AMC
USA(J)MO
USAMO
trigonometry
geometry
abstract algebra
geometric transformation

I think the plan is now to base unofficial UK squad training here, rather than at NRICH or via the mailing list. Thus optional practice exams and accompanying discussion, or random diversions, will be posted here.

I have noticed that http://people.bath.ac.uk/masgcs/ukimo2005/ is still available, so those of us who weren't in the squad two years ago could simply do the exams there, reading January for October, etc. setting us up nicely for Trinity after March.

Alternatively we could go further afield. There are nice PDFs of Yugoslav and Turkish TSTs floating around on the internet, and questions on Mathlinks (although unfortunately the PDF feature doesn't work).

The third option is to just ask Ceri (who agreed to if asked) to stick together some non-official exams for us, but this seems a little pointless given all the resources here.

Anyway, I'm thinking the October exams from the UK 2004 training regime. If anyone has any better ideas, feel free to shout me down.

I think that sounds good.

Another option (or really an extension to that idea) would be to ask Geoff if he could put the IMO2004 and earlier years pages back online- they were taken offline so the questions from those years could be used in the correspondence course last summer- though that's only necessary if or when we ran out of resources from the IMO2005 page. Other than that, there are lots of USAMO pdfs somewhere on the internet which I think are about the right difficulty for our training purposes.

If I can find last summer's Trig Drill around, then it might be worth our while just learning that as well, since random and extensive knowledge of formulae is always useful.

I think it would be fine to do some of the online TSTs. Why don't you just pick one, Tom, and link to it from this page. Then, anyone who attempts it can talk about it when they finish. Also, if you find/think of a good geometry problem suitable for starting the marathon we proposed (c. BMO2 level) then post it in a separate thread. Also, perhaps it would be possible for some of the photos from Hungary to be published?
Dominic

I think Alison was going to set up a bloggy thing where photos were going to be dumped.

If it doesn't happen for a couple of days, however....

In the meantime:

Ok. As well as the old 2004 October exams, which we shall regard as secondary, the external exam for this month is the Yugoslav TST 1996.

http://www.matf.bg.ac.yu/~matic/competitions/zadaci/selection_1996.pdf

It's just a single 3-question 4 1/2 hour exam (although the actual exam says "180 minutes" but I think we can safely extend this to 4 1/2). I hope it works ok. I'll probably try it tomorrow or Sunday if I can kill off my courseworks and AS module learning today.

Fine. Why don't we discuss it on Monday.
Dominic

[quote="Ilthigore"]I think Alison was going to set up a bloggy thing where photos were going to be dumped.[/quote]

I hope you're going to inform us as well :D

But till then, you can post [i]me[/i] the photos :P (Dominic, you were the one, who took pictures of the Hungarian Snap with me, weren't you?)

Daniel

Yes. When I have uploaded them, I will forward them to you.
Dominic

I'll get that done over the weekend :) Nice that this is happening so soon

*is Ian*

Wow, we finally have our own forum on mathlinks. About time.

Keep up the good work folks.

Just to remind everyone of the Geometry Marathon I started in this section. Perhaps someone else would like to post the second question, since it seems apparent that the first solver will not do so.
Dominic

Wow!  Impressively keen.  I'll try to do the Yugoslav TST.  If anyone wants to look at the 2001 IMO shortlist, I'm trying to do that as well.

Jack

I have just completed the Yugoslav TST. I think it is a very useful level (perhaps a little bit easy, but not too much so, still certainly hovering between BMO2 and IMO).

I'll avoid saying anything else until Monday.

In the meantime, I think I'll follow Jack's lead. Are shortlists supposed to be in difficulty order (C1=easy combinatorics, C8=hard combinatorics), or is it just random? I've found a nice little PDF at [url]http://www.ajorza.org/math/mathfiles/imoshortlist/jorza/imo2001.pdf[/url], although if anyone knows of one with smaller text...

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15624[/url] has shortlist 2001 as a pdf (and it works iff you are logged on).  Yes, they are supposed to be in increasing order of difficulty.  C2, which later became q4, is instructive and not too hard.

Edit: on further inspection, it seems these are the same pdf.

Jack

[quote]If I can find last summer's Trig Drill around, then it might be worth our while just learning that as well, since random and extensive knowledge of formulae is always useful.[/quote]

I now have a PDF of the UK trig drill. I have emailed it to those who have already posted on this thread whom I thought would want it. If anyone else wants me to forward it to them feel free. I am also trying to get hold of some of our old FSTs, which should make good training material.

I have found Zhivko's functional equation notes also: they are quite extensive but lots of the problems are also very difficult. Sadly functional equations rarely come up these days but I'll send them to the same people as you send the trig to.

FST questions for 2006/5/3 are in the national olympiads forum, but I suppose that you have them already. NST 2005(?) is there also, if memory serves correctly.

I'll try do the Yugoslav TST some point this week, though my GCSE mocks may well get in the way of that.
[i]edit: I can't access the link to the TST- maybe I'll do the 'secondary' old training exam then[/i]

Is unipotent the same as an involution? Thanks for the notes.

I have emailed Tim the Yugoslav PDF (which appears to currently not work). If anybody else needs it, PM me.

I did the TST on Saturday, and since it's Monday now I suppose I'll say that I think I have solutions for Q1 and Q2, but didn't make any useful (in my opinion anyway) progress on Q3. I expect there's an important method for those sequence-with-evil-position-to-term-formula questions :P

There is a very useful method.
For those who have now done the exam, [hide="hidden"]See Engel Chapter 9. :)[/hide]

Joseph forwarded me some more stuff today, including a few NSTs and this, which (given who I believe is the writer has already given his approval to our regime in this thread) I post as our next exam.

Could someone with slightly more patience than me provide a translation of Q1? English is preferred, though Latin is acceptable as an alternative...

Let there be a symbol $\delta$ with $\delta \neq 0 = \delta^{2}$ (oh golly gosh, this must be some kind of cheat), so $(a+b\delta)(c+d\delta)=ac+\delta(ad+bc)$. Ha ha, how terribly hilarious. Let P(x) be a polynomial of [b]immense stature[/b] with real coefficients. Prove this: P has a repeated* real root (this is funny) if and only if it has a nonreal root of the form $x\delta$ where $x$ is real.

*In 1337, "mlt1pl" - lit. "multiple"

Not quite.  $\mathbb{R}[\delta]$ is the set of all numbers of the form $a+b\delta$ (in this case).  Ah, this was such a fun exam, although I misread question 3 and so was tragically unable to prove it.

Jack

I e-mailed it to my Dad's school and he is going to show it to his sixth formers to see if they can translate it.

I wonder if any of them will try to google the Democratic Republic of Oundle.

I was snooping around the inequality forum and I came across this user:

http://www.mathlinks.ro/Forum/search.php?search_author=jivko777

This guy seems familiar... and I think some of his posts in Algebra were really good (as you would expect).


conics
parabola
function
rotation
analytic geometry
graphing lines
slope

Didn't know how hard this was, and where I should post it, but being from Sweden I've got no real idea of what the different difficulties stand for, so Iäll just post it here. Move it around if you want to.

Okay, you've got a piece of wood. It's lower edge resembles a x^2 curve.

Now you've also got some kind of 90 degrees angle. The legs or whatever they are called are infinite. As is the curve.

In the cross of the angle, you've got a very sharp knife.

You put the wood on an even larger piece of paper.
You put the angle below the piece of wood, the two legs touching the edges. You move the angle from your left to your right, always making sure that the legs touch the wood.

Now there will be a cut in the paper. This cut should resemble some kind of function. Which?

[quote="deadfish"]In the cross of the angle, you've got a very sharp knife. [/quote]

Which angle?  I don't really understand how the cut is being made.

To me, it sounds like he's asking for the locus of points having the property that the tangents drawn from the point to the parabola are perpendicular.

@OP: Why complicate the question with making cuts in wood?

I did'nt know the words for all that stuff. (And still don't)

Choose a point on the X^2 curve, and another point will have a tangent in a right angle to the first point's.

When the chosen point moves, the point where the angle is will move as well, but resembling what function?

Feel free to translate into understandable english.

I think what you're asking for is a function which covers all points such that if two lines are drawn from the point, tangent to $y=x^{2}$, they will meet at a right angle.

[hide="this is what I got"]Pick a point $P(a,b)$ in the plane.  Extend a horizontal ray from this point and rotate it 360 degrees, seeing where it intercepts $y=x^{2}$.  If the ray is rotated $\theta$, the slope of the line is $tan\theta$.  Therefore, using the point-slope formula we have a line $y = tan\theta (x-a)+b = x^{2}$.  Solve for $x$.

$x^{2}-x\tan\theta+a\tan\theta-b = 0$.  Use the quadratic formula.

$x=\frac{\tan\theta\pm\sqrt{\tan^{2}\theta-4a\tan\theta+4b}}{2}\\$

The slope at that point is $2x$, and we want this equal to $\tan\theta$

$\tan\theta = \tan\theta\pm\sqrt{\tan^{2}\theta-4a\tan\theta+4b}$

$\tan^{2}\theta-4a\tan\theta+4b = 0$.  Once again use the quadratic formula to solve for $\tan\theta$.

$\tan\theta = \frac{4a\pm\sqrt{16a^{2}-16b}}{2}= 2a\pm 2\sqrt{a^{2}-b}$

Because of the $\pm$, there are two values here.  Call them $\alpha$ and $\beta$.  Multiply the two values.

$\tan\alpha\tan\beta = (2a+2\sqrt{a^{2}-b})(2a-2\sqrt{a^{2}-b})$

We want $\alpha$ and $\beta$ to differ by 90 degrees.  The product of the tangents of two angles which differ by 90 degrees is always -1.  Therefore:

$-1 = 4a^{2}-4(a^{2}-b)$

$-1=4a^{2}-4a^{2}+4b = 4b$

$\therefore b=\frac{-1}{4}\forall a\in \mathbb{R}$.

This yields the function $\boxed{y=-\frac{1}{4}}$[/hide]

[quote="deadfish"]I guess you put it better than I did.  :oops:

As for the solution, I'm not familliar with whatever that was in the last line before the solution, but it seems to be correct.

Thank you.[/quote]$\forall a\in \mathbb{R}$ means "for all a which is a real number".  Basically what I was saying was for every value of a, the point $P(a,-\frac{1}{4})$ is on the graph.  This is the function $y=-\displaystyle\frac{1}{4}$.

Let $F$ be the focus of the parabola, $T,T'$ be points on the parabola, $t,t'$ be lines tangent to the parabola at $T,T'$.
If $F_{t}$ is the reflection of $F$ across $t$, then $F_{t}$ lies on the directrix of the parabola. I can't recall the proof of this right now, sorry.
$t\perp t' \leftrightarrow\overline{FF_{t}}\perp\overline{FF_{t'}}\leftrightarrow\triangle FF_{t}F_{t'}$ is right-angled. Since $t,t'$ are perpendicular bisectors of $\triangle FF_{t}F_{t'}$, they meet at the circumcenter. In a right triangle, the circumcenter lies on the hypothenuse.
Therefore, tangents meeting perpendicularly meet on the directrix.

[color=darkred][b]Preliminary study.[/b] Consider the parabola $y^{2}=2px\ ,$ with the focus $F(\frac{p}{2},0)$ and the directress line $y=-\frac{p}{2}\ .$ Denote two any  points $M(\frac{m^{2}}{2p},m)$ and $N(\frac{n^{2}}{2p},n)$ which belong to the given parabola. The equation of the line $MN$ is $2px-(m+n)y+mn=0$ and the equations of the tangents in the points $M$ , $N$ to the parabola are : $\|\begin{array}{ccc}MM & : : & my=px+\frac{m^{2}}{2}\\\\ NN & : : & ny=px+\frac{n^{2}}{2}\end{array}\ .$ Denote the intersection $L(u,v)\in MM\cap NN\ .$ Prove easily that $\boxed{\ \begin{array}{c}m+n=2v\\\\ mn=2pu\end{array}\ }\ .$[/color]

[color=darkblue][b]Applications.[/b]
$1.\blacktriangleright$ [u]Ascertain the geometrical locus of the point $L(u,v)$ so that $LM\perp LN\ .$[/u] In this case, the product of the slopes $\frac{p}{m}$ , $\frac{p}{n}$ of the tangents $MM$ , $NN$ respectively is equally to $-1\ ,$ i.e. $mn=-p^{2}$ what means $2pu=-p^{2}\ ,$ i.e. $u=-\frac{p}{2}\ .$ In conclusion, the required locus of the point $L$ is the directress line of the parabola.
$2.\blacktriangleright$ [u]Ascertain the geometrical locus of the point $L$ for which the line $MN$ passes through a fixed point $A(r,0)$ which belongs to the symmetry axis of the parabola.[/u] In this case, $A\in MN\ ,$ i.e. $|\begin{array}{ccc}\frac{m^{2}}{2p}& m & 1\\\\ \frac{n^{2}}{2p}& n & 1\\\\ r & 0 & 1\end{array}|=0$ $\Longleftrightarrow$ $mn=-2pr$  $\Longleftrightarrow$ $2pu=-2pr\ ,$ i.e. the required locus of the point $L$ is the line with the equation $x=-r\ .$ In the particular case when $r=\frac{p}{2}\ ,$ i.e. $A: =F$ the required locus of the point $L$ is the directress line of the parabola.

[b]Exercises.[/b]
$1.\blacktriangleright$ Ascertain the geometrical locus of the point $L$ for which $m(\widehat{MLN})\in \{45^{\circ},135^{\circ}\}\ .$
$2.\blacktriangleright$ Ascertain the geometrical locus of the point $L$ for which the distance of the focus $F$ to the line $MN$ is constant.[/color]

I understood pretty much nothing of the two last posts, but I do not see an answer... Or do I just not see it?

[color=darkred][b]Albert Eistein :[/b] if exists at least the One who see and understand, then the all is O.K. ![/color]


AMC USA(J)MO USAMO ARML inequalities MIT college

EDIT: Removed because the person who found the link apparently wasn't supposed to reveal the results.

Now it's up and official. [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/2009usamoqual.shtml[/url]. 17 in Georgia! Congrats all.

EDIT: I'll set something up at MathFest so all of you can work on old Olympiad tests.

Tom

Do you mean at MathFest, you will organize a gathering in the future? The way you worded it sounds like we'll be doing Olympiad tests during the day of MathFest.

Sadness.  For the fourth(?) time this year, I won't be able to attend a math competition cuz of music stuff.  And MathFest is always so fun.  Oh well, cya guys at varsity state!

Yeah, FIRST Championships falls on this date as well, so I won't be able to attend.

Hmm, at least two USAMO qualifiers won't be on the ARML A-Team this year.

Also, we have only 1 less qualifier than TJ. Pretty good.

EDIT: We had 11 qualifiers last year, while TJ had 6. So, I guess the above statement is less impressive. However, GA has 1 more qualifier than Exeter, which is an improvement from 6 less last year.  (And before someone from TJ/Exeter attacks this, yes I do realize that I am comparing an entire state to a single boarding school, I am just trying to determine a relative measure of improvement).

[quote="Boy Soprano II"]Psh, I still haven't forgiven TJ for getting the Cauchy-Schwarz Inequality backwards at the song contest in my freshman year.[/quote]

I remember that. Even I noticed that they had the sign backwards. (Back then, it was a big deal for me.)

Also . . .

SEVENTEEN!  YES!  This is up about 50% from last year!  This year, both the Georgia ARML A team and the Walton state math team are more prestigious than making USAMO. :)   I bet you can sympathize with the [url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam2006results.html]MIT Putnam team organizers[/url], Mr. Fulton.

I will be glad to talk to anybody about the USAMO.  I can even prepare something, if you like.

I don't know about "more prestigious". I mean, have you seen my guys?  :D And with there being [b]17[/b], it feels like a good ARML year for us. At least Harrison, Adam, and the rest of the Hampshire folks should think so!

I'm not entirely certain that the prestige of being in a group of people is solely dependent on the number of people who are in that group.
Then again, if Miles says so, it must be true.  :wink:

No, but it can be determined by inclusion.

More, specifically, let $ \leqslant$ be a partial ordering on a set $ \mathfrak{C}$ of ordered pairs of the form $ \mathcal{A} \equal{} (A_0,A_1)$, where $ A_0$ is a collection of people competitively selected from a collection $ A_1$.  By abuse of notation, we may identify $ A_0$ with $ \mathcal{A}$ and refer to $ A_1$ as $ \text{pop}\,\mathcal{A}$.  We call $ \leqslant$ a [i]prestige measure[/i] if the relation
\[ \bigl[ (\mathcal{A} \cap \text{pop}\, \mathcal{B}) \subseteq (\mathcal{B} \cap \text{pop}\, \mathcal{A}) \bigr] \implies \mathcal{B} \leqslant \mathcal{A}
\]
holds.

More generally, if $ \mathfrak{A}$ and $ \mathfrak{B}$ are [i]classes[/i] of possible samples each drawn from a fixed populations, we say that $ \mathfrak{A} \geqslant \mathfrak{B}$ if $ \mathcal{A} \geqslant \mathcal{B}$ for some $ \mathcal{A} \in \mathfrak{A}$ and $ \mathcal{B} \in \mathfrak{B}$.  (If $ \mathfrak{A} \leqslant \mathfrak{B}$ and $ \mathfrak{B} \leqslant \mathfrak{A}$, then we say that $ \mathfrak{A}$ and $ \mathfrak{B}$ are [i]equivalent with respect to prestige[/i].)

Now, if you let $ \mathfrak{A}$ be the set of possible Georgia ARML A teams for this year, and you let $ \mathfrak{B}$ be the set of possible lists of USAMO qualifiers for this year, you can see that my statement follows directly from the definition, and likewise if you let $ \mathfrak{A}$ be the set of possible Walton math teams for state. :wink:   Of course, as these groups are actually determined, it may come to pass that the relative prestige of these different groups will depend on the particular prestige ordering used, i.e., there may be no univesally agreed-upon prestige order for these sets.

[quote="gtf"]Sorry, my mind is still on Spring Break! There will be a room and materials available during MathFest for those of you that qualified for the USAMO. Perhaps we can get Miles/Santhosh/Billy or one of the others to talk about the test for the first-timers.

We'll work something out for the Walton folks so they can attend this.[/quote]

So I'm sad to say but I'm going to be at MIT CPW (admit weekend) .... and thus will go through high school having _never_ gone to a MathFest. Bah! Anyway, apologies, but I'll be unable. (I'm not very qualified either. 12 then a 6? My scores are going _down_ - you shouldn't learn what _to_ do from me :P)

Looking at the content of the past 3 posts, I'd guess that Miles would be everyone's best bet for USAMO advice. :)

Nonetheless, I advise first-timers not to stress out too much, and use all of the time if possible. Just keep picking at the problem(s) that you understand. If you can figure out roughly what the problem is asking you to do, you're doing fairly well.

Yea, I agree with Mr. Fulton. I mean, those Walton people are [i]weird[/i]. Especially that short guy. Jeez.

By the way Mr. Fulton, I can stay at MathFest to about 3, so that's better. I'll just stick that in here so I won't forget to tell you.

I agree with Xantos.

But Santosh should still speak.

And I wish I could understand what Miles wrote...

Actually, the Claus/Sitan/Other-Intelligent-Middle-Schoolers Rule (or whatever it is called) prohibits Walton from using Edward at the State tournament.
So, I am going to make two [b]conjectures[/b][size=9]*[/size]:
[hide="Conjecture#1"]Walton's state team will be Henry, Claus, Allen, and Abubakar. (Mr. Fulton will prove xor disprove this within the next two weeks.)[/hide]
[hide="Conjecture#2"]The prestige of making the Walton State team is now greater than or equal to the prestige of qualifying for USAMO. Equality occurs if one of the members barely made the cut. (Miles can probably prove xor disprove this). [/hide]

[size=9]*I am not responsible for any losses or damages directly or indirectly occurring from the making of these conjectures. <Insert more legal jargon that makes me not liable for anything here.>[/size]

Georgia AIME Statistics

[url]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2009-aa/AIME%20state%20stats/GA_STATS.HTML[/url]

Wah?  Does that mean I got an 8 on the AIME?  Or does that mean I guessed one right and didn't realize it?

Cuz I'm 100% positive I got a 10.

But then, if I had gotten an 8, wouldn't my index me a measly 212 with my 10 score, which isn't enough?

Or did tiebreaks between indices only take place between the people with 7?

Hmm, does this mean I got an 11, which is weird, but cool.

Can someone more knowledgeable (which would be anyone) clarify?

PS: I [i]wonder[/i] who the 7th grader with a 10 was? :roll:

So the questions are:

Who made the 12's?
What 9th grader made the 11?
What HS'ers made the 10's?
Who made the 9's?

I think I got one of the 12s and Tim Wu got the other one.

[quote="Boy Soprano II"][quote="Math Geek"]But then, if I had gotten an 8, wouldn't my index me a measly 212 with my 10 score, which isn't enough?[/quote]
It you took the AMC 12, then an index of 201 was enough.  In fact, if you took the 12 as underclassman, an AIME score of 8 was sufficient.  I suggest you should argue with Claus about who got the higher score, or just ask your coach what your reported score was. :)[/quote]

According to some complaints in the AMC forum, having taken AMC 10 seemed to disqualify you from the special invitation. Also, another post from AMC finance stated they only consider the numerically higher score which, in your case, is 132 on AMC 10. Therefore, the most likely case is that you underestimated your score (or guessed correctly!? :ninja: ) and actually got an 11.

P.S.: yeah, I think I got 12.

P.P.S.: I think Eric Chen (11) from Brookwood got one of the 9s.

If I'm not mistaken (i had to erase the answer on one of my scantron boxes multiple times) I got one of the 10s

Let's keep this organized:

AIME Score: people
12's: Miles, Tim
11's: Sitan, Allen, Henry
10's: Edward, Oliver, Zihao
9's: Billy, Santhosh, <11th>, <11th>
8's: Claus, <9th>, <10th>, Eric, <12th>

EDIT: Sorry if I messed up anything, feel free to fix it in your post.

Henry and I both got a 11. The 9th grader would probably be Sitan then.

Let's not forget Claus.  I might have gotten an 8.  I should have gotten a 10 and deserve neither an 11 nor an 8, especially not the former.

So Eric thought he was of an 8. Therefore the missing senior must be... Vladimir. Although it could be Zihao Li [has he done any competitiony things?]. I'm inclined to have confidence in Vladimir's score though.

Our school's AMC sponsor e-mailed me yesterday...Zihao Li got 10 on the AIME, putting him just above the index with a 103 AMC12B

Grade| #with 8+AIME | #USAMO Qualifiers
07 | 1 | 1
09 | 3 | 2
10 | 3 | 3
11 | 3 | 4
12 | 7 | 7

If an upperclassman qualified with a 7 AIME, then their AMC score would have to be at least 201-70=131.  Only 5 distinct students in GA satisfy this, 4 of which/whom??? are accounted for.

If an underclassman didn't qualify with an 8 AIME, then their AMC12 score would have to be less than 201-80=121 and their AMC10 score would have to be less than 215-80=135. This however is a lot of people.

From the stats listed and discussions on this forum, only 1 score is "unclaimed".
AIME Score: people
12: Miles, Tim
11: Sitan, Allen, Henry
10: Edward, Oliver, Zihao
9: Billy, Santhosh, Robert, Eric C.
8: Claus, <9th>, Abubakar, Eric M., Vladamir

I'm a 9th grader with an 8...am I the missing one? :O

Not anymore.

AIME Score: people
12: Miles, Tim
11: Sitan, Allen, Henry
10: Edward, Oliver, Zihao
9: Billy, Santhosh, Robert, Eric C.
8: Claus, Arvind, Abubakar, Eric M., Vladamir

yay! i liked it better shrouded in mystery tho XD

congrats to all the usamo qualifiers! :D

thx arvind!
i feel rather embarrassed to admit that I qualified for USAMO with a 136.5 on the AMC...and a [hide][b]7[/b][/hide] on the AIME.

Well, even if you didn't admit it, people could figure it out anyway. Don't worry, several people in the nation qualified like that. Otherwise, the floor would be an 8.

I got a 126 on the 10A and an 8...so...din qualify.  :noo:

Next year Arvind. I have faith that you will qualify.


algebra
polynomial
analytic geometry
function
induction
blogs
rational numbers

Find all polynomials $W$ with real coefficients possessing the following property: if $x+y$ is a rational number, then $W(x)+W(y)$ is rational.

Setting $x=y$, we get that $W(x)$ is rational for all rational $x$. If it has degree $n$, it's uniquely determined by $n+1$ points and equals it's Lagrange interpolation polynomial [*]. Since we can choose these points to have both coordinates rational, the Lagrange interpolation polynomial and thus $W$ itself is rational.

[*]: For given points $(a_i,b-_i)$, $i=1,2,3,...n$, it is defined via $\sum_{i=1}^n b_i \prod_{j=1, i \neq j}^n \frac{x-a_j}{a_i-a_j}$.
It's an easy check that it runs through all those points $(a_i,b_i)$ and has rational coefficients only if the $a_i,b_i$ are rational.

The previous post gives a necessary, but not sufficient condition.

The answer is that the only such polynomials $ W$ are of the form $ x\to ax\plus{}b$ where $ a$ and $ b$ are rational. Indeed these are the only continuous functions $ W$ from the reals to the reals such that if $ x\plus{}y$ is rational then $ W(x)\plus{}W(y)$ is rational.

[b]Proof[/b]:
Fix a rational number $ q$. Then $ x\to W(x)\plus{}W(q\minus{}x)$ takes only rational values (since $ x\plus{}(q\minus{}x)\equal{}q$ is rational), and is continuous. Therefore it is constant.

Thus, if $ x\plus{}y$ is rational, then $ W(x)\plus{}W(y)\equal{}W(x\plus{}y)\plus{}W(0)$. If we let $ Z(x)\equal{}W(x)\minus{}W(0)$, this says that if $ x\plus{}y$ is rational, then $ Z(x)\plus{}Z(y)\equal{}Z(x\plus{}y)$. In particular, $ Z$ is additive on the rationals. It follows by an easy induction that $ Z(nx)\equal{}nZ(x)$ for any integer $ n$ and rational number $ x$. Thus, $ nZ(\frac{m}{n})\equal{}Z(m)\equal{}mZ(1)$, so $ Z(q)\equal{}qZ(1)$ for all rational $ q$. Thus $ W(x)\equal{}W(0)\plus{}x(W(1)\minus{}W(0))$ for all $ x$ in the set of all rationals. Since the rationals are dense, this is true for all $ x$.
[b]QED[/b]

Luke
See my puzzle blog at [url]http://bozzball.blogspot.com[/url][/url]


If $ x \plus{} y \equal{} 8$ and $ xy \equal{} 15$,what's the value of $ x^{2}\plus{}6xy\plus{}y^{2}$ ?

A) 64
B) 109
C) 120
D) 124
E) 154

[hide="Solution"]$ x\plus{}y\equal{}8$ implies $ (x\plus{}y)^2\equal{}x^2\plus{}2xy\plus{}y^2\equal{}64$. $ x^2\plus{}6xy\plus{}y^2 \equal{} (x^2\plus{}2xy\plus{}y^2)\plus{}4xy \equal{} (x\plus{}y)^2\plus{}4xy \equal{} 8^2 \plus{} 4\times 15 \equal{} 64\plus{}60 \equal{} 124$. The answer is $ \boxed{D}$.[/hide]


Yea. Same goes for like the USAMTS problems. People were trying to sneak answers all over the place.

That'll get people to want to sneak answers more. Wouldn't it?

Hmm... I don't think so.

The big problem is, when I see a problem posted, I have no idea it's illegal... Same with most other people. For example, why should I care about the Polish MO? I'm not in Poland!

[quote="Jumbler"]Why not make a list of the problems that are not suppose to be answered, so that everybody knows. For example, the problems of Polish MO, which was not after deadline, and had to be locked, but somebody might have replied earlier.

Thanks[/quote]

Because that's not a trivial task and admins/moderators have more pressing issues to deal with.  We also assume that the community catches most of these posts and reports them to us otherwise we might have to spend more time preventing cheating.  If you want to create such a list and keep it updated, that's fine.  If you make it on the Wiki it's more likely to stay updated since everyone can edit it.

I'm not trying to sound like a jerk, but it shouldn't be AoPS's responsibility to prevent cheating.  Obviously, we take some measures during contest windows (locking AMC forum, etc.) but these actions are due to the fact that that forum is designated for that specific topic.  It's like saying the telephone companies should monitor the lines for terrorists.  That's what we have the NSA for.  :rotfl:

The AMC Forum doesn't look locked to me, even though it is a contest window uptil the 21st.

mysmartmouth: then give the competitions enough money to make up something like NSA on their own...

Hmm, maybe they lock it only for AMC 12, or perhaps it was AIME, but I remember it being locked last year...or perhaps they only lock it the day of the contest...was it locked this year at all?


Slove this equation: $ \frac {x(5 \minus{} x)}{x \plus{} 1} \displaystyle \left(x \plus{} \frac {5 \minus{} x}{x \plus{} 1}\right) \equal{} 6$

$ \frac {x(5 \minus{} x)}{x \plus{} 1} \displaystyle \left(x \plus{} \frac {5 \minus{} x}{x \plus{} 1}\right) \equal{} 6 \Leftrightarrow   \\ \Leftrightarrow 5 x^3 \minus{} x^4\minus{}6 \plus{} 13 x \minus{} 11x^2\equal{}0 \Leftrightarrow  \\ \Leftrightarrow (2\minus{}x)(x\minus{}1)(x^2\minus{}2x\plus{}3)\equal{}0$

From here $ x \in \{2; 1\}$

[quote="Ahiles"]$ \frac {x(5 \minus{} x)}{x \plus{} 1} \displaystyle \left(x \plus{} \frac {5 \minus{} x}{x \plus{} 1}\right) \equal{} 6 \Leftrightarrow \\
\Leftrightarrow 5 x^3 \minus{} x^4 \minus{} 6 \plus{} 13 x \minus{} 11x^2 \equal{} 0 \Leftrightarrow \\
\Leftrightarrow (2 \minus{} x)(x \minus{} 1)(x^2 \minus{} 2x \plus{} 3) \equal{} 0$

From here $ x \in \{2; 1\}$[/quote]
Thank Ahiles very much.


Three non-negative integers are written on a blackboard. A move is to replace two of the integers $k,m$ by $k+m$ and $|k-m|$. Determine whether we can always end with triplet which has at least two zeros

The problem was 'almost'  ;)  discussed before: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=poland+2002&t=3912  Unfortunately, there was one typo in problem which completely changed the answer to the problem and solution.

But this problem is asking if we can get a triplet with two zeroes which is a different problem. Could anybody please post a solution? Thank you.
Also, I do not think this is a combinatorics problem.

can anyone post a solution.
I have one idea
It should be checked as 0, n, n + 1


logarithms
function
algorithm
real analysis
real analysis unsolved

I was thinking about this problem for a while now, and I would like you to see what you think about it. First, denote with $ e_n,\, n \in [0,9]$ a number that has all the digits like $ e$ (the base of the natural logarithm), but the digit $ n$ left out. Is it possible to write any of these numbers using known functions and constants. As an example, $ e_8\equal{}2.712124590452354...$.

What motivation would you have for suspecting that any of these numbers have special forms?

Well, there is no reason to tie this to $ e$ alone. It can be connected to any constant. I saw some equations at [url]http://mathworld.wolfram.com[/url] that include $ \pi$ and operator $ \oplus$ (xor). So, since this is on a digit scale, I thought that maybe there is some equality on a digit scale of this sort. I'm not searching for a solution to a specific problem, but you opinion weather this could be possible (and a bit of math with it, if possible)..

Found it: [url]http://mathworld.wolfram.com/PlouffesConstants.html[/url] - fascinating!

Binary XOR makes some kind of sense to me; the spigot algorithms for $ \pi$ also exist on a binary (base $ 16$) level.  Binary is in some sense the most fundamental and important number base.  It's the choice of base $ 10$ that makes me wonder.


function
integration
absolute value
real analysis
real analysis unsolved

Let $ f: [0,1]\minus{}>R$ be a function s.t. $ |f(x)\minus{}f(y)|<\equal{}|x\minus{}y|$ and $ \int_0^1f(x)dx\equal{}0$. prove that $ |\int_0^xf(t)dt|<\equal{}\frac{1}{2}x(x\minus{}1)$ for any $ x$ in $ [0,1]$.

You meant $ \left|\int_0^x f(t)\,dt\right|\le \frac12 x(1\minus{}x)$. In fact, we can drop the absolute value because $ f$ can be replaced with $ \minus{}f$.

Let $ g(x)\equal{}f(x)\plus{}x\minus{}\frac12$. It should be clear that the function $ g$ is nondecreasing and $ \int_0^1 g(x)\,dx\equal{}0$. Its antiderivative $ G(x)\equal{}\int_0^x g(t)\,dt$ is equal to $ 0$ at $ x\equal{}0,1$ and is concave up in between. Therefore...

mlok

Could you post the details that $ G(x)$ is concave  ?
thx

It's because $ G'\equal{}g$ is nondecreasing.


find all function such that
$ \frac{f(x\plus{}3)}{f(x\plus{}2)}\equal{}x\plus{}3$

[hide=" one solution"]

$ f(x) \equal{} x!$ works.

[/hide]

I have not solved the problem. All I have done is noticed that $ (x\plus{}3)! \equal{} (x\plus{}3)\cdot(x\plus{}2)!$.

yes but i have already got that :)
so we need a solution

We use induction to show that $ f(x)\equal{}a\cdot x!$ for any constant $ a$ is the only solution.  (Note that this is only for natural numbers; I suspect there will be some kind of Cauchy-like equation for all reals.)

Base case:  Let $ f(1)\equal{}a$.  Then obviously it works.

Inductive step:  We have that, for all $ k\leq n$, that $ f(k)\equal{}a\cdot k!$.  Let $ x\equal{}n\minus{}2$.  Then
$ f(x\plus{}3)\equal{}f(x\plus{}2)(x\plus{}3)\equal{}a(x\plus{}2)!(x\plus{}3)\equal{}a(x\plus{}3)!$

Re-write as $ f(x) = xf(x - 1)$. Now the function is completely defined by it's values over $ [ - 1,0)$, where the value of $ f(0)$ should be defined appropriately*.

For example, consider $ f_{[ - 1,0)}(x) = 1$. Then:

\begin{align*} f_{[ - 1,0)}(x) & = 1 \\
f_{[0,1)}(x) & = x \\
f_{[1,2)}(x) & = x(x - 1) \\
f_{[2,3)}(x) & = x(x - 1)(x - 2) \\
f_{[N,N + 1)}(x) & = x(x - 1)\ldots (x - N) \end{align*}

And similarly we can define $ f$ for intervals $ [N,N + 1)$ for $ N < - 1$ by inducting backwards. Of course, $ f_{[-1,0)}$ doesn't have to be "nice" -- it can even be pathological.

* [i]- You must have $ f(0) = 0$ if $ f( - 1)$ is defined and finite. There is no restriction if there's a pole at -1, although $ f(0) = \lim_{x \to 0} xf(x - 1)$ is a nice way to pick the particular value and will allow you to express the function everywhere neatly in terms of the Gamma function.[/i]


calculus
integration
calculus computations

hi!!

calculate this sum:

$ \sum_{i\equal{}1}^{n}(\frac{1}{i})$

thanx.

The simplest exact formula for that sum is $ \sum_{i\equal{}1}^n \frac1i$. You are asking for something that does not exist, and we can't help you with that.

There isn't a simple exact formula for the $ n$th harmonic number.

Wikipedia has some information about integral approximations and asymptotics, though.
http://en.wikipedia.org/wiki/Harmonic_number


Prove that for any $ \delta\in[0,1]$ and any $ \varepsilon>0$, there is an $ n\in\mathbb{N}$ such that $ \left |\frac{\phi (n)}{n}-\delta\right| <\varepsilon$.

Fix any $ \delta\in[0,1]$ and any $ \epsilon>0$.

We have $ \prod_{p\ \mbox{is prime}} (1\minus{}\frac{1}{p}) \equal{} 0$ that is equivalent to
\[ \sum_{p\ \mbox{is prime}} \minus{}\log(1 \minus{} \frac{1}{p}) \equal{} \infty.\]
This is divergent series whose terms tend to 0. So, we can select a subsequence of primes $ p_1, p_2, \dots$ such that
\[ \sum_{k\equal{}1}^{\infty} \minus{}\log(1 \minus{} \frac{1}{p_k}) \equal{} \minus{}\log\delta.\]
Let $ m$ large enough integer such that
\[ \left| \sum_{k\equal{}1}^m \log(1 \minus{} \frac{1}{p_k})  \minus{} \log\delta\right| < \min\{\frac{\epsilon}{2\delta},\frac{1}{2}\}.\]
Since $ |\exp(x\plus{}t)\minus{}\exp(x)|<\exp(x)2|t|$ for $ |t|<\frac{1}{2}$, we have
\[ \left| \prod_{k\equal{}1}^m \frac{p_k\minus{}1}{p_k}  \minus{} \delta\right| < \epsilon.\]
i.e., for $ n\equal{}\prod_{k\equal{}1}^m p_k$, we have
\[ \left| \frac{\phi(n)}{n}  \minus{} \delta\right| < \epsilon.\]
Q.E.D.


trigonometry
algebra unsolved
algebra
system of equations

Solve the system of equations
  \[ \begin{cases} \tan x - \tan y = y -x \\ \sqrt{y+1} - 1 = \sqrt{x- \sqrt{y+8}}\end{cases} \]

Why did you post the statement here when you knew exactly that it was wrong?

[quote="ali"]tg x + x is an increasing function. This implies that x=y
then this gives x=y=8.[/quote] nttu is right. $\tan x+x$ is not an increasing function.

I have some doubts that this system is resolvable.

hello, with $x=y+2-2\sqrt{y+1}+\sqrt{y+8}$ we get
$\tan(y+2-2\sqrt{y+1}+\sqrt{y+8})-\tan(y)+2-2\sqrt{y+1}+\sqrt{y+8}=0$
One solution is $y \approx 1.143069793$
Only with numerical methods we can find the other solution.
Sonnhard.


inequalities
number theory solved
number theory

Find all pairs of integers (m,n) such that 1!*3!*...*(2m-1)!=n!. I found it in an AMM and I don't know how to solve it.

For all m \geq 4, there exists a prime p between m and 2m - 2.
Hence p \leq n < 2n.
Maybe this helps? Just an idea, I didn't really think about it, not yet  :)

My computer collapses while I was typing my first answer grrrrrrr  :(

To continue Arne's idea.

For n >= 7 we have :

(*) 1!*3!*...*(2m-1)! > (4m)!

So if 1!*3!*...*(2m-1)!=n! then n > 4m so by Bertand's postulate there is a prime factor in n! > 2m-1, this is not possible.

So we only have to check by hand m < 7. m=1,2,3 works at least.

(*) can probably be proved with Stirling inequality ( \sqrtn (n/e)^n < n! < e \sqrtn (n/e)^n) with a lot of courage  :D

For (*) you just need to prove (2m+1)!*(2m+3)!>(4m+1)(4m+2)(4m+3)(4m+4) which should be weak enough


IMO Shortlist
combinatorics
Extremal Graph Theory
graph theory
permutation

An $ n \times n, n \geq 2$ chessboard is numbered by the numbers $ 1, 2, \ldots, n^2$ (and every number occurs). Prove that there exist two neighbouring (with common edge) squares such that their numbers differ by at least $ n.$

Does anyone have a solution to this problem?  I have worked on it for some time and not found a solution.

We can prove more general result(*):
Let the cells of an  $n \times n$ board be filled with an integer, such that the difference of numbers in any two neighbouring (with common edge) squares is at most $d$. Then  there exists a number that appears in the board at least  $\frac{n}{d}$  times.
From this result, if  the difference of numbers in any two neighbouring (with common edge) squares is at most $n-1$. Then  there exists a number that appears in the board at least $\left\lceil{\frac{n}{{n - 1}}}\right\rceil=2$ times, contradiction.

Proof of (*):We point out the biggest number of each rows, and the biggest number of each columns, suppose $m$ is the smallest one of these pointed out numbers.
Without loss of generality, we can suppose $m$ is the biggest number of row $i$. For any column $j$, suppose the biggest number of column $j$ is $M$, ${a_{ij}} = p$,  then $M \geqslant m\geqslant p$ , because the difference of numbers in any two neighbouring (with common edge) squares is at most $d$. So in column $j$ there exist a number of $\left[ {m,m+d-1}\right]$ between $p$ and $M$.
So there are at least $n$'s numbers of  $\left[ {m,m+d-1} \right]$ appear in the board, one of these number must appears at least  $\frac{n}{d}$  times.

consider the subset {1,2,3,...,(n/2)^2} as A and other numbers as B. A and B has more than n common edges. But only n-1 numbers in B that differ less than n from numbers in A ( from (n/2)^2+1 to (n/2)^2+n-1 ). So there must be a common edge of A&B that the two numbers share this edge has difference more than n-1.

To prove A and B has more than n common edges. We can consider A in 4 different situation.
1. A has x full rows and y full columns. (xy>0)
2. A has x full rows and 0 full columns.
3. A has 0 full rows and y full columns.
4. A has 0 full rows and 0 full columns.
for each square, up and bottom edges are row edges, left and right edges are column edges.
1. row edges of A >= n-y
    column edges of A >= n-x
    edges of A >= 2n-x-y
    ∵ number of squares of A in full rows and full columns are: n*x+n*y-xy<=(n/2)^2, and 0<x<=n, 0<y<=n
    x+y<n/2
    so, edges of A >= 2n-x-y >= 3n/2
2. row edges of A =n
    so, edges of A >=n
3. column edges of A =n
    so, edges of A >=n
4. if x row has squares in A and y columns has squares in A.
    Then, row edges of A >=y, and columb edges of A >=x.
    So, edges of A>=x+y
    ∵x*y>=(n/2)^2
    So, x+y>=n
    edges of A >=n

Sorry for my bad English expression.

Here's an alternate way to use the idea in love_sc1's solution, where it suffices to partition the board into two consecutive intervals of numbers with at least $n$ borders. Let $S_k$ be the set of squares containing a number at most $k$. We consider the maximal $k$ such that $S_k$ contains neither a full row or a full column.

WLOG $S_{k+1}$ contains a full row, and $k+1$ is in row $r$, column $c$. The missing square in row $r$ gives us one vertical border between $S_k$ and its complement. We also know that $S_k$ has no full columns, so the other $n-1$ columns besides column $c$ are missing at least one square but do have a square in row $r$. We get at least one horizontal border in each of those $n-1$ columns. That's at least $n$ borders total.

[quote="Ramchandran"]Is there another solution to this problem?

The solution above seems vague and non-intuitive.

Why is this particular statement from the above solution true ?  -
"So in column $j$ there exist a number of  $\left[ {m,m+d-1}\right]$ between $p$ and $M$."[/quote]

Because “such that the difference of numbers in any two neighbouring (with common edge) squares is at most $d$.”

Can someone please post another solution or clarify the solutions posted here already. Even the official solution from the IMO compendium is not explained well and unnatural.

[quote=Takeya.O]We write $1$~$n^2$ step by step.We consider the number of empty squares which neighbor on written squares.If this number is not less than $n$ at a moment,the proof is completed.We consider the moment some row(or some column) is completed(written numbers).

[u][b]Case1:[/b]only some row is completed[/u]
Since any column isn't completed,there is at least one empty square on each column which neighbors on written square.Thus the number of empty squares which neighbor on written squares is not less than $n$.

[u][b]Case2:[/b]only some column is completed[/u]
This is the same as Case1.

[u][b]Case3:[/b]some row and some column are completed simultaneously[/u]
Just before some row and some column are completed simultaneously,there is at least one empty square on each column which neighbors on written squares in the same way as Case1.Thus the number of empty squares which neighbor on written squares is not less than $n$.

From Case1,2,3,the proof is completed.$\blacksquare$

:coolspeak:[/quote]

Amazing solution .

[quote=Takeya.O]We write $1$~$n^2$ step by step.We consider the number of empty squares which neighbor on written squares.If this number is not less than $n$ at a moment,the proof is completed.We consider the moment some row(or some column) is completed(written numbers).

[u][b]Case1:[/b]only some row is completed[/u]
Since any column isn't completed,there is at least one empty square on each column which neighbors on written square.Thus the number of empty squares which neighbor on written squares is not less than $n$.

[u][b]Case2:[/b]only some column is completed[/u]
This is the same as Case1.

[u][b]Case3:[/b]some row and some column are completed simultaneously[/u]
Just before some row and some column are completed simultaneously,there is at least one empty square on each column which neighbors on written squares in the same way as Case1.Thus the number of empty squares which neighbor on written squares is not less than $n$.

From Case1,2,3,the proof is completed.$\blacksquare$

:coolspeak:[/quote]

.We consider the number of empty squares which neighbor on written squares.If this number is not less than $n$ at a moment,the proof is completed.We consider the moment some row(or some column) is completed(written numbers)
Why if this number is not less than n at moment, the proof is completed ?

[quote=fwolth]The idea is to consider an interval of $ n\minus{}1$ consecutive integers which divides $ \{1,2,\ldots,n^2\}$ into integer sequences of length $ \binom{n}{2}$ and $ \binom{n}{2}\plus{}1$. $ n\minus{}1$ integers cannot contain a full row or column, so the two intervals must border one another, which gives the desired result. Not very rigorous, but it's only an outline.

I tried finding a nicer argument with something like expected value, coloring, or paths, but it i'didn't work out.[/quote]

The $n-1$ integers cannot contain a full row of column, but it can certainly divide the $n \cdot n$ board into two non bordering sections. It can trap one corner...

[quote=Plops]I would like to also ask Lukariman's question. Why if there are more than $n$ unwritten squares in Takeya's solution is it implied that some two adjacent numbers differ by at least $n$?[/quote]

I too ask the same question

I got my mistake so please ignore whatever I wrote below as I am not able to delete the post

This is wrong,
I don't know how to share grid here so I will just write the number in position without grid.
1.   2.    3.
6.   4.    5.
8.   9.    7.

No numbers on edge differ by n,i.e., 3

The Solution can be found in Chapter 4 of Olympiad Combinatorics by Pranav sriram. Example 11

By doing basic combinatorics you can see that the total number of pairs that are possible - the pairs which have
atleast n difference < the total number of neighboring squares in the chessboard

The left side of the gives the total number of pairs which are allowed in any two neighboring vertices
So this means as each and every pair is distinct according to the question therefore
This proves that it is not possible to put these number into the chessboard without having the difference of any two squares atleast n.


algebra polynomial quadratics function MATHCOUNTS

this is the only contest that is somewhat decent that we take (well NATS is ok i guess).

Hmm you guys should totally take the nationals test next year - i wasn't aware that a school could compete in both until after round 1 of last year, which is why i was stuck in regionals last year - nationals is way better, although I failed round one because I couldn't read my handwriting. Hm I should totally write my work neatly like last year.

ok so this thread is for discussing the problems/ how you did.

question: why are kyle and owen and suyoung not on the leaderboard for regionals?

No discussion yet I think. Some people can take the test this week still.

Yes; discussion is permitted the Monday after the competition week, in this case, November 9th.  :)

[quote="athunder"]Hurdler, i was wondering if you (or anyone else) can post a solution to number 6 on nationals. I keep expanding but get no where. I also used this formula but still no luck. #7 on nationals...nobody in the country got right. An assumption from looking at the leaderboard.
Problem 6: Given that $ \omega^{7} \equal{} 1$ Compute the following:
$ (2\omega \plus{} \omega^{2})(2\omega^{2} \plus{} \omega^{4})(2\omega^{3} \plus{} \omega^{6})(2\omega^{4} \plus{} \omega^{8})(2\omega^{5} \plus{} \omega^{10})(2\omega^{6} \plus{} \omega^{12})$

[hide="Key insight, given by hurdler"]$ \Re(\omega) \equal{} \frac {\omega \plus{} \overline{\omega}}{2}$[/hide][/quote]

okay factoring out the $ \omega^21\equal{}1$ gives $ (2\plus{}\omega^1)(2\plus{}\omega^2)(2\plus{}\omega^3)(2\plus{}\omega^4)(2\plus{}\omega^5)(2\plus{}\omega^6)$. These along with the number $ 2\plus{}1\equal{}3$ are the roots of $ (x\minus{}2)^7\minus{}1\equal{}0$. Taking the constant term of this polynomial gives $ 3(2\plus{}\omega^1)(2\plus{}\omega^2)(2\plus{}\omega^3)(2\plus{}\omega^4)(2\plus{}\omega^5)(2\plus{}\omega^6)\equal{}129$, so $ (2\plus{}\omega^1)(2\plus{}\omega^2)(2\plus{}\omega^3)(2\plus{}\omega^4)(2\plus{}\omega^5)(2\plus{}\omega^6)\equal{}\boxed{43}$.

lol I had fun bashing with conjugate pairs

it's not so bad, just 27 terms... $ \prod_{i \equal{} 1}^3(5 \plus{} \omega^i \plus{} \overline{\omega^i})$

still dnkywin's solution is very nice... hm

Hmmm yea that is nice. Taylor Han's solution was to consider the polynomial
$ \prod_{i\equal{}1}^6 (x\minus{}\omega^i)\equal{}\frac{x^7\minus{}1}{x\minus{}1}$
Then you win :O
I got a pretty funny score this round...It's congruent to a quadratic non-residue mod 4 and I'm not anywhere to be seen on the website, like usual.

darn I knew I should've learned how to use cyclotomic polynomials

man that's not even a tricky way

T__T

[hide="Well if discussion is allowed..."]We find the amount of numbers $ n$ from $ 3$ to $ 6027$ inclusive that have all even digits and are divisible by three.
[i]Case 1: $ 3\le n\le 5999$.[/i] The second through fourth digits can be $ 0,2,4,6,8$. Note that the first digit can be $ 0,2,4$, which forms exactly one complete residue set $ \bmod3$. Hence for any choice of the second through fourth digits, there is exactly one choice for the first digit that makes the number divisible by three. However, we are counting $ n \equal{} 0$ as well, so this case yields $ 5^3 \minus{} 1 \equal{} 124$ possibilities.*
[i]Case 2: $ 6000\le n\le6027$.[/i] It's not hard to list out $ 6000,6006,6024$.
In all, there are $ 124 \plus{} 3 \equal{} \boxed{127}$ possibilities.

*One can also use generating functions (the relevant one here is $ (1 \plus{} x^2 \plus{} x^4)(1 \plus{} x^2 \plus{} x^4 \plus{} x^6 \plus{} x^8)^3$) and use the roots of unity filter (but it is equivalent here because $ 1 \plus{} x^2 \plus{} x^4 \equal{} 0$ for the two primitive third roots of unity), or expand and see how many exponents are divisible by three.[/hide]

Too bad for failing this...

[b]Edit:[/b] Yay for MATHCOUNTS practice, Ubey got a perfect. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1680756#1680756]Here[/url] is a horrible solution to #7 if anybody cares.

[quote="athunder"]thank you dnkywn, your solution was great(i got most of it, except for the most important part lol) :)

quote]

which was the most important part? i can hopefully explain it

[quote="athunder"]Well basically dnkwin said this:

$ 3(2 \plus{} \omega^{1})(2 \plus{} \omega^{2})(2 \plus{} \omega^{3})(2 \plus{} \omega^{4})(2 \plus{} \omega^{5})(2 \plus{} \omega^{6}) \equal{} 129$

How did he get 129?[/quote]

so he evaluated the polynomial $ \prod_{i\equal{}0}^{6}(x\minus{}\omega^i) \equal{} x^7\minus{}1$ at $ x\equal{}2$, giving $ \minus{}128\minus{}1 \equal{} \minus{}129$. Then he divided by the $ i\equal{}0$ term (which is $ \minus{}2\minus{}1$). so he kinda got rid of signs, but w/e

Hey, did anyone find a nice unbashy way to do #4 on nationals? Because I totally carelessed on it and ended up with a stupid 6 :mad: (I fail at combo...) and tied for second in my school...(first place got a beastly 11)

[quote="gauss1181"]Hey, did anyone find a nice unbashy way to do #4 on nationals?[/quote]

Just to clarify, #4 on nationals was the same as #6 on regionals.


This will be my first time doing science contests, but I have done multiple math team contests as well.  I want to do some research in math, but am not sure where I can start off.  Does anyone have any ideas of what a good topic could be for a high-school student?  So far, I've googled for some unsolved problems in math, but they all seem to be things like a college grad-student would be doing.  In terms of math, I have done single and multivariable calculus and will be in differential equations this fall.  I can also do most AIME problems / easy-level USAMO problems decently.

Thanks

This is a good question.  I had some luck contacting professors at local universities, but the positions that were available involved helping with a very small part of a larger project.  I'd like to know about some good topics to research as well.

First you should pick a topic that you're interested in and do some preliminary research on it.  Then you can contact professors at a local university and show them what you have been working on.  That way, you will find a good match.  Even if the professors that you contact directly don't share your specific interests, they usually have connections and are willing to help you find someone else who does.  This method worked very well for me.

to think of an idea for a topic do you simply read math books and when you come across an unsolved prolbem say "hey thats cool"?

It doesn't have to be an unsolved problem. Sometimes by looking into known techniques for solving problems you can get ideas for a project. For example, look into something called Gaussian Quadrature.

Do you get any ideas like "I wonder what would happen if _______" or "Is this somehow related to ________"? If so, you might have an idea for a project!

I just read a very interesting proof  and I think I will research extensions of this postulate or maybe that Quadrature you just suggested. However, do you think it is worth it to obtain a basic NT background (i am already familiar with NT but I think perhaps I should go through a book thorougly to learn about things like Gaussian integers. This might make me more attractive to proffesors right?

Just to be clear, Gaussian Quadrature has nothing to do (as far as I can tell) with Gaussian Integers or number theory. But yeah, Gaussian integers are really cool too. I think t0rajir0u has a good post about them somewhere in his [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=634]blog[/url]. In fact, his blog is filled with lots of interesting ideas.

Gaussian integers are only a very special case of number rings, which probably is the better direction to go  :)


Assuming we only have natural numbers?

And are you saying for all possible permutations of ${k_i}$, ie $n!$ different terms in each summation?

[b]a)[/b] I don't think it's correct,for example for $m=n$ left hand will be $\frac{1}{m!}$ but the right hand is $\frac{m^m}{(m!)^2}$

I think the sum is supposed to be taken over all n-tuples of non-negative integers that sum to m.

For example, if n = 2 and m = 3, then sum over the pairs (0,3), (1,2), (2,1) and (3,0).  If n = 3 and m = 2, then sum over the triples (0,0,2), (0,2,0), (2,0,0), (1,1,0), (1,0,1), and (0,1,1).  Then the formula seems to work.

oooooops!! maybe you are right nsato I thought $0!=0$ :oops:


Prove Fn = (F[size=59]n-1[/size]-1)^2 + 1 using mathematical induction.

Sorry. It should read prove that Fn = (Fn-1-1)^2 + 1 using mathematical induction, where Fn is the nth Fermat number.

Wait... do I understand correctly that you want to show:
$F_{n}= (F_{n-1}-1)^{2}+1$ with $F_{n}= 2^{2^{n}}+1$
via induction¿
That's nonsense as long as you don't want to reduce inductive reasoning to usage of naturals.

But in general: most problems saying "prove $A$ not using $B$" are stupid as long as you are not looking for a nicer, shorter or more elementary proof (definitely not the case here).

Here goes one proof using induction  :lol: :
We see that $F_{n}= 2^{2^{n}}+1 = (2^{2^{n-1}})^{2}+1 = (2^{n-1}+1-1)^{2}+1 = (F_{n-1}-1)^{2}+1$.
No we are supposed to use induction at least one time, so let's also show that $\prod_{k=1}^{n-1}F_{k}= F_{n}-2$ (will  be left to the reader) ;)


Ok here is my situation i just started all this all this math contest thing (ive done UIL but i dont count that) the thing is i can not find anything that is going on right now or really soon that i can join.... and im very frustrated about it because a want to get started as soon as posible. Can anybody tell me anything thats going on that i can join? I dont really care how small. ( I live in Texas for regional stuff)

The only one I can think of coming up is ARML which is in early June. However, I think in many, if not most, cases, it may already be too late to join a team. The AMC tests, which begin in early Jan/late Feb are the main ones that participates in. But in the meantime, you can still prepare right now by studying on your own, going through past problems whicn you can find here, or go to a summer camp in math.

I suggest you take a look at [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php#tx]here[/url].


geometry
circumcircle
geometric transformation
rotation
geometry proposed

Let $ABCD$ and $AEFG$ be similar inscribed quadrilaterals, whose vertices are labeled
counter-clockwise. Let $P$ be the second common point of the circumcircles of the quadrilaterals beside $A$. Show that $B, E$ and $P$ are colinear.
Actually this is an easy problem and I have a nice and short solution for it, but I want to know solution using rotations and homotethy.

The source is Austria 2002, op02 problem Austria 7. My solution was 2 lines long:

With directed angles modulo 180°, the concyclic points A, E, G and P yield < APE = < AGE, while the concyclic points A, B, D and P yield < APB = < ADB, but from the similarity of our quadrilaterals ABCD and AEFG we obtain < AGE = < ADB, so that < APE = < APB, and the points B, E and P are collinear.

A solution using spiral similitude (sketched):

Since our quadrilaterals ABCD and AEFG are directly similar, there exists a spiral similitude mapping the quadrilateral ABCD to the quadrilateral AEFG. This similitude maps the points B and C to the points E and F, and leaves the point A fixed, so that this point A must be the center of this similitude. Hence, we can state: The point A is the center of the spiral similitude mapping the points B and C to the points E and F.

Now, a well-known theorem states that, in order to find the center of the spiral similitude mapping two given points X and Y to two other given points X' and Y', one has to take the point of intersection W of the lines XX' and YY', and to draw the circumcircles of triangles XYW and X'Y'W. These circumcircles intersect at the point W and also at another point, which is the center of the spiral similitude mapping the points X and Y to the points X' and Y'. Applying this to the points B and C and the points E and F, we see that, if W is the point of intersection of the lines BE and CF, then the center of the spiral similitude mapping the points B and C to the points E and F is the point of intersection of the circumcircles of triangles BCW and EFW different from W. But we know that the center of the spiral similitude mapping the points B and C to the points E and F is the point A. Thus, the point A is the point of intersection of the circumcircles of triangles BCW and EFW different from W. This shows that the point W lies on the circumcircles of the quadrilaterals ABCD and AEFG, so that W = P, and this yields that the point P lies on the lines BE and CF.

Sorry for being so short, I don't have the time for writing too many details.

  Darij

Yes, my solution is the same. Now, I'll carefully read your solution using spiral similitude. Thank you very much. :)


induction
rotation
geometry
rectangle
Gauss
modular arithmetic
arithmetic series

[quote="now a ranger"]I don't understand why $1+2+3+...+n=\frac{n(n+1)}{2}$

I just use it as an understood fact. How do you know this is true? By a proof?  :maybe:[/quote]

There are quite a few proofs to show that $\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$; induction works, as well the proof shown on the left of the AoPS website. (It shows a right triangle formed by rows of squares with one, then two, then three and so on; the triangle is rotated $180^\circ$ to create a rectangle with the original image that has side lengths of $n$ and $n+1$.)

[hide="induction"]Base case: $n=1$, clearly $1=\frac{1\cdot2}{2}$.
Assume $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$. Adding $n+1$ to both sides makes the LHS $\sum_{k=1}^{n+1}k$. The RHS becomes
$\frac{n(n+1)}{2}+n+1$
$(n+1)(\frac{n}{2}+1)$
$(n+1)(\frac{n+2}{2})$
$\frac{(n+1)(n+2)}{2}$, and the induction is complete.
[/hide]

[quote="funcia"][quote="cincodemayo5590"][quote="funcia"]not a proof exactly

1 2      3....
n n-1 n-2....
 add the terms 1 2 3 4………n is 2* the sum so divide it by 2

example:
1 to 10
1   2...5
10 9....6
so they are all 11 and there are 5 of them(10/2=5!!)
so 5*11
but if you just wanted a formula you do 10*(11/2)
i hope i am making myself clear... :([/quote]

That is how Gauss originally did this problem, and is similar to the rectangle diagram method.  :)[/quote]

though he was obviously better when it comes to explanations i think.....
anyway is there some book or something about Gauss? i keep hearing things about him(not just on aops of course)
and was wondering how everyone knows so much about him...
did he write an autobiography?[/quote]
What I know about Gauss I learned from reading THE BIG PICTURE in the AoPS volumes and from seeing his name in the forums.

[hide="proof by hockeystick"]
$\binom{1}{1}+\binom{2}{1}+...+\binom{n}{1}=\binom{n+1}{2}=\frac{n(n+1)}{2}$[/hide]

[hide="proof by hockeystick"]
$\binom{1}{1}+\binom{2}{1}+...+\binom{n}{1}=\binom{n+1}{2}=\frac{n(n+1)}{2}$[/hide]
ooh thats a kewl method i haven't seen that before

Ask a Ranger, if you don't know the proof for this you probably don't know the proof for the arithmetic series formula- its basically the same method posted by funica and used by gauss
just thought i'd say that :D

[quote="now a ranger"]I don't understand why $1+2+3+...+n=\frac{n(n+1)}{2}$

I just use it as an understood fact. How do you know this is true? By a proof?  :maybe:[/quote]
[hide]
well, when you have an arithmetic sequence summed together, the first and last has the same sum as the second and secondlast.
for example, $1+2+3+4+5+6=(1+6)+(2+5)+(3+4)$
so in other words, it is the sum of the first and last, times half of how many numbers there are (because every two numbers make a sum.
so
$1+...n=(n+1)+(n-1+2)+(n-2+3)+...(n-\frac{n}{2}+\frac{n}{2}+1)=(n+1)(\frac{n}{2})$
it also works for an odd number of addends because the middle term is half of the sum of the first and last terms.
[/hide]

[quote="bpms"][hide="proof by hockeystick"]
$\binom{1}{1}+\binom{2}{1}+...+\binom{n}{1}=\binom{n+1}{2}=\frac{n(n+1)}{2}$[/hide][/quote]

You might as well just give the counting proof, which is much easier to understand.

[b]Problem:[/b]  How many ways are there of picking two elements out of $n+1$ elements?

Solution:  Well, one way is to pick any of them (there are $n+1$) ways to do this, and then to pick any other (there are $n$).  But we can pick a given subset $\{ a, b \}$ in two ways - either by picking $a$ and then $b$, or by picking $b$ and then $a$ - so we have to divide by $2$.  Hence there are $\boxed{ \frac{(n+1)n}{2}}$ ways.

But another way is to arrange them all in a row, from first to last.

- Pick the $1^{st}$ one first.  Then the other $n$ elements can be the second member of the subset.
- Pick the $2^{nd}$ one first.  Then the $n-1$ elements after that can be the second member (we don't count the $1^{st}$ element because we already counted that subset).
- Pick the $3^{rd}$ one first.  Then the $n-2$ elements after that can be the second member (same deal).

And so forth.  In total, there are

$\boxed{ n+(n-1)+(n-2)+...+3+2+1 }$

Distinct two-element subsets.

This is a powerful technique discussed in Engel, and it is given the amazing name [b]counting the same thing in different ways.[/b]  Since both expressions count the same thing, we conclude that

$\boxed{ \frac{n(n+1)}{2}= 1+2+3+...+n }$

:)

This is how I understand the sum of an arithmetic sequence: :maybe:

S = 1+2+3+...+(n-1)+n
S = n+(n-1)+...+3+2+1

2S = (1+n)+[(2+(n-1)]+...+[(n-1)+2]+(n+1)
2S = n(1+n)
S = n(1+n)/2

I hope this can help you! :P

The most elegant proof is by inducion, as it was previously mentioned.

Although, consider the arithmetic progression:

$S=1+2+3+...+n$

In order to find the sum of infinite terms, there exists the following type:

$S_{n}=\frac{n}{2}(a_{1}+a_{n})$

Substituting in the above formula, we have:

$S_{n}=\frac{n}{2}(1+n)$ and we are done .

I personally think the diagram method that is shown on the left of this site is the most elegant, since it involves no words at all and is quite short.  :)

This can be derived w/o hockeystick if we first asasume it's a quadratic. Assume it's in the form $a{n}^{2}+b{n}+c$. Clearly, $c=0$, as the sum of the first 0 #s is 0. Now, $S(n+1)=S(n)+{(n+1)}^{2}$, so $a{(n+1)}^{2}+b{(n+1)}=a{n}^{2}+b{n}+{(n+1)}$. Rearranging, we get $a{n}^{2}+{(2a+b)}{n}+a+b=a{n}^{2}+{(b+1)}{n}+1$. Now, solving for a and b, we get $a=\frac{1}{2}$, and $b=\frac{1}{2}$. So, we simplify to get $S(n)=\frac{(n)(n+1)}{2}$

[quote="perfect628"]asasume it's a quadratic.[/quote]

To do this rigorously you need to prove that the method of finite differences works.

[hide="alternate way to explain funcias way"]The average of each pair of terms is $\frac{n+1}2$, and there are $n$ terms, so their sum is $\frac{n(n+1)}2$.

This makes it easier to account for when $n\equiv1\pmod2$ (the middle term wouldn't be paired with anything).[/hide]


modular arithmetic
induction
number theory unsolved
number theory

Find all groups of positive integers $ (a,x,y,n,m)$ that satisfy $ a(x^n \minus{} x^m) \equal{} (ax^m \minus{} 4) y^2$ and $ m \equiv n \pmod{2}$ and $ ax$ is odd.

Sorry to revive an old topic. But the problem is hard. Can someone solve it?

HINT :
$ax^m-4$ divides $x^{n-m}-1$
by induction $ax^m-4$ divides $4^{k-1}x^{n-mk}-a^{k-1}$
so $ax^m-4$ divides $4a^{k-2}(a-1)-4^{k-1}x^{n-mk}(4-x^m)$
hence $ax^m-4$ divides $ax^m+4$
thus we get $x\leq 4$ checking all case we get $x=1$ is only solution.

[quote="dinoboy"][quote="subham1729"]
by induction $ax^m-4$ divides $4^{k-1}x^{n-mk}-a^{k-1}$
[/quote]

Why is this?[/quote]
$ax^m-4$ divides $a.x^{n-m}-4x^{n-2m}$ and also $ax^{n-m}-a$
substracting we get it for $k=2$ similarly going on we get for all k such that $n\geq mk$

[quote="dinoboy"]Sorry, I quoted the wrong line. What I meant is how did you get this:
[quote="subham1729"]
so $ax^m-4$ divides $4a^{k-2}(a-1)-4^{k-1}x^{n-mk}(4-x)$
[/quote][/quote]

Put $(k-1)$ in place of $k$ and substract both and multyply by $4$

[quote="dinoboy"]That'll get you a $4 - x^m$ term or something, not $4-x$...[/quote]
oops, now edited

Your solution is still completely wrong, if you expand properly you simply get $ax^m - 4$ divides $ax^m - 4$ which is useless...

To actually solve this problem the fact that $y^2$ is a perfect square is crucial.

I am very interested in a solution to this difficult problem, so come on please post some solution! :)


geometry
trigonometry
rotation
function
parallelogram
perimeter
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14629[/img]

 :(

We begin with co-ordinate geometry, and then interpret the result.
In a rectangular system with origin $ P$,
we can define any line $ l$ through $ P$ with a suitable choice of $ s$ such that
$ l : y\cos{s} \minus{} x\sin{s} \equal{} 0$
so that for points $ A(k,p) B(l,q)$ and $ C(m,r)$ we get the required distance as
$ d : \leq | (p \plus{} q \plus{} r)\cos{s} \minus{} (k \plus{} l \plus{} m)\sin{s}| \equal{} \sqrt {(p \plus{} q \plus{} r)^{2} \plus{} (k \plus{} l \plus{} m)^{2}}|\sin{[\arctan{\frac {p \plus{} q \plus{} r}{k \plus{} l \plus{} m}} \minus{} s]}|$
but we can arbitrarily interchange horizontal and vertical axes, maximizing $ d$ when
$ \tan{s} \equal{} \frac {p \plus{} q \plus{} r}{k \plus{} l \plus{} m}$ (here we have rotated the axes so the above sine function takes a $ 1$ value instead of $ 0$).
To construct it, look for a point on $ l$ so that a co-ordinate system centred at that point would give
$ s \equal{} 0$
or
$ p \plus{} q \plus{} r \equal{} 0$.

Conclusion:

$ l$ is $ PG$ extended.

(Here is proof:
Consider projections $ U, V, W,$ of $ A, B, C,$ respectively on any line through centroid $ G$. There must be a side of this triangle that the line does not cut internally, say $ BC$.  Let $ X$ be the middle of $ BC$ and $ Y$ be the projection of $ X$ unto the line. Then $ \frac {VB \plus{} WC}{2} \equal{} YX$ by proportion. Also, $ AU \equal{} 2 YX$ so $ AU \plus{} CW \plus{} BV \equal{} 0$, precisely $ p \plus{} q \plus{} r \equal{} 0$.)

Draw three parallelograms PBDC,PAEC,PBFA.
d is the line through P satisfying the sum S=AA'+BB'+CC' where X' is the orthogonal proj of X on d.
S=AA'+BB'+CC'=AA'+DD'=BB'+EE'=CC'+FF'.
* If d doen't cut the perimeter of triangle ABC. Denote M be the midpoint of AD. S=2MM', which implies d passes through P and perpendicular to PM.
* If d cuts the perimeter of ABC.
We have known that if an arbitrary line through P cuts the perimeter of triangle ABC, then it must cut two of three sides of ABC.
Suppose d cuts two segment lines [AB],[AC]. Then, BC separates A,D. So, $ S\equal{}AA'\plus{}DD'\leq AD$, which gives us d is the perp from P to AD. ( easy to solve when the ortho proj of P on AB doen't lie on [AB]. :wink: )
Similarly, if d cuts two segment lines [BA],[BC], d is the perp from P to BE. If d cuts two segment lines [CA],[CB], d is the perp from P to CF.
Compare the length of AD,BE,CF. The max length of them is the max value of the sum S=AA'+BB'+CC', related to the construction of d.


Hello everybody... :lol: I need some physics books to download for IPHO...And also physics tests... Please give some links to download physics books and tests... Thanks

yes, how would one study for IPhO?

There are obviously alot less resources in physics than the math olympiads

http://www.phy.ntnu.edu.tw/apho2001/

http://www.geocities.com/links2ipho/problems.html

http://www.komal.hu/verseny/feladat.cgi?a=honap&h=200612&t=fiz&l=en


"I want exactly one pound of flour," said the customer to the  miller.
"Sorry," said the miller. "My scales are faulty. One arm is longer than the other."
"Do you have some lead pellets and a one-pound weight?" asked the customer.
The miller provided these things and the customer was able to weigh a pound of flour.
How did she do it?

[i]from "Fantastic book of Math Puzzles" by Margaret C. Edmiston[/i]

[u]Please hide Your solution . Write inside Hide ... /Hide[/u]

[hide="Solution?"]Assume (without loss of generality?) that with some combination of lead pellets and the one-pound weight, the balance will report an equilibrium. We search for the equilibrium. After obtaining it, remove the one-pound weight, and replace with flour until equilibrium is reached. The flour that was added must be equal to one pound.[/hide]


function
logarithms
ratio
probability
probability and stats

This result was a little surprising to me.

By Blumenthal's 0-1 law, if $ f$ is a positive function and $ B$ is a brownian motion with $ B(0)\equal{}0$, then there is a constant $ c\in [0,\infty]$ such that
\[ \limsup_{t\to 0^\plus{}} \frac{B(t)}{f(t)} \equal{} c
\]
almost surely.

It seems that it's just a consequence of the Law of Iterated logarithm. Am i wrong?

You are. $ f$ may be highly irregular and you may just easily miss all the points where the iterated logarithm bound is attained ;) In other words, the upper limit of the ratio is not the ratio of the upper limit of the numerator to some simple quantity depending on the denominator...

do you have any insight on how one might prove the result, fedja?

You mean, how to prove Blumenthal's zero-one law for the Brownian motion or what? :? You see that it, indeed, yields the result immediately, don't you: the event that the upper limit is above any particular level belongs to $ \mathcal F_0^\plus{}$ and all such events have probability either $ 0$ or $ 1$. Now just consider this probability as a function of the level. At a certain point it'll have a jump from $ 0$ to $ 1$. This point is your $ c$.

[quote="fedja"]You mean, how to prove Blumenthal's zero-one law for the Brownian motion or what? :?[/quote]

actually i meant using LIL but i see that i read your first post wrong now :)


You are organizing 3 blue books and 5 red books on a shelf with 9 spaces for books.

(a) Are the 3 blue books distinguishable from each other given the problem's information?
(b) How many ways are there to arrange the 3 blue books and 5 red books on the shelf?

[hide]a) Usually, from my experience, it's assumed that the blue books aren't distinguishable from each other.

b) Proceeding, we consider the "empty space" as a book to arrange. Our answer is
$ \frac{9!}{5!\cdot 3!\cdot 1!}\equal{}\fbox{504}$[/hide]

Could you please explain your second equation to me? I don't understand..

It's a formula in probability.

$ \frac{n!}{m_1!m_2!...}=\# of combinations$

n=total number of objects.

each m is the number of each type of objects. (5 for the red books, 3 for the blue ones, etc.)

Note that the sum of all m's must equal n.

What is the $ 1!$ in the denominator for?

Just to be neat. It's the same reason we write $ ...\cdot1$ in expansions of factorials.


Let $n$ be an integer such that $n > 3$.  Suppose that we choose three numbers from the set $\{1, 2, \ldots, n\}$.  Using each of these three numbers only once and using addition, multiplication, and parenthesis, let us form all
possible combinations.
(a) Show that if we choose all three numbers greater than $\frac{n}{2}$, then the values of these combinations are all distinct.
(b) Let $p$ be a prime number such that $p \leq \sqrt{n}$.  Show that the number of ways of choosing three numbers so that the smallest one is $p$ and the values of the combinations are not all distinct is precisely the number of positive divisors of $p - 1$.

The possible combinations of $ (a,b,c)$ are $ a\plus{}b\plus{}c, c\plus{}ab, b\plus{}ca, a\plus{}bc,(b\plus{}c)a,(c\plus{}a)b,(a\plus{}b)c,abc$. If $ a<b<c$, then $ a\plus{}b\plus{}c < c\plus{}ab < b\plus{}ca < a\plus{}bc$, $ (b\plus{}c)a<(c\plus{}a)b<(a\plus{}b)c<abc$, $ b\plus{}ca<(c\plus{}a)b$, $ a\plus{}bc<(c\plus{}a)b$. So some combinations are equal iff $ a\plus{}bc \equal{} a(b\plus{}c)$.

(a) If $ \frac{n}2<a<b<c\le n$, then $ a>\frac{c}2$. So $ a(b\plus{}c\minus{}1) > \frac{c}2(b\plus{}b) \equal{} bc$, i.e., $ a(b\plus{}c) > a\plus{}bc$. Thus all combinations are distinct.

(b) Let $ p<b<c$ be the chosen numbers. Also let $ d$ be the positive integer such that $ b\equal{}p\plus{}d$. We have $ p\plus{}bc \equal{} p(b\plus{}c)$. Therefore $ p\plus{} (p\plus{}d)c \equal{} p(p\plus{}c\plus{}d)$ and we get $ c \equal{} p\plus{}\frac{p(p\minus{}1)}d$. But $ b<c$, so $ p\plus{}d < p\plus{}\frac{p(p\minus{}1)}d$, i.e., $ d^2<p(p\minus{}1)$. Therefore $ d<p$. But $ d$ must divide $ p(p\minus{}1)$, so $ d$ divides $ p\minus{}1$. Clearly, for any divisor of $ p\minus{}1$, we get a satisfying triple. So we are done.

Босинг брат ман хам шундай ишладим


superior algebra
superior algebra solved

Let A be a ring s.t. x 2 =0 =>x=0. Let M={a in A | a 2 =a}. Prove that if a and b are in M then a+b-2ab is also in M.

[i]Romanian olympiad/2'nd round/1998/12'th grade[/i]

Are you sure that this is the statement of the problem? I think the ring must be commutative.

Actually, the condition x 2 =0 => x=0 is enough to prove that ab=ba for all a and b in M (that's the idea of the problem).

Oh, sorry. I have a terrible day. Indeed, if in a rind we have x^2=0 implyes x=0 then any idempotent element is in the center ot the ring. Indeed, let z^2=z and let any x in the ring A. Then (zxz-xz)^2=(zxz-zx)^2=0, so xz=zx=zxz. And in fact, if the ring is finite, then the number of elements of M is a power of 2.

That's exactly what I did, but looks like you saw it immediately! :D It took me some time.

Just so I can move this problem:
Because $a,b \in M$ and from what grobber and harazi proved, $(a+b-2ab)^2=a^2+b^2+4(ab)^2+2ab-4a^2b-4ab^2$ = $a+b+4ab+2ab-4ab-4ab=a+b-2ab$. So $a+b-2ab \in M$.


Let $ a_1, a_2, ..., a_n$ are permutaion of$ \{1,2,...,n\}$ Find the maximum and minimum value of the $ a_1 a_2\plus{}a_2 a_3\plus{} a_3 a_4\plus{}...\plus{}a_n a_1$

by AM-GM

$ \sqrt1\plus{}\sqrt2\plus{}...\plus{}\sqrt{n}\equal{}\sqrt{a_1}\plus{}\sqrt{a_2}\plus{}...\plus{}\sqrt{a_n}\ge{a_1}a_2\plus{}a_2a_3\plus{}...\plus{}a_na_1\ge\sqrt[n]{(a_1a_2...a_n)^2}\equal{}\sqrt[n]{(n!)^2}$ $ \rightarrow$ $ \text{max}\{a_1a_2\plus{}a_2a_3\plus{}...\plus{}a_na_1\}\equal{}\sqrt1\plus{}\sqrt2\plus{}...\plus{}\sqrt{n} , \text{min}\{a_1a_2\plus{}a_2a_3\plus{}...\plus{}a_na_1\}\equal{}\sqrt[n]{(n!)^2}$

Is it????

sorry I'm a new member here and only 7th grade...


logarithms
inequalities unsolved
inequalities

Let a,b,c,d,e are non-negative real numbers which satisfy a+b+c+d+e=1
Find the maximum value of P=abc+bcd+cde+dea+eab

The problem is on a magazine in China called 数学通讯(I don't know how to say in English)1996-8
And the solution is from Li Xiaolong from Liaoning,China.A perfect solution!

Another solution from Zheng Yingdi
$(a+b+c+d+e)^3=\sum\frac{1}{3}(a^3+b^3+c^3)+\sum3a(b-c+d-e)^2+24(abc+bcd+cde+dea+eab)$
$\geq25(abc+bcd+cde+dea+eab)=25P$
So $P\leq\frac{1}{25}$
Then the same.

Someone guess that if $x_1,x_2,\cdots,x_n(n\geq5)$,$x_1+x_2+\cdots+x_n=1$
and $n>k>n-\ln n-\frac{\ln^2n}{2n}-\frac{2\ln^3n}{3n^2}-\frac{9\ln^4n}{8n^3}$,
then $x_1x_2\cdots x_k+x_2x_3\cdots x_{k+1}+\cdots+x_nx_1\cdots x_{k-1}\leq n^{1-k}$.
But I don't know how to prove it.
Maybe there isn't a very simple proof.


I've been looking at Intermediate Algebra and Complex/Numbers Trig.

Looking at the course descriptions, I seem to know most of the basic concepts covered in those courses. Would taking these courses still be of great benefit to me in future problem solving and I should really think about still trying to do them? Or would it be a better idea to just do olympiad courses (and in those courses I seem ready for them but the material would be almost all new to me) and only do intermediate courses if I have extra time later?

Also, would going through AoPS 2 make up for some of the material in classes like those?

In case this is relevant somehow, I get around 120-130 on AMC 12 and ~ 8 on AIME.

I'd start by looking at the [url=http://www.artofproblemsolving.com/Classes/IntermAlg/PostTest.pdf]Intermediate Algebra post-test[/url].  If you can do most of those problems without much difficulty, then you're probably ready for Olympiad-level coursework.  If not, then that's probably the best class for you.

I should add that you're exactly the target audience for WOOT.

AoPS Volume 2 does have a lot of this material as well.

The Interm Complex/Trig is an AMAZING class. That's how I got really good with complex numbers. :yup:

I really want to take that class but its not offered. :(!


ratio
geometry
3D geometry
tetrahedron
octahedron
pyramid
geometry unsolved

A regular tetrahedron and a regular octahedron have edges of the same length. What is the ratio between their volumes?

I think it should be $\frac{V_{\hbox{tetrahedron}}}{V_{\hbox{octahedron}}}=\frac 14$. For many of the polyhedra we usually work with (prysms, pyramids, etc.), the following remarkable formula works for the volume: $V=\frac 16\cdot h\cdot(A_1+4A+A_2)\ (*)$, where $h$ is the altitude, $A_1,A_2$ are the areas of the "top" and "bottom" faces, and $A$ is the area of the section through the polyhedron equidistant from the "top" and "bottom" faces.

Apply $(*)$ to the construction [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21736]here[/url], and the result follows.


geometry
rectangle
combinatorics unsolved
combinatorics

Find all the couples of naturals $(m,n)$ such that it's possible to paint white or black all unit-square of a rectangle $m \times n$ such that for each unit-square, the number of unit-square which have the same colour than it and which have at least a common vertex with it (it-self included), is even.

It looks very familiar. I think it's from last year's Iranian TST (I remember solving something very similar).

Anyway, it's very simple:

Construct the graph having the unit squares as vertices and in which two vertices are connected iff the corresponding squares have at least one common vertex (but do [b]not[/b] coincide) and the same color. The hypothesis says that every vertex in this graph has odd degree, so we must have an even number of vertices. This means that one of $m,n$ must be even. Conversely, suppose $m$ (the number of rows) is even, for instance. Color the first two rows in black, then the next to rows in white, then the next two in black again, and so on.


1. $ \sum_{k \equal{} 1}^n \frac1{\sin kA \plus{} \sin (k \plus{} 1)A}$

2. $ \sum_{k \equal{} 1}^n \frac1{\sin kA \cdot \sin(k \plus{} 1)A}$

2) $ sin((k\plus{}1)A \minus{} kA) \equal{} sin[(k\plus{}1)A]cos(kA) \minus{} cos[(k\plus{}1)A]sin(kA) \equal{} sin A$

$ S sinA \equal{} cotkA \minus{} cot(k\plus{}1)A$

Rest is easy


If $x \equiv y \mod 4$ then show that $2^{x}\equiv 2^{y}\mod 5$.

[hide]If $x \equiv y \mod 4$ then $y \in \{0, 1, 2, 3 \}$. So $x \in \{ 4n, 4n+1, 4n+2, 4n+3 \}$ with $n \in \mathbb{Z}$

By replacing, we get

$2^{4n}\equiv 1 \pmod 5$, $2^{4n}$'s last digit being 6.
$2^{4n+1}\equiv 2 \pmod 5$, with $2$ as the last digit of $2^{4n+1}$
$2^{4n+2}\equiv 4 \pmod 5$, last digit 4
$2^{4n+3}\equiv 8 \pmod 5$, last digit 8

[/hide]

$x\equiv y\pmod{\varphi (m)}\implies a^{x}\equiv a^{y}\pmod m,\ \gcd (a,m)=1$

I think that all you need to note is that $2^{4}\equiv 1 \bmod 5$, so powers of $2$ cycle through $\bmod 5$ in periods of $4$.

what's that boxedexe? Euler's theorem?

Heres how I did it:
$x \equiv y \mod 4 \text{ so }\exists k \in \mathbb{Z}\text{ such that }x = y+4k$
Without loss of generality $\text{let }x \ge y \Longrightarrow k \ge 0$
$2^{4}\equiv 1 \mod{5}\text{ so }2^{4k}\equiv 1 \mod{5}$
also:
$2^{y}\equiv 2^{y}\mod{5}$, so multiply two equations to get
$2^{y}\cdot 2^{4k}\equiv 2^{y}\cdot 1 \mod{5}$
$2^{y+4k}= 2^{y}\mod{5}$
$2^{x}= 2^{y}\mod{5}$

I was hoping to do something more elementary than Euler's theorem, like what me or icx did, but was hoping maybe there'd be a more elegant way of proving it - as it feels a little clumsy.

since 5 is prime, we can use Fermat's Little Theorem instead of Euler's generalization.

Yes.  And you would have $x-y=4k\implies a^{x}\equiv a^{4k+y}\equiv a^{y}\pmod 5,\ \gcd (a,5)=1$.


[color=orange] I was wondering what excatly is nuclear physics? I understand it the study of nuclei and the forces of protons and neutrons. But could it be applied to making a nuclear weapon? [/color]

Why would you want to know ???? :)

I think that knowing Nuclear Physics is essential to somebody who wants to build such a weapon.

yes i want to make one i baught some U238 so someone tell me how to make one please before i decside to kill something else


AIME Statistics are now available on the AMC website.

http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2004-aa/04aimestats.html

I don't know when the USAMO list will be out, but would anyone like to make predictions on the cutoff now that the scores can be seen? (and it is quite obvious how the AIME I was a lot easier than II)

[quote="MithsApprentice"](and it is quite obvious how the AIME I was a lot easier than II)[/quote]

Thanks for sharing the link to the statistics.

Well, the mean and modal scores on BOTH AIMEs seem to be down this year compared to previous years, and yet the peak seventh-grade-and-below score (which is the score I track the most from year to year) is higher on the AIME II than on the AIME I, so for the right person, getting to the cut-off on the AIME II was not insuperably hard. And the number of perfect papers on the AIME I this year was also unusually high. There is always some randomness in what kind of problem best fits what kind of test-taker.

The USAMO list is out. :-D

And I'm on it! Yayyyyyy

Where can the list be found?

Congratulations, also =)

Actually, I'll answer my own question, in case anyone else is wondering..
http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamoqualstate.html

Congrats to those who made it =)

[quote="timhlu"]So the floor was 8....[/quote]

I don't know..it seems like the floor was 7

It depends on your grade.. for a freshman, it may have been 6 or 7.

I thought the floor was fixed for the lower grades...

It is fixed, and I believe it's a 7 because I got a 6...

-interesting_move

I got a 1 w00t, lol, but my school got most improved school in NJ or Bergen county or something for the AMC, so thats good, i guess, only one person got to AIME last year, this time 4 of us got to the AIME


Alcumus
algorithm
videos
pigeonhole principle
probability
MATHCOUNTS
geometry

Alcumus is now available again.  Most of the changes to Alcumus will be invisible to experienced users, as the most significant changes are in the Alcumus algorithm that selects problems.  Here are some of the main changes you will see:

* You will be able to choose which topics you want to study.  You can still let Alcumus choose your topics for you, and we recommend you do so most of the time, but if you want to focus on a single topic, you can.

* We have eliminated the 60-question "diagnostic mode".  Alcumus will switch in and out of diagnostic mode throughout your work with it, but this will be largely invisible to you, and won't affect your capabilities.

* You'll have immediate feedback after each question regarding how close you are to completing the primary topic of the question.

* You'll have more access to the videos.

* We have eliminated the old "points" and "badges" structure.  We will develop a new one to be released in the fall, and will start another thread to solicit suggestions for it.

Over the next couple months, we will add many more counting problems to the existing topics.  We expect to release an Algebra component by year-end that will be aligned with our Introduction to Algebra book and our Algebra 1/Algebra 2 classes.

Enjoy!

I just answered a few questions correctly, and then the exact same questions kept appearing.  I think this is a glitch.  Also, the solution pages say that I am in diagnostic mode.

Having looked at your report, I think it has something to do with your having answered basically all the problems in the system.  But we should at least have it mix things up a bit -- we'll look into it.

Has anyone else had this behavior?

I checked out the videos section, but it seems like I only have a few of them available. Do I have to wait until they rest are uploaded or do I have to keep on doing more problems?

Keep doing problems; you should also be able to see videos relating to a problem you have just done (there should be links below the solution if there are videos relevant to that problem).

Yeah, same here, I got all of my videos taken away.

But I think I'm doing better on the problems now.  They seem a little easier...........

Also, what about that "Basic Pigeonhole"??

The [url=http://www.artofproblemsolving.com/Wiki/index.php/Pigeonhole_Principle]Pigeonhole Principle[/url] (@maybach).

A few things:

1. I can't go to the video in the solution page.
2. Why do I only have one chance to answer?
3. When rating problems, all of the stars say the same thing (Extremely Easy, Nothing, Very Poor).

[quote="AIME15"][quote="rrusczyk"]Has anyone else had this behavior?[/quote]

Yep.  Exact same as GoldenFrog1618.[/quote]

I think it's the same cause.

i was using the select a question from current topic feature when i received the same problem twice (with several problems in between, so it's not a big problem)

the first time, i gave up
the second time, i solved it, but it still said that i gave up

however, the percentage on the side increased

because of the increase in percentage, i'm assuming that Alcumus considers the problem correct but just showed the previous screen?

edit: another slightly related problem

i received a problem i haven't done yet (yet as in after alcumus came up)
i entered the correct answer (confirmed from solution)

however, the screen said i answered incorrectly, and claims my answer was 528 (when in reality the answer was 1199976, so...)

my percentage did increase like last time

Hello, vallon22, thanks for pointing this problem out. It should be fixed now. It's mostly a display issue. As you astutely noticed, the percentages did go up, so your progress should not have been affected by the bug.

Thanks for everyone for keeping us posted on any bugs you find on Alcumus!

:D

I went on alcumus yesterday and passed a topic and got 4 new topics to study. But today when I got on I lost all my work from yesterday. Is this a bug or something?

Sometimes when I get a string of problems correct (5+), my Alcumus scores will [b]drop[/b] about a point...

I have the same exact problem! Except the progress bar also decreases along with my percentages. Also, my percentage dropped, but i got about 70 questions right out of 132 incorrect/giveup. Is this a slight glitch or not?

My progress in a topic went from 45% yesterday to 11% today is this a glitch but alcumus is remembering my questions and answers now.

Q: How many new questions were just added? And how many total are there now? I know I've completed about 1270, and many have completed much more.

I like the idea that Introduction to Algebra is also coming out, but are Intermediate topics also coming out sometime? Are there any plans to include Intermediate Algebra/Counting & Probability in Alcumus in the future?

There are plans, but they're in the pretty distant future.  We'll fill out the introduction space first.

[quote="HappyLand"]Should we time ourselves while doing Alcumus problems? Or just spend one hour on each ;)[/quote]

Alcumus is just practice with lots of problems, so I'd say you don't have to time yourself. Usually people time themselves in actual contest problems, like MATHCOUNTS, AMC, and AIME.

[quote="remember"]when you get a second try at the problem because your first answer was wrong, does that count as getting a problem incorrect?[/quote]

if you get your second try right, i'm pretty sure that it counts as a correctly done problem, although your alcumus score may reflect that first incorrect try somehow

[quote="Mikario"]I can't see any of the videos...[/quote]

If you're just starting Alcumus, you may still be in the Diagnostic Phase.  This phase gives you a wide range of problems, so as to determine your level of math.  Consequently, you may not view videos or gain/lose karma.  I forget how long the Diagnostic Phase lasts, but I think that it's about 60-100 problems.

did these glitches start from the night of 10/26/09 at about 10 o'clock p.m.? Because when I was on, my alcumus score turned to 0.00, there were no lessons available, and my percentage in each section was gone.
And after 5 minutes, the whole website said it was unavailable.

Apparently, my Alcumus score tends to increase when I get many wrong in a row.

[quote="epicfailiure"]Whenever I get something wrong, my alcumus score adds to itself, and when I get something right, it jumps significantly, but then drops very soon again.

For example, I got a question wrong, it added 5 to the score. Then I got two questions right in a row, and it went down by 11.

Can someone reply about this problem or glitch?[/quote]

Hi,

Thank you for your question. As you have not completed many problems yet, the Alcumus system will make what appear to be somewhat random adjustments to your score. This should only continue until you have done a sufficient number of problems that Alcumus has a good idea of where you are at.

Not yet. I think the staff is going to add Algebra problems in a few months (I'm not sure exactly when.)

[quote="AMCrocker"]I saw the Alcumus problems, and it only had a few subjects, like counting and prob. And the Venn Diagram. My question is, are there going to be algebra problems on there? :roll:[/quote]

Alcumus currently covers our Introduction to Counting & Probability book.  We expect to add the subjects from our Introduction to Algebra book in the next few months.

[quote="the cliu"]Could you also add some Intro to geometry (and possibly Number Theory)?
Geometry is something I really need to (and want to, for that matter) to work on.
Thanks :)[/quote]

This inquiry has been stated before. The makers Alcumus would eventually add as much as they can. As for now, just enjoy the nice free software already in Alcumus.


geometry
incenter
circumcircle
geometric transformation
homothety
geometry proposed

Given a triangle $ ABC$. A circle passes through vertices $ B$ and $ C$ and intersects sides $ AB$ and $ AC$ at points $ D$ and $ E$, respectively. Segments $ CD$ and $ BE$ intersect at point $ O$. Denote the incenters of triangles $ ADE$ and $ ODE$ by $ M$ and $ N$, respectiely. Prove that the midpoint of the smaller arc $ DE$ lies on line $ MN$.

Let $ E\^{D}N=N\^{D}C=x$, $ D\^{E}N=N\^{E}B=y$ and $ P\^{D}E=P\^{E}D=z$.
Then, we have $ M\^{D}P=y$ and $ M\^{E}P=x$.
Let $ K(\neq E)$, $ L(\neq D)$ the intersection points of the circuncircle of the triangle $ MDE$ with the lines $ EP$, $ DL$.
We have:
$ M\^{K}E=M\^{D}E=y+z=K\^{E}N$, so $ MK$ and $ EN$ are parallels.
$ M\^{L}D=M\^{E}D=x+z=L\^{D}N$, so $ LM$ and $ ND$ are parallels.
$ L\^{K}E=L\^{D}E=z=K\^{E}D$, so $ LK$ and $ ED$ are parallels.
Then, there exists and homotety which transforms the triangle $ LMK$ into the triangle $ DNE$.
The centre of that homotety is $ KE\cap LD=\{P\}$.
In particular, $ M$, $ P$ and $ N$ are collinears.


tan^2(x) + cos(2x) = 1 - cos(2x)tan^2(x)

What are the steps in order to proove it?  I think it's hard  :o

Many thanks in advance!!!

c
z

[quote="shadysaysurspammed"][hide]L.H.S:-

$\frac{sin^2x}{cos^2x}+cos^2x-sin^2x$

$=sin^2x(1-cos^2x)/cos^2x+cos^2x$

$=\frac{sin^4x}{cos^2x}+cos^2x$

$Similarly$,just simplify the R.H.S,all you have to use is $cos2x=cos^2x-sin^2x$ :) [/hide][/quote]

Hey thanks, that was quick!  2 questions though:

1) I don't get the part to go from sin^2(x) (1-cos^2(x)) to sin^4(x) , which identity is this coming from?

2) On the R.H.S, when I substitute cos(2x) with cos^2(x) - sin^2(x) and try substituting the 1 with cos^2(x) + sin^2(x) I get:

cos^2(x) + sin^2(x) - cos^2(x)tan^2(x) - sin^2(x)tan^2(x),

Here, I get stuck.  Thanks for your help.

hmm...looks like you need to practice more trigo identities...

$sin2^\theta+cos^2\theta=1$

$\implies sin^2\theta=1-cos^2\theta$

That should clear your first doubt..

Now,for the second doubt,don't make that substitution.

[hide="R.H.S"]R.H.S:-

$1-(cos^x-sin^2x)(\frac{sin^2x}{cos^2x}$

$=1-(sin^2x-\frac{sin^4x}{cos^2x})$.[/hide]

Now just substitute $1-sin^2x=cos^2x$ and you are done.[/hide]
Since you seemed pretty confused,I solved it completely but remember,everyone here will encourage you to think more rather than directly spoon feeding the solution.Since you are new,thought i should let you know :)

Thank You so much!  I'll put more effort next time!


quadratics
Vieta
algebra
algebra unsolved

Let $p, q, r$  be positive real numbers, not all equal, such that some two of the equations \begin{eqnarray*} px^2 + 2qx + r &=& 0 \\ qx^2 + 2rx + p &=& 0 \\ rx^2 + 2px + q &=& 0 . \\ \end{eqnarray*} have a common root, say $\alpha$. Prove that

$a)$ $\alpha$ is real and negative;

$b)$ the remaining third quadratic equation has non-real roots.

I type "a" as "alpha" :roll:
Suppose that the first 2 equations have one same root. So
$pa^2+2qa+r=0,qa^2+2ra+p=0$ that is
$pqa^2+2q^2a+qr=0,pqa^2+2pra+p^2=0$
if $q^2-pr=0$, then a=-q/p<0, with $p^2-qr\ge0$ we must have the equal held, so p=q=r, contradiction.
so $a=\frac{p^2-qr}{2(q^2-pr)}$, is a real.By Vieta's Theorem it's negative of course.
notic $q^2-pr\ge0$ by looking into the delta, so $p^2-qr<0$, we get the proof.□

I did it more or less similar to Skyward_Sea.
Assume that two equations hav one  similar root say $pa^2+2qa+r=0$ and $qa^2+2ra+p=0$. $(*)$
Using it we can write,
$\frac{a^2}{2r^2-2pq}=\frac{a}{p^2-qr}=\frac{1}{2q^2-2pr}$
thus $(p^2-qr)^2=4(r^2-pq)(q^2-pr)$
 $(r^2-pq) and (q^2-pr)$ are either both positive or both negative. $(**)$
For positive we are done.
For both negative we get that $\frac{q}{p}=\frac{r}{q}$  because a is complex and the roots of the eqs(*)
$(\frac{-2q \pm\sqrt{4q^2-4pr}}{2p} and \frac{-2r\pm\sqrt{4r^2-4pq}}{2q})$ gives us real parts and imaginary parts  equal to corresponding values.
$\rightarrow q^2=pr$ but it is a contradiction by (**).
Thus, both -ve assumption is wrong. It implies a is real.
All other parts are similar to the above answer.

a) If two of the equations have a common root $\alpha$, and if $\alpha \in \mathbb{C} \setminus \mathbb{R}$, then, since the two equations have real coefficients, each will also have $\overline{\alpha}$ as a root. Thus both equations are scalar multiples of $(x-\alpha)(x-\overline{\alpha}) = 0$, hence their corresponding coefficients are proportional - but this leads to $p=q=r$, which is forbidden. Therefore $\alpha$ is real, and by Viète's relations, must be negative.

b) The three discriminants are $q^2 - rp$, $r^2 - pq$ and $p^2 - qr$, and two of them are non-negative. Both cannot be null, since this again would lead to $p=q=r$, so at least one is positive. But then the third needs be negative - for example if $q^2 \geq rp$ and $r^2 > pq$, then $q^2r^2 > (rp)(pq) = p^2rq$, and so $p^2-rq < 0$.

This question is wrong, as 2 of these equations CANNOT have a common root if $p,q,r$ are all $distinct$, $real$ and $non-zero$. I have a proof that shows this:

Your convlusion that it must have omly one root is false

[quote=Wizard_32]Can you be clearer[/quote]

I will give you an example of how you are wrong.
Considered the equations.
$2x^2 -3x +1$ and $x^2 - 2x + 1$

They have a common root namely $1$ now check that your claim is false.

What we are saying is your proof is invalid.

[quote=Wizard_32][quote=AnArtist]

I will give you an example of how you are wrong.
Considered the equations.
$2x^2 -3x +1$ and $x^2 - 2x + 1$

They have a common root namely $1$ now check that your claim is false.[/quote]
The two equations are of the form $px^2 + 2qx + r =0$ and  $qx^2 + 2rx + p=0$ in which the coefficients must vary cyclicly.[/quote]

Ok take another example

$3x^2 - 2x -1$ and $-x^2 - 2x +3$ see that the co efficients vary cyclically.

[hide=nice one] lemma:- if $P(x)$ and $Q(x)$ be two distinct have a root in common ( say $\alpha$) then $P(x)-Q(x)$ and $P(x)+Q(x)$ has also root $\alpha$\\
proof is just easy to show by just taking out the factor of $(x-\alpha)$ common out!\\
WLOG, we just inspect for initial two equations and rest is just symmetrically same.\\
so we have $p(\alpha)^2+2q(\alpha)+r=0$\\
  $q(\alpha)^2+2r(\alpha)+p=0$\\
consider $f(x)=(p-q)x^2+2(q-r)x+r-p$, then $f(\alpha)=0$ and it is quadratic so we seek to find discriminant\\
$D=4(q^2+r^2+p^2-qr-pr-pq)>0 \forall p,q,r \in \mathbb{R^{+}}$\\
hence $f(x)$ has all its root real and hence $\alpha$ is real\
also $\alpha=\frac{-2q\pm \sqrt{4q^2-4pr}}{2p}<0 \forall p,q,r \in \mathbb{R^{+}}$\\
hence a part is done $\blacksquare$\\
b) $rx^2+2px+q=0$\\
$D=4(p^2-rq)$\\
also from initial two equations\\
$q^2>pr , r^2>qp\implies qr>p^2$ hence we have $D<0$ so third equation has no real roots
which proves the part b $\blacksquare$ [/hide]