Find the volume of a parallelepiped with verticies: (0,0,0) (3,0,0) (0,5,1) (3,5,1) (2,0,5) (5,0,5) (2,5,6) and (5,5,6).

http://en.wikipedia.org/wiki/Parallelepiped

Define:

a = (3,0,0)
b = (0,5,1)
c = (2,0,5)

We can rewrite our list of vertices as 0, a, b, c, a+b, b+c, c+a, a+b+c

The volume of this can easily be expressed in terms of a, b, and c as abs(det(a^T b^T c^T)), or:

$ \text{abs} \; \left | \begin{array}{ccc} 3 & 0 & 2 & 0 & 5 & 0 & 0 & 1 & 5 \end{array} \right | = 75$

If you're not familiar with the determinant formula, let's reason it out. We know that shearing this thing in a direction parallel to any one of a, b, or c does not change the volume. So, the volume of the figure "spanned" by a, b, and c would not change if we projected b and c onto the y-z plane (we sheared out the component parallel to a, which is along the x-axis). So, it's volume is the same as the volume "spanned" by vectors:

a = (3,0,0)
b = (0,5,1)
c' = (0,0,5)

We use a similar argument to shear out the z-components, since they are in the direction of c'. That is, we project a and b onto the x-y plane. We now get:

a = (3,0,0)
b' = (0,5,0)
c' = (0,0,5)

But, we know the volume of this is just 3*5*5=75


trigonometry
geometry
trig identities
Law of Sines

[hide="Proof with Extended Law of Sines"]By the Extended Law of Sines, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$, where $R$ is the circumradius.  Therefore, $a=2R\sin A$ and $b=2R\sin B$.

$a=\frac{c(\sin A-\sin B)}{\sin C}+b$
$2R\sin A=2R(\sin A-\sin B)+2R\sin B$

Dividing by $2R$, we obtain $\sin A=\sin A-\sin B+\sin B$, which is clearly true for all angles $A$ and $B$.

Notice that it's not necessary to know that $\frac{a}{\sin A}=2R$.  You could also let $x=\frac{a}{\sin A}$, and the proof would work the same way.[/hide]

And by request from mrcrank,  :)

[hide="Proof without Extended Law of Sines"]Let $x=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$.  Therefore, $a=x\sin A$ and $b=x\sin B$.

$a=\frac{c(\sin A-\sin B)}{\sin C}+b$
$x\sin A=x(\sin A-\sin B)+x\sin B$

Dividing by $x$, we obtain $\sin A=\sin A-\sin B+\sin B$, which is clearly true for all angles $A$ and $B$.[/hide]

I just added another proof above using $x$ instead of using $2R$.  (The statement $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ is known as the Extended Law of Sines.)  The two proofs are almost exactly the same.  :)

Here's a proof that may be easier to understand:

[hide]$a=\frac{c(\sin A-\sin B)}{\sin C}+b$
$a\sin C=c\sin A-c\sin B+b\sin C$
$a\sin C+c\sin B=c\sin A+b\sin C$

By the Law of Sines, $a\sin C=c\sin A$, and $c\sin B=b\sin C$, so we're done.[/hide]


is about the fundamental difficulties of the human endeavor of reconciling quantum field theory with general relativity mathematically.

My guesses:

The mystery behind gravity (makes sense!)
Teenage Mutant Bninja Gerbils
The Monkeys Bit Gerald
Tinytim might be gigantic


geometry
number theory
prime numbers
greatest common divisor

Got some questions, which should test some of you, others you'll eat in your sleep...

[quote="Question 30"]The sum of n positive integers is 19. What is the maximum possible of these n numbers?[/quote]

I got [hide]3, 3, 3, 3, 3, 4, which gives 972[/hide], any chance of getting bigger?

[quote="Question 25"]The vertices of a cube are each labelled with integers numbered from 1-8. A face-sum is the sum of the labels of the four vertices on a face of the cube. What is the maximum number of equal face-sums in any of these labellings?[/quote]

NB: Can either be 2, 3, 4, 5, or 6.

Would be great if you could help me, thanks!

[quote]Question 30 wrote:
The sum of n positive integers is 19. What is the maximum possible of these n numbers?[/quote]
maximum possible product?
i think that is it.

and i don't get question 25 =\

Don't worry man, I just got it, the answer is 6.

[hide]Faces are 9136, 8641, 2368, 2548, 2359, 1236, bit hard to see without a diagram, I'll try and upload one later on.[/hide]

---

Next question:

[quote="Question 29 (no diagram)"]]In a regular polygon there are two diagonals such that the angle between them is 50 degrees. What is the smallest number of sides of the polygon for which this is possible?[/quote]

Yep..i got 6 for that as well.

However, i made so many damn mistakes. I got questions 4 and 9 wrong...like WTF...i must be stupid or something. However i think i did well in the latter sections of the exam (except 26-30) and i forgot my eraser in my pencil case, so i was frustrated while i was doing the test.

Anyway, does anyone know the cut-off mark for a distinction? I would be quite lucky to get that..

Question 23. Five bales of hay are weighed two at a time in all possible combinations. The weights, in kilograms are:
110, 112, 113, 114, 115, 116, 117, 118, 120, 121.

What is the weight, in kilograms, of the heaviest bale?

A) 58
B) 59
C) 60
D) 61
E) 62

I got 4 wrong too, lol, I was trying to take to breeze through them, and mis-took 1/4, for the one that was split in 4, but was shaded in half...

[quote]Question 23. Five bales of hay are weighed two at a time in all possible combinations. The weights, in kilograms are:
110, 112, 113, 114, 115, 116, 117, 118, 120, 121.

What is the weight, in kilograms, of the heaviest bale? [/quote]

Can't be 58, can't be 59, can't be 60.

The numbers I found were 54, 56, 58, 59, [b]62[/b]

(54, 56) (54, 58) (54, 59) (56, 58) (56, 59), (54, 62) (58, 59) (56, 62) (58, 62) (59, 62)

Answer is E

Intermediate for me, though the thing is, some questions are across Intermediate and Senior divisions, hell, even the Junior category

Yay i got 62 as well..

Lol.. some of my friends got Q4 wrong as well..

Quesiton 15
A train travelling at a constant speed takes a quarter of a minute to pass a signpost and takes three quarters of a minute to pass completely through a tunnel which is 600m in length. The speed of the train, in kilometers per hour, is:

A) 50
B) 56
C) 64
D) 72
E) 80

Edit: These questions are from the Intermediate (grades 9-10) paper.

That one, I blew real hard, worse than Paris Hilton could :lol:

For the record, I put down 64, huge error, a friend said it was 72.

Grrr.

Huh? 72? damnit.. i think i totally misread the question..i just picked 72, did the working and found that the meters/minute for 72km/hr couldnt work out to finish 600m in 3/4minutes. So, in the final minutes, i just put a wild random guess, 56...
I hate it when i misread questions..i try to do the first 20 questions as efficiently as possible to work the harder ones, but now i find out that i stuffed up like 3-4 questions in the easier parts...

For question 30, i got 768.. but it was 972..!! eheh..oh well at least i tried

Might not be 72, a friend said, we'll wait and see...

I had 768 too, changed to 972 really late, lucky to get it.

[quote="Q27"]Each of a students in a class writes a different 2-digit number on the whiteboard. The teacher claims that no matter what the students write, there will always be at least three numbers on the whiteboard whose digits have the same sum. What is the smallest number of students in the class for the teacher to be correct?[/quote]

last five answers were: 32, 35, 11, 18, 972.

[quote="jgao"]I got a high distinction.[/quote]

How many marks did you think you got?

Because, I would say that if you are so sure you got a high distinction, you could've got higher.

[quote="sen"]last five answers were: [b]32[/b], 35, 11, 18, 972.[/quote]

I so had it. I had 144+25=169

I needed the last integer from both, and I would've had it. Instead I chose the one after. Hard to fathom. 36, and I knew it was 32.

[quote="jgao"][hide=" :) "] I got a high distinction.[/hide]

EDIT: Which one of you are from NZ, and is plalnning to try out for camp. if so Can you suggest some materials that might help. thnx  :D[/quote]

Do you mean the Olympiad Camp?

for quesiton 30 type quesitons, you want to use numbers as close to $e$ as possible, because it will yield the greatest product, and hence 2s and 3s have to be used but one must keep in mind that $2^{3}<3^{2}$ and $2+2+2=3+3$ so use 3s as much as osisble until you have to use 2s.

Yah, had 768, 2 was my first thought, then I saw 3, and thought, voila!

--

You got #19 right too I'd imagine?

What was it? It was a weird Q.

Hey :P

Oh dear i so screwed this up.. but the train one I got:

[hide]It says the train moves past a signpost, and that to pass one it takes 1/4 min. That means that the length of the train is roughly 1/4 min. It takes 3/4 of a min to fully clear a 600 m tunnel. Because it says that to pass a signpost (which isn't very large) it takes 1/4 min, subtract 1/4 min from 3/4.
And you get 2/4, which equals 1/2 min. To get one minute, multiply 600 m by 2, which is 1.2km, and then multiply that to get the hours, which turns out to be 72 km. [/hide]

I got stumped on that for a while :p

All the others I so screwed.. oh man that one where there's two squares and the points rotate.. I did the area calculation and got 2 pi. Sad.

On another note, who could do that 450 grams question? The one where it goes:

A bag weighs 450g, which is made of flour and salt. How much flour do you need to add to make the proportion of salt at 90% of what it used to be?

My firend said you had to assume quantities, butI don't think so..

[hide]50g[/hide] of flour is what I got.

Yeah, that's what my friend said, but do you know [i]how [/i]to do it? I.e. working out?

Yeh, i didnt get that question either, because I couldnt understand it.

[quote="Maths_Crazy"][quote="mym.robert"]Question 20
I have 450 grams of salt and flour mix. How many grams of flour should I add to reduce the percentage of salt in the mixture to 90% of what it was.

A) 50
B) 10
C) 30
D) 45
E) 60

[hide]There are 225g of flour, and 225 grams of salt.
To me, 50 grams seemed pretty promising and worked it out. This makes the flour= 275grams and salt is still 225grams. Therefore I now have 500 grams of the unbalanced mixture.
BEFORE mixing more flour, salt was 50% and flour was 50%. Now, salt is 45%. So, now I have 90% of the flour, since 10% of 50 is 5, and 50-5= 45. 100%-10%=90%, therefore A is the answer.[/hide][/quote]

I worked it out that way, and just to make sure:
[hide]The question was a lil' ambiguous - 450g of salt and flour mix, doesn't say how much of each. So even if the salt is 450g, adding 50g means that it is only 9/10 (90%) of what it was[/hide][/quote]

Hehe, didnt even notice that, thanks.

Anyway,
Question 22.
The nine squares of a 3x3 grid painted on a wall are to be colored red, white and blue so that no row or column contains squares of the same color. One such pattern is shown in the diagram. How many different patterns can be made?

Diagram:
RWB
BRW
WBR

A) 15
B) 6
C) 9
D) 12
E) 24

It's different to your way, I could've even been totally in the wrong comapss direction, yet I still got the same answer...

---

I think the answer for Q22 is [hide]12
Let R=1, W=2, B=3
possible first line combos: 123, 132, 213, 231, 312, 321
And there are two different combinations for the other two lines, dictated the colour (number) that's gone before.[/hide]

Hopefully I am right.

[quote="sen"]q22 is 12,

there is $\frac{x}{450}\cdot \frac{9}{10}=\frac{x}{500}$ so 50 was added.[/quote]

:)

cool, thanks :D

I got 12 for question 22 too, the way i worked it out was that there is one colour that always occupies one diagonal: so

[hide]
-wrb-
-bwr-
-rbw-

= where white occupies the topleft-botright diagonal. if you switch
red and blue, that's another solution. then u make white occupy
the topright-botleft diagonal, and switch blue and red again to make another.
So that makes 4 solutions as white occupying a diagonal.
Switch to red and blue, and thats 12.
[/hide]

did anyone by the way manage ot sneak out the question sheet? My teacher collected them at the end of the exam, he was so slow he only gave us writing paper 10 min in the exam.. :p

what do you guys think the cutoff for a medal will be?

Hmm..isnt the medal only for people who get full marks? Anyway, we had 1 student from our school recieve a medal last year and he was in year 9...crazy.
Anyway, i think this competition was slightly harder than last year's one..only in the 1-20 sections..so maybe the cut-offs will be less demanding, but i wouldnt think so. Does anyone know the usual cut-offs for distinction and high distinction?

the medals are for 1 in every 10000 competitors, prizes are no more than 1 in every 300 competitors i think.

[hide="answer for 19"]B)[/hide]

Even though I'm not from Australia :D .

[quote="Maths_Crazy"]21. How many positive integers less than 72 have the property that the highest common factor of the number and 72 is equal to 1?

A 12
B 30
C 36
D 18
E 24

I didn't understand this bit:
- the highest common factor of the number and 72 is equal to 1  :?:[/quote]

if number is x, than $(x, 72)=1$
answer is 24 btw.

[quote="Maths_Crazy"]21. How many positive integers less than 72 have the property that the highest common factor of the number and 72 is equal to 1?

A 12
B 30
C 36
D 18
E 24

I didn't understand this bit:
- the highest common factor of the number and 72 is equal to 1  :?:[/quote]
It means "how many prime numbers are lower than 72"
A prime number and 72 don't have any common factors besides 1.

[quote="Totally Zealous"]It means "how many prime numbers are lower than 72"
A prime number and 72 don't have any common factors besides 1.[/quote]
It doesn't have to be prime, just coprime. For example, $51$ isn't prime, but the greatest common factor is $1$.

[quote="Maths_Crazy"][quote]Does anyone know the usual cut-offs for distinction and high distinction?[/quote]

Last year, in New Zealand, a prize was 83, I got 75, high distinction, most of my friends got in the 50s, and got distinctions.[/quote]

hmm thats comforting to know, because i know i got at least 60, maybe even 70, which means i have a chance to get a distinction, which is good enough for me.

[quote]Do you mean the Olympiad Camp?[/quote]

Yep. :lol:


$6$ intelligent people are told that either a RED or GREEN color will be pasted on their foreheads and obviously the exact color on individual's forehead is not known to him/her.

Now they are all invited into a room and told that if they see at least one of each colour , he/she is supposed to stand on one leg.

And when they determine their color, they have to sit down.

After they are invited into the room, all of them stand on one leg.
After a while, all of them sit.

Can you tell me how many RED and and how many GREEN colors are there?

[quote="daine"][hide]2 or 4 green.[/hide][/quote]

how?

[quote="samath"]If there are:

1 green:  When everybody else remains standing, the person in the green realizes he must have green.  But nobody else can sit.  NO
2 green:  Person A, wearing green, realizes that exactly one other person is wearing green, and since that person is standing, he must be green.  Also, the other person must sit down.  Everybody else, wearing red, realizes that if they had been wearing green, exactly 3 people would have been wearing green, and neither or the originals could sit down, so everyone sits down.  YES
3 green:  Let's say person A is green: he sees 2 greens and 3 reds.  If he is red, then the greenswould have sat down; therefore he is green.  Similarily, all reds see that there cannot be 4 greens, and realize they are red, so sit down.  YES
4 green: Equivalent to 2 green: YES
5 green: Equivalent to 1 green: NO

Therefore, there could be 2, 3, or 4 greens.[/quote]

hmmm... when I told this answer, I am told that this is incorrect.
Specifically I am told that [i][b]"I a consider that all are equally intelligent ; and require the same time for analysis; I will arrive at the unique answer. Remember that all jump together"[/b][/i]

What are we missing here?

[hide]3 green, 3 red, Because they must all sit at the same time they must all see the same things, though red and green may be switched. [/hide]


analytic geometry
graphing lines
slope
geometry
function
integration
ceiling function

Plot the points $ P_i(a_i,a_i^2)$ in the coordinate plane, within the unit square $ S$ of vertices $ O(0,0), A(0,1), B(1,1), C(1,0)$, and reindex them in descending order of the slopes of the rays $ OP_i$, i.e. in descending order of the values of the $ a_i$'s. Now, it is well-known that $ \frac {1} {2} | a_ia_j^2 \minus{} a_ja_i^2 | \equal{} \frac {1} {2} a_ia_j| a_i \minus{} a_j |$ is the area of triangle $ \Delta OP_iP_j$ (also see the \href{http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem}{Shoelace Theorem}).

Clearly the triangles $ \Delta OP_iP_{i \plus{} 1}$, $ 1 \leq i \leq n \minus{} 1$, have pairwise disjoint interiors. Notice that the points $ P_i$ lie on the arc of the graph of the function $ f(x) \equal{} x^2$. Therefore the sum of their areas is strictly less than $ \displaystyle \frac {1} {2} \minus{} \int_{0}^{1} x^2 \textrm{d}x \equal{} \frac {1} {2} \minus{} \frac {1} {3} \equal{} \frac {1} {6}$, the area between the first bisector and the graph. Moreover, $ a_ia_{i \plus{} 1}(a_i \minus{} a_{i \plus{} 1}) > 0$, so
\[ 0 < \sum_{i \equal{} 1}^{n \minus{} 1} a_ia_{i \plus{} 1}(a_i \minus{} a_{i \plus{} 1}) < \frac {1} {3},\]
therefore (at least) one of the expressions is less than $ \frac {1} {3(n \minus{} 1)}$. If we want the bound to be $ \frac {1} {2007}$, that requires $ n \geq 670$.

For a better estimate, the defect area (between the triangles and the graph) can be computed to be
\[ D \equal{} \frac {1} {6} \left ( a_n^3 \plus{} (a_{n \minus{} 1} \minus{} a_n)^3 \plus{} \cdots \plus{} (a_1 \minus{} a_2)^3 \right ) \geq \frac {1} {6n^2} a_1^3, \ \textrm {so}\]

\[ \sum_{i \equal{} 1}^{n \minus{} 1} [\Delta OP_iP_{i \plus{} 1}] \equal{} \frac {1} {2} a_1^3 \minus{} \frac {1} {3} a_1^3 \minus{} D \leq \frac {n^2 \minus{} 1} {6n^2} a_1^3 \leq \frac {n^2 \minus{} 1} {6n^2},\]
therefore (at least) one of the expressions is less than $ \frac {n \plus{} 1} {3n^2}$. If we want the bound to be $ \frac {1} {2007}$, that still requires $ n \geq \left \lceil \frac {669 \plus{} \sqrt {669^2 \plus{} 4\cdot 669}} {2} \right \rceil \equal{} 670$.

But there is room for improvement (equality in this last estimate does not yield equal areas), so we are back to some algebraic manipulations, in order to lower the number to $ 669$.

See a second posting at [url]http://www.mathlinks.ro/viewtopic.php?t=299651[/url]

Here's a proof I posted a while ago:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145906[/url].

By the way, in the statement of the problem in the current thread, there is an inconsistency: 699 was used once, while 669 was used later.  I think 669 was meant throughout.


[quote="silouan"]Μας έβαλαν την εξής άσκηση (που μόνο ένα άτομο την έλυσε)
Προσπαθείστε την αξίζει τον κόπο
Δίνεται τρίγωνο ABC και σημείο Μ πάνω στην BC (στο εσωτερικό αυτής) .Νδο
$\left|\vec{MA}\right|\cdot\left|\vec{BC}\right|\leq \left|\vec{MB}\right|\cdot\left|\vec{AC}\right|+\left|\vec{MC}\right|\cdot\left|\vec{AB}\right|$[/quote]

Ξεκινάω από δεύτερο μέλος και έχω:

$\left|\vec{MA}+\vec{AB}\right|\cdot\left|\vec{AC}\right|+\left|\vec{MA}+\vec{AC}\right|\cdot\left|\vec{AB}\right| \geq \left|\vec{MA}\cdot\vec{AC}+\vec{AB}\cdot\vec{AC}\right|+\left|\vec{MA}\cdot\vec{AB}+\vec{AB}\cdot\vec{AC}\right|\geq\left|\vec{MA}\cdot\vec{AC}+\vec{AB}\cdot\vec{AC}-\vec{MA}\cdot\vec{AB}-\vec{AB}\cdot\vec{AC}\right|= \left|\vec{MA}\left(\vec{AC}-\vec{AB}\right)\right|=\left|\vec{MA}\cdot\vec{BC}\right|$

Τριγωνική ανισότητα και τέλος  :D

PS: Να υποθέσω ότι το άτομο ήσουν εσύ;   :P

[quote="stergiu"][quote="mostel"][quote="silouan"]Μας έβαλαν την εξής άσκηση (που μόνο ένα άτομο την έλυσε)
Προσπαθείστε την αξίζει τον κόπο
Δίνεται τρίγωνο ABC και σημείο Μ πάνω στην BC (στο εσωτερικό αυτής) .Νδο
$\left|\vec{MA}\right|\cdot\left|\vec{BC}\right|\leq \left|\vec{MB}\right|\cdot\left|\vec{AC}\right|+\left|\vec{MC}\right|\cdot\left|\vec{AB}\right|$[/quote]

Ξεκινάω από δεύτερο μέλος και έχω:

$\left|\vec{MA}+\vec{AB}\right|\cdot\left|\vec{AC}\right|+\left|\vec{MA}+\vec{AC}\right|\cdot\left|\vec{AB}\right| = \left|\vec{MA}\cdot\vec{AC}+\vec{AB}\cdot\vec{AC}\right|+\left|\vec{MA}\cdot\vec{AB}+\vec{AB}\cdot\vec{AC}\right|\geq\left|\vec{MA}\cdot\vec{AC}+\vec{AB}\cdot\vec{AC}-\vec{MA}\cdot\vec{AB}-\vec{AB}\cdot\vec{AC}\right|= \left|\vec{MA}\left(\vec{AC}-\vec{AB}\right)\right|=\left|\vec{MA}\cdot\vec{BC}\right|$

Τριγωνική ανισότητα και τέλος  :D

PS: Να υποθέσω ότι το άτομο ήσουν εσύ;   :P[/quote]

[color=red] ΣΧΟΛΙΟ[/color]

 Αν $\vec a, \vec b , \vec c$ είναι διανύσματα , νομίζεις ότι

    $|\vec a |\cdot | \vec b+\vec c| = | \vec a \cdot (\vec b+\vec c|$ ;
 Κοίτα μήπως είναι παγίδα !!![/quote]

Mε προλάβατε κύριε Στεργίου...
Δεν ισχύει $|\vec{u}|\cdot |\vec{v}|=|\vec{u}\cdot\vec{v}|$ αλλά ισχύει η εξής σχέση: $|\vec{u}|\cdot|\vec{v}|\geq |\vec{u}\cdot \vec{v}|\geq \vec{u}\cdot\vec{v}$

Κ. Στεργίου, Νίκο, έχετε απόλυτο δίκαιο και οι δύο! Ομολογουμένως η ισότητα ισχύει στην περίπτωση της παραλληλίας! Απλώς μου φύγε το $\geq$ ! Ευχαριστώ για τη διευκρίνιση, τη διορθώνω αμέσως!

Γεια χαρα Σιλουανέ,τα χαιρετισματα μου στην Φανη και τον Νικο.Α,ο Κων/νος τι κανει?

ΥΓ.Κeep up with the good work.(αν και περιμενα να γνωριζεις απο πιθανοτητες :rotfl: )

Katarxin as einai $H$ to ixnos tou ipsous apo to $A$ panw stin pleura $BC$. Tote exoume diadoxika: (ftiakste ena proxeiro sxima gia na deite ta parakatw).

$\overrightarrow{MB}\cdot\overrightarrow{AC}=\overrightarrow{MB}\cdot \pi\rho o\beta_{\overrightarrow{MB}}\overrightarrow{AC}=\overrightarrow{MB}\cdot\overrightarrow{HC}\ \ (1)$

kai

$\overrightarrow{MC}\cdot\overrightarrow{AB}=\overrightarrow{MC}\cdot \pi\rho o\beta_{\overrightarrow{MC}}\overrightarrow{AB}=\overrightarrow{MC}\cdot\overrightarrow{HB}\ \ (2)$

Apo tis $(1)$ & $(2)$ exoume:

\begin{eqnarray}\overrightarrow{MB}\cdot\overrightarrow{AC}+\overrightarrow{MC}\cdot\overrightarrow{AB}&=& \overrightarrow{MB}\cdot\overrightarrow{HC}+\overrightarrow{MC}\cdot\overrightarrow{HB}\nonumber \\ &=& \overrightarrow{MB}\cdot\overrightarrow{HC}+(\overrightarrow{BC}-\overrightarrow{BM})\cdot (\overrightarrow{CB}-\overrightarrow{CH}) \nonumber \\ &=& 2\overrightarrow{MB}\cdot\overrightarrow{HC}+\overrightarrow{BC}\cdot(\overrightarrow{CB}-\overrightarrow{CH}+\overrightarrow{BM}) \nonumber \\ &=& 2\overrightarrow{MB}\cdot\overrightarrow{HC}+\overrightarrow{BC}\cdot\overrightarrow{HM}\nonumber \\ &=& 2\overrightarrow{MB}\cdot\overrightarrow{HC}+\overrightarrow{BC}\cdot \pi\rho o\beta_{\overrightarrow{BC}}\overrightarrow{AM}\nonumber \\ &=& 2\overrightarrow{MB}\cdot\overrightarrow{HC}+\overrightarrow{BC}\cdot\overrightarrow{AM}\nonumber \end{eqnarray}

Etsi vrikame kai tin akrivi timi tou arxikou ginomenou ap'opou (pairnontas metra) exoume:

$|\overrightarrow{MB}\cdot\overrightarrow{AC}+\overrightarrow{MC}\cdot\overrightarrow{AB}|=|2\overrightarrow{MB}\cdot\overrightarrow{HC}+\overrightarrow{BC}\cdot\overrightarrow{AM}| \ \ (3)$

Episis :

$|\overrightarrow{MB}\cdot\overrightarrow{AC}+\overrightarrow{MC}\cdot\overrightarrow{AB}|\leq |\overrightarrow{MB}\cdot\overrightarrow{AC}|+|\overrightarrow{MC}\cdot\overrightarrow{AB}|\leq \left|\overrightarrow{MB}\right|\cdot\left|\overrightarrow{AC}\right|+\left|\overrightarrow{MC}\right|\cdot\left|\overrightarrow{AB}\right| \ \ (4)$

kai apo tin alli $|2\overrightarrow{MB}\cdot\overrightarrow{HC}+\overrightarrow{BC}\cdot\overrightarrow{AM}|\geq |2\overrightarrow{MB}\cdot\overrightarrow{HC}|+|\overrightarrow{BC}\cdot\overrightarrow{AM}|\geq |\overrightarrow{BC}\cdot\overrightarrow{AM}| \ \ (5)$

Apo tis $(4)$ & $(5)$ kathws kai apo tin sxesi $(3)$, prokiptei to zitoumeno. (Xrisimopoiiste tin $(5)$ me tin antistrofi fora, xrisimopoiiste meta tin $(3)$, kai meta parte tin $(4)$).

Asfalws iparxei kai pio eukolos dromos, apla me ton parapanw tropo fainetai kai me ti akrivws einai isos o arithmos $\overrightarrow{MB}\cdot\overrightarrow{AC}+\overrightarrow{MC}\cdot\overrightarrow{AB}$

Alexandros

Y.G. Silouan wraio to thema!

[quote="Laplacian"]Esy re cretaman den ksereis tpt ma tpt apo pithanotites na voithiseis ligoulaki?
Gia olympiada etoimazomai...[/quote]

Distixws Laplacian me tis pithanotites tou eidous pou zitas den ta paw katholou kala! Exw na ta dw apo to panepistimio kai den mporw na pw oti ta simpathousa para poli! Giati de dokimazeis sto "College Playground" na kaneis tin erwtisi sou? Ekei pithanon na se voithisoun oi "pio eidikoi".

Alexandros

εκει κι αν εχουν κολλησει...

Φέρνουμε $ME \| AB$ και $MD \| AC$ .

Τότε $\frac{AE}{AC}= \frac{BM}{BC}$ και $\frac{AD}{AB}= \frac{CM}{BC}$

και $\left|\vec{AD}\right| = \left|\vec{EM}\right|$.

Λαμβάνοντας τα παραπάνω υπ'όψιν η ανισότητα γράφεται

$\\ \left|\vec{MA}\right|\cdot\left|\vec{BC}\right|\leq \left|\vec{MB}\right|\cdot\left|\vec{AC}\right|+\left|\vec{MC}\right|\cdot\left|\vec{AB}\right| \Leftrightarrow \\ \\ \\ \left|\vec{MA}\right| \leq \frac{\left|\vec{MB}\right|}{\left|\vec{BC}\right|}\cdot\left|\vec{AC}\right|+\frac{\left|\vec{MC}\right|}{\left|\vec{BC}\right|}\cdot\left|\vec{AB}\right| \Leftrightarrow \\ \\ \\ \left|\vec{MA}\right| \leq \left|\vec{AE}\right|+\left|\vec{AD}\right| \Leftrightarrow \\ \\ \\ \left|\vec{MA}\right| \leq \left|\vec{AE}\right|+\left|\vec{EM}\right|$

που είναι η τριγωνική ανισότητα για το τρίγωνο $AEM$ .

Pe παιδιά δεν περίμενα τέτοια ανταπόκριση (10 posts!!!!!) και βάλατε και τρεις διαφορετικές λύσεις !!!!
Εγώ την έλυσα αλλιώς  :wink:
Δίνω μόνο το hint
$\left|\vec{BM}\right|=m\left|\vec{BC}\right|$ και
$\left|\vec{MC}\right|=(1-m)\left|\vec{BC}\right|$

 :wink:

Το ίδιο γράφω αλλά γεωμετρικά  :wink:


A train travels its own length in 3 seconds. It approaches a station which is 336 feet long. From the time the front of the train reaches the station until the end of the train is even with the other end of the station, 10 seconds elapse. What is the length in feet of the train?

[quote="Aequilipse"][quote="nkouevda"][hide]From the time the front of the train reaches the front of the station until the front of the train reaches the end of the station, 7 seconds elapse. Therefore, I do $\frac{336 \cdot 3}{7}= 144.$[/hide][/quote]
How do you calculate that $7$ seconds elapse? can you explain your equation?[/quote]

Label the beginning of the train point x. Now, imagine x entering the tunnel and going out of the tunnel. However, for the entire train to be out, x must be a train length away from the enterance. So, the train travels the distance of a tunnel + a train.

tunnel + train = 10, train = 3, tunnel = 7

Train speed is only 32.73 mph. It is super slow.


Construct a point inside a triangle such that cevians of it cut on the sides equal disjoint segments.

T.Y.
M.T.

Do you mean the points of equal cevians?. i.e the foci of the Steiner circumellipse

[url]http://www.xtec.cat/~qcastell/ttw/ttweng/definicions/d_Bickart_p.html[/url]
[url]http://www.xtec.cat/~qcastell/ttw/ttweng/resultats/r163.html[/url]

No. The disjoined parts of the  sides have equal lengths.The cevians are AX,BY,CZ.

    XC = YA = ZB   (or the other).

    T.Y.
    M.T.

Let $ \overline{XC} = \overline{YA} = \overline{ZB} = m.$ Then by Ceva's theorem we have

$ \frac {\overline{BX} \cdot \overline{ CY } \cdot \overline{AZ}}{m^3} = 1 \Longrightarrow m^3 = (a - m)(b - m)(c - m) \ \Longrightarrow$

$2m^3 - m^2(a + b + c) + m( ab+bc+ca) - abc = 0 \Longrightarrow m$ is not constructible.


$ S(x) \equal{} 1 \plus{} x \minus{} x^{2} \minus{} x^{3} \plus{} x^4 \plus{} x^5 \minus{} x^6 \minus{} x^7 \plus{} x^8 \plus{} .....$ for $ 0 < x < 1$
Find number of x which satisfy $ S(x) \equal{} \frac {\sqrt{2} \plus{} 1}{2}$

[quote="apratimdefermat"]$ S(x) \equal{} 1 \plus{} x \minus{} x^{2} \minus{} x^{3} \plus{} x^4 \plus{} x^5 \minus{} x^6 \minus{} x^7 \plus{} x^8 \plus{} .....$ for $ 0 < x < 1$
Find number of x which satisfy $ S(x) \equal{} \frac {\sqrt {2} \plus{} 1}{2}$[/quote][hide]$ S(x)\equal{}(1\plus{}x)(1\minus{}x^2\plus{}x^4\minus{}x^6\plus{}x^8\minus{}x^{10}\plus{}\ldots)\equal{}(1\plus{}x)\frac{1}{1\plus{}x^2}$ if $ |\minus{}x^2|<1$.
$ \frac{1\plus{}x}{1\plus{}x^2}\equal{}\frac{\sqrt{2}\plus{}1}{2} \iff 2(1\plus{}x)\equal{}(\sqrt{2}\plus{}1)(x^2\plus{}1) \iff \\
0\equal{}(\sqrt{2}\plus{}1)x^2\minus{}2x\plus{}\sqrt{2}\minus{}1$
$ x\equal{}\frac{2\pm\sqrt{4\minus{}4(\sqrt{2}\plus{}1)(\sqrt{2}\minus{}1)}}{\sqrt{2}\plus{}1}\equal{}\frac{2\pm0}{\sqrt{2}\plus{}1}$
$ \sqrt{2}\plus{}1>2$, so $ x<1$, so the solution is okay. Therefore, there is one solution of $ x$.[/hide]

[hide]
We have $ S(x)\minus{}1\equal{}(x\minus{}x^2) \plus{} (\minus{}x^3\plus{}x^4) \plus{}...$
This is a geometric progression with $ a_0 \equal{} x(1\minus{}x)$ and $ r\equal{}\minus{}x^2$

$ \Rightarrow S_{\infty}\equal{}\frac{a}{1\minus{}r} \equal{} \frac{x(1\minus{}x)}{1\plus{}x^2}$

Now we need to solve $ \frac{x(1\minus{}x)}{1\plus{}x^2}\equal{}\frac{\sqrt{2}\minus{}1}{2}$

Solving for $ x$ gives $ x \equal{} \sqrt{2}\minus{}1$[/hide]

[hide="Slightly different"]
$ S(x)\equal{}(1\minus{}x^2\plus{}x^4\minus{}x^6\plus{}...)\plus{}(x\minus{}x^3\plus{}x^5\minus{}x^7\plus{}...)
\equal{}\frac{1}{1\plus{}x^2}\plus{}\frac{x}{1\plus{}x^2}
\equal{}\frac{1\plus{}x}{1\plus{}x^2}$

Rest is same as the others.[/hide]


Can you activate Japanese on this forum? Thanks.

Um, what do you mean 'activate'?  I don't think it's a matter of flipping a switch.  We won't be implementing any other languages any time soon.

I read somewhere that you need to activate a language in order to use it...oh well, thanks anyways.

dude tare...its like impossible to do that...i mean a lot of the stuff we say can't be translated...like a translator wouldnt recognize "lol" or "sup" ya know...besides, its gonna be too wierd and I know for a fact that a computer translator cannot detect sarcasm or anything or the sort...it'd realli mess up this place...

No... a forum has different language packs. So the forum stuff you see - on the top, "FAQ - Search - Usergroups - Profile" etc would be translated into Japanese. The actual messages of the users, uh, stay the same.

But I hardly see a point to implement more languages...

And you can type in any language you want.

AoPS 是一个非常具有实用性的数学网站。

Mmm...how come what you wrote doesn't bug out and TheSouthpaw's Japanese text bugged out? For example, なぜこのテキストはうまく表示されないのでしょうか？

good point...i understood the chinese...but not the japanese

My browser showed both just fine. After I ran Tare's sentence through Babelfish, it read

"Why, this text probably won't show either?"

(This post was for Tare.)

no...it displayed it...i juss didnt know what it meant...

Right, it meant "Why does this text not show up right?"

So I guess it depends on the OS, browser, encoders, and the versions...OK, thanks!

P.S. What's "babelfish"?

[quote="Osiris"]http://babelfish.altavista.com A translation website.[/quote]
Cool! It's like google language tool but it's better! (quantatively, not necessarily qualitatively...wow, tons of vocab words :) )

...I didn't know pears had eyebrows :)

Babelfish is the reason why I can (partially) understand what rippledance and others are saying when they use Chinese characters.

I think to be able to read foreign languages like that you just have to download some little plugin thing for your browser.. you should probably be able to find out about it in your browser settings somewhere.

hmm...I somehow can't find it on altavista...


Given that a>b>c>d, determine min of

f(x) = |x-a| + |x-b| + |x-c| + |x-d|

Note that for all $b\ge x\ge c$, \[f(x)=a+b-c-d.\] Let $y>0$. If $x=b+y$, then the first absolute value decreases by $y$, while the final three increase by $y$. Siilarly, if $x=c-y$, then the last absolute value decreases by $y$ and the other three increase by $y$.

Hence, the minimum value is $a+b-c-d$.

It will be much easier if you use a number line. for example, |x-a|means the distance between x and a.


let $a,b,c>0$ and $\frac{1}{a^2+1}\ + \frac{1}{b^2+1}\ + \frac{1}{c^2+1}\ =1$ prove that this inequality
$ab+bc+ca \leq\ \frac{3}{2}$

can you show me . (I want to see the solution)

Indeed, great inequality! Here is my solution:

Denote $x=\frac{1}{a^2+1}$, $y=\frac{1}{b^2+1}$, $z=\frac{1}{c^2+1}$, then $x+y+z=2$ and $a=\sqrt{\frac{1-x}{x}}$, etc. So, the inequality can be rewritten as:
$\sum \sqrt{\frac{(1-x)(1-y)}{xy}} \leq \frac{3}{2}$
which is equivalent to:
$\sum \sqrt{\frac{(2-2x)(2-2y)}{xy}} = \sum \sqrt{\frac{(-x+y+z)(x-y+z)}{xy}} \leq 3$
Now substitute $p=-x+y+z$, $r=x-y+z$, $s=x+y-z$, then $x=\frac{r+s}{2}$, etc. and we rewrite our inequality as:
$\sum \sqrt{\frac{4pr}{(r+s)(s+p)}} \leq 3$

From Cauchy-Shwarz inequality for numbers: $\sqrt{4pr}$, $\sqrt{4rs}$, $\sqrt{4sp}$ and $\sqrt{\frac{1}{(r+s)(s+p)}}$, $\sqrt{\frac{1}{(s+p)(p+r)}}$, $\sqrt{\frac{1}{(p+r)(r+s)}}$ we get:

$\sum \sqrt{\frac{4pr}{(r+s)(s+p)}} \leq \sqrt{4(pr+rs+sp)(\frac{1}{(r+s)(s+p)}+\frac{1}{(s+p)(p+r)}+\frac{1}{(p+r)(r+s)})}$
We need to prove $\sqrt{4(pr+rs+sp)(\frac{1}{(r+s)(s+p)}+\frac{1}{(s+p)(p+r)}+\frac{1}{(p+r)(r+s)})} = \sqrt{4(pr+rs+sp)\frac{2(p+r+s)}{(r+s)(s+p)(p+r)}}=\sqrt{\frac{8(pr+rs+sp)(p+r+s)}{(r+s)(s+p)(p+r)}} \leq 3$
which after squaring both sides is:
$\frac{8(pr+rs+sp)(p+r+s)}{(r+s)(s+p)(p+r)} \leq 9$
and after clearing the denominator and expanding:
$6prs \leq p^2r + pr^2 + r^2s + rs^2 + s^2p + sp^2$
which is pure AM-GM.

Well

Let $a^2+1=x, b^2+1=y, c^2+1=z$ then we have

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2$

and we must prove

$2ab + 2ac + 2bc \leq 3$

which is equivalent to

$(a+b+c)^2 \leq (a^2+1) + (b^2+1) + (c^2+1)$

$(a+b+c) \leq \sqrt{x+y+z}$

$\sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1} \leq \sqrt{x+y+z}$

i think the last inequality is very famous and easy to prove.

the last inequality.because:
$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \leq \sqrt{\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}} \sqrt{x+y+z}$

to [b]TomciO[/b]
what about this approach?

$ \sum\frac {1}{1 \plus{} a^2} \equal{} 2$ is equivalent to the following:

$ 1 \equal{} \sum\frac {a^2}{1 \plus{} a^2}\geq\frac {(a \plus{} b \plus{} c)^2}{3 \plus{} a^2 \plus{} b^2 \plus{} c^2}$, or

$ ab \plus{} bc \plus{} ca\leq\frac {3}{2}$

that's it, I guess.


calculus
integration
geometry
geometric transformation
percent
induction

Hi all.
Continue ur discussions here and from now on any 'wrongly answered' questions should be discussed exclusively at [url=http://www.mathlinks.ro/viewtopic.php?t=187040]this[/url] place :)

Best of Luck :)

Yes Best of Luck to all and most importantly myself  :D
My expectations of the paper : Physics -- Easy
                                           Chemistry -- very very easy
                                           Maths -- Tough
 :)

All the very best to ppl all over India writing the Open test..
I'd prefer to differ from Madness..
Going by FIITJEE standards, I would like to predict,
PHYSICS- Very very tough
CHEMISTRY- Very easy
Mathematics- good...

Unfair u say that bcos u dont like chem  :mad:  :mad:

My expecation
Physics:Atleast 5 questions answered wrong
Chemicstry:Atleast 6 answered wrong and another answer elightening us with the fact that electron lies inside the nucleus
Maths:Two match the followings and  3 MCQ's  answered wrong
 :D  :P  :P  :P

Pathetic performance  :o  Its become a hatrick of such performances now  :( And how come that the assertion and reasoning ques abt $ \ce{SO2}$ is wrong? and that ques abt white ppt of $ \ce{BiCl3}$ with water too?  :huh: and he's written $ \ce{SF4}$ doesnt have a 180 degree bond. And this time he has given many none of the above (esp for that 12.6 mm sum) which he usually appr and gives an answer  :mad: . So i appr and wrote 12.6 And i appr and got 28 % for the cyclohexane while the answer is 29 % . Wth? And that isotherm match is also outrageous :mad:

The above ques in chem itself total to a 30 mark loss for me  :mad:  :mad:  And to add to it all, no one is even online to discuss now. Wth  :mad:

hi did veryyyyyyyyyyyyyyyyyyyyyyyyyyyy!!!!!!!!!!!!!11
pathetically.................
not willing to comment on my marks..........
 :stretcher:  :stretcher:  :stretcher:

[quote="madness"]Yes and the solns were more pathetic in the 1 st paper i think  :D  And even i didnt get 70 in 2nd paper  :o  And u will get 93 percentile i think  :)[/quote]

Will probably be more because not so many people have done extraordinarily in this paper and more people (who aren't so into IIT) are writing this.

And adhennae "Even I". You haven't been scoring 200's and 250's to say "Even I".

@Rohit: I got the vernier sum wrong too... 12.6 was a close approximation damn it.. But the cycle hexane sum I got 29%, I got something more than 28 so I decided to opt for 29%... Bloody, apparently a equivalents of SO2 will give a/2 of both products. I was damn irritated after seeing the solutions...  But the best one was the isotherm match.. And the BiOCl question.. We just discussed that yesterday, and I saw it from the GMP book of FIITJEE.. I should have waited till I saw the solutions before giving the feedback...

Dont even want to predict my scores... This is downright pathetic..

Inquilab Zindabad, Open Test Muradabad

[quote="Cyber-Shreyas"]I got 160. Highly upset by this pathetic score. :( After a cool 101 in paper I, I couldn't even manage a meagre 70 in paper II. How Pathetic. Paper II chem paper was so pathetic and that I have proved my computer skills by scoring a binary mark in it. (1).

I presume I will cross 90 percentile but in this paper it matters nothing.[/quote]
Hey math gini ur total shd be 171

[quote="madness"]dei thala laam abidi thaan pesuvom da  :P  IIT la katterein da na yaarunu. IIT la naan than phy topper. And do you knw prasad got 196 after a freaking 85 in phy while i am lesser but more than him in math and chem and he is considered to be equally good in all da  :D . And yes i atleast have a 97.5 percentile once to say that  :wink:[/quote]

I got 82 In physics. :D 50 + 32. If I had got atleast 20 in Chem paper II, I could have raised my std to 180. :(

And you all chotus, I see my mark only in terms of percentage not percentile.

Edit: Fool Rockesh, I said, I didn't get 70 in paper II. Read properly. How much you are gettting?

[quote="Cyber-Shreyas"]FIITJEE has already given soln dont worry, before you go an refer to it and come and post it here

x1+x2+x3 = 6n. :D (x1<=x2<=x3) Ob!

But substituting n=1 is the best way atleast in an obj paper.[/quote]

Exactly multinomial and i never see fiitjee solns  :P
And ur new pic is horrible da shre  :huh:

Exactly. Shre, revert back to the old one. At least it was spunky da fool of a took

[quote="srinitrs"][quote="Cyber-Shreyas"][quote="pardesi"]no da that's to early  may be ur son/daughter has to have a look at it only :P[/quote]

Babykkae babya???  :oops:

Anyways I used a very nice method to solve the Garland 6n  + 3 problem. I put n=1 (ob) and then did counted the no.of different trichlorobenzene compounds that existed... I am good at IUPAC, thats why. But ofcourse I got it wrong obviously cause 6+3 = 9 and I should have done trichlorononane. (FYI, this is a RSM question. Its not fair that ppl who NV RSM can just do it just like that)

Note:

0percentile is possible mathematically. 0% of India got Strictly less than you
100 is impossible. 100 percent of india cant get more than you because you yourself cant get more than you.[/quote]
very good very good but the best gulsa method is that put n=0 and eliminate 2 options and put n=1/2 and eliminate  the other option there it is the answer!!!! check it out!!!! :D[/quote]

how can put n=.5?????????

why not ???there is no restriction that n must be an integer only 6n must be an integer no so if we put n=1/2 we get 3 and 3 type colours which is ieasy to solve!!!! :D you get the answer on the spot!!! :lol:

But 3n shd be da fool so put n = 1/3  :D

[quote="madness"]But 3n shd be da fool so put n = 1/3  :D[/quote]
whatever both our methods are right!!!

who said i put 1/3?  :huh:  I solved using multinomial da  :D

discussions move dto [url=http://www.mathlinks.ro/viewtopic.php?t=186505]Metathread[/url]

That was a question I solved correctly along wit another dots problem and after  I came back, inspite of the different method, remember Baby n my answers matched?? Bluff da Shre..:D

[quote="Cyber-Shreyas"][quote="pardesi"]no da that's to early  may be ur son/daughter has to have a look at it only :P[/quote]

Babykkae babya???  :oops:

Anyways I used a very nice method to solve the Garland 6n  + 3 problem. I put n=1 (ob) and then did counted the no.of different trichlorobenzene compounds that existed... I am good at IUPAC, thats why. But ofcourse I got it wrong obviously cause 6+3 = 9 and I should have done trichlorononane. (FYI, this is a RSM question. Its not fair that ppl who NV RSM can just do it just like that)

Note:

0percentile is possible mathematically. 0% of India got Strictly less than you
100 is impossible. 100 percent of india cant get more than you because you yourself cant get more than you.[/quote]
Yes da it is a RSM question and we solved it in class and I remember baby telling the correct answer by solving for n=1 and n=2 and then by induction getting it as 1/(3n(n+1)+1). :lol:
So I directly wrote the ans without wasting time on that :D  :D

Hi friends I am new here... I scored 173 PCM(98-33-42). Yes the solutions are horrible, to say the least. Enlighten me on one aspect, are the Rank projections published by FIITJEE correct?? :oops:


Let $u_0$ be an arbitary natural number ( $\{0,1,\cdots\}$) and $a$ an arbitary odd natrual number.
the sequence $(u_n)$ is defined as follows:
\[ u_ {n+1} = \left\{ \begin{array}{ll} \frac12 u_n, &\textnormal{if }u_n \textnormal{ even}\\ u_n +a, \quad & \textnormal{if }u_n\textnormal{ odd} \end{array} \right.  \]
For which $a$ is $(u_n)$ periodic from a certain $n=n_0$ on? (like e.g. $1,6,4,2,3,2,3,2,3,2,3,\cdots$)

It will get periodic everytime:
For $u_n > a$ we have that $\frac{u_n}{2}<u_n$ and $\frac{u_n+a}{2}<u_n$, so that, depending on the parity of $u_n$ after one or two steps we get a lower number. So the sequence contains infinetely many $u_n \leq a$. But the number of positive integers $\leq a$ is finite, so after some time, we reach some of these integer again, meaning that it gets periodic (because $u_{n+1}$ depends on $u_n$ only).


number theory
relatively prime
number theory unsolved

Solve the Diophantine equation
(x − y − z)(x − y + z)(x + y − z) = 8xyz
where x, y, and z are relatively prime.

Write equation as :

$(y+x+z)((x-y-z)^2 - 4yz))$

So either $x+y+z=0$  either $4yz = (x-y-z)^2$

In second case, this gives that $y$ and $z$ (and $x$ by symetry) have same sign, wlog $\geq 0$.
Since x,y,z are relatively prime this gives that x,y,z are pair-wise relatively prime and that $x =X^2, y=Y^2, y=Z^2$
.....

I find as solutions : $(-y-z, y,z)$, $(\pm (Y \mp Z)^2, \pm Y^2, \pm Z^2)$


find with proof  if there exists a number m such that m has at least 2007 prime factors and m divides $ n^{m \minus{} 1} \minus{} 1$ where n is an integer

[hide="Fast"]To satisfy the first condition, let $ m \equal{} \prod_{j \equal{} 1}^{2007} p_j \equal{} 2 \cdot 3 \cdot 5 \cdot \ldots$

To satisfy the second condition, note that $ n^{m \minus{} 1} \minus{} 1 \equal{} (n \minus{} 1)(n^{m \minus{} 2} \plus{} n^{m \minus{} 3} \plus{} \ldots \plus{} n \plus{} 1)$ and let $ n \equal{} m \plus{} 1$.
[/hide]

but the n number is given you can not replace it with whatever you want sorry for my english

I think the OP means for all integers n, in which case no, there isn't, just consider m=n.

Oh I see. I haven't thought about this much, but I don't think m=n is a counterexample to the question; namely, if we are given a number n, we want to know if there exists at least one m such that that holds true.


I am now going to start up my own contest on this forum.  And I will post the top scores later, answers need to be PMed to me.  The deadline is by Christmas Day.  I will score with 5 points for a correct answer, if the problem is left blank, then 2 points, and if it is wrong, then 0 points.  Round 1 has 10 problems.  Good luck!

[b]Round 1[/b]

[b]1.[/b] If M is 40% of Q, Q is 15% of P, and N is 50% of P, then M/N = ...

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[b]2.[/b] How many base 10 four-digit numbers, N = [u]a[/u] [u]b[/u] [u]c[/u] [u]d[/u], satisfy all three of the following conditions?
([i]i[/i]) 4,000 :le: N < 7,000; ([i]ii[/i]) N is a multiple of 5;
([i]iii[/i]) 3 :le: b < c :le: 7.

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[b]3.[/b] The area of the triangle bounded by the lines y = x, y = -x and y = 18 is ...

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[b]4.[/b] Seven points on a circle are numbered 1, 2, 3, 4, 5, 6, and 7 in clockwise order.  A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points.  If the bug begins on point 6, after 1995 jumps it will be on the point numbered ...

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[b]5.[/b]Two rays with common endpoint O form a 30 :^o: angle.  Point A lies on one ray, point B on the other ray, and AB = 2.  The maximum possible length of OB is ...

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[b]6.[/b] If log[size=75]24[/size]p = log[size=75]9[/size]q = log[size=75]64[/size](p + 2q), what is p/q?

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[b]7.[/b] Two parallel chords in a circle have lengths 40 and 56, and the distance between them is 24.  The chord parallel to these chords and midway between them is of length :sqrt:a where a is ...

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[b]8.[/b] How many positive integer solutions (x, y) are there that satisfy 1/x + 1/y = 1/36?

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[b]9.[/b] For how many three-element sets of positive integers {a, b, c} is it true that a * b * c = 2730?

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[b]10.[/b] If :sqrt: (4 + :sqrt: x) =5, then x = ...

*If you have any questions or suggestions regarding this, contact me at WarpedKlown51435 on AIM or email me at soccerboyyp35@yahoo.com or yanman12@hotmail.com*

I would like to participate, unfortunately I know where almost all of the problems come from; it would be a bit too unfair for me to join in ...


We've seen many students ask about how to do graphics in LaTeX.  We finally have a good answer.  Maria Monks (former USAMTS gold medalist) has built an excellent guide for the AoPSWiki to the graphics package Asymptote. Check it out [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_About]here[/url].

This is great!  However, are we allowed to attach a .asy file when we submit our USAMTS solutions?

Yes, and we'd like the actual graphic file that is produced, as well.

[quote="elephantmaster"]Are we allowed to modify the preamble of the solutions.tex file to include the line:
[code]\usepackage{asymptote}[/code]?[/quote]

I'd assume so. They'll probably change the file to include that anyway.


Find all paris of positive integers $ (m,n)$ satisfying the equality

                            $ (m\minus{}n)^2(n^2\minus{}m)\equal{}4m^2n$

 :)

$ nm=0$ implies the solution $ n=m=0$. So let $ 0\not= nm$ and $ n\not= m$ from now on.

Let $ t=\gcd(m,n)$, then $ \left(\frac{n}{t}{-\frac{m}{t}\right)^2\left(\frac{n}{t}n-\frac{m}{t}\right)=4\left(\frac{m}{t}\right)^2\frac{n}{t}}$ from which we conclude that $ \left(\frac{n}{t}\right)^3n\equiv 0\bmod{\frac{m}{t}}$ and $ \left(\frac{m}{t}\right)^3\equiv 0\bmod{\frac{n}{t}}$ hold. The second equivalence gives $ n=\pm t$ and the first $ t\equiv 0\bmod{\frac{m}{t}}$, i.e. we can write $ t=t_1t_2,m=t_1^2t_2$ and substitution yields the identity:

$ (\pm 1-t_1)^2(t_2-1)=\pm 4 t_1$

The functions $ f_{\pm}(x)=\frac{(\pm 1-x)^2}{\pm 4x}$ defined over $ \mathbb{Z}^*$ fullfill $ f_+(x)=f_-(x)$. Moreover, we check that $ |f_{\pm}(x)|>1$ for $ |x|\geq 6$ and $ \frac{1}{f_{\pm}(x)}\in\mathbb{Z}\Longleftrightarrow x\in\{\mp 1,\pm 2\}$ with corresponding function values $ f_+(-1)=f_-(1)=-1,f_+(2)=f_-(-2)=\frac{1}{8}$.

The identification $ t_2-1=\frac{1}{f_{\pm}(t_1)}$ gives the additional solutions $ (n,m)=(-2,2),(18,36)$

P.S produced a typo  :blush: , however, it does not change the proof (it should read $ f_\plus{}(x)\equal{}f_\minus{}(\minus{}x)$)

ok but...
there is one more solution in positive integers...
so you must be missing something

:!:


The sum of the perimeters of three squares is 24 inches. What is the smallest possible sum of their areas in inches squared?

how would you do this problem?

[hide="Solution"]$4a+4b+4c=24\iff a+b+c=6$, and we're asked for the minimum of $a^{2}+b^{2}+c^{2}$. We can use the relation between the quadratic mean and the arithmetic mean:

$\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\geqslant\frac{a+b+c}{3}$

That yields $\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\geqslant 2\iff a^{2}+b^{2}+c^{2}\geqslant 12$, and the equal sign holds only if $a=b=c=2$.[/hide]


Find all positive integers $ k$ with the property that there are four distinct divisors of $ k$, $ k_1,k_2,k_3,k_4$, such that $ k$ divides $ k_1\plus{}k_2\plus{}k_3\plus{}k_4$.

[hide]Let $ k_1>k_2>k_3>k_4$.  Then if $ k_1 \le \frac{k}{2}$, then $ k_1\plus{}k_2\plus{}k_3\plus{}k_4<2k \implies k_1\plus{}k_2\plus{}k_3\plus{}k_4\equal{}k$.

If $ k_1\equal{}k$, it suffices to find three divisors which sum to $ k$.

So essentially we want to solve the following two equations in distinct positive integers (separately).
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}1$
The only solution to this one is $ 2, 3, 6$.  Clearly one of the three numbers must be $ 2$: otherwise, equality cannot hold.  Then the other two add to $ \frac{1}{2}$, so one of them must be $ \frac{1}{3}$, and we have our only solution.

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}\equal{}1$.
Let $ a<b<c<d$.  Because $ \frac{1}{3}\plus{}\frac{1}{4}\plus{}\frac{1}{5}\plus{}\frac{1}{6}\equal{}\frac{3}{4}\plus{}\frac{1}{5}<1$, we have $ a\equal{}2$.  Because $ \frac{1}{6}\plus{}\frac{1}{7}\plus{}\frac{1}{8}<\frac{1}{2}$, we have $ b \ge 5$.  From here it just feels like casework.  The case $ b\equal{}3$ yields $ (2, 3, 7, 42)$, $ (2, 3, 8, 24)$, $ (2, 3, 9, 18)$.  The case $ b\equal{}4$ yields $ (2, 4, 5, 20)$, $ (2, 4, 6, 12)$.  The case $ b\equal{}5$ yields no solutions, because then $ 5|c, 5|d$, and we get a contradiction.

So we get the solutions $ (a, b, c, d)\equal{}(1, 2, 3, 6), (2, 3, 7, 42), (2, 3, 8, 24), (2, 3, 9, 18), (2, 4, 5, 20), (2, 4, 6, 12)$.  So $ k$ has the property if and only if it's divisible by one of $ 6, 42, 24, 18, 20,$ or $ 12$.  That reduces to being divisible by either $ 6$ or $ 20$.
[/hide]


algebra
polynomial
induction
algebra unsolved

There is a broken computer such that only three primitive data $c$, $1$ and $-1$ are reserved. Only allowed operation may take $u$ and $v$ and output $u \cdot v + v.$ At the beginning, $u,v \in \{c, 1, -1\}.$ After then, it can also take the value of the previous step (only one step back) besides $\{c, 1, -1\}$. Prove that for any polynomial $P_{n}(x) = a_0 \cdot x^n + a_1 \cdot x^{n-1} + \ldots + a_n$ with integer coefficients, the value of $P_n(c)$ can be computed using this computer after only finite operation.

Proof:Lemma:For all $n\in Z$,we can compute $x+n$ from $x$ using this computer after only finite operation.
the proof of lemma:It's obvious that $n=0$,
If $n<0$,let $u=x,v=1$,$uv+v=x+1$,so like this after $n$ operations we have $x+n$.
If $n<0$,let $u=x,v=-1$,$uv+v=-x-1$;then let $u=-x-1,v=1$,$uv+v=-x$,like this after $-n$ operations we have $-x-n-1$;then let $u=-x-n-1,v=-1$,$uv+v=x+n$.So we get the lemma.
We use induction to prove the origin problem.
$n=0$,by lemma,we can compute any $P_0(c)=a_0\in Z$ from $c$ using this computer after only finite operation.
Then we suppose that $n=k$ the conclusion is correct.
For $n=k+1$,first we can get $y=a_0c^{n-1}+a_1c^{n-2}+\ldots+a_{n-1}-1$ by the assumption.
Then let $u=y,v=1$,$uv+v=a_0c^n+a_1c^{n-1}+\ldots+a_{n-1}c$.
Then by lemma we can compute $P_n(c)=a_0c^n+a_1c^{n-1}+\ldots+a_{n-1}c+a_n$ from $a_0c^n+a_1c^{n-1}+\ldots+a_{n-1}c$ using this computer after only finite operation.So it's also correct for $n=k+1$ and we know the conclusion is correct for any nonegative integer $n$.


trigonometry
geometry
circumcircle
perpendicular bisector
geometry unsolved

$ BK\cap AD \equal{} {Q}$ $ \implies$  $ CL\cap AD \equal{} {Q}$

if $ BK\parallel DL$  $ \implies$  $ \frac {|DC|}{|BD|} \equal{} 2$

But $ D$ is any point at $ |BC|$

No, it isn't any point, it is [b]the[/b] point on BC such that A,M,N,K,L all lie on the same circle.

it's obvious that $ KL \parallel{} MN$.
also $ A,M,N,L,K$ lie on the same circle which implies that $ \angle LNM \equal{} \angle NMK$ which implies $ DM \equal{} DN$.so $ D$ is the intersection of the perpendicular bisector of $ MN$ and $ BC$.
since BK//DL iff BD/DC=2 so we must prove that BD/DC=1/2.
u can see that in this case $ \frac {CD}{BD} \equal{} \frac {\frac {b^2 \minus{} c^2}{4a} \plus{} CT}{BH \plus{} \frac {b^2 \minus{} c^2}{4a}}$.($ H$ is the projection of $ A$ on $ BC$ and $ T$ midpoint of $ BC$)
which can be equal to $ 2$ iff $ 2.BH \plus{} \frac {b^2 \minus{} c^2}{4a} \equal{} \frac {a}{2} \implies 2.\frac {a^2 \plus{} c^2 \minus{} b^2}{2a} \plus{} \frac {b^2 \minus{} c^2}{4a} \equal{} \frac {a}{2} \implies 3b^2 \minus{} 3c^2 \equal{} 2a^2$
so we must prove the last equality.
now use the equality
$ (\frac {1}{3}.S_{\Delta DMN})^2 \equal{} (S_{\Delta MLN})^2 \equal{} (\frac {1}{2}.\sin \angle A.NL.ML)^2$
but i 'm looking for a better proof...

Thank you very much! For me, your proof is better than better! But could you please explain how does tha last equation lead to $ 3b^2\minus{}3c^2\equal{}2a^2$...?

plz tell me if i'm wrong.
we know that $ A,L,M,N,K$ lie on the same circle iff $ P_{D}^{(AMN)}\equal{}\frac{2}{3}.DM^2$.
let $ O$ be the circumcenter of $ AMN$ then we'll have $ P_{D}^{(AMN)}\equal{}DO^2\minus{}(\frac{R_{\Delta ABC}}{2})^2$.
we define $ R_{\Delta ABC}\equal{}r$.
let $ H_a$ be the projection of $ A$ on $ BC$ and $ H$ be the orthocenter of $ \Delta ABC$,and $ T$ be the midpoint of $ BC$.
it's obvious that :
$ DO\equal{}\frac{AH_a}{2}\plus{}\frac{AH}{4}$ and $ r^2\equal{}(\frac{AH}{2})^2\plus{}(\frac{a}{2})^2$.
also we have $ DM^2\equal{}(\frac{AH_a}{2})^2\plus{}(\frac{a}{4})^2$
by using the equalities we get:
$ \frac{2}{3}.(\frac{AH_a}{2})^2\plus{}\frac{2}{3}.(\frac{a}{4})^2\equal{}(\frac{AH_a}{2}\plus{}\frac{AH}{4})^2\minus{}(\frac{(\frac{AH}{2})^2\plus{}(\frac{a}{2})^2}{4})$

$ \implies \frac{2}{3}.(\frac{AH_a}{2})^2\plus{}\frac{2}{3}.(\frac{a}{4})^2\equal{}(\frac{AH_a}{2})^2\plus{}(\frac{AH}{4})^2\plus{}\frac{AH_a.AH}{4}\minus{}(\frac{(\frac{AH}{2})^2\plus{}(\frac{a}{2})^2}{4})$

$ \implies \frac{2}{3}.(\frac{a}{4})^2\equal{}\frac{1}{3}.(\frac{AH_a}{2})^2\plus{}(\frac{AH}{4})^2\plus{}\frac{AH_a.AH}{4}\minus{}(\frac{AH}{4})^2\minus{}(\frac{a}{4})^2$

$ \implies \frac{5}{3}.(\frac{a}{4})^{2}\equal{}\frac{1}{3}.(\frac{AH_{a}}{2})^{2}\plus{}\frac{AH_{a}.AH}{4}$

$ \implies \frac{5}{3}.(\frac{a}{4})^{2}\equal{}\frac{1}{3}.(\frac{AH_{a}}{2})^{2}\plus{}\frac{AT^2\minus{}(\frac{a}{2})^2}{4}$

note that in the above equality we used this $ AH.AH_a\equal{}P_{A}^{(BC)}$.we continue:

$ \implies \frac{8}{3}.(\frac{a}{4})^2\equal{}\frac{1}{3}.(\frac{AH_{a}}{2})^{2}\plus{}\frac{\frac{1}{2}.(b^2\plus{}c^2)\minus{}(\frac{a}{2})^2}{4}$

$ \implies \frac{11}{3}.(\frac{a}{4})^2\equal{}\frac{1}{3}.(\frac{AH_{a}}{2})^2\plus{}\frac{1}{8}.(b^2\plus{}c^2)$

$ \implies \frac{11}{2}.a^2\equal{}2.AH_a^2\plus{}3.(b^2\plus{}c^2)$

$ \implies \frac{11}{2}.a^2\equal{}(b^2\minus{}CH_a^2)\plus{}(c^2\minus{}BH_a^2)\plus{}3.(b^2\plus{}c^2)$

$ \implies \frac{11}{2}.a^2\plus{}CH_a^2\plus{}BH_a^2\equal{}4.(b^2\plus{}c^2)$

$ \implies \frac{11}{2}.a^2\plus{}(CH_a\plus{}BH_a)^2\minus{}2.BH_a.CH_a\equal{}4.(b^2\plus{}c^2)$

$ \implies \frac{11}{2}.a^2\plus{}a^2\equal{}4.(b^2\plus{}c^2)\plus{}2.BH_a.CH_a$

$ \implies \frac{13}{2}.a^2\equal{}4.(b^2\plus{}c^2)\plus{}2.(\cos \angle B.c).(\cos \angle C.b)$

$ \implies \frac{13}{2}.a^2\equal{}4.(b^2\plus{}c^2)\plus{}2.(\frac{a^2\plus{}(c^2\minus{}b^2)}{2a}).(\frac{a^2\minus{}(c^2\minus{}b^2)}{2a})$

$ \implies \frac{13}{2}.a^2\equal{}4.(b^2\plus{}c^2)\plus{}2.(\frac{(a^2)^2\minus{}(c^2\minus{}b^2)^2}{4a^2})$

$ \implies \frac{13}{2}.a^2\equal{}4.(b^2\plus{}c^2)\plus{}\frac{a^4\minus{}b^4\minus{}c^4\plus{}2b^2c^2}{2a^2}$

$ \implies \frac{13}{2}.a^2.2a^2\equal{}4.2a^2(b^2\plus{}c^2)\plus{}a^4\minus{}b^4\minus{}c^4\plus{}2b^2c^2$

$ \implies 12a^4\equal{}8a^2b^2\plus{}8a^2c^2\minus{}b^4\minus{}c^4\plus{}2b^2c^2$

now how can we deduce $ 2a^2\equal{}3b^2\minus{}3c^2$?
since $ 12a^4\equal{}8a^2b^2\plus{}8a^2c^2\minus{}(b^2\minus{}c^2)^2$ and we must prove that $ 4a^4\equal{}9.(b^2\minus{}c^2)^2$ so by combining these equalities we get
$ 12a^4\equal{}8a^2(b^2\plus{}c^2)\minus{}\frac{4}{9}.a^4$

$ \implies 3a^2\equal{}2(b^2\plus{}c^2)\minus{}\frac{a^2}{9}$

$ \implies 28.(\frac{a}{3})^2\equal{}2.(b^2\plus{}c^2)$

$ \implies 14.(\frac{a}{3})^2\equal{}b^2\plus{}c^2$.

now u can easily get $ b^2\equal{}\frac{10}{9}.a^2,c^2\equal{}\frac{4}{9}a^2$ (because we had $ \frac{2}{3}.a^2\equal{}b^2\minus{}c^2$)

but we can easily find a solution $ a,b,c$  with $ b>c,a\plus{}c>b,b\plus{}c>a$  for the equation $ 12a^4\equal{}8a^2(b^2\plus{}c^2)\minus{}(b^2\minus{}c^2)^2$
which doesn't satify $ b^2\equal{}\frac{10}{9}.a^2,c^2\equal{}\frac{4}{9}a^2$.
take for example
$ a\equal{}1,c\equal{}\frac{2}{3}\minus{} 10^{\minus{}2}\equal{}0,656666...$
$ b\equal{}\sqrt{(\frac{2}{3}\minus{}10^{\minus{}2})^2\plus{}4\minus{}2.\sqrt{1\plus{}4(\frac{2}{3}\minus{}10^{\minus{}2})^2}}\equal{}1,062916056...$
so we can find a triangle with sides $ a,b,c$ with $ A,L,N,M,K$ on the same circle but without $ BK \parallel{} DL$.
so your problem is wrong unfortunately :(
(sorry if it isn't so!)

in my first post in this topic i had thought that the problem is true that's why in the end i didn't write the computations since if your problem was true then we will arrive certainly at the equality $ 2a^2=3b^2-3c^2$ after computing.to be more clear...
in place of $ (\sin \angle A)^2$ we place $ 1- (\cos \angle A)^2$ similarly for $ (\sin \angle MDN)^2$ and then we can easily write the cosinus in terms of the sides of the triangles $ \Delta DMN$ and $ ABC$.also we must write the sides $ DM^2$ and $ ML^2$ in terms of the sides of the triangles $ ADB$ and $ DMN$.after doing these we must write the sides of the triangle $ DMN$ and $ \Delta ADB$ in terms of $ a,b,c$.(warning,there will be a lot of computations...)
but by doing this , i think,now, that we'll arrive at the same equation $ 12a^4=8a^2(b^2+c^2)-(b^2-c{}^2){}^2^$ not at $ 2a^2=3b^2-3c^2$!
anyway,check plz my previous post and tell me if there is sth wrong with....

[quote="Spribo"]What does $ P_{D}^{(AMN)}$ mean?[/quote]
the power of the point $ D$ wrt the circle passing through three points $ A,M,N$

Thank you! I can't see anything wrong, so I may really be wrong. Well, the original problem was to prove DM=DN, but after proving that, it seemed to me that BK//DL, so I tried to prove or disprove it but without success...

never mind,but ... plz, the next time u post another topic like this,  tell us if u are sure of the statement or not...
thx :)

[quote="Spribo"]Let ABC a triangle with AB<AC, M the midpoint of AB, N of AC. Let D a point on BC. Let K the barycenter of ADB and L of ADC. Prove that if A,M,N,K,L all lie on the same circle, then BK//DL.[/quote]
A,M,N,K,L all lie on the same circle. Let P be the midpoint of BC, H be the foot of the altitude from A.
$ BK\parallel DL \iff \frac{BH}{BO}\equal{}\frac{1}{3}$  :D


limit
logarithms
number theory unsolved
number theory

we define $A(n,k)=LCM(n,n-1,...,n-k+1)$
a)If $f(n)$ is the biggest $k$ such that $A(n,1)<A(n,2)<...<A(n,k)$ prove that there exist $c$ such that $f(n)<c\sqrt{n}$
b)Prove that  $\lim _{n\rightarrow \infty}f(n)=+\infty$.

a) Let $m=[\sqrt{n/2}].$ We have $2(m-1)*(m-1)<n$ and $2(m-1)*(m-1)+m-1<n.$ Therefore $f(n)\le n-2(m-1)^{2}<8\sqrt n$.
b)I think it is very hard.

b) Let $n-f(n)=q_{1}*q_{2}*...*q_{k},q_{i}=p_{i}^{a_{i}}$,were $q_{i}$ powers of primes. We have $f(n)\ge q_{i}$. Therefore $f(n)>k_{0}$,were $k_{0}^{k_{0}}<n$. It give $f(n)>\sqrt{\ln n }$ for big n.


Just a general remark, which you can take for granted (and if you are a math teacher, as you said somewhere, it seems strange that you have doubts about it):

$a^{b^c}$ can never mean $(a^b)^c$ because the latter case would always be written as $a^{bc}$ (therefore the second case would simplify to $2006^{2005!}$)

What you say is true, but what I have written is still valid.

For example, under the second condition $3^{2^{3}}=9^{3}=3^{2*3}$ like you said. Just solve the problem anyways.

No zellex we're looking at the second condition in the original problem. You're describing the first condition. I think that, if left unspecified, what you're writing is what you assume, but the second part of the problem is different.


calculus
derivative
logarithms
induction
real analysis
real analysis solved

You have $\frac{d^n}{dx^n}(x^n\ln x) = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1} \ln x) + (n-1)!$, which by induction gives $\frac{d^n}{dx^n}(x^n\ln x) = a_n \cdot ln(x) + b_n$ with $a_n=n! , b_n= n! \cdot \sum_{k=1}^n \frac{1}{k}$

I have found
\[\frac{d^n}{dx^n}(x^n\ln x)=n!\left(\ln x+\sum_{k=1}^n\frac{1}{k}\right)\]

[quote="kunny"]Let $n$ be positive integer.
Find the following derivative.
\[\frac{d^n}{dx^n}(x^n\ln x)\][/quote]

If by recurrence you suppose $f_n =\frac{d^n}{dx^n}(x^n\ln(x))$,
if $n=1$, then $f_1=\ln(x)+1$
Moreover, if $n\geq 1$ then $f_{n+1} =   \frac{d^{n}}{dx^{n}} \frac{d}{dx}(x^{n+1}\ln(x))$
thus $f_{n+1} = \frac{d^{n}}{dx^{n}}( (n+1)x^n\ln(x) + x^n)$
then $f_{n+1}=(n+1)f_{n} + n!$ by linearity of the derivative.
We get then $f_n = n! \cdot \ln(x) +c_n$ where $c_n$ is an integer.
To be more precise, $c_1=1$ and $c_n= n\cdot c_{n-1} + (n-1)!$.
Then, you put $b_n = c_n/n!$ you verifiy the relation
$b_n - b_{n-1}= \frac{1}{n}$, you sum this relation
and you get $b_n =  \sum_{k=1}^{n} \frac{1}{k} $.
And then, $c_{n} = n! \cdot \sum_{k=1}^{n} \frac{1}{k}$.

Fabrice


An electron from $ \text{P680}$ is used to replenish the electron loss in $ \text{P700}$ of plant cells. A photon of light strikes P700, igniting an electron to the acceptor, $ \text{NADP}^\plus{}$. Is the electron from any atom in the porphyrin ring?

What are "P680" and "P700"? What is the meaning of "igniting an electron to the acceptor"?

P680 - Photosystem II
P700 - Photosystem I

in the thylakoid membrane of chloroplasts.

When a photon of light hits an electron, it excites an electron to a higher energy level, that can be given off to an acceptor.

electron is from h2 not porphyrin---i believe

@LevatorNasolabialis: what molecule is in your figure? i thought chlorophyll had magnesium, not manganese.

just out of curiosity, what do P680 and P700 stand for (not Photosystem I and II, but the numbers 680 and 700)? Are they the names of the proteins where the electron transport chain occur?

EDIT: nvm. i looked it up. its the wavelength that the chlorophyll absorbs best.

[quote="LevatorNasolabialis"]Maybe this will help[/quote]
that's certainly not Chl....
@mwpl--in order of discovery-PS 1 is P700(700 is light's wavelength)
PS 2-P680

[quote="rituraj007"]
@mwpl--in order of discovery-PS 1 is P700(700 is light's wavelength)
PS 2-P680[/quote]

yes, i know. i remember it by the fact that its backwards. well, not the wavelengths, but the order in which the photosystems occur in the sequence of the light reactions.

they r numbered on order of their discovery

...which is the reverse of their actual order  :lol: .


calculus
integration
calculus computations

Could anyone please tell me the difference between $\int$ and $\oint$. Thanks

The first symbol is general; you're never wrong to use it. The second symbol is used specifically for a line integral around a closed curve, and usually appears in situations that call for theorems about such integrals.


This is from the USSR Olympiad Problem Book.

Find a six-digit number whose product by 2, 3, 4, 5, or 6 contains the same digits as did the original number.

And if you haven't already, you should commit it to memory :)

[quote="Gyan"][quote]you should commit it to memory [/quote]
Why? Can't you just remember how to divide by 7? :D[/quote]It sounds like you already know the answer.. Would you like to share?

[quote]It sounds like you already know the answer.. Would you like to share?
[/quote]

[hide]
Fairly well know fact from number theoy ,,
Just divide (1/7)  = .142857  142857 .....

Take the number 142857   which has this property.

(In short it comes from the fact that  7 is prime and   10^6-1 is divisible by 7  you may also try the 18 digit repeating decimal in 1/19 and see how the digits behave when you multiply by 1,2,3,4 ,,,,)   Or look for  repeating decimals in reciprocals of  even larger prime numbers ... and you can go much higher than 2,3,4,5,6 ....

Added later:
When you think about it: (try to do the division by hand)  it will be obvious that   1/7, 2/7 3/7 ...6/7  .. the pattern will repeat   once the "remainder" becomes 1 and .. hence   the above trick work,
[/hide]

[quote="Gyan"][quote]you should commit it to memory [/quote]
Why? Can't you just remember how to divide by 7? :D[/quote]

Thats the bit I was meaning when I said to remember it :P

MMM :   I   knew that.. in any case, not to waste this post,  here is some more fun observation:

 a b c d e f    digits from  above - not only have this woderful property mentioned above, but also:

a b c + d e f  = 999
a b +  c d + e f = 99

( a b c means just write a b c in decimal notation)

Enjoy

Of course, when one thinks about it, it is obvious, why it will be so...


The problems states: "Prove that there are infinite primes $ p$ such that $ \minus{}3$ is a quadratic residue $ \mod{p}$".

I feel it is very easy, but I can't solve it u_u ...

This is equivalent to the claim that there are infinitely many primes dividing a value of the polynomial $ x^2 \plus{} 3$ at an integer.  There is a standard "Euclidean" proof of this; are you familiar with the Euclidean proof of the infinitude of the primes?


1. Suppose P(x) = $ x(x \plus{} 1) \ldots (x \plus{} k \minus{} 1)$, for some integer $ k$. Let $ p$ be some prime number. Find, in terms of $ k$, the greatest integer $ e$ such that $ p^e | P(x)$ for all integers $ x$.

2. Suppose $ P(x) \equal{} x^{a_1} (x \plus{} 1)^{a_2} \ldots (x \plus{} k \minus{} 1)^{a_k}$, for some integer $ k$. Pet $ p$ be some prime number. Find, in terms of $ a_i$ and $ k$, the greatest integer $ e$ such that $ p^e | P(x)$ for all integers $ x$.

[hide="1, hint"] Why does it suffice to take $ x \equal{} 1$? [/hide]


vector
algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Recall that P3 is the vector space of polynomials of the form p(z) = a + b*z + c*z^2, where a, b, c are in the Complex field. Can the linear operator d/dz be diagonalized over this vector space? If so,
how is it done? If not, why not?

If I understand correctly the linear operator with respect to the basis $\{1, z, z^{2}\}$ would be $\left(\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 2\\ 0 & 0 & 0\\ \end{matrix}\right)$ so the minimal polynomial would be $x^{3}$, and thus has repeated roots. Therefore it is not diagonalizable.


Ok, so my sister got this extra credit problem on her homework...here goes:

There are 3 bags of marbles; red, blue, and red and blue (mixed, not 2 colored marbles). The labels are wrong. What are the [i]least[/i] amount of marbles you have to draw in order to find what the correct labels are.

The teacher says one, my family said 3. I'm thinking now about you know that all the labels are wrong, but I haven't finished thinking. Thoughts and opinions?

Edit: Figured it out. If all of them have to be wrong (which was the part we didn't realize) then if you pick a blue marble out of the pouch labled blue, it can't be blue, so it has to be mixed. The other ones are red and mixed. So the pouch labeled mixed is red and the pouch labeled red is blue.

Figured it out all by my eigth grade self  :D

[quote="i_like_pie"][quote="`Patrick"]Edit: Figured it out. If all of them have to be wrong (which was the part we didn't realize) then if you pick a blue marble out of the pouch labled blue, it can't be blue, [b]so it has to be mixed.[/b] The other ones are red and mixed. So the pouch labeled mixed is red and the pouch labeled red is blue.[/quote]
It can be either mixed or red.[/quote]

Why would there be blue marbles in a bag of red?

Anyway,
[hide]
Drawing from the blue or red bag does not work, here's why.

Draw from the red bag.

Case 1: You get a red marble.
Since this bag isn't red as the label is wrong, but it contains red marbles, it must be mixed.
The blue bag must be red in that case and it follows that hte mixed bag is blue.

You need 1 marble

Case 2: You get a blue marble.
You know this bag isn't red, but it could be blue or mixed. The blue bag could be mixed or red depending on this bag and the mixed bag could be red or blue depending on the last two bags.

You can easily swap red with blue in the last example.

Instead, you have to draw from the mixed bag.

Case 1: Marble is red
This bag isn't mixed and it contains red so it must be red. The blue bag must therefore be mixed as it can't be blue or red and the red bag must be blue.

1 marble needed

Case 2: Marble is blue
This bag isn't mixed and it contains blue so it must be blue. The red bag must therefore be mixed as it can't be blue or red and the blue bag must be red.

1 marble needed.

So the answer is 1 if you always draw from the mixed bag[/hide]

her answer isnt wrong
If you choose the mixed bag, whichever color it is is the color of the bag. So if you chose a red from the mixed that is red. Then that means that the bag that says blue must be mixed and red is mixed. It works both ways.

Edit: Figured it out. If all of them have to be wrong (which was the part we didn't realize) then if you pick a blue marble out of the pouch labled blue, it can't be blue, so it has to be mixed. The other ones are red and mixed. So the pouch labeled mixed is red and the pouch labeled red is blue.

Figured it out all by my eigth grade self  :D[/quote]

Yeah but if you choose a red from the blue bag it can be mixed or red the answer has to work all the time. Only my answer works all the time.


A rhombus is given with one diagonal twice the length of the other diagonal.  Express the side of the rhombus in terms of $K$, where $K$ is the area of the rhombus in square inches.
(A) $\sqrt{K}$
(B) $\frac{1}{2}\sqrt{2K}$
(C) $\frac{1}{3}\sqrt{3K}$
(D) $\frac{1}{4}\sqrt{4K}$
(E) none of these is correct

Hint:

[hide]
Let $d_1=x$ and $d_2=2x$. Then

\[A=K=\frac{d_1d_2}{2}\]
\[K=x^2\]

Since we are dealing with a rhombus the diagonals bisect each other at right angles so we form 4 congruent right triangles with legs of $x$ and $\frac{x}{2}$ and the side of the rhombus is their hypotenuse. You should be able to handle it from here. :)

[/hide]

[hide]The area of a rhombus is given by half the product of the diagonals so x*2x=k/2  4x^2=k   x= :sqrt: k/2 x being shorter diagonal. Now using the pythogorean theorem we can find the side by doing  ((:sqrt: k)) 2  + (( :sqrt: k)/2) 2 =s where s is the side so s^2 = k + k/4      s^2 = 5k/4 s =   (:sqrt: 5k)/2

[/hide]

is that right?


In another post, I asked how could we proceed to factorize the polynomial $x^{7}+x^{2}+1$ in the following way

$x^{7}+x^{2}+1 = (x^{2}+x+1)(x^{5}-x^{4}+x^{2}-x+1)$.

but no one answered.
Now, can someone provide an explanation on the above factorization?
Thanks.

$x^{7}+x^{2}+1=x(x^{6}-1)+x^{2}+x+1$. when I saw the problem, I thought that plugg $x=\omega$ into this equation, where $\omega$ is one of imaginary roots such that $x^{3}=1$.

You can use the same method for a polynomial as this one:$x^{3p+2}+x^{3q+1}+x^{3r}$.


i live near delhi...
did any1 living in or near delhi got ur performance card :?:
pls reply soon bcoz i haven't got mine...

lets give our NAME/CLAss/marks in only inao sr and make a list

i venkat/11/------------- 50/100


I am an eighth grader who competed in the MATHCOUNTS Nationals. I have read Volume 1 of the Art of Problem Solving nad am currently reading Volume Two. My goal is to at least make it to the USAMO.

Will it be better for me to take a subject class (i.e. Intermediate Algebra) or an Independent Study class?

The Independent Study curriculum is far broader than any one subject class and provides you with one-on-one and small group mentoring time that, from what we have seen, definitely helps students develop more quickly.

Students who take the Independent Study also get to take a "companion course," so you would get to take Intermediate Algebra at no additional charge.

The Independent Study is in its first week, so if you are interested in registering, it's a good idea to do so as soon as possible.

[quote="NoSoupForYou"]I am interested in taking Intermediate Algebra and AIME Problem Series in addition to Intermediate Independent Study A. Is it a good idea to take all 3 classes at once?[/quote]

The Independent Study is not a small amount of work.  You can expect to spend around 8-15 hours per week on it.

We will offer another AIME Problem Series in the Winter/Fall.  You could take it then.

You could take the AIME Problem Series in addition to the Independent Study and companion class if you are trying to create a 20 hour/week program for yourself.  From what I've seen, you're certainly capable of it.  I just want to be sure you understand the dedication involved in taking so much on at once.


I got an email a couple days ago about it, even though I have no idea how they got my email in the first place (at first I thought it was a scam to get $350 out of me :blush:).  Since then, I've had the sense to search Google, but I still know little more than that it's a math competition.  Can anyone offer some basic information, especially in regards to how it works in Texas?  Is it worth going to?  Prestige factor (college apps lol) and difficulty?

Taken directly from the ARML website:

The American Regions Mathematics League's annual competition brings together the nation's finest students. They meet, compete against, and socialize with one another, forming friendships and sharpening their mathematical skills. Since its inception in 1976, ARML has snowballed, burgeoned, and mushroomed into a national program, involving almost 2000 students and teachers from almost every state. Simply put, ARML is the World Series of mathematics competitions. The contest is written for high school students, although some exceptional junior high students attend each year. The competition consists of several events, which include a team round, a power question (in which a team solves proof-oriented questions), an individual round, two relay rounds (in which a contestant solves a problem and passes his/her answer to another team member, who uses this answer to solve another problem), and a super relay.

As per the original email:

ARML is a math competition, but unique in the fact that it is

(1) nationally prestigious (starting to become globally so),
(2) contains both question-and-answer problems and proof-oriented problems, and
(3) gathers hundreds of math nerdy types together so that they can hang out and have a good time.

The Texas team travels to the Iowa City site, I believe (I see them there every year), and they tend to do very well. Being a part of a top-tier ARML team is definitely something that would look well on a college application.

That is, unless you have an aversion to armldillos...

They probably got your email from AMC or some other competition you did, or maybe your coach.  They have ways.

Being selected for Texas ARML is actually quite an honor.  (By the phrasing of your post, I'm assuming you've been selected.)  It means they believe you're one of the top 45 math students in the state, and deserve to be on what is essentially the all-star team.

By all means, please go to ARML.  It's not just the prestige of competing for the national championship, or of being on one of the best math teams in the country, or the problem style (which is extremely good, as I'm sure generating would tell you *wink wink*) you should go for.  Reason #3 is what really makes it different.  You'll meet and get to know a great group of kids during the ARML season, and then you'll go and compete alongside them.  You'll have the time of your life.

[quote="Sly Si"]...or the problem style (which is extremely good, as I'm sure generating would tell you *wink wink*)[/quote]

Yes, well, let's just say I'm working on it  :)

The chance to meet, make friends with, and learn from those people is one of the best reasons to do ARML.  One day, you may become the one people look up to--I speak from personal experience.

[quote="daermon"][quote="SweetAnesthesia"]Yah, sounds like fun :lol:

But after reading some of the other threads... [i]dang[/i].  So many godly-smart people; makes me want to sleep on my math textbook (cause we all know osmosis works) :P[/quote]
dont be overwhelmed by all the smart people at AoPS- this place has some of the best math people in the world (i'm not one of them :) )
i found out about AoPS last year at ARML (6th grade, got a 0 :D )
now i'm hoping to get a 1 or 2[/quote]
[i]Participating[/i] in ARML in sixth grade is pretty commendable all on its own.  I doubt I could’ve found the surface area of a cone in sixth grade (I sure know I couldn’t do it in seventh :roll:).  I'm a frosh now but none the less slow; just ask my friends :P

Calculators are allowed on the team and power rounds and prohibited on the individual and relay rounds.

I don't actually think that "participating in ARML" is worth anything to colleges since most teams are not very competitive. Even making one of the better teams is probably not worth much if you didn't win anything individually. But as other people have mentioned, it's the experience that matters.

Yes.  Even if you don't win anything, you still have a good time.


Nice proof, zaya_yc!
[quote="zaya_yc"]
; proved ;a=b=c=d
  :)[/quote]
Are you sure?

This appears as problem 2.1.5 of Secrets in Inequalities and Problem 2.21 of Sang Tao bat dang thuc. My solution is the same as Zaya's one.


modular arithmetic
induction
abstract algebra
number theory
relatively prime

A while ago, I came across this proof of Fermat's Little Theorem which I thought was interesting, so I decided to post it.

[hide="Theorem"]

Fermat's Little Theorem states: If $ a$ is an integer that is comprime to a prime number $ p$, then $ a^{p\minus{}1}\equiv 1\pmod{p}$.

[/hide]

[hide="Proof"]
Assume the set $ G\equal{}{1,2,...,p\minus{}1}$ under multiplication taken modulo $ p$ forms a group.

Assume $ a\in G$ and $ k$ is the order of $ a$ such that, $ a^k\equiv 1\pmod{p}$.

By Lagrange's Theorem, $ k$ divides the order (cardinality) of $ G$, $ p\minus{}1$.

Therefore, $ p\minus{}1\equal{}km$ for some positive integer $ m$.

So, $ a^{p\minus{}1}\equiv a^{km}\equiv (a^k)^m\equiv 1^m\equiv 1\pmod{p}$.

Lastly, we must prove that $ G$ is invertible:

Assume $ b\in G$ is relatively prime to $ p$.

By Bezout's identity, there exist integers $ x$ and $ y$ such that $ bx\plus{}py\equal{}1$.

In modulo $ p$ this identity states $ bx\plus{}0\equiv 1\pmod{p}$, therefore $ x$ is the inverse of $ b$. (note: $ py\equiv 0\pmod{p}$ because it is divisible by $ p$)

Therefore, $ G$ is a group and thus proving the theorem.

[/hide]

Either you or someone else has posted this to Wikipedia  :) [url=http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem]here is the page devoted to F[size=75]L[/size]T proofs[/url].

There are some other nice proofs on that page as well. The modular proofs are pretty standard, and the bracelet proof was new to me (it was a cute combinatorial argument).

What many people might not be familiar with the last one, which I suggest as reading. It's a formalization of what might otherwise seem heuristic logic.

Thanks for the link, I've only seen the page dedicated to the theorem itself but not the page on the various proofs. Your right, the last proof seems pretty non-standard, I haven't seen anything like it. Can you clarify how it works? I'm not very well aquainted with dynamical systems. I liked the proof I posted because it's concise. As for the bracelet proof using combinatorics, its kinda wierd. It seems like this theorem envoked alot of abstract methods of proof which is just as important as the theorem itself.

Thanks for the reponses.

The first proof I saw was by induction and the binomial theorem. I had also seen the one using modular arithmetic. But I never saw the bracelet proof before that's on wikipedia. That one is my favorite, since it combines combinatorics with an algebraic theorem, and the proof is very simple.

If you're going to invoke Lagrange's Theorem, you're really proving a rather general statement about the multiplicative group of a finite field.  That's using stronger tools than this specific problem really requires.

The bracelet proof relies on a combinatorial notion that some people find awkward, but which should seem natural as long as you trust combinatorial proofs:  [i]if an expression counts something, it's an integer.[/i]

[b]Example:[/b]  Prove that the product of $ n$ consecutive integers is divisible by $ n!$.

Proof:  $ {k \plus{} n \choose n}$ is an integer.  :)


Find all function $f: \Re \to \Re$ such that

$(f(x) + f(z))(f(y) + f(t)) = f(xy - zt) + f(xt + yz)$

for all $x,y,z,t \in \Re$

[hide="One function"]Square[/hide]
Any others?  :ninja:

first set $t=y=0$ and figure that $f(0)=0$
then set $z=t=0$ and reduce the problem to
\[ f(x)\cdot f(y) = f(x\cdot y) \]
with logs and cauchy one can prove that $f(x)=x^\alpha$ with $\alpha\in\mathbb R$

wait actually it int proven than $f$ can be any function$x\mapsto x^\alpha$, it is just proven, that any function $f$ solving the equation has that condition ...

now i got it ...
10000th User u found the only solution.

since $f(x)=x^\alpha$ we have $f(1)=1$
now setting $x=y=z=t=1$ we get $2^2 = f(2)$
there for $f(x)=x^2$

[quote="peeta"]
\[ f(x)\cdot f(y) = f(x\cdot y)  \]
with logs and cauchy one can prove that $f(x)=x^\alpha$ with $\alpha\in\mathbb R$[/quote]

Could you give some explanation about the cauchy thing? I've heard of it as being usefull in solving this kind of problem but what is it?

[quote="peeta"]now i got it ...
10000th User u found the only solution.[/quote]

$f \equiv 0$

http://www.kalva.demon.co.uk/imo/imo02.html

the cauchy functional equation says:
\[ f(x)+f(y)=f(x+y) \]
has $f(x)=ax$ as its only solution with $a$ being any real number ...

actually (by Athur Engel) all continous solutions of
\[ f(x)f(y)=f(xy) \]
are $f(x)=|x|^\alpha$, $f(x)=\text{sgn} x \cdot |x|^\alpha$, and $f(x)=0$ and $f(x)=1$ ...

What relates cauchy functional equation
$f$ must be continuous...
IT'S A VERY IMPORTANT CONDITION


For some reason, I can't view the Asymptote codes on blogs. If I hover over the graphic, it'll say "Click here to view the code." When I click it though, nothing pops up. Mozilla 3.0.1 and XP here.

Same here.
Opera (newest 7.0?) windows XP

I have the same problem.

Firefox 2.0, OS is Fedora 8.

I experience a similar problem with $ \text{\LaTeX}$ codes that are too long for the hovering thing, but I can let my cursor hover over short codes to view the code that pops up in a blue box thing.


$ \frac{1}{a^{\beta}(b\plus{}c)}\plus{}\frac{1}{b^{\beta}(a\plus{}c)}\plus{}\frac{1}{c^{\beta}(a\plus{}b)}\ge\frac{3}{2}$

where $ a,b,c >0\ and\ abc\equal{}1$

for$ \beta\in(x\ge2: x\in N)$

Follows Holder Inequality we have
$ LHS = \sum {\frac {\left(\frac {1}{a}\right)^{\beta}}{b + c}\ge \frac {\left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c}\right)^\beta}{3^{\beta - 2}2(a + b + c)}=\frac {(ab + bc + ca)^\beta}{3^{\beta - 2}2(a + b + c)} \ge \frac {3^{\beta - 2}(ab + bc + ca)^2}{3^{\beta - 2}2(a + b + c)}\ge\frac {3abc(a + b + c)}{2(a + b + c)} = \frac {3}{2}}$

 q.e.d.

[quote="Euclid_great"]$ \frac {1}{a^{\beta}(b \plus{} c)} \plus{} \frac {1}{b^{\beta}(a \plus{} c)} \plus{} \frac {1}{c^{\beta}(a \plus{} b)}\ge\frac {3}{2}$

where $ a,b,c > 0\ and\ abc \equal{} 1$

for$ \beta\in(x\ge2: x\in N)$[/quote]

Let $ a\equal{}\frac{1}{x}$,           $ b\equal{}\frac{1}{y}$,              $ c\equal{}\frac{1}{z}$

we have $ abc\equal{}1\Longleftrightarrow xyz\equal{}1$

then the given inequality is equilevant to

$ \frac{x^{{\beta}\minus{}1}}{y\plus{}z}\plus{}\frac{y^{{\beta}\minus{}1}}{z\plus{}x}\plus{}\frac{z^{{\beta}\minus{}1}}{x\plus{}y}\geq\frac{3}{2}$

From Andreescu inequality in general form we have

$ \frac{x^{{\beta}\minus{}1}}{y\plus{}z}\plus{}\frac{y^{{\beta}\minus{}1}}{z\plus{}x}\plus{}\frac{z^{{\beta}\minus{}1}}{x\plus{}y}\geq\frac{(x\plus{}y\plus{}z)^{{\beta}\minus{}1}}{3^{{\beta}\minus{}3}2(x\plus{}y\plus{}z)}\equal{}\frac{(x\plus{}y\plus{}z)^{{\beta}\minus{}2}}{2\cdot3^{{\beta}\minus{}3}}\geq\frac{3}{2}$

because $ x\plus{}y\plus{}z\geq3\sqrt[3]{xyz}\equal{}3$


complex analysis
complex analysis unsolved

Find the image of the following region:

(see attached image)

under the action of
$ f(z)\equal{}e^{\bar{z}}$
(e raised to z conjugate)

I am lost on this one.  Thanks for any advice with this one.

Beyond $ |x|\equal{}\frac{4\pi}{5}$, there's a block of at least $ 2\pi$ in the $ y$ direction, which means the exponential wraps all the way around. That means that the image contains everything with $ |z|>\exp\left(\frac{4\pi}{5}\right)$ and everything with $ |z|<\exp\left(\minus{}\frac{4\pi}{5}\right)$. In the annulus between those, we have the region between two spiral curves, which cross at 1.


algebra
polynomial
limit
integration
logarithms
calculus
algebra proposed

Prove that for each $ n$ there exist $ (P,Q)$ deg  $ P \equal{} 3n$. deg $ Q \equal{} 2n$ such that $ 3Q'P \minus{} 2P'Q \equal{} cons$

$ P(x) \equal{} ax^{3n}$
$ Q(x) \equal{} bx^{2n}$
$ 3Q'P \minus{} 2P'Q \equal{} 6bnx^{5n \minus{} 1} \minus{} 6nax^{5n \minus{} 1}$
If we chose $ bn \equal{} an\Longleftrightarrow a \equal{} b$ then
$ 3PQ' \minus{} 2QP' \equal{} 0$
What is your problem?

a pardon.   :oops:  conts $ <> 0$ i think its not that triavial

[hide="?"]
Since $ P,Q$ can be scaled, WLOG let $ Q(x)$ have positive leading coefficient and the constant be 1. The equation
\[ 3Q^\prime (x)P(x) - 2Q(x)P^\prime (x) = 1
\]
is equivalent to
\[ P^\prime (x) = \frac {3Q^\prime (x)}{2Q(x)}P(x) - \frac {1}{2Q(x)}
\]
For $ x$ greater than all roots of $ Q$, let ${ f(x) = \frac {P(x)}{Q(x)^{3/2}}}$ so that
\[ f^\prime (x) = \frac {P^\prime (x)}{Q(x)^{3/2}} - \frac {3Q^\prime (x)P(x)}{2Q(x)^{5/2}} = - \frac {1}{2Q(x)^{5/2}}
\]
But $ \lim_{n\rightarrow \infty} f(x)$ exists. If we let the limit be $ L$, then we must have
\[ f(x) = L + \int_x^\infty \frac {dt}{2Q(t)^{5/2}}
\]

\[ P(x) = LQ(x)^{3/2} + Q(x)^{3/2}\int_x^\infty \frac {dt}{2Q(t)^{5/2}}
\]
for all $ x$ greater than all roots of $ Q$.

This seems difficult, since dealing with 5/2 powers of nonsquare polynomials typically leads to logarithms or even nonelementary integrals, yet $ Q$ cannot have any repeated roots.

There is a solution where $ n=1$. For example, if we let $ Q(x)=x^2+1$, then $ P(x)=-\frac{1}{3}x^3-\frac{1}{2}x$ is a solution.
[/hide]

i would much appreciate for your solution  :) though its not complete. plus. there seems a mistake here: u cant assume (WLOG) that the leading coeff is $ 1$ but not the cons $ a$. .thank you anyway.

This one is very important to me! Ive been working on it n 've only found solutions for $ n\equal{}1,2,3$


Find the values of n $ \leq$ 7 for which n!+1 is a perfect square.

[hide=" solution"]

$ n\equal{} 4 \implies n!\plus{}1 \equal{} 25 \equal{} 5^{2}$

$ n\equal{} 5 \implies n!\plus{}1\equal{}121\equal{}11^{2}$

$ n\equal{}7 \implies n!\plus{}1 \equal{} 5041 \equal{} 71^{2}$

So there are $ 3$ values of $ n$: $ n\equal{} 4, 5$ and $ 7$.

[/hide]

Actually,I knew how to solve it in this way.I wanted a solution without actual calculation.

Still,thanks. :)


let c be a positive real number . complete the following representation of shifts in the graph of y= f(x)

vertical shift c units upward
vertical shift c units downward
horizontal shift c units to right
horizontal shift c units to left

Hmm... isn't this just too homework-like type of problem to be here in [b]Intermediate Topics?[/b]  :roll:


what is y(y-3x) if y=0 and x=3?
i typed in 0 and i got it wrong

You may have typoed it, read the question wrong, etc. Next time, at the conclusion of the game, click "review" and see what you actually typed and what the question was.

If you did this, and even if you didn't, there is also the possibility that the answer AoPS has is incorrect. This is not unusual at all. When reviewing problems, select "report error" if this occurs.  :)

Your answer is correct. Try to report the question next time, but the majority of the time if you report a problem, it is not fixed. Maybe implement some other kind of report question thing? Maybe when you report a question, ask them what they think the answer is, and if a lot of users enter the question wrong, and put that answer, change it automatically.


(-3/x+3y)+(4/x-2y)=9 find the point(s) of intersection.

i keep coming across problems like these... i know you are supposed to make m=1/(x+3y) and n=1/(x-2y) but i dont know where to go from there...pls and thanks =]

The points of intersection with what? The axes?

Please do not double-post! The same problem at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=131971[/url].


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Verify the Galois correspondence explicitly for $Q(\sqrt 2, \sqrt 3)$, the splitting field for $x^4-10x^2+1\in Q[x]$, and its Galois group. In other words, write down the lattice of subfields of $Q(\sqrt 2, \sqrt 3)$ and the lattice of subfroups of its Galois group, and show which intermediate field corresponds to which subgroup.

Thanks a lot!

Set $y=x^2$ and solve your equation, after that identify the decomposition field and its generators, take an element of the Galois group en see what are the possible image for the generators, all that will give you information to find the Galois groups and subgroups will give you subfield
(I've changed my precedent post which contain a mistake sorry)


Consider $n\times n$ chessboard. How many ways we can put 3 knights on the board such that no two knights are threatining each other? Positions are different if they are symmetric or we can get another position by rotating the board.


Find all functions$ f: \Bbb{R}_{0}^{ \plus{} }\rightarrow \Bbb{R}_{0}^{ \plus{} }$ with the following properties:

(a) We have$ f\left( xf\left( y\right) \right) \equal{}f\left( x\plus{}y\right)$ for all x and y belong$ \Bbb{R}_{0}^{ \plus{} }.$

[quote="angelstt"]Find all functions$ f: \Bbb{R}_{0}^{ \plus{} }\rightarrow \Bbb{R}_{0}^{ \plus{} }$ with the following properties:

(a) We have$ f\left( xf\left( y\right) \right) \equal{} f\left( x \plus{} y\right)$ for all x and y belong$ \Bbb{R}_{0}^{ \plus{} }.$[/quote]

Set $ x\equal{}0$ in the equation and you get $ f(y)\equal{}f(0)$ $ \forall y$ hence the solution $ \boxed{f(x)\equal{}c}$ constant $ \forall x\ge 0$, which indeed is a solution.


topology
advanced fields
advanced fields solved

I was studying topology and it seems that I have been confused with the basic things. In the book I read it was said that "a topologist is a person who sees no difference between a doughnut and a coffee cup. However, using the fundamental group he can find a difference between those."

Now I think that a doughnut and a coffee cup are homeomorphic to each other. On the other hand, Mathworld says that "The fundamental groups of homeomorphic spaces are isomorphic." contradicting that was said in my topology book.

Can anyone show me how to use the fundamental group to show some difference between those objects?

Coffeecup: http://www.clever.ro/images/coffeecup.jpg
Doughnut: http://www.erights.org/smart-contracts/donut-lab/images/donut-lab.gif
;)

And yes, both have first fundamental isomorphic to $\mathbb{Z}$ (when seen as solids). Also the part of the $\mathbb{R}^3$ not part of it (the complement) has that fundamental group. Similar the group is $\mathbb{Z}^2$ when only considering the surface...
Just out of couriosity: what's the name of the book¿

Topologia II, It is a Finnish book written by Jussi Väisälä.

Silly my. I got confused with the Finnish words "munkki" (a doghnut with no hole inside) and "munkkirinkilä" (a doghnut with hole inside). Great. I should study more Finnish.


$ lim_{n\minus{}\minus{}>inf}(\frac{a_n}{1}\plus{}\frac{a_{n\minus{}1}}{2}\plus{}\frac{a_{n\minus{}2}}{2^2}\plus{}...\frac{a_1}{2^{n\minus{}1}})$
$ liman\equal{}a$
$ n\minus{}\minus{}>oo$

what is your problem requested for ?

$ \boxed{\lim_{n\to \infty}(\frac {a_n}{1} \plus{} \frac {a_{n \minus{} 1}}{2} \plus{} \frac {a_{n \minus{} 2}}{2^2} \plus{} ...\frac {a_1}{2^{n \minus{} 1}}) ;\quad
\lim_{n\to \infty}a_n \equal{} a}$

Need to find the first limit and given by the second one? :roll:

The answer is simply $ 2a$.

Note that if $ a_n \equal{} a$ is a constant sequence, this result is trivial. In the case of general convergent sequence $ a_n$, we know that it becomes nearly constant for large $ n$.

Based on this observation, control the difference between $ \sigma_n \equal{} a_n \plus{} \cdots \plus{} a_1 2^{1\minus{}n}$ and $ 2a(1 \minus{} 2^{\minus{}n})$ by splitting its sum into two parts:

$ |\sigma_n \minus{} 2a(1 \minus{} 2^{\minus{}n})| \leq \sum_{N \leq k \leq n} 2^{n\minus{}k} |a_k \minus{} a| \plus{} \sum_{1 \leq k < N} 2^{n\minus{}k} |a_k \minus{} a|$


inequalities
LaTeX
inequalities proposed

x,y,z positive real numbers,such that x+y+z=1 and min(x,y,z)=z

1/z^3>=72/(2-(x^2+y^2+z^2))^2

excuse me for this form of publication

No matter Anni. Here is LaTeX form

For $ x,y,z > 0$ we have $ x \plus{} y \plus{} z \equal{} 1$ and $ \min (x,y,z) \equal{} z$.  Prove that $ \frac{1}
{{z^3 }} \geqslant \frac{{72}}
{{\left( {2 \minus{} (x^2  \plus{} y^2  \plus{} z^2 )} \right)^2 }}$.

thank you the sa.
can someone give to me a solution.it's so easy

can anyone solve thissssssssssssssssssssssssssssssssssssssssss?????????????????!!!!!!!!!!!!!!!!!!11111

Homogenising yields $ (\sum x^2 \plus{} 4 \sum xy)^2 \ge 72 (\sum x)z^3$. You can also take $ 75$ instead of $ 72$. At this point, just expand the LHS and notice that the inequality is symmetric in $ x$ an $ y$. Moreover, ever term except $ z^4$ is greater than or equal to $ xz^3$, $ yz^3$, respectively.


What does a question saying a square inscribed in a right triangle give to you?

Like if you have a triangle and they tell you it is isosceles you know there are 2 equal sides.

If you know a square is inscribed in a right triangle, what do you know about the relationship between the triangle and the square?

If you know the lengths of the sides of the triangle, you can find the side length of the square.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124587[/url]

If the square is tangent to the right angle of the triangle, Then its side length is $\frac{\sqrt2}{2}$ times the bisector of the right angle.  If it has a side along the hypotenuse, rather than each of the legs, then this might not hold true.