geometry
calculus
trigonometry
ratio
cyclic quadrilateral
trig identities
Law of Sines

Hi guys

I have a small problem now. I am taking the course in Further Mathematics SL in IB, which is the hardest of the four courses availible. I am doing it as self-study, since I'm the only student in my school taking it. I'm good (maybe not for your standards) in abstract algebra, statistics, calculus, discrete mathematics, etc. However, I'm terrible at geometry, because Norway doesn't have very much geometry and AP calculus and IB maths HL, haven't given me any introduction.

I have a topic about Eucledian Geometry, and I'm able to do only some of the questions in the begining of the section. They also put up a lot of stuff that have never been proved or shown to me before as presumed knowledge. Although, I'm able to prove it myself, I don't have the practice involve. It would be fine, if the book actually had a facit, but it doesn't. My teacher says that I should use Geogebra to construct the figures, and in that way I will be able to see how the angles relate. Do any of you have any idea, how I'm going to learn the geometry?

Is there any better books that could help me to teach the material?

The common response on AoPS would be to use:

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=100]AoPS Volume 1[/url]
[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=102]AoPS Volume 2[/url]
[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=204]AoPS Intro to Geometry[/url]

I have [url=http://www.amazon.com/Geometry-Revisited-Mathematical-Association-Textbooks/dp/0883856190/ref=sr_1_11?ie=UTF8&s=books&qid=1228342731&sr=1-11]this book[/url] its pretty good.

[i]Mod edited to put link on "this book" instead of posting a looong URL in post.[/i]

If you're looking for elementary Euclidean geometry, starting from what's typically taken as the beginning, I suggest starting with our [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=204]Introduction to Geometry[/url], which you can probably get easiest in Norway from [url=http://highperception.com]Highperception[/url].

If you're looking for more advanced text, the book hunter34 links will give you some solid, advanced material.

Thanks for all of the answers. I really appreciate you answering.

Well, I don't need a book that explains me the law of sines, or trigonometry, or something like that. However, I don't need to explain to me abstract geometry either.

To make you understand what I'm actually going to study, I will tell you what topics I have.
A: Using circle Theorems
B: Concyclic Points
C: Similarity
D: Intersecting chords Theorem
E: The equation of a circle
F: Concurrency in a triangle
G: Further theorems (apollonius, Stewarts, Ptolomy
H: Proportionality in right angled triangles
I: harmonic Ratios
J: Proportional division
K: Concurrency and Ceva's Theorem
L: Menelaus' theorem
M: Euler's line and the nine point circle

I actually don't have that much problems with understanding the theorems. The problem is when I'm going to use the thorems together in a question.

This is a question I wasn't able to do without asking for help, but I have done it now. it's the fifth question of section.
A5. PQ and RS are two perpendicular chords of a circle, centre O. Prove that <POS and <QOR are supplementary.
B8 OABC is a parellogram. A circle, at centre O and radius OA is drawn. BA produced meet the circle at D. Prove that DOCB is a cyclic quadrilateral
B24 PQ and RS are tangents to a circle, centre O, from an external point P. PS is perpedicular to PQ and meets OR, produced at S. QR produced meets PS produced at T. Show that triangle STR is isoceles.

This should help you understand what I need to learn. What book is my best bet.


Find 3 integers that form a geometric progression if their sum is 21 and the sum of their reciprocals is 7/12.

[hide]You could solve it the conventional way, but I don't think that is really necessary... I mean, you have xr = y, and yr = z, so xr^2 = z, so x+ xr + xr^2 = 21, and so x(1+r+r^2) = 21, and you know that for r^2 + r + q, to have integer solutions, q = (n)(n+1), so notice that since r^2+r+1 = 21/x, for x to be an integer, that 1 - 21/x has to equal (n)(n+1), so you try x = 1,3,7,21, and for 1, that quantity is -20. So, r^2+r-20, and (r+5)(r-4) = 0. So r = 4. So, you have (1,4,16). Tada![/hide]


induction
number theory
relatively prime

Choose any (n+1) element subset from {1,2,3,..., 2n}. Show that this subset must contain two integers that are relatively prime.

[hide="Solution"]To ensure that there are  elements that are coprime, we must have 2 consecutive positive integer elements. Consider the worst case scenario when choosing $ n$ elements from the set: none of them are coprime. There are 2 such subsets: $ \{1, 3, ..., 2n \minus{} 1 \}$ and $ \{2, 4, ..., 2n \}$. Adding another element to any of these sets will yield a pair of coprime elements.[/hide]

True... But wouldn't you have to prove that there are no more than two subsets that have no coprime numbers?

After a while of thinking, I think I have come up with a more rigorous solution:

[hide="solution"] Let an arbitrary integer q be any number in the set {1,2,3,...2n-1}. Then q+1 must be in the set {2,3,4,..., 2n}. We can pair each number q with it's partner q+1 to create n sets, each of which contain two relatively prime integers. Since we are asked to select a subset containing n+1 elements, all the elements must be from the n subsets we created. By the pidgeon-hole principle, two of those elements must of been in the same two-integer-relatively-prime subsets we created. Thus, they would be relatively prime. [/hide]

Unfortunately, there are 2n-1 sets ({1,2}, {2,3}, ..., {2n-1,2n}), not n sets, so your solution isn't quite right (unless if I am misunderstanding what you said), although the basic idea is pretty much correct.
Considering the n sets {1,2}, {3,4}, {5,6}, ..., {2n-1, 2n}, would work.

I think you have misunderstood :wink: . There are n sets because 2n/2 = n. each number q is paired with each number q+1, creating n sets.

I think in his solution he meant $ \{1, 2, 3, \ldots 2n\minus{}1\}$ to be $ \{1, 3, 5, \ldots 2n\minus{}1\}$ and $ \{2, 3, 4, \ldots 2n\}$ to be $ \{2, 4, 6\ldots 2n\}$.  With these sets, his solution works the way it's supposed to.


In a triangle $ ABC$:

$ D$ is the point of trisection of $ BC$ nearer to $ B$ than $ C$.
$ E$ is the point of trisection of $ CA$ nearer to $ C$ than $ A$.
$ F$ is the point of trisection of $ AB$ nearer to $ A$ than $ B$.

Triangle $ PQR$ is the triangle formed formed by the pairwise intersection points of $ AD$, $ BE$ and $ CF$. Prove that the area of $ PQR$ is one seventh the area of  triangle $ ABC$.

[hide="Solution"]Let AD intersect CF at P, AD intersect BE at Q, and BE intersect CF at R. Then
$ [PQR] \equal{} [ABC] \minus{} [FBD] \minus{} [AFE] \minus{} [EDC] \minus{} [FPD] \minus{} [ERF] \minus{} [DQE]$
Note that $ [FBD] \equal{} [AFE] \equal{} [EDC] \equal{} \frac {2}{3} \cdot \frac {1}{3}[ABC] \equal{} \frac {2}{9}[ABC]$
Also, by Menelaus, $ \frac {CD}{CB} \cdot \frac {BF}{FA} \cdot \frac {AP}{PD} \equal{} \frac {3}{4} \implies \frac {AP}{PD} \equal{} \frac {3}{4} \implies \frac {PD}{AD} \equal{} \frac {4}{7}$.
Note that
$ [ABD] \equal{} \frac {1}{3}[ABC]$
$ [AFD] \equal{} \frac {1}{3}[ABD]$
$ [FPD] \equal{} \frac {4}{7}[AFD]$
So $ [FPD] \equal{} \frac {4}{63}[ABC]$, similarly $ [ERF] \equal{} [DQE] \equal{} \frac {4}{63}[ABC]$
So $ [PQR] \equal{} \frac {1}{7}[ABC]$[/hide]
See [url]http://en.wikipedia.org/wiki/Routh's_theorem[/url] and [url]http://en.wikipedia.org/wiki/One-seventh_area_triangle[/url]

[hide="My solution"]Using areal coordinates:

The line $ AD$ has equation $ y\equal{}2z$
The line $ CF$ has equation $ x\equal{}2y$

Solving these as simultaneous equations along with $ x\plus{}y\plus{}z\equal{}1$ gives the coordinates of their intersection $ P$ as $ (\frac{4}{7},\frac{2}{7}, \frac{1}{7})$

Working similarly for the other two gives:

$ Q\equal{}(\frac{1}{7}, \frac{4}{7}, \frac{2}{7})$

$ R\equal{} (\frac{2}{7}, \frac{1}{7}, \frac{4}{7})$

So using $ \frac{[PQR]}{[ABC]} \equal{} det \left(
\begin{array}{ccc}
x_p & x_q & x_r \\
y_p & y_q & y_r \\
z_p & z_q & z_r \end{array}
\right)$

$ \frac{[PQR]}{[ABC]} \equal{} \frac{4}{7}(\frac{16}{49}\minus{}\frac{2}{49}) \plus{} \frac{2}{7}(\frac{4}{49}\minus{}\frac{4}{49})\plus{} \frac{1}{7}(\frac{1}{49}\minus{}\frac{8}{49})$

$ \frac{[PQR]}{[ABC]} \equal{} \frac{56}{343} \minus{} \frac{7}{343} \equal{} \frac{49}{343} \equal{} \frac{1}{7}$[/hide]

Thanks for the links.

This is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1394470#1394470]1952 AHSME #49[/url].


Can someone do a deminstration using Einstein's Field equation?

[quote="mna851"]Dude, it's advanced general relativity.  First off, you're not going to be able to find too many people who know it very well.  Second off, even if there was a member on the forum that knew it, they couldn't explain it to you in a single AoPS post.[/quote]
Understandable


number theory
relatively prime
number theory unsolved

prove the among sixteen consecutive intgers it is always possible to find which is relatively prime to all the rest.

What we need to prove is that among sixteen consecuative integers, there is at least one which cannot be divided by 2, 3, 5, 7, 11, or 13. Let any number which is divisible by any of these primes belong to the set A. In the sequence there are 8 numbers divisible by 2, 5 or 6 numbers divisible by 3, 3 or 4 numbers divisible by 5 and 2 numbers divisible by 7, 11 or 13. Of course some elements have been overcounted.  Considering the minimum overcount in each case, one observes that A can at most contain 15 elements. There must herewith be an element which is relatively prime to all the other.


Στη γεωμετρία του Κοντογιάννη είδα αυτή την άσκηση :

http://www.mathlinks.ro/viewtopic.php?t=247420

 Δεν μου άρεσε ιδιαίτερα η λύση του βιβλίου γιαυτό και  την έβαλα και στο αγγλικό φόρουμ. Ρίξτε μια προσεκτική ματια και πεστε μου αν η λύση που προτείνεται στο σύνδεσμο είναι πλήρης , γιατι δεν έχω χρόνο να κοιτάξω με σχήμα όλες τις λεπτομερειες.
 Αλλά και κάθε άλλη λύση είναι ευπρόσδεκτη !

    Μπάμπης

Κυριε Στεργιου νομιζω οτι η λυση στο διεθνες  forum ειναι πολυ καλα τεκμηριωμενη και η αποδειξη με το ριζικο κεντρο πολυ ωραια.Τελικα αποδεικνυεται βεβαια οτι τα ισοσκελη τραπεζια $ ABDE ,BCEF,CDFA$ ειναι ορθογωνια με ισες διαγωνιους που διχοτομουνται.Ας το κοιταξει ομως και κανεις αλλος μηπως μου ξεφυγε κατι.

Ayti den exei pesei se IMO? Etsi nomizw... :?:


logarithms
algebra
linear equation
calculus
calculus computations

1.)
x^2/(y^(2)-5) (dy/dx) = 1/(2y)
y(1) = 6^(1/2)

i ended up getting ln(y^(2)-5) = (1/x) + c
but now can't solve for c. did i do something wrong?
i need to find the final specific equation for y=?

2.)
(du/dt) = e^(2u+2t)
u(0) = -15

for this one i got (e^(2u))/2 = (e^(2t))/2 + c
and again can't solve for c.
i need to find the final specific equation for y=?

any help?

**OK - I got the first one - but the second one, still so lost! Can anyone help me separate this one?!**

You shouldn't have any trouble solving for $ c$ in the first- but you didn't integrate correctly on the $ x$ side.

The same applies to the second. There shouldn't be any problem solving the linear equation for $ c$, but you didn't solve the differential equation correctly.

$ \frac {x^2}{y^2 \minus{} 5}\frac {dy}{dx} \equal{} \frac {1}{2y}$, so you seem to be missing a negative:

$ \ln(y^2 \minus{} 5) \equal{} \minus{} \frac {1}{x} \plus{} C$, or raise both sides by $ e$: $ y^2 \minus{} 5 \equal{} Ce^{ \minus{} 1/x}$ (note this isn't the same $ C$). Can you solve and plug in from here?

Exactly: $ e^{\ln{(f(x))}}\equal{}f(x)\equal{}\ln{(e^{f(x)})}$ And in general, if you have an equation relating $ a$ to $ b$ then $ a(n_1)\equal{}n_2$ implies $ a$ is the dependent variable (output=n_2) and $ b$ is the independent (input=n_1). You can just plug values in for the variables 'a and b' and find C without isolating the other variables.


Hello!
   I was wrestling with the concept of an indirect proof. What happens if the bounds on the proposition to be proved (if p then q) are too great? Let me illustrate this with an example. NOTE: THIS IS NOT A PROOF AT ALL AND IS USED ONLY TO ILLUSTRATE MY POINT!!!!!

 I will try to prove that "if an angle is in the first quadrant, it is an angle from 0 to 135 degrees." (I know, this doesn't make sense, but I'm trying to illustrate my point, so please bear with me :) )

Assume that the conclusion is not true, i.e. that if an angle is in the first quadrant, it is an angle between 135 and 360 degrees (non-inclusive). This is not true by defintion of a first quadrant angle, so that the assumption then is false and the conclusion is then true (according to the method of indirect proof)! But we know that the conclusion "it is an angle from 0 to 135 degrees" isn't always true (i.e. an angle 91 degrees, etc.)!

Can someone please resolve this ASAP as this question is really perplexing/bothering me?!

I thank you in advance! :)

Your statement is: Angles in the first quadrant range from 0 to 135 degrees
The converse of that statement is: Angles in the first quadrant do not range from 0 to 135 degrees
The converse is not that angles range from 135 to 360 degrees.

Furthermore, you're referencing the definition of the first quadrant in the middle of a proof of the supposed definition of the first quadrant.

Thanks for the quick reply!

Can anyone else answer this, too?

As for the "proof", it's not meant ot be a proof or in any way, shape or form imitate one - it's just kinda like a filler thing. The "proof" is not important - only the point is.

For the "The converse is not that angles range from 135 to 360 degrees", what do you mean? Does this affect what I am trying to ask here? Just for your information, I meant (and please, don't fuss over the details, just try to get my point and answer it, please :) ) 135 to 360 degrees (exclusive) and all angles coterminal to those angles....

When you assert that any angle in [0,135] is in quadrant one, and then try to prove it by contradiction, you have to assume the converse. The converse, in this case, is that not all angles in [0,135] are in quadrant one. The converse of the statement is not that angles outside of [0,135] are in quadrant one.

[quote="grn_trtle"]... to prove it by contradiction, you have to assume the [b]converse[/b].[/quote]

No, you need to assume either the [i]negation[/i] or the [i]negation of the contrapositive[/i], which are equivalent.

[b]@ theax:[/b] You will not run into this difficulty if you understand what proof by contradiction is trying to do. Basically, a statement takes the form of (p is true) implies (q is true). The negation is (p is true) does not imply (q is true). If you show that the negation is false, you are equivalently showing that the original statement is true.

In doing so, consider using more suggestive wording as to what you're trying to prove. Here's an example of what I'm talking about, in terms of your example:

Statement: If an angle is in the first quadrant, it is an angle from 0 to 135 degrees.
Contrapositive: If an angle does not measure between 0 and 135 degrees, it is not in the first quadrant.
Above reworded: There does not exist any angle in the first quadrant whose measure is not in [0, 135]

The above three statements are equivalent. To prove by contradiction, you need to invert the statement and show that the negation is false. I'll negate all three of them -- you decide for yourself which version is easiest to work with.

Negation 1: If an angle is in the first quadrant, it does not necessarily measure from 0 to 135 degrees.
Negation 2: If an angle does not measure between 0 and 135 degrees, it may still possibly be in the first quadrant.
Negation 3: There exists at least one angle in the first quadrant whose measure is not in [0, 135].

Ah, sorry, I mix up the words "converse" and "contrary" sometimes  :lol:

Thanks very much for the speedy reply! (This question has puzzled me for a
while now...)

BUT the problem is that the proposition isn't true for ALL cases. Take, for example, 91 degrees, which isn't a first-quadrant angle, but according to the "theorem" ("if an angle is in the first quadrant, it is an angle from 0 to 135 degrees." ) ... OH!!!!! I see what you mean! I am really sorry for all this confusion, but I actually worded the problem wrong!

Here is my dilemma - (I can't think of a solid example now, so...) let's say we have a circle (a pie chart, as we are going to make it) and divide them into three parts. The first part is A, second B, and third C.

Let's then say that you have a proposition (if p, then q) such that q includes A and B (and A is true, B and C are false). By method of indirect proof, we will prove that C is false and therefore the proposition (if p, then q) is true. BUT the problem is that this proposition isn't true in all scenarios! (and a proposition, if proven true, should!)

If we pick something in B and apply the "theorem" to it, it should be right, but it isn't! To really summarize my point: what happens if the proposition OVERSTATES what you can say and you can't say (is true)?!

Again sorry for the confusion!  :D

Don't apologize for asking a question if it is legitimate.

I'm not sure what you are trying to do here. Are A, B, and C independent statements? From what I gather -- you can't conclude (A&B) because B is a false statement (A&B is true only if B is true, but B is false). Also, what does disproving C have anything to do with A and B?

[quote="theax"]If we pick something in B and apply the "theorem" to it, it should be right, but it isn't! To really summarize my point: what happens if the proposition OVERSTATES what you can say and you can't say (is true)?[/quote]  No, it understates it.  Here's a cleaner example of the phenomenon that you're describing:
"Every multiple of 4 is an even integer."  (Or, equivalently, "for every integer $ x$, if $ x$ is a multiple of 4 then $ x$ is even.")
This is a true statement.  One can prove it by contradiction (show that no odd integer is divisible by 4).  What it isn't is [i]sharp[/i]: we could tighten it up a lot.  It [i]understates[/i] the truth, because it says that every member of some small set belongs to some larger superset; this leaves a lot of elements (those in the big set but not in the small set) about which it says nothing.  In your original example, "every first-quadrant angle is in $ [0, 135]$" is true; it says something about angles in $ [0, 90]$ and (via the contrapositive) something about angles in $ (135, 360)$, but is says nothing at all about angles in $ (90, 135]$.  (It is possible you are confusing the converse and the contrapositive -- there are lots of true statements whose converses are false, and vice versa.  The converse of your statement is false: it does say something about angles in $ (90, 135]$, but the thing it says is wrong.


Let x,y,z,n,k be the positive numbers such that n>k,x>y.
Prove that $ (x^k\minus{}y^k)^n<(x^n\minus{}y^n)^k$

Since x>y>0, we have $ 0< 1\minus{}\frac{y}{x}<1$

Hence we have $ \left[ 1 \minus{} \left(\frac{y}{x} \right)^k \right]^{\frac{1}{k}} < \left[ 1 \minus{} \left(\frac{y}{x} \right)^n \right]^{\frac{1}{k}} <  \left[ 1 \minus{} \left(\frac{y}{x} \right)^n \right]^{\frac{1}{n}}$

In other words $ (x^k \minus{} y^k)^{\frac{1}{k}} < (x^n \minus{} y^n)^{\frac{1}{n}}$

or $ (x^k\minus{}y^k)^n < (x^n \minus{} y^n)^k$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi,

A characteristic subgroup of G is a subgroup that remains fixed by automorphisms of G.

So, the question is, what supgroups of the p-group
G= Z_p + Z_(p^3), are characteristic? I know that if H is a homomorphism from G to G then the subgroups that are characteristic are of the form {x in G| (p^i)*x=0} for i = 1,2 or 3, and {(p^i)*x in G | x in G}. However that is a total of 4 proper subgroups. The subgroups fixed by the homomorphisms from G to G are obviously fixed by the isomorphisms from G to G (aka the automorphisms), but do the isomorphisms fix some other subgroups which the homomorhisms don't?

[This post was partially based on a misreading of the original question]

Let the elements be written as ordered pairs $(a,b)$, with $a\in \mathbb{Z}_p$ and $b\in \mathbb{Z}_{p^3}$.
Among the elements of order $p$, we can distinguish the ones of the form $(0,ap^2)$; these are divisible by $p^2$, while the others are not. There's one group to add to the list.
Similarly, the group of elements divisible by $p$ is also a characteristic subgroup not already on your list.

Now, we try to show that these are all. If a characteristic subgroup contains $(a,b)$ with $a\neq0$, it also contains $(-a,b)$ and thus $(2a,0)$. If $p>2$, this subgroup contains $(a,0)$. If such a group contains $(a,b)$ with $b\neq0$ it also contains $(a,-b)$ and therefore contains the group generated by $2b$. If $p>2$, this is the same as the group generated by $(0,b)$, and every characteristic subgroup is a product of subgroups of the two parts. We have shown that six of these products are characteristic subgroups; we need to look at the other two.
Case 1: $\{(a,0)\}$. Since $(x,y)\mapsto (x,p^2x+y)$ is an automorphism which does not fix this group, this group is not characteristic.
Case 2: $\{(0,b)\}$. Since $(x,y)\mapsto (x+y,y)$ is an automorphism which does not fix this group, this group is not characteristic.

That's it for $p>2$; I'll leave the special case $p=2$ to you.

Thank very much for your response.

However, why do you say that the group with elements divisible by p and P^2 are not on my list? Arent these the subgroups {(p^i)*x in G | x in G} i=1,2 ?

Just to clear up, if (a,b) with a not 0 is in a characteristic subgroup then (-a,b) is in there because of the automorphism
f: (a,b) to (-a,b) ? Why is (2a, 0) there?

In your next sentence, why is the group generated by 2b?

And what 6 products? Are you saying 6 groups are fixed, meaning 2 in addition to my given 4?

Sorry for all the questions

Actually, if you had looked closer, you would have seen that there are six groups on your list: the trivial group, $G$, $pG$, $p^2G$, $\{x: px=0\}$, and $\{x: p^2x=0\}$.

[quote="mark1234"]Just to clear up, if $(a,b)$ with $a \neq 0$ is in a characteristic subgroup then $(-a,b)$ is in there because of the automorphism
$f: (a,b) \mapsto (-a,b)$? Why is $(2a, 0)$ there?[/quote]
Yes. $(2a,0)$ is the difference of two elements of the group, so it is in there.

[quote="mark1234"]In your next sentence, why is the group generated by $2b$? [/quote]
It isn't; it contains that group. I'll go back and clarify my wording.

Yes, I'm saying that six groups are fixed, between my list and yours. The rest of the argument is a proof that there aren't any more (in the case $p>2$).

I am sorry, I meant 4 proper subgroups, of course the trivial group and G itself is characteristic, thank you again for clearing that up.


Lets have a nice talk about the book written by Richard Dawkins, G-d delusion!
I assume many of you guys have read the book,
I would like to discuss some of his points here,,, since I see them as purely logical issues.
any comments before we begin?
 :D

I have read part of the book anyway. I thought it was interesting. It was harsh though...

One argument that I found compelling is:
An atheist is not sure if god does or does not exist for sure. The thing that makes an atheist different from a agnostic is that he finds it more probable that God does not exist rather than that God existing.

(He might of phrased it in a different way, but thats the idea anyway)

Yeah. We could postulate that a teacup is floating around Pluto. This is like God. An agnostic says that there is no hard proof one way or the other, so they will not make a decision about the existence of the teacup. The atheist makes the leap and says that the idea of the teacup is so absurd and improbable, that we might as well say that it is false.

I think his point is that any rational human being will be like the atheist on anything except on the point of God. It is really the believers and agnostics that are being different, despite what the media would like us to believe.

As someone who calls himself an agnostic, I have decided not to make a decision not only because of the lack of proof but also because the decision means nothing to my daily life. Any decision I make could have the side effect of angering the *real* god, so it's completely useless to decide.

[quote="HilbertThm90"]Yeah. We could postulate that a teacup is floating around Pluto. This is like God. An agnostic says that there is no hard proof one way or the other, so they will not make a decision about the existence of the teacup. The atheist makes the leap and says that the idea of the teacup is so absurd and improbable, that we might as well say that it is false. [/quote]I'd say, "Why are we thinking about this?"

As to why people believe in religion, etc, many people I've talked to have had experiences which convinced them of the existence of a god.

I still consider those that have had an "experience" to be choosing the belief, because they turn a deaf ear to any other explanation. There are no situations that I know of, in which some miraculous thing happened, that I can't point to and go, oh, that's why that happened. The believer is unwilling to hear that, though.

What about Jesus Christ?
Do you think that the stories saying about he turned water to wine, walked on water etc. are not true?
I personally find it hard to believe in a God like a person but more like a source of a higher energy that creates life. A bit like the idea of karma.

I think the key word there is "story." Every culture has traditions of stories, many of which are equally fantastical. Why should we believe one set over another?

Please, there are already two discussions about these arguments running. Use them!

I think that we should shift from The God Delusion, to The End of Faith by Sam Harris. The God Delusion concerns itself with God's existence. Sam Harris is an atheist, but The End of Faith is an argument against religion from a purely ethical standpoint. He could care less about whether or not God exists. He makes the claim that organized religion is unethical and any supporter of it (or a non-supporter that is tolerant of religions) is engaging in unethical activity.


rectangular hyperbola :$ XY=c^{2}$ where $ c^{2}$ = $ \frac{a^{2}}{2}$

How do I show the focii and directices are $ \pm(\sqrt{2}c,\sqrt{2}c)$ and $ X+Y$=$ \pm\sqrt{2}c$

Well, you could either

i) rotate and apply normal procedures.

ii) show that those points and those lines have the normal focus-directrix property:

\[ \frac{d(F,P)}{d(\mathcal{D},P)}=e\]
where $ d(A,B)$ respresents the distance between $ A$ and $ B$, $ P$ is a point on the hyperbola, $ F$ is a focus, $ \mathcal{D}$ is the corresponding directrix, and $ e$ is the eccentricity.


given n unit vecto $x_{1},x_{2}......x_{n} \in R^{n}$ so that among any three there are two othogonal ones .prove that  $|\sum\limits_{ i=1}^{n}x_{i}| \leq 2\sqrt[4]{n^{3}}$

Please continue discussion about this problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=49217 .

  darij


for positive real $ x,y,z$:
$ \frac{1}{x^{2}\plus{}xy\plus{}y^{2}}\plus{}\frac{1}{y^{2}\plus{}yz\plus{}z^{2}}\plus{}\frac{1}{z^{2}\plus{}zx\plus{}x^{2}}\geq \frac{9}{(x\plus{}y\plus{}z)^{2}}$

[quote="Dumel"]for positive real $ x,y,z$:
$ \frac {1}{x^{2} \plus{} xy \plus{} y^{2}} \plus{} \frac {1}{y^{2} \plus{} yz \plus{} z^{2}} \plus{} \frac {1}{z^{2} \plus{} zx \plus{} x^{2}}\geq \frac {9}{(x \plus{} y \plus{} z)^{2}}$[/quote]
It's old and easy by Schur! :)

[quote="Dumel"]for positive real $ x,y,z$:
$ \frac {1}{x^{2} \plus{} xy \plus{} y^{2}} \plus{} \frac {1}{y^{2} \plus{} yz \plus{} z^{2}} \plus{} \frac {1}{z^{2} \plus{} zx \plus{} x^{2}}\geq \frac {9}{(x \plus{} y \plus{} z)^{2}}$[/quote]

Use"Iran 96" can kill it  :)

After expanding and using Muirhead's inequality we obtain very quickly solution.
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231097

[i]Oh , easy problem

 $ 4(x^2 \plus{} xy \plus{} y^2 )( xy \plus{} yz \plus{} zx) \ \leq \ (x^2 \plus{} xy \plus{} y^2 \plus{} xy \plus{} yz \plus{} zx)^2 \equal{} (x \plus{} y)^2 ( x \plus{} y \plus{} z)^2$

Apply the famous ineq Iran 96 : $ \left( \frac{1}{(x\plus{}y)^2} \plus{}\frac{1}{(y\plus{}z)^2} \plus{}\frac{1}{(x\plus{}z)^2} \right) (xy\plus{}yz\plus{}zx) \ \geq \frac{9}{4}$, we have :

$ \rightarrow LHS \ \geq \ \frac {4(xy \plus{} yz \plus{} zx)}{(x \plus{} y \plus{} z)^2 } \sum \frac {1}{(x \plus{} y)^2 } \ \geq \ \frac{4(xy \plus{} yz \plus{} zx)}{(x \plus{} y \plus{} z)^2 } \frac{9}{ 4(xy \plus{} yz \plus{} zx) } \ \equal{} \frac {9}{(x \plus{} y \plus{} z)^2}$[/i]


[i]Όσοι 'μεγάλοι' βάζουν απάντηση , να την κρύβουν για να βοηθάμε τους μικρότερους να σκεφτούν.[/i]
[b]ΘΕΜΑ 1ο [/b]

 Ένας φυσικός αριθμός έχει $27$ ψηφία , όλα ίσα με 1.Ποιο θα είναι το υπόλοιπο της διαίρεσης του αριθμού αυτού με το $27$; ΚΑλό είναι να μη γίνει η διαίρεση!

Hint για όσους δεν μπορέσουν να βρουν λύση βρίσκεται παρακάτω.

[hide]
Αν $a$ είναι ένας φυσικός αριθμός και $b$ είναι το άθροισμα των ψηφίων του $a$, τότε οι $a$ και $b$ διαιρούμενοι με $n=3$
ή $n=9$ είναι ισοϋπόλοιποι. Επίσης ισχύει, αν $ab$ είναι παράγοντας του $c$, τότε το $a$ είναι παράγοντας του $c$ και το $b$ είναι παράγοντας του $c$.
[/hide]

[b]ΘΕΜΑ 2ο[/b] (3η Γυμνασίου)

Αν ισχύει $3\sin x+4\cos x = 5$, να βρεθεί η $\tan x$

[b]ΘΕΜΑ 3ο[/b]

Ποιος αριθμός είναι μεγαλύτερος;

$a = \sqrt{2005}+\sqrt{2006}$
$b = \sqrt{2004}+\sqrt{2007}$

[hide] για το πρώτο μέλος θέτω β+γ=χ  γ+α=ψ  α+β=ω άρα α=ψ+ζ-χ/2 β=χ+ζ-ψ/2 γ= χ+ψ-ζ/2 κι με αντικατάσταση κι αρκετές πράξεισ θα βγει στο δευτερο μέλος αντρέσκου[/hide]

Θέμα 3ο

Υψώνουμε στο τετράγωνο και αφαιρούμε κατά μέλη  και έχουμε
$a^{2}-b^{2}=2005+2006+2\sqrt{2005\cdot 2006}-2004-2007-2\sqrt{2004\cdot 2007}=2(x+1)(x+2)-2x(x+3)=2>0$

άρα $a>b$ (όπου $x=2004$)

Κωνσταντίνος

[quote="socrates"]Για $a,b,c>0$ νδο $\frac{1}{2}(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}) \geq \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}$

 :)[/quote]

Λύση
Το δεξί μέλος βγαίνει εύκολα αφου
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{(a+b+c)^{2}}{2(ab+bc+ca)}\geq \frac{3}{2}$  από Andreescu-Cauchy .
Τώρα για το δεξί μέλος από την γνωστή
$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq 4(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$ αρκεί νδο
$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\geq\frac{1}{2}(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$
.
Θέτουμε $\frac{a}{b}=x$ κτλ (tote $xyz=1$ kai exoume ndo
$2(x^{2}+y^{2}+z^{2})\geq x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

ή
$2(x^{2}+y^{2}+z^{2})\geq x+y+z+xy+yz+zx$
που ισχύει αφού
$x^{2}+y^{2}+z^{2}\geq \frac{(x+y+z)(x+y+z)}{3}\geq\frac{3(x+y+z)}{3}=x+y+z$

kai $x^{2}+y^{2}+z^{2}\geq xy+yz+zx$

Καταπληκτική λύση Pontios! Πραγματικά πολύ έξυπνη!

Απλώς δε με βόλευε να χρησιμοποιήσω διαφορά τετραγώνων όπως ήταν δοσμένη ...  :wink:

[quote="mostel"][b]ΘΕΜΑ 4ο[/b]

Έστω $x_{1},x_{2},x_{3}$ οι ρίζες της εξίσωσης:

$x^{3}+3x^{2}-7x+1$

Να υπολογιστεί: $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$[/quote]

Epeidi mporei na min eixes graftei tote sto forum, se parapempw (kai olous tous endiaferomenous) se ena palaiotero post sto opoio sizitisame gia tis simmetrikes parastaseis me aformi ena sistima tou RAKAR kai to opoio einai to http://www.mathlinks.ro/Forum/viewtopic.php?t=110880

[hide] Apo ekei kai pera xrisimopoiwntas tin tautotita Newton pou grafoume ekei (kala den einai anagi na tin ksereis gia na vgaleis auti tin askisi alla genika xreiazetai an gia paradeigma eleges na vrethei to $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$) vriskoume tin parastasi $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ se sinartisi me ta simmetrika poliwnima $x_{1}+x_{2}+x_{3}$ kai $x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3}$ (den xreiazetai to $x_{1}x_{2}x_{3}$) ta opoia vriskoume me tin seira tous kanontas xrisi twn tipwn Vieta kai exoume katharisei.

Sigekrimena $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=(x_{1}+x_{2}+x_{3})^{2}-2(x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3})$

kai apo tous tipous Vieta exoume $x_{1}+x_{2}+x_{3}=-3$ kai $x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3}=-7$ opote $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=23$
[/hide]
Alexandros

Όντως Αλέξανδρε, δεν ήμουν εγγεγραμμένος εκείνη την περίοδο στο forum. Σε ευχαριστώ πολύ για τις επεξηγήσεις που έδωσες όπως και για το thread σχετικά με την ταυτότητα του Newton. Η αλήθεια είναι πως τώρα απέκτησα μία πιο σφαιρική εικόνα των ταυτοτήτων!

Έστω $n>4$ περιττος φυσικος αριθμος.
ΝΔΟ ο $z=n^{4}-10n^{2}+9$ διαιρειται με το $384$.

 :)

Έχουμε:

$n^{4}-10n^{2}+9=(n^{2}-3)^{2}-(2n)^{2}=(n^{2}-2n-3)(n^{2}+2n-3)=(n-1)(n+1)(n-3)(n+3)$

Όμως ο $n$ είναι περιττός, άρα έχει τη μορφή: $2k+1$. Έτσι έχουμε:

$(n-1)(n+1)(n-3)(n+3)=(2k+1-1)(2k+1+1)(2k+1-3)(2k+1+3)=2k(2k+2)(2k-2)(2k+4)=2k\cdot2\cdot2\cdot2(k-1)k(k+1)(k+2)=2^{4}(k-1)k(k+1)(k+2)\quad(k\in\mathbf{N}, k\geq2)$

Παρατηρούμε ότι $k-1$,$k$,$k+1$,$k+2$ είναι διαδοχικοί αριθμοί και μάλιστα $k\geq2$, έτσι ο αριθμός $(k-1)k(k+1)(k+2)$ είναι πολλαπλάσιο του $8$. Επίσης μέσα στους τέσσερις αυτούς όρους σίγουρα ένας θα είναι πολλαπλάσιο του $3$. Δηλαδή $(k-1)k(k+1)(k+2)=3\cdot8x=24x$, όπου $x$ ένας φυσικός. Έτσι θα έχουμε: $2^{4}(k-1)k(k+1)(k+2)=2^{4}\cdot24x=16\cdot24x=384x\ldots$

να δειξετε οτι καθε δυναμη ενος ακεραιου n (βαση και εκθετης ειναι φυσικοι) μπορει να γραφτει ως αθροισμα n διαδοχικων περιτων αριθμων.αν d ο Μ.Κ.Δ  αριθμων αυτων τοτε ποια σχεση συνδεει τη βαση και τον d ?  :)

Για την ασκηση του Κωστα.[hide]ευκολα βρισκεις οτι αν n^a=2kn+n^2 αρα κ=[n^(a-1)-n]/2 και τοτε n^a=(2k+1)+(2k+3)+.......(2k+2n-1) :) [/hide]

ωραια, υπαρχει και το δευτερο ερωτημα ,...

Αν καταλαβα καλα θες να βρω μια σχεση που να συνδεει τον d=(2κ+1,2κ+3,.........,2κ+2n-1)=1 με τον a.Eιναι πολυ απλο! a=d.a=1.a :rotfl:[hide]

δηλ δεν μπορει να ειναι διαφορο του 1 το d ?[hide][/hide]

[hide]Συφνωμη αν κανω καποιο λαθος αλλα ετσι καταλαβα.Αν d=(2k+1,2k+3,.....2k+2n-1) τοτε ισχυριζομαι οτι d=1.Που ειναι το λαθος μου?[/hide]

[quote="Αρχιμήδης 6"]Ασκηση
Δειξτε οτι μεταξυ του n και n! υπαρχουν 3[n/2] πρωτοι για n>=4.[/u][/quote]

O $n$ μάλλον είναι φυσικός.. επίσης αν όπου $n=5$ θα έχουμε $\frac{3\cdot5}{2}=7.5$. Εδώ κάνουμε στρογγυλοποίηση; Για περιττούς αριθμούς δηλαδή. Και επίσης, λογικά εννοείς τουλάχιστον $\frac{3n}{2}$ πρώτοι.

Αυτά.

 :D

Arximedes 6 εχεις δικιο.μηπως η τελευταια ειναι δυσκολη για γυμνασιο ?  :)

ΜΜΜΜ........Ενταξει δεν θελω να το παιξω εξυπνος αλλα να σου ξεκαθαρισω οτι αυτο το θεμα δεν θα επεφτε ουτε λυκειο(στην Ελλαδα) κατα την γνωμη μου ετσι αυτουσιο.Και αυτο το λεω για αλλο λογο και οχι για δυσκολια του προβληματος(θα καταλαβεται τι εννοω αν το προσπαθησεται!)Παντος δεν νομιζω οτι υπαρχουν πολλοι στο ελληνικο φορουμ που να πηγαινουν γυμνασιο επομενως προσπαθηστε την ολοι.Δεν ειναι δυσκολο παντος. :)Εξαλλου τα <<κινεζακια>> νομιζω οτι μπαινουν απο νωρις στα βαθια..... :D

[quote="Αρχιμήδης 6"]Για n=5 το 5!=120.Πραγματι μεταξυ του 5 και 120 υπαρχουν 3[5/2]=6 πρωτοι.Με το [χ] συμβολιζω το ακεραιο μερος του χ και οριζεται ως ο μεγαλυτερος ακεραιος που δεν υπερβαινει το χ.[/quote]

Δημήτρη υπάρχουν περισσότεροι από 7 πρώτοι μεταξύ του $5$ και του $120$. Για αυτό ανέφερα πιο πάνω "τουλάχιστον"...

Δηλαδή από κόσκινο του Ερατοσθένη οι πρώτοι αριθμοί είναι οι εξής:

$5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113$

[quote="Αρχιμήδης 6"]Οπως σωστα ανεφερες εννοουσα τουλαχιστον.Την ελυσες μηπως? :)[/quote]

Όχι.. φαίνεται πολύ καλή πάντως !

Prove that between n and n! there exist more or equal of 3[n/2] primes.Can sb solve this problem?

Να μια ιδέα
Από το θεώρημα Bertrand βρίσκουμε ότι υπάρχουν $k$ πρώτοι μεταξύ του $n$ και του $2^{k}m$
άρα αρκεί να δείξουμε ότι

$2^{3[\frac{n}{2}]}\leq n!$  το οποίο νομίζω ότι μπορεί να δειχθεί επαγωγικά .

Σωστα!Δεν ηταν και πολυ δυσκολη! :)

[quote="mostel"][color=red][b]ΑΣΚΗΣΗ number_goes_here[/b][/color]

Να αποδειχθεί ότι κάθε συνάρτηση μπορεί να γραφεί ως άθροισμα μίας περιττής και μιας άρτιας[/quote]
ενδιαφερον αποτελεσμα ....ειδα μια αποδειξη απο ενα βιβλιο η οποια δεν ειναι για Γυμνασιο  :P  μπορει να δωσει μια λυση καποιος ?

Έστω μια συνάρτηση $f: \mathbb R \to \mathbb R$.
Αν $g(x)=\frac{f(x)+f(-x)}{2}$ τότε $g(-x)=g(x)$.
Επίσης αν $h(x)=\frac{f(x)-f(-x)}{2}$ τότε $h(-x)=-h(x)$.
Όμως ισχύει ότι $f(x)=g(x)+h(x)$ και συνεπώς το ζητούμενο αποδείχτηκε.

Οκ, ίσως δεν ήταν για γυμνάσιο, άλλα για πρώτη λυκείου σίγουρα! Αρκεί να ορίσεις ποια συνάρτηση είναι περιττή και ποια άρτια και να κάνεις το τρικάκι που χρησιμοποίησε παραπάνω ο Στέλιος..

@stedes
[hide] άντε και με το καλό προαγωγή στο λύκειο  :pilot:  :P[/hide]

Καλησπέρα σε όλους σας
Την πέμπτη βγαίνουν τα αποτελέσματα του Θαλή
μια μικρή άσκηση παιδιά
Τι είναι προτιμότερο στο Τζόκερ
Πέντε δελτία με 6 5 από 45 ή εξι δελτία με 5 από 45 (σταθερά το 1 στα 20)
πιθανοθεωρητικά πιο προτιμότερο
κι μετά να εξετάσετε κι οικονομικά τί συμφέρει θα πληρώσουμε :)  :D  :)  :wink:


abstract algebra
superior algebra
superior algebra unsolved

Let $ E$ be a free module over the ring $ R$, with basis $ {v_1,...,v_n}$. Let $ F$ be a module, and $ w_1,...,w_n$ be elements of $ F$. Show that there exists a unique homomorphism $ f: E\to F$ s.t. $ f(v_i)\equal{}w_i$ for $ i\equal{}1,2,...,n$.

Hi,

This boils down to noticing that "f(v_i)=w_i and f a morphism" determines f completely!

@quasar987: this doesn't explain anything, it just repeats the claim.
@countdownmath: uniqueness comes from the given fact that the $ v_i$ generate $ E$, and for existence set $ f(\lambda_1 v_1 \plus{} ... \plus{} \lambda_n v_n) \equal{} \lambda_1 w_1 \plus{} ... \plus{} \lambda_n w_n$.


Let $ A$ be an $ n*m$ real matrix with rank $ m$ and $ B,\ C$ be $ m*n$ matrices.
If the marices $ B,\ C$ satisfy $ AB\equal{}AC$, then prove that $ B\equal{}C$.

Let's start with $ x$ and $ y$ being $ m\times 1$ columns. If $ A$ is an $ n\times m$ matrix with rank $ m,$ that means that the $ m$ columns of $ A$ are independent, which is the same as saying that $ Ax\equal{}0$ implies $ x\equal{}0.$ Then, $ Ax\equal{}Ay$ implies $ A(x\minus{}y)\equal{}0$ implies $ x\equal{}y.$

Now if $ B$ and $ C$ are $ m\times p$ matrices, then each of $ B$ and $ C$ is made up of a number of $ m\times 1$ columns stacked together. If $ AB\equal{}AC,$ then from above, each column of $ B$ equals the corresponding column of $ C,$ and hence $ B\equal{}C.$


calculus
function
integration
calculus computations

Let a an element of  $]1,+\infty[$ and $f$ a derivable function on
$[1,a]$ and $f'$ is continous on $[1,a]$

Prouve that:

$\int^{a}_1 [t]f'(t)dt=[a]f(a)-\sum_{k=1}^{[a]}f(k)$

Do you know Stieltjes integration?

Because this is just integration by parts with $u=[x]$, $v=f(x)\in C^1$


What is the tens digit in the sum $ 7! \plus{} 8! \plus{} 9! \plus{} \cdots \plus{} 2006!$?

$ \textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

[hide]$7!=5040$
$8!=40320$
$9!=362880$
$10!=3628800$

From $10!$ and on, we see that the ten's digit is simply $0$. Therefore, we add $4+2+8=4\rightarrow\boxed{C}$[/hide]

we want to find the smallest integer $n$ such that $n!$ is divisible by $100$ so that we can ignore it and everything above it.

$100=2^2\cdot5^2$ so the smallest integer $n$ is $5\cdot2=10$.

our answer is the tens digit of $7!+8!+9!=40+20+80=40$ which is $\boxed{4}$


Find all function $f: R^+\rightarrow R^+$ satisfying $f(x+y) \ge f(x)+yf(f(x)) \forall x,y\in R^+$

I don't think any exist.

First, note that if $a < b$, we can write $b = a + y$ for some $y \in \mathbb{R}^+$, and substituting in $a$ for $x$ yields $f(b) = f(x + y) \ge f(x) + yf(f(x)) > f(x)$, so $f(a) < f(b)$ and thus $f$ is strictly increasing. Next, note that $f(x + 1) \ge f(x) + f(f(x)) > f(f(x))$, so $f(x) < x + 1$. Thus we can write $f(x) + yf(f(x)) \le f(x + y) < x + y + 1$; letting $y \to \infty$, we find that $f(f(x)) \le 1$. Suppose $f$ is unbounded. Then we can find $t$ such that $f(t) > 1$ and $x$ such that $f(x) \ge t$, in which case $f(f(x)) \ge f(t) > 1$, a contradiction. Hence $f$ is bounded, i.e. $\exists M$ such that $\forall x \in \mathbb{R}^+, f(x) \le M$. However, if we take $y = \frac{M}{f(f(1))}$ in the given condition, we find that $f(1 + y) \ge f(1) + yf(f(1)) > \frac{M}{f(f(1))} f(f(1)) = M$, a contradiction.


Let $ a,b,c$ be positive reals such that $ ab\plus{}bc\plus{}ca\equal{}1$. Then

$ \frac{1}{a\plus{}b\plus{}c\minus{}abc}\leq\frac{3\sqrt{3}}{8}$

[quote="Ligouras"]Let $ a,b,c$ be positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Then

$ \frac {1}{a \plus{} b \plus{} c \minus{} abc}\leq\frac {3\sqrt {3}}{8}$[/quote]
$ a\plus{}b\plus{}c\minus{}abc\equal{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}abc \geq \frac{8}{9}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \geq \frac{8}{ 3\sqrt{3} }$

Done !


I don't get this circumscribing funkyswibing stuff. I just read this problem that said "blahblahblah Quadrilateral ABCD is circumscribed about Circle O. blahblahblah" I could understand that the circle is outisde of the quadrilateral. Then i read this problem that said "blahblahblah Quadrilateral ABCD is circumscribed around Circle O. blahblahblah" and the circle's somehow outside the quadrilateral still. I don't get it. I think it's vice versa (circle inside ABCD). BTW, will a question ever be worded with "inscribed about" or "inscribed around"? Just wondering. Thanks! :D

btw, if x inscribes y, then y is in x, and if x circumscribes y, then x is around y right? k, just making sure. this circumscribe inscribeiswibe funkyswibe gets kinda funky sometimes.  :|

The terminology used is "inscribed in" and "circumscribed about".  If x is inscribed in y, then x is inside y.  If x is circumscribed about y, then x is around y, I believe.


In Kedlaya's Geometry Unbound, there are "theorems" and "facts", which are several other results which are not named. Can we use them at IMO, without proving?

if they are well known , u can say only the name of them ,and then use them.

or u can write one of the links of mathlinks, on ure paper, which contains it . :D


probability MATHCOUNTS calculus geometry

Please help me with this one. I am trying to figure it out because this was my first time I encountered with this kind of probability:

FIND THE GIVEN PROBABILITY OF: 1. P(z < 1.25)

Thanks in advance!

Removed from Mathcounts because Mathcounters don't knkow statistics and certainly don't know z-values.

You can use a z-value table, or you can use calculus, in which case this is going to move further, but I'll let one of the good GS mods do that. I didn't think too much before I took it out of MC.

The probability is approximately .3944. You can look these things up in a table like this:
[url]http://www.stat.ucla.edu/~dinov/courses_students.dir/STAT10_Fall01/STAT10_Fall01/z-table.JPG[/url]

Yeah, my z-table was from 0 to x. The area less than 0 is then .5, so you just add that on to my probability to get what you have.


function
calculus
algebra
polynomial
algebra proposed

Let $a,b,c>0$ and $f(x)=x^5$ and $g(x)=ax^4+bx^2+c$ .Prove that the grafs of the functions has only one common point

Not difficult, indeed, but it's a nice problem.
$ax^4+bx^2+c=x^5 \Longleftrightarrow a+\frac{b}{x^2}+\frac{c}{x^4}=x$
Obviously $x>0$. Denote $t=\frac{1}{x^2}$. Then  $x=\frac{1}{\sqrt{t}}$ and the equation becomes $ct^2+bt+a=\frac{1}{\sqrt{t}}$.
The function in the left side is increasing on $[-\frac{b}{2c}, \infty)$, hence also on $(0, \infty)$ (as $b,c>0$), while the function in the right side is decreasing on $(0, \infty)$. It follows that their graphs have at most one common point.
On the other hand, since the degree of the original equation is odd, it surely has at least one real root.
It follows that the equation has exactly one real root.

My solution is similar to yours but I have some diffrences and I use more calculus.

To have the two equations common it should be $x>0$ after dividing by $x^5$ we

have $1=\frac{a}{x}+\frac{b}{x^3}+\frac{c}{x^5}$. Let the function

$h(x)1-\frac{a}{x}-\frac{b}{x^3}-\frac{c}{x^5}$,   $x>0$

$h'(x)=\frac{a}{x^2}+\frac{3b}{x^4}+\frac{5c}{x^6}>0$ .So $h$ is increasing

in $A= (0,+\infty)$ .There exist at most one root but it exist ?Yes because

with limits easily we find that $h(A)=(-\infty,1)$ so the $0$ is contained in $h(A)$

and we are done

I intended to completely avoid calculus (the theorem about the odd degree polynomials is an algebric one).
However, your solution seems to be more natural for those who already learned calculus.


let positive numbers a,b,c,d .Show that
$\sum$ $\frac{1}{a\sqrt{5ab+bc+2ca}}$ $\geq$ $\frac{\sqrt{2}}{\sqrt{abcd}}$

I think prove that it same way prove that:$\sum\frac{1}{a\sqrt{a+b}}\ge \frac{3}{\sqrt{2abc}}$. :roll:


[quote="quantum leap"]I just had the Ole miss math tournament today, and I was kinda surprised how they were endorsing aops.  For example, the pro-their school propaganda packets ;) they were handing out included those brochure things about the usamts, and they also gave out two 50 dollar gift cards for aops products as door prizes.  I know that although this is their first year holding it, their competition had been advertised in the "national contests" page on this site.  I was wondering whether aops sponsors all (or most or however many) of the contests that are listed on this site and if the ppl running the contests get aops material to give out in return.  I know this site supports all kinds of math contests, but i was wondering kinda to what extent  :huh:[/quote]

We sponsor some contests in this manner, but not all the ones listed on our site.  Glad to hear they were spreading the word.  I hope the contest went well!

From the HMMT website, describing their tests:"The General Test is designed to be more accessible than the Subject Tests to students with less math background ...The difficulty is somewhat comparable to a Mu Alpha Theta test or tests given in many Southeastern math tournaments."

At first i was kinda like :mad: but after this test, knowing what AIME and ARML are like, I can understand what they mean.  I guess Ole Miss'll post  the questions up later so anyone can check them out.  Dunno if anyone else knew what aops was,(woulda been nice to talk to another AoPSer there) but there were only about 30 or so people there.


geometry
rhombus
trigonometry
projective geometry
Brazilian Math Olympiad

$ABCD$ is a rhombus. Take points $E$, $F$, $G$, $H$ on sides $AB$, $BC$, $CD$, $DA$ respectively so that $EF$ and $GH$ are tangent to the incircle of $ABCD$. Show that $EH$ and $FG$ are parallel.

A solution using some basic projective geometry:

Since the quadrilateral ABCD is a rhombus, the lines AB and CD are parallel; thus, they meet at an infinite point. Denote this infinite point by $U_{\infty}$. Similarly, the lines BC and DA meet at an infinite point, which we call $V_{\infty}$.

Now, consider the "hexagon" $EFV_{\infty}HGU_{\infty}$. Agreedly, you need some abstraction to regard it as a hexagon, but anyway all of its sidelines EF, $FV_{\infty}=BC$, $V_{\infty}H=DA$, HG, $GU_{\infty}=CD$ and $U_{\infty}E=AB$ are tangent to the incircle of the quadrilateral ABCD. So the sidelines of this hexagon $EFV_{\infty}HGU_{\infty}$ are all tangent to one circle, i. e. the hexagon is circumscribed. Thus, by the Brianchon theorem, the main diagonals EH, FG and $V_{\infty}U_{\infty}$ of this hexagon are concurrent. In other words, the lines EH and FG intersect on the line $V_{\infty}U_{\infty}$.

But since $U_{\infty}$ and $V_{\infty}$ are two infinite points, the line $V_{\infty}U_{\infty}$ is the line at infinity, and thus we obtain that the lines EH and FG intersect on the line at infinity, i. e. the lines EH and FG are parallel. $\blacksquare$

  Darij

this one was quite easy with trigonometry
too easy to number 3

See the nice Yetti's proof here : http://www.mathlinks.ro/Forum/viewtopic.php?t=110204

Let $E'$ be the point on $AB$ such that $E'H \| FG$ and let $M$ be the intersection of $FH$ and $AC$. Since the corresponding sides of triangles $\triangle{AHE'}$ and $\triangle{CFG}$ are parallel, $E'G$, $FH$ and $AC$ are concurrent at $M$ which implies that $E'$ is the intersection of $GM$ and $AB$. By Brianchon's theorem $EG$, $FH$ and $AC$ are also concurrent at $M$ which implies that $E$ is the intersection of $GM$ and $AB$. Hence $E'=E$ and $EH \| FG$.


geometry
3D geometry
arithmetic sequence
absolute value
number theory unsolved
number theory

Find all integers x,y,z s.t. x^2+y^3=z^6.

Bomb

If a solution exists then there are infinitely many solutions for if $(x, y, z)$ is a solution then $(t^3x, t^2y, tz)$ is a solution as well.

but that still doesn't find ALL the possible solutions...

unless trivial cases, the only solutions imply the existence of 3 cubes in arithmetic progression, so we would have $a^3+b^3=2c^3$ I think this has been previously discussed

Perhaps a thread would justify your comment... :)

BOmb

it's funny bomb, cuz you just gave the solution to it:

http://www.mathlinks.ro/Forum/viewtopic.php?t=45946

But you still need to show that the given equation is equivalent to the existence of three cubes in arithmetic progression :)

Clearly if any two of x,y,z have a common factor
then they also share a common factor with the third, so we can divide
until the three numbers x,y,z are coprime.  Of all such solutions,
let's consider the one with the smallest absolute value of z.

Now, the equation implies  y^3 = (z^3)^2 - x^2, so we have

                  y^3  =  (z^3 + x)(z^3 - x)                     (1)

If y is odd then the two factors on the right side are mutually
coprime, so they must each be cubes, meaning there are coprime
integers a,b such that

        z^3 + x  =  a^3      and        z^3 - x  =  b^3

Adding these two gives  a^3 + b^3 = 2z^3.  On the other hand, if y is
even, then the two factors on the right side of (1) share a common
factor of 2, so one of them is twice a cube and the other is four times
a cube, meaning there exist coprime integers a,b such that

        z^3 + x  =  2a^3      and        z^3 - x  =  4b^3

Adding these two and dividing by 2 gives z^3 + (-a)^3 = 2b^3., as required.
This and other thread solves problem.

Bomb

I've checked the website [url]http://www.mathpages.com/home/kmath213.htm[/url]... It missed out another trivial solution $(x, y, z) = (\pm k^3, 0, k)$, where $k$ is an arbitrary integer. This solution is obtained from the case when $y$ is even as outlined in the solution at this website.


inequalities
real analysis
integration
function
Support
real analysis unsolved

If $ m(E)<\infty$, show that $ L^{\infty}(E)\subset L^{p}(\mathbb{R})$ for any $ 1\leq p<\infty$, and that ${ \left\|f\right\|_{p}\leq (m(E))^{1-1/p}}\left\|f\right\|_{\infty}$ for any $ f\in L^{\infty}(E)$.

Given $ f\in L^{p}$, $ 1\leq p<\infty$ and $ g\in L^{\infty}$, prove that $ fg\in L^{p}$ and that $ \left\|fg\right\|_{p}\leq\left\|f\right\|_{p}\left\|g\right\|_{\infty}$.

I'm assuming Holder's inequality is needed but I can't get the terms right.

Your proposed inequality $ \parallel{}f\parallel{}_p\leq \parallel{}f\parallel{}_\infty m(E)^{1 \minus{} 1/p}$ is wrong. Take $ f \equal{} 1$ a.e in $ E$, then $ \parallel{}f\parallel{}_p \equal{} m(E)^{1/p}$, $ \parallel{}f\parallel{}_\infty \equal{} 1$, but $ m(E)^{1/p}$ could not less than $ m(E)^{1 \minus{} 1/p}$ for any measurable set $ E$.

I suppose that you wanna prove $ \parallel{}f\parallel{}_p\leq \parallel{}f\parallel{}_\infty m(E)^{1/p}$. We take a look at the case $ 1 < p < \infty$. (the case $ p \equal{} 1$ can be seen from 0) below):
0) Note that $ f\in L^p$. This comes from $ |f(x)|\leq \parallel{}f\parallel{}_\infty$ a.e in $ E$.
1) Take $ f\in L^\infty(E)$, then $ f(x)\leq \parallel{}f\parallel{}_\infty$ a. e in $ E$ and we consider $ f$ determine on $ \mathbb R$ so that $ supp f\subset E$. You know why?
2) Now $ \parallel{}f\parallel{}_p^p \equal{} \int\limits_{\mathbb R}|f(x)|^pdx\leq \parallel{}f\parallel{}_\infty\int\limits_{E}|f(x)|^{p \minus{} 1}dx$
By $ \text{H\"{o}lder}$ inequality:
$ \int\limits_{E}|f(x)|^{p \minus{} 1}dx \equal{} \int\limits_{E}|f(x)|^{p \minus{} 1}\times1dx\leq \left(\int\limits_{E}|f(x)|^p\right)^{1/q}\times\left(\int\limits_{E}dx\right)^{1/p} \equal{}$
$ \equal{} \parallel{}f\parallel{}_p^{p \minus{} 1}\times m(E)^{1/p}$

So $ \parallel{}f\parallel{}_p\leq \parallel{}f\parallel{}_\infty m(E)^{1/p}$

The second inequality follows immediately from $ |g(x)|\leq \parallel{}g\parallel{}_\infty$ for a.e in $ E$.

Thank you.  How would I show that $ L^{\infty}(E)\subset L^{p}(\mathbb{R})$?  What I have so far is that if $ f\in L^{\infty}(E)$, then $ \left\|f\right\|_{\infty}\equal{}A<\infty$ so $ |f|\leq A<\infty$ on $ A$.  Hence $ \int\left|f\right|^{p}\leq\int\left|A\right|^{p}<\infty$ on $ E$.  But that doesn't mean it's integrable on $ \mathbb{R}$, right?

One more question, when you write
$ \int\limits_{E}|f(x)|^{p\minus{}1}dx \equal{}\int\limits_{E}|f(x)|^{p\minus{}1}\times1dx\leq\left(\int\limits_{E}|f(x)|^{p}\right)^{1/q}\times\left(\int\limits_{E}dx\right)^{1/p}$, how are you defining $ q$ and where is it coming from?

$ q$ is such that $ \frac1q \plus{} \frac1p \equal{} 1$, i.e. $ q \equal{} \frac {p}{p \minus{} 1}$.

This gives the transition from $ \left(\int\limits_{E}|f(x)|^{p}\right)^{1/q}$ to $ \parallel f\parallel _{p}^{p \minus{} 1}$ as $ \left(\int\limits_{E}|f(x)|^{p}\right)^{1/q} \equal{} \left(\int\limits_{E}|f(x)|^{p}\right)^{1 \minus{} \frac1p} \equal{} \left(\int\limits_{E}|f(x)|^{p}\right)^{\frac {p \minus{} 1}{p}}$.

Also, in the inequality you quoted, $ |f(x)|^{p\minus{}1}$ then indeed becomes $ |f(x)|^p$ as $ (p\minus{}1) q \equal{} (p\minus{}1) \frac{p}{p\minus{}1} \equal{} p$.

(Note he said he would look at the case $ 1 < p$, implying $ q$ is not ill-defined by this expression.)

[quote="JRav"]show that $ L^{\infty}(E)\subset L^{p}(\mathbb{R})$[/quote]

If $ f|_E\in L^\infty(E)$, then certainly $ \|f\|_{L^p(E)}<\infty$ (you can get a quantitative version from Holder) but you have no control on what happens outside of $ E$.  In particular $ f$ can be big enough on $ \mathbb{R}\setminus E$ that $ f|_E\in L^\infty(E)$ and $ f\not\in L^p(\mathbb{R})$.  Obviously the reverse inclusion doesn't hold either.  So what do you really mean?

If $ m(E)<\infty$, show that, as sets, $ L^{\infty}(E)\subset L^{p}(\mathbb{R})$, for any $ 1\leq p<\infty$, and that $ \left\|f\right\|_{p}\leq (m(E))^{1\minus{}1/p}\left\|f\right\|_{\infty}$ for any $ f\in L^{\infty}(E)$.  In particular, if $ f\in L^{\infty}[0,1]$, then $ \left\|f\right\|_{1}\leq\left\|f\right\|_{p}\leq\left\|f\right\|_{\infty}$ for any $ 1<p<\infty$.

This is everything, word for word.  I agree with your comment and that is why I pressed the issue in my second post.  Is pluricomplex correct that there is a typo and the exponent should read $ 1/p$ instead of $ 1\minus{}1/p$?

I think $ L^{\infty}(E)\subset L^{p}(\mathbb{R})$ here means that for any $ f\in L^{\infty}(E)$, if we consider $ f$ as a function determine on $ \mathbb R$ by means $ supp f\subset E$, then from the claim $ \parallel f\parallel _{p}\leq \parallel f\parallel _\infty m(E)^{1/p}$, it implies $ f$ must belongs to $ L^{p}(\mathbb{R})$.
($ supp f$ here is support of $ f$)

[quote="JRav"]Is pluricomplex correct that there is a typo and the exponent should read $ 1/p$ instead of $ 1 \minus{} 1/p$?[/quote]

yes, e.g. by scaling (or by setting $ f\equal{}1$ as he said earlier)


We are given[hide] (x/3-3/x)^2=0.  Since the squareroot of 0 is 0, x/3-3/x=0.  Using algebra, x/3=3/x.  Therefore x^2=3^2=9.  This means x=3, or x=-3.  The problem asks for x^4, which for either of our x's is 81.[/hide]

[hide]From the first equation, we get

(x/3-3/x) = 0

x/3 = 3/x

x = 9/x

x^2 = 9

x = 3

x^4 = 3^4 = 27[/hide]

[quote="MCrawford"]Given $\displaystyle \left( \frac{x}{3} - \frac{3}{x} \right)^2 = 0$, find the value of $x^4$.[/quote]

$x^4 = 81$

Potter, check your arithmetic at the end.

[color=white]

I did exactly what potter did, except I calculated 3^4 correctly  ;)

[/color]

[hide]if you start it off like you would in a normal ... thing, you would times both by the denominators. so i got (x 2 -9)/ (3x)=0 but of this is true, then x 2 -9 have to be 0 because that's logical. then, you add both sides by 9 then square root which gives you x=3. then you do it to the fourth power, which now gives you the answer:
x 4=81

TADA!!!!!
 :D [/hide]

yeah calculating correctly always helps.   Never fear i always make mistakes on it

[hide]
I used a different, but (I think) quicker method.
Because a value squared is zero, that value must equal 0.
So we know that
(x/3)-(3/x)=0
So, x/3=3/x
Each of those sides must equal 1, otherwise they can not be equal, so x=3
If x=3, then x^4= [b]81[/b][/hide]

[hide]$\displaystyle \displaystyle \left( \frac{x}{3} - \frac{3}{x} \right)^2 = 0$

$\displaystyle \displaystyle \left( \frac{x^2-9}{3x}\right)^2 = 0$

$x^2=9$

$x^4=81$[/hide]

[hide](x/3 - 3/x)^2 =0
So x/3 - 3/x = 0
x/3 = 3/x
x^2 = 9
(x^2)^2 = x^4 = [b]81[/b][/hide]
interestingly enough, this question appeared on the 2004 Georgetown (contest in Texas) and i put the same answer but got the question wrong.... hmmm


probability
AMC
AMC 10
AIME
USA(J)MO
USAMO
AMC 12

[quote="chuchen"]1.  How many 5 digit numbers exists in which no two consecutive digits are the same?
2.  How many four digit numbers are there in which the four digits are in decreasing order?
3.  Of all the people in world, 30% own dogs and 40% own cats and 20% of dog owners also own cats.  If a person was randomly selected what is the probablity that they would own a dog or a cat?____a dog and a cat?____
4.  The probablity that Spike will pass his math test is 80%.  The probability that Spike will pass his history test is 70%.  Calculate:
a.  p(he will pass both)        b.  p(he will fail both)
c.  p(he'll pass exactly one)  d.  p(he will fail at least one)
5.  The probability of scoring high enough on the AMC10 to qualify for the AIME is 20%.  10% of those who fail still manage to qualify for the AIME by scoring well enough on the AMC12.  Those who qualify by the AMC10 have a 15% chance of advancing past the AIME to the USAMO, while 25% of those who qualify through the AMC12 advance to the USAMO.  What is the probability of a student advancing to the USAMO?[/quote]
[hide="1"]So we have teh blanks _ _ _ _ _. The first blank you have 9 choices for it. The second one then also has 9. Then it goes to 8, then 9, then finally 8. So its 46656. [/hide]
Can somebody check mine? I'm not sure of mine.

actually there are five blanks:

_ _ _ _ _.

the first slot has 9 and since no others can't be consecutive then the others are 8, so $9 \times 8^{4}= 36864$...

at least i think thats right :wink:

[quote="usaha"]wouldnt this be 9^5?[/quote]

no because it says that the next digit cannot be the same

so therefore it has to be $9 \times 8^{4}$

[quote="usaha"]Wait, can you explain ur reasoning again?  lets say its a two digit number, then the answer would be 81 which is 9*9, not 9*8[/quote]

wait actually you are right but i think you are right for the wrong reasons....

the first digit can have numbers 1-9, the second digit can have 0-9, excluding the first digit, therefore leaving 9..so the answer is really $9^{5}$

Ya, thats what i was thinking... I was just stating an easier problem that you can brute force to show you that any question of that type is 9^x where x is the number of digits.

and number 2 is [hide]10C4=210[/hide]

[hide="3"] Since 20% of the 30% of the owners who have dogs also have cats, we can say that (.2)(.3)=.06=6% of the people own dogs and cats.

Thus 30-6=24% of the people only own dogs and 40-6=34% of the people only own dogs.

Thus, 24+34=58% own either a cat or a dog (which is part of the answer).

We already figured out that there is a chance of 6% that someone owns both a cat and a dog (this is the answer to the other part).  [/hide]
[hide="4"] 4a.  This probability is (.7)(.8)=.56=56%

4b.  This probability is (1-.7)(1-.8)=(.3)(.2)=.06=6%

4c.  Say he fails the mathematical test and passes the history test.  This probabilty is (.7)(1-.8)=.2(.7)=.14=14%
Say he fails the history test and passes the mathematical test.  This chance is (.8)(1-.7)=.8(.3)=.24=24%.  When you add these up, you get 38%, which is our answer.

4d.  The probabilty that he fails at least 1 is equal to 1-p(pass both)=1-56%=44% [/hide]
[hide="5"] Since 20% of the students who get through by AMC 10 have a 15% chance of making it to USAMO, only (.15)(.2)=.03=3% of these people make it to USAMO.

Now, there are 80% of the people who don't make it to AIME through AMC 10.  Thus, 10% of the 80% of people=8% make it through AMC 12.  25% of these people have the chance of making it to USAMO, thus (.25)(.08)=.02=2%.

 Thus, our total is 2%+3%=5% [/hide]


Maybe this was posted before, but searching such a thing is a pain in the neck...
If one types [code]$ \ $[/code], this results in some error (and if one tries to edit, the \ is gone).

If you use it in any meaningful context, it works OK: $ js;\ \ \ \ \ hdljfhld$. What exactly would one need the $ \$ \ \backslash \ \$$ for? :?

Creating spaces ;)
Well, there may be more severe bugs with the same reason.

[quote="ZetaX"]Creating spaces ;)[/quote]
Use the LaTeX command \hspace{}


Let $ u(n)$ denote the number of ways in which a set of $ \mbox{n}$ elements can be partitioned into non-empty subsets. For example, $ u(3)\equal{}5$, corresponding to:

$ \{a,b,c\}$

$ \{a,b\}, \{c\}$

$ \{a,c\},\{b\}$

$ \{a\},\{b,c\}$

$ \{a\},\{b\},\{c\}$

I thought it might have been a Catalan Recurrence praying that $ u(4) \equal{} 14$ but $ u(4) \equal{} 15$ :(

How do I do this question?


I was wondering, how can I shade a portion of a graph? I have the code:
[code][asy]
size(200);
import graph;

Label f;
f.p=fontsize(6);

xaxis(-7, 7, Ticks(f, 10, 2));
yaxis(-7, 7, Ticks(f, 10, 2));

dot("(0, 5)", (0, 5), black);
dot("(0, -5)", (0, -5), black);
dot("(5, 0)", (5, 0), black);
dot("(-5, 0)", (-5, 0), black);

draw((5, 0)--(0, -5));
draw((0, 5)--(5, 0));
draw((-5, 0)--(0, 5));
draw((0, -5)--(-5, 0));

[/asy]
[/code]

and I'm wondering how to shade the interior of the square. (this is the graph of $ |x|\plus{}|y|\le5$)

Thanks in advance :)

Try this.

[code]
[asy]
size(200);
import graph;

real a= 7, h=5;
pair b= (0,h), c= (0,-h), d= (h,0), e= (-h,0);
pen pp= black;
path r= c--e--b--d--cycle;

Label f;
f.p=fontsize(6);

xaxis(-a,a,Ticks(f,10,2));
yaxis(-a,a,Ticks(f,10,2));

dot("(0, 5)",b,pp);
dot("(0, -5)",c,pp);
dot("(5, 0)",d,pp);
dot("(-5, 0)",e,pp);

draw(r);
fill(r,yellow);

[/asy]

[/code]

I declared some new variables to make your code more efficient. Shading in a portion of the graph is done by using the fill command on a specific path, or in this case, the path r I defined earlier.

Wow!! It really works! Now, I can shade in graphs very easily . Thanks a lot!


the book "geometry revisited" by coxeter is good material for geometry.

Would the book "Introduction to geometry" by the same author be good?

http://www.amazon.com/Introduction-Geometry-2nd-H-Coxeter/dp/0471504580/ref=pd_sim_b_4/102-2718426-6433766

Thanks guys!

Coxeter's "Geometry Revisited" is a classic but is overhyped.


linear algebra
matrix
geometry
geometric transformation
reflection
trigonometry

Describe a transformation that carries the triangle with vertices $ R \equal{} (1, 2)$, $ P \equal{} (6, 7)$, and $ Q \equal{} (10,5)$ onto the triangle with vertices $ K \equal{} (0,0)$, $ L \equal{} (7,\minus{}1)$, and $ M \equal{} (9,3)$.

Where does the transformation send:
(a) (7, 4)
(b) (3, 4)
(c) (4, 4)

[quote="AndrewTom"]Peterhi, have you tried plotting the two triangles to see what's going on? I think that might give you some ideas. Hope this is of some help.[/quote]

Yes, I've plotted them, but I'm still unable to solve this, in large part because I don't have a systematic method for attacking this type of problem.

Any solution or methodological insights you could share would be most appreciated.

Thanks.

I first drew the two triangles on squared paper and saw that one gets from triangle $ PQR$ to triangle $ KLM$ by translating each point $ 1$ to the left and down $ 2$ and then reflecting in the line $ y\equal{}\frac{1}{3}x$, that is in the line $ KM$.

I used the matrix for reflection in the line $ y\equal{}x\tan \theta$ but one could also solve between $ y\equal{}\frac{1}{3}x$ and the perpendicular through the required point.

For example, for (a) $ (7, 4)$, one first translates to $ (5, 5)$ and then finds that the perpendicular through this point is $ y \equal{} \minus{}3x \plus{} 20$. (I got this by substituting $ x\equal{}5$ and $ y\equal{}5$ into $ y \equal{} \minus{}3x\plus{}c$ to find $ c\equal{}20$.)

Solving between this and $ y\equal{}\frac{1}{3}x$, they meet at $ (6, 2)$.

For (b), I get $ (\frac{14}{5}, \frac{\minus{}2}{5}) \equal{} (2.8, \minus{}0.4)$

and for (c) $ (\frac{18}{5}, \frac{1}{5}) \equal{} (3.6, 0.2)$.

I hope that this is of more help to you than my previous contribution.

Maybe this is not directly related to this problem but there is a theory behind it. Look up under [i]isometries[/i] and [i]glide-reflections[/i]; there are full discussions on how to solve these types of problems painlessly.


limit
integration
calculus
calculus computations

Evaluate $\lim_{n \rightarrow \infty} \frac{1}{(n+1)^2} \sum_{k=1}^{n} k^2 (a_k^2 + b_k^2)$

I dunno exactly about that :? . I've working on it for about a couple of hours. its remain me with Paserval's identity. so I try to define $a_k$  and $b_k$ as in Paserval's identity, and I got the limit is zero. This limit is from my friend. I must ask him for this lack of information. Sorry

Is $\sum$ inside the limit? Or it would be $0$...

I thinks I made some mistakes in my calculation  :| . Actually, while calculating this limit i looked at the $a_k^2+b_k^2$ expresion in the problem which is like the other one on paserval's indentity, $\frac{a_0}{2} + \sum_{k=1}^{\infty} (a_k^2 +b_k^2) = \frac{1}{\pi} \int_{- \pi}^{\pi} [f(x)]^2 \, dx$ or maybe from trigonometric series . But my long-nasty calculation is flaw. If this asumsion is right, maybe  Feje'r's theorem will work and so this limit is for those who have taking calculus III course (so not for me :P ). WHOA


geometry
parallelogram
trigonometry
Pythagorean Theorem
geometry proposed

1. In the square $ABCD$ is given  a  point $E$, and  out of the square a point $F$, such that $\triangle{ABE}=\triangle{BCF}$.Find the angles of the triangle $ABE$, if  $EF$ equal to the side of the square and $\angle{BFD}=90^0$. $\clubsuit$

Since $\angle BFD=\angle BCD= 90^\circ$ the quadrilateal $BFCD$ is inscriptible. Hence, $180^\circ=\angle DCF+\angle DBF=90^\circ+\angle BCF+45^\circ+\angle CBF=315^\circ-\angle BFC$, hence $\angle BCF=135^\circ$.
Is also easy to show that $\triangle BFE=\triangle EAB=\triangle FBC$ since they have all edges equal. Hence, $CFEB$ is parallelogram hence $EA=EC$ hence $E$ is on $DB$ hence  $\angle EBA=45^\circ$.

I don't realise where is the mistake. :blush:  the triagles seems to be degenerated.. :blush:

I also got the same. Tiks do you hava a solution?

[quote="xirti"]Since $\angle BFD=\angle BCD= 90^\circ$ the quadrilateal $BFCD$ is inscriptible. Hence, $180^\circ=\angle DCF+\angle DBF=90^\circ+\angle BCF+45^\circ+\angle CBF=315^\circ-\angle BFC$, hence $\angle BCF=135^\circ$.
[/quote]
$xirti$ your mistake is here.You get that $\angle BCF=135^\circ$,but how I have seen you havn't used the condition that$EF=AB$,without this condition the point $F$ can be any point in the arc $BC$.
To M4RIO:Of cours I have solution(otherway I wouldn't post this problem in this section),but I think it is too early to post my solution,try to solve it yoursalf ;) .

Since $\angle BFD=\angle BCD= 90^\circ$ the quadrilateal $BFCD$ is cyclic.

From here we have $\angle DBC = \angle DFC = 45 ^ { \circ }$  therefore $\angle BFC = \angle AEB = 135 ^ { \circ }$. $(1)$

Let $\angle EAB = \angle BCF = \omega$ and $\angle ABE = \angle CBF = \theta$.

From $(1)$    ,    $\omega + \theta = 45 ^ { \circ }$. Also $\angle EBF = \angle ABE + \theta - \theta = 90 ^ { \circ }$ so the triangle $\triangle EBF$

 is orthogonal and isosceles. We now from pythagorean theorem , if $AB =a$ we have  $EB=BF = \frac{a}{ \sqrt{2}}$ .

From sines law $\frac{a}{ \sin{ A \hat {E} B }} = \frac{ \frac{a}{ \sqrt{2}}}{ \sin{ \omega }}$ therefore $\omega = 30 ^{ \circ}$ and $\theta = 15 ^{ \circ}$ .

[quote="Tiks"]
$xirti$ your mistake is here.You get that $\angle BCF=135^\circ$,but how I have seen you havn't used the condition that$EF=AB$,without this condition the point $F$ can be any point in the arc $BC$.[/quote]

I wrote : $\triangle BFE=\triangle EAB=\triangle FBC$ because they have all edges equal: $BF=BF$, $BE=FC$, $EF=BC$. I still can't see the mistake.

As Ramanujan wrote, the triangle $BFE$ is orthogonal and isoscelles, but I proved that it is similar with $EAB$ which has an angle of measure $135^\circ$, hence contradiction.

[quote="xirti"]

It is $AB=BC \, , \, CF = AE \, , \, EB = FB$ so maybe Tiks should write  $\triangle{BAE}=\triangle{BCF}$ .

:)[/quote]

That works and makes the problem much easier.

[quote="Ramanujan"][quote="xirti"]

I wrote : $\triangle BFE=\triangle EAB=\triangle FBC$ because they have all edges equal: $BF=BF$, $BE=FC$, $EF=BC$. I still can't see the mistake.

[/quote]

[quote="Tiks"]such that $\triangle{ABE}=\triangle{BCF}$[/quote]

It is $AB=BC \, , \, CF = AE \, , \, EB = FB$ so maybe Tiks should write  $\triangle{BAE}=\triangle{BCF}$ .

:)[/quote]
In original version it was written that the triangle $BAE$ equal to the triangle $BCF$,and you should prove that it is possible only when $AB=BC \, , \, CF = AE \, , \, EB = FB$. ;)

[quote="Ramanujan"][quote="Tiks"]
In original version it was written that the triangle $BAE$ equal to the triangle $BCF$,and you should prove that it is possible only when $AB=BC \, , \, CF = AE \, , \, EB = FB$. ;)[/quote]

[quote="xirti"]Since $\angle BFD=\angle BCD= 90^\circ$ the quadrilateal $BFCD$ is inscriptible. Hence, $180^\circ=\angle DCF+\angle DBF=90^\circ+\angle BCF+45^\circ+\angle CBF=315^\circ-\angle BFC$, hence $\angle BCF=135^\circ$.

Is also easy to show that $\triangle BFE=\triangle EAB=\triangle FBC$ since they have all edges equal. Hence, $CFEB$ is parallelogram hence $EA=EC$ hence $E$ is on $DB$ hence $\angle EBA=45^\circ$.

 [/quote]

Maybe this is a proof ?   ;)[/quote]
Mayby  :D ,but ther is also a trivial variant when $AB=BF$ :P  ;) .

[quote="Tiks"]
Mayby ,but ther is also a trivial variant when $AB=BF$ :P  ;) .[/quote]

Yes i know but i leave it for you  :P


1. For function $ f: \mathbb{R} \to \mathbb{R}$ given that $ f(x^2 \plus{} x \plus{} 3) \plus{} 2 \cdot f(x^2 \minus{} 3x \plus{} 5) \equal{} 6x^2 \minus{} 10x \plus{} 17$. Calculate $ f(2009) \equal{} ?$.

http://www.mathlinks.ro/viewtopic.php?p=1477953#1477953

2. $ x$ and $ y$ are real numbers.
A-)if it is known that $ x \plus{} y$ and $ x \plus{} y^2$ are rational numbers, can we conclude that $ x$ and $ y$ are also rational numbers.
B-)if it is known that $ x \plus{} y$ , $ x \plus{} y^2$ and $ x \plus{} y^3$ are rational numbers, can we conclude that $ x$ and $ y$ are also rational numbers.

http://www.mathlinks.ro/viewtopic.php?p=1477954#1477954

3. $ a,b,c$ are sides of triangle $ ABC$. For any choosen triple from $ (a \plus{} 1,b,c),(a,b \plus{} 1,c),(a,b,c \plus{} 1)$ there exist a triangle which sides are choosen triple. Find all possible values of area which triangle $ ABC$ can take. (ps: Sorry for bad english...)

http://www.mathlinks.ro/viewtopic.php?p=1477961#1477961


Hello geniuses ...
I've alittle question about permutations
WHY 0!=1 ??
0!=1  .. it's too difficult to understand
please I need to answer

0!=1

By $_n C_r=\frac{n!}{r!(n-r)!}$, we have $ _nC_n=\frac{n!}{n!(n-n)!}=\frac{n!}{n!0!}=\frac{1}{0!}$.

Since $_nC_0=_nC_n=1$, if we suppose that $0!=1$, then this is consistent with our problem.

In this way, $0!=1$ is defined for convinience.

well since a permutation can tell you how many ways you can rearrange something. You can say that the number of ways you can rearrange nothing is one way.  :D

Let's work backwards:

$5! = 5\times 4\times 3\times 2\times 1=120\\ \\
4! = \frac{5!}{5}=\frac{120}{5}=24 \\ \\
3! = \frac{4!}{4}=\frac{24}{4}=6 \\ \\
2! = \frac{3!}{3}=\frac{6}{3}=2 \\ \\
1! = \frac{2!}{2}=\frac{2}{2}=1 \\ \\
0! = \frac{1!}{1}=\frac{1}{1}=1 $

Here's another way to think about this.  If I asked you to add all of the elements of an empty set of numbers, you probably would say the answer was 0.  That's correct.  If I add one element to the set, the new result is the old result plus the new element ($0+x$, where $x$ is the new element).

If I asked you to multiply all of the elements of an empty set of numbers, the answer is 1.  That might be surprising, but if I add one element to the set, the new result is the old result times the new element ($1\times x$, where $x$ is the new element).

In how many ways can you arrange 0 objects on a shelf?

[quote="bahamut"]Thanks mr.Kummy for your help
and thank you mr.dangvh
mr.Kummy your sulotion is very excellent but can you give me the perfect Mathimatical proof ????  :) :lol:  :lol:  :lol:[/quote]

No. There is no mathematical proof, because $0!=1$ is the definition,the same as $(x+1)^0=1$. ;)

To Kunny and Jimmy:
From that expression, you forgot to mention that x cant equal -1, otherwise to Jimmy's explanation, the denominator would be 0, and would be therefore undefined

Actually $0^0$ is defined as $1$ because $\lim_{n\to 0} n^n = 1$.  But you're right, you can't divide by zero.

Since this is a middle school forum, $\lim_{n\to 0} n^n = 1$ means that as $n$ approaches $0$, $n^n$ approaches $1$.

Well, there's one way to arrange zero objects...


Obviously, the set of all positive even numbers, E, plus the set of all positive odd numbers, O, equals the set of all positive integers, I.

How do I show this algebraically?

Let E=2n and O=2n-1 where n=1,2,3,4...

Obviously, you can't just add the two, (E + O = I).

(2n) + (2n-1) = 4n-1.

The answer should be n, not 4n-1!

Is there a way to do something like this?
Do I need to use n1 and n2 instead of a single n?

Any thoughts would be welcome...

Thanks.

Or in other words:

If E={2n:n belongs to the positive integers}
and O={2n-1:n belongs to the positive integers}

then is it possible to get the following:

E union O={n:n belongs to the positive integers} ?

And how do I show that?

Thanks.


[quote="shobber"]Given:
\[ S = 1 + \frac{1}{1 + \frac{1}{3}} + \frac{1}{1 + \frac{1}{3} + \frac{1} {6}} + \cdots + \frac{1}{1 + \frac{1}{3} + \frac{1}{6} + \cdots + \frac{1} {1993006}}  \]
where the denominators contain partial sums of the sequence of reciprocals of triangular numbers (i.e. $k=\frac{n(n+1)}{2}$ for $n = 1$, $2$, $\ldots$,$1996$).  Prove that $S>1001$.[/quote]
We have
\[ \frac{1}{k_n}=\frac{2}{n(n+1)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right) \]
Thus \[ \sum\limits_{n=1}^m\frac{1}{k_n}=\sum\limits_{n=1}^m2\left(\frac{1}{n}-\frac{1}{n+1}\right)=2\left(1-\frac{1}{m+1}\right)=2\frac{m}{m+1} \\\Rightarrow S=\sum\limits_{m=1}^{1996}\frac{1}{\sum\limits_{n=1}^m\frac{1}{k_n}} =\sum\limits_{m=1}^{1996}\frac{m+1}{2m}=\frac{1}{2}\sum\limits_{m=1}^{1996}\left(1+\frac{1}{m}\right)=998+\frac{1}{2}\sum\limits_{m=1}^{1996}\frac{1}{m} \]
It remains to show, that $\sum\limits_{m=1}^{1996}\frac{1}{m}>6$. I'm too lazy to show this, but it is definitely not difficult (unless it's wrong
 ;)  ).

[quote="Werwulff"]
It remains to show, that $ \sum\limits_{m \equal{} 1}^{1996}\frac {1}{m} > 6$. I'm too lazy to show this, but it is definitely not difficult (unless it's wrong
 ;)  ).[/quote]It is indeed easy.

$ 1\plus{}\frac12\plus{}\frac13\plus{}\frac14\plus{}\ldots\plus{}\frac1{1996}>1\plus{}(\frac12)\plus{}(\frac13\plus{}\frac14)\plus{}(\frac15\plus{}\frac16\plus{}\frac17\plus{}\frac18)\plus{}\ldots\plus{}(\frac1{513}\plus{}\frac1{514}\plus{}\ldots\plus{}\frac1{1024})>1\plus{}(\frac12)\plus{}(\frac14\plus{}\frac14)\plus{}(\frac18\plus{}\frac18\plus{}\frac18\plus{}\frac18)\plus{}...\plus{}(\frac1{1024}\plus{}...\plus{}\frac1{1024})\equal{}1\plus{}\frac12\plus{}\frac12\plus{}\frac12\plus{}...\plus{}\frac12\plus{}\frac12\equal{}6$. :D

Let $a_n=1+\frac13+\frac16+\ldots+\frac2{n(n+1)}$. Note that $S=\sum_{i=1}^{1996}\frac1{a_i}$. Now we can easily calculate $a_n$ with partial fraction decomposition and telescoping series:
$$a_n=\sum_{i=1}^n\frac2{i(i+1)}=\sum_{i=1}^n\left(\frac2i-\frac2{i+1}\right)=2-\frac2{n+1}=\frac{2n}{n+1}.$$
Then:
$$S=\sum_{i=1}^{1996}\frac1{a_i}=\sum_{i=1}^{1996}\frac{i+1}{2i}=\sum_{i=1}^{1996}\left(\frac1{2i}+\frac12\right)=\frac12\sum_{i=1}^{1996}\frac1i+998$$
so it suffices to show that $\sum_{i=1}^{1996}\frac1i>6$. Since:
$$1+\frac12+\ldots+\frac1{1996}>1+\frac12+\frac12+\frac14+\frac14+\frac14+\frac14+\frac18+\ldots+\frac1{1024}+\frac1{1024}=6$$
we're done.


[quote="shobber"]Function $f(x)=\sqrt{x^2+1}-ax$, $a>0$.
(1) Solve the inequality: $f(x) \leq 1$.
(2) Prove that when $a \geq 1$, $f(x)$ is a monotone function of $[0, + \infty)$.[/quote]

$\sqrt{x^2+1}-ax\leq 1$

$x^2+1\leq a^2x^2+2ax+1$

$(a^2-1)x^2+2ax=x((a^2-1)x+2a)\geq 0$

$x\geq 0$ and $x\geq\frac{2a}{1-a^2}$ or $x\leq 0$ and $x\leq\frac{2a}{1-a^2}$

$x\geq\frac{2a}{1-a^2}$ iff $0\leq a<1$

$x\leq\frac{2a}{1-a^2}$ iff $a>1$

Im stuck

Are you sure you can square the inital inequality?

[quote="shobber"]Function $f(x)=\sqrt{x^2+1}-ax$, $a>0$.
(1) Solve the inequality: $f(x) \leq 1$.
(2) Prove that when $a \geq 1$, $f(x)$ is a monotone function of $[0, + \infty)$.[/quote]

i believe that the best way to do part 1 is to intersect y=1, and test values to find the correct intervals for x in terms of a

10000th User: I think that you forgot $ax+1 \geq 0$.


When friction acts between two surfaces, there is clinging at the atomic level.

When one surface moves over the irregularities of the other surface, the atoms snap, and the bumps deform to give rise to [b]waves[/b] and [b]atomic motions[/b].

What kind of [b]waves[/b] are these?

I'm kind of guessing here, but since you can sometimes hear friction (think sandpaper on wood), I would think longitudinal.  But that's really nothing more than a guess.

JRav, can you try finding out from somewhere which kind of [b]waves[/b] are these?
Need the answer quite urgently.

Those are longitudinal waves, I think.  It's like having a bunch of masses tied by springs (atoms), and you pull one of the mass, then release it.  You get a longitudinal wave.

I have another question related though.  If you have an infinite line of masses m, tied to adjacent blocks by springs of stiffness k, what is the speed of a disturbance is transferred?

I am of the same opinion. Earthquakes are also a product of friction between the plates that float on molten magma.


A cylinder is inscribed in a sphere. The diameter of the cylinder is 13 cm. Its height is 13 cm. Find the volume of the sphere.
______________________________________________________
Math is a subset of life. :P

[hide]The diameter and height make a right triangle, with the hypotenuse being the diameter of the sphere.

$13^{2}+13^{2}=x^{2},$ so $x=13\sqrt{2}$

The radius of the sphere is $\frac{13\sqrt{2}}{2}.$

The volume is $\frac{4}{3}\pi\left(\frac{13\sqrt{2}}{2}\right)^{3},$ or $\frac{2197\pi\sqrt{2}}{3}.$

Approximately $3253.671\ldots$[/hide]

[b]i_like_pie[/b]
Great, you solved it  as I solved it and your answer is true.
I think your grade must be in advanced section.


I was wondering today whether it is possible to use the gamma/factorial functions to solve things such (over the reals, rather than the integers)
like
$\binom{52}{k}= 30$.
I know that no integer satisfies this, and I'm not sure if any rational number does either.
Does it involve an inverse gamma function (do they even exist?)?

Combinations are not defined for non-integer values of $k$.

They are, however, defined for non-integer values of $n$.  To be precise,

${x \choose k}= \frac{x(x-1)...(x-(k-1))}{k!}$

Where I am using $x$ to emphasize the fact that we have a polynomial in $x$, which is therefore defined for all real (and for that matter, complex) values of $x$.  The question of finding $x$ given $k$ and ${x \choose k}$ is therefore solving a polynomial of degree $k$.


Is there any number in the Fibonacci sequence whose six last digits (in decimal system) are all $ 9$?

We denote $F(0)=0,F(1)=1$ and $F(n+2)=F(n+1)+F(n)$ for all $ n\in\mathbb{Z}$
Then $(F(n),F(n+1))$ mod $1000000$ has only limited number of orders (i.e. at most $1000000^2$). That means $F(n)$ mod $1000000$ is periodic. And $F(-2)=-1$, therefore exist a $n>0$ such that $1000000|F(n)-F(-2)$ and its last six digits are all $9$


integration
function
analytic geometry
graphing lines
slope
algebra
domain

I have always hated separable variable equations. Let's solve a really simple ODE by this method: $x' = x$ on some open interval $I$. The method gives us the solutions $Ce^{t}$ where $C\neq 0$ and $x = 0$. Now, how do we show that these are actually all solutions without referring to the uniqueness and existence theorem (also of course not using a different method like integrating factor)?

In other words, I want to show that we cannot have a solution where $x(t)$ is 0 at some point and not 0 at some other point in $I$. It should be a simple contradiction, using that at a point $t$ where $x(t) \neq 0$ we have locally one of the solutions $Ce^{t}$ by the reasoning of separable variables.  However, I really cannot do it! I am trying to show that the set of points $x(t) = 0$ must be open, but I can't figure out how to do it.

Can anyone help me?

We could do something like [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=proof&t=83256]this[/url].

Kent's solution uses the integrating factor method, whilst spix's uses the Mean Value Theorem.

That is not what I want. It seems like it should be possible to construct a contradiction immediately from the fact that at a point $t$ where $x(t) \neq 0$ we have locally one of the solutions $Ce^{t}$. The problem I get is that we can't choose the same $C$ for every point, so I can't get the contradiction that $x(t)$ is bounded away from 0.

When I asked a teacher today how to do it he just drew a bunch of circles and said it was easy to show $x$ would have to be bounded away from 0 because of the above, but I can't figure out how to it :(

You're not going to get away from the existence half of the existence-uniqueness theorem; after all, separation of variables works just fine for equations that don't have unique solutions. The lemma referenced in the other thread is the core of the proof of that theorem.

Here's a version I just did for homework, with a different method for one-sided uniqueness:

[hide="Separation of variables"]Consider the initial value problem $u'=f(u)$, $u(0)=u_{0}$. Suppose also that $f$ is continuous and positive on $[u_{0},\infty)$ (We need some condition on $f$ for uniqueness.
Define $ g(u)=\int_{u_{0}}^{u}\frac1{f(v)}\,dv$. We have $g'(u)=\frac1{f(u)}$. If $U$ is a differentiable function with $g(U(t))=t$ for $t$ in some interval $I$ containing zero and $U(0)=u_{0}$, the chain rule gives that $1=g'(U(t))\cdot U'(t)=\frac{U'(t)}{f(U(t))}$ and therefore $U'(t)=f(U(t))$. This makes $U$ a solution of the initial value problem. The inverse function states precisely that such a differentiable function exists and is unique in some open interval containing zero as long as the slope $f(u_{0})$ is nonzero. That condition is satisfied, and we have existence.
For uniqueness in the IVP, suppose that $w$ is a solution to the IVP on some $I$. Since $g'(w(t))w'(t)=1$ for all $t$ in the domain of $w$ by the differential equation, $g(w(t))=t+c$ for some $c$. Since $w(0)=u_{0}$, $c=0$. Since $u$ is a two-sided local inverse of $g$, $u(t)=u(g(w(t)))=w(t)$, and the solution $u$ is locally unique. Repeating this process at each point, the set $\{x: u(x)=w(x)\}$ is open in the intersection of domains. It is clearly closed, so it must be the entire intersection, and the solution is globally unique up to how far it is extended.[/hide]

Thinking about it, here's another uniqueness proof, which should work for this problem:
[hide="Uniqueness"]Let $f$ be a positive continuous function on $(a,\infty)$, and suppose $\lim_{u\to a^{+}}f(u)=0$; extend $f$ to $[a,\infty)$ naturally. If $\int_{a}^{a+1}\frac1{f(u)}\,du=\infty$, solutions to the initial value problem $u'=f(u),u(0)=u_{0}$ are unique for all $t<0$ whenever $u_{0}>a$.
Let $g(u)=\int_{u_{0}}^{u}\frac1{f(u)}\,du$ for all $u>a$.
By the previous part, solutions are unique on $[t,\infty)$ given $u(t)$ for any $t<0$ with $u(t)>a$. This determines $u(0)$ as well: $g(u(0))=g(u(t))-t$. Since $g$ is 1-1, there is at most one $u(t)$ satisfying this equation. Consequently, $u(t)=g^{-1}(g(u_{0})+t)=g^{-1}(t)$ if $u(t)>a$.
For uniqueness to fail, $u$ must become $a$ somewhere. If it does, then we can find a sequence of points $t_{n}$ with $t_{n}$ bounded and $f(t_{n})=a+\frac1n$ by continuity of $u$. By the above, $g(a+\frac1n)=t_{n}$. Since the improper integral $\int_{a}^{a+1}\frac1{f(u)}\,du$ diverges, $\lim_{u\to a^{+}}g(u)=-\infty$, and this contradicts boundedness of $t_{n}$.[/hide]


linear algebra
matrix
linear algebra unsolved

Let $ A$ be a matrix and it's elements are nonagative .Find the condition of $ A$ such that $ A^{\minus{}1}$ has nonagative elements.

consider products of elementary matrices, particularly permutation matrices and row-multiplying matrices...


Let x,y,z>0,k<2,prove that:
      sum_cyc[sqrt(x*y/(x^2-k*x*y+y^2))]<=3/sqrt(2-k).

[quote="fjwxcsl"]Let $x,y,z>0,k<2$,prove that:
      $\sum_{cyc}\sqrt{\frac{xy}{x^{2}-kxy+y^{2}}}\leq \frac{3}{\sqrt{2-k}}$.[/quote]

I hope it`s right.

Let  a,b,c are the sides of an ariangle,3/2<=k<2,prove that:

sum_cyc[a/sqrt(b^2-k*b*c+c^2)]<=3/sqrt(2-k).


AMC USA(J)MO USAMO AIME algebra polynomial geometry

I didn't make it to USAMO this year, but I want to give it one more try
in 2005. My score on AIME was 6, and I realized that I need to know
more combinatorics, number theory, and geometry.

I am getting:

Polynomials
enumerative combinatorics by Richard Stanley

for now.

I picked them because they're recommended books from AMC site.
but I read reviews on them, and they sound pretty hard, especially,
enumerative combinatorics.

Are they good choices for now, or no?

If yes, what should i get next? (any supplements that go along w/ the books?)

If no, what would be a better choice?

There is a difference between olympiad material and university textbooks. I assume that you've already learned all the high school stuff you need for contests, so I recommand problem collections (preferably categorized by problem type) as opposed to new theory stuff. Unless that is, your goal is to explore more math as opposed to conquering contests.

My first Olympiad book was Problem Solving through Problems, by Larson. I read this first because it happened to be in our school library, but it is nevertheless a very good book for beginners (with one flaw - about 2/3 of its problem don't have solutions, but you can always post them on the forum if you get stuck). The first one I bought is Mathematical Olympiad Challenge, which I highly recommand.

For pre-olym, it's again really all about practice. The AMC Contest books may be a place to start, or you might want to take a look at the [i]Problems Problems Problems [/i]series by U of Waterloo, which features problems from Canadian competitions. The Canadian contests act as a good bridge between pre-olymp and beginner-olymp.

Olympiads often require number theory and combinatorics, the latter of which is only touched upon during high school. Number theory is never formally taught. The geometry required in olympiads is also never taught.

So there are a few topics you should go through, and there are many more.

Of course, teaching yourself takes longer than doing an AoPS course or a summer program.


algebra
polynomial
inequalities
calculus
derivative
real analysis
real analysis unsolved

let $c_0+c_1x+..+c_nx^n$ be a real polynomial. Suppose that all
roots are real and stricly negative. Show that $c_0, c_1, .., c_n$
is a log concave sequence.

Let's make it a little bit tighter: $c_k^2\geq (n-k+1)/(n-k)\cdot  (k+1)/k \cdot c_{k-1} c_{k+1}$  :D

For the original one.

We may assume $c_n=1$, then $c_i$ are symmetrical polynomials of $x_1,...,x_n$.
Let's open the brackets in $A=\sigma_{k-1}\sigma_{k+1}$ and $B=\sigma_k^2$. We will obtain summands like $x_{i_1}^2...x_{i_l}^2x_{j_1}...x_{j_{2k-2l}}$.
We see that first summand can be obtained in $A$ in $C_{2k-2l}^{k-l+1}$ ways, and in $B$ in $C_{2k-2l}^{k-l}$ ways. Q.E.D.

What is the proof for the sharp estimate?

For the sharp estimate:
  For $k=1$ it easily follows from Cauchy-Schwarz and identities. Next, apply the inequality you have found for $k=1$ to successive derivatives of the polynomial, which also have strictly negative roots.


trigonometry
analytic geometry
graphing lines
slope
vector
function

I havent posted a problem for a while, so here is one of our former easy homework problems. The answer is not ugly at all. I hope you all enjoy solving it :D  :D

We fix two charges, of magnitude $ Q_1$ and $ Q_2$, at the ends of a wire, which is turned into a half circle, of radius $ R$. we put a small charge $ q$ of mass $ m$ on the wire, and this charge can move frictionlessly along the wire, in a half circle. Somewhere on the wire the charge $ q$ will be in equilibrium. Find the frequency of small ocsillations, around this equilibrium position.

The component of the electric field along the wire is $ E \equal{} \frac {1}{16\pi\epsilon_0R^2}\left( Q_1\frac {\cos\frac {\theta}{2}}{\sin^2\frac {\theta}{2}} \minus{} Q_2\frac {\sin\frac {\theta}{2}}{\cos^2\frac {\theta}{2}} \right)$

This gives equilibrium position as $ \theta_0 \equal{} 2\tan^{ \minus{} 1}\sqrt [3]{\frac {Q_1}{Q_2}}$

We should get angular frequency by using slope $ E'(\theta_0)$, but the answer is beyond ugly :(

There has to be an easier way..

As I said before, the answer is not ugly. Just the way till you get it is ugly and lengthy. You can either linearise the force acting on the small charge, or do it the way said, with $ E'$. :D

[hide="Partial Solution"]I'm assuming that $ Q_1$ and $ Q_2$ are of the same sign...

Let $ F_1$ be the force of $ Q_1$ on $ q$ and $ F_2$ be the force of $ Q_2$ on $ q.$ Also, taking a look at the posted diagram, let $ \theta \equal{} \angle qOQ_2,$ where $ O$ is the centre of the semi-circle. By Coulomb's Law, we have
\[ F_1 \equal{} \frac {q\cdot Q_1}{8\pi\epsilon_0\cdot R^2\cdot (1 \minus{} \cos\theta)}
\]

\[ F_2 \equal{} \frac {q\cdot Q_2}{8\pi\epsilon_0\cdot R^2\cdot (1 \plus{} \cos\theta)}
\]
Now, for the charge to be in equilibrium, we need the horizontal components of each of the forces to be equal. When $ q$ is at position $ \theta,$ we get that
\[ F_1 \cos \left(\frac {\theta}{2}\right) \equal{} F_2 \cos \left(90 \minus{} \frac {\theta}{2}\right) \equal{} F_2 \sin \left(\frac {\theta}{2}\right) \iff \cos\theta \equal{} \frac {Q_2^2 \minus{} Q_1^2}{Q_2^2 \plus{} Q_1^2}
\]
Now, since $ \tau \equal{} I\alpha,$ we can state that
\[ mR^2 \cdot \frac {d\theta^2}{d^2 t} \equal{} (\vec{F_1} \plus{} \vec{F_2})\cdot \vec{u}
\]
where $ \vec{u}$ denotes the unit tangent to the semi-circle at the point where the charge is. For orientational purposes, let $ \hat{r} \times \vec{u}$ always point out of the page, where $ \hat{r}$ is the radial vector of the semi-circle. Now, note that $ \vec{F_1} \cdot \vec{u} \equal{} F_1 \cos\left(\frac {\theta}{2}\right)$ and $ \vec{F_2} \cdot \vec{u} \equal{} F_2 \cos\left(\frac {\theta}{2}\right).$ Therefore, we are left with the differential equation
\[ mR^2 \cdot \frac {d\theta^2}{d^2 t} \equal{} (F_1 \plus{} F_2)\cos\left(\frac {\theta}{2}\right)
\]
If we let $ \phi$ be the angle displacement between equilibrium and the current particle's position, we make the substitution
\[ \theta \equal{} \phi \plus{} \cos^{\minus{}1}\left(\frac {Q_2^2 \minus{} Q_1^2}{Q_2^2 \plus{} Q_1^2}\right)
\]
[/hide]
This is where I am right now, and I find the calculations to be a bit untidy. However, I still have to replace $ \sin\phi$ with $ \phi,$ etc. I'll work on this a bit later  :) Are there any errors so far?

Only that there should be one more $ R$,
$ mR^2\frac{d^2\theta}{dt^2}\equal{}(\vec{F}_1\plus{}\vec{F}_2)\cdot\vec{u}R$,
but I think you've included that in your answer :)

Anyway, getting an expression for the answer seems easy, the problem is with the trigonometric manipulations :|

I'm guessing we may have to approximate a lot.

Yes Cyber-Shreyas you are right we have to approximate a lot.
But hey, it is a number one quality of a physicist to know when and how to approximate. :D

Karth: I didnt include the signs of the charges deliberately. You know casework, is part of solving a physics problem... :D

Come on guys, its not such a hard problem. Show till the point you have managed to get... :D

[quote="Thaakisfox"]
Karth: I didnt include the signs of the charges deliberately. You know casework, is part of solving a physics problem... :D [/quote]

Well, my way works when the opposite charges on the wire are of the same sign. If they are of opposite sign, obviously, there's no oscillation.

I see there hasent been a full solution yet.
Karth: try to express the position of equilibrium with $ \tan$ function as shreyas did. Then it will be a bit more simple. Try linearising the equation of motion by using Taylor series approximation.

Here is the answer so you see its not that complicated: :D

[hide="answer"]
\[ \omega\equal{}\sqrt{\dfrac{3q}{32\pi\varepsilon_0 m R^3}\left(\dfrac{Q_1}{\cos{\frac{\varphi_0}{2}}}\plus{}\dfrac{Q_2}{\sin{\frac{\varphi_0}{2}}}\right)}\]

Where $ \tan{\dfrac{\varphi_0}{2}}\equal{}\sqrt[3]{\dfrac{Q_2}{Q_1}}$ denotes the position of equilibrium.[/hide]


$ f(x)\equal{}\frac{1}{\sin x}\minus{}\frac{1}{x}$ for $ 0<x\le\frac{\pi}{2}$

Prove that $ f(x)>0$ and $ f(x)>f(y)$ for $ x>y$

hello, defining $ h(x)=x-\sin(x)$ and we have $ h^'(x)=1-\cos(x)$, since $ h(0)=0$ and $ h^'(x)>0$ we get that $ h(x)$ is strictly monotonoulsy increasing and we have $ x-\sin(x)>0$ for all x with $ 0<x\le\frac{\pi}{2}$, and $ x-\sin(x)>0$ is equivalent with $ \frac{1}{\sin(x)}>\frac{1}{x}$.
Sonnhard.

hello, to the second part, we have $ \frac {f(x) \minus{} f(y)}{x \minus{} y} \equal{} \minus{} \frac {\cos(\xi)}{\sin^2(\xi)} \plus{} \frac {1}{\xi^2} > 0$  with $ 0 < \xi\le\frac {\pi}{2}$ since $ \sin(x) > x$ we have $ \frac {\sin^2(x)}{x^2} > 1 > \cos(x)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]$ \frac {f(x) \minus{} f(y)}{x \minus{} y} \equal{} \minus{} \frac {\cos(\xi)}{\sin^2(\xi)} \minus{} \frac {1}{\xi^2}$[/quote]
i don't understand

$ \xi\equal{}?$

(also known as "mean value theorem": [url=http://en.wikipedia.org/wiki/Mean_value_theorem]click[/url])


A triangle has two acute angles, A and B.  Show that the triangle is right-angled if, and only if,
(sinA)^2 + (sinB )^2=sin(A + B).

Let $C$ be the third angle. Then $\sin C = \sin (A+B)$

if $C=90^{o}$, then it's obvious:
$\sin C=1$ and $\sin^{2}A+\sin^{2}B = \sin^{2}A+\cos^{2}A = 1$

Inversely, we have:

$\sin^{2}{A}+\sin^{2}{B}= \sin (A+B) \Rightarrow$
$\sin^{2}{A}+\sin^{2}{B}= \sin A \cos B+\sin B \cos A \Rightarrow$
$\sin^{2}{A}-\sin A \cos B+\sin^{2}{B}-\sin B \cos A = 0 \Rightarrow$

$\boxed{\sin A (\sin A-\cos B)+\sin B (\sin B-\cos A) = 0}\ \ (1)$

Of course $\sin A, \sin B>0$.
So there are two cases for the coefficients $(\sin A-\cos B),(\sin B-\cos A):$
[b][u]Either[/u][/b] they are both zero [b][u]or[/u][/b] they have different signs (and appropriate values)

So, let's find their product:

$P = (\sin A-\cos B) (\sin B-\cos A) =$

$\sin A \sin B-\sin A \cos A-\sin B \cos B+\cos A \cos B =$

$(\cos A \cos B+\sin A \sin B)-(\sin A \cos A+\sin B \cos B) =$

$\cos(A-B)-\frac{\sin{2A}+\sin{2B}}{2}=$

$\cos(A-B)-\frac{2\sin{(A+B)}\cdot \cos{(A-B)}}{2}=$

$\cos(A-B)\cdot \left(1-\sin{(A+B)}\right)$

It is $P \ge 0,$ because $0 < A,B < 90^{o}\Rightarrow-90^{o}<A-B<90^{o}\Rightarrow \cos (A-B)>0$

If $P=0 \Rightarrow \sin(A+B)=1 \Rightarrow \sin C=1 \Rightarrow C=90^{o}$

On the other hand, if $P>0$, the only case such that $(1)$ holds is when $(\sin A-\cos B)=(\sin B-\cos A)=0$.
So we have again $C=90^{o}$


Let $ ABC$ be a triangle such that $ AB \ne CB$. Let $ D$ be the intersection of the bisector of the $ \angle B$ with the the bisector of $ \overline{AC}$. Prove that $ D$ can not be in the triangle.

They intersect on the circumcircle. Q.E.D.


function
parameterization
probability
expected value
real analysis
real analysis unsolved

I wanted to use matlab solve the ODE:y''=-(1/m)*[m*g+(Cy)*y'*(/y'/)],y(0)=0,y'(0)=Vo*sinA,Vo>0,0<A<90 degrees
I wrote the codes as
     >> sol =dsolve('Dy1=y2','Dy2=-(1/m)*(m*g+(Cy)*Dy1*abs(Dy1))','y1(0)=0','y2(0)=Vo*sinA','t')

(m,g,Cy,Vo,A are all known constants.)

It appeared to say:
"??? Error using ==> dsolve
Error, (in dsolve/IC) The 'implicit' option is not available when giving Initial Conditions."

>>

Anyone can tell me how to correct my codes? :(

[quote="Semigroups"]

     >> sol =dsolve('Dy1=y2','Dy2=-(1/m)*(m*g+(Cy)*Dy1*abs(Dy1))','y1(0)=0','y2(0)=Vo*[b]sin(A)[/b]','t')

(m,g,Cy,Vo,A are all known constants.)
[/quote]

It would be nice to define a symbolic variable t, using

syms t

Thanks for the suggestion, but I think the problem is that what does matlab mean by saying that sentence?

This is just a guess...  matlab has problems when you use symbolic variables ( i don't know how to say it in English), example;

( this is a mistake i often make ), say you want to define a function, that has certain parameter

f=inline('... (x-mu)^2...','x'),

where mu is a given parameter ( expected value in this case),

even if you have defined it previously  to be

mu=0;

matlab will completely screw it up and you will get something that doesn't make any sense, so
I point out, to figure out what's wrong in your case, it's best to use help.

Off topic: I often like to say; even if you don't what you have to do, Matlab will help you do it! :lol:


a machine for making pizza puts

salami on every 18th pizza
pinapple on every 21st pizza
peppers on every 20th pizza

how many pizzas have toppings on them?
how many pizzas have pepper and salami but not pinapple on them?

how many pizzas does your machine make? :)
if you don't mention this, the answer is $+\infty$ to both questions...

[quote="ZetaX"]I think, he asks for the probability of that things ;)[/quote]
even for the probability, there must be a certain number of Pizza's made...
otherwise it's undefined $(\frac{\infty}{\infty})$ :huuh:

I don't think infinite probabilities are undefined...:
probability = $\lim_{n \to \infty} P(n)/n$ where $P(n)$ is the expected number of pizzas of the desired type made when you make $n$ pizza's.
But it's identically the same as the probability of making such a pizza when making one pizza ;)