AMC
AIME
AMC 10
AMC 12
USA(J)MO
USAMO
group theory

There seems to be a rumor that someone has a 150/6. Is this true? The person who originally claimed this said something to this accord "I heard that (insert name) knew someone who had a 150/6", so it sounds a little sketchy to me. If this person does exist, and qualifies through index, would that automatically set the floor at 6? Or will they raise it higher if too many people got 6s?

You mean make the AMC 10 scores only worth $\frac{5}{6}$ of the AMC 12?  :P

If you notice, the page that recommends that is very obsolete (it says there are 250 qualifiers :rotfl: ). Last year, one main issue was that people did not understand that the AMC 10 might be better. This year, however, they clearly outlined the selection procedure in the USAMO Teacher's Manual.

[quote="Sly Si"]The statement about recommending that students take the AMC12 has been removed.  It's been replaced by a collection of statistics intended to aid students in choosing which exam to take.[/quote]

Its still there. The page introducing USAMO:

http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml

bottom line: [i]In advising students who desire to be selected for the USAMO whether to take the AMC 12 level contests or the AMC 10 level contests, in general, it will be to their advantage to take the AMC 12 level contests.
[/i]

Oh well, of course if we-who-are-hoping-so-bad the-cutoff-includes-us people don't end up making the USAMO, ultimately, we have no one to blame but our own stupidity at math.  :(

[quote="mysmartmouth"]OK, now just as a backup case, what if the floor is above 210?  What's the highest AMC12 score of somebody with a 7, that we know of?[/quote]

Hamster1800 said something about getting 144/7 for his composite in another topic.

But I really wonder if anyone got 145.5/7 or 150/7.

David Jia - 150/7
Vincent Dang - 150/6

Both AMC12.

Yeah, umm... both from Florida too. Yep we fail.

Yes, but the USAMO page is clearly out of date.  It also says there will be only 250 qualifiers.  In the teachers' manual, it's been changed.

Where can we find the teachers manual explaining exactly how USAMO qualification will work? I was still basing my decisions about which AMC to take on the old "recommendations".

I also didn't realize that AMC10+10*AIME could count as your index as well (thought indexes only worked with AMC12). If I put my AMC12 B instead of AMC10 A score down, will they still look at my AMC10 score?

[quote="Ubemaya"]My proctor person didn't read the manual thing, so I got confused and put down my AMC 12B score, which was 124.5, instead of my 10A score, which was 150. Is a 204.5 enough to make it?[/quote]
Your AMC score that you put down is used only for identification purposes. The AMC office will automatically tabulate your ranking using both of your scores and use whatever is more advantageous for you.

You could have easily bubbled in that you got a 150 on the AMC 12, but that wouldn't change anything, except maybe the AMC office suspecting that someone is impersonating you.

Oh my gosh!!! :coolspeak:

For a while, I thought I failed my life by getting a 6 in AIME!!!
Since I am only in 8th grade, I have some chance of making into USAMO now!!! THANK YOU VINCENT DANG :D

[quote="sunchips"][quote="Hamster1800"]If the index is below a 210, and such a person exists, then yes, 6 will be the floor. At that point, it is possible that CAMC will have to go to tiebreakers if way too many people get 6, but then you have problems (like last year), where people who took AMC 12 are disadvantaged against the people who took AMC 10 because the tiebreaker is your "Index" (where AMC10+10*AIME is an index). You could also have a person with a 135/6 for example and another with a 123/7, and the 135/6 would have a better tiebreaker, so if the situation arises, they will likely allow everyone with a 6 on AIME in 10th and below go.[/quote]

On AMC website, they [i]SPECIFICALLLY[/i] [b]recommend[/b] that students take AMC12 instead of AMC10, saying there are many advantages!!! What kind of advantages might there be if [b][i]SUCH[/i][/b] an event occurs?..... i mean.... we're all in the same grade here!

Wouldn't it be more fair ultimately to weight the AMC10 and AMC 12 scores differently if this situation occurs, to counter the advantage problem?[/quote]

the whole point of AMC 10 is to provide the top lower graders an easy access to USAMO, so that they'll get familiar with the questions earlier than their peers. if the weighting is altered, there's really no point of AMC 10

... Which isn't really problematic, because the implementation of the AMC 10 is kinda dumb, IMO.

So they will definitely be using the AMC10 score as part of the 10th grade and below index if necessary?

Sounds that way to me.  And this time there's no good argument against it, besides "it's not fair".

[quote="bookaholic"]Ugh, but it isn't fair. I took both AMC-12s despite still being eligible for the AMC-10 this year, and if I don't qualify for the USAMO because somebody's 136.5 on the AMC-10 was counted as being worth more than my 135 on the AMC-12, I will be pretty angry. I suppose it's my own fault for getting such a borderline score on the AIME, but I still don't see why they have to treat the AMC-10 equally when they could just as easily take proportionate numbers of qualifiers from the AMC-10 and AMC-12.[/quote]
Well you won't need to worry about that since you got a 7...but if they reject my 117/6, uh...I'd get angry.

[quote="bookaholic"]Ugh, but it isn't fair. I took both AMC-12s despite still being eligible for the AMC-10 this year, and if I don't qualify for the USAMO because somebody's 136.5 on the AMC-10 was counted as being worth more than my 135 on the AMC-12, I will be pretty angry. I suppose it's my own fault for getting such a borderline score on the AIME, but I still don't see why they have to treat the AMC-10 equally when they could just as easily take proportionate numbers of qualifiers from the AMC-10 and AMC-12.[/quote]

Maybe... you could study harder and score higher on the AIME then?  :)  But yeah, I'd agree that it's not fair - if they're going to take a certain AIME score as the floor, they should take everyone with that score. Although honestly, they shouldn't be doing this expansion in the first place and causing all these problems. Ever since the expansion last year, the amount of whining and complaining has spiked on the aops forum (not really talking about you in particular). Although I believe that this 3rd expansion will fix the problems, so it's all good. I'm pretty sure they will take everyone who gets the floor cutoff.

With that being said, I don't think the cutoff will be 6. Even though there is a 150-6 (from Florida, as a matter of fact, we also had a 150-7), it's unlikely that they will take 700+ qualifiers.

[quote="bookaholic"]Ugh, but it isn't fair. [/quote]

Agreed; it isn't.  But "it's not fair" will win you very few arguments, whereas "you're contradicting yourself" will win you many, assuming you argue against rational people.

I'm just curious... where are you getting the 700+ number xxazurewrathxx?

Guys, you have to realise that how many people are at the floor has NOTHING to do with what the floor is. I believe they are planning to take around the top 320 indices. Of those, the lowest AIME score becomes the floor, regardless of whether there are 0 people there or 2000 people there. In the case of too few people, they will likely take a few more indices until the floor is low enough (although this will pretty much never happen). In the case of too many people, they have the tiebreaker system in effect so that your USAMO index (meaning AMC+10*AIME regardless of which AMC you took) is the tiebreaker.

So, if the index is sub-210, the person with the 150/6 gets in, and you all have a chance, but if your tiebreaker is really bad, then you might not get in (e.x. a 97.5/6 will probably not get in). However, if there are a few too many over the floor (e.x. there are 520 qualifiers and they were only planning on 500), they'll probably not care and let them all in, like they did last year.

I don't think the AIME this year was that much harder that it would lower it to a 6. Last year, there were over 2000 people who got 6+ on the AIME, 722 of whom were underclassmen (10th grade and below). I don't think that they will allow 1000 USAMO qualifiers, after already expanding it to 500. A tiebreaker might be used, but I think even if 150/6 gets in, it would be easier to just set the floor at 7 since Vincent is an outlier.

Of course, I'm not an authority on the matter so don't get your hopes down yet.  :)

If the 150/6 gets in, THEN SIX IS THE FLOOR. It doesn't matter if there's 2000 people with a six that are underclassmen. SIX WOULD BE THE FLOOR. Now, if 2000 people have a six, then it is likely 320 people have >210 index, so the 150/6 won't get in, but the fact that there are 2000 sixes has nothing to do with the floor

Example (super contrived)
1x 150/15
2x 150/12
5x 144/7
312x150/6 (you can see why it's super contrived)
2000x ABC/6

six would still be the floor.

You may be right, but I think what might end up happening is that a lot of the people who got 7s and 8s on the AIME will end up with a lower index than the 150/6=210. In that case, would the AMC  actually accept 6 as the floor and take 2000 people? I suppose there might be a tiebreaker, but I'm not so sure a tiebreaker would be fair.

My point is, just that 6 is the floor doesn't mean everyone with a 6+ necessarily makes it. The tiebreaker system currently in place says that the higher indices take priority (150/6 > 138/7 even if the 138 was on AMC12). If the number is 2000, as such in my super contrived example, then this tiebreaker will be used. If CAMC can afford to make a second expansion this year (like they did last year), they will and try to avoid using this tiebreaker since the point of the AMC is to get people to learn math, and by making the easier test give you an advantage is somewhat contrary to this goal.

What I personally think makes more sense is to use the index as a tiebreaker for people at the floor (if necessary), and allow people with a score higher than the floor qualify regardless of their index. Compare the harder test first, then the easier test. There could be a situation where that kind of tiebreaker wouldn't be able to cut out enough people... but its not very likely. I still don't understand why they treat the AMC 10 and 12 equally... some kind of scale makes more sense, since one test is genuinely harder than the other.

Idea: take a proportional number of qualifiers from the 12 and the 10 based how how many made the floor. :idea:

[quote="pieterminate"]Idea: take a proportional number of qualifiers from the 12 and the 10 based how how many made the floor.[quote]

No..that would still be unfair-what if all the people that took the 12 got only high scores

for example-there were 4 peopLE that took the 10-scores 144,138,132,120
there were 2 people that took the 12- scores 150 141

So your telling me-lets say we take 1/2 of both-that the person who got the 141 on the amc 12 would get left out in favor of the person who got the 138 on the 10?

It isn't fair, but its more fair than setting the two equal for sure. The same could be said about taking a proportionate amount of people from each AIME... all the "smart" people could have taken the AIME I.

I like that idea!

Art of Owna, the problem you present is plausible, but statistically unlikely given that the 12 is harder than the 10.  Basically, this method recognizes that we have a certain group of people, namely those who made the floor.  This group is deemed "eligible for USAMO", but is too large to qualify them all.  Furthermore, they are split into two subgroups, the 10 and 12 takers, and indices of the two subgroups are incomparable.  So we take people from each subgroup based on the size of the subgroup.  This ignores the fact that more talented students are more likely to take the 12, but we can get away with that because the entire group is "eligible" or "good enough to qualify".

I still don't think it's quite as good as just making the floor flexible instead of pinning it to the lowest AIME score among the index qualifiers, but it would work.

I also think that's a good idea. I'm not in favor of a tie breaker at all, but if it has to be used, then it should be for people at the floor, not above it. I know a guy that got a 9 on the AIME last year and was initially left out of the USAMO because of the <10 tiebraker.

The AMC encourages taking the AMC12 over the AMC10, so it wouldn't be fair for them to introduce procedures that favor the AMC10 over the AMC12.

[quote="Treething"]Why does it matter so much? If you do not qualify because you took the AMC12 as a freshman/sophomore and got a lower score than you would've if you took the AMC10, it's your own stupidity that got you into this (lol) sticky mess in the first place. "It's not fair" generally does not apply in these cases.[/quote]

Uhhh no. There are good reasons to take the AMC12... Some people didn't know about the rule change (making the "index" tiebreaker official) and assumed that the AMC corrected the disadvantage incurred on AMC12-takers instead of just writing the disadvantage into their rules. Also, not everyone gets to take both the A and the B, so some might have opted to take the AMC12 to safely qualify for the AIME, since just a few stupid mistakes on the AMC10 could screw you over.

Anyway, more important than being "unfair," the index tiebreaker policy that was almost used last year is a bad way to select students based on their merit. I hope the AMC does not do it this year.

BTW, it "matters so much" because you can make red MOP in 9th grade, and lots of 9th graders are on the borderline for USAMO qualification.

yea..the tiebrake system is kind of unfair..but lets say there are like 1000 people that get 6s?  Surely, you wound take all of them..the only way left is to exclude everyone that got 6 and make the cutoff 7 (even if someone that made it through the index got a floor value of 6 or that with the cutoff 7, there wont be the full 500 or so people)..im not exaclty recommending this though..

Ugh, it's not the end of the world.

I've studied enough for the alternative AIME, if I do well, great, I have a decent chance of making MOP, if I don't do well, whatever, I have three more chances to make the IMO team ;p .

And what is the ultimate goal?

[quote="probability1.01"]
BTW, it "matters so much" because you can make red MOP in 9th grade, and lots of 9th graders are on the borderline for USAMO qualification.[/quote]

If one's intention is to make MOsP, maybe he/she should study enough to get off the borderline and into the safety zone  :wink: . I'm sure you would know all about that safety zone.

In recent years I've noticed that the USAMO qualification has been relaxed so much it really should have ended all complaints. Instead, people have noticed that the system is flexible and decided to complain even more about it.

[quote="xxazurewrathxx"][quote="probability1.01"]
BTW, it "matters so much" because you can make red MOP in 9th grade, and lots of 9th graders are on the borderline for USAMO qualification.[/quote]

If one's intention is to make MOsP, maybe he/she should study enough to get off the borderline and into the safety zone  :wink: . I'm sure you would know all about that safety zone.[/quote]

Well I'm not saying whether they have a right to blame things on the new policy, but there is actually quite a large difference between say an AMC 12 score of 120 and an AMC 10 score of 150. This isn't quite like the stupid mistakes issue in that people were probably not aware that a tiebreaker favoring the AMC 10 could come up.

I'd probably agree with that. That's why I think (don't hit me) that the floor should just be raised up one instead of using the tiebreaker. I know there will be less qualifiers, but it would simplify things as well and it wouldn't be "unfair" though I'm sure many would beg to differ.


[hide="solution"]
Because A is the largest angle, then $ x\plus{}9$ must be the greatest side, so set up inequalities so that it's bigger than the other two sides.
$ x\plus{}9>3x$
$ x<\frac{9}{2}$

$ x\plus{}9>x\plus{}4$
$ 9>4$ (which doesn't help us)

And by triangle inequality
$ 3x\plus{}x\plus{}4>x\plus{}9$
$ x>\frac{5}{3}$

So combining those,
$ \frac{5}{3}<x<\frac{9}{2}$

$ \frac{9}{2}\minus{}\frac{5}{3}\equal{}\boxed{\frac{17}{6}}$[/hide]

That works, but how do you guarantee that the difference is the greatest possible?

He made $ n$ as large as possible and $ m$ as small as possible.


parameterization
number theory
Diophantine equation
number theory proposed

We consider the parameters  $ (i,j,k,n)\in\mathbb{N}$.Find all positive integers $ (m_{1},m_{2},m_{3})$  such that:

$ im_{1}+jm_{2}+km_{3}=n$ and $ i+j+k=n.$

A trivial solution is $ (1,1,1).$

its looks like a kind of [url=http://en.wikipedia.org/wiki/Diophantine_equation#Linear_Diophantine_equations]linear Diophantine equation[/url], except with an extra variable.

I'd say first find solutions $ u,v$ such that $ iu\plus{}jv \equal{} gcd(i,j)$ in the usual Diophantine way. ([url=http://www-groups.dcs.st-and.ac.uk/~martyn/teaching/1003/1003linearDiophantine.pdf]here[/url] is a detailed explanation)

then find solutions $ p,q$ such that $ gcd(i,j)p \plus{} kq \equal{} gcd(i,j,k)$. This will give you an infinite number of solutions $ (up,vp,q) \equal{} (m_1',m_2',m_3')$ such that $ im_1'\plus{}jm_2'\plus{}km_3' \equal{} gcd(i,j,k)$.

Multiplying throughout by $ \frac{n}{\gcd(i,j,k)}$ should give you all the solutions $ (m_1,m_2,m_3) \equal{} \left(\frac{upn}{\gcd(i,j,k)}, \frac{vpn}{\gcd(i,j,k)}, \frac{qn}{\gcd(i,j,k)}\right)$ to your original equation.

the other part is showing that these are all the solutions, the link I gave has a proof for completeness of solutions for standard linear Diophantine equations, try extrapolating that to incorporate the extra variable. (I'm fairly confident that it should work)


Well I've been thinking of making a mock usamo, right now I have 3 problems, a #1, a #2-3, and a #4-5, or I could just combine them into only one day and make them #1,2,3. I have a couple other ideas for problems, so maybe I'll have a full 6 problems or like a 5 problem USAMO.

Post here if you have any interest in taking 1-2 days of a mock olympiad OR have some good, hard problems of your own that you might want to combine with mine. Also, post if you want to lend any other sort of assistance, like grading or checking over problems (in either case, you need to be an experienced problem-solver).

This would be great. Not to sound like I'm rushing you or anything but when do you think you will have the problems out?

Hmm well it looks like I'm deciding between just 1 day of olympiad (which I could finish right now), or the whole thing or even a 5-problem USAMO like the old days, or even hoping that someone else will contribute 1-2 problems with me.

Right now I have a decent idea for one other problem (well right now it might be a little too easy, it would probably be okay but I might work on it a little more), and then other than that I just have like 3 other ideas for problems, although none of those have been formulated that well. Right now I'm practicing (USAMO 1998 day 2) and I have a WOOT class at 7:30, but I might be able to produce 1-3 more problems by tonight (it's actually quite fun to think of them, but takes time) and have them ready tomorrow or something. I also might even have like day 1 ready tomorrow and wait a couple days for day 2.

Yeah that would be cool. Thanks so much for your efforts.

I'm interested in taking it (though I almost definitely will bomb it)

[quote="mathisfun12345"]I'm taking the USAMO. But I didn't prepare because I'm going to bomb it anyweays.[/quote]
That kind of attitude is precisely why you're gonna bomb it anyway.

Okay I have a pdf with 3 good problems I could post. I also have a slightly easier (but still not very obvious/straightforward) problem I could add somewhere.

Hmm okay I have a pretty good 1 day of an olympiad, should I just post that? Or I could try to think of another problem and add in a slightly easier problem I thought of to make a 5-problem one? Or I could even do something weird and have like 4 problems to do in 5 hours?

Post something. doesn't really matter what the format is. you can post the three problem olympiad, and if you think of another problem, just change the format.


inequalities
number theory proposed
number theory

Let $ d(n)$ be the number of positive divisors of the natural number $ n$.   Find all $ n$ such that  $ \frac {n} {d(n)}\equal{}p$   where $ p$ is a prime number

Daniel

Suppose $ n \equal{} p^rp_1^{r_1}p_2^{r_2}\cdots{p_k^{r_k}}$, where $ p,p_1,p_2,\cdots,p_k$ are distinct primes (WLOG, we assume $ p_1<p_2<\dots<p_k$.  Then,
\[ n \equal{} p\cdot{d(n)} \equal{} p(r\plus{}1)\left(r_1\plus{}1\right)\left(r_2\plus{}1\right)\cdots\left(r_k\plus{}1\right)\,.\]
Therefore,
\[ p^{r\minus{}1}p_1^{r_1}p_2^{r_2}\cdots{p_k^{r_k}} \equal{} (r\plus{}1)\left(r_1\plus{}1\right)\left(r_2\plus{}1\right)\cdots\left(r_k\plus{}1\right)\,.\]

If $ k\equal{}0$, then $ p^{r\minus{}1} \equal{} r\plus{}1$.  Hence, $p=2$ and $r=3$, or $ p\equal{}3$ and $ r\equal{}2$; i.e., $ n\equal{}8$ or $n=9$.  From now, we assume that $ k \geq 1$.

If $ p_1\geq3$, then $ p_i^{r_i} > r_i\plus{}1$ for all $ i$.  Thus, $ p^{r\minus{}1} < r\plus{}1$.  This occurs when $ r\equal{}1$ or $ (p,r) \equal{} (2,2)$.  However, $ r\equal{}1$ leads to $ p_1\equal{}2$, a contradiction.  Hence, $ p\equal{}r\equal{}2$.  That is, $ p_1\equal{}3$.  Now,
\[ 2\cdot3^{r_1} \leq 3\left(r_1\plus{}1\right)\,,\]
which occurs if and only if $ k\equal{}1$ and $ r_1\equal{}1$.  Therefore, $ n\equal{}12$.

If $ p_1\equal{}2$ and $ r\equal{}1$, then $ 2^{r_1\minus{}1} \leq r_1\plus{}1$.  Therefore, $ r_1 \equal{} 1$, $ r_1 \equal{} 2$, or $ r_1\equal{}3$.  However, $ r_1\equal{}1$ leads to a contradiction.  If $ r_1\equal{}2$, then $ p_2\equal{}3$. That is,
\[ 2\cdot{3^{r_2\minus{}1}} \leq r_2\plus{}1\,.\]
The above inequality forces $ r_2\equal{}1$ and $ k\equal{}2$.  Hence, $ n\equal{}12p$.  Now, if $ r_1\equal{}3$, then $ k\equal{}1$.  Therefore, $ n\equal{}8p$.

If $ p_1\equal{}2$ and $ r>1$, then $ p^{r\minus{}1} \geq r\plus{}1$ and $ p_i^{r_i} > r_i\plus{}1$ for all $ i>1$.  Therefore,
\[ 2^{r_1} \leq r_1\plus{}1\,.\]
Thus, $ r_1 \equal{} 1$ is the only solution.  This makes $ k\equal{}1$ and $ p^{r\minus{}1} \equal{} r\plus{}1$ (which occurs only when $ p\equal{}3$ and $ r\equal{}2$).  Therefore, $ n\equal{}18$.

In summary,
1) If $ p\equal{}2$, then $n=8$ or $ n\equal{}12$;
2) If $ p\equal{}3$, then $ n\equal{}9$, $ n\equal{}18$, or $ n\equal{}24$;
3) If $ p>3$, then $ n\equal{}8p$ or $ n\equal{}12p$.

EDIT: Thanks yunxiu for notifying an omission.

[quote="lambruscokid"]Let $ d(n)$ be the number of positive divisors of the natural number $ n$.   Find all $ n$ such that  $ \frac {n} {d(n)} \equal{} p$   where $ p$ is a prime number

Daniel[/quote]
hi
$ n\equal{} p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots{p_k^{a_k}} \equal{}\prod p_i^{a_i}$

so $ n\equal{}(1\plus{}a_1)(1\plus{}a_2)........(1\plus{}a_k)\equal{}\prod(1\plus{}a_i)$ that mean:

$ p\equal{}\frac {n} {d(n)} \equal{}\frac{\prod p_i^{a_i}} {\prod(1\plus{}a_i) }$

$ p\equal{}\prod \frac{p_i^{a_i}} {1\plus{}a_i}$ ==> $ \prod p_i^{a_i} \equal{}p*\prod(1\plus{}a_i)$

so $ p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots{p_k^{a_k}} \equiv 0[(1\plus{}a_1)(1\plus{}a_2)........(1\plus{}a_k)]$

and by using Fermat if $ p_i\wedge(1\plus{}a_i) \equal{}1$ so $ p_i^{a_i} \equiv 1[1\plus{}a_i]$

so $ p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots{p_k^{a_k}} \not\equiv 0[(1\plus{}a_1)(1\plus{}a_2)........(1\plus{}a_k)]$ contradiction !!

so $ p_i\wedge(1\plus{}a_i) \not\equal{}1$ that mean $ p_i\equal{}1\plus{}a_i$

==> $ p\equal{}\prod p_i^{a_i\minus{}1}$ ==> p is prime number so $ a_i\minus{}1\equal{}1$so $ a_i\equal{}2$ that mean $ pi\equal{}3$

so $ p\equal{}3$ and $ n\equal{}p_i^{a_i}\equal{}3^2\equal{}9$

finally  $ (n.p)\equal{}(9.3)$ :|

It is not correct.
Batominovski's one is perfect   so check it.

Daniel

[quote="Batominovski"]
In summary,
1) If $ p\equal{}2$, then $ n\equal{}12$;
2) If $ p\equal{}3$, then $ n\equal{}9$, $ n\equal{}18$, or $ n\equal{}24$;
3) If $ p>3$, then $ n\equal{}8p$ or $ n\equal{}12p$.[/quote]

$n=8$ is a soluton.


I have a rather brief question.  When an gas expands reversibly, I know that we can write w = $P \Delta V$ = $P_{\text{ext}}\Delta V$.  However, what happens when the gas expands irreversibly?  Do we use $P$ (of the gas) or $P_{\text{ext}}$, if the two are unequal, when calculating work?  I tend to think $P_{\text{ext}}$, but I certainly can not prove why this should be the case.  Any thoughts?

Thanks!

The easiest way to answer this questions is to go back to the derivation of the work = pressure x change in volume.  For which side are the underlying assumptions of the equation still valid?  (Not for the irreversibly expanding gas.)

The detailed answer deals with which assumption(s) are no longer met for teh irreversibly expansing gas side.[/code]


geometry
perimeter
cyclic quadrilateral
geometry proposed

[b]Problem.[/b] [i]Let $UV$ be a diameter of a semicircle.  $P,Q$ are two points on the semicircle with $UP<UQ$.  The tangents to the semicircle at $P$ and $Q$ meet at $R$.  If $S=UP\cap VQ$, then prove that $RS\perp UV$.[/i]

Because $UV$ is a diameter, we have $\angle{SPV}=90^{\circ}$ and $\angle{SQU}=90^{\circ}$. Call $N$ the intersection between $UQ$ and $PV$ and $M$ the intersection between $SN$ and $UV$. Then the quadrilateral $PSQN$ is cyclic. This means that $\angle{PQN}\equiv \angle{PSN}$. But $\angle{PQN}=\angle{VPU}$, therefore $\angle{USM}=\angle{UVP}$ $\Longleftrightarrow \angle{VPU}=\angle{UMS}$, which means $SR\perp UV$.

[quote="sinankaral53"]You are wrong.. you proved SN is perpendicular to UV ... if your statement is true you must prove that
"R" lies on SN.. which means R , S , N are collinear
How you got "R" lies on SN ?[/quote]

We consider the non-convex cyclic quadrilateral $UPVQ,$ as the special case of the non-convex cyclic hexagon $UPP'VQQ',$ of which $P\equiv P'$ and $Q\equiv Q'.$

So, by Pascal’s theorem we have that the points $N,$ as the intersection point of the opposite sidelines $UQ,$ $VP$ and $R,$ as the intersection point of the opposite sidelines $PP',$ $QQ'$ $($ the tangent lines of the semicircle, at $P,$ $Q,$ respectively $)$ and $S,$ as the intersection point of the opposite sidelines $UP,$ $VQ,$ are collinear.

Because of now, the point $N,$ is the orthocenter of the triangle $\bigtriangleup SUV,$ we conclude that $SR\perp UV$ and the proof is completed.

Kostas Vittas.

PS. As in a convex hexagon, we say as opposite, two sidelines of a non-convex one, when between the end of the one sideline and the start of the other $($ to the order of drawing the hexagon’s perimeter $),$ two it’s successive sidelines, are inserted.


The equation x^10 + (13x - 1)^10 = 0 has 5 pairs of complex roots a1, b1, a2, b2, a3, b3, a4, b4, a5, b5. Each pair ai, bi are complex conjugates. Find ∑ 1/(aibi).

Actually, we can solve the equation explicitly.

$(13x-1)^{10}=-x^{10}=e^{i(\pi+2k\pi)}x^{10}$ yields

$13x-1=xe^{i{(2k+1)\pi\over 10}}\iff{1\over x_{k}}=13-e^{i{(2k+1)\pi\over 10}}, k=\overline{0,9}$

Conjugates of $x_{k}$ for $k=0,1,2,3,4$ are $x_{9-k}$. We have

$\frac{1}{x_{k}\overline{x_{k}}}=\left(13-\cos{(2k+1)\pi\over 10}\right)^{2}+\sin^{2}{(2k+1)\pi\over 10}=170-26\cos{(2k+1)\pi\over 10}$.

Therefore the desired sum is

$S=5\cdot 170-26\sum_{k=0}^{4}\cos{(2k+1)\pi\over 10}$.

However, $\cos{\pi\over 10}+\cos{3\pi\over 10}+\cos{5\pi\over 10}+\cos{7\pi\over 10}+\cos{9\pi\over 10}=0$ (the first and the fifth terms add up to zero, the second and the fourth too, the third IS zero), which yields $S=5\cdot 170=850$


There was the following question.

There are 12 dimes, to distributed among 5 people, one of whom is Thomas. Assuming Thomas gets at least 50 cents, how many ways are there to do this?

[hide="MySolution"] I broke this problem down a little.

We know that Thomas has at least 5 of them, so it's just for one case, how many ways to distribute 7 dimes among 4 people.

That is simply C(10,7), since we are basically putting 3 dividers among 7 objects, since the last divider is already set on the far end, and the dividers have to be placed such that there's no empty set.

If Thomas has 60 cents, we have 6 dimes distributed among 4 people.
C(9,6).

So, this is until Thomas has all 12 dimes.

So, C(10,7) + C(9,6) + C(8,5) + C(7,4) + C(6,3) + C(5,2) + C(4,1) + C(3,0) =  330.[/hide]

Tell me if my thinking is wrong.

[hide]Say Thomas got 5+x dimes,  each other person get a, b, c, d dimes.
so

5+x+a+b+c+d=12, a+b+c+d+x=7

Consider a sequence of 7 D

we add 4 breakers that the number of D before the first breaker belongs to the first person,....
there are
(7+4)C4 such sequence, so our answer is 11C4=330, which agrees with White_Horse's answer.[/hide]

I don't see anything wrong with your solution, that's probably how I would have done it.

I think I have an easier way.

Now, we know that Thomas has 5 dimes, so there are 7 dimes remaining to be distributed among any of the 5 people (including Thomas). Once we have taken into account Thomas's restriction of having 5 or more dimes, we don't need to do any special casework for his dimes.
So it's simply the number of ways to distribute 7 dimes among 5 people, which by your formula is 11 nCr 7 = 330.


function
floor function
calculus
integration
inequalities

$ \lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor \plus{}\lfloor \sqrt{x}\rfloor \plus{}1\equal{}x$

[hide="err.."]
floor(x) can only decrease the value of x.
$ x\plus{}\sqrt{x}\plus{}1 \le x$
$ \sqrt{x} \le \minus{}1$
so no solutions
[/hide]

[quote="abacadaea"]
floor(x) can only decrease the value of x.
$ x \plus{} \sqrt {x} \plus{} 1 \le x$
[/quote]

It's true that floor(x) will only decrease the value of x if it's not integral, but shouldn't it be $ \ge$ instead of $ \le$?

[quote="abacadaea"][hide="err.."]
floor(x) can only decrease the value of x.
$ x \plus{} \sqrt {x} \plus{} 1 \le x$
$ \sqrt {x} \le \minus{} 1$
so no solutions
[/hide][/quote]

It's true that $ \lfloor x\rfloor\le x$,but I think your sign of inequality is reverse...

[hide="in that case"]
since floor(x) can only decrease x by a value less than 1, floor(x)>x-1, so floor(rt{x}*floor(rt{x}))>floor(x-rt{x})>x-rt{x}-1
floor(rt{x}*floor(rt{x}))+floor(rt{x})+1>x
x>x
no solution[/hide]

that should be right.

[hide="There are solutions"]Lemma: $ \sqrt {x \plus{} 1} \minus{} \sqrt {x} < \frac {1}{2k}$ if $ x \geq k^2$.
Proof: Move the $ \sqrt {x}$ over and square both sides. This reduces to $ 2 < 1/k^2 \plus{} 2\sqrt {x}/k$. Note $ \sqrt {x} \geq k$, so this is clearly true.

First of all, the LHS is an integer, so $ x$ must be one also.

We show that $ x \equal{} k^2 \plus{} 2k$ is a solution for all positive integers $ k$. Note $ \lfloor \sqrt {x} \rfloor \equal{} k$. We need to know $ \lfloor \sqrt {k^4 \plus{} 2k^3} \rfloor$, but note that $ (k^2 \plus{} k \minus{} 1)^2 < k^4 \plus{} 2k^3 < (k^2 \plus{} k)^2$. So it is $ k^2 \plus{} k \minus{} 1$. Adding that to $ k \plus{} 1$ gives $ k^2 \plus{} 2k$, so it checks out.

We also show that $ x \equal{} k^2 \plus{} 2k \minus{} 1$ makes the LHS too high. Calculating $ \lfloor \sqrt {k^4 \plus{} 2k^3 \minus{} k^2} \rfloor$, we can use the same bounds as above to get $ k^2 \plus{} k \minus{} 1$. Then the LHS is $ k^2 \plus{} 2k$ as before, which is greater than $ k^2 \plus{} 2k \minus{} 1$.

Suppose $ \lfloor \sqrt {x} \rfloor \equal{} k$ for a positive integer $ k$ (positive since $ x \geq 1$ by the given equation). Note that $ \lfloor \sqrt {x} k \rfloor \leq \lfloor \sqrt {x \minus{} 1} k \rfloor \plus{} 1$ for $ x > k^2$, by using the lemma. This means that if we decrease $ x$ by $ 1$ when it isn't a square number, the LHS decreases by at most $ 1$. We  have shown that $ x \equal{} (k \plus{} 1)^2 \minus{} 2$ makes the LHS too high, and so decreasing $ x$ by $ 1$ means the LHS decreases by at most how much the RHS decreases (1). This means that none of $ k^2$ through $ (k \plus{} 1)^2 \minus{} 2$ can be solutions.

Meaning that the only solutions are $ x \equal{} k^2 \plus{} 2k$ for positive integers $ k$.[/hide]

Let, $ n \le \sqrt{x} < n\plus{}1 \ \ (n \in \mathbb{N})$.
So, $ n^2 \le x < n^2\plus{}2n\plus{}1$.
Therefore, $ \lfloor \sqrt{n^2x} \rfloor \equal{} x\minus{}n\minus{}1 \Rightarrow x\minus{}n\minus{}1 \le \sqrt{n^2x} < x\minus{}n$
$ \Rightarrow \minus{}n\minus{}1 \le \sqrt{n^2x}\minus{}x < \minus{}n \Rightarrow n < x\minus{}\sqrt{n^2x} \le n\plus{}1$

And,
$ n^2 \cdot n^2 \le n^2x < n^2(n^2\plus{}2n\plus{}1) \Rightarrow \sqrt{n^4} \le \sqrt{n^2x} < \sqrt{n^4\plus{}2n^3\plus{}n^2}$
$ \Rightarrow n^2 \le \sqrt{n^2x} < n(n\plus{}1)$

Thus,
$ n^2\plus{}n < x < n^2\plus{}2n\plus{}1$
$ \therefore$  $ x \equal{} n^2\plus{}n\plus{}c$  or  $ x\equal{}n^2\plus{}2n\minus{}c$, $ x\equal{}n^2\plus{}2n$
($ c\in\mathbb{N}$)

If $ x \equal{} n^2\plus{}n\plus{}c$,
$ \lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor \plus{}\lfloor \sqrt{x}\rfloor \plus{}1 \equal{} \lfloor n\sqrt{n^2\plus{}n\plus{}c} \rfloor \plus{} n \plus{} 1$
Therefore,
$ \lfloor n\sqrt{n^2\plus{}n\plus{}c} \rfloor \equal{} n^2\plus{}c\minus{}1 \Rightarrow n^2\plus{}c\minus{}1 \le n\sqrt{n^2\plus{}n\plus{}c} < n^2\plus{}c$
$ \Rightarrow n^4\plus{}2n^2(c\minus{}1)\plus{}(c\minus{}1)^2 \le n^4\plus{}n^3\plus{}cn^2 < n^4 \plus{} 2cn^2\plus{}c^2$
$ \Rightarrow cn^2\minus{}2n^2\plus{}(c\minus{}1)^2 \le n^3 < cn^2\plus{}c^2$
$ \therefore$ Incorrect

If $ x \equal{} n^2\plus{}2n\minus{}c$,
$ \lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor \plus{}\lfloor \sqrt{x}\rfloor \plus{}1 \equal{} \lfloor n\sqrt{n^2\plus{}2n\minus{}c} \rfloor \plus{} n \plus{} 1$
Therefore,
$ \lfloor n\sqrt{n^2\plus{}2n\minus{}c} \rfloor \equal{} n^2\plus{}n\minus{}c\minus{}1 \Rightarrow n^2\plus{}n\minus{}c\minus{}1 \le n\sqrt{n^2\plus{}2n\minus{}c} < n^2\plus{}n\minus{}c$
$ \Rightarrow n\plus{}1\minus{}\frac{c\plus{}1}{n} \le \sqrt{n^2\plus{}2n\minus{}c} < n\plus{}1\minus{}\frac{c}{n}$
$ \Rightarrow n^2\plus{}2n\minus{}\frac{2}{n}\plus{}\frac{1}{n^2}\minus{}1\plus{}\frac{c^2}{n^2}\minus{}\frac{2c}{n}\plus{}\frac{2c}{n^2}\minus{}2c \le n^2\plus{}2n\minus{}c < n^2\plus{}2n\plus{}1\plus{}\frac{c^2}{n^2}\minus{}\frac{2c}{n}\minus{}2c$
$ \Rightarrow \minus{}\frac{2}{n}\plus{}\frac{1}{n^2}\minus{}1\plus{}\frac{c^2}{n^2}\minus{}\frac{2c}{n}\plus{}\frac{2c}{n^2}\minus{}c \le 0 < 1\plus{}\frac{c^2}{n^2}\minus{}\frac{2c}{n}\minus{}c$
$ \therefore$ Incorrect

If $ x \equal{} n^2\plus{}2n$,
$ \lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor \plus{}\lfloor \sqrt{x}\rfloor \plus{}1 \equal{} \lfloor n\sqrt{n^2\plus{}2n} \rfloor \plus{} n \plus{} 1$
Therefore,
$ \lfloor n\sqrt{n^2\plus{}2n} \rfloor \equal{} n^2\plus{}n\minus{}1 \Rightarrow n^2\plus{}n\minus{}1 \le n\sqrt{n^2\plus{}2n} < n^2\plus{}n$
$ \Rightarrow n\plus{}1\minus{}\frac{1}{n} \le \sqrt{n^2\plus{}2n} < n\plus{}1$
$ \Rightarrow n^2\plus{}2n\minus{}1\minus{}\frac{2}{n}\plus{}\frac{1}{n^2} \le n^2\plus{}2n < n^2\plus{}2n\plus{}1$
$ \Rightarrow \minus{}1\minus{}\frac{2}{n}\plus{}\frac{1}{n^2} \le 0 < 1$
$ \therefore$ Correct

Thus, $ x\equal{}n^2\plus{}2n \ \ (n\in\mathbb{N})$

Thank you,Mellow and Kouichi :)

[hide="my solution"]
Because the LHS is integer,we can see that $ x$ is also an integer.
Let $ x \equal{} n^2 \plus{} c$ $ (n\in N,c \equal{} 0,1,2,...,2n)$.
Thus $ \lfloor \sqrt {x}\rfloor \equal{} n$,and $ \sqrt {x}\lfloor \sqrt {x}\rfloor \equal{} \sqrt {n^4 \plus{} cn^2}$.We consider it by two cases:

(1)When $ c$ is even
if $ c\equal{}0$,we see that it doesn't work,so let$ c \ge2$.
Now we easily check the inequality $ n^2 \plus{} \frac {c}{2} \minus{} 1 < \sqrt {n^4 \plus{} cn^2} < n^2 \plus{} \frac {c}{2}$ hols,so we have $ \lfloor \sqrt {x}\lfloor \sqrt {x}\rfloor \rfloor \equal{} n^2 \plus{} \frac {c}{2} \minus{} 1$,and the equality is equiv to
$ (n^2 \plus{} \frac {c}{2} \minus{} 1) \plus{} n \plus{} 1 \equal{} n^2 \plus{} c\Longleftrightarrow c \equal{} 2n$.

(2)When $ c$ is odd
in the same way as (1),by the inequality $ n^2 \plus{} \frac {c \minus{} 1}{2} < \sqrt {n^4 \plus{} cn^2} < n^2 \plus{} \frac {c \plus{} 1}{2}$,we have $ \lfloor \sqrt {x}\lfloor \sqrt {x}\rfloor \rfloor \equal{} n^2 \plus{} \frac {c \minus{} 1}{2}$,and the equality is equiv to
$ (n^2 \plus{} \frac {c \minus{} 1}{2}) \plus{} n \plus{} 1 \equal{} n^2 \plus{} c\Longleftrightarrow c \equal{} 2n \plus{} 1$,ruled out by $ c\le 2n$.

finally,we have the solutions $ \boxed{x \equal{} n^2 \plus{} 2n}(n\in N)$
[/hide]


abstract algebra
group theory
algebra solved
algebra

as I said, this could hardly be called a problem. computing the sequence reveals that a_5 > 99/100 and we are done.

AS well as most of the other problems given in that contest (except, of course, the problem with the morphism). I also had a problem proposed for that contest, but they considered it too hard and gave only a stupid question related to it. The problem was:
  Let p a prime dividing n>2. Prove that the sentences below are equivalent:
 a) In any abelian group G of order n the equation x^n/p=e has exactly n/p solutions;
 b) p^2 does not divide n.

what a nerve!!!!
they signed that problem !!!!!

also the same person  signed a number of other problems, which were well-known, or ...

PS someone told me a super joke: (only for romanians :D) everyone is going to remember this olympiad as "chiteshiada 2004" :D

No, I told them to sign it, since they mutilated my problem! It was so trivial!!! And this problem was so nice... :( . Too bad. Nice joke!

Just  a_5 very funny
Why so easy for local MO Bucharest 2004 ?

because someone made a joke about all this olympiad.

in the same grade 10th grade, there are the following problems

find monotone f such that f(f(x))={x} for all x in [0,1] !! (I could hardly call this an olympiad problem).

then is the problem 4, which I corrected in the presented file, but which sounded like this in the actual contest:

Given n points on the unit circle, and n disks of radius 1 centered in those points. Find the intersection of the disks.

This is the problems students had to solve. Obviously the following solution is correct:

" the intersection of the disks is the set of points which belong to all disks".

No further comment.


[18:37] * Welcome to AoPS For The Win! Alpha Edition. Please review terms of service regularly.
[18:38] * USA pride detests ellipses
[18:38] * USA pride loves ellipses
[18:38] * mathelete was banned for losing THE GAME!
[18:38] * USA pride detests finding ellipses areas using integration
[18:38] jjfun1: heh
[18:38] * jjfun1 leaves again
[18:38] k00lperson: need help.
[18:38] KevinChen_Wang: another
[18:38] k00lperson: in alcumus
[18:38] Minamoto: gg
[18:38] mathelete: *me was banned for flooding!
[18:38] * USA pride loves finding ellipses areas using s-m*s-miIpi
[18:38] k00lperson: How many positive multiples of 7 that are less than 1000 end with the digit 3?
[18:38] * mathelete was banned for flooding!

[quote="mightymathnut"]okay...
 am I supposed to give an example? :starwars:  :help:  :starwars:[/quote]
yes
it can be short or long. it should usually be funny or wierd.

for example:

[23:39] * Welcome to AoPS For The Win! Alpha Edition. Please review terms of service regularly.
[23:39] TheAwesomo: sup indian phale
[23:39] TheAwesomo: cd
[23:39] jack4585: CD
[23:39] jack4585: awsome
[23:40] mathelete: repeat after me: it's a scooby-dooby-dooby scooby-dooby metal-d
[23:40] mathelete: sing along with me: it's a scooby-dooby-dooby scooby-dooby metal-d
[23:43] mathelete: ba-ba-ba-ba-ba be ba-pa ba-ba pa
[23:43] mathelete: ba-pa ba-ba pa
[23:43] mathelete: be ba-pa ba-ba pa
[23:44] mathelete: ba-pa ba-ba ba-ba-ba-ba-ba
[23:44] mathelete: be ba-pa ba-ba pa
[23:44] xxrxxhxx: gj sakk
[23:44] xxrxxhxx: above 1200 again
[23:44] xxrxxhxx: lollll
[23:44] sakkorig: =P


linear algebra
matrix
linear algebra unsolved

Let $ A$ be 2x2 matrix with eigenvalues $ \lambda_1$ and $ \lambda_2$ such that $ |\lambda_1| < 1$and $ |\lambda_2| < 1$. Prove that  $ (E \minus{} A)^{ \minus{} 1} \equal{} E \plus{} A \plus{} A^2 \plus{} \ldots \plus{} A^n \plus{} \ldots$.

Assuming that your matrix $ E$ is the identity matrix $ I$, (you may need to use your fifth problem to) make sure that infinite sum RHS and also the inverse LHS make sense. (Note the restriction on the eigenvalues). Just multiply RHS by $ I\minus{}A$, and then use the uniqueness of the inverse of invertible matrices to obtain the result.


Often you need an answer to questions like "How do I ...?", "Why doesn't it work when...?", "Why do I get errors..."? It may be that the problem is something simple like leaving out a }, a typo or a clash of packages but for anyone else to see the problem you need post a sample document called a [i]minimal example[/i]. This minimal example should be a short, but [b]complete[/b], LaTeX document. There is a very good description of such a minimal example at http://www.minimalbeispiel.de/mini-en.html which I repeat (slightly edited) here:

[quote]Creating an minimal example  is a way of tracking down and fixing bugs, as well as of finding the cause of certain behaviour. Minimal examples must be as small and simple as possible, yet complete, i.e. they accord to the base structure of a LaTeX document and can be LaTeXed without further additions.

A minimal example should be complete and working,[list][*]so that you cannot accidentally omit information important to diagnosing the problem;
[*]so that the person responding can just copy-and-paste the code to try it out.[/list]
A minimal example should be small, [list][*]so that responders does not have to concern themselves with long unnecessary lines of code;
[*]because it will narrow down possible causes of misbehaviour;
[*]because short examples are well suited for posting;
[*]because many problems, such as missing brackets, faulty syntax, forgotten switches, etc. will easily be found in the process of creating it;
[*]because short documents will help you not to lose track of what's going on. [/list]
Even beginners should have no problems creating minimal examples, since it only takes a little patience and common sense. In most cases, you will find the cause of your problems yourself, thereby making posts unnecessary.[/quote]
You will also find more at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=minxampl]How to make a “minimum example”[/url] and [url=http://www.dickimaw-books.com/latex/minexample/]Creating a LaTeX Minimal Example[/url].

So A minimal example is the smallest, most concise document that still created the same error?

Thanks! Now I understand. (I don't know why I didn't)

stevem, it seems like your final link "Creating a LATEX Minimal Example" is broken.

Some short Googling brings up [url=http://www.dickimaw-books.com/latex/minexample/minexample-screen.pdf]this[/url], which is identical to that document. (In fact, going one level up in the broken link's URL to http://theoval.cmp.uea.ac.uk/~nlct/latex/minexample/ gives a redirect to a page which links to that PDF as a version for online reading.)

Thanks for the sleuthing. I'll change the link to the web page [url=http://www.dickimaw-books.com/latex/minexample/]http://www.dickimaw-books.com/latex/minexample/[/url].


Given a connected graph $ G$,It is known that degree of each vertex is more than $ 1$.Prove that:There are 2 adjacent vertices $ X$ and $ Y$ such that if we delete $ X$ and $ Y$, $ G$ will still be connected.

That has always seemed strange to me when one posts a problem in Unsolved Section and simultaneously considers the problem to be easy.
How can you know it is easy, if you don't know its solution?

But this time actually the problem is indeed pretty easy.

[b]Solution:[/b]

Consider a spanning tree $ T$ of the graph $ G$. And let $ a$ be the leaf node in the tree that is maximally distant from the root of $ T$(if such a vertex is not unique, then choose any of them) . Suppose that $ b$ is a vertex adjacent to $ a$.

If $ b$ has a degree less than $ 3$, we are allowed to delete $ a,b$ and obviously the condition of connectedness will be still satisfied.

Assume that $ b$ has a degree more than $ 2$, and let $ c\neq a$ be a vertex adjacent to $ b$. Clearly $ c$ is a leaf node, otherwise, $ a$ would not be the vertex maximally distant from the root.

If there is an edge between $ a$ and $ c$ we are allowed to delete them both and the connectedness will still hold.

Assume that $ c$ and $ a$ are not "strictly" connected, but then due to the statement - $ c$ must have another adjacent vertex in $ G$, which is not $ b$. And the same condition holds for every vertex adjacent to $ b$, except its "parent" and $ a$.

So we are allowed to delete $ a,b$ again. Indeed, assume that after this step condition of connectedness fails. Clearly the whole graph $ T$,without $ b$ and its "sons",(call it $ T'$) is still connected. But as it has been established, every leaf node of $ T$, that is simultaneously a son of $ b$, has an adjacent vertex in $ T'$. That finishes the proof, I guess.

Coz,our techer gave this problem as a warm-up exercise which we need to prove more complicated problems.And he said that this problem is easy in considerable degree.So please let discuss more serious things.Easy or hard it is relative things.It doesn't make a sense to discuss about it.

Indeed it is much more simple. Consider a maximal path in the graph. Let X and Y be the extremities of the path. You are done.

why are these extrimities connected with an edge?

Sorry, I didn't see the word "adjacent" ! It made the problem easier. What a pity.


$n\in Z^{+}$ , $1\leq n \leq 2007$ , and $t(n)$ denotes sum of digits of $n$ when $n$ was writed in base $2$ . How many $n$ hold $3\leq t(n)$?

Obviously that number is 2007 -
the number of n such t(n)=1 (a)-
the number of n such t(n)=2 (b)

Because: 2^11 =2048 > 2007 > 2^10 =1024
(a) is equal to the ways of chosen 1 number out of 11 (2^0 to 2^10)

Similarly Because: 2^11 + 2^1 =2049 > 2007 > 2^10 + 2^9 =  1536
(b) is equal to the ways of chosen 2 different numbers out of 11

Then the number is 2007 - 11 - 55 = 1941

[quote="mszew"]An interesting spinoff of that problem:

T(k) equals the amount of different naturals numbers n such as
1<=n<=2007 & the sum of the digits of n in base ten equals k

Is it true that T(29-i)=i^2 for natural i and 1<=i<=10?[/quote]

Yes, it is true. I checked it with a computer, now I'm gonna try to prove this.


1. Prove that $ |Re(z)| < |z|$.

2. If $ z^{2}\equal{}\bar{z^{2}}$, show that z is either real or pure imaginary?

[quote="bos1234"]1. Prove that $ |Re(z)| < |z|$. [/quote]

You mean $ \le$.

[hide="Method 1"] Triangle inequality.  $ z \equal{}\Re(z)\plus{}\imath\Im(z)$. [/hide]
[hide="Method 2"] $ |\Re(z) | \equal{} |z|\cos\arg(z)\le |z|$ [/hide]

[quote="bos1234"]If $ z^{2}\equal{}\bar{z^{2}}$, show that z is either real or pure imaginary?[/quote]

[hide="Method 1"] Casework.  If $ z \equal{}\bar{z}$ then $ \Im(z) \equal{} 0$; otherwise if $ z \equal{}\minus{}\bar{z}$ then $ \Re(z) \equal{} 0$.  Note that we have the identities $ z\minus{}\bar{z}\equal{} 2\imath\Im(z)$ and $ z\plus{}\bar{z}\equal{} 2\Re(z)$. [/hide]
[hide="Method 2"] $ z^{2}\equal{}\overline{z^{2}}\implies 2\arg(z) \equal{}\minus{}2\arg(z)\plus{}2\pi k\implies\arg(z) \equal{}\frac{2\pi k}{4}$ [/hide]

How did you do it using triangle inequalities? $ |z_{1}\plus{}z_{2}|\le |z_{1}|\plus{}|z_{2}|$

why is $ |\Re(z)| \equal{} |z|\cos\arg(z)\le |z|$ considered less than |z|?

thanks for the help on both qns

#1: [edit: derrr]

#2: $ \cos\theta\leq 1$ for any $ \theta$

$ |z| \equal{} |a\plus{}ib| \equal{} |\sqrt{a^{2}\plus{}b^{2}}cis (\theta)| \equal{} |\sqrt{a^{2}\plus{}b^{2}}|\cdot |\cos\theta\plus{}i\sin\theta| \equal{}\sqrt{a^{2}\plus{}b^{2}}$

(because $ |\cos\theta\plus{}i\sin\theta| \equal{} 1$) and

$ \sqrt{a^{2}\plus{}b^{2}}\geq |a| \equal{} |Re(z)|$.

[quote="bos1234"]How did you do it using triangle inequalities? $ |z_{1}\plus{}z_{2}|\le |z_{1}|\plus{}|z_{2}|$[/quote]

:oops:  Sorry, I keep calling a different concept the triangle inequality.  I meant the last line of Carcul's post.

[quote="bos1234"]why is $ |\Re(z)| \equal{} |z|\cos\arg(z)\le |z|$ considered less than |z|?[/quote]

$ |\cos\theta |\le 1$

[quote="Hamster1800"][quote="MellowMelon"]#1: Let $ z \equal{} a\plus{}bi$, meaning we want $ |a|\leq |z|$. Then we have $ |z\minus{}bi|\leq |z|\plus{}|\minus{}bi|$. But $ z\minus{}bi \equal{} a$, so $ |a|\leq |z|\plus{}|\minus{}bi|$. Drop the $ |\minus{}bi|$ term and you're done.

#2: $ \cos\theta\leq 1$ for any $ \theta$[/quote]

You can't just drop a term in an inequality like that. $ 2\leq 1\plus{}3$, but $ 2\not\leq 1$[/quote]
That's just me flipping my inequality signs. :D You can if it's the other way, but yeah, not there.


[quote="LoboSolitario"]The chords $ AB$ and $ CD$ has no intersection inside of a circle with radius $ R$ and the lenghts of these chords are $ \dfrac{R}{2}\cdot\sqrt {10 + 2\sqrt {5}}$ and $ R\cdot\sqrt {2 - \sqrt {3}}$, respectively. Find the angle between lines $ AC$ and $ CD$.

I got at least 3 different results for the same position of chords, just changing $ A$ for $ C$ and $ C$ for $ A$, $ C$ for $ D$ and $ D$ for $ C$, etc. So i'm not sure if it's correct. :maybe:[/quote]
[hide="idea"]
Find a trig function that's value is $ \sqrt{2-\sqrt{3}}$[/hide]

[quote="JRav"]Would you mind posting what you already did so maybe someone can catch your error?[/quote]

Sure.

Set $ R\equal{}1$. Let $ n$ be the number of sides of a regular polygon (star or convex). Hence,

$ \dfrac{R}{2}\cdot\sqrt {10 \plus{} 2\sqrt {5}}\equal{}\dfrac{1}{2}\cdot\sqrt {10 \plus{} 2\sqrt {5}}\Longrightarrow n\equal{}5$ (star pentagon).

$ R\cdot\sqrt {2 \minus{} \sqrt {3}}\equal{}\sqrt {2 \minus{} \sqrt {3}}\Longrightarrow n\equal{}12$ (regular convex dodecagon).

By construction we have the figure below

One more situation.

I think there are infinite solutions, because we still can change the position of chords.


analytic geometry
geometry
ratio
articles
vector

We have coplanar points $ P$, $ N$, and $ Q$ with $ Q$ lying on $ P$ and $ N$.

I don't get something... why is it intuitive that the weight on $ Q$ is equal to the sum of the weights on $ P$ and $ N$?
I always thought of the lines segments as see-saws where there is a fulcrum between two points. I do not readily associate the fulcrum with a weight.

here is an adequate explanation:

the fulcrum must hold up the mass of the object on both ends of the lever

i know thi sleads to other questions but mass point is not actually a model, it just helpps you assign weights o various parts of triangles.

I do believe the lever equation is quite readily accessible to you aufha, \[ \minus{}\ell_1m_1\equal{}\ell_2m_2\] (the negative so you can work with an origin) and i believe that for each individual point you sum the $ \ell_i m_i$. It's been a while.

Keep in mind though that if the system is in equilibrium (i.e not moving), something must be resisting good ol' gravity now eh?

Hey  :D

So why is it clear that the "origin" or "fulcrum" has a weight and that its weight is equal to the sum of the weights of the two points on both of it's sides?

Mass points (or barycentric coordinates) can be derived and proven rigorously from area ratios alone without any recourse to physical intuition.  If the physical intuition isn't helping you, stick to the area ratios.

But the idea here is that we're replacing a line with a point centered at its center of mass with the same total mass as the entire line (which here is the sum of the masses of two vertices).  The notion of replacing an object with a point at its center of mass is a basic physical model.

This [url=http://en.wikipedia.org/wiki/Center_of_mass]article[/url] maybe can help you.

I think that much confusion was probably from the analogy to a lever, which mislead people thinking about balance and such stuff.

How do we use area ratios to prove mass points facts? How do you relate "weights" on points to areas?

Three weights on the vertices of a triangle $ ABC$ determine three centers of mass $ X, Y, Z$ of the corresponding sides.  It turns out that the way these centers are calculated is equivalent to the statement of [url=http://en.wikipedia.org/wiki/Ceva%27s_theorem]Ceva's theorem[/url] (verify this!), so the cevians $ AX, BY, CZ$ intersect at a common point called the [b]center of mass[/b] of $ ABC$ (with respect to the weights we assigned); call it $ P$.  This point is just the weighted average $ t_1 A \plus{} t_2 B \plus{} t_3 C$ of the vertices, and the coordinates $ t_1, t_2, t_3$ are called the [i]barycentric coordinates[/i] of $ P$.  These coordinates actually refer to area ratios of the triangles into which $ P$ divides $ \triangle ABC$: see [url=http://mathworld.wolfram.com/BarycentricCoordinates.html]barycentric coordinates[/url] and [url=http://mathworld.wolfram.com/HomogeneousBarycentricCoordinates.html]homogeneous barycentric coordinates[/url].

I'm reading the articles but am missing something important each time. Why do we want to normalize the barycentric coordinates?

Does it have anything to do with the reasons why we sometimes normalize vectors?

Normalizing makes each barycentric coordinate a measure of an area ratio - the ratio of the area of a sub-triangle to the entire triangle.  Since any statement you can make about a triangle continues to hold if you scale it so that its area is whatever you want it to be, it is convenient in many cases to essentially treat that area as $ 1$.


function
algebra
polynomial
Rational Root Theorem

How does one factor third degree equations? :?

By using the rational root theorem, factor by grouping, and I'm sure theres more tricks..

Or just brute force with the cubic equation.

Thanks guys, using rational root thereom helps a bit.

Also just remember that a cubic equation may no go your way sometimes.  So you may not find an exact value of a root in a long time.

This will maybe help: http://mathworld.wolfram.com/CubicEquation.html
http://mathworld.wolfram.com/CubicFormula.html

Rational root theorem + synthetic is the easiest way if you ask me... aside from using a TI 86/89.  But assuming no calculator synthetic is good.

See my post before, maybe that links will help you

[quote="math92"]I have a Ti-89, but i don't know how to use the feature to solve third degree equations.  Can anyone tell me how?[/quote]

do you have the tiitanium?

in not, plug the equation into the graph funtion and find where it crosses the x-axis.

if so, press f2 and then solve.  plug in the equation and then put ,x.
so an example is solve(0=x^3+x^2+x+1,x)


geometry
inradius
trigonometry
geometry proposed

Let $AH$ be altitude of triangle $ABC$, the excircle $(I_c)$ touch $BC, AB$ at $P, Q$ respectively, so $PQ$ meet $AH$ at $J$. Prove that:
1) $AJ = r$
2) $[I_bJI_c] = [JBC]$
Where: $r$=inradius of $ABC$ and $I_b, I_c$ are excenters of $ABC$

$\blacksquare\ 1^{\circ}.\ 2s=a+b+c;\ QA=s-b;\ PB=QB=s-a;\ m(\widehat {HPQ})=\frac B2$.
Apply the Menelaus' theorem to the transversal $\overline {PQJ}$ and the triangle $ABH:$

$\frac{PB}{PH}\cdot \frac{JH}{JA}\cdot \frac{QA}{QB}=1\Longrightarrow$ $JA=QA\cdot \frac{JH}{PH}\Longrightarrow JA=(s-b)\tan \frac B2\Longrightarrow\boxed {\ JA=r\ }$

$\blacksquare\ 2^{\circ}.\ [JBC]=\frac 12\ BC\cdot JH=\frac 12 a(h_a-r)=\frac{2S-ar}{2}\Longrightarrow$
$[JBC]=\frac{r(2s-a)}{2}=\frac{r(b+c)}{2}\Longrightarrow [JBC]=\frac{r(b+c)}{2}\ \ (1)$
$I_bI_c=AI_b+AI_c=\frac{s-c}{\sin \frac A2}+\frac{s-b}{\sin \frac A2}=\frac{a}{\sin \frac A2}=4R\cos \frac A2\Longrightarrow \boxed {\ I_bI_c=4R\cos \frac A2\ }$
$m(\widehat {JAI_c})=m(\widehat {JAB})+m(\widehat {BAI_c})=(90^{\circ}-B)+\left(90^{\circ}-\frac A2\right)=C+\frac A2\Longrightarrow$
$[I_b J I_c ]=\frac 12\cdot I_b I_c \cdot AJ\cdot \sin \widehat {JAI_c }=2Rr\cos \frac A2\sin \left( C+\frac A2 \right)\Longrightarrow$
$Rr(\sin B+\sin C)=\frac{r(b+c)}{2}\Longrightarrow [I_b J I_c ]=\frac{r(b+c)}{2}\ \ (2)$
From the relations $(1)$ and $(2)$ results $\boxed {\ [I_bJI_c]=[JBC]\ }$

[quote=sprmnt21]Let M be the ,nearest to C , intersection of CQ with the incircle (I). It is easy to see that CMI ~ CQI_c. Let F be the intersection  bet ween I_cA and PQ, by angle chasing one can prove that <CFI_c = 90° then CFI_c ~ IAI_c.  From those relation we have that r_c/r = I_cC/IC = I_cF/AF. From the last, since AH//I_cP it holds that I_cF/AF = I_cP/AJ = r_c/AJ then the thesis.[/quote]

Sorry for reviving the thread, but why does CMI ~ CQI_c necessarily hold?


Let $f: N\to N$ and $\forall n\ge 1$ we have $2+f(n+1) = f(n) +(n+f(n))^2$. If $f(1)\ne 1$ prove that $f(1)+1$ is a divisor of $f(f(1))$.

I give you a hint
[hide]
$a_n=f(n)+n$
[/hide]


geometry
3D geometry
search
analytic geometry

Consider a cuboid of dimensions $ 150,  324 , 375$cm. It is divided into cubes of dimensions $ 1 , 1 , 1$cm. When a diagonal of the cuboid is drawn, find the number of these cubes it passes through.

P.s. The diagonal is not the face diagonal but the one that passes inside the cuboid

Here are some past threads you may find useful:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=331220286&t=194642
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1375466271&t=158123

I found these by searching for "cuboid diagonal."

the links you provided are hard to understand ...
can anyone provide me with a simpler solution.....?

Cmon guys......
what is the motivation to use PIE???

I am almost certain the person who posted PIE was either joking or in a crankish mood.

[hide="Thorough explanation"]150 = 3*50, 324 = 3*108, 375 = 3*125.

Consider a 50*108*125 cuboid, and the diagonal of it. Now, as we go along the diagonal, at first our coordinates are (0, 0, 0); at the end it is (50, 108, 125). So each time, say, the x-coordinate goes from less than 37 to more than 37, we enter a new cube (and 37 is just a random number, it can be anything from 1 to 49); same for the y-coordinate and the z-coordinate.

So this way we have counted 49+107+124 = 280 "entering-new-cube" changes, plus the original cube at (0, 0, 0). However a lot of times the x- and y-coordinate change from less than n to more than n at the same time, so instead of two "entering-new-cubes", we get only one. Similar for x- and z- together, or y- and z- together.

Now, (50, 108) = 2, so the x- and y- coordinates flow over an integer at the same time twice: (25, 54) and (50, 108). Similarly (50, 125) = 25; (108, 125) = 1. But of course (50, 108) doesn't count, so we deduct one from each.

Finally, check that it never happens that all three coordinates flow over at the same time.

So this is 280 + 1 - 1 - 24 - 0 = 256.

Finally, the big cuboid is made of 27 of these, and the diagonal is the same as the diagonals through three of these, so the answer is 768.
[/hide]

Disclaimer: check my arithmetic.

You said the inclusion-exclusion solution was a joke, and then you gave a solution using inclusion-exclusion?   :huh:

Oh, that's what it stands for? Sorry, my experience with English proofs is obviously not sufficient. And here I was thinking it was a crank.  :blush:


Να λυθεί στους θετικούς ακεραίους:

$ 2^x\minus{}1\equal{}xy$

Κωστα δες εδω..
http://www.mathlinks.ro/Forum/viewtopic.php?t=241181

Για όποιον θέλει κάτι πιο δύσκολο:

Να βρεθούν τα $ n$, ώστε:
$ 2^n\plus{}1\equal{}kn^2$

Ωραια ασκηση Δημητρη.
Ειναι: $ 2^{2n} \equiv 1 (modn)$,
Εστω $ p$ ο ελαχιστος πρωτος διαιρετης του $ n$ και $ d$ η ταξη της πεπερασμενης πολλαπλασιαστικης ομαδας $ (modp)$
Τοτε $ d|2n, p \minus{} 1$. Αν $ d$ περιττος τοτε $ d|n, p \minus{} 1$ αρα αφου $ d < p$ αρα $ d \equal{} 1$. Αν $ d \equal{} 2s$ τοτε $ s|n, p \minus{} 1$ απο οπου ομοια προκειπτει $ s \equal{} 1, d \equal{} 2$
Αν $ d \equal{} 1$ ειναι $ 2 \equiv 1 (modp) \Rightarrow p \equal{} 1$ ατοπο. Αρα $ d \equal{} 2, 4 \equiv 1 (modp) \Rightarrow p \equal{} 3$
Αρα $ n \equal{} 3m$ και $ 8^m \equiv \minus{} 1 (modn)$ (1).
Εστω $ p'$ ο ελαχιστος πρωτος διαιρετης του $ n$ που ειναι μεγαλυτερος του 3 (Αυτος υπαρχει αν ο $ n$ δεν ειναι δυναμη του 3)
Τοτε αν $ d'$ ειναι η ταξη $ (modp')$ τοτε ομοια με παραπανω προκειπτει $ d' \equal{} 3$ η $ d' \equal{} 6$ (Οι τιμες $ d' \equal{} 1,2$ ευκολα απορριπτονται). Ετσι λαμβανουμε $ 64 \equiv 1 (modp')$ η $ 8 \equiv 1 (modp')$, απο τις 2 αυτες ισοτιμιες προκειπτει $ p' \equal{} 7$, τοτε ομως απο την (1) προκειπτει $ 8^m \equiv \minus{} 1 (mod7)$ το οποιο ειναι ατοπο. Αρα $ n \equal{} 3^t$

Θεωρουμε $ a_k \equal{} 2^{3^k} \plus{} 1$, τοτε $ a_k \equal{} a_{k \minus{} 1}(4^{3^{k \minus{} 1}} \plus{} 2^{3^{k \minus{} 1}} \plus{} 1) \equiv 3a_{k \minus{} 1} (mod9)$. Αρα αν η μεγιστη δυναμη του 3 που διαιρει τον $ a_k$ ειναι $ 3^m$, τοτε η μεγιστη δυναμη του 3 που διαιρει τον $ a_{k \plus{} 1}$ ειναι $ 3^{m \plus{} 1}$, και αφου $ a_1 \equal{} 9$ προκειπτει οτι για $ k \geq 2$ ειναι $ a_k \neq 0 (mod(3^{2k}))$. Αρα οι μοναδικες λυσεις ειναι για $ k \in \{0,1\}$ οι $ n \equal{} 1$ και $ n \equal{} 3$.

PS: Thanx για τη διορθωση.

[quote="NickNafplio"]
Ευκολα (μπορει να γινει με απλη επαγωγη) βλεπουμε οτι καθε δυναμη του 3 επαληθευει τη σχεση.[/quote]

Δε νομίζω  :P  Μόνο $ n\equal{}1$ και $ n\equal{}3$ είναι οι λύσεις  :wink:

Thanx για τη διορθωση. Edited.

PS: Κοιτα τι μπορει να παθει κανεις οταν κανει πραξεις χωρις χαρτι και στιλο :P

Ελπίζω τουλάχιστον να είχες πάπυρο και μολύβι... :)  :P


function
logarithms
calculus
integration
calculus computations

Solve $ y'' y \equal{} c$ for some nonzero constant c.

There is a technique for this:

Let $ v\equal{}\frac{dy}{dt}$ and then consider $ v$ to be a function of $ y.$

We get that $ y''\equal{}\frac{dv}{dt}\equal{}\frac{dv}{dy}\,\frac{dy}{dt}\equal{}v\,\frac{dv}{dy}.$

Hence we have $ yv\,\frac{dv}{dy}\equal{}c.$ This is a separable equation, and has the solutions

$ v^2\equal{}2c\ln y\plus{}k_1$ for some constant $ k_1.$ Solving for $ v$ and replacing $ v$ by $ \frac{dy}{dt},$ we get the separable equation

$ \frac{dy}{dt}\equal{}\pm\sqrt{2c\ln y\plus{}k_1}.$

This is a first order separable equation. But in trying to solve it, I run up against the integral

$ \int\frac{dy}{\sqrt{2c\ln y\plus{}k_1}}$

and that doesn't look particularly likely to have a closed-form answer.

Alright, thanks Kent.

To be honest, I'm not that great with calculus; I just stumbled upon this equation when working on a physics problem I made up in class. Thanks for the help, though.


Let's take the positive real numbers $\ x_1, x_2, x_3,...,x_n, m$, such that
$\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}=2$.Find $\ m$ such that $\sqrt{x_1+m}+...+\sqrt{x_n+m}$ is a natural number.

It's been 2 1/2 weeks with no attempts seen.

I'm "going to try to get some partial credit" on this problem by working on the case where n = 1.  (For all other n, I have not tried, and I probably will not.)

The case where n =1:
-----------------------------

sqrt(x_1) = 2

x_1 = 4

sqrt[(x_1) + m]  is a natural number,  m > 0

sqrt(4 + m) is a natural number

Then, for this case, m must be a natural number greater than or equal to 5.

Let k be a natural number >= 3.

sqrt(4 + m) = sqrt(k^2)

4 + m = k^2

Subtracting,

m = k^2 - 4,  k >= 3


We have:

Let p an odd prime and h a nonnegative integer. If one of the following conditions is satisfied:

1) $h|p-1$
2) h is even and $h|p+1$
3) $\phi(h)|h$
4) if h=rs, r,s satisfy 1) or 2), or 3) and $\gcd(r,s)=1$
5) h=$d^n$, where d satisfies 1) or 2) or 3) or 4)

then $p^h\equiv1(h)$.

Conversely: if $p^h\equiv1(h)$ then h satisfies one of the five conditions?

Uber

Conversely is wrong. Consider c(n), defined in $\phi(n)|n$.

Could you at last post a link where you defined/discussed the $c(n)$ function.
(I've seen it quite recently).

http://www.mathlinks.ro/Forum/viewtopic.php?t=80221
but I agree with {x} that you should link it in future.

I'm sorry, Rust, but I don't find an example that shows my hypotesi is wrong.

Let h=42,p=11. We have ${11^{42}=1(mod 42}$, but h can not be expression by yoor method.

Your problem have not convenient solution.
If $h|(a^h-1)$ for all (a,h)=1, then $c(h)|h, c(n)|\phi(n)$.
In your case a=p fixed, and solution is $c(h,p)|h, c(n,p)|c(n)|\phi(n).$

ok... now I'm very sad... :(
Is it possible to answer at the following question: There exists an odd prime p such that conversely is true?


I am trying to relearn algebra after 10 years and its a bit foggy to me, and word problems seem to throw a monkey wrech into things for me.  So if someone can see what I am doing wrong please let me know.  Here is the problem:

If Y is your before-tax income, suppose that the government takes away $ \frac {1}{4}$Y - 10 as taxes.  What is your pre-tax income if your after-tax income is $ \$$280?

Logically I would set up the equation as:
Before-Tax Income - Taxes = After-Tax Income

So when I plug the numbers in I get:
Y - ($ \frac {1}{4}$Y - 10) = 280

And here is how I am trying to solve it:
Y - ($ \frac {1}{4}$ $ \cdot$ $ \frac {Y}{1}$ - 10) = 280

$ \Rightarrow$ Y - ($ \frac {Y}{4}$ - 10) = 280

$ \frac {Y}{1}$ - $ \frac {Y}{4}$ - 10 = 280

I convert the two fractions to have the same denominator:
($ \frac {Y}{1}$ $ \cdot$ $ \frac {4}{4}$) - $ \frac {Y}{4}$ - 10 = 280

$ \Rightarrow$ $ \frac {4Y}{4}$ - $ \frac {Y}{4}$ - 10 = 280

$ \Rightarrow$ $ \frac {3Y}{4}$ - 10 = 280

$ \frac {3Y}{4}$ = 280 + 10

$ \Rightarrow$ $ \frac {3Y}{4}$ = 290

3Y = 290 * 4

$ \Rightarrow$ 3Y = 1160

Y = $ \frac {1160}{3}$

$ \Rightarrow$ Y = 386.66666666667

The real answer should be $ \$$360 but I have no clue what I am doing wrong to get that answer.

I have however found that if I set up the equation like this I get the correct answer:

Tax = Before-Tax Income - After-Tax Income
$ \frac {1}{4}$Y - 10 = Y - 280

So if anyone can tell me what I am doing wrong with the first equation not to get the same answer please let me know.  I have spent over 2 hours trying to figure this one problem out lol.

Thanks!

When you opened the parenthesis to combine the y's you didn't distribute the negative to the negative ten,
so instead of ending up adding the 10 to the opposite side of the equation and getting 290 you would get 270=3y/4, which leads to y=360

So when I got to this part:

Y - ($ \frac {Y}{4}$ - 10) = 280

I should have done this?

Y + -1($ \frac {Y}{4}$) + -1(-10) = 280

$ \Rightarrow$ Y - $ \frac {Y}{4}$ + 10 = 280

Thanks so much for the help! Its the same problem I had back in High School.  I would always make stupid mistakes like this trying to eliminate steps and not show my work and end up having the whole problem ruined because of one error.  Well hopefully if I keep at it and learn from my mistakes I will eventually get the hang of it.  I love problem solving till I get stuck in a rut like this and can't figure out what went wrong.  Thanks again!  Problem solved.   :D

That's correct.

You might say the whole reason the parentheses are there is so that you distribute the "-".


rotation
conics
parabola
geometry open
geometry

Let $G$ be the graph of $y=ax^2$. $G$ is rotated about the origin to obtain the parabola $H$. We know that $(20,40)$ and $(24,-7)$ lie on different arms of $H$. Compute $|a|$ in simplest form.

Enjoy, feel free to post partial results (solutions up to a point where you get stuck).


Which program should i use for geometry drawings?

Zirkle und Linear ([url=http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/doc_en/]Compass and Ruler[/url]) is the best I know and it is in several languages. The problem is that there is no way to translate their pictures to PiCTeX macros, but you can convert them (with some help) in jpg or eps.

Best,

Geometer's Sketchpad. Unfortunately it ain't free. There is another topic on the forum about these programs, so I think you should search for it.

I'd reccomend [url=http://www.mit.edu/~ibaran/kseg.html]KSEG[/url] for a free program.  There are versions for both Linux and Windows.  It has a very nice, point based interface, but the documentation is limited and I still haven't figured out a lot of things.  Also can convert to other image formats.


geometry
circumcircle
geometric transformation
homothety
ratio
geometry unsolved

In a triangle ABC with circumcentre O and centroid M, lines OM and AM are
perpendicular. Let AM intersect the circumcircle of ABC again at A′. Let lines BA′ and AC intersect at D and let lines CA′ and AB intersect at E. Prove that the circumcentre of triangle ADE lies on the circumcircle of ABC.

clearly $ \overline{AM}\equal{}\overline{MA'}$. Call N the midpoint of AC, so we have $ MN \parallel A'C$ and then $ \overline{EB}\equal{}\overline{AB}$ and equally that $ \overline{CD}\equal{}\overline{AC}$. Then we can obtain AED from ABC with an homothety with center A and ratio 2, then the circumcenter of AED is the symmetric of A wrt O (that lies on the circumcircle of ABC).


In what range is $\tan x$ convex? Concave? What about $\tan^2 x$

[quote="Rushil"]see the derivative. What's so tough???[/quote]
To see the signs of the second derivative...
Maybe he is asking the elementary proofs?
This may help: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34665]http://www.mathlinks.ro/Forum/viewtopic.php?t=34665[/url]

[quote="indybar"]What's the derivative? That's what I'm not sure.[/quote]
Calculus.
Derivative of $f(x)$ at $x_0$ is: ${\lim_{dx \to 0}}{\frac{f(x_0+dx)-f(x_0)}{dx}}$

I'm sorry I supposed you'd be knowing the definiton of derivative.i didn't realise this was non-calculus!!!

[quote="Rushil"]For convexity etc. you need to see the second derivative
${ f'(x} = ( \sec{x} )^2 $

$ f''{x} = 2 ( \sec{x} )^2 \tan{x} $

For convexity, second derivative $> 0 $ thus $ \tan {x} > 0$ Solving this inequation, we get the required domain.[/quote]

Thanks.


linear algebra
matrix
geometry
geometric transformation
rotation
trigonometry
vector

[url=http://imageshack.us][img]http://img262.imageshack.us/img262/5192/matrissss11sr6.gif[/img][/url]
Shot at 2007-07-17

[url=http://imageshack.us][img]http://img470.imageshack.us/img470/540/matrissss12cm4.gif[/img][/url]
Shot at 2007-07-17

Two approaches:

i)  It is obvious by inspection.  Since both matrices are rotation matrices, you are simply saying that rotation through an angle $ \theta$ repeated $ n$ times is the same as a rotation through an angle $ n \theta$ once.

ii)  Or, just diagonalize and note that $ A^{n}=P D^{n}P^{-1}$.  If your not sure how to do this, note that the eigenvalues should be $ \cos(\theta) \pm i \sin(\theta)$ and the corresponding (unnormalized) eigenvectors are any nonzero multiple of

$ \left[ \begin{array}{c}1\\\noalign{\medskip}i\end{array}\right]$ and $ \left[ \begin{array}{c}1\\\noalign{\medskip}-i\end{array}\right]$

Hope that helps!

Let $ A=\left( \begin{matrix}0 & \theta \\-\theta & 0 \end{matrix}\right)$ , then
$ e^{A}=\sum_{k=0}^\infty \frac{A^{k}}{k!}=\left(\begin{matrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{matrix}\right)$

Now we just use properties of exponents to get that $ \left(\begin{matrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{matrix}\right)^{n}=\left(e^{A}\right)^{n}=e^{nA}= \left(\begin{matrix}\cos(n \theta) & \sin (n\theta) \\-\sin (n\theta) & \cos (n\theta) \end{matrix}\right)$

Ibrahimsenturk,

     If you are going to post a new question, you need to start a new thread.  Nonetheless I can answer your second question here as well.

Only the trivial vector $ x=0$ will satisfy that equation since $ 3$ is not an eigenvalue of $ A$.  But $ 2$ is an eigenvalue of $ A$, did you mean to write that?

You can use induction for 1). Note:

$ \left(\begin{matrix}\cos(n\theta)&\sin(n\theta)\\-\sin(n\theta)&\cos(n\theta)\end{matrix}\right)\left(\begin{matrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{matrix}\right)=\left(\begin{matrix}\cos((n+1)\theta)&\sin((n+1)\theta)\\-\sin((n+1)\theta)&\cos((n+1)\theta)\end{matrix}\right)$


I think Russia has  a very good team,but Turkey is very strong to loose!!! :ninja:

[quote="Albanian Eagle"]I voted Germany  :)[/quote]
I think its difficult for Germany to win the cup :)

[quote="Gary Lineker"]Football is a simple game; 22 men chase a ball for 90 minutes and at the end, the Germans win.[/quote]

You can never say that Germans don't have much chance... Just remember WC 02, for instance, when nobody took them seriously.

its not like it was my favorite at the beginning, but at this point...

Is there any team that can block the superb strikers (Villa & Tores) and an awesome Goalkeeper Ikar Kasillas of Spain?There's Fabrigas also.
But there might be a final between Turkey and Russia. :lol:

I'm quite sure Germany's gonna take the cup this year.

Looking the situation I voted for Spain.  I hope it will win.  :)

Why am I the only one who has faith in italy?

Yay.Spain woooooooooooooooooooooooooooooooooon. :first:  :clap:

CAMPEÓÓÓN ESPAÑA CAMPEÓN!!!:winner_first:

hehehe... just because I speak Spanish

CAMPEÓÓÓN ESPAÑA CAMPEÓN!!!:winner_first:

BWAHAHAHAHAHAAAA!!! :winner_first:


calculus
integration
trigonometry
function
logarithms
calculus computations

1. Evaluate $ \int_{2}^{7} \frac {1}{(x + 1) \sqrt {x + 2}} dx$.

2. Evaluate ${ \int \frac {1}{2 \cos^2 x - 1}}dx$.

Tried substituting $ u = x + 1$ and $ x + 2$ and even some trigonometric functions for the first, to no avail. For the second, substituting $ u = \tan x$ gives $ \int \frac {du}{1 - u^2}$, but I can't see what to do after that (barring inverse hyperbolic functions).

For the second one, wont this work out??

$ \int \frac{dx}{a^{2}\minus{}x^{2}} \equal{}\frac{1}{2a}log(\frac{a\plus{}x}{a\minus{}x})\plus{}c$

[quote="knol"]1. Evaluate $ \int_{2}^{7} \frac {1}{(x + 2) \sqrt {x + 1}} dx$.

2. Evaluate ${ \int \frac {1}{2 \cos^2 x - 1}}dx$.

Tried substituting $ u = x + 1$ and $ x + 2$ and even some trigonometric functions for the first, to no avail. For the second, substituting $ u = \tan x$ gives $ \int \frac {du}{1 - u^2}$, but I can't see what to do after that (barring inverse hyperbolic functions).[/quote]

for the first one,let$ t=\sqrt{x+1}$,we have $ x+2=t^2+1,dx=2tdt$.The given integral can be rewritten as $ \int_{\sqrt{3}}^{\sqrt{8}}\frac{2}{1+t^2}dt=2\arctan \sqrt{8}-2\arctan \sqrt{3}$

For the second one, having got $ \int \frac {1}{1 \minus{} u^{2}}\ dx$, partial fraction gives the integrand as $ \frac {1}{2}(\frac {1}{1 \minus{} u} \plus{} \frac {1}{1 \plus{} u})$, so the integral is $ \frac {1}{2} \ln (\frac {1 \plus{} u}{1 \minus{} u}) \equal{} \frac {1}{2} \ln((\frac {1 \plus{} \tan x}{1 \minus{} \tan x}) \plus{} c$.

Sorry, I mixed up the denominator for the 1st integral. It is fixed now.

hello, for the first integral i have got $ \ln\left(\frac{3}{2}\right)$.
Sonnhard.


Find the value of Q+R if

(3x+50)/(x-3)=Q+[R/(x-3)]

[hide]Multiply both sides by $ (x\minus{}3)$ to get $ 3x\plus{}50\equal{}Q(x\minus{}3)\plus{}R$.
$ 3x\plus{}50\equal{}Q(x\minus{}3)\plus{}R \Rightarrow 3x\plus{}50\equal{}Qx\minus{}3Q\plus{}R$
$ Q\equal{}3$ and $ \minus{}3Q\plus{}R\equal{}50$
$ \minus{}3(3)\plus{}R\equal{}50 \Rightarrow R\equal{}59$
$ Q\plus{}R\equal{}3\plus{}59\equal{}\boxed{62}$
[/hide]

[quote="sdbro11"][hide]Multiply both sides by $ (x \minus{} 3)$ to get $ 3x \plus{} 50 \equal{} Q(x \minus{} 3) \plus{} R$.
$ 3x \plus{} 50 \equal{} Q(x \minus{} 3) \plus{} R \Rightarrow 3x \plus{} 50 \equal{} Qx \minus{} 3Q \plus{} R$
$ Q \equal{} 3$ and $ \minus{} 3Q \plus{} R \equal{} 50$
$ \minus{} 3(3) \plus{} R \equal{} 50 \Rightarrow R \equal{} 59$
$ Q \plus{} R \equal{} 3 \plus{} 59 \equal{} \boxed{62}$
[/hide][/quote]
How did you get Q = 3?

If you wanted to be especially tricky, you could plug in $ x\equal{}4$, and get the answer immediately.

[quote="Altheman"]If you wanted to be especially tricky, you could plug in $ x \equal{} 4$, and get the answer immediately.[/quote]
How did you come up with that?

You want Q+R, the coefficient on Q is 1, so it would be nice to have the coefficient on R be 1 also...that is if 1/(x-3)=1...x=4

Yeah, how does Q come out to be 3?

If you plug 4 in for x you're not really solving it though, you're just making up your own eq.
So if x=4 then 3(4)+5 = Q + R

Hmm... two  completely different answers.

No, you get the same answer.  It's $ 3(4)\plus{}50$, not $ 3(4)\plus{}5$.

$ Q\equal{}3$ because when you get $ 3x\plus{}50\equal{}Qx\minus{}3Q\plus{}R$, the coefficients of $ x$ on both sides of the equation have to be equal, just like how $ 50\equal{}\minus{}3Q\plus{}R$.

Oops, i made a typo here, the real equation was +5 not +50, I made a typo in the OP. Ok, I see it now, thanks.

[quote="sdbro11"]No, you get the same answer.  It's $ 3(4) \plus{} 50$, not $ 3(4) \plus{} 5$.

$ Q \equal{} 3$ because when you get $ 3x \plus{} 50 \equal{} Qx \minus{} 3Q \plus{} R$, the coefficients of $ x$ on both sides of the equation have to be equal, just like how $ 50 \equal{} \minus{} 3Q \plus{} R$.[/quote]

So are you just relying on your instinct?
I don't see how you got that still...
I know that your answer is correct, but the precedure...why does the coefficients of $ \ x$ on both side have to be same?
Can someone explain?

If you move the $ 3x$ over to the right side of the equation, you get $ 50\equal{}Qx\minus{}3x\minus{}3Q\plus{}R$, or $ 50\equal{}(Q\minus{}3)x\minus{}3Q\plus{}R$.
Since there is no $ x$ term on the left side of the equation, there must also be no $ x$ term on the right side, meaning that the coefficient of $ x$ must be 0.
$ 0\equal{}Q\minus{}3 \Rightarrow Q\equal{}3$

Does this make sense?

Alternately, you could just... y'know... use polynomial division.

$ \frac{3x \plus{} 5}{x \minus{} 3} \equal{} \frac{(3x \minus{} 9) \plus{} 14}{x \minus{} 3} \equal{} 3 \plus{} \frac{14}{x \minus{} 3}$

@ t0rajir0u - Good idea, but you messed up a bit in the numerator of the original fraction.

[quote="stack3d"]Oops, i made a typo here, the real equation was +5 not +50, I made a typo in the OP. [/quote]

The idea is the same either way, though ;)

You can't really call it a "rule", but yes, it must always be true. If you have $ n\cdot x \equal{} 0\cdot x$, then $ x \equal{} 0$.

[quote="DonkeyKong"][quote="sdbro11"]Since there is no $ x$ term on the left side of the equation, there must also be no $ x$ term on the right side, meaning that the coefficient of $ x$ must be 0.
[/quote]
I kind of get it, but is that a rule?[/quote]

The idea is more general than that.  If two polynomials, say $ ax^2 \plus{} bx \plus{} c \equal{} dx^2 \plus{} ex \plus{} f$, then the coefficients must be equal; that is, $ a \equal{} d, b \equal{} e, c \equal{} f$.  This is because

$ (a \minus{} d) x^2 \plus{} (b \minus{} e) x \plus{} (c \minus{} f) \equal{} 0$

and the only polynomial that is identically zero is the polynomial with all coefficients $ 0$.


The average (mean) of 20 numbers is 30, and the average of 30 other numbers is 20. What is the average of all 50 numbers?

A. 23
B. 24
C. 25
D. 26
E. 27

[hide="Answer"]The sum of the first set of numbers is $20\cdot30=600$.  The sum of the second set of numbers is $30\cdot20=600$.  The sum of all the numbers is therefore $600+600=1200$.  Divide by 50 to get the average.  $\frac{1200}{50}=24$  Choice B is 24, so the answer is $\boxed{B}$.[/hide]


Any suggestions you might have for improval of this topic/document are welcomed. I especially am waiting for suggestions from those which have started learning $\LaTeX$ from 0 (like say yetti? :) ) and what they think that it is best to improve.

[quote="mathgeek2006"]If you could post a command index, that would be helpful.[/quote]

I hope that by [b]you[/b], you don't mean specifically [b]Valentin[/b]. Mainly because there are literally several thousands, and is way a lot of work.

Nevertheless, I have been thinking about doing one of those by my own. Unfortunatelly I have been to lazy to give the non trivial step from desire to action, because is a big task. Nevertheless, if I find a few people, say, five or six, who are willing to write nicelly the description of one or two commands a week, I volunteer to compile them and edit them to do a document of this kind, and we can post the document here, and update it once every few weeks. Well, it is just an idea.

Best regards,

[quote="mathgeek2006"]If you could post a command index, that would be helpful.[/quote]See [i]A Guide to LaTeX[/i], 3rd edition, pages 459-547.

In other words, (a) this index already exists, and (b) there are way too many to post.

You can find lists of commands on the net for example at http://tex.loria.fr/general/latex2e.html#SEC166 or at http://ubista.ubi.pt/~dfis-wg/sgml/tex/lc3/ltx-2.html

There are other such lists, but there's nothing to match a book like the one DPatrick mentions. The latest edition of Kopka & Daly's A Guide to LaTeX costs $43 but used copies of the third edition go for around $25.

May be we could do a short guide on how  (and what) to include figures in a latex document.  I say that because of the high popularity of the posts about that.  Having a general document about that could be really usefull.  I'm ready to do my share of work if you think it's a good idea.

what do you think?

ledurt

p.s: by "you", I mean all of you on this forum

If you create such a guide, I will put it so that everyone sees it :)

We have such a guide as part of our LaTeX tutorial:

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]
then click on "Pictures"

Admittedly, it is pretty threadbare, but it's a start.

[quote="Valentin Vornicu"](this is intended for new users of the board:) We have enabled LaTeX [u]with the usual tags dollar  and  double dollar[/u] on the board.

I have attached a [url=http://www.mathlinks.ro/Forum/download.php?id=146]document[/url] that should be used as a quick tutorial/introduction in $\LaTeX$. You can test your $\LaTeX$ abilities in the [url=http://www.mathlinks.ro/Forum/index.php?f=224]Test Forum[/url] :)

Now here's what you must do if you already master the basics in LaTeX :D

Each LaTeX text should look like this

[code]the LaTeX code follows: $\$ \frac 12 - \frac 1{23} + \int^1_0 f^2(x_1) dx = \cos x + \sin \frac 1x $\$ [/code] and the output will be

[quote]the LaTeX code follows: $ \frac 12 - \frac 1{23} + \int^1_0 f^2(x_1) dx = \cos x + \sin \frac 1x $ [/quote]

Be carefull not to use \begin{...} \end{...} commands too much, because they can cause errors.

Any other questions you can post in [url=http://www.mathlinks.ro/viewforum.php?f=123]this forum[/url] (not in this topic, this is only for announcements).[/quote]

I downloaded TeXaide using the link
http://www.dessci.com/en/products/texaide/
typed the above math expression and copied it on to this message box.  It pastes as
\frac{1}{2} - \frac{1}{{23}} + \int\limits_0^1 {f^2 (x_1 )dx = Cosx + Sin\frac{1}{x}}

I kept two dollar symbols  on both sides.  I got this,
$\frac{1}{2} - \frac{1}{{23}} + \int\limits_0^1 {f^2 (x_1 )dx = Cosx + Sin\frac{1}{x}}$

I think it is good to use TeXaide to save time and using TeXaide one need not remember the syntax also.

thanx

A rather comprehensive source of reference for LaTex Symbols could be found:

http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf

 :lol:  :lol:

[quote="bchui"]:(  I wonder if we can find a Comprehensive LaTex User Manual or Book from the web which is FREE for Download?  :blush:[/quote]Try [url]http://www.tug.org/begin.html[/url].  If you use LaTeX with any regularity, purchasing [i]A Guide to LaTeX[/i] is well worth it.  [i]The LaTeX Companion[/i] is probably only worth it for very advanced users.

I believe the best way to learn $ \text{\LaTeX} $ is to scroll over that people's $ \text{\LaTeX} $ and see a pattern...it's easy to learn like that.

Masoud Zargar

or \;, \:, \quad, \qquad
$ i\quad like\quad food$

$\color{red}\text{thank you very helpful}$

What do you need to enable colours for LaTeX code?

[code]To make beautiful text, type $\text{This is the power of LaTeX} $.[/code]
Look what you get:
To make beautiful text, type $\text{This is the power of LaTeX} $.

To make pretty colors:
[code]$\textcolor{red}{\text{Hi there}}\textcolor{blue}{\text{Did I mention you can type in a variety of colors?}}$[/code]
And
$\textcolor{red}{\text{Hi there}}\textcolor{blue}{\text{Did I mention you can type in a variety of colors?}}$

In general, type
[code]$\textcolor{color}{text}$[/code]

[quote="AwesomeToad"]I need help posting in math jam.
1)How do I start my post?
2)How do I create spaces between words?
3)Must I end my post with a dollar ($ \$$) sign[/quote]
Latex is not required. Hope this helps.

[quote="ssilwa"]Latex is not required. Hope this helps.[/quote]

Hello ssilwa,

Thanks for the help, but now I would like to close any questions I have asked on AoPS prior to 2011.

Andrew

[quote="hshiems"][color=#00BF00]Can you use LaTex in Microsoft Office or Google Drive? If so, tell me how to do it/provide links.[/color]

P.S. I'm pretty sure Wolfram Alpha recognizes LaTex code.[/quote]
Wolfram|Alpha does recognize basic LaTeX code like 2^4, but it has trouble with things like nested fractions, summations, multinomials, and about half of the LaTeX symbols among many others.

Really? 'Cause this works:

[latex]\sum^{7}_{x=0} \frac{2}{1+\frac{2}{1+\frac{2}{1+\binom{x}{1+\binom{x}{1}}}}}[/latex]
It is probable that you missed a bracket or two when submitting. They can be hard to count.

Nope. Sometimes it works; most of the time it doesn't. I didn't miss a bracket, since I tested it by putting it in TeXeR first.


Find the number of possible bridge auctions.

A bridge auction consists of multiple bids in order. There are 37 possible bids: pass, double, and numerical bids (1-35)
A numerical bid can be made as long it is higher than all proceeding numerical bids
A double can only be made after certain circumstances:
x
xpp
xd
xdpp
xppd
xppdpp
where x is a numerical bid, p is pass and d is double
passes can be made at any time. If there are 3 passes in a row, the auction is terminated. Exception: If the first bid is pass, this does not count towards that limit. So the shortest possible auction is pass pass pass pass

This is the mathematical equivalent of asking somebody to dig a hole to China.

well the answer does seem unresonably large and hence difficult to find an exact answer as a number without a computer

how about if bids are restricted to the 1 level (only 5 numerical bids)?


[quote="pianoforte"]A cone with height 20 and radius 10 has an upside-down cone inside with height 10 and radius 5. What is the volume within the larger cone that is not taken up by the smaller cone?[/quote]

[hide]Just subtract the volume of the smaller cone from the larger cone:

$\frac{100\pi*20}{3}-\frac{25\pi*10}{3}$

$\frac{2000\pi-250\pi}{3}$

$\frac{1750\pi}{3}$

$570\pi$[/hide]


algorithm
number theory
Euclidean algorithm

i'm guessing they should both be + or both -. i'll do both +, the - case is equivalent.

[hide]$(2+\sqrt{3})(2-\sqrt{3})=1$, so $(2+\sqrt{3})^n (2-\sqrt{3})^n=1$ for all $n$.

Since $(2+\sqrt{3})^n=5042+b\sqrt{3}$, $(2-\sqrt{3})^n=5042-b\sqrt{3}$ so $5042^2-3b^2=1$.

Isolating $b$, $3b^2=5042^2-1=(5041)(5043)$. $5043$ is the one divisible by 3, by inspection, so divide it out to get $b^2=(5041)(1681)$.

Using the Euclidean algorithm $3*1681-5041=2$. Since neither of 5041 and 1681 are even, they are relatively prime. Therefore each must be a perfect square.

The squares can be estimated by $5041=(70+x)^2=4900+140x+x^2$ and $1681=(40+y)^2=1600+80y+y^2$. $x=y=1$ solve this.

So $b=41*71=2800+40+70+1=\boxed{2911}$.[/hide]

Well, signs are wrong, since $(2-\sqrt{3})^n$ is always of the form $a-b\sqrt{3}$ where $b>0.$ Some corrections must be made.

Okay, that might explain it. My solution is now finished.


License plates are issued that contain four digits followed by
one letter. If the letters O and I cannot be used, how many
different license plates are possible?

There are 10 choices for each digit, and 24 for the letter (since 2 letters-O and I-can't be used).  Thus, the answer is 10*10*10*10*24=240000.


geometry
power of a point
radical axis
geometry proposed

[quote="hophinhan"][b](TST 2004/4)[/b] Let ABC be a triangle, and let D be a point in its interior. Construct a circle (O1) passing through B and D, and a circle (O2) passing through C and D, such that the point of intersection of (O1) and (O2) other than D lies on the line
 AD. Denote by E, F the points where (O1), (O2) intersect side BC, respectively, and by X, Y the intersections DF intersection AB,DE intersection AC, respectively. Prove that XY is parallel to BC.[/quote]
Let $ AB\cap (O_1)=\{M\}, AC\cap (O_2)=\{N\}$.
A lies on the radical axis of $ (O_1)$ and $ (O_2)$ then $ AM.AB=AN.AC$
So BMNC is cyclic quadrilateral. We get $ \angle ANM=\angle ABC$
But $ \angle YDM=\angle ABC$ then $ \angle YDN=\angle YNM$ thus YNDM is a cyclic quadrilateral.
Similarly, XMDN is a cyclic quadrilateral.
Then YXNM is a cyclic quadrilateral.
$ \Rightarrow \angle YXM=\angle ANM=\angle ABC$
$ \Rightarrow XY//BC$ (QED)


I am running PCA on the correlation matrix for a collection of variables, and can successfully extract the eigenvectors, eigenvalues, and generate the principal component scores.  I have noticed in many multivariate textbooks that it is common to show a plot with the 2 orthogonal straight lines for PC-1 and PC-2 overlayed on top of an x-y scatterplot of the standardized values for two variables.  Question is: are the straight lines representing PC-1 and PC-2 really the eigenvectors plotted against one another or fitted lines?  Also, is there a transformation like dividing by SQRT(eigenvalue) that must be done to elements of the eigenvectors in order to get the 2 orthogonal straight lines?

Answer is as follows.  For the textbook-type plots which plot the centered $x_{1}$ vs. $x_{2}$ and a straight line going through PC-1 and PC-1, the value of the principal component for each data sample is represented in the form  $y = x_{i1}e_{1,1}+x_{i2}e_{2,1}$, where $x_{i1}$ is the value of $x_{1}$ for sample $i$, $x_{i2}$ is the value of $x_{2}$ for sample $i$, $e_{1,1}$ is the eigenvector element for $x_{1}$, and $e_{2,1}$ is the eigenvector element for $x_{2}$.

First, set $y_{i}= 0$ to get

$0 = x_{i1}e_{1,1}+x_{i2}e_{2,1}$.

Next, subtract ${ix}_{2}e_{2,1}$ from both sides to get

$x_{i2}e_{2,1}= x_{i1}e_{1,1}$.

Solve for $x_{i2}$ to get

$x_{i2}= \frac{x_{i1}e_{1,1}}{e_{2,1}}$.

To make the straight line for PC-1, simply loop over values of $x_{i1}$ at fixed intervals (over the range of centered $x_{1}$ values) to get the $x_{2}$ values.

Similarly, PC-2 is defined as

$y = x_{i1}e_{1,2}+x_{i2}e_{2,2}$,  (notice column 2 is used for eigenvectors)

so solving for $x_{i1}$ gives

$x_{i1}= \frac{x_{i2}e_{1,2}}{e_{2,2}}$

Next, loop over values of $x_{i2}$ at fixed intervals over the range of the centered values.  This will essentially overlay the fitted values of the PC-1 and PC-2 over the scatter plot of centered $x_{1}$ and $x_{2}$ values.

lep


let $ KN$ and $ KL$ be tangent to $ \odot C$ at point $ L$ and $ N$, and $ M$ is a point on the ray $ KN$, $ \odot O$ is the circumcircle of $ \triangle KML$ and $ \odot O$ meets $ \odot C$ at $ P$, and $ NQ \bot ML$ at $ Q$, prove that $ \angle MPQ \equal{} 2\angle LMK$

does anybody have any ideas? thanks a lot

notice that △PHM～△PLK
 P Q G M lie in a circle and we are done


Source: Mexico

prove that a 6 x 6 square cannot be covered completely by the "L-shaped" tetris blocks

edit: pluralized "tetris block" or else it's too obvious

[hide]Draw a 6x6 grid. Color rows 1, 3, and 5 one color and rows 2,4, and 6 another.  The L-shaped tetris block covers 3 spaces of one color and 1 space of another no matter how you place it.  There are 36 spaces to cover.  Each L-shaped block covers 4 spaces, so we need 9 blocks to cover the whole grid.  Each block covers $1\pmod{2}$ spaces of each color.  9 blocks then cover $9\pmod{2}\equiv 1\pmod{2}$ spaces of each color.  But we need to cover $18 \pmod{2}\equiv 0\pmod{2}$ of each color, and we can't. QED.[/hide]


algebra
polynomial
calculus
integration
algebra unsolved

For every non-negative integer n one can write

$(1+4\sqrt[3]{2}-4\sqrt[3]{4})$
with $a_{n}$
$b_{n}$and $c_{n}$ uniquely determined integers. Show that if$c_{n} =0,$
then $n =0$.

Let $f(x)$ be a monic polynomial of degree 1991 with integral coffcients.

Let $g(x)=(f(x))^2-9$
Show that $g(x)$has at most 1995 distinct realroots.


what math course are you talking this year?????

Geometry

This message is not too small.

[quote="akalra1"][quote="Ignite168"]Pre-Algebra[/quote]

rnt u goin into 9th?

cuz i did prealg in sixth[/quote]

Everyone's not as smart as you  :noo:  :(

no i mean that my district teaches prealg to everyone if they want to in sixth grade (i live in fairfield)

My old school had this option where you get to take Pre-alg in 5th, but the year I was in sixth grade, they screwed the schedule and stopped switching math courses. :mad:  :(  So I had to take algebra in seventh.

[quote="Phelpedo"]Ignite was joking.[/quote]
He's taking Algebra II.

no i'm going to 8th......... and m y school Worthington Hooker (math teacher) Beth Klinger teaches you nothing... she so annoying.. :(

Hooker is actaully one of the better schools in New Haven.

My school forces everybody to take Pre-Alg in 7th.

i know we score the top on the cmt's but we recently got a

IT'S SOOOOOOOOOOOOOOOOOOO ANNOYING...

...What?

And CMT's are a joke. I got a full score on my essay, full scored the DRP section, and only missed 6 questions total on the entire test.


How would I find what a,b,c are given:
$a^2+b^2+c^2=10$ and
$2a+3b+4c=8$

Is there a general technique to find (a,b,c) given:

$a^2+b^2+c^2=x$
$ma+nb+pc=y$

$x,y,m,n,p$ coefficients

Thanks!

do you mean integer solutions? otherwise, there are infinite solutions
otherwise, there are specific situations where you use inequalities to show that one of them attains a maximum value when the a=b=c

I meant all real, and I was wondering about the equality case....

Eliminating $a$ we get

$13b^2+24(c-2)b+20c^2-64c+24=0$ which is a quadratic equation .

The discriminant after some simplifying job gives us

$29c^2-64c-66\leq 0$

Well , surely there are many answers for $c$ here . If the question doesnt state that $a,b,c$ have to be integer , than there are infinity many answer  :)


MATHCOUNTS
number theory
least common multiple

uh, i guess this isn't really related to mathcounts, but it is used alot in it, i think. so, how do use mod? and when would you use it?

http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28395

Here is a time when you would use mods.

There is a rock collection. When you group them by 7, 6 is left over. When you group them by 5, 4 is left over. When you group them by 3, 2 is left over. WHat is the least amount of rocks in the collection?

so how would u use mod in that problem? (sorry, im really dumb)
lol

[hide="Like this"]Add one rock and you will have a multiple of $3$, $5$, and $7$. $LCM(3,5,7)=105$, so the answer is $104$. However, you can also express it like this: $x+1\equiv 0\bmod{7}$, $\equiv 0\bmod{5}$, $\equiv 0\bmod{3}$. Really, it isn't necessary for this problem but it gives you a basic idea.[/hide]

ok....
i guess i sorta get it, but could u elaborate a little more? and also show step by step how 2 solve the problem using mod.
thanks!

[quote="236factorial"]
There is a rock collection. When you group them by 7, 6 is left over. When you group them by 5, 4 is left over. When you group them by 3, 2 is left over. WHat is the least amount of rocks in the collection?[/quote]

[hide="my explanation"]I'll try to explain, but I'm not so sure It'll help.

Ok think of mods in terms of remainders. So when you right the 1st sentence in mod form you get ---$x=6 (mod 7)$
 the next one---$x=4 (mod 5)$
last ---------------$x=2 (mod3)$

you want x=0 (mod 7,5,3) (is that the way to write it??) you need to add one to each of these equations to get that.

so you take the lcm of 7,5,3 (so the x is the same) which is 105 and you subtract 1 (because you added 1 previously)

So the answer is [size=150]104[/hide][/size]


calculus
derivative
calculus computations

Let $ f(x) \equal{} ke^x \minus{} x$

1) For what value(s) of k does this equation have only one real root.

2) For what value(s) of k does this equation have two distinct real roots.

Please give complete formal solution.

1) f(x) has exhactly one real root when k<=0 or k=1/e

k=0. is a trivial case.
let k<0. f(x) is +infinity at x=-infinity; and f(x) = -infinity when x=+infinity; Hence, by bolzano's theorem, f(x) has atleast one real root.

Now, f'(x)=ke^x - 1; this is never zero, for real x, when k<0

Hence, f(x) cannot have more then one roots, for if it has, then, by mean value theorem, it wil have to have a zero derivative, which I proved is not possible.

Hence, if k<=0, f has exhactly one real root.

Now, let k>0. f(x) is + infinity at both x=+infinity and x=-infinity. and, f'(x) has a zero at x=-ln(k).Also, this is the only zero of f'(x). And, f''(x) is always positive.

It can be shown that f(x) should have two roots, or no root. Hence, for a single root, the two roots should be identical, i.e., it should be a double root, and hence, the derivative at the root vanishes. Hence the root has to be at x=-lnk.

So, f(-lnk)=0; this gives, k=1/e

Since we have examined all possible values these are the only values of k for which f has exhactly one root.

2)f has exhactly two distinct roots when, 0<k<1/e

f has the only local minima at x=-lnk. If x has two distinct zeroes, it must be negative at some points near the neighbourhood of the roots hence, f(-lnk)=lnek<0

i.e., 0<k<1/e


Corey's harvest, worth its weight in gold, consists of 3000 bananas. The market place where the stash can be cashed in is 1000 miles away. However, Corey must walk to the market, and can only carry up to 1000 bananas at a time. Furthermore, being a camel, Corey eats one banana during each and every mile she walks (so Corey can never walk anywhere without bananas). How many bananas can Corey get to the market?

This is a pretty common math problem.
I am trying to find an algebraic solution to the question. How would you do that?

There was an identical problem posted in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119963]Careers in Mathematics Forum[/url], which was given during a job interview.

In terms of how many Corey can get to the market, it will depend on the traveling method.
But the real and preferred question is finding the method that optimizes the number of bananas one can bring to the market. Or better yet, what is the optimal amount of bananas that can be brought to the market?

According to the employer, based on the given link, the optimal amount, which I'm not sure of myself since I haven't tried the problem yet, is:
[hide]$ 567$[/hide]

m pretty sure the maximum number of bananas that he can get to the market is 533 1/3. To see my justification, see this link:

http://www.puzzle.dse.nl/math/bananas_us.html

[quote="alanchou"]So he eats 2/3 of a banana at some point?[/quote]
Yeah I guess that the person assumed that the eating rate was constant, so for every feet the camel went, it would eat $ \frac{1}{5280}th$ of a banana.