calculus
derivative
algebra
polynomial
geometry
geometric transformation
combinatorics proposed

Let $A=\{a_1,a_2,...,a_n\}$ be set of positive integers. Define $A^*=\{a_i+a_j|1\leq i<j\leq n\}$. The important thing to mention here is if, for example, $A=\{1,2,3,4\}$ then $A^*=\{3,4,5,5,6,7\}$ - note that we write $5$ twice (because $5=1+4=2+3$).

Now, for some $A\not=B$ we have $A^*=B^*$. Prove that $|A|=|B|=2^k$.

Try to solve it by yourselves, and I will post the solution in a few days.

It's been discussed at least twice here :).

grobber, could you give a link to any of these discussions?i don't remember seeing it(but then you are my 'senior' in this forum :)  so you know best!)
anyway brilliant or not here is a solution: let $\displaystyle f_A(x)=\sum_{a \in A} x^a,f_B(x)=\sum_{b \in B}x^b$.then the condition simply says $f_A^2(x)-f_A(x^2)=f_B^2(x)-f_B(x^2)$.since the sets are finite this is an equality for polynomials.firstly $|A|=|B|$ since $|A^*|=|A|(|A|-1)/2,|B^*|=|B|(|B|-1)/2$ and $x \rightarrow x(x-1)/2$ is monotonic on the integers,so $f_A(1)=f_B(1)$. now differentiating we get
  $2(f_A(x)f_A'(x)-xf_A'(x^2))=2(f_B(x)f_B'(x)-xf_B'(x^2))$,so we have plugging $x=1,\ f_A'(1)=f_B'(1)$ or that $f_A(1)=1=f_B(1)$.
now we repeat the same thing over and over again and repeatedly plug $x=1$(basically induction). the term involving the $(k+1)^{st}$ derivative comes only from the expression $f_A'(x)f_A^{(k)}(x)-2^{k-1}x^kf_A^{(k)}(x^2)$;the other terms involve lesser derivatives of $f$ and when we plug in $x=1$,they inductively are equal to the corresponding terms on the RHS,so we get $(f_A(1)-2^{k})f_A^{(k+1)}(1)=(f_B(1)-2^{k})f_B^{(k+1)}(1)$.but since $A\neq B$ the polynomials are not equal and so for some $k,\  f_A^{(k)}(1)\neq f_B^{(k)}(1)$(otherwise the power series expansions of $f_A,f_B$ about $x=1$ agree and hence so would $f_A,f_B$).but then we have $f_A(1)=2^k=f_B(1)$.

Needless to say, what you have posted here, seshadri, is the only solution to this problem that I have ever seen, and it's probably the only one anyone has ever seen :).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25354]Here[/url]'s one of the threads.

well i haven't seen any of the posts in the algebra forum,no wonder i missed this....
and i didn't know that it was called Selfridge's theorem-i didn't know that it was a theorem anyway...
but the technique is not totally new.i believe that this method is one of the oldest technique of algebraic combinatorics.
but thanks for the thread, grobber!

[quote="seshadri"]-i didn't know that it was a theorem anyway...
[/quote]

 :)  :)  :)   :lol:

What is 'theorem,' then?  ;) What I meant was that this is (to the best of my knowledge) a result of Selfridge.  :P

[quote="vess"][quote="seshadri"]-i didn't know that it was a theorem anyway...
[/quote]

 :)  :)  :)   :lol:

What is 'theorem,' then?  ;) [/quote]

of course! :)
what i meant was:i did not know that it was a 'named' result.

Sorry about repeating, but I tried searching and found nothing.

Anyway, I hoped that somebody knows some other solution to this problem, but it seems that this is the only one till now.

[quote="Bojan Basic"]Now, for some $A\not=B$ we have $A^*=B^*$. Prove that $|A|=|B|=2^k$.[/quote]
Where is a brilliant non-trivial example? ;) :)

I'm not sure I understand your question, could you please be more specific?

More about this, I heard that not only implication but equivalency also holds (what means that for any given [tex]k[/tex] we are able to construct [tex]A\not=B[/tex] such that [tex]|A|=|B|=2^k[/tex] and [tex]A^*=B^*[/tex]). Could anybody prove this?

[quote="Bojan Basic"]More about this, I heard that not only implication but equivalency also holds (what means that for any given [tex]k[/tex] we are able to construct [tex]A\not=B[/tex] such that [tex]|A|=|B|=2^k[/tex] and [tex]A^*=B^*[/tex]).[/quote] That was my question ;)

Actually, I've seen this problem stated for $A$ and $B$ being multisets as well (i.e. they can have repeated elements), and I have also seen an example in a book, but $A,B$ had repeated elements in that example. Do you have an example of sets $A,B$, Myth?

Arrrgh! I was so close... :) I went along the same lines and got stuck somehow :). Thanks, vess!

Let's call these constructions (and their translations) 'trivial.' How about non-trivial constructions? (maybe that's what Myth meant by saying 'non-trivial example')  ;)

It is clear that any construction for $k-1$ can be extended to a construction for $k$ in an obvious manner. Call a construction $(A,B)$ [i]$k$-primitive[/i] if $|A| = |B| = 2^k$ and $(A,B)$ is not extended from any $(k-1)$-construction. For every $k$, is there a $k$-primitive construction? As far as I know this is an open problem even for $k=3$, though I'm not at all sure about this.

Grobber, perhaps the constructions you saw for multisets are primitive? Could you please post them?

A weaker question (that I still couldn't solve): do there exist, for every $k$, two sets $A,B$ of sizes $2^k$ such that no proper subsets $A_1 \subsetneq A, B_1 \subsetneq B$ satisfy $A_1^* \equiv B_1^*$? Give it a try.

--Vesselin

I'll post that example eventually, but I'll have to either think about it for a while, or find the book where I saw it :). It'll be here in a while.

[quote="grobber"]Actually, I've seen this problem stated for $A$ and $B$ being multisets as well (i.e. they can have repeated elements), and I have also seen an example in a book, but $A,B$ had repeated elements in that example.[/quote]
Well, as much as I can see the solution written here (which seems to be the only one which exists so far) makes no difference between $A$ and $B$ being multisets and not, am I wrong?

Indeed, the solution here is not affected by this. However, when I wrote that I was talking about the example for $|A|=|B|=2^k$, not about the proof that $|A|=|B|=2^k$. I never said the solution depended on $A,B$ being multisets or not, did I? :)

[quote="vess"]Let's call these constructions (and their translations) 'trivial.' How about non-trivial constructions? (maybe that's what Myth meant by saying 'non-trivial example')  ;)
--Vesselin[/quote]

the following are all also 'trivial' constructions :
if we have sets $A,B$,then $A_1=A\cup (B+n);B_1=B\cup (A+n)$,then $(A_1,B_1)$ 'extend' $(A,B)$.

Yes, of course, for every sufficiently large $n$, $A_1 = A \cup (B+n), B_1 = B \cup (A + n)$ extend $(A,B)$. This was what I meant by 'trivial' (the construction that I wrote is, of course, of this form). By 'these constructions' I meant such type of constructions, not just the translations of the particular one that I wrote.

However, I find it incredibly difficult to construct sets $A,B$ such that for no proper subsets $A_1 \subsetneq A, B_1 \subsetneq B$ does the pair $(A_1,B_1)$ satisfy $A_1^* \equiv B_1^*$. Any ideas?


Prove that if the equation $ x^3\plus{}ax^2\plus{}bx\plus{}c\equal{}0$ has 3 distinct real roots, then the equation
$ x^3\plus{}ax^2\plus{}(\minus{}a^2\plus{}4b)x\plus{}(a^3\minus{}4ab\plus{}8c)\equal{}0$ has too 3 distinct real roots.

This is very fundamental.

$ x^3\plus{}a x^2\plus{}b x\plus{}c\equiv (x\minus{}\alpha ) (x\minus{}\beta ) (x\minus{}\gamma )$

$ x^3\plus{}a x^2\plus{}b x\plus{}c\equiv x^3\plus{}(\minus{}\alpha \minus{}\beta \minus{}\gamma ) x^2\plus{}(\alpha  \beta \plus{}\gamma  \beta
   \plus{}\alpha  \gamma ) x\minus{}\alpha  \beta  \gamma$

$ a\equal{}\minus{}(\alpha \plus{}\beta \plus{}\gamma) \land b\equal{}\alpha  \beta \plus{}\gamma  \beta \plus{}\alpha  \gamma \land
   c\equal{}\minus{}\alpha  \beta  \gamma$

the second problem comes direct from above.


find the nth term for this sequence
1, 2, 10, 67, 467, 3268, 22876

This topic has been posted in a BUNCH of places.  I don't know what you are trying to achieve...


Niech $n$ oraz $k$ będą stałe oraz niech $f(s)$ dla $0\leq s\leq kn$ oznacza liczbę ciągów $k$-elementowych $a_1,...,a_k$ spełniających $0\leq a_1,...,a_k\leq n$ oraz $a_1+...+a_k=s$. Udowodnić, że wielomian
\[ f(0)+f(1)x+f(2)x^2+...+f(kn)x^{kn}=0 \]
nie ma pierwiastków rzeczywistych różnych od $-1$.

Łatwo zauważyć, że $f(s)$ to współczynnik przy $x^s$ dla wielomianu $(x^n+x^{n-1}+...+x+1)^k=\left(\frac{x^{n+1}-1}{x-1}\right)^k$ a ten nie ma pierwiastków różnych od $-1$.

I dlatego właśnie określiłem zadanie mianem prostego. Ale faktycznie było za łatwe.


I don't think this is right....

[hide]

57%

[/hide]

someone tell me if this is anywhere close.....

not sure...

[hide] I set up this equation where x is the total amount of cereal and a is the percent of cereal left after the removal:

remaining marshmallows / (non-marshmallow cereal + remaining marshmallows) = [a/100*(x*70/100)]/(x*30/100 + a/100*(x*70/100)) = 30% = 30/100

Solving yields a = 18.36, so that the percent of marshmallow removed is ~~81.6%.

seems weird... am i doing something wrong[/hide]

I think barnacle's solution is correct...

I have a solution.

Let the marshmallow portion be .7, let the entire cereal be 1.
Let how much marshmallow taken out be x.
Remember, after taking out x, the marshmallow count will be .3 that of the whole cereal. Marshmallow is the left side and the whole cereal is on the right.

.7 - x = .3(1-x)
.7 - x = .3 - .3x
.4 = .7x
x = 4/7

This tells us the amount of marshmallow taken out in comparison to the whole amount of cereal. To find the amount as compared to the original amount of marshmallow, we compute

$\frac{\frac{4}{7}}{\frac{7}{10}}$ or $\frac{40}{49}$, approximating as 81.6%.

Lets say there is 210 total things.
147 are marshmallows.
63 are not
63=7/10x
x=90
210-90=120 marshmallows were taken out
120/147=4000/49%

[quote="236factorial"]I don't get your last line, but otherwise it's correct.[/quote]Do you mean this part?[quote="nat mc"]
210-90=120 marshmallows were taken out
[/quote]
There are 90 left after the marshmallows were taken out.  210 to start with.  210-90=120 marshmallows taken out

Or do you mean this part?
[quote="nat mc"]120/147=4000/49%[/quote]
Divide by 3 on the numerator and denominator to get 40/49.  Multiply by 100%

[quote="G-UNIT"][hide] 4 /7 = 57.14% [/hide][/quote]

bahhhhh!, i forgot it was in ratio to the original amount of marshmallows :wallbash_red: ....

G-UNIT, you were very close, you just had to divide by 7/10!!!!

[quote="colts18"][quote="GoBraves"][quote="math92"][hide="clik dis"]i am not sure but i did some work and got to 4/7 which is equal to about [b]57.14%[/b][/hide][/quote]

Ok..If I am being unfair, let me know but did you actually do the work or did you just copy G-UNITs answer because

1. The chances of you getting the same WRONG answer as someone else is very low
2. This isnt the first time this has happened
3. You never show work, could you tell me how you got to 4/7?[/quote]
Actually, i logged on as math92 (i knkow his password) and i posted that post.  But still he should show more work.[/quote]

Why didn't you use your own account??
GoBraves, you have three good points there.

You are not the right person to make such a statement, its unfair to both of them, please edit your post

[quote="nat mc"]Like brother like sister I guess, they both
:spam::spam::spam: :spam: :spam: :spam:[/quote]
 :o  :mad:

haha it's all in good humor.  no need for anyone to get mad, im sure he was just kidding around too.

nonetheless, people may be offended by such comments, and I too agree it should be removed

everyone should calm down. whether math92/colts18 are the same person or brother/sister or friends relationship, leave that to admins because math92 is already flagged by Valentin, so don't go screwing around or you will likely take a nice and long vacation, so be cool  :gleam:

[quote="236factorial"]A box of lucky charms has 70% marshmallows (oh I wish that was the case!). After Lucy took out a certain amount of marshmallows, the remaining cereal was only 30% marshmallows. What percent of the original amount of marshmallows were taken out?[/quote]

236 factorial said it :D


[quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a^2 \plus{} b^2 \plus{} c^2\geq 3$.Prove that:
$ \frac {a^5 \minus{} a^2}{a^5 \plus{} b^2 \plus{} c^2} \plus{} \frac {b^5 \minus{} b^2}{a^2 \plus{} b^5 \plus{} c^2} \plus{} \frac {c^5 \minus{} c^2}{a^2 \plus{} b^2 \plus{} c^5}\geq 0$[/quote]
It seem similar to an IMO problem(i remember it is problem 3 in IMO 2005,sorry if if i have mistakes)


Why don't we try to get this active again? Do people from Ohio still go on here?

there are no contests in the summer... @?!

Still, there were no posts since the end of May :o
I guess this is a bad time, especially since people are at Mathpath and other stuff... I'll try in September/October.

"Other stuff" includes AwesomeMath starting a week from Monday :jump: !


IMO 2003 Japan is the biggest IMO ever ( at least that's what the Japanese tell us on their official site ).

Any opinions?

it seems that the majority of people really guessed the answer :D


What are 2 critical numbers and how do differ?

I'm not sure what you're asking, but if this is calculus-related, then try posting this [url=http://www.artofproblemsolving.com/Forum/index.php?f=66]here[/url].

He's just asking for a definition.
Well, without going into the calculus of it all, I'll try to define it (this is not rigorous):
Given a function f(x) (doesn't necessarily have to be continous), the critical numbers of f(x) are either:

The X values for which we get out a local maximum or minimum.

OR

If f(x) is non-continous, the point at which the graph shifts from one piece to another (think piece-wise defined).

[url]http://mcraefamily.com/MathHelp/CalculusTheorem3aCriticalNumbers.htm[/url]
That should clarify, if you've had a bit of calc.


number theory
Diophantine equation
number theory unsolved

Hello.
"Find all numbers that are simultanously triangular and quadrangular, that is, find all pairs $ (a,b)\in\mathbb{N}$ such that $ \frac{a(a\plus{}1)}{2}\equal{}b^2$.

For example, $ 36$ and $ 1225$ work. Are the only ones?

$ \sum_{k \equal{} 1}^n k \equal{} m^2 \iff \frac{n(n \plus{} 1)}{2} \equal{} m^2$
Now let $ n \equal{} \frac {a \minus{} 1}{2}, m \equal{} \frac {b}{2}$, so
\[ a^2 \minus{} 2 b^2 \equal{} 1\]
So this problem is amount to the Pell Equation.

Hense the first few minimal pair of $ (a, b)$ are $ (3, 2)$.
Once the fundamental solution is found, all remaining solutions may be calculated algebraically as
\[ a_n \plus{} b_n\sqrt {2} \equal{} (3 \plus{} 2\sqrt {2})^n\]
Thus $ (a, b)$ are
$ (3, 2)$
$ (17, 12)$
$ (99, 70)$
$ (577, 408)$
$ (3363, 2378)$
$ (19601, 13860)$
$ (114243, 80782)$
$ (665857, 470832)$
$ (3880899, 2744210)$
$ (22619537, 15994428)$

Now $ (n, m) \equal{} \left(\frac {a \minus{} 1}{2}. \frac {b}{2}\right)$, so $ (n, m)$ are
$ (1, 1)$
$ (8, 6)$
$ (49, 35)$
$ (288, 204)$
$ (1681, 1189)$
$ (9800, 6930)$
$ (57121, 40391)$
$ (332928, 235416)$
$ (1940449, 1372105)$
$ (11309768, 7997214)$

cf.
a : [url=http://www2.research.att.com/~njas/sequences/A001108]a(n)-th triangular number is a square: a(n+1) = 6*a(n)-a(n-1)+2, with a(0) = 0, a(1) = 1.[/url]
b : [url=http://www2.research.att.com/~njas/sequences/A001109]a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.[/url]

And this Diophantine Equation is [url=http://mathworld.wolfram.com/SquareTriangularNumber.html]Square Triangular Number[/url]


[url]http://hubblesite.org/newscenter/archive/releases/2007/17/[/url]

Another one for dark matter.
Score
$\text{Dark Matter}\ 1$
$\text{MOND}\ 0$
 :rotfl:

Actually...

Dark Matter: 2
MOND: 0
My Dark Matter Theory: 0 :(

My idea doesn't really make much sense anymore, now that we've actually "seen" dark matter... :roll: Oh, well. It was fun while it lasted.

:lol: Your theory was interesting, the coffee cup analog was a cool way to put it.

MOND does not seem to be popular, I guess people like contemplating the mysterious connotation associated with "dark" matter.

Yeah. I don't think people really like messing with Newtonian dynamics, either. MOND makes things so messy...what about Occam's razor? MOND is definitely not the simplest solution.

MOND does make things weird, but so does Quantum Mechanics with it own math and that works.
Does Occom's Razor always yeild the answer?

Though, I still stick to Classical Newtonian Mechanics.  :)

I don't think Occam's Razor is always right, but what if it is? Maybe we only THINK classical mechanics and quantum mechanics work. The real answer to all of physics--both macro and micro scale--could be something so obvious and simple that it's been right in front of physicists' noses for decades. I have no idea what it would be, though, and that's assuming Occam's Razor is always correct (which it isn't).

Anyway, for me MOND is just so weird that it's counterintuitive. Why would gravity not decrease proportionally as one gets far away from the center of it, even for very small amounts? There has to be a better answer, and for now, dark matter is it.

Good point. :)
That description kind of fits the present conflict with String Theory, it is so convulted that many are doubting it validity.

Why are there so many extra dimensions in string theory? I never really understood that. They're not where the dark matter is, so what's their function? Is there even a way to prove their existence, besides the fact that they "fit the equations"?

:rotfl:
That is why it is now being refuted.  Essentially they add dimensions until the math works out.
String theorists also attempt to apply stings to everything, yes including dark matter, black holes, and gravity.

What would a black hole string look like? :huh: How can one make a point into a two-dimensional string? And how could strings be applied to dark energy?

Well, when I asked a string theorist of ISSYP last year, he did not really give me details about strings and dark matter-energy, he just said that String Theory has an intepretation of it.

About black holes, he said that in String Theory a black hole (or atleast one type) can be described by a string (or loop) being around the event horizon, but still not many details.


$ 6)$ $ x$ varies jointly as $ z$ and $ y$ and inversely as the square root of $ w$. If $ x\equal{}2$ when $ z\equal{}7$, $ y\equal{}3$, and $ w\equal{}9$, what is $ z$ when $ x\equal{}5$, $ y\equal{}14$, and $ w\equal{}16.$

$ 13)$ Simplify:$ \frac{2000^{3}\plus{}64}{2000^{3}\minus{}6(2000)^{2}\plus{}24(2000)\minus{}32}$

[hide="1"]since $ x$ is jointly proportional to $ y$ and $ z$ and inversely proportion to the root of $ w$, we have
$ \frac{x\sqrt{w}}{zy}$ plugging in the  values we have $ \frac{x\sqrt{w}}{zy}\equal{}\frac{6}{21}$
now plugging in the new values, we have $ \frac{5\sqrt{16}}{14z}\equal{}\frac{6}{21}$ solving, we get $ z\equal{}5$..unless i made a computation error[/hide]

[hide="2"]is there some mistake. i tried factoring it in the form $ (x\plus{}y)^{3}$ but it's not working, unless i made some mistake[/hide]

When it says that $ x$ varies jointly as $ y$ and $ z$ and inversely as the $ \sqrt{w}$ doesn't that translate to $ \frac{xyz}{\sqrt{w}}$

[hide] $ \frac{(2)(7)(3)}{\sqrt{9}}\equal{} 14$ which gives us $ \frac{(5)(14)(z)}{\sqrt{16}}\equal{}\frac{70z}{4}$ Then $ z \equal{}\frac{4}{5}$ [/hide]

[quote="SnipedYou"]When it says that $ x$ varies jointly as $ y$ and $ z$ and inversely as the $ \sqrt{w}$ doesn't that translate to $ \frac{xyz}{\sqrt{w}}$[/quote]
If $ a$ varies inversely to $ b$, then it's $ ab$. Jointly is $ \tfrac{a}{b}$.

yes, i like pie is right. when $ x$ is jointly proportional with two other variables u can really say that those two other variables are both inversely proportional to each other and x is directly proportional to them. inversely proportional means has a constant product. e.g. $ a$ and $ b$ are inversely proportional, thus their product $ ab$ is constant.

has anyone tried the second problem?

Oops I forgot :blush: Well here's my attempt at 13 (which is probably wrong too)

[hide] $ 2000^{3}\plus{}64$ can be factored into $ (2000\plus{}4)(2000^{2}\minus{}8000\plus{}16)$ Now we can do some canceling. We have $ \frac{(2000\plus{}4)(2000^{2}\minus{}8000\plus{}16)}{2000^{3}\minus{}6(2000)^{2}\plus{}24(2000)\minus{}32}$ Which can be reduced to $ \frac{2004}{2000^{3}\minus{}32}$ There's probably an error in there.
[/hide]


Can some tell me what the symbol [b]>>[/b] means please? (From a physics problem.)

Can you use it in an expression? I suspect that it means implies but I am not sure.

[url]http://en.wikipedia.org/wiki/Inequality[/url]
It means much greater than. (There is no exact mathematical criteria for what [i]much greater[/i] than means, but it typically implies that there is a difference of at least several orders of magnitude.)


calculus
integration
trigonometry
calculus computations

calculate this
$ \int \frac{(1\plus{}\sin{x})e^x}{1\plus{}\cos{x}} dx$

$ \int \frac{(1\plus{}\sin{x})e^x}{1\plus{}\cos{x}} dx\equal{}e^{x}\tan\left(\frac{x}{2}\right)\plus{}C$.

hello
[hide="solution"]
write $ sinx\equal{}2sin\frac{x}{2}cos\frac{x}{2}$
        $ 1\plus{}cosx\equal{}2cos^2\frac{x}{2}$
$ \int \frac{(1\plus{}\sin{x})e^x}{1\plus{}\cos{x}} dx\equal{}\int e^{x}(\frac{sec^2{\frac{x}{2}}}{2}\plus{}tan\frac{x}{2}$
$ \equal{}e^{x}tan\frac{x}{2}$
[/hide]
thank u


search
combinatorics unsolved
combinatorics

Graph Airlines $ (GA)$ operates flights between some of the cities of the Republic of Graphia. There are at least three $ GA$ flights from each city, and it is possible to travel from any city in Graphia to any city in Graphia using $ GA$ flights. $ GA$ decides to discontinue some of its flights. Show that this can be done in such a way that it is still possible to travel between any two cities using $ GA$ flights, yet at least $ 2/9$ of the cities have only one flight.

[quote="JBL"]It's from Turkey 2002.  Here are two other places (one very recent) where it has appeared on the forum, neither with any solutions:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1842627630&t=316753
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1842627630&t=192924[/quote]

thanks for links but there is no solution in this topics


superior algebra
superior algebra unsolved

Suppose that $ G$ is finite and nilpotent group and $ p$ is a prime number such that $ p\mid  \| G \|$.Prove that $ p \mid \| Z(G) \|$.

$ G$ is the direct product of it's sylow subgroups. then we may assume that $ G$ is a finite $ p$-group. but then the claim is clear since nontrivial $ p$-groups have nontrivial center (a consequence of the class equation).


Prove that {1,2,3,...,100} can be partitioned into 5 sets such that no 2 elements in a set differ by a perfect square.

My sets are:
[hide]
The numbers ending in 1 and 4
The numbers ending in 2 and 5
The numbers ending in 3 and 6
The numbers ending in 7 and 9
The numbers ending in 8 and 0

The difference between any two numbers in the same set ends in 3, 7, 2, 8, or 0.  But no square less than 100 ends in any of those digits, so no difference is a square.
[/hide]


calculus
integration
function
derivative
geometry
trigonometry
real analysis

Can we find a function - or does there exist a function $f$ , which has derivative $\forall x \ in I$ , I is an iterval , but $\int_{a}^{b}f'(x)dx$ does not exist ?($a,b \ in I$).

 Or the existence of derivative is sufficient for the integral to exist ? I suspect that $f'$ must be something more , for the integral to exist , but is this really true and why?
 Thanks - Babis

suppose the area of the $f'$ in the interval $(a,b)$ tend to infinity

or is undefinied in some point $x$ in the interval

the improper integral tends to infinity and a lot of cases

It depends on what kind of integral we're looking at. The derivative can easily fail to be absolutely integrable, even if improper integrals exist.
As for improper Riemann integrals- how many bad points are you willing to avoid?

Well , the problem concerns high school analysis.I was wondering , if we must allways consider a function to have continoous derivative , if we want $f'$ to be integratable on [a, b]. The newton theorem states f' must be integratable ,but I thought that $f'$ has this propetry.It sees no, so I was seeking an example. If you have any , please send it.
 [u] Babis[/u]

The function $f(x)=x^{2}\sin(\frac1{x^{2}})$ is differentiable on $[-1,1]$. It's not Riemann integrable because the derivative is unbounded.

jmerry didn't explicitly mention that he intends to define $f(0)=0;$ that falls under the heading of "removable discontinuities deserve to be removed." The differentiability at zero is less than obvious, but it is differentiable there.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=132593]Here[/url] you will find the answer to all of your questions (see fedja's example).

Just take any continuous function on $(a,b)$ whose integral fails to exist. It is the derivative of one of its primitives ...

Really thanks !

 It was a little difficult for me to read again all this  nice theory about intregrals , after 30 years fom the university years! Your help was great! You saved me much time. To all problems I will give the condition that $f'$ is continous , so we'll have not problems with my students in high school!
  Thank you again!
              Babis


Let $ f(x)\equal{} ax^{2}\plus{}bx\plus{}c$ be a quadratic such that $ a,b,c\in\mathbb{R}^{\plus{}}$ and $ a\plus{}b\plus{}c\equal{}1$. Prove that if given a sequence of positive real numbers $ x_{1},x_{2},........,x_{n}$ such that their product is $ 1$ then:

$ f(x_{1})f(x_{2})......f(x_{n})\geq 1$.

Using induction I'll prove that :
$ p(n) : f(x_{1})f(x_{2})...f(x_{n})\geq\left (f(P_{n})\right)^{n}$ ,where $ P_{n}\equal{}\sqrt[n]{x_{1}x_{2}...x_{n}}$
$ p(1)$ is obvious , $ p(2)$ is true by Cauchy-Schwartz.
Supose that $ p(n)$ is true we must prove that $ p(n\plus{}1)$ is true:
From $ p(n)$ we get that :
$ f(x_{1})f(x_{2})...f(x_{n})f(x_{n\plus{}1})\geq\left(f(P_{n})\right)^{n}f(x_{n\plus{}1})$ so it's enough to prove that $ (f(P_{n}))^{n}f(x_{n\plus{}1})\geq\left(f(P_{n\plus{}1})\right)^{n\plus{}1}$ ,but this is just Holder's inequality.
Therefore $ f(x_{1})f(x_{2})...f(x_{n})\geq\left (f(P_{n})\right)^{n},\forall n\in\mathbb N^{*}$ .In particular for this problem $ P_{n}\equal{}1 ,\forall n\in\mathbb N^{*}$ and the result follows.


A map's key equates $ 1$ inch with $ 30$ miles. If two cities are exactly $ 25$ miles apart, how many inches apart are they on the map? Express your answer as a common fraction.

Since we know that 1 inch= 	30 miles,  and 5 miles= 	$ \dfrac{1}{6}$ [size=150], [/size]$ \dfrac{1}{6} * 5$ equals $ \dfrac{5}{6}$


logarithms
floor function
geometric series

Compute $ \sum_{n \equal{} 1}^{6237}\lfloor \log_2 n \rfloor.$

An interesting problem: Let $ a_k$ be the coefficient of $ x^k$ in the expansion of $ \sum_{i\equal{}1}^{99} (x\plus{}1)^i$. Find the value of $ \lfloor a_4/a_3 \rfloor$.

[hide="1"]
We know that $ 2^{12}<6237<2^{13}$, so for values from $ 4096$ to $ 6237$ ($ 2142$ numbers) we get $ 12$.

Let's consider $ \sum_{n\equal{}1}^{4095}\lfloor\log_2 n\rfloor$. For $ n\equal{}1$, it's $ 0$. For $ n\equal{}2,3$, it's $ 1$. For $ n\equal{}4,5,6,7$, it's $ 2$. For $ n\equal{}8,9,10,11,12,13,14,15$ it's $ 3$.

Noticing this pattern, we find that that sum is equivalent to $ \sum_{n\equal{}1}^{11}n{2^n}$.

Let $ S\equal{}1\cdot2^1\plus{}2\cdot 2^2\plus{}3\cdot2^3\plus{}\cdots\plus{}11\cdot2^{11}$.

$ 2S\equal{}1\cdot 2^2\plus{}2\cdot 2^3\plus{}\cdots\plus{}11\cdot2^{12}$.

(I fudged a little here to get a geometric series, but accounted for it by subtracting off $ 2^0$ and $ 2^1$).
$ S\minus{}2S\equal{}\minus{}S\equal{}2\plus{}(2^0\plus{}2^1\plus{}2^2\plus{}2^3\plus{}\cdots\plus{}2^{11})\minus{}2^0\minus{}2^1\minus{}11\cdot 2^{12}$.

$ \minus{}S\equal{}2\plus{}2^{12}\minus{}4\minus{}11\cdot2^{12}\equal{}\minus{}10\cdot 2^{12}\minus{}2$

$ S\equal{}10\cdot 4096\plus{}2\equal{}40962$.

Then $ 40962\plus{}2142\cdot 12\equal{}\boxed{66666}$   :o
[/hide]

[hide="2"]
The coefficient of $ x^k$ in $ (x \plus{} 1)^i$ is $ \binom{i}{k}$, so what we're dealing with here is $ a_k \equal{} \sum_{i \equal{} 1}^{99}\binom{i}{k}$. For $ i\leq k$ this is $ 0$, so what we have is $ \sum_{i \equal{} k}^{99}\binom{i}{k} \equal{} \binom{100}{k \plus{} 1}$ by Hockey Stick.

Therefore, $ \frac {a_4}{a_3} \equal{} \frac {\binom{100}{5}}{\binom{100}{4}} \equal{} \frac {100!}{95!5!}\cdot\frac {96!4!}{100!} \equal{} \frac {96}{5}$.

Our answer is, then, $ \boxed{19}$.
[/hide]


algebra
polynomial
inequalities
inequalities proposed

Thank you, silver1989 and Fitim.
The proof of silver1989 is similar to mine.
I will re-read the Fitim's (I don't have time now, sorry). But do you have another proof not expanding ?

[b]Ineq. 3.[/b] $(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2}) \ge 8 \left(\frac{abc(a^{2}+b^{2}+c^{2})}{ab+bc+ca}\right)^{2}$
[b]Ineq. 4.[/b] $3(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2}) \ge (ab+bc)^{3}+(bc+ca)^{3}+(ca+ab)^{3}$


Let $ a,b,c$ be positive real numbers satisfying $ abc\equal{}1$. Prove that \[ 3(a\plus{}b\plus{}c)\ge\sqrt{4a\plus{}5}\plus{}\sqrt{4b\plus{}5}\plus{}\sqrt{4c\plus{}5}\]

Let  $ f: D\to\mathbb R$ , $ f(x) =\sqrt{4x+5}$ ,$ f$ is concave
From Jensen's inequality we have that :
$ \sum f(a)\leq 3f(\frac{a+b+c}{3}) = 3\sqrt{\frac{4}{3}(a+b+c)+5}\leq 3(a+b+c)$ $ \Leftrightarrow$
$ (a+b+c)^{2}-\frac{4}{3}(a+b+c)-5\geq$ $ \Leftrightarrow$ $ a+b+c\geq 3$ which it's obvious.

Another solution:
The given inequality is equivalent to:
$ \sum(\sqrt{4a+5}-3a+\frac{7}{3}\cdot ln(a))\leq 0$
Let $ f: D\to\mathbb R$ $ f(x)=\sqrt{4x+5}-3x+\frac{7}{3}\cdot ln(x))$
Because $ f''(x)<0$ and $ f'(1)=0$ we are done  :D

Um...okay. Great, thanks. :| But there are solution(s) using elementary and basic inequalities. :maybe: Actually I don't consider Jensen's inequality as elementary because it uses calculus(which I haven't pretty much studied :blush: ). Anyway the following inequality is also true:

Let $ x,y,z>0$. Prove that \[ \sqrt{x^{2}\plus{}3xy}\plus{}\sqrt{y^{2}\plus{}3yz}\plus{}\sqrt{z^{2}\plus{}3zx}\le 2(x\plus{}y\plus{}z)\]

[hide]$ \sum x^{2}\ge\sum xy$
$ \Rightarrow 4\sum x^{2}\plus{}8\sum xy\ge 3\sum x^{2}\plus{}9\sum xy$
$ \Rightarrow 4(x\plus{}y\plus{}z)^{2}\ge 3\sum (x^{2}\plus{}3xy)\ge (\sum\sqrt{x^{2}\plus{}3xy})^{2}$ by Cauchy[/hide]

Both are very nice solutions! (Although I'm not sure if I understand Svejk's last one pretty much :blush: . So I'm still looking for another solution of my first inequality...)

Hi SplashD, can you prove this one:
\[ \sqrt{3x^{2}\plus{}xy}\plus{}\sqrt{3y^{2}\plus{}yz}\plus{}\sqrt{3z^{2}\plus{}zx}\le 2(x\plus{}y\plus{}z)\]

[quote="nayel"]I don't consider Jensen's inequality as elementary because it uses calculus(which I haven't pretty much studied :blush: ). [/quote]

It's not that Jensen's inequality itself is calculus - all it really does is extrapolate from the elementary (not the calculus) definition of convexity.  The part that could be considered calculus is knowing when a function is convex, but certain functions can easily be proven convex from the elementary definition.

[quote="nayel"]But there are solution(s) using elementary and basic inequalities.[/quote]is Cauchy allowed?
We know $ a+b+c\geq 3$ from AM-GM.
\begin{eqnarray*}\sqrt{4a+5}+\sqrt{4b+5}+\sqrt{4c+5}&\leq&\sqrt{3}\sqrt{4(a+b+c)+15}\\ &\leq&\sqrt{a+b+c}\sqrt{4(a+b+c)+5(a+b+c)}\\ &=&3(a+b+c)\end{eqnarray*}
not so different from Svejk's solution though


I know this is probably the wrong place to post this, but I don't know where else.  I have a TI30 scientific calculator.  However, whenever I ask for Sin it gives the value of Cosine and vice versa (the same is true for arcsin and arccos). is is possible to reprogram such a calculator?  thanks  :D

well unless you have a program button then you can't

you can program calculators using BASIC with TI-83s and up


search
MATHCOUNTS
factorial
geometry
calculus
AMC 8
AMC 10

Hi, I'm curious, as a grown-up example of a student who skipped a grade (from fifth to seventh) in school, how many of the rest of you have a declared grade skip? In other words, how many of you are in a school grade higher than the grade you would be placed in considering only your age? How many grades have you skipped?

And, most importantly, when you register for an AMC test, or a regional talent search achievement test, or for Mathcounts, or for some other math contest, do you register according to the grade you are in from the point of view of the school's records, or according to the grade you would be in from the point of view of your age?

I'm wondering how common the phenomenon is of multiple grade-skipping among you math-eager young people. Here in Minnesota, I know of a local example of a young man (that is, a boy) who is no more than eight years old, taking an accelerated Algebra I & II course this year, who is listed in that course's records as an eighth-grader. If I'm doing my math right, that is something like a quintuple grade skip. I don't know his family well enough (yet) to ask how they sign him up for talent search testing, which is something I presume they do for him. I also don't know if he is signed up for this year's AMC-8, but he sure ought to be. And, of course, someone ought to invite him to join this forum . . .

Thanks for any observations any of you can provide,

I am currently skipping two grades, [b]only in math[/b]. I started skipping in fifth since California did not allow skipping grades let alone a partial grade-skip. Instead I was assigned to a cource about "How to be successful in life" in fifth grade. (I'm sure it's experimental since I've never heard of it before nor after I took the cource; Probably because I've got a 100 on STAR 4 years running :) (I moved to the U.S. in second grade))
After I came to Massachusetts, my 6th grade math teacher (no longer a teacher; quit after she had me :twisted:; actually, she had a baby so she quit) placed me in eighth (which might've been a mistake since I skipped a lot of material like factorials and box-and-whisker graph and such). Now I am in eighth and is currently taking tenth Honors over at a high school which is across a huge field. (My middle school, another middle school, and the high school shares the same field)

I tend to take full advantage of this and register for all that I can. For example, MATHCOUNTS and AMC 10. The odd thing is since I dropped a grade when I first came here (since I didn't know any English), so I qualify for the grade below me, my original grade, and the one above. (because of the grade I'm currently in, my age, and my math class's grade, in that order)

I tend to see a lot of imaturity among those who skip grades and vice versa. This is obviously because the kids around them have more maturity then him/her, and this difference in mental age can affect the child a lot as they grow older. I'd say that you shouldn't skip grades once you're halfway through Elementary, even before. Or else you'll get this obnoxious selfish impudent kid with no friends [size=59]a.k.a. M*x*m[/size]
My two cents on skipping grades.

Skipping grades in math does not count as "skipping a grade". Generally, children may skip a grade in elementary school (and thus skip an entire grade, not just take advanced courses in one subject area), though there was an IMOer at my school a few years ago that graduated one year early, so I guess that counts as skipping a grade (senior year).

Well there's a pretty obvious difference between those who partly skip grade and take a couple of subject over at the high school and those who don't because we literary go back-and-forth between our schools. I also see the "Indulgence"syndrome even when a child is only skipping one grade.

I guess this is a special case, but still there are a lot of them so I think this should be into consideration. I know of three people who skips only math in AoPS, and I'm sure there's a lot more people in the world who do "partial skips."

Well, if you mean skipping grades in only one subject, then I am sure almost everyone here is skipping grades in math.

By the way, what is the "Indulgence" syndrome?

[quote="Tare"]I named it "Indulgence" syndrome since I forgot the scientific name of it, but it's the syndrome when people, usually children, becomes less mature than others because they are spoiled by someone such as their parents, their classmates, etc.[/quote]

hmmm....i dunno. they might be immature at the beginning, but i think they usually grow out of it (personal experience, and one rather good friend)

[quote="Tare"]I tend to take full advantage of this and register for all that I can. For example, MATHCOUNTS and AMC 10. The odd thing is since I dropped a grade when I first came here (since I didn't know any English), so I qualify for the grade below me, my original grade, and the one above. (because of the grade I'm currently in, my age, and my math class's grade, in that order)[/quote]

I'm trying to be clear about this. As far as I know, anyone can take a math contest at an unusually LOW age. I know of summer math programs that have minimum age requirements, but I don't know of any contests that do. Thus my oldest son started taking part in a junior high mathematics contest we have locally when he was of fourth-grade age (mostly to fill out the team). He was the youngest kid in the whole state in that program--unless someone equally young was designated as belonging to a higher grade. He is on the league records as a fourth-grader, because I don't declare a grade skip for him (yet). People who homeschool can be vague about grades.

But I wonder how many unusually young participants in math contests declare their grade as a grade above the grade placement for their age. For example, there are AMC-8 statistics

http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2002-8a/02stats8.html

that go all the way down to fourth grade, but I wonder if some of the nominal eighth graders who take AMC-8 are actually one or two years younger than most other eighth-graders. What grade would you declare for yourself in signing up for AMC-10?

Has anyone reading this thread done a talent search test

http://www.ditd.org/Cybersource/record.aspx?sid=12649&scat=902&stype=110

for example an out-of-level SAT I test? If so, was your test result reported as matching the grade you would be in by age, or matching some other grade?

Inquiring minds want to know . . .

Well... in math I skipped grades... in orchestra too... but nothing else...
and about skipping whole grades...
one of my friend just skipped half of kinderg. and half of 1st grade... which doesn't really say anything about him being smart.

I know that you can still qualify even if you're lower than the grade the the competition was meant for (you could even take AMC 12 in first for all I know), but you wouldn't get the [b]opportunity[/b]to do so if you don't go to high school/middle school. Sometimes you have to even [b]make [/b]that opportunity. For example, my school didn't offer MATHCOUNTS so I went and set up a MATHCOUNTS "meetings" which doesn't really help us with math (since the only teacher available was a 6th math teacher who majored in "acting") but it will at least provide some oppotunity for the more fortunate who will come to my middle school. If I stayed back that fateful year from saying "yes", then I probably wouldn't have known about AMC, even this website! It's all about whether you know the competitions exists rather than whether or not you qualify.
...I sure wish I've known of MATHCOUNTS earlier :cry: , and I'm sure there are even less fortunate around me. So when I say "qualify" I mean actually signing up and doing it, not just qualify for it and do nothing.

P.S. My friend (who's about a dozen :wink: ) took SAT and he was in the grade that he was currently in, not by his age.

if you start school early. like kindergaten and stuff, it doesnt really matter that much. like you can tell all your friends once youve reached like the 7th grade, that you were super smart and skipped a grade. yup yup

[color=cyan]In general, you are supposed to indicate the level of your education, not the level of your age.  That is, if you have a six-year-old taking in 12th grade, she should indicate her grade as 12, not as kindergarten.[/color]

Never skipped a full grade.  I did start a year early compared to most people in my state, though, because I lived in California as a kid.

I think that by the time you're in high school, "skipping grades" in individual subjects is no longer meaningful.  That's because in high school, people have a lot more choice over what courses they want to take, and it's less unusual to find kids who don't adhere to a specific guideline.  For instance, in my middle school most kids took prealgebra in 7th grade and algebra in 8th grade.  In high school, on the other hand, there is a 7th grader and a sophomore in geometry, a freshman and a senior in AP Bio, etc.

I skipped a grade; 2nd grade that is.  It's pretty cool, but there are drawbacks.  I'm almost always 1 to 1.5 years younger than my peers.  But it's also advantageous in that I "started" life earlier than they did.

-interesting_move

[quote="Ravi B"]When I was little, I skipped first and third grades.  I'm glad I did, or I would have been bored in school (still was, but less so).[/quote]

Thanks to you and to the others who revitalized this thread today. So now you participate in math contests, at a grade level two grades higher than your age placement would be, right?

Just wondering . . .

Well, I'm an adult now.  But, yes, when I was a kid, I participated in math contests two years ahead of my age.  For example, I made it to the US Olympiad at ages 14 and 15 (11th and 12th grades), but never went beyond that.  Perhaps if I had those two extra years, I could have went farther...  There's advantages and disadvantages to skipping grades.

It is kind of nice to be able to declare a grade fitting for age, but I'm not sure exactly how ethical/permitted that is, especially when most people can not declare a lower grade.  It would been nice to have an extra year in MathCounts and AMC/AIME/USAMO/(IMO?).  Maybe judging by age would be more fair, but then you have people who are just older or just younger than the threshold, so neither way is quite fair.  I suppose grade-skippers just all have that extra challenge to face.

Sadly enough, I don't think I got much out of skipping a grade, since I'm still taking math quite a few grades higher, as most of us are, I would think (who else would be such math geeks? :-P).

In a larger sense, though, these contests don't matter that much, because in the end you're going to have basically the same opportunities in research/business/whatever.

I skipped first grade. I don't suggest it; I think it was a big mistake.  (Although I do have the benefits of not having to take the new SAT's)

Skipping puts you behind socially. In elementary school, other kids don't like the idea of having someone once a grade lower than them in their class. As a result, people who skip are often ostracized, impeding the development of social skills and potentially making their school lives miserable. I got lucky in that I had and have always grown earlier than those my age, so I could fit in physically with the rest of my grade; however, other people aren't as lucky, so shorter stature and an underdeveloped appearance give classmates something else to pick on them with.

Skipping did not make me any less bored in school. It is a matter of learning speed; if they raise your grade level so that the rest of the class is "caught up" with you, but you learn faster than the rest of the class, you are doomed to be bored within a couple of weeks.

It's not really worth it: you only gain one year, which in the long run isn't much, and you lose so much more from your school experience. To add insult to injury, you're still bored too! I don't recommend it at all. Instead, I suggest looking for alternative schooling if possible (schools like AAST if available, which run on an accelerated and more in-depth program). But even if this isn't a possibility, skipping grades is not the solution.

[quote="Nukular"]Skipping did not make me any less bored in school. It is a matter of learning speed; if they raise your grade level so that the rest of the class is "caught up" with you, but you learn faster than the rest of the class, you are doomed to be bored within a couple of weeks. [/quote]

Right.  I went to quite a good school for lower and middle school, and they were kind enough to skip me through 3rd and 5th grade math, but the fundamental problem wasn't that I knew too much, it was that I learned it too fast.  That isn't a problem that skipping grades will really deal with.

Skipping four years of college math solved that problem pretty well for me.

[quote="yif man12"]my teachers told my parents i could have skipped from 4th to 6th, but since i was already the shortest and smallest in my class..... (still am, by the way)[/quote]

lol, I'm the shortest in my grade too.

I never really skipped a grade, but I started skipping math in 1st and english in 2nd.  Now I'm up one grade in english honors.  and just two grades up in math. :(

I've never skipped a grade, but supossedly I'm "accelerated" in math and Spanish, which means that I take math two years ahead of me, and can take Spanish II in 9th grade.

But skipping math doesn't really help me.  Our Algebra class was soooooooooo stupid this year.  I mean our midterm was harder than our final!  Even a person just taking 8th grade math just answer some questions.  All skipping math does for me is make me look really really smart.

But anyone who's bored in math shouldn't worry.  There will always be something to challenge us in AoPS, no matter what our age.  Unless you've like written 1,000 world reknown papers on math by the time your 18.

I skipped in math and got into 'honors' programs otherwise...my mom skipped two grades and said she had too much trouble adjusting and regretted, so I was never encouraged to.  But even skipping in math is a problem sometimes.  I never took algebra II but I picked it up in my extremely slow-paced precal class.  On the other hand, at least 3 people on my school math team never took geometry which is a mistake, especially since one of them never bothered to study it independently either.

Um, I guess I'm just a year ahead of my class since I'm in the advance program...but my school won't let me double up on math courses, which sucks.  My mom...her parents MADE her skip grades.  I think she graduated HS when she was 14/15, and graduated college when she was about 17.  My mom hated my parents because she really liked her HS...and she only had a couple of years.  I like learning stuff independently, however, that only happens when I'm motivated (which isn't often).

I've never skipped a grade, but I am 2-5 grades ahead in math. (depending on where you start from and whether you count not taking AB as a skip: in my area, a considerable percentage of students take Al II in 9th grade, but Al I is 'On-Level.')

I do have a few friends that skipped a grade, though, but they don't seem to have any problems in school.

[quote="beta"]I skipped a grade in China( from 2nd to 4th), not sure if that counts.[/quote]

I think that would count as a grade skip, especially if you are younger than most other people in your grade now in the United States. Is grade skipping common in China? One of my best buddies fifteen years ago was a guy from Hunan who had entered university at age fourteen, I think, and was in Minnesota studying for a graduate degree at that time, I think barely twenty-some years old.

I agree with Nukular, as my grade-skipping experience sounds similar.  Grade skipping sidesteps the issue.

They offered me to skip 2 grades, but my father said no cuz he did when he was small and he had been bullied for it so he said I couldn't

that's been 10 years of boring school since then :(

My school wanted me to skip from 7th to 9th but my dad said absolutly not.  His father did and had always told him it was a mistake.  So I am still in my "age grade"  unless you count my going to take algebra 1 as an 8th grader where it is normally a 9th grade course here.  I luckly have nice teachers who allow me to work ahead and last year allowed me to do the work that the 8th honors were doing even though I was still in the 7th grade class.

I skipped two grades, but in France, it always means for all subjects.
Now, I'm 16 and this year was my first and last IMO, as next year I'll be at college (well... not exactly, but more or less so... something special in France). But this is the only problem... In fact, I think it was a very good thing (except for IMO)... Or I'd still be doing boring math at high school  :D

I skipped 2nd and 6th grades. I don't think it really mattered so much, and I got exposed to lots more math than if I didn't. For example, right now I have finished all the way up to Multivariable Calculus in math classes, which is pretty good for a 14 yr old (I don't know about the ppl here though). I think skipping younger grades or grades where you lived in a "podunk" (thank you Mr. Rucscyzk, you saved my SAT) are perfectly acceptable and actually better for your development.  :D

This is an interesting issue for us, because my son started kindergarten "early" -- that is he barely missed the cutoff, but was allowed to start at "almost 5yo" anyhow.

A few years ago, we moved him to an ungraded private school.  That makes declaring grades all the more complicated. :)

For math competitions and other academic areas, we declare his grade according to the grade he would be in based on when he started Kindergarten, and when he will presumably move to high school (so the grade that is high for his age).   For summer camp where they bunk kids by grade, we also use this grade.

But for sports, we declare his grade according to his birth date (the grade he would have been in if he had started kindergarten at "almost 6yo" according to our state cutoffs).  Since he is not especially strong at sports, this doesn't seem unethical to us.   He's also small for both his age and grade, no matter which grade we declare.

For kids enrolled in normal (graded) schools I don't think there's really any choice as to which grade to declare.   For students in a higher grade for math, though, I think it is usually appropriate to declare the overall grade, and not the math grade.

(I did not skip any grades in school, but did take math 1 year ahead, starting in 8th grade.  In that case I qualified for contests intended for students in my general grade, not my math grade, though in some cases I could do both.)

For Mathcounts, you can't participate before 6th grade, and you can only compete for 3 years.  I specifically contacted MathCounts about how to declare grades for the kids in our ungraded school, and they said to base it not on age as much as on when they will be leaving our ungraded school and entering 9th grade, as we'd basically want them to be eligible for the last 3 years at our school.

For kids who would remain homeschooled or in ungraded situations beyond 8th grade, I suspect they do have rules based on age, but I'm not sure.

Some contests may have rules pertaining to the declaration of grades for homeschoolers -- it is worth checking with them.

Issues of age come up about the IMO, as well.
There were proposals (not implemented)
to limit students to 3 IMO's.
There were also some concerns about countries
where high school can extend to age 19 or grade 13
(e.g. Germany, Canada).
It's also not unknown for strong students
to use their final year of high school as a platform
to train for the IMO or enter the Westinghouse,
rather than going directly to university. In a sense
this is sandbagging, though they probably
learn more than they would at college during that
time.

tokenadult,  you asked about starting college
early in China.  Since there are national entrance
exams for college, one can, in principle,
bypass high school entirely.  I get the impression
that it's not uncommon for people in rural areas
with bad or nonexistent high schools to prepare
directly for the college admission exams.
There's no reason this can't happen at 14-15.

I never really skipped a grade... nobody has in our school... since the heads of the lower, middle, and upper (it's a K-12 private school) are pretty oldfashioned... it takes a whole lot of convincing to put the advanced math students from 7th grade into Hon. Geometry in 8th grade... that's what I did... so I did skip a grade in math...

[quote="tokenadult"]Has anyone reading this thread done a talent search test

http://www.ditd.org/Cybersource/record.aspx?sid=12649&scat=902&stype=110

for example an out-of-level SAT I test? If so, was your test result reported as matching the grade you would be in by age, or matching some other grade? [/quote]

For the talent search programs I know about, students in school are required to take the appropriate test and have their results interpreted according to the grade in which they are registered (to the  potential disadvantage of grade-skipped kids).  Again, I'm not sure of the rules for homeschoolers.

[quote="tokenadult"]Has anyone reading this thread done a talent search test

http://www.ditd.org/Cybersource/record.aspx?sid=12649&scat=902&stype=110

for example an out-of-level SAT I test? If so, was your test result reported as matching the grade you would be in by age, or matching some other grade? [/quote]

I've done the Midwest Talent Search twice, and both times I've gotten two score reports - One was just by everyone who took it, including high-school students (from the test making company), and there was one by the grade that I was in (from the college that ran the talent search).

skipped a grade of math, skipped a grade of english, did a course during the summer b/c the school had some rule about it so that put me two yrs ahead in math, and i started spanish a year early. well, skipping thoses classes helped me not be bored. so that gives me only two years in math, one year in english, one year in spanish. haha, i feel stupid  :blush:

for some reason our school wouldn't let me or my brother go ahead to be taking algebra 1 in 6th grade.  We both did perfectly on this algebra prognosis test and they still wouldn't let us take '8th honors' in 6th grade.... But for another guy they did!... It make no sense and makes me kind of sad, mad, angry

sorry to rub it in, but i would be mad too. I would ask my parents to talk to the teacher, though.

I've skipped 2 grades in math (If you actually count skipping pre-algebra as skipping a grade)(In other words, I was in algebra 1 back in 7th)
Being at TJ, I am currently very fortunate. However, some people in the area had much more opportunity than me (especially those that attended Longfellow MS). Basically, my 1st and 2nd grade educations were a waste, but then I went to a GT center in 3rd. That went pretty well until 6th grade, at which point both GT 6th grade teachers retired, and we got two new one. I can say, without exagerating, that I, as a 6th grader, was far smarter than my math/science teacher. I then went to Irving MS, which was pretty mediocre compared to other schools in fairfax.

There are two things here that annoy me:
1. I was talking to a TJ senior that went to my elementary school, and apparently, up until my year (or possibly the year before that), they let one or two kids per year take algebra 1 in 6th grade. I'm assuming this oppurtunity not being brought to my knowledge had to do with the shifting of teachers.

2. The people that went to Longfellow, as well as a few other schools in the area, had the oppurtunity to get into calculus in 9th grade. I didn't.

Okay, I'm done complaining now  :D

I skipped 5th because when I took the mathcounts test when I was in 4th grade, I could have made the team if I was in 6th grade.

Actually, I have never skipped a grade, but, as I am homeschooled, I was able to do lots of extra work and do two extra math grades in my spare time.


conics
parabola
symmetry
analytic geometry
graphing lines
slope
calculus

OK here's the situation.. (If I'm not giving enough information, please let me know - I wasn't sure)

An object is being projected into the air (ignore air resistance) from height zero at an angle 'theta'.  If the object must land 5 metres ahead (horizontal displacement) and can never reach a height of 2.5 metres or more, what is the maximum angle of launch?

I got an answer, but I don't know if it's right:[hide]tan(theta)<4, so theta<76 degrees?[/hide]

Equation of a parabola (with vertical symmetry axis) passing through the points (0, 0), (2.5, 2.5), (5, 0) is $y = -0.4x^2 + 2x$, as can be calculated by substituting these 3 coordinate pairs into $y = ax^2 + bx + c$ and calculating a, b, c. Slope of the parabola tangent at x is the derivative $y' = -0.8x + 2$ and at x = 0, the slope is $\tan \theta_{\text{max}} = y'(0) = 2,\ \theta_{\text{max}} \doteq 63.5^\circ$.

[quote="Michael Niland"]Another solution using the discriminant of a quadratic polynomial

Relative to X-Y axes through point of projection the equation of the path of the projectile is

Y = X.tanA - g.X^2/{2.V^2.(CosA)^2}............. (1)  where V is speed of projection and A is angle of projection.

The range R is given by R=2.V^2.SinA.CosA/g from which we find that g/2.V^2 =SinA.CosA/R .Now writing this expression for g/2.V^2 into equation1

the path of the particle becomes Y = X.tanA -X^2.TanA/R and if Y cannot exceed a height h then the quadratic inequality

X^2.TanA/R -X.TanA +h >=0 must hold for all values of X and A.This happens iff the discriminant is negative that is if

(TanA)^2 <=4.h.TanA/R or TanA<=4h/R and thus the greatest possible value for TanA is 4h/R.[/quote]I'm sorry but I found your solution hard to read. To make it easier for others, I edited your solution with $\LaTeX$

[quote="Michael Niland"]Another solution using the discriminant of a quadratic polynomial

Relative to $x,y$-axes through point of projection the equation of the path of the projectile is

$y=x\tan A-\frac{gx^2}{2v^2\cos^2A}$............. (1)

where $v$ is speed of projection and $A$ is angle of projection.

The range $R$ is given by $R=\frac{2v^2\sin A\cos A}{g}$ from which we find that $\frac{g}{2v^2}=\frac{\sin A\cos A}{R}$. Now writing this expression for $\frac{g}{2v^2}$ into equation (1)

the path of the particle becomes $y=x\tan A-x^2\frac{\tan A}{R}$ and if $y$ cannot exceed a height $h$ then the quadratic inequality

$x^2\frac{\tan A}{R}-x\tan A+h\ge0$ must hold for all values of $x$ and $A$.This happens iff the discriminant is negative that is if

$\tan^2A\le\frac{4h\tan A}{R}$ or $\tan A\le\frac{4h}{R}$ and thus the greatest possible value for $\tan A$ is $\frac{4h}{R}$.[/quote]


Consider f(n-1) (denoting the sample space). Its last 2 endings maybe

HH
HT
TH
TT

So we can derive f(n) by adding a T to f(n-1) or a H to f(n-1) except where the last 2 digits of f(n-1) are HH.

Since the last 2 digits are HH, the previous digit must be T.
So We must find the number of cases with THH in f(n-1), which turns out to be equal to f(n-4). So converting this to probability...

$ P(N) \equal{} P(N \minus{} 1) \minus{} (1 \minus{} p)p^3P(N \minus{} 4)$

To solve this let $ P(N) \equal{} a^n$

we get $ a^4 \equal{} a^3 \minus{} (1 \minus{} p)p^3$

Edit: Where did you get this sum da?

Whats wrong with the logic? I think I've got all cases??? Give your answer.

Raga: I think I've got the mistake.

Answer is

$ F(n) \equal{} pF(n\minus{}1) \plus{} (1\minus{}p)pF(n\minus{}2) \plus{} p^2(1\minus{}p)F(n\minus{}3)$

Put $ F(n) \equal{} a^n$

$ a^n \equal{} pa^{n\minus{}1} \plus{} p(1\minus{}p)a^{n\minus{}2} \plus{} p^2(1\minus{}p)a^{n\minus{}3}$
$ a^3 \equal{} pa^2 \plus{} p(1\minus{}p)a \plus{} p^2(1\minus{}p)$

which I'm sure you can obviously solve.  :rotfl:  :rotfl:

Here s the final answer:

the required probability is $ (0.75)^{n\plus{}1} (2 \plus{} \alpha^{n\plus{}1} \plus{} \beta^{n\plus{}1})$where $ \alpha$ and $ \beta$ are the roots of $ 3x^{2} \plus{} 2x \plus{} 1 \equal{}0$.


limit
trigonometry
algebra unsolved
algebra

[quote="Amir.S"]find:

$\lim_{n\rightarrow \infty}\sin \frac 1{2}+\sin\frac 1{2^{2}}+\dots+\sin\frac 1{2^{n}}$[/quote]
cotgx-cotg2x=$\frac1{sin2x}$

[quote="lordWings"]Nice! Do you know the source?

At least we're sure that the limit of $s_{n}=\sin \frac 1{2}+\sin\frac 1{2^{2}}+\dots+\sin\frac 1{2^{n}}$ exists,
because $s_{n}$ is increasing and bounded by the known sequence $t_{n}=\frac 1{2}+\frac 1{2^{2}}+\dots+\frac 1{2^{n}}<1$,
since $0<\sin\frac 1{2^{n}}<\frac 1{2^{n}}.$[/quote]

actually I proved the limit exsist as you did.But I couldn't find it.

can anyone find it?

[quote="lordWings"][quote="ali666"]cotgx-cotg2x=$\frac1{sin2x}$[/quote]
Nice! This would be useful if we wanted to evaluate
$\frac 1{\sin 2}+\frac 1{\sin 2^{2}}+\dots+\frac 1{\sin 2^{n}}=\cot 2-\cot 2^{n+1}$,
and in this case I'd say that the limit doesn't exist.[/quote]
 :blush:

Following ali666's idea,
$\sin\alpha=s(2\alpha)-s(\alpha)$,
where
$s(\alpha)=\lim_{n\to\infty}\sin\frac\alpha{2}+\sin\frac\alpha{2^{2}}+\dots+\sin\frac\alpha{2^{n}}$.
The limit always exists, for every $\alpha\in\mathbb{R}$, because the sequence of finite sums, at least after a certain term, is monotonic and bounded.
Can we find a closed expression for $s$?

How does that help?
I'm starting to think that we won't find a closed expression.


The area of a circle circumscribed about a regular hexagon is $ 2\pi$. What is the area of the hexagon?

[hide="Solution"]
Since the area of the circumscribed circle is $ 2\pi$ we notice that the radius of the circle is $ \sqrt{2}$.

A hexagon is made up by 6 equilateral triangles, whose side lengths are equal to the radius of the circumscribed circle.

The area of an equilateral triangle with side length $ s$ is $ \dfrac{s^2\sqrt{3}}{4}$. Therefore the area of the hexagon is,
\[ A\equal{}6\cdot\left(\dfrac{2\sqrt{3}}{4}\right)\implies A\equal{}\boxed{3\sqrt{3}}\]
[/hide]


geometry
circumcircle
algebra unsolved
algebra

let $ z_1,z_2,z_3$ is complex number such that $ |z_1|\equal{}|z_2|\equal{}|z_3|$=R
Prove that:
$ |z_1 \minus{} z_2| . |z_2 \minus{} z_3| \plus{} |z_2 \minus{} z_3| . |z_3 \minus{} z_1| \plus{}|z_3 \minus{} z_1| . |z_1 \minus{} z_2| \leq 9.R^2$

[quote="math man"]let $ z_1,z_2,z_3$ is complex number such that $ |z_1| \equal{} |z_2| \equal{} |z_3|$=R
Prove that:
$ |z_1 \minus{} z_2| . |z_2 \minus{} z_3| \plus{} |z_2 \minus{} z_3| . |z_3 \minus{} z_1| \plus{} |z_3 \minus{} z_1| . |z_1 \minus{} z_2| \leq 9.R^2$[/quote]

Consider triangle $ ABC\ : \ A(z_1)\ ;\ B(z_2)\ ;\ C(z_3)$

$ Inq < \equal{} > ab \plus{} bc \plus{} ca \le 9R^2$

Call $ O$ is circumcenter of $ \triangle ABC$ :

$ 0 \le 9.\vec{OG}^2 \equal{} (\vec{OA} \plus{} \vec{OB} \plus{} \vec{OC})^2 \equal{} 3R^2 \plus{} 2\sum \vec{OA}.\vec{OB} \equal{} 3R^2 \plus{} \sum (OA^2 \plus{} OB^2 \minus{} AB^2) \equal{} 9R^2 \minus{} a^2 \minus{} b^2 \minus{} c^2$

$ \equal{} > ab \plus{} bc \plus{} ca \le a^2 \plus{} b^2 \plus{} c^2 \le 9R^2$


logarithms
Vieta
algebra
binomial theorem
algebra proposed

If $ a,b,c$ are the roots of $ x^3\minus{}x\minus{}1\equal{}0$, compute:

$ \frac{1\minus{}a}{1\plus{}a}\plus{}\frac{1\minus{}b}{1\plus{}b}\plus{}\frac{1\minus{}c}{1\plus{}c}$

[quote="moldovan"]If $ a,b,c$ are the roots of $ x^3 \minus{} x \minus{} 1 \equal{} 0$, compute:

$ \frac {1 \minus{} a}{1 \plus{} a} \plus{} \frac {1 \minus{} b}{1 \plus{} b} \plus{} \frac {1 \minus{} c}{1 \plus{} c}$[/quote]

We have $ 1\equal{}a^3\minus{}a\equal{}b^3\minus{}b\equal{}c^3\minus{}c$ , than use it repeatedly

the sum $ \equal{}\frac{a^3\minus{}2a}{a^3}\plus{}\frac{b^3\minus{}2b}{b^3}\plus{}\frac{c^3\minus{}2c}{c^3}$
$ \equal{}3\minus{}2(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})$
$ \equal{}3\minus{}2(\frac{a^2\minus{}1}{a}\plus{}\frac{b^2\minus{}1}{b}\plus{}\frac{c^2\minus{}1}{c})$
$ \equal{}3\minus{}2(a\plus{}b\plus{}c)\plus{}2(\frac{ab\plus{}bc\plus{}ca}{abc})$
$ \equal{}3\minus{}2(0)\plus{}2(\frac{\minus{}1}{1})$
$ \equal{}1$

$ x^3\minus{}x\minus{}1\equal{}(x\minus{}a)(x\minus{}b)(x\minus{}c)\Longrightarrow \ln |x^3\minus{}x\minus{}1|\equal{}\ln |x\minus{}a|\plus{}\ln |x\minus{}b|\plus{}\ln |x\minus{}c|$

$ \Longrightarrow \frac{1}{x\minus{}a}\plus{}\frac{1}{x\minus{}b}\plus{}\frac{1}{x\minus{}c}\equal{}\frac{3x^2\minus{}1}{x^3\minus{}x\minus{}1}$, yielding $ \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\equal{}2.$

$ \therefore \frac{1\minus{}a}{1\plus{}a}\plus{}\frac{1\minus{}b}{1\plus{}b}\plus{}\frac{1\minus{}c}{1\plus{}c}\equal{}\minus{}1\cdot 3\plus{}2\left(\frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\right)\equal{}\minus{}3\plus{}2\cdot 2\equal{}\boxed{1}.$

1996 Canada, #1:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77670[/url]

[hide="Solution"]$ \frac{1\minus{}a}{1\plus{}a}\equal{}\minus{}1\plus{}\frac{2}{1\plus{}a}$.

$ \sum_{\text{cyc}}\minus{}1\plus{}\frac{2}{1\plus{}a}\equal{}\minus{}3\plus{}2\sum_{\text{cyc}}\frac{1}{1\plus{}a}$.

Let $ f(x)\equal{}x^3\minus{}x\minus{}1$. If $ x\equal{}\frac{1}{1\plus{}a}$, then $ a\equal{}\frac{1}{x}\minus{}1$, so $ \frac{1}{1\plus{}a}$ is a root of $ f\left(\frac{1}{x}\minus{}1\right)$.

It's not hard to see that $ x^3\minus{}2x^2\plus{}3x\minus{}1$ has the same roots as $ f\left(\frac{1}{x}\minus{}1\right)$. (Binomial theorem, then multiply by $ \minus{}x^3$).

From Vieta's, the sum of the roots of $ g$ is $ 2$, so the answer is $ \minus{}3\plus{}2\cdot 2\equal{}4\minus{}3\equal{}\boxed{1}$.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that every subgroup of a cyclic group is cyclic.

Let $ G$ be a cyclic group and $ H$ a subgroup. Pick a generator for $ G$, call it $ g$ then order each element based on powers of multiples (if we write additively) of $ g$. Ie $ g < 2g < 3g...$ etc. This gives an isomorphism to either $ \mathbb{Z}$ or $ \mathbb{Z}/n\mathbb{Z}$ where $ n$ is the order of the group. We will now denote $ g$ as $ 1$. let $ H$ be a subgroup of $ G$, under the ordering, there is a unique smallest element of $ H$, call it $ h$. Suppose now $ h$ does not generate $ H$, then there is an element $ f$ that is not a multiple of $ h$. Now, they must not share a common factor, otherwise it would be in the subgroup $ H$, so $ (h,f) \equal{} 1$. Then by Bezout's identity, you can find multiples of h and f that have a difference of $ 1$ but because $ H$ a group, $ 1$ therefore is in the group which means $ H \equal{} G$. Thus $ h$ generates $ H$. q.e.d.


[quote="kidwithshirt"](Sprint) A best-of-five series ends when one team wins three games. The probability of team A defeating team B in any game is  1/6 . What is the probability that team A will win the series? Express your answer as a fraction in lowest terms.

 (Sprint) We select  6  numbers at random, with replacement, from the set of integers from  1  to  600  inclusive. What is the probability that the product of the  6  numbers is  even  ? Express your answer as a commom fracton in lowest terms.

(Target) Find the probability that four randomly selected points of the geoboard shown will be vertices of a square who sides have integer length. Give your answer as a common fraction. The distance between two adjacent vertical or horizontal dots is 1.
          .  .  .  .  .  .
          .  .  .  .  .  .
          .  .  .  .  .  .
          .  .  .  .  .  .
          .  .  .  .  .  .
          .  .  .  .  .  .

What kind of level are they? I say they are pretty hard, probly national lvl?[/quote]
[hide="1"]A's combos: win three games: AAA
win four games: AABA, ABAA, BAAA
win five games: BBAAA, BABAA, ABBAA, BAABA, ABABA, AABBA
There are $\frac{5!}{3!2!}=10$ ways to get a team A to win, and we have all ten. So adding up the probabilities we get
$\frac{1}{216}+\displaystyle\left(\frac{5}{1296}\right)\times3+\displaystyle\left(\frac{25}{7776}\right)\times6=\boxed{\frac{23}{648}}$
Is that correct?[/hide]
[hide="2"]
We find the probability the product is odd. This is $\displaystyle\left(\frac{1}{2}\right)^6$ and the probability it is even is $1-$ that answer, $\boxed{\frac{63}{64}}$. [/hide]
[hide="3"]I'm getting kicked off the comp so i'm gonna guess $\boxed{55}$[/hide]

I would say national
[hide="number 1"]
Ways that A wins the series
3 wins. $\frac{1}{216}$
3 wins one loss.
The last must be a win, so we can arrange the loss in 3 places.
Thus it is $\frac{15}{6^4}=\frac{5}{432}$
3 wins two losses.
Again, a win must be last, so we can arrange the losses in $\binom 42=6$ ways.
$\frac{150}{6^5}=\frac{25}{6^4}=\frac{25}{1296}$.
Summing all of them, $\frac{46}{1296}=\frac{23}{648}$
[/hide]
[hide="number 2"]
The only way for it to be odd is if all of the numbers are odd.
There is a $\frac 12$ chance for each number, since there is replacement.
$\frac 12^6=\frac{1}{64}$
Thus the answer is $\frac{63}{64}$[/hide]
[hide="number 3"]
There are $25+16+9+4+1=55$ squares.
There are $36$ points, and thus $\binom{36}{4}$ ways to choose the points.
So the answer=$\frac{1}{1171}$ ( I used a calculator to evaluate)[/hide]


Let $ a,b\ge 0$ and $ a\plus{}b\equal{}1$. Prove that:
$ \sqrt{a^2\plus{}b}\plus{}\sqrt{b^2\plus{}a}\plus{}\sqrt{1\plus{}ab}\le  3$.

Easy!

$ \sqrt {a^2 \plus{} b} \plus{} \sqrt {b^2 \plus{} a} \plus{} \sqrt {1 \plus{} ab}$
$ \equal{} \sqrt {3}\sqrt {(a \plus{} b)^2 \plus{} (a \plus{} b) \plus{} 1 \minus{} ab}\leq \sqrt{3}\sqrt{3\minus{}ab}\leq 3$.

my solution uses Cauchy-Schwarz:

$ \sqrt{a^2\plus{}b}\plus{}\sqrt{b^2\plus{}a}\plus{}\sqrt{1\plus{}ab}\leq3$
$ (\sqrt{a^2\plus{}b}\plus{}\sqrt{b^2\plus{}a}\plus{}\sqrt{1\plus{}ab})^2\leq9$
Because of Cauchy-Schwarz-Inequality
$ LHS\leq(1\plus{}1\plus{}1)(a^2\plus{}b\plus{}b^2\plus{}a\plus{}1\plus{}ab)\leq9$ /: 3
$ a^2\plus{}b^2\plus{}a\plus{}b\plus{}ab\plus{}1\leq3$
because $ a\plus{}b\equal{}1$
$ a^2\plus{}ab\plus{}b^2\leq1$ /+ab
$ (a\plus{}b)^2\leq1\plus{}ab$ which is true, as $ a\plus{}b\equal{}1$ and $ a,b\geq0$
Equality holds if $ ab\equal{}0$ and if $ a^2\plus{}b\equal{}b^2\plus{}a\equal{}1\plus{}ab$ , that is if $ a\equal{}1$ and $ b\equal{}0$ or rather if $ a\equal{}0$ and $ b\equal{}1$


I have this code that a professor posed to use in class.  He found it on the internet and has been unable to solve it.  The only clue he gave us was A = 1.

Here it is.  Both parts are decoded seperately:

Part 1: [66-9-13-18-1-19-95-12-3-14-5-12-15-7]
Part 2: [1-12-14-56-18-21-1-23-15-5-20-18-13-18-22-9-25-15-21-15-15-14-4-20]


ratio
complex numbers
geometry proposed
geometry

Let $ABCD$ and $A'B'C'D'$ be two squares in the same plane and they are oriented in the same way. Let $A'',B'',C''$ and $D''$ be the midpoints of $[AA'], [BB'], [CC']$ and $[DD']$. Show that the quadrilateral $A''B''C''D''$ is also a square.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

The small letters denote the complex numbers corresponding to the capital letters.

Assume we have $\frac{c-d}{a-d}=i,\frac{c'-d'}{a'-d'}=i$, so $\frac{c''-d''}{a''-d''}=\frac{\frac{c+c'}2-\frac{d+d'}2}{\frac{a+a'}2-\frac{d+d'}2}=i$, meaning that $A''D''C''$ is a right isosceles triangle. We do the same for the others, and we are done.

I said "assume" in the first sentence because those ratios could also be $-i$. That doesn't matter. What matters is that they are both $i$ or $-i$ at the same time (because the two squares have the same orientation).


f(x) is defined in the positive integer, and satisfies (1) and (2)

(1) if n is a positive integer, than $ f(f(n))\equal{}4n\plus{}\frac{3}{2} (3\plus{}(\minus{}1)^n)$

(2) if n is a positive integer, than $ f(2^n)\equal{}2^{n\plus{}1}\plus{}3$

find f(322).

Thanks.

Perhaps you wanted to find meant find $ f(319)$?

Because we can, with some effort, find that $ f(n) \equal{} 2n \plus{} \frac {3}{2}\cdot\left(( \minus{} 1)^n \plus{} 1\right)$ for $ n$ of the form $ n \equal{} g(k,x) \equal{} 2^{k \plus{} x} \plus{} 2^{k \plus{} 1} \minus{} \frac {( \minus{} 1)^k \plus{} 3}{2}$, for $ x,k\geq 0$, if I am not mistaken.

Furthermore $ f(g(k,x)) \equal{} g(k \plus{} 1,x)$. These relationships can be derived by starting with (2) and applying induction. Applied to finding $ f(319)$, we have $ g(5,3) \equal{} 319$ and thus $ f(319) \equal{} g(6,3) \equal{} 638$.

This can be seen much more easily for finding $ f(319)$ by starting from $ f(2^3)$.

But there are no $ (k,x)$ yielding $ g(k,x) \equal{} 322$! Furthermore, $ f(f(n))$ cannot be a power of $ 2$ for any positive $ n$, rendering (2) useless except as a first step. And furthermore we cannot find $ n$ such that $ f(f(n))$ is $ 322$, because we have a contradiction in the parity of $ n$. This seems to imply that it is impossible to directly derive $ f(n)$ for all $ n$, unless we simply use the formula we found for the special cases. I have not yet looked at other methods extending this formula to arbitrary $ n$, because I am hoping that you simply made a typo...


inequalities
function
pigeonhole principle

Let f(X,Y) be the function $\displaystyle f(X,Y) = \frac{(2a+1)X - Y + aXY + a}{(2+a)Y - aX + XY + 1}$ where $a$ is positive,

and let $S$ be a set of  $n$  positive real numbers, where $n > a + 2$ .

Prove that we can found two  distinct numbers $X,Y \in S$ such that   $\displaystyle 1 < f(X,Y) \leq \frac{a.n+1}{n-2-a}$

 :cool:  Do you like pigeons ? :cool:

Sorry, I made a mistake :blush:  !!  The linequality has to be :

$ a < f(X,Y)  \leq \frac{an+1}{n-2-a}$

 :cool:  Sorry again  :cool:

Well if no one else is going to try it, I'll post a solution, since this thing has been frustrating me for a helluva long time.

The first part is straightforward:
The equation can be factored as follows
$F(X,Y)=\frac{a(X+1)(Y+1)+(a+1)(X-Y)}{(X+1)(Y+1)-(a+1)(X-Y)}$ and for $x>y$, F(X,Y)>0

And now the part thats been bugging me for a fricken long time... :evil:
I only came to this solution after noticing a pattern using the calculator (was trying to disprove it, I was getting a different result on the RHS b/c of a different pigeonholing)...so someone please explain how the solution can be more easily noticed, or show a simpler one.

Now we pigeonhole on $(\frac{0}{n-1},\frac{1}{n-2}],[\frac{1}{n-2},\frac{2}{n-3}]....$ So now there must exist $X,Y \in s$ such that $X-Y \le \frac{k}{n-1-k}-\frac{k-1}{n-k}=\frac{n-1}{(n-k-1)(n-k)}$

Letting $X=\frac{k}{n-1-k}$ and $Y=\frac{k-1}{n-k}$, $(Y+1)(X+1)=\frac{(n-1)^2}{(n-k-1)(n-k)}$

BTW, well see that the maximum value of F(X,Y) is independent of k, so we wont have to worry about minimizing $(Y+1)(X+1)$
So now $F(X,Y)\le\frac{a(n-1)^2+(a+1)(n-1)}{(n-1)^2-(a+1)(n-1)}=\frac{an+1}{n-2-a}$

There has got to be a nicer method, so someone PLEASE POST if u know it...since F(X,Y) has many nice properties like F(x,y)=F(1/y,1/x), however I kept getting stuck working with that...And if there isn't a nicer solution  :( then how do u think of this craziness in the time given for a contest?

Sorry GeoMar, your proof does'nt work, because you have $k < n$ so $X-Y \leq (n-1)/2$, but in fact the minimum of $X-Y$ can be so big I want ..
 :cool:  Try again  :cool:

I dont see the problem still... the smallest X-Y can be as large as you like, that is true... but two values of s will fall in the same interval... for example the intervals for n=6 would be as follows...

$\displaystyle(0,\frac{1}{4}],[\frac{1}{4},\frac{2}{3}],[\frac{2}{3},\frac{3}{2}],[\frac{3}{2},\frac{4}{1}],[\frac{4}{1},\infty)$
so there must exist an X,Y
such that $X-Y\le \frac{k}{n-1-k}-\frac{k-1}{n-k}$ for $k \in(1,5) $  This isnt necessarily the smallest X-Y...
but i guess the RHS can be approach infinity... is that a problem?

i editted out some mistakes

Hi GeoMar, had you got a nice day ? Your method is nice but  I'm still unconvinced !
It's not very clear, but It seems that you use two sentences :
(1)$ 0 < U-V < X-Y  ==> f(U,V) < f(X,Y)$ , and , (2) There are $X,Y,k$ such that $X,Y \in S$ , $1 \leq  k < n-1$ and  $X,Y \in [\frac {k-1}{n-k}, \frac {k}{n-k-1}]$

Am I right ? If yes, you have to prove  them ...

:cool: I will post later my solution .. :cool:

Ahh I see...
I'm sorry for being unclear and incomplete, but im a beginner at this...I'll try to explain a bit better
Also, I said some stuff that didn't make sense in the original post, but I think I got my ideas across

(2) came from that there are n values in s and there are n-1 intervals (for k=1,2,3...n-1) (although k cant =n-1 it can approach it)

And (1) i didn't explain, and I dont even think that it is correct.
What i guess I have to prove is that F(X,Y) is the maximum for any two values in the in the interval for which X,Y were in.  But I'm not sure how this can be done, perhaps setting $x=\frac{k-\alpha}{n-1-k}$ and $Y=\frac{k-\beta}{n-k-1}$ working that out?
but I'm not sure and if I have to do that...it becomes too messy

So I would really like to see another solution, because this just gets yuckier and yuckier.

My general problem is :

Let $f$ be a bijective increasing  function $f : ]a,b[ \rightarrow T$ , where $a < b$ are real numbers and $T$ is  sub-set of  $R$ (real numbers) .

Let $g$ be the inverse of $f$ , $g : T \rightarrow ]a,b[$ and let $S$ a finite sub-set of  $T$ , with $n+1$ elements .

If  $a < u < b$  and  $u + \frac{b-a}{n} < v$ , then there are $X,Y \in S$ such that $f(u) < f(u + g(X)-g(Y)) < f(u + \frac{b-a}{n})$ .

Could you prove it ?  :D  Take$ f:]-a,1[ \rightarrow ]0, \infty [ $such that $ f(x) = \frac{a+x}{1-x} $ :lol:  Here you are ? :lol:

:cool: You can produce a lot of nice problems from the general form :cool:


AMC
AIME
geometry
linear algebra
matrix
LaTeX
rotation

Hi,

I'm going to post AIME problems (or problems in that level) in this thread now and then. Any of you who want to practice with me on this thread can post the answers to the problems. Please do not post any problems though. I like to keep track of the number of problems I did.  :)

[color=blue]Problem 1[/color]

The triangle $ABC$ has $\angle A = 60^\circ$, $\angle B = 45^\circ$. The bisector of $\angle A$ meets $BC$ at $T$ where $AT = 24$. If the area of the triangle $ABC$ is expressed in form $a + b \sqrt{c}$ where $c$ is not divisible by squares of any prime, compute $a+b+c$.

[color=blue]Problem 2[/color]

Find $c$ if $a,b,$ and $c$ are positive integers which satisfy $c = (a+bi)^3 - 107i$, where $i = \sqrt{-1}$.

[color=blue]Problem 3[/color]

Find the value of $a_2+a_4+a_6+ \cdots + a_{98}$ if $a_1,a_2,a_3 \cdots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+ \cdots + a_{98} = 137$.

[color=blue]Problem 4[/color]

The integer $n$ is the smallest positive multiple of 15 such that every digit is either 0 or 8. Compute $\frac{n}{15}$.

[color=blue]Problem 5[/color]

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

[color=blue]Problem 6[/color]

Let $f(x) = |x-p| + |x-15| + |x-p-15|$, where $0 < p < 15$. Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x \leq 15$.

[color=blue]Problem 7[/color]

What is the largest 2-digit prime factor of the integer $n = \binom{200}{100}$.

[color=blue]Problem 8[/color]

The numbers 1447, 1005, and 1231 have something in common: each is a 4-digit number beginning with 1 that has exactly two identical digits. How many such numbers are there?

[color=blue]Problem 9[/color]

A point $P$ is chosen in the interior of $\Delta ABC$ so that when lines are drawn through $P$ parallel to the sides of $\Delta ABC$, the resulting smaller triangles have areas 4,9, and 49, respectively. Find the area of $\Delta ABC$.

[color=blue]Problem 10[/color]

When a right triangle is rotated about one leg, the volume of the cone produced is $800 \pi cm^3$. When the triangle is rotated about the other leg, the volume of the cone produced is $1920 \pi cm^3$. What is the length (in $cm$) of the hypotenuse of the triangle?

[color=blue]Problem 11[/color]

What is the sum of the solutions of the equation

$ \sqrt[4] x = \frac{12}{7 -  \sqrt[4] x}$?

[color=blue]Problem 12[/color]

Evaluate the product

$(\sqrt 5 + \sqrt 6 + \sqrt 7)(\sqrt 5 + \sqrt 6 - \sqrt 7)(\sqrt 5 - \sqrt 6 + \sqrt 7)(- \sqrt 5 + \sqrt 6 + \sqrt 7)$

[color=blue]Problem 13[/color]

If $\tan x + \tan y = 25$ and $\cot x + \cot y = 30$, what is $\tan(x+y)$?

[color=blue]Problem 14[/color]

Determine $3x_4+2x_5$, if $x_1,x_2,x_3,x_4,$ and $x_5$ satisfy the system of equations given below:

\[2x_1+x_2+x_3+x_4+x_5 = 6\\
x_1+2x_2+x_3+x_4+x_5 = 12\\
x_1+x_2+2x_3+x_4+x_5 = 24\\
x_1+x_2+x_3+2x_4+x_5 = 48\\
x_1+x_2+x_3+x_4+2x_5 = 96\]

[color=blue]Problem 15[/color]

What is the largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?

[color=blue]Problem 16[/color]

The pages of a book are numbered 1 through $n$. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in the incorrect sum of 1986. What was the number of the page that was added twice?

[color=blue]Problem 17[/color]

One commercially available ten button lock may be opened by depressing - in any order - the correct five buttons. The sample shown at right has {1,2,3,6,9} as its combination. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?

[color=blue]Problem 18[/color]

In $\Delta ABC$, $\tan(\angle CAB) = \frac{22}{7}$ and the altitude from $A$ divides $BC$ into segments of lengths 3 and 17. What is the area of $\Delta ABC$?

[color=blue]Problem 19[/color]

Find $x^2+y^2$ if $x$ and $y$ are positive integers such that

\[xy+x+y = 71\\
x^2y+xy^2 = 880\]

[color=blue]Problem 20[/color]

For how many positive integers $n < 1000$ such that $\lfloor {\log_2 n} \rfloor$ is positive and even? As usual, $\lfloor {z} \rfloor$ denotes the greatest integer function of $z$.

[color=blue]Problem 21[/color]

The square below is magic. It has a number in each cell. The sums of each row and column and of the two main diagonals are all equal. Compute $x$. ($*$ are places with no number or variable defined in the problem)

\[x\ 19\ 96\\
1\ *\ *\\
*\ *\ \ *\]

[color=blue]Problem 22[/color]

Find the value of $m$ if $m = \sqrt{1 + 28 \cdot 29 \cdot 30 \cdot 31}$.

[color=blue]Problem 23[/color]

10 points lie on a circumference of the circle. How many distinct convex polygons can be formed by connecting some or all of the points?

[color=blue]Problem 24[/color]

For some digit $d$, we have $0.{\overline{d25}} = \frac{n}{810}$, where $n$ is a positive integer. Find $n$.

[color=blue]Problem 25[/color]

We have five consecutive positive integers whose sum is cube and sum of the three middle is a square. What is the smallest possible middle integer?

[color=blue]Problem 26[/color]

The function $f$ holds $f(x) + f(x-1) = x^2$ for all $x$ values. Given that $f(19) = 94$, find the remainder when $f(94)$ is divided by 1000.

[color=blue]Problem 27[/color]

$k$ is a unique integer such that:

$\lfloor {\log_2 1} \rfloor + \lfloor {\log_2 2} \rfloor + \lfloor {\log_2 3} \rfloor + \cdots \lfloor {\log_2 k} \rfloor = 1994$. Find $k$.

[color=blue]Problem 28[/color]

The roots of $x^3+3x^2+4x-11 = 0$ are $a,b,c$. Let's say there exists another cubic polynomial with roots $a+b, b+c, c+a$. This polynomial is in form $x^3+rx^2+sx+t = 0$. What is $t$?

[color=blue]Problem 29[/color]

Starting with a unit square, a sequence of square is generated. Each square in the sequence has half the side-length of its predecessor and two of its sides bisected by its predecessor's sides as shown. Find the total area enclosed by the first five squares in the sequence.

For image, please click:

http://www.kalva.demon.co.uk/aime/aime95.html

(Do not look at the answer!)

[color=blue]Problem 30[/color]

Assume that $a,b,c,$ and $d$ are positive integers such that $a^5 = b^4, c^3 = d^2,$ and $c-a = 19$. Determine $d-b$.

Cool.

Extend altitude from C to AB, and a perpendicular from T to E on AB.

So, $TE = 12$. $AE = 12\sqrt{3}$.

Isosceles triangles, 30-60-90, 45-90-45... WHOA.

Anyways, $CF = 12\sqrt{3}$.

So, 291, after calculations. Darn if my calculations are wrong...

This is an AIME problem?...

Whoaness...

This one's just here for practice. Don't look if you don't want to:

[hide]Problem:

Here's my problem. (AIME)

In triangle ABC, a point P exists such that the triangles formed by drawing parallel lines through P to each of the sides are of areas 4, 9, and 49.

What is [ABC]?[/hide]

[hide="Answer"]Bisecting $\angle A$ gives two $30^\circ$ angles. Draw the perpendicular from $T$ to $AB$ to get a 30-60-90 triangle and an isosceles right triangle, labeling the intersection as point $D$, so we have $TD=12$ and $AD=12\sqrt{3}$, and we see that $AB=12+12\sqrt{3}$. Now, draw the perpendicular from $C$ to $AB$, labeling the intersection as point $E$, and we have 30-60-90 $\triangle ACE$, with $CE=12\sqrt{3}$ and $AE=12$. Therefore, the area of $\triangle ABC=\frac{1}{2}(12)(12\sqrt{3})+\frac{1}{2}(12\sqrt{3})^2=216+72\sqrt{3}=a+b\sqrt{c}$, so $a+b+c=291$.[/hide]

[quote="Silverfalcon"]
[color=blue]Problem 1[/color]

The triangle $ABC$ has $\angle A = 60^\circ$, $\angle B = 45^\circ$. The bisector of $\angle A$ meets $BC$ at $T$ where $AT = 24$. If the area of the triangle $ABC$ is expressed in form $a + b \sqrt{c}$ where $c$ is not divisible by squares of any prime, compute $a+b+c$.
[/quote]

My solution:

[hide]Note that $\angle ACB = 180^\circ - 60^\circ - 45^\circ = 75^\circ$.  This means that the triangle $ACT$ has interior angles $\angle TAC = 60^\circ$, $\angle ACT = 75^\circ$, and $\angle ATC = 75^\circ$.  $ATC$ is therefore an isosceles triangle, with $AT = AC = 24$.

Now apply the Law of Sines to find

$AB = \frac{24 \sin 75^\circ}{\sin 45^\circ}$

$AB = 24\sqrt{2}\left(\frac{\sqrt{3} + 1}{2\sqrt{2}}\right)$

$AB = 12(\sqrt{3} + 1)$

So we now have

$[ABC] = \frac{1}{2} (AB)(AC) (\sin \angle BAC)$

$ = \frac{1}{2}(24)(12)(\sqrt{3} + 1)(\sin 60^\circ)$

$ = 216 + 72\sqrt{3}$

Therefore,

$a + b + c = 216 + 72 + 3 = \fbox{291}$

$QED$[/hide]

[quote="Silverfalcon"]
[color=blue]Problem 2[/color]

Find $c$ if $a,b,$ and $c$ are positive integers which satisfy $c = (a+bi)^3 - 107i$, where $i = \sqrt{-1}$.[/quote]

My solution:

[hide]
Expanding the cube and simplifying gives

$c = a^3 - 3ab^2 + (3a^2b - b^3 - 107)i$

If $c$ is to be a real number, we must have

$3a^2b - b^3 - 107 = 0$

Rearranging and factoring gives

$b(3a^2 - b^2) = 107$

Noting that $a,b$ are positive integers and $107$ is prime, we must have either

$3a^2 - b^2 = 107$
$b = 1$

or

$b = 107$
$3a^2 - b^2 = 1$

The second set of equations must be discarded, as they yield a non-integer $a$.  The first set of equations gives $(a,b) = (\pm 6, 1)$.  $a$ must be positive, so we have $(a,b) = (6,1)$.  Remembering that we have

$c = a(a^2 - 3b^2)$

we can now plug in our solution for $(a,b)$ to obtain

$c = (6)(33) = \fbox{198}$

$QED$[/hide]

shouldnt it be $i = \sqrt{-1}$, not $i=-1$. That might be a little different after that then.

I did problem 2 somewhere before, recently...

[hide="Answer"]$c=(a+bi)^3-107i \Rightarrow (a+bi)^3=a^3-3ab^2+3a^2bi-b^3i$
$3a^2bi-b^3i=107i \Rightarrow b(3a^2-b^2)=107 \Rightarrow a=6$, $b=1$
Therefore, we have $c=216-18+108i-i-107i=216-18=198$.[/hide]

[quote="zanttrang"][quote="Silverfalcon"]
where $i = -1$.[/quote]

Oh, and I believe you meant $i = \sqrt{-1}$ ;)[/quote]

Yes. I edited now. I think I made typo when I was writing the original wording of the problem (probably forgot the square root part)

3 more problems..  :P

(I'll try to add new problems every day... Hopefully at that rate, I'll be able to cover most of the AIMEs by this summer.  :D )

Problem 4:

[hide]To be divisible by three, the number must have a multiple of three eights. As in, three eights, six eights, and so on. And to be divisible by 5, it must end in a zero. Thus the number is 8880 when divided by 15 yields 592. Yeesh, that was like a MATHCOUNTS countdown round problem.[/hide]

On #3, I'm assuming that common difference means that $a_1+1=a_2$, $a_2+1=a_3$, etc.

[hide="3"]$a_1+(a_1+1)+(a_1+2)+...+(a_1+97)=137 \Rightarrow 98\cdot a_1+\frac{97(98)}{2}=137 \Rightarrow 98\cdot a_1=-4616 \Rightarrow a_1=-\frac{2308}{49}$
$a_2+a_4+...+a_{98}=49\cdot a_1+49^2=-2308+2401=93$[/hide]

[hide="4"]The sum of the digits will be divisible by $3$, and will end in either a $5$ or a $0$. Since all digits are either $0$ or $8$, the number obviously ends in a $0$, and must include 3 $8$'s to sum to a multiple of $3$. Therefore, the smallest possible multiple of $15$ consisting only of $8$'s and $0$'s is $8880$, and $\frac{8880}{15}=592$.[/hide]

[hide="5"]All odd numbers can be found by taking the difference of consecutive squares, so we have eliminated the numbers $1,3,5,...,999$. Next, if we try taking the difference between $(n+2)^2$ and $n^2$, we get $4n+4$, so this set will include all multiples of $4$. $(n+3)^2-n^2=6n+9$, which consists only of odd numbers, and we have already eliminated all odd numbers. Similarly, $(n+4)^2-n^2=8n+16$, which is multiples of $8$, all of which have been covered by the multiples of $4$. Therefore, the only numbers that cannot be expressed as the difference between two perfect squares are numbers in the form $2k$, where $k$ is any odd integer, so there are $250$ numbers of this kind in the given interval, and $750$ can be expressed in this way.

This also proves that, with $n$ numbers in a given set, $\frac{n}{4}$ of those numbers cannot be expressed as the difference between two perfect squares.[/hide]

[b]EDIT:[/b] [quote="sirorange"]Yeesh, that was like a MATHCOUNTS countdown round problem.[/quote]

I'd guess it was #1 or #2, possibly on one of the easier AIME's.

I think you mean #3. ;) That's a nice proof, but I have one question: how do you get that $2N-49=a_1+a_2+...+a_{98}$?

[quote="JesusFreak197"]On #3, I'm assuming that common difference means that $a_1+1=a_2$, $a_2+1=a_3$, etc.

[hide="3"]$a_1+(a_1+1)+(a_1+2)+...+(a_1+97)=137 \Rightarrow 98\cdot a_1+\frac{97(98)}{2}=137 \Rightarrow$
$98(a_1+\frac{97}{2})=137 \Rightarrow a_1=-47$

From here, we can see that $a_{48}=0$, so the sum $a_1+a_2+a_3+...+a_{95}=0$, and $a_{96}+a_{98}=48+50=98$[/hide]

[/quote]

[hide="problem with your solution"]You made a calculation error.  If you look back over your equations, you'll find that $48a_1 = -4616$.  So, your method no longer works out nicely.  We can proceed by simple brute force:

$a_2 + a_4 + ... + a_{98} = \frac{98a_1}{2} + 1 + 3 + ... + 97$

$= \frac{-4616}{2} + \frac{(98)(98)}{(2)(2)}$

$=-2308 + 2401$

$= \fbox{93}$[/hide]

[quote="themonster"][hide="4"]
Let $N=a_2+a_4+...+a_{98}$.  Then $2N-49=a_1+a_2+a_3+...a_{98}=137$ $\Longrightarrow$ $\boxed{N=98}$
[/hide][/quote]

[hide="problem with your solution"]Check your math and you'll find that actually, $N = \fbox{93}$ ;)[/hide]

I have a question on 23
[quote]Problem 23

10 points lie on a circumference of the circle. How many distinct convex polygons can be formed by connecting some or all of the points?
[/quote]

I originally thought the answer was (10,3)+....+(10,10) also but if you label the points going in order 1,2,3...10
then would connecting 1,2,3 be the same polygon as 2,3,4?
It says how many DISTINCT

[quote="JesusFreak197"]How would connecting $(1,2,3)$ be the same as connecting $(2,3,4)$ unless they are evenly spaced? It never says that they are evenly spaced.[/quote]It doesn't say they aren't evenly spaced.  When I did this question I got it but in a lot of problems, it does assume they are evenly spaced. (Which would make the problem harder, too.)

O really? I've always thought that if it doesn't specify, it's NOT evenly spaced. Especially if its in a polygon, cuz regular polygons are so easy to work with.

It's actually not too hard if they're all evenly spaced. Once you figure out a pattern, you can apply it to every polygon from 3 sides to 10 sides. (Although, there's only one 10-sided polygon in there anyway...)

Since this was #1, I assumed that the points are evenly spaced.

Well, what was the answer? I don't think my solution is assuming that the points are evenly spaced, but it seems to be the simplest option...

What's the answer to 26? Me and Jesus are disagreeing :D

Oh, I'm an idiot. I forgot that it's missing every other odd number. :roll:

So, mine should have come out the exact same as krusty's. It is now fixed. :)

It is evenly spaced
but
the real AIME specified that having different vertices constitutes a different polygon
(It was a little NOTE under the problem)

[quote="amirhtlusa"]$ \displaystyle (\sqrt 5 \plus{} \sqrt 6 \plus{} \sqrt 7)(\sqrt 5 \plus{} \sqrt 6 \minus{} \sqrt 7)(\sqrt 5 \minus{} \sqrt 6 \plus{} \sqrt 7)( \minus{} \sqrt 5 \plus{} \sqrt 6 \plus{} \sqrt 7)$

if i recall correctly:
$ (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)$ simplifies to:
$ \minus{} a^4 \plus{} 2a^2b^2 \minus{} b^4 \plus{} 2b^2c^2 \minus{} c^4\Rightarrow$
the answer is :
$ \boxed{34}$[/quote]

This question seemed odd, to me. I just multiplied out all the brackets and simplified it. The first pair of brackets leads to $ 4\plus{}2\sqrt{30}$ while the second pair leads to $ \minus{}4\plus{}2\sqrt{30}$, which means that the final result is $ 4 \times 30 \minus{} 16 \equal{} 104$.

If you have something new to add, you can.


function
Euler
calculus
calculus computations

Does anybody know where I can download a computer program that will repeat Euler's method over and over and over again? It'd be even better if the program produced a graph.

Thanks.  :)

I believe theres a computer forum somewhere on mathlinks.

The TI-86 and TI-89 both have such a program built in. They also produce piecewise-linear graphs of the solution.

I have a program written for TI-83/84's that does Euler's Method, but it would take me a while to type it up. You may be able to find a similar one on the internet.

We're talking about Euler's method for approximating the solution to an ODE, right?

Out of curiosity, what is this needed for?

How long do you think it would take to program it?

Maple can do it also, so I assume Mathematica should be able to, and if Matlab can't, you could write a program for that.

[quote="Aunt Sally"]How long do you think it would take to program it?[/quote]

Shouldn't take very long; I had to implement this in Maple on one of my finals awhile back.

Assume you're solving $y'=g(x,y);\,y(x_0)=y_0$ on the interval $[x_0,x_1].$ Let's suppose we want to use $n$ steps in the interval.

Here's a version of the program that's almost (but not quite) in BASIC:

[code](Do some stuff to set up a plotting window.)
(Input n, x0, x1, y0.)
(If appropriate, write a subroutine/external function to compute g(x,y).)

LET dx=(x1-x0)/n
LET x=x0
LET y=y0

FOR j=1 to n
     PLOT (x,y);  (leave the "pen" down)
     LET dy = g(x,y)*dx
     LET x = x + dx
     LET y = y + dy
NEXT j

(pick up "pen")
END[/code]
If you don't write a subroutine to compute $g(x,y),$ you can just put that formula in line in the appropriate place.

The plotting details (including the commands for leaving the "pen" down and picking it up) will vary from language to language and implementation to implementation; so also the preliminary details of setting up the plotting window, and whether you want additional amenities, like axes and tick marks.

The thing is, as soon as you have this, you can improve upon Euler's method - after all, it has only first order accuracy, which is very poor for numerical purposes. The "improved Euler" or second order Runge-Kutta method requires only one additional line:

[code]FOR j=1 to n
     PLOT (x,y);  (leave the "pen" down)
     LET k1 = g(x,y)
     LET k2 = g(x+dx,y+k1*dx)
     LET x = x + dx
     LET y = y + (k1+k2)*dx/2
NEXT j[/code]
And the fourth-order Runge-Kutta method is only slightly more complicated than that:

[code]FOR j=1 to n
     PLOT (x,y);  (leave the "pen" down)
     LET k1 = g(x,y)
     LET k2 = g(x+dx/2,y+(k1*dx)/2)
     LET k3 = g(x+dx/2,y+(k2*dx)/2)
     LET k4 = g(x+dx,y+k3*dx)
     LET x = x + dx
     LET y = y + (k1+2*k2+2*k3+k4)*dx/6
NEXT j[/code]


ΚΑΛΗΣΠΕΡΑ ΣΕ ΟΛΟΥΣ
ΝΕΟΤΑΤΟ ΜΕΛΟΣ ΚΑΙ ΜΑΛΛΟΝ ΛΙΓΟ ΤΣΑΚΩΜΕΝΟΣ ΜΕ ΤΑ ΜΑΘΗΜΑΤΙΚΑ......
ΑΛΛΑ ΟΛΟ ΚΑΙ ΠΡΟΣΠΑΘΩ.ΚΑΤΙ ΕΙΝΑΙ ΚΑΙ ΑΥΤΟ!
ΘΑ ΗΘΕΛΑ ΑΝ ΜΠΟΡΕΙΤΕ ΝΑ ΜΟΥ ΠΕΙΤΕ ΤΗ ΛΥΣΗ ΤΗΣ ΠΑΡΑΚΑΤΩ ΑΣΚΗΣΗΣ.

10 επιβάτες που θεωρούνται διακεκριμένοι, βρίσκονται μέσα σε ένα λεωφορείο και απομένουν 3 στάσεις (ΣτΑ, ΣτΒ, ΣτΓ)

Α. Με πόσους τρόπους μπορούν να αποβιβαστούν όλοι οι επιβάτες από το λεωφορείο με δεδομένο ότι σε κάθε στάση μπορούν να αποβιβάζονται από κανένας έως και 10 επιβάτες;
Β. Με πόσους τρόπους μπορούν να αποβιβαστούν όλοι από το λεωφορείο αν πρέπει να κατέβουν από 4 σε δύο οποιεσδήποτε στάσεις και 2 στην άλλη;
Γ. Με πόσους τρόπους μπορούν να αποβιβαστούν όλοι από το λεωφορείο αν πρέπει να κατέβουν 4 στην ΣτΑ ή 4 στην ΣτΒ (όχι όμως από 4 και στις δύο ταυτόχρονα);

Δεν θα ληφθεί υπόψη η σειρά αποβίβασης των επιβατών σε κάθε στάση

ΣΑΣ ΕΥΧΑΡΙΣΤΩ :lol:

Λοιπόν, η άσκηση σου δεν είναι ιδιαίτερα δύσκολη, αλλά μάλλον βαρετή (και ιδιαίτερα πολύπλοκη αν δε ξέρεις συνδυαστική).  Αρχικά μπορείς να παρατηρήσεις ότι για κάθε τρόπο με τον οποίο βγαίνουν οι επιβάτες (όπως 4 στη ΣτΑ, 4 στη ΣτΒ, 2 στη ΣτΓ), αν δημιουργήσεις πίνακα και εφαρμόσεις τη θεμελιώδη αρχή της συνδυαστικής απαρίθμησης, θα βρεις 10!
Αυτό ήταν το Β ερώτημα.
Για το Α ερώτημα αρκεί να βρεις το πλήθος τρόπων που μπορούν να βγουν οι επιβάτες που ισούται με το πλήθος των λύσεων της διοφαντικής εξίσωσης $ x1 \plus{} x2 \plus{} x3 \equal{} 10$ με $ x1, x2, x3$ φυσικούς, δηλαδή 10 ανά 2 δηλαδή $ \frac{10!}{2!(10\minus{}2)!}$ . Πολλαπλασιάζοντας το πλήθος των τρόπων με τις περιπτώσεις σε κάθε τρόπο βρίσκεις $ \frac{(10!)^{2}}{2!(10\minus{}2)!}$, που είναι η απάντηση στο ερώτημα Α.
Το ερώτημα Γ λύνεται με την αρχή της συμπερίληψης-εξαίρεσης και την εφαρμογή των παραπάνω με ένα παραλαγμένο τρόπο. Το πλήθος των τρόπων είναι 10 ανά 2 πλην (10-4) ανά 1 και όλο αυτό επί 10!, δηλαδή με $ 10!(\frac{10!}{2!(10\minus{}2)!}\minus{}{\frac{(10\minus{}4)!}{1!(10\minus{}4\minus{}1)!}})$

Αν έχεις οποιαδήποτε απορία στείλε μου τη σε αυτή τη σελίδα.

Όσο πιο πολύ σκέφτομαι αυτή την άσκηση, τόσο πιο πολύ μου αρέσει. Δε θέλει ιδιαίτερες γνώσεις αλλά είναι κομψή περίληψη των κυριότερων στοιχείων της συνδυαστικής.