trigonometry
algebra
polynomial
complex numbers
algebra unsolved

Let a,b,c be complex numbers satisfying

(a+b)(a+c)=b, (b+c)(b+a)=c, and (c+a)(c+b)=a.

Prove that a,b,c are all reals.

Bomb

[color=blue][b]Problem.[/b] Let a, b, c be complex numbers satisfying

(a + b) (a + c) = b;
(b + c) (b + a) = c;
(c + a) (c + b) = a.

Prove that these numbers a, b, c are all reals.[/color]

Cool problem! Here is my solution, though presumably not the shortest possible:

First, we will rule out some degenerate cases.

[i]Degenerate case 1:[/i] One of the numbers a, b, c equals 0. Let's WLOG say that a = 0. Then, the equation (c + a) (c + b) = a becomes c (c + b) = 0. Hence, either c = 0, or c + b = 0. Now, if c + b = 0, then b + c = 0, so that the equation (b + c) (b + a) = c yields c = 0. Thus, in both cases, we have c = 0. Hence, c = a = 0. Consequently, the equation (b + c) (b + a) = c becomes $b^2=0$, so that b = 0, and thus a = b = c = 0, so the numbers a, b, c are reals.

[i]Degenerate case 2:[/i] Two of the numbers a, b, c are equal. Let's WLOG assume that a = b. Then, the equation (a + b) (a + c) = b becomes 2a (a + c) = a, so that either a = 0 or 2 (a + c) = 1. We won't consider the case a = 0, since it was already studied in Degenerate case 1; hence, we are left with 2 (a + c) = 1. Thus, $a+c=\frac12$, so that $c=\frac12-a$. Now, the equation (b + c) (b + a) = c becomes $\left(a+\left(\frac12-a\right)\right)\cdot 2a=\frac12-a$, what rewrites as $a=\frac12-a$ and thus yields $a=\frac14$. Since a = b, we thus have $b=\frac14$, and $c=\frac12-a=\frac12-\frac14=\frac14$. Hence, all the numbers a, b, c are real.

So, we have solved the problem for the degenerate cases. Hence, we can now, for the rest of our solution, consider only the non-degenerate case. This means that we can postulate

- that none of the numbers a, b, c equals 0,
- and that no two of the numbers a, b, c are equal.

Now, the equation (a + b) (a + c) = b yields

(b + c) (c + a) (a + b) = (a + b) (a + c) (b + c) = b (b + c).

We denote k = (b + c) (c + a) (a + b); then, this becomes k = b (b + c). Similarly, k = c (c + a). Hence, c (c + a) = b (b + c). In other words, $c^2+ca=b^2+bc$, so that $ca-bc=b^2-c^2$. In other words,

c (a - b) = (b + c) (b - c).

Similarly,

a (b - c) = (c + a) (c - a);
b (c - a) = (a + b) (a - b).

Multiplying these three equations, we obtain

abc (b - c) (c - a) (a - b) = (b + c) (c + a) (a + b) (b - c) (c - a) (a - b).

But since no two of the numbers a, b, c are equal, their pairwise differences b - c, c - a, a - b are all nonzero, and thus we can divide this equation by (b - c) (c - a) (a - b). What we obtain is

abc = (b + c) (c + a) (a + b).

In other words, abc = k.

Now, compare this with the equation k = b (b + c) and the similarly obtained equations k = c (c + a) and k = a (a + b). We see that

$k^3=k\cdot k\cdot k=b\left(b+c\right)\cdot c\left(c+a\right)\cdot a\left(a+b\right)$
$=abc\cdot\left(b+c\right)\left(c+a\right)\left(a+b\right)=k\cdot k=k^2$.

Since $k\neq 0$ (in fact, k = abc, and none of the numbers a, b, c equals 0), we can divide this by $k^2$ and obtain k = 1. Hence, b (b + c) = k = 1, so that $b+c=\frac{1}{b}$, or, equivalently, $c=\frac{1}{b}-b$. Similarly, $a=\frac{1}{c}-c$ and $b=\frac{1}{a}-a$.

Now, the following nice final argument is due to NoName (my own argument used analysis):

In the following, for a real number x, we denote by sgn x the following value: $\text{sgn }x=\left\{\begin{array}{cc}1&\text{if }x>0\\0&\text{if }x=0\\-1&\text{if }x<0\end{array}\right.$.

We assume, for the sake of contradiction, that the numbers a, b, c are not all reals. WLOG, let the number a not be real. Then, this number a has a nonzero imaginary part: $\text{Im }a\neq 0$. Then, $\text{sgn }\text{Im}\left(\frac{1}{a}-a\right)=-\text{sgn }\text{Im }a$ (in fact, the imaginary part of $\frac{1 }{a}$ already has its sign opposite to the sign of the imaginary part of a, and the subtraction of a obviously doesn't change this sign). Since $b=\frac{1}{a}-a$, we thus have $\text{sgn }\text{Im }b=-\text{sgn }\text{Im }a$. Similarly, $\text{sgn }\text{Im }c=-\text{sgn }\text{Im }b$ and $\text{sgn }\text{Im }a=-\text{sgn }\text{Im }c$. Hence, we get $\text{sgn }\text{Im }a=-\text{sgn }\text{Im }c=-\left(-\text{sgn }\text{Im }b\right)=\text{sgn }\text{Im }b=-\text{sgn }\text{Im }a$. But this is an obvious contradiction, as we know that $\text{Im }a\neq 0$ and thus $\text{sgn }\text{Im }a\neq 0$. Hence, our assumption that the numbers a, b, c are not all reals was wrong, i. e. the numbers a, b, c must all be reals. And the problem is solved.

  Darij

Nice problem. Darij, here is another approach. If we substract the second equation from the first one we obtain \[ (a+b)(a-b)=b-c.  \] Analougously we have the other ones.If two of them are equal then all of them are equal and the conclusion becomes obvoious. Now, let us suppose that all three numbers are different. After multiplying the equalities above, we get \[ b(b+c)=c(c+a)=a(a+b)=1.  \] If one of them is real, then clearly all of them are real.
So if we suppose that that none of these three numbers is real, we get a contradiction also. The conclusion follows.;)

Actually, it can be shown that the solutions of the system are the triples $(0,0,0),(\frac14,\frac14,\frac14)$ and $(2\cos\frac{2\pi}{7},2\cos\frac{4\pi}{7},2\cos\frac{6\pi}{7})$ (and permutations)

Enescu can you show that, please  :jump:

Hellooo anybody.. proove or disproove Enescu`s statement  :?

No need to shout! Here's a sketch:
Let $s=\sum{a},q=\sum{ab},p=abc$. Then $a,b,c$ are the roots of $f(x)=x^3-sx^2+qx-p.$
Write the equations as
\[ sa+bc=b,\quad sb+ca=c,\quad sc+ab=a. \]
Adding them yields $q=s-s^2$. Multiplying by $a,b,c$ and adding again gives $3p=-3s^3+s^2+s.$
Write them again as $(s-c)(s-b)=b,$ etc. Multiplying and performing standard computations lead to
\[ s(4s-3)(s+1)^2=0. \]
Thus, $f$ equals one of the polynomials $x^3$, $x^3+x^2-2x-1$, $(x-\frac14)^3$.

I forgot to mention why the roots of $x^3+x^2-2x-1$ are $2\cos\frac{2k\pi}{7},k=1,2,3.$
Let $\alpha=\cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$;then $\alpha^7=1$, hence
\[ \alpha^6+\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0, \]
which implies
\[ \alpha^3+\frac{1}{\alpha^3}+\alpha^2+\frac{1}{\alpha^2}+\alpha+\frac{1}{\alpha}+1=0. \]
Now, let $\alpha+\frac{1}{\alpha}=\beta$. Then
\[ \alpha^2+\frac{1}{\alpha^2}=\beta^2-2,\quad \alpha^3+\frac{1}{\alpha^3}=\beta^3-3\beta. \]
Replacing in the previous equation yields
\[ \beta^3+\beta^2-2\beta-1=0. \]
Finally, observe that $\beta=\alpha+\frac{1}{\alpha}=\cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}+\cos\frac{2\pi}{7}-i\sin\frac{2\pi}{7}=2\cos\frac{2\pi}{7}.$

Well, Enescu, you [b]really[/b] killed it !  :)  :)


trigonometry
real analysis
real analysis unsolved

[hide="Answer"]We have $ S \equal{} \sum_{n\equal{}1}^{\plus{} \infty} \frac{\sin(nx)}{n} \equal{} \frac{\pi \minus{} x}{2}$. In our case x = 1, and so your sum equals $ \frac{\pi \minus{} 1}{2}$.

To derive the above result you can use $ e^{i x} \equal{} \cos x \plus{} i \sin x$.[/hide]


For the set {2,3,4,5}, is the mean equal to the mode?

No.  There is no mode, and the mean is 3.5.


Let $a$, $b$, $c$ be three positive reals. Prove the inequality

$\frac{\left(b+c\right)\left(2a^2+b^2+c^2\right)}{a\left(b^2+c^2\right)}+\frac{\left(c+a\right)\left(2b^2+c^2+a^2\right)}{b\left(c^2+a^2\right)}+\frac{\left(a+b\right)\left(2c^2+a^2+b^2\right)}{c\left(a^2+b^2\right)}$
$\geq 12$.

  Darij


This is from page 157, example 8.12 in
AoPS "Introduction to Number Theory" (by Matt Crawford).

Question: In how many zeros does 58! end when expressed in
base 12.

I do not understand the explanation in the book. More detailed help
will be highly appreciated.

[hide]equivalently, it is saying, find the largest integer n such that 12^n|58!. There are 54 factors of 2 in 58! and 27 factors of three, so n=27[/hide]

[hide="Solution"]
$ 12\equal{}2^2*3$

There are $ 29\plus{}14\plus{}7\plus{}3\plus{}1\equal{}54$ 2's, and there are $ 19\plus{}6\plus{}2\equal{}27$ 3's.

Either way, we will have 27 0's at the end.

Answer: 27[/hide]

rofl AIME what'd u put first
and why is this in the mathcounts section?

[quote="theprodigy"]and why is this in the mathcounts section?[/quote]

What do you mean?

[quote="HiDN428"]i think he meant that we don't need it for MC.

W8 do we though?[/quote]

Yes, you need to know this for MATHCOUNTS...

I'm pretty sure you don't need to do equations like this in bases for mathcounts. (regardless of how simple it is)

It never hurts to know more than you "need to."

math154 is right, but I'm pretty sure you need to know this for mathcounts.  This is quite a bit simpler than last year's state target #8, and that included bases too.


Hey guys, I thought of a good question to ask:

What is the probability of picking two cards from a standard deck of 52 cards such that one card is black and one is red, and the value of the red card is greater than the value of the black card?

[quote="keta"]You know, instead of posting just to say what section this belongs to, why don't you actually post an answer? Yes, I'm still quite unfamiliar with the layout of the sections and the average level of problems here, but there is simply no need to state that this should be in a different place when it already is here. Would you rather I post the exact question in intermediate again?[/quote]

There certainly is a reason...so that a moderator of this forum will be alerted and can move it to Intermediate if he/she agrees with me! No need to post the same question again...if a mod agrees that this is an intermediate level question, it will be moved there. I simply posted in order to alert a mod of my opinion. No offense was intended.

Hi,

My attempt :

Total no. of ways to choose a black and a red card = $C(26,1) *C(26,1)$.
Now, out of these, either the red and the black card have equal values, or they are different (in which case, one is greater than the other).

The no. of ways of choosing a black and a red card, such that their values are equal
 = $52$, as for one black card, there are two red cards.

Now, of the choices in which the red card and black card differ in value, one of them is greater than the other. The no. of choices in which red > black is same as that in which black > red.
So, the no. of choices in which red > black = $\frac{1}{2} ( C(26,1) *C(26,1) - 52 )$

Thus, the probability of red > black
  = $\frac{C(26,1) *C(26,1) - 52}{2 * C(26,1) *C(26,1)}$
  = $6 / 13$

****************************************************************************************
Sambit


inequalities
function
geometry
geometric transformation
reflection
inequalities unsolved

It is clear that for the minimum, $ \{x_1,...x_{20}\}\equal{}\{1,...,{20}\}$. Now, $ 20$ must be adjacent to $ 1$ and $ 2$, $ 1$ must have $ 19$ as its other neighbor etc. Otherwise switching certain numbers makes the function smaller. Continuing this, the minimal solution (unique, apart from reflection) would be  \[ 11,9,13,7,15,5,17,3,19,1,20,2,18,4,16,6,14,8,12,10\] and the same or its cyclics for the second one (note that $ x_1x_{20}\equal{}10\cdot 11$ is maximal).

Note that this is not a rigorous proof... or maybe it is  :maybe:


linear algebra
matrix
inequalities
linear algebra unsolved

Let $ A$ be an $ n\times n$ $ (0,1)$-matrix with row sums $ r_i$ for $ 1\le i\le n$. Prove that
\[ \text{Per} (A)\le \prod_{i\equal{}1}^n r_i!^{\frac{1}{r_i}}\]

Just to get a sense of what's being asked:

A cheap estimate would be to say that you have terms taking a 1 from each row, so $ \text{Per}\,(A)\le\prod_{i\equal{}1}^nr_i.$

This cheap estimate is an overestimate, because it counts terms which aren't permutations. The cheap estimate would say that $ \text{Per}\,(J)\le n^n,$ when in fact $ \text{Per}\,(J)\equal{}n!.$

The inequality being asked for would show that $ \text{Per}\,(J)\le n!.$ Hence in that case, the inequality is "tight." At all levels, since $ (r_i!)^{\frac1{r_i}}< r_i$ for every $ r_i>1,$ the proposed inequality is always sharper than the cheap estimate.

And equality happens for a bunch of other matrices, the ones that are direct sums of matrices $ J_m$ where $ m|n$


n>0 is a natural number
Let denote a(n) is the number of  1 of n in base 2.
(Ex: 5=101(2),thus  a(5)=2; 14=1110(2),thus a(14)=3 )
The inequality: a(sqr(n))>=a(n) ,for all n
right or wrong ?

Wrong. The first counterexample is $n=23$.

I have a problem and I can't solve:is exist a natural number n obeys: 2.a(n^2)=a(n) ?
May someone help me?


[quote="ragnarok23"][quote="ch1n353ch3s54a1l"][quote="ragnarok23"]A really stupid hacker wants to hack the Starbuck's computer mainframe, and because of certain issues and restrictions, he can only hack it by guessing. Through some certain business and dealings, he has learned the following things:
-The password consists of 7 letters and numbers- no symbols or anything else
-It is not case sensitive
-the only numbers CAN (This does not mean they are) be in the third and fifth characters and can only be odd.
-The first, second and third characters are not vowels (y is not a vowel).
-The fifth character, if it is a letter, is a vowel
-The fourth character is a letter in the range of a-m
-The last character is some letter in the range of n-z.
Assuming the hacker guesses every possible combination before he guesses the correct password, he never repeats a combination and it takes him exactly 3 minutes to guess a combination, then how many hours will it take him to guess the password? Include time to guess the password.
Hope I did not mess up anything :D
Probably a target, #1, because of the hideous numbers.[/quote]

Wait..."-The fifth character, if it is a letter, is a vowel" what do you mean by if it is a letter? Doesn't it have to be because of "-the only numbers CAN (This does not mean they are) be in the third and fifth characters and can only be odd."[/quote]
What that means is that it doesn't have to be a letter, it can be a number.....so basically that spot can either be a letter or a number and you have to take those choices into consideration.[/quote]
Oh I thought it said first and third letters >.> :dry: -_-'

[hide]So I got 21,21,26,13,10,21,13. So then $21\times 21\times 26\times 13\times 10\times 21\times 13\times \frac{1}{20}=\boxed{20346417}$ hours...
I wish I had his hacking skills >.>[/hide]

I think there are 26 possibilites for the 6rd character.....it can be a vowel, can't it?

[quote="ch1n353ch3s54a1l"]

[hide]So I got 21,21,26,13,10,21,13. So then $21\times 21\times 26\times 13\times 10\times 21\times 13\times \frac{1}{20}=\boxed{20346417}$ hours...
I wish I had his hacking skills >.>[/hide][/quote]

there are 26 possibilities for character #6
bringing us to a grand total of
$\fbox{25190802}$ hours
which is about $1049616$ days
or about $2876$ years
have fun with that

why are you guys typing weird numbers instead of latex?

[quote="jclarkemathL314159"][hide]
There are:
21 possibilities for the 1st character
21 possibilities for the 2nd character
31 possibilities for the 3rd character
13 for the 4th
10 for the 5th
26 for the 6th
13 for the 7th

So there are 21x21x31x13x10x26x13=600703740
So it will take 600703740x3=1802111220 minutes, which is about 3426 years.

Yeah, this hacker is pretty stupid.[/hide]

Actually, what if the hacker guesses the password correctly on his first try, or any try other than the last for that matter?[/quote]

How does it matter? All the question is asking (essentially) is how long it would take to guess every possible password combination. And btw, the question asks for hours.

[quote="jclarkemathL314159"][hide]
There are:
21 possibilities for the 1st character
21 possibilities for the 2nd character
31 possibilities for the 3rd character
13 for the 4th
10 for the 5th
26 for the 6th
13 for the 7th

So there are 21x21x31x13x10x26x13=600703740
So it will take 600703740x3=1802111220 minutes, which is about 3426 years.

Yeah, this hacker is pretty stupid.[/hide]

Actually, what if the hacker guesses the password correctly on his first try, or any try other than the last for that matter?[/quote]
[quote="ragnarok23"]
Assuming the hacker guesses every possible combination before he guesses the correct password
[/quote]

[quote="math92"][quote="Klebian"]That is latex, genius.[/quote]
 :blush: sorry[/quote]

hehe.

According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88819]this[/url] topic, it has something to do with the server and they're moving severs soon.

The latex looks the same on my computer :yes:. It isn't the messed up text anymore.

[quote="pascal_1623"][hide]
I got 21 for the first, then 21, 26, 13,10, 21, 13.
Since three minutes is 1/20 of an hour and you want the number of hours, you divide the product by 20 and get 20346417.[/hide][/quote]

That is a lot of hours, I hope that I won't have to deal with this when I grow up.


http://www.mathlinks.ro/Forum/viewtopic.php?t=18182 .

  darij


Let $ G$ be a planar graph all of whose faces are triangular,

and suppose that the vertices of $ G$ are colored with three colors (e.g. $ 1,2,3$).

Show that the number of faces whose vertices are colored with all three colors is [i]even[/i] .

In my thought, in proof of [i]Sperner lemma [/i], parity of number of edges in an edge of big triangle, of which endpoints are 1 and 2, is odd.

but in this problem, what is parity of '1-2' edges in boundary of $ G$ ? we can't know it?

if  we know parity of '1-2' edges in boundary of $ G$  , this problem is trivial.

[i]OR[/i], am I incorrectly know meaning of planar graph ?

please, help me.  :)

Yes, I've got the statement (deleted something irrelevant): the infinite facet should also be triangular.

Here's my solution (maybe, not the best one).

1. Erase all edges connecting vertices with the same colour.

2. Claim: In this way we get a graph whose facets have either three or even number of edges.
Proof: If some facet is not a triangle, this means we've erased everything inside, but left everything on the border. So there are no adjacent vertices on the border having the same colour. Moreover, every facet of the original graph, which had one side on the border, should have another side on the border as well. (Otherwise, we would delete the whole triangle; would be better to draw pictures, but too lazy for that.)  Take one of such facets. Its vertices should be coloured (if we follow them on the border) a-b-a, say, 1-3-1. After the last vertex of these three, say, $ v'$, we should have some other vertex, say, $ v''$. Now take the former facet adjacent to $ v'v''$. It should have another edge on boundary, so we have something like 1-3-1-?-1. Going on in such a manner, we get alternating 1's, so we conclude with an even number of edges.

3. Now we should prove that we are left with an even number of triangles, which is really easy, so I leave you to prove this (just count everything).


who knows femmarts last theorem?
who solved it?

[url=http://en.wikipedia.org/wiki/Fermat%27s_last_theorem#Elliptic_curves_and_Wiles.27_proof]Andrew Wiles[/url] gave a correct proof in 1995.

Fermat's Last Theorem states,

There are no positive integer solutions to the equation $ a^n\plus{}b^n\equal{}c^n$ when $ n\in Z>2$ and $ a,b,c\in Z^\plus{}$.

The proof is extremely advanced and requires a very high level of specialized knowledge. In light of this the following link is essenially the proof without the subtle stuff, [url]http://www.ams.org/notices/199507/faltings.pdf[/url]. Enjoy.


Well I think it would be obvious to see that who framed this question.

The BCCI looking at the high demand for Dhoni in the Indian premier League has created 7 clones of Dhoni which along with the original make 8. Now each of the Dhonis come on stage (one by one) and the bidding takes place.. The BCCI states a max-limit of Rs.20 as the overall amount to be spent by all teams together. However, the original Dhoni being very money-minded gives a condition, the minimum price for each Dhoni should be less than or equal to the price for the previous Dhoni and no Dhoni can go for Rs.0.

In how many ways can the franchisees buy the Dhonis.

--- Courtesy Shreyas

THis is simple sum...

x1+x2+x3+x4+x5+x6+x7+x8=20
and x1<x2<x3<x4<x5<x6<x7<x8<edit::20(may be) :oops:

so it is coefficient of x^20 in expansion of something....

[quote="valeriummaximum"]THis is simple sum...

x1+x2+x3+x4+x5+x6+x7+x8=20
and x1<x2<x3<x4<x5<x6<x7<x8....

so it is coefficient of x^20 in expansion of something....[/quote]

Less than equal to.

Coeff in what???  The multi is not that simple.


Let $ (O)$ be a circle centered at $ O$and fixed  diameter at $ AB$.Let $ \Delta$ be a straight line which touches $ (O)$ at $ A$.Choose a point $ M$ on the circle $ (O)$,different from $ A$ to $ B$.The tagent line with $ (O)$ through $ M$ meets $ \Delta$ at $ C$.Let $ (I)$ be the circle through $ M$ and touches $ \Delta$ at $ C$.let $ CD$ be a diameter of $ (I)$.Prove that:
1)$ DOC$ is an isosceles triangle
2)The line through $ D$ and perpendicular to $ BC$ always passes through a fixed point when $ M$ moves on the circle $ (O)$

pls lock this topic because this problem is on Mathematical and Youth this month.
see here [url]http://www.mathlinks.ro/viewtopic.php?t=168836[/url]


[quote="perfect628"]someone's gonna come out of the blue and blow everyone away.[/quote]

like me.

Well I hate to bring everyone down, but I don't see Massachusetts finishing in the top 5 next year. We had our best shot last year, and if anyone's read my complaining, we (I) kinda blew it.

But that's not to say it won't happen, I didn't originally have high hopes for the Clarke math team, but Mr. Frost seems to know what he's doing even when working with less experienced kids (SEEING THAT WE WON THE FIRST IMLEM MEET YES THAT'S RIGHT), and if he goes to nationals again, I would expect nothing but big improvement from whatever kids are going.

I'm just gonna sit back, try for the 46, and if I get it, hope that Kevin Wen doesn't get it too.

Okay, maybe not sit back, but...

Anyone ever been in a tiebreaker?

Chapters 2005: Sergei and Noah both got 45 and Sprint #29 wrong, Noah came out on top.

If you really want my opinion, Kevin Wen will not win, and the poll is biased.

You mean biased in the way that two of the people are from Acton and the person who made this poll apparently knows a lot about the Acton team?  hmm...

Apparently I'm expected to win this time.

Kinda cool

Both me and Kevin Wen got perfects at the IMLEM meet, so I've got no clue how it'll turn out

No offense-IMLEM's not that tough to be me (perfect) on, especially the first one. Now, back to the point-no one on this poll will win. Someone will come out of nowhere. I'm hoping for the non Asian community to come back strong this year. (What are the names on that poll? That's the bias...) Last two years:George+Elizabeth in Nat. CD(neither Asian). Sorry if this is offensive: but it shouldn't be...

Sidenote: I just put capital letters EVERYWHERE they were supposed to be. This is a first for me. I'm very proud of myself.

[quote="perfect628"]

Sidenote: I just put capital letters EVERYWHERE they were supposed to be.[/quote]

Unfortunately you capitalized too many letters there.

[quote="perfect628"]I'm hoping for the non Asian community to come back strong this year.[/quote]

wtf

[quote="perfect628"]No offense-IMLEM's not that tough to be me (perfect) on, especially the first one.[/quote]

wait so remind me how many perfect scores did you get last year?

Yeah, as Mr. Frost said, no one should ever get an IMLEM question wrong, because they say what will be on the test

[quote="CatalystOfNostalgia"]

[quote="perfect628"]No offense-IMLEM's not that tough to be me (perfect) on, especially the first one.[/quote]

wait so remind me how many perfect scores did you get last year?[/quote]

I know I'm a moron-this proves nothing. And the sidenote didnt count in the capitalization thing...

You did [i]not[/i] just start a sentence with the word 'and.'

Oh and also the polls aren't exactly reliable...I had like the 3rd most votes on my own poll last year when I put myself on, look at what happened to me. Some people vote for others (in some cases multiple times) in order to suck up to them. *coughJORIANcough*

[quote="frost13"][quote="CatalystOfNostalgia"]
But that's not to say it won't happen, I didn't originally have high hopes for the Clarke math team, but Mr. Frost seems to know what he's doing even when working with less experienced kids (SEEING THAT WE WON THE FIRST IMLEM MEET YES THAT'S RIGHT), and if he goes to nationals again, I would expect nothing but big improvement from whatever kids are going.[/quote]

thanks for the compliment....sort of :-)  "seems to know what he's doing"??  Exactly how many times in a row do I need to get to Nats before that says "[u]knows[/u] what he is doing"?

I'll do my best.  Alan and Peijin if you are reading this get back to work solving problems, reading this isn't helping you win states!!!!

Catalyst, while you hate the competition you'll be happy to know we had more perfect scores on ASMA #2 than either First Colony or Fort Settlement!!  Go Clarke!![/quote]

Hmm, I highly doubt Peijin is going to read this, he's really inactive now.

About the ASMA... no thanks to some supposedly smart and crazy people *cough* Andrew *cough*

[quote="CatalystOfNostalgia"] Some people vote for others (in some cases multiple times) in order to suck up to them. *coughJORIANcough*[/quote]

At least he's gone now...

Wow, someone actually voted for me.
(I didn't vote for myself)

[quote="entertainerment"]Alkjash from acton has a good chance. David Fink also might get lucky and beat him.[/quote]

Who the heck is David Fink???

From what I've gauged, people from Acton are highly overconfident. Clearly, Lexington will take 2-3 nats team spots, and some other random person. (I'd like to hope that one of the 7-8th graders from my school has a chance, but we'll see...maybe next year for one of the 7th graders though.)

[quote="entertainerment"]Alkjash from acton has a good chance. David Fink also might get lucky and beat him.[/quote]

Who are you?

[quote="perfect628"] ... are highly overconfident. Clearly, Lexington will take 2-3 nats team spots, [/quote]

:roll:

u never know about whats gonna happen. I'd like to say that some Sharon kid will pull it off, but that's wishfull thinking. Probably a Lexington kid who'll go on to annoy everyone by owning in GBML next year. Lexington does tend to own in math.

About sharon kids Frank Zhou might win in Mathcounts.

How old is this Frank Zhou? Many people see some kid every day who is the best in their school, then are amazed when they get owned in the real competition. (personal experience-I was the one whom everyone thought was a genius)

Can we add Jason Li to the list now too?

[quote="Moldytape"]I voted for Allen because I (admittedly) think that he's better than me. I should probably practice more on States because I only had like two 3/2 hours of team practice for chapter and I did like, what, 3 full chapter competitions to prepare? Now I'm planning on joining to practice with the Clarke practice and actually do all the state comps. As for this years state's, I hope that it's hard like last years because I am NOT a perfectionist. Harder test = lower cutoff = profit.[/quote]

It's Alan :wink:

I did all of the states from 1990-2007 last year, so my state prep probably will involve national tests...

wait, if you think I'm better than you, then that probably means someone will pwn us all

dude, if my score is like when I did the '95 states in 6th grade, I would seriously cry.

[quote="Moldytape"]I am NOT a perfectionist.[/quote]
Maybe you should be. Then you can be like me and not make it back to nats in 8th grade.

[quote="CatalystOfNostalgia"]Wow, perfect628 you made it in 7th grade and not in 8th? That must suck.[/quote]

Alright you two, enough bickering.

it's actually kinda funny watching them do that

wait, how did I get so many votes on me recently?  I don't think I'm that good


I downloaded a 617MB ProTeXt package which had MikiTeXt Ghostview and all that. Problem is, I can't seem to get an output value when I go into TeXnic Center.
Also, for the Configuration Wizard for TeXnic, I skipped that...
Don't know if that really matters...
Is it supposed to work? :huh:
Help! :help:

The Configuration Wizard tells TeXnic Center where your installation of $ \text{\LaTeX}$ is, among other things.  It's sort of important.

As t0rajir0u says, the configuration wizard is absolutely vital. You can rerun the wizard by starting up TexnicCenter, Build, Define Output Profiles, Wizard button in the bottom left corner. See paragraph 7 of [[LaTeX:Downloads#TeXnicCenter]] for what to do with it.


hi friends
me and my friends wanna establish an [url=http://www.poker-on-net.com]online poker[/url] site
i know you must have lots of money to do it [which we dont have yet]
this is why we are building a demo-site,with no cash system, to try to impress the people with the money...
wanna work with us?
mail me if you do...

wow that sounds so cool. too bad i cant program at all


Not everyone is talented. Not every talented are competitors. If you feel as a talented competitor, CONGRATS! You have reached your start place.

Get details about the competitions conducted world wide in no time.

Visit our Website http://www.competearoundtheworld.com and participate in your favorite competition and experience the pleasure of winning.

We also encourage people who like to report about their competition.

What are you waiting for?

Come, test your talent and get your hands on winning.

I am more of a competitor than you

I dont need to prove it

um what exactly are we trying to prove, i visited the website and it makes no sence to me.  It doesnt seem to be at all math related could some1 explain.

This is spam, I assume. especially since he only has 1 post.


algorithm
combinatorics proposed
combinatorics

Let $n>0$ be an integer. We say that the positiv integer $k$ satisfies the condition $C_n$ if there exist $2k$ positive integers $a_1,b_1,a_2,b_2, \cdots , a_k,b_k$ pairwise distincts such that the sums $a_i+b_i$ are pairwise distincts and strictly less than $n$.

1) Prove that if $k$ satisfies the condition $C_n$ then $k \leq  \frac {2n-3} 5$.

2) Prove that 5 satisfies the condition $C_{14}$.

3) We assume that $\frac {2n-3} 5$ is an integer. Prove that $\frac {2n-3} 5$ satisfies the condition $C_n$.

Pierre.

I seem to have empirically found the algorithm which solves 3) (and thus 2)). Write the numbers $1,2,\ldots,k=\frac{2n-3}5$ in a vertical column. Since $k$ is odd, $\frac{k+1}2$ is an integer which is in the list. Now start from that integer and to the right of the integers already written start writing the numbers $k+1,k+2,\ldots,2k$ (when you reach the bottom of the list continue from the top, as if it were on a circle). Example ($9$ satisfies condition $C_{24}$):

$\begin{array}{cc}1&15\\2&16\\3&17\\4&18\\5&10\\6&11\\7&12\\8&13\\9&14\end{array}$


Source: Alternate AIME 2003

Corresponding terms of two arithmetic sequences are multiplied together to give the sequence 1440, 1716, 1848, ... Find the eigth term of this new sequence.

Solution in spoiler:

[hide]Let the sequences be:

x, x+d, x+2d

y, y+e, y+2e

xy = 1440

(x+d)(y+e) = 1716

(x+2d)(y+2e) = 1848

xy = 1440

xy + xe + yd + de = 1716

(or xe + yd + de = 1716 - 1440 = 276)

xy + 2xe + 2yd + 4de = 1848

(or 2xe + 2yd + 4de = 408)

Multiplying the second equation by two yields:

2xe + 2yd + 2de = 552

Subtracting the first gives

-2de = 144

de = -72

Subbing this back in, we find that xe + yd = 348.

We want the 8th term, which is

(x+7d)(y+7e) = xy + 7xe + 7yd + 49de

xy = 1440

7(xe + yd) = 348*7 = 2436

49(de) = 49*-72

So (x+7d)(y+7e) = 348[/hide]

Note that one of the common differences is positive, while the other is negative. Also, x>y but |e| > |d|

Notenote: edit

great solution. i didnt think of it that way.

my solution (spoiler):

[hide]

Let the 2 arithmetic sequences by in the forms:

x, x+d, x+2d...

y, y+e, y+2e...

so that we have xy=1440, (x+d)(y+e)=xy+xe+dy+de=1716, and (x+2d)(y+2e)=xy+2xe+2dy+4de=1848.  And our goal is to find (x+7d)(y+7e).  We see that xy+xe+dy+de=1440+xe+dy+de=1716 and thus xe+dy+de=276.  Now we substitute xe+dy+de=276 and xy=1440 into xy+2xe+2dy+4de=1848 to get 1440+2(276)+2de=1848 and so de=-72.  Expanding (x+7d)(y+7e), we get xy+7xe+7dy+49de=1440+7(276)+42(-72)=348.[/hide]

hope it's right!

-interesting_move

[quote="Syntax Error"]actually, i think interesting move is right[/quote]

Lol.  I was wondering why I put "de" instead of "4de"...

Anyway, some notes on the thing.

In order for the sequence to go up and then down, x has to be bigger than y, but e has to be more negative than d is positive in order to eventually offset x's lead and make it come down.

i had to copy both of your solutions on word to examine the differences. but when i found the mistake, i went back to refer, but it was already changed. oh well. good job though

[quote="interesting_move"]my solution (spoiler):
[-spoiler]
Let the 2 arithmetic sequences by in the forms:
x, x+d, x+2d...
y, y+e, y+2e...
so that we have xy=1440, (x+d)(y+e)=xy+xe+dy+de=1716, and (x+2d)(y+2e)=xy+2xe+2dy+4de=1848.  And our goal is to find (x+7d)(y+7e).  We see that xy+xe+dy+de=1440+xe+dy+de=1716 and thus xe+dy+de=276.  Now we substitute xe+dy+de=276 and xy=1440 into xy+2xe+2dy+4de=1848 to get 1440+2(276)+2de=1848 and so de=-72.  Expanding (x+7d)(y+7e), we get xy+7xe+7dy+49de=1440+7(276)+42(-72)=348.[-/spoiler]

hope it's right!

-interesting_move[/quote]
The - sign within the [] is not needed...

[quote="Tare"]The - sign within the [] is not needed...[/quote]

Yes.  It was originally [spoiler] but then I added the - because I got tired of having to highlight interesting_move's solution so much (every time I'd scroll down or change windows or something, the highlight would go away) so I edited it to make the spoiler temporarily go away.

I changed it back; I forgot to before.  Sorry!

Great job, Syntax Error and Mathfanatic.

A word on problems such as this. In many contest problems, especially problems involving systems of equations or polynomials, the trick to finding the answer is not to actually solve all of the variables, but to find products of variables that when plugged into our preliminary equation will yield the desired result. NOtice that in this problem, neither solution actaully solved for the sequences and their common difference, but rather solved for products and sums. THis is a great strategy to keep in mind and can prove useful later on.

After all, this is much better than solving for 8 variables!!

Yes, that's a very good note.  I tried to do this problem at first by solving for the variables themselves, which proved to be quite tedious.  So then I just listed some relationships between all the variables in order to get what I needed for (x+7d)(y+7e).

-interesting_move

yup yup. well now we know a good way to solve problems like these


inequalities
function
calculus
derivative
symmetry
6th edition

Please send your solution for this inequality.
So,hard (at least for me) I couldn't solve it.

Me either. I tried this problem by homogenizing(spelled correct?) and then expand to use Muirhead. I failed horribly.
Inequaliy is my strongest weakness. :(  I need more practice. :(

[quote="lightrhee"]I tried this problem by homogenizing(spelled correct?) and then expand to use Muirhead. I failed horribly.[/quote]

So did I.

same to me :P

so let's wait for the soln from the proposer

from the past edition i think this problem is from harazi! am i wrong? :D

well, darij and me were collaborating about that inequality, and we quickly found some solution, greatest parts of it were found by darij. unfortunately, i don't have the time to post the solution, but maybe i'll do that later.

Peter

[quote="Makaveli"]I believe you can do it by putting ab=x/(x+y+z) and so on[/quote]
Did you solved, Makaveli?

If so, please post your solution  ;)

PS: After expanding (I did this by writing a program which does exactly that), you probably get: \[\left. 3 [3,3,2]+2 [6,2,0]+2 [6,1,1]+[5,2,1] \geq [5,3,0]+[4,2,2]+[4,4,0]+5 [4,3,1] , \right.\] which is... hhmmm.. very hard to prove to say the least ($[x,y,z]$ means $ \sum_{\textrm{sym}}a^{x}b^{y}c^{z}$)

[quote="mahbub"]from the past edition i think this problem is from harazi! am i wrong? :D[/quote]You are wrong. Keep guessing :D

Muirhead can't work because (0, 1, 1) and its permutations are equality cases, if I'm not mistaken.

[quote="k2c901_1"]Are we allowed to post the solutions before the current edition has ended?[/quote][b]Only[/b] for the rounds that are finished. As Round 1 has finished, you can discuss about it ...

Hi! My idea was to look at the function $f(x,y,z) ( x=a,y=b,z=c)$, which is equal to the expression inequality for which we have to prove. Then we can study this function for conditional extremum with condition $xy+yz+zx=1$ But to solve it calculating of derivatives is not so nicely :|

It looks like I forgot our original solution, but I thought about it again and found something, which is the same I guess:

First, we have

$2\frac{1+a^2b^2}{(a+b)^2} - (3ab + ac + bc) - 2c^2 + (a^2+b^2)$ $=\frac{2(ab+bc+ca)^2+2a^2b^2 - ( 3ab + ac + bc )(a+b)^2 - 2c^2(a+b)^2 + (a+b)^2(a^2+b^2) }{(a+b)^2}$ $=\frac{a^2bc + ab^2c + a^4 - a^3b - ab^3 + b^4 - b^3c - a^3c}{(a+b)^2}$ $=\frac{(a-b)^2(a^2+ab+b^2-ac-bc)}{(a+b)^2}$.

Summing that equality, we get that our inequality is equivalent to

$\sum \frac{(a-b)^2(a^2+ab+b^2-(a+b)c)}{(a+b)^2} \geq 0$.

From here, one can use standard methods to conclude - I'm too lazy to complete it right now. I hope I didn't miscalculate.

Peter

[quote="perfect_radio"][quote="Makaveli"]I believe you can do it by putting ab=x/(x+y+z) and so on[/quote]
Did you solved, Makaveli?

If so, please post your solution  ;)

PS: After expanding (I did this by writing a program which does exactly that), you probably get:
\[ \left. 3 (3,3,2)+2(6,2,0)+2(6,1,1)+(5,2,1) \geq (5,3,0)+(4,2,2)+(4,4,0)+5(4,3,1) , \right.  \] which is... hhmmm.. very hard to prove to say the least ($(x,y,z)$ means $\displaystyle \sum_{\textrm{sym}} a^xb^yc^z$)[/quote]

I used Maple :D. I also ended up at the same inequality.
However, I'm convinced that $3 (3,3,2)+2(6,1,1)+(5,2,1) \geq (4,2,2)+5(4,3,1)$ is true, which can be simplified to $3 (2,2,1)+2(5,0,0)+(4,1,0) \geq (3,1,1)+5(3,2,0)$. If this is indeed true, we can easily solve the problem.

Peter you know all the solution about this problem. It is proved by expanding?

???

Above, I gave a solution which obviously didn't expand the whole thing. And as far as I took a look at the expanded version, I don't see how one can easily conclude that way. But I don't know what your question aims at.

Peter

BTW, the inequality trivially implies the famous Iran 1996 inequality - this may be the reason that it is very hard:

We have $\frac{1+a^2b^2}{(a+b)^2}\leq \frac{1+\frac 14 ab(a+b)^2}{(a+b)^2} = \frac 1{(a+b)^2} + \frac 14 ab$.

Summing this up, we get that

$\sum \frac 1{(a+b)^2} \geq \sum \frac{1+a^2b^2}{(a+b)^2} - \frac 14 \geq \frac 52 - \frac 14 = \frac 94 = \frac 9{4(ab+bc+ca)}$.

But this is exactly the Iran 1996 inequality.

Peter

[quote="Peter Scholze"]BTW, the inequality trivially implies the famous Iran 1996 inequality[/quote]I was about to post this just now :D

You know, Valentin, that at first I said to you this must be a stupid inequality, with no idea. After that, when trying to solve it, I realized it is stronger than Iran and then I knew it must have been Vasc who created it.  ;)

Well, is this inequality a piece of Vasc's creation? ;)

Yes it is. A very nice one I might add ;)

[quote="mahbub"]well i don't know if anyone noticed one thing, that is if u introduce tan here, i mean like a=tan x, ..., then x+y+z=90. but i did not see any soln from this point. is it possible to solve from here??[/quote]

I had this idea but I found no ideas to go on with the solution after this subsitution though the expression simplifies a bit.

what kinda simplification u made? can u show me?

I haven't solved it, but I have found a nice equivalent inequality. (which may be easier to solve, I don't know since I didn't solve it.)

Let's use the transformation $x=ab,y=bc,z=ac$ to get rid of the difficult constraint. Then we have, reversely $a=\sqrt{\frac{xy}{z}},b=\sqrt{\frac{xz}{y}},c=\sqrt{\frac{yz}{x}}$, with the constraint $x+y+z=1$.

This allows us the beautiful identity $1+a^2=\frac{z+xy}{z}=\frac{1-x-y+xy}{z}=\frac{(1-x)(1-y)}{z}$.
Also, this also allows us to put $(a+b)^2 = \frac{x}{yz}(y+z)^2$.

Combining the above, it's natural to write the inequality now as $\sum_{cyc}\frac{(1+a^2)(1+b^2)}{(a+b)^2} \ge \frac52 + \sum_{cyc}\frac{a^2+b^2}{(a+b)^2}$ which (after some simplification) reduces to:

 \[ \sum_{cyc}\frac{(1-y)(1-z)}{x} \ge \frac52 + \sum_{cyc}\frac{y^2+z^2}{(y+z)^2}\ \ \ \ \ \ \ (\text{for } x+y+z=1)  \]

This reduces further to: $\left(\sum_{cyc}\frac{xy}{z}\right)+\frac12 \ge \sum_{cyc}\frac{y^2+z^2}{(y+z)^2}$

Which I found very nice. :)
Or, in the old variables: $a^2+b^2+c^2+\frac12 \ge \sum_{cyc}\frac{a^2+b^2}{(a+b)^2}$ under the original conditions. This seems so easy... I hope I was at least close. :)

[quote="Peter Scholze"]Well, subtracting $-\frac 32 = -\frac 32(ab+bc+ca)$ you easily get the equivalent form

$\sum (a-b)^2\geq \sum \frac{(a-b)^2}{(a+b)^2}$.

This is the same form used by harazi and similar to the form I took to give the final argument by breaking of symmetry $a\geq b\geq c$.[/quote]

can anyone show me how to continue from this form?
I obtained this form early on and then made no progress...

Well, I guess it's no problem for you to get the form I used some posts ago to prove the inequality:

$\sum \frac{(a-b)^2(a^2+ab+b^2-ac-bc}{(a+b)^2}\geq 0$.

Then just follow my proof there.

I guess this is kind of late (I've been busy with university application essays lately), but if anyone is interested, here is my solution. I basically did expand + muirhead + further brute force. I hope that this is correct.

 Problem 1:Let $a$, $b$, $c$ be positive real numbers such that $ab+bc+ca=1$. Prove that
\[ {1+a^2b^2 \over (a+b)^2}+{1+b^2c^2 \over (b+c)^2}+{1+c^2a^2 \over (c+a)^2} \geq {5\over 2} \]

Solution: The inequality is equivalent to
\[ \sum_{sym}{(ab+bc+ca)^2+a^2b^2 \over (a+b)^2} \geq 5 \]
\[ \Longleftrightarrow \sum_{sym}{2ab\over (a+b)^2} + \sum_{sym}{c^2} \geq 5 \]
\[ \Longleftrightarrow (\sum_{sym}{2ab(a+c)^2(b+c)^2)(\sum_{sym}ab)+(\sum_{sym}a^2)(\prod_{cyc}(a+b)^2) \geq 5(\sum_{sym}ab)(\prod_{cyc}(a+b)^2)} \]

Expanding and cancelling like terms (a LOT of algebra), the inequality becomes:
\[ 2\sum_{sym}a^6b^2+2\sum_{sym}a^6bc+\sum_{sym}a^5b^2c+3\sum_{sym}a^3b^3c^2 \geq \sum_{sym}a^5b^3+\sum_{sym}a^4b^4+5\sum_{sym}a^4b^3c+\sum_{sym}a^4b^2c^2 \]
which is equivalent to,
\newline\[ (\sum_{sym}a^6b^2-\sum_{sym}a^5b^3)+(\sum_{sym}a^5b^2c-\sum_{sym}a^4b^3c)+4abc(\sum_{sym}a^{10/3}b^{5/3}-\sum_{sym}a^3b^2)+2abc(\sum_{sym}a^2b^2c-\sum_{sym}a^{5/3}b^{5/3}c^{5/3}) \]
\[ +2abc(\sum_{sym}(a^{5/3})^3+\sum_{sym}a^{5/3}b^{5/3}c^{5/3}-2\sum_{sym}(a^{5/3})^2(b^{5/3}))+(\sum_{sym}a^6b^2+\sum_{sym}a^3b^3c^2-\sum_{sym}a^4b^4-\sum_{sym}a^4b^2c^2) \geq 0 \]

Now, the first 4 terms of the above expression are non-negative due to Muirhead's inequality. The 5th term is non-negative due to Schur's inequality. Therefore, it is sufficient to show that the last term is non-negative, or
\[ (\sum_{sym}a^6b^2-\sum_{sym}a^4b^4)-(\sum_{sym}a^4b^2c^2-\sum_{sym}a^3b^3c^2) \geq 0 \]
\[ \Longleftrightarrow \sum_{sym}a^2b^2(a-b)^2(a+b)^2 \geq \sum_{sym}a^2b^2(a-b)^2c^2 \]

Since the inequality is symmetric, we can assume that $0\leq a \leq b\leq c$.

Case 1: If $a$, $b$, $c$ are the side lengths of a triangle, then $a+b\geq c$, $a+c \geq b$ and $b+c \geq a$, and the above inequality is obviously true.

Case 2: If $a$, $b$, $c$ can not make a triangle, then since $c$ is the largest value, $a+b \leq c$. Let $k$ be the positive real number such that $k=c-a-b$, then substituting in $c=a+b+k$ and simplifying, the above inequality becomes,
\[ b^2(a+b+k)^2(a+k)^2(2b+k)(2a+2b+k)+a^2(a+b+k)^2(b+k)^2(2a+k)(2a+2b+k) \]
\[ \geq 2a^2b^2(a+b)k(a-b)^2 + a^2b^2k^2(a-b)^2 \]

Now, $b^2(a+b+k)^2(a+k)^2(2b+k)(2a+2b+k)\geq b^2(4(a+b)k)(a+k)^2(2b+k)(2a+2b+k) \geq b^2(2(a+b)k)a^2(b-a)(b-a) = 2a^2b^2(a+b)k(a-b)^2$. As well, $a^2(a+b+k)^2(b+k)^2(2a+k)(2a+2b+k) \geq a^2(4(a+b)k)(b+k)^2(2a+k)(2a+2b+k) \geq a^2(b-a)kb^2k(b-a) = a^2b^2k^2(a-b)^2$. Adding these two inequalities, we see that the desired inequality is true.
 Q.E.D.

I see you apply muirhead... muirhead doesn't keep equaily for nonsymmetric equaliy cases (like 1,1,0) so I'm afraid your solution is wrong...

Actually, muirhead handles this in some cases. For example, for all of the following inequalities, equality is maintained at (1,1,0).

\[ (\sum_{sym}a^6b^2-\sum_{sym}a^5b^3)\geq 0 \]
\[ (\sum_{sym}a^5b^2c-\sum_{sym}a^4b^3c)\geq 0 \]
\[ 4abc(\sum_{sym}a^{10/3}b^{5/3}-\sum_{sym}a^3b^2)\geq 0 \]
\[ 2abc(\sum_{sym}a^2b^2c-\sum_{sym}a^{5/3}b^{5/3}c^{5/3})\geq 0 \]

And these were the ones i used.

[b]Generalization[/b] If $ a, b, c$ be [color=red]real[/color] numbers such that $ bc \plus{} ca \plus{} ab \equal{} 1,$ then

$ \frac {1 \plus{} b^2c^2}{(b \plus{} c)^2} \plus{} \frac {1 \plus{} c^2a^2}{(c \plus{} a)^2} \plus{} \frac {1 \plus{} a^2b^2}{(a \plus{} b)^2} \geq \frac {5}{2},$

with equality if $ a \equal{} 0,b \equal{} c \equal{} \minus{} 1.$

[b]Proof[/b] $ \sum{\frac {1 \plus{} b^2c^2}{(b \plus{} c)^2} }\minus{}\frac {5}{2}$

$ \equal{}\frac {(b \minus{} c)^2(c \minus{} a)^2(a \minus{} b)^2}{2(b \plus{} c)^2(c \plus{} a)^2(a \plus{} b)^2} \plus{} \sum{\frac {\left[a(b \plus{} c)\left(b^2 \plus{} c^2 \minus{} 2a^2\right) \plus{} (b \minus{} c)^2\left(a^2 \plus{} bc\right)\right]^2}{6(b \plus{} c)^2(c \plus{} a)^2(a \plus{} b)^2}}$

$ \geq0.$


In the race Math $ 600$, drivers must complete $ 200$ laps of $ 3$ miles each. Dave drove the first $ 150$ laps at $ 180$ mph and the last $ 50$ laps at $ 200$ mph. What was Dave's average speed for the race, in miles per hour?

It took Dave 150 minutes to drive the first 150 laps, and 45 minutes for the last 50 laps, so 600 miles in 3.25 hours, so 184.6153847 mph.

huh? you don't need to go into decimals.
Can't you just do:

$ (150 \cdot 180) \plus{} (50 \cdot 200) \equal{} 37000$ Then, you just divide $ \dfrac{37000}{200} \equal{} \boxed{185}$

The total distance is $600$ miles. The total time is $\left(\frac{150}{180} +\frac{50}{200}\right) \cdot 3=\frac{5}{6}+\frac{1}{4}=\frac{13}{8}$ so we have $\frac{2400}{13}.$


Consider $ABC$ a triangle with $AB=AC$, $D$ midpoint of $(BC)$ , $P\in(AD)$ with $CP^{2}=DP\cdot DA$. Let $M$ be the midpoint of $PC$. Find the angles of $ABC$ if angle $\angle AMB=90^{\circ}$

Let BM intersect PD at E. Connect B&P,D&M, $DM\parallel BP, \frac{BP}{DM}=\frac{BE}{EM}=\frac{PE}{ED}=2$
$ABDM$ lies on a circle, so we have$\angle{BAD}=\angle{BCD}=\angle{PBC}\Rightarrow BE^{2}=EP\cdot{EA}$
So $\Rightarrow PC^{2}-BE^{2}=DP^{2}-DE^{2}=DP\cdot{DA}-(DP-DE)(DA-DE)$
${\Rightarrow DP^{2}=DE(DP+DA)=\frac{1}{3}DP(DP+DA)}\Rightarrow{DA=2DP}$
so then$CP^{2}=2DP^{2}\Rightarrow \angle{CPD}=\angle{DCP}=45^\circ$
$\Rightarrow \angle{ABC}=\angle{ACB}=\arctan{2}$ :)

Well done! It can be solved using vectors, too!


Let $ x,y,z$ be a positive reals,such that $ xy \plus{} yz \plus{} zx \equal{} 1$.
Prove that
$ x(1 \minus{} y^2)(1 \minus{} z^2) \plus{} x(1 \minus{} y^2)(1 \minus{} z^2) \plus{} x(1 \minus{} y^2)(1 \minus{} z^2)\leq\frac {4\sqrt {3}}{9}$.
I'm sorry if it has been posted before.I've tried to find it,but i failed.
Best regards.

$ L.H.S. \equal{} (x \plus{} y \plus{} z) \minus{} [(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)] \plus{} xyz(xy \plus{} yz \plus{} zx) \plus{} 3xyz \equal{} 4xyz$
$ AM \minus{} GM ; \frac {xy \plus{} yz \plus{} zx}{3} \geq (xyz)^{\frac {2}{3}}$
$ \therefore xyz \leq \frac {\sqrt {3}}{9}$
$ \therefore L.H.S. \leq \frac {4\sqrt {3}}{9}$

[quote="apollo"]$ L.H.S. \equal{} (x \plus{} y \plus{} z) \minus{} [(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)] \plus{} xyz(x \plus{} y \plus{} z) \plus{} 3xyz \equal{} 4xyz$
[/quote]
I guess it must be
$ xyz(xy\plus{}yz\plus{}zx)$,instead of $ xyz(x\plus{}y\plus{}z)$?
My proof is the same.

[quote="Erken"][quote="apollo"]$ L.H.S. \equal{} (x \plus{} y \plus{} z) \minus{} [(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)] \plus{} xyz(x \plus{} y \plus{} z) \plus{} 3xyz \equal{} 4xyz$
[/quote]
I guess it must be
$ xyz(xy \plus{} yz \plus{} zx)$,instead of $ xyz(x \plus{} y \plus{} z)$?
My proof is the same.[/quote]
Sorry,I have edited it.


Source: Mathmetical Olympiad Treasures
Hey guys,
I really need help on this kind of problems (i searched every math books that i've got but with no result.. :( )

Prove that the equation
$x^{2}+(x+1)^{2}=y^{2}$ has infinitely many solutions in positive integers.

On the solution, it defined sequences
[hide]
$x_{n+1}=3x_{n}+2y_{n}+1$,
$y_{n+1}=4x_{n}+3y_{n}+2$.
[/hide]
but i have no clue where those sequences came from...

[quote="Rust"]It is equavalent to Pel's equation: $z^{2}-2y^{2}=-1,z=2x+1$. First solution $x=0,y=1.$ Minimal solution $z^{2}-2y^{2}=1$ is $z=3,y=2$, therefore all solution give by formul $z+\sqrt 2 y=(1+\sqrt 2 )(3+2\sqrt 2)^{n}$. It may be expressed by recurent formuls:
$z_{0}=1,y_{0}=1,z_{n+1}=3z_{n}+4y_{n},y_{n+1}=2z_{n}+3y_{n}$ or
$x_{0}=0,y_{0}=1,x_{n+1}=3x_{n}+1+2y_{n},y_{n+1}=3y_{n}+4x_{n}+2$.[/quote]

Thanks a lot!
transforming the given equation into pell's equation was the key move!


Prove that if all the terms with the digit 9 in their denominator are removed from the harmonic series, such as 1/9, 1/19, 1/92, 1/199, etc, the resulting series is know to converge to a sum between 22.4 and 23.3.

[size=150][b]Please do not answer here![/b][/size]

A solution to this problem would more or less also solve problem 4 of the current running 2. round of the Bundeswettbewerb Mathematik.

@Moderators: please lock this!


hi im new to this forum but my friends told me about this site and how its great for helpin ppl in math..but i need to be helped to help tutoring this kid from church..she just asked me these 5 questions out of no where and its above my level. but shes nice and has respected me greatly so im wantin to help her out. thx.

1)
If adding consecutive integers starting with 1 on a nonscientific calculator with an eight-digit display, how many integers would have to be added to exceed the capacity of the calculator?

2)
A semicircle is inscribed in an equilateral triangle as shown. Express the radius in temrs of the side S.
it shows the circle's diameter on the base of the triangle and the circle's edge seems to be 1/2 way up the tip with the other bottom half of the circle not shown. (sry no pic)

3)
Arrange the digits 0 through 9 so that the first digit is divisible by 1, the first two digits are divisible by 2, and the first three digits are divisible by 3. Continue until the first nine digits are divisible by 9 and the ten-digit number is divisible by 10.

4)
What is the maximum number of acute angles that any convex polygon can have?

5)
Find a possible next three terms in the following pattern:
1,1,1,3,5,9,17,31,__,__,__,......

id really kno how to do it so i can explain...sry not that good with these stuff im only in geometry and this girl from algebra is asking...its optional to help. just wanna help ppl out. thx[/img]

[hide="5"]
Notice that each term is the sum of the last three
SO the next three are
57, 105, 193[/hide]


geometry
incenter
symmetry
trigonometry
geometric transformation
reflection
geometry proposed

Let I be the incenter of triangle ABC.Point D on the side AB is such that BD=BC and DC=DA.Let  $ DM \perp AI$ ,$ M \in AI$.
Prove that AM=MI+IC  .

_________________________________________
Azerbaijan Land of Fire  :ninja:

See that $ ACID$ is cyclic and $ DI \equal{} CI$ due to symmetry and, if $ a, b$ and $ c$ are the sides lengths of the triangle, $ AD \equal{} CD \equal{} c \minus{} a$. Apply Ptolemy's in $ ACID$: $ AI\cdot CD \equal{} (AC \plus{} AD)\cdot DI$, or $ AI\cdot (c \minus{} a) \equal{} DI\cdot (c \minus{} a) \plus{} b\cdot DI$, or $ (AI \minus{} DI)\cdot AD \equal{} DI\cdot AC$, and we claim that, this last equality solves the problem:
Take $ P$ symmetrical of $ I$ about $ M$, then see that $ \triangle PDI \sim \triangle ADC$, from their similarity getting $ AI \equal{} 2\cdot MI \plus{} DI$, done.

[i][b]Note:[/b] Pls, disregard the circle drawn on the sketch, was made just to get $ BC\equal{}BD$, thanks![/i]

Best regards,
sunken  rock

[quote="Pirkuliyev Rovsen"] [color=darkred]Let $ I$ be the incenter of triangle $ ABC$ . Let $ D\in [AB]$ be a point such that $ BD\equal{}BC$

and $ DC\equal{}DA$ . Denote  $ D\in AI$ for which $ M \perp AI$ . Prove that $ AM\equal{}MI\plus{}IC$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ A\equal{}2x$ . Observe that $ m(\widehat {ACD})\equal{}2x$ , $ m(\widehat {ACI})\equal{}3x$  , $ MA\equal{}AD\cdot\cos x$ and the quadrilateral $ ACID$ is

cyclically. Therefore,  $ AM\equal{}MI\plus{}IC\Longleftrightarrow 2\cdot AD\cdot \cos x\equal{}IA\plus{}IC\Longleftrightarrow$ $ 2\sin 2x\cos x\equal{}\sin 3x\plus{}\sin x$ , what is truly.[/color]

Actually, I developed too much the subject solution:
Due to symmetry, $ DI \equal{} CI \ ( \ 1 \ )$, next see that $ ACID$ is cyclic, so $ \angle ACD \equal{} \angle AID$; now take $ P$ as reflection of $ I$ in $ M$, see that $ \triangle APD$ is isosceles, hence $ AP \equal{} PD \ ( \ 2 \ )$, but $ PD \equal{} DI$ and, from (1) and (2), $ AP \equal{} CI$, hence $ AI \equal{} AP \plus{} 2\cdot MI$, done.

Best regards,
sunken rock


calculus
integration
trigonometry
calculus computations

Show that $ \int_0^\pi e^{ \minus{} 2t} \sin^3(t) \ dt \equal{} \frac {48}{65} \left( \dfrac{e^{2\pi} \plus{} 1}{e^{2\pi} \minus{} 1} \right) \int_0^{\pi} e^{ \minus{} 2t} \sin^2 (t) \ dt$ given that $ \sin^3(t) \equal{} \sin(t) \left(1 \minus{} \cos^2(t)\right)$

The use of $ \sin ^ 3 t \equal{} \frac 14 (3\sin t \minus{} \sin 3 t),\ \sin ^ 2 t \equal{} \frac 12 (1 \minus{} \cos 2 t)$ solves more quickly.

Let $ I_p \equal{} \int_0^{\pi} e^{ \minus{} 2t}\sin pt\ dt\ (p \equal{} 1,\ 2,\, \cdots)$, using Integral by parts twice,

$ \boxed{I_p \equal{} \frac {p}{p^2 \plus{} 4}\{1 \plus{} ( \minus{} 1)^{p \plus{} 1}e^{ \minus{} 2\pi}\}}.$

$ \therefore \int_0^{\pi} e^{ \minus{} 2t}\sin ^ 3 t\ dt \equal{} \frac {1}{4}\int_0^{\pi} e^{ \minus{} 2t}(3\sin t \minus{} \sin 3t)\ dt$

$ \equal{} \frac {3}{4}I_1 \minus{} \frac {1}{4}I_3$, then

$ \int_0^{\pi} e^{ \minus{} 2t}\sin ^ 2 t\ dt \equal{} \frac {1}{2}\int_0^{\pi} e^{ \minus{} 2t}(1 \minus{} \cos 2t)\ dt$

$ \equal{} \frac {1}{8}(1 \minus{} e^{ \minus{} 2\pi})$

[quote="kunny"]$ \boxed{I_p = \frac {p}{p^2 + 4}\{1 + ( - 1)^{p + 1}\}e^{ - 2\pi}}.$[/quote]

Shouldn't this be $ I_p = \frac {p}{p^2 + 4}\{1 + ( - 1)^{p + 1} e^{ - 2\pi} \}$?

\begin{align*} I_{p} & = \int_0^{\pi} e^{ - 2t} \sin(pt)\ dt \\
& = \left[\sin(nt)( - \dfrac{e^{ - 2t}}{2})\right]_0^\pi + \frac {p}{2} \int_0^{\pi} \cos(pt) e^{ - 2t}\ dt \\
\implies \dfrac{2I_{p}}{p} & = \int_0^{\pi} e^{ - 2t} \cos(pt)\ dt \\
& = \left[ \cos(pt) \left( - \dfrac{e^{ - 2t}}{2} \right) \right]_0^\pi - \dfrac{p}{2} \int_0^\pi e^{ - 2t} \sin(pt)\ dt \\
& = \dfrac{1}{2} \left[( - 1)^{p + 1}e^{ - 2\pi} + 1\right] - \frac {p}{2} I_{p} \\
\implies I_{p} \left(\dfrac{4 + p^2}{p}\right) & = \left(1 + ( - 1)^{p + 1}e^{ - 2\pi} \right) \\
\implies I_{p} & = \dfrac{p}{4 + p^2} \left(1 + ( - 1)^{p + 1}e^{ - 2\pi} \right) \\
\end{align*}

Substituting this in for your expression for the LHS and making the substitution $ u = \dfrac{\pi}{2} - t$ on the RHS shows the result.


Yeah, that's what I got. I thought of that problem when I got an assignment from school to make an educational board game.  :D

[quote="ch1n353ch3s54a1l"]..But if the numbers are all 1/5, then there's a 0% chance that you roll a "roll again"...[/quote]
You keep rolling until you get a number, not "roll again." The probability of stopping on each of the 5 numbers is equal, so it's $\frac{1}{5}$.

[quote="kyyuanmathcount"]There was a problem of that sort in a team round.  It was like, "If Bob rolls a 1,2,or 3, he eats sweetened cereal, If he rolls 4 or 5, he eats unsweetened, If he rolls 6, he rolls again.  How many days in a year (no leaps) is he expected to eat sweetened cereal?"[/quote]
[hide] $\boxed{219}$. $\frac{3}{5}\cdot365=219$, if I did the arithmetic right. :D [/hide]

It was actually how many more days will he eat sweetened cereal than unsweetened. Although it's very easy to get from lotrgreengrapes's solution.

[quote="kyyuanmathcount"]There was a problem of that sort in a team round.  It was like, "If Bob rolls a 1,2,or 3, he eats sweetened cereal, If he rolls 4 or 5, he eats unsweetened, If he rolls 6, he rolls again.  How many days in a year (no leaps) is he expected to eat sweetened cereal?"[/quote]

[hide]3/5 of the time he eats sweetened. 2/5 of the time he eats unsweetened. 365/5=73 represents the 1/5 difference in days.[/hide]

[quote="SplashD"]I don't remember that problem. What year was it from?[/quote]

2001 National Team Round

[quote="SplashD"]You're playing a board game and it's your turn. You roll the six-sided dice.

The probability that you roll a 1, 2, 3, 4, or 5 are all (1/5)

What are the six sides of this dice? (Name the marks that are on all six sides)[/quote]

That's still an iffy problem, because there's still a probability that you roll "roll again"...
I don't think that's a good problem.

[hide="proof"]let's just begin by saying every side has a probability of (1/6) of being rolled

if you keep rolling 'roll again' you would get (1/6)^infinity=0, so then there's only 5 equal possibilities left.[/hide]

[hide="better proof"]let x=an arbitrary number between 1 and 5 inclusive

to calculate the probability of rolling x, add "the probability of rolling x"+"the probability of rolling 'roll again' once and then x"+"the probability of rolling 'roll again' twice and then x"...

(1/6)+(1/6)^2+(1/6)^3+(1/6)^4...=(1/6)/(5/6)=(1/5)[/hide]

How's that for you, Klebian? :P

But when you roll "roll again", it's still rolling it.

He's talking about wording about how it sounds like rolling once, not rolling until you get a number.

That's why I created the board game scenario

[hide]
Since each has a chance of 1/5, there is a 1 probability. So, the other one is roll again??[/hide]
Ummm, this confused me a lot. If you roll again, then do you just roll again without getting a certain side of the cube?


Hi,

How to insert the references in a book using latex command?

I mean if I want to say that some theorem is taken from a book, which listed in bibiliography.

Theorem: [6] - this is the one

Use \cite. For example:
[code]\documentclass{article}
\begin{document}
This book \cite{mittelbach} contains everything you need to know about LaTeX
\begin{thebibliography}{9}
 \bibitem {mittelbach} Mittelbach F. Goossens M. et al (2004) \textit{The \LaTeX{} Companion}. Reading MA:\ Addison Wesley
\end{thebibliography}
\end{document}[/code]
For more information [url=http://theoval.sys.uea.ac.uk/~nlct/latex/novices/cite.html]See Creating a Bibliography[/url] and the Bibliography section on page 73 of [url=http://tug.ctan.org/pkg/lshort-english]The Not So Short Introduction to LaTeX2e[/url].


AMC USA(J)MO USAMO Harvard college MIT USAMTS

Please introduce yourself here!

I am a 9th grader and this is my first year participating in MA math competitions. I go to Acton-Boxborough Region HS. So far I've went to the monthly math meets and took the MAML. My biggest goal in math this year is to make USAMO, and looks like I still need a lot more work.

Hi,

You should come to the Harvard-MIT Math Competitions. We are organizing it at MIT this year, in February.

Andrei

Hello. I'm from Michigan. The reason I wanted to join this community was because I wanted to get to know more people, I know some people from Massachusetts, and Massachusetts has some really good math people and I wanted to leech off their smartness :lol: ! Anyways, thanks for letting me in beta!

So you introduce yourself in the MA forum but not the MI?  :?

I'm a senior from Central Mass, I joined AoPS last year to get ready for the USAMO.

Hi, my name is Geehoon Hong.
I'm a senior at Benjamin N. Cardozo High School at Bayside, NY,
but I'll be attending MIT, so I want to get to know more Massachusetts people :)

Hi, I also go to Acton-Boxborough Regional High School, like beta. I'm currently a 9th grader there, and have been participating in MML meets on our school math team (well now the MAML meet). I joined AoPS and started math contests such as AMC, USAMTS, and the MAML Olympiad  this year, as well as doing Mathcounts on our local junior high school team last year in 8th grade.

Hi. I'm Peter Zhang, and I go to Phillips Academy.

hi i'm j ehr from western/central mass (i dont want to put more presonal info than that online but if you know me it should be sufficient).  I joined a while ago for some reason i dont remember but im part of western mass arml and i've gone to the hmmt for the past three years on the evil geniuses for a better tomorrow.

Hello. I'm a 9th grader in worcester county.  :starwars:

Hi.....

I'm going into 8th grade at Jonas Clarke Middle School, I'm going to Lexington High next year.

I want to be on Eastern MA ARML this year, and hope to make the USAMO (and get a positive score), and plan to go back to MC nats, after a sub-standard showing in 2006. Hopefully, with 3 returnees (jclarkemathL314159, perfect628, daine), we'll pwn everyone next year.

I'm from Michigan but i used to live in Massachusetts.

Like Pineappleperson I used to live in Massachusetts (I went to Newton North High School), but now I live in Illinois.  :|

I'm Sterling from Central MA. I'm part of the Western Mass ARML team. I don't remember when and why I joined, but that doesn't matter.

Hi

I'm currently a Junior at Bedford High School
our school doesn't participate in that many competitions. I got the school to start giving the AMC/AIME last year.

I guess I should check here to find out about all the good stuff happening in MA

[quote=usernameyourself]my name is usernameyourself and i live in charlotte nc yay[/quote]

yay indeed

[quote=ASweatyAsianBoie][quote=Vin.9704][quote=noahdavid]my name is noah david and i live in newton ma[/quote]

same[/quote]

Ur name is Noah David? [img]http://www.simpleimageresizer.com/_uploads/photos/26b6e016/THONKEmote_1_24.png[/img][/quote]

th0nk

[quote=pradhyaya]][quote=klip7][quote=VKU427]I am vyom and I go to stony brook school in Westford Massachusetts
I am in seventh grade[/quote]

Wait.. I'm from blanchard too[/quote]

I'm from Blanchard too. Well I just finished 8th grade so...Westford Academy I guess.[/quote]
I'm from Blanchard too.

[quote=ASweatyAsianBoie][quote=Vin.9704][quote=noahdavid]my name is noah david and i live in newton ma[/quote]

same[/quote]

Ur name is Noah David? [img]http://www.simpleimageresizer.com/_uploads/photos/26b6e016/THONKEmote_1_24.png[/img][/quote]

lolol

[quote=htzhang]hi i'm hilary i'm a freshman from abrhs

hi teengle19 and anniewang314 hehe:)[/quote]

hi hil hil :)

I'm Noah David and I'm in 8th grade at Acera school in Melrose.

[quote=Jerry122805]facts ... time passes quickly doesn't it?[/quote]

Hi


algebra
polynomial
Rational Root Theorem
algebra unsolved

Let $ P(x)$ be a polynomial with rational coefficients such that $ P^{\minus{}1}(\mathbb{Q}) \subseteq \mathbb{Q}$. Show that $ P$ is linear.

[quote="moldovan"]Let $ P(x)$ be a polynomial with rational coefficients such that $ P^{ \minus{} 1}(\mathbb{Q}) \subseteq \mathbb{Q}$. Show that $ P$ is linear.[/quote]

[hide="main ideas"]

Assume that there are $ P$ isn't linear satisfying the condition.
[b]Step 1:[/b] Multiply $ P(x)$ by non-zero integer to form $ Q(x)$ which has integer coefficients and positive leading coefficients. Clearly for any large enough $ p$ $ Q(x)\equal{}p$ has real roots.

[b]Step 2:[/b] Let $ p$ be prime and doesn't divide the coefficients of $ Q(x)$. Clearly there are infinitely many such primes.

[b]Step 3:[/b] Then by Rational Root Theorem , the rational roots of $ Q(x)\equal{}p$ is of the form $ \frac{1}{ps_i}$ or $ \frac{1}{s_i}$, where $ s_i$ are factors of the leading coefficient of $ Q(x)$.

[b]Step 4:[/b] Plug in $ x\equal{}\frac{1}{ps_i}.........(1)$ and $ x\equal{}\frac{1}{s_i}...........(2)$. For Case (2) clearly there are finite number of $ p$ s satisfying it. For Case (1) we multply both sides by a power of $ p$ and then get a finite set of non-constant polynomials with $ p$ as the variant and thus there are only a finite number of $ p$ s satisfying them. So for some large enough $ p$, $ Q(x)\equal{}p$ has no rational roots, contradiction, done.
[/hide]

Actually it's more like a solution but with small details removed.  :)

See also [url]http://www.mathlinks.ro/viewtopic.php?p=1614467#1614467[/url]

By the way, why bother writing $ P^{ \minus{} 1}(\mathbb{Q}) \subseteq \mathbb{Q}$, when clearly then we have $ P^{ \minus{} 1}(\mathbb{Q}) \equal{} \mathbb{Q}$.


What is the result when the sum of fifteen and ninety is divided by the positive difference of thirty-eight and twenty-three?


Ring Theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ f: A\to B$ be a ring homomorphism, $ S$ a multiplicative subset of $ A$, and $ T\equal{}f(S)$. Show that $ T^{\minus{}1}B\cong B\otimes_AS^{\minus{}1}A$ as $ A$-algebras.

As with all tensor product isomorphisms like this, we want to find homomorphisms in both directions.  Given an element $ b/f(s)$, what is a reasonable choice in $ B\otimes_A S^{ \minus{} 1} A$?

For the other direction, first write down an intuitive bilinear map $ B\times S^{ \minus{} 1} A \to T^{ \minus{} 1} B$.  (where should $ (b,a/s)$ go?)

Of course, you need to check that these maps are inverses, and in fact homomorphisms of $ A$-algebras.  For the first one, you will need to make sure that it's well-defined.

If you're having trouble with some specific part of that, let us know.


induction
number theory proposed
number theory

let:
$ (1\plus{}\sqrt{2})^n\equal{}p\plus{}\sqrt{2}*q$
prove that:
$ 2^k|q$ <=> $ 2^k|n$

[quote="Allnames"][quote="tuandokim"]let:
$ (1 \plus{} \sqrt {2})^n \equal{} p \plus{} \sqrt {2}*q$
prove that:
$ 2^k|q$ <=> $ 2^k|n$[/quote]
It is obvious since induction  :lol:[/quote]
Why?????
thanks


I have a few questions regarding proof-writing.
First of all: when you are trying to prove a triangle statement, must you always divide the statement into two cases, proving it first for acute triangles and than for obtuse triangles?

Second of all: when you are asked to prove an equation or expression and let's say you manipulate it until you get a true statement (e.g. $ x^2\geq 0$), can you just write "and all the steps are reversible, and we are done" as your proof, or must you rewrite the entire thing starting with you're final statement and manipulating it back to the statement we are trying to prove?

Thank you!

[quote="t0rajir0u"][quote="mihail911"]when you are trying to prove a triangle statement, must you always divide the statement into two cases, proving it first for acute triangles and than for obtuse triangles?[/quote]
That depends on the nature of the statement.  The distinction between acute and obtuse triangles can occur because a certain point $ P$ is placed in two different ways in the two cases, in which case you probably need to, but if you use trigonometry (for example), which handles both acute and obtuse angles correctly in a single triangle, you don't need to make this distinction.  (You can also get away with ignoring casework by using directed angles in some cases.)

[quote="mihail911"]Second of all: when you are asked to prove an equation or expression and let's say you manipulate it until you get a true statement (e.g. $ x^2\geq 0$), can you just write "and all the steps are reversible, and we are done" as your proof, or must you rewrite the entire thing starting with you're final statement and manipulating it back to the statement we are trying to prove?[/quote]
You don't need to write "all the steps are reversible" if you make sure to include $ \Leftrightarrow$ between all of your statements, but it's stylistically neater to prove forward from a known statement.[/quote]

Wait, you have to do the double arrow (don't know how to do in latex) between all statements?

If you want to start from the problem statement and work to a true statement, yes.  What you're saying is that "the following statements are equivalent" and that the last one is true.

Be careful with this, though. Working backwards doesn't always yield the same result.


$ ABC$ is a triangle($ BC=22$, $ CA=16$, $ AB=14$) with incentre $ I$ and $ D$, $ E$, $ F$ the midpoints of $ BC$, $ CA$, and $ AB$.

A line through $ D$ parallel to $ AI$ meets the longer of the other two sides at $ D'$. Similarly construct $ E'$, $ F'$ ( a line through $ E$ parallel to $ BI$ etc)

$ a.$ Find (with proof) the measures of $ AD'$, $ BE'$, $ CF'$
$ b.$ Why are the lines $ DD'$, $ EE'$, $ FF'$ concurrent?

a) $ AI$ intersect $ BC$ at $ G$

$ \frac{GC}{AC}=\frac{BG}{AB}=\frac{BC}{AB+AC}$
So $ GC=\frac{BC.AC}{AB+AC}=\frac{176}{15}$

$ \frac{AD'}{AC}=\frac{GD}{GC}=\frac{GC-CD}{GC}$
So $ AD'=\frac{AC(GC-CD)}{GC}=1$

We can find the measures of  $ BE', CF'$ similarly

b) $ ED$ and $ AB$ are parallel; $ DD'$ and $ AG$ are parallel
So $ \angle EDD'=\angle BAG$

$ FD$ and $ AC$ are parallel; $ DD'$ and $ AG$ are parallel
So $ \angle FDD'=\angle CAG$

We obtain $ \angle FDD'=\angle EDD'$
So $ DD'$ is the bisector of $ \angle EDF$
Similarly, we have  $ EE'$ is the bisector of $ \angle FED$
                             $ FF'$ is the bisector of $ \angle EFD$
$ DD', EE', FF'$ are 3 bisectors of the triangle $ EFD$
So $ DD', EE', FF'$ are concurrent


Find the smallest number (other than 1) in the sequence
${1 , 11 , 111 , 1111 , 11111, ...}$ that is a perfect square (like 25 or 36 or 144) or prove that none of them are perfect squares.

[hide]
Let $a_{n}$ be the number $11...1$ with [b]n[/b] ones

Suppose that there exists some $x$ and some $n>1$ such that $x^{2}=a_{n}$

Of course $x$ is odd

$x^{2}-1^{2}= a_{n}-1 = 10a_{n-1}$

$(x+1)(x-1) = 2\cdot 5\cdot a_{n-1}$

But this is a contradiction, because the left side is a multiple of 4 but the right side is not[/hide]

[hide]

The numbers are all $11\equiv 3 \mod 4$, (except for 1), but 3 is not a quadratic residue mod 4.

[/hide]

$111\ldots111$ can be written in the following form:

$111\ldots108+3$. But $111\ldots108$ is a multiple of 4. As a consequence, $111\ldots111$ has the form: $4k+3$. As a result, it can't be a perfect square.


using Stewart's Theorem, prove that in any triangle with sides a, b, c and medians m, n, p we have m^ 2+n^2+p^2=3/4(a^2+b^2+c^2)

[b](1)[/b] Please don't SHOUT by using capitals in the subject of your topic, and also please give a [i]meaningful[/i] name to your topic, for instance "Triangle and medians", or "Lengths of the medians", or "Help with Stewart's theorem", rather than just "HELP" or "I DON'T KNOW WHERE TO START".

[b](2)[/b] If you know the Stewart theorem, then you can easily compute the lengths of the medians m, n, p and get

$m^2=\frac{2b^2+2c^2-a^2}{4}$;
$n^2=\frac{2c^2+2a^2-b^2}{4}$;
$p^2=\frac{2a^2+2b^2-c^2}{4}$.

Now add these equations together and see what you obtain  ;) . So much about "where to start". Feel free to ask again if you have troubles with completing the solution.

  darij

thank you very much for your help and i will remember not to use capitalized letter.  Thank You very much  :o)


algebra
polynomial
superior algebra
superior algebra unsolved

Find \[ \min (\sqrt{2}+\sqrt{3}, \mathbb{Q}). \]

Done like in the other threads about minimal polynomials.
Or:
$\sqrt{2}+\sqrt{3}=x$
$2+2 \sqrt{6} + 3 = x^2$
$2 \sqrt{6}= x^2-5$
$24 = x^4-10x^2+25$
$x^4-10x^2+1=0$
done.


$ a$, $ b$, $ c$ are positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$.
Prove that $ \frac{a^{2}}{b}\plus{}\frac{b^{2}}{c}\plus{}\frac{c^{2}}{a} \geq 3(a^{2}\plus{}b^{2}\plus{}c^{2})$.

[quote="paracompact"]$ a$, $ b$, $ c$ are positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$.
Prove that $ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a} \geq 3(a^{2} \plus{} b^{2} \plus{} c^{2})$.[/quote]
$ c\equal{}min(a,b,c)$
this ineq <=> $ N(a\minus{}b)^2\plus{}M(a\minus{}c)(b\minus{}c)\ge 0$
in there
$ N\equal{}\frac{a\plus{}b}{ab}\minus{}\frac{2}{a\plus{}b\plus{}c}$
$ M\equal{}\frac{b\plus{}c}{ac}\minus{}\frac{2}{a\plus{}b\plus{}c}$
easy to prove $ M,N \ge 0$
so ineq was proved  :D

Another proof can be obtained if we use one nice results,that is:
$ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\geq 3\sqrt{\frac{a^4\plus{}b^4\plus{}c^4}{a^2\plus{}b^2\plus{}c^2}}$.
So from this it will be necessary to prove that
$ (a^4\plus{}b^4\plus{}c^4)\geq (a^2\plus{}b^2\plus{}c^2)^3$,or
$ (a^4\plus{}b^4\plus{}c^4)(a\plus{}b\plus{}c)(a\plus{}b\plus{}c)\geq (a^2\plus{}b^2\plus{}c^2)^3$,which is Holder's inequality.

See hear:
http://www.mathlinks.ro/viewtopic.php?p=515092#p515092
http://www.mathlinks.ro/viewtopic.php?p=868668#868668   :wink:


http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=1%2B1%3D2&t=55563

Why do people keep asking this ?

does it need to prove?

i think we should accept with this fact that:1+1=2

then you can accept as well the fact that every even number $\geq 2$ can be written as the sum of two primes...
It isn't about guessing whether this statement is true or not - it should be true- but it is important to check if all underlying axioms are consistent with our feelings about natural numbers and their properties.

We note that f(x)=x"+1" is defined as the successor function of (x"+0") when x is a non-negative integer since 1 is defined as the successor of 0.

We also note that "0" is defined as the identity for addition, and so x+0 is defined as the identity function.

We also note that "2" is defined as the successor of "1".

Thus, we have 1+1=S(1)+0=S(1)=2

[i]"God gave us the Natural Numbers. The rest is the work of Man."[/i] - Kronecker


parameterization
function
complex numbers
complex analysis
complex analysis unsolved

Suppose that $ \{a_n(z)\}_{n \equal{} 1}^\infty$ is a sequence of complex numbers depending continuously on a parameter $ z,$ and that

$ \sum_{n \equal{} 1}^\infty a_n(z)$

converges uniformly to an analytic function $ f(z)$ in some open region $ \Omega$ (and that region is the largest open region in which the series converges uniformly). Suppose further that the series is Abel summable in a larger region $ \Omega'$; that is, if $ 0 < r < 1$ then

$ \sum_{n \equal{} 1}^\infty r^na_n(z)$

converges uniformly to $ g(r, z)$ in some open region $ \Omega' \supsetneq \Omega$ and $ g(r, z)$ converges to a function $ g(z)$ on $ \Omega'$ as $ r \to 1$.

Is it true that $ g(z)$ is equal to the analytic continuation of $ f$ onto $ \Omega'$? (Note that I didn't say $ g$ must be analytic (or even continuous) on all of $ \Omega'$. Maybe it will be, but it doesn't seem entirely obvious to me...)

Well, we have that $ g$ is an extension of $ f$; the standard Abel theorem says that they agree where they're both defined.

On the other hand, I see no good reason for $ g$ to be analytic. An example:
$ a_n(z)\equal{}\begin{cases}\frac1n z^n& |z|<1\\ \frac1n\cdot\frac{z^n}{|z|^n}& |z|\ge 1\end{cases}$

$ f$ extends analytically to a branch of $ \minus{}\log(1\minus{}z)$, while $ g(z)\equal{}\minus{}\log(1\minus{}\frac{z}{|z|})$ outside the circle.


For any 2 convex polygons $S$ and $T$, if all the vertices of $S$ are vertices of $T$, call $S$ a sub-polygon of $T$.

[b]I. [/b]Prove that for an odd number $n \geq 5$, there exists $m$ sub-polygons of a convex $n$-gon such that they do not share any edges, and every edge and diagonal of the $n$-gon are edges of the $m$ sub-polygons.

[b]II.[/b] Find the smallest possible value of $m$.

The solution use complex numbers? Who can solve?


I am thinking of taking the SAT II Biology test at the end of the year, after I complete Freshman Biology I Honors.  JUst from what we've done in class already, I think I will be able to do the test by the end of the year, because the material is the same.  I will not take another biology class throughout high school, so should I take the SAT II at the end of this year?  Will I be prepared?

if you feel that you're ready, then by all means. After all, you have nothing to lose.

Since this is the only time you're taking biology, I would definitely take it at the end of the year.  Keep in mind that ARML usually falls on the last SAT date.

Oh, wow, I totally forgot about ARML.  But isn't there usually an alternate SAT test for ARMLers?

Yeah, ARML is tricky in relation to taking SAT II tests. The first day of ARML training for the Minnesota team is the May SAT date, and ARML meet day each year is the June SAT date. Let's see, the College Board says

http://www.collegeboard.com/student/testing/sat/calenfees.html

there are SAT II test dates in October, November, December, January, March, May, and June. Those October and March dates are looking good, or at least better than skipping ARML.

You can just reschedule the June SAT if you are going to ARML. That's what I did for my SAT 2's.

As far as score reporting goes, should I report to my potential colleges?  I don't even know where I want to go yet.  How does all that work?

What steps did you have to follow to reschedule the SAT II to fit in ARML?


Problem description:
[code]
Farmer John and Bessie the cow have embarked on one of those 'active'
vacations. They spend entire days walking in the mountains and then,
at the end of the day, they tire and return to their vacation cabin.

Since climbing requires a lot of energy and they are already tired,
they wish to return to the cabin using a path that has the least
difference between its highest and lowest elevations, no matter how
long that path is. Help FJ find this easy-to-traverse path.

The map of the mountains is given by an N x N (2 <= N <= 100) matrix
of integer elevations (0 <= any elevation <= 110) FJ and Bessie are
currently at the upper left position (row 1, column 1) and the cabin
is at the lower right (row N, column N).  They can travel right, left,
toward the top, or toward the bottom of the grid.  They can not travel
on a diagonal.

PROBLEM NAME: mtwalk

INPUT FORMAT:

* Line 1: The single integer, N

* Lines 2..N+1: Each line contains N integers, each of which specifies
        a square's height. Line 2 contains the first (top) row of the
        grid; line 3 contains the second row, and so on.  The first
        number on the line corresponds to the first (left) column of
        the grid, and so on.

SAMPLE INPUT (file mtwalk.in):

5
1 1 3 6 8
1 2 2 5 5
4 4 0 3 3
8 0 2 3 4
4 3 0 2 1

OUTPUT FORMAT:

* Line 1: An integer that is the minimal height difference on the
        optimal path.

SAMPLE OUTPUT (file mtwalk.out):

2

OUTPUT DETAILS:

xx...   Taking the path shown here, the minimum height is 0 and the
.xx..   maximum is 2, so the difference is 2.
..x..
..x..
..xxx
[/code]

Any idea how to solve it :?:

well, i have not thought that much but i think i am in right path!

look the elevation can be only <=110. So first set only 0s. and check if there is any way. if not set 1s. check... set 110 and check! simple.

Order: for each checking: O(N^2)
Number of checking: O(111).

Worst case Order: 111*100^2.

It is very low. so it should work!

i got better way. it will just reduce running time. u make a binary search on the height. and then make BFS.

Mahbub

[quote="yangzhe1990"]I think we should enumerate the min elevation, and then cross out points whose elevation is lower, and then dijkstra.
so it cost O(n^3).  You may use heap to reduce the runtime to O(n^2logn).[/quote]

Heap won't reduce to NlogN. The graph is nowhere near sparse (E is on same order as V^2, so ElogV = V^2logV x_x).

But N^3 is definitely fast enough anyways.

well... i think i had misread the problem. but that idea can be applied as well...

this time i want to keep track of the height difference. so how many nodes do we have? approximately: 2N^2.

now make a min_heap. 2N^2(log(2N^2)).

then extract min and construct graph.

i would like to use the idea of MST here. i will use the union-set. that will reduce the connecting fidning time to O(1) each time.

So running time: 2N^2(log(2N^2))


Is it possible to have images float inside a paragraph in LaTeX the same way that CSS does with float:right?

Thanks

Xevarion,

Remember that Latex is an edition program, originally though and created to edit mathematics and scientific text for printed version, and in a printed version you cannot have floating text or images. On the other hand, in the last years there have been created some latex packages that are usufull tocreate presentation with many of the capabilities of Power Point, and some that this program doesn't have. May be one of those packages are usufull for that, but I doubt it.

Best,

[quote="Xevarion"]Is it possible to have images float inside a paragraph in LaTeX the same way that CSS does with float:right?[/quote]Yes.  There are two packages -- [b]wrapfig[/b] and [b]picins[/b] -- that can do this.

Go to http://www.ctan.org to dowload the packages and get documentation.


I have been reading up on previous avatar help topics, but I still don't get it. (jose had this same problem) do you just copy a picture and paste it? How do you upload it? help!

tspa

First, save the file you would like to upload as a picture (.jpg if you don't know much about pictures).  Then, go to your profile.  There is a buttom at the top of the page.  Go to the avatar control panel.  Find where it says upload avatar.  Click the browse buttom and go to the location on your computer where the picture file is located.  Click whatever the OK button is (maybe something like submit...).

Is this specific enough?

if your avatar is on the internet you right click it and pick properties and that will have the URL of the avatar and you put that in your profile for your avatar


Set up an equation for the circle whose centre lies on the straight line given by x+y=0 and which passes through the points of intersection of the two circles given by (x-1)^2 + (y+5)^2 = 50 and (x+1)^2 + (y+1)^2 = 10

[hide]
The set of points equidistant from two points is equal to the perpindicular bisector of the segment between them.  Therefore, all points that are equidistant from the two intersections must pass through their perpindicular bisector and therefore through the line connecting the center of the circles.  Then the center of the circle we want is the intersection of $x+y=0$ and the line between $(1,-5)$ and $(-1,-1)$ with equation $2x+y+3=0$, which is at $(-3,3)$.[/hide]


I'm not sure if the following theorem is well-known or if indeed it is known as van Obal's Theorem. However it seems very useful and it would nice to clear up whose theorem it is once and for all:

If $ABC$ is a triangle and three cevians $AD$, $BE$, and $BF$ meet at $P$, then: \[ \frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC} \].

I think it is called the van Aubel theorem.
See

http://www.mathlinks.ro/Forum/viewtopic.php?t=15764
http://www.mathlinks.ro/Forum/viewtopic.php?t=15211
http://www.mathlinks.ro/Forum/viewtopic.php?t=46502
http://www.mathlinks.ro/Forum/viewtopic.php?t=31797
http://www.mathlinks.ro/Forum/viewtopic.php?t=56226

  darij

http://mathworld.wolfram.com/vanAubelsTheorem.html

I believe van Obal's Theorem is different from van Aubel's Theorem. van Aubel is the Napoleonic triangles theorem for quadrilaterals.


An album for humerous pictures pertaining to math!

Actually, I have no idea what "nonie" is. But if you try searching for this picture on the internet, you can see that it is popular enough and has been posted on enough sites so that it cannot be considered stealing from anyone.

:rotfl: , what grade did you get on that?


geometry
similar triangles
congruent triangles
combinatorics proposed
combinatorics

Here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=26792

It can be shown that the integers $n$ for which every triangle can be decomposed into $n$ pairwise similar triangles are $n=4$ and all the integers $n \geq 6$.

Now, here is a really nice result :

Prove that the integers $n$ for which any triangle can be decomposed into $n$ pairwise congruent triangles are  :P  :P  :P  (to be found  :D )

Pierre.

Hmm.. A first glance tells me "perfect squares", but I don't know if those are all of them :).

Do you have a nice solution, Pierre?
My arguments are quite straighforward. Suppose $n$ is not a perfect square. We are to find a triangle which has lineary independent angles over $\mathbb Q$ (it will implies that all small triangles will be similar to the big one), moreover we need $x$, $y$, $z$, $x\sqrt n$, $y\sqrt n$, $z\sqrt n$ are also linear independent other $\mathbb Q$. Then it is impossible to tile the big triangle with its small copies.

Now, I'm not home so I cannot log as easy as usual.
I will post a solution in some days.

Pierre.


Determine $ m \in \mathbb{R}$ so that these equations have a common root:

$ x^2\minus{}(m\plus{}3)x\plus{}m\plus{}5\equal{}0$
$ x^2\minus{}(3m\minus{}1)x\plus{}2m\plus{}6\equal{}0$
$ x^2\minus{}(2m\plus{}4)x\plus{}3m\plus{}7\equal{}0$

[hide="solution"]If we add the first and last equation, and subtract twice the middle equation, we can get all terms without an $ x$ to cancel and that allows for some easy factoring.

$ x^2\minus{}(m\plus{}3)x\plus{}(m\plus{}5)\plus{}x^2\minus{}(2m\plus{}4)x\plus{}3m\plus{}7\minus{}2(x^2\minus{}(3m\minus{}1)x\plus{}2m\plus{}6)$

$ \equal{}2x^2\minus{}3mx\minus{}7x\plus{}4m\plus{}12\minus{}2x^2\plus{}6mx\minus{}2x\minus{}4m\minus{}12$

$ \equal{}3mx\minus{}9x\equal{}0$

$ (3m\minus{}9)x\equal{}0$

But $ x\equal{}0$ leads to $ m\equal{}\minus{}5$ which is inconsistent.  Thus $ m\equal{}3$ and the $ x$ value that works for this is $ x\equal{}2$ as is easily checkable.[/hide]


function
inequalities
inequalities unsolved

$a,b,c$ are positive number and $a+b+c=6$. Find the least value of the expression;

$\large S= \sqrt{a^2+ \frac{1}{b+c} } + \sqrt{b^2+ \frac{1}{a+c} } + \sqrt{c^2+ \frac{1}{a+b} }$

 :arrow:  :lol:

[quote="zhaobin"]first:
$\sqrt{a^2+ \frac{1}{b+c} } + \sqrt{b^2+ \frac{1}{a+c} } + \sqrt{c^2+ \frac{1}{a+b} } \ge \sqrt{(a+b+c)^2+ \left( \sqrt{\frac{1}{b+c}}+ \sqrt{\frac{1}{a+b}}+ \sqrt{\frac{1}{c+a}}\right)^2}$
second:
$\sqrt{\frac{1}{b+c}}=\sqrt{\frac{1}{6-a}}$
use the function $\sqrt{\frac{1}{6-x}}$ is convex  on $(0,6)$:)[/quote]

I am misunderstand. You continue help me, please

can you tell me where you misunderstand?? :?:

[quote="Nameless"]Maybe he misunderstand in $\sum\frac{1}{\sqrt{b+c}}\ge\frac{3}{2}$
We have $\sum\frac{1}{\sqrt{b+c}}\ge\frac{9}{\sum\sqrt{b+c}}\ge\frac{9}{\sqrt{6(a+b+c)}}$[/quote]
\[ \sum\{b+c}\ge\frac{9}{\sum\frac{1}{\sqrt{b+c}=6 \]

[quote="Nameless"]Maybe he misunderstand in $\sum\frac{1}{\sqrt{b+c}}\ge\frac{3}{2}$
We have $\sum\frac{1}{\sqrt{b+c}}\ge\frac{9}{\sum\sqrt{b+c}}\ge\frac{9}{\sqrt{6(a+b+c)}}$[/quote]
\[ \sum\{b+c}\ge\frac{9}{\sum\frac{1}{\sqrt{b+c}}=6 \]

[quote="konlsy"][quote="Nameless"]Maybe he misunderstand in $\sum\frac{1}{\sqrt{b+c}}\ge\frac{3}{2}$
We have $\sum\frac{1}{\sqrt{b+c}}\ge\frac{9}{\sum\sqrt{b+c}}\ge\frac{9}{\sqrt{6(a+b+c)}}$[/quote][/quote]
what happen? why can't i type LAX?
please delete them

[quote="Nameless"]Maybe he misunderstand in $\sum\frac{1}{\sqrt{b+c}}\ge\frac{3}{2}$
We have $\sum\frac{1}{\sqrt{b+c}}\ge\frac{9}{\sum\sqrt{b+c}}\ge\frac{9}{\sqrt{6(a+b+c)}}$[/quote]

You use [b]BCS[/b] inequality. I understanded.
$( \sqrt{a+b} + \sqrt{b+c} + \sqrt{c+a} )^2 \leq (1+1+1)(2a+2b+2c)$
and
$\sqrt{a^2+ \frac{1}{b+c} } + \sqrt{b^2+ \frac{1}{a+c} } + \sqrt{c^2+ \frac{1}{a+b} } \ge \sqrt{(a+b+c)^2+ \left( \sqrt{\frac{1}{b+c}}+ \sqrt{\frac{1}{a+b}}+ \sqrt{\frac{1}{c+a}}\right)^2}$
Zhaobin used [b]Mincopxki[/b]
inequality
 :lol:  :P

Can use Cosi inequality to prove this problem?

$\large S= \sqrt{a^2+ \frac{1}{b+c} } + \sqrt{b^2+ \frac{1}{a+c} } + \sqrt{c^2+ \frac{1}{a+b} }$
 prove that :
 $\sum\frac{1}{\sqrt{a+b}} \geq \frac{3}{2}$(1)
 we have   cauchy ineq's by :$\sqrt{a+b} \leq \frac{4+b+c}{4}$
 hence (1)  ->
  $\sum\frac{1}{4+a+b} \geq \frac{3}{8}$
 by cauchy-Schwarz  ineq , we have :

$\sum\frac{1}{4+a+b} \geq \frac{9}{12+2(a+b+c)}= \frac{3}{8}$

[quote="ehku"]This is a problem in a magazine in VietNam.
Robinho, you shouldn't post it.
We stop here ! OK ?[/quote]

Ok, I agrre. I not send it to magazine so you don't worry.
I thanks everybody. :P  :lol:


Divisibility rules: For 11: The difference between the sum of alternating digits is divisible by 11.
For 9: Sum of digits is divisible by 9

[quote="314.math"]The $ 9$-digit number $ aab,aba,ba3$ is a multiple of $ 99$ for some pair of digits $ a$ and $ b$.  What is $ b \minus{} a$?

[hide]$ 4$[/hide]

Thanks in advance.

[EDIT] Oops, I double posted.  Could a mod please delete one of them?  Thanks.[/quote]
[hide="Solution"]
Since the number is a multiple of 99, $ aab,aba,ba3\equiv 0 \pmod{99}$ .
\[ \begin{array}{l}0\pmod{99}\equiv 0\pmod{9} \\
0\pmod{99} \equiv 0\pmod{11}\end{array}
\]
For any number the sum of digits is congruent to the number itself mod 9.
\[ a \plus{} a \plus{} b \plus{} a \plus{} b \plus{} a \plus{} b \plus{} a \plus{} 3\equiv 0\pmod{9}
\]

\[ 5a \plus{} 3b \plus{} 3\equiv 0\pmod{9}
\]
Similarly, any number is congruent to the sum of digits with alternating signs, starting with a positive sign on the far right, mod 11.
\[ a \minus{} a \plus{} b \minus{} a \plus{} b \minus{} a \plus{} b \minus{} a \plus{} 3\equiv 0 \pmod{11}
\]

\[ 3\left(b \minus{} a\right) \plus{} 3\equiv 0\pmod{11}
\]
Recognizing that $ a\, \&\, b$   are each single digits, the congruence mod 11 and the congruence mod 9 are the same number:
\[ 5a \plus{} 3b \plus{} 3\equiv 3\left(b \minus{} a\right) \plus{} 3 \pmod{9}
\]

\[ 5a\equiv \minus{} 3a \pmod{9}
\]

\[ 8a\equiv 0 \pmod{9}
\]

\[ \minus{} a\equiv 0 \pmod{9}
\]

\[ \implies a \equal{} 9
\]
If $ a \equal{} 9$  then $ 5(9) \plus{} 3b \plus{} 3\equiv 0 \pmod{9}\implies 3b \plus{} 3\equiv 0\pmod{9}$
\[ 3b \plus{} 3\equiv 0\pmod{9}\implies b\equiv \minus{} 1\pmod{9}
\]

\[ b\equiv \minus{} 1\pmod{9}\implies b \equal{} 8
\]

\[ aab,aba,ba3 \equal{} 998,989,893
\]

$ {b \minus{} a} \equal{} {8 \minus{} 9} \equal{} \minus{} 1$
[/hide]


In Ms. Smith's class, each student averages one day absent out of thirty. What is the probability that out of any two students chosen at random, one student will be absent while the other is present? Express your answer as a percent rounded to the nearest tenth.

The probability that one student is absent is $ \frac{1}{30}$, while the probability that the other student is present is $ \frac{29}{30}$. Remember we have to multiply by 2, because it could be vice-versa, so our final answer is $ \frac{1}{30} \times \frac{29}{30} \times 2 \approx 6.4\%$.


function
calculus
derivative
calculus computations

Is the following true?

For a fixed curve $C$ and fixed straight line $L$, let $P$ be a point on $C$ such that the distance from $P$ to $L$ is less than that of any other points on $C$. Then the tangent to $C$ at $P$ is parallel to $L$.

Is $C$ guaranteed to be smooth enough to have a tangent line? How many dimensions? The answers depend on these questions.

If it's a two-dimensional problem, make $C$ (locally) the graph of a function and $L$ a horizontal line. What can you say about the derivative at a minimum or maximum?


[quote="Demetres"]Δίνονται 101 θετικοί ακέραιοι ώστε αν αφαιρέσουμε οποιονδήποτε από αυτούς οι υπόλοιποι 100 μπορούν να χωριστούν σε δύο ομάδες των 50 ώστε κάθε ομάδα να έχει το ίδιο άθροισμα.

Να δειχθεί ότι όλοι οι αριθμοί είναι ίσοι.[/quote]

Υπόδειξη
    [hide] 1) βγάλε τον $ a_1$. Από την υπόθεση οι άλλοι έχουν άθροισμα κάποιον άρτιο.
             2) βγάλε τον $ a_2$. Από την υπόθεση οι άλλοι έχουν άθροισμα κάποιον άρτιο.
             3) αρα  $ a_1 \minus{} a_2$ είναι άρτιος
             4) οι αριθμοί είναι όλοι άρτιοι ή όλοι περιττοί.
             5) προσθέτωνυτας σε όλους +1, αν χρειαστεί, μπορούμε να υποθέσουμε ότι όλοι είναι άρτιοι.
             6) αφαιρώντας τον πιο μικρό από όλους μπορούμε να υποθέσουμε ότι κάποιοι είναι μηδέν και οι υπόλοιποι (αν υπάρχουν) είναι άρτιοι
             7) μπορούμε να διαιρέσουμε δια 2 όλους τους αριθμούς. Η αρχική ιδιότητα διατηρείται.
             8)  Όμως πάλι όλοι οι αριθμοί είναι άρτιοι ή όλοι περιττοί. Αλλά δεν είναι περιττοί γιατί κάποιοι είναι 0.
             9) Διαιρούμε δια 2, ξανά και ξανά. Κάποτε θα πέσουμε σε περιττούς, 'ατοπο, εκτός αν όλοι οι αριθμοί είναι 0.
[/hide]

Φιλικά,

Μιχάλης Λάμπρου

Ωραία λύση κ. Λάμπρου! Να σημειώσω ότι το πρόβλημα φυσικά ισχύει για οποιουσδήποτε $ 2n\plus{}1$ ακέραιους (όχι απαραίτητα θετικούς!) και το πιο εντυπωσιακό είναι (μόλις το διάβασα) ότι ισχύει για οποιουσδήποτε $ 2n\plus{}1$ πραγματικούς!!!

[quote="bilot4"]Να σημειώσω ότι το πρόβλημα φυσικά ισχύει για οποιουσδήποτε $ 2n \plus{} 1$ ακέραιους (όχι απαραίτητα θετικούς!) και το πιο εντυπωσιακό είναι (μόλις το διάβασα) ότι ισχύει για οποιουσδήποτε $ 2n \plus{} 1$ πραγματικούς!!![/quote]

Έχω την εντύπωση πως κάπου είχα δει μια στοιχειώδη λύση αλλά δεν την θυμάμαι  :( . Την έλυσα με γνώσεις γραμμικής άλγεβρας αλλά δεν θυμάμαι αν αυτήν την λύση είχα δει ή όχι.


modular arithmetic
least common multiple
number theory
Diophantine equation

With $ \$1000$ a rancher is to buy steers at $ \$25$ each and cows at $ \$26$ each. If the number of steers $ s$ and the number of cows $ c$ are both positive integers, then:

$ \textbf{(A)}\ \text{this problem has no solution}\qquad\\
\textbf{(B)}\ \text{there are two solutions with }{s}\text{ exceeding }{c}\qquad \\
\textbf{(C)}\ \text{there are two solutions with }{c}\text{ exceeding }{s}\qquad \\
\textbf{(D)}\ \text{there is one solution with }{s}\text{ exceeding }{c}\qquad \\
\textbf{(E)}\ \text{there is one solution with }{c}\text{ exceeding }{s}$

Yes, I think he must spend it all. I got something different.
[hide="My Solution"]
We need to find the solutions to the Diophantine equation $ 25s \plus{} 26c \equal{} 1000$. We can starting by modding out $ 25$. We obtain $ c \equiv 0 \pmod{25}$. Therefore, $ c \equal{} 25k$ for $ k \in \mathbb{Z}$. Substituting that into the equation and solving for $ s$, we obtain

$ 25s \plus{} 26\cdot 25k \equal{} 1000\implies 25s \equal{} 1000 \minus{} 26\cdot 25k\implies s \equal{} 40 \minus{} 26k.$

So the solutions to the Diophantine equation are $ (s,c) \equal{} (40 \minus{} 26k, 25k)$. We can generate each solution by plugging in a value of $ k$. Note that $ k$ must be positive because $ c$ is positive. Also note that the only solution in which $ s$ and $ c$ are both positive integers is when $ k \equal{} 1\implies (s,c) \equal{} (14,25)$. Therefore, the answer is $ \fbox{(E)}$.
[/hide]

[quote="PowerOfPi"]Does he have to spend all the money? If so, the answer is $ \mathbf{(A)}$. This is becuase the least common multiple of 25 and 26 is over 1000. Also, They can't all be steers because c must be positive.[/quote]

Wow, I thought $ 25\times26$ is over $ 1000$. It is $ 25\times26\equal{}650$. So one solution is 14 steers and 25 cows. It is the only one because the next common multiple of 25 and 26 is 1300.

So, $ \mathbf{(E)}$.

[quote="Ihatepie"]He could also buy 40 steers and 0 cows. I wish that the problem writers weren't so ambiguous.[/quote]

Actually, no, because it says "If the number of steers $ s$ and the number of cows $ c$ are both positive integers, ...". So $ c$ can't be zero.


geometry
parallelogram
ratio
geometric transformation
homothety
geometry unsolved

Given $ \triangle ABC$ with point $ D$ inside. Let $ A_0\equal{}AD\cap BC$, $ B_0\equal{}BD\cap AC$, $ C_0 \equal{}CD\cap AB$ and $ A_1$, $ B_1$, $ C_1$, $ A_2$, $ B_2$, $ C_2$ are midpoints of $ BC$, $ AC$, $ AB$, $ AD$, $ BD$, $ CD$ respectively. Two lines parallel to $ A_1A_2$ and $ C_1C_2$ and passes through point $ B_0$ intersects $ B_1B_2$ in points $ A_3$ and $ C_3$respectively. Prove that $ \frac{A_3B_1}{A_3B_2}\equal{}\frac{C_3B_1}{C_3B_2}$.

Let lines $A_1C_1$ and $AA_1$ cut $BB_0$ at $E,V,$ respectively. Parallel from $A_1$ to $C_1C_2$ cuts $BB_0$ and $C_1A_2$ at $U,Q.$ Since $C_1A_2 \parallel A_1C_2 \parallel BB_0$ and $C_1A_1 \parallel A_2C_2 \parallel AC,$ it follows that $C_1A_1C_2A_2$ is a parallelogram $\Longrightarrow$ Segments $A_1A_2$ and $C_1C_2$ bisect each other. Therefore, $C_1$ is the midpoint of $QA_2$ $\Longrightarrow$ Median $A_1C_1$ of $\triangle QA_1A_2$ bisects the parallel segment $UV \parallel QA_2$ $\Longrightarrow$ $E$ is the midpoint of $UV.$ Consequently,  pencil of lines $A_1(U,V,E,C_2)$ is harmonic. From $B_0C_3 \parallel A_1U,$ $B_0A_3 \parallel A_1V,$ $B_0B_2 \parallel A_1C_2$ and $B_0B_1 \parallel A_1E,$ it follows that pencils $A_1(U,V,E,C_2)$ and $B_0(C_3,A_3,B_1,B_2)$ are similar. Therefore, cross ratio $(C_3,A_3,B_1,B_2)$ is harmonic  $\Longrightarrow$ $\frac{_{{\overline{B_1A_3}}}}{^{{\overline{B_1C_3}}}}=-\frac{_{{\overline{B_2A_3}}}}{^{{\overline{B_2C_3}}}}.$

Maybe there is a typo : [quote]Let lines $A_1C_1$ and $AA_1$ cut $BB_0$  at $ E,V ,$ respectively[/quote]
 it should be $A_1A_2$ cut $BB_0$ at $V$ . :maybe:

let BB_0 intersect C_1C_2 , A_2A_1 , C_1A_1 , A_2C_2 at points X , Y , P , Q
Easy to see that PQ || C_1A_2 || A_1C_2 , so YD = DX , make homotety with center B_2 and k = B_2B_1/B_2D and let Y -> Y' , X -> X' , so Y'B_0 = B_0X' and B_2B_0/B_0Y' = B_2A_3/B_0A_3 = B_2B_0/B_0X' = B_2C_3/C_3B_1 . done


Hi everybody.
I am mad on Bantumi. Can you play it with me for me please. If you don't know how to play, I can post all the rules for you, okay?
Thanks.

oh you mean the thing where you have

(  ][4][4][4][4][4][4][4][  )
(  ][4][4][4][4][4][4][4][  )

where the number represents hte nmber of stone and you have to pick the stuff up and get as many in your pile as possible.

there might be an online version like on yahoo or something

[quote="moogra"]Oh yeah I remember this game. Once I palayed online and was winning so the other guy quit.
at any rate, I'm not really good, does this game have any strategy?[/quote]
yes strategy but only on the human level, i.e. it has been analytically "solved" by computers.

No way!!! They're finding "winning strategies" for, like every game there is, like Connect Four, Checkers, Chomp, and a whole bunch of others.
(Except chess  :P )

I used to be good at mancala, but there really wasn't a strategy with me, you just tried to win.  Maybe that's a good strategy...

Do you mean play online or what?

Yeah! I heard that near where I live this guy proved that Chomp can have a winning strategy for his science project or something.

If you don't know what Chomp is, [url]http://en.wikipedia.org/wiki/Chomp[/url] should help you some.

The strategy stealing argument that Chomp is a first player win is well known. Finding the actual strategy is an open problem.

No, but I think he derived the strategy too.

Talk about a science project...   :surrender: