It is not very attractive. 2.5 years in Algebra Unsolved Problems proved it! :)

Suppose $x>1$. Then $x^{3}+3x-3+\ln(x^{2}-x+1)>x$. It follows $y>x>1$, and consequently $1<x<y<z<x$ - a contradiction.

Suppose $x<1$. Then \[x^{3}+3x-3+\ln(x^{2}-x+1) \leq x^{3}+3x-3+(x^{2}-x)=x^{3}+x^{2}+2x-3<x,\] and the same contradiction again.

Therefore the only answer is $(1,1,1)$.


algebra
function
domain
complex analysis
polynomial
rotation
complex numbers

consider the power series $ f(z) = \sum_{k\ge 0} z^{2^k}$. it's convergent in $ \Delta = \{|z| < 1\}$.
what's the largest set $ \Omega\subset \mathbb{C}$ such that there exist an holomorphic function $ F: \Omega\to\C$ such that $ F|_\Delta = f$?
(i think such an $ F$ can be called the holomorphic extension of $ f$, but i'm not sure)

The key words are [i]Fabry's theorem[/i] though this particular case of it is much more elementary than the general version...

The unit circle is a natural boundary for this function.

Here's an argument for this case. (As fedja says, more elementary than the general theorem).

Suppose that there exists a connected $ \Omega$ on which such an extension is possible, and suppose that $ \Omega$ is strictly larger than $ \Delta.$ Then there exists $ z_0\in\Omega$ with $ |z_0|\equal{}1,$ and there exists $ \delta>0$ such that $ B(z_0;\delta)\subset\Omega.$ Choose an $ m$ such that $ 2\pi 2^{\minus{}m}<\frac{\delta}2.$

Write $ f(z)\equal{}\sum_{k<m}z^{2^k}\plus{}\sum_{k\ge m}z^{2^k}\equal{}g(z)\plus{}h(z).$

$ g(z)$ is a polynomial, hence entire, and we don't need to concern ourselves with it any further. We know that $ h(z)$ can be holomorphically extended to $ \Omega.$

Now for any integer $ j$ let $ h_j(z)\equal{}h(e^{2\pi i j2^{\minus{}m}}z).$ We note that in fact, $ h_j(z)\equal{}h(z)$ for $ |z|<1.$ We've just rotated the picture.

That means that $ h_j$ can be holomorpically extended to a domain that includes $ B(z_0;\delta),$ which means that $ h$ can be holomorpically extended to a domain that includes $ B(e^{2\pi ij2^{\minus{}m}}z_0,\delta).$

So we now have that if $ f$ can be holomorpically extended to $ F$ on $ \Omega$ where $ \Omega\setminus \Delta$ is nonempty then we must actually have that $ F$ can be holomorphically extended to a set which contains $ \Delta\cup\bigcup_{j\equal{}0}^{2^m\minus{}1}B(e^{2\pi ij2^{\minus{}m}}z_0,\delta).$

Since the balls overlap, this in turn includes the set $ |z|<r,$ for some $ r>1.$ That means that $ F$ is given by a power series centered at zero with radius of convergence at least $ r.$ But we already have the power series for $ F$ - it's the power series for $ f,$ and that series has radius of convergence exactly $ 1.$

So $ \Omega$ cannot contain any points outside of $ \Delta.$ The function $ f$ cannot be holomorphically extended at all.

my idea was a bit different...
call $ p_n(z) \equal{} z \plus{} z^2 \plus{} \dots \plus{} z^{2^{n \minus{} 1}}$: you can write $ f(z) \equal{} p_n(z) \plus{} f(z^{2^n})$.

now you notice that $ p_n$ does not vanish on $ (2^n \minus{} 1)$-th roots of unity (it's enough studying the imaginary part of that sum evaluated on one of them, if i'm not mistaken). on the other hand, if you consider such a root $ \omega$, you should have $ 0 \equal{} f(\omega) \minus{} f(\omega^{2^n}) \equal{} p_n(\omega)$, contradiction.


Is there a function defined on the positive integers with values in the same set such that f(f(n))=1993*n^1945 for all n? It is taken from Vietnam olympiad (guess when?) and is very nice.

It is a hard problem .
 Let M be set : {m \in N | (m,1993)=1 and m #p^1945 ,with all p  \in N}
   + ak  \in M
  for all n in N* exist the pair ( b ,k,r) b,k,r  \in N n=1993^b.ak^1945^r
f(n) = f(1993^b*a{k}^1945^r =
   +  1993^1945^(b+1) * a{k+1}^1945^(r+1) , if k is odd
   +  1993^b*a{k-1}^1945^r , if k is even

Yes, I consider it a very difficlut problem. Thanks for your solution.

Could you please explain it again..
i am sorry..i did not get it.. :)


[hide]
It is probably easier to deal with $x$-coordinates and $y$-coordinates separately.

Think about the horizontal component of the segment $AB$ (i.e., $x_2-x_1$) and find its length. Then find the appropriate $x$-coordinate of $C$.

Then work on $y$-coordinates in the same manner.[/hide]

[quote="lkryptonicl"]I have a weird question. Say you are given two points A(3,4) and B(6,-2). There is a point on AB, C, which divides the segment into a 3:4 ratio. How do you find the coordinates of C?[/quote]
[hide]
I would start by finding the length of the segment $(3\sqrt{5})$. Then the equation of the line passing through AB is $y=-2x+10$. Since the ratio is 3:4, we know that there are seven parts. Each part is $\frac{3\sqrt{5}}{7}$ units long. This multiplied by three is the distance away from A $(\frac{9\sqrt{5}}{7})$, and then using that, we find the point from A that is on the equation of the line. From the equation of the circle with the points equidistant from A, we see $\frac{405}{49}=(x-3)^2+(y-4)^2$ and solving the system (painfully) I got $x\approx.1437280611$ and $y\approx 9.712543878$[/hide]

Edit: o.0 looks like i posted too late

How can anything with a $y$-coordinate greater than 4 be on that segment?

Here's what I would do (this is what frt was talking about):
[hide]The length of the horizontal component is $6-3=3$, so the x-coordinate for the point will be $3+3(\frac{3}{7})=\frac{30}{7}$

Using the same method for the $y$-coordinate, you get $\frac{10}{7}$, so your answer should be $(\frac{30}{7},\frac{10}{7})$[/hide]

:fool:  :dry: that was dumb.

Would my method work at all (without the possible computation and organization errors)?

Or you can use vectors, which simplifies the calculation pretty much.

First you find the coordinates of $\overrightarrow{AB}$, which is just $\langle x_B-x_A, y_B-y_A\rangle$ (note that we calculate $B-A$ and not the other way round), which in the given case yields $\overrightarrow{AB}=\langle 3,-6\rangle$.

Next, you know that $AC$ is ${3\over 7}$ of $AB$, so you use the neat operation of multiplying a vector by a number. This yields a new vector which is collinear with the given vector, its length is the product of the given number and the length of the given vector, and its direction depends on the sign of the given number: if it's positive, the direction is the same as the direction of the given vector, and otherwise it's opposite.

Therefore, we can write $\overrightarrow{AC}={3\over 7}\overrightarrow{AB}={3\over 7}\langle 3,-6\rangle=\langle {9\over 7},-{18\over 7}\rangle$ (you just multiply both coordinates by the number in front).

Next, we must find the actual coordinates of $C$, which are always measured in respect to the origin $O$. By definition, a vector from the origin to a given point has the same coordinates as the given point. Therefore, we just need to find $\overrightarrow{OC}$, and the easiest way is by $\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}$, since both vectors are already known. We get $\overrightarrow{OC}=\langle 3,4\rangle + \langle {9\over 7},-{18\over 7}\rangle=\langle {30\over 7},{10\over 7}\rangle.$ (You just add the corresponding coordinates up.) Hence the coordinates of $C$ are $({30\over 7},{10\over 7}).$

I know not about vectors yet, but what i would do is find the midpoint, $M$, between segment $AB$. Then I would again find a midpoint either between $AM$ or between $MB$, it will depend on over who I want the 3:4 ratio.

Am I right or am I completely lost on the topic?

[quote="DanK"]That's pretty much what I did, I just didn't write them as vectors...[/quote]

Yup, and that's the theory behind the practice. Also, in my opinion, it's a bit "less messy" to calculate both coordinates at the same time than splitting the problem in two (but that can be just the matter of taste - both procedures are correct and equivalent).

Well, splitting it like that is the right idea, but you can't use midpoints, because it needs to be split into 7 segments...  2 midpoints would give you a $3: 1$ ratio...

Couldn't the distance formula be used, distance divided by 7, and then multiply that length by 3, and by 4???

So,
[hide="demonstration"]
$A(3,4) and B(6,-2)$

and

$\sqrt{(6-3)^2+(-2-4)^2} = \sqrt{9+36} = \sqrt{45} = 3\sqrt{5}$

so,

$3\sqrt{5}*3 = 9\sqrt5$ and $3\sqrt{5}*4 = 12\sqrt{5}$

and the distances would be $9\sqrt{5}$ and $12\sqrt{5}$
[/hide]

Yes, those are the correct distances, but that doesn't give you the coordinates of the point that divides it into segments of those lengths.  The easiest way to find that is working with x and y separately.

:dry: good point.....

but if a variable is put in for C, and the distance formula is used again for CB, then can't C be found?

Could someone explain vectors in greater detail?


[quote="shaaam"]Given that $ a, b, c$ are non-negative reals, prove that

\[ \frac{a}{\sqrt{a\plus{}2b}}\plus{}\frac{b}{\sqrt{b\plus{}2c}}\plus{}\frac{c}{\sqrt{c\plus{}2a}}\leq \frac{9}{8} \sqrt{a\plus{}2b\plus{}3c}\]

 :)

P.S: When does the inequality hold?[/quote]
See here for a stronger one: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=383369

[quote=shaaam]Given that $ a, b, c$ are non-negative reals, prove that

\[ \frac{a}{\sqrt{a\plus{}2b}}\plus{}\frac{b}{\sqrt{b\plus{}2c}}\plus{}\frac{c}{\sqrt{c\plus{}2a}}\leq \frac{9}{8} \sqrt{a\plus{}2b\plus{}3c}\][/quote]
Given that $ a, b, c$ are non-negative reals, [url=https://artofproblemsolving.com/community/c6h383369p2126398]prove that[/url]
$$\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\leq \frac{9}{8} \sqrt{a+2b+2c}$$

Given that $ a, b, c$ are non-negative reals. Prove that
$$\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\leq \frac{9}{8} \sqrt{a+2b+c}$$
$$\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\leq  \sqrt{a+2\sqrt{bc}}$$
$$\frac{a}{\sqrt{a+2b}}+\frac{bc}{\sqrt{(b+2c)(c+2a)}}\leq  \sqrt{a+b+c}$$


Karen wants to increase her physical stamina. Each morning she does push-ups and increases the number completed by five per day. After four days, she has done a total of 90 push-ups. How many did Karen do on the fourth day?

$ x\plus{}x\plus{}5\plus{}x\plus{}10\plus{}x\plus{}15\equal{}90$
$ 4x\plus{}30\equal{}90$
$ 4x\equal{}60$
$ x\equal{}15$
for first day
$ x\plus{}15\equal{}15\plus{}15\equal{}30$
for the 4th day

I don't really know any formulas for this, but what I do, is $ 90/4 \equal{} 22.5$ and I round it down to 20. That is the 2nd day. So now we have $ (20 \minus{} 5) \plus{} 20 \plus{} (20 \plus{} 5) \plus{} (20 \plus{} 10) \equal{} 15 \plus{} 20 \plus{} 25 \plus{} 30 \equal{} 90$ You are looking for the fourth day's amount so the answer is $ \boxed {30}$

Edit: Whoops, beaten by 1 minute.

Yeah, that's what I actually do during a FTW match. 90/4= 22.5, add 2.5 for third day, and another 5 for the fourth, arriving at 30.


Say I had a really annoying equation in my solution. Is "guess and check" a valid method to check for a solution to the equation? It is possible to solve the equation, it would require about 8 lines of really messy algebra. Would points be taken off?

I believe it is, but you must show that those are the only solutions to the equation as well then.


We only have to prove
$ a+b+c\ge \sum_{cyc}\frac{a(b+c)(b+c-a)}{a^{2}+bc}$
Or
$ \sum_{cyc}\frac{a(a^{2}-b^{2}-c^{2}+ab+ac-bc)}{a^{2}+bc}\ge 0$
Or
$ \sum_{cyc}\frac{a(a+2b+c)(a-b)+a(a+b+2c)(a-c)}{a^{2}+bc}\ge 0$
Or
$ \sum_{cyc}(a-b)\left(\frac{a(a+2b+c)}{a^{2}+bc}-\frac{b(2a+b+c)}{b^{2}+ca}\right)\ge 0$
Or
$ \sum_{cyc}(a-b)(a^{2}-b^{2})\left(\frac{c(a+b+c)-2ab}{(a^{2}+bc)(b^{2}+ca)}\right)\ge 0$
Now, I think I don't need to do the rest. The equality occurs when $ (a,b,c)\backsim(1,1,1)$ or $ (a,b,c)\backsim(1,1,2)$ and all permutations

[quote="behemont"][quote="ywl"]condition is equivalent to a,b,c sides of triangle.
try a=y+z,....etc.[/quote]

Can you explain the equivalence?[/quote]
the expresion is equivalent to $ s(s-a)(s-b)(s-c)\geq 0$ now some algebra will give you that they are the sides of a triangle..

what do you think about the soul of rock's proof?  :roll:

[quote="behemont"][quote="ywl"]condition is equivalent to a,b,c sides of triangle.
try a=y+z,....etc.[/quote]

Can you explain the equivalence?[/quote]
It is equivalent to $ (a+b+c)(-a+b+c)(a-b+c)(a+b-c)\geq 0$.

[quote="zaizai-hoang"]$ \bigstar$ Let $ a,b,c$ be positive real numbers such that: $ 2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})\ge a^{4}+b^{4}+c^{4}$. Prove that:
\[ \frac{b+c}{a^{2}+bc}+\frac{c+a}{b^{2}+ca}+\frac{a+b}{c^{2}+ab}\le \frac{3(a+b+c)}{ab+bc+ca}\]
[/quote]
$ \sum_{cyc}\frac{b+c}{a^{2}+bc}\leq\frac{3(a+b+c)}{ab+ac+bc}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{1}{a}-\frac{b+c}{a^{2}+bc}\right)\geq\frac{a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}-abc(a+b+c)}{abc(ab+ac+bc)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a-b)(a-c)}{a(a^{2}+bc)}\geq\sum_{cyc}\frac{bc(a-b)(a-c)}{abc(ab+ac+bc)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a-b)(a-c)(b+c-a)}{a^{2}+bc}\geq0.$
Let $ a\geq b\geq c.$ It remains to prove that $ \frac{a+b-c}{c^{2}+ab}\geq\frac{a+c-b}{b^{2}+ac},$ which true because
$ a+b-c>a+c-b$ and $ b^{2}+ac\geq c^{2}+ab\Leftrightarrow(b-c)(b+c-a)\geq0.$ :)

yeah, this is also my solution  :blush: It is easy  :P


algebra polynomial function quadratics geometry induction AMC

[hide="Solution"]Notice how the first figure has $1+4$ squares, the second has $1+4(1+2)$ squares, the third has $1+4(1+2+3)$ squares, and etc.

Thus, the 100th figure will have $1+4(1+2+\cdots+100)=20201$ squares.[/hide]

Whats with the broken lines in figure 3?

I needed some polynomial work.

[hide]The second differences come out to be all fours. Thus the function for figure n is a quadratic, and the function is $f(n)=2n^2+2n+1$. The 100th figure would have $2(100^2)+2(10^2)+1=20201$ squares. [/hide]

This is quite simple.  The left and right side squares add up to $2n^2$ squares (Note: $\sum_{i=1}^n(2i-1)=n^2$).  The middle column has $2n+1$ squares.  Therefore, we add all the squares and get $f(n)=2n^2+2n+1$.  It could also be done using area in terms of the $n$th figure.

EDIT: I had an x in f instead of an n. :P
Masoud Zargar

[quote="now a ranger"]How did you derive the polynomial equation?

Is this combinatorics?[/quote]

Alright you have, $f(0)=1, f(1)=5, f(2)=13, f(3)=25$

This is a quadratic because all the second differences are the same ($f(1)-f(0)=4, f(2)-f(1)=8, f(3)-f(2)=12$ are the first differences, then $8-4, 12-8$ are the second differences, since these are the same it is a quadratic). Quadratics are of the form $ax^{2}+bx+c$. So, plug in $0, 1, 2, 3$ in for x, and solve for $a,b,c$. This gives you $f(x)=2x^{2}+2x+1$

And if you wanted to be nitpicking, you would have to prove that this quadratic function works via induction, [b]boxedexe[/b]'s method, or a variety of other ways. On a written proof, I wouldn't just use the finite differences without a proof that the formula works since you don't know if the next values will agree or not; but on a contest where only the answer is required (AMC, AIME, etc) it'd be perfectly fine--whatever saves you time is the best method.

here's a simple, intuitive way to find a formula for this sequence:
[hide]
If you paint a chessboard pattern on each picture, you'll see that there are $n^2$ squares of one colour and $(n+1)^2$ squares of the other colour. So, the total is $n^2+(n+1)^2$.
[/hide]

[hide="Visual PIE"]Around figure $n$, draw a large square with $2n+1$ squares on each side.  (In other words, fill in the four corners.)  You have $(2n + 1)^2$ small squares in this figure.

You now must remove one "triangle" from each of the four corners.  Each of these triangles have $1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$ squares in them.  Thus, in total, you're removing $2n(n+1)$ squares when you remove all four corners.

So, there are $(2n+1)^2 - 2n(n+1) = 2n^2 + 2n + 1$ squares left in the figure.[/hide]

[hide=Another View]
Rotate all of the figures 45 degrees as in the attachment.
Now it is obvious the formula is $(n+1)^2+n^2$.
Plugging in $n=100$ gives the answer $\boxed{\textbf{(C)} 20201}$
[/hide]

[hide]I noted that since figure $x$ has $x^2+(x+1)^2$ squares, so Figure 100 had $100^2+101^2=10000+10201=20201$ squares.[/hide]
EDIT: Oops, I just realized that other people already posted this, only $x^2+(x+1)^2$ was simplified to $2x^2+2x+1.$

I had a kind of longer solution.

[hide]I basically thought of it as a square with a certain number of blocks subtracted for each of the four corners. I wrote it out.

0: $1^2 -4*0$
1: $3^3 -4*1$
2: $5^3 -4*3$
3: $7^3 -4*6$

We can see a pattern for the very last number. For index 0 it's 0. For index 1 it's 1. For index 2 its 3 = 1 + 2. For index 3 it's 6 = 1 + 2 + 3. I just assumed that it would continue like this. Since $\sum_{i=0}^n i = (1/2)n(n+1)$, we can figure out the nth index.

0: $1^2 -4*0$
1: $3^3 -4*1$
2: $5^3 -4*3$
3: $7^3 -4*6$
... .......
n: $(2n+1)^2 -4*\sum_{i=0}^n i = (2n+1)^2 -4*(1/2)n(n+1)$

Just replace n with 100 and you get $20201$.
So the answer is $\boxed{C}$.
[/hide]

[quote=Hu_bryan]I know it is not necessary, but one can also use differential equations

As the figure n increases, the number of squares being added on each time is increasing by 4. Thus we can say:

y' = 4n + 1
dy/dn = 4n + 1
 ∫dy = ∫(4n + 1) dn
y = 2n^2 + n + C          // C in this situation is 1, the number of squares in figure 0
y = 2n^2 + n + 1          // Plug in 100 for n
y = 20101[/quote]

Actually, using differential equations doesn't work; the integral represents integrating over a continuous interval, not over increments.
If you were to set up a differential equation for the continuous version of the function, you would have $y'=4n$ in a continuous form and integrating would yield $y=2n^2+C$.
You could use an initial condition of $y(0)=1$ to let $C=1$.
This yields an approximation of the result, as $2(100^2)+1$ is closest to $B$ and $C$.
This can be resolved by noting that the function under-approximates the original incremental approach.

[hide=Solution]Note that we have $1 + 4 \cdot 0, 1 + 4 \cdot 1, 1 + 4 \cdot 3, 1 + 4 \cdot 6$, and so on. The pattern is that in figure $n$, the amount is $1+4 \cdot \frac{n(n+1)}{2} \implies 20201 \implies \boxed{\text{(C)}}$.[/hide]

In first its [b] 1,5,13...,[/b]so as i observed its increasing in format of  [b]1+1(4)+2(4)............n(4).[/b]
 [b] so for 100 it will be 1+4+8+16....400 so simply leaving 1 is coming an A.P. series so we will apply the sum of an A.P. formula - n/2(1st term + last term )  in 4+8+16...400
which is coming as 20200 the add the 1 so its 20201 (C).[/b]

It is quite simple.

ans = 20201
general form = 2n^2   + 2n + 1
solution :
                       *  Ignoring the line in mid first.
                       * As we know that sum of first n odd numbers if n^2
                       * Hence, n^2   + n^2   = 2n^2
                       * Now as we know that the line in mid is 2n+1
so : 2n^2  + 2n  +  1


An amusement park has a giant double ferris wheel. The double ferris wheel has a 30 meter rotating arm attached at its center to a 25 meter main support (10 meters off the ground). At each end of the rotating arm is attached a ferris wheel measuring 20 meters in diameter rotating counterclockwise. The rotating arm takes 6 minutes to complete one full revolution, and each wheel takes 4 minutes to complete a revolution about that wheel's hub. At time t=0 the rotating arm is parallel to the ground and your seat is at the 3 o'clock postition of the rightmost wheel. SEE ATTACHED FIGURE(Not to Scale).

a) Find a formula for h=f(t), your height above the ground in meters, as a function of time in muinutes. [Hint: Your height above the ground equals the height above your wheel hub above ground plus your height above that hub.]

b) Is f(t) periodic? If so, what is it's period?

c) Approximate the least value for t such that h is at a maximum value. What is this maximum value?

Thanks for your help.

Before anyone answer, this is very like a homework so please do not answer unless the poster shows some attempt.

@golfmaster925: What work have you done to this problem so far? Care to show? We're here to help you and guide you through the work, not do the problems.  :)

I spent 2 hours on it yesterday. It's not a homework problem, its a challenge problem that I recieved for fun. This is what I came up with:

b) The function must be periodic since 2 revolutions of the big wheel = 3 revolutions of the small wheel. The period is 12 minutes because it will take them that amount of time to reach the starting point again.

I also figured out some basic things about each of the wheels, assuming that you were attempting to graph there cycles seperately.

[u]Large Wheel: [/u]
min= 10m
max= 40m
amp= 15m
midline= 25m
frequency= 1
period= 6min

[u]Small Wheel:[/u]
min= ((sin 45)*10)+10= ((radical 2/2)*10 +10)= approx. 17.07
max= ((sin 135)*10)+50= ((radical 2/2)*10 +50)= approx. 57.07
amp= 37.07
midline= 40 m
frequency= 2/3
period= 4min

I also graphed the entire problem by hand, but it didnt look to be periodic in any way except for every 12 minutes, so I was unsure on how to come up with a formula.

If you can point me in the right direction that would be great. I only know Precalculus material, so if you can keep it simpler that would be great. Thanks.

Find a formula for the height of the center of the small circle (which is easy enough).  Then find a formula for the height of the seat [i]relative to the center of the small circle[/i].  Then add them.  :)

Ok, so the formula for the height of the center of the small circle would be: 7.5sin(x)+25 . The period would be 6 minutes or just 1 for the formula. Would you then add 10 to the formula I just found to get the formula for the entire problem?


A piece of paper with the shape of a square lies on the desk. It gets dissected step by step into smaller pieces: in every step, one piece is taken from the desk and cut into two pieces by a straight cut; these pieces are put back on the desk then.
Find the smallest number of cuts needed to get $100$ $20$-gons.

Proceed as follows:

Each cut increases the total number of sides by at most 4.
Cutting an $m$-gon leads to an $n$-gon and an $r$-gon where wlog $3 \leq r \leq n \leq m + 1$.  Thus, the only way to increase the number of sides of a given shape is to cut off triangular corners.
This means that given 100 20-gons at the end, we can put a lower bound on the number of triangles that resulted from our process -- 100*16, in fact -- one for each corner after the 4th (since we can get quadrilaterals without creating new triangles) of each 20-gon.  So we must have at least 100*20 + 100*16*3 = 6800 edges once we are done.  This requires at least (6800-4)/4 = 1699 cuts.

Is this acheivable?  Well, use 99 cuts of the original paper to create 100 quadrilaterals.  Then, we need 16 cuts on each of the quadrilaterals, each taking off just one corner, and this indeed gives us 1699 cuts.  So this is in fact the minimum.  Indeed, any process that you follow will use this many cuts, provided you don't create excess polygons with more than 3 sides.

This answer generalizes very easily to any collection of polygons with "not too many" triangles.  Just, watch out for paper cuts.


ratio
AMC
AMC 10
AMC 12
Pythagorean Theorem
geometry

Circles with centers $ A$ and $ B$ have radii 3 and 8, respectively. A common internal tangent intersects the circles at $ C$ and $ D$, respectively. Lines $ AB$ and $ CD$ intersect at $ E$, and $ AE \equal{} 5$. What is $ CD$?
[asy]unitsize(2.5mm);
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=3;

pair A=(0,0), Ep=(5,0), B=(5+40/3,0);
pair M=midpoint(A--Ep);
pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];
pair D=B+8*dir(180+degrees(C));

dot(A);
dot(C);
dot(B);
dot(D);
draw(C--D);
draw(A--B);
draw(Circle(A,3));
draw(Circle(B,8));

label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,SE);
label("$E$",Ep,SSE);
label("$D$",D,NW);[/asy]$ \textbf{(A) } 13\qquad \textbf{(B) } \frac {44}{3}\qquad \textbf{(C) } \sqrt {221}\qquad \textbf{(D) } \sqrt {255}\qquad \textbf{(E) } \frac {55}{3}$

[hide]$\angle AEC\cong \angle BED$ by vertical angles, and $\angle ACE\cong \angle BDE$ since both are right angles. Then $\Delta AEC$ and $\Delta BED$ are similar, with side ratio $\frac 38$.

By Pythagorean Theorem, $CE=4$, which makes $DE=\frac{32}{3}$. So $CD=\frac{44}{3}\Rightarrow \boxed{B}$.[/hide]

[hide="Answer"]Note that $\triangle ACE$ is a right triangle, so $CE=4$. $\triangle ACE\sim \triangle BDE$, so $DE=\frac{32}{3}$, and $CD=\frac{44}{3}\Rightarrow \boxed{B}$.[/hide]

I drew in the line parallel to $AD$ through $B$ to get a similar triangle. I always do it the hard way :wallbash:

This was a 23? I scoff at its easy-ness.

Draw in tangent to center lines and use similar triangles.

[quote="PenguinIntegral"]This was a 23 [b]on the AMC 10[/b]? I scoff at its easy-ness.

Draw in tangent to center lines and use similar triangles.[/quote]

AMC10 problems do look fairly easy if you're really good at math. It makes sense. I was surprised to see that as well, but then comparing it to the AMC12, having it as a #16 isn't [b]too[/b] unreasonable. ;)

[hide=solution]$CD=CE+ED$. Note that $CE=\sqrt{AE^{2}-AC^{2}}=\sqrt{5^{2}-3^{2}}=4$. Also, $\triangle ACE\sim\triangle BDE$ so $AC/CE=BD/DE\rightarrow\cfrac{3}{4}=\cfrac{8}{DE}\rightarrow DE=\cfrac{32}{3}\rightarrow CD=4+\cfrac{32}{3}=\boxed{\textbf{(B)} ~\cfrac{44}{3}}$

To be blunt, you're basically postfarming. I can't understand what you're saying as you provide mindless numbers and no explanations. Please at least attempt to explain yourself. If you can't do that, what's the point of sending your post to the community? It's an answer with absolutely no validation, and we're not frantically trying to determine our score because this is 2021, not 2006.

[hide=Solution]Connect $AC$ and $BD$. We have $AC=3$ and $BD=8$, meaning that $EB=\frac{40}{3}$. Applying the Pythagorean Theorem, we get $DE=\frac{32}{3} \implies \frac{32}{3} \cdot \frac{11}{8} = \frac{44}{3} = \boxed{\text{(B)}}$.[/hide]

[quote name="OlympusHero" url="/community/p21651518"]
To be blunt, you're basically postfarming. I can't understand what you're saying as you provide mindless numbers and no explanations. Please at least attempt to explain yourself. If you can't do that, what's the point of sending your post to the community?
[/quote]
Don't discourage people like this. At least he is doing math, unlike the majority of recent posts in C&P.

[quote=rdspear]Writing out an equation with no explanation or logic is not "doing math".[/quote]

why are so saying mean things to me AGAIN? :(

this is making me so sad :(

[hide=#12:][quote name="rdspear" url="/community/p21673488"]
Writing out an equation with no explanation or logic is not "doing math".
[/quote][/hide] to be fair it is math, just not expressed very well

[quote=pog]Maybe, instead of being concerned at the fact that someone could dare to be "mean" to you, concern yourself with why they [i]are[/i] criticizing your behavior in the first place.[/quote]

:(

ok :(

i make better solution now :(

because u guys hurt my felings a lot :(

[hide=Solution]
We draw radii $AC$ and $BD$ to get that $3^2+CE^2=5^2$ and so $CE=4$.

We use similar triangles to get the answer of $\frac{8+3}{3}4=\boxed{\frac{44}{3}}$
[/hide]

i did power of a point E with respect to the circle with radius 3

Join AC,BD and the key points are Pythagoras theorem and similarity


"Using" how?  $+,-, \times, \div$?  Digit concatenation?

No no, I think the author of this topic wants you to write primes with only ten numbers (0;1;2;3;4;5;6;7;8;9). You are not allowed to use $+;-; \times ; \div$

[quote="wxydx00"]How many primes can be made by using numbers:1,2,3,4,5,6,7,8,9
each number can be used once,also you can don't use some of them.[/quote]

do you mean i cannot use some of them and i must use all of them? then if 1,2,3,4,5,6,7,8,9 are to be the digits of that prime, the answer is 0. because the number will then be divisible by 3.

I think "can don't use some of them" means "are allowed to not use some of them."  So... if the author meant digit concatenation, then the list will be long.  Every prime under $100$ except for $11$, for example, can be made.

[quote="wxydx00"]How many primes can be made by using numbers:1,2,3,4,5,6,7,8,9
each number can be used once,also you can don't use some of them.[/quote]

I mean that 1,2,3,4,5,6,7,8,9 can be only used once.
the answer is:
3,5,7,13,17,19,23.....
and
11,131....is not the answer.
The question is just how many, not ask you to point out all of them

Maybe it is a boring answer, but I want to make sure whether there is some beautiful solution of this problem...
My English is quite poor,sorry:)

First of all, this is not an algebra problem.  Second of all, why on earth would you expect this to yield a "nice" answer or solution?  It's just extremely ugly.  For example, there are 97 primes which use each of the digits 1, 2, 3, 4, 5 and 7 exactly once.


Hi People,

I am trying to have the statments (as accurate as possible) from the USAMOs. I have the book with the statements from 1972 to 1986, and I had found in the official site of USAMO the statements from 1989 up to 2003. I also found in the web the 2004, but I haven't been able to find the statement sheets for 1987 and 1988, and I mean, official ones. I have a good library available, if you can only send me some reference about when and where were they published, or if you have them in digital format and can send them to me (djimenez@walla.com) would be very appreciated.

I know that John Scholes have his version on his webpage, but he changes the statments "a little" (sometimes A LOT) to avoid any responsability for copy rights violation. As it is for private use, I believe there will be no problem if you have them and send them to me.

Best,

David

Why do you need the original problem statements ?  :)

Well, the answer is pretty simple. Let's put the example of the site of Scholes (kalva), he says he "changes the wording" of the problems, but sometimes he either gives an important key, or totally changes the problem. Introduces aspects that changes the soul of the problem and sometimes while refrasing, he actually comes up with a wrong problem.

It is not that I want to undervalue the excelent job that John Scholes has done, I understand his limitations, and I also appreciate the huge collection that he has done, but if I can get the original statements for any other method, I would like to.

Now, why do I want them, well, I enjoy to spend some time solving them. Also, I am now out of my country, doing my Ph.D., but when I was there I used to be one of the trainers of the Iberoamerican Team (and this year is the first time that there is going to be a team for the IMO) and I would like to have a BIG collection to train.

Best Regards,

David

BUMP.

Especially the correct wording of USAMO 1987/4 would be welcome, the version on Kalva seems bugged (an infimum isn't a minimum, and one of the given inequalities is not needed).


Let $x,y,z,t \ge 0$ and $x+y+z+t=1$. Show that:
\[ (1-\sqrt{x})(1-\sqrt{y})(1-\sqrt{z})(1-\sqrt{t}) \ge \sqrt{xyzt}  \]

so we have to show that :
$1+%Error. "sumx" is a bad command.
+\sum xy+\sum xyz +xyzt \geq \sqrt{xyzt}(1+\sum \sqrt{x}+\sum \sqrt{xy}+\sum \sqrt{xyz}+\sqrt{xyzt})$
or that $1+\sum x +\sum xy+ sum xyz \geq \sqrt{xyzt}(1+\sum \sqrt{x}+\sum \sqrt{xy}+\sum \sqrt{xyz})$
which can be shown by use of AM-GM or muirhead :)


Where would a simple number theory problem go? I put it in High School Basics for now, but should it be elsewhere?

Depends if it is Middle School level, High School or Olympiad Level.  I don't know the problem so I can't tell if you what level it is. :D


if we replace $ \frac{n}{n\plus{}1}$ by $ \frac45$ then we have an easy inequality


The only real difference is that they have a different set of problems. Also, the AIME Problem Solving Seminar takes all of its material from the AIME A Problem Series.

[quote="13375P34K43V312"]What's the difference between the AIME Problem Series' A and B, if much at all?[/quote]
They have two series at different times of year, since passing the AIME is so hard,  so people can practice by taking the class twice, and do different problems.

These responses are essentially correct.  The two classes differ mainly in that different problems are presented.


geometry
geometric transformation
reflection
function
inequalities
number theory proposed
number theory

Let $ p$ be an odd prime.
Let $ a$ and $ b$ be two integers such that none of $ a$, $ b$, $ a\plus{}b$ are divisible by $ p$.
We call an integer $ n$ good, if it satisfies the following:
(1)  $ 1\le n\le p \minus{} 1$
(2)  $ an \bmod p \le (p \minus{} 1)/2$
(3)  $ bn \bmod p \le (p \minus{} 1)/2$

Show: There exist at least $ (p \minus{} 1)/6$ good integers.

(Corrected on October 18, 2007)

[quote="gjw"]Let $ p$ be an odd prime.
Let $ a$ and $ b$ be two integers that are not divisible by $ p$.
We call an integer $ n$ good, if it satisfies the following:
(1)  $ 1\le n\le p \minus{} 1$
(2)  $ an \bmod p \le (p \minus{} 1)/2$
(3)  $ bn \bmod p \le (p \minus{} 1)/2$

Show: There exist at least $ (p \minus{} 1)/6$ good integers.[/quote]

That's wrong : Take $ a \equal{} 1$, $ b \equal{} 2$ and $ p \equal{} 3$ and we have $ 0 < (p \minus{} 1)/6$ good numbers

[quote="pco"]That's wrong : Take $ a \equal{} 1$, $ b \equal{} 2$ and $ p \equal{} 3$ and we have $ 0 < (p \minus{} 1)/6$ good numbers[/quote]

Yes, I forgot to state the condition that $ a\plus{}b$ must not be divisible by $ p$.
Sorry for this.
I have corrected the statement above.

(The original statement was actually wrong for all odd primes $ p$:
Setting $ a\equal{}1$ and $ b\equal{}p\minus{}1$ would yield a counterexample.)

[b]P(re).S.[/b] [i]I could have summarised the proof in a slightly crispier and more efficient form, but it reflects my initial brute force thought process, which might be of some interest.[/i]

First some notation. For any integer $ m$ define the sets

$ \Omega(m)=\{n=1,\ldots,p-1\,|\,1\leq nm\bmod{p}\leq\frac{p-1}{2}\}$ and
$ \Omega_0(m)=\{n=1,\ldots,\frac{p-1}{2}\,|\,1\leq nm\bmod{p}\leq\frac{p-1}{2}\}$

In order to check the claim about the number of good integers let $ x$ be a solution of $ ax\equiv b\bmod{p}$. In particular, $ x\not\equiv 0,-1\bmod{p}$ is valid. Then the number of good integers with regard to $ (a,b)$ is given by:

$ |\{n=1,\ldots,p-1\,|\,1\leq an,bn\bmod{p}\leq\frac{p-1}{2}\}|\\[0.2cm]
=|\{an=1,\ldots,\frac{p-1}{2}\,|\,1\leq xan\bmod{p}\leq\frac{p-1}{2}\}|\\[0.2cm]
=|\Omega_0(x)|\\[0.2cm]$

since $ an$ runs through all residues modulo $ p$ different from zero. Now we can either apply the following lemma to complete the proof or $ x\equiv 1\bmod{p}$ holds, which trivially fulfills the claim due to $ |\Omega_0(1)|=\frac{p-1}{2}>\frac{p-1}{6}$.

[b]Lemma[/b]: For $ m\not\equiv 0,\pm 1\bmod{p}$ we have $ \frac{p-1}{6}\leq |\Omega_0(m)|\leq\frac{p-1}{3}$

[i]Proof of lemma[/i]: First note, that $ |\Omega(m)|=\frac{p-1}{2}$ holds, since $ nm$ runs through all residues modulo $ p$ different from zero. From $ n\in\Omega_0(m)\Longleftrightarrow p-n\in\Omega(m)\setminus\Omega_0(p-m)$, we conclude $ |\Omega_0(m)|+|\Omega_0(p-m)|=\frac{p-1}{2}$. Because of the properties $ \Omega_0(m)=\Omega_0(m+p)$ and $ \frac{p-1}{6}+\frac{p-1}{3}=\frac{p-1}{2}$ and $ |\Omega_0(2)|=[\frac{p-1}{4}]$, it suffices to verify the claim for all $ m=3,\ldots,\frac{p-1}{2}$ and $ p\geq 7$.

Let $ p\leq kp<\frac{p-1}{2}m<(k+1)p$ and denote the integer part function by $ [\ldots]$, i.e. $ k=[\frac{(p-1)m}{2p}]$. Define $ b_0=0$ and for $ i=1,\ldots,k+1: b_i=\max\{t=1,\ldots,\frac{p-1}{2}\,|\,tm\leq ip\}=\min\{[\frac{ip}{m}],\frac{p-1}{2}\}$ and $ a_i=\max\{t=1,\ldots,\frac{p-1}{2}\,|\,tm\leq (i-1)p+\frac{p-1}{2}\}=\min\{[\frac{(2i-1)p-1}{2m}],\frac{p-1}{2}\}$.

We clearly observe $ 0=b_0<a_1<b_1<\ldots <a_k<b_k<a_{k+1}\leq b_{k+1}=\frac{p-1}{2}$ and making the substitution $ p=2mt+r$ with $ t\in\mathbb{N}$ and $ 1\leq r<2m$ we compute for $ i=1,\ldots,k$:

(1)  $ b_i-b_{i-1}=2t+[\frac{ir}{m}]-[\frac{(i-1)r}{m}]\in\{2t,2t+1,2t+2\}$ and
(2)  $ a_i-b_{i-1}=t+[\frac{(2i-1)r-1}{2m}]-[\frac{2(i-1)r}{2m}]\in\{t,t+1\}$ and
(3)  $ b_i-a_i=t+[\frac{2ir}{2m}]-[\frac{(2i-1)r-1}{2m}]\in\{t,t+1\}$

In particular, we have for all $ i=1,\ldots,k\,: \,b_i-a_i\not=a_i-b_{i-1}\Longleftrightarrow b_i-b_{i-1}=2t+1$. From this we conclude $ \frac{a_i-b_{i-1}}{b_i-b_{i-1}}\geq\frac{t}{2t+1}\geq \frac{1}{3}$, which obviously holds for $ i=k+1$ as well, since $ b_{k+1}>a_{k+1}\Longrightarrow a_{k+1}-b_k,\in\{t,t+1\}\,,\,b_{k+1}-a_{k+1}\leq t+1$. Therefore, we have $ |\Omega_0(m)|=\sum_{i=1}^{k+1}(a_i-b_{i-1})\geq\frac{1}{3}\cdot \sum_{i=1}^{k+1}(b_i-b_{i-1})=\frac{p-1}{6}$, which verifies the left-hand inequality of the lemma.

As for the right-hand inequality of the lemma note, that we obviously have $ |\Omega_0(m)|\leq(k+1)(t+1)$. Moreover, we have $ \frac{p-1}{2}-|\Omega_0(m)|=\sum_{i=1}^{k+1}(b_i-a_i)\geq\sum_{i=1}^k(b_i-a_i)\geq kt$.

If $ kt\geq\frac{p-1}{6}$ or $ (k+1)(t+1)\leq\frac{p-1}{3}$ holds, then we can conclude $ |\Omega_0(m)|\leq\frac{p-1}{3}$. Hence, we can assume the opposite inequalities $ kt <\frac{p-1}{6}$ and $ (k+1)(t+1) >\frac{p-1}{3}$. This implies $ (k+1)(t+1)>2kt\Longrightarrow k+t\geq kt\Longrightarrow t=k=2 \vee t=1\vee k=1$.

The case $ t=k=2$ requires $ 25<p<28$, which does not have a solution among the primes.

The case $ k=1$  yields $ m\leq 4$. Direct calculation gives $ |\Omega_0(3)|=\frac{p-1}{2}+[\frac{p-1}{6}]-[\frac{p}{3}]$ and for $ p\geq 11$ we get $ |\Omega_0(4)|=[\frac{p-1}{8}]+[\frac{3p-1}{8}]-[\frac{p}{4}]$, which both fulfill the assertion; for the proof note, that for, e.g., $ m=3$ we have $ r\in\{1,5\}$ and after substitution and transformation $ |\Omega_0(3)|=2t+\frac{r-1}{2}+[\frac{r}{3}]=\{\begin{array}{ll}\frac{p-1}{3} & \mbox{if }r=1\\[0.2cm] \frac{p-2}{3} & \mbox{if }r=5\end{array}$

The case $ t=1$ yields $ k<\frac{p-1}{6}<k+1\Longrightarrow p\equiv 5\bmod{6}$ and $ k=[\frac{p-1}{6}]=\frac{p-5}{6}$, which implies $ p\geq 11$ and $ |\Omega_0(m)|\leq \frac{p+1}{3}$. If the inequality is strict, then we are finished. Therefore, we can assume equality and we have to show that this will create a contradiction. An immediate consequence is that ${ a_{k+1}}=b_{k+1}$ and for all $ i=1,\ldots,\frac{p+1}{6}: \,a_i-b_{i-1}=t+1=2$ and for all $ i=1,\ldots,\frac{p-1}{5}: \,b_i-b_{i-1}=2t+1=3$ must hold. In particular, we have $ \frac{p-5}{6}p<\frac{p-1}{2}m\leq\frac{p-5}{6}p+\frac{p-1}{2}\Longrightarrow m=\frac{p-2}{3}$.

However, we get the contradiction $ a_1=1,b_1=3$, which completes the proof of the lemma.


calculus
integration
calculus computations

[quote="PeterP"]0.
The graphs of $ y \equal{} \sqrt [3]{1 \minus{} x^7}$ and $ y \equal{} \sqrt [7]{1 \minus{} x^3}$ are symmetric in the line $ y \equal{} x$.[/quote]

Yep, I did that as well.  :)

Any other solutions? (I'd like to see non-graph ones)

Here's a different presentation of the same proof:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=942620&#942620[/url].


For Part A, I got $ \text{C}_3\text{H}_8 \plus{} 5\text{O}_2 \rightarrow 3\text{CO}_2 \plus{} 4\text{H}_2\text{O}$.  Using the Ideal Gas Law, I got $ 27.4$ L $ \text{O}_2$ for Part B.  But I'm stuck with parts C and D.  Could someone please help?  Thanks!

(c) Hess Law.

(d) Considering a constant heat capacity, $ Q \equal{} mc \Delta T$.


Find 6 points in the plane such that any two of them are at an integer distance apart, no three of them are collinear, and no four lie on the same circle.

make a hexagon by gluing 6 congruent triangles of sides 68,85,87.
Can you find 7 :D

Nice solution!  For seven points, I found this reference:
[url]http://www.maa.org/editorial/mathgames/mathgames_05_02_07.html[/url]


Is anybody here interested in the pyraminx, a variation of the Rubik's cube?

[url=http://en.wikipedia.org/wiki/Pyraminx]Wikipedia[/url]
[url=http://www.mud.ca/puzzler/JPuzzler/JPuzzler.html]Virtual pyraminx[/url]
[url=http://www.worldcubeassociation.org/results/events.php?eventId=pyram&regionId=&years=&show=100%2BPersons&single=Single]World Records[/url]

[hide="My Story"]I have had a pyraminx for a very long time (originally bought almost 20 years ago, before I was born), but I never knew how to solve it until last year. After getting the pyraminx into working condition (it barely moved and constantly got stuck), I started figuring out how to solve it (no internet, just logic). It took me almost a day to get the basic ideas down, and then another couple of days to invent algorithms and perfect the technique. Once I had a method, I started practicing. At first I could barely do it in less than a minute, but I quickly improved and got my time down to a constant 20 seconds. After this, you can't improve very fast anymore. It took a couple of weeks after my first sub-20 to get my first sub-10. Very soon after that, I got a lucky 5-second solve. And that's basically where I am right now. I haven't practiced for quite a while now, but I can still get sub-10 without much effort.[/hide]
Personal best (lucky): 3.14 seconds (what a coincidence: pi!)
Personal best single solve: 4.92 seconds
Personal best average of 5: 5.80 seconds (this is where I got the 4.92 non-lucky and a sub-4 lucky)
Personal best average of 10: 6.79 seconds (this should be much close to 5)
Personal best average of 100: 8.12 seconds

If you own one or have ever tried solving one, vote in the poll and share your method!

It looks very interesting! I'm pretty sure that everyone wants a break from the original Rubik's cube, and why not another puzzle in another shape? It also looks like something littler kids would play, because the Rubik's cube was too difficult. Perfect for me  :lol:

Average of 10: [b]6.71 PB[/b] :D
Average of 5: 6.40 :(

I have a 3-by-3-by-3, and I instantly looked at the solutions manual.  :wallbash_red:

On i_like_pie's pyraminx, with i_like_pie timing... I don't remember. I think it was about 25 seconds or so. But then again I only tried a few times. Oh, and i_like_pie, what happened to your claimed 2.71 second solve? :P

[quote="i_like_pie"][quote="xscapezaer"]Oh, and i_like_pie, what happened to your claimed 2.71 second solve? :P[/quote]
That was too lucky to even mention. It was a proper scramble and it looked well-scrambled, but a 5-move solve is much too fast to count.

Recently:

Average of 10: 6.73 (just 0.02 above best)[/quote]

Whoa I can't even solve one (haven't tried but still). That's pretty beast. That's faster than my 2x2 time.

http://www.worldcubeassociation.org/results/events.php?eventId=pyram&regionId=&years=&show=100%2BPersons&average=Average

Go to a competition and get the National Record (or North American)


calculus
integration
logarithms
trigonometry
calculus computations

\[ \int {\log (\sqrt {1 \plus{} x}  \plus{} \sqrt {1 \minus{} x} } )dx\]

$ \int \ln (\sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x})dx$

$ \equal{}\allowbreak x\ln \left( \sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}\right) \minus{}\int \frac{\frac{1}{2\sqrt{1\plus{}x}}\minus{}\frac{1}{2\sqrt{1\minus{}x}}}{\sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}}x\,dx\allowbreak$

$ \equal{}\allowbreak x\ln \left( \sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}\right) \minus{}\int x\left( \frac{\frac{1}{2\sqrt{1\plus{}x}}\minus{}\frac{1}{2\sqrt{1\minus{}x}}}{\sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}}\frac{\sqrt{1\plus{}x}\minus{}\sqrt{1\minus{}x}}{\sqrt{1\plus{}x}\minus{}\sqrt{1\minus{}x}}\right) dx\allowbreak$

$ \equal{}\allowbreak x\ln \left( \sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}\right) \plus{}\int \frac{x^{2}}{\sqrt{1\minus{}x^{2}}}\,dx\allowbreak$

$ \equal{}\allowbreak x\ln \left( \sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}\right) \minus{}\frac{1}{2}x\sqrt{1\minus{}x^{2}}\plus{}\frac{1}{2}\sin ^{\minus{}1}x\plus{}C$


equation
algebra
polynomial
Diophantine equation
Rational roots
IMO Shortlist

Prove that $ \forall n > 1, n \in \mathbb{N}$ the equation \[ \sum^n_{k\equal{}1} \frac{x^k}{k!} \plus{} 1 \equal{} 0\] has no rational roots.

$ p$ is a  prime. First we have to prove that  $ p^k$ does not divide $ k!$.To show that put $ r$ so that $ p^r =< k < p^{r+1}$.
 the highest power of $ p$ dividing$ k!$ is $ p^r$(p adic valuation), such that $ r = [k/p] + [k/p^2] + [k/p^3] + ... + [k/p^r] ≤ k/p + k/p^2 + ... + k/p^r = k(1 - 1/p^r)/(p - 1) < k$.

Now suppose there is a rational root $ x$ of the polynomial. if we take  $ x = a/b$, where$ a$and$ b$ are coprime (a^b=1) integers. so $ a^n + na^{n-1}b + ... + (n-1)! a b^{n-1} + n! b^n = 0$. So an must be divisible by$ b$. But a and b are coprime, so$ b$ must be $ 1$, in other words, the root must be an integer.

 take $ p$ a prime dividing $ n$. Then p must divide $ a^n$ and thus $ a$. But we have$ - n! = n! (a^n/n! + a^{n-1}/(n-1)! + ... + a^1/1! )$. Each term inside the parentheses is divisible by$ p$, because $ p^k$ divides $ a^k$ but not $ k!$ So a higher power of $ p$ divides the $ rhs$ than the$ lhs$. Contradiction!

Notice that:
$\sum_{k=0}^{\infty}\frac{x^k}{k!}=1+x+...=e^x$ $\Longleftrightarrow$ $e^x-1=\sum_{k=1}^{\infty}\frac{x^k}{k!}$
Furthermore lets take: $\lim_{n\to\infty}\sum^n_{k=1} \frac{x^k}{k!}=\sum_{k=1}^{\infty}\frac{x^k}{k!}$
It follows that $\sum_{k=1}^{\infty}\frac{x^k}{k!}+1=0$ but we know that $\sum_{k=1}^{\infty}\frac{x^k}{k!}=e^x-1$
 $e^x-1+1=0$
$e^x=0$
There are no such real numbers $x$ such that this statement is true which implies that there are no real roots which implies that there are no rational roots.
And we are done!


Prove the following: There exsists a constant c > 0 such that given any set P of points in the plane, the number of triples spanning an isosceles triangle is at most cn^(7/3).

Do you have an elementary solution for this one?
I mean that this result has been proved by Pach and Sharir using Szemeredi-Trotter theorem, which is not quite elementary....

Moreover, Pach and Tardos (2002) improved it to $c \cdot n^{\frac {11e-3} {5e-1} + \epsilon}$, for some positive constant $c$ depending on $\epsilon$.
As far as I know, it is the current best upper bound (note that $ \frac {11e-3} {5e-1} = 2,1364...$)

Pierre.

I post it hoping that somebody could post an elementary one.  :D


[quote="Farcus"]so the tan(theta) = 2.7361

how do i figure out the sec(theta) and cos(theta)? I know i have to use pythagorean identity but i dont know how[/quote]

[hide]Well, $\sec^{2}\theta=\tan^{2}\theta+1$, and $\sec \theta=\frac{1}{\cos \theta}$.

I think you can figure it out from there.[/hide]


Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ \frac{a}{2b^2\plus{}ca}\plus{}\frac{b}{2c^2\plus{}ab}\plus{}\frac{c}{2a^2\plus{}bc} \ge \frac{3}{a\plus{}b\plus{}c}\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$ or any cyclic permutations. :)

[hide="@Lam"]Good luck to you, my friend! :)[/hide]

[quote="can_hang2007"]Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ \frac {a}{2b^2 \plus{} ca} \plus{} \frac {b}{2c^2 \plus{} ab} \plus{} \frac {c}{2a^2 \plus{} bc} \ge \frac {3}{a \plus{} b \plus{} c}
\]
Equality holds if and only if $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ or any cyclic permutations. :)

[hide="@Lam"]Good luck to you, my friend! :)[/hide][/quoteusing Cauchy Schwarz ,then setting a=x+y;b=x+z;z=y+z
$ \sum {(x \minus{} z)^2 (z \minus{} y) \ge 0}$

I'm very busy so i'll post the solution soon :)

[quote="tronghieu"]$ (x \minus{} z)^2(z \minus{} y) \plus{} (z \minus{} y)^2(y \minus{} x) \plus{} (y \minus{} x)^2(x \minus{} z) \ge 0$[/quote]
This inequality is not true, try $ z \equal{} 0,x \equal{} y \equal{} 1$

[quote="tronghieu"]I'm very busy so i'll post the solution soon :)[/quote]
I don't know the reason why so many people said that... If you don't have time, how can you get the time to post the "non-complete" solution and say that you don't have time?

[quote="can_hang2007"]

[quote="tronghieu"]I'm very busy so i'll post the solution soon :)[/quote]
I don't know the reason why so many people said that... If you don't have time, how can you get the time to post the "non-complete" solution and say that you don't have time?[/quote]

Take it easy can_hang2007 :wink:

[quote="can_hang2007"]Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ \frac {a}{2b^2 \plus{} ca} \plus{} \frac {b}{2c^2 \plus{} ab} \plus{} \frac {c}{2a^2 \plus{} bc} \ge \frac {3}{a \plus{} b \plus{} c}
\]
Equality holds if and only if $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ or any cyclic permutations. :)

[hide="@Lam"]Good luck to you, my friend! :)[/hide][/quote]

[hide="@Can"]Thank you very much, good friend! :)[/hide]


How many solutions does the equation$x+y+z=100$ have on the set of positive integers?

since $x,y,z$ are positive , this means $x,y,z\geq 1$ so we are to find number of solution for

$x'+y'+z'=97$ where $x',y',z'\geq 0$ are all integer .

Hence there are $\binom{99}{2}=3851$ solutions

Number of solution equation in natural
x(1)+x(2)+...x(k)=m
expressed by binomial coefficients C(m-1,k-1).


Find all functions $ f: \mathbb{N}\to \mathbb{N}$ such that:
\[ f(f(n)-n)\plus{}f(f(n)+n)=8n
\quad \forall m,n\in\mathbb N\]

[quote="quangpbc"][color=green]No body?  :D . Pco, where are you ?[/color][/quote]

Hello Quangpbc.

I have not found the solution and I'm searching.

I found the trivial soluton $ f(n)\equal{}2n$ and I found that $ n<f(n)<4n$ $ \forall n$.

That's all for the moment.

[color=green]No problem, pco and another members. I will return here later ( 9 days ), i hope i will see the solutions .

Best regards,

Quang.[/color]

[quote="praneeth"][quote="pco"][quote="quangpbc"][color=green]No body?  :D . Pco, where are you ?[/color][/quote]

Hello Quangpbc.

I have not found the solution and I'm searching.

I found the trivial soluton $ f(n) \equal{} 2n$ and I found that $ n < f(n) < 4n$ $ \forall n$.

That's all for the moment.[/quote]

Using n < f(n) < 4n for every n in N, cant we say that f(n) must be of degree 1.[/quote]

Sure, it would be true if $ f(n)$ was a polynomial. But nothing said that $ f(n)$ must be a polynomial.

So no.

[b]quangpbc[/b] is Vietnamese
So we understand that $\mathbb{N}$ is set of natural numbers ( i.e $\mathbb{N}=\{0,1,2,...\}$  in this problem :)

And dear [b]Mr.Patrick[/b] , maybe you will continue  solving  this problem :lol:


number theory
greatest common divisor
least common multiple

Find the smallest positive integer which occurs in all of the arithmetic sequences:
8, 15, 22, ...
3, 14, 25, ...
11, 24, 37, ...

[hide]The first two sequences coincide with each other every multiple of 77 (GCD 7,11).  The first intersection is 36, so the two sequences will have the same value at numbers writable as 36+77n, where n is an integer.  For the third series to also coincide here, the number must leave remainder 11 when divided by 13.  36=2(13)+10, and 77=6(13)-1, so each succesive intersection of the first two series has a remainder one less than the previous one when divided by 13.  We need to go from 10 to 11=-2, so we need 36+12(77)=960.[/hide]

Thanx! Just a note: GCD (7, 11) = 1 and LCM (7, 11) = 77  :)


hi all,
have a look at this site:

http://www.math.u-bordeaux.fr/ALGANT/

its quite good,they give you a heavy scholarship of 21000 euros per year(for non EU students),
I came to bordeaux through this prog.

regards
Rohith :)


I always struggled at knowing how many days are in a month and leap years and stuff like that. This really kills me because easy questions that I would otherwise know how to solve (1st ten in chapter sprint and some CD) are now making my score fail.

Help?

Leap years are years that are divisible by 4

You can use the "knuckle-method" to find the days in a month
say O is a knuckle and I is the skin between them
your fist would look like:
OIOIOIO

O=Big I=Small
Count it this way from first finger to pinky:
OIOIOIO
JFMAMJJ when you count back,
you start at the pinky knuckle (I'm looking at my left hand) with August like this:
OIOI OIO
   DNOSA

Sorry I tried making them lined up.
Just remember that February is 28 and on leap years it's 29
Other months are this way:
Big-31, Small-30

Good luck!

[quote="xoangieexo"]Leap years are years that are divisible by 4
[/quote]

For our purposes, this works (since I don't think you'll do MC in 2100 xD), but it's all years divisible by $ 4$ that can't be expressed as $ 400a\plus{}100k$ for $ k\equal{}1,2,3$ and integer $ a$.

[quote="math154"][quote="xoangieexo"]Leap years are years that are divisible by 4
[/quote]

For our purposes, this works (since I don't think you'll do MC in 2100 xD), but it's all years divisible by $ 4$ that can't be expressed as $ 400a \plus{} 100k$ for $ k \equal{} 1,2,3$ and integer $ a$.[/quote]

oh yeah oops... and I don't think they ask about leap years in mc.

You can use your knuckles for the months.
As for the leap years, it is every 4 years, but not every 100 years, but also every 400 years.
So 2100 isn't a leap year
But 2000 and 4000 are
And so on

How I do it:

As many people have said, the "knuckle" method is the best.  Turn your hand into a fist.  Notice how You have a high level, then low, then high, etc.  Call your first knuckle January, and because its high, it has 31 days.  Low ones have 30 days (exception for February).  Wen you run out of knuckles, start over.

Leap Years:  2008 was a leap year.  Since MathCounts questions don't ask for leap years in like 2200, etc., I just remember 2008, and minus 4, 2004, etc. and adding 4, etc. are leap years.

yeah, but it will be convenient if the year was like, 2100


Δείτε μια όμορφη που προέκυψε : Έστω $ a,b,c\geq 0$ με $ a\plus{}b\plus{}c\equal{}2$ . Να αποδείξετε ότι

$ (a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)\leq 2$

Sil
Δώσε μια γεωμετρική ερμηνεία στον $ R^3$ η άσκηση γίνεται [b]πολύ ενδιαφέρουσα!!![/b]

[quote="silouan"]Δείτε μια όμορφη που προέκυψε : Έστω $ a,b,c\geq 0$ με $ a \plus{} b \plus{} c \equal{} 2$ . Να αποδείξετε ότι

$ (a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)\leq 2$[/quote]

Ωραια ασκηση

Εχουμε:
$ LHS \equal{} \sum_{cyclic}{a^4b^2 \plus{} b^4a^2} \plus{} 2a^2b^2c^2 \leq \sum_{cyclic}{a^4b^2 \plus{} b^4a^2} \plus{} 3a^2b^2c^2 \equal{} (a^2 \plus{} b^2 \plus{} c^2)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$.

Ειναι ομως: $ a^2b^2 \equal{} \frac {1}{4}4abba \leq ab\frac {1}{4}(a \plus{} b)^2 \equal{} ab\frac {1}{4}(2 \minus{} c)^2 \leq ab$, με ισοτητα για $ a \equal{} b, c \equal{} 0$ η $ ab \equal{} 0$

Αρα:
$ (a^2 \plus{} b^2 \plus{} c^2)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \equal{} \frac {1}{8}4(a^2 \plus{} b^2 \plus{} c^2)(2a^2b^2 \plus{} 2b^2c^2 \plus{} 2c^2a^2) \leq \frac {1}{8}4(a^2 \plus{} b^2 \plus{} c^2)(2ab \plus{} 2bc \plus{} 2ca) \leq \frac {1}{8}(a \plus{} b \plus{} c)^4 \equal{} \frac {16}{8} \equal{} 2$,
με ισοτητα για $ (a,b,c) \in \{(0,1,1), (1,0,1), (1,1,0)\}$

Quite Easily Done :P

Edit: Για τη γεωμετρικη ερμηνια, εχω σκευτει το εξης αλλα ισως να υπαρχει και κατι καλυτερο:
Νδο το γινομενο των αποστασεων των τριων προβολων καθε σημειου του επιπεδου με εξισωση $ x \plus{} y \plus{} z \equal{} 2$ με θετικες συντεταγμενες πανω στα επιπεδα $ xOy, yOz, zOx$ απο την αρχη των αξονων $ O$ δεν ειναι μεγαλυτερο απο $ \sqrt {2}$.

πάρε τα σημεία Α(2,0,0) , Β(0,2,0) και Γ(0,0,2) ενα σημείο Μ κινείται στο εσωτερικό ή και τις πλευρές του ισοπλεύρου ΑΒΓ προβάλεις το Μ στα επίπεδα ΑΟΒ, ΒΟΓ, ΓΟΑ και έστω Δ,Ε,Ζ οι αντίστοιχες προβολές τότε τα τρίγωνα ΟΔΕ, ΟΕΖ. ΟΖΔ έχουν γινόμενο πλευρών (πχ ΟΔ.ΟΕ.ΔΕ) το πολυ $ \sqrt{2}$(κοκ γενικεύσεις με γ-μ, αρμ-μ...)


AMC
USA(J)MO
USAMO
inequalities solved
inequalities

Let a,b and c be positive real numbers.Prove that:
(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)>=(a+b+c)^3

Not new...

http://www.mathlinks.ro/Forum/viewtopic.php?t=5383
http://www.mathlinks.ro/Forum/viewtopic.php?t=5726
http://www.mathlinks.ro/Forum/viewtopic.php?t=12202

  Darij


conics
ellipse
geometry
geometric transformation
reflection
analytic geometry

A trapezoid $ABRS$ is inscribed in an ellipse with large axis $a$ s.t. $AB\|RS$. The reflections of $A,B$ in $a$ are $P,Q$ respectively. Show that $P,Q,R,S$ are concyclic.

Using coordinates, let the ellipse be $x^2/a^2 + y^2/b^2 = 1$ (1), let line $AB$ be $y = mx + t_1$, and line $RS$ be $y = mx + t_2$.  Then line $PQ$ is $y = -mx - t_1$.

Then $P$, $Q$, $R$, and $S$ lie on the ellipse (1), as well as the "conic"
\[(y + mx + t_1)(y - mx - t_2) = -m^2 x^2 + y^2 - m(t_1 + t_2)x + (t_1 - t_2)y - t_1 t_2 = 0. \quad (2)\]

Take the appropriate linear combination of (1) and (2) which results in $x^2$ and $y^2$ having the same coefficient.  This is also a conic with no $xy$ term, so it represents a circle.

Thanks for the solution. I guess it's doesn't look bad at all, but could you give it a try without analytic geometry?

My solution :

[b]Lemma[/b]:

Let $ A, B, C, D $ lie on a conic $ \mathcal{H} $ such that $ A, B, C, D $ are concyclic .
Let $ A^*, B^* $ be the reflection of $ A, B $ in the axis of $ \mathcal{H} $, respectively .

Then $ CD \parallel A^*B^* $ .

[b]Proof[/b]:

Let $ C^*=AC \cap \odot (AA^*BB^*), D^*=BD \cap \odot (AA^*BB^*) $ .

From Reim theorem we get $ CD \parallel C^*D^* $ . ... $ (\star) $
From Pascal theorem ( for $ C^*D^*BB^*A^*A $ and $ CDBB^*A^*A $ )
$ \Longrightarrow A^*B^*, CD, C^*D^* $ are concurrent , so combine with $ (\star) $ we get $ CD \parallel A^*B^* $ .
____________________________________________________________
[b]Back to the main problem[/b]:

Let $ \odot (QRS) $ cut ellipse again at $ P^* $ .

From the lemma we get $ P^*Q \parallel PQ \Longrightarrow P \equiv P^* \Longrightarrow P, Q, R, S $ are concyclic .

Q.E.D


Find all functions $ f: \mathbb R\rightarrow\mathbb R$ satisfying $ f(f(x)\minus{}y^2)\equal{}(f(x))^2\minus{}2f(x)y^2\plus{}f(f(y))$ for every $ x,y\in\mathbb R$

The function is a surjective one or a constant ($ f(x)\equal{}0$ for every real $ x$).
Let's substitute $ f(0) \equal{} a$.

1. $ x \equal{} 0$ and $ y \equal{} 0$ $ \Rightarrow f(a) \equal{} a^2 \plus{} f(a)$ $ \Leftrightarrow a \equal{} 0$.

2. $ y \equal{} 0$ $ \Rightarrow f(f(x)) \equal{} f^2(x) \plus{} a \equal{} f^2(x)$.

3. $ x \equal{} 0$ $ \Rightarrow f( \minus{} y^2) \equal{} a^2 \minus{} 2ay^2 \plus{} f(f(\pm y)) \equal{} f(f(\pm y)) \equal{} f^2(\pm y)$.

4. $ y \equal{} 1$ $ \Rightarrow f( \minus{} 1) \equal{} f^2( \minus{} 1) \equal{} f^2(1)$.

If $ f( \minus{} 1) \equal{} 1 \equal{} f^2(1)$ $ \Rightarrow f(f(x) \minus{} 1) \equal{} f^2(x) \minus{} 2f(x) \plus{} f^2(1) \equal{} (f(x) \minus{} 1)^2$ and we conclude from this that $ f(x) \equal{} x^2$.
If $ f( \minus{} 1) \equal{} 0 \equal{} f^2(1)$ $ \Rightarrow f(x) \equal{} 0$.

[quote="BG Yoda"]
If $ f( \minus{} 1) \equal{} 1 \equal{} f^2(1)$ $ \Rightarrow f(f(x) \minus{} 1) \equal{} f^2(x) \minus{} 2f(x) \plus{} f^2(1) \equal{} (f(x) \minus{} 1)^2$ and we conclude from this that $ f(x) \equal{} x^2$.
If $ f( \minus{} 1) \equal{} 0 \equal{} f^2(1)$ $ \Rightarrow f(x) \equal{} 0$.[/quote]

How?  :|

[quote="BG Yoda"]The function is a surjective one or a constant ($ f(x) \equal{} 0$ for every real $ x$).
Let's substitute $ f(0) \equal{} a$.

1. $ x \equal{} 0$ and $ y \equal{} 0$ $ \Rightarrow f(a) \equal{} a^2 \plus{} f(a)$ $ \Leftrightarrow a \equal{} 0$.

2. $ y \equal{} 0$ $ \Rightarrow f(f(x)) \equal{} f^2(x) \plus{} a \equal{} f^2(x)$.

3. $ x \equal{} 0$ $ \Rightarrow f( \minus{} y^2) \equal{} a^2 \minus{} 2ay^2 \plus{} f(f(\pm y)) \equal{} f(f(\pm y)) \equal{} f^2(\pm y)$.

4. $ y \equal{} 1$ $ \Rightarrow f( \minus{} 1) \equal{} f^2( \minus{} 1) \equal{} f^2(1)$.

If $ f( \minus{} 1) \equal{} 1 \equal{} f^2(1)$ $ \Rightarrow f(f(x) \minus{} 1) \equal{} f^2(x) \minus{} 2f(x) \plus{} f^2(1) \equal{} (f(x) \minus{} 1)^2$ and we conclude from this that $ f(x) \equal{} x^2$.
If $ f( \minus{} 1) \equal{} 0 \equal{} f^2(1)$ $ \Rightarrow f(x) \equal{} 0$.[/quote]
That's wrong. I need another solution  :)

Obviously $ f(x)\equal{}0$ is one solution. To find another one, we assume there exists $ f(b)\neq0$.
Let $ f(0)\equal{}a$.
$ P(0,0): f(a)\equal{}a^2\plus{}f(a)$. So $ a\equal{}0$ and $ b\neq0$.
$ P(x,0): f(f(x))\equal{}f^2(x)$ -----------------(1)
We see $ f(f(b))\equal{}f^2(b)>0$. Now we replace $ b$ with $ f(b)$. So we have $ f(b)>0$.
$ P(b,y): f(f(b)\minus{}y^2)\equal{}f^2(b)\minus{}2f(b)y^2\plus{}f^2(y)$. If we let $ u\equal{}f(b)\minus{}y^2$. we see that $ f(u)\minus{}f^2(y)$ can take any value in $ (\minus{}\infty,f^2(b)]$.
We notice that $ f(f(f(y)))\equal{}(f(f(y)))^2\equal{}f^4(y)$.
$ P(u,f(y)): f(f(u)\minus{}f^2(y))\equal{}f^2(u)\minus{}2f(u)f^2(y)\plus{}f(f(f(y)))\equal{}(f(u)\minus{}f^2(y))^2$.
Since $ f(u)\minus{}f^2(y)$ can take all negative numbers, we have
$ f(x)\equal{}x^2$ for $ x<0$.
Assume $ x<0$ and look back (1), we have
$ f(x^2)\equal{}f(f(x))\equal{}f^2(x)\equal{}x^4$.
Hence, we also have $ f(x)\equal{}x^2$ for $ x>0$.

The conclusion: $ f(x)\equal{}x^2$ or $ f(x)\equal{}0$.


linear algebra
matrix
algebra
function
domain
linear algebra unsolved

Is it true that any non invertible matrix is a product of nilpotent matrices ?

as i know it's totally fake.
a non singular matrix has its determinant not zero. and knowing that in a product of matrices, the determinant of the product is equal with the product of the determinants of the matrices that are being multiplied, it means that if you can write a nonsingular matrix as the product of nilpotent matrices, each nilpotent matrix from the product must have its determinant not zero.
but a nilpotent matrix has the determinant 0 beacause raising that matrix at a certain power you will get the zero matrix.  so it's a contradiction.
unless the elements of the matrices come from a domain with divizors of 0...or unless i misunderstood the problem.  :P

of course I made a mistake. I've edited. So what do you think about it ?

The answer is yes.I've just seen a proof in a book and I'll try to post it in the following days.

I'll put the principal steps leading to the result:
1)Show that every matrice from $GL_d(k)$ can be written as the product of matrices from $T_n^-(K) \cup T_n^+(K)$.
2)Show that every matrice from $T_n^+(K),($respectively$ T_n^-(K))$ having its last column(line) nul can be written as the product of 2 nilpotent matrices,more precisely $M=N*J_n,$where $N$ is a nilpotent matrice and $(J_n)_{i,j}=1$ when $j=i+1$ and $0$ otherwise.

$ T_n^+(K)=$ the set of superior triangular matrices.I'll come with more details if it's needed.

Something tells me it's more than a simple need of details...This problem is a real monster, killer! Alekk, have you figured out how it works?  ;)


inequalities
induction
inequalities proposed

If $\pi(n)$ is the numbers of primes form 0 to n and $n>54$ proove that
\[ \pi(n)<\frac{n}{2}-11 \]

Do I misunderstand something or is this trivial? It follows from the fact that the only even prime number is $2$ and hence $p(n)<\frac{n}{2}+1$. But we can find enogh composite odd numbers in $[1; 53]$ to compensate the difference $(\frac{n}{2}+1)-(\frac{n}{2}-11)=12$.

well I can't understand very well your demonstration(don't kill me)...I prooved it by MI(mathematical induction)


Givin that
$x+\frac{1}{x}=10$, find $x^{9}+\frac{1}{x^{9}}$.

This is the most straightforward problem I ever made up, but the most annoying.  I got the idea from a State Competition Sprint in 2000-something, but it was a lot easier for that one (I think it was 2003, give or take a year).


They are f and g two real functions of real variable x and y, satisfying the equation: $f(x+y)+f(x-y)=2f(x)g(y)$, for all x and y. Proves that if $f(x)$ is not identically null and $|f(x)|\le1$ for all x, then $|g(y)|\le1$ for all y.

we prove your result by contradiction supose that existe $t$ such $|g\left(t\right)|\gneq{1}$.
$\forall{x}\in\mathbb{R}$,we have $|f\left(x\right)|\le{1}$ then exist $sup(|f\left(x\right)|=\alpha$.
form the eguation we get $|f\left(x\right)|\le{\alpha\over|g\left(t\right)|}$
wich absurd with $\alpha$ is sup :lol:

[quote="fermat3"]we prove your result by contradiction supose that existe $t$ such $|g\left(t\right)|\gneq{1}$.
$\forall{x}\in\mathbb{R}$,we have $|f\left(x\right)|\le{1}$ then exist $sup(|f\left(x\right)|=\alpha$.
form the eguation we get $f\left(x\right)\le{\alpha\over|g\left(t\right)|}$
wich absurd with $\alpha$ is sup :lol:[/quote]
thx a lot man! :lol:


Let $a_{i}$, $i=1,\dots, n$ be real numbers such that $\max_{i,j}(|a_{i}-a_{j}|) = d$. Prove that

$(n-1)d \leq \sum_{i<j}\parallel a_{i}-a_{j}\leq \frac{n^{2}}{4}d$

Daniel

Well, it is just the absolute value.

Prove that, $(n-1)d \leq \sum_{i<j}|a_{i}-a_{j}|\leq \frac{n^{2}}{4}d$.

Now you can try it!


[b]Let[/b] $ A\equal{}[a_{ij}]_{nxn},B\equal{}[b_{ij}]_{nxn},$ [b]are two  Unita matrices.[/b]
Prove that :
1) $ |\sum_{1 \leq i,j \leq n}{a_{ij}}| \leq n$
2) $ |\det(A\plus{}B)| \leq 2^n$

[hide]1) Let $ U \equal{} \left[\begin{array}{c} 1 \\
1 \\
... \\
1 \end {array} \right]$
 Use Cauchy-Schwars inequality we have :
  $ | < AU,U > | \leq \parallel{}AU\parallel{}\parallel{}U\parallel{} \equal{} \parallel{}U\parallel{}^2 \equal{} n$ ( by $ \parallel{}AU\parallel{} \equal{} \parallel{}U\parallel{}$, $ A$ is Unita matrix)
Then we have : $ |\sum_{1 \leq i,j \leq n}{a_{ij}}| \leq n$
2)We have
  $ |\det(A \plus{} B)| \equal{} |\det(A)\parallel{}\det(I_n \plus{} A^{*}B)| \equal{} |det(I_n \plus{} C)| \equal{} |\prod_{k \equal{} 1}^{n}{(1 \plus{} \lambda_k)}| \leq \prod_{k \equal{} 1}^{n}{(1 \plus{} |\lambda_k|)} \equal{} 2^n$
By $ |\det(A)| \equal{} 1,C \equal{} A^{*}B$ is Unita matrix by $ C^{*} \equal{} B^{*}A \equal{} B^{ \minus{} 1}A \equal{} (A^{ \minus{} 1}B)^{ \minus{} 1} \equal{} (A^{*}B)^{ \minus{} 1} \equal{} C^{ \minus{} 1}$
 and if $ \lambda_k$ is a eigenvalue and have eigenvector $ x_k$ of $ C$ when we have $ \parallel{}Ax_k\parallel{} \equal{} \parallel{}x_k\parallel{}$ or $ |\lambda_k| \parallel{}x_k\parallel{} \equal{} \parallel{}x_k\parallel{}$ or $ |\lambda_k| \equal{} 1$(note : $ C^{*}C \equal{} CC^{*} \equal{} I_n,C \equal{} \frac{1}{2}(C \plus{} C^{*}) \plus{} \frac{1}{2i}i(C \minus{} C^{*})$)[/hide]
 :blush:  Sorry English of me so bad ! I remember this is problem have in a exam of DHSP HN


In square $ ABCD$, points $ E$ and $ F$ are midpoints $ AD$ and $ DC$ respectively. $ AB \equal{} 2$ cm. What is the number of square centimeters in the area of $ \triangle BEF$? Express your answer as a common fraction.

[asy]size(4cm,4cm);
pair A,B,C,D,E,F;

A=(0,0);
B=(1,0);
C=(1,1);
D=(0,1);

draw(A--B--C--D--A);

E=(1/2)*D+(1/2)*A;
F=(1/2)*C+(1/2)*D;

draw(E--F--B--E);

label("A",A,SW);
label("B",B,SE);
label("C",C,NE);
label("D",D,NW);
label("E",E,W);
label("F",F,N);[/asy]

[b]Another Solution[/b]

Let point $ G$ is midpoint $ BC$.
$ \triangle ABE \equiv \triangle CBF$ and $ Area(\triangle DEF)\equal{}\frac{1}{4}Area(DEGC)$.

Thus, $ 2^2 \minus{} \left(\frac{2^2}{4} \times 2 \plus{} \frac{1}{4} \times \frac{2^2}{2}\right)  \equal{} 4 \minus{} \left(2\plus{}\frac{1}{2}\right) \equal{} \boxed{\frac {3}{2}}$.

Or...you can subtract the sum of the areas of $\triangle ABE$, $\triangle BCF$, and $\triangle DEF$ from the area of square $ABCD$.

$4-\left(1+1+\frac{1}{2}\right)=\boxed{\frac{3}{2}}$


im pretty sur i got a 4 or 5 on this year's AIME, and i was just wondering how hard i need to work if i want to get a 9 next year.

also, i know a 4 or 5 compared to other AOPS ppl isnt that good, but compared to the several thousand who took AIME, how many people do you think i beat?
im still happy with a 4 though.. i have 3 more years to improve

well, if you look on the AMC website, you can find statistics for previous AIME's and the # of people who got a certain score. The 2005 AIME statistics will probably be out in a few weeks.

I'm aiming for 12 next year, how should I start to practice? I have a year to do it.

I spent so much time this year... and I still did badly. If I can get my practice scores to a 15, I think I can manage a 10 on the real one.

I took the AIME in 9th grade, got a 3. Then in 10th grade after studying, I got a 2. This year, with no studying, I got a 10. It seems that big improvements are possible. Perhaps it is only because this year's test was easier than last year's. Still, I think a 400% increase is marked imrpovement.

hmmm...the real concept you should get out of the AIME and other math tests is how to attack the problem. If a problem has such and such concepts, then you should use a certain method to solve each type of problem. this AIME wasn't a gimme nor was it extremely hard. They (AMC) might've thought last years tests were hard and somewhat balanced it out with an easier group of tests this year.

On the subject of next year, i'm just going to learn more problem solving strategies. And do more problems so i'll be able to recognize them.

I should also get some sleep. On the night before EVERY math competition(mathcounts, AMC, NC State math competition) i get an average of 3 hrs. of sleep. :(  :P

One person at collegeconfidential increased his AIME score from 0 to 7 in a single year with some studying, but also some gaming.

If that was 0 - last year, 7 - this year...I can't say I'm surprised...


Is it possible to have a positive integer $ x$ so that $ 3^x$ when added to $ 3^6 \plus{} 3^7$ becomes a perfect square number?

$ 3^x \plus{} 3^6(4) \equal{} y^2$

$ 3^x \equal{} (y\minus{}54)(y\plus{}54)$

We see that each expression in parenthesis must be a power of 3.

$ y\minus{}54 \equal{} 3^m$

$ y\plus{}54 \equal{} 3^n$

With this we find that $ y\equal{}\frac{3^m\plus{}3^n}{2}$.

I can't find anything else but maybe you or someone else can finish it..

[hide="solution"] We claim the answer is no.

Assume that there exist some (x, y) such that $ 3^7\plus{}3^6\plus{}3^x\equal{}y^2$. Examing $ \pmod{8}$, we must have $ 3\plus{}1\plus{}3^x\equiv{y^2}\pmod{8}$. All squares are congruent to 0, 1, or 4 (mod 8). Also, all powers of 3 are 1 or 3 (mod 8), so our sum is congruent to one of 5 or 7 (mod 8), neither of which are quadratic residues (mod 8). Thus, there is no such x.[/hide]

oh, I overthought it I guess. But where'd you get the idea to look at mod 8 specifically?

I first tried (mod 4) becuase 3 is -1 (mod 4), but that didn't work. Since 9 is 1 (mod 8), I just decided to work in (mod 8), which luckily worked out. Also, there aren't too many squares (mod 8), so it's a decent modulo to work in.


calculus
Euler
integration
limit
geometry
3D geometry
sphere

Use the principle of conservation of mechanical energy to find the velocity with which a body must be projected vertically upward, in the absence of air resistance, to rise to a height above the earth's surface equal to the earth's radius, R.
[b]Book's solution:[/b]
Let the center of the Earth be the origin. Then the initial distance of the body is r_1=R and its final position is r_2=2R. Let v_1 be the initial velocity. v_2 is the final velocity, of the body of mass m is zerp since 2R is the maximum height the body rises. Using conservation of energy, we have:
KE_1+PE_1=KE_2+PE+2
1/2m.v_1^2-G.m.m_E/r_1=1/2m.v_2^2-[b]G.m.m_E/r_2[/b]
where m_E is the earth's mass
Note that potential energy is negative(Why PE_2 is negative? I think it's positive since the object is above the origin and all of its K.E changes into P.E at the time it reaches its maximum height).The rest of this solution is just "plug and chug"
Can anybody explain it for me? I know this is an easy problem but I need to make my understanding clearly.By the way, how can I use a different color, besides black, to outstanding some headlines or key words in my post?

Hello
[color=red][hide="First problem"][/color]
Ya when the body goes up its $ KE$ decreases and as a conclusion $ PE$ increases .So the change in Gravitational energy should be $ \plus{}ve$ not the gravitational energy at the point above the surface.
So instead of saying $ PE_2$ should be $ \plus{}ve$ u should say the change in the gravitation potential energy $ [U_2\minus{}U_1]$  should be +ve.
Well
$ U(r)\equal{}\minus{}\frac{GmM_e}{R}$
if u have doubt from where this $ \minus{}ve$ sign comes?Plz go for the derivation of this formula.Hopefully u will be more clear.
  [/hide]
[color=green][hide="second problem"][/color]
use this code::
[color=name of color]write the text[/color]
[/hide]
I m sorry if  my answer doesn't satisfied u
Thank u.

Hi Kabi
In my post above, I forgot to mention to one thing:The change in gravitational force is: G.m.M_E/2R -G.m.M_E/R= -G.m.M_E/2R
Now looking at your formula above, I'm sure that I'm mistaken in the sign of the gravitational force on the body when it's below the origin. I thought when an object is below the origin, then its P.E must be negative.However, it's always positive, right? Probably that's the only mistake I made! Thank you for your help! :)

Hello
I m sorry but u r wrong.The gravitational potential energy is $ \minus{}ve$ only when we go inside the earth or below the origin.
Actually the situation totally changed when go below the origin.
Go for the derivation in some good book or net.If u don't find it tell me I will pm my derivation.Just reply me.I'll do that.
Thank u.

The sign of potential energy depends on the choice of its origin, i.e. where do you arbitrate the PE is zero. For gravitational attraction, usually it's when the attractive bodies are at infinite distant apart ( $ R\rightarrow \infty$), and the gravity does a positive work to bring they at a finite distance apart, decreasing it's PE from zero to negative value.

[quote="ghjk"]Use the principle of conservation of mechanical energy to find the velocity with which a body must be projected vertically upward, in the absence of air resistance, to rise to a height above the earth's surface equal to the earth's radius, R.
[b]Book's solution:[/b]
...

Note that potential energy is negative(Why PE_2 is negative? I think it's positive since the object is above the origin and all of its K.E changes into P.E at the time it reaches its maximum height).The rest of this solution is just "plug and chug"
Can anybody explain it for me? I know this is an easy problem but I need to make my understanding clearly.By the way, how can I use a different color, besides black, to outstanding some headlines or key words in my post?[/quote]

You're not the only one who's ever been confused by this.

In order to understand what's going on a bit of calculus is required, and that is beyond the scope of [b]Introductory Physics[/b].

For a conservative force field, like gravity, the force on a free body is given by:

$ \vec{F} \equal{} \minus{} \nabla U, \text{ where U is the potential energy function.}$

That gives us the following differential equation,

$ \minus{} \frac {\partial U}{\partial r} \equal{} \minus{} \frac {GMm}{r^2}$.

The minus sign on the left hand side of the above expression is a consequence of the Euler - Lagrange equation from the calculus of variations. The minus sign on the right hand side is due to the direction of gravity, it is always directed toward the center of the Earth (or whatever massive body you are considering.

Canceling the minus signs leaves us with, $ \frac {\partial U}{\partial r} \equal{} \frac {GMm}{r^2}$.

To solve this you integrate both sides of the following integral equation:

$ \int_{U(R)}^{U(\infty)} \mathrm{d}U \equal{} \int_{R}^{\infty} \frac {GMm}{r^2} \,\mathrm{d}r$

$ \implies U(\infty) \minus{} U(R) \equal{} \lim_{h\to \infty} \left. \minus{} \frac {GMm}{r}\right|_{r \equal{} R}^{h} \\
 \\
U(\infty) \minus{} U(R) \equal{} \lim_{h\to \infty}\left(\frac {GMm}{R} \minus{} \frac {GMm}{h}\right) \equal{} \frac {GMm}{R}$

Now you could define $ U(\infty) \equal{} 0$ and this would mean that $ U(R) < 0$, whatever $ R$ happened to be. To do so would be a bit silly since $ \infty$ is much farther away than any point  you could start from -- and therefore has higher potential energy by definition -- furthermore, $ U(\infty) \minus{} U(R)$ is the amount energy required to move an object of mass $ m$ from a point a distance $ R$ from the center of $ M$ to a point infinitely far away.

Given that is the case, the energy required to move $ m$ from $ R_e$ to $ 2R_e$ is

$ U(2Re) \minus{} U(R_e) \equal{} \left( U(\infty) \minus{} U(R_e)\right) \minus{} \left(U(\infty) \minus{} U(2R_e)\right) \\
 \\
\equal{} \frac {GMm}{R_e} \minus{} \frac {GMm}{2R_e} \equal{} \frac {GMm}{2R_e}$

The only way to get this energy is to change the kinetic energy of $ m$.

Thus $ \Delta KE \equal{} \frac {GMm}{2R_e} \implies \frac12 m\left(v_{2}^2 \minus{} v_{1}^2\right) \equal{} \frac {GMm}{2R_e}$

Hello
Here is the derivation what happen when we go below the origin.
Let the outermost sphere(blue one) be the Earth's origin with radius $ R_e$.We want to find the Potential at the point $ P$ which lie on the innermost sphere(black sphere) whose radius is $ r$.We can see the innermost sphere had divided the sphere in two parts.Therefore the total potential at point $ P$ would be sum of potential due to inner part and outer part.
The mass of the inner part, $ M'$ =${ \frac{M_e}{\frac{4}{3}pi {R_e}^3}}\frac{4}{3}pi r^3$
                                             =$ \frac{M_er^3}{{R_e}^3}$                 ...............$ [equation 1]$
The potential at $ P$ due to inner part ,$ V_1$
=$ -\frac{GM'}{r}$
=$ -\frac{GM_er^2}{{R_e}^3}$ ......................$ [using equation 1]$

To get the potential at the point $ P$ due to outer part of sphere, we divide this part in concentric shells(made of green and pink sphere).
The mass of the shell b/w radii x and x+dx ,$ dm$
=$ \frac{M_e}{\frac{4}{3}pi {R_e}^3} 4pix^2dx$
=$ \frac{3M_ex^2dx}{{R_e}^3}$
The potential due to this shell ,$ V_2$
=$ -\frac{Gdm}{x}$
=$ -3\frac{GM_exdx}{{R_e}^3}$
Thus the potential due to outer part
=$ \int_{r}^{R_e}-\frac{3GM_exdx}{{R_e}^3}$
=$ -\frac{3GM_e((R_e)^2-r^2))}{2{R_e}^3}$

The total potential at point $ P$
=$ V_1$+$ V_2$
=$ -\frac{GM_e}{2{R_e}^3}(3{R_e}^3-r^2)$
or we can say that potential energy on  a body of mass $ m$ is kept at the point $ P$or$ U(r)$
=$ -\frac{GM_em}{2{R_e}^3}(3{R_e}^3-r^2)$
Thank u.

I thought of one further comment I'd like to make on this topic.

One can in fact define $ U(\infty) = 0$.

Then $ U(\infty) -U(R) =\frac{GMm}{R} \implies U(R) = -\frac{GMm}{R}$.

Gravitational potential is negative everywhere. Gravity is a potential well.

If that's so then why is gravitational potential always written as if it were positive when one expresses it as $ mgh$ rather than using Newton's law of gravitation?

If we consider the difference in potential energy given by $ U(R_e + h) - U(R_e)$ then

$ U(R_e + h) - U(R_e) = -\frac{GMm}{R_e + h} +\frac{GMm}{R_e}$.

Now we can rewrite this expression as

${ U(R_e + h) -U(R) = -\frac{GMm}{R_e}\cdot\frac{1}{\left(1 + \frac{h}{R_e}\right)} +\frac{GMm}{R_e}}$.

The McLaurin series expansion of $ \frac{1}{(1 + x)}$ is $ \quad\frac{1}{(1 + x)}= 1 - x + x^2 - x^3 + x^4 - x^5 + ...$

If we rewrite the above difference in terms of its McLaurin series we get,

$ U(R_e + h) - U(R_e) = -\frac{GMm}{R_e}\left(1 -\frac{h}{R_e} +\left(\frac{h}{R_e}\right)^2 -\left(\frac{h}{R_e}\right)^3 + \left(\frac{h}{R_e}\right)^4 -\left(\frac{h}{R_e}\right)^5 \\+\ldots\right) + \frac{GMm}{R_e}$.

Applying the distributive property,

$ U(R_e + h) - U(R_e) = -\frac{GMm}{R_e} +\frac{GMmh}{R_{e}^2} -\frac{GMmh^2}{R_{e}^3} +\frac{GMmh^3}{R_{e}^4} - \frac{GMmh^4}{R_{e}^5}\\ +\frac{GMmh^5}{R_{e}^6} -\ldots  + \frac{GMm}{R_e}$.

$ \Rightarrow U(R_e + h) - U(R_e) = \frac{GMmh}{R_{e}^2} -\frac{GMmh^2}{R_{e}^3} +\frac{GMmh^3}{R_{e}^4} - \frac{GMmh^4}{R_{e}^5} +\ldots -$.

Now if we note the fact that $ \frac{h}{R_e} << 1$ then we can neglect every term in the series except the first one.

$ \Rightarrow U(R_e + h) - U(R_e) \approx \frac{GMmh}{R_{e}^2}$

Now we notice that $ \frac{GM}{R_{e}^2} = g$ so that,

$ U(R_e + h) - U(R_e) \approx mgh$.

However, if we apply conservation of energy to the problem at hand without any approximation, in view of the fact that potential energy is negative everywhere,

\[ \frac{1}{2}mv^2(2R_e) -\frac{GMm}{2R_e} = \frac{1}{2}mv^2(R_e) -\frac{GMm}{R_e} \\ \\
\Rightarrow \frac{1}{2}m\left(v^2(2R_e) - v^2(R_e)\right) = \frac{GMm}{2R_e} -\frac{GMm}{R_e} = -\frac{GMm}{2R_e}\]

Therefore,  $ \frac{1}{2}m\left(v^2(2R_e) - v^2(R_e)\right) < 0$.

\[ \frac{1}{2}m\left(v^2(2R_e) - v^2(R_e)\right) = \frac{1}{2}m\left(v(2R_e) - v(R_e)\right)\left(v(2R_e)+v(R_e)\right) \\ \\ \frac{1}{2}m\left(v(2R_e) - v(R_e)\right)\left(v(2R_e)+v(R_e)\right) < 0 \\ \\ \Rightarrow v(2R_e) - v(R_e) < 0 \\ \\ \therefore v(R_e) > v(2R_e)\]

This is consistent with the idea of free fall from a height of $ 2R_e$ to the surface at $ R_e$.


calculus
integration
real analysis
probability
function
real analysis unsolved

[quote="jmerry"]The convolution of two gaussians is a gaussian, and variances add. $f(y)$ should be a multiple of $e^{-\frac32 y^{2}}$. In this case, the appropriate scaling factor is $\frac3{\sqrt{2\pi}}$.[/quote]

Could you explain a bit more please.

Let $N_{\sigma^{2}}(x)= \frac1{\sigma\sqrt{2\pi}}e^{-\frac{x^{2}}{2\sigma^{2}}}$ be the probability density function for a normal distribution of mean zero and variance $\sigma^{2}$.
Then $N_{\sigma_{1}^{2}}*N_{\sigma_{2}^{2}}= N_{\sigma_{1}^{2}+\sigma_{2}^{2}}.$ The rest is just a matter of fitting constants.


I am trying to construct a complex function that is analytic on the interior of a unit disk but not so on the boundary. The set of points on the boundary where the function is not analytic should be dense.

Any ideas? Would it be possible to represent such a function as a power series? Any hints/ideas will be appreciated.

Your question is a little imprecise. After all, if we take something like

$f(z)=\sum_{n=0}^{\infty}z^{n}=\frac1{1-z}$ on the unit disk,

then the series isn't analyting anywhere on the boundary; it doesn't even converge. However, $f$ can be analytically continued to a function which is analytic everywhere except $z=1.$ So perhaps what you mean is that the function cannot be analyically continued to a set which contains a dense set on the boundary of the disk.

Lacunary power series are very good for this sort of thing. Consider:

$\sum_{k=1}^{\infty}\frac1{k^{2}}z^{2^{k}}.$


Chelsea Cricket begins hopping on a number line at 0. She wants to
get the point at 1, but she can hop only $ \frac{1}{4}$ of the distance. Each hop tires her out so that she continues to hop $ \frac{1}{4}$ of the remaining distance. How far has she hopped after five hops? Express your answer as a common fraction.

Each hop, as a fraction of the original unit interval, is given by $ \left(\frac{1}{4}\right)^{n}$ where $ n$ is the hop number.  So we get
$ \frac{1}{4}\plus{}\frac{1}{16}\plus{}\frac{1}{64}\plus{}\frac{1}{256}\plus{}\frac{1}{1024}\equal{}\frac{256}{1024}\plus{}\frac{64}{1024}\plus{}\frac{16}{1024}\plus{}\frac{4}{1024}\plus{}\frac{1}{1024}\equal{}\frac{341}{1024}$

Alternatively, one can note that these are the second-sixth terms of a geometric series and use the sum formula:
$ S_{n}\equal{}\sum_{k\equal{}0}^{n}ar^{k}\equal{}a\frac{1\minus{}r^{n}}{1\minus{}r}$
In this case, $ r\equal{}\frac{1}{4}$ and $ a\equal{}1$, and remembering to subtract out 1 (why?), we get the same answer.

I don't understand this solution.  Isn't the length of each hop $\frac{1}{4}(1-\text{length hopped already})$?  Is there any other solution for this problem?

Wait, but it says:
[quote="GameBot"]Each hop tires her out so that she continues to hop $ \frac{1}{4}$ of the remaining distance.[/quote]

So for example, the first hop is $\frac{1}{4}$ of the distance, then the second hop is $\left(1-\frac{1}{4}\right)\frac{1}{4}=\frac{3}{16}$ of the distance since it is $\frac{1}{4}$ of the [b]remaining distance[/b], not $\frac{1}{4}$ of the distance that she hopped previously.

So you just keep on adding the distance that the Cricket already hopped then multiplying it by $\frac{1}{4}$ to get the distance that she hopped on that turn.  But is there a quicker way to do it than just keep on adding and multiplying?  Because that can get nasty with those big numbers.  :(

We can set up the following recursion:

The geometric series for how far she hops on each turn is:

$\frac{1}{4}, \frac{3}{16}, \frac{9}{64}, \frac{27}{256}, \frac{81}{1024}...$

Common ratio is $\frac{3}{4}$

Sum of first five terms: $\frac{1}{4} \cdot \frac{1-\frac{3^{5}}{4^{5}}}{\frac{1}{4}}=\frac{781}{1024}$


If $ a^{3}\plus{}b^{3}\plus{}c^{3}\equal{}1$  , than prove that $ a^{6}\plus{}b^{6}\plus{}c^{6}\ge 1$   where $ a,b,c$  are  Real numbers.
But I think it is wrong.

[quote="ridgers"]If $ a^{3} \plus{} b^{3} \plus{} c^{3} \equal{} 1$  , than prove that $ a^{6} \plus{} b^{6} \plus{} c^{6}\ge 1$   where $ a,b,c$  are  Real numbers.
But I think it is wrong.[/quote]
Of course it is wrong. $ a^3\equal{}\frac{1}{3}, b^3\equal{}\frac{1}{3},c^3\equal{}\frac{1}{3}$..

I had this question in my school math olympiad and I wrote that the question is wrong and gave this example $ a^{3}\equal{}\frac{1}{2}, b^{3}\equal{}\frac{1}{3},c^{3}\equal{}\frac{1}{6}$

[quote="ridgers"]If $ a^{3} \plus{} b^{3} \plus{} c^{3} \equal{} 1$  , than prove that $ a^{6} \plus{} b^{6} \plus{} c^{6}\ge 1$   where $ a,b,c$  are  Real numbers.
But I think it is wrong.[/quote]

This inequality is obviously wrong. The right wersion would be:

If $ a^{3} \plus{} b^{3} \plus{} c^{3} \equal{} 1$, prove $ a^{6} \plus{} b^{6} \plus{} c^{6}\ge \frac{1}{3}$ where $ a,b,c$ are Reals.


algorithm
number theory
prime factorization
relatively prime

[b]Please provide hints only.[/b]

Prove that for any prime number $ p$, $ 2^{p\minus{}1}(2^p \minus{} 1)$ generates a perfect number.

This is false; the correct condition is that $ 2^p \minus{} 1$ is prime.

First of all, do you know how to compute the sum of the divisors of any number?  If you do, just compute that sum.  Otherwise, you will need to figure out how to do that.  Here are two sets of hints.

Hint set 1:
- What is the sum of the divisors of a power of a prime?
- What is the sum of the divisors of a product of two coprime integers in terms of the divisors of those integers?

Hint set 2:
- Write out the prime factorization of an arbitrary number $ n$.  What does an arbitrary divisor of $ n$ look like?
-  What does the sum of all such divisors look like?  Can you factor that sum?  (Look at small examples.)

if $ 2p\minus{}1$ is prime, then what can you do with the $ 2^{p\minus{}1}$?
and, proofs don't belong here.

uhh they kinda do
easy ones anyway

here is a algorithm type thing you can use to find the sum of the factors of a number,z, in the form $ p_1^{k_1}p_2^{k_2} \ldots p_n^{k_n}$
$ \prod^n_{x\equal{}1}\sum^{k_x}_{y\equal{}0}p_x^y$
i think
or alternatively,
$ z \cdot \prod^n_{x\equal{}1}\sum^{k_x}_{y\equal{}0}p_x^{\minus{}y}$
i think

so for a perfect number to occur, you want the last part to be 2
...
disproof for your first conjecture:
$ 2^{10} \cdot 2^{11}\minus{}1$
has factors
1 23 89 2047
2 46 178 4094
4 92 356 8188
...
point is, there are extra factors that make this number not perfect

[quote="isabella2296"][b]Please provide hints only.[/b]

Prove that for any prime number $ p$, $ 2^{p \minus{} 1}(2^p \minus{} 1)$ generates a perfect number.[/quote]

This is a product of two relatively prime numbers. ($ 2^{p \minus{} 1}$ only has factors of 2, while $ 2^p \minus{} 1$ has none).

Since they are relatively prime, the sum of factors of the whole thing can be found by taking the sum of factors of each part separately and multiplying them together. (Ask yourself why this works and prove it if necessary.)

Taking the sum of factors of the $ 2^{p \minus{} 1}$ part yields $ (1 \plus{} 2 \plus{} 4 \plus{} \dots \plus{} 2^{p \minus{} 1}) \equal{} 2^p \minus{} 1$, which is exactly the second part. So what do you want the sum of factors of the second part to be? (Remember, you want the sum of the factors of the whole thing to be twice the original expression.)

http://www.mathlinks.ro/viewtopic.php?p=560965

Nice problem with primes and perfect numbers.

It isn't right.
You haven't understood the problem, the only solution is 6.

[hide="My solution"]
Those 2 primes are $ 6k\minus{}1$ and $ 6k\plus{}1$. So the perfect number is $ 6k$
One solution-6, easily found.
If $ k>1$, then some (not necessarily all) divisors of $ 6k$ are $ 1$, $ k$, $ 2k$, $ 3k$ and $ 6k$.
Their sum is $ 12k\plus{}1>12k$, so there are not other solutions.[/hide]

[quote="isabella2296"]Okay, I think I have it!

[hide="Is This Right?"]

Say we have a prime number in the form $ 2^k \minus{} 1$. Let $ a \equal{} 2^{k \minus{} 1}(2^k \minus{} 1)$. We want to show that the sum of $ a$'s factors is $ 2a$.

The sum of the factors of $ p$ is $ p \plus{} 1 \equal{} 2^k$. We then have:

$ \text{sigma}(a) \equal{} \text{sigma}(2^{k \minus{} 1}) \cdot \text{sigma}(p) \equal{} (2^k \minus{} 1)2^k \equal{} 2n$

QED
[/hide]

Thanks everyone![/quote]
no offense, but your solution doesn't make any sense whatsoever.

[quote="isabella2296"]Okay, I think I have it!

[hide="Is This Right?"]

Say we have a prime number in the form $ 2^k \minus{} 1$. Let $ a \equal{} 2^{k \minus{} 1}(2^k \minus{} 1)$. We want to show that the sum of $ a$'s factors is $ 2a$.

The sum of the factors of $ p$ is $ p \plus{} 1 \equal{} 2^k$. We then have:

$ \text{sigma}(a) \equal{} \text{sigma}(2^{k \minus{} 1}) \cdot \text{sigma}(p) \equal{} (2^k \minus{} 1)2^k \equal{} 2n$

QED
[/hide]

Thanks everyone![/quote]

Great solution!

Flaw 1: You might want to introduce the variable $ p$. (Minor.) Maybe at the beginning, "Say we have a prime number [b][i]p[/i][/b]..."

Flaw 2: You said $ 2n$ when the original variable was $ a$ :P

Flaw 3: The official notation for sum of divisors is \sigma = $ \sigma$.

Flaw 4: You have to prove that the two factors $ 2^{k\minus{}1}$ and $ 2^k\minus{}1$ are relatively prime. That is the criterion for saying that $ \sigma(xy)\equal{}\sigma(x)\sigma(y)$.

Great solution anyway!
Next step: Prove the converse.

[i]Prove that if a perfect number is in the form $ 2^{k \minus{} 1}(2^k \minus{} 1)$ then $ 2^k \minus{} 1$ [b]must[/b] be prime[/i].
My hint helps here.

[quote="t0rajir0u"][quote="Bugi"]You haven't understood the problem, the only solution is 6. [/quote]
isabella wasn't answering your question, but the original question.

[quote="Poincare"]no offense, but your solution doesn't make any sense whatsoever.[/quote]
The solution is fully correct.  What's confusing about it?[/quote]
what does the sigma thing mean?

"Sigma" means the sum of the factors.

@Yongyi: Thanks! Working on Flaw 4.

[quote="isabella2296"]Okay, I think I have it!

[hide="Is This Right?"]

Say we have a prime number in the form $ 2^k \minus{} 1$. Let $ a \equal{} 2^{k \minus{} 1}(2^k \minus{} 1)$. We want to show that the sum of $ a$'s factors is $ 2a$.

The sum of the factors of $ p$ is $ p \plus{} 1 \equal{} 2^k$. We then have:

$ \text{sigma}(a) \equal{} \text{sigma}(2^{k \minus{} 1}) \cdot \text{sigma}(p) \equal{} (2^k \minus{} 1)2^k \equal{} 2n$

QED
[/hide]

Thanks everyone![/quote]
oh, ok! now i get it thanks to yongyi!! yay!! its an awesome solution; ingenious yet easy to understand!

Fixing Flaws 1-3...

[hide="Here It Is"]

Say we have a prime number $ p$ in the form $ 2^k \minus{} 1$. Let $ a \equal{} 2^{k \minus{} 1}(2^k \minus{} 1)$. We want to show that the sum of $ a$'s factors is $ 2a$.

The sum of the factors of is $ p$ is $ p \plus{} 1 \equal{} 2^k$. We then have:

$ \sigma(a) \equal{} \sigma(2^k \minus{} 1) \cdot \sigma(p) \equal{} (2^k \minus{} 1)2^k \equal{} 2a$

QED
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Oh, and thanks Poincare!


hei alle!

Finst det altså ikkje nordisk oppgåver på nettet? ingen som har funne det?

Det fins ingen side med samla nordisk-oppgåver som eg har sett iallfall. Men det er ikkje så viktig å øve akkurat på dei, for oppgåvene er ganske like i dei fleste konkurransar. Eg fant ei oppgåve på eit av setta som låg på abelkonkurransen.no:

Beskjeder kodes ved hjelp av nuller og enere. Kun sekvenser med maksimalt to påfølgende nuller og enere er tillatt. (For eksempel er sekvensen 011001 tillatt, mens 011101 ikke er. Hvor mange tillatte sekvenser fines med nøyaktig 12 sifre???

Denne oppgåva kan sikkert løysast veldig vanskelig med å telle opp alle kombinasjonane på ein lur måte. Men alltid i slike oppgåver er det en måte å tenkje på  som gjer alt litt enklare. Nokon forslag???

ja, eg har sett på den eg også, men eg anar ikkje kva slags måte som er riktig å bruke, finn det kanskje ut en dag...

Er det riktig?
[hide] Derfiner $a_n$ som antall slike tall med $n$ siffer. Lett å sjekke at $a_1 = 2$ og $a_2 = 4$
Hvis vi skriver til $1$ eller $0$ til slutten av tallene i $a_n$ vil vi få noen av tallene fra $a_{n+1}$.
Vi gjør da følgende: Ta et tall fra mengden $a_n$ og hvis tallet slutter på $0 (1)$ da setter du $1 (0)$ til høyre. Ved å gjøre det er vi sikre at vi ikke får $000$ eller $111$. Slik får du $a_n$ tall fra mengden $a_{n+1}$.
Så tar vi tallene fra mengden $a_{n-1}$. Hvis tallene slutter på $1 (0)$ så skriver vi til høyre $00 (11)$. Nå har vi fått alle mulige tall som oppfyller betingelsene og har lengde $a_{n+1}$.
Det vi fikk nå er en rekursiv rekke definert slik : $a_1=2$ og $a_2 = 4$, $a_{n+1}= a_n + a_{n-1}$ (Ligner litt på Fibonacci)
Vi regner videre:
$a_3=6$
$a_4 = 10$
$a_5 = 16$
$a_6 = 26$
$a_7 = 42$
$a_8 = 68$
$a_9 = 110$
$a_{10} = 178$
$a_{11} = 228$
$a_{12}= 406$
Svaret er da $406$
Ja forresten ved induksjon kan vi vise at $a_n = 2 F_{n+1}$ hvor $F_n$er n'te Fibonacci tallet dvs. $F_1=F_2=1$ og $F_{n+1}= F_{n}+F_{n-1}$
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Her er min måte. Den er kanskje heilt lik Aslan sin, men den ser litt anleis ut.
Vi ser berre på tal som startar på 0, så kan vi gange med 2 etterpå pga symmetri.

For tal med 3 siffer, er det berre 3 måtar, 011,010,001. Så skal vi leggje til nye siffer til venstre. Kall tal som sluttar med 2 like for A-tal og tal med 2 siste ulike for B-tal. 011 er eit A-tal og dei to andre B-tal. Når vi skal få 4 siffer, får ein to nye for kvart B-tal, for der kan ein både leggje til 0 og 1. Det eine av desse blir eit B-tal og det andre eit A-tal. Med A-tala kan ein berre leggje til eit tal, ellers vil ein få 2 like.

Kall antal A-tal med n siffer for $A_n$ og B-tal for $B_n$. Pga ressonementet ovanfor blir $A_n=B_{n-1}$ og $B_n=A_{n-1}+B_{n-1}$ Her blir både $A_n$ og $B_n$ Fibbonaci - følger og det kan lett sjekkast at $A_{12}+B_{12}=233$ når $A_3=1,B_3=2$. Etter å ha ganga med 2 blir svaret 466. Då spørst det kem av meg og aslan som har rekna feil.

Aslan har feil, $110 + 178 \not= 228$...men metodene er jo egentlig like...

Ja jeg må lære meg å gange og legge sammen tall tror jeg  :blush:   :P


abstract algebra
calculus
integration
algebra
function
domain
superior algebra

Ring $ R$ is semiprime if and only if for any left (or right) ideal $ I$ in $ R$,$ I^2\subseteq 0$ implies that $ I\subseteq 0$. How show example ring P that is semiprime and is not noetherian?

Do you have an example of a non-Noetherian ring?  The standard examples are integral domains, hence semiprime.


superior algebra
superior algebra solved

Let [b]$R$[/b] be a commutative ring with $a \in R$, and let [b]$B$[/b] be an ideal of [b]$R$[/b].

Let $A$ = {$x \in R: ax \in B$}.

Show that A is an ideal of $R$.[/b][/list][/code]

For any $x\in A$ we have $(ax)\in B$, thus $a(rx)=r(ax) \in B$ (it follows from the fact that $B$ is ideal) for any $r\in B$, which means that $rx$ is also in $A \Rightarrow$ $A$-ideal


How do I determine whether the redox reaction
\[ H^ \plus{} \plus{} MnO_4^ \minus{} \plus{} [Cr(NH_2 CONH_2 )_6 ]_4 [Cr(CN)_6 ]_3 \rightarrow Cr_2 O_7^(2 \minus{} ) \plus{} CO_2 \plus{} H_2 O \plus{} NO_3^ \minus{} \plus{} Mn^(2 \plus{} )
\]
takes place in acidic or basic solution?  Also, what are the two half-reactions I can write form it?

Thanks in advance!

You have H+ in your equation, so this takes place in acidic solution (also, permanganate ion is converted into Mn2+ in acid and into MnO2 in basic solution). For your second question, start by finding the oxidation numbers.


In professional football it is possible to score 6 points (a touchdown), 3 points (a field goal), or 2 points (a safety). If a touchdown is scored it is possible to score an additional point (the point after). What is the largest total score that cannot be achieved in football.

The only score that cannot be achieved in football is 1 point, hence it must be the largest.

This is because of Chicken McNugget, $ 2(3) \minus{} 2 \minus{} 3 \equal{} \boxed{1}$.

The Chicken McNugget Theorem states that for any two relatively prime positive integers x and y, the greatest integer that cannot be written with any combination of the two numbers is $xy-x-y$.
Check it out: [url]http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem[/url]


Shouldn't the sign be directed in other side - I mean $\leq$ instead of $\geq$ ?

It is clear that with opposite sign inequality is wrong.

hm ? when you multiply some things you get:

$\sum_{symm}(a+\sqrt{bc})(a+b)(a+c)$ while on the other side $3(a+b)(b+c)(c+a)$ and I think right side is greater because $a+b+c \geq \sqrt{ab} + \sqrt{bc} + \sqrt{ca}$ so when you multiply everything out and use Muirhead you got the ineq which is true for reversed sign.

What am I missing ?  :?

[quote="manlio"]For $a,b,c$ positive real numbers prove that

$\frac{a+\sqrt{bc}}{b+c}+\frac{b+\sqrt{ca}}{c+a}+\frac{c+\sqrt{ab}}{a+b} \geq 3 $[/quote]

The inequality is not true in either direction. Check a = 1, b = 2, c = 2 versus a = 1, b = 16, c = 16. But here are two correct inequalities that Manlio could have meant:

First inequality:

$\frac{a+\sqrt{bc}}{b+c}+\frac{b+\sqrt{ca}}{c+a}+\frac{c+\sqrt{ab}}{a+b} > 2.5$.

Second inequality:

$\frac{a+2\sqrt{bc}}{b+c}+\frac{b+2\sqrt{ca}}{c+a}+\frac{c+2\sqrt{ab}}{a+b} \geq 3$.

The second inequality was proven on http://www.mathlinks.ro/Forum/viewtopic.php?t=6420 (with the notations $a^2$, $b^2$, $c^2$ instead of a, b, c).

Proof of the first inequality:

Since $\sqrt{bc} \geq \frac{2bc}{b+c}$, $\sqrt{ca} \geq \frac{2ca}{c+a}$ and $\sqrt{ab} \geq \frac{2ab}{a+b}$ by GM-HM, it is enough to show that

$\frac{a+\frac{2bc}{b+c}}{b+c}+\frac{b+\frac{2ca}{c+a}}{c+a}+\frac{c+\frac{2ab}{a+b}}{a+b} > 2.5$.

But this is clear from

$\frac{a+\frac{2bc}{b+c}}{b+c}+\frac{b+\frac{2ca}{c+a}}{c+a}+\frac{c+\frac{2ab}{a+b}}{a+b}-2.5$
$=\frac{1}{2\left( b+c\right) ^{2}\left( c+a\right) ^{2}\left( a+b\right)^{2}}\cdot \left( 22a^{2}b^{2}c^{2}+2a^{4}bc+2b^{4}ca+2c^{4}ab\right.$
$\left. +6a^{3}b^{2}c+6a^{3}c^{2}b+6b^{3}c^{2}a+6b^{3}a^{2}c+6c^{3}a^{2}b+6c^{3}b^{2}a\right.$
$\left. +bc\left( 2b^{2}+3bc+2c^{2}\right) \left( b-c\right) ^{2}+ca\left(2c^{2}+3ca+2a^{2}\right) \left( c-a\right) ^{2}\right.$
$\left. +ab\left( 2a^{2}+3ab+2b^{2}\right) \left( a-b\right) ^{2}\right) > 0$

(since the numbers a, b, c are all positive). This shows that the inequality is very weak, but the constant 2.5 cannot be improved.

  Darij