rotation
number theory
least common multiple

Point $ A$ is on the edge of a circular disk. Every day at noon, the disk is rotated exactly $ 150^\circ$ in a counter-clockwise direction. What day of the week will it be the next time point $ A$ is at the same position that it was at 10am on Saturday?

Well, we have to find $ \text{lcm}(150,360)$, which is $ 1800$.  Divide this by $ 150$, to get $ 12$ days.  Thus, the day will be a $ \boxed{\text{Thursday}}$.

[quote="ernie"]Well, we have to find $ \text{gcd}(150,360)$, which is $ 1800$.  Divide this by $ 150$, to get $ 12$ days.  Thus, the day will be a $ \boxed{\text{Thursday}}$.[/quote]

Minor correction: it should be the least common multiple. :P

Alternatively, $ 150d\equal{}360k\implies5d\equal{}12k,\gcd(5,12)\equal{}1\implies d\equal{}12m$ for integer $ k,m$, and the next time occurs in $ 12$ days (the rest is the same as ernie's solution).

Sorry to revive this, but the ans is Wednesday for some reason. Is the answer wrong, or are our calculations flawed? Someone please explain this problem for me and others who might want to know why the answer is Wednesday.

Thanks.

It turns once on Saturday...

Remember, we start counting on Saturday before noon, not Saturday after noon.


Euler
geometry
circumcircle
projective geometry
geometry proposed

ABCD is a convex quadrilateral satisfying $ \angle A\equal{}\angle B\equal{}\angle C>90^o$. Prove that the Euler line of triangle ABC pass through D.

Let $ K \equiv AB \cap CD,$ $ L \equiv BC \cap DA.$ Let $ M, N$ be midpoints of $ AB, BC.$ $ \angle A \equal{} \angle B \equal{} \angle C$ implies that the perpendicular bisectors of $ AB, BC$ pass through $ L, K,$ respectively. $ G \equiv AN \cap CM$ is the centroid of the $ \triangle ABC,$ $ O \equiv LM \cap KN$ its circumcenter and $ D \equiv AL \cap KC.$ By Pappus theorem, $ G, O, D$ are collinear.


calculus
derivative
function
inequalities solved
inequalities

Prove that $ \forall x \geq 0 $ we have:
$ 1+ \frac{x}{3} - \frac{x^2}{9} \leq \sqrt[3]{1+x} \leq 1+ \frac{x}{3}- \frac{x^2}{9}+\frac{5}{81}x^3$

cheers! :D :D :cool:

A direct calculus is boring, but derivatives solve this problem.
   For the left hand, take f(x)=(1+x)^1/3-1-x/3+x^2/9.Then its derivative is 1/3*(x+1)^(-2/3)-1/3+2x/9 and this is clearly increasing, so it attains its minimum in 0, where it becomes 0. So the derivative is nonnegative and f is increasing. Finish.
   For the right hand, I suppose you meant (1+x)^1/3<=1+x/3-x^2/9+5/81x^3 (otherwise it is trivial). Take the function f(x)=1+x/3-x^2/9+5/81x^3-(1+x)^1/3. Its derivative is at least 5/27x^2-2x/9+1/3>=0 so f is increasing.

You are right harazi a solution without derivates could be quite boring! :)) 10x for the solution! :D


trigonometry
ratio
complex analysis
complex analysis unsolved

show that $ f(z)\equal{}\sum_{n\equal{}1}^{\infty}\frac{cos(nz)}{n!}$ converges uniformly on compact subsets of C

It can also be said that it converges "locally uniformly" (i.e. every point has a neighborhood in which the convergence is uniform). These two notions are equivalent on locally compact spaces such as $ \mathbb C$.

The Weierstrass M-test applies on any bounded set, because $ |\cos (nz)|\le e^{n|z|}$.

this was what i did but i was confused and not too sure. so we have $ |cos nz|\leq e^{n|z|}$ and since any compact set can be enclosed in $ |z|\leq R$ for sufficiently large R we have $ |cos nz|\leq e^{nR}$, so the series is majorized by $ \sum_{n\equal{}1}^{\infty}\frac {e^{nR}}{n!}$ which converges for example by ratio test. is that correct?


If we have $ m$ monkeys and $ n$ nuts where $ m>n$, and given that no monkeys can have more than one nut, how many ways are there to distribute the nuts to the monkeys?

Since no monkey gets more than one nut, exactly n monkeys get a nut, hence the number of ways is m choose n.

Exactly.  The generating function is $ (1 \plus{} x)^m$; one copy for every monkey.  If you want to generalize to the case where, say, no monkey can have more than $ k$ nuts, you'll want $ (1 \plus{} x \plus{} ... \plus{} x^k)^m$ and an appropriate generalization of the binomial theorem.

For the first nut, we have $ m$ possibilities, for the second, we have $ m\minus{}1$ possibilities.... for the $ n$th nut we have $ m\minus{}n\plus{}1$ possibilities.  However, because we are counting each nut as distinct, we have to divide by $ n!$, to get
$ \frac{(m)(m\minus{}1)(m\minus{}2)...(m\minus{}n\plus{}2)(m\minus{}n\plus{}1)}{n!}$.  This is equal to, of course $ m \choose n$.


What do you predict the cutoff index will be? I'm guessing about a 200 or so... since this year's combo was significantly harder than last years. Im guessing AMC 12 scores were about -7 from last year, and AIME scores at least one less, making the cutoff about 205 or so.

And I hope I'm right, my index is 212.5.

Um...Excuse my ignorance, but how is the index calculated?

A cutoff at 210 or lower would be awesome! Nukular, I'm pretty sure that 212.5 will get you in...but these are based on the indicies from 2000.

(I have an index of 217.5)

Ouch I think I have index of 199.5.  Oh well I still did much better than expected:)


[u][b]Problem.[/b][/u] Let $ A,B$ and $ C$ be three $ 2\times 2$ matrix which are commutative, i.e $ AB \equal{} BA, BC \equal{} CB, CA \equal{} AC$,  and such that $ \det(AB \plus{} BC \plus{} CA)\leq 0$. Prove that $ \det(A^2 \plus{} B^2 \plus{} C^2) \geq 0$.

[u][b]Solution.[/b] [/u] Let $ P(x)\equal{}\det(A^2\plus{}B^2\plus{}C^2\plus{}x(AB\plus{}BC\plus{}CA))$. We are easily find that $ P(x)\equal{}ax^2\plus{}bx\plus{}c$ with $ a\equal{}\det(AB\plus{}BC\plus{}CA)$ and $ c\equal{}\det(A^2\plus{}B^2\plus{}C^2)$.
Since $ P(2)\equal{}4a\plus{}2b\plus{}c\geq 0$ and $ P(\minus{}1)\equal{}a\minus{}b\plus{}c\geq 0$, $ 6a\plus{}3c\equal{}P(2)\plus{}2P(\minus{}1)\geq 0$. Hence $ c\geq \minus{}2a\geq 0$.

[size=150][b]NOTE:[/b] We can not explain why [/size] $ P(\minus{}1)\geq 0$

Please help me to explain or find another solution for this problem.

Why did you start another topic since you had already asked this question here http://www.mathlinks.ro/viewtopic.php?t=189991 ?


If $ sin({\alpha}) \plus{} sin({\beta}) \plus{} sin({\gamma}) \equal{} 0 and cos({\alpha}) \plus{} cos({\beta}) \plus{} cos({\gamma}) \equal{} 0$
then prove that
 $ sin({3\alpha}) \plus{} sin({3\beta}) \plus{} sin({3\gamma}) \equal{} 3\cdot sin({\alpha \plus{} \beta \plus{} \gamma})$.

Can any body prove it without using complex number ?
 :coolspeak:

hello, your term
$ \sin(\alpha)+\sin(\beta)+\sin(\gamma)-3\sin(\alpha+\beta+\gamma)$
is equivalent to
$ 3\cos(\alpha)^2\sin(\alpha)-3\cos(\beta)\cos(\gamma)\sin(\alpha)-\sin(\alpha)^3
+\3\cos(\beta)^2\sin(\beta)-3\cos(\alpha)\cos(\gamma)\sin(\beta)-\sin(\beta)^3-
3\cos(\alpha)\cos(\beta)\sin(\gamma)+3\cos(\gamma)^2\sin(\gamma)+3\sin(\alpha)\sin(\beta)\sin(\gamma)-\sin(\gamma)^3$
inserting
$ \sin(\gamma)=-\sin(\alpha)-\sin(\beta)$
and
$ \cos(\gamma)=-\cos(\alpha)-\cos(\beta)$
in this term gives zero after expanding all.
Sonnhard.


Start with the set $ \{3,4,12\}.$ In each step you may choose two of the numbers $ a,b$ and replace them with $ 0.6a \minus{} 0.8b$ and $ 0.8a \plus{} 0.6b$ respectively. Is it possible to reach $ (a)$ or $ (b)$ in finitely many steps:
$ (a)\qquad\{4,6,12\}$
$ (b)\qquad\{x,y,z\}\text{ where }|x \minus{} 4|,|y \minus{} 6|,|z \minus{} 12|\text{ are all less than }\frac {1}{\sqrt {3}}?$

I don't really know where to use the invariance principle.

Sorry if this is too easy for this forum :oops:

Thanks in advance.

It should read "$ 0.6a \minus{} 0.8b$" instead of $ 0.6a \minus{} 8b$, but...

[hide="Hint"]What do $ 0.6$ and $ 0.8$ remind you of...[/hide]

If you only wish to see hints and not solutions then do not click on the hidden content, as I have posted the whole solution.

[hide="Part a)"]$ (0.6a\minus{}0.8b)^2\plus{}(0.8a\plus{}0.6b)^2 \equal{} a^2 \plus{} b^2$ so the sum of squares of the elements is an invariant.

The sum of squares of the original set is $ 3^2\plus{}4^2\plus{}12^2\equal{}169$ and the sum of squares of the target set is $ 4^2\plus{}6^2\plus{}12^2\equal{}196$ so the goal is unreachable.[/hide]

[hide="Part b)"]it is very easy to tackle this part using co-ordinate geometry. Let the elements of the set be co-ordinates $ (x,y,z)$.

Since in the original set $ a^2\plus{}b^2\plus{}c^2\equal{}13^2$ the point $ (a,b,c)$ lies on the sphere with centre $ (0,0,0)$ and radius $ 13$ and thus has distance $ 13$ from the origin.

In the target set we have $ \mid x\minus{}4 \mid, \mid y\minus{}6 \mid, \mid z\minus{}12 \mid < \frac{1}{\sqrt{3}}$ so the square of each term is less than $ \frac{1}{3}$

Thus we have $ (\mid x\minus{}4 \mid)^2 \plus{} (\mid y\minus{}6 \mid)^2 \plus{} (\mid z\minus{}12 \mid)^2 < 1$.

The point $ (\mid x\minus{}4 \mid, \mid y\minus{}6 \mid, \mid z\minus{}12 \mid)$  therefore is less than one unit away from the point $ (4,6,12)$.

Since the point $ (4,6,12)$ has distance 14 from the origin, the point $ (\mid x\minus{}4 \mid, \mid y\minus{}6 \mid, \mid z\minus{}12 \mid)$ is more than 13 units from the origin.

Therefore since the point $ (a,b,c)$ is always exactly 13 units from the origin, $ (a,b,c)$ can never equal $ (\mid x\minus{}4 \mid, \mid y\minus{}6 \mid, \mid z\minus{}12 \mid)$ so the goal is unreachable.

Sorry if this was hard to understand, because the solution always makes sense in my head but I have trouble typing it up in a clear manner.[/hide]

By the way I was to creat a topic asking what is this Invariance Principle but I found this topic can someone till me in a short way?  :)

Basically the invariance principle is trying to find a property that does not change when something goes through a transformation i.e an "invariant".

A simple example is this: Consider the numbers $ (1,2,3...,12)$ At each step you may replace any two numbers by their positive difference. Prove that the last number cannot be $ 1$. The "invariant" here is that the sum of all the remaining elements always retains the same parity. Since the sum was even at the beginning, it will be even at the end, so the last number must be even.

The proof that the parity is indeed an invariant is simple: If you pick two even numbers then their difference is even so you have removed 2 even numbers and added one even number, meaning that in total you have reduced the sum by an even amount so the parity is the same. Use similar arguments for the case where you pick 2 odd numbers and the case where you pick one odd and one even.

Sorry about the error in my solution. Anything which says "the point $ (\mid x\minus{}4 \mid, \mid y\minus{}6 \mid, \mid z\minus{}12 \mid)$" should actually read "the point $ (x,y,z)$"

[hide=" Part b, little bit different"]

$ |x\minus{}4|\le \frac{1}{\sqrt3}$, so $ \frac{\minus{}1}{\sqrt3}\le x\minus{}4\le \frac{1}{\sqrt3}$

$ 4\minus{}\frac{1}{\sqrt3}\le x\le 4\plus{}\frac{1}{\sqrt3}$, and same with y,z

The least value of $ x^2\plus{}y^2\plus{}z^2$ is for least $ x,y,z$, or $ 4\minus{}\frac{1}{\sqrt3}$ and respective.

Then $ minvalue\equal{}197\minus{}\frac{44}{\sqrt3}$, which is more than 169, so in any case, the sum of squares is bigger than 169, but the invariance tells us that it must remain 169, so the answer is no

This is problem from Engel, solution there is the same as CrazyFK's... I found non-geometry access more natural, although they are almost the same

[/hide]


induction
modular arithmetic
number theory unsolved
number theory

Hello !!

$ n \in \mathbb{N}$

let : $ x_n\equal{}\frac{n(n\plus{}1)}{2}$ and $ y_n\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$ and $ z_n\equal{} x_n^2$

1/ prove that $ (\forall n \in \mathbb{N})$ $ x_n \in \mathbb{N}$ and $ y_n \in \mathbb{N}$ and $ z_n \in \mathbb{N}$

2/ let $ d_n \equal{} pgcd((2n\plus{}1),3)$

a/calculate $ d_n$  according to $ n$

b/deduct  $ pgcd (x_n;y_n)$

3/ calculate : $ \Delta_n \equal{} pgcd ( n, n\plus{}2 )$

4/deduct  $ pgcd ( z_{n\plus{}1}, z_n )$

5/ prove that : $ ( \forall n \in \mathbb{N}^*\minus{}\{1\} ) pgcd (z_n,z_{n\plus{}1} , z_{n\plus{}2} )\equal{} 1$

[quote="*210*"]Hello !!

$ n \in \mathbb{N}$

let : $ x_n \equal{} \frac {n(n \plus{} 1)}{2}$ and $ y_n \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$ and $ z_n \equal{} x_n^2$

1/ prove that $ (\forall n \in \mathbb{N})$ $ x_n \in \mathbb{N}$ and $ y_n \in \mathbb{N}$ and $ z_n \in \mathbb{N}$

2/ let $ d_n \equal{} pgcd((2n \plus{} 1),3)$

a/calculate $ d_n$  according to $ n$

b/deduct  $ pgcd (x_n;y_n)$

3/ calculate : $ \Delta_n \equal{} pgcd ( n, n \plus{} 2 )$

4/deduct  $ pgcd ( z_{n \plus{} 1}, z_n )$

5/ prove that : $ ( \forall n \in \mathbb{N}^* \minus{} \{1\} ) pgcd (z_n,z_{n \plus{} 1} , z_{n \plus{} 2} ) \equal{} 1$[/quote]

And for what date, exactly, should you give back this highly complex olympiad homework ?

Hello  Mr Patrick

je sais que ,comme vous êtes  un grand mateux vous trouvez l'exo très facile .
Cependant , j'aimerai bien avoir vos  idées pour pouvoir résoudre l'exercice
Thank you !!!

[quote="*210*"]
$ n \in \mathbb{N}$

let : $ x_n \equal{} \frac {n(n \plus{} 1)}{2}$ and $ y_n \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$ and $ z_n \equal{} x_n^2$

1/ prove that $ (\forall n \in \mathbb{N})$ $ x_n \in \mathbb{N}$ and $ y_n \in \mathbb{N}$ and $ z_n \in \mathbb{N}$
[/quote]
First method :
$ x_n$ : $ \forall n\in\mathbb{N}$, either $ n$ is even, either $ n\plus{}1$ is even, hence the result.
$ y_n$ : $ \forall n\in\mathbb{N}$, either $ n$ is even, either $ n\plus{}1$ is even. And $ 3$ divides either $ 2n$, either $ 2n\plus{}1$, either $ 2n\plus{}2$, hence the result.
Second method ( :) ) Use induction to show that :
$ x_n\equal{}\sum_{k\equal{}1}^nk$ and so is a finite sum of integers, and so is an integer.
$ y_n\equal{}\sum_{k\equal{}1}^nk^2$ and so is a finite sum of integers, and so is an integer.
[quote="*210*"]
2/ let $ d_n \equal{} pgcd((2n \plus{} 1),3)$

a/calculate $ d_n$  according to $ n$
[/quote]
$ 3 | 2n\plus{}1\iff n\equal{}1\pmod 3$, so $ d_n\equal{}3$ if $ n\equal{}1\pmod 3$ and $ d_n\equal{}1$ otherwise.
[quote="*210*"]

b/deduct  $ pgcd (x_n;y_n)$
[/quote]
$ y_n\equal{}x_n\frac{2n\plus{}1}{3}$. So if $ 3 | 2n\plus{}1$, $ pgcd (x_n;y_n)\equal{}x_n$, otherwise $ pgcd (x_n;y_n)\equal{}\frac{x_n}{3}$
You can also write $ pgcd (x_n;y_n)\equal{}\frac{x_nd_n}{3}$
[quote="*210*"]

3/ calculate : $ \Delta_n \equal{} pgcd ( n, n \plus{} 2 )$
[/quote]
$ \Delta_n \equal{} 2$ if $ n$ is even and $ \Delta_n \equal{} 1$ otherwise (if $ n$ is odd)
[quote="*210*"]
4/deduct  $ pgcd ( z_{n \plus{} 1}, z_n )$
[/quote]
$ pgcd ( z_{n \plus{} 1}, z_n )\equal{}(n\plus{}1)^2$ if $ n$ is even, and $ pgcd ( z_{n \plus{} 1}, z_n )\equal{}(\frac{n\plus{}1}{2})^2$
You can also write $ pgcd ( z_{n \plus{} 1}, z_n )\equal{}(\Delta_n\frac{n\plus{}1}{2})^2$
[quote="*210*"]
5/ prove that : $ ( \forall n \in \mathbb{N}^* \minus{} \{1\} ) pgcd (z_n,z_{n \plus{} 1} , z_{n \plus{} 2} ) \equal{} 1$ [/quote]
If $ n\equal{}2p$, $ pgcd ( z_{n \plus{} 1}, z_n )\equal{}(2p\plus{}1)^2$ and $ z_{n\plus{}2}\equal{}(p\plus{}1)^2(2p\plus{}3)^2$ hence the result.
If $ n\equal{}2p\plus{}1$, $ pgcd ( z_{n \plus{} 1}, z_n )\equal{}(p\plus{}1)^2$ and $ z_{n\plus{}2}\equal{}(p\plus{}2)^2(2p\plus{}3)^2$ hence the result.

And, I repeat, this homework is not in the good forum. It's obviously not an olympiad problem.


geometry
inradius
incenter
perimeter
geometry solved

let ABC be a triangle find the point D on BC  such that the inradius of triangles ABD
and ACD are equal

$AD=\sqrt{p\cdot(p-a)}$ where p -semiperimeter!

Let $r_1$ be the common radius, $O_1$ and $p_1$ the incenter respectively the perimeter of ABD.
The same $O_2$ and $p_2$ for ACD. Then  :
1. $r_1(p_1+p_2)=pr$, hence $\frac{r_1}{r}=\frac{p}{p+AD}$;
2. $O_1O_2=a+AD-p$;
3. Since $\triangle{IO_1O_2}$ and $\triangle{IBC}$ are similar:
   $\frac{a-p+AD}{a}=\frac{r-r_1}{r}=\frac{AD}{p+AD}$
and it follows that $AD=\sqrt{p\cdot (p-a)}$


Suppose T is a permutation on n-set {1,2,3...n}. Prove that the number of permutation T such that there exist no three numbers i<j<k with T(i)<T(k)<T(j) is equal to the nth Calatan number.[/list]

[hide="hint"]use the recursive definition of the catalan numbers.[/hide]

My approach is based on a bijection with a certain set of permutations, is this similar to what you have in mind?

for such a permutation T consider the position of n. $ t(n)\equal{}k$
it is obvious that the set A $ t(r)\le k\minus{}1$ is of the form $ \{s,s\plus{}1,...,n\minus{}1\}$. and the set B $ t(r)\geq k\plus{}1$ is $ \{1,2,...,s\minus{}1\}$ so as the number of permutations in A and B is $ C_{s\minus{}1}$ and $ C_{n\minus{}s}$ we get $ C_n\equal{}\sum_{s\equal{}1}^n C_{s\minus{}1}C_{n\minus{}s}$ so the numbers $ C_k$ satisfy the same recursion as the catalan numbers. it is easy to check that $ C_1\equal{}1$ so we are done.

I can't tell you if this interpretation of the catalan numbers is original, but you should know that a great number of interpretations are known till now.


A carnot heat engine has a theortical efficiency of $ 1 \minus{} \frac{T_c}{T_h}$

I take the same reservoirs and build another engine (Otto)

1-2 adiabatic expansion from volume $ V_1$ to $ V_2$
2-3 constant volume cooling until the gas has temperature $ T_c$
3-4 adiabatic compression until volume $ V_1$
4-1 constant volume heating until temperature $ T_h$

This engine has an efficiency of $ 1 \minus{} (\frac{V_1}{V_2})^{\gamma\minus{}1}$

i compare both efficiences, they must be equal, then $ \frac{T_c}{T_h} \equal{} (\frac{V_1}{V_2})^{\gamma\minus{}1}$
but $ T_h \equal{} T_1$, $ T_c\equal{}T_3$ and $ V_2\equal{}V_3$
that is, the points 1 and 3 are both on an adiabatic curve, contradiction. Why that?

For the same reservoirs, Carnot's cycle gives  the maximum efficiency.


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Hmm, good ascii art. :)
Copy and paste in notepad for best view.

that's a huuuuuge crotch (facing to the right)

[quote="Klebian"]Um... that's her foot.
My opinion:
Good work, but some of her torso is a bit disproportionate.[/quote]

Her?

...oops.

:rotfl:  :rotfl:
that was the funniest thing today

Daesun, how could you not realize that it's a girl?  She's got some MAJOR cleavage.


The sides of $ \triangle PQR$ are in the ratio 7:8:9. $ QN$ is the angle bisector drawn to the shortest side $ PR,$ dividing it into segments $ PN$ and $ RN$. If the length of $ PR$ is 70, then the difference between $ PN$ and $ RN$ is $ \frac{m}{n},$ where $ (m,n) \equal{} 1$. What is $ m\plus{}n$?

[hide]

Since QN is an angle bisector,

$ \frac {PQ}{PN} \equal{} \frac {QR}{RX}$

$ PN \equal{} \frac {70\cdot90}{170}$

$ RN \equal{} \frac {70\cdot80}{170}$

$ PN \minus{} RN \equal{} \frac {70}{17}$

$ m \plus{} n \equal{} 87$[/hide]


[quote="Jean-Louis"]let's have y = x^x^x...^x (x times), with y = 2.
So, 2 = x^x^x...^x
I see 2 possible solutions for x (x = 2 and x = 4)

Which one is correct ?[/quote]

Neither. I think what you are trying to say is that if you have $  x^{x^{x^{x}}}$ with $ x$ number of exponents, what are your solutions for $ y$. Thus:

if $ x = 1, y = 1$
if $ x = 2, y = 2^{2}= 4$
if $ x = 3, y = 3^{3^{3}}= 19683$

and so on. If not, please clarify your problem a little more.


Determine whether it is true that:
For any positive integers a,b, $ 4a^{2}b^{2}\plus{}4ab^{2}\minus{}a$ is not a perfect square.

$ 4a^{2}b^{2}\plus{}4ab^{2}\minus{}a \equal{} (ya)^{2}$
then
$ 4b^{2}(a\plus{}1) \equal{} y^{2}a\plus{}1$ (1)
Chose a=7 then
$ 32b^{2}\minus{}7a^{2}\equal{} 1$
I think this equation has infty solution(Pell's equation) because the equation
$ 2a^{2}\minus{}7b^{2}\equal{} 1$  has a solution (2,1) so  has infty solution.

The equation is $ 16b^{2}\minus{}7a^{2}\equal{}\minus{}1$, not the $ 16b^{2}\minus{}7a^{2}\equal{}1$.

And have a look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=159987


Let $ \textbf{V}$ and $ \textbf{W}$ be vector spaces, and let $ S$ be a subset of $ \textbf{V}$.   Define $ S^{0}\equal{}\{T\in\mathcal{L}(\textbf{V},\textbf{W}): T(x)\equal{}0\mbox{ for all }x\in S\}$.  Prove the following:
(a) $ S^{0}$ is a subspace of $ \mathcal{L}(\textbf{V},\textbf{W})$
(b) If $ S_{1}$ and $ S_{2}$ are subsets of $ \textbf{V}$ and $ S_{1}\subseteq S_{2}$, then $ S_{2}^{0}\subseteq S_{1}^{0}$
(c) If $ \textbf{V}_{1}$ and $ \textbf{V}_{2}$ are subspaces of $ \textbf{V}$, then $ (\textbf{V}_{1}\plus{}\textbf{V}_{2})^{0}\equal{}\textbf{V}_{1}^{0}\cap\textbf{V}_{2}^{0}$

(a) is definitional. $ 0\in S^0$ for sure, and if $ T_1, T_2\in S^0$ and $ \alpha, \beta$ scalars then $ (\alpha T_1 \plus{} \beta T_2)\in S^0$ by just applying an arbitrary $ x\in S$ and seeing that it is still zero.

$ (\alpha T_1 \plus{} \beta T_2)(x) \equal{} \alpha T_1 (x) \plus{} \beta T_2(x) \equal{} 0 \plus{} 0 \equal{} 0$

Is $ L: \equal{}\mathcal{L}(V,W)$ the set of homomorphisms form V to W?

(a)It is easy: $ f,g \in S^0 \Leftrightarrow (f\plus{}\lambda g)(x)\equal{}f(x)\plus{}\lambda g(x)\equal{}0\Leftrightarrow f \plus{} \lambda g \in S^0$
(b)If $ f$ is zero over $ S_2$, it might be the same over $ S_1 \subseteq S_2$...
(c)If f is zero over $ V_1$ and $ V_2$ it has to be the same, by linearity, over $ V_1\plus{}V_2$. For the other inclusion, just consider that $ V_1 \subseteq (V_1\plus{}V_2) \Rightarrow  (V_1\plus{}V_2)^0 \subseteq V_1^0$

^^$ \mathcal{L}$ is the set of linear transformations from $ V\rightarrow W$.


A cube of side length 3 is divided into 27 unit cubes(that is cubes with side length one)
Twenty five of these cubes are coloured with blue paint and the remaining 2 are coloured with yellow paint. How many distinct colourings are possible.

You should be specific
is the cube fixed on the table or it can be rotated as we want ?
do you consider totwo reflected colourings to be same or different?


Let $ a,b,c \ge 0$ and $ a,b,c$ distinct. Prove that

$ \sum \frac{4a^2\plus{}5b^2\plus{}2c^2\minus{}4ab\minus{}4ac\minus{}2bc}{(b\minus{}c)^2}\geq 23$

______________________
Panagiote Ligouras

By the summation sign, do you mean:
$ \frac {4a^2 \plus{} 5b^2 \plus{} 2c^2 \minus{} 4ab \minus{} 4ac \minus{} 2bc}{(b \minus{} c)^2} \plus{} \frac {4b^2 \plus{} 5c^2 \plus{} 2a^2 \minus{} 4ab \minus{} 4bc \minus{} 2ca}{(c \minus{} a)^2} \plus{} \frac {4c^2 \plus{} 5a^2 \plus{} 2b^2 \minus{} 4bc \minus{} 4ca \minus{} 2ab}{(a \minus{} b)^2}\ge 23$
?? :blush:

PS: The inequality seems to be too big for words!?

/**[edited]**\ :roll:  :oops:

[quote="Potla"]By the summation sign, do you mean:
$ \frac {4a^2 \plus{} 5b^2 \minus{} 2c^2 \minus{} 4ab \minus{} 4ac \minus{} 2bc}{(b \minus{} c)^2} \plus{} \frac {4b^2 \plus{} 5c^2 \minus{} 2a^2 \minus{} 4ab \minus{} 4bc \minus{} 2ca}{(c \minus{} a)^2} \plus{} \frac {4c^2 \plus{} 5a^2 \minus{} 2b^2 \minus{} 4bc \minus{} 4ca \minus{} 2ab}{(a \minus{} b)^2}\ge 23$
?? :blush:

PS: The inequality seems to be too big for words!?[/quote]

HI!!

The term is $ \plus{}2c^2$

[quote="Ligouras"][quote="Potla"]By the summation sign, do you mean:
$ \frac {4a^2 \plus{} 5b^2 \minus{} 2c^2 \minus{} 4ab \minus{} 4ac \minus{} 2bc}{(b \minus{} c)^2} \plus{} \frac {4b^2 \plus{} 5c^2 \minus{} 2a^2 \minus{} 4ab \minus{} 4bc \minus{} 2ca}{(c \minus{} a)^2} \plus{} \frac {4c^2 \plus{} 5a^2 \minus{} 2b^2 \minus{} 4bc \minus{} 4ca \minus{} 2ab}{(a \minus{} b)^2}\ge 23$
?? :blush:

PS: The inequality seems to be too big for words!?[/quote]

HI!!

The term is $ \plus{} 2c^2$[/quote]
Thanks, I edited it.

[quote="Agr_94_Math"]Well it is not a tough inequality.
U can reduce the no. of terms by introducing new variables.
Will post my solution later. Too lazy to properly post it in latex.[/quote]

Dear "Agr_94_Math", aspect your solution with joy :wink:

[quote="Agr_94_Math"]Well it is not a tough inequality.
U can reduce the no. of terms by introducing new variables.
Will post my solution later. Too lazy to properly post it in latex.[/quote]
Another way: it is sufficient to prove it for, say, $ c\equal{}0$, as we have.
$ LHS\equal{}\sum\frac{(2a\minus{}b\minus{}c) ^2\plus{}(2b\minus{}c)^2}{(b\minus{}c)^2}\equal{}\sum\frac{((a\minus{}c)\plus{}(b\minus{}c)) ^2\plus{}(b\plus{}(b\minus{}c))^2}{(b\minus{}c)^2}$


p\mid ab (testing the latex on my system)

There is a test forum for this, I just recently used it [url]http://www.mathlinks.ro/index.php?f=224[/url].


function
inequalities
limit
calculus
integration
real analysis
real analysis unsolved

hello every body
f is a function defined on [0,1] with f(1)<0<f(0).
suppose there exist a function g continous on [0;1] and the function f+g is increasing
prove that there existe c on [0;1] such that f(c)=0
sorry for my bad english

Since $ h\equal{}f\plus{}g$ is an increasing function, then $ h$ has lateral limits and:
$ h(t\minus{}) \leq h(t) \leq h(t\plus{})$ for all $ t\in (0,1)$
For the continuity of $ g$, we have: $ h(t\minus{}) \equal{} f(t \minus{}) \plus{} g(t)$ and $ h(t\plus{}) \equal{} f(t \plus{}) \plus{} g(t)$.
Then
$ f(t\minus{}) \leq f(t) \leq f(t\plus{})$ for all $ t\in (0,1)$
Let $ S$ be the supremum of $ t\in [0,1]$ such that $ f(t)>0$.
Suppose that $ S\in(0,1)$, then we have $ f(S\minus{})\geq 0$ and $ f(S\plus{})\leq 0$.
So, for the anterior inequality, $ f(S\minus{})\equal{}f(S)\equal{}f(S\plus{})\equal{}0$.
The cases $ S\equal{}0$, $ S\equal{}1$ are similars to this.

please jonfev what do you mean by t- and t+
and do we have [0,1]=(0;1) or what?
other question: where are you used the fact f(1)<0<f(0) ?
please respond me as soon as possible

\[ \mathop {\lim }\limits_{x \to x_0  \plus{} } f\left( x \right) \equal{} f\left( {x_0  \plus{} } \right)
\]

ok now i see ,thnx a lot, but where are you used f(1)<0<f(0) ?

Alain,
$ (0,1)$ = $ [0,1]\backslash \{0,1\}$
You need $ f(0)>0$ for the definition of $ S$
You will need the fact $ f(1)<0$ in the case $ S\equal{}1$
In my post, $ S\in [0,1]$, so we have three cases: $ S \equal{} 0$, $ S\in (0,1)$ or $ S\equal{}1$.
I have posted the second one, you just need to prove that $ f(S\minus{})\geq 0$ and $ f(S\plus{})\leq 0$.
You can proceed as follows:
By definition of $ S$: for all $ t> S$, $ f(t)\leq 0$.
In particular,
\[ 0\geq f(S\plus{})\geq f(S)\geq f(S\minus{})\]
By definition of $ S$ and the fact that $ f(S)\leq 0$, we must have a sequence $ S_n\uparrow S$ with $ f(S_n) > 0$. So, we conclude that $ f(S\minus{})\geq 0$.

The other two cases are solved in similar way.

thnx a lot jonfv
but please where can I find this sort of exercices (speaking about continuity without derivation and integration) ?


Extremal combinatorics
combinatorics
Set systems
imo 1991
relatively prime
IMO

Let $ S \equal{} \{1,2,3,\cdots ,280\}$. Find the smallest integer $ n$ such that each $ n$-element subset of $ S$ contains five numbers which are pairwise relatively prime.

Let's look at this another way. We aim to find the least number of integers we can [i]remove[/i] from S to leave a set which does not contain 5 pairwise coprime integers.

Let $ S_p \equal{} \{p^k | p^k \leq 280\}$
And more generally $ S_{p_1,p_2,...,p_l} \equal{} \{n \equal{} p_1^{q_1}p_2^{q_2}....p_l^{q_l}| n \leq 280 \}$
And finally $ S_1 \equal{} \{1\}$
It is clear that these sets taken over all collections of primes and 1 forms a partition of S.

If there are representatives from five sets $ S_p$, there are five mutually coprime integers, so all but four of the sets $ S_p$ must be completely removed. It is clear that (with the exception of 1, which can clearly be removed first) if $ p>q, |S_p| \leq |S_q|$ so it is worth removing $ S_p$ before removing $ S_q$ (if we are striving for a minimum).
Furthermore, if we remove two sets $ S_p, S_q$, we must also (to stop there from being 5 mutually coprime integers) remove $ S_{p,q}$ and so on, and these have similar ordering relations.
So once we have removed everything we need to keeping sizes of sets removed to a minimum, we are left with $ S_2,S_3,S_5,S_7$ and the multi-index sets not coprime to all of these.
In other words, we have the set of multiples of 2,3,5 and 7, which can be calculated (by lots of inclusion-exclusion) to have cardinality $ 216$. Therefore, a subset of $ S$ of size $ 217$ must contain 5 coprime integers.

[hide=Solution]Firstly note that $\lfloor\sqrt{280}\rfloor = 16$. Therefore, we only need to check primes $p\le 16$ to check if a number from $1$ to $280$ is prime.

I claim the answer is $\boxed{217}$. First, let's look at a subset of $S$ that consists only of multiples of $2,3,5,7$. It is easy to calculate by PIE that the total number of elements in this set is $$\left\lfloor\tfrac{280}{2}\right\rfloor+\left\lfloor\tfrac{280}{7}\right\rfloor+\left\lfloor\tfrac{280}{3}\right\rfloor+\left\lfloor\tfrac{280}{5}\right\rfloor-\left\lfloor\tfrac{280}{14}\right\rfloor-\left\lfloor\tfrac{280}{6}\right\rfloor-\left\lfloor\tfrac{280}{10}\right\rfloor-\left\lfloor\tfrac{280}{15}\right\rfloor-\left\lfloor\tfrac{280}{21}\right\rfloor-\left\lfloor\tfrac{280}{35}\right\rfloor+\left\lfloor\tfrac{280}{70}\right\rfloor+\left\lfloor\tfrac{280}{105}\right\rfloor+\left\lfloor\tfrac{280}{42}\right\rfloor+\left\lfloor\tfrac{280}{30}\right\rfloor-\left\lfloor\tfrac{280}{210}\right\rfloor=216.$$ Now, note that no matter what $5$ element subset we consider of this subset we just made, consisting of multiples of $2,3,5,7$, we will always have at least $2$ of those elements that exist in the subset that share a factor greater than $1$ by the Pigeonhole Principle since $\left\lfloor\tfrac{5}{4}\right\rfloor+1=2$. Thus, we know $n>216$ from this. We also know that there are $212$ composite numbers and $4$ prime numbers (obviously just $2,3,5,7$) in this subset of $S$. Now we do casework on the numbers $11,13$ (we don't have to check higher numbers because remember $p\le 16$!) to find the other composite numbers. We find that all of numbers, $121,143,169,187,209,221,247,253$ fills in the rest of the composite numbers from $1$ to $280$. This gives a total of $8$ numbers when counted. So we can now count up the total as $220$ composite numbers and the remaining must be prime so $59$ prime numbers ($1$ not included as prime or composite). Now suppose $X$ is a subset of our set $S$ such that $|X|=217$ ($|X|$ represents its cardinality or the number of elements in the subset $X$). Also suppose that there is no $5$ element subset of $X$ such that all $5$ elements are relatively prime to each other. This means we would have to have at most $4$ prime numbers and at least $213$ composite numbers to make this subset $X$. This means the set $S-X$ (everything that is in $S$ but not in $X$) has at most $220-213=7$ composite numbers. Now consider the following $8$ sets and notice that at least one of these must be a subset of $X$!: $$\{2^2,3^2,5^2,7^2,13^2\}$$$$\{2\times31,3\times29,5\times23,7\times19,11\times17\}$$$$\{2\times47,3\times43,5\times41,7\times37,13\times19\}$$$$\{2\times41,3\times37,5\times31,7\times29,11\times23\}$$$$\{2\times23,3\times19,5\times17,7\times13,11^2\}$$$$\{2\times43,3\times41,5\times37,7\times31,13\times17\}$$$$\{2\times37,3\times31,5\times29,7\times23,11\times19\}$$$$\{2\times29,3\times23,5\times19,7\times17,11\times13\}$$And obviously each of these sets has $5$ elements that are all relatively prime numbers, as desired.[/hide]

Sorry for the bump, but why is this problem difficult enough to be a #3? To me this problem looks more like something you would find early-mid AIME than IMO level. Is it just because IMO 25 years ago was much easier than what it is nowadays?

The answer is $\boxed{217}$. Partition $S$ into sets $A_i$ such that$$A_i = \{a \in \mathbb N \mid p_i \text{ divides } a \text{ and } p_j \nmid a \ \forall j < i\},$$where $p_i$ denotes the $i$th prime. Observe that $|A_1|+|A_2|+|A_3|+|A_4| = 216$ by a PIE computation. Notice that by definition, any two numbers that fall in different sets are relatively prime. Furthermore, if we pick 217 numbers, by Pigeonhole two of them must fall in the same set because $|A_i|$ is strictly decreasing as $i$ increases (and thus, 186 is the maximum possible sum of the cardinalities of any four of these sets). Therefore we have the result.

is my method correct here
since we wanted to find the smallest n for all subset
Worst Case is when n divisible by 2
so you need to pick at least (140) so it left by things not divisible by 2
Again Worst Case is when n is divisible by 3
but u need to count the inclusion when 2 doesn't divide n  there is 93 multiple of 3 in the set
but there is 46 number divisible by 2 subtracting we get 47
Why worst case since they are the one that appear the most in the set
Worst case when n divisible by 5
there is 56 number multiple of 5  but we need to find the pure one it is not divisible by 2 not divisible by 3 using PIE
56 = Pure+(|divisible by 10|+|divisible by 15|-|divisible by 30|) = Pure + (|28|+|18|-9)
Pure = 19
Worst case when n divisible by 7 but doesn't have (2,3,5) as factor
just bash since quite small
$7*1,7^2,7*11,7*13,7*17,7*19,7*23,7*29,7*31,7*37= 10$
$140+47+19+10 = 216$
is the required which there exist already (4) by adding one more number since the one that being left are already relative prime to 2,3,5,7 (216+1) done


I can calculate some lines of tangency, but how do you make lines of tagentcy that aren't (if you will) perpendicular to the center of the circle?

Sorry, but I don't understand you: how can a line be perpendicular to a point?  :maybe: If what you are looking for is drawing a tangent line to a given circle through a given point, then the olympiad package has the corresponding routine:
[code]
import olympiad;
size(150);
pair P=(3,4), O=origin;
draw(Circle(O,3));
draw(tangent(P,O,3,1)--P);
draw(tangent(P,O,3,2)--P);
[/code]
yields
[asy]size(150);
pair P=(3,4), O=origin;
draw(Circle(O,3));
draw(tangent(P,O,3,1)--P);
draw(tangent(P,O,3,2)--P);[/asy]
In the tangent command the first argument is the pair representing the point from which you want to draw the tangent line, the second is the pair representing the center of the circle, the third is the circle radius and the fourth tells which one of the two possible tangents you need. The command returns the point at which the tangent line touches the circle.

So this olympiad thing is a seperate package?
(And when i meant perpendicular i meant perpindicular to the x and y axis)
[asy]import graph;
draw(Circle((0,0),20));
draw((-40,20)--(40,20));[/asy]
(Like that)

Yes, almost everything is a sep[b]a[/b]rate thing :). Also, if you know how to draw a horizontal tangent line, you can easily rotate it by any angle:
[asy]import graph;
draw(Circle((0,0),20));
draw(rotate(30)*((-40,20)--(40,20)));[/asy]

Ha ha  :oops: i don't proofread do i?
So you really don't need the olympiad package?
I mean as long as you stay in that level of asymptote, you can just roatate it.
But olympiad has it's benifits right? :maybe:

The olympiad package was designed to facilitate coding the constructions that occur frequently in geometry problems. In principle, all can be done from scratch using only the basic Asymptote routines. The packages are there just to let you use some common constructions without too much hassle but there is nothing in them you couldn't write yourself given time and practice.

[asy]import graph;
draw(Circle((0,0),20));
draw(rotate(30)*((-40,20)--(40,20)));
draw(rotate(270)*((-40,20)--(40,20)));[/asy]
Is it possible for them to meet without using olympiad package or do you have to calculate it?

What do you mean? They met already :? If you need to mark/label the point that is at the intersection, use [b]intersectionpoint()[/b] command:
[asy]import graph;
draw(Circle((0,0),20));
path P=rotate(30)*((-40,20)--(40,20)), Q=rotate(270)*((-40,20)--(40,20));
draw(P); draw(Q);
pair A=intersectionpoint(P,Q);
dot(A); label("$A$",A,SE);[/asy]

I should have clarified.  :oops:
I mean make the interception point exactly meet with no extra sticking out the two lines. (Like your AX where x is the displayed end of one of the tangents)
Anyway
[asy]import graph;
draw(Circle((-20,0),20));
draw(Circle((20,0),20));
draw(Circle((0,-20),20));
draw(Circle((0,20),20));
draw((-20,20)--(-20,-20));
draw((20,20)--(-20,20));
draw((20,20)--(20,-20));
draw((-20,-20)--(20,-20));[/asy]
How would you make the regions where two circles meet be yellow and how do you make the color of the circles be green on the outside. (Thats the way it is on my homework.)
Also, how would you draw a close up of a square where two quarter circles meet, with the rest quarter circles inclulded? (Aka the region bounded by (on my figure) (0,0),(20,0),(20,20)(0,20))

[quote="Dojo"]I should have clarified.  :oops:
I mean make the interception point exactly meet.
[/quote]
The simplest way is to use the Olympiad package to draw the tangent lines from the prescribed point. It is possible to truncate the lines (or any other curves) to the intersection point as well:
[asy]import graph;
draw(Circle((0,0),20));
path P=rotate(30)*((-40,20)--(40,20)), Q=rotate(270)*((-40,20)--(40,20));
P=firstcut(P,Q).before;
Q=firstcut(Q,P).after;
draw(P);draw(Q);[/asy]
As to circles, there are several ways. The simplest one is to use the [b]Arc()[/b] command that draws only a part of a circle:
[asy]import graph;
draw(Arc((20,0),20,-90,90),green);
fill(Arc((20,0),20,90,180)--Arc((0,20),20,270,360)--cycle,yellow);
draw(Arc((20,0),20,90,270));
draw(Circle((0,-20),20));
draw(Circle((-20,0),20));
draw(Circle((0,20),20));
draw((-20,20)--(-20,-20));
draw((20,20)--(-20,20));
draw((20,20)--(20,-20));
draw((-20,-20)--(20,-20));[/asy]
Can you finish by yourself from here?


I don't known Jensen! :blush:
Can you post other solution?


trigonometry
geometry
inequalities
algebra
system of equations
triangle inequality

Given a, b, c are real numbers.
Solve the following system of equations:
$x^2+xy+y^2=a^2$,
$y^2+yz+z^2=b^2$,
$z^2+zx+x^2=c^2$.

[quote="Altheman"][quote="Abhishek"]Solution:

[hide]As the equations are homogenous, put y=mx & z=nx.
You will have to solve for m,n,x.

But I have an other idea.
Consider a triangle ABC. Consider a point P in it such that
LAPB = LAPC = LBPC =120 degrees.

Now call PB as y, PC as x, PA as z.
Apply Cosine rule to each triangle. U would see u get the system of eqns.
(a,b,c are the sides of triangle ABC.)

Now area of PAB + PBC + PCA = Area of ABC.

Apply Area formula for triangles on LHS and we get xy+yz+zx
(As we know Area of ABC)

Now adding the three eqns. given, we can get x^2+y^2+z^2.

Now use (x+y+z)^2 formula to get x+y+z.

Now adding  1 & 2, we get

x^2+y^2+z^2 + y(x+y+z) = a^2+b^2

From this we can solve for y.

||| ly we can solve for x & z.[/hide][/quote]

how do you know that a,b,c satisfy the triangle inequality?, furthermore, how do you know that they are real?[/quote]

The problem gives that they are real, but yes, they don't necessarily satisfy the triangle inequality.  But that is the motivation for the problem, I think.

[quote="Abhishek"]Actually it will have real solutions iff a,b,c satisfy triangle inequality.

For any given real x,y,z, draw 3 lines making 120 degrees from a pt. P of corresponding lengths and form the triangle.
You would see that a,b,c would be the sides of triangle ABC and hence they obey the triangle inequality.

Going thro this argument the other way round, we can prove that the system has real solns. iff a,b,c obey triangle inequality.[/quote]
**************************************************************************
Don't forget that a, b, c can be [b]negative numbers[/b] also.
Your statement could be true if we refine it as: |a|,|b| and |c| obey the triangle inequality, and provided that x, y and z are real numbers.

[b]Edit[/b]
In fact, the condition of triangle inqualities is not necessary and sufficient.
Take for an example.
$x^2+xy+y^2=61$,
$y^2+yz+z^2=31$,
$z^2+zx+x^2=28$.
You can check that $(x,y,z)=(-4,-5,6)$ is a solution of the sytem of equations.
With your argument, how can you construct the triangle having a side with negative value?

Sorry for not mentioning modulus func.

I thought it was trivial as the notion of lengths is associated with non -ve nos.


$ ABC$ is an acute triangle and $ H$ is his orthocenter .
let: $ AH$x$ BH$x$ CH\equal{}3$ and $ AH^2\plus{}BH^2\plus{}CH^2\equal{}7$
$ 1)$ find the radius $ R$ of the circumcircle  of the triangle $ ABC$.
$ 2)$ find the lengths of the triangle $ ABC$ which the area of $ ABC$ is $ maximum$.

I do not believe in spoon feeding.I want you to solve the problem yourself.

take hints:

4R^2(cos^2A+cos^2B+cos^2c)=7 gives:

4R^2/2{3+cos2A+cos2B+cos2C}=7

or 4R^2{2-4cosAcosBcosC}=7

this equation and the one supplied by you will solve the value of R
{8R^3(cos^2Acos^2Bcos^c)=3}

now for the second part

area of any triangle is given by abc/4R

It will be max only when abc is max

note that a+b+c/3>or equal to (abc)^(1/3), equality holds whena=b=c

therefore abc will be max when a=b=c (here a,b,c are sides of a triangle)

You have the other two equations. I have given you enough hints for the time being


AMC USA(J)MO USAMO geometry Support rotation inequalities

http://www.maa.org/news/073107chinaolympiad.html
This is very interesting that US is going to participate in the China Girls Olympiad. I wonder if U.S is going to send a team to the Chinese Math Olympiad next year.

cool.

did they get a full ride to amsp?

That's pretty interesting.  I had no idea there existed competitions just for girls.

So AMP as the MOP for girls now  :wink: although Sherry Gong's not there. I guess the coaches felt she was sufficiently well prepared after IMO gold medal and all.

@archimedes1 - neither did I (for the US anyway), but my impression is that this is the first year we're participating, so I don't think it means we're too far out of it.

Wow, the western team is all from the Bay Area.

Well, of the 66 USAMO qualifiers from California, 47 were from SFBA. Of the 19 others, 18 were definitely male, and the only one I'm not sure of is someone with a Korean name whom I don't know. I haven't looked around all of the other western states - I know there were two female USAMO qualifiers from Colorado, and at least one from Washington - but the Bay area may well be the bulk of the eligible pool.

About China Girls Olympiad, it started in 2001. Chinese math society organized this competition with the intension to stimulate girls' interests in math. The competitions in the past years initially had each province in china sending team(s) (usually one team with 4 girls, occasionally, some provinces sent 2 teams in the past). The numbers of teams were usually in the range of mid-30s to upper 30s. Last couple of years, Chinese math society began to invite teams from other countries to send teams to compete in this event. I believe Russia participated last year. This year, Russia, USA, Canada, Australia and some other countries are invited to send team(s).

With the support from IBM, MAA and others, US decided to send two teams in its first year of participation. All team members (except Sherry, most likely due to overlapping of her IMO competition schedule, I am very happy for her gold metal in this year’s IMO) went to AMP for three weeks of training. The selection of the team members was based on the 2006 USAMO result. According to the information I had, all top eight girls accepted the invitation. At least one girl was accepted into RSI and had to give up RSI opportunity (since making the decision to be one of the team members of the USGMO happened before RSI result). Among the eight team members, 4 are from California, one from NH, one form MA, one form FL and one form IL. Four of them were MOPpers (three red MOPpers). Hopefully, this competition experience in China will give them more motivation to invest some real effort in the future preparation of USAMO. Hopefully they can do well in this first US participation to China Girls Olympiad, so that the supporting organizations will have continued interests in supporting future US girls’ teams. All the team members got free ride at AMP. However, people should not consider AMP as girls’ MOP; rather, their participation at AMP is just an accessory to AMP.

The China Girls Olympiad competitions rotate in different cities in China each year. Each competition has eight questions divided into two mornings of four hours each. I wish all our team members good luck and have a great memorable positive experience.

Wow when I was doing one of the tests, I thought that it was all 8 questions in four hours. No wonder I ran out of time so quickly.

I would argue that the girls team got, in some sense, special attention at AMSP. Track 3 was basically designed to train the girls for the competition because like all the problems are either inequalities or geometry.

P.S. qvc122, are you somebody that I would know?

[quote="qvc122"]About China Girls Olympiad, it started in 2001. [/quote]

The first one was in 2002.

[quote="qvc122"]
According to the information I had, all top eight girls accepted the invitation. At least one girl was accepted into RSI and had to give up RSI opportunity (since making the decision to be one of the team members of the USGMO happened before RSI result). [/quote]

Yes, that was unfortunate. I believe it was Wendy Hou (from FL) that had to give up going to RSI, but she seemed happy enough with AMP.

[quote="qvc122"]
Hopefully, this competition experience in China will give them more motivation to invest some real effort in the future preparation of USAMO. [/quote]

Hmm, what's that supposed to mean? Surely you're not suggesting that they didn't put any effort into preparing for the USAMO?

I, for one, think this is an excellent opportunity.  :)

[quote="xxazurewrathxx"][quote="qvc122"]
According to the information I had, all top eight girls accepted the invitation. At least one girl was accepted into RSI and had to give up RSI opportunity (since making the decision to be one of the team members of the USGMO happened before RSI result). [/quote]

Yes, that was unfortunate. I believe it was Wendy Hou (from FL) that had to give up going to RSI, but she seemed happy enough with AMP.
[/quote]
YES
SHE THREATENED CCY WITH 5 PLASTIC KNIVES

What do you expect?  CCY threatened just about everyone with plastic knives (including nsato :rotfl: )!

@ Altheman: [quote]P.S. qvc122, are you somebody that I would know?[/quote]
Unlikely.

I found this amusing:

So I was like hey I want to find olympiad inequality stuff. So naturally, I type 'inequality olympiad' into google. The page about the GMO is the 6th result.

[quote="xxazurewrathxx"]Err, who's CCY?[/quote]

ccy is a wonderful and active member of this forum. He enjoys attacking small children with plastic knives, hacking calculators, and screaming very loudly.

Any questions?

The US team members include two 10th graders, five 11th graders and one 12th grader. The selection was based on 2006 USAMO score (not 2007) since the US teams were formed in Feb, 2007, before the 2007 USAMO test. Iris Xu is probably the only female red MOPper but not the only female MOPper. Sherry Gong is a black MOPper multiple times. Some of the girls on the US team were red MOPpers when they were in their 9th grade. Iris Xu would qualify for the team next year if US continues to send the team(s).

[quote="BFG"]They said the girls team was selected based on the USAMO score. What was the cut off USAMO score? All the girls on the team are 11th graders or above.  What are the other factors of team selection? Grade is also a factor? Note that Iris Xu (9th grade) was the only girl this year qualified for MOP but she was not on the team.[/quote]
Not only was Iris not the only female MOPper, there was also another female red MOPper... at any rate, if you had read qvc's post, it says quite clearly that it was 2006 usamo.

[quote="BFG"]Note that Iris Xu (9th grade) was the only girl this year qualified for MOP but she was not on the team.[/quote]

You forgot Lynnelle Ye, who I think is from California.

Just a little update:
http://www.msri.org/specials/gmo

What are their ranks? They said that there were 40 teams.

Who knows, but they did pretty well nonetheless.


symmetry
linear algebra
matrix
geometry
geometric transformation
reflection
linear algebra unsolved

$\mathcal{K}$ is the subset of $\mathcal{M}_{n}(\mathbb{R})$, such that $\forall A \in \mathcal{K}$, $A$ is symmetric and stochastic. $\mathcal{K}$ is clearly convex.
What are the extremal points of $\mathcal{K}$?

See the fifth message in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=stochastic&t=45600]this[/url] thread. A simple twist on that proof (we have the extra assumption of symmetry here) shows that the matrices we're looking for are the symmetric permutation matrices.

I hope I didn't overlook something.

grobber,

I'm not convinced that a "simple twist on that proof", would work, try it, and let me know :maybe:

This is what I had in mind:

That proof works by showing that a non-permutation matrix cannot be an extremal point of the set of doubly stochastic matrices. This is done by fincing a "cycle" of cells in the matrix, i.e. an ordered set $(a_{1},\ldots,a_{2t})$ of cells such that $a_{i}\in(0,1)$ and $a_{2k-1},a_{2k}$ lie in the same row, $a_{2k},a_{2k+1}$ lie on the same column. Then $a_{1}$ decreased by a small amount $\varepsilon>0,\ a_{2}$ is increased by $\varepsilon$, and so on. Afterwards, we perform the same moves, except that this time we replace "decreased" with "increased" everywhere.

In our case, we do the exact same thing, except that this time we work on both the cycle $(a_{i},\ i=\overline{1,2k})$ and on its reflection in the main diagonal. That way, the matrices we obtain are symmetric, but everything else is the same: after performing those two operations on a non-permutation symmetric doubly stochastic matrix $A$, we end up (if $\varepsilon>0$ is small enough) with two symmetric doubly stochastic matrices $B,C\ne A$ such that $A=\frac{B+C}2$.

[quote="grobber"]See the fifth message in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=stochastic&t=45600]this[/url] thread. A simple twist on that proof (we have the extra assumption of symmetry here) shows that the matrices we're looking for are the symmetric permutation matrices.
[/quote]
grobber,
For $n=3$, $A=\frac{1}{2}\begin{pmatrix}0&1&1\\ 1&0&1\\ 1&1&0 \end{pmatrix}$ seems to be an extremal point of $\mathcal{K}$, and doesn't fit your description.
Or am I missing something here :blush:

I'm glad I made that mistake; it made the problem more interesting :).

I think I've got it this time, but I'm afraid I don't have time to post the proof. It's not that big a deal anyway; apply the same ideas as in my previous post, but more carefully :).

To each symmetric stochastic $n\times n$ matrix $A=(a_{ij})$ associate a graph $G(A)$ on $n$ vertices with labeled edges as follows: $i$ is connected to $j$ iff $a_{ij}>0$, and in this case the label of the edge between $i$ and $j$ is $a_{ij}$. We allow loops, i.e. edges from a vertex to itself. It will be more convenient to describe the graphs associated to the extremal points. The result is this:

$A$ is extremal in $\mathcal K$ iff each connected component of $G(A)$ falls into one of the following three categories: (a) a loop with label $1$; (b) an edge with label $1$ connecting two vertices; (c) an odd cycle with all edges having label $\frac{1}{2}$.

If this is still not it, sorry.

It looks right. However I'm not sure why you use graph and vertices :?:

Here is a simpler (or at least more natural) way of describing the extreme points: They are the symmetric part of a permutation matrix ([i]i.e. [/i]$\frac{1}{2}(P+P^{T})$) for a permutation whose disjoint-cycle representation consists only of transpositions and odd cycles.
I have a solution which is rather long (2 to 3 pages), I'll be interested in a shorter solution.


Truly a nice one! Let's put [tex] x=\frac{a}{b},\;y=\frac{b}{c},\; z=\frac{c}{a}[/tex]. The inequality becomes, after some minor computations [tex]\frac{(1-x)^2}{x}+\frac{(1-y)^2}{y}+\frac{(1-z)^2}{z}\geq
3(1-x)^{\frac{2}{3}}(1-y)^{\frac{2}{3}}(1-z)^{\frac{2}{3}}[/tex], which follows immediately from the AM-GM inequality, because [tex]xyz=1[/tex].


A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?

$ \textbf{(A)}\ \frac{1}{10}\qquad
\textbf{(B)}\ \frac{1}{9}\qquad
\textbf{(C)}\ \frac{1}{3}\qquad
\textbf{(D)}\ \frac{4}{9}\qquad
\textbf{(E)}\ \frac{5}{9}$


I'm 16 and in 11th grade right now and considering applying to Awesome Math. I'm wondering if I'm too old, however. Can anyone give me an opinion on this?

When I was a student there, I was your age. I had a great time and the (small) age gap didn't really make a difference for me there.


Prove that

$ \lim_{x\rightarrow 1}\frac{\sqrt [a]{x}\minus{}1 }{\sqrt [b]{x}\minus{}1 }\equal{}\frac{b}{a}\;\forall\;{a,b}\in\mathbb{N}\neq 1$

using only algebraic methods.

[hide]
For $ x\neq 1$,
\[ \frac{\sqrt[a]{x}\minus{}1}{\sqrt[b]{x}\minus{}1}\equal{}\frac{f(x)}{g(x)}\]
where
\[ f(x)\equal{}\sum_{k\equal{}0}^{b\minus{}1}x^{k/b},\ g(x)\equal{}\sum_{k\equal{}0}^{a\minus{}1}x^{k/a}\]
since $ x\minus{}1\equal{}g(x)\left(\sqrt[a]{x}\minus{}1\right) \equal{}f(x)\left(\sqrt[b]{x}\minus{}1\right)$. The result follows.
[/hide]

$ lim_{x\to1}\frac{x^{a}\minus{}1}{x\minus{}1}\equal{}a$ for all a different from 0.

[quote="gotztahbeazn"]Can you explain your last step? I fail to see how the last statement proves the limit...[/quote]
$ f(x)$ has $ b$ terms that each tend to 1, and $ g(x)$ has $ a$ terms that each tend to 1.

But you use for $ a,b\in N$ but them in R


let $ a$,$ b$,$ c$ $ \in {\mathbb R} ^{*}$ , such that $ ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c\plus{}1\equal{}0$ , prove that :

$ a^2\plus{}b^2\plus{}c^2 \geq 2(ab\plus{}bc\plus{}ca)$

We set $ ab\plus{}bc\plus{}ca\equal{} A$

From $ ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c\plus{}1\equal{}0$

we have $ A\plus{}1 \equal{} \minus{}(a\plus{}b\plus{}c)$

and $ (A\plus{}1)^{2} \equal{} a^{2}\plus{}b^{2}\plus{}c^{2} \plus{}2A$

therefore after simpifying $ A^{2}\plus{}1 \equal{} a^{2}\plus{}b^{2}\plus{}c^{2}$

We now have to prove that $ A^{2}\plus{}1 \geq 2A$ which is obvious.


function
calculus
advanced fields
advanced fields unsolved

Show that if $ B\in\mathcal{B}(\mathcal{H})$ commutes with self-adjoint operator $ A$ then $ B$ commutes with every operator of the form $ F(A)$ where $ F$ is bounded Borel function on $ \mathbb{R}$.

As usual with the functional calculus of borel functions.

Yes, the definition is usual. In case here are some links:

http://en.wikipedia.org/wiki/Functional_calculus
http://en.wikipedia.org/wiki/Borel_functional_calculus

Does anyone know how to solve this problem?


ARML AMC AIME USA(J)MO USAMO MATHCOUNTS geometry

I attend a small school that does not participate in any of these contests.

Is there any way I can participate in the ARML, AIME, USAMO, etc?  :cry:

i think anyone can go to ARML. not sure about the others, but definitely not mathcounts, since its basically a school vs. school competition.

You can take the AMC 8, AMC 10, and the AIME even if you are in a small school.  All you need to do is talk with the AMC coordinator in your state, tog have him / her send a teacher or religious leader a copy of the test.  Either that, or you can go to a large school in your area to take the test.  In my area, several schools take the test in one location, and those who are home-schooled can take it there as well.

[quote="davidyko"]Presumably you know that you need to take the AMC 8 to qualify for the 10/12 (if you're in middle school), the AMC 10/12 to qualify for AIME, and the AIME to qualify for USAMO.

I'm not sure if [b]anyone[/b] can go to ARML...I'm not sure how it works, I just got invited last year.[/quote]

Umm you do not need to take the AMC8 to qualify for anything.

I would look for some universities near you that offer the tests and take them there. There's a list on the AMC Website.

[quote="davidyko"]Fine, you can be invited to take the AMC 10 by scoring well on the AMC 8. But I was wrong, it's not a prerequisite for middle school students.[/quote]

There's no such thing as an "invitation" to take the AMC10, it is an open contest, they don't send out invites.

And they don't send out invites for ARML either, that depends on where you live and you need to contact your local coordinator.

did anyone else notice that this thread was started in 2003? =o
(lock please)


email
superior algebra
superior algebra unsolved

Dear Forum,

Ok.. I have posted this in the sci.math newsgroup. I thought of posting this here too. Can one give an example or show otherwise that there exists two fields $ F_1$ and $ F_2$ which are algebraically closed and both subfield of a bigger field $ K$ and such that their compositum $ F_1F_2$ is not an algebraically closed field?

Sincerely,
Jose Capco

I suspect that the algebraic closures of $ K[x]$ and $ K[y]$ in the algebraic closure of $ K[x,y]$ will provide such an example.  I doubt that we can find a root of $ z^2 \minus{}(x^2\plus{}y^2)$ in the compositum.  I'll let you know if I think of some way to prove this...

Indeed, LydianRain's suspicion is true, at least if $ K$ is algebraically closed (though his direct counterexample doesn't work for $ \mathrm{char}k\equal{}2$) - see

Shreeram Abhyankar, [i]On the compositum of algebraically closed subfields[/i]. Proc. Amer. Math. Soc., 7:905-907, 1956.

I can PM/email (both of) you the article if you are interested.

  darij

I wonder if there is "often" such alg closed $ F_1$,$ F_2$, that $ F_1 \otimes_Q F_2$ is still alge closed.

Yeah, I found the paper a little while after posting the problem. I have written a short summary of Abhyankar's method and we have made some interesting discussion in sci.math newsgroup. The thread can be found [url=http://groups.google.com/group/sci.math/browse_frm/thread/c8e2c5ddf3381646/37951617151165d7]here[/url].

Im quite new to this forum and Im beginning to enjoy it here, I'll try to be more active here as well. I was quite pleased by the responses here and the willingness and helpfullness of darij to email us the article. Its not something you say in english.. but...  Please be take care of me :)

Sincerely,
Jose Capco


geometry
symmetry
quadratics
function
calculus
derivative
calculus computations

i think i saw this in one of the aops forums long ago.the solution wasnt given so i thought id post it again in the hope that someone would solve it.

Amount of destruction caused by a bomb at a given point is given by 100/r^2,where r is distance fromm point of impact.
a point is considered destroyed if net damage at the point is 1.
what is the max area you can destroy with 2 bombs?

thnx in advance :lol:  :lol:

I think that rotational symmetry will tell you that we may simply consider the bombs to be symmetrically placed at $x=\pm b$ for some $b>0$ on the x-axis, and consider the y-axis to be damage. Put

\[f(x)=\frac{100}{(x+b)^{2}}+\frac{100}{(x-b)^{2}}= \frac{200(x^{2}+b^{2})}{(x^{2}-b^{2})^{2}}\]

so that $f(x)=1\Leftrightarrow 200(x^{2}+b^{2})=(x^{2}-b^{2})^{2}$ which is quadratic in $x^{2}$ (and hence has four solutions in x), namely

\[x_{\pm}^{2}=b^{2}+100\pm 20\sqrt{b^{2}+25}\]

and it happens that $x_{+}^{2}>0$ for all $b>0,$ but $x_{-}^{2}=0$ for $b=0,\pm 10\sqrt{2}$ a quick check tells us that $x_{-}^{2}\geq 0$ iff $|b|>10\sqrt{2}$ or $b=0$. The four solutions are the locations on the x-axis at which the damage is 1, they occur left to right in this order: $-x_{+},-x_{-},x_{-},x_{+}$ and the middle two only exist (as real solutions) when the bombs are far enough apart that the addative overlap between the bombs is allowed to drop below 1 damage. The length of the aera destroyed is accordingly given by

\[L(b) = \left\{\begin{array}{cc}2x_{+}= 2\sqrt{b^{2}+100+20\sqrt{b^{2}+25}},&\mbox{ if }0\leq b\leq 2\sqrt{2}\\2(x_{+}-x_{-}) = 2\sqrt{b^{2}+100+20\sqrt{b^{2}+25}}-2\sqrt{b^{2}+100-20\sqrt{b^{2}+25}}, & \mbox{ if }b\geq 2\sqrt{2}\end{array}\right. \]

since the maximum occurs at the same location if we square the function and divide by 4, consider now

\[\frac{1}{4}L^{2}(b) = \left\{\begin{array}{cc}b^{2}+100+20\sqrt{b^{2}+25},&\mbox{ if }0\leq b\leq 2\sqrt{2}\\2(b^{2}+100)-2b\sqrt{b^{2}-200}, & \mbox{ if }b\geq 2\sqrt{2}\end{array}\right. \]

differentiate w.r.t. b to get

\[\frac{1}{2}L(b)L^{\prime}(b) = \left\{\begin{array}{cc}2b+\frac{20b}{\sqrt{b^{2}+25}}=2b\left(1+\frac{10}{\sqrt{b^{2}+25}}\right) ,&\mbox{ if }0\leq b\leq 2\sqrt{2}\\4b+\frac{-4b^{2}+400}{\sqrt{b^{2}-200}}, & \mbox{ if }b\geq 2\sqrt{2}\end{array}\right. \]

now the first piece of our derivative is only zero if $b=0$ (clearly the minimum since the bombs would be on top of each other) and the second piece gives the only other critical numbers, namely $b=\pm 10\sqrt{2}$ (since it is undefined there), and is indeed produces the maximum! We have that $L ( 10\sqrt{2}) = 20\sqrt{6}$ and the function is even continuous at this value. Here's the plot:

[url=http://ezimagecenter.com/out.php/i186_dabombs2d.JPG][img]http://ezimagecenter.com/out.php/i186_dabombs2d.JPG[/img][/url]

Thanks for the nasty problem!

[color=green]Question[/color]: Does this linear model (linear in that I only considered the damage along a single geometric dimension, namely the x-axis) solve the problem asked? I had considered that modeling bombs as affecting a plane (as in air-dropping bombs on the battlefield), and that WLOG one bomb would then be at the origin and  the other bomb being at, say at some point (p,q) but by radial symmetry of the damage pattern, rotating the other bomb to, say (a,0) is the same senario. Or what of 3-d? suppose the target of the bombs were some asteriod in space, perhaps... end rant.

[color=red]EDIT[/color]: Repost with 1 error(s) with multiplicity six fixed (I had posted $2\sqrt{2}$ where $10\sqrt{2}$ ought to have been. emoticons are so very lacking at this moment [unrelated].)

I think that rotational symmetry will tell you that we may simply consider the bombs to be symmetrically placed at $x=\pm b$ for some $b>0$ on the x-axis, and consider the y-axis to be damage. Put
\[f(x)=\frac{100}{(x+b)^{2}}+\frac{100}{(x-b)^{2}}= \frac{200(x^{2}+b^{2})}{(x^{2}-b^{2})^{2}}\]
so that $f(x)=1\Leftrightarrow 200(x^{2}+b^{2})=(x^{2}-b^{2})^{2}$ which is quadratic in $x^{2}$ (and hence has four solutions in x), namely
\[x_{\pm}^{2}=b^{2}+100\pm 20\sqrt{b^{2}+25}\]
and it happens that $x_{+}^{2}>0$ for all $b>0,$ but $x_{-}^{2}=0$ for $b=0,\pm 10\sqrt{2}$ a quick check tells us that $x_{-}^{2}\geq 0$ iff $|b|>10\sqrt{2}$ or $b=0$. The four solutions are the locations on the x-axis at which the damage is 1, they occur left to right in this order: $-x_{+},-x_{-},x_{-},x_{+}$ and the middle two only exist (as real solutions) when the bombs are far enough apart that the addative overlap between the bombs is allowed to drop below 1 damage. The length of the aera destroyed is accordingly given by
\[L(b) = \left\{\begin{array}{cc}2x_{+}= 2\sqrt{b^{2}+100+20\sqrt{b^{2}+25}},&\mbox{ if }0\leq b\leq 10\sqrt{2}\\ 2(x_{+}-x_{-}) = 2\sqrt{b^{2}+100+20\sqrt{b^{2}+25}}-2\sqrt{b^{2}+100-20\sqrt{b^{2}+25}}, & \mbox{ if }b\geq10\sqrt{2}\end{array}\right. \]
since the maximum occurs at the same location if we square the function and divide by 4, consider now
\[\frac{1}{4}L^{2}(b) = \left\{\begin{array}{cc}b^{2}+100+20\sqrt{b^{2}+25},&\mbox{ if }0\leq b\leq 10\sqrt{2}\\ 2(b^{2}+100)-2b\sqrt{b^{2}-200}, & \mbox{ if }b\geq 10\sqrt{2}\end{array}\right. \]
differentiate w.r.t. b to get
\[\frac{1}{2}L(b)L^{\prime}(b) = \left\{\begin{array}{cc}2b+\frac{20b}{\sqrt{b^{2}+25}}=2b\left(1+\frac{10}{\sqrt{b^{2}+25}}\right) ,&\mbox{ if }0\leq b\leq 10\sqrt{2}\\ 4b+\frac{-4b^{2}+400}{\sqrt{b^{2}-200}}, & \mbox{ if }b\geq 10\sqrt{2}\end{array}\right. \]
now the first piece of our derivative is only zero if $b=0$ (clearly the minimum since the bombs would be on top of each other) and the second piece gives the only other critical numbers, namely $b=\pm 10\sqrt{2}$ (since it is undefined there), and is indeed produces the maximum! We have that $L ( 10\sqrt{2}) = 20\sqrt{6}$ and the function is even continuous at this value. Here's the plot:

[url=http://ezimagecenter.com/out.php/i186_dabombs2d.JPG][img]http://ezimagecenter.com/out.php/i186_dabombs2d.JPG[/img][/url]

Thanks for the nasty problem![/quote]


(1) The sequence {a_n} is defined by a_1 = 2^0.5 and a_(n+1) = (2+a_n)^0.5, n > 1. Determine lim_(n->00) 4^n(2-a_n).

(2) Determine \sum_{k=1}^{infinity} k^2 / 2^k and

(3) \sum_{k=1}^{infinity} k^3 / 2^k.

for the 1st one you could do a little bit of generalization! :D

if a1 is  in [-2,2] (for example) you write a1=2cos(a) where a is in [0,pi]
and using induction you get that: an=2 cos(a/2^n)

and now the limit:
you write it like so: lim 2^(2n+1)(cos(a/2^n)-1)=lim -2^(2n+2)sin^2(a/2^(n+1))=-a^2
so:
2cosa=a1=>a=arccos(a1/2)
so lim=-(arccos (a1/2))^2

cheers! :D :D

This is for the second problem. The third one can be approached in a very similar way.

First of all, it's well-known that the sum  \sum {k=1 to n}k*x^k is easy to calculate (we just multiply it by 1-x and see what happens). Because of this, the series  S1=\sum {k=1->00}k/2^k is easy to calculate (I won't do it here). We want S2= \sum {k=1->00}k^2/2^k.

We take S1+S2= \sum {k=1->00}(k+1)*k/2^k= 2*\sum {k=1->00}(1+2+..+k)/2^k =  2*\sum {k=1->00}(k/2^k)*(1+1/2+..)=4* \sum {k=1->00}k/2^k = 4*S1, so S2=3*S1. Is it right? I might have made a mistake.

For 2nd an 3rd:

Let we have (1-x)-1 = Sum x^k. Take derivation of two side, we have
                   (1-x)-2 = Sum kx^(k-1),
So
                   x(1-x)-2 = Sum kx^k.
Again take deivation
                   (1+x)(1-x)-3 = Sum k^2x^k-1
=>               x(1+x)(1-x)-3 = Sum k^2x^k

Put x=1/2 we get  Sum k^2/2^k = 6.


algebra
polynomial
function
algebra unsolved

Let P,Q,R be polynomials with real coefficients satisfying P^4 + Q^4 = R^2.
Prove that there exists real numbers p,q,r and a polynomial S such that P = pS, Q = qS and R = rS^2 !

[i]IMO longlist 1987, problem 3 by Poland[/i]

Do you think it will be useful to solve  V^2 + W^2 = R^2 where V,W,R are polynomials for solving P^4 + Q^4 = R^2 ?

Thank's

Moubinool

Hi Moubinool,

you are right about the source. How did you find it out ?  :D

There is a mighty theorem which helps to tackle such kind of problems.

[b]Theorem[/b] (Mason)
Let $a(x)$, $b(x)$ and $c(x)$ be pairwise coprime polynomials and $a+b+c=0$. Then $\max(\deg a,\deg b,\deg c)\leq N(abc)-1$, where $N(p)$ is the number of distinct roots of $p$.

[b]Corollary 1[/b] (Devenport)
Suppose $f$ and $g$ are coprime polynomials of nonzero degree. Then \[\deg(f^{3}-g^{2})\geq \frac{1}{2}\deg f+1.\] [b]Corollary 2[/b]
Suppose $f$, $g$, $h$ are coprime polynomials and one of them is nonconstant. Then $f^{n}+g^{n}=h^{n}$ implies $n\leq 2$.

[b]Corollary 3[/b]
Suppose $a$, $b$ and $c$ are natural numbers, $2\leq a\leq b\leq c$. Then the equation \[f^{a}+g^{b}=h^{c}\] has pairwise coprime solutions only for the following triples of $(a,b,c)$:
$(2,2,c)$, $(2,3,3)$, $(2,3,4)$ and $(2,3,5)$.

[b]Corollary 4[/b]
Suppose $x(t)$ and $y(t)$ are rational functions, $m,n\geq 2$. Then the equation \[x^{n}-y^{m}=1\] has solution only for case $m=n=2$.

-----------------
At least, I immediately got a solution for initial problem using the theorem.

[quote="Myth"]

Then $\max(\deg a,\deg b,\deg c)\leq N(abc)-1$, where $N(p)$ is the number of distinct roots of $p$.

[/quote]
What do you mean with roots of $p$?

[quote="Myth"]Do you know what is roots of polynomial?[/quote]
Sorry for the stupid question  :blush:

Indeed amazing!
... And such a mighty theorem requires a mighty proof?

The proof is quite elementar.
Let $ f\equal{}a/c$, $ g\equal{}b/c$. Then $ f\plus{}g\plus{}1\equal{}0$, so $ f'\equal{}\minus{}g'$. Thus
\[ a/b\equal{}f/g\equal{}\minus{}\frac{fg'}{gf'}\equal{}\minus{}\frac{g'/g}{f'/f}\equal{}\minus{}\frac{Ng'/g}{Nf'/f},\]
where $ N\equal{}\prod (x\minus{}x_j)$, $ x_j$ are all pairwise distinct roots of $ abc$ (note that $ \deg N\equal{}N(abc)$).
Since $ g(x)\equal{}\prod (x\minus{}x_j)^{k_j}$ ($ k_j\in \mathbb Z$), we conclude that $ g'/g\equal{}\sum \frac{k_j}{x\minus{}x_j}.$ The same is for $ f'/f$.
It means that $ Ng'/g$ and $ Nf'/f$ are usual polynomials of order $ \leq N(abc)$. Now using the fact that $ a$ and $ b$ are pairwise coprime we conlude $ \deg a\leq \deg(Ng'/g)\leq N(abc)$.

[quote="Myth"]There is a mighty theorem which helps to tackle such kind of problems.

[b]Theorem[/b] (Mason)
Let $ a(x)$, $ b(x)$ and $ c(x)$ be pairwise coprime polynomials and $ a \plus{} b \plus{} c \equal{} 0$. Then $ \max(\deg a,\deg b,\deg c)\leq N(abc) \minus{} 1$, where $ N(p)$ is the number of distinct roots of $ p$.
 [/quote]
[hide="To Myth"]
Могу ли я узнать как это теорема доказывается, или хотя бы получить ссылку на то место, где это доказательство может быть найдено.Я уже пытался найти и само доказательство, и место где оно может быть найдено, но безуспешно.[/hide]

[quote="greenvert"][hide="To Erken"]
Ну вроде вот тут есть :
[url]http://www.ega-math.narod.ru/Lang/HypotABC.htm[/url]
Там в принципе то же, что и Myth написал, но подробнее.
[/hide][/quote]
[hide="To greenvert"]Огромное спасибо :)  [/hide]

Another application of this theorem is the following problem:
Assume that there exist polynomials $ P,Q,R\in\mathbb{C}[x]$,such that
$ P^{a}\plus{}Q^{b}\equal{}R^{c}$,where $ a,b,c\in\mathbb{N}$.
Prove that
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}>1$.

[quote="Erken"]Another application of this theorem is the following problem:
Assume that there exist polynomials $ P,Q,R\in\mathbb{C}[x]$,such that
$ P^{a} \plus{} Q^{b} \equal{} R^{c}$,where $ a,b,c\in\mathbb{N}$.
Prove that
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} > 1$.[/quote]

[b]Corollary 3[/b] from my first post in this thread?


There's something about this I don't understand.

For example 14x - 73 = 91. I have to solve for x. You're supposed to do the inverse steps, right? First I'm supposed to add, then devide? Well that's not giving me the right answer for x. I don't understand why.

When I do an addition one, like 2x + 21 = 71. I can easily subtract, then devide to reach the answer; 25.

Please help, thanks.

[hide="Solution"]
$ 14x\minus{}73\equal{}91$
$ 14x\equal{}164$
$ x\equal{}\frac{82}{7}$
[/hide]

Rather than focusing on some algorithm, focus on isolating the variable

[quote="leroyjenkens"]Well that's not giving me the right answer for x. I don't understand why.[/quote]

How can you be sure it's not the right answer for $ x$? Did you double check your answer by substituting in the value of $ x$ and see if the equation checks out? Remember, just because it's a decimal or an irrational value, it doesn't mean it's incorrect.

[quote="RisingStar"][hide="Solution"]
$ 14x \minus{} 73 \equal{} 91$
$ 14x \equal{} 164$
$ x \equal{} \frac {82}{7}$
[/hide]

Rather than focusing on some algorithm, focus on isolating the variable[/quote]
Hmm, I got that answer, which turns into 11.714285714285... You simplified it though, which I didn't do.
The reason I thought it wasn't right was because when I multiplied it times 14, I didn't get exactly 91. I thought I was doing something wrong.

BTW, what do you mean by isolating the variable?
[quote]How can you be sure it's not the right answer for ? Did you double check your answer by substituting in the value of  and see if the equation checks out? Remember, just because it's a decimal or an irrational value, it doesn't mean it's incorrect.[/quote]
Well I multiplied 11.71 times 14 and didn't get 91. I even multiplied 11.714285 times 14 and it still didn't get exactly 91.

Well I got another one that I don't understand. This one I can't even get close to the right answer:

7 = 4 - 2y. Do I add 7 + 2 or 7 + 4? I've done both and then devided and still didn't get the right answer.

Thanks.

[quote="leroyjenkens"]
The reason I thought it wasn't right was because when I multiplied it times 14, I didn't get exactly 91.[/quote]

This is because the numbers you are dealing with have an infinite amount of decimal places (numbers after the decimal point). So no matter how many numbers you put in your calculator after the dot, you won't get exactly the number you want. That's why we use fractions, so we can avoid this problem, not to mention they look neater.

[quote="leroyjenkens"]7 = 4 - 2y. Do I add 7 + 2 or 7 + 4? I've done both and then devided and still didn't get the right answer.[/quote]

As RisingStar said before me, the goal is to get the $ y$ all by itself. That is, we want something like "$ y\equal{}$" when we're done.

You first may want to change the minus sign to a plus sign just for clarity:

$ 7\equal{}4\minus{}2y$
$ 7\equal{}4\plus{}(\minus{}2)\times y$

Now we subtract $ 4$ from both sides:

$ 7\minus{}4\equal{}4\minus{}4\plus{}(\minus{}2)\times y$
$ 3\equal{}(\minus{}2)\times y$

Now divide by $ \minus{}2$:

$ \frac{3}{\minus{}2}\equal{}\frac{\minus{}2}{\minus{}2}\times y$

$ \frac{\minus{}3}{2}\equal{}y$
$ \boxed{y\equal{}\frac{\minus{}3}{2}}$.

Depending on how much algebra experience you have, you may feel that the above is either skipping some steps or way too detailed. Sadly (if you don't like math), you just have to practice until you can get it each time.

[quote="leroyjenkens"]BTW, what do you mean by isolating the variable?[/quote]

Having everything except for the variable itself (with no coefficient except for 1) on one side.

[quote]
Well I multiplied 11.71 times 14 and didn't get 91. I even multiplied 11.714285 times 14 and it still didn't get exactly 91.
[/quote]

You don't have to get [i]exactly[/i] 91. Remember, since you are truncating the decimal, your final answer will be a little different as well.

[quote]
7 = 4 - 2y. Do I add 7 + 2 or 7 + 4? I've done both and then devided and still didn't get the right answer.
[/quote]

Add like terms. So  you would subtract 4 from the 7, leaving

$ 3\equal{}\minus{}2y$.

Moved to Classroom Math.

Isolating the variable means getting the variable alone on one side of the equation. In your last example you can't just add and subtract and hope to get the right answer.
You have:
7 = 4 - 2y   First subtract 4 from both sides to get rid of the +4 to get:
3= -2y        Then you have to isolate the varible by dividing by -2 to get:
$ \frac {3}{ \minus{} 2} \equal{} y$

Hope that helps!

[quote="jvenezuela716"][quote="leroyjenkens"]
The reason I thought it wasn't right was because when I multiplied it times 14, I didn't get exactly 91.[/quote]

This is because the numbers you are dealing with have an infinite amount of decimal places (numbers after the decimal point). So no matter how many numbers you put in your calculator after the dot, you won't get exactly the number you want. That's why we use fractions, so we can avoid this problem, not to mention they look neater.

[quote="leroyjenkens"]7 = 4 - 2y. Do I add 7 + 2 or 7 + 4? I've done both and then devided and still didn't get the right answer.[/quote]

As RisingStar said before me, the goal is to get the $ y$ all by itself. That is, we want something like "$ y \equal{}$" when we're done.

You first may want to change the minus sign to a plus sign just for clarity:

$ 7 \equal{} 4 \minus{} 2y$
$ 7 \equal{} 4 \plus{} ( \minus{} 2)\times y$

Now we subtract $ 4$ from both sides:

$ 7 \minus{} 4 \equal{} 4 \minus{} 4 \plus{} ( \minus{} 2)\times y$
$ 3 \equal{} ( \minus{} 2)\times y$

Now divide by $ \minus{} 2$:

$ \frac {3}{ \minus{} 2} \equal{} \frac { \minus{} 2}{ \minus{} 2}\times y$

$ \frac { \minus{} 3}{2} \equal{} y$
$ \boxed{y \equal{} \frac { \minus{} 3}{2}}$.

Depending on how much algebra experience you have, you may feel that the above is either skipping some steps or way too detailed. Sadly (if you don't like math), you just have to practice until you can get it each time.[/quote]
Thanks, that was really helpful.
But if you didn't change the 4-2 to 4+(-2), then you wouldn't be doing the inverse.
You subtracted from a equation that was originally subtraction, so there was no inverse (until you changed the original to addition).
[quote]In your last example you can't just add and subtract and hope to get the right answer. [/quote]
The way I understand it is that to get the value of the variable, you have to do the opposite of what is in the equation. So for subtraction, you have to add, and for adding you have to subtract. Same for multiplication and division. But that's not correct?

[quote="shentang"]Yes that is correct, but you should know which numbers to add, subtract, multiply and divide. [quote="leroyjenkens"]7 = 4 - 2y. Do I add 7 + 2 or 7 + 4? I've done both and then devided and still didn't get the right answer.[/quote]Like in that problem I don't understand why you added 2 or 4 to 7[/quote]
Well the equation is a subtraction equation, so doing the opposite would be adding, right?

[quote="7h3.D3m0n.117"][quote="leroyjenkens"]Well the equation is a subtraction equation, so doing the opposite would be adding, right?[/quote]

No, you must analyze the terms. Because 4 is positive, you need to subtract 4 from both sides to balance the equation. You basically need to "eliminate" 4 on the RHS, and to do that, you must subtract 4.[/quote]

Why is that one different from 14x - 73 = 91? I got that one right by doing the inverse; adding. Does the answer being on the right side make the equation different than if it was on the left side? Like if the equation was 91 = 14x - 73 instead of 14x - 73 = 91?

Yes, as 7h3.D3m0n.117 said, you have to isolate the variable by subtracting 4. Remember, the sign applies to the term immediately following. There is actually a plus in front of the 4, which means that the inverse is, infact, subtraction.
Secondly, I have seen the same solutions like 3 times. STOP POSTING THE SAME SOLUTION.
 @leroyjenkens: When you have something like this problem in the future, post your solution as well, so we can be sure we're not just giving you the answer, and we can just fix your method.  :lol:

[quote="leroyjenkens"]Why is that one different from 14x - 73 = 91? I got that one right by doing the inverse; adding. Does the answer being on the right side make the equation different than if it was on the left side? Like if the equation was 91 = 14x - 73 instead of 14x - 73 = 91?[/quote]

You'll note that 73 is proceeded by a subtraction sign. To eliminate 73 from the LHS, you must add 73. Now what if the problem was

$ 73\plus{}14x\equal{}91$?

This is weird, when I tried to do the equation 7 = 4 - 2y, I kept getting the wrong answer, but when I switched it around to 2y - 4 = 7, I immediately got the right answer.
This is how I did it:

2y - 4 = 7 + 4 = 11 and then 2/11 = 5.5. So 5.5 times 2 - 4 = 7. So I got the right answer that way. Is my method incorrect?
I want my method to be correct so I don't mess up more difficult problems.
[quote]Yes, as 7h3.D3m0n.117 said, you have to isolate the variable by subtracting 4. Remember, the sign applies to the term immediately following. There is actually a plus in front of the 4, which means that the inverse is, infact, subtraction. [/quote]
Can you please explain what you mean by isolating the variable? I'm new to that.

So whenever I have a subtraction problem, the inverse would be subtraction if they're both positive numbers? What if one of them is negative?
[quote]@leroyjenkens: When you have something like this problem in the future, post your solution as well, so we can be sure we're not just giving you the answer, and we can just fix your method.[/quote]
Yeah, sorry, I should've been doing that all along. I did it above, so you can see how I got my answer.
[quote]You'll note that 73 is proceeded by a subtraction sign. To eliminate 73 from the LHS, you must add 73. Now what if the problem was [/quote]
What do you mean by eliminating the 73 from the lefthand side? ( if that's what LHS means)

And if the problem was 73 + 14x = 91, do I subtract the 14 or the 73 from 91? Does it matter which side the coefficient is on?

Thanks for the replies, guys.

Isolating the variable means getting it on its own on one side of the equation. This is done by undoing the operations on the variable.

4-2y is different than 2y-4. In the first one, 4 is positive and 2y is negative. In the second one, it has been reversed.
I think that your problem is that you don't like negative coefficients for your variables. You must remember than the sign applies to the term immediately following it, and no other.

$ 7\equal{}4\minus{}2y$

$ 11\equal{}\minus{}2y$        adding 4

$ \minus{}5.5\equal{}y$         dividing by -2

[quote="leroyjenkens"]And if the problem was 73 + 14x = 91, do I subtract the 14 or the 73 from 91? Does it matter which side the coefficient is on?[/quote]

Remember, consider the [i]terms[/i]. The terms on the LHS (left hand side) are 73 and 14x. You must perform addition/subtraction [i]only[/i] on like terms. Therefore, 14x can only go with another term with "x" in it. Therefore, ones with "y" or "$ x^2$" do not qualify.

Now since you don't have a variable ($ x$) on the RHS, you must subtract the ones without variables. Therefore, you would subtract 73.

[quote]Isolating the variable means getting it on its own on one side of the equation. This is done by undoing the operations on the variable. [/quote]
Like 2y - 4, you would devide 2y by 2 to "isolate" the variable?
[quote]4-2y is different than 2y-4. In the first one, 4 is positive and 2y is negative. In the second one, it has been reversed.
I think that your problem is that you don't like negative coefficients for your variables. You must remember than the sign applies to the term immediately following it, and no other.
[/quote]
I'll keep that in mind when I see subtraction. It's just weird to think of it that way.
[quote]Remember, consider the terms. The terms on the LHS (left hand side) are 73 and 14x. You must perform addition/subtraction only on like terms. Therefore, 14x can only go with another term with "x" in it. Therefore, ones with "y" or "" do not qualify.

Now since you don't have a variable () on the RHS, you must subtract the ones without variables. Therefore, you would subtract 73[/quote]
Ok I think I get it now. Thanks for the help guys.
I enjoy math, I'm just not very good at understanding new things about it that I learn. It makes me feel good to learn something new, it just seems to take forever for me to finally get it.

Well, Generally, if you have an equation of the form

$ Ax\plus{}B\equal{}C$ where A,B,C are constants, then $ x\equal{}\frac{C\minus{}B}{A}$

So in that, $ C\equal{}91$, $ B\equal{}\minus{}73$, and $ A\equal{}14$. So, plugging in,

$ x\equal{}\frac{72}{7}$.

We got the above by doing $ Ax\plus{}B\equal{}C \implies Ax\equal{}C\minus{}B$ by

subtracting $ B$, then we divide by $ A$ to get $ x\equal{}\frac{C\minus{}B}{A}$.

Don't memorize the forumla, understand how to get it.

We just want to isolate $ x$, or in other words, on one side have

$ x$, and the other something not involving $ x$.

Yeah, this is a little harder for Classroom Math, but w/e...

If you are like this --> :huh: then I'll try to explain again

[quote="BOGTRO"]Well, Generally, if you have an equation of the form

$ Ax \plus{} B \equal{} C$ where A,B,C are constants, then $ x \equal{} \frac {C \minus{} B}{A}$

So in that, $ C \equal{} 91$, $ B \equal{} \minus{} 73$, and $ A \equal{} 14$. So, plugging in,

$ x \equal{} \frac {72}{7}$.

We got the above by doing $ Ax \plus{} B \equal{} C \implies Ax \equal{} C \minus{} B$ by

subtracting $ B$, then we divide by $ A$ to get $ x \equal{} \frac {C \minus{} B}{A}$.

Don't memorize the forumla, understand how to get it.

We just want to isolate $ x$, or in other words, on one side have

$ x$, and the other something not involving $ x$.

Yeah, this is a little harder for Classroom Math, but w/e...

If you are like this --> :huh: then I'll try to explain again[/quote]
Yeah I understand now. Thanks. And thanks to everyone else who helped.

You guys know any good websites where I can practice my algebra and maybe learn some of the more advanced algebra?

I recommend the [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=200]Art of Problem Solving Introduction To Algebra[/url]


The ratio of length to width to height of a rectangular prism is 3:2:1. If the surface area of the prism is $ 198 \text{m}^{2}$, how many cubic meters are in its volume?

The answer key says the answer is $ 162$, and I think I know why...you did everything correctly, you just forgot to multiply it by $ 6$. I'll finish it for you:

$ 27*6\equal{}162$ :)

Why does the volume need to be multiplied by 6???
BOGTRO's explanation is correct so far, but he forgot that the sides were 3x,2x, and x.  If x=3 (as BOGTRO solved), the sides are 9,6, and 3 so the volume is 9*6*3 or [b]162[/b]

You need to multiply by 6 because:
$ 3x\times2x\times x = 6x^3$ not $ x^3$


Prove that if $ p > 1,x > 0,$

$ x^p \minus{}1 \ge p(x \minus{}1)$

It's Bernoulli's inequality (just substitute $ x$ with $ x\plus{}1$ and see the proof here: http://en.wikipedia.org/wiki/Bernoulli%27s_inequality).


geometric series
number theory unsolved
number theory

Luca Pacioli claimed that

M=1+2+2^2+2^3+2^4+...+2^26

is a prime number. Show, WITHOUT using knowledge of geometric series, that this is FALSE by finding at least two distinct prime factors of M.

Work $\bmod 7$. The residues of the powers of $2$ are $1, 2, 4$. $M \equiv 9(1+2+4) \equiv 0 \bmod 7$.


Support MATHCOUNTS trigonometry geometry

Hi,

I want the honest opinions on this matter because it's very important to me. :)

I recently read the thread and heard some people mentioning (especially middle schoolers) that my contest, COP, should be open to everyone, not just member of Art of Problem Solving.

I didn't say "yes" to that matter because I wasn't sure how much support I could get by opening it into all people, and furthermore, how am I going to send/receive the answers?

Having a website is a best way to do this but since I'm little bit busy on making one, I want to hear some opinions how should I take care of this matter.

The reason this post has title "Test" is because I want to "test" something in here.

1) First, build a thread for registrating through E-mail. Way this work is I basically ask for who you are, what school you in, state, grade, gender (so I can later may have statistics on female and male intelligence on this  :D ), and basically information.
2) My idea would be post the problems in Getting Started or Basics Forum (depending whether it's Round 1 or COPSD) like usual.
3) As soon as it's posted, I mail the contestants who were registrated to E-mail the URL of the thread in Art of Problem Solving.
4) When it's close to the deadline, I send the assurance mail again.
5) Do this same for other rounds.

I think this might just work out if I have some supports.  The reason I go for Mathcounts Forum is because I think middle schoolers who are in Mathcounts can really benefit from this (although some questions have trig which you don't need), most questions are in grasp of them if they covered some geometry and algebra.  If you're interested or want to give opinion this, please do so.

Just wondering, can any of the users who can access to the Parent Forum spread this word to that forum as well? I can't access to there (obviously) but I want to hear some advices or ideas from the parents.

Thank you and Regards,

Silverfalcon

I posted the link to this thread in the parent's forum.

Naga

Treething,

Silverfalcon is not a parent.  You can infer the rest.

Thank you, Naga.

To Treething, here are few notes on speaking in the international forum:

You can't just call someone retard because they don't know something that you know. For instance, you can't call newbie a retard because he/she doesnt' know Latex. Since you posted many threads and once a moderator, you should give respect to other people and be kind to them. In this case, would I be happy if I was 50 year old man who just came to this site and saw your thread, posted by a student in a middle school, calling myself a retard? ;)

Silverfalcon


complex analysis
function
topology
complex numbers
complex analysis unsolved

Anyone helps me with this, please:

"Let $U$ be a connected open set in $C$ (the complex numbers) and let $(f_{n})$ be a sequence of holomorphic functions on $U$ converging uniformly on compact subsets of $U$, and let $f=lim f_{n}$.   Suppose that each $f_{n}$ is everywhere nonzero on $U$ , $f$ is not a constant, and $f_{n}(U) \subset V$ ,  with $V$ a connected open set in $C$.   Show that $f(U) \subset V$."

thanks

FS

No, and the hypothesis $f_{n}\ne 0$ is not needed either.

Suppose $w_{0}: =f(z_{0})\notin V$ for some $z_{0}\in U$. Since $f$ is nonconstant (and holomorphic), there is a small circle $C$ around $z_{0}$ on which $f\ne w_{0}$. Let $\epsilon=\min_{z\in C}|f(z)-w_{0}|$. For sufficiently large $n$ we have $|f-f_{n}|<\epsilon$ on $C$. [url=http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem]Rouche's theorem[/url] yields a contradiction.

you're right mlok i dont need that $f_{n}$ has no zeros, thank you so much, i got it perfectly; the contradiction holds in the fact that you get $w_{0}=f_{n}(a)$ for some $a\in U$


This is the fourth time this has been posted here...


linear algebra
matrix
algebra proposed
algebra

Assume that you do not understand very basic calculus.  Then you may use the famous Abel Summation Formula.

Or you can just split the sum thus:

$\begin{matrix}a &+& a^{2}&+& a^{3}&+& \dots &+& a^{n-1}\\ &+& a^{2}&+& a^{3}&+& \dots &+& a^{n-1}\\ & & &+& a^{3}&+& \dots &+& a^{n-1}\\ & & & & & & & \vdots &\\ & & & & & & &+& a^{n-1}\end{matrix}$

and treat each row as a separate geometric series.

I've always remembered it using Farenhajt's way.


If a + b + c = abc but abc is not equal to 0. show that

$ 4 \equal{} \frac {(1 \minus{} b^{2})(1 \minus{} c^{2})}{bc} \plus{} \frac {(1 \minus{} a^{2})(1 \minus{} c^{2})}{ac} \plus{} \frac {(1 \minus{} a^{2})(1 \minus{} b^{2})}{ab}$

[hide]$ RHS \equal{} \frac{1}{ab} \plus{}\frac{1}{bc} \plus{}\frac{1}{ca} \minus{} (\frac{a\plus{}c}{b}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{a\plus{}b}{c}) \plus{}ab\plus{}bc\plus{}ca$

$ \equal{}\frac{a\plus{}b\plus{}c}{abc} \minus{} (ac\minus{}1\plus{}bc\minus{}1\plus{}ab\minus{}1) \plus{}ab\plus{}bc\plus{}ca$

$ \equal{}4$[/hide]

Comment: This is definitely algebra, but not something a normal 8th grade algebra class would assign for homework :!:

I commend your algebra teacher for making this beautiful problem.

[hide="do not look if you want to solve it yourself"]Let the sum be $ S$. We want to prove that $ S\equal{}4$. Multiplying the binomials together and canceling out common factors from numerators and denominators gives

$ S\equal{}\frac{1}{bc}\minus{}\frac{b}{c}\minus{}\frac{c}{b}\plus{}bc\plus{}\frac{1}{ac}\minus{}\frac{a}{c}\minus{}\frac{c}{a}\plus{}ac\plus{}\frac{1}{ab}\minus{}\frac{a}{b}\minus{}\frac{b}{a}\plus{}ab$

Since $ a\plus{}b\plus{}c\equal{}abc\neq 0$, we have $ \frac{1}{bc}\plus{}\frac{1}{ac}\plus{}\frac{1}{ab}\equal{}1$ (I divided both sides by $ abc$), so

$ S\equal{}\minus{}\frac{b}{c}\minus{}\frac{c}{b}\plus{}bc\minus{}\frac{a}{c}\minus{}\frac{c}{a}\plus{}ac\minus{}\frac{a}{b}\minus{}\frac{b}{a}\plus{}ab\plus{}1$

Since $ a\plus{}b\plus{}c\equal{}abc\neq 0$, we have $ 1\plus{}\frac{b}{a}\plus{}\frac{c}{a}\equal{}bc$, $ \frac{a}{b}\plus{}1\plus{}\frac{c}{b}\equal{}ac$, and $ \frac{a}{c}\plus{}\frac{b}{c}\plus{}1\equal{}ab$ (I divided both sides by $ a$, $ b$, and $ c$ respectively). We then have

$ S\equal{}\minus{}\frac{b}{c}\minus{}\frac{c}{b}\plus{}1\plus{}\frac{b}{a}\plus{}\frac{c}{a}\minus{}\frac{a}{c}\minus{}\frac{c}{a}\plus{}\frac{a}{b}\plus{}1\plus{}\frac{c}{b}\minus{}\frac{a}{b}\minus{}\frac{b}{a}\plus{}\frac{a}{c}\plus{}\frac{b}{c}\plus{}1\plus{}1$

$ S\equal{}1\plus{}1\plus{}1\plus{}1\equal{}4$.

Therefore $ 4\equal{}\frac{(1\minus{}b^2)(1\minus{}c^2)}{bc}\plus{}\frac{(1\minus{}a^2)(1\minus{}c^2)}{ac}\plus{}\frac{(1\minus{}a^2)(1\minus{}b^2)}{ab}$.[/hide]

I have a feeling that you go to Phillips Exeter Academy. :P


function
vector
real analysis
real analysis solved

Find all $F$:$R^4 \rightarrow R$ such that
i) $F(1,0,0,1)=1$
ii) $F(ka,b,kc,d)=kF(a,b,c,d)$
iii) $F(a,b,c,d)=-F(b,a,d,c)$
iv) $F(a+e,b,c+f,d)=F(a,b,c,d)+F(e,b,f,d)$

where $a,b,c,d,e,f,k \in R$

$F(a,b,c,d)=ad-bc$. Consider $F$ as a function of two 2-dimensional vectors $x=(a,c)$ and $y=(b,d)$. Then we have $F(e_1,e_2)=0,\,F(kx,y)=kF(x,y),\,F(x,y)=-F(y,x)$, $F(x_1+x_2,y)=F(x_1,y)+F(x_2,y)$ (here $e_1=(1,0),\,e_2=(0,1)$). So, $F$ is antisymmetric polylinear function of a couple of two-dimensional vectors. It is very well-known and not hard that such function is a multiple of a determinant.


geometry
circumcircle
trigonometry
geometry solved

Let O be the circumcenter of a triangle ABC, denote Da, Db, Dc the distance from O to BC, AC, AB. Prove that Da + Db + Dc = R + r, when ACB is acute and Da + Db - Dc = R + r, when C is obtuse angle.

maybe someone has a nice proof without using trigonometry ?

Or maybe one could tell me, where I could find a synthetic solution ?

Thank you

[quote="wellknown"]Let O be the circumcenter of a triangle ABC, denote Da, Db, Dc the distance from O to BC, AC, AB. Prove that Da + Db + Dc = R + r, when ACB is acute and Da + Db - Dc = R + r, when C is obtuse angle.[/quote]

Actually, the problem is stated a bit incorrectly. In fact, in order to have Da + Db + Dc = R + r, not only the angle ACB must be acute, but all three angles of triangle ABC must be acute, i. e. triangle ABC must be acute-angled.

A synthetic proof of the formula Da + Db + Dc = R + r for acute-angled triangles ABC can be found at

  http://www.cut-the-knot.org/proofs/carnot.shtml

(website by Alexander Bogomolny). The formula Da + Db - Dc = R + r for a triangle ABC with an obtuse angle at the vertex C can be shown analogously.

  Darij

[quote="darij grinberg"] Actually, the problem is stated a bit incorrectly. In fact, in order to have Da + Db + Dc = R + r, not only the angle ACB must be acute, but all three angles of triangle ABC must be acute, i. e. triangle ABC must be acute-angled.
[/quote]
OK, that was actually what I meant, but i didn't express myself clearly. I thank you very much for this link and your comments.


superior algebra
superior algebra unsolved

[quote="spx2"]If $ \sigma$ is a cycle with $ |\sigma| \equal{} n$ , in what conditions is $ \sigma^r$ also a cycle ?[/quote]
the condition is $ \gcd (n,r)\equal{}1$, can you prove it?

if what you trying to say is, it is a cycle of [b]length n[/b], then the condition is exactly as given by JC_math.

For cycle of arbitrary length, then any $ r$ works

[quote="JC_math"][quote="spx2"]If $ \sigma$ is a cycle with $ |\sigma| \equal{} n$ , in what conditions is $ \sigma^r$ also a cycle ?[/quote]
the condition is $ \gcd (n,r) \equal{} 1$, can you prove it?[/quote]

to be honest no, but I read the solution(if I remember well from Fraleigh solution manual)


calculus
integration
function
LaTeX
trigonometry
symmetry
calculus computations

Evaluate the definite integral of the following function between the limits 1/a and a ,where a > 0

f(x) = (sqrt(1 + x^4))/x^2

Excuse me for not being latex - savvy  :| [/code]

Mathematica thinks the answer can only be expressed in terms of hypergeometric functions.

[quote="kunny"]$ \int_{\frac {1}{a}}^{a}\frac {\sqrt {1 \plus{} x^4}}{x^2}\ dx \equal{} \int_{\frac {1}{a}}^a \sqrt {1 \plus{} \frac {1}{x^4}}\ dx$, Let $ \frac {1}{x^2} \equal{} \tan \theta$.[/quote]

I tried a similar substitution,but arrived at the following one

 integral of  [b](sin(theta))^(-3/2)[/b] between appropriate limits of [b]theta[/b]How do you solve that?

The problem falls in the category of elliptical integrals. So probably we need to employ some standard technique to solve it.

I was wondering if it could be solved by an easier approach.

One comment: the substitution $ u\equal{}\frac1x$ transforms the integral into itself. That means, by symmetry, that the original problem could have been written as $ 2\int_1^a\frac{\sqrt{1\plus{}x^4}}{x^2}\,dx$

That means that to solve this for an arbitrary $ a$ means precisely to find the antiderivative $ \int\frac{\sqrt{1\plus{}x^4}}{x^2}\,dx.$

And that runs into what JBL mentioned: not expressible in elementary terms.

By Integral by parts, $ \int_{\frac {1}{a}}^a \frac {\sqrt {1 \plus{} x^4}}{x^2}\ dx \equal{} \left[ \minus{} \frac {1}{x}\sqrt {1 \plus{} x^4}\right]_{\frac {1}{a}}^a \plus{}\int_{\frac{1}{a}}^a \frac {1}{x}\cdot \frac {4x^3}{2\sqrt {1 \plus{} x^4}}\ dx$

$ \equal{} a\sqrt {1 \plus{} \frac {1}{a^4}} \minus{} \frac {1}{a}\sqrt {1 \plus{} a^4} \plus{} 2\int_{\frac {1}{a}}^a \frac {x^2}{\sqrt {1 \plus{} x^4}}\ dx$

$ \equal{} 2\int_{\frac {1}{a}}^a \frac {x^2}{\sqrt {1 \plus{} x^4}}\ dx$.

Kunni,How do you evaluate the last integral?