The "closed form" of this recurrence can be written in terms of the roots of the polynomial $ x^3 \equal{} x \plus{} 1$.  Unfortunately, the roots of this polynomial are annoying to write out and essentially require the [url=http://en.wikipedia.org/wiki/Cubic_equation#Roots_of_a_cubic_function]cubic formula[/url].  I discuss the general idea [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url].

[quote="mr.affable"]I understand that the first terms of the perrin sequence are markes as:

P(0) = 3   P(1)= 0   P(2)= 2

and is defined by the equation P(N) = P(N-2) + P(N-3)

but what is the equation for finding the Nth term in the sequence, without previous knowledge of the previous 2nd and 3rd terms?[/quote]

The characteristic equation is:

\[ n^3\minus{}n\minus{}1\equal{}0
\]

Then find the roots.  There is only 1 and its irrational so for simplicity call it $ r$.  So $ r^N$ is a root of the equation.  We need two more and they are $ N\cdot r^N$ and $ N^2\cdot r^N$.  Then we have

\[ P(N) \equal{} a r^N \plus{} b (N\cdot R^N) \plus{} c (N^2\cdot R^N)
\]

Then plug in the initial values and we get 3 equations and 3 variables so you can solve for $ a, b$ and $ c$.

That would be the case if $ x^3 \equal{} x \plus{} 1$ had a single [i]repeated[/i] root, but the two complex roots also count.  (Otherwise, how do you explain the behavior of the recursion $ P(N) \equal{} \minus{} P(N \minus{} 2)$?)

so, is the Perrin sequence anything like the Fibonacci sequence, where I can generate any n'th term simply by using a formula.  I know the quotient of f(n)/f(n-1) approaches the value of 1+sqrt5 / 2 as n approaches infiniti within the fibonacci sequence.  Is this also the case in the Perrin sequence?

It is. The ratio of consecutive terms will approach one of the roots of $ x^3\minus{}x\minus{}1\equal{}0$, just based on the characteristic equation.

Now, since the two initial conditions are real, it is obvious that all terms of the sequence will be real, and thus their ratio will always be real. This tells us that the ratio of the roots will approach one of the [i]real[/i] roots of $ x^3\minus{}x\minus{}1\equal{}0$, and there is only one such real root.

This root is $ x\equal{}\frac{1}{3} \left(\frac{27}{2}\minus{}\frac{3 \sqrt{69}}{2}\right)^{1/3}\plus{}\frac{\left(\frac{1}{2} \left(9\plus{}\sqrt{69}\right)\right)^{1/3}}{3^{2/3}} \approx 1.324717957244746 \ldots$

A solution using generating functions can also be done:
We'll rewrite that recursion by index shifting:
$ P(n + 3) = P(n+1) + P(n)$
Multiply both sides by $ x^n$ and sum over $ n$:
$ \sum_{n\ge0} P(n+3) x^n = \sum_{n\ge0} P(n+1) x^n +\sum_{n\ge0} P(n) x^n$
We'll want to call $ G(x) = \sum_{n\ge0} P(n) x^n$ because this way, the coefficients of G(x) give us the values of P(n).
Now we want to write the other two terms in terms of G(x).
For the left, we will multiply by $ \frac{x^3}{x^3}$ and pull a factor of $ \frac{1}{x^3}$ out of the denominator, so the sum (LHS) is now
$ \frac{1}{x^3}\left(\sum_{n\ge0} P(n+3) x^{n+3}\right)$.
This is nearly the correct form for the G(x), but it is index shifted. So we re-index.
$ \frac{1}{x^3}\left(\sum_{n\ge3} P(n) x^{n}\right)$
Doing this, however, will miss the $ n=0, 1$, and $ 2$ cases. These, however, are $ 3, 0x$, and $ 2x^2$ respectively (given as $ P(0), P(1)x, P(2)x^2$)
Therefore, we will have the LHS equal to $ \frac{1}{x^3}\left(\left({\sum_{n\ge0} P(n) x^n\right) - 2x^2 - 3\right)}$

This means that the left hand side will reduce to $ \frac{1}{x^3}(G(x) - 2x^2 - 3)$

By a similar manipulation, the first term on the right hand side reduces to $ \frac{1}{x}(G(x) - 3)$
Now, we've got $ \frac{G(x) - 2x^2 - 3}{x^3} = \frac{G(x) - 3}{x} + G(x)$.
Solving for $ G(x)$, we have
$ G(x) = \frac{x^2 - 3}{x^3 + x^2 -1}$

Now we break this down into partial fractions.
This is where we have to, unfortunately, invoke the cubic formula.
After we've done that, we can basically recombine into power series and get the closed form.
But, as the posters above me have already shown, it's going to be ugly.

Thanks DiscreetFourierTransform, I can't really say it makes it easier :) , but your explanation certainly does help me understand it better.  And yes, using the cubic formula looks like a very arduous process.


geometry
induction
Mafia
geometric transformation
reflection

i have found a few problems that really stump me, i could really use the answers and an explanation, if you can answer any of the questions below, please icq, aim, or msn me preferably, or post here, they really are confusing to me!

(By the way, i know all the answers are whole numbers)

34 - ?
24 - 10
16 - 7
8 - 5
4 - ?

Numbers separated with a '-'. (1-2)
-----------------------------

1
2
10
140
5740
???

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

On a small planet not far away lived a strange race. A long time ago all the members of that species had one of three eye colors: blue, green or brown. In some point in history a famous war took place, initiated by the blue and brown eye faction. They genocided all green eyes, thus eliminating the green eye color from their race, and reducing the population of the planet to a few hundred. Soon after that war the factions were disbanded, and eye color was from that point on considered an inappropriate topic. Some rules were later established for the benefit of the whole society, so that something like this could never, ever happen again. The rules were as follows:

1. Nobody is allowed to talk about eyecolors.
2. If anyone gets to know his own eyecolor he must commit suicide at midnight in the town square.

Now I must explain the way this species was thinking, so you understand their reactions later on. You must understand that they all were really strict about those rules. In the beginning some people happened to accidently see their eyecolor in some reflective surface or puddle of water, and every last one of them faithfully comitted suicide the night after. After a few incidences all mirroring surfaces were removed from the cities, to minimize the risk. Also, this species was a very logical thinking. Although they still lacked knowledge in some areas, they thought clearly and logically, and trusted in the results they got, even if it led to their own death. Last, the way it worked after the rules were established was that everyone knew the eyecolor of everyone else on the planet, but not his own. Of course, on a small planet like this, they all saw everyone everyday, so a suicide was immediately noted.
Well, the people on this little planet were living for a few hundred years in peace and harmony, when something bad happened:

A little spacecraft crashed down in the desert near one of the large cities, leaving a big crater. Immediately the friendly people tried to rescue the astronauts, but it was too late. The only thing they recovered from the craft was a book about mathematics. In the same night a big meeting was called on, and everyone came together in a large hall. The book was read to the whole population, because it offered new concepts that were so impressive and striking that everyone had to know about them. But, knowledge isn't always a good thing. The knowledge in that book made the people aware of facts they didnt know about before, and made them realize a few things. In the following nights everyone was really nervous.

Then, in the 250th night after the dreaded meeting, all brown eyes committed suicide, thus eliminating the whole species, and leaving the planet an empty brick in the wide universe.

Blue eyes present during the meeting?
Number of nights in which suicides happened?
Are there more blue or brown eyes in the meeting?
Which mathematical concept made them kill themselves?

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

You traced the target computer across the globe. You finally end up with two coordinates.
One lies at [80 00 00 N 007 45 00 E] and the other at [70 00 00 S 007 45 00 E].
While you plan your next attack, you suddenly get curious... and want these questions answered:

1. How far away are these two towns from each other?
2. If a car drives 200 km/h, how long will it take you to drive from one town to the other?

Just full numbers. (e.g. 12, 104, 1006)

Distance in km?
Drive time in hours?

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

And Thanks in advance!

1) twenty-four = 10
sixteen = 7
eight = 5

You should be able to figure out the rest.

[quote="white_horse_king88"]This makes no sense.[/quote]

His problems, or my answers?

Since this is a math problem, shouldn't it be in one of the math forums, i. e. Getting Started, Intermediate, etc.

For the first problem, the answer is  (it's in spoiler):

[hide]10.[/hide]

[quote="white_horse_king88"]What topic would it qualify in? Mathematical Riddles? I think that's why the fun and games forum was created in the first place.[/quote]

Lol, good point. And I do find a problem with aliens committing suicide a tid-bit amusing.

[quote="h_s_potter2002"][quote="white_horse_king88"]His problem about the alien invaders - that was so pointless. Do a math problem or something - why waste precious time pondering how many aliens committed suicide?[/quote]

It's probably for homework or something.  I can't think of any people who would care about how many aliens comitted suicide.[/quote]

 :(

As for the alien problem, the mathematical concept is the Riemann Hypothesis! jk jk

The true concept is [hide]induction[/hide].

This problem is actually very flawed but I won't get into that.

how is it flawed? anyway, the problem is confusing.  very confusing.  But i like it, but still you should change the aliens into mafia people!!!

i am the mafia and i will kill you.
woh she (mafia).  woh hway sha nee.
soy la mafia y te voy a matar.
je suis (mafia) et je vais? (kill you)

:D math camp joke

I still don't understand the alien problem. Induction???

lol.. guess what? i learned how to prove by induction here... *gasp from audience*  is it cause blue eyes can see their reflection in brown eyes? i don't know...

Random guess: All brown eyed people can see their reflections in blue eyed people, so if you can't see your reflection ur blue eyed - either ways I don't give a chicken whether any alien decided to commit suicide.

First imagine that there is only 1 brown eyed alien alive, and he found out there was at least one brown eyed alien on the planet, then he would see everyone else with blue eyes, thus leading him to realize that he is the one brown eyed alien, so he now know his eye color and commits suicide.
Now imagine there are only 2 brown eyed aliens alive, and they can see each other having brown eyes, thus they wait 1 day to see if each other commits suicide, but since they both are waiting on each other, but since nothing happens, they both realize that they themselves are brown eyed because the other brown eyed person didnt commit suicide yet, so they both commit suicide on day 2.

Now you guys can build on that if you want


Given a circle $\omega$ and its center $P$. Let $AB$ be the diameter of $\omega$. $X$ is a point on $\omega$ different from $A$ and $B$. $M$ is a point on $AB$ such that $XM \perp AB$. Let the line tangent to $\omega$ at $X$ intersect $AB$ at $D$. $N$ be any point on segment $XM$ different from $M$. Prove that circle with diameter $ND$ (call $\omega_{1}$) is orthogonal to $\omega$ (tangent line from $P$ to $\omega_{1}$ touch $\omega_{1}$ at the intersections of these two circle).

Are u serius??This is not a problem for advansed section.
Two circles are orthogonal if and only if any line passing through the center of one of them cuts the two circles at A,B,C,D so that these point are harmonic conjugates.
Here u have to prove that D,A,M,B are h.c.
This is obvious-just similar triangles.

Sorry for my stupid :blush:  I don't know that theorem.

Oh it's ok then!Good choice for reading it...


I didn't know about USAMTS until now, and I know that Round 1 is already over.  Can I still register for it, but only participate in rounds 2-4?  Thanks.

(I realize that my maximum score would only be a 75, which seems pretty bad.)

Yes, you can. Just follow regular instructions for joining. Also, Round 2 is due in a week :(

Yes, you can. You can join at any round.


In each square of the figure below one places a different number from $ 1$ through $ 9$ such that the sum of the five entries in the horizontal row is equal to the sum of the five entries in the vertical column. The positions of $ 3$, $ 5$, and $ 7$ are given. Find all possibilities for the number in the darkened square.

[hide]Let $ M$ be the number in the middle
Let $ R$ be the sum of the 3 squares at the right
Let $ B$ be the sum of the 2 squares at the bottom

$ 5\plus{}7\plus{}M\plus{}B \equal{} 3\plus{}M\plus{}R \Rightarrow$

$ R \equal{} B\plus{}9$

But $ 1\plus{}2\plus{}\cdots \plus{}9 \equal{} 45$, so $ 3\plus{}5\plus{}7\plus{}L\plus{}M\plus{}B\equal{}45 \Rightarrow R\plus{}M\plus{}B \equal{} 30$

So we have $ M\plus{}2B\plus{}9 \equal{} 30 \Rightarrow M\plus{}2B \equal{} 21$

Then $ M$ must be odd. But there are only two remaining odd numbers $ 1$ and $ 9$

For $ M\equal{}1 \Rightarrow 2B\equal{}20 \Rightarrow B\equal{}10\Rightarrow \boxed{B\equal{}2\plus{}8 \text{ or } B\equal{}4\plus{}6}$

For $ M\equal{}9 \Rightarrow 2B\equal{}12 \Rightarrow B\equal{}6\Rightarrow \boxed{B\equal{}2\plus{}4}$[/hide]


It has been determined that dialing 9-1-1 in the case of no emergency for five minutes is about as unhealthy as eating a cheeseburger with two packets of mayonnaise. It has also been determined that dialing the fire department in the case of no emergency for sixteen minutes is as unhealthy as eating two cheeseburgers and a packet of mayonnaise. Finally, a survey conducted by males over the age of 25 has shown that dialing 9-1-1 is twice as unhealthy as dialing the fire department.

Eating six packets of mayonnaise is as unhealthy as dialing 9-1-1 for how many minutes?

[hide]Dialing 9-1-1 is twice as unhealthy as dialing the fire department.
And I'm guessing that's for the same number of minutes.
So 9-1-1 for 5 min = c + 2m
fire dept for 16 min = 2c + m

9-1-1 for 80 min = 16c + 32 m
fire dept for 80 min = 10c + 5m
So 16c + 32 m = 2(10c + 5m)
16c + 32m = 20c + 10m
22m = 4c

11m = 2c
5.5 m = c

And 9-1-1 for 5 min = c + 2m
                               = 7.5 m

So 9-1-1 for 5 min = 7.5m, and we're looking for 6m.
That's $\frac45$ of 5 min.

Answer is 4 minutes![/hide]

What a dumb problem.  :rotfl:

At least make them somewhat realistic if possible...


geometry
geometric transformation
reflection
perpendicular bisector
geometry unsolved

Let ABC be an isosceles triangle (AB = AC). Consider a point D on the side AC such that $ C\hat{B}D \equal{} 60^{o}$ and a point E on the side AB such that $ B\hat{C}E \equal{} 50^{o}$. Determine the angle $ B\hat{D}E$.

The given conditions are insufficient for a unique solution. Probably, condition $ \angle CAB \equal{} 20^\circ$ is missing. :?:

Yeap...I am sorry....I forgot to type the hypothesis $ \widehat{CAB} \equal{} 20^{\circ}$.

$ \angle ABC \equal{} \angle BCA \equal{} 80^\circ$. Reflect $ BD$ in the perpendicular bisector of $ BC$ into $ CF,$ $ F \in AB.$ $ BD, CF$ intersect at $ G.$ $ \triangle GBC,$ $ \triangle GDF$ are both equilateral, $ BG \equal{} BC.$ $ \triangle BCE$ has angles $ \angle EBC \equal{} 80^\circ, \angle BCE \equal{} 50^\circ \Longrightarrow$ the remaining angle is $ \angle CEB \equal{} 50^\circ,$ this triangle is isosceles, $ BE \equal{} BC$. Then $ BE \equal{} BG,$ $ \triangle EBG$ is isosceles. $ \angle EBG \equal{} 80^\circ \minus{} 60^\circ \equal{} 20^\circ$, the remaining angles are $ \angle EGB \equal{} \angle GEB \equal{} 80^\circ.$ Then $ \angle GEF \equal{} 180^\circ \minus{} 80^\circ \equal{} 100^\circ.$ $ FD \parallel BC \Longrightarrow$ $ \angle AFD \equal{} \angle ABC \equal{} 80^\circ$. $ \triangle GDF$ is equilateral $ \Longrightarrow \angle DFG \equal{} 60^\circ.$ Then $ \angle GFE \equal{} 180^\circ \minus{} (80^\circ \plus{} 60^\circ) \equal{} 40^\circ.$ $ \triangle GEF$ has angles $ \angle GEF \equal{} 100^\circ, \angle GFE \equal{} 40^\circ \Longrightarrow$ the remaining angle is $ \angle FGE \equal{} 40^\circ,$ this triangle is isosceles, $ EG \equal{} EF.$ $ \triangle GDF$ is equilateral $ \Longrightarrow DF \equal{} DG.$ Quadrilateral $ DFEG$ is a kite, the diagonal $ DE$ bisects the angle $ \angle GDF \equal{} 60^\circ \Longrightarrow \angle BDE \equal{} \angle GDE \equal{} 30^\circ.$

[quote="Duncreggan"][color=darkred]Let $ ABC$ be a triangle for which $ AB = AC$ and $ A=20^{\circ}$ . Consider the points $ \{\begin{array}{ccc}
D\in (AC) & , & m(\widehat {CBD}) = 60^{\circ}\\\
E\in (AB) & , & m(\widehat {BCE}) = 50^{\circ}\end{array}$ . Ascertain $ m(\widehat {BDE})$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Construct the point $ F\in (AC)$ for which $ m(\widehat {CBF})=20^{\circ}$ .
Prove easily that $ BC=BE=BF=DF=EF$. In conclusion, $ \boxed {m(\widehat {BDE})=30^{\circ}}$ .[/color]


Prove that $ sin(18^o) \equal{} \frac{\phi}{2}$ where $ \phi$ is the gold number $ \frac {\sqrt {5} \minus{} 1}{2}$

[hide="Hint"]
Prove that if $ x \equal{} sin(18^o)$ then $ 4x^2 \plus{} 2x \equal{} 1$

Actually, this is a problem in Prasonof's geometry.
[/hide]

[hide="Solution"]
Let AC be the bisector of an equilateral triangle with base AB and $ \hat{C} \equal{} 36^o$.

The triangle ACD is an isosceles one and ABC ~ BDA.

Therefore, $ CD \equal{} AD \equal{} AB \equal{} 2sin(18^o)BC$ and $ DB \equal{} 2sin(18^o)AB \equal{} 4sin^2(18^o)BC$

let  $ x \equal{} sin(18^o)$ then

$ BC \equal{} CD \plus{} DB \equal{} 2xBC \plus{} 4x^2BC \Leftrightarrow 1 \equal{} 2x \plus{} 4x^2 \Leftrightarrow$

$ x \equal{} \frac { \minus{} 2 \minus{} 2\sqrt {5}}{8} < 0$ or $ x \equal{} \frac { \minus{} 2 \plus{} 2\sqrt {5}}{8} \equal{} \frac{\phi}{2}$

so, as $ x \equal{} sin(18^o) > 0$, it is

$ x \equal{} \frac{\phi}{2}$  :D
[/hide]

You are wrong. $ \sin 18^\circ$is not $ \frac {\sqrt {5} \minus{} 1}{2}$ but $ \frac {\sqrt {5} \minus{} 1}{4}.$

Here is my solution:

The quadrilateral $ ABCD$ is inscribed in the cicumcircle of the regular pentagon $ ABCDE$ with side length 1, W.L.O.G.

Ptolemy's theorem gives $ AB\cdot CD \plus{} BC\cdot AD \equal{} AC\cdot BD$, let $ x$ be the diagonal length of the penatagon, we have

$ 1\cdot 1 \plus{} 1\cdot x \equal{} x\cdot x$, yielding $ x \equal{} \frac {1 \plus{} \sqrt {5}}{2}.$ Then remark the isoceles $ ACD$ with $ AC \equal{} AD \equal{} x,\ CM \equal{} \frac {1}{2}$ here, $ M$ is the mid

point of $ CD \equal{} 1.$ Since $ \angle{CAM} \equal{} 18^\circ,\ \therefore \sin 18^\circ \equal{} \frac {CM}{AC} \equal{} \frac {1}{2x} \equal{} \boxed{\frac {\sqrt {5} \minus{} 1}{4}}$.

You are right kunny, I misculculated $ 2\cdot4$. :blush:  I have corrected my post.

None the less, your solution is impresive.  :lol:


How difficult would it be for an 8th grader to get in to AMP, how many qs would he need to get right on admission test, do they look at ur contests from previously in the year?

[quote="CatalystOfNostalgia"]I hear they accepted very many people, including those with low scores, which I believe because there's no way certain people there solved 7-8 problems.

I'd say as long as you can get 5 problems, you'll be accepted, probably more if you're older, but as you're in 8th grade, you'll be fine. I'm pretty sure they don't look up your contest scores, because the test A is before any of the AMC's, but they do ask for them when you're choosing tracks. Also, when they say that they consider your application test score when helping you choose your track, I don't know if this is completely true, because my roommate (or apartmentmate) and I basically wrote down what we wanted and Mr. Zuming almost immediately said it was fine.[/quote]
Well they won't stop you from going into whatever track you choose (at least I don't think), but they will discourage you [i]immensely[/i] if, for example, you got a 2 on the AIME and want to be in track I.

Classes are divided into 'tracks,' in which you will stay for all three weeks. You pick your track, with the guidance of various people at AMP (and your parents), right when you arrive.

There are six tracks, [i]generally[/i], the lower the number the harder. Attatched are track listings and descriptions from 2007.

How difficult would it be for me to get in?

It depends on your level of math, if you are quite familiar with proofs, it would be quite easy to get in, even if they are wrong.  Last year, i was a 6th grader, I got in quite easily.  They look at how well you can express your ideas in proof form.  They dont care if it is correct or not.  Last year, i submitted 6, yet only 4 were correct.
It should be quite easy if you can get 4+ on the AIME

But im only in 8th grade, i barely made AMC... how hard is it 2 get 4+ on AIME?


Can someone please help me with this problem.
Hints are more appreciated than a full solution but i will be thankful for any kind of help.

Let $V$ and $W$ be vector spaces, and let $T$ and $U$ be nonzero linear transformations from $V$ to $W$. If $R(T)\cap R(U) =\{0\}$, prove that $\{T,U\}$ is a linearly independent subset of $L(V,W)$.

where $L(V,W)$ is the set of all linear transformations from $V$ to $W$ and $R(T)$ is the range of $T$.

Suppose $aT+bU$ is the zero function. Choose $x$ such that $T(x)\ne 0.$ This is possible because $T$ is not the zero function. Then $U(bx)=bU(x)=-T(ax)=T(-ax)\in R(T)$ Since $R(T)\cap R(U)=\{0\},$ we must have that $bU(x)=0$ from which it follows that $aT(x)=0.$ Since we assumed $T(x)\ne 0,$ we have $a=0.$

Now find $y$ such that $U(y)\ne 0.$ The same argument with the letters switched then shows that $b=0.$ Hence assuming $aT+bU=0$ implies $a=b=0,$ which is the definition of independence.

thanks.
is this basically how you go about proving that a set of functions are independent?

Well, disjoint (except for zero) ranges is an awfully strong condition; you might be asked to prove independence with less to go on than that. But do rely on the definition of independence. Assume a linear combination is zero, now what can you learn about the coefficients?


geometry
3D geometry
pyramid
algebra
linear equation
similar triangles
calculus

I need help on solving this related rates problem:

The base of a pyramind-shaped tank is a square tank with sides of length 12 feet, and the vertex of the pyramind is 10 feet above the base.  The tank is filled to a depth of 4 feet, and the water is flowing into the tank at the rate of 2 ft^3 per minute.

Find the rate of change of the depth of water in the tank. V=BH/3

Could anyone explain and show me the steps to solving this questions?

Thanks :ewpu:

I call the height of water from bottom $h$, and volume of water inside the tank $V$.You're given that $\frac{dV}{dt}= 2$. Question is $\frac{dh}{dt}= ?$. Write down height in terms of water volume, ie $h(V)$, and using chain rule, conclude that

$\frac{dh}{dt}= \frac{dh}{dV}\frac{dV}{dt}$

could you explain what the last part mean?

here is what i did:

dv/dt=1/3 12*12 dh/dt

2=48 dh/dt

dh/dt=1/24

i didn't use the 4 given in the question, and i don't think i did it right.

Intiutively, you can say the rate of change of height will depend on height, so you're quite wrong.

$V = \frac{12^{2}}{3}10-\frac{(12 \times (10-h)/10)^{2}}{3}(10-h)$
$V=48\left(10-\frac{(10-h)^{3}}{100}\right)$
Differentiating both sides, we find $dV = 1.44 (10-h)^{2}dh$. So $\frac{dh}{dV}= \frac{1}{1.44 (10-h)^{2}}$.

$\frac{dh}{dt}= \frac{1}{1.44 (10-h)^{2}}2$
At $h=4$, it's $\frac{2}{1.44 \times 36}$

[quote="juicybooty911"]you got v=144/3 *(something huge)

i can't figure out what that huge bunk is?[/quote]

Well, my calculator says I got ~0.0385803.
Where exactly have I lost you?

you lost me at the first v= equation, i got 144/3, but what's all that after it?

i tried a different way:

since i have a right triangle inside, i can say that its base is X and its height h and set up a proportion

X/h=6/10  (6 came from teh base divide by 2)
X=6h/10=3h/5

so

V=1/3Bh= 1/3 (2x)^2 h

plug in X

dV/dt=1/3 (6h/5)^2 h
dV/dt=1/3 (36hh/25) h
dV/dt=12/25 h^3
dV/dt=(12/25) 3h^2 dh/dt
dV/dt=23.04 dh/dt  (plug in 4 for h)
2/23.04=dh/dt

dh/dt=.08681 ft/min

did i do somthing wrong?

Hmm, first of all you can only use $V = \frac{bh}{3}$ if you are filling the pyramid with vertex down, or that will not be the equation for volume. Then you would get the answer in the previous post. However, if you are filling the pyramid with vertex up, the volume is not as simple of an expression and you would have to go through the calculations with a different volume equation.

EDIT: Salviati was using the second case, in which the vertex is up, as implied by the problem statement.

I believe the question is filling the pyramid with vertex up.

what do you mean by different equations? up and down are not the same? whoa...

For the filling of a vessel of whatever shape,

$\frac{dV}{dt}=A(y)\frac{dy}{dt}$

where $y$ is the depth to which it has been filled and $A(y)$ is the surface area at that depth.

That's what you should focus on: computing $A(y).$ This will be the area of a square, and you should be able to find the sides of that square by either a linear equation or a similar triangles argument.

ok, i got .03858 after doing it again

except this time i turned the thing upside down so i can use the 1/3bh equation.  Instead of plugging in 4, i plugged in 6.


x^n+y^n=z^n

Is there a proof that x, y, and z must be unique? (for equations of this type to work at all, not really getting into fermat stuff) edit: actually I just need a proof that they can't all be the same... I think I got it. Still appreciate anything!)

This sort of equation can have finitely many or infinitely many integer solutions, depending on the equation. A homogeneous polynomial equation of degree $ >2$ will often have infinitely many different (not proportional) integer solutions if it has any, while a nonhomogeneous equation will usually only have finitely many.

As an example, consider elliptic curves $ y^2\equal{}x^3\plus{}ax\plus{}b$, with $ a$ and $ b$ integers and not both zero. No such curve has infinitely many integer solutions. Some of these have no rational solutions, some have finitely many rational solutions (all rational solutions are integers in this case), and some have infinitely many rational solutions. Homogenizing with a third variable $ z$ would make these rational solutions integer solutions: $ y^2z\equal{}x^3\plus{}axz^2\plus{}bz^3$ can have infinitely many truly different integer solutions.


function
induction
algebra unsolved
algebra

Find all the functions f from N to N such that:
f(f(n))+f(n+1) =n+2 holds for all $n/in{N}$

[quote="Rust"]By induction $f(n)=[\frac{\sqrt 5-1}{2}n]+1$.[/quote]
Why do you have this formula -correct beautifull result ? :)
How do you think ?

[quote="Nameless]But why $f(f(n))=[a[an]+a]+1$[/quote]
By induction, because $f(n)=1+[an]\le n$ and (by induction) for all k<=n $f(k)=[ak]+1$.


Determine all real functions $ f,g$ that for every $ x,y$ $ \in$ $ R$ :

$ f(x)\minus{}f(y)$ $ \equal{}$ $ (x\minus{}y)g(x\plus{}y)$

$ (x,y) \rightarrow (\minus{}x,x)$, $ f(\minus{}x)\minus{}f(x)\equal{}\minus{}2x g(0)$
$ (x,y) \rightarrow (x,0)$, $ f(x)\minus{}f(0)\equal{}x g(0)$
$ f(\minus{}x)\minus{}f(0)\equal{}\minus{}x g(0)$
$ f(x)\equal{}x g(0)\plus{}f(0)$ then $ f(x)\equal{}ax\plus{}b$ with $ g(0)\equal{}a$ and $ f(0)\equal{}b$

$ (x,y) \rightarrow (5x,x)$, $ f(5x)\minus{}f(x)\equal{}4x g(6x)$

$ (x,y) \rightarrow (4x,2x)$, $ f(4x)\minus{}f(2x)\equal{}2x g(6x)$

$ f(5x)\plus{}f(4x)\minus{}f(x)\minus{}f(2x)\equal{}6x g(6x)$
$ a5x\plus{}a4x\minus{}ax\minus{}a2x\equal{}6x g(6x)$
$ 6x g(6x)\equal{}a6x$
$ x g(x)\equal{}ax$ for $ x \neq 0$
$ g(x)\equal{}a$ then $ \boxed{g(x)\equal{}a}$

and $ \boxed{f(x)\equal{}a x \plus{}b}$ for any real numbers $ a$ and $ b$

[quote="mszew"]$ (x,y) \rightarrow ( \minus{} x,x)$, $ f( \minus{} x) \minus{} f(x) \equal{} \minus{} 2x g(0)$
$ (x,y) \rightarrow (x,0)$, $ f(x) \minus{} f(0) \equal{} x g(0)$
[/quote]
are you sure about last step?

infact , we can reduce given equ to f(x)=xg(x)+c and
xg(x)-yg(y)=(x-y)g(x+y)
last equation can have solution g(x)=a,or g(x)=x, so f could be x^2+c or ax+c.

I agree with ith_power pointing out that substituting y = 0 gives xg(x) and not xg(0) on the right-hand-side but as regards the next post, I substituted y = 0 to get the first equation and then y = -x to get g(x) + g(-x) = 2 g(0), so that g is an odd function if g(0) = 0. I can see that both g(x)=a and g(x) = x are solutions but I can't see how to show this.


[quote="dna"]if $x.y.z$positive:
$\sum\frac{x}{(x^2+y^2)^{1/2}}\ge2$.[/quote]

> 1, not $\geq 2$, and solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=17549 .

  Darij


arithmetic sequence
number theory
relatively prime
number theory proposed

An increasing arithmetic progression consists of 100 hundred positive integers . Is it possible that every two of them are relatively prime ? If not, show your proof .

As recently proven, there exists also a arithmetic progression of primes of any length.
So yes, they all can be relativly prime.
An easier prove would be to consider the numbers $1, 1+n!, 1+2 \cdot n!, 1+ 3 \cdot n!,..., 1+n \cdot n!$. Since $\gcd(1+an!,1+bn!)=\gcd(1+an!,(a-b)n!=1$, these are all coprime.


AMC
AIME
analytic geometry
Euler
graph theory

Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the mmediately preceding pair, and in which (i, j) and (j, i) do not both appear for any i and j. Let D40 be the set of all dominos whose coordinates are no larger than 40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of D40.

Dont look at these unless you have spent at least 30 minutes on this and have gotten nowhere:

REALLY big hint: [hide] Use graph theory [/hide].

Bigger hint: [hide] euler tours [/hide].

[hide] This is simply the complete graph of forty verticies, where no line can be crossed in two directions. Thus, we need to check if our graph is Eulerian. As it turns out, the order of the outer verticies is 39 (one to every vertex), and so the graph is not Eulerian. The best we can do is to omit one side edge, which leaves two verticies with even orders, and then we can establish a Eulerian path from one vertex to the other. Thus, our answer is 40C2 - 1.[/hide][/hide]

There is one error in your method:

[hide] For a graph to have an eulerian path, either all vertices have even degree or exactly two have odd degree. You still leave 38 vertices with odd degree.

[/hide]

Ahhh!

Can't remember my graph theory correctly...I had assumed it was the other way around.

I suppose that in this case, we remove all direct halfway paths (i.e. from n to n+20) except for one, and so we now have all even verticies except for two, which we can use as the Eulerian path.

Is the answer then 40C2 - 19?


Compute \[\frac{2^{10}-2^8}{2^7-2^6}.\] Express your answer as a fraction.

What the heck? How'd you do the top part?

Lol 10kth user! Thanks!

G-unit you should explain a bit more if possible.Thanks.


quadratics
algebra
quadratic formula
algebra unsolved

solve equation in $ R$:

$ x^2 \plus{} 5x \minus{} 10 \equal{} \sqrt {60 \minus{} 24x \minus{} 5x^2}$

Square both sides to get $ x^{4} \plus{} 10x^{3} \plus{} 10x^{2} \minus{} 76x \plus{} 40 \equal{} 0$

Factorise the left-hand-side to get $ (x^{2} \plus{} 4x \minus{} 10)(x^{2} \plus{} 6x \minus{} 4) \equal{} 0$

Use the quadratic formula to get $ x \equal{} \minus{} 2 \pm \sqrt {14}$ and $ x \equal{} \minus{} 3\pm\sqrt {13}$.

Check by substituting into original equation to see which two of these work and which two are extraneous solutions due to squaring.

Left-hand-side is positive for $ \minus{}2\plus{}\sqrt{14}$ and $ \minus{}3\minus{}\sqrt{13}$ and negative for the other two, so these are the only two real solutions.


A basketball team has won seven of its first fifteen games. How many wins in their next ten games will raise their winning percentage for the season to 60%?

A.2  B.6	   C.7	D.8	E.9

[hide=" click to reveal hidden content"]

$ \frac{7\plus{}x}{15\plus{}10} \equal{} \frac{60}{100}$

$ \implies x\equal{}8$

[/hide]


The area of the triangle is equal with half of the half-circles area.  Calculate angle B.

[url=http://f.imagehost.org/view/0812/Got_Damn][img]http://f.imagehost.org/0812/Got_Damn.jpg[/img][/url]

Half of the half-circle area $ \equal{}\frac{r^2 \pi}{4}$, let the center of the circle be $ O$,

Then area of $ \triangle ABC\equal{}2[COB]\equal{}2(r \sin \angle B)(r \cos \angle B)$
$ \equal{}r^2 \sin \angle 2B$

So $ \sin 2B\equal{}\frac{\pi}{4} \Rightarrow B\equal{}\frac{1}{2}sin^{\minus{}1} \frac{\pi}{4}$

I've done much the same:

Let angle $ CAB$ be $ \theta$.

Then $ \sin\theta = \frac{AC}{2r} and \cos\theta = \frac {BC}{2r}$

Therefore area of triangle $ =\frac{1}{2}\times 2r\sin \theta \times2r\cos \theta = \frac{\pi^{2}}{4}$

Therefore $ r^{2}\sin 2\theta = \frac{\pi^}{4}$

Therefore $ \sin 2\theta = \frac{\pi}{4}$

Therefore $ \theta = \frac{1}{2} \arcsin (\frac{\pi}{4})$.

Let the centre be O, and $ \angle COB$ be $ (180 - x)$ degrees

$ (1/2) \times r^2 \sin x + (1/2) \times r^2 \sin (180 - x) = 1/4 \times r^2 \pi$

$ \sin x = 1/4 \times \pi$

then $ x/2 = \angle B$


Για την πρώτη, όλα στο πρώτο μέλος, διαιρώ με το $15^{x}$ και θεωρώ τη συνάρτηση

$f(x)=\left(\frac{1}{15}\right)^{x}+\cdots+\left(\frac{14}{15}\right)^{x}-1-\left(\frac{16}{15}\right)^{x}-\cdots-\left(\frac{20}{15}\right)^{x}$

της οποίας η παράγωγος είναι $<0$ άρα η συνάρτηση $f$ είναι γνησίως φθίνουσα κι επειδή η $x=1$ είναι προφανής ρίζα (αριθμητική πρόοδος $1+2+\cdots+14=105$ και $15+16+\cdots+20=105$) άρα είναι μοναδική. Όμορφη και έξυπνη άσκηση.

Αλέξανδρος

poly ωραια αλεξανδρε....

Τι λετε, κανει για πανελληνιες?
Για βιβλιο...?

 :)

Γεια σου Θανάση . Όπως πάντα ωραίες ,<<γλυκές>> ασκησούλες.
Για την Β)
Έστω $z=x+yi$ όπου χβτγ μπορούμε να υποθέσουμε ότι $x,y\geq 0$
Η ζητούμενη παράστaση είναι ίση με $\sqrt{2}y+2x$
Άρα μένει να βρούμε το ελάχιστο και το μέγιστο της $\sqrt{2}y+2\sqrt{1-y^{2}}$
Μετά ο μηδενισμός της παραγώγου είναι $y=\frac{1}{\sqrt{3}}$
και μετά το ελάχιστο επιτυγχάνεται στο y=1

[quote="socrates"][color=indigo][b]Solve the equation $a^{x}+a^{\frac{1}{x}}=\lambda$, where $a>1$ and $\lambda >2a$. For example, $9^{x}+9^{\frac{1}{x}}=84\Longleftrightarrow x\in \{2,\frac{1}{2}\}$.[/b][/color][/quote]
[color=darkred][b][u]Lemma.[/u] For $a>1$ the function $f\ : \ (0,\infty )\rightarrow \mathcal R\ ,\ f(x)=a^{x}+a^{\frac{1}{x}}$ is decreasing on $(0,1]$ and increasing on $[1,\infty )$.[/b][/color]

[color=darkblue][b][u]Proof.[/u] Denote $b=\sqrt a>1$ and the functions $\{\begin{array}{c}h\ : \ (0,\infty )\rightarrow (1,\infty )\ ,\ h(x)=b^{x}\\\\ \phi \ : \ (0,\infty )\rightarrow [2,\infty )\ ,\ \phi (x)=x+\frac{1}{x}\\\\ \psi\ : \ (0,\infty )\rightarrow \mathcal R\ ,\ \psi (x)=x-\frac{1}{x}\end{array}$. Observe that these functions are positive and bijectively.

Therefore, $f(x)=b^{2x}+b^{\frac{2}{x}}=b^{x+\frac{1}{x}}\cdot(b^{x-\frac{1}{x}}+b^{\frac{1}{x}-x})$, i.e. $f=h(\phi )\cdot \phi [h(\psi )]$. Prove easily that

the functions $h$ and $\psi$ are increasing on $(0,\infty )$, but the function $\phi$ is decreasing on $(0,1]$ and increasing on $[1,\infty )$.

In conclusion, the function $f$ is decreasing on $(0,1]$ and increasing on $[1,\infty )$.

[u]Remark.[/u] For any $x>0$ , $f(x)\ge 2a$ and for any $\lambda >2a$ the equation $f(x)=\lambda\ ,\ x>0$ $\Longleftrightarrow$ $x\in\{x_{1},x_{2}\}$, where $x_{1}x_{2}=1$.[/b][/color]

[quote="socrates"][color=darkred][b]For any  $z \in \mathcal C$ , $|z|=1$ $\implies$ $\sqrt{2}\leq |1+z|+|1+z^{2}|\le 4$[/b][/color][/quote]
[color=darkblue][b][u]Proof I.[/u][/b] Prove easily that for any $\omega\in\mathcal C$ , $|\omega |=1\Longleftrightarrow |1+\omega^{2}|=|\omega+\overline\omega |$ and in this case $|\omega+\overline\omega |=r\in [0,2]$.
Denote $\omega\in\mathcal C$ for which $\omega^{2}=z$. From $|z|=1$ obtain $|\omega |=1$. Thus, $f(z)\equiv |1+z|+|1+z^{2}|=$ $|1+\omega^{2}|+|1+\omega^{4}|=$
$|\omega+\overline \omega |+|\omega^{2}+\overline\omega^{2}|=r+|r^{2}-2|$, where $|\omega+\overline \omega |=r\in [0,2]$. If $r\in (\sqrt 2,2]$, then $f(z)>\sqrt 2$ is evidently.
Suppose $r\in [0,\sqrt 2]$. Therefore, $f(z)\ge \sqrt 2$ $\Longleftrightarrow$ $2-r^{2}\ge \sqrt 2-r\ge 0$ $\Longleftrightarrow$ $r^{2}-r+\sqrt 2-2\le 0$ $\Longleftrightarrow$
$(\sqrt 2-r)(r+\sqrt 2-1)\ge 0$, what is truly. We have equality iff $r=\sqrt 2$, i.e. $z\in \{\pm i\}$.
[b]Remark.[/b] $|1+z|+|1+z^{2}|\le 1+|z|+1+|z|^{2}=4$. For $z: =-z$ obtain the initial exercise !

=======================================================================================

[b]Lemma 1.[/b] $\boxed{\ z\in\mathcal C\ ,\ |z|=1\implies |1+z^{2}|=|z+\overline z|\ }$.

$\{\begin{array}{c}|1+z^{2}|^{2}=(1+z^{2})(1+\overline z^{2})=1+z^{2}+\overline z^{2}+(z\overline z)^{2}=2+z^{2}+\overline z^{2}\\\ |z+\overline z|^{2}=(z+\overline z)(\overline z+z)=z^{2}+\overline z^{2}+2z\overline z=2+z^{2}+\overline z^{2}\end{array}$ because $|z|=1\Longleftrightarrow z\overline z=1$.

[b]Lemma 2.[/b] $\boxed{\ a\in\mathcal R\ ,\ |a|\le 1\implies \sqrt{1+a}+|a|\sqrt 2\ge 1\ }$.

$1\blacktriangleright a\in [0,1]\implies$ $\{\begin{array}{c}|a|\sqrt 2\ge 0\\\\ \sqrt{1+a}\ge 1\end{array}\|$ $\implies$ $\sqrt{1+a}+|a|\sqrt 2\ge 1$ with the equality iff $a=0$.
$2\blacktriangleright$ $a\in[-1,-\frac{\sqrt 2}{2}]\implies$ $\sqrt{1+a}\ge 0\ge 1+a\sqrt 2=1-|a|\sqrt 2$ $\implies$ $\sqrt{1+a}+|a|\sqrt 2>1$.
$3\blacktriangleright$ $a\in[-\frac{\sqrt 2}{2},0)$. Thus, $\sqrt{1+a}+|a|\sqrt 2\ge 1$ $\Longleftrightarrow$ $\sqrt{1+a}\ge 1+a\sqrt 2\ge 0$ $\Longleftrightarrow$ $1+a\ge 1+2a\sqrt 2+2a^{2}$ $\Longleftrightarrow$
$1\le 2\sqrt 2+2a$ $\Longleftrightarrow$ $a\ge \frac{1-2\sqrt 2}{2}$, what is truly because $a\ge-\frac{\sqrt 2}{2}>\frac{1-2\sqrt 2}{2}$, i.e. $\sqrt 2>1$.

[b][u]Proof II.[/u][/b] Denote $z=a+bi\ ,\ a^{2}+b^{2}=1$. Observe that $|1+z|=|1+a+bi|=\sqrt{(1+a)^{2}+b^{2}}=\sqrt{2(1+a)}$ and from the first lemma $|1+z^{2}|=|z+\overline z|=|2a|=2|a|$. Thus, $|1+z|+|1+z^{2}|\ge \sqrt 2$ $\Longleftrightarrow$ $\sqrt{1+a}+|a|\sqrt 2\ge 1$ , what is truly from the second lemma because $a^{2}+b^{2}=1\implies |a|\le 1$. We have equality iff $a=0$, i.e. $z\in\{\pm i\}$.

[b]A more easy similar exercise.[/b] $z\in\mathcal C\ ,\ |z|=1\implies |1+z|+|1+z^{2}|+|1+z^{3}|\ge 2$ ([u]M. Becheanu, L. Panaitopol[/u]).

[b][u]Request.[/u][/b] Can anyone give another proof using only identities or inequalities over $\mathcal C$without sinking as $z=a+bi$ or $z=\cos\phi+i\cdot \sin\phi$  ?![/color]

[color=darkblue][b]Very nice the your proposed problems, Socrates ! Thank you for it.[/b][/color]

You are welcome, Mr Virgil!!

Very nice and instructive your observations, as well..

I ll post some more very soon....

 :)

ΚΟΝΨΗ ΓΕΩΜΕΤΡΚΗ ΛΥΣΗ ΤΩΝ ΜΙΓΑΔΙΚΩΝ

[quote="Virgil Nicula"][color=darkblue][b]Very nice proof, [u]R_boris[/u],[/b] but you solved in a particular case $\mathcal{R}e(z)\ge 0$, where $2\cdot\mathcal R e(z)=z+\overline z$.

Indeed, $BD\ge \sqrt 2$ $\Longleftrightarrow$ $|z^{2}+z|\ge \sqrt 2$ $\Longleftrightarrow$ $|z+1|\ge \sqrt 2$ $\Longleftrightarrow$ $(z+1)(\overline z+1)\ge 2$ $\Longleftrightarrow$ $2+2\cdot\mathcal{R}e(z)\ge 2$ $\Longleftrightarrow$ $\mathcal R e(z)\ge 0$.

[b]Remark.[/b] This problem is equivalently with the folowing :[/color]

[quote][color=darkred][b]Let $ABC$ be an $A$- isosceles triangle ($b=c$) with the circumcircle $w=C(O,R)$.
Denote the reflection $B'$ of the vertex $B$ w.r.t. the circumcenter $O$. Prove that $B'A+B'C\ge R\sqrt 2$.[/b][/color][/quote]
See http://www.mathlinks.ro/Forum/viewtopic.php?t=152962[/quote]
YES you' re right.If$Re(z)<0\Rightarrow Re(-z)>0\Rightarrow|1-(-z)|+|1+z^{2}|\ge \sqrt{2}\Rightarrow|1+z|+|1+z^{2}|\ge \sqrt{2}$ but$Re(z)<0\Rightarrow |1-z|>|1+z|$additioning the above inequality's we have the result

Λοιπον συνεχιζω με μερικες ακομα ασκήσεις..
Ελπιζω να σας δροσισουν........

 :lol:

Α) Solve in $\mathbb{Z}\times \mathbb{Z}$ the equation $x^{5}-xy^{2}+y^{2}-1=0$

B)Consider a triangle $ABC$ and the interior bisectors of its angles $AA_{1}$,$BB_{1}$ and $CC_{1}$ ($A_{1},\ B_{1}, \ C_{1}$ are on the respective sides $BC$ etc )
        If $ \vec{AA_{1}}, \ \vec{BB_{1}},\ \vec{CC_{1}}$ form a triangle find the angles of $\triangle{ABC}$

 :)

[quote="silouan"]Η πρώτη είναι άκρως δροσιστική  :P  .Η δεύτερη sorry αλλά με παραπέμπει σε υπολογιστικά πράγματα οπότε με 40 βαθμούς θεωρείται ακατάλληλη .

Λοιπόν για την πρώτη .Γράφεται $(x-1)(x^{4}+x^{3}+x^{2}+x+1-y^{2})=0$
*Αν x=1 δίνει άπειρες λύσεις της μορφής $(x,y)=(1,y)$

*Αν $x\neq 1$ τότε $(2y)^{2}=4x^{4}+4x^{3}+4x^{2}+4x+4$
και για $x\in (-\infty ,-1)\cup\left( 3,+\infty\right)$  έχουμε

$(2x^{2}+x)^{2}<(2y)^{2}<(2x^{2}+x+1)^{2}$ άρα δεν έχουμε λύσεις .

Άρα $x=-1$ ή $x=0$ ή $x=2$ ή $x=3$
από όπου παίρνουμε τα ζεύγη $(x,y)=(-1,1),\ (-1,-1),\ (0,1),\ (0,-1),\ (3,-11),\ (3,11)$

QED[/quote]

well done!!

ταχυτατος και ευθυβολος.....

 :lol:

[quote="Virgil Nicula"][b]A more easy similar exercise.[/b] $z\in\mathcal C\ ,\ |z|=1\implies |1+z|+|1+z^{2}|+|1+z^{3}|\ge 2$ ([u]M. Becheanu, L. Panaitopol[/u]).[/quote] [url=http://www.mathematica.gr/forum/viewtopic.php?p=184725#p184725]εδώ[/url]


Can someone show me a proof that $e^{-x^2}$ does not have an elementary primitive?  Or a general method (if it exists) of prooving that a function does not have an elementary primitive?
Thanks.

Do you read Russian? (I know a web site where the full proof is posted but it is in Russian).

Try  [url]http://www.math.niu.edu/~rusin/known-math/93_back/elementary.int[/url] then. It is a sketch of a proof but you may be able to understand it...

OK I found an even better reference: [url]http://math.hunter.cuny.edu/ksda/papers/churchill.pdf[/url]


function
integration
algebra
domain
limit
real analysis
real analysis unsolved

Find all continous  functions $ f: [0,a] \minus{} > [0,\infty]$ with the property $ f(t) \leqq (\int_{0}^{t}f(s)ds)^{2}$, $ \forall t \in [0,1]$.

You probably wanted the domain of $ f$ to be $ [0,1]$, for the problem wouldn't make sense for $ a<1$, and for $ a>1$, the rest of the domain would be irrelevant.

In this case, let $ g(t)\equal{}\int_0^t f(s)\ ds$ so $ g^\prime (t)\leq g(t)^2$. Also $ g(0)\equal{}0$ and $ g$ increasing, so if $ g(a)\equal{}0$ for any $ a\in [0,1]$, then $ g(t)\equal{}0$ for every $ t\in [0,a]$.

We have the trivial solution $ g\equiv 0$ giving $ f\equiv 0$. Otherwise, let $ b$ be the largest real number satisfying $ g(b)\equal{}0$. For $ t\in (b,1]$, $ g$ is nonzero. In this case, let $ h\equal{}x\plus{}\frac{1}{g}$ so that
\[ h^\prime \equal{}1\minus{}\frac{g^\prime}{g^2}\geq 0\]
meaning $ h$ increasing. However, $ \lim_{t\rightarrow b^\plus{}} h(t)\equal{}\infty$ which is a contradiction. So we necessarily have $ f(x)\equal{}0$ for all $ x\in [0,1]$.


Bob is 5 years old.  Bobito is half as old as four times the age of Bobella.  Bobella is twice as old as Bob.  How old is everyone?

[hide="click if colts rock"]
bob-5
bobella-10
bobito-20[/hide]

Extension:  Bobitoella is 4 times as young as half of Bobellaito.  Bobellaito is half as old three times Bob's age.  How old is Bobitoella and Bobellaito?

[hide="my answer"]
Bobitoella=15 years old
Bobellaito=7.5 years old[/hide]


[quote="Jonhs"]Let $x;y>0$ satisfying $x+y=1$.Find min of:
$P=\frac{x+2y}{\sqrt{1-x}}+\frac{y+2x}{\sqrt{1-y}}$[/quote]
$P\geq3\sqrt2\Leftrightarrow\sum_{cyc}\frac{1+x}{\sqrt x}\geq3\sqrt2\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{1+x}{\sqrt x}+\frac{1}{\sqrt2}\cdot\left(x-\frac{1}{2}\right)-\frac{3}{\sqrt2}\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{\left(\sqrt{2x}-1\right)^{2}\left(\sqrt x+2\sqrt2\right)}{\sqrt x}\geq0.$ :)

The following inequality stronger:
Let $x$ and $y$ are positive numbers satisfying $x+y=1$. Prove that
$\frac{x+7y}{\sqrt{1-x}}+\frac{y+7x}{\sqrt{1-y}}\geq8\sqrt2.$ :wink:

[quote="Jonhs"]Let $x;y>0$ satisfying $x+y=1$.Find min of:
$P=\frac{x+2y}{\sqrt{1-x}}+\frac{y+2x}{\sqrt{1-y}}$[/quote]
the problem can be written by finding min of
$S=\sqrt{x}+\sqrt{y}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}$
we have $S \geq \frac{2}{\sqrt{x}+\sqrt{y}}+(\frac{2}{\sqrt{x}+\sqrt{y}}+\sqrt{x}+\sqrt{y}) \geq \frac{2}{2\sqrt{\frac{x+y}{2}}}+2\sqrt{2}=3\sqrt{2}$

[hide]this is problem 3(51) in one of the newest magazine in VN  :wink: [/hide]


calculus
derivative
function
algebra
functional equation
calculus computations

1. $ x \epsilon N^ \plus{} , f(0) \equal{} 0,$

$ f(x) \minus{} f(x \minus{} 1) \equal{} x \plus{} 1$

$ f^{\prime}(6) \equal{} ?$

2. $ x \epsilon Z^ \plus{} , f(0) \equal{} 0$

$ f(x) \equal{} x^2 \plus{} f(x \minus{} 1)$

$ f^{\prime}(1) \plus{} f^{\prime}(0) \equal{} ?$

Derivatives don't make sense for functions on the integers. Both questions are nonsense.

Notation note: If you use \prime for a derivative, put it in an exponent. $ f^{\prime}$ looks much better.

[quote="jmerry"]Notation note: If you use \prime for a derivative, put it in an exponent.  looks much better[/quote]

Why not use $ f'$ instead?

[quote="queen"]
$ f(x) \equal{} x^2 \plus{} f(x \minus{} 1)$

$ f^{\prime}(1) \plus{} f^{\prime}(0) \equal{} ?$

[/quote]

$ f(x) \equal{} x^2 \plus{} f(x\minus{}1) \\ \\
f'(x) \equal{} 2x \plus{} f'(x\minus{}1) \\ \\
f'(1) \minus{} f'(0) \equal{} 2(1)$

 :maybe:
But I doubt if this is correct.

$ f'(1)$ and $ f'(0)$ are supposed to be constants. So, in both cases the answer should be $ 0$.

[quote="Rajiv"]Why not use f' instead? [/quote]I do.

The larger point is that functions defined on the integers don't have derivatives. We need intervals in $ \mathbb{R}$ or something similar to make sense of the derivative.
Interpolation doesn't help, either. A function interpolating the known values and satisfying the functional equation could have any value for its derivative at a chosen point.

[quote="jmerry"]
Interpolation doesn't help, either. A function interpolating the known values and satisfying the functional equation could have any value for its derivative at a chosen point.[/quote]

I haven't studied interpolation, so I beg pardon... :(


AMC
AMC 10
AMC 12
AIME
probability
summer program
Mathcamp

What is the averge AMC10/12 score for someone in middle school. I remember getting like 12 out of the first 15 before I went to sleep last night. I am going into 7th grade, almost exactly 12 1/2 years old. Is this good?Bad?

i'd think an average middle school AMC 10 score would be around 70-80ish right? i got 85  :D , 3rd best at my school on the 2004 test. The AMC website has the stats. (80.3 for 2005 AMC 10A and 80.4 for AMC 10B)

[quote="paladin8"]12/15 on the AMC10 would be pretty decent, you'd probably have to take a look at the last 10 problems to be sure of what kind of score you're looking at. 12/15 on the AMC12 would be really good, enough to qualify for AIME already, but again, look at the last 10 and see what you can do.

I didn't take it until 8th grade, when I got 113 on the AMC10, with qualification at 121.5. I'm sure other people here could share their scores so you could "compare" in a sense. And if you're going into 7th grade, I think you're already well ahead of people in my year (well me at least, I didn't know about the AMC until a couple of weeks before I took it); then again, 3 years ago everything was completely different because AoPS wasn't a full-blown math community yet.

Basically, there's very little chance you can have a "bad" score on the AMC at 7th grade because MOST students are unaware of it (though the number that are is growing rapidly due to AoPS).[/quote]

Getting 12 problems right on an AMC 12 qualifies for AIME!?!??!?

I can do that!

How can you have a score of 70 if there are only 25 problems?

Each problem is worth 6 points.

I got around 112 in 8th grade, but that was on the AMC 12.

Anyways, since this is the AoPS forums, you're probably gonna find a pretty high "average". Keep in mind that the true average is probably really low, cuz alot of people take it.

The average score for middle school testtakers on the AMC 10 and 12 is higher than the overall average, but that's mainly because the middle schoolers who are taking it have been selected by their school or teacher, so it's more of a selective process.

If you only answered 12 of the first 15 and then no others, that'd get you a 104.5, which is very respectable. I'm assuming that you didn't get anything wrong and omitted the other 13 problems.

Yea, comparing middle schoolers on the AMC 10 won't really shouw the big picture.  Even taking it in 7th grade is a big thing.  And if you keep visiting AoPS, and doing problems, I'm sure you could make it to AIME.

If you want to compare yourself with other middleschoolers though, take the AMC 8, or MathCounts.

[quote="probability1.01"][quote="MysticTerminator"]dude I remember getting a 106 on the AMC-10. and I was annoyed. b/c the cutoff was 116.[/quote]

I can't remember what I got.  I heard Alex got a 116 some time ago on the AMC 10... cutoff was 117 :(.

Anyhows as long as you work at it I think just about anybody can make AIME.[/quote]

OUCHYPOOS

I bet my sister could never ever ever make AIMEs. mainly b/c she's a chihuahua

[quote="probability1.01"]
Anyhows as long as you work at it I think just about anybody can make AIME.[/quote]

I wouldn't be too sure about that... if you mean AoPS frequenters (or maybe even just those who have read the books) then yes, perhaps, but it's not because nobody works that 10,000 out of 400,000 make the AIME.

AMC scores in middle school--

123 AMC 10 5th
102.5 AMC 12 6th
112.5 AMC 12 7th
136 AMC 12 8th

probability--I'd have to agree with paladin on this one.

Also, there are lots of people who qualify for the AIME, but don't work enough.  (like me).  But yeah, I'm gonna devote this entire year to WOOT and other Olympiad level junk.  Orznorz.  Whatev.

[quote="MysticTerminator"]dude I remember getting a 106 on the AMC-10. and I was annoyed. b/c the cutoff was 116.[/quote]

When was that Arnav? In 5th grade?  :P

ORZNORZ...  :rotfl:

That was teh most annnoying thing eva...

[quote="bubala"]
123 AMC 10 5th
102.5 AMC 12 6th
112.5 AMC 12 7th
136 AMC 12 8th[/quote]

that is some high scores..... WOW. I didnt get above 130 in AMC10 until 8th grade, and never gotten above 130 in AMC12 ever. Very Respectable. I will be amazed even more if you beat my AIME scores in that ages.

[quote="white_horse_king88"][quote="MysticTerminator"]dude I remember getting a 106 on the AMC-10. and I was annoyed. b/c the cutoff was 116.[/quote]

When was that Arnav? In 5th grade?  :P[/quote]

ooh no!!! guess again!!!!! :lol::lol::lol:

[quote="white_horse_king88"]ORZNORZ...  :rotfl:

That was teh most annnoying thing eva...[/quote]

agreed, so why is it still around ... just say no!!!

[quote="xxreddevilzxx"]I will be amazed even more if you beat my AIME scores in that ages.[/quote]

ooh sounds like a challenge!
HEAD TO HEAD!
HEAD TO HEAD!

[quote="MysticTerminator"][quote="white_horse_king88"]
ORZNORZ...  :rotfl:

That was teh most annnoying thing eva...[/quote]

agreed, so why is it still around ... just say no!!!

[/quote]

Apparently, Zhai Zhai's roomate at MOP said "orznorz" all the time.  (And omgnore).  I picked it up from him at mathcamp, so ORZNORZ lives on!!!

[quote="reddevilz"][quote="bubala"]

123 AMC 10 5th
102.5 AMC 12 6th
112.5 AMC 12 7th
136 AMC 12 8th

[/quote]

that is some high scores..... WOW. I didnt get above 130 in AMC10 until 8th grade, and never gotten above 130 in AMC12 ever. Very Respectable. I will be amazed even more if you beat my AIME scores in that ages.[/quote]

The reason I did well was because I was in high school math courses and so I knew a bunch of random junk.  After 7th grade I decided to start actually working a couple hours per day because I knew I had USAMO qualifying potential.  Still, I was too lazy lazy lazy.  Now I find out you AoPS people work 5 hours a day on random junk!  Yeah, I should start being un-lazy, un-lazy, un-lazy.

[quote="bubala"][quote="MysticTerminator"][quote="white_horse_king88"]
ORZNORZ...  :rotfl:

That was teh most annnoying thing eva...[/quote]

agreed, so why is it still around ... just say no!!!

[/quote]

Apparently, Zhai Zhai's roomate at MOP said "orznorz" all the time.  (And omgnore).  I picked it up from him at mathcamp, so ORZNORZ lives on!!!
[/quote]

great. now kill it.

[quote="MysticTerminator"][/quote][quote="bubala"][quote="MysticTerminator"][quote="white_horse_king88"]
ORZNORZ...  :rotfl:

That was teh most annnoying thing eva...[/quote]

agreed, so why is it still around ... just say no!!!

[/quote]

Apparently, Zhai Zhai's roomate at MOP said "orznorz" all the time.  (And omgnore).  I picked it up from him at mathcamp, so ORZNORZ lives on!!!
[/quote][quote="MysticTerminator"]

great. now kill it.[/quote]
With what? The nearest large blunt object?

In 7th grade I got a 144 on AMC 10, then got a 8 on AIME. I believe treething has the same scores, or nearly the same.

[quote="PenguinIntegral"][/quote][quote="MysticTerminator"][/quote][quote="bubala"][quote="MysticTerminator"][quote="white_horse_king88"]
ORZNORZ...  :rotfl:

That was teh most annnoying thing eva...[/quote]

agreed, so why is it still around ... just say no!!!

[/quote]

Apparently, Zhai Zhai's roomate at MOP said "orznorz" all the time.  (And omgnore).  I picked it up from him at mathcamp, so ORZNORZ lives on!!!
[/quote][quote="MysticTerminator"]

great. now kill it.[/quote][quote="PenguinIntegral"]
With what? The nearest large blunt object?[/quote]

that would suffice!!! :lol:

[quote="xxreddevilzxx"][quote="PenguinIntegral"][quote="xxreddevilzxx"]..... I will be amazed even more if you beat my AIME scores in that ages.[/quote]

What do I have to beat?[/quote]

haha..... that wasnt really a challenge, as Arnav said. I got 1 as a 7th grader, 2 as eighth. But, considering the fact that back then there were not much of resources as AoPS around, that is a respectable score.[/quote]
... I got 8 on the AIME in 8th without AoPS...

[quote="krustyteklown"][/quote][quote="xxreddevilzxx"][quote="PenguinIntegral"][quote="xxreddevilzxx"]..... I will be amazed even more if you beat my AIME scores in that ages.[/quote]

What do I have to beat?[/quote]

haha..... that wasnt really a challenge, as Arnav said. I got 1 as a 7th grader, 2 as eighth. But, considering the fact that back then there were not much of resources as AoPS around, that is a respectable score.[/quote][quote="krustyteklown"]
... I got 8 on the AIME in 8th without AoPS...[/quote]

That doesn't change the fact that it's a very respectable score if you got higher than it.

[quote="MysticTerminator"][quote="white_horse_king88"]I bet if you sneak into HaiTao (Zhai Zhai's RM)'s room at night and modify his memory and teleport him back in time before MOP, then ORZNORZ will be erased.

Am I the only one who didn't find out about AMC's until like 9th grade? LOL.[/quote]

wtf you made MOP in your first year of doing AMC? sheesh...[/quote]

I'm going to agree with Arnav.

[quote="JesusFreak197"][quote="white_horse_king88"]Am I the only one who didn't find out about AMC's until like 9th grade? LOL.[/quote]

I didn't find out about it until 8th grade, after it had already happened.[/quote]

Same. Looking forward to convincing my HS to do it this year. :roll:

[quote="paladin8"]I found out about it 8th grade about a week before the B test happened... it's pretty amazing how much AMC participation among MS students has grown with the advent of AoPS. It also means that USAMO/MOP/IMO Qualification will become increasingly more difficult (joy! ;)). But maybe I can avoid the brunt of that wave of AoPSers :P.[/quote]

I also found out about the same time as you, hehe. "A" test already happened.

[quote="JesusFreak197"][quote="bubala"]Also, there are lots of people who qualify for the AIME, but don't work enough. (like me).[/quote]

Same here. When I took the AMC-10 this last year, I had never seen any AMC test before, and I got a 134.5. :P I guess it  comes from my experience with MC, mostly. Then I got a 3 on the AIME, though, so by the time you get there you need to be studying to get farther. :P[/quote]
That's strange. I took the AMC-10 for the first time this past school year, and got a 125, then went on to get a 6 on the AIME...then again, I like a few hard questions more than many short ones. Perhaps I'll flourish in the USAMO if I'm good enough.  :D

a lot of ppl find out in different ways... it really depends on the school... which really sucks becuase if you are super smart but unfortunately your school doesnt really care, then you could be missing out on amc/aime during high school

for middle school i think a lot of aops-ers are advanced in math and take high school courses at a high school  and thats how i found out about amc   i barely missed the cutoff on amc-12   i got a 99 (ithink) cutoff is 100  :mad:   well i've got all of high school to improve! :D


inequalities
trigonometry
function
inequalities unsolved

[color=darkblue]Let $ x, y >0$ and $ x\plus{}y\equal{}1$. Find minimum of $ A\equal{}\frac{x}{\sqrt{1\minus{}x}}\plus{}\frac{y}{\sqrt{1\minus{}y}}$.[/color]

This inequality can be proved by many ways. I will suggest the first solution:
Using Chebyshev inequality, we have:
$ \frac{x}{\sqrt{1\minus{}x}}\plus{}\frac{y}{\sqrt{1\minus{}y}} \geq \frac{(x\plus{}y)}{2}(\frac{1}{\sqrt{1\minus{}x}}\plus{}\frac{1}{\sqrt{1\minus{}y}})$
By using Cauchy Schwartz inequality, we have the following result:
$ \frac{1}{\sqrt{1\minus{}x}}\plus{}\frac{1}{\sqrt{1\minus{}y}} \geq \frac{4}{\sqrt{1\minus{}x}\plus{}\sqrt{1\minus{}y}}$
But the Bunhiacovsky inequality helps us:
$ \frac{4}{\sqrt{1\minus{}x}\plus{}\sqrt{1\minus{}y}} \geq \frac{4}{\sqrt{2(2\minus{}x\minus{}y)}}$
Now we get the last result, minimum of the expression is $ \sqrt{2}$.
The equality holds if and only if  $ x \equal{}y \equal{}\frac{1}{2}$

[color=darkblue]Thank you. I sloved this problem by let $ x\equal{}sin^2\alpha, y\equal{}cos^2\alpha$ but it's long.[/color]

[quote="thanhnam2902"][color=darkred]Let $ x, y > 0$ and $ x + y = 1$. Find minimum of $ A = \frac {x}{\sqrt {1 - x}} + \frac {y}{\sqrt {1 - y}}$ .

 I solved this problem by substitution $ x = sin^2\phi$ and $ y = cos^2\phi$ , but [b]it's long[/b].[/color][/quote]

[size=150][color=darkred][b]It is no long , [u]Thanhnam[/u] ![/b][/color] [/size]

[color=darkblue]With your nice substitution obtain : $ A = \frac {\sin^3\phi + \cos^3\phi}{\sin\phi\cos\phi} = f(\sin\phi + \cos\phi )$ , where $ \{\begin{array}{c} f\ : \ [1,\sqrt 2]\ \rightarrow\ \mathbb R \\
\ f(t) = \frac {t(3 - t^2)}{t^2 - 1}\end{array}$

because $ \sin\phi\cos\phi = \frac {t^2 - 1}{2}$ and for $ (\forall)\ x\in[0,\frac {\pi}{2}]$ we have $ t\in[1,\sqrt 2]$ . Prove easiky that $ f'(t) < 0$ , i.e. the function $ f$ is decreasing.

Therefore, $ \begin{array}{cccccccc} \phi & \blacktriangleright & \| & 0 & \ldots & \frac {\pi}{4} & \ldots & \frac {\pi}{2} \\
 \\
= = & = = & \| & = = & = = & = = & = = & = = \\
 \\
t & \blacktriangleright & \| & 1 & \nearrow & \sqrt 2 & \searrow & 1 \\
 \\
f(t) & \blacktriangleright & \| & \infty & \searrow & 1 & \nearrow & \infty\end{array}$ . Thus, $ (\forall )\ t\in [1,\sqrt 2]$ we have $ f(t)\ge f(\sqrt 2) = \sqrt 2$ , i.e. $ A\ge \sqrt 2$ .

We have equality if ans only if $ t = 1\ \Longleftrightarrow\ \phi = \frac {\pi}{4}\ \Longleftrightarrow\ x = y = \frac 12$ .[/color]

Apply the inequality of convexity $ mf(x)\plus{}nf(y)\ge f(mx\plus{}ny)\ ,\ m>0\ ,\ n>0\ ,\ m\plus{}n\equal{}1$ for the convex function

$ f(t)\equal{}\frac {1}{\sqrt t}$ and $ m: \equal{}y$ , $ n: \equal{}x$ . Obtain : $ \frac {y}{\sqrt x}\plus{}\frac {x}{\sqrt y}\ge\frac {1}{\sqrt {y\cdot x\plus{}x\cdot y}}\equal{}\frac {1}{\sqrt {2xy}}$ , i.e. $ \boxed {\ \frac {x}{\sqrt {1\minus{}x}}\plus{}\frac {y}{\sqrt {1\minus{}y}}\ge\frac {1}{\sqrt {2xy}}\ }$ .

Since $ 2xy\le \frac {(x\plus{}y)^2}{2}\equal{}\frac 12$ obtain $ \frac {1}{\sqrt {2xy}}\ge \sqrt 2$ . In conclusion, $ \frac {x}{\sqrt {1\minus{}x}}\plus{}\frac {y}{\sqrt {1\minus{}y}}\ge\sqrt 2$ with equality iff $ x\equal{}y\equal{}\frac 12$ .

$ \sqrt{1\minus{}x}\equal{}\sqrt{y}$...

$ \frac{x}{\sqrt{y}}\plus{}\frac{y}{\sqrt{x}}\ge \sqrt{2}\cdot\sqrt{x\plus{}y}$

$ \frac{x^2}{y}\plus{}2\sqrt{xy}\plus{}\frac{y^2}{x}\ge 2(x\plus{}y)$

$ x\equal{}u^2$, $ y\equal{}v^2$

$ \frac{u^4}{v^2}\plus{}2uv\plus{}\frac{v^4}{u^2}\ge 2 u^2\plus{}2v^2$

$ \frac{(u\minus{}v)^2}{u^2v^2}\cdot (u^4\plus{}2u^3v\plus{}u^2v^2\plus{}2uv^3\plus{}v^4)\ge 0$


Show that $4\leq xyz\leq \frac{112}{27}$ on the curve $x+y+z=5$ and $xy+yz+xz=8$,
by solving it as an extremevalue problem.

(Its sort of maximizing the volume of a box with restrictions of side length and area)
Try to describe the constraint region.

The constraint curve is a circle. It is centered at the point $\left(\frac53,\frac53,\frac53\right)$ and has radius $\frac{\sqrt{6}}3.$ An example of a point on this circle: $\left(\frac53,\frac{5-\sqrt{3}}3,\frac{5+\sqrt{3}}3\right).$

As for the optimization problem, we use Lagrange multipliers. Let $f(x,y,z)=xyz,g(x,y,z)=x+y+z,$ and $h(x,y,z)=xy+xz+yz.$ We set $\nabla f=\lambda\nabla g+\mu\nabla h,$ or

$yz=\lambda+\mu(y+z)$
$xz=\lambda+\mu(x+z)$
$xy=\lambda+\mu(x+y).$

Adding these three equations, we get

$xy+xz+yz=3\lambda+2\mu(x+y+z)$ which simplifies to

$8=3\lambda+10\mu$ or $\lambda=\frac{8-10\mu}3.$

Now subtract the first two equations to get

$(y-x)z=\mu(y-x)$ so either $x=y$ or $z=\mu.$

Similarly, from the 2nd and 3rd equations, we can conclude that either $y=z$ or $x=\mu.$

One possibility would be $x=y=z,$ but putting that into the constraint equations leads to no solution. (Geometrically, this line is perpendicular to the plane of the constraint circle and passes through its center.)

That leaves us with $x=y=\mu$ with $z$ different (and similar possibilities). Put that into the third equation to get

$\mu^{2}=\lambda+2\mu.$ Now put $\lambda=\frac{8-10\mu}3$ to get

$3\mu^{2}-10\mu+8=0.$ This has solutions $\mu=2$ or $\mu=\frac43.$

This, and permutations of this, gets us two sets of critical points.

First case: $(2,2,1),(2,1,2),$ or $(1,2,2).$ At that point, $xyz=4.$

Second case: $\left(\frac43,\frac43,\frac73\right),\left(\frac43,\frac73,\frac43\right),$ or $\left(\frac73,\frac43,\frac43\right).$ At that point, $xyz=\frac{112}{27}.$

The constraint curve is a compact set, so $f$ must assume a maximum and a minimum. Since we have found all of the critical points, the values we have found must include both the maximum and the minimum. It follows then that on this circle, $4\le xyz\le\frac{112}{27}.$

$x,\ y,\ z$ are the [color=blue]positive real roots [/color] of the cubic equation $f(t): =t^{3}-5t^{2}+8t=k\Longleftrightarrow f\left(\frac{4}{3}\right)\leq k\leq f(2)$, we are done.


induction
number theory proposed
number theory

guys,have a go at this one.hope u come up with sumthing shorter and sweeter (the motto (while solving problems of course is
KISS- "K"eep "I"t "S"hort & "S"weet )
as for the problem :D
The Fibonacci numbers are defned by F0 = F1 = 1 and F_n+2 =
F_n+1 + F_n for n  \geq  0. Let A and B be natural numbers such that A^19 divides B^93 and also B^19 divides A^93. Prove: for all natural numbers n  \geq  1
the number (A^4 + B^8)^[F_(n+1)] is divisible by (AB)^[F_n].
cheerss

A hint:

Prove that A | B5 and B | A5.
Use induction :)

well my proof uses induction as well (but it's kinda dirty,if u get what i mean)
 :D

My induction is quite sweet - the fact that A|B5 and B|A5 simplifies things a lot.


Seriously what is it with getting owned by Oklahoma teams? I mean, it wasn't even the GOOD oklahoma this time.

Not sure, but I think this is what people call "bad luck" nowadays.

Bah more like a curse. Anyone know what ranking we are now?

119? they only rank the top 25, at least LSU and OK lost.

What do you think? Better hurry up to $ 194$.

Bah just because 2x13 is not 213 or, by what is posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231250]here[/url] 312.

Are you kidding me? THat failz. Although there's more than enough time to bounce back especially since we are playing Texas (2?)

Excuse me, but Texas is now ranked #1 after defeating Oklahoma.  (I'm from Texas, btw.)

[quote="mewto55555"]Wow did you just change the equation to make me look like a total noob?[/quote]

That's actually pretty funny. I thought you were acting like a total noob.

Moogra, I can't help it xD

HUnter make a new topic. THis is about Mizzou. We fail.

Texas 56, Missouri 31.
As the saying goes, there's always next year.

Be that as it may, with some n00b QB Mizzou will just do worse. Every weekend Rams win, Mizzou loses....

the world is messed up when rams beat teh cowboys.  next im going to get a friend.

:rotfl:  :rotfl:  :ninja: People don't realize that most of sports is luck.

Hm so if I am like Chuck Norriz im just lucky?

But nothing happens when I click on the link! :huh:

[quote="mewto55555"]http://en.wikipedia.org/wiki/Chuck_Norris

Better?[/quote]

Missed a "t"

He pwns? Google, "chuck norris jokes". He's been elevated to just below omnipotent deity.

What does "Gotcha" mean? I searched it in the dictionary but couldn't find it... Pardon my bad English. :maybe:


In triangle ABC the point X lies on BC.
(i) Suppose that
angle BAC = 90,  and that X is the midpoint of BC, and that angle BAX is one third of angleBAC. What can you say(and prove!) about triangle ACX?
(ii)Suppose that angle BAC = 60±, that X lies one third of the way from B to C, and that AX bisects angle BAC. What can you say (and prove!) about triangle ACX?

1st is easier than 2nd :D

i)[hide]equilateral[/hide]
II) what does 60± mean?
You do know that AC=2AB.

sorry :oops:  +_ is degrees okay :D  and AC=2AB is NOT given. :(

AC=2AB is deduced from the fact that AX bisects BAC and CX=2BX.  Not sure if this requres some sort of trig...

how did you get 2BX=CX? in the second one? how?

...Well I'm assuming that what you said applies to the problem.  You said "...X lies one third of the way from B to C..." ...

Let $BX=x$, $AB=y$, and $AC=z$.  Since $X$ is one-third of the way from $B$ to $C$, $CX=2x$.  By Angle Bisector Theorem,
\[\frac{AB}{BX}=\frac{AC}{CX}\\
\frac{y}{x}=\frac{z}{2x}\\
2xy=xz\\
2y=z\]
Therefore, $AC=2y$.  Draw an altitude from $C$ to $AB$.  Let us call the foot of the altitude $Y$. Since $\angle BAC=60$ and $\angle AYC=90$, $\triangle ACY$ is a $30$-$60$-$90$ triangle.  Since the hypotenuse of the triangle $AC$ equals $2y$, the shortest side $AY$ equals $y$.  Therefore, $AY$ equals $AB$ and $\angle ABC=90$.  Since $\angle BAC=60$ and $\angle ABC=90$, $\angle ACB=30$.  Since $\angle BAC=60$ and $AX$ bisects $\angle BAC$, $\angle XAC=30$, and triangle $ACX$ is isosceles.


algebra polynomial function logarithms calculus derivative domain

Find the remainder term $R_4(x)$ of a function $f(x)=\ln x$ by Taylor Expansion about $x=2$ . Determine an upper limit for $|R_4(x)|$ for $1\leq x\leq 3$ .

I can find $R_4(x)=\frac{f^{(4)}(c)(x-2)^4}{4!}$ for $c\in [1,3]$ . But how to find the upper limit ? Anyone help ?

Since you don't know what $c$ is, you have to consider the worst case.

$f^{[4]}(x)=-\frac6{x^4}.$

We're expanding about $2$ and $c$ will be between $x$ and $2.$ with $x$ in the interval $[1,3]$, the best we can say about $c$ is that it is in the inteval $(1,3).$

Now look at that formula for the fourth derivative? What's the worst case? Since it's a decreasing function, it's at its largest when $c=1.$

So: $|f^{[4]}(c)|\le 6,\,|x-2|\le 1,$ so $|R_4(x)|\le \frac{6\cdot 1^4}{24}=\frac14.$

We can do better on the right hand half of your interval. If $x\in[2,3],$ then $|f^{[4]}(c)|\le|f^{[4]}(2)|=\frac38.$

For $x\in[2,3],$ we have $|R_4(x)|\le \frac{\frac38\cdot 1^4}{24}=\frac1{64}.$

thanks Prof .

Since we have two cases , (i) $x\in [1,2]$ , max error is $\frac{1}{4}$ , (ii) $x\in [2,3]$ , max error is $\frac{1}{64}$ . So from this two cases , can we say that the max error is still $\frac{1}4$ ? And is it in general that if we wish to compute the max error from taylor remainder term , we have to consider two cases and see which one is bigger ?

From $R_4(x)=\frac{1}{4!}f^{(4)}(c)(x-2)^4$ we have $|R_4(x)|=\frac{1}{4!}\cdot \left(-\frac{6}{c^4}\right)(x-2)^4=\frac{1}{4}\cdot \frac{1}{c^4}(x-2)^4.=\frac{1}{4}\left|\frac{x-2}{c}\right|^4,$ thus we have

$|R_4(x)|=\frac{1}{4}\cdot \left(\frac{1}{|c|}|x-2|\right)^4.$ Since $|x-2|$ means that the distance between $2$ and $x$ on a number line, for $1\leq c\leq 3,\ 1\leq x\leq 3,\ \frac{1}{|c|}$ is maximized when $c=1$ and the distance $|x-2|$ is maximized when $x=1\ or\ x=3,$ yielding $|R_4(x)|\leq \frac{1}{4}\cdot \left(\frac{1}{1}\cdot 1\right)^4=\boxed{\frac{1}{4}}.$

[quote="kunny"]From $R_4(x)=\frac{1}{4!}f^{(4)}(c)(x-2)^4$ we have $|R_4(x)|=\frac{1}{4!}\cdot \left(-\frac{6}{c^4}\right)(x-2)^4=\frac{1}{4}\cdot \frac{1}{c^4}(x-2)^4.=\frac{1}{4}\left|\frac{x-2}{c}\right|^4,$ thus we have

$|R_4(x)|=\frac{1}{4}\cdot \frac{1}{|c|}|x-2|^4.$ Since $|x-2|$ means that the distance between $2$ and $x$ on a number line, for $1\leq c\leq 3,\ 1\leq x\leq 3,\ \frac{1}{|c|}$ is maximized when $c=1$ and the distance $|x-2|$ is maximized when $x=1\ or\ x=3,$ yielding $|R_4(x)|\leq \frac{1}{4}\cdot \frac{1}{1}\cdot 1^4=\boxed{\frac{1}{4}}.$[/quote]

so is there any significant for us to find the other side of the interval like what mention by Prof Kent Merryfield in #2
where in his last sentence he wrote the max error to be $\frac{1}{64}$ ?

Basically Im confused with whether we should consider $\frac{1}{4}$ as the answer (since it is bigger) or $\frac{1}{64}$ as the answer ( more accurate ?) ??

[quote="shyong"][quote="kunny"]Sorry for that I made a typo. I have just edited. I think that the answer is $\frac{1}{4}$ for $1\leq c\leq 3.$

kunny[/quote]

But isnt that in Taylor Series , we must have the $c$ to be between $x_0$ and $x$ (which in this case $c$ is in between $2$ and $x$) ??[/quote]

I don't quite understand this comment... but you're given that $c \in [1,3]$ and that's all you know about $c$. So the only sure thing you can say is that it is off by at most $\displaystyle \frac{1}{4}$. If you were given $c \in [2,3]$ (which you are not), then it's possible to reduce the error to $\frac{1}{64}$.

Try writing out the third degree Taylor polynomial. Compare $P_3(1)$ with $\ln{1} = 0$.

No, you have to pick the larger term. When you're estimating error, you always want to estimate from above.

[quote="shyong"][quote="paladin8"]
I don't quite understand this comment... but you're given that $c \in [1,3]$ and that's all you know about $c$. So the only sure thing you can say is that it is off by at most $\displaystyle \frac{1}{4}$. If you were given $c \in [2,3]$ (which you are not), then it's possible to reduce the error to $\frac{1}{64}$.

Try writing out the third degree Taylor polynomial. Compare $P_3(1)$ with $\ln{1} = 0$.[/quote]

actually , the question didnt gives $c\in [1,3]$ . Thats just some mistake i made in my solution . I should have written $c$ is between $x_0$ and $x$  ( isnt that what it should be stated in Remainder term of taylor series?)

In this cases , $x_0$ was picked as $2$ so that $c$ lies between $x$ and $2$ . So we have two cases , either $2\leq c\leq x$ or $x\leq c\leq 2$ .

So from what you had mention above , if among the two cases , where  one we get $1/4$ and another we get $1/64$ , we can choos the SMALLER error (1/64) as the maximum error of our remainder term ??[/quote]

Hmm, but basically $c$ is the value of $x$ in the given domain that maximizes $R_4(x)$, right? So we do know that $c \in [1,3]$. We want to find the value of $x \in [1,3]$ that maximizes $f^{(4)}(x)$ and that value is $c$. It was shown above that $c = 1$, so $R_4(x) \le \frac{1}{4}$. I think the whole $\frac{1}{64}$ thing was just to comment on how there is a much better bound on the interval $[2,3]$ than on $[1,2]$. It wasn't directly relevant to the question at hand.


geometry
geometric transformation
rotation
symmetry
trigonometry

I'm trying to win an arguement with my dad.  He thinks an equilateral triangle (all 60 degrees) is not point symmetrical and I think it is point symetrical.
Can anyone help explain?
Thanks!

EDIT: Incorrect explanation. Attachment removed.
Gosh, is it point symmetrical, I know it's line symmetrical.

It is point symetrical.

It has to be...all of the sides are of equal length.

Let's use the Pythagorean Theorem.

$a^2 + b^2 = c^2$

If:

a=a,
b=b,
then,
c=c.

Same with this:
c=c,
a=a,
so,
b=b.

Thanks!  That's what I thought too.  Think my dad was trying to make it more complicated than that.

But 236factorial and hwenterprise that is call line symmetric, not point symmetric?  :huuh:
I think point symmetric is put pencil in center of equilateral triangle and spin 120 degrees, same triangle. Spin again 120 degrees, same triangle. Spin one more time 120 degrees, same triangle you started.

[EDIT]Oh I'm sorry. I make mistake. My symmetric is called rotation symmetric not point symmetric but 236factorial and hwenterprise are line symmetric. [b]Equilateral triangle is not point symmetric[/b]

Alright, I'm confused now.
Is an equilateral triangle point symmetrical or not?  I understand symmetry but what's point symmetry?

i think point symetry is similar to regular symetry... :?

i dont think you understand symmetry, in fact, symmetry is harder than you think

but point symmetry is not that hard if explained correctly

a figure has point symmetry if every pair of corresponding points of the image is joined by a segment passing through exactly one point and this point is the center of symmetry and this center of symmetry bisects every single one of these segments

examples of figures exhibiting point symmetry:

square, because the center of the square is the center of symmetry

letter N

letter S

the graph $y=x^3$ is point symmetric because the graph is point symmetric at the origin

the graph $y=\sin x$

now, is the equilateral triangle point symmetric or not? i think you should know the answer by now

Okay, I see now. By the way, I doubt classroom math people know about the periodic sine funcion.  :roll:

you are correct but ill just leave it there so in case he knows it he can graph it with his graphing calculator ;)

[quote="ritchjp"]I believe it WOULD be point symmetrical. The point would be the intersection of all the perpendicular bisectors of  the sides . Not only that, since it is an equilateral triangle, all the angular bisectors would define the same point. Since that point is equally distant from all the sides and all the angles, it would make the triangle symmetrical about that point.[/quote]You are wrong. Put triangle upside-down, now not the same triangle. So not point symmetric. I made mistake before. You thinking of rotation symmetry

you are correct 10000th user. I found this web page that may be of interest to jamers.

  http://www.teachersnetwork.org/dcs/math/symmetry/Point/


[url]http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_12A_Problems/Problem_14[/url]
why wouldnt it just be 1/2 because the probablility of having even number of dots is the same as odd?

i believe it also has to do with the fact that the total number of even dots on a dice is greater than the total number of odd dots, even though there are 3 odd and 3 even faces.
e.g. 2+4+6=12
1+3+5=9.
there is a greater probability of picking an even dot because there are more even faced dots.

do you see what I'm talking about?

Well, if you remove it from an odd face, there is more chance of getting an even, and if you remove it from an even, you hae a less chance. So, the probabilities depend on what happened before, so it's not just 1/2.

removing a dot from the odd face has the same probability as removing a dot from the even. when removed, they come out as the other.
1 3 5 becomes 0 2 4
2 4 6 becomes 1 3 5

[quote="abacadaea"]removing a dot from the odd face has the same probability as removing a dot from the even[/quote]This is false. Each dot has an equal chance of being chosen. There are 12 dots on even faces, while there are only 9 dots on odd faces, so there is a 4/7 probability of removing a dot from an even face, while there is only a 3/7 chance of removing a dot from an odd face.


Jack drives at 40 mph for an hour, then at 50 mph for an hour.  What is his average speed?

Since Jack drove both speeds for the same amount of time, we can do,
$ 50\plus{}40\equal{}90$. What we do after that is we divide it by 2. So it looks like this:
$ 50\plus{}40\equal{}90$ $ \rightarrow$ $ \dfrac{90}{2} \equal{} 45$

Actually, you would use [url=http://en.wikipedia.org/wiki/Harmonic_mean]harmonic mean[/url] to obtain an answer of $ 48$.

No, for equal times, it is simply the arithmetic mean. It's harmonic for equal distances.


HINT: factor, you find out that $a(a^{4}-a^{2}+1)=2$. Factor. :)

HAPPY FIRST POST :welcomeani:

[quote="MellowMelon"][hide="Partial solution"]AM-GM $a^{5}$ and $a$ to get that $a^{5}+a \geq 2a^{3}$ with equality only if $a^{5}= a$. But that's only true if $a$ is 1 or 0, neither of which satisfy the problem conditions. So $a^{5}+a > 2a^{3}$.

Thus $2 = a^{5}-a^{3}+a > 2a^{3}-a^{3}= a^{3}$. So $2 > a^{3}$. That means $a^{6}< 4$.[/hide]

Because I am dreadfully bad at inequalities I haven't figured out how to do the other part, and I need to stop procrastinating on doing something else. :([/quote]
a good solution  !!! :D
Sorry I dont know much about E so I can not explain my idea completely
We have $a^{6}+1 =(a^{2}+1)(a^{4}-a^{2}+1)$
and $a(a^{4}-a^{2}+a)=2$( $a \neq 0$)
so $a^{6}+1 =(a^{2}+1).\frac{2}{a}$
 because $a^{2}+1 \geq 2a$
So we have $a^{6}+1 \geq 4$ -->
$a^{6}\geq 3$
 But that's only true if $a=1$ and $2 = a^{5}-a^{3}+a > 2a^{3}-a^{3}= a^{3}$.
--> $a^{6}>3$
this is my solution. :D
P/s: thank to Arvind_sn  :)

The left bound is a special case of India 1994 #2:

Given $x, r \in \mathbb{R}$ such that

$x^{5}-x^{3}+x = r$

Prove that

$x^{6}\ge 2r-1$

Edit:  The technique MellowMelon's solution suggests an additional extension for $x \in \mathbb{R}^{+}$, which is of course

$2r-1 \le x^{6}\le r^{2}$

[hide="Solution"] Note that $2r = 2x^{5}-2x^{3}+2x$, so equivalent we wish to show that

$x^{6}-2x^{5}+2x^{3}-2x+1 \ge 0 \Leftarrow$
$(x-1)^{2}(x^{4}-x^{2}+1) \ge (x-1)^{2}(x^{4}-2x^{2}+1) \ge 0$

QED. [/hide]

The next  :P
We have$n\in N, n \geq 2, x \in R+$
$x^{n}=x^{n-1}+x^{n-2}+...+x+1$
Prove that: $2 > x> 2-\frac{1}{n}$


geometry
area of a triangle
Heron\u0027s formula
number theory unsolved
number theory

It easy to get that $(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})=0$
assume that$\sqrt{a}-\sqrt{b}+\sqrt{c}=0$
then $a+c+2\sqrt{ac}=b$
$ac$ is a square.
if $(a,c)=d>1$
then$d|a,d|b,d|2\sqrt{ac},then d|b$ contraduction
then$(a,c)=1$
then$a,b$ are both square.

[quote="Hawk Tiger"]It easy to get that $(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})=0$
[/quote]

From where did you get the idea that $a^{2}+b^{2}+c^{2}-2(ab+bc+ca)$ is equal to this in quotation, is it from Heron's formula for the area of the triangle: $\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{4}\sqrt{a^{4}+b^{4}+c^{4}-2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}$ or from somewhere else?

[quote="Jumbler"]Let $a,b,c$ be a positive integer. If $gcd(a,b,c)=1$ and $a^{2}+b^{2}+c^{2}=2(ab+bc+ca)$, then prove that $a,b,c$ are perfect square.[/quote]
$a^{2}+b^{2}+c^{2}=2(ab+bc+ca)\Rightarrow(a-b-c)^{2}=4bc$
if $gcd(b,c)=d$then$d^{2}|4bc=(a-b-c)^{2}\Rightarrow{d|a}\Rightarrow{d|gcd(a,b,c)=1\Rightarrow{ gcd(b,c)=d=1}}$
therefore a,b are perfect square.


ckck  this equation have 8 roots... not only one....

its not wrong write $\sqrt{144}$

dont need to edit.......

Yes. But the only root is $0$. Try it yourself.

reals should have been specified, in short.

Normally when you take square roots, only the positive value is allowed.

[quote="Thales418"]the problem dont say this

we need to give a complete solution

in a test if you just put 0 your solution is wrong

if the problem dont say how the solution look like, we need a general solution[/quote]
true, but isn't this from mathcounts or something?

Its a test that you only need answers :roll:

http://www.mathcounts.org should give you a descriptoin


I have attached the code and the complied output (in pdf format).

Notice that the proof is incomplete. That's because it looks too ugly for me to continue.

First of all I don't like that fact that every first sentence of a paragraph is indented while all the others isn't. I would like every sentence not indented...

Secondly, I have a hard time deciding what I should name for this math. problem.
Now this problem (inprogress.tex) is not from any text or whatsoever. It is something that I come up with due to this similar question in my math assignment 1:
let $\{ A_\alpha \} _{\alpha \in J}$ be an indexed collection of sets. \\
let B be a set. Then

\[ B \cup \left( \bigcap_{\alpha \in J} A_\alpha \right) = \bigcap_{\alpha \in J} \left( B \cup A_\alpha \right)  \]

So I really shouldn't call the problem in "inprogress.tex" a lemma. Maybe I should call it "my problem 1.1"?

For your first problem, put in the preamble:[code]\setlength{\parindent}{0in}[/code]

[b]PS:[/b] I don't know why do you start a new paragraph every time you want to start a new line!!  :| . \\ is enough!

oops you are right... \\ is enough. Thanks for mentioning. Now my proof looks better :lol:


function
inequalities proposed
inequalities

Let $f$ be a convex functions three times diferentiable with $f'''\geq 0$ and $a,b,c,d$ positive real numbers. Prove that:
$f(a)+f(b)+f(c)+f(d)+2f(\frac{a+b+c+d}{4}) \geq  f(\frac{a+b}{2})+f(\frac{a+c}{2})+f(\frac{a+d}{2})+f(\frac{b+c}{2})+f(\frac{b+d}{2})+f(\frac{c+d}{2})$


inequalities solved
inequalities
maximum value

If $a + b = c, b + c = d, c + d = a$, and $b$ is a positive integer. Find the largest possible value of $a + b + c + d$.

Easily, we have $a+b = c, d = b+c= a+2b, a = c+d = 2a+3b$.
It follows that  $a= -3b, c = -2b, d = -b$.
Thus $a+b+c+d = -5b  \leq -5$, since $b \geq 1$, with equality iff $b = 1, a=-3, c = -2, d = -1$.

Pierre.


Given are the negative integers a, b, x, y with the assumption that a, b have no common factor. Prove that ab−a−b represents the greatest number which cannot be written in the form ax + by.

I think this is commonly known as the "Chicken McNugget Theorem" for some odd reason. Anyway, there are many proofs around AoPS that can be found searching this.

[quote="Iversonfan2007"][quote="towersfreak2006"]I think this is commonly known as the "Chicken McNugget Theorem" for some odd reason. [/quote]

yea i don't get that either it's kinda random[/quote]
It may not be that random at all. This is what I found out, http://mathworld.wolfram.com/McNuggetNumber.html

Just look at each residue class mod either of a and b.

Here is a proof that $ax+by\neq ab-a-b$, but I'm not sure of how to finish it by showing that $ab-a-b$ is the [i]greatest[/i] such integer.

Suppose that for some $a,b,x,y\in\mathbb{N}$, we have $ax+by=ab-a-b$.  Then we would have \begin{eqnarray*}by\equiv a-b\bmod a&\implies& b\left(y+1\right)\equiv 0\bmod a \\ ax\equiv b-a\bmod b &\implies &a\left(x+1\right)\equiv 0\bmod b \end{eqnarray*} This means that for some $m,n\in\mathbb{N}$ we have \[a\left(nb-1\right)+b\left(ma-1\right)=ab-a-b\implies m+n=1.\] However, this cannot be so since $m,n\in\mathbb{N}=\left\{1,2,\dots ,n,\dots\right\}$.  Therefore we have a contradiction meaning it is impossible for $ax+by=ab-a-b$.

Any insight as to how we might show this is the greatest integer?

[quote="zanttrang"][quote="amcavoy"]
Suppose that for some $a,b,x,y\in\mathbb{N}$, we have $ax+by=ab-a-b$.  Then we would have \begin{eqnarray*}by\equiv a-b\bmod a&\implies& b\left(y+1\right)\equiv 0\bmod a \\ ax\equiv b-a\bmod b &\implies &a\left(x+1\right)\equiv 0\bmod b \end{eqnarray*} [/quote]
I don't understand this part.  Wouldn't it be $by \equiv-b \mod a$ and $ax \equiv-a \mod b$?[/quote]
$-b\equiv a-b\bmod a$ and likewise for $\bmod b$.  Either way will lead to the same result;  I just wrote it that way because that's how I'm used to dealing with negative residues.

[quote="amcavoy"]
$-b\equiv a-b\bmod a$ and likewise for $\bmod b$.  Either way will lead to the same result;  I just wrote it that way because that's how I'm used to dealing with negative residues.[/quote]
Oh, right - duh   :wallbash:  It's certainly been too long since I've done modular arithmetic...

$gcd(a,b)=1$, so basically if you have $0, a, 2a, 3a,...(b-1)a$, then you have a complete residue class mod b. Furthermore, these are all different. Now, it's pretty clear that you can obtain any value greater than (b-1)a (since you have every residue and you can just keep adding b), and that the last residue you get is (b-1)a (mod b), so the largest element is $(b-1)a-b=ab-a-b.$


symmetry
inequalities
inequalities unsolved
Belarus

[quote="N.T.TUAN"]What will occur if $a=b=-1$?  :D[/quote]
Your right I forgot to type that a 'n b are positive
Now its corrected..

[quote="Zamfirmihai"][quote="FOURRIER"]a and b are POSITIVE real numbers,prove:
 [img]http://www.gnux.be/latex/data/632c549d85dd17cf280dc10569715992.png[/img][/quote]

I think it works if you "separate" the therms and then aplying Cauchy.

I mean:

a^2 + b +3/4= a^2 + sqrt (b^2) + (1/2)^2 + (sqrt2 / 2)^2 and be careful when regrouping the therms[/quote]
$\left(a^{2}+\left(\sqrt b\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt2}\right)^{2}\right)\left(b^{2}+\left(\sqrt a\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt2}\right)^{2}\right)=...$
Zamfirmihai, continue, please. Thank you.

I have just solved this problem basically. :lol:

$\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+a+\frac{3}{4}\right)\geq \left(a+b+\frac{1}{2}\right)^{2}\geq 4\left(a+\frac{1}{4}\right)\left(b+\frac{1}{4}\right).$

[quote="arqady"][img]http://www.gnux.be/latex/data/6ebaf6a45195143a13a940baa5985b74.png[/img][/quote]

How did you get that please  :lol: ? and thanks  :D

your solution can be completed by substitution $x+y=u$ and $xy=v$ and then it becomes a cubic in u and can be diffrentiated to find critical points.
:NT TUAN please reaD my PM to you
anuj

[quote="kunny"]Is my solution right? :)[/quote]
kunny, why $\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+a+\frac{3}{4}\right)\geq \left(a+b+\frac{1}{2}\right)^{2}$ $?$

[quote="arqady"]$\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+a+\frac{3}{4}\right)\geq\left(a^{2}+a+\frac{3}{4}\right)\left(b^{2}+b+\frac{3}{4}\right)\Leftrightarrow$
$\Leftrightarrow(a+b)(a-b)^{2}\geq0.$ :wink:[/quote]
Yeah  :lol: , I got this after developping  :roll:
but  I was meaning what did you apply:Cauchy?
Or is it a classical FORMULATE??

[quote="kunny"]tangent line![/quote]
I don't understand, what do you mean. Detale, please. Thank you.
[quote="FOURRIER"]...but I was meaning what did you apply:Cauchy?
Or is it a classical FORMULATE??[/quote]
I think, it is accident. But nice accident. :)

[quote="arqady"][quote="kunny"]tangent line![/quote]
I don't understand, what do you mean. Detale, please. Thank you.
[/quote]

idea: The symmetry of $a,\ b$ of our inequlity gives the equality of the given inequality; $a=b=\frac{1}{2}.$

Here is my complete solution.

[color=blue]Problem[/color] Let $a,\ b$ be positive numbers.Prove that $\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+a+\frac{3}{4}\right)\geq \left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right).$

Solution:

For $a>0,\ b>0,$ we have $\left(a-\frac{1}{2}\right)^{2}\geq 0\Longleftrightarrow a^{2}\geq a-\frac{1}{4},$ which can be also obtained by considering the tangent line of $y=x^{2}$ at the tangent point $\left(\frac{1}{2},\ \frac{1}{4}\right).$ Thus $a^{2}+b+\frac{3}{4}\geq a+b+\frac{1}{2},$ similarly by symmetry of $a,\ b$ we have $b^{2}+a+\frac{3}{4}\geq b+a+\frac{1}{2},$ since the both sides of these inequality are positive, multiplying the inequalities for both sides gives $\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+a+\frac{3}{4}\right)\geq \left(a+b+\frac{1}{2}\right)^{2}.$ Use the inequality $(x+y)^{2}\geq 4xy$ for $x>0,\ y>0,$ yielding
$\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+a+\frac{3}{4}\right)\geq \left(a+b+\frac{1}{2}\right)^{2}\geq 4\left(a+\frac{1}{4}\right)\left(b+\frac{1}{4}\right),$ which is to be proved.

[quote="arqady"]I think, it is accident. But nice accident. :)[/quote]
9WD, What do you mean?  :rotfl:

[quote="FOURRIER"][quote="arqady"]I think, it is accident. But nice accident. :)[/quote]
What do you mean?  :rotfl:[/quote]
This inequality was in my some olympiad and I solved it. :roll:


Thấy các bạn trường lhp cũng nhiều quá. Mà sao các nhân amser đâu rồi ta. Không hiểu học hành nhiều quá có biết tới mathlinks không vậy. Hai bác mod để khoảng hai tuần mà không ai trả lời hắng xóa nha!!

Amser chính hiệu đây.
Nhưng bây giờ có khi phải đổi thành Hanoier chứ nhí


WELCOME BACK!!! :D

The forums have been dead without you hehe.

Good job at IMO :)

How was your trip?

[quote="lightrhee"]Yeah! I'm back! :)
I came back last Tuesday, but I didn't have time to use my computer until now.
How is everyone(including CanMoo :moose: ) doing?[/quote]

Wonderful! We haven't had any posts here in ages  :sleeping:

And thank you for spelling CanMoo correctly! (I personally prefer to add a hyphen: Can-Moo, but that doesn't make any difference). In case you don't already know, Richard and his IOI gang changed its game to Canmuu (which triggers the pronounciation /can-m-you/). But we math people are sticking with Can-Moo and we shall not let it be abused!!!

[quote="billzhao"]And thank you for spelling CanMoo correctly! (I personally prefer to add a hyphen: Can-Moo, but that doesn't make any difference). In case you don't already know, Richard and his IOI gang changed its game to Canmuu (which triggers the pronounciation /can-m-you/). But we math people are sticking with Can-Moo and we shall not let it be abused!!![/quote]
Yay, flame war over the moose :moose:
I guess there are two solutions:
1. It's called Canmoo
2. It's called Can-moo
The reason we had it down as Canmuu is because we didn't want it to get mixed up with the American mascot, which was a cow. Also, Canmuu looks more native Canadian.
By the way, check the previous thread to see the super-leader of the Canadian IOI delegation. I still have the nametag.

Sorry. I was busy for a while.

Richard won a gold medal in IOI? Congraturation! :D  And don't forget to bring CanMoo :moose:  to the winter camp.

Was anyone trying to steal the moose during IOI? I'm quite sure that someone tried, since there were 9 attempts to steal CanMoo during IMO.

1) A South-American(I believe that he was a Venezuelan) tried to grab the moose and hide him at the airport.

2) One of the New Zealander grabbed the moose and ran. I caught him. :D

3) The New Zealander tried again. He had the moose for a longer time than his first try, but I still got him. This is the time that I warned him about my black belt.

4) The South-American tried again. He pretend to take a picture, snatch the moose and ran.

fifth to eighth attempts was on the same day, the day Emily was coming.

5) The New Zealander successfully took the moose when no Canadians were around. It was hidden under a chair. I stole it back when no New Zealanders were around. :lol:

6) Four of the Mexican team members tried. They also pretended to take a picture. Two of them grabbed me and the others ran off with the moose. I got really mad this time.

7) The New Zealander tried again! I kicked the cup he was holding, which was filled with water. He got wet. :lol:

8) Valentin also tried to steal the moose! He stole the moose without being noticed. I saw him putting something down and walk away, but didn't know that it was the moose. Valentin thought I saw the whole thing and gave the moose back.

9) The New Zealander again! I was chasing him for a long time, but I still got him.

[quote="lightrhee"]8) Valentin also tried to steal the moose! He stole the moose without being noticed. I saw him putting something down and walk away, but didn't know that it was the moose. Valentin thought I saw the whole thing and gave the moose back.[/quote]Lucky shot :D Keep it safe till Slovenia guys  :coool:  :diablo:

[quote="lightrhee"]
don't forget to bring CanMoo :moose:  to the winter camp.

Was anyone trying to steal the moose during IOI? I'm quite sure that someone tried, since there were 9 attempts to steal CanMoo during IMO.
[/quote]
I won't forget about bringing Canmoo :moose: to winter camp. Right now, it's sitting proudly on top of my bookshelf wearing the Canadian flag and offcourse, the super-leader nametag from the IOI.

Nonody tried to steal the moose at the IOI :D. As a result, we came up with the postulate  that programmers are more civilized than mathematicians. However, we nearly changed Canmoo's name and set off a flame war on AOPS (check the IOI newsletters)

Edit: Now I just realized that the Newzealanders tried to steal Canmoo a whopping 5 times. If I knew this at the time, I really should have sent the kiwi to a speed where no kiwi had gone before. Oh well....guess we'll get our revenge in Slovenia. :ninja:

[quote] Four of the Mexican team members tried. They also pretended to take a picture. Two of them grabbed me and the others ran off with the moose. I got really mad this time.
[/quote]
Well, for all of us, you overact, we weren´t going to keep it for ever. I´m sorry for making you mad but if we didn´t try to stole every pet we saw, we weren´t mexicans (acourding to our olympiad traditions). However, we kidnap the Southafrican pet "Chubuchef" and it was returned with a new T-shirt. Hope they like it.

Haha I'm sure everything was in good fun friends :)

[quote="lightrhee"]
3) The New Zealander tried again. He had the moose for a longer time than his first try, but I still got him. This is the time that I warned him about my black belt.
[/quote]

Ah hahahaha, that black belt sure is something.  Ah hahahahahahahaha....

[quote="lightrhee"]7) The New Zealander tried again! I kicked the cup he was holding, which was filled with water. He got wet. :lol: [/quote]

You got some water on me too!  :mad:

I actually witnessed first-hand the cup flying out the New Zealander's hand and spilling water everywhere. Great action there ... scared me a bit too  :oops:.

[quote="billzhao"]

I actually witnessed first-hand the cup flying out the New Zealander's hand and spilling water everywhere. Great action there ... scared me a bit too  :oops:.[/quote]

lol, I could just see them running back, "WHOA!  David REALLY is a black belt!!!"

Godo times, good times...

wow... this forum is significantly less dead now that David is back...

How's school going for everyone?

Waterloo is awesome :D  Best university EVAR!!!


How many pairs $ (a,b)$ of integers such that

$ a^2b^2 \equal{} 4a^5 \plus{} b^3$　are there?

[quote="yunxiu"]There are 8 solutions.[/quote]

Your answer is correct,yunxiu^^
Would you write how you solved it?

If $ab=0$, than $a=b=0$,now we suppose $ab \ne 0$, $(a,b) = k$, $a = k{a_1}, b = k{b_1}$,$({a_1},{b_1}) = 1$, than we have $ka_1^2b_1^2 = 4{k^2}a_1^2 + b_1^3$, so ${a_1}\left| {b_1^3} \right. \Rightarrow {a_1} = 1$, and $kb_1^2 = 4{k^2} + b_1^3 \Rightarrow b_1^2\left| {4{k^2}} \right. \Rightarrow {b_1}\left| {2k} \right.$. Suppose $2k = c{b_1}$, we have $kb_1^2 = 4{k^2} + b_1^3 \Rightarrow c{b_1} = 2{c^2} + 2{b_1} \Rightarrow c\left| {2{b_1}} \right.$, suppose $2{b_1} = cd$, so $cd = 4c + 2d\Rightarrow c-2|4c$,and we have $\left( {c - 2,c} \right) = (c - 2,2)\left| 2 \right.$, so $c - 2 =  \pm 1, \pm 2, \pm 4, \pm 8$.
If $c - 2 = 8$, than $c = 10 \Rightarrow d = 5 \Rightarrow {b_1} = 25 \Rightarrow k = 125$,so $(a,b) = \left( {{5^3},{5^5}} \right)$;
If $c - 2 = 4$, than $c = 6 \Rightarrow d = 6 \Rightarrow {b_1} = 18 \Rightarrow k = 54$,so $(a,b) = \left( {2 \times {3^3},{2^2} \times {3^5}} \right)$;
If  $c - 2 = 2$, than $c = 4 \Rightarrow d = 8 \Rightarrow {b_1} = 16 \Rightarrow k = 32$,so $(a,b) = \left( {{2^5},{2^9}} \right)$ ;
If  $c - 2 = 1$, than $c = 3 \Rightarrow d = 12 \Rightarrow {b_1} = 18 \Rightarrow k = 27$, so $(a,b) = \left( {{3^3},2 \times {3^5}} \right)$ ;
If  $c - 2 =  - 1$, than $c = 1 \Rightarrow d =  - 4 \Rightarrow {b_1} =  - 2 \Rightarrow k =  - 1$,so $(a,b) = \left( { - 1,2} \right)$;
If  $c - 2 =  - 2$, than $c = 0 \Rightarrow b = 0$, contradiction;
If  $c - 2 =  - 4$, than $c =  - 2 \Rightarrow d = 2 \Rightarrow {b_1} =  - 2 \Rightarrow k = 2$,so $(a,b) = \left( {2, - 4} \right)$;
If  $c - 2 =  - 8$, than $c =  - 6 \Rightarrow d = 3 \Rightarrow {b_1} =  - 9 \Rightarrow k = 27$,so $(a,b) = \left( {27, - {3^5}} \right)$。
So there are $8$ solutions
$(a,b) = (0,0),\left( {{5^3},{5^5}} \right),\left( {2 \times {3^3},{2^2} \times {3^5}} \right),\left( {{2^5},{2^9}} \right),\left( {{3^3},2 \times {3^5}} \right),\left( { - 1,2} \right),\left( {2, - 4} \right),\left( {{3^3}, - {3^5}} \right)$。

[color=#FF0000][moderator edit: Very nice, replying after about three years. :?][/color]


Points $A$, $B$, $C$, and $D$ lie on a line, in that order, with $AB=CD$ and $BC=12$. Point $E$ is not on the line, and $BE=CE=10$. The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find $AB$.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=507[/img]

\[ \text {(A) } 15/2 \qquad \text {(B) } 8 \qquad \text {(C) } 17/2 \qquad \text {(D) } 9 \qquad \text {(E) } 19/2  \]

[hide="Solution"]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=508[/img]

The altitude $EF$ is equal to $\sqrt{10^2-6^2}=8$. Let $AB=CD=x$. Then \[ AE^2 = AF^2+EF^2=(6+x)^2+8^2. \] Since the perimeter of $\triangle BEC$ is $32$, we know that \[ AE + ED + AD = 2\sqrt{(6+x)^2+8^2}+(12+2x)=64. \] Solving this equation gives $AB=x=\boxed {9}$.
[/hide]

[hide="Answer"]We know also that $AE=DE$. $2AE+2AB+12=64\Rightarrow AE+AB=26$, and the altitude from $E$ to $AD$ has length $8$ by Pythagorean. $(6+AB)^2+8^2=AE^2\Rightarrow AE^2-AB^2=12AB+100$, so $(AE+AB)(AE-AB)=26(AE-AB)=12AB+100\Rightarrow 13AE=19AB+50$. Add $13AB$ to both sides, and we have $338=32AB+50\Rightarrow AB=9\Rightarrow \boxed{D}$.[/hide]


can somebody please explain to me what number "e" is? Thanks

Please take a look at  http://en.wikipedia.org/wiki/E_%28mathematical_constant%29 . If there are more concrete questions, you can ask.