Να βρείτε όλους τους θετικούς ακεραίους , οι οποίοι είναι ίσοι με το $ 13 \minus{}$ πλάσιο του αθροίσματος των ψηφίων τους.

 Μπάμπης

ψαχνωντας βρηκα αυτο το πολυ ωραιο ξεχασμενο προβλημα :)
η λυση μου ειναι η εξης:
εστω οτι ο αριθμος ειναι ο $ xy...zk = x10^{n - 1} + y10^{n - 2} + .... + k = 13(x + y + ....z + k)$με $ 9 > = x,y,z....,k > = 0$και n ο αριθμος των ψηφιων του.
εχουμε ως δεδομενο με παραγοντοποιηση οτι
$ (10^{n - 1} - 13)x + (10^{n - 2} - 13)y + .... - 12k = 0$
$ = > k = \frac {(10^{n - 1} - 13)x + (10^{n - 2} - 13)y + ....z10^{n - (n - 1)}}{12}\leq9$
$ < = >$
$ (10^{n - 1} - 13)x + (10^{n - 2} - 13)y + ....z10^{n - (n - 1)}\leq108$
ομως για $ n\geq4 = > (10^{n - 1} - 13)x\geq987x\geq987 > 108$παιρνωντας δεδομενο οτι το χ δεν ειναι μηδεν
αρα ατοπο.αρα $ n\leq3$
διαπιστωνουμε επισης οτι $ 13/x10^{n - 1} + y10^{n - 2} + .... + k = > x10^{n - 1} + y10^{n - 2} + .... + k\geq13$
αλλα και οτι
$ x10^{n - 1} + y10^{n - 2} + .... + k = 13(x + y + ....z + k)\leq117n$
τοτε παιρνουμε $ n\geq\frac {1}{9} = > n\geq1$
η περιπτωση n=1 καθιστα αδυνατο τον αριθμο να ειναι πολ/σο του 13 και
για n=2
παιρνουμε $ 10x + y = 13x + 13y < = > - 3x = 12y$=>x,y eterosimoi.ατοπο
επομενως $ n = 3$
για n=3,
$ 100x + 10y + z = 13x + 13y + 13z(1) < = > y = 29x - 4z\leq9$
$ < = > x\leq\frac {9 + 4z}{29}\leq\frac {45}{29} = > x = 1$
για χ=1 η (1) γραφεται $ y + 4z = 29$
διοφαντικη με λυσεις $ y = 1 - 4t,z = 7 + t$,t ακεραιος
πρεπει$ 0 < z,y\leq9$
$ = > t = 0, - 1, - 2 = > (y,z) = {(1,7),(5,6),(9,5)}$
οι ζητουμενοι αριθμοι μας λοιπον ειναι οι $ 117,156,195$  :D QED


trigonometry
function
complex analysis
complex analysis unsolved

Let $ f_1(z)\equal{}\sin{z},f_n(z)\equal{}f_{n\minus{}1}(\sin{z})$. Is the family of holomorphic functions $ f_n,n\equal{}1,2,3,\ldots$ normal in $ \{z \in \mathbb{C} : |z|<1 \}$ ?

No, $ f_n(0)\equal{}0$ but $ f_n(0.9i)\to\infty$

Could you please explain the following : take a sequence $ f_n(z)\equal{}z\plus{}n$. It is clear that it diverges locally uniformly yo $ \infty$ in $ \mathbb{C}$. That i understand.
In the book they claim that it neither converges nor diverges locally uniformly in $ \bar{\mathbb{C}}$

Why is it so ? Why doesn't it converge locally uniformly to the constant function, which is $ \infty$ everywhere.

[quote="eugene"]in $ \bar{\mathbb{C}}$[/quote]
The sequence $ z\plus{}n$ does not go to $ \infty$ uniformly in any neighborhood of $ \infty$.

Oh, yeas, indeed. For some stupid reason i forgot to look at the neighborhood of $ \infty$.
You are saying that if $ |f_n(z)| > M(R)$ for all $ n > N(R)$ for some $ N(R),M(R)$ and all $ |z| > R$. But of course this neighborhood contains $ z$ with $ z\equal{}\minus{}n$ say with big enough $ n$, which is a contradiction.

How about everywhere black in the plot below?

[img]http://img409.imageshack.us/img409/9944/iteratedsinzng6.jpg[/img]

Julia set of $ f(z)\equal{}\lambda \sin z$ has been studied quite a bit. There is even a Youtube video.
[youtube]-oJaE-iBIvA[/youtube]


If you choose "other," please post what the "other" is.

[color=indigo][moved from bay area: please reserve threads like these to the fun factory][/color]

Sadly, it only allowed me to put nine options.

COUGHGAGWHEEZE

im sorry, please, do continue.

and if you dont know its best to not know

please?

tellz me plz

plz

my 10th grade friends started laughing at "69"

i feel left out :(

plz tell me

ummmmm.......68<69<70!

theres always wikipedia

[quote="Joe10112"]explain thouruohly plz

plz?

like the actual meaning

i cant go on wikipedia

or on google[/quote]

It is seriously a very sick, disgusting joke and inappropriate for anyone here. Let's not spread the contamination.

eww please don't talk about that. it's gross especially when your bio teacher refers to it *chokegagdie*

anway... on a more pure note... my fave number is $ 6$ =] because a lot of things in my life relate to $ 6$ =] although i'm not a devil child... the devil stole my number. he's the copy cat  :D  :rotfl:

I like $ 6$ too. I also like $ 56$ a lot. $ 15$, $ 56$, and $ 89$ are the three most awesome two-digit numbers, with $ 13$, $ 21$, $ 22$, $ 82$, and $ 88$ being the next-coolest two-digit numbers. If you ask for a cool triplet of two-digit numbers, $ 21$, $ 55$, and $ 89$ form a symmetrical pattern on the number pad while being Fibonacci numbers at the same time, and one of them, $ 89$, is one of those three epicly-awesome numbers. Isn't that cool?

(I'm sounding even more nerdy than I normally am, right?)

okayz

i wont ask then

i will not ask 4 "69"

I like $ \boxed{16.7}$.  Hey, my favorite number isn't a whole number!  I'M A TRUE AoPSER!  :lol:

Well, gee, you're leaving out an awful lot of numbers, don't you think? There's 1, and 1.1, and 1.01, and 1.001, and 1.91, and a whole lot more. My favorite number is $ \phi$.

I beg to differ about r15s11z55w89y21's (beg me if I spelled it wrong o_0) "15 and not 42" image....

Favorite number = 842310 GO FIGURE!

$ \text{My favorite number would have to be }{\color{cyan}\pi \color{black}+\color{red}i}$

My favorite numbers are roots of $ x^7\minus{}1$.

[quote="AIME15"]Where's 1337?[/quote]
exactly, where is 1337? And what about pi?!

Favorite number: lim x->(+0), 1/x

Shoot, that converges towards infinity.

[quote="r15s11z55y89w21"][quote="dragon96"][quote="mousy"]

Favorite number = 842310[/quote]

Is that the game number that you got your highest score?[/quote]

I don't even think we got to that number yet...[/quote]

You know, besides 42, none of the numbers are really favorites of AoPSers. If I made a poll, I would have:

42
pi
1337
i
e
56154
56
15
96 (7th graders only)
other

[quote="eggylv999"]$ \minus{} 146e^{\pi i}$[/quote]

Which would be 146?
While you're at it, why not write $ \minus{}146e^{\pi e^{\frac{\pi}{2}i}}$? :D

My favorite number is pi, which is not a interger. :lol:

$ \pi$.

I'm surprised no one has said pi until TPAM.

Oh man... 69... I don't want to hear that number again after what happened on the bus last year.

If you're looking for numbers comprised of Fibonacci numbers, $ 558921$ pwns all.

Mine is 6 for some weird reason.....It was like that ever since I was about 2 years old.

6 is the 1st perfect number and the only number such that its proper factors multiply to the number itself

5 is coool   thats why it's in my username

2. The only even prime. The only number where $ n^n=n\times n=n+n$. When writing square roots, 2 is so special that it doesn't need to be written for someone to understand its power. The base that computers work in.

My favourite number is the sum of all real numbers that are not expressible as the sum of arbitrarily but finitely many algebraic numbers.

[quote="PowerOfPi"]$ n^n = n\times n = n + n$.[/quote]
LaTeX \fail.


Suppose that $ABC$ is a triangle and $H \in BC, D \in AB, E \in AC$. Also $AH,CD$ and $BE$ are concurrent and $AH$ is an altitude of $\triangle ABC$. Prove that $\angle EHA=\angle AHD$.

$DE$ intersects $AH$ and $BC$ at $M,N$. We have $(NDME)=-1$ ( it's well-known ). Also $HM \perp HN$ so $HM$ is the bisector of $\angle MHN$


$\\ 2x^{2}+1 > x^{2}\Rightarrow \frac{x}{2x^{2}+1}< \frac{x}{x^{2}}= \frac{1}{x}$ and now you can take $\delta = \frac{1}{ \epsilon }$ .

Ramanujan, could you explain why we choose $\delta = \frac{1}{\epsilon}$ in more detail? Thanx.

[quote="bythecliff"]How can I show that $\lim_{x\to\infty}{\frac{x}{2x^{2}+1}}= 0$?

I can't just state it. I need to show it using $\delta, \epsilon$, or the like. How can I do this?

THanx[/quote]

$\lim_{x\to\infty}{\frac{x}{2x^{2}+1}}= 0$ if and only if
$\forall\epsilon>0\, \exists\delta >0\mbox{ suchthat }|x|>\delta\Rightarrow \left| \frac{x}{2x^{2}+1}-0\right| <\epsilon$

choose $\delta =\frac{1}{\epsilon}$ so that $|x|>\delta\Rightarrow \left| \frac{x}{2x^{2}+1}-0\right| =\frac{x}{2x^{2}+1}< \frac{1}{x}=\epsilon$ QED


A cyclist has won a competition. Then the reporters asked him what was the speed of it, then, he replied:

"If I had been walking 4 km / ha more than I would have returned an hour earlier than when he returned, knowing that the distance is 120 km."

So what was the speed of the cyclist?

Thanks, my answer was 20 too.
Thanks man ;)


In a party, each person knew exactly $ 22$ other persons. For each two persons $ X$ and $ Y$, if $ X$ and $ Y$ knew each other, there is no other person who knew both of them, and if $ X$ and $ Y$ did not know each other, there are exactly $ 6$ persons who knew both of them. Assume that $ X$ knew $ Y$ iff $ Y$ knew $ X$. How many people did attend the party?
[i]Yudi Satria, Jakarta[/i]

The answer is $ n \equal{} 100$.

Consider an arbitrary person $ A$. Let $ F(A)$ be the set of friends of $ A$ and $ S(A)$ be the set of strangers of $ A$. Then $ |F(A)| \equal{} 22$ and $ |S(A)| \equal{} n \minus{} 23$. Consider a person $ S\in S(A)$. Since $ S$ does not know $ A$, $ S$ and $ A$ must have $ 6$ friends in common. Thus, each person $ S\in S(A)$ knows exactly $ 6$ people in $ F(A)$. Hence, the number of the relationship $ F(A) \minus{} S(A)$ is $ 6(n \minus{} 23)$. On the other hand, every two people in $ F(A)$ do not know each other (since they all know $ A$), therefore, every people in $ F(A)$ knows exactly $ 21$ people in $ S(A)$. This means that the number of relationship $ F(A) \minus{} S(A)$ is $ 21\cdot 22$. From the above argument, we deduce that $ 6(n \minus{} 23) \equal{} 21\cdot 22$, which yields $ n \equal{} 100$, QED.


By using identity cot x - cot 2x = cosec 2x, prove that
cot x - cot 4x > 2 for 0 < x <  :pi:/4

studing $x \mapsto \cot{x} - \cot{4x} -2$ no?  :blush:

[hide]

$\cot{x} - \cot{4x} = (\cot{x}-\cot{2x}) + (\cot{2x}-\cot{4x}) = \csc{2x} + \csc{4x}.$
But since $\csc{x} \ge 1$, we know that the expression is at least 2. We also know that $\csc{x}=1$ at $x=\frac{\pi}{2}, \frac{3\pi}{2}$ so $\csc{2x}$ is never 1 on the interval. Then suppose $\csc{4x}=1$. So $x=\frac{\pi}{8}$. Then $\csc{2x} > 1$ so the expression is still greater than 2.[/hide]


function
trigonometry
limit
calculus
calculus computations

I am stuck with a)

Consider the function $ f(x) =\sin(1/x)$

a) Find a sequence of x-values that approach $ 0$ such that $ \sin(1/x) = 0$.
[Hint: Use the fact that $ \sin(\pi) =\six(2\pi) =\sin(3\pi) = ... =\sin(n\pi) = 0.$].

b) Find a sequence of x - values that approach $ 0$such that $ \sin(1/x) = 1$. [Hint: Use the fact that $ \sin(n\pi/2) = 1$ if $ n = 1,5,9,...$].

c) Find a sequence of x-values that approach 0 such that $ \sin(1/x) =-1$.

d) Explain why your answers to any two parts (a)-(c) show that $ \lim_{x\to 0}\sin(1/x)$ does not exist.

a) can be seen easily if you think a little bit. $ \sin k\pi \equal{} 0$ for all $ k\in\mathbb{Z}$.  Now just solve the equation $ \frac{1}{x}\equal{} k\pi$ with respect to $ x$ and prove that this value goes towards zero if $ k\to\infty$.

So

a)

$ x \equal{}\frac{1}{k\pi}$

$ \sin\left(\frac{1}{\pi}\right) \equal{} 0.313$
$ \sin\left(\frac{1}{2\pi}\right) \equal{} 0.158$
$ \sin\left(\frac{1}{3\pi}\right) \equal{} 0.106$

So all of these values are approaching $ 0$.

Did I do it right?

b)

$ x \equal{}\frac{n\pi}{2}\equal{}\frac{1}{x}$
$ x \equal{}\frac{2}{n\pi}$.

$ \sin\left(\frac{2}{\pi}\right) \equal{} 0.59$
$ \sin\left(\frac{2}{5\pi}\right) \equal{} 0.127$
$ \sin\left(\frac{2}{9\pi}\right) \equal{} 0.0706$

So these also approach $ 0$.

???

For (a) you don't want the value of the function to go to $ 0$, you want it to always be zero. Look at where $ x$ is. If $ x \equal{}\frac{1}{k\pi}$ then $ f(x) \equal{}\sin(k\pi)$, not $ \sin\left(\frac{1}{k\pi}\right)$. The thing you want to go to $ 0$ is the $ x$ value. So the first thing you wrote is correct, but plugging it back into the function you flipped it upside down.

Likewise for (b) you want the value of the function to always be $ 1$, yet the sequence you pick goes to zero.

:( I still don't have a sequence of x values. is it $ \pi, 2\pi, 3\pi$?

$ f(x)\equal{}\sin\left(\frac{1}{x}\right)$

Using the sequence of $ x$ values $ x\equal{}\frac{1}{k\pi}$. (Technically they should probably be notated as a sequence $ x_{k}\equal{}\frac{1}{k\pi}$.

You get a sequence of $ f(x)$ values $ f\left(\frac{1}{k\pi}\right)\equal{}\sin\left(\frac{1}{1/k\pi}\right)\equal{}\sin(k\pi)\equal{}0$ for all $ k$.


Haley has enlarged a 3" by 5" picture so that both the lenght and width are tripled.  By what number must the original are of the pricture be multiplied to get the enlarged area?

[hide]
$x(3\cdot5)=3\cdot3\cdot3\cdot5$
$x=\frac{3\cdot3\cdot3\cdot5}{3\cdot5}$
$x=9$[/hide]

[hide]9 times. Each side is tripled so the area is $3^2$ times more (9). i didnt really need to calculate the actual area  :D[/hide][/hide]

[quote="risinglegend"][hide="im hiding your answer"]9 times[/hide]. is it correct?[/quote] Yes it is correct but you are supposed to tell us why its correct and HIDE your answer so other people can have a chance at the problem unless someone is asking for help.

It is useless to just post an answer and not a solution to a problem other people have already solved and explained. IT's not useless if you post a solution


Нереально онай гой!!!

[quote="Ulanbek_Kyzylorda KTL"]Let O be the center if the excircle of triangle ABC opposite to A. Let M be the midpoint of AC and let P be the intersection of lines MO and BC. Prove that if <BAC=2<ACB, AB=BP.[/quote]

Let's $ \angle ACB \equal{} \gamma$. $ AO \cap BC \equal{} A_1$
We have
 $ AO\equal{}\frac{AC \cdot sin(90^o\plus{}\gamma)}{sin(90^o\minus{}3\gamma)}$
 $ A_1O\equal{}\frac{A_1C \cdot sin(90^o\minus{}\gamma)}{sin(90\minus{}3\gamma)}$
So
$ \frac{A_1O}{AO}\equal{}\frac{A_1C}{AC}\equal{}\frac{AA_1}{AC}$

By Menelaus on $ \triangle AA_1C$ and line $ OPM$ we get $ \frac{CM}{AM} \cdot \frac{A_1P}{CP} \cdot \frac{AO}{A_1O}\equal{}1$
or
$ \frac{A_1O}{AO}\equal{}\frac{A_1P}{CP}$
Therefore
$ \frac{A_1P}{CP}\equal{}\frac{AA_1}{AC}$
therefore $ AP$ is bisector of $ \angle A_1AC$ $ \rightarrow$ $ \angle BAP\equal{}3\gamma$
$ \angle APB \equal{}180^o\minus{}\angle PAB \minus{}\angle ABP\equal{}3\gamma$
SO $ AB\equal{}PB$

[color=darkred][size=125][b]Nice problem ! Thank you.[/b][/size] [/color]

[quote="Ulanbek_Kyzylorda KTL"]Let $ \triangle\ ABC$ , the midpoint $ M$ of $ [AC]$ and $ P\in MI_a\cap BC\ .$ Prove that $ A \equal{} 2C\ \Longleftrightarrow\ BP \equal{} AB\ .$[/quote]

Denote $ D\in \overline {AII_a}\cap BC$ . Apply the [b]Menelaus[/b]' [u]theorem[/u] to the transversal $ \overline {I_aPM}$ for $ \triangle\ ADC$ and obtain

$ \frac {PD}{PC} \equal{} \frac {I_aD}{I_aA} \equal{} \frac {ID}{IA}\ .$ Thus, $ A \equal{} 2C\Longleftrightarrow\ IP\parallel AC\ \Longleftrightarrow\ \triangle ABI\equiv\triangle IPD\Longleftrightarrow AB \equal{} BP\ .$


calculus
integration
function
calculus computations

I think this is the right way to do it, correct me if I am wrong.
First differentiate both side, so by FTC we have
${f(x)}^{2006}=\int_{0}^{x}f(t)dt+1 \\ \implies 2006{f(x)}^{2005}f'(x)=f(x)\\ \implies{f(x)}^{2004}f'(x)=1/2006$
Now, it is a first-order seperable differential equation. with $f(x)=1$, for $x=0$
so by integrate both side, we have
$\frac{{f(x)}^{2005}}{2005}=\frac{x}{2006}+C\\ \implies f(x)=(\frac{2005x}{2006}+C)^{\frac{1}{2005}}$
By subsititute the initial condition in, we have $C=1$
so $f(x)=)=(\frac{2005x}{2006}+1)^{\frac{1}{2005}}$
so $f(2006)=(2006)^{\frac{1}{2005}}$

Your answer is correct. :)
The answer is $2006.$

[quote=galacta]But we first need to ensure that f(x) is differentiable.This isn't provided in the conditions![/quote]

$f(x)^{2006}=\int_0^x f(t) \ dt+1\quad (*)$

The RHS is a continuous function of $x$ by fundamental theorem of calculus (FTC). Hence the LHS is continuous too. Now $f$ is positive on $[0,\infty).$ Therefore continuity of $f(x)^{2006}$ implies that $f(x)$ is continuous. Since $f$ is continuous, applying FTC once again, we get that RHS of $(*)$ is differentiable. Hence $f(x)^{2006}$ is differentiable. Now, $f(x)>0$ for all $x\ge 0.$ Hence $\log (f(x)^{2006})=2006\log f(x)$ is differentiable. Thus, $f(x)=e^{\log f(x)}$ is differentiable too. :)

[quote=integrated_JRC][quote=galacta]But we first need to ensure that f(x) is differentiable.This isn't provided in the conditions![/quote]
No. We need only continuity of $f$. From there, it follows from the fact the integral of any continuous function is differentiable. As the RHS is differentiable, so is the LHS. :)
———————
Correct me if I’m wrong.[/quote]

the LHS is the 2006th power of f(x),so in elementary calculus,we first need to solve out f(x) by take the 2006th root on every side.Then we see the differentiability of f(x) by the chain rules.I THINK there is also a theorem talking about the differentiability of implicit function which can used in this problem directly instead of solve f(x) out.But we need more advanced tool to prove this theorem.


function
integration
calculus
calculus computations

Last 3 weeks I gave the following problem to my students and  :huh: no one can solve it correctly.

Does there exist continuous function $ f : [0,1] \to \mathbb R$ satisfying
\[ \int_0^1 {f\left( x \right) \,dx }  = \frac{1}{3} + \int_0^1 {\left( {f\left( {x^2 } \right)} \right)^2  \,dx } .
\]

[hide]$ f(x) = \sqrt{x}$.[/hide]

wouldn't f(x)=a, where $ a^2\minus{}a\plus{}1/3\equal{}0$ work?

Putting $ x^2 \equal{} y$ in the last integral, we get that $ \int_0^1 \left( \frac {f(x)}{\sqrt 2 \sqrt [4]{x}} \minus{} \frac {\sqrt [4] x}{\sqrt 2} \right)^2 \, dx \equal{} 0$ :)

It's worth mentioning that I'm talking improper integrals here, i.e. as $ a \to 0$, $ \int_a^1 \frac {f^2 (y)}{2 \sqrt y} \, dy$ approaches the second integral from below.

[quote="0714446459923"]wouldn't f(x)=a, where $ a^2 \minus{} a \plus{} 1/3 \equal{} 0$ work?[/quote]

Yes, if that equation had real solutions ;)

JBL:Yes, if that equation had real solutions.

But that equation had no real solutions.The solution perfect_radio given is the exclusive solution.

how do you prove that that is the only solution?

Sorry. The solution perfect_radio given is the exclusive real and continuous solution.
The proof Perfect_radio had given 1n the 3th floor.If you want more simple or evident explain,I suggest that you may putting x=y^2 in the first integral.

[quote="bookworm_vn"][quote="chernfey"]I suggest that you may putting x=y^2 in the first integral.[/quote]

Oh, one can easily see that but how to proceed further  :roll:[/quote]

Hello,bookworm_vn,I think you really need to see perfect_radio's proof seriously.
f>=0,continuous ,in [a,b],and int[a,b]f(x)dx=0  => f=0 in [a,b].
If you still intangible,I all say nothing.

The full solution, just for the sake of completeness:

[hide]Set $ x \equal{} y^2, dx \equal{} 2y\, dy$ in the first integral, so we have
\[ \int_0^1 2y \cdot f( y^2 ) \,dy \equal{} \frac {1}{3} \plus{} \int_0^1 {\left( {f( {y^2 } )} \right)^2 \,dy }.
\]
Rewriting $ \frac {1}{3} \equal{} \int_{0}^{1} y^2 \,dy$ and bringing everything over to the right gives
\[ \int_0^1 f(y^2)^2 \minus{} 2yf(y^2) \plus{} y^2 \,dy \equal{} 0,
\]
or if we set $ g(y) \equal{} f(y^2) \minus{} y$
\[ \int_0^1 g(y)^2 \,dy \equal{} 0.
\]
Since $ f$ is continuous, $ g^2$ is a continuous, nonnegative function.  It's a standard exercize in a first proof-based calculus class to show that the integral of a nonnegative continuous function over some interval can be zero if and only if the function is uniquely zero.  Thus $ g(y)^2 \equal{} 0$ and so $ f(y^2) \minus{} y \equal{} 0$ for all $ y \in [0, 1]$.  It follows that $ f(x) \equal{} \sqrt {x}$ is the unique function satisfying the desired equation.[/hide]


If $ \frac{1}{a} \plus{} \frac{1}{b} \equal{} \frac{2}{b}$ and $ ac > 0$.Prove that $ \frac{a\plus{}b}{2a\minus{}b}\plus{}\frac{c\plus{}b}{2c\minus{}b}\ge4$

if it is in that way that is written here ... than this inequality is equivalent to :

$ \frac {c \plus{} b}{2c \minus{} b}\ge2$

where $ a\equal{}b$

and then it is very easy  :)  :D

[quote="Obel1x"]If $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {2}{b}$ and $ ac > 0$.Prove that $ \frac {a \plus{} b}{2a \minus{} b} \plus{} \frac {c \plus{} b}{2c \minus{} b}\ge4$[/quote]

Im sorry its : If $ \frac {1}{a} \plus{} \frac {1}{c} \equal{} \frac {2}{b}$ and $ ac > 0$.Prove that $ \frac {a \plus{} b}{2a \minus{} b} \plus{} \frac {c \plus{} b}{2c \minus{} b}\ge4$

i made this problem the easiest one in mathlinks.ro  :(  :oops:

We have $ b\equal{}\frac{2ac}{a\plus{}c}$; substituting, the given inequality is equivalent to $ \frac{a\plus{}3c}{2a} \plus{} \frac{c\plus{}3a}{2c} \geq 4 \Leftrightarrow \frac{c}{a}\plus{}\frac{a}{c} \geq 2$, which follows from homogeneity (WLOG $ a,c > 0$) and AM-GM.


calculus
integration
number theory
Diophantine equation

Hey guys,

I'm stuck on the following problem:

Find the integral solutions to the Diophantine equation 317x+241y=9.

I subtracted 317x to get 241y=9+317x and then put the equation in mod 317 to get 241x is congruent to 9(mod 317).

That's pretty much where I get confused. Is there a way to find a multiple of 217 that is congruent to 9(mod 317) without random guessing and checking?

Thanks,
mathgeek93

hello, i hope this will help you
http://www-groups.dcs.st-and.ac.uk/~martyn/teaching/1003/1003linearDiophantine.pdf
Sonnhard.

Using Theorem $ 4.6$ in the reference given by Sonnhard and finding lots of multiples of $ 317$ and $ 241$ on a calculator, I get $ 317 \times 35 \plus{} 241 \times ( \minus{} 46) \equal{} 9$ and so $ x_{0} \equal{} 35, y_{0} \equal{} \minus{} 46$ and the general solution is

$ x \equal{} 35 \plus{} 241t, y \equal{} \minus{} 46 \minus{} 317t$ for all $ t \in \mathbb{Z}$.


inequalities
probability
expected value
algebra proposed
algebra

Show that , for any $z_1,z_2...z_n\in \mathbb{C}$, there are $a_1,a_2...a_n\in\{\pm 1\}$such that:
\[|\sum(a_iz_i)|^2\leq(\geq)\sum (|z_i|^2)\]

The probabilistic method works well here.  Let $a_1$, ..., $a_n$ be independent, random elements of $\{ \pm 1 \}$.  Let's compute the expected value of the left-hand side of the desired inequality:
[tex]\begin{eqnarray*}
E({\bigl\vert \sum_i a_i z_i \bigr\vert}^2)
  & = & E(\sum_{i, j} a_i z_i a_j \overline{z_j}) \\
  & = & \sum_{i, j} z_i \overline{z_j} E(a_i a_j) \\
  & = & \sum_i {\left\vert z_i \right\vert}^2.
\end{eqnarray*}
[/tex]
Hence, there exists a specific choice of $a_i$ such that
[tex]
{\bigl\vert \sum_i a_i z_i \bigr\vert}^2
  \geq  \sum_i {\left\vert z_i \right\vert}^2
   =  \sum_i {\left\vert a_i z_i \right\vert}^2.
[/tex]
Similarly, there exists a specific choice of $a_i$ such that
[tex]
{\bigl\vert \sum_i a_i z_i \bigr\vert}^2
  \leq  \sum_i {\left\vert z_i \right\vert}^2
  =  \sum_i {\left\vert a_i z_i \right\vert}^2.
[/tex]

Could you please tell more about this probabilistic method. Is it only based on the fact that you sum up all possible sums, or is it more than that(not necessary on this problem)?

The version of the problem I was given used $\epsilon$ instead of $a$ and assumed WLOG that $\sum |z_i|^2 =1$.

Let $z_k=a_k+b_ki$ for all $1\le k\le n$. Now, notice that the problem condition is equivalent to
\[(a_1^2+a_2^2+\dots+a_n^2)+(b_1^2+b_2^2+\dots+b_n^2)=1,\]
and the problem is asking us to prove that there exist $\epsilon_1$, $\epsilon_2$, $\dots$, $\epsilon_n$ such that
\[ \left|\sum_{k=1}^n \epsilon_k z_k \right| \le 1,\]
and squaring both sides of the equation, this is equivalent to proving that
\[(\epsilon_1a_1+\epsilon_2a_2+\dots+\epsilon_na_n)^2+(\epsilon_1b_1+\epsilon_2b_2+\dots+\epsilon_nb_n)^2\leq 1.\]
And since $\epsilon_i^2a_i^2$ is always equal to $a_i^2$, we can re-express this sum as
\[\sum a_i^2 + 2\sum \epsilon_i\epsilon_ja_ia_j+\sum b_i^2 + 2\sum \epsilon_i\epsilon_jb_ib_j,\]
and since we're given that $(a_1^2+a_2^2+\dots+a_n^2)+(b_1^2+b_2^2+\dots+b_n^2)=1$, this gives us that this is equal to
\[1+2\sum \epsilon_i\epsilon_ja_ia_j+2\sum \epsilon_i\epsilon_jb_ib_j.\]
Now note that if we sum this over all possible $n$-tuples of the different $\epsilon$'s, the coefficient of $a_ia_j$, $\epsilon_i\epsilon_j$ is $-1$ with $\frac{1}{2}$ probability and $1$ with $\frac{1}{2}$ probability, meaning that the average value of $\epsilon_i\epsilon_ja_ia_j$ over all $n$-tuples of the $\epsilon$'s for all $(i,j)$ is $0$. Similarly, the average value of $\epsilon_i\epsilon_jb_ib_j$ over all $n$-tuples of the $\epsilon$'s for all $(i,j)$ is $0$. \\\\
This gives us that the average value of
\[\left| \sum_{k=1}^n \epsilon_k z_k \right|^2=1+2\sum \epsilon_i\epsilon_ja_ia_j+2\sum \epsilon_i\epsilon_jb_ib_j,\]
is $1+0$, or $1$. This implies that
\[\left|\sum_{k=1}^n \epsilon_k z_k \right| \le 1,\]
for some $n$-tuple of $\epsilon$'s, which means that
\[\left|\sum_{k=1}^n \epsilon_k z_k \right| \le 1,\]
for the same $n$-tuple of $\epsilon$'s, finishing the problem.

The square of the LHS is just
\begin{align*}
& \left(\sum_{k=1}^n \varepsilon_k z_k\right)\cdot\overline{\left(\sum_{k=1}^n \varepsilon_k z_k\right)} \\
=& \sum_{i=1}^n\sum_{j=1}^n \varepsilon_i\varepsilon_j z_i \overline{z_j} \\
=& \sum_{i=1}^n \varepsilon_i \varepsilon_i |z_i|^2+\sum_{i=1}^n\sum_{j\ge 1, j\ne i} \varepsilon_i\varepsilon_j z_i \overline{z_j} \\
=& 1+\sum_{i=1}^n\sum_{j\ge 1, j\ne i} \varepsilon_i\varepsilon_j z_i \overline{z_j}.
\end{align*}
By randomly choosing $\varepsilon_i$, the expected value of the second sum is $0$, so some choice works. $\blacksquare$

De-homogenize so that $\sum |z_i|^2 = 1$, and use $\epsilon_i$ instead of $a_i$. Then the LHS equals
\begin{align*}
\left| \sum_{k = 1}^n \epsilon_kz_k \right| ^2 &= \left(\sum_{k = 1}^n \epsilon_kz_k \right) \left( \sum_{k = 1}^n \epsilon_k \overline{z_k} \right) \\
&= \sum_{k = 1}^n \epsilon_k^2 |z_k|^2 + \sum_{i < j} (\epsilon_i \epsilon_j z_i \overline{z_j} + \epsilon_i \epsilon_j \overline{z_i}z_j) \\
&= 1 +  \sum_{i < j} \epsilon_i \epsilon_j (z_i \overline{z_j} + \overline{z_i}z_j).
\end{align*}
When we sum this over all $2^n$ possible combinations of $\epsilon_1, \epsilon_2, \ldots, \epsilon_n$, each of the $\epsilon_i$ is $-1$ half the time and $1$ the other half of the time so the $\sum_{i < j} \epsilon_i \epsilon_j (z_i \overline{z_j} + \overline{z_i}z_j)$ terms will "vanish" and the sum is just $2^n$. Hence for some choice of $\epsilon_i$, the sum is at most $1$.


AMC
USA(J)MO
USAMO
inequalities solved
inequalities

nobody solve ???? Please help me !!!  :(  :(

[quote="Chocolate M&M"]Let a, b, c, d are real number such that $\large a^2+b^2+c^2+d^2 \leq 1$ . Find maximum of :

$\large A=(a+b)^4+(a+c)^4+(a+d)^4+(b+c)^4+(b+d)^4+(c+d)^4$[/quote]

We have $\large (a+b)^4 \leq (a-b)^4+(a+b)^4=2(a^4+b^4+6a^2b^2)$
Similarly, we have
$\large A \leq 6(a^4+b^4+c^4+d^4+2a^2b^2+2a^2c^2+2a^2d^2+2b^2c^2+2b^2d^2+2c^2d^2)=6(a^2+b^2+c^2+d^2)^2 \leq 6$
that equality holds when $\large a=b=c=d=\frac{1}{2}$   :D


Professor Gamble buys a lottery ticket, which requires that he pick six different integers from 1 through 46, inclusive. He chooses his numbers so that the sum of the base-ten logarithrms of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property - the sum of the base-ten logarithrms is an integer. What is the probability that Professor Gamble holds the winning ticket?

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

[hide]Let $x_i$ be the numbers on his ticket and $y_i$ be the numbers on the winning ticket. We must have $\log x_1 + \log x_2 + \cdots + \log x_6 = \log (x_1x_2x_3x_4x_5x_6) \in \mathbb Z$ and $\log (y_1y_2y_3y_4y_5y_6) \in \mathbb Z$ as well. This means that both $x_1x_2\cdots x_6$ and $y_1y_2\cdots y_6$ must be powers of 10. The highest power of ten possible is $10^6$. We can also get $10^5$ (in 2 ways) and $10^4$, but it is impossible to get $10^3$, $10^2$, $10^1$, or $10^0$, so our answer is $\frac{1}{4} \implies \boxed {B}$.[/hide]


Find all natural numbers $ n$ with exactly $ 8$ natural divisors which sum is equal to $ 3780$ (including 1 and $ n$).

$ n$ has exactly $ 8$ divisors $ \Leftrightarrow n\equal{}pqr, p^3 q, p^7$, where $ p,q,r$ are different prime numbers.Also $ 3780\equal{}2^2*3^3*5*7$
[b]Case 1:[/b]$ n\equal{}pqr$, $ (p\plus{}1)(q\plus{}1)(r\plus{}1)\equal{}2^2*3^3*5*7$, then we have one of them, say $ p\equal{}2$ as $ 8 \nmid R.H.S.$
So $ (q\plus{}1)(r\plus{}1)\equal{}2^2*3^2*5*7$, we have $ q, p\equal{}1(mod \;\; 4)$ as $ 8 \nmid R.H.S.$
Now let $ q<r$, then $ q\plus{}1<\sqrt{2^2*3^2*5*7}$,$ q \le 34$, by trial (as possible values of $ q$ are $ 5,13,17,29$), we get $ n\equal{}2*13*89\equal{}2314$ or $ n\equal{}2*29*41\equal{}2378$

[b]Case 2:[/b]$ n\equal{}p^3 q$, so $ (p^3\plus{}p^2\plus{}p\plus{}1)(q\plus{}1)\equal{}2^2*3^3*5*7$ , in this case we get $ p<(2^2*3^2*5*7)^{\frac{1}{3}}$, so $ p \le 10$, by trial we get $ n\equal{}2^3*251\equal{}2008$

[b]Case 3:[/b]$ n\equal{}p^7$, then $ p^7<2^2*3^3*5*7$, so $ p\equal{}2 or 3$ which is impossible as $ 2,3|3780$,

So the only solutions are $ n\equal{}2008, 2314, 2378$


If $\displaystyle\sum_{n=1}^{k}\displaystyle \left(\frac{1}{\sqrt{2n-1}}-\frac{1}{\sqrt{2n+1}}\right)=\frac{10}{11}$, what is the positive integer value of $k$?

[hide]Just from writing down the first few terms, we notice that we have a telescoping series.  The sum of the first $k$ terms is thus:

$1-\frac{1}{\sqrt{2k+1}}=\frac{10}{11}$
$\sqrt{2k+1}=11$
$2k+1=121$
$k=\boxed{60}$[/hide]


Prove that, \[ (a\plus{}3b)(b\plus{}4c)(c\plus{}2a) \geq 60abc \]for all real numbers $ 0\leq a\leq b\leq c$.

[quote="Moonmathpi496"]Prove that,
\[ (a \plus{} 3b)(b \plus{} 4c)(c \plus{} 2a) \geq 60abc
\]
for all real numbers $ 0\leq a\leq b\leq c$.[/quote]
$ (a \plus{} 3b)(b \plus{} 4c)(c \plus{} 2a) \equal{} 25abc \plus{} 2a^2b \plus{} 6ab^2 \plus{} 3b^2c \plus{} 12bc^2 \plus{} 4c^2a \plus{} 8ca^2 \ge 60abc \iff$
$ ab(2a\plus{}6b) \plus{} bc(3b\plus{}12c) \plus{} ca(4c\plus{}8a) \ge 35abc \iff$
$ 2\frac{a}{c}\plus{}6\frac{b}{c} \plus{} 3\frac{b}{a}\plus{}12\frac{c}{a} \plus{} 4\frac{c}{b}\plus{}8\frac{a}{b} \ge 35\iff$
By AM-GM:
$ LHS \ge 35 \sqrt[35]{\frac{a^{10}b^9c^{16}}{a^{15}b^{12}c^8}} \ge 35 \equal{} RHS \iff$
$ c^8 \ge a^5b^3$ which is obvious since $ c \ge a,b$
Equality iff $ a\equal{}b\equal{}c$.

[quote="Moonmathpi496"]I am interested to know if there is any easy and efficient method to find the coefficient to apply AM-GM. I know that there is a method named Balancing Coefficients.

Did you use balancing coefficient method?[/quote]
I don't know what the balancing coefficient method is, since i've never heard about it :blush: The trick is to guess that the ineq only has equality when $ a\equal{}b\equal{}c$. I thought there was at least one more, so that was why I multiplied out. But you should just have the equality in mind in this case. So while $ 2a \plus{} b \ge 2\sqrt{2ab}$ is true, it won't help us much since $ 2a\plus{}b \neq 2\sqrt{2ab}$ when $ a\equal{}b$. So you would choose $ 2 \cdot a \plus{} b \ge (2\plus{}1) \sqrt[2\plus{}1]{a \cdot a\cdot b} \equal{} 3\sqrt[3]{a^2b}$. Hope you understood :)


Let a, b, c be real numbers satisfying a < b < c, a + b + c = 6 and ab + bc + ca = 9.
Prove that 0 < a < 1 < b < 3 < c < 4.

well just by some fill-in you can easily find that they're all positive so the 0< part is pretty easy

then a<b<c is also given, so no need worrying bout that.

So we got: 0 < a < ? < b < ? < c < ?

At least 2 of them must be lower than 3 because of ab + bc + ca = 9

so 0 < a < ? < b < 3 < c < ? or 0 < a < ? < b < ? < c < 3

suppose c<3, then a+b>3, then ab + c.(b+a) = 9 results us that ab < 0, which cannot be, so 0 < a < ? < b < 3 < c < ? is the correct one.

If one of them (c thus) was bigger than 4, then the other two have sum lower than 2, which makes ab + c.(b+a) = 9, and which ends us up in a quadratic equitation saying a=b=1 (which contradicts a<b<c

so 0 < a < ? < b < 3 < c < 4

now last thing to prove is that a < 1 < b

suppose they were both smaller:
then c > 4, contradiction

suppose both larger:
say a = 1+d, b= 1+e
d+e+c = 4      (d+e<1 as c>3)
so work the second one out:
ed+dc+ec+c = 5
and this is obviously impossible with 1 > d,e > 0 and 4 > c > 3
contradiction

thus it is possible no other way then that 0 < a < ? < b < 3 < c < 4

I solved it this way:

Let t = abc. Then a, b, c are the roots (= zeros) of f(x) = x3 - 6x2 + 9x - t = 0.
We have f'(x) = 3(x - 1)(x - 3) so f(3) < 0 < f(1). Hence a < 1 < b < 3 < c.
Since f(1) = 4 - t > 0, t = abc < 4, and since f(3) = -t < 0, abc > 0.
As b, c are positive, a > 0.
Also, f(x) is strictly increasing on (3, +oo) and f(4) = f(1) = 4 - p > 0 so c < 4.


You'll have to go to the complex numbers, but there do indeed exist solutions.  (In fact, they're all just scalings and permutations of a single solution $(1, \omega, \omega^{2})$ where $\omega$ is one of the complex third-roots of 1.

[quote="Tim_Lou"]I don't think that
$x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

is possible.

can anyone think of a example of (x,y,z) that satisfy the equation?[/quote]

The third roots of unity work. :)


It's pretty easy proving limits when having for example $ \lim_{x\to4}(2x\plus{}1)\equal{}9.$

But I'm getting stuck when tryin' to prove that $ \lim_{x\to2}x^2\equal{}4.$

We have $ 0<|x\minus{}2|<\delta$ & $ |x^2\minus{}4|<\epsilon\implies|x\plus{}2\parallel{}x\minus{}2|<\epsilon.$ What's the next step?

Create an upper bound for $ |x\plus{}2|$, say $ M$, so that $ |x\plus{}2|<M$. Choose $ \delta\equal{}\frac{\epsilon}{M}$ :)

Here's what I would do

$ |x^2 \minus{} 4| < \epsilon$

$ \minus{} \epsilon < x^2 \minus{} 4 < \epsilon$

$ \minus{} \epsilon \plus{} 4 < x^2 < \epsilon \plus{} 4$

since all three are greater than zero for sufficiently small epsilon
$ \sqrt { \minus{} \epsilon \plus{} 4} < x < \sqrt {\epsilon \plus{} 4}$

$ \sqrt { \minus{} \epsilon \plus{} 4} \minus{} 2 < x \minus{} 2 < \sqrt {\epsilon \plus{} 4} \minus{} 2$

Since the magnitude of lower bound of x-2 is greater than the magnitude of the upper bound we select the lower bound as our largest window

$ |x \minus{} 2| < 2 \minus{} \sqrt { \minus{} \epsilon \plus{} 4}$

and there you go

$ \delta \equal{} 2 \minus{} \sqrt { \minus{} \epsilon \plus{} 4}$ for any sufficiently small epsilon

That works as long as you're not building up foundations of calculus. (Typically, showing square roots exist uses the continuity of $ x^2$.)

So you want to show $ |x\minus{}2\parallel{}x\plus{}2|<\epsilon$ for all $ x$ satisfying $ |x\minus{}2|<\delta$, for some positive $ \delta$ depending on $ \epsilon$.

If $ \epsilon$ is small, $ |x\minus{}2|$ will close to 2, so $ |x\plus{}2|$ will be close to 4. In particular, we should be able to find an open interval around 2 for which $ |x\plus{}2|<5$. We can: $ |x\minus{}2|<1$ works.

So choosing $ \delta \equal{}\min \left\{ 1,\frac{\epsilon}{5}\right\}$ satisfies $ |x\minus{}2\parallel{}x\plus{}2|<\epsilon$ for all $ x$ such that $ |x\minus{}2|<\delta$.

[quote="scorpius119"]Typically, showing square roots exist uses the continuity of $ x^2$.[/quote]
Not really, it's easy to define each decimal in the root


An object is propelled upward with an initial velocity of 20 meters per second.  The distance s (in meters) of the object from the ground after t seconds is s = -4.9t^2 + 20t.

(a) What is the maximum height?

(b) What is the minimum height?

QUESTION:

Where did you get 200/49 and 100/49?

$\frac{200}{49}\rightarrow{\frac{20}{4.9}=\frac{200}{49}}$

$\frac{100}{49}\rightarrow{\frac{1}{2}}\cdot\frac{200}{49}=\frac{100}{49}$

ya I converted the decimal into a fraction and all I did there was complete the square..

[hide="solution"]its just $\frac{-b}{2a}$ or $\frac{-20}{(2)(-4.9)}$ which equals $\frac{200}{(2)(49)}=\frac{100}{49}$

the minimum is obviously $0$[/hide]

shouldnt this be in classroom math?  this is an algebra 1 thing.

no, algebra 1 canbe in hsb if its easy enuf


Find $ f''(x)\minus{}5f'(x)\plus{}3f(x)$, if    $ f: \mathbb{R}\rightarrow\mathbb(\minus{}\frac{\pi}2, \frac{\pi}2)$,   $ f(x)\equal{}arctgx$


Let $a,b,c$ are positive number . Prove:
$\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}\leq (a+b+c)$

Yeah. Here is what I come up with [edit according to Albanian Eagle]
Let $a+b+c=1$ Find min and max of:
$\frac{a^{2}}{b+c+1}+\frac{b^{2}}{b+c+1}+\frac{c^{2}}{a+c+1}$

$\frac{a^{3}}{b+c+1}+\frac{b^{3}}{b+c+1}+\frac{c^{3}}{a+c+1}$

$\frac{a^{4}}{b+c+1}+\frac{b^{4}}{b+c+1}+\frac{c^{4}}{a+c+1}$

:(

there should be a misprint because this is trivial... :maybe:

Er,I still don't understand....
$\sum \frac{a^{k}}{b+c+1}\leq \sum a^{k}\leq \sum a=1$ when $k\geq 1$
and a=1,b=0,c=0 the equality holds.
Is there any mistake?

[quote="Gibbenergy"]Let $a,b,c$ are positive number . Prove:
$\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}\leq (a+b+c)$

Yeah. Here is what I come up with [edit according to Albanian Eagle]
Let $a+b+c=1$ Find min and max of:
$\frac{a^{2}}{b+c+1}+\frac{b^{2}}{c+a+1}+\frac{c^{2}}{a+b+1}$

$\frac{a^{3}}{b+c+1}+\frac{b^{3}}{c+a+1}+\frac{c^{3}}{a+b+1}$

$\frac{a^{4}}{b+c+1}+\frac{b^{4}}{c+a+1}+\frac{c^{4}}{a+b+1}$

:([/quote]


Help me figure out how many different possible combinations of a 20x20x20 Rubik's Cube

[hide]
We have to look at the different types of pieces that make up the cube

Corner Pieces
Edge Pieces
Center Pieces

Corner Pieces are straightforward: there are $ 8$

Edge pieces can be subdivided into 9 subcategories based on how close to the corner they are. (20 sided cube, minus two for corners, and symmetry).

Center pieces can be subdivided into 45 categories of pieces (due to symmetries, this value is 1/8 of the total 18^2 adjusting for diagonal pieces)
9 are diagonal pieces, and the other 36 are other center types

Now we must enumerate each type

Amount of pieces
Corner: 8
Edge Subcategory type: 12 x 2 = 24  (12 edges to a cube. 2 on each edge)
Center Diagonal type: 6 x 4 = 24 (6 faces, four on each face)
Center reg type: 6 x 8 = 48

Number of orientations:
Corner: 3
Edge: 2
Center: 1

throw it into one long expression, and you have

$ (8!\cdot3^{8})(24!\cdot2^{24})^{9}(24!\cdot 1^{24})^{9}(48!\cdot 1^{48})^{36}$

If you put a picture on each face, suddenly the orientations on the center face skyrockets to 4, and you have to adjust accordingly.
[/hide]

Or we could just say that there are a ****load of possible combinations.

That number is not correct, you don't take into account that there are (a lot of) combinations that are impossible to reach.


Hi

I just started USAMTS
What software do you prove questions?
And what software do you use to create images?
Thanks :D

*What software do you use to format your proofs?

[url=http://www.texniccenter.org]TeXnicCenter[/url] can turn LaTeX into PDFs.
[url=http://www.wolfram.com]Mathematica[/url] could potentially help solve problems.

Use a $ \text{\LaTeX}$ editor. You see the sidebar (its to your left  :D) where it says LaTeX? That's a quick tutorial of LaTeX. A common one here is TeXnicCenter (google it.)

To make images, anything will work. Even microsoft paint (just save it as a jpg file). I like using Microsoft Visio 2003 because of the easy access to all the shapes and even variables (letters in Monotype Corsiva font  :P). Then, at the top, write

\usepackage{graphicx}

and where you want to insert the picture

\includegraphics[scale=(insert scale)]{insert file name}

which types of files can you put (.doc, .gif,...) and how would you put in an asymptote code?

You should preferably use some sort of Integrated $ \text{\LaTeX}$ Environment to format your proofs. These will typically output your $ \text{\LaTeX}$ into .pdf or other files by themselves and help you out with accessing the function library. Also, to use asymptote or other add-ons, you simply type in the path to the asymptote info you download from somewhere (I think, can't remember).

I know I was terribly vague in explaining, but I use [url=http://kile.sourceforge.net/]Kile[/url] for my $ \text{\LaTeX}$ needs. It only works for the KDE environment though (Linux, Mac).


It can be also solved with the following nice lemma:
In the triangle $ ABC$ let $ D, E, F$ be the points of tangency of the incircle with $ BC, CA, AB$. Let $ X, Y, Z$ be the points on incircle. Then $ XD, YE, ZF$ are concurrent iff $ AX, BY, CZ$ are concurrent.
Proof is an application of a trigonometric Ceva, I can post it in case anybody is interested, it was probably on the forum already.

Let $ l, k, m$ be the tangents at $ A_{2}, B_{2}, C_{2}$ and we denote by $ P, Q, R$ intersection of lines $ k, m$, $ m, l$, $ l, k$ . As Little Gauss has already mentioned  line $ l$, line $ B_{1}C_{1}$ and line $ BC$ concurr at one point and similary for $ B_{2}$, $ C_{2}$. Since $ AA_{1}, BB_{1}, CC_{1}$ concurr in one point from the Desaurges theorem the intersections $ B_{1}C_{1}\cap BC$, $ C_{1}A_{1}\cap CA$, $ A_{1}B_{1}\cap AB$ are collinear (one can omit the Desaurges here and see that these intersections lie on radical axis of circumcircle of $ ABC$ and circumcircle of $ A_{1}B_{1}C_{1}$). Once more from the Desaurges theorem we conclude that lines $ AP, BQ, CR$ concurr in one point. But note that the circumcircle of $ ABC$ is an incircle of $ PQR$ so from the lemma we are done.

Actually, this lemma is called [b]Steinbart's Theorem[/b]  :) !

Nice!

Let $A_0B_0C_0$ be a triangle whose intouch triangle is $ABC$. Let $T_C=A_1B_1 \cap AB$ and define other analogues of it. Clearly, $\ell=T_AT_BT_C$ is the tripolar of $H$ and these three points are collinear. Now, by the radical axis theorem, one can see that $T_CC_2$ is tangent to $(ABC)$. Therefore, the pole of $\ell$ in the circumcircle $(O)$ of $ABC$ is the intersection point of the three lines $C_0C_2,B_0B_2,A_0A_2$. We see that by the Steinbart theorem, the lines $AA_2,BB_2,CC_2$ are concurrent. The result holds.

[quote=TomciO]It can be also solved with the following nice lemma:
In the triangle $ ABC$ let $ D, E, F$ be the points of tangency of the incircle with $ BC, CA, AB$. Let $ X, Y, Z$ be the points on incircle. Then $ XD, YE, ZF$ are concurrent iff $ AX, BY, CZ$ are concurrent.[/quote]
Shouldn't this lemma have some more assumptions,? Let's take three cevians concurrent in $P$ inside the incircle and not lying on any of $AD$, $BE$, $CF$. Let $AP$ intersect incircle in $X$ and $X'$. Let one of intersection points of $BP$ with incircle be $Y$ and of $CP$ be $Z$. Now triples $AX $, $BY $, $CZ $ and $AX' $, $BY $, $CZ $ can't be both concurrent.

[quote=Little Gauss]Let $ BC\cap B_{1}C_{1}=X, CA\cap C_{1}A_{1}=Y, AB\cap A_{1}B_{1}=Z$.
It is easy to show that $ BCB_{1}C_{1}$ is cyclic, so radical center of $ (ABC), (BCB_{1}C_{1}), (B_{1}C_{1}A_{2})$ is $ X$.
$ \Rightarrow$ $ XA_{2}$ is tangent to $ (ABC)$ at $ A_{2}$.
WTS : $ \frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=1$.
[/quote]

Hello, how does  $\frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=1$ prove the concurrency? Isn't Ceva supposed to be on the sides?

[quote=jayme]Dear Mathlinkers,

for the Steinbart's theorem, you can see and more on

http://jl.ayme.pagesperso-orange.fr/Docs/Les%20points%20de%20Steinbart%20et%20de%20Rabinowitz.pdf

Sincerely
Jean-Louis[/quote]

do you have this article in english version???

[quote=pohoatza]$ AA_{1}$, $ BB_{1}$, $ CC_{1}$ are altitudes of an acute triangle $ ABC$. A circle passing through $ A_{1}$ and $ B_{1}$ touches the arc $ AB$ of its circumcircle at $ C_{2}$. The points $ A_{2}$, $ B_{2}$ are defined similarly. Prove that the lines $ AA_{2}$, $ BB_{2}$, $ CC_{2}$ are concurrent.[/quote]

[hide=Solution]
[b]Lemma[/b] Let $A_1,B_1,C_1$ be points on $\odot{ABC}$ of a triangle $ABC$.Then $AA_1,BB_1,CC_1$ concur iff $\left(\dfrac{BA_1}{A_1C}.\dfrac{CB_1}{B_1A}.\dfrac{AC_1}{C_1B}\right)=1$

[i]Proof[/i]   Follows from Trig Ceva.$\square$
               --------------------------------------------
Let $T_C=A_1B_1\cap AB$ and $T_A,T_B$ defined similarly.Note that $TC_2$ is tangent to $\odot{ABC}$ via radical axis theorm on $\odot{ABC},\odot{ABA_1B_1},\odot{A_1B_1C_2} \implies T_CAC_2\sim T_CC_2B \Rightarrow \dfrac{C_2A}{C_2B}=\dfrac{T_CA}{T_CC_2}=\left(\sqrt{\dfrac{T_CA}{T_CB}}\right)$.Thus since $T_AT_BT_C$ is the orthic axis so $T_A,T_B,T_C$ collinear hence by our lemma and Menelaus we are done.$\blacksquare$

[/hide]

[quote=Little Gauss]Let $ BC\cap B_{1}C_{1}=X, CA\cap C_{1}A_{1}=Y, AB\cap A_{1}B_{1}=Z$.
It is easy to show that $ BCB_{1}C_{1}$ is cyclic, so radical center of $ (ABC), (BCB_{1}C_{1}), (B_{1}C_{1}A_{2})$ is $ X$.
$ \Rightarrow$ $ XA_{2}$ is tangent to $ (ABC)$ at $ A_{2}$.
WTS : $ \frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=1$.
But $ \frac{BA_{2}}{A_{2}C}=\frac{XA_{2}}{XC}=\frac{\sqrt{XB\cdot XC}}{XC}=\sqrt{\frac{XB}{XC}}$,
$ \frac{BX}{XC}\frac{CB_{1}}{B_{1}A}\frac{AC_{1}}{C_{1}B}=1 \Rightarrow \frac{XB}{XC}=\frac{B_{1}A}{CB_{1}}\frac{C_{1}B}{AC_{1}}$.
$ \Rightarrow \frac{BA_{2}}{A_{2}C}\frac{CB_{2}}{B_{2}A}\frac{AC_{2}}{C_{2}B}=\frac{AC_{1}}{C_{1}B}\frac{BA_{1}}{A_{1}C}\frac{CB_{1}}{B_{1}A}=1$.

Is it right?

BTW, Problems in this olympiad are very interesting for me. Is there any official homepage of it?[/quote]

Ya its correct


Let $ A \equal{} \{1;2;...;25\}$. Prove that: for all $ B\subset A$ in which $ B$ have 17 elements always exit $ a\neq b$ in $ B$ satisfy $ ab$ is a square number.

Obviously $ B\equal{}\{2,3\}$ is a counterexample.

We can except $ \{7,14,21,11,22,13,17,19,23\}$ (9 terms) and have at least 8 terms in B form $ 2^k3^m5^n$.
B had not terms $ 2*3*5\equal{}30, (k,m,n)\not \equal{}(1,1,1)\mod 2$, therefore 8 terms in these form had 2, suth that $ k_1\equal{}k_2\mod 2, m_1\equal{}m_2\mod 2, n_1\equal{}n_2\mod 2$.

OK Rust! Your solution is better than me!


Is there a file with pictures for this test? I found a pdf by anirudh on the wiki for 2006 10A. I was wondering about if the 10B has another file with pictures, not just words saying external image.

As I have said many times before, this is in violation of the AMC's copyright.  We have removed the link from the wiki.

However, since the file is not hosted on our site, there's not much we can do about it.


$ x,y,z > 0$ . Prove that: $ \sum \frac {\sqrt {y}}{\sqrt {x \plus{} y}} \le \frac {3}{\sqrt {2}}$

[hide="Hint"]$ \sum_{\mathbf{cyc}} \frac {1}{\sqrt {1 \plus{} \frac{x}{y}}}$

Replace with $ \frac{x}{y} \equal{} a$, $ \frac{y}{z} \equal{} b$, $ \frac{z}{x} \equal{} c$, $ abc\equal{}1$

Then our sum becomes $ \frac{1}{\sqrt{1\plus{}a}} \plus{} \frac{1}{\sqrt{1\plus{}b}} \plus{} \frac{1}{\sqrt{1\plus{}c}}$

Define $ P_{[n]} [ \ldots]$ be the nth power mean of the contained list. Define $ L\equal{} (1\plus{}a,1\plus{}b,1\plus{}c)$, and let $ S$ be our sum. We can write:

$ S \equal{} \frac{3}{\sqrt{P_{\left[ \minus{}\frac 12 \right]}[L]}}$

Power Mean kills it[/hide]


geometry
conics
parabola
analytic geometry
graphing lines
slope

$F$ is the focus of a parabola $y=ax^2$. A line through $F$ meets the parabola at $P$ and $Q$. If $PF=p$, $QF=q$, then find the value of $\frac{1}{p}+\frac{1}{q}$.

[hide]The focus must be at $(0, \frac{1}{4a})$.  Given that that value is constant, I know it's $4a$, but here's a proof.

Suppose a line of slope $m$ passes through the focus.  Then, it has equaiton $y=mx+\frac{1}{4a}$  At the intersections, we also have $y=ax^2$.  Hence, \[ ax^2=mx+\frac{1}{4a} \implies ax^2-mx-\frac{1}{4a}=0 \] \[ \implies x=\frac{m \pm \sqrt{m^2+1}}{2a} \implies ax^2=\frac{2m^2+1 \pm 2m\sqrt{m^2+1}}{4a} \]  The distance from a point on the parabola to the focus is the same as the distance to the directrix.  The directrix has equation $y=-\frac{1}{4a}$.  Thus the distance is just the y-coordinate plus $\frac{1}{4a}$.  So the two distances to the directrix are then \[ \frac{2m^2+2 \pm 2m\sqrt{m^2+1}}{4a}=\frac{m^2+1\pm\sqrt{m^2+1}}{2a} \]  We know $\frac{1}{p}+\frac{1}{q}=\frac{p+q}{pq}$.  Thus, \[ \frac{p+q}{pq}=\frac{\frac{2m^2+2}{2a}}{\frac{(m^2+1)^2-(m^2(m^2+1))}{4a^2}}=\frac{m^2+1}{a}\frac{4a^2}{m^2+1}=\boxed{4a} \] [/hide]


function
algebra
domain
inequalities
calculus
calculus computations

Given a function $ f: I \to R$ defined on an interval I, let $ f*(t) \equal{} \mathop {\sup }\limits_{x \in I} (tx \minus{} f(x))$.

Show that the set $ I*$ of $ t \in R$ for which $ f*(t) \ne \infty$ is either empty or consists of a single point, or is an interval of the line, and in this last case the function $ f*(t)$ is convex on $ I*$

can we get some continuity of f?

if $ f(x_n) \to \minus{}\infty$ for some $ (x_n)$, then $ f * (t) \equal{} \sup_{x \in I} tx\minus{}f(x) \equal{} \infty \forall t \to I*$ empty

therefore if I* is not empty, we must have at least $ \inf_{x\in I} f(x) > \minus{} \infty$

looking at the statement i feel like it might not be true?

The statement of the problem doesn't say anything about f being continuous (it's worded rather vaguely, actually).

So what of the case where f is bounded below? We can give an upper bound for f*(t) in  this case - provided I is bounded, which I assume. When is I* degenerate? Whether f is bounded above or not seems to be irrelevant.

There seems to be something [url=http://www.cc.gatech.edu/~lebanon/notes/convexFunctions.pdf]here[/url]under the name 'Fenchel conjugate' and also [url=http://books.google.com/books?id=byF4Xb1QbvMC&pg=PA106&lpg=PA106&dq=convex+conjugate+supremum&source=web&ots=zHpFHrPVem&sig=u274cKPo50ZMpXaU--J_eLL98mc&hl=en&sa=X&oi=book_result&resnum=10&ct=result#PPA107,M1]here[/url].

So linear transformations in R are an example of functions with Legendre transforms defined on a degenerate set - PROVIDED we take 'an interval' to mean R. It appears these transforms are usually defined on R (with f taking on the value infinity outside its domain) so let's work with that understanding. I wish there was an expert on this though.

okay i guess, assume now that it is not empty and is not a single point

then $ \sup_{x \in I} (t_1x \minus{} f(x)) \equal{}f*(t_1) < \infty, \sup_{x \in I} (t_2x \minus{} f(x)) \equal{}f*(t_2)< \infty, t_1<t_2$

$ t_1<t<t_2 \to t_1x \minus{} f(x)< tx \minus{} f(x) < t_1x \minus{} f(x)$

$ \to \sup_{x \in I\cap R^\plus{}} \left(t_1x \minus{} f(x)\right)\le \sup_{x \in I\cap R^\plus{}} \left(tx \minus{} f(x)\right) \le\sup_{x \in I\cap R^\plus{}} \left(t_2x \minus{} f(x)\right)$

not sure how to control the negative x that are in I though

You are trying to show connectedness, I presume. For -x, wouldn't you reverse the inequalities? (Can we say something about whether I* is connected or closed).

I edited my post while you were writing to suggest a change in how we define f*.

$ \sup_{x \in I\cap R^\minus{}} \left(t_0x \minus{} f(x)\right)\ge \sup_{x \in I\cap R^\minus{}} \left(t_1x \minus{} f(x)\right) \ge\sup_{x \in I\cap R^\minus{}} \left(t_2x \minus{} f(x)\right)$

defining $ f_i^\minus{} \equal{}\sup_{x \in I\cap R^\minus{}}\left(t_ix \minus{} f(x)\right),f_i^\plus{} \equal{}\sup_{x \in I\cap R^\plus{}}\left(t_ix \minus{} f(x)\right)$

$ f*(t_1)\equal{}\max (f_1^\minus{},f_1^\plus{}) \le \max (f_0^\minus{},f_2^\plus{}) < \infty$

which makes $ f*(t)$ an interval. so at least the first claim is done.

showing convexity requires something from what you said above i thinks


function
real analysis
real analysis unsolved

Let $f: [0,1]\to R$ be continuous over $[0,1]$ and $f(0)=f(1)=0$. Prove that exist convex continuous function $g: [0,1]\to R$ such that $g(0)=g(1)=0$ and $g(x)\geq f(x)$ for all $x\in [0,1]$.

Shouldn't that be $g(x)\le f(x)?$

Here's a construction:

For each $m\in\mathbb{R},$ let $h(m)=\min_{x\in[0,1]}(f(x)-mx).$

Now define $g(x)=\sup_{m\in\mathbb{R}}(mx+h(m)).$

Since $mx+h(m)\le f(x)$ for each $m,$ it follows that $g(x)\le f(x).$

Now pick two points $x$ and $y$ and some $t\in(0,1).$

For each $m,$

$(1-t)(mx+h(m))+t(my+h(m))=m[(1-t)x+ty]+h(m)$

Now take the sup over $m$ of both sides. The sup of the right hand side is $g[(1-t)x+ty],$ by definition. On the left hand side.

$\sup_{m}[(1-t)(mx+h(m))+t(my+h(m))]$

$\le \sup_{m}(1-t)(m(x)+h(m))+\sup_{m}t(m(y)+h(m))$

$=(1-t)\sup_{m}(m(x)+h(m))+t\sup_{m}(m(y)+h(m))=(1-t)g(x)+tg(y).$

Hence $g[(1-t)x+ty]\le(1-t)g(x)+tg(y)$ and $g$ is convex.

Wow :)  a realy nice problem and a beautiful solution Kent Merryfield
Is the original statement also true?

[quote="ciprian"]Is the original statement also true?[/quote]
Of course not. Just let $f$ be some function with $f(c)>0$ for some $0<c<1.$ To have $g(x)\ge f(x),$ we'd have to have $g(0)=0, g(c)>0,g(1)=0,$ and that violates convexity.

Of course g must be <=0 in [0,1]:) Sorry for my nonsense :(
But concave? The same argument works? just switching min with max and sup with inf?


1)Explain which compound is the weaker base:

[hide]Since the lone pair of the first is in conjugation so I suppose it should be the weaker base but if we consider acid/base wise , then the second one is more acidic hence less basic.[/hide]

2) Rank the following amines in the increasing basic nature;
[hide]
Doubtful about para and ortho isomers. Also I want to know what will happen if the substituent groups attached are Cl and NO2 respectively instead of CH3[/hide]

1. The strongest acid is oxalic acid, and so it is the weakest base. Remember, in dicarboxylic acids the second proton is always more relutant to leave the molecule.

2. The strongest base is cyclohexylmethylamine: there is no ressonance possiblity for the nitrogen lone pair.

With an electron withdrawing group like -NO2 in the ring, basicity will decrease, since the nitrogen lone pair will be more conjugated with the ring.

Ans 1 -edit , overlooked carcul's post :D

Ans -2
is it ->
4>3>2>1

4 - no resonance
3 - +M, +I
2 - +I
1 - Ortho Effect


Prove that $2n \choose n$ is divisible by $n+1$.

The identity $(n+1) \binom {2n}{n-1} = n \binom {2n}n$ shows that $(n+1)|n \binom {2n}n$ and, since $\gcd(n, n+1) = 1, (n+1)|\binom {2n}n.$

We know that $\frac 1{n+1} \binom{2n}n$ is the number of triangularisations of a $n$-gon (the $n$-th Catalan number).
This number is trivially an integer.

[quote="Peter"]Prove that $ 2n\choose n$ is divisible by $ n\plus{}1$.[/quote]

One may slightly generize the argument of mathmanman.

Proposition. Let $ N$ and $ k$ are positive integers with $ N\geq k$ and $ \gcd(N\plus{}1, k\plus{}1)\equal{}1$. Then, $ N\choose k$ is divisible by $ k\plus{}1$.

Proof. It is well-known that
\[ (N\plus{}1){N\choose k}\equal{} (k\plus{}1){{N\plus{}1}\choose{k\plus{}1}}.\]
Hence, $ (N\plus{}1){N\choose k}$ is divisible by $ k\plus{}1$. Since  $ \gcd(N\plus{}1, k\plus{}1)\equal{}1$, this implies that $ N\choose k$ is divisible by $ k\plus{}1$.

As you know,$C_n=\frac{1}{n+1}\binom{2n}{n}$ is called as [b]Catalan number[/b].

$\binom{2n}{n}-\binom{2n}{n-1}=\frac{(2n)!}{n!n!}-\frac{(2n)!}{(n-1)!(n+1)!}$
$=\frac{(2n)!}{n!(n+1)!}=\frac{1}{n+1}\binom{2n}{n}$ is positive integer.The proof is completed.$\blacksquare$ :coolspeak:


MATHCOUNTS geometry Euler AMC AIME probability algebra

[quote="Ignite168"][quote="bpms"][quote="jhredsox"]Does anyone have a list from easiest to hardest for chapters, states, and nats?
-jorian[/quote]
The chapters from hardest to easiest. would be in order 04, 05, 03, 02, 06, 01 in my opinion.
For states it gets a little more complicated. Looking through my old ones I have come up with this order from hardest to easiest:
06, 05(killer target) ,99, 04, 01, 03, 02, 96, 00, 98, 93, 94, 97.
For nationals, I think this is pretty accurate:
05 (killer overall), 06, 04, 03, 01, 02, 96(last ten sprints), 97, 00, 98, 99 (easier than state of this year), 95, 94, 93, 91, 92, 90,[/quote]

'06 States was TONS easier than '05.

'99 was a lot harder than most of the nationals. I got a 17 on it while I got much higher on 05.[/quote]
There were some good questions on the 06 sprint, and I got a 20 on the 05 sprint when I did it officially in sixth grade so It must've been easy (though I got 3 right on the target that year)...

well thank you bpms for all of your hard work in this area

and ignite

bpms, u went to nats, right?
-jorian

why are all these new tests harder than the old ones?
-jorian

ty for this
so....they generally have been getting harder?
-jorian

89 was the hardest year, although it [i]was[/i] in the old format :wink:

90 was by far the easiest(1st one in current format)

[quote="13375P34K43V312"][quote="bpms"][quote="13375P34K43V312"]04 nats was WAY harder than 06.[/quote]
They are of comparable difficulty but IMO 06 was slightly harder.
To jhredsox, I think the caliber of the students has increased, or mathcounts was getting too high scores. Perhaps a combination...
EDIT: Yes, I went to nats.
EDIT 2: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98871[/quote]

IMO 2006??

I think 2006 state was harder than 2004 state, but the 2004 nat test was harder.

The 2001 State was...ugh...#5 and #6 on target could easily have come on nationals. I found 01 nat easier than 01 state.

I think this year (2007), the countdown cutoff will be around 40, with the average top 57-er having a score of like 24/12 or so. It will be a little easier than last year's.

Anyway, the 1984 nat masters round was really hard.[/quote]

IMO=in my opinion (in this instance)

Somehow, I think this year will be uber hard, like harder than 2005, dunno why

I had that feeling too, but this one's stronger.

I don't see how your feelings could have any basis at all.

[quote="13375P34K43V312"]Yeah, I guess we just have to see how everything plays out.

I personally am hoping for a hard test, since I modestly say that I believe that I know more "real" math than most mathcounts national competitors, and I might be able to use higher-level shortcuts (such as Fermat-Euler or Vandermonde) on a hard Mathcounts test.[/quote]
Fermat Euler=Euler's Totient?
Anyays, Euler's Totient only applies to the cases where you want the last 2 or maybe 3 digits and the binomial theorem dominates for those. For example, Euler's Totient would help with an old problem find the last three digits of $9^{2004}$, but the binomial theorem dominates. As for vandermonde, it is powerful, but I don't think I have ever seen it used before in mathcounts.

[quote="Phelpedo"][quote="bpms"]Looking through my old ones I have come up with this order from hardest to easiest:
06, 05(killer target) ,99, 04, 01, 03, 02, 96, 00, 98, 93, 94, 97.
[/quote]

You know this is just 93-2006 -{95} in skewed order.[/quote]
Well how else could it be organized :roll:. The only one I am missing is 95...
EDIT: Now that I look at it, vandermonde is completely useless...

[quote="13375P34K43V312"]Yeah you're right, since vandermonde is proven using a combinatorial argument + casework, it would probably just be easier to go through the casework  :D

there have been recent last3/last2 digit problems, however.[/quote]
Did you read what I wrote
[quote="bpms"]Fermat Euler=Euler's Totient?
Anyays, Euler's Totient only applies to the cases where you want the last 2 or maybe 3 digits and the binomial theorem dominates for those. For example, Euler's Totient would help with an old problem find the last three digits of $9^{2004}$, but the binomial theorem dominates. As for vandermonde, it is powerful, but I don't think I have ever seen it used before in mathcounts. [/quote]

How would you use the binomial theorem for that problem?

To perfect 628, How did you do at nats last year?

Hard tests are WAY better in my opinion, and 07 shoulkd be one of the hardest ever. I hope it's THE hardest ever.

[quote="jhredsox"]wait. why do u hope its the hardest ever?
-jorian[/quote]
Careless mistakes have less weight and can be afforded more.

Yeah, with an easy test, it's just a matter of who makes the least careless mistakes but with a harder test, you can better determine who has the most knowledge.

[quote="jhredsox"]thats what i was thinking in the back of my head. though, i hope there's not too much geometry on the test.

what's the hardest geometry area mathcounts will go into?
-jorian[/quote]
I love geometry at the mathcounts level :). The hardest mathcounts problems aren't what you should be worried about if you want to get better at geometry. What you should be looking at is medium level AIME problems.

ummmmmmmmmmm.........can u tell me where i find these?
-jorian

[quote="star99"]for me the geometry problems are much easier than other problems, especially probability, and anything to do with counting/permutation.  and those show up a heck lot on MC[/quote]

Yeah. Geometry on mathcounts, or even in general, is much easier than counting/permutation/probability.

bpms: Can you please stop offering advice like that? I talk to Jorian outside of AoPS and he hasn't taken Geometry yet. MathCouts Nationals problem he doesn't get most of the time and comments like that will only discourage him more.

@ignite:  what you just said would probably hurt him a lot more.

everyone one on aops should know that there are tons of people on the forum better than they are at math, and thats why they are here, to learn to be that good.

Sorry, but I was referring to the whole, "Do problems that are really hard for you to get better". I'd say the hardest mathcounts problem on geometry ever period would be
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11092
I have no clue how someone would get that in 3 min.
Or even 6 for that matter.

wow, bpms, that's a real hard question

and btw, the geometry portion for me is really hard compared to the probability portion and the algebra
-jorian

[quote="pianoforte"]I hate that question. I've seen it so many times and I still keep getting different answers for that one.[/quote]I still don't know how one would get it in 3 min. Read Ravi B's solution it's good (not surprising seeing as he is a beast at problems :D)

isn't that an inefficient way to test difficulty?
-jorian

Maybe.  Do you know of a better method?  I like to solve the problems anyway to be well prepared to coach, so I might as well time myself doing so.

a good way is to do what u do now, and then record the percentage of people who got it right (don't do sprint!, they might not get to it)
-jorian

[quote="Ravi B"]Thanks for the compliment.  While I was preparing to coach the NY team last year, I solved that question in 3 minutes 20 seconds.  (I record my times to measure the difficulty of questions.)  It probably would have taken me a lot longer except that I had seen the question a couple of years ago when I posted that solution.[/quote]
Really? Well, any solution under 5 minutes is okay seeing as the pair to that problem can be done in under a minute with a calculator.

Well you could interpret its "difficulty" many ways.

There could be 29 ultra easy questions, which take like 10 minutes total (i mean for sprint), then 1 uberly hard question that only like the top people at nationals can solve.

Then, there could be a test with 30 questions that are all like State #15 on the sprint.


geometry
geometric transformation
reflection
geometry solved

give us procedure to draw 3 cevaian line with equal size in the traingle .(with compass and ruler).

That's the h3ll of a problem! Actually, in [url=http://forumgeom.fau.edu/FG2003volume3/FG200307index.html]Paul Yiu, [i]On the Fermat lines[/i], Forum Geometricorum 3 (2003) p. 73-81[/url], paragraph 6, item (14), you can find a construction of the two focii of the Steiner inellipse of triangle ABC. If you reflect the centroid G of triangle ABC in these two focii, you get two new points, which are the focii of the Steiner circumellipse of triangle ABC. These two new focii have the property that the cevians of each of these focii have equal length. And these points are the only points with this property.

Please don't ask me about the proof of this construction. See also [url=http://groups.yahoo.com/group/Hyacinthos/messages/211?viscount=100&expand=1]the Hyacinthos messages #211, #212, #215, #217, #222, etc.[/url].

  Darij


[quote="Maverick"]If a, b, c are positive numbers with a + b + c = 1, then prove

[tex]\frac{1}{bc+a}+\frac{1}{ca+b}+\frac{1}{ab+c}\geq \frac{27}{4}[/tex].[/quote]

Indeed easy, but very nice (I have TeXed the formula in order to better appreciate its beauty  :D ).

[tex]\frac{1}{bc+a}+\frac{1}{ca+b}+\frac{1}{ab+c}=\frac{1}{bc+a\left( a+b+c\right) }+\frac{1}{ca+b\left( a+b+c\right) }+\frac{1}{ab+c\left( a+b+c\right) }[/tex]
[tex]=\frac{1}{\left( bc+ca+ab\right) +a^{2}}+\frac{1}{\left( bc+ca+ab\right) +b^{2}}+\frac{1}{\left( bc+ca+ab\right) +c^{2}}[/tex]
[tex]\overset{\text{by HM-AM}}{\geq }9\cdot \frac{1}{\left( \left( bc+ca+ab\right) +a^{2}\right) +\left( \left( bc+ca+ab\right) +b^{2}\right) +\left( \left( bc+ca+ab\right) +c^{2}\right) }[/tex]
[tex]=9\cdot \frac{1}{3\left( bc+ca+ab\right) +\left( a^{2}+b^{2}+c^{2}\right) }=\frac{27}{4}\cdot \frac{4}{9\left( bc+ca+ab\right) +3\left( a^{2}+b^{2}+c^{2}\right) }[/tex]
[tex]\overset{\text{since }bc+ca+ab\leq a^{2}+b^{2}+c^{2}}{\geq }\frac{27}{4}\cdot \frac{4}{8\left( bc+ca+ab\right) +4\left( a^{2}+b^{2}+c^{2}\right) }[/tex]
[tex]=\frac{27}{4}\cdot \frac{4}{4\left( a+b+c\right) ^{2}}=\frac{27}{4}\cdot \frac{4}{4}=\frac{27}{4}[/tex].

Equality if and only if $a=b=c=\frac13$.

  Darij

Or alternatively, by Cauchy,

$\sum \frac{1}{bc+a} \geq \frac{9}{a+b+c+ab+bc+ca} \geq \frac {9}{a+b+c+\frac{(a+b+c)^2}{3}} = \frac{27}{4}$ with equality iff $a=b=c=\frac{1}{3}$.

Whoa... I made the exact same inequality a while ago   :shock:

It's posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13175]here[/url] with even more solutions.

Alternatively,

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

$\displaystyle  = \frac{\sum (b+c)}{(a+b)(b+c)(c+a)} = \frac{2}{(a+b)(b+c)(c+a)}$

And by AM-GM, $\displaystyle {(a+b)(b+c)(c+a)} \leq \left(\frac{(a+b)+(b+c)+(c+a)}{3}\right)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$

Yet another solution,

[tex] \displaystyle \sum \frac{1}{bc+a} = \sum \frac{1}{bc+a(a+b+c)} = \sum \frac{1}{(a+b)(a+c)} [/tex]

From Holder,

[tex] \displaystyle \left( \sum \frac{1}{(a+b)(a+c)} \right)^{1/3} \left( \sum (a+b) \right)^{1/3} \left( \sum (a+c) \right)^{1/3} \geq (1+1+1) [/tex],

and since [tex]\displaystyle \sum (a+b) = \sum (a+c) = 2[/tex], this is equivalent to the original inequality.


There seems to be NO link to the Problem of the day from the Index page. Even on the resources page, there is no link to it. However, there is one on the Mathlinks Homepage.  Please see to this.

It is an experimental feature so for now the links are only presented there where they are now.

OK. Dont forget to add a prominent one when you decide to (if) make it permanent!  :lol:


Triangles $ABC$ and $XYZ$ are similar, with $A$ corresponding to $X$ and $B$ to $Y$. If $AB=3$, $BC=4$, and $XY=5$, then $YZ$ is:

$ \textbf{(A)}\ 3\frac 3 4 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 6\frac 1 4 \qquad \textbf{(D)}\ 6\frac 2 3 \qquad \textbf{(E)}\ 8$

[hide]Becuase ABC and XYZ are similar, their respective sides must be proportional.  Therefore,

$\frac{AB}{XY} = \frac{BC}{YZ}$
$\frac 35 = \frac4{YZ}$
$3(YZ) = 20$
$YZ = \frac{20}3$ = D[/hide]

answer

--[hide] (D) [i]yz/bc=xy/ab[/i]
[/hide]

[/i]

[hide="we just learned this today..."]
create a proportion:
ab/xy=bc/yz
substitute in the values:
3/5=4/yz
cross multiply:
3(yz)=20
yz=6 2/3
or
D
[/hide]

[hide]
$\frac{AB}{XY}=\frac{BC}{YZ}$
Plug in the values:
$\frac{3}{5}=\frac{4}{YZ}$
$3(YZ)=20$
So $YZ=6\frac{2}{3}$
$\boxed{\boxed{D}}$[/hide]

[hide]
The anwer is D. A corresponds to X and B corresponds to Y. therefore AB corresponds to XY. If AB=3, and XY=5, then we know the ratio between the two are 3:5.

therefore

4 x 5/3 = 20/3 = 6+2/3
[/hide]

[hide]As the length of AB corresponds to the length of XY, we find that the ratio of sides in triangle ABC to triangle XYZ is 1:5/3. BC corresponds to YZ, so 4*5/3=20/3, or [b]D)6 2/3[/b].[/hide]

That was an AMC 12 problem?

[quote="easyas3.14159..."]That was an AMC 12 problem?[/quote]

Well it's number 2, and the first 5 are usually a piece of cake (usually...). You'd be surprised how many people get these types of problems wrong.

[quote="4everwise"]Triangles $ABC$ and $XYZ$ are similar, with $A$ corresponding to $X$ and $B$ to $Y$. If $AB=3$, $BC=4$, and $XY=5$, then $YZ$ is:

\[ \text{(A)}\ 3\frac 3 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 6\frac 1 4 \qquad \text{(D)}\ 6\frac 2 3 \qquad \text{(E)}\ 8  \][/quote]

[hide]without drawing the pic, you can tell that ab and xy correspond
3/5=4/x
20=3x
x=20/3= 6 2/3
[/hide]


trigonometry
analytic geometry
ratio
quadratics
email
geometry
3D geometry

It seems that problem solving marathon is a good way to keep a forum active, especially for this one. So, here we go!

[b]Problem One[/b]

The initial kinetic energy of the ball is 6J,  :theta: =30, Find the kinetic energy of the ball when it touches the incline again.

ok this problem is either VERY good, and challenging., in that case i like it.
or it's very simple and you 'va forgotten to give more info about the problem--because nothing about the hight differenct is known, if it was, then the problem would be only simple case of conservation of energy
please state wether the problem is correct...or not. :)

assuming the problem is correct, is the answer:
$6(1+\sqrt{3})$?

Edit: I deleted this because my answer was completely off.

here is what i did:
we know that if we launch a projectile motion with an angle $\theta$ on the horizantal surface, the time it takes the projectile to reach the ground would be:
$t=\frac{2v_0\sin{\theta}}{g}$. but since the are not on a horizantal surfact the effective $g$ would be $g\sin{\theta}$ therefor the time it takes for the ball to reach the surface again, in our problem would be:
$t=\frac{2v_0\sin{\theta}}{g\sin{\theta}}=\frac{2v_0}{g}$
if the diffrence of hieghts of the ball before and after moving is: $h$. then based on similar tirangles, the ball would move a lentgh of $2h$. (if this is confusing. let the point that ball left the ramp be $A$ and the point that came back be $B$, then $AB=2h$)
also we know this:
$2h=v_0\cos{\theta}. t=\frac{2v_0^2\cos{\theta}}{g}$
now based on conservation of energy:
$\frac{1}{2}mv_0^2+mgh=\frac{1}{2}mv_2^2$ first thing you can do is to cancell out $m$'s, after substituting for $h$, you'll get:
$\cos{\theta}v_0+\frac{1}{2}v_0=\frac{1}{2}v_2^2$
we know that $\theta=\frac{\pi}{6}$
so we get:
$\frac{\sqrt{3}}{2}v_0+\frac{1}{2}v_0=\frac{1}{2}v_2^2$
therfore you get: $(1+\sqrt{3})v_0^2=v_2^2$ therefore
$E_{k2}=(1+\sqrt{3})E_{k1}=6(1+\sqrt{3})$

i did all this, but i have this feeling that something is wrong ;)

If you define your coordinate system's x-axis to be along the inclined plane the problem becomes much easier, I'll post my full solution later.

[quote="amirhtlusa"][quote="shobber"]there is a trick to solve this problem...
[hide]
14J[/hide][/quote]
could you explain. ? ;)[/quote]
$\tan{\alpha}=2\tan{\beta}$

:huh:
ok, can you explain, so we can move on...? :|
since there is nothing given about the hight difference, first thing you have to do is to find the height diffrence based on the given values...
then you have to use the conservation of energy to solve the problem.
how the heck, did you get 14???? :ninja:

There is no need for any dirty tricks, I think. I did exatcly what amirhtlusa described:

[quote]since there is nothing given about the hight difference, first thing you have to do is to find the height diffrence based on the given values...
then you have to use the conservation of energy to solve the problem. [/quote]
and I got 14 Joules:

[hide="step by step"]
Regular xy coordinate system, but y is positive as we go downwards. Denote v the original velocity of the ball. Then:

x = v.t
y = (1/2).g.t^2

Ball hits the incline when (at time t):

y/x = tan 30
sqrt(3).y = x
(sqrt(3)/2).g.t^2 = v.t
t = (v/g).(2/sqrt(3))

That makes height difference h = y(t) - y(0)

h = (1/2).g.t^2 = (1/2).g.((v/g).(2/sqrt(3)))^2
h = (2/3).(v^2)/g

Which makes potential energy decreacse

deltaE = m.g.h
deltaE = (2/3).m.(v^2) = 4/3 * ((1/2).m.v^2)

But the term ((1/2).m.v^2) is the original kinetic energy of the ball! That implies kinecit energy increase of (4/3) * 6 J = 8 J, total kinetic energy is therefore 6 + 8 = 14 Joules.

(sorry for not using the TEX typesetting)
[/hide]

For a body doing a projectile motion, there is always:
\[\tan{\alpha}=2\tan{\beta}\]

[quote="kubus"]There is no need for any dirty tricks, I think. I did exatcly what amirhtlusa described:

[quote]since there is nothing given about the hight difference, first thing you have to do is to find the height diffrence based on the given values...
then you have to use the conservation of energy to solve the problem. [/quote]
and I got 14 Joules:

[hide="step by step"]
Regular xy coordinate system, but y is positive as we go downwards. Denote v the original velocity of the ball. Then:

[b]x = v.t
y = (1/2).g.t^2 [/b]

Ball hits the incline when (at time t):

y/x = tan 30
sqrt(3).y = x
(sqrt(3)/2).g.t^2 = v.t
t = (v/g).(2/sqrt(3))

[/hide][/quote]
that is not true...

What is not true?

If you mean the x and y coordinate calculations - they are independent, there is no acceleration in x, therefore x = v.t. There is no initial velocity in y direction, but acceleration of magnitude g downwards (i.e. towards positive y), that means velocity vy = g.t and position (vertical position) y = (1/2).g.(t^2) = g.t.t/2.

Or is it something else which is not clear?

shouldn't be:
$x=v\cos{30}t$
$y=\frac{1}{2}g\sin{30}t^2$
actually it depends on the coordinates,
what was your coordinate system??

[hide="my solution"]
The first ball (M) bounces up with velocity v. Looking from the reference frame of M, m is approaching with speed 2v. Since it is much lighter, it just bounces back with velocity 2v. That is 3v seen from the ground. Now m sees mu approaching with speed of 4v. It also bounces back with 4v, that is 3v+4v = 7v seen from the ground. 7 times larger velocity means 49 times larger kinetic energy = 49 times largegr potential energy at the top of the trajectory = maximum height of 49H.

This is the same as when you have a ball bounce from the earth. Why don't you always calculate the energy and momentum conservation laws for the ball and earth but just assume the ball bounces back with exactly the same velocity? In fact, the earth DOES get an impulse, but since its mass is so large, the velocity is too small and it takes up almost no energy (which depends on v squared).
[/hide]

PS. I'm not going to be online for about two weeks, have fun here ... I'll leave you another one:

There are 3 dogs (A, B, C) standing in vertices of an equilateral triangle with sides of length $a$. In one moment, they all start to chase each other: A runs exactly in the direction towards B, B towards C and C towards A. They all have constant magnitude of velocity $v$. How long does it take before they all meet?

wow. thats actually a really good way to do the problem

[quote="mna851"]Won't the dogs never meet or am I missing something.[/quote]
They do meet. The problem says that [b]runs exactly in the direction towards[/b]. Which means, the direction of the velocity of every dog changes. They are not doing motion along a straight line path.

this way isnt very elegant.. but i think it might work:

first.. notice that the dogs are always in an equilateral triangle wrt eachother when they are running.

now look at the picture.. this is a picture of what happens in time $dt$. the dogs go $dx$, and the sides of the triangle decreases size by $ds = s_1-s_2$

if you know $ds$ in terms of $dx$, then you know $\frac{ds}{dt}$ in terms of $v$.

then its easy to do..

however.. i havent found a nice way to find $ds$. any seggustions?

OOOO now that I understand the problem ill attempt it.

[b]Try to solve it without calculus!!![/b]
There [b]is[/b] a way to do without calculus, but very very difficult. I remember my teacher solving this problem but i cannot recall how he did it...
So please try it.

instead of looking at $ds$ for $\frac{ds}{dt}$, maybe looking at $dr$ would work?

I conjecture that if one dog moves $dx$ towards another dog, it moves $dr = dx \cos(30)$ towards the center, the point they all wind up at.  See the picture for my reasoning. It uses a 90-90-0 right triangle :) , so I'm not confident that it works.

If this is the case, then $\frac{dr}{dt} = \frac{dx}{dt} \cos(30) = v \cos(30) = v \frac{\sqrt{3}}{2}$

The dogs are travelling a distance $\frac{a}{\sqrt{3}}$, at a velocity of $v\frac{\sqrt{3}}{2}$, so the answer should be $\frac{2a}{3v}$, right?

maybe it will help:
assume one dog remains rest.

Due to the symmetry of the system, we assume that all dogs meet at the same time. So, $t$ can be the time required for dog A and dog B to meet.

Let us imagine the whole action in the reference frame of dog A. If we sketch $v_{B} - v_{A}$ for different times, dog B follows a somewhat arc-like (of a circle) path towards A (this is an assumption), where the arc is inside the triangle and AB is a chord. (see sketch)

Looking at the angles of the resultant vector at the beginning and the ending points, $\angle XAB = \angle XBA = 30 \deg$.
Thus, $\angle OAB = \angle OBA = \angle AOB = 60 \deg$, where O is the centre of the circle.
Thus triangle AOB is equilateral.

Let $s = AB = OB \mbox{(radius)}$
Then $arc AB = \frac{1}{3} \pi s$ (B's path)

The magnitude of the velocity of dog B is always the same and can be calculated. (since angle between $v_{B}$ and $v_{A}$ is $120 \deg $)
\[v_{R} = \sqrt{v^{2} + v^{2} - 2v^{2}cos120 \deg}\]
\[= \sqrt{3}v\]
So, $t = \frac{s}{v} = \frac{\frac{1}{3} \pi s}{\sqrt{3}v} = \frac{\pi s}{3 \sqrt{3} v}$
where $s$ is the side of the triangle, and $v$ is the velocity of the dogs.
Well, it seems I have some unexplained assumptions but they just make sense. Could anybody explain (prove) the assumptions or, otherwise, give a better solution?.

I understand that my solution above is wrong.

As the bugs move, they make an equilateral triangle with the same centroid at all times as repeated above. In the reference frame of the rotating triangle, the velocity component towards the centre is always $v cos 30$ (Think of the angle between the direction of a dog and the segment joining it with the centroid.)
At the same time, the displacement in this direction is also $s cos 30$. Thus,
\[ t = \frac{s cos 30}{v cos 30}\]
\[ t = \frac{s}{v}\]

Let's revive the Marathon!!!!

Problem: Hydrogen gas is stored at a high pressure in a small, spherical container. The gas is introduced into a light balloon and its pressure becomes equal to the external atmospheric pressure. Is it possible that the balloon could lift the container in its final state? Assume that the temperature of the gas remains constant.

[color=orange](P115)[/color]

[b]NO ![/b]

[b]Rushil:[/b] Please, as a moderator you should not encourage this type of
thread! It is horrible so post problems after each other like this. It gets to messy.

Actually one should split each problem into a new thread.
And in the future only post one problem per thread.
It makes it easier to search the forum, and the members can easily be more active
and up to date, especially new members.

[b]color[/b], what's the point of posting comments like this one on forsaken two-year old threads? We have established a marathon with rules, so I don't see why this old thread has any relevance...

Ok sorry, I am quite new in the section.
I was searching for resistors and saw this thread, and that's why i did not
notice its date.

So if you already have adapted its good  :lol: .
But why continue with the marathon form?


On circle $ O$, points $ C$ and $ D$ are on the same side of diameter $ \overline{AB}$, $ \angle AOC \equal{} 30^\circ$, and $ \angle DOB \equal{} 45^\circ$. What is the ratio of the area of the smaller sector $ COD$ to the area of the circle?
[asy]unitsize(6mm);
defaultpen(linewidth(0.7)+fontsize(8pt));

pair C = 3*dir (30);
pair D = 3*dir (135);
pair A = 3*dir (0);
pair B = 3*dir(180);
pair O = (0,0);
draw (Circle ((0, 0), 3));
label ("$C$", C, NE);
label ("$D$", D, NW);
label ("$B$", B, W);
label ("$A$", A, E);
label ("$O$", O, S);
label ("$45^\circ$", (-0.3,0.1), WNW);
label ("$30^\circ$", (0.5,0.1), ENE);
draw (A--B);
draw (O--D);
draw (O--C);[/asy]$ \textbf{(A)}\ \frac {2}{9} \qquad \textbf{(B)}\ \frac {1}{4} \qquad \textbf{(C)}\ \frac {5}{18} \qquad \textbf{(D)}\ \frac {7}{24} \qquad \textbf{(E)}\ \frac {3}{10}$

[hide="Solution"]$ \frac{180\minus{}45\minus{}30}{360} \equal{} \frac{105}{360} \equal{} \frac{7}{24} \implies \textbf{(D)}$[/hide]


It would seem like there would be some people that would want to do both. Ah, well, there is always multi-tasking! I do that every class anyway, so nothing new here...

You can indeed sit in both classrooms.  Open a second browser, log in, go to the classroom.  There is a link just below the classroom window that you'll have to click in the second browser to get into a second classroom.


Let $ a,b,c>0$. Prove that:

$ \dfrac{a}{2a\plus{}b}\plus{}\dfrac{b}{2b\plus{}c}\plus{}\dfrac{c}{2c\plus{}a}\leq 1$

[hide]Let $ \frac {b}{a} \equal{} x$, $ \frac {c}{b} \equal{} y$, and $ \frac {a}{c} \equal{} z$, then $ xyz \equal{} 1$. We have to show
$ \frac {1}{2 \plus{} x} \plus{} \frac {1}{2 \plus{} y} \plus{} \frac {1}{2 \plus{} z} \le 1$.
Clearing denominators, this reduces to
$ xy \plus{} yz \plus{} xz \ge 3$
Which is true by AM-GM.
Or we could clear denominators right away and get
$ \frac{c}{a}\plus{}\frac{a}{b}\plus{}\frac{b}{c} \ge 3$ true by AM-GM.
[/hide]

First of all, see easily that $\displaystyle\sum_{cyc} \dfrac{b}{2a+b}=\displaystyle\sum_{cyc} \dfrac{b^2}{2ab+b^2}\ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1$.
Then we have $\displaystyle\sum_{cyc}\dfrac{a}{2a+b}=\dfrac{1}{2}\displaystyle\sum_{cyc}\dfrac{2a}{2a+b}=\dfrac{1}{2}\displaystyle\sum_{cyc}\left(1-\dfrac{b}{2a+b}\right)\le \dfrac{1}{2}(3-1)=1$.

[quote=sqing][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=100583&hilit=Czech+Slovak]Czech and Slovak1999:[/url]
For arbitrary positive real numbers  $ a,b,c$ prove the inequality\[ \frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2a}\plus{}\frac{c}{a\plus{}2b}\ge 1.\][/quote]

$\sum_{cyc}\frac{a}{b+2c}\geq 1$
$\iff$
$\sum_{cyc}\frac{a^2}{ab+2ca}\geq 1$

By Titu,

$\sum_{cyc}\frac{a^2}{ab+2ca}\geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$

By Rearrangement,

This is $\geq 1$ :D


What is the slope of a line parallel to $ 2x\plus{}4y\equal{}\minus{}17$? Express
your answer as a common fraction.

lets just ignore -17.

than the equation becomes 4y=-2x, and y=$ \minus{}\frac{1}{2}$.

$ \minus{}\frac{1}{2}$ * something is -1, so the answer is 2.

answer : 2

@FantasyLover, I think you did this problem in thinking that the problem asked for the slope of the line perpendicular to the line of $ 2x+4y=-17 $, which is why you set the slope times "something" equal to $-1$.

We know that the slope line $A$ perpendicular to line $B$ is the same as the slope of line $B$, all we have to do is find the slope of line $B$ by turning its equation into slope-intercept form $y=mx+b$ where $m$ is the slope of the line:

$ 2x+4y=-17 $

$4y=-2x-17$

$y=\frac{-2}{4}x-17$

Therefore, the slope of this line is $-\frac{2}{4}$, or $\boxed{-\frac{1}{2}}$.


vector
trigonometry
linear algebra
matrix
linear algebra unsolved

Let $E$ be a $n$-dimensional vector space over $\mathbb{C}$. Prove that any $f \in L(E)$ we can represent as a linear combination of no more then $4$ unitary matrices.

eugene wich oral, I mean wich concours.

Lemma:  ($h\in L(E)$, hermitian $h=h^{*}$, $|||h|||\leq 1$ )
is equivalent
(there exist $u\in U(E)$, $h=\frac{1}{2}(u+u^{*})$)

U(E) is unitary group

I think I've seen it before, but I don't think that it was on the forum.

Write $A$ (I replaced your $f$ with $A$) as $\frac{A+A^*}2+\frac{A-A^*}2$ (Moubinool's lemma was basically a hint given in that problem :)). $\frac{A+A^*}2$ is a Hermitian matrix. Assume it's diagonal. Some real multiple $t\frac{A+A^*}2,\ t\in\mathbb R\setminus\{0\}$ of it will have norm $\le 1$, so its diagonal entries will be real numbers in the interval $[0,1]$. We can denote them by $\cos\alpha_i,\ i=\overline{1,n}$. Now let $U$ be the diagonal matrix with entries $\cos\alpha_i+i\sin\alpha_i$. It's unitary, and $\frac{U+U^*}2=t\frac{A+A^*}2$, which means that $\frac{A+A^*}2=\frac{U+U^*}{2t}$ (note that this means that $\frac{A+A^*}2$ is a linear combination of two unitary matrices; it doesn't matter whether it's diagonal or not; I just assumed it was because it seemed easier to write down).

We do something similar for $s\frac{A-A^*}2,\ s\in\mathbb R\setminus\{0\}$, except that this time we denote its eigenvalues (which are purely imaginary) by $i\sin\beta_i,\ i=\overline{1,n}$. We then have $s\frac{A-A^*}2=\frac{V-V^*}2$, where $V$ is the diagonal matrix with entries $\cos\beta_i+i\sin\beta_i$.

One can better do: 2 unitary matrices are enough.
  Let A in Mn(C).By polar decomposition A=UH where U is an unitary
matrix and H is an hermitian nonnegative matrix.There exists an unitary matrix P and a real >=0 diagonal matrix D s.t. H=P*DP.
A=UP*DP; it remains to show the result for D.
  Let u=max(Djj).For all j there exists a real Sj s.t. Djj=ucos(Sj)=
u/2exp(iSj)+u/2exp(-iSj).
Let U1=diag(exp(iSj)) and U2=diag(exp(-iSj)).  U1 and U2 are unitary matrices such that D=u/2U1+u/2U2.


function
limit
pigeonhole principle
algebra unsolved
algebra

Let $ N$ denote the set of natural numbers. Let $ f$ takes $ N$ to $ N$ be a bijective funtion and assume

that there exist a finite limit

$ \displaystyle\lim_{n\to \infty }\frac {f(n)}{n}\equal{}L$

What are the possible values of $ L$?

Nice one.

[hide="Hint"]
If $ L \equal{} 0.9$, then for $ n$ large enough we must have $ f(n)/n < 1$. Is this possible?
[/hide]

[hide="Solution"]
We can certainly have L=1 by taking f(n)=n. We prove that $ L$ cannot take any other value.

If $ f(n)/n \to L < 1$, then by the definition of a limit we must have $ f(n) / n < 1$ for all $ n$ larger than some $ N$. So we only have a finite number of $ k$'s such that $ f(k) > k$: the only candidates are $ k$'s such that $ k < N$.

We can therefore find an $ M$ so large so that for any $ k$ with $ f(k)>k$ we have $ M > f(k)$. We can also assume $ M > N$. But now the interval $ [1,2,\ldots,M]$ is mapped onto itself by $ f$, and $ f(M\plus{}1)$ must also belong to that interval since $ M>N$ and anything larger than $ N$ becomes smaller when $ f$ is applied. This contradicts the pigeonhole principle.

The case $ L > 1$ can be handled similarly, or by observing that replacing $ f$ with $ f^{\minus{}1}$ (which is also a bijection) causes $ L$ to be replaced by $ 1/L$.
[/hide]


Let (O) be a semircircumference. diameter BC . A a variable point on (O) , let H be the foot of the perpendicular from A to BC, Q belows to BA (in the extension though A) such that AQ=AH . show that the loci of Q is a circumference . Identify the center and the lenght of the radius.

Sorry planegeometry I confused the problem with another one, here is the correct enuntiation , the problem is very easy

Let (O) be a semircircumference. diameter BC . A a variable point on (O) , Q belows to BA (in the extension though A) such that AQ=AC . show that the loci of Q is a circumference . Identify the center and the lenght of the radius

loci of Q is a circumcircle  with D its center ,where D is the midpoint of arcBC  the radius^2=DB^2=DC^2=1/2BC^2

[quote="Tiger100"]Let $ B$ , $ C$ be two fixed point on a given circle $ w$ and a mobile point $ A$ which belongs to one of the arcs which are

defined in the circle $ w$ by the chord $ [BC]$ . Define the point $ Q\in AB$ for which $ A\in BQ$ and $ AQ\equal{}AC$ . Show that

the loci of the point $ Q$ is a circle with the center $ D$ in the midpoint of the choosen arc and its radius is equally to $ DC$ .[/quote]

[quote="Virgil Nicula"][quote="Tiger100"]Let $ B$ , $ C$ be two fixed point on a given circle $ w$ and a mobile point $ A$ which belongs to one of the arcs which are

defined in the circle $ w$ by the chord $ [BC]$ . Define the point $ Q\in AB$ for which $ A\in BQ$ and $ AQ \equal{} AC$ . Show that

the loci of the point $ Q$ is a circle with the center $ D$ in the midpoint of the choosen arc and its radius is equally to $ DC$ .[/quote][/quote]

Thanks for your nice extension mr Virgil, I see that the proof is exactly the same.


Prove that we cannot find a continuous function $f:RxR->R$ such that for any continuous function $g:R->R$ there is $t$ with $g(x)=f(t,x)$ for all $x$. What if we replace $R$ with $[0,1]$.

For the first part:

Given a continuous function $f:\mathbb R^2\to\mathbb R$, define $g$ to be continuous on the set $\bigcup_{n\in\mathbb Z}[2n,2n+1]$ s.t. $\min_{[2n,2n+1]}g>\max_{[n,n+1]\times[2n,2n+1]}f$, and then extend it to the whole real line (this is obviously possible).

As for the second question, could you please phrase it exactly as it should be? Which instance of $\mathbb R$ do we replace with $[0,1]$? Is $f$ defined on $[0,1]^2$, or $\mathbb R\times [0,1]$, or does $g$ take values in $[0,1]$, or what? :)

Just replace each time the symbol $R$ with $[0,1]$. :)

[quote="ilyasu1"]For both parts define $g(x) = f(x,x)+1$.
So if $g(x) = f(x,t)$, then $g(t) = f(t,t) = f(t,t)+1$.[/quote]

Very nice solution.

Except that for the second question it doesn't work, since both $g$ and $f$ must take values in $[0,1]$ :).

Indeed, this was a very silly mistake  :blush: .

But what about this approach:

let $t_n$ be a number such that $f(x,t_n) = \sin(2\pi n x)$.
So $t_n \in [0,1]$. So $t_{n_j} \to t$, i.e. there's a converging subsequence.
Consider $f(0,t)$. $f(0,t_{n_j}) \to f(0,t) = 0$, and yet $f(\pi / 4n_j, t_{n_j}) \to f(0,t) = 1$,
since both $(0,t_{n_j})$ and $(\pi /4n_j, t_{n_j})$ converge to $(0,t)$.

I hope that this time there are no errors.


A square-table of dimensions $ n\times n$, where $ n\in N^*$, is filled arbitrarly with the numbers $ 1,2,...,n^2$ such that every number appears on the table exactly one time. From each row of the table
is chosen the least number and then denote by $ x$ the biggest number from the numbers chosen. From each column of the table is chosen the least number and then denote by $ y$ the biggest number from the numbers chosen. The table is called [i]balanced [/i]iff $ x \equal{} y$. How many balanced tables we can obtain?

The problem is from BMO 2001 Shortlist.

Please someone solve this...........


Let $ABC$ be a triangle with $\angle A=120$. The angle bisectors of $\angle A, \angle B, \angle C$ meet the sides $BC,CA,AB$ at $D,E,F$. Show that $\angle EDF = 90$.

posted before: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=114998]foot of angle besector[/url]


It's probably the same for all.

Is there any problem with it? I consider it as a good option, because it actually does something, rather than asking you still. And if you want to go to the forum, you can do it with one click.

Ok, thanks. I don't mind it, but I thought maybe my computer is messed up or something.

[quote="PowerOfPi"]After I post a message, a screen appears asking me whether I want to see my message or go back to the forum. After about 5 seconds, it automatically goes to my message. Is this the same with everybody, or is it just my computer?[/quote]
That doesn't happen with me.

Neither with me.  This probably won't work, but try clearing your cookies.  It works for everything :P


here are 1000 men standing in a circle numbere 1 to 1000 on their back. the person with number 1 is given a sword. he kills 2 and gives sword to 3. 3 kills 4 and passes the sword to 5. the process continues so on....................THE PROCESS CONTINUES TILL ONLY 1 PERSON IS LEFT. wat is the number of the last remaining person.

I am assuming that at the end every "round", person #1 starts off the "next round" again. By this is is pretty clear that person 1 will be the last one left. Or am I going wrong anywhere?

Well, that's true for the first few rounds, but do it yourself, and you'll quickly see that that is not the case.

Ok, so how about this: if #1 starts first every round, it is impossible to it to be out that round, isn't that true? Or is it like they are all like arranged in a circle?

EDIT: never mind...I over-read the question for the word circle.  :blush:

What happens is, 1 doesn't start all of the rounds.

1 will suddenly cease to start the round when the number of people in the previous round is odd.I have reason to belive that this will happen; the reason is that 1000 is not a power of 2.

What a violent problem!!! Involves killing and violence....hmmm....

Anyway, since person 1 kills person 2, and person 3 kills person 4...etc etc, person 1 starts the round, so he is the only one still alive :)

[quote="TheDeity"]I tested this out for 2, 4, 6, 8, and 10 and all of them ended with the first person as the last one standing, so I'm going to say it is the first person. :P[/quote]

Well, I tested 10, and I ended up with 5 as the last one standing, so...

In that scenario, 2, 4, 6, 8, 10, 3, 7 die in that iorder, and 9 has the sword, and 1 is right next to him, so...

@the cliu: person 1 doesn't always start the round, as we pointed out, so your logic is incorrect.

no the first person does not always start...

have a close observation and u find tha he gets killed in the 4th round.

and fishythefish ,  961  is wrong.

I guess I'll throw out a guess.  When I first read this problem, I thought the answer was in the form $ 2^n \plus{}1$.  So is it $ \boxed{513}$?

Well, I wasn't expecting too much. I used Microsoft Excel to list all the people, then knocked out all of the evens, and so on. In the cases where the first person was killed before they started the round, I took every other person starting with the second guy. Anyways, I ended up getting 961, but I didn't have too much faith in that answer. (Does this attempt make sense?)

I copied fishthefish's idea with Excel, but I think that 961 gets killed in the same round that 1 gets killed.
It got confusing after a couple of rounds though so I might have messed up the numbers.
I eventually got 977.

yes you are right tornado.adv4

actualy this is called as josephus problem.   u guys can wiki it if u want some more information on it

Who says that 1 starts each round?  Number one is not anything special, other than the fact that he starts the entire process.  Then he is as normal as number 527.

Anyway, for the Josephus problem for every 2, the answer lies in the binary system:  Write the number in binary.  Then take the first digit and move it to the end.  It's that simple!

For example, 10 in binary is 1010, then move the 1 to the end you get 0101 or 101 or 5 in decimal.

For 1000, in binary this is 1111101000.  Move the first digit to the end to get 1111010001, which in decimal is 977.

Every second person is killed, versus every third or fourth, ...

[quote="tenniskidperson3"]
Anyway, for the Josephus problem for every 2, the answer lies in the binary system:  Write the number in binary.  Then take the first digit and move it to the end.  It's that simple!
[/quote]

Can you explain why this works?

Thanks!

so if every third person was killed, you write it in trinary (whatever) and move the first digit to the end?

And if every third person is killed and you have a number like 123 in trniary, then you get 231, which is obviously not right?

EDIT: Hm never mind. This will only work consistently for every second person. So then whats the strategy if its every third person? And how do you prove the binary thing?

No, it doesn't generalize; it's just written that way to make it easy to remember.  I have worked on finding a strategy for every 3, and so far it has been unfruitful.

To prove that it does work, we show that it works for each power of 2 and then use induction.

For each power of 2, we use induction.  It works for 1, because the algorithm gives 1.  To do the inductive step, with $ 2^{k \plus{} 1}$ people, the first people to be eliminated are all even numbered people, and 1 ends up with the knife again.  Thus relabel all the people that are left from $ 2n \minus{} 1$ to $ n$.  Thus 1 has the knife and there are $ 2^k$ people left, use inductive step.  Obviously the algorithm gives 1 for each power of 2, so this case is true.

Now we tackle the case when the number of people is not a power of 2.  Call when the person who has the knife killing the person ahead of him and passing on the knife a step.  Now suppose there are $ 2^k \plus{} p$ people where $ 0\leq p\leq 2^k \minus{} 1$ and right now, person 1 has the knife.  The algorithm gives that the person to survive is $ 2p \plus{} 1$.

So do induction on $ p$.  It works for $ p \equal{} 0$, as we've already shown.  So now we need to show that if it works for $ p$, then it works for $ p \plus{} 1$.  Suppose that for $ p$, person $ 2p \plus{} 1$ survives.  What happens first is that person 1 kills person 2 then gives the knife to person 3.  If we relabel all the numbers $ m$ as $ m \minus{} 2$ except 1 is relabeled as $ p$, we have a situation where person 1 has the knife and there are $ 2^k \plus{} p$ people left.  So obviously person $ 2p \plus{} 1$ will survive.  Now this person, with the old labeling, is actually person $ 2p \plus{} 1 \plus{} 2 \equal{} 2p \plus{} 3 \equal{} 2(p \plus{} 1) \plus{} 1$.  This works for all numbers $ 0\leq p\leq 2^k \minus{} 2$.  For $ p \equal{} 2^k \minus{} 1$, we've already shown that $ 2^{k \plus{} 1}$ works, so the inductive step is complete and we're done.


inequalities
number theory unsolved
number theory

Find all positive integers $ n$ such that for all odd integers $ a$, if $ a^2<\equal{}n$ then $ a$ divides $ n$.

We claim that the solution space is $ \{1,2,\ldots,9,12,15,18,21,24,30,45\}$. As for the proof:

Choose the unique $ m\geq 0$ such that $ (2m\plus{}1)^2\leq n<(2m\plus{}3)^2$ holds, i.e. $ n$ has the odd integer divisors $ 1,3,\ldots,2m\plus{}1$. This yields the inequality: $ (2m\plus{}3)^2>\prod_{i\equal{}1}^m(2i\plus{}1)$. This requires $ m<3$, otherwise we would conclude $ (2m\plus{}3)^2>(2m\plus{}1)(2m\minus{}1)(2m\minus{}3)$ yielding the contradiction $ 0>8m^3\minus{}14m\minus{}16m^2\minus{}6\equal{}(8m^2\minus{}14)(m\minus{}2)\minus{}34\geq (72\minus{}14)(3\minus{}2)\minus{}34>0$.

$ m\equal{}2$ yields $ 25\leq n <49$ and $ 15|n$, i.e. the solutions $ n\equal{}30,45$
$ m\equal{}1$ yields $ 9\leq n<25$ and $ 3|n$, i.e. the solutions $ n\equal{}9,12,15,18,21,24$.
$ m\equal{}0$ yields $ 1\leq n<9$ and all such $ n$ are solutions.