Two identical cylinders of infinite length, both with diameter $9 \sqrt{2}$ cm, lie on a plane and intersect at right angles. A rectangular prism is to be placed entirely inside the intersection region. What is the maximum volume, in $cm^{3}$, of the rectangular prism?

There was originally a diagram with this question as well. However, it's not hard to visualise it so for the sake of challenge, no diagram.

[hide="Solution"]Since the diameter is $9\sqrt{2}$, The largest value of the diagonal of the prism's side is $9\sqrt{2}$. Therefore, the side length is 9 cm and the volume is 729 $cm^3$[/hide]

That's right.. I imagined a cube as the largest rectangular prism. :thumbup:


AMC
USA(J)MO
USAMO
AMC 10
AMC 12
AIME
AIME II

Vote above.

[edit] Sorry for the misspelling of "the"!

[color=red]Moderators: Could you remove the duplicate answer choices? Thanks![/color]

[quote="blootzboy"]where's the poll...[/quote]

Sorry, technical problems ... should be up now.

Given AMC10 scores are much higher than AMC12 scores and AIME II seems to be harder than AIME I, my prediction is there will be split USAMO indexes for this year.

kfeng40 and Professor Merryfield predicted the [b]USAMO Index based on AMC10[/b] scores will be much higher than [b]USAMO Index based on AMC12[/b].

...which would be completely terrible for me >_< 129 on AMC 10B (bad day >_>) and 8 on AIME II. Hopefully not. My gut feeling, which is probably worth nothing, is that the index for both the first and the second cut will be very similar; although the second cut may be higher by a bit, I don't think it'll be [b]MUCH [/b] higher.

well i really hope the floor is high so the second index will be really low or not used..

Don't worry about the duplicate answer choices. I can't remove them - it's some sort of bug. All the votes will be tabulated correctly.

If there is no split index I think it will be lower than last year's index (204). If there is, then maybe high 200's for aime I.

You meant high 100's, I assume?  Or low 200's?

[quote="funtwo"]my question is: on the aime there was only one place to bubble in 1 amc score. will the other amc score be known and taken into account for usamo selection (in case you dont make it via your bubbled score, but you make it via the other score)? if the amc10 index is really much higher than the amc12 index, this might happen to some people who bubbled their amc10 score[/quote]

The score bubbled is irrelevant. From a post by an AMC officer, if one takes more than one test, then they automatically use the higher one (irrespective of whether it's AMC 10/12) in the qualification. So if you scored higher on AMC 10, you would be considered in the second selection (even if you would already qualify via the first selection).


What are some of the weird stubstitue teacher you've have?

I had one that that I think was a professional basketball player, one who thought that $\cong$ meant not equal to, and one who said that she came from mars and was over 100 years old.

But what I hate about days when you have substitute teachers is that the teachers always leave the stupidest things for us to do, and the substitue teacher just sits at the desk eating his/her donuts and coffee.  And almost all substitute teacher are horrible at math (meaning they can't understand how to solve for x if x+1=2).

One substitute teacher of mine told us to all call 1-800-FOCUSNATION and he told us all to focus. he was really cool and always wore a soccer shirt and green pants.

And this other sub once had us watch a movie and he fell asleep until the bell rang at the end of class which woke him up.

We had one sub that didn't say a word to us the entire 90 minutes.  She just sat there eating McDonalds and didn't even make an action to acknowledge that we existed.

I've had a sub on several occasions who claims that she took Calculus in the sixth grade.  She was also the yo-yo champion of California or something like that.  When she was in her twenties she "wrote e-mail" (like the whole system).  She did all her programming in C, but she seems to have forgotten every single bit (she didn't even remember what a loop was).  But now she's gone onto bigger and better things: substituting for a bunch of kids who really don't care about any of her odd achievements.

Some subs I've had.

One of the subs is a magician. After every class, he performs a magic trick for the students, and it is really cool.

Another one is an odd one. She can't speak well, but she claims she was in a car crash. She also claims that she was struck by lightning into a coma. Over this last summer, she was arrested for marijuana possession.

hmm

one of my sub teachers in elementary was in jail, then went out.

in first grade and third grade there were these subs that were there for almost the whole year. it was really annoying.

On multiple occasions, I had a sub that actually was struck by lightning.  Twice.

on of my other subs, he was in the army. boy was he a grump.

I've had this one sub many many times in many many different classes. And I swear, that guy is a braniac. He's like... 60 years old, but really smart.. and he knows everything... chem, bio, physics, calc, french, lit, history...

o.0 amazing... he's actually a substitute TEACHER...  :)

why do people even [i]care[/i] about what kind of subs we had before?

remember this is F&G

Besides, I care :)

I didnt have any weird subs :(

Y :varepsilon: AH, Y :varepsilon: AH, Y :varepsilon: AH, I KNOW THAT THIS IS f&g  :varepsilon: CTION BUT I'IVI  JUST WON :varepsilon:  :gamma: ING WHY P :varepsilon: OPL :varepsilon:       :gamma:  :varepsilon:    INT :varepsilon:  :gamma:  :varepsilon: ST :varepsilon:

99.999999999999999% of the time, the substitute teacher SUX!!!!!

and 10.0101010101% of the time, the substitute teacher is actually a good teacher (don't have to be likable though)

*sigh* why can't the teachers just stay healthy and plan their schedules so that nothing interfers with their teaching?!?!

Once I've had a sub who was a wrestling coach.

[quote="uglyman621"]*sigh* why can't the teachers just stay healthy and plan their schedules so that nothing interfers with their teaching?!?![/quote]

The same reason students don't do the same: it's unavoidable, or they have more important things to do.  Do you seriously think that teachers WANT to get sick and worry about a substitute potentially screw up their lesson plans?  Give the teachers a break, they're allowed to have lives, too.  My Spanish teacher, for example, can't come to school next week because she's getting married; sounds like a valid excuse to me, don't you think?

I've had several strange subs including one that was in my english class last year who couldn't seem to speak coherent english.

My mom who is a retired army person is also a sub and I've heard shes pretty strange though I have not had her yet as a sub in one of my classes.

good point uglyman.  actually i think that 80% is bad teacher and 20% is a fun and good teacher.

Once, last year, for French class, we had this sub who sang and danced in the middle of class for no reason.  And there was another time, we had this strict old lady who picked on everyone's pronounciation so badly that it took maybe 5 minutes for one guy to get through one sentence.  ("No no... it's a-VREEL.  Feel that long 'VREEL'!")

oh yeah another time for me is that there was a sub and he made us do all these weird things.  it was weird.

Once I had a sub who let us play computer games since we finished our work (she gave out most of the answers so..)

[quote="themonster"]I've had a sub on several occasions who claims that she [b]took Calculus in the sixth grade.[/b]  She was also the yo-yo champion of California or something like that.  When she was in her twenties she "wrote e-mail" (like the whole system).  She did all her programming in C, but she seems to have forgotten every single bit (she didn't even remember what a loop was).  But now she's gone onto bigger and better things: substituting for a bunch of kids who really don't care about any of her odd achievements.[/quote]

That's not such a fantastic achievement.......I took calculus in 5th grade.

We had a fat guy. I mean [i]really[/i] fat. One of those "super heavy McDonald's users" they talked about on "Supersize Me". He was also abnormally tall. Combined, he was huge. His bald head almost grazes the ceiling.

His face was sort of slack and he had odd glasses. One lens appeared to be thicker than the other by a significant amount. When we filtered into the class, instructions were already written on the board so he didn't say anything. Instead, he made odd motions on his stomach, where his hand, palm facing the stomach, went up and down in a rolling motion. From his view, it would look like ocean waves. He also moved his lips, mouthing words to himself, and occasionally smoothed back invisible hair.

At one point, he got up and started touching my shoulders. It was the most uncomfortable, awkward situation ever.

Heh.
Today in French, we had a sub with a heavy Russian accent.
One of the kids did a great impression of him.
And, there was this guy, who was forced to wear makeup. Like, this girl actually made him FAB-U-LOOZ (in their words).

Another time, we had a sub who literally spat in your face from across the room. I'm glad I wore a hoodie that day.

i have a substitute who was a ventriloquist. He was the best sub i have ever had as he ended class early to put on his show

I had a sub that said to just run a couple (Yes, no specify, just "a couple") laps around the field, then to go have lunch. The whole class ran 1 lap and went to lunch. We had a 35 minute extra lunch!


Let $ a$ , $ b$ , $ c$ be real and positive numbers. Prove that :

$ (\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}1)^{2}\geq 3(a\plus{}b\plus{}c)\plus{}2(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})\plus{}1$

[quote="cnyd"]I thinks sai00 did mistake,Because;
for $ a \equal{} b \equal{} c \equal{} 2$  $ \implies$  $ (\frac {5}{2})^{2} \equal{} \frac {25}{4}$
but $ 3(a \plus{} b \plus{} c) \equal{} 24$ $ ?$[/quote]

Yes...my mistake.... $ abc\equal{}1$

$ (\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}1)^2\equal{}(ab\plus{}bc\plus{}ca\plus{}1)^2\equal{}(ab\plus{}bc\plus{}ca)^2\plus{}2(ab\plus{}bc\plus{}ca)\plus{}1\ge3(a\plus{}b\plus{}c)\plus{}2(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})\plus{}1$, as $ abc\equal{}1$, $ (ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c)\equal{}3(a\plus{}b\plus{}c)$, and $ ab\plus{}bc\plus{}ca \ge a\sqrt{bc}\plus{}b\sqrt{ac}\plus{}c\sqrt{ba}\equal{}\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$, by Muirhead or $ x^2\plus{}y^2\plus{}z^2 \ge xy\plus{}yz\plus{}zx$.


I tried to prove that a^(m/n) = nth root of (a^m).  See if I did it right.

Yep, that's correct  :) .

I can't find any error with it.

I am not sure if this is very important, but i don't see any definition of what 'h' is. Usually you must define every single variable or special notation that you incorporate into your proof.

If this is an [i]Algebra I[/i]-type of proof, then I would say the material is OK. A more rigorous and fundamental proof of it requires much material that is beyond the scope of this forum.

Now, apart from that, there are exceptions to your proof. To make this simple, you should assume that $ a>0$.

[quote="varunrocks"]I would say that there is no proof because when we use a square root we imply fractional exponents.[/quote] :huh: Then your implication needs a proof. And rOadToAIME offers you that proof... Just because a 'fact' seems obvious to you, that doesn't mean it wouldn't require a proof. As simple as 'all equilateral triangles are equiangular', it needs a proof too. We just accept some commonly used facts and that's another story.


group theory
number theory
superior algebra
superior algebra unsolved

let G a group of order n (i.e $ |G| \equal{} n$). Show:

$ n \equal{} \Sigma_{m|n} \varphi(m)$

It's not very sensefull to assume we are given a group and then not using that group ;)
And for the problem, there are many ways. E.g. you could check it for prime powers and use multiplicativity. Or you could count elements of order $ m$ in $ \mathbb Z / n \mathbb Z$ (additive group).

The idea is to use the group properties in order to probe the theorem. Remember that group theory and number theory are very,very closed cousing. :lol:

Advise me that I use this theorem:

let G a group de order n. if m|n then there is $ \varphi(m)$ elelments of order m

That theorem is simply wrong, you need $ G$ to be the [b]cyclic[/b] group of order $ n$ for it to be true ;)
But then thats what I meant with my last comment above:

Every element has one specific order, and this order always divides $ n$. Thus if $ \psi(m)$ is the number of elements of order $ m$, then $ \sum_{m|n} \psi(m) \equal{}|G| \equal{}n$ as every element is counted exactly once. Untill here, we didn't use any property in $ G$ except that it's finite. But if $ G$ is cyclic, then $ \psi(m) \equal{} \varphi(m)$ by your theorem and the result follows.

can some prove this result please ?
''let G a group de order n. if m|n then there is$ \phi(m)$  elelments of order m

That's obviously false. Maybe you mean "cyclic"?
You might as well be working with $ \mathbb{Z}/n\mathbb{Z}$, and there this is very close to the definition of $ \phi$.


OMG! I just got an email iviting me to the national camp! Yay! Who else is coming??

I guess I got invited because of Galois perfect....  :wink:

 :lol:

Where is it this year? Same place as last yr?

[quote="carpo"][quote="bryant124"]lol u just got an email??
i thought i got my forms 2 or 3 months ago...
im going...
i dont kno the weather there tho...any advice what kind of clothes i should bring?????? :ninja:  :ninja:[/quote]
Check the weather online before you go, but when I went it was really really hot (speaking from a Vancouverite's perspective). Also you should bring mosquito repellent in case you play outside.[/quote]

but its sooo hot in vancouver now... :rotfl: ...im gonna die of dehydration...just got back from central library and theres not even any wind...

and MOSQUITOES...arrghh...hate those things...sunscreen...sunscreen...i have a tan already for playing tennis for an hour yesterday...

wait...we're talking about montreal right?

and yea...see u guys in calgary...(if i survive montreal's weather...) :|

Well... I was talking about the one in Montreal...   :|

Yeah, it was really hot when I went there (it was also in Montreal). And very little rain.

thats nice...its sooo hot these days...its not even june and its burning in vancouver...

i once did a math contest on a really hot day and my pen kept slipping out of my hands...lol :rotfl:

[quote="bryant124"]but is cool indoors?
if its only hot outside and pretty much cool inside or if they have air conditioning then i can stand it...although i would love to play lots of sports outside... :([/quote]

The classroom was air conditioned but not the rooms. I had trouble with that.

Also you should try to play some sports to relax.

thats ok...we're probably only gonna be in our rooms at night...and we can open the windows to cool down...and i dont mind if its hot outside... :P  :D

It's HOT.......yes, I still remember that I can't sleep very well at night, so you really wish the weather is nice on that week, OR, bring a fan!(if you are driving to Montreal). I remember Kyle brought one and it helped sooo much when we were playing cards in our room! but then the main hallway was not that hot... I don't know why the rooms were so hot.....

Nevertheless, it's an extremely FUN experience, so ENJOY it! (and it's not as stressful as in winter camp :lol: )

Invitations are sent in several rounds, based on performance on various contests throughout the year.  That's why you all got them at different times.  I was an instructor last year, and I'll be there again this year.  I remember it being quite cool last year...but then again I was coming straight from New Jersey, where it was sweltering.  There's air conditioning in some places, but not in others - I don't remember exactly where.  Bring a variety of summer and spring/fall clothing, and also be prepared for rain - we will probably have an excursion (pre-scheduled, rain or shine) that involves walking outside.  Yes, bring mosquito repellent.  And sports are mandatory for everyone - you MUST be seen physically exercising at least once in the week, and preferably every day.  It's unhealthy to do math all day long (that's pretty hypocritical of me, actually...I'm doing research with a math prof right now).  Anyways, you guys will have a lot of fun, and hopefully learn a lot of math too! And yeah, it's certainly not as stressful as Winter Camp.  Can't wait to meet you all!

I'm sure there are badminton courts...that school has everything.  I don't recall anyone visiting them last year though.  The sports last year were soccer, basketball, and frisbee, I think.  The atmosphere is pretty laid-back, though, so if there's enough interest, I'm sure badminton could be incorporated.

COOL...i luv sports...im just not sure if im gonna be able to last long in the hot weather...im really short of energy lately...haha :rotfl:

haha...WindingFunction...i luv ur signature,,,Two lemmas make a dilemma...that made me laugh really hard...

[quote="bryant124"]haha...WindingFunction...i luv ur signature,,,Two lemmas make a dilemma...that made me laugh really hard...[/quote]

Sadly, I cannot claim credit for it, but I like the quote very much myself.   :D

[quote="Foundation"]It'll be nice to meet many new people. Will there be badminton courts there? I'm debating whether I should bring my badminton raquet or not. I was wondering what kind of sports were avaliable.[/quote]
I think you can loan a raquet at the court. i can't bring too much with me so i won't bring a raquet

o...is there a weights training room there?

I haven't seen a weight room first-hand, and I don't think past campers have used it, but, again, I wouldn't be surprised if there was one.  Keep in mind, though, that this is a math camp! We only have an hour or two a day designated for sports, and if you have all these semi-exotic requests, the camp organizer might become overworked.  I'll almost certainly be lucky enough not to have to deal with this at all, so I won't get upset at anyone for asking about it - I'm just concerned about the sanity of poor Daniel Gatien.  :lol:

Yeah, Mr. Gatien will be quite busy answering all our emails about transportation already. We don't want to give him too much trouble. I'm going to just bring a racquet or two and birdies for fun if I have room and play soccer and basketball if they're available. Thanks for your replies WindingFunction.

I've figured out who everyone is except bryant124.  So 'fess up bryant...who are you?

That makes perfect sense.  I wonder why I couldn't guess.  Must have been too tired out from you all.  :wink:


Since this problem is too easy for the college playground and moderately hard for most people in high school (I think), I will allow pre-olympiad level people to solve this.

What is the area bounded by the curves $y^2=12x$ and $y^2=24x-36$?

Masoud Zargar

i think you should put it in" Calculus computations and tutorials" !!!

Okay, sorry.  Would I mod please move it there?  Thank you.

Masoud Zargar

I suppose no one's solving this problem...well here's teh answer to the ones who seek...

We say
$12x=24x-36\\ x=3\\ y=(+-)\sqrt{12x}=(+-) 6$

Since we know that we have symmetry about the x axis, we solve for the first quadrant area, and later double it.  We dra a horizontal strip of width $dy$ and length $x_2-x_1$. The area $dA=(x_2-x_1)dy$, where $x_1$ and $x_2$ are found by solving each equation for x.
$x_1=\dfrac{y^2}{12}$ and $x_2=\dfrac{y^2+36}{24}$.
By integration, we get
\[ A=\int_{0}^{6}(x_2-x_1)dy\\ =\int_{0}^{6}\left (\dfrac{y^2}{24}+\dfrac{36}{24}-\dfrac{y^2}{12}\right) dy\\ =\int_{0}^{6}\left (\dfrac{3}{2}-\dfrac{y^2}{24}\right) dy\\ =\dfrac{3y}{2}-\dfrac{y^3}{72}|_{0}^{6}\\ =6 \]

Since we dealt with symmetrical figues, we have to do $6\cdot 2$.
$\therefore A=12\text{ units}^2$

Masoud Zargar

You should have waited till some mod moved this to the appropriate forum!!! You had just posted the problem!!!!!

Oh!  Well, I thought it woudl be better to answer this one and then post another one.  :oops:

Masoud Zargar

Learn to post in the right forum.  In particular, this means you should take the time to read the descriptions of the various forums before you post in them.

Sure, I'll do that from now on. It's just that I didn't know there was a calculus section in the high school section.

[quote="boxedexe"]It's just that I didn't know there was a calculus section in the high school section.[/quote]

There isn't.

What about the Calculus Computations and Tutorial?  Rushil said that that woudl be a suitable place.

EDIT: I see, so it leads me to the college playground forums.  If so, why is there a link from the high school forums to the college playground?

Also, you shouldn't post solutions to your own problems unless it's been a long time and/or lots of people have failed and want a solution.  No one even tried, and it looks like you only gave it TWO hours before posting the solution.

[quote="Altheman"]this is just really basic calculus...if you know calc, it is easy, if not, you won't solve it[/quote]
I agree


function
advanced fields
advanced fields unsolved

Find the cardinality of $ X\equal{}\{A \subseteq \aleph_{100}|\;\; |A|\equal{}\aleph_0\}$.

The cardinality is $ \aleph_{100}^{\aleph_0}$.
Let $ S$ be the set of the subsets of $ \omega_{0} \times \omega_{100}$. $ |X| \equal{} |S|$ since $ \aleph_{100} \times \aleph_0 \equal{} \aleph_{100}$. The set of functions $ f: \;\; \aleph_{0} \rightarrow \aleph_{100}$ is clearly a subset of S and $ \aleph_{100}^{\aleph_0}\le |X|$ is proved.
For the other side of the disequation it is enough to show that if $ x \in X$ it exists a function $ f: \;\; \aleph_{0} \rightarrow \aleph_{100}$ such that x is the range of f.
Cantor-Bernstein implies the assumption.


vector
induction
geometry proposed
geometry

Let $ABC$ be a triangle with the altitudes $AA'=h_{A},BB'=h_{B},CC'=h_{C}$ and $A_{1}A_{2}\cdots A_{n}$, $n\geq 3$ a convex polygon. Denote $\overrightarrow{v_{k}}$ the vector of lenght $A_{k}A_{k+1}$ perpendicular on $A_{k}A_{k+1}$, orientated to the exterior of the polygon.
 a) $\frac{a}{h_{a}}\cdot\overrightarrow{AA'}+\frac{b}{h_{b}}\cdot\overrightarrow{BB'}+\frac{c}{h_{c}}\cdot\overrightarrow{CC'}=\overrightarrow{0}$;

 b) $\overrightarrow{v_{1}}+\overrightarrow{v_{2}}+\,.\,.\,.\,\+\overrightarrow{v_{n}}=\overrightarrow{0}.$

a)Put $\overrightarrow{x}=\tex{LHS}$, then easy see that $\overrightarrow{x}\cdot \overrightarrow{AB}=\overrightarrow{x}\cdot \overrightarrow{AC}=0$. Therefore $\overrightarrow{x}=\overrightarrow{0}$.
b)From a) and induction on $n$.


geometry
circumcircle
trigonometry
geometric transformation
rotation
complex numbers
geometry proposed

Regular triangles $ABC$ , $CDE$ , $EHK$ (vertices are given counterclockwise) with pairwise common verices C and E are located in a plane so that $\vec{AD}=\vec{DK}$ . Prove that the triangle $BHD$ is also regular .

P S (I would like to see a solution with complex numbers)

[quote="silouan"]P S (I would like to see a solution with complex numbers)[/quote]

Can anyone find a solution like this ? :maybe:

Show very easy and very quickly with the complex numbers !

[hide="Proof 1 (synthetic)."][color=darkblue]
Denote the intersections $X_{1}\in AK\cap BE$ and $X_{2}\in AK\cap CH\ .$
$BCE\equiv ACD\ (C\--\ s.a.s.)$ $\Longrightarrow$ $\angle EBC\equiv\angle DAC$, $\angle CEB\equiv\angle CDA$, $BE=AD$ $\Longrightarrow$ the quadr. $ABCX_{1}$, $ECDX_{1}$ are cyclically.
$EHC\equiv EKD\ (E\--\ s.a.s.)$ $\Longrightarrow$ $\angle EHC\equiv\angle EKD$, $\angle ECH\equiv\angle EDK$, $HC=KD$ $\Longrightarrow$ the quadr. $EKHX_{2}$, $EDCX_{2}$ are cyclically.
The points $D$, $X_{1}$, $X_{2}$ belong to intersection between the line $AK$ and the circumcircle of the triangle $ECD$. Thus, $X_{1}\equiv X_{2}\equiv X$ $\Longrightarrow$ $BE=AD=HC=DK$, the quadr. $ABCX$, $ECDX$, $EKHX$, $EDCX$ are concyclically and $m(\angle AXB)=$ $m(\angle BXC)=$ $m(\angle CXD)=$ $m(\angle DXE)=60^{\circ}\ .$
The quadr. $EKHX$ is cyclically $\Longrightarrow$ $\angle XEH\equiv \angle XKH$, i.e. $\angle BEH\equiv\angle DKH$ $\Longrightarrow$ $BEH\equiv DKH$ $\Longrightarrow$ $BH=DH\ .$
The quadr. $EDCX$ is cyclically $\Longrightarrow$ $\angle XED\equiv\angle HCD$, i.e. $\angle BED\equiv\angle HCD$ $\Longrightarrow$ $BED\equiv HCD$ $\Longrightarrow$ $BD=HD\ .$
In conclusion, $BD=DH=HB$, i.e. the triangle $BDH$ is equilateral.[/color]

[/hide][hide="Proof 2 (with the complex numbers)."][color=darkblue]
Denote $X(x)$- the point $X$ with the affix $x$ and the complex number $\omega=\cos\frac{\pi}{3}+i\cdot\sin\frac{\pi}{3}$ which characterizes the rotation counterclockwise with $60^{\circ}$ degrees. Know $\boxed{\ \omega^{2}=\omega-1\ }$ and $\omega^{3}=-1$. Fasten the points $\underline{A(a)}$, $\underline{C(c)}$ and $\underline{D(d)}$. Therefore : $k=2d-a$ , $\boxed{\ b=c+(a-c)\cdot\omega\ }$ , $e=c+(d-c)\cdot\omega$ and
$h=k+(e-k)\cdot\omega=$ $k\cdot(1-\omega )+e\cdot\omega =$ $(2d-a)\cdot(1-\omega )+[c+(d-c)\cdot\omega ]\cdot\omega =$ $(2d-a)\cdot (1-\omega )+c\cdot\omega+(d-c)\cdot (\omega-1)=$
$(2d-a-d+c)+(a-2d+c+d-c)\cdot\omega$ $\Longrightarrow$ $\boxed{\ h=(d+c-a)+(a-d)\cdot\omega\ }\ .$
The triangle $BDH$ is equilateral $\Longleftrightarrow$ $h-d=(b-d)\cdot\omega$ $\Longleftrightarrow$ $(c-a)+(a-d)\cdot\omega=[c+(a-c)\cdot\omega-d]\cdot \omega$ $\Longleftrightarrow$
$(c-a)+(a-d)\cdot\omega=(c-d)\cdot\omega+(a-c)\cdot (\omega-1)$ $\Longleftrightarrow$ $(c-a)+(a-d)\cdot\omega =(c-a)+(c-d+a-c)\cdot \omega$, what is truly.

[b]Remark.[/b] The proposed problem is equivalently with the following implication (vertices are given or all counterclockwise or all clockwise) : \[\boxed{\ \begin{array}{c}\omega^{2}-\omega+1=0\\\ c=b\cdot\omega+a\cdot (1-\omega )\\\ e=d\cdot\omega+c\cdot (1-\omega )\\\ k=h\cdot\omega+e\cdot (1-\omega )\\\ 2\cdot d=a+k\end{array}\Longrightarrow h=b\cdot\omega+d\cdot (1-\omega )\ }\ .\] [/color][/hide]

Thank you vey much mr Nicula . :)
Also a nice solution is the following
Rotation by $60$  around $C$ takes triangle CAD into triangle CBE . and rotation by $60$ around $H$ takes triangle HBE into HDK


Find the greatest integer $ n \in \mathbb{N}^*$ for wich the inequality is true:

\[ \frac{1}{\sqrt[3]{4}\plus{}\sqrt[3]{2} \plus{} 1} \plus{} \frac{1}{\sqrt[3]{9} \plus{} \sqrt[3]{6} \plus{} \sqrt[3]{4}}\plus{} \dots \plus{} \frac{1}{\sqrt[3]{(n\plus{}1)^2}\plus{}\sqrt[3]{n(n\plus{}1)}\plus{} \sqrt[3]{n^2}}< 1 \plus{} \sqrt[3]{6\sqrt{3}\minus{}10} .\]

[hide="Hints"]1. use $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$
2. Prove that $ \frac {1}{(n \plus{} 1)^{\frac 23} \plus{} (n(n \plus{} 1))^{\frac 13} \plus{} n^{\frac 23}} \equal{} (n \plus{} 1)^{\frac 13} \minus{} n^{\frac 13}$[/hide]


When a rectangular wooden board is cut, the saw blade removes one-eighth of an inch of wood which becomes unused sawdust. A 5''
by $ x$'' board is cut 4 times perpendicular to the length and the result
is a set of boards that are each 5'' by 6.5''. What is the value of $ x$?

We do \[ \frac{1}{8}\cdot4\plus{}6.5\equal{}\boxed{7}\]

Actually, no.

Since it's cut 4 times, there are 5 boards. Then there are $ 4\times\frac{1}{8}\equal{}\frac{1}{2}$. The answer is $ 5\times6.5\plus{}0.5\equal{}\boxed{33}$.

[quote=math.is.gud]ooooooohhhhh[/quote]

[quote=math.is.gud]i get it[/quote]

A) Don't post twice in a row

B) Try not to post or bump threads unless if you have something meaningful to add to the question.


geometry
3D geometry
superior algebra
superior algebra unsolved

How many figures can be obtained by intersecting the infinite dimensional cube
$|x_k|\leq 1$  $k=1,2,....,$ with a two-dimensional plane ?

I don't know what is meant by figures here, but I would say:
all (convex) n-gons and a circle/ellipse

Is it correct¿


algebra
polynomial
geometry
topology
superior algebra
superior algebra solved

Give an example of an irreducible polynomial $f \in \mathbb{R}[x,y]$, whose zero set $Z(f)$ in $A_{\mathbb{R}}^{2}$ is not irreducible.

As far as I know this is false. On the other hand it's listed in Hartshorne (page 8), so it  is probably correct :)

I was really convinced than in an U.F.D., such as $\mathbb{R}[x,y]$ irreducibles were prime. Then some other theorem of affine algebraic sets says: "an algebraic set is irreducible iff its ideal is a prime ideal".

So, where did I go wrong? :)

What about $f\left(x,y\right)=\left(x^{2}-1\right)^{2}+y^{2}$ ?

The zero set of $f$ over $\mathbb{R}$ is the intersection of the line $y=0$ with the circle $x^{2}-1=0$, and this intersection consists of the two points $\left(-1,0\right)$ and $\left(1,0\right)$, thus not being an irreducible algebraic set since each of these two points is an algebraic set itself. On the other hand, the polynomial $f$ is irreducible, as can be easily seen: Assume that $f$ is reducible. Then, $f=f_{1}f_{2}$ with $f_{1}$ and $f_{2}$ being polynomials from $\mathbb{R}\left[x,y\right]$. Denote by $\deg_{y}g$ the degree of any polynomial $g$ in the variable $y$. If $\deg_{y}f_{1}=0$, then $f_{1}$ is a polynomial of the variable $x$ only; thus, it has a complex root $x_{1}$ such that $f_{1}\left(x_{1},y\right)=0$ for every $y\in\mathbb{C}$; since $f=f_{1}f_{2}$, this yields that $f\left(x_{1},y\right)=f_{1}\left(x_{1},y\right)f_{2}\left(x_{1},y\right)=0f_{2}\left(x_{1},y\right)=0$ for every $y\in\mathbb{C}$; but this is impossible, since $f\left(x_{1},y\right)=\left(x_{1}^{2}-1\right)^{2}+y^{2}$ cannot have the same value for all $y$. Hence, $\deg_{y}f_{1}$ cannot be $0$. Thus, $\deg_{y}f_{1}\geq 1$. Similarly, $\deg_{y}f_{2}\geq 1$. Since $\deg_{y}f_{1}+\deg_{y}f_{2}=\deg_{y}\left(f_{1}f_{2}\right)=\deg_{y}f=2$, this yields that $\deg_{y}f_{1}=\deg_{y}f_{2}=1$. Hence, for every constant $x_{0}\in\mathbb{R}$, the factorization $f=f_{1}f_{2}$ induces a factorization of the polynomial $f\left(x_{0},y\right)\in\mathbb{R}\left[y\right]$. But setting $x_{0}=0$, we see that the polynomial $f\left(x_{0},y\right)=\left(0^{2}-1\right)^{2}+y^{2}=1+y^{2}$ is irreducible over $\mathbb{R}$, so we get a contradiction, and it is proven that $f$ is irreducible.

[quote="Jacobson"]"an algebraic set is irreducible iff its ideal is a prime ideal".[/quote]

This is only true over algebraically closed fields, and $\mathbb{R}$ is not one of these.

  darij

The real problem lies in the difference between $\mathbb A^{2}_\mathbb R$ (the scheme) and $\mathbb A^{2}(\mathbb R)=\mathbb R^{2}$ (the set of $\mathbb R$-valued points). Usually, algebraic geometry should be performed on the former, whereas the latter should be considered just a set (which may carry some topology, not necessarily the Zariski one). Correspondingly, $V(f)\subset\mathbb A^{2}_\mathbb R$ [i]is[/i] irreducible. Hartshorne's book is somewhat old-fashioned, modern textbooks like Liu give appropriate emphasis to the point functor. owk


a) Do there exist natural numbers m, n such that 7/17= 1/m + 1/n?
b) Let p be a prime number. For which natural number k do there exist natural numbers m, n such that k/p=1/m + 1/n.

2. Yes. For example $ n \equal{} mp,p \equal{} km \minus{} 1$ is prime. In these case must be $ k|p \plus{} 1$ and solution $ m \equal{} \frac {p \plus{} 1}{k},n \equal{} mp$.
1. Had not solution, because $ 7\not |18$.

$ \frac{1}{m}\plus{}\frac{1}{n}\equal{}\frac{k}{p}$
$ (m\plus{}n)p\equal{}kmn$
So we can assume $ m\equal{}pa$.
$ pa\plus{}n\equal{}kan$
$ pa\equal{}(ka\minus{}1)n$ (1)

Case 1: $ n\equal{}pb$
(1) implies $ \frac{1}{a}\plus{}\frac{1}{b}\equal{}k$
Hence $ k\equal{}1,a\equal{}b\equal{}2,p$ is arbitrary

Case 2: $ p \nmid n$
Therefore $ ka\minus{}1\equal{}pb$
$ a\equal{}bn$
$ m\equal{}pbn$
Thus (1) implies
$ pb\plus{}1\equal{}kbn$
$ b(kn\minus{}p)\equal{}1$
Hence $ b\equal{}kn\minus{}p\equal{}1$
So: we have $ m\equal{}pn$ and $ kn\equal{}p\plus{}1$.
That means $ k$ is a divisior of $ p\plus{}1$, $ n\equal{}\frac{p\plus{}1}{k}$, $ m\equal{}pn$


Hi,

Where can I find sprint round papers from previous years? Pl let me know!

Thanks!

The 2009 sprints can be found on the mathcounts website. You will have to trade for the others.

I dont have anything to trade. this is my 1st time participating in Mathcounts. Our school sprint round is next week.

Anyone have the previous years papers that they can give without a trade?

I can e-mail or PM tests to you.

How about a 2008 Chapter Sprint?

When I was practicing for MathCounts last year, I also did a lot of AMC8s and a few AMC10s (these can be found under resources at mathlinks)
The problems are very similar to MathCounts problems, so you should try those if you are looking for something to practice


For natural number $ n$, define $ a_n\equal{}2^2\plus{}4^2\plus{}6^2\plus{}\cdots \plus{}(2n)^2,\ b_n\equal{}1^2\plus{}3^2\plus{}5^2\plus{}\cdots \plus{}(2n\minus{}1)^2$.

Find $ \lim_{n\to\infty} \frac{\sqrt{a_n}\minus{}\sqrt{b_n}}{\sqrt{n}}$.


These seem to fit the basic section fairly well.

1) A straight line is drawn through a circle of radius 5 such that it cuts the circle into two pieces. To one side of the line, the length of arc is 5 * pi. What is the area of the portion of the circle on the other side of the line?

2) A cylindrical water pitcher is 13 inches tall and has a radius of 5 inches. A solid cylinder, half as tall as the pitcher and with a radius of 2.5 inches is placed inside the pitcher. By what fraction is the amount of water the pitcher can hold reduced?

3) the price of ground coffee beans is d dollars for 8 ounces and each ounce makes c cups of brewed coffee. In terms of c and d, what is the dollar cost of the ground coffee beans required to make 1 cup of brewed coffee?

Since these are SAT difficulty problems, they were fairly easy to solve.
These are as hard as SAT math problems usually gets.

[hide="1"]
The circumference of the circle is $10\pi$
If the arc is $5\pi$, then it is half of the circumference, and thus the line that makes the arc is a diameter of the circle.
Thus, the other part of the circle is half the area, or $12.5\pi$ units squared.[/hide]

Sure:
[hide="2"]
Volume of Pitcher:
$13 \cdot 5^{2}\pi$
$325\pi$
Volume of Obstruction:
$6.5 \cdot 2.5^{2}\pi$
$40.625\pi$

Initial volume - Obstruction Volume = Diminished Volume
$284.375\pi$

Ratio of Diminished V to Original V:
$\frac{284.375\pi}{325\pi}$
$0.875$
This is equivalent to $\boxed{\frac{7}{8}}$

[/hide]
[hide="3"]
If you make one cup of coffee, you need one ounce of beans.
If you pay a full dollar for eight ounces of beans, you only need to pay 1/8 of a dollar for one ounce of beans.
If c = 1, d = 1/8.
The dollar cost is 1/8 of a dollar, or
$\boxed{12.5}$ cents.
[/hide]


search
number theory proposed
number theory

hallo,
Let $p$ be a prime, greater that 5 and $A=\{p-n^{2}: n \in N$,$n^{2}< p \}$. Prove that ther exists $x,y \in A$ s.t. $x \not= 1$ and $x|y$.  :)

Posted at least 3 times before, I would apreciate it if someone searchs it (don't have much time at the moment).
Some ISL or ILL I think, but not sure.


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

Let $ A$ be a ring. Show that the following are equivalent:
i) $ A$ is Jacobson (i.e. $ J(A/p)\equal{}0$ for all $ p\in Spec(A)$, where $ J$ denotes Jacobson ideal).
ii) every finitely generated $ A$-algebra $ B$ which is a field is finite over $ A$.

remark that (i) is equivalent to $ rad(\mathfrak{a})\equal{}jac(\mathfrak{a})$ for every ideal $ \mathfrak{a}$, and that both properties are stable under taking quotients. thus for (ii) => (i) it's enough to prove that $ rad(0)\equal{}jac(0)$ (well, $ rad(A)$ and $ jac(A)$ are more common, which is a bit inconsequent).

take $ f \notin rad(0)$. then $ A_f \neq 0$, so that we may choose some maximal ideal $ \mathfrak{m}$ of $ A_f$. now $ A \to A_f \to A_f / \mathfrak{m}$ is of finite type, hence finite by (ii). if $ \mathfrak{n}$ is the kernel of $ A \to A_f / \mathfrak{m}$, we have a finite embedding $ A/\mathfrak{n} \to A_f / \mathfrak{m}$ of an integral domain into a field. thus $ A/\mathfrak{n}$ is a field, i.e. $ \mathfrak{n}$ is a maximal ideal. since $ A \to A_f / \mathfrak{m}$ maps $ f$ to a unit, we have $ f \notin \mathfrak{n}$. thus $ f \notin jac(0)$.

for (i) => (ii) we probably have to generalize noether's normalization lemma?

Proof of (i) => (ii): Let $ A\rightarrow B$ be a finitely generated $ B$-algebra where $ B$ is a field. Wlog assume that $ A$ sits in $ B$, hence $ A$ is an integral domain. Denoting the quotient field of $ A$ by $ Q(A)$, the morphism $ Spec(B)\rightarrow Spec(Q(A))$ is of finite type, hence splits through $ \mathbb A^n_{Q(A)}$ for some natural $ n$ and such that $ F: Spec(B)\rightarrow\mathbb A^n_{Q(A)}$ is integral, by Noether normalization theorem. Since $ F$ is of finite type, it comes from an integral morphism $ F': Spec(B)\rightarrow \mathbb A^n_{A_f}$ for some non-zero $ f\in A$. Hence $ \mathbb A^n_{A_f}$ is a field or, in other words, $ n\equal{}0$ and $ A_f$ is a field. Thus the standard affine subset $ D_f$ of $ Spec(A)$ contains only one point, say $ x$, which is closed in $ D_f$. Since for Jacobson rings it can be tested if a point is closed can be tested with respect to every open neighborhood of $ x$,  $ x$ is closed in $ Spec(A)$. This implies that the dense set $ D_f$ is closed, hence equal to $ Spec(A)$, i.e. $ f$ is a unit in $ A$.
Altogether $ Spec(B)\rightarrow Spec(A)$ is integral and of finite type implying it is a finite morphism.


[quote="Interval"]Burger Queen offers 4 types of burgers, 5 types of beverages, and 3 types of desserts.  If a meal consists of 1 burger, one beverage and one dessert, how many possible meals can be chosen?[/quote]

lol...burger queen

[hide]there are 4 ways to choose a burger, 5 ways to choose a drink and 3 ways to choose a dessert so its $4\cdot 5\cdot 3=60$[/hide]

[quote="Interval"]Burger Queen offers 4 types of burgers, 5 types of beverages, and 3 types of desserts.  If a meal consists of 1 burger, one beverage and one dessert, how many possible meals can be chosen?[/quote]
[hide]$3\cdot4\cdot5=\boxed{60}$[/hide]

[hide="same solution but more detailed"]
First you pick 1 burger, then a beverage, then a dessert. There are 4 ways to pick 1 burger. Then when you go get a beverage, there are 5 possibilities. There are 3 ways to get desserts.
Because there are 4 possibilities, then 5, then 3 for each, then you multiply them. $4*5*3=\boxed{60}$[/hide]

Here's a general method for solving these kinds of problems.

Say you've got $n$ different items in the quantities of $a_{1}, a_{2}, \ldots a_{n}$, and to form a combination of these items (a "meal," in this instance) you need $b_{1}$ from $a_{1}$, $b_{2}$ from $a_{2}$, and so on.

Then, the number of combinations is equal to $\prod_{k=1}^{n}\binom{a_{k}}{b_{k}}= \binom{a_{1}}{b_{1}}\cdot \binom{a_{2}}{b_{2}}\cdots \binom{a_{n}}{b_{n}}$, where $\binom{X}{Y}$ is "X choose Y" and $\binom{X}{Y}=\frac{X!}{Y! \cdot (X-Y)!}$.

This is finding the number of ways to pick the items from one group, multiplied by the number of ways to pick the items from the second group and so on, so that all possible combinations are accounted for.

I thank you very much.  All replies are great math notes.


trigonometry
algebra
difference of squares
special factorizations

Problem: solve for x:

$(5+2\sqrt{6})^{\sin{x}}+(5-2\sqrt{6})^{\sin{x}}=2\sqrt{3}$

I have made a little progress:
[hide]
That can be reduced to

$(\sqrt{2}+\sqrt{3})^{2\sin{x}}+(\sqrt{2}-\sqrt{3})^{2\sin{x}}=2\sqrt{3}$

But while I know difference of squares, I cannot figure how to break down the general form

$(a+b)^{2c}+(a-b)^{2c}$[/hide]

Any ideas?

Find the positive square root of 5+ 2 sqrt(6)
This equals sqrt(3) + sqrt(2)      1)
Similarly find that of 5 - 2 sqrt(6)
This equals sqrt(3) - sqrt(2)       2)
 Add 1) and 2) to get 2 sqrt(3)

Then sin(x) = 1/2

and this is easy to solve for x

But you still have that $^{\sin{x}}$ exponent to deal with, so you can't just add them... right?

Replace sin(x) by u

Then we are solving  1)

 (5 + 2 sqrt6)^u + (5-2 sqrt6)^u = 2sqrt3

for u

I say that u = 1/2

  5 + 2 sqrt6= (sqrt3 +sqrt2)^2

 So (5 + 2 sqrt6)^(1/2) = sqrt3 + sqrt2

   5 - 2 sqrt6 =(sqrt3 - sqrt2)^2
 (5 - 2 sqrt6)^(1/2) = sqrt3 - sqrt2

  Then we have  2)

 (5 + 2 sqrt6)^(1/2)  + (5 - 2 sqrt6)^(1/2) = 2 sqrt3

Compare 1) and 2)
u = 1/2

 Then solve sin(x) = 1/2

for x

Put $u=(\sqrt{3}+\sqrt{2})^{2\sin x},v=(\sqrt{3}-\sqrt{2})^{2\sin x}$. Then

$u+v=2\sqrt{3}, uv=1$

Hence $u,v$ are the solutions of $t^{2}-2t\sqrt{3}+1=0$, and those are $t_{1,2}=\sqrt{3}\pm\sqrt{2}$

For $(\sqrt{3}+\sqrt{2})^{2\sin x}=\sqrt{3}+\sqrt{2}$ we get $\sin x={1\over 2}$

For $(\sqrt{3}+\sqrt{2})^{2\sin x}=\sqrt{3}-\sqrt{2}$ we get $\sin x=-{1\over 2}$ [u][this solution is lost by MathMaven53]
[/u]
Therefore the solutions are $x=\pm{\pi\over 6}+2k\pi\lor x=\pm{5\pi\over 6}+2k\pi, k\in\mathbb{Z}$


Από γνωστή Σέρβικη ανισότητα (όποιoς θέλει έχω και την απόδειξη) το lhs είναι μεγαλύτερο ίσο του:

$ \sqrt\frac {3(a \plus{} b \plus{} c)}{2}$

Μετά απλά AM-GM στη συνθήκη.

[quote="Protonios"]Από γνωστή Σέρβικη ανισότητα (όποιoς θέλει έχω και την απόδειξη)
[/quote]
H αποδηξή μου γι'αυτή.
$ RHS \ge \frac{(a\plus{}b\plus{}c)^2}{\sum a \sqrt{b\plus{}c}}$.
Ετσι έχουμε νδο:

$ \frac{(a\plus{}b\plus{}c)^2}{\sum a \sqrt{b\plus{}c}}\ge \sqrt{\frac{3(a\plus{}b\plus{}c)}{2}} \Longleftrightarrow 2(a\plus{}b\plus{}c)^3 \ge 3(\sum_{sym}a^2b\plus{}2 \sum ab\sqrt{c\plus{}a}\sqrt{c\plus{}b})$.
Aλλά $ \sum ab\sqrt{c\plus{}a}\sqrt{c\plus{}b} \le \frac{\sum ab(a\plus{}b\plus{}2c)}{2}$.
Eτσι $ LHS \le 3(2\sum_{sym}a^2b\plus{}6abc)$ το οποίο εύκολα βγένει μικρότερο από το rhs.

Μήπως είναι διαφορετική η λύση σου Κώστα???

εγω παντως τ πηρα αλλιως...πρωτα μια andreescu, οπως κ εσεις,μετα μια b.c.s. για τον παρονομαστη τ κλασματοσ π προεκυψε κ μετα α.ν.δ.ο. $ a\plus{}b\plus{}c \geq ab \plus{}bc\plus{}ca$ κατι που προκυπτει απο schur και cauchy χρησιμοποιωντας τν σχεση που δινεται..... :wink:

Προκύπτει και με jensen στη συνάρτηση $ f(x) \equal{} \frac {x}{\sqrt{1 \minus{} x}}$ :wink:


Frank bought a dog one day for 2.50 dollars and spent one hour per day for nine days trying to sell it. Food for the dog cost him 25 cents, and he spent a total of four hours in cleaning and caring for the dog. He sold the dog for 3.75 dollars.  How much did he profit per hour for the time spent on the whole transaction?   (Round to the nearest cent.)

A.8 cents/hour	B.10 cents/hour C. 11 cents/hour   D. 14  cents/hour  E.	29 cents/hour

This problem is missing information.

EDIT: 200th post

[quote="sri340"]Frank bought a dog one day for \$2.50 and spent one hour per day for nine days trying to sell it. Food for the dog cost him 25 cents, and he spent a total of four hours in cleaning and caring for the dog. He sold the dog for \$3.75.  How much did he profit per hour for the time spent on the whole transaction?   (Round to the nearest cent.)

A.8 cents/hour	B.10 cents/hour C. 11 cents/hour   D. 14  cents/hour  E.	29 cents/hour[/quote]

Here is the problem. (I got this by quoting then editing)
\$ should be \\$ in LaTeX.

[hide]The answer is A. Profit $1/ 13 hrs.[/hide]


Two fuel tanks, one a cylinder and one a cone, are full of jet fuel. The cylindrical tank holds 500 gallons more than the conical one. If 200 gallons are pumped from each tank the cylindrical one will contain twice as much fuel as the conical one. How many gallons of jet fuel does the cylindrical tank contain when it is full?

Cylindrical= x

Conical=y where x and y are their volumes.

x=500+y
x-200=2(y-200)

We want x. From the first equation, y=x-500. Substituing:

x-200=2(x-500-200)
x-200=2x-1400
3x=1600

x=1600/3?


algebra
polynomial
function
induction
inequalities
algebra solved

Let n <> 0 be an integer. Prove that 2*cos(nx) is a polynom of degree n of 2*cos(x).

German IMO Selection Test 1979, problem 5

Tchebyshev polynomials express cos (nx) as a function of cos (x) with degree n. So if cos (nx) = T_n (cos(x)), where T_n(cos(x)) is the nth Tchebyshev polynomial, 2 cos (nx) = 2 T_n (cos(x)). Correct?

Base: 2cos(0x)=1, 2cos(x) are polynoms of 2cos(x).

2cos((n+1)x)=2cos(x)*2cos(nx)-2cos((n-1)x) - polynom by induction

Hi Myth,

you gave the solution which is to be found in the official booklet which is the most natural approach. Concerning StRyKeR I am not familiar with Tchebyshev polynomials but only with Tchebyshev inequality. Are they somehow related ?

I don't think Chebyshef polynomials are related to the famous Chebyshef inequality ...

cos(nx)/2n-1 is a Chebyshev polynom of second type. There Chebyshev polynomes of first type.
Chebyshev polynomes and Chebyshev inequlity aren't related to each other.


I think it would be a great idea to make something like a collection of the entire solved problem section, downloadable. Begginer's and Advanced.
It is on-line, but it is difficult to read the entire  collection  from the net.
Also, with teleport pro, it would be hard to get.

that is obviously out of the question, at least for now.

I do want to put up at some time in the future a file with all the problems statements, and the links to the solutions.

Sorry, but why is it out of the question?
I mean, is it a problem with the free space on the server?
I saw that the entire mathlinks takes about 15megs :?

ok, maybe it's not so obviously out of the question :), but it is still out of the question :) :P


function
calculus
derivative
analytic geometry
graphing lines
slope

g(x)=x^2 - 5x + 6
a. Substitute and simplify the expression (g(x+h)-g(x))/(h)
b. Explain what happens to the expression when h gets "really-really" close to 0.

confused.

Hint, what they are having you do is find the difference quotient for that fucntion, just plug in what they say to plug.
Now as for when it approaches $0$, then you get the derivative of the function.

The difference quotient find is the slope of a line going through two points of the function, limiting the displacement $h$ to 0 gets you the instantaneous rate of change or derivative.

i guess i am confused on what h is. is that just the variable?

No, it is the horizontal displacement, or distance along the x-axis.
Since you are adding $h$ to $x$ you are getting another $y$ value on the function.
The formula for slope is $\frac{\text{the difference in}\ y}{\text{the difference in}\ x}$, in the case of the difference quotient, $h$ is the difference in $x$ and to find the difference in $y$ you need to plug in the $x$ values.

h is just a variable.  (As Astro suggests, you can think of it as ``horizontal displacement'' from x if you are thinking of graphing y = g(x+h).)

In the first part, you are writing (g(x+h)-g(x))/(h) in terms of x and h, treating x and h as two different variables.

In the second part, you are asked to describe how this expression from the first part behaves if h is very, very small.

[hide]
Algebraically, you can do

$\frac{(x+h)^{2}-5(x+h)+6-[x^{2}-5x+6]}{h}$

$\frac{x^{2}+2xh+h^{2}-5x-5h+6-x^{2}+5x-6}{h}$

$\frac{2xh+h^{2}-5h}{h}$

$2x-5+h$

as h gets really close to zero, you're left with

$2x-5$, which not coincidentally is the derivative (slope) of your original function. Derivatives are (arguably) the beginnings of calculus.
[/hide]


Bobby gave Fred as many coins as Fred had. Then, Fred gave Bobby as many coins as he then had. After that, they both had 20 coins. How much did each have in the beginning?

[hide="hint"]Work backwards to find the answer.[/hide]

[hide="Answer"]B is Bobby and F is Fred.  At the end they have 20 each.  Before that, Freddoubled the amount of coins Bobby had.  That means before the second transaction, $B=\frac{20}{2}=10$ and $F=20+10=30$.  Before that, Bobby doubled how many Fred had so the original amounts are $F=\frac{30}{2}=15$ and $B=10+15=25$.[/hide]


[quote="buzzer11"]construct a set containing 5 integers a, b, c, d, and e such that no subset of the set has a sum of elements of a multiple of five.

sorry, dont have the answer to this. I need help on it.[/quote]

I don't think that's possible.

apparently, it is...  :maybe:

Ill put up the solution that the book gives me if you want, but its hard to understand.

No, it isn't.

http://www.mathlinks.ro/Forum/viewtopic.php?t=103079

thats strange...

its in "Challenging problems in algebra" by Alfred S. Posamentier and Charles T. Salkind...

Your question says to prove that that there are no subsets whose sum is divisible by 5.

The book's solution is to prove there is at least one subset whose sum is divisible by 5.

"[b]1-2[/b] Find five positive whole numbers [i]a, b, c, d, e[/i] such that there is no subset with a sum divisible by 5."

Maybe I misunderstood the problem... now that I think of it, i probably did. sorry guys!  :wink:

Yes that is the question and you were supposed to see that you can't find 5 such integers.

:P I [i]thought[/i] it was a strange problem...


articles AMC AIME conics USA(J)MO USAMO algebra

Greetings,

For those who have registered for my contest, you should've all received the first article "Problem Solving Strategies" with commonly used strategies in four difficult AIME problems. The second article will instead be devoted into a specific topic.  But, this time, I'll take a poll to decide which topic are people most interested in and write the article on it. Before you cast your vote, please take a look to read the description.

[hide="Conics"]
Level: AMC to AIME
Conics article will skip the basic terminology that can be found on regular textbook and instead, focus on the actual problem solving of them. Conics is not a very common topic that is taught in high school curriculum so this article will concentrate solely to prepare students for conics problems in math contest. The problems will be taken from the actual math contest as well as modifications of them. Anyone who is not familiar with conics is recommended to read a section of conics before reading this article. Majority of problems will be more helpful to AIME because the only problems shown up on AMC are in last 5 or 6 problems. [/hide]

[hide="Cevas Theorem"]
Level: AIME to USAMO
Ceva's Theorem is a very popular topic, and most people are aware of it. Thus, instead of focusing on its proofs (which can be easily read from Wikipedia), this article will focus on some tough problems that may or may not require Ceva's Theorem. Also, Menelaus' Theorem will be mentioned in this article because they are essentially same in some perspective. It is recommended that a reader is familiar with this theorem prior to reading the article. [/hide]

[hide="Polynomial Relationship"]
Level: AIME to USAMO
This broad article (broad as in not as strict as other two) will focus specifically on polynomials that are too complex to be dealt with simple Viete's Formula or Newton's Sums. The article will aim to concentrate on strategies to work on polynomials beyond usual strategy. There is no specific recommendation as this article will be much easier to approach than others but it will [i]not[/i] be easy. [/hide]

This poll will only be up till the end of May.

Just in case that anyone who didn't vote yet, the poll is only valid till the Saturday of this upcoming week! Vote!

So far, the polynomial relationship is ahead of other two topics!  :D

Although it's not quite the end of May yet (only couple hours away), I'm going to end the poll now. The topic for the second article will be POLYNOMIAL RELATIONSHIP. Thank you to all who have casted your vote.

I do not honestly know when it'll be available (I hope before the contest starts). It takes me sometime to find good problems and solutions and it takes another great amount of time for me to write the article so yeah.

To prevent further voting, the topic is now locked and un-stickied.


inequalities
TST
USA
three variable inequality

alright.. the idea of this is to suppose that:
2/a+ 3/b + 6/c < 6
2/b+ 3/c + 6/a < 6
2/c + 3/a + 6/b <  6  and get to a false inequality and then at least two of the inequalities :
2/a+ 3/b + 6/c >= 6
2/b+ 3/c + 6/a >= 6
2/c + 3/a + 6/b>= 6  are true!

cheers!

No , then at least one of the inequalities is true .. Not TWO.

guess you are right.. sorry!  :D

thank you iandrei! bye!

Hey, I solved this problems last night.This is my solution:
Without loss of general, we suppose a=max{a,b,c}
Put S= 2/a + 3/b + 6/c
T = 2/b + 3/c + 6/a
P = 2/c + 3/a + 6/b
Applying AM-GM, we have
S^2 >= 18(1/ab + 2/ac + 3/bc) = 36(1/ab + 1/bc + 1/ca)
+ 18/b(1/c-1/a) (1)
Since a + b + c>=abc, we have 1/ab+ 1/bc +1/ca >=1 (2)
Combine (1),(2), we have s^2>=36 then s>=6
If bc<=1, we have P> 3/a + 4(3/bc)^1/2 > 4.3^1/2>6
If bc>=1, we prove that P+ T >=12
We have P+T = 5/c + 8/b + 9/a
Since a+ b + c>= abc, and bc>1, we have a<=(bc-1)/(b+c)
Then P+ T >= 5/c + 8/b +9(bc-1)/(b+c)
We prove 5/c + 8/b +9(bc-1)/(b+c) >=12, mutiplying bc(b+c),
we get f(b)= (9c^2 -12c +5)b^2 + 4c(1-3c)b + 8c^2>= 0
Quadratic f(b) have D'=-36c^2(c-1)^2<0
then f(b).(9c^2 -12c +5)>=0
but 9c^2 -12c +5= (3c-2)^2+1>0
then f(b)>=0
P+T >=12 then P or T>=6

A1lqdSchool, can  you  please  check  your  computations  please??
thanks

Here is a perhaps more motivated solution that also generalizes the problem a bit.

First of all, it is pretty easy to smooth so that we may assume $a+b+c = abc \implies \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}= 1$. Let [b]A[/b] be the matrix that cyclically shifts a vector's components to the right, so that [b]A[/b] <x, y, z> = <z, x, y>. It is not difficult to verify that the points [b]r, Ar, AAr[/b] form an equilateral triangle, and the plane containing it is normal to the vector <1, 1, 1>.

Now consider the set S of all vectors [b]r[/b] satisfying [b]r[/b] $\cdot$ [b]Ar[/b] = 1. It can be verified that [b]s[/b] = <1, 1/2, 1/3> is in this set, and also [b]t[/b] = <1/a, 1/b, 1/c> must be in this set.

Let [b]u[/b] = $\left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$. Note that any vector in S can be decomposed into the form [b]u(r)[/b] + [b]v(r)[/b], where [b]u(r)[/b] is in the direction of <1, 1, 1> and [b]v(r)[/b] is orthogonal to <1, 1, 1>. We rewrite [b]r[/b] $\cdot$ [b]Ar[/b] = 1 as

1 = ([b]u(r)[/b]+[b]v(r)[/b])([b]u(r)[/b] + [b]v(Ar)[/b])
= |[b]u(r)[/b]|^2 - |[b]v(r)[/b]|^2/2

(Notice that [b]v(Ar)[/b] is a 120-degree rotation of [b]v(r)[/b] in the plane of [b]r, Ar, AAr[/b]).

Now suppose for the sake of contradiction that [b]t[/b] has a dot product with two of [b]s, As, AAs[/b], WLOG [b]s[/b] and [b]As[/b], that is less than one. Then the direction of [b]v(t)[/b] must be within 120 degrees of the direction of one of [b]v(s), v(As)[/b], so WLOG it is within 120 degrees of [b]v(s)[/b].

For ease of notation, let $U_{t}$ = |[b]u(t)[/b]| and $V_{t}$ = |[b]v(t)[/b]|, and define $U_{s}$ and $V_{s}$ similarly. We then have:

1 > [b]ts[/b] = ([b]u(t)[/b] + [b]v(t)[/b])([b]u(s)[/b] + [b]v(s)[/b])
= [b]u(t)u(s)[/b] + [b]v(t)v(s)[/b] $\ge U_{t}U_{s}-\frac{V_{t}V_{s}}{2}$

$\left( 1+\frac{V_{t}V_{s}}{2}\right)^{2}> U_{t}^{2}U_{s}^{2}$

$\left( 1+\frac{V_{t}V_{s}}{2}\right)^{2}> \left( 1+\frac{V_{t}^{2}}{2}\right) \left( 1+\frac{V_{s}^{2}}{2}\right)$.

This is a contradiction of Cauchy. Thus at least two of the following are true:

[b]t(AAs)[/b] $\ge 1$
[b]t(As)[/b] $\ge 1$
[b]ts[/b] $\ge 1$

which is equivalent to the problem.

[quote="hxtung"]Let $ a, b, c$ be positive reals such that $ a \plus{} b \plus{} c\geq abc$. Prove that at least two of the following three inequalities:

$ \frac {2}{a} \plus{} \frac {3}{b} \plus{} \frac {6}{c}\geq 6$,

$ \frac {2}{b} \plus{} \frac {3}{c} \plus{} \frac {6}{a}\geq 6$,

$ \frac {2}{c} \plus{} \frac {3}{a} \plus{} \frac {6}{b}\geq 6$

are true.[/quote][quote="A1lqdSchool"]Let $ a \equal{} \frac {1}{x} ,b \equal{} \frac {1}{y},c \equal{} \frac {1}{z}$, we have $ yz \plus{} zx \plus{} xy\geq 1$, rewrite the three inequalities as :

$ 2x \plus{} 3y \plus{} 6z\geq 6 , 2y \plus{} 3z \plus{} 6x\geq 6,2z \plus{} 3x \plus{} 6y\geq 6$.

WLOG assume that $ 2x \plus{} 3y \plus{} 6z < 6 , 2y \plus{} 3z \plus{} 6x < 6$, then

$ 144 > [(2x \plus{} 3y \plus{} 6z) \plus{} (2y \plus{} 3z \plus{} 6x)]^2 \equal{} (8x \plus{} 5y \plus{} 9z)^2$

$ \equal{} 9(y \minus{} 3z)^2 \plus{} 16(2x \minus{} y)^2 \plus{} 144(yz \plus{} zx \plus{} xy)\geq 144\Longrightarrow$contradiction.

Problem is solved.[/quote]Another proof is based on

$ (2x \plus{} 3y \plus{} 6z)(2y \plus{} 3z \plus{} 6x) \minus{} 36(yz \plus{} zx \plus{} xy)$

$ \equal{} 12x^2 \plus{} 6y^2 \plus{} 18z^2 \minus{} 15yz \plus{} 6zx \minus{} 14xy$

$ \equal{} \frac {(12y \minus{} 15z \minus{} 14x)^2}{24} \plus{} \frac {23(3z \minus{} 2x)^2}{24}\geq0$.

By the way, if $ x, y, z$ be positive reals such that $ xyz\geq 1$, then at least two of the following three inequalities are true :

$ 2x \plus{} 3y \plus{} 6z\geq m , 2y \plus{} 3z \plus{} 6x\geq m,2z \plus{} 3x \plus{} 6y\geq m$,

where $ m \equal{} \sqrt [3]{486 \plus{} 117\sqrt {39}} \equal{} 10.675\cdots$ is the best possible.

Similar problem see http://www.mathlinks.ro/Forum/viewtopic.php?t=185491

[quote=hxtung]Let $ a$, $ b$, $ c$ be positive reals such that $ a+b+c\geq a\cdot b\cdot c$. Prove that at least two of the following three inequalities:
$ \frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6$; $ \frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6$; $ \frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$ are true.[/quote]
[hide=Jichen:][size=50]p/3530903621[/size][/hide]
Let $ a$, $ b$, $ c$ be positive reals such that $ a+b+c\geq abc$. Prove that


Those triangles are similar by SAS. The triangle to the right is has side lengths of 1/2 the triangle on the left. Thus the the length of AB is half of 14, so the length is 7 cm.

[quote="GameBot"]The side lengths of both triangles to the right are given in centimeters. What is the length of segment $ AB$?

[asy]pair A,B,C,D,E,F,G;

A=(0,0);
B=12*dir(0);
C=20*dir(120);
D=8+B;
E=D+6*dir(0);
F=D+10*dir(120);

draw(A--B--C--cycle);
draw(D--E--F--cycle);

label("A",F,N);
label("B",E+(1,0));
label("6",.5*(A+B),S);
label("14",.5*(B+C),NE);
label("10",.5*(A+C),SW);
label("\small{$120^{\circ}$}",A,NE);
label("3",.5*(D+E),S);
label("5",.5*(D+F),SW);
label("\tiny{$120^{\circ}$}",D,NE);[/asy][/quote]

You can see that the triangle to the right is similar to the triangle on the left by SAS similarity.
You can also see that the ratio of the triangle to the right to the triangle to the left is $ \frac{1}{2}$
So if the corresponding side to $ AB$ is 14, and the ratio is $ ]frac{1}{2}$, then $ AB$ should be $ 7$


geometry
geometric transformation
function

Hi everyone,
I found a problem in vietnamese forum, but I don't understand vietnamese. Can anyone translate it please ?? Just make a translation and post it here or send it to me as a Private message.
Click [url=http://www.ddtoanhoc.net/forum/index.php?act=ST&f=22&t=2822]here[/url] to see the problem.
Thanks  :lol:

Does there exist a function $s:\mathbb{Q}\rightarrow\{-1, 1\}$ such that if $x$
and $y$ are distinct rational numbers satisfying $xy = 1$ or $x + y \in \{0, 1\}$,
then $s(x)s(y) = -1$?

Bài 3:: Tồn tại hay không hàm số $s:\mathbb{Q}\rightarrow\{-1, 1\}$ có tính chất: $\forall x,y \in \mathbb{Q}$ thỏa mãn: $xy = 1$ hoặc $x + y \in \{0, 1\}$ thì $s(x)s(y) = -1$

Problem 3:: Does there exist a function $s:\mathbb{Q}\rightarrow\{-1, 1\}$ having property: $\forall x,y \in \mathbb{Q}$ satisfying: $xy = 1$ or $x + y \in \{0, 1\}$ then $s(x)s(y) = -1$

[quote="orl"]analysis, stop it!!! If you want to translate something, then translate the following [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=147370#p147370]file.[/url][/quote]
too difficult to read!

em mu` anh van nen khong hieu gi` lun lam` sao gio` day

[color=blue]Mù Anh Văn mà lại vào diễn đàn này hả em? Thật không đó[/color]


A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose endpoints are (3,17) and (48,281)? (Include both endpoints of the segment in your count.)

$\textbf{(A)}\ 2 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 16 \qquad
\textbf{(E)}\ 46$

[hide="Answer"]Slope of the line is $\frac{281-17}{48-3}=\frac{264}{45}=\frac{88}{15}$. So, the line will go through a lattice point every 15 points to the right, so we have $4\Rightarrow \boxed{B}$.[/hide]

[hide]We see that the slope of the segment is $\frac{88}{15}$ and since the right triangle (with the segment as the hypotenuse) has the base as $45$, then we have $3$ latice points every time we move $15$ to the right ($x values=18,33,48$) but we also must remember to count $3$, therefore we have a total of $\boxed{4}$ lattice points.[/hide]


Here is an interesting problem from ARML 1982:

Let's say we are given 4 bags of coins in which all the coins in a bag weigh the same, and the coins weigh 1,2 or 3 ounces. Take 1 coin from bag 1, 3 from bag 2, 9 from bag 3, and 27 from bag 4. The coins weigh 95 ounces. Determine the weight of a coin from each of the 4 bags.

A variety of answers is appreciated, as my answer for this problem is fairly complicated ;)

Well just guessing and checking, it's easy to see $\{2,1,1,3\}$ is a solution.

[hide]the marbles in the fourth bag must weigh 3 ounces because if they weighed 2 ounces and the marbles in all the other bags weighed 3, the total weight would only be 93 ounces.
THe marbles in bag 3 must then weigh 1 ounce because if they weighed any more the marbles would weigh more than 93 ounces.
The marbles in bag 2 must then weigh 1 ounce in order to keep the total weight under 93 ounces.
So the marbles in bag 1 must weigh 2 ounces.

Bag 1= 2 oz.
Bag 2= 1 oz.
Bag 3= 1 oz.
Bag 4= 3 oz. [/hide]

does anyone have a mathematical equatione way to solve this? If I changed the problem slightly the guess and check way you guys are using would be too complicated a way to use.

[hide="Can I use mods?"]Let x, y, z, w be the weights of the coins in bags 1, 2, 3, 4, respectively.
x+3y+9z+27w=95
Consider mod 3. x must be 2.

3y+9z+27w=93
y+3z+9w=31
Consider mod 3. y must be 1.

3z+9w=30
z+3w=10
Consider mod 3. z must be 1.

3w=9
w=3.[/hide]

I didn't say you couldn't use mods, I personally used bases to solve this problem.

I had two approaches toward this problem. The first was with smart guesing and checking, and the second was with mods.

Personally, I found the mods to take a longer time than the guessing and checking.  ;)

[hide="Guess and Check"]The one fact I used: Coins in bag 4 must weigh 3 ounces. If they weigh 2, it is impossible for the overall weight to be 95. If they weigh 4, it's more than 95.

After a minute or two of working with numbers, we find the solution is $\{2,1,1,3\}$ [/hide]

[hide="Mods"]First, we notice that the number of marbles taken in each bag is a multiple of 3, with the exception of bag 1. This is a strong sign to use modulus 3.  :)

Let the marbles in bag 1 weigh a ounces, in bag 2, b ounces, and c, d, for the remaining two bags.

$a+3b+9c+27d=95$, Taking modulus 3 of both sides, $a\equiv 2(mod 3)$

Now, we plug in the value for a and solve once again for b, c, and d. The answer is $\{2,1,1,3\}$
[/hide]

Please post your solution with number bases. I can't think of a number base approach. Nice problem, by the way.


Has anyone bought these books and made good progress because of them?
I just want to know what to expect once I recieve them, apparently shipment to Canada can take up to 15 days I think.  :(

if u read CAREFULLY and do a lot of work on it, u can progress really FAST! :D i did too

If this book is so recommended, are there any others that help as much?

umm... i cant remmeber the name of the book, but it is by Paul Zeitz, the former IMO coach. His book helps alot (if you finish AoPS 1/2)

Art and Craft of the Problem Solving by Paul Zeitz is an excellent book.

To hiitsme, I wouldn't recommend that to you now though. Get aops book first and then see how you can do. If you can do both versions (which are really tough) you should be ready because going over through aops 2 can help you pass to USAMO and trust me, it happened in this site.

Not me though.

yea, i forgot to mention, Art and Craft book by Zeitz is recommended along with AoPS2 or after AoPS2. sorry.

hmm......

Any others?

I think billzhao recommended "problem solving through problems" by Larson, any comments on that 1?

You seem to know alot about these books, how good were you when you finished some these books? Did you see drastic improvements?
How many aime questions can you do (estimate)?

um... it is NOT just finishing the book, but getting the jest of everything... and being able to apply those mathematical concepts to the problems like AIMe... i can now do about 809 AIMEs and 12-13 without ANY MISTAKES... (which tend to never happen... :( )

Obviously just reading the books won't make you better. 12-13 aime is very, very impressive. what grade are you in btw?

9. Hmmm, it'd make me feel pretty stupid if you were younger than me. Still, 12-13 for your age is very nice. When'd you start studying math?

Man, you two make me feel slow. I'm a senior and I think I got 4 right on this past AIME.

Anyhow, I'm looking into purchasing the two books as a part of my high school math club's thing. THat way, we all get to use it, and it is on the fundraising tab.
Regardless, I obviously have room for improvement. Since I'm graduating this summer, I'm not looking to improve on tests or anything, just for my general knowledge and cause it'll make me feel less stupid.
I've always had a knack for math, just never really had resources available to me, so I feel if I applied myself, I can understand it.

Any suggestions?

You might all look around the "Other Problem Solving Topics" forum -- there have been lots of similar conversations here in the past.


[b]Three concentric circles are drawn with radii of 8 cm, 12 cm, and 20 cm. If a dart lands randomly on this target, what are the probabilities of it landing in each region?[/b]

Just find the area of each ring and the area of the whole board. If we're assuming that the dart lands on the board, and any particular point is equally probable, then the answer is just the ratio of the area of the relevant region to the area of the whole board.

It would be much more interesting if there was a probability distribution of some kind, rather than uniformity. :D

Also, this is probably not in the right forum.


Let $ E$ be a system of $ n^2 \plus{} 1$ closed intervals of the real line. Show that $ E$ has either a subsystem consisting of $ n \plus{} 1$ elements which are monotonically ordered with respect to inclusion or a subsystem consisting of $ n \plus{} 1$ elements none of which contains another element of the subsystem.

[url=http://en.wikipedia.org/wiki/Dilworth%27s_theorem]Dilworth's Theorem[/url]. Interestingly, it was published the same year this problem was posed...

this problem reminds me of the Erdos-Szekers theorem so maybe we can use the same approche


Find the value of $p$ such that the quadratic equation $x^2+(1+i)x+(p-i)=0$ has a real root. Then find the real root of the equation.

(Note: $i=\sqrt{-1}$)

EDIT- $p$ is a real number.

[hide]p $= i$
roots $= 0, 1 + i$

Method:

Plugging into the discriminant gives:

$SQRT(2i - 4p - 4i) = SQRT(6i - 4p)$

So we want this value to be $1 + i$ to cancel out the $-b$ factor. $4p$ needs to be $4i$ so $p = i$.

Discriminant is thus $SQRT (2i) = 1 + i and -1 - i$

The real root is thus $0$ and the new complex root is $1 + i$[/hide]

by quadratic formula for the root to be real

$\frac{-(1+i)\pm \sqrt{((1+i)^2-4*(p-i))}}2$
has to be real
so
$\sqrt{\sqrt{4*p^2+9}+2p}-1$ or $-\sqrt{\sqrt{4*p^2+9}+2p}-1$
(the coefficient of the i) has to be real
the negative radical cannot equal 1, and the positive radical has solution $-2$
Thus the root is $2/2=1$

$\boxed{p=-2, root=1}$

breez,
i think p has to be integer, or at least real
if it is complex there are infinte solutions for p, such that x is real

can anyone confirm/refute that i'm right?

Yeah, $p$ is a real number. Sorry for the ambiguity.
qwertyq got the right answer :)

By the way, discriminant and quadratic formula don't always work when the coefficients are not real numbers. It turns out that they still give the correct answer in this problem, but there's a "safer" way, too. Can you find it?

We can shuffle this equation $x^2+x+ix+p-i=0$ into \[ x^2+x+p=i-ix=i(1-x) \]
If $x \in \Re$ and $p \in \Re$, then $(x^2+x+p) \in \Re$ and $(1-x) \in \Re$. So we have i times a real number equals a real number, which can't happen.

[quote="blahblahblah"]That is a good way of looking at it, but you're not thinking hard enough.

You are saying that if $x$ is real, then $i(1-x)$ is always (pure) imaginary.  But there's one value of $x$ for which that isn't true...[/quote]

Oh yeah...forgot about that. Anyway that's a pretty safe way of solving the problem. X must equal one and p must equal -2.


Find all eight solutions for x^8=1.

Use spoilers.

All eight answers are [hide]1, -1, i, -i,  :sqrt: i, - :sqrt: i, i :sqrt: i, -i :sqrt: i[/hide]

[hide]First take the square root on both sides: X^8=1  x^4= :pm: 1.  Then take the square root again on both sides, remember,  :sqrt: -1=i.  Thus x :^2: = :pm: 1, :pm:  i.  Then take the square root again but we haven't defined square root of i as another letter.  Anyway  :pm:  :sqrt: -i= :pm:  :sqrt: ((-1)i)= :pm: i :sqrt: i.  That leaves x :equiv:  :pm: 1,  :pm: i,  :pm:  :sqrt: i, and  :pm: i :sqrt: i[/hide]

Can you give a more explicit form of sqrt(i), etc.?

wouldnt this problem be a simple application of de moivre's theorem?

Try squaring one of your last 4 solutions -- tell me what you get.  Then see if you do in fact have 12.

I get [hide]e^(i*:pi:/2)[/hide]which squared gives [hide]-1[/hide]

Wait, now you've officially confused me.  What is equal to e^(i:pi:/2)?

Wait,  :sqrt: i = ei* :pi: /4, and (i+1)*(:sqrt: 2 /2) = ei* :pi: /4, so there are only 8 solutions.

There you go.  That was what I was trying to point out in my earlier comment.  Note that z8 - 1 = 0 is an 8th degree polynomial, and thus it must have 8 roots.

[quote="ComplexZeta"]Yes, you can solve it with De Moivre's Theorem as well as many other ways.[/quote]
We're doing factoring (again for the third time) and my math teacher mentioned that although he didn't mention what it is and mathworld didn't really help, can you give a brief explanation of what that is?

DeMoivre's Therem states that e^i*x = cos x + i*sin x. I'm not sure that I even know of an elementary (i.e. non-calculus) proof of this. I suppose most pre-calc teachers just have to present it as a matter of faith. It's really useful for a number of things though.

Umm.. thats not De Moivre's theorem.

De Moivre's theorem states that (a cis :theta:1) (b cis :theta:2) = ab cis (:theta:1+:theta:2). From there you can get to the more usually seen formula of (a cis :theta:)n = an cis n:theta:.

If you want to use it to find all solutions to the above:

[hide]

Right hand side = 1 = 1 cis (2n:pi:). So taking 8th roots of both side using De Moivre's theorem gives x = 1 cis (2n:pi:/8) = cis (n:pi:/4). You then substitute n=-3 through 4 or whatever set of 8 in a row you want, and then expand that into an a+ib form if you like (cis x is cos x + i sin x).

[/hide]

[/hide]

what gauss said was euler's formula, no?  Is that what it's called?

yeah well DeMoivre is basically a corollary of that. (duh - ae^ix*be^iy=abe^i(x+y) is clearly obvious once you transform into e^i:theta: notation)


f(x) = ax+b, where a and b are positive.  given f(f(x)) = bx +a, show a+b=1

does anyone know? i got up to ax+b=x, then i dont know what to do next...

B(ax+b)+a=bx+a
abx+b^2 +a=bx+a
abx+b^2=bx
ax+b=x

Is this right so far? Where do I go next?

f(x) is ax+b. Thus, f(f(x)) would be a(ax+b)+b. What I would do is set that equal to bx+a and then evaluate.

Ok. I did that. Then I got bx + a = xa^2 + b.

xa^2 + b - a - bx = 0
x(a^2 - b) + b - a = 0
x(a^2 - b) = a - b

What's next?

Arrg.  :( I dont think this is working... I know this is probably the wrong way to go when proving something, but get a or b on one side in a+b=1 and then see if the equation you just posted, with the substitution, is possible in all cases (0=0 or something).

oh great, never mind about substituting, I'll get back to you.


Find all solutions to $ c^2 \plus{} 1 \equal{} (a^2 \minus{} 1)(b^2 \minus{} 1)$, in integers $ a$, $ b$, and $ c$.
[hide="Hint"]
Modulo 4
[/hide]

if one of $ a,b,c$ were zero then we would have $ a \equal{} b \equal{} c \equal{} 0$ which is a solution,now let $ a,b,c\neq 0$.looking at the equation modulo 4,we get that $ a \equal{} 2a_1,b \equal{} 2b_1,c \equal{} 2c_1$ where $ a_1,b_1,c_1$ are integers and we have:

$ c_1^2 \equal{} 4a_1^2b_1^2 \minus{} a_1^2 \minus{} b_1^2$

now looking at the last equation modulo 4 we get that $ a_1 \equal{} 2a_2,b_1 \equal{} 2b_2,c_1 \equal{} 2c_2$ where $ a_2,b_2,c_2$ are integers as we have:

$ c_2^2 \equal{} 16a_2^2b_2^2 \minus{} a_2^2 \minus{} b_2^2$

anyway if we continue this algorithm we will end up with a contradiction,because in n-th step we will have:

$ a_{n} \equal{} 2a_{n \minus{} 1},b_{n} \equal{} 2b_{n \minus{} 1},c_{n} \equal{} 2c_{n \minus{} 1}$ hence we must have $ a \equal{} 2^na_n,b \equal{} 2^nb_n,c \minus{} 2^nc_n$ where $ a_n,b_n,c_n$ are integers,which isn't possible if $ a,b,c\neq 0$...

so the only solution is $ (a,b,c) \equal{} (0,0,0)$.

There is also a generalized form of this exercise.
$ c^n \plus{} 1 \equal{} (a^n \minus{} 1)(b^n \minus{} 1)$, where $ n$ is odd and $ n>2$
Are there any non-zero integer solutions?


ratio
geometry
circumcircle
geometry unsolved

[quote="delta"]Let $\triangle ABC$ be an equilateral triangle $AC=BC$. The incircle touches $AB$ at $M$, $BC$ at $D$ and $AC$ at $K$. $AD$ intersects the incircle at $S$.
a) if $KS$ intersects $AB$ at $P$ find $\frac{AP}{AM}$.
b) if $CS$ intersects $AB$ at $T$ find $\frac{AT}{AM}$.[/quote]
Dear Delta.

I see that the solution is not difficult, if the triangle $\bigtriangleup ABC,$ is an equilateral one. I don’t understand the statement of $AC = BC.$

If $\bigtriangleup ABC,$ is equilateral, we have that the point $S,$ is the midpoint of the segment $AO,$ where $O,$ is it’s circumcenter ans so,

$KS\parallel CM$ $\Longrightarrow$ $\frac{AP}{AM}= \frac{AS}{AO}= \frac{1}{2}$ $,(1)$

From the right triangle $\bigtriangleup AMO,$ taken as it’s transversal the segment line $CT,$ by Menelaus’s theorem,

we have $\frac{TA}{TM}\cdot \frac{CM}{CO}\cdot \frac{SO}{SA}= 1$ $\Longrightarrow$ $\frac{TA}{TM}= \frac{2}{3}$ $\Longrightarrow$ $\frac{AT}{AM}= \frac{2}{5}$ $,(2)$

Kostas Vittas.

I am really sorry. [b]equilateral[/b] is a mistake. Shud be [b]isosceles[/b] $AC=BC$. :blush:. If possible I would like a moderator to change the title too!

I am surprised because of I ascertain that we have again the same ratios.

That is the point $P,$ is the midpoint of the segment $AM$ and so $\frac{AP}{AM}= \frac{1}{2}$ $,(1)$ $($ not so difficult to prove $).$

Also if I am not mistaken, it is $\frac{AT}{AM}= \frac{2}{5}$ $,(2)$ $($ I have been proved this result, by a difficult way $).$

I will check my notes more carefully and if there is not something wrong,  I will post here next time the proofs I have in mind.

Kostas Vittas.

[b][size=109]PROOF OF THE RESULT 1.[/size][/b] $\left( \frac{AP}{AM}= \frac{1}{2}\right).$

We denote as $Q,$ the intersection point of $AD,$ $KM.$

As well known, the point $Q,$ is harmonic conjugates of $A,$ with respect to the point s $S,$ $D.$ $($ because of the segment lines $AK,$ $AM,$ are tangents of the circle $(O),$ at points $K,$ $M,$ respectively $).$

That is the pencil $K.AQSD,$ is in harmonic conjugation.

Because of now, $AB\parallel KD,$ we conclude that the point $P,$ is the midpoint of the segment $AM$ $($ A parallel line to one of the rays of a pencil in harmonic conjugation, intersects the other three it's rays at three points, formed two equal segments $).$

Hence, $\frac{AP}{AM}= \frac{1}{2}$ $,(1)$ and the proof of the [b][size=109]result  1,[/size][/b] is completed.

Kostas Vittas.

[b][size=109]PROOF OF THE RESULT 2.[/size][/b] $\left( \frac{AT}{AM}= \frac{2}{5}\right).$

We denote as $R,$ the intersection point of  the segment lines $BC,$ $KS.$

In the triangle $\bigtriangleup ABD,$ taken as it’s transversal the segment line $PSR,$ by Menalaus theorem, we have that $\frac{PA}{PB}\cdot \frac{RB}{RD}\cdot \frac{SD}{SA}= 1$ $,(1)$

From $(1)$ $\Longrightarrow$ $\frac{SD}{SA}= \frac{PB}{PA}\cdot \frac{RD}{RB}= \frac{3}{1}\cdot \frac{RD}{RB}$ $,(2)$

Also in the triangle $\bigtriangleup ABD,$ taken as it’s transversal the segment line $TSC,$ we have that $\frac{TA}{TB}\cdot \frac{CB}{CD}\cdot \frac{SD}{SA}= 1$ $,(3)$

From $(3)$ $\Longrightarrow$ $\frac{TA}{TB}= \frac{SA}{SD}\cdot \frac{CD}{CB}$ $,(4)$

From $(2),$ $(4)$ $\Longrightarrow$ $\frac{TA}{TB}= \frac{1}{3}\cdot \frac{RB}{RD}\cdot \frac{CD}{CB}= \frac{1}{3}\cdot (B,D,R,C) = \frac{1}{3}\cdot \frac{1}{(B,D,C,R)}$ $,(5)$

We denote as $L,$ the intersection point of $AR,$ $DK$ and we have the follows equal double and simple ratios.

$(B,D,C,R) = (P,S,K,R) = (R,K,S,P) = (L,K,D) = (A,P,B)$ $($ because of $LD\parallel AB$ $).$

$\Longrightarrow$ $(B,D,C,R) = (A,P,B) = \frac{BA}{BP}= \frac{4}{3}$ $,(6)$

From $(5),$ $(6)$ $\Longrightarrow$ $\frac{TA}{TB}= \frac{1}{3}\cdot \frac{3}{4}= \frac{1}{4}$ $,(7)$

From $(7)$ $\Longrightarrow$ $\frac{TA}{TA+TB}= \frac{1}{1+4}$ $\Longrightarrow$ $\frac{AT}{AB}= \frac{AT}{2(AM)}= \frac{1}{5}$

Hence, $\frac{AT}{AM}= \frac{2}{5}$ and the proof of the [b][size=109]result 2,[/size][/b] is completed.

Kostas Vittas.

It is so much easier to prove the [b][size=109]result 2,[/size][/b] by considering the triangle $\bigtriangleup APK,$ taken as it's transversal the segment line $TSC.$

So, by Menelaus theorem we have that $\frac{TA}{TP}\cdot \frac{SP}{SK}\cdot \frac{CK}{CA}= 1$ $,(1)$

From $(1)$ $\Longrightarrow$ $\frac{TA}{TP}= \frac{SK}{SP}\cdot \frac{CA}{CK}= \frac{KD}{AP}\cdot \frac{AB}{KD}= \frac{4}{1}$

$\Longrightarrow$ $\frac{TA}{TA+TP}= \frac{AT}{AP}= \frac{4}{1+4}= \frac{4}{5}$

Hence, $\frac{AT}{AM}= \frac{2}{5}$ and the proof of the result [b][size=109]result  2,[/size][/b] is completed.

Kostas Vittas.


Sue is four times as old as Tom is now, and she is $ \frac{8}{3}$ as old as Tom will be in 7 years. How many years old was Tom three years ago?

Let $ t$ be Tom's age and let $ s$ be Sue's.

Then,

$ 4t\equal{}s$ and
$ s\equal{}\frac{8}{3}(t\plus{}7)$.

So, $ 4t\equal{}\frac{8}{3}t\plus{}\frac{56}{3} \implies \frac{4}{3}t\equal{}\frac{56}{3} \implies t\equal{}14$.

So, $ t\minus{}3\equal{}11$.


MATHCOUNTS
geometry
algebra
polynomial
binomial theorem
AMC 10

[hide="Big hint for 2"]$(a-b)^2=a^2-2ab+b^2$
$(a^2+ab+b^2)(a-b)=a^3-b^3$[/hide]

[hide="Big hint for 3"]$3333_4=3(4+1)(4^2)+3(4+1)=15(16^1)+15(16^0)$.

The only problem is, $A$ and $B$ aren't digits, they'll both be $f$. :huh: I guess they just mean the $15$'s, but that really doesn't make sense.[/hide]

thank you! here's what i got using your hint i'm not sure it's right

[hide]2. 1300
[/hide]
[hide]3. 30[/hide]

is that right?

Both of those are the answers I got.

I keep forgetting how to do problems like #1. :huh:

I'm pretty sure number 1 is not Mathcounts level, but anyway.

Let $x^{99} + \ldots + x + 1 = N$.

We know $N(x-1) = x^{100} - 1$.

We also know then, that:

$\frac{N}{x+1} = \frac{x^{100}-1}{(x-1)(x+1)} = \frac{x^{100}-1}{x^2-1}$

But oh, we know that that $x^{100}-1$ divided by $x^2-1$ has no remainder, and thus $N$ divided by $x+1$ has no remainder either.

[quote="LuCy4EvA"]i need some help on these questions:

1. find the remainder when $X^{99}+X^{98}+X^{97}...X+1$ is divided by $X+1$

2. If, $XY=10$, and $X-Y=10$. Find $X^3-Y^3$

3. If $3333$ in base $4$ is equal to AB in base $16$ where $A$and $B$ are digits, find the value of A+B.

i'm not sure number 2 is possible.[/quote]

  These are some cute problems!

[hide="Question 1"]

  Note that:

$x^n= [(x+1)-1]^n$

  This of course demonstrates the binomial theorem and when we divide the expanded form of $[(x+1)-1]^n$ by $x+1$ we see that $x+1$ is a factor of every term except the last, $(-1)^n$.  If $n$ is odd then $x+1$ will leave a remainder of $(x+1)-1=x$ when divided into that expression.  Whereas is $n$ is even, it will leave a remainder of $1$ when divided into that expression.  In the sequence:

$x^{99} + X^{98} + ... X^1 + X^0$

  There are 50 odd terms and 50 even terms for a total remainder of 50x + 50.

  $\frac{50x+50}{x+1}=50$

  Meaning the remainder is 0.

[/hide]

[hide="Question 2"]

  The key to solving this problem is knowing how to manipulate equations.  We first begin by noticing our unknown expression can be factored as a difference of squares.

$x^3 - y^3 = (x-y)(x^2+xy+y^2)$

  From our given information we know the first expression, $x-y$ , is equivalent to 10 and the second term of our second expression, $xy$ , is equivalent to 10.  This looks like:

$10(x^2 + 10 + y^2)$

  We are still left with however, the terms $x^2$ and $y^2$.  We need to somehow manipulate our given equations $xy=10$ and $x-y=10$ to obtain $x^2$ and $y^2$.  I personally went about this by looking at the equation $x-y=10$ and knew if I were to square it, I would obtain the terms $x^$ and $y^2$.  Lets try this out:

$(x-y)^2 = x^2 - 2xy + y^2$

  Which means:

  $x^2+y^2=(x-y)^2 + 2xy$

  Now we have an equation that consists of only the equations we have values for:

$x^2 + y^2 = (10)^2 + 2(10) = 120$

  So now our original equation becomes:

$x^3-y^3= 10 ( 120 + 10 ) = 10(130) = \boxed{1300}$ [/hide]

[hide="Question 3"]  We could first convert this number to base 10 and then to base 16, but there is an even easier way.

$3333_4 = 3(4^3) + 3(4^2) + 3(4^1) + 3(4^0)$

  Notice now that 16 and 4 have a direct exponential relationship $16=4^{1/4}$ so we re-write this number in terms of powers of 16 ( what base 16 is ).

$3333_4 = 3(16^{1/2})^3 + 3(16^{1/2})^2 + 3(16^{1/2})^1 + 3(16^{1/2})^0$

  When we simplify this we get:

  $15(16^1) + 15(16^0)=FF_16$

  The sum of these two digits is: $15 + 15 = \boxed{30}$ [/hide]

wow these help a bunch! thank you.

here's another problem

find the area of a regular octagon inscribed in a square of side two.

i'm not sure what it would like if you have an octagon inside of a square.

[hide="Faster way but not so lucid way for 1"]Observe that when you long-divide the first polynomial by x+1, the first term of the result is x^98. Now see that this removes both the x^99 and x^98 terms from the dividend. Now see that it does this for all the other terms down to x+1, which is obviously divisible by x+1. [/hide]

Oh, of course. :roll: It's like finding the sum of the coefficients of a polynomial, except there you plug in $1$.

I have a hint for the octagon problem but I'm not sure how to explain it without a picture. :huh:

If I remember correctly, that last was an AMC 10 problem....[hide="How to put the octagon inside the square"]Chop off the corners of the square.[/hide]

[hide="solution for octagon problem"]Using what Solafidefarms said, say that the length of a side of the octagon is $x$. Then the legs of the right triangles on the corners is $\frac{x}{\sqrt{2}}$, because $x$ is the hypotenuse of a 45-45-90 triangle. That simplifies to $\frac{x\sqrt{2}}{2}$. Since there are two of them of them on the side of the square, and the length of the side of the octagon on the square is $x$, you can set up an equation $x\sqrt{2}+x=2$, solving the equation yields $2\sqrt{2}-2$. You can find the area of the octagon by finding the area of the right triangles and subtracting that from the area of the square. The area of the square is 4. The area of one triangle is $\frac{(2\sqrt{2}-2)^2}{2} = 6-4\sqrt{2}$. Finding the area of all four is $24-16\sqrt{2}$, So the area of the octagon is $4-24+16\sqrt{2}=\boxed{16\sqrt{2}-20}$[/hide]

[quote="cj"][quote="LuCy4EvA"]i need some help on these questions:

1. find the remainder when $X^{99}+X^{98}+X^{97}...X+1$ is divided by $X+1$[/quote]

[hide="my simple answer"] You just need to plug in x=-1 and you'll get the answer. This is called the remainder theorem[/hide][/quote]

Haha, duh.  :blush:

Wow I cant believe I didnt see x was congruent to -1 mod x+1..... whitehorseking's solution made it look like the problem was much more difficult than it was..... :lol:

[quote="JesusFreak197"]Oh, of course. :roll: It's like finding the sum of the coefficients of a polynomial, except there you plug in $1$.

[/quote]

I would've used some synthetic division, since there is a noticible pattern.


Find the integer solutions of equation:
$ x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} y^2$

[quote="Αρχιμήδης 6"]Find the integer solutions of equation:
$ x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} 0$[/quote]

x = -1 ? :lol:

Sorry  :oops:  i want to be perfect square...

Λάθος προσέγγιση! Ευχαριστώ τον Δημήτρη (Αρχιμήδης6) για την υπόδειξη του λάθους!

Aν για τετραδα ακεραιων $ a,b,c,d$  ισχυει οτι $ ab\equal{}cd$ και $ (a,b)\equal{}(c,d)\equal{}1$ λες τοτε οτι $ |a|\equal{}1$ η $ |a|\equal{}c$ η $ |a|\equal{}d$ η $ |a|\equal{}cd$.Δεν ισχυει κατι τετοιο...π.χ $ 3.35\equal{}15.7$

H paragontopoihsh mas dinei $ (x^2\plus{}1)(x\plus{}1)\equal{}y^2$ apo edw vlepoume oti $ x>\equal{}\minus{}1$. Gia $ x\equal{}\minus{}1,0,1$ exoume luseis. An x artios, tote megistos koinos diaireths twn $ x^2\plus{}1,x\plus{}1$ einai 1, ara k ta 2 ginontai teleia tetragwna ara $ x^2\plus{}1\equal{}r^2\equal{}>x\equal{}0$ An x perittos tote mkd=2 ara $ x^2\minus{}2z^2\equal{}\minus{}1$ kai $ x\plus{}1\equal{}2w^2$
H luseis ths prwths einai $ x_i\plus{}z_i\sqrt2\equal{}(1\plus{}\sqrt2)^{2i\plus{}1}$ kai prepei na exoume x tetoio wste na isxuei kai h deuterh...Exw thn entupwsh oti auto isxuei mono gia $ x\equal{}7$, alla den mporesa na vrw mia apodei3h akoma...

Σωστα salonikios αλλα απο δω και περα αρχιζει η δυσκολια της ασκησης.Βρηκα μια δικη μου λυση της ασκησης...θα την στειλω αφου λυθει. :wink:

Αν και οι εξισώσεις Pell περιέχουν συσσωρευμένη πληροφορία νομίζω ότι είναι δύσκολο να συνεχιστεί η λύση του Ηλία  :(

Και δυστυχώς έχω την εντύπωση ότι αυτή η άσκηση δεν λύνεται εντελώς elementary .
Η λύση που έχω βρει βασίζεται στο θεώρημα που λέει ότι η καμπύλη $ y^2\equal{}x^3\plus{}ax\plus{}b$ έχει πεπερασμένο πλήθος λύσεων .
Δημήτρη ποια είναι η λύση σου ?

Σιλουανε θα την εχω στηλει αν οχι σημερα μεχρι αυριο.Παντος η δικη μου προσεγγιση ειναι elementary...Δεν την περιμενα και ετσι την ασκηση.... :mad:

$ x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} y^2 \equal{} > (x \plus{} 1)(x^2 \plus{} 1) \equal{} y^2$
Aν $ x$ αρτιος μοναδικη λυση η $ x \equal{} 0$
Aν $ x$ περιττος τοτε $ x \plus{} 1 \equal{} 2A^2$ , $ x^2 \plus{} 1 \equal{} 2B^2$.
Aντικαθιστωντας την πρωτη στην δευτερη παιρνουμε την εξισωση $ A^2 \plus{} (A^2 \minus{} 1)^2 \equal{} B^2$
1)Aν $ A$ περιττος τοτε:
$ A^2 \equal{} u^2 \minus{} v^2$ & $ A^2 \minus{} 1 \equal{} 2uv$ $ (u,v) \equal{} 1$ σχεση(a). $ u$ περιττος $ v$ αρτιος.
Αφου $ A^2 \equal{} u^2 \minus{} v^2 \equal{} > A^2 \plus{} v^2 \equal{} u^2$ αρα $ A \equal{} k^2 \minus{} l^2$ $ v \equal{} 2kl$ $ u \equal{} k^2 \plus{} l^2$ με $ (k,l) \equal{} 1$
Η (a) γινεται $ (k^2 \minus{} l^2)^2 \minus{} 1 \equal{} 2(k^2 \plus{} l^2)2kl \equal{} > (k^2 \minus{} l^2)^2 \equal{} 4kl(k^2 \plus{} l^2) \plus{} 1$ Eστω $ k \plus{} l \equal{} c$, $ k \minus{} l \equal{} d$ τοτε $ c^4 \minus{} d^4 \minus{} 2c^2d^2 \plus{} 2 \equal{} 0$
$ D \equal{} 8(d^4 \minus{} 1) \equal{} z^2$ αρα $ z \equal{} 4z_1$ οποτε $ (d^2)^2 \minus{} 2z_1^2 \equal{} 1$
MΕΛΕΤΗ ΤΗΣ $ X^2 \minus{} 2Y^2 \equal{} 1$ ME $ X$ TEΛΕΙΟ ΤΕΤΡΑΓΩΝΟ.
Εστω $ (x_n,y_n)$ οι λυσεις της εξισωσης τοτε $ (x_n \minus{} y_n$ριζα$ 2) \equal{} (1 \minus{}$ριζα$ 2)^{2n}$ & $ (x_n \plus{} y_n$ριζα$ 2) \equal{} (1 \plus{}$ριζα$ 2)^{2n}$
αρα $ x_n \equal{} [(1 \plus{}$ριζα$ 2)^{2n} \plus{} (1 \minus{}$ριζα$ 2)^{2n}]/2 \equal{} \sum_{k \equal{} 0}^{n}C(2n,2k)2^k$.Επειδη το $ x_n$ ειναι τελειο τετρ. kai επειδη παντα για καθε $ n$  παρατηρω κι αποδεικνειω οτι $ x_n \equal{} a_n^2 \plus{} ( \minus{} 1)^n$ με $ a_n \equal{}[$ριζα2$ /2][(1 \plus{}$ριζα2$ )^{2n} \minus{} (1 \minus{}$ριζα2$ )^{2n}]$ αρα δεν εχει λυσεις...
Αν A Αρτιος τοτε υστερα απο πραξεις καταληγουμε σε μια εξισωση της μορφης $ (e^2 \plus{} 2f^2)^2 \minus{} 2(2f)^2 \equal{} 1$ με $ e$ περιττο. για την οποια λυπουμε αλλα βρηκα λαθος στην λυση μου...θα την κοιταξω σχολαστικα

$ x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} y^2 \Leftrightarrow (x^2 \plus{} 1)(x \plus{} 1) \equal{} y^2$
πρέπει $ x \plus{} 1 \ge 0 \Leftrightarrow x \ge \minus{} 1$
για $ x \equal{} \minus{} 1$ είναι $ y^2 \equal{} 0 \Leftrightarrow y \equal{} 0$ άρα $ (x,y) \equal{} ( \minus{} 1,0)$

για $ x \ge 0$:
έστω ότι οι αριθμοί $ x^2 \plus{} 1, x \plus{} 1$
έχουν κοινό παράγοντα k μεγαλύτερο του 2 και μικρότερο του μικροτέρου τους τότε
$ x \plus{} 1 \equiv a \plus{} 1 (mod k)$ και $ x^2 \plus{} 1 \equiv a^2 \plus{} 1 (mod k)$
(a η μικρότερη λύση της ισοτιμίας $ x \equiv a (mod k)$ ) άρα
$ a \equiv \minus{} 1 (mod k) \Leftrightarrow a^2 \equiv 1 (mod k)$ και $ a^2 \equiv \minus{} 1 (mod k)$ που είναι άτοπο γιατί k > 2

άρα: (όλοι οι δυνατοί συνδυασμοί διαιρετέων)
είτε $ x \plus{} 1 \equal{} x^2 \plus{} 1 \equal{} y \Leftrightarrow x(x \minus{} 1) \equal{} 0, x \equal{}$ άρτιος $ \Leftrightarrow x \equal{} 0 \Leftrightarrow (x,y) \equal{} (0,\pm1)$
είτε $ x \plus{} 1 \equal{} x^2 \plus{} 1 \equal{} 2\frac {y}{2} \Leftrightarrow x(x \minus{} 1) \equal{} 0, x \equal{}$ περιττός $ \Leftrightarrow x \equal{} 1\Leftrightarrow (x,y) \equal{} (1,\pm2)$
είτε $ x \plus{} 1 \equal{} 1$ και $ x^2 \plus{} 1 \equal{} y^2$ άρα $ (x,y) \equal{} (0,\pm1)$
είτε $ x^2 \plus{} 1 \equal{} 1$ και $ x \plus{} 1 \equal{} y^2$ άρα $ (x,y) \equal{} (0,\pm1)$
είτε $ x \plus{} 1 \equal{} 2$ και $ x^2 \plus{} 1 \equal{} \frac {y^2}{2}$ άρα $ (x,y) \equal{} (1,\pm2)$
είτε $ x \plus{} 1 \equal{} 2(\frac {y}{2})^2$ και $ x^2 \plus{} 1 \equal{} 2$ άρα $ (x,y) \equal{} ( \minus{} 1,0)$ ή $ (x,y) \equal{} (1,\pm2)$

άρα οι ακέραιες λύσεις της εξίσωσης είναι οι
$ (x,y) \equal{} ( \minus{} 1,0), (x,y) \equal{} (0,\pm1), (x,y) \equal{} (1,\pm2)$

Δεν μου φαίνεται ότι παίρνεις όλες τις περιπτώσεις... Δεν παίρνεις γαι παράδειγμα την περίπτωση που $ x \plus{} 1 \equal{} 2n^2, x^2 \plus{} 1 \equal{} 2m^2$ όπου $ y \equal{} (2nm)^2$. Σε μια τέτοια περπίπτωση βγαίνει και η λύση $ x \equal{} 7, y \equal{} 20$ την οποία προανέφερε και ο Ηλίας :wink: .

Φίλε maniopa νομίζω ότι αν ήταν τόσο εύκολη θα είχε είδη λυθεί . :wink:
Εδώ λέμε οτι μάλλον δεν λύνεται elementary ...

Aν καταφερουμε να αποδειξουμε με elementary τροπο οτι  δεν υπαρχει φυσικος $ n>1$ ωστε $ [(\sqrt{2}\plus{}1)^{2n}\minus{}(\sqrt{2}\minus{}1)^{2n}]/\sqrt{2}$ να ειναι τελειο τετραγωνο φυσικου τοτε η εξισωση μας εχει απλα λυθει!(εικαζω οτι δεν εχει λυσεις για $ n>1$) :)

Δημήτρη, η παράσταση που έδωσες είναι ίση με

$ \displaystyle\sum_{k\equal{}1}^{n} \binom{2n}{2k\minus{}1}2^k$. Μήπως βοηθάει στο να δείξουμε ότι δεν έιναι τέλειο τετράγωνο?

Αλέξανδρος

Υ.Γ. Για $ n\equal{}2$ εως $ n\equal{}1000$ τo έλεγξα με υπολογιστή και πράγματι δεν είναι τέλειο τετράγωνο.

Nαι το ηξερα αυτο cretanman αλλα σιγουρα δεν ηξερα οτι αληθευει μεχρι και για $ n\equal{}1000$ Σε ευχαριστω παρα πολυ!!!  :)  :idea:

Εστω οτι $ a_n \equal{} \sum_{k \equal{} 1}^{n}\binom{2n}{2k \minus{} 1}2^k$ τοτε $ a_n \equal{} 6a_{n \minus{} 1} \minus{} a_{n \minus{} 2}$ με $ a_1 \equal{} 4$ και $ a_2 \equal{} 24$. Επομενως αρκει νδο κανενας ορος της ακολουθειας δεν ειναι τελειο τετραγωνο για $ n > 1$.

Δημήτρη η ακολουθία αυτή μοιάζει πολύ με τις περίφημες ακολουθίες Lucas (η fibonacci είναι ειδική περίπτωση αυτής). Όμως μπορεί το ζητούμενο (για $ n > 2$ κανείς όρος δεν είναι τέλειο τετράγωνο) να μην είναι τετριμμένο. Σε παραπέμπω σε ένα paper του Ν. Τζανάκη από το Πανεπιστήμιο Κρήτης (είναι φημισμένος στον τομέα των διοφαντικών εξισώσεων με πολλές δημοσιεύσεις στο περίφημο Journal of Number Theory και cites σε πολλά βιβλία και papers ξένων ερευνητών) στη σελίδα

http://www.math.uoc.gr/%7Etzanakis/Papers/LucasActa_061006.pdf

που αναφέρεται ακριβώς σε τέτοια προβλήματα και στα οποία γίνεται χρήση Αλγεβρικής Θεωρίας Αριθμών. Θα το κοιτάξω κι άλλο το πρόβλημα! Εαν παραμείνει ανοικτό, θα περάσω να βρω τον ίδιο, να του το αφήσω για να το δει και να μου πει την γνώμη του.

Αυτό που προσπαθούσα να κάνω ήταν να "κόψω" $ mod n$ για κατάλληλο $ n$ έτσι ώστε αυτό που θα μείνει, να μην είναι τετραγωνικό υπόλοιπο $ mod n$ κι έτσι να καταλήξω σε άτοπο. Μάταια όμως. Μου έμειναν πολλές περιπτώσεις και συγκεκριμένα οι $ n \equal{} 12k \plus{} u, u \equal{} 0,1,5,6,9$ καθώς για τα υπόλοιπα $ u$ το έχω δείξει. Οι υπόλοιπες  περιπτώσεις (για διαφορετικό $ mod$) ήταν πολύ χειρότερες ...

Αλέξανδρος

Τωρα σε ευχαριστω Αλεξανδρε πιο πολυ! :wink: Εγω απεδειξα οτι ο $ a_{4k\plus{}3}\equal{}2mod6$ αρα δεν ειναι τελειο τετραγωνο αλλα για τις αλλες περιπτωσεις δεν ξερω...εχω και μια αλλη ιδεα να υπλογισω τα τετραγωνα , που φρασω τον $ a_n$ ,συναρτηση του $ n$ ...

To $ u \equal{} 5$ απορρίφθηκε διότι όλοι οι αριθμοί της μορφής $ n \equal{} 3r \plus{} 2$ απορρίπτονται. Συνεχίζουμε για τα υπόλοιπα...

[b][u][color=red]EDIT[/color][/u][/b] Κατάφερα να πετάξω έξω και το $ u\equal{}6$ καθώς όλοι οι αριθμοί $ n$ της μορφής $ n\equal{}4r\plus{}2$ απορρίπτονται.

Αλέξανδρος

Εστω οτι για καθε $ n$ ισχυει οτι $ (b_n \minus{} 1)^2 < a_n < b_n^2$
τοτε παρατηρω οτι για $ n > 6$:
αν $ n \equal{} 0,2,4mod5$ τοτε $ b_n \equal{} 2b_{n \minus{} 1} \plus{} b_{n \minus{} 2} \minus{} 1$
αν $ n \equal{} 1,3mod5$ τοτε $ b_n \equal{} 2b_{n \minus{} 1} \plus{} b_{n \minus{} 2}$
Δεν το εχω αποδειξει(λογω περασμενης ωρας δεν το προσπαθω) αλλα κατι τετοιο σημαινει οτι για καθε ορο $ a_n$ υπαρχει $ b_n$ ωστε $ (b_n \minus{} 1)^2 < a_n < b_n^2$ και αρα ο $ a_n$ δεν ειναι ποτε τελειο τετραγωνο. :)

Ειμαι σιγουρος οτι ισχυει :) ...Παρατηρησα και απεδειξα επαγωγικα οτι ${ lim(b_{n + 1}/b_n})^2 = lim(a_{n + 1}/a_n) = c$ oπου $ c$ μια σταθερα για την οποια ισχυει $ c^2 - 6c + 1 = 0$ και ειναι η $ 3 + 2\sqrt {2}$.Απεδειξα δηλαδη οτι η εξισωση $ x^2 - 8y^4 = 1$ οτι δεν εχει ακεραιες λυσεις παρα μονο τις τετριμμενες...


calculus
derivative
function
algebra
polynomial
real analysis
real analysis unsolved

Given $n>1$ find all real functions, n times derivables such that $ f(x) f'(x) f\"(x)... f^{n}(x)=0$ for all $x$.

Remember : I note $f(A)=\{f(a)|a\in A\}$
First of all, there are  two easier assertions (try to prove   :D ):
(I)  $ g'(x_0)\neq 0 \Longrightarrow \displaystyle \exists u > x_0 ,0\not\in g(]x_0,u])$
(II) $(h' \ continue)\&(u>x_0)\&(0\not\in h'(]x_0,u])\Longrightarrow \displaystyle \exists{v\in ]x_0,u]},0\not\in h(]x_0,v])$

Suppose that there is $x_0$ such that $f^{(n)}(x_0) \neq 0$
From $(I)$ and $(II)$,  it's easy to see that there is $u_{n-1}\geq...\geq u_k\geq...\geq u_0 > x_0$ such that
$\ 0\not\in f^{(n-1)}(]x_0,v]),...,0\not\in f^{(k)}(]x_0,u_k]),...,0\not\in f(]x_0,u_0])$

Manifestly, from the initial equality we obtain

$\forall x\in ]x0,u_0] , f^{(n)}_{(x)}=0 \Longrightarrow \forall x\in ]x0,u_0] , f^{(n-1)}_{(x)}=k(const.)  \Longrightarrow f^{(n)}_{(x_0)}=0$ , so contradiction with the initial supposition.

Finally $\forall x , f^{(n)}(x)= 0$, so $f$ is a polynomial of at most $(n-1)$ degree

 :cool: What's the solution from $X$ (big danger for the  French industry) ?!  :cool:


Find all natural numbers $X$ such that the product of the digits of $X$ equals $X^{2}-10X-22.$

There's idea of my solution:
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Let $X$ have $n>1$ digits. Then $X^2-10X-22\geq 10^{2n-2}-10^n-22$. Maximum of product of digits is $9^n$. But we can prove that for $n\geq 3$ is $X^2-10X-22\geq 10^{2n-2}-10^n-22> 9^n$. So we have to consider just 1-digit and 2-digits numbers.

1-digit numbers are easy, for them must $x^2=x^2-10x-22$ which is impossible

For 2-digit numbers is product of digits maximaly 81, but for $X\geq 17$ is $X^2-10X-22> 81$ so it's just about considering numbers from 10,11,...,16. You can find out, that it sits only for 12.
[/hide]


vector
function
linear algebra
linear algebra unsolved

Prove that the dual of the infinite-dimensional vector space(over any field) isn't isomorphic to this space.

(The dual here is supposed in it's purely algebraic sense, i.e. the set of all linear functionals on the vector space).

Let $\{e_{\alpha}\}_{\alpha\in A}$ be a Hamel basis for the given vector space $V.$ We are assuming that $A$ is an infinite set. Then an arbitrary element of $V$ can be written uniquely as

$\sum_{\alpha\in A}f(\alpha)e_{\alpha}$

where $f: A\mapsto F$ is any function of finite support.

An arbritrary element $\Lambda$ of the dual of $V$ can be associated uniquely to an arbitrary function $g: A\mapsto F.$ The functional is then given by

$\Lambda\left(\sum_{\alpha\in A}f(\alpha)e_{\alpha}\right) = \sum_{\alpha\in A}g(\alpha)f(\alpha)$

What is left is a cardinality argument: arguing that there are more functions than there are finitely supported functions. I think this is going to split into some casework depending on the cardinality of $A$ and the cardinality of $F.$

This problem is related to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97458]that one[/url].

If $V$ is a $F$-vectorspace with an infinite basis $A$ then, as [b]Kent Merryfield[/b] noted,  the dual space $V^*$ is isomorphic to $F^A$. So the dimension of $V^*$ is $|F|^{|A|}$ which is greater than $|A|=dim V$.

Note, that it is possible to have $|V^*|=|V|$ (e.g. when $|F|=2^{|A|}$).