calculus
integration
real analysis
derivative
complex analysis
function
calculus computations

Evaluate $ \int_0^\infty e^{\minus{}x^2} cos(2xt) dx$

The answer is $ \frac{\sqrt{\pi}}{2} e^{\minus{}t^2},$ I know some how that I need to use the fact of $ \int_0^\infty e^{\minus{}x^2} dx\equal{}\frac{\sqrt{\pi}}{2}$ and apply the Lebesgue Dominated Theorem, but I could not figure it out.

Please help. Thanks alot.

This is the Fourier transform of a Gaussian. I know at least three ways of doing this. One of those requires a complex variables justification; I'll leave that one alone. Here's another one:

Define $ F(t)\equal{}\int_{0}^{\infty}e^{\minus{}x^2}\cos(2xt)\,dx$

Take the derivate with respect to $ t.$ (That's where you're going to use the LDCT, in justifying taking the derivative under the integral sign.)

$ F'(t)\equal{}\minus{}\int_0^{\infty}2xe^{\minus{}x^2}\sin(2xt)\,dx.$

Now integrate that by parts:

$ F'(t)\equal{}\left.e^{\minus{}x^2}\sin(2xt)\right|_{x\equal{}0}^{x\equal{}\infty}\minus{} \int_0^{\infty}2te^{\minus{}x^2}\cos(2xt)\,dx.$

That is, $ F'(t)\equal{}\minus{}2tF(t).$

This is a first order linear (and also separable) ODE and is easy to solve: $ F(t)\equal{}F(0)e^{\minus{}t^2}.$

The fact you speak of - that is, the integral of the Gaussian itself - suffices to determine $ F(0)$ and give the appropriate initial condition for that differential equation.

:D
$ I\equal{}\frac{1}{2}\int_{\minus{}\infty}^{\infty}e^{\minus{}x^2}\cos(2xt)\,dx\equal{}\frac{1}{4} \int_{\minus{}\infty}^{\infty}e^{\minus{}x^2}\left( e^{2\imath xt} \plus{} e^{\minus{}2\imath xt} \right)\, dx$

$ \equal{}\frac{1}{4}\int_{\minus{}\infty}^{\infty} e^{\minus{}x^2\plus{}2\imath xt}\plus{}e^{\minus{}x^2\minus{}2\imath xt} \, dx\equal{}\frac{e^{\minus{}t^2}}{4}\int_{\minus{}\infty}^{\infty} e^{\minus{}\left(x\minus{}\imath\,t\right)^2}\plus{}e^{\minus{}\left(x\plus{}\imath\,t\right)^2} \, dx$

$ \equal{}\frac{e^{\minus{}t^2}}{4}\left(\sqrt\pi\plus{}\sqrt\pi\right)\equal{}\frac{\sqrt\pi\,e^{\minus{}t^2}}{2}$

to solve the last $ 2$   integrals, put $ y\equal{}x\minus{}\imath\,t$ and $ y\equal{}x\plus{}\imath\,t$, respectively.
in both cases, you end up with $ \int_{\minus{}\infty}^{\infty} e^{\minus{}y^2} \,dy\equal{}\sqrt\pi$

[quote="misan"]to solve the last $ 2$   integrals, put $ y \equal{} x \minus{} \imath\,t$ and $ y \equal{} x \plus{} \imath\,t$, respectively.[/quote]
That's exactly the method requiring complex variables justification that I said above I would leave alone for now. Those may look like substitutions in an integral, but they aren't really substitutions; they're acts of shifting a complex line integral sideways to a different contour. They are valid, of course; the justification quotes Cauchy's Theorem (that is, the line integral of a holomorphic function about a closed contour is zero.)

So that's two of the three methods I referred to above. My third method involves characterizing the functions in $ \mathbb{R}^2$ that are both radial (depending only on $ r\equal{}\sqrt{x^2\plus{}y^2}$ ) but also expressible as a product of a function of $ x$ times a function of $ y.$


For a triangle ABC:

If ra=2, rb=3, rc=6, find the missing parts r=?, K=?, R=?, a=?, b=?, c=?

where r is the radius of the circle tangent to the triangle, K is the area of the triangle, R is the radius of the circle surrounding the triangle and a,b,c are the sides opposite the vertices A,B,C.

So far I have r=11, K= square root of 396, a= 27*sq.rt 11/ 11, b=16* sq.rt 11/ 11, c=6*sq.rt 11/ 11

I can't figure out R and quite frankly all of the other numbers seem a bit off???

abc=4RK
bc=2Rh
K=sr (s= 1/2 perimeter of triangle)
K= (s-a)ra etc.

Yes- your $r$ is way off, and that's trouble for everything. Remember that the incircle is smaller than any excircle.

$\frac1r=\frac1{r_a}+\frac1{r_b}+\frac1{r_c}=\frac36+\frac26+\frac16=1$, so $r=1$.
$K^2=rr_ar_br_c=1\cdot2\cdot3\cdot6=36$, so $K=6$.
This gives $s=6$, $s-a=3$, $s-b=2$, $s-c=1$, so $a=3,b=4,c=5$
Now $R=\frac{abc}{4K}=\frac{60}{24}=\frac52$.


function
Asymptote
ratio
limit
calculus
calculus computations

Hey can anyone help me with this limit problem. Its on my webwork and its the only one i cant figure out.

Lim       sqrt (x^2 + 3x) / (4-4x)
x> -inf

i found -1/4 but then I found -inf, both answers are wrong and according to the webwork, this function has an horizontal asymptote so it obviously rules out my answer of -infinity.

How do I even do a problem like this?

As $ x\to\minus{}\infty,$ $ \sqrt{x^2\plus{}3x}$ looks a lot like $ \sqrt{x^2}.$ But $ \sqrt{x^2}$ is not $ x.$ (Which is how you would have gotten the incorrect answer $ \minus{}\frac14.$) Instead, it's $ |x|,$ which is $ \minus{}x$ for negative $ x.$ The fraction looks like $ \frac{\minus{}x}{\minus{}4x}$ and the limit is $ \plus{}\frac14.$

Here's my take on this problem:

Since $ x$ approaches $ \minus{} \infty$, we can take $ x$ to be negative.

Thus, by dividing both the numerator and denominator by $ \sqrt {x^2}$, which is equal to $ \minus{} x$, we get $ \frac {\sqrt {x^2 \plus{} 3x}}{4 \minus{} 4x} \equal{} \frac {\sqrt {1 \plus{} \frac {3}{x}}}{ \minus{} \frac {4}{x} \plus{} 4}.$ Now we have the fraction where both numerator and denominator have limits defined at $ \minus{} \infty$; thus we can express the limit at $ \minus{} \infty$ as the ratio of those two limits. Since $ \lim_{x \to \minus{} \infty}\sqrt {1 \plus{} \frac {3}{x}} \equal{} 1$ and $ \lim_{x \to \minus{} \infty}( \minus{} \frac {4}{x} \plus{} 4) \equal{} 4$, the limit of their ratio at $ \minus{} \infty$ is equal to $ \frac {1}{4}$.


trigonometry
logarithms
function
quadratics
complex analysis
algebra
quadratic formula

$e^{ix}=\cos{x}+i\sin{x}$
The rest is left as an excercise to the reader.

Well, actually, it's not that simple.  The inverse sine function is multivalued in the complex plane.  In particular, the principle value of $\arcsin z$ involves taking branch cuts along $\vert\Re (z)\vert>1$.

Now, assuming you're working with a reasonable branch of the inverse sine, you can use the techniques magnara described to finish it off.

I'm still confused, can someone show me how to evauate $\sin^{-1}{(2)}$?

Unfortunately, $\sin^{-1}(2)$ is not well defined, e.g, there are infinitely many numbers such that $\sin z=2$.  Now, $\text{Sin}^{-1}(2)$ (the principal value) IS well defined; if no one else posts a method, I will later.

[quote="Magnara"]What is $\frac{e^{ix}-e^{-ix}}{2i}$?[/quote]

It's clearly $\sin x$.  But the idea is that in the complex plane, which we need to go to to make sense of something like $\sin^{-1}(2)$, there are multiple numbers such that $\sin z=2$.  So we make use of something called a branch cut to create a single valued branch where we can make sense of something like $\sin^{-1}(2)$, or $(-1)^{\frac {1}{2}}$, etc.

blah, keep in mind that terms like "principal value" and "branch cut" don't mean anything to people who haven't taken Complex Analysis yet. You have to try to answer the question on the level and in the language that the person asking can understand.

For example, if they are asking a question about $\sin^{-1}{(x)}$ it might be safe to assume they know about one-to-one functions and inverses - so you can couch your explaination in those terms, and then make an extension into principal values and branch cuts from there.

The first step to answer your question Jimmy is to prove to yourself that $\sin {(x)} = \frac{e^{ix} - e^{-ix}}{ix}$; and then ask the question, what does $\sin^{-1}{(2)}$ really mean?

Well, assuming that $\sin z=\frac {e^{iz}-e^{-iz}}{2i}$,

it's clear that via the laws of exponents that $\sin(z+2\pi)=\sin z$.

So this takes care of the multivalued part, e.g, it make sense to say that $\sin^{-1}(z)=y$ if $\sin y=z$, but there's problems with that definition. So what we do is create a branch cut, that is, we specify that the argument of a complex number must be within a certain interval (usually a multiple of $2\pi$).  After doing this, we can say that $\sin^{-1}(2)$ specifies a unique value, in the same sense as there are two numbers such that $x^2=4$ but we say that $\sqrt x=2$.

I understand that by using the identity $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$, we can see that  $\sin{(\frac{\pi}{2}-i\ln{\sqrt{3}+2})}=2$, but how would you find the principle value of say $\sin^{-1}{(3)}$?  What I mean is that given the principle value I can verify it, but how would one find the principle value?

Here's the answer (if it helps):
[hide]$\frac{\pi}{2}-i\ln{(2\sqrt{2}+3)}$[/hide]

I answered my own question, $\frac{e^{ix}-e^{-ix}}{2i}=3 \Rightarrow e^{ix}-e^{-ix}-6i=0$  I then multiplied both sides of the equation by $e^{ix}$, thus $e^{2ix}-1-6ie^{ix}=0$.  Let $u=e^{ix} \Rightarrow u^2-6iu-1=0 \Rightarrow u=(2\sqrt{2}+3)i$. Then set this equal to $e^{xi}$, thus $x=\frac{\pi}{2}-\ln{(2\sqrt{2}+3)}i$.  Thanks for your help everyone.

In general if we seek to find $\sin^{-1}{(k)}$ for some real value $k$ with $|k| \ge 1$, we seek to find a complex number $\theta$ with $0 \le \Im (\theta) < 2\pi$ such that $\sin {(\theta)} = k$.

This means that, $\frac{e^{i\theta}-e^{-i\theta}}{2i} = k$. Letting $z = e^{i\theta}$, we have:
$z - 1/z = 2ik$
$z^2 - 2ikz - 1 = 0$

By quadratic formula, $z = ik \pm \sqrt{1-k^2}$, or $z = i(k \pm \sqrt{k^2-1})$.

Now replacing $z$ by $e^{i\theta}$ (and choosing the positive sign in order for $\theta$ to be in the right region) we have:
$e^{i\theta} = i(k + \sqrt{k^2-1})$
$i\theta = \ln {(i)} + \ln {(k + \sqrt{k^2-1}) = \frac{\pi}{2}i + \ln {(k + \sqrt{k^2-1})}}$
$\theta = \frac{\pi}{2} - i \ln {(k + \sqrt{k^2-1})}$.


[hide]the mean of 2 numbers is 25, so the sum has to be 25x2 or 50.  Plus, the range is 10, so one has to be 10 more than the other.  Given this information, we could see that:

x+(x+10)=50     2x+10=50     2x=40      x=20

since one number is 20, then the other is 20+10, or 30[/hide]

[quote="mathblast"]HELP!!!

The mean of two numbers is 25
The range is 10
What are the two numbers

How do you figure this problem out????[/quote]

[hide="faster"]You could also figured out two numbers differ by 10 but have mean of 25:

so two numbers should be $25 \pm \frac{10}{2}$, which are $20$ and $30$.

(One number is 5 more than 25, the other one is 5 less than 25, which gives the average)[/hide]

[hide="Simple Problem solution"] Basically, we know that

$x-y=10$
$\frac{x+y}{2}=25$

So just multiply the second equation by 2 to get

$x+y=50$

$x-y+x+y=50+10$
$2x=60$
$x=30$
$y=20$

So our solutions are [b]x=30,y=20[/b].[/hide][/hide][/quote]


how use the Latex in FORUM.... i would write any problem, but I dont know how....Help me please... I dowload the TeXnicCenter,

See section ''Latex help'' on the top of forum.  :wink:

In particular, go to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url]. You don't need TexnicCenter if you just want to post LaTeX on the forum.

Try [url=http://www.artofproblemsolving.com/Wiki/index.php/Latex]here[/url] as well.


The median of the integer set n, n+3, n+4, n+5, n+6, n+8, and n+12 is 15. What is the mean of the set?

[hide]Because there are 7 terms the 4th is the median.  This means that n+5=15.  Then n=10 so the mean is 108/7[/hide]

yup, that's what i got.....just take the numbers, and find the median of that set. Since it says the median is 15 , and the number is n-5, it is n-5=10. Then n=5. Plug it into the rest of the set, and divide by the number of numbers in the set(7) to get your answer.


summer program
MathPath
MATHCOUNTS
geometry
calculus
AMC
USA(J)MO

What books are missing from our book list [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Books.php]here[/url].

We'll be updating it soon.  So far, I'll be adding Andreescu's 102 Combinatorial Problems, Problem Solving Through Problems, Music of the Primes, the Hungarian Problem Book III, 500 Mathematical Challenges.  What else should be there?  (I may not add 'em all, but I'll add as many as I have time for).  Also, where should your suggestion be in the list (beginning, intermediate, advanced; text, problems, or general interest)?

Thanks!

Simon's list is [url=http://www.amazon.com/exec/obidos/tg/listmania/list-browse/-/2J1BZG9FH9R8L/qid%3D1053886525/sr%3Dlmlf-10/ref%3Dbr%5Flmlf%5F%5F10/102-0505827-1800914]here.[/url]

Generatingfunctionology is a very important one.  Plus, it's free online [url=http://www.math.upenn.edu/%7Ewilf/DownldGF.html]here.[/url]

There are some good Number Theory books out there.  One I'd recommend is George E. Andrews' [i]Number Theory[/i], found [url=http://www.amazon.com/exec/obidos/tg/detail/-/0486682528/qid=1061754238/sr=2-1/102-0505827-1800914?v=glance&s=books]here.[/url]

I am in the process of updating my website list. I'm going to include descriptions, etc.  If you're considering updating the websites or contests page, I can send you something that already has descriptions and hyperlinks.

Thanks much - we'd be happy to have that list when you're finished.

I would add From Zero to Infinity and The Pleasures of Counting, both books I learned about on the MathPath site.

[quote="claudius0425"]The Elements of Abstract and Linear Algebra, by Edwin H. Connell, from whence I gleaned all of my basic abstract algebra knowlege (doesn't do high level Galois theory).
Avaliable online in several formats at:
http://www.math.miami.edu/~ec/book/[/quote]

By the way, the forum about the algebra text at that same miami.edu Web site also uses the phpBB forum software, but the color scheme looks MUCH better and is MUCH more readable. Maybe this forum should adopt the same look.

"More Mathematical Challenges", Tony Gardiner, is a collection of problems from the UK Junior  Mathematical Olympiad.  (Perhaps, a bit similar in style and level to State or National MATHCOUNTS Team rounds.)  (via Cambridge University Press)

"Mathematical Puzzling", A. Gardiner, has a broader variety of creative puzzles than above book.  These 'fun' problems are not specifically contest problems.  (I suspect that A. Gardiner is for Anthony or Tony Gardiner.)  (via Dover)

Advanced : "Challenging Problems in Geometry 1" and "Challenging Problems in Geometry 2" by Alfred S. Posamentier and Charles Salkind.  I have the two 1970 volumes; but Dover might have combined them in 1990.  I also have seen "Challenging Problems in Algebra 1" and "Challenging Problems in Algebra 2" listed; but I don't own them.  Both are a bit dry, but great depth for improving advance geometry skills.

For high school level recreational reading, I recommend anything by Eli Maor, including:  "e: The Story of a Numer"; "Trigonometric Delights"; and "To Infinity and Beyond: A Cultural History of the Infinite".

For calculus students, William Dunham's "Journey Through Genius" is a brilliant blend of history and mathematics!

For USAMO and Putnam level (perhaps AIME as well), I recommend any of Ross Honsberger's books available from MAA:  "Mathematical" Gems or Morsels or Plums ... etc.  Each brief chapter presents an elegant problem and solution.  Often the problem originated from Putnam or Olympiad competitions.

To purchase used books at often reasonable prices, I have used these websites:  http://www.abebooks.com/  or  http://www.biblio.com/  .  Bookstores from across the US list their books on either or both sites with prices, shipping, and condition noted.  "Buyer beware", but I have had no problems.

Good reading!


limit
search
calculus
calculus computations

hello folks  :D

evaluate $ \lim_{n\to\infty}\;\;\sum_{k\equal{}1}^n\;\;\sqrt{\dfrac{1}{n^2}\plus{}\left( \sqrt{1\minus{}\left(\dfrac{k}{n}\right)^2 \;\;}\;\minus{}\; \sqrt{1\minus{}\left(\dfrac{k\minus{}1}{n}\right)^2 \;\;}\right)^2 \;\;}$

We have come full circle:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1408318761&t=155455


I have a question kind of like rvictoria's. How do you convert from square yards to square feet? Any help would be appreciated.

I do not understand that can u be a little more clear?

$1yd=3ft$
$1yd^{2}=3^{2}ft^{2}$

So to convert from square yards to square feet you multiply by nine.

Generally to convert something you obey the following rules:
If

$length_{1}=a(length_{2})$ then
$area_{1}=a^{2}(area_{2})$ and
$volume_{1}=a^{3}(volume_{2})$

Example: $1km=1000m$ so $1km^{2}=(1000)^{2}m^{2}$

1 square yard is like 3 feet by 3 feet, because 1 yard is 3 feet. So 1 square yard is equal to 9 feet


I made up a problem that was a bit too difficult, and so I made this problem up because it was similar and easier. I figured that if I understood this problem better that I would be able to solve the more difficult one. It stills seems quite difficult to solve this in an organized way. Of course we could use the brute force method and it would take forever. I have some other questions regarding the same situation but I just wrote the main question that I'm having some problems with.

10 people show up for game night at a friend’s house and put their hat on the hat rack as they come in the door. There is an explosion outside and they all jump up and rush out the door to see what it was, grabbing a hat without looking as they fly by the rack.

1. What is the probability that at least one of them has their own hat?

Let's generalize, and say there are $ n$ people at the party.  The probability that a person DOES NOT pick up his own hat is $ \frac{n\minus{}1}{n}$.  Since there are $ n$ people, the probability that no one picks up his own hat is $ \left(\frac{n\minus{}1}{n}\right)^{n}$.  Note that as $ n\rightarrow\infty$, then this probability approaches $ \frac{1}{e}$.

The question asked is the complement to what the probability is that no one gets their own hat. This question is answered by [url=http://en.wikipedia.org/wiki/Derangement]derangements[/url].

Thanks for the replys, so would you say that it's 1-(1334961/10!)?

A note on disambiguation:
There are many "hat problems". I know this one as the "Vienna hat problem"- which means that it has in fact been asked many times before, and with hats.

[quote="MellowMelon"]The question asked is the complement to what the probability is that no one gets their own hat.[/quote]I'm aware of this, I wanted the OP to have to think a tiny bit. :P

[quote="JRav"]Let's generalize, and say there are $ n$ people at the party.  The probability that a person DOES NOT pick up his own hat is $ \frac {n \minus{} 1}{n}$.  Since there are $ n$ people, the probability that no one picks up his own hat is $ \left(\frac {n \minus{} 1}{n}\right)^{n}$.  Note that as $ n\rightarrow\infty$, then this probability approaches $ \frac {1}{e}$.[/quote]

The events of picking up hats are not independent. (It's quite easy to check that this formula fails for the $ n\equal{}2,3$ cases.)

I have a couple of pages full of numbers trying to show examples with up to 5 people and hats and have looked for the pattern and haven't found one. I have a couple of other pages of trying to compute the probability using combinations with probabilities and this is what I have so far.

I have (10 choose 1)*1/(10*9) + (10 choose 2)*1/(10*9*8) and so on until (10 choose 10) except skipping (10 choose 9) because you can't have one person with the wrong hat and all the others correct.

I ended up with 1/9 + 1/16 + 1/42 + 1/144 + 1/600 + 1/2880 + 1/15120 + 1/80640 + 1/3628800 = .20645 on my calculator, didn't feel like finding a common denominator.


function
limit
calculus
derivative
topology
calculus computations

We have a function $ f: \mathrm{R}^2\rightarrow \mathrm{R}$, given by $ f(x,y)=(\mathrm{distance}((x,y),[-1,1]\times\{0\}))^2$. Where is $ f$ continuous and where differentiable?

I do not have solutions, therefore I would like to know, if the following solution is valid:

$ f(x,y)=
\{
\begin{array}{ccl}
	(x-1)^2+y^2 &; & x\geq 1\\
	y^2 &; & -1<x<1\\
	(x+1)^2+y^2 &; & x\leq -1\\
\end{array} \;$

[b]Continuity:[/b]

Obviously all the subfunctions are continuous. Hence we have to study just the border cases: points in $ \{(1,y),y\in \mathrm{R}\}$ and $ \{(-1,y),y\in \mathrm{R}\}$. But since
$ f(1,y_0)=\lim_{\begin{array}{cc}
	x \uparrow 1\\
        y \rightarrow y_0
\end{array} \;}f(x,y)=y^2_{0}$, $ f$ is continuous also on the first set and similarly on the second. Therefore $ f$ is continuous on $ \mathrm{R}^2$.

[b]Differentiability:[/b]

$ \frac{\partial{f}}{\partial{x}}=\{
\begin{array}{ccl}
	2(x-1) &; & x\geq 1\\
	0 &; & -1<x<1\\
	2(x+1) &; & x\leq -1\\
\end{array} \;$

$ \frac{\partial{f}}{\partial{y}}=\{
\begin{array}{ccl}
	2y &; & x\geq 1\\
	2y &; & -1<x<1\\
	2y &; & x\leq -1\\
\end{array} \;$

With the same reasoning as above we conclude, that $ \frac{\partial{f}}{\partial{x}}$ and $ \frac{\partial{f}}{\partial{y}}$ are also continuous on $ \mathrm{R}^2$.

Continuity of $ f$ and both partial derivatives gives differentiability of $ f$ on $ \mathrm{R}^2$.

[quote="t0rajir0u"]I don't understand your notation, but whatever object you're trying to measure the distance to, your function shouldn't be differentiable there; consider the analogy with the one-variable case.[/quote]
The square distance is though.

What Kalle is trying to explain: the correct function (the distance) should be the square root of the function tmrfea wrote down.

As a function from $ [0,\infty)$ to $ [0,\infty),$ the square root function is continuous everywhere and differentiable everywhere except zero. Hence, from the work tmrfea has already done, the correct distance function should be continuous everywhere (that's always true for $ g(x)\equal{}d(x,F)$ for any closed set $ F$ in any metric space), and differentiable everywhere except on $ [\minus{}1,1]\times\{0\}.$

Note that for $ |x|\le1,$ we have $ f(x,y)\equal{}|y|.$ There's no way that can be differentiable for $ y\equal{}0.$

Thank you all for your answers.

But $ f(x,y)$ is defined as a square distance of $ (x,y)$ from $ [\minus{}1,1]\times \{0\}$.
[quote]$ f(x,y)\equal{}(\mathrm{distance}((x,y),[\minus{}1,1]\times\{0\}))^{2}$[/quote]


geometry
geometric transformation
dilation

0.67 grams of potassium dichromate (VI), $K_{2}Cr_{2}o_{7}$, was dissolved in one litre of water to form a solution W. Solution W was diluted with water to five times its original volume to form solution X. Solution X was diluted with water to ten times its original volume to form solution Y.

What is the concentration of Y in g/L?
(Relative atomic mass: K, 39.1; Cr, 52.0; O, 16.0)

A. 0.0067
B. 0.0134
C. 0.0447
D. 0.1340

the number of mole of $K_{2}Cr_{2}O_{7}$ wont change under dilution with water . The original concentration is $0.67g/L$ .
using $M_{1}V_{1}=M_{2}V_{2}$ where $M_{1}$ is original concentration , $V_{1}$ is original volume etc we apply this formula two times to get the conc of Y be

$\frac{0.67}{5(10)}=0.0134g/L$

Hmm ... weird .. i din use the RMM of potassium dichromate at all  :roll:

Maybe the RMM's are there to confuse ppl.  :lol: Nice solution.

indeed it is a nice solution...
i just want to know what is this Malaysian quiz..some kind of national contest or something...
can you post the whole list of problems...
or at least the interesting ones?

I will try to post all the problems tomorrow if possible.. hehe. But I have to do a translation because half of the quiz is in Malay. This is a national level quiz for students in Malaysian Form 5 (Year 11\Grade 11).


[quote="warut_suk"]The 4th term of a geometric sequence is $108$ and 7th term is $2916$. What's the first term?[/quote]

[hide]
$ar^3=108$
$ar^6=2916$
$ar^3=\frac{2916}{r^3}$
$108r^3=2916$
$r^3=27$
$a=\boxed{4}$
[/hide]

[hide]The 4th term must have been multiplied by x 3 times to obtain the 7th term.

$108x^3=2916$
$x=3$

Then the third term is 108/3 or 36, the second term is 36/3=12, the first term is [color=orange]4 [/color][/hide]

[hide]THe first term in the sequence is $a$, the fourth term is $ar^3$, and the seventh term is $ar^6$, now:

$a=x$
$ar^3=108$
$ar^6=2916$
$ar^3 \cdot r^3=ar^6$
$r^3=27$
$\frac{ar^3}{r^3}=a$
$\boxed{a=4}$[/hide]

[hide]$a+ar+ar^2...$

$ar^3=108$

$\text{Factor 108 to get } 9\cdot12=3\cdot3\cdot4\cdot3=3^3 \cdot 2^2$

$ar^6=2916$

$ar^6=ar^3 \cdot r^3$

and then i don't know what to do

[size=75][color=green][mod edit]Substitute $ar^3=108$, so $r^3=27$. $a=108/r^3=4$ and we are done.[/color][/size][/hide]


algorithm
maxtrix
rounding
IMO Shortlist

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

Is this correct?
Assume the array is m rows x n columns. Let's assume for now that none of the numbers are integers. We can subtract any integer amount from any of the numbers without altering the problem. Thus, we can assume all the numbers are between 0 and 1, exclusively. In this case, the sum of all the rows will be less than n, and the sum of all columns will be less than m. We can make any of the numbers equal to either 0 or 1. Note that the sum of all the rows must equal the sum of all the columns, so the sum of the row totals equals the sum of the column totals. So if the sum total in row k is $ a_{k}$, we know there will be $ a_{k}$ 1's in that row and the rest 0s. Starting with the first row, place the $ a_k$ 1's in all the leftmost columns that currently need it, this must work in the end since the total number of 1's need for the rows is the same as the number needed for the columns.
Now, if s of the numbers in a row are integers then they will turn to a 0 but will not be needed for that row because the row sum will be less than n-s, so we are fine in this case.

How will you do in this case with "0" and "1",dgreenb801 ?
Collumn sums: 2 2 3 4 0
Row sums: 5 4 1 1

The column sum for the last column must be greater than 0 unless they are all integers, in which case the row sum would have to be less than 5, as I showed before.

[quote="dgreenb801"]Starting with the first row, place the $ a_k$ 1's in all the leftmost columns that currently need it, this must work in the end since the total number of 1's need for the rows is the same as the number needed for the columns.[/quote]  This is extremely sketchy, and it does not obviously rule out something like what No Reason wrote.  Since the problem is true, no one can actually produce a counter-example, but I'm certainly not convinced by the rigor of what you've written.  (I mean, you've given an algorithm that almost certainly works, but your proof of its correctness is just an assertion.)

Hmm, I'm not really sure of how you would rigorously prove something that seems so obvious. Possibly we could assume the process doesn't work, then either we need more 1s or we run out of 1s, in either case it's a contradiction as the row sums don't equal the column sums?
Anyways, if there are any integers, we can just leave the 0s where they are and the row and column they are in will just need that much less.

[quote="dgreenb801"]Hmm, I'm not really sure of how you would rigorously prove something that seems so obvious.[/quote]
Can't prove it $ \Rightarrow$ not obvious.

If your proof is an algorithm, you must do two things:
1. Show the algorithm does what it's supposed to.
2. Show the algorithm terminates.
Leaving out one of these steps as "trivial" is at your own peril.

My personal suggestion is to go back to the drawing board; patching up what you have I think is actually harder than finding some of the other solutions to this problem.

This may not work but I think it does.

First, note that we only have to consider the fractional parts of all of the numbers in the grid, so that every number is in [0,1). Say the rectangle is $ m\times n$ ($ m$ rows, $ n$ columns). Along the side of length $ m$, let the sums of the numbers in the rows be $ a_{1},a_{2},\ldots,a_{m}$. Similarly, along the side of length $ n$, let the columns have sums $ b_{1},b_{2},\ldots,b_{n}$. We wish to show that given $ \sum a_{i}\equal{}\sum b_{i}$, we can place 0's and 1's in the rectangle so that we get the desired sums. We do this by induction.

The claim is clear when $ m\equal{}n\equal{}1$. Now, consider adding a row or a column to a board, so that it's, say $ m\times(n\plus{}1)$. We now have $ a_{1},a_{2},\ldots,a_{m}$ and $ b_{1},b_{2},\ldots,b_{n\plus{}1}$. Note that there are at least $ b_{n\plus{}1}$ nonzero values among $ a_{1},a_{2},\ldots,a_{m}$, since if not, the sum of the values in the $ (n\plus{}1)$-st column will be at most $ b_{n\plus{}1}\minus{}1$, because there are at most this many nonzero values, and the values in these boxes are all less than 1 (in the original board). Now, take the corresponding rows to $ b_{n\plus{}1}$ of the nonzero $ a_{i}$, and put a 1 in the box of the $ n\plus{}1$-st column for each. What's left is an $ m\times n$ board, with non-negative sums for the rows and $ b_{1},b_{2},\ldots,b_{n}$ for the sums of the columns. Also, we can check that $ b_{1}\plus{}b_{2}\plus{}\cdots\plus{}b_{n}$ is equal to the sum of the sums of the rows, since $ a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{m}\equal{}b_{1}\plus{}b_{2}\plus{}\cdots\plus{}b_{n\plus{}1}$, and we subtracted the same amount from each sum in filling out the $ (n\plus{}1)$-st column. Thus, by the inductive hypothesis, we can fill in the rest of the grid.

Are you sure you can apply the inductive hypothesis? Haven't you made the row sums nonintegral?

By the way, your solution suggests that if there are no nonzero entries, we can take any row or column and arbitrarily relabel it with 0s and 1s to get the right sum, and still be okay. However,
[code]0.5  0.5  1
0.5  0.5  1[/code]
In this board, taking the second row and relabeling it as 1,1,0 fails. So something has to have gone wrong.

Well for the first point, I don't think so, because I fill in the rightmost box in a given row with a 0 or 1, and what's left will be integral because the whole thing was originally integral.

And for the second, I was assuming that everything in the grid is less than 1, because we can take the fractional parts of every entry - thus the 1's would be 0's. I agree that this is a little weird, though, I'll have to think about it more but I'm not sure I see the hole just yet.

OK, I see the problem now. You need to allow rows with all 1's as well, because at some point you will need to start a row or column with a string of consecutive 1's (even if it's one 1). Before, I was assuming that if $ a_{i}\equal{}n\plus{}1$, you're still allowed to put a 0 in this row, but this makes it so that in the $ m\times n$ grid, you have that $ a_{i}$ is too big. But this isn't hard to deal with; first take all the rows such that $ a_{i} \equal{} n \plus{} 1$, and fill them in accordingly, then you can essentially ignore them by deleting these rows and adjusting (decreasing) all of the $ b_{i}$'s by 1 for each row that we fill in with all 1's. Then, you should be able to fill in the rightmost column arbitrarily (avoiding the zero rows).

I think [b]dgreenb801[/b]'s argument can be slightly modified to work... the idea is that there's a lot of freedom in filling out $1$'s.

First, we can take out the rows and columns that only have integers. It's also clear that we only care about the fractional part of each square, so we can assume that all squares contain a nonnegative number strictly less than $1$, and every row and column has at least one number that is not an integer (in fact, there must be at least two or else the sum of the row or column would be non-integral). Let $a_i\ge1$ be the sum of the numbers in row $i$ and $b_j\ge1$ be the sum of the numbers in column $j$. Also let $r_i$ be the number of non-integers in row $i$ and $c_j$ be the number of non-integers in column $j$, so because all the numbers in squares are nonnegative and strictly less than $1$, we have $r_i\ge a_i+1\ge2$ and $c_j\ge b_j+1\ge2$ for all $i,j$.

Now we see that taking the floor of a number is simply changing it to $0$ and taking the ceiling is changing it to $1$. We want to change the non-integers to $0$'s and $1$'s and end up with exactly $a_i$ $1$'s in row $i$ and $b_j$ $1$'s in column $j$ for all $i,j$. In row $i$, there are exactly $r_i$ numbers we change to $0$ or $1$, and in column $j$, there are exactly $c_j$ numbers we change to $0$ or $1$ (call such a number [i]available[/i]). Note that $\sum r_i=\sum c_j$ and $\sum a_i=\sum b_j$.

We prove that this is always possible by strong induction on $m+n$. I think the base cases are when $mn=0$, which is hopefully trivial (it should be since it means we've finished changing everything). We will take out column $1$, in which we want to place $b_1$ $1$'s (with $c_1\ge b_1+1$ available numbers to choose from). Without loss of generality, we can rearrange the rows so that rows $1$ through $c_1$ have their column $1$ squares available. Say we make the first $b_1$ of these $1$ and the others $0$.

[b]Case 1:[/b] $r_i\ge a_i+2$ for $b_1<i\le c_1$. If we take out column $1$, then for $1\le i\le b_1$, $a_i,r_i$ are both reduced by $1$, for $b_1<i\le c_1$, $r_i$ is reduced by $1$ (the squares in column $1$ and row $i$ here are the available squares in column $1$ which have been made $0$), and for $c_1<i\le m$, $a_i,r_i$ are unaffected. Hence $r_i$ will remain at least $a_i+1$ for all rows, and the columns from $2$ to $n$ are unaffected, so this case follows from the inductive hypothesis.

[b]Case 2:[/b] $r_i\ge a_i+2$ for $b_1<i\le x<c_1$, and $r_i=a_i+1$ for $x<i\le c_1$ (we can assume without loss of generality that the rows are ordered like this). For $x<i\le c_1$, this means that row $i$ has $r_i=a_i+1$ available numbers, and since the squares in column $1$ and row $i$ have been made $0$, taking out column $1$ reduces $r_i$ by $1$ to $a_i$. Since we need $a_i$ squares in row $i$ to be $1$'s, though, we need to make all $r_i$ squares in row $i$ not in column $1$ equal to $1$. If we do this for all $x<i\le c_1$ (we are essentially taking out these rows), then it's clear that for each column $2\le j\le n$, the respective $b_j$ and $c_j$ are both reduced by the same amount (more precisely, both decrease by the number of squares column $j$ has available in rows $x<i\le c_1$). Hence after taking out rows $x<i\le c_1$, $c_j$ is still at least $b_j+1$. Furthermore, taking out these rows does not affect any other rows, so we can use the same analysis as in case $1$ to conclude that $r_i$ will remain at least $a_i+1$ for all rows not taken out. By the inductive hypothesis, we can place $1$'s into the remaining array as desired.

Sorry, what I wrote is completely wrong... if $a_i$ ever becomes zero, then we need to take row $i$ out, which screws everything up. :oops: I really need to get better at combo...

Consider the bipartite graph corresponding to the table and put the numbers on corresponding edges.
In an arbitrary graph we call an edge with non-integer number on it temporary and also call the subgraph consisting of the temporary edges temporary.

We apply series of operations to numbers on edges so that operations have three propeties (*):
1.If the number on an edge before and after an operation is $x$ and $y$ we have $\lfloor x \rfloor \le y \le \lceil x \rceil$
2. At least one non-integer number becomes integer or equivalently number of temporary edges decrease.
3.Sum of numbers on neighboring edges of any vertex become unchanged.
It is not hard to see that by these properties the procedure will end and at last we will have the desired numbers.

Procedure is as follows(**):Suppose we have a bipartite graph $G$ so that sum of numbers on neighboring edges of any vertex is an integer.
Call its temporary graph $H$.The degree of any vertex in $H$ can't be $1$.
If all degrees are zero we stop and we are done.
If not ,we will have a cycle in $H$ .let $x_{1},...,x_{2k}$ be numbers on the cycle's edges.
let $f(x)=min{\{ |x-\lfloor x \rfloor|,|x-\lceil x \rceil| \} }$ and $\varepsilon=min { \{ f(x_{1}),...,f(x_{2k}) \} }$.
Now we shift these numbers by $\varepsilon$ so that (*) hold.Step is done and we goto (**) again.

Is it right?

Again graph theory comes to the rescue. I will slightly rearrange your argument, for increased clarity.

You consider the complete bipartite graph $G$ of vertices $R \cup C$, where the edge between a row in $R$ and a column in $C$ is labeled by the number filling their intersecting cell. By eliminating the edges labeled with integer values, you obtain your[i] temporary [/i]subgraph $H$. No vertex $v$ in $H$ may have degree $1$, since that would contradict the sum of the labels of the edges incident with $v$ being an integer. Also eliminate from $H$ all isolated vertices (of degree zero), if any. Since now all degrees of the vertices in $H$ are at least $2$, $H$ will contain a cycle; since $H$ remains bipartite, the cycle is of even length. You define your value $\varepsilon > 0$ (no edge in $H$ is labeled with an integer). Now [b]alternately[/b] shift the values of the labels by $\pm \varepsilon$, starting with the one realizing the minimum represented by $\varepsilon$ (for which the label now will become integer). This ensures that all labels remain within the floor and the ceiling of their original values, and moreover, that the sums of the labels of the edges incident with any vertex stay integer, since[b] exactly[/b] two labels are affected for any row or column participating in the cycle, and since those edges are adjacent, the effect of the alternating shift cancels (or no labels are affected for the rows and columns not participating in the cycle). Finally, we noticed that at least one edge is now labeled with an integer, which makes that returning to creating a new temporary subgraph $H$ will strictly decrease its number of edges. The process can only stop when $H$ is empty, which is what was required. [color=#0000FF]Congratulations![/color]

Thanks mavropnevma.You filled details I didn't mention:cycle has even length and we must shift alternately and why we don't have degree 1 and why sums become unchanged.I see your proof is more coherent and readable than mine :)

this problem also can be solved using Hall's theorem

Perhaps I'm missing something, but can't we just do:

[hide]
Subtract the floor of each entry. Now, consider it as filling a bunch of 1's and 0's into a matrix whose row and column sums are known, and the sum of all the row sums equals the sum of all column sums. Also, no row sum is greater than or equal to the number of columns and no column sum is greater than or equal to the number of rows.

WLOG we have no columns or rows with sum 0; otherwise we can just fill them in with all 0's and remove them from consideration.

Induct on the number of columns. The induction hypothesis is that we can fill it in with 1's and 0's given that the sum of the row sums equals the sum of the column sums and no row sum is greater than or equal to the number of columns or column sum greater than or equal to the number of rows.

The base cases are an $m \times 1$ dude and an $m \times 2$ dude. $m \times 1$ is clearly doable since it must be all 0's (due to integer sums).

To solve on $m \times 2$ induct on the number of rows. The induction hypothesis is the same as the above.

 $1 \times 2$ and $2 \times 2$ are clearly easy to do (for the latter just list out all the cases it takes like 5 sec to figure each one out).

Assume $k \times 2$ is possible for all such possibilities. Consider row $k+1$. Now, consider the two column sums: they sum to $k+1$ (since no row is 0), and are both at least $1$ since they are at most $k$. WLOG let the left column have the greater sum (or possibly equal). Then, fill in row $k+1$ as 1 0. This leaves the maximal column sum as less than or equal to $k-1$, and the sum of the row sums equal to the sum of the column sums, so by the induction hypothesis, we can fill in the rest accordingly.

Now, given that we can fill in a $m \times n$, which satisfies all our conditions, we show we can fill in an $m \times (n+1)$

Sort the columns by sum, with the greatest in position $n+1$. Let column $n+1$ sum to $x$; in column $n+1$ just fill in 1's in the $x$ rows which have greatest row sums. The remaining rows and columns satisfy the conditions of the induction hypothesis; they clearly satisfy the first condition (row sum sums=column sum sums), as well as the third (no column sums to the number of rows or more). To show they satisfy the second, all we have to do is show that at most $x$ rows have sum $n$, but this is easy since the sum of the columns is at most $x(n+1)=nx+x<nx+n=n(x+1)$, which is the minimum the sum would be if at least $x+1$ had sum $n$.

So, our induction is complete.[/hide]

[hide="An example of how to do this"]
Consider like:

[code]
? ? ? ? ? 4
? ? ? ? ? 2
? ? ? ? ? 1
? ? ? ? ? 3
2 2 2 1 3[/code]

Now sort the rows/columns in increasing order.

[code]
? ? ? ? ? 1
? ? ? ? ? 2
? ? ? ? ? 3
? ? ? ? ? 4
1 2 2 2 3[/code]

Fill in the columns from right to left according to which sum is the largest remaining just like our induction:

[code]
? ? ? ? 0 1
? ? ? ? 1 2
? ? ? ? 1 3
? ? ? ? 1 4
1 2 2 2 3[/code]

This is equivalent to:

[code]
? ? ? ?  1
? ? ? ?  1
? ? ? ?  2
? ? ? ?  3
1 2 2 2 [/code]

Again and again to get:

[code]
? ? 0 0  1
? ? 0 0  1
? ? 1 1  2
? ? 1 1  3
1 2 2 2 [/code]

or

[code]
? ? 1
? ? 1
? ? 0
? ? 1
1 2 [/code]

Eliminate the 0 for

[code]
? ? 1
? ? 1
? ? 1
1 2 [/code]

Now the bottom row to get us to a 2x2:

[code]
? ? 1
? ? 1
0 1 1
1 2 [/code]

[code]
? ? 1
? ? 1
1 1 [/code]

which is easy:

[code]
1 0 1
0 1 1
1 1 [/code]

We can now build back up to get:

[code]

1 0 0 0 0
0 1 0 0 1
0 0 1 1 1
0 1 1 1 1
[/code]

and then rearrange to

[code]
1 1 1 0 1
0 1 0 0 1
0 0 0 1 0
1 0 1 0 1[/code]
which clearly works.

[/hide]

[quote="Zhero"][quote="Mewto55555"]Now, given that we can fill in a $m \times n$, which satisfies all our conditions, we show we can fill in an $m \times (n+1)$

Sort the columns by sum, with the greatest in position $n+1$. Let column $n+1$ sum to $x$; in column $n+1$ just fill in 1's in the $x$ rows which have greatest row sums.[/quote]
This doesn't work because some of the 1's you fill in could be in squares that you're not allowed to modify (i.e. if a square contains an integer, you cannot do anything with it).[/quote]

Good point, but that's easily fixed; if a row of $m$ has $k$ 0's and there are $n$ columns, then it can sum to at most $n-k-1$ instead of $n-k$; similarly with columns. So, just fill in the others and you reach a situation where you win; all you have to do is slightly modify the inductive hypothesis.

EDIT: No, math154 provided a counterexample to troll this line of argument.

someone who cares to check. The above if correct is a totally new solution.

I don't find anything wrong with the reading nor the solution. :-D

For each square, define its [i]weight[/i] to be the minimum distance between it and an integer. (Formally, $w(x) = \min(\{x\}, 1 - \{x\})$ where $\{x\}$ is the fractional part of $x$.) Our goal is to make all squares have weight zero while preserving the row and column sums.

Suppose there are $N$ squares with nonzero weight, where $N \ge 1$. Consider the bipartite graph $G$ with rows and columns as vertices, and two vertices (one row and one column) connected by an edge of weight $w$ if the intersection of the row and column has weight $w > 0$. There are $N$ edges in $G$. Notice that it is impossible for a vertex to have degree exactly $1$, as that implies the corresponding row or column has non-integer sum. Thus, each non-isolated vertex has degree at least $2$, implying the existence of a cycle. The graph being bipartite ensures the cycle has even length. Let $w$ be the minimum weight in the cycle. Then by alternating adding and subtracting $w$ to the squares corresponding to edges of the cycle, we preserve the row and column sums. Furthermore, we can choose to start the alternating add/subtract such that an edge with weight $w$ will have weight $0$ after the transformation. Thus, the number of edges (and the number of squares) with nonzero weight decreases by at least $1$, so after finitely many steps we can decrease $N$ to $0$, as desired.

Example: the cycle of squares with values 1.5, 2.3, 3.6, 1.8 has edge weights 0.5, 0.3, 0.4, 0.2. The minimum weight is 0.2, attained for the 1.8 square. Now alternate adding/subtracting 0.2 to get 1.3, 2.5, 3.4, 2.0, and 2 is an integer (with 0 weight).

[quote=vjdjmathaddict][size=100][size=100]Can someone please check this algoritmic solution I doubt the validity.
We do the filling greedily. WLOG we assume the rows and columns to be arranged in decreasing order

Let the sum of numbers in row $i$ be $N_i$, and those in column $i$ be $M_i$. Firstly look at the table as an
$n*n$ matrix.
We say that we switch on a number if we change it into least integer the element $x_{ij}$, and we switch of the element if we change into greatest integer.
Now the algorithm
Initially switch off every element
[b][u]Step 1[/u][/b]
Switch on one by one each elements in each row $i$(suppose) starting from column 1 till you reach $N_i$.
[b][u]Step 2[/u][/b]
If at any moment switching on $x_{j1}$(remember the rows and columns are arranged in decreasing order) increases the sum more than $M_1$ go to the nearest column where we switch on the elements and continue from step 1.Continue the same for other columns.
[b][u]Claim[/u][/b]

By the end of the process the desired configuration is reached

Suppose the desired configuration is not reached.
Now there are two bad things that could happen
(1) The row sums are reached before column sums are reached.
(2) The column sums are reached before row sums are reached.

Suppose (1) happens
Note that the sum of rows is same as that of the sum of the columns.
That means there must be some numbers in some column $i$ such that the sum of numbers in some column must exceed $M_i$ but this contradicts the algorithm.
Suppose (2) happens then similar to the argument above the sum of number in some row must exceed $N_i$ impossible by algorithm. Also since elements are finite the algorithm terminates.[/size][/size][/quote]

How do you handle $N_1 = 5, N_2 = 1, N_3 = 0, M_1 = 2, M_2 = 2, M_3 = 2$?

Here's an approach that doesn't use graph theory, but the machinery of [b]linear algebra[/b] instead.
Assume for the sake of contradiction that we've done the maximal number of rounding we can do, and we still have non-integer terms. Then, there are at least two of them in each row/column. Replace these by variables.
Now, write for each row/column an equation relating the unknowns and numbers in the grid to the integer sum.
This gives a system of $m+n$ equations. However, we know that there is at least one redundant equation in here as the sum of the $m$ row equations is equal to the sum of the $n$ column equations. Thus, we have strictly less than $m+n$ linearly independent equations.
However, we have at least $\max{(2x,2y)}\ge x+y>x+y-1$ variables, so our system has nontrivial null space. In particular, some of our variables are free, so we can swap one of them for the nearest integer, contradicting our maximality assumption.

[quote=Wizard_32][quote=ISL 1998 C1]A rectangular array of numbers is given. In each row and each column, the sum of all numbers is an integer. Prove that each nonintegral number $x$ in the array can be changed into either $\lceil x\rceil $ or $\lfloor x\rfloor $ so that the row-sums and column-sums remain unchanged. (Note that $\lceil x\rceil $ is the least integer greater than or equal to $x$, while $\lfloor x\rfloor $ is the greatest integer less than or equal to $x$.)[/quote]
Let the array be $m \times n.$ For any given row $\mathcal{R}$, let the elements be $y_i$ for $1 \le i \le n.$ Call the [i]value[/i] of $\mathcal{R}$ the number of terms on which we use $\lceil \circ \rceil$ so that the sum of the row remains constant.

A simple calculation shows that $\text{value} (\mathcal{R})=\{y_1\}+\cdots+\{y_n\} \in \mathbb{N}$. From this note that the $0<\text{value}<n.$ The only issue now is to choose which elements in each row so that the choice corresponds with the columns.

To see this, draw a bipartite graph with the rows and the columns as the two parts. Label each vertex by its value.

Consider the simple algorithm of choosing the first point (i.e. the left most point) from the first part, then connecting it to as many vertices as its value. This is possible since $\text{value}<n=\text{no. of vertices in the second part}.$ For instance:
[asy]
unitsize(3);
int y=20; //y-gap between parts
int x=15; //x-gap between dots

draw((0,y)--(0,1), arrow=Arrow(HookHead), blue);
draw((0,y)--(x,1), arrow=Arrow(HookHead), blue);

draw(circle((0,0),1)); label("$2$",(0,-5)); fill(circle((0,0),1));
draw(circle((x,0),1)); label("$1$",(x,-5)); fill(circle((x,0),1));
draw(circle((2x,0),1)); label("$3$",(2x,-5)); fill(circle((2x,0),1));

draw(circle((0,y),1)); label("$2$",(0,5+y)); fill(circle((0,y),1));
draw(circle((x,y),1)); label("$2$",(x,5+y)); fill(circle((x,y),1));
draw(circle((2x,y),1)); label("$1$",(2x,5+y)); fill(circle((2x,y),1));
draw(circle((3x,y),1)); label("$1$",(3x,5+y)); fill(circle((3x,y),1));

[/asy]
Repeat this procedure for the points in the first part moving from left to right. If the value of a vertex from the second part is achieved at some point, delete it. Keep following this algorithm.
[asy]
unitsize(3);
int y=20; //y-gap between parts
int x=15; //x-gap between dots

draw((0,y)--(0,1), arrow=Arrow(HookHead), blue);
draw((0,y)--(x,1), arrow=Arrow(HookHead), blue);

draw((x,y)--(0,1), arrow=Arrow(HookHead), red);
draw((x,y)--(2x,1), arrow=Arrow(HookHead), red);

draw((2x,y)--(2x,1), arrow=Arrow(HookHead), magenta);

draw((3x,y)--(2x,1), arrow=Arrow(HookHead), purple);

draw(circle((0,0),1)); label("$2$",(0,-5)); fill(circle((0,0),1));
draw(circle((x,0),1)); label("$1$",(x,-5)); fill(circle((x,0),1));
draw(circle((2x,0),1)); label("$3$",(2x,-5)); fill(circle((2x,0),1));

draw(circle((0,y),1)); label("$2$",(0,5+y)); fill(circle((0,y),1));
draw(circle((x,y),1)); label("$2$",(x,5+y)); fill(circle((x,y),1));
draw(circle((2x,y),1)); label("$1$",(2x,5+y)); fill(circle((2x,y),1));
draw(circle((3x,y),1)); label("$1$",(3x,5+y)); fill(circle((3x,y),1));

[/asy]
This algorithm would stop with each vertex satisfied since the sum of the values of the two parts are equal. Hence we are done. $\square$[/quote]

From this solution it is clear to me that the algorithm will stop.But then in which cells will we apply the ceiling function???

Sorry, but bumping it to get an answer of my above question

how the heck did you do those asymptote codes

[quote=rachelshi][asy]
unitsize(3);
int y=20; //y-gap between parts
int x=15; //x-gap between dots

draw((0,y)--(0,1), arrow=Arrow(HookHead), blue);
draw((0,y)--(x,1), arrow=Arrow(HookHead), blue);

draw((x,y)--(0,1), arrow=Arrow(HookHead), red);
draw((x,y)--(2x,1), arrow=Arrow(HookHead), red);

draw((2x,y)--(2x,1), arrow=Arrow(HookHead), magenta);

draw((3x,y)--(2x,1), arrow=Arrow(HookHead), purple);

draw(circle((0,0),1)); label("$2$",(0,-5)); fill(circle((0,0),1));
draw(circle((x,0),1)); label("$1$",(x,-5)); fill(circle((x,0),1));
draw(circle((2x,0),1)); label("$3$",(2x,-5)); fill(circle((2x,0),1));

draw(circle((0,y),1)); label("$2$",(0,5+y)); fill(circle((0,y),1));
draw(circle((x,y),1)); label("$2$",(x,5+y)); fill(circle((x,y),1));
draw(circle((2x,y),1)); label("$1$",(2x,5+y)); fill(circle((2x,y),1));
draw(circle((3x,y),1)); label("$1$",(3x,5+y)); fill(circle((3x,y),1));

[/asy]

[/quote]

FTFY

[i]Rough and sloppy solution for storage, which I think is similar to OldMath's solution.[/i]

[b]Solution[/b].    Consider the bipartite graph with a vertex for each row and column in the graph, and edges weighted by the corresponding array entry. Remove all edges with integers weights.

    Repeat the following procedure:
   [list=1]
[*]Consider some cycle in the graph (exists by Euler as vertices have even degree).
[*]Choose the edge $e$ in this cycle whose weight is closest to an integer, with a distance $x$.
[*]Go around each edge in the cycle, alternating between adding and subtracting $x$. Do this such that $e$'s weight becomes an integer.
[*]This leaves the sum of edge-weights at each vertex the same.
[*]After doing this, for each edge that has become an integer, we apply either the floor or ceiling function to the corresponding array entry, and then remove the edge from the graph.
[/list]
Since this procedure removes edges, we can repeat until there is no edges left, at which point there will be no non-integer elements in the array, and the final row and column sums will stay the same (since we kept the sum of edge-weights at each vertex constant).

[b]Remark[/b]. [i]I had a look in the IMO Compendium after doing this problem, and I think the proof they use there is the same as this proof, modulo graph theory.[/i]

Let the distance between a square and an integer be $\Theta$. Let $\Theta_{\text{min}}$ be defined as the [i]strength[/i] of a square. The main goal of the problem is to make the strength of all the squares equal to $0$, this however needs to be done while preserving the sums of the row and the coloumns.
Now suppose that there are $X$ squares that have non-zero strength. Where we have $X \ge 1$. Now we employ graph theory to solve the problem. Consider a bipartite graph $\Omega$ where we have the rows and coloumns as vertices and we have one row and one coloumn that is connected by an edge having strength $\nabla$. If the intersection of the row and column have strength of $\nabla > 0$. Thus there are $X$ edges in $\Omega$. Observe the fact that it is impossible to have a vertex of degree $1$, if this happens, then it implies that the corresponding row or coloumn sum is not an integer. This implies the existense of a cycle. Since $\Omega$ is bipartite, the length of the cycle will be even. Let $\nabla$ be the minimum strength of the cycle, then we do alternate adding and subtracting with $\nabla$ to the squares with corresponding edges of the cycle. This way we preserve the row and the coloumn sum. We can also choose to start the alternation addition or subtraction in such a way that an edge having a strength of $\nabla$ will be having a strength of $0$ after the process. Hence the number of edges having strength $\nabla > 0$ decreases at least by $1$. Hence after a finite number of steps we can decrease $X$ to $0$, which is what the problem desired for. $\blacksquare$

Let a "nonintegral cell" be a cell with a nonintegral number. We will use induction on $n$, the number of nonintegral cells. The base case is when $n=0$. In this case, doing nothing satisfies the condition.

Note that every nonintegral cell has another nonintegral cell in the same row and another in the same column. Otherwise, the row and column sum wouldn't be an integer.

If $n>0$, let $x_1$ be some nonintegral cell. For odd $i>1$, let $x_i$ be a nonintegral cell in the same row as $x_{i-1}$ and for even $i>1$, let $x_i$ be a nonintegral cell in the same column as $x_{i-1}$ such that $x_i\neq x_{i-1}$.

We know this is possible because every nonintegral cell has another nonintegral cell in the same row and another in the same column. Let $k$ be the smallest value of $k$ such that there exists $j< k$ such that either of the following are true:
-$k$ is even, and $x_k$ and $x_j$ are in the same row (we know $x_k\neq x_j$, since if they were equal then $x_{k-1}$ would be in the same column as $x_j$, satisfying the condition below, a contradiction)
-$k$ is odd, and $x_k$ and $x_j$ are in the same column ($x_k\neq x_j$ by the same logic as above)
Note this must be possible since there are a finite number of rows and columns.
Moreover, after $k$ has been chosen, let $j$ be the highest value less than $k$ such that one of the above is true. Then, let $a_j,\ldots, a_k$ be a finite sequence of cells with $a_j,\ldots, a_k=x_j,\ldots, x_k$.

We claim that in each row and column, there is either $1$ cell $a_i$ with $i$ odd and $1$ cell $a_i$ with $i$ even or there is no cell $a_i$. First we will show that each row with any cell $a_i$ has a cell of each parity.

To do this, we will first show that $j$ and $k$ have different parity. Assume $k$ is even. Assume towards a contradiction that $j$ is even. Then, $a_k$ and $a_j$ are in the same row. However, $a_{j+1}$ is also in this row, so $j$ is not maximal, a contradiction. By similar logic, $k$ and $j$ cannot both be odd.

WLOG, assume $k$ is even and $j$ is odd. Then, they are in the same row. In other rows, if $i\neq k, j$, then either $a_{i-1}$ or $a_{i+1}$ must be in the same row as $a_i$.
For columns, for odd $i$, $a_{i+1}$ is in the same column as $a_i$ and $a_{i-1}$ is in the same column as $a_i$ for even $i$.

We will now show that no row or column has more than two of the $a_i$.

Assume towards a contradiction that $x$ and $y$ are both even and $a_x$ and $a_y$ are in the same column with $y>x$. Then, $a_{y-1}$ is also in this column, and $y-1$ is odd. Also, $x<y-1$, $y-1$ is odd, and $a_x$ and $a_{y-1}$ are in the same column. Therefore, $k$ is not minimal, a contradiction.

Assume towards a contradiction that $x$ and $y$ are both odd and $a_x$ and $a_y$ are in the same column with $y>x$. Then, $a_{x+1}$ is also in this column. So, $y>x+1$, $y$ is odd, and $a_y$ and $a_{x+1}$ are in the same column. If $y\neq k$, then $k$ is not minimal, a contradiction. If $y=k$, then $x+1>j$, so $j$ is not maximal, a contradiction. Apply the same logic to rows, and we find that any row or column with any of the $a_i$ has one $a_i$ with $i$ even and one $a_i$ with $i$ odd.

Let $x\in\mathbb{R}$ such that $x+a_i$ is an integer for some $i$ and $|x|$ is minimized. Add $x$ to all of the cells $a_m$ with $m\equiv i\pmod{2}$. Subtract $x$ from all of the cells $a_m$ with $m\not\equiv i\pmod{2}$. Because every row and column has the same number of odd- and even-indexed cells from $a_j,\ldots,a_k$, this will not affect row and column cells. Since $|x|$ is minimal, this will not affect the floor or ceiling of any of the cells that don't become integers.

So, at least 1 non-integral cell will become an integer. So, by induction, we can take the floor or ceiling of every other non-integral cell such that the row and column sums stay the same.

tldr: Because each non integral cell has another non integral cell in the same row and column, we can make a "cycle" of non integral cells such that we can add some constant to half of them and subtract some constant from the other half such that the row and column sums stay the same. Then, we can use induction to turn all the cells into integers.

Assume WLOG that every number in the grid is in the interval $[0,1)$. We will create an algorithm that works on grids that aren't only integers yet that changes the grid such that each number in the grid is still in the interval $[0,1]$, each row and column sum is unchanged, no cell that is already an integer is modified, and the number of integers in the grid strictly increases. By repeatedly applying the algorithm, the number of integers in the grid will be a strictly increasing monovariant until the entire grid is filled with only $0$s and $1$s where the cells that were previously $0$s stay $0$s, when we will be clearly done.

For the algorithm, first pick any noninteger in the grid and let Catloe be on it. Note that if a row or column has at least one noninteger, it must have at least two nonintegers. So, select another noninteger in the row, and let Catloe walk to it in a straight line. Now, select another noninteger in the column Catloe is in, and let Catloe walk there. This repeats until Catloe reaches a cell he was already in. Catloe has completed a loop that only turns on nonintegers. Let the turns, in order, be $t_1,t_2,\dots,t_n$. Note that $n$ has to be even.

Define a number's [i]awaness[/i] to be its distance to the closer of $0$ or $1$. Choose the $t_i$ with the lowest awaness(ties can be broken arbitrarily). Assume WLOG that it is closer to $0$. Then we reduce the cell $t_i$ to $0$, and go around the loop and alternate increasing and decreasing each cell by $t_i$(so increase $t_{i+1}$ by $t_i$, decrease $t_{i+2}$ by $t_i$, etc, increase $t_{i-1}$ by $t_i$). Clearly, nothing gets out of bounds, no integer is modified, the number of total integers is increased, and the row and column sums are preserved(for every pair in a row/column, one is decreased while the other is increased). Therefore, our algorithm works, and we are done.

Quite a beautiful problem, actually.

If there are no non-integers then we are done. Otherwise, we claim that there exists squares $s_1,s_2,\dots,s_{2k}$ such that $s_0=s_{2k},s_{1}=s_{2k+1}$, and for $0\le i\le 2k-1$,
[list=i]
[*] The numbers written in each of these squares is non-integer.
[*] The squares $s_{i}$ and $s_{i+1}$ are either on the same row, or the same column.
[*] The squares $s_{i}$ and $s_{i+2}$ are neither on the same row, nor the same column.
[*] The squares $s_1,s_2,\dots,s_{2k}$ are pairwise distinct.
[/list]
First, we note that each row and each column which does contain non-integers has at least two of them. If one only has one non-integer, then the sum must be non-integer, which is a contradiction.

Using that fact, we may start from any non-integer square $s$, and then move onto a non-integer square of the same row, which must exist. Then, we may move onto a non-integer square of the saw column, which also must exist. Repeating this, we must either eventually reach a square we've reached before. Suppose the first time this happens, we reached $s_1$ for the second time.

Now, let the path we took land on the squares $s_1,s_2,\dots,s_{2k},s_1.$ Connect squares with adjacent indices $\pmod {2k}$. Note that any such must land on an even number of squares, because at each square there is one horizontal and one vertical segment, so there are equal amounts of horizontal and vertical segments, and therefore an even number of segments total, meaning that even number of squares.

Now, none of the squares $s_1,s_2,\dots,s_{2k}$ are the same because that violates the fact that the second $s_1$ is the first time a square is stepped on twice. Our path's algorithm dictates that $\text{(i)}$, $\text{(ii)}$ and $\text{(iii)}$ are followed.

Therefore, our claim is proven. Now, of these squares $s_1,s_2,\dots,s_{2k}$ and selected one of the ones that is closest to an integer, WLOG let it be $s_1$. Now, move this to the closest integer, move $s_{i+1}$ by the same amount that $s_i$ was moved, but in the opposite direction. One can check that each row's and column's sum is preserved, and there is at least one more integer square.

Repeat this until there are no more non-integer squares, at which point we are done.

Let there be $m$ rows and $n$ columns. Let each of the elements in the array be in the range $(0,1]$. We now have the much simpler problem of assigning $0,1$ to elements of an array such that the rows and columns sum to known target quantities. In addition, some of the elements are forced to be $1$(namely the integer ones). Now we just take the following greedy algorithm, take the column whose target sum $k$ is least, then we are forced that the total target sum is $kn$, so by pigeonhole there must exist at least $k$ rows which have nonzero target sum.  Observe that the fixed $1$ elements do not matter because we can just take them and then we will still have enough good rows left. After this procedure is done, literally just delete the column and adjust the target sums for the rows accordingly. Repeating finishes.

Notice that we can subtract the floor parts of each element, and simplify the task to the following:
	- We are given an $m \times n$ grid
	- We have row totals $a_1, a_2, \dots, a_m$ where all $0 \leq a_i < n$
	- We have column totals $b_1, b_2, \dots, b_n$ where all $0 \leq b_i < m$
	- Prove that we can put ones and zeroes to satisfy the above conditions.

Now, we can prove the claim by induction. The base case when $m = 1$ is trivial, so now we will solve the problem for $m+1$ rows where the problem for $m$ rows is true.

First, note that if any of the columns has a total sum $b_i = 0$, then we can just delete the column since we know it must be all zeroes anyways.

Pick the row that has the largest total $a_i$. Note that $ma_i \geq S$, where $S$ is the total number of $1$'s needed.
This implies that $a_i$ is at least the number of $b_j$'s that are equal to $m-1$.

Then, this means that if we put the $1$'s in the $i$th row such that the $1$'s are placed in the columns with the greatest totals required (the columns with the greatest $b_j$), then deleting that row (since we've finished filling it) leaves us with a problem satisfying the constraints but for the $m \times n$ case. This proves the induction. $\blacksquare$

First, smooth so that all numbers have integer part $0$. Then, delete any rows or columns that have all $0$s, to obtain a new $m \times n$ grid. Upon imposing a coordinate system, the grid can be understood as the graph $K_{m, n}$, where edges of the graph that have weight $0$ are deleted. Since this graph $G$ is bipartite, there exists a cycle of even length. Let $e$ denote edge whose weight is closest to $0$ or $1$, and let $w$ denote it's weight. Let $w' = \min(w, 1-w)$. Then starting with $e$, alternate between adding and subtracting $w'$ to the weights in the cycle, where the value of $w$ is changed to one of $0$ and $1$, and after that delete all integral edges. Do this for all cycles. Repeat this process until all weights are made either $0$ or $1$. Clearly, the total sum in each row and column is preserved, and all numbers are changed to $0$ or $1$. Thus we are done.

oops what am i doing

somehow no solve oops

motivation is like to change one thing, see how it changes another thing, and then just keep going like that

and this just solves

smh my head

though it is essentially the same concept as JMO 2024/2


Θα μου πει κάποιος τη γνώμη του και για το 3 θέμα της Γ Λυκείου; Το έχει λύσει κανείς με διαφορετικό τρόπο από αυτόν που παραθέτει η ΕΜΕ στην ιστιοσελίδα της;;

Πάντως το πρώτο μου φάνηκε αρκετά εύκολο.Πιστεύω θα το έχει λύσει η πλειοψηφία.

Φιλε a.c! ο τροπος της ΕΜΕ ειναι και ο πιο απλος...και ισως μοναδικος..και αυτο γιατι αναγκαια κατευθυνεσαι στην ευρεση της ακολουθειας και υστερα τα πραγματα ειναι απλα...Γενικα ειναι δυσκολο να εχεις πλυθωρα διαφορετικων τροπων για τοσο απλες ασκησεις... :wink:

Συμφωνώ με τον Αρχιμήδη που τώρα έμαθα ότι λέγεται Δημήτρης... αλλά υπάρχει άλλος ένας ακόμη τρόπος. Ίσως είναι πιό σύντομος και elegant αλλά δεν είναι ο ενδεδειγμένος, με την έννοια ότι δεν μπορείς να αντιμετωπίσεις παρόμοιες ασκήσεις με τον τρόπό αυτό.
Εύκολα μπορείς να δείξεις ότι
$ a_{n\plus{}l}\minus{}a_n\equal{}l^2\plus{}2ln$
Παρατηρώ ότι $ a_n \cdot a_{n\plus{}1}\minus{}a_n\equal{}{a_n}^2\plus{}2n\cdot a_{n}\equal{}a_{n\plus{}a_n}\minus{}a_n$
Συνεπώς $ a_n\cdot a_{n\plus{}1}\equal{}a_{n\plus{}a_{n}}\Longrightarrow q.e.d.$

[quote="Nick Rapanos"]Συμφωνώ με τον Αρχιμήδη που τώρα έμαθα ότι λέγεται Δημήτρης... αλλά υπάρχει άλλος ένας ακόμη τρόπος. Ίσως είναι πιό σύντομος και elegant αλλά δεν είναι ο ενδεδειγμένος, με την έννοια ότι δεν μπορείς να αντιμετωπίσεις παρόμοιες ασκήσεις με τον τρόπό αυτό.
Εύκολα μπορείς να δείξεις ότι
$ a_{n \plus{} l} \minus{} a_n \equal{} l^2 \plus{} 2ln$
Παρατηρώ ότι $ a_n \cdot a_{n \plus{} 1} \minus{} a_n \equal{} {a_n}^2 \plus{} 2n\cdot a_{n} \equal{} a_{n \plus{} a_n} \minus{} a_n$
Συνεπώς $ a_n\cdot a_{n \plus{} 1} \equal{} a_{n \plus{} a_{n}}\Longrightarrow q.e.d.$[/quote]

 :yup: :)

Μα ειναι η ιδια φιλοσοφια... :maybe:

koita auth einai h lysh enos sumathiti mou kai nomizo oti einai pio syntomh kai den einai idia filosofia otan vgazei typo gia deikti ths morfhs $ n\plus{}l$...alla de xreiazetai na epektathoume afou opos eipes ki esy einai dyskolo na vreis diaforetikous tropous gia eukoles askhseis.apla mias kai zitithike eipa na grapso auton...


function
logarithms
algebra
domain
functional equation

A real-valued continuous function satisfies for all real $x$ and $y$ the functional equation

$f(\sqrt{x^{2}+y^{2}}) = f(x)f(y)$.

Prove that $f(x) = [f(1)]^{x^{2}}$.

[hide="hint"]
Prove it for all numbers of the form $2^{\frac{n}{2}}$ for integers $n$.
[hide="further hint"]
Show that any positive rational number can be written as a sum of powers of $2$. Or any positive real, if we allow infinite such terms. [/hide][/hide]

[quote="Farenhajt"]Or we can just

[hide="do this:"]
Substitute $g(t)=\ln f(\sqrt{t})$ to get $g(x^{2}+y^{2})=g(x^{2})+g(y^{2})$. By Cauchy, the only continuous solution is $g(t)=ct$, hence $f(\sqrt{t})=e^{ct}\implies f(x)=e^{cx^{2}}=[f(1)]^{x^{2}}$[/hide][/quote]

OK, but you need to justify $f(\sqrt{t}) > 0$ and say something about $t < 0$.

Point taken.

1. Plugging $y=-z$ in, we get $f(z)=f(-z)$, hence $f$ is even and we can restrict the domain to $\mathbb{R}_{0}^{+}$;

2. Putting $z=\sqrt{x^{2}+y^{2}}\iff y=\sqrt{z^{2}-x^{2}}$, we get $f(\sqrt{z^{2}-x^{2}})={f(z)\over f(x)}$. Assume there exists $b$ such that $f(b)=0$. If $f$ is non-constant, then there must exist smallest of such $b$'s. But then from $f(x)f(\sqrt{b^{2}-x^{2}})=0$ we find that there exists $b'<b$. Hence $f$ is constant, i.e. $f(x)\equiv 0=[f(1)]^{x^{2}}$. Otherwise, $f(x)\neq 0$ for all $x$. Plugging $x=y={a\over\sqrt{2}}$ in, we get $f(a)=f^{2}\left({a\over\sqrt{2}}\right)>0$.

[quote="Farenhajt"]Substitute $g(t)=\ln f(\sqrt{t})$ to get $g(x^{2}+y^{2})=g(x^{2})+g(y^{2})$. By Cauchy, the only continuous solution is $g(t)=ct$, hence $f(\sqrt{t})=e^{ct}\implies f(x)=e^{cx^{2}}=[f(1)]^{x^{2}}$[/quote]

what do you mean when you say "By Cauchy"?


group theory
abstract algebra
superior algebra
superior algebra unsolved

This is probably dumb question. I expect the answer to be no

Nonabelian simple groups, such as $ A_5$. There are maximal subgroups, but no nontrivial normal subgroups.

The trivial group is a maximal normal subgroup of a simple group, as I see it (and I don't see why you wouldn't agree).

$ \mathbb Q / \mathbb Z$ doesn't have maximal (normal) subgroups.

It was ambiguous language- I read it as maximal and normal, which is different.


inequalities
LaTeX
inequalities proposed

Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
      $ \frac {1}{2a^2 \plus{} 2a \plus{} bc} \plus{} \frac {1}{2b^2 \plus{} 2b \plus{} ca} \plus{} \frac {1}{2c^2 \plus{} 2c \plus{} ab}\geq\frac {1}{ab \plus{} bc \plus{} ca}$
@quykhtn:[hide]I think you have long proof and you can wrong for your proof if you using Iran 96 for it :wink: [/hide]

After homogenising, the inequality is equivalent to
$ \sum_{cyc} \frac{ab+bc+ca}{(2a+b)(2a+c)} \ge 1$ or
$ \sum_{cyc} \frac{b}{2a+b}+ \sum_{cyc}\frac{c}{2a+c} \ge 2.$
Applying Cauchy yields
$ \sum_{cyc} \frac{b}{2a+b} \ge \frac{ (\sum_{cyc}a)^2}{\sum_{cyc}b(2a+b)} = 1$ and
$ \sum_{cyc} \frac{c}{2a+c} \ge\frac{ (\sum_{cyc}a)^2}{\sum_{cyc}c(2a+c)} = 1$,
proving the inequality.

ps: how to center $ \text{\LaTeX}$-formulas?

[quote="2424"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
      $ \frac {1}{2a^2 \plus{} 2a \plus{} bc} \plus{} \frac {1}{2b^2 \plus{} 2b \plus{} ca} \plus{} \frac {1}{2c^2 \plus{} 2c \plus{} ab}\geq\frac {1}{ab \plus{} bc \plus{} ca}$(*)
$ (*) < \equal{} > \sum_{cyc}{\frac {1}{(2a \plus{} b)(2a \plus{} c)}}\geq\frac {1}{ab \plus{} bc \plus{} ca}$
Apply Cauchy-Schwarzt, we have
  $ LHS\geq\frac {4(a \plus{} b \plus{} c)^2}{\sum{(b \plus{} c)^2(2a \plus{} b)(2a \plus{} c)}} \equal{} P$
And $ P \geq RHS < \equal{} > \sum{ab(a \minus{} b)^2}\geq0$(true!)
So,we are done!


:idea: The historical tradition of the indian classical music dates right back to the evolution of the cosmos. The holy vedas have described that period as"when there was neither truth nor the gory pangs of falsitude;when the world had not yet germinated and there was no   space:then the SHRISTI consisted of NAAD, the divine note which is forever unchanging ,  which never was produced by vibrational mode ,and which shall never cease to be .

The pythagorean school of thought categorizes music as one of the elements of mathematics .Though this idea is now discarded ,yet many esteemed mathematicians continue to be good at music.SHEER COINCIDENCES :?: The indian classical music is closely linked to maths:our forefathers had discovered the exact no. of raags or musical compositions using combinatorics and they had also found out the exact ranges of frequencies which  most  mellifluously chant their  concordant notes at our ears.

Tradition has thrown some light on the varied aspects of our music.It is said that musicians could kindle oil lamps ,summon dark clouds ,charm in beautiful and innocent fauna :such was the power of their music.It may or may not be true,but its varied effects on human health:mental ,moral and spiritual can be witnesseed in toto.

Use your discretion ,fair reader and make use of your discretion and judgement.May you revel through the vagaries of indian classical music and return satisfied.May god shower his choicest blessings upon you.!!

I have to disagree with some of the historical assertions here, but I do like the sound of Indian music.

Hello,
    It seems like you are talking of Hindusthani music,Anant with trems like 'raag' and 'naad'.There is another domain down south-Carnatic music which too is steeped in tradition.But,yes,one need not link music to math or science.It is best treated as an art and the historical credibilites of musicians as mathematicians or scientists are debatable.However,as an art,Indian music,both Hindusthani and Carnatic, though different in form, are truly wonderful.

:roll: Dear sathej the substratum from which carnatic and hindustani music arise are the same.NAAD ,as a matter of fact is more serenly ,nay in an unprofaned manner is expressed in its carnatic counterpart.The latter is deemed to be more intellectual in spirit.

regarding the fact of leaving maths and music alone i would like to say that in a world where scientists have started alluding the properties of subatomic particles to human thoughts (quantum physics) relating maths and music is not difficult. COGITO ERGO SUM. Music only extends one's conciousness to higher planes of thought. :)
FURTHER READING:(the tao of physics,the turning point :both by fritjoff kapra(an american author.)

Hello Anant,
    Though the 'substratum' consisting of the Sapta Swaras and other essentials are same,rendering methodology(called Sampradaya down south) and Ragam and Talam aspects are poles apart. For instance, few ragams(we call them ragam while North Indians call it Raag) are of course similar like Shankarabharanam.While Hindustani music begins with training in Shankarabharanam,Carnatic music starts off with Mayamalavagowla.And rendering of the same Raga is different in both forms.
 As of quantum physics relating to mind,they are only allusions and are yet to be put on a sound footing.And moreover, relating music and math is not 'required'. Just back from a kutcheri here of TM Krishna.Was outstanding.The great Vidwan Umayalpuram K Sivaraman accompanying him on the Mridhangam commented that it was a heavenly gathering.True music cannot be explained.Here down south, mere mathematical exhibition of Swaras termed Kanakku is denounced as jugglery. 'Advaitham' of the singer and the listener are stressed.In that state of bliss,why link math and music?You transcend to an unknown and undescribable experience of sorts.
PS:1)I learn the Carnatic violin(strictly Carnatic Sampradaya) from Shri Parur MA Sundareswaran.
2)I have read the Tao Of Physics by Capra and have had a brief look at the other book.


How to demonstrate that there are only five Platonic solids?

Thank you for helping. :P

[quote="edriv"]Some hints:
Suppose that a face has at least 6 sides. Then, using angles, prove that three faces can't concur in one vertex. And of course there can't be less than three faces concurring in a vertex, so the faces can have only 3,4,5 sides.
If each face has three sides, than the number of faces concurring at one vertex is at most 5.
If each face has four sides, than the number of faces concurring at one vertex is at most 3.
If each face has five sides, than the number of faces concurring at one vertex is at most 3.

Prove that each of these cases yield a platonic solid and you're done![/quote]

I will follow your hint to work out the solution. Thank you edriv. :P

[url=http://en.wikipedia.org/wiki/Platonic_solid]Here[/url] you can also find a nice topological proof (Section 3.2).


does anyone know who was selected for the traveling team to the IPhO? The training camp ended yesterday, but there's nothing on the website about who was selected. Does anyone know anyone at the camp who gave them some feedback?

The members of the traveling team are:

Eric Mecklenberg: junior at Hawken School, Gates Mills, OH Anson Hook: senior at Princeton High School, Princeton, NJ. Elena Udovina: senior at the Hathaway Brown School, Shaker Heights, OH. Jeffrey Middleton: senior at LBJ High School, Austin, TX. Yi Sun: sophomore at the Harker School in San Jose, CA. Daniel Whalen: (Alternate) sophomore at Phillips Academy, Andover, MA.

(from the website, which was just updated)

Physics is for nerds!  Only a nerd would do something mathy.  We need to find an example nerd.  Hmmm... a nerd spends 5 hours a day doing math or science... and preferably likes to talk to himself... hey!  Why is everyone looking at me?? :-D

Physics is for the nerds who end up for some reason getting lots and lots of money from the Nobel Prize. :-D

Physics is for nerds.  OK, well college is for nerds too!  OK, well those 6-figure salaries are for nerds too!!  Wait -- is it too late to become a nerd? :-D

The comment's only offensive if you think "nerd" is an insult...But who would think that? :)

[quote="probability1.01"]Physics is for nerds!  Only a nerd would do something mathy.  We need to find an example nerd.  Hmmm... a nerd spends 5 hours a day doing math or science... and preferably likes to talk to himself... hey!  Why is everyone looking at me?? :-D[/quote]

Or at me.


vector
conics
parabola
geometry
power of a point
radical axis
geometry proposed

In the following By construction ,we mean construction via compass and ruler .
These problems are a little hard and worth thinking on ,so i hope you enjoy them.
[color=blue]Problems[/color] For given two circles like $s_1$ &$s_2$ construct the line $\ell $ such that
1.$\ell $ be parallal to a given line and make equal segments in the circles
2.$\ell$ be parallal to a given line and the sum (or difference ) between the segments which are made in the circles be equal to a given constant
3.$\ell$ passes through a given point ,and make equal segments in the circles.
4.$\ell$ passes through a given point and sum(or difference) between then segments in the circles be eqaul to a given constant.

For the first two:

For (1) draw a line through the center of $s_1$ parallel to the given line $\ell$, and another line through the center of $s_2$ perpendicular to $\ell$. Draw a circle of the same radius as $s_1$ and centered at the intersection point between the two lines. If this circle cuts $s_2$ in two points, then the construction has a solution, and the solution is the line connecting the two points where the drawn circle cuts $s_2$.

For (2), when we're looking for a specific difference, the construction above can be modified to yield the result, since if,  for example, the chord determined in $s_2$ must be larger than the one in $s_1$ with $t$, then we can translate $s_2$ along a vector of length $t$ (and parallel to the given line), and perform a construction analogous to the one above, except that this time we work with the region obtained by intersecting $s_2$ and the translate of $s_2$ instead of $s_2$ itself.

If, on the other hand, we're looking for a specific sum $t$ of the lengths of the two chords, then we could first draw a line perpendicular to $\ell$ through the center of $s_2$, and another line $d$, parallel to this one, at distance $\frac t2$ from it. Next, translate $s_1$ along a vector parallel to $\ell$ s.t. the center of the translate lies on $d$. If this circle cuts $s_2$, then a line we're looking for is a line parallel to $\ell$ passing through one of the intersection points.

I hope it's correct. It looks so on my sketch :).

The lines intercepting chords of equal length on the circles envelope the parabola with focus the midpoint F of OO' (centers of the circles) and tangent at the vertex the radical axis L of the circles. Thus, to solve 1 - or 3 -, we have to find the lines tangent to a given parabola and parallel to a given direction - or going through a given point -
1) The perpendicular at F to the given direction intersects L at T on the required line.
3) If U is the given point and if the circle with diameter UF intersects L at M and M', the required lines are UM and UM' (if they intersect the circles)


what is alcumulus? i heard about it somewhere...

It's a codename for another of our projects. When the time comes you will know the details (just like AoPS FTW).

Wow the board is gaining alot of new great features.  :)

gr8 u have codenames and keep that under wraps  :)

[quote="Valentin Vornicu"]It's a codename for another of our projects.[/quote]

Wow, some things are [i]so[/i] secret that you have to resort to codenames? woah.


[quote="nhlebris"]Ok I know this doesn't seem that complicated but I need to figure this out really quick. I'm doubting myself. Here is the problem

There are six girls renting an apartment paying equal rent. They owe 1400 each. The lanlord will only accept two checks. Molly put in a check for 4500 and Nicole put in a check for 3900 for a total of 8400. Molly was to take care of Julie and Mel, and Nicole was to get money from Tami and Ivory. Nicole recieved 1500 from Tami and 1400 from Ivory. She gave Tami back 100.  Molly received 1500 from Julie and 1500 from Mel. She gave them each back 100. How much does Nicole owe Molly to make everything equal? It's 300 right?[/quote]
[hide]Since there are 6 girls, each owing 1400 dollars, that's $1400*6=8400$, which was paid.
Molly takes care of Julie and Mel so she should have paid $1400*3=4200$, but she paid 4500 dollars.
Molly gets 3000 dollars from Julie and Mel but then give each back 100 dollars so Molly paid $4500-(1400*2)=1700$.
Nicole takes care of Tami and Ivory so she should have paid $1400*3=4200$, but she paid 3900 dollars.
Nicole gets 2900 dollars from the girls and gives Tami back 100 dollars so Nicole paid $3900-(1400*2)=1100$.
Therefore, Nicole should pay $\frac{1700-1100}{2}=\$300$.[/hide]


If there are [i][b]n[/b][/i] players in a knock-out style tournament, prove there needs to be [i][b](n-1)[/b][/i] games played to determine a champion.

[hide="Not sure how to do with induction..."]If there are $ n$ players, then there will be one such player who will win all his games to be the champion. There are $ n \minus{} 1$ players, however, who will be eliminated to determine the one champion and because there is one elimination per game, there will be $ n \minus{} 1$ games.[/hide]

induction?

for n=2, there is obviously 1 game (plz dont tell me i need to prove this!!)
assume  n=k is true, then for n=k+1, 1 game is needed to have k player left (and this needs k-1 matches), o for k+1 matches, you need k-1+1=k=(k+1)-1, which agrees with the problem

Yeah, stevenmeow, that's what I got.

Is it in Vol 2? I got it from another book, and the solution they had was a bit different:

[b]Base case:[/b] When n = 2, there is 1 game.

[b]Case 1:[/b] n is even (n=2k)

There are k games, and k winners. Induction hypothesis: these k winners need (k-1) games to determine the champ. k + (k-1) = 2k - 1 = n - 1.

[b]Case 2:[/b] n is odd (n=2k+1)

There are k losers from the first k games, and the (k+1) left need to play k games (by the induction hypothesis). k + k = 2k = n-1.


Natural numbers k and n have divisors a1,a2,a3p....and b1,b2,b3...bq. If the sum of the divisors of each number are equal, and the sum of the reciprocals of the divisors of each number are equal, prove n=k.

[hide]
[hide="The conjecture is true for primes."]
Consider $n\in \mathcal{P}: =\{2,3,5,7,11,\ldots\}$. Sum of divisors is $1+n$. Then \[ 1+n = b_1+b_2+\cdots+b_q  \] and \[ 1+\frac{1}{n} = \frac{1}{b_1}+\frac{1}{b_2}+\cdots+\frac{1}{b_q}\Longleftrightarrow n=\frac{1}{\frac{1}{b_1}+\frac{1}{b_2}+\cdots+\frac{1}{b_q}-1}.  \] Consider the case $q=1$. Then $k=1$ and $n$ doesn't exist. So consider next $q=2$. This gives us $n=b_2$ and $n=\frac{1}{\frac{1}{b_2}}=b_2$ which works. If $q\ge 3$, we get $n=b_2+b_3+\cdots+b_q$ and $n=\frac{1}{\frac{S_{q-2}}{b_2b_3\cdots b_q}}$ where $S_{q-2}$ is the second symmetric sum of $b_2,\ldots ,b_q$. Solving the equation gives us $-(q-2)b_2b_3\cdots b_q = \text { something positive }$. Therefore there are no solutions for $q\ge 3$, so $q=2\implies b_2=k=n$, as claimed. $\boxed {}$
[/hide]

[hide="The rest follows inductively."]
Each positive integer $n>1$ can be expressed as the product of primes (for $n=1$ the conclusion is trivial to prove). So if we take any elements, not necessarily distinct, $m_1,m_2,\ldots,m_i$ of $\mathcal{P}$, all $n$ can be attained by their product. So it follows inductively. I think. Maybe... :?
[/hide]
[/hide]

[hide="Hint"]
For any positive integer $n$,
\[ n= \frac{\sum_{d|n}d}{\sum_{d|n} \frac{1}{d}}  \]
[/hide]

1/b1+1/b2+1/b3...+1/bq=(1/b+1/bq)+(1/b2+1/b(q-1))+....
and that sums to be (bq+b(q-1)+b(q-2)+...b)/n
and likewise for k
however since we know that bq+b(q-1)+...+b=ap+a(p-1)+a(p-2)+...+a, numerators of both are equal, this implies that 1/n=1/k, ergo k=n

So is my solution incorrect? If so, why?

Unless i have misunderstood, it needs more strengthening since (1+p1)(1+p1) isnt equal to 1+p1+p1^2 (sum of factors of p1^2), though i think it should work if you induce a separate case for numbers which are divisible by a perfect square


Does anyone know how to get the present value of annuity notation in Latex?

It looks "something" like this $ a_{\overline{n}|i}$.  I've seen this done in texts but it comes out much neater than that.  Any ideas are appreciated.

For the curious: This is the present value of an annuity that makes payments at the end of each of the periods.  There are $ n$ periods at the periodic interest rate $ i$.

Try this (for a document, not on the forum)
[code]\def\annuity#1#2{_{\vbox{\hrule height .1pt
\kern 1.5pt
\hbox{$\scriptstyle {#1}\kern 1pt$}}
\vrule width .1pt #2}}[/code]
Then use it with
[code]$a\annuity{n}{i}$[/code]You can change the various dimensions if you wish to play with the way it looks.

[size=75](This is code edited from other forums to get what you wanted).[/size]


MATHCOUNTS
search
function
number theory
relatively prime
prime numbers
totient function

Find the sum of all positive rational numbers that are less than $10$ and that have denominator $30$ when written in lowest terms.

[hide="Solution"]There are 8 fractions which fit the conditions from 0 to 1: $\frac{1}{30},\frac{7}{30},\frac{11}{30},\frac{13}{30},\frac{17}{30},\frac{19}{30},\frac{23}{30},\frac{29}{30}$

The sum of these numbers is 4. Notice that there are also 8 terms from 1 to 2, and we can obtain them by adding 1 to each of our first 8 terms. For example, $1\frac{19}{30}$. Seeing this pattern, our answer is $4(10)+8(1+2+3+\cdots+9)=\boxed{400}$ [/hide]

That seems a bit easy even for a number one.

It seems like one of those annoying mathcounts problems...
Why do these problems still exist on aime's?? :mad:

[hide]
The number in the numerator can range from $1$ to $299$. The sum of all these numbers is equal to $\frac{1+299}{2} \times 299$

But since the number must be in lowest terms, we must remove all the numbers that contain either a factor of $2$, $3$, or $5$.

Using PIE, the sum of all such numbers is equal to:

$150 \times \displaystyle \left((149+99+59) - (49+19+29) + 9 \right) = 150 \times 219$.

The desired sum is then equal to $\frac{150\times (299-219)}{30} = \boxed{400}$.
[/hide]

Principle of inclusion-exclusion.  Search, it's been asked on the forum many times.

tnx but it still doesn't help. how did he found the sum?
and the search give nothing, is it a bug?

No, it should work.  It's probably because of the server shift.  Give it some time.  Maybe tomorrow I'll explain how he used PIE in this problem, if no one beats me to it.  It's too late right now.

[quote="4everwise"][hide="Solution"]There are 8 fractions which fit the conditions from 0 to 1: $ \frac {1}{30},\frac {7}{30},\frac {11}{30},\frac {13}{30},\frac {17}{30},\frac {19}{30},\frac {23}{30},\frac {29}{30}$

The sum of these numbers is 4. Notice that there are also 8 terms from 1 to 2, and we can obtain them by adding 1 to each of our first 8 terms. For example, $ 1\frac {19}{30}$. Seeing this pattern, our answer is $ 4(10) \plus{} 8(1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} 9) \equal{} \boxed{400}$ [/hide][/quote]
Sorry for revive, but the first part of this solution (i.e., finding the 8 fractions), can be circumvented by noting that $ \phi(30)\equal{}8$, and that the 8 relatively prime numbers below 30 will pair up into 4 groups that sum to 30. Then we see that the first 8 fractions sum to 4, and can proceed similarly.

[quote="Brut3Forc3"][quote="4everwise"][hide="Solution"]There are 8 fractions which fit the conditions from 0 to 1: $ \frac {1}{30},\frac {7}{30},\frac {11}{30},\frac {13}{30},\frac {17}{30},\frac {19}{30},\frac {23}{30},\frac {29}{30}$

The sum of these numbers is 4. Notice that there are also 8 terms from 1 to 2, and we can obtain them by adding 1 to each of our first 8 terms. For example, $ 1\frac {19}{30}$. Seeing this pattern, our answer is $ 4(10) \plus{} 8(1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} 9) \equal{} \boxed{400}$ [/hide][/quote]
Sorry for revive, but the first part of this solution (i.e., finding the 8 fractions), can be circumvented by noting that $ \phi(30) \equal{} 8$, and that the 8 relatively prime numbers below 30 will pair up into 4 groups that sum to 30. Then we see that the first 8 fractions sum to 4, and can proceed similarly.[/quote]

How did you know they will pair up into groups that sum to 30? I think a similar idea is used in the PIE solution (the $ 150$ factor).  What makes it work?

[quote="amacfie"]How did you know they will pair up into groups that sum to 30?[/quote]  If $ 0 < a < m$ and $ (m, a) \equal{} 1$ then $ (m, m \minus{} a) \equal{} 1$ and so $ a, m \minus{} a$ pair up as desired.  (With special consideration for the case $ m \equal{} 2$.)

[quote="cdymdcool"][quote="Brut3Forc3"][quote="4everwise"][hide="Solution"]There are 8 fractions which fit the conditions from 0 to 1: $ \frac {1}{30},\frac {7}{30},\frac {11}{30},\frac {13}{30},\frac {17}{30},\frac {19}{30},\frac {23}{30},\frac {29}{30}$

The sum of these numbers is 4. Notice that there are also 8 terms from 1 to 2, and we can obtain them by adding 1 to each of our first 8 terms. For example, $ 1\frac {19}{30}$. Seeing this pattern, our answer is $ 4(10) \plus{} 8(1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} 9) \equal{} \boxed{400}$ [/hide][/quote]
Sorry for revive, but the first part of this solution (i.e., finding the 8 fractions), can be circumvented by noting that $ \phi(30) \equal{} 8$, and that the 8 relatively prime numbers below 30 will pair up into 4 groups that sum to 30. Then we see that the first 8 fractions sum to 4, and can proceed similarly.[/quote]

What do you mean by $ \phi(30) \equal{} 8$?[/quote]I mean [[Euler's totient function]], otherwise known as the phi function. $ \phi(n)$ (or sometimes $ \Phi(n)$) counts the number of positive integers less than (or equal to $ n$ which are relatively prime to $ n$. It's pretty clear that in this problem, we require numerators relatively prime to 30, so the phi function works very well here.

This is also a revive, but we can still directly apply totient function to 300, as it shares the same prime factors as 30. Thus the answer can be found in similar fashion.


advanced fields
advanced fields unsolved

For any (planar) triangle T, is there is a 3-coloring of the (infinite) plane with no monochromatic copy of T?  ;)

This is a conjecture in Euclidean Ramsey theory. You can see Graham's talk for more detail:

http://www.msri.org/publications/ln/msri/2003/introdcgeom/graham/1/


Let $ X_n$ be be a sequence of iid random variables with mean $ 0$ and variance $ 1$. Show that $ Y_n\equal{}max\{\frac {|X_1|}{\sqrt{n}}, . . . , \frac {|X_n|}{\sqrt{n}}\}\to 0$     in distribution.

By CLT, I could show that $ \frac {1}{\sqrt{n}}(X_1\plus{}X_2\plus{}. . . \plus{}X_n)\to N(0,1)$ in distribution. And I also know that $ P(Y_n\leq c)\equal{}[P(\frac {|X_n|}{\sqrt{n}}\leq c)]^n$               for any real $ c$.
I don't know how to proceed further. Can anyone help me? Also I would be more glad if someone could give a technique to solve the problems where max. or min. involved.
Thanks.

[quote="limsup"]Let $ X_n$ be be a sequence of iid random variables with mean $ 0$ and variance $ 1$. Show that $ Y_n \equal{} max\{\frac {|X_1|}{\sqrt {n}}, . . . , \frac {|X_n|}{\sqrt {n}}\}\to 0$     in distribution.

By CLT, I could show that $ \frac {1}{\sqrt {n}}(X_1 \plus{} X_2 \plus{} . . . \plus{} X_n)\to N(0,1)$ in distribution. And I also know that $ P(Y_n\leq c) \equal{} [P(\frac {|X_n|}{\sqrt {n}}\leq c)]^n$               for any real $ c$.
I don't know how to proceed further. Can anyone help me? Also I would be more glad if someone could give a technique to solve the problems where max. or min. involved.
Thanks.[/quote]
assume that $ c \equal{} 1, X \equal{} X_1$, we have $ P(|X| \leq \sqrt {n})^n \equal{} (1 \minus{} P(|X| > \sqrt {n}))^n \geq 1 \minus{} nP(|X| > \sqrt {n})$. Put $ Y_n \equal{} X^2I[|X| > \sqrt {n}]$, we have $ Y_n \to 0$, bounded by $ X^2$ then $ EY_n \to 0$ and $ nP(|X| > \sqrt {n}) \to 0.$

Thanks for your reply.
Why do you assume that $ c\equal{}1$?  Wouldn't your argument work for any $ c>0$? I could see that it works for $ c>0$ also.


Let $ E$ be a set of positive real numbers and define $ \sum_{x \in E} x \equal{} \sup_{F \in \mathcal{F}} s_{F}$ where $ \mathcal{F}$ is the collection of finite subsets of $ E$ and $ s_{F}$ is the sum of the elements in $ F$. Show that $ \sum_{x \in E} x < \infty$ only if $ E$ is countable.

Let $ E_n \equal{} \{ x \in E \, | \, x \geq 1/n \}$. Then $ |E_n| < \infty$, since $ \sum_{x \in E} x$ is finite. Then $ E \equal{} \bigcup_{n\equal{}1}^{\infty} E_n$, so $ E$ is countable.


vector
function
calculus
calculus computations

Do solutions of a second-order homogeneous differential equation form a 2-dimensional vector space?

You did not say what exactly is a "2nd order differential equation" and what constitutes its "solution"... but the short answer is yes. A solution of a decent 2nd order ODE is determined by the initial values of the function and its derivative. The set of all initial values is 2-dimensional.

I think that the relevant theorem is:

Let y1(t) and y2(t) be two linearly independent solutions to a linear homogeneous second-order differential equation, then all solutions are in the form:

c_1*y1+c_2*y2

So if these were somehow vectors, then they would form a basis...


In a certain population the ratio of the number of women to the number of men is 11 to 10. If the average (arithmetic mean) age of the women is 34 and the average age of the men is 32, then the average age of the population is

$ \textbf{(A)}\ 32\frac{9}{10}\qquad\textbf{(B)}\ 32\frac{20}{21}\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 33\frac{1}{21}\qquad\textbf{(E)}\ 33\frac{1}{10} $

[hide="Answer"]We can put this in simplest case by saying that there are 11 women and 10 men. Therefore, we have $\frac{34\cdot 11+32\cdot 10}{21}=33\frac{1}{21}\Rightarrow \boxed{D}$.[/hide]

According to my 1983 AHSME, the answer choices for problem #9 should be:

$ \textbf{(A)}\ 32\frac{9}{10}\qquad\textbf{(B)}\ 32\frac{20}{21}\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 33\frac{1}{21}\qquad\textbf{(E)}\ 33\frac{1}{10} $


function
trigonometry
real analysis
real analysis unsolved

Here is a further problem I'm stumped on.

Suppose 0<a<1, but a is not equal to 1/n for any natural number n. Find a
function, f, which is continuous on [0,1] and which satisfies f(0)=f(1),
but which does not satisfy f(x)=f(x+a) for any x.

It is not clear how to construct such a function.

Construct any $ a$-periodic function $ g$ such that $ g(0)\ne g(1)$ and add $ [g(0)\minus{}g(1)]x$ to it.

binator, I think that now you should prove that it is not possible if $ a\equal{}\frac{1}{n}$  :)

binator has already posted that problem ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184684]here[/url]); that's why this is [i]another[/i] intermediate value problem (although actually this has nothing to do with the intermediate value theorem, while the other question does).

[i]Construct any -periodic function  such that  and add  to it.[/i]

What exactly is your function g ? How is it related to f ?

Any $ a$-periodic continuous function that comes to your mind. The only requirement is that $ g(0)\ne g(1)$. If nothing comes to your mind within 10 minutes, [hide="Look here"]take $ g(x)\equal{}\cos\frac{2\pi}ax$[/hide] As to relation, "add $ [g(0)\minus{}g(1)]x$ to it" means $ f(x)\equal{}g(x)\plus{}[g(0)\minus{}g(1)]x$.

[i]Any -periodic continuous function that comes to your mind. The only requirement is that . If nothing comes to your mind within 10 minutes, [/i]

Thankyou so much for your reply. Its so obvious when the key step is filled in.


Source - AMC12 B 2004

I don't quite understand the solution to this problem, someone help please:

in $\triangle ABC, AB=BC$ and ${BD}$is an altitude. Point $E$ is on the extension of $AC$ such that $BE=10$. The values of tan $\angle CBE$, tan $\angle DBE$, and tan $\angle ABE$ form a geometric progression, and the values of cot $\angle DBE$, cot $\angle CBE$, and cot $\angle DBC$ form an arithmetic progression. What is the area of $\triangle ABC$?

The solution starts off with:
[hide]Let $\angle DBE =\alpha$, and $\angle DBC=\beta$. Then $\angle CBE = \alpha - \beta$ and $\angle ABE = \alpha + \beta$, so $tan(\alpha-\beta)tan(\alpha+\beta)=tan^2(\alpha)$.
This is the part that I don't get. Why is $tan(\alpha-\beta)tan(\alpha+\beta)=tan^2(\alpha)$?
[/hide]

Thanks

It's because the tangents form a geometric progression. The product of first and third term will be the square of the middle term. (Basically using the principle of geometric mean)

ooooooooooo
completely forgot about the geometric progression.....
thanks