This problem be constructed from 2 lemma (I think it' very useful):
Lemma 1(I think it' very useful): Let $ ABCD$ is a cyclyc quadrilatel in $ (O)$. $ AB$ intersects $ CD$ at $ E$, $ AD$ intersects $ BC$ at $ F$, and $ AC$ intersects $ BD$ at $ R$. We have:
a) $ P_{E}^{2}\plus{}P_{F}^{2}\equal{} EF^{2}$
b) $ R$ is orthorcenter of $ \Delta OEF$.
Lemma 2 Let $ ABCD$ is a cyclyc quadrilatel in $ (O)$. $ AB$ intersects $ CD$ at $ E$, $ AD$ intersects $ BC$ at $ F$, the tangent at $ A$ and $ C$ intersects each other at $ M$, the tangent at $ B$ and $ D$ intersects each other at $ N$. We have: $ M, N, E, F$ are conlinear.
(This lemma can prove by easy way if we use Pascal theory)
Solution for the problem: Call the intersection of $ (O)$ with $ AB, BC, CD, DA$ are $ M, N, P, Q$ respectively. Use two lemma we get the result.

I think it is easy  :)
    when I saw this problem I solved it in ten minutes :lol:
   I saw a very easy problem in last January(or before this exam):
   Let ABCD be a quadrilateral and a circle W is inscribed in it.Let K is the tangent point of AB to W,L is the tangent point of BC to W,M is the tangent point of CD to W,N is the tangent point of DA to W.Show that AC,BD,KM and LN are concurrent.
   In the problem of Turkey TST this problem and inversion finish the proof. :lol:

I think I saw Darij post this before using something called Newton's theorem. It says that if W,X,Y, and Z are the points of tangency of the circle to AB, BC, CD, and DA, then K is also the intersection of WY and XZ.
Anyways, if this is true then ZX is the polar of Q wrt O and WY is the polar of P wrt O so since K is the intersection of both polars, PQ is the polar of K wrt O, which completes the proof.
EDIT: 1000th post!
Also, a nice proof is given here [url]http://www.cut-the-knot.org/Curriculum/Geometry/CircumQuadri.shtml[/url].

I solved with trigonometry

Let $ N$ is tangent point of  $ DC$ to $ K$

Let $ M$  is tangent point of $ AD$  to $ K$

Let $ L$  is tangent point of $ BC$  to   $ K$

Let $ W$ is tangent point of $ AB$ to $ K$

We know $ ML \minus{} WN \minus{} BD \minus{} AC$ are concurrent;

$ \angle OPC \equal{} \angle OPW \equal{} \angle ONW \equal{} \angle OWN \equal{} \alpha$

$ \angle MQO \equal{} \angle OQL \equal{} \angle OLM \equal{} \beta$

$ \angle PQC \equal{} x,\angle QPC \equal{} y$  Let $ OX\perp PQ$

$ \implies$  $ |OX| \equal{} \frac {rsin(\alpha \plus{} y)}{sin\alpha}$

$ \angle EON \equal{} \theta$  $ \implies$  $ |OE| \equal{} \frac {rsin\alpha}{sin(\alpha \plus{} \theta)}$

in $ \Delta AQP$;  $ sin\beta sin(\alpha \plus{} y) \equal{} sin\alpha sin(\beta \plus{} x)$

in $ \Delta NOL$ ; $ sin\beta sin(\alpha \plus{} \theta) \equal{} sin \alpha sin(x \plus{} y \minus{} \theta \plus{} \beta)$

$ \implies$  $ y \equal{} \theta$

$ \implies$  $ |OX|.|OE| \equal{} r^{2}$ $ \blacksquare$

[b]A step by step proof[/b]

Let the circle touch $AB$, $BC$, $CD$, $DA$ at $S$, $N$, $M$, $L$, respectively.
Let $R$ be the intersection of $LN$ ile $SM$.
Sine Theorem for $\triangle RDM$ and $\triangle RLD$ gives \[ \begin{align} \dfrac{RD}{DM} = \dfrac{\sin \angle RMD}{\sin \angle DRM} = \dfrac{\sin \angle RLD}{\sin \angle LRD} \tag {1} \end {align*}\]
Sine Theorem for $\triangle RBN$ and $\triangle RBS$ gives \[\begin{align} \dfrac{RB}{BN} = \dfrac{\sin \angle RNB}{\sin \angle BRN} = \dfrac{\sin \angle RSB}{\sin \angle SRB} \tag {2} \end {align*}\]
Since $\sin \angle RMD = \sin \angle RSB$ and $\sin \angle RLD = \sin \angle RNB$,  after multiplying $(1)$ and $(2)$, we have \[\begin{align} \dfrac{\sin \angle SRB}{\sin \angle BRN} = \dfrac {\sin \angle DRM}{\sin \angle LRD}  \tag {3} \end {align*}\] Then, $\angle SRB = \angle DRM$. So $D,R,B$ are colinear.
Similarly, $A,R,C$ are colinear. Thus, $R=K$.

Stewart for isosceles $\triangle QLN$ gives \[\begin{align} QN^2 - KL \cdot KN = QK^2  \tag {4} \end {align*}\] Stewart for isosceles $\triangle PMS$ gives \[\begin{align} PM^2 - KM \cdot KS = PK^2  \tag {5} \end {align*}\] Since power of $K$ is $KL\cdot KN = KS \cdot KM$, substracting $(5)$ from $(4)$ gives \[\begin{align} QK^2 - PK^2 = QN^2 - PM^2   \tag {6} \end {align*}\] Also $QN^2 = OQ^2 - ON^2$ and $OP^2 = OP^2 - OM^2$. So \[ \begin{align} QK^2 - PK^2 = OQ^2 - OP^2  \tag {7} \end {align*}\]  This is equivalent to $PQ \perp OK$. Let $OK \cap PQ = \{H\}$.
Since $HPMO$ is cyclic, $\angle OPM = \angle MHO$. We also have $\angle OPM = \angle KMO$. So $\angle OHM = \angle KMO$. This is equivalent to $OK\cdot OH = OM^2$. $\blacksquare$

[quote=orl]Quadrilateral $ ABCD$ has an inscribed circle which centered at $ O$ with radius $ r$. $ AB$ intersects $ CD$ at $ P$; $ AD$ intersects $ BC$ at $ Q$ and the diagonals $ AC$ and $ BD$ intersects each other at $ K$. If the distance from $ O$ to the line $ PQ$ is $ k$, prove that $ OK\cdot\ k \equal{} r^2$.[/quote]


se considera in plan C(O;R) si o dreapta d exterioara cercului. Sa se determine locul geometric al centrelor cercurilor tangente la C(O;R) si dreapta d.

intuitiv cred ca e o portiune dintr-o hiperbola, practic habar n-am.

Mie imi ies doua parabole.

[i]Cazul 1.[/i] Cercul (de centru $P$) e tangent la $d$ si tangent extern la cercul dat.

Il luam pe $O$ drept origine, iar axele le alegem astfel incat $d \| Oy$. Unitatea de masura = raza cercului dat.
Fie $x=t$ ecuatia lui $d$.
Noi trebuie sa avem $\textrm d (d,P) = OP-1$ (seamana foarte mult cu definitia parabolei, nu?), adica $x-t = \sqrt{x^{2}+y^{2}}-1$. Il trecem pe $1$ in stanga, ridicam la patrat si obtinem $y^{2}= 2x(1-t)+(1-t)^{2}= 2(1-t) \left( x+\frac{1-t}2 \right)$, o parabola. E destul de clar ca orice punct de pe aceasta parabola e centrul unui cerc tangent la $d$ si tangent extern la cercul dat.

[i]Cazul 2.[/i] Cercul (de centru $P$) e tangent la $d$, iar cercul dat e tangent intern la acesta.
E cam la fel, numai ca obtinem alta parabola.

si eu cred la fel
pentru cazul 1
construiesc dreapta $d'$ paralela cu $d$ astfel incat $d$ este "intre" $d'$ si $\mathcal{C}$, iar distanta dintre $d$ si $d'$ este chiar $R$. fie $T$ centrul cercului care este tangent la $\mathcal{C}$ si $d$, $r$ raza acestuia, si $X$ respectiv $X'$ proiectiile lui $T$ pe $d$ si $d'$. aceasta conditie (de tangenta) este echivalenta cu :
$TO=r+R$
si
$TX=R$.
din constructia lui $d'$ va rezulta $TX'=R+r$, adica $d(T,O)=d(T,d')$, deci parabola.
pentru cazul 2 analog, doar ca il luam pe $d'$ intre $\mathcal{C}$ si $d$.


function
real analysis
integration
limit
real analysis unsolved

[i](i) Define the set $ C(\mathbb{T}): \equal{} \left\{f\in C(\mathbb{R})|f(t \plus{} 1) \equal{} f(t)\mbox{ for all } t\in\mathbb{R}\right\}$ of all 1-periodic continuous functions on $ \mathbb{R}$. Describe how $ C(\mathbb{T})$ van be regarded as a subspace of $ (C[0,1],\|..\|_\infty)$. Verify that this subspace is closed, and hence also a Banach space.[/i]

My answer: for $ f\in C(\mathbb{T})$, look at the restriction $ \hat{f}: \equal{} f|_{[0,1]}$. Then $ \hat{f}$ contains all information about $ f$ because of 1-periodicy (i.e. for every $ x\in\mathbb{R}$, we can (re)construct $ f(x)$ from $ \hat{f}$). Now we have $ \hat{f}\in C[0,1]$.
But as for closedness, there actually is nothing to check I guess: conversely, every $ g\in C[0,1]$ can be continued in a 1-periodic fashion to all of $ \mathbb{R}$ becoming an element of $ C(\mathbb{T})$, and in this way the two spaces are the same. In other words, if I write $ \hat{C}(\mathbb{T}) \equal{} \{\hat{f} \equal{} f|_{[0,1]}|f\in C(\mathbb{T})\}$, then $ \hat{C}(\mathbb{T}) \equal{} C[0,1]$. Is this reasoning correct?

[i](ii) Verify that the convolution $ (f\star g)(t): \equal{} \int_0^1 f(t \minus{} s)g(s)\mbox{d}s$ of two 1-periodic continuous functions $ f,g\in C(\mathbb{T})$ is again a 1-periodic continuous function.[/i]
For all $ t\in\mathbb{R}$ we have $ (f\star g)(t \plus{} 1) \equal{} \int_0^1 f(t \plus{} 1 \minus{} s)g(s)\mbox{d}s \equal{} \int_0^1 f(t \minus{} s)g(s)\mbox{d}s \equal{} (f\star g)(t)$. My concern is that here I only used the 1-periodicy of $ f$, and not of $ g$?

Thanks for the help.

[quote="BlueVelvet"]conversely, every $ g\in C[0,1]$ can be continued in a 1-periodic fashion to all of $ \mathbb{R}$ becoming an element of $ C(\mathbb{T})$, and in this way the two spaces are the same. In other words, if I write $ \hat{C}(\mathbb{T}) = \{\hat{f} = f|_{[0,1]}|f\in C(\mathbb{T})\}$, then $ \hat{C}(\mathbb{T}) = C[0,1]$. Is this reasoning correct?[/quote]I realize this is not correct; the values at 0 and 1 are not necessarily equal. So I think instead we have $ \hat{C}(\mathbb{T})=\{f\in C[0,1]\ |\ f(0)=f(1)\}$. To show closedness, it the following correct?
Take a Cauchy sequence $ f_n\in \hat{C}(\mathbb{T})$, so $ f_n(0)=f_n(1)$ for all $ n$. Since $ (C[0,1],\|..\|_\infty)$ is complete, there is a limit $ f\in C[0,1]$. The convergence $ f_n\to f$ is uniform, so certainly pointwise. This means $ f(0)=\lim f_n(0)=\lim f_n(1)=f(1)$, so $ f\in \hat{C}(\mathbb{T})$.


algebra
polynomial
function
Vieta
complex numbers

Given that -3+sqrt(7) and 1+2i are two of the roots of the polynomial function P(x)=3x^5+8x^4+ax^3+bx^2-74x-40, and that a and b are integers, find the other three roots.

Would 2 of the roots be -3-sqrt(7) and 1-2i, but how would I find the third?

Yes, that is known as the Complex Conjugate Root Theorem - If $ a \plus{} bi$ is a root of a polynomial with real coefficients, then so is $ a \minus{} bi$

[hide]Maybe you could multiply the conjugate roots and get nice numbers like 2 and 5, and then use polynomial division to factor those out.[/hide]

Too tired to think properly. Sorry...

[quote="xpmath"]Perhaps Vieta's would help... you know the sum of the other 4 roots.[/quote]

And the product, as well.

I wouldn't know any other way to do it.


Let $\displaystyle 1=d_{1}<d_{2}<...<d_{k}=n$ be the divisor of $n$.
Find the smallest possible of $n$ such that:
$\displaystyle d_4=4d_3-5d_2$ and $d_5=2d_2d_3d_4-1$.

For the begining you could try to prove that [tex]d_2, d_3, d_4, d_5[/tex] are prime numbers.

I don't think so Maja,at least that idea comes from my proof.
First,we can see that $d_2$ is a prime number.If $(d_2,d_4)=1$ then $d_2 \cdot d_4$ is a divisor of $n$.But $d_5 > d_2 \cdot d_4>d_4$.
So $d_4$ is divided by $d_2 \Rightarrow 4d_3$ is divided by $d_2$.
Take $d_4=b \cdot d_2$.
If $d_3$ is a multiple of $d_2$ then $d_3=a \cdot d_2$.
Since $d_3>a>d_1=1$ then $a=d_2 ; d_4>b>a=d_2 $then $b=d_3$.
Hence $d_2^3=4d_2^2-5d_2$.This equation have no root $d_2>1$.
That means $d_2|4 \Rightarrow d_2=2$
Since $d_3$ isn't a multiple of $d_2$ then $d_3$ is a prime number too.
If $(d_3,d_4)=1$ then $d_3 \cdot d_4$ is a divisor of $n$ and $ d_5>d_3 \cdot d_4>d_4$.
Hence $d_4$ is divided by $ d_3 \Rightarrow d_3|5 \Rightarrow d_3=5 \Rightarrow d_4=10 \Rightarrow d_5=199$.
So the least value of $n$ is $1990$. :)

My mistake  :blush:
I also got $d_2=2, d_3=5$ in my "proof" (which is a little bit longer than yours), but it is realy a mistery why I thought that it can not be.

Anyway, thanks for solution!


Each of 6 points in space is connected to the other 5 points by line segments. Each segment thus formed is colored green or purple. Show that it is impossible to color all the segments without forming a triangle in which all three segments are the same color. (AoPS Volume 1)

We wacht one point $A$. That point is connected with five others so of that five three must be same color by piginhole principle. Let's say that $AB,AC,AD$ are coloured with same let's say green. Then if one segment $BC,CD$ or $BD$ is coloured green we have monochromatic triangle but if all of them are coloured purple then triangle $BCD$ is monocgromatic. :D

Did you just copy that from AoPS volume 1? :P

:D I have known this problem for years :)


Henry's little brother has 8 identical stickers and 4 sheets of paper, each a different color. He puts all the stickers on the pieces of paper. How many ways are there for him to do this, if only the number of stickers on each sheet of paper matters?

The stated solution used casework, but it also said that the solution equals $ \binom{11}{3}$, but doesn't explain why.
I know this probably has to do with balls and urns, but I've thought about it and can't figure out how to make it work.

Can someone explain the balls and urns solution?

Call the number of stickers on the first sheet of paper $ a_1$, the number on the second sheet of paper $ a_2$, the number on the second sheet of paper $ a_3$, and the number on the second sheet of paper $ a_4$

Since only the number of stickers on the piece of paper matters, and there are 8 total stickers, the number of ways to do this is in 1-1 correspondence with the number of nonnegative integer solutions to \[ a_1\plus{}a_2\plus{}a_3\plus{}a_4\equal{}8\].

If we define $ b_i: \equal{}a_i\plus{}1$ this is also in 1-1 correspondence with the number of positive integer solutions to \[ b_1\plus{}b_2\plus{}b_3\plus{}b_4\equal{}12\].

Express 12 as $ (1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1)$. To get an ordered quadruplet in the form $ (b_1, b_2, b_3, b_4)$, we need three of these 11 plus signs to be changed into commas. So the answer is $ \binom{11}{3}$.

we have 8 Strikers that we need to divide into 4 groups. Lets add 3 partitions to 8 stickers  (partitions are identical). Now if we do all arrangements of these 11 things, the partitions will divide the 8 stickers into 4 groups and thus give us all possible ways of pasting them on 4 pages. This arrangement can be done Pn 11!/(8!x3!) ways. This is equal to  11[size=200]C[/size]3

I have a question similar to that of andersonw.

I tried to use the Hockey-Stick Identity, where $n=8$ and $r=4$. So:

$ \dbinom {n+r+1}{n+1} = \dbinom {13}{9} = 715 $

What doesn't this work?

This relies on something called Balls and Urns.

Look to this page for more reference: [url]https://www.artofproblemsolving.com/Wiki/index.php/Distinguishability[/url]

Is there a way to do this without listing out the possibilities?

I don't think so, but it is good casework practice.

Actually, as referenced above, you can use Balls and Urns to find that the answer is equal to $ \binom{11}{3}$, or $\boxed{165}$.

Hint:
[hide]Use the ‘Balls and Boxes’ formula, which is $\binom{n + k - 1}{k - 1}$, where $n$ is the balls and $k$ is the boxes.[/hide]
Using the formula, we can understand why it's 11 Choose 3.

The reason you can use Balls and Urns or whatever it is called is because  you have indistinguishable balls and distinguishable boxes, which allows for the $\binom{n + k - 1}{k - 1}$.

[quote=mathdate]we have 8 Stickers that we need to divide into 4 groups. Lets add 3 partitions to 8 stickers  (partitions are identical). Now if we do all arrangements of these 11 things, the partitions will divide the 8 stickers into 4 groups and thus give us all possible ways of pasting them on 4 pages. This arrangement can be done Pn 11!/(8!x3!) ways. This is equal to  11[size=200]C[/size]3[/quote]

[quote=andersonw]The solution equals $ \binom{11}{3}$, but doesn't explain why.
[/quote]

OK, [hide=here's]There are 8 "balls" and 3 dividers to make slots for them. That's a total of 11 symbols. Go back to the word-arranging thing and you'll have $\binom{11}{3}$.[/hide] a primitive and quite simple method to do these sorts of questions. No formulas included, just simple combination basics!  :thumbup:

[quote=exmath89]The formula is actually $\dbinom{n+r-1}{r-1}$.[/quote]

that's not true. ahaanomegas has the correct formula.

[quote=dysfunctionalequations]

If we define $ b_i: \equal{}a_i\plus{}1$ this is also in 1-1 correspondence with the number of positive integer solutions to \[ b_1\plus{}b_2\plus{}b_3\plus{}b_4\equal{}12\].

Express 12 as $ (1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1)$. To get an ordered quadruplet in the form $ (b_1, b_2, b_3, b_4)$, we need three of these 11 plus signs to be changed into commas. So the answer is $ \binom{11}{3}$.[/quote]

I don't see why you have to switch in your third paragraph. Can't you still do the same reasoning?

4 urns, so basically you can divide 3 dividers to get four sections(urns) like o|o|o|o and since there are 8 stickers(balls) then there are (8+3)! / (8! * 3!) ways or 11c3 ways. 165.

These problems are always the hardest...
I missed the answer by flipping 1...

Here's why:
[hide=Why it's 11C3]Use the stars & bars method.
Each sticker represents a star and each paper represents a bar.
For example, a possible arrangement would be:
*****||***|
This corresponds to the arrangement (5, 0, 3, 0).
There are eleven spots to place items, and 3 of them are bars.
Thus, the answer is 11c3.[/hide]

Here's a simplification of boing123's explanation:
[hide=Solution]Imagine we have to place 3 dividers to make 4 groups (for the sheets of paper), like this: _S_SS_SSSSS. Because there are 3 dividers and 8 stickers, we just count the number of ways to order them, which gives us the Hockey Stick Identity, leading us to the final answer of $\boxed{165}$.[/hide]

Not a coincidence. See [hide]Chapter 13 in Intro to Counting and Probability, 13.4 [hide]I used [hide]stars & bars: 8 stickers and 4 sections : XX sec 1|XX sec 2 |XX sec 3 |XX sec 4 Then there is $\binom{8+4-1}{4-1}$ since 4-1=3 |'s are used.[/hide][/hide][/hide]

[quote=bobert1]Without considering the color of the paper, we can put the stickers on the papers in the following combinations:
 \begin{align*}
& (8,0,0,0) \\
& (7,1,0,0) \\
& (6,2,0,0) \\
& (6,1,1,0) \\
& (5,3,0,0) \\
& (5,2,1,0) \\
& (5,1,1,1) \\
& (4,4,0,0) \\
& (4,3,1,0) \\
& (4,2,2,0) \\
& (4,2,1,1) \\
& (3,3,2,0) \\
& (3,3,1,1) \\
& (3,2,2,1) \\
& (2,2,2,2).
\end{align*}
However, we must take into account the fact that the sheets of paper are different colors. To find the total number of arrangements, we find the number of arrangements for each grouping:
 For $(8,0,0,0),$ there are $\dfrac{4!}{3!}=4$ ways.

 For $(7,1,0,0),$ we have $\dfrac{4!}{2!}=12$ ways.

 For $(6,2,0,0),$ there are $\dfrac{4!}{2!}=12$ ways.

 For $(6,1,1,0),$ there are $\dfrac{4!}{2!}=12$ ways.

 For $(5,3,0,0),$ we have $\dfrac{4!}{2!}=12$ ways.

 For $(5,2,1,0),$ there are $4!=24$ ways.

 For $(5,1,1,1),$ there are $\dfrac{4!}{3!}=4$ ways.

 For $(4,4,0,0),$ we have $\dfrac{4!}{2!2!}=6$ ways.

 For $(4,3,1,0),$ there are $4!=24$ ways.

 For $(4,2,2,0),$ there are $\dfrac{4!}{2!}=12$ ways.

 For $(4,2,1,1),$ we have $\dfrac{4!}{2!}=12$ ways.

 For $(3,3,2,0),$ there are $\dfrac{4!}{2!}=12$ ways.

 For $(3,3,1,1),$ there are $\dfrac{4!}{2!2!}=6$ ways.

 For $(3,2,2,1),$ we have $\dfrac{4!}{2!}=12$ ways.

 For $(2,2,2,2),$ there are $\dfrac{4!}{4!}=1$ way.

In total there are $\boxed{165}$ different ways.[/quote]

You almost verbatim copied the Alcumus official solution.

[hide=Post #29 by smileapple][url=aops.com/community/user/367773][b]smileapple[/b][/url] · Yesterday at 11:33 PM [url=aops.com/community/p20691961](view)[/url][color=transparent]helo[/color]
Hello? Hello?

-----------
[color=#5b7083][aops]x[/aops] 29[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@smileapple][img]https://avatar.artofproblemsolving.com/avatar_367773.jpg?t=1605332113[/img]
AoPS User[/tip] wut

[hide=Alcumus Solution]Ignoring the different colors of paper, we can put the stickers on the sheets of paper in the following groups:\begin{align*}
& (8,0,0,0) \\
& (7,1,0,0) \\
& (6,2,0,0) \\
& (6,1,1,0) \\
& (5,3,0,0) \\
& (5,2,1,0) \\
& (5,1,1,1) \\
& (4,4,0,0) \\
& (4,3,1,0) \\
& (4,2,2,0) \\
& (4,2,1,1) \\
& (3,3,2,0) \\
& (3,3,1,1) \\
& (3,2,2,1) \\
& (2,2,2,2).
\end{align*}For each of these combinations, we will list how many distinct ways there are to put the groups of stickers on the different sheets of paper.

$\bullet$ For $(8,0,0,0),$ there are $\dfrac{4!}{3!}=4$ ways.

$\bullet$ For $(7,1,0,0),$ we have $\dfrac{4!}{2!}=12$ ways.

$\bullet$ For $(6,2,0,0),$ there are $\dfrac{4!}{2!}=12$ ways.

$\bullet$ For $(6,1,1,0),$ there are $\dfrac{4!}{2!}=12$ ways.

$\bullet$ For $(5,3,0,0),$ we have $\dfrac{4!}{2!}=12$ ways.

$\bullet$ For $(5,2,1,0),$ there are $4!=24$ ways.

$\bullet$ For $(5,1,1,1),$ there are $\dfrac{4!}{3!}=4$ ways.

$\bullet$ For $(4,4,0,0),$ we have $\dfrac{4!}{2!2!}=6$ ways.

$\bullet$ For $(4,3,1,0),$ there are $4!=24$ ways.

$\bullet$ For $(4,2,2,0),$ there are $\dfrac{4!}{2!}=12$ ways.

$\bullet$ For $(4,2,1,1),$ we have $\dfrac{4!}{2!}=12$ ways.

$\bullet$ For $(3,3,2,0),$ there are $\dfrac{4!}{2!}=12$ ways.

$\bullet$ For $(3,3,1,1),$ there are $\dfrac{4!}{2!2!}=6$ ways.

$\bullet$ For $(3,2,2,1),$ we have $\dfrac{4!}{2!}=12$ ways.

$\bullet$ For $(2,2,2,2),$ there are $\dfrac{4!}{4!}=1$ way.

In total, there are$$4+12+12+12+12+24+4+6+24+12+12+12+6+12+1=\boxed{165}$$ways for Henry's brother to put the stickers on the sheets of paper.

Notice that this answer equals $\binom{11}{3}.$ Is this a coincidence?[/hide]

[hide=My Solution]Use stars and bars, because we are getting 8 indistinguishable stickers into 4 distinguishable papers. This means that the answer will just be 8+4-1c4-1=11c3=165.[/hide]

Why is the solution so long :dry:

AND I got a similar question right after this one:
Henry's little brother has $8$ identical stickers and $4$ identical sheets of paper. How many ways are there for him to put all of the stickers on the sheets of paper, if only the number of stickers on each sheet matters?

Why does Henry's little brother need to think so hard on how he places his stickers?

Anyways, it is still very annoying, counting. Are you sure there's another alternative solution?

- If this ever appears on a math competition, I'm doomed because I am very bad at counting all possibilities.


Let $a,b,c$ be non-negative numbers such that $a+b+c=2$.
Find the maximum value of
$E=\frac{bc}{a+r}+\frac{ca}{b+r}+\frac{ab}{c+r}$,
where $r$ is a given positive number.

Hi, Vasc. It's nice, but I think It's not too difficult. Isn't it?  :lol:

Here is my solution I only prove for $k=2$. In the other case, The prove is only diffirence a bit.

$\sum\frac{ab}{c+2} \le \frac{1}{2}$

$\Leftrightarrow 2\sum a^2b^2+4\sum ab(a+b) + 8abc \le abc+2(ab+bc+ca)+16$

$\Leftrightarrow 2(ab+bc+ca)^2+14(ab+bc+ca) \le 16+21abc$

$\Leftrightarrow (ab+bc+ca-1)(ab+bc+ca+8) \le 21abc$

Apply the inequality :

$(a+b-c)(b+c-a)(c+a-b) \le abc \Rightarrow ab+bc+ca-1 \le \frac{9}{8}abc$

And notive that $ab+bc+ca \le \frac{4}{3}$

We are done. The equal takes with $a=b=c=\frac{2}{3}$ or $a=b=1,c=0$.

Nice solution for $r=2$ !

But what is the maximum value?

It's $max(\frac{1}{k},\frac{4}{2+3k})$ If my computing is not wrong.


The expression $ a^3 \minus{} a^{ \minus{} 3}$ equals:

$ \textbf{(A)}\ \left(a \minus{} \frac {1}{a}\right)\left(a^2 \plus{} 1 \plus{} \frac {1}{a^2}\right) \qquad\textbf{(B)}\ \left(\frac {1}{a} \minus{} a\right)\left(a^2 \minus{} 1 \plus{} \frac {1}{a^2}\right)$
$ \textbf{(C)}\ \left(a \minus{} \frac {1}{a}\right)\left(a^2 \minus{} 2 \plus{} \frac {1}{a^2}\right) \qquad\textbf{(D)}\ \left(\frac {1}{a} \minus{} a\right)\left(a^2 \plus{} 1 \plus{} \frac {1}{a^2}\right)$
$ \textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
$ a^3\minus{}a^{\minus{}3}\equal{}a^3\minus{}\frac{1}{a^3}\equal{}\left (a\minus{}\frac{1}{a} \right ) \left (a^2\plus{}1\plus{}\frac{1}{a^2} \right )$, or $ \boxed{\textbf{(A)}}$.
[/hide]


Let  :tri:  ABC  be in condition:

 sinA= (sinB+sinC)/(cosB+cosC)    (This is a fraction)

 What is the kind of this triangle?[/hide]

The question has already been answered on the other subforum.


This belongs in Intermediate, not HSB.

Anyway, it's basically roots of unity.

[quote="ashrafmod"]$\cos (\frac{2\pi}{21})+\cos (2.\frac{2\pi}{21})+\cos (4.\frac{2\pi}{21})+\cos (5.\frac{2\pi}{21})+\cos (8.\frac{2\pi}{21})+\cos (10.\frac{2\pi}{21}) =\frac{1}{2}$[/quote]
Let $z=e^{i\frac{2\pi}{21}}$, and $\zeta_{k}=\cos\frac{2k\pi}{21}=\frac{z^{2k}+1}{2z^{k}}$. Note that by simplifying the following, you obtain the required result.
\[\zeta_{1}+\zeta_{2}+\zeta_{4}+\zeta_{5}+\zeta_{8}+\zeta_{10}=\frac{1}{2}. \]
Masoud Zargar

[hide="Solution"] Let $\omega = e^{i \frac{2\pi}{21}}$ be a primitive twenty-first root of unity.  Then we want to evaluate

$R = \Re \left( \omega+\omega^{2}+\omega^{4}+\omega^{5}+\omega^{8}+\omega^{10}\right)$

Notice that these are the first half of the primitive roots $\bmod 21$.  The other units have the same real part; in other words,

$\Re \left( \omega^{k}\right) = \Re \left( \omega^{21-k}\right)$

So that

$2R = \Re (z)$

Where $z = \omega+\omega^{2}+\omega^{4}+\omega^{5}+\omega^{8}+\omega^{10}+\omega^{11}+\omega^{13}+\omega^{16}+\omega^{17}+\omega^{19}+\omega^{20}$ is precisely the sum of the primitive roots $\bmod 21$.

What's missing?  At this point, we have removed exactly the multiples of $3$ and $7$; in other words,

$z = \left( \sum_{k=0}^{20}\omega^{k}\right)-\left( \omega^{0}+\omega^{3}+...+\omega^{18}\right)-\left( \omega^{0}+\omega^{7}+\omega^{14}\right)+1$

The first expression is the sum of all of the twenty-first roots of unity, the second the sum of all of the seventh roots of unity, and the third the sum of all of the third roots of unity.

Now, since $x^{n}-1 = 0$, by Vieta's, all of those expressions are equal to zero.  Therefore $z = 1$ and $\Re(z) = 2R = 1 \Leftrightarrow R = \frac{1}{2}$. QED. [/hide]

Edit:  [hide="Oh... of course..."] There is a theorem which states that the sum of the primitive roots $\bmod n$ is $\mu(n)$.  Of course, $\mu(21) = 1$.   :blush: [/hide]


I have 2 equations.
One is multiple lines, and I used \begin{eqnarray}
the other one is one line, and I used\begin{equation}

The output shows the right formats of each equation.
But the problem is: The two equations have different alignments. It looks like:
v1=xxxxxxxxxxxxxxxxxxxxxxxx         (1)
            v2=*******   (2)

equation (2) is like centered.

How to deal with this?

Thanks

eqnarray is obsolete as it lays out equations badly (see [url=http://tug.org/pracjourn/2006-4/madsen]Avoid eqnarray![/url]) so replace it with the align environment (which requires the amsmath package).
BUT the equation environment does not centre the label as you have it, so post a complete minimal example to find out why it is doing this.

Thanks! Align and split  two commands work!

Hi weng,

perhaps have a quick look at [url=http://texblog.net/latex-archive/maths/eqnarray-align-environment/]eqnarray vs. align[/url].
Concerning mathematical typesetting, here's a list of tutorials and documentation: [url=http://texblog.net/latex-link-archive/mathematics/]Mathematics with LaTeX[/url].

Stefan


$ \texttt{if a,b,c are three positive numbers, show that: }$
\[ \frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\geq \frac{9}{4(a+b+c)}
\]
$ (a-b)^2\geq 0\\[8pt]
a^2+b^2 \geq 2ab\\[8pt]
(a+b)^2 \geq 4ab\\[8pt]
\texttt{then}\\[8pt]
\frac{c}{(a+b)^2}\leq \frac{c}{4ab}\\[8pt]
\texttt{so in that way we got: }\\[8pt]
\frac{a}{(b+c)^2}\leq \frac{a}{4bc}\\[8pt]
\frac{b}{(a+c)^2}\leq \frac{b}{4ac}\\[8pt]
\texttt{then i sumed the inequalities}\\[8pt]
\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\leq \frac{b}{4ac}+\frac{c}{4ab}+\frac{a}{4bc}\\[8pt]
=\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\leq \frac{b^2}{4abc}+\frac{c^2}{4abc}+\frac{a^2}{4abc}\\[8pt]
\texttt{from the very known inequality}\\[8pt]
a^2+b^2+c^2\geq ab+bc+ac\\[8pt]
\frac{a^2+b^2+c^2}{abc}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.....\beta\\[8pt]
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}....\alpha\\[8pt]
\alpha .... \beta\\[8pt]
\frac{a^2+b^2+c^2}{4abc}\geq \frac{9}{4(a+b+c)}$
so i got both LHS and RHS of the inequality but i can't demonstrated it
i need help

Well, your solution isn't correct because yo have acquired:
$ \frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\leq  \frac{a^{2}+b^{2}+c^{2}}{4abc}\geq  \frac{9}{4(a+b+c)}$ which is not sufficient to prove our inequality.

Proof:
Rewrite the inequality in the form:
$ \frac{x+z-y}{2y^{2}}+\frac{x+y-z}{2z^{2}}+\frac{y+z-x}{2x^{2}}\geq \frac{9}{2(x+y+z)}$
Where:
$ x=a+b \\ y=b+c \\ z=c+a$
It's equal to:
$ \frac{(x+z)}{y^{2}}+\frac{(x+y)}{z^{2}}+\frac{(y+z)}{x^{2}}-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq \frac{9}{x+y+z}$
Now multiply by $ x+y+z$ and what will be to prove is:
$ (\frac{(x+z)}{y})^{2}+(\frac{(x+y)}{z})^{2}+(\frac{(y+z)}{x})^{2}\geq 12$
The last inequality is true, beacause we have the inequality beetwen quadratical and arithmetical mean, transforming the last inequality it suffices to prove:
$ \sqrt{\frac{(\frac{(x+z)}{y})^{2}+(\frac{(x+y)}{z})^{2}+(\frac{(y+z)}{x})^{2}}{3}}\geq 2$
${ \sqrt{\frac{(\frac{(x+z)}{y})^{2}+(\frac{(x+y)}{z})^{2}+(\frac{(y+z)}{x})^{2}}{3}}\geq
\frac{(\frac{(x+z)}{y}))+(\frac{(x+y)}{z}))+(\frac{(y+z)}{x}))}{3}}\geq \frac{2+2+2}{3}=2$
beacause we have $ \forall_{x,y\in R_{+}} \ \frac{x}{y}+\frac{y}{x}\geq 2$
:wink:

Without loss of generality, suppose that $ a\plus{}b\plus{}c\equal{}3$.  Now observe that \[ \frac{(9\minus{}2a)(a\minus{}1)^2}{(4(a\minus{}3)^2}\ge 0\iff \frac{a}{(3\minus{}a)^2}\ge \frac{1}{2}\cdot (a\minus{}1)\plus{}\frac{1}{4}\] Then \[ \sum \frac{a}{(b\plus{}c)^2}\equal{}\sum \frac{a}{(3\minus{}a)^2}\ge \sum \frac{1}{2}\cdot (a\minus{}1)\plus{}\frac{1}{4}\equal{}\frac{3}{4}\equal{}\frac{9}{4(a\plus{}b\plus{}c)}\] as desired.

Could you please explain why we can assume that $ a \plus{} b \plus{} c \equal{} 3$ without losing the generality?  :huh:
Do you mean we can assume that, bacause the same method can be applied for all $ k\in R_{ \plus{} }$ such that: $ a \plus{} b \plus{} c \equal{} k$  :?:

If we replace $ a,b,c$ in the original inequality with $ \lambda a, \lambda b, \lambda c$, for $ \lambda$ a positive real number, both sides are multiplied by $ \frac{1}{\lambda}$ so the direction of inequality will not change under this operation.  Thus, for any $ a,b,c$, we can choose a $ \lambda$ such that $ a\plus{}b\plus{}c\equal{}3$ (or any other positive real that is convenient) and the direction of the equality will not change.  This is called homogenizing the inequality, I believe.

[quote]If we replace a,b,c in the original inequality with \lambda a, \lambda b, \lambda c, for \lambda a positive real number, both sides are multiplied by \frac{1}{\lambda} so the direction of inequality will not change under this operation. Thus, for any a,b,c, we can choose a \lambda such that a+b+c=3 (or any other positive real that is convenient) and the direction of the equality will not change. This is called homogenizing the inequality, I believe.[/quote]
Hi teki teki by homogenizing the inequality do you mean to choose a convinient value as in Limits by definition exercises where you choose a $ \delta$ convinient. Is that what you mean??

By using Cauchy-Shwarz inequality we have:

$ (\sum(\frac{\sqrt{a}}{b\plus{}c})^{2})(\sum(\sqrt{a})^{2})\geq(\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b})^{2}\geq (\frac{3}{2})^{2}\equal{}\frac{9}{4}$

(Last inequality is true because of well known Nesbitt's inequality)

[quote="asintota"]Hi teki teki by homogenizing the inequality do you mean to choose a convinient value as in Limits by definition exercises where you choose a $ \delta$ convinient. Is that what you mean??[/quote]

I don't see what this has to do with limits... he is just making a simplifying assumption. Let B be the statement that must be proven, and A be a simplified statement. He shows that if A, then B. Then he only needs to prove A, which is easier to prove than B.


The number $ 2008!$ (factorial $ 2008$) means the product of all the integers $ 1, 2, 3, 4,
. . . , 2007, 2008$. With how many zeroes does $ 2008!$ end?

the number if zero equals the biggest x that $ 10^x$ divides 2008!. Obviously there are more factors of 2 in 2008! than factors of 5, so we only need to count number of factors of 5.

among 1-2008, 401 numbers are multiple of 5, 80 numbers are multiple of 25, 16 numbers are multiple of 125, 3 numbers are multiple of 625. Thus, the number of factor of 5 in 1*2*3*...*2008 is 401+80+16+3=500

[quote="greentreeroad"]there are more factors of 2 in 2008! than factors of 5, [b]so we only need to count number of factors of 5.[/b][/quote]

Could you explain the bold part please.

Number of zeroes equivalent to the power of 10, which equals 2 times 5. But since there are many more 2's than 5's just count 5's (the extra 2's wont do anything to add 0's).

[hide="hint"]
Number of prime $ p$'s in $ n!$ is equal to $ \lfloor\frac{n}{p}\rfloor+\lfloor\frac{n}{p^2}\rfloor+\cdots+\lfloor\frac{n}{p^k}\rfloor$ where $ p^k$ is the highest power of p dividing $ n!$[/hide]
[hide="answer"]$ 500$[/hide]
[hide="solution"]
Well, we see that the no. of 5's dividing 2008 is 500.
As # of 5's dividing n! is definitely greater than the # of 2's :lol:
Hence # of ZERO's at the end of $ 2008!$ is 500.[/hide]


I want an academically challenging school.

Perhaps I'm somewhat (okay extremely biased): HOPKINS!!!!  If you want more info on Hopkins, just let  me know and I'll try to answer your questions.

OK, please cancel out Chase Collegiate, as I find this school extremely easy.

Are these colleges or parochial schools?  If you mean colleges, then go with Hopkins as I said earlier.

If you mean parochial schools, well then I have no idea.

Dude, you missed the best school on you list: Glastonbury High School!

Note: All schools listed are [b]PRIVATE[/b] schools.

What about Choate? Ur going to Taft though, right? lulz. im a bit late.  :lol:  :wink:

[hide]Hopkins is overrated  :D [/hide]


Gnorizei mipos kaneis na mou diatiposei ti leei to topologiko Limma tou Zorn?To xriazomai gia tin apodiksi tou Limmatos tou Whitney ...

Έψαξα λίγο στο internet και βρήκα κάποια πράγματα στη [url=http://en.wikipedia.org/wiki/Zorn%27s_lemma]Wikipedia[/url]. Έχει και 2 references, εάν δε σε πληρεί η Wikipedia, που μπορείς να δεις.

sta agglika to eixa vrei kai egw alla me diskolevei para poli na to katanoisw etsi kai giafto zitisa ama kapoios to gnorizei na to anaferei(sta ellinika) wste na to katalavw! thnx pantws ;)

oxi distixos den einai afto(nomizw dld) loipon tha grapsw pou to xriazomai mipos kai i ekfrasi topologiko limma einai lathos....

kanw ena eidiko thema epano stin megistiki sinartisi tou Hardy - Litltlewood kai panw stous idiazontes oloklirotikous telestes , gia olo afto xriazete ena lima tou whitney pou leei oti ama exeis ena klisto sinolo F mporeis na pareis mia oikogenia mpallwn B1,B2,...,Bk , ... tetoia wste : 1. Bk ksenes metaksi tous , 2. U4Bk = A = F simpliroma , 3. 8Bk tomi me to F na einai diaforo tou kenougia kathe k

Pano stin apodiksi pouxw , pernei ena simio x panw sto A kai thetw d(x) tin apostasi tou x apo to F , kai theorei tin sferiki perioxi B(x,d(x)/4) kai mia ipooikogenia aftis pou einai megisti estw B(x1,d(x1)/4) , ... ,B(xk,d(xk)/4) kai einai ksenes metaksi tous kai leei oti mia tetoia ipooikogenia iparxei apo to Lemma tou Zorn kai den anaferei kati allo ......


counting
triangle inequality
combinatorics
Triangle
IMO Shortlist
IMO Longlist

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

[quote=orl]Let $n$ be a positive integer. Find the maximal number of non-congruent triangles whose sides lengths are integers $\leq n.$[/quote]

So $a \le b \le c \le n$ and obviously $a+b>c$. Let $P_n(k)$ denotes the number of such triplets with $c=k,k=1,2,.....,n$. For $k$ even, $P_n(k)=k+(k-2)+(k-4)+...+4+2=\frac{k^2+2k}{4}$, and for $k$ odd, $P_n(k)=\frac{(k+1)^2}{4}$
So $P_n(1)+P_n(2)+.....+P_n(n)=\frac{n(n+2)(2n+5)}{24}$ when $n$ is even, and $\frac{(n+1)(n+3)(2n+1)}{24}$ when $n$ is odd.


When Monique visited Australia, the American dollar was equivalent to $ 1\frac{1}{2}$ Australian dollars.  She purchased a sweater priced at 30 Australian dollars.  What was the equivaelnt number of American dollars?

[quote="GameBot"]When Monique visited Australia, the American dollar was equivalent to $ 1\frac {1}{2}$ Australian dollars.  She purchased a sweater priced at 30 Australian dollars.  What was the equivaelnt number of American dollars?[/quote]

1 dollar = 1.5 Australian dollars, so the sweater is $ 30 \div 1.5 \equal{} \boxed{20}$ American dollars.


How to identify all the ions if the solution contains:

(1) $ \text{SO}_4^{2\minus{}}, \text{PO}_4^{2\minus{}}, \text{I}^\minus{}, \text{CO}_3^{2\minus{}}$

(2) $ \text{Ag}^\plus{}, \text{Pb}^{2\plus{}}, \text{Cu}^{2\plus{}}, \text{Cr}^{3\plus{}}, \text{Ba}^{2\plus{}}$

Those are all ions so I'm not really sure what you want.  Are you looking for the names of the ions?  In the first list, we have sulfate, phosphate, iodine, and carbonate.

I think he wants the qualitative tests :)

Some remarks first: $ \ce{PO4^2\minus{}}$ should be $ \ce{PO4^3\minus{}}$, phosphate ion; and $ \ce{I\minus{}}$ is iodide ion, not iodine.

Also, do you want qualitative tests or quantitative separation of the ions?

For part b) I think i know how to identify silver ion and lead ion. Here goes:
Put 6M HCl solution into the ions solution which may contain silver and/ or lead, you will see a white precipitate.
Centrifuge the solution, pour the supernant into another test tube, now you have a white precipitate (ppt) of silver and lead, right?
Add 1 ml of distilled water into the ppt, heat the test tube in the hot water bath with stirring for about 3 mins.
If Lead is present in the origional ions solution, it will be dissoved in hot water bath;    whereas the silver will not.
Confirmation of lead: pour off the supernant to another test tube. To this liquid, add 6 M of H2SO4 ( sulfuric acid), if there is a ppt of lead sulfate, it will prove that lead is present.
Confirmation of silver: from the ppt ( after boiling), add ammonia to the ppt till the ppt dissolves, then add 6M HCl. if there is a ppt form of HCl, it will prove that silver is present.

I know it is confusing but if you have any question, i am willing to answer. and i will think about how to identify the rest.  :)

[quote="Carcul"]Some remarks first: $ \ce{PO4^2 \minus{} }$ should be $ \ce{PO4^3 \minus{} }$, phosphate ion; and $ \ce{I \minus{} }$ is iodide ion, not iodine.

Also, do you want qualitative tests or quantitative separation of the ions?[/quote]
Thanks a lot, Carcul, for pointing out the typo! Yes, I want qualitative tests.

And for i_thirst_for_knowledge, thanks for your answers, they are great. And I'm thinking of adding $ \ce{H2SO4}$ to the liquid after we add $ \ce{HCl}$, then gives $ \ce{Ba^2\plus{}}$.

After that I'm thinking of using $ \ce{K4[Fe(CN)6]}$ to identify $ \ce{Cu^2\plus{}}$ and ether to identify $ \ce{Cr^3\plus{}}$.

There are generally dry tests and wet tests for anions. Ofcourse there are anions like $ \ce{SO4^{2\minus{}}}, \ce{PO4^{3\minus{}}}$ which don't have dry tests (as far as i know).

For wet tests, one needs to prepare the extract.

Preparation of extract: To a small amount of the salt, thrice the amount of $ \ce{Na2CO3}$ is added followed by 25-30 ml of distilled water and heated for 10-15 minutes and filtered. The filtrate is called sodium carbonate extract.

1. $ \ce{SO4^{2\minus{}}}$ : a. The extract is neutralised by dil. $ \ce{HCl}$ and $ \ce{BaCl2}$ solution is added. A white precipitate of $ \ce{BaSO4}$ is obtained.
$ \ce{Ba^{2\plus{}}} \plus{} \ce{SO4^{2\minus{}}} \rightarrow \ce{BaSO4}$
b. The extract is neutralised by acetic acid and lead acetate is added. A white precipitate of $ \ce{PbSO4}$ is obtained.
$ \ce{Pb^{2\plus{}}} \plus{} \ce{SO4^{2\minus{}}} \rightarrow \ce{PbSO4}$

Will post the rest later  :)

[color=green]Kent Merryfield, i don't think this lab would require an atomic absorption spectrometry. this quantitative analysis would be done in micro scale, using common chemical. that's what i think.  :) [/color]

[quote="i_thirst_for_knowledge"]shobber, i have a question. isn't chromate form precipitate when you add sulfuric acid to the solution?
i know BaSO4 is s white ppt but chromate sulfate is also a ppt, right?[/quote]
I think that $ \ce{Cr2(SO4)3}$ should be soluble. At least from what I did today, I detected $ \ce{Cr^3\plus{}}$ from the supernant (by adding $ \ce{H2O2}$ and $ \ce{NaOH}$) after adding sulfuric acid. But I don't know if my method of detecting $ \ce{Cr^3\plus{}}$ is alright......

[quote="Kent Merryfield"]
With respect to all of those cations ... your lab wouldn't happen to have an atomic absorption spectrophotometer, would it?
[/quote]
We don't have the atomic absorption spectrophotometer here. See, I'm not a chemistry major so I just do the most basic experiments, in the test tubes.


Yeah. And I was the 26th (with 15 others) top scoring girl on AIME I. Whee~

But with the top 7th grader being Canadian, you can say that you're the 3rd in the US.

tied for 29th in AIME1.... for sophs.

now we call THAT a mediocrity.

I don't think that making USAMO should be considered mediocre under any standards.

[quote="nr1337"]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2005-aa/05aimestats.html

While I'm still disappointed with my score, at least I can say that I was the 4th top scoring 7th grade on the  AIME I. :D[/quote]
Only 4th? Wow, sounds like you need to step it up a notch and stop slacking :P What is your USAMO index?
Great job, everyone -- especially all the pre-high school students who participated in the AMCs and AIME. Extra kudos to everyone who qualified for the USAMO. ;)

I scored below average for 8th graders (I think).  But what's weird is that I did better compared to everyone than to just 8th graders alone.

So who qualifies for the USAMO?  It looked like there were alot less high scores on the AIMEII.

[quote="ZennyK"]What is your USAMO index?[/quote]

Ugh.  170.5  *Many* stupid mistakes on the AMC12 dropped my index a lot...

Maybe next year... practice practice practice.... My USAMO index in practicing is usually around 230, I guess it's just one of those things about test-taking.


Source: Alternate AIME 2003

Find the largest possible integer n such that sqrt[n] + sqrt[n+60] = sqrt[m] for some non-square integer m.

Sqrt(n) + Sqrt(n+60) = Sqrt(m)

Squaring, we have

2n + 60 + 2*Sqrt(n^2 + 60n) = m
2*Sqrt(n^2 + 60n) = m - 2n - 60

Squaring again,

4n^2 + 240n = m^2 + 4n^2 + 3600 - 4mn - 120 m + 240n
0 = m^2 + 3600 - 4mn - 120m
n = (m^2 - 120m + 3600)/(4m) = m/4 - 30 + 900/m

Since we want to maximize n, we must maximize m (not sure how to prove that we must maximize m, can anyone enlighten me?) such that n is an integer. Thus, m/4 and 900/m must be integers, so 4|m and m|900. Let us try the largest factors of 900. m != 900 because 900 is a perfect square, as 900 = 30^2. m != 450 because 4 does not divide 450. m = 300 does work, however, since 4|300 and 300 is not a perfect square. Substituting, we find that n = 48.

How to prove that we want to maximize m:

Let f(m) = m/4 -30 + 900/m.

It can be found by differentiation that the minimum of f(m) is at m=60, as m increases from there, f(m) is increasing. If m starts at 60 and gets smaller, f(m) is also increasing, with an asymptote at m=0. We can therefore check the largest non-square m less than 900 that is divisible by 4, which is 300 and yields n=48. We must also check the smallest positive non-square m divisible by 4, which is 12, and this also yields n=48. Therefore, n can be no larger than 48.

[quote]How to prove that we want to maximize m:

Let f(m) = m/4 -30 + 900/m.

It can be found by differentiation that the minimum of f(m) is at m=60, as m increases from there, f(m) is increasing. If m starts at 60 and gets smaller, f(m) is also increasing, with an asymptote at m=0. [/quote]

Wait, I am a bit confused. You state that there is a local minimum (and in fact, a minimum for m >= 0), but how does that prove that we must maximize m?

[quote]Wait, I am a bit confused. You state that there is a local minimum (and in fact, a minimum for m >= 0), but how does that prove that we must maximize m?[/quote]

For all m > 60, f(m) is increasing, so we want to take the largest possible m. I forgot to say that we are also trying to minimize m, since when m starts at 60 and gets smaller, f(m) is also increasing (there is an asymptote at m=0). We therefore want to take the largest <i> and </i> smallest positive values of m (sorry about that), which are 300 and 12. Both give n=48.

Ah, ok. I understand better now. Thanks.

By the way, concerning your following statement:

[quote]We must also check the smallest positive non-square m divisible by 4, which is 12[/quote]

Shouldn't it be 8 = 4 * 2 and not 12 = 4 * 3, since 8 is not a perfect square (8 = (2*Sqrt2)^2)?

Well, it would be, except 900 is not divisible by 8.

I found out something rather interesting, but I don't know if this will get anywhere, but:

you can rewrite n=a*k^2
and                 n+60=a*p^2

Substituting for a into the 2nd equation and then solving for n, you get that n=60*k^2/(P^2-k^2). It just so happens that the maximum n occurs when p^2-k^2 is the largest odd factor of 60 that satisfies the given conditions (5 in this case, since if p^2-k^2=15, m would be a perfect square.)

My question is: is there a reason for this?

p^2 - k^2 = 5 is an interesting theory, and if it were true one could only speculate as to the reason for its truth. It would truly be a testament to the fact that mathematics is beautiful. Unfortunately, p^2 - k^2 = 20.

wat r u talking about?

p^2-k^2=5, where p=3, k=2

n = 48, a = 3, k = 4, p = 6. That's Wat I R Talking About.

I assume you are assuming a to be the largest possible quadratfrei integer value. If k = 2, p = 3, a = 12. 12 contains 4 = 2^2.


[quote="ngoclam1088"]
\[ {x}^{5} \plus{} \frac {x}{\sqrt {x^2 \minus{} 2}} \minus{} 2008 \equal{} 0\]
[/quote]

Are you sure that the problem statement is correct ?

There is a unique solution to this equation, which is about $ \sqrt 2\sqrt{1\plus{}\frac 1{2008^2}}\sim 1.414213739...$ and I'm not sure there is a normal way to find a closed form for this result.  :huh:

[quote="Bugi"]There are 2 real and 4 complex solutions to this:

Real are $ 1.41421...,4.57622...$[/quote]

You're right, I missed this second real number.


Their ratio of speed is 1:2. When they first met, they travelled a total of 2 times the distance between the shopping center and his home. So they met at $\frac{2}{3}$  of the total distance from the shopping center to his home. The same things occurs for the second time (they meet at $\frac23$ of the distance between where they first met and home). So the answer is
$ 36 \cdot \frac13 \cdot \frac23  = 8$

I think it is $32$ minutes, that is, $4$ minutes from his house.

[quote="riddler"]Bill and his dog walk home from the shopping center. It takes Bill 36 minutes
and his dog walks twice as fast. They start together, but the dog reaches home
before Bill and returns to meet Bill. After meeting Bill, the dog walks home,
again at double speed, and then turns back to meet Bill again. Bill starts at
noon to walk home. How many minutes later does he meet the dog for the
second time?

Ive got the answer but any cool solutions :gleam:[/quote]

Just guessing here ,

first the dog uses 18min to walk home and return to meet Bill 18*1/3= 6min after .Then the dog walk back again in 1/2*(36-18-6)=6min and return to Bill after 1/3*(36-18-6-6)=2min later . So they will meet again after 18+6+6+2=32min

I understand how you guys have got $32$ but the $2$nd time means after he has come from his house to Bill not going home after that.
So the answer is $24$.

I got mine the same way as shyong.

I think the question could have been more clear, as both of us apparently misread it.

I also thought the same as you guys in the $1$st place but it was a multiple choice question so then I understood what they meant.

I thought the question is asking how many minutes after they first met did they meet for the second time :P

24 is correct.

When the dog reached the house the first time, the man had walked 18 minutes. Then the dog walked back, and met the man 2/3 of the way from his house to the halfway point (since dog walked twice as fast, and 2x + 1x = 3x). The dog goes 2/3 of halfway (or 1/3) in 1/3*18 or 6 minutes. So 18+6=24.
But that's assuming when they started, they've already met.


Ring Theory
calculus
integration
algebra
function
domain
abstract algebra

Is a subring of an artinian ring again an artinian ring ?

No: With the criterion that an integral domain is Artinian iff it is Noetherian of dimension 0 one of the two conditions might fail to descend. For one of the properties take $ K[X]$ sitting in the quotient field $ K(X)$.


Let  a 1987-gonon and the numbers 1,2,...,3974 .
In each side of the polygonon we put 3  of these numbers (two numbers in the vertices and one in the interior of the side ) .
No number is putting two times .
There is a way to put the numbers such that in each side the sum of the 3 numbers is the same ?

EDITED: Yes, you're right.. i think i was out of my mind  :blush: .. ok then...

Suppose we have a $p$-gonon, where $p$ is prime, it's easy to show that, labelling the vertex with $a_{1},a_{2},\cdots,a_{p}$, the sum $\sum_{i}a_{i}$ must be  $\equiv 0\pmod p$.

Thus we can proceed in this way.. Let's say that $a_{1}\equiv 1\pmod p, a_{2}\equiv 2\pmod p$ and so on.

We are now allowed to choose the $a_{i}$'s in this way:

$\{\begin{array}{c}{a1}a_{i}=i \mbox{ if i is odd}\ a_{i}=i+p \mbox{ if i is even}\end{array}$

It's now easy to check that all the odd numbers are selected. Moreover we have

$a_{p}+a_{1}=a_{1}+a_{2}-2=a_{2}+a_{3}-4=\cdots=a_{p-1}+a_{p}-2*(p-1)$

Since all the sums on the sides differ one to another by $2$, it sufficies to put the $p$ even numbers left, starting with the biggest in the side whose vertices are $a_{1}, a_{p}$ and proceeding clockwise.

[hide="click to reveal"]
An example that this works for example for $p=7$ is to put the numbers in this order  $1,12,9,10,3,8,11,6,5,4,13,2,7,14$  :)
[/hide]

I think you are somewhere wrong because there exist  such labeling of the numbers .


calculus
integration
function
real analysis
real analysis unsolved

Let $ (f_n)$ a bounded R-integrables in $ [a, b]$ functions succession.

Let $ F_n (x) \equal{} \displaystyle\int^x_a f_n (t) dt, \, a\leq x \leq b.$ Prove there exists a subsuccession of $ (F_n)$ that converges uniformly en $ [a, b]$.

-------

My biggest complication is that $ [a,b]$ is not enumerable :(. Any hint is welcomed.

The usual English word is "sequence", not "succession".

That first sentence doesn't work as English in that order, and the precise meaning matters. If it's that $ f_n$ is a sequence of bounded integrable functions, the result is false by the example of $ f_n(x)\equal{}nx^{n\minus{}1}$ on $ [0,1]$. If it's an equi-bounded* sequence of functions (that is, there's a bound $ M$ that works for all $ n$ and all $ x$ simultaneously), this has a chance.
*The usual modifier for this notion is "uniformly"; "equi" only comes in when there's more than one dimension things need to be uniform in. I'm bending those rules a bit, so this isn't quite standard terminology.

Oh, sorry about that. I'll keep it in mind.

You are right, that is exactly what I meant. It hadn't thought in counter-examples, but that seems to work. I'll review the exercise.

Ty ^^

You were also right about the exercise, it's an equi-bounded sequence.


The tessellation shown consists of congruent obtuse triangles arranged in a regular hexagon. What is the number of isosceles triangles of any size in the figure?

[asy]size(100); real side = 1; real angle = 120;

path myTriangle(real side = side, real angle = angle)
{
real newangle = 90-angle/2;
return (0,0)--(side,0)--(side/2,side/2*tan(newangle*pi/180))--cycle;
}

path[] hexagonSector(real side = side,real angle = angle)
{
path inTriangle = myTriangle(side,angle);
real newangle = 90-angle/2;
pair reflectvertex = (side/2,side/2*tan(newangle*pi/180));
pair farvertex = (side,0);
path[] out = inTriangle ^^ reflect((0,0),reflectvertex)*inTriangle ^^ reflect(farvertex,reflectvertex)*inTriangle ^^ reflect(farvertex,  reflect((0,0),reflectvertex)*farvertex)*reflect(farvertex,reflectvertex)*inTriangle;
return out;
}

path[] tilePieces(path[] piece,real offset)
{
path[] out;
path[] newdude = rotate(offset)*piece;
for(int i = 0; i<6;++i)
{
newdude = rotate(60)*newdude;
out.append(newdude);
}
return out;
}

draw(tilePieces(hexagonSector(),30));[/asy]

Do equilateral triangles count as isosceles triangles?

There are 24 small isosceles triangles.
There are 6  equilateral triangles.

I think thats all, so if you see any more, tell me.
I got 30.


Let S be a set of n terms $ (n>3)$ satisfies that : if we remove an arbitrary term in S, the rest can be divide into two set whose sum of all terms are equal to the other. Prove that if n is odd, then all terms in S are equal, and if n is even, there isn't any set which stisfies the condition.

S = {1,1,1,1,3}

seems to be a counter-example to the problem statement.

Well, if the problem statement assumes all are distinct, it is asking us to prove a contradiction which seems a bit silly. Is the problem statement right?

Also, what does 'n terms' really mean? n integers, n reals?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=290685
and follow the links.  (Please don't tell me that that's not exactly the same problem before following and reading the links.  The problem as stated here is false.)


I'm not sure how easy or difficult people will find this, but I'll stick it here by default.

Box 1 contains 1 ball of color 1, 2 balls of color 2, 3 balls of color 3, up to n balls of color n.

Box 2 contains n balls of color 1, n-1 balls of color 2, n-2 balls of color 3, down to 1 ball of color n.

A ball is randomly chosen from Box 1 and placed into Box 2. A ball is then randomly chosen from Box 2. Which color ball is most likely to be chosen?

[hide]Obviously, if the first-chosen ball is of color $ 1$ or of any color from $ 3$ to $ n$, the most likely color of second-chosen ball is $ 1$. If the first-chosen ball is of color $ 2$, it's easy to show that the probability of picking the ball of the same color at the second turn doesn't exceed the probability of picking the ball of color $ 1$ in the former case. Hence the answer is $ 1$.[/hide]

EDIT: Thanks to MellowMelon

You sure? Box 2 contains $ n$ balls of color $ 1$ and only $ 1$ ball of color $ n$. You can correct your arguments to show that the most likely color will be $ 1$.

Now try to find $ p(k)$, the probability that color $ k$ is chosen from the second box.

[hide="Solution"]For each color $ k$, there are two cases:

Case 1: A ball not of color $ k$ is picked from box $ 1$. This has probability $ 1\minus{}\frac{2k}{n(n\plus{}1)}$. Then the probability of picking $ k$ from box $ 2$ is $ \frac{2(n\plus{}1\minus{}k)}{n(n\plus{}1)\plus{}2}$. Multiplied, we get $ (1\minus{}\frac{2k}{n(n\plus{}1)})\cdot\frac{2(n\plus{}1\minus{}k)}{n(n\plus{}1)\plus{}2}$.

Case 2: A ball of color $ k$ is picked from box $ 1$. This has probability $ \frac{2k}{n(n\plus{}1)}$. Then the probability of picking $ k$ from box $ 2$ is $ \frac{2(n\plus{}2\minus{}k)}{n(n\plus{}1)\plus{}2}$. Multiplied we get $ \frac{2k}{n(n\plus{}1)}\cdot\frac{2(n\plus{}2\minus{}k)}{n(n\plus{}1)\plus{}2}$.

Then we add. For purposes of clarity, let $ A \equal{}\frac{2}{n(n\plus{}1)}$ and $ B \equal{}\frac{2}{n(n\plus{}1)\plus{}2}$. Then

\[ p(k) \equal{} (1\minus{}Ak)(n\plus{}1\minus{}k)B\plus{}Ak(n\plus{}2\minus{}k)B\]
\[ p(k) \equal{} B(n\plus{}1\minus{}k)\plus{}AkB \equal{} B(n\plus{}1\minus{}k\plus{}Ak)\]
\[ p(k) \equal{}\frac{2(n\plus{}1\minus{}k\plus{}\frac{2k}{n(n\plus{}1)})}{n(n\plus{}1)\plus{}2}\]

You'll notice that as $ k$ increases, the numerator of this fraction decreases, implying that $ 1$ is indeed the answer.[/hide]

Actually, I wasn't paying attention and misordered the boxes  :roll: you're right, of course. My intended point was: the color which is represented by the highest number of balls in box 2.

Farenhajt's solution is nice because it catches the real spirit of the problem, which is that the answer actually depends extremely weakly on the contents of Box 1 and only slightly more strongly on the contents of Box 2.  We were given some very explicit information about the contents of these boxes (the number of appearances of $ i$ in Box 1 is $ i$; in Box 2 it is $ n\plus{}1\minus{}i$).  My follow-up question is, how much of this information is irrelevant to the answer?  How much can we reduce our knowledge of the contents of Box 1 and Box 2 and still be able to solve this problem correctly?

Actually, let me add some more: what if we draw 2 balls from Box 1 and add them to Box 2?  Is there any number of balls we can draw from Box 1 that will change the answer?  What is the smallest such number of balls?


rotation
geometry
3D geometry
search
\/closed

If I was to encounter a post that was well, spam, should I rate it a 1, or not rate it at all?

There's no point in rating true commercial spam. Those posts get deleted, and those users get banned.

Is there a way to find out which/what my posts were rated?

MY rating rotates between 1.9 and 2.0 and I don't know which, by whom, or what posts were rated.

Rating has no effect on some people. But for some reason, I still care about my rating. :|

I think this to be sort of weird, because I can see what other posts were rated.

@dragon96: moi aussi

Is a user's ratings just the average ratings of all their posts?Cuz 2.0 seems sorta low...

Please, search for previous topics in which all of the questions were answered.

@pascal12: that's because people usually rate just bad posts (ask yourself, how many 1's and how many 6's have you given to other people)

[quote="Bugi"]@pascal12: that's because people usually rate just bad posts (ask yourself, how many 1's and how many 6's have you given to other people)[/quote]

I use ratings to tell people when they have done something good (i.e. a brilliant proof) or when they have done something bad (waste someone else's time with a stupid problem). Generally I give a lot more 6's than 1's, and nothing in between.

@Differ: and that's very good if you do like that. Unfortunately, this forum has far more people who target just bad posts.

[quote="PowerOfPi"]No, you can't see which posts were rated or who rated them.

Non-commercial spam, I think, should be rated 1 because if you hover over one cube, it says "Poor (Spam)".

EDIT: @AwesomeToad: By which posts were rated, I meant new ratings. You can see the post's overall rating.[/quote]

I can't see the post's overall rating. I still don't get how the rating changes from 1.9 to 2.0 to 1.9 to 2.0 .... For example, yesterday I got on in morning and my rating was 1.9. Then 15 minutes later it was 2.0. Then I came in afternoon and it was 1.9 again.

EDIT: I know I sound overly concerned about post rating, but this is weird and I'm just wondering.

[quote="PowerOfPi"]Your posts only have one set of six cubes, because you can't rate your own post. The only set should display your overall rating for the post. I rated your last post a 5. See if you can see that.[/quote]

1) What do you mean by "one set of 6 cubes"? I know I'm being really annoying now, but I don't really get it.
2) No, I cannot see anything for that post and my rating is still 1.9


Τώρα είδα τα θέματα! Αρκετά όμορφα. Ειδικά το 2 και το 3 μου άρεσαν πολύ. Έχω μια πολύ όμορφη λύση για το 2 (χρησιμοποιώ το τετράπλευρο Miquel). Θα προσπαθήσω να βρω χρόνο για να βάλω τις λύσεις μου. Πάω να ρίξω μια προσεκτκιή ματια...

Πιστεύω πως τα θέματα ταιριάζουν με την Ελληνική ομάδα!Ξέρει κανένας πως πήγαμε?

Το 3ο θέμα χαρακτηρίζεται εξαιρετικά δύσκολο και λέγεται ότι το έλυσε μία μόνο Κινέζα .
Αν είναι έτσι , τότε θα είναι το δυσκολότερο πρόβλημα όλων των εποχών σε ΙΜΟ .

Κινέζα e?
Μπραβο!!!
Θα μας βαλουν τα γυαλια αυτες οι γυναικες... :roll:

ki omos filoi mou auto mallon mas voithaei os ellhnikh omada!!!!!!

ego an hmoun diagonizomenos tha xairomoun na yparxei provlhma pou to xei lysei mono ena atomo, parolo pou prokeitai gia psema sthn sygekrimenh periptosh.
kai leo oti prokeitai gia psema giati sthn imo symmetexei kai o jurie, opote yparxoun toulaxiston dyo atoma pou to exoun lysei (allo an o filos jurie einai xamhlon tonon)!

opos kai na xei mallon einai yper mas. vazo to mallon dioti en genei to oti yparxei ena provlhma pou den to exei lysei sxedon kanenas kai poly perisootero ligoi einai autoi pou exoun sygentrosei merikous pontous apo to idiio thema, mas voithaei para poly san ellhnikh omada, alla yparxei kai mia periptosh megalhs sysorreushs stis mikres vathmologies kai opote symvainei kati tetoio exoume polla aproopta. h gnomh mou einai oti den tha yparxei auth h periptosh, dioti to proto thema htan vato gia poly kosmo kai to deutero xarakthrizetai os eukolo pros mesaio gia omades eksoikeiomenes me thn eukleideia geometria kai kathara mesaio gia olous tous ypoloipous.

opos kai na xei, pisteuo oti oi dikoi mas kai o kathenas ksexorista diaskedasan me ta themata o kathenas gia to diko tou logo.den nomizete oti ta themata eixan kati attractive mesa tous??

auta

~nick

ps. gia to provlhma 2, auto pou mou kinhse thn periergeia emena kai me odhghse sth lysh einai to oti to shmeio E einai to shmeio Miquel gia to plhres tetrapleuro BPMC. meta einai gonstotato theorhma oti ta ixnh ton katheton apo to shmeio miquel pros tis pleures kai tis diagonies tou plhrous tetrapleurou einai shmeia syneutheiaka. apo ekei exoume ED=EB pou einai arketo gia thn oloklhrosh ths apodeikshs (kati goniakes sxeseis mono) .

molis vro xrono tha valo th lysh kai gia to 1.

Ο Iurie έλυσε μόνο το 1ο θέμα και αυτό μάλιστα (το 1β) στα τελευταία 15 λεπτά , όπως λέει ο ίδιος !!!!!
http://www.mathlinks.ro/Forum/viewtopic.php?t=159844

MONO AN MPOROYSE KAPOIOS NA TA EXEI EDO GIA EUKOLIA
PS
RE NIKO AKOMA EKEI EISAI ??

nai edo eimai kai den auth th stigmh den exo xrono, alla auto gia to jurie den to iksera......

oxi pos me peirazei an de grapsei kala o jurie ,isa-isa.....alla denksero isos prpei na eimaste epifilaktikoi me tetopies prosopikothtes. merikes fores lene kai psemata stis olbiades (vlepe MARKO JEVREMOVIC) alla nomizo oti den einai tetoios o jurie!

pantos ena einai sigouro. an den egrapse o jurie, sigoura euthynetai to jet lang.epishs den boro na pistepso oti den elyse to 2 o jurie (ton exo dei na lynei geometria kai ksero pos syberiferetai..einai apla kataplhktikos!). re paidia mipos enoei oti ta elyse ola se ena tetarto kai apla ton paidepse to 1???

auto to telftaio metaksy sovarou kai asteiou....

ps. gia tous dikous mas kseroume tipota?

Οι χώρες που πρότειναν τα θέματα στη φετεινή ΙΜΟ :

1: New Zealand
2: Luxembourg
3: Russia
4: Czech Republic
5: United Kingdom
6: Netherlands

Ξέρει κανείς πως τα πήγε η ελληνική ομάδα  :maybe:  :?:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159844&start=80

Στη μέση της σελίδας έχει ένα σχόλιο του Vornicu για τα ελληνικά αποτελέσματα.

Ευχαριστώ πολύ  :)

Official results for Greek team

GR1. SILOUANOS BRAZITIKOS              12 pts
GR2. ALKISTIS MAVROEIDI                   10 pts
GR3. DIMITRIOS PAPADIMITRIOU         15 pts
GR4. NIKOLAOS ROMANIDIS                 24 pts
GR5. MICHAIL SAVVAS                         14 pts
GR6. ANASTASIOS VAFEIDIS                14 pts

Εξαιρετικές επιδόσεις από τα παιδιά .Μπορεί να φέρουμε και 5 μετάλλια .  Μπράβο σε όλους .

[quote="artisti"]Official results for Greek team

GR1. SILOUANOS BRAZITIKOS              12 pts
GR2. ALKISTIS MAVROEIDI                    9 pts
GR3. DIMITRIOS PAPADIMITRIOU         15 pts
GR4. NIKOLAOS ROMANIDIS                 24 pts
GR5. MICHAIL SAVVAS                         14 pts
GR6. ANASTASIOS VAFEIDIS                14 pts

Εξαιρετικές επιδόσεις από τα παιδιά .Μπορεί να φέρουμε και 5 μετάλλια .  Μπράβο σε όλους .[/quote]

sigxaritiria ki apo mena
alla eiste bebaioi oti to medal cut off tha einai tetoio oste na paroume 5 metallia??
makari

[size=150][color=red]ΠΑΡΑ ΠΟΛΥ ΚΑΛΑ !!![/color][/size]

 Όπως λένε τα  cut offs , παίρνουμε ένα ασημένιο (αργυρό) και τρία χάλκινα. Η ομάδα πήγε πολύ ωραία :lol:Άπό μικροατυχίες ίσως χάσαμε ένα - δύο ακόμα μετάλλια.Δεν πειράζει όμως !

                               Συγχαρητήρια σε όλους  !!!

 [u]Μπάμπης[/u]

Sygxarhthria ki apo emena sta paidia!! eksairetikes epidoseis gia thn ellada, fetos. Mono pou afto pou thelo na epishmano einai oti exoume ena ashmenio, tria xalkina kai dyo efhmes mnhes (den prepei na to ksexname ki as mhn einai metallio). kai symfono apolyta me ton kyrio babh. apo mikroatyxies xasame alla dyo metallia, ki oxi apo eleipsh mathimatikon gnoseon!

auta apo emena...

kai pali sygxarhthria sta paidia!

The IMO 2007 Jury has voted for the medal cut-offs:
[color=yellow]Gold 29 [/color]
[color=indigo]Silver 21 [/color]
[color=brown]Bronze 14 [/color]

Η Ελληνική ομάδα παίρνει :

1. Μπραζιτίκος Σιλουανός   εύφημη μνεία
2. Μαυροειδή Άλκηστις   εύφημη μνεία
[color=brown]3. Παπαδημητρίου Δημήτριος  χάλκινο μετάλλιο [/color]
[color=indigo]4. Ρωμανίδης Νικόλαος  αργυρό μετάλλιο[/color]
[color=brown]5. Σάββας Μιχαήλ  χάλκινο μετάλλιο
6. Βαφείδης Αναστάσιος  χάλκινο μετάλλιο
 [/color]

Τα αποτελέσματα βρίσκονται στη διεύθυνση :

http://www.imo2007.edu.vn/index.php?module=ViewInforTableForm.php&act=viewresult&ct=GRE

[quote="artisti"]The IMO 2007 Jury has voted for the medal cut-offs:
[color=yellow]Gold 29 [/color]
[color=indigo]Silver 21 [/color]
[color=brown]Bronze 14 [/color]

Η Ελληνική ομάδα παίρνει :

1. Μπραζιτίκος Σιλουανός   εύφημη μνεία
2. Μαυροειδή Άλκηστις   εύφημη μνεία
[color=brown]3. Παπαδημητρίου Δημήτριος  χάλκινο μετάλλιο [/color]
[color=indigo]4. Ρωμανίδης Νικόλαος  αργυρό μετάλλιο[/color]
[color=brown]5. Σάββας Μιχαήλ  χάλκινο μετάλλιο
6. Βαφείδης Αναστάσιος  χάλκινο μετάλλιο
 [/color][/quote]

Αφού τα αποτελέσματα είναι έτσι , τότε όλα τα παιδιά πρέπει να γυρίσουν με χαμόγελο , περήφανα και ευτυχισμένα!
 Ξανά συγχαρητήρια και στα παιδιά και στους γονείς τους αλλά και σε όλους τους συντελεστές .
 Μακάρι η φετινή εμφάνιση να δώσει σε όλους τη δύναμη να συνεχίσουμε ως χώρα με μεγαλύτερο μεράκι την προσπάθεια και να φτάασουμε ακόμα ψηλότερα !

[u]Μπάμπης[/u]

ΘΕΡΜΑ ΣΥΓΧΑΡΗΤΗΡΙΑ ΚΙ ΑΠΟ ΜΕΝΑ
πάντα τέτοια κι κυρίως πολυ καλύτερο μέλλον εύχομαι
μπράβο σε όλους κι όλες για αυτή την καλή παρουσία

Θερμά συγχαρητήρια σε όλη την ομάδα!! Φέτος ήταν καλή χρονιά για μας αφού βρεθήκαμε στην 34 θέση ενώ τα προηγούμενα χρόνια (πλήν Αθήνας 2004) είμασταν κάτω του 50. Και πάλι εύγε!

Συγχαρητήρια σε όλα τα παιδιά για τις διακρίσεις τους! Είναι πολύ μεγάλη η επιτυχία της χώρας μας μετά από χρόνια!

ΜΠΡΑΒΟ ΣΕ ΟΛΟΥΣ ΣΑΣ!

Αλέξανδρος

Συγχαρητήρια και απο εμένα στα παιδιά για τα μετάλλια που έφεραν στην Ελλάδα  :clap:

nickolas

η σειρά μας είναι 33η στον κόσμο !
Και 7η στην Ευρωπαική Ένωση !!

[color=red][b]ΑΠΑΡΑΙΤΗΤΟ ΣΧΟΛΙΟ[/b][/color]

Θεωρώ χρήσιμο να τονίσω ότι η εύφημη μνεία είναι [b]μεγάλη  τιμητική διάκριση [/b], εφάμιλλη μεταλλίου , για τον λόγο ότι αποδεικνύει την ολυμπιακή πάστα του μαθητή και την ικανότητα να λύνει με ιδιαίτερα ευφυή τρόπο τέτοια προβλήματα. Υπό αυτό το πρίσμα , όλοι μας οι μαθητές  της ελληνικής αποστολής είναι μεταλλιούχοι.
 Σκεφτείτε ακόμα την περίπτωση που κάποιος μαθητής θα συγκέντρωνε 14 μονάδες , άρα θα έπαιρνε χάλκινο, χωρίς να λύσει ολόκληρο κανένα θέμα.Βέβαια αυτή η παρατήσηση  δεν αφορά κανένα από τους τρεις μαθητές της ομάδας μας ,αφού  και αυτοί  μάζεψαν τις 14 ή 15 μονάδες λύνοντας  και ολόκληρα θέματα.
 Από κάθε άποψη λοιπόν η ελληνική ομάδα σημείωσε μεγάλη επιτυχία. Οι αρμόδιοι ας αναλύσουν αυτή την επιτυχιία και ας κάνουν ό,τι καλύρερο μπορούν , ώστε οι διακρίσεις να συνεχιστούν.

 [b][color=red]Ξανά συγχαρητήρια !!![/color][/b]

[u]Μπάμπης[/u]

Συγχαρητήρια για τις τρομερές επιτυχίες τις ομάδας. Μπράβο παιδιά. :)

ΣΥΓΧΑΡΗΤΗΡΙΑ ΠΑΙΔΙΑ!!!
Σας ζηλεύω μια και εγώ εχω περασει μονο Θαλη...... :oops:
Κατι αλλο που μπορει να ειναι off-topic
Ο Νκος ο Ραπανος δεν στελεχωνε την ομαδα φετος???

Συγχαρητήρια σε όλους και από 'μένα!

Καλή επιστροφή στα πάτρια εδάφη  :wink:

Συγχαρητήρια παιδιά, κατ' αρχήν για την προσπάθεια, και φυσικά για τα αποτελέσματα. Εύχομαι καλή συνέχεια σε όσους απο σας έχετε ακόμα μια (και δυο) δυνατότητες συμμετοχής στην IMO. Χαίρομαι ιδιαίτερα για το Νίκο (Ρωμανίδη) γιατί δικαιώθηκε για την περσινή του ατυχία.

~

[i]Aviator:[/i] Ο Νίκος (Ράπανος) την περίοδο αυτή επέλεξε να πάει στο [url=http://www.cee.org/rsi/]RSI[/url] και όχι στην IMO.

Συγχαρητήρια σε ολα τα παιδια της Ελλας που αγωνιστηκαν για το καλυτερο της χωρας μας και να ξερουν οτι αρκετες φορες δν φενεται η αξια του διαγωνιζομενου απο τους διαγωνισμους.Ειναι και η τυχη......!Οι ολυμπιαδες ειναι απλα η αρχη,το εναρκτηριο λακτισμα!Τελος να μην ξεχνουν οτι η αποσταση μεταξυ ευφημης μνειας και χρυσου δν ειναι και τοσο μεγαλη οσο φαινεται.... :wink:

Συγχαρητήρια κι απο εμένα σε όλους  :)

[color=red][b]ΜΙΑ ΑΠΑΡΑΙΤΗΤΗ ΠΡΟΣΘΗΚΗ[/b][/color]

 Από αυτό το φόρουμ ,και από όλους μας, έχουν γίνει πολλές συζητήσεις για τον ρόλο της ΕΜΕ στους διαγωνισμούς.Οι προτροπές  και οι παραινέσεις μας για καλύτερη προετοιμασία της ομάδας μας φαίνεται ότι αυτή τη φορά δεν πέρασαν απαρατήρητες. Μαθητές της ομάδας που παρακολούθησαν τα τελευταία μαθήματα - έστω αυτά - μου εξέφρασαν την ικανοποίησή  τους για το επίπεδο , την αρτιότητα και την αποτελεσματικότητα των μαθημάτων.Όλοι τους τα βρήκαν πολύ χρήσιμα.
 Θεωρώ λοιπόν , ως απλό μέλος της ΕΜΕ , να εκφράσω τα συγχαρητήριά μου στους συναδέλφους που παρέδωσαν αυτά τα μαθήματα , αλλά και σε όλους τους ανθρώπους που εμπλέκονται στη διαδικασία των διαγωνισμών και των μαθημάτων, σε όλα τα παραρτήματα της ΕΜΕ .
 Η καλή εμφάνιση εύχομαι να αποτελέσει το εφαλτήριο για καλύτερη οργάνωση και προετοιμασία των μαθητών μας, σε όλα τα επίπεδα και όχι μόνο στην τελική φάση.Η θέσπιση του [b]θερινού μαθηματικού σχολείου [/b]από το παράρτημα της Ημαθίας , είναι ένα πολύ καλό ξεκίνημα που αξίζει θερμής αποδοχής κάτι βέβαια που ήδη συμβαίνει , όπως ευγενικά μου μετέφερε ο πρόεδρος του παραρτήματος Κώστας Παπαδόπουλος. Μακάρι η προσπάθεια να συνεχιστεί και να επεκταθεί - σε άλλον ίσως νομό - και σε παιδιά από την Ε'  Δημοτικού ως την Α' Γυμνασίου.
  Δεν παραβλέπω βέβαια και άλλες δραστηριότητες στα μαθηματικά που ξεκίνησαν στην πατρίδα μας και  που θα συμβάλλουν στην προβολή και τη διάδοση των μαθηματικών και του ολυμπιακού πνεύματος , αλλά για αυτό θα μιλήσω σε άλλο μήνυμα.Ακόμα είναι νωρίς. Για την ώρα τα συγχαρητήρια μου στοχεύουν στην ΕΜΕ , για την άξια προβολή της χώρας μας στο διεθνές μαθηματικό πεδίο.
 Εύχομαι καλή συνέχεια και ανοδική επίδοση στα χρόνια που ακολουθούν !

                                      [b]ΜΠΑΜΠΗΣ[/b]


Μπορεί κανείς να γράψει την απόδειξη οτι μεταξύ του
του ν και του 2ν υπαρχει πάντα ενας πρώτος αριθμός.

sef καλως ηρθες.το θεωρημα που ζητας εχει λιγο μεγαλη αποδειξη.ας δωσει καποιος ενα link στο φιλο μας γιατι ειναι ομως   :lol: φανταστικη!

Να ένα pdf  :)

[url=http://en.wikipedia.org/wiki/Bertrand%27s_postulate]Bertrand's postulate[/url]
[url=http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate]Proof of Bertrand's postulate[/url]

Φαίνεται να έχει ενδιαφέρον και η σελίδα στην οποία το βρήκα

 :arrow:     [url]http://www.lacim.uqam.ca:16080/~plouffe/[/url]

Ευχαριστώ για τη βοήθεια, ξεχασα να αναφερω αρχικά
οτι ψάχνω την απόδειξη που έδωσε ο Ramanujan σ αυτό
το θεώρημα

Νομίζω ότι ζητάς αυτό

[url]http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/Cpaper24/page1.htm[/url]

Κοίτα και το


$ x,\ y$ satisfies the following three inequalities:

$ x>2,\ y\geq \frac{x}{x\minus{}2},\ x\plus{}y\leq 6$.

Find the maximum and minimum values of $ 3x\plus{}2y$.

[hide]
15 and 16?

im pretty sure x must be within 3 and 4 or else the conditions fail,, I just dont know how to show this
[/hide]

[hide="solution"]
First combine the following; $ y\le 6 \minus{} x$ and $ y \le \frac {x}{x \minus{} 2} \equal{} 1 \plus{} \frac {2}{x \minus{} 2}$ to get

$ x^2 \minus{} 7x \plus{} 12 \le 0 \Rightarrow x \in [3,4]$

Now under a restrtiction $ x \plus{} y \equal{} k$, the expression $ 3x \plus{} 2y$ will clearly be maximised when $ x$ is maximised and minimised when $ x$ is minimised
($ \because 3x \plus{} 2y \equal{} 3x \plus{} 2(k \minus{} x) \equal{} x \plus{} 2k...$)

With the sum $ x \plus{} y \equal{} 6$, $ Max(x) \equal{} 4$ we have $ (x,y) \equal{} (4,2)$. This complies with all of our restrictions so we need not go any further. $ Max(3x \plus{} 2y) \equal{} 3(4) \plus{} 2(2) \equal{} 16$

For the minimum we have $ y \ge 1 \plus{} \frac {2}{x \minus{} 2} \Longrightarrow 3x \plus{} 2y \ge 3x \plus{} 2 \plus{} \frac {4}{x \minus{} 2}, x\in [3,4]$

Now we use AM-GM
$ 3x \plus{} 2 \plus{} \frac {4}{x \minus{} 2} \equal{} 8 \plus{} 3(x\minus{}2) \plus{} \frac {4}{x \minus{} 2} \ge 8 \plus{} 2 \sqrt{4\cdot 3} \equal{} 8 \plus{} 4\sqrt{3}$

Min, Max $ \equal{} \boxed{4\sqrt {3} \plus{} 8, 16}$
 [/hide]

Of course, we could also use graphing, but ocha's solution is very nice  :D .


Let  $ a,$ $ b,$ $ c,$ $ x$ and $ y$ are positive numbers such that $ ay \plus{} bx \plus{} \sqrt3(ab \minus{} xy) \equal{} 0$ and
$ a^{2}\plus{}x^{2}\equal{}b^{2}\plus{}y^{2}\equal{}(x\minus{}y)^{2}\plus{}c^{2}.$ Prove that $ c\equal{}a\plus{}b.$

[hide="Solution"]Since $ a,b,x,y$ are positive, $ \sqrt{3}(xy\minus{}ab) \equal{} ay\plus{}bx > 0$. So $ xy > ab$...

We have
\[ 4(ab \minus{} xy)^{2} \equal{} (ay \plus{} bx)^{2} \plus{} (ab \minus{} xy)^{2} \equal{} (a^{2} \plus{} x^{2})(b^{2} \plus{} y^{2}) \equal{} (a^{2} \plus{} x^{2})^2
\]
Since $ xy > ab$,

$ 2(xy \minus{} ab) \equal{} a^{2} \plus{} x^{2} \equal{} b^{2} \plus{} y^{2} \equal{} (x \minus{} y)^{2} \plus{} c^{2}$

So

$ a^{2} \plus{} x^{2} \minus{} 2(xy \minus{} ab) \plus{} b^{2} \plus{} y^{2} \equal{} (x \minus{} y)^{2} \plus{} c^{2}$

Therefore $ (a \plus{} b)^{2} \plus{} (x \minus{} y)^{2} \equal{} (x \minus{} y)^{2} \plus{} c^{2}$. $ c$ positive implies $ c \equal{} a \plus{} b$.

EDIT: It turns out that $ xy > ab$ must hold...[/hide]


$1\geq x^{3}\geq 0$, so $0\leq (1-x^{3})\leq 1$
it's obvious now that the minimum of $A$ is $0$.

PROVE That the Maximum value of A is:(5.^(5/3))./(4*(3.^(4/3)).
and occur for:the root of the:(x.^3=9/10.8).

Guys, if you have too much time, play some stupid RPG, read some book, but don't spam this forum. Please.

  darij

it is obvious that  $A=x^{5}(1-x^{3})^{3}\geq 0$ with $x \in [0,1]$
so
when $x=0, A=0$
so the min of $A$ is $0$


Wait...you just enrolled today? If there hasn't been any classes yet, there wouldn't be a message board. When you click on "Forum", scroll down until you get to a subtopic called "Online classes". Once the classes start, there will be a forum there called "AMC 12 Problem Series", which is the message board.

[quote="CyanRedYellow"]Hey, I just enrolled in the amc 12 class today, and I'm pretty clueless on how this whole online-class thing works. So...when I go to my classes, I can get to the page that has the transcript and stuff, but when I click on the link to the message board, it tells me that the forum I selected does not exist. What am I doing wrong?[/quote]

Nothing.  You haven't been added to the proper usergroup yet.  You will be this morning.  Also, you'll receive an email this morning with more information about the class.

Okay, so I can see the stuff there now, thank you!


you should think about these a bit before you post them, they seem like homework, look at the section in your book that talks about this, but really, this is pretty simple, arccot and cot are inverses, and arcsin 1/sqrt(2) is also trivial

the second one, think that the bottom does not matter for solving roots (unless it makes it undefined), so the top is all you need to worry about, and that is quite basic too


given the array of integers $ a_{n}$ :
$ a_{1}\equal{}2;a_{2}\equal{}7$
$ \minus{}\frac{1}{2} < a_{n\plus{}1}\minus{}\frac{{a_{n}}^2}{a_{n\minus{}1}} \leq \frac{1}{2}$ for every $ n\geq2$
prove that:$ a_{n}$ is odd and show the array satisfies the condition

You are wrong, $ a_{n \plus{} 1} \equal{} [\frac 12 \plus{} \frac{a_n^2}{a_{n \minus{} 1}}]$ and $ a_3 \equal{} 50$ - even.

[quote="Rust"]You are wrong, $ a_{n \plus{} 1} \equal{} [\frac 12 \plus{} \frac {a_n^2}{a_{n \minus{} 1}}]$ and $ a_3 \equal{} 50$ - even.[/quote]
are you sure?
because $ a_{3}\equal{}25$ is odd

By induction, $ a_{n\plus{}1}\equal{}3a_n\plus{}2a_{n\minus{}1}$.

can you tell me why your array satisfies  this
$ \minus{}\frac{1}{2} < a_{n\plus{}1}\minus{}\frac{{a_{n}}^2}{a_{n\minus{}1}} \leq \frac{1}{2}$

$ |a_{n \plus{} 1} \minus{} \frac {a_n^2}{a_{n \minus{} 1}}| \\
\equal{} |3a_n \plus{} 2a_{n \minus{} 1} \minus{} \frac {a_n(3a_{n \minus{} 1} \plus{} 2a_{n \minus{} 2})}{a_{n \minus{} 1}}| \\
\equal{} |2a_{n \minus{} 1} \minus{} \frac {2a_na_{n \minus{} 2}}{a_{n \minus{} 1}}| \\
\equal{} |2\frac {a_{n \minus{} 1}^2 \minus{} a_na_{n \minus{} 2}}{a_{n \minus{} 1}}| \\
\equal{} |\frac {a_{n \minus{} 1}^2 \minus{} a_na_{n \minus{} 2}}{a_{n \minus{} 2}}|*|\frac {2a_{n \minus{} 2}}{a_{n \minus{} 1}}| \\
\le \frac12*1$
Here we use
$ |\frac {a_{n \minus{} 1}^2 \minus{} a_na_{n \minus{} 2}}{a_{n \minus{} 2}}|\le\frac12$ by induction assumption and
$ |\frac {2a_{n \minus{} 2}}{a_{n \minus{} 1}}|\le1$ from $ a_{n \minus{} 1} > 3a_{n \minus{} 2}$.


Let $ A_1,A_2,...$ be a sequence of infinite sets such that $ |A_i \cap A_j| \leq 2$ for $ i \not\equal{}j$. Show that the sequence of indices can be divided into two disjoint sequences $ i_1<i_2<...$ and $ j_1<j_2<...$ in such a way that, for some sets $ E$ and $ F$, $ |A_{i_n} \cap E|\equal{}1$ and $ |A_{j_n} \cap F|\equal{}1$ for $ n\equal{}1,2,... .$

[i]P. Erdos[/i]

I don't understand this. What if $ A_i \cap A_j \equal{} \emptyset$ for all $ i \neq j$?


A sequence $ < a_{n} > $ is defined in terms of a sequence $ < b_{n} > $ by .. $ a_0 = b_0 , a_n = b_n -\ \alpha b_{n-1} , n >1 $ and $ | \alpha | < 1 $. Express $ < b_n > $ in terms of $ < a_n > $ and show that if  $ a_n \rightarrow L , then\ b_n \rightarrow \frac{L}{1- \alpha} $

[quote="blahblahblah"]$b_n=\sum^n_{i=0} \alpha^ia_{n-i}$ by a simple induction.

For the second part, given $\epsilon>0$, there exists $n$ so that $|a_N-L|<\epsilon$ for $n>N$.

Then pick $m$ large enough that

$|\alpha^{m-n}|<\epsilon\cdot\max_{i\in\{1,2,\ldots,m-n\}} |a_i|$.

holds.

It should now be clear that

$b_m<\frac {\alpha^{m-n+1}-1}{\alpha-1}|L+\epsilon|+n\epsilon$

As $m\to\infty$, we get the desired result.  The other side of the inequality is trivial to derive from the above results, and the limit of $b_n$ follows.[/quote]

The first part is effectively trivial. But since i don't know anything about series etc. Can you explain it in sinpler terms. Also do you have any other simpler method!!!

No, I don't have a simpler method, because this is just standard epsilon delta manipulation.  Possibly I am missing something, but I don't see an easier way.

I use liberally in there the standard result that $a_i$ converges to $L$ iff for all $\epsilon>0$, there exists $n\in\mathbb{N}$ such that $m>n$ implies $|a_m-L|<\epsilon$.

can you plz explain the part about picking $m$

It's obvious.  We have a finite set of numbers that are nonzero (if they are zero, they are small enough anyways).  Hence eventually $|\alpha^t|$ is smaller than any of them for large enough $t$, since $|\alpha|<1$.


Let $ P   \in {\mathbb R}[x]$, $\mbox{\rm degree}[P]\ge 1\; , $ and denote $ F(x)=\ds P\left(ax^4+bx^3+cx^2+dx \right). $ If $ F $ is even polynomial ,i.e. $ F(-x)=F(x)\; ,\; \forall x\in {\mathbb R} ,$ prove (or disprove) that  $ P(x) $ is even  or  $b=d=0 \; .$
If $   \displaystyle abcd \ne 0 $ , it's possible for $ F $ to be an even polynomial  ?

If $P$ is injective $F(x)=F(-x)$ =>
$ax^4+bx^3+cx^2+dx=ax^4-bx^3+cx^2-dx$ $\forall x \in R$ => $b=d=0$


While doing problems form V.V. Prasolov's „Problems in Plane Geometry (Zadachi po planimetrii)“, I ran into this problem:

[quote="Prasolov"]20.13. Given finitely many points on the plane such that any line passing through two
of the given points contains one more of the given points. Prove that all the given points
belong to one line.[/quote]

After seeing the problem (and its solution), I thought about the possibility of obtaining a general result. That is why I conjectured:

[b]Conjecture[/b]: [color=darkblue]Finitely many points are given in n-dimensional (hyper)space such that any (n-1 dimensional) hyperplane containing n points contains at least one more point. Then all given points lie in the same (n-1) dimensional hyperplane.[/color]

Prasolov's example is special case n=2. Can a general proof be given (if the conjecture is true, of course)?

In addition to my first post:

1. Is the condition that all points lie in a plane (in the original Prasolov's problem) really necessary? The argument used in the solution given in the book seems to hold for higher dimensions of space in which the points lie

[hide="The solution from the book"][quote=Prasolov]Suppose that not all of the given points lie on one line. Through every pair of
given points draw a line (there are finitely many of such lines) and select the least nonzero
distance from the given points to these lines. Let the least distance be the one from point
A to line BC, where points B and C are among given ones.
On line BC, there lies one more of the given points, D. Drop perpendicular AQ from
point A to line BC. Two of the points B, C and D lie to one side of point Q, let these be
C and D. Let, for definiteness, CQ < DQ.

Then the distance from point C to line AD is smaller than that from A to line BC which
contradicts to the choice of point A and line BC.[/quote][/hide]

2. In the light of question 1. and assuming that my conjecture is true: is the condition that points lie in a space with a dimension higher by one than the dimension of the hyperplanes formed necessary?

If i'm right it's the [url=http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem]Sylvester–Gallai theorem[/url] or [url=http://mathworld.wolfram.com/SylvestersLineProblem.html]Sylvester's Line Problem[/url]. There are some links at this sites, may be they will help you.

Thank you very much. In fact, "my" generalization (in slightly different form) was conjectured in 1951 (T. Motzkin, proven by S. Hansen in 1965.), as stated in

Generalizations of the Sylvester Problem,M. Edelstein (Mathematics Magazine, Vol. 43, No. 5 (Nov., 1970), pp. 250-254)

[url]http://www.jstor.org/stable/2688739[/url]

50 years too late :)


inequalities
algebra
polynomial
inequalities proposed

Let $ a,b,c$ be real numbers. Find the maximum value of constant $ k$ so that the following inequality holds
$ a^4\plus{}b^4\plus{}c^4\plus{}k(a^3b\plus{}b^3c\plus{}c^3a) \ge 0.$

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} k\left(a^3b \plus{} b^3c \plus{} c^3a\right) \ge 0$ holds for all real numbers $ a,b,c$ if and only if $ \minus{} 1\leq k\leq k_1 \equal{} 1.4894\cdots$,

where $ k_1$ is the unique real root of irreducible polynomial $ 7k^5 \plus{} 17k^4 \plus{} 16k^3 \plus{} 16k^2 \minus{} 64k \minus{} 128.$

With equality if $ k \equal{} k_1,a \equal{} x_3 \equal{} \minus{} 1.8030 \cdots,b \equal{} x_2 \equal{} 0.84846\cdots,c \equal{} x_1 \equal{} 1.9545\cdots,$

where $ x_1,x_2,x_3$ are all real roots of irreducible polynomial

$ 40824x^{15} \minus{} 204120x^{14} \plus{} 326592x^{13} \plus{} 34992x^{12} \minus{} 915624x^{11} \plus{} 1658232x^{10}$

$ \minus{} 1697436x^9 \plus{} 1174824x^8 \minus{} 585630x^7 \plus{} 216702x^6 \minus{} 60642x^5 \plus{} 13050x^4 \minus{} 2187x^3$

$ \plus{} 279x^2 \minus{} 24x \plus{} 1.$

[quote="arqady"]See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112423[/quote]
Your proof is very nice, arqady. Thank you very much.


calculus
integration
floor function
trigonometry
calculus computations

\[ \int\limits_0^{100} {\left\lfloor {\tan ^{ \minus{} 1} x} \right\rfloor } dx\]
Please show all the steps clearly, I need to understand..the substitution of variables..

This has got nothing to do with subsitution of variables.

Just break up $ [0,100]$ into $ [0, \tan 1]$ and $ [\tan 1, 100]$ and integrate on those separately.


ARML
USAMTS
geometry
geometric transformation
reflection
MATHCOUNTS
AMC

Tryouts for the 2009 Chicago ARML Teams will be held on Thursday, February 26, 2009, from 4 to 10 pm (with a break for dinner) at Evanston Township High School in Evanston, IL.

Tryouts are open to any student who has not yet completed high school and lives in the northern 2/3 or so of Illinois.

For more information, see http://chicagoarml.org/

By the way, it's traditional to eat dinner at [url=http://maps.google.com/maps?f=q&source=s_q&hl=en&geocode=&q=3727+Dempster+St,+Skokie,+IL&sll=42.009509,-87.675705&sspn=0.219386,0.44632&ie=UTF8&t=h&layer=c&cbll=42.040738,-87.721638&panoid=nkpV-pFlix2bVzj39ngO1A&cbp=12,140.66294184473654,,0,5&ll=42.040608,-87.721646&spn=0.006852,0.013947&z=17&iwloc=addr]Hubs Restaurant[/url]

(sorry if you were supposed to do this Coach Greenspan, but I thought I'd get the word out a tiny tiny bit early?)

lol, just a tiny bit earlier.

I'll be there!  I just barely make the requirements for above 39 degrees latitude though. :P

[quote="gauss1181"]Hmmm..I'm not sure whether I could try out or not. Do you have to register in advance and inform your school or something, or could you just attend the tryouts yourself? Are the tryouts free? And is dinner actually provided free of charge?[/quote]

I think you just go. For free. Except no free food. Hubs isn't a terribly fancy restaurant, but it's not McDonald's either, so plan on like $ \$$10 ish?

[quote="gauss1181"]Other than chicagoarml.org, are there any other sites for more info?[/quote]

http://www.artofproblemsolving.com/Forum/index.php?f=348 isn't very "informational"

http://www.arml.com/index.php

http://www.artofproblemsolving.com/Wiki/index.php/ARML

http://www.artofproblemsolving.com/Wiki/index.php/Chicago_ARML

[quote="gauss1181"][quote="suma_milli"][quote="gauss1181"]You really think you can make the Chicago ARML team, suma? :maybe:
Or are you just going there for fun or just to see me??? LOL :D[/quote]

I think I might possibly make D-Team at least,  but yeah it'll be fun too :P[/quote]
Wait so there are multiple Chicago ARML teams? What's the D-team?[/quote]

chicagoarml.org/dteam/

Huh?

There's no way you can be on D-team...tryouts ALREADY happened January 31 :(  :wink:

No that's from last year. See at the bottom it says last updated December 2007?

lol that's what I was confused about at first too.

[quote="gauss1181"]Didn't they say they select only 9th and 10th graders for D-team but students of any age up to high school age can be on teams A, B, and C?[/quote]
 :maybe:

lol just maybe... :maybe:

Yes I poked/nudged guassy to go to ARML tryouts :P

I really can't imagine you in real life.   :huh: I can imagine a person wearing a huge Mr.Green costume and a short asian girl like you said.... but still..

[quote="suma_milli"]
I really can't imagine you in real life.   :huh: I can imagine a person wearing a huge Mr.Green costume and a short asian girl like you said.... but still..[/quote]
 :maybe:  :huh:

Spam is good for your health.
On topic, uh...see the rest of you ARML people on teh 26th?

Are we allowed to discuss the ARML hw or does it all have to be done individually?

so that means we can check answers as long as we already did the problem right?

"You are expected to attempt every problem, and make at least some effort to get help if you get stuck. Good sources for help are: current and former ARML team members, your math team coach, or any of the D-team coaches. We especially encourage you to use the D-team email group, <name removed>, to share questions, ideas, and hints about the problems. (The only thing that is discouraged is simply posting an entire solution to a problem.)"

- D-team page from Coach Karafiol.

I emailed Coach Greenspan today to see about an email list or group for the C-team.  There is usually one for ABC, but I think we are doing C a bit differently this year.

Yes, you can discuss the problems.  Please do not do it in the open on a forum like this, but I am happy to set up a closed group for the problems.  What we'd like is for you to see the different techniques of solving.  We'd like to see organized solutions.  We'd like to see mature mathematical proofs if needed.

If you do not understand a problem, then please ask.  We are trying to help you along your mathematical journey.  Some of the problems are easy. Some seem like case work.  Some involve topics you may not have covered.  Please hurt your brain trying to solve the problem before asking.  You can only truly know the right path after walking down the wrong one.

I'd like you to tear the problems apart.  Figure out why certain numbers were used.  Could the problem be extended?  How could you combine a problem with another topic to make a better problem?  Push each question to the corners of mathematics and see what is there.  ARML does not use very deep topics for the questions, but it finds recesses and alcoves within basic concepts.

What we do not want is sharing solutions.  It is unfair to your teammates if you are not working out your brain.

coach fritz -- see HW for my email address or pm me if needed

[quote="sunehra"]In my problem set it says you're allowed to talk with members of the team and it doesn't specifically say that you can't get help from anyone. The only thing it tells you not to do is use a calculator.[/quote]

Oh so you made the team sunehra, which one?

y'all can PM me for hw help if you want.

@coaches: I promise I won't cheat :]

I made the D-Team (as I mentioned in a previous post).

I made the A-Team.  Me and Ravi were competing. He stayed one ahead of me the whole time, and he finished one ahead of me. Josh Dement made A/B, too.

[quote="suma_milli"]So Ravi made the A team as an 8th grader? Wow..[/quote]
And Ravi is a seventh grader....

[quote="suma_milli"]Oops I keep forgeting because all of us 8th graders keep talking about him.

..... :|

So Ravi made A/B as a [i]7th grader[/i]?  Even more beastingness.[/quote]
yes... :o  :o  :o

bobby shen made texas gold as a 7th grader


Square ABCD into the circle of radius R. Prove for any point of the circle M equality
$ AM^4 \plus{}BM^4 \plus{}CM^4 \plus{}DM^4 \equal{} 24R^4$

Let $ A\equal{}(1,1)$, $ B\equal{}(1,\minus{}1)$, $ C\equal{}(\minus{}1,\minus{}1)$, and $ D\equal{}(\minus{}1,\minus{}1)$. $ M\equal{}(x,y)$ where $ x^2\plus{}y^2\equal{}2$.

The rest is a computation with the distance formula...

Yes but the equation is homogeneous so we might as well suppose that $ R\equal{}\sqrt{2}$.

By [b]Pithagorean theorem[/b] from $\triangle AMC\implies AM^2+CM^2=AC^2\iff AM^2+CM^2=4R^2\ (\ 1\ )$ and from $\triangle BMD\implies BM^2+DM^2=4R^2\ (\ 2\ )$.
Squaring the 2 equalities and adding them side by side: $\sum AM^4 +2(AM^2\cdot CM^2+BM^2\cdot DM^2)=32R^4\ (\ 3\ )$.
But, expressing in 2 different ways the area of $\triangle AMC\implies AM\cdot MC=AC\cdot MX$ (where $X$ is the pedal of $M$ onto $AC$); squaring it it yields $\left(AM\cdot CM\right)^2=4R^2\cdot MX^2$, and similar for the other product: $\left(BM\cdot DM\right)^2=4R^2\cdot MY^2$, where $Y$ is the pedal of $M$ onto the diameter $BD$.
Adding the last two equalities: $\left(AM\cdot CM\right)^2+\left(BM\cdot DM\right)^2=4R^2(MX^2+MY^2)$, but $(MX^2+MY^2)=XY^2=OM^2=R^2$, $O$ being the center of the square, with the latter equalities, from $(3)$ we get the desired relation.

Best regards,
sunken rock


real analysis
calculus
integration
function
real analysis unsolved

Assume wlog that your function is non-negative. If you denote $ a_n \equal{} \lambda(x: g(x)\in[n,n \plus{} 1])$, then the statement can be reformulated as:

If for any positive $ b_n$ s.t. the series $ \sum_{n \equal{} 1}^\infty a_n b_n$ converges,  the series $ \sum_{n \equal{} 1}^{\infty} n a_n b_n$ converges as well, then there is $ n_0$ s.t. $ a_n \equal{} 0$ for $ n\ge n_0$.

Does this series reformulation make your life easier? Just tell if not, I will give other hints.

Otherwise, you can try to use Banach-Steinhaus theorem.

Hello Sir,

Thanks for your help. I do not know how to procede with the given hint, so could you please help more.

[quote="santug"]Let g be a Lebesgue measurable function on R such that g*f is Lebesgue integrable for every fuction in L(R).Show that g is essentially bounded[/quote]

Hint: Suppose there are pwdj sets E_n, 0 < m(E_n) < oo, such that g > 2^n on E_n for each n. Can you then find f = sum c_n*X(E_n) leading to a contradiction?

Ok, further hint: if all $ a_n$ are non-zero, then you can take $ b_n\equal{}1/(n^2 a_n)$, so that the series $ \sum a_n b_n$ converges, while $ \sum n a_n b_n$ does not.

Now try to extend this to the case where there are some $ a_n$ equal to zero, but still infinitely many of them non-zero.


Fie a,b,c numere pozitive . Sa se arate ca:
(a^3+1)(b^3+1)(c^3)>=(a^2b+1)(b^2c+1)(c^2a+1)

[quote="Andrei Mihnea Stefan"]Fie a,b,c numere pozitive . Sa se arate ca:
$\ (a^{3}+1)(b^{3}+1)(c^{3}+1)\geq\ (a^{2}b+1)(b^{2}c+1)(c^{2}a+1)$[/quote]
banuiesc ca asta e forma corecta...

Desfacand parantezele, $\sum_\textrm{cyc}b^{3}c^{3}+\sum_\textrm{cyc}a^{3}\geq \sum_\textrm{cyc}a^{3}b c^{2}+\sum_\textrm{cyc}a^{2}b$.

$AM-GM$: $2 a^{3}b^{3}+b^{3}c^{3}\geq 3 \sqrt[3]{a^{6}b^{9}c^{3}}= 3 a^{2}b^{3}c$.

Adunand toate inegalitatile obtinem $\sum_\textrm{cyc}b^{3}c^{3}\geq \sum_\textrm{cyc}a^{3}b c^{2}$.

Din inegalitatea rearanjamentelor obtinem $\sum_\textrm{cyc}a^{3}\geq\sum_\textrm{cyc}a^{2}b$ si e gata problema.

O alta solutie este folosind holder $(a^{3}+1)(a^{3}+1)(b^{3}+1)\geq (a^{2}b+1)^{3}$ \c{s}i analoagele. :)

In acest caz, inegalitatea e foarte cunoscuta..
[i]Solutie[/i]
Deoarece inegalitatea este simetrica, putem presupune ca avem ordinea $\ a\leq\ b\leq\ c$.Folosim spargerea.Pentru asta, sa aratam prima data ca $\ (a^{3}+1)(b^{3}+1)\geq\ (a^{2}b+1)(ab^{2}+1)$.Aceasta devine $\ a^{3}+b^{3}\geq\ a^{2}b+ab^{2}$ si este adevarata, conform Inegalitatii lui Muirhead.Pana acum am demonstrat ca $\ LHS\geq\ (a^{2}b+1)(ab^{2}+1)(c^{3}+1)$.Inecrcam sa demonstram ca $\ (ab^{2}+1)(c^{3}+1)\geq\ (b^{2}c+1)(c^{2}a+1)$, sau $\ c^{3}+ab^{2}\geq\ ac^{2}+b^{2}c$.Aceasta din urma e echivalenta cu $\ (c-a)(c-b)(b+c)\geq\ 0$.Si demonstratia se incheie. :)

Ai grija cu presupunerea faptului ca $a \geq b \geq c$. In cazul de fata n-ai voie s-o faci, ci doar $c = \max \left\{ a,b,c \right\}$.

[quote="stancioiu sorin"]Aceasta devine $\ a^{3}+b^{3}\geq\ a^{2}b+ab^{2}$ si este adevarata, conform Inegalitatii lui Muirhead.[/quote]

Sorine... o sa se supere Cezar pe tine  :D  Cum sa folosesti Muirhead?

m-am gandit la asta, sa stii...e echivalenta cu $\ (a-b)^{2}(a+b)\geq\ 0$.Si nu mai folosim Muirhead. :D

1)Generalizarea poate merge cu inductie...
2)  Am mai gasit o inegalitate asemanatoare cu cea a lui Andrei.E vorba de
$\ (1+x^{3}y)(1+y^{3}z)(1+z^{3}x)\geq\ (1+x^{2}yz)(1+y^{2}zx)(1+z^{2}xy)$

Atunci facem ca mateivld, cu Holder, ca in postul #4. :)


MATHCOUNTS
mathleague.org
geometry
3D geometry
sphere
ARML

hi everyone,
thanks to those of you who came out to the contest today, and congratulations to all the winners.  i was really glad we were able to pull off another contest before mathcounts season starts.  based on the surveys i got it looks like a lot of you are interested in having another (harder) contest on march 14, so i'll do everything i can to make that happen.  you can watch our website (http://mathleague.org/jhevents.php) for more information, or sign up on our mailing list at http://mathleague.org/mailman/listinfo/mailinglist_mathleague.org if you want to be informed as soon as we have details confirmed.  good luck to everyone at chapter mathcounts..

-Tim

Yay so how did everybody do?

I epic-failed written round...an arithmetic error on sprint, a logic fail on sprint, and a formula fail on target.  42 score.  Tie for 4th in 7th grade.

I also got 42 score, missing numbers 1,4,6, and 19 on the Sprint. How did you know you tied for 4th?

Well my score tied for 4th.

43 was 3rd :)

I missed the cone and sphere one on target, and 6 and 8 (i think) on sprint.  Arithmetic errors fail.

based on the surveys i got it looks like a lot of you are interested in having another (harder) contest on march 14, so i'll do everything i can to make that happen.

Harder? As in how hard?
Well March 14 is in time for State MC, that's ok.

Dude, dragon, Hopkins had a "Nerd Day" celebration last year on Pi Day, except it was EPIC PHAIL. PHAIL. BY THE WAY, IT WAS PHAIL. OH, DID I MENTION IS WAS RIDICULOUS PHAIL. THEY HAD NERD DAY EXCEPT IT WAS PHAIL. OH YEAH, IT WAS PHAIL.

Ok.

james4l: Apparently state/nats level.

AIME15: Dang, nice on the CD.

And my score is too embarrassingly low to post. If I had gotten about the same score as my teammates, we should have gotten at least 2nd place team.

Slightly off topic, but has the location of the February 14 contest been determined yet?

You mean march 14, right?

I don't think so.  At least it doesn't show up on http://mathleague.org/jhevents.php.

Oh sorry, I meant the high-school contest for ARML team selection. That's supposed to be on Feb. 14, which is pretty soon.

Oh bleh, that interferes with my MATHCOUNTS chapter...

Is there another one in March? (and I need to get top 3, no excuses)

The February 14th contest is at Stanford.

[quote="dragon96"]And it also happens to be Pi Day. Unfortunately, no school ever celebrates Pi Day...[/quote]

My middle school did. I had a cool 7th grade math teacher.

I went there. I got 1st place in 7th grade. 28 on sprint round with 2 arithmetic errors. >.> Perfect target, and 9 on the team.

Lol I got owned in the countdown.

Did anyone feel like the sprint round questions weren't really arranged in increasing difficulty?

I had a subtraction error, a forgotten formula, and a logic/misread fail.

But I got a perfect score!!!!!!!!!!!!!!!!! 42

I agree... number 30 was pretty easy, compared to 25-29, in my opinion.

Target problem #8 (the one with the dartboard) was a target problem on 1993 state sprint.

I found the last two target problems too easy.

Yeah, they were easier than most of the countdowns.