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MIT
college
algebra
system of equations

[quote="chemrock"][b][i]RESULTS ARE ANNOUNCED .
GOT SELECTED IN CHEMISTRY BUT SAD THAT I COULDN'T MAKE THE CUT IN PHYSICS
[/i][/b][/quote]

Me too in chemistry  :lol:   :D
Was not expecting in Physics though

got selected in chem. dint write physics( wouldnt have got selected neways). who else has got selected?

hey congrats 2 all of u.. i made it in phy.. couldnt clear d cut for chem

is there any prescribed syllabus... more like it.. topics...  which come for INPho...
wt kind of a paper is it???

and can we get hands on previous year papers by any how???

best of luck !!!

I am selected for physics but couldn't make it for chem.
Does anybody how the paper of inho is like how many question?
Anybody selected in bothe chem and physics?
what was the likely cutoff for chem?

Didn't get through both.: ( Thought I would clear atleast physics. Some of my friends got through and seemingly none of them expected more than 135.. Either they were lying or that was really the cut-off in which case, I dunno. Didn't attempt astronomy.

Oh! Well... Congrats to everybody who got through.

Edit: Can those who got through NSEP tell me how much they were expecting?

Pathetic results.. Din make both.. Am dead disappointed.. I donno how, but still seemingly an easy 150+ dusn seem to b enuf in phy.. Shre, I still think our papers were screwed somewhere in the middle..  :(  It jus cant get worse than this.. The future has something more in store.
Congrats fer all those who got thru in the olympiads...
n baby, good job.. Well ensured that only Andy students made it to Incho from ps..  :rotfl:

MY NAME IS ROHIT KANNAN :bruce:

edit
P.S. Pardesi see the pm i sent you  :wink:
Is it all right now Shreyas?

[quote="Cyber-Shreyas"][quote="pinnaka"]is there going to be a practical part in inpho?[/quote]

INPHO is only a written round, pure subjective round. http://en.wikipedia.org/wiki/Indian_National_Physics_Olympiad

However there is a practical round for the top notch among Inpho. Can you please tell me how much you got?

@Madness: Please delete that. its not good to reveal personalized info about others when they themselves don't want to reveal it.[/quote]

the marks are not yet revel.. but I have done very well in the subjective part And I expetcted to score about 180-195. by the way the paper was easy

[quote="madness"][quote="pardesi"]got through astronomy.don't know hwy couldn't get through physics...the cutoffs don't seem more than 120.
two friends of mine got through all three  :o[/quote]

Is one of your friends Abhishek Kar???[/quote]
yes Abhishek Kar and Sandeep Samal
i also know Varun and vivek(only by their names) got through NSEC and NSEP
YEs i agree  with GV the results are pathetic...those who expected good(after seeing the officail solutions [url=http://www.mathlinks.ro/viewtopic.php?t=179067]here[/url] didn't get through  :huh:
all of us here expecting 130+ didn't get through

[quote="thebadri"]but this year 12th standards are allowed right? some brochure from tifr said that india would be participating in the IOAA and invited registrations.. any idea about selection? considering we only wrote the astronomy senior paper, would the next round also be common for astronomy senior and astrophysics?[/quote]

I am not very sure...but our team leader had told us that we will give the same INAO (senior) paper + some other probs...
(just a speculation... anyways whatever the paper be, you will have to write it :lol: )

But the camp will be different in the sense that lectures for both the olympiads will be different although accomodation may be the same...
(IIT bombay hostel this year)

12th class people are only eligible for IOAA..
11th class people have a choice between IAO and IOAA.
(though the choice might end the year India stops going to IAO which is very likely in the near future)

ha. thanks for the information.. i presume they'd select 30 odd..? because the IOAA website had said only 12th graduates were allowed to write it..

No no...
there were also people who were 13 years old ...(although they dint win medals)

the age limit is 20 years and any one below the age not entered into university is eligible (not necessarily 12th graduates)

[quote="pradeepmathias"]

I know a guy expecting 140 in phy obj and 200+ in chem... didn't get thru. If you ask me, his papers weren't evaluated properly...!

And I was expecting 169 in chem...

[/quote]

i know a guy who was expecting 150+ in phy and dint make it. but he expected much lesser in chem and he got in. another guy was expecting 195( official solutions) in chem dint make it.
[quote="pardesi"]

one thing their mechanism what so ever was this time was a complete failure.
it failed like hell
[/quote]
pardesi, you're rite. they took so long to give the results and they get it all wrong :mad:
btw do we get the marks in the result sheets, at least we'll get to know then whether it was a mistake on their part or ours

varunarasu
Yes we do get the marks in the sheet,But they(marks)  have to be wrong otherwise they(those who corrected) will  be wrong :wink:

how many of u guys have recieved your selection letters for the inos
i have not yet recieved my letter yet
they mentioned that a prize in the form of a book is also gonna be awarded to the nse clearants
anybody who got through nsec please report about the book and letter

[quote="pardesi"]well got my astro letter...no books :D
books only for top 25 people in the entire process(NSE,INO,I OTC)[/quote]

In physics and chem, top 1% of nation get books no?

Top 1% is the whole lot  :D   :D   :D
But my friends and me didnt get books

no i don't think so ....then i should have got it last time in NSEA ...i was almost perfect :D

Paradesi ku chinna vayasu laye semma talent  :D  :D  :D

Treat fer crossing 1600 posts  :D   :D
And btw you have to cross 2500 posts for becomming A Birch and Swinnerton Dyer  :o

ennu ore rendu AITS result vandha podhum


We have a probability game, where we have $ N$ number of events, each of which outcome can be $ A,B$ or $ C$. We do/will NOT know real probabilities afterwards: only the discrete outcome ($ A, B$ or $ C$) of each event.

Player 1 forecasts these events with certain probabilities (not only guess what is the outcome, but gives probabilities for each outcome option), and Player 2 as well with own probability estimates.

How we can know how accurate Player 1 and Player 2's predictions were (relation to reality) and how to measure the accuracy?

I heard that one can start solving the problem by modeling the process by multinomic distribution and takes Dirichlet's distribution as a priory. Can anyone say how this approach works?

I also heard that Akaike information criterion solves the problem.


inequalities
trigonometry
geometry unsolved
geometry

Show that for all cute triangle $ABC$ (ie.. $A,B,C<\pi/2$) then
$\sum \sin B\sin C+4\sum\sin^{2}B/2\sin^{2}C/2\leq 4\prod \sin A/2$


counting
distinguishability
AMC
AIME
complementary counting
AIME II

[quote="007math"]Since you can distinguish students, won't you have to multiply the answer by $ 3!$ to get a total of $ 120$ ways?

I'm guessing there is a way to do this without listing, and maybe using complementary counting.  Can someone post a solution this way?[/quote]

But this problem doesn't address that you can distinguish students.
I just take it as the total number of ways that three bodies can occupy
three seats out of eight seats, such that none of them are seated next
to each other.

For all I know they could be placing identical triplets (same clothing, etc.)
without labels on them into three of the eight seats.

If the question had stated three specific names of the students, or
stated that the students were distinguishable from each other, I
would readily see your different solution instead of mine.

Such as this:

"Of the 8 chairs in a row, students A, B, and C occupy three of them.
If none of them sit next to each other, in how many ways can this
be done?"

I think the phrasing of the question makes it partly ambiguous.

Well, my logic was that students are people, and every person is different, therefore making every student different.

I am known for over thinking the simplest of questions though...

[hide="Complementary Counting?"]

*Before I begin, note that I originally interpreted the students as distinct so different arrangements of the same setup will be counted.

There are really only a couple ways where the students can violate the rule: when all three of the students are together or two of the students are together.

[b]Case[/b]: all of the students are seated next to each other.

If we use this to notation to represent all eight seats: _ _ _ _ _ _ _ _

then we can consider the three students ***, as one block or _ _ _  *** _ _

Basically, there are 6 different ways to move this 'block' around and hence

6 * 3! = 36 different positions for the students.

[b]Case 2[/b]: Two of the students are seated next to each other.

Subcase 2a: the two seated students are along the edge so: ** _ _ _ * _ _

is one possible configuration. There are 2 * 5 = 10 ways to get the 3 students to line up and 6 ways to select the order so 60 ways.

Subcase 2b: The two seated students are along the interior so : _ _ ** _ * _ _

There are 5 ways to get the two students to do this and 4 spots for the third person and 6 ways to order them so

5*4*6 = 120.

Now we take the total which is $ 6\times \dbinom{8}{3}$ and subtract our 216 from it. This ends up to be 336 - 216 =[b]120 [/b]

 [/hide]

[quote="Arrange your tan"][quote="mr. math"]Of the 8 chairs in a row, students occupy three of them. If none of them sit next to each other, in how many ways can this be done?[/quote]

I don't have a relatively sophisticated method using combinatorics formulas.

Let x be a student.

..................# of positions for the right-most student

x_x_x _ _ _  (4)
x_ _x_x_ _   (3)
x_ _ _x_x _  (2)
x_ _ _ _x_x  (1)

_x_x_x_ _  (3)
_x_ _x_x_  (2)
_x_ _ _x_x  (1)

_ _x_x_x_ (2)
_ _x_ _x_x (1)

_ _ _x_x_x  (1)

(4 + 3 + 2 + 1) + (3 + 2 + 1) + (2 + 1) + (1) =
10 + 6 + 3 + 1 =
20

$ 20$ ways[/quote]

Isn't is $ 20 \cdot 3!$ since there are $ 3!$ ways to arrange them? I got $ 20 \cdot 3! \equal{} \boxed {120}$ as my answer.

I appear to get something different  :maybe:

So, we have an initial count of 8*7*6 = 336 ways assuming that order matters.

Now we can use PIE. The number of ways in which A and B are sitting together is 7*2!*6 = 84 (We can think of A and B as a superperson, therefore, there are just 7 seats now for 2 people, which can be done in 7*6 ways, and the two people in the superperson can be arranged in 2! ways.)

The cases where A and C, and the cases where B and C sit together is also 84 by symmetry. So our count so far is 3*84 = 252.

Now we count the cases where all three of them sit together. The superperson now has 3 people in it, so we now only have 6 seats to arrange them in, which can be done in 6 ways, and then there are 3! ways to arrange the people in the superperson. Which gives us 6*3! = 36 ways.

From PIE,

#(A+B) + #(A+C) + #(B+C) - 2#(A+B+C)

(here #(A+B) for instance denotes the number of ways to seat so that A and B sit next to each other).

So we have 252 - 2(36) = 180. And 336-180 = [b]156[/b].

Where am I going wrong?

[quote="limac"]
From PIE,

#(A+B) + #(A+C) + #(B+C) - [b]2#(A+B+C) [/b]
[/quote]

If I'm not mistaken, from PIE we should get

#(A+B) + #(A+C) + #(B+C) - [b]#(A+B+C)[/b]

So instead, we have $ 252 \minus{} 36 \equal{} 216$.  Thus, $ 336 \minus{} 216 \equal{} \boxed{120}$, which is what everyone else got.

Here's my solution. (I'm assuming the kids are indistinguishable.)

x=a student
|=an empty chair

So first, there are the three students: xxx. Because they can't be next to each other, we have x|x|x. We now have 4 spots (notice that opposite sides of an empty chair are the same) to put our 3 remaining chairs. There are three cases:

All 3 chairs go in the "same" spot: There are $ 4$ spots, so there are $ 4$ ways for this to happen.
Two chairs go in one spot and the remaining one in another: There are 4*3 ways for this to happen.
All chairs go in different spots: There are $ \binom{4}{3} \equal{} 4$ ways.
The answer is $ 4 \plus{} 12 \plus{} 4 \equal{} 20$.

If we treat the chairs between the students as "filled", we see that there are 6*5*4=120 ways of seating them (if indistinguishable students just divide by 6).

Sorry, but I don't really see why the answer is 6*5*4. (I know it is the correct answer.) Can you elaborate?

We have 8 chairs, but two of them (the chairs between each student) cannot be sat in, so its just $ 6_P3\equal{}6\cdot5\cdot4\equal{}120$

Cant we use bijections?

Here is one proper arrangement (with X's as kids and _'s as chairs)

[u]X[/u]  _ _  [u]X[/u] _ _  [u]X[/u]   _

now if we remove one chair between each student, we will have a way to arrange the kids without restrictions. There will be 6 chairs left and 3 kids to arrange in them, and we wont have restrictions on how to do this.

Now if the kids are indistinguishable, our answer is 6C3=20

If the kids are distinguishable (which is more likely) then the answer is 6x5x4=120

ok, I'm really trying to understand why the 6*5*4 thing works, but I don't. I mean, if a proper order of the students was
[u]x[/u] _ _ [u]x[/u] _ _ [u]x[/u] _
and we took out two chairs between the students to get
[u]x[/u] _ [u]x[/u] _ [u]x[/u] _
the first student would still be restricted to the first two seats and the second student to the second two seats, like the first and the second student couldn't occupy the first two seats because they would be next to each other.
And why don't we have to account for the number of ways that we can pick two seats to be empty?

Once we have a proper arrangement:

x _ _ x _ _ x _

and we take one chair out:

x _ x _ x _

now we have a method to count the ways to arrange the kids WITHOUT restrictions. Since we took the chair out between the students, we no longer have restrictions, so the question is reduced to "how many ways can three *distinguishable* kids be arranged in six chairs?"

Does that make sense? The2009 AIME II #6 could be solved by bijections. If you dont get this explanation, look at the solutions for that question. Also, there is a section in AoPS vol2 that deals with bijections

Heres another way to think about this by setting up an equation:
Assume without the loss of generality that the three kids $ a_1,b_1,$ and $ c_1$ sit with $ a_1$ before $ b_1,$ and $ b_1$ before $ c_1$

let $ a$ be the number of chairs from the beginning of the row to $ a_1$, let $ b$ be the number of chairs from $ a_1$ to $ b_1$, let $ c$ be the number of chairs from $ b_1$ to $ c_1$, and let $ d$ be the number of chairs from $ c_1$ to the end of the row. Then we have
$ a \plus{} b \plus{} c \plus{} d \equal{} 5$

But b and c must be at least one, so our equation reduces to
$ a \plus{} b \plus{} c \plus{} d \equal{} 3$
Which by balls and urns has a total of $ \binom {3 \plus{} 4 \minus{} 1}{4 \minus{} 1} \equal{} 20$ solutions. But we need to multiply that by $ 3!$ since the kids are distinguishable, so our answer is $ 120$

Hello sirs thanking you for the wonderful solution. Verily elegant

I understand now! Thanx alot!
Here is my explanation for people who didn't understand, like me.

If we take away the two chairs that are between the students, which are the chairs that cannot be sat in, we have six chairs to choose from, and three chairs are to be filled by three students. And now the students can sit in any of the six chairs that they want, they are not restricted.

The reason that they are not restricted is because

If we choose three seats for the three students to sit in, out of the six chairs, like so...
[u]x[/u] [u]x[/u] [u]x[/u] _ _ _
...then we can just put the two seats, the two that the students couldn't choose from, back inbetween the students...
[u]x[/u] _ [u]x[/u] _ [u]x[/u] _ _ _
...because the two chairs are still there, we are just ignoring them.

If we give the three students six chairs to sit in, no matter what arrangement they choose, we can put the two chairs back in between the students, one inbetween the first two students, and one inbetween the second two students, so that they are not next to eachother. For example, if the three students choose to sit like so...
[u]x[/u] _ _ [u]x[/u] [u]x[/u] _
...we can put one chair between the first and second student, and one chair between the second and third student, and they are no longer sitting next to eachother...
[u]x[/u] _ _ _ [u]x[/u] _ [u]x[/u] _

First we let the three students choose their seats out of the six chairs, and then we put the two chairs back between the students to ensure that they are not sitting next to eachother.


Let $ n$  be an odd integer greater than one. Prove that the sequence:
\[ \dbinom{n}{1},\dbinom{n}{2}, \ldots , \dbinom{n}{\frac{n\minus{}1}{2}}\] contains an odd number of odd numbers.

I have a question.

Around what level of difficulty is that series of books?

I think most of it is geared towards olympiad training, but I think the intro problems are around AMC 10/12 in difficulty.

[hide="Hint"]Show that the sum is odd.[/hide]

[hide="Solution"]
Let $ S$ be the Sum, so

$ S\equal{}\dbinom{n}{1},\dbinom{n}{2},\ldots ,\dbinom{n}{\frac{n\minus{}1}{2}}\equal{}\dbinom{n}{n\minus{}1},\dbinom{n}{n\minus{}2},\ldots ,\dbinom{n}{\frac{n\plus{}1}{2}}$

We know that $ S\plus{}S\plus{}\binom{n}{0}\plus{}\binom{n}{n}\equal{}\sum_{i\equal{}0}^n \binom{n}{i}\equal{}2^n$

Which means that $ S\plus{}S\plus{}\binom{n}{0}\plus{}\binom{n}{n}\equal{}2^n\implies 2S\equal{}2^n\minus{}2\implies S\equal{}2^{n\minus{}1}\minus{}1$ which is odd.
[/hide]


calculus derivative integration logarithms function algebra domain

If $ f'(x)\equal{}\frac{2}{x}$ and $ f(\sqrt{e})\equal{}5$, then what is the value of $ f(e)$?

Hint: What is the integral of $ \frac {1}{x}$ with respect to x?

$ f(x) \equal{} \int \frac{2}{x} \; dx \plus{} C \equal{} 2\ln|x| \plus{} C$

$ 5\equal{} f(e^{\frac{1}{2}}) \equal{} 2\ln( e^{\frac{1}{2}} ) \plus{} C \equal{} 2 \cdot (1/2) \ln(e) \plus{} C \equal{} 1 \plus{}C$

And so $ C \equal{} 5\minus{}1\equal{}4$, meaning our expression for $ f(x)$ is now $ 2\ln(x) \plus{} 4$. Hence $ f(e) \equal{} 2\ln(e) \plus{}4 \equal{} 6$.

If $ f(x)$ is a function from $ \bf R$ to $ \bf R$, differentiable everywhere except at $ x \equal{} 0$, with $ f'(x) \equal{} \frac{2}{x}$ and $ f(\sqrt{e}) \equal{} 5$, what can we say about $ f(\minus{}e)$?

;)

AP Calculus kids: you should be aware that JBL isn't playing entirely fair here. Don't be too trusting.

Kent, I can't remember if I learned this particular "trick question" from you or from jmerry.  I wish my calculus students were savvy enough that I could explain this to them some time, but I think it would result in more confusion than enlightenment :(

This is an important point for AP students to understand. It appeared in an AP Free Response question in 2006. [url=http://www.collegeboard.com/prod_downloads/student/testing/ap/calculus_ab/ap06_frq_calculus%20ab.pdf]#5[/url]

I believe less than 2% of students got this question completely correct.

An [url=http://apcentral.collegeboard.com/apc/members/repository/ap07_calculus_DE_domain_fin.pdf]article[/url] for AP teachers related to this topic.

[quote="JBL"]If $ f(x)$ is a function from $ \bf R$ to $ \bf R$, differentiable everywhere except at $ x \equal{} 0$, with $ f'(x) \equal{} \frac {2}{x}$ and $ f(\sqrt {e}) \equal{} 5$, what can we say about $ f( \minus{} e)$?

;)[/quote]

I see no reason why $ f( \minus{} e)$ shouldn't be $ 6$ since $ 2\ln|x| \plus{} C$ is an even function. If the antiderivative were simply $ 2\ln(x) \plus{} C$ then $ \minus{} e$ would lie outside the function's domain.

Yes, $ f(\minus{}e)$ can be $ \minus{}6$. You did, however, make a silly computation error.

[quote="JBL"]Kent, I can't remember if I learned this particular "trick question" from you or from jmerry.[/quote]I can't say either, but I'm pretty sure that one of us learned it from the other, so it doesn't really matter. In the quick search I tried, I found one of my posts and none of Kent's, but there are likely other instances I missed.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=953067#953067]Here[/url] is the link.

Yes, I think that's probably the post that I remember.  Thanks for the links, gauss!

The answer to my question, for those keeping score, is that actually we can have $ f(\minus{}e)$ equal to any real number.  Specifically, the functions that satisfy the conditions I listed are defined by $ f(x) \equal{} 2 \ln x \plus{} 4$ when $ x > 0$, $ f(0) \equal{} C_1$ and $ f(x) \equal{} 2\ln (\minus{}x) \plus{} C_2 \,(\equal{} 2 \ln |x| \plus{} C_2)$ when $ x < 0$, for arbitrary real numbers $ C_1$ and $ C_2$.  The fact that we can choose the constants for the two branches separately is an alternative phrasing of the content of jmerry's and gauss's links: solutions to the differential equation $ \frac{df}{dx} \equal{} \frac{2}{x}$ only exist on the intervals $ (0, \infty)$ and $ (\minus{}\infty, 0)$ -- there is no solution that exists on all of $ \bf R$.  To get a function that has the derivative we want on all of $ \bf R$, we can patch together any two solutions (and plug the hole at 0 arbitrarily) -- in particular, the initial value that we have can't tell us anything about what goes on on the other side of the discontinuity.

[quote]I wish my calculus students were savvy enough that I could explain this to them some time, but I think it would result in more confusion than enlightenment [/quote]

If you're referring to why we cannot infer what $ f(\minus{}e)$ is based on the given information, I don't think it's that tricky to explain. My book seems to have fine clarity on the issue:

[quote]Recall that the most general antiderivative on a given interval is obtained by adding a constant to a particular antiderivative. We adopt the convention that when a formula for a general indefinite integral is given, [b]it is valid only on an interval.[/b] [/quote]

[asy]draw( (-10,0)--(10,0), Arrows );
draw( (0,-10)--(0,10),Arrows);

size(200);

real f(real x) {return log(x)+4 ;}
real g(real x) {return log(-x)+8 ;}

draw(graph(f,0.1,8));
draw(graph(g,-8,-0.1));

label( "\small{$0$}",(0,0),SW);
label( "\small{$x$}",(10,0),S);
label( "\small{$y$}",(0,10),W);
label( "\small{$f(x)$}",(5,7));
label( "\small{$g(x)$}",(-5,7));[/asy]

In the above example, $ f(x) \equal{} \ln(x) \plus{}4$ and $ g(x) \equal{} \ln(\minus{}x) \plus{} 8$ have the same derivative, even though their constants do not match.

Hey,
[quote="jmerry"]
A short version of what I think:
$ \ln(x)$ is an antiderivative of $ \frac1x$ on $ (0,\infty)$.
$ \ln( - x)$ is an antiderivative of $ \frac1x$ on $ ( - \infty,0)$.
$ \ln|x|$ is not an antiderivative of $ \frac1x$ on $ ( - \infty,\infty)$.
[/quote]
I agree, however isn't the following true,
$ \ln|x|$ is an antiderivative of $ \frac1x$ on $ ( - \infty,0){\cup}(0, \infty)$.
[quote="JBL"]
If $ f(x)$ is a function from $ \bf R$ to $ \bf R$, differentiable everywhere except at $ x = 0$, with $ f'(x) = \frac {2}{x}$ and $ f(\sqrt {e}) = 5$, what can we say about $ f( - e)$?
[/quote]
[hide="APSStudent wrote:"]
[b]1. The problem statement, all variables and given/known data.[/b]
Consider the general problem,
If $ {f(x)}$ is a function from $ {\bf{R}}$ to $ {\bf{R}}$, differentiable everywhere except at $ {{x} = {0}}$, with,
$ {{{f}^{\prime}}{(x)}} = {\frac {a}{x}}$
where,
$ {f(\alpha)} = {\beta}{\,}{\,}{,}{\,}{\,}{{\alpha} > {0}}$

What can we say about,
$ {f(\gamma)}{\,}{\,}{,}{\,}{\,}{{\gamma} < {0}}$

[b]2. Relevant equations.[/b]
$ {\lVert{x}\rVert} = {\begin{Bmatrix}{{x}{\,}{\,}{,}{\,}{\,}{{x} \geq {0}}}\\[2ex]
{{ - x}{\,}{\,}{,}{\,}{\,}{x < 0}}\end{Bmatrix}}$

$ {\lVert{x}\rVert} = {\sqrt {({{x}^{2}})}}$

For,
$ {\log_{b}{x}}{\,}{\,}{,}{\,}{\,}{{b} > {0}}$
$ D = (0, \infty)$
$ R = ( - \infty, \infty)$

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
$ {{{f}^{\prime}}{(x)}} = {\frac {a}{x}}$
$ {{\int_{}^{}}{\biggl[{dy}\biggr]}} = {{\int_{}^{}}{\biggl[{\frac {{a}{dx}}{x}}\biggr]}}$
$ {y} = {{a}{\int_{}^{}}{\biggl[{\frac {dx}{x}}\biggr]}}$
$ {y} = {{a}{({{\ln{\lVert{x}\rVert}} + {{C}_{1}}})}}$

Let, $ {{C} = {{a}{{C}_{1}}}}$
$ {y} = {{{a}{\ln{\lVert{x}\rVert}}} + {(C)}}$
$ {y} = {{{a}{\ln{\lVert{x}\rVert}}} + {C}}$

The motivation for the following step comes from realizing that although,
$ {\lVert{x}\rVert} = {\begin{Bmatrix}{{x}{\,}{\,}{,}{\,}{\,}{{x} \geq {0}}}\\[2ex]
{{ - x}{\,}{\,}{,}{\,}{\,}{x < 0}}\end{Bmatrix}}$
in this differential equation problem we are not just simply concerned with the values $ {x}$ and $ {y}$ can take on as a consequence of the absolute value function, but also the values that $ {C}$ can take on as a consequence of the absolute value function for each particular solution to the differential equation. That being said, we need to consider how the constant $ {C}$ changes as a consequence of the absolute value function.

That is, from the definition of the absolute value function let,
$ {y} = {\begin{Bmatrix}{{{a{\ln{({x})}}} + {{C}_{A}}}{\,}{\,}{,}{\,}{\,}{{x} > {0}}}\\[2ex]
{{\text{undefined}}{\,}{\,}{,}{\,}{\,}{{x} = {0}}}\\[2ex]
{{{a{\ln{({ - {x}})}}} + {{C}_{B}}}{\,}{\,}{,}{\,}{\,}{{x} < {0}}}\end{Bmatrix}}$

Where for convenience in notation let,
$ {{{f}_{{}_{A}}}{(x)}} = {{{a{\ln{({x})}}} + {{C}_{A}}}}$
and
$ {{{f}_{{}_{B}}}{(x)}} = {{{a{\ln{({ - {x}})}}} + {{C}_{B}}}}$

If we consider the particular solutions to this differential equation,
$ {({{x}_{1}}, {{y}_{1}})} = {({\alpha}, {\beta})}{\,}{\,}{,}{\,}{\,}{{\alpha} > {0}}$
and
$ {({{x}_{2}}, {{y}_{2}})} = {({\gamma}, {{y}_{2}})}{\,}{\,}{,}{\,}{\,}{{\gamma} < {0}}$

We see specifically that $ {({\alpha}, {\beta})}$ is a particular solution to the particular equation $ {{{f}_{{}_{A}}}{(x)}}$ since $ {{\alpha} > {0}}$ and that $ {({\gamma}, {{y}_{2}})}$ is a particular solution to the particular equation $ {{{f}_{{}_{B}}}{(x)}}$ since $ {{\gamma} < {0}}$.

Since we have the initial condition $ {{({{x}_{1}}, {{y}_{1}})} = {({\alpha}, {\beta})}}$ we can proceed to solve for the particular solution to $ {{{f}_{{}_{A}}}{(x)}}$ since $ {{\alpha} > {0}}$.
In other words, solving for $ {{C}_{A}}$,
$ {{{f}_{{}_{A}}}{(x)}} = {{{a{\ln{({x})}}} + {{C}_{A}}}}$
$ {({{y}_{1}})} = {{{a{\ln{({({x}_{2})})}}} + {{C}_{A}}}}$
$ {({\beta})} = {{{a{\ln{({(\alpha)})}}} + {{C}_{A}}}}$
$ {{C}_{A}} = {{\beta} - {a{\ln{\alpha}}}}$

So that,
$ {{{f}_{{}_{A}}}{(x)}} = {{{a{\ln{(x)}}} + {({\beta} - {a{\ln{\alpha}}}})}}$
$ {{{f}_{{}_{A}}}{(x)}} = {{{a{\ln{(x)}}} + {{\beta} - {a{\ln{\alpha}}}}}}$

Simplifying,
$ {{{f}_{{}_{A}}}{(x)}} = {{a{\ln{({\frac {x}{\alpha}})}}} + {\beta}}$

Since we have the initial condition $ {{({{x}_{2}}, {{y}_{2}})} = {({\gamma}, {{y}_{2}})}}$ we can proceed to solve for the particular solution to $ {{{f}_{{}_{B}}}{(x)}}$ since $ {{\gamma} < {0}}$.
In other words, solving for $ {{C}_{B}}$,
$ {{{f}_{{}_{B}}}{(x)}} = {{{a{\ln{({ - {x}})}}} + {{C}_{B}}}}$
$ {({{y}_{2}})} = {{{a{\ln{({ - {({{x}_{2}})}})}}} + {{C}_{B}}}}$
$ {{y}_{2}} = {{{a{\ln{({ - {({\gamma})}})}}} + {{C}_{B}}}}$
$ {{C}_{B}} = {{a{\ln{({ - {\gamma}})}}} - {{y}_{2}}}$

However, not knowing the value of $ {{y}_{2}}$ we cannot explicitly calculate the value of $ {{C}_{B}}$.
In other words, what we can say about $ {f(\gamma)}$ is that since we do not know the value of $ {{C}_{B}}$, then $ {f(\gamma)}$ can be made to take on any arbitrary value dependent on the choice of $ {{C}_{B}}$.
Specifically,
$ {f(\gamma)} = {{a{\ln{({ - {\gamma}})}}} + {{C}_{B}}}$
[/hide]
[/hide]
[quote="JBL"]
The answer to my question, for those keeping score, is that actually we can have $ f( - e)$ equal to any real number.  Specifically, the functions that satisfy the conditions I listed are defined by $ f(x) = 2 \ln x + 4$ when $ x > 0$,
$ f(0) = C_1$
and $ f(x) = 2\ln ( - x) + C_2 \,( = 2 \ln |x| + C_2)$ when $ x < 0$, for arbitrary real numbers $ C_1$ and $ C_2$.  The fact that we can choose the constants for the two branches separately is an alternative phrasing of the content of jmerry's and gauss's links: solutions to the differential equation $ \frac {df}{dx} = \frac {2}{x}$ only exist on the intervals $ (0, \infty)$ and $ ( - \infty, 0)$ -- there is no solution that exists on all of $ \bf R$.  To get a function that has the derivative we want on all of $ \bf R$, we can patch together any two solutions (and plug the hole at 0 arbitrarily) -- in particular, the initial value that we have can't tell us anything about what goes on on the other side of the discontinuity.
[/quote]
I agree with your points, however why is that we can let,
$ {f(0)} = {{C}_{1}}$
for some arbitrary (real) constant, $ {{C}_{1}}$?

The problem with the above statement is that $ {{y} = {{{a}{\ln{\lVert{x}\rVert}}} + {C}}}$ is undefined at $ {f(0)}$, since $ {{x} = {0}}$ lies outside of the domain of $ {\ln{\lVert{x}\rVert}}$ or for that matter $ {\log_{b}{(x)}}$ for any $ {{b} > {0}}$.

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]


Prove also that if a_1*...*a_n=1, then
   sum sqrt(((a_i)^2+a_i)/2)<=sum a_i.
  This one is not so hard.
   And now another question. Which of the two inequalities is stronger?

We can prove it in the same way:

sqrt((ai^2+ai)/2) + sqrt((ai+1)/2) <= ai + 1
So
Sum sqrt((ai^2+ai)/2) <= Sum ai + n - Sum sqrt((ai+1)/2)
Now, by AM-GM Sum sqrt((ai+1)/2) >= n and we are done.
I'm thinking now for the second question.

Namdung

That can also be solved using Cauchy:
  sum sqrt(x_i*(x_i+1)/2)<=sqrt[ (sum x_i)*(sum (x_i+1)/2)] and the last quantity is smaller than sum x_i, since sum x_i>=n.


Let a,b,c >o,$ a \neq b \neq c$.Prove that:
$ a^2 \plus{}b^2\plus{}c^2 \plus{} 48abc(ab \plus{}ac\plus{}bc)\geq a^3 \plus{}b^3 \plus{}c^3\plus{}16abc$

[quote="talaaiha"]Let a,b,c >o,$ a \neq b \neq c$.Prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 48abc(ab \plus{} ac \plus{} bc)\geq a^3 \plus{} b^3 \plus{} c^3 \plus{} 16abc$[/quote]
Your problem is incorrect.
Counter example:
$ a\equal{}2,b\equal{}1,c\rightarrow 0$.


What is the probability that a randomly selected two-digit number is a multiple of 14? Express your answer as a common fraction.

Two-digit multiples of 14: 14(1) through 14(7)
Total of 7 two-digit multiples of 14.

The two digit numbers range from 10-99 inclusive, so there are 90 two digit numbers total.

Probability is then $ \frac{7}{90}$.


AMC
Hi
USAMO
geometry
3D geometry
prism
parallelogram

Legs $L_1, L_2, L_3, L_4$ of a square table each have  length $n$, where $n$ is a positive integer.  For how many ordered  4-tuples $(k_1, k_2, k_3, k_4)$ of nonnegative integers can we cut  a piece of length $k_i$ from the end of leg $L_i \; (i=1,2,3,4)$ and still have a stable table?

(The table is [i]stable[/i] if it can be placed so that  all four of the leg ends touch the floor. Note that a cut leg of length 0 is permitted.)

Proposed new wording for question 4 - it's a limerick!

I once had an n-height square table
That I wanted to stand on a gable.
So I hired a plumber
Who cut a whole number
From each leg, so the table stayed stable.

With the given information, and assuming each leg is distinct before cutting, how many ways could the plumber have cut the legs of my table?

[quote]With the given information, and assuming each leg is distinct, how many ways could the plumber have cut the legs of my table in terms of n?[/quote]

But each leg doesn't have to be distinct after each cut.

By the way, myth is right -- K1 + K3 = K2 + K4.
Another way to look at it is if l1, l2, l3, and l4 are the legs after the cuts

l1 + l3 = l2 + l4

Then its all about finding a formula...I'll post it once the USAMO formula opens up again.

JB

[quote="Myth"]Am I missing something? We are to find the number of 4-tuples $(k_1, k_2, k_3, k_4)$ satisfying $k_1+k_3=k_2+k_4$, $0\leq k_i\leq n$?[/quote]

That's absolutely correct, but how did you come up with that so fast? It took me a few minutes to get around that fact. Then it took me quite a few more to convince myself (with the help of the table I was working on, which imagined its legs being cut off) it's true.

Hmm... What is the problem? I just wrote the condition that four points lies in the same plane, i.e. they are complanar. :? Indeed, it immediately kills the problem, since we can easily count all such quadruples.

[quote="jackb1115"][quote]With the given information, and assuming each leg is distinct, how many ways could the plumber have cut the legs of my table in terms of n?[/quote]

But each leg doesn't have to be distinct after each cut.[/quote]

I meant that the legs were distinct before the cut.

Hmm... I think your arguments have the same core with my arguments. Actually, I simply used the fact that, clearly, a plane intersects regular square prism on parallelogram (sorry for my english).

BTW, why did you change your flag?

I took this approach:

First, we count the number of ways of cutting the same length off each leg, which is clearly (n+1).

We note that making three integral cuts determines the length of the fourth cut, and furthermore that the fourth cut must have integral length.  Also, the shortest cut must be across from the longest cut.

For the cases where we do not cut the same length from each leg, we assume WOLOG that k1 is the shortest cut, making k3 the longest cut, and if there are two shortest cuts, let k2 be the other one.  Later we will multiply our count here by 4 so that we count when each of k2, k3, and k4 are the shortest cuts (and we don't count any cases more than once).

Then 0 <= k1 <= k2 < k3 <= n; we want the number of solutions to this, which is equivalent to

1 <= a1<a2<a3 <= n+2, when a1=k1+1, a2=k2+2, a3=k3+2.

This is just (n+2 choose 3).

So our total is n+1 + 4(n+2 choose 3)

here's how i did number 4, it's essentially the same as everyone else's method:

let Lx be the length of leg x, and Px be the endpoint of leg x.
say that L1 is directly across from L3(diagonally). in order for the table to be stable, the line that passes through P1 and P2 must be paralell to the line that passes through P3 and P4. the condition L2 - L1 = L3 - L4 ensures that the 4 points are coplanar

let the difference L2 - L1 = m
since the problem states that each leg can either be cut entirely, not cut, or cut integrally, m must be an integer on the domain [0,n]
if there are two adjacents legs with length n, then there are (1+n-m) different cuts that will produce this difference.

the pair of legs opposite these two must have the same difference, but can still be cut in any of the (1+n-m) ways.

thus we take the following sum:

sum of (1+n-m)^2 from m=0 to m=n

since we have restricted m to be greater than or equal to zero, we have neglected to count the cases in which L1 is greater than L2. thus we multiply the sum by two, for there are just as many cases in which L2>L1 as those in which L1>L2. but what about when m=0? at m=0 the expression only needs to be counted once, thus to avoid overcounting, we subtract the number of cases in which m=0 (there are (1+n)^2 such cases.

if we examine the sum of (1+n-m)^2 from m=0 to m=n, we note that the first term will be (1+n)^2 and the last term 1^2. so this is essentially the same as the sum of the first (1+n) squares. plugging into the formula x(x+1)(2x+1)/6 and performing the said process to count accurately, we have:

2*(n+1)(n+2)(2n+3)/6 - (1+n)^2

= (n+1)(n+2)(2n+3)/3 - (1+n)2

=[(n^2+3n+2)(2n+3)-3(n^2+2n+1)]/3

=(2n^3+3n^2+6n^2+9n+4n+6-3n^2-6n-3)/3

 =(2n^3 + 6n^2 +7n + 3)/3

I interpreted "Note that a cut leg of length 0 is permitted" to mean "Cutting off a leg by a length of 0 is permitted" but that a length of 0 after cutting would leave no table leg, and would be non-permissible.

I also put in my solution that if it was meant to be interpreted "A leg of length 0 after cutting is permitted" then all the "n" in my solution could be replaced by "n+1".  Would they take off a mark for that?

(My solution above is with the accepted interpretation - but on the contest, I put
n+4(n+1 choose 3) instead of
n+1+4(n+2 choose 3)

I did this differently from the way CodeBlue did: I took the sum of the number of 4-tuples meeting the condition and having maximal element $i$, for $i = 0$ to $n$. To find these, I took the sum of the number of tuples $(k_1,k_2,k_3,k_4)$ for $m = n$ to $2n$, such that $k_1 + k_3 = m$. Since the maximum element had to be i, I set $k_2 = n$. Then $k_4$ is fixed, and we get that the number of these tuples is $2n - m$. There is one overlap here when you rotate (when $k_1 = m - n$ and $k_1 = n)$, so get rid of one of them, and you get $2n - m - 1$ tuples. Multiply this by four, then sum them and add one for $m = 2n$. You get, for $j$ between $n$ and $2n-1$, $4 \sum (2n - j - 1) + 1$, which is equivalent to, for $j$ between $1$ and $n$, $4 \sum j + 1$. Then replace n with i, and we take the sum of this for $i$ between $0$ and $n$. Eventually, I get the same answer as CodeBlue87 did (some reassurance that I did it right).

[quote="MithsApprentice"]Could you explain why it's obvious that "coplanar 4-tuple" obviously implies

[quote="Myth"] $(k_1, k_2, k_3, k_4)$ satisfying $k_1+k_3=k_2+k_4$, $0\leq k_i\leq n$[/quote]

? I eventually used an argument stemming from the fact that if there were to be a leg coming from the center of the table, it would have to be the average of k_1 + k_3 and k_2 + k_4, which led (with a bit of work) to the same conjecture you cited.[/quote]

Should be immediately obvious :)

I did $k_1+k_4=k_2+k_3$, did cookies and kids, and got the answer to be \[ 2\sum\limits{k=1}^{n}{k^2}+(n+1)^2=\frac{2n^3+5n^2+5n+2}{3} \]
I allowed a leg of length 0.

How about this solution:

Let $ C_i$ be the number of sets of cuts $ \{k_1,k_2,k_3,k_4\}$ that result in the longest leg being of length $ i$.  Then $ C_i\equal{}C_{i\minus{}1}\plus{}x$, and we will evaluate $ x$ to develop a recursion.  Note that if a set of cuts has non-zero cuts, then if that set is counted in $ C_i$, subtract $ 1$ from each of the cuts to obtain a set of cuts that is counted in $ C_{i\minus{}1}$.  For example, if $ \{2,3,4,5\}$ was counted in $ C_5$, then $ \{1,2,3,4\}$ would have been counted in $ C_4$ because subtracting the same length from each leg preserves the quality of the legs being coplanar, and there is a bijection.

Now we must evaluate those sets of cuts that don't fall in this bijection, which are clearly those where one leg is completely cut off.  After some simple geometric reasoning that I won't include here, we see that one corner has a leg of length of zero, and the opposite leg, called the initial leg, will have length $ i$  If the remaining legs have length $ y$ and $ z$, $ (y,z)\equal{}(0,i),(1,i\minus{}1),...,(i\minus{}1,1),(i,0)$, so $ i\plus{}1$ options.  The initial corner can be any of the four legs, but each of the four permutations of the set of cuts $ {0,0,i,i}$ is repeated twice, so we have $ 4(i\plus{}1)\minus{}4\equal{}4i$ total additional sets of cuts.  We conclude that $ C_i\equal{}C_{i\minus{}1}\plus{}4i$, and note that $ C_0\equal{}1$.  Now we can create generating functions.  $ F(x)\equal{}\sum_{i\equal{}0}^\infty C_is^i$.  Also, $ G(x)\equal{}\sum_{i\equal{}1}^\infty 4i$. From the recursion, we have
\[ F(x)\equal{}C_0\plus{}xF(x)\plus{}G(x)\Longrightarrow F(x)\equal{}\frac{1\plus{}G(x)}{1\minus{}x}\]
This final equation is easy to analyze.  Simply use $ G(x)$ as the first term of a geometric series with constant multiplier of $ x$.  This gives $ C_i\equal{}2i^2\plus{}2i\plus{}1$.  The total number of sets of cuts for a table of legs of length $ n$ is the sum of all the $ C_i$, $ 0\le i\le n$, from which we deduce the answer $ \frac{2n^3 \plus{} 6n^2 \plus{}7n \plus{} 3}{3}$.

Why so much overkill?! Here's the first complete non-overkill solution in the thread.

As noted, it's immediately obvious that the condition is equivalent to $ k_1\plus{}k_3\equal{}k_2\plus{}k_4$.

To count these, fix the sum to be $ s$; clearly $ 0\le s \le 2n$. For $ s\le n$, we have that there are $ s\plus{}1$ (ordered) ways to partition $ s$ into two parts; consider $ (s\minus{}i,i)$ with $ i$ ranging from $ 0$ to $ s$. $ s>n$ is in one-to-one correspondence with $ s<n$; replace $ s$ partitioned into $ x\plus{}y$ with $ 2n\minus{}s$ and $ (n\minus{}x)\plus{}(n\minus{}y)$. Because we're choosing the partitions of $ s$ with replacement, our final sum is $ 1^2\plus{}2^2\plus{}3^2\plus{}\ldots\plus{}n^2\plus{}(n\plus{}1)^2\plus{}n^2\plus{}\ldots\plus{}3^2\plus{}2^2\plus{}1^2$. Using the formula for the sum of $ n$ squares, this comes out to $ \frac{2n^3 \plus{} 6n^2 \plus{}7n \plus{} 3}{3}$ and hence this is our answer, as desired.

What a weird problem. Hardest part is probably noticing what the condition means.

We claim that the condition is equivalent to the claim that $a+d=b+c=k$ for some $0\leq k\leq 2n,$ where $a\leq b\leq c\leq d$ are the leg lengths. This is because $\vec{A}+\vec{D}=\vec{B}+\vec{C}$ where $A,B,C,D$ are the vertices of the square in that order. Looking at the component parallel to the legs finishes.

    Obviously for $0\leq k\leq n$ there are $(k+1)^2$ tables, which corresponds to $\frac{(n+1)(n+2)(2n+3)}{6}.$

    Now for $n+1\leq k\leq 2n,$ send $a\to n-a'$ and $d\to n-d'.$ Then $a'+d'=2n-k,$ so there are $(2n-k+1)^2$ tables, which corresponds to $\frac{n(n+1)(2n+1)}{6}.$

    Add it all up to get
    \[\frac{2n^3+6n^2+7n+3}{6}.\]

[hide = Solution]
The main part is deciphering the condition. The condition is equivalent to the averages of the lengths of opposite legs being equivalent, or

\[\frac{(n - k_1) + (n - k_3)}{2} = \frac{(n - k_2) + (n - k_4)}{2}\]

which would mean that

\[k_1 + k_3 = k_2 + k_4.\]

Now let this common value be $v$ (that is, $k_1 + k_3 = k_2 + k_4 = v$). We consider two cases

$\textbf{Case 1:}$ $0 \le v \le n$

Since there are $v+1$ solutions to $a+b = v$ with $0 \le a,b \le n$, the total number of ordered quadruples for a particular $v$ is $(v+1)^2$. Summing over all $v$ from $0$ to $n$, we get that the total number of quadruples for this case is

\[1^2 + 2^2 + \cdots + (n+1)^2 = \frac{(n+1)(n+2)(2n+3)}{6}.\]

$\textbf{Case 2:}$ $n+1 \le v \le 2n$

this time, we consider the part of each leg that remains after we cut it. Consider the number of solutions to $a+b = 2n - v$ with $0 \le a,b \le n$. This will have $2n - v + 1$ solutions, so the total number of quadruples for a particular $v$ is $(2n - v + 1)^2$. Summing over all $v$ from $n+1$ to $2n$, we get that the total number of quadruples for this case is

\[n^2 + (n-1)^2 + \cdots + 1^2 = \frac{n(n+1)(2n+1)}{6}.\]

Summing the total for both cases, we get an answer of

\begin{align*}
\frac{(n+1)(n+2)(2n+3) + n(n+1)(2n+1)}{6} & = \frac{(2n^3 + 9n^2 + 13n + 6) + (2n^3 + 3n^2 + n)}{6}\\
& = \frac{4n^3 + 12n^2 + 14n + 6}{6}\\
& = \boxed{\frac{2n^3 + 6n^2 + 7n + 3}{3}}. \square
\end{align*}
[/hide]

Assume the $L_i$ are numbered clockwise. The table is stable if and only if the new ends of the tables are coplanar. Given three choices of table leg lengths removed, $k_1,k_2,k_3$, the fourth must be uniquely determined by intersecting the plane with the fourth leg. But also, the midpoint the segment between the table leg endpoints of $L_1$ and $L_3$ and the midpoint of the segment between the table leg endpoints of $L_2$ and $L_4$ must coincide because the table is square. That is, $((n-k_1)+(n-k_3))/2 = ((n-k_2)+(n-k_4))/2$, or $k_1+k_3=k_2+k_4$. For a particular integer $0\le a\le n$, there are $(a+1)^2$ choices of $(k_1,k_2,k_3,k_4)$ with $k_1+k_3=k_2+k_4=a$ and for $n+1\le a\le 2n$ there are $(2n+1-a)^2$ choices. So overall there are
\[(n+1)^2 + 2\sum_{i=1}^n i^2 = (n+1)^2 + 2\cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)+3(n+1)^2}{3} = \]
\[\frac{n+1}{3}\cdot (n(2n+1)+3(n+1)) = \frac{n+1}{3}\cdot(2n^2+4n+3) = \frac{(n+1)(2n^2+4n+3)}{3}.\]

This problem is a bit ambiguous.
1. The problem doesn't state the legs are all parallel.
2. The problem doesn't specify whether a cut of length $n$ is allowed. I interpreted it as a cut of length $n$ not being allowed, ending up with a different answer.

My solution is similar to post 18. 2021 Fall AMC 10A #17 has basically the same idea.

Define $a_i = L_i - k_i$ for all integers $i \in [1, 4]$.

Linearity of Height implies that it is necessary and sufficient to have $a_1 + a_3 = a_2 + a_4$, where $L_1$ and $L_3$ are diagonally opposite wrt the square.

Now, it's clear that the number of satisfactory $4$-tuples is just $$\sum_{i = 1}^{n + 1} i^2 + \sum_{j = 1}^{n} j^2 = \frac{(n+1)(n+2)(2n + 3)}{6} + \frac{n(n+1)(2n+1)}{6} = \frac{2n^3 + 6n^2 + 7n + 3}{3}$$ which finishes. $\blacksquare$

[b]Remark:[/b] This solution is probably too terse for the actual USAMO; I'm just lazy.

Notice we must have $k_1+k_3=k_2+k_4$ where $0\le k_i\le n.$ The number of solutions to this equation is $$\sum_{i=0}^{n}{(i+1)^2}+\sum_{j=1}^{n}{j^2}=\frac{2n^3+6n^2+7n+3}{3}.$$ by casework on whether $0\le k_1+k_3\le n$ or $n+1\le k_1+k_3\le 2n.$ $\square$

Restate the problem in terms of $k_i$:
[quote=Rephrased Problem]Find the number of ordered tuples $(k_1,k_2,k_3,k_4)$ of integers such that $k_1+k_3=k_2+k_4$ and $0\le k_i\le n$ ($i=1,2,3,4$).[/quote]
Let this common value be $t$, with $t$ fixed. If $0\le t\le n$, there are $t+1$ ways to choose $k_1$ and $t+1$ ways to choose $k_3$, then the tuple $(k_1,k_2,k_3,k_4)$ is uniquely determined. If $n<t\le2n$, then we have $t-n$ choices for $k_1$ and $t-n$ choices for $k_3$. Then the answer is:
$$\sum_{t=0}^n(t+1)^2+\sum_{t=n+1}^{2n}(t-n)=\frac{(n+1)(n+2)(2n+3)}6+\frac{n(n+1)(2n+1)}6=\frac{2n^3+6n^2+7n+3}3.$$

Set the table in the $xy$ plane such that WLOG it’s coordinates are $(1,1),(-1,1),(-1,-1),(1,-1)$, corresponding to $L_1,L_2,L_3,L_4$, respectively.

The problem statement is equivalent to there being a plane passing through the four ends of the legs, and since a plane is determined by three points, we will fix $L_2,L_3,L_4$.
The coordinates of these points are $(-1,1,n-k_2),(-1,-1,n-k_3),(1,-1,n-k_4)$. The plane passing through these points is $(k_4-k_3)x+(k_2-k_3)y+2z=2n-k_2-k_4$.

Then at $x=y=1$, we have $z=n-k_2+k_3-k_4$.
Going back to our problem statement, it is necessary and sufficient to have $n-k_1=n-k_2+k_3-k_4\implies k_1=k_2-k_3+k_4$.

We use generating functions. Our problem becomes finding the sum of the coefficients of the terms in the expansion of $(1+x+x^2+…+x^n)^2(1+x^{-1}+x^{-2}+…+x^{-n})$ that have an exponent between $0$ and $n$, inclusive. This expands to $(1+2x+3x^2+…+(n+1)x^n+nx^{n+1}+…+x^{2n})(1+x^{-1}+…+x^{-n})$, and from here our desired value is $\sum_{k=0}^{n+1} k^2 +\sum_{k=0}^{n} k^2=\frac{n(n+1)(2n+1)}{6}+\frac{(n+1)(n+2)(2n+3)}{6}=\boxed{\frac{2n^3+6n^2+7n+3}{3}}$

Notice that $k_1+k_3=k_2+k_4$ because all four of the leg ends must lie on the same plane. Notice that when $k_1+k_3=k_2+k_4=x\le n$, there are $(x+1)^2$ ways to pick $k_i\le n$. This gives us $\frac{2n^3+9n^2+13n+6}{6}$ ways. Notice that when $k_1+k_3=k_2+k_4=x>n$, there are $(2n-x+1)^2$ ways to pick $k_1\le n$. This gives us $\frac{2n^3+3n^2+n}6$ ways. In total, there are $\frac{2n^3+6n^2+7n+3}3$ ways.

The main claim is the following:

[b][color=#f00]Claim.[/b][/color] The table is stable if and only if $k_1+k_3=k_2+k_4$.

[i]Proof.[/i] Observe that for the four vertices of the legs to be coplanar, they must form a parallelogram because the quadrilateral formed is the projection of a square. Thus, equating coordinates yields the result. $\blacksquare$

Now, we can simply split into cases based on the value of $k_1+k_3$ to get the answer $$2\sum_{i=1}^n i^2 + (n+1)^2 = \frac{(n+1)(2n^2+4n+3)}3.$$

[hide=solution]The table is [i]stable[/i] if $k_1+k_3=k_2+k_4$. So let's call the common value $k$ so that $k_1+k_3=k_2+k_4=k$. Let $c_k$ be the number of ways to make the table stable for each value of $k$. We want $\sum_{k=0}^{2n}c_k$.

Note that each table leg is at least 0 and at most $n$, hence we'll break this into two sums so that it's easier to handle:
\[\sum_{k=0}^n c_k+\sum_{k=n+1}^{2n}c_k.\]
If $0\le k\le n$, then there are $k+1$ ways to partition $k_1$ and $k_3$, and another $k+1$ ways to partition $k_2$ and $k_4$, for $(k+1)^2$.

If $n+1\le k\le 2n$, each of the $k_i$ is at most $n$ and at least $0$. There are $(2n-k+1)^2$ ways to partition the $k_i$ in this interval. So we want to compute
\[\sum_{k=0}^n (k+1)^2+\sum_{k=n+1}^{2n}(2n-k+1)^2.\]
Expanding, you can evaluate it as $\boxed{\frac{(n+1)(2n^2+4n+3)}{3}}$ using  \[\sum_{k=0}^n k=\frac{n(n+1)}{2}\text{ and }\sum_{k=0}^n k^2=\frac{n(n+1)(2n+1)}{6}.\]
$\blacksquare$[/hide]

We must have $k_1 + k_3 = k_2 + k_4$. So, since $0\leq k_1, k_2, k_3, k_4\leq n$ we do casework on the possible values of $k_1+k_3$ and arrive at our answer of \[2(1^2 + 2^2 +\cdots +n^2)+(n+1)^2 = \frac{2n^3+6n^2+7n+3}{2}.\]

I claim that there are exactly $\frac{2n^3+6n^2+7n+3}{3}$ ways to cut the legs of the table.

We use 3D-coordinate geometry. WLOG, let the side of the square table be $s$. In order to have a "stable" table, we must have $(0,0,k_1)$, $(s,0,k_2)$, $(s,s,k_3)$, and $(0,s,k_4)$ be on the same plane. (We consider $(0,0,0)$ to be the end of leg $L_1$ before the cuts.) Taking the three legs $L_1$, $L_2$, $L_3$, we find that the equation of the plane containing the three ends of those legs is
\[\frac{k_1-k_2}{k_1}x+\frac{k_2-k_3}{k_1}y+\frac{s}{k_1}z=s,\]
and plugging in $(0,s,k_4)$, we have that $k_4=k_3+k_1-k_2$, meaning that the question is equivalent to

[b](USAMO 2005/4, Rephrased)[/b]
  Find all ordered quadruples of nonnegative integers $(k_1,k_2,k_3,k_4)$ such that $k_i\leq n$ and $k_1+k_3=k_2+k_4$.

Let the sum be $x$. Note that if $x\leq{}n$, then there are $x+1$ ways to partition $x$ into two nonnegative integers less than $n$. If $x>n$, we have that there are $2n-x+1$ ways to partition $x$ into two nonnegative integers less than $n$. Using this, we find that there are
\[1^2+2^2+\dots+n^2+(n+1)^2+n^2+\dots+2^2+1^2=\frac{2n^3+6n^2+7n+3}{3}\]
possible ordered quadruples, finishing the problem.

[quote name="Myth" url="/community/p213023"]
Am I missing something? We are to find the number of 4-tuples $(k_1, k_2, k_3, k_4)$ satisfying $k_1+k_3=k_2+k_4$, $0\leq k_i\leq n$?
[/quote]

When $x \le n$, $k_1+k_3=x$ has $x+1$ solutions, and $k_2+k_4=x$ also has $x+1$ solutions, so total is $(x+1)^2$ per $x$
Total for this case: $1^2+2^2+...+(n+1)^2=\frac{(n+1)(n+2)(2n+3)}{6}$
Otherwise, $k_1+k_3=x$ has $2n+1-x$ solutions, and $k_2+k_4=x$ also has $2n+1-x$ solutions, so total is $(2n+1-x)^2$ per $x$
Total for this case: $n^2+(n-1)^2+...+1^2=\frac{n(n+1)(2n+1)}{6}$
The totals sum to $\frac{2n^3+6n^2+7n+3}{3}$

0 MOHS so hard

Note that the points $(0, 0, n-k_1)$, $(0, 1 n-k_2)$, $(1, 1, n-k_3)$, and $(1, 0, n-k_4)$ are coplanar. Realize that the points are coplanar if and only if  $k_1+k_3 = k_2+k_4$. Thus we want the number of solutions $(k_1, k_2, k_3, k_4) \in \{0, 1, \dots, n\}^4$ such that $k_1+k_3=k_2+k_4$. Employ gen func. Then we want the sum of the squares of the coefficients of \[ (1+x+\dots+x^n)^2, \] which has coefficients $1$, $2$, ..., $n+1$, $n$, $n-1$, ..., $1$. The sum of their squares is \[ \frac{n(n+1)(2n+1)+(n+1)(n+2)(2n+3)}{6} = \frac{2n^3+6n^2+7n+3}{3}, \] which is the answer.

The feet of the legs are coplanar iff $k_1+k_3=k_2+k_4$. Thus casework on this equal value gives our answer of
\[1^2+2^2+\ldots+n^2+(n+1)^2+n^2+\ldots+2^2+1^2 = \boxed{\frac{2n^3+6n^2+7n+3}{3}}. \quad \blacksquare\]

Let the point where $L_i$ meets the table top be $V_i$, and the point where it touches the ground after being cut be $F_i$. Clearly, the slope of $V_1V_2$ must be equal to that of $V_4V_3$, giving us $n-k_1-n+k_2 = n-k_4-n+k_3 \implies k_2-k_1 = k_3-k_4$, and the number of 4-tuples satisfying this is clearly $2 \cdot \sum_{k = 1}^{n} (n+1-k)^2 + (n+1)^2$, which is easy to simplify.


inequalities
search
inequalities proposed

Very nice inequality. I solve it by contradiction method. The solution bases on the following result

If $ a+b+c=3$ then
\[ 3(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})+7a^{2}b^{2}c^{2}\le 16.\]

Is it also your solution, Virgin?

A very interesting idea is that Virgin's Inequality is totally equivalent to my inequality (actually, VASC and I also proposed this inequality, I just proposed it before him, so both VASC and me appear as its author in my book)

\[ \sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}\ge \frac{3}{2}.\]

You can see, Virgin's one is symmetric, but my own is cyclic.

More interesting, the solutions to both of them are [b]totally the same[/b]. Thanks to Virgin to show me this relationship.

The following inequality holds for $ p \ge \frac{49+9\sqrt{17}}{32}=2.691...$:
\[ \sqrt{\frac{a}{b+pc}}+\sqrt{\frac{b}{c+pa}}+\sqrt{\frac{c}{a+pb}}\ge \frac{3}{\sqrt{1+p}}. \]

[b][u]Extension[/u] (similarly to the last Cartoaje's inequality).[/b] $ \left\{\begin{array}{c}p\in N^{*}\ ,\ \{x,y,z\}\subset (0,\infty )\\\ p(x+y+z)+1+p^{3}=xyz\end{array}\right\|$ $ \implies$ $ \sum\frac{1}{\sqrt x}\ge \frac{3}{\sqrt{1+p}}$.

[quote="Virgil Nicula"]$ \left\{\begin{array}{c}\{x,y,z\}\subset (0,\infty )\\\ 3(x+y+z)+28=xyz\end{array}\right\|$ $ \implies$ $ \sum\frac{1}{\sqrt x}\ge \frac{3}{2}$.[/quote]

I found a very nice solution, but just for the case $ x,y,z\in [0,8]$. It is dedicated to Mircea Lascu and his current Article on Gazetta Magazine.

We consider the following equivalent problem

\[ 3(x+y+z)+7=16xyz \ (x,y,z\in [0,2]).\]

Prove that $ \sqrt{yz}+\sqrt{zx}+\sqrt{xy}\ge 3\sqrt{xyz}$.

Among three numbers $ x-1,y-1,z-1$ there exist two numbers having the same sign. Assume that
\[ (y-1)(z-1) \ge 0 \Rightarrow 1+yz \ge y+z.\]
Let $ t=\sqrt{yz}$ then

\[ 3(x+1+yz)+7 \ge 16xyz \Rightarrow 10+3t^{2}\ge x(16t^{2}-3).\]
\[ \Rightarrow \frac{1}{\sqrt{x}}\ge \sqrt{\frac{16t^{2}-3}{3t^{2}+10}}.\]

By AM-GM, we have

\[ \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\ge \frac{1}{\sqrt{x}}+\frac{2}{\sqrt[4]{yz}}\ge \sqrt{\frac{16t^{2}-3}{3t^{2}+10}}+\frac{2}{\sqrt{t}}=A.\]

It is easy to conclude that $ A\ge 3$ if $ t\le 2$.

[quote="Vasc"]The following inequality holds for $ p \ge \frac {49 \plus{} 9\sqrt {17}}{32} \equal{} 2.691...$:
\[ \sqrt {\frac {a}{b \plus{} pc}} \plus{} \sqrt {\frac {b}{c \plus{} pa}} \plus{} \sqrt {\frac {c}{a \plus{} pb}}\ge \frac {3}{\sqrt {1 \plus{} p}}.
\]
[/quote]

This is a very hard inequality of Vasc again unsolved. :wink:

I would like to see a solution, if possible. :roll:

[quote="hungkhtn"][quote="Virgil Nicula"]$ \left\{\begin{array}{c}\{x,y,z\}\subset (0,\infty ) \\
\ 3(x \plus{} y \plus{} z) \plus{} 28 \equal{} xyz\end{array}\right\|$ $ \implies$ $ \sum\frac {1}{\sqrt x}\ge \frac {3}{2}$.[/quote]

I found a very nice solution, but just for the case $ x,y,z\in [0,8]$. It is dedicated to Mircea Lascu and his current Article on Gazetta Magazine.

We consider the following equivalent problem
\[ 3(x \plus{} y \plus{} z) \plus{} 7 \equal{} 16xyz \ (x,y,z\in [0,2]).
\]
Prove that $ \sqrt {yz} \plus{} \sqrt {zx} \plus{} \sqrt {xy}\ge 3\sqrt {xyz}$.

Among three numbers $ x \minus{} 1,y \minus{} 1,z \minus{} 1$ there exist two numbers having the same sign. Assume that
\[ (y \minus{} 1)(z \minus{} 1) \ge 0 \Rightarrow 1 \plus{} yz \ge y \plus{} z.
\]
Let $ t \equal{} \sqrt {yz}$ then
\[ 3(x \plus{} 1 \plus{} yz) \plus{} 7 \ge 16xyz \Rightarrow 10 \plus{} 3t^{2}\ge x(16t^{2} \minus{} 3).
\]

\[ \Rightarrow \frac {1}{\sqrt {x}}\ge \sqrt {\frac {16t^{2} \minus{} 3}{3t^{2} \plus{} 10}}.
\]
By AM-GM, we have
\[ \frac {1}{\sqrt {x}} \plus{} \frac {1}{\sqrt {y}} \plus{} \frac {1}{\sqrt {z}}\ge \frac {1}{\sqrt {x}} \plus{} \frac {2}{\sqrt [4]{yz}}\ge \sqrt {\frac {16t^{2} \minus{} 3}{3t^{2} \plus{} 10}} \plus{} \frac {2}{\sqrt {t}} \equal{} A.
\]
It is easy to conclude that $ A\ge 3$ if $ t\le 2$.[/quote]

This solution is OK but it works only for the case $ x,y,z\in [0,8]$ so it is not a solution for the inequality

 \[ \sqrt {\frac {a}{b \plus{} 3c}} \plus{} \sqrt {\frac {b}{c \plus{} 3a}} \plus{} \sqrt {\frac {c}{a \plus{} 3b}}\ge \frac {3}{2}.
\]

with $ a\plus{}b\plus{}c\equal{}3$ :(

That is old, Manilo. Try to search in the forum. Both Vasc and I posted this problem before and there is a very nice solution of Arqady by contradiction method.

Yes, the link is

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=863565#863565

but I am searching for a lot of months a direct solution :wink: that only can_hang2007 has found.

A lot of congratulation for you, can_hang2007.

[quote="hungkhtn"]Very nice inequality. I solve it by contradiction method. The solution bases on the following result

If $ a \plus{} b \plus{} c \equal{} 3$ then
\[ 3(a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2}) \plus{} 7a^{2}b^{2}c^{2}\le 16.
\]
Is it also your solution, Virgin?[/quote]

How does this give a contradiction? You still have $ \equal{}$...?

I guess I'm asking how you precisely phrase the argument.


function
analytic geometry
graphing lines
slope

Here's some AMC-level questions (from the 2001 NV Prize Exam).  Each question has a number which corresponds to where you'd probably find a problem like this on an AMC10.

1. For each positive integer n, let a_n = 1/n - 1/(n+1).  What is the sum of the first 2001 terms of this sequence?  (3)

(1) 2000/2001
(2) 2001/2000
(3) 2001/2002
(4) 2002/2001
(5) none of these

2. What is the period of the funcion f(x) = (sin x)(sin(:pi: - x))? (12)

(1) :pi:/2
(2) :pi:
(3) 3 :pi: /2
(4) 2 :pi:
(5) 1

3. One line has slope (5-sqrt(24))^(-1) and a different line has slope 5 + sqrt(24).  What can be concluded about the lines? (15)

(1) They are parallel
(2) They are perpendicular
(3) Their union includes a 45 degree angle
(4) Their union includes a 60 degree angle
(5) None of the above

and from part II:

4. Prove: tan :theta:  - cot :theta:  + 2 cot 2 :theta: = 0. (n/a)

5. I am twice as old as my son.  28 years ago I was 9 times as old as my son was.  How old is my son now? (10)

Taking a break from homework...

[color=white]1.  a_n could be found by 1/n - 1/(n+1), so if you substitute n = 1 and n = 2, you can see that :

a_1 = 1 - 1/2 = 1/2
a_2 = 1/2 - 1/3 = 1/6

The sum of the first two terms of the sequence is 4/6, or 2/3.  There is a pattern reoccuring that the sum of a_n + a_(n-1)+....+ a_1 will be n/(n+1).  Therefore, the sum of the first 2001 terms will be (3), 2001/2002.

5. Let x be the age of the father now.  Let y be the age of the son now.

x=2y
(x-28)=9(y-28)

x=64, and y=32.  The son is 32 years old now.[/color]

Oasis, it appears you are hiding your text by turning it white.  Instead, I suggest you use "Spoiler text," the button for which is available just above font color.

Number 4:

[hide]

cot2 :theta: = (1-tan^2 :theta: )/2tan :theta:

Rewrite as:

(tan^2 :theta: -1) / tan :theta: + (1-tan^2 :theta: )/tan :theta:  = 0

[/hide]

rlee wrote:Also, for #1:
[hide]
Oasis's solution is fine, but it may be much simpler to see that the series telescopes. (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... (1/2000 - 1/2001) + (1/2001 - 1/2002). All terms cancel except first and last, and so the answer is 1 - 1/2002 = 2001/2002
[/hide]

Yes, I was thinking that there was a easier way to that question because I vaguely recall a question very similar to that is in the 2002 or the 2001 AMC 10.  I did solve for it this way when I was practicing for the AMC, but didn't remember it off the top of my head...

JBL, I'll remember that!

I'll take the leftovers.

(2) I just solved it by looking at it graphically in my head, but an algebraic description is that [hide]sin( :pi: -x)=sin(x), so the function reduces to sin :^2: (x). The period of sin(x) is 2 :pi: , but since sin(x)=-sin( :pi: +x), sin(x) repeats itself every  :pi: except for alternating signs. Since x :^2: =(-x) :^2: , sin :^2: (x) has a period of  :pi: [/hide].

(3) [hide](5- :rt24: ) :^-1: = 1/(5- :rt4: ). Multiplying numerator and denominator by the "conjugate" 5+ :rt4: to rationalize the denominator, we have (5+ :rt24: )/(5 :^2: -24) = 5+ :rt24: , or the same slope as the second line. Hence, the two lines are parallel.[/hide]


superior algebra
superior algebra unsolved

Let $ \rho : G \to GL(V)$ be a representation and let $ \varphi : G \to \mathbb{C}$ be some complex function. The Fourier transform $ \hat{\varphi}(\rho)$ is defined by $ \hat{\varphi}(\rho) \equal{} \sum_{g \in G} \varphi (g) \rho (g)$. Prove the Fourier inversion formula: $ \varphi (g) \equal{} \frac{1}{|G|} \sum_{\sigma} [\dim(V_{\sigma}) trace(\sigma(g^{\minus{}1})\hat{\varphi}(\sigma)) ]$ , where the summation is over all irreducible representations $ \sigma : G \to GL(V_{\sigma})$ of $ G$.

[quote="Thunder33"]Let $ \rho : G \to GL(V)$ be a representation and let $ \varphi : G \to \mathbb{C}$ be some complex function. The Fourier transform $ \hat{\varphi}(\rho)$ is defined by $ \hat{\varphi}(\rho) \equal{} \sum_{g \in G} \varphi (g) \rho (g)$. Prove the Fourier inversion formula: $ \varphi (g) \equal{} \frac {1}{|G|} \sum_{\sigma} [\dim(V_{\sigma}) trace(\sigma(g^{ \minus{} 1})\hat{\varphi}(\sigma)) ]$ , where the summation is over all irreducible representations $ \sigma : G \to GL(V_{\sigma})$ of $ G$.[/quote]

Recall:
(1) $ \sum_{\chi \in Irr(G)} \chi(1)^2 \equal{} |G|$
(2) $ \sum_{\chi \in Irr(G)} \chi(1)\chi(s) \equal{} 0$ if $ s \neq 1$.

Now, since everything in sight is linear in $ \varphi$, it will suffice to check the equality when $ \varphi(g) \equal{} \delta_{gh}$ for some fixed $ h \in G$. In this case, if $ \sigma$ is a representation of $ G$, then $ \hat{\varphi}(\sigma) \equal{} \sigma(h)$. This gives:
$ \frac{1}{|G|} \sum_{\sigma} \dim(V_{\sigma}) Tr(\sigma(g^{\minus{}1})\hat{\varphi}(\sigma)) \equal{} \frac{1}{|G|} \sum_{\chi \in Irr(G)} \chi(1) \chi(g^{\minus{}1}h)$
If $ g \equal{} h$, then by (1), the sum is 1. If $ g \neq h$, then by (2), the sum is 0.


[quote="mysmartmouth"]If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x+y+z$ is

$\text{(A) }18\qquad\text{(B) }19\qquad\text{(C) }20\qquad\text{(D) }22\qquad\text{(E) }24$[/quote]

[hide]Dividing the first 2 equations, we get $\frac{z}{y}=2$.  Multiply this by the third equation to get $z^{2}=144$ so $z=12$.  This mean $x=4$ so $y=6$.  22[/hide]

[hide]if we multply all of them and square root, we get abc=something...then divide that something by ab=something to get c, etc[/hide]

Very sorry to revive, but is it possible to use that $ (x\plus{}1)(y\plus{}1)(z\plus{}1) \equal{} xyz \plus{} xy \plus{} xz \plus{} yz \plus{} x \plus{} y \plus{} z \plus{} 1$?  I see that the methods posted above are much better, but I was just wondering.  Again, sorry to revive, but I didn't want to start a new thread since there was already one.

I can't see how something like SFFT would work here; perhaps someone else can.

I don't know if it would, I just noticed how we were given those values and $ xyz$ is easily calculated.  But again, I do realize that the other ways are the fastest.

The only thing I'm a little wary about is that you would have to solve anyways for the $ (x\plus{}1)(y\plus{}1)(z\plus{}1)$?  Or am I misreading the approach?

If xy=24, then x can be 3,4,6, or 8 as well as y.

If you try and substitute those numbers into the other equations, then you will find your answer.

Take 3 and 8 for example...

(3)(8)=24     so x=3 and y=8

(3)z=48 divide 48 by 3 which equals 16 so z=16

(8)(16)=128....which doesnt work. If you switch the x and y with 8 and 3, it still wont work.

Eventually after you try substituting x with 4 and y with 6, you will find your answer.

(4)(6)=24   x=4  y=6

(4)(z)=48 divide 48 by 4 which equals 12 making z=12

(6)(12)=72

[color=red]4+6+12=22 making your answer D.[/color]

[quote="darkspartan931"]what else could they be?[/quote]

Real numbers. All you're given is that $ x,y,z$ are positive.
Assuming they're integers is an approach that just happens to work out here (this being an AMC10 #10) but it rarely will in difficult problems. It also leads you to a guess-and-check kind of solution, which is a lot more work than making the simple observations that people above made.

Not necessarily. Remember that the sum of conjugate irrational numbers and the products can both be integers and therefore, you can't conclude that operands in a sum are integers based on the fact that the sum itself is an integer.

[quote="darkspartan931"]by the way....they have to be integers because the answers are integers :|[/quote]

This doesn't work out in the problem, but say $ (a,b,c) \equal{} (5 \plus{} \sqrt{2}, 9 \minus{} \sqrt{2} \plus{} i, \minus{}i)$. Here, $ a \plus{} b \plus{} c \equal{} 14$, an integer even though the original numbers were irrational and imaginary.

[quote="007math"][quote="darkspartan931"]by the way....they have to be integers because the answers are integers :|[/quote]

This doesn't work out in the problem, but say $ (a,b,c) \equal{} (5 \plus{} \sqrt {2}, 9 \minus{} \sqrt {2} \plus{} i, \minus{} i)$. Here, $ a \plus{} b \plus{} c \equal{} 14$, an integer even though the original numbers were irrational and imaginary.[/quote]

In *this* problem, it states that $ x, y$, and $ z$ are positive, so you
cannot use $ y \equal{} 9 \minus{} \sqrt{2} \plus{} i$ and $ z \equal{} \minus{}i$ as a part of counterexample
because these are not positive as the problem required.

[quote="Arrange your tan"][quote="007math"][quote="darkspartan931"]by the way....they have to be integers because the answers are integers :|[/quote]

This doesn't work out in the problem, but say $ (a,b,c) \equal{} (5 \plus{} \sqrt {2}, 9 \minus{} \sqrt {2} \plus{} i, \minus{} i)$. Here, $ a \plus{} b \plus{} c \equal{} 14$, an integer even though the original numbers were irrational and imaginary.[/quote]

In *this* problem, it states that $ x, y$, and $ z$ are positive, so you
cannot use $ y \equal{} 9 \minus{} \sqrt {2} \plus{} i$ and $ z \equal{} \minus{} i$ as a part of counterexample
because these are not positive as the problem required.[/quote]

Then we can just have $ x \equal{} 5, y \equal{} 9 \minus{} \sqrt2, z \equal{} \sqrt2$.  $ x, y, z$ are positive and real,so it works out with $ x \plus{} y \plus{} z \equal{} 14$.

$ \frac{x}{y} \equal{}\frac{2}{3},\  \frac{y}{z} \equal{}\frac{3}{6} ,\ x: y: z\equal{}2: 3: 6,\ x\equal{}2t,\ y\equal{}3t,\ z\equal{}6t\\
x\cdot y\equal{}24\Rightarrow 6t^{2}\equal{}24\Rightarrow t\equal{}2\Rightarrow x\equal{}4,\ y\equal{}6,\ z\equal{}12\Rightarrow x\plus{}y\plus{}z\equal{}22$

[quote="darkspartan931"]what else could they be?[/quote]
In general, the solution for positive $ x, y, z$ and $ a, b, c$ to $ xy \equal{} a$, $ yz \equal{} b$, and $ xz \equal{} c$ is
$ x \equal{} \sqrt {\frac {ac}{b}}, y \equal{} \sqrt {\frac {ab}{c}}, z \equal{} \sqrt {\frac {bc}{a}}$

So $ x \plus{} y \plus{} z$ would equal $ \frac {ab \plus{} bc \plus{} ac}{\sqrt {abc}}$

The question is, given that $ \frac {ab \plus{} bc \plus{} ac}{\sqrt {abc}}$ is an integer, does that imply that $ x \equal{} \sqrt {\frac {ac}{b}}, y \equal{} \sqrt {\frac {ab}{c}}, z \equal{} \sqrt {\frac {bc}{a}}$ must all be integers?

[hide]I believe the answer is yes. But a proof of it probably doesn't belong in this forum, so I've created a separate thread for it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=321893]here[/url].[/hide]

lol just start guessing, I got it on the first try

[hide=sol] $$(xy)(xz)(yz)=24\cdot 48 \cdot 72\Longrightarrow xyz=288$$
$$xyz\div xy=z\Longrightarrow z=12$$
$$xyz\div zx=y=6$$
$$xyz\div yz=x=4$$
$$x+y+z=12+6+4=\boxed{22}$$[/hide]

It's basically a simple symmetric system. It's no different from \begin{align} x+y&=24 \\ x+z &=48 \\ y+z&=72 \end{align} (in format, of course).

[quote=mysmartmouth]If $ x$, $ y$, and $ z$ are positive with $ xy \equal{} 24$, $ xz \equal{} 48$, and $ yz \equal{} 72$, then $ x \plus{} y \plus{} z$ is

$ \textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24$[/quote]

[b][hide]Solution:
x=4
y=6
z=12[/hide][/b]

[quote=mysmartmouth]If $ x$, $ y$, and $ z$ are positive with $ xy \equal{} 24$, $ xz \equal{} 48$, and $ yz \equal{} 72$, then $ x \plus{} y \plus{} z$ is

$ \textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24$[/quote]

[hide=Solution]xy=24, xz=48, yz=72, z/y=2 So, z^2 is 144 and z is 12. So, x=4, and y=6. Summing everything up, we get 4+6+12=22. The answer is (D)[/hide]

The fastest way to do it is to divide the first and second equation and multiply it by the third. Then everything will fall in place.

I just took the square root after multiplying all 3 equations.


geometry
perimeter
incenter
geometric transformation
homothety

See the figure below :

[color=red](edited)[/color]

[b]Hint [/b]

 We have $ CP \equal{} \tau \minus{} \beta, BP \equal{} \tau \minus{} \gamma$

Take AI, extend it to meet BC at X and use length chasing to show that it is parallel

[quote="AstroPhys"]How can $ \text{IP} \| \text{AM}$?[/quote]

Yes , you are right and sorry. I hope now to be correct.

[quote="stergiu"][b]Hint [/b]
We have $ CP \equal{} \tau \minus{} \beta,\ BP \equal{} \tau \minus{} \gamma$[/quote][color=blue]Nice hint![/color] :)

[img]11784[/img]

Let $ \tau$ be the semi-perimeter of the triangle

Let $ D$ be the point where the incircle touches the side $ BC$. Then

$ BD\equal{}\tau\minus{}b \equal{} CP$ and
$ CD \equal{} \tau\minus{}c \equal{} BP$

So the $ M$ is the midpoint of $ DP$

Now we use the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106337]theorem[/url]:

[quote="shobber"]The incircle of triangle $ ABC$ touched side $ BC$ at $ D$. Let the midpoint of $ BC$ be $ M$. Show that $ MI$ bisects $ AD$ where $ I$ is the incentre of triangle $ ABC$[/quote]

Let $ N$ be the midpoint of $ AD$. Then $ MI$ is passing through $ N$. From the triangle $ DAP$ we can see that $ MN\parallel AP$

Let $ B'$ be a point on the extension of the ray $ CB$, such that $ BB' \equal{} BA$. Similarly:
Let $ C'$ be a point on the extension of the ray $ BC$, such that $ CC' \equal{} CA$

$ \triangle BB'A$ is isosceles. So let $ \angle BB'A \equal{} \angle BAB' \equal{} \phi$. The angle $ CBA$ is external for the triangle $ BB'A$.

So $ B \equal{} 2\phi \equal{} > \phi \equal{} \angle BB'A \equal{} \frac {1}{2}B \equal{} > B'A \parallel BI$, making the triangles $ DBI,DB'A$ similar

Similarly $ C'A \parallel CI$

Now let $ A'$ be a point such that $ B'A' \parallel BA$ and $ C'A' \parallel CA$

Clearly the triangles $ ABC,A'B'C'$ are similar (having B=B' and C=C') and $ A$ is the incenter of $ \triangle A'B'C'$ (it belongs to two angle bisectors)

also, $ P$ is the midpoint of $ B'C'$ because $ PB \plus{} BB' \equal{} PB \plus{} BA \equal{} PC \plus{} CA \equal{} PC \plus{} CC'$

We have homothety with center $ D$.
Any point of $ A,B,C,I,M$ maps to the corresponding $ A',B',C',A,P$

So $ AP \parallel IM$


limit
trigonometry
function
calculus
real analysis
real analysis unsolved

Find

 :huh:

 $ \lim_{x \to+\infty}$  $ x^{2}-xsin(x)+cos(x)+sin(x)cos(x)$

is it +infinity?bcoz the last two terms are bounded and the second term has limit 1.so x^2 shall dominate the others.

Yes it's $ +\infty$ but can you post your solution?

Thank you!

[quote="batata"]is it +infinity?bcoz the last two terms are bounded and [color=blue]the second term has limit 1[/color].so x^2 shall dominate the others.[/quote]
[color=blue]It isn't true[/color]
[quote="giannis"]$ \lim_{x \to+\infty}$  $ x^{2}-xsin(x)+cos(x)+sin(x)cos(x)$[/quote]
We have for example for $ x>1$:
$ x^{2}-x\sin x+\cos x+\sin x\cos x\geq x^{2}-x\sqrt{x}+1+2=x^{2}\left(1-\frac{1}{\sqrt{x}}\right)+3$ and it has limit $ \infty$\for $ x\to\infty$, so our limit is $ \infty$ too.

My solution but can anybody tell me if i am right?

we have $ L=\lim_{x \to+\infty}$  $ x^{2}-xsin(x)+cos(x)+sin(x)cos(x)=\lim_{x \to+\infty}$  $ x^{2}(1-\frac{sin(x)}{x}+\frac{cos(x)}{x^{2}}+\frac{sin(2x)}{2x^{2}})$

but for $ x>0$ we have  $ |\frac{sin(x)}{x}|\leq \frac{1}{|x|}=\frac{1}{x}$ $ \Longleftrightarrow$ $ -\frac{1}{x}\leq \frac{sin(x)}{x}\leq \frac{1}{x}$

but $ \lim_{x \to+\infty}\frac{1}{x}=\lim_{x \to+\infty}-\frac{1}{x}=0$ so $ \lim_{x \to+\infty}\frac{sin(x)}{x}=0$

With the same way $ \lim_{x \to+\infty}\frac{cos(x)}{x^{2}}=0$ and $ \lim_{x \to+\infty}\frac{sin(2x)}{2x^{2}}=0$

so $ L=+\infty (1-0+0+0)=+\infty$

[quote="giannis"]My solution but can anybody tell me if i am right?[/quote]
you're right  :wink:

Yes, that works. This is about size, and dividing by the large term is a good way of getting at that.
In big-O/little-o notation, the function is $ x^{2}+O(x)$ or $ x^{2}+o(x^{2})$, and the $ x^{2}$ term dominates.

[quote="giannis"][color=darkred]Find [/color]$ L=\lim_{x \to\infty}\left(x^{2}-x\sin x+\cos x+\sin x\cos x\right)$.[/quote]
[color=darkblue][b]Proof.[/b] Observe that $ f(x)=x^{4}\cdot\frac{x-sin x}{x^{3}}+g(x)$, where $ \lim_{x\to\infty}x^{4}=\infty$, $ \lim_{x\to\infty}\frac{x-\sin x}{x^{3}}=\frac{1}{6}\ \ (*)$

and the function $ g(x)=\cos x+\sin x\cos x$ is bounded, more exactely $ |g(x)|\le 2$.

Since $ \infty\cdot\frac{1}{6}=\infty$ and $ \infty+\mathrm{''bound''}=\infty$, then $ L=\lim_{x\to\infty}f(x)=\infty$.

[b]Remark.[/b] I used for $ (*)$ the [b]l"Hospital rules[/b] : $ \lim_{x\to\infty}\frac{x-\sin x}{x^{3}}=$ $ \lim_{x\to\infty}\frac{1}{3}\cdot\frac{1-\cos x}{x^{2}}=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$.[/color]

That's just wrong, Virgil. L'Hopital's rule is entirely counterproductive here.

Wow such complications for $ f(x) \geq x(x-1)-2 \to \infty$ ...  :huh:

[quote="jmerry"][color=darkred][b]That's just wrong, Virgil[/b]. l'Hospital's rule is entirely counterproductive here.[/color][/quote]
[color=darkblue][b]O.K., V. Vornicu[/b] (the your proof if more nice than [b]Giannis[/b]') and [b]Jmerry[/b] !
[b]Jmerry[/b], you are too much roughly with me ! Here I wanted solve otherwise.
The [b]Giannis[/b]' proof is standard - [b][u]first of all, the most rapid term pull out as factor[/u][/b] ![/color]

[quote="jmerry"]If I gave something like this to my calculus students, giannis' solution is what I would want to see. I'd actually expect less explanation, but it's exactly the right idea.[/quote]

Exactly !

 Moreover , Giannis is a high school student (he will be in class 12 in September) and his solution is based on the standard marerial  which we have in our school book for the  central uni - examination .

 The romanian textbook in analysis(class 11) is more rich in theorems than our book in Greece(CLASS 12). Giannis can not use Val.V's theory for a solution in his exams. I write all these things to inform  you about some differences in the theory which some text books contain.

 [u]Babis[/u]

[quote="Virgil Nicula"][quote="giannis"][color=darkred]Find [/color]$ L \equal{}\lim_{x\to\infty}\left(x^{2}\minus{}x\sin x\plus{}\cos x\plus{}\sin x\cos x\right)$.[/quote]
[color=darkblue][b]Proof.[/b] Observe that $ f(x) \equal{} x^{4}\cdot\frac{x\minus{}sin x}{x^{3}}\plus{}g(x)$, where $ \lim_{x\to\infty}x^{4}\equal{}\infty$, $ \lim_{x\to\infty}\frac{x\minus{}\sin x}{x^{3}}\equal{}\frac{1}{6}\ \ (*)$

and the function $ g(x) \equal{}\cos x\plus{}\sin x\cos x$ is bounded, more exactely $ |g(x)|\le 2$.

Since $ \infty\cdot\frac{1}{6}\equal{}\infty$ and $ \infty\plus{}\mathrm{''bound''}\equal{}\infty$, then $ L \equal{}\lim_{x\to\infty}f(x) \equal{}\infty$.

[b]Remark.[/b] I used for $ (*)$ the [b]l"Hospital rules[/b] : $ \lim_{x\to\infty}\frac{x\minus{}\sin x}{x^{3}}\equal{}$ $ \lim_{x\to\infty}\frac{1}{3}\cdot\frac{1\minus{}\cos x}{x^{2}}\equal{}\frac{1}{3}\cdot\frac{1}{2}\equal{}\frac{1}{6}$.[/color][/quote]

$ \lim_{x\to\infty}\frac{1\minus{}\cos x}{x^{2}}\equal{}\frac{1}{2}$?

sorry i read it as xsin(1/x).but u were exactly right.THANKs

[quote="giannis"]My solution but can anybody tell me if i am right?

we have $ L =\lim_{x\to+\infty}$  $ x^{2}-xsin(x)+cos(x)+sin(x)cos(x) =\lim_{x\to+\infty}$  $ x^{2}(1-\frac{sin(x)}{x}+\frac{cos(x)}{x^{2}}+\frac{sin(2x)}{2x^{2}})$

but for $ x > 0$ we have  $ |\frac{sin(x)}{x}|\leq\frac{1}{|x|}=\frac{1}{x}$ $ \Longleftrightarrow$ $ -\frac{1}{x}\leq\frac{sin(x)}{x}\leq\frac{1}{x}$

but $ \lim_{x\to+\infty}\frac{1}{x}=\lim_{x\to+\infty}-\frac{1}{x}= 0$ so $ \lim_{x\to+\infty}\frac{sin(x)}{x}= 0$

With the same way $ \lim_{x\to+\infty}\frac{cos(x)}{x^{2}}= 0$ and $ \lim_{x\to+\infty}\frac{sin(2x)}{2x^{2}}= 0$

so $ L =+\infty (1-0+0+0) =+\infty$[/quote]

This is also the standard solutions for Japanese high school students. I like it very much! :lol:


$ f (x) \equal{} x \plus{} 3$ and $ g(x) \equal{} x^2 \minus{}6$, what is the value of $ f (g(2))$?


limit
logarithms
integration
calculus
calculus computations

Find the limit \[\lim_{n\to \infty}\frac{\sqrt[n+1]{(2n+3)(2n+4)\ldots (3n+3)}}{n+1}\]

The other classical method:

If $a_{n}= \frac{\sqrt[n+1]{(2n+3)(2n+4)\ldots (3n+3)}}{n+1}$, then
\[b_{n}= \ln a_{n}= \frac1{n+1}\cdot \left( \sum_{k=0}^{n}\ln (2n+3+k)-(n+1) \ln (n+1) \right) . \]
By Cesaro-Stolz we get that
\begin{eqnarray*}\lim_{n \to \infty}b_{n}&=& \lim_{n \to \infty}\left( \ln[(3n+4)(3n+5)(3n+6)]-\ln[(2n+3)(2n+4)(n+2)]+\ln \frac{(n+1)^{n+1}}{(n+2)^{n+1}}\right) \\ \ &=& \ln \frac{27}{4e},\end{eqnarray*}
so $\lim_{n \to \infty}a_{n}= \frac{27}{4e}$.
(Maybe I made some computational mistakes since I don't get the same answer.)


The k-th symmetric mean is defined as follows:
\[ \sigma _k (a) \equal{} \sqrt[k]{{\left( {\frac{{a_1 a_2 ...a_k  \plus{} a_1 a_2 ...a_{k \minus{} 1} a_{k \plus{} 1}  \plus{} ....... \plus{} a_{n \minus{} k \plus{} 1} a_{n \minus{} k \plus{} 2} ...a_n }}
{{\left( \begin{gathered}
  n \hfill \\
  k \hfill \\
\end{gathered}  \right)}}} \right)}}
\]
Prove that if $ k<l$
then,
\[ \sigma _k (a) > \sigma _l (a)
\]
My solution is pretty ugly...... :blush:
Please provide a neat solution. :D

This is known as [url=http://www.artofproblemsolving.com/Wiki/index.php/Maclaurin%27s_Inequality]Maclaurin's Inequality[/url].


linear algebra
matrix
linear algebra unsolved

X and B_0 are real square matrices -nxn.  we define inductively the sequence B_i=B_(i-1)X-XB_(i-1). prove that if X=B_(n^2) then X=0.

By the same method one can prove a much stronger result:
Proposition: if $\exists{k}\geq1$ s.t. $B_{k}=0$ then $B_{0}X=XB_{0}$
Proof: It's sufficient to prove : let $k\geq1$ s.t. $B_{k+1}=0$; then $B_{k}=0$. I use notations of my precedent post.
Indeed $B_{k}X=XB_{k}$ and $trace(B_{k})=0$. $trace(B_{k}^{r})=trace(\sum_{i=0}^{k}(-1)^{i}comb(k,i)X^{i}B_{0}X^{k-i}B_{k}^{r-1})=$
$\sum_{i=0}^{k}(-1)^{i}comb(k,i)trace(B_{0}X^{k}B_{k}^{r-1})=0$. Thus $B_{k}=0$.

Currently I make many errors .
1)  In my first post (june 16) I proved only that if $B_{k}=X$ then $X$ is nilpotent or $X^{n}=0$.
Thus if $k\geq{2n}$ then, in the formula giving $B_{k}$, the index $i$ or the index $k-i$ is $\geq{n}$; thus if $\exists{k}\geq{2n}$ s.t. $B_{k}=X$ then $X=0$, what is stronger that the required result.
2) In my second post (june 18) I proved only that if $B_{k+1}=0$ with $k\geq1$ then $B_{k}$ is nilpotent.


I just finished the school target round today and got first in my school.
How bout everyone else?

I don't think we're allowed to discuss it...and I haven't taken it yet :P

Oh. Wait when will everyone be done? Is there a date?

The school round dates are determined by each school. The school can have them anytime from the beginning of the school year to practically the day before the chapter competition. The chapter competition dates vary greatly too, so there is no set date. It might be safe to discuss after the state competitions. Ask rrusczyk for clarification.

This test should not be discussed until after the last chapter test has been given. A lot of schools may give this test as the last test prior to chapter.  Donot discuss the problems.

without discussing the actual content, can we discuss our own results?

just outta curiosity, whats the harm of informing about difficulty?

It would actually be in your best interest to edit those scores out. It might tell people about the difficulty if they know your math skill, etc.

@nikeballa96: Well, they might prepare differently. For example, the test might be easy/hard which would make people practice making less careless mistakes/practice harder problems and stuff like that.

So we're not allowed to say anything about 09 Mathcounts until Feb 28.

feb 1. chapters are feb 1-feb 28, thus school competitions must be done by feb 1, thus we can discuss SCHOOL on the first, CHAPTERS in the 28th, and STATES...idk when.

Um no, nikeballa96, bad logic. :P Some places don't have chapter until the 28th, so schools in those regions might not take the school test until the 28th (in the most extreme case).


we have 64 1x1x1 small cubes and we want to stack them up into a 4x4x4 larger cube size. among these 64 cubes, half of them are totally painted with color blue and half of them are painted with color red. so in other words, 32 cubes are painted blue on each of its 6 faces and 32 cubes are painted red on each of its 6 faces too. what is the maximum number of red 1x1 faces can be seen on the outside of the larger 4x4x4 cube?

[quote="DarkKnight"][hide]
I think it's 72[/hide][/quote]
and your reason is...?

[hide]Well, in a 4x4x4 cube, there are 24 edge cubes (2 faces) and of course 8 corner cubes (3 faces). In total, that adds up to 32 cubes and thats how many red cubes there are. So obviously, you use all the red cubes as edges and corners. 24*2=48. 3*8=24. 24+48=72.[/hide]


Να υπολογιστεί το ολοκλήρωμα :

$ \int \frac{cos(y^2)}{y} dy$

Τι ειναι το ΠΚ?

Πανεπιστήμιο Κρήτης

Μετά από πολυήμερη απουσία, είχε χαλάσει ο υπολογιστής ,επανέρχομαι
Είναι γνωστό ότι το $ \int \frac{sinx}{x\minus{}a}dx$ δεν υπολογίζεται στοιχειωδώς
Τώρα $ \int \frac{cosx}{x}dx\equal{}\frac{1}{2}\int \frac{cos(x^2)}{x^2}d(x^2)\equal{}\frac{1}{2}\int \frac{cosy}{y}dy\equal{}\frac{1}{2}\int \frac{sinw}{w\minus{}\pi /2}dw$

Μα δεν έχουμε να κάνουμε με το $ \int \frac{cosy}{y}dy$ αλλά με το $ \int \frac{cos(y^2)}{y}dy$ :maybe:

Ναι αλλά αν στο $ \int \frac{cosy}{y} dy$ θέσουμε $ y\equal{}u^2$ τότε $ \int \frac{cosy}{y} dy\equal{}\int \frac {cos(u^2)}{u^2} 2u du\equal{}2 \int \frac {cos(u^2)}{u} du$ οπότε αν το ζητούμενο ολοκλήρωμα υπολογιζόταν στοιχειωδώς θα υπολογιζόταν στοιχειωδώς και το $ \int \frac{cosy}{y} dy$ άτοπο... :maybe:

[quote="bouzoukman"]Με αφορμή αυτή την άσκηση θα θέσω μία ερώτηση που πάντα είχα απορία...

Πότε ένα ολοκλήρωμα θεωρούμε ότι δεν υπολογίζεται στοιχειωδώς,δηλαδή πότε είμαστε σίγουροι ότι δεν ξέρουμε και δεν μπορούμε αναλυτικά και με κανόνες να το υπολογίσουμε??? :wink:

Πάντα είχα απορία πότε ένα ολοκλήρωμα καταλαλήγουμε να το κατηγοριοποιούμε στα ολοκληρώματα που δεν υπολογίζονται στοιχειωδώς...[/quote]

simfwna me enan kathigiti mou (sto AUTH) e ama ipologizotane tha to eixame vrei  :rotfl:

[quote="SorrowMan"]

simfwna me enan kathigiti mou (sto AUTH) e ama ipologizotane tha to eixame vrei  :rotfl:[/quote]

Βασικά κι εγώ σκέφτηκα ότι για να μην το έχουν βρει τόσοι και τόσοι μεγάλοι μαθηματικοί λογικά δεν θα υπολογίζεται...

[quote]Πότε ένα ολοκλήρωμα θεωρούμε ότι δεν υπολογίζεται στοιχειωδώς,δηλαδή πότε είμαστε σίγουροι ότι δεν ξέρουμε και δεν μπορούμε αναλυτικά και με κανόνες να το υπολογίσουμε???[/quote]
Δεν το έχω πολυψάξει το θέμα αλλά κάτι βρήκα [url=http://en.wikipedia.org/wiki/Differential_Galois_theory]εδώ[/url].Γενικά ένα ολοκλήρωμα λέμε ότι δεν υπολογίζεται στοιχειωδώς αν δεν μπορεί να εκφραστεί συναρτήσει στοιχειωδών συναρτήσεων κι όχι αν δεν υπάρχουν συγκεκριμένες μεθοδολογίες υπολογισμού... :wink:


AoPS is pleased to announce that we have a new MATHCOUNTS book for sale!

It is called [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=2004]Practice Competitions for MATHCOUNTS[/url] and is written by the Massachusetts state team coach Josh Frost (whom you may know better as [b]frost13[/b] on AoPS).  The book consists of 4 [b]all-new[/b] complete practice MATHCOUNTS contests, including Sprint, Target, and Team rounds, and has complete detailed solutions to all the problems.

[quote="frost13"][quote="ernie"]What level are the tests?[/quote]

The short answer is that they are all state+ level.  My 8 kids who all qualified for state this year all scored lower on the tests in the book than they did at state this year or really even on the average of all the old state tests.  Each test has a mix of problems ranging in difficulty as most tests do.  I never write tests for the chapter level as we start preparing for states from day 1, so they are all at least state level.

Mr Frost[/quote]

Sounds very interesting. :)  Are there any really tough questions that require a lot of thought?

The answer to that will differ for everyone.  My team averaged 4-5 points lower on my tests than they did on the real state this year.  They averaged a 144 on the AMC10 and a 37.5 on the tests in my book.  Are there a lot of easy problems?  Sure, number 1 through 10 should be easy for people really working on MATHCOUNTS, but still challenging for beginners (my 3 6th graders averaged about 15-18 on the tests).

They also include team rounds which are hopefully harder.  In fact there was one problem I had to convince the editors at MATHCOUNTS that my answer was correct on since they kept getting something different :lol:

how long did u take you to make/publish the book, like when did u start writing the book..just wondering? :oops:

I started last summer and spent several hours a week for a couple months to write the tests.  Then once I had the problems I wrote the detailed solutions.  Then it took several months going back and forth with MATHCOUNTS to make edits to make sure wording wasn't ambiguous, grammar and punctuation was correct, solutions were clear and that formatting was pretty :lol: Probably 6 - 8 months from start to finish.

does this book include tests we haven't done at mc practice? or did you test them all out on us?

[quote="frost13"]Students from my school were the Guinea Pigs for all the tests in the book, although they didn't necessarily see all the solutions in the book.[/quote]
hmm, I've heard that some of your students create mathcounts mock tests. Does this book have any of them? (like the Shen Test, Waffle's Mock Mathcounts, etc.)

Hehe. You'll definitely have lots of buyers. I bet there are a lot of ppl in the world like my coach. He buys almost every book on aops.  :wink:

No, these are all my tests.  My students do write some good problems, but none are ready for publication :-) In fact I encourage everyone to try writing their own tests, especially those going to Nationals.  The tests always end up having several problems that are impossible, or that are ambiguous.  Writing this really made me appreciate the work MATHCOUNTS does.  When you write problems you have a particular solution in mind a lot of the time, but wording the problems so that no other interpretation can be used is not easy.

It took giving the tests to kids, editing based on their comments, and then 3 more rounds of edits to get these finalized for the book!!

[quote="waterbird"]How much would a person who was aspiring to make it to nationals benefit from this book? Would you consider this book very helpful if you were preparing your students for states with it?

Also, you would consider these problems more difficult than a regular state sprint test, right?[/quote]

As stated above, these are some of the tests I used in preparing my team for states this year where they averaged a 41.5 on the individual test.  They are definitely meant for state preparation.  As for how good a judge I am of that, I have been coaching MC for 11 years now, my teams have been 1st in the state the last 9 years (2 different states) and I have always had at least one kid make Nationals (2 kids 5 times).

umm so u wouldnt recommend this book for anyone preparing for school, chapter, nationals, etc.

As I stated in an earlier post, I don't write tests to prepare for chapter, I prepare them for states.  So if you are hoping to make Nationals, this is what the book was written for.

As for the questions about whether I'd recommend it for people preparing for the other 3 levels, I would think the answer obvious, and my guess there was some wise-guy sarcasm in there, but here goes.  These are the tests I used this year before chapter.  All 8 of my kids scored at least 41 and finished in the top 13 in a very tough chapter that had 4 USAMO qualifiers and 3 of the top 4 in one of the tougher states(MA)(6 of my kids were in the top 14 in MA).  So yes, the tests are good for preparing for all levels.  My goal is to write tests so that if the questions are easy like chapter you have the skills you need to be successful, but if the tests are hard like state, you are not thrown and do just as well at state.

I hope that's clear.

Are these questions harder than most State/National Level problems?

Read the thread.  They are supposed to be a little harder than states.

I'm currently persuading my mom to buy this book(It looks cool!), and she told me that I could buy it with my own money, not hers.

 :mad: I only have 30 in hand, because of my invitation to 4 consecutive birthday parties.

Can we see a sample...cuz we cant just buy it wthout lookin at some samples :wink:

I agree. We should at least see like a question from each section of the rounds. :arrow:

Actually, Mr. Frost created those problems.... :|

Maybe a sample problem please?

Guys, I am not posting "sample problems", what would be the point.  The tests are just like other MC tests.  They start off easy and they get harder as you go through the test.  If I posted the hardest problem in the book, then when you got the book you'd be disappointed that it was too easy.  If you are looking for more practice TESTS (not just random problems) to prepare for states next year, this is what the book was made for.  There are also solutions to every problem, from the easiest to the more challenging, so that all levels of student, including those without a serious coach, can learn how to do MATHCOUNTS style problems.


I am trying to prove this by contradiction in the following way

What I have so far is: if 7(2)^-1/2 is rational then it must be expressed as p/q where q>0 and, and p and q must not both be even otherwise you could simplify the fraction.

then,

(7(2)^-1/2)^2 = (p/q)^2
49(2)=p^2/q^2

which means p^2 = 49(2)(q^2), and because there is a 2 in there you can say that p^2 is even and thus p is even.
Therefore p can be written as 2k, and the above expression as
(2k)^2 = 49(2)(q^2),

which means that
(q^2) =4k^2/49*2, thus  (q^2) =2k^2/49, and q=(2)^-1/2(7)(k)

And this is where I am stuck, I dont know how to show that q is even. That would contradict the initial assumption and finish the prove....

Any suggestions????

[quote="Piffo"]I am trying to prove this by contradiction in the following way

What I have so far is: if 7(2)^-1/2 is rational then it must be expressed as p/q where q>0 and, and p and q must not both be even otherwise you could simplify the fraction.

then,

(7(2)^-1/2)^2 = (p/q)^2
49(2)=p^2/q^2

which means p^2 = 49(2)(q^2), and because there is a 2 in there you can say that p^2 is even and thus p is even.
Therefore p can be written as 2k, and the above expression as
(2k)^2 = 49(2)(q^2),
[/quote]

$ 4 k^2\equal{}49 \cdot 2 \cdot q^2$
$ 2k^2\equal{}49 \cdot q^2$
then using your same argument as before: q is even which is a contradiction both p and q are even.

haha, that is true..Feeling a bit idiotic here...

Thanks for pointing it out :)

Lemmas:

- The product of a rational and irrational number is irrational.
- The inverse of an irrational number is irrational.

The irrationality of $ 7(2)^{\minus{} \frac{1}{2} }$ then follows from the irrationality of $ \sqrt{2}$, which is a well-known proof along the same lines.


Excess saturate sodium fluoride solution is added to a solution of aluminim sulfate, i felt like it impossible to predict the exact product.....

a solution of hydrogen peroxide is exosed to strong sunlight, why aren't the product O2+H2?

Also, why doesn't NH4Cl aqueous in the reaction "ammonia gas is added to hydrogen chloride gas"?

1) Aluminium fluoride is a possible product: check the corresponding Ks value.

2) Hydrogen peroxide is expected to disproportionate into oxygen and water. Formation of O2 and H2, even if thermodynamically favoured, is too unfavourable from a kinetic point of view to be of real importance.

3) What does that sentence mean?

for part C if you add a gas to a gas there is no solution for the NH4Cl to dissolve in. Thus I believe that the product will be in solid phase.

Yes, that's correct: ammonium chloride forms as a solid.


[color=darkblue]Let two circles $ (O; R)$ and $ (O'; R)$, let $ (O; R)$ meet $ (O'; R)$ at $ A$ and $ B$. Let $ I$ a point lie out of two circles. Let $ (d)$ is a line through point $ I$ and $ (d)$ meet $ (O; R)$ at $ M$ and $ N$. Let $ MB$ meet $ (O'; R)$ at $ M'$ and let $ NB$ meet $ (O'; R)$ at $ N'$. Prove that $ \angle{OAO'} \equal{} \angle{NAN'} \equal{} \angle{MAM'}$.

[b][u]Note:[/u][/b] $ (O; R)$ and $ (O'; R)$ are two circles same radius.[/color]

Dear thanhnam and Mathlinkers,
iwhy are you considering the points I and I'?
The result is "the angle of two circles".
An angle chasing gives the result.
Sincerely
Jean-Lois

It is obvious because AN=AN' AM=AM'  AO=AO'


limit
induction
inequalities
calculus
calculus computations

$ 0<x_1<1$ and $ x_{n\plus{}1}\equal{}x_n(1\minus{}x_n).$ Prove that $ \lim_{n \rightarrow \infty }nx_n\equal{}1.$

[hide]My attempt: $ \forall n\geq 2$ , $ 0<x_n \leq \frac 14$ and $ x_{n\plus{}1}\minus{}x_n\equal{}\minus{}x_n^2<0 \Longrightarrow x_{n\plus{}1}<x_n \Longrightarrow \lim_{n \rightarrow \infty }x_n\equal{}0$ and we know $ nx_n<(1\plus{}x_n)^n \Longrightarrow \lim_{n \rightarrow \infty }nx_n \leq 1$ .[/hide]

Ehsan2004, this is a quite easy exercice. Well, fisrt of all, we can prove by an easy induction that $ x_{n}\in (0,1)$. Hence the sequence $ (x{}_{n}{}_{n\geq 1}$ is bounded. Since $ \frac{x_{n+1}}{x_{n}}=1-x_{n}<1$ it follows that $ (x_{n})_{n}$ is decreasing, so by Weierstrass criterion it follows that our sequence is convergent, so there exists $ \lim_{n\to\infty}x_{n}=l\in [0,1]$. By passing to the limit in our reccurence, we get $ l=l(1-l)$, so $ l=0$.
Now, by applying Stolz-Cesaro theorem we get \[ \lim_{n\to\infty}nx_{n}=\lim_{n\to\infty}\frac{n}{1/x_{n}}=\lim_{n\to\infty}\frac{(n+1)-n}{1/x_{n+1}-1/x_{n}}=\lim_{n\to\infty}\frac{x_{n}x_{n+1}}{x_{n}-x_{n+1}}=\lim_{n\to\infty}\frac{x_{n}^{2}(1-x_{n})}{x_{n}(1-1-x_{n}}=1,\]
because $ \lim_{n\to\infty} x_{n}=0$.

As you have already shown $ x_{n}$ is decreasing and $ x_{n}\rightarrow 0$ $ \Leftrightarrow$
$ \frac {1}{x_{n}}$ is increasing and $ \frac {1}{x_{n}}\rightarrow \infty$

So from Stolz-Cesaro lemma we have:

$ \lim_{n\rightarrow \infty}nx_{n} \equal{} \lim_{n\rightarrow \infty} \frac {n}{\frac {1}{x_{n}}} \equal{} ^{S C} \lim_{n\rightarrow \infty} \frac {n \plus{} 1 \minus{} n}{\frac {1}{x_{n \plus{} 1}} \minus{} \frac {1}{x_{n}}} \equal{} \lim_{n\rightarrow \infty} \frac {x_{n \plus{} 1}x_{n}}{x_{n} \minus{} x_{n \plus{} 1}} \equal{} \lim_{n\rightarrow \infty} \frac {x_{n \plus{} 1}}{x_{n}} \equal{} 1$
(for the 4 th step I  used the recurence relation : $ x_{n \plus{} 1} \minus{} x_{n} \equal{} x_{n}^2$)

$ x_n$ decreasing to zero does not imply $ \lim_{n\rightarrow\infty}\frac{x_{n+1}}{x_n}=1$ nor does it allow the application of Stolz Cesaro, e.g., the sequence $ x_n=e^{-an}$ fulfills your properties for $ a>0$, but the limit is $ e^{-a}$.

I guess a proof has been provided in some other post already. First note, that $ f(x)=x(1-x)\leq\frac{1}{4}$ and it is monotone increasing between $ 0$ and $ \frac{1}{2}$.

Claim 1: $ \forall n>1: 0<x_n<\frac{1}{n+1}$. The LHS inequality is obvious and the RHS inequality is proved by induction with $ x_2=f(x_1)\leq \frac{1}{4}<\frac{1}{3}$ and $ x_{n+1}=f(x_n)\leq f\left(\frac{1}{n+1}\right)=\frac{n}{(n+1)^2}<\frac{1}{n+2}$.

Claim 2: $ \forall n>1,k\geq 0: x_{n+k}\geq\frac{1}{x_n^{-1}+k(1-x_n)^{-1}}$. This is shown by induction on $ k$ with $ k=0,1$ obvious. Direct calculation applying the induction hypothesis yields:

$ \begin{array}{rl}x_{n+k+1}= & x_{n+k}(1-x_{n+k})\geq \frac{x_n^{-1}+k(1-x_n)^{-1}-1}{(x_n^{-1}+k(1-x_n)^{-1})^2}\\[0.25cm]
 = & \frac{(x_n^{-1}+k(1-x_n)^{-1}-1)(x_n^{-1}+(k+1)(1-x_n)^{-1})}{(x_n^{-1}+k(1-x_n)^{-1})^2(x_n^{-1}+(k+1)(1-x_n)^{-1})}\\[0.25cm]
 = & \frac{(x_n^{-1}+k(1-x_n)^{-1})^2+kx_n(1-x_n)^{-2}}{(x_n^{-1}+k(1-x_n)^{-1})^2(x_n^{-1}+(k+1)(1-x_n)^{-1})}\\[0.25cm]
 \geq & \frac{1}{x_n^{-1}+(k+1)(1-x_n)^{-1}}\end{array}$

Claim 1 shows that $ \lim_{n\rightarrow\infty}x_n=0$ holds and that $ nx_n<1$ is bounded. Moreover, we have $ (n+1)x_{n+1}=(n+1)x_n(1-x_n)>(n+1)x_n\left(1-\frac{1}{n+1}\right)=nx_n$, i.e. $ nx_n$ is an increasing sequence. Since it is bounded it has some limit $ l\leq 1$. Using claim 2 we have  $ l\geq (n+k)x_{n+k}\geq\frac{n+k}{x_n^{-1}+k(1-x_n)^{-1}}$. Considering the limit $ k\rightarrow\infty$ we see that $ l\geq 1-x_n$ and since $ x_n$ converges to zero, we must have $ l\geq 1$ and therefore $ l=1$.

sorry, [b]Svejk[/b], I only looked at your last limit $ \lim\frac{x_{n\plus{}1}}{x_n}$, which triggered my email. In this particular case, the last limit is equal to $ \lim (1\minus{}x_n)\equal{}1$ using the recursive definition of the sequence. Well, then you obviously can apply Stolz Cesaro (nicely spotted !)

thanks very much, but what is the Stolz -Cesaro theorem ?????? :blush:

have a look at  [url]http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf[/url] , it covers some modifications as well