If you posted something and you don't like it and you just don't want to edit it, are you able to delete it by yourself or only a mod/admin can delete it?

If it is not the last post, then only mods or admins can delete your post.
If the last post happens to be first post, no - a mod or admin can.
If it is the last post, you can delete.

[quote="jmerry"]You can delete one of your posts if (a) it is a reply, not the first post of the thread, (b) it is the last post in the thread and nobody has replied to it, and (c) it has been less than a day, so you're still in the editing window. Other posts can only be deleted by moderators and administrators.[/quote]

How do you delete it? Do you edit then erase everything?

There's a specific delete button that shows up.


Let $ x,y,z \geq 0$ with $ x \plus{} y \plus{} z \equal{} 1$,prove
\[ x^ky^s \plus{} y^kz^s \plus{} z^kx^s \leq \frac {k^ks^s}{(k \plus{} s)^{k \plus{} s}}
\]
holds for $ k\ge{1},s\ge{1},(k,s)\neq(1,1).$


geometry
circumcircle
geometric transformation
reflection
geometry proposed

In the plane is given a circle, a point $A$ on it and a point $M$ in it. A chord $BC$ is passing through $M$. Prove that the circles passing through middle-points of all triangles $ABC$ is touch a fixed circle.

Since the rircumradius R of $\triangle ABC$ is fixed, the radius $\frac{R}{2}$ of its 9-point circle (N) is also fixed, hence we only have to show that the locus of N is a circle. Let $A_{1}, B_{1}, C_{1}$ be the midpoints of BC, CA, AB. Let $O_{1}$ be the circumcenter of $\triangle AC_{1}B_{1}$ and $M_{1}$ the intersection of AM with $C_{1}B_{1}.$ The triangles $AC_{1}B_{1}\sim \triangle ABC$ are centrally similar with similarity center A and similarity coefficient $\frac{1}{2}.$ Since O, M are fixed, $O_{1}, M_{1}$ are also fixed. Let $A_{1}$ be the midpoint of $C_{1}B_{1}.$ Since $O_{1}M_{1}$ is fixed and $\angle O_{1}A_{2}M_{1}= 90^\circ$ is right, the locus of $A_{2}$ is a circle with diameter $O_{1}M_{1}.$ Since N is a reflection of $O_{1}$ in $C_{1}B_{1},$ i.e., a reflection of $O_{1}$ in $A_{2},$ the locus of N is a circle with fixed center $M_{1}$ and fixed radius $M_{1}O_{1}.$


The equation $ x^2\minus{}kx\minus{}12\equal{}0$ has only integer solutions for certain positive integers $ k$. What is the sum of all such values of $ k$?

For this equation to have integer solutions, $ k$ must be equal to the following:

$ 1$

$ 4$

$ 11$

Thus, the answer is $ \boxed{3}$.

[quote="GameBot"]The equation $ x^2 \minus{} kx \minus{} 12 \equal{} 0$ has only integer solutions for certain positive integers $ k$. [b]What is the sum[/b] of all such values of $ k$?[/quote]

if the integers you put are correct, then our answer would be $ 1\plus{}4\plus{}11\equal{}16$

our answer is $ 16$

Yes my integers are correct.

$ x^2 \minus{} x \minus{} 12$ factors into $ (x \minus{} 4)(x \plus{} 3)$, which has integer solutions.

$ x^2 \minus{} 4x \minus{} 12$ factors into $ (x \minus{} 6)(x \plus{} 2)$, which has integer solutions.

And finally, $ x^2 \minus{} 11x \minus{} 12$ factors into $ (x \minus{} 12)(x \plus{} 1)$, which also has integer solutions.

So, as vallon said, the answer is $ 1 \plus{} 4 \plus{} 11$, or $ \boxed{16}$.


I don't know if this is the right forum to put this, but...

How do you make the cover art for the introduction and intermediate series?(especially intermediate counting and probability. I know that they don't make rubix cubes like THAT!)

Look at the beginning, they use KaleidoTile Software.
The Intro to Algebra book was designed by Vanessa Rusczyk.
:D


calculus
integration
inequalities
trigonometry
calculus computations

Prove:\[ \int_{0}^{\frac{\pi}2}x \left(\frac{\sin nx}{\sin x}\right)^4dx \le \frac{n^2 \pi^2}{8},n \in Z \]


How many subsets of the set $ \{1,2,3,4,\cdots,30\}$ have the property that the sum of the elements of the subset is greater than 232?

I'm trying to do this on my own, but I'm not having an incredible amount of luck with proceeding, so if anybody could offer the slightest hint, it would be appreciated.


A sculptor works with large wooden cuboids (rectangular prisms), cuts them into three smaller cuboids and then paints them black, white and grey. The volume of the black cuboid is always twice the volume of the white one and the volume of the white cuboid is always twice the volume of the grey one.
The diagram shows how this works. The cuts are always parallel to the face marked with w and h. We will always denote the surface areas of the separated cuboids by A1 (blue) , A2 (white) and A3 (grey).

[img]http://img127.imageshack.us/my.php?image=untitledea5.jpg[/img]

a) The sculptor is given a block for which l = 1, w = 6 and h = 3.
Show that with these values the ratio of A1 to A2 is 3:2
b) Show that for any cuboid the sculptor cuts, A1 is always less than 2A2
c) The sculptor wants a cuboid with h = w, which will give A1 to 3A3 . what would be the value of l in terms of w?
d) The sculptor decides she would like to have A1 to A3 = 4 : 3. If this is to be the case, find a formula linking l, w and h and give an example of the dimensions of a cuboid, before it is cut, for which this will work.

I've already solved a, which is easy.

For (B):
The black block has twice the volume of the white block, and we know that the smaller the size of an object, the greater the surface area to volume ratio. The result that $A1 \leq 2A2$ follows from this right?

I don't know how that works with this question because obviously the sculptor can cut A1 as much as he wants

$h$ and $w$ are constant for all three cuboids cut from the original cuboid?

Based on the text, I believe so I'm just wondering whether I could use volume of them but the text it says denote the surface areas A1, A2, A3 so does that mean I have to always use surface areas to work it out?

Don't worry about abc, I've solved them using only algebraic expression, I was heading a wrong direction when I realized how stupid I was but now I still have problem with d. Could someone help me out with this question?
I know that 3A1 = 4A3 but I just cant get the answer and the dimensions

This is how the cuboid looks like:
[url=http://imageshack.us][img]http://img114.imageshack.us/img114/3088/untitledea5pc3.jpg[/img][/url]


geometry
rectangle
parallelogram
analytic geometry
calculus
calculus computations

Using Green's Theorem, find the outward flux of F across the closed curve C.

F =(x2 + y2)i + (x - y)j; C is the rectangle with vertices at (0, 0), (2, 0), (2, 7), and (0,2).

Well Green's Theorem states
$ \oint_{C}\textbf{F}\cdot\ \text{d}\textbf{s}\equal{}\iint_{A}\left(\frac{\partial N}{\partial x}\minus{}\frac{\partial M}{\partial y}\right)\ \text{d}A$
so once you parametrize $ C$, which isn't hard since it's 4 straight lines, you can evaluate the line integral.

the point given or vertices given will not draw a rectangular shape unless the last vertice is (0,7)

Well then $ C$ isn't a rectangle.  It's still a parallelogram so it isn't hard to parametrize it.  It's still 4 straight lines.

do you think the question is correct because the answer given are not right it will not be a rectangle unless the last value is (0,7)

If the book says it's a rectangle, then the coordinates are wrong.  I would try both calculations and see which answer agrees with the one in the back of the book.


conics
ellipse
parabola
geometry
cyclic quadrilateral
geometry theorems

A.)Can quadrilateral ABCD be transform to cyclic ABCD by affine transformation?
B.)Can pentagon ABCDE be transform to regular pentagon ABCDE by affine transformation?

In general, pentagons cannot be transformed into regular pentagons (I'm not sure whether all convex pentagons can be transformed into cyclic pentagons), since affine transformations preserve parallelism.

Convex quadrilaterals can be, though (clearly if one point is inside the triangle formed by the other three, then we can't transform to a cyclic quadrilateral): let $X=AB\cap CD$; using directed segments, affine transformations preserve $k_1=\overline{XA}/\overline{XB}$ and $k_2=\overline{XC}/\overline{XD}$. But we can take $X,A,C$ wherever we want since they are not all collinear. In particular, by a suitable transformation we can set $\overline{X'A'}=\sqrt{|k_1|}$  and $\overline{X'C'}=\sqrt{|k_2|}$, whence $\overline{X'A'}\cdot\overline{X'B'}=\mathrm{sgn}(k_1)=\mathrm{sgn}(k_2)=\overline{X'C'}\cdot\overline{X'D'}$ (note that $k_1,k_2$ have the same sign by convexity).

Another note: it is not true that for every convex pentagon, there exists an affine transformation mapping it to a cyclic pentagon. We first observe that affine transformations map ellipses to ellipses. We next observe that any five points determine a conic section. Therefore, if we take 5 points on a parabola (e.g., $(0,0)$, $(1,1)$, $(-1, 1)$, $(2, 4)$, $(-2, 4)$), no affine transformation would be able to map that pentagon to an ellipse, (or else we may invert this transformation to map an ellipse to a parabola, which is absurd.)


In the diagram you must move from Start to End by moving up, down, or right to an adjacent box. You may not pass through a box more than once. Begin at 0 and $ \emph{subtract}$ the integer in the next box on the path. From that result, $ \emph{subtract}$ the integer in the next box on your path and so on, until you arrive at End. What is the value of the maximum result that can be obtained?
[asy]import olympiad; import graph; import geometry_dev; size(250); defaultpen(linewidth(0.8));
void drawRectangle(Label center, pair blCorner, real sidex, real sidey){
  draw((blCorner)--(blCorner + (sidex,0))--(blCorner + (sidex,sidey))--(blCorner + (0,sidey))--cycle);
  label(center,(blCorner + (sidex/2,sidey/2)));
}
  drawRectangle(Label("START"),origin,3,1);
  drawRectangle(Label("0"),(3,0),1,1);
  drawRectangle(Label("$-2$"),(4,0.5),1,1); drawRectangle(Label("1"),(4,-0.5),1,1);
  drawRectangle(Label("0"),(5,0),1,1); drawRectangle(Label("6"),(5,1),1,1); drawRectangle(Label("$-5$"),(5,-1),1,1);
  drawRectangle(Label("3"),(6,0.5),1,1); drawRectangle("4",(6,-0.5),1,1); drawRectangle("$-3$",(7,0),1,1);
  drawRectangle("$-1$",(8,0.5),1,1); drawRectangle("$-6$",(8,-0.5),1,1); drawRectangle("END",(9,0),3,1);[/asy]

At each step, you want to subtract the smallest integer possible (unless there is something better, which doesn't occur in this problem). So our answer is $ 0 \minus{} ( \minus{} 2) \minus{} 0 \minus{} ( \minus{} 5) \minus{} 4 \minus{} ( \minus{} 3) \minus{} ( \minus{} 6)\minus{}(\minus{}1) \equal{} \boxed{13}$.


Three coins are tossed. What is the probability of getting 2 heads and 1 tail? Express your answer as a common fraction.

There are $ 2^3$ ways three coins can be tossed. There are $ 3$ ways to get 2 heads and 1 tail. (You can get the tail on the first, second, or third flip.) The answer is thus $ \frac{3}{8}$.

Assuming that the coin is unbiased, then the probability of getting heads or tails is $ \dfrac{1}{2}$.

Now we can use Binomial Distribution.

$ P(2H,1T) \equal{} \binom{3}{2}\left( {\frac{1}{2}} \right)^2\left({\frac{1}{2}}\right)^1 \equal{} \dfrac{3}{8}$

another way is to use complimentary counting.

There are only 4 different outcomes when you flip 3 coins;
all 3 heads, 1 head and 2 tails, 2 heads and 1 tail, and all tails.

First, the chances of getting a coin flip is $\frac{8}{8}$, or $1$, the basic amount.
Then, the chances of getting all heads or all tails is $\frac{1}{8}$, getting all heads and tails is $2 * \frac{1}{8} = \frac{1}{4}$.
Finally, we don't know the chances of getting 1 head and 2 tails or 2 heads and 1 tail, but the probability of getting them are equal, so we denote it as $2x$ (2 for both outcomes)

complimentary counting: $1 - \frac{1}{4} - 2x = 0$

we get $2x = \frac{3}{4}$
and therefore the answer (x) is $\boxed{\frac{3}{8}}$.


function
number theory unsolved
number theory

[quote="jbmorgan"]Find all functions $ f: \mathbb{N}\rightarrow\mathbb{N}$ such that

$ 2(f(m^2 \plus{} n^2))^3 \equal{} f^2(m)f(n) \plus{} f^2(n)f(m)$ for all distinct natural numbers $ m,n$[/quote]
$ f(n)\equal{}c$,where $ c$ is a constant number

Can you prove that there are no other solutions?


Inam asamie bachehaye teame iran too IMO 2008 ke too spain bargozar mishe :
$ 1)$ Kasra Ahmadi (Allame Hellie Tehran)
$ 2)$ Milad Bakhshizadeh (Allame Tabatabayie Tehran)
$ 3)$ Mohammad Jahangoshayi (Shahid Soltanie Karaj)
$ 4)$ Nima Hamidi (Shahid Soltanie Karaj)
$ 5)$ Amir Sepehri (Shahid Beheshtie Zanjan)
$ 6)$ Mehdi Yazdi (Shahid Sadooghie Yazd)

Omidvaram emsal too IMO teame khoobi dashte bashim :)


Jane has four decks of cards, red, blue, green, and yellow. Each deck has 15 cards, numbered 1-15. Jane picks two cards without replacement. What is the probability that the first card that Jane picks is a red card with a prime number and the second card is a non-prime? Express your answer as a common fraction.

[hide]
There are 60 cards.

For the first card, there are 6 primes that are red.
Thus a $\frac{6}{60}=\frac{1}{10}$ chance.

For the second card, there are 23 primes left, and so there are 36 non primes.
Thus a $\frac{36}{59}$ chance.

$\frac{1}{10}\cdot\frac{36}{59}=\frac{18}{295}$[/hide]

LOL. I got to this problem when doing it with like 30 seconds left. I made a stupid mistake and missed it. But honestly, this must be like the easiest #26 in nationals history or something...

[hide]First portion:

There are 6 prime numbers that are red out of 60 total cards. The probability of the first thing coming true is ${1}/{10}$

Second portion:

There are 9 total primes per color, for a total of 36 primes. There are 59 cards to choose a card from (without replacement). So the probability of the second thing coming true is ${36}/{59}$

Multiply the two portions to get ${18}/{295}$[/hide]

Lesson: Do not skip the first words of a problem or you'll be forever confused. I didn't see the part that said 4 decks of cards.

Maybe states/nats will be this easy this year. :P

You have 6/60 chance for the first one.
A 36/59 chance for the second one. Multiply those together and you get....[hide]36/590. [/hide]


calculus
integration
function
vector
real analysis
topology
Functional Analysis

if f: [a,b]-> X (some banach space) is riemann integrable, does that mean that || f ||_x is riemann integrable as well?

$f$ is Riemann integrable on $[a,b]$ iff for any $\epsilon>0$ the interval $[a,b]$ can be partitioned into subintervals $I_{k}$ such that $\sum_{k}\mathrm{osc}_{I_{k}}(f)|I_{k}|<\epsilon$. (Here $\mathrm{osc}_{I}(f)=\sup_{x,y\in I}\parallel f(x)-f(y)\parallel$. Post-composition of $f$ with any Lipschitz map preserves this property.

[i]Most pointless remark of the week:[/i] the notion "Riemann integrable" makes sense for maps into an abstract metric space, when no reasonable definition of "Riemann integral" is available.

I'm not that sure: unlike $\mathbb R$, in infinite-dimensional Banach spaces you can have a limit of Riemann sums with large sums of oscillations times lengths. Let, for instance, $X$ be a Hilbert space with an orthonormal basis $\{e_{t}\}_{t\in\mathbb R}$ and let $f(t)=e_{t}$. Then the norm of each Riemann sum for the interval $[0,1]$ and any partition does not exceed the square root of the length of the longest interval in the partition, so this terrible function is Riemann integrable to $0$ if one understands Riemann integrability of a vector-valued function as existence of a limit of Riemann sums. This conclusion will remain true if you change $e_{t}$ to $2e_{t}$ here or there but in this way you can get $\|f\|$ even not Lebesgue measurable! I should confess that I'm not completely sure what is the commonly accepted definition of Riemann integrability of vector-valued functions, i.e., whether it is just existence of a limit of Riemann sums or the oscillation condition cited by mlok. My point was only that those two are no longer equivalent.

Now I think it's the existence of Riemann sums... Indeed, [url=http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:0764.28008&format=complete]this paper[/url] concerns, among other things, the spaces $X$ such that every $X$-valued Riemann integrable function is a.e. continuous. There is also [url=http://www.worldscibooks.com/mathematics/5905.html]a recent book[/url] on Riemann-style integrals of Banach space-valued functions.

thanks guys. it was recently suggested to me that a function such as this maybe a suitable counterexample:
$f: [0, 1] \rightarrow \ell^{2}$ such that $\int_{0}^{1}f = 0$ but $\parallel f \parallel = \chi_{Q}$ and therefore not riemann integrable.

This works
$\{r_{n}\}_{n=1}^\infty = \mathbb{Q}\cap [0, 1]$ and $\{e_{n}\}_{n=1}^\infty$ a collection of "orthonormal basis" of infinite dimension. i.e. $e_{1}= (1, 0, 0, ...), e_{2}= (0, 1, 0, 0, ...)$

$f : [0, 1] \rightarrow \ell^{2}$ such that, $f(x = r_{n}) = e_{n}$ and$f(x = otherwise) = 0$


$\log_{2} 64 = x$ $\Longrightarrow$ $2^x = 2^6$ $\Longrightarrow$ $x = 6$.

You just guessed that how did you get it?

[hide]We have $\log_{2}64=x \Longrightarrow 2^x=64 \Longrightarrow 2^6=64, x=\boxed{\boxed{6}}$[/hide]

As shown by Andreas and ragnarok23, when you get to the step $2^x=64$, you're basically saying "2 to the what power equals 64?" Since $2^6=64$, 6 is the answer.

Hes writing $64$ as $2^y$. So then you can equate the exponents.

lol its the way the aops volume 1 book teaches it  :P

[quote="lillyiggie"]Log base 2 (64) = x
So thats 2 to the x = log 64? but what does x equal[/quote]

[hide]$log_{2} 64=x \Longrightarrow 2^x=64 \Longrightarrow x=6$[/hide]

[quote="math92"][quote="Andreas"][quote="José"]$log_{2} 64=x \Longrightarrow 2^x=64 \Longrightarrow x=6$[/quote]
 :spam:[/quote]
if he copied your answer then its spam, if he came up with that on his own it isn't spam.[/quote]

I did it by my own


A= [6  4]
     [-2 -1]

B=[2  -3]
     [4   4]

tr((AB)^-T)= ?  Can you give any hints of what to do with the negative transpose[/list]

That notation makes no sense to me. No one is interested in anything called a "negative transpose". On the other hand the inverse of a transpose - which is also the transpose of an inverse - is often an interesting item. Do you perhaps mean $ ((AB)^{\minus{}1})^T,$ which is the same as $ ((AB)^T)^{\minus{}1}?$

Also, that "tr" you have in front. That would usually mean the trace of a matrix. Is that what you intend?

well yes the tr stands for trace of a matrix ive been trying to do this matrix and this is what a did I multiply A.B and then do the inverse of AB is that corrrect?  After doing that I did the transpose right?

((AB))= [28   -2]
             [-8    2]

((AB^-1))=[7/10    -1/20]
                 [-1/5      1/20]

is this correct?


A and B can build a wall in 24 hrs. After A worked 7 hrs, B helped A and both together finished the rest of the work in 20 hours. How long does it takes to each one to make the work?

[quote="lingomaniac88"]Let $a$ be the amount of time in hours it takes A to build the wall and $b$ be the amount of time in hours it takes B to build the wall.

So in 1 hour, A can do $\frac{1}{a}$ of it, and in 1 hour, B can do $\frac{1}{b}$ of it.  I hope this helps you get started.[/quote]

using this you get:
[hide]
$\frac{1}{a}+\frac{1}{b}= \frac{1}{24}$
since they worked together for $20$ hrs, they worked on
$\frac{20}{24}$ or $\frac{5}{6}$ of the job.
that means $a$ did $\frac{1}{6}$ in $7$ hrs, or $\frac{1}{42}$
in one hour. thus $a$ can do the job in 42 hrs.
use $\frac{1}{a}+\frac{1}{b}= \frac{1}{24}$ to solve for
$\frac{1}{b}$ which is $\frac{1}{56}$ so $b$ does the
job in 56 hrs
[/hide]

[quote="Interval"]Okay...excellent replies.  I did not know that a fractional equation set up was needed to answer this question.

Is there a formula or equation for questions dealing with working alone and working together?[/quote]
Not really. You just need to know when to use these kinds of fractions in these kinds of problems. Try this. A pipe can drain a tank in 8 hours. Another pipe can fill the same tank in 6 hours. John who was watching the tank accidentally left both pipes on. How many hours will it take for the tank to be filled with water?


inequalities
geometry
trigonometry
geometry proposed

(A.Abdullayev, 9--11) Prove that the triangle having sides $ a$, $ b$, $ c$ and area $ S$ satisfies the inequality
\[ a^2\plus{}b^2\plus{}c^2\minus{}\frac12(|a\minus{}b|\plus{}|b\minus{}c|\plus{}|c\minus{}a|)^2\geq 4\sqrt3 S.\]

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=199673 (with not quite the nicest solution).

  dg

Let $ a \leq b \leq c$.  Hence, we need to show that
\[ b^2 \minus{} a^2 \minus{} c^2 \plus{} 4ac \geq 4\sqrt {3}S\,.\]
However, since $ 2ac\cos(B) \equal{} a^2 \plus{} c^2 \minus{} b^2$ and $ 2S \equal{} ac\sin(B)$, the inequality is equivalent to
\[ ac\big(2 \minus{} \cos(B)\big) \geq \sqrt {3}ac\sin(B)\,.\]
Thus, it suffices to show
\[ \cos(B) \plus{} \sqrt {3}\sin(B) \leq 2\,,\]
but this can be done using Cauchy-Schwarz Ineq, or expanding $ \cos\left(B \minus{} 60^\circ\right)$.  In addition, the equality holds iff the second largest angle is equal to $ 60^\circ$, which, in this case, is $ \angle{B}$.


Can someone move all of this guy's problems to a middle school forum?


I know this might not be the right place to ask for help on computer, but it is related to USAMTS. My story is here: half an hour ago I tried to transfer my Round 4 solutions to my new computer, but for some freak reasons all the related files disappeared in both my old and new computers. Afterwards I searched through the disks and found nothing, and later discovered that my recycle bin setting was "Always Delete Completely". So it's very likely that I deleted my files accidentally and, due to my dumb automatical setting, permanently. I don't really want to re-type everything again since I'm so poor at LaTeX. If anyone can tell me how to recover files deleted completely like this, I would greatly appreciate. Thanks in advance.

I know there's some kind of computer program that can recover files that have been "permanently" deleted, but since it costs money, it might not be the best option.

You could go to http://www.download.com, search on "lost files", and download a file-recovery program for a free trial. That trial should last long enough to recover the accidentally erased files.

However, there is a risky paradox involved. Any downloaded file is written to supposedly empty space on your disk, such as the space that contains your lost file. If you download a file-recovery program, it could overwrite the lost file that you want it to recover. It is a risk you have to take, unless you can download it onto your new computer and use it to remotely scan your old computer.

Erin Schram

To both of you, thanks for your helpful response. I've just re-typed my solutions for this round - it's not really a good experience - since it's my only choice. Some of my friends also introduced to me the software you mentioned, but considering its cost and my inability to use these modern technology, I decided to write it again... :huh: hopefully I will not make this mistake again. Anyways thank you very much. :)

For future reference:

[quote="ZzZzZzZzZzZz"]...but since it costs money, it might not be the best option.[/quote]

That's not true: there are several free file-recovery programs. You can find some browsing the Internet for a while. Scan the program for viruses to avoid running a fake spyware program. To avoid (or at least minimize) writing to the disk you need to recover from, install the program to a flash disk.


Veronica High School has 85 seniors, each of whom plays on at least one of the school's three varsity sports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 are on teh baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball and football teams; and 13 are on both the baseball and lacrosse teams. Compute the number of seniors playing all three sports given that twice this number are the members of the lacrosse team.

[hide]
I get 11.
I think the formula for this is a+b+c-ab-ac-bc+abc, where ab is someone doing first two, etc.
74+26+2x-18-17-13+x=85
100+3x-48=85
52+3x=85
3x=33
x=11
[/hide]


quadratics
number theory
greatest common divisor
relatively prime

Please write your solution and as the instructions say if necessary to "prove" your solutions:
Thanks in Advance  :D

heres the link for the problems, please note that you need acrobat or some form of pdf. reader to view the file
[url]http://www.math.ksu.edu/%7Esoibel/2002-7-8.pdf[/url]

in my opinion this should be moved down to intermediate
[hide="1"]
Assume we can find such squares.
$ 4m \plus{} 3 \equal{} x^2 \plus{} y^2$
One of the ones on the right must be even, the other odd.  WLOG x is even.  x=2a.
$ 4m \plus{} 3 \equal{} 4a^2 \plus{} y^2$
Rearrange.
$ 4(a^2 \minus{} m) \equal{} 3 \minus{} y^2$
Consider the equation mod 4.
$ 0\equiv 3 \minus{} y^2 \mod 4$
But quadratic residue mod 4 is always 0 or 1, so this equation has no solutions.

[/hide]

For 2, couldn't you theoretically choose all 1001 even numbers, of which no 2 are relatively prime?

[hide="3"]
$ \frac {(100!)^{50}}{\binom{100}{1}\binom{100}{2}\cdots}$
probably unhelpful, but it looks like it might go somewhere
[/hide]

[hide="4"]Place four points such that it forms a square with side length 1. Then place a point over any one side (outside the square) that it forms an equilateral triangle of side 1 with two of the square's vertices. The final point goes 1 unit "below" it in the square, or such that it is "above" the opposite side of the square and forms an equilateral triangle.
Good job if you actually understand that.[/hide]

That doesn't work for question 4, since the point above is only next to 2 other points.

[hide="question 1"]

$ k^2$ and $ n^2$ are congruent to either 0 or 1 mod 4.

Therefore, $ k^2 \plus{} n^2$ is congruent to 0,1, or to mod 4, for all integers k and m.

But 4m+3 is congruent to 3 mod 4, so it can't be of the form k^2+m^2.

[/hide]

[hide=" question 2"]

This question is simply wrong, just take 2,4,6,8,...,2002.

The question is really supposed to be 1002 numbers

In this case, pair off the numbers (1,2) (3,4) (5,6) ... (2001,2002)

There are 1001 pairs, so 2 numbers lie in the same pair by pigeonhole. 2 consecutive integers have gcd 1, since

(k+1)-k=1.

Therefore, since there is at least 2 numbers which make such a pair, we are done, since their gcd is 1.

[/hide]

[hide="question 3"]

$ 1!2!3!...49!51!52!...100! \equal{} (100^{1}*99^{2}*98^{3}*...*51^{50})*(49^{51}*48^{52}*...*1^{99})$

we may remove any terms with even powers, since they dont affect whether the number is a square or not.

$ (100*98*96*...*52)*(49*47*...*1) \equal{} 2^{25}*(50*49*...*26)*(49*47*...*1)$

removing all of the doubled terms again

$ 2^{25}*(50*48*...*26)*(25*23*...*1) \equal{} 2^{25}*2^{13}*(25*24*23*...*13)*(25*23*...*1)$

removing all of the doubled terms again

$ 2^{(25 \plus{} 13)}*(24*22*20*18*16*14)*(11*9*7*5*3*1) \equal{} 2^{38}*(24*22*20*18*16*14)*(11*9*7*5*3*1)$

removing the 2^ term, which is a square,

$ (24*22*20*18*16*14)*(11*9*7*5*3*1) \equal{} 2^{6}*(12*11*10*9*8*7)*(11*9*7*5*3*1)$

removing all of the doubled terms again, and the 2^ term,

$ (12*10*8)*(5*3*1) \equal{} 2*2*3*5*2*2*2*2*5*3*1 \equal{} 2^{6}*3^{2}*5^{2}$

which is a perfect square.

[/hide]

[hide="question 4"]
take an equilateral triangle with side length one. There are 3 points.

Take another equilateral triangle that's exactly one cm higher than the first one, with sidelength one, and the same orientation. Details are easy to fill out from here.

[/hide]

sorry but i could'nt understand number 3, can someone please simplify the wording or is it just my lack of understanding into what is happening, I could'nt read much more than into the 2nd line until i started getting confused.

[quote="rofler"]That doesn't work for question 4, since the point above is only next to 2 other points.[/quote]
I'm pretty sure it touches three points.... Actually, my configuration is exactly what you did, except worded differently. (And horribly.)


So my competition is on the 21st, today is the 14th, (I live in Washington state btw) and I want to know if you guys know of any formulas or math tricks that might help out there, like SFFT, Balls and Urns,etc.

Edit: Oh, and what score would I need on written to get into Countdown (top 10) and what score to get top 4

Wow.. Washington? That's where Darryl Wu lives!! I think...  :maybe:  But your score would really have to be at least 43+ or something. But at least you have countdown. I can't really help you with tricks, because you probably know more than me.   :blush:

43? Unless this year is easier then normal, you definitely do not need to score that high. Darryl Wu will probably get like perfect but their are still 3 spot open. Just practice a lot of problems and you will naturally learn tricks. I learned balls and urns on my own.

Would a score of mid to upper 30s get me to countdown?

I can usually do good at Countdown, I got 6th or 7 seat and beat everyone up to 2nd seat at chapter.[/quote]

hmm...probably don't need 43, mid-upper 30s should get u in cd but u have to get top 4 through the countdown unless u get 40+

and unless u were in darryl wu's chapter u know u're going to have an extra guy in front of u...so only 2 spots :P

I'm sorry. I meant top 4. But there should be some REALLY good people in Washington. Don't worry Romeo. I bet your one of them!  :wink:  But just listen to the scores other people tell you. I probably don't have as much experience  because I am a 7th grader. :(


geometry
perimeter
geometric transformation
rotation

3)  In 1736, the cousin of the famous Gabriel Fahrenehit, Otto Fahrenheit, invented his angle measuring scale.  No rotation was 32 and half a rotation was 212.  When Otto Fahrenheit measured the angles of a hexagon with his not-so-famous Fahrenheit protractor and then added them what was the sum of the angles?

4)  The net (2-d drawing of a 3-d shape) of a Green house is given.  The 20 by 20 square is the floor of the building.  IF the air system circulates 8 cubic feet per second, how many minutes will it take to circulate the entire volume of the building (answer to the nearest minute) (pic below)

[hide]Answers:
According to the answer key...here they are...but I don't think they are correct

1) 7 $\frac{1}{24}$
2) 1080
3) 912
4) 29
[/hide]

[quote="ckck"][quote="Walk Around The River"][quote="ckck"][hide]use the formula A=rs. Since its an isosceles triangle, we have a 5-12-13 triangle. So the height is 12. Quickly finding the area of the triangle which is 60, and the semi-perimeter is 18, we can find the radius. The radius is $\frac{10}{3}$.[/hide][/quote]

no, you found the radius of an inscribed circle.

[hide]using the height of the triangle being 12, as ckck correctly explained, we can do one of two things.

1. use the formula for radius of a circle circumscibed around a triangle with sides a, b, and c.

$r=\frac{abc}{4(area)}$
so $\frac{13\cdot13\cdot10}{4\cdot60}=\boxed{\frac{169}{24}}$

or 2. we can be creative.

draw a radius from the center of the circle, which lies on the altitude of the triangle to one of the base angles.  We then have a right triangle with legs 5, 12-r, and r.  Using pythagorean, we gets $169-24r+r^{2}=r^{2}$

So 169=24r and $r=\boxed{\frac{169}{24}}$

EDIT: yah, as the answer key says $\frac{169}{24}=7\frac{1}{24}$[/hide][/quote]
Oops, my bad. I thought the circle was inscribed. Well, I was too tired to see it. Oh well...[/quote]

Without looking at the diagram you knew there was a 5-12-13 triangle inside a circle? Wow!

Ah, thanks Walk Around the River!  I'm still confused on the other questions though  :huh:

[quote="Phoenix257"]1)  What is the radius of the circle? (pic below)

2)  HEPTAGO is a star polygon with its 7 vertices equally spaced around a circle.  If Sam walks along the sides of the star polygon and turns at each vertex, how many degrees will she turn in coming back to her original position? (pic below)[/quote]
[hide="1, a lower tech method"]Ok, using pythag the altitude has length 12.
Now draw the radius from circles center to any of the vertices.
Call this length x.
$(12-x)^{2}+25=x^{2}$
$144-24x+25=0$
$169=24x$
$x=\frac{169}{24}$[/hide]

Can anyone get the second question ? (the star one)

[hide="#2"]Since all vertices are located on the circle and are equally spaced, each "outside" angle must be congruent to one another. Therefore, if we were to find the measure of one angle, suppose $PTA$, we could find the total number of degrees she turns by multiplying by seven. We notice that since P, A, and T are all on the circle, the angle $PTA$ is $\frac{1}{2}$ of the central arc $PA$. The measure of this arc is $\frac{360}{7}$. Therefore the angle $PTA=\frac{360}{14}$ and the total [b]$\frac{360}{7}$[/b].[/hide]

I was thinking that too but somehow the answer is [hide]1080[/hide] :huh:

EDIT: i didn't read the question right :oops:
i think you can solve it by assuming that you get a regular heptagon in the middle, which is a perfectly okay assumption

[hide]A possible solution to the second question...

Mark the intersection of GA and HE as $X$.  The angles that are like $\angle GXH$ are equal to $\frac{arc GH+arc EA}{2}= \frac{360}{7}\cdot \frac{3}{2}$.  There are seven of these angles, so they add up to 540 degrees.  Angles like $\angle PTA$ are half arc $PA = \frac{360}{7 \cdot 2}$.  There are 7 of these angles so they add up to 180 degrees.  Together, that's $540+180 = 720$.  Maybe the answer key is wrong?

[/hide]


If $ a,b,c$ are nonnegative real numbers, then

$ 2\sum a^6\plus{}16\sum a^3b^3 \ge 9\sum a^2b^2(a^2\plus{}b^2)$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ 2\sum a^6 \plus{} 16\sum a^3b^3 \ge 9\sum a^2b^2(a^2 \plus{} b^2)$.[/quote]
We have
\[ a^6\plus{}b^6\plus{}16a^3b^3\minus{}9a^2b^2(a^2\plus{}b^2) \equal{}(a^3\minus{}b^3)^2 \minus{}9a^2b^2(a\minus{}b)^2 \equal{}(a\minus{}b)^2[(a^2\plus{}ab\plus{}b^2)^2\minus{}9a^2b^2] \ge (a\minus{}b)^2[9\cdot a^2b^2\minus{}9a^2b^2]\equal{}0\]
Hence
\[ 2\sum a^6\plus{}16\sum a^3b^3\minus{}9\sum a^2b^2(a^2\plus{}b^2) \equal{}\sum [a^6\plus{}b^6\plus{}16a^3b^3\minus{}9a^2b^2(a^2\plus{}b^2)] \ge 0\]
We have done. :)

A stronger inequality is the following:

$ \sum a^6 \plus{} 8\sum a^3b^3\plus{}2abc(\sum a)(\sum ab) \ge 5\sum a^2b^2(a^2 \plus{} b^2)$.

[quote="Vasc"]A stronger inequality is the following:

$ \sum a^6 \plus{} 8\sum a^3b^3 \plus{} 2abc(\sum a)(\sum ab) \ge 5\sum a^2b^2(a^2 \plus{} b^2)$.[/quote]

Is there any mistake, Vasc.

Of course. This is corrected statement.

A stronger inequality is the following:

$ \sum a^6 \plus{} 8\sum a^3b^3 \plus{} 2abc(\sum a)(\sum ab) \ge 5(\sum a^2)(\sum a^2b^2)$.


geometry
geometric transformation
reflection
calculus
trigonometry
Pythagorean Theorem

A man is 6 miles east and 5 miles south of his home. He is also 3 miles north of a river, which is 8 miles south of his home. What's the least amount of miles he may travel to fetch water from the river and then return home?

[hide]
Distance from him and river : 3mi
Distance from river and home : 10mi by pythagorean theorem
Total distance : 13mi
[/hide]

[quote="mathgeek2006"][hide]12.58529852..........[/hide][/quote]

How did you get that?

[hide] the answer is ten (there is a really special way to do this involving using the river as a mirror[/hide]

Good job you guys, a lot of critical thinking involved. I'll post the correct answer tomorrow...  :)

This is a good one ,I tried using a lot of laws of reflection [hide]
12.77 miles[/hide]

[hide]
I got 12.529964...

River is a mirror, reflect home, make right triangle with legs 6 and 11. Find hypotenuse.
[/hide]

[hide]First he travels 3 miles south to the river. then he goes along the hypotenuse of the triangle (with sides 6 and 8) which is 10 miles. so he travels 13 miles.[/hide]

does someone want to give in the final answer

Hey everybody:

 Using Calculus I got an answer of 12.53 miles.
 Hamster1800 had the best solution.

Calculus has no place in this forum and is utterly unnecessary to solve this problem.

If you post an answer, please explain how you got your answer.  This thread contains many guesses -- mostly wrong.  But without explanations, nobody can help you correct your mistakes and nobody can learn from what you did right.  Please post explanations, not just answers!

The man must travel 6 miles west to get home.  He must travel 3 miles south and then 8 miles north to get home.  Using the river as a mirror, the man's minimum travel distance is the hypotenuse of a right triangle with legs of 6 and 11.

So, assuming the man can access the river (flowing east-west) at any point along its length, the minimum distance is $\sqrt{6^2+11^2}=\sqrt{157}\approx 12.53$.

I got an answer of thirteen because he has to go 3 miles down and then then travel across the hypothenuses which is ten.

People are trying to minimize distance to the river, which is incorrect. While this is not terrible, it is not the minimum SUM of the distanced from him to the river to the house. What we want, is for him to hit the river at the angle, so that when he leaves the river, he goes at the same angle, but on the other sides. some higher level math is nice here to show why it works to reflect the house, but let's look at it like this.

When you reflect the house over the river, it moves down and you can create a line to the contact point on the river. Now, instead of being 8 miles north of the river, the house is 8 miles south of the river. Obviously, any point of contact on the river will be "on the way" to his house, so we want the shortest distance: a straight line. This straight line will intersect the river. This point of intersection is where he should get water from the river.

Does any one want to give in an answer or Am i correct at 12.7

Yes, the answer is approximately 12.53. Can calculus be used to solve any problem? Just curious.

[quote="nat mc"]It has already been given.  You just put the guy beneath the river, and draw a straight line to his house.  Legs of 6 and 11, make the hypotenuse and the answer $\sqrt{6^2+11^2}=\sqrt{157}\approx12.53$[/quote]

Oh that explains it very well.  :)  :10:

4everwise -- Calculus can't be used to solve [i]any[/i] problem, but it is powerful when dealing with areas under curves, rates, and particularly in this kind of problem: optimization (that is, maximizing or minimizing quantities).

Most math competition students I know despise calculus simply because it is less elegant or insightful than a wiser approach with more basic math concepts.  As seen here, the reflection using the river as a mirror is an excellent approach, and calculus (while giving the same answer as this approach), is just brute force.  A computer could do the calculus approach, but could not conceive of the mirror idea by itself.

Hi everyone:

 The only reason I used calculus was this: I believe that the best way to check an answer to a problem is by solving it a different way. I'm a strong believer in being able to solve a problem with more than one solution. I agree (and I mentioned this in my last post on this topic) that Hamster1800 had the best solution. It was creative
yet simple to understand. I only used calculus as a check.

I completely agree with you!  I'm not bashing your solution  :) .  The spirit of checking a problem multiple ways is good, but I dont want people who haven't heard of calculus to start giving up on optimization problems and what not... and also not to assume that just because a problem seems like it requires calculus that calculus is the only way to solve it.

so many interesting examples of Archimedes solving areas under curves exactly without using calculus..

Let's not curve the topic into something several years ahead of this forum!  :D
I have never learned calculus, neither precalculus. I should start studying that  :lol:

Don't you just hate it when a problem goes into triginometry and calculus stuff? I always get lost. :(

Anyway, shouldn't the answer be 13 how ccy did it? I read the explanation but I don't quite understand what you mean by mirror and a load of other things. Please explain.

I'm learning pre-calculus next school year and didn't know what it was for until now.  :(

Anyway, let's get this thread back on track.


[quote="hash_include"]hey all.
i jus wrote FT1 today at home and wanted to know how every1 else has done so i can check wat a "decent" mark is.

So ppl, plz post ur marks if youve written it too!

I got 215 (120 + 95)

FYI it was out of 522 (270 + 252) for all those who've not written it. :)[/quote]

Thala!!

Well, not sure about these

Chappli - 252
Prasad - 246
Shreyas - 557

No da ,thats my no of posts.

haven't written it ...not even thinking of writing it :P

rock got 162 in paper 1 and it seems 118 in paper2 making it 280!!!  :D
Me too not written!  :D

Havent written it yet. Was in the process of cleaning my room and now all my FIITJEE papers are missing. Must be the work of a over neatness concious mother of mine :D

But hell, my room actually looks organized...

I HATE IT

P.S - Sorry about the rant, but yeah, will post here once I take the test

I got 256, Prasad 242.. You swapped the units digit ,shre...

Yes I got 161 in paper 1 and 120 in paper 2 so totally 281:
Ofcourse central marks in all of them and felt chemistry was easy out of the three :D  :D

Chemistry:108
Physics:86
Maths:87
Central marks in all of them :D  :D  :D

I thought paper-2 was really good and I spent 1hr and 15mins. in that paper for maths and still ended up getting 33 which was awful and then rolled over chemistry da. :D


A fair coin and a 6-sided number cube with numbers 1 through 6 on its faces are simultaneously tossed. What is the probability that a ``head'' and an even number appear, respectively, on the top faces of the coin and the cube?

Each of these independent events comes up with $ \frac{1}{2}$ probability, so the answer is with multiplication.

$ \frac{1}{2} \cdot \frac{1}{2} \equal{} \frac{1}{4}$


Yes, I like that one. Please come one and send me your results.

Yes, It is my result when I solve the ChinaTST 2005. I'm really like the not homogeneous inequalities. i'll sent you the tex file next day about my work. I hope you'll enjoy it!

[quote="The soul of rock"]Sent to pvthuan, I hope you'll like this
Prove that for all x,y >0, we always have
$\frac{4}{(x+1)^{4}}+\frac{4}{(y+1)^{4}}+\frac{x^{2}y^{2}}{x^{2}y^{2}+1}\ge 1$[/quote]
It's true for all reals $x\neq-1$ and $y\neq-1$. :wink:

It's just a corollary of the known inequality: If $a$ and $b$ are positive real numbers, then \[\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2} \ge \frac{1}{1+ab}.\]

[quote="TheIronChancellor"]And how can you deal with the last one?[/quote]
With the most natural way:
$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}-\frac{1}{1+ab}=\frac{ab(a-b)^2+(ab-1)^2}{(a+1)^2(b+1)^2(ab+1)}$.


Consider a standard twelve-hour clock whose hour and minute hands move continuously. Let $m$ be an integer, with $1 \leq m \leq 720$. At precisely $m$ minutes after 12:00, the angle made by the hour hand and minute hand is exactly $1^\circ$.
Determine all possible values of $m$.

[hide]The hour hand moves 1/12th the speed of the minute hand.  $1/6+5x$ where x is the hour 12:00=0 1:00=1,...  would form one degree with the hour hand if the hour hand did not move.  However, the hour hand does so an infinite series is set up.  Everytime the minute hand catches up to form one degree if the hour didn't move, the hour hand moves at 1/12th the minute hand's speed and prevents it.  This means that the expression above is multiplied by $12/11$.  $2/11+60/11x$.  60 is 5 mod 11, so x=4 works.  The resulting term means that at 22 minutes past 4:00, the minute hand and hour hand forms a one degree angle.  That's 262 minutes after 12:00.[/hide]

[hide="solution"]consider the angles starting at 12:00, going clockwise, after $m$ minutes, the minute hand will make an angle of $6m$, and the hour hand will make an angle of $\frac{m}{2}$, we then have that $|6m-\frac{m}{2}|=\pm 1+360k$ ${\{m,k}\} \subset \mathbb{Z}$
\begin{eqnarray} 11m=\pm 2+720k\\ 11m=\pm 2+5k+11*65*k\\ 11(m-65k)=\pm 2+5k \end{eqnarray}
we get the solutions: $k=4$, and $m-65k=2$ or $\boxed{m=262}$ or
$k=7$, $m-65k=3$ then $\boxed{m=458}$
by CRT these are the unique solutions mod 55 (the numbers being k and m-65k) if we add 55, the minutes are not in the given range so that is all the solutions[/hide]


Evaluate :
\[ S_n \equal{} \arctan{\frac {1}{2}} \plus{} \arctan{\frac {1}{2 \cdot 2^2}} \plus{} \dots \plus{} \arctan{\frac {1}{2 \cdot n^2}}
\]
.

Is it true that :

$ \arctan{(2k \plus{} 1)} \minus{} \arctan{(2k \minus{} 1)} \equal{} \arctan{\frac {1}{2 \cdot k^2}}$ ?? If yes , prove it :)

I think it is true, but I feel like I'm missing something in my proof:

$ \arctan(2k\plus{}1)\equal{}a$, $ \arctan(2k\minus{}1)\equal{}b$

$ \tan(a\minus{}b)\equal{}\frac{\tan(a)\minus{}\tan(b)}{1\plus{}\tan(a)\tan(b)}\equal{}\frac{(2k\plus{}1)\minus{}(2k\minus{}1)}{1\plus{}(2k\plus{}1)(2k\minus{}1)}\equal{}\frac{1}{2k^2}$

Now just take the $ \arctan$ of both sides.

... well after this the evaluation is almost............ trivial  :P

Interesting: $ \lim_{n\rightarrow \infty}{S_n}$??

 :)

What is the motivation for identity used?

Well, in terms of arguments, it's given by

$ \frac {1 \plus{} (2k \plus{} 1)i}{1 \plus{} (2k \minus{} 1)i} \equal{} \frac {(1 \plus{} (2k \plus{} 1)i)(1 \minus{} (2k \minus{} 1)i)}{4k^2 \minus{} 4k} \equal{} \frac {2k^2 \plus{} i}{2k^2 \minus{} 2k}$

These sort of identities should be expected; they appear, for example, in [url=http://en.wikipedia.org/wiki/Machin-like_formula]Machin-like formulae[/url].


[quote="ductrung"]For any positive real numbers $a,\ b$ and $c$,
$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{81abc}{2(a+b+c)^3}$[/quote]

In [b]"Toantuoitho[\b] magazine in Vietnam
it stronger ..

We can make $x=\frac{a}{a+b+c}$, $y=\frac{b}{a+b+c}$ and $z=\frac{c}{a+b+c}$. And now we have to prove that (In this passage we must use the fact that $x+y+z=1$)

$\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x} \geq \frac{81xyz}{2}$.

But we know that:

$\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x} \geq 3\frac{\sqrt[3]{xyz}}{\sqrt[3]{(x+y)(y+z)(z+x)}}$.

Let $P=\sqrt[3]{xyz}$.

We can observe too that $\frac{(x+y)+(y+z)+(z+x)}{3} \geq \sqrt[3]{(x+y)(y+z)(z+x)}$, and so:
$\frac{3}{2} \leq \frac{1}{\sqrt[3]{(x+y)(y+z)(z+x)}}$.

And now we have that:

$\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x} \geq \frac{9}{2}P$ and we´ll have $\frac{9}{2}P \geq \frac{81}{2}P^3$, if and only if $\frac{1}{3}\geq P$, and this last ineq is true because $\frac{x+y+z}{3} \geq \sqrt[3]{xyz} = P$

[quote="socrates"][quote="ductrung"]For any positive real numbers $a,\ b$ and $c$,
$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{81abc}{2(a+b+c)^3}$[/quote]

I think the stronger ineq holds, too:

$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{27abc}{2(a+b+c)(ab+bc+ca)}$

Happy new year :)[/quote]

[i]Unless[/i] my computer program is buggy, I don't think it's true. (Funny that it's exactly the original ineq I attempted until I found a counter example using computer.) Otherwise, please be so kind to post a solution.

Happy New Year !!!

I have another nice solution.

$\sum_{cyclic} \frac{a}{a+b} \geq \frac{81abc}{2(a+b+c)^3}$

$\Leftrightarrow (\sum_{cyclic} a+b ) ( \sum_{cyclic} \frac{a}{a+b} )\geq \frac{81abc}{(a+b+c)^2}$

Here, by $Cauchy-Swartz$ $Ineq$, we get $(\sum_{cyclic} a+b ) ( \sum_{cyclic} \frac{a}{a+b} )\geq (\sum_{cyclic} \sqrt{a})^2$

Therefore, It is enough to prove that $(a+b+c)^2 (\sqrt{a}+\sqrt{b}+\sqrt{c})^2 \geq 81 abc$.

This is self-evident by using $AM-GM$.

$Q.E.D.$

[quote="ductrung"]
[i]Unless[/i] my computer program is buggy, I don't think it's true. (Funny that it's exactly the original ineq I attempted until I found a counter example using computer.) Otherwise, please be so kind to post a solution.

Happy New Year !!![/quote]

Your computer's statement [i]must[/i] be true...

Anyway here goes my solution, please check its validity... :P

$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} = \frac{ac}{ac+bc}+\frac{ab}{ab+ac}+\frac{bc}{bc+ab}\geq \frac{(\sqrt{ac}+\sqrt{ab}+\sqrt{bc})^2}{2ab+2bc+2ca}$

So it remains to show that

$(\sqrt{ac}+\sqrt{ab}+\sqrt{bc})^2 \cdot (a+b+c)\geq 27abc$

which is true due to AM-GM :

$(\sqrt{ac}+\sqrt{ab}+\sqrt{bc})^2\geq (3\sqrt[3]{abc})^2$

as well as

$a+b+c\geq 3\sqrt[3]{abc}$

 :)

[quote="socrates"]
Your computer's statement [i]must[/i] be true...
[/quote]

Hello socrates,

So simple and totally valid. I regret deeply that I discarded it so quickly (such an easy problem) 'cause the computer says so - it is a rounding error! Anyway, please excuse me and thanks for your reply.


ARML
AMC
USA(J)MO
USAMO
Stanford
college
analytic geometry

Hi everyone,
i'm working on getting a bunch of information out about ARML tonight including practice schedules, how to pay, etc., but i need to pass this on right away.  You can go to http://www.southwest.com right now and buy a flight to Vegas for ARML for 117 bucks if you want to fly.  If you're worried about being on the team, i can assure you that i will accept you onto the team if you want to go ahead and purchase a flight.  Remember, you will still need to pay the registration fee, but for those of you who were planning to fly, this may be your best chance to get a good deal.  Sale ends at midnight apparently, so don't waste any time.  :)  Again, my flight details are Southwest flight 215 from SJC-LAS on May 28 and Southwest flight 270 from LAS-SJC on May 30, and you need to arrange to be on the same flight as me if you're flying..

-Tim

yeah, i canceled my old flight and bought a new one at this rate!  :)

Er, my mom just checked on southwest.com, and the lowest rates she can find are 92 and 99 (there and back, respectively), which after tax amounts to over 200 bucks.  Did you get some sort of discount, and when did you change the flight?

EDIT: Found the promotion.  Booked the flight :)

Hi everyone,

This letter is for everyone - students, parents, and coaches - planning to go to ARML this year.  First, if you haven’t yet filled out an application form, do it now.  http://mathleague.org/armlapp.php.  There is still time.  This goes for every person who is going to ARML regardless of age – I filled one out for myself just an hour ago.  As for organizing the teams, it looks like we have the interest and the transportation capacity to take four full teams to ARML this year, so my goal will be to do just that.  A and B teams have been chosen according to the formula I outlined at last Saturday’s contest, and they are listed below alphabetically.  Anyone else who wants to go to ARML may as well, subject of course to space constraints as described in the next paragraph.  I mentioned earlier that the cost of the trip would be $ \$$250; I’m going to knock that down to $ \$$245, because assuming you went to three tryouts you’ve already paid $ \$$30, and $ \$$30 off last year’s cost brings us to $ \$$245.  Of course, I promised at the beginning of the year that ALL your tryout entry fees would come back to you; so if you attended more than three tryouts feel free to take an additional $ \$$10 off the cost for each additional tryout you went to (eg. if you went to six tryouts your cost is only $ \$$215)..

If you want to secure a spot on this year’s ARML team, you need to bring a check for the amount of your registration fee - payable to Great Plains Math League - to the practice on May 2 (see below).  I may be able to accept payments next week assuming there is still space on one of our teams.  If on Saturday we have more interest in our lower two teams than I can accommodate, I will accept students in order of their ARML selection indices as defined last weekend.  If those two teams remain unfilled after Saturday, I will accept students onto those teams in the order their checks are received.  If students selected for the A or B team are unable to participate, I will move students up from lower teams based on some objective combination of selection index and USAMO performance..

You will also need to submit a notarized permission slip, which can be found at http://mathleague.org/permission.pdf.  You have two options to get this turned in:  you can hand it to me with your check at the practice on May 2, or you can mail your permission slip to PO Box 111, Sparta WI 54656.  I will not collect any permission slips personally after this Saturday’s practice.  Permission slips don’t have to be in until May 20th, but please get yours in as soon as possible, because the person checking the PO Box in Wisconsin will be bugging you incessantly about your permission slip until it shows up.  You will also be bugged in the same manner if you filled out an application form but haven’t committed to going or not going to ARML by next week..  :)

Because the vast majority of you indicated on your application forms that leaving early on Thursday would not be a problem, I anticipate having the bus leave from Stanford at 2pm on Thursday May 28th, heading straight for Las Vegas.  If this is impossible for you, let me know when you submit your check and I will hold your check until a firm decision has been made regarding departure time.  I will be glad to return any such checks in the event this presents an irreconcilable schedule conflict; aside from that, we will be unable to issue refunds if you end up not being able to go to ARML.  We will probably leave Las Vegas around 5pm on the 30th, putting us back to Stanford shortly after midnight.  Again, if you are thinking of flying and are willing to pay the extra cost, make sure you coordinate with me so that you’re on my flight..

Finally, a word about practices.  We’ll be running afternoon practices from 2:00 to 5:00 on May 2, 10, and 24, and from 2:00 to 8:30 (with dinner served) on May 16.  These will be in the math building at Stanford (you all know where that is I’m sure), and we will be giving more details at the first practice.  Suffice it to say practices will be a little different this year, so you’ll have to show up to see what we mean..

And now on to our alphabetical listing of A and B team qualifiers:

A team
Arul	Vishal
Chang	Alan
Chen	Yanping
Chen	Lewis
Chu	Timothy
Liu	Andrew
Liu	Bowei
Lu	Anthony
O'Dorney	Evan
Saxena	Amrit
Shweh	Kevin
Wang	Bayley
Wein	Alex
Yang	Kevin
Yang	Patrick

B team
Boyle	John
Burks	Michael
Chen	WilliamZ
choi	dongho
Day	Cynthia
Gu	Albert
Heh	Kevin
Ho	Tony
Kouevda	Nikita
Kulkarni	Archit
Lau	Matthew
Sung	Felix
Swernofsky	Joseph
Young	Brandon
Zhang	Brian

Eh...what if we can't exactly remember how many tryouts we went to? :P

ah yes.  i guess in that case you could send an email to operations at mathleague dot org and someone can look that up for you.  same thing if you're not sure if you filled out your application correctly/at all..

BTW i'm assuming everyone got the letter above via email as well; if you didn't and think you should have, send an email to operations and we'll try to figure out what went wrong..

For those flying, is the bus expense taken off the price or not?

I did not get the email.

In response to Silas' post below: Yes, got the new one.

I have not been receiving any emails either, and I have signed up numerous times over the past few months.

Same here, even though signed up for team and qualified for A team.

Same here.  I signed up for ARML, then signed up for mailing list twice with the same email account, no emails yet.

The email was just resent to a smaller mailing list (just this year's team).  Did you get it this time?

probably better if anyone who didn't get the email contact mathleague directly rather than posting here - we do want to keep this one non-spammy..  :)

Hey,

Just wondering, for the AIME scores used in the ARML index, is the team based on last years AIME scores or this years AIME scores?

Thanks,
Timothy Chu

[quote="Scrambled"]is the team based on last years AIME scores[/quote]

I wish it were  :(   This year's scores are used in the index.


Where can I find an exhaustive list of permitted attachment types? I can't find it in FAQ.

Here you are: http://www.mathlinks.ro/Forum/faq.php?mode=attach
The 0-s there mean that there is no limit for the upload of one single file (other than your own total quota).


number 1 solution was given to me by someone else that I talk to :lol:

In a trapezoid the bases are 5 inches and 8 inches, and the legs are 4 inches and 6 inches. If the legs are extended to meet in a point, how much must the shorter leg be extended?

Label the length extended from the leg whose side is $4, x$. If we also extend the leg whose side is $6$ until it meets the extended length $x$, we notice that we form $2$ similar triangles. So we have \[\frac{x}{x+4}=\frac{5}{8}\Rightarrow 8x=5x+20 \Rightarrow x=\frac{20}{3}\]

nice explanation  :)

so you could also find the extension needed for the longer leg also if you wanted

[quote="i_like_pie"]Trapezoids don't have medians...[/quote]

Yes they do.

As he solved
if we label extension of longer side as $a$
then  $\frac{5}{8}=\frac{a}{6+a}$
so $a=10$ inches[/b]


[quote]ΤΟ ΣΕΜΙΝΑΡΙΟ ΑΝΑΛΥΣΗΣ ΚΑΙ ΚΑΤΑΣΚΕΥΗΣ ΜΑΘΗΜΑΤΙΚΩΝ ΠΡΟΒΛΗΜΑΤΩΝ
(ΣΥΝΤΟΝΙΣΤΗΣ : ΕΥΓΕΝΙΟΣ ΑΓΓΕΛΟΠΟΥΛΟΣ) ΘΑ ΓΙΝΕΙ ΚΑΝΟΝΙΚΑ ΤΗΝ
ΠΑΡΑΣΚΕΥΗ 26 IΑΝΟΥΑΡΙΟΥ 2007
ΣΤΙΣ 17:00, ΚΤΙΡΙΟ Ε, 2ος ΟΡΟΦΟΣ,
ΣΤΗ ΑΙΘΟΥΣΑ ΣΕΜΙΝΑΡΙΩΝ ΤΟΥ ΤΟΜΕΑ ΜΑΘΗΜΑΤΙΚΩΝ (ΣΤΗΝ ΠΕΡΙΠΤΩΣΗ ΠΟΥ
ΑΥΤΟ ΔΕΝ ΕΙΝΑΙ ΕΦΙΚΤΟ ΛΟΓΩ ΤΩΝ ΚΙΝΗΤΟΠΟΙΗΣΕΩΝ ΤΩΝ ΦΟΙΤΗΤΩΝ
ΕΧΕΙ ΕΞΑΣΦΑΛΙΣΤΕΙ ΑΛΛΗ ΑΙΘΟΥΣΑ ΣΤΟ ΚΤΙΡΙΟ ΔΙΟΙΚΗΣΗΣ).

ΟΙ ΣΥΜΜΕΤΕΧΟΝΤΕΣ ΚΑΛΟ ΘΑ ΕΙΝΑΙ ΝΑ ΕΡΘΟΥΝ ΛΙΓΟ ΝΩΡΙΤΕΡΑ ΑΠΟ ΤΙΣ 17:00
ΠΡΟΚΕΙΜΕΝΟΥ ΝΑ ΕΞΑΣΦΑΛΙΣΤΕΙ Ο ΚΑΛΥΤΕΡΟΣ ΔΥΝΑΤΟΣ ΣΥΝΤΟΝΙΣΜΟΣ.

ΣΑΝ ΘΕΜΑ ΤΟΥ ΣΕΜΙΝΑΡΙΟΥ ΕΧΕΙ ΚΑΤ' ΑΡΧΗΝ ΠΡΟΤΑΘΕΙ ΤΟ ΘΕΩΡΗΜΑ ΤΟΥ FERMAT (x^n +
y^n = 1).[/quote]
Είπα ν' ανοίξω νέο topic αφού υπάρχουν ήδη  3-4 άτομα που ενδιαφέρονται.Εδώ θ' ανακοινώνουμε τα θέματα των επόμενων σεμιναρίων καθώς το site τους δεν ανανεώνεται συχνά.

akougete endiaferon, to seminario einai anoikto gia olous?

φυσικα και ειναι ανοικτο για τον καθενα,  :)

ελπιζω ο Θανος να μην μας την κανει ,,, οποιος θελει να παει ας ζητησει λεπτομεριες του κτιριου (ειναι δυσκολο να το βρεις )  :D

Ποια αίθουσα συγκεκριμένα είναι  αυτή που γίνεται η συζήτηση??
Είναι μία από αυτές που γράφουμε και γίνονται μαθήματα??

ειναι στην σχολη Σ.ΕΜΦΕ(εφαρμοσμενων μαθηματικων και φυσικων επιστημων
τομεας μαθηματικων -- κτηριο Ε.
2ος οροφος
αιθουσα σεμιναριων
μαλλον δεν θα καταφερω να ερθω :(
[hide]θα δεις και στην εισοδο τους καταληψειες τις σχολης ! δε θα εχεις προβλημα να μπεις , ειναι καλα παιδια  :) [/hide]

:blush:
 :blush:

βασικά είχα στείλει email την Τετάρτη και μου απάντησε την Πέμπτη αλλά δεν το είχα δει :(
αυτή τη φορά όμως θα είμαι εκεί!

δεν πειραζει  :)
@thatin: εσυ θα μιλησεις ?

[quote="kostas"]ελπιζω ο Θανος να μην μας την κανει ,,, [/quote]

simpairana pws to ikseres...
nai

το ηξερα ,  :P

Εγώ όμως, θα έρθω.Την προηγούμενη φορά,χμμ,ε,χμμ,...,ε,thatiiiiin;;;;;; :mad:
[hide]Αλήθεια,ένα pm που σου έχω στείλει,το είδες;[/hide]

[quote="thatin"]telika de tha ertheis?[/quote]
οχι, αλλα οποιος θελει ας πει τι κανατε και ειπατε ,παρακαλω :)

@Fischerman
όχι ακόμα!
ε, τι? κάνατε μια ωραιότατη εισαγωγή!

παρουσιάστηκε η απόδειξη που δεν παρουσιάστηκε στο προηγούμενο και ο Αγγελόπουλος μίλησε για τριτοβάθμιες καμπύλες και ειδικότερα για τις εφαπτομένες καμπής τους

την επόμενη παρασκευή θα γίνει

askisi gia to spiti:
aposeikste to avatar mou

Το σεμινάριο αν ξέρετε θα γίνεται κάθε Παρασκευή?

ειναι καθε Παρασκευη την ιδια ωρα

Την προηγούμενη Παρασκευή (2/2) δεν έγινε το σεμινάριο.

kserei kaneis an tha ginei auti tin paraskeui?

Δεν έχουν στείλει mail.

emena pote de mou exei erthei mail...
ama erthei pes mou plz
exeis pm...!

συνεχιζει να γινεται αυτο το σεμιναριο ? αυτη την Παρασκευη θα γινει?

[quote]Ε Λ Ε Υ Θ Ε Ρ Ο    Σ Ε Μ Ι Ν Α Ρ Ι Ο
Ανάλυσης και Κατασκευής Μαθηματικών προβλημάτων
http://www.math.ntua.gr/~probsolv/

(τρίτος χρόνος λειτουργίας, 2007-2008)
Συντονιστής : Ευγένιος Αγγελόπουλος

Το Σεμινάριο απευθύνεται σε φοιτητές οποιουδήποτε έτους. Στο παρελθόν το έχουν παρακολουθήσει και διδάσκοντες Μέσης Εκπαίδευσης. Είναι ανοιχτό σε συναδέλφους του Τομέα Μαθηματικών του ΕΜΠ (ορισμένοι έχουν στο παρελθόν παρουσιάσει θέματα).
Τα θέματα που αναπτύσσονται (βασικά από τους φοιτητές τους ίδιους) είναι εκτός ύλης, προσεγγίσιμα όμως με στοιχειώδεις γνώσεις. Απώτερος στόχος είναι η οργάνωση θεμάτων σε προβλήματα-εργασίες, με αλλεπάλληλα βήματα, έτσι ώστε να καθοδηγείται ο λύτης στο να χρησιμοποιεί τις γνώσεις του για να προσεγγίσει το νέο και άγνωστο.
Η λέξη ελεύθερο υποδηλώνει ότι δεν υπάρχει καμία υποχρέωση συμμετοχής και καμία υποχρέωση παρουσίας (εκτός από την ηθική υποχρέωση να παρουσιάσει κάποιος την ομιλία που έχει αναλάβει). Αντίστοιχα, δεν υπάρχει και καμιά ανταμοιβή, βαθμολογική ή άλλη, πέρα από την ικανοποίηση της περιέργειας και την εν γένει απόλαυση που η ίδια η συμμετοχή προσφέρει.

Κάθε Παρασκευή, ώρα 17:00

Αίθουσα σεμιναρίων, κτίριο Ε (ΣΕΜΦΕ), 2ος όροφος
(Πολυτεχνειούπολη, Ζωγράφου)

Έναρξη :		Παρασκευή 23 Νοεμβρίου 2007[/quote]

OXI :D .Βρήκα αυτό που έψαχνα :lol: http://eclass.uoa.gr/courses/MATH206/


logarithms
geometry
complex analysis
algebra
polynomial
function

Hi, I was wondering if anyone can help me with this inverse Laplace Transform using the complex inversion formula.  Thanks.

Question: Evaluate $ \displaystyle\mathcal{L}^{\minus{}1}\left\{\ln \left(1\plus{}\frac{1}{s^2}\right)\right\}$ by the inversion formula.

use a 3 keyhole contour

What can you say about the branch-cut geometry of the holomorphic extension of $ \text{Log} \frac{P_n(z)}{Q_m(z)}$ to the left half-plane $ \textnormal{Re}(z)<0$ with $ P$ and $ Q$ polynomials (of sufficient degree to make the inversion valid) when both the zeros and poles are arranged along the imaginary axis like your function?  How will the function value along (horizontal) branch-cuts at each pole and zero differ depending on the order of each pole and zero?


Which one of the following statement explain why the emission spectrum of atomic hydrogen is a line spectrum?

A.an electron can possess any amount energy
B.the electron in hydrogen atom can only occupy fixed energy levels.
C.Transition of an electron from a higher energy level to a lowerenergy level emits light
D.An emission spectrum consists of discrete lines which have fixed frequencies.

I would say B, but I have to review my stuff.

The correct answer is B.

The theory of fixed energy levels was constructed to answer the question as to why line spectrums were found for hydrogen (and everything else).  The uniqueness of the frequency is the basis for spectral chemical analysis.

We may find new evidence that calls this theory into questions, but as far as I know it seems correct so far.


[quote="sri340"]AB
                                       +BA
                                     ______
                                       CD4
In the correctly worked addition problem above, each letter represents a different nonzero digit. What is one possible value of the two-digit number represented above as AB ?

(A) 74       (B) 31       (C) 86           (D) 47         (E) 95[/quote]

By adding the units digits, we see that $ a \plus{} b \equiv 4 \mod{10}$, this means that $ a \plus{} b \equal{} 4$ or $ a \plus{} b \equal{} 14$.  Note how our sum is a 3 digit number, because of this there is no way for $ a \plus{} b \equal{} 4$ (try plugging in some numbers and see why).  Because $ a \ne b$, we can simply guess & check.  If $ a \equal{} 6$ and $ b \equal{} 8$ we get $ 68 \plus{} 86 \equal{} 154$, since $ a \ne b \ne c \ne d$, $ ab \equal{} 86$.

Also, because you are given choices, simply add the two digits given and see if they equal $ 14$.  This eliminates A, B, and D.  Checking 86 and 95, we see only 86 works.

Another way is to note that
$ 10a\plus{}b\plus{}10b\plus{}a\equal{}11(a\plus{}b)$
The only number in the hundreds that is divisible by $ 11$ and ends in $ 4$ is $ 154$
Hence $ ab$ is $ 86$


counting
distinguishability
geometry
geometric transformation
rotation

One problem in the first volume of Aops asks,
In how many ways can four idential red chips and two identical white chips be arranged in a circle?

I failed to solve the problem correctly, and I understood the solution the solutions manual gave me.  However...I was wondering why and where my method failed.

Here is my solution:
We take a simpler situation, and pretend we are ordering the chips in a line.  There are six chips, so there are 6! ways the chips can be arranged.  However, since the four red chips are indistinguishable, each possible arrangement is repeated 4! times.  So we must divide 6! by 4!.  We do the same thing with the white chips, so we have $ \frac{6!}{4!2!}$.  This equals 15, the number of possible arrangements if we were concerned with a line.  I'm pretty sure I am right up to now (although this may be an unnecessary step...).

However, we are dealing with a circle.  Since rotations of the circles don't matter, we divide by n, the amount of chips on the circle...so my final solution is
$ \frac{15}6$.  This is obviously wrong because it is not an integer number.

Once again, I know a different approach to the problem that will get me to the right answer, but I really want to know what is wrong with this logic.  I am not looking for an alternate solution, but rather a correction.

Thanks

[quote="choidavid91"]However, we are dealing with a circle.  Since rotations of the circles don't matter, we divide by n[/quote]

You can only do that if all of the objects are distinguishable (I think).

I'm not too sure about this logic myself.

When you convert a string of beads into a bracelet, you're overcounting different configurations by different amounts. I really can't find a clear way to explain precisely why this occurs, but if you list out all 15 configurations (it's not too bad), you should at least be able to see how this happens.


limit
real analysis
real analysis unsolved

Let $f,g : \mathbb{R}\to\mathbb{R}$. $f$ has Darboux property and for all $x\in\mathbb{R}$

$\lim_{h \to 0}\frac{f(x+h)-g(x)}{h}=0$

exists and is finite.

Prove that $f=g$ ... i just can't solve it  :(

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88958]some hints in romanian[/url]

$\lim_{h\to0}f(x+h)=g(x)$ so $f$ has finite lateral limits which implies that it hasn't disconituities of the first order but also $f$ has Darboux property which means that it hasn't discontinuities of the second order thus $f$ is continous.

What is the Darboux property? And if the limit is 0, doesn't that imply its finite?

That $=0$ after the limit is probably a typo.

The Darboux property is also called the intermediate value property. For any value $t$ between $f(a)$ and $f(b)$ there is $c\in(a,b)$ such that $f(c)=t$.

Yes, it's a typo  :maybe: , i'm sorry.

Thanks for the links and indication.


algebra
polynomial
function
quadratics
AMC
AIME
rational function

Sorry if this is a little unclear, but:

[hide="Solution"]Denote the three equations, from top to bottom, as $(1)$, $(2)$, and $(3)$. Notice that $(2)-(1)=11$ and $(3)-(2)=111$ are both sequences of consecutive odd integers. Let's denote these sequences as $(4)$ and $(5)$, respectively. Now, notice that $(5)-(4)$ gives $2x_1+2x_2+\cdots+2x_7=100$, which we can use to find the desired sequence. Our answer is $(3)+(5)+100=\boxed{334}$.[/hide]

[hide]Let the left-hand sides of the equations be A, B, C, and D (from top to bottom).

Notice that $z^2 + (z+2)^2 = 2(z^2 + 2z + 2) = 2((z+1)^2 + 1)$

Applying this to the coefficients of the equations, $(A+C)/2 = B + (x_1 + x_2 + ... + x_7)$ and that $(B+D)/2 = C + (x_1 + x_2 + ... + x_7)$

Plugging values of A, B, and C into the first equation, $x_1 + x_2 + ... + x_7 = 50$, so plugging this into the second equation gives:

$(12+D)/2 = 123 + 50$, $\boxed{D = 334}$[/hide]

[quote="Farenhajt"][hide]The corresponding coefficients with $x_k$ are $k^2,(k+1)^2,(k+2)^2$, and we're looking for a linear combination of equations which will yield the coefficient $(k+3)^2$ with $x_k$. By the method of undetermined coefficients we have

$ak^2+b(k+1)^2+c(k+2)^2=(k+3)^2$

Expanding and equating the coefficients with the corresponding powers of $k$, we get

$a+b+c=1$
$2b+4c=6$
$b+4c=9$

and the solution is $a=1, b=-3, c=3$

Hence the desired sum is $1\cdot 1-3\cdot 12+3\cdot 123=334$[/hide][/quote]

can you explain this a little bit more?? It seems like a powerful system solving technique with patterns and I'd like to learn it

[quote="Farenhajt"][quote="pkerichang"]can you explain this a little bit more?? It seems like a powerful system solving technique with patterns and I'd like to learn it[/quote]

[hide="Detailed explanation"]Well, the general idea is this: We have $3$ equations with $7$ variables, hence there's no unique solution, so the idea of solving it should be discarded as too cumbersome. On the other hand, we can try to find numbers $a,b,c$ such that when we multiply the first equation by $a$, the second by $b$ and the third by $c$, and then add all that up, we get exactly the desired expression as LHS. (The "shop" term for the resulting equation is [i]linear combination[/i] of the given equations.)

Therefore, in the first equation we'll get terms of the form $ak^2x_k$, in the second of the form $b(k+1)^2x_k$, and in the third $c(k+2)^2x_k.$ In the same time, the free terms will be $a\cdot 1, b\cdot 12, c\cdot 123$.

Now adding up must yield the terms of the form $(k+3)^2x_k$ for EVERY $k.$ So we write

$ak^2+b(k+1)^2+c(k+2)^2=(k+3)^2$

This is where actual power technique comes handy, and it's called [i]the method of undetermined coefficients[/i] (at least I think that's the correct english term :)): We don't know what are the values of $a,b,c$, but we know that they must satisfy the above equation for every $k$. Let's expand and group the LHS and expand the RHS:

\begin{eqnarray*}ak^2+b(k^2+2k+1)+c(k^2+4k+4) &=& ak^2+bk^2+2bk+b+ck^2+4ck+4c\\ &=& (a+b+c)k^2+(2b+4c)k+(b+4c)\end{eqnarray*}

$(k+3)^2=k^2+6k+9$

So we get

$(a+b+c)k^2+(2b+4c)k+(b+4c)=k^2+6k+9$

Now the important point: These are two polynomials in $k$ and the [u]only[/u] way they could be equal for EVERY $k$ (this "every" is very crucial) is that all the corresponding coefficients (leading, linear and constant) are equal. Therefore we must have

$a+b+c=1$
$2b+4c=6$
$b+4c=9$

This is the system of linear equations which isn't difficult to solve by any known method. For example, subtracting the third equation from the second we get $b=-3.$ Now substituting that into the third we get $-3+4c=9\implies 4c=12\implies c=3$. Then the first becomes $a-3+3=1\implies a=1.$

So we have found $a,b,c$ such that we get what we want: If we multiply the first equation by $1$, the second by $-3$ and the third by $3$ and add that all up, we get the desired expression. So we must calculate the free term we'll obtain in the process. As we said, it must be $a\cdot 1+b\cdot 12+c\cdot 123=1-36+369=334.$

Hope this helped :)

NB. The method of undetermined coefficients is applicable in many different situations concerning polynomials (notably in calculating remainders in polynomial division and splitting a rational function into simple fractions), so having it in mind could sometimes be a life-saver.[/hide][/quote]

Thank you for this useful approach to solve systems of equations (the coefficient technique I know already, just not the name).  Are there any circumstances where it's not applicable???

As far as I know, it can be used only for polynomials and rational functions (but of course, the scope of problems involving these is immense).

I'm not sure what do you mean, shady... Can you post some more details from their solution?

[quote="Farenhajt"]$ak^2+b(k+1)^2+c(k+2)^2=(k+3)^2$[/quote]

Just a note, at this point you can just substitute $k=0,-1,-2$ to get \[ b+4c=9, \] \[ a+c=4, \] and \[ 4a+b=1, \] from which you get the same solution $(1,-3,3)$. :)

Sure, but it's always better to work with variables, since that way you make sure that the method is applicable. (Remember that trigonometric sum? :))

[quote="Farenhajt"]Remember that trigonometric sum? :)[/quote]

:D Ok, sure.

A free term is the number on the RHS of an equality. Let's say you have these two:

$x=12$
$y=7$

When you multiply the first by $a$ and the second by $b$, you get

$ax=12a$
$by=7b$

Now if you add the LHS's, you must also add the RHS's:

$ax+by=12a+7b$

Does this help?

And yes, this method can be used to solve the problem you posted, but I'll let someone else to show it (it's not too difficult).

thanks that has helped. ill will try to do the problem i posted with this method, later right now i have to head to class.  :blush:  :(

[quote="maokid7"]i have a few questions.

what are the free terms. i do not understand y u used these values.

my other question is, would you be able to use that method on this problem.

$a_1+8a_2+27a_3+64a_4=1$
$8a_1+27a_2+64a_3+125a_4=27$
$27a_1+64a_2+12a_3+216a_4=125$
$64a_1+125a_2+216a_3+343a_4=343$

show that
$a_1(x+1)^3+a_2(x+2)^3+a_3(x+3)^3+a_4(x+4)^3\equiv(2x+1)^3$[/quote]

[hide]To show this, all you need to do is show that the two cubics are equal for at least 4 different values of $x$, because if they were not equal, their difference would be nonzero and of degree 3 or less, so it would have at most 3 roots. But finding 4 values of $x$ is easy, since the cases where $x$ is 0, 1, 2, and 3 are given.[/hide]

wow you guys sure made a big deal of an easy problem :D jk not really
I just have an easy solution because when I first saw the problem I had just finished doing this problem: which is in aops vol 2
[quote="maokid7"]i have a few questions.

what are the free terms. i do not understand y u used these values.

my other question is, would you be able to use that method on this problem.

$a_{1}+8a_{2}+27a_{3}+64a_{4}=1$
$8a_{1}+27a_{2}+64a_{3}+125a_{4}=27$
$27a_{1}+64a_{2}+12a_{3}+216a_{4}=125$
$64a_{1}+125a_{2}+216a_{3}+343a_{4}=343$

show that
$a_{1}(x+1)^{3}+a_{2}(x+2)^{3}+a_{3}(x+3)^{3}+a_{4}(x+4)^{3}\equiv(2x+1)^{3}$[/quote]
anyway, all you do is notice that $f(x)=\sum (x+k)^{2}*x_{k}$. F is obviously a quadratic equation. We have $f(1)=1,f(2)=12,f(3)=123$. Now look at differences: the first ones are $11,111$. The second level of differences must be a constant (because this is a quadratic), and we have $111-11=100$ must be this value. So we can easily find $f(4)$ using only basic arithmetic: $100+111=211$ is the next first level difference, so $f(4)=211+f(3)=\boxed{334}$

[quote="shadysaysurspammed"]Farenhajit,could you solve this question using the coeffoicient method?I do have the solution but i just can't understand it..

$1.2+2.3+3.4.....$.Find sum.

Now in this ,They've equated the given thing to $A+Bn+C(n+1).....$
Why?[/quote]

I don't see why you need the A, B, C thing but here is a solution:

$\sum_{n=1}^{k}{n+\frac{n+1}{10}}$
$\frac{1}{10}\sum_{n=1}^{k}{11n+1}$
$(\frac{1}{10})(11\binom{k+1}{2}+k)$
$\sum=\frac{11k^{2}+13k}{20}$

[quote="lingomaniac88"][quote="Farenhajt"][hide]The corresponding coefficients with $x_{k}$ are $k^{2},(k+1)^{2},(k+2)^{2}$, and we're looking for a linear combination of equations which will yield the coefficient $(k+3)^{2}$ with $x_{k}$. By the method of undetermined coefficients we have

$ak^{2}+b(k+1)^{2}+c(k+2)^{2}=(k+3)^{2}$

Expanding and equating the coefficients with the corresponding powers of $k$, we get

$a+b+c=1$
$2b+4c=6$
$b+4c=9$

and the solution is $a=1, b=-3, c=3$

Hence the desired sum is $1\cdot 1-3\cdot 12+3\cdot 123=334$[/hide][/quote]

Uh... what did you do?[/quote]

Farenhajt notes that the coefficient of every $x_{k}$ in the system is, in the first equation $k^{2}$, in the second equation $(k+1)^{2}$, and in the third equation $(k+2)^{2}$.

The goal of the problem is to find the value of the expression where every coefficient is $(k+3)^{2}$.  Thus, we multiply the first equation by some real, add it to the second equation multiplied by some real, then to the third equation multiplied by some real, to get that desired $(k+3)^{2}$.

To do this we need to find reals $a, b, c$ that satisfy the equation Farenhajt specified, as coefficients of a polynomial of $k$.  We equate coefficients to get his system and solve.

I hope this is a little clearer.

first we know that
if f(x)=ax^2+bx+c then (f(x1)-f(x2))/(x1-x2)=(f(x1+x2)-f(0))/((x1+x2)-0)
it is not so difficult to prove it....

and then we look at this question
we have f(x)=x1(x+1)^2+x2(x+2)^2+........x7(x+7)^2
and f(0)=1 f(1)=12 f(2)=123 and we want to find f(4)

then it is easy.....

I'm sorry that I've made a mistake above
that we want to find f(3) instead of f(4)

Nobody mentioned this was an AIME problem. I think like 1987 or something like that.

Sorry for the revive, but no one has posted the following solution yet:
[hide=Summation overkill]Rewrite all of the equations as follows:
$\sum_{i=1}^{7}i^2x_i = 1$
$\sum_{i=1}^{7}(i+1)^2x_i = 12$
$\sum_{i=1}^{7}(i+2)^2x_i = 123$
And we would like to find the value of $\sum_{i=1}^{7}(i+3)^2x_i$. We can rewrite the desired quantity as $\sum_{i=1}^{7}i^2x_i + 6\sum_{i=1}^{7}ix_i + 9\sum_{i=1}^{7}x_i = 1 + 6\sum_{i=1}^{7}ix_i + 9\sum_{i=1}^{7}x_i$. The second equation rewrites to $2\sum_{i=1}^{7}ix_i + \sum_{i=1}^{7}x_i = 11$ and the third equation rewrites to $2\sum_{i=1}^{7}ix_i + 2\sum_{i=1}^{7}x_i = 61$. Solving this system of equations gives us $\sum_{i=1}^{7}ix_i = -\frac{39}{2}$ and $\sum_{i=1}^{7}x_i = 50$. So our answer is $1 + 6\left(-\frac{39}{2}\right) + 9(50) = \boxed{334}$.

[hide=Solution]We want a cool solution to post on the aops forums, so we try to obliterate everything at the same time :P Thus, we want some linear combination of $x^2, (x+1)^2, (x+2)^2$ to give $(x+3)^2$. Trying random stuff and equating some coefficients gives that the first minus three times the second plus three times the third gives the desired (try it yourself if you haven't; its kinda magical!), so the answer is $1-36+369=\boxed{334}$.[/hide]

[hide=Is this legal]
Use finite differences
$E_2-E_1=11$
$E_3-E_2=111$
$(E_3-E_2)-(E_2-E_1)=100$
Since the second difference in a quadratic is constant, we know $(E_4-E_3)-(E_3-E_2)=100$
So $E_4-E_3=211$
$E_4=334$
[/hide]

Edit: it is legal because we can say that the LHS are $E(1)$, $E(2)$, $E(3)$ of the function $E(n)=\sum_{k=0}^6 x_k(k+n)^2$. Clearly this is quadratic wrt $n$.

[quote=smartguy888]Where did we get the $(x+3)^2$ from?[/quote]

I'm going to assume you're responding to my post, because its the most recent that uses that terminology.

To be honest, using $x$ there is a bit confusing.

[hide=maybe this gives away some stuff so dont read it if you haven't solved this yet]The point is that the value we're looking for can be expressed as $\sum_{i=1}^{7} (i+3)^2x_i$, so we want to use some combination of $i^2, (i+1)^2,$ and $(i+2)^2$ to create $(i+3)^2$ to obtain every term needed.[/hide]

tbh I think finite differences is overkill

Also, you automatically know that $(x+3)^2$ can be represented as a linear combination of $x^2, (x+1)^2, (x+2)^2$ because those 3 polynomials form a basis of all polynomials with degree at most $2$.

[hide=Solution]We have $a^2, (a+1)^2, (a+2)^2$ and want $(a+3)^2$, so take $x(a^2)+y(a+1)^2+z(a+2)^2=(a+3)^2$. This means $x+y+z=1, 2y+4z=6, y+4z=9$. Multiplying the last equation by $2$ gives $2y+8z=18$, so $4z=12 \implies z=3$. This means $y=-3$ and $z=1$, so the answer is $1 \cdot 1 - 3 \cdot 12 + 3 \cdot 123 = \boxed{334}$.[/hide]


I am getting stuck on the following problem, I have tried to use the map $ p_{*} : \pi_1(S^n)\to \pi_1(\mathbb{R}P^n), \mbox{and} \,\,\,p_{*} : H_{n}(S^n)\to H_{n}(\mathbb{R}P^n)$   but it  seems  not working  :(

Let $ n\geq2$ and  $ p : S^n\to \mathbb{R}P^n$  be the standard covering map. Prove that $ p$ is NOT nullhomotopic

Any idea?

Here $ S^n$ is the universal cover for $ \mathbb{R}P^n$, and so the covering map induces an isomorphism of $ \pi_k$ for $ k > 1$. In particular $ p_* : \pi_n(S^n) \to \pi_n(\mathbb{R}P^n)$ is not the zero map.

thanks Nukular

Your solution is great, I also figured out one solution for my own :

Let $ \tilde {p}$ be the lifting map from $ S^n \to S^n$ of $ p : S^n\to \mathbb{R}P^n$, by the uniqueness of the lifting map I have $ \tilde {p}\equal{}Id_{S^n}$ so I have $ Id_{S^{n}}$ homotopic to constant map. This is a contradiction since $ S^n$ is not contractible

1234567a: your solution is almost the same as Nukular's. The reason you know the lifting map $ \tilde{p}$ exists is because $ \pi_1(S^n)\equal{}0$.  So in fact, if $ \tilde{X}\rightarrow X$ is a covering map, then by the lifting criterion, for $ n\geq 2$, $ \pi_n(\tilde{X})\cong\pi_n(X)$ precisely because we can lift any map from $ S^n\rightarrow X$ into its covering space and do all our homotopy stuff in the universal cover!


I am finding following Engel book to be quite difficult. I have finished AoPS 2, can score about a 9 or 10 consistently on the AIME (provided that I don't make any careless errors), have no olympiad experience, and am taking WOOT. When providing solutions to the questions, for example, what is the motivation to find the solutions? Though the solution itself is important, how can we figure out the motivation to solve the problem?

try Problem Solving Through Problems it's nicer! I'm doing it right now  :)

The Larson is much easier at the beginning, but gets hard fast  :wink:

[quote="PenguinIntegral"]What chapter are you doing, specifically?

The first few are easy, but some of that NT stuff at the end blows my mind...[/quote]

Dude... when I read the invariance chapter, i'm like what on earth is this?

Also, which is better for Olympiad preparation? Larson or PSS?

Well PSS is more difficult and longer...but if you can't do it, there's no point in bulling through it.

[quote="Phelpedo"]I'd imagine that he has to look at the solutions for most of the problems, and upon doing so the motivations for these solutions seems rather obscure.

-If you have to look up solutions for most/all of the problems (especially in chapter 1), you should probably start easier...you gain very little by simply looking at the solution and a lot more by wrestling with the problem, understanding why certain things work and don't work, even if you never get the solution.

I will say that Engel's solutions are rather minimal, which can be good because you still have to figure out large parts for yourself.[/quote]
The motivation to decompose some large number into a sum of fourth powers still boggles me to this day. But I honestly agree with you, Karth should probably start easier. If he's in the same spot I was about a year ago and am in now where everything was/is way too hard or way too easy, I don't have any good suggestions.

I have to second PSS for a large part of pre-olympiad prep. Larson and ACops skim way too much off the top.

Would you mind posting some problems you are having difficulty with, Karth?


the equation of a projectile can be written as $ y \equal{} xtan\theta \minus{} \frac {1}{4h}x^2(1 \plus{} tan^2\theta) \; where \frac {V^2}{2g} \equal{} h , g > 0$
(i)show that the point $ (X,Y)$can be hit by firing at two different angles $ \theta_1$ and $ \theta_2$ if $ X^2 < 4h(h \minus{} Y)$
(ii) show that no point ABOVE the x axis can be hit by firing at two different angles such that $ \theta_1 < \pi/4$ and  $ \theta_2 < \pi/4$
i have completed part one but i have no idea how to approach part two

Treat $ \tan \theta$ as a variable and solve for it (it's a quadratic equation), you then get
\[ \tan \theta \equal{} \frac {2hx\pm \sqrt {4h^2x^2 \minus{} x^4 \minus{} 4hx^2y}}{x^2}
\]
Now show that  if $ x > 0$ then $ \frac {2hx \plus{} \sqrt {4h^2x^2 \minus{} x^4 \minus{} 4hx^2y}}{x^2}\geq 1$ (or it's undefined) and therefore it corresponds to $ \theta \geq \frac {\pi}{4}$.


algebra
functional equation
IMO
IMO 2003

His result for IMO 2003: 777 776. He missed perfect score by one point. That's really unlucky for him.

What did he do wrong on problem 6? Why did he scored 6 points? Does anyone knows?

i belive that (from what he said) he didn't acutally solve the problem - it was more like a very stupid marking scheme!

example "for not proving that ... " -1p !!! etc.

I remember that Mihai Manea from Romania got 41/42 points .. that is pretty unlucky!  :(  .. this happened a few years ago.. I don't remember quite well when.. but it happened!  :( ... I guess that all contestant who get 41/42 are unlucky.. one point away from perfectness!  :(

cheers :D

i was on the team that year (2001) and yeah mihai was good. the 6 which you get at an IMO can from a 3-4 to a 7, from my point of view.

I myself got three 6-s last year, and one of them was a clear 6 (at the functional equation I forgot to mention that the solutions found verify the original equation - much as all the Romanian team did) and the other two, were 7-s ( i repeat in my opinion).

I heard that in the year Mihai Manea took 41 pnts, 4 42 pnts the organizers gave a notebook. So he didn't get one
 :(


哥们你在哪个城市呀? 几年级?

国内有好多奥数网站讨论问题和发表材料更方便.如"数学竞赛论坛","奥数之家论坛".可在GOOGLE打中文搜索.
我在"奥数之家"主持专栏,那儿也有不少国内外的材料,欢迎光临.www.aoshoo.com./bbs1

[quote="georgew_bush"]:!:
我是一名中学生，是一位巨时网民，但始终没能找到好的数学资源网站
希望mathlinks可以改变
谢谢大家[/quote]
你就是tale吗？


[quote="perfect_radio"]Find all $f : \left( 0, \infty \right) \to \mathbb R$ such that $f(x) \leq \ln x$ for all $x > 0$ and $f(xy) \leq f(x)+f(y)$ for all $x, y > 0$.[/quote]

simple problem.

f(x) <=lnx
for x=1 we have f(x) <=0

f(xy) <= f(x) + f (y) <= lnx + lny; relation (1)
for x=y=1 we have: f(1) <= 2*f(1) <= 0 so rezults that f(1) = 0

Now taking y=1/x in (1) we have:
f(1) <= f(x) + f(1/x)<= 0;but f(1)=0 so f(x) + f(1/x) =0 so f(1/x)= -f(x)   we say to that relation (2)

in relation (1) we take x=y=1/x rezults:

f(1/(x^2)) <= 2*f(1/x) <= ln(1/(x^2)); but from relation (2) rezults:

f(1/(x^2)) <= -2*f(x) <= - ln(x^2)  we multiply with (-1) so it rezults:

f(1/(x^2)) >= 2*f(x) >=2*lnx so f(x) >= lnx; but the problem says that f(x) <= lnx so we finally got that f(x) = lnx;

[/hide]

two things: 1) you have to always check your solution works
2) there are more solutions, such as, 0.5 ln x

[quote="perfect_radio"]Nice solution, Zamfirmihai.

[quote="Altheman"]1) you have to always check your solution works[/quote]
True.
[quote="Altheman"]2) there are more solutions, such as, 0.5 ln x[/quote]
False.[/quote]

yes,  I forgot to check.

mersi

uhh...? wow, am i being stupid or something?, 0.5 ln x works...

no, it doesn`t work!

f(x)<=lnX =>

0.5 * lnx <= lnx  rezults  0 <= lnx . but if you chose x=1/e rezults
lnx = -1 rezults 0 <= -1. false


Has someone tell me about square residues?
Thank so much.

i suppose it is quadratic residues  :)

it is related to number theory

please check [url]http://en.wikipedia.org/wiki/Quadratic_residue[/url]

You might also want to visit the following links:
1. http://mathworld.wolfram.com/QuadraticResidue.html
2. http://primes.utm.edu/glossary/page.php?sort=QuadraticResidue
3. http://www.mathreference.com/num-mod,res.html
4. http://marauder.millersville.edu/~bikenaga/numbertheory/quadratic-residues/quadratic-residues.html
5. http://www.iec.csic.es/~ismael/QRs.htm

Or rather, try out your search in google:
  http://www.google.co.in/search?q=Quadratic+residue&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a


Let Fb be Feuerbach point,and $ M_{c},M_{c},M_{c}$  be midpoints of sides AB,BC,CA respectively.Let ⊙I be the incircle of triangle ABC. The three sides  are tangent to the circle I at D,E,F respectively.
Then the Simson lines for Fb for triangle MCMaMb and DEF are the same,and the line is parallel to the Euler line of triangle DEF.
Let $ H_{c},H_{c},H_{c}$ be feet of three vertices C,A,B on three sides AB,BC,CA respectively.Then Simson Line $ H_{2}H_{4}$ is parallel to Euler line GH of triangle DEF.
  In the figure,the Simson lines are green,and Euler line is blue.
[img]http://bbs.mathchina.com/usr1PvjRWKew/5/68/CEB4C3FCC3FB_1164420633.jpg[/img]
I solved this problem by using conjugate ratio [url]http://webs.uvigo.es/adg2006/talks_files/ling.pdf[/url].

Because the Feuerbach point $ F_{b}$ of triangle $ ABC$ is the Anti-Steiner of the Euler line of triangle $ DEF$ w.r.t triangle $ DEF$, thus the Simson line for $ F_{b}$ w.r.t triangle $ DEF$ is parallel to the Euler line of triangle $ DEF$.


...this is derivation of the pythagorean theorm.

E ) X+Y=Z.

answer ; e

Could you please explain in further detail? This isn't my question, but I want to know the solution as well. Thanks.

im not really a teacher type... but w/e

in the derivation of pythagorean theorm, there are three squares around the triangle.

and in the problem, those triangles are 1/2 of the squares.

in derivation of pythagorean theorm, its $ x^2\plus{}y^2\equal{}z^2$,

and in this problem, it becomes $ \frac{x^2}{2}\plus{}\frac{y^2}{2}\equal{}\frac{z^2}{2}$, which is also right.

also food for thought
most the time multiple choice answers are e :wink:  :wink:


topology
ratio
function
real analysis
real analysis solved

hi all,

here is a problem that seems intuitive but rather hard (i find) to formalize:
consider the set E of all the continuous (with uniform norm) fonctions from D to S, where D is the unit disk and S the unit circle. Prove that E is arc connected.

Do you mean $\forall\, x,y\in E$ $\exists$ homotopy $H:[0,1]\to E$ s.t. $H(0)=x$ and $H(1)=y$?

[quote="grobber"]Couldn't we take the homotopy (from $f$ to $g$) $\phi(t)(x)=$ the point dividing the positively directed arc $\widehat {(f(x),g(x))}$ in the ratio $t:1-t$? By positively directed arc I mean the directed arc going from $f(x)$ to $g(x)$ in a counterclockwise direction.

It looks Ok to me right now, but it's 4 AM, so I'm sure you'll be understanding, should it turn out to be just another stupid mistake of mine :).[/quote]
I think it doesn't work, because it is possible $H(t)\not\in E$. Unfortunately, I don't know how to describe appropriate example using my poor english. :(

Yeah, I think I also see a pretty obvious counterexample (it's not a concrete counterexample, but I think I can see why the construction above can fail).

Just take $y(u)\equiv (1,0)$ and $x(u)$ as it is described on figure, $t=\frac{1}{2}$.
Let $\epsilon>0$ be small number. You see that for $u=(w,\epsilon)$ point $H(t)(u)$ lies in neighbourhood of $(1,0)$ and for $u=(w,-\epsilon)$ point $H(t)(u)$ lies in neighbourhood of $(-1,0)$.
Ok?

Ok, I'll have another shot at it. I'm sure there will be someone to dismiss this one too, but I think that learning through mistakes is the best way of learning :D.

Since path connectedness is an equivalence relation, it's enough to show that every function is path connected to the constant function. Moreover, since any two constant functions are obviously path connected, it's enough to show that $f$ is path connected to the constant function $g$ whose value is $f(0)$ (the value of $f$ in the origin; I'll use complex number language). Now take $H(t)(x)=f((1-t)x),\ \forall x\in D,\ \forall t\in [0,1]$.

Is it totally stupid? :)


inequalities
trigonometry
quadratics
Euler
algebra
inequalities unsolved

Oh,I think It's very well-Known.
$x=\cos{A},y=\cos{B},z=\cos{C}$

if A+B+C=180 then
cos^2 A+cos^2 B + cos^2 C + 2cosAcosBcosC=1
and cosA+cosB+cosC<=3/2

But we have to show that if $x^2+y^2+z^2+2xyz=1$ then there exist $A$, $B$, $C$, such that $A+B+C=180^{\circ}$ and $x=\cos{A}$, $y=\cos{B}$, $z=\cos{C}$.

This is not the same. At first, I suppose, we have to prove that $x,y,z \leq 1$. Don't you think so?

well, x,y,z <=1 are quite obvious because
x^2<x^2+y^2+z^2+2xyz=1

[quote="Remike"]But we have to show that if $x^2+y^2+z^2+2xyz=1$ then there exist $A$, $B$, $C$, such that $A+B+C=180^{\circ}$ and $x=\cos{A}$, $y=\cos{B}$, $z=\cos{C}$.[/quote]

Indeed. In other words, we have to prove:

[b]Theorem 1.[/b] If x, y, z are three positive reals such that $x^2+y^2+z^2+2xyz=1$, then there exists an acute-angled triangle whose angles A, B, C satisfy x = cos A, y = cos B and z = cos C.

[i]Proof.[/i] The proof is not as easy as some people think. First we start by noting that, since x, y, z are positive, we have $x^2 < x^2+y^2+z^2+2xyz = 1$, so that also x < 1. Similarly, y < 1 and z < 1. Since x < 1 and y < 1, there exist two acute angles A and B such that x = cos A and y = cos B.

Now, define an angle C = 180° - A - B. Since both angles A and B are acute, we have A < 90° and B < 90° and thus C = 180° - A - B > 180° - 90° - 90° > 0°. On the other hand, clearly, C < 180°. Thus, the angle C lies between 0° and 180°. Hence, with C = 180° - A - B, we have

Now we are going to prove that z = cos C. In fact, since A is an acute angle, $\sin A = \sqrt{1-\cos^2 A} = \sqrt{1-x^2}$ (since x = cos A), and similarly, $\sin B = \sqrt{1-y^2}$.

$\cos C = \cos\left(180^{\circ}-A-B\right) = \cos\left(180^{\circ}-\left(A+B\right)\right) = -\cos\left(A+B\right) = -\left(\cos A \cos B - \sin A \sin B\right)$
$= \sin A \sin B - \cos A \cos B = \sqrt{1-x^2}\sqrt{1-y^2}-xy = \sqrt{\left(1-x^2\right)\left(1-y^2\right)}-xy$.

On the other hand, the equation $x^2+y^2+z^2+2xyz=1$ can be rewritten as

$z^2 + 2xy\cdot z + \left(x^2+y^2-1\right) = 0$,

what is a quadratic equation for the number z, and its two solutions are

$z_1 = \frac{-2xy+\sqrt{\left(2xy\right)^2-4\left(x^2+y^2-1\right)}}{2}$;
$z_2 = \frac{-2xy-\sqrt{\left(2xy\right)^2-4\left(x^2+y^2-1\right)}}{2}$.

Actually, $z = z_2$ is impossible since $z_2$ is negative, while z was assumed to be positive. Hence, we must have $z = z_1$. Thus,

$z = \frac{-2xy+\sqrt{\left(2xy\right)^2-4\left(x^2+y^2-1\right)}}{2} = \frac{-2xy+\sqrt{4\left(1-x^2\right)\left(1-y^2\right)}}{2}$
$ = \frac{-2xy+2\sqrt{\left(1-x^2\right)\left(1-y^2\right)}}{2} = \sqrt{\left(1-x^2\right)\left(1-y^2\right)} - xy$.

Comparing this with the equation $\cos C = \sqrt{\left(1-x^2\right)\left(1-y^2\right)}-xy$ obtained above, we see that z = cos C. Also, the equation C = 180° - A - B yields A + B + C = 180°. Furthermore, we know that the angles A and B are acute; moreover, since C is an angle between 0° and 180° and its cosine is positive (cos C = z, and z is positive by assumption), it is also acute, and thus, the angles A, B, C are the three angles of an acute-angled triangle. Thus, Theorem 1 is proven.

Theorem 1 justifies the substitution x = cos A, y = cos B and z = cos C, with A, B, C being the angles of an acute triangle, for positive reals x, y, z satisfying $x^2+y^2+z^2+2xyz=1$.

BTW, we had a discussion of the same inequality at http://www.mathlinks.ro/Forum/viewtopic.php?t=21387 .

[b]PS.[/b]

[quote="Sailor"]Belive it or not, this property can be used to solve this [i]unsolved [/i] problem:
http://www.mathlinks.ro/Forum/viewtopic.php?t=23446[/quote]

Hmm, do you mean the identity $\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A\cos B\cos C = 1$ ? Well, maybe, I don't know (I have never been interested about that problem).

  Darij

This equality can be derived from the first cosine theorem.

I have the same idea with cosA,cosB,cosC.We use only that cosA+cosB+cosC=1+r/R=<3/2,using the Euler Inequalitie. ;)

We can use approach ABC to solve this problem
When we use this we have:a=b
Then we only have to prove this inequality with two variables such as:a and c or b and c.Easy?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3398963#p3398963]With condition x^2+y^2+z^2+2xyz=1:[/url]
$1< x+y+z \leq \frac{3}{2}.$

[quote="Remike"]Let $x$, $y$, $z$ be positive real numbers such that $x^2+y^2+z^2+2xyz=1$. Prove the inequality $x+y+z \leq 3/2$.[/quote]
[hide="Non-Trigo"]
We will prove that $x+y+z\leq \frac{3}{2}$ diven $x^2+y^2+z^2+2xyz=1$ and $x,y,z\in R^+$
now, if $x\geq y\geq z$
then $3x^2+2x^3\geq x^2+y^2+z^2+2xyz=1$
$\implies 3x^2+2x^3\geq 1$ if $x \leq \frac{1}{2}$ then $3x^2+2x^2\leq 1$ which is not possible hence
$x\geq \frac{1}{2}$
and $3z^2+2z^3\leq x^2+y^2+z^2+2xyz\implies 3z^2+2z^3\leq 1$ if $z\geq \frac{1}{2}\implies 3z^2+2z^3\geq 1$hence $z\leq \frac{1}{2}$
which means if $x\geq y \geq z$ then we can surely say that $x\geq \frac {1}{2}$ and $z\leq \frac{1}{2}$
but we can't say anything about $y$ i.e. it may be more that $\frac{1}{2}$ or less than $\frac{1}{2}$
now, forget about the condition $x\geq y\geq z$
we can say that two of $(x-\frac{1}{2})$, $(y-\frac{1}{2}$),$(z-\frac{1}{2})$ will bear the same sign
hence $(x-\frac{1}{2})(y-\frac{1}{2})\geq 0\implies (2x-1)(2y-1)\geq 0\implies 2xy+\frac{1}{2} \geq x+y$
hence $2xy + \frac{1}{2}+z \geq x+y+z\implies$ it is enough to prove
$\frac{3}{2}\geq 2xy+\frac{1}{2}+z$
i.e.,$1 \geq 2xy+z$
we will assume the contrary
i.e. $2xy+z\geq 1$
but $1=x^2+y^2+z^2+2xyz\geq 2xy+z(z+2xy)$
if $z+2xy\geq 1\implies z(z+2xy)\geq z \implies 2xy+z(z+2xy)\geq 2xy+z\implies 2xy+z(z+2xy)\geq 1$(as we have assumed $z+2xy\geq 1$
but we have already proved $2xy+z(z+2xy)\leq 1$ hence $2xy+z\geq 1$ is not possible hence $2xy+z\leq 1$
Hence Proved. :)[/hide]


hey i wanted to type up my algebra midterm in latex in the texnic center thing but i wanted to shorten the margins so i can fit more wording on a line

so does ne1 kno the latex for that? changing the margins?

I have it set to this for my document:

\textwidth 6.5in
\oddsidemargin 0in
\evensidemargin 0in

which works pretty well, but you can change it to whatever.   Those \()sidemargin commands are for the left side and the right side apparently depends on it--at least that's how I understand it, but I don't know much about $\LaTeX$.  And the odd, even are just odd and even pages, and they should probably be the same unless you are putting together a book.  Also, the \()sidemargin commands also accept negative values, so that should help if you are making the \textwidth bigger.


Are there integers $m$ and $n$ such that $5m^2 -6mn+7n^2 =1985$?

$ (5m \minus{} 3n)^2 \plus{} 26n^2 \equal{} 5.1985$
But $ (\frac { \minus{} 26}{397}) \equal{} \minus{} 1$ imply that it has no solution.

Solving the equation w.r.t. $ m$ we obtain $ m \equal{} \frac {3\pm\sqrt {9925 \minus{} 26n^2}}{5}$, which means $ 9925 \minus{} 26n^2$ must be a perfect square.
Let $ k^2 \equal{} 9925 \minus{} 26n^2$. So we have $ k^2\equiv_{13}6$, and it's easy to see $ 6$ is not a quadratic residue modulo $ 13$, therefore the equation has no solutions on integers.