abstract algebra
email
conics
parabola
ratio
geometry
calculus

How did u ppl do?

and... are ppl who got 68-71 writing the repechage? (i got 71)

I'm not sure we're allowed to discuss the problems (though I got them all ^_^)!!

Also, about 200 people were allowed to do the repechage (at least) as I only got 66 on COMC...  :(  :(

Well... we CAN discuss it after today, can't we?

ya~
i got 71 and was invited to write it

[quote="TCAtect"]Heh, I got a 67 and was invited to write it. I know a friend who got a 64 and was also invited to write CMOR. I had some problems with the last question. Really couldn't explain it properly.  :P

Got a question though, are only the top 10% scorers invited to write the CMO? I heard that about 80 get invited each year, and since roughly 200 people were invited to write the CMOR, it sounds like only around 20 people get selected for the CMO. Is this true?[/quote]

i've seen #10 (although i don't need to do it)
i am not sure y u saying it's hard to explain......it's a NT question.....
use module or some other basic techniques u can come up with a nice and short solution

Yeah, I agree with Vincent_Five, 10 was a pretty easy problem... (i hadnt seen it before...)

Also, I had a really long solution for 9 with 4 cases; did anyone have a short solution to 9???

ya

i machine gunned number 9 using parabolas. the easiest solution is sub x=1 and x=-1. you then compare them after adding

[quote="vincent_five"]
i've seen #10 (although i don't need to do it)
i am not sure y u saying it's hard to explain......it's a NT question.....
use module or some other basic techniques u can come up with a nice and short solution[/quote]

Just heard that today too. Saw a senior use modulo to solve the problem, solution quite elegant too. For some reason, I just never thought of using modulo. :(

can someone post if they get an email about their score. and by the way, i only got 61 but still got the repechage. so its probably 60 or above.

[quote="Eric Kim"](for 9) i found a really short solution using addition and subtraction of inequalities... but not sure it's right...[/quote]

Quite a few people I know used the same thing. Did you sub in x=0, 1, and -1?

i used parabola as well (some expansion and compression thing)
(0,1)(1,-1)&(-1,-1)
or
(0,-1)(1,1)&(-1,1)

When are we allowed to see the questions

Hmmm...did everyone here receive a hard copy of the CMO invitations?

Apparently, no one in our school received an invitation yet. Not even Dmitri or John (with scores 80 and 72). Weird...perhaps it is some conspiracy cooked up by our math dept. head.

Well, I received my invitation (for CMO) through a letter (addressed to me)...

Hmm...so you did "directly" receive your invitation? I thought that they sent the invites to schools or something.

Crap, that reminds me, I don't think I wrote the correct postal code on my COMC answer sheet...

woot i got my invitation for cmo :)  not apmo though hope i get one :maybe:

[quote="TCAtect"]Hmm...so you did "directly" receive your invitation?[/quote]

No, not directly... it was sent through the school, and my math teacher gave it to me...

[quote="vincent_five"]i think ur school will receive a package of invitation letters and the head of math will distribute them to u guys.......

anywayz, who else is writing APMO[/quote]
I am writing both CMO and APMO. Lucky to be a Canadian. Does anyone how many are invited to write the APMO?

Did anybody find out the problem set for this year?(2008). I couldn't find out the problem set for this year, even in the link provided above. Can you show me how to find out these problems? By the way, can non-citizen students like me take it?

Hey can anyone post the 2008 problems or post a link? Do you have even earlier ones? 2009, 2010 are available oneline.

hey bigsams, do you mind if you post a link to CMOR 2009? I can't find one myself.

(btw, this is for my friend)

These are the only ones I could find:
[url=http://www.cemc.uwaterloo.ca/cmor_files/2009/2009CMOR.pdf]CMOR 2009[/url]
[url=http://cemc.uwaterloo.ca/cmor_files/2010/2010CMOQR.pdf]CMOR 2010[/url]
[url=http://cemc.uwaterloo.ca/cmor_files/2011/2011CMOQR.pdf]CMOR 2011[/url]

And 2011 will be up tomorrow. The link for it is on the other topic. Maybe we should like archive the ones we have, just for future use?

EDITED... added the 2011 link

Thunder, how are you preparing for CMO?
it's my first time (don't say that's what she said)
and i think i could consult you since you are so good at math.

Thanks

Yeah Thunder's post has the links I am using. I was hoping that someone could post earlier ones, but those people are mostly in university now and probably don't visit AoPS much.
Does anyone know at what time 2011 will be up? Midnight? lol

@Bigsams

You must be [u]too[/u] excited for CMOR haha, I was too last year :)
Don't overkill though.


My teacher assigned us a problem that is way beyond our level(Junior in High School), well, we havent learned it  and I'm told that its a college level problem, dealing with abstract algebra.
Anyways, without further ado, the problem. "What is the remainder of 3 to the 98th power(98 is an exponent to 3, just to clarify) when it is divided by 5."

PLease, any assistance with equations and formulas as well as explanations in fairly understandable terms would be greatly appreciated. I would like to solve this myself, but some assistance to set the problem up and give me some guidance to the answer would be awesome.

Thank you very much,

Scarp

3^1 mod 5 = 3
3^2 mod 5 = 4
3^3 mod 5 = 2
3^4 mod 5 = 1
3^5 mod 5 = 3
3^6 mod 5 = 4
3^7 mod 5 = 2
3^8 mod 5 = 1

all 3^(multiples of 4) have a remainder of 1 when divided by 5.
so 3^ 96 is 1, 97 is 3, 98 is 4

That's the simple way to do it without knowing what fermat's little theorem is.

This problem uses modular arithmatic.

What we want to find is $3^{98}\pmod{5}$.

Start writing out the first few terms, mod 5.

3, 4 (actually 9), 2 (actually 27), 1 (actually 81), 3, etc.

As you can see, this pattern repeats every 4 terms, so the first term is the same as the 5th, the second is the same as the 6th, 10th, etc.

So know, we need to find what 98 is the same as.

$98\equiv2\pmod{4}$, so it is the second term in the sequence, or $\boxed{4}$.

If you don't understand modular arithmetic, you might not understand this, but it will still work if you get the idea of looking for a pattern.

Basically, $k\equiv j\pmod{m}$ means that $m|k-j$, or $m$ is a factor of $k-j$.

NOTE:  this is only true for integers!

[quote]"What is the remainder of 3 to the 98th power(98 is an exponent to 3, just to clarify) when it is divided by 5." [/quote]

Fermat's Little theorem states n^(p-1) = 1 mod p where p is prime and n is not divisible by p. ( a special case of Euler's Formula, which you can look up if you want). your problem wants 3^98 mod 5. we know 3^4 = 1 mod 5, so 3^98 = ((3^4)^24)(3^2) = 3^2 mod 5.  3^2 = 4 mod 5, which is our answer.

if you call this a college level problem, your teacher's insane....... :maybe:

I think the teacher meant college level in that number theory is a college class, but your right it is pretty simple if you've done any.


Why isn't my AMC 10A score mailed back yet?  The school complained, but aren't they supposed to be here by now???

Hmm...our school's hasn't come back either.  Maybe in a week or so would be my guess.

-interesting_move

[quote="WarpedKlown1335"]Why isn't my AMC 10A score mailed back yet?[/quote]

I remember Mr. Dunbar, the AMC director, said that mail delay is sometimes unavoidable at this time of year. And he has just posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=71140&highlight=#71140]a reply on another thread[/url] explaining the processing done at AMC headquarters. 10A results are only just now going out in postal mail.

I am in suspense about a summer program application, so I can feel sympathy,  but I'm trying not to get too worried. Take a deep breath and relax.

Yannik, it's only been...what...three weeks?  USAMTS takes several months, Mandelbrot takes at least 6 weeks (although most of that is just waiting), etc.  Plus, AMC A is probably the biggest competition around.  Give it a little time.


geometry
circumcircle
geometric transformation
reflection
ratio
trigonometry
geometry proposed

In acute triangle $ ABC,$ points $ P,Q$ lie on its sidelines $ AB,AC,$ respectively. The circumcircle of triangle $ ABC$ intersects of triangle $ APQ$ at $ X$ (different from $ A$). Let $ Y$ be the reflection of $ X$ in line $ PQ.$ Given $ PX>PB.$ Prove that $ S_{\bigtriangleup XPQ}>S_{\bigtriangleup YBC}.$ Where $ S_{\bigtriangleup XYZ}$ denotes the area of triangle $ XYZ.$

It's well-known that $\triangle{XPQ}\sim\triangle{XBC}$ and $\triangle{XBP}\sim\triangle{XCQ}$ (all similarities here are direct). First we prove some easy configuration facts: $\angle{XPQ}=\angle{XBC}$ forces $X,A$ to lie the same side of $BC$, whence $\angle{PXB}=\angle{QXC}$ forces $X,A$ to lie on the same side of $PQ$.

Let $\alpha=\angle{PXB}=\angle{QXC}$ and $\beta=\angle{XBP}=\angle{XCQ}$ so that $\alpha<\beta<90^\circ$ (since $\triangle{ABC}$ is acute and $PX>PB$). For convenience, define the similarity ratio $r = PQ/BC = XP/XB = \sin\beta/\sin(\alpha+\beta)$.

For points $U,V,W$, let $d(U,VW)$ denote the directed distance from $U$ to $VW$ (directed so that $X,A$ are "above" $PQ$ and $BC$). Then $d(Y,BC) = d(X,BC) - XY\cos\alpha$ and $XY = 2\cdot d(X,PQ) = 2r\cdot d(X,BC)$. It follows that
\begin{align*}
\frac{2[YBC]}{2[XPQ]} = \frac{2[YBC]}{2r^2[XBC]} = \frac{| BC\cdot d(Y,BC)|}{r^2 BC\cdot d(X,BC)}
&= \frac{|d(X,BC) - 2r\cdot d(X,BC)\cos\alpha|}{r^2\cdot  d(X,BC)} \\
&= \frac{\sin(\alpha+\beta)}{\sin^2\beta}|\sin(\alpha+\beta)-2\sin\beta\cos\alpha| \\
&= \frac{|\sin(\alpha+\beta)\sin(\alpha-\beta)|}{\sin^2\beta} \\
&= \frac{|\sin^2\alpha - \sin^2\beta|}{\sin^2\beta}<1,
\end{align*}as desired.

Interesting problem :)

Let $ PQ $ and $ BC $ meet at $ T $. And circle $ (TXY) $ meet $ PQ,BC $ at $ R,S $
(1) First, we show that $ Y $ is below $ BC $. (Actually, $ Y $ is on $ BC $ when $ PX=PB $)
 $ R $ is the midpoint of arc $ XY $. And $ \triangle{XRS} \sim \triangle{XPB} \sim \triangle{XQC} $.
Which means $ XR >RS $ so, $ Y $ is below $ BC $.

(2) Now we show the problem.
$ \frac{S_{\bigtriangleup YPQ}}{S_{\bigtriangleup YBC}}= \frac{{PQ} \times {RY}}{{BC} \times {SY}}=\frac{{XR} \times {RY}}{{XS} \times {SY}} >1 $.

So, done :D


Apparently, I'm in an awkward situation.

I'm aware that AMC takes place in Feb 1st and 16th

Apparently, I'm living in Hong Kong and will be going to the US in Feb 14th.

I certainly wont be able to take it on Feb1st, but how about the 16th of Feb?

I know it is extremely late to order test papers for AMC but do u think my new public HS will allow me to take AMC in such short notice?  Its a rather rural HS and I'm worried that it won't have enuff finance to hold the competition.  I've called the HS already, but I wasnt able to contact the math department.

"Late Registration Contest B (Jan. 10,-Feb. 7) $52"

So if I do manage to get in contact with some1 who is in charge or knows about the test, will I be able to take it after informing that I would like to participate in it?

Ya.. Its my utmost laziness of not making the calls earlier on in Dec that takes the blame.  I really would wish to take the exam.  For UKMT (this is a british version of AMC), the school just ordered loads of copies of the exam and anyone who wished to take it, takes it.  Is this the same for AMC?  Or do onli those who ordered gets to participate?

Your best chance is to keep contacting your high school, and to try to get them to administer the AMC B on the 16th.

Unfortunately, if they are a small school, they may only be giving it on the 1st.

Actually, if it's a really small school, you might even be the only person to take it, making it easier for you to take the AMC.


I have been trying to obtain a TI-84+ Silver Edition emulator for my computer, but I'm stuck on the process of downloading the ROM img. I following the instructions on ticalc.org (http://www.ticalc.org/programming/emulators/romdump.html#5) for the TI-83+ downloading the ROM img, but when I try to run the romdump.exe, nothing happens. Can somebody tell me what I'm doing wrong/alternate websites, etc.?

umm, can someone help me out? Maybe I posted in the wrong topic... if so, could a mod move it or something, please? Thanks

Well, if I remember correctly, it's illegal to upload ROM images online, so you'll have to get it from your calculator.

I don't remember much else about the entire process, but you might try calculator-specific forums like those at http://www.maxcoderz.org or http://www.detachedsolutions.com, though.

yeah, the article on ticalc gave instructions on getting the ROM img. from my calculator, but it said i needed to run an executable file that was supposed to be included on the file it told me to download, but when i ran it nothing happened, and the calc would not send me the ROM img. when i ran the program that was supposed to do it. help? anyone?


We know that for a natural number M > 1 and a real number $0 \leq q < 1$, there exist unique $b_i \in \{0,1,2,...,M-1\}$ s.t

\[ q = \sum_{i=0}^\infty b_i M^{-i}\]

but what I tried to generalize the M-base to:

for any number $0<p<1$ and $0<q<p$, there exist unique $b_i \in \{0,1,2,...,\text{int}(1/p)\}$  where int(x) defines integer part of number x, s.t:

\[ q = \sum_{i=0}^\infty b_i p^{i}\]

I don't know whether this is true or not. Can you prove or disprove it?
If you know any websites on this topic, please let me know, I appriciate it.
Is there any genelization for M-base numbers?

[quote="vuhung"]for any number $0<p<1$ and $0<q<p$, there exist unique $b_i \in \{0,1,2,...,\text{int}(1/p)\}$  where int(x) defines integer part of number x, s.t:
\[ q = \sum_{i=0}^\infty b_i p^{-i}\]
I don't know whether this is true or not. Can you prove or disprove it?
[/quote]

vuhung,

As you are posting it, it is clearly not true, because if $0<p<1$ then $p^{-i}>1$, and therefore there is no way that you can obtain a positive number less than 1 as the result of the serie you are posting.

On the other hand, if you change $p^{-i}$ for $p^i$, then, in I believe is not true in general, but I still don't know exactly how.

By the way, the sum should start on $i=1$ in the first case, and in $i=2$ in the second.

Best,

[quote="djimenez"]
vohung,

As you are posting it, it is clearly not true, because if $0<p<1$ then $p^{-i}>1$, and therefore there is no way that you can obtain a positive number less than 1 as the result of the serie you are posting.

On the other hand, if you change $p^{-i}$ for $p^i$, then, in I believe is not true in general, but I still don't know exactly how.

By the way, the sum should start on $i=1$ in the first case, and in $i=2$ in the second.

Best,[/quote]

I am sure it is your typo but in Vietnamese, "vohung" and "vuhung" have the same meaning :)

Btw, I changed to

\[ q = \sum_{i=0}^\infty b_i p^{i}\]

and as I stated, this is an "open question", I could not prove or disprove it....

[quote="vuhung"]I am sure it is your typo but in Vietnamese, "vohung" and "vuhung" have the same meaning :)
[/quote]

vuhung,

Just in case, the mistake (already edited) was totally honest, I didn't mean to confuse terms. I confess to be totally ignorant with respect to asian languages, and just misread the name.

I will continue trying to se what can I do with your problem.

Best,

[quote="jmerry"]It's not true in general because these expansions are not unique. For example, in base $\frac{1+\sqrt5}2,$ $.100$ and $.011$ represent the same number. Uniqueness can only happen if the largest allowed decimal $.aaaaa\dots$ is less than or equal to 1. By your rules, this only happens at the integers.

You can always construct at least one expansion by the greedy algorithm.[/quote]

Yep, you are right! :first:

But what would happen on the other extreme. There is no unicity, but existence? I there always a representation of this kind for any number? Probably it does... yes, I think it does!

Best,

[quote="jmerry"]It's not true in general because these expansions are not unique. For example, in base $\frac{1+\sqrt5}2,$ $.100$ and $.011$ represent the same number. Uniqueness can only happen if the largest allowed decimal $.aaaaa\dots$ is less than or equal to 1. By your rules, this only happens at the integers.

You can always construct at least one expansion by the greedy algorithm.[/quote]
Thanks jmerry, now I have to find another way to solve my problem :) I think there is another way to genralize M-base numbers. I will post if found any.


How can i find The imaginary part of (1-2i)^2-i.

This is what i have done so far:
(1-2i)^2 (1-2i)^-i

=(1-4i+4)(1-2i)^-i

If you don't mind, using $\LaTeX$ to format your posts makes them much easier to read.  Anyway, do you mean $(1-2i)^{2}-i$ or $(1-2i)^{2-i}$?

mets, that's not true (in fact, basically everything can be written as a real plus an imaginary)
Just typing it into a ti-83 gives $-(1.64...)-(.202...)i$...
http://en.wikipedia.org/wiki/Exponentiation#Computing_complex_powers
Explains the steps how to compute complex powers.
(It's pretty textbook)


The Wall Street Journal published a study today claiming that mathematician is the best career when considering such factors as working conditions and income potential.  In fact many of their top ranking careers draw heavily on mathematics and problem solving.

   1. Mathematician
   2. Actuary
   3. Statistician
   4. Biologist
   5. Software Engineer
   6. Computer Systems Analyst

You can read more commentary and find links to the article from the Metroplex Math Circle site:

[url]http://metroplexmathcircle.wordpress.com/2009/01/06/wsj-survey-says-mathematician-best-job/[/url]

Similar studies have been done over the years.


vector trigonometry calculus integration

Hello everybody!
Here's a (easy) problem:
A small snail is sliding from the top of a ball (the motion is frictionless). Show that the snail will fall away from the ball when it covers a lenght on the ball corresponding to an angle theta so that cos(theta)=2/3. (See figure attached)
I hope i was clear enough! :D

The action of the force of gravity on the snail in combination with the curve of the ball gives the snail a small horizontal acceleration, as long as the snail is in contact with the ball.  After that it stops accelerating.

At whatever point the snail and the ball part, the horizontal acceleration must stop, although the vertical acceleration continues.

The snail's trajectory when it leaves the ball will be at a tangent to ball surface.

The question is, when does the snail have enough horizontal speed to free itself from the ball, given it is also accelerating downward?

Since the snail is sliding not dropping, the vertical acceleration is *not* the simple Newton gravity law of acceleration.  The acceleration is being retarded by the opposing surface of the ball.

Yet the vertical force on the snail *is* constant and conforms to the gravity equation F = GMM/d^2. (d = distance to earth's centre doesn't change significantly.).

My guess is that the snail cannot depart from the ball until the resultant vector is >= the tangent.   This is a simultaneous equation type problem.

The answer will be different if the snail were a ball rolling (even with negligible friction) since some energy will be being stored as angular momentum.

They seperate if the force between them is 0. So we will have:
\[mg \cos{\theta} = m \frac{v^2}{R}\]
Use the conservation of energy , we will have:
\[R(1-\cos{\theta})mg=\frac{1}{2}mv^2\].
Solve the system, we get $\cos{\theta} = \frac{2}{3}$.

another question: if the bigger ball is not fixed, what is $\theta$?

I could not solve my own problem, and I do not think it's doable (at any sort of easy level). A small angle approximation cannot be made for obvious reasons (the angle is large)!

There is a way to solve the general harmonics motion differential equation without making the small angle approx but it is not elementary. So blah.

[quote="Spoon"]Nice shobber :). Assuming gravity is g, how much time does it take for this to happen?[/quote]
I find a calculus solution. So here we go:

Assume it turns a very small angle $d\theta$

Then $s=R \sin{d\theta}=R d\theta$

Use the conservation of energy, we will have $mv^2=2mgR(1-\cos{d\theta})$
Simplify, we get: $v=\sqrt{2gR(1-\cos{d\theta})}$

$t= \int{dt}= \frac{R}{\sqrt{2gr}} \int {\frac{d\theta}{\sqrt{1-\cos{d\theta}}}}$

I am sure it is not correct since it will have $R$ in the final answer...But I think whatever calculus way we use, we shall always try to get $\int f(\theta) d\theta$.

I am still trying to find a basic solution without solution...

[quote="Spoon"]Nice shobber :). Assuming gravity is g, how much time does it take for this to happen?[/quote]

The initiation of the motion of the snail is not clear to me. It starts out with zero energy (as I saw to be silently assumed in the solutions), that means, zero speed. That means, it doesn't slide down. Therefore, the snail must get a very small impulse coming down the ball, and we assume that this impulse is very small and then solve the problem. When we limitly approach the impulse to zero, the critical take-off angle limitly goes to whatever arccos(2/3) is. But the issue with time is different. If we approach to zero with the impulse, the time needed for the snail to take off goes to infinity.

[quote="kubus"]The initiation of the motion of the snail is not clear to me. It starts out with zero energy (as I saw to be silently assumed in the solutions), that means, zero speed. That means, it doesn't slide down. Therefore, the snail must get a very small impulse coming down the ball, and we assume that this impulse is very small and then solve the problem. When we limitly approach the impulse to zero, the critical take-off angle limitly goes to whatever arccos(2/3) is. But the issue with time is different. If we approach to zero with the impulse, the time needed for the snail to take off goes to infinity.[/quote]
Yes! This is precisely why I could not solve it! Because as far as the initial conditions for the differential equation are concerned, there is no initial velocity! hehe, which means the snail NEVER moves. And when dealing with "time", you cannot just assume a tiny "nudge", because the magnitude of this nudge will affect your result. In fact, it should affect the result to the original question, even though as Kubus pointed out it was "silently assumed".


A triangle with side lengths in the ratio $ 3: 4: 5$ is inscribed in a circle of radius $ 3.$ What is the area of the triangle?

$ \textbf{(A)}\ 8.64 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 5\pi \qquad \textbf{(D)}\ 17.28 \qquad \textbf{(E)}\ 18$

[hide] Since the triangle has side lengths in the ration of 3:4:5, we can say that it is a right triangle (since 3,4,5 is a Pythogorean Triplet).  We know the radius of the triangle is 3, so the radius is 6.  Now, since the triangle is right, we know that the hyptonuse is the diameter of the circle.  Say that 5x=6.  Thus, x=1.2.  Therefore, 3x=3.6 and 4x=4.8.  Now, this means that the length of the other sides are 4.8 and 3.6.  Finally, we multiply these sides and divide by 2 to get 8.64, so the answer is A. [/hide]

[hide]The triangle must be inscribed in a semicircle because it is right.

The ratio of the diameter to the hypotenuse is $6: 5$. The area of a 3-4-5 triangle is $6$.

The ratio of the areas is the ratio of the sides squared, so $A=6\cdot\left(\frac65\right)^{2}=8.64\Rightarrow\boxed{\text{A}}$[/hide]

This was in AMC10A
Answer is [b]8.64[/b]

Yeah, I can reiterate every answer already posted on the AMC 12A too...


inequalities
calculus
derivative
algebra solved
algebra

I think I found a really nice way of solving Inequalities with the property (any inequality can be reduced to g(x1,x2,x3.....,xn)>=0):
If from  g(x1,x2,x3.....,xn)>= it is also true that
g(\lambda x1,\lambda x2,\lambda x3.....,\lambda xn)  >=0
with  \lambda >0 then there is a very easy way to prove them.
Also note that after some algebra, a lot of IMO-type inequalities have this property

It uses a really nice idea and I think that after that, some inequalities become not challenging. It is like Shturm's way of proving ineqs ( but it doesn't require so much algebra and it doesn't lead to even harder inequalities). I showed it to my Olympiad Teacher at School so he said it is allright.
:)
:D

can you give some examples on some inequalities?

simple and IMO ineqs?

cheers! :D

Sorry, I don't have a lot of time, but it is something like this:
g(x1,x2,x3.....,xn)>=0
If from g(ax1,ax2,ax3.....,axn)>= it is also true that
g( ax1, ax2, ax3....., axn) >=0
with a>0
We take "a" as being the maximum of them.
So we have g(b1,...bn)>=0 , where bi=(ai)/(amax).
Now all bi have 0<bi<=1.
With some calculus, this problem becomes bookwork.
The idea is to look at bi as a separate variable, and bj, with i!=j as constants. Take derivates and see where equality it is positive or negative.
The best example is IMO2001 ineq.
Although this could not be so useful for easy problems , for a IMO second question it worths spending 1/2 h to find a solution that surely exists.

i'm not sure of that - in fact I disagree - what you are saying it not only wrong, but it's against all the principles of the IMO competitions. Their goal is to nurture talent and cleverness, rather than a good memory, and long time practice.

       :( :(  :(
   :(                :(
:(     :(    :(      :(
:(                        :(
:(       :( :(         :(
:(    :(        :(   :(
  :(                   :(
     :(              :(
          :( :( :(

PS Although the last two can get you far away, in order to reach high limits,  you must also posses the first two ones :D

some equations (a lot of \sqrt ) maybe very difficult to take derivitive, let alone prove its positive

The notation seemed to be confusing...
Is the following what you're meaning?
Example: g(x,y)=x^2-xy+y^2>=0
Then
dg/dx=2x-y
dg/dx=0 iff y=2x
Then g(x,2x)=3x^2>=0
Similar for y

Then this method could be very impractical for inequalities with more terms, and for this inequality, AM-GM works better.

I made some mistakes when writing it.
But for the given inequality, take consider x being the maximum.
Taking g((x/x),(y/x))>=0 we get g((x/x),(y/x))= 1 - (x/y)+(x/y)^2>=0
Taking (x/y)=t gives us 1 - t + t^2>=0. 0<=t<=1. Now it is easy.
Sorry Valentin, or you angry or what?
I don't say this is the best way to solve an IMO iequality. But to solve other kind of inqualities, it is really good.
And by the way, is somebody here who could write the MONGOLIAN Inequality? It is something about the averages of 3 and 2 consecutive integers.
? by DUST


Help me solve this problem.
You have two strings(length<=5000) consists only of two letters 'a' and 'b'. If 'a' and 'b' are adjancent you can switch their places. You should compute min nuimber of switches you can get second string from the first.

Well, if you just consider the positions of the 'a's, then they have to match up from left to right, so just move the leftmost a to the correct position, then the second leftmost, etc..


Γεια σας Κύριε Μπάμπη τι κάνετε ? Να η λύση μου.
By AM-GM we have $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\frac{1}{\sqrt[3]{a^{2}b^{2}c^{2}}}$

and
$1+a=1+\frac{a}{2}+\frac{a}{2}\geq 3\sqrt[3]{\frac{a^{2}}{4}}$

and similarly $1+b\geq 3\sqrt[3]{\frac{b^{2}}{4}}$  and
 $1+c\geq 3\sqrt[3]{\frac{c^{2}}{4}}$

. Multiplying and we find the minimum  :wink:

Σας άρεσε ?

[quote="silouan"]Γεια σας Κύριε Μπάμπη τι κάνετε ? Να η λύση μου.
By AM-GM we have $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq 3\frac{1}{\sqrt[3]{a^{2}b^{2}c^{2}}}$

and
$1+a=1+\frac{a}{2}+\frac{a}{2}\geq 3\sqrt[3]{\frac{a^{2}}{4}}$

and similarly $1+b\geq 3\sqrt[3]{\frac{b^{2}}{4}}$  and
 $1+c\geq 3\sqrt[3]{\frac{c^{2}}{4}}$

. Multiplying and we find the minimum  :wink:

Σας άρεσε ?[/quote]

 Nice solution Silouan !!!We have only to observe that the expression really attends the minimum $\frac{81}{4}$ , when $a=b=c= 2$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $H$ be a normal subgroup of $G$, $\mathcal{K}$ is a conjugacy class of $G$ contained in $H$ and $x\in \mathcal{K}$. Prove that $\mathcal{K}$ is a union of $k$ conjugacy classes of equal size in $H$, where $k=|G:HC_{G}(x)|$.

I believe this is part of #19 from the 3rd edition of Dummit and Foote. Here's a write-up for your question, which I happened to have done sometime last year. I hope the details make sense. If you need help with the rest of the question, feel free to ask.

I've posted the solution as an attachment, albeit as an image file. I presume you'll be able to view it.

[quote="oldman"]How do you define $G^{[x]_H}$ any way? Thanks![/quote]

I guess I used different notation from Dummit and Foote --- whereas they use $G_s$ for the stabilizer of $s$ in $G$, I used (from class) $G^s$. With regard to your question, then $G^{[x]_H}$ refers to the stabilizer of an element $x$ from $[x]_H$. The latter, as defined in my solution, is the set $\{xhx^{-1}:h \in H\}$.

Hope this helps clarify that! :)

[quote="grobber"]A generalization:

Given a finite transitive action (i.e. a finite group acts on a finite set) of $G$ on $S$, and given a normal subgroup $H\lhd G$, the action of $H$ on $S$ has $k$ orbits of equal size, where $k=[G:H\cdot\mbox{Stab}_G(x)]$ for some $x\in S$.

The fact that the sizes of the orbits under the action of $H$ on $S$ are equal follows from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47737]this other problem you've posted[/url].

The size of $S$ is precisely $k'=[G:\mbox{Stab}_G(x)]$, so we want to show that the size of the orbit of $x$ under the action of $H$ is $\frac{k'}k=[H\cdot\mbox{Stab}_G(x):\mbox{Stab}_G(x)]$. This follows immediately from the fact that the orbit of $x$ under the action of $H\cdot\mbox{Stab}_G(x)$ is the same as the orbit under the action of $H$, and the stabilizer of $x$ in $H\cdot\mbox{Stab}_G(x)$ is $\mbox{Stab}_G(x)$.

P.S.

I suppose your groups were finite in this exercise? :?[/quote]

Thanks, I did like this
As you observed, $G$ can acts on orbits of $H$ in $S$ and this acts still transitive. The stabilizer of this action is $H\,Stab_G(x)$. Therefore $k=[G:H\,Stab_G(x)]$.


This is the first time I have taken the Putnam.
When/Where/How do we get our results back.
Also, Do we get a breakdown of what we scored
on each question or just a total score and rank?

 Thanks.

The results come back in mid-March. A sheet with your individual score is mailed to the faculty supervisor at your college - nothing gets sent to you individually.

All you ever learn is your total score and rank. You will not be told what the question-by-question breakdown is.


probability
combinatorics unsolved
combinatorics

$6$ distinct points have been chosen on a circle. We accidentally name them $A,B,C,D,E,F$. What is the probability that the triangle $ABC$ does not intersect triangle $DEF$ ?

The two triangles do not intersect precisely when $A,B,C$ are consecutive points.  There are $6$ sets of 3 consecutive points, and $\binom{6}{3}= 20$ sets of 3 points, so the probability is $6/20 = 0.3$.


[b](1) find the function f(x) If  f:R+->R+,

If f(f(x)) = 3x+1/(x+3)[/b]

From the hypothesis:
\[ f(3x \plus{} \frac {1}{x \plus{} 3}) \equal{} 3f(x) \plus{} \dfrac{1}{f(x) \plus{} 3} \quad [1]
\]
( both equals $ f(f(f(x))$)

Also note that $ y \equal{} 3x \plus{} \frac {1}{x \plus{} 3} > 0 \wedge x > 0 \iff x \equal{} g(y) \equal{} \minus{} \frac32 \plus{} \frac16 y \plus{} \frac16 \sqrt {69 \plus{} 18y \plus{} y^2}$. This and [1] give rise to a formula for recursively evaluating the value of $ f(y)$ in terms of previously known value of $ f(x)$.

Because $ g(y) > 0 \iff x > \frac13$, we need all the values of $ f(x)$ in the interval $ (0,\frac13]$ in order to generate every value of the function. Next, I will describe all $ f(x)$.

Given $ h : (0,\frac13] \rightarrow \mathbb{R}^ \plus{}$ be any function. The function $ f(x)$ extended  from $ h(x)$ is defined recursively as:
\[ f(x) \equal{} \begin{cases} h(x) \text{, \quad if }0 < x\leq \frac13 \\
3f(g(x)) \plus{} \dfrac{1}{f(g(x)) \plus{} 3} \text{, \quad otherwise} \end{cases}
\]
This family of functions gives us all possible solution of $ f(x)$.


calculus
integration
calculus computations

maybe this has been asked before...  :maybe:

in how many ways,  [u][i][b]without[/b][/i][/u] using the fact

$ \left ( \int_{ \minus{} \infty}^\infty e^{ \minus{} x^2}\;dx \right )^2\; \equal{} \; \int_{\theta\equal{}0}^{2\pi}\int_{r\equal{}0}^\infty re^{ \minus{} r^2} \;dr\;d\theta$

 can you evaluate the integral

$ \int_0^\infty e^{ \minus{} x^2}\;dx$

Call
\[ A(t) \equal{} \left(\int_{0}^{t} e^{ \minus{} x^2} dx\right)^2 , B(t) \equal{} \int_0^1\frac {e^{ \minus{} t^{2}(1 \plus{} x^2)}}{1 \plus{} x^{2}} dx
\]
Differentiating with respect to $ t$ we have,
\[ A'(t) \equal{} 2 \int_{0}^{t} e^{ \minus{} x^2} dx \cdot e^{ \minus{} t^2}
\]
and
\[ B'(t) \equal{} \int_{0}^{1} \minus{} 2t e^{ \minus{} t^{2}(1 \plus{} x^2)} dx
\]

\[ \equal{} \minus{} 2 \int_{0}^{1} t e^{ \minus{} t^2 x^2} dx \cdot e^{ \minus{} t^2}
\]

\[ \equal{} \minus{} 2 \int_{0}^{t} e^{ \minus{} u^2} du \cdot e^{ \minus{} t^2}
\]

\[ \equal{} \minus{} A'(t)
\]
So for some C we need,
\[ A(t) \equal{} \minus{} B(t) \plus{} C.
\]
Letting $ t \rightarrow 0 \plus{}$ the LHS tends to 0 and the RHS tends to $ \minus{} \int_{0}^{1} \frac {dx}{1 \plus{} x^2} \plus{} C \equal{} \minus{} \pi/4 \plus{} C$ so $ C \equal{} \pi/4.$
Now let $ t \rightarrow \infty$. We have,
\[ \left(\int_{0}^{\infty} e^{ \minus{} x^2} dx\right)^2 \equal{} 0 \plus{} \frac {\pi}{4}
\]
Or,
\[ \int_{0}^{\infty} e^{ \minus{} x^2} dx \equal{} \frac {\sqrt {\pi}} {2}.
\]

That was amazing   :)

Now can you prove that:

$ \int^{x}_{0}e^{ \minus{} u^{2}}\text{d}u \equal{} \frac {\sqrt {\pi}}{2} \minus{} \frac {\frac {1}{2}e^{ \minus{} x^{2}}}{f(x)},\;\; x > 0$

where $ f(x) \equal{} x \plus{} \frac {1}{2x \plus{} \frac {2}{x \plus{} \frac {3}{2x \plus{} \frac {4}{x \plus{} \cdots}}}}$


Find all natyral numbers $ n$ and $ k$ such that the $ n^{n}$ have $ k$ digits ,and $ k^{k}$ have $ n$ digits (in base 10).

[b]ee"[/b]

$ \begin{cases} 10^{k\minus{}1} \le n^n < 10^k \\
10^{n\minus{}1} \le k^k < 10^n\end{cases}$

If $ n \ge 10$, we have $ k^k<10^n \le n^n$, so $ k<n$, but this leads to $ n^n<10^k \le n^k \Rightarrow n<k$, contradiction.
So by checking $ n\equal{}1, 2, ..., 9$, we know that $ (n,k)\equal{}(1,1),(2,2), ..., (7,7)$ are the only solutions.

Well done , very nice solution

[b]ee"[/b]

Sorry, I made a mistake in my solution above.After checking $ n \equal{} 1,2,...,9$, the solutions should be $ (1,1),(8,8)$ and $ (9,9)$.


counting distinguishability LaTeX factorial

How many 4 letter words can be formed from the letters of KITTENS?

[quote="Art of Owna"]Is it [hide]480..if im right ill post a solution[/hide][/quote]

I'm not sure. I'm working without an answer here. Post your solution anyway. I'm working with two cases but am unsure of how to set them up.

[hide]well, theres K I E N S T T

So lets say how many are there when theres only 1 of each letter
6*5*4*3 (there are only 6 differnt letters to choose from, 5..and so on)=360

Lets say there are 2 of one kind of letter (in this case T)
So we set the 2 T's and when we need 2 other letters.  so 5*4*1*1, but the T's can go in 4C2 places so 5*4*6=120

120+360=480

I checked it using another way..so I think Im right..
You can just say how many there are total (Ts are distinguishable)
7*6*5*4
But T's arent distinguishable so we counted the Txxx case twice
and the TTxx case twice too
so 7*6*5*4-4*5*4*3-6*5*4=480[/hide]

just do casework.

Case 1:  no T's

Case 2:   1 T

Case 3: 2 T's

[hide]

casework:

No Ts in word:

$5*4*3*2 = 120$ words

1 T in word

$4$ places the T can go. $5*4*3$ for the other three spots
$(5*4*3) * 4 = 240$ words

2 T in word

$6$ ways the Ts can be placed. $5*4$ choices for the other letters.
$(5*4) * 6 = 120$ words

total words = $120+240+120 = \boxed{480}$
[/hide]

it is simply a matter of casework....sry my latex is messed up

[hide][b]No t's- [/b]

The remaining letters are k, i, e, n, s. Thus, there are five factorial=[b]120[/b] ways.

[b]one t-[/b]

Since you have to have a t, there are four places to set it. The remaining three spots have 5x4x3=60 ways. Thus, there are 4x60=[b]240[/b] ways.

[b]two t's- [/b]

There are four choose two= six ways to set the t's. Then the remaining two spots have 5x4=20 ways. Thus, there are 6x20=[b]120[/b] ways.

The number of four letter words is 120+240+120=[b]480[/b] ways.  [/hide][/b]

You dont need to split the one t and no t (even though its easier to realize and organize if you do)

ALl you need to know is when theres 2 of one letter (only T in this case) and 1 of every letter
there are 6 distinguishable letters so 6*5*4*3 (when all 4 are different)+6*5*4 (when TTxy where x and y are differnt letters)=480

that works too.....probably a much faster way there

[hide]Case 1: Both T's are included, Then there are ${5\choose 2}4!/2!=120$.
Case 2: Only 1 T is included, Then there are ${5\choose 3}4!=240$.
Case 3: No T's included, Then there are ${5\choose 4}4!=120$.
480 in total.[/hide]

[quote="GlenthemanN"][hide="woudlnt it be.."]
$420$?

There are $7$ letters, thus number of 4 letter arrangements is $7P4 = 840$

But there are two t's so we have to distinguish between them. There are $2!$ ways to arrange the t's for every permutation so the answer is

$\frac{7P4}{2!}= \frac{7!}{2!3!}= \frac{5040}{12}= \boxed{420}$

I am almost certain, I dont know how you guys all get 480, I did this and a few questions identical to this one in my textbook and this is the way to go about it..
[/hide][/quote]

You subtracted the ones that were (T1)(A)(B)(C) wehre A and B are leters and (T2)(A)(B)(C), those yes you counted twice since they are both the same TTAB, but ones that have both Ts, you didnt over count twice

[quote="GlenthemanN"]I'm not sure what you even said, but this is a question from the text I mentioned, which is pretty much identical to the original posts question:
[quote]
What is the number of four-letter words that can be formed using the letters in each of the following words?

i) GOSSOON
[/quote]

Using the same logic as above there are $7P4 = 840$ ways to form a 4 letter word from 7 letters. But O repeats 3 times and S repeats twice. There are $3!$ permutations of the O's in the same three positions, and $2!$ for the S's in the same two positions. So our answer would be
$\frac{7P4}{2!3!}= \frac{7!}{2!3!3!}= 70$

The answer in the back of the book says 70.[/quote]

I dont think that book answer is right.

[quote="p4fn2w"]I was taught the same thing as GlenthemanN in my algebra II class, so I am confused here.

After doing a little experimenting, I think that method of finding permutations with repeated elements only works if you permute the entire string of words. If you are looking for permutations with only 4 letters out of the original 7-letter word, then it gets complicated and you need a different approach.

I'll try to look more into this...[/quote]
I dont even understand that method... :blush:
but if the 4 letter words permutations for GOSSOON
it would be
so OOOSSGN
4 different : 4*3*2*1
two Os only with no other repeats: 3C2*4!/2!
two Ss only with no other repeats: 3C2*4!/2!
three Os: 3C1*4C3
two Os and two Ss:4C2
so add them up and you get 114?

p4fn2w,GlenthemanN.. What book did you learn the thing about the permutaitons?  Is it in a school textbook because im pretty sure thatway is wrong.

Umm for an example, how many two letter words can you get out of ABCDD

Listing out there are
AB
AC
AD
BA
BC
BD
CA
CB
CD
DA
DB
DC
DD

if you conunt that is 13
but with your way of 5P2/2! (for 2 letter words so P2 and divide by 2! because D is repeated twice) taht is only 10, not 13...

Yeah, it was from my algebra II class text book. But I only remember problems where you have to permute an entire string of letters. I never remembered a problem in the book where you have to permute partial elements of the whole set.

So for example, GOSSOON has (7!)/(2!3!) permutations of 7-letters only, but if you want to look for 6-lettered permutations or less, then you need an entirely different approach.

The reasoning behind this is that GOSSOON has 7 elements, so there are 7! ways to permute this. But since there are 3 O's, the O's can permute 3! = 6 ways, so you are overcounting the case six times since the O's are indistinguishable which is why you divide it by 6. The same goes for 2!.

But if you arent using all the letters of GOSSOON, then there will be some cases where the O's wont be able to permute 6 ways. If you are constructing a 4-lettered word from GOSSOON, then there is a case where you are using only 1 O, so you dont divide anything; where you are using 2 O's, so you only divide by 2!; and where you are using all 3 O's, so in those subcases you divide by 3!. You have to split it into cases, I think.

I hope that makes sense..

Yeah I realize the problem with the way ive gone about it. I see that it only works to do that when you permutate the whole string..

The book is called "Finite Mathematics" By John J. Del Grande, W. Bisset, and Edward Barbeau Jr.

There actually werent any examples in the section about repetitions and choosing n lettererd words from words with more than n letters, I just figured it would work for the question and it did according to the back of the book...

if we find the two letter permutations of the word "zoo" and let the first o be $o_{1}$ and the seconds $0_{2}$ then we have
zz
zo
zo
oz
oo
oo
oz
oo
oo
= 9 permutations, but we can elinate 5 because oo repeats 4 times, oz 2 times and zo 2 times
so there is only 4 distinct permutations
zz
zo
oz
oo

From my above method this gives $3P2 \div 2! = 3$ but we know the answer is 4.. so it definatly doesnt work.

EGh!

There shouldn't be 9 permutations there.  "oo" will not repeat 4 times.  Like you said call them $o_{1}$ and $o_{2}$. Then you get $o_{1}o_{2}$ or $o_{2}o_{1}$.  There are only two ways to do that.  Also "zoo" only has one "z" so there is no "zz" term.  So there are only 3 ways to get two letter permutations from "zoo":
zo
oz
oo

Yes, so to sum up, the method that GlenTheMann mentioned only works when you're taking permutations of the entire string, that is, you're looking at the number of permutations of the string (n!) and then looking at all the permutations of the repeated letters (a!b!c! etc.) so to look at the number of permutations with repeated elements of the entire string you're calculating $\frac{n!}{a!b!c!}$ where $a,b,c$ are the number of specific repeated elements. Also, the book 'finite mathematics' that you're referring to was the OAC (grade 13) textbook for the course called, evidently, "finite mathematics". The book itself is, IMO, not bad at all for a highschool text (with maybe the exception of the chapters on statistics and the normal distribution).


let Sn be the number of permutation (a1,a2,a3....,an)of (1,2,3....,n)
such that 1 =<Iak-kI=<2 for all k .show that (7/4)Sn-1<Sn<2Sn-1 for n>6

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=3710[/url]

I don't think a full treatment of the solution is given here, just some talk about it.

can you frove that
                          Sn=(Sn-1+Sn-2+Sn-3+Sn-4)-Sn-5   ?

thanks

Continue in the given link, don't double-post.


geometry
3D geometry
sphere
Pythagorean Theorem
geometry solved

When three spherical balls, each of radius 10 cm, are placed in a hemispherical dish, it is noticed that the tops of the balls are all exactly level with the top of the dish. What is the radius, in centimetres, of the dish?

Source : Australian Math Competition, Intermediate, 1999

Let $A, B, C$ be the centers of the 3 congruent spherical balls $(A), (B), (C)$, touching each other and forming an equilateral triangle with the side $AB = 2r = 20\ \text{cm}$. Let $H$ be the orthocenter of this equilateral triangle. Let $D$ be the center of the sphere $(D)$ of the hemispherical dish centered in the common tangent plane of the spheres $(A), (B), (C)$. The distance of the center $A$ from the vertical axis $DH$ of the hemispherical dish is equal to

$AH = AB \frac{\sqrt 3}{3} = \frac{2r \sqrt 3}{3}$

Using Pythagorean theorem for the right angle triangle $\triangle AHD$,

$DA = \sqrt{DH^2 + AH^2} = \sqrt{r^2 + \frac{4r^2}{3}} = r \sqrt{\frac 7 3}$

The tangency point $T$ of the spheres $(A), (D)$ is on their center line $DA$ and the radius of the sphere $(D)$ is

$R = DT = DA + AT = r \sqrt{\frac 7 3} + r = \left(1 + \sqrt{\frac 7 3}\right) r \doteq 2.5275\ r = 25. 275\ \text{cm}$

Didn't think of the Pythagoras's theorem there... my hat is off to a brilliant solution.  :)


induction
abstract algebra
modular arithmetic
quadratics
Divisibility Theory
number theory

A number of the type $m^2+9$ can, with exception of $p=3$, only have prime divisors $p \not \equiv -1 \mod 4$:
If $q \equiv -1 \mod 4$ and $q \neq 3$, we have $3$ invertible $\mod q$ and get $m^\prime^2 \equiv -1 \mod q$ with $m^\prime \equiv m \cdot 3^{-1} \mod p$. But now we easily derive a contradiction with Fermat's little theorem: $1 \equiv m^\prime^{p-1} = (m^\prime^2)^\frac{p-1}2 \equiv (-1)^\frac{p-1}2 \equiv -1 \mod p$.

Claim: exactly those $n$ being a power of two work.
If $n$ has any odd prime divisor $q$, then $2^q-1 | 2^n-1$, so we just have to disprove the possibility of $2^q-1|m^2+9$ for $q$ any odd prime.
Since $q>1$, $2^q-1 \equiv -1 \mod 4$, thus there is a prime $p \equiv -1 \mod 4$ dividing $2^q-1$. If we would have $p=3$, then $2^2 \equiv 1 \mod p$ yielding $1 \equiv 2^q \equiv 2^{q \mod 2} \equiv 2^1 \mod p$, impossible. Thus $p>3$, but now $p$ has to divide $m^2+9$, impossible by the above.

We are left to prove that all $n=2^k$ work and do so by induction on $k$:
$k=1$ gives the number $3$ which obviously works with $m_1=3$.
If we have found $m_k$ with $m_k \equiv -9 \mod (2{}^{2{}^k}-1)$, we see that $2{}^{2{}^{k+1}}-1=(2{}^{2{}^k}+1)(2{}^{2{}^k}-1)$ and that these two factors are coprime. By the chinese remainder theorem we find $m_{k+1} \equiv m_k \mod (2{}^{2{}^k}-1)$ and $m_{k+1} \equiv 2{}^{2{}^{k-1}} \mod (2{}^{2{}^k}+1)$ and this one works.

[quote="ZetaX"]$ m_{k \plus{} 1} \equiv 2^{2^{k \minus{} 1}} \mod (2^{2^k} \plus{} 1)$[/quote]

Does it work for $ m_{k \plus{} 1}^{2}\equiv {\minus{}9}\mod(2^{2^{k}}\plus{}1)$ ? :maybe:

the only prime in form of 4k+3 which can devide m^{2}+9 is 3. and so n must be a power of 2; so we should find for which n 2^{2^{n}} devides m^{2}+9.

Lemma 1 : all prime factors of 2^{2^{n}} module 4 is 1.

Lemma 2: a|b^{2} +1 iff all odd prime factors of a module 4 is qual to 1 and 4 does not divide a.

so we can easily get that 2^{2^{n}}=3p, so that by using Lemma 1,2 there exist an m such that p devides m^{2} +1. finally we get 3p devides 3.m^{2} +3 and so (3m)^2 +9. so we done.

we will prove by induction on $n$ :

let $2^{2^{n}}-1|m^{2}+9$ for some $m\in \mathbb{Z}$.it's clear that $\upsilon _{3}(2^{2^{n}}-1)=1$.
$so 2^{2^{n}}-1=3c$. now we are going to prove there exist $m'\in \mathbb{Z}$ so that $2^{2^{n+1}}-1|m'^{2}+9.$
let $m'=3(\frac{m}{3})+3tc=3k+3tc$. now what we want here is a $t\in\mathbb{Z}$ such that :
 $2^{2^{n+1}}-1=(2^{2^{n}}-1)( 2^{2^{n}}+1)|m'^{2}+9 =9((k+tc)^{2}+1)\rightarrow c.2^{2^{n}}+1|((k+tc)^{2}+1$.
it's clear that $c|RHS.$ by Chinese Remainder Theorem it is also clear that there exist such a $t$ so that $k+tc\equiv 2^{2^{n-1}} mod 2^{2^{n}}+1.$ (note that $(c,2^{2^{n}}+1 )=1 )$.
so $TIWWWTP$. (This Is What We Want To Prove :P  ).

[quote="Peter"]Determine all positive integers $n$ for which there exists an integer $m$ such that $2^{n}-1$ divides $m^{2}+9$.[/quote]

[hide="My Solution"]
If $n=1$, we are done. Else, let $p$ be an odd prime divisor of $2^{n} - 1$. Assume there exists an integer $m$ such that $2^{n} - 1$ divides $m^2 + 9$. This means $m^2 \equiv -9 \pmod{p}$. This means $\left(\dfrac{-9}{p}\right) = \left(\dfrac{-1}{p}\right) = 1$, so $p \equiv 1 \pmod{4}$. Since all prime divisors of $2^n - 1$ are $1 \pmod{4}$, we know $2^n - 1 \equiv 1 \pmod{4}$, so $2^n \equiv 2 \pmod{4}$. This means $n = 1$, contradiction. Therefore, our only solution is $\boxed{n = 1}$.
[/hide]

actually this forgets the prime 2.

The above is clearly false; the argument does not work for the prime $3|2^n - 1$ when $n$ is even (in particular if it is a power of two).

We can only define the quadratic residue or Lengendre symbol when gcd(9, p)=1.
We have to show when p and 9 have a G.C.D.
if gcd(9, p) is not 1, we can find p=3 (is only prime that divides 2^n-1), it means n=2.
when, n=2, we can find out that m can be any number if m is divided by 3.

Sorry... My English grammer is not good...

[hide="A20"]
First, we see that $n=1,2,4$ works. Now, for $n>4$, we want to find numbers $n$ such that $-9$ is a quadratic residue modulo $2^n-1$. If $3\not|2^n-1$, then $a^2\equiv-9\mod2^n-1$ means there exists $b$ such that $b^2\equiv -1 \mod2^n-1$, which is clearly impossible, since $2^n-1 \equiv 3\mod4$. So, $3|2^n-1 \implies 2|n$. Now, if $3|n$, then $a^2\equiv-9\mod2^n-1$ means there exists $b$ so that $b^2 \equiv -1\mod \dfrac{2^n-1}9$, which is impossible since $\dfrac{2^n-1}9$ must be $3\mod4$. So $3\not|n$. We now wish to find $m$ so that $m^2\equiv-9\mod2^n-1$. Since $3|m$, we get that, for $3p=m$, that $p^2+1 \equiv 0\mod\dfrac{2^n-1}3$. This is possible whenever $2^n-1$ has no other 3 mod 4 divisors other than 3.
Since $a|n \implies 2^a-1 | 2^n-1$, we get that the only prime divisor of $n$ is 2, and indeed $2^{2^x}-1$ can be factored as $(2^2-1)(2^2+1)(2^4+1)(2^8+1)\cdots$, so we get that $n=2^x$ for non-negative integral $x$ $\boxed{}$[/hide]


I need help not answers to #7. The other questions serve as a reference.

Stop creating new threads about this!

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259406
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259373
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259455
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259516
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259529
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259294


Prove that, for every natural k, the number (k^3)! is divisible by

(k!)^(k^2 +k+1).


SMO 2007, Open #4

In each of the following 7-digit natural numbers: \[1001011,5550000,3838383,7777777,\]every digit in the number appears at least three times. Find the number of such 7-digit natural numbers.

[hide]Since there must be at least 3 occurances of the same digit, there cannot exist more than two digits.
The first digit isn't zero so there's nine choices.
First, assume there's two digits (Case 1).
$9*(\binom{6}{3}+\binom{6}{4})*9=81*35$
This is true since there's 9 possible choices for the second type of digit and it can either exist 3 or 4 times.  By placing those digits in place, the placement of the remaining 2 or 3 original digits is forced.
Now, let's assume there's only one digit (Case 2).  Here, there's only 9 ways to do that.
Thus, the total number of ways is 81*35+9=2835+9=2944[/hide]


induction
number theory
prime numbers
number theory proposed

This is my 600 post! so i wanted to post a problem that is one of my favorites:

let $n>3$ be an odd integer. Prove there is a prime $p$, such that $p$ divides $2^{\phi(n)}-1$ but $p$ doesn't divides $n$.

I have some thoughts on it. :)

Suppose $n=p_1^{a_1}...p_k^{a_k}$, then $\varphi(n)=p_1^{a_1-1}(p_1-1)...p_k^{a_k-1}(p_k-1)$, so $\varphi(n)=2^{l}r$, where $l\geq k$.
[b]Case 1.[/b] Suppose $r\geq 3$. We have now $2^{\varphi(n)}-1=2^{2^lr}-1=(2^{2^{l-1}r}+1)...(2^{2r}+1)(2^r+1)(2^r-1)$. All these factors are pairwise coprime, $so 2^{\varphi(n)}-1$ has at least $l+1>k$ prime factors :)
[b]Case 2.[/b] Suppose $r=1$. It means that all $a_i=1$ and $(2^{2^{l-1}}+1)...(2^{2}+1)(2^1+1)$ contains exactly $p$ prime factors. It follows that all $2^{2^i}+1$ are perfect powers of prime numbers, which are not coprime with $p$. But $2^{2^5}+1$ is not a perfect power. So $l\leq 5$. We know $2^1+1$, $2^2+1$, $2^4+1$, $2^8+1$, $2^{16}+1$ are prime numbers, so we obtain $n=(2+1)...(2^{2^{l-1}}+1)$, $2\leq l\leq 5$ (since $n\neq 3$ we exclude $l=1$). But in this case $\varphi(n)=2^{2^l-1}>2^{l}$ -- contradiction.

Nice Myth! i didnt know this solution and it is so simple, and creative! Thanks for sharing it with us!

hello pascual. this problem appeared in my tst for nationals in mexico. i gave a partial solution  my general idea was the same as myth's. when i saw the full solution i was under the impression that there couldnt be a diferent solution for this problem, i cant imagine a diferent way to attack this problem. so could you please show me your other solution please? :blush:  your the master here jaja :P

The official solution for the problem was the same as Myth's one

Ok, I will post it this afternoon!

LOL BTW Myth is the Master here i think! :D  even knowing i would like to be!!!! jaja

I am glad they used my solution for their olympiad  :P

Lemma 1:

let $p$ be an odd prime, such that it divides $a-1$, then $p^b|a^m-1$ if and only if $p^b|m(a-1)$.

This lemma is in the theorem section, and it appears as "Hensels Lemma", but i dont think it is its name.

Let $n=p_1^{t_1}p_2^{t_2}...p_k^{t_k}$, then $p_i^{r_i}|2^{\phi(n)}-1$ if and only if it divides $\phi(n)(2^{p_i-1}-1)$ by lemma 1. so we get, if we suppose that all prime factors of $2^{\phi(n)}-1$ are factors of $n$, their max product is at most:

$(2^{p_1-1}-1)(2^{p_2-1}-1)...(2^{p_k-1}-1)\phi(n)$

so we will be done if we showed $(2^{p_1-1}-1)(2^{p_2-1}-1)...(2^{p_k-1}-1)\phi(n)<2^{\phi(n)}-1$

this can be showed by induction on the number of prime factors, using that $(2^{p-1}-1)(p-1)p^{t-1}<2^{(p-1)p^{t-1}}-1$ and also induction on $t_1+t_2...+t_k$ will do the job.

sorry but i dont understand your aplication of the lemma in the problem. note that that im not saying its wrong im just saying i dont understand it. :blush:  could you  please explain just that part please.

[quote="Omid Hatami"]Of Course you used their solution.[/quote]
What do you mean :(

hmm a variant came up in our selection test! for what n does $2^{\phi(n)}-1$ divide $n^n$ - basically the same idea.


For any positive real numbers $ a,b$ and $ c$,
\[ \sum_{cycl}\frac{a}{a+b}\le\frac{3}{2}\left( 1+K\cdot\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}\right) \le \sum_{cycl}\frac{a}{b+c}\]
where $ K=1/2$ for the left-hand side and $ K=21/32$ for the right-hand side.

[quote="fjwxcsl"]we have

(1) sum(a/(a+b))<=3/2*[1+sqrt(3)/9*sum((a-b)^2/(a+b+c)^2))];

(2) 3/2*[1+2/3*sum((a-b)^2/(a+b+c)^2)]<=sum(a/(b+c)).

See here:

http://www.irgoc.org/bbs/dispbbs.asp?boardid=19&id=2614&star=1#2614[/quote]

Yeah, the inequalities can be sharpened. But the original constants admit a simple pen-and-paper solutions.

@fjwxcsl: egaugnal ruoy gnidnatsrednu srehto gnitcepxe tonnac


geometry
angle bisector
geometry unsolved

The points $K$ and $L$ lie on the sides $AB$ and $AC$, respectively, of the triangle $ABC$
such that $BK = CL$. Let $P$ be the point of intersection of the segments $BL$ and $CK$.
$M$ be an inner point of the segment $AC$ such that the line $MP$ is the parallel to the
bisector of the $\angle BAC$. Prove that $CM = AB$

We need to prove $AK=ML$.Let $BK=a,KA=b,AL=c,LC=a$ by Menelaus: $\frac{CL}{CA}\cdot \frac{AK}{KB}\cdot \frac{BP}{PL}=1$.Therefore, $\frac{BP}{PL}=\frac{a+c}{b}$.So, we can write $BP=(a+c).k,PL=b.k$.Let $X$ be intersection of angle bisector of $A$ and line $BL$.According to the interior angle bisector theorem: $\frac{BX}{XL}= \frac{a+b}{c}$.Hence, $BX=(a+b).k,XL=c.k$.Since $\frac{AL}{XL}= \frac{ML}{PL}$, we find that: $ML=b$.

(solution: Fatih EGİ)

[quote="Yosh..."][color=darkred]The points $K$ and $L$ lie on the sides $AB$ and $AC,$ respectively, of the triangle $ABC$ such that $BK = CL.$ Let $P$ be the point of intersection of the segments $BL$ and $CK,$ and let $M$ be a point lie on line $AC$ such that the line $MP$ is the parallel to the angle bisector of the $\angle BAC.$
Prove that $CM = AB$ [size=150][u]Nice problem![/u][/size] [/color][/quote]
[color=darkblue][b]Proof.[/b] Construct the point $D$ so that $ABDC$ is a parallelogram. Denote $S=BD\cap CK$
We have: $\frac{PS}{PC}=\frac{BS}{CL}=\frac{BS}{BK}=\frac{DS}{DC}\implies DP$ is angle bisector of $\angle BDC\implies M\in DP$
So $\widehat{CMD}=\widehat{BDM}=\widehat{CDM}\implies CM=CD=AB$[/color]

[color=darkred][b]Remark 1.[/b] Let $ABC$ be a triangle and let $M\in AB$, $N\in AC$ be two points so that the sideline $BC$ don't separate these points and $MB=NC$
Then the geometrical locus of the middlepoint of the segment $[MN]$ is a parallel line to the bisector of the angle $\widehat{BAC}$.

[b]Remark 2.[/b] Let $ABC$ be a triangle and let $M\in AB$, $N\in AC$ be two points so that the sideline $BC$ separates these points and $MB=NC$.
Then the geometrical locus of the middlepoint of the segment $[MN]$ is a perpendicular line to the bisector of the angle $\widehat{BAC}$.[/color]

See here for some other solutions: http://www.mathlinks.ro/Forum/viewtopic.php?p=415147#415147

[quote="scarface"]So, we can write $BP=(a+c).k,PL=b.k$.
Hence, $BX=(a+b).k,XL=c.k$[/quote]

Why we can conclude that "k" is same?

Thx..

$\frac{BP}{PL}=\frac{a+c}{b}\Longrightarrow \frac{BP}{a+c}=\frac{PL}{b}=k$.Here, $k$ is constant of the proportion.So we can write, $BP=(a+c).k, PL=b.k$Then, $BL=BP+PL=(a+b+c).k...(*)$ According to the interior angle bisector theorem: $\frac{BX}{a+b}=\frac{XL}{c}=t$.Here, $t$ is constant of the 2nd proportion.Hence, $BX=(a+b).t, XK=c.t$ and $BL=BX+XL=(a+b+c).t...(**)$. from $(*)$ and $(**)$, we yields $k=t$.


Let's say America didn't go into Iraq and didn't go after Saddam Hussein.  How do you think everything would be different?

I don't really see the value of a hypothetical like this.  It adds nothing to any consideration of what to do about the world [b]as it is[/b], and is just another way to play the blame game.  Stop being childish.

I don't think anyone really cared overly much about Iraq before the war, so it would have been the same as life before 2000

Oh, and republicans would have congress.

Terrorists based in Iraq would have killed everyone with WMD's.

Realistially?
We wouldn't have soldiers in Iraq.
Addam Hussein would be in power.
Etcetera.


function
limit
number theory open
number theory

maybe this is a well-known open problem, maybe it is easier than i think: given a set A of finitely many primes. prove that there are only finitely many $n$ such that any prime dividing $n$ or $n+1$ belongs to A.

i could hardly make any progress on it. i would especially like to see explicit bounds on the number or the size of the $n$ satisfying that property.
the original thing i was thinking of(which immediately implies the problem above): does there exist a function $f: \mathbb{N}\to \mathbb{N}$ such that $f(xy)=f(x)f(y)$ and $\lim_{n\to \infty} \frac{f(n+1)-f(n)}{n}=0$.

that one now does imply the 3n+1-conjecture for sufficiently large numbers.

Peter

[quote="Peter Scholze"]the original thing i was thinking of(which immediately implies the problem above): does there exist a function $f: \mathbb{N}\to \mathbb{N}$ such that $f(xy)=f(x)f(y)$ and $\lim_{n\to \infty} \frac{f(n+1)-f(n)}{n}=0$.
[/quote]

don't $f(x)=1$, $f(x)=0$ and $f(x)=x$ all fullfil these requirements? :?

As for the other problem, n and n+1 cannot have any prime in common in their factorization, so basicly what you need to do is checking if $p_1^{q_1}\cdots p_n^{q_n}$ can be written in the form of $n\cdot(n+1)$ in an infinite amount of ways, providing you may choose all $q_i$.

So I think we can rewrite the problem in a nicer wording: for given $n$, $\exists m: m\cdot(m+1) = k\cdot n\cdot(n+1)$ with either $k|n$ or $k|n+1$?

I didn't solve it yet though, but it looks much too nice to be an unknown open problem :cool:

sorry for the second thing, i wanted to forbid "trivial" solutions such as the ones you gave(let's take $f(2)>f(3)$ as an extra condition).
btw, can the moderators delete one of the two topics? i posted it twice...sorry!

Peter

[quote="Peter VDD"]for given $ n $, $ \exists m: m\cdot(m+1) = k\cdot n\cdot(n+1) $ with either $ k|n $ or $ k|n+1 $?[\quote]

this is not quite true...the condition for $k$ should be that all its prime divisors divide either $n$ or $n+1$.

Peter

[quote="Peter Scholze"]sorry for the second thing, i wanted to forbid "trivial" solutions such as the ones you gave(let's take $f(2)>f(3)$ as an extra condition).[/quote]

Ah, I think I start getting the relation you're investigating between the 2 problems...

I thought for reals we have that $f(x)\cdot f(y)=f(x\cdot y)$ implies $f(x)=x^k$, but that's $R\rightarrow R$.
So if we consider only the lattice points of the function created by a non-negative integer value...
that means we must have $f(x)=x^k\ (k\in\mathbb{Z})$ right? I'm not really familiar with this though. :(

so if $f(2)>f(3)$ that is a contradiction... since $k<0$ and $f(x>1) \not\in \mathbb{N}$ for $k\in\left]-\infty;0\right[$.

hmm...maybe you should not trust on some results you heard of but on your own mind. and that one tells you that either $f(x)=0$ or $f(1)=1$ (very easy), and if $f(1)=1$, we may arbritarily choose $f(p)$ for all primes $p$. now, we have $f(\prod p_i^{a_i})=\prod f(p_i)^{a_i}$. all these functions give solutions.
it is now some work to find how the second problem implies the first. i didn't check everything, but i think if there are infinitely many $n$ with that property, then $f$ has to be the identity at all primes of $A$. thus, let's add the condition in the second problem that $f$ is not the identity for all primes in $A$. so i'll state it again to make the whole thing clear:
given an arbritary finite set of primes $A$. then there exists a nonconstant function $f:\mathbb{N} \to \mathbb{N}$ with the following properties:
1) $f(xy)=f(x)f(y)$
2) $\lim_{n\to \infty} \frac{f(n+1)-f(n)}{n}=0$
3) $\exists p\in A:\;  f(p)\neq p$.

Peter

A problem generalizing your first problem was posted some time ago. If my memory serves right, I think the topic was started by sam-n and it stated that given $k$ and a finite set of primes, $A$, there are only finitely many solutions to the equation $x-y=k$ s.t. the prime factors of $x,y$ are in $A$. It was unsolved for quite some time, until Harazi posted a solution which I didn't understand (and made no efforts to understand, because, even though it looked cool, it dew on too much advanced stuff :)). However, nobody managed to solve it in a more elementary way, so it might just be that this is the way to solve it.

I'll look for the topic.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4724&highlight=xy+k]Here[/url] it is. I read it, and it's not that tough to understand, but really tough to find :).

uhm...yeah. thanks grobber, but i came to the same point as harazi and knew there was such a result in that direction(couldn't remember the wording, but anyway). thus i was sure that the problem is true. however, i want a short nice elementary solution(or at least only using results i really understand).

Peter

btw, sam-n's generalization is implied by my second problem as well...the more i look at it, it has more nice consequences. i'll try to find a proof using that theorem harazi used.

Peter


I realize that the only reason I'm good at/ like math below the precalc level is because of math competitions and fun problems. I don't particularly enjoy college math, though, and I was wondering if anyone know of any good problem solving books that address the following subjects:

Calculus
Multivariable Calculus
Differential Equations
Linear Algebra
Real Analysis

Thanks. I'm lookig for books other than past Putnam's, though.

problem solving through problem,L.C.Larson....very nice and suitable book for anybody.and also it has many other topics.

Intermediate Real Analysis published by Springer has lots of problems and topics to learn and enjoy.  It begins with very simple, yet delightful problems, and it builds up gradually.

Masoud Zargar


function
limit
real analysis
real analysis unsolved

Do you mean the Mean Value Theorem? or the relaxed constraint where the function $ f$ only needs to be continuous on $ [a,b]$, $ f(a)=f(b)$ and for all $ x\in (a,b)$ we have $ \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\in [-\infty, \infty]$?

[quote="Orif"]Prove the generalization of Rolle's theorem[/quote]

Let $ f: [a,b]\mapsto \mathbb{R}$ be continous on $ [a,b]$ and differentiable on $ (a,b)$ define the function,
$ g(x) = f(x)-m(x-a)-f(a)$ where $ m=\frac{f(b)-f(a)}{b-a}$.
Then show that $ g(a)=g(b)$ and conditions for Rolle's theorem are satisfied.

So, $ g'(c) = 0$ for some $ c\in (a,b)$.
Meaning,
$ f'(c) = \frac{f(b)-f(a)}{b-a}$.


calculus
derivative
LaTeX
limit
logarithms
function
calculus computations

can someone help me with this review sheet. its due on monday, and i have been sick for the last week and am completely lost. thank you.

Find the limitsof the following:

lim(x->2) for: ((x^2)-4), x<2
(x-2), x>2

lim(x->3) for: (x)/(3-x)

lim(x->9) (x-9)/((sqrt of x)-3)

lim(x->2) for ((x^3)-8)/((x^2)-4)

lim(x-> -3) for thirdroot(10+3x)

lim(x->0) for (ln(x+2)^X)/(x))

Using the long method, find the derivative of f(x)=(x^3-5x^2+7)

$\LaTeX$ed:

[quote="kataklysm08"]Find the limits of the following:

$\lim_{x \rightarrow 2}x^{2}-4$
$\lim_{x \rightarrow 3}\frac{x}{3-x}$
$\lim_{x \rightarrow 9}\frac{x-9}{\sqrt{x}-3}$
$\lim_{x \rightarrow 2}\frac{x^{3}-8}{x^{2}-4}$
$\lim_{x \rightarrow 3}\sqrt[3]{10+3x}$
$\lim_{x \rightarrow 0}\frac{\ln((x^{2})^{x}}{x}$

Using the long method, find the derivative of $f(x)=x^{3}-5x^{2}+7$[/quote]

Perhaps this should be moved to a different forum? (I'm not a moderator; I don't know)

These should be in the Calculus Computation Tutorials forum.

Basically, for most of these limits, you can simply substitute the value that x approaches into the function.
However, if you do this on some, you'll end up dividing by zero.
You'll need to use algebra to rewrite the equations in such a way that division by zero doesn't occur.

The last limit is a bit tricker than the rest.

For the derivatives one, it's simply $f'(x) = 3x^{2}-10x$ by the power rule.

For the last one I think they want the difference quotient method, of if that is what they mean the "long way".

I think this should be in the calculus forum.  For the derivative, I think the "long method" you are referring to is:
$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0}\frac{(x+h)^{3}-5(x+h)^{2}+7-(x^{3}-5x^{2}+7)}{h}$. See what you can do from there.

Correction for the first problem in $\LaTeX$.
[quote]$\lim_{x \rightarrow 2}\begin{cases}x^{2}-4 & x < 2\\x-2 & x > 2\end{cases}$[/quote]

Bad $\LaTeX$, how are piecewise coded?  :maybe:

[color=green][Edited by moderator; copy it to see what the code looks like.][/color]

[quote="7h3.D3m0n.117"]1. $\lim_{x \rightarrow 2}x^{2}-4$
2. $\lim_{x \rightarrow 3}\frac{x}{3-x}$
3. $\lim_{x \rightarrow 9}\frac{x-9}{\sqrt{x}-3}$
4. $\lim_{x \rightarrow 2}\frac{x^{3}-8}{x^{2}-4}$
5. $\lim_{x \rightarrow 3}\sqrt[3]{10+3x}$
6. $\lim_{x \rightarrow 0}\frac{\ln((x^{2})^{x}}{x}$

Substituting values in most of the limits require (sometimes) division by 0, which is not allowed in calculus. However, use the Conjugate Theorem to work around some of the limits (like number 3). If all else fails, use L'Hospital's Rule (where you calculate the derivitives of the limit in order to solve the limit itself).[/quote]

so 1. would be 0?

I'll do number 6.
The numerator is $\ln((x^{2})^{x})$ which can be rewritten as $x \ln x^{2}$
We get $\frac{x \ln x^{2}}{x}$, so the quotient simplifies to $\ln(x^{2})$, which has a limit at $-\infty$ as $x$ goes to $0$

For number one  :maybe:
[quote]lim(x->2) for: ((x^2)-4), x<2
(x-2), x>2 [/quote]
Am I reading it wrong or it is a piecewise function?
Then making it not have a limit.

Technically, a piecewise function can have a limit if the limit from the right is equal to the limit from the left.

So:
$\lim_{x \rightarrow 2}x^{2}-4 = 0$
and $\lim_{x \rightarrow 2}x-2 = 0$

So the limit at $x = 2$ exists.

i called my friend. she said that for the ones that lead to division by zero, i need to average the terms before and after the given term. is this correct?

:rotfl:
After all that, I still read the problem wrong.  Though I read it as a piecewise correctly.

For the ones that you get zero, either manipulate the function or put in very close numbers to the limit to find it out.

Learn something known as the Squeeze thm. too.
That's a helpful one.
I liken it to making a limit sandwich.

Pick two functions that f(x) and g(x) such that f(x) > original function for all x, and g(x) < original function for all x.

Then find the limits of those functions where you want to, and it is guaranteed that if the limit of the original function exists, it lies between f(x) and g(x)

Moderators, this forum should be locked.

Japanese Communities Modeartor

kunny

This is rule! The problems of differentiation, derivative, limit etc are not allowed to post High School Forum, even in Intermidate or Olympiad forum.
In Japan, Japanese students usually study those problem at the age of 17~18, except for some junior high school students at the age of 14, 15.
Anyway your problem should be posted in  [color=red]Calculus Computation and Tutorials[/color] :)

kunny

[quote="kataklysm08"]4. $\lim_{x \rightarrow 2}\frac{x^{3}-8}{x^{2}-4}$

so if i plugged in 1.95 as my lower limit and 2.05 as my upper limit, with a .01  pitch in my calculator, i could then take the values for 1.99 and 2.01, average them, and get the right answer?[/quote]

Yes, you can do that, or you can use L'Hospital's Rule:

$\lim_{x \rightarrow 2}\frac{x^{3}-8}{x^{2}-4}$

$\lim_{x \rightarrow 2}\frac{3x^{2}}{2x}$

substitution would lead to $\frac{12}{4}= 3$

[quote="kunny"]Should be locked! :mad:

You should read this!

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=26438[/url][/quote]

Well that's for the Intermediate Forum. I'm sure a mod can move this to a Calculus forum.

[color=blue]To Japanese high school students and junior high school students![/color]

[color=red]You can't use L'Hopital's theorem in Japanese university entrance examination. [/color]

Japanese Communities Moderator

kunny

why do you care so much? does helping a junior in high school precalculus offend you?

guess what buddy, im not taking the japanese university entrance exam, im just trying to learn efffing math.  this is for AMERICAN learning institutions, so i dont think your opinion is required.

How did this ever end up in the complex analysis forum?

Sorry for not noticing this earlier. It is an appropriate - if perhaps not all that interesting - topic for the "Calculus Computations and Tutorials" forum, and it will be moved.


AMC USA(J)MO USAMO ARML AIME quadratics inequalities \/closed

If I had a choice between the AMC/AIME algebra class or the AMC-12 problem series (during the summer), which would you suggest?

Having had you in class I would suggest that you are better off taking the algebra class.  I personally feel that students who can already make it to the USAMO will get way more out of our subject courses.

Some of the algebra course will be review for you.  Take a look at the Post Test.  If you can solve all the problems, skip the class.

Unfortunately the Post Test does not test for everything covered in the class (it would be hard to do that).  The class covers dozens of problems on the ARML or AIME level and a fair amount (10% of the class) of Olympiad level algebra problems (after building steadily in each lesson).

I am currently writing script for difference equations that I will be adding to the class for this coming session.  The course will also cover uses of the quadratic equation that lead to many Olympiad ideas and substitution methods that help solidify problem solving right up to Olympiad level.  There is also a full day on functional equations that gets up the level of Olympiad problems.

The Olympiad level problems are on the whole not as challenging as those in the Inequalities class you are taking.  If I were you I would see the algebra class as having the potential to solidify ideas, close a gap or two in knowledge, provide practice for the AMC12 and AIME, and provide a couple of dozen practice Olympiad problems.


calculus
integration
trigonometry
calculus computations

First, we show that

$\int_{0}^{\frac{\pi}{2}}\cos^{n}x \cos (n+2k)x \, dx = 0$

for all $n = 0, 1, \cdots$ and $k = 1, 2, \cdots$. This can be achieved by mathematical induction.

(1) $n = 0$, then $\int_{0}^{\frac{\pi}{2}}\cos^{n}x \cos (n+2k)x \, dx = \int_{0}^{\frac{\pi}{2}}\cos 2kx \, dx = 0$
(2) Suppose it holds for $n$. Then $\int_{0}^{\frac{\pi}{2}}\cos^{n+1}x \cos (n+1+2k)x \, dx$ $= \int_{0}^{\frac{\pi}{2}}\frac{1}{2}\cos^{n}x \{ \cos (n+2+2k)x+\cos (n+2k)x \}\, dx = 0$,
so it also holds for $n+1$.

Now we apply the same method we used in (2) to the original integral. Then

$\int_{0}^{\frac{\pi}{2}}\cos^{n+1}x \cos (n+1)x \, dx$ $= \int_{0}^{\frac{\pi}{2}}\frac{1}{2}\cos^{n}x \{ \cos (n+2)x+\cos nx \}\, dx$ $= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\cos^{n}x \cos nx \, dx$.

So we have

$\int_{0}^{\frac{\pi}{2}}\cos^{n}x \cos nx \, dx = \frac{1}{2^{n}}\int_{0}^{\frac{\pi}{2}}\, dx = \frac{\pi}{2^{n+1}}$.


function
algebra
domain
calculus
calculus computations

what's the faster way to identify if a function is bounded or not?

Let D be the domain of a function f(x). For f to be bounded it means you can find a positive number M such that $ \forall x \in D \,\, |f(x)| \leq M$. Geometrically, this means you can sandwich the graph of f(x) between two horizontal lines. So perhaps looking at the graph is the faster method. Of course, some experience also plays a role.


[quote="spider_boy"]Prove that $\{n\sqrt7\}<1-\frac{1}{6n}$  where $n$ is a natural number and $\{x\}$
is fractional part function.
[hide]The original question was $\frac{1}{2n}<\{n\sqrt7\}<1-\frac{1}{6n}$.
I have solved (only) the first part by using some number theory,so I posted the problem
in this section.[/hide][/quote]

Let $(a+1)^2 > 7n^2 =a^2+t \geq a^2 \Rightarrow$
$\{\sqrt{7}n\}=\sqrt{a^2+t}-a$
So we have to prove
$\sqrt{a^2+t}-a<1-\frac{1}{6n} \Leftrightarrow$
$t<2a(1-\frac{1}{6n})+(1-\frac{1}{6n})^2$
but $\frac{2a}{6n} <1 \mbox{ as } a^2< 9n^2) \Rightarrow$
if $t \leq 2a-1$ then we prove the statement.

In the another case $t=2a \Rightarrow$ we have to prove
$a^2+2a<a^2+2a+1 - \frac{a+1}{3n} +\frac{1}{36n^2} \Leftarrow$
$\frac{a+1}{3n} <1 \Leftarrow (a+1)^2=7n^2+1<9n^2=(3n)^2$

So we'v proved the statement in any case.

V:)


The message is too small. Please make the message longer before submitting.

All the letters in your topic title are uppercase (A, B, C, ...). Please make some of them lowercase (a, b, c, ...)

 :mad:

ANYBODY THERE????

I'm here...
And Altheman, do you go to New Trier? Because I think I know you from NTMT  :wink:

[quote="Quevvy"]I'm here! If you know me......... I got a perfect score on the AMC 8![hide=":first:"]Yeah![/hide][/quote]
Wow. I just realized you hid something under that emoticon. Pretty sneaky.

I think we need to pool our money to buy him plane tickets to come up here tomorrow.

[quote="Altheman"][quote="probability1.01"]No. Let $A = \{ x | x \mbox{ is from Illinois}\}$. It is not difficult to show that $A$ is empty.[/quote]

yes...to match the awesomeness...

let $S$ be the set of people, let $f: S\rightarrow\{0,1\}$ such that $f(x)=1$ if $x$ does arml, otherwise $f(x)=0$...why is it that $f(\text{Alex Zhai})=0$?[/quote]

Hmmm... I just assumed Alex did Chicago ARML. Way to stick it to those Chicago guys, Alex.

Just to clarify, I was on the national championship Mathcounts team with Alex three years ago. But I'm from the southwest part of Illinois, so I latched on to the Missouri ARML team instead. Sure, it means coming in near-last place every year. But it also saves 8 hours in travel time per practice.

[quote="Aryth"]Ah, Quevvy's at my school. Haha, small world.  Not really.

[b]AMC[/b]
[b]2005= 8:[/b] 20 (oh. crap.)
[b]2006= 10A:[/b] 144 [b]AIME I:[/b] 6 [b]8:[/b] 25
[b]2007= 10A[/b]: 144 [b]12B:[/b] 117

[b]MC[/b]
School: 1st
Regionals: 1st
State: 1st
Nationals: Soon to be 228th.  :P[/quote]

Wait, you're in 8th grade? Those are really nice 7th grade scores...


function
Support
real analysis
advanced fields
advanced fields unsolved

Could you give me and example of a smooth map R -> R that has a dense set of critical values.

Just take any enumeration $ q_n$ of the rationals and create any smooth function $ f(x)$ such that $ f(n) \equal{} q_n$, $ f'(n) \equal{} 0$.

To elaborate on that, consider something like
\[ \phi(x) \equal{} \begin{cases} e^{ \minus{} \frac {x^2}{1 \minus{} x^2}}\left(1 \minus{} e^{ \minus{} \frac {1}{x^2}}\right) & 0 < |x| < 1 \\
1 & x \equal{} 0 \\
0 & \text{otherwise}\end{cases}\]
It is $ C^\infty$, has compact support on $ [ \minus{} 1,1]$, and has a critical value at $ x\equal{}0$.  Then $ f(x) \equal{} \sum_{n \equal{} 1}^\infty q_n \phi (x \minus{} n)$ is a function of the type suggested by fedja.

Yeah. Thank you so much for such a detailed answer.


The 2-digit numeral $AB$, with $A \not= 0$, represents a prime if the numeration base is 8, 10, or 12.  Find $AB$.

[hide]Well, 8a + b = Prime, 10a + b = Prime, and 12a + b = Prime. So, if a = 1, the primes are 2 apart, and for a = 2, they are 4 apart, so with that pattern, I listed some primes with difference 2, 4, 6, 8... to try. 2 and 4 are impossible right away, so try 6. 8 * 3 = 24, and 24 + 7 = 31, which is the first term in three primes 6 apart. They are 31, 37, and 43. So, a = 3, and b = 7, for 37.[/hide]


A 48-foot-wide building has a roof that rises 4 feet for each 12-foot horizontal change. If the roof has a 2-foot overhang, what is the distance from the peak, P, to the edge of the overhang, O?

[asy]draw((0,0)--(48,0));
draw((-6,-3)--(24,12)--(54,-3));
label("P", (24,12), N);
label("O", (-6,-3), S);
label("48\ ft", (24,0), S);
label("2\ ft", (-3,-1.5), NW);[/asy]

From the peak to the junction is 24 horiz = 8 vert = 8rt(10) total distance.

Answer: 2+8rt(10)


function
floor function
algebra unsolved
algebra

find the number of real solutions

a=[a/2]+[a/3]+[a/5]

where [.] implies greatest inter function

Suppose $ \{.\}$ is the fractional part function.
In other words, for all real $ x$, $ \{x\}\equal{}x\minus{}\lfloor x\rfloor$
The given equation is equivalent to:
$ a\equal{}\frac{a}{2}\minus{}\{\frac{a}{2}\}\plus{}\frac{a}{3}\minus{}\{\frac{a}{3}\}\plus{}\frac{a}{5}\minus{}\{\frac{a}{5}\}$
$ \iff a\equal{}\frac{a}{2}\plus{}\frac{a}{3}\plus{}\frac{a}{5}\minus{}\{\frac{a}{2}\}\minus{}\{\frac{a}{3}\}\minus{}\{\frac{a}{5}\}$
$ \iff a\equal{}\frac{31a}{30}\minus{}\{\frac{a}{2}\}\minus{}\{\frac{a}{3}\}\minus{}\{\frac{a}{5}\}$
$ \iff \frac{a}{30}\equal{}\{\frac{a}{2}\}\plus{}\{\frac{a}{3}\}\plus{}\{\frac{a}{5}\}$
$ \iff a\equal{}30(\{\frac{a}{2}\}\plus{}\{\frac{a}{3}\}\plus{}\{\frac{a}{5}\})$

We know that for all integers $ x$, $ n$:
$ n\{\frac{x}{n}\}$ is the remainder when $ x$ is divided by $ n$
Suppose $ i,j,k$ are the remainders when $ a$ is divided by $ 2,3,5$ respectively.
So the equation becomes:

$ a\equal{}15i\plus{}10j\plus{}6k$

Note that $ 15i\plus{}10j\plus{}6k\equiv i\mod 2$
And $ 15i\plus{}10j\plus{}6k\equiv j \mod 3$
And $ 15i\plus{}10j\plus{}6k\equiv k\mod 5$

So for all triples $ (i,j,k)$ fix $ a$ at $ 15i\plus{}10j\plus{}6k$ which gives a solution.

Since there are $ 2,3,5$ possible values for $ i,j,k$ respectively, there are $ 2\cdot 3\cdot 5$ triples $ (i,j,k)$
Hence there are $ \boxed{30\text{ solutions}}$


Find all solutions to the pair of equations

SinX/SinY =sqrt2 ,         TanX/TanY= sqrt3

$\frac{SinX}{SinY}=\sqrt{2} , \frac{TanX}{TanY} =\sqrt{3}$

You can make that look even nicer with the \sin and \tan commands...

$\frac{\sin{X}}{\sin{Y}} = \sqrt{2}$ and $\frac{\tan{X}}{\tan{Y}} = \sqrt{3}$

[hide]
$\frac{\tan{X}}{\tan{Y}} = \frac{\sin{X}}{\sin{Y}}\cdot\frac{\cos{Y}}{\cos{X}} = \sqrt{2}\cdot\frac{\cos{Y}}{\cos{X}} = \sqrt{3}$

So $\cos{X} = \sqrt{\frac{2}{3}}\cos{Y}$ and $\sin{X} = \sqrt{2}\sin{Y}$.

But $1 = \cos^2{X}+\sin^2{X} = \frac{2}{3}\cos^2{Y}+2\sin^2{Y} = \frac{2}{3}+\frac{4}{3}\sin^2{Y}$

so then

$\sin^2{Y} = \frac{1}{4} \Rightarrow \sin{Y} = \pm \frac{1}{2} \Rightarrow Y=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$

and we get corresponding $X$ values of

$X = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.

I'm not sure, but that seemed like a very round-a-bout way to get at an obvious answer.[/hide]


The National Marriage Council wishes to invite $n$ couples to form 17 discussion groups under the following conditions:

(1) All members of a group must be of the same sex; i.e. they are either all male or all female.

(2) The difference in the size of any two groups is 0 or 1.

(3) All groups have at least 1 member.

(4) Each person must belong to one and only one group.

Find all values of $n$, $n \leq 1996$, for which this is possible.  Justify your answer.

Sorry to bring this post back up but the current solution seems very vague (in fact, it's not quite a solution at all).

Also, I'm interested in the problem and I want to bring this post back to life so that it can have another chance of someone finding and posting a solution. :)

It probably hasn't been solved or gotten a detailed solution because it's rather ugly casework without any neat insights to be had.

Suppose there are $k$ groups of males and $17 - k$ groups of females; WLOG $k \geq 9$. Also suppose that all the groups have $m$ or $m+1$ members. Then $mk \leq (m+1)(17-k)$, or $(2m+1)(2k-17) \leq 17$. Since $m \geq 1$, we have $2k - 17 < 6$, so $9 \leq k \leq 11$.

$k = 11$? Then only $m = 1$ works. We have at least 11 males and at most 12 females. So $n = 11, 12$ work.

$k = 10$? Then only $m = 1$ or $m = 2$ works. For $m = 1$, we have at least 10 males and at most 14 females. So $n = 10$ through $14$ work. For $m = 2$, we have at least 20 males and at most 21 females. So $n = 20, 21$ work.

$k = 9$? Then $m$ can be as high as 8. For a given $m$, we have at least $9m$ males and at most $8(m+1)$ females. This gives the range of solutions seen in post #2 which I won't recopy here. It turns out that this set of solution subsumes all the previous cases, so that is the complete list of answers.


integration
calculus
calculus computations

Show that $ \left(\alpha - \frac {1}{\alpha} - x\right)(4 - 3x^2)$ has just one maximum and just one minimum, and that the difference between them is $ \frac {4}{9}\left(\alpha + \frac {1}{\alpha}\right)^3$.

What is the least value of this difference for different values of $ \alpha$?

[hide="My Solution"]

Let $ f(x) = (c - x)(4 - 3x^2) = 3x^3 - 3cx^2 - 4x + 4c$, where $ c = \alpha - \frac {1}{\alpha}$.

$ f'(x) = 9x^2 - 6cx - 4$, $ D/4 = ( - 3c)^2 - 9\cdot ( - 4) = 9(c^2 + 4) > 0$, thus we can write $ f'(x) = 9(x - \alpha )(x - \beta)\ (\alpha < \beta)$.

Since $ f(x)$ has a local maximum at $ x = \alpha$ and  a local minimum at $ x = \beta$,

The difference of extrema : $ d = f(\alpha) - f(\beta) = \int_{\beta}^{\alpha} f'(x)\ dx$

$ = - 9\int_{\alpha}^{\beta} (x - \alpha)(x - \beta)\ dx = \frac {3}{2}(\beta - \alpha)^3$.

Now $ f'(x) = 0\Longleftrightarrow x = \frac {c\pm \sqrt {c^2 + 4}}{3}$, yielding $ \beta - \alpha = \frac {2}{3}\sqrt {c^2 + 4}$.

$ \therefore d = \frac {3}{2}\left(\frac {2}{3}\sqrt {c^2 + 4}\right)^3 = \frac {4}{9}\left(\alpha + \frac {1}{\alpha}\right)^3$. Q.E.D.

Then $ d$ is minimal $ \Longleftrightarrow \beta - \alpha$ is minimal $ %Error. "Lonogleftrighatrrow" is a bad command.
c = 0$, yielding $ d_{min}=\frac{32}{9}$.[/hide]

My Solution:

Let $ f(x) = (c - x)(4 - 3x^2) = 3x^3 - 3cx^2 - 4x + 4c$, where $ c = \alpha - \frac {1}{\alpha}$.

$ f'(x) = 9x^2 - 6cx - 4$, $ D/4 = ( - 3c)^2 - 9\cdot ( - 4) = 9(c^2 + 4) > 0$, thus we can write $ f'(x) = 9(x - \alpha )(x - \beta)\ (\alpha < \beta)$.

Since $ f(x)$ has a local maximum at $ x = \alpha$ and  a local minimum at $ x = \beta$,

The difference of extrema : $ d = f(\alpha) - f(\beta) = \int_{\beta}^{\alpha} f'(x)\ dx$

$ = - 9\int_{\alpha}^{\beta} (x - \alpha)(x - \beta)\ dx = \frac {3}{2}(\beta - \alpha)^3$.

Now $ f'(x) = 0\Longleftrightarrow x = \frac {c\pm \sqrt {c^2 + 4}}{3}$, yielding $ \beta - \alpha = \frac {2}{3}\sqrt {c^2 + 4}$.

$ \therefore d = \frac {3}{2}\left(\frac {2}{3}\sqrt {c^2 + 4}\right)^3 = \frac {4}{9}\left(\alpha + \frac {1}{\alpha}\right)^3$. Q.E.D.

Then $ d$ is minimal $ \Longleftrightarrow \beta - \alpha$ is minimal $ \Longleftrightarrow c = 0$, yielding $ d_{min} = \frac {32}{9}$.


Let $ x,y,z\geq 0$ be such that $ x^2\plus{}y^2\plus{}z^2\equal{}1$.

Prove that $ 1\leq \frac{z}{1\plus{}xy}\plus{}\frac{y}{1\plus{}xz}\plus{}\frac{z}{1\plus{}xy}\leq \sqrt{2}$

[quote="outback"]Let $ x,y,z\geq 0$ be such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$.

Prove that $ 1\leq \frac {z}{1 \plus{} xy} \plus{} \frac {y}{1 \plus{} xz} \plus{} \frac {z}{1 \plus{} xy}\leq \sqrt {2}$[/quote]
Fix it please.

[quote="outback"]Let $ x,y,z\geq 0$ be such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.

Prove that $ 1\leq \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ca} \plus{} \frac {c}{1 \plus{} ab}\leq \sqrt {2}$[/quote]
Firstly, we prove that:
$ (a \plus{} b \plus{} c)^2 \leq 2(1\plus{}bc)^2 (1)$
Indeed, we have :
$ (1) \Leftrightarrow 2(ab\plus{}bc\plus{}ca) \leq 1\plus{}4bc\plus{}2b^2c^2 \Leftrightarrow 2a(b\plus{}c) \leq a^2\plus{}(b\plus{}c)^2\plus{}2b^2c^2$
$ \Leftrightarrow (b\plus{}c\minus{}a)^2\plus{}2b^2c^2 \geq 0$ (true)
Therefore,
$ \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ca} \plus{} \frac {c}{1 \plus{} ab}\leq \frac{\sqrt{2}a}{a\plus{}b\plus{}c}\plus{}\frac{\sqrt{2}b}{a\plus{}b\plus{}c}\plus{}\frac{\sqrt{2}c}{a\plus{}b\plus{}c}\equal{} \sqrt{2}$
Other, we have :
$ a\plus{}abc \leq a\plus{}\frac{a(b^2\plus{}c^2)}{2}\equal{}1\minus{}\frac{(a\minus{}1)^2(a\plus{}2)}{2} \leq 1$
therefore,
$ \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ca} \plus{} \frac {c}{1 \plus{} ab}\geq a^2\plus{}b^2\plus{}c^2\equal{}1$
We are done .

$ 1\leq \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ca} \plus{} \frac {c}{1 \plus{} ab}$

$ (a\plus{}b\plus{}c)^2\equal{} \left (\sum_{cyc} \sqrt a \sqrt{1 \plus{} bc} \cdot \frac{\sqrt a}{\sqrt{1\plus{}bc}} \right )^2$$ \le \left (\sum_{cyc} \frac{a}{1\plus{}bc} \right ) \left ( \sum_{cyc} a (1 \plus{} bc) \right )$

$ \sum_{cyc} \frac{a}{1\plus{}bc} \ge \frac{(a\plus{}b\plus{}c)^2}{a\plus{}b\plus{}c\plus{}3abc} \ge 1$

$ p^2 \minus{} p \minus{} 3r \ge 0$


If $a$ and $b$ are integers such that $x^2-x-1$ is a factor of $ax^3+bx^2+1$, then $b$ is

(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

[hide]If we take the third factor to be (mx+n), we get mx^3+(n-m)x^2-(m+n)x-n.  We thus know that m+n=0 and n=-1, so m=1, and (n-m)=b=-2.  A.[/hide]

[hide="Answer"]Factoring out, we have $(ax-1)(x^2-x-1)$ so that we end with a $+1$ on the end and $ax^3$ at the beginning. Therefore, we have $ax^3-(a+1)x^2-(a-1)x+1$. We have no $x$ factor, so $a-1=0\Rightarrow a=1$, and therefore $b=-(a+1)=-2$. $\boxed{A}$[/hide]


How many ways are there to place two kings on a 8x8 chessboard so that they do not attack each other?
(you know hte drill, kings attack all adjacent squares. and they are opposite color kings.)

[hide]n = 64^2 - (36 * 8 + 24 * 5 + 4 * 3) = 3676

Not sure about this..seems too easy...[/hide]

Whoops. Here is revised solution:

[hide]n = 64 * 63 - (36 * 8 + 24 * 5 + 4 * 3) = 3612[/hide]

Looks good to me, assuming "the two kings are of different colors" means that they are distinct.


The product of any two of the positive integers 30, 72, and $N$ is divisible by the third.  What is the smallest possible value of $N$?

[hide]Needs factors of 2, 3, and 5. For it to be divisible by 72, it needs 2 more 2's and a 3. So the answer is $60$[/hide]

You can prime factor both 30 and 72, so you get 5X3X2, and 2^3 and 3^2. I then just used trial and error with these numbers to get 5X2^2X3=[b]60[/b].


Given that $w$ and $z$ are complex numbers and $w^2+z^2 = 7$ and $w^3+z^3=10$, find the greatest value of $w+z$
Source: 1983 AIME #5

[hide="I tried..."]
Since $w^2+z^2 = 7$, we have $(w+z)^2=7-2wz$
Factoring $w^3+z^3=10$, we get $(w+z)(7-wz)=10$.
Solving for $wz$ in the first equation and plugging it into the second gives you as a final equation
$(w+z)^3+7(w+z)-20=0$ so I'm thinking that the greatest real solution would be the answer. The "only" problem is that I can't find any real solutions to the equation...and I don't see how $w$ and $z$ being complex numbers affect the question.[/hide]

[quote="Hellstar"][hide="I tried..."]
Since $w^2+z^2 = 7$, we have $(w+z)^2=7-2wz$
Factoring $w^3+z^3=10$, we get $(w+z)(7-wz)=10$.
Solving for $wz$ in the first equation and plugging it into the second gives you as a final equation
$(w+z)^3+7(w+z)-20=0$ so I'm thinking that the greatest real solution would be the answer. The "only" problem is that I can't find any real solutions to the equation...and I don't see how $w$ and $z$ being complex numbers affect the question.[/hide][/quote]
[hide]
The first line should be $(w+z)^2=7+2wz$.
You should get $-(w+z)^3+21(w+z)-20=0$, whose solutions are $w+z=4,1,-5$.[/hide]

This "classic" problem was posted about 8 times in AoPS. Here is the link. Please use search button.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367523#p367523


What is the number of permutations of length $N$,consisting all the numbers between $1$ and $N$
such that in each permutation all the numbers isn't on their places?

I think your meant was $D_n=n!\sum_{r=0}^n\frac{(-1)^n}{r!}$


Hey does anyone here think they qualified for the USAMO?

based on [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142291[/url] I did.. I hope.

omfg I CANNOT wait for that email to come.. I'll do whatever my mom tells me to do for the rest of the day.

then again, maybe not :lol:

Nice job (especially for an 8th grader)
if I didnt make any stupid mistakes (or just like 1 stupid mistake), I wouldve qualified but I didnt... :mad: but theres always next year...even though I thought I could qualify for this year.

I think Carmel High School had like 2 qualifiers or so?

EDIT:only one..only two have ever made it from our school

i didn't make it. my index was 175 and i got a 4 on the AIME.

Thanks.

I was in despair because I got a 7 on the AIME II and people were mentioning 8 or 9 as the USAMO floor.. but yeah.. just got my email and like *spazzzz*

I think about five people from our school qualified, including me :huh:

EDIT: heh, [i]three[/i] people from our school.. six people from west side

the qualifiers are posted: [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007usamoqual.shtml[/url]

at least u made it. i was in despair cuz i made a careless mistake on #2. and i could've gotten #4. and i made a careless mistake on #16 on AMC 10.

hmm only 10 people from Indiana made it..thats worse than last year and with the expansion...o well..nice job to tyler zou-seems like hes the only one in our school that made it (even though about 5 others were probalbay 1 question frm amking it)

I guareentee next year Indiana will have like 20 USAMO qualifiers (double this year)..

"The message is too small.. sldkfjskadfj"

whatever.

Art of Owna, how come?

Looking at all the 7th, 8th and 9th graders that almost qualified this year/or wouldve qualified if it werent for dumb mistakes (I know next year you can make the same dumb mistakes-but theres a potential for 20)

COnsidering there are 11 frm thie year
Only 1 is a 12th grader so next year we'll probalby have a guareenteed 10.  I can probalby name 10 more that can qualify next year if something bad or a lot of careless msitakes do not take place.

i hope i can get an 8 next year. in 9th grade, don't you get to go to this really good math camp for free if you get top 30 for 9th graders?(on AIME)

I hope I can get 12+ for next year just because I feel like letting myself down...

Actually I hope for a 10..I probalby can get a 10 as long as I make NO careless or even non-careless mistakes (especially with one year)

uh dude -_-... its called MOSP, and you can only go if you're top 500... unless you're talking about a scholarship for ninth graders, then i have no idea (to yuqing :D)

guess what? i'm stupid. no wonder i don't even know it's called MOSP. (To [color=red][size=200]LYNDON[/size][/color])

[quote="mathcrazed"]that's like... not really plausible lol, david benjamin and nathan benjamin only got 2 or 3, but then again, Paul does like practice every day,.... they have Zuming Feng over there!![/quote]
Yea, Feng is really good. Paul improved a lot this past year, I have conversed with him a little bit.

Dude Phil, fix your signature..

lol nathan was just kidding about getting perfect ... no duh XD

Let me clarify my earlier post.  I thought that nathan benjamin got a perfect score (lyndon told me this) so I thought when lyndon said he got 2 or 3..I was a little confused.

[quote="mathcrazed"]ya well... i joke actually nathan jokes lol[/quote]

uh really?  he doesnt seem like the humorous type.  (dont take that offensively)

and lol he just bsed your dad?  uhh..probalby not..

[quote="mathcrazed"]no my dad knew he was joking...[/quote]

is taht kid half asain?  squinty eyes..I know hes half jew..but he seems asian. even though his last name is american.

and the year that he won, sergei is actually his half brother!! ( i think )... LoL, cause his dad has two wives, one is divorced now

Oh wow... you people... amuse me... to such an extent that I'm actually going to post on AoPS and address some stuff.

We are half-Taiwanese, half-Russian (in response to the above, Russian Jew). Our last name is American-sounding because our dad changed his last name when coming to America, for the sake of not having a ridiculously hard to spell/pronounce last name.

Sergei is [i]not[/i] our half-brother. That would be [i]far[/i] too simple. He is our half-brother's mother's second-husband's brother's son. I'll leave it to you to figure out how that works.

And no, Nathan and I are not the most serious people out there. :P

I never knew the Benjamins weren't american.. :D

hehe sorry.

[quote="indianamath"]woah...i never knew david was an AoPSer![/quote]
If you define AoPSer to be someone with an AoPS account, then yes.

By any other definition, no. :)

[quote]I never knew the Benjamins weren't american..
hehe sorry.[/quote]
No offense taken. :) Meh, we were born in America if that counts... although I certainly have much less nationalistic pride than most Americans... especially with regards to the current regime.

wow I got a 10 on USAMO

It sucks for Nathan because he's a freshamn and red MOP cutoff was 9 and he got 8


Let [tex]A_0B_0C_0[/tex] be a triangle and P a point. Define a new triangle whose vertices [tex]A_1[/tex] ; [tex]B_1[/tex] ; [tex]C_1[/tex]
as the feet of the perpendiculars from P to [tex]B_0C_0[/tex] ; [tex]C_0A_0[/tex] ; [tex]A_0B_0[/tex] , respectively. Similarly,
define the triangles [tex]A_2B_2C_2[/tex] and [tex]A_3B_3C_3[/tex] . Show that [tex]A_3B_3C_3[/tex] is similar to [tex]A_0B_0C_0[/tex] .

Denote angles $PA_0B_0$, $PA_0C_0$ and so on. Then using fact that $PB_1A_0C1$ (and all remaining) is inscribed, we immediately obtain values for angles $PA_1B_1$ and so on. Repeat the same for point $P$ and triangle $A_1B_1C_1$, and further for $P$ and $A_2B_2C_2$. Finally you will find expressions for angles of $A_3B_3C_3$ via angles of $A_0B_0C_0$.

[Sorry, Myth, for de-facto repeating your post. I didn't see it when I posted the mine...]

Well-known theorem, but I don't find a proof online, so let me post one:

We will use directed angles modulo 180.

Since $\measuredangle A_0B_1P = \measuredangle A_0C_1P = 90^{\circ}$, the points $B_1$ and $C_1$ lie on the circle with diameter $A_0P$. Hence, $\measuredangle PB_1C_1 = \measuredangle PA_0C_1$. In other words, $\measuredangle PB_1C_1 = \measuredangle PA_0B_0$. Similarly, $\measuredangle PC_2A_2 = \measuredangle PB_1C_1$, $\measuredangle PA_3B_3 = \measuredangle PC_2A_2$. Thus, $\measuredangle PA_3B_3 = \measuredangle PA_0B_0$. Similarly, $\measuredangle PB_3A_3 = \measuredangle PB_0A_0$. Hence, the triangle $PA_3B_3$ is directly similar to the triangle $PA_0B_0$. Analogous reasoning shows that the triangle $PB_3C_3$ is directly similar to the triangle $PB_0C_0$ and that the triangle $PC_3A_3$ is directly similar to the triangle $PC_0A_0$. Hence, the triangle $A_3B_3C_3$, which can be subdivided into the triangles $PA_3B_3$, $PB_3C_3$ and $PC_3A_3$, is directly similar to the triangle $A_0B_0C_0$, which can be subdivided into the triangles $PA_0B_0$, $PB_0C_0$ and $PC_0A_0$.

$\blacksquare$

By the way, this generalizes to n-gons; for instance, for quadrilaterals, the quadrilateral $A_4B_4C_4D_4$ is similar to the quadrilateral $A_0B_0C_0D_0$.

  Darij

I have a feel that it is from Kvant...  :?

Darij, it is ok!
Your remark about quadrilaterals was unknown for me. At least I never thought about it.

[quote="Myth"]I have a feel that it is from Kvant...[/quote]

Well, it is [url=http://kvant.mccme.ru/1976/08/gif/76_08-40.gif]Kvant problem 356[/url], but actually it is well-known. It's Theorem 1.92 in chapter 1 9 of

H. S. M. Coxeter, S. L. Greitzer, [i]Geometry Revisited[/i], Toronto - New York 1967.

The generalization to arbitrary n-gons (not just quadrilaterals) is straightforward:

[i]For any n-gon K and any point P in its plane, denote by $T_P\left(K\right)$ the n-gon formed by the orthogonal projections of the point P on the sidelines of the n-gon K. Then, for any n-gon K and any point P in its plane, the n-gon $T_P^n\left(K\right)$ is directly similar to the n-gon K.[/i]

The proof is almost the same as the one for the triangles.

  Darij

[quote="darij grinberg"][quote="Myth"]I have a feel that it is from Kvant...[/quote]

Well, it is [url=http://kvant.mccme.ru/1976/08/gif/76_08-40.gif]Kvant problem 356[/url][/quote]
It means that I saw this problem in another place, since I saw it long time ago and I didn't have access to Kvant problems. Though, it is possible that it was in some article in Kvant in 90th. :?