[quote="aev5peru"]HOLA A TODOS...FOREROS  :D
Aca les pongo un problema weno es algo q creo q se cumple...espero q lo resuelvan..digo q creo q se cumple por q aun no me sale  :rotfl:  :(
SUERTE!!
L1 es paralelo a BC, L1 pasa por A, D es un punto en la recta BC, P es el punto medio de AD, y B1 y C1 son las intersecciones de BP y CP con L1 respectivamente, ademas B2 y C2 son las intersecciones de PB y PC con AB y AC respectivamente, Sea Q el punto de interseccion de A2B1 con A1B2, y D' el punto de corte de AQ con BC, demostrar que BD = D'C[/quote]
upssss..notaran que faltan el punto A2 y A1..esos puntos son C2 y C1 respectivamente..ademas....no es necesario q P sea punto medio de AD puede ser cualquier punto en el segmento AD.
SUERTE

Todavía no está claro el enunciado, por ejmeplo en:
[quote="aev5peru"]ademas B2 y C2 son las intersecciones de PB y PC con AB y AC respectivamente[/quote]

B2=b y C2=C... además los puntos A,B y C no están bien definidos (bueno uno supone que están por ahí formando un triangulo pero igaul se debe decir)

Esta es mi interpretación del problema:
En el triángulo $ ABC$ sea $ D$ un punto en el lado $ BC$. Sea $ P$ un punto en el segmento $ DA$ y $ l_1$ una recta paralela a $ BC$ que pasa por $ A$. Nombramos las siguientes intersecciones:
$ C_1 \in CP \cap L_1$
$ C_2 \in CP \cap BA$
$ B_1 \in BP \cap L_1$
$ B_2 \in BP \cap CA$
$ Q \in C_1B_2 \cap C_2B_1$
$ D' \in AQ \cap BC$

Si es así?
Por cierto, lindo problema  :wink: .

[hide="Solución"][b]Definición:[/b] (Razón cruz) Sean $ A$, $ B$, $ C$ y $ D$ puntos colineares. La razón cruz de los pares de puntos $ (A,B)$ y $ (C,D)$ es.
\[ \Re(A,B;C,D) \equal{} \frac {AC/CB}{AD/DB}
\]
[b]Lema de la Razón cruz[/b]: Sean $ a$, $ b$, $ c$ y $ d$ cuatro rectas concurrentes. Para las rectas $ \ell_1$ y $ \ell_2$ se definen $ A_i \equal{} a \cap \ell_i$, $ B_i \equal{} b \cap \ell_i$, $ C_i \equal{} c \cap \ell_i$ y $ D_i \equal{} d \cap \ell_i$ para $ i \equal{} 1,2$. Entonces:
\[ \Re(A_1,B_1;C_1,D_1) \equal{} \Re(A_2,B_2;C_2,D_2)
\]
.

(Si alguine no conoce el lema puede probarlo usando areas)

[b]Solución del problema:[/b] Supongamos que la figura está en el plano proyectivo. Como en este plano todo par de rectas se cortan (también las paralelas) podemos decir que $ X \in l_1 \cap CB$
También defino:
$ A_1 \in AC \cap C_2B_1$
$ A_2 \in AB \cap C_1B_2$
Tenemos:
$ \minus{} (BD'/D'C) \equal{} \frac {BD'/D'C}{BX/D'X} \equal{} \Re(B,C;D',X)$
Usando el lema:
 $ \Re(B,C;D',X) \equal{} \Re(C_2,A_1;Q,B_1) \equal{} \Re(C_2,A;A_2,B) \equal{} \Re(P,B_1;B_2,B)$
Por el otro lado:
$ \minus{} (CD/DB) \equal{} \frac {CD/DB}{CX/XB} \equal{} \Re(C,B;D,X)$
Usando el lema:
$ \Re(C,B;D,X) \equal{} \Re(B_2,B;P,B_1)$

$ \Re(P,B_1;B_2,B) \equal{} \frac {PB_2/B_2B_1}{PB/BB_1} \equal{} \frac {B_2P/PB}{B_2B_1/B_1B} \equal{} \Re(B_2,B;P,B_1)$
De esto concluimos $ \minus{} (BD'/D'C) \equal{} \minus{} (CD/DB)$ que equivale a $ BD \equal{} D'C$

Q.E.D.[/hide]

Sea $ L_{\infty}$ el punto del infinito de $ \ell_1,BC$  y defínase las intersecciones $ T \equiv B_2C_2 \cap PQ,$ $R \equiv \ell_1 \cap B_2C_2,$ $ S \equiv \ell_1 \cap PQ$ y $ M \equiv AT \cap BC.$ Basta observar que en el cuadrivértice $ B_1C_1C_2B_2$ se tiene que $ (P,Q,T,S) \equal{} (C_2,B_2,T,R) \equal{} \minus{} 1$ $ \Longrightarrow$ $ (P,Q,T,S) \equal{} (D,D',M,L_{\infty}) \equal{} \minus{} 1$ $ \Longrightarrow$ $ M$ es punto medio de $ DD'$ y $ (C_2,B_2,T,R) \equal{} (B,C,M,L_{\infty}) \equal{} \minus{} 1$ $ \Longrightarrow$ $ M$ es punto medio de $ BC.$ Consiguientemente si $ M$ es punto medio de $ BC$ y $ DD',$ se sigue pues que $ BD \equal{} CD'.$


Let $a_1, a_2, ...,$ and $b_1, b_2, ...$ be arithmetic progressions such that $a_1 = 25$, $b_1 = 75$ and $a_{100} + b_{100} = 100$. Find the sume of the first $100$ terms of the progressions $a_1 + b_1, a_2 + b_2, ...$.

Source: AHSME 1975

[hide]Notice that $a_{1}+b_{1}=100$ and $a_{100}+b_{100}=100$.  Therefore, the difference between successive terms in progression $a$ is the opposite of the difference in progression $b$, and $a_{n}+b_{n}=100$ for any $n$.  The answer is $10000$.[/hide]


geometry
circumcircle
perimeter
geometry solved

Let $ABC$ be an arbitrary triangle. A circle touches the sides $AB$ and $AC$ of triangle $ABC$ at the points $P$ and $Q$, respectively, and also is tangent internally to the circumcircle of the triangle $ABC$. Prove that the midpoint of the segment $PQ$ is the center of the incircle of the triangle $ABC$.

This was a problem on the IMO1993 Shortlist  :) .
Indeed, it is a generalization of IMO 1978 problem 4, since that IMO problem had the extra (unnecessary) condition that $AB=AC$.

Let us apply inversion with center A and radius 1.
Circumcircle of ABC will become B'C', the initial circle will become escribed excircle wich tangents B'C'. Thus AP' = PAB'C'/2 = (1/AB+1/AC+BC/AB.AC)/2 = PABC/2AB.AC ==> AP = 2Ab.AC/PABC = 2r/sin A. On the other hand if we reconstruct p.P for incircle we obtain the same value!

i don't understand why AP' = P_{AB'C'}/2
and, what's your point P here? is it the tangent point desribed in the original problem?

P' is the image of P ==> P' is a point of tangent of escribed excircle with line AB'. PAB'C' is the perimeter of AB'C'.

AP'=PAB'C'/2 is a well known fact.

can you explain the 'fact' in more details?
or can you tell me what is the original theorem of the fact and where can i find it?

Let \omega be escribed excircle of ABC, \omega tangents segment BC in p.A1 and lines AB and AC in points C1 and B1 respectively.
It is FACT: BA1=BC1, CA1=CB1, AB1=AC1 ==>
PABC = AB+AC+BC = AB+AC+BA1+CA1 = (AB+BC1) + (AC+CA1)=AC1 + AB1 = 2AC1.

thank you all the same.
i understood it later by myself. i just did not understand your english at first. sorry for my poor english.

Problem: The $ A \minus{}$mixtilinear incircle of $ \triangle ABC$ touches $ C(O,R)$, $ CA$, and $ AB$ at $ T$, $ E$, and $ F$, respectively. Prove that $ I$ is the midpoint of $ EF$.

Proof: Let $ BI$ and $ CI$ intersect the circumcircle again at $ B_1$ and $ C_1$. Let $ Y\in B_1T\cap CC_1$ and $ Z\in C_1T\cap BB_1$. Clearly, $ B_1$ and $ C_1$ are the midpoints of small arcs $ CA$ and $ AB$, respectively. It is a well-known result that $ T$, $ F$, $ C_1$; $ T$, $ E$, $ B_1$ are triples of collinear points. Now $ (C_1FZT) \equal{} B(C_1FZT) \equal{} (C_1AB_1T) \equal{} C(C_1AB_1T) \equal{} (YEB_1T)$. Thus $ C_1Y$, $ FE$, and $ ZB_1$ are concurrent. But $ I\in CC_1\cap BB_1$ so $ I\in EF$. Since $ AE \equal{} AF$, and $ \angle FIA \equal{} \angle EIA$, $ I$ is the midpoint of $ EF$, as desired.


A person has 12 CDs of the following types:

A - 3
B - 4
C - 2
D - 1
E - 2

If 5 CDs are randomly selected, what is the probability that either 3 A's or 3 B's will be selected.

I tried (C(3,3)C(9,2) + C(4,3)C(8,2))/C(12,5) = 37/198, but the textbook had an answer of 13/66.

What am I doing wrong?

Thanks.

[quote="mz94"][hide="Hint"]your method is correct. however, you fail to count the different orders in which the CDs can be selected[/hide][/quote]

I probably missed something; what's wrong with qszwdxefc's method? Since he/she didn't count ordering in the denominator, shouldn't he/she also remove an ordering term in the numerator, which is what he/she did?

Thanks, but using permutations results in 37/1980, which is still incorrect.

Wait, you should get the same answer regardless of permutation/combination.

The CDs are probably indistinguishable.

And I think he probably typed the answer wrong.  I just did it and get the same 37/198 with permutations.

Is there any way we could be missing a two in the numerator, as that would give us the "correct" answer?

A two in the numerator won't fix it. $ 2(\frac{37}{198})\equal{}\frac{37}{99}$, not $ \frac{13}{66}$.

Sorry for being vague, I meant adding a two to the numerator (making $ \frac{39}{198} \equal{} \frac{13}{66}$). :blush:

Hmm that seems to work... but where does the +2 come from? :huh:

Hey guys guess what, the textbook made an error :mad:.

Thanks anyways.

bleh sorry for failing i didnt really read through the whole solution. MEHHHHH


geometry
perimeter
ratio
3D geometry
tetrahedron
parallelogram
rhombus

If M is the midpoint of AD,N-midpoint of BC and K-midpoint of IE and K,M and N lie on one line,then proov that AB=CD.

[quote="lym"]It's wrong.[/quote]
It can't be wrong.
We have that I is the center of the circle,ACxBD=E.
Please check it on some program to see that is true.

Come on!I don't believe that noone can solve it.

[b]Lemma:[/b] Given triangle $ ABC$. Let $ G,H$ be two points on rays $ AB, AC$, respectively such that $ GB \equal{} HC \equal{} x$. Denote $ D \equal{} BH\cap CG, J,N$ the midpoints of $ BC, GH$. The A-bisector of triangle $ ABC$ intersects $ BH$ at $ K, JN$ intersects $ BH$ at $ I$. Then $ DI \equal{} IK$.
[b]Proof:[/b]
We know that $ JN//AK, JN$ bisects the perimeter of triangle $ ABC$ and denote $ L$ the intersection of $ JN$ and $ AC$ then $ LC \equal{} \frac {AB \plus{} AC}{2}$.
We have $ \frac {BK}{BH} \equal{} \frac {AB}{AB \plus{} AH} \equal{} \frac {AB}{AB \plus{} AC \plus{} x}. (1)$
On the other side, applying Menelaus's theorem: $ \frac {BD}{DH}.\frac {CH}{CA}.\frac {GA}{GB} \equal{} 1$ then $ \frac {BD}{DH} \equal{} \frac {CA}{GA}$
$ \Rightarrow \frac {BD}{BH} \equal{} \frac {CA}{CA \plus{} GA} \equal{} \frac {CA}{AB \plus{} AC \plus{} x} (2)$
Applying Menelaus's theorem again we obtain $ \frac {IB}{IH}.\frac {LH}{LC}.\frac {JC}{JB} \equal{} 1$ thus $ \frac {IB}{IH} \equal{} \frac {LC}{LH}$
$ \Rightarrow \frac {IB}{BH} \equal{} \frac {LC}{LC \plus{} LH} \equal{} \frac {AB \plus{} AC}{2(AB \plus{} AC \plus{} x)} (3)$
From $ (1),(2)$ and $ (3)$ we claim $ BI \equal{} \frac {1}{2}(BK \plus{} BD)$ therefore $ I$ is the midpoint of $ KD$.(QED)
[b]Back to our problem:[/b]
Let $ J$ be the intersection of $ AB$ and $ CD$. Since $ O$ lies on J-bisector of triangle $ AJD$ then $ OJ//HI$.
Applying the lemma above we get the intersection of $ FO$ and $ IH$ is the midpoint of $ OF$. Our proof is completed.

I have some questions:
1.Why $ JN\parallel{}AK$?
2.How $ LC \equal{}\frac{AB\plus{}AC}{2}$?

By the first question we must use the theorem of Thales,right?

Actually livetolove212 proves the reverse of the problem, but we have to assume it is OK, by the uniqueness of the construction.
A shorter way to prove the above lemma (using the same notations) would be:
1) Apply Menelaos to $ \triangle AGC$ with the transversal $ BDH$ and get $ \frac {CD}{DG} \equal{} \frac {AB}{AH}$ ( 1 );
2) From angle bisector theorem we have $ \frac {BK}{KH} \equal{} \frac {AB}{AH}$ ( 2 ), from (1) and (2) $ D$ and $ K$ divide the diagonals $ CG$ and $ BH$ respectively at the same ratio, hence the midpoint $ (I)$ of the segment $ KD$ lies on $ JN$.

To prove of this statement we shall first prove the following
[b][size=134][color=darkblue]Lemma[/color][/size][/b]: [i][color=darkblue]Let $ ABCD$ be a quadrilateral, $ E$ and $ F$ points on $ (AB)$ and $ (CD)$ respectively such as $ \frac {AE}{BE} \equal{} \frac {DF}{FC}$, $ L$ and $ M$ points on $ (BC)$ and $ (AD)$ such as $ \frac {BL}{LC} \equal{} \frac {AM}{MD}$. Then there exists $ K \in EF \cap ML$ and $ \frac {MK}{KL} \equal{} \frac {AE}{BE}$ and $ \frac {EK}{KF} \equal{} \frac {BL}{LC}$[/color][/i]

To achieve this goal: join $ B$ and $ C$ with $ M$, the parallel through $ E$ at $ AD$ intersects $ BM$ at $ I$, while the parallel trough $ F$ at $ AD$ intersects $ CM$ at $ J$. It's easy to see that $ \frac {EI}{FJ} \equal{} \frac {AM}{MD}$ and that $ \frac {IM}{IB} \equal{} \frac {JM}{JC}$, hence $ IJ \parallel{} BC$. From $ \triangle KIE \sim \triangle KJF$ we get $ \frac {KI}{KJ} \equal{} \frac {EI}{FJ} \equal{} \frac {AM}{MD} \equal{} \frac {BL}{LC}$, hence $ M, \ K$ and $ L$ are collinear, lemma being proven.

If $ L$ and $ M$ are midpoints of $ BC$, $ AD$ respectively, then $ I$ and $ J$ are midpoints of $ EG$ and $ FH$ and $ K$ becomes the intersection of the diagonals $ EF$ and $ GH$, and the claim from 2) is solved as well. ( For an easy understanding, see the attachment).

[color=red]Remark: As it can be seen from the proof of the lemma, the property holds for a tetrahedron as well.[/color]

For barcelona: I presume there is no need to further explain your queries - the first one is sketched down, while the second one resulting from easy calculations as $ JL \parallel{} AK$.

Best regards,
sunken rock

Well i don't understand this part:
[quote="livetolove212"]
Let $ J$ be the intersection of $ AB$ and $ CD$. Since $ O$ lies on J-bisector of triangle $ AJD$ then $ OJ//HI$.
Applying the lemma above we get the intersection of $ FO$ and $ IH$ is the midpoint of $ OF$. Our proof is completed.[/quote]
In my problem there is no points like:O,F or H.
So can anybody show how to apply that lemma.

[quote][b][size=100][color=DarkBlue]A circumscribed quadrilateral $ ABCD$ is given, with incircle $ (I)$ and $ AB = CD$ and let be the point $ E\equiv AC\cap BD.$ Prove that the line connecting the midpoints $ M,\ N,$ of the side-segments $ BC,\ AD$ respectively, bisects the segment $ EI,$ where $ I$ is the center of $ (I).$[/color][/size][/b][/quote]
[b]PROOF.[/b] - Let be the point $ F\equiv AB\cap CD$ and based on the below [b]Lemma 1[/b], we have that $ MN\parallel FI$ $ ,(1)$

Also, based on the [b]Lemma 2[/b], we have that $ F'E\parallel FI$ $ ,(2)$ where $ F'$ is symmetric point of $ F,$ with respect to the midpoint $ M$ of $ BC.$

From $ (1),\ (2)$ we conclude that the line segment $ MN$ bisects the segment $ EI$ and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA 1. - A triangle $ \bigtriangleup ABC$ is given and let be the points $ P,\ Q$ on its side-segments $ AB,\ AC$ respectively, such that $ BP = CQ.$ Prove that the line connecting the midpoints $ M,\N,$ of $ BC,\ PQ$ respectively, is parallel to the angle bisector of $ \angle A.$[/color][/size][/b]

[b][size=100][color=DarkBlue]LEMMA 2. - A triangle $ \bigtriangleup ABC$ is given and let be the points $ P,\ Q$ on its side-segments $ AB,\ AC$ respectively, such that $ BP = CQ.$ prove that the point $ K\equiv BQ\cap CP,$ lies on the line parallel to the angle bisector of $ \angle A,$ though the point $ A',$ as the symmetric point of $ A$ with respect to the midpoint $ M$ of the side-segment $ BC.$[/color][/size][/b]

Kostas Vittas.

PS. I will post here later the proofs of the above [b]Lemmata[/b] I have in mind.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA 1. - A triangle $ \bigtriangleup ABC$ is given and let be the points $ P,\ Q$ on its side-segments $ AB,\ AC$ respectively, such that $ BP = CQ.$ Prove that the line connecting the midpoints $ M,\N,$ of $ BC,\ PQ$ respectively, is parallel to the angle bisector of $ \angle A.$[/color][/size][/b][/quote]
[b]PROOF OF THE LEMMA 1.[/b] - (We use the same [b]schema[/b], as mentioned before by [b]sunken rock[/b] ) - Let be the points $ E,\ Z,$ such that $ BPNE,\ CQNZ$ to be parallelograms.

So, from $ MB = MC$ and $ BE\parallel = PN = NQ\parallel = CZ,$ based on the [b]Thales theorem[/b], we conclude that the points $ E,\ M,\ Z,$ are collinear and $ ME = MZ$ $ ,(1)$

Also, from $ NE\parallel = BP = CQ\parallel = NZ$ $ \Longrightarrow$ $ NE = NZ$ $ ,(2)$

From $ (1),\ (2)$ we conclude that the line segment $ MN$ is the angle bisector of the $ \angle N$ of the isosceles triangle $ \bigtriangleup NEZ.$

Because of now, $ NE\parallel AB$ and $ NZ\parallel AC,$ we conclude that the line segment $ MN$ as the angle bisector of $ \angle N,$ is parallel to the line segment $ Ax,$ as the angle bisector of $ \angle A$ ( because of the homologous rays of these angles, have the same direction ) and the proof of the [b]Lemma 1[/b] is completed.

Kostas Vittas.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA 2. - A triangle $ \bigtriangleup ABC$ is given and let be the points $ P,\ Q$ on its side-segments $ AB,\ AC$ respectively, such that $ BP \equal{} CQ.$ prove that the point $ K\equiv BQ\cap CP,$ lies on the line parallel to the angle bisector of $ \angle A,$ though the point $ A',$ as the symmetric point of $ A$ with respect to the midpoint $ M$ of the side-segment $ BC.$[/color][/size][/b][/quote]
[b]PROOF OF THE LEMMA 2.[/b] - We draw the linew through the points $ P,\ Q$ and parallel to $ AC,\ AB$ respectively and let $ L$ be, their point of intersection.

We have now the configuration of the parallelogram $ ABA'C,$ where $ A'$ is the symmetric point of $ A$ with respect to the midpopint $ M$ of the side-segment $ BC$ and then, we conclude that the points $ K,\ L,\ A'$ are collinear as well. ( see the [b]Lemma 2[/b] in the post #7# [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107997][b][color=Blue]Here[/color][/b][/url] ).

Let be the points $ X\equiv BA'\cap QL,\ Z\equiv CA'\cap PL$ and it is easy to show that $ LXA'Z$ is a rhombus, because of $ A'Z\parallel \equal{} XL\parallel \equal{} BP \equal{} CQ\parallel \equal{} ZL\parallel \equal{} A'X$.

So, we have that the line segment $ A'K$ is the angle bisector of $ \angle A'$.

We conclude now, that $ A'K\parallel AN,$ where $ AN$ is the angle bisector of $ \angle A,$ because these angles have their rays parallel each other with opposite direction and the proof is completed.

Kostas Vittas.

In fact,we can get a more generally conclusion:in ABCD,AB insection CD on F,AC insection BD on E,I is the inceter of ABF,G,H,m is the midpoint of AC,BD,IE.Then AB=CD if and only if GHM colinear.
Proof:We know AI//GH if and only if AB=CD from above,for GHJ is the Newton Line of AEDFBC,so J is the Midpoint of EF,
then GHM colinear if and only if AI//GH.


topology
real analysis
real analysis unsolved

Let $V$ be a normed space and $f: A\subset (M,d) \to V$ be continuous at $x_{0}\in A$.
Show that if $f(x_{0})\neq 0$, $f$ is nonzero in a neighborhood of $x_{0}$.

Suppose $f$ is continuous at $x_{0}\in~A$ and $f(x_{0})$ is nonzero. Given any $\epsilon>0$ we can find a integer $M$ such that $\frac{1}{n}<\epsilon$ whenever $n>M$ for integer $n$; so that $\parallel f(x)-f(x_{0})\parallel <\frac{1}{n}<\epsilon$ whenever $d(x,x_{0})<\delta$ and $x\in~A$. Consider the sequence of neighborhoods $N_{\frac{1}{n}}(f(x_{0}))=\{x\in~A~|~\parallel f(x)-f(x_{0})\parallel <\frac{1}{n}<\epsilon\}$. As $n\rightarrow\infty$; $\frac{1}{n}\rightarrow~0$ so that $f(x)\rightarrow~f(x_{0})$ for all neighborhoods $N_{\frac{1}{n}}(f(x_{0}))$ whenever $n>M$. Since $f(x_{0})$ is nonzero, $f$ is nonzero for all such neighborhoods.

Let's see..  $V$ is Hausdorff, so $\exists U \ni f(x_{0})$ such that $0 \notin U$. Now $U'=f^{-1}(U)$ is an open nbh of $x_{0}$, such that $f(x)\neq 0 \forall x\in U'$.


Did  anybody in Romany  - or anywhere - find any simple solution to BMO - 2007 , probl 2 , different than the official solution and which is this solution ?

Thank you ! -  Babis

There was this guy from the Romanian BMO Team... I forgot his name... Besides substituting $y : = f(x)$, he also tried $y : = 2 \cdot f(y)-f(x)$. He then combined the two relations and the conclusion followed easily.

he is nicolae papara (rom 6)
his solution was very nice and simple, indeed :
1) take $y=f(x)$. it follows that $f(2y)=f(0)+(2y)^{2}$.
2) take $y=2f(y)-f(x)$. using 1), the relation becomes $f\big(2f(x)-2f(y)\big)=f(0)+\big(2f(x)-2f(y)\big)^{2}$ (just make the computations, it`s easy).
3) $f\equiv 0$ is solution. take $x_{0}$ such that $f(x_{0})\neq 0$, and note that
$8f(x_{0})y=2f\big(f(x_{0})+y\big)-2f\big(f(x_{0}-y)\big)$,
and take $y=\frac{x}{8f(x_{0})}$. use 2) and the problem is over.

Thank you !

 Really nice approaches!

 babis


This problem was, in my opinion, bit too easy for AIME #7. I'm not sure though. :)

Let $x,y,z$ be a positive real numbers such that following holds true:

\[xyz = 1\\
x + \frac {1}{z} = 5\\
y + \frac {1}{x} = 29\]

If that is the case, then find $z + \frac {1}{y}$.

From the first equation, $\frac{1}{z}=xy$ and from the third, $y=29-\frac{1}{x}$. Substituting this into the second equation we get  $x(30-\frac{1}{x})=5$ so $x=\frac{1}{5}$  then solve for y and z to get $y=24$ and $z=\frac{5}{24}$ thus $z+\frac{1}{y}=\frac{1}{4}$

Great.

Next time, please put your answer in spoiler though.

Thank you.


Suppose that $ a$ and $ b$ are different fixed positive integers.
Show that the coupled equations $ xy \plus{} zw \equal{} a$ and $ xz \plus{} yw \equal{} b$ cannot have infinitely many solutions for integers $ x$, $ y$, $ z$ and $ w$.

By adding and subtracting the equations we get $ (x\plus{}w)(y\plus{}z) \equal{} a\plus{}b$ and $ (x\minus{}w)(y\minus{}z) \equal{} a\minus{}b$, hence by multiplying, $ 0 < | (x^2\minus{}w^2)(y^2\minus{}z^2) | \equal{} | a^2\minus{}b^2 |$. Therefore $ 0 < | (x^2\minus{}w^2)| \leq | a^2\minus{}b^2 |$ and $ 0 < | (y^2\minus{}z^2)| \leq | a^2\minus{}b^2 |$.
But $ 0 < | A^2 \minus{} B^2 | \leq | C |$ has finitely many integer solutions for fixed $ C$.

Notice that if we allow $ a\equal{}b$, then we can take $ y\equal{}z\equal{}1$, arbitrary $ x$, and $ w\equal{}a\minus{}x$.

That's correct.

It should also be apparent that the number of solutions is at most $ d(a\plus{}b) d(a\minus{}b)$ where $ d(n)$ means the number of divisors of $ n$.


arkadaslar sadece merak ettigim icin soruyorumda siz kac yasindasiniz.(eger istemiyorsaniz cevaplamayin)

neden herkes yasını söyledikten sonra forumdan çekilmeyi tercih ediyor anlamıyorum neyse ben de 15 yaşındayım.

Ben de 17 yasindayim.

Saraybosnadan selam :)


Functional Analysis
real analysis
real analysis unsolved

Let $ X$ be a Banach space and $ F$ be a closed subset. Suppose that $ X$ is obsorded by $ F$. Prove that $ int(F) \neq \emptyset$

[quote="kingkong"]$ X$ is obsorded by $ F$[/quote]
The correct spelling is A-B-S-O-R-B-E-D.
Do you know the Baire category theorem? Your problem follows from it in one line...


modular arithmetic
number theory
greatest common divisor

For each nonnegative integer $n$ we define $A_n = 2^{3n}+3^{6n+2}+5^{6n+2}$. Find the greatest common divisor of the numbers $A_0,A_1,\ldots, A_{1999}$.

[i]Romania[/i]

$A_0=35$, but $A_1$ is not a multiple of $5$ so it is either $7$ or $1$. Since $5^6=3^6=2^3=1 \pmod 7$, we have that $7$ divides every number so the gcd is $7$.

[hide="2500'th post "]

Note that $A_0=35$, so we evaluate $A_1\pmod{5}$ and $A_1\pmod{7}$.  $A_1\pmod{5}$ is equal to $3+3^8=4$ by fermat's little theorem so we disregard the factor of $5$.  Now, I will prove that all terms in the sequence are $0\pmod{7}$:

$2^{3n}+3^{6n+2}+5^{6n+2}\equiv 0\pmod{7}$ is what we desire.  We now get $2^{3n}+9^{3n+1}+5^{6n+2}\equiv 0\pmod{7}$ as desired or $2^{3n}+2^{3n+1}+(-2)^{6n+2}\equiv 0\pmod{7}$.  This is the same as $2^{3n}*3+2^{6n+2}\equiv 0\pmod{7}$ as desired.  Note that $2^{3n}\equiv 8^n\equiv 1\pmod{7}$ and $2^{6n+2}\equiv (2^{3n+1})^2\equiv 4\pmod{7}$ giving $2^{3n}*3+2^{6n+2}\equiv 3+4\equiv 0\pmod{7}$ hence completing our proof.

The answer is henceforth $\boxed{7}$. [/hide]

for n=0, we get A0=35. if n=1, we get A1=397194. so the greatest common divisor of A0, A1, A2,.....,A1000 is either 1 or 7, assume the answer is 7, so we will prove it.
we know that 2^3n is equal to 8^n which is congruent to 1 mod 7, and 3^6n+2 is equal to 9^3n+1 which is congruent to 2^3n+1 mod 7, similarly we get 5^6n+2 is congruent to 4^3n+1 mod 7. we have 2^1 congruent to 2 mod 7, 2^2 is congruent to 4 mod 7, 2^3 is congruent 1 mod 7, 2^4 is conruent to 2 mod 7, etc. so 2^3n+1 will be congruent to 2 mod 7, analogously, we get 4^3n+1 will be congruent to 4 mod 7, so the answer is 7.

Notice that $A_0=2^0+3^2+5^2=1+9+25=35.$ Then, the greatest common divisor must be one of the factors of $35,$ namely $1,5,7$ or $35.$

Next, notice that $A_1=2^3+3^8+5^8=8+3(3^7)+5(5^7)\equiv 8+3\cdot 3+5\cdot 5 \equiv 42\equiv 0 \mod 7.$ This means that $A_1$ is a multiple of $7.$ However, $A_1= 2^3+3^8+5^8\equiv 8+6561 \equiv 3+1=4\mod 5.$ As a result, since $5\nmid A_1,$ the greatest common divisor must not be a multiple of $5,$ so it needs to be $1$ or $7$ from before.

Then, we will show that for all integers $n\ge 0,$ $7\mid A_n.$ Our base case is $A_0 =35\equiv 0 \mod 7,$ which we calculated before.

Next, assume that $2^{3n}+3^{6n+2}+5^{6n+2}\equiv 0\mod 7.$ We then have\begin{align*}A_{n+1}&=2^{3(n+1)}+3^{6(n+1)+2}+5^{6(n+1)+2}\\
&=2^{3n+3}+3^{6n+8}+5^{6n+8}\\
&=2^{3n}2^3+3^{6n+2}3^6+5^{6n+2}5^6\end{align*}
By FLT, $3^6\equiv 1 \mod 7$ and $5^6\equiv 1\mod 7.$ So, we then have
\begin{align*}A_0&\equiv 2^{3n}(8)+3^{6n+2}(1)+5^{6n+2}(1)\\ &\equiv 2^{3n}+3^{6n+2}+5^{6n+2} \equiv 0 \mod 7.\end{align*}
by our inductive hypothesis. So, $7\mid A_{n+1},$ meaning that $7$ divides all elements $A_n$ and so $\gcd(A_0,A_1,\dots, A_{1999})=7.$ $\square$


[quote="ProtestanT"][quote="sonny"]What integer n has the property such that $n^{72}<5^{96}<{(n+1)^{72}}$?[/quote]
[hide]
$n^{3}<5^{4}<(n+1)^{3}$
$n^{3}<625<(n+1)^{3}$
$8^{3}<625<9^{3}$
$n=8$
[/hide][/quote]

I don't know if that is the correct answer but your solution looks right  :)

[quote="wii"]I don't know how you can do it without a calculator, but it shows that 8^72 < 5^96 < 9^72[/quote]

You CAN do it without a calculator!
[hide="Solution (with guess and check)"]Raise both sides to the power of $\frac{1}{24}$ to get:

$n^{3}<5^{4}<(n+1)^{3}$

$n<5\sqrt[3]5<n+1$

You know that $1<\sqrt[3]5<2$, so you could guess and check to get $n=8$. [/hide]He beat me to it...

[quote="funcia"][quote="wii"]I don't know how you can do it without a calculator, but it shows that 8^72 < 5^96 < 9^72[/quote]
how do you do it with?[/quote]
He probably guessed and checked until you get an answer.

[quote="funcia"][quote="wii"]I don't know how you can do it without a calculator, but it shows that 8^72 < 5^96 < 9^72[/quote]
how do you do it with?[/quote]
Remember that this is an old state sprint problem. But with a calc you can use logs.

To vishalarul: I wouldn't simplify your radical so much unless I had $5^{1/2}$ memorized to the thousandth place, just to be safe, but if you like to live on the dangerous side, go ahead :)
And as i remember it, this was #30 on 1997 state sprint round....and btw, if you ever have a logarithm with a base that is neither $e$ nor $10$ then you can use the change of base rule which says $\log_{x}{y}=\frac{\log{y}}{\log{x}}$

Here's another way to solve the problem (and a more general way to look at these types of problems just in case the exponents aren't so friendly)
[hide]$n^{72}<5^{96}<{(n+1)^{72}}$
$72\log{n}<96\log{5}<72\log{(n+1)}$
$n<5^{96/72}<n+1)$
$n<(625)^{1/3}<n+1$
Every mathlete who has read the Diaz Bible knows that $9^{3}=729$ and that $8^{3}=512$
$\therefore n=8$[/hide]

[quote="xscapezaer"]Seriously, vishalarul, you know there're no calcs on sprint round. sheesh. how exactly do you "know" that 1 < 5^(1/3)< 2?[/quote]

1^(1/3)=1

8^(1/3)=2

Conclusion follows.


[quote="behdad.math.math"]prove that infinite numbers n exist that we can write each of n,n+1 and n+2 as the sum of perfect squares of two integers.[/quote]
[hide="Solution"]
We let $ n \plus{} 1$ be a perfect square, say $ k^2$.  Then, we have that $ k^2 \equal{} k^2 \plus{} 0$ and $ k^2 \plus{} 1$ are both sums of two squares.  It suffices to show that there are infinitely many $ k$ so that $ k^2 \minus{} 1 \equal{} k^2 \minus{} 2k \plus{} 1 \plus{} (2k \minus{} 2)$ is a sum of squares.  Yet, if $ 2k \minus{} 2 \equal{} l^2$ for some $ l$, then we have that $ k^2 \minus{} 1 \equal{} (k \minus{} 1)^2 \plus{} l^2$.  We obtain the set of solutions: $ n \equal{} k^2 \minus{} 1 \equal{} l^2 \plus{} (\frac {l^2}{2})^2$ for even $ l$. [/hide]


Yellow Boy Chapter 1 (An epic novel)

Yellow Boy
-------------------------------------

Chapter 1: The Awkward Arrival
-------------------------------------
The story begins when I was traveling on the airplane one day. My friend Gabe and I were going to the unknown land called the United States,San Francisco to be exact, all because we were forced by our parents to escape our fate of the foot-massaging business in China.
A while after dinner Gabe pointed at a man and said in a chinky chinese voice,

"I wanna eat him, he looks like some good chocolate".

I said to him, "You stupid turtle asse, its not even a good joke, that's guy is one of those African people you see in Roots"

Gabe looked at me like I've lost my nuts and said, "You b'astard guy, don't underestimate me." The man saw us looking at him and said, "Watchu looking at, you lil' chinese biches?"

Gabe immediately said in broken English: "Hi chocolate man!"

I saw the man's eyes bulge out like bull testicles and the most ghetto expression ive ever seen. But fortunately, none of us got hurt and we successfully continued the trip with our heads on our shoulders.

Whoosh, the airplane came down to the earth as I gazed upon this new peice of land that looks very weird and a bit disturbing as well, its name was America. I got out of the plane and as I was entering the lobby some girls with yellow colored hair,looking like my morning excrete, were checking for passports and stuff like that. When it came to my turn, I needed to show them my Chinese passport.

But crap, at that moment I just all my pockets, but I coulden't find it! The girl got impatient and roared,

"Please show your identification! Or else we can't let you pass!"

I suddenly felt a rectangular hard spot in my pants and finally realized, "Oh shiet, the passport is INSIDE my pants." I tried not to explode in this awkward situation and tried to explain, but they wouldn't understand my broken Chinglish.

At last, I had no choice but to take off my pants. At that instant, an old chinese lady came up to me and said in chinese:

"You little stupidy shiz, keep you pants ON!" And then she ACTUALLY smacked it with a cane. At that same moment I didn't feel any pain, but that moment what I felt was much more magnificient and cunning than pain, it was as if my spirit has been set free and there was a rocket engine propelling me through space. I farted.

Silence. After 2 seconds, the old woman fell and fainted. I quickly showed the girl, who was half stunned, my passport and quickly ran through the exit, like the time fat pigs were chasing me in Beijing.

Then I remembered about Gabe, and ran back to get him. He was nowhere to be seen. After I informed the police about the case, apparently when he was in the airport bathroom, he misused the American toilet and had his foot stuck in the hole. When I went to get

him he looked like a mess, but after half an hour we finally made it through the airport.

We did not have a cell phone at the time and therefore noone could pick us up. The only currency we carried were chinese dollars, and

at the time I was unsure if they could be used in America.

To get to the destination we were going, we needed a taxi. We finally halted a cab and the driver was a nice middle-aged mexican man.
He said:

"Where are you headed amigo?"
We couldn't understand him but we showed him the address. He said, "oh! Long distance! you pay a lot when you get there ok?"
I said ok and we started to have the funnest ride of our lives.

On the street, we saw various brands of Japanese cars that we have seen in China before. We both would ocassionally say thoughout the

trip: "ooo, Honda, made in Japan, very fast. OOO, mitsubishi, made in Japan, very very fast." Near the end of our trip as we were

closer to our destination, the Mexican driver stopped the car, and said: "This is the end of our trip". We looked at the taxi-meter, it

said "247.89". We both shouted:

"WHAT?! So expensiv!"

Then he said:

"Haha. Price-Meter, made in Japan. Very fast."

At that point we didnt know what to do. We couldn't use our chinese money. So I said to Gabe in chinese:

"Let's run for it". He replied: "1,2.."

On 3 we both ran out of the taxi running rapidly north. When we were about to become out of sight, we heard a car driving behind that

and I suddenly heard Gabe cry, "OOOWW! My assie!" I turned around and saw him flying aross the road like chinese gone wild. Then at

that moment I knew what happened. He mexican driver has gone mad! Before I could hide in the nearest Bush, I felt a force of 58,000

newtons hit my behind. Then. nothing.

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

It was quite suprising that I woke up the next day. The strange thing was that I didn't know where I was. I looked around the room, it

was white and I was sleeping in a king sized bed. But the weird thing was.. there's a lump of something under the sheet right next to

me.

TO BE CONTINUED

a few points:

(1) Proof-read for grammatical and/or spelling mistakes.
(2) Racism is very highly looked down upon in AoPS.
(3) So is cussing.


calculus
integration
trigonometry
function
logarithms
real analysis
real analysis unsolved

I'm posting this because I had looked at integrals in mathlinks.ro and I found lots of interesting ideas. When I was a student a teacher of mine asked me and all my colleagues for an integral that was most notable for us. I liked this experiment and I want here to post 1-3 integrals by user - that were most notable to you. I know for experienced people like misan, kunny, Virgil Nicula, Carcul, N.T.TUAN (sorry if I not mentioned some good solver) it will not be an easy task, because they had thousands of interesting integrals, but I believe here we may collect really beautiful integrals.

For me for example - my favorite integrals are:
http://www.mathlinks.ro/Forum/viewtopic.php?t=167800
http://www.mathlinks.ro/Forum/viewtopic.php?t=168714
http://www.mathlinks.ro/Forum/viewtopic.php?p=935835#935835
http://www.mathlinks.ro/Forum/viewtopic.php?t=168335

$ \int e^x \sin x dx$

$ \displaystyle \int \sum_{n \equal{} 0}^{\infty}(\minus{}1)^n x^{2n}dx \equal{} \arctan(x)$

The second one led me to discover that $ 4\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n}{2n \plus{} 1} \equal{} \pi$ which was when I decided that I wanted to study math.

hi,

there are innumerable interesting and beautiful integrals such as the ones you are seeking. i have seen a whole variety of them on this forum posted by a lot of people before, including myself. yes, i am aware of a handful of integrals that i have seen smart students stumble on, but of-course, you will always find someone elsewhere (like there are a few people on this forum) that can solve those integrals using some clever and unorthodox techniques. [ includes the people you mentioned plus didilica, sos440, perfect radio, dr doe, lordykelvin, jmerry, and many more. ]

anyhow, here are 10 integrals that i am remembering now, and are presented for your entertainment, LoL ! (i have given here the answers to 5 of them, the rest you should try and see if you can do them on your own first! ) i also think i have also posted them on the forum before. i like integrals a lot myself, but not that much, as my interest primarily lies in the theory of functions of several complex variables.

by the way, these are fairly difficult single integrals to solve, mind you. [ i also know a handful of double and triple integrals that are hard to solve...  :| ]

$ \textbf{(i)}\;\;\;\quad\int_0^{\frac{\pi}{2}}\;\cos^{\minus{}1}\;\left(\dfrac{\cos\,x}{1\plus{}2\cos\,x}\right)\;dx\;\equal{}\;\boxed{\frac{5\,\pi^2}{24}}$

$ \textbf{(ii)}\;\;\;\quad\int_{\minus{}\infty}^{\infty}\;\dfrac{dx}{\pi^2\plus{}(1\minus{}x\plus{}e^x)^2}\;\equal{}\;\boxed{\frac{1}{2}}$

$ \textbf{(iii)}\;\;\;\quad\int_0^1\;\dfrac{\tan^{\minus{}1}\,\sqrt{\;2\plus{}x^2\;}}{\sqrt{\;2\plus{}x^2\;}\cdot(1\plus{}x^2)}\;dx\;\equal{}\;\boxed{\frac{5\,\pi^2}{96}}$

$ \textbf{(iv)}\;\;\;\quad\int_0^{\frac{\pi}{2}}\;\dfrac{x^2}{x^2\plus{}\ln^2\,(2\cos\,x)}\;dx\;\equal{}\;\boxed{\frac{\pi}{8}\left(\;\ln(2\pi\,e)\minus{}\gamma\;\right)}$

$ \textbf{(v)}\;\;\;\quad\int_0^{\infty}\;\dfrac{x^8\minus{}4x^6\plus{}9x^4\minus{}5x^2\plus{}1}{x^{12}\minus{}10x^{10}\plus{}37x^8\minus{}42x^6\plus{}26x^4\minus{}8x^2\plus{}1}\;dx\;\equal{}\;\boxed{\frac{\pi}{2}}$

$ \textbf{(vi)}\;\;\;\quad\int_0^\infty\;\dfrac{\tan^{\minus{}1}\,(3x)\cdot\tan^{\minus{}1}\,(5x)\cdot\tan^{\minus{}1}\,(7x)}{x^3}\;dx$

$ \textbf{(vii)}\;\;\;\quad\int_0^1\;\dfrac{(4x\plus{}1)\left(1\plus{}x\plus{}x^2\right)^{\frac{3}{2}}}{(4x^2\plus{}1)\cdot(3\plus{}2x\plus{}x^2)^2}\;dx$

$ \textbf{(viii)}\;\;\;\quad\int_0^{\infty}\;\dfrac{1}{\sqrt{\;(1\plus{}x)\cdot(3\plus{}x)\cdot(5\plus{}x)\cdot(7\plus{}x)\;}}\;dx$

$ \textbf{(ix)}\;\;\;\quad\int_1^4\;\dfrac{\tan^{\minus{}1}\,\left(\dfrac{\sqrt{\;(x\minus{}1)\cdot(16\plus{}x^2)\;}}{x\minus{}4}\right)}{\sqrt{\;16\plus{}x^2\;}}\;dx$

$ \textbf{(x)}\;\;\;\quad\int_0^1\;\dfrac{\sin^{\minus{}1}\,\left(\dfrac{1\plus{}x}{3\plus{}2x}\right)}{\sqrt{\;(2x\minus{}x^2)\cdot(3x^2\plus{}10x\plus{}8)\;}}\;dx$

My favorite integral is

$ \int_0^1x^x\,dx$

Its solution is based on $ e^x\equal{}\sum_{k\equal{}0}^\infty\frac{x^k}{k!}$ and integration by parts.

Las week, I composed some integral for my students, here is the file http://anhngq.files.wordpress.com/2007/09/gt2-01.pdf.

Thanks to your responses to this topic - very good integrals of all levels!
  misan's integrals seems to be very difficult at least for me - math is only my hobby . (I've finishted not mathematical high school and I've finnished IT in economical university.) They are very good indeed.
  bookworm_vn - your integrals are interesting they are for normal people I think most of them are a little under mine and of mine colleagues level when I was student (but not forget I took part to an competition for students - not listed in mathlinks and had second rank for my university - because I supported the winner and made stupid mistake). There are some interesting irrational integrals - well done job as a teacher.
  Hetidemek's integral is posted many time in mathlinks - it is interesting.
  hadjimurad - :-) he is on two ends - interesting but classical or hard and interesting - second one seems to be very good - like misan's integrals - if I don't spend time with additional training and thinking I cannot solve it.
  It will be good if more people send their integrals. As a whole I like simple and genious things ... like me ;-)

Indefinite integrals: http://anhngq.files.wordpress.com/2007/09/gt2-01.pdf
Proper integrals: http://anhngq.files.wordpress.com/2007/09/gt2-02.pdf
Mean value theorems: http://anhngq.files.wordpress.com/2007/09/gt2-03.pdf
Other topics: http://anhngq.files.wordpress.com/2007/09/gt2-04.pdf
Improper integrals: http://anhngq.files.wordpress.com/2007/09/gt2-05.pdf

[quote="borislav_mirchev"]Very good job, bookworm_vn!
Thank you for sharing your experience.
What is the speciality of your students?[/quote]

Just the problems of Calculus I for all students in my department (over 1,100 students), I mean these problems are at low level. For the best/outstanding students, I will give an other problem set.


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
geometry
geometric transformation

Hello everyone, here are the details of the newest (unofficial) AoPS contest:

[u]Contest Format:[/u]
Iversonfan2005 and myself are writing/sponsoring a [b]multiple choice[/b] math contest called the Masters Invitational Math Contest.  It is on a middle to high school level, but anyone can participate.  There will be awards on for a high-scorer on every grade level, and an award for the top three scorers overall.

[u]Difficulty:[/u]
The contest will be [b]50 questions[/b], multiple choice, ranging from National MathCounts to upper level AIME difficulty.  We would like every AoPSer to participate, but do not be discouraged by harder problems towards the end.  Although we encourage middle-schoolers to participate, middle schoolers are not expected to be able to anwer the higher numbered questions.

[u]Scoring:[/u]
Each correct answer will earn 5 points, a question left blank will earn 1 point, and an incorrect answer will earn 0 points.  Scores will range from 0 - 250.

[u]Time Limit:[/u]
Due to our inability to accurately track time, all participants will have [b]3 weeks [/b]from the time of contest posting to turn in their answers.

[u]Participation:[/u]
If you want to participate, just reply to this thread indicating so.  Include your grade and state!  The contest will be posted in a week or two.  This was posted in the SC forum because I moderate there, but it is for EVERYONE to participate in!

[b][u]IMPORTANT EDIT[/u][/b]
[u][b]The PDF will be posted on April 16th.  All answers will be due exactly 3 week later, on May 7th. [/b][/u] We added two weeks due to USAMO and AP exams.  Also, due to the added time, the exam has gone from 40 questions to 50 questions.  Please note these changes!

Rule Change 4/13/06 - I've heard a lot of complaining about younger middle school students pulling down state scores.  So, we've changed the rules to AMC/Coastal Carolina style:  A state/country is allowed unlimited participants.  However, the state/country score will consist of the top 3 scores from that state/country added together.

Rule Clarification 4/14/06 - All information required is your AoPS username and grade.  Name and state are optional, and tests will be graded without a real name or state.  However, you will not be placed on a regional team, and your certificate will have your AoPS username on it.  Sorry for the confusion!

Rule Document Changed 4/14/06 - Rules changed to reflect above notices.  Please re-view new rules document, entitled Version 2.0.

I am interested :)

EDIT: I saw this as a global announcement, but is it just for SCers?

I definitely want to participate.  My only question is how to turn in our answers.  Do we just pm one of you once we have them?

Yes, just PM [b]iversonfan2005 [/b]or [b]mysmartmouth[/b] to turn in answers.  More details will follow when the contest is released.

Ok, I will make a list of participants by grade.  Also, when you sign up to participate, include your grade so I can add you to the list.  This is going to be a nice, clean organzied contest, and results will be distributed in the crisp MathCounts way.

Updated List
[u]6th Grade & Below[/u]
Hypatia 2 - MD
Demetrius - MD
jea7 - MD

[u]7th Grade[/u]
ragnarok23 - CA
PenguinIntegral
tyy
kyyuanmathcount - MI
slamchess - IN
inyoung - VA
Fanatic - IN
easyas3.14159... - CA
goldendomer

[u]8th Grade[/u]
biffanddoc - DE
Zenith Xan - VA
ch1n353ch3s54a1l
ehehheehee
236factorial - NY
Klebian - TX
Captain Sugar - OK
robinhe - MI
Ignite168 - CT
baseballfreak2
k8reindeer - IL
pianoforte - KS
baseballfreak2
swirly3
fireemblem13 - TX
azndemon43 - CA
davidyko - CA
mathgeniuse^ln(x)
Xaero - OH
Pitter
tmaseda - MD
crystal_serina_granger
froglover - MD
angelgirl
pineappleperson

[u]9th Grade[/u]
nat mc - IN
jb05 - NC
chess64 - MD
Xantos C. Guin - GA
apeman5291 - SC
me@home
4everwise
matt276eagles - PA
origamiidiot - SC
Andrew619 - FL
surge - FL
chuckzeduck - CO
Mathlete9366
solafidefarms - GA
EFuzzy - MD
h_s_potter2002 - OH
GoBraves - NC
bob123 - CA
Meghrabi - Morocco
jrshoch - CT
E^(pi*i)=-1 - MD
chesspro - CA
LordoftheMorons - VA
keeyounglee - VA
Pineappleperson - MI
rs9999 - SC
Russian Ninja - GA
krsattack - NY
Pitter - MD
Boy Soprano II - GA

[u]10th Grade[/u]
DanK
Hokkage - CA
kimby_102 - CA
K81o7 - NY
Adunakhor - NY
dark_sunburst - SC
eryaman - TX
ExterAY -  FL
jhcreinhardt - AL
filletwho
SamE - CO
Pakman2012 - FL
Altheman - IL
Jeff219
The Zuton Force - MA
diana25 - Romania
Karth - India
10000th User
cincodemayo5590 - IL
lkryptonicl - MA
tiredepartment - China
nateharman1234 - IL
shadysaysurspammed - India
Zellex - CA
calculuslover800 - WA
nrz - TX
MathDancer - MD
aahmadi - KT

[u]11th Grade[/u]
deej21
elemmnop - AL
MysticTerminator - NC
arjunlandes - DC
scorpius119 - PA
gotztahbeazn - IL
ShoeFactory - CA
arkmastermind - SC
inteluser - FL
perfectnumber628 - NY
djshowdown2 - IL
Abhishek - India
pkerichang - CA
WarpedKlown1335 - FL
dangvh - FL
klocraft - MI
agolsme
playmor10is - India
xxreddevilzxx - WA
quantum leap - TN
gljirain - VA

[u]12th Grade[/u]
Hikaru79 - Canada
amirhtlusa - WA
daveed - MD
eXistenZ - Israel
Aislynn - MD
M4RI0 - Peru

[b]117? Participants So Far![/b]

PLEASE POST YOUR STATE/COUNTRY IF YOU WISH TO BE INCLUDED ON YOUR REGIONAL TEAM!

8th grade for me (I didnt feel like editing)

Hm, I might as well... I'm in 11th grade.

Please add a link to this contest in your signature, and help us spread the word!

[quote="Xantos C. Guin"]Everyone from GA is a 9th grader. Now there will be competition for #1 in GA.[/quote]Karth (10) is actually from GA.  He may be competing for India, though.

So who here is competing for Georgia or India?

[quote="Karth"]Er, I live in Georgia, but am from India... do we compete for our own country?  :D

EDIT: I'll just compete for Georgia, since I live there. But if many people are competing for India, I'll just compete for India  :P[/quote]

I believe you have to compete for the state/country you live in, or no state/country.

So unless you live in India and GA simutaneously, you are competing for the state you live in.

This is why I'm competing for GA not Antarctica.

GA:
Xantos C. Guin
solafidefarms
Boy Soprano II
Russian Ninja

hey I just wanted to add that I'm from CT, already said I'm in 9th grade, and my name is Jason Hoch (JRSHoch, JRSH are initials  ;) )

I will do it

From Virginia, 11 th grade

WHat time will it be posted? It's today, right?

I just got a PM from Iversonfan (im presuming that he sent it out to everyone) which said to check here for the posted contest... but i dont see it anywhere.

ok everyone its on a different thread on the SC forum. cool.

OK, due to bad proofreading, we had a couple of mistakes on the test.  So to keep things nice and clean, the test will be posted somewhere else (See link below).  Furthermore, I will lock this thread to keep all posting in one place.

[b]If you want to join now, that's fine, JUST PM YOUR INFO WITH YOUR ANSWERS!

PLEASE ENSURE YOU HAVE THE LATEST VERSION OF THE TEST.[/b]

CURRENT TEST VERSION AVAILABLE HERE:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=485044#p485044

All information is now at this location!
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86656


Evaluate $p_{67}(67) - p_{65}(67)$, where $p_k(n)$ is the number of partitions of $n$ with summands no larger than $k$.

[hide]This seems so easy that I think I must be making a mistake:

Wouldn't it be all partitions of 67 that use summands larger than 65?  If so the only two are 66,1 and 67, so the answer is 2.[/hide]


analytic geometry
search
floor function
combinatorics unsolved
combinatorics
Chip firing

There are $ 2009$ pebbles in some points $ (x,y)$ with both coordinates integer. A operation consists in choosing a point $ (a,b)$ with four or more pebbles, removing four pebbles from $ (a,b)$ and putting one pebble in each of the points
\[ (a,b\minus{}1),\ (a,b\plus{}1),\ (a\minus{}1,b),\ (a\plus{}1,b)\]

Show that after a finite number of operations each point will necessarily have at most three pebbles. Prove that the final configuration doesn't depend on the order of the operations.

This is a standard lemma in the theory of [url=http://scholar.google.com/scholar?q=chip-firing+game&hl=en&btnG=Search]chip-firing games[/url].

[quote="cyshine"]There are $ 2009$ pebbles in some points $ (x,y)$ with both coordinates integer. A operation consists in choosing a point $ (a,b)$ with four or more pebbles, removing four pebbles from $ (a,b)$ and putting one pebble in each of the points
\[ (a,b \minus{} 1),\ (a,b \plus{} 1),\ (a \minus{} 1,b),\ (a \plus{} 1,b)\]
Show that after a finite number of operations each point will necessarily have at most three pebbles. Prove that the final configuration doesn't depend on the order of the operations.[/quote]
Let $ n \equal{} 2009$.
Let the pebbles lie on coordinate points $ p_1\equal{}(x_1,y_1),p_2\equal{}(x_2,y_2),p_3\equal{}(x_3,y_3), \cdots, p_n\equal{}(x_n,y_n)$.
Now define $ f(p_i) \equal{} (x_i)^2 \plus{} (y_i)^2$, and $ A \equal{} f(p_1) \plus{} f(p_2) \plus{} \cdots \plus{} f(p_n)$. It is apparent that $ A$ increases with $ 4$ whenever a move is made.
Assume that the problems statement is false. Then we can always make another move, and after $ m$ moves $ A$ is at least $ 4m$. The maximal distance from $ (0,0)$ is at least $ \sqrt{\frac{A}{n}} \ge \sqrt{\frac{4m}{n}} \to \infty$ as $ m \to \infty$. So for any $ d$ we can find a number of moves $ m$ and a pebble $ p_k$ such that the distance from $ p_k$ to $ (0,0)$ is at least $ d$. Let the original maximal distance (before any moves were made) be $ d_0$. Then the pebble must have been on at least $ d\minus{}d_0$ different points. It is easy to see that if there has been a pebble on to adjecent points then there will always be at least one pebble on one of them. So there must be at least $ \left \lfloor \frac{d\minus{}d_0}{2} \right \rfloor$ points. But $ \left \lfloor \frac{d\minus{}d_0}{2} \right \rfloor \to \infty$ as $ d \to \infty$, and there has to be infinitly many points. Contradiction.
So after a finite number of moves there are at most three pebbels on each point.

After the $n-$th operation, let $S_n$ be the set of points $(x_i,y_i)$ such that $(x_i,y_i)$ contains at least one pebble. Let $E_n= \sum_{(x_i,y_i) \in S_n} p_i(x_i^2+y_i^2)$. Observe that $E_{n+1}=E_n-4(x_t^2+y_t^2)+(x_t-1)^2+y_t^2+(x_t+1)^2+y_t^2+x_t^2+(y_t-1)^2+x_t^2+(y_t+1)^2= E_n+4$ and notice that the sum of the coordinates at every step is invariant, so if $2009[\sum_{(x_i,y_i) \in S_n} x_i+y_i]^2$ is bounded, which implies that $E_n$ is bounded, otherwise since $E_n \leq 2009[\sum_{(x_i,y_i) \in S_n} x_i+y_i]^2$ would be unbounded, a contradiction. Therefore, the operations cannot occur indefinetely.

Moreover, since the process ends, by Chip Firing Theory we know that the final position is unique, so we are done.

$\blacksquare$


[size=150]Use intermediate calculus level to show:[/size]
[img]http://img15.imageshack.us/img15/350/28664013.jpg[/img]

It's not really calculus that you need here. Two lines lie in the same plane if and only if they intersect. Your two lines intersect if and only if there exists and $ \tau_1$ and $ \tau_2$ such that

$ x_1 \plus{} a \tau_1 \equal{} x_2 \plus{} a_2 \tau_2$

$ y_1 \plus{} b_1 \tau_1 \equal{} y_2 \plus{} b_2 \tau_2$

$ z_2 \plus{} c_1 \tau_1 \equal{} z_2 \plus{} c_2 \tau_2$

or

$ a_1 \tau_1 \minus{} a_2 \tau_2 \equal{} x_2 \minus{} x_1$

$ b_1 \tau_1 \minus{} b_2 \tau_2 \equal{} y_2 \minus{} y_1$

$ c_1 \tau_1 \minus{} c_2 \tau_2 \equal{} z_2 \minus{} z_1$

So, it becomes a question of determining whether the vectors $ (a_1,a_2,a_3)$, $ (a_2,b_2,c_2)$ and $ (x_2 \minus{} x_1,y_2 \minus{} y_1,z_2 \minus{} z_1)$ are linearly independent, which is exactly what the determinant tells you.

By the way, I would recommend not starting the same thread in two different forums on AOPS.


trigonometry
geometry
cyclic quadrilateral
trig identities
Law of Cosines
algebra
difference of squares

Prove Brahmagupta's formula, which states that for a quadrilateral $ WXYZ$,

$ [WXYZ]^{2}\equal{} (s\minus{}WX)(s\minus{}XY)(s\minus{}YZ)(s\minus{}WZ)$
$ \text{ }\minus{}WX\cdot XY\cdot YZ\cdot WZ\cdot\cos^{2}{\frac{X\plus{}Z}{2}}$

where $ s \equal{}\frac{WX\plus{}XY\plus{}YZ\plus{}WZ}{2}$.

[hide="Proof"]
This should really be in Pre-Olympiad?, but anyway.
I'd rather just simplify Brahmagupta's formula, so it states that the area of a cyclic quadrilateral can be expressed as $ \sqrt{(S-p)(S-q)(S-r)(S-s)}$, where $ S$ is the semiperimeter.

I'll just draw a diagram so it's easier to see what's going on.

[img]10174[/img]

Now, just divide the thing into two triangles, $ BDC$ and $ ADB$. Then, you just find the area of the cyclic quad is just $ \dfrac{rs\sin C}{2}+\frac{pq\sin A}{2}$. Also, $ \sin A=\sin C$, because $ ABCD$ is a cyclic quadrilateral. When expressed, it becomes:

\begin{align*}[ABCD] &=\dfrac{rs\sin C}{2}+\frac{pq\sin A}{2}\\ [ABCD]^{2}&=\frac{\sin^{2}A(pq+rs)^{2}}{4}\end{align*}

Then, we multiply it out to simplify it.

\begin{align*}4[ABCD]^{2}&= (1-\cos^{2}A)(pq+rs)^{2}\\ &= (pq+rs)^{2}-\cos^{2}A(pq+rs)^{2}\end{align*}

from before. Then just apply law of cosines and substitute to get $ 2\cos A(pq+rs)=p^{2}+q^{2}-r^{2}-s^{2}$.

Use difference of squares now to get $ 2\cos A(pq+rs)=(p+r)+q^{2}-s^{2}$.

Now, a lot of messy simplification results in $ (p+q+r-s)(p+q+s-r)(p+r+s-q)(q+r+s-p)$.

This simplifies into (with more messy simplification)

\[ 16[ABCD]^{2}= 16\left(\frac{p+q+r+s}{2}-p\right)\left(\frac{p+q+r+s}{2}-q\right)\left(\frac{p+q+r+s}{2}-r\right)\left(\frac{p+q+r+s}{2}-s\right)\]

Divide by 16 and take the square root to get

Note that we have $ \frac{p+q+r+s}{2}$, which is our semiperimeter. Therefore, we actually have:

\[ 16[ABCD]^{2}= 16(S-p)(S-q)(S-r)(S-s)\]

Divide by sixteen and take the square root to get $ [ABCD] =\sqrt{(S-p)(S-q)(S-r)(S-s)\}}$, and we are done. $ \Box$
[/hide]


Two circles have a common external tangent of length 8 and a common internal tangent of length 6. Find the product of the circles' areas.

[hide]Let the radii be $x$ and $y$, with $y>x$.  We can tell that for the internal tangent, we can complete a triangle to make a right triangle with legs $(x+y)$ and $6$, and hypotenuse is the length between the circles' centers.  With the external tangent we do something similar to get a right triangle with legs $(x-y)$ and $8$, hypotenuse again is the distance between the centers. So: $6^2+(x+y)^2=8^2+(x-y)^2 \Rightarrow 4xy=28 \Rightarrow xy=7$.  We are looking for $x^2\pi \cdot y^2 \pi=(xy)^2 \pi^2=49 \pi^2$.[/hide]


probability geometry trigonometry rotation AMC

A region $ S$ in the complex plane is defined by \[ S \equal{} \{x \plus{} iy: \minus{} 1\le x\le1, \minus{} 1\le y\le1\}.\] A complex number $ z \equal{} x \plus{} iy$ is chosen uniformly at random from $ S$. What is the probability that $ \left(\frac34 \plus{} \frac34i\right)z$ is also in $ S$?

$ \textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78$

You made a typo :] It's $ \minus{} 1\le y \le 1$. (I think).

Um....

I think you should have had

$ 3/2\le x\minus{}y \le 3/2$ and $ 3/2\le x\plus{}y \le 3/2$. I'm pretty sure you can't get $ 4/3i$ from $ S$.
I got $ E$ after doing the same thing you did afterwards, though.

[quote=stuck][hide="Solution"]
In DeMoivre's form, we have that
$ (\frac {3}{4} \plus{} \frac {3}{4}i) \equal{} \frac {3 \sqrt {2}}{4} (\cos{\frac {\pi}{4}} \plus{} isin{\frac {\pi}{4}})$

This means that for a point $ z \in S$, by multiplying we rotate that point $ \frac {\pi}{4}$ degrees and multiply it by $ \frac {3 \sqrt {2}}{4}$. So the permissible points are the original square multiplied by the reciprocal ($ \frac {2 \sqrt {2}}{3}$) rotated by $ \frac {\pi}{4}$. The area of this square inside the larger square is $ \frac {28}{9}$, so
$ \boxed{D}$
[/hide][/quote]

wait this is actually very pro but you can also easily solve this problem through [b]geometric probability[/b].
[hide=Solution]
Our total region has area $4$. Expanding $3/4(1+i)(x+iy)$, we get $-4/3 \leq x-y \leq 4/3$ and $-4/3 \leq x+y \leq 3$. The desired region has area $28/9$ so the answer is $\boxed{\textbf{(D)}\ \frac79}$

[quote=stuck][hide="Solution"]
In DeMoivre's form, we have that
$ (\frac {3}{4} \plus{} \frac {3}{4}i) \equal{} \frac {3 \sqrt {2}}{4} (\cos{\frac {\pi}{4}} \plus{} isin{\frac {\pi}{4}})$

This means that for a point $ z \in S$, by multiplying we rotate that point $ \frac {\pi}{4}$ degrees and multiply it by $ \frac {3 \sqrt {2}}{4}$. So the permissible points are the original square multiplied by the reciprocal ($ \frac {2 \sqrt {2}}{3}$) rotated by $ \frac {\pi}{4}$. The area of this square inside the larger square is $ \frac {28}{9}$, so
$ \boxed{D}$
[/hide][/quote]

Why did you have to multiply by the reciprocal?
And I thought the area inside the square was an octagon?

[quote=aznlord1337]Multiply the number to get the following conditions:

$ 4/3 \leq x\minus{}y \leq 4/3$
$ 4/3 \leq x\plus{}y \leq 4/3$.

Consider the square of area four with vertices $ (1,1), (1,\minus{}1), (\minus{}1,\minus{}1), (\minus{}1,1)$. The four lines cut off four triangles of equal area from each corner of the square, and a quick computation reveals that the area of each triangle is $ 2/9$. So the valid area bounded by the lines is $ 4\minus{}4(2/9) \equal{} 28/9$. $ 28/36 \equal{} 7/9$, $ D$.[/quote]

[quote=RunpengFAILS]aznnlord's solution is correct. (its 3/4 not 3/2)[/quote]

Is it 3/4 or 4/3?


Let$S=\{1,2,..,n\}$.A partition of S into three set$A,B,C$with$|A|,|B|,|C|>\frac{n}{4}$Prove that there exist 3 elements$a\in A,b\in B,c\in C$such that there is one of them equal the sum of another elements

Have a look http://www.mathlinks.ro/Forum/viewtopic.php?highlight=contest&t=3781 you can see a long solution for above problem

i think this problem is very nice but it is too hard,too. :D
anyway does it have a nice solution or not?


trigonometry
function
inequalities
calculus
calculus computations

The problem should be moved to Intermidate.

Anyway, Here is a hint:

What's $ \frac {8}{2}?$

$ \frac{d}{dx} sin^8x\plus{}cos^8x\equal{}8sin^7xcosx\plus{}8cos^7x(\minus{}sinx)$
$ \equal{}8sinxcosx[sin^6x\minus{}cos^6x]$
Set $ \frac{d}{dx} \equal{} 0$.
This leaves us with a couple of cases:

1)$ sinx\equal{}0$
$ x\equal{}n\pi \forall n \in Z$
2)$ cosx\equal{}0$
$ x\equal{}n\pi\plus{}\frac{\pi}{2} \forall n \in Z$
3)$ sin^6x\minus{}cos^6x\equal{}0$
$ (sin^3x\minus{}cos^3x)(sin^3x\plus{}cos^3x)\equal{}0$
$ (sinx\minus{}cosx)(sin^2x\plus{}sinxcosx\plus{}cos^2x)(sinx\plus{}cosx)(sin^2x\minus{}sinxcosx\plus{}cos^2x)\equal{}0$
From the first and third brackets:
$ sinx\equal{}\pm cosx$
$ x\equal{}\frac{(2n\plus{}1)\pi}{4} \forall n \in Z$
From the second and fourth brackets:
$ (1\plus{}sinxcosx)(1\minus{}sinxcosx)\equal{}0$
$ (1\minus{}sin^2xcos^2x)\equal{}0$
$ 1\equal{}sin^2xcos^2x$
Does not exist, since $ sin^x$ and $ cos^x$ are both in the interval $ [\minus{}1, 1]$, but are never both $ 1$ or $ \minus{}1$.

After checking all the roots, it turns out that the maximum is $ 1$, as expected.

$ \sin ^{8}x\plus{}\cos ^{8}x\equal{}\frac{1}{32}\left[ \left( \cos 4x\plus{}7\right) ^{2}\minus{}32\right]$

$ \frac{1}{32}\left[ \left( \minus{}1\plus{}7\right) ^{2}\minus{}32\right] \leq \sin ^{8}x\plus{}\cos ^{8}x\leq \frac{1}{32}\left[ \left( 1\plus{}7\right) ^{2}\minus{}32\right]$

$ \frac{1}{8}\leq \sin ^{8}x\plus{}\cos ^{8}x\leq 1$

:o   kunny
[quote="kunny"]...

Anyway, Here is a hint:

What's $ \frac {8}{2}?$[/quote]

more generally consider $ f(x) \equal{} \sin^{2n}x \plus{} \cos^{2n}x$
it's well defined function on $ [0,\frac {\pi}{2}]$ and is periodic with period $ T \equal{} \frac {\pi}{2}$
thus it's enough to find maximum value on the interval $ [0,\pi/2]$
note that the derivetive of $ f$ has roots $ root's \equal{} 0, \pi/4, \pi/2$
it's enough to find value of the function  $ f$ at the point's $ 0,\pi/4, \pi/2$
thus maximum value is equal to $ 1$.  and the minimum $ 2\left( \frac {1}{\sqrt {2}} \right)^{2n}$
in the case $ n \equal{} 1$ we have $ max \equal{} min \equal{} 1$ or $ \sin ^{2}x \plus{} \cos^{2}x \equal{} 1$ :lol:

[quote="vijaymenon"]what is the greatest value of [b](sin x)^8 + (cos x)^8[/b][/quote]
$ (\sin^2 x)^4 \plus{} (\cos^2 x)^4 \le (\sin^2 x \plus{} \cos^2 y)^4 \plus{} 0^4 \equal{} 1$ by Karamata, because $ x \to x^4$ is convex.

of course. it's also true without karamata . fior example :
$ 1 \equal{} (\sin^{2}x \plus{} \cos^{2}x)^{n} \geq \sin^{2n}x \plus{} \cos^{2n}x$
 and equality hold's for example for $ x \equal{} 0$
last inequality is trivial. For example expand  $ (\sin^{2}x \plus{} \cos^{2}x)^{n} \equal{} binom \equal{} \sin^{2n}x \plus{} ("...." > 0) \plus{} \cos^{2n}x \geq \sin^{2n}x \plus{} \cos^{2n}$

[quote="IndoChina"]$ \sin ^{8}x \plus{} \cos ^{8}x \equal{} \frac {1}{32}\left[ \left( \cos 4x \plus{} 7\right) ^{2} \minus{} 32\right]$

$ \frac {1}{32}\left[ \left( \minus{} 1 \plus{} 7\right) ^{2} \minus{} 32\right] \leq \sin ^{8}x \plus{} \cos ^{8}x\leq \frac {1}{32}\left[ \left( 1 \plus{} 7\right) ^{2} \minus{} 32\right]$

$ \frac {1}{8}\leq \sin ^{8}x \plus{} \cos ^{8}x\leq 1$[/quote]

That's correct.

Here is my solution:

$ \sin ^ 8 x\plus{}\cos ^ 8 x\equal{}(\sin ^ 4 x\plus{}\cos ^ 4 x)^2\minus{}2\sin ^ 4 x\cos ^ 4 x$

$ \equal{}\{(\sin ^ 2x\plus{}\cos ^ 2x)^2\minus{}2\sin ^ 2 x\cos ^ 2 x\}^2\minus{}2\sin ^ 4 x\cos ^ 4 x\ \Longleftarrow t\equal{}\sin ^ 2 x\cos ^ 2x$

$ \equal{}(1\minus{}2t)^2\minus{}2t^2\equal{}2t^2\minus{}4t\plus{}1\equal{}2(t\minus{}1)^2\minus{}1\equiv f(t)$, since $ t\equal{}\frac{1}{4}\sin ^ 2 x\Longrightarrow 0\leq t\leq \frac{1}{4}$,

$ f(0)\leq f(t)\leq f\left(\frac{1}{4}\right)$, yielding $ \frac {1}{8}\leq \sin ^{8}x \plus{} \cos ^{8}x\leq 1$.

I see no need to invoke anything specific about sine and cosine other than the Pythagorean identity. Let $ u=\cos x$ and $ v=\sin x.$

Maximize $ u^8+v^8$ subject to the constraint $ u^2+v^2=1.$

Better, let $ s=u^2$ and $ t=v^2.$ Maximize $ s^4+t^4$ for nonnegative $ s,t$ subject to $ s+t=1.$

How about Lagrange multipliers?
\[ \begin{array*}4s^3&=\lambda\\4t^3&=\lambda\\s+t&=1\end{array*}\]
The only critical point happens when $ s=t=\frac12,$ at which point $ s^4+t^4=\frac1{16}+\frac1{16}=\frac18.$

But ... what about the endpoints? $ s=0,t=1$ and $ s=1,t=0$ both result in a value of $ s^4+t^4=1,$ so that's the maximum and $ \frac18$ is the minimum.

We could also get that minimum by Hölder's inequality, noting that the dual exponent of $ 4$ is $ \frac43.$

$ s+t\le (1^{4/3}+1^{4/3})^{3/4}(s^4+t^4)^{1/4}$

$ (s^4+t^4)^{1/4}\le(s+t)\cdot 2^{-3/4}$

$ s^4+t^4\le \frac18(s+t)^4$

I wanted you to remark the half of $ 8$, or, $ \sin ^8 x\equal{}(\sin ^ 4 x)^2.$


$ a,b,c>0$, and $ a\plus{}b\plus{}c\equal{}1$
Prove that   $ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\le\sqrt{3}$

[quote="Bayasgalan"]$ a,b,c > 0$, and $ a \plus{} b \plus{} c \equal{} 1$
Prove that   $ \sqrt {a \plus{} \frac {(b \minus{} c)^2}{4}} \plus{} \sqrt {b} \plus{} \sqrt {c}\le\sqrt {3}$[/quote]

http://www.mathlinks.ro/viewtopic.php?p=995272#p995272


probability
geometry
combinatorics unsolved
combinatorics

$2$ points are selected with equal probability of a line segment with length $1.$ What is the probability that you can make up a triangle out of the $3$ resulting line segments?

I think we can change the problem to every segment has to be less than $\frac{1}{2}$, but I have no idea how to calculate the probability.
Could you show me please? :lol:

[hide="Hint"]Draw a graph with the position of the first pt as x axis and the other as y axis, then find out the feasible area.[/hide]
[hide="Answer"]$\frac{1}{4}$[/hide]

http://www.mathlinks.ro/Forum/viewtopic.php?t=84708

If you're interested, I've now killed the generalization I mentioned there.
The result is (unless I'm wrong) : $1-\frac n{2^{n-1}}$.

[quote="modeler"][quote="OHO"]Why the probability is the area?

So there is no difference if one of the segments can be $\frac{1}{2}$ long?[/quote]

There may be a segment with length $\frac{1}{2}$, but the probability for this to happen is 0, so we need not consider this case.[/quote]No. I mean if we change the problem
from
find a probability that you make up line segments with length $<\frac{1}{2}$
to
find a probability that you make up line segments with length $\leq \frac{1}{2}$,
will the same answer still be the same?

Yes.

Because we are currently considering a continuous range of values for the length, one point has exactly 0 area under it in a probability density curve.


Find, in terms of $ n$ ($ n \in \mathbb{R}$), the ratio of the coefficient of $ x^3$ to that of $ x^2$ in the expansion of $ (3\plus{}2x)^n$ in ascending powers of $ x$, where $ |x|<1.5$. [hide="(Method)"] Consider $ 3^n(1\plus{}2/3x)^n\equal{}3^n[1\plus{}2/3nx\plus{}\frac{n(n\minus{}1)}{2}(2/3x)^2\plus{}\cdots]$; the ratio is easily found to be $ \frac{2n\minus{}4}{9}$[/hide] Given that $ n\equal{}2.5$ and $ x>0$, find the least value of $ r$ such that the coefficient of $ r$ is negative.

Sorry, the question is asking for the 'least value of $ r$ such that the coefficient of $ x^r$ is negative'.
Let me attempt the question again. [hide="Solution"]
$ 3^{2.5}\left[\frac {2.5(1.5)\cdots(3.5 \minus{} r)}{r!}\right](2/3x)^r \equal{} 3^{2.5}\left[\frac {(1.875)( \minus{} 1)^{r \minus{} 3}(0.5)\cdots(r \minus{} 3.5)}{r!}(2/3)^r \right]$. Observe that for $ r \equal{} 4$, we will have a $ \minus{} 1$ on the numerator.
[/hide]

But by doing this, am I assuming in any way that the least value of $ r$ is greater than 3?


What's the easiest way to prove that

$ \tiny\frac{1}{sinA}\ \plus{} \ \frac{1}{sinB} \ \plus{} \ \frac{1}{sinC} \ \geq \ 2\sqrt{3}$

[quote="whomer"]What's the easiest way to prove that

$ \tiny\frac {1}{sinA}\ \plus{} \ \frac {1}{sinB} \ \plus{} \ \frac {1}{sinC} \ \geq \ 2\sqrt {3}$[/quote]

$ \frac{1}{sinA}\plus{}\frac{1}{sinB}\plus{}\frac{1}{sinC}\geq 3\sqrt[3]{\frac{1}{sinAsinBsinC}}\equal{}\frac{3}{\sqrt[3]{sinAsinBsinC}}\geq\frac{3}{\frac{sinA\plus{}sinB\plus{}sinC}{3}}\equal{}\frac{9}{sinA\plus{}sinB\plus{}sinC}\geq\frac{9}{3sin\frac{A\plus{}B\plus{}C}{3}}\equal{}\frac{9}{3sin60^{0}}\equal{}\frac{9}{3*\frac{\sqrt{3}}{2}}\equal{}2\sqrt{3}$

$ 0<A,B,C<180^{0}$-----> $ \frac{1}{3}sinA\plus{}\frac{1}{3}sinB\plus{}\frac{1}{3}sinC\leq sin{\frac{A\plus{}B\plus{}C}{3}}$

Ok, that was the proof I knew, thanks anyway zaya

[quote="whomer"]What's the [b]easiest way[/b] to prove that

$ \sum \csc A \geq \ 2\sqrt {3}$[/quote]I don't know what if it is easiest to you :wink:
But, sometimes Jensen's Inequality can make your life easy when you are trying to solve a trigonometric inequality.
Let, $ f(x) \equal{} \csc x$
Now,check that, $ f''(x) \equal{} \csc^3 x \plus{} \cot^2 x\csc x$.
So, $ f''(x)\geq 0$ when, $ x\in [0,\pi]$
Now,
\[ \frac {1}{3}\sum \csc A \geq \csc \left ( \frac {\sum A}{3} \right ) \equal{} \csc \frac {2 \pi}{3} \equal{} \frac {2}{\sqrt {3}} \iff \sum \csc A \geq \ 2\sqrt {3}
\]


Maybe easy, but hard for me.

Find all natural solutions of
\[ n^{3}\minus{}10^{4}m \equal{} 2006\]

$ n^{3}\equal{}2006\plus{}10^{4}m\equiv6\mod16$. There are no solutions.

[quote="Lazarus"]
$ n^{3}\equiv 11\mod25$, which is impossible.[/quote]

$ n\equiv 21\mod{25}$?

In fact, there is a unique value of $ n\in\\mathbb{Z}_{10^{2006}}$ such taht $ n^{3}\equiv\frac{10^{2006}\minus{}1}{9}$.  If $ n\in\mathbb{Z}$ satisfies this congruence, then we set $ m\equal{}\frac{n^{3}\minus{}\frac{10^{2\minus{}\minus{}6}\minus{}1}{9}}{10^{2006}}$.  Therefore, there are infinitely many solutions.


algebra
polynomial
search
function
Galois Theory
superior algebra
superior algebra unsolved

Let $ a\in \mathbb{F}_p$ with $ a\neq 0$, and let $ K$ be the splitting field of the polynomial $ x^p\minus{}x\minus{}a$ over $ \mathbb{F}_p$. Prove that $ [K: \mathbb{F}_p]\equal{}p$ and that the Galois group $ G(K/\mathbb{F}_p)$ is cyclic.

I'm usually not one to make angry "OMG USE THE SEARCH FUNCTION" posts, but it's clear that you didn't even try here.  This exact question was posted 2 weeks ago, [i]with the same title[/i]: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265094[/url]

Come to think of it, even the text is the same.  What's the story? EDIT: and why do you guys seem not to know that groups of prime order are cyclic?


ratio
arithmetic sequence
geometric sequence

if it is given that the numbers a,b, and c form an arithmetic sequence, do you know that a<b<c? Or is it that c can be the smallest of the three? And does b have to be the middle number?

if it is given that a,b, and c form an increasing arithmetic sequence--- same questions as above

thanks a lot

Yes, b must be in the middle (given the sequence is a,b,c).

Remember that an arithmetic sequence can go negative, so 3,2,1 and 1,2,3 are both arithmetic sequences, so a/c are neither greatest/least unless specified.

However, if it is given that $ a,b,c$ form an [b]increasing[/b] arithmetic sequence, then the common difference is positive, so $ a<b<c$.

[quote="10000th User"][quote="Utter Loser"]I know a common ratio of 1 is not considered a geometric sequence, but is a common difference of 0 also considered an arithmetic sequence?[/quote]Unless I'm mistaken but I don't see why a sequence like $ a,a,a$ with common ratio of 1 is not considered geometric. But, a $ 0,0,0$ sequence is pretty much ambiguous because there is no way to tell what the common ratio is.[/quote]

I meant for a sequence like 4,4,4. Is it considered arithmetic??

[quote="Utter Loser"]I meant for a sequence like 4,4,4. Is it considered arithmetic??[/quote]Sure, why not? It satisfies 4-4=4-4, which is usually the $ b\minus{}a\equal{}c\minus{}b$ criteria for arithmetic.

[quote="Utter Loser"][quote="10000th User"][quote="Utter Loser"]I meant for a sequence like 4,4,4. Is it considered arithmetic??[/quote]Sure, why not? It satisfies 4-4=4-4, which is usually the $ b \minus{} a \equal{} c \minus{} b$ criteria for arithmetic.[/quote]

Then the only reason 4,4,4 isn't considered geometric is because the sum of geometric sequence's formula won't work then, or is it just because 4,4,4 is arithmetic, and a sequence can't be arithmetic and geometric???[/quote]Please don't mix up the questions asking several things in one sentence.

1) The sequence is arithmetic because $ 4\minus{}4\equal{}4\minus{}4$ and hence it is arithmetic.

2) The sequence is geometric because $ \frac{4}{4}\equal{}\frac{4}{4}$ and hence it is geometric.

3) The formula for the sum of the terms of this geometric sequence does not work because the formula assumes that $ |r|<1$.

4) Just because the formula for the sum fails to work that does not imply that the sequence is not geometric.

5) A sequence can be both arithmetic and geometric at the same time. Likewise, a sequence can neither be arithmetic nor geometric as well.


Hello!

I am wondering if any kind person would post the solutions for the AMC 10 A for 2009

for #9, 10, 16

thanks so much for your help
 :lol:

Check the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2009]Resources section[/url]. Click on the number of a problem to go to that problem's thread.

oh wow i only saw the "S" before

thanks a lot!!

See also the official answers
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/09-2-1012-Answers.shtml
at our website.

-- Steve Dunbar
MAA Director, American Mathematics Competitions


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[color=blue][size=200][b]Welcome everybody to the next IMO 2009 in Germany!!![/b][/size][/color]

The International Mathematical Olympiad celebrates its 50th anniversary, i.e. half a century. Except for [url=http://www.mathlinks.ro/viewtopic.php?t=15614]IMO 1981[/url] which was restricted to various local IMO-style olympiads it took place every year.

[url=http://www.mathlinks.ro/viewtopic.php?t=15623]Here is an introduction[/url] if you ever wondered what IMO, ISL and ILL actually means. More information on the history and selection process of the [url=http://www.mathlinks.ro/viewtopic.php?t=21653]German IMO Team can be found here.[/url]

And finally [url=http://www.imo2009.de/]the web page of the IMO 2009 in Bremen.[/url] To brush up your geography knowlege [url=http://maps.google.de/maps?f=q&hl=de&geocode=&q=bremen&ie=UTF8&ll=53.074228,8.811035&spn=14.127293,48.164063&t=h&z=5&om=1]a link to Bremen on google maps.[/url]

The host of the next [url=http://www.imo2009.de/index.html]50th International Mathematical Olympiad 2009[/url] will be in [b]Germany.[/b] It will be hosted at the rising [url=http://math.jacobs-university.de/]Jacobs University Bremen.[/url] It will also be the 20th anniversary after the last IMO in Braunschweig, Germany in 1989.

Now we need to prove to the world that we cannot only organize soccer world cups!  :D

[url=http://www.math.uni-rostock.de/~gronau/]Dieter Gronau,[/url] Germany's long-serving (deputy) team leader is among the winners of the [url=http://www.amt.edu.au/erdgronau.html]Paul Erdös Award for 2008.[/url] All prize winners including Bruce Henry, Australia and Leou Shian, Taiwan are shown [url=http://www.amt.canberra.edu.au/wfnmcann08.html]here.[/url] In the past the [url=http://www.amt.edu.au/wfnmc.html]World Federation Of National Mathematics Competions (WFNMC)[/url] awarded a prize to [url=http://www.amt.canberra.edu.au/wfnmcaw.html]Wolfgang Engel and Arthur Engel.[/url]

Now the organisers are starting to add some information to the page for the [url=http://www.imo2009.de/imo/index.php?option=com_content&view=frontpage&Itemid=1&lang=en]International Mathematical Olympiad (IMO) 2009, Bremen.[/url]

According the [url=http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=12&Itemid=20&lang=en]schedule[/url] the jury members are due to arrive this Friday, July 10th and contestants next Monday, July 13th. On Sunday, July 19th the last jury meeting will be held and the 50th IMO Anniversary celebration will take place. Apparently [url=http://www.math.ucla.edu/~tao/tags.html]at this time[/url], July 17-20, fields and IMO gold medal winner [b]Terence Tao[/b] will be in Bremen as well. On Wednesday, July 22nd, will be the day of departure.

Total costs of the IMO are estimated at 1.6 million euros and are mostly covered by [url=http://www.bmbf.de/en/]Federal Ministry of Education and Research.[/url] Other [url=http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=6&Itemid=6&lang=en]sponsors[/url] include the state of Bremen and private sponsors as ArcelorMittal GmbH Bremen and BTC AG (Business, Technology, Consulting).

Roughly 1,100 people will be involved in the IMO. [url=http://www.imo-official.org/year_reg.aspx]Here[/url] you can see the total numbers by function and participating country:

[b]Total number of countries sending delegates:[/b] 108
[b]Total number of contestants:[/b] 575
[b]Total number of team leaders:[/b] 105
[b]Total number of deputy leaders:[/b] 102
[b]Total number of observers category A:[/b] 57
[b]Total number of observers category B:[/b] 39
[b]Total number of observers category C:[/b] 16

Some more information about the [url=http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=12&Itemid=20&lang=en]programme:[/url]

[b]Friday, July 10th: [/b]

- arrival of jury members

[b]Saturday, July 11th and Sunday, July 12th:[/b]

- jury meetings

[b]Monday, July 13th:[/b]

- jury members conduct jury meeting
- deputy team leader and contestants arrive

[b]Tuesday, 14th July, 1am to 1.30pm:[/b]

- Opening ceremony with a video message by Federal Chancellor Angela Merkel and other greetings
- presentation by mathematician [url=http://en.wikipedia.org/wiki/Albrecht_Beutelspacher]Albrecht Beutelspacher[/url] on „Calculating without a calculator“,
- parade of the participating IMO teams

[b]Wednesday, July 15th and Thursday, July 16th:[/b]

- [url=http://www.mathlinks.ro/resources.php?c=1&cid=16]actual IMO exams[/url]

[b]Friday, July 17th and Saturday, July 18th:[/b]

- excursions in the area of Bremen for contestants, delegations members are involved in coordination and marking of exam papers

[b]Sunday, July 19th:[/b]

- 50th IMO Anniversary celebration including short talks by the following notable mathematicians:

-- [url=http://en.wikipedia.org/wiki/B%C3%A9la_Bollob%C3%A1s]Béla Bollobás[/url]
-- [url=http://en.wikipedia.org/wiki/Timothy_Gowers]Timothy Gowers[/url]
-- [url=http://en.wikipedia.org/wiki/L%C3%A1szl%C3%B3_Lov%C3%A1sz]László Lovász[/url]
-- [url=http://en.wikipedia.org/wiki/Stanislav_Smirnov]Stanislav Smirnov[/url]
-- [url=http://en.wikipedia.org/wiki/Terence_Tao]Terence Tao[/url]
-- [url=http://en.wikipedia.org/wiki/Jean-Christophe_Yoccoz]Jean-Christophe Yoccoz[/url]

- final round of talks and meeting/chatting opportunities among contestants and top mathematicians

[b]Monday, July 20th: [/b]

- excursions for contestants and delegation members

[b]Tuesday, July 21st:[/b]

- closing ceremony with greetings by Federal Research Minister Annette Schavan and Bremen's Science Senator Jürgens-Pieper
- medals/prizes are awarded
- IMO flag is passed to the next host country Kazakhstan
- pictures and interviews at the research expedition of Germany
- farewell dinner

[b]Wednesday, July 22nd:[/b]

- departure of jury members, deputy leaders, observers and contestants

[i]German Team:[/i]

And finally a picture of the [url=http://imo-official.org/year_reg_team.aspx?code=GER]German IMO Team[/url] can be found [url=http://www.stifterverband.org/presse/pressemitteilungen/2009_07_02_matheolympiade/mathematikolympiade_2009_deutsches_team.jpg]here.[/url] From left to right the members are:

    *  Christoph Kröner (19) vom Gymnasium Stein, Stein
    * Malte Lackmann (18) von der Klaus-Groth-Schule, Neumünster
    * Martin Merker (18) vom Carl-Zeiss-Gymnasium, Jena
    * Jens Reinhold (17) vom Helmholtz-Gymnasium, Bielefeld
    * Lisa Sauermann (16) vom Martin-Andersen-Nexö-Gymnasium, Dresden
    * Bertram Arnold (17) vom Georg-Cantor-Gymnasium, Halle (Saale)

On the very right you can find German IMO leader Dr. Eric Müller and the team mascot "Mathematigerin". The picture was taken in front of the [url=http://www.mfo.de/]Mathematical Institue in Oberwolfach.[/url]

[i]Contact details of organizers:[/i]

[b]Dr. Harald Wagner:[/b]

CEO, Bildung und Begabung e.V. (Research and Gift), Science Centre
Ahrstraße 45
D-53175 Bonn
Tel. 0228/95915-10
During the IMO: Tel. 0171/87 64 884, Email: hw@imo2009.de

[b]Hanns-Heinrich Langmann:[/b]

International contacts, Bildung und Begabung e.V. (Research and Gift), Science Centre
Ahrstraße 45
D-53175 Bonn
Tel. 0228/95915-20
Email: hl@imo2009.de

[b]Thomas Vogt:[/b]

Press contacts, Deutsche Mathematiker-Vereinigung (German Mathematician Association), media office mathematics, Technische Universität Berlin (Technical University Berlin)
Straße des 17. Juni 136
10623 Berlin
Tel.: 030/314-78 788
Fax: 030/314-78 787
Mobile Phone: 0170/20 33 263
E-Mail: vogt@math.tu-berlin.de
URL1: http://www.mathematik.de
URL2: http://www.dmv.mathematik.de

[quote="leepakhin"]Does anyone know if the talks will be recorded?[/quote]

I do not know but I suggest to ask people from the [url=http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=24&Itemid=32&lang=en]Local Organizing Office[/url] for it, and maybe they will consider the idea if they have not already done so.

It could be posted on [url=http://videolectures.net/]video lectures[/url] or on youtube (google videos) to save money for the organizers.

A copy from the [url=http://www.imo2009.de/imo/index.php?option=com_content&view=article&id=30&Itemid=39&lang=en]latest IMO news, 26 June, 2009:[/url]

[b]John Templeton Foundation sponsors IMO video documentary[/b]

The John Templeton Foundation (U.S.A.) has just awarded a grant to Bildung und Begabung e.V., the association organising the IMO in cooperation with Jacobs University, to produce a video documentary of the IMO 2009. Bildung und Begabung e.V. has commissioned Oliver Wolf, a journalist and psychologist to create this film. He and his colleagues, up to two camera teams, will be present on campus for the full time of the IMO, they will wear name tags and shirts like the members of the organising team. So you may expect to be filmed once in a while for that documentary.

The project takes advantage of the pivotal role of the 50th IMO anniversary in the world mathematics community, which brings outstanding mathematicians together to discuss their IMO experiences, then have these mathematicians interact with this year’s IMO participants. Among the top level mathematicians, all of them former IMO medallists, are three winners of the Fields Medal, the equivalent of a Nobel Prize for mathematicians.

The video documentary, of professional quality, will be made of the IMO and of the mathematicians’ interactions with the public and with the students.

John Templeton Foundation’s funding will allow this experience to be shared with many more students who don’t actually attend the IMO themselves. Many countries have gifted students, perhaps students with genius potential, but no opportunity for them to compete in a team or attend the IMO.  The DVD will benefit this population in several ways.  It will give them examples of problems and contest materials to work on. It will provide for them models of how to work on difficult mathematical problems.  It will likewise provide mathematicians in these countries with models for work with young people.

All IMO 2009 participants will receive a complimentary copy of the DVD before the end of this year.

[b]New publication “German Teams at the IMOs 1959-2008” available[/b]

Just in time for the 2009 IMO, a publication will be available with a compilation of all German IMO participants from 1959 to 2008:

W. Engel, H.-D. Gronau, H.-H. Langmann, & H. Sewerin (2009). The German Teams at the International Mathematical Olympiads 1959-2008. Bad Honnef: K.H. Bock (96 pp.).

Additionally, statistics on all participating countries and their results are provided as well as a list of the most successful participants, awards given to IMO participants, and the regulations for the IMO.

Complimentary copies of the booklet will given to all Team Leaders, Deputy Team Leaders, and IMO Officials.

[b]The 50th International Mathematical Olympiad will be held from 14 – 21 July in Bremen. [/b]

The most talented young mathematicians from all over the world will be trying to find elegant solutions to difficult mathematical problems in their bid to win gold, silver, and bronze medals. Jacobs University Bremen will be the venue. The organisers are expecting around 600 students from 104 countries. Together with the team leaders, team guides, observers and the team of organisers around 1100 people will be involved in the nearly fortnight long event.

After several qualifying rounds six ace German mathematicians have made it to the forthcoming final.

The six members of the German team are:

[i]Bertram Arnold (17) from Georg-Cantor-Gymnasium, Halle (Saale)
Christoph Kröner (19) from Gymnasium Stein, Stein near Nürnberg
Malte Lackmann (18) from der Klaus-Groth-Schule, Neumünster near Kiel
Martin Merker (18) from Carl-Zeiss-Gymnasium, Jena
Jens Reinhold (17) from Helmholtz-Gymnasium, Bielefeld
Lisa Sauermann (16) from Martin-Andersen-Nexö-Gymnasium, Dresden
[/i]
The jury will be chaired by Hans-Dietrich Gronau, professor of mathematics in Rostock.

The youngest German olympian, Lisa Sauermann from Dresden, is just 16 years old and is in 10th grade at school. But this will be her third International Mathematical Olympiad. The same goes for 18 year old Malte Lackmann from Schleswig-Holstein. Although the other four maths experts are new to the olympiad, they, too, have raked in prizes in federal and national competitions.

"Germany is entering a very strong team ", said Dr. Harald Wagner, Executive Director of Bildung und Begabung e.V., organiser of the 2009 IMO in Bremen. "But it will remain exciting to the end." For in previous years the Chinese and Russians have scooped most of the gold medals. But even comparatively small countries such as South Korea, Vietnam and Romania, for example, enter very strong candidates.

"This year's outstanding achievements in the mathematical olympiad at federal and national level have shown once again that many German school students enjoy and are good at maths", said Wolfgang Lück, mathematics professor and president of the German Federation of Mathematicians. "They should all be role models for us – just as enthusiasm for maths is stronger and more accessible in other cultures than it is in Germany. I wish the German ace mathematicians cool heads and the best of luck in the competition!"

[b]Schedule [/b]

[i]From Saturday, 11.7.09[/i]: Jury meetings

[i]Tuesday, 14.7.09, 11:00 - 13:30[/i]
Opening ceremony: open to the press, in German and partly in English, with photo opportunity.

Venue: Pier 2, Gröpelinger Fährweg 2, 28237 Bremen
With a video message from German Chancellor Angela Merkel, speeches, a presentation by Albrecht Beutelspacher, mathematician from Gießen ("Calculating without a calculator") and the parade of the IMO teams

[i]Wednesday, 15.7.09, and Thursday, 16.7.09:[/i] Examinations (the actual contest)

[i]Sunday, 19.7.09, 14.30 [/i]
Anniversary celebration: open to the press, in English, with photo opportunity
Venue: Musical Theater, Richtweg, 28195 Bremen
With former IMO medallists Prof. Béla Bollobás, Prof. Timothy Gowers, Prof. László Lovász, Prof. Stanislav Smirnov, Prof. Terence Tao and Prof. Jean-Christophe Yoccoz with short lectures, discussion session and opportunity for the participants and the top mathematicians to meet.

[i]Tuesday, 21.7.09, 11:00 - 13:30 [/i]
Awards and closing ceremony (open to the press, in English, with photo opportunity)

Venue: Die Glocke, Domsheide 6-8, 28195 Bremen
With speeches from, among others, German Research Minister Annette Schavan and Bremen's Senator for Science, Renate Jürgens-Pieper. Presentation of medals and handover of the IMO flag to Kazakhstan, the host country in 2010, followed by photo session and interview opportunities.

[b]Administration, History, Sponsoring, Press Contacts: [/b]

The International Mathematical Olympiad is a mathematics competition at the highest level for high school students, hosted in a different country in July every year. In each of two four and a half hour long examinations the participants, no more than 20 years old, solve three challenging mathematical problems selected by an international jury over the previous days.

The 2009 IMO is organised by the association Bildung und Begabung e.V. (Bonn) in cooperation with Jacobs University Bremen. The Federal Ministry of Education and Research bears the largest share of the total budget of 1.5 million Euros while the Federal State of Bremen, Stifterverband für die Deutsche Wissenschaft and both main sponsors ArcelorMittal and BTS are also making considerable contributions. Other sponsors are: Atlantic Hotel Sail City, Bremer Landesbank, JOKE Event AG, Nordmetall Stiftung, and Stadtwerke Bremen (SWB). The organisers gratefully acknowledge additional financial support from BEGO Bremer Goldschlägerei, Bremer Messegesellschaft HGV, Die Sparkasse Bremen AG, Hannover Rückversicherung AG, NORDMILCH AG, Verkehrsverbund Bremen/Niedersachsen GmbH (VBN), GEOtec Zeichen- und Kunststofftechnik GmbH, Mathematikum e.V., Zirgon GmbH.

Interviews with jury members and delegation leaders as well as participants from Germany and abroad can be arranged by appointment for the following times: 16.7. from 15.00 - 16.30 and 18.7. from 16.00 to 17.30. Contact to the German team or the jury can also be arranged prior to the contest.


The idea of this thread is to get various special propoerties / special numbers  that can help form an outline guide for helping understand / solve number theory problems....

Squares...

I accidentally found that squares follow a pattern...if you now squares till 25 here is a easy way to figure out till 50....

20 ^ 2 = 400
21 ^ 2 = 441
22 ^ 2 = 484
23 ^ 2 = 529
24 ^ 2 = 576
25 ^ 2 = 625 = > here a new pattern starts...
26 ^ 2 = 676 = > 24^2 + 100
27 ^ 2 = 729 => 23^2 + 200
28 ^ 2 = 784 => 22^2 + 300

and on it goes till 50!
50 ^ 2 = 2500 => 0^2 + 2500

now the pattern continues after 50 in a different way..but thats a different story...

another interesting pair of numbers & squares...

4^2 = 16
subtracting 1 from 4 and 1 from 16
we have 3 on LHS and 15 on RHS...
Now 34 ^ 2 = 1156 => insert the 15 between the 1 & 6
334 ^ 2 = 111556 = > for every 3 at the right hand corner insert a 15 at the center!

7^2 = 49
subtracting 1 from 7 and 1 from 49
we have 6 on LHS and 48 on RHS...
Now 67 ^ 2 = 4489 => insert the 48 between the 4 & 9
667^ 2 = 444889 = > for every 6 at the right hand corner insert a 48 at the center!

Any more interesting numbers properties like these???


search
number theory unsolved
number theory

prove that there are infinitely many positive integers $ x,y$ and $ z$ solutions to the equation :$ x^2\plus{}y^5\equal{}z^3$ with $ xyz$ not equal to $ 0$.

u can put $ x\equal{}2^{15t\minus{}5}$ , $ y\equal{}2^{6t\minus{}2}$ and $ z\equal{}2^{10t\minus{}3}$

[quote="ZetaX"]Can't you  calculate yourself ;)[/quote]
of course, i know but i'm asking for the method that Rust find to this answer  , and I think that we can kill the problem if we search $ x,y$ and $ z$ as  $ 2^m,2^n$ and $ 2^p$ which it is given by khashi70.

Consider $ x^m\plus{}y^n\equal{}z^k$, were $ m,n,k$ relatively prime.
Let $ x\equal{}d^{nr}a,y\equal{}d^{mr}b$, then $ x^m\plus{}y^n\equal{}d^{mnr}(a^m\plus{}b^n)$ if $ d\equal{}a^m\plus{}b^n, z\equal{}d^l$ we get solution if $ mnr\plus{}1\equal{}lk$. It may be satisfyed, because (m,n,k) relatively prime.
We can consider $ z\equal{}d^{nr}a,y\equal{}d^{kr}b, d\equal{}a^k\minus{}b^n, x\equal{}d^l,nkr\minus{}1\equal{}lm$ too.

[hide]$ (3,\minus{}1,2)$ is a solution +
if $ (x,y,z)$ is a solution then  $ \forall \alpha \in \mathbb{Z^*}$ $ (\alpha^{15} x,\alpha^6 y , \alpha^{10} z)$ is a solution too
so $ (3 \alpha^{15},\minus{}\alpha^6 , 2 \alpha^{10})$ is a solution for all $ \forall \alpha \in \mathbb{Z^*}$ [/hide]


If $a , b$are real numbers and $(a + \sqrt{a^2 + 1} ) \cdot (b + \sqrt{b^2 +1} ) = 1$ , prove that
            $a + b = 0$

 [u]Babis[/u]

[hide]$\left( a+\sqrt{a^2+1}\right) \left( b+\sqrt{b^2+1}\right)=1\Rightarrow a+\sqrt{a^2+1}=\sqrt{b^2+1}-b\Rightarrow a+b=\sqrt{b^2+1}-\sqrt{a^2+1}\Rightarrow (a+b)\left(\sqrt{a^2+1}+\sqrt{b^2+1}\right)=(b+a)(b-a)$

Suppose $a+b\neq 0$. Then $b=a+\sqrt{a^2+1}+\sqrt{b^2+1}$. Since $\sqrt{a^2+1}>-a$ and $\sqrt{b^2+1}>b$, then $b=a+\sqrt{a^2+1}+\sqrt{b^2+1}>b$, a contradiction. So $a+b$ must be 0.[/hide]

I will give a solution a bit strange. Let $P(x)=x+\sqrt{x^2+1}$
Easily we find that
$P'(x)>0$ so $P$ is increasing so ''1 to 1'' .We have $(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})$ => $a+\sqrt{a^2+1}=-b+\sqrt{b^2+1}$

and so $P(a)=P(-b)$ and because $P$ is ''1 to 1'' we have that $a=-b$ and $a+b=0$

and we are done.  :D  :)


Two percent of the Earth's surface is permanently covered by ice. If 3.9 million square miles of the surface are covered by ice, how many square miles are in the surface area of the Earth? Express your answer in scientific notation with two decimal places of accuracy in the mantissa.

We can write this as $ .02\equal{}3.9/x$, where $ x$ is the total number of square miles in millions.  Solving gives $ x\equal{}195$, so the earth has 195 million square miles, or $ 1.95\times 10^{8}$.


Give $a,b,c$ are positive integers such that:$a+b+c=1$.Prove that:

$\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}\geq\frac{3}{2}$

actually $\frac94\geq \sum\frac{a}{a+bc}\geq 2$
but it is very easy :D

$\frac{a}{a+bc}= \frac{a}{bc-b-c+1}= \frac{a}{(1-c)(1-b)}= \frac{a}{(a+b)(b+c)}$

We do the same thing with the others. We get:
$\frac{9}{4}\geq \frac{(a+b+c)(a(b+c)+b(c+a)+c(a+b))}{(a+b)(b+c)(c+a)}\geq 2$

which is easy after multiplying

I am sorry, $a+b+c=3$.$a,b,c$are positive reals.Prove that:
$\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}\geq\frac{3}{2}$
Thank you.

[quote="Frey"]I am sorry, $a+b+c=3$.Prove that:
$\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{c}{c+ab}\geq\frac{3}{2}$
Thank you.[/quote]
It's wrong. :wink:
See here for positive numbers:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21689

Oh,no.It's right.
$a,b,c$ are positive reals.

There is a some cunfusions in here.
$a,b,c$ are positive reals that also means $a,b,c$ are positive number. While don't show clearly then the word "positive number" is "positive reals "


function
algebra
functional equation
algebra unsolved

Find all function $ f$:$ R$ to $ R$ such that:
(i)$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ for every two real number $ x,y$
(ii)$ f(\frac{1}{x})\equal{}\frac{f(x)}{x^2}$ for all $ x \not\equal{} 0$

Very nice!!!!!

From (i) we can take that:
$ f( \minus{} x) \equal{} \minus{} f(x)$.
Setting $ y \minus{} \minus{} > 1$ and $ f(1) \equal{} c$ the given functional equation becomes:

$ 1)$ $ f( \minus{} x) \equal{} \minus{} f(x)$
$ 2)$ $ f(x \plus{} 1) \equal{} f(x) \plus{} c$.
$ 3)$ $ f\left(\frac {1}{x} \right) \equal{} \frac {f(x)}{x^2}$.

From $ 2$ we take:
$ f(\frac {1}{x} \plus{} 1) \equal{} f\left(\frac {1}{x} \right) \plus{} c \equal{} \frac {f(x)}{x^2} \plus{} c$ $ \boxed{1}$.

But $ f(\frac {1}{x} \plus{} 1) \equal{} f\left(\frac {1}{\frac {x}{x \plus{} 1}} \right) \equal{} \frac {f\left(\frac {x}{x \plus{} 1} \right)}{\frac {x^2}{(x \plus{} 1)^2}}$

But $ \frac {x}{x \plus{} 1} \equal{} 1 \minus{} \frac {1}{x \plus{} 1}$.

$ f \left(\frac {x}{x \plus{} 1}\right) \equal{} f(1 \minus{} \frac {1}{x \plus{} 1}) \equal{} f(1) \plus{} f\left( \minus{} \frac {1}{x \plus{} 1} \right) \equal{} c \minus{} f(\frac {1}{x \plus{} 1}) \equal{} c \minus{} \frac {f(x \plus{} 1)}{(x \plus{} 1)^2} \equal{} c \minus{} \frac {f(x) \plus{} c}{(x \plus{} 1)^2}$

So $ f\left(\frac {1}{x} \plus{} 1 \right) \equal{} \frac {c \minus{} \frac {f(x) \plus{} c}{(x \plus{} 1)^2}}{\frac {x^2}{(x \plus{} 1)^2}}$ $ \boxed{2}$

Now from $ \boxed{1},\boxed{2}$ we take $ \frac{c(x \plus{} 1)^2 \minus{} f(x) \minus{} c}{x^2} \equal{} \frac {f(x)}{x^2} \plus{} c$.

$ \boxed{f(x) \equal{} cx}$ where $ c \equal{} f(1)$

 :)


The longest side of a 30 degree-60 degree-90 degree triangle is equal to the longest side of a 45 degree-45 degree-90 degree triangle. What is the ratio of the perimeter of the 30 degree-60 degree-90 degree triangle to the perimeter of the second triangle, to the nearest hundred.


algebra
polynomial
integration
calculus
calculus computations

I'm fond of this question very much: Let ${\mathcal P}_{n}$ be the space of all polynomials of degree $\leq n$ and $\alpha\neq0$. Show that for every $p\in M_{n}$ there exists $q\in M_{n}$ and constant $C$ so that
$\int p(t)e^{t\alpha}dt=q(t)e^{t\alpha}+C$

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[quote="mlok"]This is calculus, specifically integration by parts $n$ times.[/quote]

okey but I have a very nice sol. without calculating? Guess why?