I've branched off this topic for discussion of the qualifying scores.  The announcement is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133657]here[/url].

[quote="usaha"]wow, i feel bad for someone in my class.  He got a 115.5 on the amc 10a, and he wouldve got 117, but thinking he needed 120 to pass, he guessed on the last quetsion and he guessed wrong.  So if he didnt guess, he wouldve had 117... but hes probalby gonna pass the amc 12B[/quote]This is why that I feel that strategies that involve answering a certain pre-determined number of questions are not wise.  Answer the questions that you can solve (or are pretty sure that you have solved), and leave blank the ones that you don't know (or guess if you can eliminate at least two wrong answers).

It isn't everything, but it is the biggest high school competition there is.  I think it is like Mathcounts is to middle school.

[quote="bpms"][quote="Karth"]Hmm... has this ever happened before?[/quote]
Probably, consider the sheer number of takers.
Also, in past the qualifying score was an 84 on the 12.
$6x+2.5(25-x) \geq 84$
$3.5x \geq 21.5$
$x \geq 7$
7 correct answers and you can take the AIME...

When are the statistics going to be up?[/quote]
I doubt anyone answered only 7 questions correctly and then left the rest of it blank.

Statistics are probably not going to be up until at least (around) the time schools get their school reports.

[quote="white_horse_king88"]That really defeats the purpose of the AMC. Only 7?  :rotfl:

Anyway, some people's predictions were right on the money. A few points lower than the typical 100/120 index - but not significantly so. That means there are people out there that are now more encouraged to try more problems and that some of those are actually getting the ones they are reaching for correctly. So the new points system is good, I think.[/quote]
The first ones must've been hard for 84 to be top 5% percent.

The one on which the cutoff was 84 (either 2000 or 2001, I don't remember which) was absurdly difficult, I'm told (though I was a couple of years too young to have taken it myself).  Also, the scoring at the time only gave 2 points for a blank, which would mean you need 9 right to qualify.  And it may not have even been that easy to get 9.  I'll have to try to dig that test up sometime and see if it's that hard.

As for this year's, I can't really say too much until I take the tests.  But not having seen the questions, I was guessing the 12 cutoff would be higher (i.e. 100) and the 10 cutoff slightly lower.

Edit: The 2001 cutoff was 84.  I took the AMC10 that year, so it didn't affect me.
http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2001-12a/01-1012stats.html

[quote="Kent Merryfield"]Where did 84 come from? The AMC-12 (as opposed to the AHSME) only goes back to the year 2000. The lowest qualifying index I can remember was a 94. I think that was in the year 2000. And the scoring rules in 2000 weren't 0-2.5-6, they were 0-2-6. The 0-2.5-6 scale was introduced in 2002. The minimum number of correct questions to score 94 in a 0-2-6 system is 11.

For 1999 and earlier the test was the AHSME. In the 1990's, it was a 30-question test, and was scored on a 0-2-5 scale.[/quote]

I'm guessing this:

http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2001-12a/01-1012stats.html

And yeah, it was 0-2-6 for a year or two, though test takers really don't know of that since it was like 6 years ago.

[quote="now a ranger"]It's too late to feel bad. I don't feel any pressure even though I missed the cut by 1.5/3 points on the A. Since I have a state math tournament a week after, I can have other things to do. AMC/AIME/USAMO isn't EVERYTHING.[/quote]
you are completely right man, we need to forget the past and focus on the present and future


geometry
circumcircle
geometric transformation
reflection
incenter
trigonometry

Dear Livetolove,
you have rediscovered the P-Hagge's circle.
Nice result for you
Sincerely
Jean-Louis

Dear [b]Jayme[/b],
Thanks you for the name of this circle. I don't know it was [b]P-Hagge[/b]'s circle. But what about this problem?
Let $ A_3, B_3,C_3$ be the reflections of $ A_1,B_1,C_1$ across the midpoints of $ BC, CA, AB$, respectively. Prove that $ H,A_3,B_3,C_3$ are concylic.

[hide]Two problems are equivalent.[/hide]

Dear Livetolove and Mayhlinkers,
If my reference are well, your new problem has been presented at
TST Chine (2006).
Perhaps can you precise this reference.
Sincerely
Jean-Louis

Thanks you again dear [b]Jayme[/b],
Reference:
http://www.mathlinks.ro/viewtopic.php?p=550561#550561

And here are other results:

Let $ (A_2B_2C_2)\cap HA, HB, HC$ again at $ A_4, B_4, C_4$. Prove that $ A_2A_4, B_2B_4, C_2C_4$ are concurrent.
$ (A_3B_3C_3)\cap HA,HB,HC$ again at $ A_5,B_5,C_5$. Prove that $ A_3A_5, B_3B_5, C_3C_5$ are concurrent.

Dear Livetolove and Mathlinkers,
as you can see on my website, an article intitled "The P-Hagge's circle" is coming next.
You will see the proof of your first concurrence.
For the second, I only proof that the points "5" are concyclic with H, but I don't treat the concurrence...
Sincerely
Jean-Louis

Dear [b]Jayme[/b],
For the second concurrency, you can use Ceva-sine theorem or note that $ (A_3B_3C_3)$ is the Q-[b]Hagge[/b]'s circle of triangle $ ABC$ ($ P$ and $ Q$ are isogonal conjugate wrt $ ABC$)

This is the last result I've discovered:
Denote $ G$ the reflection of $ H$ across the center of $ (A_2B_2C_2). AG, BG, CG$ intersect $ (A_2B_2C_2)$ again at $ A_6, B_6, C_6$. Prove that $ A_4A_6, B_4B_6, C_4C_6$ are concurrent.

Dear Livetolove and Mathlinkers,
for the concurrence you use the Ceva's theorem... Why not a synthetic proof?
For your nice last discoverie, it can be proved with the theorem of the six pedals (see :http://www.cip.ifi.lmu.de/~grinberg/geometry2.html).
Sincerely
Jean-Louis

The last can be solve by using Ceva-sine theorem too. Note that $ AG, BG, CG$ are concurrent and $ \Delta ABC\sim \Delta A_4B_4C_4$. You can try.

Here are other problems involved:
[b]Problem 1:[/b]
Given triangle $ ABC$ with circumcircle $ (O)$, orthocenter $ H$.Denote $ A_1, B_1, C_1$ points on $ (O)$ such that $ AA_1//BB_1//CC_1, A_2,B_2,C_2$ the reflections of $ A_1,B_1,C_1$ across the lines $ BC, CA, AB$, respectively. Prove that $ H,A_2,B_2,C_2$ are concylic.
[b]Problem 2:[/b]
Given triangle $ ABC$ with circumcircle $ (O)$, orthocenter $ H$.Denote $ A_1, B_1, C_1$ points on $ (O)$ such that $ AA_1//BB_1//CC_1, A_2,B_2,C_2$ the reflections of $ A_1,B_1,C_1$ across the midpoints of $ BC, CA, AB$, respectively. Prove that $ H,A_2,B_2,C_2$ are collinear.

Any solution? Because I want to ignore an infinity point.

Some other interesting results:
[b]Problem 3: [/b]Given triangle $ ABC$ and an arbitrary point $ P$ on circumcircle $ (O)$. Denote $ A', B',C'$ be the reflections of $ P$ across the lines $ BC, CA, AB$. Prove that $ H,A',B',C'$ are collinear. (Steiner line)
[b]Problem 4:[/b] Given triangle $ ABC$ and an arbitrary point $ P$ on circumcircle $ (O)$. Denote $ A', B',C'$ be the reflections of $ P$ across the midpoints of $ BC, CA, AB$. Prove that $ H,A',B',C'$ are concyclic.
[b]Problem 5:[/b] Given triangle $ ABC. P$ and $ Q$ are isogonal conjugate wrt $ \Delta ABC$. Let $ O_1, O_2$ be the circumcenters of P-Hagge's circle and Q-Hagge's circle. Prove that $ O_1O_2//PQ$

Dear livetolove212
Can you post your solutions for problem 5?
It should be a really nice problem.
And the origin problem can be also found as China tst 2006 quiz2 Problem2.

[b]Problem 6:[/b] Given triangle $ ABC$, the Nine-point circle $ (E)$ and an arbitrary point $ P$ on $ (E)$. Let $ Q$ be the isogonal conjugate point of $ P$ wrt $ \Delta ABC$. Prove that the center $ I$ of Q-Hagge circle lies on (E) too. Moreover, $ I$ is the reflection of $ P$ across point $ E$.

[b]Solution of problem 5:[/b]
[b]Lemma:[/b] Given triangle $ ABC$, orthocenter $ H$, circumcenter $ O$ and an arbitrary point $ P$ inside. $ AP, BP, CP$ intersect $ (O)$ again at $ A_1, B_1, C_1$. Denote $ A_2,B_2,C_2$ the reflections of $ A_1,B_1,C_1$ across the midpoints of $ BC, CA, AB$, respectively, $ O_2$ the center of $ (A_2B_2C_2)$. Then $ OO_2HP$ is a parallelogram.
[b]Proof: [/b]
Denote $ A',C'$ the reflections of $ A,C$ across the midpoint of $ AC$, respectively.
Then $ \Delta A'B_1C'$ is the reflections of $ \Delta A_2B_2C_2$ across the midpoint of $ AC$.
Since $ AA'//A_2C, AA' = A_2C, A_2C//BA_1, A_2C = BA_1$ we get $ AA'A_1B$ is a parallelogram. $ (1)$
Similarly, $ BCC'C_2$ is a parallelogram. $ (2)$
Let $ X,Y,Z$ be the midpoints of $ AA_1, BB_1, CC_1$, respectively then $ X,Y,Z$ lies on $ (OP)$.
But from $ (1)$ and $ (2), X, Y$ are the midpoints of $ BA', BC'$.
A homothety $ H^{\frac {1}{2}}_B: A'\rightarrow X, B_1\rightarrow Y, C'\rightarrow Z$ therefore $ \Delta A'B_1C'\rightarrow \Delta XYZ$
A symmetric with center is the midpoint of $ BC$
$ S: B\rightarrow B', H\rightarrow H'$ the orthocenter of triangle $ AB'C$, $ O_2\rightarrow O'_2$
$ \Rightarrow O_2H\rightarrow O'_2H'$
It’s easy to see that $ \angle ABH = \angle CBH' = \angle CBO$ thus $ B,O,H'$ are collinear.
Hence $ H^{\frac {1}{2}}_B: H'\rightarrow O, O'_2\rightarrow O''_2$ the circumcenter of triangle $ XYZ$.
$ \Rightarrow OP\equiv OO''_2//H'O'_2//HO_2$ and $ OP = 2OO''_2 = H'O'_2 = HO_2$
So $ O_2OPH$ is a parallelogram. We are done.
[b]Return to our problem:[/b]
From the lemma above, $ OO_1// = QH, OO_2// = PH$ thus $ PQ// = O_1O_2$
Our proof is completed.
[b]Remark:[/b]
(1) An interesting result: The radius of Q-Hagge circle is equal to $ OP$.
(2) Problem 6 is only a corollary of the lemma above, note that the center of Q-Hagge circle is always a reflection of $ P$ across the center of Nine-point circle.

Dear Mathlinkers,
Hagge circle has many corollaries. Here are three interesting problems I've found:
[b]Problem 7:[/b] Given triangle $ ABC$ and the circumcenter $ O$. $ P$ and $ Q$ are isogonal conjugate wrt triangle $ ABC$ such that $ O,P,Q$ are collinear. Prove that P-Hagge circle and Q-Hagge circle are tangent.
[b]Problem 8:[/b] Given triangle $ ABC$ and circumcircle $ (O,R)$. A circle which has center $ O$ and radius $ r < R$ cuts $ BC, CA, AB$ at $ A_1,A_2,B_1,B_2,C_1,C_2$, respectively. Let  $ E, F$ be the Miquel points of triangle $ ABC$ wrt $ (A_1,B_1,C_1)$ and $ (A_2,B_2,C_2)$ respectively. Prove that E-Hagge circle and F-Hagge circle are congruent.
[b]Problem 9:[/b] Denote $ O,I,H,N$ the incenter, circumcenter, orthocenter and [b]Nagel[/b] point of triangle ABC. Prove that $ HN \equal{} 2OI$

Any solution?

Problem 8 is only a corollary of this problem:
[b]Problem 10:[/b] Given triangle $ ABC$ and an arbitrary point $ O$ inside. A circle which has center $ O$ intersects $ BC, CA, AB$ at $ A_1,A_2,B_1,B_2,C_1,C_2$, respectively. Denote $ E,F$ the Miquel point of triangle $ ABC$ wrt $ (A_1,B_1,C_1)$ and $ (A_2,B_2,C_2)$. Prove that $ OE \equal{} OF$ and $ E, F$ are isogonal conjugate wrt $ \Delta ABC$.
Other corollary:
[b]Problem 11:[/b] Let $ K$ be the reflection of $ H$ wrt the center of P-Hagge, $ G$ be the centroid of triangle $ ABC$. Prove that $ Q,G,K$ are collinear and $ QG \equal{} \frac {1}{2}GK$
[b]Problem 12:[/b]  + When $ P\equiv G$ the centroid of triangle $ ABC$. Prove that $ P$ lies on Q-Hagge.
+ Prove that the center of O-Hagge $ \equiv O$ ($ O$ is the circumcenter of triangle ABC)

[quote="livetolove212"]Given triangle $ ABC$ with circumcircle $ (O)$, orthocenter $ H$ and an arbitrary point $ P$ inside. $ AP, BP, CP$ intersect $ (O)$ again at $ A_1, B_1,C_1$, respectively. Let $ A_2, B_2,C_2$ be the reflections of $ A_1,B_1,C_1$ across the lines $ BC, CA, AB$, respectively. Prove that $ H, A_2, B_2, C_2$ are concyclic.

[b]Note: [/b]When $ P\equiv I$ ($ I$ is the incenter), $ (A_2B_2C_2)$ is the [b]Fuhrmann[/b]'s circle of triangle $ ABC$.[/quote]
First we have $ A_1B_1C_1\sim A_2B_2C_2$
Then $ (C_2A_2,C_2B_2)\equiv (C_1B_1,C_1A_1)(mod \Pi )$
We proff
$ (HA_2,HB_2)\equiv (C_1B_1,C_1A_1)$
$ \Leftrightarrow (HA_2,BC)\plus{}(BC,CA)\plus{}(CA,HB_2)\equiv (C_1B_1,C_1C)\plus{}(C_1C,C_1A_1)$
$ \Leftrightarrow (BC,A_1A_0)\plus{}(BC,CA)\plus{}(B_0B_1,AC)\equiv (AB_1,AC)\plus{}(BC,BA_1)$
$ \Leftrightarrow (BA_1,A_1A_0)\plus{}(BC,CA)\plus{}(B_0B_1,AB_1)\equiv 0$
$ \Leftrightarrow (AB,AA_0)\plus{}(AA_0,BB_0)\plus{}(BB_0,BA)\equiv 0$
$ \Leftrightarrow(AB,BA)\equiv 0$
done.
$ A_0,B_0,C_0$ are symmetric of $ H$ by $ BC,CA,AB$

[quote="livetolove212"]
This is the last result I've discovered:
Denote $ G$ the reflection of $ H$ across the center of $ (A_2B_2C_2). AG, BG, CG$ intersect $ (A_2B_2C_2)$ again at $ A_6, B_6, C_6$. Prove that $ A_4A_6, B_4B_6, C_4C_6$ are concurrent.[/quote]
Can you post your solutions for this problem :wink: ?
It is very nice, but i can't solve  :(

[quote="livetolove212"][b]Solution of problem 5:[/b]
[b]Lemma:[/b]
A symmetric with center is the midpoint of $ BC$
$ S: B\rightarrow B', H\rightarrow H'$ the orthocenter of triangle $ AB'C$, $ O_2\rightarrow O'_2$
$ \Rightarrow O_2H\rightarrow O'_2H'$[/quote]
I think it must be :"A symmetric with center is the midpoint of $ AC$ "  :)
This topic is very very good, thanks you very much.

[quote="conan6032"]
Can you post your solutions for this problem :wink: ?
It is very nice, but i can't solve  :([/quote]

I'm busy now so I will give you a note: "Using Ceva-sine theorem". If you still can't solve this problem, I will post the solution.
Good luck!
N.V.L


Bobert decided to apologize to Caddlebot for pushing him off the Leaning Tower of Treething, so he decided to give him a present. It was a man-eating cat.

The first day, the man-eating cat takes one bite out of Caddlebot.
The second day, the man-eating cat takes two-thirds of a bite out of Caddlebot.
The third day, the man-eating cat takes four-ninths of a bite out of Caddlebot.
The fourth day, the man-eating cat takes eight-twentysevenths of a bite out of Caddlebot.

Every day, the man-eating cat eats two-thirds as much as it did on the previous day.

How much of Caddlebot does the man-eating cat eat in total?

[hide]This is just a infinite geometric series, so we can just do the problem with the forumla $\frac{a}{1-r}$.  IN this case, a is 1, and r is $\frac{2}{3}$.  Plugging in the values into the forumla, we get 3 bites total.[/hide]

I don't know if this is a bit hard for getting started, but anyway:
[hide]This is just an infinite geometric sequence.
1/(1-2/3)=1/(1/3)=3.  The cat will take [b]3 [/b]bites.[/hide]


John writes a number with 2187 digits on the blackboard, each digit being a 1 or a 2. Judith creates a new number from John's number by reading his number from left to right and wherever she sees a 1 writing 112 and wherever she sees a 2 writing 111. (For example, if John's number begins 2112, then Judith's number would begin 111112112111.) After Judith finishes writing her number, she notices that the leftmost 2187 digits in her number and in John' number are the same. How many times do five 1's occur consecutively in John's number?

(A) 182 (B) 183 (C) 184 (D) 185 (E) 186

OMG, i took the pascal contest too. and i really want to know the answers to the last 5 questions. plzzzz, where did you get the questions?
by the way, i put B for #24, for #25, i think i put C.


Στο βιβλιο θεωριας αριθμων της ΕΜΕ υπαρχει η παρακατω ασκηση:

Γενικευση Κινεζικου Θεωρηματος:
Ν.δ.ο οι γραμμικες ισοτιμιες $ x \equiv a_i (modn_i)$, εχουν κοινη λυση αν και μονο αν $ gcd(n_i, n_j)|(a_i \minus{} a_j)$ για καθε $ i$ διαφορο του $ j$. Αν υπαρχει λυση, αυτη ειναι μοναδικη $ (mod(lcm(n_1, n_2, ..., n_k)))$

Ειναι πολυ ευκολο να δειξουμε οτι για να υπαρχει λυση πρεπει $ gcd(n_i, n_j)|(a_i \minus{} a_j)$, το αντιστροφο ομως με δυσκολευει... στη λυση του βιβλιου γραφει οτι η αποδειξη ειναι ομοια με αυτην του απλου Κινεζικου Θεωρηματος, οποτε μαλον θα πρεπει να βρουμε μια λυση και μετα να αποδειξουμε οτι αυτη ειναι ισοτιμη με τις υπολοιπες λυσεις $ (mod(lcm(n_1, n_2, ..., n_k)))$, ομως επηδη μεχρι τωρα δεν εχω διαβασει και πολλα πραγματα ουτε εχω λυσει πολλες ασκησεις απο Θεωρια Αριθμων δεν μπορω να σκευτω κατι. Θα ηθελα αν μπορει καποιος να βοηθησει...

Ευχαριστω

Ουτε ενας; :(

Είμαι καινούργιος στο forum. Ελπίζω να μην έχω παραβεί κάποιους κανόνες.

Σχετικά με το θεώρημα:

Το πρώτο βήμα που θα συνιστούσα, είναι να δείξεις ότι η ισοτιμία  $ x \equiv a \bmod n$ είναι ισοδύναμη με το σύστημα $ x \equiv a \bmod p_1^{r_1}, \ldots , x \equiv a \bmod p_k^{r_k}$ όπου $ n \equal{} p_1^{r_1}, \cdots , p_k^{r_k}$ σε γινόμενο πρώτων παραγόντων.

Το δεύτερο βήμα είναι να μετατρέψεις το υπάρχων σύστημα ισοτιμιών σε ένα ισοδύναμο σύστημα όπου όλα τα $ n_i$ είναι δυναμεις πρώτων αριθμών και να ελέγξεις ότι αν στο πρώτο σύστημα ισχύει ότι $ \gcd(n_i,n_j)|(a_i\minus{}a_j)$ (για κάθε $ i \neq j$) τότε αυτό ισχύει και στο νέο σύστημα.

Τέλος να δείξεις ότι το νέο σύστημα έχει λύση.


modular arithmetic
number theory
prime factorization

What is the smallest positive integer that has exactly 18 positive integer factors and leaves a remainder of 9 when divided by 31?

[hide] For a positive integer $ x$, we can calculate its number of positive factors by using its prime factorization $ 2^a3^b5^c7^d\dots$. If you know its prime factorization, then the number of factors is $ (a\plus{}10(b\plus{}1)(c\plus{}1)\dots$. In this case the number of factors is $ 18$, so $ 18\equal{}(a\plus{}1)(b\plus{}1)(c\plus{}1) \dots$. We note that a solution of this is for one of the variables to be $ 5$, another to be $ 2$, and the rest $ 0$. We next note that $ 2^5\equiv1\pmod(31)$ and $ 3^2\equiv9\pmod(31)$. Clearly, this integer $ x$ with the prime factorization of $ 2^53^2$ iscongruent to $ 9\bmod(31)$, and after using a quick check to not that no other cases that are small enough work, our minimum integer is $ 32\times{9}\equal{}\boxed{288}$.[/hide]

Yes, this checks, the others being $ 2^{17}, 2^{8}\cdot3,$ and  $ 2^{2}\cdot3{2}\cdot 5 \equal{} 180$ but leaves a remainder of $ 25$ when divided by $ 31$.

[quote="gaussintraining"] We note that a solution of this is for one of the variables to be $ 5$, another to be $ 2$, and the rest $ 0$.[/quote]
but also two variables may be equal $ 2$ and one may be $ 1$... why didn't you check this case? :)

I know. But I noticed when working through the problem that $ 2^5\equiv1\pmod{31}$. Also, I did make a quick check of other cases
[quote="gaussintraining"]and after using a quick check to note that no other cases that are small enough work[/quote]


A family $ \aleph$ of subsets of a set $ X$ is said to have the finite intersection property
if $ F_1 \cap F_2 \cap \ldots \cap  F_n$ is non-empty for any $ n \in \mathbb{N}$ where $ F_i \in \aleph$, i.e. finite intersections in $ \aleph$ are not null.
Prove that a topological space is compact if and only if the intersection of all subsets of $ X$ belonging to $ \aleph$ is not null for all $ \aleph$ having the finite intersection property.

This is a reasonably manageable alternative formulation of compactness, but I found it rather nice, so I thought I'd post it.

Ashwath

Yes, it is equivalent to following assertion

''Every open cover has a finite sub-cover''.


modular arithmetic
function
floor function
algebra

Here is another question that I need help on. I actually thought about it during my problem series class tonight but I think I'll get better answer by posting here.

Say, you get a mod equation as below:

$\frac{7}{6} \alpha \equiv 7 \mod 11$

Nice thing is that 11 is a prime number so I guess I can simply divide by 7/6? I never really understood how this would work when the number multiplied with a variable, $\alpha$ in this case, is just rational number, not necessarily integer.

Thanks!

Someone correct me if I'm wrong (I have a feeling I am)

I think I remember reading somewhere that, in that equation, $a$ is the integer that when multiplied by $7$ and divided by $6$ is $\equiv 7\pmod{11}$.  I think you're allowed to do that (it has a meaning) but in general is not a good notation to use.

[quote="JesusFreak197"]So, that would mean that $\alpha \equiv 6\bmod{11}$, right?[/quote]

I suppose. I don't see why it will NOT work but I've never seen a floor function with fractions.  :ninja:

[quote="chess64"]Floor function? I don't see any floor function...you mean congruency, don't you? :huuh: Floor functions are $\lfloor x \rfloor$. ;)[/quote]

You're right. I must've typed wrong word when I did this (since I was doing algebra problem that time too --> probably had a floor function in it)


Could any Japanese user explain what everyday life in Japan is like?  Thanks in advance.

Your post doesn't make any sense. You got to be kidding! What do you mean by what it's like in Japan. Are you asking if people living in Japan grow horns on their heads? Or how much time they spend on the bathroom?

You might want to ask the user undefined117, because I definately know he has lived in Japan (or is currently living in).

I did spend a week in Tokyo as part of my vacation, and I discovered several things:

1.) The climate is always hot.

2.) The rail lines are hell. Everyone uses a rail line to get to their destination. School girls, business men, tourists, etc.

3.) Every man who goes to work wears a suit.

4.) 100 Japanese Yen = 1 USD

5.) "Underground" markets are common - where venders sell goods at market stalls.

6.) The Japanese people are [i]very[/i] nice. They've helped us constantly when we were lost in the rail lines and many of them spoke fluent English.

mathwizarddue, find somewhere else to ask people to do your homework.  This thread is locked, and all future threads of a similar sort will be locked as well.


Four distinct lines $L_1,L_2,L_3,L_4$ are given in the plane: $L_1$ and $L_2$ are respectively parallel to $L_3$ and $L_4$. Find the locus of a point moving so that the sum of its perpendicular distances from the four lines is constant.

We get a parallelogram, whose sides extended bound 9 areas.

If the constant is exactly the sum of the distances that L1,L2 and L3,L4 are apart, then the locus is exactly the center region only.

If the constant is larger, we get a regular octagon with an affine transformation applied to it.  The transformation is exactly the transformation required to bring a square to the parallelogram (defined by L1, .. L4) that we have.

Sorry for reviving this but does anyone have a solution? I don't understand the one posted above.

Denote our parallelogram formed by our lines $ABCD$, and let $x$ be the distance between $L_1$ and $L_3$, and $y$ the distance between $L_2$ and $L_4$. For a distance $D$, let $P$ denote the locus of points whose sum of distances is equal to $D$. We note $3$ cases.
$\newline$
Case $1$, $D < x+y$. This cannot be satisfied, so $P$ is empty. $\newline$
Case $2$, $D = x+y$. This is satisfied for all points in and on $ABCD$, so $P$ here is the boundary and interior of $ABCD$ $\newline$
Case $3$, $D > x+y$. Then note that $P$ is a centrally symmetric octagon. This immediately follows from this: consider an isosceles triangle $ABC, AB = AC$ with $G$ on $BC$. Then note that the sum of the distances from $G$ to $AB$ and $AC$ is constant, since the area of $ABC$ can be written in terms of the sum of the two distances.
$\newline$
Hence, our locus is just a centrally symmetric octagon, as shown in the diagram below. $\quad \square$

[asy]
import graph; size(15cm);
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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
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pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);

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draw((-7.55104938369913,3.827564196120023)--(0.772267672117848,7.021461198022683), linewidth(2) + blue);
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dot((5.613701122765273,-5.290072197015603),linewidth(4pt) + dotstyle);
dot((13.937018178582253,-2.0961751951129455),linewidth(4pt) + dotstyle);
dot((14.465434341079671,2.298576985247421),linewidth(4pt) + dotstyle);
dot((-2.7042738026251647,-4.289935622883647),linewidth(4pt) + dotstyle);
dot((9.09024259750829,6.021324623890725),linewidth(4pt) + dotstyle);
dot((-8.079465546196548,-0.5671879842403441),linewidth(4pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]


Cast mi je i zadovoljstvo da mogu nas podforum proglasiti otvorenim!  :lol:
Znam da nismo svi zadovoljni imenom, ali eto... ja sam Valentinu poslao oba imena u onom PMu, pa sad sto nas je ovako udesio... nije do mene  :roll:
Bilo bi fino da neko od nas dobije ovdje moderatorske "privilegije", da malo sredimo forum... Ovaj topic je mjesto i za prijedloge i sugestije...

P. S. Sad kad vec imamo ovaj svoj podforum, ne bismo smjeli dopustiti da nam zamre (kao nekim nacionalnim podforumima (npr. nasim zapadnim sosedima s kojim su nas ponovno ujedinili  :) , bez uvrede tovarisi...).

Ipak ex juga kao naslov - ma svejedno...

I sta radi raja iz komsiluka u slobodno vrijeme? Ima li ko kakav dobar zadatak?

Ma evo, ima finih zadataka na proslogodisnjem shortlistu... To je ono cime se ja trenutno bavim. A sta ima u susjedstvu?  :D

Pozdrav iz komsiluka!  :lol:

Zar je lanjski shortlist vec objavljen? Gdje ga mogu naci? EDIT: nasao :)

Ovdje se odmara od skole i radi matematika. Pomalo se zabavljam smisljanjem zadataka ovih dana.

Samo nemaj smisljati funkcionalne  :rotfl:  drzi se nejednakosti: u par poteza napravis fin zadacic za koji ostatku svijeta treba dva dana da rijese (imam takvih vec gomilu kod kuce, prestao sam ih vise i pokazivati drugima-istraumirani su  :lol: )

Ona funkcionalna jeste bila glupost, al ne smisljam ja njih. A sto se nejednakosti tice nisam toliko dobar da bi ih lako smisljao. Kad pokusavam smislit nejednakost onda poznate mnozim, dijelim al rijetko dobijem ista sto nije trivijalno za dokazati. Lakse je recimo smisliti planimetrijski zadatak ili diofantsku jednadzbu.

Jel kod vas u Hrvatskoj ovo MEMO predstavlja test za izbor ekipe za Medjunarodnu matematicku olimpijadu.

[b]Hsiljak [/b] sto te bolan nema na pripremama?

Jel mi ko probo rijesit onu geometriju sto sam stavio na Proposed and own problems (ono beautiful iz Irana  / mada nije ni blizu beautiful al eto reko nek raja makar pogleda).

[quote="delegat"][b]Hsiljak [/b] sto te bolan nema na pripremama?

Jel mi ko probo rijesit onu geometriju sto sam stavio na Proposed and own problems (ono beautiful iz Irana  / mada nije ni blizu beautiful al eto reko nek raja makar pogleda).[/quote]

Ma imao sam obaveza kolega, sta ces... A geometrija ti je Boze sacuvaj :lol:

Nije, MEMO je Srednjoeuropska olimpijada, evo tu malo informacija: http://www.oemo.at/info/memo.php (siguran sam da ćeš skužiti toliko na njemačkom).

[quote]Ma imao sam obaveza kolega, sta ces... A geometrija ti je Boze sacuvaj [/quote]
Dobro kolega Danas smo imali nesto s Damirom Neki test i tako
Eno stavio sam i onu funkciju sto je Dino bio dao na testu na Algebra proposed and own problems pa ako neko nema preca posla nek je rjesava (i za nju sam stavio da je [b]beautiful[/b])

[quote]Nije, MEMO je Srednjoeuropska olimpijada, evo tu malo informacija: http://www.oemo.at/info/memo.php (siguran sam da ćeš skužiti toliko na njemačkom[/quote]

Ma kakav ja i njemacki
Vjeruj mi da prvi put cujem za to takmicenje (bar pod ovim nazivom)
Kada bude ta olimpijada tj. u kom mjesecu i kad je prvi puta odrzana Ko je pokretac ovoga?
Jel to bude ko mediteranska u svakoj zemlji posebno il pak bude zemlja domacin takmicenja?
Kolko bude zadataka i kolko svaka zemlja moze imati ucesnika?
Kako se ovo zove na engleskom (mislim takmicenje) i imaju li gdje zadaci?

Ovak, natjecanje se održava na jednom mjestu... ove godine je to u Eisenstadtu u Austriji, 20.-26.9., a dalje će se valjda (?) održavati po kružnom principu. Zemlje koje sudjeluju su za sad: Austrija, Slovenija, Hrvatska, Švicarska, Poljska, Češka i Slovačka (možda se sljedeće godine priključe i Njemačka i Mađarska). Natjecanje je nastalo iz Austrijsko-poljskog natjecanja, a ovo mu je prva godina postojanja. Bit će šest natjecatelja iz svake zemlje (ne smiju biti 4. razred srednje a kod nas se plasiralo dvoje iz trećih, troje iz drugih i jedan iz prvih razreda i ne smiju sudjelovati na IMO-u u bilo kojem trenutku prije MEMO-a). Sastoji se od pojedinačnog dijela (5 sati, 4 zadatka) i ekipnog dijela (4.5 sati, 4 zadatka). Mislim da se na engleskom isto kaže MEMO, ali nisam siguran... Evo, za sada toliko.

Dobro hvala ti.
Ovo je jel tako prvi put da se organizuje ovo takmicenje (bar sam tako razumio na osnovu tvog pisanja)?
Sta je onda ono Donova mathematic olympiad. Ucestvuje li i Hrvatska na tome i ko je uopce organizator tog takmicenja.
(Kako se sjecam prosle godine je Srbija ucestvovala dok mi iz Bosne nismo).

Pozdrav tebi i ostalim "jugoslovenima" :)

[quote="goc"]muljas mali, ne mnozis jer bi inace rijesio nejednakost na memo testu onako kako sam vas i naucio, da lijepo sve pomnozis i gotovo...e, ma sve cu ti oprostit, al ta nejednakost me ganja ;)[/quote]

Jel ti to grizes sebe il mene? Daj se malo smiri covjece. Ja sam sjebo i ja sam kriv. ccc....

a grizem se. malo mi zao dodje, samo to. nije frka mali, imas ti jos vremena. samo lipo citaj bolesna rjesenja po mathlinksu od cike harazija,arqadya,vijetnamaca i takovih likova ;)

Normalno da imam vremena. Pa nije matematika samo u natjecanjima. Mozda mi je prije bilo bolesno stalo do uspjeha na natjecanjima, al sad sam malo ohladio od toga. Ne da mi se ni visit na mathlinksu stalno, vec mi je kompa na vrh glave. Preko ljeta cu se malo primit knjizevnosti, koliko dugo nisam postenu knjigu procitao.

pa nisam ni reko da moras na natjecanja. rjesavaj zadatke i citaj bolesna rjesenja za svoj gust. meni to bar tako dodje...
a zao mi dodje zato sto ti rezultati na natjecanjima donose hrpu pogodnosti tipa stipendije i olaksanja u skoli itd...

Iskreno, ne znam što je Donova MO, ali mislim da Hrvatska nije nikad sudjelovala na tome, čuo bih valjda... Da, ovo je prvi put da postoji MEMO natjecanje, ali na neki način ima svoju prošlost kroz Austrijsko-poljsko natjecanje.

Joj, dajte se smirite goc i djuro... Djuro nije uspio ove godine, ali uspjet će sljedeće. Trebao si ga vidjeti jučer popodne na tulumu i ujutro na ekipnom ... bio je lud. Kao ekipa smo (Djuro, Ines, ja) imali 18 bodova, a drugi su imali 2 boda. Dobro, jest da je to bila ekipa prvašića, ali dobro...  :D Bio je ful dobar.

[quote="delegat"]
Sta je onda ono Donova mathematic olympiad. Ucestvuje li i Hrvatska na tome i ko je uopce organizator tog takmicenja.
(Kako se sjecam prosle godine je Srbija ucestvovala dok mi iz Bosne nismo).
[/quote]

Donova MO bi ustvari trebao biti Danube Cup (iliti Dunavski Kup). to je takmicenje koje se organizuje od 2005. (cini mi se), a pozivnice valjda dobijaju sve zemlje kroz koje protice Dunav  :)  Prve godine su ucestovale ekipe Srbije i Crne Gore, Bugarske, Rumunije i Moldavije, a za 2006. ne znam.

Pozdrav svima i nadam se da cemo opstati i druziti se ovdje naukom Pozdrav svima, Slovencima, Hrvatima, Bosancima, Crnogorcima, Kosovarima, Srbima, Makedoncima, Madjarima, Romima i svima ostalim da ne nabrajam ( pade me na pamet on pjesma " Sto nas ima masala")

Izgleda da ce i "nas"  forum  zamreti ...
Al' ajde, kada su kod vas ostalih izborna za olimpijadu i balkanijadu ?
Kod nas je izborno za nedelju dana pa cu znati da li cu obici jos jednu "former " jugoslovensku republiku ...

Pa dobro, vecina national foruma na MLu nisu pretjerano aktivni... Sta ces.

Sto se naseg (BH) izbornog tice, sredina maja je najvjerovatniji termin (kao i inace). Poznato je da je nama mrsko ici na Balkanijadu, pa mozemo drzavno odrzati tako kasno.

P.S. cestitke za hundred points... jos jednom si opravdala nadimak  :)

Donova je Danube cup, koji je probno odrzan te godine, sa ciljem da se posle prosiri na sve zemlje kroz koje protice Dunav. Bili smo mi, Rumuni, Moldavci i Bugari, tako da sam bio srecan sto sam izvukao srebro :D Koliko sam upucen, ovo takmicenje nije zazivelo, nije se odrzalo nijedne godine potom.


function
algorithm
number theory
Euclidean algorithm
algebra solved
algebra

A function on two variables (from Z+^2 --> Z+) is defined:
f(x,x) = x+2
f(x,y)=f(y,x)
f(x,x+y)/f(x,y) = x+y/y.
Find a closed expression for f(x,y) in terms of x,y.

Let us take and work backwards from any pair given to us, assuming WLOG a<b and compute f(a,b).

Let us use [tex]\frac{f(x,x+y)}{f(x,y)}=\frac{x+y}{y}[/tex]

[tex]\frac{f(a,b)}{f(a,b-a)}=\frac{b}{b-a}[/tex]

By the Euclidean Algorithm we know that [tex]b=ka+r_1[/tex] with [tex]0 \le r_1 \le a[/tex] for some unique k.  By using the above property k times we arrive at:

[tex]\frac{f(a,b)}{f(a,b-ka)}=\frac{b}{b-ka}[/tex]

Since [tex]f(x,y)=f(y,x)[/tex] we know [tex]f(a,b-ka)=f(b-ka,a)[/tex]

Now we continue with the Euclidean algorith knowing there exists a unique m such that [tex]a=m(b-ka)+r_2[/tex] with [tex]0 \le r_2 \le b-ka[/tex]

We know that [tex]\frac{f(b-ka,a)}{f(b-ka,a-m(b-ka))}=\frac{a}{a-m(b-ka)}[/tex]

From the above two steps we have [tex]\frac{f(a,b)}{f(b-ka,a-m(b-ka))}=\left( \frac{b}{b-ka} \right) \left( \frac{a}{a-m(b-ka)} \right)[/tex]

Now we continue this process until we arrive at a point where both components are the same.  By the Euclidean Algorithm used this must occur at gcd(a,b).

At this point we have [tex]\frac{f(a,b)}{f(gcd(a,b),gcd(a,b))}=\left( \frac{b}{gcd(a,b)} \right) \left( \frac{a}{gcd(a,b)} \right)[/tex]

Now from the first property we know that [tex]f(gcd(a,b),gcd(a,b))=gcd(a,b)+2[/tex]

Plugging this into the equation above we get our solution that:
[tex]\frac{f(a,b)}{(gcd(a,b)+2)}=\left( \frac{b}{gcd(a,b)} \right) \left( \frac{a}{gcd(a,b)} \right)[/tex]

Thus [tex]f(a,b)=\frac{ab \left( gcd(a,b)+2 \right)}{ \left( gcd(a,b) \right)^2}[/tex] in a closed form as desired.

Here is the example [tex]f(35,100)[/tex] worked out:

[tex]\frac{f(35,100)}{f(35,30}=\frac{100}{30}[/tex], so [tex]f(35,100)=\left(\frac{100}{30}\right)\left(f(30,35)\right)[/tex]

Since [tex]\frac{f(30,35)}{f(30,5)}=\frac{35}{5}[/tex], [tex]f(35,100)=\left(\frac{100}{30}\right)\left(\frac{35}{5}\right)\left(f(5,30)\right)[/tex]

[tex]\frac{f(5,30)}{f(5,5)}=\frac{30}{5}[/tex] implies [tex]f(35,100)=\left(\frac{100}{5}\right)\eft(\frac{35}{5}\right)\left(f(5,5)\right)[/tex]

Thus, [tex]f(35,100)=\frac{100*35*f(5,5)}{5*5}=\frac{100*35*7}{5*5}=980[/tex]


inequalities
calculus
search
function
inequalities unsolved

[quote="Marius Mainea"]If $ a,b,c > 0$ then
\[ \sqrt {\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}} \plus{} \sqrt {\frac {b^3}{b^3 \plus{} (c \plus{} a)^3}} \plus{} \sqrt {\frac {c^3}{c^3 \plus{} (a \plus{} b)^3}}\ge 1
\]
[/quote]
To standardize  $ a \plus{} b \plus{} c \equal{} 1$
Let$ f(t) \equal{} \sqrt {\frac {a^3}{a^3 \plus{} (1 \minus{} a)^3}}$ we easy prove $ f"(t) \ge\ 0$
By Jensen we have
$ f(a) \plus{} f(b) \plus{} f(c) \ge\ 3f(\frac {a \plus{} b \plus{} c}{3} \equal{} 1$
We have done!

It's very old  :)
$ \sqrt[]{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}} \geq\ \frac{2\minus{}a}{3}$ $ \Leftrightarrow  \frac{(a\minus{}1)^2(a\minus{}3)^2}{27(a^2\minus{}3a\plus{}3)} \geq\ 0$

An other way
$ \sqrt[]{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}} \geq\ \frac{a^2}{a^2\plus{}b^2\plus{}c^2}$
is also true

[quote="Allnames"]An other way
$ \sqrt []{\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}} \geq\ \frac {a^2}{a^2 \plus{} b^2 \plus{} c^2}$
is also true[/quote]

  Try $ c\to0$ , $ a\equal{}b\to\frac{1}{2}$

[quote="nguoivn"]It's very old  :)
$ \sqrt []{\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}} \geq\ \frac {2 \minus{} a}{3}$ [/quote]

   Try $ a\to 0$

Oh, very sorry. I'll check again my calculus :D
You can see a solution in "Sang tao bat dang thuc". of PKH :D

[quote="Allnames"][quote="Marius Mainea"][quote="Allnames"]An other way
$ \sqrt []{\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}} \geq\ \frac {a^2}{a^2 \plus{} b^2 \plus{} c^2}$
is also true[/quote]

  Try $ c\to0$ , $ a \equal{} b\to\frac {1}{2}$[/quote]
My proof is true ,Marius Mainea,i think you should check your example carefully
Proving it is quite simple[/quote]

 Yes ,sorry.

[quote="Marius Mainea"][quote="onlylove_math"][quote="Marius Mainea"]If $ a,b,c > 0$ then
\[ \sqrt {\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}} \plus{} \sqrt {\frac {b^3}{b^3 \plus{} (c \plus{} a)^3}} \plus{} \sqrt {\frac {c^3}{c^3 \plus{} (a \plus{} b)^3}}\ge 1
\]
[/quote]
we easy prove $ f"(t) \ge\ 0$
[/quote]

  Why$ f^{\prime\prime}(x)\ge 0$?[/quote]
I sure $ f"(x)\ge\ 0$ . If not $ f"(x) \le\ 0 \rightarrow$your problem 's wrong  :lol:

[quote="Marius Mainea"]If $ a,b,c > 0$ then
\[ \sqrt {\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}} \plus{} \sqrt {\frac {b^3}{b^3 \plus{} (c \plus{} a)^3}} \plus{} \sqrt {\frac {c^3}{c^3 \plus{} (a \plus{} b)^3}}\ge 1
\]
[/quote]
By Holder we obtain: $ \left(\sqrt {\frac {a^3}{a^3 \plus{} (b \plus{} c)^3}}\right)^2\sum_{cyc}a^3(a^3\plus{}(b\plus{}c)^3)\geq(a^2\plus{}b^2\plus{}c^2)^3.$
Hence, it's remains to prove that $ (a^2\plus{}b^2\plus{}c^2)^3\geq\sum_{cyc}a^3(a^3\plus{}(b\plus{}c)^3),$ which is obvious.

[quote="Allnames"]No you cant say that,my friend,How can you do with this problem
Let $ a,b,c \ge\ \frac {3}{4}, \sum a \equal{} 1$Prove $ \sum \frac {a}{1 \plus{} a^2}\le \frac {9}{10}$
Here is Poland 1991[/quote]
Sorry my friend but my ideal 's right . I say that because I believe problem need prove which 's allways true  :)

And with your problem you should check carefully

Because with $ a,b,c \ge\ \frac {3}{4} \rightarrow \sum a \ge\ \frac {9}{4} > 1$ ?????  :lol:

If don't have$ a,b,c \ge\ \frac {3}{4}$ I think it completely similar

With$ f(x)\equal{}\frac {x}{1 \plus{} x^2}$ we easy prove $ f"(x) \le\ 0$ We have done! :lol:

[quote="Allnames"]No you cant say that,my friend,How can you do with this problem
Let $ a,b,c\ge \frac {3}{4}, \sum a \equal{} 1$Prove $ \sum \frac {a}{1 \plus{} a^2}\le \frac {9}{10}$
Here is Poland 1991[/quote]
$ \frac{a}{a^2\plus{}1}\le \frac{18}{25}a\plus{}\frac{3}{50}$

[quote="onlylove_math"][/quote][quote="Marius Mainea"][quote="onlylove_math"][quote="Marius Mainea"]If $ a,b,c > 0$ then
\[ \sqrt {\frac {a^3}{a^3 + (b + c)^3}} + \sqrt {\frac {b^3}{b^3 + (c + a)^3}} + \sqrt {\frac {c^3}{c^3 + (a + b)^3}}\ge 1
\]
[/quote]
we easy prove $ f"(t) \ge\ 0$
[/quote]

  Why$ f^{\prime\prime}(x)\ge 0$?[/quote][quote="onlylove_math"]
I sure $ f"(x)\ge\ 0$ . If not $ f"(x) \le\ 0 \rightarrow$your problem 's wrong  :lol:[/quote]
Your idea is not correct, onlylove_math. ;)
For example, the broaden of Thien's example is
For all $ a,b,c \ in R$ such that $ a+b+c=1$. Then
\[ \sum \frac{a}{a^2+1} \le \frac{9}{10}\]
This inequality is true and I have posted it once time in the forum, you can search it. :)
But the function $ f(x) =\frac{x}{x^2+1}$ is not always convex (or concave). You can check it ;)

[quote="can_hang2007"][/quote][quote="onlylove_math"][/quote][quote="Marius Mainea"][quote="onlylove_math"][quote="Marius Mainea"]If $ a,b,c > 0$ then
\[ \sqrt {\frac {a^3}{a^3 + (b + c)^3}} + \sqrt {\frac {b^3}{b^3 + (c + a)^3}} + \sqrt {\frac {c^3}{c^3 + (a + b)^3}}\ge 1
\]
[/quote]
we easy prove $ f"(t) \ge\ 0$
[/quote]

  Why$ f^{\prime\prime}(x)\ge 0$?[/quote][quote="onlylove_math"]
I sure $ f"(x)\ge\ 0$ . If not $ f"(x) \le\ 0 \rightarrow$your problem 's wrong  :lol:[/quote][quote="can_hang2007"]
Your idea is not correct, onlylove_math. ;)
For example, the broaden of Thien's example is
For all $ a,b,c \ in R$ such that $ a + b + c = 1$. Then
\[ \sum \frac {a}{a^2 + 1} \le \frac {9}{10}
\]
This inequality is true and I have posted it once time in the forum, you can search it. :)
But the function $ f(x) = \frac {x}{x^2 + 1}$ is not always convex (or concave). You can check it ;)[/quote]
Thank can_hang2007 & Allnames! I have mistaked  :D .But I believe my proof for Marius Mainea 's problem is true  :lol:
And with Poland 1991 i think it 'll be solve by pqr or CHY  :lol:

[quote="onlylove_math"][/quote][quote="can_hang2007"][/quote][quote="onlylove_math"][/quote][quote="Marius Mainea"][quote="onlylove_math"][quote="Marius Mainea"]If $ a,b,c > 0$ then
\[ \sqrt {\frac {a^3}{a^3 + (b + c)^3}} + \sqrt {\frac {b^3}{b^3 + (c + a)^3}} + \sqrt {\frac {c^3}{c^3 + (a + b)^3}}\ge 1
\]
[/quote]
we easy prove $ f"(t) \ge\ 0$
[/quote]

  Why$ f^{\prime\prime}(x)\ge 0$?[/quote][quote="onlylove_math"]
I sure $ f"(x)\ge\ 0$ . If not $ f"(x) \le\ 0 \rightarrow$your problem 's wrong  :lol:[/quote][quote="can_hang2007"]
Your idea is not correct, onlylove_math. ;)
For example, the broaden of Thien's example is
For all $ a,b,c \ in R$ such that $ a + b + c = 1$. Then
\[ \sum \frac {a}{a^2 + 1} \le \frac {9}{10}
\]
This inequality is true and I have posted it once time in the forum, you can search it. :)
But the function $ f(x) = \frac {x}{x^2 + 1}$ is not always convex (or concave). You can check it ;)[/quote][quote="onlylove_math"]
Thank can_hang2007 & Allnames! I have mistaked  :D .But I believe my proof for Marius Mainea 's problem is true  :lol:
And with Poland 1991 i think it 'll be solve by pqr or CHY  :lol:[/quote]
[quote="Marius Mainea"]If $ a,b,c > 0$ then
\[ \sqrt {\frac {a^3}{a^3 + (b + c)^3}} + \sqrt {\frac {b^3}{b^3 + (c + a)^3}} + \sqrt {\frac {c^3}{c^3 + (a + b)^3}}\ge 1
\]
[/quote]
The idea of proving
\[ \sqrt{\frac{a^3}{a^3+(b+c)^3}} \ge \frac{a^2}{a^2+b^2+c^2}\]
is nice, but I dont like it.
Here is another solution by Holder's Inequality
By Holder's Inequality, we have
\[ LHS^2 \cdot (\sum a^3(a^3+(b+c)^3)) \ge (\sum a^2)^3\]
It suffices to prove that
\[ (\sum a^2)^3 \ge \sum a^6+\sum a^3(b+c)^3\]
\[ Leftrightarrow 3\sum a^2b^2(a^2+b^2)+ 6a^2b^2c^2 \ge 2\sum a^3b^3 +3abc\sum ab(a+b)\]
which is true because
\[ 3\sum a^2b^2(a^2+b^2) \ge 6\sum a^3b^3\] (AM-GM Inequality)
and by Schur's Inequality, we have
\[ 4\sum a^3b^3 +6a^2b^2c^2 = 4(\sum a^3b^3 +3a^2b^2c^2)-6a^2b^2c^2 \ge 4abc\sum ab(a+b) -6a^2b^2c^2 \ge 3abc\sum ab(a+b)\]
We have done.
@onlylove_math: my technique is CYH not CHY. CHY does not have any meanings.

A nice proof with CBS is [url=http://www.mathlinks.ro/viewtopic.php?search_id=573793642&t=206757]here[/url]

[quote="Allnames"]No you cant say that,my friend,How can you do with this problem
Let $ a,b,c\ge \frac {3}{4}, \sum a \equal{} 1$Prove $ \sum \frac{a}{1 \plus{} a^2}\le \frac {9}{10}$
Here is Poland 1991[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=397704&hilit=Poland+inequality
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=169923&hilit=Poland+inequality
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=54&t=180&hilit=Poland+inequality

Here is another solution by Holder's Inequality
By Holder's Inequality, we have
\[ LHS^2 \cdot (\sum a^3(a^3+(b+c)^3)) \ge (\sum a^2)^3\]
It suffices to prove that
\[ (\sum a^2)^3 \ge \sum a^6+\sum a^3(b+c)^3\]
\[ \Leftrightarrow 3\sum a^2b^2(a^2+b^2)+ 6a^2b^2c^2 \ge 2\sum a^3b^3 +3abc\sum ab(a+b)\]
which is true because
\[ 3\sum a^2b^2(a^2+b^2) \ge 6\sum a^3b^3\] (AM-GM Inequality)
and by Schur's Inequality, we have
\[ 4\sum a^3b^3 +6a^2b^2c^2 = 4(\sum a^3b^3 +3a^2b^2c^2)-6a^2b^2c^2 \ge 4abc\sum ab(a+b) -6a^2b^2c^2 \ge 3abc\sum ab(a+b)\]

Very nice and interesting sol.  :roll:

[quote=sqing][quote=Marius Mainea]If $ a,b,c>0$ then \[ \sqrt{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}}\plus{}\sqrt{\frac{b^3}{b^3\plus{}(c\plus{}a)^3}}\plus{}\sqrt{\frac{c^3}{c^3\plus{}(a\plus{}b)^3}}\ge 1\][/quote]
If $ a,b,c>0 ,n\geq 2$ then \[ \sqrt{2}\geq \sqrt{\frac{a^n}{a^n\plus{}(b\plus{}c)^n}}\plus{}\sqrt{\frac{b^n}{b^n\plus{}(c\plus{}a)^n}}\plus{}\sqrt{\frac{c^n}{c^n\plus{}(a\plus{}b)^n}}\ge 1\][/quote]

$n = 2$，$$\sum {\sqrt {\frac{{{a^2}}}{{{a^2} + {{\left( {b + c} \right)}^2}}}} }  > \sum {\sqrt {\frac{{{a^2}}}{{{a^2} + 2{{\left( {b + c} \right)}^2}}}} }  \ge 1$$


lat a,b are positive integers and p is prime number such that $(a,p)=(b,p)=1$.Prove that for all $k,l \in N^{*}$ there exist $n \in N^{*}$ such that: $n^{l}a^n-b \equiv 0 (mod p^{k})$ :)  :)
Sorry for my mistake

What is $p$? Is it given or does it has to work for all $p$?

Pierre.

Sorry to you.I've edited it  :blush:  :blush:


abstract algebra
superior algebra
superior algebra unsolved

Let $ R$ be a ring.  Show that a free module is isomorphic to $ \oplus_{I} R,$ where $ I$ is an index set.

The definition I am using is:

A module $ F$ is free if there exists a set of elements $ e_i$, $ i \in I$ such that every element $ f \in F$ can be uniquely be
written as $ f \equal{} a_1e_1\plus{}....\plus{}a_ne_n$
and the sum is finite.

It should be $ f \equal{} a_1e_1 \oplus \cdots \oplus a_ne_n.$

Consider the map $ \bigoplus_{i\in I}R\to F$ given by $ (a_1,a_2,\ldots,a_n,0\ldots,0,0,\ldots)\to a_1e_1 \plus{} \ldots \plus{}a_ne_n$. What you write in the first post implies that the map is surjective and injective.

How can I show that the map is a homomorphism, and I cant see that it is surjective and injective. Please help. Thanks

Oh I see it is actually clear that the map is bijective since every element $ f$ can be uniquely written as
$ f \equal{} a_1e_1\plus{}...\plus{}a_ne_n$.

But how can I show that it is a homomorphism?

[quote="peteryellow"]But how can I show that it is a homomorphism?[/quote]
There are two axioms for an $ R$-module homomorphism, and both are trivial to verify. (as long as you remember that the action is componentwise on $ \bigoplus_{i\in I} R$)

It is trivial that the map is a homomorphism:

Is the following true?

$ \phi(a_1e_1,....,a_ne_n,0,0,...)\phi(b_1e_1,....b_ne_n,0,....)
 \equal{} (a_1e_1\plus{}......a_ne_n)(b_1e_1\plus{}....\plus{}b_ne_n) \equal{} \phi(a_1e_1,...a_ne_n,0,0,...b_1e_1,...b_ne_n,0,0,......)$

Is it true? Thanks for your help. :)

What you've written doesn't make sense for a number of reasons.  First of all, $ e_1$ is not an element of $ R$, so your expression involving $ \phi$ is strange.

More importantly, you are trying to multiply two elements of $ F$, which doesn't make any sense, as $ F$ is not a ring, it is an $ R$-module.  What you need to show is that $ \phi(X \plus{} Y) \equal{} \phi(X) \plus{} \phi(Y)$, and $ \phi(rX) \equal{} r \phi(X)$, for all $ X$ and $ Y$ in $ \bigoplus_{i\in I} R$.  To do this you will need to look closely at the definition of $ \phi$, and the definitions of the $ R$-module structures on $ F$ and $ \bigoplus_{i\in I} R$.

Also note that the notation $ (a_1,a_2,\ldots,a_n,0\ldots,0,0,\ldots)\to a_1e_1 \plus{} \ldots \plus{} a_ne_n$ is a little sloppy, as $ I$ is not necessarily a countable set.  But I would address the above concerns first.

[quote="peteryellow"]CAn you please shoe that $ \phi$ is an isomorphism, what about that $ \phi$  is bijective does it follow from the definition as I said, or is it also wrong?[/quote]

The simplest way to show that $ \phi$ is bijective is to construct an inverse map $ \psi: F \to \bigoplus_{i\in I} R$.

As a sample, let's go through the whole proof when $ |I| = \{1\}$.

We define $ \phi : R \to F$ by $ \phi(r) = re_1$ and $ \psi : F \to R$ by $ \psi(re_1) = r$.  We have to check that $ \psi$ is well-defined.  But if $ re_1 = r' e_1$ then $ r = r'$ by your definition of a free module, so $ \psi(re_1) = r = r' = \psi(r'e_1)$.

We show that $ \phi$ is a homomorphism.  If $ r, r'\in R$, then $ \phi(r + r') = re_1 + r'e_1 = (r + r')e_1 = \phi(r) + \phi(r')$, verifying the first axiom for $ R$-module homomorphisms.

Also, $ \phi(r' \cdot r) = \phi(r' r) = (r' r) e_1 = r' \cdot (r e_1)$. (here we have used the $ R$-module axioms for $ R$ in the first step, and for $ F$ in the last step)  This is $ r' \cdot \phi(r)$, so we have verified the second axiom for $ R$-module homomorphisms.

Now, we check that $ \psi$ is a two-sided inverse for $ \phi$.  $ (\psi \circ \phi)(r) = \psi(r e_1) = r$, and $ (\phi \circ \psi)(r e_1) = \phi(r) = r e_1$, so $ \psi \circ \phi = id$ and $ \phi \circ \psi = id$.  Hence $ \phi$ is a bijective homomorphism, so it is an isomorphism.

Make sure that you understand every single step of this proof before thinking about direct sums!  As you get into more advanced algebraic structures (modules vs. rings, infinite direct sums vs. finite direct sums, etc.) it gets more and more important to be very careful and rigorous.  I get the sense that you have entered into this problem without being 100% sure of the definitions you are using.  Anyway, I'll see about posting a solution to the complete problem later, as instruction.

Thanks for your help, it will be nice if you can post a solution to the problem.


function
integration
calculus
derivative
topology
real analysis
real analysis unsolved

Let $f\in L^{1}(R)$ and define Hardy's associated function by $M_{f}(x)=\sup_{y>0}{\frac{1}{2y}\int_{x-y}^{x+y}{|f(t)|dt}}$. Prove that for some constant $C>0$ and all $A>0$ the set of those $x$ with $M_{f}(x)>A$ has measure at most $\frac{C}{A}\parallel f\parallel _{1}$.

Harazi, here http://www.mathlinks.ro/Forum/viewtopic.php?t=84774&highlight=function+hardy
you may find such a constant $C$ and a reference to the proof as well.

An intuition towards the proof:

For each point at which the maximal function is $>A,$ you have an interval $I$ such that the average of $|f|$ on $I$ is greater than $A.$

Suppose all those intervals were disjoint. Then we'd have

$\sum|I|=\frac1A\sum A|I|\le \frac1A\sum\int_{I}|f|\le \frac1A\int|f|.$

Now, the intervals aren't disjoint, so we can't to exactly that. But it gives us something to shoot for. Can we get disjoint intervals of this type to cover at least some fixed fraction of the set $\{M_{f}>A\}?$ If we could, we could use the idea above.

That's a covering lemma, and one that's nowhere near as fancy as the Vitali lemma. So our proof will ultimately rest on some kind of covering lemma (and that in turn means that there's something geometric in this theorem, something that depends on us working in $\mathbb{R}^{n}$ and not some arbitrary metric/measure space.)

In turn, if you ask me to prove Lebesgue's theorem on the differentiation of the integral, my preferred proof will go through this, the Hardy-Littlewood maximal theorem.

notice that one can also take $M_{f}(x) = \sup_{I_{x}}\int_{I_{x}}f(t)dt$ where the supremum is over all the compact intervals that contain $x$.

[quote="Kent Merryfield"]there's something geometric in this theorem, something that depends on us working in $\mathbb{R}^{n}$ and not some arbitrary metric/measure space.[/quote]
I think that the doubling condition -- $\mu(B(x,2r))\le C\mu(B(x,r))$ with $C$ independent of $x$ and $r$ -- would make this work in a metric measure space $(X,\mu)$. It is a geometric condition, of course, - for example, it implies that the Hausdorff dimension of $X$ is finite.

Yes, that doubling condition is sufficient, and the constant in the Hardy-Littlewood maximal theorem would depend only on that constant $C.$


Fix $b>1$. For real x, define $b^{x}= \sup B(x)$ where $B(x) = \{ b^{t}: t \in \mathbb{Q}, t \leq x \}$. Prove that $b^{x+y}= b^{x}b^{y}$ for all real $x$ and $y$.

Note: It was first proven for the case where $x,y \in \mathbb{Q}$.

I'm going to take a stab at it..

If the theorem was proven for rationals then you should think about the following simpler problem:

If A and B are bounded sets of positive real numbers and C is the set of all ab where a is in A and b is in B then sup C = sup A * sup B.

This seems to be the solution.

Is it true that $B(x+y) = \{ b^{s}b^{t}: s,t \in \mathbb{Q}, s \leq x, t \leq y \}$? In other words, if $q \in \mathbb{Q}$ and $q \leq x+y$, can $q$ always be written as a sum $s+t$ where $s,t \in \mathbb{Q}, s \leq x, t \leq y$?

I think we need the above to be true for the simpler problem you proposed to indeed be the solution. That's my only worry.

Yes. Choose any rational $s$ strictly between $q-y$ and $x$.


trigonometry geometry circumcircle inradius

Prove that in triangle $ABC$, $\cos A+\cos B+\cos C = \frac{R+r}{R}$ where $R$ is the circumradius and $r$ the inradius of the triangle.


function
limit
real analysis
real analysis unsolved

[quote="harazi"]Let $f: [0,1]\to R$ be an increasing function and define $g(x)=\lim\sup_{h\to 0}{\frac{f(x+h)-f(x)}{h}}$. Prove that the outer measure of the set of $x$ such that $g(x)>1$ is smaller than or equal to $f(1)-f(0)$.[/quote]

Let $m^{*}$ be the outer measure of the set $S = \{ x : g(x) > 1\}$. Then for any point $x \in S$ and arbitrary small $h$, there exists an interval $[x+h,x]$ such that $f(x+h)-f(x) > 1$. These intervals compose a Vitali covering of $[0,1]$, thus we can use Vitali's theorem to conclude that there exists a finite number of disjoint intervals of this cover $[x_{n}+h_{n}, x_{n}]$, $n=1,...,N$ such that the measure $m(\bigcup_{n=1}^{N}[x_{n}+h_{n}, x_{n}] ) > m^{*}-\epsilon$ for arbitrary $\epsilon > 0$. Thus we have

$f(1)-f(0) \geq \sum_{n=1}^{N}(f(x_{n}+h_{n})-f(x_{n})) > \sum_{n=1}^{N}h_{n}> m^{*}-\epsilon$, which in turn implies that $m^{*}\leq f(1)-f(0)$.


function
number theory unsolved
number theory

Let $k$ be a positive integer and $N_k$ be the number of sequence of length $2001$ , all member of which are elements of the set ${\{0,1,2,...,2k+1}\}$, and the number of zeroes among these is odd. Find the greatest power of $2$ which divides $N_k$. ;)

Let $f(x)$ be the generating functions of the numbers of sequences with a given number of zeroes: that is, the coefficient $f_r$ of $x^r$ in $f(x)$ is the number of sequences with exactly $r$ zeroes.  The question is asking for $f_1 + f_3 + \cdots + f_{2001}$.  We'll get this as $\frac12(f(1) - f(-1))$.

Now $f_r = \binom{2001}{r} (2k)^{2001-r}$ and so $f(x) = \sum_{r=0}^{2001} f_r x^r = \sum_{r=0}^{2001} \binom{2001}{r} (2k)^{2001-r} x^r$ and this is just $(x+2k)^{2001}$.  The sum wanted is $\frac12(f(1) + f(-1))$ which is $\frac12(2k+1)^{2001}-(2k-1)^{2001})$.  One of $2k\pm1$ is divisible by 4 and the other is exactly divisible by 2 and no higher power.  So the power of 2 in $N_k$ is just $2^{2000}$.


Let be $ a,b,c,d,e,f\in \mathbb{R}_\plus{}$ such that $ d^2\plus{}e^2\plus{}f^2\equal{}3$ . Prove that :

$ \frac{a}{d}\plus{}\frac{b}{e}\plus{}\frac{c}{f}\ge \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$

[quote="alex2008"]Let be $ a,b,c,d,e,f\in \mathbb{R}_ \plus{}$ such that $ d^2 \plus{} e^2 \plus{} f^2 \equal{} 3$ . Prove that :

$ \frac {a}{d} \plus{} \frac {b}{e} \plus{} \frac {c}{f}\ge \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}$[/quote]
Easy Holder.  :wink:

suppose that,$ a \geq b \geq c$ and$ f\geq e \geq d$ then by Chebychev,
$ S\equal{}\frac{a}{d}\plus{}\frac{b}{e}\plus{}\frac{c}{f} \geq \frac{1}{3}(a\plus{}b\plus{}c)\left(\frac{1}{e}\plus{}\frac{1}{d}\plus{}\frac{1}{f}\right)$
By Holder
$ \left(\frac{1}{e}\plus{}\frac{1}{d}\plus{}\frac{1}{f}\right)^2(e^2\plus{}d^2\plus{}f^2)\geq 27$
then, $ \frac{1}{e}\plus{}\frac{1}{d}\plus{}\frac{1}{f} \geq 3$
and it's easy to prove that $ a\plus{}b\plus{}c\geq \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$
then $ S\geq \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$

[quote="peine"]suppose that,$ a \geq b \geq c$ and$ f\geq e \geq d$ then by Chebychev[/quote]
This step is wrong :wink:
Here is my proof:
$ LHS \ge \frac{(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^2}{d\plus{}e\plus{}f}\ge \frac{3(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca})}{\sqrt{3(d^2\plus{}e^2\plus{}f^2}}  \equal{}RHS$
We have Q.E.D.Equality holds when $ a\equal{}b\equal{}c,,,d\equal{}e\equal{}f\equal{}1$

Given $ d$, $ e$ and $ f$ we have to prove an inequality which is homogeneus in $ a$, $ b$ and $ c$ so we may suppose that $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\equal{}3$.
From this it follows that $ \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\ge\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$, with equality only if $ a\equal{}b\equal{}c$.
Also we have that from $ d^2\plus{}e^2\plus{}f^2\equal{}3$ it follows that $ d\plus{}e\plus{}f\le 3$, with equality only if $ d\equal{}e\equal{}f\equal{}1$.

Now
$ 3\plus{}\dfrac{a}{d}\plus{}\dfrac{b}{e}\plus{}\dfrac{c}{f}\ge\dfrac{a}{d}\plus{}d\plus{}\dfrac{b}{e}\plus{}e\plus{}\dfrac{c}{f}\plus{}f\ge$

$ 2\sqrt{a}\plus{}2\sqrt{b}\plus{}2\sqrt{c}\equal{}3\plus{}\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\ge 3\plus{}\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$

Then $ \dfrac{a}{d}\plus{}\dfrac{b}{e}\plus{}\dfrac{c}{f}\ge\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$ with equality only if all variables are $ 1$.

:)


(1/3)<a<b<12 then find the integer values of 2/a+3/b

Looking through your posts, you are continually posting the same problems in multiple forums. Please post your questions once, and once only.


algebra
polynomial
inequalities
algebra proposed

Let $ n$ be a natural number having at least $ k\plus{}1$ different prime factors . Let $ P(x) \in \mathbb{R}[x]$ be a polynomial with its degree not exceeding $ k$ . Prove that if $ P(1),P(2),...,P(n) > 0$ then the following inequality holds for every $ 1 \le i \le n$ :

$ P(1) \plus{} P(2) \plus{} ... \plus{} P(n) > 2P(i)$


Prove that if p,q are prime numbers for which

(p^2 + 7pq + q^2)^(1/2) + (p^2 + 14pq + q^2)^(1/2)
is natural number, therefore p=q

 :mad:


Consider the following inequality: [tex]\frac{(a_1+a_2+\ldots+a_n)^n}{a_1a_2\ldots a_n}+kn^2-n^n\geq k(a_1+a_2+\ldots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_1}+\ldots+\frac{1}{a_n}\right)[/tex]. I believe that the biggest possible value of k is [tex]k=(n-1)^{n-1}[/tex]. This inequality is true for [tex]n=3,4[/tex]. I hope to find an answer for the general case. ;)

Can you explain where that [tex](n-1)^{n-1}[/tex] comes from?  So far I've only deduced that this inequality is a comparison of the relative strengths of AM-GM and Cauchy, the latter winning considerably, as the inequality is equivalent to

[tex]\displaymode{
\frac{\left(\sum a_{i} \right)^n}{\prod a_{i}} - n^n \ge k \cdot \left( \left( \sum a_{i}\right)\left( \sum \frac{1}{a_{i}} \right) - n^2 \right)
}[/tex]

Does it help to set any of [tex]\displaymode{\sum a_{i}, \prod a_{i}, \sum \frac{1}{a_{i}}}[/tex] equal to 1?

[quote="pcalin"]Consider the following inequality: $\displaystyle \frac{(a_1+a_2+\ldots+a_n)^n}{a_1a_2\ldots a_n}+kn^2-n^n\geq k(a_1+a_2+\ldots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_1}+\ldots+\frac{1}{a_n}\right)$. I believe that the biggest possible value of $k$ is $k=(n-1)^{n-1}$. This inequality is true for $n=3, 4$. I hope to find an answer for the general case. ;)[/quote]

:)


Prove that for any positive integer $n$, \[\left\lfloor \frac{n+1}{2}\right\rfloor+\left\lfloor \frac{n+2}{4}\right\rfloor+\left\lfloor \frac{n+4}{8}\right\rfloor+\left\lfloor \frac{n+8}{16}\right\rfloor+\cdots = n.\]

Interesting.
Notice that $\left \lfloor {\frac{n+2^k}{2^{k+1}}}\right \rfloor=\left \lfloor {\frac{n}{2^k}}\right\rfloor -\left \lfloor \frac{n}{2^{k+1}}\right \rfloor$

I want to know if there is any other solution than the one given by Hawk Tiger!!!!!

Here is a rather contrived proof (but still a proof nonetheless):

Consider the binary representation of $ n$: $ n\equal{}2^{u_0}\plus{}2^{u_1}\plus{}\cdots\plus{}2^{u_t}$.  Then it is easy to see that

$ \left\lfloor\frac{n\plus{}2^{k\minus{}1}}{2^k}\right\rfloor\equal{}2^{u_i\minus{}k}\plus{}2^{u_{i\plus{}1}\minus{}k}\plus{}\cdots\plus{}2^{u_t\minus{}k} \plus{}(0)\text{ if }u_{i\minus{}1}<k\minus{}1\text{ and }\plus{}(1)\text{ if }u_{i\minus{}1}\equal{}k\minus{}1$

for $ u_i\ge k$.  Thus, the sum in question is equivalent to

$ (2^{u_0\minus{}1}\plus{}2^{u_0\minus{}2}\plus{}\cdots\plus{}1)\plus{}(2^{u_1\minus{}1}\plus{}2^{u_1\minus{}2}\plus{}\cdots\plus{}1)\plus{}\cdots\plus{}(2^{u_t\minus{}1}\plus{}2^{u_t\minus{}2}\plus{}\cdots\plus{}1)\plus{}t\plus{}1$

$ \equal{}2^{u_0}\minus{}1\plus{}2^{u_1}\minus{}1\plus{}\cdots\plus{}2^{u_t}\minus{}1\plus{}t\plus{}1$

$ \equal{}n$

I am not sure but can someone (maybe peter) check out if this problem appeared in the 1968 IMO?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213453&search_id=335968213]here[/url].  The second solution is essentially Hawk Tiger's and the first is essentially the cosinator's.

[hide="Hint"] use Induction in the intervals $[2^{n},2^{n+1})$[/hide] and then represent $n$ as $n=2^{n} +m$ so that m is less than $2^{n}$.

Here's a way to see why the identity is true and also why Hawk Tiger's identity is true. Consider all the numbers less than or equal to $n$. We count which of these have $v_2$ equal to 0, 1, 2, and so on. Let $k$ be fixed;we are counting the number of nonnegative $m$ such that $2^{k+1}m + 2^k \le n$. This is simply $\left\lfloor\dfrac{n+2^k}{2^{k+1}}\right\rfloor$. Summing over all $k$, we get the desired result. For Hawk Tiger's identity, note that the first term is just the number of multiples of $2^k$ and we subtract the second term which is the number of multiples of $2^{k+1}$


algorithm Gauss calculus integration logarithms

Find the first instance where there are 10 natural numbers in a row that are all composite.

Is there a nice way to do this, or is it just guess and check?

considering there really isn't a nice formula for primes, i would say your best bet is guess and check

That's right. Did you use guess and check?

There's so specific algorithm for primes, but I do remember a problem on an AoPS competition asking for the first decade where all the years were composite.

So trial and error is the only way.

What about 114, 115, 116, 117, 118, 119, 120, 121, 122, 123? I guess you could solve it by finding the first instance when $\pi(n)= \pi(n+9)$ Where n is composite. But guess and check is probably best.

Oh, right...I guess that does work...I am such an idiot...

I thought that they had to be in the same decade as someone else thought...

GOOD JOB.

Oh yeah. That is right :rotfl:  When I did the problem, I made the same mistake as mathgeniusln^x did, even thought I wrote it :rotfl:

Gosh, this is one of the questions that I do not like...doable, but you still make a mistake....

well gauss derived a formula that approximates the distribution of primes right. i think it was something like $\pi(x)=$integral$( \frac{dx}{\ln x})$. does that help much?

I think thats too inaccurate for small values of x.

I remember I read in this book some equation to find prime numbers but now I forgot...

You could write a lot of numbers and use the sieve. Or you could look at a list of the first couple hundred numbers if you are too lazy. This method is quicker than it seems.


geometry
parallelogram
MATHCOUNTS
conics
ellipse
absolute value

Find the equation whose graph is as shown at the right. (MATHCOUNTS 1989)

(The graph is a parallelogram with x-intercepts $\pm 6$ and y-intercepts $\pm 4$.)

In the marathon, an answer but no solution was posted.

[hide="hint"]absolute value[/hide]

[hide="hint 2"]contains both $|x|$ and $|y|$ because it's a quadrilateral[/hide]

Off topic but what's the equation of a square to begin with?
This should just be an equations topic.

[hide="236!"]The equation of a square would be $|x|+|y|=k$ where $\pm k$ are the x- and y-intercepts and it's centered at $(0,0)$.[/hide]

[quote="anirudh"]Can you just roughly redraw the graph on Paint or Photoshop? I think I need the graph.[/quote]

I think it's clear enough...you can draw your own graph. Draw a point at $(0,4), (0,-4), (6,0), $ and $(-6,0)$. Connect the points and you get your parallelogram.  ;)

it doesn't look right.

[img]http://4aunlimited.co.nr/AoPS/parallelogram.PNG[/img]

The equation of a parallelogram with diagonals on the axes is very much like that of an ellipse :D

[hide="my answer"]
$\frac {|x|}{6} + \frac {|y|}{4} = 1$
[/hide]

Correct, but can you explain your solution please? In the AoPS book, it says "trial and error yields $4|x|+6|y|=24$, or $2|x|+3|y|=12$." or something like that. :?

[quote="Hamster1800"]I don't want to ruin it for other so I'll hide it.
[hide="my solution"]
The form of my equation came from the form of the equation for an ellipse with center (0,0):

$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$

where a is half of the length of the horizontal axis and b is half the length of the vertical axis.

The way this works, is it makes the numerator positive, so each of the four quadrants have congruent figures. So, by morphing that into straight lines for the four quadrants (using absolute values instead of squaring), we have the general equation for a parallelogram with center (0,0) and diagonals that are horizontal and vertical:

$\frac {|x|}{a} + \frac {|y|}{b} = 1$
(a and b are the same things)

Why are we not squaring the denominators? Because the numerators have not been squared so we don't need to. If you did square it, it would create a larger parallelogram than we want.

Plugging in a and b give my solution:

$\boxed{\frac{|x|}{6} + \frac{|y|}{4} = 1}$

[/hide][/quote]

Very nice, thankyou. Much better than the AoPS book solution at least. :D


Suppose we have two events: A and B. P(A)>0 and P(B)>0.

If P(A|B) > P(B) then show that P(B|A) > P(A)

I've tried this problem by replacing P(A|B) to be P(AB)/P(B) and using Bayes' Rule to change the forms around. Each time, however, I wound up back to where I started.

Can anyone figure this out?

Thanks!
-PD

Forget Bayes rule; keep it simple.

You have $ P(A\mid B)>P(A).$

As you said, we can use $ P(A\mid B)\equal{}\frac{P(A\cap B)}{P(B)}.$ Put that in, and multiply by $ P(B)$ to get the symmetric form

$ P(A\cap B)>P(A)P(B).$

I'll let you finish it from there.

Wait! I don't have P(A|B) > P(A)!

(1) I start off with: P(A|B) > P(B)

(2) From there, I am trying to prove that P(B|A) > P(A). (2)

If we multiply both sides of (1) with P(B), wouldn't we get P(AB) > P(B)*P(B)? How would that help?

Thanks! I'll take the advice of K.I.S.S. (Keep it short and simple)
-PD

It doesn't help- this is false as stated.

Example: $ P(A)\equal{}\frac56,P(B)\equal{}\frac12,P(A\cap B)\equal{}\frac13$; we get $ P(A|B)\equal{}\frac23>\frac12$ and $ P(B|A)\equal{}\frac25<\frac56$.

Just to let everybody know, this problem no longer needs to be solved. My professor made a typo on this problem!

-PD


If x, y and z are positive real numbers such that x+y+z=xyz
prove the inequality:
(2x+2y+2z)( (x+y)ˇ2/z +(y+z)ˇ2/x + (z+x)ˇ2/y )>=sqrt27(x+y)(y+z)(z+x)

[quote="delegat"]If $x,$ $y$ and $z$ are positive real numbers such that $x+y+z=xyz$
prove the inequality:
$(2x+2y+2z)\left(\frac{(x+y)^{2}}{z}+\frac{(y+z)^{2}}{x}+\frac{(z+x)^{2}}{y}\right)\geq\sqrt{27}(x+y)(y+z)(z+x)$[/quote]
Let $x+y+z=3u,$ $xy+xz+yz=3v^{2}$ and $xyz=w^{3}.$ Then our inequality equivalent to
$2\cdot\sum_{cyc}(x^{3}y+x^{3}z+2x^{2}y^{2})\geq\sqrt{27}\cdot\sum_{cyc}(x^{2}y+x^{2}z+\frac{2}{3}\cdot xyz)\Leftrightarrow$
$\Leftrightarrow2(x^{2}+y^{2}+z^{2})(xy+xz+yz)+4(xy+xz+yz)^{2}-10(x+y+z)xyz\geq$
$\geq\sqrt{27}((x+y+z)(xy+xz+yz)-xyz)\cdot\sqrt{\frac{xyz}{x+y+z}}\Leftrightarrow$
$\Leftrightarrow18u^{2}v^{2}-10uw^{3}\geq(9uv^{2}-w^{3})\cdot\sqrt{\frac{w^{3}}{u}}\Leftrightarrow$
$\Leftrightarrow(18u^{2}-9\sqrt{uw^{3}})v^{2}-10uw^{3}+w^{3}\cdot\sqrt{\frac{w^{3}}{u}}\geq0.$
But $18u^{2}-9\sqrt{uw^{3}}\geq0$ and $v^{2}\geq\sqrt{uw^{3}}.$ Hence remain to prove that
$(18u^{2}-9\sqrt{uw^{3}})\cdot\sqrt{uw^{3}}-10uw^{3}+w^{3}\cdot\sqrt{\frac{w^{3}}{u}}\geq0\Leftrightarrow$
$\Leftrightarrow18t^{2}-19t+1\geq0,$ where $t=\sqrt{\frac{u^{3}}{w^{3}}}\geq1.$
But $18t^{2}-19t+1=(t-1)(18t-1)\geq0,$ which trivial.
The following inequality true too:
Let $x,$ $y$ and $z$ are positive real numbers such that $x+y+z=xyz.$ Prove that
$(x+y+z)\sum_{cyc}\frac{(x+y)^{2}}{z}+4x^{2}y^{2}z^{2}\geq\sqrt{27}(x+y)(y+z)(z+x).$


Could you tell me when will be the date of romaninian national mathematics olympiad final round this year ?

Because I am interested about the problems

thx


五位數中，至少有一個位數是3，且此數為3的倍數者，總共有多
少個？

[color=brown]Translate:
[/color]
In all 5-digits integers .how many possible numbers which are multiple of 3 and contains at least one digit of 3.

[hide="ans."]
is 12504
[/hide]

Taking the digits $ \bmod3$, we find that
$ 0,3,6,9\equiv0\pmod3$
$ 1,4,7\equiv1\pmod3$
$ 2,5,8\equiv2\pmod3$

Since there must be at least one three, there must be at least one "zero" in each case.

The possible five-digit numbers are of the forms:
We use complementary probability for the first four cases (i.e. find the numbers divisible by three without restrictions, and subtract the numbers without any threes divisible by three).

$ 0,0,0,0,0$: $ 3\cdot4^4\minus{}2\cdot3^4\equal{}606$ numbers.
$ 2,1,0,0,0$: $ \left(\frac{5!}{3!}\cdot3\cdot3\cdot4^3\minus{}\frac{4!}{2!}\cdot3\cdot3\cdot4^2\right)\minus{}\left(\frac{5!}{3!}\cdot3\cdot3\cdot3^3\minus{}\frac{4!}{2!}\cdot3\cdot3\cdot3^2\right)\equal{}5904$ numbers.
$ 1,1,1,0,0$: $ \left(\frac{5!}{3!2!}\cdot3^3\cdot4^2\minus{}\frac{4!}{3!}\cdot3^3\cdot4\right)\minus{}\left(\frac{5!}{3!2!}\cdot3^3\cdot3^2\minus{}\frac{4!}{3!}\cdot3^3\cdot3\right)\equal{}1782$ numbers.
$ 2,2,2,0,0$: $ \left(\frac{5!}{3!2!}\cdot3^3\cdot4^2\minus{}\frac{4!}{3!}\cdot3^3\cdot4\right)\minus{}\left(\frac{5!}{3!2!}\cdot3^3\cdot3^2\minus{}\frac{4!}{3!}\cdot3^3\cdot3\right)\equal{}1782$ numbers.
$ 2,2,1,1,0$: We know that the "zero" must be a $ 3$. Thus, the amount for this case is $ \frac{5!}{2!2!}\cdot3^2\cdot3^2\cdot1\equal{}2430$ numbers.

Adding up, we get $ 606\plus{}5904\plus{}1782\plus{}1782\plus{}2430\equal{}\boxed{12504}$.

 :ewpu:  :wacko:

[quote="math154"]Taking the digits $ \bmod3$, we find that
$ 0,3,6,9\equiv0\pmod3$
$ 1,4,7\equiv1\pmod3$
$ 2,5,8\equiv2\pmod3$

Since there must be at least one three, there must be at least one "zero" in each case.

The possible five-digit numbers are of the forms:
We use complementary probability for the first four cases (i.e. find the numbers divisible by three without restrictions, and subtract the numbers without any threes divisible by three).

$ 0,0,0,0,0$: $ 3\cdot4^4 \minus{} 2\cdot3^4 \equal{} 606$ numbers.
$ 2,1,0,0,0$: $ \left(\frac {5!}{3!}\cdot3\cdot3\cdot4^3 \minus{} \frac {4!}{2!}\cdot3\cdot3\cdot4^2\right) \minus{} \left(\frac {5!}{3!}\cdot3\cdot3\cdot3^3 \minus{} \frac {4!}{2!}\cdot3\cdot3\cdot3^2\right) \equal{} 5904$ numbers.
$ 1,1,1,0,0$: $ \left(\frac {5!}{3!2!}\cdot3^3\cdot4^2 \minus{} \frac {4!}{3!}\cdot3^3\cdot4\right) \minus{} \left(\frac {5!}{3!2!}\cdot3^3\cdot3^2 \minus{} \frac {4!}{3!}\cdot3^3\cdot3\right) \equal{} 1782$ numbers.
$ 2,2,2,0,0$: $ \left(\frac {5!}{3!2!}\cdot3^3\cdot4^2 \minus{} \frac {4!}{3!}\cdot3^3\cdot4\right) \minus{} \left(\frac {5!}{3!2!}\cdot3^3\cdot3^2 \minus{} \frac {4!}{3!}\cdot3^3\cdot3\right) \equal{} 1782$ numbers.
$ 2,2,1,1,0$: We know that the "zero" must be a $ 3$. Thus, the amount for this case is $ \frac {5!}{2!2!}\cdot3^2\cdot3^2\cdot1 \equal{} 2430$ numbers.

Adding up, we get $ 606 \plus{} 5904 \plus{} 1782 \plus{} 1782 \plus{} 2430 \equal{} \boxed{12504}$.

 :ewpu:  :wacko:[/quote]

nice work of analysis... :lol:  :roll:


trigonometry
inequalities proposed
inequalities

if x,y<=1 and -1<=x,y,prove that

A)
\[ \frac{\sqrt{1-y^2}}{y}+\frac{\sqrt{1-x^2}}{x}<1 \]
                                                                                Luca Victor Iliesiu

______________________________
B) is coming soon(very soon)

A counterexample is $x = y = \frac{1}{2}$.

Well, since we know that $-1 \le x, y \le 1$, we can let them to be cosines. ;)

So take $x = \cos \alpha$ and $y = \cos \beta$. We want to prove that $x\sqrt{1 - y^2} + y\sqrt{1 - x^2} \leq 1$ or $\cos \alpha \sin \beta + \sin \alpha \cos \beta \leq 1$, or $\sin(\alpha + \beta) \leq 1$, which is trivial.


AMC
USA(J)MO
USAMO
inequalities
induction
inequalities proposed

[quote="iura"]Sorry if this problem was before on forum

Given a sequence $(a_n)$ of reals with $a_{i+j} \leq a_i+a_j$ for any $i,j \geq 1$.
Prove that

$a_1+\frac {a_2}2+\cdots+ \frac {a_n}n \geq a_n$[/quote]

when $n=1$,it is quite obvious
now assume $n<k$,it 's right,and we will prove it 's dtill right for $n=k$,
for $n<k$ ,we can get $(k-1)$ ineq,add them

This is the third time this problem appears.

First time USAMO 1981-5
Second time China Math Olympiad 1993-6
Third time APMO 1999-2

I will write more clear explanations.

By induction.
For $n=1$ the statement holds.
Suppose it holds for all values 1, 2, ..., $n-1$. Let's prove it for $n$.
We have the following inequalities: $a_k+a_{n-k}\geq a_n$ for $k=1,2,...,n-1$. Using the induction proposition we conclude that $\frac{a_1}{1}+...+\frac{a_k}{k}+a_{n-k}\geq a_n$, moreover $a_n\geq a_n$. Summing up all these inequalities we obtain $\frac{na_1}{1}+...+\frac{na_{n-1}}{n-1}+a_n\geq na_n$.
That's all.


calculus
integration
function
trigonometry
logarithms
calculus computations

Find the minimum value of the following function $f(x) $ defined at $0<x<\frac{\pi}{2}$.

\[f(x)=\int_0^x \frac{d\theta}{\cos \theta}+\int_x^{\frac{\pi}{2}} \frac{d\theta}{\sin \theta}\]

$\int \sec x \,dx = \int \sec x \frac{\sec x + \tan x}{\sec x + \tan x} \ dx$

$u=\sec x + \tan x\Rightarrow du = (\sec x \tan x +\sec^2 x)$

Thus the integral becomes $\int \frac{du}{u}=\ln|u|=\ln|\sec x + \tan x|$.

$\int \csc x \,dx = \int \csc x \frac{\csc x + \cot x}{\csc x + \cot x} \ dx$

$u=\csc x + \cot x\Rightarrow du = (-\csc x \cot x -\csc^2 x)$

Thus the integral becomes $-\int \frac{du}{u}=-\ln|u|=-\ln|\csc x + \cot x|$.

Using these identities, we see that $f(x) = \ln|\sec x +\tan x| + \ln|\csc x + \cot x|$.

$\frac{d}{dx} f(x) = \frac{\sec x \tan x +\sec^2 x}{\sec x + \tan x} - \frac{\csc x \cot x + \csc^2 x}{\csc x + \cot x } = 0$

Simplifying this gives

$\frac{2(\sin x - \cos x)}{\sin(2x)} = 0 \Rightarrow \sin x = \cos x \Rightarrow x = \frac{\pi}{4}$.

This value is a minimum because $f''\left(\frac{\pi}{4}\right)$ is positive.

And finally, $f\left(\frac{\pi}{4}\right) = \ln\left(\sqrt{2}+1\right)+\ln\left(\sqrt{2}+1\right) = \ln\left(2\sqrt{2}+3\right)$.


kenken puzzles are awesome... they are similar to sudoku, but betterly awesome

nim. its is a highly combinatorial game :)

Shouldn't this thread be in Math and Strategy Games?

Yeah.

Also, the 24 game is pretty fun.  But hard.

[quote="Nick31"]I'm looking for various games involving maths, strategies and whatnot. I know chess has been studied by some of the greats like Euler, De Moivre and so on. I guess Go is the same. What other games to mathematicians play ?

also, what pc versions are the best for chess, go and so on ? Is chessmaster still the best out there ?[/quote]
No definitely not!
Chessmaster is now a complete beginner compared to the best programs.

[b]Rybka 3[/b] is clearly the star of the stars, even being able to give Pawn handicaps to human IMs and still win!! Rybka dominates easily other programs as also of course to all the best top human GMs.

 Other hyper strong programs are Zappa Mexico II, Hiarcs 12, Shredder 11, Fritz 11, Naum 3.1, Deep Sjeng 3, Deep Junior 10.1, Toga II 1.4.1.
 Each of these programs on a common, even obsolete, PC(not to speak about todays QUAD or Octal machines) can beat any day the best of the best human top GMs like Anand or Kramnik or Carlsen etc....


A rectangular block of cheese (16 cm by 12 cm by 7 cm) is covered with wax. If the block is cut into 1 cm cubes, what percent of the cubes will have wax on at least one side? Leave your answer rounded to the nearest tenth.

$ 14*10*5$ cubic cm would have no wax at all. That fraction is $ \frac{(14)(10)(5)}{(16)(12)(7)}\equal{}\frac{25}{48}$

So the fraction of the cubes that have at least 1 side with wax is $ 1\minus{}\frac{25}{48}\equal{}\frac{23}{48}$, which is approximately $ \boxed{47.9}$ percent.

[quote="gauss1181"]$ 14*10*5$ cubic cm would have no wax at all. That fraction is $ \frac {(14)(10)(5)}{(16)(12)(7)} \equal{} \frac {25}{48}$

So the fraction of the cubes that have at least 1 side with wax is $ 1 \minus{} \frac {25}{48} \equal{} \frac {23}{48}$, which is approximately $ \boxed{47.9}$ percent.[/quote]

Hmmm.... I agree with ya


[quote="nateharman1234"][hide] 5 choices for the first, 4 for the second, and 4 again for the third, 5x4x4 = 80 codes[/hide][/quote]
Correct. :)

[hide="solution"]

We shouldn't count them directly! :rotfl:  We can instead notice that thre are 5 choices for the first digit, 4 for the second, and 4 for the third, since it can be the same as the first. We multiply through and get 80.[/hide]

[quote="i_like_pie"]How many different 3-digit security codes are possible using the the digits 1-5, if the second digit can't be the same as the first, and the third digit can't be the same as the second?[/quote]

[hide]The first number could be any number one through five, so there are $5$ different choices. The second can't be the same as the first, to there are $4$ different choices for the second number. The third number cannot be the same as the second but [i]can be the same as the first[/i], so there are $4$ choices for this number.

Thus, the answer is $5*4*4=80$[/hide]

[quote="i_like_pie"]Extension: How many x-digit security codes can be created using the digits 1-5 if adjacent digits can't be the same?[/quote]

[hide]The first digit can be $1$ to $5$, so there are $5$ choices. The second has $5-1=4$ choices. The third has $5-1=4$ choices, and so on. Thus, the answer is:

$5*4*4*4*... = 5(4^{x-1})$[/hide]


If $ (a^{2}\minus{}b^{2})\sin \theta\plus{}2ab\cos \theta\equal{}a^2\plus{}b^2$.Find $ \tan \theta$ in terms of $ a$ and $ b$.

Divide both sides by $ \cos \theta$.

Square both sides (having used $ \sec^{2} \theta \equal{} 1 \plus{} \tan^{2} \theta$).

This gives a quadratic in $ \tan \theta$.

Then apply the quadratic formula.

Hope this is helpful.

it ís so easy
set $ t\equal{}tan(x/2)$ then we have $ sin(x)\equal{}\frac{2t}{1\plus{}t^2}$ and $ cos(x)\equal{}\frac{1\minus{}t^2}{1\plus{}t^2}$
replace them in  problem we have:
\[ (a^2\plus{}b^2)2t\plus{}2ab(1\minus{}t^2)\equal{}(a^2\plus{}b^2)(1\plus{}t^2)\]
\[ \Leftrightarrow t^2(a\plus{}b)^2\minus{}2(a^2\minus{}b^2)t\plus{}(a\minus{}b)^2\equal{}0\]
\[ \Leftrightarrow t\equal{}\frac{(a\minus{}b)}{(a\plus{}b)}\]
We have \[ tan(x)\equal{}\frac{2t}{1\minus{}t^2}\equal{}\frac{a^2\minus{}b^2}{2ab}\]. the end. :rotfl:

Observe that the maximum value of LHS is $ a^{2}\plus{}b^{2}$.Equality occurs when $ \frac{a^2\minus{}b^2}{\sin \theta}\equal{}\frac{2ab}{\cos \theta}$.
So $ tan \theta\equal{}\frac{a^{2}\minus{}b^{2}}{2ab}$

hello, rewriting your equation in the following form
$ \frac {a^2 \minus{} b^2}{a^2 \plus{} b^2}\sin(\theta) \plus{} \frac {2ab}{a^2 \plus{} b^2}\cos(\theta) \equal{} 1$
then there exists an angle $ \phi$ with
$ \sin(\phi) \equal{} \frac {a^2 \minus{} b^2}{a^2 \plus{} b^2}$
and
$ \cos(\phi) \equal{} \frac {2ab}{a^2 \plus{} b^2}$
after a known additional theorem we have
$ \cos(\theta \minus{} \phi) \equal{} 1$ and
$ \tan(\theta) \equal{} \tan(\arccos(1) \plus{}\phi)$.
Sonnhard.

Thank you all,for ur help.
@Raghav-Very short and nice solution!!

[quote="Dr Sonnhard Graubner"]
$ \cos(\theta \plus{} \phi) \equal{} 1$
Sonnhard.[/quote]
Shouldn't it be $ \cos(\theta \minus{} \phi) \equal{} 1$

hello, yes, you have right, i have corrected it. Thank you for your hint.
Sonnhard.


Aqui estan los problemas mas interesantes de la  version de este año! (fue en junio)

[b]Problema 3[/b]
(propuesto por P.  Restrepo)

Sea $d(n)$ la suma de divisores de un numero. Demuestre que $d(n)$ nunca es una potencia de 5 mayor que 1.

[b]Problema 5[/b]
(propuesto por S. Saavedra)

Una Jotax es una figura formada por 3 casillas que corresponden a una $L$ (de 3 casillas para ariiba y 2  de lado) sin la casilla de la "esquina" (la que tiene otras dos casillas adyacentes).

Hallar todas las dimensiones de un tablero para que pueda ser diseccionado por jotax! (cada casillita de las Jotax mide 1*1)

[b]Problema 6[/b]
(propuesto por E. Leon)

Se tiene un plano tringular, formado por vertices de triangulos equilateros adyacentes de lado 1 (algo asi como una colmena) Se pinta un triangulo con vetices en estos puntos sin puntos sobre sus bordes. si el triangulo tiene lado $L$ demuestre que tiene $\frac{L^2-1}{2}$ puntos en su interior.

[quote="Pascual2005"]Aqui estan los problemas mas interesantes de la  version de este año! (fue en junio)

[b]Problema 3[/b]
(propuesto por P.  Restrepo)

Sea $d(n)$ la suma de divisores de un numero. Demuestre que $d(n)$ nunca es una potencia de 5 mayor que 1.
[/quote]
Tomamos un $n > 1$ de la forma : $n = p^i$, $p \in \mathbb{P}$.
Sabemos que la formula para la suma de divisores de un numero es :
$\prod \frac{p^{i+1}-1}{p-1}$ (suponemos : $=5^k$).
Entonces : $\frac{p^{i+1}-1}{p-1}=5^j$.
En modulo 5, tenemos : $p^{i+1}=1$, por eso : $i+1 = 4k$.
Factori[b]z[/b]amos : $(p+1)(p^2+1)(\sum_{m=0}^n (p^4)^m)=5^j$.
Entonces, $p+1$ es una potencia de 5.
Es decir : $p = 5^x - 1$, que esta divisible por 4 : contradiccion! ($p \in \mathbb{P}$).
En conclusion, no hay $n \ne 1$ que verifica la relacion.  :D

Sorry if my spanish is not understandable.  :(
It would be very kind of you if you could correct my mistakes. :)

Oh, and, nice problem. ;)

EDIT: Thanks José, if that's my only mistake, that's cool then. :)

It isn't "factorisamos", it's "factorizamos".


[quote="amcandio"]Express 1 / 2 as the sum of all different and all fractions of the form 1 / (n ^ 2) with n a natural number.  :)[/quotthis seems a bit confusing.can you clarify this, please?

Find a subset $ S$ of $ \mathbb{N}$ such that $ \sum_{n \in S} \frac {1}{n^2} \equal{} \frac {1}{2}$.

I'm assuming the problem is supposed to be what TZF stated.
Motivated by the Euler product identity for $ \zeta(2)$, we can take $ S$ to be the set of integers whose prime factors are only 2 and 3, except for the number 1. The sum for this is
\[ \left(\sum_{i\equal{}0}^\infty \frac1{4^i}\right)\left(\sum_{i\equal{}0}^\infty \frac1{9^i}\right)\minus{}1\equal{}4/3\cdot 9/8\minus{}1\equal{}1/2\]

I think the poster wants a [b]finite[/b] set $ S$  :)


MATHCOUNTS
rotation
geometry
geometric transformation
reflection

An equilateral triangle is placed in its starting position on a table with its starting position on a table with its vertices labeled as shown. On each move, it can either be reflected over a vertical axis through its top vertex (switching the location of the bottom two vertices), or it can be rotated clockwise 120 degrees about its center. How many different sequences of 5 moves will return the triangle to its original starting position?

I am not totally sure about my answer cause I am not as beastly as those other people on the internet, but I got 243. The reason for this is :

Since on any move, you are going to go back to same position. No matter how many times you try, you still end up with the correct position. Therefore, there are 3 choices for the first move (120 degrees: right or left, and flip) and 3 choices for the second and so on... 3^5!!!!!

The possible sequences all involve two reflections and three rotations. We just need to find the number of ways in which we can conduct these five manipulations. This is $ 5!/2!3! \equal{} 10$.

The answer is $ \boxed{10}$

johnathonchou711: Not quite. You can only rotate clockwise. And each rotation is $ 120$ degrees. You can only get back to your original position with three rotations. Likewise, if you reflect, you need two to get back to your original orientation. Thus all orders of five manipulations will have three rotations and two reflections.

Ok, it is 5 after all.

T = Turn and R = Rotate.

RRTTT
TRRTT
TTRRT
TTTRR
RTRTR

Are correct sequences. Others don't work.

Actually I think you can only solve this by listing. Like the problem with the guy numbering his fingers at last year's nats.


One line divides a plane into two regions. Two lines can divide a plane into, at most, four regions. What is the maximum number of regions possible using four lines?

We find a pattern. For the nth additional line we add n to the number of regions for n-1.

1 line: 2
2 lines: +2 = 4
3 lines: +3 = 7
4 lines: +4 = 11


It should be $x^5-x-1<0$
So it should be $x>x^5$ <=> $1>x^4$so $1>|x|$ and we are done

[quote="Altheman"]i think it is $x<1.1673...$ since there is only one real root, and by testing, we get that it is true to the left( i used my calc to solve it) i dont think that there is an efficient way to solve for roots of quintic polynomials[/quote]

I forgot to type that $x\in N$. Now, could you prove that $x\leq1$  :?:

there was something in my algebra 2 text book where you find the lower bounds of the roots of an equation
but anyway, if you know the bounds is like 1.1..., then you want to find the closest integer that is less than that, so it is obviously 1


induction
number theory
prime factorization

[color=olive]a) Show that if a, b, and c are integers with (a,b)=(a,c)=1, then (a,bc)=1.
b) Use mathematical induction to show that if a[size=75]1[/size], a[size=75]2[/size],.....,a[size=75]n[/size] are integers, and b is another integer such that (a[size=75]1[/size],b)=(a[size=75]2[/size],b)=...=(a[size=75]n[/size],b)=1, then(a[size=75]1[/size]a[size=75]2[/size]...a[size=75]n[/size], b)=1[/color]

Thanks for your help.

For the first part, assume for contradiction that there exists a prime $p$ such that $p|a$ and $p|bc$. By Euclid's Lemma, either $p|b$ or $p|c$. In either case, it contradicts the problem statement that $gcd(a,b)=gcd(a,c)=1$.

For the first one....

[hide]The prime factorization of a has no numbers in common with the prime factorization of b or c. When you multiply b and c, you don't add any new numbers to the prime factorization, so the prime factorization of a has no numbers in common with the prime factorization of bc, so (a,bc)=1. [/hide]


In the set $ S_n \equal{} \{1, 2,\ldots ,n\}$ a new multiplication $ a*b$ is defined with the following properties:

[b](i)[/b] $ c \equal{} a * b$ is in $ S_n$ for any $ a \in S_n, b \in S_n.$
[b](ii)[/b] If the ordinary product $ a \cdot b$ is less than or equal to $ n,$ then $ a*b \equal{} a \cdot b.$
[b](iii)[/b] The ordinary rules of multiplication hold for $ *,$ i.e.:

[b](1)[/b] $ a * b \equal{} b * a$ (commutativity)
[b](2)[/b] $ (a * b) * c \equal{} a * (b * c)$ (associativity)
[b](3)[/b] If $ a * b \equal{} a * c$ then $ b \equal{} c$ (cancellation law).

Find a suitable multiplication table for the new product for $ n \equal{} 11$ and $ n \equal{} 12.$


trigonometry
function
vector
ratio
advanced fields
advanced fields unsolved

My prof. said that we will have to integrate the Serret-Frenet equations with constant torsion and constant curvature to get the general form of the curve for my test.

Much edited post:

This was my attempt: I'll use $\kappa (s)=\kappa_{0}$ and $\tau (s)=\tau_{0}$: Since $\dot{T}=\kappa_{0}\vec{n}$ we have $\ddot{T}=\kappa_{0}\dot{n}$, and from the second equation $\dot{n}=-\kappa_{0}T+\tau_{0}B$, so $\ddot{T}=\kappa_{0}(-\kappa_{0}T+\tau_{0}B)$. Then from $\dot{B}=-\tau_{0}n$ combined with the first one we get:
$\dddot{T}=-(\kappa_{0}^{2}+\tau_{0}^{2})\dot{T}$.

Now by integrating twice we get $\dot{T}=-(\kappa_{0}^{2}+\tau_{0}^{2})x(s)$, this is the differential equation $\frac{d^{2}x}{ds^{2}}+(\kappa_{0}^{2}+\tau_{0}^{2})x(s)=0$. This is solved by $x(s)=C_{1}e^{\sqrt{-(\kappa_{0}^{2}+\tau_{0}^{2})}s}+C_{2}e^{-\sqrt{-(\kappa_{0}^{2}+\tau_{0}^{2})}s}$, but since $\kappa_{0}$ and $\tau_{0}$ are reals the square root is negative. Thus $x(s)=C_{1}\cos\left(\sqrt{\kappa_{0}^{2}+\tau_{0}^{2}}s\right)+C_{2}\sin\left(\sqrt{\kappa_{0}^{2}+\tau_{0}^{2}}s\right)$.

I guess my question is now: Is it reasonable that with constant torsion and constant curvature the equation is always of that form given? And should the constants $C_{1}$ and $C_{2}$ be determined somehow. That determines the amplitude of these functions, and so it seems that it should change the curvature? This should be a vector-valued function in $\mathbb{R}^{3}$, is the answer (assuming it is correct) just $C_{1}\cos\left(\sqrt{\kappa_{0}^{2}+\tau_{0}^{2}}s\right)+C_{2}\sin\left(\sqrt{\kappa_{0}^{2}+\tau_{0}^{2}}s\right)\vec{e}_{1}+s\vec{e}_{2}+s\vec{e}_{3}$? Or is it $C_{1}\cos\left(\sqrt{\kappa_{0}^{2}+\tau_{0}^{2}}s\right)+C_{2}\sin\left(\sqrt{\kappa_{0}^{2}+\tau_{0}^{2}}s\right)\vec{e}_{1}$?

New theory: $x(s)=C_{1}\cos\left(\sqrt{\kappa_{0}^{2}+\tau_{0}^{2}}s\right)\vec{e}_{1}+C_{2}\sin\left(\sqrt{\kappa_{0}^{2}+\tau_{0}^{2}}s\right)\vec{e}_{2}+s\vec{e}_{3}$

Any takers on this one? Still completely unsure... I guess I can pinpoint one of the major points of confusion to, do I still solve the differential equation the same way even though it is vector-valued and not scalar valued, and if so (since it probably is) what does that do to the answer? Hence the three solutions given all different from eachother using the same answer.

There is probably a cleaner way to do this, but my memory is not serving me that well, so I'll plod along as follows:

First, for some curve $\alpha$ that is not necessarily unit speed, we note that $\alpha' = vT, \alpha'' = \frac{dv}{dt}T+\kappa v^{2}N, \alpha''' = \text{stuff }+\kappa \tau v^{3}B$, so we can calculate $\tau$ as $\frac{(\alpha' \times \alpha'') \cdot \alpha'''}{\|\alpha' \times \alpha''\|^{2}}$.

Lemma. Given a unit speed curve $\alpha(t)$, $\alpha$ is a cylindrical helix if and only if $\frac{\tau}{\kappa}$ is constant, where we take "cylindrical helix" to mean that if we let $\beta(t) = T(t)$ the tangent vector (so that $\beta(t)$ is always in $S^{2}$), then $\beta$ actually lies in some plane.

Proof. Recall that a given curve lies in a plane if and only if the torsion vanishes, so we compute $\tau_{\beta}$, the torsion of $\beta$. We have that $\beta' = \kappa N, \beta'' = \frac{d\kappa}{dt}N+\kappa(-\kappa T+\tau B)$, so $\beta' \times \beta'' = \kappa^{2}\tau T+\kappa^{3}B$. Next, $\beta''' = \text{stuff}N-3 \kappa \frac{d\kappa}{dt}T+(2 \frac{d\kappa}{dt}\tau+\frac{d\tau}{dt}\kappa) B$, so in the formula above, we see that $0 = (\beta' \times \beta'') \cdot \beta'''$ if and only if $0 = \kappa^{4}\frac{d\tau}{dt}-\kappa^{3}\tau \frac{d\kappa}{dt}= \kappa^{5}\frac{d}{dt}\left( \frac{\tau}{\kappa}\right)$, which establishes the lemma.

Thus, curves with $\tau, \kappa$ identically constant are certainly cylindrical helices, as described above. In particular, the tangent vector lies in some plane, i.e. there exists a vector $u$ and constant $c$ such that $T(t) \cdot u \equiv c$ identically. Now, consider the curve $\gamma(t) = \alpha(t)-ctu$. We have that $\gamma \cdot u = \alpha \cdot u-ct$; if we differentiate this, we obtain $T \cdot u-c = 0$, so $\gamma \cdot u$ is a constant and hence $\gamma$ lies in a plane normal to $u$. Now, we finally use that $\tau, \kappa$ are constant (as opposed to just using that their ratio is constant) by noting that $\gamma' \cdot \gamma' = 1-c^{2}$, $\gamma'' = \kappa N$, i.e. $\gamma$ has the same curvature as $\alpha$, so $\gamma$ is a unit-speed plane curve with constant curvature. Namely, the plane curvature of $\gamma$ is $\pm$ the space curvature, and as it is continuous, it cannot "jump" between the two values. Thus, the plane curvature is constant. We can now use results from plane curves, such as the standard result that now tells us $\gamma$ runs around a circle. (Essentially, we can construct the center $\gamma+\frac{1}{k}N$, where $k, N$ are the plane curvature and normal. Differentiate this to get 0 to see that this point we constructed is a constant, whereupon $\gamma$ has constant distance from this constant point. We can now write $\gamma$ in terms of $\cos, \sin$ of some function $\theta$ and check that $\theta$ is linear to see that $\gamma$ traces out the circle.)

Thus, $\alpha$ is a circle in some plane plus some linear term in the vector normal to that plane, i.e. a circular helix.

Thanks. I took my test today and wrote something that was vaguely correct (I think). It came out to be a circular cylindrical helix. I think I messed up a sign, though, because the torsion was positive, but I had a negatively oriented curve. Oh well, if that was really the only mistake, then I'm in good shape.


limit
induction
algebra unsolved
algebra

Let ${a_{n+1}=f(a_n)},f(x)=\frac{1}{2-x}$.  If $a_1\in R$, then
1) $1<a_1<2$ exist k: $a_k>2$,
2) if $a_1>2\Longrightarrow a_2<0$,
3) if $a_1<0\Longrightarrow 0<a_2<\frac 12 <a_3<1.$
Let ${\frac 12 <a_i<1\Longrightarrow 1-a_{i+1}=\frac{1-a_i}{1+(1-a_i)}}<1-a_i$, therefore sequence increase and limit=1.

Solve the recurrence equation, we have $a_n=1+\frac{a_1-1}{(1-a_1)n+a_1}$

how do you solve the recurrence equation?

By induction, or calculate $A^n$, were Ax=(ax+b)/(cx+d) if A=(a,b)(c,d).


function
superior algebra
superior algebra solved

Lets consider the sum f:R\{-3,-2} ->R and f(x)=1/(x^2+5x+6);
a) compute the n-th derivate;
b) compute the sum sum(i=1,n) f^(k) (i);

where f^(k) is the k-th derivate of the function!

cheers! :D :D


Define $ A @ B$ to be $ \frac{A}{B}\plus{}A\times B$. What is the value of $ 20 @ (4 @ 2)$?

First determine 4@2. Plugging in:
4/2 + 4x2 = 10

Then determining 20@10:
20/10 + 20x10 = 202


We have acute angle with apex P and point A in this angle. Find on shoulders angle points X and Y, that |PX|=|PY| and sum |AX|+|AY| is the smallest

well, I don't think this is a calculus problem :|

Sorry for mistake Leonard. :P  Can anyone resolve it for me?

This is a problem from an ongoing math contest  :!:  :!:  :!:    (this year's Polish Mathematical Olympiad)

[Moderator edit: discussion to be continued at http://www.mathlinks.ro/Forum/viewtopic.php?t=169314 ]


Could  anyone  prove  the  Maclairin' s theorem.

 Let  a  convex  agle  xOy  and  let  A  and B  are  points  at    Ox  and Oy   . If  OA + OB = constant
  then  the  circumradus  circles  of  the  triangle  OBA   have  a common  point  P different  to    O  .

[quote="gdeskos"]Could  anyone  prove  the  Maclairin' s theorem.

 Let  a  convex  agle  xOy  and  let  A  and B  are  points  at    Ox  and Oy   . If  OA + OB = constant
  then  the  circumradus  circles  of  the  triangle  OBC   have  a common  point  P different  to    O  .    [/quote]

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=367254#p367254[/url]


Two numbers, x and y, have a geometric meean of 12.5 and an arithmetic mean of 12. Find x 2 +y 2

We create two equations:

$\sqrt{x * y} = 12.5$
and
$\frac{x + y}{2} = 12$

We find that xy = 156.25 from the first equation.
We also find taht x + y = 24.

x+y = 24.

(x+y)2 = 242

x2 + 2xy + y2 = 576.

x2 + y2 + 2(156.25) = 576

[b]x2 + y2 = 263.5[/b] There's your answer.