geometry
circumcircle
geometric transformation
rotation
geometry proposed

[hide="Solution"]
Let $ T$ be the midpoint of arc $ DE$ on the circumcircle of $ \triangle EID$.  Notice that $ \angle DOE \equal{} 2\angle BID \equal{} 2\angle BAD$, so $ \angle DOT \equal{} \angle BAD$.  This, combined with the fact that $ DO \equal{} OT$ gives us the spiral similarity: $ \triangle BAD\sim \triangle TOD$, which yields that $ (AO, BT) \equal{} \angle ADB$.  We then observe that $ \triangle ABD\sim \triangle AEC$ with $ AB \equal{} AD$, which gives the rotation: $ \triangle ABE\cong \triangle ADC$, meaning that $ DC \equal{} BE$.  We have that $ \angle TDC \equal{} \angle TEB$ and $ TD \equal{} TE$, yielding the rotation: $ \triangle TEB\cong \triangle TDC$.  This gives that $ \angle TBI \equal{} \angle TCI$, so $ TIBC$ is cyclic, giving that\[ 90 \minus{} \angle ADB \equal{} \frac {\angle BAD}{2} \equal{} \frac {\angle DOT}{2} \equal{} \angle TED \equal{} \angle TIC \equal{} \angle TBC\]so we have that $ (BT, AO) \plus{} (BT, BC) \equal{} 90$, from which we conclude that $ AO\perp BC$. [/hide]

EDIT: Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=841255#841255

I don't believe that my exact solution was posted in the above link, but the idea was the same.

[quote="April"]Let be given a triangle $ ABC$. Construct outside it two similar isosceles triangles $ ABD$, $ ACE$, where $ AB \equal{} AD$ and $ AC \equal{} AE$. Let $ I$ be the intersection of $ BE$ and $ CD$; and $ O$ the circumcenter of triangle $ IDE$. Prove that $ AO$ is perpendicular to $ BC$.[/quote]
Here is my solution
We have $ A,D,B,I$ lie on a circle and $ A,E,C,I$ lie on a circle and $ DC \equal{} BE$
Let $ AC\cap (ADI) \equal{} H;DA\cap (AIE) \equal{} K$

We have
$ OB^2 \minus{} OC^2 \equal{} (BE^2 \minus{} P_{B|(O)}) \minus{} (CD^2 \minus{} P_{C|(O)}) \equal{} (CD^2 \minus{} \overline{CI}.\overline{CD})\minus{}(BE^2 \minus{} \overline{BI}.\overline{BE}) \equal{} \overline{DA}.\overline{DK}\minus{}\overline{EA}.\overline{EH} \equal{} DA^2 \minus{} EA^2 \plus{} \overline{AD}.\overline{AK} \minus{} \overline{AE}.\overline{CH} \equal{} AB^2 \minus{} AC^2 \plus{} \overline{AD}.\overline{AK} \minus{} \overline{AE}.\overline{AH}$

Easy to see $ H,K,E,D$ lie on a circle so $ \overline{AD}.\overline{AK} \minus{} \overline{AE}.\overline{AH} \equal{} 0$

So $ OB^2 \minus{} OC^2 \equal{} AB^2 \minus{} AC^2$ so $ OA\perp BC$
As desired! :D


It basically states that the expansion of $ (a \plus{} b)^n$ is related to combinatorics in that the coefficients follow the Pascal's Triangle.

$ (a \plus{} b)^n \equal{} \sum_{i \equal{} 0}^n \binom ni a^{n \minus{} i}b^i$

For example, if $ n \equal{} 5$, then

$ (a \plus{} b)^5 \equal{} a^5 \plus{} 5a^4b \plus{} 10a^3b^2 \plus{} 10a^2b^3 \plus{} 5ab^4 \plus{} b^5$

This follows from the fact that in the product

$ (a \plus{} b)(a \plus{} b)(a \plus{} b)\cdots (a \plus{} b)$

there are $ \binom ni$ ways to choose $ i$ a's and $ n \minus{} i$ b's.

[quote="mathemagician1729"]Can someone introduce the Binomial Theorem and its properties and useful qualities in MC?[/quote]

its actually not that useful in mc, there are few questions that binomial theorem will be handy on, and you can brute force all of them.

hmm..since when did i start giving out advices.... :roll:

I've actually seen a question that was a DIRECT application of the Binomial Theorem. It went something like this:

Find the coefficient of $ x^3$ in the expansion of $ (x \plus{} 1)^6$.

[size=59]Actually, that was a problem from my mock test.[/size]


Let $\{a_i\}$ be a seqaunce of integer numbers such that :

$(a_m,a_n)=a_{(m,n)}$

Prove that there exist a seqaunce like $\{b_i\}$ such that

$a_n=\prod_{i \mid n} b_i$

Standing 5 topics below this: http://www.mathlinks.ro/Forum/viewtopic.php?t=66566


[quote="jarod"]Prove that : $\forall x,y,z \in R$ :
  $(x^2+y^2+z^2)^3 \geq (x^3+y^3+z^3)^2+(xy^2+yz^2+zx^2)^2+(x^2y+y^2z+z^2x)^2$[/quote]

Hmmm your Inequality is WRONG for real numbers !

 Suppose that x=y=z=1 ok then you will get a Contradiction.

[quote]\sum|y^3z^3|+3x^2y^2z^2\ge\sum (x^3y^2z+x^2yz^2)\Leftrightarrow \sum a^3+3abc\ge \sum(a^2b+a^2c) [/quote]
 I think they are not equivalent

it suffices to show $2 \sum x^4y^2 + 6x^2y^2z^2 \ge 2\sum_{cyc} x^3y^3 + 2 \sum x^3y^2z$
which is true since $x^4y^2 + x^2y^2z^2 \ge 2x^3y^2z$ (and similar ineq)
and $\sum x^4y^2 \ge 2\sum_{cyc} x^3y^3$ (amgm)


topology
advanced fields
advanced fields theorems

Is it usually taken for granted that if $ \left\langle X,\mathcal{T}\right\rangle$ is a toplogical space, $ \forall x\in X\exists U\in\mathcal{T}$ such that $ x\in U$ (i.e. every point of $ X$ is a member of some open set)?

For example, the statement "A [i]connected[/i] and [i]pathwise locally-connected[/i] space is [i]path-connected[/i]" is tue only if we make that assumption

one of the axioms for $ \tau$ being a topology on $ X$ is that $ X\in \tau$.
so, yes, it is "usually taken for granted".


An archer pulls back his arrow with a force proportional to the displacement, and must exert a 200N force to hold the 30g arrow in position ready to shoot. The string of the blow(assumed inextensible) is 1.6m long, and the center of the string is pulled back 30cm. The mass of the string and bow are negligible.Compute:
a) The potential energy stored in the bow
b) The tension in the string
c) The speed with which the arrow leaves the string

1) We can think of the bow string as a spring.  Then $ F\equal{}kx\equal{}200\equal{}.3k\Rightarrow k\equal{}666.67$.  Then $ U\equal{}\frac{1}{2}kx^{2}\equal{}\frac{1}{2}(666.67)(.3)^{2}\equal{}30\text{J}$

Your answers are right for a) and c) question, respectively(according to the provide answer from the book). However, the answer for question b) is: 267 N, which I really don't understand how to get it(I think your result is right in this question too).


I thinks problems should read:
Let $ a,b,c,d$ be four numbers in non constant arithmetic progression.
(if series is constant then $ a\equal{}b\equal{}c\equal{}d$, and product is $ a^4$ which is perfect square)


induction
combinatorics unsolved
combinatorics

Let $ n$ be a positive integer. Determine the size of the largest subset of $ \{\minus{}n,\minus{}n\plus{}1,...,n\minus{}1,n\}$ which does not contain three elements $ a,b,c$ (not necessarily distinct) satisfying $ a \plus{} b \plus{} c \equal{} 0$

Do you mean to say - Any three elements when selected randomly from the subset (of that particular set you had mentioned) , the sum of those 3 numbers shouldn't be equal to zero, is that what you meant?

tdl: All of these problems have already been posted here: http://www.artofproblemsolving.com/Forum/index.php?f=133

[hide="please check my solution"]We claim that the answer is $ 2k$ for $ n\equal{}2k$, which we will prove by induction. The base case is obvious, assume this is true of $ 2k$. Consider the set $ \{\minus{}2n\minus{}2,\minus{}2n\minus{}1,\minus{}2n,\minus{}2n\plus{}1,\ldots,2n\minus{}1,2n,2n\plus{}1,2n\plus{}2\}$.

Assume that we have a satisfying subset of size at least $ 2n\plus{}3$. So we need to take at least 3 among $ \{\minus{}2n\minus{}2,\minus{}2n\minus{}1,2n\plus{}1,2n\plus{}2\}$. WLOG, we take $ \minus{}2n\minus{}2,\minus{}2n\minus{}1$.

(i) We take four of them.
We must take $ 2n\minus{}1$ of $ (1,2n),(2,2n\minus{}1),\ldots,(n\minus{}1,n\plus{}2),n,\minus{}n,(\minus{}n\plus{}1),(\minus{}n\minus{}2),\ldots,(\minus{}1,\minus{}2n)$ (at most one of each pair). WLOG we take $ n\minus{}1$ of $ (1,2n),(2,2n\minus{}1),\ldots,(n\minus{}1,n\plus{}2)$ (exactly one of each pair). We can't take $ n\plus{}1,\minus{}n\minus{}1$. We must take $ n,\minus{}n$. We can't take $ 2n$ because $ (\minus{}n)\plus{}(\minus{}n)\plus{}2n\equal{}0$ and we can't take $ 1$ because $ (\minus{}2n\minus{}2)\plus{}1\plus{}(2n\plus{}1)\equal{}0$, a contradiction.

(i) We take $ \minus{}2n\minus{}2,\minus{}2n\minus{}1,2n\plus{}1$.
We must take at least $ 2n$ of $ (\minus{}2n,\minus{}2n\plus{}1,\ldots,2n\minus{}1,2n)$. Exactly one of each following pairs is taken: $ (1,2n),(2,2n\minus{}1),\ldots,(n\minus{}1,n\plus{}2),(n,n\plus{}1),(\minus{}1,\minus{}2n),(\minus{}2,\minus{}2n\plus{}1),\ldots,(\minus{}n,\minus{}n\minus{}1)$. We can't take $ n\plus{}1$, so we take $ n$. Now we can't take $ n\plus{}2$, so we take $ n\minus{}1$. Continue this, we must take $ 1,2,\ldots,n$. But $ (2n\plus{}1)\plus{}1\plus{}(\minus{}2n\minus{}2)\equal{}0$, contradiction.

(ii) We take $ \minus{}2n\minus{}2,\minus{}2n\minus{}1,2n\plus{}2$.
We must take exactly one of each of the following pairs: $ (1,2n),(2,2n\minus{}1),\ldots,(n,n\plus{}1),(\minus{}1,\minus{}2n\minus{}1),(\minus{}2,\minus{}2n)\ldots(\minus{}n,\minus{}n\minus{}2)$ (we can't take $ \minus{}n\minus{}1$). We can't take $ n\plus{}1$, so we take $ n$. We can't take $ n\plus{}2$ so we take $ n\minus{}1$, continuing, we take $ 1,2,\ldots,n$. Now we can't take $ \minus{}2n$, but we can't take $ \minus{}2$ either, a contradiction.

Therefore our induction is complete. The construction is obviously by taking all odd numbers.[/hide]


inequalities
function
calculus
derivative
inequalities unsolved

[quote="silouan"]Γεια σου Κύπριε Αδελφέ .
Here is my solution .
Consider the function $ln(\frac{1-x}{x})$  and it is convex at the interval ($0,\frac{1}{2}$) so aplly Jensen's inequality .[/quote]

  Is it convex? I think the sign of the inequality should be reverse. :D

[quote="Lovasz"]But $f''(x)=\frac{1-2x}{x^2(1-x)^2}?$[/quote]
Yes f''(x)>0  so my solution workes , do you have another opinion ??

Sorry, I calculate second derivative wrong. :(


Let $X$ be the set of real numbers greater than $1$. Define $f: \ X\rightarrow X$ and $g: \ X\rightarrow X$ by $f(x)=2x$ and $g(x)=\frac{x}{x-1}$
Show that given any real numbers $1 < A < B$ we can find a finite sequence $x_{1}=2,x_{2},...,x_{n}$ such that $A<x_{n}<B$ where $x_{i}=f(x_{i-1})$ or $x_{i}=g(x_{i-1})$

[quote="Rust"]It is not true if B<=2. Because $f(x)>2,g(x)>2 \ \forall x\in (1,2)$.
In this case n=1. If B>2 it is true, because we can choose $x_{i}=min(f(x_{i-1},g(x_{i}))\le 2$.[/quote]
why it is not true if B<=2 ? For example take A=1.1 and B=1.9
Now we have $x_{1}=2$, $x_{2}=f(x_{1})=4$,  $x_{3}=g(x_{2})=\frac{4}{3}$ and $1.1<\frac{4}{3}<1.9$

I am sorry, i think that must be all $x_{i}\in (A,B) \ \forall i\le n$.

For prove we can consider matice $F=(\frac{2 \ \ 0}{0 \ \ 1})$ and $G=(\frac{p \ \ 0}{q \ \ r}), \ p=1=q,r=-1.$ All multiply f(x),g(x) is equavalent multiply matrices F and G. For example $f(g(x))=A(\frac{x}{1})=\frac{ax+b}{cx+d}$, were matrice $A=FG.$
Therefore sequence $x_{i}=f(x_{i-1}) \ or \ x_{i}=g(x_{i-1})$ can be represent $x_{n}=A(\frac{2}{1})$, were $A=F^{i_{1}}G^{j_{1}}....F^{i_{n}}G^{j_{n}}, i_{k},j_{k}$ equal  0  or  1.
We have $A=(\frac{a \ \ 0}{c \ \ d}), a=2^{k}, d=(-1)^{l}$, were k is number $\{i_{m}|i_{m}=1\}$ or
k=\sum_{m=1}^{n}i_{m}, \ l=\sum_{m=1}^{n}j_{m}.
For prove this statement sufficiently to show, that for any integer c we can choose $i_{m},j_{m}$ suth that $(\frac{a \ \ 0}{c \ \ d})=F^{i_{1}}G^{j_{1}}...F^{i_{n}}G_{^{}}{j_{n}}$.
It is possible, because $A_{1}A_{2}=(\frac{a_{1}a_{2}\ \ \ \ \ \ \ 0}{a_{2}c_{1}+d_{1}c_{2}\ \ d_{1}d_{2}})$.


Given that $ a^2c \plus{} b^2a \plus{} c^2b \equal{} 1$
Prove that:
\[ ab \plus{} bc \plus{} ca \leq \sqrt {a \plus{} b \plus{} c}
\]
P.S: There exists a beautiful solution.
And also, i am sorry if it is too easy to be in this section.

Can I solve it? Here is the simpliest solution in my opinion
[hide="solution"]$ (a \plus{} b \plus{} c)(b^2a \plus{} c^2b \plus{} a^2c)\ge (ab \plus{} bc \plus{} ca)^2$[/hide]

[quote="shaaam"]Given that $ a^2c + b^2a + c^2b = 1$
Prove that:
\[ ab + bc + ca \leq \sqrt {a + b + c}
\]
P.S: There exists a beautiful solution.
And also, i am sorry if it is too easy to be in this section.[/quote]
Thank you very much for your gift, Shaam. With it, I will have many lucky things in new year (right now, we are in Tet Holidays at Vietnam which is our greatest festival in New Year) :)
Here is my solution for it:
Applying Cauchy Schwarz Inequality, we have
\[ (ab+bc+ca)^2=( \sqrt{a}\cdot \sqrt{ab^2} +\sqrt{b}\cdot \sqrt{bc^2} +\sqrt{c} \cdot \sqrt{ca^2})^2 \le (a+b+c)( ab^2+bc^2+ca^2)=a+b+c\]
Therefore
\[ ab+bc+ca \le \sqrt{a+b+c}.\]
:)

WOW, you are so fast... And here is the beautiful solution i am talking about:

From Cauchy-Schwarz:
\[ (a \plus{} b)(ab \plus{} c^2) \geq b(a \plus{} c)^2
\]
therefore,
\[ a \plus{} b \geq \frac {b(a \plus{} c)^2}{ab \plus{} c^2}
\]
so we have:
\[ 2\sum_{cyc} a \geq \sum_{cyc} \frac {b(a \plus{} c)^2}{ab \plus{} c^2}
\]
and after applying Cauchy-Schwarz again:
\[ \left(\sum_{cyc} \frac {b(a \plus{} c)^2}{ab \plus{} c^2}\right)(\sum_{cyc} ab^2 \plus{} bc^2) \geq 4(ab \plus{} bc \plus{} ca)^2
\]

\[ \sum_{cyc} \frac {b(a \plus{} c)^2}{ab \plus{} c^2} \geq 2(ab \plus{} bc \plus{} ca)^2
\]
so we have:
\[ 2(a \plus{} b \plus{} c) \geq \sum_{cyc} \frac {b(a \plus{} c)^2}{ab \plus{} c^2} \geq 2(ab \plus{} bc \plus{} ca)^2
\]

\[ ab \plus{} bc \plus{} ca \leq \sqrt {a \plus{} b \plus{} c}
\]
I used Cauchy-Schwarz twice, and since Can_Hang likes Cauchy-Schwarz, i decided to dedicate it to him... But it was too easy to him  :lol:

EDIT: I am so sorry.... i didn't notice that, it is direct consequence of Cauchy-Schwarz...  :blush:  :blush:  :blush:

VERY VERY SORRY....

Here is the new Problem. i hope it is not as trivial as before:

Given that $ a,b,c \geq 0$, Prove that:

\[ a^3\plus{}b^3\plus{}c^3\plus{}\sqrt{3}\cdot \left(\frac{abc(a\plus{}b\plus{}c)}{\sqrt{a^2\plus{}b^2\plus{}c^2}} \right) \geq a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)
\]

ineq $ \Leftrightarrow\sum a^3 + 3abc - \sum a^2(b + c) + \sqrt {3}abc(\frac {\sum a}{\sqrt {\sum a^2}} - \sqrt {3})\geq 0$
${ \Leftrightarrow\sum(\frac {a + b - c}{2} - \frac {\sqrt {3}abc}{\sqrt {3\sum a^2}(\sqrt {3\sum a^2} + \sum a})}(a - b)^2\geq 0$
Suppose $ a\geq b\geq c\Rightarrow S_c\geq S_b\geq S_a$
$ S_b + S_a = c - \frac {2\sqrt {3}abc}{3\sum a^2 + \sqrt {3\sum a^2}\sum a}$
We only prove that $ 3\sum a^2 + \sqrt {3\sum a^2}\sum a\geq 2\sqrt {3}ab,Right$


calculus
integration
geometry
analytic geometry

[quote="Rushil"]Given that the electric field in the entire space is $ \vec{E} =\ ax \ \hat{i} $

[b]Find the charge distibution in space which produces this field and explain how and why such a charge distribution produces such an electic field.[/b]

[hide]Differential form of Gauss's law[/hide][/quote]

$\rho = \epsilon \nabla \cdot \vec{E} = \epsilon a$

This result implies that charge is uniformly distributed over the whole space.

That part is real easy.!!!!! :P

The main question (which even Griffiths has said to ponder over) is how such a uniform charge distribution produces such an electric field??? :D

Griffiths is David J. Griffiths, author of Introduction to Electrodynamics.

This problem is from Chapter 2, problem 50. And to be more specific, the question asks:

"How do you account for the fact that the field points in a particular direction, when the charge density is uniform.?"

I think the answer is that we can't distinguish the direction of the E field at this point, because there are many solutions for E that will satisfy this condition. (e.g. $\vec{E}=ay\ \hat{y}$ produces the same density as does $\vec{E}=az\ \hat{z}$). I think because it's uniform, the direction is arbitrary and can be chosen as such. For more of a mathematical reason, we do not have the proper boundary conditions to satisfy the uniqueness and existense theorem, thus a unique solution cannot be found, and thus inifinitely many E fields satisfy this charge density.

... sounds interesting, Spoon ... I understand that many solutions apply to that charge distribution, but does that mean that it is "physically forbidden" to have such charge distribution? (does that mean that if the universe is infinite, net charge is 0?) If not, what charge would you measure?

[quote="kubus"][quote]
That part is real easy.!!!!! Razz

The main question (which even Griffiths has said to ponder over) is how such a uniform charge distribution produces such an electric field??? Mr. Green
[/quote]

That was going to be my question, Rushil;)

BTW, who is Griffiths? And, you still haven't answered my question about the "200 problems" book, could you pls give me some advice on that?[/quote]

I did .I PMed you. Didn't you get it!!!

Spoon, I understand what point you are trying to make but in this case the field isn't arbitrarily pointed. Its along the x direction and increasing!!!!

If you do this in your course you are bound to have discussed this with your friends and teachers. What do they say?

Do you know griffiths' answer? Hasn't anyone ever contacted him???

Also regarding the uniquness thm, we aren't solving Laplace's eqn. We are given the field , not the charge distribution.If we knew the charge and had to find the field, won't uniqeness thm. come in then? ?

[size=150]Note :-Gauss's law in integral form also gives the same answer.... So no role of dirac delta fn!!!![/size]

[quote="Rushil"]Spoon, I understand what point you are trying to make but in this case the field isn't arbitrarily pointed. Its along the x direction and increasing!!!![/quote]
We know this, that is not what the question is asking.

[quote="Rushil"]Do you know griffiths' answer? Hasn't anyone ever contacted him???[/quote]
Griffiths does not provide solutions to any problems for his book, something many students (and professors) think is really bad about the book (myself included). Others on the other hand, like the fact that no solutions are available. Griffiths claims in the preface that the groups are split pretty evenly so he doesn't see enough demand for the solutions to merit him providing them.

Contacted him? For answers? That's a good one. All right, I won't get all sarcastic now, so I will just say... No I haven't and if someone did I'm pretty sure 1) he wouldn't read what you sent and 2) If he did, he certainly wouldn't provide you with the solution to what could potentially be your homework problem or open book test question...

[quote="Rushil"]Also regarding the uniquness thm, we aren't solving Laplace's eqn. We are given the field , not the charge distribution.If we knew the charge and had to find the field, won't uniqeness thm. come in then? ?[/quote]
It is naive to think that the uniqueness and existence theorem should be directly related to Laplace's equation in any way. On the contrary, the uniqueness and existence theorem is extremely robust and can be applied to countless different differential equation situations.

Again, you misunderstood what the question is asking. In this situation, yes we know the E field and we know that it's $\vec{E} =\ ax\ \hat{x}$. We've established that. Now, the question is asking us to consider them problem in reverse! Given the fact that the charge density is uniform, would it be possible to determine the E field? The quick answer is No.

This would require us to solve the coupled integral equation $\nabla\times\vec{E}=0$ and $\epsilon\nabla\cdot\vec{E} = \rho$. This set of equations, like any other differential equation is unique only given initial condictions to satisfy the "constant of integration". In our case here, the constant is in fact a Guage Transformation, which is more complex than just a constant. I won't go into that here, but let's suffice it to say that given this differential equation, we lack the necessary boundary conditions to determine $\vec{E}$. The entire purpose of this question is to show that $\rho$ alone does not ALWAYS uniquely determine $\vec{E}$.

[quote="Rushil"][size=150]Note :-Gauss's law in integral form also gives the same answer.... So no role of dirac delta fn!!!![/size][/quote]
I don't know what you mean here...

So basically you want to say that the first part of the question ends once we determine the charge distribution which(acc. to Gauss's law) creates this electric field. The 2nd part of finding the electric field due to this distribution is unsolvable without the boundary conditions.

I think I do agree with you on this viewpoint!!! It's a very intricate relationship indeed!!!

My remark regarding Dirac Delta fn. was meant to indicate that there can be no discrepancy due to the presence of Dirac delta fn.

So basically , the 'paradoz' stands resolved?????? :?:

The integral form of Guass's Law depends on the boundary conditions, which in this case do not exist. Could you show how you think it should be applied? I am pretty sure it only works one way here...

And it's spelled 'Paradox' :)

I considered a Gaussian box("pillbox" as Griffiths says) and applied Gauss's law in integral form on it.It's really easy if you see so.

I son't quite understand what you mean by "it's a Paradox"

Just to know - which university/college do you go to????

Yes I know you use a pillbox with the integral form of Gauss's Law, I'm asking if you could show what you did, i.e. show your work and your solution.

If you consider a box of lenght $ L $ and area of face $A$ at a distance of $x$ from the coordinate origin , we get that $ - ax \times A + a(x+L) A = \frac{q_{enc}}{\epsilon_o} $

Thus, we have ....

oh no i get it I have to assume that charge density is constant to get the result.

Sorry!!!!!!!!!


function
number theory unsolved
number theory

I was just wondering what is the asymptotic upper bound on the divisor function (number of divisors of an integer)?

Under the title "Growth of the Function" http://en.wikipedia.org/wiki/Euler's_totient_function


geometry
calculus
trigonometry
ratio
area of a triangle
Heron\u0027s formula
trig identities

$ \triangle ABC$ has $ AB \equal{} 6$, $ BC \equal{} 7$, $ CA \equal{} 8$.  Points $ P$, $ Q$, and $ R$ are on $ AB$, $ BC$, and $ CA$ respectively with $ BP \equal{} BQ$ and $ AP \equal{} AR$.  Find the maximum area of $ \triangle PQR$.

Note: I am yet to find a solution without calculus, but I'm fairly certain there is one.

After using the law of cosines to find the angles of A, B, and C I defined BQ as X.

BQ=PB=X
AP=AR=6-X
RC=2+X
CQ=7-X

I then used the law of sines to find that
PQ=X(Sin B)/(Sin BPQ)
PR=(6-X)(Sin A)/(Sin APR)

I used the law of cosines to find that

PQ= sqrt((93/32)X^2-(45/4)X+225/8)

Using my calculator to program Heron's formula and experimenting with values gives me a maximum value of approximately  5.3265 when X=3.04

Does anybody have an exact solution to this problem?

Let $ AP\equal{}AR\equal{}x$. Then we have that $ BP\equal{}BQ\equal{}6\minus{}x, QC\equal{}x\plus{}1,$ and $ CR\equal{}8\minus{}x$.

[b]Lemma[/b]: $ \triangle ABC$ has points $ P, Q, R$ on sides $ AB, BC,$ and $ CA$ respectively such that $ \frac{AP}{AB}\equal{}p, \frac{BQ}{BC}\equal{}q,$ and $ \frac{CR}{AC}\equal{}r$. Then $ [\triangle PQR] \equal{}[\triangle ABC]\cdot [pqr\plus{}(1\minus{}p)(1\minus{}q)(1\minus{}r)]$.

Applying this to our problem, we have that $ [\triangle PQR]\equal{}[\triangle ABC]\cdot\left[\frac{x\cdot(6\minus{}x)\cdot(8\minus{}x)\plus{}(6\minus{}x)\cdot(x\plus{}1)\cdot x}{6\cdot7\cdot8}\right]$

This is the same as $ \frac{9(6x\minus{}x^2)}{336}\equal{}\frac{3(6x\minus{}x^2)}{112}$

So we wish to maximize $ 6x\minus{}x^2,$ which occurs when $ x\equal{}3$.

Thus the ratio of areas is $ \frac{27}{112}$.

To find the area of triangle $ ABC$, we use heron's: $ \sqrt{\frac{21\cdot9\cdot7\cdot5}{2^{4}}}\equal{}\frac{21\sqrt{15}}{4}$

Thus $ [\triangle PQR]\equal{}\frac{27}{112}\cdot \frac{21\sqrt{15}}{4}\equal{}\frac{81\sqrt{15}}{64}$

There's probably some calculation error in my solution, but you get the idea.


geometry
3D geometry
prism
geometry theorems

Can somebody give me a hint what implies the Poincare theorem?

I know it is something about an inscribed and described $ n \minus{}$polygon... but the main statement is incomprehensible to me  :oops:

Do you mean [url=http://sbseminar.wordpress.com/2007/07/16/poncelets-porism/]Poncelet's porism[/url]?

Exactly  :)  Thank you [b]puuhikki[/b] :P

Is there a geometric proof by the way ? :blush:

The general theorem (due to Poncelet) is as follows:

Suppose circles $\omega_i$ belong to the same pencil and $A_0$ is a point on $\omega_0.$ Tangent to $\omega_1$ from $A_0$ cuts $\omega_0$ again at $A_1,$ tangent to $\omega_2$ from $A_1$ cuts $\omega_0$ again at $A_2,$ etc., the tangent to $\omega_{i+1}$ from $A_i$ cuts $\omega_0$ again at $A_{i+1}.$ Suppose that for some $n,$ the points $A_n$ and $A_0$ coincide. Then for any $B_0$ on $\omega_0,$ the similarly constructed point $B_n$ coincides with $B_0.$


Let $ a,b,c$ be nonnegative numbers such that $ a^3\plus{}b^3\plus{}c^3\equal{}3$

Prove that $ a^4b^4\plus{}b^4c^4\plus{}c^4a^4\le 3$

We have
     $ ab \leq \frac {a^3 \plus{} b^3 \plus{} 1}{3} \equal{} \frac {4 \minus{} c^3}{3} \equal{} > a^4b^4 \leq \frac {4a^3b^3 \minus{} a^3b^3c^3}{3}$
  We have to prove that
             $ 4(a^3.b^3 \plus{} b^3.c^3 \plus{} c^3.a^3) \minus{} 3a^3b^3c^3 \leq 9$
  Let $ a^3 \equal{} x,b^3 \equal{} y,c^3 \equal{} z$ we have Schur


I think I have the answer for this one (I observed a pattern), but I didn't finish the proof (I didn't show those I've seen are the only solns):

Find all pairs$(x;y)$ of integer numbers with:
$x|y^3+1$ and $y|x^2+1$.

What pattern did you see?
I only noticed that if you take (x,y) to be a solution of the system, then
(x, (x 2 +1)/y) and ((y 3+1)/x,y) are also solutions.
I tried to use some descending techniques. No luck yet.

Here's what I think (the pattern I saw):

There are some 'primitive solutions (x,y)=(n^2+n+1,n^2+1) for all natural n, and then all other solutions are obtaind from these by the transformations you observed. This is, of course, only what I think. I have no proof or any idea for a proof.

If x = y then (x, y) = (1, 1) is a solution.
If y = 1 then x = 1.
If y = 2 then x = 1 or x = 3.

Let (x, y) be a solution with x not equal to y and y > 2.
Then (x2 + 1)(y3 + 1)/(x2y) is natural.
Hence (x2 + y3 + 1)/(x2y) is natural.
Hence x2 + y3 + 1 \geq x2y ---> x < y.
Now x2 + 1 = yz, with z natural and x > 1.
If z \geq x, then yz \geq (x + 1)x > x2 + 1, contradiction.
Hence z < x, and z divides x2 + 1.
Since z = (x2 + 1)/y,
y3 + 1 = [(x2 + 1)3 + z3]/(z3).
Hence x2 divides z3 + 1.
Hence z|(x2 + 1) and x2|(z3 + 1), z < x.
So z \leq 2. Then we have the cases z = 1, z = 2.
An easy case check gives (3, 5) and as the only additional solution.

Sorry  :?:?:?

I solved the problem y|(x2 + 1), x2|(y3 + 1) above, not Grobbers problem.

The propblem I posted comes from that file with NT questions from Hojoo Lee. I found quite a few solutions for this, but they all respect the pattern I spoke about. I really think this might be it, but I don't know how to prove that these are the only solutions.

I also tried a descending argument, like amfulger, but nothing worked...

Well Hojoo Lee joined the forum :)
So maybe he knows the solution :D

Hey! This forum is becoming really popular! :D Now that I think abt this, I don't know if I want to know the proof... Maybe just to know if this answer is correct, and then I could try to show it's correct.

prove y=k^2+1 & find all x.solve this one by sequence :D

[quote="dna"]prove y=k^2+1 [/quote]
How :?:
[quote="dna"] find all x.solve this one by sequence [/quote]
And further...

it is some how too long but the idea is sequene & you must seperate it in three  cases 1) y| x+1 2) y|x^2+x+1 3)exist P: P|x+1 &P|y
then easily solve original problem by sequence. :D

write your solution more detail,dna  :mad: Every mathlinkers want a detail solution and they don't need the hint.ok :mad:


algorithm
algebra
polynomial
number theory
greatest common divisor
relatively prime

It is well known that the largest n for which ax+by=n has no integer solutions is n=ab-a-b. But how about for the more general equation a_1*x_1+a_2*x_2+...+a_k*x_k = n?

Even in the case n = 3, I don't believe there is a "nice" formula.  Of course, there must in some sense be a formula, since for every triple of numbers without common divisor there is such a number.  But you can't just write it down in terms of the a_i with any kind of ease.

Right.  I don't believe there is a formula for n = 3 either.  But there are efficient algorithms that find the answer for n = 3.  See the references in http://mathworld.wolfram.com/CoinProblem.html .  (Actually, that link suggests that these references provided a formula for n = 3, but I believe that is an overstatement.)

For any fixed number of variables, Ravi Kannan showed that there is a polynomial-time algorithm to solve the problem.

By the way, I think you meant the xi should be nonnegative integers, not just integers.  Also the gcd of the ai should be 1.  Otherwise, even your formula for n = 2 is incorrect.

There isn't a formula for n=3, but, interestingly enough, if the (pairwise? -- I'm not sure if this is necessary) gcd of (a,b,c) is 1, then the largest nonachievable value of abx+bcy+caz (x,y,z nonnegative integers) is known and not all that impossible to find I think.

Right.  Here is a more general result that supplies the formula for that case and more.  Let a, b, and c be three integers with gcd 1.  Let g(a, b, c) be the largest integer not achievable with a, b, and c.  Let D be the gcd of a and b.  Then Johnson's formula says that g(a, b, c) = Dg(a/D, b/D, c) + (D - 1)c.  Basically that formula lets you assume that the three numbers are pairwise relatively prime.

Applying Johnson's formula (once or even twice) shows that if a, b, and c are pairwise relatively prime, then g(ab, ac, bc) = 2abc - ab - ac - bc.  (Yes, you need the numbers a, b, and c to be pairwise relatively prime; otherwise gcd(ab, ac, bc) would be bigger than 1.)


When  10^20 - 1999  is written as a single whole number, the sum of its digits is

A144    	B.153   	C. 156  	D. 162    E.171


Find all functions$ f: \mathbb N^{2} \rightarrow \mathbb N$ such that $ f(x,y) \plus{} f(y,x) \equal{} f(x^{2},y^{2}) \plus{} 1$ holds for all natural numbers $ x$ , $ y$.

I can't see any contradiction coming from defining $ f(1,a), f(1,b), f(a,b), f(b,a)$ for all nonsquare $ a,b$ arbitrarily, and then letting all the other values be generated from that. ($ f(1,1) \equal{} 1$ is the only requirement)


I was thinking of books about every level:

[u]Introductory[/u]
Algebra - Intro. to Algebra
Number Theory - Intro. to Number Theory
Counting and Probability - Intro. to Counting and Probability
Geometry - Intro. to Geometry

[u]Intermediate[/u]
Algebra - Interm. Algebra
Number Theory - ???
Counting and Probability - Interm. Counting and Probability
Geometry - Geometry Revisited

[u]Olympiad[/u]
Art and Craft of Problem Solving
Problem Solving Strategies

What book is good for intermediate number theory and is there anything missing in the list above?

104 number theory problems from the training of the IMO team (yes, that's the title)

wouldnt that book be EXTREMELY hard though?


limit
trigonometry
search
real analysis
real analysis unsolved

Let $f$ is continuous on $[a,b]$,and for any $x \in (a,b)$,the limit of $\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{h}$ is exist(and defined it as $g(x))$,if $f(b)>f(a)$,prove there is a $x\in (a,b)$  such that $g(x)>0$

g(x)=2f'(x)

Integrate both sides over [a,b] and argue by contradiction, say.

How did you prove that $f^\prime$ exists?

[quote="§outh§tar"]f'=g/2 by definition[/quote]
Only if $f'$ exists. Consider $f(x)=|x|$, for which $f'(0)$ does not exist, yet $g(0)=0$.

You are quite right mlok. It seems my Calc BC teacher lied to me.  :blush:

I will try to fix this.

Q: If g as defined is non-zero at a point x can f still be non-differentiable at x? My knowledge of limits is close to nonexistent.  :oops:

let $f(x) = \begin{cases}ax & x\geq 0\\bx & x < 0\end{cases}$. Then $g(0)$ exists and $g(0) = a+b$. This example is a simple extension of mlok's.

Is it true that for every $x\in\mathbb{R}$ g(x) has lateral derivates?..

Later edit: Excuse me :oops:

[quote="scytale"]Is it true that for every $x\in\mathbb{R}\ g(x)$ has lateral derivates?[/quote]
No it isn't true. Consider $g(x)=x^{2/3}$ at the origin.

Or worse: $g(x)=x\sin\left(\frac1x\right)$ with $g(0)=0.$

What do you mean by lateral derivates?
No one want to solved the  original problem?
I have solved that,It is a nice problem I think.
So I will post the solution.
In fact we should only prove there is a $ x \in (a,b)$,such that $ g(x) \ge 0$
(because,You can search a $ k > 0$,such that $ f(b) \minus{} f(a) > k(b \minus{} a)$,then consider $ h(x) \equal{} f(x) \minus{} kx$)
find every $ f(a) < c < f(b)$
define $ x \equal{} \sup \{y: f(y) < c \}$
Then you can prove:
$ f(x) \equal{} c$,and also there is a increasing sequence converges to $ x_n$,such that$ f(x_n) < c$ by the defination of $ x$.
then:
$ g(x) \equal{} \lim_{n \to \plus{} \infty} \frac {f(2x \minus{} x_n) \minus{} f(x_n)}{x \minus{} x_n} \ge 0$
We complete the proof.

In fact,we  can prove such $ x$ is uncontable.


function
trigonometry
calculus
calculus computations

A. if for the function h(x) is"

h"(x)+h(x)=0 for all $ x\in R$
    and h(0)=1 , h'(0)=0,
    show that h(x)=cosx

B. if for the function f(x) is

f"(x)=f(x), for all $ x\in R$
    and f(0)=f'(0)=1
    find the function f(x)?

A. If $ h(x)\equal{}\cos{x}$, then $ h'(x)\equal{}\minus{}\sin{x}$ and $ h''(x)\equal{}\minus{}\cos{x}$.  Then $ h''(x)\plus{}h(x)\equal{}\minus{}\cos{x}\plus{}\cos{x}\equal{}0$.  In addition, $ \cos{0}\equal{}1$ and $ \minus{}\sin{0}\equal{}0$.
If you know how to solve differential equations, this is a basic second-order homogeneous equation with constant coefficients.

hello,
B. with $ y(x) \equal{} e^{\lambda x}$ we get $ \lambda^2 \cdot e^{\lambda x} \equal{} e^{\lambda x}$. From here we get $ \lambda_{1,2} \equal{} \pm1$ and our solution
has the form $ y(x) \equal{} f(x) \equal{} C_1e^{ \minus{} x} \plus{} C_2e^x$.
Sonnhard.

In the last post, Sonnhard showed you how to solve that differential equation so I'll solve the first one for you as a differential equation:

Characteristic equation is $ r^{2}+1=0=(r-i)(r+i)\Rightarrow \lambda=0,\ \mu=1$

Then the solution is $ e^{0\cdot t}(A\cos{t}+B\sin{t})=A\cos{t}+B\sin{t}=f(t)$.

Then $ f'(t)=-A\sin{t}+B\cos{t}$ and $ f''(t)=-A\cos{t}-B\sin{t}$ and using the initial conditions:
$ f(0)=A=1$ and $ f'(0)=B=0$

And subbing in, we have
$ f(t)=\cos{t}$, as desired.

[quote="JRav"]A. If $ h(x) \equal{} \cos{x}$, then $ h'(x) \equal{} \minus{} \sin{x}$ and $ h''(x) \equal{} \minus{} \cos{x}$.  Then $ h''(x) \plus{} h(x) \equal{} \minus{} \cos{x} \plus{} \cos{x} \equal{} 0$.  In addition, $ \cos{0} \equal{} 1$ and $ \minus{} \sin{0} \equal{} 0$.
If you know how to solve differential equations, this is a basic second-order homogeneous equation with constant coefficients.[/quote]

this is backwards -- you have the ODE and the boundary conditions and you want to show the [i]uniqueness[/i] of $ f(x)\triangleq \cos x$ as a solution, not show the existence of a solution (which is more or less trivial to verify as you did above)

That's why I solved the equation and showed that $ \cos{x}$ is the solution in my second post.


combinatorics solved
combinatorics
games

The two mathematicians Lagrangia and Galois  :D  play the following game:
They select and take from the set {0,1,2,3,...,1024} 512, 256, 128, 64, 32, 16, 8, 4, 2, 1 numbers away alternately. Lagrangia starts and takes 512 numbers away, then Galois 256 numbers etc.  Finally there are two remaining numbers a,b (a<b). Galois pays Lagrangia the following amount of money: abs(b-a).
Lagrangia wants to obtain as much money as possible. And vice versa Galois wants to loose at least money as possible.

Exlain how much money Lagrangia can earn at most ! Assume that they all try their best.

I think that the best way for Lagrangia is to divide make the numbers as far away from each over as possible.
I mean he should take the first time 1,3,5,.......1023.
Then do in such a way, that the numbers become as far away as possible.
But Galois :D should take 1,1023,3,1021.........

Let initial set be {0,1,...,22n}.

Galois.
We will show strategy for Galois such that his loss will be no more than 2n.
Suppose before k-th turn of Lagrangia (k\geq n) there are N=22n-2k+1 in the set: a1<a2<...<aN, and aN-a1\leq 22n-k. For k=0 it is true. Let us show that after (k+1)-th turn of Galois this property will satisfy.
After Lagrangia turn set will contain 2M-1=22n-2k-1+1 numbers (M=22n-2k-2+1) b1<b2<...<b(2M-1) and b(2M-1)-b1\leq 22n-k ==>
(1) b(2M-1)-bM\leq 22n-k-1
or
(2) bM-b1\leq 22n-k-1.
If (1) is true then Galois expunge (M-1) numbers b(M+1),b(M+2),...,b(2M-1).
If (2) is true then Galois expunge (M-1) numbers b1,b2,...,b(M-1).
In any case new set will contain before new Lagrangia (k+1)th turn M=22n-2k-2+1 numbers c1<...<cM and cM-c1\leq 22n-(k+1).

Thus after n turns of Lagrangia and Galois set will contain exactly 2=22n-2n+1 numbers a1<a2 and loss of Galois will be a2-a1 \leq 22n-n=2n.

Lagrangia.
Now we will show strategy for Lagrangia such his winnings will be no less then 2n.
First turn of Lagrangia must expunge all odd numbers so new set will be {0,2,4,...,22n}. After Galois turn set will be contain N=22n-2+1 numbers a1<a2<...<aN and a(i+1)-ai \geq 2=22 for all i. Lagrangia must again expunge all odd-positioned numbers, then after Galois turn set will be contain numbers b1<b2<...<bN' (N'=22n-4+1) and b(i+1)-bi \geq 22. And so on: each turn Lagrangia must expunge odd-positioned numbers ==> after n-th turn of Galois set will contain 2 numbers c1<c2 and c2-c1 \geq 2n.

Thus assured winnings of Lagrangia is nothing more than 2n.


[quote="vijaymenon"]but i see that -1/6 also satisfies the equation...[/quote]

Are you sure $ \minus{}\frac{1}{6}$ satisfies the equation? :?:

[quote="stephencheng"][quote="vijaymenon"]but i see that -1/6 also satisfies the equation...[/quote]

Are you sure $ \minus{} \frac {1}{6}$ satisfies the equation? :?:[/quote]

Um, I see is not.

oops i very much sorry made a mess of it,...very sorry :(  :(

hello, by factoring the equation
$ (12x\minus{}1)(6x\minus{}1)(4x\minus{}1)(3x\minus{}1)\equal{}5$
we get
$ (2x\minus{}1)(12x\plus{}1)(36x^2\minus{}15x\plus{}4)\equal{}0$ and from here we get the solutions
$ x_1\equal{}\minus{}\frac{1}{12}$, $ x_2\equal{}\frac{1}{2}$ ,$ x_3\equal{}\frac{5}{24}\plus{}\frac{1}{24}i\sqrt{39}$, $ x_4\equal{}\frac{5}{24}\minus{}\frac{1}{24}i\sqrt{39}$.
Sonnhard.

Just paste the problem into http://www.wolframalpha.com, a so called computational knowledge engine. It will find the solutions for you.


Let $x_1,...,x_N$ be a finite set of points in an euclidean space of dimension $n$ and $N\leq 2n$. Prove the existence of 2 hyperplanes $H_1,H_2$  which are not parallel and whose union contains the $N$ points.

First of all, we can assume that the points are in general position, i.e. every $n+1$ of them are affine-independent. Otherwise, for any $n+1$ affine-dependent points, we can take a hyperplane determined by $n$ of them, and it will intersect the one determined by the other $n$.

Now take any $n+2$ points among the $2n$ which were given. Radon's Theorem says that we can partition this set of size $n+2$ into two subsets with intersecting convex hulls. If both sets have size at least $2$, then just complete each one of them to a set of size $n$ by using points which are not among the $n+2$. The hyperplanes determined by the two groups of $n$ points each will intersect.

If, on the other hand, one of the sets has size $1$, and the other one has size $n+1$, then the set of size $1$ is a point inside a simplex determined by the other $n+1$ points. The line determined by this point and one of the $n+1$ points intersects the hyperplane determined by the other $n$, and we can complete this line to a hyperplane by adjoining the $n-2$ points which have been left out, thus getting two intersecting hyperplanes.


Prove that as $x$ goes to infinity, $1^{n}+2^{n}+3^{n}...+x^{n}=\frac{x^{n+1}}{n+1}$.  This solution requires no knowledge of caculus other than knowing how $x+1=x$ when $x$ goes to infinity.
EDIT: Look at my second post for the properly-worded problem.

Well... $x^{2}= O\left(e^{|x|}\right)$ as $|x| \to \infty$.

[quote="The QuattoMaster 6000"][quote="Kurt Gödel"][quote="The QuattoMaster 6000"]Prove that as $x$ goes to infinity, $1^{n}+2^{n}+3^{n}...+x^{n}=\frac{x^{n+1}}{n+1}$.  This solution requires no knowledge of caculus other than knowing how $x+1=x$ when $x$ goes to infinity.[/quote]

Both sides of your equation go to infinity, so it doesn't really make sense. It's like saying that $x^{2}= e^{x}$ if $x$ is large enough :wink:[/quote]
Sorry, bad wording.  Forget my first post.  This is what I mean:
Prove that as $\epsilon$ goes to 0, $(\epsilon)^{n+1}(1^{n}+2^{n}+3^{n}...+(\frac{a}{\epsilon})^{n})=\frac{a^{n+1}}{n+1}$ where $a$ is an arbitary real number.[/quote]

This still doesn't quite make sense because we don't know that $\frac{a}{\epsilon}$ is an integer, and when it isn't...  :?:  :!:

I think a better restatement would be
\[\lim_{m\rightarrow \infty}\frac{1^{n}+2^{n}+\cdots+m^{n}}{m^{n+1}}=\frac{1}{n+1}\]
for integral $m$, of course.


MATHCOUNTS Vieta geometry probability perimeter conics parabola

Traditional marathon except...

You dont have to solve the old problem to post up a new one (this will save time)

Please post up your answers when you have reached a solution.

Rules...
1. NO irrelevant posts
2. Tell us where the question comes from
3. Number your problems
4. Give the number of the problem when you post the solution
5. HAVE FUN!!!!

If this officially means the other marathon is dead, then...

Problem 1 (as far as I know, I made up the numbers, the idea comes from somewhere else)

Find the sum $ \binom{4}{0}\plus{}\binom{5}{1}\plus{}\binom{6}{2}\plus{}\binom{7}{3}\plus{}\binom{8}{4}\plus{}...\plus{}\binom{15}{11}$.

6. Post the answer, NOT SOLUTION, with you question
--------------------------------------

[size=200]Question 2[/size]

Acme T-shirt Company charges a $ \$$50 set up fee plus $ \$$9 per shirt...
Beta T-shirt Company charges no set up fee, but charge $ \$$14 per shirt.
What is the minimum number of shirts for which a customer saves money by using Acme?

2001 Chapter Target

[hide="for cheaters or people who already have an answer  :lol: "]
11 shirts[/hide]

I'll post solution after a few people post theirs

[hide="Question 1"]Let the number of shirt bought be $ x$.
Then we want $ 50 \plus{} 9x < 14x$
Solving for $ x$, $ x > 10$.  The minimum is thus $ 11$ shirts bought.[/hide]

[hide="Problem 1"]
Draw Pascal's triangle.  Notice that $ \binom {4}{0}, \hdots$ are all on the same diagonal.  This is known as the hockey stick (?).  The sum of $ \binom {n}{0} \plus{} \binom{n \plus{} 1}{1} \plus{} \hdots \plus{} \binom{n \plus{} k}{k} \equal{} \binom {n \plus{} k \plus{} 1}{k}$
Thus, the desired answer is $ \binom {16}{11} \equal{} \binom{16}{5} \equal{} 4368$[/hide]

GautamR, posting answers, not solutions, is a direct violation of etiquette on the forums, so everyone should ignore rule #6.

[quote="GautamR"]ok sorry - wanted to give a challenge.....[/quote]

How exactly is not posting a solution more of a challenge?

ok...

Revised Rule 6

NO one can post up an answer if their solution (approach to the answer) is already posted above (what Fraenkel said)

[quote="Fraenkel"]Add a rule that no one shall repeat the same answer given by someone else in a previous post unless they have a different solution, this will save ALOT of space and time. (Perhaps also if the mod can delete those posts it would be great in the long run.

[size=200]Question 3[/size]

If there are six AOPSers seven different books, and twelve different pencils? If you choose one of each category, how many different combinations can you obtain?[/quote]

[hide="question 3"]
EH, it's just $ 6\times 7\times 12\equal{}504$
[/hide]

[size=200]Problem 4[/size]

Find the sum of the reciprocals of the roots of $ 4x^{2}\plus{}13x\minus{}43$, as a common fraction if necessary.

[quote="Fraenkel"]Add a rule that no one shall repeat the same answer given by someone else in a previous post unless they have a different solution, this will save ALOT of space and time. (Perhaps also if the mod can delete those posts it would be great in the long run.

[size=200]Question 3[/size]

If there are six AOPSers seven different books, and twelve different pencils? If you choose one of each category, how many different combinations can you obtain?[/quote]

Easy - if I hadnt been delayed by that "No consecutive posts for 10 minutes" thing, I would have posted sooner

[hide]

6 AOPSers * 7 books * 12 pencils
=504[/hide]

[hide="Solution to #4"]Let the roots be $ c_1$ and $ c_2$. The sum of the reciprocals will be $ \frac {1}{c_1}\plus{}\frac{1}{c_2}\equal{}\frac{c_1\plus{}c_2}{c_1*c_2}$
By Vieta's, $ c_1\plus{}c_2\equal{}\frac {\minus{}13}{4}$ and $ c_1*c_2\equal{}\frac{\minus{}43}{4}$
Thus, the desired sum is $ \frac {13}{43}$[/hide]

[size=200]Problem 5[/size]
Given that one whup and two cliks is equivalent to 4 watts of energy, three cliks and one pent is equivalent to 6 watts of energy, and two whups and four pents is equivalent to 8 watts of energy, how many cliks are need to generate the same amount of energy as one whup?

To GautamR, you just violated your own rule.  Posted the same solution as alanchou.

[quote="mathcrazed"][hide="Solution to #4"]Let the roots be $ c_1$ and $ c_2$. The sum of the reciprocals will be $ \frac {1}{c_1} \plus{} \frac {1}{c_2} \equal{} \frac {c_1 \plus{} c_2}{c_1*c_2}$
By Vieta's, $ c_1 \plus{} c_2 \equal{} \frac { \minus{} 13}{4}$ and $ c_1*c_2 \equal{} \frac { \minus{} 43}{4}$
Thus, the desired sum is $ \frac {13}{43}$[/hide]

[size=200]Problem 5[/size]
Given that one whup and two cliks is equivalent to 4 watts of energy, three cliks and one pent is equivalent to 6 watts of energy, and two whups and four pents is equivalent to 8 watts of energy, how many cliks are need to generate the same amount of energy as one whup?

To GautamR, you just violated your own rule.  Posted the same solution as alanchou.[/quote]

[hide="problem 5"]
So
$ w\plus{}2c\equal{}4$
$ 3c\plus{}p\equal{}6$
$ 2w\plus{}4p\equal{}8$

From the third equation, $ w\plus{}2p\equal{}4$, so $ c\equal{}p$.  Subbing into the second equation, $ 4c\equal{}6$, so $ c\equal{}1.5$.  Substituting into the first equation, $ w\plus{}3\equal{}4$, so $ w\equal{}1$ and the answer is $ \frac{1}{1.5}$ or $ \boxed{\frac{2}{3}}$.
[/hide]

I'm guessing you just made it up as you went along...

[size=200]Problem 6[/size]
This problem probably showed up somewhere

Find the sum of the last two digits in $ 7^{2001}$.

I didnt see that.

I was waiting for the 10 minutes to pass.

Sorry

-----------------------
I didnt see that.

I was waiting for the 10 minutes to pass.

Sorry

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

How exactly do I do the tens???

[hide="Answer to Problem 6"]

We have to keep finding patterns in the units and tens digits...

units ---- [b]7[/b], 7*7=4[b]9[/b], 9*7=6[b]3[/b], 3*7=2[b]1[/b], 1*7=[b]7[/b] and so on (repeating "block" = 4)

tens--- a bit more difficult im not sure if im right here...
after 0-- , 4, 4, 0, 0, 4, 4, ... (repeating block = 4)

2001/4 = something with 1 as remiander... units is 7
2001-1(for the original 0)/4 = something remainder 0...tens is 4

answer (I hope) is 7+4

=11

[/hide]

[size=200]Problem 7[/size]

For positive integer $ k$, when 60 is divided by $ k^2$, the remainder is 6.  What is the remainder when 100 is divided by $ k$.

2001 Chapter Target

[hide="answer"]
1
[/hide]

[hide="Problem 6"]$ 7^{2001} \equiv 7^{2001\minus{}1998} \mod 10$
This is because $ x^{\phi(n)}\equiv 1 \mod n$ for $ (x,n)\equal{}1$
[hide="The Velocity of a Condor at 216 Feet in a Dive for Prey"]Unfortunately, we can't discuss this in the Mathcounts forum (I think).[/hide]
The answer is thus $ 4\plus{}3\equal{}7$.
Another way is to find the pattern, but that might take a while.[/hide]

[size=200]Problem 7[/size]

Find the sum of the digits of the sum of the reciprocals of the solutions to the equation $ x^2\minus{}60x\plus{}5\equal{}0$.

Yep, I'm just making things up as I go :D.

Sorry, but not really.  Try looking at the probability of the same thing if there is only 1 coin, then only 2 coins, then only 3 coins and so on....

By looking at a simpler case you may notice something...

This problem comes up so much!
Not MC level though...

[hide]
Let's call the number of ways to flip $ n$ coins without 2 consecutive heads $ a_n$.

Now, let's split the 10-flips into two cases:
Case 1:
The first flip is heads. The second flip must be a tails (otherwise there would be HH).
Then, the other 8 flips can be made in $ a_8$ ways.

Case 2:
The first flip is tails. The other 9 flips can be made in $ a_9$ ways.

Thus, we have the recursion $ a_n \equal{} a_{n\minus{}1} \plus{} a_{n\minus{}2}$.
Starting with $ a_1 \equal{} 2$ and $ a_2 \equal{} 3$, we plug into the recursion:
2,3,5,8,13,21,34,55,89,144

So the answer is $ \frac{144}{2^{10}} \equal{} \boxed{\frac{9}{64}}$.

You could also have found $ a_{10}$ by noticing that it is equal to the 12th fibbonacci number and plugging into Binet's formula, but that's totally over MC level...

[/hide]

Nobody would want to evaluate Binet's formula for $ n\equal{}12$ anyways.

And we should move on soon, or never

Problem $ \sqrt{256}\equal{}4^2\equal{}F_6\plus{}F_3$

The [[Fibonacci Sequence]] is defined as $ F_0\equal{}1$ and $ F_1\equal{}1$, and $ F_{n\minus{}1}\plus{}F_n\equal{}F_{n\plus{}1}$. A pair of consecutive Fibonacci Numbers are picked at random. What is the probability that their sum is a multiple of 3?

I got that too, using mods and stuff

Ok, problem 17

Given that $ f(x)\equal{}x^{4}\minus{}10x^{3}\plus{}17x^2\plus{}22x\plus{}24$ factors as $ (x^2\minus{}3x\plus{}2)(x^2\minus{}7x\plus{}12)$ (I hope), find all $ x$ such that $ f(x)\equal{}0$.

[hide]Just factor the quadratics: $ (x \minus{} 2)(x \minus{} 1)(x \minus{} 3)(x \minus{} 4)$

$ x \equal{} 1,2,3,4$
Btw, that is the wrong factoring. The $ x^2$ term is way off.[/hide]

[b][size=150]Problem 18[/size][/b]

The diameter of a circle with center $ O$ is extended to a point $ A$.  The tangent lines to the circle containing point $ A$ are drawn.  Given that the distance from $ A$ to the closest point on the circle is $ 3$ and the distance from $ A$ to a tangent point is $ 9$, find the radius of the circle.

[hide="my solution"]From basic geometry, we know that any line tangent to the circle is perpendicular to a radius. If we draw the circle out, we see a right triangle. Labeling the radius as x and applying the pythagorean theorem, we get $ x^2 \plus{} 9^2 \equal{} (x \plus{} 3)^2$
Solving for $ x$ yields $ x \equal{} 12$[/hide]

Hoping that the above is right...

A teacher takes a class of 9 to a movie theater. (The student's names are A, B, C, D, E, ...The teacher randomly arranges seating. However, the students will be satified if and only if:

A-D are sitting together. (e.g. DCAB... )
D is NOT sitting next to E or F.
H and I are sitting together.

What is probability that the students will be satisfied?

[quote="darkmatter47"]
A-D are sitting together. (e.g. DCAB... )
D is NOT sitting next to E or F.
H and I are sitting together.

What is probability that the students will be satisfied?[/quote]

I'm not sure what you mean... Is there someone sitting [i]between[/i] A and D? Does the same apply for D and E and D and F?

I meant A and B and C and D are sitting together, but not neccessarily in that order.

[hide]
First, we count the number of seating arrangements without the second condition.
We can think of HI as a group and ABCD as a group, so the number of such arrangements is $ (5!)(4!)(2) \equal{} 5760$.

Then, we count the number of the arrangements we just counted that aren't desirable.

Case 1: The group ABCD is sitting on either side of everyone else (ABCDxxxx or xxxxABCD)
If ABCD sits on a side, then, D must be the one closest to everone else. The ABC can be ordered in any way, so $ 3!$ ways.
Then, there are two ways to choose the person to sit next to D (E or F), and $ 3! \cdot 2$ to order the rest.
So the total number of such seatings is $ (2)(3!)(2)(3!)(2) \equal{} 288$.

Case 2: The group ABCD is sitting in one of the 3 middle positions. (xABCDxxx, xxABCDxx, xxxABCDx)
D can sit on either side of the ABCD, and the others can be ordered in $ 3!$ ways.
There are two ways to choose the person sitting next to D (E or F), and $ 3! \cdot 2$ to order the rest.
For a total of $ (3)(2)(3!)(2)(3!)(2) \equal{} 864$ seatings.

So there are $ 5760 \minus{} 864 \minus{} 288 \equal{} 4608$ desirable seatings, out of a total of $ 9! \equal{} 362880$ possible seatings, so the probability desired is $ \frac{4608}{362880} \equal{} \boxed{\frac{4}{315}}$[/hide]

[b][i][u]Problem [/u][/i]/[b] $ \lfloor \sqrt{400\plus{}1}\rfloor$

Compute

\[ 2\plus{}6\plus{}10\plus{}14\plus{}\cdots \plus{}154\].

Do not use a calculator.

[hide]That expression is equal to $ 2(1\plus{}3\plus{}...\plus{}77)$, which is 2 times the sum of the first 39 odd numbers. Which is 3042.[/hide]

Problem 21.
Billy and Bob decide to flip coins for no good reason. What's the probability that out of 10 coins they flip, there are 5 heads and 5 tails?

Um, I think this is the answer:
$ \frac {\binom{10} {5}} {2^(10)}\equal{}\frac {63} {256}$

[quote="xpmath"][hide]That expression is equal to $ 2(1 \plus{} 3 \plus{} ... \plus{} 77)$, which is 2 times the sum of the first 39 odd numbers. Which is 3042.[/hide]

Problem 21.
Billy and Bob decide to flip coins for no good reason. What's the probability that out of 10 coins they flip, there are 5 heads and 5 tails?[/quote]
I think you could use Pascal's triangle for this.
[hide=" Using Pascal's triangle"]
Tenth row of Pascal's triangle:
1 10 45 120 210 252 210 120 45 10 1
So here it shows that.
There's 1 way for all heads.
There's 10 ways for 9 heads
There's 45 ways for 8 heads.
There's 120 ways for 7 heads.
Etc.
So it said that there's 252 ways for five heads and five tails.
Out of the possible $ 2^10$ total ways
Which gets you $ \frac{252}{1024}$
Which simplifies down to $ \frac{63}{256}$
So yeah, I agree with Math Geek's answer.
[/hide]


In a scalen triangle ABC the tangent to its circumcircle (with center O) at C meets AB in M. The line perpendicular to OM in M intersects BC and AC in P, Q respectively. Prove that MQ=MP.

The perpendicular from $C$ to $OM$ (call it $\ell$) is the polar of $M$ wrt the circle, so $\ell, CM,CA,CB$ form a harmonic quadruple. $PQ\|\ell$, so the other three lines of the harmonic quadruple cut two equal segments on $PQ$, Q.E.D.

Another solution: The butterfly theorem (in its general form - Theorem 1 in [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i]) says that [i]if A, B, C, D are four points on a circle, and O is the center of this circle, and if the lines AB and CD meet at a point M, then the perpendicular to the line OM through the point M intersects the lines BC and DA at two points P and Q such that the point M is the midpoint of the segment PQ[/i]. Now, in your problem, it is enough to apply this to the four points A, B, C, C, and the proof is complete.

  Darij

Thanks darij
I knew butterfly theorem in its usual form(unforunatly)


[quote="Totally Zealous"]If triangle PQR has sides 40, 60, and 80, then the shortest altitude is K times the longest altitude. Find the value of K.[/quote]

[hide]I guess it's $\boxed{2}$[/hide]

[hide] Let's call the area of the triangle $A$. The altitude to the 80 side is $\frac{A}{40}$, from the area formula. The altitude to the 40 side is $\frac{A}{20}$. So, the answer's $2$.[/hide]

Altitude to 80 is $x/40$. Altitude to 40 is $x/20$. So K=2.


[quote="KingSmasher3"]The AIME is a challenging compeition for many people. How did you do?
Obviously, older students generally do better than younger ones, so dived your score by your grade level.  :wink:[/quote]

Using your scale, higher than a 2 is very unlikely. And as far as scoring above a 3(using your scale) is concerned, it's not very plausible.

Even if I got a 15 on AIME, since I'm in 8th grade, on your scale I can't even get a 2. o_o

You can look at the General Statistics from past years, like [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2009-aa/09-AIMEstats.shtml]2009[/url], [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2008-aa/08-AIMEstats.shtml]2008[/url], ... (more available [url=http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml]here[/url]), and see scores based on grades.

This makes me feel stupid cuz I'm less than 1  :(  :P

The highest I can think of is Michael Ma (the 4th grader who qualified for USAMO), who earned a 9 on the AIME.

And even if you do something like (AIME score) - (grade), I think he still wins.

There's no place on the poll for less than 0.5 (which would include me :( ).

I'm hoping to qualify for USAMO next year, though. I was fairly inexperienced when I took the AIME last year: I had never even heard of the AMC until the January before.

Yeah, come on.  :mad:

8th grade = 3.

$ \frac{3}{8} \equal{} 0.375 < 0.5 \equal{}$lowest classification. hmf. :)


function
real analysis
real analysis unsolved

hi all,

let u_n a sequence of real numbers. Prove that there
exists a function C^(oo) so that f^(n)(0)=u_n where
f^(n) is the n-th derivative.

$f(x) = f(0) + f'(0) {x \over 1!} + f''(0) {x^2 \over 2!} + f'''(0) {x^3 \over 3!} + ...$ which is sometimes convergent and sometimes divergent. taking $u_n = f^{(n)}(0) = (n!)^2$ we clearly see that it's convergent only at $x=0$ and we cannot differentiate it one term after another. therefore i think this statement isn't true.

This is Borel Theorem, classical problem in oral examination at
Ecole Normale Suprieure  or Ecole Polytechnique X, or "leon d'Agrgation de Mathmatiques"

For  $\epsilon >0$ for any $p\in N$ construct a familly of functions
$g_p:[-1;1] -->R$ $C^{\infty}$ such that
$g_p^{(p)}(0)=1$, for any k  $\in N-{(p)}$ ,  $g_p^{(k)}(0)=0$
and for any $k\in {0,1,...,p-1}, ||g_p^{(k)}|| \leq \epsilon$

The function is $f=\sum u_n.g_n$ is the solution


What's the minimum value of $f\text{:}\mathbb{R}\rightarrow\mathbb{R}$ defined by : $f(x)=\ln{\left(1+\sqrt{1+x^{2}}\right)}$.

Is there an solution for it or... that's all  :? ?

Are we using calculus here then it can be easily shown that f(x)is increasing in [0,infinity).Hence the answer.

Oh!
After posting I see that
1+sqrt(1+x^2)>=1+sqrt(1+0^2)   (i.e. for x=0)
ln(1+sqrt(1+x^2))>=ln(1+sqrt(1+0^2))=ln(2)
Hence proved

Since $x^2\geq 0$, we have $1+\sqrt{1+x^2}\geq 2$, the equality holds when $x=0$.
Then $y=\ln t\ (t>0)$ is increasing function, we have $f(x)\geq \ln 2$,yielding the desired minimum value $\ln 2$.

Oh! sorry, you have just posted the solution,varun. :blush:

My solution is exactly same as yours.

kunny


http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2004-aa/04aimeanswerforms.html

I've noticed on 2004 AIME answer sheet on the link above, they ask your current citizenship status. I'm in process for getting my green card, and most likely I will receive my green card on June 6th, 2005. So can I take USAMO this year since I'm just off by 2 month?

i had the same situation last year. I am getting mine at April of 2005. then, i wasnt sure what to do, but I cleared it with Mr rusccyk and Steve Dunbar, the AMC coordinator. I dont think they enforce the rule of "Green Card - At Least" on USAMO anymore.

Definitely call the AMC and ask them.  Their number is (402) 472-2257.

Last year, I obtained a clarifying statement from our Policy Committee.  The new policy is (from the USAMO Teachers' Manual, reproduced on the web from source at http://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/05TM-USAMO/html-05U/04-05usamo-text.html
 is:

"U. S. citizens and students legally residing in the United States and Canada (with qualifyng scores) are eligible to take the USAMO."

Steve Dunbar
AMC Director

[quote="AMCDirector"]
"U. S. citizens and students [b]legally residing [/b]in the United States and Canada (with qualifyng scores) are eligible to take the USAMO."

Steve Dunbar
AMC Director[/quote]

What do you mean by "legally residing", if someone is applying for green card, does it qualify for "legally residing".  How about the kids of visiting graduate students or scholars on temp visa?  Do they qualify?

[quote="Silverfalcon"]Doesn't legally residing means:

You have come United States legally and you're legal alien? :)

I learned that in my civics class. I'm not sure if it's same thing or not. :D[/quote]

This is what I think.

As long as you're in the process of applying for citizenship, I think that it's all right to take the USAMO.

[quote="tetrahedr0n"]From what I understand, a student that legally resides in the US (ie, not an illegal alien) can take it. Being in the process of applying for citizenship is not necessary.[/quote]
I can confirm that; I still have not even applied for citizenship. All the AMC asked me last year before USAMO was whether I was a US citizen, to which I answered "No" and was still allowed to take the exam.

I wasn't aware that this was a new policy. I sure hope it stays this way.  :P


|| x/y - y/x || = (x^2)/y + y/(x^2)

I found a solution for the integer x and y values x= -3 y= 9 and x= -2 y = -4   So,

x+y= {-6, 6 } Is there any more solutions?


Prove that $ \sqrt{n^2\plus{}2}$ is irrational for all integers $ n$.

Let $ \sqrt{n^2\plus{}2}\equal{}\frac{q}{p}$, where $ p,q$ are relatively prime positive integers.
Then  $ n^2\plus{}2\equal{}\frac{q^2}{p^2} \Rightarrow p\equal{}1$ and $ n^2\plus{}2\equal{}q^2$.
$ \Rightarrow (q\plus{}n)(q\minus{}n)\equal{}2$, as $ (q\plus{}n)$ and $ (q\minus{}n)$ are both odd or both even, this is impossible. Contradiction.

Another profe is:
If $ n \equal{} 2k$ then it must be $ 4k^2 \plus{} 2 \equal{} j^2$ ,where $ j$ is even,so
we got $ 4k^2 \plus{} 2 \equal{} 4m^2$ or $ 2k^2 \plus{} 1 \equal{} 2m^2$ contradiction..
If $ n$ is odd,then $ n^2 \plus{} 2 \equal{} 8k \plus{} 3$
So it must be $ 8k \plus{} 3 \equal{} j^2$ with $ j$ odd.So $ j^2 \equal{} 8m \plus{} 1$
From these two we got that $ 8k \plus{} 2 \equal{} 8m \leftrightarrow 4k \plus{} 1 \equal{} 4m$ contradiction...

How do you get $ p\equal{}1$?
(sorry if my question is silly)

Additional Question :
Prove that $ \pi \equal{} 3.14159265....$ is irrational.

$ n^2\plus{}2$ is an integer, so $ \frac{q^2}{p^2}$ is also an integer and since $ (p,q)\equal{}1$, $ p\equal{}1$.

Question 2 (about $ \pi$): see

http://www.lrz-muenchen.de/~hr/numb/pi-irr.html
http://www.mcs.csuhayward.edu/~malek/Mathlinks/Pi.html

[quote="Dimitris X and Arrange your tan"]Another proof is:

Let n, k, j, m, K, and M belong to the set of integers.

If n is even, and $ n \equal{} 2k$, then it must be $ 4k^2 \plus{} 2 \equal{} j^2$ ,
where $ j$ is even, and j is of the form $ 2m$, so
we got $ 4k^2 \plus{} 2 \equal{}
4m^2$ or $ 2k^2 \plus{} 1 \equal{} 2m^2$ contradiction..

If n is odd and n = 2k + 1, then $ n^2 \plus{} 2 \equal{}$
$ (2k \plus{} 1)^2 \plus{} 2 \equal{}$
$ 4k^2 \plus{} 4k \plus{}3 \equal{}$
$ 4k(k \plus{} 1) \plus{} 3 \equal{}$
$ 8K \plus{} 3$.

So it must be $ 8K \plus{} 3 \equal{} j^2$, with $ j$ odd, and $ j$ is of the form $ 2m \plus{} 1$.
So $ j^2 \equal{} (2m \plus{} 1)^2 \equal{}$
$ 4m^2 \plus{} 4m \plus{} 1 \equal{}$
$ 4m(m \plus{} 1) \plus{} 1 \equal{}$
$ 8M \plus{} 1$.

From these two we got that
$ 8K \plus{} 3 \equal{} 8M \plus{} 1$
$ 8K \plus{} 2 \equal{} 8M \leftrightarrow 4K \plus{} 1 \equal{} 4M$ contradiction...[/quote]

(These added steps, spacing, changes should help a
reader better follow the turns the original proof took.)

If you're going to go about improving people's writing, you might as well learn enough LaTeX so as to make helpful suggestions.  For example, go read about the align* environment for properly formatting chains of equations.

[quote="JBL"]If you're going to go about improving people's writing, you might as well learn enough LaTeX so as to make [MORE]** helpful suggestions.  For example, go read about the align* environment for properly formatting chains of equations.[/quote]

** This must have been an oversight to leave out this word
(or the like), else anyone can see all of the helpful suggestions
 I put in the box as compared to the original.

No, I'm mainly satisfied to get some more mathematical steps
just typed in here (as opposed to "improving people's writing"),
Latex style or not.

I can be consistent and stay away from Latex altogether if
I want and just type in the steps.


modular arithmetic
inequalities
pigeonhole principle
number theory
prime numbers
number theory proposed

Which prime numbers can be written as sum of $ 2$ perfect squares.

[hide="My Proof"]
Let $ p$ be a prime number such that $ p = a^2 + b^2$ for some positive integers $ a$ and $ b$.

Then
$ \begin {eqnarray*}p-1 &=& a^2 + b^2 - 1 \\
&=& (a+1)(a-1) + b^2$.

We will assume $ p$ is odd, since $ p=2$ is a trivial case, where we can clearly see $ 2 = 1^2 + 1^2$ by inspection. If $ p$ is odd, then either $ a^2$ is odd and $ b^2$ is even, or $ a^2$ is even and $ b^2$ is odd. In other words, for $ a$ and $ b$, one of them must be odd and the other must be even.

Thus assume $ a$ is odd without loss of generality, then either $ (a+1)$ or $ (a-1)$ must be a multiple of $ 4$ (i.e. $ a = 3 \Rightarrow a+1=4$, $ a=5 \Rightarrow a-1=4$, $ a=7 \Rightarrow a+1 = 8$, etc.). Since $ b$ is even, $ b^2$ can be rewritten as $ (2n)^2 = 4n^2$ for some positive integer $ n$. It is clear that $ b^2$ is also a multiple of $ 4$.

Therefore, a factor of $ 4$ can be pulled out from $ (a+1)(a-1) + b^2$, which implies $ p-1$ is divisible by $ 4$.

We will remember this conclusion as $ p \equiv 1 \pmod 4$.

Now let's assume $ p$ is a prime number again, and that $ p \equiv 1 \pmod 4$.
Then we know that $ x^2 \equiv -1 \pmod p$ has solutions for some $ x$ by Theorem 2.12 in Section 2.1.

Let $ f(u,v) = u + xv$ for some integers $ u$ and $ v$ such that $ 0 \le u \le K$ and $ 0 \le v \le K$, where $ K = [\sqrt{p}]$ (Greatest integer less than or equal to $ \sqrt{p}$).
Note that $ K \le \sqrt{p} \le K+1$ since the root of a prime is a non-integer.
Since $ u$ and $ v$ can each be up to $ K - 0 + 1$ values (because the inequality is inclusive), we have $ (K+1)^2$ possible combinations of pairs $ (u,v)$. But $ K+1 > \sqrt{p} \Rightarrow (K+1)^2 > p$, so the number of pairs is greater than $ p$. In this case, then consider $ f(u,v) \pmod p$. Since $ (K+1)^2 > p$, this implies there is not enough residue classes to each contain one and only one $ f(u,v)$. Therefore, the pigeonhole principle applies, which tells us that there must be a residue class that contains $ f(u,v)$ for two distinct pairs $ (u_1,v_1)$ and $ (u_2,v_2)$.

By the definition of a residue class that contains $ f(u,v)$ for two distinct pairs $ (u_1,v_1)$ and $ (u_2,v_2)$,
$ f(u_1,v_1) \equiv f(u_2, v_2) \pmod p \\
u_1 + x v_1 \equiv u_2 + x v_2 \pmod p \\
u_1 - u_2 \equiv -x (v_1 - v_2) \pmod p$

Let $ a = u_1 - u_2$ and $ b=v_1 - v_2$.
We then get
$ a \equiv -xb \pmod p$, which we can multiply by itself to get
$ a^2 \equiv (-xb)^2 \pmod p \equiv x^2b^2 \pmod p$. Recall that we stated in the beginning that $ x^2 \equiv -1 \pmod p$. So we end up with $ a^2 \equiv -b^2 \pmod p$. Thus $ p | (a^2 + b^2)$ by definition of congruence.

We can check to see that $ a^2 + b^2 > 0$ since the pairs $ (u_1, v_1)$ and $ (u_2, v_2)$ are distinct from each other (implying $ a,b \neq 0$).
Note also that $ -K \le a \le K$ since $ a = u_1 - u_2$, where $ u_1 \le K$ and $ u_2 \ge 0$. Similarly, $ -K \le b \le K$. Since $ K < \sqrt {p}$, this gives us $ |a|< \sqrt{p}$ and $ |b| < \sqrt{p}$. We can square each of the two equations to get the inequalities $ a^2 < p$ and $ b^2 < p$. Adding them together, we get $ a^2 + b^2 < 2p$. Combining the two statements we have shown, we get $ 0 < a^2 + b^2 < 2p$.

However, recall from earlier that $ p| a^2 + b^2$ but $ p$ is the only multiple of $ p$ in the interval $ (0, 2p)$. Therefore, $ p$ must equal $ a^2 + b^2$.

In the end, we have proved that $ p \equiv 1 \pmod 4$ if and only if $ p$ can be written as a sum of two squares. Therefore, any prime number that is congruent to $ 1 \pmod 4$ can be written as a sum of two squares.
[/hide]

It seems like you do a little more work than is necessary; in the first part, $ x^2 \equiv 0, 1 \bmod 4 \implies a^2 \plus{} b^2 \equiv 0, 1, 2 \bmod 4$.  As for one of my favorite approaches to the second, see [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]Toolkit 2[/url] here.


My school's math curriculum is pretty shallow and I'm trying to teach myself how to prove trig identities. These are giving me grief and I'm hoping for some help with proving them:

$ \frac {\sin (x\minus{}y)}{\sin (x\plus{}y)} \equal{} \frac {\tan x \minus{} \tan y}{\tan x \plus{} \tan y}$

$ \frac {\sin (x\minus{}y)}{\sin y} \plus{} \frac {\cos (x\minus{}y)}{\cos y} \equal{} \frac {\sin x}{\sin x \cos y}$

$ \frac {\tan (x\plus{}y) \minus{} \tan y}{1 \plus{} \tan (x\plus{}y) \tan y} \equal{} \tan x$

(1) Write the RHS in terms of sine and cosine.
(2) Expand the LHS using the sin & cos addition formulas and combine fractions.
(3) Put everything in terms of sin and cos. (Or you can be clever and notice the expression for the tangent addition formula on the LHS).

I'll show the first one.

Using the sum/difference formula for $ \sin (A \pm B)$,
$ \frac{\sin(x\minus{}y)}{\sin(x\plus{}y)} \equal{} \frac{\sin x \cos y \minus{} \cos x \sin y}{\sin x cos y \plus{} \cos x \sin y}$
Dividing the numerator and denominator by $ \cos x \cos y$ (which is valid as long as $ x, y \neq \frac{\pi}{2} \plus{} n \pi$) gives
$ \displaystyle{\frac{\frac{\sin x}{cos x} \minus{} \frac{\sin y}{\cos y}}{\frac{\sin x}{cos x} \plus{} \frac{\sin y}{\cos y}} \equal{} \frac{\tan x \minus{} \tan y}{\tan x \plus{} \tan y}}$

The others are similar. Use a trig identity to expand, then apply some algebraic operation to combine or simplify the result.

Expanding the third one is not the best way to go.

[hide="Hint"]
Recall that $ \tan(a\pm b)\equal{}\frac{\tan a\pm\tan b}{1\mp \tan a\tan b}$.
[/hide]


Discuss.

Which rounds were easy, which were hard? I personally thought 3 was the hardest (based on other people from my school who scored 0 TOTAL on that round), then round 1 came next. Round 2 was like super easy (like always).

I did rounds 1, 4, 5. Imaginary numbers kill me. Thus... I lose 4 points.

I think alkjash and illusiat did the exact opposite... 2, 3, and 6. Oh well.

idk didnt take it but I got a 19 on USAMTS u? also should have got a 15 on RIML but got 2 point of on a SUPER technicality. a ration a/b was 2310 and we were supposed to write 2310/1  :o

Rounds 3, 5, 6, the geo/trig rounds...I got an 18. I thought the rounds were pretty good...except I always get scared when I get something like 14040 as an answer. I'm actually somewhat surprised I didn't screw up anything, especially on trig...

[quote="Yongyi781"]The lesson I learned is that I should FORCE MY TEAM MEMBERS TO LET ME PICK THE PROBLEMS.

Meh. I really didn't want to do #1, since everyone knows how bad I am at complex numbers, but everyone just forced me to do it. Their excuse was that "you're the smartest person in the school! You should do the hard problems."

The net result was that I scored very low.[/quote]

It happens when you're smart like that. You end up taking the harder problems.

[quote="Pineappleperson"]i hate how AB is so jacked this year...I actually have to take mml seriously now b/c a 14 could get you kicked off the team   :([/quote]

SO JACKED

Yep, I can't stress enough how bad I am at complex numbers... :noo: And they still make me do that round :noo:

[quote="ac-king"]Go AB! :D

but yeah... I epically failed... 14.

I got a 2 on the factoring round and then took an alg 2 test on factoring with harder problems and less time and I got a 100 O.o.

Maybe I'm just bad at mml in general.[/quote]

Alan Chiao?

And this is a topic about the MAML II in March, not the second MML.

[quote="Yongyi781"]No, this is about the second MML. I should know, since I created this topic!![/quote]

gg corwin

guyz z0mggg i'm so excited for the mml meet on thursday lollz!!!!

Look! I posted two consecutive posts!

[hide]
See [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1091]here[/url] for details.[/hide]

[quote="CatalystOfNostalgia"]i'm so excited for tomorrow!!!!![/quote]

Too bad I don't get to see you. We're in a different division :wallbash:
Then again, we would be in like last place if we were in your division :|


the quickest way is to specify a number such as 2,
f(2)=3/1=3
f(-2)=-1/-3=1/3
hence A is correct

if there're no selections given, we can solve it in this way:

f(-x)=(-x+1)/(-x-1)=-(x-1)/[-(x+1)]=(x-1)/(x+1)=1/f(x)

you'll notice that the condition [b]x 2 doesn't equal 1[/b] hasn't been used,because it only ensures that the denominators may not be zero.

[hide]The answer is 1/f(x) or A.  f(x)=(x+1)/x-1.  f(-x)=(-x+1)/-x-1.  We don't recognize this as one of solutions yet so we multiply both sides by -1.  f(-x)=(x-1)/x+1=1/f(x).[/hide]

I'm pretty pathetic

I spent 10 minutes using equation to find out the answer...

well,,

then I realized that it was a function

well,,

I used 2 and 3 to check if the function worked

well,, A is the answer......


integration
limit
real analysis
real analysis unsolved

Let $ f: [0,1]\to \mathbb{R}$ continuous and $ g_{n}: [0,1]\to R, g_{n}(x)=\int_{0}^{1}min(x,t^{n})f(t)dt,n\in \mathbb{N}^{*}$.
Compute $ \lim_{n\to \infty}g_{n}(x)$

The limit is $ 0$.

$ f$ is continuous on $ [0,1]$ and hence bounded.

$ \biggm|\int_{0}^{1}\min(x, t^{n})f(t)dt\biggm|\leq \int_{0}^{1}\min (x, t^{n})Mdt$

$ \leq \int_{0}^{1}t^{n}Mdt=\frac{M}{n+1}$.

It follows that

$ 0\leq |g_{n}(x)|\leq \frac{M}{n+1}$.

The desired result follows.


integration
function
real analysis
real analysis unsolved

Show that an orthonormal sequence $ \{\phi_n : n\geq 1\}$ in $ L^2(a,b)$ is complete iff

$ \sum_1^\infty\big|\int_a^x\phi_n(t)dt\big|^2 \equal{} x \minus{} a$ for all $ x\in (a,b)$.

I think I have the forward direction, since $ f \equal{} \chi_{(a,x)}\in L^2(a,b)$, so you just work out the norm of the projections and get precisely what you want.

I see no way of getting that set complete just from that equality. Maybe the problem is I'm trying to be too constructive and there is some contradiction from assuming it isn't?

Suppose that $ \{\phi_n\}$ satisfies the following condition: $ \sum_n\left|\int_Ig\phi_n\right|^2\equal{}\int_I|g|^2$ for all $ g\in B$ where $ B$ is a dense subspace of $ L_2(I),$

Would that be sufficient for $ \{\phi_n\}$ to be complete?

If you can answer that, can you then use the stated condition to show this property for all step functions? After all, the set of step functions is dense in $ L^2(I).$

Ah. Thanks. That was obviously the natural thing to try. I don't know why I never seem to think of it.  :mad:


A $4v$ battery is connected to a $10$ ohm resistor as shown.  The circuit is placed in a magnetic field directed into the page that is detereorating at a rate of $150 \  \text{tesla}/\text{s}$  Calculate the current in the circuit and state the direction.

[hide="solution"]
Let's first summarize what's given:

\[emf_0 = 4\ V\]
\[R = 10\ \Omega\]
$L = 0.12\ m $, the length of each side of the square loop
$A = L^2 = (0.12)^2 = 1.44\ \times \ 10^{-2} \ m^2$ , the area of the loop
\[\frac{dB}{dt} = - 150\ Tesla/s\]

Now, first consider the magnetic flux, $\Phi_B = \int B \cdot dA$. But since the area vector is in the direction of the normal vector to the square, i.e. B is parallel to A, the dot product vanishes and we get $\Phi_B = BA$

Furthermore, from Faraday's Law:

\[emf_{in} = - \frac{d\Phi_B}{dt} = - \frac{d (BA)}{dt} = - A \frac{dB}{dt}\]
\[emf_{in} = - L^2 \frac{dB}{dt} = 2.16 V\]

Furthermore, by Ohm's Law, we have $emf = I R$.
Here $emf = emf_{total} = emf_0 + emf_in = 6.16 V$
Therefore, $ i = \frac{emf_{tot}}{R} = \frac{6.16}{10} = 616\ mA$

Now to determine the direction of the current, we'll use Lenz's Law which says the system will try to oppose a change in flux, i.e. the decreasing flux will be counteracted by creating an induced magnetic field in the direction of B (into the page) so by the right hand rule, the current i must travel clockwise.
[/hide]

Hope that helps.


um...if i enroll for the Intro to C&P class like this week, and the class starts in Oct., but the book is included as a part of the class price, does that mean, I will get the course book in Oct, or several days after i place my order for enrollment in the class...???

If you enroll now, you will get the book when you enroll; that is, it will be sent right away.

The only class for which this is not the case is Intermediate Counting & Probability, as the book will not be published until September.


quadratics
superior algebra
superior algebra theorems

Hi,

I am trying to get a better understanding of quadrics, quadratic forms , and symmetric bilinear forms

as i understand it roughly : in char nontwo, symmetric bilinear forms and quadratic forms correspond, and quadratic forms correspond to quadrics (if the quadric is no subspace)

This is a question of mine : is it correct that in an n dimensional space over a finite field( char may be two)
there are only 2 nonsingular quadratic forms when n is even
and only 1 nonsingular quadratic form when n is odd

It is my goal in general to understand orbits, isometries, projective orthogonal groups (preserving the form and preserving it up to scalar multiplication is not the same right?) better

I am interested in any technical details as well concerning quadratic form/bilinear form , and characteristic two

[quote="fredbel6"]
This is a question of mine : is it correct that in an n dimensional space over a finite field( char may be two)
there are only 2 nonsingular quadratic forms when n is even
and only 1 nonsingular quadratic form when n is odd[/quote]

up to similarity

how many orbits are there then

O

i see i have at least one mistake here ....
when n is odd, there are two i guess

If it is relevant I am aware of the fact than any symmetric bilinear form in charnontwo can be diagonalised, and that the determinants being square play a role

plz help

there are the two natural orbits (square and nonsquare determinant). in even dimension you can diferentiate them also by the dimension of the largest totally isotropic subspace( that is a subspace on which the quadratic form is zero).

look up any book on geometric algebra (artin's) or for example ashbacher's gp theory book.


$ a$, $ b$, $ c$ are positive reals such that $ ab\plus{}bc\plus{}ca\equal{}1$
Prove the inequality:

$ a\plus{}(1/a)\plus{}b\plus{}(1/b)\plus{}c\plus{}(1/c)\geq (a\plus{}b)(3ab\plus{}1)\plus{}(b\plus{}c)(3bc\plus{}1)\plus{}(c\plus{}a)(3ca\plus{}1)$

Nice inequality .
we have that
$ 3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}9abc \equal{} 3(a\plus{}b\plus{}c)\minus{}9abc \equal{} 3(ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a))$
so the inequality is equivalent to

$ 9abc\plus{}\frac{1}{abc}\geq 4(a\plus{}b\plus{}c)$ or we have to prove that

$ 1\plus{}9a^{2}b^{2}c^{2}\geq 4abc(a\plus{}b\plus{}c)$ or $ (ab\plus{}bc\plus{}ca)^{3}\plus{}9a^{2}b^{2}c^{2}\geq 4abc(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$ (1)

But schur inequality says that $ (x\plus{}y\plus{}z)^{3}\plus{}9xyz\geq 4(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)$
Put $ x \equal{} ab$ $ y \equal{} bc$ $ z \equal{} ca$ and you will take the desired (1)


The fenced area of a yard is a 15-foot by 12-foot rectangular region with a 3-foot by 3-foot square cut out, as shown. What is the area of the region within the fence, in square feet?

[asy]draw((0,0)--(16,0)--(16,12)--(28,12)--(28,0)--(60,0)--(60,48)--(0,48)--cycle);
label("15'",(30,48),N);
label("12'",(60,24),E);
label("3'",(16,6),W);
label("3'",(22,12),N);[/asy]

The area of the whole piece is 180 and the area of the taken piece is 9 thus 180-9 = [b]171[/b]


The sphere has radius 10 cm with a circle in the middle, and the smaller circle's center is 6 cm from the center of the sphere. What is the area of the smaller circle in terms of pi?

NOTE: the problem might be a little confusing without the diagram, but it's understandable. Try your best.
(there wasn't much info on the diagram anyways)

See diagram

[hide]
Looking at the right triangle, you can see the radius is 8, so $64\pi$?
[/hide]

I agree that it's

[hide]60pi[/hide]
for the same reason,
but what if the 2nd circle was in a plane that wasn't parallel to the plane with the first circle...how does that figure?

the 2nd circle is parallel to the plane of the first circle.  :)

[quote="Treething"]- then you just draw another diameter circle parallel to the smaller circle
[color=orange]
[Think of something more creative to say than swearing - Billy][/color][/quote]

he's my god too >:|


What is the maximum number of  distinct regions that $ 50$ lines and $ 60$ circles can divide a rectangle into?

I do not wish to count the lines and circles; is there a formula?

Hi,I hope this works something
$ n$ lines divide a circle into ${ \frac{n(n+1)}{2}} +1$ regions.

[hide="Method"]Compute the maximum number of intersection points. Doable by adding circles and lines iteratively. Use that two lines can intersect at one point, and circles can intersect any circle or line at two points.

Consider the planar graph in which each intersection point, intersection of a line or circle with a corner of the rectangle, and the rectangle corners are the vertices, and the lines and circles form the edges. (Graph may not be simple) You already know the number of vertices of this graph.

You can compute the number of edges by noting that each line and circle starts off as one edge, and with each intersection point they are split in a way that adds two more. However the first split on a circle doesn't add an edge, so take out an ovecount of $ 60$.

Now apply $ V \minus{} E \plus{} F \equal{} 2$ and you're home. There are some details to be ironed out, like the case that a circle intersects the rectangle edge.[/hide]


linear algebra
matrix
inequalities
linear algebra unsolved

Supposed that an $ nxn$ matrix $ A$ statisfying $ A$ is not equal to $ I_n$ where $ I_n$ is the identity matrix of order $ n$, and for a positive integer $ k$ we have $ {A}^{k\plus{}1}\equal{}I$, prove that $ A^k$ is not equal to $ I$.

Isn't that obvious? $ A^k\equal{}I$ implies $ A^{k\plus{}1}\equal{}A\equal{}I$. Contradiction.

Oh.. I think it must use Schur Decomposition. I'm sorry.
Is there anyone who have any problems about Schur Decomposition ? Please post your problems here. Thanks in advance


[hide]This is a well known factorization:

\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\][/hide]

For the derivation, consider a polynomial $P(x)$ with roots $a,b,c$ and work from there.

There is also a nice solution using determinants.

By the way, this identity is called "Dorin's favorite identity."


What number can be subtracted from both the numerator and denominator of $ \frac{19}{24}$ so that the resulting fraction will be equivalent to $ \frac34$?

$ \frac{3}{4}\equal{}\frac{15}{20}$ so the number is $ \boxed{4}$.

We can interpret this as if there are $ 19$ of thing one and $ 5$ of thing two. In the end, we desire there to be $ 1$ of thing two for every $ 3$ of thing one--that is, there are $ 15$ of thing one, and the desired answer is $ 19 \minus{} 15 \equal{} \boxed4$.


[quote="dickchimney"]Chào Vũ Hùng. ?ồng chí đang h?c ở đâu đấy :D[/quote]
Mình học ở Tokyo, không phải nghành chi?nh là toa?n nhưng râ?t yêu toa?n, và cũng hay phải động vào toa?n nên mơ?i mò vào mathlinks.
Trong này i?t người Việt nhỉ

[quote="pigfly"][quote="vuhung"]
Trong này i?t người Việt nhỉ[/quote]
Sao lai it'?Nhieu` lam' ma`.De nghi 2 anh doi lai quoc tich di,ai doi` lai Ukraine voi' Nhat,bo' tay luon :D[/quote]

 Nhưng mà cái c? Ukraine rõ đẹp :D . ?ổi thành Vn thì cũng không còn bí hiểm nữa :(

[quote="pigfly"]
Sao lai it'?Nhieu` lam' ma`.De nghi 2 anh doi lai quoc tich di,ai doi` lai Ukraine voi' Nhat,bo' tay luon :D[/quote]
Anh cu?n ơi, ca?i cờ đo? gọi là Country Flag chơ? hông phãi là quô?c tịch đâu. Cờ Việt Nam nhìn đỏ qua? cho?i mă?t. Vơ?i lại mình sô?ng ở đâu thì để cờ nươ?c đo? cho diễn đàn no? phong phu? .... :)

[quote="vuhung"]
Anh cu?n ơi, ca?i cờ đo? gọi là Country Flag chơ? hông phãi là quô?c tịch đâu. Cờ Việt Nam nhìn đỏ qua? cho?i mă?t. Vơ?i lại mình sô?ng ở đâu thì để cờ nươ?c đo? cho diễn đàn no? phong phu? .... :)[/quote]
Minh` so'ng o dau thi ghi dia chi vao Location chu' :? ?Con` co` la "select ountry" ma`.Ro rang la` the hien quoc tich ma`.Ma` nen goi la em Cun' nhe' :D
@anh Dung: Anh gioi the phai de quoc tich VietNam cho nguoi ta biet nguoi Viet cung gioi lam chu  :D

[quote="pigfly"]
Minh` so'ng o dau thi ghi dia chi vao Location chu' :? ?Con` co` la "select ountry" ma`.Ro rang la` the hien quoc tich ma`.Ma` nen goi la em Cun' nhe' :D
[/quote]
Country và Nationality kha?c nhau hoàn toàn mà!

[quote="pigfly"]@anh Dung: Anh gioi the phai de quoc tich VietNam cho nguoi ta biet nguoi Viet cung gioi lam chu  :D[/quote]

 Anh chuyen roi nhung ma khong phai vi cau noi tren cua em dau nhe ;)

chao moi nguoi.
chuc mung nam moi.chuc moi nguoi van su nhu y.
 chi vuhung la con gai a?hay qua.
Tren Mathlinks toan con trai.Hoi buon

[quote="dickchimney"][quote="nonIMO"]National forums by number of posts:
1.Deutsch - 181 [i] :first: [/i]
2.Français -  97  :D
3.Chinese -  96  :what?:
4.Română -  76  :?
5.Spanish -   55  :(
6.Russkij -    47   :mad:
7.Vietnam -  37  :huh:
8.Korean -   12  :wacko:
9.Magyar -    0  :no:[/quote]

 Ghét chưa, anh em buôn chuyện đê :D[/quote]
Em dong y' hoan` toan` voi' anh Dung :P ,anh em ta phai buon chuyen manh len chu,ko the de tinh trang heo hut nhu the nay duoc :D.Anh Dung lam phat dau tien nao!
@lqdschool:ban hoc o truong Le Quy Don ,Da Nang a`?Sao ban co nhieu de TST the,cam on nhieu ve de TST nhe' :love:

[quote="dickchimney"]
 Cún!! Anh vẫn chưa biết danh tính thật của chú!! :?[/quote]
Em tiet lo nhe,anh co nho Mai Thanh Nga tren H-A-O ko? ;)  :)

Nhớ rồi :D. Vì thế nên cứ g?i chú là Nga nhỉ ;)
 Mai Thanh Nga là ngư?i iu à?? :blush:  :P

Hom nay da 27 Tet roi,chuc moi nguoi co mot cai Tet that vui ve! :) [quote="dickchimney"]Nhớ rồi :D. Vì thế nên cứ g?i chú là Nga nhỉ ;) [/quote]
 Y' kien tuyet voi! :lol:
[quote="dickchimney"]Mai Thanh Nga là ngư?i iu à?? :blush:  :P[/quote]
Ko co dau anh a`:D ,duoc the thi da may lam roi :P .Ma anh em ta hay dung do nhay het o HAO roi ML,vui that! :)

Cha soi noi qua, nam moi chuc moi nguoi manh khoe, thanh cong va that that giau co nhe!

[quote="dickchimney"]Vũ Hùng là con gái à ?? Mình cũng có mấy cô bạn h?c ở Nhật, không h?c Toán nhưng cũng thích toán !!   :blush: :D  Eh, thấy ngư?i quen thì phải chào nhé, nhưng mà nếu không phải thì thôi :blush:  :P
 Cún!! Anh vẫn chưa biết danh tính thật của chú!! :?[/quote]
Vũ  Hưng, làm quả giả con gái để nhập nhằng oánh cả con đen thôi. Bên này này bác quen tụi sinh năm bao nhiêu? Nếu biết là h?c trư?ng nào thì thế nào em cũng lần ra ngay

hello VietNam;I live in Cuba;but i have many friends in Vienam;my nick name"kiemkhach" is name of my friend ;he is Vietnamese :D

[quote="vuhung"]
Vũ  Hưng, làm quả giả con gái để nhập nhằng oánh cả con đen thôi. Bên này này bác quen tụi sinh năm bao nhiêu? Nếu biết là h?c trư?ng nào thì thế nào em cũng lần ra ngay[/quote]

 Tụi sinh năm 83, h?c trư?ng nào thì chịu :P .
 Bác Kiếm Khách chắc lại bốc phét rồi. :D

Hi! ca nha` minh` la`thanh` vien moi rat vui duoc lam quen .

[quote="dickchimney"] Tụi sinh năm 83, h?c trư?ng nào thì chịu :P .
 Bác Kiếm Khách chắc lại bốc phét rồi. :D[/quote]

Vậy bạn co? biê?t

Hằng, Yên Ba?i
Tuâ?n Hùng, BKHN, câ?p III Yên Hòa,
Quang Duy, Hà Tây, Ba?ch Khoa HN cũ
Long, Bă?c Ninh, học ?H xây dựng HN cũ
Nhã, Huê?
Hà, Linh ?à Nẵng,
?ư?c Giang BK TPHCM
?ư?c Phương, BK TPCHM

Mình kể từ bă?c chi? nam rồi, không biê?t cô? kiê?m được người quen không?

lau lau post bai ma khong biet moi nguoi la ai
xin phep co y kien moi nguoi xung danh day du de tien noi chuyen
okie? :D

[quote="dickchimney"]Híc, lâu không vào đây nên không trả l?i bạn Hưng được, thất lễ quá :( .  Mình là mình biết có mỗi Hằng Yên Bái thôi  :D[/quote]
Em Hằng Yên Ba?i là của hiê?m bên này học Toa?n đâ?y Dick Chim ạ. Tụi kha?c toàn học mâ?y ngành ư?ng dụng thôi à.

Mình cũng 1, 2 lần đa?nh bo?ng bàn vơ?i em này. Kha? lă?m. I?t nhâ?t là vô địch bo?ng bàn nữa giải Lưu Học Sinh VN toàn Nhật Bản 2003! congratz.

[quote="vuhung"]
Em Hằng Yên Ba?i là của hiê?m bên này học Toa?n đâ?y Dick Chim ạ. Tụi kha?c toàn học mâ?y ngành ư?ng dụng thôi à.

Mình cũng 1, 2 lần đa?nh bo?ng bàn vơ?i em này. Kha? lă?m. I?t nhâ?t là vô địch bo?ng bàn nữa giải Lưu Học Sinh VN toàn Nhật Bản 2003! congratz.[/quote]

 Thế à, hồi trước em cứ tưởng bạn ý sang h?c điện tử viễn thông gì gì đấy!! Thế cũng tốt, con bé đấy thông minh lại yêu Toán, h?c chắc sẽ khá lắm!! :)
 Hê,  thế bác Hưng h?c gì mà em thấy bác chơi cả mấy bài v? tích phân Lebesgue ghê vậy :D ??
 P.S:Bác Hưng g?i nick em kiểu đấy thâm quá :D.

[quote="dickchimney"]
 Thế à, hồi trước em cứ tưởng bạn ý sang h?c điện tử viễn thông gì gì đấy!! Thế cũng tốt, con bé đấy thông minh lại yêu Toán, h?c chắc sẽ khá lắm!! :)
 Hê,  thế bác Hưng h?c gì mà em thấy bác chơi cả mấy bài v? tích phân Lebesgue ghê vậy :D ??
 P.S:Bác Hưng g?i nick em kiểu đấy thâm quá :D.[/quote]
Hi Dick Chim,

Tên ba?c như vậy thì cư? thê? mà đọc ra chư? mình đâu co? thâm nho nhọ chêm giề, khà khà.
Mâ?y ca?i bâ?t đẳng thư?c Lebesgue thì giải cho vui thôi, không phải là chuyên nghành của mình. Thực ra thì ca?ch đây 2 năm, mình cũng động  nhiều đều Sobolev Spaces nên cũng biê?t qua về ti?ch phân Lebesge -- thu? thực là chỉ biê?t qua thôi, vì không đọc no? thì không hiểu sâu được Sobolev Spaces.

Con be? Hằng a?, được đâ?y. Mình nhường cho cậu! Cô? lên nhe? không chị em ở Nhật này lâ?y chồng sơ?m lă?m.

[quote="vuhung"]
Tên ba?c như vậy thì cư? thê? mà đọc ra chư? mình đâu co? thâm nho nhọ chêm giề, khà khà.
Mâ?y ca?i bâ?t đẳng thư?c Lebesgue thì giải cho vui thôi, không phải là chuyên nghành của mình. Thực ra thì ca?ch đây 2 năm, mình cũng động  nhiều đều Sobolev Spaces nên cũng biê?t qua về ti?ch phân Lebesge -- thu? thực là chỉ biê?t qua thôi, vì không đọc no? thì không hiểu sâu được Sobolev Spaces.

Con be? Hằng a?, được đâ?y. Mình nhường cho cậu! Cô? lên nhe? không chị em ở Nhật này lâ?y chồng sơ?m lă?m.[/quote]

 Hè, em mạn phép đoán chắc là bác h?c điện điện tử gì đó thì mới phải h?c Sobolev Spaces, hehe. Bản thân em cũng chưa được h?c cái này!
 Bé Hằng lấy chồng hay chưa thì cũng chả đến lượt em, hic!!

[quote="dickchimney"]
Con be? Hằng a?, được đâ?y. Mình nhường cho cậu! Cô? lên nhe? không chị em ở Nhật này lâ?y chồng sơ?m lă?m.[/quote]

 Hè, em mạn phép đoán chắc là bác h?c điện điện tử gì đó thì mới phải h?c Sobolev Spaces, hehe. Bản thân em cũng chưa được h?c cái này!
 Bé Hằng lấy chồng hay chưa thì cũng chả đến lượt em, hic!![/quote]

Mình học tin, dick đoa?n xem mình học nghành hẹp nào mà động đê?n Sobolev Spaces :).

Con ga?i học toa?n thì tô?t nhâ?t ki?nh nhi viễn chi, nghĩa là tôn  trọng ca?c chị nhưng yêu đương thì đừng lolz

[quote="vuhung"]
Mình học tin, dick đoa?n xem mình học nghành hẹp nào mà động đê?n Sobolev Spaces :).

Con ga?i học toa?n thì tô?t nhâ?t ki?nh nhi viễn chi, nghĩa là tôn  trọng ca?c chị nhưng yêu đương thì đừng lolz[/quote]

 Bác h?c electro-comunication à, hay là information system??

[quote="dickchimney"][quote="vuhung"]
 Bác h�?c electro-comunication à, hay là information system??[/quote][/quote]
Mình học information technology :)

[quote="mathrus"]hello everybody!

lan dau vao` day choi, han hanh lam` quen![/quote]
Ha^n ha.nh ha^n ha.nh ^^

[quote="analysis"]minh` nguoi viet, xin chao` vu hung va` chao` ca nha`

cai hinh` nho? cua? vu hung nhin muot qua'[/quote]

  Ờ nhìn vào ai cũng nghĩ vuhung là một cô gái dễ thương  :blush:

.Cho hoi co phai anh Vuhung co phai la nguyen vu hung ngay xua rat noi tieng tren bao toan khong?

Cho em  hoi ,cac bac co biet mot so nguoi noi tieng ngay xua tren bao toan bay gio the nao khong.Vd:hoang ngoc minh(phu tho)tran tuan anh(khanh hoa)...

Hoang Ngoc Minh thi minh biet day , ong dang hoc Bach khoa HN va co 1 co ban gai cung la sinh vien Bach khoa .

Vay con Pham Gia Vinh Anh(HCV42/42 2001),Vu Ngoc Minh (HCV 2000,2001),Dao Hai Long(HCV 92,93),Ngo Dac Tuan(HCV 95,96).

em cung muon bik cac nguoi noi tieng do hien h dang la`m j ne` noi tieng ma`
 thanh tik cac noi bat thi` ca mu`ng


geometry
geometric transformation
reflection
circumcircle
geometry unsolved

(B.Frenkin) Construct the triangle, given its centroid and the feet of an altitude and a bisector from the same vertex.

[b]Notations: [/b]

$ ABC$ is the triangle, $ G,P,Q$ the given points respectively. $ P,Q$ are opposite $ A$
$ X.Y.Z\implies$ the points $ X,Y,Z$ lies on a line in that order where the order is required.

[b]Construction: [/b]

The following points are easily constructable with the properties with them.

$ D: P.D.G\ \&\ PG \equal{} 4GD$
$ E: P.Q.E\ \&\ DP \equal{} DE$
$ A: AP\perp PQ\ \&\ A.G.E$
$ F: A.Q.F\ \&\ FE\perp PQ$
$ S: SE\perp PQ\ \&\ SA \equal{} SF$
$ B,C: SA \equal{} SB \equal{} SC\ \&\ P.Q.B.C$

i have another construction  :D i'm not good at constructions, but i think this is easy... i use the same notation from the previous solution...

let $ G'$ be the foot of the perpendicular from $ G$ to line $ BC$ (or line $ PQ$, it's the same)... we know that $ GG' \equal{} \frac {1}{3}AP$, so we construct $ A$ in the same semiplane of $ G$ wrt line $ BC$, so that $ AP \equal{} 3GG'$. then we can construct the midpoint $ M$ of side $ BC$ and let $ l$ be the perpendicular line to $ BC$ passing through $ M$ (i.e, the perpendicular bissector). also, we have that the reflection of $ AP$ across $ AQ$ passes through the circumcenter, so, its intersection with $ l$ gives the circumcenter of triangle $ ABC$. from this, $ B$ and $ C$ are obtained by intersecting this circle and line $ BC$...


Does it have to do with multiplying by $2^{20}$ to get a 21 digit number?[/hide]

Oh, I remember this one was a Sprint.

Ah, this is a mean problem.  Easy with logs, but easy if there is an easy button.

Where do we find the easy button? (iow what's the easy solution)

[quote="Ignite168"]Determine the number of digits in the number $5^{20}$.[/quote]
[hide]
$5^{20}*2^{20}$ is $10^{20}$
$2^{20}$ is a little more than $10^{6}$
$\frac{10^{20}}{10^{6}}=10^{14}$
$5^{20}$ is a little less than $10^{14}$
so instead of 15 digits,
there are $14$ digits
[/hide]

[hide]
$5^{5}= 3125$
$3.125^{2}$ is a little bit less than $10$, so $(5^{5})^{2}$ is a little bit less than 10000000. Therefore $5^{2}0$ is a little less than $10^{1}4$, so it has 14 digits.
[/hide]

That was the button I was talking about good job.


if  j  is  chosen at random from thr set ( 4,5,6)  and k is chosen at random from the set (10,11,12), what is the probability that the product of j and k is divisible by 5 ?

the probability that the product will not be divisible by 5 is:
$ \frac {2}{3}$ x $ \frac {2}{3}$ = $ \frac{4}{9}$
Subtracting $ \frac {4}{9}$ from 1 gives the probability is $ \frac {5}{9}$


Suppose that [i]a[/i] and [i]b[/i] are two integers with GCD 1.
Prove that if [i]p[/i] is any odd prime which divides a^2 + b^2,
then p ≡ 1 (mod 4)

I have trouble solving this... anyone with some mouth-opening solutions?

assume $p=4k+3$
you know that $a^{2}=-b^{2}$ mod $p$ so $a^{4k+2}=-b^{4k+2}$ mod $p$
from fermats little theroem you have $a^{4k+2}=1$ and $b^{4k+2}=1$ mod $p$ because you know that neither $a$ nor $b$ are divisible by $p$ because if one is then so is the other and then $gcd(a,b) \geq p$
now you get $1=-1$ mod $p$ or $p|2$ which is not true...

[hide="Equivalently"] $a^{2}\equiv-b^{2}\bmod p \Leftrightarrow \left( \frac{a}{b}\right)^{2}\equiv-1 \bmod p \Leftrightarrow \left( \frac{-1}{p}\right) = 1 \Leftrightarrow p \equiv 1 \bmod 4$ [/hide]
[hide="Or"] If $p | a^{2}+b^{2}$ in the integers then $p | (a+bi)(a-bi)$ in the Gaussian integers.  Assume $p$ is prime in the Gaussian integers; then it divides either $a+bi$ or $a-bi$, but in fact if it divides one it divides the other, contradiction; hence $p$ is composite in the Gaussian integers, so some $c+di | p \implies p = c^{2}+d^{2}$ after a bit more work. [/hide]


geometry
geometric transformation
dilation

Well seemingly a normal person would write the following answers  :D

1. The answer would be $ \ce{MeC(OH)CD2CD2CD3}$

2. The answer would be $ \ce{PhC(OH)(CD2CD2CD3)C(Ph)D2)}$

But as u have asked this question, I feel

1. The answer could be $ \ce{MeC(OH)CHDCD2CD3}$  (maybe ir shows isotope effect or H shift)        (or) $ \ce{MeC(OD)CH2CD2CD3}$ :D

Is it right?

no also note that the solvent is $ \ce{H2O}$ which i forgot to give in the question

Erm a doubt  :maybe:  does this show isotope effect at all???  :)

definitely an acid base reaction will take place but it would also depend on the conc of the compound as h20 can also act as an acid(and wud preferably react if solution is dilute)


Let $a, b, c$ and $n$ be positive integers such that $(a +bc)(b + ac) = 5^n.$ Prove that $n$ is even.

:lol: Well, we can sopose that gcd(a,b)=1 and a>b (because it is clear fora=b). We write
a+bc=$5^x$ and b+ac=$5^y$. Easely we find that $5^x$ divides (a+b)(c+1) and (a-b)(c-1). But 5 cannot divide both (a+b) and (a-b), also 5 cannot divide c-1and c+1 at the same time. So we have to see two cases:
1) $5^x$ divides a+b and c-1. We have $5^x$=a+bc or $5^x$=a+b+(c-1)b. From here we get c=1 and a+b=$5^x$ and the result is clear.
2) $5^x$ divides a-b and c+1. We have $5^x$=a+bc or $5^x$=a-b+(c+1)b. From here we get a=b=1 and the result is clear


Put the equation in standard form for an ellipse

16x^2-32x+25y^2+50y=359

The answer I guessed on and was correct was 1/25(x-1)^2+1/16(y+1)^2=1

how would you get that answer step by step thanks...

You seem to have forgotten to divide by $400$ on the RHS  :o