC^2$


Three One inch squares are placed with their bases on a line. the center square is lifted out and rotated $45$ degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point B from the line on whic hthe bases of the original squares were placed?

A. 1

B. $\sqrt{2}$

C. $\frac{3}{2}$

D $\sqrt{2}+\frac{1}{2}$

E. 2

[quote="akkaren"]Possibly I'm being an idiot, but it only works (answer-wise) if point B is the top vertex, of all the reasonable points I can think of...[/quote]

yeah that's the question.

sorry about that

[hide="Solution"] So, the square's turned sideways, wedged between the other two. At the top of the side squares, draw a one inch line perpendicular to the diagonal of the middle square. Two right triangles are formed by the diagonal, the sides of the triangle, and the one-inch line. Their angles are 45 degrees and so their legs are all the same, 1/2. Since the height of the side squares is 1, the distance from the bottom vertex of the square to the one inch line is 1-1/2=1/2. Add the diagonal of the square and you get answer D. [/hide] I hope I didn't mess that up.


Fina all pairs $ (x,y)$ of real numbers for which $ x^2 \minus{} xy \plus{} x \minus{} y \equal{} 0$ and $ 4x^2 \plus{} xy \plus{} y^2 \equal{} 10$. (give answer in exact form).

[hide]I thought this was the only interesting question on the contest, and they marked my answer wrong because I didn't rationalize  :mad: [/hide]

[hide="Hint"]$ x^2 \minus{} xy \plus{} x \minus{} y \equal{} 0 \iff (x\plus{}1)(x\minus{}y)\equal{}0$[/hide]


probability
combinatorics proposed
combinatorics

The three forum members Maverick, alekk and iandrei play the following game: Arbitrarily they take a subset with k elements from the set{1,2,3,...,1986} away, k \leq 1986 is a natural number. Suppose that every subset can occur with the same probability. Generate the sum of all elements from k and divide it by 3. Maverick, alekk, iandrei win respectively according the remainder. If the remainder is 0, Maverick wins. alekk according to 1 and iandrei according to 2.
Determine the values for which the game is fair !

Note: A game is fair when every forum member has the same probability to win the game. The probability is defined as quotient of the number of appropriate cases divided by the number of possible cases !

This is one of the oldest problem on that section. Please Orlando, could you be more precise in the statement :
1) They may choose the same subset I presume?
2) You compute the sum of what? The sum of all the 3k elements from the chosen sets?
3) I think the question is to determine the values of k such that the game is fair, am I right?

Pierre.


Can someone please help me? I have some problems on my computer, a couple of days ago it started displaying a screen with auto-protected results that has a virus scan thing and there are also ads that always appear, some aurora ads. I don't know why this is happening and how to fix this. Can anyone please help?

Did you find NaviHelper.dll in your C:\WINDOWS\system32 ? If so, google it, you can find the solution easily. It's popular these days.

Sounds like you have spyware and also perhaps virus problems.  You might try downloading free software like AdAware from [url=http://www.lavasoft.de/]Lavasoft[/url].

If it's a big deal and you can't figure out how to fix it, as a last resort you could always back up all your important files and reformat the C drive and reinstall Windows.  This is kind of a pain because it takes a while to reinstall all the programs you cleared.

Or get Spybot S&D: http://www.safer-networking.org/en/download/

Don't reformat unless you can't fix it and the local computer shop can't either.

Also look into a program called Hijack This

be sure you learn how to use it before you do...

[quote="hello"]I have the Lafsoft Spybot program and I ran it and cleaned the viruses that shows I think but the auto-protected results keeps appearing.[/quote]
What does the auto-protect things say?  Do they say some thing like "Tracking cookies blocked." or "registry change detected."?  If it does, then you must have Ad-Watch loaded which is like an auto detection software.  Don't worry about it, because its normal.  But if you get annoyed, just unload Ad-Watch.

No, it does a virus scan and mostly adware appears, also there are pop up ads that appear.

In that case, it's definitely spyware, because viruses generally don't make ads.   Yea I would recommend reformatting because it is soooo hard to remove every single spyware on your computer.  And be careful, spywares can reveal your personal information.

You should do a system restore. Just go to start, all programs, accessories, system tools, system restore. Choose a date before your computer was malfunctioning, and restore it to that date.

It's most likely spyware - as was said viruses dont usually  err.. make ads. I wouldnt do a reformat for just some spyware, definitely not. System Restore is an OPTION.. but its also a pain, unless you havent changed ur comp since the last system restore, and you can only do it with XP.. and some people dont have Restore turned on or the last restore point was like 6 months ago. Usually creates one when you install a program. Either way, a free spyware scanner like Lavasoft Ad-Aware or Spybot Search & Destroy will probably clear up the problem. I also like Microsoft Anti-Spyware (Beta) - its not complete but it has a very clean interface and from what ive read in reviews it is very powerful. If none of the above fails, before resorting to a reformat, and maybe even before a system restore you could use Hijack This... but you either need to 1. know how to interpret everything you see OR 2. post it on a message board till someone will read it for you and tell you what to do. Its not a spyware scanner, it... just displays all the stuff on the computer that does certain stuff.

[quote="hello"]What does system restore mean?[/quote]
A System Restore is when you restore your computer's hard drive to a later date, turning everything back to normal. This clears up many PC problems.

[quote="Xantos C. Guin"][quote="pop up I received once"]Do You Hate Pop-Ups!? If so click here.[/quote]

Obviously a scam.
[/quote]

It doesn't realize that it's a pop-up to begin with  :rotfl:

[quote="Cyberman"]Just use LIVE-CD Knoppix a Linux version OS. Basically you download the .iso and burn it on a cd and put it in the computer, restart the computer and go into the linux like windows back in the day would.

It's a pretty nice feature if you use DSL or something like that.[/quote]
Yeah that's a good suggestion, he sounds someone who would know how to work knoppix.

Anyway...
[quote="hello"]they tried puttin service pack 2, i dont know, i still have all this crap on my comp, i dont want to reformat, lol, maybe i should.[/quote]Who's they? And... You don't have service pack 2?????? By your post it isn't clear. In any event, nothing you do is going to make a bit of difference if you don't have service pack 2. Also, if you don't, that means you haven't been installing microsoft updates. That's another no-no. These days you MUST keep your computer up-to-date by the hour, literally. Turn on auto updates, always install them instantly.

Do you have the Windows firewall on? If not, turn it on! (If you didn't have SP2, then turn it on when you get SP2). I read a study where they determined that it took the average unfirewalled windows computer 12 minutes upon connecting to the internet before it had spyware. You DONT have to click anything to get spyware, all you have to have is an open port. So PLEASE make sure you have a firewall, make sure it's working, and make sure you always keep your computer up-to-date.

It is very common these days for a virus to disable 2 things immediately, those are 1) Your windows installer, and 2) your anti-virus. This guarantees that you won't be able (without extremely advanced computer knowledge) to remove the virus. And if you try reinstalling, you won't be able to.

You have to get your virus scan working (and make sure it's up-to-date also!) before you can fix the other problems. If you can't get it working, then I would say you'll likely have to pay to have someone try to fix it, get a very knowledgable friend to help you, or reformat your computer.

Once you get the computer working correctly, I would suggest you run a virus scan at least once a week, and also run SpyBot and Ad-aware once a week as well.

Linux is awsome, you just need to learn the commands.  And it is 100x more secure than Windows, partly because most people dont use linux so computer harckers dont hack linux or make viruses that target linux.  But yea, try knoppix because in knoppix you can still access Windows formatted harddrives.

[quote="hello"]whats the firewall thing? I am bad in comp so i dont know much.[/quote]

Google firewall and get one as soon as possible. It's just a security system consisting of a combination of hardware and software that limits the exposure of a computer or computer network to attack from crackers; commonly used on local area networks that are connected to the internet :D

Service Pack 2 has a built in firewall, so you don't need to download any third party ones.

I think we also need to give Hello a little background in here:

Spyware: Any software (software: ANY program) on your computer which relays information to another computer that is not yours through the internet - through a PORT (read on) on your computer.

Port: Gateway to the internet. Basically, like a door to your computer, and theres I think something like 10000 of them on your computer. Works both in and out.

Firewall: Program that works in the background on your computer, designed to block all ports (metaphor: lock all doors) on your computer - so no spyware gets in, and no information from already existing spyware gets to the comp its trying to. Unfortunately, the firewall cannot block all ports, or you could not surf the web, or do anything to do with the internet, which some programs (WHICH COULD BE SPYWARE - even a program that you use could secretly be a 'spy) need. Some programs need to connect to the internet, whetehr for innocent reasons or not, and using internet explorer or aol is one of those programs, and for that, certain ports need to be opened. And, programs cant share ports, the doorway gets kinda crowded. So overrall, even w/ a firewall, you will have some ports open. Because internet explorer (or aol or anything to surf the web) needs ports open, anything you see in internet explorer could plant something in your computer. A popup ad that you click could, without you knowing it, cause some spyware to be downloaded to your computer through an open port that internet explorer needs open. So simply, you get spyware. Lesson: never click on a popup. Also, only go to websites that you are 100% sure (and not because it says it is, remember, people LIE) are safe - even w/out you clicking anything, going to a website can start a download, even give a hacker access to your computer through the internet. INTERNET EXPLORER OPTIONS, AND OTHERS: In addition to firewalls, Internet Explorer and other internet browsers like it have security options, which prevent SOME things from being downloaded to your computer w/ or w/out you knowing it. You have to set these settings yourself, get someone who is very knowledgable in computers to help you. Unfortunately, Internet Explorer and all browsers have security holes in them which allow some trickier spyware to get through anyway. Nothing you can do about that except switch browsers (Firefox is good) and get rid of spyware when it gets to your computer.

To Be continued tommorrow - too tired to finish this I'll edit and finish.

Added "tommorrow"-

There are many firewalls (programs that are firewalls) out there. Windows XP comes with one, and with Service Pack 2, a better one (go to microsoft.com and look around for downloading Service Pack 2) . To turn on the firewall in Windows, google "XP firewall" or something like that (im not gonna go into the steps now). Windows firewall though isnt the best, and many people opt to have an additional, third party firewall installed. Norton and Mcafee suites both come with firewalls, and I believe both have a standalone firewall for sale. ZoneAlarm is also a very popular one. Again, do NOT download a firewall which is advertised in a popup, or is advertised to you in an email.

Service Pack (& SP2): Basically, a package of many security and performance updates for a particular operating system (Windows XP is an example of an operating system). The latest, Service Pack 2, as stated above, can be found on Microsoft's website (http://www.microsoft.com).
~~~~~~~
P.S. Also, Hello, at this point I'd recommend a full reformat.. your laptop is pretty bad at this point and I doubt you have a system restore point thatll get you safe, nor an antispyware program that will solve your problems.

Overall, that was pretty well said...

[quote="ToBe"]Port: Gateway to the internet. Basically, like a door to your computer, and theres I think something like 10000 of them on your computer. Works both in and out.[/quote]
The TCP port range varies with OS but in Windows it goes up to 65535. (which is the same number of rows you can have in an excel worksheet! Crazy.)

Oooh sorry yeah.. thats a lot of ports, I just forgot the number... thanks. I know it varies by OS, but I think Hello has XP from what's been said - if Hello had Linux or Mac this problem wouldnt happen =D. *hopes this post is not considered spam*

[quote="anirudh"]3 things:

1. DO NOT DO SYSTEM RESTORE. It still keeps the spyware on. A spyware is like a like or something in an important file. THat file doesn't get editted during system. You just get exposed to risk.

2. The firewall in Windows is not safe enough. You should get a router TrueSPIFirewall or ZoneAlarm, or better-Both.

3. You should download SUSE Linux, (it's better), it looks a littel like Windows, and it's safer. Personal is available at http://www.novell.com , and Professional is available for download at http://www.opensuse.com , but if you are going to use opensuse, first install zonealarm, cuz in 3 days, I got attacked by 10,000 hackers, but ZoneAlarm blocked it. You can get a TrueSPIFirewall at http://www.netgear.com , and you can get ZoneAlarm Free at http://www.zonelabs.com . These are safe websites. If you don't want Linux, then just reformat your hard drive (ask your dad how to do htat, and if he doesn't know, google it).[/quote]

:\

1. Did you read the thread?  someone who's not very good at computers shouldn't bother with linux.

2. System Restore is absolutely safe.  It backs up files and restores your computer to a previous state.  If you know when the ads started coming, try restoring to a date before that time.

[b]Before you try any of these, try this.[/b]
go to microsoft.com
click on Microsoft antispyware
[hide="link"](http://g.msn.com/mh_mshp/98765?http://www.microsoft.com/downloads/details.aspx?FamilyID=321cd7a2-6a57-4c57-a8bd-dbf62eda9671&displaylang=en&&HL=Windows+AntiSpyware+(Beta)&CM=PopularPicks&CE=popDownloads)[/hide]
Do the stupid 'Genuine Advantage' check
Download and install the file.
Run the Antispyware icon now on the desktop
Scan for spyware.  If found, select delete for all of them.

[TEMPORARY FIX (if you can't get ms antispyware at the time.]
Go Start > Run.
Type msconfig. Enter
Go to the Startup Tab.  Uncheck all the stuff there.
Reboot
Ignore the warning.
(to revert, just ckeck them back again)

Hope this helps.

Also, please hide your thingie. It messes up the layout.

Ways to fix it:

:arrow: I tried to uninstall Mathematica by system restore and it didn't uninstall it. The files were still on C:. Therefoe, it won't do it with Spyware.

 :arrow: Go to ZoneLabs adn download the ZoneAlarm Pro Trial. Then run it's spyware check. It always finds one when I check it

:arrow: ONe more thing. You miht have botzor. Just go to my computer and then search your local drives for botzor.exe

:arrow: If you don't find it, then go to run and type in regedit. THen click on mycomputer in the tree and go to edit>find. Type in botzor .exe. If you find it(it should be 001), then right click hte entry and press delete. Accept th econfirmation.

arrow: If you do have it, then go to http://microsoft.com and read how to recover froim it.

NOTE: ALL MY STEPS HAVE ONLY BEEN TESTED ON XP. I JUST REMEMBERED THAT BOTZOR CREATES ADS(BECAUSE I JUST GOT RID OF IT FROM MY MACHINE)
I hope these help   :)

[quote="anirudh"]Trust me. I tried to uninstall Mathematica by system restore and it didn't uninstall it. The files were still on C:. Therefoe, it won't do it with Spyware.[/quote]

 First of all, do you understand the principles of System Restore??  System Restore does not intend to remove the files from C, but to restore the registry and to restore user files.

 The registry, not the files themselves controls the startup and loading of programs on the computer.  Therefore, simply restoring the registry to an earlier state would be sufficient in disabling spyware.  If you don't believe me, go create a system restore point and go download spyware to try it out.  (This challenge is provided in an AS-IS basis.  Liability for damages is not expressed or implied)

 Second of all, Why were you trying to uninstall software by System restore?  System Restore stores information about previous states of the computer, but not as extensively as an installation log.

 Third of all (yes, i have nothing to do), it does not hurt to try System Restore when you have spyware, though MS antispyware is better.  Just back up some important documents somewhere, and you can use System Restore.  If you aren't satisfied, simply run it to revert the changes.

If you still are skeptical, go consult google for some detailed destriptions about the registry and System Restore.

ut all you have to do is remove botzzor from the registry

How do you know that he has botzor? Botzor isnt unique in making popup ads, it's just one of many that do.


calculus
derivative
calculus computations

You need to use the chain rule twice.  Evaluate both second partial derivatives on the left side and multiply by the variable indicating.  If you do it correctly, terms should cancel out and you should be left with the expression on the right side.


ratio
geometry
parallelogram
geometry proposed

A triangle $ ABC$ has an obtuse angle at $ B$. The perpindicular at $ B$ to $ AB$ meets $ AC$ at $ D$, and $ |CD| \equal{} |AB|$.
Prove that $ |AD|^2 \equal{} |AB|.|BC|$ if and only if $ \angle CBD \equal{} 30^\circ$.

[quote="n^4 4"]Does anyone have a nice solution to this problem?  I am most curious.[/quote]

I'm curious too. Anyone?

owlie

If $\widehat{DBC}=30^\circ$, then $\widehat{ABC}=120^\circ$, hence $b^2=a^2+c^2+ac=(AD+c)^2$, or $AD(AD+2c)=a(a+c)$ $(1)$.
Construct the equilateral triangle $BCK$, with $K$ on $AB$, $CF\perp AB$, $F$ the midpoint of $BK$, take $L$ the symmetrical of $D$ about $C$ and get $BD\parallel CF \parallel KL$, but $(1)$ is equivalent to $ AD\cdot AL = AK \cdot BK$, or $\frac{AK}{AL}=\frac{AD}{BK}$, but $\frac{AK}{AL}=\frac{AB}{AD}$, hence $\frac{AD}{BK}=\frac {AB}{AD}$, or $AD^2=ac$; from the way of construction, it's obvious that the problem is iff type.

Best regards,
sunken rock

If $AD^2=AB.BC$ then let $E$ on ray $AB$ such that $AD^2=AB.AE$ so $AE=BC$ and $ED\perp AC\Rightarrow ED^2=ED.EA$
Construct parallelogram $ABCF,EF\cap BC=M$ so $AF=BC=AE,CF=AB=CD$
$\frac{CM}{CF}=\frac{BM}{BE}=\frac{AF}{AE}=1\Rightarrow CM=CF=CD$
Let $w$ be the circle through $M,D,F$ with center $C$
Because $ED\perp CD$ so $w$ is tangent to $ED$ at $D$ so $ED^2=EM.EF$
$\Rightarrow ED.EA=EM.EF$ so $MBAF$ is an inscribed quadrilateral
We have $\widehat{EFA}=\widehat{MBE}=\widehat{FAE}\Rightarrow EF=EA=AF$
Thus $\widehat{DBC}=90^{0}-\widehat{MBE}=90^{0}-\widehat{FAE}=30^{0}$


if there's a dartboard with numbers __ and __ and there's an infinite number of darts thrown, what are the numbers if the highest impossible number to get is 59?

See the [url=http://en.wikipedia.org/wiki/Coin_problem#n_.3D_2]chicken mcnugget problem[/url].


geometry
geometric transformation
reflection
trapezoid
power of a point
radical axis
geometry unsolved

AM, AN are two tangent, HM = HN , AH' = A'H',
prove A', B, C are in a same line.

[img]http://phsco.org/sketch01.jpg[/img]

Denote  (O), (P) the circles with tangents $ m \equiv AM,\ n \equiv AN.$ Reflect m, n in $ x \equiv HH'$ into $ m' \equiv A'N,\ n' \equiv A'M.$ Let n' meet (O) again at K and let m' meet (P) again at L. AMNA' is isosceless trapezoid with $ MN \parallel AA'$ and AM = A'N. $ \angle KBM \equal{} \angle KMA \equiv \angle A'MA \equal{} \angle ANA' \equal{} \angle LBN$ and M, B, N are collinear $ \Longrightarrow$ K, B, L are collinear. $ \angle NMK \equiv \angle BMK \equiv NMA' \equal{} \angle MNA \equal{} \angle BLN \equiv \angle KLN$ $ \Longrightarrow$ KMLN is cyclic, $ A'K \cdot A'M \equal{} A'N \cdot A'L$ $ \Longrightarrow$ A' is on the radical axis BC of the circles (O), (P).


geometry
3D geometry
sphere
Pythagorean Theorem

What kind math computations is for 4th grade?

Addition, subtraction, division and multiplication

The answer to that question depends on whether you mean classroom math, extra math, contests, etc. If you just mean common classroom math, probably all operations with rational numbers. This refers to multiplication, addition, subtraction, division, some basic exponents, and more. You probably also have basic properties of numbers (commutative, distributive, associative), primes/composites, negative numbers, fractions, etc.

Even though there are not many competitions for 4th graders, you should probably know the basics for the following subjects:

1) Integer square roots, perfect squares (1, 4, 9, 16, 25,...).

2) The pythagorean theorem (with integer sides)

3) Common right triangle types $ (3\minus{}4\minus{}5, 30^\circ\minus{}60^\circ\minus{}90^\circ, 5\minus{}12\minus{}13)$

4) Formulae for area of common 2D shapes (polygons and circles) and volume/surface area of 3D objects (spheres, cylinders, cones, polyhedra)

5) Basic algebra (up to 2 variables) in the integers

6) Common measurements, customary and metric

And some other stuff. Note: You really don't need to know most of that stuff for classroom math, but it is important for harder contests. You can always post questions on the forum, look at the AoPS Wiki, or ask a teacher for help with these topics.

Thank you so much for the reply.I mean the conteste.

contests?
i think simple algerbra skill and what salahcool said.


Yes, it is true, for in case you were wondering, but it's pretty ugly too.

You solved it ?why you say that it is ungly?

Let $x=a+b+c$, $y=ab+bc+ca$ and $z=abc=1$.

Then the given inequality is equivalent to $6\sqrt{x}<x^2+y$, or $6<x^\frac{3}{2}+\frac{y}{\sqrt{x}}$.  By the AM-GM, $x^\frac{3}{2}+\frac{y}{\sqrt{x}}\ge2\sqrt{xy}$, $x\ge3\sqrt[3]{z}=3$ and $y\ge3\sqrt[3]{z^2}=3$.  It is easy to confirm that equality cannot be achieved on all of these inequalities simultaneously, so the inequality is strict and the result follows.

Because the problem itself is ugly (quite a weak inequality too) and because my solution was pretty ugly too. (The solution Diarmuid gave is shorter.)

Arne your solution will be ungly as you said  but could you post your solution?

Homogenise to \[6\sqrt{abc(a + b + c)} \leq (a + b + c)^2 + (ab + bc + ca).\]
Square. Get a very ugly expression with only a few negative terms and use Muirhead. (I had to think only for 1 minute but I admit - it's extremely ugly :))
(Of course you can take shortcuts while expanding... That's what I did.)


MATHCOUNTS
AMC
USA(J)MO
USAMO
summer program
MathPath
videos

About 5 hours a day for MATHCOUNTS and I aim to get top 150 at Nats. My other hobbies are soccer, chess, reading, and robotics/programming (not much). I wish I was doing something else this summer, but I don't have anything else to do.

Currently, I don't do much with mathcounts. I'm going to a chess camp this summer starting Monday, so I don't have a lot of time. I plan to do some stuff with mathcounts, if I can force myself to, after the chess camp ends. I aim to make nats this year.

About 4 hours of [b]math[/b], spent learning new topics, not doing old tests (not right now, anyway).  Goals: Make Nats (hard) if so, make countdown (medium), positive score on USAMO or top 12 on USAJMO.

@BOGTRO: I thought you were going to MathPath?

EDIT: I forgot to say my hobbies: video games, running, playing piano, sports besides beach sports

[quote="tennis123"]How long do you do MathCounts every day and what are your goals? Also, what other hobbies do you have? Are you doing anything else this summer?[/quote]

A few hours (3-4.) My goal is to make Nationals and be the top 50 - I think I have a shot at making Nationals but I dunno about top 50. My other hobbies are C++, failing chess, etc. I'm also working on my website.

What happened, BOGTRO?

[quote="AIME15"]About 4 hours of [b]math[/b], spent learning new topics, not doing old tests (not right now, anyway).  Goals: Make Nats (hard) if so, make countdown (medium), positive score on USAMO or top 12 on USAJMO.

@BOGTRO: I thought you were going to MathPath?[/quote]

Lol, i find it funny how making nats from california is harder than making nats countdown.

I study decently (maybe 2 hours a day?). I think I have a semi-sorta-decent chance at making nats this year, only because this year I can only think of 2 people who are guaranteed a spot on the team (athunder, if he makes the school team :P , & thechineesenoob) and a semi-semi-semi.............-semi-decent chance at making top 57 at nats.


LaTeX
real analysis
real analysis unsolved

Suppose the b_{n} is a sequence of nonnegative numbers that converges to 0, and x_{n} is a real sequence that satisfies abs(x_{n}-a) <= b_{n} for large n.
Prove that x_{n} converges to a.
Sorry, this is my first time using latex  :oops:

I don't think that
Chose $ b_n \equal{} 0,\forall n$
$ a_{n}\to\ a \minus{} 1$
two sequence satisfy condition
What do you mean ?

In order to make latex work in the forum, you must put dollar signs around the text you want in math mode, i.e. putting dollar signs on both sides of b_n gives $ b_n$.

Choose any $ \varepsilon > 0$.  Then, since $ b_n \to 0$, we can find $ N$ such that $ b_n < \varepsilon$ for all $ n \geq N$, and $ N$ is "large" for purposes of the second line.  Then, conclude.


topology
function
superior algebra
superior algebra unsolved

Let $ X$ be a topologiocal space, $ Y$ be a Hausdorff topological space, and $ f,\ g$ be two continuous maps from $ X$ to $ Y$.

(1) Show that a subset $ \{x\in{X} | f(x) \equal{} g(x)\}$ of $ X$ is closed.

(2) Suppose that there exists a subset $ A$ of $ X$ such that $ \overline{A} \equal{} X$ and $ f|_A \equal{} g|_A$, where $ \overline{A}$ is the closure of $ A$ and $ f|_A,\  g|_A$ are the restrictions of $ f,\ g$ to $ A$, respectively. Show that $ f \equal{} g$.

My attempt at (1):

Let $ Y$ be an ordered set and give $ Y\times Y$ the order topology, so it is Hausdorff.  Consider the set $ \Delta^{\minus{}}\equal{}\{y_{1}\times y_{2}|y_{1}>y_{2}\}$.  Let $ (y_{1},y_{2})\in\Delta^{\minus{}}$ so $ y_{1}>y_{2}$.  If $ y_{2}$ is the immediate predecessor of $ y_{1}$ then $ y_{1}\times y_{2}\in [y_{1},\infty)\times (\minus{}\infty,y_{2})\equal{}(y_{2},\infty)\times (\minus{}\infty,y_{1})\subset\Delta^{\minus{}}$.  If $ y_{1}>y>y_{2}$ for some $ y\in Y$ then $ y_{1}\times y_{2}\in (y,\infty)\times (\minus{}\infty,y)\subset\Delta^{\minus{}}$.  Hence $ \Delta^{\minus{}}$ is open.  Since the map $ (f,g): X\stackrel{\Delta}{\rightarrow}X\times X\stackrel{f\times g}{\rightarrow}Y\times Y$ is continuous, the preimage $ (f,g)^{\minus{}1}(\Delta^{\minus{}})\equal{}\{x\in X| f(x)\geq g(x)\}$ is closed.

Do something similar for $ \{x\in X|f(x)\leq g(x)\}$ to show it is closed.  Then an intersection of two closed sets is closed so $ \{x\in X|f(x)\equal{}g(x)\}$.

2. Suppose $ f\ne g$. Then there exists $ x\in \overline{A}$ such that $ f(x)\ne g(x)$. Because $ X$ is Hausdorff, so is $ A$. Therefore we can find an open set $ U\subset \overline{A}$ such that $ x\in U$ and $ f(U)\cap g(U) \equal{} \emptyset$. As $ A$ is a dense subset of $ X$, we can find $ x^\prime \in A\cap U$ such that $ f(x^\prime)\ne g(x^\prime )$.

JRav: Is every Hausdorff space an ordered set?

The converse to (2) is a very fun problem: If $ X$ is a Hausdorff space and $ A\subset X$ such that $ \overline{A}\neq X$, show that there exists a Hausdorff space $ Y$ and continuous functions $ f,g: X\to Y$ such that $ f\neq g$ but $ f|_A \equal{} g|_A$.

(1) Let $ A \equal{} \left\{x \in X \mid f(x) \equal{} g(x)\right\}$. We have to prove that $ \overline{A} \subseteq A$ or, in other words, if $ x \in X$ is such that for every open neighbourhood $ W_{x}$ we have $ W_{x} \cap A \neq \emptyset$, then $ f(x) \equal{} g(x)$.
Suppose $ f(x) \neq g(x)$. Then there exist two open neighbourhoods $ U_{x}, U_{y}$ of $ f(x), g(x)$ respectively, such that $ U_{x} \cap U_{y} \equal{} \emptyset$. But $ f^{\minus{}1}(U_{x})$ and $ g^{\minus{}1}(U_{y})$ are open sets of $ X$ and so $ f^{\minus{}1}(U_{x}) \cap g^{\minus{}1}(U_{y})$ is an open neighbourhood of $ x$. Now let $ m \in (f^{\minus{}1}(U_{x}) \cap g^{\minus{}1}(U_{y}))$. We have $ f(m) \in U_{x}$ and $ g(m) \in U_{y}$ but since $ U_{x} \cap U_{y} \equal{} \emptyset$, we have $ f(m) \neq g(m)$ which means $ m \notin A$. Then $ (f^{\minus{}1}(U_{x}) \cap g^{\minus{}1}(U_{y})) \cap A \equal{} \emptyset$, contradiction.


function
algebra
polynomial
algebra proposed

1)find all functions $f:\mathbb Z\rightarrow\mathbb Z$,such that:
         $f(m+f(f(n)))=-f(f(m+1))-n$  for all x,y in integer
2) find all functions $f:\mathbb R\rightarrow\mathbb R$, such that:
          $f(x^5-y^5)=x^2*f(x^3)-y^2*f(y^3)$   for all x,y in real
two problems aren't difficult and i have two long solutions,so i want to have the other solution :D  :D  :D

For 2) :
Let $x=y$, it comes $f(0)=0$.
Thus for all real $x$, let $x=a^5$ then
$f(x) =f(a^5-0)= a^2f(a^3) = -(-a^2f(a^3))=-f(0-a^5) = -f(-x)$, so that $f$ is odd.
Moreover for all $x$, we have $f(x^5) = x^2f(x^3).$

It follows that for all $x,y$, using that $x=a^5,y=(-b)^5$ :
$f(x+y)=f(a^5-b^5)= a^2f(a^3)-b^2f(b^3) = f(a^5) - f(b^5) = f(x) + f(y).$
Then $f$ satisfies the Cauchy functional equation.

It is well-known that for all integer $n$ and real $x$, we have :
$f(n)=nf(1)$ and $f(nx) = nf(x)$.

Now, let $a$ be any real number. For all $n$, we have :
$f((a+n)^5) = a^2f(a^3)+5nf(a^4)+10n^2f(a^3)+10n^3f(a^2)+5n^4f(a)+n^5f(1) = P(n)$ .
In the other hand :
$f((a+n)^5) = (a+n)^2f((a+n)^3) = (a^2+2an+n^2)(f(a^3)+3nf(a^2)+3n^2f(a)+n^3f(1)) = Q(n).$

Now, the polynomials $P(X)$ and $Q(X)$ are equal for all integer $n$, so that they are equal for all $X$. Thus, we may identify the coefficients. The coefficient of $n^4$ gives $5f(a) = 2af(1)+3f(a)$, so that $f(a)=af(1).$
Thus $f(x) = xf(1)$ for all $x$.

Conversely, any function of the form $f(x)=c \cdot x$ where $c$ is a given constant, is clearly a solution of the problem.

Pierre.

For 1) :
Assume that $f$ is a solution. Let $f^2 (x) = f(f(x)).$
Then, for all integers $m,n$, we have $f(f^2(m)+f^2(n)) = -f^2(f^2(m)+1) - n.$
The LHS is symmetric in $m,n$ so that the RHS has to be too. It follows that $f^2(f^2(m)+1)+n = f^2(f^2(n)+1)+m$,
that is $m-n = f^2(f^2(m)+1) - f^2(f^2(n)+1).$
But, the initial equation leads to $f^2(f^2(m)+1) = f(-f^2(2)-m) = f(-k-m)$, where $k=f^2(2).$ Then, for integers $m,n$ we have :
$m-n = f(-k-m)-f(-k-n).$
Choosing $n=-k$ and $p=-m-k$, we obtain $f(p)=f(0)-p$ for all $p$.
Moreover $f(f(p)) = f(f(0)-p)=f(0)-(f(0)-p) = p.$

Conversely, if $f(n) = c - n$ where $c$ is some integer constant, then it is straightforward to verify that $f$ is solution if and only if $f(n) = 1-n.$

Pierre.

with the problem one you have the wrong answer because $f(n)=-1-n$ for all n in integer. may be with constant $c$ you calculate wrong :D  :D


algebra
polynomial
number theory solved
number theory

I found these problems on the web. They are gathered by             Torsten Sillke on 2001-11-20. They are called mathematical impossibilities.

Problem A:
  Prove that the product of four consecutive
  positive integers cannot be a perfect square.

Problem B:
  Prove that the product of three or more consecutive
  positive integers cannot be a perfect square.

Problem C:
  Can both x^2 + y and x + y^2 be squares for x and y natural numbers?

Problem D:
  Why is  n(n+2) = m(m+1)  impossible for positive integers m, n?

Problem E:
  Can p+q and p-q be squares for p and q primes?

Problem F:
  Show that n! cannot be a prefect square for any integer n>1.

Problem Z:
  Let P a monic polynomial in Z[X].
  We suppose that :
  - the degree of P is even,
  - for all positive integer n, P(n) is a square in Z.
  Can we say that P is a square in Z[X] ?

You should ignore problems A and B since A is solved by me in the "Perfect squares return :)" topic and B is a particular case of the problem I propose there which in turn is a particular case of the Erdos- Selfridge Theorem: "The product of consecutive positive integers in never a power".

Problem F is well known: By Bertrand there is a prime p between n/2 and n for n>2 and we have p|n! and p 2 does not divide n!, so n! is not a square.

for problem F. it is well known that there is a prime number between [n/2] and n for any n greater than 12. Obviously this prime appears only once in the decomposition of n!. The other cases (n<12) are easy (are they? :D) to check.

Problem D:
(n+1) 2=m 2 +m+1, so
(2n+2) 2 =(2m+1) 2 +3.
The equation a 2 +3=b 2  has only the solution a=1 and b=2, so m=0 - a contradiction to m is positive.

Maybe Torsten was kidding about impossibilities.

[quote="Valentin Vornicu"]wierd I found you posting before I did :)[/quote]
It happens.  :D

Problem E:
Yes: p=q=2.
No - if you exclude this case.
We have p>q.
If p+q=a 2  and p-q=b 2, then 2q=a 2-b 2. But a 2[u]=[/u]b 2[u]=[/u]{0 or 1} (mod 4), so 4|2q, so 2|q, so q=2.
So a 2 -b 2 =4, so (a-b)(a+b)=4, so a=2, b=0, so p=q=2.

[quote="Valentin Vornicu"]by the way mihai, be aware that the postulate is valid only starting from k=6, that is from 12! :D[/quote]
1.5<2<3
2<3<4
2.5<3<5
3<5<6
3.5<5<7
4<5<8
4.5<5<9
5<7<10
5.5<6<11
6<7<12
......

You must be pointig to some other theorem.

and now let us leave aside the easy problems and give a nice solution to a good one: problem no. C (that sounds good :D)

suppose x2+y =a2 and x+y2=b2, where a and b are positive integers themselves.

From the two equalities, taking into account that  x,y \geq 1 we have that x <a and y <b thus x \leq a-1 and y \leq b-1 which means that adding them up and multiplying that

a2 + b2 = x(x+1)+y(y+1) \leq (a-1)a +(b-1)b = a2 + b2 -a-b contradiction!!!  :?

[quote="amfulger"] You must be pointig to some other theorem.[/quote]
sorry mihai you are right. I mistakenly thought about the extended version of the postulate, which states that there are actually two primes between k and 2k :)

Z.

We can find "integer part" of square root of arbitrary polynomial, i.e. if $P(x)=x^{2n}+a_{2n-1}x^{2n-1}+\cdots +a_0$ then $P(x)=(x^n+b_{n-1}x^{n-1}+\cdots +b_0)^2+c_{n-1}x^{n-1}+\cdots+c_0$. In our case $a_i \in \mathbb N$, $(i=0,1,\ldots,2n-1)$ thus  $b_j$, $c_j \in \mathbb Q$, $(j=0,1,\cdots,n-1)$. Let $M \in \mathbb N$ such that $Mc_j$, $Mb_j \in \mathbb N$ for all $j$. Consider $R(x)=M^2P(x) \Rightarrow R(m)$ is perfect square for all $m \in \mathbb N$ and $R(x)=U^2(x)+V(x)$ where $U, V \in \mathbb Z[X]$. Since $\deg V < \deg U$ it is obvious $V(x)$ must equal 0. Thus $P(x)= \frac{U^2(x)}{M^2}= \left( \frac{U(x)}M \right)^2$. If $\frac{U(x)}{M}$ has non-integer coefficients then $\left( \frac{U(x)}{M}\right)^2$ has non-integers coefficients too because $\frac{U(x)}{M}$ is a monic polynomial (well known statement?).


Then you might want to know an interesting identity:

[hide]
$\prod_{k=1}^{\lfloor(n-1)/2\rfloor}\tan\left(\frac{k\pi}{n}\right)=\begin{cases}\sqrt{n}&\text{if }n\text{ is odd}\\ 1&\text{if }\text{if }n\text{ is even}\end{cases}$
[/hide]

Edit: Oh wait, sorry.  I guess that doesn't help much.

That is a very interesting identity.  What's the proof?

I think the proof is beyond me.  :blush: I learned that one from mathworld.

[hide]
Well, if one enters the expression I gave into a calculator, etc., the result is $-1$...

Maybe it would be best to rewrite the expression in terms of $\sin$'s and $\cos$'s?

[/hide]

$\Rightarrow \tan 12\tan 24\tan 48\tan 96$

$\Rightarrow \frac{\sin 12\sin 24\sin 48\sin 96}{\cos 12\cos 24\cos 48\cos 96}$

$\Rightarrow \frac{2\sin 6\cos 6(2\sin 12\cos 12)(2\sin 24\cos 24)(2\sin 48\cos 48)}{\cos 12\cos 24\cos 48(-\sin 6)}$

$\Rightarrow-16\cos 6\sin{12}\sin{24}\sin48$

[u][b]** From Transformation and Products and Sums **[/b][/u]
$\Rightarrow-4(2\sin 24\cos 6)(2\sin 12\sin 48)$

$\Rightarrow-4(\sin{30}+\sin{18})(\cos{36}-\cos{60})$

$\Rightarrow-\frac{4(2\sin 18+1)(2\cos 36-1)}{4}$

$\Rightarrow-(2\cos 72+1)(2\cos 36-1)$

$\Rightarrow-(2(2\cos^{2}{36}-1)+1)(2\cos 36-1)$

$\Rightarrow-(4\cos^{2}36-1)(2\cos 36-1)$

Substitute $\cos{36}=\frac{\sqrt{5}+1}{4}$

$\Rightarrow-\left(4\left(\frac{\sqrt{5}+1}{4}\right)^{2}-1\right)\left(2\left(\frac{\sqrt{5}+1}{4}\right)-1\right)$

$\Rightarrow-\left(\frac{3+\sqrt{5}}{2}-1\right)\left(\frac{\sqrt{5}+1}{2}-1\right)$

$\Rightarrow-\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{4}$

$\Rightarrow-\frac{4}{4}\implies\boxed{-1}$

Ya, I had to find the transformation and products of sum as they were not in the top of my head and I knew the value for $\sin 18$ so i was easily able to figure out the value for $\cos 36$..just in case if someone was wondering.

This might be tedious ... but $\sin 12^\circ$ can be directly computed
[hide]because you can figure out the value of $\sin 72$ by using a $36-72-72$ triangle (bisect one of the base angles) and then you can get $\sin 12^\circ$ using the subtraction identity $\sin(72^\circ-60^\circ)$.

Also, all multiples of $3^\circ$ are constructable using a compass and straightedge.[/hide]

[quote="Zakary"]Evaluate $\tan({12}^{\circ})\tan({24}^{\circ})\tan({48}^{\circ})\tan({96}^{\circ})$.[/quote]
[hide]$\frac{\sin 12 \cdot \sin 24 \cdot \sin 48 \cdot \sin 96}{\cos 12 \cdot \cos 24 \cdot \cos 48 \cdot \cos 96}$
Multiply by $\frac{\sin 12}{\sin 12}$
The bottom becomes $2^{-4}\cos 192$
$\frac{\cos 12}{16}$[/hide]
Just thought that that was interesting...


algebra
polynomial
geometry
3D geometry
factorization
sum of cubes
algebra unsolved

Let $n$ be an integer greater than $3$. Prove that all the roots of the polynomial
$P(x) = x^{n}-5x^{n-1}+12x^{n-2}-15x^{n-3}+\cdots+a_{0}$
cannot be both real and positive.

Sum of the roots is 5, the sum of their cubes $5^{3}-3\cdot 5\cdot 12+3\cdot 15=-10$. Sum of cubes of positive reals cannot be negative, am I right?

Wow, what's a nice solution.. GOod job, Kondr..  :D


Two blondes are sitting in a street cafe , talking about their children. One says that she has three daughters . The product of their ages equals 36 and the sum of the ages coincides with the number of the house across the street . The second blonde replies that this information is not enough to figure out the age of each child. The first agrees and adds that the oldest daughter has he beautiful blue eyes. The the second solves the puzzle. You might solve it too!

We want at least two triples of integers such that their product is $ 36$, sum is equal, but precisely one has a greatest element.  These are $ (1,6,6)$ and $ (2,2,9)$.  The latter is the answer.

Who says the latter is the answer? The second-eldest daughter could have been born 9 months after the eldest daughter, and at that time have the same age. Or they could have been non-identical twins born just minutes after each other. There is still not enough information, and we still cannot distinguish between the two possibilities.

This is a modified version of Cenuss taker problem.

TRICK QUESTION!

They are both blondes, need I say more? :P


Let a,b,c be positive real and a+b+c=1
Prove that: $\sqrt{a^{2}+bc}+\sqrt{b^{2}+ca}+\sqrt{c^{2}+ab}\leq \frac{3}{2}$

[quote="superacidduck"]Let $a,b,c$ be positive real and $a+b+c=1$
Prove that: $\sqrt{a^{2}+bc}+\sqrt{b^{2}+ca}+\sqrt{c^{2}+ab}\leq \frac{3}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63645


Suppose that $ n$ is the product of three consecutive integers and that $ n$ is divisible by $ 7$. Which of the following is not necessarily a divisor of $ n$?

$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42$

We have $ n \equal{} a(a \plus{} 1)(a \plus{} 2)$

$ n$ will always be divisble by 2 and 3 (can you see why?), and in this case, also a multiple of 7.

Thus 6 (2x3), 14(2x7), 21(3x7) and 42(2x3x7) all must divide into n, making our answer $ \boxed{\text{D}}$

Well, I did it a little bit differently.

I divided all the answers divisible by seven by seven to get;

(A.) 6    (B.) 2   (C.) 3   (D.) 4   (E.) 6

Answers A and E have to both be correct, because they are both the same and two answers can't be wrong.

The answer to D is (2)(2), so if B is wrong, then D must also be wrong, but they can not both be wrong, so B must be true.

This leaves C and D. C must be true, because out of three consecutive integers, one of them must be divisible by three, because if the first number is 1 (mod 3), then the third number must be divisible by three. If the first number is 2 (mod 3), then the second number is divisible by three, and if the first number is equal to 3 (mod 3), then it is obviously divisible by three.

The product of three consecutive integers is only divisible by four if the smallest number of the three is even, because that would make two of the three numbers even, and 2 powers of 2, which is four.

@Wickedestjr: That would take a while to do on the actual AMC... You should look for a more time-effective way to do it.

The simplest way I can think of is to just find the smallest value of $ n$ and test it.
Let $ n\equal{}5\cdot6\cdot7\equal{}2\cdot 3\cdot 5\cdot 7$.
$ n$ is divisible by $ 6\equal{}2\cdot 3$.
$ n$ is divisible by $ 14\equal{}2\cdot 7$.
$ n$ is divisible by $ 21\equal{}3\cdot 7$.
$ n$ is divisible by $ 42\equal{}2\cdot 3\cdot 7$.
So it's obviously $ \boxed{\text{D}}$.

This may seem like a lot of work at first glance, but it's pretty quick on paper...

Actually, Wicked's method isn't that slow, although it does abuse multiple choice. He could have skipped from A and E cannot be the answer to D is the answer, though. If a number is divisible by 6, it is divisible by 2 and 3.

Basically, all you need to see is that in three consecutive integers, at least one must be even and one must be divisible by three.

Trtle's method is a bit flawed because if you chose, say, 12*13*14 you would get that n is divisible by all the choices. That, and it's brute force.

[quote="Nerd_of_the_Ages"]
Trtle's method is a bit flawed because if you chose, say, 12*13*14 you would get that n is divisible by all the choices. That, and it's brute force.
[/quote]

I said "...find the smallest value of $ n$" for this very reason.
But you're right, it's brute force :lol:


[quote="mdk"]If the range of f(x,y,z)=x/(x+y)+y/(y+z)+z/(z+x) for positive x,y
is (A,B) find A+B[/quote]
$\LaTeX$ your formulas
[quote="mdk"]If the range of $f(x,y,z)=\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}$ for positive $x,y$
is $(A,B)$ find $A+B$[/quote]
[hide="Some crazy thoughts"]Since $z$ can be negative, we can make $y+z\rightarrow0^{+}$, so $B\rightarrow+\infty$. We can also make $z+x\rightarrow0^{+}$ so that $A\rightarrow-\infty$. Therefore, $A+B$ is indeterminate. Now, if $z$ has to be positive, then it is just Nesbitt's inequality.[/hide]

[hide="notice the holes"]
at $x=-z$ and $y=-z$[/hide]


n is a positive integer. Prove that we cannot find an integer m and an integer r > 1 such that n(n + 1)(n + 2) = m^r.

we have $(n^{2}+2n)(n+1)=m^{r}$ , now assume that $(n^{2}+2n,n+1)=d$
$d|n+1 \longrightarrow d|n^{2}+2n+1$ but $d|n^{2}+2n \longrightarrow d|1 \longrightarrow d=1$
now we have $n^{2}+2n=x^{r}$ and $n+1=y^{r}\longrightarrow n^{2}+2n+1=y^{2r}$
now assume that $y^{2}=z$ now we have $z^{r}-x^{r}=1$ , now it's contradiction because $r>1$ and $x,z$ are natural numbers :)


Sorry if this sounds stupid but:

Whats the difference between
$ \angle BAC \equal{} \angle DAC\
and\
\angle BAC \cong \angle DAC$

I know one means is equal to and the other is congruent but what is the difference between the 2 terminologies?

Typically it really doesn't matter, because the two statements are virtually the same.
However, usually if you use an equal sign, it would be $ m\angle BAC \equal{} m\angle DAC$. Congruency would still be $ \angle BAC\cong\angle DAC$.

$ \angle BAC\cong\angle DAC$ means that $ m\angle BAC\equal{}\angle DAC$ by definition.

Yes but some people happen to be very strict about it...just saying

I think taht if two angles are the same measure, its better to say the first one.

The question wasn't "which angle should I use?"

It was asking the difference between congruency and equal angle measures.
Each statement validates the other, by definition of angle congruency.

However, you have to be careful when you write proofs, because you might have found that 2 angles are congruent, but you must write that down first, before you can say that their measures are equal. (I believe that's what you meant, math154?)

@fishythefish: Yes, I did mean that, but was too lazy to write it all down. :P


search
number theory unsolved
number theory

$ p$ is a prime number
$ 1 \plus{} \frac{1}{2} \plus{} \frac{1}{3} \plus{} \frac{1}{4} \plus{} ...\plus{} \frac{1}{p\minus{}1} \equal{} \frac{m}{n}$

prove that $ p|m$

Thank You!


Find the largest integer that divides the following three numbers with the same remainder in each case $54437, 20093, 52997$.

If you'd just given me more time :(

[hide="Here it is"] Obviously, we want the $gcd$ of their pairwise differences.  Since two of the differences sum to the third, we only need to take the $gcd$ of two of them.  I will take

$gcd(54437-52997, 54437-20093) = gcd(1440, 34344) = gcd(1440, 1224) = gcd(216, 1224) = gcd(216, 144) = gcd(72, 144) = \boxed{ 72 }$.

Where I've repeatedly applied the Euclidean Algorithm.

(Well, oops, I did some subtraction wrong on the real thing anyway.  :( ) [/hide]


search
trigonometry
inequalities unsolved
inequalities

Let $a,b,c,d$ be the real numbers statisfies \[(a^{2}+1)(b^{2}+1)(c^{2}+1)(d^{2}+1)=16\] prove that \[-3 \leq ab+ac+bc+ad+bd+cd-abcd \leq 5\] ??? Why the "search" always return "1000 match" result? how to use search on this forum.

thx before.

[quote="bolpoin"]
??? Why the "search" always return "1000 match" result? how to use search on this forum.
thx before.[/quote]
I join your question.
[quote="bolpoin"]Let $a,b,c,d$ be the real numbers statisfies \[(a^{2}+1)(b^{2}+1)(c^{2}+1)(d^{2}+1)=16\] prove that \[-3 \leq ab+ac+bc+ad+bd+cd-abcd \leq 5\] [/quote]
 I remember, that the solution is based on following idea:
Let $ab+ac+bc+ad+bd+cd-abcd=S.$
Hence, $(S-1)^{2}=((1-ab)(cd-1)+(a+b)(c+d))^{2}\leq$
$\leq((1-ab)^{2}+(a+b)^{2})((1-cd)^{2}+(c+d)^{2})=16.$

Thank argady. Almost same with mine, except i'm start with $(S+3)(S-5) \leq 0$ yeah that's just style matter. :D

I'm curious to see the solution with trigonomety subtitution. anybody know that kind of solution? i'm just not to fluenty with trigonometry stuff  :lol:


geometry
rotation
rectangle
similar triangles

A semicircle with diameter $d$ is contained in a square whose sides have length $8$. Given the maximum value of $d$ is $m- \sqrt{n}$, find $m+n$.

[hide="Solution"]It's best to draw a diagram as you go along.

We know that the diameter must touch two sides of the square such that the semicircle is tangent to two sides of the square.

We let the vertices of the square be $A,B,C,$ and $D,$ and the diameter touch $BC$ and $CD$ at $B_1$ and $C_1$, respectively. The points of tangency on $AD$ and $AB$ will be $D_1$ and $A_1$, respectively. The center of the semicircle will be $O$.

We now draw $A_1O$ and $D_1O$ and drop a perpendicular from point $C_1$ to $OD$, and call that point $E$. We also know that $m \angle BOD=180^\circ$, $m \angle A_1OD_1=90^\circ$, and $m\angle BOA_1= m \angle D_1OC_1$ because a circle is symmetric. Therefore, $m \angle EOC_1 = 45^\circ$ and $\triangle EOC_1$ is a right isoceles triangle.

Now label all the radii you see. $a_1O=C_1O=r$. $EC_1= \frac{r}{\sqrt{2}}$.

Finally, we notice that the length of the square equals $A_1O+EC_1=r+\frac{r}{\sqrt{2}}$

Therefore, $r\left( 1+\frac{1}{\sqrt{2}} \right)=8$
Simplifying, $r=\frac{8}{1+\frac{1}{\sqrt{2}}}=18-8\sqrt{2}$
$d=32-16\sqrt{2}=32-\sqrt{512}$

Our answer is therefore $32+512=544$

[/hide]

Let $d=2\sqrt{2}l,$ remark the diagonal of the square, we have $2l+\sqrt{2}l=8\sqrt{2}\Longleftrightarrow l=8(\sqrt{2}-1).$ so we get $d=2\sqrt{2}\cdot 8(\sqrt{2}-1)=32-16\sqrt{2}=32-\sqrt{512},$yielding  $m=32,n=512.$
The answer is $m+n=544.$

Prove the semicircle has maximum area if it's symmetric over the diagonal of the square now.  =).

That's a little harder.

I solved this problem a (slightly) different way.

First of all, in the solution 4everwise posted, I don't think you can have both the diameter touch the square and be tangent in two other places.
Also, how do you prove that you found the greatest diameter?

Start with a semicircle with radius r centered at the origin. Now rotate it, and given theta as the measure of roation, we write the side lengths of the rectangle which surrounds the semicircle. We want to find a square, which happens when r + sin(theta) = r + cos(theta) or sin(theta)=cos(theta). Thus the side of the square is r + r*sqrt(2)/2 which equals 8. We solve for r and then diameter in the same way as before.

Sorry, but I got lost (although I think I got this right on the real test, with making a lot of assumptions of course :blush: ) Could someone include a diagram because some of the stuff is too hard to follow.

*EDIT*

Nevermind, I found what I did on the real test. By the way(I remember now...), when I did this problem, I put down the radius instead of the diameter  :wallbash_red:

You have to have the diameter touch in two places and have it be tangent in two. The diameter touches at its endpoints, and then for the semi-circle to have maximum area it must be as large as possible, which requires logically that it be tangent at two other places. If you got the right answer, then your semicircle does that, too.

Yeah, it's just that I couldn't redraw my diagram (well, I got kinda confused at WarpedClown's thing where he said that the semicircle is symmetric about the diagonal of the square)..at first, I looked at the wrong diagonal....which led me to think that my diagram was wrong...but now I see what he meant and everything is fine  :P

That was actually directed at me@home. I probably should have quoted him.

I still can't really see how to get the semicircle tangent in four places. But my answer was correct...
I probably am being really stupid in some way, could you try to draw the diagram?

Not that I can post on here.

I don't see how it wouldn't be tangent at four points. For one thing, if it isn't, I don't know how you can solve it, because you can't use all of the similar triangles and pre-determined values (a lot of them come out to be things like $r$ or $r\sqrt{2}$).

It's possible to have it tangent at 4 points, but it's not given. It would be best if somebody proved it in their solution.

I don't feel like it though, as my method is rather tedious. :P

This problem was posted in the Intermediate Geometry Marathon. My solution to it has the diagram with the semicircle touching at 4 points ;)


Your question doesn't really make sense. The answer could be anything.

Maybe this belongs to the Braintesters section.

No, this is just not a well formulated question. I can give you many sequences that fulfil this, and I mean normal natural sequences.

If you're given $m$ terms of sequence ($a_1$, $a_2$, ... $a_m$), there's a way you can always compute the explicit formula for the sequence.

$a_n=a_1f_1(n)+a_2f_2(n)+...+a_mf_m(n)$, where $f_p(x)=\lfloor\frac{1}{|x-p|+1}\rfloor$

The trick is, $f_p(x)=1$ for $x=p$ and $f_p(x)=0$ for $x\ne p$.

For example, $a_3=a_1f_1(3)+a_2f_2(3)+a_3f_3(3)+...+a_mf_m(3)$, but since $f_3(3)=1$ and $f_p(3)=0$ ($p\ne 3$), $a_3=(1)(a_3)=a_3$, so the formula is true.

[quote="infinity"]Y1=3, Y2=11, Y3=24

a) Find Y6
b)Find a rule for Yn
c)Find Y40[/quote]

I think he mean's that  $Y_n$ is arithmetic sequence . $Y_{n+1}=Y_n + r$
very hard problem  :rotfl:

[quote="Arne"][quote="MathsCrack"]

I think he mean's that  $Y_n$ is arithmetic sequence .[/quote]

That would surprise me since the first terms are not in arithmetic progression.[/quote]
mmm yes the arnt arithmetic sequence  :blush:

But the termswise difference could be an arithmetic sequence for example.

(geez, incredible how much people from Belgium that all of a sudden get to this site :) welcome welcome!)

You can create an infinite number of possible sequences.  I believe that you can only have one of each degree though...
Usually, you're expected to take the simplest one, however, and that in this case would be degree 2.
But, in fact, you could make it an exponential series... :D


in ham yek masale tarkibiyate ziba:
$9$ khane az $100$ yek jadvale $10.10$ mariz hast.dar har vahed zamani har khane ke hade aghal $2$ khane mojavere zelei ash mariz bashad mariz mishavad.sabet konid bimari dar kole jadval faragir nemishavad.
solve & enjoy. :)

navardayii baz ham ghodratman zaher shod...
hamishe mohit mariza ya sabete ya kamtar mishe.

bale,daghighan hamin eede kar mikone, khoobe:)

[quote="amin_68"]masale ba'adi:
man adady az majmoe $A=\{1,2,...,144\}$entekhab mikonam.shoma bayad adade mara biyabid.shoma har dafe mitoonid yek zir majmoe az majmoe $A$ entekhab konid va az man beporsid ke aya adadam dar majmoe entekhabi shoma hast ya na?
man har dafe agar be shoma kheyr begooyam,$1000$ toman va agar bale begooyam $2000$ toman migiram.minimom poole lazem baraye yaftane adade man ra biyabid.
solve & enjoy. :)[/quote]
noktash aadade fibonachi hast:
$144=55+89$
ebteda daste $55$ tai ra midahim pas ba yek dolar be $89(F{}_n{}_-{}_1)$ va ya ba $2$ dollar be $55(F{}_n{}_-{}_2$ miresim pas haddeaksar ba $9$ dollar be $F_3$ miresim va dar nahayat ba $2$ dolar be javab miresim
$\Rightarrow$ $2+9=11$ dolar
esbate inike in algorithm minimum ast niz ba esteghrae do moghadame hast:
$\{1,...,F_n\}$ $n\geq 3$ , ba $n-1$ dolar:
baraye $n=3,4$ ke hokm vazeh ast
$n-1,n\Rightarrow n+1$   $\{1,...,F_n_+_1\}$
zir majmooe $A$ ra ba $m$ ozv dar nazar migirim:
[b]halate aval:[/b]$2$ dolar => $m$ ozv
[b]halate dovom:[/b]$1$ dolar => $F_n_+_1 - m$ ozv
dar halate aval gharar midahim $m\geq F{}_n{}_-{}_1$ => $2$ dolar baraye halate aval + $(n-2$farze esteghra) = $n$
dar halate dovom gharar midahim $m<F{}_n{}_-{}_1$ => $F_n_+_1 -m\geq F_n$ => $1$ dolar baraye halate docom + $(n-1$farze esteghra)
pas in algorithm minimum ast
bebakhshid age bad tozih dadam
chon aslan halo hosele nadashtam vaghti dashtam ino post mikardam
vali char choobe algorithm va javabe an va ideaye esbatesh hamine:
az $F_n$ ba $2$ dolar be $F{}_n{}_-{}_2$ miresim
pas az $F_12$ ba haddeaksar $9$ dolar be $F_3$ miresim va dar nahayat ba $2$ dolar be javab miresim
dar halate kolli niz baraye majmooe $\{1,...,F_n\}$ ba $(n-1)$ dolar hokme masale mohaghagh mishavad
baraye esbat ham az esteghrae do moghadame estefade mikonim
 ;)  :D

[quote="amin_68"]merci babak jan vali vaghti aghaye alipour in masale ro hal mikard man ham sare kelas boodam. :)[/quote]
pas age bad tozih dadam khodet zahmat bekesho dorostesh kon
merC


The sum of any two consecutive triangular numbers is a perfect square.

Let the two consecutive triangular numbers be $ \frac{n(n\minus{}1)}{2}$ and $ \frac{n(n\plus{}1)}{2}$, where $ n\ge 2$.

$ \frac{n(n\minus{}1)}{2}\plus{}\frac{n(n\plus{}1)}{2}\equal{}\frac{n^2\plus{}n\plus{}n^2\minus{}n}{2}\equal{}\frac{2n^2}{2}\equal{}n^2$, which is a perfect square.

$ \text{QED}$


Let $ a;b;c \in\ R$ and $ abc\equal{}1 , a^3 > 36$ , Prove that :

$ \frac{a^2}{3}\plus{}b^2\plus{}c^2 > ab\plus{}bc\plus{}ca$

P/s : This is a problem of grade 8 . And grade 8 only study AM-GM & C-S  :lol:

Suppose that $ b,c>0$ because otherwise $ ab\plus{}bc\plus{}ca<\minus{}ab\plus{}bc\minus{}ca$ and in this case $ \minus{}b>0$ and $ \minus{}c>0$.

$ a^3>36 \Rightarrow \frac {1} {b^3c^3} >36 \Rightarrow 1>36b^3c^3$

Suppose that $ c$=min$ (b,c) \Rightarrow 1>36b^3c^3 \ge 36c^6 \Rightarrow \frac {1} {c^3}>6 \Rightarrow \frac {a^3} {c^3}>6^3 \Rightarrow a>6c$

$ \bullet$ 1st case: $ b \ge 6a$

$ \Rightarrow \frac {b} {6} \ge a>6c \Rightarrow \frac {b^2} {6} \ge ab; \frac {a^2} {3} >\frac {6ac} {3}\equal{}2ac; \frac {b^2} {2} \plus{} \frac {c^2} {2} \ge bc;$

$ \Rightarrow \frac {a^2} {3} \plus{} \frac {4b^2} {6} \plus{}\frac {c^2} {2}>ab\plus{}bc\plus{}2ac>ab\plus{}bc\plus{}ca \Rightarrow \frac {a^2} {3} \plus{}b^2\plus{}c^2>ab\plus{}bc\plus{}ca$

$ \bullet$ 2nd case: $ b<6a$

$ \frac {a^2} {6}>\frac {6ac} {6}\equal{}ac$

It is enough to prove that: $ \frac {a^2} {6} \plus{}b^2\plus{}c^2>ab\plus{}bc$

But, from $ a>6c \Rightarrow ab>6bc \Rightarrow \frac {ab} {6}>bc$ so $ \frac {7ab} {6}>ab\plus{}bc$

We will show that: $ \frac {a^2} {6}\plus{}b^2>\frac {7ab} {6} \Leftrightarrow \frac {a^2} {6}\minus{}\frac {7ab} {6}\plus{}b^2>0$

$ \Delta \equal{} \frac {49b^2} {36}\minus{}\frac {4b^2} {6}\equal{}\frac {25b^2} {36}$

Since $ \frac {1} {6}>0$ and max$ (b_1,b_2)$=max$ (\frac {b} {36},\frac {b} {6})\equal{}\frac {b} {6}<a \Rightarrow \frac {a^2} {6}\minus{}\frac {7ab} {6}\plus{}b^2>0$

We have the equivalent inequality:
$ \frac{a^2}{4} \minus{}a(b\plus{}c) \plus{} (b\plus{}c)^2 \plus{}\frac{a^2}{12}\minus{}3bc \geq 0$
which leads to
$ (\frac{a}{2}\minus{}b\minus{}c)^2 \plus{} \frac{a^3\minus{}36}{12a} \geq 0$ which is true because $ a^3 >36 >0$ and $ abc\equal{}1$

[quote="moldovan"]Suppose that $ b,c > 0$ because otherwise $ ab \plus{} bc \plus{} ca < \minus{} ab \plus{} bc \minus{} ca$ and in this case $ \minus{} b > 0$ and $ \minus{} c > 0$.

$ a^3 > 36 \Rightarrow \frac {1} {b^3c^3} > 36 \Rightarrow 1 > 36b^3c^3$

Suppose that $ c$=min$ (b,c) \Rightarrow 1 > 36b^3c^3 \ge 36c^6 \Rightarrow \frac {1} {c^3} > 6 \Rightarrow \frac {a^3} {c^3} > 6^3 \Rightarrow a > 6c$

$ \bullet$ 1st case: $ b \ge 6a$

$ \Rightarrow \frac {b} {6} \ge a > 6c \Rightarrow \frac {b^2} {6} \ge ab; \frac {a^2} {3} > \frac {6ac} {3} \equal{} 2ac; \frac {b^2} {2} \plus{} \frac {c^2} {2} \ge bc;$

$ \Rightarrow \frac {a^2} {3} \plus{} \frac {4b^2} {6} \plus{} \frac {c^2} {2} > ab \plus{} bc \plus{} 2ac > ab \plus{} bc \plus{} ca \Rightarrow \frac {a^2} {3} \plus{} b^2 \plus{} c^2 > ab \plus{} bc \plus{} ca$

$ \bullet$ 2nd case: $ b < 6a$

$ \frac {a^2} {6} > \frac {6ac} {6} \equal{} ac$

It is enough to prove that: $ \frac {a^2} {6} \plus{} b^2 \plus{} c^2 > ab \plus{} bc$

But, from $ a > 6c \Rightarrow ab > 6bc \Rightarrow \frac {ab} {6} > bc$ so $ \frac {7ab} {6} > ab \plus{} bc$

We will show that: $ \frac {a^2} {6} \plus{} b^2 > \frac {7ab} {6} \Leftrightarrow \frac {a^2} {6} \minus{} \frac {7ab} {6} \plus{} b^2 > 0$

$ \Delta \equal{} \frac {49b^2} {36} \minus{} \frac {4b^2} {6} \equal{} \frac {25b^2} {36}$

Since $ \frac {1} {6} > 0$ and max$ (b_1,b_2)$=max$ (\frac {b} {36},\frac {b} {6}) \equal{} \frac {b} {6} < a \Rightarrow \frac {a^2} {6} \minus{} \frac {7ab} {6} \plus{} b^2 > 0$[/quote]
[quote="great math"]We have the equivalent inequality:
$ \frac {a^2}{4} \minus{} a(b \plus{} c) \plus{} (b \plus{} c)^2 \plus{} \frac {a^2}{12} \minus{} 3bc \geq 0$
which leads to
$ (\frac {a}{2} \minus{} b \minus{} c)^2 \plus{} \frac {a^3 \minus{} 36}{12a} \geq 0$ which is true because $ a^3 > 36 > 0$ and $ abc \equal{} 1$[/quote]
Thanks! :)


geometry
function
analytic geometry
trigonometry
vector
trig identities
Law of Cosines

[b]Dear Friends[/b],
I found this problem in a book. It made me crazy to solve. And now I need your helps. Thanks a lot.

The problem is as follows:

“Let be given a circle (O) and a point A outside the circle (O). A straight line (d) passing through the point A intersects the circle (O). Let M be an intersection point of (d) and (O). Let D and E respectively are the feet of the internal and external bisectrix of the angle A of the triangle OAM. Find the locus of D and E”.

This is not a usual geometric locus. The locus is a function $r(\theta)$

We apply a coordinate system with $O(0,0)$
Suppose that the radius of the circle is $1$ and the point $A$ is $A(a,0)$ with $a>1$

$M = (\cos \theta, \sin\theta)$

[hide="From the Law of cosines in the triangle OAM we get:"]
$AM^2 = a^2+1^2-2a cos \theta$
$AM^2 = a^2-2a cos \theta+cos^2 \theta +(1-cos^2 \theta)$
$AM^2 = (a-cos \theta)^2 +sin^2 \theta$[/hide]$AM = \sqrt{(a-cos \theta)^2 +sin^2 \theta} \ \ (1)$

[hide="Point D"]
The point $D$ is the one on the segment $OM$ which satisfies the equation $\frac{OD}{DM} =\frac{OA}{AM}$,

set $r=OD\Rightarrow DM=OM-OD = 1-r$

$\Rightarrow  \frac{r}{1-r} =\frac{a}{AM} \Rightarrow  r = \frac{a}{a+AM}$

Hence the point $D$ lies on the curve $\boxed{r(\theta) = \frac{a}{a+\sqrt{(a-cos \theta)^2 +sin^2 \theta}}}$

This curve is closed ([color=green][u]see graph, green curve[/u][/color])[/hide]
[hide="Point E"]For the point $E$ we have again $\frac{OE}{EM} =\frac{OA}{AM} \ \ (2)$

but $E$ is out of the segment $OM$

a) $AM<OA \Rightarrow  EM = OE-OM \ (M$ between $O,M)$
b) $AM>OA \Rightarrow  EM = OE+OM \ (O$ between $E,M)$
c) $AM=OA \Rightarrow$There is not point $E \ (AOM$ is isosceles $)$

In the first case the angle of $E$ is $\theta$.

In the second case the vectors $\vec{OM},\vec{OE}$ have opposite directions, so the angle of $E$ is $-\theta$

we set again $r=OE$

a) $EM = OE-OM \Rightarrow  EM=r-1$

$(2) \Rightarrow  \frac{r}{r-1} =\frac{a}{AM} \Rightarrow  r = \frac{a}{a-AM}$

b) $EM = OE+OM \Rightarrow  EM=r+1$

$(2) \Rightarrow  \frac{r}{r+1} =\frac{a}{AM} \Rightarrow  r = \frac{a}{AM-a}$

In this case we should not forget that the angle is $-\theta$

So we can say that, at any case the locus of $E$ is the curve $r(\theta) = \frac{a}{a-AM}$

and from $(1)$ we get $\boxed{r(\theta) = \frac{a}{a-\sqrt{(a-cos \theta)^2 +sin^2 \theta}}}$

The locus of $E$ consist of two branches.

The first [color=blue](blue)[/color] is when $AM<OA$
The second [color=red](red)[/color] is when $AM>OA$

$AM=OA \Leftrightarrow AOM \text{ isosceles }\Leftrightarrow \cos \theta = \frac{OM}{2OA} = \frac{1}{2a} \Leftrightarrow$
$\theta_0 = \cos^{-1}(\frac{1}{2a})$

So, if $\theta \in [-\pi,\pi)$ then the first case is when $\theta \in (-\theta_0,\theta_0)$[/hide]

In the attached graph it is $a=1.1$

Dear [b]pointos[/b],
Thank you very much for your replies. Let me think about your solution for a while. If I still have troubles, I will ask you for some more.


geometry
trapezoid
analytic geometry
vector

Polyhedron $ABCDEFG$ has six faces.  Face $ABCD$ is a square with $AB=12;$ face $ABFG$ is a trapezoid with $\overline{AB}$ parallel to $\overline{GF},$ $BF=AG=8,$ and $GF=6;$ and face $CDE$ has $CE=DE=14.$  The other three faces are $ADEG, BCEF,$ and $EFG.$  The distance from $E$ to face $ABCD$ is 12.  Given that $EG^2=p-q\sqrt{r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$

Here is my solution:

[hide]Set up $ABCDEFG$ in a 3-D coordinate system. Define $A(0,0,0)$, $B(12,0,0)$, $C(12,12,0)$, and $D(0,12,0)$. Let $H$ be the midpoint of $\overline{CD}$, and let $J$ be the point on plane $ABCD$ such that $EJ = 12$. H is therefore $H(6,12,0)$. Because triangle $CDE$ is isosceles, it can be determined that $EH = \sqrt{EC^{2}-CH^{2}}= \sqrt{196-36}= \sqrt{160}$. Triangle $EJH$ is a right triangle with hypotenuse $\overline{EH}$; therefore, $JH = \sqrt{EH^{2}-EJ^{2}}= \sqrt{160-144}= 4$. Because the figure is symmetric with respect to the plane $x = 6$, we can assume that the x-coordinate of point E is 6. Furthermore, since the distance from $J$ to $\overline{CD}$ is 4, the distance from $J$ to $\overline{AB}$ is 8, and the y-coordinate of point E is 8. Since $\overline{EJ}$ is perpendicular to plane $ABCD$, the z-coordinate of point E is 12.

We have now determined that $A = (0,0,0)$, $D = (0,12,0)$, and $E = (6,8,12)$. However, it is given that ADEG is a plane. For A, D, E, and G to lie in the same plane, the triple product of vectors AD, AE, and AG must be 0 (as in, the volume of the parallelopiped formed by those vectors is 0). Let $G = (a,b,c)$. Therefore,

\begin{eqnarray*}AG \cdot ( AD \times AE ) &=& 0 \\ <a,b,c> \cdot ( <0,12,0> \times <6,8,12> ) &=& 0 \\ <a,b,c> \cdot <(12)(12)-(0)(8),(0)(6)-(0)(12),(0)(12)-(12)(6)> \ &=& 0 \\ <a,b,c> \cdot <144,0,-72> \ &=& 0 \\ 144a-72c &=& 0 \\ 144a &=& 72c \\ 2a &=& c \end{eqnarray*}

Furthermore, since $FG = 6$ and the polyhedron is symmetrical with respect to the plane x = 6, the x-coordinate of F must be 9 and the x-coordinate of G must be 3. Therefore, $a = 3$ and $c = 2(3) = 6$. It is given that $AG = 8$, so, applying the distance formula: $3^{2}+y^{2}+6^{2}= 8^{2}\Rightarrow y = \sqrt{19}$. Hence, $G = (3,\sqrt{19},6)$ and $E = (6,8,12)$.

Applying the distance formula one last time,
\begin{eqnarray*}EG^{2}& = & (6-3)^{2}+(8-\sqrt{19})^{2}+(12-6)^{2}\\ & = & 9+8^{2}-2(8)(\sqrt{19})+(\sqrt{19})^{2}+36 \\ & = & 128-16\sqrt{19}\end{eqnarray*}.

And so $p+q+r = 128+16+19 = \boxed{163}$[/hide]

Hmm, I did this problem assuming that J was outside ABCD and I couldn't find any contradictions, can someone show why J cannot be outside ABCD? (I drew a polyhedron with J outside ABCD and it looks fine)

Are there two possible answers depending on if ABCDEFG is convex or not?