Two dogs, each travelling at 10 ft/sec, run toward each other from 500 feet apart. As they run, a flea flies from the nose of one dog to the nose of another at 25 ft/ sec. The flea flies between these dogs in this manner until it's crushed when the gods collide. How far did the flea fly?

Have fun!  :trampoline:

[hide]Let's see, we can make this problem alot simpler by doing one thing, realizing that all we want is the amount of time it took for the dogs to reach eachother, because that's the amount of time the fly will have ran. So the two dogs together run 20 feet in a second, so it takes them 500/20=25 seconds, so the flee is running for 25 seconds, so we want 25*25=[b]625 feet[/b][/hide]

Not all that difficult. . .

I posted this in the marathon. The total distance it travels is $\text {rate} \cdot \text {time}$. Since it's flying back and forth for 25 seconds, and its rate is 25, the answer is 625.

Ok. I was implying that some people would find it hard because I am in 8th grade and can't really base it on what I felt, since it's a middle school forum.

Well I'm in eighth grade as well, so. . . .

It says nothing about the dogs colliding, just something about [b]gods[/b] colliding :roll: ...

[quote="tarquin"]I like funny problems about animals.[/quote]

I like funny animals talking about their problems, like a depressed capybara!

There is a funny story about this problem being given to the mathematician John von Neumann, considered by many to be the most brilliant man who ever lived.  The poser of the problem expected him to overthink it and try to sum the infinite series, but von Neumann solved it almost immediately.  But when the poser said, "You must have known the trick", von Neumann responded, "What trick?  I simply summed the infinite series".

Yeah I kinda did the infinite series thing, until I noticed....


modular arithmetic
number theory unsolved
number theory

Let $ n \equal{} \prod_{1\leq i \leq k} p_i^{\alpha_i}$, $ n$ odd, $ p_i$ primes, $ \alpha_i \geq 1$. Since $ \gcd(x \minus{} 1,x \plus{} 1) \leq 2$ for any integer $ x$, it means that from $ x^2 \equiv 1 \pmod{n}$, i.e. $ (x \minus{} 1)(x \plus{} 1) \equiv 0 \pmod{n}$, follows there exists a partition $ I \cup J \equal{} \{1,2,\ldots,k\}$, $ I \cap J \equal{} \emptyset$, such that $ x \equiv 1 \pmod{\prod_{i\in I} p_i^{\alpha_i}}$ and  $ x \equiv \minus{}1 \pmod{\prod_{j\in J} p_j^{\alpha_j}}$, with unique solution $ x$ modulo n, from the Chinese Remainder Theorem. But there are exactly $ 2^k$ such partitions, hence that many solutions.


Hello everybody.

This is a thrilling game of politics, economy, life living in a city except: theres nothing to start with.

A group of settlers move into a forest clearing, where they discover rich soil, and a rich bank with unlimited money. What they don't know is that this place is oftenly touched by God. (me) In this game, you will make your own town, its laws, taxes and if you ever need to kick someone out for not doing something, just ask god. I do not care if you have a mayor or no type of politics at all. I do not care if you develop your city name or make an ultimate ruler. There is only one rule. What happens, happens. If I say: BAM you're dead. Then you're dead and we chuck you out of the city.

Some suggestions:
*Make some sort of university.
*Have doctors or dentists.
*Start your own buisness if you want to.
*Start political campaigns.

I was lying. There are some other humanly rules:
1. You must eat-> or you die.
2. You must drink-> or you die.
3. You must do what god says.

Thats it!
You don't need to say /in or anything. You can join whenever you want to.

I will be the first 'citizen'. Notice how I post:

Dojo walks in and cuts down the trees to build house.
Dojo uses tools made from rocks to construct his house.
Dojo notices the fertile land around and picks some berries to eat.

I would join, except that if god is playing, it would be more than a little unfair.

Chen-Hsi walks in and cuts down the trees to build house.
Chen-Hsi uses tools made from rocks to construct his house.
Chen-Hsi notices the fertile land around and picks some berries to eat.

AIME15 walks in and cuts down trees to build house.
AIME15 uses tools made from rocks to construct his house.
AIME15 noticed the fertile land and plants crops around AIME15's house.  AIME15 then picks berries to eat.
AIME15 uses surplus wood to build fence around AIME15's house and crops.

AHHHHHHH!!!!!!
Okay this is not gonna work. So here is whats gonna happen.
There will be a grid from 1-50. Each 'plot' is a number which you must purchase with a loan from a bank.
okay continue :D

Sorry you guys, restarting the game.

Okay the rules:
1. The bank has unlimited money in it.
2. You must eat/drink every day, by either growing your own crops or later buying from the store.
3. The current only source of water is the pure, fresh water that flows though the land.

Okay do not build anything yet, just make an initial sign-up list that can have more people come in and out:

1. chenhsi
2. AIME15
3. james4l

I dunno, whats the objective of your life?

i will join. better explaintion of game.

[b]join[/b]. how long is a day in this place? cuz some ppl can't get on AoPS every single normal day.

ISN'T ANYONE GOING TO FOLLOW UP ON THIS???  :mad:

Definitely not a real day....maybe 4-5?  Or a specified amount of posts?

eggylv999 becomes god and floods AoPS city.

[size=200][b][i][u]THE END.[/size][/u][/i][/b]

1. This isn't AoPS city.
2. Stop spamming.
3. Dojo should not let you join.

Thank you.  That is all.


In an experiment, a scientific constant $ C$ is determined to be $ 2.43865$ with an error of at most $ \pm 0.00312$.  The experimented wishes to announce a value of $ C$ in which every digit must be significant.  That is, whatever $ C$ is, the announced value must be the correct result when $ C$ is rounded to that number of digits.  The most accurate value the experimented can announce for $ C$ is

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.4 \qquad \textbf{(C)}\ 2.43 \qquad \textbf{(D)}\ 2.44 \qquad \textbf{(E)}\ 2.439$

[hide="My Solution"]
The answer is clearly $ \boxed{(E)}$ because it has the most accurate value with the appropriate rounding.
[/hide]

EDIT:  :blush:, I forgot to account for tolerance.

The actual value is between $ 2.44177$ and $ 2.43553$ which both round to $ 2.44 \boxed {D}$


geometry
circumcircle
cyclic quadrilateral
projective geometry
geometry proposed

Let a triangle $ABC$ and let $H$ its orthocenter and $BB_{1}$ its altitude .
Let also $M$ the midpoint of $BC$ .Let the perpendicular from $H$ to $AM$ intersect $AM$ at $K$ .Prove that KH is bisector of the angle $CKB_{1}$

Let X the intersection of HK and AC.
$HKB_{1}A$ is cyclic, so, $B_{1}KX = HAB_{1}= 90-C$!

So, we have to prove that K, H, A and C are concyclic, because $CBB_{1}= 90-C$!

To prove this, we will use a very nice fact, and the problem follows easily!

Lemma:

Let D and E the intersections of AH and CH with BC and BA!
We have that DE, HK and AC are concurrent!

Proof:

Let W the circle with center in the midpoint of AC and ray AC/2!
Let Y the intersection of DE and AC. The polar of Y with respect to W is HB, so, the polar of B with respect to W is YH.
We know that the polar of a point is perpendicular to the segment trought the point and the center of the circle!
So, $YH$ is perpendicular to $BM$! and the result comes..

_

Now, we have to prove that $YH.YK = YA.YC$,
but $YH.YK = YB_{1}.YM$, because $HKMB_{1}$ is cyclic,
$YD.YE = YA.YC$, because $ACDE$ is cyclic
and $YD.YE = YB_{1}YM$ (nine-point circle)!

We denote as $AA_{1},$ $CC_{1},$ the other two altitudes of $\bigtriangleup ABC$ and as $X,$ the intersection point of $AC,$ $HK.$

We consider the vertex $A$ of $\bigtriangleup ABC,$ as a point outwardly of the circumcircle $(M)$ of the cyclic quadrilateral $BCB_{1}C_{1},$ centered ( as clear ) at the midpoint $M$of $BC.$

So, The segment line $HK,$ is the polar of $A$ with respect to $(M),$ as the line passes through the point $H$ $($ the intersection point of $BB_{1},$ $CC_{1}$ $),$ and perpendicular to $AM$ $($ the line connecting the point $A,$ with the center of $(M)$ $).$

So, the point $X,$ is the harmonic conjugate of $A,$ with respect to $B_{1},$ $C$ and hence, the pencil $K.AXB_{1}C,$ is in harmonic conjugation.

Because of now, $AK\perp KX,$ we conclude that the segment line $KX$ bisects the angle $\angle B_{1}KC$ and the proof is completed.

Kostas Vittas.

[color=indigo][i]Let me show you my solution[/i].

Denote by $X$ the intersection of $HK$ and  $AC$ and by $A_{1}$ the foot of the $A-height$. Then we have:

$HKMA_{1}$ is cyclic so $\angle{A_{1}MK}=\angle{KHA}=\angle{KB_{1}X}\Longrightarrow MKB_{1}C : cyclic$

So $\angle{B_{1}MC}=\angle{B_{1}KC}$, therefore $2 \cdot (90-C)= 90-C+\angle{XKC}\Longrightarrow q.e.d.$

Nick Rapanos[/color]

[color=darkred][b][u]Lemma (w-k).[/u][/b] For $\triangle ABC$ and $P$ denote : $\{\begin{array}{c}D\in AP\cap BC\ ,\ E\in BP\cap CA\ ,\ F\in CP\cap AB\\\\ L\in EF\cap BC\ ,\ S\in LP\cap AC\end{array}$. Then the division $(A,E,S,C)$ is harmonically.[/color]

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] Denote : $C_{1}\in AB\ ,\ CC_{1}\perp AB$ ; $L\in B_{1}C_{1}\cap BC$ ; $S\in LH\cap AC$.

From the w-k property $LH\perp AM$, i.e. $K\in LH$ and the above lemma obtain that the division $(A,B_{1},S,C)$ is harmonically, i.e. $\frac{B_{1}A}{B_{1}S}=\frac{CA}{CS}$.

Because $KA\perp KS$ (from a w-k characterization of a harmonical division) results that the ray $[KS$ is the bisector of the angle $\widehat{CKB_{1}}$.

[u]Remark.[/u] "[b]w-k[/b]" means "[b]well-known[/b]".[/color]

Thank you all for trying my problem  :)
My solution is similar to vittasko's one and that was the idea I constructed the problem .
But I must tell that the solution of my friend nick rapanos <<kills>>  the problem
Great solution Nick thank you for sharing with us  :)

Hope that you enjoyed the problem.


calculus
AMC
AIME
USA(J)MO
USAMO
probability
number theory

i'm going to be junior and im done with precal
(ill take calculus bc+statcis)
i have both AoPS1 and2 but i dont know if i should do vol1 or not
cuz it can be waste of time(though it was really helpful last summer)

this summer i really want to use my time effectively

one of you recommened focusing on AIME so i will

i already enrolled AMC12class so i think im fine preparing for AMC

but for AIME instead of just solving practice problems
is there any good book i can use ?????

im looking forward taking the USAMO if i make AIME

any suggestion????????please???

btw i don have anyone whos super genious in math near me
so i'll have to teach myself..................(other than people in this website =)

i think it really depends if you have fully mastered the material...doing well in pre-calc in school doesn't really translate to doing well in competition math...even though AoPS volume 1 may have material that you may find repetitive or boring (proportions, exponents, ectr.), it's made to give you as indepth of an explanation as possible...if you have already done introduction to number theory, counting, probability, ectr., then i would recommend doing AoPS volume 2 because it is made for higher levels of competition (AIME, USAMO)...if u still haven't, then i would recommend starting from scratch at volume 1...it's worth the time and effort spent!

Start with the subjects you're not completely comfortable with in Volume 1.  Work the problems, particularly the ones with needles next to them.  If you have a lot of trouble with them, read the relevant chapters carefully.  If you're not having much trouble with them, then move on to Volume 2.


calculus
integration
trigonometry
function
derivative
logarithms
calculus computations

Can you find

[1]$ \int(tan^{2}x\plus{}tanx)e^xdx$

[2]$ \int \frac{ xe^x}{x^2\plus{}2x\plus{}1}dx$

[3]  $ \int_{ \minus{} 2008}^{2008}x^{2007}ln(e^x\plus{}sinx\plus{}2)dx$

hello, write your first integral in the form:
$ \int e^x\left(\tan(x) \minus{} 1\right)dx \plus{} \int e^x\left(\tan^{2}(x) \plus{} 1\right)dx$ and integrate the first one
by parts. The solution is $ e^x\left(\tan(x) \minus{} 1\right) \plus{} C$.
Sonnhard.

hello, by the same procedure we get
$ \int \frac{xe^x}{(x\plus{}1)^{2}}dx\equal{}\int \frac{e^x}{x\plus{}1}dx\minus{}\int \frac{e^x}{(x\plus{}1)^{2}}dx\equal{}\frac{e^x}{x\plus{}1}\plus{}\int \frac{e^x}{(x\plus{}1)^{2}}dx\minus{}\int \frac{e^x}{(x\plus{}1)^2}dx\equal{}\frac{e^x}{x\plus{}1}\plus{}C$
Sonnhard.

Can you post your solution of the third one?

Sorry ...  :(  the third integral is  $ \int_{ \minus{} 2008}^{2008}t^{2008}ln(cost \plus{} 2)dt$

Solution

Donate the function $ f(x) \equal{} \int_{ \minus{} x}^{x}t^{2007}ln(cost \plus{} 2)dt$

Then $ I \equal{} f(2008)$ .It's enough to find $ f$

The first  derivative is

$ f'(x) \equal{} (\int_{ \minus{} x}^{x}t^{2007}ln(cost \plus{} 2)dt)' \equal{} ( \minus{} \int_{0}^{ \minus{} x}t^{2007}ln(cost \plus{} 2)dt)' \plus{} (\int_{ 0}^{x}t^{2007}ln(cost \plus{} 2)dt)' \equal{} x^{2007}ln(cosx \plus{} 2) \minus{} x^{2007}ln(cos(x) \plus{} 2) \equal{} x^{2007}ln1 \equal{} 0$ so $ f(x) \equal{} c \equal{} f(0) \equal{} 0$

Then $ I \equal{} 0$

[quote="giannis"]Sorry ...  :(  the third integral is  $ \int_{ \minus{} 2008}^{2008}t^{2007}ln(cost \plus{} 2)dt$
[/quote]

Well for an odd function f, $ \forall a \in \mathbb{R}$, the integral$ \int_{\minus{}a}^{a} f(x) \, dx \equal{} 0$

Its clear that $ t^{2007}\ln(\cos t \plus{} 2)$ is an odd function :D

Yes ... Nice idea  :wink:  but how about this

${ \int^{\frac{1}{e}}_{e} \frac{lnx}{x^2+1}}dx$

[quote="giannis"]Yes ... Nice idea  :wink:  but how about this

${ \int^{\frac {1}{e}}_{e} \frac {lnx}{x^2 + 1}}dx$[/quote]

With $ x = e^t$ substitution,

$ \int^{\frac {1}{e}}_{e} \frac {lnx}{x^2 + 1}dx = \int^{-1}_{1} \frac {te^t}{e^{2t} + 1}dt$

But $ \frac{te^t}{e^{2t} + 1}$ is a odd function :D


Prove or disprove that there exist two tetrahedra $T_1$ and $T_2$ such that:
(i) the volume of $T_1$ is greater than that of $T_2$;
(ii) the area of any face of $T_1$ does not exceed the area of any face of $T_2$.

There are two such tetrahedra. This is a trivial consequence of the following fact: there are tetrahedra of volume arbitrarily small whose faces have areas arbitrarily large. For example, consider an equilateral triangle so big as you want and take a point on the perpendicular to the plane of the triangle in its center, so close to the center of the triangle as needed, then the resulting tetrahedra has a volume as small as you want and the areas of the faces are as big as you desire.


[quote="GameBot"]How many integers between $ 100$ and $ 500$ do not contain an even digit?[/quote]

Number of such numbers is $ \ 2\times 5\times 5\equal{}50$


algebra
polynomial
arithmetic sequence
arithmetic series
AMC

[color=cyan]A hotel has its floors numbered in the standard order; that is, 11, 12, 14, 15 (which is the top floor).  The hotel manager wishes to add an additional floor onto the top of his hotel.  He decides that the best way to number the new floor is by using a cubic polynomial with sequential values 11, 12, 14, and 15, and taking as the next floor number whatever the cubic polynomial would give as the next value in sequence.  What is this floor number?

(Sorry, that was rather poorly phrased.)[/color]

[quote]He decides that the best way to number the new floor is by using a cubic polynomial with sequential values 11, 12, 14, and 15, and taking as the next floor number whatever the cubic polynomial would give as the next value in sequence. [/quote]

Would someone mind explaining to me what a "sequential value" of a polynomial is?

[quote="Fierytycoon"]Ah. I see. Thanks. But it could be such that, say, f(any number) = 11, f(any number+1) = 12, ..., where any number is not restricted to being 1, right?[/quote]

Well, that would be really nice, but no.

It has to be a cubic polynomial, and the one you mentioned is a linear function.  We need something like f(x) = ax:^3: + bx:^2: + cx + d

[quote="Fierytycoon"]Ah. I see. Thanks. But it could be such that, say, f(any number) = 11, f(any number+1) = 12, ..., where any number is not restricted to being 1, right?[/quote]

Yeah. Thats right.. the values can occur anywhere in the sequence, not just at the start. But it shouldn't change anything, just think of shifting the x axis across..

[color=cyan]Yeah, I knew this was nto the best-formulated problem.  I just meant that the x's in f(x) that gave you 11, 12, 14, and 15 had to be consecutive terms of an arithmetic sequence, and you need to find what happens at the next term in the sequence.  Although TripleM maybe gave you a good idea why the problem might let you do something nicer than look at the general form of 5 consecutive terms of an arithmetic sequence. . .[/color]

It can, because that's exactly the same thing. I mean to say, if f(n) = 11, f(n+1) = 12, f(n + 2)=14, f(n+3)=15, then the polynomial f((x+n(k-1))/k) will do what you want for whatever k you like, and is obviously of the same degree; so the difference of the arithmetic series doesn't matter at all.

Anyhow, just take third differences, which are constant for a cubic, and extrapolate the next term, which is 13.


You've got some pieces, whose shape is shown in the attached picture. Prove that the rectangles you can build with it have at least one side which is multiple of 4.
By the way, how to prove $2^kn-7$ with k and n positive integers makes an infinite number of perfect squares with any k?

i'm assuming the second one is a differnt Q.
that 's not true.
let $n=k=3$, you get: $2^3*3-7=17$. it's not a perfect square :huuh:
or $n=3, k=5$ you get $89$ :huuh:
or $n=5, k=3\Rightarrow 33$ :huuh:

I didn't express myself well.
For the second question, it is:
given K, you can find infinite n so that

$2^kn -7$ is a perfect square
Bye

For the first problem, to make a rectangle, we must get rid of the jagged edges.  Given two of those shapes, we can connect them in two different ways so that the jagged edges connect, not so that the straight edges connect.  The first way forms a $3\times 4$ rectangle.  The second way forms a $2 \times 4$ rectangle with unit squares sticking out of the top left and bottom right corners.  To prove davi24's problem, we must prove two things.  First, we must prove that any large rectangle formed by any number of $3\times 4$ rectangles has at least one side that is a multiple of $4$.  Second, we must prove that the figure formed by the second method of connecting the jagged edges cannot be formed into a rectangle by adding any number of pieces.

I don't think you have to do all this to solve it. In the paper with the first question there's a little help: paint only the "cells" with odd coordinates... But I don't know how this could help me.


Θεωρούμε τον $ F$-διανυσματικό χώρο $ V$ με διάσταση $ 2$. Έστω $ \left\{e_{1},e_{2}\right\}$ μια βάση του $ V$. Θεωρούμε τους ενδομορφισμούς $ f$ και $ g$ του $ V$ με $ f(e_{1})\equal{}f(e_{2})\equal{}e_{1}\plus{}e_{2}$ και $ g(e_{1})\equal{}e_{2},g(e_{2})\equal{}e_{1}$. Να αποδείξετε ότι $ f\circ g\equal{}g\circ f$ και $ f^{2}\equal{}2f$

[quote="Γιώργος"]Θεωρούμε τον $ F$-διανυσματικό χώρο $ V$ με διάσταση $ 2$. Έστω $ \left\{e_{1},e_{2}\right\}$ μια βάση του $ V$. Θεωρούμε τους ενδομορφισμούς $ f$ και $ g$ του $ V$ με $ f(e_{1}) \equal{} f(e_{2}) \equal{} e_{1} \plus{} e_{2}$ και $ g(e_{1}) \equal{} e_{2},g(e_{2}) \equal{} e_{1}$. Να αποδείξετε ότι $ f\circ g \equal{} g\circ f$ και $ f^{2} \equal{} 2f$[/quote]

Επειδή δεν βλέπω να τη λύνει κανείς θα βάλω τη λύση έτσι για να κλείσει και αυτό το post.

[hide]i) Σύμφωνα με τη θεωρία αρκεί να δείξουμε ότι $ (f\circ g)(e_{1}) \equal{} (g\circ f)(e_{1})$ (α)

 και $ (f\circ g)(e_{2}) \equal{} (g\circ f)(e_{2})$ (β).

Έχουμε $ (f\circ g)(e_{1}) \equal{} f(g(e_{1})) \equal{} f(e_{2}) \equal{} e_{1} \plus{} e_{2}$ (1) και

 $ (g\circ f)(e_{1}) \equal{} g(f(e_{1})) \equal{} g(e_{1} \plus{} e_{2}) \equal{} g(e_{1}) \plus{} g(e_{2}) \equal{} e_{1} \plus{} e_{2}$ (2)

Από (1) και (2) έπεται η σχέση (α).

Έχουμε ακόμα $ (f\circ g)(e_{2}) \equal{} f(g(e_{2})) \equal{} f(e_{1}) \equal{} e_{1} \plus{} e_{2}$ (3)

και $ (g\circ f)(e_{2}) \equal{} g(f(e_{2})) \equal{} g(e_{1} \plus{} e_{2}) \equal{} g(e_{1}) \plus{} g(e_{2}) \equal{} e_{1} \plus{} e_{2}$ (4).

Από (3) και (4) ισχύει η σχέση (β). Επομένως ισχύει $ f\circ g \equal{} g\circ f$

ii)
 $ f^{2}(e_{1}) \equal{} (f\circ f)(e_{1}) \equal{} f(f(e_{1})) \equal{} f(e_{1} \plus{} e_{2}) \equal{} f(e_{1}) \plus{} f(e_{2}) \equal{} 2(e_{1} \plus{} e_{2}) \equal{}2f(e_{1})$.

$ f^{2}(e_{2}) \equal{} (f\circ f)(e_{2}) \equal{} f(f(e_{2})) \equal{} f(e_{1} \plus{} e_{2}) \equal{} f(e_{1}) \plus{} f(e_{2}) \equal{} 2(e_{1} \plus{} e_{2}) \equal{} 2f(e_{2})$.

Ισχύει λοιπόν
$ f^{2}(e_{1}) \equal{} 2f(e_{1})$

$ f^{2}(e_{2}) \equal{} 2f(e_{2})$   και επειδή $ e_{1},e_{2}$ είναι βασικά διανύσματα, θα έχουμε ότι $ f^{2} \equal{} 2f$ :P [/hide]

E ntax i askisi ine full elementary...isos i orologies tromazun.. :P
An $ A$ o pinakas tis$ f$ kai $ B$ tis $ g$ ,ine

A={{1,1},{1,1}}  kai

B={{0,1},{1,0}}

opote ine eukolo n dume oti

$ A*B \equal{} B*A$  kai  $ A^2 \equal{} 2*A$ .......


function
calculus
derivative
integration
modular arithmetic
symmetry
Functional Analysis

Hello,
I'd like to get some help/hints on the following question:

Let X be a linear space of continuous functions f on [-1,1] (C[-1,1]) such that except for a finite number of points on [-1,1], the derivative of f exists, continuous and it (the derivative) is square integrable on [-1,1].
That is such f that $ \int_{ \minus{} 1}^{1} |f'(t)|^2 dt < \infty$.

Define for f in X
$ \| f\| _X \equal{} ( \int_{ \minus{} 1}^{1} |f'(t)|^2 dt \plus{} \int_{ \minus{} 1}^{1} |f(t)|^2 dt) ^{\frac {1}{2}}$.

Let $ \phi$ be defined on X by $ \phi (f) \equal{} f(0)$.

Let H be the completion of X. In the Hilbert space H, find g such that
for any f in X,
$ \phi (f) \equal{} \langle f,g \rangle$.
(Here the inner product is the one one that corresponds to the defined norm in X.)

Thanks in advance.

Let's look at this through Fourier coefficients:

For $ f\in X,$ we can write $ f(t)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}\widehat{f}(n)e^{\pi int}.$

From that (and Parseval's formula, and the effect of derivatives on Fourier coefficients) we have that
\[ \|f\|_X^2\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}(1\plus{}\pi^2n^2)|\widehat{f}(n)|^2\]
This can be extended to
\[ \langle f,g\rangle_X\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}(1\plus{}\pi^2n^2)\widehat{f}(n)\overline{\widehat{g}(n)}\]
In this language, the "Dirac delta" is $ \phi(f)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}\widehat{f}(n).$

In that setup, the $ g$ we need to answer the problem is given by $ \widehat{g}(n)\equal{}\frac1{1\plus{}\pi^2n^2}$ or
\[ g(t)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}\frac{e^{\pi int}}{1\plus{}\pi^2n^2}\]
It is possible to write this $ g$ in closed form - I'll leave that to someone else.

I just realized that I defined
\[ \|f\|_X^2\equal{}\frac12\int_{\minus{}1}^1|f'(t)|^2\,dt\plus{}\frac12\int_{\minus{}1}^{1}|f(t)|^2\,dt\]
which isn't quite what kar said. The consequences shouldn't be too serious, but there might be some stray extra factors of $ 2$ or $ \sqrt{2}$ floating around.

Kent Merryfield, thank you!
Actually after I posted the question, I've found something similar solved (in a different, more "constructive" way), but your solution seems "easier" (although I didn't computed this g in closed form yet.).

So thanks again, I'll look more carefully in your solution and maybe try to solve it in the other way too...

I get $ g(t)\equal{}\frac{\cosh(t\minus{}1)}{\sinh(1)}$ on $ [0,2),$ continued to be periodic of period $ 2.$ I might be off by a constant multiple.

The idea: since $ \widehat{g}(n)\equal{}\frac1{1\plus{}\pi^2n^2},$ then $ \widehat{g''}(n)\minus{}\widehat{g}(n)\equal{}\minus{}1.$  From that we get that $ g''(t)\minus{}g(t)\equal{}\minus{}2\delta$ (considered as a distribution). The rest is just solving the differential equation.

Kent, would you mind explaining how to solve the differential equation? I think I may have made a misstep in my education by not taking an ODE class  :wink:

All right. Except for what happens at $ t \equal{} 0$ and by extension all the places where $ t\equiv 0\pmod{2},$ we have $ g'' \minus{} g \equal{} 0.$ The solutions to that are linear combinations of $ e^t$ and $ e^{ \minus{} t}.$

We do have that $ g$ is continuous (Weierstrass M-test on the series), so we need to have $ g(0^ \plus{} ) \equal{} g(2^ \minus{} ).$ It's that symmetry that leads me to $ a\cosh(t \minus{} 1).$ (Since $ g$ is an even function on $ [ \minus{} 1,1]$ and periodic of period $ 2,$ it follows that its graph is also symmetrical about $ t \equal{} 1.$)

Now we need to get $ \minus{} 2\delta(t)$ from the second derivative. If $ g(t) \equal{} a\cosh(t \minus{} 1),$ extended to be periodic, then $ g'(0^ \plus{} ) \equal{} a\sinh( \minus{} 1) \equal{} \minus{} a\sinh(1),$ while $ g'(0^ \minus{} ) \equal{} g'(2^ \minus{} ) \equal{} a\sinh(1).$ Then the (weak) second derivative at zero is the jump in $ g'$ multiplied by $ \delta,$ which is to say $ \minus{} 2a\sinh(1)\delta,$ which is supposed to equal $ \minus{} 2\delta.$ Hence $ a \equal{} \frac1{\sinh(1)}.$

There's one more thing to check: could we add some constant to this? For that I compute

$ \widehat{g}(0) \equal{} \frac12\int_{ \minus{} 1}^1g(t)\,dt \equal{} \int_0^1g(t)\,dt\stackrel{?}{ \equal{} }1.$

But since $ \int_0^1\cosh(t \minus{} 1)\,dt \equal{} \sinh(1),$ this matches and there's no constant to add.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Here is a problem given at the "Dan Barbilian" contest last year in Campulung-Muscel:

Let A belong to M_3(C) so that det(A+3*A^t)=0. Show that det(A+A^t)=-8/3det(A)

where A^t is the transpose matrix of A, this means that A^t is the matrix where the columns of A have been replaced by the lines!

cheers! :D :D

Denote A'=A t.
Denote by P(x)=det(A+xA'). It is clear that P is a polynomial in X of degree  \leq 3, and the coefficient of the term of degree 3 is det(A')=det(A).
We have det(A+3A')=0, so P(3)=0.
det(M)=det(M'), so det(A'+3A)=det(A+3A')=0, so P(1/3)=0.

If det(A)=0, then P has degree  \leq 2, and P(0)=det(A+0A')=det(A)=0, so P has 3 distinct roots (0,3,1/3), so P=0, so P(1)=det(A+A')=0=-8/3det(A).

If det(A)<>0 then A is invertible, P has degree 3 and (X-3)(X-1/3)|P, so P(X)=det(A)(X-3)(X-1/3)(X-t), where t is the third root of P.
We have det(A)=P(0)=-t*det(A) and det(A)<>0, so t=-1, so P=det(A)(X-3)(X-1/3)(X+1).
det(A+A')=P(1)=det(A)*(-2)*(2/3)*2=-8/3*det(A).

Nice problem :D


I was wondering where you take these competitions, i know that AMC10 is taken at school but where are the others taken, and if taken somewhere else, do you take it in a room where others are taking it. This is directed to AIME and USAMO.

the AIME and USAMO are for the most part taken at the same site you took the AMC at.

And if multiple people qualified from your school, you will take test in the same room.  Otherwise, you take the test by yourself.

So you wouldn't get to travel anywhere, well unless you make IMO.  :mad:

well, what about trophies and such?

[quote="simo14"]no trophies.. reward=fun. u can travel to mop.[/quote]

only if u make it that far lol :P

u get these nice plaques for state amc winner. -.-

MOP is the [url=http://en.wikipedia.org/wiki/Mathematical_Olympiad_Program]Math Olympiad (Summer) Program[/url].

[quote] u get these nice plaques for state amc winner. -.-[/quote]

how do they decide state winners in case of ties?

[b]serialk11r[/b][quote]Aren't there so many AMC10 perfects that they just give like a medal or pin or something?[/quote]

depends on ur state i guess...some states have like no one getting perfects, some have like 10 people :thumbup:

[quote="worthawholebean"]As far as I can tell, the plaques are actually for *regional* winners.[/quote]

[quote="Quickster94"]I'm not sure about the state plaque, searching on the forum leads me to many posts where people say they never got plaques as highest scorers, but I got one for top AMC 10-A in my state last yr (and some1 in the 10B beat me), I had a score of 120, which says a lot about my state :D  And, I can't find anything about state plaques on the AMC website, could some1 from the AMC clarify this?  or perhaps it is up to state discretion?

I'm also curious to what happens in the case of many many perfect scores (perhaps only the states small enough to not have that problem give plaques?)[/quote]

quickster94: yeah same here. i had the highest AMC10A score [130.5 ><] in WI, and i got a plaque. :// & someone in 10b beat me too i think.
but else they'd have to make like. 20 plaques for texas

aw man i made perfect amc10 but i only got a small pin. and it said school winner, so i suppose this isnt what y'all are talking about

Last year, I got a 114 on AMC 10 and got the school winner pin.  School Winner is not the same as Regional Winner :P


Given the equation \[ 4x^3 \plus{} 4x^2y \minus{} 15xy^2 \minus{} 18y^3 \minus{} 12x^2 \plus{} 6xy \plus{} 36y^2 \plus{} 5x \minus{} 10y \equal{} 0,\] find all positive integer solutions.

[quote="orl"]Given the equation
\[ 4x^3 \plus{} 4x^2y \minus{} 15xy^2 \minus{} 18y^3 \minus{} 12x^2 \plus{} 6xy \plus{} 36y^2 \plus{} 5x \minus{} 10y \equal{} 0,
\]
find all positive integer solutions.[/quote]

$ 4x^3 \plus{} 4x^2y \minus{} 15xy^2 \minus{} 18y^3 \minus{} 12x^2 \plus{} 6xy \plus{} 36y^2 \plus{} 5x \minus{} 10y \equal{}4x^{3}\minus{}8x^{2}y\plus{}12x^{2}y\minus{}24xy^{2}\plus{}9xy^{2}\minus{}18y^{3}\minus{}12x^{2}\plus{}24xy\minus{}18xy\plus{}36y^{2}\plus{}5x\minus{}10y\equal{}4x^{2}(x\minus{}2y)\plus{}12xy(x\minus{}2y)\plus{}9y^{2}(x\minus{}2y)\minus{}12x(x\minus{}2y)\minus{}18y(x\minus{}2y)\plus{}5(x\minus{}2y)\equal{}(x\minus{}2y)(4x^{2}\plus{}12xy\plus{}9y^{2}\minus{}12x\minus{}18y\plus{}5)\equal{}(x\minus{}2y)((2x\plus{}3y)^{2}\minus{}6(2x\plus{}3y)\plus{}5)\equal{}(x\minus{}2y)(2x\plus{}3y\minus{}1)(2x\plus{}3y\minus{}5)\equal{}0$

$ x\minus{}2y\equal{}0$--> $ x\equal{}2y$ --> $ (x;y)\equal{}(2a;a)$    ;   $ a\in Z$
$ 2x\plus{}3y\minus{}1\equal{}0$ --> $ x\equal{}\frac{1\minus{}3y}{2}$ ---> $ (x;y)\equal{}(\minus{}3b\minus{}1;2b\plus{}1)$  ;     $ b\in Z$
$ 2x\plus{}3y\minus{}5\equal{}0$ --> $ x\equal{}\frac{5\minus{}3y}{2}$ ---> $ (x;y)\equal{}(\minus{}3c\plus{}1;2c\plus{}1)$  ;     $ c\in Z$


set theory
advanced fields
advanced fields unsolved

(i apologize if it is stupid)
When we build theory to describe our real world, we assume it's completeness (absolute truthness). It is nothing but a logic convenience. As to math, we also assume arithemetic system (denoted by N() in the following text) is complete (actually all structures. that's part of the "semantic definition" of structure, i.e this is our strategy, we invent such concept to study). However is this strategy right? If it is right, then whenever we find a proposition P, we should tell its truthness no matter it can or cannot be proved by the present axioms. Is this the case? It seems not. Then what's the point to view N() as a completely determined "thing"? What will our attitude be like towards those propositions to whose truthness we have no "concrete resonse" to believe (some propositions implies the inconsistence of N() so we assume its falseness)? Is there or should there always be resonse (which we have not yet found) for us to prefer one of Q and -Q? If we will adopt both of it and its negation then what's point to view N() as a completely determined thing?

To state the above question in another way, lets consider all turing machine and imagin we are building a theory to describe it (mainly whether a machine on an instance halt or not). Our knowledge can be make presice by two r.e sets A and B, each contains some instances which we believe to halt (if it's in A) and un-halt （in B). Suppose we miss take a un-halt example u halt (i.e u$ \in$ A), and get A' and B' (note that B' contains not only u but also trivil changed "u", i.e ingeneral instances "implied" by "concrete reasons") however how will we find out we are wrong. If we take a halt instance as an un-halt one then by finite steps we know we are wrong. If $ u\in A$ implies (via some "right" rules) some halt instance to be un-halt we also know we are wrong……In general for some instances we have concrete reason to believe their "haltness" or "un-haltness" (by concrete reason we mean, except the above reason, for example if a machine is found to code "all machines" then we know it wont halt for all instances). But what about the other instances? Shall we persuing its truthness by finding other concrete reasons (quite ambiguous) since we have assume all instances must be halt or un-halt?

My personal view is we dont have to.  And there is no concrete reasons that can make us prefer one of Q and -Q. Because mathematic study everything "useful" but concrete. (logic is ultimately from reality, we assume the consistence of arithemetic which can not be essentially prooved.)

What on Earth do you mean by "complete"?

the theory describing a structure (induced by this structure) is complete if every proposition in this theory is either right or wrong

That is consistent.  Complete can have different meanings.  Usually It means that there are no truths that can not be proven or It says all truths.

1. Statement 1 is unprovable.

If 1 is proven then it is inconsistent.  If 1 is not proven then 1 is true.  You now have a true statement that can not be proven. If it is consistent then it is incomplete.

1.  A can not prove 1.

A can not prove 1 but another system of axioms B can.  There will be another truth that B can not prove.

I still don't really understand your question.  The question of whether a particular theory is consistent has to be asked within the context of another theory in which you perform the proof of consistency (if it is possible).  It doesn't have any meaning out of context.

I'm sorry for the ambiguity in my expression.

It seems that before 20 centry we don't distinguish between structure and the logic language describing them. (And we found "finite" language cannot grasp some structure.) But now we do, and we assume those structures are nature objects we study and thus we assume (implicitly) that structures are (should be) completely determined. Or in another word we call those things structures and assume they are completely determined. And thus Th(M) must be a complete theory. So, when you say if a fact "L" can not be proved in a system A (logic language or set theory language or so) but it can be proved in B, you are actually assuming we have concrete reason (induced by this structure) to believe the statement "L" (whatever it is, i.e any statement induced by this structure) must be either true or false.

You may immediately argue that this is of course because of the definition of structure. However let's consider the theory we build to describe turing machine (as in the first post paragraph 2). You would think it absolutly right to assume that a turing machine will either halt or un-halt on an instance. However see the argument in that paragraph. (I'm not saying this is false but questioning on a consequence of this "assumption")

We never grasp some completely determined structures (whatever we use, they are all essentially "finite language"). It is just a convenience, a strategy (to assume the structure we study is determined), nothing's wrong. But, when we met a statement cannot be proved in the present system (all of them), shall we keeping finding for those concrete reasons that can convince us of the truth (falseness) of the statement or shall we develope both systems in which the statement is true and false respectively. sometimes we can indeed find those concrete reasons, for example when this statement imply the inconsistence of arithemetic, but this isn't always the case is it?

well, I think my question can be viewed as a doubt about Platonism. The narrow definition of Platonism is the determinacy (or in phrase in the question, completeness) of arithmetic, and the generalized Platonism is the determinacy of set theory. I don't mean that I do doubt about the determinacy of arithmetic, it is crazy, but I'm wondering the reason that all people so strongly believe this------the determinacy of arithmetic.

 I thought it was a mere assumption (logical convenience). i.e is it possilble that some day we met an arithmetic proposition that we couldn't find evidence to believe either its truth or false. Or can we even answer this question even if it is possible?


this is nguyen anh tuan 's inequaliti  :lol:

no one can ?
 :mad:
realy; it ís hard problem

$\frac{1}{48}$ is betest but i cannot show this

Sorry, but $\frac{1}{48}$ is not the best constant . I have solved it already and if yu want, i will post my solution :)

@mathboy: mình nghĩ bạn nên thay đổi cách viết đi, đừng nên viết "no one can" hay cái gì đó, vì làm như thế rất làm mất lòng người khác và vừa rồi bạn đã làm cho Dj ... :( Mình chỉ góp ý thôi chứ không có ý gì khác đâu! :)

The following problem was used in the shortlist problem in VMEO3 but wasn't used in offical exam:

Find the maximum real value $k>0$ such that $\forall x,y,z>0: xy+yz+zx+xyz=4$ then we have :

$\frac{x+y+z}{xy+yz+zx}\le 1+\frac{(x-y)^{2}+(y-z)^{2}+(z-x)^{2}}{k}$

And you can't share this problem,Can

@: Anh nói vậy thôi chứ các em cứ trao đổi về bài này thỏai mái đằng nào hè nàydiễn đàn chả làm cuốn tập hợp tất cả các bài và lời giải của shortlist VMEO

I have a problem :
for : $xy+yz+zx+xyz=4$
Show that :
          $\frac{x+y+z}{xy+yz+zx}\leq 1+\frac{(x-y)^{2}+(y-z)^{2}+(z-x)^{2}}{16}$
but it is easy
@nthd:  em thấy bài này trong một quyển vở của anh Lê Đình Huy

[quote=mathboy_cntHDVN]problem1:
$ab+bc+ca+abc=4$
prove that :
$\frac{a+b+c}{ab+bc+ca}\leq1+\frac{1}{48}\sum(a-b)^{2}$[/quote]

The minimal $k$ for which the inequality
$$\frac{a+b+c}{ab+ac+bc}\leq1+k(a^2+b^2+c^2-ab-ac-bc)$$
is true for all non-negative $a$, $b$ and $c$ such that $ab+ac+bc+abc=4$ is
$$k=\max_{x\geq0}\frac{x(x+1)^2}{4(x+2)^2(x^2+x+1)}=0.04048...$$


prove that if $\gcd(a,b)=1$ and $ab=k^{2}$ for some integer $k$, then $a=r^{2}$ for some integer $r$ and $b=s^{2}$ for some integer $s$.

[hide="Challenge"] Can you do this problem without talking about prime factorizations at [b]all[/b]? [/hide]

suppose that p is a prime

then p|k^2 implies p|k, or p|k by definition of a prime, so obviously, p^2|k^2

p can only divide one of them...but it that one twice, this is true for all primes

ok i failed

lol

anyway, is there something with division algorithm...i ono

[u]Challenge[/u]

Induction. For all x<=a,y<=b, x and y not equal to a and b simultaneously, assume that if gcd(x,y)=1, xy=m^2, then x and y are both squares.

First, a cannot equal b, since they would both equal k, and gcd(a,b)=k.

Now, one of a,b must be representable as x(n^2) for some largest n>1, and x. To show this, it is obvious that any integer can be represented in this form, with n>=1, by just setting n=1.

-->But both a and b can't have n=1.<--

Therefore, wlog, a=x(n^2) b=y
x(n^2)y=k^2
xy=(k/n)^2

x,y are both squares.

If anyone can prove the statement inside the arrows, the challenge is done.

[hide]Assume the contrary. WLOG then $a$ has some prime factor with a odd degree. The product $ab$ also has this prime factor with an odd degree since $\gcd(a,b)=1$. But perfect squares have only prime factors with even degrees. Contradiction.[/hide]

[quote="rofler"]t is obvious that any integer can be represented in this form, with n>=1, by just setting n=1.

-->But both a and b can't have n=1.<--
...
If anyone can prove the statement inside the arrows, the challenge is done.[/quote]

Unfortunately, the statement inside the arrows is the content of the problem.

I'm not sure my challenge is possible ...  :|

:mad:

...

lol

i spent just a bit too long on that...lol

This property doesn't hold in most rings without unique factorisation.
Just an example: $(2+2\sqrt{-5})(3-3\sqrt{-5})=36=6^{2}$, but the two factors on the left are no squares and are coprime (in the sense of $\gcd(a,b)=1$).

However, $ab=c^{n}$ and $\gcd(a,b)=1$ implies $a=x^{n}, b=y^{n}$ in some number rings if the class number is coprime to $n$ (this is up to units: $(-1) \cdot (-1) = 1$ in integers, but $-1$ is not a square).


[quote="SplashD"][quote="nealwu"]How many ways can you cover a unit square with two smaller squares (with side length strictly less then 1)?[/quote]

I don't that's possible, there's like this whole proof.

Hmm, I wonder who you're trying to impersonate?[/quote]
Yeah, it's impossible...

[quote="nebula42"][quote="b-flat"]I'm pretty sure this problem is impossible...

Err, why the username nealwu when he already has a different user name?[/quote]
That's not me.  :wink:  :roll: That's my classmate Delong. He can't think of any good names or something like that, so he uses mine, which is starting to get pretty annoying.  :wink:
Anyway, about the problem, I think the proof had something to do with showing that you can't cover the full perimeter of the larger square, but proofs aren't MathCounts level, so I'll move it to Intermediate.[/quote]

Wait what? You're Neal Wu?
When did this happen?

Okay, I think he should just change the username to something else.

I think it is impossible.

[quote="srulikbd"]it's obvious it's impossible but how do you prove it?[/quote]

By sheer obviousness.

[quote="10000th User"]By contradiction maybe? We could assume the smaller squares don't overlap and fall entirely within the unit square.

[hide="My guess without overlap"]Suppose we can cover the unit square with two squares whose sides are r and s. Since the whole unit square is covered,

$r^{2}+s^{2}=1$

But, $(r+s)^{2}=1$.

$2rs=0$, contradiction...[/hide]

Here is where I have some doubts on my argument but for some reason it 'looks OK' when we try to cover it with squares that can overlap.

[hide="My guess with overlap"]This time not by contradiction. Same notation except that r and s must be at least 1/2 for the squares to overlap while covering the square.

The overlapping area is a square and its area is at least $(r+s-1)^{2}<1$. Then the area covered by the squares without the overlap should be $(r^{2}+s^{2})-(r+s-1)^{2}=1-2(1-r)(1-s)$. Since the side lengths are strictly less than 1, $1-2(1-r)(1-s)<1$.[/hide][/quote]

I'd say this is pretty good, however, for the part $1-2(1-r)(1-s)<1$, if both r and s are greater than one, it is also true, but that isn't possible in the problem.

Also, where you say "the area covered by the squares without the overlap should be” that makes it sound like you are not including the overlapped area at all (yet your equation includes it).  A better wording might be "the area of the unit square that is covered” or something similar so you know that you are including the overlap (but just once).

On the first one, you could also note that $s+r<1$ is possible, but it would also not be possible to even have a total area of one.  The proof would go something like:

$s+r<1$
$(s+r)^{2}<1$
$(s+r)^{2}-2sr<1$
$s^{2}+r^{2}<1$

None of this is really important, but it would just help to make a more thorough proof  :)

[quote="blahblahblah"]Two smaller squares can definitely cover all four corners of a larger square (think of the diagonals!)[/quote]
Hmm... I never thought of that.  :oops:
So does the question mean that the smaller squares have to be completely inside the larger square or not?

[hide="problem"]
Are the smaller squares limited to the inside of the unit square?  Not that it should make a difference but that possibility still needs to be taken into account for a rigorous proof.
[/hide]

Yes, of course the squares can be outside.  So for example, one of the smaller squares can cover two adjacent sides with its diagonal and the other square cover the other 2.
So far, the attempted proofs are kind of "blatantly false" or "proof by optimism"-T2.  This problem is actually kind of hard (Olympiad level).  It's from MOP Test 1 Problem 3.  It has no trivial solution (Proof by sheer obviousness=bad=0), but the idea of proof by perimeter is VERY correct.
[b]P.S. Also, Delong challenges Neal to find an elegant solution[/b] :rotfl:

It is possible if and only if there is a square that has area of the original square. You have a 5*5 square, and you have a 5*5 and a 3*3 to covre it. Use the 3*3, and then the 5*5 and voila! If you mean not ovrlapping, it is impossible.

Dang, but they are overlapping.  I think got a 0 for this problem.  Come to think of it, I think I got a 0 on that test.  Dang.

But unlike the rest of you, I know how to do it now! HAHAhAha *falls to the ground and starts twitching*

[quote="ychwyo"]This problem is actually kind of hard (Olympiad level).  It's from MOP Test 1 Problem 3.  It has no trivial solution [/quote]

Shouldn't be in Intermediate, then.  Someone move?

[hide="Hint on the proof by perimeter"] What is the maximum amount of the perimeter that a square with side length strictly less than $1$ can cover? [/hide]

[quote="10000th User"]By contradiction maybe? We could assume the smaller squares don't overlap and fall entirely within the unit square.

[hide="My guess without overlap"]Suppose we can cover the unit square with two squares whose sides are r and s. Since the whole unit square is covered,

$r^{2}+s^{2}=1$

But, $(r+s)^{2}=1$.

$2rs=0$, contradiction...[/hide]

Here is where I have some doubts on my argument but for some reason it 'looks OK' when we try to cover it with squares that can overlap.

[hide="My guess with overlap"]This time not by contradiction. Same notation except that r and s must be at least 1/2 for the squares to overlap while covering the square.

The overlapping area is a square and its area is at least $(r+s-1)^{2}<1$. Then the area covered by the squares without the overlap should be $(r^{2}+s^{2})-(r+s-1)^{2}=1-2(1-r)(1-s)$. Since the side lengths are strictly less than 1, $1-2(1-r)(1-s)<1$.[/hide][/quote]
is your first proof is rigorous? cause maybe on one side they are together but what about the others sides?
[hide="my solution"]there are 2 cases one square is overlap the vertex of the great square, or the edge: 1)the remaining shape canno't be a square because it have 6 sides. 2)the remaining shape canno't be a square cause it have 8 sides. is it ok?[/hide]

10000th user:  The overlapping area is not necessarily a square.

srulikbd:  The two squares only have to [b]cover[/b] the square of side length $1$.  It doesn't matter how many sides their union has.

I didn't talked about their union...I say what the shape that one square leave, which isn't a quadrilateral, and that why it of course canno't be a square...
[quote="nebula42"][quote="ehehheehee"]Neal said something about not being able to cover the entire perimeter...

[hide]To cover the entire entire perimeter you must cover one side of the unit square

Since the two squares we are using to cover that one side have shorter sidelengths than the side we need to lay them so that they are adjacent and have their boundries exactly on the side of the unit square.

Since their side lenghts are shorter than 1 the squares would not be able to touch the opposite side.  So it is impossible to completely cover one side and the opposite side at the same time.[/hide][/quote]
Hmm... I think we can even say that you can't cover the unit square with three squares of side lengths strictly less than $1$.
[hide="Proof"]Since each of the smaller squares has side length less than $1$, we can easily see that one of the squares can only cover one corner. Thus with three squares we can only cover three of the four corners, so it is impossible to completely cover the unit square with three strictly smaller squares.[/hide]
We can even generalize the problem to this:
If we have a unit $n$-gon, then the minimum number of strictly smaller $n$-gons needed to completely cover the larger $n$-gon is $n$. (The proof is similar to the proof for the unit square.)

Haha, Delong, I owned your problem.  :P[/quote]
do you sure? how do you cover equilateral triangle?(with equilateral triangles) I think you need 4 triangles...

srulikbd, what t0rajir0u was saying is that "cover" does not mean "cover exactly with no overlap."  The fact that taking away one square doesn't leave a quadrilateral is irrelevant -- the question is whether taking away one square leaves something that would fit inside another square.  Similarly, one can cover a triangle with three triangles shrunk by a factor of 2/3 simply by sticking one in each corner.  This creates overlap, of course, but that's not a problem.  It also doesn't cover any points outside the triangle, but that wouldn't be a problem, either.


inequalities
induction
graph theory
combinatorics proposed
combinatorics

Let's consider the complete graph whose vertices are the persons. Color red the edge $ A \minus{} B$ if $ A$ and $ B$ are knowing each other, and blue otherwise.
Thus, we are looking for the maximum number of vertices of such a graph knowing that there is no red $ K_n$ and no blue $ K_3$.
This maximum is $ R(n,3) \minus{} 1$, where $ R(.,.)$ is the Ramsey number.

Therefore, the problem asks to prove that $ R(n,3) \leq \frac {n(n \plus{} 1)}2$.

This upper bound follows easily from the well-known inequality $ R(a,b) \leq R(a \minus{} 1,b) \plus{} R(a,b \minus{} 1)$, which gives $ R(a,b) \leq \binom {a \plus{} b \minus{} 2}{b \minus{} 1}.$

Pierre.

Written in simple words.
We will use induction on n and thus assume it is true for n-1.
Choose a person. he will be friends with people in A and not with people in B. it is clear that then every two persons in B are friends. Thus |B| is at most n-1. Now because our person knows every one in A by the induction hypothesis A consists of at most $ \frac{(n\minus{}2)(n\plus{}1)}{2}$ people. thus we get a maximum of $ \frac{(n\minus{}2)(n\plus{}1)}{2}\plus{}1\plus{}(n\minus{}1)\equal{}\frac{(n\minus{}1)(n\plus{}2)}{2}$ for our set.

[quote=pohoatza]Let $ n$ be a nonzero positive integer. A set of persons is called a $ n$-balanced set if in any subset of $ 3$ persons there exists at least two which know each other and in each subset of $ n$ persons there are two which don't know each other. Prove that a $ n$-balanced set has at most $ (n \minus{} 1)(n \plus{} 2)/2$ persons.[/quote]

Alternatively, work with the complementary graph $G^{*}$; which is triangle-free and has no independent set of size $n$.
We prove that $|V(G)|<\binom{n+1}{2}$ by induction on $n \ge 2$; base case being obvious. The point is to delete no more than $n$ vertices from $G^{*}$ to reduce to the case where the induced sub-graph no longer has independent sets of size $n-1$ and apply induction hypothesis.

Pick the vertex $v$ with maximal degree. Let $\{u_1, \dots, u_k\}$ be the neighbours of $v$. Observe that $N(v)$ must be an independent set else $G^{*}$ has a triangle. Thus, $k \le n-1$. Now delete $v$ along with $u_1, \dots, u_k$ from $G^{*}$ and suppose we still have an $(n-1)$ independent set $S$ in the induced subgraph. Then $v$ has no neighbours in $S$ (since all of them were deleted prior to constructing $S$). However, appending $v$ to $S$ yields an $n$-independent set in the original graph; a contradiction! Hence the induction hypothesis works!

Ramsey kills it...... Another problem is posted in Bosnia tst

Ramsey kills it
$R(r,s)\le R(r–1,s)+R(r,s–1)$
$R(r,s)\leq \binom {r+s–2}{r–1}$
Here $(r, s) =(n, 3)$
$R(r,s)\leq  \binom {n+1}{n–1}$
$R(n,3)\le \frac{n(n+1)}{2}$ and thus...

@Paragdey12 This is highly of topic, but seeing your recent posts in HSO forum (the above post, [url=http://artofproblemsolving.com/community/c6h201580p9797538] #1 [/url],  [url=http://artofproblemsolving.com/community/c6h1525174p9779108] #2 [/url], [url=http://artofproblemsolving.com/community/c6h31799p9798617] #3 [/url] for example), I have no idea how what you're writing is either helpful or provides a full solution (or atleast sketch of a solution) - it seems (for example check #3) that you're simply writing random theorems related to the problem.

Please don't spam threads in HSO with useless posts like this.

Note that $\frac{(k-1)(k+2)}{2}+1=\binom{k+1}{2}$. Thus, the contrapositive of the problem is asking us to show that a triangle free graph with at least $\binom{k+1}{2}$ has an independent size of at least $k$. This follows directly from Ramsey's theorem applied to $R(3,k)$, but we'll outline a proof here for completeness.

We proceed by induction on $k$ with the base case of $k=1$ being obvious. Now suppose the problem is true for $k-1$, and suppose we have a graph $G$ on $\binom{k+1}{2}$ vertices that is triangle free. Let $v$ be any vertex. If the degree is at least $k$, then by the triangle free, the neighbors of $v$ form an independent set of size $k$. So WLOG, suppose that the degree of $v$ is at most $k-1$. Thus, there are at least $\binom{k+1}{2}-1-(k-1)=\binom{k}{2}$ vertices not connected to $v$, and by the inductive hypothesis, they have a $(k-1)$-independent set. Adding in $v$ solves the problem.

Create a graph in which two persons are connected if they don't know each other and replace $n$ with $k$.

Induct on $k$. The result is trivial at $k=1$.

For the induction step, suppose otherwise. By induction hypothesis, there exists an independent set $S$ of size $k-1$. As $\frac{(k-1)(k+2)}{2}-(k-1)=\binom{k}{2}$, there are more than $\binom{k}{2}$ non-elements of $S$. Let the set of non-elements of $S$ be $T$ and for each $v\in S$, let $T_v$ be the set of neighbors of $v$ in $T$. Observe that each element of $T$ must be in some $T_v$, as otherwise the result is clearly false. Consider a particular $T_v$. Observe that $T_v$ is an independent set because there are no triangles. If there are two or more vertices $u$ in $T_v$ such that there is no $v\ne x\in S$ with $u\in T_x$, we could replace $v$ with those two vertices to generate an independent set of size $k$, absurd. Thus over all vertices in $T$, at most $k-1$ (the number of $T_v$s) do not have edges to multiple elements of $S$. This implies the average cardinality of a $T_v$ is greater than
\[\frac{2\cdot (\binom{k}{2}-(k-1))+(k-1)}{k-1} = k-2+1=k-1,\]
so some $T_v$ is an independent set with size larger than $k-1$, contradiction.

Phrase this in graph theory terms, replace $n$ with $k$.

Assume that $k$ is minimal, so there exists an independent set of size $k-1$, let that set be $S$. For every $v\in\S$, there is at most one vertex $w$ adjacent to $v$ such that no other vertex in the independent set is adjacent to $w$. This is because if two such vertices $w, x$ existed, then get rid of $v$ and add in $w,x$ into the independent set. This means there exists at most $k-1$ vertices that are adjacent to exactly one vertex in the independent set, so every other vertex is adjacent to at least $2$.

Observe that the degree of each vertex $v$ is at most $k-1$, because if it was greater, then since the graph is triangle free, we can choose all the vertices adjacent to $v$ in our independent set, for an independent set of size $k$. Let $n$ be the number of vertices. The number of edges adjacent to a vertex in $S$ is at least $k-1 + 2(n-(k-1)-(k-1))$, since there are at most $k-1$ vertices not in the independent set adjacent to exactly one $v\in S$, and the other $n-(k-1)-(k-1)$ vertices not in the independent set are adjacent to at least $2$ $v\in S$. However, there are at most $(k-1)^{2}$ edges with one vertex in $S$, so
\[(k-1)^{2} \geq (k-1) + 2(n-2(k-1)) \Rightarrow 2n\leq (k-1)((k-1) - 1 + 4) \Rightarrow n\leq \frac{(k-1)(k+2)}{2}\]
This is our desired result.

Rephrase the question as prove that when we have at least $\frac{n(n+1)}{2}$ persons it is impossible to have a $n$-balanced set

We shall proceed by induction on $n$

Base case $n=2$

Observe that $n \leq 2$ or else we will immediately form a triangle. At the same time , $\frac{(2-1)(2+2)}{2}=2$

Hence our base case holds

Now, suppose that it holds for some $n=y$ , consider the $n=y+1$ case

Consider the person with the maximal degree, let it be $Zhong Xina$

If $Zhong Xina$ has $\geq y+1$ neighbours, then we are done since the set of neighbours must form an independent set or else we have a triangle.

So, suppose that $Zhong Xina$ has $\leq y$ neighbours

This means that we have at least  $\frac{(y+1)(y+2)}{2}-y-1=\frac{(y)(y+1)}{2} $

By induction hypothesis, this means that in the $\frac{(y)(y+1)}{2}$ vertices, we have at least one $y$ sized independent set

Note that the vertices in this $y$ sized independent set all do not share any edge with $Zhong Xina$ since it is not its neighbour

Hence the vertices and $Zhong Xina$ form a $y+1$ sized independent set

Hence, by  induction, our claim is true

Is ZhongXina a reference to the famous AoPSer JohnCena

We prove the contrapositive statement by induction on $k$: Every triangle-free graph with at least $\binom {k+1}{2}$ vertices has an independent set with size at least $k$. If some vertex has degree at least $k$, we are done. Otherwise, remove some vertex and its neighbors and induct down (we can now add this vertex to the independent set from the IH).

It suffices to show that any graph with at least ${k+1 \choose 2}$ edges that is triangle-free has an independent set of size $k$.

We proceed by induction; assume for the sake of contradiction the result is false. Note that every vertex in the graph has degree at most $k-1$, else the set of all neighbors of that vertex would constitute such a set. But now delete that vertex and all its neighbors; we are left with at least ${k \choose 2}$ edges, so the remaining graph has a $k-1$-independent set.

On the other hand, including that original vertex creates a desired set, contradiction.


calculus
integration
geometry
trigonometry
function
calculus computations

It means that there is no "net" area.  If the bounds on the definite integral are the same, this can be seen because there is clearly no area under a point.

If one has something like $\int_{0}^{2\pi}\sin{x}dx$, one can see that the area between $\sin{x}$ and the y-axis "cancels" out because the "positive" area (that above the y-axis) is equal in magnitude to the "negative" area (that below the y-axis).

That's kind of how I think about it.  Was that helpful?

err..

sorry for the poor question

lets me rewrite

if we have a function $f$  if we get $f' =0$  we can get minimum, and maximum or another interesting points.

if we get  $f'' =0$  we can study the curvature of the function etc..

but if the have te integral iqual to zero,, how can it help in the graph?

in nothing?!

It means that the average value of the function over the interval of integration is zero. And if the function is continuous, it means that the function must have at least one zero within the interval of integration.


How do you get the inverse of
f(x) =  4 + 2x + e^x ?

[quote]How do you get the inverse of
$ f(x)\equal{}4\plus{}2x\plus{}e^{x}$ ?[/quote]

this is hard for me but i can give a hint
try to make $ x$ the subject or in other words find $ x$ in terms of $ q$ in this equation first $ 2x\plus{}e^{x}\equal{}q$ then you can find the inverse easily.

[quote="Icy"]Actually, just replace $ x$ and $ y$, then solve for $ y$.

For example,
$ y \equal{} 4\plus{}2x\plus{}e^{x}$ becomes $ x \equal{} 4\plus{}2y\plus{}e^{y}$.

Solving for $ y$, in this case, is...difficult. You could leave it in its implicit form I guess...[/quote]
I quite agree, I got as far as $ \ln{\left(x\minus{}4\minus{}y\right)}\equal{}y$ and couldn't go from there... :(

Where did you get this problem?  The function you gave does not have an inverse expressible in terms of elementary functions.

t0rajir0u,

It's a regular "webwork" assignment. (In this case it's a practice assignment for no marks). I have no clue how to find a numerical answer for f^-1 (7) = 0.

[quote="Jean-Louis"] f^-1 (7) = 0.[/quote] What do you mean?

[quote="Jean-Louis"]I have no clue how to find a [b]numerical[/b] answer for f^-1 (7) = 0.[/quote]

This is a hint, and it is a hint telling you not to use algebra.  Rather, graph $ y \equal{} f(x)$ and $ y \equal{} 7$ and find their point of intersection with the help of a calculator.

t0rajir0u,

we're apparently not supposed to use calculators for the assignment.
I plugged in f^-1 (7) = 0. It turns out it's the correct answer.

f(x) = 4 + 2x + 3e^x
7 - 4 - 2x - 3e^x = 0
3 = 2x + 3e^x
3 = 2(x + 3/2 e^x)
ln(3) = ln(2) + ln(x + 3/2 e^x)
e^ln(2/3) = x + 3/2 e^x
3/2 = x + 3/2 e^x
So, x = 0, as (0) + 3/2 e^0 = 3/2

Does this make sense?

[quote="Jean-Louis"]I plugged in f^-1 (7) = 0. It turns out it's the correct answer.

f(x) = 4 + 2x + 3e^x[/quote]

This is different from the function you posted.  What is the problem?  If it's to verify that $ f^{\minus{}1}(7) \equal{} 0$, this is obvious from the fact that $ f(0) \equal{} 7$ and $ f(x)$ is a sum of monotonic functions.


Let $ a,b,c$ are positive real numbers such that $ a\plus{}b\plus{}c \equal{} 3$. Prove that:
\[ \frac{1}{\sqrt{a^{2}\minus{}3a\plus{}3}}\plus{}\frac{1}{\sqrt{b^{2}\minus{}3b\plus{}3}}\plus{}\frac{1}{\sqrt{c^{2}\minus{}3c\plus{}3}}\le 3\]

Yes, we can. It is some trip when proving tangent line inequalities, from my work with mr PvThuan. I'll present the very sort solution.
Without loss of generality, we assume that $ x\le y\le z$ and let
$ f(x) \equal{}\frac{1}{\sqrt{x^{2}\minus{}3x\plus{}3}}$
By taking derivate, we have $ f$ is increasing on $ [0,1.5]$ and decreasing on $ [1.5,3]$ so $ \max_{x\in[0,3]}f(x) \equal{} f(1.5)$.
[b]1)Case[/b] $ min\{x,y,z\}\ge\frac{1}{2}(\minus{}1\plus{}\sqrt{5})$ is trivial by getting tangent line

[b]2)Case[/b] $ x \equal{} min\{x,y,z\}<\frac{1}{2}(\minus{}1\plus{}\sqrt{5})$, and we have

  Case $ z >\frac{1}{4}\left(6\plus{}\sqrt{6}\right)$ is trivial
  Case $ z <\frac{1}{4}\left(6\minus{}\sqrt{6}\right)$ then $ y >\frac{1}{4}\left(6\plus{}\sqrt{6}\right)$
  And now, by Jensen inequality, $ f(y)\plus{}f(z)\le 2f\left(\frac{y\plus{}z}{2}\right)$, then, the rest is easy because
  \[ \max_{x\in[0,3]}f(x)\plus{}2f\left(\frac{3\minus{}x}{2}\right)\equal{}3f(1)\equal{}3\]

P/s:Sorry but I'll delete this post after 2 day so please don't post anything more here. I want stop posting in Mathlinks with number of post is 365. Thanks

very nice proof, rocky and pvthuan. My solution and your are rather like each other. I also using Underfined Coefficient Technique and Jensen's inequality.
Have any proof for it ?!

Can any mod delete all my posts in this topic. Due to zaizai_hoang's post, I can't delete mine. Thanks

P/s: This solution is mine only


Let $m\in N$ and $E(x,y,m)=(\frac{72}x)^m+(\frac{72}y)^m-x^m-y^m$, where $x$ and $y$ are positive divisors of 72.
a) Prove that there exist infinitely many natural numbers $m$ so, that 2005 divides $E(3,12,m)$ and $E(9,6,m)$.
b) Find the smallest positive integer number $m_0$ so, that 2005 divides $E(3,12,m_0)$ and $E(9,6,m_0)$.

a) $2005$ divides those for $m=400k$ where $k \in \mathbb{N}$
   just count these $E...$ and use Fermat's Little Theorem

for problem b) you will use the lemma:
             $(a^{m}-b^{m},a^{n}-b^{n})=a^{(m,n)}-b^{(m,n)}$  (if $(a,b)=1$)
so we get $m=400*q+r (1 \leq r \leq 400)$
so $E(3,12,m)=24^m+6^m-3^m-12^m \equiv 24^r+6^r-3^r-12^r(401)$
     $E(9,6,m)=8^m+12^m-9^m-6^m  \equiv 8^r+12^r-9^r-6^r(401)$
     so that $24^r+8^r-3^r-9^r \equiv 0(401) \rightarrow (3^r+1)(8^r-3^r) \equiv 0(p)$
 so use the lemma we easy to show that $r=400$  so $m=400*q$
we find all r such that $3^r+1 \equiv 0(401)$
easily to prove that $r=200$ is the unique
so $m=400*q+200$
now it easy to see the minimum $m$.
if it has some mistake ,please correct my mistake.Thanks :?  :?


Let's convert words into symbols first.
$ \text{is}$ means $ \equal{}$, and $ \text{of}$ means $ \times$.
Call our "what number" $ x$.  We can now set up the equation:
\[ 84\equal{}40\%\times x\implies 84\equal{}.4x\implies x\equal{}\boxed{210}\]


geometry
geometric transformation
reflection

$ P$ and $ Q$ are reflections of $ (2,\minus{}3)$ across the $ x$-axis and $ y$-axis, respectively. Find the length of $ \overline{PQ}$ in simplest radical form.

Point $ P$ is $ (2,3)$, while point $ Q$ is $ (\minus{}2,\minus{}3)$.  Use the distance formula.

$ \sqrt{(2\minus{}(\minus{}2))^2 \plus{} (3\minus{}(\minus{}3))^2)} \equal{} d$

$ \sqrt{16 \plus{} 36} \equal{} d$

$ \sqrt{52} \equal{} d$

$ d \equal{} \boxed{2\sqrt{13}}$


Prove that
(nC0)+(nC1)+(nC2)+................+(nCn)=2^n

***it is said to work with the right side..)

Notice that by the Binomial theorem,

$ 2^n \equal{} (1 \plus{} 1)^n \equal{} \binom{n}{0}1^n \plus{} \binom{n}{1}1^{n \minus{} 1}*1 \plus{} ... \plus{} \binom{n}{n \minus{} 1}1* 1^{n \minus{} 1} \plus{} \binom{n}{n}1^n \equal{} \binom{n}{0} \plus{} \binom{n}{1} \plus{} ... \plus{} \binom{n}{n \minus{} 1} \plus{} \binom{n}{n}$

[hide="More fun proof"]
Consider a coin being flipped $ n$ times. We'll count the number of possible outcomes.
Let's do some casework. If there are no heads and all tails, we have $ \binom{n}{0}$ ways for that to happen (let's say we're choosing the number of heads out of the total number of trials).
If there's one head, $ \binom{n}{1}$.
If there's two, $ \binom{n}{2}$, and so on.

We can see that the number of outcomes is $ \sum_{k\equal{}0}^{n}\binom{n}{k}$.

But, of course, there being two outcomes at each point (heads or tails) this is really just $ 2\cdot 2\cdot 2\cdots\equal{}2^n$.

These count the same thing, so they're equal.
[/hide]


Can somebody help me with the achievement checklist question. I have attached the question as a file.

Calculus posts should go in the calculus forum.


Show that the n digits after the decimal point in $ (5 \plus{} \sqrt{26})^n$ are all equal.

[quote="mathophile593"]Subtract (5-sqrt(26))^n?[/quote]

Well, rather, subtract $ (\sqrt{26}\minus{}5)^n$. Then all the digits after the decimal point are like... 9s, because $ \sqrt{26}\minus{}5$ is so small (as in, $ \sqrt{26}\minus{}5 \equal{} \frac{1}{\sqrt{26}\plus{}5} < \frac{1}{10}$).

[quote="mathwizarddude"]Excuse me, but I'm a little lost by your solutions; how would this show that the n digits after the decimal point in $ (5 \plus{} \sqrt {26})^n$ are all equal?[/quote]

[hide="explaination"]$ K \equal{} (5 \plus{} \sqrt {26})^n \plus{} (5 \minus{} \sqrt {26})^n \equal{} 2\displaystyle\sum_{k \equal{} 0}^{\lfloor\dfrac{n}{2}\rfloor}5^{n \minus{} 2k}\cdot26^k$ which is an integer for all $ n\in\mathbb{N}$.

Since $ |(5 \minus{} \sqrt {26})^n| \equal{} \dfrac{1}{(5 \plus{} \sqrt {26})^n} < \dfrac{1}{10^n}$, $ (5 \plus{} \sqrt {26})^n$ is within $ \dfrac{1}{10^n}$ of $ K$.

So, $ K \minus{} \dfrac{1}{10^n} < (5 \plus{} \sqrt {26})^n < K \plus{} \dfrac{1}{10^n}$.

Or equivalently, $ (K \minus{} 1).\underbrace{999\ldots999}_{n \ 9's}0 < (5 \plus{} \sqrt {26})^n < K.\underbrace{000\ldots000}_{n \minus{} 1 \ 0's}1$.

Thus, the last $ n$ digits are ether all $ 0$'s or all $ 9$'s.[/hide]


when are they going to post the AIME answers?

No discussion of the AIME.  Please do not even mention the AIME until the AMC Forum is unlocked.


limit
real analysis
real analysis theorems

I can't find what something like $ \lim_{h \rightarrow 0}f(a \plus{} h) \equal{} L$ exactly mean.

Is it that we can find a $ \delta > 0$ for each $ \epsilon > 0$ so that if $ |h| < \delta$ then $ |f(a \plus{} h) \minus{} L| < \epsilon$ ?

But if we just now that $ a$ is an accumulation point then this definition is ofcourse wrong.

I think $ \lim_{h \rightarrow 0}f(a \plus{} h) \equal{} L$ is the same as $ \lim_{x \rightarrow a}f(x) \equal{} L$.


does anyone have ap calculus free-response tests older than 2002 and does anyone have the solutions guidelindes for 2002 and 2003?

[quote="1/(ln x)"][quote="texas137"]http://www.moems.org/calcSolo.htm[/quote]

Is this books just solutions? Do they include the free-response questions?[/quote]

Of course, both questions and solutions for both the AB and BC exams for the years indicated. But only the free response; no multiple choice.


Find the smallest possible value of $a + b + c$ where $a$, $b$ and $c$ are different positive integers that satisfy the following equation:
\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{7}{10}\, .\]

Could you explain your answer?  For example, where did you get those numbers?  How do you know they are smallest?

well this will be one of the problems where answer comes to my mind first and then try to make sure by proving.

[hide]
I took a glance at it and then I thought it would be better if a,b,and c are close to each other. So I plugged in numbers around 4,5 and I was able to get the answer less than a minute.

Proving)

let a<b<c

1<a<4 because 1/1 is greater than 7/10 and 1/4 + 1/5 + 1/6 is less than 7/10

if a was 3, b was 5, and c was 6, we can get 14 as a sum.

Now, we want to know if it is a minimum.

if a was 2, 1/b + 1/c = 2/10 -> c is at least 10

a(2) + b(bigger than 2) + c(bigger than 10) > 14

if a was 3, in order for b+c is less than 11 was b being 4 and c being 5 or 6, which clearly doesnt make 1/a + 1/b + 1/c adds up to 7/10

Q.E.D.
[/hide]

(Edited by moderator to put hide tags in correct position.)

Okay, good explanation.

Bonus question: What are all the (unordered) triples $\{a,b,c\}$ that satisfy the conditions?


LaTeX
limit
logarithms
algebra unsolved
algebra

I will help you latex:
let sequence $ a_n$ with $ \{  \begin{array}{l}    a_1 = 1 \\    a_{n+1}+1=a_n+\frac{1}{a_1+a_2+...+a_n}   \end{array}  $
Find $ \lim_{n\to \infty} \frac{a_n}{\sqrt{2\ln{n}}}$


geometry
geometric transformation
trigonometry
function
calculus
integration
calculus computations

Hey, i only have the basic concepts to intergration and these two questions really have me stumped.
1) Determine the area between;
y=sin2x   and    y=sin(x - pi/4)
between  x=0   and  x=pi

2)When no vertical translation is considered, the sine curve can be represented by the function y=a.sin.(bx+c)  If the area contained between the curve, the x asix and ordinates at x= 0 and x=pi is 4 units^2 investigate the values of a,b and c
thankyou in advance.Janet.

for 1 it's you evaluate this integral
\[\int_{0}^\pi | \sin{2x}-\sin{(x-pi/4)}|dx \]
modulus signs are there since you need the area not the integral. Study the interval and function sign , then split the integration interval, and remove the modulus sign. Then it's only basic integrals.


ok.... this is how i would do it...

[hide="Informal proof"]First.... We need to proof algebraically that
$a^3+b^3+c^3=(a+b+c)^3+3(a+b)(b+c)(c+a)$

In order for $a^3+b^3+c^3$ to be equal to $(a+b+c)^3$...
$3(a+b)(b+c)(c+a)$ must equal zero, and for that to happen...
either  $a=-b,b=-c,$ $or$ $c=-a$

Now in any of those there cases, there would be two numbers cancelling each other out... which means that only one number will be powered, making the statement true.For instance...

$a^3+b^3+c^3=(a+b+c)^3+3(a+b)(b+c)(c+a)$ If and only if $a=-b$ (could also be $b=-c$ or $c=-a$) so...
$(-b)^3+b^3+c^3=(-b+b+c)^3$  Lets cancel out terms...
$c^3=(c)^3$ which is of course true for any value of c.

And also $c^{2n+1}=(c)^{2n+1}$ will be true for any value of c and n.[/hide]

If someone can confirm this, I would really apreciate it; tell me about any mistake I've done if that is the case.

[quote="Jhonny Ruiz"]ok.... this is how i would do it...

First.... We need to proof algebraically that
$a^3+b^3+c^3=(a+b+c)^3+3(a+b)(b+c)(c+a)$

If someone can confirm this, I would really apreciate it; tell me about any mistake I've done if that is the case.[/quote]

uhhh...$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$

So $a+b=0$ or $b+c=0$...

If $b+c=0$, then  $b^{2n+1}=-c^{2n+1}$,hence  $a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}$

and also if $c+a=0$ or $a+b=0$


lol thats physics

F=magnitude of universal gravitation

anyways $ F \equal{} \frac {GMm}{r^2} \implies Fr^2 \equal{} GMm \implies \frac {Fr^2}{Gm} \equal{} M$

answer : $ \frac {Fr^2}{Gm}$

oh theres a mistake. G is multiplied to the whole thing.

...thats same thing

$ a\left(\frac {x}{y}\right) \equal{} \frac {ax}{y}$

i'm not exactly sure, but i don't think this belongs in classroom math
be sure that the difficulty of your problem is appropriate for the forum you're posting in

[quote]lol thats physics [/quote]
physics definitely does not belong in classroom math

Well, maybe it could. He's not asking anyone to do the physics, just solve for a variable. If you didn't know that this was physics, you wouldn't have said that. At most, this belongs in HSB.

okay this seems pretty easy
we have $ F  \equal{}  GMm/R^2$
now we make M as the subject of the whole xpression
so we get
$ FR^2/GM$

Yeah...See FantasyLover's post... :|  :|  :|

he explained the confusion of g! :maybe:

Yes, but look at his first one...Better yet, read all the posts before you start posting in threads.


Prove or disprove it.
Let $ A,B\in C^{n\times n}$, if $ rank(A)\equal{}n$, then there exist $ s,t \in R$ such that $ det[(sA^*\plus{}tB^*)(tA\plus{}sB)]\neq 0$.

That seems very similar to what you posted here, don't you think?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259891


conics
ellipse
parameterization
greatest common divisor
Diophantine Equations

Find all pairs $(x, y)$ of positive rational numbers such that $x^{2}+3y^{2}=1$.

I observe that $O(1,0)$ is such a rational pair. I now consider this as an ellipse in cartesian 2-space. If a point $A(x,y)$ on it is rational, then $y/x-1$ will be rational (unless x=1, but this case has already been addressed), so the line $AO$ has rational gradient. Conversely, if I take any line $y=mx-m$ through $O$ with $m\in\mathbb{Q}$, it meets the ellipse again at $(m^{2}+1)x^{2}+2m^{2}x+m^{2}-1=0$ where $x=\frac{-2m^{2}\pm \sqrt{4m^{4}-4(m^{2}+1)(m^{2}-1)}}{2(m^{2}+1)}=\frac{-m^{2}\pm 1}{m^{2}+1}$, where $y=mx-m=\frac{-2m^{3}-m \pm m}{m^{2}+1}$ which is clearly a rational point.

By arguing both ways I have shown that all rational points on the given ellipse are of the form above, so the solution is \[(x,y)=\left(\frac{-m^{2}\pm 1}{m^{2}+1},\frac{-2m^{3}-m \pm m}{m^{2}+1}\right)\ \ \forall m \in \mathbb{Q}\]

[quote="Moonmathpi496"]Nice solution!:)
if they asked to find int solution.The problem would be 100 times easier...But actually i dont know how to find rational solution.
How to do it?Using cartesian geometry?Or is there any general way?
After all,how can i learn it?

Thank u
Moon[/quote]

In fact, there exists an elementary approach making use of no analytic geometry.
Let $ (x, y)$ be a solution to our equation. If some component is zero, we ascertain $ (x, y) \equal{} (\pm 1, 0)$. Of course that $ (\pm x, \pm y)$ are also solutions, so let's assume now $ x, y > 0$. Then write $ x \equal{} \frac{p}{q}$, $ y \equal{} \frac{r}{s}$ with $ p, q, r, s \in \mathbb{N}$ and $ (p, q)$, $ (r, s)$ coprime pairs.
We rewrite the equation as $ p^2s^2 \plus{} 3r^2q^2 \equal{} q^2s^2$. Notice that $ q^2$ divides $ p^2s^2$ is implying $ q^2 | s^2$, and then $ s^2 | 3r^2q^2$ yields $ s^2 | 3q^2$. So $ 3q^2 \equal{} ms^2 \equal{} mnq^2$, $ \{m, n\} \subset \mathbb{N}$, hence either $ s^2 \equal{} 3q^2 \Rightarrow \sqrt{3} \equal{} \frac{s}{q}$, known to be false, or $ q \equal{} s$. It has just been inferred that the equation is reduced to $ p^2 \plus{} 3r^2 \equal{} q^2$.
The methods from now on are varied, however I prefer the idea to factor as $ (p \minus{} r)(p \plus{} r) \equal{} (q \minus{} 2r)(q \plus{} 2r)$. By the Four-Number Theorem, there exists an unique quadruple of integers $ (a, b, c, d)$ such that $ a > 0$, $ b, d$ are of greatest common divisor $ 1$ and $ p \minus{} r \equal{} ab$, $ p \plus{} r \equal{} cd$, $ q \minus{} 2r \equal{} ad$ and $ q \plus{} 2r \equal{} bc$ hold. Thus $ p \equal{} \frac{cd \plus{} ab}{2}$, $ r \equal{} \frac{cd \minus{} ab}{2}$, $ q \equal{} \frac{bc \plus{} ad}{2}$ and $ r \equal{} \frac{bc \minus{} ad}{4}$. Equating the two formulas for $ r$ yields $ 2(cd \minus{} ab) \equal{} bc \minus{}ad$, or equivalently $ (2c \plus{} a)d \equal{} (2a \plus{} c)b$. Now since $ (b, d) \equal{} 1$, $ 2c \plus{} a \equal{} bv$ and $ 2a \plus{} c \equal{} dv$ for some integer $ v$. By solving the system, $ a \equal{} \frac{(2d \minus{} b)v}{3}$ and $ c \equal{} \frac{(2b \minus{} d)v}{3}$. Replacing in the previous relations finally leads us to $ p \equal{} \frac{v}{6} \cdot (4bd \minus{} b^2 \minus{} d^2)$, $ q \equal{} \frac{v}{6} \cdot (2b^2 \plus{} 2d^2 \minus{} 2bd)$ and $ r \equal{} \frac{v}{6} \cdot (b^2 \minus{} d^2)$.
In the end, for convenience rename the parameters $ m : \equal{} b$, $ n : \equal{} d$ and replace them in the formulas for $ p, q, r$ and then $ x, y$ to ascertain the general form of the solution
$ (x, y) \equal{} \left(\pm \frac{4mn \minus{} (m^2 \plus{} n^2)}{2(m^2 \plus{} n^2 \minus{} mn)} , \pm \frac{m^2 \minus{} n^2}{2(m^2 \plus{} n^2 \minus{} mn)} \right)$, for all $ (m, n) \in \mathbb{Z}^2 \backslash \{ 0, 0 \}$.

Let x=p/q, y=r/s （p,q,r,s∈N） and （p,q）,（r,s） are coprime pairs.By above PhilAndrew's solution, p^2＋3r^2=q^2 holds.From now, my method is different.

（q＋p）（q－p）=3r^2

g=G.C.D（q＋p,q－p） is ≦2. Because if g is ≧3, g | 2p and ∃g' （≧2）| g s.t. g' | p.So g' | q holds, this is contradiction.Then g is ≦2.Apparently q＋p, q－p are same parity.

Case1）q＋p, q－p are odd
Case1-1）3 | q＋p
We can write q＋p=3m^2, q－p=n^2 with（m,n） are coprime odd pairs, and 3m^2＞n^2.Then p=（3m^2－n^2）/2, q=（3m^2＋n^2）/2, r=mn.So x=（3m^2－n^2）/（3m^2＋n^2）, y=2mn/（3m^2＋n^2） with （m,n） are coprime odd pairs, and 3m^2＞n^2.These satisfy the equation.

Case1-2）3 | q－p
We can write q＋p=n^2,q－p=3m^2 with （m,n） are coprime odd pairs, and 3m^2＜n^2.Then p=（n^2－3m^2）/2, q=（3m^2＋n^2）/2, r=mn.So x=（n^2－3m^2）/（3m^2＋n^2）, y=2mn/（3m^2＋n^2） with （m,n） are coprime odd pairs, and 3m^2＜n^2.These satisfy the equation.

By Case1-1） , Case1-2）, We write the solution, x=|（3m^2－n^2）/（3m^2＋n^2）|, y=2mn/（3m^2＋n^2） with （m,n） are coprime odd pairs---α.

Case2）q＋p, q－p are even
Clearly q＋p and q－p are distinct in mod 4.

Case2-1）q＋p≡0, q－p≡2 mod 4
Case2-1-1）3 | q＋p
We can write q＋p=3・2^a・m^2, q－p=2n^2 with a≧2,（m,n） are coprime odd pairs, and 3・2^（a－1）・m^2＞n^2.Then p=3・2^（a－1）・m^2－n^2, q=3・2^（a－1）・m^2＋n^2, r=2^｛（a＋1）/2｝・mn.So x=（3・2^（a－1）・m^2－n^2）/（3・2^（a－1）・m^2＋n^2））, y=（2^｛（a＋1）/2｝・mn）/（3・2^（a－1）・m^2＋n^2） with a≧2, （m,n） are coprime odd pairs, and 3・2^（a－1）・m^2＞n^2.

Case2-1-2）3 | q－p
Similarly to above Case, x=｛2^（a－1）・n^2－3m^2｝/｛2^（a－1）・n^2＋3m^2｝, y=（2^｛（a＋1）/2｝・mn）/｛2^（a－1）・n^2＋3m^2｝ with a≧2, （m,n） are coprime odd pairs, and 3m^2＜2^（a－1）・n^2.

Case2-2）q＋p≡2, q－p≡0 mod 4
Case2-2-1）3 | q＋p
Similarly to above Case, x=（3m^2－2^（a－1）・n^2）/｛3m^2＋2^（a－1）・n^2｝,
y=（2^｛（a＋1）/2｝・mn）/｛3m^2＋2^（a－1）・n^2｝ with a≧2, （m,n） are coprime odd pairs, and 3m^2＞2^（a－1）・n^2.

By Case2-1-2）, Case2－2－1）, we write the solution, x=|（3m^2－2^（a－1）・n^2）/｛3m^2＋2^（a－1）・n^2｝|, y=（2^｛（a＋1）/2｝・mn）/｛3m^2＋2^（a－1）・n^2｝ with a≧2, （m,n） are coprime odd pairs---β.

Case2-2-2）3 | q－p
Similarly to above Case, x=｛n^2－3・2^（a－1）・m^2｝/｛n^2＋3・2^（a－1）・m^2｝, y=（2^｛（a＋1）/2｝・mn）/｛n^2＋3・2^（a－1）・m^2｝ with a≧2, （m,n） are coprime odd pairs, and 3・2^（a－1）・m^2＜n^2.

By Case2-1-1）, Case2-2-2）, We write the solution, x=|｛3・2^（a－1）・m^2－n^2｝/｛3・2^（a－1）・m^2＋n^2｝|, y=（2^｛（a＋1）/2｝・mn）/｛3・2^（a－1）・m^2＋n^2｝ with a≧2, （m,n） are coprime odd pairs---γ.

Therefore the solution are α,β,γ form.

Hey, IIthigore.

y=mx－m meets the ellipse at （3m^2＋1）x^2－6m^2・x＋（3m^2－1）=0 where x=（3m^2±1）/（3m^2＋1）, y=（－m±m）/（3m^2＋1）?

So the solution are x=（3m^2－1）/（3m^2＋1）, y=（－2m）/（3m^2＋1） with m＜－1/3^1/2.


Let $ a$ and $ b$ are positive numbers such that $ a^9 \plus{} b^9 \equal{} 2.$ Prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{a}\geq2
\]

Here is my solution . The inequality is equivalent to

$ (a^3 \plus{} b^3)^9\geq 256a^9b^9(a^9 \plus{} b^9)$  and setting $ \frac {a}{b} \equal{} k$ we have to prove that

$ (k^3 \plus{} 1)^9\geq 256k^9(k^9 \plus{} 1)$ or equivalently

$ (k \minus{} 1)^4(k \plus{} 1)(k^2 \minus{} k \plus{} 1)(k^2 \plus{} k \plus{} 1)^4(1 \plus{} 12k^3 \plus{} 70k^6 \plus{} 12k^9 \plus{} k^{12})\geq 0$  which is clearly true
 :)

Arqady which is your solution?

C.S. & Tangent line gives the proof! :lol:

It's false! :blush:

My proof:
After silouan's reasonings we obtain $ \frac {a^2}{b} \plus{} \frac {b^2}{a}\geq2\Leftrightarrow
(k^3 \plus{} 1)^9\geq 256k^9(k^9 \plus{} 1)\Leftrightarrow$
$ \Leftrightarrow(k^3\plus{}1)^8\geq256k^9(k^6\minus{}k^3\plus{}1)\Leftrightarrow f(k)\geq0,$
where $ f(k)\equal{}8\ln(k^3\plus{}1)\minus{}9\ln k\minus{}\ln(k^6\minus{}k^3\plus{}1)\minus{}\ln256.$
But $ f'(k)\equal{}\frac{24k^2}{k^3\plus{}1}\minus{}\frac{9}{k}\minus{}\frac{6k^5\minus{}3k^2}{k^6\minus{}k^3\plus{}1}\equal{}3k^2\left(\frac{8}{k^3\plus{}1}\minus{}\frac{3}{k^3}\minus{}\frac{2k^3\minus{}1}{k^6\minus{}k^3\plus{}1}\right)\equal{}$
$ \equal{}\frac{9(k^3\minus{}1)^3}{k(k^9\plus{}1)}.$ Id est, $ f(k)\geq f(1)\equal{}0.$
What your proof, kunny? Thank you.

Sorry, it was $ 2\sqrt[9]{2}<a\plus{}b\leq 2$. My solution was false! :blush:

kunny

Let $ x\equal{}\frac{a^2}{b}$, etc. Hence $ a^3\equal{}x^2y$, etc.

So the problem is equivalent to proving that

$ x\plus{}y\ge 2$ when $ (xy)^3(x^3\plus{}y^3)\equal{}2$. Note that $ x\equal{}y\equal{}1$ is a possible values for this condition, so we are not discussing an empty problem.

Suppose for sake of contradiction that $ x\plus{}y<2$, then

$ 2\equal{}x^3y^3(x^3\plus{}y^3)\le\frac{(x\plus{}y)^9}{2^8}<2$ which is a contradiction hence we are done.

[the identity follows by factoring]

The inequality $ (a^3\plus{}b^3)^9\geq 256a^9b^9(a^9\plus{}b^9)$ is equivalent to
\[ (a^6\plus{}2a^3b^3\plus{}b^6)^4\geq 256a^9b^9(a^6\minus{}a^3b^3\plus{}b^3)\]
If we let $ X\equal{}a^3b^3$ and $ Y\equal{}a^6\minus{}a^3b^3\plus{}b^6$, then AM-GM gives $ (3X\plus{}Y)^4\geq 256X^3Y$.

[quote="Altheman"]
$ 2 \equal{} x^3y^3(x^3 \plus{} y^3)\le\frac {(x \plus{} y)^9}{2^8} < 2$ which is a contradiction hence we are done.
[/quote]
But why $ x^3y^3(x^3 \plus{} y^3)\le\frac {(x \plus{} y)^9}{2^8}$ is true? :wink:

scorpius119, beautiful!  :lol:


I'm taking over AOPS Planet from Bugi who suggested a remake.

I'll take at most 10 people (subject to change).

Players (so far):
[b]jain_harshi (6)[/b]
[b]AIME15 (4)[/b]
coolts
[b]SuperMathBoy (1)[/b]
hongsquare
[b]ernie (2)[/b]
[b]eggylv999 (3)[/b]
[b]BOGTRO (5)[/b]

If you signed up in the last pages of the first thread, I will [have] send a confirmation pm.

Numbers assigned are based on order of joining/confirming. (don't worry they won't give anyone an advantage)

WOOT, I AM HERE. GO STARWARSSSSSSSSSSSSSSSSSSSSSSSSS :starwars:  :starwars:  :starwars:

Keep it small, and make sure everyone is active.

I feel we have enough people. Sign ups are still open, but the game starts on the 12th (yay Friday!!!!).

Good.  Let's make this a quick, fun game.

Everyone pm me a color to use on the map (I want them confidential so you won't know who is who).

We will still be starting Friday. If I don't get a color pm from the rest of you by the time I get on the computer on Friday, your color will be assigned. (irrelevant to the game but you probably won't like what I choose for you  :P )

Game starts now!!!

jain_harshi (6)
AIME15 (4)
coolts (7)
SuperMathBoy (1)
ernie (2)
eggylv999 (3)
BOGTRO (5)
crazypianist1116 (8)

Each of you has been chosen as the leader of an anonymous tribe of cavemen. You're goal is to advance through the ages and be the sole ruler of the planet. What stands in your way: each other, neutral barbarians, and a couple sup3rcrash3r controlled nations (they start stone age too).

Command your forces. Everyone has 10k people. There will be NO SUCH THING as "mewto credits" or "sup3r money". There will be no currency at all, unless you decide to establish it. Trade your needs with other tribes. You can back stab all you want, but trades must be fair.

A special event can happen to someone every three days (for now, everyone will have them. Later, it might become random).

Formulas (plot your food and stuff (only food and pop. growth for now (sorry, no latex))):
Getting food (f) based on fertility of your land (e (1-10)) and number of workers (w): f =  (ew)/100
Rate of eating food (f) based on population (p): f = p/100
Population growth (g) based on current population (p) and surplus of food (f): c = p/10 + s/100

Starting tips:
-try to grab as much land as you can for now
-use [most of] your people as workers to get a food surplus
-mine, chop trees... OH WAIT!!! No tools! Haha. Figure it out.
-you can move diagonally on the map
-don't move away from water if you don't want to have some of your people die of thirst (again, figure it out)
-your first plot of land has 7 fertility

Warring: I don't expect much player/player warring for now, but each person has 100 health. You all have sticks and stones. They do 1 damage. More on this later. "You" can also fight. "You" have 10000 health and also have sticks and stones. Don't expect to kill anyone so soon.

So, this starts as soon as I post the map (maybe at 7 or 8 pm). I recommend you have 9k workers getting food and sending the 1k to go get more land.

Last thing. PM me all your moves. The thread is for diplomacy and trading (once we get that far).

HAY?

WHEN DID THE MOD SAY YOU CAN HAVE ALLIANCES?!?!

[quote="sup3rcrash3r"]
Last thing. PM me all your moves. The thread is for diplomacy and trading (once we get that far).[/quote]

Implies alliances are allowed.  Also, why can't different tribes have alliances...

My post "HAY?" implies but does not forbid you from having alliance.

Just don't tell anyone your color.

EDIT: actually you can... but not recommended to yet

Posting so I dont have 4666 posts.

My color is

[hide]
[hide]
[hide]
arent u gonna quote this yet/

not tellin because its not recomended
[/hide][/hide][/hide]

This is too complicated, I drop out.

[b]DROP OUT, burns all land with the newfound fire knowledge[/b]

hmm...so what will become of the land that got burned? given to someone?

http://lunarwars.net/

basically the same thing except more complicated and manegeable.

More about Lunar Wars: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249676[/url]

*DROP OUT*

BURNS AIME'S LAND ONCE AGAIN.

NEED NEW MOD


$P,Q$ is midpoint of $AC, BD$ diagonals of convex quadrilateral $ABCD$, $PQ$ intersects with $AB, CD$ at $N,M$ respectively.Prove that circumcenters of triangle $NAP, NBQ, MQD, MPC$are concurrent.

I am not sure but I guess this is the same as the 2nd problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=25772 .

  darij


A pentagon $ABCDE$ is inscribed in a circle of radius $10$ such that $BC$ is parallel to $AD$ and $AD$ intersects $CE$ at $M$. The tangents to this circle at $B$ and $E$ meet the extension of $DA$ at common point $P$. Suppose $PB=PE=24$ and $\angle BPD$ is 30 degrees. Find the length of $BM$

[hide="Solution"]The right triangles $POB,PEB$ are equal (they have a common hypotenuse $OP,\ OB=OE=radius,\ PB=PE=24$)
So, $\angle POB = \angle POE$

From the circle, $\angle BCE = \frac{1}{2} \angle BOE = \angle POE$

 :idea: But $CB \parallel MP $ hence $\angle BCE = \angle PME$

So we have $\angle POE = \angle PME$

This means, that, if we construct the circle passing through points $P,O,E$, then $M$ must be on this circle ($M$ and $O$ are in the same side of $PE$). :roll:

In other words, point $M$ lies on the circumcircle of the triangle POE.

But $POE$ is a right triangle, so the center of its circumcircle is the midpoint $K$ of $OP$: $KO=KP=KE$

:o

Since the common hypotenuse of $POB,POE$ is the diameter of the circle $(K,KP)$, point $B$ also lies on this circle.

Hence, $\angle MKB = 2\angle MPB = 60$ ;)

The isosceles triangle $MKB$ is equilateral, so $MB=KB=>\ MB=KO$

The final step is to find $KO=\frac{OP}{2}$, but it's easy since $OP^2=OB^2+PB^2=10^2+24^2=100+576=676=26^2=>$
$OP=26=>$
$MB=KO=\frac{26}{2}=13$[/hide]


Could someone explain how proving $ \frac{\vec{A}\plus{}\vec{B}\plus{}\vec{C}}{3}$ and showing that each of the line to the medians are in the ratio $ 2: 3$ with respect to the centroid proves that they concur? I'm having trouble visualizing/understanding why this is true.

[quote="Tenoreoz"]Could someone explain how proving $ \frac {\vec{A} \plus{} \vec{B} \plus{} \vec{C}}{3}$ and showing that each of the line to the medians are in the ratio $ 2: 3$ with respect to the centroid proves that they concur? I'm having trouble visualizing/understanding why this is true.[/quote]
Well, you only really need one of them to show that the medians concur. I will use the ratio (it is actually $ 2: 1$)
[hide="Solution"]
Consider medians $ CF$ and $ BE$ of triangle $ ABC$ which intersect at $ G$.  We notice that $ F$ is the midpoint of $ AB$ and $ E$ is that of $ AC$.  Hence, $ FE\parallel BC$, so $ \angle FEG \equal{} \angle GBC$ and $ \angle GFE \equal{} \angle GCB$.  Therefore, we obtain $ \triangle FGE\sim \triangle CGB\implies \frac {FG}{CG} \equal{} \frac {FE}{BC} \equal{} \frac {1}{2}$.  Thus, $ 2FG \equal{} CG$.  Now, allow the median $ AD$ to be the final median of $ \triangle ABC$ and allow it to hit $ CF$ at $ G'$.  We can also show that $ FG' \equal{} 2CG'$ (using the exact same method as we used to show that $ FG \equal{} 2CG$.)  This gives that $ FG \equal{} 2CG$ and $ FG' \equal{} 2CG'$ where $ G$, $ G'$, $ F$, and $ C$ are collinear.  Hence, we realize that $ G' \equal{} G$ and the medians are thus concurrent.  [/hide]


function
calculus
derivative
calculus computations

[quote="dule_00"]I heard that there is defintion for $x^{th}$ derivate of function where $x\in \mathbb{R}$ . Anyone who isn't laisy to explain it to me will be welcome?[/quote]

Hope this helps:

[url]http://www.math.montana.edu/frankw/ccp/calculus/deriv/derlimit/learn.htm[/url]

You're probably thinking of "fractional calculus", which studies ways of meaningfully interpreting partial iterates of the differential operator. The Wikipedia article has some info: http://en.wikipedia.org/wiki/Half-derivative , but it will likely not satisfy you if you are looking for more in-depth coverage. Perhaps google "fractional calculus". Note however that this is not quite the kind of basic stuff that belongs in "Computations and Tutorials", IMO.


AMC USA(J)MO USAMO MIT college Harvard Princeton

Just out of curiosity, does putting "USAMO qualifier" on a college application make ur application stand out more to a somewhat significant degree? I mean I realize that obviously any national olympiad is better than none, but I was just wondering if it would actually make them look more closely at ur application and stuff.

I think that it depends from university to univeristy. For example I think that it would make some difference for MIT, Caltech, but it would make no difference for say Harvard :)

just my 0.02 $.

Some schools (notably MIT and Caltech) ask specifically about AMC and AIME scores on the face of the application, so of course they will be interested in USAMO qualifiers. Harvard, Princeton, Stanford, etc. are also well aware that USAMO qualification is a rare qualification for a math-interested young person.

[quote="rrusczyk"]I disagree - it certainly will make a difference at Harvard.  It won't get you in for sure, but it will make a difference.  All the top schools pay attention to the programs which identify the very top students in various fields.  For example, within a week of being publicly named a USAMO winner my senior year of high school, I received an invitation from the dean of academic affairs at Princeton (to which I had already matriculated) to visit her when I arrived on campus.  So, the non-MITs definitely pay attention, even if you don't tell them.  (Though admittedly, they probably won't be scouring the whole USAMO list!)

Just take a look at the number of USAMO qualifiers at Harvard.  It's not cause-and-effect, probably, but there's certainly a significant correlation.[/quote]I was thinking just about "USAMO Qualifier" and not "USAMO no. 1" :)

Of course it does not hurt one to mention there "usamo qualifier", one can just benefit from that, especially in some colleges (like MIT there it will certainly matter).

As far as the international admission is concerned, only MIT is interested in IMO medals, other univ. just reject people with flawless records (maximum GPA in high-school, over 1750 in SATs and IMO medals) because they don't like their essays :)

I'm no college admissions agent, but I do think that being an USAMO qualifier... even last place, will boost your chances.  I mean, there are only so many hundred qualifiers in the nation.  Heck!  There are 900 1600s every year and there are so many less USAMO qualifiers.

[quote="Valentin Vornicu"]I was thinking just about "USAMO Qualifier" and not "USAMO no. 1" :)
[/quote]

I will note that I was just a USAMO qualifier at the time I was *accepted* to Princeton! (And I wasn't #1 -> 'USAMO winner' denoted top 8 back then, and denotes top 12 now; I was actually #4.)

I imagine foreign admissions is a real crapshoot, even more than US ones (far fewer slots available - but I'm just guessing).  Grad schools, on the other hand, are packed with foreign students.

Only about 50 11th graders qualify for the USAMO.  So it's a relatively rare person who can write "USAMO qualifier" on their college application.


Two trains leave from the station at the same time, moving in opposite directions. One train averages 75 kilometers per hour for the first two hours and 60 kilometers per hour after that. The second train averages 70 kilometers per hour. After how many hours will the trains be 745 kilometers apart? Express your answer as a mixed number.


Let $ p$ to be a norm over $ C([a,b])$ for which $ C_p([a,b]) \equal{} (C([a,b]), p )$ is a Banach space. Assume that $ p(x_n) \rightarrow 0$ implies $ x_n(t) \rightarrow 0$ for all $ t \in \mathbb Q \cap [a,b]$.

Show that $ p$ is equivalent to the usual norm of $ C([a,b])$.

I´m trying to solve this question for almost four days and I got nothing so far :(

Thanks ^^

Hint: Apply the closed graph theorem to the identity operator.

thanks :)

I haven´t realised how simply is it :P


inequalities
calculus
integration
function
real analysis
real analysis unsolved

Hello, This is my first post in calculus section  :) :

[color=blue]Problem[/color]Let $f$ be a differentiable function on the interval $[0,1]$ and $0 < f'(x) \leq 1$ for $x \in [0,1]$.
Let $f(0)=0$ prove that : \[[\int_{0}^{1}f(t)dt ]^{2}\geq \int_{0}^{1}f(t)^{3}dt\]


Given the triangle $ ABC$ ( $ \angle B \ge \angle C$) and $ AM$ is its median. Let $ P$ be a point on $ AM$ such that $ PB \equal{} PM$ and $ D$ is the midpoint of $ BM$. The line through $ D$ and perpendicular to $ PB$ intersects $ PB$ at $ G$ and $ AB$ at $ E$. The line through $ D$ and perpendicular to $ PC$ intersects $ PC$ at $ H$ and $ AC$ at $ F$. Prove that $ GH$ and $ EF$ are parallel.

Here is my counter-example,because your is not clear :wink: :
[img]http://img263.imageshack.us/img263/6131/counterexamplepv9.png[/img]

[quote="mathangel"]I think that it 's not correct.

Look at the figure:[/quote]

sorry, $ P$ is the point satisfying $ MP\equal{}MB\equal{}MC$ and $ D$ is the intersection of the line through $ P$ and perpendicular to $ BC$ and side $ BC$

Then it 's not difficult.

Let $ R$ and $ S$ be intersections of $ BP$ and $ AC$ and of $ CP$ and $ AB$

Applying the Ceva's Theorem to the triangle $ ABC$ and 3 intersecting lines: $ AM$, $ BR$ and $ CS$ we have:

$ \frac{SA}{SB}\frac{MB}{MC}\frac{RC}{RA} \equal{} 1$

Then : $ \frac{SA}{SB} \equal{} \frac{RA}{RC} \leftarrow RS//BC$

Using Thales' Theorem: $ \frac{PS}{PC}\equal{}\frac{PR}{PB}$

We also have:

$ \frac{GE}{GD}\equal{}\frac{PS}{PC}$ and $ \frac{HF}{HD}\equal{}\frac{PR}{PB}$

Therefore : $ \frac{GE}{GD} \equal{} \frac{HF}{HD}$

From the converse of Thales' Theorem, $ EF//HG$ (QED  :) )

Figure: (I don't know why it 's not clear enough ???)


Three circles of equal radius r are tangent. Their centers are collinear
from left to right. Triangle ABC is constructed.

A is the center point of the circle on the left
B is the center point of the circle on the right.
C is a point on the circle on the right where the tangent line thru A
touches the circle.
D and E are points on the middle circle where the line AC intesects the
middle circle.

Find the length of the line segment DE in terms of r.

I got $ED=2r\sqrt{\frac{13}{17}}$.
However, it would probably be better if someone else checked before I post my solution.

You are correct with a little more added.

hmm.. name the center of the middle circle P... and the midpoint of DE. Q

apparently PQ is perpendicular to DE.. and P is the midpoint of AB, so $PQ=\frac{r}{2}$

use pythagorean theorem.. $DE=2*\sqrt{r^{2}-(\frac{r}{2})^{2}}=r\sqrt{3}$