(I posted this in the Introductory Physics forum too, but it might be more appropriate here)

These questions drove my physics professor nuts. Any help you can give would be greatly appreciated. Thanks!

Imagine fixing the position of one long, straight wire on a table top and placing another (free to move) long, straight wire on the table (parallel to the first). Put antiparallel currents in the wires. Then it is not hard to see that the magnetic force acting on the freely movable wire is directed away from the fixed wire, and hence this wire will move. Since W = F*D (this probably isn't the best form to use here, since the magnitude of force is variable, but its direction is constant), any displacement in the direction of the force will result in work being done. However, it is a fundamental fact that the magnetic force can do no work (as it always acts perpendicular to the velocity of a charged particles). How can the above argument be reconciled with this fact?

Also, my textbook (Serway and Jewett, Physics for Scientists and Engineers, page 846) mentions that utility companies transport energy at low currents and high potential differences (versus low potential differences and high currents) to minimize the I^2 R loss. In this case, R is fixed as the resistance of the wires in the powerlines. However, by Ohm's law the voltage is proportional to the current, so a high voltage seems to imply a high current (also, power can be expressed as V^2/R, so a high voltage should give a high power loss). Any ideas about what the text meant? Maybe they mean fix the resistance as high as possible, and crank up the voltage as well... ok really I have no idea.

And finally, how can two magnetic north poles facing one another repel? (Specifically, why is a superconductor expelled from a solenoid). Since the magnetic force is always perpendicular to the field lines, how can the force have radially repulsive component? I've been thinking this might have something to do with magnetic field fringing; if it does, is it difficult to calculate this force (i.e. find how much of the magnetic field is fringing)?

Once again, thanks for any help or insight you can provide.

Ok, for the first question. The magnetic field does not do work; its some "other" force that does the work.

We imagine the wire to be stiff and unable to bend. Now, there are two forces acting on the electrons. One force is due to the magnetic field (and perpendicular to the velocity). The second force is the force of the wire against the electron; the electron cannot move in a circular path because of this second force.

We now take a look at the wire (minus the electrons). This wire has no force due to the magnetic field. However, the wire exerts a force on the electrons (as stated above). Therefore, the electrons exert a force back on the wire (newton's law). This is the force which causes the wire to move. (This all happens because of the magnetic field, but, it is not the magnetic field itself doing the work).

Disclaimer: This could all be hopelessly incorrect.

Question 2 -
Consider the whole circuit where the power source, the place where power is needed (the transformer, I guess) and other resistors (the wire) in series. Now, as in a series circuit, to calculate the power lost at the wire, we need to know either the potential difference accross only the wire ($V_{wire}$) or the current through the wire($I$) if we know the resistance $R_{wire}$.
But the voltage you are thinking of decreasing is the voltage accross the whole circuit that contains the transformer too, which is the wrong approach.

Briefly, that is what I see the problem as. Perhaps, you can also figure out the mathematical details and show that a decrease in voltage would only increase power loss. I can but I am not really sure if I am complicating things.


calculus vector derivative function integration geometry

Usually Calc III refers to Multivariable Calculus.

yes it is mostly multivariable, along with rigorous polar transformation, and usually includes some vector analysis...

is AP calc that you take in HS considered calc I + calc II or just calc I?

AP Calculus BC is Calc I and Calc II. AB is just Calc I I would assume.

[quote="MysticTerminator"]Yeah, you're right. People from my high school who take AB Cal go on to Cal 2 afterwards. People who take BC Cal go on to Cal 3.[/quote]
which school?  Is it a big one?

um no not really...I go to CHCCS (Chapel Hill schools). I think they're supposed to be good. I don't really know.

Does anyone know what are differential equations used for?

Basically anything that involves rates of change..

very elementary calculus is taught in pre-calc, like derivatives of power functions, exponential, etc. and definite integrals using the midpoint method and stuff... im doing differentiation in calc right now and its much much harder than the pre-calc basics.

btw, is Advanced Calc supposed to be Calc III and multivariable? its just the calc after AB and BC

man, calc 3 is the hardest math class i've ever taken, ever...(the fact that i dont do homework doesnt help either)

on the last test (2 parts, each part 50 points), i got a 42% on part one....

Hehe...lucky ol' me.  I'm taking precalc!  It's so hard to keep awake in my class...zzzz.

-interesting_move

[quote="SPiKEzZ"]I think Calculus AB is I and BC is II with a little Calculus I. I think a bit of Calculus I is taught in Pre-Calculus. Calculus III I think is Multivariable Calculus, and I think after that is Differential Equations.[/quote]

Grr.  In my [stupid] precalc class, we never learned any calculus topics like limits or differentiation.  Precalc was basically a review of algebra I/II, geometry, and trig.


function
limit
induction
algebra open
algebra

Let $ k$ be a real number. Find all functions $ f:\mathbb{R}\to\mathbb{R}$ such that $ f$ is continuous on $ \mathbb{R}$ and $ f(x) \equal{} f(x^{2}\plus{}k)\;\forall x\in\mathbb{R}$.

Note: If $ k > 0$ then we have a problem in Putnam.

If $ x^{2}\plus{}k\equal{}x$ had roots $ z_{1}\le z2$, then for $ x\equal{}x_{0}>z_{2}$ we can bound sequence $ x_{n\plus{}1}^{2}\plus{}k\equal{}x_{n}\to z_{2}<x_{n\plus{}1}<z_{n}$.
Because $ \lim x_{n}\equal{}z_{2}$ and f(x) continiosly, we get $ f(x_{0})\equal{}f(x_{1})\equal{}...\equal{}f(z_{2})$. By anologue $ f(x_{0})\equal{}f(z_{2})$ if $ z_{1}<x_{0}<z_{2}$ and $ f(x_{0})\equal{}f(z_{1})$ if $ x_{0}<z_{1}$, therefore $ f(x)\equal{}const$.

If that equation hasn't real root  then...?  :maybe:

In these case ($ k>\frac{1}{4}$) for any continiosly $ g(x): [0,k]\to R$ with $ g(0)\equal{}g(k)$ we can construct $ f(x)$, suth that $ f(x)\equal{}g(x)\ \forall x\in[0,k]$.

With $ 0<k <\frac{1}{4}$

Because $ f(x) \equal{} f(\minus{}x)$ so we must consider $ x\ge 0$.

The equation $ x^{2}\minus{}x\plus{}k \equal{} 0$ has two roots $ 0\le a_{1}< a_{2}$.

For any $ a > 0$,consider the sequence $ (x_{n}): x_{1}\equal{} a,x_{n}\equal{} x_{n\minus{}1}^{2}\plus{}k$

a,$ a_{1}< a < a_{2}$.By induction we have $ a_{1}< x_{n}< a_{2}$ and $ (x_{n})\searrow$

b,$ 0 < a < a_{1}\Rightarrow 0 < x_{n}< a_{1}$ and $ (x_{n})\nearrow$

c,$ a > a_{2}\Rightarrow x_{n}> a_{2},x_{n}\searrow$

Then for all cases $ \exists\lim_{n\to\infty}(x_{n}) \equal{} b$ ($ b \equal{} a_{1}$ or $ a_{2}$) so $ f(a) \equal{} f(b)$

Hence $ f(x) \equal{}\text{constant},\forall x\in\mathbb{R}$

$ k \equal{}\frac{1}{4}\to$ VNTST 1982

Different cases, i think it is simirlary  :blush:

[code]edited[/code]

[quote="quangpbc"]

Different cases, i think it is simirlary  :blush:[/quote]
I don't think that, you can find in

Functional Equations and How to Solve Them (Problem Books in Mathematics)
by Christopher G. Small
with case $ k>0$.

I will trying solve the last cases ...  :arrow: Can you share this book for me, N.T.TUAN  :D

[quote="Rust"]If $ x^{2}\plus{}k \equal{} x$ had roots $ z_{1}\le z2$, then for $ x \equal{} x_{0}> z_{2}$ we can bound sequence $ x_{n\plus{}1}^{2}\plus{}k \equal{} x_{n}\to z_{2}< x_{n\plus{}1}< z_{n}$.
Because $ \lim x_{n}\equal{} z_{2}$ and f(x) continiosly, we get $ f(x_{0}) \equal{} f(x_{1}) \equal{} ... \equal{} f(z_{2})$. By anologue $ f(x_{0}) \equal{} f(z_{2})$ if $ z_{1}< x_{0}< z_{2}$ and $ f(x_{0}) \equal{} f(z_{1})$ if $ x_{0}< z_{1}$, therefore $ f(x) \equal{} const$.[/quote]

I think your solution is right with $ 0<k<\frac{1}{4}$ and you must consider three cases as my solution  :)


[quote="flip2004"][b] How many real  roots has the equation  $ \displaystyle  \sum\limits_{k=0}^{n}{n\choose k}^p X^k = 0 \: \; \; ,\; \;  p \in {\mathbb N}^*  $ ? [/b]   [/quote]
The answer is that for all  $  p \in {\mathbb N }^{*} $ the equation has only real roots. The solution is based on following [i] nice [/i]  proposition :
[b] Theorem.[/b] [i]Suppose that the polynomials (with real coefficients)
\[\left\{\begin{array}{lcl}
A(X)&=&a_0X^n+a_1X^{n-1}+....+a_{n-1}X+a_n  \\
  B(X)&=&b_0X^n+b_1X^{n-1}+....+b_{n-1}X+b_n
\end{array}\right.
\] have only real roots. If $B(x) $ has all roots positive (or negative),  then the polynomial
\[F(x)=a_0b_0X^n+a_1b_1X^{n-1}+....+a_{n-1}b_{n-1}X+a_nb_n  \]
has also only real roots.[/i]
[b]I do'nt remember whom belongs this theorem, (please any reference).[/b]
In our case $ a_k={n\choose k} $ , $b_k={n\choose k}^{q-1} \; ,\; q\in \{2,3,...\},$ and  $ A(X)=\sum\limits_{k=0}^{n}{n\choose k}X^k=(1+X)^n \; .$

[quote="flip2004"][quote="flip2004"][b] How many real  roots has the equation  $ \displaystyle  \sum\limits_{k=0}^{n}{n\choose k}^p X^k = 0 \: \; \; ,\; \;  p \in {\mathbb N}^*  $ ? [/b]   [/quote]
The answer is that for all  $  p \in {\mathbb N }^{*} $ the equation has only real roots. The solution is based on following [i] nice [/i]  proposition :
[b] Theorem.[/b] [i]Suppose that the polynomials (with real coefficients)
\[\left\{\begin{array}{lcl}
A(X)&=&a_0X^n+a_1X^{n-1}+....+a_{n-1}X+a_n  \\
  B(X)&=&b_0X^n+b_1X^{n-1}+....+b_{n-1}X+b_n
\end{array}\right.
\] have only real roots. Then the polynomial
\[F(x)=a_0b_0X^n+a_1b_1X^{n-1}+....+a_{n-1}b_{n-1}X+a_nb_n  \]
has also only real roots.[/i]
[b]I do'nt remember whom belongs this theorem, (please any reference).[/b]
In our case $ a_k={n\choose k} $ , $b_k={n\choose k}^{q-1} \; ,\; q\in \{2,3,...\},$ and  $ A(X)=\sum\limits_{k=0}^{n}{n\choose k}X^k=(1+X)^n \; .$[/quote]
Wow,extremely nice theorem,do you have a proof?


The first three towers in a sequence are shown. The $ n$th tower is formed by stacking $ n$ blocks on top of an $ n \times n$ square of blocks. How many blocks are in the 99th tower?

[asy]pair A;
for(int i = 1; i<4; ++i){
A = ((i-1)*(i+4)/2,0);
for(int k = 0; k <= i; ++k){
draw(A+(k,0)--A+(k,i)); draw(A+(0,k)--A+(i,k));
draw(A+(i-1,i+k)--A+(i,i+k));
}
draw(A+(i-1,i)--A+(i-1,2*i));draw(A+(i,i)--A+(i,2*i));
}[/asy]

If the nth tower is formed with an n by n square and n more blocks, it must contain n^2+n=n(n+1) blocks. Since n is 99, there are 99(100)=9900 blocks.


My idea is that for people here that have microphones, the winner of the previous round types something (no longer than a short paragraph) that the other people need to say. People will submit their voices and there will be a poll of which one is the funniest, and the one with the most votes wins. This would be very similar to the contests in the graphics and arts design section. I would post an example, but they're all on my brother's computer, which he is using (I can definitely post up a funny example by monday).

Anyone interested in this (I probably won't have time to organize this once school starts, though)?

It will be once I get a chance to post my examples  :D

how do you submit your voices? record it and attach it to a file?


Is there a closed form for $\prod^{n}_{k=0}\binom{n}{k}$?

[quote="tjhance"]that isn't close form thought is it?  :maybe:[/quote]
No it isn't.

He is implying that it can't be simplified, I believe.

$\Big ( \prod_{k=0}^{n}\frac{1}{k!}\Big ) = \Big ( \prod_{k=0}^{n}\frac{1}{(n-k)!}\Big ) \Rightarrow \prod_{k=0}^{n}\binom{n}{k}= (n!)^{(n+1)}\Big (\prod_{k=0}^{n}\frac{1}{k!}\Big )^{2}$

Slightly more concise, but I don't think you can get the product of factorials in closed form.

i just show that dont have a explicit formula to simplify this


geometry
perimeter
inequalities
triangle inequality

Let $ABC$ be a triangle such that $AB < AC$.

(a) Prove that for every point $M \in (AB)$ there is only one point $N \in (AC)$ such that the perimeter of $AMN$ is equal to $AB$.

(b) Prove that there is at least one point $N \in (AC)$ such that, for all $M \in (AB)$, the perimeter of $AMN$ is $> 2 AB$.

[i]Dan Popescu[/i]

[hide="(a)"]Assume for the sake of contradiction that for some $M\in (AB)$ there are two different points $N_{1},N_{2}\in (AC)$ with $N_{1}$ on $\overline{AN_{2}}$ such that the respective perimeters of $AMN_{1}$ and $AMN_{2}$ are equal to $AB$.  Then the perimeter of $AMN_{1}$ is equal to the perimeter of $AMN_{2}$.  Therefore, $AN_{1}+MN_{1}=AN_{2}+MN_{2}$.  Since $AN_{2}=AN_{1}+N_{1}N_{2}$, we have $MN_{1}=N_{1}N_{2}+MN_{2}$.  But by the Triangle Inequality, we have $MN_{1}<N_{1}N_{2}+MN_{2}$, a contradiction.  Therefore, for every $M\in (AB)$ there is only one point $N\in (AC)$ such that the perimeter of $AMN$ is equal to $AB$.[/hide]

[hide="(b)"]Suppose we have two different points $D_{1},D_{2}\in (AB)$ with $D_{1}$ on $\overline{AD_{2}}$.  By the Triangle Inequality, we have $ND_{1}<ND_{2}+D_{1}D_{2}$.  Since $AD_{1}+D_{1}D_{2}=AD_{2}$, we have $AD_{1}+ND_{1}<AD_{2}+ND_{2}$.  Adding $AN$ to both sides of this inequality, we obtain $AN+AD_{1}+ND_{1}<AN+AD_{2}+ND_{2}$, so the perimeter of $AD_{1}N$ is less than the perimeter of $AD_{2}N$.  Therefore, if the perimeter of $ABN$ is greater than $2AB$, then for all $M\in (AB)$, the perimeter of $AMN$ is greater than $2AB$.

Let $X$ be on $\overline{AC}$ such that $AB=AX$.  $X$ exists because $AC>AB$.  If $N$ is on $\overline{CX}$, then the perimeter of $ABN$ is $AB+AN+BN>AB+AX+BN=2AB+BN>2AB$.  Therefore, the perimeter of $AMN$ is greater than $2AB$ for all $M\in (AB)$.[/hide]

[quote="perfect_radio"][hide="Remark regarding (a)"]Don't forget to mention why $N$ actually exists.[/hide][/quote]

I don't think $N$ necessarily exists.  If $M$ is closer to $B$ than it is to $A$, then the perimeter is always greater than $AB$.

Here is my solution for part (a):

Assuming for the sake of contradiction that there are two such N, N1 and N2. Then AB = AM + MN1 +N1A = AM + BM, so MN1 + N1A = BM. Then you have AB < AC, so AB = AM + MN1 + N1A and AB = AM + MN2 + N2A and AC = AN1 + N1C but AC = AN2 + N2C as well, so you have two inequalities:

AM + MN1 + N1A < N1A + N1C and

AM + MN2 + N2A < N2A + N2C

Now you cancel and get:

AM + MN1 < N1C and

AM + MN2 < N2C

When you subtract, the two, you will get:

AM + MN1 - AM - MN2 < N1C - N2C

Simplify to get MN1 - MN2 < N1C - N2C

Now, you have MN1 + N1A = BM, so you add N1A to each sides and subtract N2A from each sides and get:

(MN1 + N1A) - (MN2 + N2A) < (N1C + N1A) - (N2C + N2A)

and you have BM - BM < CA - CA which would mean 0 < 0 which is a contradiction. There there is only one N.

I don't know how to prove that there actually exists a N though. I agree with matt276eagles that there does not necessarily exits N for each M.


题目：有一个六位数，当分别用2、3、4、5、6乘它后，得到的数还是个六位数，而且是原来的六个数字组成，仅仅是改变了原来的数字顺序。
问：是哪六个数字

要求给出结果和过程

希望有多种不同解法！

设它们由小到大依次是A，B，C，D，E，F。
A=(abcdef),则a=1(否则F=6A>=6*200000>999999)
显然，A、B、C、D、E、F首位各不相同，所以组成X(X为ABCDEF中任意一个）的6个数字各不相同，分别为A，B，C，D，E，F的首位。于是f不等于1。
若f为偶数，则E=5A,E末位为0，但A，B，C，D，E，F的首位均不为0，矛盾
若f=5，则B=2A，B末位为0，但A，B，C，D，E，F的首位均不为0，矛盾
若f=3，则ABCDEF末位依次为3、6、9、2、5、8，各不相同，所以A由3、6、9、2、5、8组成，但A中有a=1，矛盾
若f=9，则ABCDEF末位依次为9、8、7、6、5、4，各不相同，所以A由9、8、7、6、5、4组成，但A中有a=1，矛盾
所以f=7，所以ABCDEF末位依次为7、4、1、8、5、2。
所以A由1、2、4、5、7、8组成，且a=1，f=7。
接下来可以用枚举法，也可以通过讨论依次推出e、d、c、b的值。
最后可以得到：A=142857，B=285714，C=428571，D=571428，E=714285，F=857142

That's a nice question!!(Pardon me but something is wrong and I can't type in Chinese now..)


function trigonometry limit algebra domain topology calculus

$lim_{(x,y)\rightarrow(2,1)}\frac{sin^{-1}(xy-2)}{tan^{-1}(3xy-6)}$

What??? Okay, I get that if I can prove that it is continous at (2,1) that the function's limit = the function, but the function is undefined at (2,1). I believe (due to taking small intervals approaching that point) that the answer is zero but how do I prove that?

Thanks!

The limit is $\frac13.$

Let $u=xy-2.$ Then we're looking at $\frac{\sin^{-1}u}{\tan^{-1}(3u)}.$

You can show (by power series or L'Hôpital's rule, or by a couple of other means) that

$\lim_{u\to 0}\frac{\sin^{-1}u}{\tan^{-1}(3u)}=\frac13.$

Then, $u(x,y)=xy-2$ is a continuous function whose value at $(2,1)$ is zero. Theorems about composing limits with continuous functions then lead us to

$\lim_{(x,y)\to(2,1)}\frac{\sin^{-1}(2x-1)}{\tan^{-1}(3xy-6)}=\frac13.$

I had a feeling if I found:
$\lim_{u\to 0}\frac{\sin^{-1}u}{\tan^{-1}(3u)}$
That I would get the answer...thanks for reminding me about L'Hospital's.

There's another consideration here. The function $f(x,y)=\frac{\sin^{-1}(xy-2)}{\tan^{-1}(3xy-6)}$ is not defined on a (punctured) neighborhood of $(2,1).$ The function fails to be defined whereever $xy-2=0,$ which is a curve that passes through $(2,1).$

What we do about it depends on what exact definition you have adopted for the meaning of limit.

I prefer the following: Suppose $D$ is a subset of metric space $X$ and $f: D\mapsto Y$ (where $Y$ is another metric space.). Then I'm willing to say $\lim_{x\to y}f(x)=L$ provided the following holds:

1. $y$ is a limit point of the domain $D,$ and

2. $\forall\,\epsilon>0,\,\exists\,\delta>0$ such that if $x\in D$ and $0<d(x,y)<\delta,$ then $d(f(x),L)<\epsilon.$

If you're willing to accept this as your working definition of limit, then you will accept my contention that the limit above is $\frac13.$

But if your definition of limit omits the "$x\in D$" in #2, thereby implicitly requiring that $f(x)$ be defined for all $y$ such that $0<d(x,y)<\delta,$ then you would not accept this situation as the existence of a limit.

A good portion of  what your second post contained went right over my head...however, my textbook allows a limit of 2 variables to be defined if the values approaching those points are defined when really close to that point but not exactly that point and L'Hospitals gives 1/3.

[quote]... if the values approaching those points are defined when really close to that point but not exactly that point ...[/quote]
That's what my second post was about: does "when really close" mean [b]every[/b] "really close" point, or just all of those really close points that are actually in the domain of the function? That's the difference between two possible meanings of limit.


What are some maths jobs for BSc gradutes  that recquire minimal direct personal contact? online tutoring is one. Any others?

What's the motivation for the lack of personal contact ?
It makes quite a difference in the options out there.

Language issues would be handled differently to mobility.


Let  N = 1234567898765432AB ,  where A and B are digits, possibly equal.  Of the many different possible values of N, HOW MANY are divisible by 18?

A.Two	B.Three	C.Four	D.Five E.	Ten

Easy Question
For N to be divisible by 18
B should be even, sum of digits should be divisible by 9.

so the possible values of A,B are:
(1,0)
(2,8)
(4,6)
(6,4)
(8,2)

So the answer is [b]five[/b].


function
real analysis
real analysis unsolved

Give an example of a function $f: \Re\to\Re$ that is continuos in every point of$\Re - \mathcal l Q$ and discontinous in every point of $\mathcal l Q$.

$f(x) = 0$ if $x\in R\setminus Q$ and $f(x) = \frac{1}{q}$ if $x=\frac{p}{q}$ where $p,q \in Z$ and $(p,q)=1$.

that`s what I found too.... do you know any other function that is in accordance with the hypothesis?


Three circumferences have radii $ r,r_1,R$ and are tangent externally, two by two. The internal common tangent of the first two circumferences intercept the circumference with radius $ R$ in points $ A$ and $ B$ . Calculate the length of the chord $ \overline{AB}$

I have indicated a solution using pythagoras, but the amount of algebraic manipulation is enormous.

I was wondering if there was a simpler solution than mine that reduces the amount of manipulation necessary. The answer seems simple, there must be a way...

[hide="Answer"]$ \frac {4R\sqrt {rr_1}}{r \plus{} r_1}$[/hide]

Denote the circles $ (X,r_1), (Y,r_2), (O,R).$ Circles $ (X), (Y)$ are externally tangent at $ I,$ circles $ (X), (O)$ at $ J$ and circles $ (Y), (O)$ at $ K.$ Inversion with center $ I$ and power $ \varrho^2 \equal{} 4r_1r_2$ (where $ \varrho$ is the inversion circle radius) takes $ (X), (Y)$ to parallel lines $ x, y \perp XY,$ such that distance of $ I$ from $ x, y$ is $ 2r_2, 2r_1,$ respectively. Circle $ (O)$ goes to a circle $ (O', r_1 \plus{} r_2)$ tangent to $ x, y$ at $ J', K'.$ The common internal tangent of $ (X), (Y)$ at $ I$ cuts $ (O)$ at $ A,B$ and $ (O')$ at $ A',B'.$ It cuts $ J'K'$ at the midpoint $ M'$ of $ A'B'.$ Power of $ M'$ to $ (O')$ is $ M'A' \cdot M'B' \equal{} M'J' \cdot M'K'$ $ \Longrightarrow$ $ A'B' \equal{} 2 \sqrt {2r_2 \cdot 2r_1} \equal{} 4 \sqrt {r_1r_2}.$ The inversion center $ I$ is similarity center of $ (O), (O')$ $ \Longrightarrow$

$ AB \equal{} A'B' \cdot \frac {R}{r_1 \plus{} r_2} \equal{} \frac {4R \sqrt {r_1r_2}}{r_1 \plus{} r_2}.$


algebra
polynomial
modular arithmetic
induction
number theory unsolved
number theory

Since the multiplicative order of $ 2$ modulo $ 7^n$ is $ 3\cdot 7^{n\minus{}1}$, we have that
i) $ 2^{7^n}$ is a zero of the cyclotomic polynomial $ \Phi_3(x)\equal{}x^2\plus{}x\plus{}1$ modulo $ 7^{n\plus{}1}$;
ii) $ 2^{7^n}$ is a zero of the polynomial $ f(x)\equal{}\Phi_3(x)\Phi_{21}(x)\equal{}x^{14}\plus{}x^7\plus{}1$ modulo $ 7^{n\plus{}2}$.

Lemma. $ 2^{7^n}$ is a zero of the polynomial $ g(x)\equal{}\frac{(1\plus{}x)^7\minus{}1\minus{}x^7}{7}\equal{}x^6 \plus{} 3x^5 \plus{} 5x^4 \plus{} 5x^3 \plus{} 3x^2 \plus{} x$ modulo $ 7^{n\plus{}2}$.

Proof. The statement directly follows from Bezout identity for the polynomials $ f(x)$ and $ g(x)$:
\[ (6x^2 \plus{} 6x \plus{} 7)\cdot f(x) \plus{} (\minus{}6x^{10} \plus{} 12x^9 \minus{} 13x^8 \plus{} 9x^7 \minus{} 4x^6 \plus{} 2x^5 \minus{} 4x^4 \plus{} 2x^3 \plus{} x^2 \minus{} 2x \plus{} 1)\cdot g(x) \equal{} 7\cdot \Phi_3(x).\]

Theorem.
\[ 3^{7^{n}} \equiv 2^{7^{n}} \plus{} 1 \pmod{7^{n \plus{} 2}}.\]

Proof by induction on $ n$.

For $ n\equal{}0$, the statement is trivial. Suppose it holds for some $ n$.
Then
\[ 3^{7^{n\plus{}1}} \equiv (2^{7^n}\plus{}1)^7\equiv 2^{7^{n\plus{}1}} \plus{} 1 \plus{} 7\cdot g(2^{7^n})\pmod{7^{n\plus{}3}}.\]
By Lemma, we have
\[ 7\cdot g(2^{7^n})\equiv 0\pmod{7^{n\plus{}3}}\]
implying that
\[ 3^{7^{n\plus{}1}} \equiv 2^{7^{n\plus{}1}} \plus{} 1 \pmod{7^{n\plus{}3}}\]
that completes the proof.

[quote="maxal"]Since the multiplicative order of $ 2$ modulo $ 7^n$ is $ 3\cdot 7^{n \minus{} 1}$, we have that
i) $ 2^{7^n}$ is a zero of the cyclotomic polynomial $ \Phi_3(x) \equal{} x^2 \plus{} x \plus{} 1$ modulo $ 7^{n \plus{} 1}$;
ii) $ 2^{7^n}$ is a zero of the polynomial $ f(x) \equal{} \Phi_3(x)\Phi_{21}(x) \equal{} x^{14} \plus{} x^7 \plus{} 1$ modulo $ 7^{n \plus{} 2}$.

Lemma. $ 2^{7^n}$ is a zero of the polynomial $ g(x) \equal{} \frac {(1 \plus{} x)^7 \minus{} 1 \minus{} x^7}{7} \equal{} x^6 \plus{} 3x^5 \plus{} 5x^4 \plus{} 5x^3 \plus{} 3x^2 \plus{} x$ modulo $ 7^{n \plus{} 2}$.

Proof. The statement directly follows from Bezout identity for the polynomials $ f(x)$ and $ g(x)$:
\[ (6x^2 \plus{} 6x \plus{} 7)\cdot f(x) \plus{} ( \minus{} 6x^{10} \plus{} 12x^9 \minus{} 13x^8 \plus{} 9x^7 \minus{} 4x^6 \plus{} 2x^5 \minus{} 4x^4 \plus{} 2x^3 \plus{} x^2 \minus{} 2x \plus{} 1)\cdot g(x) \equal{} 7\cdot \Phi_3(x).\]
Theorem.
\[ 3^{7^{n}} \equiv 2^{7^{n}} \plus{} 1 \pmod{7^{n \plus{} 2}}.\]
Proof by induction on $ n$.

For $ n \equal{} 0$, the statement is trivial. Suppose it holds for some $ n$.
Then
\[ 3^{7^{n \plus{} 1}} \equiv (2^{7^n} \plus{} 1)^7\equiv 2^{7^{n \plus{} 1}} \plus{} 1 \plus{} 7\cdot g(2^{7^n})\pmod{7^{n \plus{} 3}}.\]
By Lemma, we have
\[ 7\cdot g(2^{7^n})\equiv 0\pmod{7^{n \plus{} 3}}\]
implying that
\[ 3^{7^{n \plus{} 1}} \equiv 2^{7^{n \plus{} 1}} \plus{} 1 \pmod{7^{n \plus{} 3}}\]
that completes the proof.[/quote]
I'm sorry but I didn't understand your solution with polynomial. Can you explain it clearly?
THX

[quote="maxal"][b]litongyang[/b], what particular point is not clear?[/quote]
Like this:
 $ 2^{7^{n}}$is a zero of the cyclotomic polynomial $ \Phi_{3}(x) \equal{} x^{2}\plus{}x\plus{}1$ modulo $ 7^{n\plus{}1}$

[quote="litongyang"]Like this:
 $ 2^{7^{n}}$is a zero of the cyclotomic polynomial $ \Phi_{3}(x) \equal{} x^{2} \plus{} x \plus{} 1$ modulo $ 7^{n \plus{} 1}$[/quote]
That follows from the facts:
\[ 2^{7^n}\not\equiv 1\pmod{7},\]

\[ 2^{3\cdot 7^n}\equiv 1\pmod{7^{n \plus{} 1}},\]
and
\[ \Phi_{3}(x) \equal{} \frac {x^3 \minus{} 1}{x \minus{} 1}.\]
So, if we plug $ x \equal{} 2^{7^n}$ into $ \Phi_{3}(x)$, it evaluates to zero modulo $ 7^{n \plus{} 1}$.

[quote="123eeee"]In fact, $ g(x) \equal{} \frac {(1 \plus{} x)^{7} \minus{} 1 \minus{} x^{7}}{7} \equal{} x^{6} \plus{} 3x^{5} \plus{} 5x^{4} \plus{} 5x^{3} \plus{} 3x^{2} \plus{} x \equal{} x(x \plus{} 1)\Phi_{3}^{2}(x)$[/quote]
Indeed. That proves that $ 2^{7^n}$ is a zero of $ g(x)$ modulo $ 7^{2n\plus{}2}$, i.e., even a stronger statement than given in Lemma.


Let $a,b,c$ be three positive reals. Prove that
$\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq\frac{3}{\sqrt[3]{abc}(1+\sqrt[3]{abc})}$
Is there a nice proof without many computations?

Yes, I remember to have solved it quite a long time ago: http://www.mathlinks.ro/Forum/viewtopic.php?t=6044 .

Thanks. One of my teammates has posted it 4 months ago.
I am interested in the inequality.
$ \frac{1}{a_{1}(1+a_{2})}+...+\frac{1}{a_{n-1}(1+a_{n})}+\frac{1}{a_{n}(1+a_{1})}\geq\frac{n}{\sqrt[n]{a_{1}...a_{n}}(1+\sqrt[n]{a_{1}...a_{n}})}$, where $ a_{1},...,a_{n}$ are positive reals.
I've checked it by a program(just test a lot of cases :D ) and I believe the inequality is right. But how to prove it?

P.S should I put it in the Unsolved problems section?

I do not think that this inequality is always true. For example, if it is true, than by taking the product of the n numbers 1 we find that Shapiro's inequality is true for all n, which is certainly false. But there are many many other questions about this: if we impose some restrictions on the product, what happens? Or, if we consider small cases for n? For any of these questions I do not have an answer no (or never!), but the topic is certainly too interesting to be quitted. So, let's concentrate...

What is even nicer is that the cyclic inequality in three variables darksheer posted has a solution using Cauchy! Try to find it... I just love Cauchy's inequality!

[quote="harazi"]I do not think that this inequality is always true.[/quote]
You're right. I am sorry that I've made a mistake in my program. And the sequence {4, 1, 4, 1,...,4, 1} is always a counterexample for every even integers n.

But I think the following inequality holds

$(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n})(\frac{1}{1+a_1}+\frac{1}{1+a_2}+...+\frac{1}{1+a_n})\geq \frac{n^2}{r(1+r)}$,

where $a_i$  are positive numbers and $r=\sqrt[n]{a_1a_2...a_n}$.


geometry
circumcircle
incenter
perpendicular bisector
angle bisector

Let $ ABC$ be a triangle,$ D$ is a point in $ AC$ such that $ BD\equal{}BA$
$ I$ and $ J$ is centre of incircles of triangle $ ABC$ and $ BDC$
Prove that $ AJ \perp BI$

Wow this is way easier than I thought...

It is easy enough to see that $ I$ is the intersection of the perpendicular bisector of $ BD$ and the angle bisector of $ \angle BCD$, so we have a well-known configuration with triangle $ BCD$ and point $ I$ as the midpoint of arc $ BD$ of the circumcircle of that triangle. In particular, we have $ ID\equal{}IJ$. Also since $ AB\equal{}BD$, $ I$ is an incenter, $ AI\equal{}ID$, so we have $ AI\equal{}JI$, so the perpendicularity is obvious.

[quote="Altheman"]Wow this is way easier than I thought...

It is easy enough to see that $ I$ is the intersection of the perpendicular bisector of $ BD$ and the angle bisector of $ \angle BCD$, so we have a well-known configuration with triangle $ BCD$ and point $ I$ as the midpoint of arc $ BD$ of the circumcircle of that triangle. In particular, we have $ ID \equal{} IJ$. Also since $ AB \equal{} BD$, $ I$ is an incenter, $ AI \equal{} ID$, so we have $ AI \equal{} JI$, so the perpendicularity is obvious.[/quote]
I don't understand.I think $ I$ isn't the intersection of the perpendicular bisector of $ BD$ and the angle bisector of $ \angle BCD$

According to my dynamic Sketch in Geogebra this problem is not true  :wink: . AJ is not perpendicular to BI , Would you mind  checking the problem again Obama ?

[geogebra]f268c0530abaf1d8b52c51f4922bbc3287ccb83a[/geogebra]


Putnam
Stanford
college
algebra
polynomial
linear algebra
matrix

So, what did you think of it? Was it easy? Hard? Mixed? More or less as you expected? Surprising and strange?

I think I did better relatively on that than I ever did on AMC or AIME. I'm not too happy with what I did (especially on section A), but it certainly could have been much worse. Of course, I won't know for sure until March, but I don't think I did too much worse than I hoped. Not likely that I did much better, either, though.   :|

From preliminary impressions, the A problems were harder than most years. You're not alone in finding them difficult. The B problems were more normal, including the one truly easy problem on this year's test.

I thought it was more difficult than usual.  I got complete, well-written solutions to A1,A5,B1,B2, and assorted sketchy solutions that I'm not expecting more than a couple of marks from.

In retrospect, A2 was definitely doable (probably the easiest question on the A paper) but I never got to it, as I spent roughly two hours on A1 (and roughly 20 minutes on A5, go figure).  A3 also seemed doable, as well - I made some progress on it, and then decided to go back to A1, reasoning that I must've missed something trivial.

As for the B paper, I blew through B1, B2, made a rather dumb mistake on B3, and completed a significant portion of B4.  I should have at least been able to finish B4, given that I finished B1 and B2 within an hour.

Given the (self assessed) quality of my solutions to A1,A5,B1,B2, I'm hoping for around 40, and I would imagine that that would be enough for top 200 this year.  Maybe I should send the markers Christmas presents, or something.

I don't really have anything to compare it to, but I did find the A section kind of hard, and the B section kind of easy.  I solved A1, A2, B1, B2, and B4.  The rest of A looked very difficult, except A3, and A3 only [i]looked[/i] approachable (it seems that it was the only problem that no one at Stanford solved).  B3 was not too bad; after hearing the solution, it sounded like something I might have come up with had I had more time.  I've barely looked at B5 or B6.

Ravi Vakil, who runs Putnam here, thought that with the exception of B1, the test overall was somewhat harder than usual.  We all agreed that B1 was pretty easy.  So it will be interesting to see how that affects the statistics.

This was my first Putnam, but I think it was harder than it has been  in recent years (based on trying the old exams these last few weeks). There is no question that the first half was the more difficult one and that B1 was the easiest problem on the exam. I almost finished A5 when I ran out of time, so I'll make sure to watch the clock more next year. It should be interesting to see the score distribution.

I did A1, A2, B1, B2, B4

but I made a really stupid mistake on B1  :blush: I left out coefficients of 2 when writing the solution and never went back to check because of how easy it was...lol.

While sitting and proctoring, I managed A1, A2, B1, B2, and B4. I could see that the the answer to B3 had to be $f(x)=cx^r$ for any $r,$ but I didn't have a convincing proof that that was all solutions.

I spent a lot of time banging my head against B5 (the harmonic polynomial) with nothing to show for it.

I thought the whole A section was unusually difficult. The very quick B1 and the relatively easy B2 partly made up for that. I really don't know what to predict about scores.

Just wanted to let you guys know that I've inserted the problems in the contest/resources section: http://www.mathlinks.ro/Forum/resources-2-23-2005.html in case you want a .pdf containing all the problems.

My goal for the test was to not get 0. I got the B1 answer but maybe not rigourous enough. I shouldn't get zero though. I got 0 questions from A. I tried to use Abstract Alegbra for B6 but I didn't realize it was a matrix but I was surprised how well it worked using transpositions. I'm not very good at Abstract Algebra considering I only got 53 and 57 on my last two tests but I used it on the Putnam which was pretty neat. I'm pretty sure I made my goal of not getting zero so that means this year's putnam was a success.

Well I'm a highschool student but just doing the problems at home I thought that the A section was quite a bit harder than the B section although I was able to solve A1, A2, I didn't think A3 was terrible although it certainly was not easy, and had seen A5 on the forum earlier (as well as in a book of mine). On the B section I got B1, B2, B4 and I had some ideas on B3 but I don't think I would have gotten the credit.

Ack. I didn't manage a single calculus question. Got 1,2,4 on both parts though. When I looked at A3, I muddled around a bit and figured I didn't know what I was doing, so I skipped it. After I finished A4, I took one look at A5 and A6 and figured that muddling through A3 was probably a better idea. Of course, I didn't get anything after that. I managed to finish B1, 2, and 4 with an hour to spare, but the remaining three questions looked practically hopeless. I don't even know enough math to attempt some of these questions...

I just hope I manage to really nail those questions that I did. Argh. It's going to be some time till I can tell if I messed up on any question. It'll only be out in several months' time right?

[quote="Yossarian"]It'll only be out in several months' time right?[/quote]
Some time around March 10 to March 15 seems to be typical. Official results for your school and summaries and national listings are mailed to the faculty member who supervises the Putnam at your college; how much he chooses to tell you, and by what means, is up to him.

Wow. That's a long time.
I sure hope I get full or almost full credit for the questions I did. It's my first Putnam, so I'm not too certain how polished the solutions must look. But I've another three Putnams to go, so I guess there's nothing to worry about now.

I got A1, A2, B1, B2, B3, B4, B6. I'm hoping that gets an honorable mention.

What do you think how the top people of the last year performed? Will Ana Caraiani retrieve a full solution for 11 problems again? How did it work out for Jacob Tsimerman?

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam2004results.html

Well, anyway it takes some month until results will be released. So let's spread the rumours! :D

Obviously I didn't do the Putnam exam but it looks quite a bit harder than last year's exam to me...

[quote="Xevarion"]Being from Princeton myself, I'm also quite interested in how Andrei Negut and Aaron Pixton did.[/quote]What was the team this year? Andrei Negut, Ana Caraiani and Aaron Pixton?  :? What about kappa?

I thought it was them. Maybe I'm wrong though.

[quote="pierrecurie"]I've heard several references to March.  Is that when they post the results (which I'm waiting for)?[/quote]

Yes.

It would certainly make sense.  Most universities, unless they have a really good reason (e.g., Jacob Tsimerman at Toronto last year) just take the top three scorers from the year before.  Princeton had 2 Putnam Fellows last year, so obviously they will be on the team again, and Andrei Negut was in the next 10 scorers, so that certainly qualifies as a very good reason.

I assume that the 5th place at the IMO wasn't enough though :)

Ana Caraiani and Aaron Pixton are both (twice, once resp.) Putnam fellows and (twice, twice resp.) IMO gold medalists, so they're untouchable, and Andrei Negut did finish in the top 15 on the Putnam last year and his best showing at the IMO was better than Kappa's.

I don't think the decision is hard to make - maybe next year it will be if all 4 of them are in the top 15.

Yeah, I guess 3 Romanians in the team is just too much to ask :D I wonder when was the last time there were less than 2 Romanians in Priceton's Putnam  team?  :?  :)

If that isn't rhetorical, either 2001 (where they didn't place in the top 10 and hence I cannot find the list of their team members) or 2000.

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam2000results.html

[quote="blahblahblah"]If that isn't rhetorical[/quote] :D  :D

A3 was pretty hard for a #3 in my opinion.

i didnt take this exam, but after trying it i found that section A was rather difficult in comparison to previous years. However, i thought that the B section was much easier.

in part A i attempted number 2,3, and 5. but was not able to develope complete solutions so i probably wouldn't have received any credit for them.

in part B i got complete solutions for numbers 1 and 3. while for number 4 i ran out of time befor i could finish my solution,this problem really didnt take too much creativity. number 2 i started to look at but had not idea how to start other than just trying small values, and that still didnt get me anywhere.

one question, what is the correct answer to number A2?

[quote="maokid7"]one question, what is the correct answer to number A2?[/quote]
See the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64432]2005 A2[/url] thread.


Are restriction enzymes found in eukaryotes? They seemed to be derived from bacteria cells for scientists to use but can they also be located in eukaryotes?

This may answer your question.

http://www.nature.com/embor/journal/v2/n4/full/embor438.html

I believe restriction enzymes are only found in prokaryotes, mainly as a form of defense from invading foregin viruses.

arent endonucleases involved in crossing over, which occurs in eukaryotes too?
Then they must be present in eukaryotes!!

yes,that's right--they are found in animal cells


Calculate:

2^3+3^5+2^4+3^6+....+2^200+3^202

how do you exactly work this problem? do you use the geometric sum formula?

can somebody help me out?

Calculate:

1+2^3+2^6+.....2^300

It's too big for my TI-86 to even calculate

Would it help if i told you to sum the 2's and 3's seperatly? Do you know the formula for a geometric sum?

actually i just did that.... i got both geometric sums separately and then i added them together?

I got roughly 3.60x10^96


function
integration
calculus
real analysis
real analysis unsolved

let $A=\{f\in C^{1}[0,1]: f(0)=0,f(1)=1,\mid f'(t)\mid \leq 1 \forall t \in [0,1]\}$ . What is the cardinality of A?

edit: i finally figured a way for putting something that looks like the mod

This is silly: the unique such function is $f(x) = x$.  If we ever have $f'(x) < 1$ then $1 = f(1)-f(0) = \int_{0}^{1}f'(x) dx < \int_{0}^{1}1dx = 1$, a contradiction, so $f'(x) = 1$ and $f(x) = x$.

It also follows from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120465

sorry, it never struck me to use the integral theorem  :oops:

Actually, it's not entirely silly; the fact that $f, g$ continuous, $f \leq g$ and there is some $c \in [a, b]$ such that $f(c) < g(c)$ implies that $\int_{a}^{b}f(x) dx < \int_{a}^{b}g(x) dx$ is not entirely trivial (although it is intuitively obvious), and it is necessary for my initial proof.


calculus
integration
trigonometry
linear algebra
matrix
calculus computations

express $ \int_{0}^{1}\frac{sin(\frac{lnx}{2})}{2(x\minus{}1)}dx$ in the form $ \sum_{k\equal{}1}^{\infty}\frac{1}{ak^2\plus{}bk\plus{}c}$

I'm wondering if
1.it is [hide="what I think it is"]a=4,b=0,c=1[/hide]
2.the answer is impossible to write in closed form

3.there's an easier way to do it

Substitute $ x\mapsto e^{\minus{}x}$ then $ I\equal{}\minus{}\frac12 \int_0^\infty \frac{\sin \frac{x}{2}}{1\minus{}e^{x}} \, dx$ .

Now use the formula $ \int_0^\infty \frac{\sin(\alpha x)}{1\pm e^{\beta t}}\, dt\equal{}\frac{1}{2\alpha}\minus{}\frac{\pi}{2\beta}\, \begin{matrix} \text{csch} \\ \coth \end{matrix} \left(\frac{\alpha\pi}{\beta}\right)$

to obtain $ I\equal{}\frac12 \left(\frac{\pi}{2}\coth \left(\frac{\pi}{2}\right)\minus{}1\right)$ .

On the other hand $ \sum_{k\equal{}1}^\infty \frac{\alpha^2}{k^2\plus{}\alpha^2}\equal{}\frac12 \left(\alpha\pi\, \coth(\alpha \pi)\minus{}1\right)$ .

Hence $ \sum_{k\equal{}1}^\infty \frac{1}{4k^2\plus{}1}$ is also $ \frac12 \left(\frac{\pi}{2}\coth \left(\frac{\pi}{2}\right)\minus{}1\right)$.


limit
real analysis
real analysis unsolved

[hide="Solution"]Divide both numerator and denominator with $ n^{\frac {3}{2}}$. You get the limit: $ 1$ (as all elements in numerator and denominator but two give zero in the limit process after the dividing).[/hide]

As $ n$ gets large, you can make the substitutions

$ \sqrt{n^3\minus{}2n^2\plus{}1} \rightarrow n^\frac{3}{2}$

$ \sqrt[3] {n^4\plus{}1} \rightarrow n^\frac{4}{3}$

$ \sqrt[4] {n^6\plus{}6n^5\plus{}2} \rightarrow n^\frac{3}{2}$

$ \sqrt[5] {n^7\plus{}3n^3\plus{}1} \rightarrow n^\frac{7}{5}$

and this gives you:

$ \lim_{n \rightarrow \infty} \frac{n^\frac{3}{2}\plus{}n^\frac{4}{3}}{ n^\frac{3}{2}\minus{}n^\frac{7}{5}}\equal{}\lim_{n \rightarrow \infty} \frac{1\plus{}n^{\frac{4}{3}\minus{}\frac{3}{2}}}{ 1\minus{}n^{\frac{7}{5}\minus{}\frac{3}{2}}}\equal{}1$

[quote="milin"]As $ n$ gets large, you can make the substitutions

$ \sqrt {n^3 \minus{} 2n^2 \plus{} 1} \rightarrow n^\frac {3}{2}$

$ \sqrt [3] {n^4 \plus{} 1} \rightarrow n^\frac {4}{3}$

$ \sqrt [4] {n^6 \plus{} 6n^5 \plus{} 2} \rightarrow n^\frac {3}{2}$

$ \sqrt [5] {n^7 \plus{} 3n^3 \plus{} 1} \rightarrow n^\frac {7}{5}$

and this gives you:

$ \lim_{n \rightarrow \infty} \frac {n^\frac {3}{2} \plus{} n^\frac {4}{3}}{ n^\frac {3}{2} \minus{} n^\frac {7}{5}} \equal{} \lim_{n \rightarrow \infty} \frac {1 \plus{} n^{\frac {4}{3} \minus{} \frac {3}{2}}}{ 1 \minus{} n^{\frac {7}{5} \minus{} \frac {3}{2}}} \equal{} 1$[/quote]

This way of working is dangerous, you take a limiet in a limiet.


greatest common divisor
Diophantine Equations

Find all pairs $(a,b)$ of positive integers that satisfy the equation \[a^{b^{2}}= b^{a}.\]

[quote="Peter"]Find all pairs $ (a,b)$ of positive integers that satisfy the equation
\[ a^{b^{2}} \equal{} b^{a}.
\]
[/quote]
Solution at http://www.kalva.demon.co.uk/imo/isoln/isoln975.html
Notice first that if we have $ a^m \equal{} b^n$, then we must have $ a \equal{} c^e, b \equal{} c^f$, for some c, where m=fd, n=ed and d is the greatest common divisor of m and n. [Proof: express a and b as products of primes in the usual way.]

In this case let d be the greatest common divisor of a and b2, and put $ a \equal{} de, b^2 \equal{} df$. Then for some c, $ a \equal{} c^e, b \equal{} c^f$. Hence f $ c^e \equal{} e c^{2f}$. We cannot have e = 2f, for then the c's cancel to give e = f. Contradiction. Suppose 2f > e, then $ f \equal{} e c^{2f \minus{} e}$. Hence e = 1 and $ f \equal{} c^{2f \minus{} 1}$. If c = 1, then f = 1 and we have the solution a = b = 1. If c ≥ 2, then $ c^{2f \minus{} 1} \geq 2^f > f,$ so there are no solutions.

Finally, suppose 2f < e. Then $ e \equal{} f c^{e \minus{} 2f}$. Hence f = 1 and $ e \equal{} c^{e \minus{} 2}$. $ c^{e \minus{} 2}$ ≥ $ 2^{e \minus{} 2}$ ≥ e for e ≥ 5, so we must have e = 3 or 4 (e > 2f = 2). e = 3 gives the solution a = 27, b = 3. e = 4 gives the solution a = 16, b = 2.

$(1,1)$ and $(16,2)$ are the only solutions. I would post the complete solution later as I am busy.

My calculator says that $\left(27,3\right)$ is also a solution.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group, and assume that $N \lhd G$ is normal in $G$, and that $|N|$ is co-prime to both $|G/N|$ and $|\mathrm{Aut} \, G|$. Prove that $N$ is a direct factor of $G$, i.e. there exists a subgroup $K < G, K \cong G/N$, such that $G = N \times K$.

Enjoy!  :lol:
--Vesselin

I think I have a "proof" :), but I'm using a theorem which I will not prove (I read it some time ago in Rotman's "An Introduction to the Theory of Groups"; I don't want to copy the proof from the book though :)): If a central subgroup $N$ of a group $G$ has index $k$, the map $x\mapsto x^k$ is an endomorphism of $G$.

First of all, let's observe that $(|N|,|\mbox{Aut}(G)|)=1$ implies that $N$ is central (i.e. it's contained in the center $Z$ of $G$): on the one hand we have $\dfrac{|N|}{|N\cap Z|}\ |\ |N|$, and on the other hand we have $\dfrac{|N|}{|N\cap Z|}=\dfrac{|NZ|}{|Z|}\ |\ \dfrac{|G|}{|Z|}=|\mbox{Inn(G)}|\ |\ |\mbox{Aut(G)}|$, so $|N|=|N\cap Z|\Rightarrow N\le Z$.

Now take, just as in the above-cited theorem, $k$ to be the index of the central subgroup $N$ of $G$, and take $K$ to be the kernel of the endomorphism of $G$ defined by $x\mapsto x^k$. Since $(k,|N|)=1$, it's clear that every coset of $G$ modulo $N$ contains one and only one element of $K$, so $|K|=k$, and the conclusion follows.


In base $R_1$ the expanded fraction $F_1$ becomes $.373737\cdots$, and the expanded fraction $F_2$ becomes $.737373\cdots$.  In base $R_2$ fraction $F_1$, when expanded, becomes $.252525\cdots$, while fraction $F_2$ becomes $.525252\cdots$.  Find $R_1 + R_2$ (written in base 10).

[hide]$.\overline{37}_a=3a^{-1}+7a^{-2}+3a^{-3}+7a^{-4}...=\frac{3a+7}{a^2-1}$

Similiarily,

$.\overline{73}_a=\frac{7a+3}{a^2-1}$

$.\overline{25}_b=\frac{2b+5}{b^2-1}$

$.\overline{52}_b=\frac{5b+2}{b^2-1}$

For $F_1$, we have:

$\frac{3a+7}{a^2-1}=\frac{2b+5}{b^2-1}$

For $F_2$, we have:

$\frac{7a+3}{a^2-1}=\frac{5b+2}{b^2-1}$

Divide the two equations on both sides:

$\frac{3a+7}{7a+3}=\frac{2b+5}{5b+2}$

$15ab+6a+35b+14=14ab+35a+6b+15$

$ab+29(b-a)=1$

Assume a<10 and b<10, then b<a and ab can be 30 or 59.  59 is a prime so, only 30=5*6 work.

a=5 and b=6.  The answer is [b]a+b=11[/b]

Question: what if a or b>10?  If a=11 and b=8, 11*8+29(8-11)=1 also work.  Does this problem have more than one answer?

EDIT:
1. No need to restrict a<10 and b<10.  In fact, a>7 and b>5
2. Therefore, a=6, b=5 does not work.

Solution: a=11 and b=8.  $R_1+R_2=11+8=19$

Question: How to prove there is no another pair of a and b so that ab+29(b-a)=1.  I still don't know if this problem has more than one solution.
[/hide]

[hide="Response to jli"]My approach was pretty much the same as yours.
Since we can have 7s in base a, a is at least 8. Similarly, b is at least 6. Therefore, your answer of a=5, b=6 is incorrect. I think your other solution is the correct one. I haven't come across any others yet.[/hide]

jli: I don't think you should assume that a, b < 10.  There's nothing wrong with bases higher than 10, but more importantly, the solution that you found does not work.  There is no such digit as 7 in base 5, so a cannot be 5.  More specifically, a must be at least 8, and b must be at least 6.

[quote="jli"]Question: How to prove there is no another pair of a and b so that ab+29(b-a)=1.  I still don't know if this problem has more than one solution.
[/quote]

If you calculate $b$ in terms of $a$ form your equation and then substitute to the equation of either fraction,

$ab + 29(b - a) = 1\ \Rightarrow b = \frac{29a + 1}{a + 29}$

$\frac{3a + 7}{a^2 - 1} = \frac{2(29a + 1)(a + 29) + 5(a + 29)^2}{(29a + 1)^2 - (a + 29)^2}$

$(29a + 1)^2 - (a + 29)^2 = ... = (29^2 - 1)(a^2 - 1) = 840(a^2 - 1)$

This means that the equation can be reduced by $a^2 - 1 \neq 0$. Then it becomes a quadratic equation, which has 0, 1, or 2 roots. The quadratic equation is

$63a^2 - 546a - 1617 = 0$

and after reducing by 21,

$3a^2 - 26a - 77 = 0$

This quadratic equation has 2 roots and only one is a positive integer:

$a = \frac{26 \pm \sqrt{26^2 + 12 \cdot 77}}{6} = \frac{26 \pm \sqrt{1600}}{6} = 11$ or $-\frac{7}{3}$


A spring is 18 inches long and a force of 10 pounds is necessary to compress it up to 16 inches long. If the force is given by $ f(x) \equal{} kx$ where $ k$ is a constant and $ x$ the displacement. What's the constant $ k$? How much work is done to compress the spring from 16 inches long up to 12 inches long?

Here's what I did:

Let's find $ k,$ then $ f(2)\equal{}2k\equal{}10\implies k\equal{}5,$ so the work is $ \int_0^4 5x\,dx.$ Is this correct?

Since we are only compressing the spring from [b]$ 16$[/b] in. to $ 12$ in.,
work = $ \int_{2}^{6}5x \mathrm{d} x \equal{} 30$ pounds.

EDIT: changed the upper bound to $ 6$ instead of $ 4$

Wait, wouldn't it be $ \int_{2}^{6}5x\mathrm{d}x\equal{}80$?

[quote="savy2020"]Yes it would be  $ \int_{2}^{6}5x\mathrm{d}x \equal{} 80$.[/quote]
I don't get very well how you got that bounds, can you explain any further please?

Obviously they'd be the limits for "$ x$"(Here '$ x$' is the compression).
As you need to find the work done in compressing the spring from $ 16inches$ to $ 12 inches$, the bounds would be the compressions (measured from the original length of $ 18inches$). in the respective cases.


function integration algebra polynomial trigonometry calculus inequalities

Let $ f,g\in C[0,1]$  where $ C[0,1]$ denotes the set of continuous functions on interval $ [0,1]$ and suppose that

$ \int_{0}^{1}{x^{n}f(x)dx}\equal{}\int_{0}^{1}{x^{n}g(x)dx}$

for all $ n\in\{0,1,2,...\}$

Prove that $ f \equal{} g$

I appreciate your help

If we put $ h\equal{}f\minus{}g$ then we have $ h$ is perpendicular to every polynomialds. Because set of polynomials is dense in $ L^{2}[0;1]$ so $ h(x)\equal{}0$ almost every where. Now $ h$ is continous so then $ h\equiv 0$.

We also use Parseval fomula in Fourier Theory to prove this problem. By this indentity we just need to show all Fourier coefficients of $ f\minus{}g$ is zero. But this is obviously because the functions $ \sin x$ and $ \cos x$ are analytic, ie.. we could obtain it by an power uniform series.

Thanks for the reply, there's one thing I didn't understand and that is

'because the set of polynomials are dense in L2 h=0 almost everywhere.'

Is this the fact from measure theory or something?

A more elementary approach:

You have to prove that if $ f$ is a continuous function such that $ \int_{0}^{1}x^{n}f(x)dx\equal{}0$ for all $ n$ then $ f\equal{}0$.

$ f$ is continuous , thus it can be approximated uniformly by polynomials, and hence

for $ \epsilon >0$ there is a polynomial $ P$ such that

$ |f(x)\minus{}P(x)|\leq \sup_{x\in[0,1]} |f(x)\minus{}P(x)|<\epsilon$.

Thus

$ \int_{0}^{1}|f(x)\minus{}P(x)|^{2}dx<\epsilon^{2}$.

It follows that

$ \int_{0}^{1}f^{2}(x)dx\plus{}\int_{0}^{1}P^{2}(x)dx<\epsilon^{2}$ and hence

$ \int_{0}^{1}f^{2}(x)dx\leq \epsilon^{2}$.

Since $ \epsilon$ is arbitrary fixed we get that $ \int_{0}^{1}f^{2}(x)dx\equal{}0$ from which it folows that $ f\equal{}0$.

Thanks for your solution didilica,
my memory of measure theory is fading so please correct me if I'm wrong:
In the last step of your solution, you have integral of f squared equal to zero, and concluded that f is zero.
Are you invoking the theorem which roughly says that the integral of non negative (?) function is zero iff the function is zero almost everywhere?
And the reason you want f squared is so that you have non negative function is that right?

One more thing,
integral of absolute value of f minus P squared was less than epsilon squared and from that follows that integral of f squared plus integral of p squared is less than epsilon squared,
now, absolute value of f minus P squared is equal to f squared plus P squared minus 2 f P isn't it?
So i can't quite see how integral of f squared plus integral of P squared is less than epsilon squared

And lastly, you have something plus something is 'strictly less' than epsilon squared implies someting is 'less than or equal to' epsilon squared.
Does this work? I've never seen this implication before.

Thanks!

Well,
1) You do not have to use measure theory concepts to prove the last implication.

We have this result: If $ f\geq 0$ is a continuous function on an interval , say $ [0,1]$, then $ \int_{0}^{1}f(x)dx \equal{} 0$ implies that $ f \equal{} 0$.

By contradiction , assume that $ f\neq 0$. It follows that there is $ x_{0} \in [0,1]$ such that $ f(x_{0})\neq 0$. Let $ f(x_{0}) > 0$.

Then, since $ f$ is continuous at $ x_{0}$ there is an interval $ I \equal{} (x_{0} \minus{} \delta, x_{0} \plus{} \delta)$ contained in $ [0,1]$ such that $ f(x) > 0$ on $ I$.

Now,

$ 0 \equal{} \int_{0}^{1}f(x)dx\geq \int_{I}f(x)dx > 0$ which is absurd.

2) Note that, since $ \int_{0}^{1}x^{n}f(x)dx \equal{} 0$ for all $ n$, then if your polynomial $ P$ is , say, $ P(x) \equal{} a_{k}x^{k} \plus{} \cdots \plus{} a_{0}$ then

$ \int_{0}^{1}f(x)P(x)dx \equal{} a_{k}\int_{0}^{1}x^{k}f(x)dx \plus{} \cdots a_{0}\int_{0}^{1}f(x)dx \equal{} 0$ based on the hypothesis of the problem.

3) That inequality should have been strict, sorry about that.

Thank you so much for your help didilica!!

[quote="sit.think.solve"]Let $ f,g\in C[0,1]$  where $ C[0,1]$ denotes the set of continuous functions on interval $ [0,1]$ and suppose that
$ \int_{0}^{1}{x^{n}f(x)dx} \equal{} \int_{0}^{1}{x^{n}g(x)dx}$
for all $ n\in\{0,1,2,...\}$
Prove that $ f \equal{} g$
[/quote]

I found that a bit more stronger is true, here's:
Let $ f,g\in C[0,1]$  where $ C[0,1]$ denotes the set of continuous functions on interval $ [0,1]$ and suppose that
$ \int_{0}^{1}{x^{n}f(x)dx} \equal{} \int_{0}^{1}{x^{n}g(x)dx}$
for all $ n\geq0$ which $ 3\mid n$. Then $ f \equal{} g$


Find all real numbers $ x$ such that

$ \sqrt{x\minus{}10} \plus{} \sqrt{x\minus{}20} \equal{} \sqrt{x\plus{}22}$.

hmm,i found $ x\equal{}28$ but i don't really like my solution :( I am wondering if there is any smarter way(some substitution maybe?) to
solve this equation.
Basically what i did was squaring twice so i got rid of the square roots and ended up with a quadratic formula.One of the roots satisfies
the conditions of the problems.

You 're right. This is the only root. And, to be honest, I've never seen a shorter way to solve it, except squaring both sides (as you did)...


In a meeting there are 6 participants. It is known that among them there are seven pairs of friends and in any group of three persons there are at least two friends. Prove that:
(a) there exists a person who has at least three friends;
(b) there exists three persons who are friends with each other.

[i]Valentin Vornicu[/i]

[hide="Solution"]
(#)We prove by contradiction.

Let A and B be friends. Friendship relation: let it be <=>. So A<=>B.
Easy to prove: at least one of them has a friend C ('cause of 7 pairs)
Let A<=>B<=>C (that doesn't mean A<=>C!!!). So A<=>C isn't true(#).
Again because of 7 pairs, one of them has a friend D. It's either A<=>D or C<=>D. Let it be C<=>D.
Again there are not any more friendships between A,B,C and D(#).
Again because of 7 pairs, one of them has a friend E. It's either A or D (#). Let it be D.
Do the same for F.

So we get: A<=>B<=>C<=>D<=>E<=>F
There are 5 friendships, so if A,B,C,D,E and F form a hexagon, and friendships are sides and diagonals, there must be at least one diagonal=> b) proved.
And the person from which 1 diagonal is drawn has 3 friends (easy!)
[/hide]

It's a very nice problem!

[hide="Solution"]
Since we have seven friendships, the total friend-count is 14. By pure pidgeonhole, one guy has 3 friends. QED for part (a).

Take that guy's three friends. Two of them must be friends. Those two, plus the person with three friends, satisfies part (b).

I like to think of the people as vertices on a graph, and friendships as edges.[/hide]

Nice solution Darmani!
I usually complicate every solution, you can see my solution for this :wink:


probability
function
integration
calculus
calculus computations

I'm just a beginning mathematics enthusiast.Could anyone tell me the meaning of "Gaussian[/b] Law[b]"related to
the 'normal[/b] distribution[b] of[/b] error[b]?'

                                                       Respecrfully,
                                                       A.Eddington[/b]

The standard normal distribution is the continuous probability distribution on the real line that has the density function

$\frac1{\sqrt{2\pi}}e^{-x^{2}/2}.$

That is, if $X$ is a random variable with a standard normal distribution (the word "standard" here means mean zero and variance one), then

$P(a<X<b)=\frac1{\sqrt{2\pi}}\int_{a}^{b}e^{-x^{2}/2}\,dx$

The normal distribution draws its importance from the [b]Central Limit Theorem[/b]. There are several different sets of hypotheses that can be used in the CLT; here's a straightforward version.

Suppose $X_{i}$ is a sequence of independent identically distributed random variables, with each $X_{i}$ having mean $m$ and variance $\sigma^{2}.$

Let $Z_{n}=\frac{\sum_{i=1}^{n}X_{i}-nm}{\sigma\sqrt{n}}$

Then $Z_{n}$ converges in distribution to a standard normal distribution as $n\to\infty.$

A binomial distribution with large $n$ can be regarded as the sum of $n$ identical independent trials, and the Central Limit Theorem applies; we may use the normal distribution to make reasonable estimates of cumulative probabilities in that case.

I'm teaching a college-level class (calculus prerequisite) in probability and random variables this semester. We finish the class with an explanation (but not really a proof) of the Central Limit Theorem. One of my contentions in that class: nothing in nature is exactly normally distributed. The normal distribution does appear to be ubiquitous in nature, but that is a consequence of the Central Limit Theorem - that many things we observe are themselves effectively the sum of many different random inputs.

See: random walks; heat equation; Maxwell-Bolzmann distribution.


Putnam
linear algebra
calculus
calculus computations

Hi, maybe this is supposed to be in another forum..

I find that olympiad-like problems improves your mathematical skills much more than practice problems in general. So I would only assume this applies to topics you find in university as well. I am wondering if there exists olympiad-[i]like[/i] problem books for topics like calculus/analysis, number theory/group theory, linear algebra, etc... I am very interesting in getting such a book. Do you have any references?

Hi Jarle10.

What about the following?

http://www.amazon.com/Putnam-Beyond-Razvan-Gelca/dp/0387257659

or the Putnam books here: http://www.artofproblemsolving.com/Books/AoPS_B_Item.php

This site looks good: http://math.arizona.edu/~savitt/teaching/09math294/

Hope this is helpful.


calculus
topology
vector
real analysis
Princeton
college
linear algebra

If anyone has any useful advice/opinion regarding Princeton's courses 217 (Honors Linear Algebra) and 218 (Analysis in Multiple variables), please feel free to comment... Should 217 be taken (normally) before 218? What prior knowledge do you need to have in order to be fairly comfortable with these courses?

Without knowing anything about you, I don't think anyone can give you solid advice on which math course is right for you. You should wait until you talk to a professor in the department before you decide on a math course for certain.

That being said, MAT 217 is strongly recommended before MAT 218. Especially if the MAT 218 course uses Spivak's [u]Calculus on Manifolds[/u] as the course description indicates it will. To understand anything in that book you need a strong background in the theory of vector spaces and linear transformations (which is the major topic of MAT 217).

Furthermore, if you're going to be a freshman, you should be cautious about skipping MAT 215 to go into MAT 217 unless you are really comfortable with the material covered in MAT 215 (i.e. most of the material covered in the first 8 chapters of [u]Principles of Mathematical Analysis[/u] by Rudin) and "definition-theorem-proof" style mathematics.

You can look at the past exams for some math courses [url=http://www.math.princeton.edu/undergraduate/Pexams.html]here[/url].

First, thank you for the advice!
Second, I should do a bit of explaining: I am reluctant about taking 218 - because of 215. In the past month, I've looked a bit into Rudin's book and a couple of exams in the website you indicated and realized that although I was familiar with most of the stuff there I felt I was a bit insecure about some (for example: I studied 2 years of analysis at school but everything in R; I'm very unfamiliar with metric spaces, or topology for that matter). Especially the last chapters of Principles... -well, they do pose a little problem. The reason I still considered 218 was simply because I talked with some Princeton students who had a similar background as mine who said you can catch up with 215 in the meanwhile.
Hmmm... well, I don't know about that. & besides, you're right: I will wait until September and do a bit more research.  :)

I imagine that MAT 215 and 218 do in fact have a lot of overlap in material. But for most students, this probably isn't a bad thing. My general opinion is that it's a greater sin to not have been exposed to something you need to know, than to be exposed to it twice.

When I was there in the mid 90s they didn't have 215, just 218. I think 215 was introduced because a.) it allowed them to "beef up" the 217 and 218 classes by having students get their feet wet in a class preceeding them, and b.) I think some faculty at the time felt that a few students coming out of 217/218 alone weren't coming to higher level courses with a strong enough analysis background. So having two courses of "advanced calculus/intro-analysis" (215 and 218) would give a more solid basis for Stein's newly introduced analysis sequence 330-332/433 and for other courses.

Is it possible to skip MAT 215 and just do 217/218? Well at one point before there was 215, it was common. Probably now, it is done by a few students. But I think for most students the introduction of 215 was for a good reason. Getting familiar with metric spaces and topology are very important. When you get to 218 and they start talking about hausdorff spaces and compact sets and open coverings - already being familiar with metric spaces and topology will come in handy!! But for some students, it is possible to learn those things on their own or while taking 218. I imagine metric spaces and basic point-set topology is covered in Folland's book (though I have never seen it).

BTW, for a good, easy-to-read (and cheap!) book on metric spaces and topology try [u]General Topology[/u] by Michael Gemignani.

Thanks for your advice. I, myself, am in the same situation. I have studied Analysis in Single Variable at a nearby university using "Elementary Analysis: The Theory of Calculus" by Kenneth Ross. It covers analysis using Real Numbers, but goes a bit into Topology-metric spaces and sets (since the Heine-Borel Theorem was an important part of the course).

Topology was my favorite part of the course, but I think I will take this course again, because:

1. I took the Analysis course during the summer and it still skipped a lot of important things that I think I can learn by myself, but I would like to take it again because I've never been exposed to proofs this rigorously before.
2. I don't think I did as well as I wish I had in that Analysis course
3. I think taking it again at Princeton would make it more interesting and give me some time to play around with some ideas, ask (stupid) questions and learn by trial and error, which my 18 years of experience has shown to be an effective way to learn.


I doubt that anyone who clicked on this actually believed it.

eh. I could email my mathcounts state team and say that I am resigning, but I won't.

[quote="bowser"]HAPPY APRIL FOOLS DAY!!!!! :] hehe you fell for it and clicked it!! YAY!!  :D  :D  :D happy April fools day![/quote]

hmm

exactly HOW many ppl that clicked on it do you think CLICKED on it because they thought it was REALLY a prize?

and remember, this is like a math website for super-smart ppl

so get a life

better reasoning is that this seems a lot like spam
people sometimes look at spam to laugh at the moron (bot) who posted it

I thought there would be an interesting discussion about banner ad 'games' in here but no....
[color=#EFEFEF]red goes the moon[/color]

[quote="perfect628"][quote="1=2"][quote="randomdragoon"]I thought there would be an interesting discussion about banner ad 'games' in here but no....
red goes the moon[/quote]
 :rotfl:

They are annoying. I got two prizes in less than one minute. The prizes have to be fake.[/quote]
NOOOOOOOOOOOOOOOOOOOO!!!!! SO MY XBOX [i]ISN'T[/i] COMING?!?!?![/quote]

 :rotfl:

[quote="mz94"]
so get a life
[/quote]

That's rude... :|


Let $ x, y, z$ be the real numbers such that
$ 9x \minus{} 9y \plus{} 7z \equal{} 2$ and $ 7x \plus{} 2y \minus{} 6z \equal{} 9$
Evaluate $ x^2\minus{}y^2\plus{}z^2$

I believe there are multiple answers for different values of $ z$.


geometry
quadratics
conics
parabola
analytic geometry
calculus
derivative

[hide]I don't see why it works but if the fence has sides 1200, 600, 1200, 600, the sum would be 1200+600+600=2400 and the area would be 720000 which is larger than 800*800=640000.[/hide]

I think you need some additional constraints on the first problem:
[hide]
Suppose we use a semicircle where the river acts as the border for the diameter.  Then 2400 is half of the circumference:
$ 2400 \equal{} \pi r$
$ r \equal{} \frac{2400}{\pi}$

The area enclosed is half of the area of the full circle:
$ A \equal{} \frac{\pi r^2}{2}$
$ A \equal{} \frac{\pi}{2} \frac{2400^2}{\pi^2}$
$ A \equal{} \frac{5,760,000}{2 \pi}$
$ A \approx 916732.5$ which is clearly a larger area.

I'm pretty sure that the optimal area in this case is going to be a circular region (as opposed to some polygon - I think you can make some intuitive arguments as to this point) where the river acts as a chord across the circle, but I'm too lazy to use Lagrange multipliers to find the optimal area.  The above is just a demonstration of the easiest such area to compute that is greater than 720,000.
[/hide]

for #1 you have a quadratic and you complete the square to find the maximum possible area.

Solutions anyone? #1 is simply the equation of a parabola and to find the vertex, you complete the square.  The y-coordinate is your answer.  Rather not explain right now because.... it's high school stuff

[hide="#1 Solution"]Let the height of the pasture be $ h$, then we have that the width is $ 2400\minus{}2h$.  We thus wish to maximize $ h(2400\minus{}2h)\equal{}\minus{}2(h^2\minus{}1200h)$ Complete the square.

$ \minus{}2(h^2\minus{}1200h\plus{}360000\minus{}360000)\equal{}\minus{}2(h\minus{}600)^2\plus{}720000$.
Our maximum value obviously occurs when $ h\equal{}600$, which gives the area as $ 720000$[/hide]

[quote="zolojetto"]i dont see how you get number one, maybe there is a larger area which obviousely there is because you got one but i thought becuase the river makes up one length we have 2400 meters of fencing to make 3 sides of the square, maybe a square isnt the largest area but then what is the largest area, so i get one side of the square is 800 and then the area is 640000.[/quote]
1200+ 600+600;600 x 1200 = 720000

I'm going to cheat:
[hide="Cheating Solution"]The volume is given by $ \pi r^2h\equal{}1$, and the surface area is $ SA\equal{}2\pi r\left(r\plus{}h\right)\Rightarrow SA\equal{}S(r)\equal{}2\pi r\left(r\plus{}\frac{1}{\pi r^2}\right)$.
Then, $ S'(r)\equal{}\frac{2\left( 2\pi r^3\minus{}1 \right)}{r^2}$.
Setting $ S'(r)$ equal to zero, we have:
$ r\equal{}\sqrt[3]{\frac{1}{2\pi}}$
Plugging this value in for $ r$ in the surface area equation, we have:
$ \boxed{SA\equal{}3\sqrt[3]{2\pi}}$[/hide]

[hide="a different way to think of #1"]
to maximize the area if there were no river, you would use the same amount of fence for the lengths and the widths. we do the same here

to maximize the area, you want to use the same amount of fence for the length as with the width. the length we have x=1200, the width is 2y=1200 y=600 so its 600*1200[/hide]

wow, never thought of a way to use derivatives to do it. Hey, nobody at the actual competition cares about HOW you do it, even if u guess. Even if u hi-jacked the answers..... (although you might have to face federal law)

But seriously, I just wanna get this straight. You set it to zero, because you are trying to find the max of the original function, right? Cuz that's when the change is zero? Would it have worked just doing $ \frac{\minus{}b}{a}$?

That would have worked, but it's $ \frac{\minus{}b}{2a}$.

You simply find the critical points of the derivative and check that it is indeed a local minimum that testing the concavity around that point ~_-... We shouldn't talk about this.

[quote="darkmatter47"]wow, never thought of a way to use derivatives to do it. Hey, nobody at the actual competition cares about HOW you do it, even if u guess. Even if u hi-jacked the answers..... (although you might have to face federal law)

But seriously, I just wanna get this straight. You set it to zero, because you are trying to find the max of the original function, right? Cuz that's when the change is zero? Would it have worked just doing $ \frac { \minus{} b}{a}$?[/quote]

Yes, Google the word "optimization". Yes, I set the derivative to 0 because I wanted to find the max/mins.

[quote="mathcrazed"]Farmer Brown wants to fence in a rectangular pasture using $ 2400$ meters of fencing.  Given that there is a river next to his pasture that may be used as a border, find the largest possible area of the pasture.

[hide="Answer"]$ 720,000$[/hide]

Much Harder (using elementary methods, that is):
Find the minimal value of the surface area of a cylinder in $ cm^2$ given that the volume of the cylinder is $ 1cm^3$.

[hide="Answer"]$ 3\sqrt [3]{2\pi}$ (you can check the algebra for me when $ r \equal{} \frac {\sqrt [3]{1}}{2\pi}$ and $ h \equal{} 2r$[/hide]

Ok I edited it. Thanks Haoye[/quote]

Aren't these all classic calculus problems?
Let x and y be lengths of the sides, where x is the side parallel to the wall.
$ x \plus{} 2y \equal{} 2400$
Find max of $ xy$.
$ x \equal{} 2400 \minus{} 2y$
$ 2400y \minus{} 2y^2$
$ y^2 \minus{} 1200y \equal{} 0$
DERIVATIVE TIME!
$ y^2\minus{}>2y$
$ \minus{}1200y\equal{}\minus{}1200$
so.
$ 2y \equal{} 1200$
$ y \equal{} 600$
$ x \equal{} 1200$
$ 600*1200 \equal{} 720000$

[quote="mathcrazed"]There's a reason I said using elementary methods... Do it without calculus -_-.[/quote]

Yes, but you already gave the elementary method (complete the square). I think an AoPS rule is not to repeat solutions. Also I believe using calculus is legal in Mathcounts? I think they only care about the answer, not how we got there.

I think it's kinda illegal in the MC forums here, since this community cares about how you get the answer.

or at least most of it...


The two-digit integers from $19$ to $92$ are written consecutively to form the large integer $N = 19202122 . . . 909192$.  If $3k$ is the highest power of $3$ that is a factor $N$, then find the exact value of k.

If you find the sum of $19, 20, ..., 92$, this will be the same as $N$ mod 3 and mod 9. The sum in question is $111\cdot37$, which is divisible by $3$ but not $3^2$...therefore the largest $k = 1$.

U get a cookie!  Good job; that's correct.


Let $ A'B'C'$ be a pedal triangle of arbitrary point $ P$ with respect to triangle $ ABC$. $ P^*$ is isogonal conjugate of $ P$ with respect to triangle $ ABC$. $ R_p$ is circumradii of triangle $ A'B'C'$. The rays $ [B'P,[C'P$ intersect circle $ (P^*,2R_p)$ at $ B'',C''$, respectively. Prove that $ BB'',CC''$ intersect on circle $ (P^*,2R_p)$.

Note that, this is generalization of the problem on the post [url=http://www.mathlinks.ro/viewtopic.php?t=267167&sid=7e75f0684576ac37920943ca9c183451]Nice problem about concurrent:D[/url]

Great work dear Linh  :lol: , all my problems always wait for your nice solution :)!

[b]Proposition:[/b] $P$ is a point on  the plane of $\triangle ABC$ and $\triangle P_AP_BP_C$ is the pedal triangle of $P$ WRT $\triangle ABC.$ $A_1,B_1,C_1$ are the midpoints of $PA,PB,PC.$ Lines $PP_A,PP_B,PP_C$ cut the pedal circle $\odot(P_AP_BP_C)$ again at $P_1,P_2,P_3.$ Then $P_1A_1,$ $P_2B_1$ and $P_3C_1$ concur  on $\odot(P_AP_BP_C).$

[b]Proof:[/b] Let $U$ be the Poncelet point of $ABCP,$ which lies on the pedal circle of $P$  WRT $\triangle ABC$ (for a proof, see post #2 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=406543]here[/url]). $\odot(P_AP_BP_C)$ cuts $BC$ again at $D.$ $R$ is the orthogonal projection of $A$ onto $PC$ $\Longrightarrow$ $\odot(URA_1C_1)$ is 9-point circle of $\triangle APC.$

$\angle DUC_1=\angle DUP_B-\angle C_1UP_B=180^{\circ}-\angle CPP_B-\angle PRP_B$

$\angle DUC_1=\angle PAC-\angle CPP_B=\angle PAC-\angle RAC=90^{\circ}-\angle APC$

Thus, $\angle DUA_1=\angle DUC_1+\angle C_1UA_1=90^{\circ}-\angle APC+\angle APC=90^{\circ}.$ Since $\angle DUP_1=90^{\circ},$ then $U \in P_1A_1.$ Likewise, $P_2B_1$ and $P_3C_1$ pass through $U.$

P.S. Note that a dilatation with center $P$ and factor $2$ produces the configuration of the proposed problem. P-Pedal circle $\odot(P_AP_BP_C)$ with radius $\varrho$ goes to the circle $\odot (P^*,2\varrho)$ centered at the isogonal conjugate $P^*$ of $P.$

[quote="buratinogigle"]There is a generalization for above

[b]Problem. [/b]Let $ABC$ be a triangle and $A',B',C'$ lies on $BC,CA,AB$, resp. $P$ is Miquel point of $A'B'C'$ (the common point of circumcircles $(AB'C'),(BC'A'),(CA'B')$). $PA',PB',PC'$ cut the circumcircle $(A'B'C')$ again at $P_a,P_b,P_c$. Let $O_a,O_b,O_c$ be circumcenter of triangles $AB'C',BC'A',CA'B'$, resp. Prove that $O_aP_a,O_bP_b,O_cP_c$ concur on $(A'B'C')$.[/quote]
Use homotety with center at P and coefficient 1/2 , then Luisgeometra's problem for O_aO_bO_c and P

[quote=Luis González][b]Proposition:[/b] $P$ is a point on  the plane of $\triangle ABC$ and $\triangle P_AP_BP_C$ is the pedal triangle of $P$ WRT $\triangle ABC.$ $A_1,B_1,C_1$ are the midpoints of $PA,PB,PC.$ Lines $PP_A,PP_B,PP_C$ cut the pedal circle $\odot(P_AP_BP_C)$ again at $P_1,P_2,P_3.$ Then $P_1A_1,$ $P_2B_1$ and $P_3C_1$ concur  on $\odot(P_AP_BP_C).$
[/quote]

Let $ T $ be the Poncelet point of $ ABCP, $ then $ T $ lies on the pedal circle $ \odot (P_AP_BP_C) $ of $ P $ WRT $ \triangle ABC $ and the 9-point circle $ \odot (A_1B_1P_C) $ of $ \triangle APB, $ so $ \measuredangle A_1TP_C = \measuredangle A_1B_1P_C = \measuredangle PBA = \measuredangle PP_AP_C = \measuredangle P_1TP_C. $ i.e. $ T \in A_1P_1. $


Prove that if a+b=1 with a>0 and b>0, then

(a+1/a)^2 +(b+1/b) is greater than or equal to 25/2

It's easy to show that it is greater than 8 but I'm not sure how to get the extra 9/2.

[quote="Gibbenergy"]Well it is just AM-GM ,isn't it?
 We have $a,b \geq 0$ then $a^2+b^2 \geq \frac{(a+b)^2}{2}$ and $\frac1a+\frac1b \geq \frac{4}{a+b}$

So we have:
$(a+\frac1a)^2+ (b+\frac1b)^2 \geq \frac{(a+b+\frac1a+\frac1b)^2}{2} \geq \frac{(1+ \frac{4}{a+b})^2}{2} =\frac{5^2}{2}= \frac{25}{2}$
Done![/quote]

It should be $a,b>0$ because he/she said that $a>0$ and $b>0$.  Since $a+b=1$ $a,b>0$, $a,b\notin\mathbb{Z}$. However, I like your solution.  Nice job! :)

EDIT:
[quote="SirErnest"]
(a+1/a)^2 +(b+1/b)
[/quote]

Did you mean (a+1/a)^2 +(b+1/b)^2 or (a+1/a)^2 +(b+1/b)?

Masoud Zargar


Let $ ABCD$ be a convex quadrilateral such that $ \hat{DAC}\equal{} \hat{DBC}$,$ M$ and $ N$ be the midpoints of $ AB$ and $ CD$,respectively, $ O \equal{} AC \cap BD$,and $ P$ and $ Q$ be the projections of $ O$ on $ AD$ and $ BC$,respectively.Prove that $ MN \perp PQ$

If $ \angle DAC \equal{} \angle DBC$ the quadrilateral ABCD is cyclic, and the problem is problem 2037 of russian journ. "Kvant" no.2/2007.
[i]The proposer P. Kojevnikov (П. Кожевников) [/i]
[b]The solution for the problem appear in no.5/2007, p.21.[/b]

It has been posted before.
http://www.mathlinks.ro/Forum/viewtopic.php?t=55836
http://www.mathlinks.ro/Forum/viewtopic.php?t=213068

See also.
http://www.mathlinks.ro/Forum/viewtopic.php?t=63589
http://www.mathlinks.ro/Forum/viewtopic.php?t=216738

Kostas vittas.


Let $S$ be a subset of $\{1, 2, 3, \cdots, 1989 \}$ in which no two members differ by exactly $4$ or by exactly $7$. What is the largest number of elements $S$ can have?

the answer is $ 905$
$ A_k \equal{} \{11k \plus{} 1,11k \plus{} 3,11k \plus{} 4,11k \plus{} 6,11k \plus{} 9\},k \equal{} 0.1,2,...180$
and it is easily to check we can select at most $ 5$ terms from $ 11$ consecutive integers
QED


I have heard about the Calculus class coming up this fall.  Will there also be a Calculus book in the near future?

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=216923]This[/url] will probablly help you (check out rrusczyk's last comment).

In the future, yes.  Not in the "near" future.  Next summer/fall, probably.


algebra
binomial theorem
combinatorics proposed
combinatorics

Evaluate

\[{\sum_{r=0}^n\frac{ {}_nC{}_r}{\displaystyle\sum_{s=0}^r\frac{{}_rC{}_s}{\displaystyle\sum_{t=0}^s {}_sC{}_t {\left(\frac{\sqrt{5}-1}{2}\right)^t}}}}\]

Let $\phi = \frac {1+ \sqrt 5} 2$.
From the binomial theorem, we have :
$\dsp \sum_{t=0}^s C_s^t (\frac { \sqrt 5 - 1} 2 )^t = ( 1 + \frac { \sqrt 5 - 1} 2 )^s =  \phi} ^s.$
Now the second floor becomes :
$\dsp \sum_{s=0}^r C_r^s (\frac 1 {\phi}) ^s = (1 + \frac 1 {\phi} )^r = \phi ^r$
Then, the desired sum is
$\dsp \sum_{r=0}^n C_n^r  (\frac 1 {\phi} )^r = ( 1 + \frac 1 {\phi} )^n = \phi ^n.$

Pierre.

Thank you for your reply, Pierre!

You are right! To tell the truth,I would like to have been generized this problem.

I have been imagined  $1+\frac{1}{1+\frac{1}{1+\frac{1}{1+x}}}$,repeating this procedure and [b]Fibonacci's sequence[/b].

kunny


function
integration
calculus
linear algebra
matrix
geometry
absolute value

Let $ q\geq 0$ be a non-identical 0 continuous function, with period 1  and such that $\int_{0}^{1}{ q(t)dt}\leq 4$. Prove that if $ x\"+q(t)x=0$ then $x$ is bounded.

How about $q(x)=-1$ ? Seems like there is some mistake in the formulation of the problem  :(   :?

It is supposed to be positive. I still don't quite believe it, although the integral condition rules out the simple examples as in alekk's problem.

Oops, I didn't notice the positivity condition  :blush:  Then it seems correct and I can even prove it with $4$ replaced by $1$. Also, it is not very hard to show that you cannot replace $4$ by any larger constant. Unfortunately, I still cannot solve the problem as stated.

P.S. I finally figured it out! :lol:  I do not have time right now to post the solution, so I'll do it later in the day or tomorrow (unless somebody beats me to it :P )

[quote="fedja"] I finally figured it out! :lol:  I do not have time right now to post the solution, so I'll do it later in the day or tomorrow (unless somebody beats me to it :P )[/quote]
OK, here goes. Note first of all that the equation $y''+qy=0$ is equivalent to the system of $2$ first order equations $Y'(t)=Q(t)Y(t)$ where
\[
Y(t)=\left(\begin{matrix}y\\y_1\end{matrix}\right), Q(t)=\left(\begin{matrix}0&1\\-q(t)&0\end{matrix}\right)
\]
and $y_1$ is the same as $y'$  in our second order linear equation. Let $\Phi(t)=\left(\begin{matrix}a(t)&b(t)\\c(t)& d(t)\end{matrix}\right)$ be the fundamental matrix of our system, i.e., the solution of $\Phi'=Q\Phi$, $\Phi(0)=I$ where $I$ is the $2\times 2$ identity matrix. We have to show that $\Phi$ is bounded on the entire real line. Due to $1$-periodicity of $Q$, we have $\Phi(t)=\Phi(\{t\})\Phi(1)^{[t]}$, so we need to show that all integer powers of $\Phi(1)$ are uniformly bounded. This will follow if we show that $\Phi(1)$ has two distinct eigenvalues whose absolute value is $1$. The first (easy) observation is that $\frac d{dt}\det \Phi(t)=\text{Tr}\Phi'(t)\Phi(t)^{-1}=\text{Tr}Q(t)=0$, so $\det \Phi(1)=1$. Therefore, it is enough to show that $\text{Tr}\Phi(1)\in(-2,2)$. Writing the equations for $\Phi$ as a system of $4$ equations for $a$, $b$, $c$, and $d$, we see that $\text{Tr}\Phi(1)=a(1)+b'(1)$ where $a$ and $b$ are the solutions of $y''+qy=0$ with $a(0)=1$, $a'(0)=0$ and $b(0)=0$, $b'(0)=1$. Now comes the main

[b]Lemma:[/b] If $q>0$ and $\int_0^1 q\,dt\leqslant 4$, then the equation $y''+qy=0$ has no conjugate points on $[0,1]$, i.e., there exists no solution $y$ of the equation that is not identically $0$ and vanishes at $2$ distinct points on this interval. This absence of conjugate points condition is actually all we need from $q$. The proof of Lemma is hidden
[hide="here"]
Assume the contrary, i.e., that a non-trivial solution $y$ can vanish at two points $t_1<t_2$. Without loss of generality, we may assume that $y$ is positive between $t_1$ and $t_2$. Let $\alpha=y'(t_1)$ and $\beta=-y'(t_2)$. Since $y''=-qy<0$ between $t_1$ and $t_2$, $y$ is concave on $[t_1,t_2]$ and, thereby, the graph of $y$ is below both of its the tangent lines at $t_1$ and $t_2$. Drawing the lines and finding the intersection point, we see  that $y\leqslant \frac{\alpha\beta}{\alpha+\beta}$. But then
\[
\alpha+\beta=y'(t_1)-y'(t_2)=\int_{t_1}^{t_2}qy\,dt <\frac{\alpha\beta}{\alpha+\beta}\int_0^1 q\,dt=\frac{4\alpha\beta}{\alpha+\beta}
\]
which is impossible for $\alpha,\beta>0$.
[/hide]
Now let us use the lemma. From the equation, it is clear that $a(t)$ decreases as long as it stays positive. But if $a(t)$ goes into the negative area, it can never come back (otherwise we would have two conjugate points on $[0,1)$). Thus, $a< 1$ on $(0,1]$. Similarly, $b$ has to remain positive on $(0,1)$, so it is concave on $[0,1]$ and therefore $b(t)<t$ and $b'<1$ on $(0,1]$. From here we conclude immediately that $a(1)+b'(1)<2$. Now, using the equation, we can also write
$a(1)=1-\int_0^1 (1-t)a(t)q(t)\,dt$
$b'(1)=1-\int_0^1 b(t)q(t)\,dt$
Thus $a(1)+b'(1)=2-\int_0^1[(1-t)a(t)+b(t)]q(t)\,dt$. But $(1-t)a(t)+b(t)<1-t+t=1$ on $(0,1)$ and, therefore, the integral is less than $4$. The end.


trigonometry
geometry
area of a triangle
Heron\u0027s formula

Is there an extension to Heron's Formula that allows to calculate a quadrilateral or any other n-gon without drawing in triangles?

Brahmagupta's Formula for quadrilaterals.  Since every polygon follows its own properties, it would be impossibel to find one that works for all n-gons, at least I think it should be with my understanding of mathematics.

$A=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2[\frac{1}{2}(A+B)]}$
for quadrilaterals, where $s$ is the semiperimeter, and $a,b,c$ and $d$ are teh side lengths.

Masoud Zargar

For a cyclic quadrilateral, just remember;
$A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$

Yup, and that is obviously true from Brahmagupta's formula. :)

Simple reasoning - the side lengths of an $n$-gon do not uniquely determine its area, hence, there is no formula based solely on the side lengths for $n > 3$.  However, an extension for a [i]cyclic[/i] $n$-gon may be possible.  Who wants to try?


i got $ \frac{4}{11}$ too.

there are total 18 numbers that contain 4, and there are 24 multiples of 4, and there are 6 both.

18+24-6=36

answer : $ \frac{4}{11}$

There are 24 multiples of 4 that are less than 100 and there are 18 numbers that have the digit 4 in them.

That means the probablility is (24+18)/100=42/99.
However, there are 6 common numbers, so the answer is 36/99=4/11

Yes, I got the same answer as you.

just a heads up, there are 19 numbers with 4 in them
here's the list if you want to count them

[hide="numbers with four in them"]
4
14
24
34
40
41
42
43
44
45
46
47
48
49
54
64
74
84
94[/hide]

i got the same answer though

If you have to select different numbers, then 0. IF you must select the same:

122-3 ways
311-3 ways

so $ \frac{6}{11^3}\equal{}\frac{6}{1331}$

[quote="vallon22"]ya, i know
i solved the problem after posting....
anyway, the new problem

there are 2 ways to get a sum of 5 with 3 numbers if order doesn't matter
3, 1, 1
2, 2, 1
there are 165 different sets of numbers you can choose, so our answer is $ \frac {2} {165}$

however, if order does matter, then we have 6 ways to get 5 by choosing 3 numbers out of the 165
3, 1, 1
1, 3, 1
1, 1, 3
2, 2, 1
2, 1, 2
1, 2, 2
therefore, our answer is $ \frac {2} {55}$[/quote]

How did you get 165?

[quote="mewto55555"]11x10x9/6=165

That's the umber of ways you can choose 3 [b]different[/b] numbers.[/quote]

So answer is 0 or 6/1331.

How would you solve if the sum would be not 5, but 18?

[hide]First of all, can you reselect the numbers more than once.  If so the solution is...

The numbers that add up to 5 are 1,2,2 1,1,3.  There are 3 ways to rearrange each group of numbers to get distinct rearrangements.  There are a total of $ 11^3$ combinations so the probability is...

                                            $ \dfrac{6}{1331}$[/hide]

[quote="Bachukas"]
How would you solve if the sum would be not 5, but 18?[/quote]

You could use brutal casework, or you could use generating functions, which does not belong in the middle school forums


symmetry
binomial coefficients
Pascal\u0027s Triangle

Need some help with the following two combinatorics problems:

1.) If  ${8 \choose x} = 28$, compute all possible values of $x$.

2.) How many positive 6-digit numbers contain the digit "1" exactly once, the digit "2" exactly twice, and the digit "3", exactly three times? (For example, 122333, or 231323)

I haven't really played aroud with this problem yet.  Yet, its pretty obvious that x has to be an integer less than 8.  So if one was lazy, or lacking much mathematical creativity they could try every integer less than 8 and greater than 0.

[quote="DPopov"]#1

[hide]
We know that there are exactly two solutions to this, unless $x=4$ which we quickly see is impossible, since $\displaystyle{8 \choose x}=\displaystyle{8 \choose 8-x}$. Thus, we recognize that this is valid for $x=\fbox{2}$ and $x=8-2=\fbox{6}$

[/hide][/quote]

It seems that you did get the right answer, but did you just automatically recognize that the answer is 2 and 6, or did you do it any other way?

#2...I'm tired, i may have a mistake

[hide]
Since there are five spots available we want 3 of them for the 3's, 2 for the 2's and the remaining one for the one. Hence there are $\displaystyle{5 \choose 3} \cdot \displaystyle{3 \choose 2} \cdot 1=10 \cdot 3=\fbox{30}$ different combos.

[/hide]

You can always use the formal definition to figure it out algebraically but I knew it right away pretty much. Also, the symmetry of two solutions is obvious.

The only way to do #1 is to happen to remember what values ${8\choose n}$ can take on, or to calculate them yourself and see which work.  For number 2, I believe that DPopov is wrong -- it should actually be [hide]60, not 30 -- where he wrote ${5 \choose 3}$ he should have written ${6 \choose 3}$ (since there are 6 digits).[/hide]

[quote="MaThS"]It seems that you did get the right answer, but did you just automatically recognize that the answer is 2 and 6, or did you do it any other way?[/quote]

As you do more and more problems with the binomial coefficients you begin to understand them much more.  Here are some of the things I would have thought of when doing this problem:

The binomial coeffiecents for $\binom{n}{k}$ are symmetric (think of Pascal's triangle) so if I find one answer I know how to get the other one.

The answer can't be 0 or 1 because I know that $\binom{n}{0}$ and $\binom{n}{1}$ are 1 and $n$, respectively.  So then we try 2 and it works.  The other number will be $8-2=6$.

To see how we found the second solution consider the identity:

\[\binom{n}{k}=\binom{n}{n-k}.\]

This is where we get $8-2=6$ for the second solution.  Now there are few different ways to show the validity of this identity.

First let's try straight algebra.  To do this we need to know that $\binom{n}{r}=\frac{n!}{r!(n-r)!}:$

\begin{eqnarray*}
\binom{n}{k} &=& \binom{n}{n-k}\\
&=& \frac{n!}{(n-k)!(n-(n-k))!}\\
&=& \frac{n!}{(n-k)!k!}\\
&=& \binom{n}{k}
\end{eqnarray*}

You could also use committee forming.  Choosing a $k$ member committee from $n$ people is the same as choosing the $n-k$ ppl not to be on the committee.

You could also use a block walking argument with Pascal's triangle....taking $k$ lefts is equivalent to taking $k$ rights when going to the $n$th row.

If you don't know much about the binomial coefficients and how they relate to Pascal's triangle I suggest that you at least try to learn a little bit about them.  You can ask more questions here on the forum and someone will help you if you still don't understand.

For number 2: Another way to think (and this should be a part of  your toolbox)

[hide]Permutation of 6, there is a set of  3 identical digits, and another set of two identical digits. so you get 6!/((3!)*(2!))[/hide]

answer to number 2 is equivalent of an answer of a question that's like this:
find the number of ways to arrange ABBCCC in different ways.
the answer is equal to
6C1 * 5C2 * 3C3 or just 6! / (3!2!1!) = 60


If $ a$, $ b$, $ c$, and $ d$ are the solutions of the equation $ x^4 \minus{} bx \minus{} 3 \equal{} 0$, then an equation whose solutions are
\[ \frac {a \plus{} b \plus{} c}{d^2}, \frac {a \plus{} b \plus{} d}{c^2}, \frac {a \plus{} c \plus{} d}{b^2}, \frac {b \plus{} c \plus{} d}{a^2}
\]is

$ \textbf{(A)}\ 3x^4 \plus{} bx \plus{} 1 \equal{} 0\qquad \textbf{(B)}\ 3x^4 \minus{} bx \plus{} 1 \equal{} 0\qquad \textbf{(C)}\ 3x^4 \plus{} bx^3 \minus{} 1 \equal{} 0$
$ \textbf{(D)}\ 3x^4 \minus{} bx^3 \minus{} 1 \equal{} 0\qquad \textbf{(E)}\ \text{none of these}$

[hide]

The sum of the roots, $ a \plus{} b \plus{} c \plus{} d$, is equal to the coefficient of $ x^3$, or $ 0$. We wish to find the polynomial with roots $ \minus{} \frac {1}{d}$, $ \minus{} \frac {1}{c}$, $ \minus{} \frac {1}{b}$, and $ \minus{} \frac {1}{a}$. Reversing the coeffecients of the polynomial gives a new polynomial with roots that are the recipricoals. So, we now have $ \minus{} 3x^4 \minus{} bx^3 \plus{} 1 \equal{} 0$. For any $ f(x)$, we can find a new polynomial $ g(x)$ with roots that are the negative values of the roots of $ f(x)$ by evaluating $ f( \minus{} x)$. So, our new polynomial is $ \minus{} 3x^4 \plus{} bx^3 \plus{} 1 \equal{} 0$. If we multiply this by $ \minus{} 1$, we get $ \textbf{(D)}$
[/hide]

[quote="lifeisacircle"][hide]
Reversing the coeffecients of the polynomial gives a new polynomial with roots that are the recipricoals. [/quote]

Can you please tell me what theorem this is so i can google it- reciprocal root theorem?
I think i have prooved it to myself
[hide]
$ P(x) \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c)(x \minus{} d) \equal{} 0$

$ G(x) \equal{} \left(x \minus{} \frac {1}{a}\right)\left(x \minus{} \frac {1}{b}\right)\left(x \minus{} \frac {1}{c}\right)\left(x \minus{} \frac {1}{d}\right)$

$ \Rightarrow G(x) \equal{} \frac {1}{abcd} (ax \minus{} 1)(bx \minus{} 1)(cx \minus{} 1)(dx \minus{} 1)$

Now it is easier to see that it is true.

[/hide][/hide]

lifeisacircle: [hide]
I think you meant $ \textbf{(D)}$
[/hide]

[hide]From vieta, $ 0 \equal{} a\plus{}b\plus{}c\plus{}d$, $ 0 \equal{} ab\plus{}\cdots$, $ b \equal{} abc\plus{}\cdots$, $ \minus{}3 \equal{} abcd$.

Since $ \minus{}d \equal{} a\plus{}b\plus{}c$, the first root becomes $ \minus{}1/d$, and similarly, the roots of the new polynomial are $ \minus{}1/d,\minus{}1/c,\minus{}1/b,\minus{}1/c$. From vieta, the constant term is

$ 1/(abcd) \equal{} \minus{}1/3$, and the coeff of $ x^3$ is $ \minus{}(\minus{}1/a\minus{}1/b\minus{}1/c\minus{}1/d) \equal{} (abc\plus{}\cdots)/(abcd) \equal{} b/(\minus{}3) \equal{} \minus{}b$, thus the polynomial is D.[/hide]


The angle of elevation from Lone Pine to the top of Mr. Whitney is 10 degrees and 50 minutes.  Van Dong Le, travleing 7.00 km from Lone Pine along a straight, level road toward Mt. Whitney, finds the angle of elecation to be 22 degrees and 40 minutes.  Find the height of the top of Mt. Whitney aboe the level of the road.

I am having trouble drawing the diagram.  Will someone please help me solve the problem?

Draw a right triangle $ABC$ with angle $B=90degree$ and angle $ACB=10degrees50min$

$M$ on $BC$ such that $CM=7$ and angle $AMB=22degrees40min$

That's the diagram. $A$ is the top of Mt. Whitney. $B$ is the bottom of the mountain. $C$ is Lone Pine and $M$ is the point 7.00km from Lone Pine (where Van Dong Le stops). Now you have to find $AB$.


The square piece of paper shown is folded over so that vertex A lies on vertex C. The paper is folded again so that vertex C lies on the midpoint of DB. Given that the length of one side of the square is 6cm, find the length in centimeters of the second fold. Express your answer as a decimal rounded to the nearest hundredth.
[asy]size(150); import geometry_dev;
draw(unitsquare);
label("$A$",(0,1),NW); label("$B$",(1,1),NE); label("$C$",(1,0),SE); label("$D$",(0,0),SW);
perpendicularmark((0,1),SE); perpendicularmark((1,1),SW); perpendicularmark((1,0),NW); perpendicularmark((0,0),NE);[/asy]

The length of the fold is simply half of the diagonal on the square.

$ \boxed{3\sqrt{2}}$

However, the problem states to express your answer as a decimal rounded to the nearest hundredth.
$3\sqrt2 \approx \boxed{4.24}$


Εστω οι a,b,c πραγματικοι και διαφοροι μεταξυ τους ωστε $ (a\minus{}b)\sqrt[3]{1\minus{}c^3}\plus{}(b\minus{}c)\sqrt[3]{1\minus{}a^3}\plus{}(c\minus{}a)\sqrt[3]{1\minus{}b^3}\equal{}0$.Να δειχθει οτι $ (1\minus{}a^3)(1\minus{}b^3)(1\minus{}c^3)\equal{}(1\minus{}abc)^3$.

Ωραία άσκηση, αλλά νομίζω πως δεν έπρεπε να μας πεις να χρησιμοποιήσουμε την ταυτότητα του Euler. Την κάνει αρκετά πιο εύκολη.

My solution:
Thetoume:
$ x\equal{}(a\minus{}b)\sqrt[3]{1\minus{}c^3},y\equal{}(b\minus{}c)\sqrt[3]{1\minus{}b^3},z\equal{}(c\minus{}1)\sqrt{1\minus{}a^3}$,exoume
$ x\plus{}y\plus{}z\equal{}0$.Ara

$ x^3\plus{}y^3\plus{}z^3\equal{}3xyz \Leftrightarrow \sum_{cyc}(a\minus{}b)^3(1\minus{}c^3)\equal{}3(a\minus{}b)\sqrt[3]{1\minus{}c^3}(b\minus{}c)\sqrt[3]{1\minus{}b^3}(c\minus{}1)\sqrt{1\minus{}a^3} \Leftrightarrow \sum_{cyc}a^3\minus{}3a^2b\plus{}3ab^2\minus{}b^3\minus{}a^3c^3\plus{}3c^3a^2b\minus{}3c^3b^2a\plus{}c^3a^3\equal{}3(a\minus{}b)\sqrt[3]{1\minus{}c^3}(b\minus{}c)\sqrt[3]{1\minus{}b^3}(c\minus{}1)\sqrt{1\minus{}a^3}\Leftrightarrow \left(\sum_{cyc}\frac{ab(1\minus{}c^3)}{(b\minus{}c)(c\minus{}a)} \right)^3\equal{}(1\minus{}c^3)(1\minus{}b^3)(1\minus{}c^3)$
Ara arkei ndo
$ \left(\sum_{cyc}\frac{ab(1\minus{}c^3)}{(b\minus{}c)(c\minus{}a)} \right)\equal{}(1\minus{}abc) \Leftrightarrow
\sum_{cyc}ab(a\minus{}b)(1\minus{}c^3)\equal{}(1\minus{}abc)(a\minus{}b)(b\minus{}c)(c\minus{}a)$,pou isxuei...
Q.E.D.
Pantws ontos ean den eleges ton Euler i askhsh tha itan polu pio duskoli.....


inequalities calculus induction AMC USA(J)MO USAMO percent

Prove that $ (n!)^2 \geq n^n$. Anyone with a non-calculus solution?

hm i think induction will work

1) $ n\equal{}1$

$ (1!)^2\equal{}1^1$ - true

2) $ n\equal{}k$

Let $ (k!)^2\ge k^k$

3) $ n\equal{}k\plus{}1$

we have to figure out whether $ (k\plus{}1)!^2\ge(k\plus{}1)^{k\plus{}1}$ or not

$ (k\plus{}1)!^2\equal{}(k!)^2*(k\plus{}1)^2 \implies (k!)^2\ge(k\plus{}1)^{k\minus{}1}$

since $ k^k$ is always greater or equal to $ (k\plus{}1)^{k\minus{}1}$(you can prove it by easy induction),

$ (k!)^2\ge k^k \ge (k\plus{}1)^{k\minus{}1} \implies (k!)^2\ge (k\plus{}1)^{k\minus{}1}$

Q.E.D.

[quote="FantasyLover"]
since $ k^k$ is always greater or equal to $ (k \plus{} 1)^{k \minus{} 1}$(you can prove it by easy induction),
[/quote]

The USAMO will dislike that.

Note that because

$ (n!)^2 \equal{} \prod_{k\equal{}1}^n {k} * \prod_{k\equal{}1}^n {(n\plus{}1 \minus{} k)} \equal{} \prod_{k\equal{}1}^n {k(n\plus{}1\minus{}k)}$
and since

$ k(k\minus{}1) \leq n(k\minus{}1)$
$ k(1\minus{}k) \geq n(1\minus{}k)$
$ k(n\plus{}1 \minus{} k) \geq n$,

we have

$ (n!)^2 \equal{} \prod_{k\equal{}1}^n {k(n\plus{}1\minus{}k)} \geq \prod_{k\equal{}1}^n {n} \equal{} n^n$

[quote="Yongyi781"]The USAMO will dislike that.[/quote]

$ k^k > (k\plus{}1)^{k\minus{}1} \iff k \geq \left( 1 \plus{} \frac{1}{k} \right) ^{k\minus{}1}$

The RHS can be easily shown to be less than $ 3$ -- see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236644]this thread[/url].

Then, the inequality is clearly true for $ k \geq 3$.

It is easy to show it is true for $ 1$ and $ 2$.

[quote="TZF"][quote="Yongyi781"]The USAMO will dislike that.[/quote]

$ k^k > (k \plus{} 1)^{k \minus{} 1} \iff k \geq \left( 1 \plus{} \frac {1}{k} \right) ^{k \minus{} 1}$

The RHS can be easily shown to be less than $ 3$ -- see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236644]this thread[/url].

Then, the inequality is clearly true for $ k \geq 3$.

It is easy to show it is true for $ 1$ and $ 2$.[/quote]He's referring to the exposition of the proof. You really don't want to write ''you can prove it by easy induction'' when that step needs some explanation or additional proof of such claims for USAMO.

fantasy lover's goals are to make USAMO, not necessarily do well on it. Therefore, who cares if is explanation isn't the best. Also, what percent of people actually write out USAMO level solutions more often then not? I'm not trying to offend, just pointing out some relevant points.

But I should point out that most (if not all) of the people who aim to make USAMO want to try their best on it and get as good a score as possible that they can get. Not necessarily make Red MOP, but at least put forth some effort.


calculus
integration
function
floor function
geometry
rectangle
derivative

Compute $ \int_2^5\int_1^3\lfloor x \plus{} y\rfloor\,dx\,dy.$

Well, I considered the inner integral and then I made the following: $ \int_{1}^{3}{\left\lfloor x \plus{} y \right\rfloor \,dx} \equal{} \int_{1 \plus{} y}^{3 \plus{} y}{\left\lfloor t \right\rfloor \,dt} \equal{} \underbrace{\sum\limits_{j \equal{} 1 \plus{} y}^{2 \plus{} y}{\int_{j}^{j \plus{} 1}{j\,dt}}}_{(*)} \equal{} 2y \plus{} 3.$

On $ (*)$ I'm not sure at all if that's correct, since nothing proves that $ y\in\mathbb Z,$ but the thing is that this actually works, because last integral yields $ 30$ which is the value of the whole double integral.

The value of the integral is 3(1/2) + 4(3/2) + 5(4/2) + 6(3/2) + 7(1/2) = 30. In order to see this, you need to tile the rectangle [1,3]x[2,5] with polygonal regions where the integrand is constant.

Ah, but the step with the $ (*),$ does it hold?

Your are right to have doubts, since for non-integer $ y$ the step (*) is not justified. The reason why you still right result is that for any real $ y$
\[ \int_y^{y\plus{}1} \lfloor t \rfloor dt \equal{} y.\]
You can either check this by calculation or note that it is true for $ y\equal{}0$, and check that both sides have derivative $ 1$ (using the fundamental theorem of calculus for the left side). So if you skip step (*), but use the above you directly get
\[ \int_{y\plus{}1}^{y\plus{}3} \lfloor t \rfloor dt \equal{} y\plus{}1\plus{}y\plus{}2\equal{}2y\plus{}3.\]