Find possitive integers a, b , c, d such that :
 $ a \leq b \leq c \leq d =n^2 $
 $ a+b+c+d = m^2 $
 $ a^2 + b^2 +c^2 +d^2 = 2021 $

I don't know what you mean , Myth !!! :?  :?  This problem is so long that it take me 45' to write down its solution . I need a better way !!!!  :(  :(

:D  :D  :D .
Firstly, use the fact that [tex] 2021 \geq n^4 \geq \frac{2021}{4} [/tex] to have [tex] n \in \lbrace {5,6} \rbrace [/tex] . Use something like that too, we have [tex] m \in \lbrace {8,9} \rbrace [/tex].
However,in the case [tex] n=6 [/tex]  we have [tex] a^2+b^2+c^2=725 [/tex] and [tex] a+b+c=m^2-36 \geq 28 \Rightarrow ab+ac+bc \leq 29 [/tex], and this is impossible.
So [tex] n=5 [/tex]  and [tex] a^2+b^2+c^2=1396 [/tex] and [tex] a+b+c=m^2-25 [/tex] .It easy to see that $ a,b,c $ are even then $ m=9$. Hence $ a+b+c = 56 $. Take $ a=2a';b=2b',c=2c' $ and then we have:
[tex] a'^2+b'^2+c'^2=349 [/tex]  and
[tex] a'+b'+c'=28 \Rightarrow 2(a'b'+b'c'+c'a')=435[/tex] .

Uh, how boring it is !. It must have another solution or this one is really a monster. ;)

Of course, it is an very easy problem, but it is so ugly :(

There are some mistakes in your post , Anh Cuong !!
 If $ n=6 ==> ab +bc +ca >29 $ .
 We have only one root :
   $ a= 10 ; b= 15; c= 20; d =36 $

Oh, I am sorry. So you must have the conclusion that:
[tex] a+b+c=45 [/tex] and [tex] a^2+b^2+c^2=725 [/tex]. Use something like [tex] 2(b^2+c^2) \geq (b+c)^2 [/tex] to have [tex] a\geq 10 [/tex].
In addition [tex] a \leq 14 [/tex]. And the rest is caculator's job.  :)


$ \triangle ABC$ is a right triangle. The altitude from C meets the hypotenuse AB at D and CD is six inches shorter than twice BD. If DC is subtracted from BD, the result is one inch more than AD. Find BD in inches.

I think you should put a space between \triangle and ABC in your LaTeX code, lol.

[asy]size(200);
pair A=(0,0), B=(4,0), C=(4,3), D=foot(B,A,C);
D(B--C--A--B--D);
D(rightanglemark(A,B,C));
D(rightanglemark(A,D,B));
MP("A",D(A),SW); MP("B",D(B),SE); MP("C",D(C),NE); MP("D",D(D),NW);[/asy]
[color=darkred]Given: $ CD+6=2BD$ and $ BD-CD=AD+1$
Find: $ BD$[/color]

[hide="Solution"]Rewrite the first known equation as $ CD=2BD-6$. Now, substitute $ CD$ into the second known equation to get $ BD-2BD+6=AD+1$, or $ 5-BD=AD$.

Using similar triangles, $ \frac{CD}{BD}=\frac{BD}{AD}$.

Substituting the expressions for $ AD$ and $ CD$ in terms of $ BD$ into that equation, we get:

\begin{align*}\frac{2BD-6}{BD}&=\frac{BD}{5-BD}\\\\
10BD-2BD^2-30+6BD&=BD^2\\\\
3BD^2-16BD+30&=0\end{align*}
And we get an irrational solution? :huh:[/hide]

i_like_pie, AB is the hypotenuse, not AC.

[hide]
Let $ BD\equal{}x$.
Then $ CD\equal{}2x\minus{}6$ and $ AD\equal{}5\minus{}x$. $ \triangle ADC \sim \triangle CDB$.
$ \frac{BD}{CD}\equal{}\frac{CD}{AD}$
$ \frac{x}{2x\minus{}6}\equal{}\frac{2x\minus{}6}{5\minus{}x}$
$ 5x^2\minus{}29x\plus{}36\equal{}0$
$ (5x\minus{}9)(x\minus{}4)\equal{}0$
$ x\equal{} \frac{9}{5}$ makes $ CD$ negative, so $ x \equal{} \boxed{4}$
[/hide]


calculus
integration
factorial
algebra unsolved
algebra

Given that $a,b,c,d\in \mathbb{N}$ , find all the solution for the following equation

$a!=b!+c!+d!$

 :)

Because obviously $a>max (b,c,d)$, $a-1\geq max(b,c,d)$ thus
$a!=a.(a-1)!\geq a.(max(b,c,d))!$, so for a>3 $a!>3.(max(b,c,d))!\geq b!+c!+d!$ and the eq cannot be satisfied.
For a=1 and a=2 is the left side lower then 3 and the right side greater or equal to 3, so a=3.
a=3, b=c=d=2 works (6=2+2+2). But didn't manage to prove there arent any other solutions for a=3 ;)

[quote="Kondr"]Because obviously $a>max (b,c,d)$, $a-1\geq max(b,c,d)$ thus
$a!=a.(a-1)!\geq a.(max(b,c,d))!$, so for a>3 $a!>3.(max(b,c,d))!\geq b!+c!+d!$ and the eq cannot be satisfied.
For a=1 and a=2 is the left side lower then 3 and the right side greater or equal to 3, so a=3.
a=3, b=c=d=2 works (6=2+2+2). But didn't manage to prove there arent anz other solutions for n=3 ;)[/quote]

ya , you are correct  :)


[quote="Jutaro"]Let $f(x)$ be a cubic polynomial with rational coefficients. If the graph of $f(x)$ is tangent to the $x$ axis, prove that the roots of $f(x)$ are all rational.[/quote]

If $x=r$ is the point where the graph of $f(x)$ is tangent to the $x$ axis, then $f(r)=f'(r)=0$ and r is at least a double root.
Polynomial division of $f(x)$ by $f'(x)$ gives $f(x)=(ax+b)f'(x)+px+q$, where $(a,b,p,q)$ are all rationals.

So we have $pr+q=0$ and r is rational.
Then polynomial division of $f(x)$ by $(x-r)^{2}$ gives $f(x)=(ux+v)(x-r)^{2}$, with $(u,v)$ rational and hence the third root is rational too.

--
Patrick

actually i do not understand the solution .please help me giving a detailed explanation and solution.

please help,

Hence all coefficients of $ f(x)$ are rationals, all coefficients of $ f'(x)$ are rationals.

So

[quote="pco"]
Polynomial division of $ f(x)$ by $ f'(x)$ gives $ f(x) \equal{} (ax \plus{} b)f'(x) \plus{} px \plus{} q$, where $ (a,b,p,q)$ are all rationals.
[/quote]

And hence $ f(r)\equal{}f'(r)\equal{}0$ ,

[quote="pco"]
So we have $ pr \plus{} q \equal{} 0$ and r is rational.
[/quote]

And then all coefficients of $ (x\minus{}r)^{2}$  are rationals, so

[quote="pco"]
Then polynomial division of $ f(x)$ by $ (x \minus{} r)^{2}$ gives $ f(x) \equal{} (ux \plus{} v)(x \minus{} r)^{2}$, with $ (u,v)$ rational and hence the third root is rational too.
[/quote]

[quote=DavyBa]The statement of this problem is incorrect. If f(x) = x^3 + 2x, then its three roots are: 0, i√2, and -i√2 which are not all rational. Its graph is tangent to the X axis in x = 0. The derivative f`(0) is not equal to 0, but the second derivative f``(0) is. The statement probably should be: "all its real roots are rational".[/quote]

Does the polynomial f(x) = x^3 + 3x have 0 as a root of multiplicity 2? Not... right?...

No, it does not have a root of 0 multiplicity 2

[quote=DavyBa]The statement of this problem is incorrect. If f(x) = x^3 + 2x, then its three roots are: 0, i√2, and -i√2 which are not all rational. Its graph is tangent to the X axis in x = 0. The derivative f`(0) is not equal to 0, but the second derivative f``(0) is. The statement probably should be: "all its real roots are rational".[/quote]

Actually, the graph is not tangent to the $x$-axis at $x=0$. This is called an "inflection point" or "saddle point", where it sort of curves out. These are identified when $f''(x)=0$.

To solve the original problem, calculus actually is not required.

[hide=Solution]
It is well known that a polynomial is tangent to the $x$-axis at $x=r$ when the multiplicity of $r$ in the polynomial is a positive even integer.

Thus, the multiplicity must be $2$. Let $f(x)=a_3x^3+a_2x^2+a_1x+a_0$. Then, the sum of the roots (counting multiplicity) is $-\frac{a_2}{a_3}$, or a rational number. Thus, the final root is just $-\frac{a_2}{a_3}-2r$, which is a rational number. Thus, all three roots of $f(x)$ are rational.

Q.E.D.
[/hide]

Yes, you are correct. I realized it myself just recently. The graph of this polynomial is not tangent to X axis since the first derivative at x = 0 is positive. It was a temporary blackout. :)
Thank you for your reply.

Also, thank you for posting the solution. May I ask you about one detail in this solution for my understanding?
Since the multiplicity of r in such special cubic polynomial is 2, it's clear that such cubic polynomial has 3 real roots: r, r, and some real root q. It's also clear that the sum 2r+q (and the product (r^2)q, etc.) of these three real roots are rational (Vieta's formulas). How does it prove that all three roots are rational?

It's clear how to prove it by using calculus. Let's examine two types of cubic polynomials whose graphs are tangent to X axis. Firstly, consider polynomials f(x) = a(x-b)^3 with rational a and b, that have root r = b with multiplicity 3. Both first and second derivatives f`(b) = f``(b) = 0, so its graph is tangent to X axis and b is the inflection point. All three roots are r = b (which is a rational number).
The second case is when root r of polynomial ax^3 + bx^2 + cx + d has multiplicity 2. Then, r is a local extremum. In this case, f(r) = f`(r) = 0. Polynomial division f(x)/f`(x) gives us the remainder px + q in which p and q are rational numbers. From f(r) = f`(r) = 0 it follows that pr + q = 0 and that r is a rational number. From this it follows by Vieta's formula that the third root is also rational. Q.E.D.


I making a personality quiz that goes like the following.

User takes a test to determine which type of person they are more like. So there will be say 5 types, each given a percentage of 100 that they are closest to at quiz end. The test has 50 questions, plus a age question and a sex question. Each test has 5 possible answers. The user clicks through radio button to fill out the quiz. No sweat so far.

My issue is in the test scoring, this is were i get confused. When making a test, each answer is weighted to fit a specific personality type. The weights are shown on screen as +6 thru 0 thru -6. (In code as 1-13, zero the default get 7 points). I have tried a few different scoring algos but non seem to be accurate. One other thing, I have to assume that a value other than zero WILL NOT be chosen when building a quiz.

1st try - Multiple count of all questions times 7 (7 is middle or 0 on scale), then divide by total points gotten from weighted answers. This ends up being something like 350/372 = 94%. This is all good till you get a score like 350/352 = 99%. In percentage you can assume that the second score is higher, but in reality the first score had the greater point spread.

2nd try - Add the highest value (up to 13) from each questions answers together for each personality type, then divide by total points gotten from weighted answers. This ends up something similar to the above try, but with higher percentages going to those personality types with answers that have no weights attached to them.

OK so now that i have written this Im really thinking that the quizzes will not work if weights are not set to other than zero (aka 7 points). Am I totally off here? Can anyone lead me in a proper direction. Thanks much for helping my keep what hair is left on my head.

Cesar

If all you are going for is most a like, then why not try just assigning questions like this. A=1 B=2 C=3 D=4 E=5 . Then you could add their answers together, then the people that match would be the closest number. Do you care if there are ties? A person might have more then one match...That should keep it simple for you. You must be a VB user... Or if you are going for a certain range maybe just do a select case statement... Total >5 and <8 then this person is this type. Get my drift? I would just keep it simple.

I may be misunderstanding what you want, but why don't you just do $ \frac{\text{points earned}}{\text{points possible}}$ (adding 6 to each answer to make the answer always nonnegative)?

Example:

Question 1:  -4
Question 2:  -3
Question 3:  4

Result:  $ \frac{(\minus{}4\plus{}6)\plus{}(\minus{}3\plus{}6)\plus{}(4\plus{}6)}{12\plus{}12\plus{}12}\equal{}\frac{15}{36}$?


limit
calculus
integration
real analysis
real analysis solved

$\{t_n\}$ is a sequence of numbers. Define $E = \{ x \in \mathbb{R}: \lim\limits_{n\to\infty} e^{i t_n x} \mbox{ exists.} \}$ and $m(E)>0$. Show that $\{t_n\}$ is convergent.

[b]Case 1.[/b] Suppose $\{t_n\}$ is bounded. If $\{t_n\}$ has two limit points $a$ and $b$ then we conclude $\lim_{n\to \infty}e^{ixt_n}=e^{iax}$ and $\lim_{n\to \infty}e^{ixt_n}=e^{ibx}$. It means $e^{i(a-b)x}=1$ for all $x\in E$,so $(a-b)x=2\pi k$ for all $x\in E$, i.e. $E$ is a countable set, so $m(E)=0$ -- contradiction.

[b]Case 2.[/b] Suppose $\{t_n\}$ is unbounded. Suppose wlog $t_{n_k}\to+\infty$. Denote $t_{n_k}$ as $t_n$. Since $m(E)>0$ then there exists segment $I=[n,n+1]$ s.t. $m(I\cap E)>0$.
It is clear that for all $x,y\in E$ and $n,m\in\mathbb Z$ number $nx+my\in E$. It follows that $m(I\cap E)=m(I)=1$. Now open http://www.mathlinks.ro/Forum/viewtopic.php?t=19082 and see my solution.

So you utilize the fact that $m(E\cap I) =1$ by using integral, and Lebesgue dominant convergence theorem!!
 It took me a while to realize that


Support
college
MATHCOUNTS
geometry
geometric transformation
reflection
Berkeley Math Circle

The Joint Mathematics Meeting is an annual conference held jointly by several big math organizations.  This year it is held in San Francisco.  What's special this year is that the first national "Who Wants to be a Mathematician" contest (sponsored by AMS, American Math Society) takes place at this meeting.  The previous WWTBAM contests were all local runs at various locations, including the one at Southern California in December 2008, hosted by Math Zoom.

The contest will be held tomorrow (Thursday Jan 14, 2010) at the Moscone Conferece Center.  Ten students from all over US were chosen to participate in this contest.  The winner of this contest receives $ \$$5000 as prize.  In fact all participats for this final event receive a cash prize.

I am glad to say that Kevin Yin, one of the students from Math Zoom, has been selected.  AMS covers the airfare and hotel for him and a parent.  I am also here today in San Francisco, going to the contest tomorrow to support him.  I will report how it goes.

I've been getting updates from a colleague of mine who is attending the meetings. It sounds like the winner was Evan O'Dorney.

Yes, I failed when I thought that distributive property worked the opposite way.

One of the problems was a ridiculous counting problem that everybody ran out of time on, and that I managed to guess right (lucky). The rest were solvable. On #2 I was able to simplify on time but apparently nobody else was.

The winner was O'Dorney, and second place was Ben Zouzmer (sp?). In the finals, apparently Zouzmer decided the only way he would get any points was by quick guessing skills, and indeed that is how he managed to get points.

Well, the second group of semifinalists, including Kevin Yin and Evan O'Dorney, were much stronger than the first group.  The problems were harder too.  To be fair to Kevin, he was the leader at the time of the last problem.  The last problem was the one with the distributive and associative properties of a newly defined operation.  And after the problem was displayed on screen, the Indian boy had a problem with his remote control, and Mike Breen, the game host, spent some time to sort it out, during which everyone answered, and he still allowed the full 90 seconds time after that.  I was thinking, OMG, there's no way anyone can get this one wrong, and Kevin actually beat Evan O'Dorney!  Alas, that was not to be  :( ...  Still, Kevin did a good job.

Anyway, Evan does deserve to win.  He was lightning fast at the final face-off; the other contestant didn't have a chance.  Last time I saw him was at the 2007 MOSP, he was just a very young kid at the time (but in Blue MOP).

I will write more details later.

The national "Who Wants to be a Mathematician" contest was held on Thursday, January 14, 2010 at the Joint Mathematics Meetings in San Francisco.  This contest was held by Ameircan Mathematical Society.  Math Zoom hosted this contest in Southern California in December 2008.  The San Francisco contest was the first time that this contest was run on a national scale.  The 10 contestant are:

- Rohit Agrawal, Junior at Wayzata High School (MN)
- Rebecca Easterwood, Senior at Shades Valley High School (AL)
- Brian Freidin, Senior at Glenbrook North High School (IL)
- Daniel Li, Senior at Thomas Jefferson High School for Science and Technology (VA)
- Kathy Lin, Senior at Los Alamos High School (NM)
- Ofir Nachum, Senior at Algonquin Regional High School (MA)
- Evan O'Dorney, Junior in the Berkeley Math Circle (CA)
- Charles Xu, Junior at Fairview High School (CO)
- Kevin Yin, Junior at San Marino High School (CA)
- Ben Zauzmer, Junior at Upper Dublin High School (PA)

The contestants were selected based on scores on the national qualifying test.  Kevin Yin, a Math Zoom student, represented Southern California.  All these students have very strong backgrounds in math contests.

The contestants were divided into two groups to have two semifinal games, where the contestants answered 8 questions, each with specified time limit and increasing point values.  The first group were Rebecca Easterwood, Brian Freidin, Daniel Li, Ofir Nachum, and Ben Zauzmer.  Ben Zauzmer was the winner of this group.  The second groups included Rohit Agrawal, Kathy Lin, Evan O'Dorney, Charles Xu, and Kevin Yin.  Evan O'Dorney won this group.  Kevin did well by leading the first 7 rounds, but Evan O'Dorney, twice IMO silver medalist, won on the last question.  Then it was the final face-off between Ben and Evan.  The final round ran like the Countdown Round in MATHCOUNTS: for each question, whoever gets the correct answer first wins the points.  Evan O'Dorney obviously had the advantage with faster reflect and more expereince, so he won with no suspense.  Evan received the top prize of $ \$$5000 for himself and $ \$$5000 for his school.  The other 9 contestants also received cash and other prizes for themselves and their schools, at least $ \$$500 each.


induction
number theory unsolved
number theory

Sequence $ a_1,a_2,......$ satisfy $ \displaystyle\sum_{d|n} a_{d}\equal{}2^n$ for all natural number $ n$.

Prove that $ n|a_n$ for all n

[quote="Baronov"]Let $ a_{n}$ be the number of sequences of length $ n$ of 0,1 which cannot be obtained by writing some smaller sequence k (k>1) times. I.e 010101 is not such sequence since it can be obtained by writing 01 3 times. It is not difficult to prove that this sequence is the same as the sequence from the problem statement(the same identity is true for it). Now prove that if we take the last symbol from such sequence and put it in the first place we get again such sequence. So by this operation the $ a_{n}$ sequences split into classes of equivalence. Each class has $ n$ sequences in it. So $ a_{n}$ is divisible by n.[/quote]

I can't understand what it means. You want to talk about binary system?? If you do, I think there are many $ a_n$ which you talked.

For example, n=4. 1001, 1011, 1100, 1000...... etc   Please explain what you mean.

$ a_{n}$ is the number of all such sequences. For n=2. $ a_{2}\equal{}2$ because the sequences 01 10 count. And the sequences 00 and 11 don't count.

OK, before I read your solution, I should write mine...
For every $ n$, main idea is to prove that if $ p^\alpha \parallel{} n$ (highest power of $ p$ which divides $ n$) then $ p^\alpha | a_n$. Then this is true for every prime $ p$ which divides $ n$, so $ n|a_n$.

For this we use induction. Suppose that for every $ k < n$, $ k|a_k$. Let $ n \equal{} p^\alpha m$ where $ p \not | m$, and $ 1 \equal{} d_1 < d_2 < ... < d_k \equal{} m$ are all positive divisors of $ m$. Then
\[ \sum_{d|n} a_d \equal{} (\sum_{d|\frac {n}{p}}a_d) \plus{} a_{p^\alpha} \plus{} a_{p^\alpha d_2} \plus{} ... \plus{} a_{p^\alpha m} \equal{} 2^{p^{\alpha \minus{} 1}m} \plus{} a_{p^\alpha} \plus{} ... \plus{} a_{p^\alpha d_{k \minus{} 1}} \plus{} a_n \equal{} 2^{p^\alpha m}.
\]
Now because $ p^{\alpha} | a_{p^\alpha d_i}$ (bacuse of the induction), and $ p^{\alpha} | 2^{p^\alpha m} \minus{} 2^{p^{\alpha \minus{} 1}m} \equal{} 2^{p^{\alpha \minus{} 1}m}(2^{p^{\alpha \minus{} 1}(p \minus{} 1)m} \minus{} 1 )$ because $ p^{\alpha \minus{} 1}(p \minus{} 1) \equal{} \varphi(p^\alpha)$, then $ p^\alpha | a_n$. That's it

[edit:]@Baronov: I read your solution, and I am very interested in it. Can you please post as complete solution as you can? Thanks
[edit 2:] Ok, I get the solution :)

Baronov, I understand what you mean.  I will try that idea.

And, it's very nice solution, spongebob. Thank you very much!


calculus
integration
modular arithmetic
linear algebra
matrix
number theory solved
number theory

let $A=(a_{ij})$ and $B=(b_{ij})$  be two n x n matrices with integral entries. suppose $a_{ij}\equiv b_{ij}\pmod{m}$ for all i,j. then prove $detA\equiv detB\pmod{m}$

it's not very formal, but I think the idea is right. I can't explain in english very well...
[hide]
since $a_{ij}-b_{ij} \equiv 0 \mod m$ we can write that $a_{ij}=b_{ij}+k_{ij}m$, $k_{ij} \in \mathbb {Z}$
So we can substitute in matrix $A$ for each $a_{ij}$ his value in terms of $b_{ij}+k_{ij}m$. To calculate $\det A$ we only
find elements equal to the ones that we find calculating $\det B$ plus some element all $\equiv 0 \mod m$
So $\det A-\det B \equiv 0 \mod m$
[/hide]

It follows from

$detA= \sum_{ \sigma \in X } \epsilon (\sigma)a_{1 \sigma(1)}...a_{n \sigma(n)}$

and the property

$a \equiv b \pmod{ m}$ and $c \equiv d \pmod{ m} \Rightarrow a c \equiv b d \pmod{ m}$


trigonometry
calculus
function
geometry
trig identities
Law of Sines

Prove that $\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$.

Why can't I post a problem's source in any forum other than this one?

Perhaps because the admins haven't made it possible for you yet? Anyway, trig and proofs do not belong in the MC forum, by order of the admins, no matter how easy, so to Getting Started this goes...

well, $\sin\frac{\pi}{4}=\sin 45^{\circ}$.  $\cos\frac{\pi}{4}=\cos 45$.
$\cos (90^{\circ}-\theta)=\sin\theta\implies\cos (90^{\circ}-45^{\circ})=\cos 45^{\circ}=\sin 45^{\circ}$.

And the proof is complete.

Masoud Zargar

But you didn't prove what they're equal to.

Just take a unit square with its diagonal, that'll do the trick ;)

lol, I thought you knew the rest...joking.

Well, you can take a unit square, and do $\sqrt{x^2+y^2}=\sqrt{2}$.  The hypotenuse is as a result $\sqrt{2}$. $\sin\theta=\frac{opp}{hpp}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$.  :)

EDIT: $\cos\theta=\frac{adj}{hpp}$.  $opp=adj=1$, so $\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$.
Masoud Zargar

We could also create a right triangle with two 45 degree angles.  The law of sines tells us that the sides opposite those angles must have equal length.  If we have a right triangle with sides $x$ and $x$ then the hypoteneuse is $x\sqrt{2}$.  From this triangle $sin(x)=cos(x)=\frac{x}{x\sqrt{2}}=\frac{1}{\sqrt{2}}$

[quote="JesusFreak197"]I'm not following your reasoning.

Try it using $\cos \frac{\pi}{2}=0$.[/quote]

How do you mean? like this?

$\cos 2(\frac{\pi}{4})=\cos \frac{\pi}{2}=0$
$\cos^{2}\frac{\pi}{4}-\sin^{2}\frac{\pi}{4}=0$
$\cos \frac{\pi}{4}=\sin \frac{\pi}{4}$

Yeah, that's more what I had in mind, but you still didn't prove that they're equal to $\frac{1}{\sqrt{2}}$. Sorry for how weird this seems; my Calculus book basically builds a lot of stuff up from the foundations, including trig functions, but I can't really put down as a given everything that it's proven so far. What's intended is a purely algebraic solution; so, although Shogia's solution works as well, I can't do that based on what's been proven so far in the book. ;) (And don't worry, I've already solved it. It's fairly straightforward if you've been following the book, but I thought it was an interesting exercise to do.)

prototype: try using another equivalent of $\cos 2x$.

Isn't it as easy as:
$cos(45)=sin(90-45)=sin(45)$
Since they are equal, we use the fact that we are in the unit circle:
(I replace $sin(45)$ and $cos(45)$ with $t$.

$t^2+t^2)=1\\ t^2=1/2\\ t=1/\sqrt{2}$

[quote="JesusFreak197"]As I said, that does work, but I'm looking for a purely algebraic approach, not geometrical in any way.[/quote]
What is your 'algebraic' definition of sine and cosine?

Algebraic? Okay. Since $\sin {x} = \cos {(\pi/2 - x)}$, we have $\sin {\pi/4} = \cos {\pi/4}$. Now you want me to prove that it is equal to $\frac{1}{\sqrt{2}}$? Well $\sin {\frac{\pi/2}{2}} = \sqrt{\frac{1-\cos {\pi/2}}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$...

I don't really understand why you want an "algebraic" approach though... :huh:

[quote="white_horse_king88"]Since when was trigonometry allowed in Getting Started? Strange... :?[/quote]
[quote="MCrawford"]Throughout the history of this site it has been difficult to keep the Getting Started Forum at the appropriate difficulty level.  For this reason I have decided to create a guide to help users understand what material is appropriate for this forum.

First, an explanation of suitable problems:

$\bullet$ Most AMC problems (between the first 5 and the last 3-5)
$\bullet$ Harder MATHCOUNTS problems
$\bullet$ The very simplest AIME problems are also reasonable
$\bullet$ Most problems from most local competitions
$\bullet$ The easier problems from current ARML (or about half of those from 10 to 25 year old ARML/NYSML contests)
$\bullet$ Most math league problems, though harder problems such as the last 3 or so individual problems on a Mandelbrot competition should go in Intermediate.
$\bullet$ Most AoPS Volume 1 material.  Volume 2 nearly always belongs in the Intermediate Topics forum and sometimes the Pre-Olympiad forum (non-trivial proofs).

[b]Note: If a problem involves a non-trivial step using more than one area of mathematics (algebra with number theory, geometry with serious algebra, etc.) then it does not belong in the Getting Started forum.  Remember that the average score on a typical AIME is 2-3 problems.  That's given the fact that AIME participants already score well on a qualifying exam.  At most 1 in 20 AIME problems belong in this forum.[/b][/quote]
So, yes, simple trigonometric questions are allowed in Getting Started because simple trigonometry have appeared in AMC test questions, local competitions and so on, BUT just use commonsense because harder trigonometry problems belong to Intermediate Forum. Use responsibly.

[quote="JesusFreak197"]
prototype: try using another equivalent of $\cos 2x$.[/quote]

Ohh. I get what you mean:

$\cos 2(\frac{\pi}{4})=1-2\sin^{2}\frac{\pi}{4}=0$

$\sin (\frac{\pi}{4})=\frac{1}{\sqrt{2}}$

The same for cosine


let $ (a,b,c)>0$ Prove or disprove  that : $ \frac{24(ab\plus{}ac\plus{}bc)}{(a\plus{}b\plus{}c)^2} \plus{} \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \geq 11$
equality occurs  when $ a\equal{}b\equal{}c$ and $ a\equal{}2b\equal{}2c$
[hide] the inequality is equivalent to : $ (a\minus{}b)^2( \frac{1}{2ab} \plus{} \frac{a\minus{}b}{6abc} \minus{}\frac{4}{(a\plus{}b\plus{}c)^2}) \geq 0$  :wink: [/hide]

[quote="anas"]let $ (a,b,c) > 0$ Prove or disprove  that : $ \frac {24(ab \plus{} ac \plus{} bc)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq 11$
equality occurs  when $ a \equal{} b \equal{} c$ and $ a \equal{} 2b \equal{} 2c$
[/quote]
I have obtained that it's true. :maybe:

BTW , I think that the following is also true : with the same condition we have :
$ \frac {2\sqrt {2}(ab \plus{} ac \plus{} bc)}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq 3 \plus{}2\sqrt {2}$  :wink:

[quote="anas"]let $ (a,b,c) > 0$ Prove or disprove  that : $ \frac {24(ab \plus{} ac \plus{} bc)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq 11$
equality occurs  when $ a \equal{} b \equal{} c$ and $ a \equal{} 2b \equal{} 2c$.[/quote]
This inequality is wrong. Try $ a\equal{}7$, $ b\equal{}6$ and $ c\equal{}17$.

you are right Vasc , anyway it was just a conjecture , and I Had a feeling  that it was true  :oops:

[quote="Vasc"][quote="anas"]let $ (a,b,c) > 0$ Prove or disprove  that : $ \frac {24(ab \plus{} ac \plus{} bc)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq 11$
equality occurs  when $ a \equal{} b \equal{} c$ and $ a \equal{} 2b \equal{} 2c$.[/quote]
This inequality is wrong. Try $ a \equal{} 7$, $ b \equal{} 6$ and $ c \equal{} 17$.[/quote]
Yes, you are right, Vasc.
This inequality is wrong because $ 27t^3 \minus{} 252t^2 \plus{} 480t \minus{} 256\leq0$ is wrong for $ 1\leq t\leq\frac {26 \plus{} \sqrt {460}}{9},$
where $ t \equal{} \frac {(a \plus{} b \plus{} c)^2}{3(ab \plus{} ac \plus{} bc)}$ and $ a> b> c.$

[quote="Vasc"]This inequality is true for $ a\le b\le c$.[/quote]
Yes, since
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {1}{2}\left( \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b}\right)
\]
with $ a\le b \le c$.


Two players alternately write integers on a blackboard as follows: the first player writes $ a_{1}$ arbitrarily, then the second player writes $ a_{2}$ arbitrarily, and thereafter a player writes a number equal to the sum of the two preceding numbers. The player after whose move the obtained sequence contains terms such that $ a_{i}-a_{j}$ and $ a_{i+1}-a_{j+1}(i\neq j)$ are divisible by $ 1999$, wins the game. Who has the winning strategy

I suppose that $ a_{1},a_{2}>0$

Let $ \{F_{n}\}$ be the Fibonacci numbers, with $ F_{1}=F_{2}=1$ and $ F_{n}= F_{n-1}+F_{n-2}, \ \forall n\geq 3$

For $ \{a_{n}\}$ we have  $ \boxed{a_{n}= a_{n-1}+a_{n-2}}\ , \ \forall n\geq 3\ (1)$

$ \hline$
$ \star$ [color=brown][size=134]We'll see that [b]j=1[/b][/size][/color]

set $ m=1999$

Suppose that there are $ x,y$ with $ x>y$ such that $ \left\{\begin{array}{l}a_{x}\equiv a_{y}\mod m\\ a_{x+1}\equiv a_{y+1}\mod m \end{array}\right\}\ (2)$

Then from $ (1)$ we get $ \boxed{a_{x+k}\equiv a_{y+k}\mod m}\ , \ \forall k$

From $ (2)$ and $ a_{n}=a_{n+2}-a_{n+1}$ we get $ \boxed{a_{x-k}\equiv a_{y-k}\mod m}\ , \ \forall k =1,2,...,y-1$

For $ k=y-1$ and for $ k=y-2$ we get:

$ a_{x-y+1}\equiv a_{1}\mod m$
$ a_{x-y+2}\equiv a_{2}\mod m$

When $ a_{x-y+2}$ appears in the sequance, the game ends. So $ i=x-y+1$ and $ j=1$

So the winner is the player who adds the numbers $ a_{x-y}$ and $ a_{x-y+1}$ to provide the number $ a_{x-y+2}$

The 1st player provides the numbers $ a_{1},a_{3},a_{5},...$, hence the odd indexed numbers
The 2nd player provides the numbers $ a_{2},a_{4},a_{6},...$, hence the even indexed numbers

[color=red]If $ x-y$ is an odd number then the 1st player is the winner
If $ x-y$ is an even number then the 2nd player is the winner[/color]

$ \hline$
[size=134][color=brown]Case 1. The sequence $ a_{n}$ is the Fibonacci sequence[/color][/size]

Suppose that $ a_{1}=a_{2}=1$. So $ \{a_{n}\}=\{F_{n}\}$

So we care about the [url=http://www.research.att.com/~njas/sequences/A001175]Pisano periods (period of Fibonacci numbers in modulo m)[/url]

[url=http://www.research.att.com/~njas/sequences/b001175.txt]Here[/url] we can see that the Pisano period for $ m=1999$ is $ 666$

So the smallest positive integer $ c$ with the properties

$ a_{c+1}\equiv a_{1}\mod m$
$ a_{c+2}\equiv a_{2}\mod m$

is the number $ c=666$

$ c+2$ is even, so [b][u]the 2nd player wins[/u][/b]
$ \hline$
[size=134][color=brown]Case 2. Generalization for any sequence $ a_{n}$[/color][/size]

We'll prove that the period $ c=666$ is always the same, independently of $ a_{1}$ and $ a_{2}$

[color=brown][size=150]Theorem[/size] $ \boxed{a_{n}= F_{n-2}\cdot a_{1}+F_{n-1}\cdot a_{2}}\ , \forall n\geq 3$[/color]

[color=blue][size=150][hide="Proof"][/size]For $ n=3$ we have
$ a_{3}=F_{1}\cdot a_{1}+F_{2}\cdot a_{2}\iff$
$ a_{3}=a_{1}+a_{2}$, true

Suppose that the theorem holds for any positive integer $ <n$. We'll prove that it holds also for $ n$

$ a_{n-1}= F_{n-3}\cdot a_{1}+F_{n-2}\cdot a_{2}$
$ a_{n-2}= F_{n-4}\cdot a_{1}+F_{n-3}\cdot a_{2}$

$ a_{n}=a_{n-1}+a_{n-2}\Rightarrow $

$ a_{n}= (F_{n-3}+F_{n-4})\cdot a_{1}+(F_{n-2}+F_{n-3})\cdot a_{2}\Rightarrow $
$ a_{n}= F_{n-2}\cdot a_{1}+F_{n-1}\cdot a_{2}$[/color][/hide]$ \hline$

Let $ m$ be a positive integer
Let $ c$ be the Pisano period of the number $ m$
That is, $ c$ is the smallest positive integer such that $ F_{n+c}\equiv F_{n}\mod m , \ \forall k$

We'll prove that is is also period for the sequence $ \{a_{n}\}$

[hide="c is one period of a_n"]$ \boxed{a_{n}= F_{n-2}\cdot a_{1}+F_{n-1}\cdot a_{2}}$

$ a_{n+c}= F_{n-2+c}\cdot a_{1}+F_{n-1+c}\cdot a_{2}\Rightarrow $
$ a_{n+c}\equiv (F_{n-2}\cdot a_{1}+F_{n-1}\cdot a_{2}) \mod m\Rightarrow $
$ a_{n+c}\equiv a_{n}\mod m$[/hide]
[hide="c is the smallest period of a_n"]We define the sequence $ \boxed{K_{n}=a_{2}\cdot a_{n+1}-a_{1}\cdot a_{n+2}}$. We'll prove that $ F_{n}=\frac{K_{n}}{K_{1}}$

$ a_{n+1}= a_{1}\cdot F_{n-1}+a_{2}\cdot F_{n}$
$ a_{n+2}= a_{1}\cdot F_{n}+a_{2}\cdot F_{n+1}$

$ K_{n}= a_{2}(a_{1}\cdot F_{n-1}+a_{2}\cdot F_{n})-a_{1}(a_{1}\cdot F_{n}+a_{2}\cdot F_{n+1})$

$ K_{n}= a_{1}a_{2}\cdot F_{n-1}+a_{2}^{2}\cdot F_{n}-a_{1}^{2}\cdot F_{n}-a_{1}a_{2}\cdot F_{n+1}$

$ K_{n}= (a_{2}^{2}-a_{1}^{2})\cdot F_{n}-a_{1}a_{2}\cdot (F_{n+1}-F_{n-1})$

$ K_{n}= (a_{2}^{2}-a_{1}^{2})\cdot F_{n}-a_{1}a_{2}\cdot F_{n}$

$ K_{n}= (a_{2}^{2}-a_{1}^{2}-a_{1}a_{2})\cdot F_{n}$

$ K_{n}= K_{1}\cdot F_{n}$

So it is $ F_{n}=\frac{a_{2}\cdot a_{n+1}-a_{1}\cdot a_{n+2}}{K_{1}}\ \ (3)$

We assume that there is a smaller period for $ \{a_{n}\}$, that is, there is $ r<c$ such that $ a_{n+r}\equiv a_{n}\mod m, \ \forall n$.

From $ (3)$ we get $ F_{n+r}=\frac{a_{2}\cdot a_{n+r+1}-a_{1}\cdot a_{n+r+2}}{K_{1}}\equiv \frac{a_{2}\cdot a_{n+1}-a_{1}\cdot a_{n+2}}{K_{1}}= F_{n}$

So $ F_{n+r}\equiv F_{n}\mod m$, contradiction, because $ c$ is the smaller period of $ \{F_{n}\}$[/hide]

[size=150]Conclusion[/size]
$ c=666$ is a period for the sequence $ \{a_{n}\}$, independently of $ a_{1},a_{2}$
We have
$ a_{c+1}\equiv a_{1}\mod 1999$
$ a_{c+2}\equiv a_{2}\mod 1999$

$ c+2$ is even, so [b][u]the 2nd player wins[/u][/b]

Well, pontios, I'm surprised why your solution looks so complicated and involves something like the Pisano period. Isn't there any easier way?


[quote="mr. math"]if a and b are distinct, odd primes, then how many distinct positive factors does 4a^2b^3 have?

how would u do this problem?[/quote]

Is that expression $4a^{2}b^{3}$ or $4a^{2b^{3}}$?

[hide="hint"]to find The number of factors of a number prime factorize it, add one to the exponents, and multiply these together.[/hide]
The first one klebian

I got that too. it's a well-known procedure... can you figure out why it works??

Has something to do with the Counting Principle.

Because the positive whole number $N=p_{1}^{a_{1}}p_{2}^{a_{2}}...p_{k}^{a_{k}}$, so any one of $N$'s factors can be written as $p_{1}^{i_{1}}*p_{2}^{i_{2}}...p_{k}^{i_{k}}$, where $i_{1}=0,1,2,...a_{1}, ... i_{k}=0,1,2,...,a_{k}$.

Therefore, to get a factor of N there are k steps. The first step is choosing $p_{1}^{i_{1}}$ and there are $a_{1}+1$ ways to do so. The second step there are $a_{2}+1$...the kth step has $a_{k}+1$ steps. So by the Counting Principle there are $(a_{1}+1)(a_{2}+1)...(a_{k}+1)$ factors of N.

[quote="nebula42"][quote="Arvind_sn"]Yes, that's what I got as well. :)  However, can anyone find a way to prove that for any number $a^{b}*c^{d}*....$ where a and c (and other prime factors) are prime, The number of divisors is b+1 times C+1 times...(so on, so on)? I have been working with that rule for a year now and I still do not have the proof.[/quote]
To generalize what bpms wrote, let's say $n={p_{1}}^{e_{1}}{p_{2}}^{e_{2}}{p_{3}}^{e_{3}}\dots$ (where the $p_{i}$ are primes). Consider the number $x ={p_{1}}^{d_{1}}{p_{2}}^{d_{2}}{p_{3}}^{d_{3}}\dots$, where $d_{i}= 0, 1, 2, \dots,$ or $e_{i}$ for any $i$. Then we can see that any number of this form is a factor of $n$, and any factor of $n$ can be expressed in this form. This means that the number of factors of $n$ is equal to the number of possibilities for $x$, which is $(e_{1}+1)(e_{2}+1)(e_{3}+1)\dots$.

For a specific example of this, see bpms's post above.[/quote]
A proof by Neal Wu, that is foolproof...

Yeah, it's actually just some fairly simple combinatorics...


calculus
derivative
function
algorithm
integration
real analysis
real analysis unsolved

Let $ V$ be an $ n$-dimensional subspace of $ C([0,1])$ consisting of smooth functions with the sup-norm. Prove that the norm of the differentiation operator $ D: V\to C([0,1])$ is at least $ 2n\minus{}2$ and that this estimate is sharp.

[quote="fedja"]Let $ V$ be an $ n$-dimensional subspace of $ C([0,1])$ consisting of smooth functions with the sup-norm. Prove that the norm of the differentiation operator $ D: V\to C([0,1])$ is at least $ 2n \minus{} 2$ and that this estimate is sharp.[/quote]
Very nice :)
The challenge to show that any such $ n$-dim subspace contains a function $ f$ such that $ |f(x)| \le 1$ with equality at at least $ n$ points in [0,1], and that the sign of $ f$ alternates at these points. The result then follows from the mean value theorem, since at least two adjacent points in this set must have a distance $ \le \frac {1}{n \minus{} 1}$.

This is related to the [url=http://en.wikipedia.org/wiki/Remez_algorithm]Remez algorithm[/url] for best $ L^\infty$-approximation.

[edited 10/15]

Yes, it is a nice solution though I had a simpler approach in mind when posting this problem: just split $ [0,1]$ into $ n\minus{}1$ equal subintervals and consider the function $ f\in V$ whose integral over each of these subintervals is $ 0$ :)


If $ 2^3 \times 4 \times 8 \equal{} 2^{k}$, what is the value of $ k$?

By base two, don't you mean in terms of 2? Base 2 is binary, like $ 1001001110$.

You can express it has base 2 but you really shouldn't. I think he meant powers of 2.

Perhaps he meant $ \log^2 {8}$.  By that, I mean that the $ 2$ is low, not high.  :blush:


I have to create a balsa wood bridge for my physics class and our instructor is giving us 40g (maximum) of balsa wood to work with. We can only use elmer's white glue to assemble the bridge and the less wood we use, the better. But we want to support a large amount of weight which will be stranded from the center of the bridge. Any ideas for construction that will allow for maxmum weight? My prof said his balsa bridge in high school held over 100 pounds.

To make this a proper Physics problem could you specify some details for those unfamiliar with Balsa? eg. Density, Young's Modulus, how bendable it is, Properties of the glue etc.?

Many designs for efficient bridge building are already known. You may want to experiment with some of them.

P.S.: Shouldn't this be in the Astronomy and Engineering forum ?


conics
ellipse
analytic geometry
graphing lines
slope

In the xy-plane, what is the length of the shortest path from $ (0, 0)$ to $ (12, 16)$
that does not go inside

1. the circle $ (x \minus{} 6)^2 \plus{} (y \minus{} 8)^2 \equal{} 25$

2. the ellipse $ (x\minus{}6)^2/4 \plus{} (y\minus{}8)^2/9 \equal{} 1$

3. both?

My method is a little long-winded (it involves equating the slopes of the tangents and other messy algebra); I'm looking for a quick solution, if any.

[hide="The Key Observation"]The ellipse is entirely contained within the circle.[/hide]
[hide="Solution"]Ignore the ellipse because it is contained entirely within the circle.

Our path will have three parts:
Segment OA: A line segment from the origin to the circle, and will be tangent to the circle where it hits
Arc AB: Follows the circle for a particular arc
Segment BC: Will fly off tangent to the circle in a line segment, and end at the destination

Observe that the center of the circle is the midpoint of the start and ending segments of the path, so our path is symmetric.

For part 1: Let Z be the center of the circle. Observe that OA^2+R^2=OZ^2=6^2+8^2
OA^2+25=100
OA = 5 rt(3)

Part 3 has the same length as well; 5 rt(3)

Part 2:
Note that measure(OZA) = measure(BZC)=60, and measure(OZC)=180.
Thus, measure(AZB)=180-60-60=60, or 1/6 of the circumference of the circle.
The circumference is 2*pi*5 = 10pi, so this part has length (5/3)pi

Thus, the shortest path has length 10*rt(3) + (5/3)*pi[/hide]


modular arithmetic
algebra
polynomial
induction
number theory unsolved
number theory

If $p >3$ is a prime, and $a, b$ are positive integers, then
\[
\binom{ap}{bp} \equiv \binom{a}{b} \pmod{p^3}
\]

Does anybody have a nice and quick way of proving this result? (and is there a name for it?)

You only have to prove that $C_{2p}^p - 2 = 0 \mod [p^3]$ or equivalentely that $(2p)! - 2(p!)^2 = 0 \mod[p^5]$.
Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=21489

Clearly, we only have to look at the case $a \geq b$.
Let's consider an urn containing $ap$ balls, with $p$ balls for each color $c_1, \cdots , c_a$.
Now, choose $bp$ balls from it. There are $C_{ap}^{bp}$ such choices.
In another hand, th total number of choices is also equal to the sume of the numbers of the form $C_p^{i_1}C_p^{i_2} \cdots C_p^{i_a}$, where the sum is over all the $a$-uples $(i_1, \cdots , i_a)$ such that $0 \leq i_j \leq p$ for all $j$ and $i_1 + \cdots + i_a = bp$. (1)

Since $C_p^k = 0 \mod[p]$ for $k=1, \cdots, p-1$, we deduce that modulo $p^3$ this sum reduces to the terms for which at most $2$ are not $0$ or $p$.
- The part of the sum for which all the $i_j$'s are $0$ or $p$ is $C_a^b$ (we only have to choose the $b$ subscripts for which $i_j = p$ to fulfill (1)).
- There cannot be only one of the $i_j$ which is not $0$ or $p$ because the sum of all of them is $0$ modulo $p$ from (1).
- If there are exactly two which are not $0$ or $p$ : then, from (1) they are respectively equal to $k$ and $p-k$ where $k=1, \cdots, p-1$. We have $a(a-1)$ ways to choose these two colors. Then, we have to choose $b-1$ colors among the $a-2$ remaining ones for which we have selected $p$ balls. The others are now determined. Thus, the part of the sum in that case is
$\dsp a(a-1)C_{a-2}^{b-1} \sum_{k=1}^{p-1} C_p^kC_p^{p-k} = a(a-1)C_{a-2}^{b-1}(C_{2p}^p  - 2)$.
Therefore, $C_{ap}^{bp} - C_a^b = a(a-1)C_{a-2}^{b-1}(C_{2p}^p  - 2) \mod [p^3]$, as claimed.

Pierre.

I have Defined $\frac{a}{b} = a b^{-1}$. We know that the usual property of fractions holds in modulo arithmetic inverses.

[color=#0040FF]Lemma 1 :[/color] Work in $\mathbb{Z}_p$. $p$ is a prime. We know that if an $n$th degree polynomial over $\mathbb{Z}[x] $ vanishes for $n+1$ integers, then all the coefficients of that polynomial is divisible by $p$, i.e the polynomial is identically zero.

So by Fermat's theorem $x^{p-1} - 1 = 0$  for $x= 1, \cdots , p-1$. So $x^{p-1}-1 = (x-1) \cdots (x-p+1) = x^{p-1} + \cdots + s_1 x + (p-1)!$. By willson's theorem $(p-1)! = -1$, so we have that $s_{p-2}x^{p-2}+ \cdots + s_{2}x^2 + s_1{x}= 0$ for $x= 1,\cdots , p-1$. So $p \mid s_i $  $\forall i \in \left \{ 1, \cdots, {p-2} \right \}$. Where the definition of $s_i$ is easy to guess.

[color=#0040FF]Wolstenholme theorem : [/color] if $p$ is a prime. Then $1 + \frac{1}{2}+ \cdots + \frac{1}{p-1} \equiv 0 \mod p^2 $.

Proof: $\sum_{i=1}^{p-1}\frac{1}{i} = \sum_{i=1}^{(p-1)/2}\frac{1}{i}+ \frac{1}{p-i}= p \sum_{i=1}^{(p-1)/2}\frac{1}{i(p-i)} $. So it suffices to prove $\sum_{i=1}^{(p-1)/2}\frac{1}{i(p-i)} \equiv 0 \mod p $. Indeed $\sum_{i=1}^{(p-1)/2}\frac{1}{i(p-i)} \equiv - \sum_{i=1}^{(p-1)/2}\frac{1}{i^2} \mod p  $. But note that each $\frac{1}{i^2}$ is congruent to unique number modulo $p$ from the set $\left \{ 1^2, \cdots , \left ( \frac{p-1}{2} \right )^2 \right \}$. So $\sum_{i=1}^{(p-1)/2}\frac{1}{i^2}\equiv \sum_{k=1}^{(p-1)/2}k^2 \equiv \frac{(p^2-1)p}{24} \mod p   $. Since if $p \ge 5$, we have $\gcd(24,p) = 1$, so the sum is congruent to $0$ modulo $p$.  Therefore the theorem is proved.

Now its easy to see that  $s_1 = (p-1)! \left ( 1 +\cdots +  \frac{1}{p-1} \right )  \equiv 0 \mod p^2$

[color=#0040FF]Lemma 2: [/color] Note that $(pk-1)\cdots (pk -p + 1) = (pk)^{p-1}+ \cdots + s_2 (pk)^2+ s_1 (kp) + (p-1)!$. Since $p \mid s_2$ by lemma 1, $p^3 \mid  s_2 (pk)^2$. Since we have just proved that $p^2 \mid s_1$, we get $p^3 \mid  s_1 (kp)$. So we get $(pk-1)\cdots (pk -p + 1) \equiv (p-1)! \mod p^3 $. Since $\gcd((p-1)!, p^3) = 1$, and $(p-1)! \mid (pk-1)\cdots (pk -p + 1) $,

we get $\binom{pk-1}{p-1} = \frac{{}(pk-1)\cdots (pk -p + 1)}{(p-1)!}\equiv 1 \mod p^3$.

[color=#0040FF]Lemma 3: [/color] Note the identity $k\binom{kp-1}{p-1} = \binom{kp}{p}$. So $ \binom{kp}{p} \equiv k\binom{kp-1}{p-1} \equiv k \equiv \binom{k}{1} \mod p^3$

Now we will prove the original problem i.e $\binom{kp}{lp} \equiv \binom{k}{l} \mod p^3$, for prime $p \ge 5$ by induction over $l$. We have already showed that it is true for $l=1$. Assume that it is true for $l=c$. We will show that it is true for $l= c+1$.

We have $\binom{kp}{cp+ p} = \frac{(kp)!}{(cp+p)! (kp-cp - p)!} = \frac{(kp)!(kp-cp)\cdots (kp-cp - p+1)}{(kp-cp)! (cp+p)\cdots (cp+1)(cp)! }$  $= \frac{(kp)!(k-c)(kp-cp-1)\cdots (kp-cp - p+1)}{(kp-cp)! (c+1)(cp+p-1)\cdots (cp+1)(cp)! }$

Note that the numbers $(kp-cp-1), \cdots , (kp-cp-p+1), (cp+p-1) \cdots, (cp+1)$ all are coprime with $p^3$. So

$\binom{kp}{cp+ p} \equiv  \frac{(kp)!(k-c)(kp-cp-1)\cdots (kp-cp - p+1)}{(kp-cp)! (c+1)(cp+p-1)\cdots (cp+1)(cp)! }$
$ \equiv  \frac{(kp)!(k-c)(kp-cp-1)\cdots (kp-cp - p+1)\left ( (cp+p-1)\cdots (cp+1) \right )^{-1}}{(c+1)(kp-cp)! (cp)! }$ $ \mod p^3$

In a manner similar to proving lemma 2: we can show that  $(kp-cp-1)\cdots (kp-cp - p+1) \equiv (p-1)! \mod p^3$, and $ (cp+p-1)\cdots (cp+1) \equiv (p-1)! \mod p^3$

Now using this in above congruence relation, and again noting the fact that $\gcd((p-1)!, p^3) = 1$, we get

$\binom{kp}{cp+ p} \equiv \frac{(kp)!(k-c)(kp-cp-1)\cdots (kp-cp - p+1)\left ( (cp+p-1)\cdots (cp+1) \right )^{-1}}{(c+1)(kp-cp)! (cp)! }$  $   \equiv \frac{(kp)!(k-c)(p-1)!\left ( (p-1)! \right )^{-1}}{(c+1)(kp-cp)! (cp)! }$
$ \equiv \frac{(kp)!(k-c)}{(kp-cp)! (cp)!(c+1)} \equiv \binom{kp}{cp}\frac{k-c}{c+1}$ $ \equiv \frac{k!(k-c)}{c!(k-c)!(c+1)}$  $ \equiv \binom{k}{c+1} \mod p^3 $ and the induction is complete.

Wikipedia claims this is in fact wolstenholme's theorem.
http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem


A square with side length 2 and a circle share the same center. The total area
of the regions that are inside the circle and outside the square is equal to the
total area of the regions that are outside the circle and inside the square. What
is the radius of the circle?

for diagram, click link. problem #25
http://www.math.ksu.edu/main/handbook/ProblemSets/AMC8Problems

solution(first part)-
25. (A) Because the circle and square share the same interior region and the area
of the two exterior regions indicated are equal.

how do you know that is true?

if the area of the square is S and the area of the circle is C and the area of the interior is I, based on the question, you have S-I=C-I. So, S=C.

haha
i remember solving this problem two years ago.
I got this one right!  :D


limit trigonometry inequalities integration ratio logarithms calculus

Compute the following limit:

$ L \equal{} \lim_{n \to \plus{} \infty} \frac{\sin \theta \plus{} \sin \frac{\theta}{2} \plus{} \sin \frac{\theta}{3} \plus{} ... \plus{} \sin \frac{\theta}{n}}{1 \plus{} \frac{1}{2} \plus{} \frac{1}{3} \plus{} ... \plus{} \frac{1}{n}}$

$ L \equal{} \theta,$ since $ \lim_{n\to\infty}\frac {\sin\frac {\theta}n}{\frac {1}n} \equal{} \theta.$ More generally, consider the following:

$ a_n$ and $ b_n$ are (eventually) positive sequences such that $ \lim_{n\to\infty}\frac {a_n}{b_n} \equal{} r > 0$ and $ \sum_{n \equal{} 1}^{\infty}a_n \equal{} \sum_{n \equal{} 1}^{\infty}b_n \equal{} \infty.$ Then:

$ \lim_{n\to\infty}\frac {\sum_{k \equal{} 1}^na_k}{\sum_{k \equal{} 1}^nb_k} \equal{} r.$

Given $ \epsilon > 0,$ ($ \epsilon < r$) find $ N$ such that for $ n > N,\ (r \minus{} \epsilon)b_n < a_n < (r \plus{} \epsilon)b_n.$

Then for $ n > N,$

$ \frac {\sum_{k \equal{} 1}^Na_k \plus{} (r \minus{} \epsilon)\sum_{k \equal{} N \plus{} 1}^nb_k} {\sum_{k \equal{} 1}^Nb_k \plus{} \sum_{k \equal{} N \plus{} 1}^nb_k} < \frac {\sum_{k \equal{} 1}^na_k}{\sum_{k \equal{} 1}^nb_k} < \frac {\sum_{k \equal{} 1}^Na_k \plus{} (r \plus{} \epsilon)\sum_{k \equal{} N \plus{} 1}^nb_k} {\sum_{k \equal{} 1}^Nb_k \plus{} \sum_{k \equal{} N \plus{} 1}^nb_k}$

As $ n\to\infty,$ the outer sides of this chain of inequalities tend to $ (r \minus{} \epsilon)$ and $ (r \plus{} \epsilon),$ respectively. Hence the sum in the middle can be eventually made arbitrarily close to $ r,$ and thus converges to $ r.$ Note why we needed the series to diverge; the result would be false without that.

(Of course, this is at heart a variation on Cesaro-Stoltz.)

A continuous variation: suppose $ f(x)$ and $ g(x)$ are both positive for all sufficiently large $ x,$ continuous on $ [0,\infty),$  that $ \lim_{x\to\infty}\frac{f(x)}{g(x)}\equal{}r>0,$ and that $ \int_0^{\infty}f(x)\,dx\equal{}\int_0^{\infty}g(x)\,dx\equal{}\infty.$ Then

$ \lim_{x\to\infty}\frac{\int_0^xf(t)\,dt}{\int_0^xg(t)\,dt}\equal{}r.$

Note that this can be settled by L'Hôpital's rule.

Thanks for the detailed explanations Professor Merryfield. I originally solved this limit by using directly Cesaro-Stolz theorem (however, the book source said the limit equals theta/2). Another way of solving this with the same theorem:

If lim a_n exists then $ \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \equal{} \lim a_n$. Now, if both lim a_n and lim b_n exist, then

$ \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{b_1 \plus{} b_2 \plus{} ... \plus{} b_n} \equal{} \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \cdot \frac {1}{\frac {b_1 \plus{} b_2 \plus{} ... \plus{} b_n}{n}}$

$ \equal{} \lim a_n \cdot \frac {1}{b_n} \equal{} \lim \frac {a_n}{b_n}$.

In our case, $ a_n \equal{} \sin \frac {\theta}{n}$ and $ b_n \equal{} \frac {1}{n}$, and so, as Professor Kent already stated,

$ L \equal{} \lim \frac {\sin \frac {\theta}{n}}{\frac {1}{n}} \equal{} \theta$.

[quote="Carcul"]If lim a_n exists then $ \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \equal{} \lim a_n$. Now, if both lim a_n and lim b_n exist, then

$ \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{b_1 \plus{} b_2 \plus{} ... \plus{} b_n} \equal{} \lim \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \cdot \frac {1}{\frac {b_1 \plus{} b_2 \plus{} ... \plus{} b_n}{n}}$

$ \equal{} \lim a_n \cdot \frac {1}{b_n} \equal{} \lim \frac {a_n}{b_n}$.[/quote]
This would only work if $ \lim a_n$ and $ \lim b_n$ were nonzero, which means it would not work for this particular problem.

Note that in our particular problem, $ \lim\frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}n\equal{}0,$ and similarly for $ b_n,$ hence the ratio of those quantities is less than fully useful.

also by aproximation for big M we can say $ \sum_{k \equal{} M}^{n} \sin \frac {\theta}{n}$~$ \theta \ln n \plus{} C \plus{} o(1)$ then
$ \sum_{k \equal{} 1}^{n} \sin \frac {\theta}{n}$~$ \theta \ln n \plus{} C' \plus{} o(1)$
and
$ \sum \frac {1}{n}$~$ \ln n \plus{} \gamma \plus{} o(1)$

[quote="Kent Merryfield"]This would only work if $ \lim a_n$ and $ \lim b_n$ were nonzero, which means it would not work for this particular problem.[/quote]

Yes, you are right, thanks. I suppose the same could be said if both limits were infinity.


The sequence $(a_n)$ define : $a_0=1, a_1=1 ,\ \ a_n=2a_{n-1}+a_{n-2}\ \ \forall n>1$
Find the minium natural number $n$ such that $1024 \mid a_n$

If $a_{n-2}$ is odd, then $a_n$ is odd. $a_0,a_1$ are odd, therefore $a_n$ are odd for all n.

I'm sorry ! I have post a wrong problems ! I correct it !
The true problem is :

The sequence $(a_n)$ define : $a_0=0, a_1=1 ,\ \ a_n=2a_{n-1}+a_{n-2}\ \ \forall n>1$
Find the minium natural number $n$ such that $1024 \mid a_n$

We can easily find that the charasteristic equation of this sequence is
 $a_n=\frac{(1+\sqrt2)^n+(1-\sqrt2)^n}{2\sqrt2}$
 And we can show that $a_n$ is odd when $n$ is odd.
 Assume that $n=2^\alpha p$ where $\alpha\ge1$ and $p$ is odd positive integer.
 From opening brackets and cancelling common terms out we get
 $a_n=(\begin{array}{c}2^{\alpha}p\\1 \end{array} )+(\begin{array}{c}2^{\alpha}p\\3 \end{array} )2+...+(\begin{array}{c}2^{\alpha}p\\2^{\alpha}p-1 \end{array} )2^{2^{\alpha-1}p-1}$
We can easily prove that $2^{\alpha+1}|(\begin{array}{c}2^{\alpha}p\\ 2k+1\end{array}).2^k$ for all $k\ge1$.
So smallest such $n$ is $10$. ;)

P.S.It is a special case of shorlisted problem of IMO 1988.

just calculating $a_n(n\leq 10)$
then it can be solved.
and we can apply induction to prove
$2^n$is a divisor of $a_n$but $2^{n+1}$isn't


Prove that for $ p\equiv3(mod4)$ with $ p$ as a prime,
$ \sum_{k \equal{} 1}^{p \minus{} 1}k^2(\frac {k}{p}) \equal{}p\sum_{k \equal{} 1}^{p \minus{} 1}k(\frac {k}{p})$

$ \sum_{k\equal{}1}^{p\minus{}1}k^2(\frac kp)\equal{}\frac 12 \sum_{k\equal{}1}^{p\minus{}1}[k^2(\frac kp) \plus{}(p\minus{}k)^2(\frac{p\minus{}k}{p}]\equal{}\frac 12 \sum_{k\equal{}1}^{p\minus{}1}(\frac kp )[k^2\minus{}(p\minus{}k)^2]\equal{}\minus{}\frac{p^2}{2}\sum_{k\equal{}1}^{p\minus{}1}(\frac kp )\plus{}p\sum_{k\equal{}1}^{p\minus{}1}k(\frac{k}{p})\equal{}p\sum_{k\equal{}1}^{p\minus{}1}k(\frac kp ).$


Each block on the grid shown in Figure is 1 unit by 1 unit.  Suppose we wish to walk from $ A$ to $ B$ via a 7 unit path, but we have to stay on the grid -- no cutting across blocks.  How many different paths can we take?[asy]size(3cm,3cm);int w=5;int h=4;int i;for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label("B", (w-1,h-1), NE);label("A", (0,0), SW);[/asy]

R : advance to the right
U : advance to the up
The route supports that I arrange two R and two U to one line.
Therefore,
$ \dfrac{(4\plus{}3)!}{4! \ 3!}\equal{}\binom{7}{3} \equal{} \boxed{35}$.


Let $a,b,c$ be distinct real numbers.The sequence $\{x_n\}$ is the arithmetic sequence whose the first three terms are $a,b,c$ in this order and the sequence $\{y_n\}$ is the geometric sequence whose the first three terms are $c,a,b$ in this order.

(1) Express $b,c$ and the common difference of $\{x_n\}$ in terms of $a.$
(2) Find $\sum_{k=1}^8 y_k$ in terms of $a.$

(3) Let $\{z_n\}$ be the sequence whose the first three terms are $b,c,a$ and the progression of differences for the sequence is an arithmetic sequence.Find the $n$-th term of $z_n$ in terms of $a.$

[quote="Altheman"][hide="part 1"]by definition of the sequences,
$a^2=b+c$, and $a+b=2c$
$2a^2-2b=a+b$
$\frac{2a^2-a}{3}=b$
a similar process shows $c=\frac{a^2+a}{3}$
then the common difference is $c-b=\frac{2a-a^2}{3}$[/hide][/quote]

are you sure ? I think it should be

[hide]
$a+c=2b$ , $bc=a^2$ $\implies (c-b)(c-4b)=0$ $\implies c=4b$ since $a,b,c$ are distinct .

Hence $b=-\frac{a}{2}$,$c=-2a$ and the common different = $c-b=-\frac{3a}{2}$
[/hide]

[hide="part 2"]
$y_n=(-2a)(-1/2)^{n-1}$

So $\sum^8_{k=1}y_k=(-2a)\frac{1-(-1/2)^8}{3/2}=-\frac{511a}{384}$
[/hide]

[hide="part 3"]
$/{z_n/}=c-b,a-c...$

$c-b=-3a/2$ , $a-c=3a$ . Hence the common different of $z_n$ is $3a-(-3a/2)=9a/2$

If we denote the progression different of $z_n$ be $z'_n$ then $z'_n$ is an arithmetric sequence with first term $c-b=-3a/2$ and common different $9a/2$

Hence $z_n=z_1+\sum^{n-1}_{k=1}z'_n=-3a/2 +3a/2\sum^{n-1}_{k=1}(3k-4)=\frac{3a(3n-2)(n-3)}{4}$
[/hide]

You seemed to make a mistake in the calculation of Part 2,3.


Point A is the intersection of y=6/x and y=(3/4)x:^2:. Point O is the origin. Point B is also on y=6/x and is sqrt(37) away from point O. Point P is on the y-axis. Triangle ABO has the same area as triangle APO. What is the coordinate of point P?

My solution is: [hide]Since A is where 6/x and 3/4x :^2: intersect, then the equations are set so that 6/x=3/4x :^2:   6=3/4x :^3:   8=x :^3:   2=x.  Then plug x into either equation to get y=3. Then graph 6/x.  Since B is  :sqrt: 37, it is  :sqrt: (1 :^2: +6 :^2: ) (remember my Henry Dudeney Problem  ?).  Since 6/x is in the first and third quadrants, B can be (1,6), (6,1), (-6,-1), and (-1,-6).

Since [ABO]=[APO] and P is on the y-axis, the area is 2S/2 or S since height is 2.  S is area of ABO.  With B at (1,6), using Pick's Theorem, we get [ABO] 3/2+4-1=9/2.  P can be (0,4 1/2) and

(0,4 1/2).  With B at (6,1), [ABO]=4/2+7-1=8.  P can be (0,8) and (0,-8).  We don't need to [ABO] with B at (-1,-6) or (-6,-1), because [ABO] with B at (-1,-6) :equiv: [ABO] with B at (1,6) and [ABO] with B at (-6,-1) :equiv: [ABO] with be at (6,1)

Ergo, P can be at (0,-8), (0,-4 1/2), (0, 4 1/2), and (0, 8)[/hide]

I apologize for tweaking it incorrectly, the original problem just gave out the coordinates of A and B so I changed it, but I forgot to say that B is in the first quadrant, sorry frodo.

Anyways can you explain what "Pick's Theorem" is anyway? I think it might be useful to know...

You got the correct answers though frodo, good job :)

Pick's Theorem explains the area of any polygon on a graph.  The formula is (Border points)/2+(inner points)-1.

ahem. i'm fairly sure the polygon has to have vertices at lattice points.

(the inner pts. and boundary pts. in your formula also refer to lattice pts., methinks)

MysticTerminator (a.k.a. Bob) is correct.


function
vector
calculus
integration
topology
real analysis
real analysis solved

I have a solution for this, but it's rather unenlightening.  Hence I'll break up the problem into two parts.

Let $C([0,1])$ be the set of continuous functions on the interval $[0,1]$.  Define our metric as $d(f,g)=\vert \vert f-g\vert \vert_{\infty}$ (the sup norm)

Let $B=\{f\in C([0,1]):\vert \vert f\vert \vert_{\infty}\leq 1\}$

a) Prove that $B$ is not compact
b) Exhibit an open cover of $B$ with no finite subcover.

If what you're looking for is an example of a metric space together with a closed bounded non-compact subset, a simple example would be this: the real line with the metric $d(x,y)=\min(1,|x-y|)$.

The whole space is bounded, so every subset is bounded, but if you cover it with open intervals of length $<1$, you can't extract a finite collection of intervals which still covers the space.

No; I already know a few such examples.  Take $x\in\mathbb{Q}$ such that $2<x^2<3$.  It's clearly bounded, if I remember correctly its closed, but its not compact.

I was just doing this in class this week, so I might as well post the example I was using.

Let $f_n$ be the piecewise linear continuous function with vertices at $(0,0),\,\left(\frac1n,1\right),\,(0,1).$

For all $n$, $||f_n||_{\infty}=1,$ so they are all contained in the closed unit ball $B$ of $C[0,1].$

We can directly calculate that if $n>m,\,||f_n-f_m||_{\infty}=1-\frac mn.$

The sequence is not a Cauchy sequence, and no subsequence of it can be a Cauchy sequence. Thus we have a sequence with no convergent subsequence.

We already know the closed unit ball is not compact. We can use this example to construct our open cover with no finite subcover.

For each $k$, let $U_k$ be the ball centered at $f_{2^k}$ of radius $\frac14.$ Then each ball contains only one of the $f_{2^k}.$ Let $U_0$ be the original set $B$ (the closed unit ball) with each of the points $f_{2^k}$ deleted. Since points of this sequence have no accumulation point, the set containing them all is already closed and the complement is open.

$U_0\cup\left(\cup_kU_k\right)$ is an open cover of $B.$ A finite subcollection of these open sets can cover at most finitely many of the $f_{2^k}$ and is thus not a cover of $B.$

In that last example, I cut down to the subsequence $f_{2^k}$ just for the convenience of using balls of the same radius. We could have used the full original sequence if we'd wanted to.

Well-known fact: in a normed vector space, the closed unit ball is compact iff the vector space is finite dimensional.

Out of curiosity, what level course was that for?

Junior level required course for math majors. In our version, the second semester of a two semester sequence. Our first semester is mostly concerned with basic epsilon mechanics; the second semester does the Riemann integral, uniform convergence, and some material (such as this) on metric spaces.


Palindromes are numbers that read the same backwards and forwards, like 5665. What is the least possible positive four-digit palindrome that is divisible by 3?

Since the palindrome has to start and end with the same number, to make it the smallest it should start and end with 1. Now we need to get the smallest value that adds up to 3 because of 3's divisibilty property. 3 wouldn't work because 4 positive numbers couldn't add up to 3. So try 6. We need to numbers that add up to 4, that are the same. 2 works!  So the number is 1221.

[quote="quaksterjohny"]3 wouldn't work because 4 positive numbers couldn't add up to 3.[/quote]

Well, we could have $ 3000$...what you should say is that the sum can't add to $ 3$ because the sum of the digits of palindromes with an even number of digits is always even.


I've only seen it twice, but now i'll look out for it and tell you if i find out. ;)

[quote="nat mc"]I love that commercial! I need help with just one name that i can never catch. It goes:

Pedro: Cisse

Jose: Kaka

Pedro: Zidane

Jose: Beckham

Pedro: Defoe

Jose: Kahn

Pedro: ?????????

Jose: Beckenbauer

Pedro: Platini

What is it??[/quote]

I thik it is Messi or something like that but I don't know.  Those are pretty sweet commercials.

[quote="Drunken_Math"]There's actually three of those I think.

One is the one when they choose their players. Another is when they actually play and a goal blocked hits a person trying to fix their car, I think. Another one was when they play and a shot goes off the top crossbar and the goalie grabs it and they argue bout whethers its a goal or not and then jose gets called in by his mom.

Those commercials are pretty cool. :D[/quote]

Yep those were pretty cool.

How did you guys know the name was Pedro again?

[quote="Drunken_Math"]There's actually three of those I think.

One is the one when they choose their players. Another is when they actually play and a goal blocked hits a person trying to fix their car, I think. Another one was when they play and a shot goes off the top crossbar and the goalie grabs it and they argue bout whethers its a goal or not and then jose gets called in by his mom.

Those commercials are pretty cool. :D[/quote]

I think there are actually two of each one minute length, but they are cut sometimes to 30 seconds.

first one- they choose teams, and go over lineups and strategies and coinflip etc all the way up to the kickoff

second one- they play. defoe blcoks a shot and it hits the car. jose sends kaka to the bench. lampard shoots and it hits the crossbar and Kahn disputes it wasnt in. then jose's mom calls him home

Yes, now that I see it again, it definitely does sound like messi, though in the second one, i don't see him playing

I have seen the one where Kahn blocks a goal, and Lampard starts celebrating with Cisse and all. Then Beckenbauer is shown once. Then Vieri comes and speaks to Ballack while Kahn keeps protesting he has made a save.

There is another one where a lady starts scolcding the players from a balcony. Jose takes the ball and starts going home.

Yeah, but I still don't understand how you were able to interpet that the name of the boy against Jose is Pedro.  I couldn't figure this out.

Did I write Vieri? I meant Vieira of course!
By the way, I have not seen the first one yet, and am not sure if that is shown here, so I can't help you. Can you find it on the net and out the link here?


calculus
integration
logarithms
geometry
limit
trapezoid
calculus computations

Let $C: y=\ln x$. For each positive integer $n$, denote by $A_n$ the area of the part enclosed by the line passing through two points $(n,\ \ln n),\ (n+1,\ \ln (n+1))$ and denote by $B_n$ that of the part enclosed by the tangent line at the point $(n,\ \ln n)$, $C$ and the line $x=n+1$. Let $g(x)=\ln (x+1)-\ln x$.

(1) Express $A_n,\ B_n$ in terms of $n,\ g(n)$ respectively.

(2) Find $\lim_{n\to\infty} n\{1-ng(n)\}$.

[quote="kunny"]let B_n be the area of the region bounded by the tangent line at (n,ln) on...[/quote]

(n, ln?)

[hide="1)"]$ A_n$ can be expressed as the difference between the area under $ C$ and the trapezoid formed by the points $ (n, \ln{n})$ and $ (n\plus{}1, \ln{(n\plus{}1)}$ with the $ x$-axis.

Using the antiderivative $ F(x) \equal{} x \ln{x} \minus{} x$ and the formula $ A\equal{}\frac{1}{2}(b_1\plus{}b_2)h$:

Area under curve: $ \int_n^{n\plus{}1} \! \ln{x}\, dx \equal{} (n\plus{}1)\ln{(n\plus{}1)} \minus{} n \ln{n} \minus{} 1$
Area of trapezoid: $ \frac{1}{2}(\ln{n}\plus{}\ln{(n\plus{}1)})$
$ A_n \equal{} (n\plus{}1)\ln{(n\plus{}1)} \minus{} n \ln{n} \minus{} 1 \minus{} \frac{1}{2}(\ln{n}\plus{}\ln{(n\plus{}1)}) \equal{} (n\plus{}\frac{1}{2})g(n) \minus{} 1$

$ B_n$ on the other hand, can be expressed as the difference between the trapezoid formed by $ (n, \ln{n})$ and $ (n\plus{}1, \ln{n}\plus{}\frac{1}{n})$ and the area under $ C$ on that interval.

Area of trapezoid: $ \ln{n}\plus{}\frac{1}{2n}$
Area under curve: same as before, $ (n\plus{}1)\ln{(n\plus{}1)} \minus{} n \ln{n} \minus{} 1$
$ B_n \equal{} \minus{}(n\plus{}1)g(n) \plus{} \frac{2n\plus{}1}{2n}$
[/hide]

[hide="2)"]The given expression is equivalent to $ 2n^2(A_n \plus{} B_n)$, but I am wondering how helpful that is.

However, I did find that mere application of L'Hopital's rule gives the answer $ \frac{1}{2}$. :D [/hide]

That's correct. :lol:

Can anyone solve question (2) without using l'Hopital for Japanese high school students?


geometry
geometric transformation
reflection
geometry unsolved

We are given a line $ g$ with four successive points $ P$, $ Q$, $ R$, $ S$, reading from left to right. Describe a straightedge and compass construction yielding a square $ ABCD$ such that $ P$ lies on the line $ AD$, $ Q$ on the line $ BC$, $ R$ on the line $ AB$ and $ S$ on the line $ CD$.

Construct circles $ (X), (Y), (Z), (T)$ with diameters $ PR, QR, RS, PS.$ $ A, B, C,D$ are on $ (X), (Y), (Z), (T).$ Perpendicular bisectors $ x, y, z, t$ of $ PR, QR, RS, PS$ cut $ (X), (Y), (Z), (T)$ at $ K, L, M, N,$ such that $ K, M$ are on one side of the line $ g$ and $ L, N$ on the other side. $ KM, LN$ are diagonals of the square $ ABCD.$ $ KL$ cuts $ (X), (Z)$ again at $ A, C$ and $ LN$ cuts $ (Y), (T)$ again at $ B, D.$ Reflections $ K', L', M', N'$ of $ K, L, M, N$ in $ g$ can be used in the same way, resulting in a square $ A'B'C'D',$ which is reflection of $ ABCD$ in $ g.$

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.AMO2008Problem3

Vo Duc Dien


What are some important reactive intermediates in Organic Chemical Reactions (or just reactions in general). I know some include carbocations, benzyne etc. I just want to make a short (or general) list of them so I can study each one individually.

Thanks.

P.S. How can we experimentally determine the existence of these species and whether they actually exist/form during the course chemical reactions.

Here are some :
carbocations
carbanions
free radicals
benzyne
carbenoids
sigma complex
carbene
nitrene

I think sigma complex can be isolated in certain cases and it can also be affirmed by colligative properties
 :)

Carbenoids and carbenes are, in practice, the same thing.

Except for the fact that carbenoids are actually metal-complexed species?

That's right, but the general outcome is the same. May I ask why are you "studying" reactive intermediates?

I am not 'studying' in the sense that I want to study their properties. I am just making a general list of the things encountered and the reactions associated with their formation (helps me remember organic reactions).


search
superior algebra
superior algebra unsolved

A group is generated by elements $ x,y$ s.t. $ x^{\minus{}1}y^2x\equal{}y^3$ and $ y^{\minus{}1}x^2y\equal{}x^3$.
Find the number of elements of this group.

$ (1)\quad y^3 \equal{} x^{\minus{}1}y^2x$ (given)
$ (2)\quad x^3 \equal{} y^{\minus{}1}x^2y$ (given)
$ (3)\quad y^6 \equal{} x^{\minus{}1}y^4x$ (from (1))
$ (4)\quad y^9 \equal{} x^{\minus{}1}y^6x$ (from (1) and (3))
$ (5)\quad y^9 \equal{} x^{\minus{}1}y^6x$ (from (1) and (3))
$ (6)\quad y^9 \equal{} x^{\minus{}2}y^4x^2$ (from (5) and (3))
$ (7)\quad y^9 \equal{} (yx^{\minus{}3}y^{\minus{}1})y^4(yx^3y^{\minus{}1}) \equal{} yx^{\minus{}3}y^4x^3y^{\minus{}1}$ (from (6) and (2))
$ (8)\quad y^9 \equal{} x^{\minus{}3}y^4x^3$ (from (7), conjugating both sides by y)
$ (9)\quad x^{\minus{}3}y^4x^3 \equal{} x^{\minus{}2}y^4x^2$ (from (8) and (6))
$ (10)\quad y^4x \equal{} xy^4$ (from (9))
$ (11)\quad y^6 \equal{} y^4$ (from (3) and (10))
$ (12)\quad y^2 \equal{} e$ (from (11))
$ (13)\quad y^3 \equal{} e$ (from (11) and (1))
$ (14)\quad y \equal{} e$ (from (12) and (13))

It then follows from (2) that x = e. So the whole group reduces to the identity element.

Somehow like this :[url]http://www.mathlinks.ro/viewtopic.php?search_id=705284132&t=189778[/url].


Let a,b,c,d in [- \pi /2, \pi /2] be such that

  sena+senb+senc+send=1  and cos2a+cos2b+cos2c+cos2d \geq 10/4

Prove that a,b,c,d are in [0, \pi /6].

I think the last inequality is >=10/3, not 10/4.


combinatorics
IMO Shortlist
graph theory
Extremal Graph Theory

Given seven points in the plane, some of them are connected by segments such that:

[b](i)[/b] among any three of the given points, two are connected by a segment;
[b](ii)[/b] the number of segments is minimal.

How many segments does a figure satisfying [b](i)[/b] and [b](ii)[/b] have? Give an example of such a figure.

Again Turan, in $G'$ we have no triangle by (i), so we have $\varepsilon '\leq {n^2\over 4} = {49\over 4}$.
Thus maximum  $\varepsilon' = 12$ so minimum $\varepsilon = {7\choose 2}-12 = 9$.

Now minimal configuration is union $K_3$ and $K_4$.

I too got 9... but is there any proof without graph theory?

$4G \le\ n^2$ by the well known mantels theorem
So here n=7 actually you we are to consider [n^2/4] so that n=7
So min is 12 and for minimum 7C2–12
There is nothing so difficult


A document is divided into 50 parts. Each of these parts is assigned to a person. Each person has her or his own cell phone. What is the minimum number of callings to be made among these 50 people so that everyone knows information in the whole document?

I think the puzzle means that when person A calls person B, they all learn what each of them knew before.
So if we have A, B, C, and D go like so:
A calls B
C calls D
B calls C
A calls D
Then each person knows all the information. (That may not be the minimum amount of calls for 4 people, but it isn't 3+2+1.)

well, it's not 1225 since 99 is an upper bound.

the reason for that is that lets say

a calls b
b calls c
c calls d.

49 calls 50.
50 calls 1.
1 calls 2.

48 calls 49.

Let's look at the 4-person example. Let's denote a to be the information assigned to the person A, b to B, etc.

So initially, this is what they know:
A: a
B: b
C: c
D: d

The first calling doesn't matter, because it's symmetric. So let's say A calls B. Here's the table now.
A: ab
B: ab
C: c
D: d

And C calls D:

A: ab
B: ab
C: cd
D: cd

Now, we can go A->C and B->D to complete the table in 4 calls:
A: abcd
B: abcd
C: abcd
D: abcd

The idea here is that we want to get from a total of 4 pieces of information to 16 pieces of information in the least number of calls. Thus we want to maximize the average amount of information transferred per call. Working binarily achieves this.

The binary method for $ 4$ people doesn't generalize:

In general, let $ x$ be the number of parts/people.  Let $ f(x)$ be the number of calls that to be made.

By the binary strategy from Yongyi781, $ f(2x) \equal{} 2f(x) \plus{} x/2$, where you split each group into two halves, have each half learn all the info in each half, and then make calls to transfer information between the halves.  This $ f$ grows exponentially, when a clear upper bound can be set at $ f(x) \equal{} 2x\minus{}3$.  (Each person calls the first person ($ x\minus{}1$ calls), who then has all the info and calls back the remaining $ x\minus{}2$ people without complete info.  Note that the last call in the first half produces $ 2$ people with all the info.)

This problem has probably been explored in the comp. sci. literature, so I'm going to guess that there are better upper bounds than $ 2x\minus{}3$ for large $ x$, but I don't know what they are.  It seems that there may be some optimizations where after some people have complete information, you can have an additional call between people with incomplete information that produces two people who do have complete information.  This would reduce the number of calls in the second half of the algorithm, since you don't need one person with complete info to make two calls.  In general, if you attempt to grow info in a binary fashion, there should only be one other pair for which this is true.

What abour 98? They all call one person and he calls them all back.

I think 97.

One special person is chosen. Then, 48 people call him and tell him all the parts except the last person's part. Next, the last person exchanges all 50 parts with the special person. Then, the special person calls the other 48 people back, with the whole document. Now everybody knows the document. $ 48\plus{}1\plus{}48\equal{}97$

96

This can be achieved as follows: choose four people (A, B, C, and D) as
the "core group".  Each person outside the core group phones a member of
the core group (it doesn't matter which); this takes n-4 calls.  Now the
core group makes 4 calls: A-B, C-D, A-C, and B-D.  At this point, each
member of the core group knows everything.  Now, each person outside the
core group calls anybody who knows everything; this again requires n-4
calls, for a total of 2n-4.


Let $ Y$ be a subspace of $ X$, then
$ dim(Y)+dim(X/Y)=dim(X)$.

-------

If $ Z=X+Y$, then $ Y$ is isomorphic to $ Z/X$.

Let $ f: X \to X/Y$. Then $ Ker f =Y\, , \, Im f=X/Y$
 :maybe:


[quote="Ligouras"]Let $ r_a, r_b, r_c$ exradii, of an acute-angled triangle $ ABC$. Let $ H$ be the orthocentre, and let $ d_a, d_b, d_c$ be the distances from $ H$ to the sides $ BC, CA,$ and $ AB$, respectively. Prove or disprove that

$ r_a \plus{} r_b \plus{} r_c \ge 3(d_a \plus{} d_b \plus{} d_c)$ .

________
 :|[/quote]
$ r_a \plus{} r_b \plus{} r_c\equal{}4R\plus{}r$
$ HA\plus{}HB\plus{}HC\equal{}2R(cos(\alpha)\plus{}cos(\beta)\plus{}cos(\gamma))\equal{}2(R\plus{}r)$
$ h_a\plus{}h_b\plus{}h_c\equal{} \frac{1}{2R}(r^2\plus{}s^2\plus{}4rR)$
$ d_a \plus{} d_b \plus{} d_c \equal{} h_a\plus{}h_b\plus{}h_c\minus{}(HA\plus{}HB\plus{}HC)$
So we must prove :   $ 3s^2 \le 20R^2\plus{}2Rr\minus{}3r^2$ , which is obviously true ( $ 3s^2 \le (4R\plus{}r)^2 , 2r \le R)$

Dear [b]Sergic[/b] is Very nice your solution. Thanks
Dear friends, Desire to have further proof (original or classical...) without the use of the trigonometry. Aspect


Putnam
factorial
induction
abstract algebra
vector
logarithms
strong induction

Show that every positive rational number can be written as a quotient of products of factorials of (not necessarily distinct) primes. For example, $ \frac{10}9\equal{}\frac{2!\cdot 5!}{3!\cdot 3!\cdot 3!}.$

I used strong induction on the prime numbers to show that every prime number could be written this way, so every natural number can be written this way, and then so can every quotient of natural numbers, as desired...

I used a similar method:
It suffices to prove that each integer can be represented this way.
Induction
Base case: n=2=2!

Assume numbers <= n have a representation of a quotient of products of primes

Prove n+1:

Case 1: n+1 is prime
n+1 = (n+1)!/n! but the bottom is n x n-1 x n-2....2 x 1.  By induction hypothesis, each term in the product has a representation.

Case 2: n+1 is composte
n+1=p_1xp_2...p_n=(p_1)!/(p_1-1)! x (p_2)!/(p_2-1)! ... (p_n1)!/(p_n-1)!, and we can replace the bottom numbers in the same way as in case 1.

Although I'm not sure what strong induction is (as opposed to induction in general), I think I solved this one the same way.  All primes can be written like this, therefore you can obtain all the natural numbers, therefore you can get all positive rationals (since rationals are just a quotient a/b of integers, b not equal to zero).

First time I've taken the Putnam exam.  It was fun. :)

[quote="Kierhammer"] ...I'm not sure what strong induction is (as opposed to induction in general...[/quote]

In strong induction, the induction hypothesis is different. In normal induction, you assume it is true for some k and show that it must be true for k+1. In strong induction you assume it is true for 1,2,3,...,k, and then show that it must be true for k+1. The induction hypothesis is "stronger".

Going back to the topic, did anyone think that this problem was too straightforward to be a putnam problem (considering that you could just do the standard sort of induction)?

I think the difficulty of the problem was realizing that all you needed to do was prove it true for any prime p.  After you saw through that it was straightforward.  It was my first time taking the putnam though so I have no idea how difficult the A1, B1s are supposed to be.

[quote="themandotcom"]

Case 2: n+1 is composte
n+1=p_1xp_2...p_n=(p_1)!/(p_1-1)! x (p_2)!/(p_2-1)! ... (p_n1)!/(p_n-1)!, and we can replace the bottom numbers in the same way as in case 1.[/quote]

Question...

If n+1 is composite, shouldn't it be a product of POWERS of primes?

Edit: Well, I guess it doesn't affect your argument....

Here's a high-concept view.  One can regard the positive rational numbers under multiplication as a $ \mathbb{Z}$-module in the obvious way using prime factorizations (in fact they are the countable direct sum of copies of $ \mathbb{Z}$).  The factorials of the primes form the rows of a lower-triangular linear transformation on this module, and lower-triangular linear transformations are always invertible in the obvious way, especially if they have $ 1$s on the diagonal.

Yes, that's how I saw it. The bit I wrote down featured a vector and an upper triangular matrix. (with factorials as the columns, of course).

For what it's worth, here's how I actually wrote down the argument:

Let $ p_1, p_2, ...$ denote the primes, and strong induct on the largest index $ i$ such that $ p_i$ appears in the prime factorization.  (This seems to me much more natural than inducting on the size of the integers themselves.)  $ i \equal{} 1$ is trivial, since any rational number with only $ 2$ in its prime factorization is of the form $ (2!)^k, k \in \mathbb{Z}$.  If you know the result for $ i \le k$, then suppose that $ v_{p_{k \plus{} 1}}(q) \equal{} r$ and consider $ q (p_{k \plus{} 1}!)^{ \minus{} r}$.

Is it OK to show that (using induction) since every prime can be written in terms of the quotient of products of factorials of primes, every rational can too?

I mean, for every rational number, both its numerator and its denominator can each be re-written as the product of primes.....

please help

thanks

my argument is simply saying that every prime can be written as the quotient of products of factorials of primes.

By standard induction, the base case is the smallest prime number, 2 = 2!.

Suppose the argument is true for the kth smallest prime, prove it's true for the (k+1)th smallest prime.

Suppose the (k+1)th prime = p.

p! = p*(p-1)*(p-2)*...*1 = p*(q1^e1)*(q2^e2)....
where the qn's are just prime numbers, each less than p, and en's are just some exponents.

By the inductive hypothesis, all the qn's can be written as quotient of products of factorials of primes.

So, p = p!/[(q1^e1)*(q2^e2)...]

I actually did strong induction on the naturals, not the primes. I first showed that the set $ S$ of all numbers that can be written as the products of...  etc. is closed under addition and inversion, so $ \mathbb{N}\in S$ $ \Rightarrow$ $ \mathbb{Q}\in S$. Then induction: $ 1\equal{}\frac{2!}{2!}$, and two cases for $ n\plus{}1$:

$ n\plus{}1$ is prime: $ \frac{(n\plus{}1)!}{n!}\in S$ since $ n!\in S$.

$ n\plus{}1$ is composite: $ n\plus{}1 \equal{} ab$, where $ a,b \in S\Rightarrow ab \equal{} n\plus{}1\in S$

For integers a and b, I took the  log of all the prime from 2 to the mth prime, where m is the largest prime divisor of a or b. which is a vector space of $ \log p_k$, where $ p_k$ is the kth prime number, and all the $ \log p_k!$ forms a basis of the vector space.  then we can prove $ \log \frac{a}{b}$ is in this space. then prove we can write $ \log \frac{a}{b}$ as linear combination of $ \log p_k!$, where all coefficient are integers. then it imples $ \frac{a}{b}$ can be written as a product of $ p_k!$ and $ \frac{1}{p_k!}$

Do you think it's trivial that if $ a$ and $ b$ follow the form, then so do $ ab$ and $ \frac{a}{b}$ ? Will I lose points if I state that these properties are true but don't actually show them? During the test I thought they were fairly obvious.

We shall prove that every natural number can be written as a quotient of product of factorials of primes [not necessarily distinct] by induction.
we have $1=\frac{2!}{2!}$ ,$2=\frac{2!.2!}{2!}$ ,$3=\frac{3!}{2!}$ ,$4=\frac{2!.3!.2!}{3!}$
now , we assume that , the result is true for $n=1,2,....,m$.
now , if $(m+1)$ is composite , then , we can write , $m+1=uv$ with [$u,v>1$] .so , $u,v \le m$
so , both $u$ and $v$ can be written as quotients of product of factorials of primes.
so , $m+1$ can be written as a quotient of product of factorials of primes.
And if $m+1$ is a prime , then , $m+1=\frac{(m+1)!}{m!} =\frac{(m+1)!}{1.2....m}$ .
now , all of $1,2,....,m$ can be expressed as the quotients of product of factorials of primes.
so , $m+1$ can also be expressed as a quotient of product of factorials of primes .
hence , by PMI , we conclude that , we can write all positive integers in the [i]'required form'[/i].
so , clearly , all rational numbers can be written in the [i]'required form'[/i] .

I thought I had a correct solution to this, but it doesn't involve induction.  Here is what I did, if someone could look it over and tell me if it works that would be greatly appreciated.

Consider any rational number, and WLOG write it as $p/q$, where $p$ and $q$ are relatively prime positive integers.  Next notice that there is some prime factorization (possibly just itself) of both $p$ and $q$.  Hence we can write

$p = p_1^{e_1}p_2^{e_2}p_3^{e_3} \cdots$ where all $p_i$ are prime

and

$q = q_1^{f_1}q_2^{f_2}q_3^{f_3} \cdots$ where all $q_i$ are prime.

Next notice that it is an algebraic fact that $q_i = \frac{q_i!}{(q_i-1)!}$, and similarly that $p_i = \frac{p_i!}{(p_i-1)!}$.  Hence we have

$p = \left(\frac{p_1!}{(p_1-1)!}\right)^{e_1}\left(\frac{p_2!}{(p_2-1)!}\right)^{e_2}\left(\frac{p_3!}{(p_3-1)!}\right)^{e_3} \cdots$
and

$q = \left(\frac{q_1!}{(q_1-1)!}\right)^{f_1}  \left(\frac{q_2!}{(q_2-1)!}\right)^{f_2}\left(\frac{q_3!}{(q_3-1)!}\right)^{f_3} \cdots$.

Hence the quotient $\frac{p}{q}$ can be expressed as a quotient of products of factorials of primes.

There is only one prime number $p$ with the property that $p-1$ is also prime.

I have an alternative, which works for all except prime numbers. If somebody can check it that would be great. First, let

$$x = \prod n_k!$$

$$y = \prod j_i!$$

Since $a,b$ are both positive integers, it comes:

$$\frac{a}{b} = \frac{x}{y} \implies a = b\cdot \frac{x}{y}$$

Suppose this holds for: $\varphi = \{1, 2, 3, ... , a - 1 \}$

So, suppose:

$$a -1 = b \cdot \frac{x}{y}$$

Consider even $a$. Since $a \ge 1$ it follows that:

$$\frac{a}{2} \le a - 1$$ Which means,

$$a = 2! \cdot \frac{x}{y}$$

Suppose $a$ was a non-prime odd. Let $q_k = \{3, 5, 7, 9 \}$

So that:

$$\frac{a}{q_k} < n-1$$ Hence,

$$a = q_k \cdot \frac{x}{y}$$ Of which, $q_k =o_l! Since it is an integer.

This works for all but prime odd. [b]Any feedback?[/b]

[hide]Let the rational numbers be p/q.If p/q is an random integer,it can be easily written as the product of factorials
.If it is a strict quotient,consider if p-1 and q-1 are integers,p/q=p！*（q-1）！/（p-1）！/q！.If not,p-x and q-y
must be prime,certainly it can be written as a product of factorials. :winner_third:   [/hide]

Neither of the preceding two solutions makes any sense to me.  The first never defines any of its notation nor hypotheses.  The second sentence of the second solution is obviously false: how can 5 be written as a product of factorials?

The induction hypothesis was for $n = \{1, 2, ... n-1\}$ also the notation is fairly simple.

$$\prod n_k! = n_1! \cdot n_2!....$$

What is wrong in my (first answer)?

You do not define any of the symbols $a, b, x, y, n_k, j_i$, nor say what role they have in the proof.  You do not say "This is a proof by induction; my induction hypothesis is ..."  Basically no line in your post has enough explanation for a person to decipher it.

@JBL, Sorry. Will you still consider checking it please? I'll describe the symbols.

$a, b$ are positive integer numbers that aren't prime. $n_k, j_i$ are positive integers. $x, y$ are factorial products. Then I used the hypothesis thatL

$$a -1 = b\frac{x}{y}$$

Meaning $a-1$ is a positive integer that can be written in that form.

Okay, so I transferred my original post. Sorry about that. Anyways, during my practice for the Putnam I came across problem B1 from the 2009 exam. I came to a solution quickly but I had trouble expressing it with the clarity necessary to receive a full 10 points. My professor advised me to formally write it up, and so I did, but he seemed to think it needed more rigor to satisfy the strict grading. Anyways, I'm hoping I could get some suggestions on what score my proof would receive and how I must improve it to receive full credit.

 Theorem: Show that any rational number may be expressed as a quotient of products of factorials of primes.

 It is sufficient to prove the theorem for any prime, since rational numbers are ratios of integers, and integers are products of primes. Now, before we proceed via infinite descent, we will introduce a brief lemma.

 Lemma: Let $k$ be a prime factor of $n!$ for composite $n$. Then, $k!<n!$.
 Proof: Since $n$ is composite, it follows that $k\leq{n-1}$. Therefore, $k!\leq{(n-1)!}<n!$.

 With this lemma, we may now prove the theorem of interest.

 Proof: Consider $p_0!$ where $p_0$ is an arbitrary prime. Now, divide $p_0!$ by the factorials of the prime factors of $(p_0-1)!$, denoted as $p_1!, p_2!,...,p_n!$, (note that the $p_i$'s aren't necessarily distinct) the result is
$$\frac{p_0!}{p_1!p_2!...p_n!}=\frac{p_0}{(p_1-1)!(p_2-1)!...(p_n-1)!}.$$
 We assert that if one repeats this process by multiplying the numerator by the factorials of the prime factors of each $(p_i-1)!$ in the denominator, and so on, that the process will terminate with each term other than $p_0$ being $2!$, at which point we may multiply the numerator or denominator by a sufficient number of $2!$'s to produce $p_0$. To see why this is true, note that $p_i-1$ is composite since $p_i$ is prime. Applying our lemma, each $p_i!$ for $p_i$ a prime factor of $(p_j-1)!$ is necessarily less than $(p_j-1)!$, therefore after each step the largest number in either the numerator or denominator has decreased, and the process must terminate as intended.

Call a rational number [i]good[/i] if it can be written in such a way. Also, let $p_n$ denote the $n$th prime.

[b]Claim:[/b] All primes are good.
We use strong induction. The bases cases $2=2!$ and $3=\frac{3!}{2!}$ are trivial. For the induction step, note that $p_n!$ can be factorized into: a product of primes strictly less than $p_n$, and $p_n$ itself. Then dividing $p_n!$ be the product of these primes gives that $p_n$ is good.

Since the set of good numbers is closed under multiplication and division, it suffices to check $1$, which is obviously good.

We first show any prime $p$ is expressible in this format via strong induction as follows:

Base Case: We can experiment with small cases to see this is true.

Inductive Step: Observe that $p = \frac{(p)!}{(p-1)!}$. Now we know $(p-1)! = (p-1)(p-2)….$. Each $i$ satisfying $1 < i < p$ can be prime factorized into factors less than $p$. From here, our inductive hypothesis tells that each of these prime factors are expressible in the desired form. Thus, we can substitute for each $i$ the appropriate rational representation via first factorizing, and then using the desired form for each prime factor. Thus completes the inductive step.

Now any rational number can have its numerator and denominator prime factorized, and we can readily substitute.


geometry
trigonometry
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines
Law of Sines

A triangle of sides $8,12,18$ is inscribed in a circle. Find the area within the circle but outside of the triangle.

[hide="Kinda ugly..."]Using Heron's formula gets that the area of the triangle is $\sqrt{1463}$.  Another formula states that the area of triangle $ABC$ equals $\frac{abc}{4R}$ where $R$ is the circumradius.  So $\sqrt{1463}=\frac{1728}{4R}=\frac{432}{R}$.  Therefore $R=\frac{432}{\sqrt{1463}}$.  The area of the circle should therefore be $\frac{186624}{1463}\pi$.  So the answer is $\boxed{\frac{186624}{1463}\pi-\sqrt{1463}}$.[/hide]

I wouldn't consider this on the level of HSB.

[hide]
We're looking for $A = \pi R^{2}-K$
By law of cosines,
$18^{2}= 8^{2}+12^{2}-2 \cdot 8 \cdot 12 \cdot \cos \theta$
$\cos \theta = \frac{-29}{48}\implies \sin \theta = \frac{\sqrt{1463}}{48}$
By law of sines,
$2R = \frac{18}{\sin \theta}\implies R = \frac{432}{\sqrt{1463}}$
$K = 0.5 \cdot 8 \cdot 12 \cdot \sin \theta = \sqrt{1463}$
$A = \frac{186624\pi}{1463}-\sqrt{1463}$
[/hide]


Stop clogging up the forum with easy questions.  I won't even bother to answer this one.

This is not an AIME problem, it is an AMC 12 one. Anyways, the answer is $\boxed{B}$ after inspection.


在直角△ABC中, AC = BC, 弧DEF的圓心為A. 如果圖中兩個陰影部分的面積相等, 那麼AD:DB?

Translate:
In a right-angled triangle $ \triangle ABC, AC \equal{} BC, arc \widehat {DEF}$ is formed as the centre $ A.$If  the two shaded areas are equal , then $ AD: DB\equal{}$?

 :icecream:

[hide]
Note that shaded area DBE is just $ [ABC]$ minus the big non-shaded area ADEC.

Also, the shaded area ECF is area of the partial circle ADF minus the big non-shaded area.

Let $ AE \equal{} r$ and $ \angle EAC \equal{} \alpha$. The shaded area equality gives us

$ \frac {1}{2}(r\cos{\alpha})^2 \equal{} \frac {1}{8} \pi r^2 \Leftrightarrow \cos\alpha \equal{} \frac {\sqrt {\pi}}{2}$

To find ratio $ AD: DB$, we realize that is equal to ratio $ [ADC]: [DCB]$.

$ [ADC] \equal{} \frac {1}{2}(r)(r\cos\alpha)\sin\frac {\pi}{4}$

$ [DCB] \equal{} [ABC] \minus{} [ADC] \equal{} \frac {1}{2}(r\cos\alpha)^2 \minus{} \frac {1}{2}(r)(r\cos\alpha)\sin\frac {\pi}{4}$

So, $ \frac {[ADC]}{[DCB]} \equal{} AD: DB \equal{} \frac {\sqrt {2}}{\sqrt {\pi} \minus{} \sqrt {2}}$.
[/hide]


For some reason I think I am just having a brain freeze and I cant solve for x. Can someone help?
[tex]{\frac{4}{3}}^x=\frac{8\sqrt{3}}{9}[/tex]

sorry, the LHS is [tex]{\frac{4^x}{3^x}}[/tex]

btw, i dont think calculators are allowed, this is part of a problem in AoPS Vol 2


I signed up for the mailing list to remind me of upcoming math jams, but I haven't received any notifications since the one in late April about the math jam on USAMO day 1 problems. Are any of you who are on the list noticing this too?

There are a couple possibilities:

1) Your email is invalid - about 5% of people who join the mailing lists put a typo in their email.

2) Our emails are landing in your Spam folder.  You'll have to tell your spam folder that we aren't spam.

#2 is likely the problem.  If you let me know what mail server you're using, we can take a look at if all our mail is being blocked at a certain server (this hasn't happened since we moved to our own server - last year when we were on a shared server, other folks on it would spam, and we'd get blocked).

I'm using AOL... which I doubt would be blocking all of your emails. But maybe I set some personal mail security setting that I forgot I set.


topology
function
real analysis
real analysis theorems

I am taking a course which relies a lot on topology, but I haven't ever taken any topology. I'd like to check off what relatively closed means, so that I am not using incorrect definitions.

Let M be a subset of the the complex plane with the usual metric. Is it true that $S \subset M$ is relatively closed iff:

1) $x \in M, \, x \in \partial S$ implies $x \in S$

2) For every sequence $(x_{j})$ of points in $S$ converging to a point $x \in M$, $x$ belongs to $S$

I mostly feel unsure about 1). Should it be something else?

(I am wondering if S being relatively closed is equivalent to each of 1 and 2 alone, not both at the same time).

$S$ is relatively closed if it is closed in the induced topology on $M$, which happens when $S=M\cap E$ for some closed set $E\subset\mathbb C$. Note that any such $E$ must contain $\overline{S}$ (the closure of $S$ in the plane topology). Therefore, $S$ is closed in $M$ iff $\overline{S}\cap M=S$ (3).

It is easy to see that (3) is equivalent to (1).
Since the plane is a first countable space, (1) is equivalent to (2).

Thank you very much mlok. You really helpd me out with this. I understand now.

For proving that 1) and 2) are equivalent I just used that $x \in \bar{S}$ iff there is a sequence in S converging to $x$. Is first-countability some sort of generalization of this?

I am also wondering about the general notion of connectivity. Path-wise connected feels much more intuitive to me than connected. I have seen that they are equivalent for open sets, so I can see why you would like to use the general notion of connectedness in such a situation, since it is a lot easier to work with technically. However, I can't help but wonder if sets that are not path-wise connected but are connected really 'look connected' if you draw them in the plane. Is it possible to give some intuitive characterization of connected sets (that are not path-wise connected)?

Your "iff" does not apply in all topologies. It works in metric topologies, but sequences don't determine everything in some larger spaces.

An example:
Let $X$ be the space of functions from $\mathbb{R}$ (or any uncountable set) to the two-element set $\{0,1\}$, with the product topology. (The basic open sets are of the form $\{f: f(x_{i})=a_{i}\text{ for }1\le i\le n\}$)
A sequence converges in the product topology if and only if it converges pointwise.
Let $S$ be the set of all functions which are equal to 0 at some finite collection of points, and 1 elsewhere. The closure of $S$ is all of $X$, but no sequence in $S$ converges to the zero function.


Is there a way to inscribe three unit circles inside another bigger circle without having to calculate a lot of coordinates?

One can obtain it wtih an inversion:
[code]import geometry_dev;
import animate;
animation A;

size(10cm);

int n=20;
for (int i=1; i < n; ++i) {
  real r=i/20;
  line L1=line(N,S), L2=shift(2*r*E)*L1;
  transform T=shift(0,-2*r);
  circle C1=circle(r*E,r), C2=T*C1, C3=T*C2;

  inversion inv=inversion((-1,0),1);
  save();
  draw(inv*L1^^inv*L2);
  draw(inv*C1^^inv*C2);
  A.add();
  restore();
}

A.movie();
[/code]
Clik


[quote="vutang"] Prove that for all real numbers $a,b,x,y$ satisfying $x+y=a+b$ and $x^4 + y^4=a^4+b^4$ then $x^n + y^n=a^n+b^n$ for all $n\in\mathbb{N}$[/quote] [hide][/hide]

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The only possible solutions of the system of equations are
$x=a,y=b$ or $x=b,y=a$
Direct checking will show that these two sets of values are solutions.
In order to show that they are the only possible solutions,
we can use the transformation
$x=u-v$ and $y =u+v$.

Alternative method
Let $t=xy$.
$x^2+y^2=(x+y)^2-2xy=(a+b)^2-2t$
$x^4+y^4=(x^2+y^2)^2-2(xy)^2$
$a^4+b^4=[(a+b)^2-2t]^2-2t^2$
Therefore, $a^4+b^4=(a+b)^4-4(a+b)^2t+4t^2-2t^2$
Simplify, $t^2-2(a+b)^2t+(2a^3b+3a^2b^2+2ab^3)=0$
Solving, $t = ab$ or $t = 2a^2 + 3ab+ 2b^2$ (rejected as $\Delta < 0$)
Finally, we have $x+y = a + b$ and $xy= ab$.
Solving x and y, $x=a,y=b$ or $x = b, y = a$.

Question to think about :idea:
Given $x+y=a+b$ and $x^5+y^5=a^5+b^5$, where a,b,x,y are real numbers, show that
$x^n+y^n=a^n+b^n$ for all natural numbers n.
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parameterization calculus derivative function

$ x,y$ are real numbers such that $ x^2\plus{}y^2\equal{}1$. Find the Min and Max value of $ P\equal{} \frac {2(x^2\plus{}6xy)}{1\plus{}2xy\plus{}2y^2}$

[hide="simplification, sort of."]$ x^2\equal{}1\minus{}y^2$. Thus $ P\equal{}\frac{2\minus{}2y^2\plus{}12xy}{1\plus{}2xy\plus{}2y^2}\equal{}\minus{}1\plus{}\frac{14xy\plus{}3}{1\plus{}2xy\plus{}2y^2}$. Thus we need to find the minimum and maximum value of $ Q\equal{}\frac{14xy\plus{}3}{1\plus{}2xy\plus{}2y^2}$.[/hide]

[hide="Hint"]Let $ \mathbf{c} \equal{} x^2 \minus{} y^2$ and $ \mathbf{s} \equal{} 2xy$.

Observe that $ \mathbf{c}^2 \plus{} \mathbf{s}^2 \equal{} 1$

Then, the given expression can be rewritten as: $ P \equal{} \frac {1 \plus{} \mathbf{c} \plus{} 6\mathbf{s}}{2 \minus{} \mathbf{c} \plus{} \mathbf{s}}$

[i]What kind of parametrization am I hinting at here?[/i][/hide]

[quote="Lawrence Wu"]i would go ahead and try to take the partial derivatives and find critical points[/quote]The fun is not to use calculus... besides that's overkill for this problem.

Fine. I'll bite, TZF.   :lol:

I'm going to expand on his hint with a larger semi-hint: think of some functions which start with the respective bolded letters.

[quote="1=2"][hide="simplification, sort of."]$ x^2 \equal{} 1 \minus{} y^2$. Thus $ P \equal{} \frac {2 \minus{} 2y^2 \plus{} 12xy}{1 \plus{} 2xy \plus{} 2y^2} \equal{} \minus{} 1 \plus{} \frac {14xy \plus{} 3}{1 \plus{} 2xy \plus{} 2y^2}$. Thus we need to find the minimum and maximum value of $ Q \equal{} \frac {14xy \plus{} 3}{1 \plus{} 2xy \plus{} 2y^2}$.[/hide][/quote]

Then???? I think ur result is slow.!! :(


Up to now i have only the approach that's similar with your solution.
 :) Hmm. I think it's a inteliggent solution


geometry
geometric transformation
rotation

A tribe of eight Native Americans sits in a circle around a campfire. If the chief must sit between his mother and his wife, how many different seating arrangements are possible? Two seating arrangements are considered the same if each individual has the same person to the right and the same person to the left in both.

First fix the chief, such that there are $ 2$ ways to arrange his mother and wife. Then, there are $ (8\minus{}3)!$ more ways to arrange the remaining people, for a total of $ 2\cdot5!\equal{}\boxed{240}$ seating arrangements.

What about rotations? wouldn't you have to divide 240 by 8?

If the chief is fixed in a certain position, then we don't need to worry about rotations.

[hide = Solution]

We bundle the mother, wife, and chief into one person. The number of ways to order the people are $\frac{6!}{6} = 120$, and multiplying by $2$ since the mother and wife are interchangeable gives us $\boxed{240}$

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