geometry
geometric transformation
reflection
3D geometry

In this game, I'll say a word, and another word that relates to it, and you must give me, in as many steps as you require, a series of links to that word. Then you may give a third word, and the person after us must then relate the second and third words, and so on.

Rules:
1) Relations must be positive. You can't say "Bears are not sandals" to relate bears to sandals. That's cheating.
2) Relations must be factual. You can't say "Humans can eat radios" to relate humans to radios. That's also cheating.

The first word is: COTTON

The second word is: WINDOWS VISTA

Good luck trying to get this in six steps or less.

COTTON is worn by humans
Humans made WINDOWS VISTA

Second word: WINDOWS VISTA
Third Word: SEATTLE

WINDOWS VISTA has the word windows in it.
There are windows in SEATTLE.

Third Word: SEATTLE
Fourth Word: THIOL OXIDASE

EDIT: 42nd Post :D

In SEATTLE they emit chemicals.
An example of a chemical is THIOL OXIDASE.

Fourth word: THIOL OXIDASE
Fifth word: TELEMETRY

[b]Thiol Oxidase[/b] is a key ingredient in the manufacturing of the [b]Telemeter[/b], which is used for [b]Telemetry.[/b]
 :?:
5th [b]Telemetry[/b]
6th [b]Sun[/b]

Wikipedia says: A telemeter is a device used to transmit measurements of a quantity to a remote location. Telemeters are generally the physical devices used in telemetry and may consist either of analog or digital transmitters.

Thiol Oxidase is an enzyme... lol.

Telemetry is used in communcations satellites, and we have sent satellites to observe the sun.

6th = Sun
7th = Linux

The sun shines on the earth
Linux is on Earth

7th=Linux
8th=Calculator

Linux is an OS, several calculators use OSes.

8th = Calculator
9th = Trachea

[quote="bluecarneal"][b]Calculators[/b] are made by humans, humans have [b]tracheas[/b](what is the plural of trachea???)

9th Trachea
10th uranium-235[/quote]

Uranium-235 was in the Hiroshima bomb that affected the tracheas, among other body parts, of many Japanese.

10th: uranium-235
11th: a bird

Oh, another note about the "factual relation" rule, it cannot be a historical fact, and it must be specific and notable enough that it does not apply to something else that is completely unrelated. (So, for example, "Linux is on Earth" is not valid, but "Linux is an OS" is.) Sorry... :(

Don't worry, above posters, you don't need to change your wordings. But just so that you feel better:

Uranium-235 is an isotope of uranium, which is a radioactive element that decays into lead. Lead, among other things, is hazardous to the respiratory system, which involves the trachea.

The uranium 235 got into animals,
mainly their tracheas,
and killed many birds.

11th- a bird
12th- square root

A flying bird looks like a square root sign.

12th - square root
13th - facsimile

To make facsimiles, you can use a copy machine. Copy machines kind of look like square root signs, the same way that flying birds (somehow) look like square root signs.

13 - facsimile
14 - tap dancing

tap dancers look alike, so they are facsimiles.

14 - tap dancing
15 - ethernet

Should be easy

Tap dancing can be quite fun, and so can a computer, which uses Ethernet.

15- ethernet
16- gold

ethernet uses gold wires.

16:gold
17:bomb

Bomb's are a color, like black. Black is a color. like gold.

Gold and Richard Ruscyzk

No, bomb and Richard Ruscyzk.

Building bombs involves using math to maximize blast radius, therefore relating to Richard Ruscyzk.

18:Richard Ruscyzk
19:Whirlpool

Richard Ruscyzk lives in a house.
Some houses have a hot tub which may be made by whirlpool.

19: Whirlpool
20: Trikadekaphobia

Triskaidekaphobia - did you mean this word, just misspelled, I believe.

WHIRLPOOLS can be dangerous, so some people are scared of them.
Sometimes people believe the number thirteen is similarly dangerous, so they have TRISKAIDEKAPHOBIA.

20: Triskaidekaphobia
21: laughter

Triskaidekaphobia can appear absurd to some (unsympathetic) people. Absurd things can cause laughter.

21: laughter
22: twenty-two

LAUGHTER is contagious, many times.
This can cause TWENTY-TWO people to laugh.

22. twenty-two
23. movies

Some MOVIES feature SHINING TURTLES.

24. shining turtles
25. Wii

[quote="mz94"]TWENTY TWO is one less than 21
21 is a title of a MOVIE(S)

23. movies
24. shining turtles[/quote]

just to point out, 22 is one less than 21???

SHINING TURTLES are shiny (duh), which implies reflected light.

The WII also reflects light

Sorry, couldn't think of a better one. Are there Wii games with shining turtles? I don't know.

25. Wii
26. termite

WIIs play games. Some games involve termites.

26. termite
27. nanotube

[quote="james4l"]evil: yes - I was hoping someone would point out the Mario - shining Koopa correlation.

TERMITES build tunnels. Tunnels look like NANOTUBES.

27. nanotube
28. diamond

I will shoot you if you don't get a direct link.[/quote]

...they're both made of carbon.

28. diamond
29. Gateway Computers

("some games feature termites" is not a good enough answer. Perhaps name a game that featured termites...)
("Rusczyk lives in a house" is not good either... anyone could live in a house. It's too general a statement. These aren't meant to be done in one or two steps, you know...)

putting 3 diamonds together forms a cube-like representation

gateway computers have the gateway logo, which is a cube

Gateway Computers
St. Louis

Gateway computers share a word in their name with the Gateway Arch, which is in St. Louis.

30. St. Louis
31. New Orleans

They are both U.S. cities.

31. St. Louis
32. Infested Terran (suggested by pythag011)

ST LOUIS is a home to mankind
Some of mankind would like to become INFESTED TERRANS
I think it's supposed to be New Orleans, but... somehow it ended up as St. Louis....... I just copy the post before me
Sorry I don't know where that came from xD

32. INFESTED TERRANS
33. OCTOPUS

I am not an OCTOPUS and I am not and INFESTED TERRAN.

OCTOPUS

COMPUTER

Octopus is an animal.
Computers have a virus called a computer worm, which is also an animal.

35. COMPUTER

36. POPCORN

3333, I notice that you revived Connect-a-word. Please don't do that. Create a new revival topic (in the FF, seeing that this is a marathon) and vote instead :P

[quote="3333"]Why is it better to create a new topic?[/quote]

Primarily because cluttering the Games forum isn't good.


Many times I'm looking for a thread, but I can't find it even though i know some keywords.  I know on some forums you can search for a term and posts that have the term come up.  Is there a similar function on Aops? If so, how do  i use it?

Use the .... [url=http://www.mathlinks.ro/search.php]search function[/url]?!  :huh:

Tip: To maximize efficency in using the search, don't search for EXACT phrases. Also, don't be afraid to play around the restrictions (author, time, etc.) a little.

Oh thanks guys :D

Just noticed the button next to memberlist now  :blush:


Exactly one ordered pair of positive integers (x,y) satisfies the equation 37x+73y=2016. What is the sum of x+y?

======================
Use Euclidean algorithm.
73= 37+36
37=36+1

73= 37+(37-1)
2×37-73=1
multiply both side by 2016.
2016(2×37-73)=2016
4032×37-2016×73=2016

(x,y)=(4032, -2016) : a (special) solution

Common solution : x= 4032 +73n , y=-2016-37n , n : integer

If n=-55 then x=17, y=19.

∴ x+y=36

you could also use mod 36 to get straight to the answer :wink:

what ever x is, 37x will leave a remainder of x when divided by 36. Likewise, whatever y is, 73y will leave a remainder of y when divided by 36. Since 2016 is divisible by 36, the remainders must add to 36.


Divison Series:
Yanks over Twins in 3
Boston over  LAA in 5

Phillies over Rox in 3
Mannywood over Cards in 4

Championships:
Yanks over Boston in 5
LAD over Phillies in 7

WS!
Yankees over LAD in 4

[quote="Jongy"]It's that time of year again.  Pick your Division Series, Champion Series, and World Series winners!
ALDS:
Yankees over Minnesota 3-0
Boston over LA Angels 3-2
NLDS:
Phillies over Rockies 3-1
Cardinals over Dodgers 3-1
ALCS:
Yankees over Boston 4-2
Cardinals over Phillies 4-2
WORLD SERIES:
Yankees over Cardinals 4-3[/quote]

Kinda hard now that Dodgers are winning 2-0... but then again, I had the Cardinals winning the World Series...


group theory
abstract algebra
linear algebra
matrix
induction
superior algebra
superior algebra unsolved

Let $ G$ be the multiplicative group generated by the real matrices $ a \equal{} \left(\begin{array}{cc}  2 & 0 \\ 0 & 1 \end{array}\right)$ and $ b \equal{} \left(\begin{array}{cc}  1 & 1 \\ 0 & 1 \end{array}\right)$. If $ H$ is the set of all matrices in G whose (main) diagonal entries are 1, then $ H$ is a subgroup that is not finitely generated.

Attempted Solution:

First, how do you prove that $ H$ is closed? The set of all matrices whose (main) diagonal entries are 1 is not closed. Second, how do prove that the existence of a finite set of generators is impossible? Contradiction?

in the subgroup of $ GL(2,R)$ consisting of upper diagonal matrices we may consider the subgroup of those matrices whose diagonal entries are 1. thus $ H$ is a group. I suggest to determine $ H$ explicitely e.g. using the determinant.

Okay, I can prove that $ H$ contains $ \left\{\left(\begin{array}{cc}1 & \frac{m}{2^n} \\ 0 & 1 \end{array} \right) | m \in \mathbb{Z}, n \in \mathbf{N} \right\}$. Furthermore, the structure of $ H$ admits a monomorphism to Q given by $ \phi\left[ \left(\begin{array}{cc}1 & q \\ 0 & 1 \end{array} \right) \right]\equal{} q$. So ISTS that no subgroup of Q that contains the elements $ \frac{m}{2^n}$ has a finite generating set. Is there a simple way to do that?

in fact, we have $ a^{ \minus{} n} b^m a^n \equal{} \begin{pmatrix} 1 & m/2^n \\
0 & 1 \end{pmatrix}$. now an easy induction on the length of a word in $ < a,b >$ shows that $ H$ consists exactly of those matrices. thus $ H$ is isomorphic to $ \mathbb{Z}[1/2] \leq \mathbb{Q}$. every element of $ \mathbb{Z}[1/2]/\mathbb{Z}$ is annihiliated by some power of $ 2$. so if it was finitely-generated, some power would annihilate the whole group, which is not the case.

In other words, a finitely generated subgroup of $ H$ has a maximum value of $ n$, but $ H$ does not.


MATHCOUNTS
quadratics
function
geometry
algebra
quadratic formula

I have a Ti-83-plus and am trying to get calculator programs to use during mathcounts target round.

I already have the quadratic formula, can anyone help me??? I can't download anything but I could manually enter the program with access to the source code.

Although most calculator formulas just distract you during a test (I have a silver edition, I use a Bank of America calculator on MathCounts). Here are a few:

http://www.ticalc.org/pub/83plus/basic/math/

[quote="bottlecap"]I already have the quadratic formula, can anyone help me??? I can't download anything but I could manually enter the program with access to the source code.[/quote]

Learn BASIC programming and program your own.

The only program that would be useful other than quadratics is n choose r programs, and those are native functions of the calculator./

Some native functions are hard to get to (MATH - < - < - Down down...?) so I programmed the ones I would need ($ \sqrt [x]{y}$) so I could save a couple seconds.  :wink:

speaking of loading apps is there a tutorial for helping me load apps on? I have a ti-84 Silver edition.

Make Cyclic quads and heron's area formula programs.


calculus
derivative
function
calculus computations

Prove that if all the second partial derivatives of $ f$ are continuous at a point $ (a,b)$ then $ f_x, f_y$ and $ f$ are all continuous at $ (a, b)$.

Do you know how to prove that a function is continuous if its derivative is continuous?

Nope ):
Prove? No idea, thought it was assumed since..well its differentiable there. Which poses the problem on how to actually prove my given question.


How many triangles with integer-length sides exist with the property that their areas and perimeters are numerically equal?

[quote="probability1.01"]Well our good friend Heron says that we must satisfy

[tex] \frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4} = a+b+c [/tex]

[tex] \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} = 4a+4b+4c [/tex]

[tex] (a+b+c)(-a+b+c)(a-b+c)(a+b-c) = 16(a+b+c)^2 [/tex]

[tex] (-a+b+c)(a-b+c)(a+b-c) = 16(a+b+c) [/tex]

Or maybe we find all the triangles in which a circle of radius 2 can be inscribed.  I don't know how to find the integer-sided ones though.

Stuck after that. :-([/quote]

You're on the right track.  Try making the substitution:

x = s-a, y = s-b, z = s-c and see what that simplifies to.  In particular, what does x+y+z equal?


uhm.. i was looking at $x^n+y^n=z^n,n>2$
now...$z>x,y$ no ?

$x^n+y^n=(x+(z-x))^n=$ $C_{n}^{0}(z-x)^n +C_{n}^{1}x(z-x)^{n-1}+...+C_{n}^{n}x^n$
we cancell out $x^n$ so we now have
$y^n=\sum_{k=0}^{n-1} C_{n}^{k} x^k (z-x)^{n-k}$
but in the RHS we have some term something like this $(z-x)^n$
but this contains $z^n$.now we have some other stuff in RHS
so it is $z^n+stuff>y^n$(as we supposed to be true in the first part)
so we have that it is not possible

now i'm sure this is not right...i just dont know where i'm mistaking
please help me find the error...because it is impossible that this should be the proof
also it seems weird to me that i didnt even considered real numbers and it worked...
wich is too weird

Come on, you know that $ \left(z \minus{} x\right)^n$ is not greater than $ z^n$... or what do you mean by "contains $ z^n$"?

[b]EDIT:[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=45697 .

  darij

perhaps the "stuff" could be negative.  or, wait, $z^n$ has to be greater than $y^n$, right?  so what's wrong with $z^n+stuff$ being greater than $y^n$?

(yeah, you know something is wrong with your proof because of course there are irrational solutions that should turn up...)

thank you for pointing up those negative terms...they make the difference
i understand

Can anyone give me a solution for Fermat last theorem?

Hey,that's really a difficult problem,and the mathematicians used 358 years to solute it.
The proof is very very long and difficult.So maybe no one can post it here.

The proof is available on the net; googling for it could help.


There was once a cool problem involving factorials and it asked how many zeroes the number ended with... but the trick was that if you counted fives (which is the wont), you got that it ended with 2 zeroes, but you had to count twos, and you would see that it actually ended with 0 zeroes.

Anyone remember the exact numbers, and preferably, what test this is from?

Thanks

but once you get to 2!, you have a factor of 2

jorian

There have been a few of these.  Try searching for "how many zeros".  Here are a couple of them:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84288[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134451[/url]

There is another recent one in the marathon on this message board.

The problem is like:

How many zeroes does \[\frac{25!}{8!*8!*8!}\]
end with?

[quote="mathcrazed"][hide]ok well, you find the number of 0's on top subtract the number of 0's on bottom.  that gives you 6- 3= 3.[/hide][/quote]

 :rotfl: remember, we argued aobut this type of problem in class??

Yes the answer is

[hide]
1
[/hide]

I can post a quick solution if you don't get it.

ya mathcrazed thought you could subtract zeros..but you cant.. because 40/8=5, which has 0 zeros, but 1-0=1, so if you just subtracted top zeros from bottom you would bet 1 not 0..your trying to subtract powers of 5 and 2 (in this case 2s-not 5s)


calculus derivative symmetry function algebra domain search

One approach could be using Lagrange multipliers, because the system you'll get taking derivatives is a system of biquadratic equations, which is quickly solvable, and it shows obvious symmetry, which makes the process of looking for $ a,b,c,\lambda$ simple. Maybe I've missed something in my calculations, but you get something like this:

$ a \equal{} b \equal{} c \equal{} \frac {1}{3},\lambda \equal{} \minus{} 25/9$, which I got from the only useful solution of biquadratic equation

$ \minus{} 45a^4 \plus{} 30a^2 \plus{} \lambda \equal{} 0$, which is $ a \equal{} \sqrt {\frac {5 \minus{} \sqrt {5}\sqrt {\lambda \plus{} 5}}{15}}$. Now after plugging in we get the minimum of the function $ 10(a^3 \plus{} b^3 \plus{} c^3) \minus{} 9(a^5 \plus{} b^5 \plus{} c^5)$ which is $ 1$.

(I've skipped the seek for extrema on the edges of domain, which in this case reduces to $ a\to 1$, $ b,c \to 0$ which gives the same result as above. I know that the solution lacks rigor, but I don't have much time ;) )

This problem was already discussed here:
http://www.mathlinks.ro/viewtopic.php?search_id=1722134015&t=161399
See last post in the topic for my proof.

And why did you thanhak5 steal [b]Myth[/b]'s avatar picture?

For me, i'd like to use this classic way

I claim for every real $ x \in (0,1)$,

$ 10x^5\minus{}9x^3 \ge \frac{25}{9}x\minus{}\frac{16}{27}$

So, $ \sum 10a^5\minus{}9a^3 \ge \sum \frac{25}{9}a\minus{}\frac{16}{27}\equal{}\frac{75\minus{}48}{27}\equal{}1$

 :P

Ronald Widjojo, that inequality is not true, it is reversed when x is greater than about .234

[quote="thanhak5"]For $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$
      Prove that   $ 10(a^3 \plus{} b^3 \plus{} c^3) \minus{} 9(a^5 \plus{} b^5 \plus{} c^5) \geq 1$[/quote]

Hello [b]Ronald Widjojo[/b]!

For all $ a,b,c$ (even if $ a\plus{}b\plus{}c\ne1$) the following identity can be easily proved by removal of parentheses:

\[ {10 \left( {a}^{3}\plus{}{b}^{3}\plus{}{c}^{3} \right)  \left( a\plus{}b\plus{}c \right) ^{2}
\minus{}9\left({a}^{5}\plus{}{b}^{5}\plus{}{c}^{5}\right)\equal{} \left( a\plus{}b\plus{}c \right) ^{5}\plus{}\frac{15}{2}
 \left( a\plus{}b \right)  \left( a\plus{}c \right)  \left( b\plus{}c \right) \left(  \left( a\minus{}b \right) ^{
2}\plus{} \left( a\minus{}c \right) ^{2}\plus{} \left( b\minus{}c \right) ^{2} \right)}\]

It clearly proves, that if $ a\plus{}b\plus{}c\equal{}1$, then

\[ {10 \left( {a}^{3}\plus{}{b}^{3}\plus{}{c}^{3} \right)
\minus{}9\left({a}^{5}\plus{}{b}^{5}\plus{}{c}^{5}\right) \geq 1}\]

Thank you for the nice inequality!

Sergey Markelov


linear algebra
matrix
linear algebra unsolved

Hi

I was wondering if there's a quick way to see this fact :

if H is a Hilbert  spacevector and if T:H->H is a normal mapping (that means $ TT^\star \equal{} T^\star T$), then the spectrum of the mapping $ T T^\star$ is exactly the set

$ \{ |\lambda|^2 , \lambda \in sp (T) \}$

Thank you

Case 1: assume $ H$ is finite dimensional.

By Schur's lemma, we may assume WLOG that $ T\equal{}\begin{bmatrix}\lambda_1&*&\dots&*\\ 0&\lambda_2&\dots&*\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&\lambda_n\end{bmatrix}$ is upper triangular.

Then by direct calculation,

$ TT^*\equal{}\begin{bmatrix}|\lambda_1|^2&*&\dots&*\\ 0&|\lambda_2|^2&\dots&*\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&|\lambda_n|^2\end{bmatrix}$

By the hypothesis of normality, this also equals $ T^*T.$

Each $ \lambda_k$ on the diagonal is a member of the spectrum of $ T,$ so the result is evident.

I'll let someone else post the infinite dimensional case.

well, if $ \lambda \in \sigma (T) \Longrightarrow |\lambda | \leq \parallel{}T\parallel{} \Longrightarrow  |\lambda |^2 \leq \parallel{}T\parallel{}^2\equal{}\parallel{}T^{\star}T\parallel{}$, so $ \sigma (T^{\star}T) \subseteq \{ |\lambda |^2 ,\; \lambda \in \sigma (T)\  \}.$ Suppose that  there exists $ \lambda \in \sigma (T)$ such that $ |\lambda |^2 \not \in \sigma (T^{\star}T)$, therefore $ T\minus{}\lambda I$ is not invertible and $ T^{\star}T\minus{}|\lambda|^2I$ is invertible. Now we know $ T$ is normal then $ T\minus{}\lambda I$ is normal too. In the other hand we have $ TT^{\star}\minus{}|\lambda|^2I\equal{}(T^{\star}\minus{}\overline{\lambda}I)(T\minus{}\lambda I)\plus{}\overline{\lambda}(T\minus{}\lambda I)\plus{}\lambda(T^{\star}\minus{}\overline{\lambda}I)$.

It's not hard to verify that $ T\minus{}\lambda I$ is not invertible if and only if there exists a sequence of $ \{ x_n \}$ s.t. $ \parallel{}x_n\parallel{}\equal{}1$ and $ \parallel{}(T\minus{}\lambda I)x_n\parallel{} \rightarrow 0$ and also from normality we know $ \parallel{}(T\minus{}\lambda I)x\parallel{}\equal{} \parallel{}(T^{\star}\minus{}\overline{\lambda}I)x\parallel{} \;\forall x \in H\ .$

Hence
$ \parallel{}(TT^{\star}\minus{}|\lambda|^2I)x_n\parallel{}\equal{}\parallel{}(T^{\star}\minus{}\overline{\lambda}I)(T\minus{}\lambda I)x_n\plus{}\overline{\lambda}(T\minus{}\lambda I)x_n\plus{}\lambda(T^{\star}\minus{}\overline{\lambda}I)x_n \leq \parallel{}(T^{\star}\minus{}\overline{\lambda}I)\parallel{} \parallel{}(T\minus{}\lambda I)x_n\parallel{}\plus{}|\overline{\lambda}| \parallel{}(T\minus{}\lambda I)x_n\parallel{}\plus{}|\lambda| \parallel{}(T^{\star}\minus{}\overline{\lambda}I)x_n\parallel{} \rightarrow 0.$
Therefore $ TT^{\star}\minus{}|\lambda|^2I$ is not invertible and it's contradiction, so we're done.

Hi Kent,
a priori your proof is false because $ T^*$ is not upper triangular but hopefully it is correct because $ T$ is in fact diagonalizable in an orthonormal basis.
In the infinite dimensional case:
 $ T$ continuous
 $ \lambda\in\sigma(T)$ iff  $ T\minus{}\lambda{I}$ has no continuous inverse.
.Hi ehsan2004,
your proof for:
    $ \sigma (T^{\star}T) \subseteq \{ |\lambda |^2 ,\; \lambda \in \sigma (T)\ \}$ is unclear.

Sorry, I screwed that up. I don't know why I thought $ TT*$ was upper triangular.

Let's try this again. Assume WLOG that $ T$ is upper triangular and write it as $ T\equal{}\begin{pmatrix}\lambda_1&a^*\\0&T'\end{pmatrix}$

Compute $ TT^*$ and $ T^*T$ $ T^*T$ has an upper left corner entry of $ |\lambda_1|^2,$ while $ TT^*$ has an upper left entry of $ |\lambda_1|^2\plus{}a^*a.$

Since $ TT^*\equal{}T^*T,$ we have that $ a^*a\equal{}0.$ But $ a^*a\equal{}\|a\|^2,$ so we concluded that $ a$ is the zero vector. We then get that the remaining square piece, $ T',$ is itself normal.

What this shows is that an upper triangular normal matrix is in fact diagonal. Since, by Schur's Lemma, every matrix is similar to an upper triangular matrix, we conclude that every normal matrix can be diagonalized. Which proves the theorem.

[quote="loup blanc"]
Hi ehsan2004,
your proof for:
    $ \sigma (T^{\star}T) \subseteq \{ |\lambda |^2 ,\; \lambda \in \sigma (T)\ \}$ is unclear.[/quote]

yeah, I got it, it should be  $ \sigma (T^{\star}T) \subseteq \{ |\lambda |^2 ,\; |\lambda|\leq \parallel{}T\parallel{} \ \}.$ But now I don't know how this can be shown $ \sigma (T^{\star}T) \subseteq \{ |\lambda |^2 ,\; \lambda \in \sigma (T)\ \}.$  :maybe:


The drawing shows a person who, starting from rest at the top of a cliff, swings down at the end of a rope, releases it, and falls into the water below. There are two paths by which the person can enter the water. Suppose he enters the water at a speed of 13.0 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 5.20 m above the water? Ignore the effects of air resistance.

Could someone please help me get started with this problem?  Thanks!

Here's the way I approach it: I know the speed he reaches at the bottom, so I know his maximum speed.  I'm assuming this is at no height.  I then go in reverse, and figure out his speed on the upward return, to a height of 5.20 meters.

Yeah, using conservation of energy should work for this problem.


calculus
integration
function
real analysis
real analysis unsolved

Find all measurable/ continuous/ differentiable functions $f: [0,1]\to [0,1]$ such that for any nonnegative measurable function $g$ we have $\int_{0}^{1}{g(x)dx}=\int_{0}^{1}{g(f(x))dx}$ (the Lebesgue measure is considered here).

It suffices to consider $g=\chi_{E}$ for measurable sets $E\subset [0,1]$. The property becomes $|E|=|f^{-1}(E)|$, that is, $f$ is measure-preserving. There is a lot to be said about measure-preserving transformations, but you can't expect to classify them all.

What about the differentiable ones? I suspect only $x$ and $1-x$ work.

If you also require $f$ to be invertible, then yes. Otherwise there are many more, although it's not very easy to come up with an explicit example. (I'll try to.)

I was wrong, as usual. The measure-preserving property implies $|f(E)|\ge |E|$ for any measurable $E\subset [0,1]$. If $f$ is also  differentiable, then $|f'(x)|\ge 1$ everywhere. This implies that $f'(x)$ has constant sign, hence $f$ is invertible, hence $|f'(x)|=1$ for all $x$.


integration
calculus
real analysis
real analysis unsolved

let $ p>1$ and $ \phi(x)\equal{}x^{\minus{}\frac{1}{p}}\chi_{(0,1]}$
and $ (r_{n})$ be a countable dense subset of $ \mathbb{R}$, put $ F(x)\equal{}\sum_{n\geq 1}2^{\minus{}n}\phi(x\minus{}r_{n})$
a) prove that $ F\in L^{1}(\mathbb{R})$
b) Prove that $ F^{p}$ is not integrable on any non-empty interval $ [a, b]$ of $ \mathbb{ R}$.

Any idea will be  highly appreciated.

The key points: $ \int\phi \equal{} \frac{p}{p\minus{}1}$ is finite, whie $ \int\phi^p\equal{}\infty.$ More specifically, for any $ \delta>0,$ $ \int_0^{\delta}\phi^p\equal{}\infty.$

To answer a), interchange sum and integral. (The fact that the integral is finite is then perhaps the only way to prove that the sum converges to a finite value for almost every $ x.$)

To answer b), note that the sum is greater than any single term of the sum, and in any interval, you can find at least one rational number.

thanks Prof. Kent (again  :) )
that was I though but I was not sure, thanks for your clarification


Ehh...i thought the $ \frac{1} {p(a_k)}$ was in the summation, so i ended up with something like $ \frac{k} {p(a_k)}\equal{}\frac{1} {a_k}\plus{}\frac{k\plus{}1} {p(a_{k\plus{}1})}$

anyone else do that?

Except it doesn't have a dummy variable in it, and convention says $ \sum a \plus{} b \equal{} b\plus{}\sum a$.

Hmm...so thats why i couldn't get it... :(

[quote="archimedes1"][quote="MathAndKnowledge"][quote="pianoforte"]Except it doesn't have a dummy variable in it, and convention says $ \sum a \plus{} b \equal{} b \plus{} \sum a$.[/quote]
I don't know what sort of convention has this.  If you really think you're right, then you'll have a hard time persuading the SAT people if you take the SAT Math Level 2 Subject Test.  I know that SAT math is extremely easy (I got a perfect on the aforementioned test), so it'd really stink if you followed your convention and fell short of an 800.
For any questions on the SAT, $ \sum a \plus{} b \equal{} \sum(a \plus{} b)$.  So abacadaea has a point; the notation used for the USAMTS problem was extremely unclear and ambiguous.  Of course, I didn't submit solutions for the last three problems on Round 2 (school work took up my time), so I don't have a real problem with this.[/quote]
I think the convention pianoforte was referring to applies only if $ b$ has nothing to do with the dummy variable.  In general, people won't be summing the same thing over if nothing is changing.[/quote]

Yeah, but there was an actual SAT problem in which $ \sum\limits_{k \equal{} 1}^5 3k \plus{} 2 \equal{} \sum\limits_{k \equal{} 1}^5 (3k \plus{} 2)$, even though $ 2$ has nothing to do with the dummy variable.

[quote="MathAndKnowledge"]
I don't know what sort of convention has this.  If you really think you're right, then you'll have a hard time persuading the SAT people if you take the SAT Math Level 2 Subject Test.  I know that SAT math is extremely easy (I got a perfect on the aforementioned test), so it'd really stink if you followed your convention and fell short of an 800.[/quote]

First of all, you can get plenty of questions wrong on subject tests and still get 800s (especially physics), but that's another issue.

Second, where specifically did you get that summation problem you cited? I highly doubt that's an SAT question. It's convention that you don't assume parentheses are where they aren't. For example, $ \sin 3\theta \plus{} 2 \neq \sin (3\theta \plus{} 2)$. This comes up all the time on the AP Calculus test, and you won't get your way with arguing the notation, as you said. I don't see why the SAT would be any different with summation.

[quote="MathAndKnowledge"][quote="pianoforte"]Except it doesn't have a dummy variable in it, and convention says $ \sum a \plus{} b \equal{} b \plus{} \sum a$.[/quote]
Yeah, but there was an actual SAT problem in which $ \sum\limits_{k \equal{} 1}^5 3k \plus{} 2 \equal{} \sum\limits_{k \equal{} 1}^5 (3k \plus{} 2)$, even though $ 2$ has nothing to do with the dummy variable.[/quote]
Just because the SAT misused the notation does not mean that we will.

I had checked over the problems beforehand for ambiguities, but I am so accustomed to the conventions of summation notation that I did not realize that some people might misinterpret it. If I had noticed the ambiguity, I would have suggested rewriting the equation as
$ \sum\limits_{j\equal{}0}^{n\minus{}1}\frac{1}{a_j} \equal{} 1 \minus{} \frac{1}{p(a_n)}$.

I will try to remember that ambiguity with summation notation while I evaluate future problems.

Erin Schram
USAMTS grader and problem evaluator


given a rectangle ABCD, with E on BC and F on AB, area of triangles AFD, FBE, and ECD are all equal to 1. what is the area of triangle FDE?

and no other information is given :(

See the attached diagram for additional notation.

$ [AFD] \equal{} 1\implies b(a \minus{} x) \equal{} 2\implies x \equal{} a \minus{} {2\over b}$

$ [CDE] \equal{} 1\implies a(b \minus{} y) \equal{} 2\implies y \equal{} b \minus{} {2\over a}$

$ [BEF] \equal{} 1\implies xy \equal{} 2\implies \left(a \minus{} {2\over b}\right)\left(b \minus{} {2\over a}\right) \equal{} 2$

The last equation expands to

$ ab \minus{} 2 \minus{} 2 \plus{} {4\over ab} \equal{} 2$

Putting $ S \equal{} ab$, we get

$ S \minus{} 6 \plus{} {4\over S} \equal{} 0\iff S^2 \minus{} 6S \plus{} 4 \equal{} 0$

The solutions of the last quadratic are

$ S_{1,2} \equal{} \frac {6\pm\sqrt {36 \minus{} 16}}{2} \equal{} \frac {6\pm 2\sqrt {5}}{2} \equal{} 3\pm\sqrt {5}$

However, the area of the rectangle can't be smaller than $ 3$, otherwise the triangles couldn't fit in. Therefore $ S \equal{} 3 \plus{} \sqrt {5}$, which finally gives $ [DEF] \equal{} \sqrt {5}$


function
linear algebra
linear algebra unsolved

Let $ \emph{W}$ be the subspace of $ \mathbb{R}^4$ spanned by (1, 0, -1, 2) and (2, 3, 1, 1).
Which linear functions $ \emph{l(x)}$ = $ c_1x_1\plus{}c_2x_2\plus{}c_3x_3\plus{}c_4x_4$ are in the annihilator of $ \emph{W}$?

$ l\in Ann(W)\Longleftrightarrow (c_1,\ldots,c_4)\in W^{\bot}\equal{} <(1,0,\minus{}1,\minus{}1),(0,1,\minus{}2,\minus{}1)>$


inequalities
pigeonhole principle
inequalities unsolved

Mr. Chen ,would you tell me how to do it? thanks!

It's ugly.
But [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=119231]the following inequality[/url] is true and nice:
Let $a,b,c\geq 1$ such that $\frac 1{a^{2}+47}+\frac 1{b^{2}+47}+\frac 1{c^{2}+47}=\frac 1{24}$.
Prove that $a+b+c\geq 15$.

We prove that $a,b,c \geq 1$ and $\sum \frac{1}{a^{2}+47}= 1/24$ implies $a+b+c \geq 15.$  (All the sums will be cyclic over the appropriate variables.)

First let $x=24/(a^{2}+47),$ $y=24/(b^{2}+47),$, and $z=24/(c^{2}+47),$ and observe that $0<x,y,z \leq \frac{1}{2}.$  Substituting into the original relations, we must prove that $x+y+z=1$ implies that $\sum \sqrt{\frac{24}{x}-47}\geq 15.$

Let $f(x)= \sqrt{\frac{24}{x}-47}$ for $0<x \leq \frac{1}{2}.$  We have

\[f''(x)=\frac{24(18-47x) \sqrt{\frac{24}x-47}}{x^{2}(24-47x)^{2}},\]

and therefore $f$ is concave up for $0<x<18/47$ and concave down for $18/47<x<1/2.$  Our proof proceeds by considering the tangent lines and secant lines to the graph of $f.$

We consider three cases.

Case 1:  Suppose that $x,y,z \leq 25/54.$  We have $\sqrt{\frac{24}x-47}\geq \frac{61-108x}5.$  To establish this we square both sides (remembering that $x \leq 25/54 < 61/108$ so all the terms are positive) and rearrange in the equivalent form
\[(1-3x)^{2}(25-54x) \geq 0,\]
which is clearly true. Equality occurs only when each variable equals $1/3$ or $25/54,$, but their sum can be $1$ only if each equals $1/3.$
(I found this inequality because a section of a graph which is concave up lies above its tangent lines, and the line $y=\frac{61-108x}5$ is the line tangent to the graph of $f$ at $x=1/3.$)  Thus in this case we have
\[\sum f(x) \geq \sum \frac{61-108x}{5}= 3\frac{61}5-\frac{108}5 \sum x =\frac{183}5-\frac{108}5=15.\]

Equality occurs when $x=y=z=1/3,$ which makes $a=b=c=5.$

Case2: Exactly one of $x,y,z$ is greater than $25/54.$  Assume WLOG that $25/54<x \leq 1/2$ and that $0<y,z \leq 25/54.$  Consider the secant line of $f$ which connects the points $(25/54, f(25/54)$ and $(1/2,f(1/2).$  Its equation is $y=\frac{86-162x}5.$  Since $f$ is concave down on this interval (see above about $f''$) , the graph lies above this secant line and we have $\sqrt{24/x-47}\geq \frac{86-162x}5.$  Equality occurs where the secant line intersects the graph--when $x=25/54$ or $x=1/2.$

Now we consider the line tangent to $f$ at $x=1/4.$  Its equation is $y=\frac{97-162x}7,$ and we have $\sqrt{24/x-47}\geq \frac{97-162x}7.$  To see this note that $y,z \leq 25/54 < 97/162,$ so we square both sides and rearrange the inequality, getting the equivalent $(1-4x)^{2}(49-96x) \geq 0,$ which is true for $x \leq 49/96 <1/2.$  Equality occurs here when $y$ and $z$ equal $49/96$ or $1/4.$

Now we combine the two inequalities—one for $x$ and the other for $y$ and $z$ to write

\[\sum \sqrt{24/x-47}\geq \frac{86-162x}5+\frac{97-192y}7+\frac{97-192z}7 \newline =\frac{6(262-189x-160y-160z)}{35}\newline =\frac{6(262-160(x+y+z)-29x}{35}\newline =\frac{6(102-29x)}{35}\newline \geq \frac{6(102-\frac{29}{2}}{35}\newline =15.\]

Since $x+y+z=1,$ we see from the conditions above that equality occurs when $\{x,y,z\}=\{1/2,1/4,1/4\},$ which is $\{a,b,c\}=\{1,7,7\}.$

Case 3: Exactly two of the variables exceed $25/54.$  (All three can’t be larger than $25/54$ since their sum is $1.$)  WLOG assume that $25/54 < x,y \leq 1/2.$  Then we must have $z=1-x-y <4/54.$  But $f(4/54)=\sqrt{277}>15,$ so in this case we clearly have $f(x)+f(y)+f(z)>15.$  There can be no equality in this last case.

This concludes the proof of the original inequality, and we see that equality holds only when $a,b,c$ are cyclically equal to $5,5,5$ or $1,7,7.$

[hide="this also works (for the nicer one)"]
suppose for the sake of contradiction that $a+b+c<15$. Substitute $x=\frac{15a}{a+b+c}$ etc. so that $x>a\geq 1$ and $x+y+z=15$ Since $\frac{1}{t^{2}+47}$ is clearly decreasing for $t>1$, we have
\[\sum_\text{cyc}\frac{1}{a^{2}+47}>\sum_\text{cyc}\frac{1}{x^{2}+47}\]
Therefore, we only need to show that the right hand side is $\geq \frac{1}{24}$ to arrive at the contradiction.

Let $f(t)=\frac{1}{t^{2}+47}$. If $f(t)$ were concave up in the interval $[1,13]$, we'd be done, but that would be too easy! In fact,
\[f^\prime^\prime (t)=\frac{2(3t^{2}-47)}{(t^{2}+47)^{3}}\]
Setting $\alpha =\frac{\sqrt{141}}{3}$, $f$ is concave up for $t\geq \alpha$ and concave down for $t<\alpha$.

We now show that we can make two of the variables equal or one of them equal to 1 without changing their sum or increasing $\sum_\text{cyc}\frac{1}{x^{2}+47}$. This is true, since by Pigeonhole, either two variables are in the concave up zone or the concave down zone. WLOG let $x\leq y\leq z$, so that either $y$ and $z$ are in the concave up zone, in which case we can apply Jensen to force $y$ and $z$ equal:
\[f(y)+f(z)\geq 2f\left( \frac{y+z}{2}\right) \]
or $x$ and $y$ are in the concave down zone, in which case we can apply Majorization to separate $x$ and $y$ as much as possible. If $x+y\geq 1+\alpha$, this will send $y$ to $\alpha$ and then we can Jensen $y$ and $z$ as in above. Otherwise, $x$ will be sent to 1. (this can be done with
\[f(x)+f(y)\geq f(x+y-\alpha)+f(\alpha) \]
for the first case and
\[f(x)+f(y)\geq f(1)+f(x+y-1) \]
for the second.)

We are now left with a triple of the form $(A,B,1)$ or $(A,A,B)$. In the first case, it remains to prove $f(A)+f(B)\geq \frac{1}{48}$ given $A+B=14$. In fact, for all $t\geq 1$,
\[f(t)\geq \frac{1}{96}-\frac{7}{48\cdot 96}(t-7) \]
which follows from
\[48\cdot 96+7(t-7)(t^{2}+47)-48(t^{2}+47)=(t-7)^{2}(7t+1)\geq 0 \]
This leaves us with the second case, $2f(A)+f(B)\geq \frac{1}{24}$ given $2A+B=15$. This is equivalent to $2f(A)+f(15-2A)\geq \frac{1}{24}$ for $A$ in the interval $[1,7]$. Now we multiply this out:
\[48((15-2A)^{2}+47)+24(A^{2}+47)-(A^{2}+47)((15-2A)^{2}+47) \]

\[=4(7-A)(A-5)^{2}(A+2)\geq 0 \]
A similar method should work for the uglier inequality where $a,b,c\geq 0$.
[/hide]

[quote="behemont"]here is a bit shorter solution...
all sums are cyclic over $a,b,c$:

CSB gives:
$\frac{1}{24}=\sum \frac{1}{a^{2}+47}\geq \frac{9}{\sum a^{2}+3*47}$
that is
$\sum a^{2}\geq \ 75$
and now $QM-AM$ gives
$a+b+c \geq \ 15$
 :D[/quote]
Except that the inequality points the wrong way on the last line, $a+b+c\leq \sqrt{3(a^{2}+b^{2}+c^{2})}$...  :|

It's not so easy because of the (1,7,7) equality case. It's probably doable with Muirhead and Schur (starting with the beginning of my solution and then substituting $p=x-1$ etc., then getting a homogenous inequality in terms of $p,q,r$) but that seems like a LOT of scary computation, and i don't see an obvious way of using Cauchy.

ooops............ :oops:

but isn't equality case also (5,5,5)?


I am really bad at tangents. I missed them on several ma0 practice tests from the past, and it really brought my score down.

16. BA, ED, and BD are tangents to circle O. BD= 8+6rt3 and angle B=60. How far would CE be from the center of Circle O?

17. In the figure tangents AB and CD are parallel and ponts A and C are on circle O. If BD is a tangent and point D is not on circle O, find the measure of angle BOD.

18.The radius of a circle is 5". Tangents drawn from an external point P form an angle of 120 degrees. How far is P from te center of the circle?

11. In a tunnel with a cross section that is semicircular in shape, a vertical 10 ft pole touches the top of the tunnel when the pole's foot is 4 feet from the side of the tunnel. What is the maximum height of the tunnel.

22. Two circles of respective radii 7 and 5 ae externally tangent to each other. How long is their common external tangent?

CLICK TO SEE MORE CLEAR PICTURE  :oops:

[quote="ckck"]Hey now a ranger, for number 22 are the two circles touching? If they are not then how far apart is their centers?[/quote]
The problem says they are externally tangent.

[hide="18"]Draw the radius to the points of tangency and connect it to the exterior point. Two 30-60-90 triangles are formed. Then since the radius of the circle is opposite of the 60 degree angle, we have the distance as $\boxed{\frac{10\sqrt{3}}{3}}$[/hide]

[hide="22"]$\sqrt{(7+5)^{2}-(7-5)^{2}}=2\sqrt{35}$[/hide]

[quote="Xaero"][hide="22"]$\sqrt{(7+5)^{2}-(7-5)^{2}}=2\sqrt{35}$[/hide][/quote]
How did you get that? :ninja:

Since the circles are externally tangent, connecting the radii to the external tangent forms right angles, since the distance from the centers of the circles are 12, we can move the tangent until we form a right triangle, which then by the pythagorean theorem, we have $2\sqrt{35}$.

I immediately thought of that, but it doesn't work. :wink:

[img]http://www.initdotd.com/~nikita/Pictures/Tangent.jpg[/img]

I think you can draw a line from P perpendicular to BR in which the length of the segment is equal to AB. Therefore, you get a right triangle.

Let the intersection of this perpendicular with $BR$ be $F.$ We know $AP=BF=5$ and $FR=2,$ but we don't know the length of $AF$ or $PB$ and therefore can't do anything else.

Why do we need $AF$ or $PB$ if the question is asking for $PF$ or $AB$? Also we can find $AF$ or $PB$ since they are the diagonals of a rectangle.

[quote="Farenhajt"]How about examining $\triangle PFR$? What's $PR$? What's $FR$? What segment is $PF$ equal to?[/quote]
$PR=12,\,FR=2,\,PF=AB$

[quote="Farenhajt"]There you go  :)[/quote]
But $\triangle PFR$ isn't right. What am I supposed to do with it? :huh:

It isn't right?  :huh:

Please confer this:

[quote="Xaero"]...draw a line from P perpendicular to BR...[/quote]

and your own words (!):

[quote="i_like_pie"]Let the intersection of this perpendicular with $BR$ be $F.$[/quote]


calculus geometry AMC USA(J)MO USAMO MATHCOUNTS trigonometry

I was fortunate enough to have a teacher for honors calculus my freshman year who was also in charge of math competitions. Quite a number of stories can be told about him and his room. He's a really great guy. In his room, there are many cabinets of old math competitions and math books covering unconventional topics. I don't think anyone had ever read a lot of those books before I got there. Taking competitions was really good for me because I had never been seriously challenged before in mathematics. So I got lots of math books and studied them until I was able to win math contests. By that time, I had learned a lot of math in general, so I was able to do more than just win competitions and actually do some serious math.

How did you get to the point where you were taking Calculus in your freshman year? What was your middle school math education like, and what have you taken since?

Do you recall any of the books specifically that your teacher had in his cabinet?

I took normal classes in middle school (i.e. prealgebra in 6th grade, algebra I in 7th grade, and geometry in 8th grade). But my middle school teacher let one other student and me go a little faster than the rest of the class because we weren't being challenged. But then we still weren't being very challenged. When I asked him to give me a recommendation to honors calculus for 9th grade, he refused, so I had to take algebra II/trigonometry, which was another one of those really boring classes where I never paid any attention but knew all the material. But I met Mr. Headley, who used to teach an analytic/geometry class until the school district decided that students learn too much in that class and that it couldn't be offered any longer (the wonders of CA public schools...). He saw that I was interested in math (and I guess fairly good at it too), and so he helped me enter his class in the second semester (the calculus part).

He has all sorts of problem books. He has the AoPS books (multiple copies of them), a bunch of AHSME books, USAMO books, Russian problem books, some books by Polya, just about all the New Mathematics Library books, some of the new olypimad booksfrom around the world, some books on geometry problems, and many more. Naturally, he has tons on textbooks too.

I took Algebra 1 in the 6th grade. Then I took these wierd math classes called Integrated 2 and Integrated 3. They're supposed to cover Geometry and Algebra 2 at the same time in a two year sequence, but when you take the class it doesn't make sense. Anyway, I'm taking calculus next year as a freshman.

A lot of people at my school are terrible at math so they can't really introduce any interesting topics. It gets really boring.

i went by the highest curriculum my school could hand, which was algebra I in the 7th (lool). when i transfered schools in the 5th grade i was supposed to be doing pre-algebra, instead i got this weird crap book with some girl with a fro playing chess. anyways. i was extremely bored of algebra for two reasons. the class sucked. ie: we covered that x+4=4+x midyear (once again a resounding lool). and two i already knew the subject. cause in mathcounts you better know it or you might as well give up. so i already knew some of geometry from mathcounts. so i decided to skip that. and do algebra II in 8th grade. at least now i might be challenged in my stupid school. (which just so happens to one of the best schools in measly Oklahoma). oh well, back to my geetar to pass time until the inevidable boring school year.(sobs)...........

Well I took algebra in the fourth grade and then for some reason I took the regular math in 5th and 6th grade which I wasn't at all challenged.  Then that all changed when I was in seventh grade when I took algII/trig honors at a magnet high school and I learned A LOT in that class, I had an awesome teacher also.  Then this year as an 8th grader, I took geometry honors at my zoned high school and it was so useless that I did precalculus next door with the geometry teacher's husband(teaches calculus).  So I really should take calculus but I'm just going to pad my GPA with an easy precalc course.  :-P

Let's see... I took Algebra I in 5th grade, then moved to a different school where for 6th grade they put me in 7th grade math.  Then in 7th grade I took geometry (had to test out of Algebra I).  Meanwhile, while I was taking geometry my math teacher game me an Algebra II/Trig book (because I he thought it'd prepare me more for AMC 10).  But in order to get credit for it, between 7th and 8th grade I had to take notes on all of the lessons and do all the usually-assigned problems and take all the chapter tests and the two semester finals from Algebra II/Trig.  Then in 8th grade I took Precalc, and I'm gonna take Calc AB freshman year.

Kinda messy.  Oh, and of course I "wasted" :roll: (according to some of my peers at school) many hours reading math books and doing mathcounts and amc problems.

Oh, and I used to do EPGY.  But I quit after Beginning Algebra because... guess what came next?  Intermediate Algebra and I'd already taken Algebra I twice and Algebra II once.  :roll: But no unconventional topics, or special math programs aside from EPGY... just lots of spare time.

algebra I and II in grade 7, geometry in 8, still have to do trig, ap calc ab in fall.

I forgot to mention I retook algebra I in the seventh grade while I took Alg II/trig because I wanted honors credit instead of regular credit.  And yeah...people here thought I was an idiot for doing math 24/7 these past two years.  *That was an exaggeration*

Btw...should I take precalculus or calculus as a freshman this fall?

hey...  well, so far, my math education has been mostly fostered by my dad who is a mathmatician and would love nothing more than for everyone else to love math... he has tutored me since i can remember, and up till 7th grade gave me summer classes, the biggest of which was when i did an online course in algebra II summer before my 6th grade year...  my elementry school was rather restricting in getting ahead in math, but i went as far as i could, and left in seventh grade for a small competive private school where I am about to go into an AP Calculus BC class in 10th grade...  im in richard's saturday class in san diego...  i've been involved in a few math competitions...  but i didnt really go to any outside math classes besides my dads and richards just this year...  and by the way of math books... dad has copies of them all downstairs, but i dont particularly remember any one of them as being wonderful...  my pre-calc teacher chose the book he used just cause his friend wrote it, so...  anyways, hope that helps!

What math education? Yeah, my parents did think I was "mentally challenged" and they still kinda do but that's besides the point. My math education started when I used baseball cards to help me learn division with decimals... I kinda never really paid attention in class... and I still don't so I never started learning anything till I reached high school. I had to use baseball cards to do decimal division... So although I never had the privilege of having my dad as a mathematician.. which would be pretty ill, I prefer the idea of learning maths to solve problems since problem solving will help you throughout your life.. solving mundane maths problems is just, well.. quite mundane.

But, if you're wondering I took the Algebra II in frosh year, then I skipped pre-calc (it's essentially the same class) and enter Calc I in soph year. Then I took Calc II the summer between my soph and jr. year and 1st sem. jr. year I did indep. study of Calc III Number theories, Comb, and some Adv. geom. Indeed I did get caught in the Calculus trap... I then studied some analysis and ODE, DE, PDE ad nauseum... It was only recently did I find out about some problem solving and become interested in it...

BTW, has anyone heard of Routh's theorem and barycentrix, sounds pretty tite to me...

Enjoy,
IW.OAV

LOL, my school must be really far behind all your schools. My school system only offers pre-algebra to seventh grade gifted and talented students. And even then, we could only go to Algebra I in eighth grade. From there, it was Geometry Frosh year, Algebra 2/Trigonometry Soph year, PreCalculus Junior year, and AP Calculus AB/BC (depending if the students wish to study more on their own for BC) Senior year. My school system was definitely retarded. The administrations have a lack of motivation to improve the curriculum. The only other math course besides the ones I listed that are not basic (e.g. courses that basically teach to add, multiply etc.) is Statistics. Our school is mostly an atheletically known school, where 25% of the kids failed the english section of a state administered basic skills test (which basically means, according to NJ, that 25% of our junior class are illiterate). You know how are school system responds to this kind of problem, they build the biggest gym in the state, which costed the taxpayers (us) 4 million dollars. Our schools gotta be one of the crappiest schools in the state. Jeez...that's why I take all my math courses at my local county college. Lol...sorry about rambling on.

My education looked like this. In first grade I was too gifted for my class so they moved me up a step. There I was doing math from 5th or 6th grade, I don't remember accurately. My math interest declined a bit in 4th grade. Then when I actually began 7th grade we had a radical change of teachers. I got this really stupid old man, who was on the verge of being blind -- thus depressed. And accidently it went over us students. He didn't have any lectures at the normal lessons, we didn't do any math only discussed our social life with friends. This went on until I began 2st grade in high school (we have nice middle school grades and three high school in Sweden) where I got this super intelligent math teacher who thought gifted students not only the standard curriculum, but also made them feel that maths is something beyond that. I spontaneously did the four years I missed in math and now I'm here. Going for my third and last year in high school. Loving math most of everything. Pretty dramatic, but unluckily I didn't have anyone who supported me in 5th grade onward. I still don't. The school I go to is miserable. So I have to study myself. Having the most interest in number theory at the moment.

all i can say is that IW.OAV's avatar rocks the house.

Sincerely,
High Inquisitor Syntax Error

same here
oh..great quote, Syntax/Sharkmasta/Handsome B Good/whatever
and did you read Harry Potter 5 by any chance?

My math education followed pretty much the same pattern...I had some really stupid teachers teaching me basic math courses which I was way too advanced for teaching me out of some really stupid books until Geometry. That  year, a really great man from NCSSM, Mr. Goebel (you may know him, he did the math team until recently) offered to tutor me. I had a great year, knocked off both Geometry with this really cool computer program that let me understand the stuff a whole lot better, and really stressed construction, and Algebra II, which we mostly did by him giving me hints on how to prove the theorems that he gave me. My best year in math ever....until the year right after that when I started having Mr. Stuart at CHHS. He introduced me to the math club in the first place. From then on, he taught me Precal and Cal BC with his normal classes, and he was a really great teacher. So....here I am now.

Art of Problem Soving Decree #3

No making fun of the High Inquisitor

Sincerely,
High Inquisitor Syntax Error

im on aim at the moment, talk to me at Handsome B Good

People don't usually get much respect from self-appointed titles. Keep that in mind.

fine, ill edit all of the posts to not have self-appointed. *yawn*

uh..what were decrees 1 & 2? And don't you think you should start with ~20 instead of 3? And ... whatever gave you the idea that I was making fun of you?

Hi.

I went to school in Ontario and never got any enrichment of any kind. In grade 9 however, I had a good math techer and got interested in math. After that I learned some stuff from the internet, went to the Canada USA Mathcamp (which is great) twice, and did the AP Calculus BC exam on my own.

-Murray

Since you brought up this old subject...

I had this really weird program back in CA in 5th where all the high scorers in the standarized test gets to go (including English, Science, and SS, not just math). In this class you get lectured about how to have a successful life and stop yourself from being a loser (which had nothing to do with math)
We also learned stuff like "how to memorize" and "how to set a goal" which didn't really teach us any facts, but more of a help with setting life's course. I'm pretty sure it was an experiment, so I'm sure not all people take the course in CA. Other than that, I have nothing interesting to say except that I skipped two grades [b]only[/b] in math.

Btw, Educational...I mean Art of Problem Soving Decree :) #1 can be found [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?p=1943]here[/url], but it doesn't mean anything because he misspelled "site" :P
I couldn't find #2 though. It was probably in the Etc.

and "I respect you Mermaid[b]man[/b]"...I mean High Inquisitor :)
Can I be on your Inquisitorial Squad? :D

[quote="rrusczyk"]What's the really cool program you mentioned?[/quote]

erg sorry I didn't see this for a couple of monts. It's called Elica. Free program; you can google it to find the homepage, i believe. anyways, it helped me a lot with the whole construction idea thing. Basically, we started out just like programming in commands to find like midpoints of lines and stuff, and we worked our way up. It was pretty cool.

OOO, I love threads like these.  Let's see.  In elementary school, I was in all the math classes common to elementary school, I usually had 100% in math and my dad wanted to advance me farther but my mom did not because that would mess up my life socially or something.  So I entered 6th grade and again was in basic 6th grade math where my math teacher tried to get me into more advanced math, but once again my mom denied that.  In 7th grade, I entered "advanced" math, Pre-Algebra.  I did real well in there like usual but at the end of the year, a lot of bad stuff happened and I pretty much lost all my motivation.  In 8th grade, I entered Algebra 1 but I did not do as well but still well enough? I got B+'s because I never really did my homework and in high school, I did Geo, Algebra 2, and then Pre-Calc. In Geo and Algebra 2, I had little motivation still but did well and in Pre-Calc, I started getting my motivation back and did pretty well.  Now I am in AP Calc AB as a senior which is advanced I suppose but not nearly as advanced as I could have been or as you guys are.  I dominate though just like I used to with the highest grade in the class  :D What a happy ending!

[quote="gauss202"]I'm curious to know what different people's math education has been like. What classes have you taken, were you in any special "new math" programs, does your teacher cover unconventional topics in class, what books did you use, have you read any books that were particularly enlightening to you, etc etc..[/quote]

Hi, this is my first post. I just turned eleven, and I'm homeschooled. I was born in Washington State and have lived most of my life in Minnesota.

The first math materials I used were the Miquon Math materials. I was living in Taiwan then, so I also used some Taiwan textbooks (just a little). After Miquon Math I used Singapore Primary Mathematics. We came back to the States in 2001. I went to the M-FACTs Saturday program at the U. of Minnesota in 2001-2002 and to the 2002 and 2003 Math Fun Fair.  :roll:

One book I really enjoyed was The Number Devil. These days I'm reading The Pleasures of Counting and From Zero to Infinity. We have a lot of other math books around the house. We just got AoPS this week.

In June 2002, I started using Stanford's EPGY distance learning program. I finished Beginning Algebra in May 2003. I also completed the Johns Hopkins Center for Talented Youth Honors Prealgebra course in December 2002. Now I'm in the University of Minnesota Talented Youth Mathematics Program (UMPTYMP) accelerated Algebra I & II course, using Algebra for College Students 6th edition by Jerome Kaufmann and Karen Schwitters. My teacher is one who does NOT agree with the text book in any way at all, so he's always giving his own ways of doing things.

I was on the Home School team of the SW Metro Division of the MN Junior High Mathematics league as an alternate in the 2001-2002 season and as a regular team member in the 2002-2003 season. I've been to the ITCEP (U. of Minnesota) summer programs Tech Week and PRIME in 2002. I took the 2002 AMC 8 math test.

i love your avatar. riemann is the best!

[quote="J031p"]LOL, my school must be really far behind all your schools. My school system only offers pre-algebra to seventh grade gifted and talented students. And even then, we could only go to Algebra I in eighth grade. From there, it was Geometry Frosh year, Algebra 2/Trigonometry Soph year, PreCalculus Junior year, and AP Calculus AB/BC (depending if the students wish to study more on their own for BC) Senior year.[/quote]

Yeah, my school in 7th grade was like that. It was a "magnet" school, basically a school only for gifted students, still it only had pre-alg for 7th. It was good in MathCounts competitions though, my team was first in the state, and one of my teammates made it to nationals, and i've done competitions since. anyway, i moved and my current school allows you to jump 2 years instead of my original 1, but i just continued at the normal pace in school and am now in pre-calc as a junior, but i've read math books when i can, my favorite is the classic Flatland.


inequalities
linear algebra
matrix
quadratics
vector
inequalities proposed

Let $S=\{1,2,...,2005\}$ and $A_1,A_2,...,A_{2005}$ be subsets of $S$ (not neccessary distinct).  For all $i,j$ denote
\[ a_{ij}=card\{A_i\cap A_j\}  \]
Prove the following inequality
\[ \sum_{i=1}^{2005}\sum_{j=1}^{2005}(a_{ij}x_ix_j)\geq 0  \]
for every real numbers $x_1,...,x_{2005}$

See also http://www.mathlinks.ro/Forum/viewtopic.php?p=362353#362353

[quote="grobber"]The matrix of this quadratic form is the gramian matrix of the characteristic vectors of the sets $A_i$, so it's positive semidefinite.[/quote]
Please do not use linear algebra theory. Elementary proof please???


:ninja: Ok, I am new here, and I REALLY need help with algebra.

 Right now I am on 'The Percent of Change', and a problem that I will present to you all is as thus~

[b]The original cost of an item and a discount rate are given. Find the sale price of each item to the nearest cent.[/b]

 12. Tennis Balls: $4.50; 25%

****

Help?

To find the new price of the item, you take the discount, multiply it by the price, then subtract that price from the original price.

\begin{eqnarray*}
4.5-(4.5\cdot.25)&=&4.5(1-.25)\\
&=&4.5\cdot.75\\
&=&\$3.38
\end{eqnarray*}

Ok, how do you find the percent of increase? Here is one such problem~

Original: 20
New: 16

 Let me see if I can get it...

Take the idea from the other post.

$ \frac{20\minus{}16}{20}\equal{}\frac{4}{20}\equal{}\frac{1}{5}$.


I know that if you have five distinct points in the 2d plane, and you join every point to every other point, there will be a maximum of 5C2 = 10 lines produced, but obviously it depends on where the points lie. I can see that the minimum number of lines will be 1 (all the points lie on a straight line). But what about the in between numbers? I can get 5, 6 and 8 but no others. Can someone please help me to work out why those numbers. I want to create a formula that will tell me which numbers are possible/impossible with n points. Does this problem have a name? It's very difficult to search for things like this on the web!

With many thanks.

Hm, this seems like an interesting problem. I don't know if this problem has already been solved but here is how I would handle the case $ n \equal{} 5$ points.

Case 1: All 5 points are non-collinear - Obviously, there are $ \binom{5}{2} \equal{} 10$ lines.
Case 2: 3 points out of 5 are collinear (1 "collinear group") - First there are $ \binom{5}{2} \equal{} 10$ lines that you could create. But this overcounts since you could pick two collinear points to create the same line. There are $ \binom{3}{2} \equal{} 3$ ways to create that line by picking two points. So, in this case, there are $ 10 \minus{} 3 \plus{} 1 \equal{} 8$ lines.
Case 3: 3 points out of 5 are collinear (2 "collinear groups") - By the same reasoning, there are $ \binom{10}{2} \minus{} \binom{3}{2} \plus{} 1 \minus{} \binom{3}{2} \plus{} 1 \equal{} 6$ lines here.
Case 4: 4 points out of 5 are collinear - There are $ \binom{10}{2} \minus{} \binom{4}{2} \plus{} 1 \equal{} 5$ lines.
Case 5: All 5 points are collinear - Obviously, there is only 1 line.

Therefore, you can create 1, 5, 6, 8, and 10 lines.
As n increases however, I think this process becomes more tedious, so there might exist a different way.

[quote="ocha"]How many lines is this? it looks like seven to me

[geogebra]48cf39ebdf9dc78ce6c2df383a27a308e56620f5[/geogebra]

Can you define a line? A line is continuous in both direction, whereas I have used segments. If three points are colinear, do they form 1 line or 2 segments?[/quote]

1) The OP is asking about lines, not segments. A line is defined as a one-dimensional figure which extends infinitely in both directions, and it is uniquely determined by 2 points. So choosing between collinear points always defines the same line.

2) You missed the line connecting the top-right point to the bottom-left point, making it 8 lines.


I have a problem with your solution. Correct me if i am wrong.

one problem, i cant help but notice that you wrote sqrt.(n+1)/(n+1)<sqrt.(n+2)/(n+2) as a step in your reasoning

which i found, by subbing a value, say 1, to be false, eg, 1/rt2 is not less than 1/rt3 !!

what do u think ?

This is not calculus.  You correctly point out an error in akech's solution -- since $\sqrt{x}$ is an increasing function, $\frac{1}{\sqrt{x}}$ is decreasing.  In order to show the desired inequality, you should start with the fact that $4(n+1)^{3}> (2n+1)^{2}(n+2)$ for all positive integers $n$.  (You'll have to use the fact that the square root function is increasing, as well.)


At 12 o'clock, the three hands of a clock (hour, minute, second) are in one line. What is the first time after noon where the second hand is the angle bisector of the angle formed by the hour and minute hands?

[quote="fourierseries"]Do you mean angle bisector as the angle bisector of the acute angle?[/quote]

Yes,  :blush:


Pegs are put in a board $ 1$ unit apart both horizontally and vertically.  A reubber band is stretched over $ 4$ pegs as shown in the figure, forming a quadrilateral.  Its area in square units is

[asy]
int i,j;
for(i=0; i<5; i=i+1) {
for(j=0; j<4; j=j+1) {
dot((i,j));
}}
draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));
[/asy]

$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 5.5 \qquad \textbf{(E)}\ 6$

[hide]Divide it into 4 triangles. $ \frac {1(2)}{2} \plus{} \frac {2(3)}{2} \plus{} \frac {1(3)}{2} \plus{} \frac {1(1)}{2} \equal{} 6$[/hide]

Pick's theorem:
$ A\equal{}i\plus{}\frac{b}{2}\minus{}1$
$ A$ $ \text{\equal{} area}$
$ i$ $ \text{\equal{} interior points}$
$ b$ $ \text{\equal{} boundary points}$

Plugging in the numbers:
$ A\equal{}5\plus{}\frac{4}{2}\minus{}1$
$ A\equal{}5\plus{}2\minus{}1$
$ A\equal{}6$
$ \fbox{E}$

[hide="Also"]Shoelace:

The vertices are $ (0,1),(1,3),(3,0),(4,1)$.
So the area is $ |3\plus{}12\plus{}1\minus{}3\minus{}1|\cdot\frac{1}{2}\equal{}\frac{12}{2}\equal{}\boxed{\textbf{(E)}\ 6}$.[/hide]

Hide solutions, people.

[quote="hypersonic"]Pick's theorem:
$ A \equal{} i \plus{} \frac {b}{2} \minus{} 1$
$ A$ $ \text{ \equal{} area}$
$ i$ $ \text{ \equal{} interior points}$
$ b$ $ \text{ \equal{} boundary points}$

Plugging in the numbers:
$ A \equal{} 5 \plus{} \frac {4}{2} \minus{} 1$
$ A \equal{} 5 \plus{} 2 \minus{} 1$
$ A \equal{} 6$
$ \fbox{E}$[/quote]

hmm ...What is exactly Pick's theorem :
 :thanks for helping :

Pick's theorem is like this way to calculate the are in problems like that, the dots one. There could be like an [b]irregular shaped [/b]thing in the dots board, and the pick formula helps you [u]estimate[/u] that area. But it could be used in [b]regular shapes[/b] to find the [b]exact[/b][b] value[/b] too.

Again, the formula is A= i+e/2-1
i=interior dots
e=exterior dots
A=area

You'll learn this in geometry.


I was wondering whether anybody could answer this question:

What would one use a one line table to establish? Be specific.
Thanks.

This question didn't make much sense to me but what it actually asked was:

What would one use a one line table to establish(Sec 14.1)? Be specific.

Section 14.1 is titled Goodness-of-Fit Tests when Category Probabilities are Completely Specified so I would guess the question would mean:

What would one use a one line table to establish Goodness-of-Fit Tests? Be specific.

Does this question make sense to you?


Try this qn

http://iit-redefined.zzl.org/viewtopic.php?f=7&t=246

I think there is a similar question to this in my physics text.  I think it helps to think of it as a bunch of series and parallel elements.

You should see a bit of symmetry if you draw $ BFCG$ inside $ AEDH$. Let the resistance of $ HP$ be $ xR$ and the resistance of $ PG$ be $ (1\minus{}x)R$ (all other sides have resistance $ R$). Now to solve this, let's apply a voltage $ V$ across $ A$ and $ P$, resulting in a total current flow of $ I_1 \plus{} I_2$; let the current flowing into $ P$ from $ H$ be $ I_1$ and the current flowing in from $ G$ be $ I_2$. Also, let the current from $ D$ to $ C$ be $ I_3$. We can figure out all the other currents in terms of these by using the junction rule. Using the loop rule a few times and the fact that $ V \equal{} (I_1 \plus{} I_2) R_{eq}$, I found that $ R_{eq} \equal{} \frac{9 \plus{} 5x \minus{} 10x^2}{12} R$.

I can't see any symmetry.

Something wrong with the answer: for x =0 we get a beautiful symmetry with

$ R_{eq} \equal{} {5 \over6} R$

that doesn't agree with your expression.

I knew I must have made a mistake somewhere ;). That case actually corresponds to $ x\equal{}1$, but I carefully worked out the arithmetic again, this time swapping $ x$ and $ 1\minus{}x$. I got $ R_{eq} \equal{} \frac{10\plus{}4x\minus{}5x^2}{12} R$. For $ x \equal{} 0$, this gives $ R_{eq} \equal{} \frac{5}{6}R$, and for $ x\equal{}1$, it gives $ R_{eq} \equal{} \frac{3}{4}R$, which is right, I think. It's interesting to note that the maximum resistance of $ \frac{9}{10}R$ is achieved when $ x \equal{} \frac{2}{5}$.

You're quite right.
I've get
\[ R_{eq} \equal{} {{9\plus{}6x\minus{}5x^2}\over{12}} R,\]
in the first assumption ( HP -> xR)


It seems to be false, gemath. What happens if $x=a,y=b,z=c$? The following is true but obvious

\[x+y+z \ge \sqrt{\frac{3\sqrt{3}abc}{a+b+c}}\]

[quote="gemath"]Given $x,y,z,a,b,c>0$ such that $x+y\ge a+b,y+z\ge b+c,z+x\ge c+a$ prove that $x+y+z\ge 6\sqrt{\frac{abc}{a+b+c}}$ :)[/quote]
Trivial :P $x=y=z=a=b=c=1$.

[quote="gemath"]That's right hungkhtn, it doesn't have equlities, I type wrongly it must be
Given $x,y,z,a,b,c>0$ such that $x+y>a+b,y+z>b+c,z+x>c+a$ prove that $x+y+z\ge 6\sqrt{\frac{abc}{a+b+c}}$[/quote]

Sorry Gemath. Your inequality is still false because you can choose $x$ close to $a$, $y$ close to $b$ and $s$ close to $c$ as much as you want. There is no difference between $\ge$ and $>$.

Trivial $a=b=c=1$ and $x=y=z\to 1$ and $x,y,z>1$. :P

Ok that's right thanks I must check it! :oops:


linear algebra
matrix
inequalities
algebra
polynomial
abstract algebra
linear algebra unsolved

Let $n,p \geq 2$ be two integers and $A$ an $n\times n$ matrix with real elements such that $A^{p+1} = A$.

a) Prove that $\textrm{rank} \left( A \right) + \textrm{rank} \left( I_n - A^p \right) = n$.

b) Prove that if $p$ is prime then \[ \textrm{rank} \left( I_n - A \right) = \textrm{rank} \left( I_n - A^2 \right) = \ldots = \textrm{rank} \left( I_n - A^{p-1} \right) .  \]

I solved it:) yay
So :
a) Suppose $rank (A)=p$ then we can say that $\alpha_1=\alpha_2=...=\alpha_{n-p}=0$ and the rest are roots of the equation $x^p=1$ so $rank(I_n-A^p)=n-p$ which solves point a.
b) we can write the eigenvalues(those different from $0$) of $A$ as  $\alpha_i=\epsilon^{k_i}$ where $\epsilon$ is a different from $1$ of $x^p=1$. So the rank of $I_n-A^m$ is given by the number of eigenvalues of $A^m$ equal to $1$ or $\epsilon^{mk_i}=1$ or $p|mk_i$ sice $p$ is prime $(p,m)=1$ so that number is constant which solves the problem.
It`s a little brief but it`s the main idea.

[quote="spix"]a) Suppose $rank (A)=p$ then we can say that $\alpha_1=\alpha_2=...=\alpha_{n-p}=0$ and the rest are roots of the equation $x^p=1$ so $rank(I_n-A^p)=n-p$ which solves point a.[/quote]

I thought of something similar, but I got only an inequality, so I didn't write the idea.

http://mathworld.wolfram.com/MatrixRank.html thank god for Internet !  :D

[quote="perfect_radio"]Let $n,p \geq 2$ be two integers and $A$ an $n\times n$ matrix with real elements such that $A^{p+1} = A$.

a) Prove that $\textrm{rank} \left( A \right) + \textrm{rank} \left( I_n - A^p \right) = n$.
[/quote]

The matrix A vanishes the polynomial $X^{p+1}-X = X(X^{p}-1)$

$gcd(X,X^{p}-1) = 1$ theorem kernel decompostion give

$R^n=ker(A)\oplus ker(I-A^p)$ take the dimension

$n=dim(ker(A))+dimker(I-A^p)$
to finish use rank theorem $n=rank(f)+dimker(f)$

you get $n=rank(A)+rank(I-A^p)$

[quote="perfect_radio"]Let $n,p \geq 2$ be two integers and $A$ an $n\times n$ matrix with real elements such that $A^{p+1} = A$.

b) Prove that if $p$ is prime then \[ \textrm{rank} \left( I_n - A \right) = \textrm{rank} \left( I_n - A^2 \right) = \ldots = \textrm{rank} \left( I_n - A^{p-1} \right) .  \][/quote]

Let's work in M(n,C)  the matrix A is trigonalisable its eigenvalues is contains in
the set of the roots of  $X^{p+1}-X$ wich are $0,e^{\frac{2ik\pi}{p}}$ with$0\leq k \leq p-1$

For any $1\leq q \leq p-1$
$I-A^{q}$ is similar to a triangular superior with diagonal $1-d_1^q,...,1-d_n^q$
where $d_j^q$ could be $0$ or $e^{\frac{2ikq\pi}{p}} \neq 1$ since
$0\leq k \leqq-1$, $0\leq q \leq p-1$ (kq is prime with p)

$rank(I-A)=...=rank(I-A^{p-1})$

Hi, i'm new here
My solution for the first part was like the official solution and for the second I proved that
$\ rank(I_n-A)=rank((I_n-A)*(I_n+A+A^2+...+A^\ t-1 )$ for $\ t\leq p-1$.
That is true because if $\ T$ is invertible then $\ rank(AT)=rank(A)$.
So it is enough to prove that $\ det(T)<>0$ if  $\ T=I_n+A+A^2+...+A^\ t-1$
If $\ x$ is an eigenvalue of $\ A$ then $\ x=x^\ p+1$ and $\ 1+x+x^2+...+x^\ t-1$ is an eigenvalue of $\ T$.
Assume $\ x<>0$ and $\ 1+x+x^2+...+x^{t-1}=0$ then $\ x^t=1$ , but we also have that $\ x^p=1$.
Because $\ (t,p)=1$ there are $\ n,m$ such that $\ p*n+t*m=1$ so $\ x=x^{p*n+t*m}=1$ but this is false because
$\ 1+x+x^2+...+x^{t-1} =0$ . So $\ T$ is invertible.
But I only received 6 points for this solution  :?

It seems kind of blurry.. how much points did you get overall?

I got 26 overall, the problems were not that hard.  :)
What is blurry about the solution?

The $<>$ confused me maybe you should use $\neq$  ;)

well i used sylvester's inequality for point b) ( for a) as well :D ) and then the fact that if p is prime the roots of $x^p=1$ are all different from the roots of $x^k=1$ if $k<p$... :roll:, except 1.

Hello, excuse this stupid question... [quote="Moubinool"]$gcd(X,X^{p}-1) = 1$ theorem kernel decompostion give
$R^n=ker(A)\oplus ker(I-A^p)$...[/quote] I didn't know the "theorem kernel decomposition" and After informations, I don't know what is $R^n$??!  :blush: Prayer not to laugh...  :oops:

First part is trivial by Sylvester inequality and for second part just note $x\equiv y (mod p) \implies A^x=A^y$ so if for some $i\neq j$ we've $rank(I_n-A^i)\neq rank(I_m-A^j)$ with $p>j>i$ then we've $rank(I_n-A^i)\neq rank(I_m-A^{j+pt})$ for all $t\in\mathbb N$ and thus $i$ can't divide $j+pt$ for all $t\in\mathbb N$ and so $i|px$ for some $x<i$ and that's absurd since $gcd(i,p)=1$ so we're getting contradiction and so done.

Because $A^{p+1}=A$ it is very easy to see that $Im(A)\oplus ker(A)=\Bbb{R}^n$. It is also obvious to see that $A(I_n-A^p)=(I_n-A^p)A=O_n$.
For the first part we want to show that the image of $\Bbb{R}^n$ under $A$ is in fact the kernel of $I_n-A^p$, which implies our relation.
If we choose a vector $u\in Im(A)$, then $(I_n-A^p)u=0$ since $(I_n-A^p)A=O_n$. Now let's choose a nonzero vector $v\in ker(I_n-A^p)$.
If $Av=0$, then we would obtain that $v=I_n v=A^pv=0$ which is a contradiction, therefore $v\in Im(A)$ .
For the second part we want to show that $ker(I_n-A^j)=ker(I_n-A)$ for all $j=2,3,...,p-1$.
In order to do this, we have to notice that if $A^kv=v$ for some vector $v$ and some positive integer $k<p$, then $Av=v$.
This holds because we look at the cycle $v,Av,Av^2,...,Av^p=v$ in which the total number of vectors must divide $p$.
Now suppose that $u\in ker(I_n-A^j)$ for some $j$, which means that $(I_n-A^j)u=0$ or $A^ju=u\Rightarrow Au=u\Rightarrow u\in ker(I_n-A)$.
It is obvious that a vector from $ker(I_n-A)$ belongs to $ker(I_n-A^j)$, so the proof is complete.
.


advanced fields
advanced fields unsolved

Is it possible to color the plane $Oxy$ with countably many colors such that no non-degenerate monochromatic triangle $ABC$ with $AB\|Ox,\ AC\|Oy$ is formed?

Yes. Let f(x) = x if x is a rational number, f(x) = 0 if x is an irrational number.
For each point (x,y) in the plane, the color of the point is f(x).

Can you provide a link to the problem posted by sam-n?

UPDATE : As JBL pointed out, my coloring does not work.

Under that coloring, $A(0,0), B(0,1)$ and $C(\sqrt{2}, 0)$ are a triangle with the property we're trying to avoid, no?

Let X be a set. A corner in $X^{2}$ is a triple ( (x,y), (x, y'), (x', y)) with $x \neq x'$,$y \neq y'$ with x, y, x', y' in X.

Let $\Omega$ be the ordered set of all countable ordinals. This set has a strange property that for all b in $\Omega$ the interval [0,b] is countable while the entire set $\Omega$ itself is uncountable.

It is possible to color $\Omega^{2}$ with countably many colors without mono-chronic corners: Let E be the set of even integers and O the set of odd integers. The union of E and O is the set of colors. For each x in $\Omega$, color the line $\{x\}\times [0,x]$ using a bijection between the line and E. Color the line $[0,x) \times \{x\}$ using a bijection between the line and O. This coloring does not form mono-chronic corners.

This means that if $\aleph_{1}= 2^{\aleph_{0}}$ then such a coloring is possible also in the Euclidean plane $\mathbb{R}^{2}$.

The existence of such coloring turns out to be equivalent to [url=http://en.wikipedia.org/wiki/Continuum_hypothesis]continuum hypothesis[/url]

Suppose the Euclidean plane is colored with countably many colors without mono-chronic corners. Let $A$ be an infinite subset of $\mathbb{R}$. We will use the coloring to show that the cardinality of $A$ is either countable infinity or the continuum. For each real number $x$, let $Y_{x}$ = { y | (x,y) is the only point on the line $\{x\}\times \mathbb{R}$ having the same color as (x,y) }.  Let $Y_{A}= \cup_{x \in A}Y_{x}$.

If $Y_{A}$ is not equal to $\mathbb{R}$, there exists a number $y_{0}\in\mathbb{R}-Y_{A}$, each point in the set $A \times \{y_{0}\}$ has different color, for otherwise a mono-chronic corner is formed, therefore the set $A$ is countable.

If $Y_{A}$ is equal to $\mathbb{R}$, then since each $Y_{x}$ is countable, we have $|\mathbb{R}| = |Y_{A}| \leq  |A|\cdot \aleph_{0}= |A|$.


1.If a,b,c are positive reals,such that $ ab\plus{}ac\plus{}bc\plus{}abc \equal{} 4$,prove that :
$ a\plus{}b\plus{}c\geq ab\plus{}ac\plus{}bc$ (I wonder whether it is true or not).
I will compile this with another one(I have posted it,but still no solver :maybe: ):
2.quote="stephen38"]Prove that $ \sum_{cyc}\left(a^{2}b^{2}\plus{}abc\right)\geq\left(a\plus{}b\plus{}c\right)\left(ab\plus{}bc\plus{}ac\minus{}1\right)$
for a,b,c positive reals and $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3$.[/quote]

1. [hide="A hint"]Use lemma : Let $ x,y,z>0$, then $ xy\plus{}yz\plus{}zx\plus{}2xyz\equal{}1$ iff $ x\equal{}\frac{a}{b\plus{}c},y\equal{}\frac{b}{c\plus{}a},z\equal{}\frac{c}{a\plus{}b}(a,b,c>0).$[/hide]

[quote="stephen38"]1.If a,b,c are positive reals,such that $ ab\plus{}ac\plus{}bc\plus{}abc \equal{} 4$,prove that :
$ a\plus{}b\plus{}c\geq ab\plus{}ac\plus{}bc$ (I wonder whether it is true or not).
[/quote]
$ ab\plus{}ac\plus{}bc\plus{}abc \equal{} 4\Leftrightarrow\frac{1}{2\plus{}a}\plus{}\frac{1}{2\plus{}b}\plus{}\frac{1}{2\plus{}c}\equal{}1.$
We obtain: $ x\equal{}\frac{1}{2\plus{}b}\plus{}\frac{1}{2\plus{}c}\minus{}\frac{1}{2\plus{}a}\equal{}1\minus{}\frac{2}{2\plus{}a}\equal{}\frac{a}{2\plus{}a}>0,$
$ y\equal{}\frac{1}{2\plus{}a}\plus{}\frac{1}{2\plus{}c}\minus{}\frac{1}{2\plus{}b}\equal{}\frac{b}{2\plus{}b}>0$ and
$ z\equal{}\frac{1}{2\plus{}a}\plus{}\frac{1}{2\plus{}b}\minus{}\frac{1}{2\plus{}c}\equal{}\frac{c}{2\plus{}c}>0.$
Hence, $ x\plus{}y\plus{}z\equal{}1,$ $ a\equal{}\frac{2x}{y\plus{}z},$ $ b\equal{}\frac{2y}{x\plus{}z}$ and $ c\equal{}\frac{2z}{x\plus{}y}.$
Thus, it remains to prove that:
$ \sum_{cyc}\frac{x}{y\plus{}z}\geq2\sum_{cyc}\frac{xy}{(x\plus{}z)(y\plus{}z)},$ which equivalent to the Schur's inequality:
$ \sum_{cyc}(x^{3}\minus{}x^{2}y\minus{}x^{2}z\plus{}xyz)\geq0.$

Thanks for your nice solutions,N.T.Tuan and arqady. :lol:
[quote="arqady"]$ ab\plus{}ac\plus{}bc\plus{}abc \equal{} 4\Leftrightarrow\frac{1}{2\plus{}a}\plus{}\frac{1}{2\plus{}b}\plus{}\frac{1}{2\plus{}c}\equal{} 1.$[/quote]
Man,how can you think something like this???

[quote="stephen38"]
[quote="arqady"]$ ab\plus{}ac\plus{}bc\plus{}abc \equal{} 4\Leftrightarrow\frac{1}{2\plus{}a}\plus{}\frac{1}{2\plus{}b}\plus{}\frac{1}{2\plus{}c}\equal{} 1.$[/quote]
Man,how can you think something like this???[/quote]
I knew it! :D


I draw a card from a standard 52-card deck.  If I draw an Ace, I win 1 dollar.  If I draw a 2 through 10, I win a number of dollars equal to the value of the card.  If I draw a face card (Jack, Queen, or King), I win 20 dollars.  If I draw a $ \clubsuit$, my winnings are doubled, and if I draw a $ \spadesuit$, my winnings are tripled.  (For example, if I draw the $ 8\clubsuit$, then I win 16 dollars.)  What would be a fair price to pay to play the game?  Express your answer as a dollar value rounded to the nearest cent.

I assumed a round in which every card is drawn exactly once and divided by 52.

Either red suit gets 1 + 2 + ... + 10 = 55 + 20 + 20 + 20 = 115
Clubs gets twice this.
Spades triple.

This seems an easier explanation than the one given.

Thus our answer is 115+115+230+345=7x115=805 per 52 games, or about 15.48 per game.

[hide="Full Solution in LaTeX"]
Let $E_1$ be the expected winning if a heart or diamond is drawn. We get:

$ E_1 = \$ \frac {115}{13} $

$ E = \frac {1}{4} (E_1 + E_1 + E_2 + E_3) = \frac {1}{4}(7E_1) = \$ \frac {805}{52} $

$ = \boxed {\$15.48} $

NOTE: When entering this on Alcumus, we are expected to enter 15.48 and not with the dollar sign (I think!)
[/hide]

How did the answer get $\frac {1} {13}$ multiplied by $(1+2+3+...)$ why isn't it $\frac {1} {52}$ times $(1+2+3+4+5+...)$ Comes from the solution given.
 :o

There are 4 cards in each rank. So you count each of the numbers once and multiply by $\frac{1}{13}$.
If you want to divide by $52$, you'd have to count each rank $4$ times.

by linearity of expectation, why is it not E1+E1+E2+E3? Why multiply by 1/4?

[hide=Solution]
There is a 1/52 chance of drawing each card. So, we need to sum the total value of all of the cards in the deck, which is 115(2)+230+345=805. Therefore the answer is about [b]15.48[/b] dollars.
[/hide]

I have an easier variant of the solution.

I thought of it this way:

The total price with the cards is $(1+2+3+...+10)$ for each of the ace-10 cards and $20+20+20$ for each face card. The average winnings of the cards without multipliers is $\frac{(1+2+3+...+10)+(20\cdot3)}{13} = \frac{115}{13}.$

The average multiplier is $\frac{1+1+2+3}{4} = \frac{7}{4}.$

Multiplying these two together, we get approximately $\boxed{\$15.48}$ for our answer.


Would $ 1$-digit numbers be considered palindromes?

Well, does the number read the same forwards and backwards? (Guess this answers your question...)

"A palindrome is a word, phrase, number or other sequence of units that can be read the same way in either direction."
"a word, line, verse, number, sentence, etc., reading the same backward as forward"
"a word or phrase that reads the same backward as forward"

"A palindrome is a number that reads the same from left to right and from right to left."

I guess it would be, looking mostly at the specific definition for numbers..... Interesting Question.

WORDS EXAMPLES:-

RADAR,MALAYALAM,MADAM.ETC...

I was asking for $ 1$-digit palindromes....I already know word examples (loin oil, racecar).  And please don't caps your text, caps generally implies yelling or something important, and I'm pretty sure you were't trying to yell at me.

Dude, you just put like 4 "useless posts" around this forum asking people to stop spamming when you are spamming.

Yes, it would be considered a palindrome. Palindromicity can be applied to ANY natural number, so any one digit number is a palindrome. Also... the three ways of a palindrome would be:
n consists of a single digit
n consists of two equal digits
n consists of three or more digits, the first and last digits are equal, and the number obtained by stripping the first and last digits off n is itself palindromic.

[quote="cgyao15"]useless posts = :spam:[/quote]

That's not a useless post, cgyao.
THIS IS AN EXAMPLE OF A USELESS POST:
[hide][quote="cgyao15"]useless posts = :spam:[/quote][/hide]

and what about 10-digit plaindromes?

[quote="Eulers_Apprentice"]yeah... i agree with joyofpi. 1 digit numbers are palindromes.
it also fits with the pattern (basic C+P):

# of 1 digit palidromes: 9
# of 2 digit palidromes: 9
# of 3 digit palidromes: 90
# of 4 digit palidromes: 90
# of 5 digit palidromes: 900
# of 6 digit palidromes: 900[/quote]

Actually, 10 1 digit palindromes counting 0!!!  :D

Well that depends if 0 is positive or negative. Either way, I dont think 0 counts since it is neither positive or negative. Sorry to rain on your parade Johnathanchou711.  :(

so what would the patten be expressed in terms of n where n is the number of digits?

uhhh i dont know how to latex-ify it but

for integer n>1

if odd $ 9\cdot10^{\dfrac{n\minus{}1}{2}}$
if even $ 9\cdot10^{\dfrac{n\minus{}2}{2}}$

[quote="cgyao15"]i don't think it works for two does it?[/quote]

$ 9\cdot10^{\dfrac{2\minus{}2}{2}}\equal{}9\cdot10^0\equal{}9\cdot1\equal{}9$

[quote="mz94"]Using logic, its a relatively simple pattern to derive. Take every 2-digit palindrome, and squeeze each number 0-9 in between them. you now get 10 times as many palindromes (each one has for x0x through x9x). The same applies for 4 digit--just squeeze another number 0-9 in between it (so x00x through x99x)

so excluding the case of 1 digit, the number of x-digit palindrome forms the pattern you have just found[/quote]

HiDN428: There's your proof.

So could negatives or decimals be considered palindromes?

For example: 73.37

there's actually no need to spam... but i think they would... :wink:

[quote="isabella2296"]-- Do not make a post saying, "Hey! That's spamming! Don't do it, __________!"
[/quote]
[quote="levans"]Please do not post messages that are meant to "teach" others about forum etiquette. Messages such as "do not revive old topics" should not be placed on the forums[/quote]
[quote="levans"]Clutters the forums, and moderators have to delete the messages. Moderators will tell the user as needed. In general, we don't really have a rule against reviving old posts.[/quote]

Seriously, if PowerOfPi is asking a question that hasn't been answered yet, then how is it considered spamming?

technically, he's reviving the old post... :blush:

i guess so, cuz it is the same number from beginning to end  :rotfl:

[quote="007math"][quote="joyofpi"]! A palindromic number MUST BE A NATURAL NUMBER. [/quote]

So I guess negatives and decimals are not palindromes.

Since I was the creator of this thread, I allow PowerofPi to ask a question. :D Now he's not in trouble. :D[/quote]

Umm...What part of this post do you not understand?


The product of two numbers is $-4$ and their sum is $12$.  What is the sum of their reciprocals.  Please explain because I don't understand how to do it.

[hide]
$ab=-4$
$a+b=12$
and we are looking for $\frac{1}{a}+\frac{1}{b}$
which is equal to
$\frac{b+a}{ab}$
which is equal to
$-3$
by substitution.

I hope that's right.

[/hide]

EDIT: What did I do? :huh:
Man, I must really be losing it.

i have no idea what you did circle squared....
I think you misinterpreted the rpobme

[hide]let a and b be the two numbers. Then we have ab= -4, and a+b=12.  Let's first see 1/a+1/b in the simplified form.  This gives us (a+b)/ab= 12/(-4)= -3.  Thus, -3 is ur answer.[/hide]

[hide]Let the two numbers be $a$, and $b$.

We want $\frac1a+\frac1b$, which when written as a single fraction, is:

$\frac1a+\frac1b=\frac{a+b}{ab}$.

We know from the problem that $a+b=12$ and that $ab=-4$.

Therefore, $\frac1a+\frac1b=\frac{12}{-4}=-3$.[/hide]

Thanks for the solution, but I just still don't get how you go from $1/a+1/b$ to $\frac{a+b}{ab}$

Can someone explain that?
Am I being dense?

All you are doing is getting a common denominator.

Steps:

$\frac1{a}+\frac1{b}=\left(\frac1a\right)\left(\frac{b}b\right)+\left(\frac1b\right)\left(\frac{a}a\right)=\frac{b}{ab}+\frac{a}{ab}=\frac{a+b}{ab}$

Did you see that bobdole's topic.  What a jerk!  He posted that a good math website was "(believe me it was inappropriate)(it started with a 'P' ande ended with 'layboy')"

Of all the dumb things to do on this forum...
Luckily the moderators are cool and deleted it before most people could see it...
I'd like to take a moment and thank the moderators for doing a great job here on the MATHCOUNTS Forum...
This question is pretty much answered, so, would a fantastic moderator lock this please?...


Give $x,y,z \geq 0$ satisfying: $x+y+z=3$.
Prove that: $(xy+yz+zx)(x^{2}y+y^{2}z+z^{2}x) \leq 9$
However: $(xy+yz+zx)(x^{3}y+y^{3}z+z^{3}x) \leq 16$  :)

[quote="Anh Cuong"]Give $ x,y,z \geq 0$ satisfying: $ x \plus{} y \plus{} z \equal{} 3$.
Prove that: $ (xy \plus{} yz \plus{} zx)(x^{2}y \plus{} y^{2}z \plus{} z^{2}x) \leq 9$
However: $ (xy \plus{} yz \plus{} zx)(x^{3}y \plus{} y^{3}z \plus{} z^{3}x) \leq 16$  :)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=230715


Whats the number of ways to seat n people around a circle, without any restrictions?
How is this generalization derived?

if the order doesn't matter, then it would be a factorial problem
n!
seeing that the first spot in the circle will have N people available, then the next spot will have n-1 available, and so on.[/code]

[quote="witofsun"]n!/n
the formula of permutation isn!, the circle case should be n!/n[/quote]

Yes for circular permutations no. of ways = (n-1)!

(n-1)!. The way that this is derived is that On a line, for obvious reasons, the possible number of ways is n!. (Because n choices for the first spot, n-1 for the second, etc.). But since we have a circle, the positions can be rotated without changing. Therefore, they can be rotated n times, so we divide by n to get (n-1)!


Let $ \\a,b,c,d>0$ and $ \\a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1.$ Prove that:
$ \\\sqrt{1\minus{}a}\plus{}\sqrt{1\minus{}b}\plus{}\sqrt{1\minus{}c}\plus{}\sqrt{1\minus{}d}\geq\ a\plus{}b\plus{}c\plus{}d$

[quote="boroo12"]Let $ \\
a,b,c,d > 0$ and $ \\
a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1.$ Prove that:
$ \\
\sqrt {1 \minus{} a} \plus{} \sqrt {1 \minus{} b} \plus{} \sqrt {1 \minus{} c} \plus{} \sqrt {1 \minus{} d}\geq\ a \plus{} b \plus{} c \plus{} d$[/quote]
$ \sum_{cyc}\sqrt{1\minus{}a}\geq\sum_{cyc}\sqrt{a}\geq a\plus{}b\plus{}c\plus{}d.$ :wink:

Why $ \sum \sqrt {1 \minus{} a} \ge \sum \sqrt{a}$?


number theory
greatest common divisor
AMC
AIME
relatively prime
AIME II

[quote="Calculus"]Is one relatively prime to itself?[/quote]

Of course, two integers are relatively prime means that their GCD is 1.

I don't think $1$ is looked at in that way, since it only has one factor.

[quote="Calculus"]Is one relatively prime to itself?[/quote]

Well..... I think by definition relatively prime integers deal with integers such that N>1. I could be wrong though.

Consulting the Mathworld Bible: http://mathworld.wolfram.com/RelativelyPrime.html

we have "Two integers are relatively prime if they share no common positive factors (divisors) except 1."  Therefore, by this definition, 1 and 1 are relatively prime.  However, it seems any math question that forced you to consider whether 1 is relatively prime to itself is poorly written, since any such ambiguity should be made clear.

Hey I just did that test...no wonder my answer was off by one for that problem. :mad:

[quote="beta"][quote="zanttrang"][quote="Calculus"]This question came up because of #11 2000 AIME.[/quote]

They wouldn't tell you that the sum of all possible sides is $m/n$ if the value ends up being $1/1 = 1$; in that case they would just tell you that it's a positive integer.  So you can infer from such cases that they don't mean for both the numerator and the denominator to be 1.[/quote]

Oh no they want 1/1, check out the solution, they want everything whether positive integer or not(that means 2/1, 5/1, 1000/1, 25/2, 2/25, 1/1000, etc)

On AIME if they say it, they mean it.[/quote]
Ohh - silly me, I was looking at #11 for the AIME II (since I was using the Contest Resources page and not Kalva).  Sorry about that!


$f(x)\geq0$ for all $x>0$ and $y>0$ and $f(x+y)=f(x)+f(y)+2\sqrt{f(x)f(y)}$
Solve for $f(x)$.

I'm totally lost... :blush:

$f(x+y) = f(x) + f(y) + 2\sqrt{f(x)f(y)}$
Since $f(x) \geq 0$, we have $\sqrt{f(x)} = g(x)$ exists and is always positive. So, substituting, we get:
$(g(x+y))^2 = g(x)^2 +g(y)^2 + 2\sqrt{g(x)^2g(y)^2} = g(x)^2 + 2g(x)g(y) + g(y)^2 = (g(x) + g(y))^2$

$\implies g(x+y) = g(x) + g(y)$ since we already know $g$ is always positive. So, we can deduce $g(x) = k_1 x$ for some constant, which then tells us $f(x) = k x^2$.

I just wanted to ask that in many places we replace the composition of 2 fuctions (e.g. $(f(x))^2 $ above) by another function and solve the functional equation for it. Though it obviously does seem correct, is it flawless???

wow, seems really easy when you get the answer.  Sigh...I just don't "see" things.

This is a standard functional eqn with a quite advanced solution

[quote="Pork_Chop8"][quote="MysticTerminator"][quote="Pork_Chop8"]$g(x+y) = g(x) + g(y)$ since we already know $g$ is always positive. So, we can deduce $g(x) = k_1 x$ for some constant.[/quote]

???[/quote]
Yes, a little left out, I agree :lol: .[/quote]

then please fill it in. it's not quite that easy.

So we want to solve $f(x+y)=f(x)+f(y)$. $f(1)=f(\frac1n)+f(\frac1n)...+f(\frac1n)$ Where we have $n$ $f(\frac1n)$s.  So  $f(1)=nf(\frac1n)$.

$f(\frac1n)=\frac1nf(1)$.  Multiply $m$ to both sides we get $\frac{m}{n} f(1)=mf(\frac1n)=f(\frac{m}{n})$.  So $f(x)=f(1)x$ and $f(1)$ can be any arbituary $k$.

Not quite!!!

All you have proved is for the rational nos. For this function to be valid $ \forall \mathbb{R} $ you have to use a continuity or monotonicity argument which is a bit too advanced for pre-olympiad level!!

No I am talking about the rigoous proof ehich uses the thoerem that every infinte sequence of rational nos.(bounded) teds to a real number.

[quote="stupidkid"]Sorry forgot to state the continuity.
But this isn't too tough, just think of the graph of the function, if it's true for all rational numbers and the graph is continuous then we can deduce that it is true for all real numbers.[/quote]

who said the function was continuous? or monotonic?


hi,

im new here but am a keen mathematician, at least i think  :lol:
i am interested to see what say to these problems that are meant to be very challenging...

[b]Q1.[/b] find all pairs $x$ and $y$ that satisfy the equation
       $1/x - 1/y = 1/2005$

[b]Q2.[/b] for what values of $a$ there are exactly two real numbers that satisfy the equation
      $x^3 + (a-1)x^2 + (9-a)x - 9 = 0$
           Find all possible answers.

[b]Q3.[/b] Prove that the inquality
        $(a+7b)(4a + b) ≥ 3a^2 + 33ab + 2b^2$
           holds for all real numbers $a$ and $b$
(it is a "greater than and equal to" sign between $(4a+b)$ and $3a$) not sure if it will come up

and one more for good luck...

[b]Q4. [/b]When a positive integer $N$ is written in base $9$, it is a two-digit number. when $6N$ is written in base $7$, it is a three-digit number obtained from the two-digit number by writing digit $4$ to its right. Find the decimal representations of all such numbers $N$.

Hope you enjoy them... These questions are the questions we make in class and work on so i am currently on number 2  :)

Hope you enjoy maths as much as I do. Does anyone program? I program in BASIC and C++ and know a bit of Java.

[u]Anyways lets see how we all go  :D[/u]

Is your inequality:
\[(a+7b)(4a+b) \geq 3a^2+ 33ab+ 2b^2\]?

$(a+7b)(4a+b) \geq 3a^2+ 33ab+ 2b^2$
$\Longleftrightarrow 4a^2+29ab+7b^2 \geq 3a^2+33ab+2b^2$
$\Longleftrightarrow a^2-4ab+5b^2 \geq 0$
$\Longleftrightarrow (a-2b)^2+b^2 \geq 0$.

Equality occor when $a=b=0$.

For question 1 , $x$ and $y$ are natural number or all real number set ? :?

[quote="keshin"][b]Q1.[/b] find all pairs $x$ and $y$ that satisfy the equation
       $1/x - 1/y = 1/2005$[/quote]

[hide]I assume that $x, y$ are integers, since otherwise there are infinitely many solutions, and they can be expressed trivially.  To solve this, multiply both sides by $2005xy$, to get $xy + 2005x - 2005y = 0$, or $(x-2005)(y+2005) = -2005^2$.  From here, find all pairs of factors of $-2005^2$ (there are eight ordered pairs), and equate these pairs with $(x-2005,y+2005)$.[/hide]

[quote][b]Q2.[/b] for what values of $a$ there are exactly two real numbers that satisfy the equation
      $x^3 + (a-1)x^2 + (9-a)x - 9 = 0$
           Find all possible answers.[/quote]

[hide]If a polynomial equation has real coefficients, its complex roots come in pairs, namely, conjugates - there can't be an odd number of them.  So if exactly two numbers satisfy the cubic, either there is a double root (i.e. the LHS is equal to $(x-r_1)^2(x-r_2)$ for some $r_1, r_2$) or the third root is not real.

Consider the latter case.  If the equation has two real and one imaginary roots, it is of the form $(x-r_1)(x-r_2)(x-a-bi)$, $b \neq 0$, so when expanded, its constant term (the one that doesn't contain $x$) will be imaginary.  But we know that its constant term is -9.  So there are no solutions here.

So let us consider the former case.  Expanding $(x-r_1)^2(x-r_2)$ and equating its corresponding coefficients with $x^3 + (a-1)x^2 + (9-a)x - 9$ yields

\[a-1 = -r_2 \qquad (1)\]
\[9-a=r_1^2-2r_1=r_1(r_1-2) \qquad (2)\]
\[-9=2r_1r_2 - r_1^2r_2 = r_1r_2(2-r_1) \qquad (3)\]

Combine (1) and (2) to eliminate $a$; we get
\[r_2+8=r_1(r_1-2) \qquad (4)\]
Note that the RHS of (3) is just the RHS of (4) times $-r_2$.  So substitute:
\[-9 = -r_2(r_2+8).\]

Solve; $r_2 = -9$ or $1$.  Thus $a = 0$ or $10$.[/hide]

For #3, shobber is correct.  ≥ is rendered as a different symbol in LaTeX.  Use \ge instead.

ok looking good. thanks for the help :) any troubles you are having?

ta
Kesh

[quote="riddler"][quote="keshin"]

[b]Q1.[/b] find all pairs $x$ and $y$ that satisfy the equation
       $1/x - 1/y = 1/2005$
[/quote]
this question has appeared in one of the math forums.... I can't remember which though....[/quote]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44307]http://www.mathlinks.ro/Forum/viewtopic.php?t=44307[/url]

Now try proving that if $P(x)$ is a polynomial with rational coefficients and $P(z) = 0$, then $P(\bar{z}) = 0$.

Or better yet, if $z = a + b\sqrt{c}$ is a root of $P(x)$, then $z = a - b\sqrt{c}$ is also a root.

Q3.
[hide]
expand the left side:

4a^2 + 29ab + 7b^2 >= 3a^2 + 33ab + 2b
subtract it all to the left side

a^2 - 4ab + 5b^2 >= 0

(a-2b)^2 + b^2 >= 0

any real number squared is a non negative number, or any number 0 and greater. I am done[/hide]

[hide="Q4"]
Let $t_9$ and $u_9$ be the tens and units digits, respectively, for $N$ in base $9$.

\begin{eqnarray*}
9t_9+u_9 &=& N\\
49t_9+7u_9+4&=&6N \end{eqnarray*}

Substituting:

\begin{eqnarray*}
49t_9+7u_9+4 &=& 6(9t_9+u_9)\\
u_9+4 &=& 5t_9 \end{eqnarray*}

Since both of the digits have to be less than $7$ we have a very limited number of choices to check.  Obviously, for the LHS to be divisible by $5$, $u_9=\{1,6\}$, with corresponding values $t_9=\{1,2\}$.  Converting $11_9$ and $26_9$ into base $10$, we get the values of $N$ to be $\{10,24\}$.

Is that right?
[/hide]


$a_1,a_2,...$ is a strictly increasing sequence of natural numbers.
Prove, that there is such $n$, that:
$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\frac{a_3}{a_4}+...+\frac{a_n}{a_{n+1}}<n-2005$

Define $b_k = 1 - \frac{a_k}{a_{k+1}}$.  Then

$n - \sum_{i=1}^{n} b_i < n - 2005$, which we can rewrite
$\sum_{i=1}^{n} b_i > 2005$

If $a_k$ increases quickly (for example, if $a_k = 2^k$) then $b_k$ has large values and the inequality is easily shown (for $a_k = 2^k$ we get $b_k = \frac{1}{2}$ and $n = 4011$) so we must prove for the worst case, which is if $a_k$ increases as slowly as possible.  For a sequence of natural numbers, this means increments by one.

Specifically, $a_1 >> 1$ and $a_k = a_{k-1} + 1$.  Then $b_k = \frac{1}{a_1 + (k-1)}$, so $\sum^{\infty} b_i$ diverges because it is a harmonic series.  Q.E.D.