geometry
circumcircle
trigonometry
quadratics
3D geometry
function
Putnam

I thought it was really fun, especially the relay races.

The only problems were:

-The teams were too big. 10 people was hard to work with 1 problem sheet.
Maybe either give multiple problem sheets or cut down team sizes to like 5-8 people.
-No time in between tests.

Other than that, it was awesome, especially the 24 tournament where we got to meet other people.

[quote="chess64"]PLEASE GIVE US MORE COPIES OF THE PROBLEMS NEXT TIME[/quote]

Yeah, I think Nathan's already heard this at least a hundred times ;)

DUDE TEAM CAROLINA WON

[quote="chess64"]PLEASE GIVE US MORE COPIES OF THE PROBLEMS NEXT TIME

oh and screw the 24 tournament what we need is a BUGHOUSE TOURNAMENT[/quote]

I agree with the more copies though everybody probably complained about that.

I would also like the lecture to be a different time frame than the tournament, whether it be 24 or bughouse or something else, so that people would be able to do both.

I did write all this on the feedback sheet but by the time I finished, they were starting to hand out awards and I couldn't get my chocolate.

Team and power were hard. :(

Are full results going to be posted online?

Team and power were awesome :) We (central MD) got 5th in power, 6th in team and 7th overall. It helped that we randomly had Jeremy Hahn on our team :D He did all of section A and I did like all of section B and E^(pi*i)=-1 did section D I think. Everyone else pretty much just talked most of the time. I don't think anyone else understood the definition :roll: Oh and one really annoying person kept asking us if we were making progress on the problems.

Hmm I didn't think we did good on the team round but I guess the other teams did worse. Once again, Jeremy Hahn saved the day :lol:

Umm I got owned on the individual though... :(

They should use the papers that they wasted on the relay slips for extra copies of the problems ;)

[hide="about the geometry test"]oh man i messed up so badly on the geometry test...the problems were not so difficult, but i made so many computation errors...ahhh i even checked all my work too, but to no avail...also the answer for problem 9 is clearly wrong... it should be $25x^{2}-36xy+13y^{2}=4$, not $25x^{2}-36xy-13y^{2}=4$...

the problems that were the 65,65,50 triangle, and the last were very good!

sadly, i did not see the solutions until later...i did something slightly different, i used the circumcircle of D (and the two points on the right), then a little angle bisector theorem (why did they take the tests away btw?)

for the last problem, i am disappointed not to see the geometric solution on the key...

basically take the midpoint of the middle segment, call this M, then M is the center of the square (easy to see), then the other point determining the left segment, the locus of that point is an appolonious circle, to minimize that angle, take a tangent from point (D?) then you get the angle is 45-30=15 as desired

also i disagree with the answer for problem 4... (the tangent circle one...circle tangent to c_1, c_2, c) right?

inversion, or not (i am familiar with that argument...), i get $\frac{5-2\sqrt{3}}{13}$

(this problem is also trivialized by descartes' circle theorem...)

(likewise, the point inside the triangle is trivialized by trig ceva...)

the find sine 36 problem was annoying...i remembered teh derivation, but i switched a sign on the quadratic for cos 72....

so i ended up with a terrible score...also problem 1 was suprisingly difficult...i messed up my system of equations and could have had the solution too...if you call the arc AB=a, etc you get the system

a+b=360-2*96, etc

then make the angle you want to find x, then use the 720 degrees to get the last angle, and a+b+c+d+e+f=360, solves it

overall, i would say the geo test was fair, a little more time would have been nice, i felt too rushed...also i wish we could challenge the answers[/hide]

obviously multiple copies of problems would have been nice...although we had the ingenuity to cut the paper into sections and write the definitions on the board, so it was not that big a deal

i am kind of pissed about the relay...i got the wrong problem for the second test (i was in seat 2, but i got 2-4)

i figured that the person in front of me got the wrong answer...nothing i could do with x=3..., i just solved it anyway, it is obvious that the polyhedron has 4 sides (say they are n-gons....then $(9x*n)/2=18x$ so $n=4$), then by the lecture, the only regular polyhedron with square faces is a cube, and then you can find x, and you solve for the distance, and the length is obvious

however, the person in the fourth seat did not do this, (i could have passed this answer if i had been in that seat...)

too bad i only realized after i was discussing with my teammates after the answers had been collected

as a final word, although i have some complaints, i really enjoyed the contest otherwise, the geometry problems were pretty clever, and the power contest problem was something that felt accessible to me, the team problems were pretty good too,

On the whole, I had lots of fun.  Apart from getting to see old friends and physically meet a few people I previously only knew through the internet (Saiem G., and, however briefly, Nathan S.  Sorry I didn't get to chat much with y'all), the actual competition was well done on the whole (barring the obvious problems with Relays and Question sheets).

Funny story about question sheets, actually.  We (Lawrenceville) had our first actual team run-through... yesterday, and the following conversation occurred almost word-for-word:

Me:  "Ok, let me look off with you."
Tom:  "Why didn't you just print 10 copies?"
Me:  "I printed 5, they might not give us one for every team member."
Tom:  "Why would they do that?"
Me:  "Just a hunch... better to practice this way just in case."

As for difficulty, my team was able to get 5 on the team round, which implies that it was a bit too easy, IMO (half of our team consisted of freshmen -- and not any particularly brilliant freshmen either.).  The power round was perfect.  I've seen some power rounds with topics like Inversion of Circles and even Generating Functions, but I had never heard of "refinement," and, furthermore, I didn't have any clue what it would apply to, which made it feel very much like unexplored territory.  Awesome stuff.

More time between tests would be nice.  IMO, it would be best if the contest ran up until 5:00 or so, so that teams could have dinner together right afterwards.  It's not such a big deal for the traveling teams (Quagga and... Quagga), but for local teams, people disperse pretty quickly unless given a reason to stay together.  Having it be near dinnertime when the contest wraps up would make it easy for teams to go have dinner together and talk amongst themselves.

Finally, and most importantly, [b]please don't force each team to send 2 people to each topic test.[/b]  It would be good if you could find some sort of workaround to that rule (since, presumably, it's there to prevent bunching on several tests).  When I was arranging the divisions for our teams, I had to put our three best members into topics we didn't really want to do, since we couldn't in good conscience boot our freshmen from Algebra or Geometry, since they wouldn't stand a chance in the other topics (This may have worked in our favor though, since Ed Zheng still managed to place 2nd in Number Theory in spite of his wanting to take Algebra...).  Again, this isn't a problem for the XTREEM traveling teams, but if you're going as a school team, chances are you don't have 10 people who are very good at each of these topics.

On the whole, for your first year, you guys did put together a great contest.  I was particularly impressed by the the quality of the questions (Barring #10 on the team round, that is.  Check out http://www.usamts.org/Solutions/Solution2_1_17.pdf  But I'm sure things like that are bound to happen.  A repeat question showed up on the Putnam this year, apparently), but it was to be expected, I suppose, since Nathan S. has a ridiculous amount of question-writing experience.  Assuming that someone steps up and organizes things after I've graduated next year (our Math Club is notorious for having lazy presidents who stop doing anything as soon as they get into their fancy college of choice...), Lawrenceville will be there again.

NB:  Chess64 was talking about results... are those posted somewhere?  Maybe they were in the packet and our coach didn't see them or something...?

Oh, I think our coach had to check something with them and they just told her the (tentative) results while she was doing it... lol :D

[quote="Altheman"][quote="chess64"][quote="Altheman"]
...i got 2-4...[/quote]

[hide]
the only polyhedron with twice as many edges as faces is a cube :)

$\therefore 9x = 6\implies x=\frac{2}{3}$, which is the answer you should have received

(that was jeremy hahn's solution; i got 2-3)
[/hide]

for the first relay apparently the second problem was hard or something so i finally got my answer after 11 minutes... and it was really boring because my question was a trivial combinatorics identity ($\binom{n}{1}+\binom{n}{3}+\cdots+\binom{n}{n}= 2^{n-1}$ for $n$ odd)[/quote]

the whole point was that i was in seat 2, not seat 4, the first person had the correct answer, just that i was not in the right place, so i had to ignore the answer, but of course the person behind me could do nothing...[/quote]

oh

ok

btw, were you on the florida team?

[quote="chess64"][/quote][quote="Altheman"][quote="chess64"][quote="Altheman"]
...i got 2-4...[/quote]

[hide]
the only polyhedron with twice as many edges as faces is a cube :)

$\therefore 9x = 6\implies x=\frac{2}{3}$, which is the answer you should have received

(that was jeremy hahn's solution; i got 2-3)
[/hide]

for the first relay apparently the second problem was hard or something so i finally got my answer after 11 minutes... and it was really boring because my question was a trivial combinatorics identity ($\binom{n}{1}+\binom{n}{3}+\cdots+\binom{n}{n}= 2^{n-1}$ for $n$ odd)[/quote]

the whole point was that i was in seat 2, not seat 4, the first person had the correct answer, just that i was not in the right place, so i had to ignore the answer, but of course the person behind me could do nothing...[/quote][quote="chess64"]

oh

ok

btw, were you on the florida team?[/quote]

ahem, quagga from florida  :D

oh and about the relay 1-2, that problem was ambiguous, it said that the people pursued and gained at y miles an hour, this is unclear because the y miles an hour can refer either to the pursued, or the gained, based on the structure of the sentence, (to me the gained part just reiterated that they were heading towards the prisoner at y mph)

[quote="Cicatriz"]Are full results going to be posted online?[/quote]
We're definitely going to email the full results to all of the coaches. I'd like to put them online, but I'm not that good with internet-related stuff, and I don't want to make promises that other people will have to fulfill.

[quote="Altheman"](why did they take the tests away btw?) [/quote]
If you mean why didn't the proctors let you keep the individual tests, it's because we never told them whether you were allowed to, and some of them thought you weren't supposed to keep them. Thanks for mentioning this -- I didn't even think of it until after the tests; next year you'll be able to keep everything.

[quote="Altheman"]also i wish we could challenge the answers[/quote]
I don't immediately see how to make this happen easily (how do we make sure everyone gets the solutions? when will the challenges take place?), but it's worth thinking about. Does anyone have suggestions?

[quote="Altheman"] although we had the ingenuity to cut the paper into sections and write the definitions on the board, so it was not that big a deal [/quote]
Were you the ones who wrote the initial definition on the board  and wrote "WTF" under it?
(At least three different teams wrote "WTF" on their chalkboards during the power test!)

[quote="eyefragment"] IMO, it would be best if the contest ran up until 5:00 or so[/quote]
If we let the contest run a bit later, we could definitely give more time for everything, which would make it less rushed and easier to stay on schedule. But we didn't want to end too late because the teams that drove probably would want to drive back that evening. Was our concern misplaced? Would you guys prefer a longer contest that had more time for everything (and between everything), even though it would end later?

[quote="eyefragment"] Finally, and most importantly, please don't force each team to send 2 people to each topic test. [/quote]
Well, this presents an interesting problem, though: different tests will always have different difficulties. For example, judging by the scores, the Number Theory test was markedly easier than any of the other tests. What if a team was full of people who loved number theory and had all 10 people take it? And another team was full of people who hated number theory, so no one took it? And then the first team beats the second team hugely on the individual total and wins, even though the first team was actually better.
Here's one option in my mind: this year, we removed the idea of upper and lower divisions, because almost no teams signed up for the lower division. Next year, I suspect we'll have more teams interested in that (because we'll expand next year, and because this year's tests were very hard). So, we could continue to enforce the number of tests rule for the upper division teams, but relax it for the lower division teams.
What are your opinions about this (the rule, this idea, other ideas)?

[quote="eyefragment"]But I'm sure things like that are bound to happen. A repeat question showed up on the Putnam this year, apparently[/quote]
Putnam 2005 A5 was posted on AoPS long before the contest by kunny. Putnam 2006 B4 (I think) was posted by Omid Hatami. Also, WindingFunction wrote some problem, and then someone posted an equivalent problem in the Calculus forum. And she wrote another problem that was on an old TJ mock olympiad (which I saw posted in the preolympiad forum). Of course those two didn't appear on the tests, since we caught them in advance, but ...

[quote="chess64"]Oh, I think our coach had to check something with them and they just told her the (tentative) results while she was doing it... [/quote]
Yes, she was concerned because you guys had told her you did well on the power test, but (as all teams surely noticed) we didn't write the power test scores on the score sheets we gave you, so she was worried that we had lost your score (how would she know that we hadn't done it for anyone, rather than just you?). So we checked to make sure we had counted your power score, and that's how she saw the results.
Again, I'll see about how we can most easily get the results our to you.

Here are some thoughts of my own:
It's interesting that even though the relays were what we had the most trouble with (what with the no row I fiasco and all), it seems like a lot of you contestants found that to be the highlight of the day. I saw some comments on the feedback forms mentioning the ARML super-relay. Would you want us to devote more time in the future to fun/intense/exciting events like that?

I'd like to hear what you thought of Conway's talk. Unfortunately I was unable to stay (I only popped my head in for a moment) so I don't know what it was about or how it went. I noticed that it was a bit hard to hear from the back -- if we have a lecture again next year, we'll give the speaker a clip-on mike. Was his talk interesting? accessible?

In the future, should we announce scores as well as places with the awards? That slows down the awards ceremony a little bit, but I remember when I went to contests like this, I was always interested in the scores. I had actually intended to have scores announced this year, but I forgot to mention it to anyone else...

Thanks again for all your comments! As some of you guessed, all of us have heard countless times that we need more copies of the tests next year. But it's not bad that so many people keep repeating it -- it lets us know how important it is to fix this for next year. By the way, as far as I can tell, the solution is just to give everyone more copies of the test. But does anyone have other ideas about how we could make the team and power tests easier for everyone to work together on?

Another question: was there enough scratch paper?

All right, I've definitely written way too much already. If you've gotten all the way to the bottom, thanks also for reading all of it!

[quote="Xevarion"]Were you the ones who wrote the initial definition on the board  and wrote "WTF" under it?[/quote]

haha that was us :lol: actually first we decided we needed to write the definition on the board, and then some people started working on it and the rest were standing there being confused, and one of them wrote WTF under it :P

[quote="JSteinhardt"]Lol it was semi-directed at both. I was just pointing out to Arnav that his team wasn't exactly weak as far as Power Rounds go. I think that an IMO gold medalist, a Black MOPper, and a strong Red MOPper is certainly a large concentration of talent.

Though when I made that comment I was under the impression that you had at least 4-5 MOPpers on your team (Jeffrey Manning and Will weren't on it?).[/quote]

Nope.  We had another J. Manning, from SC, but SC has never had a MOPper (or at least not in the last 10 years).  The only 3 NC people we had were Arnav, Yasha, and John.  We were planning to go as SC, but 3 MOPpers without a team were too good to resist.

dude also for some weird reason they passed up noah blach

but anyway I wasn't saying that we were weak at power, I was saying TJ was balanced

there is a difference

slightly watered down?

dude 3 NCers to 7 SCers is not slightly watered down

that is like OH MAN CIVIL WAR ALL OVER AGAIN

oh wait we were in the south I keep forgetting since y'know I live in chapel hill and all

[quote="joml88"]You can't really force a team to work together by how you structure the test.  You can get them to split it up (like how the Team Test is tackled by most teams) at best.

Ioanu Gradinaru (SC) went to MOP in 1996 and 1997.  So that's barely within a 10-year timeframe.  I would have expected you to have heard of her Sean.  She was a student of Rushman, won USC 3 times, and just an overall genius.

You can think of Team Carolina as either a slightly watered down NC team or a souped-up SC team.

For population comparisons:

FL = 17, 789, 964
FL + IL + AL = 34,656,357
NC + SC = 12,061,325[/quote]

Oh, wow, Ioanu was in 1996?  From the way Mr. Rushman talks, it seems like she was from like the late 1980s.  But yeah, I think she was our only MOPper, ever.

Pick better rooms next time.

We had a concrete block and hammer in our room.

That was punishment for getting their part wrong.

-sledgehammer to the face-

it wasn't awesome. pick better rooms next time.

rotfl

hahaha

wow

was this like the armoury or something

perhaps the smithy

There was also a bunch of clay and a triple wheel and multiple files and a string and pulley system and a huge batman projector

Oh, well our room was bad because we couldn't gather around a desk, and the chairs were bolted to the tables...Of course, you don't need to gather when you have MULTIPLE COPIES OF THE TEST!  But seriously, it wasn't as bad as Duke a few years back when the proctor was like talking to us and stuff during the test.

Please don't end the contest at 5. It is already long enough, and any team that wants to have dinner together can feel free to do so--forcing that to happen is just kind of weird. I mean what if somebody has to get back by a certain time or just doesn't want to get home late at night?

But the contest as it was was excellent. I will certainly come back next year. My only complaint is that the relays started at the time they were supposed to end. You should organize those better (make sure the right relays are distributed and have one proctor for each team, not a bunch of scattered proctors).

[quote="Xevarion"][quote="Pakman2012"]Pick better rooms next time.

We had a concrete block and hammer in our room.[/quote]
(a) WTF?? What room was this?
(b) What's wrong with that?[/quote]

Florida A was in McDonnell A01...

I believe it...the physics department is weird...

[quote="jb05"]But the contest as it was was excellent. I will certainly come back next year. My only complaint is that the relays started at the time they were supposed to end. You should organize those better (make sure the right relays are distributed and have one proctor for each team, not a bunch of scattered proctors).[/quote]

One proctor for each team with the teams arranged vertically? That could get annoying...  Although it probably would have been better than what ended up happening.  Which was unintentionally.  Anyways, we may be leaning towards not having relays next year...

AHHHHHHHHHHHHH i lost the 24 tornument cuz the guy that was judging walked away to get something just as i got the answer.... and then wehn he came back ... i forgot it..... :'( lol
almost top 10 in number theory cept i counted factors of 66! wrong...

O and are we allowed to form teams from multiple schools? cuz our team woulda been alot better if we werent just from my school... we had a kid get a 0 on algebra and a kid get 0 on geometry...

panda, but if i recall, you could still just tap and answer without the proctor there for game 24..


linear algebra
matrix
superior algebra
superior algebra unsolved

Let A, B be nxn matrix over a field F. Prove that if A and B are similar then $ \lambda*I_{n} \minus{} A$ and $ \lambda*I_{n} \minus{} B$ are equivalent in $ M_{n}(F[\lambda])$. Prove the converse.

Assume $ P(XI_n\minus{}A)\equal{}(XI_n\minus{}B)Q$ for some $ P,Q\in GL_n(F[X])$.
Using division with rest we get $ Q\equal{}T(XI_n\minus{}A)\plus{}R$ with some $ T\in \mathcal{M}_n(F[X])$ and $ R\in \mathcal{M}_n(F)$.
(In fact if we write $ Q\equal{} \sum_{k\ge 0}Q_kX^k$ with $ Q_k\in \mathcal{M}_n(F)$, then $ R\equal{}\sum_{k\ge 0}Q_kA^k$)

Then
$ P(XI_n\minus{}A)\equal{}(XI_n\minus{}B)T(XI_n\minus{}A)\plus{}(XI_n\minus{}B)R\\
\hspace*{0.75in}\equal{}(XI_n\minus{}B)T(XI_n\minus{}A)\plus{}R(XI_n\minus{}A)\plus{}RA\minus{}BR$.
So $ RA\minus{}BR\in\mathcal{M}_n(F)$ is right-divisible by $ XI_n\minus{}A$, which gives $ RA\minus{}BR\equal{}0$
To show, that $ A,B$ are similar we just need to show, that $ R$ is invertible. (then $ A\equal{}R^{\minus{}1}BR$)

But we find $ Q^{\minus{}1}\in GL_n(F[X])$ and using division with rest we can write
$ Q^{\minus{}1}\equal{}\tilde{T}(XI_n\minus{}B)\plus{}\tilde{R}$ with some $ \tilde{T}\in \mathcal{M}_n(F[X])$ and $ \tilde{R}\in \mathcal{M}_n(F)$.
And then
$ I_n\equal{}Q^{\minus{}1}Q\\ \hspace*{0.14in}\equal{}
Q^{\minus{}1}T(XI_n\minus{}A)\plus{}Q^{\minus{}1}R\\ \hspace*{0.14in}\equal{}
Q^{\minus{}1}T(XI_n\minus{}A)\plus{}\tilde{T}(XI_n\minus{}B)R\plus{}\tilde{R}R\\ \hspace*{0.14in}\equal{}
Q^{\minus{}1}T(XI_n\minus{}A)\plus{}\tilde{T}R(XI_n\minus{}A)\plus{}\tilde{R}R$.
So $ I_n\minus{}\tilde{R}R\in \mathcal{M}_n(F)$ is right-divisible by $ XI_n\minus{}A$, which gives $ I_n\equal{}\tilde{R}R$ and $ R$ invertible.


Let $a, b, c$ be  positive reals
Prove
\[(a+b)^2+(a+b+4c)^2\geq \frac{100abc}{a+b+c}\]
Not difficult, but well, I like this question

Cute one. Here's what I did: I wrote (a+b+4c)^2 as 4*((a+b+4c)/2)^2. I applied Cauchy and AM-GM a few times. I'm not in town and I can't write all of it. Eventually I got that the equality case is a=b=2c. Bye! :D

well,even though grobber pointed a way,i did it differently
here's the solution
put c=x/2
then after opening up the whole thing
2(a^3+b^3+x^3) + 5(a 2 x + b 2 x) +
6(a 2 b + bx 2 ) + 6 (ab 2 +ax 2 )  \geq  40abx
is the inequality to be proved.apply term by term am-gm to the quantities in the parentheses
equality case a=b=x=2c
QED

Lololol:
http://mathdb.org/notes_download/elementary/algebra/ae_A5c.pdf

Apparently a 2001 Hong Kong TST problem. (Scroll to the page 7).

i have solved this problem before
but instead of 100
we had to find the constant k(k in the place of 100)
here you also have to prove that there is no k less than 100

[hide="Solution"]We can solve this fairly efficiently by dumbassing (multiplying out the inequality and using AM-GM and/or Schur's Inequality). After multiplying both sides by $(a + b + c)$, we obtain on the left,
\[\left(2a^3 + 2b^3 + 16c^3\right) + \left(6a^2b + 6b^2a + 10a^2c + 10b^2c + 24c^2a + 24c^2b\right) + 20abc.\]
We now make the substitution, $t = 2c.$ The motivation for this substitution is simply checking values in an attempt to find the equality case. Indeed equality occurs for $a = b = 2c.$
We now rewrite as
\[\Big(\left(2a^3 + 2b^3 + 2t^3\right) + \left(6a^2b + 6b^2a + 5a^2t + 5b^2t + 6t^2a + 6t^2b\right)\Big) + 10abt\]\[\ge 40abt + 10abt = 50abt = 100abc\]
by Weighted AM-GM, as desired.[/hide]

it is enough when a=b. and this is easy by derivation


geometry
algebra
system of equations
geometry proposed

Let $P$ be the point where the two cevians intersect, and let $D,E,F$ be the intersections of $AP,BP,CP$ with $BC,CA,AB$ respectively. From now on all segments are oriented. We use Van Aubel to find $\frac{PF}{FC}+\frac{PE}{EB}=\frac{PA}{AD}=-\frac 7{15}-\frac 13=-\frac {12}{15}\Rightarrow \frac{PD}{AD}=\frac 15\Rightarrow [ABC]=40\Rightarrow?=21$. Is it correct?

[quote="grobber"]We use Van Aubel to find ...[/quote]

The solution is correct...

But what is Van Aubel? Seems the solution gets pretty short so I guess it's a strong theorem here :)

I have no idea what Von Aubel is either (I've heard the name used for a different result).  So I'll post my solution instead; it's a simple area calculation which I suspect is how Grobber's Von Aubel is proved.  Stealing his notation:

$\frac{[APF]}{[APE] + 7} = \frac{FP}{PC} = \frac{[PBF]}{[PBC]} = \frac{1}{2}$.  Similarly, $\frac{[APE]}{[APF] + 4} = \frac{EP}{PB} = \frac{7}{8}$.  Solving this system of equations gives $([APF],[APE]) = (\frac{28}{3}, \frac{35}{3})$ so that
$[AFPE] = \frac{28}{3} + \frac{35}{3} = 21$.

Let $ABC$ be a triangle and let $P$ be a point. Furthermore, let $E=BP\cap CA,\ F=CP\cap AB,\ D=AP\cap BC$. Then (all segments are oriented) $\frac{EA}{EC}+\frac{FA}{FB}=\frac{PA}{PD}$. This is Van Aubel's theorem. Notice that in the solution for this problem the notations of $A,P$ are actually interchanged (I applied the theorem for triangle $PBC$ and point $A$).

Yes, that's pretty much how it's proven: using areas.

For the van Aubel theorem, which is sometimes called the van Obel theorem, you can also look at http://www.cut-the-knot.org/triangle/VanObel.shtml and http://www.cut-the-knot.org/triangle/Gergonne.shtml . The theorem also appeared in [url=http://forumgeom.fau.edu/FG2001volume1/FG200104index.html]K. R. S. Sastry, [i]Heron Triangles: A Gergonne-Cevian-and-Median Perspective[/i], Forum Geometricorum 1 (2001) p. 17-24[/url].

  Darij

Consider a triangle and 2 lines that each go through a corner and intersects the opposing segment, such that the areas are as on the attachment. Find the "?"
[img]https://cdn.artofproblemsolving.com/attachments/8/a/5f0d909dae8abe067a14a2e34326a64d2f9c38.gif[/img]


can someone teach some basic permutaions and combinations real quick?

thanks

REALLY fast overview

Permutations is how many ways to choose a certain number of things out of a certain number of things where order matters.

Combinations are the same, except order does not matter.

P(n, r) = n!/(n-r)!
C(n, r) = n!/(n-r)!r!

Also, try using using AoPSWiki

Try the AoPSWiki:  [[Combination]]
Here's another good link:
[url]http://arml.yanco.com/20012002/armlcombinatorics.pdf[/url]

Any connection?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1033464#1033464[/url]


function
integration
calculus
probability and stats

Let $ X$ and $ Y$ be two random variables such that $ Y\equal{}g(X)$. We know that $ E\{Y\}\equal{}\int_{\minus{}\infty}^{\plus{}\infty}yf_Y(y)dy$. Under what conditions does this integral equal $ \int_{\minus{}\infty}^{\plus{}\infty}g(x)f_X(x)dx$? (Please prove their equality under those conditions.)

Hello,

g has to be a measurable and bounded application. The proof uses the Radon-Nikodym theorem.


bonjour.  :D

(lately i have been learning several languages, e.g. how to say hello, thank you, and good night....... ummm......  that is it !  :rotfl:  )

if

 $ \textbf f(x,y)\; \equal{} \;  \sum_{n \equal{} 1}^\infty \sum_{m \equal{} 1}^\infty \dfrac{1}{(mx \plus{} ny)(mx \minus{} ny)}$

then show

$ \textbf f(x,y) \plus{} \textbf f(y,x)\; \equal{} \;\dfrac{\dfrac{3}{2}\,\zeta(2)}{xy}$

[quote="Extremal"]mm.. for which $ x,y$?[/quote]

oh ! i forgot to say :)

$ mx\ne ny$

$ x,y>0$

My calculation shows that answer is something like this :
$ \frac {\pi^{2}}{12}(\frac {1}{x^{2}} \plus{} \frac {1}{y^{2}}) \minus{} \frac {\pi}{2xy}\sum_{n \equal{} 1}^{\infty}\frac {cotan(n\pi y/x) \plus{} cotan(n\pi x/y)}{n}$
Hence I think your answer must include $ x^{2},y^{2}$ Are you sure in your answer? :maybe:


The numbers $a,b,c$ are the digits of a three digit number which satisfy $49a+7b+c = 286$. What is the three digit number $(100a+10b+c)$?

Shows me not to do things in my head so often... $286-245=41$, not $31$. :roll:

49a + 7b is a multiple of 7. Now, 7 (7a+b) + c = 286 = 7×40+6. So, c = 6 & 7a+b = 40 = 7×5+5. So, a = 5 & b = 5. Hence the required number is 556.


Does anyone here use Linux?  And if so, what distro do you have?  I use SuSE.

It's an open source operating system.  Like Windows, or Mac, except you can change the code.

You made me very curious. I would also like to learn Linux.Can anyone tell me a reference or something,or maybe a book. :roll:  :)

solafidefarms, can u be more specific? I don't think I did understand all of it. :?  ;)

Okay, those are the distros I use.

To get started with Linux, go to http://www.goosee.com/puppy/download/downpage.htm and download the LiveCD and burn it to a bootable CD. Then put the CD you just burned into your CD drive and restart your computer. Be sure your BIOS is set to boot from the CD drive before the HD. Then follow the instructions. Don't worry - it hasn't done anything to your hard drive. You'll still be able to boot Window$ if you eject the CD. Anyway, PuppyLinux doesn't have a LOT of features, but it has some. You'll be able to see how it works and looks.

When you're done with that, (I hope you have faster internet than dialup...), try Knoppix(search for it). Do the same thing.

Then if you still like it, go to http://www.linux.org.

[quote="Super Naterul"]What is Linux?[/quote]
AAAAHH people don't know what Linux is?!?! :o It's an evil hacking software used by the government of Northern Madagascar to scare cats, how could you not have heard about it? It was linked to the selling of illegal weapons of math instruction, and Al-Gebra!

I'm just joking. It's an operating system (like Windows, Mac, etc.). Did you really not know that?

I use Knoppix. It has fun games.

Linux is more command based, like MS DOS.  But in the recent years, Linux became more graphical and user friendly.  Lets just say it is a lot more difficult to master than Windows, but also a lot more secure and stable.

I had Gentoo, but my brother switched to Ubuntu while I was at camp.  Oh well.

Mac > Linux > Having no OS at all >> Windows

[quote="nr1337"]Mac > Linux > Having no OS at all >> Windows[/quote]
Very well put, but I would say Mac and Linux are equal since they are both based off Unix.
$Mac \;OS=Linux>Having \;no \;OS \;at\; all>-\infty>Windows$

[quote="RC-7th"]But stupidkid, how well does Mac or Windows work with your hard drive of thirty 30 kids? :D[/quote]
LOL, eh HUGE lag...


How many terms are in the expansion of $(a+b+c+d)^{n}$ (in terms of $n$)?

[quote="ragnarok23"][hide]Is it $4^{n}$? I'll post a solution if it is right. :maybe: [/hide][/quote]

I don't think that's right, try n=2 for example

[quote="ragnarok23"][hide]Is it $4^{n}$? I'll post a solution if it is right. :maybe: [/hide][/quote]

That's correct if you don't combine terms... :P

[quote="Jashper"][hide]I don't know if $5n$ would be right, because I didn't expand the cubed version, but when squared there is 10 terms.[/hide][/quote]
Try $n=1$.  :wink:
[hide] There are always the 4 terms $a^{n}, b^{n}, c^{n}, d^{n}$. For every $n$, I think it increases by $6$, so $\boxed{4+6n}$? [/hide]

Extension: How many terms are in $(a_{1}+a_{2}+...+a_{k})^{n}$, in terms of $k$ and $n$?

Here's what I got...

[hide]You know that each term is of form $a^{x}b^{y}c^{z}d^{w}$. You also know that $x+y+z+w=n$. You are solving for non-negative integers so $\boxed{\binom{n+3}{3}}$...I'm not completely sure if that is right or not...[/hide]

Ok I did it again and got [hide]$4^{n}-{n\choose 2}+{n\choose3}\ldots+{n\choose n}=4^{n}-2^{n}+n$[/hide]

I think chinesechess4all is right. Mine definitely doesn't work.

Anyone want to try the extension? It's quite easy after you've seen chinesechess's solution.

[quote="boostage"][hide="Extension"]
With the same reasoning as ch1n353ch3s54a1l's solution: \[\binom{n+k-1}{k-1}\] [/hide][/quote]

This is correct, I think.  You can look up the generalized result on Mathworld; it's called the Multinomial Theorem.

A lot of the answers given here are [b]very[/b] far off  :huh:  Don't underestimate the value of checking a few small numerical examples before making a conjecture.


Inequalities?
Let $ 2007\minus{}c\equal{}d$ so that $ c\plus{}d\equal{}2007$. Then we have $ (c\plus{}d)(LHS)\geq a\plus{}b\equal{}1 \iff LHS \geq \frac{1}{2007}$. So we must have $ ck\equal{}\frac{a^2}{c}, dk\equal{}\frac{b^2}{d} \iff k\equal{}\frac{a^2}{c^2}, k\equal{}\frac{b^2}{d^2}$ for some real $ k$, implying $ ad\equal{}cb$.
I can't finish it that well, but simplifying LHS, we have $ \frac{a^2d\plus{}b^2c}{cd}\equal{}\frac{ad(a\plus{}b)}{cd}\equal{}\frac{a}{c}\equal{}\frac{1}{2007}$, and similarly $ \frac{b}{d}\equal{}\frac{1}{2007}$. So $ \frac{a^{2008}}{c^{2007}}\plus{}\frac{b^{2008}}{d^{2007}}\equal{}\frac{1}{2007^{2007}}(a\plus{}b)\equal{}\frac{1}{2007^{2007}}$.


The multiplicity of a particular factor of a polynomial refers to the number of times that that factor appears in the polynomial. For example $(x-1)$ appears twice in $f(x)=(x-1)^{2}(x+3)$. Thus we say that 1 is a zero of multiplicity 2. Hope that helps.

[quote="Interval"]What exactly is the idea of multiplicity in terms of polynomial functions?[/quote]
after the polynomial function is completely factored into linear factors, equal or oppsite linear factors will be contracted to $(x-a)^{k}$. The multiplicity of the zero $a$ is the power of it expressed as a factor, or $k$. The multiplicity of $a$ is $k$.

Basically, the multiplicity for a certain zero $a$ is how many times $x-a$ divides evenly into the polynomial function.

So, basically, the multiplicity of a function is the exponent, right?

multiplicities are of roots. basically, the multiplicity of a root is the exponent.
the function as a whole doesn't have a multiplicity.


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yani che irad az kojast?

tamame posthai ke tooye [b]community forum[/b]ha gozashte shode az roo post shomar rafte kenar,yeni posthai ke too masalan ghesmate [b]Iran,Romania,...[/b]  ersal shodan tooye shomare post be hesab nemian,yeni alan mehdi faghat 5 ta post too ghesmataye dge bejoz Iran gozashte...
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[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=172668[/url]


probability
function
limit
parameterization
logarithms
invariant
real analysis

Given the following 2-order stochastic difference equation
x(k+1) - m(k) x(k) + x(k-1)=0 (k=2,3,...; x(1)=100; x(2)=101)
where m(k) is a real number uniformly distributed in the interval [a,b] and it can be viewed as a stochastic process since it changes with k,

what is the sufficient and necessary condition for the convergence of the stochastic sequence x(k) (k=1,2,...)?

I assume that $ m_k$ are i.i.d. Then the answer is that for any non-degenerate $ [a,b]$ the sequence $ x_k$ is unbounded with probability $ 1$. You can show that if $ p_k$ is the density of $ x_k$, then $ p_k$ is a bounded function for $ k\ge 3$ and $ \|p_k\|_\infty\to 0$ as $ k\to\infty$ (you can get much sharper results, of course, but this is already enough to answer your question about convergence if I haven't misinterpreted it).

I agree with you on your answer. Thank you for reply.
If a coefficient is added as follows into the equation, then do you think there will be different results for the convergence of its solution? Can you show me the Suff&Nece condition for its convergence? Thank you in advance.

x(k+1)-m(k)x(k)+0.5x(k-1)=0,   (k=2,3,...; x(1)=100; x(2)=101)
m(k) in [a,b]

(Assume that m(k) are i.i.d. )

Yes, adding a coefficient less than $ 1$ changes everything dramatically. I need to think a little (or, perhaps, not so little) bit to figure out the answer to your new question and my free time is limited, so be patient for a while. There is a chance that somebody will beat me to it though :).

Well, I have nothing to add in terms of proofs, but empirically we have fast convergence to 0 when that coefficient is 1/2 or even 9/10, but when it's 99/100 everything is blowing up.  If I had to guess, though, I would say that this latter behavior is just evidence that 5000 trials is not enough to get convergence with a coefficient so close to 1.  (When the coefficient is .98, I get some trials that look like they're blowing up and others that look like they're converging.)

I do not know what you are trying to model, but let's suppose that everything stays positive:
\[ X_{k \plus{} 1} \equal{} U_k X_k \plus{} \alpha X_{k \minus{} 1}\]
with $ U_k$ an iid sequence of random variables uniformly distributed on $ [a;b] \subset [0;\infty[$. If one rewrites it as
\[ X_{k \plus{} 1} \equal{} (1 \minus{} \alpha) X_k \plus{} \alpha X_{k \minus{} 1} \plus{} V_k X_k\]
with $ V_k$ uniform on $ [A;B] \equal{} [a \minus{} (1 \minus{} \alpha);b \minus{} (1 \minus{} \alpha)]$ one easily sees that if $ [A;B] \subset [0;\infty[$ then $ \lim X_k \equal{} \plus{} \infty$ almost surely and if $ [A;B] \subset [ \minus{} (1 \minus{} \alpha);0[$ then $ \lim X_k \equal{} 0$ almost surely. Hence the interesting case is
\[ a \leq 1 \minus{} \alpha \leq b\]

@JBL: what value of $ a$ and $ b$ have you used ?

Ah, somehow I didn't even see those parameters while reading!  I used $ [0, 1]$.

Here is what I know about the positive case. If the equation is $ X_k\equal{}U_k X_{k\minus{}1}\plus{}\alpha X_{k\minus{}2}$ with $ \alpha\in(0,1)$ and some i.i.d. $ U_k\ge 0$ that are not too wild, then we can form the new random variable
\[ V\equal{}U_1\plus{}\cfrac{\alpha}{U_2\plus{}\cfrac{\alpha}{U_3\plus{}\cfrac{\alpha}{U_4\plus{}\dots}}}\]
If $ E\log V<0$, then $ X_k\to 0$ almost surely and if $ E\log V>0$, then $ X_k\to\plus{}\infty$ almost surely. Of course, this is rather useless without an efficient way to compute $ E\log V$ in terms of $ a$ and $ b$ and I haven't figured out how to do it yet. Any ideas?

Needless to say, the non-positive case is much worse. Currently I do not know how to get anything reasonably nice for it (OK, you, probably, can get an answer similar to the positive case one, but it will be completely useless because the involved quantities are next to impossible to compute).

general theorems on continuous state Markov state that under mild conditions (kind of irreducibility and aperiodicity), if this Markov chain $ (Y_k)_k$ has an invariant probability $ \mu$ then this Markov chain is also Ergodic. The Markov chain of interest is the one defined on $ E \equal{} [a; \plus{} \infty[$ by:
\[ Y_{k \plus{} 1} \equal{} U_{k} \plus{} \frac {\alpha}{Y_k}\]
where $ U_k$ are iid and uniformly distributed on $ [a;b]$. Irreducibility (for Lebesgue measure for example) and aperiodicity are not hard to prove, but how did you show that there exists an invariant probability $ \mu$. If $ \mu$ exists (and $ \ln$ is $ \mu$-integrable), then $ \mu$-almost everywhere one would have $ \frac {1}{n} \sum_1^n \ln(Y_k) \to \int_E \ln(x) \, d \mu(x)$, which gives the condition you mentioned, but I am still struggling to show that this $ \mu$ exists  :blush:

For the positive case, just assume that $ U_k$ have density bounded by $ M$ and supported on finite interval. Write the convolution equation for the density of $ Y$. The right hand side will give you a compact linear operator from the convex closed set $ \{p\in L^1: 0\le p\le M,\textstyle{\int_0^\infty p=1\}\subset L^1}$ to itself (you can even add the $ p(x)=O(x^{-2})$ condition at infinity, so summability of logarithm will be automatic). The Shauder lemma finishes the story. Another way is just to show directly that the continuous fraction in question converges with probability $ 1$ ;)

Let's talk about the positive case a bit more. What bothers me a lot is that we have this intractable sharp condition $ E\log V<0$. But, if you just look at $ EX_k$, you'll get a simple sufficient condition $ EU_k<1\minus{}\alpha$. And I have no idea how to deduce the former from the latter.  :wacko:

a side question: how would you do to compute a good approximation of this invariant measure (ie. the distribution of the continuous fraction) A few million iterations (computation time ~ 10sec) of a naive Monte Carlo estimate of $ Q_{k\plus{}1} \equal{} U_k \plus{} \frac{ \alpha}{ Q_{k\minus{}1}}$ gives a pretty good approximation (but of course the approximation is horrible near the points where the density is very small), but I am curious about other available methods. (I do not know anything about computational methods for integral equation etc...) This should not be so hard to compute that invariant distribution this it is included in $ [m;M]$ with $ m\equal{}a \plus{} \frac{\alpha}{M}$ and $ M\equal{}b\plus{} \frac{\alpha}{m}$.

[quote="alekk"]This should not be so hard to compute that invariant distribution this it is included in $ [m;M]$ with $ m \equal{} a \plus{} \frac {\alpha}{M}$ and $ M \equal{} b \plus{} \frac {\alpha}{m}$.[/quote]
Well, I'm not so sure about this: if you have the equation $ Y_k\equal{}U\plus{}\alpha Y_{k\minus{}1}$, you already have a big headache computing the corresponding infinite convolution and our problem is hardly any better. Of course, you can always just experiment, but I would prefer some more explicit answer. The problem is that I do not see one yet.  :wallbash:

Here's a random, perhaps naive, thought.  We can write the recurrence $ X_k \equal{} U_k X_{k\minus{}1} \plus{} \alpha X_{k\minus{}2}$ in the matrix form
\[ \begin{bmatrix} x_k \\ x_{k \minus{} 1} \end{bmatrix}
    \equal{} \begin{bmatrix} U_k & \alpha \\ 1 & 0 \end{bmatrix}
          \begin{bmatrix} x_{k\minus{}1} \\ x_{k\minus{}2} \end{bmatrix}.\]
So our question boils down to the convergence of a product of random iid 2x2 matrices.  Does that make life easier?  I know that there is a ton of work on the product of random matrices, but I don't know if any of it is useful to us.

Alas, it seems pretty much the same to me: we still need that invariant distribution to reduce the question to a pure computation. I looked at some of the relevant papers starting with the famous 1973 paper of Kesten. They have great results [b]under[/b] the assumption that our limit is $ 0$ (or is not $ 0$) but do not provide the slightest clue as to how to check that assumption in practice except for the brute force random simulation (they have some equivalent conditions, but they are equally useless. :( )

If you find anything, let me know :).

@Fedja: I do not understand why this is not true that the invariant measure of $ Q_{n\plus{}1} \equal{} U \plus{} \frac{\alpha}{Q_n}$ is not included in $ E \equal{} [m;M]$. If one starts in $ E$ one stays in it and the invariant distribution is unique isn't it. Where am I wrong ?

You aren't wrong. The invariant measure is compactly supported if $ a>0$. I just said that it may be rather hard to find a function even if you know that it has compact support, is infinitely smooth, etc.

Actually, when I looked at the phrase again, I realized that I probably just misunderstood your broken English. The intended meaning apparently was "It is not hard to show that the invariant measure is supported on $ [m,M]$". The meaning I got was "The invariant measure should not be so hard to compute since it is supported on $ [m,M]$". :)

thanks Fedja. Yes, I meant "The invariant measure should not be so hard to compute since it is supported on $ [m,M]$", and by "compute" I meant: find a good numerical approximation. Basically, if we are given $ a,b,\alpha$, how can we efficiently compute $ \int_{E} \ln(x) d \mu(x)$ ?

That's exactly the question that gives me a headache... I even went as far as asking a few probabilists I know. No result so far :(.

Of course, you can try to solve the integral equation by projecting it onto a finite dimensional space and solving the corresponding linear system, but this method can hardly be called "efficient".


Ok, I thought of a pretty weird question:

Take four points in general position, call them the step 1 points. Then take all possible lines connecting them, call these the step I lines. Then take all the intersections of these lines, call these the step 2 points (note that these contain the step 1 points). Then take all of the lines you can get from these, call these the step II lines...

The sequence of the number of points for the first few steps look like this: 4, 7, 13, 25, ?

Call the $n$th number in that sequence $a_{n}$. Is it always true that $a_{n+1}= 2a_{n}-1$? If not, what is the general formula?

Whoa, that is pretty awesome, but are you sure about the 25? I'm getting a much larger number, and I am taking into account concurrent lines and whatnot...

Just to clarify, I'm counting the number of distinct points. Otherwise, the problem is trivial. (Also, I used KSEG to draw my diagram, I didn't actually prove that all those lines were concurrent.)

Unfortunately, I stopped one too soon. The next number in the sequence is easily over 60, and seems to be around 100. (the diagram was fun though, sometimes there were like eight lines going through a single point!)

I guess I should have known - the sequence for the number of lines seemed to have no pattern at all.

Um, I can't install KSEG for this particular computer (some random non Qt compliance copyright issue), so I can't really check my diagram now. It's your word against my past-self's word, all of which is being transmitted via a series of tubes.

I think this is a very beautiful and hard problem: beautiful because the statament is so simple!

Try starting with four points that form a square, and then count also the points at infinity. Now concurrences and collinearities become clearer... but only for the first steps! Then it's only a mess  :rotfl:


number theory
greatest common divisor
linear algebra
matrix
Euler
superior algebra

For any integers $ a$ and $ b$, let $ \text{gcd}\left(a,b\right)$ denote the greatest common divisor of $ a$ and $ b$.

Let $ A\in\text{M}_{n}\left(\mathbb{Q}\right)$ be the $ n\times n$ matrix defined by $ a_{i,j}\equal{}\text{gcd}\left(i,j\right)$. Hereby, $ a_{i,j}$ means the entry of the matrix $ A$ in the $ i$-th row and the $ j$-th column.

Prove that $ \det A \equal{}\prod_{k \equal{} 1}^{n}\phi\left(k\right)$, where $ \phi:\mathbb{N}\to\mathbb{N}$ is the Euler totient function.

I was lasy to post a solution. I still am.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45[/url]

Let $f:\mathbb{N}\to\mathbb{C}$ be a function. Define $F:\mathbb{N}\to\mathbb{C}$ by $F(n)=\sum_{d|n}f(d)$.
Define $M_n(F)=(F(gcd(i,j)))_{i,j=\overline{1,n}}$. Then $\det M_n(F)=f(1)f(2)\ldots f(n)$.

Proof:
Let $a_{i,j}=\left\{\begin{array}{cc}1&\mbox{ if }j|i\\0&\mbox{ if }j\nmid i\end{array}\right.$
Let $A=(a_{i,j})_{i,j=\overline{1,n}}$ and $B=(a_{j,i}f(i))_{i,j=\overline{1,n}}$.

$A$ and $B$ are lower triangular matrices and $\det A=1$, $\det B=f(1)f(2)\ldots f(n)$.

Let $AB=(c_{i,j})$. Then $c_{i,j}=\sum_{k=1}^n a_{i,k}a_{j,k}f(k)$ and $a_{i,k}a_{j,k}f(k)=\left\{\begin{array}{cc}f(k)&\mbox{,if } k|i,\ k|j\\0&\mbox{,otherwise}\end{array}\right.$.
So $f(1)f(2)\ldots f(n)=\det A\cdot \det B=\det (c_{i,j})=\det(\sum_{k|gcd(i,j)} f(k))=\det (F(gcd(i,j)))$.

This proof is taken from an article of Emil Daniel Schwab published in Gazeta Matematica 9/1989.
My solution for the initial problem used a boring process of triangulation of $M_n(F)$, but it worked in some other general case described in the previous link.

[i]Proof.[/i] assume that $ B \equal{} [b_{ij}]$ such that  $ b_{ij}\equal{} 1$ if $ j|i$ and otherwise $ 0$ ,and $ \Phi \equal{} diag(\phi(1),\dots,\phi(n))$ then you can easily verify that $ B\Phi B^{T}\equal{} A$.QED

http://matematika.etf.bg.ac.yu/publikacije/pub/P13(02)/rrad01.pdf

I figured that this is a good source on reading up about determinant of gcd matricies


Let a,b,c be non-negative numbers such that a+b+c=3. Prove that
1) $abc+\frac{12}{ab+bc+ca}\ge5$
2) $abc+\frac{9}{ab+bc+ca}\ge4$

I don't think so  :wink:
Problem a) is in Vasc book pr. 47

$Solution:$ $a)$
By Shur
$(\sum{a})^{3}+9abc\geq 4(\sum{a})(\sum{bc})$ $\Rightarrow $ $9abc\geq 12\sum{ab}-27$ .
By $A(M)\geq G(M)$
$9abc+\frac{12*9}{\sum{bc}}\geq 12\sum{ab}+\frac{12*9}{\sum{bc}}-27\geq 5*9$ $\Rightarrow $

$abc+\frac{12}{\sum{ab}}\geq 5$

[quote="Little Gauss"]Let a,b,c be non-negative numbers such that a+b+c=3. Prove that
2) $abc+\frac{9}{ab+bc+ca}\ge4$[/quote]
I think, it's wrong: $a=b=1.2$ and $c=0.6.$ :wink:

I think, if $a,$ $b$ and $c$ are positive numbers such that $a+b+c=3$ then
$abc+\frac{3\alpha}{ab+ac+bc}\geq\alpha+1$ for all $\alpha\geq\frac{27+7\sqrt{21}}{18}=3.28211...$ :wink:


$ 1$, Given $ a, b, c > 0$. Prove that:$ \sqrt {\frac {a^2 \plus{} 2b^2}{a^2 \plus{} ab \plus{} bc}} \plus{} \sqrt {\frac {b^2 \plus{} 2c^2}{b^2 \plus{} bc \plus{} ca}} \plus{} \sqrt {\frac {c^2 \plus{} 2a^2}{c^2 \plus{} ca \plus{} ab}} \ge 3$

$ 2$, Given $ a, b, c > 0$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that:  $ \left(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\right)\left(\frac {1}{(1 \minus{} ab)^2} \plus{} \frac {1}{(1 \minus{} bc)^2} \plus{} \frac {1}{(1 \minus{} ca)^2}\right) \le \frac {9}{4}$

$ 3$, Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$.

Prove that:$ \sqrt {\frac {a^2 \plus{} ab \plus{} b^2}{ab \plus{} 2c^2 \plus{} 3}} \plus{} \sqrt {\frac {b^2 \plus{} bc \plus{} c^2}{bc \plus{} 2a^2 \plus{} 3}} \plus{} \sqrt {\frac {c^2 \plus{} ca \plus{} a^2}{ca \plus{} 2b^2 \plus{} 3}} \ge \frac {3}{\sqrt {2}}$
 :)

[quote="Abdek"]1.
After using AM_GM inequality We need only prove that :

$ \prod(a^2 \plus{} 2b^2) \ge \prod (a^2 \plus{} ab \plus{} bc)$

By Cs we have:

$ (b^2 \plus{} a^2 \plus{} b^2)(b^2 \plus{} c^2 \plus{} c^2) \ge (b^2 \plus{} ac \plus{} bc)^2$
$ (c^2 \plus{} b^2 \plus{} c^2)(c^2 \plus{} a^2 \plus{} a^2) \ge (c^2 \plus{} ab \plus{} ac)^2$
$ (a^2 \plus{} c^2 \plus{} a^2)(a^2 \plus{} b^2 \plus{} b^2) \ge (a^2 \plus{} bc \plus{} ab)^2$

multiplying these three inequalities we obtain the desired result ! :)

________________
Mharchi abdelmalek[/quote]
Nice solution  :P
P2:It's easy by Am-Gm :)
We have:$ (1 \minus{} xy)^2 \ge (1 \minus{} x^2)(1 \minus{} y^2)$
So:
$ LHS \le (\sum a^2b^2)(\sum \frac {1}{(1 \minus{} a^2)(1 \minus{} b^2)}) \equal{} \frac {2\sum a^2b^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)} \le \frac {9}{4}$

I just observe that the first inequality and the third one are essentially the same in ideas

For the third one, we have

\[ (a^2\plus{}c^2\plus{}(a\plus{}c)^2)(b^2\plus{}c^2\plus{}(b\plus{}c)^2) \ge (ab\plus{}2c^2\plus{}3)^2\]

We finish our proof here.

[quote="nguoivn"]$ 2$, Given $ a, b, c > 0$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that:  $ \left(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\right)\left(\frac {1}{(1 \minus{} ab)^2} \plus{} \frac {1}{(1 \minus{} bc)^2} \plus{} \frac {1}{(1 \minus{} ca)^2}\right) \le \frac {9}{4}$[/quote]
The following inequality is stronger:

If $ a,b,c$ are real numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}1,$ then
$ \frac{1}{(1\minus{}ab)^2}\plus{}\frac{1}{(1\minus{}bc)^2}\plus{}\frac{1}{(1\minus{}ca)^2} \le \frac{27}{4}.$


My custom CSS block is gone!!! When I try to replace it, it gets deleted again? My poor little bloggy looks so non-Billy now... ;)

Seriously, though, my custom CSS is gone and when I try to put new CSS in, after I click submit, it goes away again.

Complaint #2: What happened to my entry titles?

Complaint #3: I can't change the number of entries on each page of replies! It's stuck at 20... I've tried to change it to 21, 22, 30, etc. but it stays at 20. [But the blog option textbox stays at whatever I change it to...]

Yeah. For some reason, my titles keep disappearing! i have it set as black but it never shows!

Also: every time i try to change the hex code it never accepts it! (background colors)

Complaint 4: On the posts beyond the first page of replies, the original post shows up as a blank by Guest, not the actual post with me as the author. See say http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=145224&w=34&start=20 or the third page of replies.

Sorry for being so annoying :(...

I have the same problem. Even if I try to change the color of the titles, they do not appear on the blog.

Also, my CSS code is gone for some reason even though I did not touch it.

We might be having some problems with that CSS folder the CSS files are kept on. I will take a look at it, but I'm really working on the ratings right now, so you have to wait a couple more days guys  :blush:

Hi guys, i have a problem with my blog too...take a look: http://www.artofproblemsolving.com/Forum/weblog.php?w=672

why is the font of the tites so smaaaaaaall noone can read it...I tried changing them to Max in blog settings but no change... :( whats happening...my dear blog.... it was so beautiful... :(

OK, I've partially solved the problem. Try putting your own CSS now. Do not use the text block for scripts! :)

All of you which have their titles "invisible" have probably tried to edit the Custom CSS block and made some illegal declaration there. Edit it again, and make sure you have the correct file. Don't try putting Javascript or PHP code into it, as it will not be parsed, and your CSS will be corrupted.

[quote="Valentin Vornicu"]All of you which have their titles "invisible" have probably tried to edit the Custom CSS block and made some illegal declaration there. Edit it again, and make sure you have the correct file. Don't try putting Javascript or PHP code into it, as it will not be parsed, and your CSS will be corrupted.[/quote]

I don't think it helps but my custom CSS block is empty and I have never added one.

[quote="Valentin Vornicu"][quote="7h3.D3m0n.117"][quote="Valentin Vornicu"]Your blog looks fine to me ...[/quote]

Hmm. It doesn't look fine with me. I'm using Mozilla maybe it only works fine on IE?[/quote]I'm also using Mozilla Firefox (tested in IE 6, 7 too) - as a future notice, I always test stuff on these 3 browsers, Opera 9 and Firefox for Mac.

You must have the old css file in your cache. Go to the page and after it's loaded hit refresh (F5).[/quote]

Yup. It worked. Thank you!

[quote="solafidefarms"]So what is wrong in my block?

[code]#mainColumn{
background:none;
background-image:url(http://billydorminy.homelinux.com/bg.jpg);
background-repeat:repeat;
}

#a1{position:absolute;top:15px;left:300px;}
#webloginfo th{background:transparent;}
#topBG,#wrapper2,#wrapper{background:transparent; border:0 !important;}
#wrapper,.entryContent{width:80% !important;}
#mainColumn, #rightColumn, #header{background:transparent !important;}
#navigation_box{opacity:0.3;filter:alpha(100);}
#navigation_box :hover{opacity:0.8;filter:alpha(255);}
#header{border-top:0 !important;}[/code][/quote]

Ah yep, it was in my cache. Many, many, many thanks :).

No problem guys. I've fixed the contributors bug as well, and I'll see what else needs immediate fixing (I think it was editing entries?)

[quote="solafidefarms"]Can you check out the replies bugs too, where the second and later pages of replies don't have the original entry at the top, and where you can't change the number of replies on each page? Thanks![/quote]Yeah, I have just switched to some new stuff that needs urgent attention but I'll get back on this sometime next week (probably Tuesday).


find all $x,y,n\in \mathbb{N}$ such that :
${(x-y)}^n=xy$

salam   (x,y)=d   x=Xd  ,  y=Yd    d^n-2(X-Y)^n=XY  (X,Y)=1   (XY , X-Y)=1    X-Y=1    XY=d^n-2  xy=d^n              X=r^n-2   Y=s^n-2   ( r^n-2)  -(s^n-2) =1  n-2<2  n=3,n=2,n=1    x=Y(Y+1)^2  ,y=Y^2(y+1)

[color=red]Edit by Megus: It is sufficient to write without colors and enlarged fonts, don't you think?  :mad: [/color]


Evaluate: $ \left( \frac{1}{2} \plus{} \frac{1}{3} \right) \left( \frac{1}{2} \minus{} \frac{1}{3} \right)$

Or we can do difference of squares to find that $ \frac{1}{4} \minus{} \frac{1}{9} \equal{} \frac{5}{36}$

Yes but if you are in a hurry, which do you know faster, 1/2 +- 1/3 or 1/4-1/9?


geometry
3D geometry
algebra
function
domain

determine the triples of integers $(x,y,z)$ satifying the equation
$x^3+y^3+z^3=(x+y+z)^3$.

ok so i have simplified it to

$(x^3+y^3)=(x+y+z)^3-z^3$

$(x+y)(x^2+xy+y^2)=(x+y)(x^2+y^2+3z^3+2xy+3xz+3yz)$

$-xy=3z^2+3xz+3yz$

$3z^2+3z(x+y)+(xy)=0$

i was wondering if i have been doing this right and if i am can you please provide a hint about what to do next.
if i have not been  doing it correctly can you please point me in the right direction.

o and i already know that $(0,0,z),(x,0,0),(0,y,0),(0,0,0)$ all satisfy this. also if $x=-y,x=-z,y=-z$ then it is also satisfied.

[quote="maokid7"]determine the triples of integers $(x,y,z)$ satifying the equation
$x^3+y^3+z^3=(x+y+z)^3$.

ok so i have simplified it to

$(x^3+y^3)=(x+y+z)^3-z^3$

$(x+y)(x^2+xy+y^2)=(x+y)(x^2+y^2+3z^3+2xy+3xz+3yz)$

$-xy=3z^2+3xz+3yz$

$3z^2+3z(x+y)+(xy)=0$

i was wondering if i have been doing this right and if i am can you please provide a hint about what to do next.
if i have not been  doing it correctly can you please point me in the right direction.[/quote]

Your equation $(x^3+y^3)=(x+y+z)^3-z^3$ is fine.  The next equation should be
\[ (x+y)(x^2-xy+y^2)=(x+y)(x^2+y^2+3z^2+2xy+3xz+3yz). \]
After that, be very careful with your algebra.  Also, I recommend not dividing by $x + y$.

there is nothing special about x and z, we also know that (x+y) is a factor...this is actually a pretty common factorization

[quote="dondigo"]Notice, that for all $x,y,z$ $(x+y+z)^3-x^3-y^3-z^3=3(x+y)(x+z)(y+z)$
So if $(x+y+z)^3=x^3+y^3+z^3$, then $(x+y)(x+z)(y+z)=0$
The answer is $x=-y$ or $x=-z$ or $y=-z$[/quote]
how do you get that

For all real $x,y,z$
$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(x+z)(y+z)$
It's easy to check that it's true

thanks nsato for pointing out an error in my algebra  :P
and, thanks to all of you that helped.

o and nsato why do u recommend that i do not divide by $x+y$?

[quote="maokid7"]
 and nsato why do u recommend that i do not divide by $x+y$?[/quote].
if $x=-y$ then $x+y=0$ :huuh:

yes i know that, but would u be able to tell at that point that $x=-y$?

this is a common problem with factoring, say you have $x^2-4x+3=0$, when you factor, you get $(x-3)(x-1)=0$, you do not divide by $x-3$, what if $x=3$, that would be dividing by $0$

the same logic applies to your problem, you divide by $x+y$, but what if $x=-y$?, you are dividing by 0

ooh ok now i see y.
so instead of dividing i should just set the RHS equal to 0 and continue to factor.

On the dividing by (x+y), sure it equals 0 if x=-y, but can't we still divide by that to cancel, out, if we address that as a solution? ie, if I'm solving x^2+x=0, I could see from x(x+1) what the solutions are, but would I miss a solution/get a wrong answer if I address that x=0 is a solution (it's in the domain no matter what we do to our equation; plugging in yields 0) and then dividing by x to further solve the equation?

dondigo, the equality you supplied, how did you come up with that? Looks like you did some intense factoring, but you made it look really easy.

I believe stuff like that can be accomplished with an appropriate calculator  :)

i'm a big fan of the TI-89.....but dondigo is better than any calculator out there.

Such things like that are very useful as you propably know now, so they schould be known by heart :)


Find a NSC on $a,b$ positive such that $[\mathbb Q(\sqrt{a+2\sqrt{b}}:\mathbb Q]$ is
$2$

This is a bit awkward.. $t=\sqrt{a+2\sqrt b}$ has to be irrational and the root of a second degree equation with rational coefficients. Isn't this correct?


ARML USAMTS AMC USA(J)MO USAMO MATHCOUNTS

I'm wondering about how everyone found out about this forum. Having this information could help to make it more accessible to others.

I found it after a message was posted about it through one of the imo mailing lists, I forget which.

One of my e-mail buddies told me about it...I think it was Chinaboy (Martin) but I can't remember for sure.  Anyway, at that point I was googling all sorts of math problem sources, so I would have found it under the Google Sponsored Links, anyway (wow!  You type in "ARML, USAMTS, USAMO, or MATHCOUNTS" and it comes up!).

...as long as we're on the subject of "The Art of Problem Solving", I find it interesting to note that there is another book of the same name about a completely differen thing (I think).  It's at [url]http://www.amazon.com/exec/obidos/tg/detail/-/0471858080/002-6902264-8479254?vi=glance[/url]

There has been an announcement in one the Yahoo mailing lists [imo-problems] or [Math Olympiad] (don't remember which one).

By the way I recommend those two lists with a lot of nices and original problems.

I found it through a John Hopkins mailing list for people that get a certain score on the SAT before they're 13.

It wasn't me.  I don't even know you.

Besides, if you had meant me, you wouldn't have said "Top 2", you would have said "Top 10" or "Top One".  Top 2 would only be used if you were about to refer to the 2nd guy.

The reason I wrote top 2 is because they keep passing each other, so therefore ranks 1/2 would be in frequent flux. :)

[quote="Lord Venom"]I found out through one of the top 2 posters...it's not too hard to find out which one[/quote]

Oh, you must mean Simon!

Oops, didn't mean to say that.  But some people might still be puzzled over who Simon is...

[quote="rippledance"]If you read the french post in the Etc. forum, you would have immediately known who Lord Venom was referring to, Top 2 or not.[/quote]

If I remember correctly, Lord Venom wrote ``SRS'' in the French post and not Simon, so that makes it not quite so obvious, although I do post my last name in my signature. Then again, when I write my initials, I usually (but not always) use SR.

[quote="ComplexZeta"]If I remember correctly, Lord Venom wrote ``SRS'' in the French post and not Simon, so that makes it not quite so obvious, although I do post my last name in my signature. Then again, when I write my initials, I usually (but not always) use SR.[/quote]

Oh, I didn't even notice that.  Probably because SRS is also the abbreviation for my school, Sage Ridge School.

i had already been to the site before to order the books. then the father of one of my peons told me to get on the mathjam thingie. which happened to be the imo one. aargh. still, i instantly found my summer planned out for me


A cylinder-shaped tube has two different sections, in which two pistons of areas $ A_{1}>A_{2}$  are inserted. They are connected by a non-extensible thread, as shown in the figure below. The mass of the pistons and the thread together is $ m$.

There is air at normal atmospheric pressure in the external part of the tubes, which occupies a volume $ V_{0}$ at a temperature $ T_{0}$.

a) Find the pressure of the gas between the pistons.

b) Find the elevation $ \Delta h$ of the pistons produced by an increase $ \Delta T$ in the temperature of the gas, when the system reaches its equilibrium.

[img]http://img256.imageshack.us/img256/7011/prob10ri6.jpg[/img]

Solution. (Not sure if this is correct, but i will give it an attempt)

(a)Assume that the distance between the pistons is small, so the external air pressure is equal at both pistons.

Let this external air pressure be $ P_{e}$. Let the internal pressure be $ P_{i}$.

$ P_{i}\geq P_{e}$ because otherwise the pistons would be pushed further together by the external pressure.

The (upwards) force due to pressure on the upper piston is: $ F_{1}\equal{} (P_{i}\minus{}P_{e})A_{1}$.

The (downwards) force due to pressure on the lower piston is: $ F_{2}\equal{} (P_{i}\minus{}P_{e})A_{2}$

Both these forces tend to make the string more taught, and therefore these forces will act on both pistons.

The tension in the string cancels. So the only force left to consider is that due to gravity, which acts on the whole system.

Hence: $ F_{1}\equal{} F_{2}\plus{}mg$

$ \Rightarrow (P_{i}\minus{}P_{e})A_{1}\equal{} (P_{i}\minus{}P_{e})A_{2}\plus{}mg$

$ \Rightarrow P_{i}(A_{1}\minus{}A_{2}) \equal{} P_{e}(A_{1}\minus{}A_{2})\plus{}mg$

$ \Rightarrow P_{i}\equal{} P_{e}\plus{}\frac{mg}{A_{1}\minus{}A_{2}}$

(b)Let the new internal pressure of the gas be $ P_{j}$

By Kelvin's gas law:

 $ \frac{P_{j}}{T_{0}\plus{}\Delta T}\equal{}\frac{P_{i}}{T_{0}}$

$ \Rightarrow P_{j}\equal{}\frac{(T_{0}\plus{}\Delta T)P_{i}}{T_{0}}\equal{} P_{i}\plus{}\frac{\Delta T}{T_{0}}P_{i}$

We assume that the external pressure does not remain constant as the pistons move, but we assume the density of the surrounding air does. Let the external air pressure at equilibrium be $ P_{f}$.

$ \Rightarrow P_{j}\equal{} P_{f}\plus{}\frac{mg}{A_{1}\minus{}A_{2}}$

$ \Rightarrow P_{j}\minus{}P_{f}\equal{}\frac{mg}{A_{1}\minus{}A_{2}}\equal{} P_{i}\minus{}P_{e}$

$ \Rightarrow P_{j}\minus{}P_{i}\equal{} P_{f}\minus{}P_{e}$

This is confusing. $ P_{j}> P_{i}$ hence the new external pressure is greater than before, so the system has "sunk", so I have probably made a mistake.

$ \Rightarrow\frac{\Delta T}{T_{o}}P_{i}\equal{}\minus{}\rho g\Delta h$ where $ \rho$ is the density of the air.

$ \Rightarrow\Delta h \equal{}\minus{}\frac{\Delta T}{\rho g T_{0}}P_{i}$

$ \Rightarrow\Delta h \equal{}\minus{}\frac{\Delta T}{\rho g T_{0}}\left (P_{e}\plus{}\frac{mg}{A_{1}\minus{}A_{2}}\right) \equal{}\minus{}\frac{\Delta T}{\rho T_{0}}\left (\frac{P_{e}}{g}\plus{}\frac{m}{A_{1}\minus{}A_{2}}\right)$


In $ \triangle ABC$ points $ D , E , F$ are marked on sides $ BC , CA , AB$ respectively , so that

$ \frac {BD}{DC}$ $ \equal{}$ $ \frac {CE}{EA}$ $ \equal{}$ $ \frac {AF}{FB}$ $ \equal{}$ $ 2$

Show by a dissection argument that the triangle formed by the lines $ AD , BE , CF$ has an area $ \frac {1}{7}$ that of $ \triangle ABC$

[b]1).[/b]$ \mbox{Let } \{M\} = BE\cap CF; \ \{N\} = CF \cap AD; \ \{P\} = AD \cap BE,$
$ \mbox{ and if denote (XYZ) = area of }\triangle{XYZ}, \mbox{ then have: } \\
(BCF) = (ACD) = (ABE) = \frac {1}{3}.(ABC); \ \ (1)$
[b]2).[/b] $ \mbox{Apply the Menelaus theorem in } \triangle {BCF} \mbox{ at A - N - D, obtaind: }$
$ \frac {|NC|}{|NF|}.\frac {|AF|}{|AB|}.\frac {|DB|}{DC|} = 1\Rightarrow \frac {|NC|}{|NF|} = \frac {3}{4}\Rightarrow\frac {|NC|}{|FC|} = \frac {3}{7}\Rightarrow \\
\Rightarrow \frac {(DCN)}{(BCF)} = \frac {|CN|.|CD|}{|CF|.|BC|} = \frac {3}{7}.\frac {1}{3} = \frac {1}{7}\Rightarrow \\
\Rightarrow (APE)=(BMF)=(DCN)=\frac{1}{7}.(BCF)=\frac{1}{21}.(ABC). \ \ (2)$
[b]3).[/b]$ (1);(2)\Rightarrow (MNP)=$
${ =(ABC)-[(ABE)+(BCF)+(ACD)}+[(APE)+(BMF)+(DCN)]=\\
=\dots=\frac{1}{7}.(ABC).$

[quote="Rijul saini"]Hey can some one solve this
Even pure geometry can be used
please......i want an answer[/quote]

____________
this is my answer....  :wink:


function
calculus
derivative
invariant
real analysis
real analysis unsolved

Find all linear continuous maps of the space of $C^{\infty}$ real functions defined on $[-1,1]$ (with the uniform norm) that commute with the operator of derivation.

In order that the functions used be simpler, I'm going to assume that the interval $I$ we're working on is actually $[0,2\pi]$ instead of $[-1,1]$. The hypotheses don't change, since $[-1,1]$ can be carried onto $I$ through a diffeomorphism with constant derivative.

Let $T$ be the linear operator we're looking for. I claim that $T$ is multiplication by a constant.

By setting $T(f+ig)=Tf+iTg$, we can make $T$ a continuous linear operator on the [i]complex[/i] linear space $\mathcal C^\infty(I;\mathbb C)$ of infinitely differentiable complex-valued functions on $I$, normed with the supremum norm. The property of commuting with the derivative $D$ is clearly preserved by this procedure.

Civen a complex number $a$, let $e_{a}$ be the function on $I$ defined by $e_{a}(x)=e^{ax},\ \forall x\in I$. Since $De_{a}=ae_{a}$ and the only solutions to the equation $Df=af$ are the scalar multiples of $e_{a},\ T$ must carry $e_{a}$ onto a scalar multiple $c_{a}e_{a}\ (*)$ of $e_{a}$.

The linear space $\mathcal A$ spanned by $e_{ni},\ n\in\mathbb Z$ is a unital algebra which is self-adjoint, and separates all pairs of points except for the endpoints $0,2\pi$ of $I$. By Stone-Weierstrass, the closure $\overline{\mathcal A}$ of $\mathcal A$ in $\mathcal C^\infty(I;\mathbb C)$ is the set of functions which take equal values in the two endpoints of $I$. Since $\mathcal A$ is $T$-invariant, so is $\overline{\mathcal A}$. In other words, if $f(0)=f(2\pi)$, then $Tf(0)=Tf(2\pi)$. This means that $f(2\pi)-f(0)\mapsto Tf(2\pi)-Tf(0)$ is a well-defined linear map on $\mathbb C$, so it's of the form $z\mapsto cz$ for some constant $c$. In turn, this implies that all $c_{a},\ a\in\mathbb R$ appearing in $(*)$ are equal to $c$, and since the linear space spanned by $e_{a},\ a\in\mathbb R$ is dense in $\mathcal C^\infty(I;\mathbb C),\ T$ must be multiplication by $c$. $c$ is real, of course, because $T$ carries $\mathcal C^\infty(I;\mathbb R)$ into itself.


Compounds A and B on reaction in ether medium and subsequent acidification and oxidation give 2,5-dimethyl-3-hexanone. What are A and B?  :maybe:  :maybe:  :(

There are multiple possibilities, but reaction in ether medium (followed by acidification and oxidation, giving a ketone) calls to mind Grignard reagents. Write the structure of the product and make a retrosynthetic analysis: what alcohol is the immediate percursor of the ketone? What Grignard reagent and aldehyde/ketone should you mix in order to get that alcohol?

Is it CH3--CHMgBr--CH3 and 3-methyl butanal ??  :maybe:  :maybe:  :maybe:

That's right. Is there another possibility?


One invention saves 30% on fuel; a second, 45% ; and a third, 25% . If you use all three inventions at once can you save 100% ? If not, how much ?

[size=75][i]from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide[/i]
[/size]

 :rotfl:

The first invention lets you only use 70% of the fuel you usually need to use. The next allows you to use 55% of the fuel you need to use (55% of the 70%), etc. You end up using about 29%, (actually about 28.8 something). Subtracting this from 100% gives the amount you save, or about 71%.

Think of it like a discount at a store. If there is a shirt for 50 dollars, and it's on sale for 80% of the original price, it's sold for 40 dollars. If they put a 20% sale on this, it is once again 80% of the 40 dollar price, or 32 dollars. As a shortcut, you can do $ (80%)(80%)=(.8)(.8)=.64
$ If we multiply this $ .64$ by our original price, 50, we get $ (.64)(50)=32$ Just as we found out before, by doing one discount after another.

Hope this helps :lol:

wow. good job solving that, u helped me a lot !!!! :D

:clap:  :10:

Wow! Nice job explaining the problem AIME15USAMO!
Seriouly, Nice job!

No, of course you can not get 100% savings.

Multiply each of the savings by 100% and add them together. Subtract the grand total by 100%, and get the answer, being what the actual savings is.

Use common sense, too. Can you really get 100% savings on something. If something is 50%, and then 50%, its 50% OF the 50% that it was. Sorry for confusing you with the example. Hope that helped.

According to one theory, if you stand twenty feet from a wall, and walk halfway each time, in theory you will never reach the wall. If you use these inventions, you will always use this:

Say you save 50% with one 25% with another.

You would save 25% of the 50% you already saved.

Guys, no need to revive a topic that's over a month old.

@Soccerblls2   well, what you say is correct. But I think what the question needs is the % of fuel you would save if you use [b]one of each[/b] rather than [i]some[/i] of each.

@pascal i think that not mentioning "over a month old" wouldve stopped it  :|

hey pascal by posting to people to stop replying, u are just inviting them to keep replying to your reply. if u were to just leave the post alone, then people would probably just stop posting naturally.


function
real analysis
real analysis unsolved

Please give me the solutions of these problems below:
1/Let $ a>0$ and $ a\not\equal{}1$.Find,with proof,all the functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
$ f(ax)\equal{}f(x)\plus{}2f(\minus{}x),x\in\mathbb{R}$
2/1/Let $ a>0$ and $ a\not\equal{}1$.Find,with proof,all the functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
$ f(ax)\equal{}f(x)\plus{}f(\minus{}x),x\in\mathbb{R}$
3/1/Let $ a>0$ and $ a\not\equal{}1$.Find,with proof,all the functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
$ f(ax)\equal{}1,x\in\mathbb{R}$

do you mean... continuous functions? differentiable functions? onto functions?


Find $ x,y,z,t,u,v$ satisfying 4 condition:
a)
$ x^3\plus{}z\plus{}u\equal{}c_1$
$ y^3\plus{}t\plus{}v\equal{}c_2$
$ 3xy^2\minus{}2v\plus{}u\equal{}c_3$
$ 3x^2y\minus{}2u\plus{}v\equal{}c_4$

b)
$ x^3\plus{}z\minus{}u\equal{}c_1$
$ y^3\plus{}t\minus{}v\equal{}c_2$
$ 3xy^2\plus{}2v\minus{}u\equal{}c_3$
$ 3x^2y\plus{}2u\minus{}v\equal{}c_4$

Where $ c_i$ is given constants

[quote="hungnsl"]Find $ x,y,z,t,u,v$ satisfying 4 condition:
a)
$ x^3 \plus{} z \plus{} u \equal{} c_1$
$ y^3 \plus{} t \plus{} v \equal{} c_2$
$ 3xy^2 \minus{} 2v \plus{} u \equal{} c_3$
$ 3x^2y \minus{} 2u \plus{} v \equal{} c_4$

b)
$ x^3 \plus{} z \minus{} u \equal{} c_1$
$ y^3 \plus{} t \minus{} v \equal{} c_2$
$ 3xy^2 \plus{} 2v \minus{} u \equal{} c_3$
$ 3x^2y \plus{} 2u \minus{} v \equal{} c_4$

Where $ c_i$ is given constants[/quote]

For [b]a)[/b] and [b]b)[/b], given any fixed $ x$ and $ y$ , we can always find $ z,t,u,v$ satisfies the conditions. Then we can just solve it like solving a system of linear equations.

Sorry. i forgot the most important condition:

$ x,y,z,t,u,v \geq 0$

i think it's not very easy  :maybe:


[quote="Dimitris X"][quote="Xaenir"]If $ \alpha ,\beta$ be the roots of $ ax^2 \minus{} bx \plus{} c$ such that $ a,b,c \in N$ and$ \alpha ,\beta \in (0,1)$ Prove that $ abc > 25$[/quote]
Are you sure for this???
$ \alpha \plus{} \beta \equal{} \frac {b}{a} < 2$
And $ \alpha \cdot \beta \equal{} \frac {c}{a} < 1$
So we have that:
$ b < 2a$.And
$ c < a \boxed1$
By the first we extract that $ b^2 < 4a^2 \boxed 2$
If we divide $ \boxed 2,\boxed 1$,we have $ \frac {b^2}{c} < \frac {4a^2}{a} \Leftrightarrow b^2 < 4ca$
So the discriminant $ < 0$ so we have no real roots  :wink:[/quote]
[quote]
So we have that:
$ b < 2a$.And
$ c < a \boxed1$
By the first we extract that $ b^2 < 4a^2 \boxed 2$
If we divide $ \boxed 2,\boxed 1$,we have $ \frac {b^2}{c} < \frac {4a^2}{a}$"[/quote]

I think your inequalities cannot hold invoved such division :

$ p< q , A<B \implies  {A\over p } < {B\over Q}$ is non logical.
e.g. $ 1< 4 , 3 < 8  \implies {3\over 1 } < { 8\over 4}$ is wrong!!

[quote="Xaenir"]If $ \alpha ,\beta$ be the roots of $ ax^2 \minus{} bx \plus{} c$ such that $ a,b,c \in N$ and$ \alpha ,\beta \in (0,1)$ Prove that $ abc > 25$[/quote]

$ (a,b,c)\equal{}(5,5,1)$ satisfies the condition.

[quote="Dimitris X"][quote="Xaenir"]If $ \alpha ,\beta$ be the roots of $ ax^2 \minus{} bx \plus{} c$ such that $ a,b,c \in N$ and$ \alpha ,\beta \in (0,1)$ Prove that $ abc > 25$[/quote]
Are you sure for this???
$ \alpha \plus{} \beta \equal{} \frac {b}{a} < 2$
And $ \alpha \cdot \beta \equal{} \frac {c}{a} < 1$
So we have that:
$ b < 2a$.And
$ c < a \boxed1$
By the first we extract that $ b^2 < 4a^2 \boxed 2$
If we divide $ \boxed 2,\boxed 1$,we have $ \frac {b^2}{c} < \frac {4a^2}{a} \Leftrightarrow b^2 < 4ca$
So the discriminant $ < 0$ so we have no real roots  :wink:[/quote]
Sorry!!!I'm completely wrong :oops:
What a careless mistake :wallbash_red:
$ c < a \Leftrightarrow \frac {1}{c} > \frac {1}{a}$...

[quote="stephencheng"][quote="Xaenir"]If $ \alpha ,\beta$ be the roots of $ ax^2 \minus{} bx \plus{} c$ such that $ a,b,c \in N$ and$ \alpha ,\beta \in (0,1)$ Prove that $ abc > 25$[/quote]

$ (a,b,c) \equal{} (5,5,1)$ satisfies the condition.[/quote]

So does $ (4,4,1)$, unless $ \alpha$ and $ \beta$ are distinct

[quote="stephencheng"][quote="Xaenir"]If $ \alpha ,\beta$ be the roots of $ ax^2 \minus{} bx \plus{} c$ such that $ a,b,c \in N$ and$ \alpha ,\beta \in (0,1)$ Prove that $ abc > 25$[/quote]

$ (a,b,c) \equal{} (5,5,1)$ satisfies the condition.[/quote]
No...
The ineq is strict :wink:
Ocha you are wrong too...

I found a really ugly solution:
$ *$ For $ c \geq2$,we have that
$ \alpha \beta \equal{} \frac {c}{a} < 1\Leftrightarrow c < a\Leftrightarrow a\geq c \plus{} 1$
Also by the discriminant $ b^2 \geq 4ca > 4c \cdot c$
So $ b > 2c \Leftrightarrow b\geq 2c \plus{} 1$
So we have to prove that $ (c \plus{} 1)(2c \plus{} 1)c > 25$,which is true for all $ c\geq 2$

$ *$For $ c \equal{} 1$
we have that $ b\geq 2c \plus{} 1 \Leftrightarrow b\geq3$ and $ a\geq 2$
And now we take cases:
$ 1)$ for $ a \equal{} 2,b \equal{} 3,c \equal{} 1$
$ 2)$ for $ a \equal{} 3,b \equal{} 3,c \equal{} 1$
$ 3)$ for $ a \equal{} 4,b \equal{} 3,c \equal{} 1$
$ 4)$ for $ a \equal{} 4,b \equal{} 4,c \equal{} 1$
$ 5)$ for $ a \equal{} 5,b \equal{} 4,c \equal{} 1$
$ 6)$ for $ a \equal{} 5,b \equal{} 5,c \equal{} 1$
I havent check these values,but when we applying them the solutions $ \notin (0,1)$...(but it must be $ \alpha \not \equal{} \beta$,and the inequality is not strict because the triple $ (5,5,1)$ satisfies the condition)
So the inequality holds....
Please check it


I use Kile, and in one of my documents, I got the following message: ./MNT:103:Underfull \hbox (badness 10000) in paragraph

What does that mean? It didn't seem to have any effect in the document.

It's not an error message but a warning. The difference is that an error message will affect your document but a warning is drawing your attention to a possible problem, but isn't serious.
An underfull \hbox is where LaTeX has filled a line to make it both left and right justified but has put in what it considers to be too much spacing perhaps due to a wrongly placed \\. The smaller the badness the less is this extra spacing.
So carefully look at the paragraph it complains about and see how it looks. If you can't see anything wrong and cannot find any wrong new lines, then ignore the warning. Of all the warnings it is the least important.


A space diagonal of a cube is $ 10\sqrt3$ inches long. How many square inches are in the surface area of the cube?

The space diagonal is the edge multiplied by $ \sqrt3$.  So the edge length is $ 10$, and the SA is $ \boxed{600}$.

In other words, the formula for getting the surface area of a [b]cube[/b] is $6\cdot x^2$, where $x$ is the side length of the cube. The formula for getting the space diagonal of a cube is $x\sqrt3$. Since we know the space diagonal but not the side length, we can reverse this formula to get $x\sqrt3=10\sqrt3$ or $x=10$. Since we know that $6\cdot x^2=$ surface area, we plug in $10$ for $x$. Therefore, $6(10)^2=6(100)=\boxed{600}$.


Let $ a,b,c\in \left[0,1\right]$ and $ k\in \left[0,3\right]$. Prove or disprove that:

$ (1\minus{}a^k)(1\minus{}b^k)(1\minus{}c^k)\le (1\minus{}abc)^k$

[quote="dduclam"]Let $ a,b,c\in \left[0,1\right]$ and $ k\in \left[0,3\right]$. Prove or disprove that:

$ (1 \minus{} a^k)(1 \minus{} b^k)(1 \minus{} c^k)\le (1 \minus{} abc)^k$[/quote]
It is "prove", my friend. To prove this inequality, firstly, we will show that for all $ x,y,z \in [0,1]$, then
\[ (1\minus{}x^3)(1\minus{}y^3)(1\minus{}z^3) \le (1\minus{}xyz)^3\]
\[ \Leftrightarrow x^3\plus{}y^3\plus{}z^3\minus{}3xyz \ge x^3y^3\plus{}y^3z^3\plus{}z^3x^3\minus{}3x^2y^2z^2\]
Assume that $ 1 \ge x \ge y \ge z$, then we have
\[ x^3\plus{}y^3\plus{}z^3\minus{}3xyz \ge x^3(x^3\plus{}y^3\plus{}z^3\minus{}3xyz)\]
It suffices to prove that
\[ x^3(x^3\plus{}y^3\plus{}z^3\minus{}3xyz) \ge x^3y^3\plus{}y^3z^3\plus{}z^3x^3\minus{}3x^2y^2z^2\]
\[ \Leftrightarrow x^6\minus{}y^3z^3 \minus{}3x^2yz(x^2 \minus{}yz) \ge 0\]
\[ \Leftrightarrow (x^2\minus{}yz)^3 \ge 0\]
which is true.
Now, using this inequality, we have
\[ (1\minus{}a^k)(1\minus{}b^k)(1\minus{}c^k) \le (1\minus{}t^k)^3\]
where $ t\equal{}\sqrt[3]{abc}$.
And all we have to do now is to show that
\[ (1\minus{}t^k)^3 \le (1\minus{}t^3)^k\]
The rest is easy!
We have done. :)

Nice, Can! :blush:

[quote="can_hang2007"]To prove this inequality, firstly, we will show that for all $ x,y,z \in [0,1]$, then
\[ (1 \minus{} x^3)(1 \minus{} y^3)(1 \minus{} z^3) \le (1 \minus{} xyz)^3
\] [/quote]
This inequality also is starting point when I think to above inequality.  :)

[quote="FelixD"]I did it with the AM-GM inequalitiy. Then
$ (1 \minus{} a^k)(1 \minus{} b^k)(1 \minus{} c^k) \leq \left( 1 \minus{} \frac {a^k \plus{} b^k \plus{} c^k}{3} \right)^3 \leq \left(1 \minus{} (abc)^{\frac {k}{3}}\right)^3$
So it suffices to prove that
$ (1 \minus{} x^k)^3 \leq (1 \minus{} x^3)^k$
for $ x \equal{} \sqrt [3]{abc}$ and that's really easy, like Can mentioned before.[/quote]
Very nice and short solution indeed! Can you prove $ (1 \minus{} x^k)^3 \leq (1 \minus{} x^3)^k$ for me ? :)

We have $ a,b,c \in [0,1]$ and therefore also $ \sqrt [3]{abc} \equal{} x \in [0,1]$.
Then $ 1 \minus{} x^k \leq 1 \minus{} x^3 \Leftrightarrow x^{3 \minus{} k} \leq 1$ and thats obviously true. Thus,
$ (1 \minus{} x^k)^3 \leq (1 \minus{} x^3)^3 \leq (1 \minus{} x^3)^k$ and the last inequalitiy is true, because $ (1 \minus{} x^3)^{3 \minus{} k} \leq 1$.
That's it.  :wink:

[quote="FelixD"]I did it with the AM-GM inequality. Then
$ (1 \minus{} a^k)(1 \minus{} b^k)(1 \minus{} c^k) \leq \left( 1 \minus{} \frac {a^k \plus{} b^k \plus{} c^k}{3} \right)^3 \leq \left(1 \minus{} (abc)^{\frac {k}{3}}\right)^3$
So it suffices to prove that
$ (1 \minus{} x^k)^3 \leq (1 \minus{} x^3)^k$
for $ x \equal{} \sqrt [3]{abc}$ and that's really easy, like Can mentioned before.[/quote]

Nice, FelixD  :lol:


When you dissolve salt, for example, in water, the volume of the solution is the same as the initial volume of the solvent (unless supersaturated). If you were to dissolve a liquid in water (rubbing alcohol, perhaps), would the volume of the solution still be the same as the initial volume of water?

Thanks for your help.

In general, when you dissolve *anything* in a solvent, a solid or a liquid, the final volume is not the same as the initial volume of solvent. This because 1) solvent-solvent intermolecular forces are replaced in part by cation/anion-solvent or solute-solvent interactions, and these forces have unequal strenghts, and 2) there are differences between the packing of molecules in the final solution and in the pure components (solvent and solute(s)), due to differences in sizes and shapes of the molecules/ions in the mixture.

This is the reason I ask:

I have dissolved various substances into water. I add enough to float an object with density about 1.1 kg/L. For some of them, like MgSO4 $ \bullet$ 7H2O, the mass of the solution is about 1.19 times the mass of the original water. For others, like NaCl, the mass of the solution is about 1.07 times the original weight of the water and has a density of 1.1. Is it possible for the volume to decrease due to the changes in the charges? Is it possible to predict whether the volume will increase or decrease?

Sorry for sounding like such a n00b  :blush: .


[quote="BanishedTraitor"]For alcohols the only useful haloform reaction is the $ \text{Iodoform Test}$. Secondary alcohols that contain $ \minus{} CH(OH)CH_3$ as part of the structure are oxidized by iodine to the $ \minus{} COCH_3$ structure.

Personally I thought $ \alpha$ carbon is a concept attached to the presence of carbonyl groups. So I would say no, it is only necessary for the alcohol to have the structure I described above.

P.S. Actually, I believe the Iodoform Test is also valid for ethanol.[/quote]
Any methyl ketone containing group can undergo haloform especially iodoform reaction i believe!!! :ninja:  :D  :maybe:

But we were talking about alcohols undergoing haloform reactions.

[quote="Pardesi"]Is it necessary for a Alcohol showing haloform Reaction to have atleast a hydrogen attached to the alpha Carbon atom[/quote]

For an alcohol to show the haloform reaction it must necessarily have an alpha carbon with [i]three[/i] hydrogens. Hence the general structure given by BanishedTraitor (and yes, ethanol also gives the test). The reason for this lies in the mechanism:

1) $ R\minus{}CHOHCH_3 \plus{} X_2/NaHO \longrightarrow R\minus{}COCH_3$

2) $ R\minus{}COCH_3 \plus{} X_2/NaHO \longrightarrow R\minus{}COCH_2X$

3) $ R\minus{}COCH_2X \plus{} X_2/NaHO \longrightarrow R\minus{}COCHX_2$

4) $ R\minus{}COCHX_2 \plus{} X_2/NaHO \longrightarrow R\minus{}COCX_3$

5) $ R\minus{}COCX_3 \plus{} HO^\minus{} \longrightarrow RCO_2H \plus{} X_3C^\minus{}$

6) $ RCO_2H \plus{} X_3C^\minus{} \longrightarrow RCO_2^\minus{} \plus{} X_3CH$

where one can see that for the halyde carbanion to depart it must have three halogens - this because the carbanion is only stable enough to act as a leaving group when it have three halogens that can delocalize the negative charge.

ok if that's the meaning of alpha carbon then i din't meant  methyl group
i meant the carbon directly attached to OH
should it necessarily have a hydrogen

Of course it must have an hydrogen, look the mechanism I posted above! In the first step there is an oxidation of the alcohol to the methyl ketone, and that can only happen if there is at least one hydrogen on the carbon bearing the -OH group.

hey carcul, the first alpha halogenated intermediate dat you get, isnt dat the intermediate obtained in favorskii rearrangement too. den y cant favorskii reaction be a competing reaction, especially when dere is a base too.

Yes, the alpha-halo ketone is an intermediate in the Favorskii reaction. However, the haloform pathway is favoured because those methyl hydrogens are specially acidic and the step where the trihalo carbanion departs is essentially irreversible.

Sorry for reviving old threads. But can you give a mechanism as to why ethanol shows iodoform??

are all alcohols oxidised by the above mixture(primary and secondary)? or is it a specific reaction? :)

Other secondary alcohols are also expected to be oxidized, but they don't undergo haloform.

[quote="Cyber-Shreyas"][quote="Carcul"]Ethanol (and the other methyl alcohols) are first oxidized to a methyl ketone by the mixture halogen/base. The remaining mechanism is as I gave above.[/quote]

Why is that ethanol alone gets oxidized and not propanol or the like??[/quote]
Shreyas propanol also is oxidised by $ \ce{X_2}$ / $ \ce{NaOH}$ to propanal because halogens are mild oxidising agents. So ur statement there is wrong.

Then the product would be propanoic acid, since aldehydes are more easily oxidized than alcohols.


calculus
integration
trigonometry
Euler
function
parameterization
linear algebra

$I = \int {e^{ax} \cdot\cos \left( {bx} \right)dx = } \frac{1}{a}\int {\left( {e^{ax} } \right)^\prime \cdot\cos \left( {bx} \right)dx = }$

$= \frac{1}{a} \cdot \left[ {e^{ax} \cdot\cos \left( {bx} \right) - \int {e^{ax} \cdot\left( {\cos bx} \right)^\prime dx} } \right] =$

$= \frac{1}{a} \cdot \left[ {e^{ax} \cdot \cos \left( {bx} \right) + \int {e^{ax} \cdot b \cdot \sin \left( {bx} \right)dx} } \right] =$

$= \frac{1}{a} \cdot e^{ax} \cdot \cos \left( {bx} \right) + \frac{b}{a} \cdot \int {e^{ax} \cdot \sin \left( {bx} \right)dx} =$

$= \frac{1}{a} \cdot e^{ax} \cdot \cos \left( {bx} \right) - \frac{b}{{a^2 }} \cdot \int {\left( {e^{ax} } \right)^\prime \cdot \sin \left( {bx} \right)dx} =$

$= \frac{1}{a} \cdot e^{ax} \cdot \cos \left( {bx} \right) + \frac{b}{{a^2 }} \cdot \left[ {e^{ax} \cdot \sin \left( {bx} \right) - \int {e^{ax} \cdot b \cdot \cos \left( {bx} \right)dx} } \right] =$

$= \frac{1}{a} \cdot e^{ax} \cdot \cos \left( {bx} \right) + \frac{b}{{a^2 }} \cdot e^{ax} \cdot \sin \left( {bx} \right) - \frac{{b^2 }}{{a^2 }} \cdot I \Rightarrow$

$\Rightarrow I = \frac{{e^{ax} }}{{a^2 + b^2 }}\left[ {a \cdot \cos \left( {bx} \right) + b \cdot \sin \left( {bx} \right)} \right] + \mathcal{C}$

Or we can use euler formula $e^{\theta i} = \cos \theta + i \, \sin \theta$.
\begin{eqnarray*} \int e^{ax} e^{bxi} \, dx &=& \int e^{ax} (\cos bx + i \sin bx) \, dx \\ &=& \int e^{ax} \, \cos bx \, dx + i \int e^{ax} \, \sin bx \, dx \\ \end{eqnarray*}
And also
\begin{eqnarray*} \int e^{ax} e^{bxi} \, dx &=& \int e^{(a+bi)x} \, dx \\ &=& \left(\frac{1}{a+bi} \right) \, e^{(a+bi)x} + C \\ &=& \left(\frac{a-bi}{a^2 + b^2}\right) \, (e^{ax}) (e^{bxi}) + C \\ &=& \left(\frac{ e^{ax}}{a^2 +b^2} \right) (a-bi) ( \cos bx + i \sin bx ) + C \\ &=& \left(\frac{e^{ax}}{a^2+b^2}\right) (a \cos bx + b \sin bx + (a \sin bx - b \cos bx) i ) + C \\ &=& \frac{e^{ax}( a \cos bx + b \sin bx) + e^{ax} (a \sin bx - b \cos bx) i }{a^2 +b^2} + C \\ &=& \frac{e^{ax}( a \cos bx + b \sin bx)}{a^2 +b^2} + C_1 + \left(\frac{e^{ax}( a \cos bx + b \sin bx)}{a^2 +b^2} +C_2 \right) i \end{eqnarray*}
Where we write $C=C_1 + C_2 i$. So from that two result and matching the real part with other real part and imaginary part with other imaginary part we have \[ \int e^{ax} \cos bx \, dx = \frac{e^{ax}( a \cos bx + b \sin bx)}{a^2 + b^2} + C_1  \] \[ \int e^{ax} \sin bx \, dx = \frac{e^{ax}( a \cos bx - b \sin bx)}{a^2 + b^2} + C_2  \]

Consider

\[ (e^{ax}\cos bx)'=\cdots\cdots \]

\[ (e^{ax}\sin bx)'=\cdots\cdots \]

Think of this as a nonhomogeneous linear ordinary differential equations:

$y'=e^{ax}\cos bx$

The corresponding homogeneous equation is $y'=0,$ which has the solution $y=C.$ We'll add that in the end.

The nonhomogeneity - the right hand side - belongs to that class of functions for which the method of undetermined coefficients is appropriate. The guidelines for that method now suggest that we make the following guess, with adjustable parameters $A$ and $B$:

$y=Ae^{ax}\cos bx+Be^{ax}\sin bx$

Now we differentiate this (I'm pretty sure this is what kunny was getting at in his last post):

$y'=A(ae^{ax}\cos bx-be^{ax}\sin bx)+B(ae^{ax}\sin bx +be^{ax}\cos bx)$

$=(aA+bB)e^{ax}\cos bx+(-bA+aB)e^{ax}\sin bx$

Equating this to the right hand side of the original equation, and using the fact that $e^{ax}\cos bx$ and $e^{ax}\sin bx$ are linearly independent functions, we get

$aA+bB=1$
$-bA+aB=0.$

From this we get that $A=\frac{a}{a^2+b^2},\,B=\frac{b}{a^2+b^2},$ leading us to the same answers everyone else has gotten.

One look at that $2\times 2$ matrix and we realize that we are in effect talking about complex numbers whether we intended to or not; the method is closely allied to what beginner said above.

The calculus books I know would direct you to the method that metru used, but there's room for all of these other methods as well.

---

A slightly stiffer challenge along the same lines. I'll make it a definite integral, which will actually makes things slightly easier. Suppose that $a>0.$ Compute:

$\int_0^{\infty}xe^{-ax}\cos bx\,dx.$

$\int_0^{\infty}xe^{-ax}\cos bx\,dx = \Re \int_0^{\infty}x e^{(-a + bi)x} \, dx =$

Then we can integrate by parts.

$\Re [\frac{x}{-a+bi} e^{(-a + bi)x} |_0^{\infty} - \int_0^{\infty} \frac1{-a+bi} e^{(-a + bi)x} \, dx] =$

$\Re \frac{-a-bi}{a^2+b^2} \frac{x}{e^{ax}} (\cos (bx)+i \sin (bx)) |_0^{\infty} - \Re \frac1{(-a+bi)^2} \frac{\cos(bx)+i \sin (bx)}{e^{ax}}|_0^{\infty} =$

$[\frac{-a}{a^2+b^2} \frac{x \cos (bx)}{e^{ax}} + \frac{b}{a^2+b^2} \frac{x \sin(bx)}{e^{ax}} - \frac{a^2-b^2}{(a^2+b^2)^2} \frac{\cos (bx)}{e^{ax}} + \frac{2ab}{(a^2+b^2)^2} \frac{\sin (bx)}{e^{ax}}]|_0^{\infty} =$

$\frac{a^2-b^2}{(a^2+b^2)^2}$

I hope I did that right, and I don't think the final answer simplifies anymore.

Is that right Dr. Merryfield?

That is correct, FMako, and that is precisely the method I would have used.

What do you mean by completely?  Are these solutions incomplete?

We should take into consideration of the signs of $a,b$.

Everyone's answer is correct no matter what the sign of $b$ is (assuming it to be real). The only place where the sign of $a$ matters in in my supplemental problem, which does require that the integral converges; in my statement of that problem, I specified the sign in such a way as to assure convergence.

I really don't see what kunny is concerned about.

The complete answer is the following.

For $a=b=0,\ x+Const.$

For $a=0,b\neq 0,\ \frac{1}{b}\sin bx+Const.$

For $a\neq 0,b=0,\ \frac{1}{a}e^{ax}+Const.$

For $ab\neq 0,\ \frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+Const.$

kunny

We would assume that $a,b\neq 0$ for it will be simply too easy.
Let $I=\int e^{ax}\cos bx \ dx$.
By integration by parts, we have

$\\u=e^{ax}\hspace{.3in}dv=\cos bx \ dx\\ du=ae^{ax} \ dx\hspace{.3in}v=\frac{1}{b}\sin \ bx$

Hence,
$\\I=\int e^{ax}\cos bx \ dx\\ =\frac{1}{b}e^{ax}\sin bx-\frac{a}{b}\int e^{ax}\sin bx \ dx$

Using integration by parts again using

$\\u=e^{ax}\hspace{.3in}dv=%Error. "sinbx" is a bad command.
\ dx\\ du=ae^{ax} \ dx\hspace{.3in}v=-\frac{1}{b}\cos \ bx$

$\\\frac{1}{b}e^{ax}\sin bx-\frac{a}{b}\int e^{ax}\sin bx \ dx\\ =\frac{1}{b}e^{ax}\sin bx-\frac{a}{b}\left(-\frac{1}{b}e^{ax}\cos \ bx+\frac{a}{b}\int e^{ax}\cos \ bx \ dx\right)\\ \implies I=\frac{1}{b}e^{ax}\sin bx+\frac{a}{b^2}e^{ax}\cos \ bx-\frac{a^2}{b^2}I\\ I=\int e^{ax}\cos bx \ dx=\frac{be^{ax}\sin bx+ae^{ax}\cos \ bx}{a^2+b^2}+C$

EDIT: I agree with kunny, since some situations such as $a=b=0$ does not satisfy our results.

Masoud Zargar

If $a=b=0$, then taking the limit of the RHS that is in the form $0/0$. Do you know that we don't just plug $a$ and $b$ to that integral, we actually use limit. It deal with uniform convergence.


ratio
geometry
trapezoid
circumcircle
geometric transformation
reflection
complex numbers

For any a not 0, show that the points

a
-conj a
1/a
-1/conj a
1
-1

are cocyclic.

Please help!

Complex numbers $a,b,c,d$ are concyclic iff $\frac{b-a}{b-d}\cdot \frac{c-a}{c-d}\in R$, in other words $\frac{b-a}{b-d}\cdot \frac{c-a}{c-d}= \overline{\frac{b-a}{b-d}\cdot \frac{c-a}{c-d}}$. I hope this works, and I hope I didn't make any mistakes...

Thanks for the suggestion, but I've tried the cross-ratio method with no success. I have also tried using Ptolemy’s Theorem, but that doesn't seem to work either. :( Any other ideas?

Of course the cross ratio method works. $a,$ $-\bar a$ are symmetric wrt the imaginary axis, $(-1,+1,a,-\bar a)$ is isosceles trapezoid, cyclic, $-\bar a$ is on circumcircle of the $\triangle (-1,+1, a).$ For $(-1,+1,a,1/a),$

$\frac{-1-a}{-1-\frac{1}{a}}\cdot \frac{1-\frac{1}{a}}{1-a}= \frac{-1-a}{-a-1}\cdot \frac{a-1}{1-a}=-1$

The cross ratio is real, $1/a$ is also on circumcircle of the $\triangle (-1,+1, a),$ same as circumcircle of the $\triangle (+1,+1, 1/a).$  $1/a,$ $-1/\bar a$ are symmetric wrt the imaginary axis, $(-1,+1,1/a,-1/\bar a)$ is isosceles trapezoid, cyclic, $-1/\bar a$ is on circumcircle of the $\triangle (-1,+1, 1/a).$

Geometric interpetation is this: $a' =-1/\bar a$ is inversion with negative power (ordinary inversion followed by reflection in the inversion center) of $a$ in the unit circle $\mathcal U$ (because $z' =-1/\bar z$ carries $\mathcal U$ to itself, +1 to -1, and -1 to +1). Circumcircle $\mathcal C$ of the $\triangle (-1,+1, a)$ cuts the inversion circle $\mathcal U$ in diametrically opposite points, hence the inversion carries $\mathcal C$ into itself, so that $a' =-1/\bar a \in \mathcal C.$ Now use the isosceles trapezoids $(-1,+1,a,-\bar a)$ and  $(-1,+1,1/a,-1/\bar a)$.


keith weighs 20kg more than beth while henery weighs 30kg less than twice as much as beth. if keith and henry weigh the same  hoe much does beth weigh

Please stop flooding the forum.

If you're ignoring my PMs, I will have your threads deleted - regardless whether it's homework or not.

$ k\equal{}b\plus{}20$
$ h\equal{}2b\minus{}30$

If $ k\equal{}h$, then $ b\plus{}20\equal{}2b\minus{}30$

$ 50\equal{}b$

Beth is 50 kg.


One elevator leaves the $ 1^{\text{st}}$ floor of a building at exactly 9 a.m.  It takes five seconds to travel between floors, stays 10 seconds at each floor and stops at each floor on its way to the top.  Another elevator leaves the $ 12^{\text{th}}$ floor (the top floor) at exactly 9 a.m., takes five seconds to travel between floors, stays 15 seconds at each floor and stops at each floor on its way to the bottom.  What is the first floor on which both elevators will be available at the same time?

Since everything is running on 5 second increments, it will be handy to talk about time T in units of 5 seconds.  Let's let T=0 be 9:00, T=1 be 9:00:05 T=2 be 9:00:10  and so on.

Now notice that the upward elevator is closed and traveling whenever T is a multiple of 3.  The other elevator is closed and traveling when T is a multiple of 4.

UP elevator:
T=1,2  floor 2
T=4,5 floor 3
T=7,8 floor 4
T=10,11 floor 5
T=13,14 floor 6
T=16,17 floor 7
....
DOWN elevator:
1,2,3  floor 11
5,6,7 floor 10
9,10,11 floor 9
13,14,15 floor 8
17,18,19 floor 7
...

The first floor which has the same T value in both tables is floor 7, at T=17.

The first method to come to mind was to list the numbers 1-20 and circle nonmultiples of 3 and label with the floor, then box nonmultiples of 4 and label them with the floor, seeing where they lined up.

Anyone have a faster trick? :/

Say that they meet on the $ n^{th}$ floor. Then accounting for the margin of error, we have, where $ (a\sim b)$ takes on some arbitrary value from $ a$ to $ b$,
$ (5 \plus{} 10)(n \minus{} 2) \plus{} (5\sim 15) \equal{} (5\plus{}15)(11 \minus{} n) \plus{} (5\sim 20)$
$ \implies3(n \minus{} 2) \plus{} (1\sim 3) \equal{} 4(11 \minus{} n) \plus{} (1\sim 4)$
$ \implies3n \minus{} 6 \plus{} (1\sim 3) \equal{} 44 \minus{} 4n \plus{} (1\sim 4)$
$ \implies7n \equal{} 50 \plus{} ( \minus{} 2\sim 3) \equal{} (48\sim 53)\implies n \equal{} \boxed7$.