conics
algebra
polynomial
geometry theorems
geometry

What is Poncelet´s Theorem?
Please, give me a theorem and proof  :D

Try searching on Google [b]"Poncelets Theorem" proof[/b]

Poncelet's theorem is in some sense a statement about elliptic curves (curves of degree 3 without rational parametrization).  Pairs consisting of (point $P$ on conic $C_1$, line $\ell$ through $P$ tangent to conic $C_2$) define an algebraic variety which is, in fact, an elliptic curve.  Going from one such pair to the next is equivalent to addition $x_i \to x_i + a$ on the curve.
The conics satisfy the $n$-Poncelet condition, if and only if $n . a = 0$ in the group law of the curve, so that the polygon closes irrespective of starting point.

However, one can ask about elementary proofs.  I've never heard of one --- but Poncelet found the theorem without the use of elliptic curves or torsion.  For triangles and quadrilaterals in a pair of circles, one can check (with some work, it is not so easy) that the Poncelet condition is the vanishing of a polynomial in radii and mutual distance of the circles, and so is independent of the polygon.  This scheme of proof seems hard to extend to $n$-sided polygons.

Here is what I know about Poncelet's theorem. [u][b]Poncelet's theorem:[/b][/u]
[i]Let circle $\beta$ lies inside the circle $\alpha$. Let $A$ be a point on $\alpha$, draw a tangent line from $A$ to $\beta$ and let it intersect $\alpha$ at $A_{1}$ , do the same for $A_{1}$ and so on . If  for some point $A$ there is $n$ such that $A_{n}=A$ then it is truth for all points of $\alpha$.[/i]

In fact the following theorem is also right:[i] Let $O,I$ are the centers of $\alpha$ and $\beta$ respectevely.Let $A$ is a point on $\alpha$, $B$ is a point obtained after $\frac{n}{2}$ steps that are written in above theorem. $OI\cap{AB}$ does not depend from $A$. [/i]


How do I get my square root to go over the entire number instead of just part of it? $\sqrt121$

enclose your number in {}'s

\sqrt{121}
$\sqrt{121}$

That applies to most functions. for example, the code for $\frac{2x+3}{x-1}$ is

\frac{2x+3}{x-1}

To render $x^{a_{n}+1}$ we write x^{a_n+1}.


If we define [p/q]>0 in Q to mean pq>0, prove that the relation ">" is well defined - ie if (p'/q')~(p/q) then pq>0-->p'q'>0.

I don't know how to start this...

I assume the equivalence relation you're using is $ \frac{p'}{q'} \sim \frac{p}{q} \Leftrightarrow p'q \equal{} pq'$.  This problem may look intimidating, but it's not; $ pq > 0$ just means that $ p, q$ are either both positive or both negative, which is the same as the intuitive meaning of $ \frac{p}{q} > 0$.  So all you have to do is casework on the signs of $ p, q, p', q'$.


inequalities
function
complex numbers
topology

Let us consider the following metric on R:

d(x,y)= |e^x-e^y|.

Is R complete with this metric.

I tried to show that any Cauchy sequence with respect to this metric is Cauchy in R with Euclidean metric but was not successful. I feel like I am missing some basic inequalities - I tried Lagrange theorem for e^x to find some but alas...

Any help is greatly appreciated.

$ \mathbb{R}$ is not complete under this metric.

Consider the sequence $ x_n\equal{}\minus{}n.$

For $ m<n$ we have $ d(x_m,x_n)\equal{}e^{\minus{}m}\minus{}e^{\minus{}n}<e^{\minus{}m},$ and that can be made arbitrarily small.

This is a Cauchy sequence, and there is no real number for it to converge to.

For any $ A>\minus{}\infty,$ the set $ [A,\infty)$ is complete under this metric, but the whole line is not complete.

I suspect that the troubles you ran into trying to use the mean value theorem in your attempted proof had to do with not being able to bound $ e^{\xi}$ away from zero. If so, your trouble was real - part of the indication that the proof wasn't going to work at all.

[quote="Kent Merryfield"]$ \mathbb{R}$ is not complete under this metric.

Consider the sequence $ x_n \equal{} \minus{} n.$

For $ m < n$ we have $ d(x_m,x_n) \equal{} e^{ \minus{} m} \minus{} e^{ \minus{} n} < e^{ \minus{} m},$ and that can be made arbitrarily small.

This is a Cauchy sequence, and there is no real number for it to converge to.

For any $ A > \minus{} \infty,$ the set $ [A,\infty)$ is complete under this metric, but the whole line is not complete.

I suspect that the troubles you ran into trying to use the mean value theorem in your attempted proof had to do with not being able to bound $ e^{\xi}$ away from zero. If so, your trouble was real - part of the indication that the proof wasn't going to work at all.[/quote]

Thank you for the prompt reply.

But when I tried to prove that $ x_n \equal{} \minus{} n.$ is not convergent with respect to d I bumped onto the same difficulties.

It seems to me that $ x_n \equal{} \minus{} n.$ converges to 0 with respect to d because $ d( \minus{} n,0)\equal{} e^{ \minus{} n}$ as you pointed out can be made arbitrarily small.

It does not converge to 0. In fact, you can see that if $ \minus{}n \rightarrow x$ in this metric, then $ e^{\minus{}n} \rightarrow e^x$ as numbers. But $ e^x \equal{} 0$ has no solution in the real line, hence the limit does not exist.

I found a direct proof:

Let {x_n} be Cauchy w.r.t d. Then {y_n=e^(x_n)} is Cauchy w.r.t. euclidean metric, hence converges to some b>=0.

If b>0 then {x_n} converges to lnb; if b=0 then {x_n} converges to 0. Hence every Cauchy is convergent, hence R,d is complete.

[quote="Dimitrina"]if b=0 then {x_n} converges to 0[/quote]

This is wrong, as Professor Merryfield has given an example, also what I have said. Remember that $ e^0 \equal{} 1$.

More general version: suppose that $ f\colon \mathbb R\to\mathbb R$ is an injective function (not necessarily continuous). Then the following are equivalent:
(1) $ \mathbb R$ is complete in the metric $ d(x,y) \equal{} |f(x) \minus{} f(y)|$
(2) $ f(\mathbb R)$ is a closed subset of $ \mathbb R$ (in the standard metric)

The function $ f(x)\equal{}e^x$ has range $ (0,\infty)$ -- not closed.

[quote="Soarer"]It does not converge to 0. In fact, you can see that if $ \minus{} n \rightarrow x$ in this metric, then $ e^{ \minus{} n} \rightarrow e^x$ as numbers. But $ e^x \equal{} 0$ has no solution in the real line, hence the limit does not exist.[/quote]

Thank you - indeed a stupid mistake.

[quote="mlok"]More general version: suppose that $ f\colon \mathbb R\to\mathbb R$ is an injective function (not necessarily continuous). Then the following are equivalent:
(1) $ \mathbb R$ is complete in the metric $ d(x,y) \equal{} |f(x) \minus{} f(y)|$
(2) $ f(\mathbb R)$ is a closed subset of $ \mathbb R$ (in the standard metric)

The function $ f(x) \equal{} e^x$ has range $ (0,\infty)$ -- not closed.[/quote]

That is a nice way to generate new problems :)

Thus we can conclude that R with either of the following distances:
$ d(x,y)\equal{}|x^3\minus{}y^3|$
$ d(x,y)\equal{} ln(1\plus{}|x\minus{}y|)$

is a complete metric space :?:

Yes for the first one.

The second distance is not of the form $ |f(x)\minus{}f(y)|$. It is still complete, though.


modular arithmetic
number theory proposed
number theory

Let $ k_{n} > k_{n \minus{} 1} > ... > k_{1} > 1$ and $ a_{1},a_{2}, ... , a_{n}$ be integers. For all $ N$ integers, it possible to choose such i ($ n\geq i\geq 1$) for which  $ k_{i}|N \minus{} a_{i}$. Determine the smallest possible value of n.

[hide]We claim that the answer is $ n=4$.

Construction: $ k_1=2,k_2=3,k_3=4,k_4=6$, $ a_1=0,a_2=0,a_3=1,a_4=5$. Clearly $ 0\pmod 2, 0\pmod3, 1\pmod4,5\pmod 6$ covers all integers.

Proof of minimality: Assume that $ n\le 3$ is possible, then $ n=3$ must be possible. Let $ \text{lcm}(k_1,k_2,k_3)=x$. Note that $ a_1\pmod{k_1},a_2\pmod{k_2},a_3,\pmod{k_3}$ must cover all numbers in mod $ x$. There are at most $ \frac{x}{k_1}+\frac{x}{k_2}+\frac{x}{k_3}$ integers covered. So $ \frac{x}{k_1}+\frac{x}{k_2}+\frac{x}{k_3}\ge x$, i.e., $ \frac1{k_1}+\frac1{k_2}+\frac1{k_3}\ge1$. If $ k_1\ge3$, then $ \frac1{k_1}+\frac1{k_2}+\frac1{k_3}\le\frac13+\frac14+\frac15<1$. So $ k_1<3$. Thus $ k_1=2$. If $ k_2\ge4$, then $ \frac1{k_1}+\frac1{k_2}+\frac1{k_3}\le\frac12+\frac14+\frac15<1$. So $ k_2=3$. If $ k_3>6$, then $ \frac1{k_1}+\frac1{k_2}+\frac1{k_3}<\frac12+\frac13+\frac16=1$. So $ k_3\le6$. Modulo 2, 3, 4 clearly cannot cover all integers. The same holds for 2,3,5 and 2,3,6. The proof is complete.[/hide]

[hide="a related problem"][url]http://www.mathlinks.ro/viewtopic.php?t=279797[/url][/hide]

[quote="Johan Gunardi"]We claim that the answer is $ n \equal{} 4$.
Clearly $ 0\pmod 2, 0\pmod3, 1\pmod4,5\pmod 6$ covers all integers.
[/quote]

7 is not covered

sorry.. :blush:  now i think the answer is $ n\equal{}5$ (see the example in the link i gave in the previous post). it's just casework..

I think $ n>5$. Take 4 primes $ p\equal{}2,3,5,7$, and $ 0(p)$. Stayed 48 numbers mod 210 and 11 factors $ k_i\equal{}6,10,14,15,21,35,30,42,70,105,210$ and we can chose $ a_i$ for these factors for cover 48 staed numbers.

Actually the result [i]is[/i] $ n\equal{}5$. I posted my proof here: [url=http://bboyjordan.wordpress.com/2009/10/31/every-number-m2m-nm-has-all-digits/]Problem10.A conditon for being a covering system[/url]. []

Ps. I don't know for other ML users, but at least for me most of Rust's messages are quite incomprehensible  :huh:


geometry
circumcircle
Euler
power of a point
radical axis
geometry proposed

See the following posts:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16566[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41678[/url]

[quote="April"]Let $ ABC$ be a triangle. The median $ AA_1$ intersects the circumcircle $ (O)$ of triangle $ ABC$ at $ A_2$. Let $ X$ the intersection of the line passes through $ A_1$ and perpendicular to $ OA$ with the tangent line in $ A_2$ of the circle $ (O)$. Points $ Y$, $ Z$ are defined similarly. Prove that $ X$, $ Y$, $ Z$ are collinear.[/quote]
[img]http://i209.photobucket.com/albums/bb172/Ineq/thayHa.jpg[/img]
Let $ G$ is centroid of $ ABC$ and $ K$ is center Euler circle of $ ABC$
Easy to see $ XA_2 \equal{} XA_1$
We have $ P_{X|(O)} \equal{} XA_2^2 \equal{} XA_1^2$
$ KA_1\parallel{} AO$ so $ KA_1\perp XA_1$
So $ XA_1^2 \equal{} P_{X|(K)}$
so $ P_{X|(O)} \equal{} P_{X|(K)}$ so X lie on radical axis of $ (O)$ and $ (K)$
Similar we have $ Y,Z$ lie on radical axis of $ (O)$ and $ (K)$ too
As desire!:D
[hide]@April:Bài này là một trong mấy bài thầy Minh Hà cho chúng em làm đó anh :D lời giải trên là lời giải thầy cho chúng em đó ạ :D[/hide]


geometry rectangle counting distinguishability

In the Valley of the Shadow of Death, there is no light, so the demons cannot see. All paper money consists of rectangles with the corners cut off some, all, or none of the corners. Currency denominations are determined by shape. The dimensions of the rectangles are 2.5 units by 6 units, and the triangles are right triangles with legs of length 1 unit. Demons can recongize the value of the money by feeling which corners have or have not been cut off. Of course, they cannot tell the front from the back of these bills. What is the number of distinguishable denominations are possible for the currency in this land?.

Billy

Funny because I just did that sprint today with 28.
[hide]You can have no corners cut off, one corner got off, 2 corners cut off which can be done 3 ways, 3 corners cut off, and all 4 cut off. This makes 7.[/hide]

[hide]13. For tri, 1 ways for 0 cut off, 2 ways for 1, 2 ways for 2 and 1 way for 3.
For rectangle, 1 way for 0, l way for 1, 3 ways for 2, 1 way for 3 and 1 way for 4. Add and get 13!![/hide] .

The problem is only referring to rectangles with triangles cut off the corners. So you wouldn't talk about triangles with corners cut off.

And you also have to make sure that you cannot flip the shape around to achieve a shape you have already cut off

[hide="I agree with tarquin"]1 way for no corners cut off
1 way for 1 corner cut off
3 ways for two corners cut off
1 way for three corners cut off
1 way for all corner cut off

again totaling 7.[/hide]

It appears that the answer is 7.

Why are there 3 ways for 2 corners to be cut off?

1 way, the corners at the ends of the shortest sides
second way, the corners at the ends of the longest sides
3rd way, the corners at the opposite of the diagonals??????????

We can have 2 cut off of one short end, 2 cut off a long sides, or 2 cut off directly on opposite corners.

I agree.
One question though:
Can you Cut off triangles after you've cut off a triangle from the came corner?

No, read the question. It says you can only cut off right isosceles triangles with legs of 1.

OOOOOOOOOOHHHH. Look what I just saw! [hide]We can cut off 2 trianlges on 1 side if we put the hypotenuse on the long side and we don't cut of corenrs on that side. And we can cut off 1 on the ends! And if we cut off one or more corners of a long side, we can cut another in the middle![/hide]

This is scary! I see why it is in the All-Time  Greatest book!!

Billy

thats what I was trying to say!
Ha!
Hold on!
If you cut a triangle from the reactangle, then put the long side on the cut, you get a square. Now you have 5 corners! NOW cut a triangle from the cut square, and that's another one! THEN you could do it again and get two cut squares!
Oh, great....
I I thought I had seen everything...

It looks like I accidentally removed a condition from the problem. The original problem says you can only cut off the 4 corners of the rectangle. If my problem had stated that, everyone would have gotten it right. However, nobody read the question well enouhg, except DWX!

Everyone try to solve it without this condition now!(the way the problem is right now.)

[hide="Answer and picture"]
Seven.

[img]http://img.photobucket.com/albums/v259/drspyder/money.jpg[/img]
[/hide]

Given that you can cut off more than 4 corners... the answer will be really big...


function
integration
abstract algebra
real analysis
real analysis unsolved

let $c_k$ be complex numbers. One defines $S_n(x)=\sum_{-n}^{+n} c_k e^{ikx}$ and $K_m(x)=\frac{1}{m} \sum_{k=0}^{m-1} S_k(x)$.
Prove that if the $K_m$ are uniformly bounded in $L^p$, $1 < p \leq +\infty$, then there exists a function $f \in L^p$ such that $c_k = \int_{0}^{2 \pi} f(x) e^{-ikx} dx$, for all $k \in \mathbb{Z}$

wow! a very brief solution  :)  could detail a little bit more ? (anyway, mine was very different ..)

Sorry, I was in a bit of a hurry before so I was just giving a suggestion. I think I misread your question, and I was thinking about trying to prove something else. I've got a lot of work to do right now but I'll give this some thought later tonight.


[quote="ckck"]Julie begins counting backwards from 1000 by 2's, and at the same time, Tony begins counting forward from 100 by 3's. If they count at the same time, what number will they say at the same time?[/quote]

[hide]It is same as finding the intersection of lines $y=3x+100$ and $y=1000-2x$, which is $(180,640)$. So they will say $640$ at the same time.[/hide]

[quote="ckck"][quote="ProtestanT"][hide]
1000-100=900
because 2+3=5, i divide by 5
900/5=180
180*2=360
180*3=540
add back the 100
100+540=$640$
[/hide][/quote]
I don't understand why you added the 100 to the 540 instead of the 360. Then what was the whole purpose of multiplying 180 by 2?[/quote]
it is to show that he will count down 360 from 1000...
and the other person will count up 540 from 100


The arithmetic mean of the nine numbers in the set $ \{9,99,999,9999, . . . ,999999999\}$ is a $ 9$-digit number $ M$, all of whose digits are distinct. The number $ M$ does not contain the digit

$ \textbf{(A)}\ 0 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 4 \qquad
\textbf{(D)}\ 6 \qquad
\textbf{(E)}\ 8$

the sum of the elements of M is $ 9(1\plus{}11\plus{}...\plus{}111111111)$

so the mean is $ 1\plus{}11\plus{}...\plus{}111111111$

which is $ 123456789$

and that does not contain 0.

answer: A

[hide] $ 9M\equal{}9(1\plus{}11\plus{}111\plus{}...\plus{}111111111)$ so $ M\equal{}1\plus{}11\plus{}111\plus{}...\plus{}111111111$. As we can see, for each place value, the number of ones is less than 10, so there will never be a 0 appearing. Hence, our answer is $ \text{(A) 0}$.[/hide]


inequalities
Cauchy Inequality
inequalities unsolved

Let a,b,c be three positive real numbers satisying a+b+c=3.Find the minimum value of $ \frac{1}{2\minus{}a^{2}}\plus{}\frac{1}{2\minus{}b^{2}}\plus{}\frac{1}{2\minus{}c^{2}}$

[quote="duythuc_lqd"]Let a,b,c be three positive real numbers satisying a+b+c=3.Find the minimum value of $ \frac {1}{2 \minus{} a^{2}} \plus{} \frac {1}{2 \minus{} b^{2}} \plus{} \frac {1}{2 \minus{} c^{2}}$[/quote]

My solution

by Cauchy Inequality, $ (a^2 \plus{} b^2 \plus{} c^2)(1^2 \plus{} 1^2 \plus{} 1^2)\ge(a \plus{} b \plus{} c)^2$. So, $ a^2 \plus{} b^2 \plus{} c^2\ge3$.

So, by Cauchy Inequality again, $ ((2 \minus{} a^2) \plus{} (2 \minus{} b^2) \plus{} (2 \minus{} c^2))(\frac {1}{2 \minus{} a^{2}} \plus{} \frac {1}{2 \minus{} b^{2}} \plus{} \frac {1}{2 \minus{} c^{2}})\ge(1 \plus{} 1 \plus{} 1)^2$.

Since $ a^2 \plus{} b^2 \plus{} c^2\ge3$, we can know that the min is 3.

It's incorrect.$ 2\minus{}a^{2},2\minus{}b^{2},2\minus{}c^{2}$ is not sure be three positive real numbers so you can't use Cauchy.
Your proof is false

[quote="duythuc_lqd"]It's incorrect.$ 2 \minus{} a^{2},2 \minus{} b^{2},2 \minus{} c^{2}$ is not sure be three positive real numbers so you can't use Cauchy.
Your proof is false[/quote]

Then there is no minimum value because $ \frac{1}{2\minus{}a^2}$ can be made artitrarily small (very negative).


inequalities
geometry
geometric inequality
inequalities unsolved

hey, can anybody do it ? or If you solved it can you pm for me, I and you can have a paper in crux for it! :)

really is it too hard for mathlinkers ? :D  :D

ok that's a nice joke  :D  :D ,I can't edit my post!  therefore I post it again
Given $x,y,z>0,x+y+z=1$ prove that
$(2+\frac{\sqrt{3}x}{\sqrt{y^{2}+yz+z^{2}}})(2+\frac{\sqrt{3}y}{\sqrt{z^{2}+zx+x^{2}}})(2+\frac{\sqrt{3}z}{\sqrt{x^{2}+xy+y^{2}}})\ge 27$
sorry but now it looks quite nice :D!

come on this is not that hard problem that you think!
mathlinkers are able to solve 10times harder problems than this one ... you will see my friend!

ok that's right, I believe that ,too.Mathlinkers is very super :D , but it is really hard with me, therefore if anybody have solution, please post it for me learn  :)

Actually I know mathlinker are very super but it's unfortunate that nobody want to help me  :D  :D do you think that Sadi  :D

It's ugly.   :(

That's why noone wants to bother with it (and matlinkers have a good eye for aesthetics I think  :lol: )

Hey but I don't think it is more ugly than [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=119306]this one[/url]  :D  :D  :D

[quote="gemath"]Given $x,y,z>0,x+y+z=1$ prove that
$(2+\frac{\sqrt{3}x}{\sqrt{y^{2}+yz+z^{2}}})(2+\frac{\sqrt{3}y}{\sqrt{z^{2}+zx+x^{2}}})(2+\frac{\sqrt{3}z}{\sqrt{x^{2}+xy+y^{2}}})\ge 27$
 :)[/quote]

Hi all mathlinkers!! It's a long time since I have posted it, actually nobody can do ?  :(

Actually we don't need to have x+y+z=1. If this inequality is true then it will be true for any x,y,z>0.

Yes you are right Mamat, actually by computer I see it is true, but unfortunate It is too hard with me!... and you?
@I heared from a mathlinker (maybe hungkhtn), he say his method can be solved all problems of symmetric inequalities of three variable, but in this problem I think it is only a joke :D :D

Ok sorry for that :oops: .Now I will psot the rgiht solution.Notice we can take the triangle ABC with side lenghts $\sqrt{y^{2}+yz+z^{2}}$ and the similar ones.The point T is the Toricelli point and we $TA=x,TB=y,TC=z$.Now from a well known geometric inequality we know $ \sum_{cyc}\frac{TA\cdot TB}{CA\cdot CB}\geq 1$ for any point inside the triangle.So this solves the problem. :lol:

[quote="mateivld"]Ok sorry for that :oops: .Now I will psot the rgiht solution.Notice we can take the triangle ABC with side lenghts $\sqrt{y^{2}+yz+z^{2}}$ and the similar ones.The point T is the Toricelli point and we $TA=x,TB=y,TC=z$.Now from a well known geometric inequality we know $\sum_{cyc}\frac{TA\cdot TB}{CA\cdot CB}\geq 1$ for any point inside the triangle.So this solves the problem. :lol:[/quote]

 Nice solution (if it is true :) )

Good ideal mateivld, but write as you we need to prove
$(2+\frac{\sqrt{3}TA}{a})(2+\frac{\sqrt{3}TB}{b})(2+\frac{\sqrt{3}TC}{c})\ge 27$
Does it have the relation with your ineq $\sum\frac{TBTC}{bc}\ge 1$  :?:

[quote="gemath"]
Does it have the relation with your ineq $\sum\frac{TBTC}{bc}\ge 1$  :?:[/quote]
   [color=purple] It's nice and true for all M in plane.gemath do you nice geometry inequality?
  .[/color]

Maybe you don't understand me, I known that inequality true for all $M$, but my meaning is if you want to do follow mateivld you need to prove
$(2+\sqrt{3}\frac{TA}{a})(2+\sqrt{3}\frac{TB}{b})(2+\sqrt{3}\frac{TC}{c})\ge 27$
and I see this inequality doesn't relate the inequality $\sum\frac{TBTC}{bc}\ge 1$
Can mateivld or naybody show me clearly the solution! :maybe:


geometry
parallelogram
geometry unsolved

let $ ABCD$ be a qadrilateral.$ H$ and $ F$ are the midpoints of $ AD$ and $ BC$ respectively.$ AB$ and $ DC$ meet in $ O$ .take $ OM\equal{}AB$ and $ ON\equal{}DC$.construct a paralleologram $ OMKN$.prove that $ OK\equal{}2HF$ and $ OK$ and $ HF$ are parallel

[quote="ishfaq420haque"]let $ ABCD$ be a qadrilateral.$ H$ and $ F$ are the midpoints of $ AD$ and $ BC$ respectively.$ AB$ and $ DC$ meet in $ O$ .take $ OM \equal{} AB$ and $ ON \equal{} DC$.construct a paralleologram $ OMKN$.prove that $ OK \equal{} 2HF$ and $ OK$ and $ HF$ are parallel[/quote]
where is the location of M and N ?
If M and N are arbitrary , I don't think it's true

$ M$ and $ N$ are in $ OA$ and $ OD$ respectively.$ OM$ and $ AB$ are in the same direction.same occurs for $ ON$ and $ DC$

[b][size=100][color=DarkBlue]EQUIVALENT  PROBLEM. - A triangle $ \bigtriangleup ABC$ is given and let $ M,\ N$ be, the midpoints of its side-segments $ AB,\ AC,$ respectively. Let $ D,\ D'$ be, two arbitrary isotomic points on $ AB$ and also, let $ E,\ E'$ be, two ones on $ AC$ $ ($ So, we have $ AD \equal{} BD',$ $ AE \equal{} CE'$ and suppose that $ D,\ E,$ are between $ A,\ M$ and $ A,\ N,$ respectively $ ).$ Prove that $ AX\parallel \equal{} YZ,$ where $ X,\ Y,\ Z$ are the midpoints of the segments $ DE,\ D'E',\ BC,$ respectively.[/color][/size][/b]

[hide]Easy to prove, as a direct application of the [b][size=100]ERIQ theorem[/size][/b] $ ($ = [b]E[/b]qual  [b]R[/b]atios  [b]I[/b]n  [b]Q[/b]uadrilateral $ ).$[/hide]

Kostas Vittas.

[quote="ishfaq420haque"]could you kindly tell me the ERIQ theorm>?i cant catch it[/quote]
You can find two elementary proofs of this theorem, in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Collinerity of orthocenters[/url] of [b][size=100]shobber[/size][/b] (posts #10# and #11# applying the corresponded schemas from the post #14#).

Best regards, Kostas Vittas.


group theory
abstract algebra
superior algebra
superior algebra solved

Suppose every subgroup of a group $ G$ is normal. Does it follow that $ G$ is abelian ?

Consider the subgroup generated by two elements $ a$ and $ b$. If $ \langle a\rangle$ and $ \langle b\rangle$ are normal, $ a^{\minus{}1}ba\equal{}b^k$ and $ b^{\minus{}1}ab\equal{}a^\ell$ for some integers $ k,\ell$.
Now, $ a^{\minus{}\ell}\cdot a\equal{}b^{\minus{}1}a^{\minus{}1}ba\equal{}b^{\minus{}1}\cdot b^k$, so $ a^{1\minus{}\ell}\equal{}b^{k\minus{}1}\equal{}\alpha\in \langle a\rangle\cap\langle b\rangle$.

If the groups generated by $ a$ and $ b$ intersect trivially, $ a$ and $ b$ commute.
Otherwise, we can continue by noting that $ \alpha\equal{}a^{\minus{}1}\alpha a\equal{}(b^k)^{k\minus{}1}\equal{}\alpha^k$ and similarly $ \alpha^\ell\equal{}\alpha$
Testing the simplest possible nontrivial example ($ k\equal{}\ell\equal{}3$, $ |a|\equal{}|b|\equal{}4$), the quaternion group works.


calculus
integration
trigonometry
calculus computations

More generally:

Let $a_{i}$, $i=1\cdots n$ be fixed real numbers.

Calculate:

$I(a_{1},\cdots, a_{n})=\int_{0}^{\pi/2}\cdots \int_{0}^{\pi/2}\frac{dx_{1}\cdots dx_{n}}{1+\tan^{a_{1}}x_{1}\cdots \tan^{a_{n}}x_{n}}$.

You can apply Fubini's to get $I_{n}= (\frac{\pi}{4})^{n}$, right?

It seems to me that the answer is $\frac{\pi^{n}}{2^{n+1}}$.


1. Find all $ f: (1, \inf ) \to \mathbb R$ which satisfy \[f(xy)(x\minus{}y)\equal{}f(y)\minus{}f(x)\]  for all $x,y>1$.

2. Find all $ f: \mathbb R^\plus{} \to \mathbb R^+$ such that \[f\left( \frac{x\minus{}y}{f(y)} \right)\equal{}\frac{f(y)}{f(x)}\] for all $x>y>0$.

[quote="gks921217"]1. $ f: (1, \inf ) \rightarrow R, f(xy)(x \minus{} y) \equal{} f(y) \minus{} f(x) \forall x,y > 1$. Find all $ f$.[/quote]

Let $ P(x,y)$ be the assertion $ f(xy)(x \minus{} y) \equal{} f(y) \minus{} f(x)$

1) If $ \exists u > v > 1$ such that $ f(u) \equal{} f(v) \equal{} 0$, then $ f(x) \equal{} 0$ $ \forall x > 1$
================================================
Adding $ P(x,y)$, $ P(y,z)$ and $ P(z,x)$, we get a new assertion $ Q(x,y,z)$ : $ f(xy)(x \minus{} y) \plus{} f(yz)(y \minus{} z) \plus{} f(zx)(z \minus{} x) \equal{} 0$

Suppose now $ \exists u > v > 1$ such that $ f(u) \equal{} f(v) \equal{} 0$. Let then $ w\in(\frac uv,uv)$ : $ Q(\sqrt {\frac {vw}u},\sqrt {\frac {uw}v},\sqrt {\frac {uv}w})$ $ \implies$ $ f(w) \equal{} 0$ $ \forall w\in(\frac uv,uv)$
And since $ P(u,v)$ $ \implies$ $ f(uv) \equal{} 0$ and $ P(\frac uv,v)$ $ \implies$ $ f(\frac uv) \equal{} 0$, we get : $ f(x) \equal{} 0$ $ \forall x\in[\frac uv,uv]$

Taking then $ \epsilon$ such that $ u \minus{} \frac uv > \epsilon > 0$, we have $ uv > u > u \minus{} \epsilon > \frac uv$ and so $ f(u) \equal{} f(u \minus{} \epsilon) \equal{} 0$ and so, using the same method : $ f(x) \equal{} 0$ $ \forall x\in[\frac {u}{u \minus{} \epsilon},u(u \minus{} \epsilon)]$
And so, taking $ \epsilon$ as little as we want : $ f(x) \equal{} 0$ $ \forall x\in(1,u^2)$
And so (just replay the mechanism with $ u^2,v$, then $ u^4,v$, ... : $ f(x) \equal{} 0$ $ \forall x > 1$
Q.E.D.

2) The only solutions are $ f(x) \equal{} \frac ax$
==========================
Clearly these functions are solutions.
Let then $ g(x)$ any solution
The function $ f(x) \equal{} g(x) \minus{} \frac {3g(3)}{x}$ is clearly a solution and is such that $ f(3) \equal{} 0$
Then $ P(3,2)$ $ \implies$ $ f(6) \equal{} f(2)$
Then $ P(6,2)$ $ \implies$ $ f(12) \equal{} 0$

And so $ f(3) \equal{} f(12) \equal{} 0$ and so $ f(x) \equal{} 0$ $ \forall x > 0$, according to 1) above.

And so $ g(x) \equal{} \frac {3g(3)}{x}$
Q.E.D.

Hence the unique family of solutions $ \boxed{f(x) \equal{} \frac ax}$

[quote="gks921217"]2. $ f: R^ \plus{} \rightarrow R^ \plus{} , f( \frac {x \minus{} y}{f(y)} ) \equal{} \frac {f(y)}{f(x)} \forall x > y > 0$. Find all $ f$.[/quote]

Let $ P(x,y)$ be the assertion $ f(\frac{x\minus{}y}{f(y)})\equal{}\frac{f(y)}{f(x)}$

Let $ 1\plus{}f(1)\equal{}a$

$ P(1\plus{}f(1),1)$ $ \implies$ $ f(1)\equal{}\frac{f(1)}{f(1\plus{}f(1))}$ and so $ f(1\plus{}f(1))\equal{}1$ and so $ f(a)\equal{}1$

$ P(2a,a)$ $ \implies$ $ 1\equal{}\frac 1{f(2a)}$ and so $ f(2a)\equal{}1$

$ P(x,a)$ $ \implies$ $ f(x\minus{}a)\equal{}\frac 1{f(x)}$ $ \forall x>a$ and so $ f(x\minus{}2a)\equal{}\frac 1{f(x\minus{}a)}\equal{}f(x)$ $ \forall x>2a$

$ P(x,2a)$ $ \implies$ $ f(x\minus{}2a)\equal{}\frac 1{f(x)}$ $ \forall x>2a$

Comparing these two lines, we get $ f(x)\equal{}1$ $ \forall x>2a$

So $ f(2a\plus{}x)\equal{}1$ $ \forall x>0$ and so $ P(2a\plus{}x,2a)$ $ \implies$ $ f(x)\equal{}1$ $ \forall x>0$

And this indeed is a solution. Hence the result : $ \boxed{f(x)\equal{}1\forall x}$


function
inequalities
conics
parabola
absolute value
real analysis
real analysis unsolved

Could someone help me with this problem?
Let f: (-1,1) to R, and f'(0) exists. If the sequences (a_n), (b_n) converges to 0 , then define the difference quotion D_n=( f(b_n) - f(a_n) ) / ( b_n - a_n)

1) Show that lim D_n = f'(0) as n approaches infinity under the following condition
 a)   a_n < 0 < b_n
 b)  0 < a_n < b_n and  b_n / ( b_n - a_n ) < or equal to M
 c) f'(x) exists and is continuous for all x in (-11)

2) let f(x)= x^2 * sin(1/x) for x does not equal 0 and f(0)=0. f is differentiable everywhere in (-1,1) and f'(0)=0. Find a_n , b_n that tend to 0 in such a way that D_n converges to a limit unequal to f'(0)

[quote="mlok"]Subtract a constant from $ f$ to make $ f(0) \equal{} 0$ - this does not affect $ D_n$.
You can also subtract the linear function $ f'(0)x$ from $ f$ - this changes $ D_n$ by $ f'(0)$.

So we assume that $ f(0) \equal{} f'(0) \equal{} 0$. Informally, $ |f(x)|$ is small compared to $ |x|$ near $ 0$. Use this to show that $ |f(b_n) \minus{} f(a_n)|$ is small compared to $ |b_n \minus{} a_n|$ under the assumptions stated. Epsilon and delta are your friends.[/quote]

I was able to figure out part 1) thanks to my GSI's help. I just play with the inequality obtained from the hypothesis and what it means for lim D_n= f'(0) .

I still don't get part 2) so, your hint for me on part 2) is this " So we assume that...Epsilon and delta are your friends."
I don't get it when you say absolute value of f(x) is small compared to abs(x) near 0 and abs( f(b_n) - f(a_n) ) is small compared to abs(b_n - a_n) . Can you put this in term of inequality? Thanks.

Actually, my "hint" was about part 1.

In part 2 we have $ f'(0) \equal{} 0$, and we want to find $ a_n,b_n$ such that $ D_n\to L\ne 0$. The graph of $ f$ is wiggles between two parabolas $ y \equal{} x^2$ and $ y \equal{} \minus{} x^2$. It touches the top parabola when $ \sin(1/x) \equal{} 1$, i.e. $ x \equal{} \frac{1} {\pi/2 \plus{} 2\pi n}$. It touches the bottom parabola when $ \sin(1/x) \equal{} \minus{} 1$, i.e. $ x \equal{} \frac {1} { \minus{} \pi/2 \plus{} 2\pi n}$. Would these points work as $ a_n$ and $ b_n$? Give it a try.


LaTeX
logarithms
limit
calculus
summer program
MathPath
real analysis

um, couldnt it be anything because anything to the power of infinity is infinite. unless its one or smaller.  But iv done some weird root calculations and if it's to follow the pattern, (decreasing from around 3^1/3), then:

[hide]The answer is x, where x approaches 1.  (This is, of course, in saying its yth root of y where y approaches infinity).[/hide]

[quote="Robert0201"]What is [tex]\sqrt[\infty]{\infty}[/tex]?
p. s.: this whole LaTeX thing is pretty cool.[/quote]

This isn't formal, but

[tex]\sqrt[\infty]{\infty} = \infty^{\frac{1}{\infty}} = \infty^0[/tex]

[tex]= e^{\ln(\infty^0)} = e^{0 \cdot \ln \infty} = (e^0)^{\ln \infty}[/tex]

[tex]= 1^{\infty} = 1[/tex]

1. This doesn't belong in Getting Started.

2. There is no such thing as [tex]\sqrt[\infty]{\infty}[/tex]; the closest thing that one can get is

[tex]\displaystyle \lim_{x \to \infty} \sqrt[x]{x}[/tex]
[tex]=\displaystyle \lim_{x \to \infty} {x^{1/x}}[/tex]

As [tex]x \to \infty[/tex], [tex]1/x \to 0[/tex], so

[tex]=\displaystyle \lim_{x \to \infty} {x^0} = 1.[/tex]

Well, while [tex]\displaystyle\lim_{x\to\infty}x^\frac{1}{x}[/tex] does equal 1 (btw osiris had a flaw in his proof... see if you can find it), that does not mean that [tex]\infty^\frac{1}{\infty}=1.[/tex] To see this, note that as x goes to infinity, [tex]x^x[/tex] goes to infinity and [tex]\frac{1}{x}[/tex] goes to [tex]\frac{1}{\infty}[/tex] (im using infinity rather loosely), and

[tex]\displaystyle \lim_{x\to\infty}(x^x)^\frac{1}{x} = \infty^\frac{1}{\infty}[/tex],

however you can easily check that THAT limit goes to infinity, and not 1. So, [tex]\infty^\frac{1}{\infty}[/tex] is indeterminant, and so it has no value.

[quote="Robert0201"][tex]\frac{1}{\infty}[/tex] is 0 so [tex]\infty^{\frac{1}{\infty}}[/tex] is 1[/quote]

It's not quite that simple, though your logic is good.

[quote="Robert0201"][tex]\frac{1}{\infty}[/tex] is 0 so [tex]\infty^{\frac{1}{\infty}}[/tex] is 1[/quote]

zableman is correct... Your question does not discuss the rate at which the two values approach infinity and the rate is what matters in a limit... you cannot generalize [tex]\displaystyle{\sqrt[\infty]{\infty} = \lim_{x\rightarrow \infty}x^{\frac{1}{x}}}[/tex] as zableman pointed out.

[tex]\sqrt[\infty]{\infty}[/tex] alone is indeterminant and does not have a value.

[quote="Robert0201"][tex]\frac{1}{\infty}[/tex] is 0 so [tex]\infty^{\frac{1}{\infty}}[/tex] is 1[/quote]

This is a pretty common naive "understanding" among people who've had calculus-for-engineers courses rather than a rigorous real analysis course. But the statement is not fully correct. Infinity is not a number (as JBL once asked, what's the number just before infinity, or just after?). I think there is even a photograph of John Horton Conway teaching on this issue at last year's MathPath camp on the [url=http://www.mathpath.org]MathPath site[/url].

[quote="AoPS Volume 2"]DO NOT treat [tex]\infty[/tex] as a regular number.[/quote]

A lot of weird things happen if you let infinite be a number, so it is best to only use it to write infinite sums and limits as (some variable) goes to infinite.

[quote="Osiris"]1. This doesn't belong in Getting Started.[/quote]

Osiris is correct that a challenge problem that really involves taking limits properly doesn't belong in the Getting Started forum. But I will retitle the thread, and then all will be fine here, because this is a common kind of beginner mistake (treating "infinity" as just another number) that people should learn how to avoid while they are still just getting started.

Ordinarily, the best thing to do before posting a challenge problem for others to solve is to KNOW THE SOLUTION; otherwise, it's not a challenge problem but an informational question. Any informational question such that what the question is about is understandable to a beginner is welcome here, so this thread stays put after I retitle it.


How many n-character strings could be formed using the letters A,B,C,D,E such that A appears an even number of times in the formed strings?

It will probably be useful to use the fact that $ \sum_{k \geq 0} \binom{n}{2k} x^{n \minus{} 2k} \equal{} \frac{(x \plus{} 1)^n \plus{} (x \minus{} 1)^n}{2}$.


does any body know:who Paco Delucia is?

thanks... ;)

sure! i love paco!
the best flamenco guitar player in the world...

wow...

i think,he has also the best technic...

oh ,i want to know that wether he is alive or not...

Of course he's still alive! :)

Check out his website : http://www.pacodelucia.org/

well yes he is a very good flamenco player,... but i won't call him the best, but defenatly in my top3 list.

i play flamenco myself, but some times, his songs are jost to boring... ;) .

i like Juan martin better. :)

i saw Paco in concert 2 years ago..

armithlusa do you know who Tomatito is? he's another goodflamenco player..
well i don't think Paco's music is boring!

yes i know Tomatito, he is good...

i didn't mean all of Paco's music is boring (some my favorites are his songs)...what i meant can be understood from what Juan martin said:

"sometimes only few single notes sound a lot better than, long chains of fast ones...".

 :)

[quote]"sometimes only few single notes sound a lot better than, long chains of fast ones..."[/quote]

I don't want to hijack the thread here, but based on that thought, do you like the music of the late jazz pianist and composer Thelonius Monk? I like it a lot.

yeah, i do.
but i still like Flamenco Better.

its not boring,is it?

every time i here his songs i enjoy them alot. ;)


function
inequalities
induction
limit
ratio
real analysis
real analysis unsolved

Let $f : \mathbb R \to \mathbb R$ be a continuous function such that $f(f(x))+f(x) = 2x+a$, where $a \in \mathbb R$.

Find all such functions.

I'll prove it for $a\neq 0$, and $f$ increasing.

$f(x)=f(y)\Rightarrow f(f(x)=f(f(y))\Rightarrow x=y$,and since $f$ is injective and continuous, it is monotonic.
If $f(x)=x$ we easily obtain $a=0$. Contradiciton. Hence, $f(x)>x$ or $f(x)<x$ for all x.

Case 1) $f$ is increasing and $f(x)>x$. Take the sequence $a_n=fofofof..of(x_0)$, n times. We have that this sequence is increasing since $f(x)>x\Rightarrow fof(x)>f(x)\Rightarrow...\Rightarrow a_n>a_{n-1}$.
But computing $a_n$ form $a_{n+2}+a_{n+1}=2a_n+a$ we get smth of the form $a_n=a\frac n 3 +(-2)^n$ which is not increasing for large $n$.
Case 2) If $f$ is increasing and $f(x)<x$ then $g(x)=f^{-1}(x)>x$ and is increasing,  and the same aproach works.

An easier proof can be done for $a=0$ and $f$ increasing. Assume WLOG that there exist $x$ such that $f(x)>x\Rightarrow fof(x)>f(x)>x\Rightarrow 2x>2x$. Contradiction. Hence $f(x)=x$ forall $x\in \mathbb{R}$.

I'm looking forward too see a solution for decreasing $f$. :D

I had the same ideas as you. Namely, I proved that $f$ is bijective and strictly increasing. After that, I tried to work out the recurrence relation, which turned out much uglier than yours. (I mean the coefficients)

But I found a function $f$ that verifies the relation given ??! Doesn't $f(x) = x + \frac{a}{3}$ work?

How did you proved that $f$ is increasing?  :?  this is the missing part from my proof....

About the recurence realtion: denote by $b_n=a_{n+1}-\frac{an}{3}$. We get that $b_{n+2}+b_{n+1}-2b_n=0$. now denote by $c_n=b_{n+1}-b_n$. We get that $c_{n+1}+2c_n=0$, hence $c_n=(-2)^n\cdot c_1$.
Hence $a_n=\frac{an}{3}+(-2)^n \cdot c_1$. and not $a_n=\frac{an}{3}+(-2)^n$  as I written in my previous post. So, if $c_1=0$ we obtain the solution $f(x_0)=x_0+\frac a 3$.  And if $c_1\neq 0$ we obtain a contradiction mentioned in my previous post.

Since $x_0$ was chosen arbitrarly, we have $f(x)=x+\frac a 3$.

bump... $f(x) = -2x$ is a solution for $a=0$ :(

I think that you can guess what mistake I've done. (I didn't notice it because I didn't write it on a paper)

My proof was only for increasing functions. $f(x)=-2x$ is decreasing.

I can prove that if $a\neq 0$, $f$ must be increasing.
Firstly, I prove that $fof(x)\neq x$. Assume that there exist an $x_0$such that $fof(x_0)=x_0$. Then $a_{2n}$ and $a_{2n+1}$ are constant sequences, but from the recurence realtion there are of the form $a_n=\frac{an}{3}+(-2)^n \cdot c_1$ hence not constant.

So, $f(f(x))>x, \forall x$ implies that $f(x)<x+a\forall x$, but if x tens to $-\infty$ f(x) goes to $\infty$ and the inequality fails to hold. Similarly, when $f(f(x))<x$.

I'll think about the case $a=0$ and $f$ decreasing.

The last case : $a=0$ and $f$ is decreasing.
It's easy to see that $fof$ is increasing and has limits $\pm \infty$ at $\pm \infty$  and $f$ has limits $\pm \infty$ at $-\infty$ and $\infty$ respectively. From this facts follows that $f$ have only one fixed point, name it $p_0$

Denote by $g(x)=f(x)+2x$. We have that $gof(x)=g(x)$, which implies that $g(fof(x))=g(x)$ and $g(hoh(x))=x$. where $h$ is the inverse of $f$.

Chose an $x_0$ and denote by $a_n=f^{(2n)}(x_0)$  and $b_n=h^{(2n)}(x)$. Since $fof$ and $hoh$ are increasing we have that $a_n$ and $b_n$ are monotonic.
Let $x_0<p_0$ WLOG. Now we have $fof(x_0)>x_0$ or $hoh(x_0)>x_0$. Take WLOG that $fof(x_0)>x_0$. It is easy to prove by induction that $a_n<p_0$ and is increasing. Hence $a_n$ is convergent to $\ell\in \mathbb{R}$.

Since $fof(a_n)=a_{n+1}$, tending n to infinity,since $fof$ is continuous, we have $fof(\ell)=\ell$, and from the given relation $f(\ell)=\ell$, hence $\ell=p_0$.

Since $g$ is continuous, we have $g(p_0)=\lim_{x\to p_0}g(x)=\lim_{n \to \infty}g(a_n)=\lim_{n \to \infty}g(x_0)=g(x_0)$. Since $x_0$ was chosen at random, we have that $g$ is constant $=g(p_0)$. Hence $f(x)=c-2x$, the only $c$ for which the realtion holds is $c=0$.

The only solution in this case is $f(x)=-2x$.

Now the proof is complete but you have to put the pieces together in oder to write down a full solution.  :D

It's gonna take a while to thoroughly check your solution... Anyway, thanks. If there's something I don't understand, I'll let you know.

BTW, instead of $fof$ you can write $f \circ f$.

[code]f \circ f[/code]

This problem appeared in some romanian shortlist, a few years ago. Can anyone tell me who is the author of this?

I do not know the shortlisted problems. Can you post more shortlisted problems for the 12-th grade? Thanks.

What about $f$ increasing and $a=0$?  :blush: Can it be done in the same way?

Also,

[hide="Minor correction"]The given relation tells us that \[ a_{n+2} + a_{n+1} = 2 a_n + a .  \]

Letting $\displaystyle b_n = a_n - \frac{a(n-1)}{3}$, we get \[ \displaystyle b_{n+2} + \frac{a(n+1)}{3} + b_{n+1} + \frac{an}{3} = 2 b_n + \frac{2a(n-1)}{3} + a \, \, \Longleftrightarrow \, \, b_{n+2} + b_{n+1} = 2 b_n .  \]

Finally, let $\displaystyle c_n = b_{n+1} - b_n$, so \[ \displaystyle b_{n+2} - b_{n+1} = 2 b_n - 2 b_{n+1} \, \, \Leftrightarrow \, \, c_{n+1} = -2 c_n .  \]

Thus, $\displaystyle \left( c_n \right)_{n \geq 1}$ is a geomtric progression of ratio $\displaystyle -2$, so $\displaystyle c_n = (-2)^{n} c_0$.

We have \[ b_n - b_0 = \sum_{k=0}^{n-1} \left( b_{k+1} - b_k \right) = \sum_{k=0}^{n-1} c_k = \sum_{k=0}^{n-1} (-2)^k c_0 = \frac{(-2)^n-1}{-3} \cdot c_0 ,  \]
so $\displaystyle b_n = b_0 + \frac{(-2)^n-1}{-3} \cdot c_0$. Hence, \[ a_n = \frac{a(n-1)}{3} + b_0 + \frac{(-2)^n-1}{-3} \cdot c_0 .  \]

We have $\displaystyle b_0 = a_0 + \frac{a}{3}$, $\displaystyle b_1 = a_1$, so $\displaystyle c_0 = a_1 - a_0 - \frac{a}{3}$. Thus, \[ a_n = \frac{an}{3} + a_0 + \frac{(-2)^n-1}{-3} \cdot \left( a_1 - a_0 - \frac{a}{3} \right) . \][/hide]

PS: this topic is getting way too long.

[quote="xirti"]
An easier proof can be done for $a=0$ and $f$ increasing. Assume WLOG that there exist $x$ such that $f(x)>x\Rightarrow fof(x)>f(x)>x\Rightarrow 2x>2x$. Contradiction. Hence $f(x)=x$ forall $x\in \mathbb{R}$. :D[/quote]

I payed no atention to details. I hope the proof works like we say in romanian ("ca idee..") :)


Hi everyone

I am trying to to inserted a wrap figure into new environment such theorem.
I have try to the package wrapfigure but it creates a floating which I trying to avoid.
So I tries to write my macro which include figure but the problem is that I need access to
the caption command. when I labels the new figure I need to write label nuber to a file which I do not
know how to do. Then I will need to read the label with ref command.

Here is what I have done so far.

Any suggestion.

http://www.potto.org
where free textbooks roam!

\documentclass[12pt]{book}
\usepackage{amsmath, amsthm}
\usepackage{latexsym}
\usepackage{amssymb}

\newtheoremstyle{myExampl}{\baselineskip}{\baselineskip}{\itshape} {}{\sffamily}{:}{\newline}{}
\theoremstyle{myExampl}
\newtheorem{examl}{Example}[chapter]

\newcommand{\latexEx}[2]{
\begin{examl}
\label{#1}
#2
\end{examl}
}

\begin{document}
\latexEx{mom:ex:nozzle}{
\parbox{0.63\textwidth}{
Liquid flows through a symmetrical nozzle as shown in the Figure \ref{mom:fig:nozzle} with
a mass flow rate of 0.01 [$ gk/sec$].
The entrance pressure is 3[$ Bar$] and the entrance velocity is 5 [$ m/sec$].
The exit velocity is uniform but unknown.
The exit pressure is 1[$ Bar$].
The entrance area is 0.0005[$ m^2$] and the exit area is 0.0001[$ cm^2$].
What is the exit velocity?
What is the force acting the nozzle?
Assume that the density is constant $ \rho \equal{} 1000 [kg/m^3]$ and the volume in the nozzle is 0.0015 [$ m^3$]
}
\parbox{0.05\textwidth}{$ \,$}
\parbox {0.32\textwidth}{ %{\resizebox{1.5in}{!}
   %{\includegraphics{nozzle.eps}}}
        this is substitute for the figure
        this is substitute for the figure
        this is substitute for the figure
        this is substitute for the figure
        this is substitute for the figure
   \stepcounter{figure}
   \begin{raggedright}
      {Fig. \thefigure{} Schematic of a nozzle for the discussion on the forces and for
      example \ref{mom:ex:nozzle}.}
   \end{raggedright}
   }
}

\end{document}

If I understand you correctly, you want a caption and label for a figure that is not floating? If so, there's a command \captionof that does exactly this. You can use either the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=capt-of]capt-of package[/url] or the [url=http://tug.ctan.org/pkg/caption]caption package[/url].

Hi

Thank you stevem!
I think that you solve my problem.
If I already asking for in this macro.
I would like to have the text wrap around the figure
similar to wrapfigure pagage with given number of line
but from the top.

I have made the changes you suggested.
here is the text.

\documentclass[12pt]{book}
\usepackage{amsmath, amsthm}
\usepackage{latexsym}
\usepackage{amssymb}
\usepackage{capt-of}

\newtheoremstyle{myExampl}{\baselineskip}{\baselineskip}{\itshape} {}{\sffamily}{:}{\newline}{}
\theoremstyle{myExampl}
\newtheorem{examl}{Example}[chapter]

\newcommand{\latexEx}[2]{
\begin{examl}
\label{#1}
#2
\end{examl}
}

\begin{document}
\latexEx{mom:ex:nozzle}{
\parbox{0.63\textwidth}{
Liquid flows through a symmetrical nozzle as shown in the Figure \ref{nozzle} with
a mass flow rate of 0.01 [$ gk/sec$].
The entrance pressure is 3[$ Bar$] and the entrance velocity is 5 [$ m/sec$].
The exit velocity is uniform but unknown.
The exit pressure is 1[$ Bar$].
The entrance area is 0.0005[$ m^2$] and the exit area is 0.0001[$ cm^2$].
What is the exit velocity?
What is the force acting the nozzle?
Assume that the density is constant $ \rho \equal{} 1000 [kg/m^3]$ and the volume in the nozzle is 0.0015 [$ m^3$]
}
\parbox{0.05\textwidth}{$ \,$}
\parbox [t]{0.32\textwidth}{ %{\resizebox{1.5in}{!}
   %{\includegraphics{nozzle.eps}}}
        this is substitute for the figure
        this is substitute for the figure
   %\stepcounter{figure}
        \captionof{figure}[move]
      {Schematic of a nozzle }
        \label{nozzle}
   }
}

In the line [code]\parbox{0.63\textwidth}{ [/code]add the t parameter so you get [code]\parbox[t]{0.63\textwidth}{[/code]
Also you may wish to replace the line [code]\parbox{0.05\textwidth}{$\,$}[/code] by [code]\hfill[/code]

thank you for your tips on the space.

If you latex the last small file that I posted you will get two paragraphs that are
aliened at the bottom.
I am trying to get the first paragraph "climb" on the top of the second one.
Wrap the first paragraph around the top of the second one.
I have tried to use the package shapepar with the \cutout command but I get either
confused or it not working within this environment

thank you,

What is wrong with doing it manually breaking the paragraph where appropriate such as
[code]\documentclass[12pt]{book}
\usepackage{amsmath, amsthm}
\usepackage{latexsym}
\usepackage{amssymb}
\usepackage{capt-of}
\usepackage{graphicx}

\newtheoremstyle{myExampl}{\baselineskip}{\baselineskip}{\itshape} {}{\sffamily}{:}{\newline}{}
\theoremstyle{myExampl}
\newtheorem{examl}{Example}[chapter]

\newcommand{\latexEx}[2]{
\begin{examl}
\label{#1}
#2
\end{examl}
}

\begin{document}
\latexEx{mom:ex:nozzle}{
	Liquid flows through a symmetrical nozzle as shown in the Figure \ref{nozzle} with
	a mass flow rate of 0.01 [$gk/sec$].
	The entrance pressure is 3[$Bar$] and the entrance velocity is 5 [$m/sec$].
	The exit velocity is uniform but unknown.
	The exit pressure is 1[$Bar$].
	The entrance area is 0.0005[$m^2$] and the exit area is 0.0001[$cm^2$].
	\begin{center}
	   \includegraphics{nozzle.eps}
	   \captionof{figure}[move]{Schematic of a nozzle }
	   \label{nozzle}
	\end{center}
	What is the exit velocity?
	What is the force acting the nozzle?
	Assume that the density is constant $\rho = 1000 [kg/m^3]$ and the volume in the nozzle is 0.0015 [$m^3$]
}
\end{document}[/code]
which is how books often do it?

Thanks for the reply.

I like to have the figure on the side (mostly on the right).
My struggle is to put it on the side and to wrap the text around as below
example

text text text text text text text text text text text text text text text
text text text text text text text text text text text text text text text
text text text text text text text text text  [b]figure figure figure figure[/b]
text text text text text text text text text  [b]figure figure figure figure[/b]
text text text text text text text text text  [b]figure figure figure figure[/b]
text text text text text text text text text  [b]figure figure figure figure[/b]
text text text text text text text text text  [b]figure figure figure figure[/b]
text text text text text text text text text  [b]figure figure figure figure[/b]

hanks for the help.

picins is less fussy than wrapfig so you can do something like
[code]\usepackage{picins}
...
\latexEx{mom:ex:nozzle}{	Liquid flows through a symmetrical nozzle as shown in the Figure \ref{nozzle} with
	a mass flow rate of 0.01 [$gk/sec$].
	\parpic[r]{\parbox{5cm}{\includegraphics{nozzle.eps}\captionof{figure}{Schematic of a nozzle} \label{nozzle}}}
	The entrance pressure is 3[$Bar$] and the entrance velocity is 5 [$m/sec$].
	The exit velocity is uniform but unknown.
	The exit pressure is 1[$Bar$].
	The entrance area is 0.0005[$m^2$] and the exit area is 0.0001[$cm^2$].[/code]
Other options eg \parpic[rf] puts a frame round the picture - see [url=http://tug.ctan.org/tex-archive/macros/latex209/contrib/picins/picins.txt]picins.txt[/url]

Hi

I think that you solve the problem with some minor correction.
I add resizebox to control the size of the figure and
\parindent 0pt to eliminate the new paragraph appearance.
I also try to back (raise the paragraph with \vspace{-.5\parskip}
but is does not move up probably \parskip is undefined.
I try to move it up with \vspace{14pt} and it indeed moves up.
While I am using 11 font I can figure out the distance but a font change
will require manual changing.  any suggestion.

[code]

\latexEx{mom:ex:nozzle}{   Liquid flows through a symmetrical nozzle as shown in the Figure \ref{nozzle} with
   a mass flow rate of 0.01 [$gk/sec$].
   The entrance pressure is 3[$Bar$] and the entrance velocity is 5 [$m/sec$].
   The exit velocity is uniform but unknown.
   \parpic[r]{\parbox{1.5in}{\resizebox{1.0in}{!}{\includegraphics{nozzle.eps}}%
                        \captionof{figure}{Schematic of a nozzle} \label{nozzle}}}
        %\vspace{-.5\parskip}
        \vspace{-3pt}
        {\parindent 0pt
   The exit pressure is 1[$Bar$].
   The entrance area is 0.0005[$m^2$] and the exit area is 0.0001[$cm^2$].
   Liquid flows through a symmetrical nozzle as shown in the Figure \ref{nozzle} with
   a mass flow rate of 0.01 [$gk/sec$].
   The entrance pressure is 3[$Bar$] and the entrance velocity is 5 [$m/sec$].
   The exit velocity is uniform but unknown.
   The exit pressure is 1[$Bar$].
   The entrance area is 0.0005[$m^2$] and the exit area is 0.0001[$cm^2$].
   What is the exit velocity?
   What is the force acting the nozzle?
   Assume that the density is constant $\rho = 1000 [kg/m^3]$ and the volume in the nozzle is 0.0015 [$m^3$]
        }
}[/code]

No need to use \resizebox as you can use \includegraphics[scale=0.5}{nozzle.eps} with whatever fraction you want.
Also no need for {\parindent 0 pt ... } as \noindent will work.
\vspace{-.5\parskip} won't work as it is a rubber length 0.0pt plus 1.0pt. Use something like
\vspace{-0.3\baselineskip}
or use a font based height like ex
\vspace{-1ex}

Don't forget [url=http://tug.ctan.org/tex-archive/macros/latex209/contrib/picins/picins.txt]picins.txt[/url] gives lots of options such as \picskip and captions.

Again thank you for your time and help.

Below what I assembled. While it requires some manual work it will do the job.
I need to conceal the details with a macro.

thanks,

http://www.potto.org

[code]
\documentclass[12pt]{book}
\usepackage{amsmath, amsthm}
\usepackage{latexsym}
\usepackage{amssymb}
\usepackage{capt-of}
\usepackage{graphicx}

\newtheoremstyle{myExampl}{\baselineskip}{\baselineskip}{\itshape} {}{\sffamily}{:}{\newline}{}
\theoremstyle{myExampl}
\newtheorem{examl}{Example}[chapter]

\newcommand{\latexEx}[2]{
\begin{examl}
\label{#1}
#2
\end{examl}
}
\usepackage{picins}

\begin{document}
\latexEx{mom:ex:nozzle}{   Liquid flows through a symmetrical nozzle as shown in the Figure \ref{nozzle} with
   a mass flow rate of 0.01 [$gk/sec$].
   The entrance pressure is 3[$Bar$] and the entrance velocity is 5 [$m/sec$].
   The exit velocity is uniform but unknown.
   %\parpic[r]{\parbox{5cm}{\includegraphics{nozzle.eps}\captionof{figure}{Schematic of a nozzle} \label{nozzle}}}
   %\parpic[r]{\parbox{1.5in}{\resizebox{1.0in}{!}{\includegraphics{nozzle.eps}}%
   \parpic[r]{\parbox{5cm} {\includegraphics[scale=0.5]{nozzle.eps}
                        \captionof{figure}[more schematic]{Schematic of a nozzle} \label{nozzle} }}
        \vspace{-0.3\baselineskip}
   {\noindent
   The exit pressure is 1[$Bar$].
   The entrance area is 0.0005[$m^2$] and the exit area is 0.0001[$cm^2$].
   Liquid flows through a symmetrical nozzle as shown in the Figure \ref{nozzle} with
   a mass flow rate of 0.01 [$gk/sec$].
   The entrance pressure is 3[$Bar$] and the entrance velocity is 5 [$m/sec$].
   The exit velocity is uniform but unknown.
   The exit pressure is 1[$Bar$].
   The entrance area is 0.0005[$m^2$] and the exit area is 0.0001[$cm^2$].
   What is the exit velocity?
   What is the force acting the nozzle?
   Assume that the density is constant $\rho = 1000 [kg/m^3]$ and the volume in the nozzle is 0.0015 [$m^3$]
   }
}
\end{document}
[/code]


The sequence a_n satisfies $ a_{0}=1$, $ a_{1}=0,a_{2}=1, a_{3}=16$, and

$ a_{(}n+2)-a_{(}n-2)=18(a_{(}n+1)-a_{(}n-1))$ for n greater than or equal to 2. Prove that a_n is a perfect square for all n.

[quote="mdk"]The sequence $ a_{n}$ satisfies $ a_{0}=1$, $ a_{1}=0$,$ a_{2}=1$, $ a_{3}=16$, and

$ a_{n+2}-a_{n-2}=18(a_{n+1}-a_{n-1})$ for n greater than or equal to $ 2$. Prove that $ a_{n}$ is a perfect square for all $ n$.[/quote]

The sequence of the square roots, $ b_{n}$, is:

$ 1$, $ 0$, $ 1$, $ 4$, $ 17$, $ 72$, $ 305$, $ 1292$, $ 5473$.

It can be seen that this follows the recurrence:

$ b_{n+1}=4b_{n}+b_{n-1}$

The rest is a trivial induction.

Comment: we also have that: $ a_{n+3}=17a_{n+2}+17a_{n+1}-a_{n}$


The first four terms of an infinite sequence $S$ of decimal digits are $1$, $9$, $8$, $2$, and succeeding terms are given by the final digit in the sum of the four immediately preceding terms. Thus $S$ begins $1$, $9$, $8$, $2$, $0$, $9$, $9$, $0$, $8$, $6$, $3$, $7$, $4$, $\cdots$. Do the digits $3$, $0$,  $4$,  $4$ ever come up consecutively in $S$?

Yes.

If we know any four consecutive terms of $ S$, we can calculate "the whole sequence", i.e. we know the terms after these four, as well as those before. Since there are only finitely many combinations of four digits, that means that some combination will repeat itself. Obviously, somewhere between those two occurences, digits $ 1, 9, 8, 2$ come up again. By calculating the four terms before $ 1, 9, 8, 2$, we see that those are exactly $ 3, 0, 4, 4$ ($ x_4 \plus{} 1 \plus{} 9 \plus{} 8\equiv 2 \,(\cdot10)$ implies $ x_4 \equal{} 4$, $ x_3 \plus{} 4 \plus{} 1 \plus{} 9\equiv 8 \, (\cdot 10)$ implies $ x_3 \equal{} 4$, $ x_2 \plus{} 4 \plus{} 4 \plus{} 1\equiv 9 \, (\cdot 10)$ yields $ x_2 \equal{} 0$, and finally $ x_1 \plus{} 0 \plus{} 4 \plus{} 4\equiv 1 \, (\cdot 10)$ yields $ x_1 \equal{} 3$).
We conclude that digits $ 3, 0, 4, 4, 1, 9, 8, 2$ come up consecutively in $ S$.

Could you please explain a little more this?

[quote="idioteque"]that means that some combination will repeat itself. Obviously, somewhere between those two occurences, digits $ 1, 9, 8, 2$ come up again. [/quote]

Going backwards from the first occurence of that 4-digit combination, we will get to the beginning of this sequence, i.e. to $ 1, 9, 8, 2$. That means that going backwards from the second occurence of that 4-digit combination we will also get to $ 1, 9, 8, 2$, since we use the same rule to find the digits before...
This actually means this sequence is periodic.

Let's say that digits $ 0, 8, 6, 3$ (in that order) repeat themselves. Going back from the first occurence of it, we find that the members before are (I write them in the order we calculate them) $ 9, 9, 0, 2, 8, 9, 1$. We see that we came to $ 1, 9, 8, 2$, which means that $ 1, 9, 8, 2$ will also appear if we calculate terms before second occurence of $ 0, 8, 6, 3$.

Regardless of that "repeating" combination, going back from the first occurence of it, we will get to $ 1, 9, 8, 2$ (beginning of the sequence), so we know that $ 1, 9, 8, 2$ appear somewhere between the first two occurences of that repeating combination.

I hope it's somewhat clearer...


I tried doing a search but got way too many irrelevant results.
Can some one explain to me what is the IB program anyways? Because my school is probably going to offer it in the next 2 years.
What do I need to qualify (e.g. do I need any second languages?) and what is offered in the IB?
I looked at their website and I'm still confused.
Thanks.

There's info in [url=http://artofproblemsolving.com/Forum/topic-10870.html][u]this thread[/u][/url].

IB is an abbreviation for International Baccalaureate. You don't need anything to qualify.. just be hard working. You don't need to be super smart, but you need to be ready to do a lot of extra work to get the diploma. Freshman and Sophomore year is just like warmup.. and nothing changes much... I guess it depends on what school you go to. But for some reason at my school, we switch around Chemistry and Biology and IB kids take Chem sophomore year instead of junior year. Anyway.. IB is mostly extra stuff in the English part.. you have to write two "world lit" essays that get sent somewhere to be graded.. do a couple of oral presentations.. and do an extended research essay with 30+ sources. You need to be in IB for all four years of high school, since it's an ongoing thing. Also, in Junior and Senior year, you take IB tests. They're somewhat similar to AP tests.. since they happen at the same time so you can't really take some AP tests. IB tests are required, and you get graded I think from 1 to 9.. and you have to take 6 tests (at least 3 higher level). You have to get some score to pass... I forget right now, but if you pass the tests and everything, you get an IB diploma, which looks good I guess. Higher level classes are English HL, World History HL, Bio HL, Math HL.. maybe a foreign language... that's all our school offers. Oh, and also for IB you need 150 hours of community service hours (called CAS for commuity, action, service) over junior and senior year. So IB really starts at Junior year, since you do nothing extra freshmen and sophomore year. Oh yeah, and then you take a class called "Theory of Knowledge" which is this weird philosophy course, and have a presentation for that too. Uhh sorry for long explanation but yeah.. that pretty much sums it up.


whoa here is a wierd malfunction

[img]http://img3.freeimagehosting.net/uploads/08d3c974bf.jpg[/img]

as you can see darij grinberg's post is messed up...also, the resolution is messed up as the post width should be bigger

is this just me?

I have just commented on this in http://www.mathlinks.ro/Forum/viewtopic.php?t=135795 ;)

[quote="darij grinberg"]Explanation:

When you submit a post, the engine [b]first[/b] checks your syntax (particularly, whether you have more "/quote"s than "quote"s) and [b]then[/b] edits some LaTeX code. #H34N1 exploited this by putting in some LaTeX code which contained a "quote" which then dissolved itself upon LaTeX code auto-editing.

Funny bug, but I would prefer having it fixed (I can imagine it being potentially harmful).[/quote]

  darij


If kyyuanmathcount has 3 testicles, how many will his child have?

im guessing his child will have one and three quarters. (ouch that's gotta hurt)

I'm sorry I have two more than you...I shouldn't make people jealous.


MATHCOUNTS
videos
Math Kangaroo
Purple Comet

You can post any resources you find here. They can be practice problems from anywhere or concepts. I will start:

[url=http://agmath.com/]AGMath[/url]
[url=http://www.mathcountscentral.com]MathCounts Central - Tips, Practice Problems, ...[/url]
[url=http://www.mathcountsmn.org/tests.php]Minnesota Mathcounts - Problems section[/url]
[url=http://mathcounts.saab.org/mc.cgi]Saab's MathCounts Drills[/url]
[url=http://sites.bergen.org/mathcompetition/exams.asp]Bergen County Academies Math Competition[/url]
[url=http://www.kangurusa.com/clark/pdb/]Math Kangaroo Problem Database[/url]
[url=http://www.math.concordia.ab.ca/kangaroo/index.php?kn_mod=samples]Math Kangaroo (this is the Canadian version)[/url]
[url=http://www.stfx.ca/special/mathproblems/]Word Problems[/url]
[url=http://purplecomet.org/welcome/practice]Purple Comet[/url]
[url=http://www.themathleague.com/index.php?option=com_content&view=article&id=97&Itemid=79]Math League[/url]
[url=http://www.mathematica.ca/eng/preparatory.htm]Mathematica[/url]
[url=http://www.math.toronto.edu/oz/turgor/archives.php]Tounament of Towns[/url]
[url=http://www.coolmath.com/]Cool Math[/url]

At least for me, Saab's Mathcounts Drills doesn't work. I think it's supposed to be http://mathcounts.saab.org/mc.cgi

This has some past tests.

[url=http://mathacle.com/assessments/MATHCOUNTS/]Mathacle[/url]

You guys are forgetting the most obvious resource:
Art of Problem Solving

yea the little video things on the sidebar are really useful for when you are waiting for the page to load and stuff

Some tricks:

http://www.math-magic.com/index.htm

And some tricks/past tests if you look on the side bar:

http://www.academicmeet.com/index.php

This site helps too: http://www.math.ksu.edu/main/handbook/ProblemSets


$SABC$ o piramida triunghiulara regulara.Un plan $\alpha$,paralel cu baza, taie segmentele $AS$,$BS$ respectiv $CS$ dupa punctele $M$,$N$,respectiv $P$.Daca $CN$ si $BP$ se intersecteaza in $Q$,sa se arate ca dreapta $MQ$ trece printr-un punct fix din spatiu si sa se determine acesta.
[color=blue][hide="Indicatie"]Fie $T$ simetricul punctului $A$ fata de mijlocul segmentului $BC$[/hide][/color]

Se demonstreaza in 3 randuri cu proprietatea:

Un plan ce contine o dreapta paralela la un alt plan, taie planu respectiv intr-o dreapta paralela cu dreapta initiala.
(aja am facut eu la jose) :|

Eu am facut q Menelaos..si a iesit misto si asa.

da :) oricum tot mai scurta e cu proprietatea asta...se demonstreaza ca ala e paralelogram de jos si iese ^ ^


I've been reading a series of articles in the NYTimes about socioeconomic class in America today.  Please write your opinions.  Does class matter?

Just to be clear:  you want this to be, "Does class matter," not "Should class matter," in contemporary America.  Correct?

[quote="bubala"] about socioeconomic class in America today. [/quote]
Since I take "America" to mean the entire continent, here's a short description of the extent of class, er, distinction in Mexico: A maid (yes, a maid) who works in a rich neighboorhood in Mexico will feel superior to a maid who works in a not-so-rich neighborhood. But Latin America's always had more evident class distinctions than the US.

I think class matters most in shaping first impressions about someone. People's socioeconomic status often show in the way they're dressed, the cars they drive, where they live, etc, and we tend to form first impressions based on those things. But once you get to know someone well, class doesn't matter as much as the person herself.


Find all integers a , b , c : a< b<c and abc -1 is divisible by (a-1)(b-1)(c-1)

This is an IMO-92 one :

http://www.kalva.demon.co.uk/imo/isoln/isoln921.html

Pierre.


I inscribe a regular hexagon within a circle. The hexagon has sides measuring 1 . What is the area of the part of the circle not covered by the hexagon?

Let's draw a diagram:

[asy]size(5cm);
draw(unitcircle);
draw(polygon(6));[/asy]

First we find the area of the hexagon, then subtract that from the circle.

We can divide this hexagon into 6 equilateral triangles.

Each one has an area of:

$ \dfrac{1^2\sqrt{3}}{4}\equal{}\dfrac{\sqrt{3}}{4}$

Therefore, the total area of the hexagon is:

$ 6\cdot\dfrac{\sqrt{3}}{4}\equal{}\dfrac{3\sqrt{3}}{2}$

Now for the area of the circle.

If we draw the diagonals:

[asy]size(5cm);
draw(unitcircle);
draw(polygon(6));
for(int i=0; i<6; ++i)
for(int j=i+1; j<6; ++j)
draw(dir(60*i)--dir(60*j));[/asy]

We can see that the radius is 1. The area is therefore $ \pi$.
The area is:
$ \boxed{\pi\minus{}\dfrac{3\sqrt{3}}{2}}$

Yep.

I made it up for my class to do, and nobody got it right.

Ya me too! Everyone got it wrong except me.
 :ninja:  :ninja:  :ninja:
I'm a ninja!!!!!!!!


abstract algebra
group theory
superior algebra
superior algebra unsolved

construct an onto homomorphism from D5 to a group of order 2.

-representations and characters of groups

Easy,

Your kernel has to be the cyclic group of order 5 in D5. Map those elements to the identity and the others the only non-identity element.


LaTeX
analytic geometry
graphing lines
slope
geometry

Will the Latex source code for Round 3 be posted, so we don't have to create our own diagrams?  :?

If not, what's an easy way to use Latex to add diagrams to our solutions?

Thanks!  :D

I kind of decoded the diagrams for previous rounds.  It's not that hard.  You'll need to look at them to understand my babble.  Start with a point in put, then in the curly-braces a symbol or a line denoted by (x,y) with slope y/x and then the horizantal lenght in another set of curly-braces.

Although, I am having trouble drawing arcs.  I guess I could just evade the problem.  Like, writing $\theta$=m$\angle ABC$ in the paragraph instead of drawing $\theta$ in the diagram.  But, I like drawing things in for the visual.

You can take a look at the LaTeX tutorial Pictures section by clicking the 'LaTeX' at the right for how to include diagrams in LaTeX.

To make my diagrams, I use the Cabri Geometry II Demo, which you can download for free. It won't let you save or print, so I press PrintScreen and crop it and make it B&W in photoshop.


For AMC 8: Your score is the number of problems you get right.  For example, a perfect score would be 25 (because there are 25 problems total.)  There is no penalty for guessing.

For MathCounts: You get 1 point for each correctly answered Sprint problem, and 2 points for every correctly answered Target problem.  So, a perfect score would be 46.

Hope this helps!  :)

They wouldn't know if you guessed or not. If you're stuck on a question, its better to guess than to leave it blank. You have a higher chance of getting it right. Thats what he means. :wink:

AMC 8:  Numerical score from 0-25;  no penalty for guessing, 25 problems, 40 minutes

MC:  Sprint Round, 30 problems, 40 minutes; each problem worth one point
Target Round;  4 pairs of problems; 6 minutes per pair;  each problem worth two points
No penalty for guessing, but usually it is hard to get a correct answer by guessing(except those really rare multiple choice problems)

Total Score:  46

Team round doesn't count.

The scoring for AMC is mainly 6 points for each correct 0 points for wrong and 1.5 points for skipped

That is the AMC 10 and AMC 12, but for the AMC 8 it is 1 point for each correct and 0 for blank or incorrect.


Does anyone know of a good math or science summer camp in or around Upstate New York? I know there's a good astronomy camp at Alfred University, and Cornell has an engineering program. Also, where in NY can I take the AMC or be involved in other math clubs and programs?

AMC is in some pretty secluded spots, usually.

One group is called 'Albany Area Math Circle' and it meets in a little town called 'Niskayuna'. It is very close to Clifton Park.


geometry
geometric transformation
reflection

What is the point $ (x,y)$ reflected over the

1. x-axis
2. y-axis
3. origin
4. x=y


probability
function
logarithms
calculus
integration
limit
real analysis

Define the sequence $ a_0=1$ and $ a_n=a_{[n/2]}+a_{[n/3]}+a_{[n/6]}$. Study the convergence of $a_n/n$. In afirmative case, compute the limit.

The sequence $a_n/n$ has a quite strange behaviour.
Maybe answer is "no convergence"?

I have no idea on this problem. This sequence is so strange, that I really do not know how to attack it.  :(  :blush:

Another killer problem by harazi... I believe I can show that the limit exists (haven't checked some details yet though). But to find it... Have mercy on my poor soul, harazi! :lol:

No, I have not!  :mad: . You think you're the only one to make peoples mad?  ;) There are hours of estimations for that cruel problem with nested roots and I still cannot prove the damn convergence. Revenge!

Well, Moubi, if you have a solution, then post it.  ;)

I don't have solution, and probability is not my cup of tea

But the probabilist does have one? That would be so nice.

[quote="Moubinool"]A probabilist told this can be solved with probability theory[/quote]
In other words, I guess he have affirmitive answer?

Here is the proof of existence of the limit of $a_n/n$. Define the function $f(x)$ by $f(x)=6x$, $0<x<\frac16$, $f(x)=1$ for $\frac16\leq x< 1$, $f(x)=a_n$ for $n\leq x<n+1$. Then, for $x\geq 1$, we have $f(x)=f(x/2)+f(x/3)+f(x/6)$. Let now $g(y)=f(e^y)e^{-y}$, $y\in\mathbb R$.
Obviously, $g$ is bounded, non-negative and $g(y)e^y$ is non-decreasing. We need to prove that $g(y)$ has a limit as $y\to+\infty$. The equation for $f$ rewrites as $(g*\mu)(y)=0$ for $y\geq 0$ where $\mu=\delta_0-\frac12\delta_{\ln 2}- \frac13\delta_{\ln 3}- \frac16\delta_{\ln 6}$. It is easy to see that the Fourier transform of $\mu$ vanishes only at $0$. So, we can use a version of the Wiener Tauberian theorem to conclude that $g*\psi$ tends to $0$ at $+\infty$ for any $L^1$-function $\psi$ with total integral $0$. This, together with non-decreasing property of $g(y)e^y$ allows to conclude that for any $\ell>0$ the oscillation of $g$ over the interval $[x,x+\ell]$ tends to $0$ as $x\to +\infty$. It remains to note that once $c<g<C$ on some interval of length $\ln 6$, it stays between $c$ and $C$ everywhere to the right of that interval.
So, the existence of the limit can be proved with good old Fourier analysis. I wonder how probability theory helps to find the limit. Can you tell us that, Moubinool?

Through my "ultra-advanced" qualitative analysis (writing out sixty terms of the sequence) it looks as though a_n/n is converging to about 1/2. The functionv seems to have a natural tendency to plateau, and then jump by large values rather than small ones. Who knows.

[quote="Philpoleon"] The functionv seems to have a natural tendency to plateau, and then jump by large values rather than small ones. Who knows.[/quote]
Aww well would you please tell me in plain mathematical terms.By the way did you really compute 60 terms.That requires some tenacity ;)

[quote="Myth"]Really, I computed 10000 terms of $a_n/n$, but I haven't saw any convergence.[/quote]
How did you manage to do it.Thats simply amazing.The behaviour of the sequence is very erratic. :(

it does converges towards. For those who understand french here is an exam where there is a proof !

Aww alekk is it in French but i dont know any french.Is there an English version of the file you have uploaded..

I don't know whether someone will undertake the task of translating the entire exam text into English, but, as a temporary remedy, let me translate the part that relates to the computation of the limit (the proof of its existence can be found in my post #11; the exam uses a slightly different approach but the key idea is more or less the same). I'll also adjust the general statements to our particular case.

2) Put $F(s)=\int_0^\infty f(e^t)e^{-st}\,dt$ ($f$ is the same as in my post #11).

a) Prove that $F(s)=\frac {3-2^{-s}-3^{-s}-6^{-s}}{s(1-2^{-s}-3^{-s}-6^{-s})}$ for $Re s>1$

b) Prove that there exists $\ell\in\mathbb R$ such that $F(s)-\frac\ell{s-1}$ is analytic near $1$

c) (this is not from the exam, but it finishes the problem) Find $\ell$ and show that it must be the same as $\lim_{x\to+\infty}\frac {f(x)}x$ (assuming that the latter limit exists).

The really cute part is a). The rest is routine. Hope this helps... :)

Yeah that really helped.But I cant help saying that that French exam must be very "tough". :D

[quote=Moubinool]A probabilist told this can be solved with probability theory[/quote]

See George Lowther answer with Markov process

[url= https://math.stackexchange.com/questions/487957/how-prove-this-nice-limit-lim-limits-n-to-infty-fraca-nn-frac12-lo] probability approach Markov process [/url]

https://math.stackexchange.com/questions/487957/how-prove-this-nice-limit-lim-limits-n-to-infty-fraca-nn-frac12-lo
————
Here a good article on this sequence
[url= http://www.dtc.umn.edu/~odlyzko/doc/arch/sequence.family.pdf] see [/url]
http://www.dtc.umn.edu/~odlyzko/doc/arch/sequence.family.pdf
Asymptotic behavior of family sequences by Erdös , Hildebrand , Odlyzko,Pudaite, Reznick
Pacific Journal of mathematics volume 126 , number 2 , 1987 , pp 227-241
[url= https://projecteuclid.org/euclid.pjm/1102699802#info] see here [/url]
https://projecteuclid.org/euclid.pjm/1102699802#info

[url= https://projecteuclid.org/download/pdf_1/euclid.pjm/1102699802] see the article [/url]
https://projecteuclid.org/download/pdf_1/euclid.pjm/1102699802

————
Mathematics magazine November 1982 , issue 55, vol 5 ,problem section 300, problem 1158

Solution of problem 1158 in mathematics magazine January 1984 ,issue 57, vol 1, problem section page 49-50

See also
Mathematics magazine January 1985, volume 58 , No1, problem section pp 51-52

——————

ENSET 1991 M’ ( École Normale Supérieure Enseignement Technique de Cachan )
[url= https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzA5Njc5YWI5ZmExMGFkNTBhYTVkOTU0N2Q4Y2QyZTE2ZWVlNDk2LnBkZg==&rn=Q2FjaGFuOTEucGRm] ENSET 1991[/url]
https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNC9mLzA5Njc5YWI5ZmExMGFkNTBhYTVkOTU0N2Q4Y2QyZTE2ZWVlNDk2LnBkZg==&rn=Q2FjaGFuOTEucGRm

—————
Sujet de concours UPS
[url= https://concours-maths-cpge.fr/] sujets concours UPS [/url]

Put

année 1991
concours ENS Cachan
filière MP
matière mathématiques

——-

[url= https://artofproblemsolving.com/community/c7h85493p497642] see [/url]
https://artofproblemsolving.com/community/c7h85493p497642


Prove that gcd(n,k)*C(n,k)/n is an integer for all positive integers n, k.

Answer

We can find integers $ x$ and $ y$ such that $ gcd(n,\,k) = nx+ky$

$ \Longrightarrow \frac{{gcd(n,\,k)}{\binom{n}{k}}}{n}= \frac{(nx+ky)\binom{n}{k}}{n}$
${ = x\binom{n}{k}+\frac{ky}{n}}\binom{n}{k}$
${ = x\binom{n}{k}+\frac{ky}{n}}\left(\frac{n!}{k!(n-k)!}\right)$
$ = x\binom{n}{k}+y\left(\frac{(n-1)!}{(k-1)!(n-k)!}\right)$
$ = x\binom{n}{k}+y\left(\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}\right)$
$ = x\binom{n}{k}+y\binom{n-1}{k-1}\in Z$


logarithms
function
complex numbers
puzzles

Neither $e^z$ nor $\ln z$ are one-to-one functions when dealing with complex numbers, so you can't just blindly invert them.

The problem is when you have

$1 = e^{2z}$ (correct), and you deduce
$\ln 1 = 2z$ (incorrect).

In this particular case $z = i \pi$.  However, $\ln 1 = 0$.

Your reasoning fails because $e^{z} = e^{z + 2 \pi i}$.  You cannot therefore assume that $\ln e^{z} = z$.

First off, $z=i\pi$  (not that it really matters)
I think you're fine up till
(-1)^2=(e^z)^2

1=e^(2z)

ln(1)=2z

It's not that ln(1)=2z, but ln($(-1)^2$)=2ln(-1)=2z.
(yeah, i was beaten to the post :ninja: )

Actually, it's correct with one modification: 2z = ln(1) (mod 2pi*[b][i]i[/b][/i]).

So then 2z = 2pi*[b][i]i[/i][/b]*k

However, we later find that only the odd values of k work.  This fact was "lost" in the squaring of both, since it effectively doubled all exponents.

Choosing the principal log, though, which is always <= 2pi*[b][i]i[/i][/b], we avoid this problem, and end up with z = [b][i]i[/i][/b]*pi.


We have the set A = {1,2,3,...,2002} of positive integers. What is the maximum size of a subset B of A which fullfills these conditions:

1) There are no two consecutive integers in B
2) There is no integer in B that divides some other integer in B

I think the set {2001,1999,1997,...,669,666,664,...,334} with 834 elements is one of those possible sets. So intuitively the answer is 834. Anyone that could check that?


I'm not sure if the date of the School Round is specified by MATHCOUNTS, but I will make a poll.

Our school is taking the test in November.

I don't really know cause the coach is still making the schedule. Seeing from last year, it was late, like around January.

[quote="FantasyLover"]im the only one whos participating in mathcounts so there wont be a school round  :D

CountDown : 8 Posts away from Navier-Stokes Equation[/quote]

You aint the only one...  :P

[quote="isabella2296"]Yeah, we won't have a school round either...officially, there's 8 people in Math Club, because otherwise we can't have a club, but 4 are people I bribed to join(25 cents each!) so we already have a team.[/quote]
Wow, bribery seems to be a pretty good way to get kids to join math club... :P

LOL..lets see...our school had $ 46$ people in mathcounts originally...last week we had like a preliminary round to narrow it down to 20.... :rotfl:

dude, our elementary school last year had an MC program so we could participate unofficially. And, at first the people thought that mathcounts would be hard mulitiplication and division problems! I overheard these girls talking like "i know multiplication so i should get a near perfect score on MC..." i'm like wait and watch. within two weeks there were only 4 people in the club one of which didn't know much math but his dad is a physicist!

I think it's in December, but I'm not worried at all.

[quote="Jongao"]Usually, we get 10 practice problems per week before the school round in January.  Although there is like only 8 smart people in our school, so our team is "predecided." :P[/quote]

wait.

ur a middle schooler, jongao??

Well, everyone who has posted here should be in middle school, or else it's kind of pointless. We have around 100 kids, some do it for fun, others do it cause their parents forced them :P . That's why we have 5 tests to determine the top 16 or 12 and then from their it is reduced to 8. That is also why math is not environmentally friendly.

[quote="shentang"]That is also why math is not environmentally friendly.[/quote]

I would say that copying the same social studies test for 200+ kids is hardly more environmentally friendly. :P

January 28 for me (Which happens to be my friend's birthday...) :D
Or rather, January 28 for the middle school students :P

[quote="gauss1181"][quote="ernie"]My MC teacher says I don't have to do state, cuz I "automatically" made the team.  The other ppl have to take it though. :P[/quote]
HOW THE HECK IS THAT POSSIBLE?[/quote]

cuz... the ppl at my school think im beast just bcuz i made state xD

[quote="ernie"]My MC teacher says I don't have to do [b]state[/b], [...][/quote]

Such is the cause of gauss1181's amazement.

Oh lol.  I meant school.

I wish I could skip state, and go straight to nats though.  :D

Never mind, my school tryout is on December 4. I think, unless my coach isn't saving the school tryout for the last tryout.


abstract algebra
superior algebra
superior algebra unsolved

Hello,

Let G be a group which admits an element of finite order >1, and which has exactly two conjugacy classes.
Prove that G is cyclic of order 2.

(Note that if we assume G to be finite the problem is trivial)

@+

All the elements have the same order, which is prime. If this prime is $2$, we're done (the group turns out to be abelian, and the conclusion follows immediately), so assume that this prime $p$ is $\ge 3$.

Let $a$ be an element of this group. There is another element $u$ s.t. $uau^{-1}=a^2\ (*)$, because $p>2\Rightarrow a^2\ne 1$. Multiply $(*)$ to the left with $u$ and to the right with $u^{-1}\ p-1$ more times to get $a=a^{2^p}$, so $a^{2^p-1}=1$, which means that $p|2^p-1$, and it's clear from here.

Thanks Grobber, this time I had the same proof  :D


inequalities
LaTeX
algebra proposed
algebra

Let $a,b,c \in \mathbb{C}$, $a \neq b \neq c \neq a$, $|a|=|b|=|c|=1$.
$S_{m}=|a+b|^m+|b+c|^m+|c+a|^m$. Prove that if there is $p \in \mathbb{N}$ s.t. $S_{2^{^p}} \leq 3$, then $\forall n \in \mathbb{Z}$, $n \geq 0$, we have $S_{n}=3$

It smells like inequalities here.I have an idea.I'm thinking at Holder inequality.
 by the way,is this your proposal?

It is a problem proposed for RMO a few years ago, i think (i don't know by who).

[quote="cezar lupu"]Here is my proof: By Holder,we have: x^2^m+y^2^m+z^2^m>=((x^2+y^2+z^2)^2^(m-1))/3^2^(m-1)-1.
 We take x=|b+c|,y=|c+a| and z=|a+b|. We deduce that S(2^m)>=S(2) ^(2^(m-1)). By the identity
  |a+b|^2+|b+c|^2+|c+a|^2=|a|^2+|b|^2+|c|^+|a+b+c|^2 we have S(2)=2+|a+b+c|^2>=3 and because
 S(2^m)>=3 it follows that S(2^p)=3 and a+b+c=0. :P[/quote]

do you know latex? if you don't you should try and learn it :). I can't understand the powers in Holder's ineq.

anyway your proof is really nice and short compared to mine.

and what about this: $|a+b|^2+|b+c|^2+|c+a|^2=|a|^2+|b|^2+|c|^2+|a+b+c|^2$
it's the first time i see it and i can't manage to prove it. give some clues plz


Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run $ 100$ meters. They next meet after Sally has run $ 150$ meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

$ \textbf{(A)}\ 250 \qquad \textbf{(B)}\ 300 \qquad \textbf{(C)}\ 350 \qquad \textbf{(D)}\  400\qquad \textbf{(E)}\ 500$

Solution here
[hide]let the circumference of the circle be C, the distance Sally ran before the first meeting be S, the distance Brenda ran between the first and second meeting be B.
100+S=C/2
B+150=C
150+B=200+2S

we also know that the ratio of distance ran would be constant if their speed was constant, so S/100=150/B
Solving for that, we get S=15000/B
substituting that into the equation for S, we get:
150+B=200+(30000/B)
150B+B^2=200B+30000
B^2-50B-30000=0

using the quadratic equation, we get
B=200 or B=-150; the negative answer is clearly extraneous.

B+150=C
200+150=C
C=350
Therefore the length of the track is 350.
[/hide]

i did it a different way

[hide]well say it takes B and Sally  t  time to cover the whole track, since they started at opposite sides, they run only half the tract so in t/2 time, Brenda runes 100, so in t she will run 200, then the next time they meet, they've covered the whole track, so in t, sally runs 150, 150 + 200 = 350 -> C[/hide]

edit: wait lol i already did this question
wow i have bad memory hahha

If people searched for 'Brenda & Sally', this same exact question was asked at least 5 times...


[quote="Fraenkel"]A piece of wire 20 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
a) How much wire should be used for the square in order to maximize the total area?
b) How much wire should be used for the square in order to minimize the total area?[/quote]
Are there more conditions? If not,

After taking any length of wire and forming it into a square, it will always be possible to create a equilateral triangle from the remaining piece. So to maximize the area of the square we need the largest possible number less than 20. Such a number does not exist as you can keep getting closer and closer to 20, without reaching it.

Same thing about minimizing the area of the square, the wire length have to be as close to 0 as possible, and again there isn't such a number.

By total area he means the sum of the areas of the triangle and the square.

[hide="hint"]

Set as $ 3x$ the lengh of the wire for the triangle (I think it is easier than setting $ 4x$ for the lenghto of the wire for the square) and expres the total area in terms of $ x$. Remeber that a cuadratica function like $ ax^2\plus{}bx\plus{}c, a\ne0$, takes its minimun(or maximun) value in $ \minus{}\frac{b}{2a}$.

It would have a minimun if $ a>0$ or a maximun if $ a<0$, this because:

 $ ax^2\plus{}bx\plus{}c\equal{}a(x\plus{}\frac{b}{2a})^2\plus{}\frac{4ac\minus{}b^2}{4a}$
[/hide]

 :)


A magazine advertises its $ \$12$ annual subscription price as $ 70\%$ off the newsstand price. What is the number of dollars per year a reader would spend if the magazine were purchased at the newsstand price?

[quote="jjx1"]It should be $ 40. 12 / 0.3\equal{}$40[/quote]

That was negative infinite % helpful!!! >=<

Let the newsstand price be $x$.
We notice that $70\%$ off the newsstand price means that the subscription price is $30\%$ of the newsstand price.
We can set up an equation:

$0.3x=12$ ($30\%$ of newsstand price equals $\$12$).

$x=\frac{12}{0.3}=40$


function
trigonometry
real analysis
real analysis unsolved

an easy question i asked for HW yesterday... :D

given

$ \Gamma(1\plus{}z)\cdot\Gamma(1\minus{}z)\;\equal{}\;\frac{\pi z}{\sin\, (\pi\,z)}$

now show

$ \bigg|\,(\imath\,z)\,!\,\bigg|^{2}\;\equal{}\;\frac{\pi z}{\sinh\, (\pi\,z)}$

( where $ \imath\;\equal{}\;\sqrt{\;\minus{}1\;}$ )

$ |\Gamma(1\plus{}iz)|^{2}\equal{}\Gamma(1\plus{}iz)\,\overline{\Gamma(1\plus{}iz)}\equal{}\Gamma(1\plus{}iz)\,\Gamma(1\minus{}iz) \equal{}\frac{\pi i z}{\sin\pi i z}\equal{}\frac{i\;\pi z}{i\;\sinh\pi z}$


trigonometry quadratics algebra polynomial AMC

Are we supposed to find the actual numerical value of a?

[quote="darkquantum"]Suppose that sinA and cosA are the roots of
ax^2+(2a-2)x+2-a=0.
If 0 :le: A :le: 2 :pi: ,

Find a and A.[/quote]
I will call sin A =s and cos A = c because I'm that lazy. we know that s+c=(2-2a)/a, and sc=(2-a)/a. Subtracting the twice the second equation from the square of the first, we find that s:^2:+c:^2:=4(a:^2-2a+1)/a:^2:-2(2-a)/a. Multiplying through by a:^2: and noting that the s:^2:+c:^2:=1, we find a:^2:=4a:^2:-8a+4-4a+2a:^2:, or 5a:^2:-12a+4-0. This factors into (5a-2)(a-2)=0, or a=2/5 or 2. Plugging each of these back into the original equation in turn, we get the two possible equations

a=2 gives 4x:^2:+2x=0, which has roots x=0, -1/2. This cannot work because the sum of the squares of the roots is not one.

a=2/5 gives 4x:^2:/25-6x/5+8/5=0. Multiplying through by 25/2, we get the quadratic 2x:^2:-15x+20=0. The quadratic equation gives x=(15:pm::sqrt:(225-160))/4=(15:pm::sqrt:65)/4. Hmmm i dont think the sum of the squares of those either is 1. HMMMMMMMM........

well, anyways, i think that's the general idea....

a*sinA^2+(2a-2)*sinA+(2-a)=0, a*cosA^2+(2a-2)*cosA+(2-a)=0, since they're the roots. You add and get a+2-a+(2a-2)(sinA+cosA)=2+(2a-2)(sinA+cosA)=0, hence (2a-2)(sinA+cosA)=-2. In the polynomial, 2a-2 is the negative of the sum of the roots. So (2-2a)^2=2. a=1-1/sqrt(2). It's trivial to find A.


Mrs. Riley recorded this information from a recent test taken by all of her students.  Using the data, what was the average percent score for these $ 100$ students?

\begin{tabular}{|c|c|}
\multicolumn{2}{c}{}\\\hline
\textbf{$ \%$ Score}&\textbf{Number of Students}\\\hline
100&7\\\hline
90&18\\\hline
80&35\\\hline
70&25\\\hline
60&10\\\hline
50&3\\\hline
40&2\\\hline
\end{tabular}

[quote="GameBot"]Mrs. Riley recorded this information from a recent test taken by all of her students.  Using the data, what was the average percent score for these $ 100$ students?

$ \begin{tabular}{|c|c|}
\multicolumn{2}{c}{}\\\hline
\textbf{$\%${ Score}\&\textbf{Number of Students}\\\hline
100\&7\\\hline
90\&18\\\hline
80\&35\\\hline
70\&25\\\hline
60\&10\\\hline
50\&3\\\hline
40\&2\\\hline
\end{tabular}$[/quote]

Hmm...

[quote="GameBot"]Mrs. Riley recorded this information from a recent test taken by all of her students.  Using the data, what was the average percent score for these $ 100$ students?

$ \begin{tabular}{|c|c|}
\multicolumn{2}{c}{}\\\hline
\textbf{\% Score}&\textbf{Number of Students}\\\hline
100&7\\\hline
90&18\\\hline
80&35\\\hline
70&25\\\hline
60&10\\\hline
50&3\\\hline
40&2\\\hline
\end{tabular}$[/quote]

since i g2g soon, I'll let someone explain the computation here: \[ \frac{700\plus{}1620\plus{}2800\plus{}1750\plus{}600\plus{}150\plus{}80}{100}\equal{}\boxed{77}\]

Is there a trick to get this fast? It took me too long to get any points.


Show that the sequence defined by : $ a_1\equal{}0$ , $ a_2\equal{}1/2$ , $ a_{n\plus{}1}\equal{}1/3(1\plus{}a_n\plus{}a_{n\minus{}1}^3)$ converges and determine it's limit.

The mapping $ F(x,y)\equal{}\left(y,\frac{1\plus{}y\plus{}x^3}3\right)$ is a contraction from $ [0,a]^2$ to itself with the distance given by $ \max\{|x\minus{}x'|, b|y\minus{}y'|\}$ where $ a$ is slightly less than $ \sqrt{\frac 23}$ and $ b$ is slightly greater than $ 1$ (to be chosen in this order). The equation $ 1\plus{}x\plus{}x^3\equal{}3x$ has only one root in the interval $ [0,a]$, which is $ \frac{\sqrt 5\minus{}1}2$. The rest should be clear.


geometry
conics
ellipse
3D geometry
sphere
incenter

an ellipse is a generalization for circles, what about more generalization, ie constant sum from 3 points? where can I find information about it?

[quote="yagaron"]an ellipse is not a generalization of circles....

a circle is a special case of ellipses. but this does not means that ellipses are generalizations.

a n-sphere is a generalization of circles and spheres.

and what do you mean "constant sum from 3 points" ?[/quote]

I think he meant in the same sense that an ellipse is the locus of all points such that the sum of distances from the two loci is constant.

I'm pretty sure the 3-dimensional analogue is the [url=http://mathworld.wolfram.com/Ellipsoid.html]ellipsoid[/url]; but I'm not sure.

${A_{n}=\left\{\alpha_{n},\beta_{n}\right\}}$
${P=\{x,y\}}$

${\sqrt{\left(P-A_{1}\right).\left(P-A_{1}\right)}+\sqrt{\left(P-A_{2}\right).\left(P-A_{2}\right)}+\sqrt{\left(P-A_{3}\right).\left(P-A_{3}\right)}=K}$
${\sqrt{\left(x-\alpha_{1}\right)^{2}+\left(y-\beta_{2}\right)^{2}}+\sqrt{\left(x-\alpha_{2}\right)^{2}+\left(y-\beta_{2}\right)^{2}}+\sqrt{\left(x-\alpha_{3}\right)^{2}+\left(y-\beta_{2}\right)^{2}}=K}$

You cannot make a line model like ellipses. with 2 focus in a horizontal or vertical to simplify things.

but it does not have nothing especial

err.... It looks normal...

Try substituing plus operator to a multiplication, like

${\sqrt{(a-x)^{2}+(b-y)^{2}}\sqrt{(c-x)^{2}+(d-y)^{2}}\sqrt{(e-x)^{2}+(f-y)^{2}}=k}$

the result is much more beautiful.

If $k$ is small the result is 3 little quasi-ellipses in countour of the 3 points.

But a tip:

If we have 3 points, we have a triangle. lets theses 3 points be the focus of our new equation. Let a,b,c be the distance from the points to the equation. so we can make

$a+b+c =K$
$a b c =K$

OR my favorite

$a^{2}=b c$, $b^{2}=a c$, $c^{2}=b a$

These 3 equations are ellipses that passes througth 2 points of the triangle, the incenter and on excenter.

see ya


If the expression $ \begin{pmatrix}a & c \\
d & b \end{pmatrix}$ has the value $ ab\minus{}cd$ for all values of $a, b, c$ and $d$, then the equation $ \begin{pmatrix}2x & 1 \\
x & x \end{pmatrix} = 3$:

$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x \qquad\\
\textbf{(B)}\ \text{Is satisified for only 2 values of }x \qquad\\
\textbf{(C)}\ \text{Is satisified for no values of }x \qquad\\
\textbf{(D)}\ \text{Is satisfied for an infinite number of values of }x \qquad\\
\textbf{(E)}\ \text{None of these.}$

[hide]$ 2x^2\minus{}x\equal{}3\implies 2x^2\minus{}x\minus{}3\equal{}0\implies x\equal{}1.5, \minus{}1\implies\boxed{B}$[/hide]


I don't know if it is rude to open a topic like this  :blush: but as an observer who participate in this IMO as well as the score coordination, I think the coordination this year is very good. It seems that the coordinators work hard on investigating the substantial of the problems, preparing a detailed marking scheme, reading the solutions before coordination, checking essential steps and so. And they made the coordination no bargaining.

I think all the contestants of our team get the scores they deserved and in my opinion the coordination this year is very nice, even comparing to the past (this is my fourth time in IMO but second time as an observer, so I don't know much in fact  :blush: ). Is it the case for all participants?

By the way, was anything about this discussed at the last jury meeting?

according to my leader, the last jury meeting ends quickly. the jury meeting only decides on pakistan's team's problem, and that's the only important thing. and after that, the submission of information of IMO 2009 to IMO website.

[quote="wangsacl"]the jury meeting only decides on pakistan's team's problem, and that's the only important thing. [/quote]

What was the problem?

The Pakistan's problem with the marking scheme is they're debating a point over number 3. The Jury decides not to give them one point, because many countries got the same results with the Pakistan's team (maybe something like showing that both the sequence have a common difference, or sth like that, I forget  :blush: ) and get 0 point, since it is on the marking scheme.

In the marking scheme, there was one point for defining that the sequence $ s_{i\plus{}1} \minus{} s_i$ has a maximum (or a minimum) value. As many contestants (three french  :wink: ), a pakistanese one proved that the sequence $ s_{i\plus{}1} \minus{} s_i$ is bounded. As it is a sequence of integers, the desired result follows easily, but not mentioning the word maximum or minimum gave no point  :wink:
We may not agree with that way of marking, but this was the same for everyone. A silver french medal turned into bronze...

Pierre.


See the main thread for the rules. An additional rule: each person may only post once. Go.

[size=150]Deadline: Wednesday, November 12 at 7:30 PM EST[/size]

PLEASE EVERY1 POST AND DISREGARD THE EXEMPTION.

Bleh.  Another failure.  How disappointing.

Indeed. cf249 posted 4 minutes late, and Brut3Forc3 and jhredsox93 did not post at all. That is quite unfortunate. No money.

Thank you super losers who care about exemptions. I am 90% sure jh is mole now! Or exceedingly stupid, a distinct possibility.

YOU IDIOTS!
IT'S OBVIOUS: BRUTEFORCE IS[size=0]not[/size]THE MOLE

OMGUS ITS U PACMAN
Or maybe it's me! For the next quiz, put me for all the answers!!! :D

Rofl. Either jh is great mole or stupid. Probably mole becuz he makes me think hes stupid, but hes a halfway decent survivor mod.

jhredsox93 can not view this thread, can he?

He can, and we obviously didnt expect him to until after it had 5 replies  :wink:

actually, i'm really sorry lol, i forgot i had to post here