Hi, I have read this book called "Passenger to Frankfurt" by Agatha Christie and I kind of need help on it.  I'm having a test on program called AR (Accelerated Reader) and it's going to take a lot of part in my English grade. To do good on it, I think, that I need to understand the plot fully and able to talk about it.

So, I'm asking if we can talk about the book in here. I'll go first.  I'll tell about the plot of this book and if you read this book or heard about it, read through (it's really good book so even you didn't read it, you'll enjoy it!) and if something I said isn't right or want to add something please post! Or you can just post on what you think or know about the book!

THANKS.

Here it goes:

A main charater named Sir Stafford Nye is a former diplomat who was returning from his work to London, England.  While he was waiting at airport lounge, he met a mysterious woman who asked him to save her life by giving him his passport to London.  Sir Stafford Nye accepts her idea of
giving his passport because with some kind of unusual feeling and close resemblance of his sister, Pamela.  Yet, the story of Pamela or his sister ends right here.  After giving his passport, he follows what the mysterious woman tells him to do.  He drinks a glass of water with some kind of dope in it, and when he wakes up, he finds his passport and wallet are missing.
But because of who he is, he gets back to his home and talks about his unusual meeting at the airport to the Colonel Pikeway and Henry Horsham, who's in Secret Service department.  Sir Stafford Nye, however, couldn't remember well and soon, he talks his Aunt Matilda.  While talking with her, Sir Stafford Nye begins to wonder about who that mysterious woman he met and at one night, a car tried to kill him (this part isn't big).  He luckily avoids the crash but that event made him to know more about that person. One day, Sir Stafford Nye crashes with a woman on the bridge and after helping her up, she disappears.  But as soon as she disappeared, he notices the ticket to Festival Hall for following evening.
Sir Stafford Nye goes to the Festival Hall and during the play of Young Siegefried, he again faces the mysterious woman. She, however, don't talk anything much except giving him a handwritten note.  After trying to sing the note in his head, he finds that it's the note of Young Siegefried, and while thinking about it, he travels to his Aunt Matilda.  During the conversation with her at her house, he hears word "Young Siegfried" again and that word gives him an unusual feeling of it.  After returing from the Aunt Matilda's house, he goes to the Embassy's party and again, he faces that mysterious woman in name of Countess Zerkowski and after the party, he gets on a car which Countess rode from.  Inside the car, he gets to talk with her and learns that she is the woman of many identites.  Yet the important thing in here was that where she led him was the castle where Lord Altamount, a formerly well-known man in whole England, and Mr. Robinson (another strong person on politics), Henry Horsham, Colonel Pikeway were staying.  At there, he learns that he was asked to save the world (bah!! Just like in cartoons) yet in different way (now that's interesting).  What he finds is that world is in near to be destroyed by a young revolutionaires who has the ideas of Hitler-like, which is to KILL.  After learning about it, he decides to be a member of it (he actually didn't have much choice though) and soon, he departs with Countess Zerkowski, or simply Mary Ann, to the South Africa where he meets a big, fat, but strongly powered woman who owns a lot of armies named Big Charlotte. Now, that was all that were talked for Sir Stafford Nye's adventure.
The story, however, go for more.  Conferences occur at France and England, trying to save the world from young mad revolutionaires.  In England, where Lord Altamount and many other people were, they get information from Sir Stafford Nye and many other agents on the plan of the revolution as a diagram.  There were five letters and letters were F, S, A, J, and D. F for finance, S for science, A for armament.
After looking at those, they tried to find on what they  [i]actually[/i] can do but what they find is, a weak ability of them.  But after more information, they find on something call "Project B" or project Benvo which is a project that makes people permanently benveolence.  Lord Altamont himself go to meet a guy named Robert Shoreham who made the Project B but burned everything, and at there, Lord Altamont tries to convince the person to revive that project.  But unfortunately, Lord Altamont's one of aid was found traitor and one nurse who seemed to be working on that castle, who was actually Juanita, a person who stood for letter J, kills Lord Altamont.  Yet, his death didn't happen for vain.  Robert Shoreham decides to save the world and the story ends on where he decides to revive it.

I know I have skipped many parts because even at now, I didnt understand the story hundred percent.  I know Agatha Christie's books aren't all difficult like Shakespeare or Dickens but they are, in my opinion, still classics.  So, I need to have a discussion to help me understanding it so please read this and give me a hand.

P.S. Test is tomorrow!!!

 :maybe:

Please try to keep in mind that this site is NOT a homework help site. This site is mainly for doing math problems/challenging others with math problems or asking for help on (non-homework!) math questions.

Thank you for the advice, Simon. Yes, I know it's a math site and to post something that's deal with literature, I must go to the literature forum.  But the literature forum I found don't really have regular posts on my topic (actually none!) and I also know this isn't for place asking homwork.

Yet, let me just point this out.  I wasn't asking for the answer to my homwork. I was asking for the discussion of this book. Discussion is different to homework.

Hope I made my point bit clear. Still, sorry about this inconvenience.

If you want to discuss literture go [url=http://thehiddenside.proboards22.com/]here[/url].  ;)

Hey! I remember AR, from 7th grade. I got 1100 points!


Find the closed form and (at least 3) recurrence forms of the sequence.

6, 12, 9, 18, 15, 27, 24, 39, 36, 54, 51, . . .

I think the pattern is: +6; -3; +9; -3; +12; etc

We have $ a_{2n \plus{} 1} \equal{} a_{2n} \minus{} 3$ and $ a_{2n} \equal{} a_{2n \minus{} 1} \plus{} 3n \plus{} 3$. Therefore $ a_{2n \plus{} 1} \equal{} a_{2n \minus{} 1} \plus{} 3n$ which implies $ a_{2n \plus{} 1} \equal{} 3\left(2 \plus{} \frac {n(n \plus{} 1)}{2}\right)$. Thus $ a_{2n} \equal{} a_{2n \plus{} 1} \plus{} 3 \equal{} 3\left(3 \plus{} \frac {n(n \plus{} 1)}{2}\right)$. Hence $ a_n \equal{} \frac {3(\lfloor\frac n2\rfloor^2 \plus{} \lfloor\frac n2\rfloor \plus{} 5 \plus{} ( \minus{} 1)^n)}{2}$.


How many two-digit prime numbers (for example, 23) can be formed by selecting two digits (possibly the same) from the set:  {1, 2,  3,  4,  5,  6}?

A.5	B.6	C.7	D.8	E.9

[hide=" click to reveal hidden content"]

$ 8$, I think: $ 11, 21, 23, 31, 41, 43, 53, 61$.

[/hide]

@AndrewTom, $ 21$ is not a prime

so they are:

$ 11,13,23,31,41,43,53,61$

The answer is the same ($ 8$) but you included $ 21$ instead of $ 13$

Thanks, Demonsthenes, I can't imagine how I had that mistake.


Let $ 0 \le \alpha, \beta, \gamma \le \frac{\pi}{2}$ satisfy the conditions

$ \sin \alpha\plus{}\sin \beta\plus{}\sin \gamma\equal{}1, \sin \alpha \cos 2\alpha\plus{}\sin \beta\cos 2\beta\plus{}\sin \gamma \cos 2\gamma\equal{}\minus{}1.$

Find all possible values of $ \sin^2 \alpha\plus{}\sin^2 \beta\plus{}\sin^2 \gamma$.

[hide]For all $ x \in \left[0, \dfrac{\pi}{2} \right]$ we have: $ \cos 2x \ge \minus{}1$ and $ \sin x \ge 0$.

Thus, $ \sin x \cos 2x \ge \minus{}\sin x$ with equality iff $ x \equal{} 0$ OR $ x \equal{} \dfrac{\pi}{2}$.

Therefore, $ \sin\alpha\cos 2\alpha\plus{}\sin\beta\cos 2\beta\plus{}\sin\gamma\cos 2\gamma \ge \minus{}(\sin\alpha\plus{}\sin\beta\plus{}\sin\gamma) \equal{} \minus{}1$

with equality iff $ \alpha,\beta,\gamma \in \left\{ 0 , \dfrac{\pi}{2} \right\}$.

Since $ \sin(0) \equal{} 0$ and $ \sin\left(\dfrac{\pi}{2}\right) \equal{} 1$, one of $ \alpha,\beta,\gamma$ must be equal to $ 1$ and the other two must be equal to $ 0$.

Therefore, $ \sin^{2}\alpha\plus{}\sin^{2}\beta\plus{}\sin^{2}\gamma \equal{} 1$. [/hide]


This one is quite difficult I must admit :

Let a(1), a(2), ..., a(5) be real numbers such that :Sigma:a(k) = 1.
Prove that for 1 :le: i < j :le: 5 we have min(a(i) - a(j)) :le: 1/10.

Good luck :mrgreen:

Oh no, I almost forgot this one ... No try ?

see, I don't understand everything here....

people post  :Sigma: equations all the time yet don't specify a variable, starting or ending values.....

what does ":Sigma:a(k) = 1" refer to?

I think in this case the only variable is k. That sum you asked abt is a(1)^2+a(2)^2+a(3)^2+etc.

Hmm, just tried this one, it didn't seem too difficult but I may have made a mistake..

[hide]

We want to prove that two of the a(k)s differ by 1/:sqrt:10. So instead, we assume that all of the a(k)s are at least 1/:sqrt:10 apart, and prove that sum of the squares is at least 1. Then the equality case proves it.

So assume that they are all that far apart. Now we want to minimise the sum of the squares. Now I'm gonna use real informal language here  but you should be able to see what I mean.. basically if any two of the a(k)s are more than that distance apart, we can either "push" the top value down if its positive, or "push" the bottom value up if its negative so that the distance between them is exactly 1/:sqrt:10. This will decrease the sum of the squares. And any gap of more than that distance must surely have the top value positive or the bottom value negative! So we keep doing that, and end up with the sum of squares minimised when all the spaces are 1/:sqrt:10. Letting the middle value be x, the sum of the squares is 5x:^2: + 1 (rest cancel out), which is greater than or equal to 1. And it is proved.

[/hide]

could you rigorize that with a [hide]smoothing argument? just wondering, i'm not good at inequalities...[/hide]

Well it is pretty rigorous, but instead of just saying if the numbers are a(1)<a(2)<a(3)<a(4)<a(5) and a(i+1)-a(i) > 1/:sqrt:10, then if a(i+1) is positive, decreasing it will decrease the sum of squares, and if a(i) is negative, increasing it will decrease the sum of squares. I just used push cos its shorter :)


Να δείξετε ότι $ b^{a}c^{b}a^{c}\le (\frac{a\plus{}b}{2})^{a}(\frac{b\plus{}c}{2})^{b}(\frac{c\plus{}a}{2})^{c}\le a^{a}b^{b}c^{c}$όταν $ a>0,b>0,c>0$
Υπόδειξη Λογαρίθμηση Jensen ΘΜΤ
Λύση σε λίγες μέρες

η λύση είναι στο attachemant


a regular dodecagon ABCDEFGHIJKL is inpolygon of square(area of square$ \equal{}24$).
point A,D,G, and J are in middle of each sides.
find area of this dodecagon. (picture is omitted.)
(IMTS-R12-2. is this posted on "national olympiad section"? not yet..)
i've a solution but i'd like to share more sol.s

ps.quite easy

[quote="IW@IT"]My answer is $ 18$. Is it correct?? If it is correct, I'll post my solution later :)[/quote]If you're not sure if it's correct, perhaps post your solution so we can see if it's correct, otherwise you'll never know if it's correct or not. A person cannot learn proper mathematics without learning from mistakes.  :roll:

@10000th User: Thanks for the advice :lol:

My Solution
From the area, we get the sides of the square is $ 2\sqrt{6}$
The area of the dodecagon is the sum of 12 small triangles

Let the center of the dodecagon $ \equal{}O$
Consider $ \triangle OAB$
$ OA\equal{}OB\equal{}\sqrt{6}$ and $ \angle AOB\equal{}30^0$
So $ [OAB]\equal{}\frac{1}{2}sin 30^0.\sqrt{6}.\sqrt{6}\equal{}\frac{3}{2}$
The area of the dodecagon $ \equal{}12.\frac{3}{2}\equal{}18$ :)

I'm not very sure with my solution, sorry If I made any mistake

IW@IT//Yea. that's correct(pls be confident with your solution~)
10000th User//thank you for replying
IW@IT(2)//oh,no that's the same with my sol !
 My one..
$ S\equal{}12 * \frac{1}{2} a^2sin{30} \equal{} 12 * \frac{1}{2} * 6 * \frac{1}{2} \equal{} 18$


algebra
function
domain
superior algebra
superior algebra solved

Let (A,+,*) be a ring in which 1=/=0 and which has no divisors of 0. f:A->A, f(x)=x10 is a morphism of the group (A,+). Show that A is a field with 2 or 4 elements.

Some observations:

1)2=0.
f(1)=110=1, so f(-1)=-1. But f(-1)=(-1)10=1, so 2=0.

2) x6=1 for all x<>0.

x10+x8+x2+1=(x+1)10=f(x+1)=f(x)+1=x10+1.
So x8+x2=0. Since A is a domain, this implies x6=1 for all x<>0.
This implies A is a field.

3) f(x)=x10=x4 is an endomorphism of (A,+).

Some more.
If A has 2 elements, then good for him.

4. If A has more than 2 elements, from [2] in my previous post, x6=1 for all x<>0.
So (x+1)6=1 for all x<>1, so
x6+x4+x2+1=1, so x4+x2+1=0 for all x<>0 or 1.
x4+x2+1=(x2+x+1)2, so x2+x+1=0 for all x<>0,1.

We get x3+1=(x+1)(x2+x+1), so
x3=1 for all x<>0,1. So x3=1 for all x<>1.

Now f becomes the identity endomorphism and it is no longer useful.
----------------------------------------------------------------------------------
The problem reduces to:
5. Let A be a field in which 1<>0, 2=0 and for all x<>0 we have x3=1. Prove that A has 2 or 4 elements.

Proof: If A has 2 elements the A is Z2.
If A has more than 2 elements, then x2+x+1=0 for all x<>0 or 1. [6]

If A has 4 elements, then good for him.

If A has more than 4 elements, then A includes {0,1,x,x+1,y,y+1,x+y,x+y+1
xy,xy+1}, for some x,y,  and it is easy to see that all the elements of the set are distinct.

By [6], we have (x+y)2+x+y+1=0, so xy+yx=1.
x=x(xy+yx)=xxy+xyx=xy+y+xyx
x=(xy+yx)x=xyx+yxx=xyx+yx+y
Add the last two to get 1=xy+yx=0 (a contradiction).

So A has 2 or 4 elements.


Euler
geometry
geometric transformation
reflection
circumcircle
homothety
analytic geometry

Prove that $FF' || OH.$ Where $F$ is Fermat point, $F'$ is its isogonal conjugate and $O$ and $H$ are circumcenter and orthocenter of $\triangle ABC.$

Is $F'$ the second fermat point? Because if it is, then it's not true. As far as I know, If $F_1,F_2$ are the first and second Fermat points, and $I_1,I_2$ are the first and second isodynamic points (just think that $I_i$ is the isogonal conjugate of $F_i$), then $F_iI_i\|OH$.

[There was an error, so I've edited it.]

what are isodinamic points? can someone explain a little about it?

The locus of point $M$ s.t. $\frac{MB}{MC}=\frac{AB}{AC}$ is a circle having the segment bounded by the foot of the $a$-bisector and the foot of the exterior $a$-bisector. This circle is called the $a$-Apollonius circle. Similarly we define the $b$ and $c$-Apollonius circles. It can be shown that these three circles are coaxal, and the two points through all three of them pass are called the isodynamic points of the triangle $ABC$.

It's also true that the isodynamic points are the isogonal conjugates of the Fermat points, and this is, I think, more useful in this context. Another interesting property (which is crucial in proving the fact that they're the isogonal conjugates of the Fermat points) is that these two are the only points which have equilateral pedal triangles.

[quote="Omid Hatami"]Prove that $FF' || OH$ that $F$ is fermat point and $O$ and $H$ are citcumcenter and orthocenter.[/quote]

The only thing that makes me wonder is this strange "Very hard" with 9 exclamation signs. Especially this is because my experience shows that a problem you call "hard" (even without any exclamation sign) is already far beyound my brain. And now this one, really strange...

In fact, I assume that you mean by F' the isogonal conjugate of F, i. e. the first isodynamic point of the triangle.

For reasons of convenience, I rename F' as J (just since I am more used to this letter for an isodynamic point) and hence rewrite your problem as follows:

[i]Let F be the first Fermat point and J the first isodynamic point of a triangle ABC. Prove that the line FJ is parallel to the Euler line of triangle ABC.[/i]

For the solution, I will use three lemmas:

[b]Lemma 1.[/b] If $P_a$, $P_b$, $P_c$ are the reflections of a point P in the sides BC, CA, AB of a triangle ABC, then the circumcenter of the triangle $P_a P_b P_c$ is the isogonal conjugate Q of the point P with respect to the triangle ABC, and we have $P_b P_c \perp AQ$, $P_c P_a \perp BQ$ and $P_a P_b \perp CQ$.

[b]Lemma 2 (extended Napoleon theorem).[/b] If D, E, L are the centers of the equilateral triangles erected outwardly on the sides BC, CA, AB of the triangle ABC, then the triangle DEL is equilateral, and its center is the centroid G of triangle ABC.

[b]Lemma 3.[/b] For the first Fermat point F of triangle ABC, we have $EL \perp AF$, $LD \perp BF$, $DE \perp CF$.

Now, consider the reflections X, Y, Z of the point J in the sides BC, CA, AB of triangle ABC. We know that the first Fermat point F and the first isodynamic point J of triangle ABC are mutually isogonal conjugate points, so we can apply Lemma 1 and see that the circumcenter of the triangle XYZ is the isogonal conjugate of the point J, i. e. the point F, and that we have $YZ \perp AF$, $ZX \perp BF$ and $XY \perp CF$. Together with $EL \perp AF$, $LD \perp BF$, $DE \perp CF$ (from Lemma 3), this yields YZ || EL, ZX || LD and XY || DE. Hence, the triangles XYZ and DEL are homothetic. Since the triangle DEL is equilateral (Lemma 2), it follows that the triangle XYZ is also equilateral. Hence, instead of saying that the [i]circum[/i]center of the triangle XYZ is the point F, we can simply claim that the [i]center[/i] of the triangle XYZ is the point F.

Now, since the triangles XYZ and DEL are homothetic, there exists a homothety h mapping the triangle XYZ to the triangle DEL. This homothety h maps the center F of the triangle XYZ to the center of the triangle DEL, hence to the centroid G of triangle ABC (because of Lemma 2). Now let $J_1$ be the image of the point J in the homothety h. Then, since a homothety maps lines to parallel lines, we have $DJ_1 \parallel XJ$. But $XJ \perp BC$ (since the point X is the reflection of the point J in the line BC). Thus, $DJ_1 \perp BC$. But the point D is the center of the equilateral triangle constructed outwardly on the side BC of triangle ABC, and hence lies on the perpendicular bisector of this side BC. Thus, the line $DJ_1$, passing through D and being perpendicular to BC, must be the perpendicular bisector of this side BC. In other words, the point $J_1$ lies on the perpendicular bisector of the side BC. Similarly, the same point $J_1$ lies on the perpendicular bisectors of the other two sides of triangle ABC. And this shows that our point $J_1$ coincides with the circumcenter O of triangle ABC. Hence, the image of the point J in the homothety h is the point O.

So we have seen that the homothety h takes the points F and J to the points G and O, respectively. Therefore, GO || FJ. But the line GO is just the Euler line of triangle ABC, and thus we see that the line FJ is parallel to the Euler line of triangle ABC.

Qed..

See also [url=http://groups.yahoo.com/group/Hyacinthos/message/7957?expand=1]Hyacinthos message #7957[/url] for some related results.

[b]PS.[/b] I had also posted some explanations about the isodynamic points on http://www.mathlinks.ro/Forum/viewtopic.php?t=6489 .

  Darij

[quote="darij grinberg"]The only thing that makes me wonder is this strange "Very hard" with 9 exclamation signs. Especially this is because my experience shows that a problem you call "hard" (even without any exclamation sign) is already far beyound my brain. And now this one, really strange...[/quote]

Perhaps that just means you're better in geometry, darij? ;)

Believe me the problem is very difficult.Of course your geometry is excellent.
$F'$ also is a point in triangle that:
\[ <F'BC=FBA , <F'CB=FCA\]

[quote="Omid Hatami"]Believe me the problem is very difficult.Of course your geometry is excellent.[/quote]

Thanks, I'm always glad to hear this...  ;)  but I remember there are some much harder problems about the Fermat and isodynamic points.

Actually, a triangle ABC has two Fermat points, two Napoleon points and two isodynamic points. These six points altogether are called [b]FNI points[/b] (for: [b]F[/b]ermat, [b]N[/b]apoleon, [b]I[/b]sodynamic). It turns out that any two of these six FNI points are collinear with (at least) one other interesting triangle center; hence, altogether, we get $\displaystyle \binom{6}{2}=15$ interesting collinearities, the so-called [b]FNI collinearities[/b]. For a list of these collinearities, you can look at [url=http://groups.yahoo.com/group/Hyacinthos/message/6602?expand=1]my Hyacinthos message #6602[/url], or you can also consult four Forum Geometricorum papers by the late Lawrence S. Evans ([url=http://forumgeom.fau.edu/FG2002volume2/FG200209index.html]paper 1[/url], [url=http://forumgeom.fau.edu/FG2002volume2/FG200211index.html]paper 2[/url], [url=http://forumgeom.fau.edu/FG2003volume3/FG200304index.html]paper 3[/url], [url=http://forumgeom.fau.edu/FG2003volume3/FG200319index.html]paper 4[/url]).

The problem with all the FNI collinearities is that they are quite easy to show using barycentric coordinates, but most of them haven't been proven synthetically yet.

The problem you posted, namely to show that the line FF' is parallel to the Euler line of triangle ABC, is number [b]X[/b] in my list (actually, in the list, I don't say "the line is parallel to the Euler line", but I say "the line passes through the Euler infinity point"; actually, this Euler infinity point is just the infinite point (on the projective plane) which is common to all lines parallel to the Euler line). Of course, the collinearity number [b]XI[/b] is analogous. The collinearity [b]IX[/b] has a not-so-difficult synthetic proof, too. But does anybody have synthetic proofs to the other 15 - 3 = 12 FNI collinearities?

And actually, once the day will come and all 15 collinearities will be shown, the time will be ready for a synthetic proof of the [url=http://www.thejuniverse.org/geometry/LC/index.html]Lester circle theorem[/url]. So you see, there are lots of VERY hard problems in geometry left to be solved...

[quote="Omid Hatami"]$F'$ also is a point in triangle that:
\[ <F'BC=FBA , <F'CB=FCA\][/quote]

Yes, that's exactly the definition of F' as the isogonal conjugate of F.

  Darij

Dear all,
another proof has been put on my website in an article intilted "La fascinante figure de Cundy" (volume 2 (2008)).
http://perso.orange.fr/jl.ayme/
Sincerely
Jean-Louis


Show that $\vert 12^m -5^n\vert \ge 7$ for all $m, n \in \mathbb{N}$.

Easy :Suppose there exist $ k < 7$ and the equation has solution .
$ |12^m \minus{} 5^n|\equiv 1(\mod 2)$ and $ 3,5\not|12^m \minus{} 5^n$ so we must prove for $ k \equal{} 1$
Suppose $ |12^m \minus{} 5^n| \equal{} 1$
[u]Case 1[/u]$ 12^m \minus{} 5^n \equal{} 1$
But $ 5^m \plus{} 1\equiv 2(\mod 4)$ then the equation has no solution.
[u]Case 2[/u]$ 12^m \minus{} 5^n \equal{} \minus{} 1$
$ 5^n \minus{} 1\equiv ( \minus{} 1)^n \minus{} 1 (\mod 2)$ so $ n \equal{} 2p$
$ 12^m \equal{} (5^p \minus{} 1)(5^p \plus{} 1)$ has no solution!
This problem was be prove.

[quote="TTsphn"]
$ 12^m \equal{} (5^p \minus{} 1)(5^p \plus{} 1)$ has no solution!
[/quote]

This is quite unclear and seems to skip many lines of proof.

Let $N = \left| {{{12}^m} - {5^n}} \right|$, since $(N,30) = 1$, so if $N \leqslant 6$, then $N = 1$.
Because ${5^n} \equiv 5,3,4,9,1(\bmod 11)$, so ${12^m} - {5^n}\not  \equiv  \pm 1(\bmod 11)$, hence $N \ne 1$, contradiction.
So $N \geqslant 7$.

Of course, when $m = 1$, we get the solution $12^1 - 5^1 = 7$. I claim that $| 12^m - 5^n| \ge 19$ if $m \ge 2$. To see this, we are going to view the equation modulo $3120$; for $m \ge 2$ we have $12^m \pmod{3120} \in \{144, 1728, 2016, 2352\} = A$ while $5^n \pmod{3120} \in \{5, 25, 125, 625\} = B$. Since the integers in $A$ and $B$ that are closest to one another are $125$ and $144$ we get $|12^m - 5^n| \ge 19$ for all  $m, n \in \mathbb{N}$ with $m \ge 2$.


group theory
abstract algebra
pigeonhole principle
superior algebra
superior algebra unsolved

Let $ G$ be a group. Asumme that  the number of subgroups of $ G$ is fiinte, is $ G$ finite?

Case I) $ G$ has an element of infinite order.

Case II) Every element of $ G$ has finite order.

First, we see that each element has a finite order. Assume the contrary. Then there exist an element $ g \in G$ such that $ |g| \equal{} \infty$. Consider the subgroups <$ g$>, <$ g^2$>, <$ g^3$>,.... There are infinitely many such subgroups, which is a contradiction. Hence, every element has a finite order. Also, the group is finitely generated. If there are infinitely many generators, then the cyclic groups of the generators are subgroups, and there are infinitely many of them, which is a contradiction. Since the group $ G$ is finitely generated, and each of the generators have a finite order, $ G$ is finite.

[quote="Twin Prime Conjecture"]
Since the group $ G$ is finitely generated, and each of the generators have a finite order, $ G$ is finite.
[/quote]

The group $ G \equal{} \langle a,b \rangle / N( \langle a^2,b^2 \rangle )$ is finitely generated, with each generator having finite order, yet $ G$ is not finite.

A slightly different way to organize things:

Suppose $ G$ is infinite, and consider all of the singly generated subgroups $ \langle g\rangle$. Any such cyclic group has only finitely many generators; $ \phi(n)$ if it has order $ n$ and $ 2$ if it is infinite. Therefore, as $ g$ runs over all of the infinitely many elements of $ G$, each cyclic subgroup appears only finitely many times. By pigeonhole, there must be infinitely many such groups.

Look [url=http://www.mathlinks.ro/viewtopic.php?t=275696]here[/url] for still another proof of this inquiry of yours, esteemed [i]rulin[/i]. ;)


power of number
Sequence
modular arithmetic
number theory
prime numbers
IMO
IMO 1989

Prove that for each positive integer $ n$ there exist $ n$ consecutive positive integers none of which is an integral power of a prime number.

There are at most $1+\sqrt[2]{n}+\sqrt[3]{n}+\sqrt[4]{n}+...+\sqrt[\left\lfloor log_2(n)\right\rfloor]{n} \leq 1+ \sqrt n log_2(n)$ 'true' powers $m^k , k\geq 2$ in the set $\{1,2,...,n\}$. So when $p(n)$ gives the amount of 'true' powers $\leq n$ we get that $\lim_{n \to \infty} \frac{p(n)}{n} = 0$.

Since also $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$, we get that $\lim_{n \to \infty} \frac{p(n)+\pi(n)}{n} = 0$. Now assume that there is no 'gap' of lenght at least $k$ into the set of 'true' powers and the primes. Then this would give that $\frac{p(n)+\pi(n)}{n} \geq \frac{1}{k}$ for all $n$ in contrary to the above (at east this proves a bit more).

Edit: to elementarize the $\lim_{n \to \infty} \frac{\pi(n)}{n} = 0$ part:
Look $\mod (k+1)!$. Then all numbers in the residue classes $2,3,4,...,k+1$ are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length $k$, there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each $(k+1)!$ numbers, again contradicting $\lim_{n \to \infty} \frac{p(n)}{n} = 0$.

Or we can just choose $2n$ distinct primes $\{p_i\}_i,\{q_i\}_i,\ i=\overline{1,n}$, and take a positive integer $k$ such that $k\equiv -i\pmod{p_iq_i},\ \forall i\in\overline{1,n}$, which must exist by the Chinese Remainder Theorem. The numbers $k+1,k+2,\ldots,k+n$ will do the trick.

A little simpler, huh? :)

I have found the same solution as dblues and I wonder isn't it too easy for IMO #5? :D

Extension:  Replace prime with power of any integer.

[hide="Hint"]
Let $x\equiv p_1\pmod{p_1^2}, x+1\equiv p_2\pmod{p_2^2},\cdots$. [/hide]

But then any integer is a power of itself.

I mean perfect power (i.e the exponent is greater than one).

By taking (by CRT) $x_0$ a positive integer solution of the system of congruences $x+k\equiv p_k\pmod{p_k^2}$, $1\leq k \leq n$, and $x = x_0 + \prod_{k=1}^n p_k^2$, we also guarantee $x+k$ is composite for all $k$.

[quote="dblues"]Well, a much simpler solution would be to consider the set of $n$ integers $\{ [(n+1)!]^2 + 2, [(n+1)!]^2 + 3, \ldots , [(n+1)!]^2 + (n+1) \}$.

:P[/quote]

Not to nitpick, but I think that you should prove that they are all NOT perfect powers of a prime number.

I will explain [b]dblues's[/b] solution .
Let's assume that one of the numbers is the power of a prime.
Let it be $[(n+1)!]^2+i=p^k$ with $p$ prime and $2\leq i\leq n+1$.
From this we get that $i$ divides $p^k$ so $i=p^l$ with $l\geq 1$.
So $p\leq i\leq n+1$. Obviously $k=v_p([(n+1)!]^2+i)=\min(v_p([(n+1)!]^2),v_p(i))=v_p(i)=l$
$i=p^l=p^k=[(n+1)!]^2+i$ a contradiction.
$v_p(x)$ is the exponent of $p$ in $x$.

[quote="aleph_zero"]How do we get $ (n+1)!^2+i=p^k\Rightarrow i\mid p^k $??[/quote]
Because $i|(n+1)! \implies i|(n+1)!^2+i \implies i|p^k$

My solution
Let ${p_1},{p_2},...,{p_{2n}}$ be $2n$ prime intergers
Consider the system of equations
$x \equiv  - 1\left( {\bmod {p_1}} \right)$
$x \equiv  - 1\left( {\bmod {p_2}} \right)$
$x \equiv  - 2\left( {\bmod {p_3}} \right)$
$x \equiv  - 2\left( {\bmod {p_4}} \right)$
$..$.
$x \equiv  - 1\left( {\bmod {p_{2n - 1}}} \right)$
$x \equiv  - 1\left( {\bmod {p_{2n}}} \right)$
By CRT, the system have one $x$ satisfies
So, $x + i \vdots {p_{2i - 1}},{p_{2i}}$, $i = \overline {1,n} $
....

Let us take any $a\in\mathbb{N}$ and consider $n$ consecutive integers $a+1,..., a+n$ . If none of this numbers is an integral power of a prime number, then we are done. Now suppose ${p_1}^{a_1}<...<{p_k}^{a_k}$ are all the numbers (among  $a+1,..., a+n$) which are integral powers of prime numbers ( $p_1, ...,p_k$ - prime numbers). I take $N=s*{p_1}^{a_1+1}*...*{p_k}^{a_k+1} $ ( I choose $s\in\mathbb{N}$ so that $N>(a+n)$) and prove that none of $n$ consecutive integers $N!+a+1,..., N!+a+n$ is an integral power of a prime number. There are 2 cases:
1. $a+i$($n\geq i\geq 1$) isn't an integral power of a prime number. $(N!+a+i )\vdots{(a+i)}$, so $N!+a+i$  isn't an integral power of a prime number.
2.$a+i={p_j}^{a_j}$  ( $n\geq i\geq 1$ ) for some $j$,  $k\geq i\geq 1$. Suppose $(N!+a+i )=p^t$ for some prime number $p$ and  $t\in\mathbb{Z}$. Note that $(N!+a+i )\vdots{(a+i)}$ , so $p={p_j}$ . Also $N!\vdots N\vdots {({p_j}^{a_j+1})}$ and $a+i={p_j}^{a_j}$ so ${p_j}^{a_j+1}$ doesn't divide $N!+a+i$, so ${p_j}^{a_j}\geq N!+a+i$ . Contradiction.
 So $N!+a+i$ isn't an integral power of a prime number for $n\geq i\geq 1$ .

The problem is essentially asking to find a natural number $m$ such that $$m, m+1, \dots, m+n-1$$ are not powers of any prime.
Consider the list of primes $\{p_1, p_2, p_3, \dots\}$. Simply pick $m$ such that $m$ satisfies the following congruences.

\begin{align*}
  m \equiv 0 \pmod {p_1p_2} \\
  m+1 \equiv 0 \pmod {p_3p_4} \\
  \vdots \\
  m+n-1 \equiv 0 \pmod {p_{2n-1}p_{2n}} \\
\end{align*}

A solution exists by CRT. Further, each number in the sequence $\{m, m+1, \dots, m+n-1\}$ is divisible by the product of $2$ primes, so this $m$ works.

Let $p_1,p_2,p_3,\dots,p_n$ are $N$ prime numbers.By CRT we can find arbitrary large $x$ such that:
\begin{align*}
		x &\equiv p_1-1 \pmod{p_1^2} \\
		x &\equiv p_2-2 \pmod{p_2^2} \\
		&\phantom\equiv\vdots \\
		x &\equiv p_n-n \pmod{p_n^2}.
	\end{align*}
But then $x+i$ is divisible by $p_i$ but not by $p_i^2$.Now take $x$ sufficiently large in that particular residue class $\pmod{p_1p_2\dots  p_N}$ so that $x+i\ne p_i$.
$\square$

let $\{p_i\}$ and $\{h_i\}$be two sequences of prime numbers and $\{p_i\} \cap \{h_i\}=\emptyset$
Hence we are looking for $x$ the solutions to this system
\begin{align*}
x \equiv i \pmod{p_i\cdot h_i}
\end{align*}
for $i =\{1, 2,... ,n\}$ and by the Chinese Remainder Theorem it has a solution

Clearly, the product of two distinct primes cannot be the power of a prime, and if an integer is a multiple of a product of two distinct primes it also cannot be the power of a prime. The problem statement is equivalent to the following system of congruences:

\begin{align*}
x &\equiv 0 \pmod {p_1p_2} \\
x+1 &\equiv 0 \pmod {p_3p_4} \\
\cdots \\
x+n-1 &\equiv 0 \pmod {p_{2n-1}p_{2n}} \\
\end{align*}

which can be rewritten as

\begin{align*}
x &\equiv 0 \pmod {p_1p_2} \\
x &\equiv -1 \pmod {p_3p_4} \\
\cdots \\
x &\equiv -n+1 \pmod {p_{2n-1}p_{2n+1}} \\
\end{align*}.

By CRT this system has a solution $x$, which means those consecutive integers exist, as desired. $\blacksquare$

Choose the first $2n$ prime numbers, and choose $a$ satisfying $$a \equiv{-k} \pmod{p_{2k-1}p_{2k}}$$ for all $k$ (Which clearly exists by CRT). Then $a+1, \dots, a+n$ Is a valid set satisfying the desired conditions.

Take $$a=(2n)!-n$$
Then $<a+i>_{0\leq I \leq n-1}$ works :)

Let $p_1,\dots, p_{2i}$ be a sequence of $2i$ distinct primes. By CRT, there exists a $q$ such that \begin{align*} &p_1p_2\mid q+1 \\ \vdots \\ &p_{2i-1}p_{2i} \mid q+i, \end{align*} the claim follows.

By CRT there exists a positive integer $m$ such that $m \equiv -i \pmod {p_{2i-1}p_{2i}}$ for all positive integers $i$, where $p_i$ denotes the $i$th prime. Then the $n$ integers not including $m$ immediately succeeding it work.

[hide = Proof of IMO 1989] Let $p_1, p_2, \dots, p_n$ and $q_1, q_2, \dots, q_n$ be two sets of primes. Then we have that considering the congruences for some arbitrary $X$ the equations,

\begin{align*}
X & \equiv -1 \pmod{p_1q_1} \\
X & \equiv -2 \pmod{p_2q_2} \\
 &\phantom \equiv \vdots \\
X &= -n \pmod{p_nq_n},
\end{align*}

has a solution. Now this system of congruences gives us that each $X + i$ is divisible by the product of two primes $p_i$ and $q_i$ and hence will never be a prime power for any $i$. $\blacksquare$ [/hide]


LaTeX
calculus
derivative
analytic geometry
graphing lines
slope
limit

On another computer I downloaded the 3 programs on the AoPS website in order to create LaTeX documents, but again I can't get it to work.  I think I am following all of the instructions on the website correctly, but I am having problems.  This is what happens:

(1) type up the document
(2) save it
(3) press ctrl F7
(4) go to Build Output
(5) look for the DVI document in the same folder as the TeXnicCenter document is in - but can't find it
(6) change LaTeX=>DVI to LaTeX=>PDF
(7) ctrl F7
(8) Build Output
(9) View Output - get error message "cannot execute the command"
(10) look for PDF in folder but can't find it.

Sorry I keep on irritating you with questions.  Please help. Thanks.

What is a source code?  What I type up for the document in TeXnicCenter?

[quote] Save the document (press Ctrl-S or click File, then Save) as 'Hello' (don't include the quote marks in the name) in a folder of your choice. The file will appear in your folder as 'Hello.tex.' We will call this the source file.
[/quote]

Okay, so I assume you mean the whole document.  THis is the start of it (I didn't finish it here b/c of the trouble I was having):

\noindent MTH 141 URI \\
\noindent Professor Kulenovic \\ \\

\noindent \textbf{Differentiation Formulas} \\ \\

\noindent We know that the derivative is the slope of the tangent line, the instaneous rate of change, and: \\
$\displaystyle\lim_{h\to\0}{\frac{f(a+h)-f(a)}{h}}$. \\
\noindent However, to make this easier, to be able to find the derivatives of all sorts of functions, you can use differentiation formulas. \\ \\

\noindent Derivative of a Contant Function, $f(x)=c$ \\
$\displaystyle\lim_{h\to\0}{\frac{f(a+h)-f(a)}{h}}$
$\displaystyle\lim_{h\to\0}{\frac{f(a+h)-f(a)}{h}}$
\noindent The Product Rule: the derivative of a product of two functions\\
$\displaystyle(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$ \\

Also, what error messages are you getting?  (There's a box where a bunch of log messages come out during compiling - this will usually tell you what's wrong if there's no output.)

In non-message board LaTeX, you have to include

\documentclass{article}
\begin{document}

at the begining and at the end you must have

\end{document}

This is not in the source code, so I'm assuming that's the problem.  You should read the the LaTeX guide on this site or search google for LaTeX tutorials.  You should be able to find some info on the structure of a LaTeX document.


ai do' dich ho^. em cai' nay` sang Tieng' Viet voi'
In a rectangle triangle the projections of the CATHETI on the HYPOTHENUSE are 3 m and 12 m long .

em moi' doan' mo` la`
tam giac vuong , hinh` chieu' cua? 2 canh. ben len canh huyen` la` 3 va` 12 ... ko bit' co' dung' ko :-SS , tu` dien cua? em ko co' may` tu` kia :-SS

[quote="N.T.TUAN"]''rectangle triangle '' nghia la tam giac vuong a?  :o[/quote]

HTA hình như là Hoàng Tuấn Anh thì phải? :lol:


induction
number theory unsolved
number theory

[quote="Beat"][i]Solve in natural numbers the equation $x!\cdot y!=(n!)^3$.[/i][/quote]

Let $x \geq y$ (this will be useful in further solution). We have that $(n!)^3\equiv0(\mod n^3)$. From this we get that
1)
$x \geq 2n$
$y \geq n$

or 2)
$x \geq 3n$
$y<n$

So if we put minimum values of $x$ and $y$ in 1) we have:
$(2n)!n!=(n!)^3$
$(2n)!=(n!)^2$

Using induction we easily prove that $(2n)!>(n!)^2$ for $n>1$.

In 2) we get:
$(3n)!=(n!)^3$

Again, with induction $(3n)!>(n!)^3$ for $n>1$.

So we have the only solution $(x,y,n)=(1,1,1)$.

[quote="abdurashidjon"]How have you found that $(2n)!n!=(n!)^3$ can you give the solution[/quote]

I have put minimal possible values of $x$ and $y$ and after that proved that $LHS>RHS$.

This actually works good for prime number but I'm not sure about composite ones. I know that there should be one step that I missed. I think I will write this later if noone do it before me.

Still if you don't understand how it works with primes I can explain.
If you have $n$ prime, then $(n!)^3\equiv 0 (\mod n^3)$. From this we have that $x!y!\equiv 0 (\mod n^3)$, so this product must have $3$ numbers divisible by $n$. Since it is prime the least values of $x$ and $y$ can be $2n$ and $n$.

[quote="Beat"][i]Solve in natural numbers the equation $x!\cdot y!=(n!)^3$.[/i][/quote]
It is easy. Let $x\le y,y/2<p\le y$ (p - prime). We have $ord_p(x!y!)=1. \ or \ 2$, but $ord_p((n1)^3)=0, \ or \ ge 3$. It give $x=y=n=1$,

Rust can you explain your solution? I don`t understand it.
Is it sure that there will exist a prime p such that $y/2<p\le y$
And what is ord?
Please rewrite your solution. Thank you

I use, that for $y\ge 2$ exist prime p: $\frac y2<p\le p$. Therefore $ord_p (y!)=1,ord_p(x!)\le 1, (x\le y)$.
$ord_p((n!)^3)=3ord_p(n!)$. Therefore $ord_p((n!)^3)=0,p>n$ or $(ord_p((n!)^3)\ge 3$. It give contradition if y>1.

Rust uses Bertrand's postulate stating that there is a prime in $[n;2n],\ \forall n\in\mathbb{N}$.
I bet he uses $ord_p$ as the biggest power $d$ dividing some particular number.

eLeM, your solution does not work for composite numbers $n$.


induction
function
calculus
derivative
Putnam
calculus computations

Consider the function f: R->R defined by setting, for all x in R,
f(x) = 3sin(x) / (2 + cos(x)).

Two questions:

1. Let x_o be a point in the interval (0 , 2[pi]/3], and define a sequence {x_n}_n>=0 inductively, by setting x_n = f(x_n-1), for n>=1. Prove that {x_n}_n>=0 is a strictly decreasing sequence in (0 , 2[pi]/3], and that lim_n->oo x_n = 0.

2. Find a positive real exponent b such that the sequence {(n^b)(x_n)}_n>=1 converges to a real number C > 0.

If possible, please provide descriptions along with your proofs for 1 and 2. Thank you.

Does anybody have any thoughts about this problem?

I am stumped.  :maybe:

It looks nicer like this:

[quote="Cezanne123"]Consider the function $f: R\to R$ defined by
$f(x) = 3\sin(x) / (2+\cos(x))$.

Two questions:

1. Let $x_{0}$ be a point in the interval $(0 , 2\pi/3]$, and define a sequence $\{x_{n}\}_{n\geq0}$ inductively, by setting $x_{n}= f(x_{n-1})$ for $n\geq1$. Prove that $\{x_{n}\}_{n\geq0}$ is a strictly decreasing sequence in $(0 , 2\pi/3]$ and that $\lim_{n\to\infty}x_{n}= 0$.

2. Find a positive real exponent $b$ such that the sequence $\{(n^{b})(x_{n})\}_{n\geq1}$ converges to a real number $C > 0$.

If possible, please provide descriptions along with your proofs for 1 and 2. Thank you.[/quote]

For the first part of 1, I think it should be enough to analyze the derivative of $f$.

First, it helps to note that
\[f(x) = x-\frac{x^{5}}{180}+O(x^{7}) \]
as $x$ approaches $0$.  For Question 2, we can then use the method from this recent Putnam problem:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122460[/url].

If my calculations are right, we get $b = \frac{1}{4}$.


Two boats start in a race and the faster boat gains 5 feet over the other every 55 yards the faster boat travels. To the nearest foot, by how many feet will the faster boat win a half-mile race?

The faster boat gains $ 5\text{ feet}$ per $ 55\text{ yards}\equal{}165\text{ feet}$. That is equivalent to gaining $ 1\text{ foot}$ per $ 33\text{ feet}$. Knowing a $ \text{mile}\equal{}5280\text{ feet}\implies\frac{1}{2}\text{ miles}\equal{}2640\text{ feet}$, we evaluate $ \frac{2640}{33}\equal{}\boxed{80}$

(because the problem says "to the nearest foot", I sorta suspect this isn't right since it gives a whole number value, so could someone check it?)

[quote="RoFlLoLcOpT"]
(because the problem says "to the nearest foot", I sorta suspect this isn't right since it gives a whole number value, so could someone check it?)[/quote]

There was a similar problem like this, except it asked for $0.375\cdot1608$ to the nearest tenth. Obviously, $1608$ is divisible by $8$, so there is no need to round to the nearest tenth. So... I don't think these wordings matter :D


linear algebra
matrix
induction
superior algebra
superior algebra solved

Pls help me with this one!!!

Prove that if A,B are matrices M_n (R) with n natural , n>=2 so that AB=BA prove that A^k*B^l=B^l*A^k, for any l,k natural.

help me please! :D  (full solution)

bye bye

Isn't this kind of easy? I think induction works (2 variables). First we prove that A*B^n=B^n*A for any n by induction after n, and we call the sentence A^k*B^n=B^n*A^k for any n P(k). P(1) is true and we prove P(k)->P(k+1). To get from P(k) to P(k+1) we do this: for any n we have A^(k+1)*B^n=A*B^n*A^k (from P(k))=B*A*B^(n-1)*A^k= B^2*A*B^(n-2)*A^k=..=B^n*A^(k+1).

It's really easy, but maybe I misunderstood?


[quote="Svejk"]From Stoltz-Cezaro's lemma we have that:
$ \lim_{n\rightarrow\infty}n^{\minus{}\frac{3}{2}}\sum_{j \equal{} 1}^{6n}\sqrt{j}\equal{}\lim_{n\rightarrow\infty}\frac{\sqrt{6n\plus{}6}}{\sqrt{(n\plus{}1)^{3}}\minus{}\sqrt{n^{3}}}$[/quote]
I think there should be:
$ \lim_{n\to\infty}\frac{\sum_{k\equal{}1}^{6}\sqrt{6n\plus{}k}}{\sqrt{(n\plus{}1)^{3}}\minus{}\sqrt{n^{3}}}$


Given triangle $ABC$, let $M$ be the midpoint of side $AB$ and $N$ be the midpoint of side $AC$. A circle is inscribed inside quadrilateral $NMBC$, tangent to all four sides, and that circle touches $MN$ at point $X.$ The circle inscribed  in triangle $AMN$ touches $MN$ at point $Y$, with $Y$ between $X$ and $N$. If $XY=1$ and $BC=12$, find, with proof, the lengths of the sides $AB$ and $AC$.

:rotfl: Where'd you get this problem??

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=125

Scroll down to #5 for a solution.

As with the other USAMTS problems, could someone post a solution to this one for the Mathlinks-AoPS resources?

Thanks.

(btw, I don't mind if you copy a solution from the solutions you posted just as long as it is correct ;) )

Since tangents from each point are equal, we can label points of tangency, for the smaller circle, let us call the points of tangency to $AN$ as $P$ and at $AM$ let us call point of tangency $R$, $AP=AR$, thus let us assign both of them the variable $a$. Now $MR=MY$ so let us give them the variable $b$. $NP=NY$ so they have the variable $c+1$ whereas $NL$ (let us call $L$ the point of tangency for the bigger circle on $AC$)$=c$ Now by calling the point of tangency for the bigger circle on $BC$ as $Q$ and since $CL=CQ$ we can give them the variable $d$. Now give the final point of tangency of the bigger circle at $T$ which is on $AB$, and thus $MT=b+1$ and since $TB=QB$ we can give them the variable $e$. Finally after all this since $AN=NC$, then $a+c+1=c+d \Longrightarrow a+1=d$, now we can replace all $d$ with $a+1$. Now similarily, since $AM=MB$ then $a+b=b+1+e \Longrightarrow e=a-1$, from here we can replace all the $e$, now since $BC=12=a+1+a-1 \Longrightarrow a=6$, so now let's replace every $a$ with $6$, now we only have two variables left, $c$ and $b$, notice how the tangents $AL=AT$, thus $b=c$, from here, we're almost set, now notice how triangles $ANM$ is similar to triangle $ACB$, from there, let's make a proportion: $\frac{2c+1}{12}=\frac{6+c}{12+2c}$, and from there we get that $c=2.5$ and from plugging that back in we get $\boxed{AC=17, AB=19}$

[hide="Answer"]$MN=\frac{1}{2}BC$ since it connects the midpoints of $AC$ and $BC$. Since $BCNM$ is cyclic, we have $BM+CN=MN+BC=12+6=18$ and $AB+AC=36$. Drawing the radii to the tangent points of each side of $BCNM$, and labeling those points as $R$, $S$, $T$, and $X$ (already named), we use Pythagorean to see that $BR=BS$ and $CS=CT$. Therefore, $BS+CS=BC=12=BR+CT$, and we also see that $AB=AC=12$. Now, since $XY=1$ and $\triangle AMN\sim \triangle ABC$, then we have altitude $AZ$ and $SZ=2$. Using Pythagorean again, we have $(AR+BR)^2-(BS+2)^2=AZ^2$ and $(AT+CT)^2-(CS-2)^2=AZ^2$, so $(12+BR)^2-(BR+2)^2=(12+CT)^2-(CT-2)^2\Rightarrow 5BR=7CT$. $BR+CT=12\Rightarrow BR=12-CT$, so $5(12-CT)=7CT\Rightarrow 60-5CT=7CT\Rightarrow 12CT=60\Rightarrow CT=5$, $BR=7$, and we have $\boxed{AB=19}$ and $\boxed{AC=17}$.[/hide]

I'm too lazy to type it out, so here you go...

O1,O2 are all on the angle bisector of <BAC; Thus , A,O1,O2 are collinear ;By MN//BC and O1N bisect <ANM;O2C bisect <ACB→O1N//O2C；so △AO1N∽△AO2C→AN/AC=O1N/O2C=AO1/AO2=1/2；By △WO1N∽△ZO2C； so ：O1N/O2C=R1/R2（O1W/O2Z）=1/2；WZ+PQ=WN+NZ+PM+MQ=MY+MX+XN+YN=2MN=2*（1/2）BC=BC=12；Notice that：AP=AW；AQ=AZ→PQ=WZ；so WZ=PQ=6=MN；MN=MY+YN=WZ=WN+NZ→MY=NZ；Let MY =x，YN=y； we get that ：1）x+y=MN=6；2）x-y=1；solve it we get ： x=3.5;y=2.5→By YN/KC=WN/ZC=1/2=AN/AC and <ANM=<ACB; we get that :point :A,Y,K are collinear →KC=2YN=5;KB=2MY=7;then we get AB=BM+2PQ=BK+2*6=12+7=19;
 SO AC=KC+2WZ=5+2*6=17; so the answer is :< AB=19 and AC = 17> ..QED(I like this problem very much!!  :D )

[asy]
size(200);
defaultpen(linewidth(0.6)+fontsize(11));
real r = 4*sqrt(3);
pair A = (4,r), B = (0,0), C = (5,0), M = (A+B)/2, N = (A+C)/2;
pair I = incenter(A,B,C), J = incenter(A,M,N);
draw(A--B--C--A^^M--N^^incircle(A,B,C)^^incircle(A,M,N));
dot("$A$",A,dir(90),linewidth(3.3));
dot("$B$",B,SW,linewidth(3.3));
dot("$C$",C,SE,linewidth(3.3));
dot("$X$", (I.x,r/2),S,linewidth(3.3));
dot("$Y$",(J.x,r/2),dir(90),linewidth(3.3));
dot("$M$",M,dir(145),linewidth(3.3));
dot("$N$",N,dir(20),linewidth(3.3));
[/asy]
Let $AC =: b$ and $AB =: c$ for ease of typesetting.  By Pitot on quadrilateral $BMNC$,
\[
\frac c2 + \frac b2 = 12 + 6 = 18,
\]
ergo $b + c = 36$.  Furthermore, the incircle of $\triangle ABC$ is also the $A$-excircle of $\triangle AMN$, so
\[
1 = XY = MY - NY = AM - AN = \frac c2 - \frac b2,
\]
ergo $c - b = 2$.  Solving this system yields $\boxed{AB = 19\text{ and }AC = 17}$.


Find real  numbers  $a,b$ such that  for every  $x,y\in\mathbb{R}$  we have $|ax+by|+|ay+bx|=|x|+|y|.$

Putting $x=1,y=0$ and $x=b,y=-a$ we get

$|a|+|b|=1$
$|a^2-b^2|=|a|+|b|=1$

The second equation can be rewritten

$||a|-|b||\cdot(|a|+|b|)=1 \implies |a|-|b|=\pm 1$

From $|a|+|b|=1\land |a|-|b|=\pm 1$ we find all the solutions: $(\pm 1,0),(0,\pm 1)$


Tha radius of convergence for the series:

$ (x^n n^n)/(2^n n!)$

would be?

You meant $ \sum_{n\equal{}1}^{\infty}\frac{n^nx^n}{2^nn!},$ right?

A weak form of Stirling is that $ \sqrt[n]{n!}\sim \frac{n}e.$ This gets us that

$ \left(\frac{n^n}{2^nn!}\right)^{\frac1n}\sim \frac{ne}{2n}\equal{}\frac e2.$

The reciprocal of this is the radius of convergence for the power series, so the radius of convergence is $ \frac2e.$

Stronger forms of Stirling will tell us that the series diverges for $ x\equal{}\frac2e$ but converges conditionally as an alternating series for $ x\equal{}\minus{}\frac2e.$

If $ a_n = \frac{n^n}{2^n n!}$, from D'Alembert we know that the radius of convergence $ R$ of the power series $ \sum a_n x^n$ equals ${ \lim_{n\to\infty}\frac{a_n}{a_{n+1}} = \lim_{n\to\infty} \frac{n^n}{2^n n!} \frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}} = \lim_{n\to\infty} 2 \left(\frac{n}{n+1}\right)^n$

Now it's easy to see that $ R = 2e^{-1} = \boxed{\frac{2}{e}}$.

[quote="Kent Merryfield"]You meant $ \sum_{n \equal{} 1}^{\infty}\frac {n^nx^n}{2^nn!},$ right?

A weak form of Stirling is that $ \sqrt [n]{n!}\sim \frac {n}e.$ This gets us that

$ \left(\frac {n^n}{2^nn!}\right)^{\frac1n}\sim \frac {ne}{2n} \equal{} \frac e2.$

The reciprocal of this is the radius of convergence for the power series, so the radius of convergence is $ \frac2e.$

Stronger forms of Stirling will tell us that the series diverges for $ x \equal{} \frac2e$ but converges conditionally as an alternating series for $ x \equal{} \minus{} \frac2e.$[/quote]

Yes, I did mean $ \sum_{n \equal{} 1}^{\infty}\frac {n^nx^n}{2^nn!},$, but I didn't know how to write summation. Thanks.

In this part:

$ \left(\frac {n^n}{2^nn!}\right)^{\frac1n}$

where did the $ x^n$ term go?

Theorem: the power series $ \sum_na_nx^n$ has radius of convergence $ R$ where

$ \frac1R\equal{}\limsup_{n\to\infty}\sqrt[n]{a_n}.$

(With the appropriate interpretations of zeros and infinities.)

@ uglysolutions - Thanks. That's how I had worked it out, but I didn't relaize that $ \lim_{n\to\infty} \left(\frac {n}{n \plus{} 1}\right)^n$ would resolve to $ e^ { \minus{} 1}$

[quote="Kent Merryfield"]Theorem: the power series $ \sum_na_nx^n$ has radius of convergence $ R$ where

$ \frac1R \equal{} \limsup_{n\to\infty}\sqrt [n]{a_n}.$

(With the appropriate interpretations of zeros and infinities.)[/quote]

That's an interesting idea that I didn't know. What does sup stand for?

That $ \limsup$ stands for 'limit superior': see [url=http://mathworld.wolfram.com/SupremumLimit.html]this[/url].


Tiffany's current automobile averages 30 miles per gallon of gasoline.
Gasoline costs $ \$1.20$ per gallon. She is thinking about buying a new
car that averages 18 miles per gallon. If she drives 30,000 miles,
how many more dollars will she have to spend on gasoline using the
new car rather than the old one?

With her current car, she will need $ \frac{30000}{30}\equal{}1000$ gallons, and with the new one she will need $ \frac{30000}{18}\equal{}\frac{5000}{3}$ gallons. Thus, the difference in price is $ 1.20(\frac{5000}{3}\minus{}1000)\equal{}\boxed{\$800.00}$.

[quote="chess940"]That method might take to long. By that time this(  :rotfl: )could happen to you!![/quote]

Erm, it takes less than 45 seconds, which is the standard countdown time.


function
real analysis
real analysis theorems

How prove this nice result:
The continuous functions of $\mathbb{R}$ in $\mathbb{R}$ are bijection with $\mathbb{R}$.

It's been discussed before, I think.

First of all, if $c$ is the continuum cardinality, then there are $c$ constant (and thus continuous) functions. This means that there are at least $c$ continuous functions. On the other hand, a continuous function is determined by the images of the rationals, so there are at most as many continuous functions as there are sequences of real numbers. Now try to show that there are $c$ sequences of real numbers, and you're done.

To be honest, the problem sounded like absurd for me. Now I see what bénabar meant.

The result we discussed recently was related to functions with a Darboux's property. We showed that there are at least $2^c$ such functions. Right?

We discussed the one about Darboux functions recently, but this one has also been discussed.


Γειά χαρά σε ολους αυτους που προσπαθούν να αναβαθμήσουν το φόρουμ μεσα από την συνεργασία.Επανείλθα δρυμίτερος.Το λεω αυτό γιατί τον τελευταίο καιρό έχω ελαχιστο χρόνο για να ασχοληθώ με τα μαθηματικά τα οποία λατρεύω.Να λοιπόν και η άσκηση που θέλω να σας προτείνω: Να βρείτε όλους τους φυσικούς n ώστε ο n!^(n!+1) να είναι τέλειο τετράγωνο. :pilot:

Αρκεί να βρούμε πότε το $n!$ είναι τέλειο τετράγωνο .

Το $n!$ για $n>1$ δεν είναι τέλειο τετράγωνο :

Ο Bertrand λέει ότι μεταξύ των $k, 2k \, \forall k \in \mathbb{N}$ υπάρχεί πρώτος $p$ . Έστω $k=2m$ ο μεγαλύτερος άρτιος μικρότερος ή ίσος του $n$ .

 Τότε υπάρχει πρώτος $p$ με $m<p<2m$ και $p|n!$ . Επίσης είναι $2m=n$  ή  $2m=n-1$. Τότε $p \geq m+1$ άρα $2 p \geq 2m+2 >n$

συνεπώς ο $p^2$ δεν διαιρει το $n!$ δηλαδή το $n!$ δεν μπορεί να είνα τέλειο τετράγωνο.

Για $n=1$ είναι τέλειο τετράγωνο.

 :)  :cool:

Αν και ειμαι λαθος και συμφωνω με τον Ramanujan μπορεί o x να μην είναι τελειο τετραγωνο αλλά ο χ εις την χ+1 να ειναι τετραγωνο.πχ χ=3.

μήπως γιατί ο n! είναι άρτιος? :oops:

[quote="kostas"]μήπως γιατί ο n! είναι άρτιος? :oops:[/quote]

Ακριβώς !  :)

[quote="Αρχιμήδης 6"]Ευχαριστώ Ramanujan για την ωραία σου λύση.[/quote]
:)

Εγώ ευχαριστώ για την ωραία άσκηση. Υπάρχει άλλη λύση ?   :)

Πιστεύω πως η εύρεση μιας αλλης λυσης είναι μια προκληση.Προς το παρον δεν ξερω καποια,αλλα σιγουρα θα προσπαθησω να βρω.Οποιος λοιπον βρει μια αλλη λυση παιδια ας την στήλει.Προσπαθήστε!

Καποια αλλη λύση εχει να προτίνει καποιος?Πιστεύω πως η ερώτηση του Ramanujan ειναι μια προκληση!Ετσι Ramanujan? :pilot:

Φαινεται οτι η ασκηση Ramanujan εχει καποια δυσκολια στο να  αποδυχθει με αλλο τροπο.Θα την προσπαθήσω ομως κι αλλο.Κώστα εσυ πως τα πας με την ασκηση?

Την προσπάθησα τα ξημερώματα χθές και σήμερα.Είναι πραγματι σπουδαία ασκηση που μετρά την τεχνική του λύτη.Λύνεται με συνολα πρωτων ρε παιδια αλλα καπου κολλαω.θέλει πολυ προσοχη.

Γεια σε ολους.το ελληνικο αποκτα και παλι κινηση.καλη ξεκουραση και στα παιδια των πανελληνιων.και μην ανησυχητε για διαγωνισματα και αποτελεσματα.Θα ηθελα και μιλαω και εκ μερους του Αρχιμηδη 6 που το προτινε να παλεψετε αυτη την ασκηση n!=y^2 ακολουθοντας αλλη πορεια απο τη λυση το Ραμανατζαν. να σας τονισω οτι κατι τετοιο δεν (προφανως) μπορεσαν οι συναδελφοι στο advanced.


geometry
3D geometry
modular arithmetic
number theory

If $ \gcd(xyz,7) = 1$, then $ x^{3}\equiv \epsilon_{x}\mod 7$, $ y^{3}\equiv \epsilon_{y}\mod 7$ and $ -z^{3}\equiv \epsilon_{z}\mod 7$ (by Fermat's little theorem), where $ \epsilon_{x}, \epsilon_{y}, \epsilon_{z}\in \{\pm 1\}$. Let $ \epsilon = \epsilon_{x}+\epsilon_{y}+\epsilon_{z}$ for convenience. Then obviously $ 0 \le |\epsilon| \le 3$. Anyway, $ |\epsilon| = 0$ is impossible, because of yielding $ 1 \equiv 0 \bmod 2$. So in fact $ 0 < |\epsilon| \le 3$. Then $ \epsilon \not\equiv 0 \bmod 7$, and consequently $ 7\nmid (x^{3}+y^{3}+z^{3})$ (absurd).

[hide="more easily..."]Note that all cubes are congruent to $ 1,0$ or $ -1\ \mod 7$. As $ x^{3}+y^{3}\equiv z^{3}\pmod 7$, if none of $ x,y,z$ are congruent to $ 0$, two must be congruent $ \mod 7$.
$ x\equiv z\Rightarrow y\equiv 0,\quad y\equiv z\Rightarrow x\equiv 0$. If $ x\equiv y$, then $ z\equiv 2,0$ or $ -2$. Thus we conclude that one of them must be congruent to $ 0$. [/hide]

I think the last one is more illustrative.

[quote="jonysatie"]Let $ x,y,z$ be positive real number and  ....[/quote]
Please edit, "real number" to  "integer"  (or use the term "natural number")  ...

[quote="jonysatie"]Let $ x,y,z$ be positive real number and $ 7|(x^{3}+y^{3}-z^{3})$. Show that one of them is divisible by $ 7$, at least.[/quote]

It Should be ... Let $ x,y,z$ be positive integers and $ 7|(x^{3}+y^{3}-z^{3})$. Show that one of them is divisible by $ 7$, at least.

PD: I couldn't edit my first post


symmetry
inequalities
function
calculus
derivative
inequalities proposed

Let $ a,\ b$ and $ c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Find the maximum value of the expression:

\[ P\equal{}\sqrt[3]{\frac{a^3\plus{}b^3}{3}}\plus{}\sqrt[3]{\frac{b^3\plus{}c^3}{3}}\plus{}\sqrt[3]{\frac{c^3\plus{}a^3}{3}}\plus{}3\sqrt[9]{a^2b^2c^2(a\plus{}b)(b\plus{}c)(c\plus{}a)}\]

Can you solve it by only AM-GM?  :P

[quote="mbucko"]No. We can notice symmetry towards a,b,c and the fact that extremum value is when a=b=c=1.[/quote]

why no? do you have a proof that we can't prove it by AM-GM?

I don't know if I understood the question correctly, then. I thought he was asking about solutions other than the one that takes advantage from AM-GM.

Now, I see that the question is "by only AM-GM", so Mr Pain rinnegan might be asking about a solution that uses ONLY AM-GM (and nothing more).

No, I don't have a proof that we can't prove it by AM-GM.

[quote="mbucko"]

Now, I see that the question is "by only AM-GM", so Mr Pain rinnegan might be asking about a solution that uses ONLY AM-GM (and nothing more).
[/quote]

No, there is not necessary to prove it by only AM-GM. It was an additional task, because i proved it by using only AM-GM :lol:

[quote="mbucko"]No. We can notice symmetry towards a,b,c and the fact that extremum value is when a=b=c=1.[/quote]

You want to say that all symmetric inequalities reach max when a=b=c ?
Do you have proof for that?  :)

Not all. In order to learn more about the specific situation, please, turn this 3-var function into 1-var function by replacing $ b: \equal{} a$ and $ c: \equal{} a$. Then, deriviation to learn how function behaves. One can imagine how this 3-var function looks (in 3 Kartezian dimension and color).

I don't uderstand you... after replacement that you suggested, you will get:

$ P \equal{} (\sqrt [3]{\frac{2}{3}} \plus{} \sqrt[3]{2})\cdot3a$

Derivative is:

$ P' \equal{} (\sqrt [3]{\frac{2}{3}} \plus{} \sqrt[3]{2})\cdot3$ , and waht do we have with this? :?:

Pain, what is your solution? can you post it here.?

[hide="Maybe this could be interesting to"] $ x^3\plus{}y^3\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2)$[/hide]

Ah, Mbucko, that does not constitute a proof. Not all 3-variable symmetric inequalities have the equality case $ a\equal{}b\equal{}c$, furthermore, showing it to be true when $ a\equal{}b\equal{}c$ does not imply that it will be true in general.

For example, $ x(x\minus{}y)(x\minus{}z)\plus{}y(y\minus{}x)(y\minus{}z)\plus{}z(z\minus{}x)(z\minus{}y)\ge 0$ has a non-trivial equality case, $ (t,t,0), (t,0,t), (0,t,t), (t,t,t)$.

Furthermore, your method would only show it to be true when the 3 variables are equal in general, for example, you would argue that $ a^3\plus{}b^3\plus{}c^3 \le 3abc$ because it holds true when $ a\equal{}b\equal{}c$!

Please stop spamming in the inequalities forums with these non-proofs.

Cheers,

Rofler

Ah yes, I do like being called gentleman :D
Nonetheless, what you said still gave no motivation to actually solving the original problem.

Cheers,

Rofler


superior algebra
superior algebra unsolved

Suppose we have a tower of fields F < E < K.
I know that if K is a Galois extension over F then K is Galois over E and E is Galois over F.
(I believe a Galois extension is one that is both normal and separable)

The converse of the above although does not hold.  So does anyone know any examples of fields such that K is Galois over E and E is Galois over F but K is NOT Galois over F ?

Already the first statment is wrong:
Let $F=\mathbb{Q}$ and $K=F[\zeta_3, \sqrt[3]2 ]$ ($\zeta_3$ being a third root of unity) be the splitting field of $x^3-2$.
Then the subfield $E=F[\sqrt[3]2]$ is not galois over $F$. You can only prove that $K$ will be galois over $E$.

For the second take $E=F[\sqrt 2] , K=E[\sqrt[4]2]$.


quadratics
number theory
relatively prime
Diophantine equation
algebra unsolved
algebra

Show that there are infintely many pairs $(a,b)$ of relatively prime integers (not necessarily positive) such that both the equations \begin{eqnarray*} x^2 +ax +b &=& 0 \\ x^2 + 2ax + b &=& 0 \\ \end{eqnarray*} have integer roots.

Basically, we just want the discriminants of these equations to be perfect squares.

Hence, we need to show that there exist infinitely many coprime integers such that \[ a^2 - b\textrm{ and }a^2 - 4b  \] are perfect squares.

Now, consider the solutions $(x_n,y_n)$ (where $x_1 < x_2 < \cdots$) to \[ x^2 - 3y^2 = 1.  \] This is a Pell equation with smallest solution $(2,1)$. Hence, there exist infinitely many such solutions $(x_n,y_n)$. Now, take $a = 2y_n$ and $y_n^2 - 1$. Then we get \[ a^2 - b = 4y_n^2 - y_n^2 + 1 = 3y_n^2 + 1 = x_n^2,\ \ a^2 - 4b = 4y_n^2 - 4y_n^2 + 4 = 2^2.  \] So we only need to find infinitely many values of $a$ and $b$, chosen this way, which are coprime. But any divisor of $a$ and $b$ also divides $a^2 - 4b = 4$, so the only possible common divisor of $a$ and $b$ is 2. It is easy to see that $y_n$ will be even when $n$ is, so we take $n$ to be even, and we are done.

This is kind of a bad proof but can't we set a = 0 and b to a perfect square * -1 making both equations a difference of two squares?

No that can't be done coz the problem mentions that a & b are relatively prime but every integer divides 0....

[hide=Solution]
We need to find infinitely many solutions to the system
\[\begin{cases}
a^2 - b &= m^2 \\
a^2 - 4b &= n^2
\end{cases}\]
where \(a,b\) are relatively prime. Let us find infinitely many solutions to \[4n^2 - m^2 = 3a^2\]

We can take \(\gcd(m,n) = 1\) since we do not need to find all such solutions. Then
\[(2n - m)(2n + m) = 3a^2\]
and \(\gcd(2n-m, 2n + m) = 1\), so set
\begin{align*}
2n - m &= 3k^2 \\
2n + m &= l^2
\end{align*}

This leads to
\[(a,m,n) = \left(kl, \frac{l^2 - 3k^2}{2}, \frac{3k^2 + l^2}{4}\right)\]
and thus the pair
\[(a,b) = \left(kl, \frac{(3k^2 + l^2)^2 - 4(l^2 - 3k^2)^2}{48}\right)\]
works for infinitely many values of \(k,l\), and we are done.
[/hide]

even easier method is that the roots ( say e,f to equation 1 and g,h to equation 2) are infinitely many i.e prove the system ef = gh and e+f = 2(g+h) has infinitely many solutions , this is easy since the parameterization of xy = zw, for integers x,y,z,w is well known


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

let $ k \equal{} \mathbb{F}_p(t)$ and $ A \equal{} k[x,y]/(x^p \minus{} t)$. as a homework, I have to show that $ A$ is a regular ring (i.e. for every prime ideal $ m$ of $ A$, the localization $ A_m$ is regular local, i.e. the maximal ideal $ \overline{m}$ of $ A_m$ satisfies $ dim(A_m) \equal{} dim_{A_m / \overline{m}} (\overline{m}/\overline{m}^2)$). the problem is that I only have the definition of regular and no idea how to calculate both sides of the equation. it's only clear to me that $ A$ is a noetherian integral domain ;).

this would be cool :-). PIDs are always regular, since local PIDs are fields or DVRs.

but how do you know that $ A$ is a PID? (just a hint, please)

$ A$ is (isomorphic to) a polynomial ring in one indeterminate over a field. owk


A piece is placed in each unit square of an $ n \times n$ square on an infinite board of unit squares. A move consists of finding two adjacent pieces (in squares which have a common side) so that one of the pieces can jump over the other onto an empty square. The piece jumped over is removed. Moves are made until no further moves are possible. We look at the $ n \times n$ square plus the rows and the columns immediately to the left, right, above, and below it (so we're looking at an $ (n \plus{} 2) \times (n \plus{} 2)$ square). At least how many pieces go outside this square?

Anybody?

Actually I would like to see ANY kind of bound you can get... I'd be very interested.


let $ a$,$ b$ and $ c$ positives reals numbers such that  $ abc\equal{}1$

prove that  $ \frac{1}{\sqrt{b\plus{}\frac{1}{a}\plus{}\frac{1}{2}}}$+$ \frac{1}{\sqrt{c\plus{}\frac{1}{b}\plus{}\frac{1}{2}}}$+$ \frac{1}{\sqrt{a\plus{}\frac{1}{c}\plus{}\frac{1}{2}}}$ $ \geq \sqrt{2}$

[quote="greatestmaths"]let $ a$,$ b$ and $ c$ positives reals numbers such that  $ abc \equal{} 1$

prove that  $ \frac {1}{\sqrt {b \plus{} \frac {1}{a} \plus{} \frac {1}{2}}}$+$ \frac {1}{\sqrt {c \plus{} \frac {1}{b} \plus{} \frac {1}{2}}}$+$ \frac {1}{\sqrt {a \plus{} \frac {1}{c} \plus{} \frac {1}{2}}}$ $ \geq \sqrt {2}$[/quote]
$ \sum_{cyc}\frac{1}{\sqrt{2\left(b\plus{}\frac{1}{a}\right)\plus{}1}}\geq\sum_{cyc}\frac{1}{b\plus{}\frac{1}{a}\plus{}1}\equal{}1.$

[quote="greatestmaths"]if u have the link give it :)[/quote]
 [hide]http://www.mathlinks.ro/viewtopic.php?t=141304[/hide]


[quote="Carcul"][hide="Answer"]This is the Neber rearrangement, which proceeds via the intermediate formation of an azirine to give an alpha-amino ketone as the final product.[/hide][/quote]
could you explain it in detail using an example???

For example:

1) Butanone + NH2OH -----> Butanonoxime

2) Butanonoxime + TsCl/base -----> Butanonoxime tosylate

3) Butanonoxime tosylate + base/H2O -----> 1-aminobutanone


Hi all,

I was wondering: in mathematics courses (specifically graduate level classes), do the problems ever require the same type of clever/creative thinking as contests such as the AIME, USAMO, IMO, Putnam and other olympiads do? Will not being able to successfully do the contest level problems (in untimed settings) imply a lack of ability to pursue graduate level courses in mathematics and, eventually, research?

Any and all input is appreciated...I am very nervous and am wondering if I should even do research mathematics...

I'm sure that some level of mathematical creativity is required, but I doubt it will be as great as the amount required for contests.

I think that research mathematics can require a lot of creativity; you can make it as creative as you want.  In my opinion the greatest challenge in research mathematics is thinking of a good problem to solve - you are not given a specific problem as you are in contests.  Even proving a conjecture can require a deal of creativity.  Fermat's last theorem for example, took over 100 years to solve.  If it was an uncreative straightforward problem, it would have been solved instantly.
A more USAMO-like problem with actual importance is Fermat's Little Theorem.  Proving it requires creative math, and it is still very useful.
Some open problems are in recreational mathematics.  The solutions are probably fairly elementary, but require a great deal of creative thought.
If you would like to see some examples of creative research math try looking at things like Domino coverings of squares and Aztec Diamonds, Harary's generalized tic-tac-toe, and one of my favorites, the Collatz conjecture.  These will probably show you that research math is not necessarily dull (though it can be if you investigate problems which are dull to you).
Certainly, do not get discouraged if you cannot solve all contest problems.  I don't think any mathematician in the world would find all of them easy!

A USAMO or IMO or Putnam problem is known to have an answer. You have to be clever enough to find it, in half an hour or an hour and a half or some such severely limited time.

A research problem is [i]not[/i] known to have an answer. It might be false, it might be a practical dead end, it might be that you haven't asked the right question yet. The answer may be something you did not anticipate when you started asking the question. You may need to develop new nomenclature, new definitions, new notation - and then that new nomenclature may lead you to a problem you didn't know you were working on.

Half the battle - no, not half, it's more than that - lies in figuring out what to ask in the first place. [i]That [/i] is the key experience that contests don't necessarily provide - the art of asking your own questions.

And of course, you don't tie things up in a neat package in an hour and a half. There's no such time limit. You need time - mostly you need time to be wrong, because you're going to be wrong most of the time. But people who aren't quick (and because they're not quick, they weren't good on contests) can be very deep and successful research mathematicians.

Oh, there are some notorious problems in the world, such at the Goldbach Conjecture and the Collatz Conjecture, that I'm convinced exist mostly as vehicles for driving people insane. Don't stake your hopes on questions like those. As to what questions you should pursue - well, you're going to have to learn a lot more mathematics just to get there.

[quote="Kent Merryfield"]A USAMO or IMO or Putnam problem is known to have an answer. You have to be clever enough to find it, in half an hour or an hour and a half or some such severely limited time. [/quote]

A big problem, at least for me, is that I'm not "clever" enough to find the answers with arbitrary amounts of time for a lot of the problems. I know for the top contestants it's more of a speed thing, but my big issue is that I just can't do these kinds of questions, period.!

[quote]And of course, you don't tie things up in a neat package in an hour and a half. There's no such time limit. You need time - mostly you need time to be wrong, because you're going to be wrong most of the time. But people who aren't quick (and because they're not quick, they weren't good on contests) can be very deep and successful research mathematicians.[/quote]

This and the previous problem lead me to ask: do the contest problems (olympiad level) require a lot of creativity to solve in general or is this need greatly diminished when the time limit is removed? Personally, I would get the same score on a USAMO or IMO whether I had 9 hours, 9 days, or 9 weeks to work on the problem set...so if the creativity needed for research and/or grad course problem sets is higher (but just on different time scales), I think I'm going to be screwed...

[quote="Kent Merryfield"]Oh, there are some notorious problems in the world, such at the Goldbach Conjecture and the Collatz Conjecture, that I'm convinced exist mostly as vehicles for driving people insane. Don't stake your hopes on questions like those. As to what questions you should pursue - well, you're going to have to learn a lot more mathematics just to get there.[/quote]
I don't believe the two conjectures you have metioned are vehicles for driving people insane.  I believe they are both true and therefore have a solution. However, I do agree that they are very hard and one should not expect to solve them (though I bet no one thought that any one would solve Fermat's Last Theorem).  I think that even though you shouldn't expect to solve them, that there is nothing wrong with trying to solve them.  I spent some time trying to solve the Collatz conjecture, and although I didn't get very far with the actual conjecture, I did discover some interesting things about powers of 2 and 3.

Don't worry about problems; you can get better at problems.  The question you need to ask yourself is: do you love math?  If you do, then don't let your ability on contest problems stop you from pursuing a career.

Hmm, fair enough. I do love mathematics and have no plans of not pursuing it.

I am going to be taking graduate level courses from here on out, so does olympiad experience help for that, or are the skill sets for each largely orthogonal? Basically, is the level of creative thinking needed to do graduate problem sets in advanced subjects at the olympiad/putnam level or is it a fair bit lower (or higher?)


Nα βρεθουν τα ζευγη θετικων ακεραιων $ (a,b)$ για τα οποια ο αριθμος $ ab\plus{}a\plus{}b$ διαρει τον $ a^2\plus{}b^2\plus{}1$

Edit: Η λυση ειναι λανθασμενη :P

Πολύ ωραίος Νίκο!

Θα μπορούσες επίσης να πάρεις το ζεύγος με b ελάχιστο και να δείξεις ότι
$ Kb\plus{}K\minus{}a<b$ (πώς?), οπότε πάλι τελείωσες :)

Εξαιρετικά χρήσιμη μέθοδος για τέτοου είδους ασκήσεις!

Νικο στο ειπα οτι θα βαλω ασκηση και το εκανα  :wink: . Και εγω περιπου με vietta jumbing την εχω λυσει ομως προσεξε οτι αποτελει λυση και το ζευγος $ (1,2^2)$ ή και το $ (2^2,3^2)$

Λοιπον, η παραπανω λυση ειναι λαθος λογο ενος λαθους βιασυνης στις ανισοτικες σχεσεις που χρησημοποιω

Βαζω λοιπον μια σωστη λυση η οποια ειναι για $ a,b:$ μη αρνητικους ακεραιους:

Εχουμε $ \frac {a^2 \plus{} b^2 \plus{} 1}{ab \plus{} a \plus{} b} \equal{} K \in Z$ (1)
Σταθεροποιουμε τον θετικο ακεραιο $ K$ και θεωρουμε τα ζευγη $ (a, b)$ που επαληθευουν την (1), και λογο συμμετριας μπορουμε χωρις βλαβη της γενικοτητας να πουμε οτι $ a \geq b$ για ολα τα παραπανω ζευγη. Θεωρουμε τωρα το ζευγος αυτο με το ελαχιστο $ a \plus{} b$, και εστω οτι αυτο ειναι το $ (a_0, b_0)$
Η εξισωση (1) γραφεται: $ a^2 \minus{} aK(b \plus{} 1) \plus{} (b^2 \minus{} Kb \plus{} 1) \equal{} 0$ η οποια απο τους τυπους Vieta εκτος απο το ζευγος $ (a_0, b_0)$ εχει λυση και το ζευγος $ (a_1, b_0) \equal{} (Kb_0 \plus{} K \minus{} a_0, b_0)$
Ομως απο τις υποθεσεις μας εχουμε:
$ a_0 \plus{} b_0 \leq a_1 \plus{} b_0 \Rightarrow a_1 \geq a_0 \Rightarrow (b_0 \plus{} 1)K \geq 2a_0 \Rightarrow {b_0}^3 \plus{} {b_0}^2 \plus{} b_0 \plus{} 1 \geq {a_0}^2b_0 \plus{} {a_0}^2 \plus{} 2a_0b_0 \geq {b_0}^3 \plus{} {b_0}^2 \plus{} 2{b_0}^2 \Rightarrow b_0 \plus{} 1 \geq 2{b_0}^2$.
Επηδη ομως για $ b_0 > 1$ η τελευταια ανισοτητα δεν ισχυει, εχουμε $ b_0 \equal{} 0$ η $ b_0 \equal{} 1$

Για $ b_0 \equal{} 1$ εχουμε: $ {b_0}^3 \plus{} {b_0}^2 \plus{} b_0 \plus{} 1 \geq {a_0}^2b_0 \plus{} {a_0}^2 \plus{} 2a_0b_0 \Rightarrow {a_0}^2 \plus{} a_0 \leq 2 \Rightarrow a_0 \in \{0, 1\} \Rightarrow a_0 \equal{} 1$, αφου πρεπει $ a_0 \geq b_0$. Για $ (a, b) \equal{} (1, 1)$ προκειπτει $ K \equal{} 1$

Για $ b_0 \equal{} 0$ εχουμε: $ {b_0}^3 \plus{} {b_0}^2 \plus{} b_0 \plus{} 1 \geq {a_0}^2b_0 \plus{} {a_0}^2 \plus{} 2a_0b_0 \Rightarrow {a_0}^2 \leq 1 \Rightarrow a_0 \equal{} 1$, αφου το ζευγος $ (0, 0)$ απορριπτεται. Για $ (a, b) \equal{} (1, 0)$ προκειπτει: $ K \equal{} 2$

Αρα εχουμε να λυσουμε την εξισωση $ a^2 \plus{} b^2 \plus{} 1 \equal{} K(ab \plus{} a \plus{} b)$ για $ K \equal{} 1$ και $ K \equal{} 2$.
Για $ K \equal{} 1$ η εξισωση ισοδυναμα γραφεται:
$ (a \minus{} b)^2 \plus{} (a \minus{} 1)^2 \plus{} (b \minus{} 1)^2 \equal{} 0$, που εχει λυση μονο την $ (a, b) \equal{} (1, 1)$
Για $ K \equal{} 2$, παιρνουμε τη διακρινουσα της 2βαθμιας εξισωσης που προκειπτει ως προς $ a$ (λογο συμμετριας) και βρησκουμε την εξης απειρια ακεραιων λυσεων:
$ (a, b) \equal{} (k^2, (k \plus{} 1)^2)$, για καθε $ k \in Z^ \plus{}$ (και συμμετρικα).

Ετσι πολυ ωραια Νικο. Εχω παρομοια λυση.


A photographer took some pictures at a party with $10$ people. Each of the $45$ possible pairs of people appears together on exactly one photo, and each photo depicts two or three people. What is the smallest possible number of photos taken?


geometry
incenter
geometric transformation
reflection
geometry proposed

Dear everyone.

For a triangle $ ABC$, let $ M$, $ N$ be midpoints of $ AB$, $ AC$ respectively. And let $ CH$ be $ C$-altitude of the triangle ($ H$ on $ AB$). The incircle of the triangle $ ABC$ touches $ BC$, $ CA$, $ AB$ at $ D$, $ E$, $ F$ respectively.
Then the incenter of a triangle $ HMN$ is on $ EF$.

Thanks for reading my post.

Let $ A'$ be a reflection of $ A$ in $ CH,$ let $ (I), (I')$ be incircles of $ \triangle ABC, \triangle A'BC$ with common side $ a \equal{} (BC)$ and semiperimeters $ s, s'.$ Let $ (I')$ touch $ AB$ at $ F'$ and let parallel to $ EF$ through $ I'$ cut $ AB$ at $ P.$ $ \triangle A'BC \sim \triangle HMN$ are centrally similar with similarity center $ A$ and coefficient 2. Therefore, it is sufficient to show $ AP \equal{} 2AF.$ $ \triangle CAA'$ is isosceles with $ CA \equal{} CA'$ $ \Longrightarrow$ $ s' \equal{} s \minus{} AH,$ $ \angle CA'B \equal{} 180^\circ \minus{} \angle CAB$ and $ \angle A'PI' \equal{} \angle AFE \equal{} 90^\circ \minus{} \frac {\angle CAB}{2} \equal{} \angle BA'I' \equal{} \angle PA'I'$ $ \Longrightarrow$ $ \triangle I'A'P$ is isosceles with $ I'A' \equal{} I'P$ and $ A'F' \equal{} F'P.$ Thus $ AP \equal{} AA' \plus{} A'P \equal{} 2 AH \plus{} 2 A'F' \equal{} 2 AH \plus{} 2 (s' \minus{} a) \equal{} 2(s \minus{} a) \equal{} 2 AF.$

As the following diagram shows, you also need the condition that $ BC > AC$:
[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/cb57eb144c9833a1fb57dddd5e64fe48.png[/img]

Thanks. If $ BC < AC,$ then the proposition holds for the excircle $ (I)$ of $ \triangle ABC$ opposite to $ A$ tangent to $ CA, AB$ at $ E, F$ and the excenter $ J$ of the $ \triangle HMN$ opposite to $ M$: $ J \in EF.$

Sorry for late thanks  :)
When $ BC<AC$, it doesn't hold  :blush:
I have a little different solutions, but they are not beatiful than yetti's.

Slightly more general: Let $ ABC$ be a given triangle and let $ B'$, $ C'$ be the orthogonal projections of $ B$, $ C$ on $ CA$, and $ AB$, respectively. Denote by $ N$, $ P$ the midpoints of the sides $ CA$ and $ AB$, and let $ X$ be a point moving on the segment $ B'C'$. Then the incenter of $ XNP$ moves along the line $ EF$.


Define $ (x\Delta y)$ to mean $ 2x\minus{}3y$. Evaluate $ ((4\Delta 3)\Delta(5\Delta 3))$.

this problem can be broken down into three steps
the first part, the second part, and then the final $ \Delta$ in the middle

let's solve the first part of the question first
$ 4\Delta 3 \equal{} 2(4) \minus{} 3(3) \equal{} 8\minus{}9 \equal{} \minus{}1$
now we'll solve the second part
$ 5\Delta 3 \equal{} 2(5) \minus{} 3(3) \equal{} 10\minus{}9 \equal{} 1$

now we'll do the final step
$ \minus{}1\Delta 1 \equal{} 2(\minus{}1) \minus{} 3(1) \equal{} \minus{}2 \minus{} 3 \equal{} \minus{}5$

our answer is -5

if anyone sees a mistake, plz let me know


Is this from some AIME ..it is really common....and boring to do...

Let the equations be 1,2,3,4

[hide]$1-2: z-y=1$

$3-4: w-x=3$

$2-3: w-y=2$

$1-4: z-x=2$

$2-4: w-z=1$

And now find all in terms of one vaiable or do whatever you want to [/hide]
I hope i did not mess up the equations...and : means implies,not ratio.

Can we solve that by matrix? I haven't really studied matrix :blush: , but I saw chess64 once solve something similar by matrix.

I believe you could do the same thing that shady did by manipulating augmented matrices; but given the particular equations you would probably end up writing more.

Nice job, this is what I have, but you could also evaluate this using determinants or matrices.  Do not use determinants becasue it'll take you a long time!

Masoud Zargar

[quote="danthemanvt"]We could also just add the equations to get

5x+5y+5z+5w=70
x+y+z+w=14

THen since

x<y<z<w

we see that x=2  y=3  z=4  w=5[/quote]

You guys are really umm... well ... unimaginative. :D danthemanvt demonstrates the elegant, simple solution! ;)

[quote="Arne"][quote="danthemanvt"]Then since

x<y<z<w

we see that x=2  y=3  z=4  w=5[/quote]

We don't need x < y < z < w...[/quote]

I agree.

And to add what danthemanvt did, we can simply solve this equation by seeing that these equations are "symmetric." We can simply add all four and divide four to get $x+y+z+w$ and subtract this from other four to get the desired results.

[hide]We add the equations to find that $w+x+y+z=14$. Doing some quick subtration, we can find the answers. [/hide]


Find all positive intergers number $ x,y,z \in{ Z_{+}}$ such that: $ 4^{x}+4^{y}+4^{z}$ is a perfect square number.

All the integers can be written in this form:
$ {(i+2k,j+2k,2k)}\forall{k\in{Z_{+}}}$
and $ (i,j)\in\left\{{(1,1),(3,2),(5+2t,3+t)\right\}\forall{t\in{Z_{+}}}}$

Thanks a lot. Can you post your solution?

Wlog, let $ z \le y \le x$. Then plainly $ 4^{x}+4^{y}+4^{z}= t^{2}$, for some $ t \in \mathbb{N}$, iff $ 1+2^{2(y-z)}+2^{2(x-z)}= (t \cdot 2^{-z})^{2}$, and we are left with sieving for any $ (a,b,c)\in\mathbb{N}_{\;\!0}^{3}$ s.t. (1): $ 1+2^{2a}+2^{2b}= c^{2}$. Here as usual, $ \mathbb{N}= \{1, 2, \ldots\}$ and $ \mathbb{N}_{\;\!0}= \mathbb{N}\cup \{0\}$. We lose no generality taking $ a \le b$. Since then $ (2^{b}+1)^{2}= 2^{2b}+2^{b+1}+1$, necessarily $ b+1 \le 2a$. In particular, if $ b+1 = 2a$, then $ b = 2n+1$, for some $ n\in\mathbb{N}_{\;\!0}$, and $ a = (b+1)/2 = n+1$. Actually, note that $ (a,b) = (n+1, 2n+1)$ is a solution of (1). Thus $ (x,y,z) = (m, m+n+1, m+2n+1)$ and any of its permutations are solutions of the original problem, for arbitrary $ m, n \in \mathbb{N}_{\;\!0}$. Indeed, we show there are no others apart from the former ones.

Indeed, by way of contradiction, suppose $ b+1 < 2a$. Then $ \frac{1}{2}(b+1) < a < b$ and consequently $ b-a < a-1$. Set $ c = 2^{p}q+1$, where $ p \in \mathbb{N}$ e $ q \in 2\mathbb{N}_{\;\!0}\!+1$ (btw, note this is consistent). Then $ 1+2^{2a}+2^{2b}= c^{2}$ iff $ 2^{2p}q^{2}+2^{p+1}q+1 = 2^{2b}+2^{2a}+1$, which is impossibile for $ p > 1$ (recall that any integer $ \ne 0$ can be uniquely represented by a sum of distinct powers of $ 2$). Therefore $ c-1 = 2\alpha$ e $ c+1 = 2^{2a-1}\beta$, for $ \alpha, \beta \in 2\mathbb{N}_{\;\!0}\!+1$, and $ \alpha\beta = 2^{2k}+1$, provided that $ k = b-a$. So $ 2^{2a-2}\beta = \alpha+1$ and $ 2^{2a-2}\beta^{2}= 2^{2k}+\beta+1$. But $ \beta \ge 3$, then $ 2^{2k}+\beta+1 = 2^{2a-2}\beta^{2}> 2^{2a-2}(3\beta+1) > 2^{2a}+\beta+1 \ge 2^{2k}+\beta+1$ (absurd). The claim follows.

but (1,3,4) is not a solution

Solving $ 2^{X}+2^{Y}+2^{Z}=c^{2}$ whit $ X=2x,Y=2y,Z=2z$ we will get the result.

Dividing by 4 many times this is equivalent to $ 2^{i'}+2^{j'}+1={c'}^{2}$  or  $ 2^{i'}+2^{j'}+2={c'}^{2}$

The second one has no solutions infact $ 2^{{i'}-1}+2^{{j'}-1}+1=\frac{{c'}^{2}}{2}$ the first member must be odd while the second even.

So all the solutions can be written in this way: $ ({i'}+2k,{j'}+2k,2k)$ for some $ {i'},{j'}$

If   $ {c'}^{2}\leq 529$   the only solutions are $ ({i'},{j'})=(2,2)$   or   $ ({i'},{j'})=(6,4)$
Note If $ {c'}^{2}>529$ the couples        $ ({i'},{j'})=(10+4t,6+2t)$ $ \forall{t}\in{Z_{+}}$ are solutions.In this case there aren't other solutions;infact suppose $ (x,a)$ and $ ({x'},a)$(fixed a) solutions then $ 4^{x}+4^{a}+1=k^{2}$ and $ 4^{x'}+4^{a}+1={k'}^{2}\Rightarrow (2^{x}-2^{x'})(2^{x}+2^{x'})=(k-{k'})(k+{k'})$ which has no different solutions.
Dividing all the coples by two we get the result:
$ (x,y,z)=(i+k,j+k,k)$ and $ (i,j)\in\left\{(1,1),(3,2),(5+2t,3+t)\right\}$ $ \forall{t}\in{Z_{+}}$

:blush: s.tringali I'm sorry for my mistake... anyway It's also in my solution
Your solution is very good.

I know that $ (a,x\plus{}a,2x\plus{}a\minus{}1)$ which $ a,x\in N$ are some of the solutions. :blush:


geometry
geometric transformation
analytic geometry
symmetry
reflection
rotation

The graph of  $ y \equal{} f(x)$ is plotted in the coordinate plane. Find the equation of the curve when this graph is reflected as follows. Justify your answer.

(1) When this graph is reflected in the $ x$ axis.

(2) When this graph is reflected in the $ y$ axis.

(3) When this graph is reflected in the line $ y \equal{} x$.

(4) When this graph is reflected in the line $ y \equal{} \minus{} x$.

(5) When this graph is reflected through the point $ (p,\ q)$.

(1) When this graph is reflected in the $ x$ axis. $ \minus{}f(x)$

(2) When this graph is reflected in the $ y$ axis. $ f(\minus{}x)$

(3) When this graph is reflected in the line $ y \equal{} x$. $ f^{\minus{}1}(x)$

(4) When this graph is reflected in the line $ y \equal{} \minus{} x$. $ \minus{}f^{\minus{}1}(\minus{}x)$

(5) When this graph is reflected in the point $ (p,\ q)$.
... not sure what "is reflected in the point" means...

(5) I don't know how should I state the context. :(

In the plane, "reflection through a point" is the same as rotation by 180 degrees around that point.  It has that name for the following reason: to reflect something in a line, you draw the path joining the thing to the line, then extend that path through the line.  To reflect something through a point, you do precisely the same thing.  Reflection through the origin is the map $ (x, y) \mapsto (\minus{}x, \minus{}y)$, or $ (x, y, z) \mapsto (\minus{}x, \minus{}y, \minus{}z)$ in three dimensions.

[quote="JBL"][quote="Flame"]Maybe you mean central symmetry?[quote="mathwizarddude"]... not sure what "is reflected in the point" means...[/quote][/quote]In the plane, "reflection through a point" is the same as rotation by 180 degrees around that point.  It has that name for the following reason: to reflect something in a line, you draw the path joining the thing to the line, then extend that path through the line.  To reflect something through a point, you do precisely the same thing.  Reflection through the origin is the map $ (x, y) \mapsto ( \minus{} x, \minus{} y)$, or $ (x, y, z) \mapsto ( \minus{} x, \minus{} y, \minus{} z)$ in three dimensions.[/quote]

Yes, this is what I want to say, thank you!

kunny

(5) . $ 2q\minus{}f(2p\minus{}x)$

First reflect about (p,0) and then (0,q).

(I think this is correct)

You can translate the graph to (0, 0), flip it, and translate it back to get y = 2q - f(2p - x).


let a,b,c,d- positive reals

\[\begin{array}{l}2(ab+ac+bc+ad+bd+cd)+abc+cbd+cda+abd = 16 \\ proof \\ a+b+c+d \ge \frac{{2(ab+bc+cd+ad+ac+bd)}}{3}\\ \end{array}\]

Same idea http://www.mathlinks.ro/Forum/viewtopic.php?t=85836 . In that topic, you'll see general problem .


Use this property:
$a^{log_{a}x}=x$ in order to obtain the expresion of $f(x)$ and $g(x)$


conics
ellipse
trigonometry
calculus
calculus computations

[quote="RaduSorici"][i]P[/i] is a point on an ellipse and [i]l[/i] is a line through [i]P[/i] perpendicular to the tangent to the ellipse at [i]P[/i].  [i]F[/i] and [i]G[/i] are the foci of the ellipse, and [i]E[/i] and [i]H[/i] are the points on line [i]l[/i] such that [i]EF[/i] and [i]GH[/i] are perpendicular to [i]l[/i]. Prove that [i]PE*PH[/i] is constant.[/quote]

[i]l[/i] is the normal to the ellipse at P  .
Let the equation to the ellipse is
$\dfrac{{x^2 }}{{a^2 }} + \dfrac{{y^2 }}{{b^2 }} = 1\,$ and the point P is
$\,P(a\cos \theta ,\,b\sin \theta )$, If e is the eccentricity of the ellipse, F and G can be taken as $(ae, 0), (-ae, 0)$ respectively
EF and GH are parallel to tangent at P

Equations to the lines [i]EF[/i] and [i]GH[/i] are
$\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = e\cos \theta$ and

$\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = - e\cos \theta$

Now PE*PH is nothing but the product of perpendiculars from P on to these lines.
$\Rightarrow \left| {\dfrac{{\cos ^2 \theta + \sin ^2 \theta - e\cos \theta }}{{\sqrt{\dfrac{{\cos ^2 \theta }}{{a^2 }} + \dfrac{{\sin ^2 \theta }}{{b^2 }}} }}} \right|\left| {\dfrac{{\cos ^2 \theta + \sin ^2 \theta + e\cos \theta }}{{\sqrt{\dfrac{{\cos ^2 \theta }}{{a^2 }} + \dfrac{{\sin ^2 \theta }}{{b^2 }}} }}} \right|$

which can be simplified as $b^2$.


function
geometry
geometric transformation
limit
real analysis
real analysis unsolved

Please answer the following questions for me:

1) Why the Fourier transform decomposes a fuction into a continuous sum of characters (eigenfunctions for translations) ?
2) Let $ g$ be an uniformly continuous function on $ \mathbb R$. Is there a function $ f$ in $ L^1(\mathbb R)$ such that $ \widehat{f}\equal{}g$ a.e. in $ \mathbb R$.

A counterexample for the second: $ g(x)\equal{}1$.

Thank you, but if we assume that $ \lim\limits_{x\to\infty}g(x)\equal{}0$ and $ g$ is uniformly continuous, does there exist such $ f$?

Just as a hint: the answer to this question is no.


[quote="peregrinefalcon88"]can anyone tell me anything about the russian shar?

http://www.mathlinks.ro/Forum/resources.php?c=143&cid=149

i found this on mathlinks and i want to jion it next year but i know nothing about it.[/quote]

I suggest you contact [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=45777]Doctor A[/url] and tell us your findings!  :)

The Sharygin Olympiad is taking place as we speak (I think today was the second day). Usually, in January or February, Alexey Zaslavsky or Alexey Myakishev, the main organizers of the olympiad, puts on a PDF on the [url=http://tech.groups.yahoo.com/group/Hyacinthos/]Hyacinthos Forum[/url] with some problems you must solve (and send them via email) in order to be invited for the final round of Sharygin, which is held in Dubna (a city very close to Moscow), Russia.

The number of problems you have to solve in order to get accepted isn't fixed (varies from 5 to 15) and you have a lot of time to do them (I think 1 May or 15 April is the deadline). Anyway, don't worry, I don't think they reject people too much.
Here's how the first round of this year's olympiad looked like: [url]http://geometry.ru/olimp/sharygin/2009/zaochn-e.pdf[/url].

If no one puts the link to this correspondence round on MathLinks by February, I shall put it myself here, in this section.

Indeed, Doctor A (Arseniy Akopyan), an organizer himself, can provide you more details about the contest.

so all i have to do is solve the problems and send in the solutions and i have a chance at making the next round? dont have to sign up or anything?

FYI Im from Virginia in the U.S.

Yes. There were students from UK, Romania... Last year, so yes (I also thought before it's just for Russia :( )

[quote="peregrinefalcon88"]so all i have to do is solve the problems and send in the solutions and i have a chance at making the next round? dont have to sign up or anything?[/quote]
You practically sign in when you submit your solutions (there will be an email adress in the file with the problems).

Hello,

I am very interested in participating in this competition each year right up until i finish high school. But i dont really understand how the Russian school system works. Basically can you participate in the competition right up to and during the last year of high school? thankyou

[quote="pohoatza"]
Indeed, Doctor A (Arseniy Akopyan), an organizer himself, can provide you more details about the contest.[/quote]

lol...I was wondering if that was really him)

I heard you took part in it, but you have to know some russian in order to do so, don't you?


limit
real analysis
real analysis unsolved

Let [tex]f:\mathbb{R}\to\mathbb{R}[/tex] continuous and derivable. Prove that if [tex]\displaysyle\lim_{x\to\infty}((f'(x))^2+f^3(x))=0[/tex], then [tex]\displaysyle\lim_{x\to\infty}f(x)=0[/tex].

Firstly let's prove that there exists $\lim_{x\to\infty}f(x)$.Suppose on contrary $\alpha=\liminf_{x\to\infty}f(x)<\beta=\limsup_{x\to\infty}f(x)$,thus there is the sequence $(x_n)_n$,$x_n \to \infty$ such that $f'(x_n)=0$ and $f(x_{2n-1})\to \alpha$, $f(x_{2n})\to\beta$. Putting this results into problem condition we get that $\alpha^3=\beta^3$, which means that $\alpha=\beta$, which contradicts our supposion.So,we've proved that there exists $\alpha=\lim_{x\to\infty}f(x)$, it is not difficult to verify that in thios case $\lim_{x\to\infty}f'(x)=0$, so we get that $\alpha=0$.


geometry
3D geometry
ratio
perimeter
analytic geometry
absolute value
Pythagorean Theorem

please give full solutions not just answers:

1.what is the area of a circle defined by the equation:
(x-3)^2+(y+4)^2=12.1

2.In terms of pi, what is the area of an inscribed circle in a right triangle with sides 3, 4, and 5.

3.What is the area of the figure defined by the points (1,5) (1,2) (6,3) and (4,0)

4. A circle of radius 36 is divided into 12 equal sectors. One sector is
folded in such a way that it becomes a cone. What is the volume of
the cone? Express your answer as a common fraction in terms of pi.

5.A circle of radius 36 has a square inscribed in it. Another circle is
inscribed in the square which has a square also inscribed in it. This
pattern continues infinitely. What is the sum of the areas of all the
circles? Express your answer in terms of .

6.On Monday the number of absent students from a class was 1/13 of
those present. On Tuesday the number of absent students decreased
by one and became 1/20 of those present that day. How many students
are in the class?

7.What is the largest value of x such that 3^x divides 100! ?

8. A cone with radius 6 and height 3 is cut out of a truncated cone with
radii 6 and 8 and height 4. What is the volume of the resulting figure?
Express your answer as a common fraction in terms of pi.

9. Mickey started to read a book on the 1st of March. Every day he read
the same number of pages and finished the book on the 31st of March.
If on the first day he had read ¼ of his usual number of pages, but
had increased the number of pages read by one page each day, he
would still have completed the book. How many pages are in the book?

10. A car traveling at a constant rate covers the distance from A to B in 1
hour. It left A for B at the same time as a walker left B for A walking
at a constant rate. When the car met the walker it gave him a lift to A.
Then the car started again for B and arrived there in 2 hours and 40
minutes after its first departure from A. In how many hours would the
walker cover the distance from A to B?

These are the same as some of the questions in the haytest but i need help with these. Help would be appreciated!

1.  If you don't want to spoil the Haytest for yourself, I suggest you avoid this thread until you have taken the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31345]Haytest.[/url]

2.  Although this is a request for help, I think it would be best if you hide your solutions and your answers.

I'll help a little.
#10
[hide]Since the car arrived at destination $B$ two hours and forty minutes after its initial departure, it spent an hour and forty minutes in getting to the walker and taking him back to $A$. That would mean that the car took $\frac {100 min}{2} = 50$ minutes in getting to the walker. The total amount of time it takes for the car to get to $B$ is sixty minutes, so the ratio of their speeds must be $1:5$. The walker must take $5$ hours.[/hide]

[hide]
1. The equation for a circle is (x-a)^2+(y-b)^2=r^2.  Since r^2=12.1 the area of the circle is [b]12.1 :pi:  or  37.994.
[/b]
2.  The area of a triangle with a circle inscibed in it is Semi-perimeter * Radius.  We already know the area is 6 and the semi-perimeter is 6 so the radius must be one, so the answer is [b]:pi:[/b]

3. For this question I used the "Shoelace" method which is kinda hard for me to show.  But it should look like this
In absolute value bars:
1  5
1  2

6  3
4  0
1  5
You draw diagonals from each number to the one down and to the right/left of it.  So for example it's 1*2.  Now add all the products on one side to the other making (2+3+0+20) and (5+12+12+0) then subtract the two to make -4.  Divide it by to to make -2 and then absolute value it (remember it's in absolute value bars) to make [b]2[/b].

4. If you draw the picture you find the slant height length is 36, because the radius becomes the slant height.  The circumference of the circle is 72:pi: so a twelfth of it is 6:pi:.  So 6:pi: is now the circumference of the base of the cone so the diameter of the cone is 6 and the radius is 3.  So now we need to find the height.  We can use pythagorean theorem so 36^2-3^2= 1287 which is 3:sqrt: :143.  So 3:sqrt: 143 is the height so now we find the area.  ( 3:sqrt: 143 *9:pi: )/3= :sqrt: 143 *9:pi:= [b] 9:pi::sqrt: :143 or 337.94 [/b]

5.  This uses the equation for the sum of infinite geometric sequence which is a/1-r, where a is the first number, and r is the ratio between numbers.  We can find r by testing some numbers.  The radius of the first circle will be 36.  The second will be 18:sqrt:2.  so 18:sqrt: 2/36= :sqrt: 2/2.  But because they want the ratio for areas you sqare it to get 1/2.  a is just 36^2 * :pi: so that;s [b] 36^2 :pi:/ 1/2 or 36^2 * :pi: *2 or 2592 :pi:. [/b]

6.  We need to assign variables to the people who are not here are here.  Let's say x is people present.  So the peoplen not present is x/13.  So total amount of people is 14x/13.  Now, one person is here, so the absent goes down one, but the present goes up one so that makes x/13-1=1/20(x+1).  Solve for X so get 39.  So in the very beginning there were 39 people present.  1/13 of that is 3 so there are [b]42 people present[/b], or you could plug it in to 14x/13 to get 3*14=[b]42[/b]

7.  You want to find the number of 3s in 100! so it's 100/3=33, 100/9=11, 100/27=3, 100/81=1.  So 33+11+3+1=[b] 48[/b]

8. We first find the volume of the original cone which is 256:pi:/3.  Then we subtract the smaller cone's volume which is 36 :pi: .  So (256-108):pi:/3= [b] 148 :pi: /3 [/b]

9.  I don't know what they're asking so I can't do the problem.

10.  If the car can travel the whole distance in 1 hour, we know that the time it took the car to get from A to B after dropping off the hitch-hiker is 1 hour and 40 minutes.  Or 100 minutes.  So since the car goes only a certain distance twice, you know the time it takes the car to travel the each distance is 50 minutes.  Since he can go the whole way in 60 minutes, the car travels 5/6 of the distance and the hitch-hiker travels 1/6 distance.  It takes 50 minutes for the hiker to go 1/6 because he has to go the same amount of time as the car.  So 6*50 to cover the whole distance is 300 minutes.  Divide by 60 to find hours and you get [b] 5  hours [/b]
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Secret asian, some of your answers are not in the correct forms.
It's not like I know any of these questions though.

[hide]For number 3, the figure drawn with the coordinates given can be broken into two triangles, and finding the area of them shouldn't be hard as you can subtract some the coordinates obtain the heights, bases, etc.
The area, according to me, comes out to be 9.[/hide]

I'll help with #7:

[hide] There are 33 multiples of 3^1 (100/3 rounded down is 33)
There are 11 multiples of 3^2.
There are 3 multiples of 3^3.
There is 1 multiple of 3^4.
Adding these up gives n=48 (where n is the greatest value such that 3^n divides 100!). [/hide]

He forgot to say that If he would have still finished by the 31st.
#9
[hide]Let's say he usually reads $x$ number of pages each day. So there are $31x$ pages in the book, but if he had read $\frac 14 x$ on the first day and increased the previous day's pages read by $1$ for the next thirty days, he still would have completed the book. So the equation is $31x = \frac {31}4 x + 465$, solving $x = 20$. So the total number of pages is $31 * 20 = 620$.[/hide]

#1 is extremely easy, you should cover that in Algebra I.  [hide]The equation of a circle is $(a-x)^2 + (b-y)^2 = r^2$

$r$ is the radius.  So you should be able to go from there:  if $r^2$ = 12.1, the area is just $pi * r^2$[/hide]


Solve the equation:
$1.\quad\frac{1+\sin x+\sin^{2}x+\sin^{3}x+\cdots+\sin^{n}x+\cdots}{1-\sin x+\sin^{2}x-\sin^{3}x+\cdots+(-1)^{n}\sin^{n}x+\cdots}=\frac{4}{1+\tan^{2}x},\, \text{with}\; x\neq\frac{\pi}{2}+k\pi,\;k\in\mathbb{Z}$
$2.\quad\sin x+2\sin{2x}=3+\sin{3x}$

[hide="obvious hint"]
geometric series[/hide]

Could you show me the complete solution?
Thank you and $\mathbb{HAPPY\,NEW\,YEAR}$ :D

partial solution

[hide] using sum of infinite geometric series, we find that

(1 + sinx)/(1-sinx) = 4(1-sin^2 x)
letting y = sin x, we obtain the cubic :o  equation 4y^3-4y^2-5y+3 = 0.
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Yes, it's very easy and you are right. Can you solve the part $2.$ (more easily :?: ) :D