projective geometry
geometry
3D geometry
sphere
similar triangles
geometry proposed

Let extension of opposite sides of any convex quadrilateral intersect at K and L points. Let the diagonals of quadrilateral intersect at O point. Let "d" line that passes through O point and parallel to KL. Prove that the midpoint of segment inside the quadrilateral of that d line is "O" point.

Im wondering is it a well known theorem? And what is the source of this problem. I like it very very much.

Dear Mathlinkers,
let ABCD a quadrilateral, P the intersection of AD and BC, Q the intersection of AB and CD.
It is well knowed that (PA, PBD, PO, PQ) = -1 (A result of Pappus, proposition 131 livre VII), then the rest follows (Pappus corollary c'est un corollaire of the proposition 129) when d is parallel to PQ.
Sincerely
Jean-Louis

Does  it mean this is  a problem discovered by Papus?

I tried to understand your solution but I cannot. May you explain it without involving Papus theorem and only using similar triangles?

Dear Borislav and Mathlinkers,
let ABCD a quadrilateral, P the intersection of AD and BC, Q the intersection of AB and CD.

1. let Q' be the point of intersection of PO and AB
2. Accordling to Pappus, proposition 131 livre VII, 	(A, B, Q', Q) = -1
3. We consider the pencil :			(P; A, B, Q', Q) = -1
4. Let E, F the points of intersection of d and BC, of d and DA.
5. The line d passing through O and being parallel to the ray PQ,     O is the midpoint of EF.

Sincerely
Jean-Louis

It seems to me that Pappus Prop. 129 is synthetic,
    when same Pappus but Prop. 131 Book VII  is not.

    M.T.

I'm sorry but please may you give more detailed steps. And is there a proof not using Papus. If your proof use Papus - please say the exact statement you use. I don't understand the notation given by Jean-Louis. Math is not my profession maybe it is the reason.

... I saw a solution using ... sphere and pyramid...

I think there should be written a detailed solution to this problem - it is good enough.

I'm sorry for double sequential post but I don't have a solution to this problem I can understand. May someone try to post a detailed solution to this problem (or solution based of homothethy/similarities or someone else well known)?

Let us to use the wording of [b][size=100]Jean-Louis[/size][/b].

So, let be the points $ P\equiv AD\cap BC,\ Q\equiv AB\cap CD,\ Q'\equiv AB\cap PO$ and we denote as $ E,\ F,$ the intersection points of $ BC,\ AD$ respectively, from the line through the point $ O\equiv AC\cap BD$ and parallel to $ PQ.$

We draw the line through the point $ D$ and parallel to $ PQ,$ which intersects the line segments $ PO,\ BC,$ at points $ M,\ N$ respectively and we denote the point $ K\equiv CD\cap PO.$

Because of now, $ EF\parallel DN\parallel PQ,$ based on [b][size=100]Thales theorem[/size][/b], it is enough to prove that $ MD \equal{} MN.$

$ \bullet$ In the triangle $ \bigtriangleup CDN$ with the line segment $ PK$ as its transversal, applying the [b][size=100]Menelaus theorem[/size][/b],

we have that $ \frac {KC}{KD}\cdot \frac {MD}{MN}\cdot \frac {PN}{PC} \equal{} 1$ $ \Longrightarrow$ $ \frac {MD}{MN} \equal{} \frac {KD}{KC}\cdot \frac {PC}{PN}$ $ ,(1)$

From $ DN\parallel PQ$ $ \Longrightarrow$ $ \frac {PC}{PN} \equal{} \frac {QC}{QD}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {MD}{MN} \equal{} \frac {KD}{KC}\cdot \frac {QC}{QD}$ $ ,(3)$

But, from the complete quadrilateral $ PCODAB,$ we conclude that the points $ C,\ K,\ D,\ Q$ are in harmonic conjugation, as well known and easy to prove by [b][size=100]Ceva theorem[/size][/b] and [b][size=100]Menelaus theorem[/size][/b] (in every complete quadrilateral, each of its diagonals is intersected from the other two ones, at two points which are in harmonic conjugation with respect to the edges of the reference diagonal).

So, we conclude that $ \frac {KC}{KD} \equal{} \frac {QC}{QD}$ $ ,(4)$

From $ (3),$ $ (4)$ $ \Longrightarrow$ $ \frac {MD}{MN} \equal{} 1$ $ \Longrightarrow$ $ MD \equal{} MN$ $ \Longrightarrow$ $ OE \equal{} OF$ and the proof is completed.

Kostas Vittas.

Why do you repost http://www.mathlinks.ro/Forum/viewtopic.php?t=69971 in exactly the same wording?

  darij

Thank you for detailed explanation, vittasko. Now it is far better.

To darij - I'm sorry, I didn't know the problem is posted before - it is the reason.
But this posting again is not bad ... we know different approaches for solving the same problem.

I have a question.

May someone give me a link to the proof of  the following statement only by Menelaus, Ceva and maybe similar triangles?

[quote="vittasko"]
But, from the complete quadrilateral $ PCODAB,$ we conclude that the points $ C,\ K,\ D,\ Q$ are in harmonic conjugation, as well known and easy to prove by [b][size=100]Ceva theorem[/size][/b] and [b][size=100]Menelaus theorem[/size][/b] (in every complete quadrilateral, each of its diagonals is intersected from the other two ones, at two points which are in harmonic conjugation with respect to the edges of the reference diagonal).
[/quote]

[quote="borislav_mirchev"]May someone give me a link to the proof of  the following statement only by Menelaus, Ceva and maybe similar triangles?
[quote="vittasko"]
But, from the complete quadrilateral $ PCODAB,$ we conclude that the points $ C,\ K,\ D,\ Q$ are in harmonic conjugation, as well known and easy to prove by [b][size=100]Ceva theorem[/size][/b] and [b][size=100]Menelaus theorem[/size][/b] (in every complete quadrilateral, each of its diagonals is intersected from the other two ones, at two points which are in harmonic conjugation with respect to the edges of the reference diagonal).
[/quote][/quote]
In the triangle $ \bigtriangleup PCD,$ with the point $ O$ outwardly to it, based on the [b][size=100]Ceva theorem[/size][/b], we have that $ \frac {KC}{KD}\cdot \frac {AD}{AP}\cdot \frac {BP}{BC} \equal{} 1$ $ ,(1)$

In the same triangle, taken as its transversal the line segment $ BQ,$ based on the [b][size=100]Menelaus theorem[/size][/b], we have that $ \frac {QC}{QD}\cdot \frac {AD}{AP}\cdot \frac {BP}{BC} \equal{} 1$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {KC}{KD} \equal{} \frac {QC}{QD}$ $ ,(3)$

By the same way, in the triangle $ \bigtriangleup PAB$ with the point $ O$ inwardly to it and the line segment $ CQ$ as its transversal, we can prove that $ \frac {Q'A}{Q'B} \equal{} \frac {QA}{QB}$ $ ,(4)$

In the triangle $ \bigtriangleup KQQ'$ now, with the line segments $ BD,\ BP$ as its transversal, applying again the [b][size=100]Menelaus theorem[/size][/b],

we have that $ \frac {OK}{OQ'}\cdot \frac {BQ'}{BQ}\cdot \frac {DQ}{DK} \equal{} 1$ $ ,(5)$ and $ \frac {PK}{PQ'}\cdot \frac {BQ'}{BQ}\cdot \frac {CQ}{CK} \equal{} 1$ $ ,(6)$

From $ (5),$ $ (6)$ and because of $ \frac {DQ}{DK} \equal{} \frac {CQ}{CK}$ from $ (3),$ we conclude that $ \frac {OK}{OQ'} \equal{} \frac {PK}{PQ'}$ $ ,(7)$ and the proof is completed.

$ \bullet$ Note please that after $ (3)$ we can say that the pencil $ P.CKDQ$ is also in harmonic conjugation and so, because this pencil is intersected from the line segment $ BQ,$ based on the [b][size=100]Pappus theorem[/size][/b] as mentioned by [b][size=100]Jean-Louis[/size][/b], we conclude that the points $ B,\ Q'\ A,\ Q$ are in harmonic conjugation too.

The line through the point $ O$ and parallel to $ PQ,$ intersects also the pencil $ P.CKDQ$ and so, we can say that the points $ E,\ O,\ F,\ \infty,$ are in harmonic conjugation too.

In this special case we have also $ \frac {OE}{OF} \equal{} \frac {\infty E}{\infty F} \equal{} 1$ $ \Longrightarrow$ $ OE \equal{} OF.$

Kostas Vittas.

Thank you, vittasko, excellent work. I'll be happy if some day I become smart like you.


function
advanced fields
advanced fields unsolved

Show that if a network $ N$ contains no directed path from the source $ s$ to the sink $ t$, then the maximum possible total flow on $ N$ (for any flow function) is $ 0$.

I am not sure how to prove this. I know I want to use the [b]max flow min cut theorem[/b] that would show that the value of a maximum flow and the capacity of a minimum cut are both zero. So, I need some help with this one. Thanks in advance.

Here is a cut: let S be the set of all vertices reachable from s, and let T be everything else. There are no edges going from S to T by definition, and both are nonempty since there is no path from s to t. Therefore the max-flow is at most the value of this cut, which is zero.


counting
distinguishability
geometry
3D geometry
algebra
polynomial

$2n$ objects of each of three kinds are given to two persons, so that each person gets $3n$ objects. In how many ways can this be done?

assuming kinds are distinguishable, then we just have a list of 6n distinguishable objects to give to two people so that each gets 3n, or (6n)C(3n)

[quote="me@home"]we just have a list of 6n distinguishable objects[/quote]

Not correct. We have $2n$ of kind, say A, $2n$ of B, and $2n$ of C.  :P

[hide]
For the sake of being more interesting, these objects are now apples

the three types of apples are Green, Red, and Blue

We have $2n$ green apples, $2n$ red apples, and $2n$ blue apples

The first person will get anywhere from 0 to 2n of each type of apple, so the number of ways he can receive apples is

$(2n+1)^{3}$
[/hide]

What is it was $3$ people?  And yellow apples instead of blue :lol: ?

Incidentally, no one has correctly answered yet.  me@home treated indistinguishable objects as distinguishable objects; chickendude neglected the limitation that each person receive exactly $3n$ objects.

Yes I didn't really understand the question... i'm so bad on wording.
2n A's, 2n B's, 2n C's.
Person 1 gets x A's, y B's, z C's.
Then person 2 gets everything else.
x+y+z=3n

-------

Okay I finally understand the question:
solve in non-negative ints x+y+z=3n with x,y,z<=2n

-------

There are 2n+1 choices for x; then, x+y<=3n so y<=3n-x so there are 3n-x+1 choices of y. There is always one choice for z; but z<=2n means that x+y>=n. So min(3n-x,2n)>=y>=n-x. This suggests two cases:
[b]1[/b] 3n-x>=2n<=>n>=x
2n>=y>=n-x yields 2n-n+x+1=n+x+1 choices for y
To find total choices for y we sum over x. $\sum_{x=0}^{n}n+x+1=n(n+1)+n(n+1)/2+n+1$
[b]2[/b] 2n>3n-x<=>x>n
3n-x>=y>=0(>n-x) so 3n-x+1 choices for y. Sum over x, i.e. $\sum_{x=n+1}^{2n}3n-x+1=n[3n-n-(n+1)/2+1]$
[b]Total:[/b] $\boxed{n[2n+(n+1)+2]+1}$
For instance when n=3 we have
x+y+z=9, 0<=x,y,z<=6
There should be 37 solutions.
Indeed, for x=0: (3,6)...(6,3) 4 solutions
x=1: (2,6)...(6,2) 5 solutions
x=2: (1,6)...(6,1) 6 solutions
x=3: (0,6)...(6,0) 7 solutions
x=4: (0,5)...(5,0) 6 solutions
x=5: (0,4)...(4,0) 5 solutions
x=6: (0,3)...(3,0) 4 solutions
(which is 37)

Okay I think that's it...

Oh ya, btw there's a much nicer solution.
[hide="HINT"]
That above formula simplifies to $(n+1)^{3}-n^{3}$[/hide]

1729 posts! There was some story about some mathematician that said that every number had something special about it and someone said, well what about 1729, and the answer is it is the smallest number that can be written as the sum of two cubes in two different ways. It is $12^{3}+1=9^{3}+10^{3}$.

Just letting you know that.... :roll:

[quote="me@home"]1729 posts! There was some story about some mathematician that said that every number had something special about it and someone said, well what about 1729, and the answer is it is the smallest number that can be written as the sum of two cubes in two different ways. It is $12^{3}+1=9^{3}+10^{3}$.

Just letting you know that.... :roll:[/quote]

For the record, that was Ramanujan and Hardy; Ramanujan was in the hospital for some reason, and Hardy was visiting; Hardy remarked that the taxi he had taken to get there was numbered 1729, which was, according to him, "a very boring number". Ramanujan corrected him by pointing out the property you just did.

Interestingly enough, there is an entire wiki page dedicated to interesting properties of this number: http://en.wikipedia.org/wiki/1729_%28number%29

[quote="me@home"]Yeah, that was the story...

AK! It's the guy who's avatar I stole er... asked for.
Remember? That must've been like a year ago...[/quote]

Hehe, yup, I remember you :) It was quite a long time ago. I've been lurking on AoPS non-stop since then, not posting too much (in the meantime I've started University math degree), but you seem to have really flourished here; you used to have fewer posts than I did back in the day! :)

[quote="miyomiyo"]What if it was $3$ people?[/quote]
Anybody?

[i]$2n$ objects of each of three kinds are given to three persons, so that each person gets $2n$ objects. In how many ways can this be done?[/i]
Is equivalent to: $x_{i}\in\mathbb{N}_{0}$
\begin{align*}x_{1}+y_{1}+z_{1}&=2n\\ x_{2}+y_{2}+z_{2}&=2n\\ x_{3}+y_{3}+z_{3}&=2n\\ \overline{2n+2n+2n }&=\overline{6n}\end{align*}

Isn't this the committee problem? (In how many ways can you separate nk people into n committees of k each?)


Arqady could you please post your proof ? Thanks.

silouan, I found two ways: by Jensen's inequality and by Mildorf's inequality.
Try more. :wink:

Thank you arqady for the hint .
By your hint let $\sqrt[3]{a}=x$  and let $x+y+z=1$ so by weighted Jensen we
have
$\sum x\sqrt[3]{\frac{y^{3}+z^{3}}{2}}\leq \sqrt[3]{\sum \frac{x(y^{3}+z^{3})}{2}}$

Now it suffices to prove that
$2(x^{2}+y^{2}+z^{2})^{3}\geq (\sum x(y^{3}+z^{3}))(x+y+z)^{2}$

which it true by Muirhead and AM-GM (after expanding)

Yes, silouan! :lol:
Or by Mildorf's inequality: $\sqrt[3]{\frac{y^{3}+z^{3}}{2}}\leq \frac{y^{2}+z^{2}}{y+z}.$ :wink:

[quote="arqady"]
Or by Mildorf's inequality: $\sqrt[3]{\frac{y^{3}+z^{3}}{2}}\leq \frac{y^{2}+z^{2}}{y+z}.$ :wink:[/quote]

Great Arqady  :lol:  . Very nice solution . Thanks for it .
BTW , you will go to Vietnam as a Leader ?

[quote="silouan"]
BTW , you will go to Vietnam as a Leader ?[/quote]
No, silouan. At us goes only Shai Geron.

[quote="arqady"]
 At us goes only Shai Geron.[/quote]

What a pity  :(  . But I  hope I will you meet you some day  :)

[quote="Karth"][quote="arqady"]Let $a,$ $b$ and $c$ are non-negative numbers. Prove that

$\sqrt[3]{ab+ac}+\sqrt[3]{ab+bc}+\sqrt[3]{ac+bc}\le\sqrt[3]{2}\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}\right)$[/quote]

Let $a \ge b \ge c.$ the result follows by Rearrangement.[/quote]
I don't see it. :maybe:  Explain pleas. :wink:  Thank you.

Eek. It doesn't work  :( What happened... let me check...

EDIT: Sorry, Rearrangement doesn't work. But here's another quick solution  :D

[hide="Another quick solution"]

The inequality is homogenous with degree $2/3,$ so let $a = 0.$ We then have to solve the inequality

$\[2 \sqrt[3]{bc}\le \sqrt[3]{2b^{2}}+\sqrt[3]{2c^{2}}\]$

which is true by AM-GM. $\blacksquare$

[/hide]

[quote="Jan"]Karth,
Your so called 'quick solution' contains a very fundamental mistake.
I assume you read about this technique without really knowing why it is valuable.
If you use homogenisation, please make sure you understand why you can make an assumption, and what assumtions are allowed.

Here's why one can make an assumption on the variables in a homogenous inequality.
If we want to prove the expression
\[f(a_{1},a_{2},...,a_{n}) \geq g(a_{1},a_{2},...,a_{n}) \]
where both $f$ and $g$ are homogenous, we know that we get an equivalent statement if we multiply every variable by some constant factor $\lambda \neq 0$.
We know this, because
\[f(\lambda a_{1},\lambda a_{2},...,\lambda a_{n}) \geq g(\lambda a_{1},\lambda a_{2},...,\lambda a_{n}) \]

\[\iff \lambda^{k}f(a_{1},a_{2},...,a_{n}) \geq \lambda^{k}g(a_{1},a_{2},...,a_{n}) \]

\[\iff f(a_{1},a_{2},...,a_{n}) \geq g(a_{1},a_{2},...,a_{n}) \]
Proving the desired inequality is thus equivalent to proving the inequality with the same variables that are 'scaled'.
Now, we know that this equivalence exists, so we can 'choose' a particular $\lambda$ in such a way that thing turn out nicely for us. For example, we can choose a $\lambda$ such that $\lambda a_{1}+\lambda a_{2}+...+\lambda a_{n}= C$ for some desired $C$ (often we pick $C=1$) or that $a_{1}a_{2}...a_{n}=C$, or $a_{1}=C$ or whatever.

Obviously, you can't assume that $a=0$, because there are cases in which $a\neq 0$, and in those cases, you can never find a $\lambda \neq 0$ to make $\lambda a = 0$.[/quote]

I know why homogenization works... but I forgot that $\lambda \neq 0...$ :(

[quote="Phelpedo"]A lot of inequalities can be "proven" by multiplying both sides by 0 if you don't check for reversibility  :wink:[/quote]

Actually, all can, but that was just a careless error on my part  :D


Let $ X_{\alpha}$ be a nonempty topological space for each $ \alpha \in A$ and let $ X \equal{}\prod X_{\alpha}$.

1. If $ V\neq \emptyset$ is an open set in $ X$, then $ \pi_{\alpha}^{\minus{}1}(V) \equal{} X_{\alpha}$ for all but finitely many $ \alpha \in A$.

Proof: Note that the $ \alpha^{th}$ projection map: $ \pi_{\alpha}: X \to X_{\alpha}$ is an open map. So that $ \pi_{\alpha}^{\minus{}1}(V)$ is open in $ X_{\alpha}$ for each $ \alpha$. By the definition of a base for the product topology on X, we must have $ \pi_{\alpha}^{\minus{}1}(V) \equal{} X_{\alpha}$ except for $ \alpha \equal{} \alpha_{1}, ..., \alpha_{n}$.

I hope this is correct and if so then the next two are the ones I need some hints.

2. If $ b_{\alpha}$ is a fixed point in $ X_{\alpha}$, for each $ \alpha \in A$, then $ X'_{\alpha_{0}} \equal{} \{x \in X|x_{\alpha} \equal{} b_{\alpha}$ whenever $ \alpha \neq\alpha_{0}\}$ is homeomorphic to $ X_{\alpha_{0}}$.

3. If $ b_{\alpha}$ is a fixed point in $ X_{\alpha}$, for each $ \alpha \in A$, then $ A \equal{} \{x \in X|x_{\alpha} \equal{} b_{\alpha}$ except for finitely many $ \alpha \in A \}$ is a dense set in X.

The idea I have for 3 is to show that the complement of A is an open set.

For 2:

To show homeomorphism you have to construct a map in each direction. Maps to products are easy: in any category, a map to a product $ f: A \to \coprod_\alpha X_\alpha$ is determined [i] uniquely [/i] (and naturally) by a collection of maps $ f_\alpha : A \to X_\alpha$. There are also projection maps $ \pi_\alpha : X \to X_\alpha$. That's really all you have to work with in general, so start off with those. We have that $ f_\alpha = \pi_\alpha \circ f$.

1) We need a map $ X \to X_{\alpha_0}$. The map $ \pi_{\alpha_0}$ is a good candidate -- we don't have any other choice, really.

2) So we need a map in the other direction, from $ X_{\alpha_0}$ to $ X$.  A map to $ X$ is a family of maps to $ X_{\alpha_i}$; now we don't know anything about these other spaces, so we can't construct arbitrary maps from $ X_{\alpha_0}$ to them. However we have points $ b_{\alpha_i} \in X_{\alpha_i}$. The constant map taking $ X_{\alpha_0}$ to $ b_{\alpha_i}$ is something we can take, though. We use the identity map $ X_{\alpha_0} \to X_{\alpha_0}$ in addition. This collection of maps gives us a map $ f : X_{\alpha_0} \to X$. In actuality $ f$ is a map to the subspace $ X'_{\alpha_0}$. Then, by definition, ${ \pi_{\alpha_0} \circ f = id_{X_{\alpha_0}}}$. Furthermore, $ f \circ \pi_{\alpha_0} = id_{X'_{\alpha_0}}$.

So, we have our homeomorphism.

A minor remark to your proof of 1. which looks fine to me: I think you mean $ \pi_\alpha(V)$ instead of $ \pi^{ \minus{} 1}_\alpha(V)$.

As for 3) we better define $ B: \equal{} \{x \in X|x_{\alpha} \equal{} b_{\alpha}$ except for finitely many $ \alpha \in A \}$ to avoid logically problems. Otherwise we would have a circular definition $ A: \equal{}$ expression of $ A$. ;) [quote]The idea I have for 3 is to show that the complement of A is an open set.[/quote]This would imply that $ B$ is closed which is not true in general.

But to show that $ B$ is dense, let's take $ x\notin B$ and $ U$ an arbitrary open set of $ X$ containing $ x$. We must show that $ U\cap B\neq \emptyset$.
WLOG we can assume that $ U$ is an element of the basis of the product topology of $ X$, i.e.
\[ x\in U \equal{} \prod_{\alpha\in A'}U_\alpha\times\prod_{\alpha\notin A'}X_\alpha\]
where $ A'$ is a finite subset of $ A$ and open sets $ U_\alpha \subset X_\alpha$ for every $ \alpha\in A'$. Since $ x$ is not contained in $ B$, there is $ \alpha'\notin A'$ with $ x_{\alpha'}\neq b_{\alpha'}$. Then define $ y\in X$ with $ y_\alpha \equal{} x_\alpha$ for $ \alpha\in A'$ and $ y_\alpha \equal{} b_\alpha$ for $ \alpha\notin A'$. Then $ y\in U\cap B$. Hence $ B$ is dense in $ X$.


inequalities
geometry
circumcircle
incenter
geometry proposed

Let $ O,H,I$ be circumcenter, othorcenter, incenter of a triangle prove that
\[ OH^{2}\ge OI^{2}+2IH^{2}\]

Using the well-known identities:

$ OH^{2}= 9R^{2}-2p^{2}+2r^{2}+8Rr$, $ IH^{2}= 4R^{2}-p^{2}+3r^{2}+4Rr$ and $ OI^{2}=R^{2}-2Rr$

the inequality is equivalent to
\[ R\geq 2r. \]

oh nice solution it is so easier than I thought, thank you  :)


[quote="tdl"]Given a sequence $ a_1,a_2,...,a_{99}$ of one-digit numbers with the poperty that if for some $ n$ we have $ a_n \equal{} 1$, then $ a_{n \plus{} 1}\neq 2$; and if for some $ n$ we have $ a_n \equal{} 3$, then $ a_{n \plus{} 1}\neq 4$. Prove that exist two number $ k,l\in\{1,2,...,98\}$ such that $ a_k \equal{} a_l$ and $ a_{k \plus{} 1} \equal{} a_{l \plus{} 1}$.[/quote]
There are 100 pairs (a,b). If in your case all 98 pairs are distinct, then if $ (a,1)$ exist for any a, then $ a_{99}\equal{}1$ (because not exist (1,2)). If exist (a,3) for any a, then $ a_{99}\equal{}3$. Therefore there are not exist pairs $ (a,1)$ or $ (b,3)$ in $ (a_i,a_{i\plus{}1}),i\equal{}1,2,...98$. It give in the set of pairs $ (a_i,a_{i\plus{}1}),i\equal{}1,2,...98$ one of 97 pairs (exlused (1,2),(3,4) and one of (a,1),(b,3)) meeting twice.


Okay, This is a cross post - no response in other thread, so posted here....
Source:  Based on an Old world  math book

"Did your teacher give  you this problem, or is it from MNO?" I asked "It looks   rather hard."

"Nahin Uncle", said Chintoo , "Its called NMO, and actually I made it up. Its an equation, and my age (in years) is one of the roots"

"Then  it can't be that hard", I chuckled, " Polynomial equation, all coefficients are integers, your  root is an integer (I hope a positive one :) ) .. let me try x=7 .. No that gives 77"

"Do I look 7 yeas old"., Chintoo  pouted?

"Okay baba, lets us try a larger  integer,.. no that does not give a zero either, it gives 85"

"You KNOW  I am older than that... Not funny." STOP Kidding."

Ok so we stop kidding... but how old is Chintoo?

(BTW - NMO is a  math competition not unlike AMC ; "nahin" just means "no" and "uncle" is what every Indian friend's son/daughter  calls me)
(As usual hide your answers)

I don't know if this is helpful, but...
[hide="Hint"]The polynomial has to have degree of greater than 1. [/hide]

[hide="Another hint"]Suppose the second age that the uncle guesses is $k$.  Then $k-7|85-77$, so $k=8,9,11,15$.[/hide]


Find the primes x,y \in N s.t.
C(x,y)=x+y+xy+4,
where C(n,k)=n!/(k!(n-k)!)

I thought I've already prposed a solution for this one, but my answer has disapeared ?
Never mind.

Assuming x >=  y > 0

Since x is prime, we must have x | x+y+xy+4 or x | y+4.
x = 2 or x=y have no solution obviously, so x and y have same paritty i.e x = y+2 or x=y+4.

x = y+2 gives no positive solution
x = y+4 gives only x=7 and y=3 as only solution.

[quote="belenos"]I thought I've already prposed a solution for this one, but my answer has disapeared ?
Never mind.
[/quote] you  posted when I said not to post because the site was moving to a new place :D ;)


hi, i'm new here and i'd like to ask you for some help ...

i cannot figure out how to solve this one:

log z=log (2u^2-2v^3u)-1/2(log v-3log (2u))

(i have to find solution for z (z=?)).

thanks!

[quote="samath"]Well, if you're new  here, I recommend that one of the first things you do is learn LaTex.

For now:

$log z=log (2u^2-2v^3u)-\frac{log v-3log 2u}{2}$

Is that you're question, or did I mess something up?[/quote]

[hide="solution"]you want to solve for z right?

if you simplify that,
$log z= \frac{2 log (2u^2-2v^3u)}{2}-\frac{log v - log 8 u^3}{2}$
$=\frac{log (2u^2-2v^3u)^2 - log \frac{v}{8u^3}}{2}=\frac{log \frac{(2u^2-2v^3u)^2\cdot 8u^3}{v}}{2}$
$=log \sqrt{\frac{(2u^2-2v^3u)^2\cdot 8u^3}{v}}$

so ${z=\sqrt{\frac{(2u^2-2v^3u)^2\cdot 8u^3}{v}}=\boxed(2u^2-2v^3u)\cdot 2u\sqrt{\frac{2u}{v}}}$

i hope i didnt make any careless mistakes :) [/hide]


The Yankees and the Red Sox are having a playoff series. The series continues until one team has won three games. Assuming each team has an equal chance to win each game, what is the expected number of games in the series?

[hide="Solution"]Since there are a maximum of five games, there are five games in which we can choose three of them to be Yankee wins. Thus, there are $ \binom{5}{3} \equal{} 10$ ways for three Yankee wins to be spread out over "five" games.

There is $ \binom{2}{2} \equal{} 1$ possibility that the Yankees win the series in $ 3$ games.
There are $ \binom{3}{2} \equal{} 3$ possibilities that the Yankees win the series in $ 4$ games.
There are $ \binom{4}{2} \equal{} 6$ possibilities that the Yankees win the series in $ 5$ games.

We can ignore the Red Sox side because that case will yield the same calculations.

Thus, the expected value is $ \frac {1(3) \plus{} 3(4) \plus{} 6(5)}{10} \equal{} \boxed{4.5\ \text{games}}$

I am weak in counting and probability. I might have done something wrong, so correct me if I'm wrong.[/hide]

You still must account for the Red Sox side and that is not how you would calculate the expected probability.

That solution is perfect -- if the probabilities that each team wins were not symmetric, then one couldn't make the given simplification, but since they are, there's no reason not to.

Edit: this was just wrong.

at the end why do u multiply the 1 by 3 the 3 by 4 and so on?

hm the math contest this problem is from saids the answer is 33/8

Sorry, I'm having a really terrible day.  :oops:  The first part of the solution (counting how many ways the series can end in 3, 4 or 5 games) is fine.  The calculation of the expected value from those numbers is just wrong -- one shouldn't multiply by $ \frac{1}{10}$ because a particular series outcome of length 3 occurs with a different probability (1/8) than a particular series outcome of length 4 or 5 (1/16 or 1/32, respectively).

This problem can be solved directly by Bernoulli's distribution, which is said that:"If you want to find the probability of the result which happens n times and the probability for this is t and you try the event h times(h>n,of course) then you can get the result: P(A)=C(n,h).t^n(1=t)^(h-n)"
In this case the 5>=h must be>=3, so we have to consider 3 cases respectively. Plus all of 3 cases and get the result that must be P(A). Then everything else is just applying the formula for calculating expected value.But wait! Does 2 team tie? If it does, this solution is completely wrong(I don't think we have that case, don't we?)

Does anybody have a complete detailed solution?

If we combine your (123456789's) solution with JBL's suggestion, then you get
[hide]
$ 2\left(\frac{{2\choose 2} \cdot 3}{2^3} \plus{} \frac{{3\choose 2} \cdot 4}{2^4} \plus{} \frac{{4\choose 2} \cdot 5}{2^5}\right) \equal{} \frac {33}8$
[/hide]

[quote="azjps"]If we combine your (123456789's) solution with JBL's suggestion, then you get
[hide]
$ 2\left(\frac {{2\choose 2} \cdot 3}{2^3} \plus{} \frac {{3\choose 2} \cdot 4}{2^4} \plus{} \frac {{4\choose 2} \cdot 5}{2^5}\right) \equal{} \frac {33}8$
[/hide][/quote]

azjps, why do you multiply by $ 2$ at the end? Is it to account for the Red Sox side? But why must it be accounted for, if this is symmetrical?

EDIT: Never mind. I get it.

Because, say after $ 3$ games, since we want the expected number of games in the series, we want to count the total number of possible outcomes that work and divide by the number of possible games. Since we divided by all $ 2^3 \equal{} 8$ possible series of games, we also need to include when the Red Sox won three games. [that was a horrible explanation]

Edit: Whoops, didn't see your edit.

Actually, azjps' explanation made it easier for me to understand. Thanks!


There are 1000 points on a plane.  We know that at least 3 out of any 4 points are on the same   straight line. Prove that all of them with the exception of one  point the most are on the same straight line.

Well if two or more points weren't in that straight line, you could simply pick those two points plus two points in the straight line and no three of them would be in the same line, meaning that at most, one can be out of line.


geometry
circumcircle
angle bisector
projective geometry

A circle touches internally the circumcircle of the triangle ABC.  The circle also touches the sides AB and AC of triangle ABC at P and Q respectively.

Prove that the midpoint of the segment PQ is the centre of the incircle of triangle ABC.

[quote="Michael Niland"]  [color=red]Let $ ABC$ be a triangle with circumcricle $ \zeta(O,R)$ and incircle $ \omega(I,r)$.Let $ \zeta_{A}$ be the circle that internally tangent to sides $ AB,AC$ at points $ P,Q$,respectively, and to $ \zeta$ at point $ R$.Prove that $ I$ is the midpoint of seqment $ PQ$  [/color]     [/quote]
Solution:
[color=blue]At first we'll prove one famous lemma:[/color]
[color=green]Let $ \gamma$,$ \beta$ be the circles,such that $ \beta$ is internally tangent to $ \gamma$ at point $ A$.Let $ B,C\in\gamma$,such that the line $ BC$ is tangent to $ \beta$ at point $ D$.Prove that
$ AD$ is the angle bisector of $ \angle BAC$.
Proof of the lemma:
Let $ l$ be the common tangent line to $ \gamma$ and $ \beta$.Let $ E$ be the arbitrary point on $ l$.WLOG suppose that $ E,B$ lies on the same semiplane wrt $ AD$.Denote $ F=AB\cap\beta$,
then $ \angle FAD=\angle BAE=\angle BCA\Rightarrow$ since $ \angle BDF+\angle FDA=\angle BDA=\angle BCA+\angle DAC\Rightarrow$
$ \angle DAC=\angle FDB=\angle BAD$.So we've finished the proof of lemma.[/color]
[color=blue]Denote $ C_{1}=RP\cap\zeta$,$ B_{1}=RQ\cap\zeta$,now using lemma we have that $ CC_{1}$ and $ BB_{1}$ are angle bisectors of $ \angle ACB$ and $ \angle ABC$ respectively.Now apply Pascal's
theorem we have that $ P,Q,\{CC_{1}\cap BB_{1}\}=P,Q,I$ are collinear,now using that $ AP=AQ$ we get the result.[/color]

Something like a generalization to this problem was posted by me [url=http://www.mathlinks.ro/viewtopic.php?t=170192]here:[/url]           (Just take there that C=D).
Here is an image to this nice problem:
[img]http://img120.imageshack.us/img120/1579/pascalql0.png[/img]

Famous problem!

See here: http://www.mathlinks.ro/Forum/viewtopic.php?p=367377#p367377

Oh,yes i know that it is famous problem,but i didn't find the link.So thank you for your link. :)
Dear,April,maybe you want to solve one very nice geometry problem,which was posted by me one month ago,but there is no solution :(
You can find this problem [url=http://www.mathlinks.ro/viewtopic.php?t=166170]here[/url] .Thank you. :wink: Tell me if you try it.


search
MATHCOUNTS
quadratics
linear algebra

What were you guys like as children(elementary school age)?

When  did you  learn to read/write/add/subtract/multiply/divide?

What sort of stuff were you interested in?

I am curious because aops/mathlinkers tend to be a lot more intelligent/studious than  the  general population(at least so it seems to me).

I could multiply when I was four. I could add 5-digit numbers and multiply 2-digit numbers in my head when I was eight.

I remember I could read probably when I was 3? Or maybe 4.

I learned to add, subtract, multiply, divide, (double digit by double digit) by the time I was 8, I think.

In elementary school, I liked to play basketball and read books.  Math only became a real interest for me in 6th grade.

I was pretty smart for my age as a kid,
but as time went on
if you were to graph my relative intelligence versus time it would be very much like the graph of y=1/x

I still have trouble adding/multiplying today :lol:

It doesn't matter where you start, it matters where you go.

Well let's see...

--I could talk when I was 6 months old
--I could add and multiply 2 digit numbers in my head when I was 1
--I could read and write when I was 2 and a half
--I learned algebra when I was 5
--I learned math up to precalculus when I was 6
--I learned most of college math by the end of elementary school

If you really believed that... :D Anyways I don't really think any of this stuff is relevant to... anything?

I was speaking 16 languages and solving IMO problems while still in the womb. With one brain hemisphere tied behind my cortex.

I finally learned to speak fluently when I was 13, and I can only add/multiply two one-digit numbers in my head currently, I can't even do two-digit operations in my head.

I cannot read or write.

[quote="joml88"][quote="probability1.01"]
If you really believed that... :D Anyways I don't really think any of this stuff is relevant to... anything?[/quote]

[quote="agolsme"]
I am curious because aops/mathlinkers tend to be a lot more intelligent/studious than the general population(at least so it seems to me).[/quote]

I'm pretty sure that if we were to do a study we would find that AoPSers on average excelled far more quickly than the 'average' student does.  What one does in their childhood should be a big indicator of what (s)he will do later on in life.  Things can change a little bit depending on the person's individual experience but usually environments don't change that dramatically and people turn out more or less how they were expected to.

I'm almost certain there have been studies done on this type of thing before...I should look into that and see what those studies showed...[/quote]

Really? What about that Johns Hopkins talent search SET thingy... supposedly they got a lot of above average adults, but they didn't get a bunch of Nobel laureates as had been expected. Anyway, it's my opinion that what happens in childhood has a very unpredictable effect on what happens later in life (when it actually matters).

[quote="joml88"]Ok, well whatever you perspective may be, I think this topic still proves to be relevant to see if there is any apparent link between people's progression during their youth and how they perform later.[/quote]

Well another thing is... will anyone really say anything unless they could solve IMO problems when they were 5? Also, how do you know how they will perform later? Finally, it's pretty easy to teach any kid that "3 + 9 = 12", so how do you know when you've really learned how to add/subtract/multiply/divide?

The part that I think is relevant is what people were interested in. For me, I basically played outside all day or with legos. That's all I did in elementary school. What fun!

Certainly there is no guarantee that people will perform later.  A lot of people's preferences will change over time.  Life altering events might take place.

Nonetheless, genetics play a large role in how people develop.  People's genome's wont change from childhood.  So why wouldn't we see a general relationship between what people do when they are young and how they perform later? (certainly there are exceptions to this but [i]on average[/i] there should be a trend)  I do not assert that if a child learns, say, multiplication a year before everyone else they will end up performing above average later in life (i.e. I don't think that there is a correlation between the two)...I only assert that there is a positive relationship between the two (as one goes up so does the other).

I just remember that in elementary school, I was really bad at almost everything except maybe computers.

So around 2nd grade, I failed an enrollment test that had multiplication of 1-digit numbers, because I kept checking my work, not sure that $x\times y=y\times x$.

Around 4th grade, I learned long division after moving to California unable to do the division worksheets from school.

Around 6th grade, my dad taught me the "proper" way to do problems like $\frac{4}{8}=\frac{32}{x}$ (before, I was guessing and checking).

And then around 7th grade, I found the Mathcounts club at school, and then AoPS... :)

I could climb stairs, unannounced, at eight months
my sister taught me to write my name in kindergarten
in third grade of elementary school i was addicted to binary system for some reason
since fourth grade, i got an interest in atomic bombs, and nuclear reactors
in sixth grade i was amazed at the $t=\sqrt{\frac{2 x}{g}}$formula for falling objects and kept calculating it for every possible height
in first year and second year of high school ( i mean uhm, you now first and second year after elementary, the americans probably call that junior high?) I hated maths, got about 60% on weekly tests, because I really didn't like the teacher, then I got the vibe again after I only had decent teachers after that, and an amazing world opened when I was introduced to a thing called olympiads in fifth year of high school

Now however, I swing to every side since I'm at uni
Partially due to a really wrong choice, partially due to unflexibility, i lost a year (the interval in which my scores lay had a width of 90 percent..) and if i don't watch out it will happen again

I learned adding etc. at 4; I finished the first-grade book rather quickly, but the second-grade book required me to read, so guess what I did? I was in the 3rd grade math book when I turned 5.

I learned to add/subtract when I was 3 and to multiply/divide when I was around 4, and I learned algebra around age 7.

I learned to read Chinese when I was 3 and English when I was 5.

When I was 5, I was obsessed with various engineering stuff like aircraft, spacecraft, and large buildings.

i remember doing 6th grade chinese textbooks when i was 6
if only i even kept up the 1 grade a year pace, i'd be a crazy chinese problem solving machine too...

I didn't know anything till someone taught it to me and never really took the initiative to learn stuff on my own. Especially not when i was three or something. yup, kindergarten and before were carefree days of tying shoes and zipping zippers, and "This is Spot... see spot run..." I will admit I was the first one in my class to zip my zipper. does that count for anything?

I was visualizing platonic solids in the crib (no, really, who am I ?).

wow some of you guys are weird.
I couldn't read until I was 4/5 and add/subtract until then either

I could read and write by the time I was 3.  I don't remember exactly how old I was when I learned addition, subtraction, multiplication, and division, but by the time I was in kindergarten, my parents were giving me linear algebra problems to solve mentally while waiting at restaurants.  So ya, I've basically been a math nerd all my life.

I just read Science stuff all the time since I was 7 till 11ish. I used to read physics books, biology books, and always watched the Discovery Channel. Then I moved from New York to South Carolina and couldn't care less about those things ;) Of course, I started doing math stuff like end of 10th grade.

[quote="calc rulz"]I proved Fermat's Last Theorem when I was in the womb. Unfortunately, it did not fit on the margin of the....let's just say it was too large to come out  ;)[/quote]

yeah, i did that with the Reimann Hypothesis, except that it did come out ;).

I could read chapter books when I was 4 and was able to add 4 or 5 digit numbers before kindergarten. I don't remember much of my early-elementary school life except I could always finish my homework in a fraction of the time it takes everyone else to do it and ever since way back I did 1-3 hours of non-school related work every day...

I guess when older people asked you "what is 3 + 2 son?", you would - as a four year old - respond "since addition with real numbers is commutative it equals 2 + 3". :lol:

Anyways, I lcould talk at an age of 5 months but no one stimulated me to learn math and read, so nothing impressive after that.

I am homeschooled, and I remember the time that I skipped the "orange book" (kindergarten) for and did the "red book" (1st grade) in kindergarten. Since then, I have been ahead in math.

The funny thing was, my parents were feeling bad for me because I didn't talk (well enough, anyways), untill quite a while, and also because I was known to run into walls. This was also because my older brother was speaking full sentances at a young age.

I learned how to solve quadratics last winter...


Edit :blush:  sorry! didnt read it properly

However, the qestion asks for a 'positive constant' c, therefore -2 cannot be correct??

[quote="01allenh"]3. Determine a positive constant c such that the equation
xy^2 − y^2 − x + y = c
has precisely three solutions (x, y) in positive integers.

Please give me a hint.[/quote]

Hint[hide] Seperate the x and y's assuming c=0.

x(y^2 − 1) = y^2-y
x(y-1)(y+1) = y(y-1)
x = y/(y+1)

Now, solve using c>0.

If c>0, we get two cases.

x>y/(y+1)
x<y/(y+1)

Shall I post more? not sure...

[/hide]

[hide="Another, admittedly much less elegant method"]

Let $f(x,y)=xy^{2}-y^{2}-x+y$
Tabulate the values of $f(x,y)$ for positive integers $x,y$, find one that comes up three times, and prove that it can't come up anywhere else by considering what happens if you fix $x$ and increase $y$ or vice versa.[/hide]

I'm unsure of what you are trying to say. Also btw, $x(y+1)-1\geq y+1-1=y>y-1$ so the x factor (no pun intended) is always greater than the $y$.
Every dichotomy of $c$ into $ab \ (a>b)$ has b=y-1, a=x(y+1)-1
or (b,a) works iff $a\equiv-1 \ (mod \ y+1) \implies a\equiv-1 \ (b+2)\implies ab\equiv 2 \ (b+2)$

i can feel it, we are a few steps of insight away from a solution (that means i'm stuck i will think about it)

The equation of a circle that goes through 3 integral points?

[quote="cincodemayo5590"][quote="Franz Joseph"]Just a guess. :blush:

[hide]$xy^{2}-y^{2}-x+y=(y-1)[x(y+1)-1]=c$

We deduce that a number $c$ must be able to be expressed as the product of two numbers in $3$ ways if three different solutions $(x,y)$ are wanted (given that $m \times n$ and $n \times m$ are considered to be distinct). Since no such number has this property, we conclude that no number $c$ of the desired property exists.

Edit: I mean that no positive integers can be expressed in the form of $m \times n$ in three ways where $m \times n$ and $n \times m$ are considered to be different.

[/hide][/quote]

Err.. can't $p^{2}$ be expressed as $1 \cdot p^{2}, \ p \cdot p, \text{ and }p^{2}\cdot 1$? (for prime $p$)

If i get anything, i'll post shortly... workin on it...  :P[/quote]

Oh yeah, how embarrassing :blush: .

Any solution does fine, sounds like youve got an interesting one though?!

[quote="jimhu"]does this problem [b]have[/b] to be solved using number theory?[/quote]

Well, considering it asks for positive [b]integer[/b] solutions...

seamusoboyle:  Not a circle.

[hide]
We know that $c=xy^{2}-y^{2}-x+y=(y-1)[x(y+1)-y]$. Since $c$ can only have three positive unique factorizations, we can conclude that it is the square of a prime ($p^{2}\cdot 1, \ p \cdot p, \ 1 \cdot p^{2}$)

Then, the three intger positive integer solutions are these:

1)  $y-1=1 \iff y=2 \implies x=\frac{p^{2}+2}{3}$

2)  $y-1=p \iff y=p+1 \implies x=\frac{2p+1}{p+2}$

3)  $y-1=p^{2}\iff y=p^{2}+1 \implies x=1$.

The third case will always give integer solutions, so we don't need to worry about it. The first two cases give the criteria for $p$:  $\frac{p^{2}+2}{3}\text{ and }\frac{2p+1}{p+2}\in \mathbb{Z^{+}}$.

But, $\frac{2p+1}{p+2}= 1+\frac{p-1}{p+2}$ which is non-integral for all primes except $p=1$. Checking, we see that $\frac{1^{2}+2}{3}=1$ which is an integer, so the three solutions would be:
$(1,2), \ (1,2), \text{ and }(1,2)$, which definitely are not unique.

Thus, we can conclude that no such $c$ exists that satisfies the given requirements.[/hide]

You could have a $c$ that it leads to an incosistant set of equations for all but 3 factorisations of it


Start with Volume 1 or Algebra (don't know which comes first, probably Vol. 1), then do the other. Next, do Number Theory and Intro to C&P (I personally think Alcumus is a good way to do this, but may not be a complete substitute), in any order. Finally, end the introduction series with Geo. and it would be good to review Volume 1 again. Then go on to the Intermediate, and Volume 2 next (you could review it before starting the intermediate series).

MC Nationals goes up to the end of the Intro series, in general. Keep doing problems along the way to see if you have improved.

EDIT: You might be able to skip ahead a little depending on what level you are at and when you want to finish what book, but, if time allows, I would go through things you already know anyway quickly to make sure you know it.

[quote="monkeygirl12"]I am probably going to be on a mathcounts team this tear and I am wondering what would be a good topic to study (for example number theory, algebra etc.).
And also what AoPS book should I study (I currently have Introduction to Number Theory AoPS Volume 1+2 and Algebra and am willing to buy more) ?[/quote]

Study all topics thorougly they will help you much later on.

Before doing AoPS Vol.1 you should do most of the Intro series.
NT doesn't go much on Mathcounts but is interesting and should be studied.:)
You can wait though

Actually, quite a few MATHCOUNTS problems involve Number Theory - for example, #2, 3, 11, 12, 17, 19, 22, 27, and 29 on this year's State Sprint Round involved some Number Theory.

To the OP: I recommend studying Combinatorics and Number Theory the most, as they are the least taught subjects used in competition math.

Is there a section in Intro about Diophantine equations?

I hate those...

Linear Diophantine equations are covered in the Introduction to Number Theory course, but not in the textbook.


integration
calculus
ratio
real analysis
real analysis unsolved

evaluate :D

$ \dfrac{1}{1\plus{}\dfrac{1^2}{1\plus{}\dfrac{1^2}{1\plus{}\dfrac{2^2}{1\plus{}\dfrac{2^2}{1\plus{}\dfrac{3^2}{1\plus{}\dfrac{3^2}{1\plus{}\dfrac{4^2}{1\plus{}\dfrac{4^2}{1\plus{}\dfrac{5^2}{1\plus{}\dfrac{5^2}{1\plus{}\dfrac{6^2}{1\plus{}\dfrac{6^2}{1\plus{}\dfrac{7^2}{1\plus{}\dfrac{7^2}{1\plus{}\ldots
}}}}}}}}}}}}}}}$

here is a hint:  :)

show that the above can be written as $ \boxed{4\,\int_0^\infty \dfrac{x\,e^{\minus{}\sqrt 5\;x}}{\cosh\,x}\;\;\textbf dx}$
and then attempt to solve the integral.

Doing some quick and dirty calculations, I found that the integral you gave was equal to
 \[ 2 \Psi_1({\varphi})\] where $ \varphi$ is the golden ratio and $ \Psi_1$ is the trigamma function.

I know nothing about continued fractions and would be interested in seeing a proof the equivalence of this fraction and the integral you provided


For which of the following values of $ k$ does the equation $ \frac{x\minus{}1}{x\minus{}2}\equal{}\frac{x\minus{}k}{x\minus{}6}$ have no solution for $ x$?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$

[quote="worthawholebean"]For which of the following values of $ k$ does the equation $ \frac {x \minus{} 1}{x \minus{} 2} \equal{} \frac {x \minus{} k}{x \minus{} 6}$ have no solution for $ x$?

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$[/quote]

Once expanded with cross multiplying, $ x^{2}\minus{}7x\plus{}6 \equal{} \minus{}kx\plus{}2k$ results.
$ \minus{}5x\plus{}kx\equal{}2k\minus{}6$
$ x(5\minus{}k)\equal{}2k\minus{}6$
$ x\equal{}\frac{2k\minus{}6}{5\minus{}k}$
K cannot be $ 5$

Question, so if $k = 1$, you can't divide both sides by $x-1$, because it might be $0$, right?

If $k=1$, then you can factor out $x-1$ from both sides. This means $x=1$ is a solution

@above so $x-1=0$ for that to hold. To divide by an expression one must first cover the possibility that that expression equals $0$, because otherwise we would be dividing by $0$ which is not possible. (Also, Notice that if we plug in $x=k=1$ in the original expression, we get a valid result)

[quote=Zanderman]@bove, what?
If $k = 1$
$(x-1)(x-6) = (x-2)(x-1)$, if we divide both sides by $x-1$, we get $x-6 = x-2$, which is impossible...[/quote]

Can't divide, since tautology


Nu am gasit nicaieri programa pe 2006 (sper ca am cautat suficient de bine) daca o are cineva si are bunavointa sa o atasez ii multumesc anticipat. Am auzit ca intr-a 10 a nu se mai fac polinoame :|

O gasesti printre ultimele fisiere de pe olimpiade.ro si au sters-o de pe edu.ro , inca nu sunt foarte lamurit de ce  :?. Dar am atasat-o mai jos. Da, polinoamele sunt doar la nationala, ca si geometria in spatiu.

La a 11-a se zice in programa de "formula lui Binet-Cauchy"...... stie cineva ce zice formula asta? :blush:

Chiar as vrea si eu programa la clasa a-11-a, daca se poate? :)


AMC
AIME
probability
trigonometry
ratio
geometry
trig identities

1 480
2 017
3 308
4 087 (corrected from 077, thanks Sara)
5 315
6 489
7 083
8 098

Solutions:
1. Draw a picture. We want all the lattice points in a 30x30 except for those in two 30x15 triangles, which have 210 lattice points each. Answer, 900 - 420 = 480

2. Two ways to get an odd sum: {all odd}, {2 even, 1 odd}. Since there are five odds and four evens, the only way that everyone gets an odd sum is if one person gets {all odd} and the other two get {2 even, 1 odd}.

First guy: P({all odd}) = 5/9*4/8*3/7 = 15/126
Second guy: P({2 even, 1 odd - the first time}) = 3*4/6*3/5*2/4 = 3/5
Third guy: P({2 even, 1 odd - the second time}) = 1, since these will be the only ones left

These could be arranged in three different ways (which guy gets all odd?), so in all the probability is 3*15/126*3/5 = 3/14 => 17

3. Similar triangles. PQA ~ PCD, so 735/(735+112+x) = PQ/PC = AQ/DC. Also, QBR ~ CDF, so 112/x = QR/CR = QB/CD = (CD-AQ)/(CD) = 1 - AQ/CD = (112+x)/(735+112+x).

Solve. x = 308.

4. Draw a graph, with the x-axis representing the first mathematician and the y-axis the second. We have $\frac{(60-m)^2}{60^2} = 60\%$, so $m = 60 - 12 \sqrt{15}$ => 87

5. Let $N = a_n 7^n + \ldots + a_1 7 + a_0$, where $a_i < 7$. This is the base 10 form. Thus N in base 7 looks like $a_n a_{n-1}\ldots a_1 a_0$. We want 2*(base 10 N) to look like (base 7 N), that is,
$2(a_n 7^n + \ldots + a_1 7 + a_0)$ looks like $a_n a_{n-1}\ldots a_1 a_0.$

So $2(a_n 7^n + \ldots + a_1 7 + a_0) = 10^n a_n + ... + 10a_1 + a_0$.
$a_0 + 4a_1 = 2a_2 + 314a_3 + (\textrm{some big number})a_4 + \ldots$
so $a_3, a_4, \ldots = 0$ (otherwise RHS will be very big)

$a_0 + 4a_1 = 2a_2$
$a_2$ is at most 6 from our previous restriction. If $a_2 = 6$, then $a_0 + 4a_1 = 12$; we want $a_1$ as big as possible (since it is the sevens place whereas $a_0$ is only the ones) so $a_1 = 3$. Now N = 630 in base 7, or 315 in base 10.

6. How many ways are there to make three segments form a triangle, with a fourth segment to spare? Well, there are 10C3 triangles, and 10C2 - 3 = 42 remaining segments, so 10C3 * 42.

There are 45C4 ways to choose four segments from 10C2 = 45 of them.

10C3*42/(45C4) = 16/473 => 489.

7. Let DQ = ER = FP = x. EFR ~ QER, so 1/(1-x) = (1-x)/x, so x = (3-sqrt(5))/2. EQ = ED-ER = 1 - x = (sqrt(5)-1)/2.

Apply law of cosines on EQR, recalling that angle REQ is 60 degrees, to find $QR^2 = 7 - 3 \sqrt{5}$. So the ratio of areas is $\left(\frac{2}{QR}\right)^2 = 7+3\sqrt{5}$ => 7^2 + 3^2 + 5^2 = 83.

#1 haha i was going to write why i disagreed with keone's answer, then i realized my mistake.. anyway, here's how to find the # of lattice points =b

since x and y must be positive integers, only look at the stuff within the square with verticies (1,1,) (1,30) (30,1,) (30,30)

for the triangle under the line y=(1/2)x:
x=1: 0 lattice points under the line
x=2: 0 (because you do not count points under the line)
x=3: 1
x=4: 1
...
x=30: 14

#4 i have the same equation
(60-m)^2/60^2 = 3/5
5(60^2 - 120m + m^2) = 3(60^2)
5*3600 - 600m + 5m^2 = 3*3600
5m^2 - 600m = -2*3600
m^2 - 120m + 3600  = 3600 - 2*3600/5 = 2160
(m - 60)^2 = 2160
m - 60 = +- 6*sqrt(60) = +-12*sqrt(15)
m = 60 - 12*sqrt(15)
so 87

For #4, I wouldn't even expand that square on the left side.  Instead, multiply both sides by 60^2 and take the square root:

(60-m)^2/60^2 = 3/5
(60-m)^2 = 3/5*60^2
60-m = sqrt(3/5)*60
m = 60-60sqrt(3/5) = 60 - 60sqrt(15)/5 = 60-12sqrt(15).  So you are right, 087 is the answer.

my answers match with 1-6, havent done 7,8 yet


linear algebra
matrix
vector
linear algebra unsolved

Let $A$ be a $n\times n$ matrix with complex elements and let $A^\star$ be the classical adjoint of $A$. Prove that if there exists a positive integer $m$ such that $(A^\star)^m = 0_n$ then $(A^\star)^2 = 0_n$.

[i]Marian Ionescu, Pitesti[/i]

Just a quick sketch:

If $A^*$ (or $A$) is not invertible, then it has rank $\le 1$. Every matrix of rank $\le 1$ can be represented as $uv^t$, where $u,v$ are column vectors with entries in the respetive field. Now $A^*^k=u(v{}^tu){}^{k-1}v^t=(v{}^tu){}^{k-1}A^*$ (because $v^tu$ is just a scalar, so we can move it around). If $A^*=0$, we're done: its square is $0$ as well. Otherwise, $A^*^k=0$ and the preceding relation imply $v^tu=0\Rightarrow A^*^2=(v^tu)A^*=0$, and again, we are done.

I can't figure out why is this true:

[quote="grobber"]If $A^*$ (or $A$) is not invertible, then it has rank $\le 1$.[/quote]

Wasn't that $n-1$???

The other results were known to me, though.

In general, for a non-invertible matrix, you can only conclude that the rank is $\le n-1$, of course. But in this setting, what I wrote is true: if $A$ is non-invertible, then $A^*$ has rank $\le 1$. Try it. It's important to figure stuff out by yourself :).

P.S.

Also, I hope I didn't make any mistakes, and it is actually true :).

You mean that the rows of $A^\ast$ are multiples of each other? I can't prove that, hence I feel pretty stupid.

[quote="perfect_radio"]$A^\ast$ [/quote]

What does this mean ?

[quote="perfect_radio"]You mean that the rows of $A^\ast$ are multiples of each other? [/quote]
It's a consequence of Sylvester's inequality. If $\ rank(A)=n-1$ (the other cases are trivial) then $\ 0=rank(AA^*)\geq n-1+rank(A^*)-n$.

What are you all saying? If $A$ is a real matrix with first row (1,1,2), second row (0,1,1) and third row (1,0,1), then both $A$ and $A^{\ast}$ have rank 2.

Or I do not understand what is adjoint? Maybe, it is a matrix of $(n-1)\times(n-1)$ minors with some signs? I would not call it so. At least, self-adjoint operator is such operator $A$, for which $(Ax,y)=(x,Ay)$ for all vectors $x,y$.

$A^\star$ reffers to the [url=http://planetmath.org/encyclopedia/MatrixAdjoint.html]the classical adjoint or the adjugate[/url] of $\displaystyle A$. I included the word "classical" so that people from other countries know what it means.

[quote="grobber"] Every matrix of rank $\le 1$ can be represented as $uv^t$, where $u,v$ are column vectors with entries in the respetive field.[/quote]

How did we have that?
 I didn't understand this part :( please help me

If $A$ has rank $1$ then $A=QI_1D$ where $I_1=diag(1,0,0,...,0)$ the rest is obvious.


Prove that for any $ n\in \mathbb N$ the number $ 1\plus{}\dfrac{1}{3}\plus{}\dfrac{1}{5}\plus{}\ldots\plus{}\dfrac{1}{2n\plus{}1}$ is not an integer.

[hide="Hint"]Think about the largest power of three.[/hide]

[hide="start"]
induction:

1) $ n\equal{}1$

$ 1\plus{}\frac{1}{3}$ is not integer.

2) $ n\equal{}k$

assume that $ 1\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{2k\plus{}1}$ is not integer, but it can be expressed as fraction $ \frac{b}{a}$ where a and b are relatively prime.

3) $ n\equal{}k\plus{}1$

$ 1\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{2k\plus{}1}\plus{}\frac{1}{2k\plus{}3}\equal{}\frac{b}{a}\plus{}\frac{1}{2k\plus{}3}\equal{}\frac{2bk\plus{}3b\plus{}a}{2ak\plus{}3a}$

idk how to go from there...
[/hide]

[quote="DreamTeam"]Prove that for any $ n\in \mathbb N$ the number $ 1 \plus{} \dfrac{1}{3} \plus{} \dfrac{1}{5} \plus{} \ldots \plus{} \dfrac{1}{2n \plus{} 1}$ is not an integer.[/quote]

[hide="This may help you"]
[url]http://www.mathlinks.ro/viewtopic.php?t=230324[/url][/hide]

Let $ m$ be the greatest integer such that $ 3^m\le n$. Hence all other numbers are of the form $ 3^k\cdot l$, where $ k<m$ and $ l$ is not divisible by 3. So if we take the lowest common denominator, the denominator is of the form $ 3^m\cdot j$; and all numerators, except the $ \frac1m$, are divisible by 3. So if we sum up, the numerator is not divisible by 3 and the denominator is divisible by 3, and hence the value is not an integer.

How about using Betrand's postulate? For any $n \geq 1$, there is a prime $p$ such that $n+1  < p < 2n+2$. From the bounds, it is clear that $p$ is odd; hence the term $\frac{1}{p}$ appears somewhere in the required sum.

If we put all the terms under a common denominator:

$1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n+1} = \frac{3 \dots (2n+1) + 1 \cdot 5 \dots (2n+1) + \dots }{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n+1)}$,

Then all but one of the terms in the numerator are divisible by $p$. This means that the numerator is NOT divisible by $p$ while the denominator is, and therefore it cannot be an integer.

[hide=Solution? Can someone verify]
If the sum is not an integer, then $v_2\left(\sum_{k=0}^{n} \frac{1}{2n+1}\right)<0$.
Note that $v_2\left(\sum_{k=0}^{n} \frac{1}{2n+1}\right) = v_2(\min(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \cdots , \frac{1}{2n+1})) = v_2(\frac{1}{2n+1}) < 0$. $\square$
[/hide]
edit: this does not work since the 2-adic valuations are not necessarily distinct, I'll maybe fix this later

Remark: for the $n=2k+1$ it is trivial, because $1+\frac{1}{3}+\frac{1}{5}+\ldots +\frac{1}{2n+1}=\frac{2a}{2b+1}$


Find $ x$ and $ y$ from the equation:

$ (x+y)^\log_{10} x$ = $ \frac{3}{2}y$

One solution is $ x\equal{}1$, $ y \equal{}\frac{2}{3}$.

Is it the only one?

[quote="AndrewTom"]One solution is $ x \equal{} 1$, $ y \equal{} \frac {2}{3}$.

Is it the only one?[/quote]

How did u find the solutions?{explain please)

Isn't $ x \equal{} 6, y \equal{} 4$ also a solution?

[hide="My Method"]
$ (x\plus{}y)^{log_{10}x} \equal{} \frac{3}{2}y$

$ (x\plus{}y)^{log_{10}x} \equal{} 10^{log_{10}\frac{3}{2}y}$ ... using the property $ n \equal{} a^{log_an}$

one solution can be

$ x\plus{}y \equal{} 10 , x \equal{} \frac{3}{2}y \\ \\ \Rightarrow \ x \equal{} 6, y \equal{} 4$
[/hide]

Andrew's solution came from $ (x\plus{}y)^{log_{10}x} \equal{} \frac{3}{2}y \equal{} 1$

You're right, Nora.91,

I used $ a^{0}\equal{}1$.

[quote="AndrewTom"]You're right, Nora.91,

I used $ a^{0} \equal{} 1$.[/quote]

Only two real solutions to (x,y)?

No any others?


A plane is traveling between City A and City B, which are 2000 miles apart. City A is due north of City B, and there is a strong, constant wind blowing due south. At constant speed, it takes a plane 5 hours to go from A to B with the wind at its tail, but 8 hours to go back when facing this headwind. What is the speed of the wind?

find the rate of both ways

with wind = 2000/5 = 400 mph

against wind = 2000/8 = 250 mph

find the difference between the two rates (150)

then divide by 2, thus, ans = 75 mph

u have 2 find the average of the 2 speeds- 325

then find the difference between that and the speed w/ or w/o wind

and u get 75


If $ a,b,c$ are nonnegative real numbers such that no two of which are zero and $ (a\plus{}b\plus{}c)(a^3\plus{}b^3\plus{}c^3)\equal{}1,$ then
$ \frac{b^2c^2}{a^2\plus{}bc}\plus{}\frac{c^2a^2}{b^2\plus{}ca}\plus{}\frac{a^2b^2}{c^2\plus{}ab} \le \frac{1}{2}.$

My dear friend, it is very nice in my opinion!!!

I will try. :)

Can anyone give us a nice proof? I used Cauchy-Schwarz and Schur, but I dont really like this proof.


number theory
prime numbers
arithmetic sequence

let $ p$ and $ q$ prime numbers such that : $ 3<p<q$
let $ a\equal{}p\plus{}q$.
[hide]you can see that $ a$ isn't prime because it's even . [/hide]

1.find the biggest  divisor  for  $ a$ and  # $ a$
2.prove that  all divisor for $ a$ are  small than $ q$.
3.prove that  all prime divisor  for $ a$ are  small than $ p$.

$ q>\frac{a}{2}$, so no divisors of $ a$ can be $ >q$. This implies 2 and 3.

[quote="Brut3Forc3"]$ q > \frac {a}{2}$, so no divisors of $ a$ can be $ > q$. This implies 2 and 3.[/quote]

And if $ q>3$?

Unless you can disprove Goldbach' conjecture, we don't know how to set an absolute bound on the biggest divisor of $ a$ (otherwise this wouldn't be an open problem).

For 3: Are you sure this is true?  The Tao-Green theorem states that we can find infinitely long arithmetic progressions of primes.  In particular, if three primes are in arithmetic progression, then the sum of the outer two is double the middle one.  For example, let $ p\equal{}17$ and $ q\equal{}29$.  Then $ p\plus{}q\equal{}46\equal{}2 \cdot 23$.

[quote="t0rajir0u"]greatestmaths is missing an additional constraint.  I have seen this problem before, but I don't remember the constraint; possibly it's $ p, q \equiv 1 \bmod 6$.[/quote]

Oh, that explains it.  But hm, if the constraint is that the two primes are congruent modulo some number, the Tao-Green theorem still tells us that c) is false, right?  For the constraint you gave, for example, p=7 and q=19...maybe it's that one is 1mod6 and the other 5mod6?

[quote="K81o7"]maybe it's that one is 1mod6 and the other 5mod6?[/quote]
No, I don't think that that works either.  Consider $ 5$ and $ 79$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="abcdefghi"]let $G$ and $G'$ be a groups, and let $H$ and $H'$ be normal subgroups of $G$ and $G'$, respectively. let $\phi$ be a homomorphism of $G$ into $G'$. show that $\phi$ induces a natural homomorphism $\phi_{*}: (G/H)%Error. "rarrow" is a bad command.
(G'/H')$ if $H'$ contains $\phi[H]$[/quote]

I think you mean,
$\phi_{*}: G/H \mapsto G'/H'$
Anyway,
Define $\phi_{*}(xH)=\phi(x)H'$  $\forall x\in G$

Is this what you mean?


geometry
3D geometry
sphere
symmetry
calculus
calculus computations

Find the dimensions of the rectangular box of largest volume that can be inscribed in a sphere of radius 1.

I'm really just unsure of the constraints that might arise when inscribing a box in a sphere, I'm fairly confident about the rest of the process.

if its in a sphere then the eight vertices will be of the form

$ (\pm a, \pm b, \pm c)$ for $ a^2\plus{}b^2\plus{}c^2\equal{}1$

so $ V\equal{}8abc$

Using kenn4000's idea, we need to maximize $ f(x,y,z) \equal{} 8xyz$ subject to the constraint $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$

Hence, by symmetry we must have that $ x\equal{}y\equal{}z\equal{} \plus{} or \minus{} \sqrt3/3$

So that the inscribed rectangular box is a cube with side length $ 2\sqrt3/3$


Show that if the equation $x^2+y^2+1=xyz$ has positive integral solutions $x,y,z$, then $z=3$.


[color=red][size=150]RULES[/size][/color]

1 - At the start of Round 1, I will announce height of the Score Board.
Ex: “Round 1: The SB is 200 units high.”

2 - You send the Mod a PM declaring how high up the Score Board you want to place your thumbtack.

      Please use this format:
      [b]Round 1: 109[/b]

3 - You earn, in points, the distance between your thumbtack and the next thumbtack below you.

4 - If 2 or more people try place their thumbtack on the same spot, they end up stabbing each other instead. Neither of the thumbtacks gets placed on the SB.

[i]Ex: A-99, B-182, C-60, D-99, E-175 [/i]

Round 1:
\begin{tabular}[t]{|c|c|c|c|}\multicolumn{7}{l}{{\bf{\large Round 1 Results}}}\\\hline{\bf Scoreboard 1}&{\bf 200}&{\bf Points}&{\bf Total}\\\hline B & 182 & 7 & 7\\\hline E & 175 & 115 & 115 \\\hline A & 99 & 0 & 0 \\\hline D & 99 & 0 & 0 \\\hline C & 60 & 60 & 60 \\\hline Scoreboard 1 & 0 & Points & Total \\\hline \end{tabular}

[i]…B earns 7 points, E earns 115 points, C earns 60 points. [/i]

[u]Your goal[/u]: To have the most points in the end, and to never be eliminated.

[u]Timing[/u]: Each round will last 2 days. Failure to send in your entry without prior notice will result in no thumbtack being placed.

[u]Players[/u]: I'll be accepting up to 15 players.

[u]Game progression[/u]: The SB’s will eventually get taller and taller. On every even-numbered round (2,4,6,etc.) the person with the lowest score will be eliminated. There may be twists some rounds, more on that later.

[u]Ties[/u]: If at the end of an elimination round there's a tie for last place, we will pause the game to do a Tie-Breaker Round. This round will be done like an ordinary round, but only those tied will participate. Points earned in TBRs will not count towards total score.

[u]Removing thumbtacks[/u]: You're allowed to change your mind up to 1 time each round. All changes must be made before the deadline, or they will not be accounted for.

      When changing your mind, use this format:
      [b]Round 1: Remove Thumbtack 109[/b]
      [b]Round 1: 182[/b]

Sign-ups end in 72 hours of this post, unless I decide to change it or 15 people sign up.

Participants:
1. Treething
[b]2. Ignite168[/b]
[b]3. life=tennis+math[/b]
[b]4. junggi[/b]
5. ShoeFactory
6. Bubka
7. Akalra1
8. Kurt Godel
[b]9. Kheg_k[/b]
[b]10. Mustafa[/b]

[hide="leaderboard"]
Ignite168-282
L+T=m- 189
Kheg_k-110
Junggi- 105
mustafa-100 [/hide]

I'm in and my only goal will be stabbing Daesun.

[quote="Xevarion"]I'm just wondering how the tiebreaker is supposed to work. If there's only one other person, you know for sure it's stupid to go anywhere other than the top.[/quote]

Why? Tiebreaker is almost the hardest part. You have to consider where your ONE opponent will put his tack. If he puts it at 50, and you put it at 49, of course you win, but would anyone put it at 50?
It's actually not stupid to go anywhere other than the top.

i m in

[color=white]blahblahblah[/color]

A wanna play!
I wanna play!

I want ta play!

I want to play!

I want to play please!

[quote="bubka"]I am in.

Let us play a round, and then I will think of improvements. Does not sound a bad game to begin with. But 2 days for a round is too long. Rather let players queue up thumbtacks.[/quote]

Nah, no fun in queueing up, it'll be 2 days a round.

[quote="Xevarion"]Klebian: regarding tiebreaker - Sorry, I misread the rules. What I'm saying is: there are two players. You lose if the other player plays below you. Why should you even make it possible for the other guy to play below you? So play at the bottom. Both players will do it and it's still a tie.[/quote]

Play at the bottom where?

You don't lose if he's below you, you lose if he scores more than you. Both players want to win, it's not that easy to win.

[quote="Xevarion"]Klebian: regarding tiebreaker - Sorry, I misread the rules. What I'm saying is: there are two players. You lose if the other player plays below you. Why should you even make it possible for the other guy to play below you? So play at the bottom. Both players will do it and it's still a tie.[/quote]

If you put 1, and he puts 10, he scores 9 and wins!

How so?

The message is too small. Please make the message longer before submitting.

You can't change the rules halfway though..

This needs a bump, I'm still waiting for a submission from life=tennis+math.

Anyone interested in replacing life=tennis+math?

Er, meh, fine. Mustafa replaces l+t=m. Read over the rules for this round and send in a submission asap.

Junggi vs Mustafa

junggi-80
mustafa-75

Kheg vs Ignite

ignite168-0
kheg_k- -78

Junggi gets 5, mustafa gets 75, kheg_k gets -78.

mustafa-347
ignite168-240
junggi-217
kheg-160

Last part of round 11

Junggi vs Kheg
Mustafa vs Ignite

[quote="Klebian"]abandoned due to inactivity[/quote]

There was no inactivity, everybody submitted their tasks (maybe kheg_k didn't) but you could've replaced him, or could've just eliminated him looking at how behind he is.

I sent in my submission, right? If someone didn't, they probably just forgot to, and that could be put right by a reminderatory pm.


Question:Give an example of a 1-dimensional subspace U of R3.

Answer: Any line through the origin is a one-dimensional subspace of R3.
$ U\equal{}\{t\cdot (a,b,c)\,\mid\,\,t\in\mathbb{R}\,,\,(a,b,c)\neq (0,0,0)\}$

what are a,b,c ?are they real numbers? Can you give me an example please.

Many Thanks

of course $ a,b,c$ are real. we're talking about $ \mathbb{R}^3$ here.

thanks for your help.

to answer

Question:Give an example of a 1-dimensional subspace U of R3.

U={t.(1,0,0) s.t t is real} is this correct?


It says only 'california university' in website of the show..
What college is that actually?

Well, his brother works for LAPD, right? So what colleges are in the area?

the thing is filmed at caltech, but in the show they call it cal-sci.

(wow, never thought that random bit of trivia would ever be useful!  :D )


$Ii(k)=1$ if $Xk$ belong to $bi. =0, otherwise. Bi=\frac{\sum_{k=1}^N Ii(k)}{N}$
Prove that $E${$Bi$}$= P(X$belong to$bi).E${}is the expect value.
Prove that $Var${$Bi$}$=\frac{1}{N}P(X$belong to$bi)[1-P(X$belong to$bi)]. Var${} is the variance.


The quadrilateral $ ABCD$ is cyclic. Points $ E$ and $ F$ are chosen at the diagonals $ AC$ and $ BD$ in such a way that $ AF\bot CD$ and $ DE\bot AB.$ Prove that $ EF \parallel BC.$

$EF\parallel BC$ $\Longleftrightarrow$ the quadrilateral $EADF$ is inscribed $\Longleftrightarrow$ $\widehat {EAF}\equiv \widehat {EDF}$ $\Longleftrightarrow$  $\widehat {CAF}\equiv \widehat {EDB}$ $\Longleftrightarrow$ $90^{\circ}-m(\widehat {CAF})=90^{\circ}-m(\widehat {EDB})$ $\Longleftrightarrow$ $\widehat {ACD}\equiv \widehat {ABD}$ $\Longleftrightarrow$ the quadrilateral $ABCD$ is inscribed, what is truly. [hide="Remark."] The both Rogue's geometry problems from the $KYIV$ festival must appear at the section "Pre-Olympiad" and not here.[/hide]

Thank you for solution, Virgil Nicula!
May be, this problem is really too easy for this section.

Here is the inverse problem:

The quadrangle $ABCD$ is convex. Points $E$ and $F$ are chosen at
the diagonals $AC$ and $BD$ in such a way that $AF\bot CD$ and
$DE\bot AB.$ Prove that if $EF\parallel BC$ then $ABCD$ is cyclic.


trigonometry
function
Asymptote
complex analysis

The complex potiential $F= A$ $sin z$ for a flow inside the semi-infinite region $\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$, $y \geq 0$

Write the equations of the streamlines

(Admittedly from the archives...)

If $F=A \sin z$ is the complex potential, then the streamlines are given by $\Im F = K$ where $K$ is constant.  Writing $z=x+iy$ gives $F=A \sin x \cosh y+i A \cos x \sinh y$.  Therefore the streamlines are $\cos x \sinh y = K_{new}$ or equivalently $y=arcsinh (\frac{K_{new}}{\cos x})$.  These curves comprise a family of even functions on $-\frac{\pi}{2}\leq x \leq \frac{\pi}{2}$ with asymptotes at the endpoints that are "u"-shaped in nature pointing upwards.

(Small joke, these can not be "steam lines" since the flow must be incompressible  :lol: )


[quote="13375P34K43V312"]In triangle XYZ, A and B are on XY, C and D are on YZ, and E and F are on ZX such that ABCDEF is a cyclic hexagon with circumcenter P. If AB=CD=EF, show that P=I.

[hide="hint 1"]PAB, PCD, PEF are congruent[/hide]
[hide="hint 2"]hint 1 implies that the altitudes from P to AB, CD, EF are equal[/hide][/hide][/quote]

The solution follows really easily from your hints, but you forgot to define $ I$. I'm assuming it's the incenter.

[hide]Like 1337 said, $ \triangle PAB$, $ \triangle PCD$, and $ \triangle PEF$ are all congruent by SSS congruency.

Hence the altitudes from $ P$ to each of the sides must be equal. This implies that $ P$ is equidistant from each side. Since $ I$ is the unique point equidistant from each side, $ P\equal{}I$.[/hide]


I want to do a math fair project for the coming school year, but have to do a lot of background reading into advanced math in order to be able to understand certain areas of math well enough to come up with a good project idea. However, the formal language used by many of these books/websites (such as MathWorld) is very hard for me to understand, as all of the math that is taught in my school is very informal and rather basic. What are some good books/websites for me to transition into easily understanding more-formalized math?

Math research can be a great motivation to learn new topics. However, I understand your problem. The formal mathematical language tends to be concise and unambiguous. I would suggest that you try reading some basic proofs of simple theorems (pythagorean, $ e$'s irrationality,...). This will expose you to different methods of proof as well as mathematical language. Another way to help may be to read less formal websites, [url]http://www.mathforum.org[/url] is great for this. Also, try to read the forums here on AOPS, you might find a great source for ideas and help improve your math language and computational skills as well. Use AOPS to ask a question if you don't understand, I'm positive that many people, some of which are extremely adept at mathematics, can give you help. To help you out a little better: What do you plan on doing for your project? What level of math have you studied? What level of math do you need to study in order to do this project?
Reply here or PM me if you need help.

I'm not much of a mathematician, so, I don't really have a broad base to speak from.  But I believe this website's own textbooks should have some information which is a very good introduction into formal/advanced mathematics.  I did a project once, investigating infinite regression, which is a pretty interesting topic to people who ordinarily don't delve much into math.  Hope this helps....


Let $ \displaystyle x_{1},x_{2},...\: ,x_{n}$ be nonnegative real numbers with
\[ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n}\leq\frac {1}{2}
\]
Show that
\[ (1 \minus{} x_{1})(1 \minus{} x_{2})\cdot\cdot\cdot(1 \minus{} x_{n})\geq\frac {1}{2}
\]

This is indeed well-known.It is also called Weierstrass's inequality.

[url]http://mathworld.wolfram.com/WeierstrassProductInequality.html[/url]


For three sequences $(x_n),(y_n),(z_n)$ with positive starting elements $x_1,y_1,z_1$ we have the following formulae:

\[ x_{n+1} = y_n + \frac{1}{z_n} \quad y_{n+1} = z_n + \frac{1}{x_n} \quad z_{n+1} = x_n +  \frac{1}{y_n} \quad (n = 1,2,3, \ldots)\]

a.) Prove that none of the three sequences is bounded from above.

b.) At least one of the numbers $x_{200},y_{200},z_{200}$ is greater than 20.

A BWM problem? As Darij used to say, it is probably in Engel's book....

Bingo! :D
This is problem 9-24.

Pierre.

[hide="a)"]
$\sum_{cyc}{x_{n+1}}=\sum_{cyc}{x_{n}}+\sum_{cyc}{\frac{1}{x_n}}$
Since all are positive, $\sum_{cyc}{\frac{1}{x_n}}>0$ so the overall sum is increasing.
Now I know there is a simple way to prove there are each increasing...[/hide]

[edit:]
[hide="b)"]
By Am-Gm: $x_1+\frac{1}{x_1}\geq 2\sqrt{1}=2$.
So, $x_2+y_2+z_2=x_1+\frac{1}{x_1}+y_1+\frac{1}{y_1}+z_1+\frac{1}{z_1}\geq 6$$\therefore \text{there exists one of } x_2, y_2, z_2, \geq 2$. Also, one of $\sum_{cyc}x_1 , \sum_{cyc}\frac{1}{x_1}$ is $\geq 3$. We just use recursively (I haven't figured out the details...[/hide]


Fatal error: Allowed memory size of 8388608 bytes exhausted (tried to allocate 184320 bytes) in /data/home/mathlinks/www/includes/template.php(127) : eval()'d code on line 112

greetz :D ;)

ooh, nice, another one for the collection :)

Fatal error: Allowed memory size of 8388608 bytes exhausted (tried to allocate 184320 bytes) in /data/home/mathlinks/www/includes/functions_post.php on line 381

too many users on site :P can't help that. I WILL (at some point in the future) hopefully have time to get my butt down to the server and do some of the stuff I need to do.

Unfortunately it's olympiad time + college time and I don't have much to spare :D.

I will give it a shot this Monday, ok? :)

sure

besides, since when is having too many users a NEGATIVE thing for any site :?

;)

well if you get there, just put off the memory limit, I can't imagine they host more important sites than this one :D

they only have one other site, which is on the same comp as this one. it's the main site for a College in bucharest, and sometimes a lot of students access it to get their homework done, and talk among themselves. That is when usually it get's crowded in the memory banks :).

Oh i see

Here that's no problem, all homeworks are passed to eachother via msn, same as for all talking, so our school's servers are never overcrowded.

(well in fact they're always overcrowded, they're just way too light -- a 4ghz processor for 100 client terminals :( )


The product of two consecutive positive even integers is 624.  What is the sum of the two integers?

[quote="james4l"]Notice that we can express the even numbers as x-1 and x+1 where x is an odd number.

x^2-1=624
x^2=625
x=25
The sum is 2x=50.[/quote]

That's elegant


Let
$ a \equal{} 1$
$ b \equal{} 2$,
$ c \equal{} 3$
$ d \equal{} 4$

then $ bacd \equal{} 2134$  and  $ dabc \equal{} 4123$

Let
first letter$ \equal{} U_{1}$
second letter$ \equal{} U_{2}$
third letter$ \equal{} U_{3}$
fourth letter$ \equal{} U_{4}$
bth letter$ \equal{} U_{n}$

In $ dabc$
$ U_{1}\equal{} d$ we know that $ d$ comes after the letters $ abc$ so we say that $ U_{1}\equal{} 3$
because $ abc$ is made of $ 3$ letters

$ U_{2}\equal{} a$ what comes before $ a$ ? no other letter comes before $ a$, so we say that value of $ U_{2}\equal{} 0$
[color=red]this is where i need help with:the line below[/color]

$ U_{3}\equal{} b$ what comes before $ b$ obviously its $ a$ but we dont say value of $ U_{3}\equal{} 1$ because $ a$ is repeated before $ b$ in $ dabc$ so we say value $ U_{3}\equal{} 0$

$ U_{4}\equal{} c$ what comes before $ c$ ? it is $ ab$ this is made of $ 2$ letters we have to write value of $ U_{4}\equal{} 2$ [color=red]But we dont [/color]because both the letters $ a$ and $ b$ is in $ dabc$ repeated before $ c$ so we say value of $ U_{4}\equal{} 0$

$ U_{3}$ and $ U_{4}$ is where i have difficulty with. i dont know How do i write this repeat thing in mathematical language? i know the maths language for $ U_{1}$ and $ U_{2}$ just not for $ U_{3}$ and $ U_{4}$ ([color=green]Repeated ones[/color])

this is how i find the values for none repeated ones ($ U_{1}$ and $ U_{2}$).
$ U_{1}\equal{} d$ as i said above $ d \equal{} 4$ so i just do $ d\minus{}1 \equal{} 4\minus{}1 \equal{} 3$ same for $ U_{2}\equal{} a\minus{}1 \equal{} 1\minus{}1 \equal{} 0$

$ dabc \equal{} U_{1}\plus{}U_{2}\plus{}U_{3}\plus{}U_{4}\equal{} 3\plus{}0\plus{}0\plus{}0 \equal{} 3$

How do I find the values of $ U_{z}$ that repeats themselves ?

i hope i have explained clearly and posted this in the right place.

The definition of $ U_{i}$ appears to be the number of letters occurring after the letter in the $ i$th position that come before that letter in the alphabet. So in your example:

$ U_{1}\equal{} 3$ because the first letter is $ d$, and the $ 3$ letters that come after all come before $ d$ in the alphabet.

$ U_{2}\equal{} 0$ because the second letter is $ a$, and $ 0$ letters come before it in the alphabet.

$ U_{3}\equal{} 0$ because the third letter is $ b$, and of all the letters after it ($ c$), $ 0$ come before it in the alphabet.

$ U_{4}\equal{} 0$ because there are $ 0$ letters after it, so of them $ 0$ come before it in the alphabet.

I don't see a need for more precise mathematical language than that...

two of them you look before it and the other two you look after it ?

i am confused can you explain abit more please?

what about $ daceb$ and $ fbeadc$? how would you go about doing these?
the one we did in the post above had even number of letters (4)
what would i do for odd number of letters ($ daceb$)?

For none of them do you look before it. I'm sorry if I somehow misled you there. Always look after the letter you're considering for any letters that occur before the one you're on in the alphabet. For each one there is, add one to appropriate $ U_{i}$. If there are none, make it 0.

$ daceb$:
$ U_{1}\equal{} 3$ ($ acb$ come before $ d$ in alphabet, after $ d$ in word)
$ U_{2}\equal{} 0$
$ U_{3}\equal{} 1$ ($ b$ comes before $ c$ in alphabet, after $ c$ in word)
$ U_{4}\equal{} 0$
$ U_{5}\equal{} 0$

$ fbeadc$
$ U_{1}\equal{} 5$ ($ beadc$ come before $ f$ in alphabet, after $ f$ in word)
$ U_{2}\equal{} 1$ ($ a$ comes before $ b$ in alphabet, after $ b$ in word)
$ U_{3}\equal{} 3$ ($ adc$ comes before $ e$ in alphabet, after $ e$ in word)
$ U_{4}\equal{} 0$
$ U_{5}\equal{} 1$ ($ c$ comes before $ d$ in alphabet, after $ d$ in word)
$ U_{6}\equal{} 0$

[quote]$ daceb$:
$ U_{1}\equal{} 3$ ($ acb$ come before $ d$ in alphabet, after $ d$ in word)
$ U_{2}\equal{} 0$
$ U_{3}\equal{} 1$ ($ b$ comes before $ c$ in alphabet, after $ c$ in word)
$ U_{4}\equal{} 0$
$ U_{5}\equal{} 0$[/quote]

I get everything except ''$ U_{4}\equal{} 0$'' isnt it suposed to be $ U_{4}\equal{} 1$?

The 4th letter is $ e$, and of all the letters that come after $ e$ in that word ($ b$), none of them come before it in the alphabet. You do understand the general definition, right?

EDIT: Ok, just shoot me. Posts don't get much funnier than this...

Um. $ b$ comes before $ e$ in the alphabet.

Yes i did undrestand the definition Thanks for the very much help.

but with your definition and my own difinition $ U_{4}\equal{}1$ not $ U_{4}\equal{}0$
and no $ b$ comes before $ e$ in the alphabet.

how do i generalize those idea in the posts above? i mean like this , but for all the repeat and nonerepeat ones.
[quote]this is how i find the values for none repeated ones ($ U_{1}$ and $ U_{2}$).
$ U_{1} \equal{} d$ as i said above $ d \equal{} 4$ so i just do $ d \minus{} 1 \equal{} 4 \minus{} 1 \equal{} 3$ same for $ U_{2} \equal{} a \minus{} 1 \equal{} 1 \minus{} 1 \equal{} 0$

$ dabc \equal{} U_{1} \plus{} U_{2} \plus{} U_{3} \plus{} U_{4} \equal{} 3 \plus{} 0 \plus{} 0 \plus{} 0 \equal{} 3$[/quote]

or in other words, how do i write those idea in mathematical language?


[quote="kunny"]The sequence $\{a_n\}$ is defined as follows.

\[a_1=1,\ a_{n+1}=\frac{3n-1}{3n+5}a_n\ \ (n=1,2,3,\cdots)\]

Evaluate $\sum_{n=1}^{\infty} a_n$[/quote]

$\sum_{n=1}^{\infty} a_n=\infty$. Probably there is a typo in the statement, because it is not convergent (the general term does not go to zero.)

Best,

David, the general term does go to zero. Look again.

Noodling around on a spreadsheet, then back-calculating from the answer, once the answer is known:

The sum is $\frac53$.

The $n$th partial sum of the series is $\frac{5n}{3n+2}.$

From that you can calculate that

$a_n=\frac{5n}{3n+2}-\frac{5n-5}{3n-1} = \frac{10}{(3n-1)(3n+2)}.$

And that satisfies the recursion. So we're done.

You are right, Kent Merryfield.

Here is my solution.

$(3n+5)a_{n+1}=(3n-1)a_{n}\Longleftrightarrow (3n+5)(3n+2)a_{n+1}=(3n+2)(3n-1)a_{n}$

$=\cdots\cdots =(3\cdot 1+2)(3\cdot 1-1)a_1=5\cdot 2\cdot a_1=10$

Therefore $a_n=\frac{10}{(3n+2)(3n-1)}$

Upss!  :oops: . Sorry, for some reason I had in my calculations that the general term goes to 1, but I can't understand what I did (messy scratch, mixed with a lot of other problems)  :blush:

Best,


calculus
derivative
inequalities unsolved
inequalities

[quote="Xevarion"]Suppose $a, b, c > 0$.

Prove that
$\frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}}\ge \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$

More generally, is it true that
$\sum_{cyc}\frac{a^{p}}{b^{p}+c^{p}}\ge \sum_{cyc}\frac{a^{q}}{b^{q}+c^{q}}$ whenever $p \ge q \ge 0$?
[hide="What I tried"] I didn't actually prove this though I am confident it is true. Actually it suffices to prove it for $q = 1$ because we can just replace $p$ by $rq$ and then $a^{q}, b^{q}, c^{q}$ by $x, y, z$. So then I thought if we take the derivative with respect to $r$ (assuming $x, y, z$ fixed), and it's nonnegative, then we win. But I couldn't prove it, even though it looks reasonably nice. [/hide][/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=harazi&t=23752
Take a look! :)

Cool, thanks! I couldn't find it by searching but I thought it seemed a natural question so it was probably old... :P
Actually I proved my original problem (the particular case) by mixing variables. This probably works for [b]harazi[/b]'s as well. But the derivative way is cleanest I think, even though calculus is generally looked down upon when it's not necessary.


geometry
circumcircle
incenter
ratio
quadratics
angle bisector
perpendicular bisector

In the triangle $ABC$ let $AD$ be the interior angle bisector of $\angle ACB$, where $D\in AB$. The circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $BCD$. Prove that $AC^2 = AD\cdot AB$.

Cannnot leave such a nice problem unanswered. Label the sides of the triangle $\triangle ABC$ as $a = BC, b = CA, c = AB$ and its angles at the vertices $A, B, C$ as $\alpha, \beta, \gamma$. Let $O$ be the common circumcenter of the triangle $\triangle ABC$ and the incenter of the triangle $\triangle BCD$. Since $O$ is the circumcenter of the triangle $\triangle ABC$, it lies on the perpendicular bisector of the side $BC$. Since the incenter of the triangle $\triangle BCD$ lies on the perpendicular bisector of its side $BC$, this triangle is isosceles with $DB = DC$, Hence, the angles $\frac \gamma 2 = \measuredangle BCD = \measuredangle CBD = \beta$ are equal. The external angle angle at the opposite vertex $D$ of this isosceles triangle is equal to the sum of these internal angles, that is, $\measuredangle CDA = \frac \gamma 2 + \beta = \frac \gamma 2 + \frac \gamma 2 = \gamma$. As the sum of the internal angles of the triangle $\triangle ADC$ is $180^o$,

$\alpha + \gamma + \frac \gamma 2 = \alpha + \frac 3 2 \gamma = 180^o$

The triangle $\triangle BOD$ is isosceles, because $OB = OC = R$ are radii of the circle $(O)$. Since $O$ lies on the bisectors of the angles $\measuredangle BCD = \frac \gamma 2$ and $\measuredangle CBD = \beta = \frac \gamma 2$, the internal angle $\measuredangle BOC$ of this isosceles triangle is equal to

$\measuredangle BOC = 180^o - \left(\frac \beta 2 + \frac \gamma 4 \right) = 180^o - \frac \gamma 2$

But the angle $\measuredangle BOC$ is the central angle of the arc $CAB$, which means that it is equal to double the angle $\measuredangle CAB = \alpha$ spanning this arc:

$\measuredangle BOC = 180^o - \frac \gamma 2 = 2 \alpha$

Hence, we obtained 2 equations for the angles $\alpha, \gamma$ and these angles can be calculated:

$\alpha + \frac 3 2 \gamma = 180^o$

$2 \alpha + \frac \gamma 2 = 180^o$

$\alpha = \gamma = \frac {360^o}{5} = 72^o$

This means that the triangles $\triangle ABC$ and $\triangle ADC$ are both isosceles with $c = AB = BC = a$ and $CA = CD$. Since the angle bisector $CD$ of the angle $\gamma = \measuredangle BCA$ cuts the opposite side $c = AB$ in the ratio of the adjacent sides $a = BC, b = CA$, we have

$\frac{AD}{BD} = \frac b a$

and since $BD = CD = CA = b$

$AD = \frac{b^2}{a}$

Consequently,

$AD \cdot AB = \frac{b^2}{a} \cdot c = \frac{b^2}{a} \cdot a = b^2 = AC^2$

In addition,

$\frac a b = \frac c b = \frac{AB}{CA} = \frac{AB}{BD} = \frac{AD + BD}{BD} = 1 + \frac b a$

and we obtained a quadratic equation for the ratio $\frac a b$:

$\left( \frac a b \right)^2 - \frac a b - 1 = 0$

$\frac a b = \frac{1 \pm \sqrt 5}{2}$

Since this ratio is positive, the negative root is not acceptable and $\frac a b = \frac{1 + \sqrt 5}{2}$ is equal to the golden ratio.

Aren't there ways of drawing the perpendicular bisector and angle bisector without having them coincide.

[quote="Valentin Vornicu"]In the triangle $ABC$ let $AD$ be the interior angle bisector of $\angle ACB$... [/quote]

Isn't it CD and not AD?

[quote="yetti"][quote="mna851"][quote]Since the incenter of the triangle  lies on the perpendicular bisector of its side , this triangle is isosceles with [/quote]

Is this a well known theorem?[/quote]
This is a simple deduction, too simple to be a theorem. The problem condition states that the point $O$ is the circumcenter of the triangle $\triangle ABC$. Therefore, the point $O$ lies on the perpendicular bisector of its side $AB$ and the triangle $\triangle BCO$ is isosceles with $OB = OC = R$, the circumradius. Hence, the angles $\measuredangle BCO = \measuredangle CBO$ are equal. The problem condition also states that the point $O$ is the incenter of the triangle $\triangle BCD$. This means that the lines $BO, CO$ are the angle bisectors of the angles $\measuredangle CBD, \measuredangle BCD$. As a result, $\measuredangle CBD = 2 \measuredangle CBO = 2\measuredangle BCO = \measuredangle BCD$, which means that the triangle $\triangle BCD$ is isosceles.[/quote]

Once you have come this far, $\measuredangle CBD = \measuredangle BCD = \measuredangle ACD$ Thus by alt seg, $AC$ is tangent to the circumcircle of $CDB$, so we are done.

[quote="kueh"]Once you have come this far, $\measuredangle CBD = \measuredangle BCD = \measuredangle ACD$ Thus by alt seg, $AC$ is tangent to the circumcircle of $CDB$, so we are done.[/quote]
Very nice, I missed that.


Given $a<>0$ and $b<>0$
Find all couple of positive integers $(a,b)$ so that there exists a bijective funtion $f$ from integers $N$  to integers $N$ ($0$ is included )  such that :
For all integers $m\geq 0$ & $n\geq 0$ we have :
$f(amn+m+n)=bf(mn)+f(m)+f(n)$

This post may be irrelevant, but the correct writing is "vieillesse" ...

Merci,je l'ai corrigé  :blush:
What about the problem ?


$Question 1.$ There are black and white checkers on some squares of a $100$ x $100$ board, with at most one checker on each square.  First, we remove every black checker that is in the same column as any white checker.  Next, we remove every white checker that is in the same row as any remaining black checker.  Prove that for some color, at most $2500$ checkers of this color remain.

$Question 2.$ Every cell of a $100$ x $100$ board is colored in one of $4$ colors so that there are exactly $25$ cells of each color in every column and in every row.  Prove that one can choose two columns and two rows so that the four cells where they interest are colored in four different colors.

[quote="xxxyyyy"]
$Question 2.$ Every cell of a $100$ x $100$ board is colored in one of $4$ colors so that there are exactly $25$ cells of each color in every column and in every row.  Prove that one can choose two columns and two rows so that the four cells where they interest are colored in four different colors.[/quote]
I'm guessing Graph theory. :)
I haven't done any though.

[quote="gotztahbeazn"]errr first off do you mean a $10x10$ board for question 2?[/quote] No, I don't think so.  :)

[quote="gotztahbeazn"]errr first off do you mean a $10x10$ board for question 2?[/quote]

No, its not $10 \times 10$

Editded.

"\times" looks much nicer than "x" for multiplication in latex.

$10\times 10$ versus $10x10$.

[quote="nodnarbo"]I find ways for qustion one where a color ends up with more than 2500 :huh:  :?:[/quote]

Yes, but the questions asks you whether it is possible to end up with both colors in excess of 2500.
($\neg(\textrm{some\,color\,has\,at\,most\,2500}) = \textrm{not\,all\,colors\,have\,more\,than\,2500}$, not $\textrm{some\,colors\,have\,more\,than\,2500}$.)

These seem a little hard for intermediate, maybe.  (At least, no one is getting them.)  Maybe move up to Pre-Olympiad?


The sequence $ c_0, c_1, . . . , c_n, . . .$ is defined by $ c_0 \equal{} 1, c_1 \equal{} 0,$ and $ c_{n\plus{}2} \equal{} c_{n\plus{}1} \plus{}c_n$ for $ n \ge 0$. Consider the set S of ordered pairs $ (x, y)$ for which there is a finite set $ J$ of positive integers such that $ x \equal{} \sum c_j , y \equal{} \sum c_{j \minus{} 1}$ ($ j$ belong to $ J$). Prove that there exist real numbers $ a, b,  m$ and $ M$ with the following property: An ordered pair of nonnegative integers $ (x, y)$ satisfies
the inequality $ m < ax \plus{} by < M$ if and only if $ (x, y)$ belong to $ S$.
Remark: A sum over the elements of the empty set is assumed to be $ 0$.


I vote for China

edit: Sorry, I should have added Other Country

[i]Moderator's Note: 'Other Country' option added. -mathfanatic[/i]

I voted for the United States of America.

Judging from past results, the winning team is [i]most likely[/i] to be China, but other countries like Russia and USA are strong competitors.

Yeah, the United States has a good chance this year.  However, Bulgaria proved last year that the unexpected can happen.

Why is everyone so confident that U.S. will win?

Well, um they are our team... I think most of the people on AoPS are from the US...so we want US to win.

Voted for the US just out of blind patriotism.

Luxembourg only sends two people... so yeah, its quite unlikely they will win. My opinion is that Vietnam will do very well. Go Romania.

i think its kinda obvious i just picked a random little country. they really do only send 2 people? i guess you do learn things in summer!

This year's team is Tiankai Liu, Tony Zhang, Alison Miller, Matt Ince, Aaron Pixton and Oleg Goldberg...Anders Kaseorg is alternate...good luck to them...

but my vote goes to China...

It seems that this is mainly a poll of who people want to win, not who they think will win. I think most people (including me) are fairly uninformed about the actual strengths and weaknesses of the different teams.

Last year Tiankai Liu wasn't there.... This year he's back gg to other IMO countries.

well, but no really, despite the fact that china has no mop, they still manage to beat us almost every year, and last last year when tianki was there, we got 4th.

[quote="MCrawford"]While China is the usual favorite, I wouldn't count out the U.S. team this year.  There may be no bigger favorite to answer 6 problems than Tiankai.  The team also boasts a gold medalist from last year's Russian team.  6 seniors makes for a lot of experience and maturity.  Two of the other team members have been alternates in the past and a U.S. alternate would likely win a silver medal in most years.

Let's also not forget the AoPS factor![/quote]

The team also has a gold medalist from last year's U.S. team.

I vote U.S.  There are a lot of veterans.  Plus, I'm going to root for my home country.

AoPS is probably too new to be a major factor this year, but I imagine it will be in 2-3 years.

P.S. I added the "other country" option.

I'm gonna vote for china. Its sorta obvious how good they r. Ofcourse i'm not dissing U.S. or ne thing, its just that, well...

[quote="Valentin Vornicu"]because I don't think any member of the US delegation is going to waste time now to do this, i'll do it :P

so basically US has won 5 gold medals, with Oleg 40p being first, and one silver through Matt Ince who got 31 (gold cut-off was at 32!).

in country's ranking US is second with 212 points, after China 220 points.[/quote]

Congrats to people who voted for China( and US).

well that's it, china won (again) ..... :) us was close though, and russia too.

[quote="Ghang Hwan"]I voted for the KOREA :D  :D  :D  :D  :D  :D  :D[/quote]You're a little bit too late :) China already won :D
We'll make another poll for Mexico though, just look in the Mexico Forum.


What five digit number when multiplied by 4 produces a five digit number whose digits are those of the original number but in reverse order?

Problem solved. Answer : 21978

Thanks to all who have considered this problem.


i was just wondering if odds is the chance of an event happening to the chance of an event not happeninig or is it the vice-versa (not happening to happening)

It depends on how you say it.  If you say "the odds of an event happening are," then it is happening:not happening.  But if you say "the odds of an event not happening are," then it is not happening:happening.

o i was just wondering because in the volume one of the aops book, i got confused on how they explained how odds worked.  i interpreted as the no:yes so like if it asked the prob of a thing winning is 7 out of 100 it said the odds of winning was 93:7 and i think it should really be 7:93

If they don't state anything about if it's odds for or odds against

ex: What are the odds that he wins?

It would be (according to AoPS I) odds against: Odds for.

If they state it: ex: What are the odds for him to win?

It means for: Agaiinst.

I think.  Odds are an iffy subject in my opinion, they usually do state if the question is for or against.

Yes it is usually stated in the problem (or, at least, I think it should be).

I learned it as success: failure, but that was just the way the textbook taught it.

For example, when betting on racehorses, odds are always expressed as failure: success.

In general, I think that odds aren't the greatest way to express answers, unless it is stated which method you are using, but I would put success: failure if it does not tell you which way to express them.

Hope this helps  :)

ex:

If the odds of something happening are 1000 to 1,
the probability of that event happening is 1/1001

Let's take the game roulette, there's a $\frac{9}{19}$ chance of you landing on red or black. However, they are offering 2:1 odds on red or black, that means it would be fair if there was a $\frac{1}{3}$ chance of landing on red or black. Instead, it's tilted in your favor. So are you saying the casino's way of expressing odds is wrong?


Nick, Janet, Huiju, and Amanda go to the World's Largest Brat Fest. Each of them is equally likely to order a brat, a hot dog, or a veggie brat. What is the probability that 2 of them order a brat, and the other two order hot dogs?

[hide]There are $ \dbinom42\dbinom22\equal{}6$ ways for the condition to hold true, and $ 2^4\equal{}16$ total ways, so the probability is $ \frac6{16}\equal{}\boxed{\frac38}$.[/hide]

[quote=Aequilipse][hide="Hm..?"]Hm, the correct answer for this on Alcumus is $\frac{2}{27}$.
Can somebody find the flaw in my [hide="work?"]$\binom42 (\frac{1}{3})^2 \cdot \binom42 (\frac{1}{3})^2 = \frac49$[/hide][/hide][/quote]

You don't need to multiply again by $\binom42$, since the other two people have to fill in the spots anyways. Or, you can think of it as $\binom22$ which is just $1$.