MATHCOUNTS
number theory
relatively prime

[quote="espark52"]how does the therom go again?[/quote]

Why don't you just google it?

Is it this?

the largest number that cannot be created by adding any number of the integers m and n is equal to mn-m-n

of course mathcounts knows that people have caught on to that theorem by now and change something it so you cannot just blindly apply the theorem

there is a proof (that I would have to dig out), but proofs aren't really put on the mathcounts section

Hmm. Is it
[hide]abc-ab-bc-ac-a-b-c??

So it would be... 900-60-90-150-6-10-15 = 569 ?

I think there might be a little bit about this in AoPS vol. 2, chapter 24, diphantine equations, but I'm too lazy to go check.[/hide]

Oh yeah, what's with "chicken mcnugget"
Who came up with it, Ronald McDonald? :rotfl:

I guess you're right...

Well you could try it with smaller numbers, like 2, 3, 6 or something.

Hang on, if I'm not mistaken, it should be larger, since there are more numbers, so the largest would be inflated.

Ignite, can you post a solution? I'm still convinced that mine is right though.

I am convinced that mine is wrong because I applied the mc nugget's theorem with two numbers that are not relatively prime where 10+10+6+6+6+6

but, it also makes since that with more numbers, you can make more numbers by adding them together.

if you can make more numbers by adding them together, the limit of what you can't make should be less

Sorry but I am also convined that you are wrong...

15+10*5+6*84 gives your number

[quote="Ignite168"]What is the largest value that you cannot pay for with unlimited 6 cent stamps, unlimited 10 cent stamps, and unlimited 15 cent stamps?[/quote]

[hide="My solution"]First of all we have to check if these are relatively prime to eachother which they are.

Let us consider modulo 6.

This means you have 0(mod 6), 4(mod 6), and 3(mod 6), respectively.

So now we have to find out the smallest number that satisfies 0(mod 6), 1(mod 6), ... 5(mod 6).

0(mod 6)~6 cents, 12 cents, 18 cents, 24 cents,...
1(mod 6)~10+15=25 cents, 31 cents, 37 cents, 43 cents,...
2(mod 6)~10+10=20 cents, 26 cents, 32 cents, 38 cents,...
3(mod 6)~15 cents, 21 cents, 27 cents, 33 cents,...
4(mod 6)~10 cents, 16 cents, 22 cents, 28 cents,...
5(mod 6)~10*2+15=35 cents, 41 cents, 47 cents, 53 cents,...

Now, we have to subtract 6 from each modulo 6 to get

0, 19, 14, 9, 4, and 29, respectively.

Each of these numbers is the largest number that cannot be achieved with any amount of stamps with their respective modulo.  0 doesn't count since we are assuming the number is positive.  Also, note that any number that is not specified with the previous six numbers that are larger than the largest number of the six numbers can be achieved.  So the answer is the largest number of the six numbers previously which is 29.

So the answer is 29.

$29$

[color=white]I do not know how I was able to come up with that solution[/color]
[/hide]

[quote="pgpatel"][quote="ehehheehee"]nope, there is no 3 number thingy (in book II), and also, it makes sense that a 3 number thingy is most like going to smaller than the two number thing

[hide="I think this is the answer, but it would be much harder if Ignite used more evil numbers"]the smallest number that cannot be made with 6 and 10 is 44, we know that 44 cannot be made with 6 and 10 but not 15 yet

44-15=29 cannot be made with 6's and 10's (odd number cannot be made with even numbers)

44-15-15=14 cannot be made with 6's and 10's

so the answer is 44[/hide][/quote]
Four sixes and two tens... And yes the numbers must be relatively prime.[/quote]

Does my answer make sense? Furthermore is it correct?

[quote="Phelpedo"]I don't think that the general formula for the 3-case will be that simple. Neil's solution looks like a good approach.[/quote]

Thank You.  I sort of learned it from my Math Teacher.


Hi there,

i'm interested in learning matlab...
Does anyone have a link where i can download matlab?

thanks...

You cannot download it. It is a commerical product. But you use Octave instead.  :)


quadratics
algebra
polynomial
logarithms
sum of roots

[quote="anirudh"][quote="xeroxia"]Find the sum of roots of equation $4^{x}-2^{x+4}+16=0$.[/quote]

This should be in HSB minimum, but
[hide]quadratics, synthetic division, polynomial division[/hide] could  help.[/quote]

The last two wouldn't help you at all.

[hide="Hint"]Set $2^{x}=a$ and rewrite the equation[/hide]

[quote="anirudh"][quote="xeroxia"]Find the sum of roots of equation $4^{x}-2^{x+4}+16=0$.[/quote]

This should be in HSB minimum, but
[hide]quadratics, synthetic division, polynomial division[/hide] could  help.[/quote]

Can you stop doing the following?
1) Repeating an answer without providing a solution after already 5-6 responses have been posted
2) Repeating the exact same solution after 5-6 responses have been posted
3) Providing hints when no one has responded that don't help you with the problem at all

[quote="me@home"]I would like to say that I too am very annoyed with anirudh's posts.
[hide="Btw, your solution is wrong"]
So from your solution, we find $2^{r_{1}}+2^{r_{2}}=16$ but we really want $r_{1}+r_{2}$. Instead, we use the same idea to find that $2^{r_{1}}2^{r_{2}}=16=2^{r_{1}+r_{2}}\implies r_{1}+r_{2}=\boxed{4}$ as wanted.[/hide][/quote]
[hide="How Did You Get..."]$2^{r_{1}}+2^{r_{2}}=2^{r_{1}}2^{r_{2}}$[/hide]

[quote="Phelpedo"]It's not true generally, but in this case it is. Sum of the roots of a quadratic is -b/a, product is c/a. The equation is

$y^{2}-16y+16$, so the two are the same.[/quote]
Does that work for cubic equations (somehow similarly)?

Here's the reason. I think it is somewhat related to the binomial theorem's explanation.
The polynomial can be factored to the product of the following with one number chosen from each row to find each part of the equation when multiplied out.
$(x+r_{1})$
$(x+r_{2})$
$(x+r_{3})$
.
.
.
$(x+r_{n})$
So with this when we want to find the coefficient of $x^{n-1}$ we know that every possible combination is a one where we chose all of the x's and one of the r's. So in the polynomial without combining like terms we get that every coefficient of $x^{n-1}$ is $r_{1},r_{2},...r_{n}$. So the sum of them is the coeffecient of $x^{n-1}$ when you combine like terms. This is where newtons sums. You can use similar reasoning to find the coeffecient of any terms  in a polynomial if you know the roots.

[quote="kyyuanmathcount"]I would like to add:

4. Giving three unconnected hints to potentailly mislead others because you haven't solved the problem yourself.

Comments such as those are considered spam (unless posted in G&FF for some reason) and, in the future, will be deleted because they have no significant mathematical content.

[hide="Solution"]We can let $2^{x}=y$.  The equation becomes $y^{2}-16y+16=0$.  The sum of the roots of this are $\boxed{16}$.  [/hide][/quote]

:wallbash:   :oops:   Sorry, don't read my solution, or at least the sum of the roots part.

[quote="bpms"][quote="anirudh"][quote="xeroxia"]Find the sum of roots of equation $4^{x}-2^{x+4}+16=0$.[/quote]

This should be in HSB minimum, but
[hide]quadratic[/hide] could  help.[/quote]
It is advised against editing what you say when you make a mistake. It is good to admit when you make a mistake and people who are new to the forum may think everyone here is perfect at math...[/quote]

Agreed  :thumbup:

Shall we get back on topic now?

Well what is back on topic? The problem is solved and the tool has been explained...
btw the symmetric sums are called newton sums but I think the method is called viete's

Maybe a lock would "bring it back on topic" so people like me stop spamming here.

[quote="xeroxia"]Find the sum of roots of equation $4^{x}-2^{x+4}+16=0$.[/quote]

[hide]OKay, let $2^{x}=a$.

Then we get $a^{2}-16a+16=0$

We see that the roots of this equation are $\frac{16 \pm \sqrt{16^{2}-4*16}}{2}$.

Both of these numbers are positive, so we know that There is a value of $x$, such that $2^{x}=a$ for those values of a.

Let us call the values of the solutions $x_{1}$ and $x_{2}$, where $x_{1}> x_{2}$

So $2^{x_{1}}=8+4\sqrt{3}$, and $2^{x_{2}}=8-4\sqrt{3}$.

So notice that we are looking for $x_{1}+x_{2}$.

So we multiply the above equations to get

$2^{x_{1}+x^{2}}=16$.

So, this means $x_{1}+x_{2}=4$

$4$[/hide]

[hide="solution"]

We can put the bases of the terms with exponents to 2 by doing this:

$2^{2x}-16*2^{x}+16=0$

and substitute and solve:

$y=2^{x}$

$y^{2}-16y+16=0$

$y = \frac{16 \pm \sqrt{256-64}}{2}=\frac{16 \pm 8\sqrt{3}}{2}=8 \pm 4 \sqrt3$

So

$2^{x}= 8 \pm 4 \sqrt3$

We can call the roots $x_{1}$ and $x_{2}$

We can multiply $8+4 \sqrt3$ and $8-4 \sqrt3$ to get

$2^{x_{1}+x_{2}}= 64-48 = 16 \rightarrow x_{1}+x_{2}= \boxed{4}$[/hide]

Why the identical solutions???? Also there is basically one way to solve this problem and it has been discussed!!... :o
I think its time for a mod to lock this... lol

[hide="Tricky"] Let the roots be $a, b$.  Consider their sum $a+b$ and note that

$2^{a+b}= (2^{a}) \cdot (2^{b})$

Which is the product of the roots of the quadratic $u^{2}-16u+16 = 0$.  This is obviously $16$; ergo,

$a+b = \log_{2}16 = \boxed{4}$. [/hide]

The solutions from t0rajir0u and me@home are showing the tricky part of this problem.
The tricky part is you are asked for sum of roots, but you can reach the solution by product of roots.


evaluate using a counting argument

${n\choose0}^2+{n\choose1}^2+{n\choose3}^2+\dots+{n\choose{n}}^2$

${n \choose k}^2 = {n \choose {n-k}}$

Now we want to choose $n$ persons from a group of $2n$, with $n$ men and $n$ women : ${2n \choose n}$ ways to do this.

We can also pick $k$ men and $n-k$ women. Summing for $k = 0 \cdot n$ ....


Sides $ BA$ and $ CA$ of $ \triangle ABC$ are extended through $ A$ to form rhombuses (rhombi?) $ BATR$ and $ CAKN$. $ BN$ and $ RC$, intersecting at $ P$, meet $ AB$ at $ S$ and $ AC$ at $ M$. Draw $ MQ$ parallel to $ AB$ (where $ Q \in BC$).

a) Prove $ AMQS$ is a rhombus

b) Prove that the area of $ \triangle BPC$ is equal to the area of quadrilateral $ ASPM$

I have a proof but it is a bit long (for part a) at least). I want to see if there is a shorter way.

May be I misunderstand your problem, but... I don't see that $ AMQS$ is a rhombus!:blush:
[img]http://i227.photobucket.com/albums/dd72/TPFC/unknown.png[/img]

$ K$ and $ T$ should be on $ AB$ and $ AC$ (extended), respectively I think.

[img]http://i227.photobucket.com/albums/dd72/TPFC/unknown-1.png[/img]
OK! With $ K,T$ on $ AB,AC$ we have:
$ \frac{AM}{AC}\equal{}\frac{AB}{AB\plus{}NC}\equal{}\frac{AB}{AB\plus{}AC}$
$ \frac{AS}{AB}\equal{}\frac{AC}{RB\plus{}AC}\equal{}\frac{AC}{AB\plus{}AC}$
Then $ AM\equal{}AS$ and $ \frac{AM}{AC}\plus{}\frac{AS}{AB}\equal{}1\Leftrightarrow \frac{AS}{AB}\equal{}\frac{CM}{AC}\equal{}\frac{MQ}{AB}$
It mean $ AM\equal{}AS\equal{}MQ$, then $ AMQS$ is a rhombus!

Part b)
We have:
$ S_{BPC}\equal{}S_{BMC}\minus{}S_{PMC}\equal{}\frac{CM}{AC}S_{ABC}\minus{}S_{PMC}$
$ \equal{}\frac{AS}{AB}S_{ABC}\minus{}S_{PMC}\equal{}S_{SAC}\minus{}S_{PMC}\equal{}S_{ASPM}$
It is OK!

Solution to part (b):

Note that $ (BPC) \plus{} (ACS) \plus{} (ABM) \minus{} (ABC) \equal{} (ASPM)$. Hence we need only prove that $ (ACS) \plus{} (ABM) \equal{} (ABC)$.

Verify that $ \frac {(ACS)}{(ABC)} \equal{} \frac {AS}{AB}$ and $ \frac {(ABM)}{(ABC)} \equal{} \frac {AM}{AC}$. Using $ \frac {AS}{AB} \plus{} \frac {AM}{AC} \equal{} 1$, we immediately get $ (ACS) \plus{} (ABM) \equal{} (ABC)$.

PS. Thanks tdl.


geometry
3D geometry
sphere
superior algebra
superior algebra unsolved

Prove that points on the bondary of a sphere,i.e points that satisfy condition:$ x^2\plus{}y^2\plus{}z^2\equal{}r^2$,for some $ x,y,z,r\in\mathbb{R}$, and points of a plane have the same cardinality [u]without falling back on using inversion[/u].

[quote="ZetaX"]Some ways:

a) It's a manifold.

b)  There are (lots of) injections from $ \mathbb R$ to the sphere/plane, and both are subsets of $ \mathbb R^3 \cong \mathbb R$. Thus all mentioned sets have the same cardinality.

c) The sphere is the 1-point-compactification of the plane.

d) Inversion ;)[/quote]

Please avoid using of d) ;)
show at least one such bijection.

[quote="Erken"]
...show at least one such bijection.[/quote]
I don't know what Inversion is, but one such bijection may be found by Stereographic projection
http://en.wikipedia.org/wiki/Stereographic_projection

You want a explicit "bijection" (inversion also lacks one point) but not the most natural one¿ Sorry, I find this rather ridiculous. I always hate things that explicitely forbid stuff, especially if that stuff is rather canonical, elementary and straightforward.
I gave proofs that do not directly give bijections but still proof the required equality (and if one really wants, one can construct an explicit bijection out if them).


Show that for any positive integer n, there exists a positive integer z with exactly n digits, such that no digit of z is 0, and such that z is divisible by the sum of the digits of z.

  Darij

darij, can you please show your solution to this. thx

In my opinion it's a really hard problem:

[url]http://www.kalva.demon.co.uk/short/soln/sh98n7.html[/url]

I published this problem on forum last autumn. First time I saw it (with solution different from mentioned above) in 1993.


I couldn't figure out how to solve two problems, any hints will be appreciated^^

1. If n is a positive integer, then n = p + a^2, where a is integer and p is prime or p = 1.
2. If n is an integer and n > 2, then there exists a prime p such that n<p<n!

I don't think the first question is a valid prompt.  n = 25 can't be expressed in that form it seems...

25 - 25 = 0
25 - 16 = 9
25 - 9 = 16
25 - 4 = 21
25 - 1 = 24
25 - 0 = 25
These are all the possible squares you can subtract from 25 and none yield a prime number.

For number 2, a hint.

[hide]

If $ n! \plus{} 1$ is not a prime number, then the factors that make it up are primes greater than $ n$.

[/hide]

An overkill method for Question 2 would be using Bertrand's postulate, which states that there exists a prime between any natural number $ n$ and $ 2n$. Because $ n>2$, $ n!>2n$, therefore the upper bound we get is actually more than we need.

[i]I don't think the first question is a valid prompt. n = 25 can't be expressed in that form it seems...

25 - 25 = 0
25 - 16 = 9
25 - 9 = 16
25 - 4 = 21
25 - 1 = 24
25 - 0 = 25
These are all the possible squares you can subtract from 25 and none yield a prime number. [/i]

[b] But
25 = -11 +36

I thought that -11 is also prime even though negative sign is attached.
Can you give me an advice for that?[/b]

The real reason I chose $ n \equal{} 25$ is because if n is a perfect square, it has only 2 viable options to test to see if it can be expressed in that form.  I replace $ n$ with $ n^{2}$ in the rest of this post (up till my last sentence).

$ n^{2} \equal{} p \plus{} a^{2} \rightarrow (n\plus{}a)(n\minus{}a) \equal{} p$.  Clearly $ n$ and $ a$ must be consecutive integers if $ p$ is to be prime.  Further, $ (n\plus{}a)$ must be prime.  The counter-example I stumbled across was $ n^{2} \equal{} 13^{2}$, because then $ a \equal{} 14$ or $ a \equal{} 12$ and the sum of $ 13 \plus{} a$ is prime for neither of these two values of $ a$.

So I now hold that $ n \equal{} 13^{2}$ is a counter-example to the statement in question.

[quote="winternights1"]The real reason I chose $ n \equal{} 25$ is because if n is a perfect square, it has only 2 viable options to test to see if it can be expressed in that form.  I replace $ n$ with $ n^{2}$ in the rest of this post (up till my last sentence).

$ n^{2} \equal{} p \plus{} a^{2} \rightarrow (n \plus{} a)(n \minus{} a) \equal{} p$.  Clearly $ n$ and $ a$ must be consecutive integers if $ p$ is to be prime.  Further, $ (n \plus{} a)$ must be prime.  The counter-example I stumbled across was $ n^{2} \equal{} 13^{2}$, because then $ a \equal{} 14$ or $ a \equal{} 12$ and the sum of $ 13 \plus{} a$ is prime for neither of these two values of $ a$.

So I now hold that $ n \equal{} 13^{2}$ is a counter-example to the statement in question.[/quote]

Thank you^^

But there is one common problem which i think is similar to the (wrong)question no.1

Prove that any [b]positive[/b] integer 'n' can be expressed as $ 2^k * i$ where 'i' is an odd whole number and k is integral
[b]EDITED[/b]

[hide="Induction"]
If $ n$ is odd, we have the required expression $ 2^0\cdot n$.

If $ n$ is even, then by the inductive hypothesis $ \frac n2$ can be expressed in the form $ 2^k\cdot i$. Then the required form for $ n$ is $ 2^{k\plus{}1}\cdot i$.[/hide]

[quote="aswdaniel"]
2. If $ n$ is an integer and $ n > 2$, then there exists a prime $ p$ such that $ n<p<n!$[/quote]
$ n<n!\minus{}1<n!$ where $ n>2$, note that $ n!\minus{}1$ is not divisible by $ 2,3,4,\cdots,n$. So it's a prime or has a prime divisor. $ n<p \le a<n!$.


Find all $(a,b)$ integers such that $\frac{a^3+1}{ab-1}$ is an integer


[quote="rofler"]Congratulations CMO 2009 winners

[b]First Place:[/b] Jonathan Schneider
[b]Second Place[/b] Danny Shi
[b]Third Place[/b] Jarno Sun

[b]Honourable Mentions[/b]
Ahmad Abdi
Robin Cheng
Nikita Lvov
Hunter Spink <----That's me :P
Chen Sun

More information is outlined here: [url]http://www.cms.math.ca/MediaReleases/2009/cmo-winners[/url][/quote]

Congrats to the Moose for getting HM. :)

[quote="sunchips"]OOHHHHHHHH!!!!!
hahaha Amazing!

Yes. A moose escaped his fate.  :lol:[/quote]

It was indeed amazing that you got HM. Good job Moose!

Yeah. Great job to all winners (including myself  :D ), especially the ever-so-legendary moose.

Congratulations to all the winners!!! :D

And especially to the Doll who pwned again
b/c there was no difficult synthetic geo :rotfl:

And yes, good job, Moose! :)

[quote="sdkudrgn88"]Aw man, Schneider is appearing everywhere in these math contests. He got into the Canadian IOI team too. (Sorry for using your last name, Jonathan, but hey, it's a habit.)

Now I have to get onto the CMO next year. This year I failed the repechage because of the last question.[/quote]

You mean, Robin Cheng.

Hey, what books did you guys pick?

I know Danny probably picked the creative writing in mathematics book :P j/k.

[quote="soledad42"][quote="rofler"]
I know Danny probably picked the creative writing in mathematics book :P j/k.[/quote]

A better speculation would be the MIT mathmatician one.[/quote]

Noobs. Both wrong. :P

[quote="cynthia1991"]I see two unfamiliar names, Nikita Lvov and Ahmad Abdi. Cool

and to Danny:

even if Hunter and Jarno guessed wrong, it seems to me that those are the only two reasonable guesses. (this tells something about you, doesn't it?) btw, how have your English classes been?[/quote]

I chose the origami book (something I can actually understand :P). English has been pretty brutal and is consuming around 90% of my time and is pretty stressful. (Amp, you know how I spend the rest 10%, riiiiiiiiight?) Are you back in Toronto?

[quote="rofler"]But Danny, as I told you in math night, Economics tells you to buy the priciest book, sell it, THEN rebuy the origami book and pocket a little extra change.[/quote]

I am not cheap enough to actually do that.

Is there a copy of the CMO report online anywhere?

Hmm, I don't think it's up yet but we got a copy at the seminar so if you want to know anything from it I could tell you.


Let a be a positive integer. Find all positive integers n such that b = a^n satisfying the condition that a^2+b^2 is divisible by ab+1.

[hide]
removing b

find all positive integers n such that

$ a^{2}+a^{2n}$ is divisible by $ a^{n+1}+1$

factoring the left hand one, we have

$ a^{2}(1+a^{2n-2})$

now its obvious that $ a^{2}$ is not a factor of $ a^{n+1}+1$ so we can ignore it

$ 1+a^{2n-2}$ must be divisible

to be divisible, $ 2n-2$ must be an odd multiple of $ n+1$ so that we have a sum of odd powers:
$ a^{odd}+b^{odd}= (a+b)(a^{odd-1}+some other stuff+b^{odd-1})$

$ 2n-2 = b(n+1)$ where b is a positive, odd integer
$ 2n-2 = bn+b$
$ n(2-b) = b+2$
$ n = \frac{b+2}{2-b}$

since n must be a positive integer, b must be 1
$ n = 3$
[/hide]


function
calculus
integration
inequalities
logarithms
derivative
real analysis

Hello,

I would appreciate any ideas / hints on how to investigate the following integral inequality:

\[ \displaystyle \int_0^{\frac{1}{2x}} \log\left(1\plus{}\epsilon \left(F(t) \plus{} F\left(\frac{1}{x} \minus{} t\right)\right)\right) dt  > 0.5\log(1\plus{}\epsilon) \]

$ F(t)$ is an arbitrary continuous function in the interval $ [0,1]$ satisfying:

\[ F(t) \plus{} F(1\minus{}t) \equal{} 1, F(t) >\equal{} 0 \]

We also have that $ 0.5 < x < 1, \, \, \epsilon > 0$

So I'm basically interested in functions $ F$ and $ \epsilon$ values satisfying and also not satisfying the above inequality.
Grateful for any insights.

Hi, I have tried the following idea. I want to prove that the derivative of the above integral is negative when we increase $ x$. That would give me that the integral is increasing when we decrease $ x$, right? Okey, so I study what happens to the integral when I have $ x + h$ for some small $ h$. So I Taylor expand several times:
\[ \log(1 + \epsilon(F(t) + F(\frac {1}{x + h} - t))) = \log(1 + \epsilon(F(t) + F(\frac {1}{x} - t - \frac {1}{x^2}h + O(h^2)))) = \log(1 + \epsilon(F(t) + F(\frac {1}{x} - t) + F'(\frac {1}{x} - t)\cdot( - \frac {1}{x^2}h + O(h^2)) + \ldots )) = \log(1 + \epsilon(F(t) + F(\frac {1}{x} - t))) + \frac {1}{1 + \epsilon(F(t) + F(\frac {1}{x} - t ))}\cdot (\epsilon(F'(\frac {1}{x} - t)\cdot ( - \frac {1}{x^2}h + O(h^2)) + \ldots )^2
\]
Now, If I study
\[ (1/h)\cdot ( \int_0^{1/(2(x + h))}\log(1 + \epsilon(F(t) + F(\frac {1}{x + h} - t))) dt - \int_0^{1/(2x)}\log(1 + \epsilon(F(t) + F(\frac {1}{x} - t))) ),
\]
that is, the derivative quotient, I end up with the following two terms(after using the Taylor expansion of the $ \log$ up there and by letting $ h \rightarrow 0$):
\[ \int_0^{1/(2(x + h))} \frac {\epsilon F'(\frac {1}{x} - t)\cdot ( - \frac {1}{x^2})}{1 + \epsilon(F(t) + F(\frac {1}{x} - t))} dt \,\, - \,\, (\frac {1}{h})\cdot \int_{1/(2(x + h))}^{1/(2x)} \log(1 + \epsilon(F(t) + F(\frac {1}{x} - t))) dt
\]
Now if I call the function in the $ \log$ in the last integral above for $ G(t,1/x)$, then the last integral there should be equal to:
\[ \frac {1}{2x^2}G'_t(\frac {1}{2x},\frac {1}{x})\log(G(\frac {1}{2x},\frac {1}{x}))
\]
Am I right or have I made some miscalculation somewhere? I would appreciate any comments on this.

Okay, I used the integral mean value theorem on the last integral above and got it to be:
\[ \minus{} \frac {1}{2x^2}\log(1 \plus{} 2\epsilon F(1/(2x)))
\]
Now the strange this is the following. I input $ F(t) \equal{} t, \, 0 \leq t \leq 1 ; F(t) \equal{} 0, t > 1$ in the original integral. Then I just integrate that and get the following expression (calling the original integral for $ I(x)$):
\[ I(x) \equal{} \left(\frac {1}{x} \minus{} 1\right)\log\left(1 \plus{} \epsilon\left(\frac {1}{x} \minus{} 1\right)\right) \minus{} \left(\frac {1}{x} \minus{} 1\right) \plus{} \frac {1}{\epsilon}\log\left(1 \plus{} \epsilon\left(\frac {1}{x} \minus{} 1\right)\right) \plus{} \left(1 \minus{} \frac {1}{2x}\right)\log\left(1 \plus{} \frac {\epsilon}{x}\right)
\]
Now if I differentiate that expression, I get something totally different compared to the derivative I got in the above general expression, when I integrated the integral immediately. I checked on Wikipedia, and the above derivation I did (in the previous post) is exactly the Leibniz integral rule. So I'm wondering why it doesn't work for this particular function? It shouldn't matter that the function is discontinuous (and it's derivative too) at one-two points, because we are integrating, right? Any ideas on what might go wrong in this case?

I take my statement back. It works out fine, I just did some miscalculation. Thanks for your attention! :)


limit
trigonometry
calculus
derivative
induction
logarithms
real analysis

[color=darkblue]$ \lim_{x \to 0} \frac {1 \minus{} cos(a_1x).cos(a_2x)...cos(a_nx)}{x}$, with $ a_1, a_2, ..., a_n \in R$.[/color]

Sorry, are you sure that the denominator is $ x$? I think that it is $ x^2$.

$ \lim_{x \to 0} \frac {1 \minus{} cos(a_1x).cos(a_2x)...cos(a_nx)}{x^2} \equal{}$

$ \lim_{x \to 0} \frac { \sum_{i\equal{}1}^n a_i \sin a_ix \prod_{j \ne i} \cos a_jx}{2x}\equal{}0$

$ \lim_{x \to 0} \frac { \sum_{i\equal{}1}^n a_i^2\cos a_i x \prod_{j \ne i} \cos a_jx}{2}\equal{}\sum_{i\equal{}1}^n \frac{a_i^2}{2}$

[quote="thanhnam2902"][color=darkblue]$ \lim_{x \to 0} \frac {1 \minus{} cos(a_1x).cos(a_2x)...cos(a_nx)}{x^2}$, with $ a_1, a_2, ..., a_n \in R$.[/color][/quote]

Hi! kuny, I have some thing wrong in my post, I have to change it as you say.

I think that again with L'hopital rule the result follows immediately, like the previous solution. But if it is not allowed  to use it, once we get the result with derivatives it is easy to think of mathematicacl induction with respect to n. For n=1 it is trivial. Sopose for n=k is true, for n=k+1 we have $ lim_{x \to 0} \frac {1 \minus{} cos(a_1x).cos(a_2x)...cos(a_{k\plus{}1}x)}{x^2}$=$ lim_{x \to 0} \frac {1 \minus{} cos(a_{k\plus{}1}x) \plus{} cos(a_{k\plus{}1}x) \minus{}cos(a_1x).cos(a_2x)...cos(a_{k\plus{}1}x)}{x^2}$=
$ lim_{x \to 0} \frac {1 \minus{} cos(a_1x).cos(a_2x)...cos(a_{k\plus{}1}x)}{x^2}$$ lim_{x \to 0} \cos(a_{k\plus{}1}x)$ + $ lim_{x \to 0} \frac {1\minus{}cos(a_{k\plus{}1}x)}{x^2}$ and the result follows immediately

[quote="thanhnam2902"][color=darkblue]$ \lim_{x \to 0} \frac {1 \minus{} cos(a_1x).cos(a_2x)...cos(a_nx)}{x^2}$, with $ a_1, a_2, ..., a_n \in R$.[/color][/quote]
[u][b]Proof.[/b][/u] Denote $ L_n \equal{} \lim_{x \to 0} \frac {1 \minus{} cos(a_1x).cos(a_2x)...cos(a_nx)}{x^2}$ , $ n\in N^*$ . Observe that $ L_1 \equal{} \frac 12\cdot a_1^2$ and

$ L_{n \plus{} 1} \equal{} \lim_{x\to 0}\ \frac {1}{x^2}\cdot \left\{\left[1 \minus{} \cos\left(a_{n \plus{} 1}x\right)\right] \plus{} \cos\left(a_{n \plus{} 1}x\right)\cdot\left[1 \minus{} \prod_{k \equal{} 1} ^{n}\cos\left(a_kx\right)\right]\right\} \equal{}$ $ a^2_{n \plus{} 1} \plus{} L_n$ .

Therefore, $ L_1 \equal{} \frac 12\cdot a^2_1$ and $ L_{n \plus{} 1} \equal{} L_n \plus{} \frac 12\cdot a^2_{n \plus{} 1}$ for any $ n\in N^*$ . In conclusion, $ \boxed {\ L_n \equal{} \frac 12\cdot\sum_{k \equal{} 1}^na^2_k\ }$ .

[u][b]A similar exercises:[/b][/u]

$ 1\ \blacktriangleright\ \lim_{x\to 0}\ \frac {1}{x^{n \plus{} 2}}\cdot\left[n!\cdot x^n \minus{} \prod_{k \equal{} 1}^n\sin\ (kx)\right] \equal{} \frac 16\cdot n!\cdot\sum_{k \equal{} 1}^nk^2$ .

$ 2\ \blacktriangleright\ \lim_{x\to 0}\ \frac 1x\cdot\left[1\minus{}\sqrt [p]{\prod_{k\equal{}1}^n\ln (e\plus{}kx)}\right]\equal{}\minus{}\frac {n(n\plus{}1)}{2pe}$ .

[quote="Virgil Nicula"][quote="thanhnam2902"][color=darkblue]$ \lim_{x \to 0} \frac {1 - cos(a_1x).cos(a_2x)...cos(a_nx)}{x^2}$, with $ a_1, a_2, ..., a_n \in R$.[/color][/quote]
[u][b]Proof.[/b][/u] Denote $ L_n = \lim_{x \to 0} \frac {1 - cos(a_1x).cos(a_2x)...cos(a_nx)}{x^2}$ , $ n\in N^*$ . Observe that $ L_1 = \frac 12\cdot a_1^2$ and

$ L_{n + 1} = \lim_{x\to 0}\ \frac {1}{x^2}\cdot \left\{\left[1 - \cos\left(a_{n + 1}x\right)\right] + \cos\left(a_{n + 1}x\right)\cdot\left[1 - \prod_{k = 1} ^{n}\cos\left(a_kx\right)\right]\right\} =$ $ a^2_{n + 1} + L_n$ .

Therefore, $ L_1 = \frac 12\cdot a^2_1$ and $ L_{n + 1} = L_n + \frac 12\cdot a^2_{n + 1}$ for any $ n\in N^*$ . In conclusion, $ \boxed {\ L_n = \frac 12\cdot\sum_{k = 1}^na^2_k\ }$ .

[u][b]A similar exercises:[/b][/u]

$ 1\ \blacktriangleright\ \lim_{x\to 0}\ \frac {1}{x^{n + 2}}\cdot\left[n!\cdot x^n - \prod_{k = 1}^n\sin\ (kx)\right] = \frac 16\cdot n!\cdot\sum_{k = 1}^nk^2$ .

$ 2\ \blacktriangleright\ \lim_{x\to 0}\ \frac 1x\cdot\left[1 - \sqrt [p]{\prod_{k = 1}^n\ln (e + kx)}\right] = - \frac {n(n + 1)}{2pe}$ .[/quote]

Hi! Virgil Nicula and kunny! I want have some problem about limit of funtion. Can you give me some problem. Or you give a link to a book. Thank you very much.


Okay;

Christina Meyer owns a used car lot in Florida. Right now her car lots had 900 2006 models, 2,000 2001 models, and over 10,000 1999 and under models. If she has exactly 33,000 cars; how many 1999 and undermodels does her car lot have?

[hide]
Okay; so she has 33,000 cars total.

33,000

She has 900 2006 models;

  33,000
[u]-     900[/u]
  32,100

She has 2,000 2001 models;

  32,100
[u]-   2,000[/u]
  30,100

She has over 10,000 1999 models and under. How many 1999 and under models does her car lot have?

[quote]30,100 1999 and under model in Christina Meyer's used car lot.[/quote][/hide]

Don't worry how easy it is because it doesn't matter to you.  If you are a high-schooler ( ;) ) then do not judge the problems in the middle-school forums.

And please stop spamming.

It's plain 33000-2000-900=30100! I think that is a sixth grade problem.

OK.  All spam and irrelevant posts about the difficulty level of this problem have been removed.

Generally, it is not necessary to question the problems posted here (especially by high schoolers because this is a [i]middle school[/i] forum).  Us mods will eventually figure it out.


[quote="Vasc"]Yes, Tiks. My solution is the same.[/quote]

I supposed so Vasc :) .


modular arithmetic
number theory
greatest common divisor

Find all integers $ n > 1$ such that $ n$ is a power of 3 and $ n\minus{}1$ is five times a power of two.

[hide]
We have $ n\equal{}3^x$ and $ n\minus{}1\equal{}5\times 2^y$, so $ 3^x\minus{}1\equal{}5\times 2^y$.
Looking at the remainders of $ 3^x$ mod 5, we see that $ 3^x\equiv 1 \pmod{5}$ iff $ x\equiv 0 \pmod{4}$, so let $ x\equal{}4x_1$. Now we have $ 81^{x_1}\minus{}1\equal{}5\times 2^y \iff (81\minus{}1)(81^{x_1\minus{}1}\plus{}81^{x_1\minus{}2}\plus{}\cdots\plus{}1)\equal{}5\times 2^y \iff 81^{x_1\minus{}1}\plus{}81^{x_1\minus{}2}\plus{}\cdots\plus{}1\equal{}2^{y\minus{}4}$.
Now, note that $ x_1\equal{}1$, $ y\equal{}4$ satisfies this, and in this case, $ n\equal{}3^4\equal{}81\equal{}5\times 2^4\plus{}1$, so this solution does indeed work. From now on, assume $ x_1>1$. This implies $ y>4$, so the right side is even, and the left side can only be even if $ x_1$ is even (if $ x_1$ was odd, there would be an odd number of odd terms). Writing $ x_1\equal{}2x_2$, and going back to the equation $ 81^{x_1}\minus{}1\equal{}5\times 2^y$, we now have
$ 81^{2x_2}\minus{}1\equal{}5\times 2^y \iff 6561^{x_2}\minus{}1\equal{}5\times 2^y \iff (6561\minus{}1)(6561^{x_2\minus{}1}\plus{}\cdots\plus{}1)\equal{}5\times 2^y \iff (160\times 41)(6561^{x_2\minus{}1}\plus{}\cdots\plus{}1)\equal{}5\times 2^y$.
The left side is divisible by 41, while the right side is not, so there can be no more solutions, and the only solution is $ n\equal{}81$.

I feel like there should be an easier way to do this, but oh well.[/hide]

[hide="My Solution"]We are trying to solve $ 3^a\minus{}1\equal{}5 \cdot 2^b$

Make the intial observation by looking mod 5 that $ a\equal{}4k$

$ 3^{4k}\minus{}1\equal{}5 \cdot 2^b$

$ 9^{2k}\minus{}1\equal{}5 \cdot 2^b$

$ (9^k\plus{}1)(9^k\minus{}1)\equal{}5 \cdot 2^b$

Now one of the terms on the left must be a power of 2, and the other must be 5 times a power of 2. Since $ gcd(9^k\plus{}1,9^k\minus{}1)\equal{}2$ we know that one of them must contain EXACTLY one factor of 2 (not divisible by 4 or 8 etc.). This comes about since they are both even, and so have at least one factor of 2, and their gcd is 2, so they cannot both have more than one factor 2.

If the term which is a power of 2 has exactly one factor of 2, then it is equal to 2. This only comes about when $ k\equal{}0$ and so $ 9^k\plus{}1\equal{}2$.

If the term which is 5 times a power of 2 has exactly one factor of 2, then it is equal to 10. This only comes about when $ k\equal{}1$ and so $ 9^k\plus{}1\equal{}10$.

So our only two possibilities are $ k\equal{}0$ and $ k\equal{}1$

$ k\equal{}0 \Rightarrow a\equal{}4k\equal{}0 \Rightarrow n\equal{}3^a\equal{}1$

$ k\equal{}1 \Rightarrow a\equal{}4k\equal{}4 \Rightarrow n\equal{}3^a\equal{}81$

And since we required $ n>1$ the only possible solution is 81. This turns out to be a genuine solution.
[/hide]


What is the largest positive integer that cannot be written as 42a + b, where a and b are positive integers and b is composite?

I remember solving this once. I'm not doing it again, but it was just tedious work trying to
[hide]find which arithmetic sequence with initial term less than 42 and common difference 42 started with the most primes.[/hide]
Not the best AIME problem IMO.


calculus
integration
calculus computations

show that integral (from 0 to 1) x^m (1-x)^n dx = (n/(m+1)) integral (from 0 to 1) x^(m+1) (1-x)^(n-1) dx. for m, n>0. otherwise, show:

integral (from 0 to 1) x^m (1-x)^n dx = (m! n!)/(m+n+1)!

What happens when you try integration by parts?


pigeonhole principle
algebra proposed
algebra

If the edges of a complete graph K_6 are colored in 2 colors.Prove the graph contains a monochromatic triangle.

[quote="salute2005"]I think there are so many ways to solve.One of them :
  Use the pigeonhole principle.There are two monochromatic triangles. :)[/quote]

Check it!

Let V_1, V_2,...,V_6 be the vertices of K_6. If a pair of edges v_iV_j and v_iv_k are of the same color then
angle v_jv_iv_k monochroamatic. Let r_i and b_i be the respective numbers of red and blue edges obtaining from v_i. Then r_i + b_i05 for all i and there are: The sum[C(r_i,2)+C(b_i,2)]>= sum[C(2,2) + C(3,2)]=24 monochromatic angles.
On the otherhand, in each monochromatic triangle, there are three monochromatic angles, while in each of the other triangles, there is one monochromatic angle. For m:the number of monochromatic triangles, since there are a total of C(6,2)=20 triangles, there are 3m+(20-m)=20+2m monochromatic angles.
Therfore 20+2m >=24  or  m>=2...

Have a look here, you will be interested...

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=monochromatic+triangles&t=5820

Pierre.


Can you create all of the integers from 37 to 67 (inclusive) with two 2's, three 3's, and combine them using addition, subtraction, Multiplication, division, and exponentiation ?

You may use any number of parentheses to control the order of operations and you don't have to use all five numbers each time.

For example, 37 = (2 . 3)^2 + 3/3

Can you create 93 in the same way?

Jean-Louis,

do not use a period for multiplication.
It looks as if you're using decimals.

Instead, you can make your example as:

37 = (2*3)^2 + 3/3 or

37 = (2 X 3)^2 + 3/3 or

37 = (2(3))^2 + 3/3 or

37 = ((2)3)^2 + 3/3 or

37 = ((2)(3))^2 + 3/3

Or someone can switch the order of the first "2" and "3" and/or place
 "3/3 +" in front of the squared expression instead for each set.

--------------------------------------------------------------

IMO, not having to use all five numbers each time makes this a somewhat
weaker puzzle.  It is still interesting and challenging relatively speaking (IMO).

A possible expression (and its slight variant) for 93:
[hide]3*2^(3 + 2) - 3 or 3*2^(2 + 3) - 3[/hide]

[quote="abacadaea"]could we do something like 52=33+22-3?[/quote]

I guess so.. :lol:

Wow!! Nice, pco!!

You must have spent bunch of time on this one.. :lol:


Four men want to cross a bridge at night.  They can cross the bridge at most two at time and must (repeat must) use a flashlight to go across.  Of course the only have one flashlight!  If the four men can cross individually in 1 minute, 2 minutes, 8 minutes, and 10 minutes respecively, what is the minimum amount of time needed to get all four men across the bridge?  Assume (for example) that the first and third men will take a total of eight minutes to get across (the length of the slower man) if they travel across the bridge together.

Don't think this is right, but...

[hide]
So, they have to cross the bridge a total of 5 times.

Call the slowest person A, 2nd slowest B, 2nd fastest C, and fastest D.

So, since the speed of the slowest person is the speed, the speed of at least one crossing has to be the speed of A or slower, at least one has to be the speed of B or slower, and at least one has to be the speed of C or slower. The rest has to be as fast as possible, which occurs if D crosses the other 2 times.

This can occur if crossing 1 has AD, 2 has A, 3 has AC, 4 has A, and 5 has AB. So the total time is:

$ 2 \plus{} 8 \plus{} 10 \plus{} 2(1) \equal{} 22$[/hide]

[hide="here is my solution"] 1 and 2 go across first=2 min, then 1 goes back=1 min, next 10 and 8 go across=10 min, and 2 goes back=2min, and finally 1 and 2 come back together=2 min.  2+1+10+2+2=17 minutes[/hide]

This should go into the brainteasers section...
[hide="To Clarify linzk3"]Call the people 1, 2, 8, and 10, based on how long they take to cross, and call the flashlight $ X$.
$ \begin{tabular}{|c|c|c|c|} \hline Side 1 & Bridge & Side 2 & Time Taken \\
\hline 1, 2, 8, 10, X & & & 0 \\
\hline 8, 10 & 1, 2, X & & 0 \\
\hline 8, 10 & & 1, 2, X & 2 \\
\hline 8, 10 & 1, X & 2 & 2 \\
\hline 1, 8, 10, X & & 2 & 3 \\
\hline 1 & 8, 10, X & 2 & 3 \\
\hline 1 & & 2, 8, 10, X & 13 \\
\hline 1 & 2, X & 8, 10 & 13 \\
\hline 1, 2, X & & 8, 10 & 15 \\
\hline & 1, 2, X & 8, 10 & 15 \\
\hline & & 1, 2, 8, 10, X & 17\\
\hline \end{tabular}$
[/hide]


Prove that if the inscribed quadrilateral with perpendicular diagonals is also a circumscribed one, then it is symmetric with respect to one of its diagonals. :?:

let a,b,c,d be sides of quadrilateral we have:
$a^{2}+c^{2}=b^{2}+d^{2}$ and $a+c=b+d (\ast)$ hence $ac=bd\Longrightarrow (a-c)^{2}=(b-d)2\Longrightarrow |a-c|=|b-d|$hence $a=b\ ,c=d$ or $a=d\ ,b=c$


Find all polynomials with real coefficents such that: $ p(x)p(2x^2\minus{}1)\equal{}p(x^2)p(2x\minus{}1)$

if p has a real root it will be in the $ [ \minus{} 1,1]$ ( why?  :wink: )
we take $ r$ as the maximum of the (different from 1) roots of  $ P$, . its easy to see then that $ P(2\sqrt {\frac {r \plus{} 1}{2}} \minus{} 1) P(\frac {r \plus{} 1}{2}) \equal{} 0$ ,
and  $ \frac {r \plus{} 1}{2} > r$ ,  $ 2\sqrt {\frac {r \plus{} 1}{2}} \minus{} 1 > r$ ( contradiction)
so $ P(x) \equal{} (x \minus{} 1)^kQ(x)$, where $ Q(x)$ not zero for real x.
Now our very first relation gives
 $ Q(x)Q(2x^2 \minus{} 1) \equal{} Q(x^2)Q(2x \minus{} 1) \leftrightarrow V(x^2) \equal{} V(x) \equal{} > V(x^{2^k}) \equal{} V(x)$, for all k naturals. ,
 where $ V(x) \equal{} \frac {Q(2x \minus{} 1)}{Q(x)}$ so $ V(x) \equal{} V(1) \equal{} 1 \equal{} > Q(2x \minus{} 1) \equal{} Q(x) \equal{} c \equal{} > P(x) \equal{} c(x \minus{} 1)^n$

for infinitely many x, V has the same value


In a round robin tournament, occasionally a cyclic 3-set occurs, that is,a set (a,b,c) of teams where a beats b, b beats c, and c beats a. If 23 teams play a round robin tournament with no ties, what is the greatest number of cyclic 3-sets that can occur>?

Am I forced to use cyclic 3-sets or can I use cyclic N-sets? Just need this clarification.

I imagine if the problem asks for cyclic $ 3$-sets, then... the problem is asking for cyclic $ 3$-sets.

[hide="solution"]
The question boils down to: How many groups of three exist in a set of twenty-three in which order does not matter?
Using the nCr function for $ _{23}C_{3}$ is equivalent to saying $ \frac{23!}{3!(23\minus{}3)!}\equal{}\boxed{1771\text{ possible cyclic 3\minus{}sets}}$
Wow, that's a lot :!:
[/hide]
@t0rajir0u: I (just) realized that cyclic N-sets with N>3 would be made up of cyclic 3-sets

[u]Solution.[/u]  I will prove an identity in terms of $ n=2m+1$ participants.  Notice that a 3-set is acyclic iff a player defeats the two other players in the set or if one of the players loses to the other two players in the set.  Let $ v_i$ and $ \ell_i$ be the number of victories and losses of players number $ i$, respectively.  Hence, the number of acyclic 3-sets is
\[ \begin{eqnarray*}C&=&\sum_{i=1}^n\binom{v_i}{2}=\sum_{i=1}^n\binom{\ell_i}{2}\\ &=&\frac{1}{2}\sum_{i=1}^n(\binom{v_i}{2}+\binom{\ell_i}{2}).\]
Notice that $ \binom{v_i}{2}+\binom{\ell_i}{2}=\frac{1}{2}(v_i^2+\ell_i^2-(v_i+\ell_i))$.  By the QM-AM inequality,
\[ \binom{v_i}{2}+\binom{\ell_i}{2}\geq (\frac{v_i+\ell_i}{2})^2-\frac{v_i+\ell_i}{2}.\]
Since $ v_i+\ell_i=n-1\forall i\in[n]$,
\[ \binom{v_i}{2}+\binom{\ell_i}{2}\geq (\frac{n-1}{2})^2-\frac{n-1}{2}=\frac{(n-1)(n-3)}{4}.\]
Id est, $ C\geq \frac{n(n-1)(n-3)}{8}$.
The number of 3-sets is $ \binom{n}{3}=\frac{n(n-1)(n-2)}{6}$, that is, the maximum number of cyclic 3-sets is
\[ \frac{n(n-1)(n-2)}{6}-\frac{n(n-1)(n-3)}{8}=\frac{n(n-1)(n+1)}{24}.\]
Notice that such a maximum can be constructed easily. $ \Box$

[u]Remark.[/u] Notice that $ \frac{n(n-1)(n+1)}{24}=\sum_{k=1}^mk^2$.

I've seen this problem two or three times before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1062351500&t=21878

[quote="MellowMelon"]Not quite. By your logic, you can get $ 4$ such $ 3$-sets with a $ 4$ person tournament. Try actually constructing the tournament that way. You'll find it can't be done.[/quote]
Yeah  :blush: , I noticed this as I was going to sleep last night.


geometry
ratio
trigonometry
geometry unsolved

[quote="Inequalities Master"]The sides $ AB$,$ BC$ and $ CA$ of the triangle $ ABC$ are tangent to the incircle of the triangle $ ABC$ with center $ I$ at the points $ C_1$,$ A_1$ and $ B_1$, respectively.Let $ B_2$ be the midpoint of the side $ AC$.Prove that the lines $ B_1I$, $ A_1C_1$ and $ BB_2$ are concurrent.[/quote]
Construct $ Bx$ parallel $ AC$
$ B_1I\cap Bx \equal{} K ; B_1I\cap C_1A_1 \equal{} N ; C_1A_1\cap Bx \equal{} H ; BN\cap AC \equal{} B_2'$
We have $ Bx\perp IK$ so $ \hat{IKC_1} \equal{} \hat{IKA_1} ( \equal{} \frac {1}{2}\hat{ABC})$
so $ (C_{1}A_{1}NH) \equal{} \minus{} 1 \Rightarrow B(C_{1}A_{1}NH) \equal{} \minus{} 1$
$ BH$ parallel $ AC$ so $ B_2'$ is midpoint of AC (As need!)

Dear Mathlinkers,
it is possible to find a synthetic solution without cross ratio.
Sincerely
Jean-Louis

[quote="Nicula Virgil"] [color=darkred][size=117][b][u]Here is a new enunciation ![/u][/b][/size] The incircle $ (I)$ and the $ A$ - excircle $ (I')$ of $ \triangle ABC$ are tangent to its sides in $ D\in BC$ , $ E\in CA$ ,

$ F\in AB$ and $ D'\in BC$ , $ E'\in CA$, $ F'\in AB$ respectively. Denote $ U\in EF\cap DI$ , $ U'\in E'F'\cap D'I'$ . Prove that $ \{U,U'\}\subset AM$ . [/color] [/quote]

[color=darkblue][b][u]Commentary.[/u][/b]This simple and nice classical problem has many synthetical ([u]without cross ratio[/u]) proofs, such said [b]April[/b] in the upper reply. Sorry, [b]Jayme[/b],

if I pester you ! I"ll present you a [b]very short[/b] metrical proof ([u]with cross ratio[/u]) which is otherwise ! For it, I"ll use the following well-known lemma:[/color]

[color=darkred][b][u]Lemma.[/u][/b] Let $ ABC$ be a triangle. For $ X\in (BC)$ , $ Y\in (CA)$ , $ Z\in (AB)$ denote $ P\in AX\cap YZ$ . Then $ \boxed {\ \frac {PZ}{PY} \equal{} \frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot\frac {AC}{AB}\ }$ .[/color]
[hide="Proof."]
$ \left\{\begin{array}{c} \frac {XB}{XC} \equal{} \frac {AB}{AC}\cdot\frac {\sin\widehat {XAB}}{\sin\widehat {XAC}} \\
 \\
\frac {AZ}{AY}\cdot\frac {\sin\widehat {PAZ}}{\sin\widehat {PAY}} \equal{} \frac {PZ}{PY}\end{array}\right\|\ \bigodot\ \implies\ \frac {PZ}{PY} \equal{} \frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot\frac {AC}{AB}$ .[/hide]
[color=darkblue][b][u]Proof of the proposed problem with a new enunciation.

$ \blacktriangleright$ [/u][/b] Observe that  $ \left\|\begin{array}{c}
IE \equal{} IF\\\\
AE \equal{} AF\end{array}\right\|$ and $ \left\|\begin{array}{c} m(\widehat {UIF}) \equal{} B \\
 \\
m(\widehat {UIE}) \equal{} C\end{array}\right\|$ . Thus, $ \frac {UF}{UE} \equal{} \frac {IF}{IE}\cdot\frac {\sin\widehat {UIF}}{\sin\widehat {UIE}} \equal{} \frac {\sin B}{\sin C} \equal{} \frac bc$ ,

 i.e. $ \boxed {\ \frac {UF}{UE} \equal{} \frac bc\ }\ (1)$ . Denote $ V\in EF\cap AM$ . Using in $ \triangle ABC$ the upper l[b]emma[/b] for $ X: \equal{} M$ , $ Y: \equal{} E$ , $ Z: \equal{} F$ , obtain

$ \frac {VF}{VE} \equal{} \frac {MB}{MC}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}$ , i.e. $ \boxed {\ \frac {VF}{VE} \equal{} \frac bc\ }\ (2)$ . From the relations $ (1)$ , $ (2)$ results $ U\equiv V$ , i.e. $ \boxed {\boxed {\ U\in AM\ }}$ .

$ \blacktriangleright$ Observe that  $ \left\|\begin{array}{c} I'E' \equal{} I'F' \\
 \\
AE' \equal{} AF'\end{array}\right\|$ and $ \left\|\begin{array}{c} m(\widehat {U'I'F'}) \equal{} B \\
 \\
m(\widehat {U'I'E'}) \equal{} C\end{array}\right\|$ . Thus, $ \frac {U'F'}{U'E'} \equal{} \frac {I'F'}{I'E'}\cdot\frac {\sin\widehat {U'I'F'}}{\sin\widehat {U'I'E'}} \equal{} \frac {\sin B}{\sin C} \equal{} \frac bc$ ,
'
 i.e. $ \boxed {\ \frac {U'F'}{U'E'} \equal{} \frac bc\ }\ (1')$ . Denote $ V'\in E'F'\cap AM$ . Using in $ \triangle AF'E'$ the upper l[b]emma[/b] for $ X: \equal{} V'$ , $ Y: \equal{} C$ , $ Z: \equal{} B$ , obtain

$ \frac {MB}{MC} \equal{} \frac {V'F'}{V'E'}\cdot\frac {AB}{AC}\cdot\frac {AE'}{AF'}$ , i.e. $ \boxed {\ \frac {V'F'}{V'E'} \equal{} \frac bc\ }\ (2')$ . From the relations $ (1')$ , $ (2')$ results $ U'\equiv V'$ , i.e. $ \boxed {\boxed {\ U'\in AM\ }}$ .

[b][u]Remark.[/u][/b] See[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=215471][color=red][b]here[/b][/color][/url] [color=darkblue]an interesting extension of this problem.[/color]


Find all real numbers $ x\geq 998$ such that $ \frac{f(\frac{x}{1998})}{f(\frac{x}{998})}$ is maximal, where $ f(x)$ is the floor function of $ x$.

Note that $ [2x]\ge2[x]$ (it is well known and easy to prove). If $ \frac{[x/1998]}{[x/998]}>\frac12$, we get $ 2[x/1998]>[x/998]$. So $ [x/999]>[x/998]$ which is impossible. So $ \frac{[x/1998]}{[x/998]}\le\frac12$. If $ x\equal{}1998$, the expression equals $ \frac12$. So the maximal value is $ \frac12$. It's easy now to find $ x$.

firstly $ \frac {[x/1998]} {[x/998]}$ is decreasing after 1998 this is because after every 998 numbers, $ [x/998]$ increases by 1 whereas $ [x/1998]$ increases after 1998 numbers. so, every time $ [x/1998]$ increases by 1, $ [x/998]$ increases by 2. if you check values less than 1998, $ \frac {[x/1998]} {[x/998]}$ will be $ 0$. so, the maximum value of the fraction occurs at $ x \equal{} 1998$, which is $ \frac {[1998/1998]} {[1998/998]} \equal{} \frac {1} {2}$


Let $ 0<p<1$. Prove that:
$ \frac{2}{e} \le p^{\frac{p}{1\minus{}p}}\plus{}p^{\frac{1}{1\minus{}p}}$

[quote="NguyenDungTN"]Let $ 0 < p < 1$. Prove that:
$ \frac {2}{e} \le p^{\frac {p}{1 \minus{} p}} \plus{} p^{\frac {1}{1 \minus{} p}}$[/quote]

$ f(x)\equal{}x^{\frac {x}{1 \minus{} x}} \plus{} x^{\frac {1}{1 \minus{} x}}$ is a strictly decreasing function for $ x\in[0,1)$

Since $ \lim_{x\rightarrow 1^\minus{}}\frac{1}{1\minus{}x}\ln(x)\equal{}\minus{}1$, we have $ \lim_{x\rightarrow 1^\minus{}}x^{\frac {1}{1 \minus{} x}}$ $ \equal{}\lim_{x\rightarrow 1^\minus{}}x^{\frac {x}{1 \minus{} x}}$ $ \equal{}e^{\minus{}1}$

So $ \lim_{x\rightarrow 1^\minus{}}f(x)\equal{}\frac{2}{e}$ and so the result.


All these creatures are poppygongs:
4, 2, 8
6, 3, 9
14, 2, 16
2, 10, 20

None of these are poppygongs:
2, 3, 4
8, 800, 80
7, 17, 27
25, 35, 45

The first numbers number is x, the second is y, and the last is Z, for example, on the first line of poppygongs 4=x, 2=y, and 8=z.

1. What is the formula to make sure 3 numbers are a poppygong?
2. Which of these are poppygongs? (choose A, B, C, or D)

.............(A).............(B).........(C).........(D)
.........7, 49, 7 :;: 3, 4, 1 :;: Both :;: Neither

[quote="vicseb"]All these creatures are                                         None of these are poppygongs:                                                       poppygongs:[/quote]
I don't quite understand this sentence.  :oops:

EDIT: Nevermind, My comp screwed up the page, I see it now.  :)

Wait, you said 2 is a poppybong and 2 isn't a poppybong. I don't get it.

No, 2 is a part of an equation which forms a number defined as a poppygong.
for example, 4=x, 2=y and 8=z so you do something like xyz, which forms a poppygong. (thats probably not the right formula)

yea, its not the continuous number source, or else i might as well just list them instead of sort them in groups. It's the group that matters to find the problem.

[hide]Part A: $x$ can be any number. In relation to $x$, $y$ is $\frac{1}{2}x$. $z$ is $2x$.
Part B: D because neither a nor b meets the requirements of my formula in Part A.[/hide]

yes, but that formula wouldn;t work for 14, 2, and 16 and also 2, 10, and 20

well at first inspection it would seem that three numbers are a poppygong is x + y = z or if xy=z.  however I dont see whats the pattern as to how to decide which to use

Ummm....do you want to post the solution, vicseb?  I wanna see the answer, but this thread is pretty much dead.

Actually, I'll admit that I'm not sure of the answer myself, appeared in this math book I was reading once... :dry:

The only relation I can find:

Given x, y, and z are Poppybongs, one of the following is true:

xy = z
x+y = z

And it's implied if they're not poppybongs, neither of those are true.

So I would say D, but I could be horribly wrong and just making stuff up  :P

EDIT: Typo  :oops:

I'm noticing that all poppygongs are even, so I'm guessing it has something to do with a 2...

[hide]i think that a poppygong is x+y = z or xy=z. my answer is (A), because 7*7 is 49. 3+4 is not 1 and 3*4 is not 1, so it isn't a poppygong.[/hide]

wow, that actually works! :lol: thanks :lol:

Captain Sugar, you made a mistake.

You said that it's a poppygon if and only if $x+y=z$ or $xy=z$.  But you said that A is a poppygon.  $7+49\ne7$ and $7\cdot49\ne7$, so it's not a poppygon.

I got it!
x+y=z or
x*y=z

x and y are always the smallest numbers

EDIT: just remembered that it can be neither

Also, any type of math can be involved in this in case that isn't the right formula. Rounding to the nearest multiple of x is included


Are you homeschooled or do you go to "government" school? :)

There are other kinds of school than homeschool or public schools.

JBL is correct that the poll choices don't allow everyone to answer. I, for one, am an adult who has already completed my schooling (all of which was operated by governmental bodies from kindergarten through law school).

hahaha I like your term "government school" :D

yeah, you should defn add parochial school. as well as boarding school. (did i spell parochial correctly?)... and private college....................... and public u's.......... as well as not in school, as tokenadult said.

Homeschooled and proud of it.

Gubment school is Neal Boortz's term for it, so by definition me and him are Boortzist.

Boarding school is private, parochial is public, law is private, and if you are an adult, vote on past experience or what you do to your children.

Billy

"law is private?"

Parochial schools are definitely [i]not[/i] public schools -- they are private and just happen to be run by religious organizations.

as jbl said. yeah, when we were paying 300 buck a month tuition for parochial school... PRIVATE!

i went to waldorf, parochial (2 different affiliations) and public.

any other waldorf kids?

Woo..  The only private schooler...
I went to a public school for 8 years before I realized that all the kids there were evil (this is [i]not[/i] a general description of people in public schools, just this one).

i go to what u call a "government" school which is simply an overcrowded school that is several hundred students over maximum capacity. fun fun.

I used to go to a public school. Went to ones from K-8th. It actually wasn't too bad. Pretty nice, and they had tough teachers too. But now I'm at a private school in Wally World.

I've been to public and private schools.  Private school I went to was nice, but it seemed somewhat too sheltered.  Public elementary school I went to was pretty good--OK teachers, good environment.  Then I moved a few states away and the public middle school there was terrible :(.  All the bad stuff: overcrowdedness, bad grades, horrific behavior, etc.  Now I'm at a selective public school, and it's generally pretty good.  Kind of like what I hear TJ to be, only much smaller.

I've only been to public schools.  My area has a really nice environment, the school doesn't have too many problems.

[color=indigo]I've been homeschooled from second grade (I'm in 7th now)[/color]    :)

I go to a public school, went to public schools all my life, and will probably go to public schools til I get out of school, but I think I want to go to all kinds of different schools just to see what they are like, but public school is cool too

[quote="white_horse_king88"]Haha. I like that title. Government schooled. It's almost as if we were trained specifically by the government to act on behalf of the government. Reminds me of the Cultural Revolution in China.[/quote]

Yeah, my mind crossed that route too.

I currently go to a Catholic Public School.

[quote="Sunny"]I currently go to a Catholic Public School.[/quote]

Um, perhaps you are mistaken?  If it's a Catholic school, it's definitely not public.  If it's a public school, it's not Catholic.

hmm, well, its not private at all.  Maybe public wasn't the right word...what would you call it?

public schools all the way! and look where i am now! w00t. i love having friends that pay $20,000 a year to go to a private school and then beating them in math competitions =P

Of course in England the term "public school" long meant "school open to the public," as contrasted with a school only for members of one family. In most places with British usage in English, the term "state school" designates what the original poster is calling a "government school" and most people answering in the thread are calling a "public school." I generally use the terms "government-operated school" and "privately operated school" when the distinction I wish to draw is a distinction of operating authority. In my categorization, schools operated by churches (whether or not they are open to the public  :)  ) are one subset of privately operated schools. I only attended government-operated schools in my childhood and early adulthood.

I don't know anything about parochial school, so perhaps it is private.

It is interesting that some goverment schoolers are beating the private schoolers. However, as a whole, homeschoolers do better than institutionalized and 'private' children. Take the spelling bee, the geography bee, the Reader's Digest Word Power Challenge(no homeschoolers that [i]won[/i] yet but close!), and SAT scores. You would think that the winners would come in the same ratio as the ratio to the other schoolers, but they are higher.

Anyway, who thinks that the first person to win 2 competitions nationally in the same year will be a homeschooler? A+Math perhaps? SOmone to come? Me? :blush: Doubtful that it will be me.

Billy

I suppose it could be privately operated, but I know for a fact that our principal makes sure we understand and let people understand we are not a private school.  We are government funded in everything we do

[color=green]Homeschooled for a couple of years, but it sure wasn't traditional homeschooling. It was heavily community based, and there were about ten of us in my little work group, which was about the equivalent of three years. That was fun. :D

Now, I go to a small, private school. I think it would be interesting to attend a government-run school, but it would not be in my best interest to stay there.[/color]

Billy -- the fact that homeschoolers are over-represented among winners of things is not, in itself, a very interesting fact.  What would be more interesting would be if there was still dominance after adjusting for demographic differences.  Homeschooler's parents must have sufficient monetary means that one of their parents can forgo working and must be sufficiently well educated to actually teach their children something.  I bet that, if you adjust for income and education differences, homeschoolers will still do somewhat better than the average student, but I strongly doubt there will be any major dominance on their part.

I agree with beating private schoolers. Haha.

Bring it on biatches.

Going to a school with others makes an environment, and can offer a ton of things homeschooling doesn't, such as highly competative sports and local math and science competitions. Schools offer accellerated programs too, so one doesn't have to be constrained by limitations.

[quote="bleumoose"]The other thing with homeschool being highly represented in competitions: you have to also remember, that at a school, there is a curriculum that has to be followed. I am worried that perhaps a lot of these homeschoolers (I'm not saying yourself, necessarily) are being "taught to the test." This is an ongoing worry in public schools, especially with the "No Child Left Behind" Act. The problem with "teaching to the test" (or to the competition) is that it cuts into time for other education, which means one isn't as well-rounded. While it's all well and good to do well in a certain competition, this shouldn't be one's main focus. I don't mean to sound accusatory, I very much respect those with the self-discipline to homeschool, I am just bringing up a point to consider.[/quote]

You read Mathew Crawford's story of school? Good read.

Being taught to the test? You do the smallest possible amount of other things when you study for (fill in the blank). The way to avoid being very little rounded is to study for competitions in all areas, or at least 2. I am actually studying for 3 right now, and last year was the first year I had ever competited. Yes you study for the test, but not to the exclusion of everything else.

I am studying geography, math, and vocabulary for competitions right now, and then I study an hour for my other subjects.

We have school normally till you get done with the day's work, which is generally at least 2 lessons in the school system. It takes about 6-7 hours a day.

[quote="solafidefarms"]Anyway, who thinks that the first person to win 2 competitions nationally in the same year will be a homeschooler?[/quote]

George Thampy (a homeschooler) came pretty close in winning first place in the national spelling bee and second place in the national geography bee in 2000. One of the kids he beat (barely) in the spelling bee was another homeschooler, an AoPSer, namely Alison.

I public schooled for 9 years and then moved to another state and switched to home schooling and much prefer it

[quote="bleumoose"]The other thing with homeschool being highly represented in competitions: you have to also remember, that at a school, there is a curriculum that has to be followed. I am worried that perhaps a lot of these homeschoolers (I'm not saying yourself, necessarily) are being "taught to the test." This is an ongoing worry in public schools, especially with the "No Child Left Behind" Act. The problem with "teaching to the test" (or to the competition) is that it cuts into time for other education, which means one isn't as well-rounded. While it's all well and good to do well in a certain competition, this shouldn't be one's main focus. I don't mean to sound accusatory, I very much respect those with the self-discipline to homeschool, I am just bringing up a point to consider.[/quote]

It makes sense to me that "teaching the test" would happen more in public schools than for homeschoolers, since public schools have to have a certain amount of people pass a standardized test, or else they won't get funding becuase of No Child Left Behind.

[quote="eryaman"]I am "government schooled".  I go to a public school in Texas, which sucks.  I would rather be homeschooled or go to a good private school, as public schools in Texas don't tend to be very good. (Texas has a <1000 average SAT score)[/quote]

California's education is not much, if any, better

Govt-schooled for 1, 5, and 6th grade...
Went to a British Intl [private] school for the second half of seventh grade..
American Intl [private] school for the rest.

I've sort of tried homeschool during the summers, but it didn't work out.

[quote="solafidefarms"]

Boarding school is private

[/quote]

i go to a boarding school and its government run. IMSA.
Not all boarding schools are private.

i used to go to a private school in texas, then i moved to China and i go to a private school cause pretty much all the international schools are private


AMC AIME USA(J)MO USAMO MATHCOUNTS algebra polynomial

That's a judgment call.  I would say start slowing things down during the last week...but don't do nothing.  Doing nothing is almost as bad as burning out.

[quote="Calculus"]About when is it time to stop studying for the aime? I read before in a post that we should try to stop studying math a week before.[/quote]
Always keep studying until on of the following things happens:
1)You pass out
2)You start to bleed
3)You are better than everyone else in the country

Penguin, you forgot 4) When you are too old.

i would say the sunday before the test...
and on monday just have fun...(wait AIME is on tuesday right?!? :P )

I recommend studying pretty intensively (say, about one practice AIME per day) up until the day before the test. The last full calendar day before the test just do fun things (other than school homework  :blush:  ) to get into a positive mood. That's assuming you've had a few weeks to study, and have lots of original AIME  tests and solutions to practice with.

The manual recommends 9 am to 12 noon.  So people in different time zones will be taking it at different times.

[quote="randomdragoon"]While we're on the topic, what should you eat for breakfast the day of the contest?[/quote]

Cereal?  :rotfl:

[quote="JesusFreak197"][quote="dakyru"][quote="PenguinIntegral"]
3)You are better than everyone else in the country[/quote]

I believe you meant everyone else in the world.   ;)[/quote]

Well, when you're better at AIME than everyone else in the country, I think that's generally good enough, since AIME is only national, not international.

BTW, I've heard that AIME I is on Tuesday, March 7th, but haven't heard anything else yet. What time is it held, or does that depend on where you're taking it?[/quote]

I don't think there is a set time like with the USAMO, just to allow more flexibility I assume. As for when to stop studying, I suggest just taking the day before off (like many have said). Take the weekend to do my mock AIME :P

Concert pianists usually eat bananas a couple hours before they perform because something in them (not sure what) helps calm the brain down and lowers the anxiety. Don't know whether that would help everyone, but I find that I can think more clearly when I'm not under pressure.

Up until 5 minutes before the test starts... :D
Just kidding

Seriously though, I would recommend using the day before the test to go over good test-taking skills and relax.

And eat pop-tarts for breakfast on the day of the AIME.
I hear it gives you special problem solving abilities.

[quote="randomdragoon"]While we're on the topic, what should you eat for breakfast the day of the contest?[/quote]

I usually eat a roll and an egg, but that's me so... :D

krustyteklown, u just used your Mod power :D

when the forum first got locked by Dpatrick, yif man12 was the last one to post.

[quote="alpha123"]I am sure we can all agree that now is the time to stop preparing for the aime (atleast the ones who took aime1  :) )[/quote]
What?  Blasphemy!  Now's the time to START preparing for next year's...  :lol:

[quote="alpha123"]I am sure we can all agree that now is the time to stop preparing for the aime (atleast the ones who took aime1  :) )[/quote]

The rest of us have two more weeks to go... of LOTS of math (seeing that it's spring break for me...).

[quote]What? Blasphemy! Now's the time to START preparing for next year's... [/quote]

well. thats only true if ur not a senior  :P

lol, i didn't visit this discussion before the aime, but i stuck to my traditional method, which seems to be generally shared, of eggs & cereal too. Just protein puts you to sleep, just sugar won't last.

But I was wondering... does anyone else think that math makes you more hungry? As soon as I started hard-core preparing for the aime; i mean, taking practice aimes every day, I found I could not stop getting hungry, no matter what i did. I've been having two lunches and two dinners for the past week & a half now. (And I only weigh 140 lbs. although I am 5'11'')

Haha I have the same feeling sometimes... When I've been thinking hard about math for a few hours, I tend to get the munchies, although that could also be due to the marij--wait umm what? I just "suddenly" lost my train of thought! But seriously math takes a lot of energy, which might be why you get hungry :)

I wonder how much energy it takes to calculate how much energy it takes calculating the amount of energy needed to calculate the amount of energy...+n
(n being the initial amount of energy calculated solving the problem)

Oh, and would that series diverge? :D

If not, does it know its Godel number?

haha! this forum has become a veritable sinkhole to my parent's food bill; one glance at all this hunger related math and im off to the fridge for some cheese!

(comes back to the computer after trip to fridge)

and if hunger's tendency to cause a mathematician to think about hunger mathematically increased in direct proportion to the strength of the hunger, i wonder if the strength of hunger would increase exponentially or only in a polynomial manner?

(heads off for more)

when is it time to stop studying for AIME?

When the proctor confiscates your AOPS book prior to AIME   :rotfl:

It's definitely time to begin studying for next year for the people who just took AIME I.

Well if you made USAMO then now is the time to study for USAMO.


function
topology
real analysis
real analysis unsolved

Prove that there is no function $f: \mathbb{R}\to\mathbb{R}$ continous only on $\mathbb{Q}.$ ;)

This is very well-known, and it has been discussed before, but eh.. :)

First notice that the set $C(f)$ of continuity points of a function $f$ is a $G_\delta$ set, i.e. a countable intersection of open sets. We can see this as folows: we define the oscillation of the function $f$ in a point $x$ as being $\omega(f,x)=\inf_{\varepsilon>0}\sup_{x_1,x_2\in(x-\varepsilon,x+\varepsilon)}|f(x_1)-f(x_2)|$. It's easy to see that $f$ is continuous in $x$ iff $\omega(f,x)=0$. Let $G_n$ be the set consisting of those $x$ for which $\omega(f,x)<\frac 1n$. We have $C(f)=\bigcap_{n\ge 1}G_n$, and, since it's clear that each $G_n$ is open, the result follows.

All we need to do now in order to obtain a contradiction is show that $\mathbb Q$ is not a $G-\delta$ set. This is equivalent to proving that $\mathbb R\setminus\mathbb Q$ is not an $F_\sigma$ set, i.e. a countable union of closed subsets of $\mathbb R$. Assume the contrary. This means that we can find closed sets $\tilde F_1,\tilde F_2,\ldots$ in $\mathbb R$ s.t. $\bigcup_{n\ge 1}\tilde F_n=\mathbb R\setminus\mathbb Q$. Let $(q_n)_{n\ge 1}$ be an enumeration of $\mathbb Q$, and put $F_n=\tilde F_n\cup\{q_n\}$. The sets $F_n$ are also closed, and, since each one of the $\tilde F_n$’s has empty interior, the $F_n$’s have empty interiors as well. We now have $\mathbb R=\bigcup_{n\ge 1}F_n$, so $\mathbb R$, a complete metric space, can be written as a countable union of nowhere dense closed subsets. This contradicts Baire’s Category Theorem, so we’re done.

[quote="cimabue"]
Then $\overline{G_n}=\mathbb R$.
But any open set is a countable union of disjoint open intervals.
Then the complement $\mathcal CG_n=\mathbb R-G_n$ is at most countable.
[/quote]

This is not true, in general. Think about the Cantor set. It's a complement of a dense open set, but it's not countable.

Can we use also use density of irrationals?

no because you can construct function that are continuous only on $\mathbb{R} - \mathbb{Q}$

To make this complete, we should prove that given any $G_\delta$ subset of $\mathbb R$, there is a function whose set of continuity points is precisely that set. I think this should work.

Let $G=\bigcap_{n\ge 1} G_n$ be a $G_\delta$ set, with $G_n$ being open sets. For $n\ge 1$, let $F_n$ be the complement of $G_n$ in $\mathbb R$. Finally, take $E_1=F_1,E_{n+1}=F_{n+1}\setminus\bigcup_{k=1}^nF_k,\ \forall n\ge 1$. We define $f_n$ to be $0$ on $\mathbb R\setminus E_n,\ 1$ on the boundary of $E_n$, and $\frac 12$ in the rationals and $\frac 14$ in the irrationals of every maximal open interval contained in $E_n$. Now we take our function $f$ to be $\sum\frac 1{5^n}f_n$.


geometry
incenter
circumcircle
inradius
trapezoid
geometric transformation
rotation

In a triangle $ABC$, it is drawn a circumference with center in the incenter $I$ and that meet twice each of the sides of the triangle: the segment $BC$ on $D$ and $P$ (where $D$ is nearer two $B$); the segment $CA$ on $E$ and $Q$ (where $E$ is nearer to $C$); and the segment $AB$ on $F$ and $R$ ( where $F$ is nearer to $A$).

Let $S$ be the point of intersection of the diagonals of the quadrilateral $EQFR$. Let $T$ be the point of intersection of the diagonals of the quadrilateral $FRDP$. Let $U$ be the point of intersection of the diagonals of the quadrilateral $DPEQ$.

Show that the circumcircle to the triangle $\triangle{FRT}$, $\triangle{DPU}$ and $\triangle{EQS}$ have a unique point in common.

Let r be the inradius and $r_0$ radius of the given circle centered at the incenter I. Let A', B', C' be the tangency points of the incircle (I, r) with the sides BC, CA, AB. The triangles $\triangle DEF \cong \triangle PQR$ are congruent and both are obtained from the contact triangle $\triangle A'B'C'$ by a spiral similarity with the center I, the same similarity coefficient $\frac{r_0}{r}$ and rotational angles $\angle DIA' = \angle EIB' = \angle FIC' = \phi$ resp. $\angle PIA' = \angle QIB' = \angle RIC' = -\phi$ of the same magnitude and opposite signs. Consequently,

$2 \phi = \angle DIP = \angle EIQ = \angle FIR =$

$\angle (DE, PQ) = \angle (EF, QR) = \angle (FD, RP) =$

$= \angle DUP = \angle QSE = \angle RTF =$

$= \angle QUE = \angle RSF = \angle DTP$

Thus the circumcircles of the triangles $\triangle FRT, \triangle DPU, \triangle EQS$ all meet at the incenter I and moreover, these circumcircles are congruent to each other.

Since $ I$ is equidistant from $ BC,CA,AB,$ then it's clear that the intersected chords $ DP,EQ$ and $ FR$ are congruent, thus $EQFR ,  FRDP$ and $ DPEQ$ are isosceles trapezoids $ \Longrightarrow$ $ QR \equal{} EF,$ $ FD \equal{} RP  ,$ $DE \equal{} PQ.$ This implies that $ \triangle DEF$ and  $ \triangle PQR$ are congruent having $ I$ as their common circumcenter. Moreover,  $ SFRT,$ $ TDPU$ and $ UEQS$ are cyclic due to $ \angle EFD \equal{} \angle QRP,$ $ \angle FDE \equal{} \angle RPQ$ and $ \angle DEF \equal{} \angle PQR.$ $ I$ is obviously center of the rotation taking $ \triangle DEF$ into $ \triangle PQR$ and since the midpoints $ M, N$ of the sides $ QR, EF$ are homologous under such rotation, the measure of the angle $ \angle MIN$ is precisely the rotational angle $\Longrightarrow$ $ \angle MIN \equal{} \angle RIF.$ But since $ IMSN$ is cyclic, we have $ \angle MIN \equal{} \angle FSR \equal{} \angle FTR \equal{} \angle FIR$ $ \Longrightarrow$ $ I \in \odot(FRT).$ By similar reasoning, we conclude that $ I \equiv \odot(FRT) \cap \odot(DPU) \cap \odot(EQS).$


inequalities
factorial
modular arithmetic
quadratics

plz can any one teach me how to get the unit digit of thus
7!+8!+9!+10!+...+2007!
[quote]if anybody ask you who i am tell them i am the greatest[/quote]

It's $ 0$.  Any number $ n!$ where $ n\geq 5$ will end in $ 0$ so the sum will end in $ 0$ also.

$ 7! \plus{} 8! \plus{} 9! \plus{} 10! \plus{} ... \plus{} 2007!\equiv r\mod{10}$

and as JRav said "Any number $ n!$ where $ n\geq$ will end in $ 0$ so the sum will end in $ 0$ also"

$ 7! \plus{} 8! \plus{} 9! \plus{} 10! \plus{} ... \plus{} 2007!\equiv 0\mod{10}$

so the answer is $ 0$

I think when you were typing your inequality in the first line, you forgot the 5 because as it stands now, it reads $ n\geq$, which doesn't really make sense.

Yep. The reason is because starting from 5! and above, it includes the two terms 2 and 5, which multiplies to ten. Usually when there are problems involving digits and factorials (especially ones where computation is impossible), this should be the first thing in your mind.

Similar to your problem is this one.  Try it:

Find all solutions $ (x,y)$ to the equation below:
$ y^{2}\equal{}1!\plus{}2!\plus{}3!\plus{}\cdots \plus{}x!$

[hide]For $ x \ge 5$, $ x! \equiv 0 \pmod{10}$. So for $ x \ge 4$, $ y^2 \equal{} 1! \plus{} 2! \plus{} 3! \plus{} \cdots \plus{} x! \equiv 1! \plus{} 2! \plus{} 3! \plus{} 4! \equiv 3 \pmod{10}$.

Since $ 3$ is not a quadratic residue in $ \pmod{10}$, the equation has no solutions for $ x \ge 4$.

Testing $ x \equal{} 1,2,3$ yeilds $ (x,y) \equal{} (1,\pm1)$ and $ (3,\pm3)$ as the only solutions. [/hide]

You can also use that Chinese Remainder thing.

For more about the Chinese Remainder Theorem, click [url=http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem]here[/url].

How might you use the Chinese Remainder Theorem?

I suppose it would just be to consider it mod 2 and mod 5, only the latter of which gives any useful information...


algebra
polynomial
linear algebra
linear algebra unsolved

Suppose $A,B\in\mathbb M_n(\mathbb F)$ which $(AB)^n=0$. Prove that $(BA)^n=0$

characteristic polynomial of AB = $x^n$ because AB is nilpotent. Since characteristic polynomial of BA is also $x^n$, by cayley hamilton we have $(BA)^n = 0$.

There is no need to use Cayley Hamilton. You see if $(AB)^n=0$ then $(BA)^{n+1}=0$ and we can prove easily if $X\in\mathcal M_n(\mathbb F)$ and $X^m=0$ then $X^n=0$


algorithm
geometry
geometric transformation

Consider a set A.
We know that 0 belongs to A.
i) If x Belongs to A then 2x+1 belongs to A;
ii) If 3x+1 belongs to A then x Belongs to A;
Is A equal to N?

A=N is equivalent to the statement that starting at some number x, doing the following algorithm we some time get 1:
if x is even, go to x/2
if x is odd, go to 3x-1

there is a similar problem which is still an open conjecture: replace 3x-1 by 3x+1 and ask again if get 1 at some time. i wonder if that problem is much easier, i don't see anything at the moment...

Peter

I think you tried to solve it orally, Peter! it's very simple and I'm sure such a good problem solver like you would solve it in 5 minutes.
Consider natural numbers ar vertices of  an infinite graph.
Connect with directed edges 3n+1 with n and n with 2n+1.
The question asks: Can we reach from 0 every vertex?
Now consider the set A={4,13,6,19,9..} (the law is clear)
Now every element from A can be reached only from another element from A and 0 is not in A
QED

thanks...in my translation of the task that corresponds to 5 not having that property: we would get from 5 to 14 to 7 to 20 to 10 to 5, so we get a circle. i was just too lazy to check if i find a counterexample.
btw why do you i'm a good problem solver? i'm doing that much stupid mistakes in the forum...but thanks for the compliment :D

Peter

Peter, I'm not stupid!
Everyone sees how many problems do you solve and how nice compliments you receive from Darij.
Among the active members of ML IMO participants I think only people like lhvietbao, andrew, maybe sasha are better than you.
besider, you are a 9th-grader!
I just wonder why we didn't meet at IMO 2003

thanks...
but actually i'm a 10th grader and wasn't at the IMO last year just because of the reason that i didn't participate at the "bundeswettbewerb mathematik", so i couldn't qualify for the TST's in that year. but however, i can still participate four times. that'll be fun enough :D
but i would be astonished of winning a gold medal this year because in that case it could be that i win the next years gold medals as well and then i would nearly be as good as Christian Reiher who won just a bronce medal more...

Peter

Well, there are many smart people on the forum. Many new IMO records can expected in the future. Iura still can win 4 further gold medals. This medal together with four further gold medals would mean a new IMO record. I think sasha even could get 5 gold medals (in the case of one additional point of Iura this also could have been possible for Iura).  :)

and iura is talking about me being that smart as well...wuuaaaaaaaahhhhhhhh :D
however, we'll see in a week how i'm doing the tests, sure i won't solve any of the problems ;)

Peter

[quote="vinoth_90_2004"]hi,
iura, how old are you? (just wondering) in other countries, does school finish just before you are 20?
im also 9th grade though im really really hopeless :( compared to people like you. i'll only do the imo 3 times anyway  :D.
vinoth[/quote] it seems that you are very confident in your abilities of making the imo team :) :D

no i didnt mean that :blush: ... i can make the team at most 3 times anyway  :P

[quote="Peter Scholze"]btw orl, you're a german, but i don't know who you are! did you participate at IMO and when?

Peter[/quote]

Hi Peter,

I did not attend IMO. I don't know you, too. But I would like to get to know you. Maybe you write me a message.

PS: I live in Griebnitzsee (Babelsberg, Potsdam). And Berlin is not too far away...  :)

[quote="iura"]Sasha can't get 5 gold medals unless he stays one year more in the school(I wouldn't do that). He's the same grade as I do.
I forgot you guys in Germany have 13 grades.  Since you have 3 more years to participate, you're like a 9th grader to me.
Anyway...[/quote]

Hi Iura,

not all federal states in Germany implement the system of 13 years. There are some federal states who have 12 years too.  :)

yeah, those guys from the former GDR mostly have 12 grades. but although i lived in GDR i've got 13 grades since Berlin has. anyway, they're planning to change that, but that'll need some years(i heard those who attend the 4th grade will have 12 years), so i can participate 3 more times.

Peter


One of Euler's conjectures was disproved in the 1960's by three American mathematicians when they found an integer $n$ which satisfies the following equation.  Find $n$. \[ 133^5+110^5+84^5+27^5=n^5 \]

[hide](a+b)^5:ge:a^5+b^5, so 133^5>84^5+27^5. So this whole expression is less than 3*133^5. as 1.25^5>3, k<133*1.25=166.25. Also, by euler, we have n^5:equiv:n mod 10, so k has last digit 4. k>133, obviously. This means k=134, 144, 154, or 164. I'm leaning toward 144, because it's nice, but I'm not sure...  oh wait. we use mod 9 analysis now. 133^5+110^5+84^5+27^5:equiv:7^5+2^5+3^5+0^5:equiv:343*49+32+0+0:equiv:1*4+5+0+0:equiv:0, so answer is 144?[/hide]

Can't you shove that whole thing into your calculator (if you have one of those "big number" ones and take the fifth root of it?

First of all, this is the AIME exam which doesn't allow calculators.  Second of all, it is almost universally better to do something without a calculator, if you are able to.


vector
function
calculus
integration
superior algebra
superior algebra unsolved

Let $ V$ be a vector space,
we know $ a \notin V$, but know nothing about $ b$. can we conclude that $ a \plus{} b \notin V$?
Thanks a lot for any help!

Not to mention that a lingering question remains: where does $ a$ live? the answer is NO.

For $ b\equal{}0$ we have $ a\plus{}b \equal{} a \not \in V$;
For $ b \equal{} \minus{}a$, we have $ a\plus{}b \equal{} 0 \in V$.

thank you, mavropnevma.
Now i make the question detailed:
$ V$ is a vector space, $ \phi, \psi$ are measurable functions so that the integral exists, e.g. $ \phi(x) \equal{} \psi(x) \equal{} x^{1/3}$.
$ x_{\plus{}} : \equal{} \max\{x,0\}$, $ x_{\minus{}} : \equal{} \minus{}\min\{x,0\}$
We know
$ a \equal{} \int_{\minus{}\infty}^{t} \phi((t \minus{} s)_{\plus{}}) \minus{} \psi((\minus{}s)_{\plus{}}) ds \notin V$
and
$ b \equal{} \int_{\minus{}\infty}^{t} \phi((t \minus{} s)_{\minus{}}) \minus{} \psi((\minus{}s)_{\minus{}}) ds$.
can we see if $ b \in V$ or not? then the case $ a\plus{}b$ is clear.
Thanks!


search
inequalities
inequalities unsolved

Search 'Ineq3' author 'manlio'
There are 3 solutions given.
If you see the inequality case, you will know why you couldn't solve it.

[quote="A. S. K."]a,b,c positive
$(a^{2}+b^{2}+c^{2})^{2}\ge 3(a^{3}b+b^{3}c+c^{3}a)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6026


Give a male name without common letters with neither Charles nor John. (Thrying to addapt it from Spanish)

[quote="NoSoupForYou"]Names that work:

Tim - 170th most popular
Guy - 252nd
Wm - 364th
Kim - 397th
Timmy - 479th
Ty - 665th
Kip - 995th[/quote]
Well, the once I know, they are not names itself but hypocoristics. Tim and Timmy comes from Timothy, Ty comes from Tyler, the others, I really don't know them!

[quote="NoSoupForYou"]Guy is a name.[/quote]Then it works!

[quote="NoSoupForYou"]What's the answer supposed to be?[/quote]In this case it is not totally applicable, because I am addapting the puzzle from Spanish, the answer I have is for Spanish, and not with Charles and John, but their contraparts in that language, Carlos and Juan, and the worst part is that the answers are not that common in Spanish!

[quote="NoSoupForYou"]With Carlos and Juan, it's much easier, because you are allowed to use 'e's. I would estimate that around 5-10% of the top 1000 American names would fit here, as opposed to 0.5% for Charles and John.[/quote]Yes, but again, it was in spanish, where most of the names do not! I was trying to addapt to the situation  ;)

[quote="NoSoupForYou"]Names that work:

Tim - 170th most popular
Guy - 252nd
Wm - 364th
Kim - 397th
Timmy - 479th
Ty - 665th
Kip - 995th[/quote]

Wm? I've never heard of that name before... :?  ;)

and Kim isn't really a guy name...at least I've only met girls named kim.

[quote="spoudyal"]Doesn't Pedro work?[/quote]Spoudyal,

Neither in the original post (Charles has e and r in common with Pedro) nor in the spanish version, because Ca[color=red][b]R[/b][/color]l[color=blue][b]O[/b][/color]s and Ped[color=red][b]R[/b][/color][color=blue][b]O[/b][/color] have two letters in common.

By the way, Kim as well, as a female name, is a hypocoristic for Kimberly!

Just in case, there was a double post, one was erased (It was the exact same post twice!)

[quote="NoSoupForYou"]I got the names from [url]http://names.mongabay.com[/url].[/quote]Very interesting webpage! My name is the sixth most popular in the US (or at least it was in 1990)


The first logic solution to this inequality .Thank you Arqady very nice solution . :D

[quote="silouan"]Thank you Arqady very nice solution . :D[/quote]
Thank you, silouan. :D  :)


LaTeX
inequalities
inequalities unsolved

[quote="Soarer"]Go to proposed forum and find the thread by remike. I suppose that's what you mean. But I really don't understand what 3.5 is. Can you write the question in latex?![/quote]
That is the power of two in the denominator

Then take $a_1=...=a_n=1$ and it's wrong.

i think Soarer is talking about the "Mongolian inequality", am I right?

that had $a_1 \geq \ldots \geq a_n$ ( or was it $a_1 \leq \ldots \led a_n$, can't remember )

can anyone explain the soln to the mong inequality for $n=5$ (for any positive $a_1,...$)? it appears in a text in russian (by Khrabrov - sorry if i messed up the name) , but i can't really understand what is said there

thank you


can you install a program by moving the necessary files manually to the Program Files folder?

No. The installer does some other stuff too, including adding registry keys and stuff.

There are a few programs that do not rely heavily on the Registry. If you are unable to install, you can try moving the program files (though using the installer is safer because it will add an easy uninstallation).


trigonometry
inequalities proposed
inequalities

[quote="TRAN THAI HUNG"]Let a,b,c >0.Prove that:
$ (a \plus{} b \plus{} c)^3 (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b) \le 27a^2b^2c^2$
 :)  :)[/quote]
Smilley, it follows from $ 9(x \plus{} y)(x \plus{} z)(y \plus{} z)\geq8(x \plus{} y \plus{} z)(xy \plus{} xz \plus{} yz).$ :wink:

It's weaker than an old and nice result of quykhtn-qa1:
$ 4a^2b^2c^2 \geq\ (a^3\plus{}b^3\plus{}c^3\plus{}abc)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$  :)
See my proof for this stronger here:

[quote="Smilley"]I remember i once solved this...but unfortunately with expanding :D
I tried but i still haven't find solution without expanding,could you write it? At least hint or something[/quote]
Here is my solution.
If a,b,c are not the three side of a triangle. It's true
If a,b,c are the three side of a triangle.
$ (a \plus{} b \plus{} c)^3 (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b) \le 27a^2b^2c^2$
$ \begin{array}{l}
  \Leftrightarrow (a \plus{} b \plus{} c)^2 16S^2  \le 27a^2 b^2 c^2  \\
  \Leftrightarrow (a \plus{} b \plus{} c)^2  \le 27R^2  \\
  \Leftrightarrow (\sin A \plus{} \sin B \plus{} \sin C)^2  \le \frac{{27}}{4} \\
  \Leftrightarrow \sin A \plus{} \sin B \plus{} \sin C \le \frac{{3\sqrt 3 }}{2}(done) \\
 \end{array}$
Anyway thanks nguoivn for the stronger one.
Can you solve this?
Find the best n:
$ \frac{{(a \plus{} b \plus{} c)^n (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)}}{{\sqrt[3]{{(abc)^{n \plus{} 3} }}}} \le 3^n$
true for all a,b,c. :lol:


This shouldn't be in the intermediate topics forum.

[hide="HINT"]
Division by zero is fun!
[/hide]

[hide]Function $ f(x) / g(x)$ is not defined where $ g(x) \equal{} 0$ , so we only have to find the solution of the second function:

$ g(x) \equal{} 2x^2 \minus{} 10 \equal{} 2(x^2 \minus{} 5)$

Solutions are:
$ A \equal{} \minus{} \sqrt {5}$
$ B \equal{} \sqrt {5}$[/hide]
Is that exercise more suitable for the high school basic part of the forum?

exact $ A\equal{}\minus{}\sqrt {5}, B \equal{} \sqrt {5}$.
OK.

[hide]The function $ \displaystyle \frac {f\left(x\right)}{g\left(x\right)}$ is undefined when $ g\left(x\right) = 0$, because then the denominator would be zero, and we all know what happens when we divide by zero! So:
\[ \& 2x^2 - 10 = 0 \\
\& \Rightarrow x^2 - 5 = 0 \\
\& \Rightarrow x^2 = 5 \\
\& \Rightarrow x= \pm \sqrt {5}
\]
Since $ - \sqrt {5} < \sqrt {5}$, $ \boxed{A = - \sqrt {5} \text{ and } B = \sqrt {5}}$.[/hide]


calculus
integration
logarithms
calculus computations

[color=darkblue][b]Ascertain[/b] $\boxed{I=\int_{0}^{1}\frac{x}{(x^{4}+1)\sqrt{x^{4}+1}}\ dx}\ .$[/color]

Here is an idea:
Let $I$ be the value of your integral:

The substitution $x^{2}=t$ implies that

$I=\frac{1}{2}\int_{0}^{1}\frac{dt}{(1+t^{2})\sqrt{1+t^{2}}}$.

Another substitution, $t=\frac{1}{y}$ shows that

$I=\frac{1}{2}\int_{1}^{\infty}\frac{ydy}{(1+y^{2})\sqrt{1+y^{2}}}$.

The final substitution $y^{2}=z$ gives the answer:

$I=\frac{1}{4}\int_{1}^{\infty}(1+z)^{-3/2}dt=\frac{1}{2\sqrt{2}}$.

[quote="Carcul"]Or we could use the substitution $x^{-4}+1 = t^{2}$ to get $P =-\frac{1}{2}\int{t^{-2}}= \frac{x^{2}}{2\sqrt{1+x^{4}}}$. Then evaluating the definite integral between 0 and 1 gives $\frac{1}{2\sqrt{2}}$.

Carcul[/quote]

Very nice.

Just substitute $x^{2}=tg\theta$ and the answer is quite trivial


[quote="José"]Good luck![/quote]

thankz ;)

for more information of Iberoamerican Teams

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83501]here[/url] (in spanish) ;)

Carlos Bravo ;)
Lima -PERU

[quote="carlosbr"]HI PEOPLE...
THIS THE PERUVIAN TEAM FOR IMO 2006 -SLOVENIA..

[b]PER1:Daniel Soncco Huarsaya[/b].(Gold ConoSur2006, Bronze IMO2005, Gold APMO2005, Gold ConoSur2005, Silver IbMO2004)
[b]PER2:Jossy Alva Gutierrez[/b]! (Bronze IMO2005, Bronze ConoSur2005)
[b]PER3:Franco Vargas Pallete[/b]. (Gold ConoSur2006, Bronze IMO2005, Silver IbMO2005, Bronze ConoSur2005)
[b]PER4:Maria Huanuco Candia[/b] (Silver ConoSur2006)
[b]PER5:Paolo Melgar Povez [/b]
[b]PER6:Aldo Quillas Paredes[/b]
------------
Mathlinkers:

PER1: Chen241290
PER2: js131090
PER3: mcphisto

Thankz ....

Carlos Bravo ;)
Lima- PERU[/quote]

I(jossy alva) also had Silver IbMO2005,with franco  :) . thanks


Let $ \triangle{ABC}$ be an isosceles triangle with $ AC\equal{}BC$ and $ \angle{ACB}\equal{}20^o$. Let $ D$ and $ E$ points on $ AC$ and $ BC$ such that $ \angle{DBA}\equal{}60^o$ and $ \angle{EAB}\equal{}70^o$. Find $ \angle{DEA}$

Hint

Let prove the following.
F on BD.
G, F, E is collinear.

Let a point P on BC s.t. $ \angle PAB \equal{} 50^\circ$. From [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=179738]this[/url] we have $ \angle PDB \equal{} 30^\circ$. Call $ G: AE \cap DB$, se we have $ \angle GAP \equal{} \angle GBP \equal{} 20^\circ$ and $ \angle GEP \equal{} \angle GDP \equal{} 30^\circ$ so the quads ABPG and GPED are cyclic from which $ \angle DEP \equal{} \angle PGB \equal{} \angle PAB \equal{} 50^\circ$ and so $ \angle DEA \equal{} \angle BED \minus{} \angle AEB \equal{} 50 \minus{} 30 \equal{} 20^\circ$


For any positive real numbers $ a,b,c,x,y,z$ we have the inequality:
  $ a^3x+b^3y+c^3z+3abc min(x,y,z)\geq ab(a+b)min (x,y)+bc(b+c) min(y,z)+ca(c+a) min(z,x)$ .

This is really very nice and I think original.

Nice yes, hard no  :)

WLOG assume $x\geq y\geq z$. Then we have to prove

$a^3x+b^3y+c^3z+3abcz \geq ab\left(a+b\right)y+bc\left(b+c\right)z+ca\left(c+a\right)z$.

But this is just the sum of

$a^3\left(x-y\right)\geq 0$ (clear, since $x\geq y$);
$a^3\left(y-z\right)+b^3\left(y-z\right)\geq ab\left(a+b\right)\left(y-z\right)$ (clear, since $a^3+b^3\geq ab\left(a+b\right)$ and $y\geq z$);
$a^3z+b^3z+c^3z+3abcz\geq ab\left(a+b\right)z+bc\left(b+c\right)z+ca\left(c+a\right)z$ (clear from Schur).

But the idea is quite interesting  ;)

  Darij


geometry
circumcircle
incenter
vector
geometry proposed

Let ABC be a triangle with circumcenter O and incenter I. Let $A_1$, $B_1$, $C_1$ be the midpoints of the altitudes of ABC. The inscribed circle of ABC touches the sides BC, CA, AB at D, E, F respectively. Show that the lines $DA_1$, $EB_1$, $FC_1$ and OI are concurrent.

We put masses a+b at C, b+c at A, c+a at B.

Group masses A(b) and B(a) to the foot $C_2$ of the altitude from C.
Then group C(a+b) and $C_2$(a+b) to the midpoint $C_1$ of $CC_2$.
A(c) and B(c) to the midpoint D of AB.

So center of masses lies on $DA_1$. And then $DA_1$, $EB_1$ and $FC_1$ are concurrent.

To prove the second part we need to prove that vectors

(b+c,c+a,a+b)
(a,b,c)
(sin 2A,sin 2B,sin 2C)

are lineary  dependent. But I think, that this is not true

So you're saying the problem is wrong? I'd love to show you my solution, but I have forgotten it! :D I hope I can remember it and make it all clear...

I'm trying to retrace my steps, so here's something which (I think) might help:
Try showing that Ia (the excenter corresponding to side BC) is on DA1.

Call Ia is center of escribed circle in angle A
AIa cuts (O) at M then M is a midpoint of IIa
It is easily to prove A1D throughs Ia
The line through Ia perpendicular to BC cuts IO at T.
A1D cuts IO at S
r/(2R)=DI/IaT=SI/ST
r/(2R-r)=SI/TI=SI/2OI
so SI = 2OI*r/(2R-r) => S is fixed point

Nice. Definitely nicer than my solution (I still can't remember it, but I remember it was uglier :D).

I didn't know that IaT=2R, or that the circumcircle of ABC cuts IIa in its midpoint, but I take your word for those (they don't seem too hard to check oput). :)

Sorry. I've replace altitudes by bissectors in my solution, so I get a solution for this problem: :(

Let ABC be a triangle with circumcenter O and incenter I. Let $A_1$, $B_1$, $C_1$ be the midpoints of the bissectors of ABC. The inscribed circle of ABC touches the sides BC, CA, AB at D, E, F respectively. Show that the lines $DA_1$, $EB_1$, $FC_1$ are concurrent.


linear algebra
matrix
vector
linear algebra unsolved

Does any body know if  latin squares always invertible. Could some one give me a direction to find papers that discuss this matter?

Thanks

Here is a non-invertible Latin square: $ \begin{bmatrix}4&3&1&2\\2&4&3&1\\1&2&4&3\\3&1&2&4\end{bmatrix}$

Thanks for the counterexample.

Now let me ask further, is there any condition to tell which latin squares are invertible?

There is almost certainly nothing simpler than just calculating the determinant. There are just too many latin squares.

Yes, but computing determinant for large size matrix is hard.

How about latin square of size $ p\times p$ where $ p$ is prime. Is it always invertible?

Is there a reason you'd suspect this? (Other than to exclude the example given).

Empirically, its true for p=2,p=3,and p=5. For p=7 there's just too many to do
any numerical experiments.

Actually after doing some research the latin square matrix that I interested in has a special name. It is Circulant matrix.
$ C$ is circulant if $ C$ is of the form
\[ C: \equal{}\begin{pmatrix}c_0&c_1&\cdots&c_{n\minus{}1}\\c_{n\minus{}1}&c_0&\cdots&c_{n\minus{}2}\\\hdotsfor{4}\\c_1&c_2&\cdots&c_0\end{pmatrix}\]
and any circulant matrix of size $ p\times p$ where $ p$ is prime is nonsingular.
I don't know if it is still true for the general latin square.

For circulants, the answer is easy. Circulants are diagonalized by the Fourier matrix, so the determinant can only be zero if one of the eigenvalues $ \sum_k c_k\omega^{jk}$ is zero ($ \omega$ is a primitive $ n$th root of unity). If $ p$ is prime, those roots of unity generate a $ p\minus{}1$-dimensional vector space over $ \mathbb{Q}$. We have $ \omega^0\plus{}\omega^1\plus{}\cdots\plus{}\omega^{p\minus{}1}\equal{}0$, and no other nontrivial rational linear combinations are; this means that if the $ c_k$ are rational and some eigenvalue is zero, either $ c_0\equal{}c_1\equal{}\cdots\equal{}c_{p\minus{}1}$ or $ c_0\plus{}c_1\plus{}\cdots\plus{}c_{p\minus{}1}\equal{}0$.

On the other question- well, I haven't found a Latin square on $ \{1,2,\cdots,7\}$ with zero determinant yet in a few hundred (random, then systematic) tries. I have found one with the smallest possible nonzero determinant $ 196$, and I'm inclined to believe that there probably are some with determinant zero; there are just so many more squares than possible determinants.

One more thing: $ \begin{bmatrix}1 & 4 & 6 & 3 & 2 & 7 & 5 \\
4 & 2 & 1 & 5 & 7 & 3 & 6 \\
2 & 1 & 5 & 7 & 3 & 6 & 4 \\
3 & 5 & 2 & 1 & 6 & 4 & 7 \\
5 & 3 & 7 & 6 & 4 & 1 & 2 \\
6 & 7 & 3 & 4 & 5 & 2 & 1 \\
7 & 6 & 4 & 2 & 1 & 5 & 3\end{bmatrix}$

I didn't find that with the general search.

[Added in edit]
There are singular Latin squares on $ \{1,2,\cdots,n\}$ for $ n\equal{}4$ and all $ n\ge 6$.
We construct them as follows:
First, build four rows $ r_1,r_2,r_3,r_4$ with $ r_1\plus{}r_2\equal{}r_3\plus{}r_4$. We start with base cases $ \begin{bmatrix}1&4&3&2\\4&1&2&3\\3&2&1&4\\2&3&4&1\end{bmatrix},$ $ \begin{bmatrix}1&5&2&4&6&3\\4&1&5&3&2&6\\2&4&1&6&3&5\\3&2&6&1&5&4\end{bmatrix},$ $ \begin{bmatrix}1&4&6&3&2&7&5\\4&2&1&5&7&3&6\\2&1&5&7&3&6&4\\3&5&2&1&6&4&7\end{bmatrix},$ and $ \begin{bmatrix}1&2&8&6&3&9&7&4&5\\5&6&1&3&7&2&4&8&9\\2&1&4&7&9&5&8&3&6\\4&7&5&2&1&6&3&9&8\end{bmatrix}$ for $ n\equal{}4,6,7,9$ respectively.

For larger $ n$, we step from $ n$ to $ n\plus{}4$ by appending the block $ \begin{bmatrix}n\plus{}1&n\plus{}4&n\plus{}3&n\plus{}2\\n\plus{}4&n\plus{}1&n\plus{}2&n\plus{}3\\n\plus{}3&n\plus{}2&n\plus{}1&n\plus{}4\\n\plus{}2&n\plus{}3&n\plus{}4&n\plus{}1\end{bmatrix}$ to these rows.

Finally, complete these rows to a Latin square; this is always possible by repeated application of Hall's marriage theorem.


Trying to determine whether surgery is needed for my wife.

So I am trying to figure this out.

7125 people in the united states, out of a population of about 298,444,215 get a particular type of cancer each year.

I have the percentage of the population.

7,125 is 0.002387% of 298,444,215 (population of US)
But what are the odds? Like for example: "The odds of getting this cancer are 100 to 1"

Anyone know how to compute that?

Thanks!!

Whoa! A post in the MN forum?!  That's almost a shame - I think we were en route to becoming the longest-unused forum on AoPS.
Well, I suppose I should answer your question now, even though it's probably been answered elsewhere already (for future reference, posts in this forum are not likely to be noticed).
Odds are given as "(percent successful) to (percent failure)".  So odds of 99-1 for an event mean that 99% of the time the event will happen, and 1% it won't.  So for your problem, 7125 out of 298,444,215 possible means odds of 7125 to (298444215-7125) = 7125 to 298437090 , which are odds of about 41885 to 1 against getting the cancer.

By the way, I don't use odds very often (except 1,3,5,7, ... :) ) so I might have forgotten completely what I'm doing and made an entirely incorrect post.  Including this line.  Ha!


geometry
rectangle
circumcircle
geometry proposed

[quote="Virgil Nicula"]In a rectangle $ MATE$ consider $ S\in (ME)$ , $ R\in (MA)$ so that $ SR\perp MT$ .

Construct the projection $ Q \in (RE)$ of $ M$ to line $ RE$ . Prove that $ QS\perp QT$ .[/quote]
It is nice.
Let  H is the intersection of MT and SR.
We have $ RQ.RE\equal{}MR^2\equal{}HR.SR$
Then S,H,Q,E are concyclic.
Since S,H,T,E are are concyclic too. So, we get  S,H,T,E ,Q are concyclic.Then $ QS\perp QT$

Alternatively, $\{ X \} = AT \cap MQ$; from $RAES$ cyclic (not a big deal to prove it) we get $\widehat{REM}=\widehat{SAR}$, but $\widehat{REM}=\widehat{RMQ}$ (from $RME$ being a right-angled triangle), hence $MAXS$ is a rectangle, which makes $XTES$ rectangle . The angles $\angle XQE$ and $\angle SHT$ being right angles, the points $X$, $T$, $E$, $S$, $H$ and $Q$ are concyclic, $XE$ and $TS$ are diameters of their circumcircle, done.

Best regards,
sunken rock

Nice problem!

Let $MQ$ meet $AT$ at $W$. Since $\angle WQE= \angle WTE=90$, $QWTE$ is cyclic, so $\angle TWE= \angle TQE$. We want to show $QS \perp QT$, which is equivalent to $QTES$ is cyclic since $\angle SET=90$, which is equivalent to $\angle TQE= \angle TSE$. Thus, if we can show $\angle TWE= \angle TSE$, we will be done.

Note that $\triangle AWM \sim \triangle RME$, thus $AW= MR \cdot \frac{AM}{ME}$. Also, $\triangle AMT \sim \triangle RSM$, thus $MS= MA \cdot \frac{MR}{AT}$. Since $ME=AT$, $AW=MS$. Thus, $\angle TWE= \angle TSE$, and we are done.

Dear Virgil and Mathlinkers,
this nice problem is a variation (with a square) of the one which we can find in Kvant, janvier 1987.
Can someone precises this reference?
Sincerely
Jean-Louis

Hello,
[quote="Virgil Nicula & sunken rock"]In a rectangle $MATE$, consider $ S\in (ME)$ , $ R\in (MA)$ so that $ SR\perp MT$ .  Construct the projection $Q$ of $M$ to line $RE$ . Additionally,  consider $\{ W \} = AT \cap MQ$ [/quote]
Orthogonality at $Q$ is now proved. What is the enveloppe of line RW ?
Best regards, Pierre.