Reid, Tiankai, Gabriel, Oaz, Dong, and Ian are praticing and they are working on the proofs to 40 questions. Each of the team members must solve at least 1 question, except Reid, who will solve at least 2. How many ways can the 6 IMO team members answer the proofs in this manner?

Billy

[hide]I'm going to take a leaf out of biffanddoc's book and see if a recently learned formula can be correctly applied by me:

OK, so there are 7 given problems that must be correctly solved. Subtract 7 from 40 to get 33. So the other thirty-three can be solved by 6 different people. Call each problem X, because it doesn't matter whether or not Gabriel solves problem #'s 1&2 or 39&40. Line up all 33 X's (don't actually do this, this is just to help explain why the formula works). To to try to dish out problems to each person, you could use five dividers and place them wherever you want in the X's. For example, if there were five X's and two dividers, and l = a divider, llXXXXX would mean 0 extra problems solved by the person to the left of the first divider, 0 extra problems solved by the person between the two dividers, and 5 problems solved by the person to the right of the divider If you have 33 X's, you have 38 spots (33+5) to put em in and choose 5 of them, so the answer is

38 Choose 5 = 501,942.

BTW I need some help understanding my own reasoning lol. [/hide]

Could someone help explain fanatic's reasoning a little? I'm kinda confused.

Could someone help ME with my reasoning? I'm a little confused.

(No sarcasm)

*sigh*
Are these people on the US IMO team?

There are 40 problems.
We know that the 5 besides Reid will solve 1 each at least.
Reid will solve at least 2.
That leaves 33 left to be distributed among the 6 people.

Here's where the tricky part is.

Say that each of the Os are the problems.
We then have
OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO (33 Os)

We need to divide them between the 6 people.
If we insert | s, as dividers, we can do so. The dividers can go anywhere. The dividers show where the problems stop for one person and go to the next.
A possible arrangement is:
||||||OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

Reid has the problems on the left of the first divider. That is none.
Tiankai has the ones between the first and second divider. THat is none.
Gabriel has those between the 2nd and 3rd divider. That is none.
Oaz has those between the 3rd and 4th divider. That is none.
Dong has those between the 4th and 5th divider. That is none.
Ian has those to the right of the 5th divider. That is 33

We now have 38 objects (5 dividers/33 problems) and we can put the 5 dividers in any of the spots.
Thus the answer is $\binom{38}{5}$


geometry
geometric transformation
reflection
similar triangles
angle bisector
geometry solved

Kindly prove that the Lemoine point of a right triangle is
    the mid-point of altitude to the hypotenuse.

    Col. Pestich.

Isn't this a trivial application of trilinears?

Why do that when there are tons of nice solutions? :)

For instance (I'm sure it can also be solved without any fancy theorems), The Lemoine point is also the intersection of all the lines connecting the midpoint of a side to the midpoint of the corresponding altitude. This is clearly the midpoint of the altitude corresponding to the hypothenuse when the triangle is right-angled.

[quote="pestich"]Kindly prove that the Lemoine point of a right triangle is
    the mid-point of altitude to the hypotenuse.[/quote]

I am sure that a synthetic proof of this is given in Honsberger's "Episodes in 19th and 20th Century Euclidean Geometry", but I don't have the book at hand right now, so I post my own proof (probably it's the same one as in Honsberger's book ;) ):

If ABC is a right-angled triangle with < BCA = 90°, and if H is the foot of the altitude from the vertex C to the hypotenuse AB, then we have to show that the symmedian point of triangle ABC is the midpoint of the segment CH.

Call S the midpoint of the segment CH, and call K the symmedian point of triangle ABC. Then, we want to show that S = K.

In fact, let M be the midpoint of the side BC of triangle ABC. Then, on the one hand, < BCA = < CHA (since < BCA = 90° and < CHA = 90°); on the other hand, < BAC = < CAH. Thus, the triangles BCA and CHA are similar. Corresponding points in similar triangles form equal angles; since the points M and S are corresponding points in the similar triangles BCA and CHA (in fact, the point M is the midpoint of the side BC of the first triangle, whereas the point S is the midpoint of the side CH of the latter one), it follows that < BAM = < CAS.

On the other hand, since the point K is the symmedian point of triangle ABC, the line AK is the A-symmedian of triangle ABC. Hence, the line AK is the reflection of the A-median of triangle ABC in the angle bisector of the angle CAB. But the A-median of triangle ABC is just the line AM (since the point M is the midpoint of the segment BC); hence, the line AK is the reflection of the line AM in the angle bisector of the angle CAB. This leads to < BAM = < CAK. Together with < BAM = < CAS, this implies < CAK = < CAS, and thus, the point S lies on the line AK. Similarly, we see that the point S lies on the line BK. Hence, the point S is the point of intersection of the lines AK and BK; this immediately yields S = K, what proves our assertion.

Yes, Grobber's proof is also nice, I am just posting this one since it seems most elementary to me.

[b]PS.[/b] Pestich - your way of naming problems is cool, but I have included some data in the "source" field so that one can better identify the problem.

  Darij

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/01.%20Problemes%20sur%20le%20point%20de%20Lemoine.pdf
p. 12…

Sincerely
Jean-Louis


Let $a,b,c>0$ such that $abc=1$.Prove that:

         $\frac{1}{2+a+ab}+\frac{1}{2+b+bc}+\frac{1}{2+c+ca}\leq\frac{3}{4}$ ;)

It is classical trick, isn't it? I mean, $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$? ;)

Yes,indeed,but have you solved it completely? ;)

I'm not sure,but I think there is a trigonometrical solution. :?  :huh:

Cezar! I usually do solve problems completely!  ;) . Unfortunately, it solves itself after the substitutions. There is almost nothing to prove. Sorry.

Cezar did you use Cebasev towards proving it??? ;)

Socrates,I don't think so.I haven't tried it yet.Did you? ;)

Yes and after making the substitution $\displaystyle a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$ , doing the calculations and setting $yz=k, xz=l \ and\ xy=m$ I got
$\sum\frac{k}{2k+l+m}\leq\frac{3}{4}$

then I Tried Cebasev..... ;)  ;)  :)

By the way,in my solution I reduced the problem to Nesbitt's inequality. ;)

By the way, could you tell me the main steps of your solution?? :)

The ineq is really nice :lol:  :lol:

After the substitution that Harazi was talking about the inequality reduces to:

             $\frac{x}{2x+y+z}+\frac{y}{z+x+2y}+\frac{z}{2z+x+y}\leq\frac{3}{4}$

 This inequality is equivalent with Nesbitt.I think it is clear now,or isn't it? ;)  :?

Yes you are right

I just posed $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{y}$ and reached the same.....

However, is under these circumstances my solution correct??? :)

[quote="socrates"]Assume $k\geq l\geq m>0$ then $\frac{k}{2k+l+m}-\frac{l}{2l+k+m}=\frac{(k-l)(k+l+m)}{(2k+l+m)(2l+k+m)}\geq0$

so $\sum(2k+l+m)\cdot \sum\frac{k}{2k+l+m}\leq3\cdot (k+l+m)$ and the conclusion follows!! :)

I was quite brief so....

Please check it! I am not sure!! ;)[/quote]

 If the calculations in the first part are right,well,I thin it is indeed a nice solution. ;)

remember $ \sum \frac{1}{1+a+ab}\ =1 $ holds for $ abc=1 $ then use this one:
$ \sum \ (\frac{1}{3}+\frac{1}{1+a+ab}+ \frac{1}{1+a+ab} )\geq\sum\frac{9}{4+2ab+2a} $ then we easily have the result


I think I have it.

$ \overrightarrow{OA}$  X  $ \overrightarrow{OB}=\overrightarrow k$

$ \frac{1}{2}$$ \overrightarrow{OA}$ X $ \overrightarrow{OB})=\frac{\overrightarrow k}{2}$

$ Area(\triangle OAB)=|\frac{\overrightarrow k}{2}|$

Now,$ \overrightarrow{OA}$ is a constant vector and  $ \overrightarrow{OB}$  being a variable vector,the area of the triangle will be a constant only if B moves on a line parallel to  $ \overrightarrow{OA}$.

Thus locus of B is a line parallel to $ \overrightarrow{OA}$.

a direct would be to take OA along the x axis and the angle between them as 8 (theta)

we directly get rsin8= constant => line parallel to x ..

[quote="pankajsinha"]If the vector product of a constant vector $ \overrightarrow{OA}$ with a variable vector $ \overrightarrow{OB}$ in a fixed plane $ OAB$ be a constant vector,then find locus of B.[/quote]

Let variable vector $ \ \overrightarrow{OB}\equal{}\overrightarrow{r}$, known vector $ \ \overrightarrow{OA}\equal{}\overrightarrow{a}\ne\overrightarrow{0}$ and their product $ \ \overrightarrow{r}\times \overrightarrow{a}\equal{}\overrightarrow{p}$

$ \implies \overrightarrow{a}\times\left(\overrightarrow{r}\times \overrightarrow{a}\right)\equal{}\overrightarrow{a}\times\overrightarrow{p}$

$ \implies \left(\overrightarrow{a}\cdot\overrightarrow{a}\right)\overrightarrow{r}\minus{}\left(\overrightarrow{r}\cdot \overrightarrow{a}\right)\overrightarrow{a}\equal{}\overrightarrow{a}\times\overrightarrow{p}$

Let $ \ \overrightarrow{r}\cdot \overrightarrow{a}\equal{}\lambda$, a variable scalar.

$ \implies\boxed{\boxed{ \overrightarrow{r}\equal{}\frac {1}{|\overrightarrow{a}|^2}\left(\overrightarrow{a}\times\overrightarrow{p}\plus{}\lambda\overrightarrow{a}\right)}}$

$ \implies$  locus of B is a line parallel to $ \ \overrightarrow{OA}.$


Este convergent sirul dat de recurenta: $ x_{n+1}=\frac{x_n ln(n+2)} {1+n x_n^2}, n>0, x_0>0 $ ?

[quote="infodata"]Este convergent sirul dat de recurenta: $x_{n+1}=\dfrac{x_n ln(n+2)} {1+n x_n^2}, n>0, x_0>0$ ?[/quote]

se arata imediaat ca $x_n>0$ prin inductie. in plus, avem ca $0 < x_{n+1} \leq  \dfrac { ln (n+2) } { 2 \sqrt n }$, de unde rezulta imediat ca $x_n$ converge la zero (am folosit ca $1+n x_n^2 \geq  2 \sqrt n x_n$). solutie data de prof. marius damian.

Gasiti un sir $y_n\rightarrow \infty$ pentru care $y_n\cdot x_n\rightarrow 1$ (sirul $y_n$ "dizolva" sirul $x_n$).


[quote="jmerry"]Currently, any new integral with limits, such as $\int\limits_0^1f(t)\,dt$ displays with the limits directly above and directly below the integral sign.
I noticed this while composing a long post on a differential equation- it's distracting.

Is this connected to the other issues of today?[/quote]Let's see now ... $\int_0^1 f(t) dt$

I typed exactly the same thing you did, but the code was changed after the fact to \int\limits instead of \int. This didn't show up in preview.

$\int_0^1 f(t)\,dt$

Well, no problem now.


You have 5x ≡ 4 (mod 3) and you want to find a value for x. You can say that 5x ≡ 1 (mod 3) and just keep adding or substracting 3 until you get a multiple of 5. Example :
1 + 3 = 4 + 3 = 7 + 3 = 10. Well 10 / 5 is an integer so you get x = 2. If there any other method of solving for a value of x?

yes, there is a good theorem for it: ax = b mod m => sometimes there is an y so that ay=1 (we call y the inverse of a mod m) and then x=yb is the requested number.

Alright thanks. I fell pretty dumb now. That was actually in solution but was indented so I skipped that line. Thanks again.


pigeonhole principle
combinatorics unsolved
combinatorics

We have $n^2$ balls and $n$ boxes . Draw $n^2$ balls by $n$ colours. Prove that we can take $n^2$ ball in $n$ box such as one box have $n$ balls and this box have most 2 colours of ball.

If I understood the problem correctly, we're given $n^2$ balls, each of which is colored in one of $n$ colors, and we must put the balls in $n$ boxes, $n$ in each box, so that each box contains no more than two distinct colors.

We'll prove it for $kn$ balls, $k$ colors and $k$ boxes, with $n$ balls in each box, by induction. If $k=1$, the result is obvious. Suppose it is true for $k-1$ and take the $kn$ balls. Since there are $k$ colors, by pigeonhole there must be a color for which there are at least $n$ balls (call it A) and a color for which there are at most $n$ balls (call it B). Take box 1 and put all the B-colored balls in it. Then put A-colored balls in, till there are $n$ balls (obviously we can do this since we have at least $n$ A-colored balls available). Box 1 is done and color B is finished; we're left with $(k-1)n$ balls, $k-1$ colors, $k-1$ boxes and we require $n$ balls in each of the other boxes. This is the $k-1$ case, so we're done.


topology
real analysis
real analysis unsolved

Prove that in a metric space every open set is the union of countable no. of closed sets & every closed set is the intersection of a countable no of open sets.

Let A be a closed subset of R & C be a compact subset of R then prove that the set A+C={a+c:a belonging to A & c belonging to C}is closed

For the first one, in order to show that every closed set $F$ is a $G_{\delta}$ set, take $G_n$ to be the union of the balls centered at the points of $F$ of radius $\frac 1n$.

For the second one, assume we have a sequence $a_n+c_n\to x$. Since $C$ is compact, there is a convergent subsequence $c_{k_n}\to c\in C$, which means that $a_{k_n}$ is also convergent. If $a$ is its limit, we have $a\in A$ and $x=a+c\in A+C$.

[quote="grobber"] $G_{\delta}$ set, .[/quote]
I am sure i knew about $G_{\delta}$ set,i mean I heard about it.Can you please tell me more about it.


geometry
rectangle
trigonometry
analytic geometry

Three squares in a row. Draw a diagonal through the rectangle formed by two squares and a crossing diagonal through the rectangle formed by three squares. Show WITHOUT USING TRIG that the acute angle formed is 45 degrees

PS: How to construct a diagram in this forum (using some commands and coordinates system???)?

It is very easy with trig, but here is with coordinates.
The intersection is (6/5,3/5)
Drop a perp from (0,1) to y=1/2x (the line going across two squares), the intersection is (2/5,1/5)
It is now easy to show using the distance formula the triangle with the three points I've typed so far is iscoceles right.

anaylitic geometry is a good solution but are there any others cause i always think its nice to know any non anylitic geometry proofs just cause they tend to be less work and more insight

[geogebra]1435edc514ec10dd78d125f144b5d018d3f7446c[/geogebra]

Proof without words.  :)


Unfortunately no one on this forum made nats, so I think this thread'll be forgotten.

[quote="mihail911"]So, state mathcounts was yesterday. i managed to get 3rd individually (tied for 2nd, lost tiebreaker), and some crazy 6th grader took first. And no one from the south made nats. You crazy northerners  :D . I'm guessing albert wu made it for sure.
btw, is it just me or are the state trophies really "cheap"? I mean, I like the chapter ones better. The trophies are n't in the traditional "30-60-90 triangle style"
good job north and make us proud at nats!!!!![/quote]

when they gave me my state trophy it was broken
as in, one of those thingies that holds the spinny part on was missing

haha i dropped my states trophy while going back to my car and it broke

do all the states have this same cheap trophy style?
i like the traditional one more.

I wouldn't know, never having seen a state trophy, never having made it to states :(

What do they look like?

[quote="pythag011"][quote="mihail911"]So, state mathcounts was yesterday. i managed to get 3rd individually (tied for 2nd, lost tiebreaker), and some [b]crazy 6th grader took first[/b]. And no one from the south made nats. You crazy northerners  :D . I'm guessing albert wu made it for sure.
btw, is it just me or are the state trophies really "cheap"? I mean, I like the chapter ones better. The trophies are n't in the traditional "30-60-90 triangle style"
good job north and make us proud at nats!!!!![/quote]

crazy 6th grader(Me)????[/quote]
lol. what was your individual score?

lol. completely forgot. well, i mean, considering i saw this person and stuff. yeah, ok, my mistake.  :lol:

[quote="Myself"]Back in the day, they gave out triangular trophies at California states, but I believe in 2004 they switched over to the plastic turning ones. I don't know about other states though.[/quote]

BOOOOOO...that was my year :(.

[quote="Myself"]Back in the day, they gave out triangular trophies at California states, but I believe in 2004 they switched over to the plastic turning ones. I don't know about other states though.[/quote]

They did in 2004? I remember seeing them 2 years ago.


superior algebra
superior algebra unsolved

Prove that $ (\mathbb{R} \minus{} \{0\},\cdot)$ is not isomorphic to $ (\mathbb{R}, \plus{} )$

First, convince yourself that $ 1$ must map to $ 0$.  Now how many solutions are there to $ x\cdot x\equal{}1$?


inequalities
LaTeX
inequalities proposed

So here is a very easy problem:

Prove:
The inequality
$\sqrt{a^2+(1-b)^2} + \sqrt{b^2+(1-c)^2} + \sqrt{c^2+(1-a)^2} \geq \frac{3 \sqrt 2}{2}$
holds for arbitrary real numbers a,b

[b]Modified by Orl[/b].

To Pierre or any other moderator; Please make this problem $ \text{\LaTeX} $ readable. Well here's my solution
[hide]
AM-QM inequality:
${\sqrt {\frac{a^2 +(1-b)^2}{2}}  + \sqrt{\frac{b^2 + (1-c)^2}{2}} + \sqrt\frac{c^2 + (1-a)^2}{2}} \geq \frac{a+1-b}{2} + \frac{b+1-c}{2} + \frac{c+1-a}{2} = \frac{3}{2}$
So the inequality above holds for abitrary real numbers a,b[/hide]

Let $\vec x=(a,1-b),\vec y=(b,1-c),\vec z=(c,1-a)$.
By using triangular inequality,we have $|\vec x|+|\vec y|+|\vec z|\geqq|\vec x+\vec y+\vec z|$,so we obtain

RHS=${\sqrt{(a+b+c)^2+\{3-(a+b+c)}\}^2}=\sqrt{2\left(a+b+c-\frac{3}{2}\right)^2+\frac{9}{2}}\geqq\frac{3}{\sqrt{2}}$.


Hello,

prove that $ 2^m \plus{}1$ is not divisible by $ 2^n \minus{}1$, where $ n>2$ and both $ m$ and $ n$ are natural numbers.

[hide="A try"]$ \frac {2^m \plus{} 1}{2^n \minus{} 1}\in\mathbb{N}$

We have that the numerator has to be greater or equal than the denominator:

$ 2^m \plus{} 1 \ge 2^n \minus{} 1$, which is $ m\ge n$.

So, $ m \equal{} n \plus{} k$ ($ k\ge 0 \in \mathbb{N}$). Replacing:

$ 2^k \plus{} \frac {2^k \plus{} 1}{2^n \minus{} 1}\in\mathbb{N}$. So, here we are at the starting point (now with $ k$ in place of $ m$)...
So $ n$ is bigger than any number. There's no $ n\in\mathbb{N}$ which satisfy the condition.
[/hide]

Let me know if it is a valid proof... :)

Hello

Well, to me honest, I cant seem to understand your proof, sorry.  :blush:

This is what I've done:

Lets assume the opposite, let $ \frac{2^{m}\plus{}1}{2^{n}\minus{}1} \equal{}k\in\mathbb{N}$.

Now we can expand this:

$ \frac{2^{m}\plus{}1}{2^{n}\minus{}1} \plus{} 1\equal{}k\plus{}1$

From there we have $ \frac{2^{m}\plus{}1\plus{}2^n\minus{}1}{2^{n}\minus{}1}\equal{}k\plus{}1$

Finally, $ \frac{2^{m}\plus{}2^n}{2^{n}\minus{}1}\equal{}k\plus{}1$

Now since $ 2^m \plus{}2^n$ is divisible by $ 2$ and $ 2^n \minus{}1$ is not, a contradiction ($ k$ is not in $ \mathbb{N}$ ).

Hence, proved.

Well, the idea is this:

If $ \frac {2^{m} \plus{} 1}{2^{n} \minus{} 1}$ has to be integer, then $ m\ge n$. Hence, we can set $ m \equal{} n \plus{} k$, $ k$ integer.
So we get that $ \frac {2^{k} \plus{} 1}{2^{n} \minus{} 1}$, has to be integer too, so $ k \ge n$.
so, $ m$ is $ \ge 2n$. Repeating it, we get $ m\ge 3n$ and so on...
We see that there's no number in $ \mathbb{N}$ with that property.

I hope now it is clear... Please judge it and then tell me what do you think :)

Anyway, compliments: your solution is really good (as far as I know)! I like it! :coolspeak:

I cannot understand this part in Gabuntu's solution
$ \frac{2^{n\plus{}k}\plus{}1}{2^n\minus{}1} \equal{}> 2^{k}\plus{}\frac{2^{k}\plus{}1}{2^{n}\minus{}1}$
You need to clarify

@Tomekk.

Sorry, but I don't understand your solution. Why does the fact that the numerator is divisible by $ 2$ and the denominator is not cause $ k$ not to be integral? Couldn't the numerator simply be a multiple of $ 2$ also divisible by the denominator? While this would make the quotient even, wouldn't we expect this because $ k \plus{} 1$ is supposed to be even?

What I did was notice the following.

Let $ \frac{2^{m}\plus{}2^{n}}{2^{n}\minus{}1}\equal{}k\plus{}1$

Factor this as the following (notice that $ m>n$).

$ \frac{2^{n} \cdot (2^{m\minus{}n}\plus{}1)}{2^{n}\minus{}1}\equal{}k\plus{}1$

Since $ gcd(2^{n}, 2^{n}\minus{}1)\equal{}1$, then it must follow that $ 2^{n}\minus{}1 | 2^{m\minus{}n} \plus{}1$.

Let $ p$ be such that it is the maximum integer that satisfies $ (p\plus{}1)n>m>pn$.

Since $ 2^{n}\minus{}1 | 2^{m\minus{}n} \plus{}1$ if $ 2^{n}\minus{}1 | 2^{m}\plus{}1$, then use this $ p$ times to get that $ 2^{n}\minus{}1 | 2^{m\minus{}pn}\plus{}1$. But since $ m\minus{}pn<n$, then $ 2^{n}\minus{}1>2^{m\minus{}pn}\plus{}1$. This is a contradiction.

Sorry, I don't think that I considered a number of cases including the $ n|m$ case or some where the final result does not hold because $ n$ is small. But the above is my solution outline.

[quote="Poincare"][quote="Tomekk"]Hello

Well, to me honest, I cant seem to understand your proof, sorry.  :blush:

This is what I've done:

Lets assume the opposite, let $ \frac {2^{m} \plus{} 1}{2^{n} \minus{} 1} \equal{} k\in\mathbb{N}$.

Now we can expand this:

$ \frac {2^{m} \plus{} 1}{2^{n} \minus{} 1} \plus{} 1 \equal{} k \plus{} 1$

From there we have $ \frac {2^{m} \plus{} 1 \plus{} 2^n \minus{} 1}{2^{n} \minus{} 1} \equal{} k \plus{} 1$

Finally, $ \frac {2^{m} \plus{} 2^n}{2^{n} \minus{} 1} \equal{} k \plus{} 1$

Now since $ 2^m \plus{} 2^n$ is divisible by $ 2$ and $ 2^n \minus{} 1$ is not, a contradiction ($ k$ is not in $ \mathbb{N}$ ).

Hence, proved.[/quote]

Nice! where'd you get this?[/quote]

Well this was a problem on some competition in my country some years ago, and I was not able to prove it for some time, so I was desperate and started trying some stuff like this  :P

[quote="RobRoobiks"]I cannot understand this part in Gabuntu's solution
$ \frac {2^{n \plus{} k} \plus{} 1}{2^n \minus{} 1} \equal{} > 2^{k} \plus{} \frac {2^{k} \plus{} 1}{2^{n} \minus{} 1}$
You need to clarify[/quote]

Hi, that's why:

$ m\equal{}n\plus{}k$, so
\[ \frac{2^{n\plus{}k}\plus{}1}{2^n\minus{}1} \equal{} \frac{2^n2^k\minus{}2^k\plus{}2^k\plus{}1}{2^n\minus{}1}\equal{}2^k\plus{}\frac{2^k\plus{}1}{2^n\minus{}1}\]

I'm sorry it wasn't clear...   :oops:

At ShowEverywhere: I am not sure if it's needed to do what you've done  :) . That's why I would be grateful if someone who is good at number theory tells us here, so I know how I should be solving these kind of problems in future. :P

I'm just curious as to why $ k$ is not in $ N$ by what you've shown (I don't know that much about NT, I just wanted to know for future reference).

No, Tomekk, that solution is not correct. That's like saying $ \frac {6}{3}$ isn't an integer because $ 2$ divides $ 6$ but it doesn't divide $ 3$.
So we do have to do what snoweverywhere did, and his solution seems correct.

Can you post a link to this problem, i want to see the source.

This problem is amazing!

I think that this is actually in AOPS V2 Chapter 23.

We could write

$ k = \frac {2^{ m - an} + 1}{2^ {n} - 1} + \sum_{i = 1}^{a} {2^{ m - in}}$

If $ r \equiv m \bmod n$, then $ r = m - an$

So it suffices to find r such that ${ 2^ {n} - 1\mid 2^ {r} + 1}$

But $ r\equiv (0,1,,.....n - 1,n)$

Hence the result follows.


algebra
polynomial
induction
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ F$ be an infinite field. Suppose that $ f \in F[x_1,x_2,...,x_n]$ is nonzero.
Show there exists $ a \equal{} (a_1,a_2,...,a_n) \in F^n$ with $ f(a_1,...,a_n) \ne 0$.

[hide="Hint"]Use induction on n using the fact that a polynomial in $ F[X]$ with infinitely many zeros is trivial together with the isomorphism $ F[X_1,\ldots,X_n]\equal{}F[X_1,\ldots,X_{n\minus{}1}][X_n]$.[/hide]

[quote="-oo-"]a simple modification of the proof shows this generalization: if $ f \neq 0$, then there is some $ a \in \overline{F}^n$ such that $ f(a) \neq 0$. in particular, you see that $ F[x_1,...,x_n]$ is jacobson.[/quote]If you mean that it generalizes to arbitrary fields $ F$ with algebraic closure $ \overline{F}$, then this is trivially implied by the first assertion without any modification of the proof (since $ \overline{F}$ is infinite). Because of that I deem the first assertion logically stronger.

ps:[quote="-oo-"]in particular, you see that $ F[x_1,...,x_n]$ is jacobson.[/quote]This is one version of Hilbert Nullstellensatz. I am wondering how you conclude it - without any machinery of commutative algebra.

the kernel if the evaluation $ f \to f(a)$ is maximal. you don't need any nullstellensatz.

ok, in fact, I meant only that rad(0) = nil(0), in which case some authors say jacobson. autors mean rad(a)=nil(a) for every ideal.

[quote="-oo-"]the kernel if the evaluation $ f \to f(a)$ is maximal. you don't need any nullstellensatz.[/quote]With this the Jacobson property is still not in sight.
[quote]ok, in fact, I meant only that rad(0) = nil(0), in which case some authors say jacobson. autors mean rad(a)=nil(a) for every ideal.[/quote]So what about the other ideals different from (0) to see the Jacobson property here?
As I said the Jacobson property of $ F[X_1,\ldots,X_n]$ cannot be shown with the assertion in the first posting or by an amendment of the proof. The heavy machinery of commutative algebra is needed here to deal with the case of an arbitrary ideal.

yes, this is clear to me ... I just meant that the intersection of all maximal ideals is $ 0$, ok?

Ah, you meant that it follows over an algebraically closed field that the Jacobson ideal of the polynomial ring in several variables is zero. This is something different to what you wrote which caused my confusion. Never mind, no problem.


What is the difference between long division and synthetic division of polynomials?

errr long division you write out all the $x's$ or whatever the variable is, along with the degree that the $x$ is raised to

also for long division, you can divide equations of degree more than $1$, in other words quadratics and cubics, etc

whereas in synthetic division, i believe you set the division up using just the coefficients, and the whole set up is different.... you'll have to ask someone more familiar with latex and tables and stuff to show you the setup

and also synthetic division i think you can only divide monic equations of degree 1....

Yeah, the problem with synthetic division is that you can only divide monic polynomials into a polynomial, cause the entire process of synthetic division is based on the free term of the monic.

I prefer long division every time because with synthetic division I have to remember how to do it. Long division is basically the same as numerical long division, which I'll never forget how to do. Also, as both people before me have said, you can only divide by first-degree polynomials.

It's the same with "FOILing." In 8th grade everyone else used the FOIL method to expand things like $(\alpha-\beta)(\gamma-\delta)$ but I hated it because (a) it's very silly (b) expanding products of polynomials is an algorithmic process that can theoretically be applied to as many polynomials of an degree as you want, and FOILing works only in a very special case. Other kids in my class just learned FOIL and didn't understand what they were doing. It's probably the same with synthetic division in precalculus classes, but I wouldn't know since I didn't take the time to take it at high school (thank god).

[quote="AntonioMainenti"]It's probably the same with synthetic division in precalculus classes, but I wouldn't know since I didn't take the time to take it at high school (thank god).[/quote]Hmmm synthetic division is taught in Algebra II at my school.  Our teacher actually goes through a quick explantion of why it works, the real problem is that the kids don't care, they just want to know how to do it so they can do it on the test.

[quote="symmetry"]Thank you all for your hints and quick reply.
I prefer long division.  Synthetic gives me a headache.[/quote]You're very kind, but trust me, we like talking about this stuff around here, so there's really no need to thank us for answering all your questions.  :)


inequalities
Cauchy Inequality
inequalities proposed

Let a,b,c be positive number such that $ \frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3} \equal{} 3$
Prove that
 $ (\sqrt {a^3 \plus{} abc} \plus{} \sqrt {b^3 \plus{} abc} \plus{} \sqrt {c^3 \plus{} abc})^2 \leq 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3a^2b^2c^2\plus{}3abc)$

[quote="8826"]Let a,b,c be positive number such that $ \frac {bc}{a^2} \plus{} \frac {ca}{b^2} \plus{} \frac {ab}{c^2} \equal{} 3$
Prove that
 $ (\sqrt {a^3 \plus{} abc} \plus{} \sqrt {b^3 \plus{} abc} \plus{} \sqrt {c^3 \plus{} abc})^2 \leq 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc)$[/quote]
Id est, we need check only the case $ a\equal{}b\equal{}c.$ :wink:

[quote="8826"]Let a,b,c be positive number such that $ \frac {bc}{a^2} \plus{} \frac {ca}{b^2} \plus{} \frac {ab}{c^2} \equal{} 3$
Prove that
 $ (\sqrt {a^3 \plus{} abc} \plus{} \sqrt {b^3 \plus{} abc} \plus{} \sqrt {c^3 \plus{} abc})^2 \leq 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc)$[/quote]
Since $ \sum \ \frac{bc}{a^2}\equal{}3$ we have : $ a\equal{}b\equal{}c$
Then your inequality 's true :wink:   :)

Why nobody solve my ineq?
Is it wrong again? Or is it very easy? :(

Let a,b,c,d be positive numbers such that $ \sum\frac{1}{a^4b^4}\equal{}4$
Prove that
 $ (\sum\sqrt{a^4\plus{}abcd})^2 \leq 2(\sum a^4\plus{}4abcd\plus{}4a^3b^3c^3d^3\plus{}\sum(\frac{abc}{d})^2)$

[quote="8826"]Let $ a,b,c$ be positive numbers such that $ \frac {1}{a^3} + \frac {1}{b^3} + \frac {1}{c^3} = 3.$ Prove that

$ \left(\sqrt {a^3 + abc} + \sqrt {b^3 + abc} + \sqrt {c^3 + abc}\right)^2 \leq 2\left(a^3 + b^3 + c^3 + 3a^2b^2c^2 + 3abc\right).$[/quote]Using Cauchy inequality, we have

$ \left(\sum{\sqrt {a^3 + abc\right)^2\leq\left(\sum{a}\right)\left[\sum{\left(a^2 + bc\right)}\right]}}$

$ = 2\left(3abc + \sum{a^3} + \sum{\frac {b^2c^2}{a}}\right) - \sum{\frac {(b - c)^2}{2bc}}\left[2a^2(b + c) - abc + bc(b + c)\right]$

$ \leq2\left(3abc + a^3 + b^3 + c^3 + 3a^2b^2c^2 \right).$


real analysis
inequalities
calculus
integration
geometry
geometric transformation
dilation

We want to show that if the set E of points of discontinuity of $ f: I \to R$ does not have measure 0 then f is not Riemann integrable. $ \omega (f;x)$ denotes the oscillation of f at the point x in I. $ E \equal{} \bigcup\limits_{n \equal{} 1} {E_n }$where $ E_n \equal{} \{ x \in I|\omega (f;x) \ge {\textstyle{1 \over n}}\}$.

Assume $ E_{n_o }$does not have measure zero.

If P is a partition of I, we can break P up into two groups of intervals A and B, where $ A \equal{} \{ I_i \in P|I_i \cap E_{n_o } \ne \emptyset ,\omega (f;I_i ) \ge {\textstyle{1 \over {2n_o }}}\}$and $ B \equal{} P \minus{} A$.

It seems to me that $ I_i \cap E_{n_o } \ne \emptyset \to \omega (f;I_i ) \ge {\textstyle{1 \over {2n_o }}}$ so I'm not sure why the author described A that way. He says that either a point of $ E_{n_o }$is in the interior of some interval $ I_i$then it is "obvious" that $ I_i$ is in A and otherwise it is on the boundary of several intervals in P, in which case the oscillation of f must be at least $ {\textstyle{1 \over {2n_o }}}$ on one of these intervals "because of the triangle inequality."

I don't see how the triangle inequality shows that either. And if the point of discontinuity were on the boundary of several intervals, wouldn't the oscillation of f on [i]all[/i] of those intervals have to be at least $ {\textstyle{1 \over {2n_o }}}$ (assuming that were true)?

It's true, but it doesn't matter. There are finitely many intervals, and we can simply ignore the finite collection of boundary points. It's also quite possible for the oscillation to be smaller on one of the intervals- consider a jump discontinuity like $ f(x)\equal{}\begin{cases}1&x\ge 0\\0&x<0\end{cases}$.

Here's how I'd do it:
Define $ E$ and $ E_n$ as in your proof, and suppose $ E_N$ has positive measure $ a$.
Let $ P$ be any partition, and let $ E_P$ be the intersection of $ E_N$ with the union of the interiors of the intervals in $ P$; $ E_P$ also has measure $ a$. The difference between the upper and lower sums for $ P$ is at least $ \sum_{J\in P}|J|\sup_{x\in \text{int }J}\omega(f;x)\ge \frac1N\sum_{J\cap E_P\text{ nonempty}}|J|\ge \frac{a}N$
This is bounded away from zero for all partitions, and $ f$ is not Riemann integrable.

I think your suggestion is essentially what the author is trying to do, although he has not yet introduced upper and lower sums (he defined the integral differently to make it suitable for normed spaces). That's a good example you gave too.

And was the "it's true" with regards to what I said about the second property of A being redundant? If so I wonder why he said it had to do with the triangle inequality.

You're right too about the points on the boundary not mattering. Doh.

EDIT: I think I understand a little more now. In your example, A would include the left interval of the partition if 0 was on its boundary but not the right right interval, even though both contain 0. I'm still not seeing how where the 1/2 comes from (if it comes from the triangle inequality).

EDIT: Alright I figured out where the triangle inequality comes from. The statement's proof is definitely not as trivial as the way it's stated would imply though. Your example is what helped me to think about it.

I have one last question jmerry. In the sufficiency proof, the author starts out by saying for each epsilon, $ E_\varepsilon \equal{} \{ x \in I|\omega (f;x) \ge \varepsilon \}$is a compact set of measure zero so there exists a finite system of intervals $ I_1 ,...,I_k$such that $ E_\varepsilon \subset C_1 \equal{} \bigcup\limits_{i \equal{} 1}^k {I_i }$ and $ \sum\limits_{i \equal{} 1}^k {I_i } < \varepsilon$. Then he says 'let us denote by $ C_2$ and $ C_3$ the unions of the intervals obtained from the intervals $ {I_i }$ by dilation with center at the center of $ {I_i }$ and scaling factors 2 and 3 respectively. It is clear that $ E_\varepsilon$ lies strictly in the interior of $ C_2$ and that the distance d between the boundaries of the sets $ C_2$ and $ C_3$ is positive.

We also note that any subset of the interval I whose diameter* is less than d is either contained in $ C_3$ or lies in the compact set $ K \equal{} I \minus{} (C_2 \minus{} \partial C_2 )$, where $ \partial C_2$ is the boundary of $ C_2$and hence $ C_2 \minus{} \partial C_2$ is the set of interior points of $ C_2$.'

*The diameter of a set is the supremum of the Euclidean distances between all pairs of points in the set

I don't understand the last paragraph. It seems to me that $ C_2$ is contained in $ C_3$ so I don't understand how what he says can be true. He has this habit of finding everything obvious.


function
linear algebra
matrix
topology
analytic geometry
real analysis
real analysis unsolved

Let ${{f: \mathbb{R}^{n}}\to\mathbb{R}^{m}}$ be a continuous differentiable function and $n>m$.
i)Prove that if $f$ is injective there is an open set $A\in R^{n}$ so that the rank of $J_{f}$ is $m$ for all $x\in A$. ($J_{f}$ is the Jacobian matrix of $f$ ).
ii)$f$ is not injective.

For i), set $A=\varnothing$.  :roll: Otherwise the statement is false, since $f$ could be constant, for example.

I don't know how to prove (ii)  :maybe:

I modified i).. sorry for the mistake...

We don't really need to make the effort of proving (a) :).

There is a generalization of the Inverse Function Theorem, called the Rank Theorem, I think (names aren't that important anyway :)), which states that if $f: \mathbb R^{a}\to\mathbb R^{b}$ is a $C^{1}$ function such that the rank of $J_{f}$ on some open subset $U$ of $\mathbb R^{a}$ is constant (equal to $k\le\min(a,b)$, say), then after a suitable homeomorphism of $\mathbb R^{b}$ onto itself, $f|_{U}$ can be regarded as taking values in the subspace of $\mathbb R^{b}$ consisting of those points whose last $b-k$ coordinates are $0$. In other words, locally, we may as well assume that $f$ takes values in $\mathbb R^{k}$.

Using this, if $p\le m$ is the maximal rank of the Jacobian of $f$ on $\mathbb R^{n}$, then in some open subset $U$ of $\mathbb R^{n}$ we can assume that $f$ takes values in $\mathbb R^{p}$, and its Jacobian has full rank $p$. Now we can apply the Implicit Function Theorem to the map $f: U\to\mathbb R^{p}$ to conclude that it cannot be one-to-one: write $\mathbb R^{n}$ as $\mathbb R^{p}\times\mathbb R^{n-p}$ and its elements as $(x_{1},\ldots,x_{p},y_{1},\ldots,y_{n-p})$, assume WLOG that the Jacobian of $f$ with respect to the $x_{i}$ is non-singular, and it follows that there is a map $g: V\subset\mathbb R^{n-p}\to W\subset\mathbb R^{p}$ ($V,W$ are open) such that $f(g(y),y)$ has the same value for all $y\in V$. This shows that $f$ cannot be one-to-one.


I cant answer your question directly, but I can prove that there are more real numbers than natural numbers. Maybe with this, you can find how to prove your problem.  :wink:

This proof requires the knowledge of cardinality.

Let $f$ be a function from the natural numbers to $(0,1)$. If we can show that $f$ is not a surjection, then our proof is complete because then NO SUCH FUNCTION can be a surjection or a bijection. Let $a_{i}$ be the $i$th decimal place of $f(i)$. (i.e.- $f(3) = .76583$ $a_{3}= 5$) The number $a_{i}$ depends on the function. Let $b_{i}={2}$ if $a_{i}= 3$, $3$ if $a_{i}$ is not ${3}$ (i.e.- $f(1) = .91 a_{1}= 9$ ; $b_{1}= 3$). Let $b = 0.b_{1}b_{2}b_{3}...$. Assume that $b$ is a member of the image of $f$. Then there exists some natural number $i$ such that $f(i) = b$. The $i$th decimal place of $f(i)$ is $a_{i}$. The $i$th decimal place of $b$ is $b_{i}$. $a_{i}$ can never equal $b_{i}$, so $f(i)$ is not equal to $b$. This is a contradiction, so $b$ is NOT in the image of $f$. Therefore, the image of $f$ does not equal $(0,1)$. Thus, $f$ is not a surjection and thus, not a bijection. $\boxed{QED}$

Corrolary: $card N < card (0,1)$ and $card N < card R$

sorry, don't know how to make natural numbers and real number symbols.

and sorry if it is very confusing. This proof requires the knowledge of cardinality, surjections, bijections, and images.

And $\mathbb{N}$ has the same cardinality as $\mathbb{Q}$ because you can put all rationals between 0 and 1 in a row:
1/1,
1/2,  2/1,
1/3,  2/3,  3/3,
etc
(if you want you can delete the terms that appear not for the first time)

Even stronger, there are as many algebraic numers as naturals. An algebraic number is a solution of a polynomial equation with integer coefficients. E.g. $\sqrt{2}$ is algebraic because it is a solution of $X^{2}-2=0$. All rationals are algebraic because a/b is a root of aX-b=0.

[quote="nebula42"][quote="Ignite168"]Are there more rational or irrational numbers?

I think a proof would be waaaaay too advanced for me as well as this forum, so I just want a little bit of the concept involved and the beginnings of an explanation. Thanks.

Also, it will probably be easier to first do it for two easier to understand sets like integers, or whole numbers, or natural numbers, etc.[/quote]
Well, Canton's diagonal proof (which is in AoPS 2 somewhere) showed that $\mathbb{R}$ has a larger cardinality than $\mathbb{N}$. You can create a correspondence between $\mathbb{N}$ and $\mathbb{Q}$ in the following way:

$1$
$2 \ 3$
$4 \ 5 \ 6$
...

$\frac11$
$\frac12 \ \frac22$
$\frac13 \ \frac23 \ \frac33$
...

This means that $\mathbb{N}$ has the same cardinality as $\mathbb{Q}$. You can then prove that the set of irrationals ($\overline{\mathbb{Q}}$?) has larger cardinality than $\mathbb{N}$ by using the diagonal proof in a modified way (which may or may not work, I'm not sure), and this shows that the set of irrationals has larger cardinality than $\mathbb{Q}$.

However (as stated in AoPS 2), you can't talk about something like "Are there more irrationals than rationals?" or "How many integers are there?"; you can only refer to the size of the sets (as in cardinality).[/quote]

Oh yeah... I learned that one too!

but I didnt remember it off the top of my head...

 :)

[quote="Ignite168"]Are there more rational or irrational numbers?

I think a proof would be waaaaay too advanced for me as well as this forum, so I just want a little bit of the concept involved and the beginnings of an explanation. Thanks.

Also, it will probably be easier to first do it for two easier to understand sets like integers, or whole numbers, or natural numbers, etc.[/quote]

Well the set of reals is uncountable (no bijection with the naturals) while the set of rationals is (bijection with the naturals, shown above). However, it is well-known that the union of two countable sets is countable, so that means the irrationals must be uncountable as well and thus there are more irrationals than rationals.

Ugh, thread was moved to Intermediate so I'm guessing I won't be understanding anything else posted in this thread.


Let $ ABC$ is a triangle, let $ (I; r)$ is inscribed circle of $ ABC$ triangle. Let $ AI, BI, CI$ meet circumcribed of $ ABC$ triangle at $ D, E, F$ respectively. Prove that $ \boxed {\frac{ID}{IA}\plus{}\frac{IE}{IB}\plus{}\frac{IF}{IC} \ge 3}$

$ \frac {DI}{AI} = \frac {sin\frac {A}{2}}{2sin\frac {B}{2}sin\frac {C}{2}}$

Then ${ \sum \frac{DI}{AI}} \ge 3\sqrt[3]{\prod \frac{DI}{AI}} = 3\sqrt[3]{\frac{1}{8 \sin{\frac{A}2} \cdot \sin{\frac{B}2} \cdot \sin{\frac{C}2}}}\ge 3$

For AM-GM and for $ \sin{\frac{A}{2}} \sin{\frac{B}{2}} \sin{\frac{C}{2}} \le \frac{1}{8}$

Thank! I have o way to done this problem, but it's long. Your solution is sort and nice.


function
calculus
derivative
number theory
prime factorization
calculus computations

The local linear approximation of a function f will always be greater than or equal to the function's value if, for all x in an interval containing the point of tangency,

A) f'<0
B) f'>0
C) f''<0
D) f''>0
E) f'=f''=0

Pretty sure it's B. There's really nothing to it, just draw a graph.

We are of course, assuming that $ f$ is several times differentiable on the interval $ [a,b]$.
Let $ g(x)\equal{}f(x_0)\plus{}f'(x_0)(x\minus{}x_0)$
We want that $ f(x)\leq g(x)$
i.e. $ f(x)\minus{}f(x_0)\leq f'(x_0)(x\minus{}x_0)$

Now, by Langrange's Mean Value Theorem, $ \forall x$ $ \exists$ $ c\in [x_0,x]$ such that
$ \frac{f(x)\minus{}f(x_0)}{x\minus{}x_0}\equal{}f'(c)$

Hence we have that $ \forall x$ $ \exists$ $ c\in[x_0,x]$ such that $ f'(c)<f'(x_0)$ if $ x>x_0$ and vice-versa.
Thus, by definition $ f''(x_0)>0$ is not possible.
Hence, $ f''(x_0)\leq 0$

The answer is C.

Prime factorization: you should read "local linear approximation" as "tangent line, if it exists." Yes you do just draw a graph, but do so correctly.

The "executive summary" of shreyas_patankar's post: If the function always lies below its tangent line, then it must be concave down. That makes its second derivative non-positive.

---

Slight quibble with the problem: we don't have an if and only if. That $ f''(x) < 0$ in a neighborhood of the point of tangency is a sufficient condition for the curve to lie below the tangent line, but not a necessary condition. One example: $ f(x) \equal{} \minus{} x^4$ near zero. We have $ f''\le0$ but we don't have $ f'' < 0.$ The problem is carefully worded (with an "if" instead of an "iff") so that it is still correctly worded, but that's still something I noticed.


function
probability
limit
trigonometry
topology
real analysis
real analysis unsolved

Find function(s) which are continues in :R but Arent diffrentiable in any point of R.:

Such functions are plentiful.

If you consider $ C[a,b]$ as a metric space with the sup norm, the set of functions that are differentiable at any point as a set of Baire first category (a countable union of nowhere dense sets). Hence in that sense, "most" continuous functions are nowhere differentiable.

If you make the continuous functions into a probability measure space via Brownian motion, the set of functions that are differentiable at any point as a set of measure zero. That is, a Brownian path is almost surely nowhere differentiable.

In an undergraduate analysis class, when I seek to provide an example, I usually use
\[ f(x)\equal{}\sum_{n\equal{}1}^{\infty}\left(\frac{3}4\right)^{n}\cos(4^{n}x).\]
This is continuous as the limit of a uniformly convergent series of continuous functions.

As for proving it is nowhere differentiable, I start with this lemma:

If $ f$ is differentiable at $ x$ and if $ y_{n}$ and $ z_{n}$ are sequences with $ y_{n}\le x<z_{n}$ and $ \lim y_{n}\equal{}\lim z_{n}\equal{}x,$ then
\[ \lim\frac{f(z_{n})\minus{}f(y_{n})}{z_{n}\minus{}y_{n}}\equal{}f'(x).\]
In particular, that limit must exist. Therefore if we can find sequences $ y_{n}$ and $ z_{n}$ for which this limit doesn't exist, then we have proved that the function is not differentiable at that point.

Another undergraduate examples include the Blancmange function. I think this example is one of the simplest examples to understand intuitively.

Let $ w$ be the function on $ \mathbb{R}$ with period 1, given by $ w(x) \equal{}\begin{cases}x &\left(0\leq x\leq\frac{1}{2}\right)\\ 1\minus{}x &\left(\frac{1}{2}\leq x\leq 1\right)\end{cases}$ on $ [0, 1]$. You'll see sawtooth-like graph when you plot this. Then define

$ f(x)\ \equal{}\ \sum_{n \equal{} 0}^{\infty}\frac{w(2^{n}x)}{2^{n}}$

This function is continuous by Weieretrass M-test. But it is also nowhere differentiable, with additional property that it actually becomes fractal.

Here is a rough explanation why $ f$ is differentiable at nowhere:

Intuitively, differentiability at a point means that we can approximate the graph of a function to a line at the point, i.e, we can draw a tangent line at the point.

In a microscopic scale, the graph of function and the tangent line would be so close that we can nearly distinguish them.

[img]http://myhome.shinbiro.com/~sosQED/blancmange3.gif[/img]

This animation shows that, however, if you zoom in to some fixed point of the plot of $ f$, you'll see only a clear self-similarity of $ f$.

Similar situations occur at any point on the graph of $ f$. So we cannot think of a tangent line, and hence $ f$ is nowhere differentiable.

Of course, one who want to prove it rigorously may do like this:
choose appropriate sequence of difference quotients which must converge to $ f'(x)$ if it exists), and then show that it does not converge.

I provide another way to construct nowhere differentiable functions. This relates to lacunary sequences. We assume to have a function $ f$ such that $ \sum a_{n}\cos\lambda_{n}t$ is its Fourier series, where $ \lambda_{n}$ is a lacunary sequences. If $ f$ is differentibale at any point then its coefficients $ a_{n}=o(\lambda_{n}^{-1})$. That means for a lacunary series, differentiablity at a point  implies differentibility on an every where dense set. By this fact we can take , namely Weierstrass function $ \sum 2^{-n}\cos2^{n}t$, this is a continuos function but nowhere differentiable.
The example which is given by Kent above is also nowhere differentiable for the same reson. :)


integration
function
probability and stats

The following data concerning the number of thousands of miles traveled between bus motor failures adapted from an experiment are below:

[img]http://i99.photobucket.com/albums/l290/janbmyffi/NewPicture.jpg[/img]

b) Test the hypothesis that the data for the fifth bus motor failure follows an exponential distribution at $ \alpha\equal{}0.10$

How would this test be done in Maple, Statistica, or Matlab? How do you find the Maximum Likelihood Estimator and the chi-squared estimator(for this problem 6.26)? How would we find $ \mu$(which in this case was 35.33) using any of this software? The detailed answer to this question is below:

[hide]Our test is that $ H_0: EXP(\mu)$
To get a the actual Maximum Likelihood Estimator (MLE) of $ \mu$ we can take the sum of the average miles in each interval and multiply them by the number of failures and then divided by the sum of the failures so:
$ \mu\equal{}\frac{10(29)\plus{}30(27)\plus{}50(14)\plus{}70(8)\plus{}90(5)\plus{}110(2)}{29\plus{}27\plus{}14\plus{}8\plus{}5\plus{}2}\equal{}35.647$
To get the actual estimate, we have to take:
$ P(a \leq X \leq b)\equal{}\int_{a}^{b}\frac{1}{\mu}e^{\frac{\minus{}x}{\mu}}dx\equal{}e^{\frac{\minus{}a}{\mu}}\minus{}e^{\frac{\minus{}b}{\mu}}$ then $ L\equal{}(1\minus{}e^{\frac{\minus{}20}{mu}})^{29}(e^{\frac{\minus{}a}{\mu}}\minus{}e^{\frac{\minus{}b}{\mu}})^{27}.........(e^{\frac{\minus{}100}{\mu}})^2$ If $ \theta\equal{}\frac{1}{\mu}$ then $ L\equal{}(1\minus{}e^{\minus{}20\theta})^{29}(1\minus{}e^{\minus{}20\theta})^{27}(1\minus{}e^{\minus{}20\theta})^{14}.....(1\minus{}e^{\minus{}20\theta})^{2}\equal{}e^{(27)\minus{}20\theta}e^{\minus{}20\theta(14)}e^{\minus{}20\theta(3)(8)}e^{\minus{}20\theta(4)(5)}e^{\minus{}100\theta}$
Since 20(27+2(14)+3(8)+4(5))+200=2180
Then $ L\equal{}e^{\minus{}2180\theta}(1\minus{}e^{\minus{}20\theta})^{83}$
Taking $ (log(L))'\equal{}0$ we have
$ \minus{}2180\plus{}83log(1\minus{}e^{\minus{}20\theta})\equal{}0$ so $ \theta\equal{}\frac{\minus{}1}{20}log(\frac{2180}{3840})$ so $ \mu\equal{}\frac{1}{\theta}\equal{}35.33$
Now figuring out the explected values of break downs if the fifth bus follow an exponential distribution, we calculate using this table:
[img]http://i99.photobucket.com/albums/l290/janbmyffi/NewPicture-1.jpg[/img]
Taking computing $ \sum\frac{(observed\minus{}expected)^2}{expected}\equal{}6.26$ Taking the Chi-squared statistic at the $ 1\minus{}\alpha\equal{}0.10$ level for 4 degrees of freedom, which is 7.87 we fail to reject the null hypothesis[/hide]

Anybody know if such analysis is available? I also have Matlab and Mathematica.

[quote="Mr.Doe"]For matlab:

help kstest

You will find there everything you need.[/quote]

Do you know how I could find the MLE or this function as shown in the solution in Matlab or any other program?

[quote="223rem"]The following data concerning the number of thousands of miles traveled between bus motor failures adapted from an experiment are below:

[img]http://i99.photobucket.com/albums/l290/janbmyffi/NewPicture.jpg[/img]

b) Test the hypothesis that the data for the fifth bus motor failure follows an exponential distribution at $ \alpha \equal{} 0.10$

How would this test be done in Maple, Statistica, or Matlab? How do you find the Maximum Likelihood Estimator and the chi-squared estimator(for this problem 6.26)? How would we find $ \mu$(which in this case was 35.33) using any of this software? The detailed answer to this question is below:
[/quote]You'll get a somewhat better estimate fro $ \mu$ with the EM-Algorithm.

Is that in Matlab? Can you give me an example of how it is used, for this problem for instance? Thanks, I don't really use Matlab, mostly maple.


In my McMurry organic chemistry textbook, it says that one method of hydrogenation of an alkyne is to use a lithium catalyst in liquid ammonia. This gives a trans double bond.

The lithium donates an electron to a pi bond in the alkyne, and the pi bond breaks to form a carbon radical and a lone pair. The carbon radical gets another electron from lithium and turns into a lone pair. At this point both carbons are sp2 hybridized. Both lone pairs each get a proton from their own ammonia molecules. I've drawn out the orbitals involved and I don't understand why the lone pairs are trans to each other. Why are they trans?

Because charges of the same sign minimize repulsion between them by maximizing their distance. So, trans orbitals give greater distance between electrons than cis orbitals.


geometry
MIT
college
MATHCOUNTS
Functional Analysis
number theory

I have to decide between Colorado School of Mines, University of Missouri at Rolla, and Texas A&M.  I really want to go to the best math school, but I might go into engineering.  I have searched their sites some and cannot come to a conclusion.  Does anybody know where I can find which ones are best for undergraduate math/science and the reputations/realities for these schools?

[quote="djimenez"]Texas A&M is the only one that I have heard, and actually I have heard a lot of good things about it.[/quote]

Oh really? Like what?

(er...if any consol ppl are reading this...hehe...no offense to you folks)

[quote="MithsApprentice"][quote="djimenez"]Texas A&M is the only one that I have heard, and actually I have heard a lot of good things about it.[/quote]

Oh really? Like what? [/quote]

OK, first, it seems like they have good groups in Functional Analysis, Algebraic Geometry, Number Theory, PDEs and Probability. If you review, since something like '97 or '98, they have been in a pretty descent position in most of the rankings (actually, pretty often are rated withing the top ten), higher in some of them than some of the Ivy Leages.

Also, I have walked around in the corridors of the school, and I have seen lots of posters about conferences and congresses, and pretty often within the main speakers there is someone from Texas A&M.

(Also, I heard from a non-totally reliable source that few years ago someone made a billionare donation to the School of Mathematics there, and that is why they had hired some big names in several areas.)

Best,

if you like math and science, then a tech school probably will be the best: MIT, CA tech, GA tech, etc

[quote="djimenez"]

OK, first, it seems like they have good groups in Functional Analysis, Algebraic Geometry, Number Theory, PDEs and Probability. [/quote]

But the poster said that he might want to move into engineering, and someone correct me if I'm wrong but A&M isn't exactly at the top in that. I wasn't aware of how good their math department was though. Thanks for that heads up.

Btw, what are the best engineering schools?

I read years ago that either Texas A & M, or UT Austin, had huge oil fields (or deposits) on their land; that this had made them fairly wealthy; and that they offered large sums of money to hire academic luminaries.  Does anyone know which school this describes?

[quote="tokenadult"]How did you come up with that list of three schools in three different states? Your answer to that question might help me think of some advice for you on choosing among those schools.[/quote]

These are the schools I have applied to.  Well, I also applied to Rice and MIT, but already know about them.  Sadly, everything has turned against me in the application process, so I don't expect to get in.

A friend of mine went to the School of Mines last year for engineering, and hated it.  He transfered out.

Does anyone else have some personal information about any of those schools? I may ask about this head-to-head comparison over on College Confidential and see what people there think.

Hi, juggle7, I asked about this on a thread at College Confidential,

http://talk.collegeconfidential.com/showthread.php?t=40988

and there are some answers there now.

Hope this helps! Good luck in your studies.

It is worth noting that the person I was talking about was a bit abnormal.  The smart, quiet, antisocial type who doesn't necessairily represent the whole population.

Thank you very much tokenadult.  You did more than I would have expected.  All the replies were very helpful.

If I remember correctly, I think A&M's, engineering school is ranked like 16th or something.  I live in Austin, where UT is the most popular school.  I like College Station's environment, it is a small town I believe.

yea college station is a small town...but everything's so...well...maroon.

I am an Engineering graduate, and all three schools are well respected in the Engineering community.  There has been little talk of the size of the schools or the different Engineering specialties each of these schools offer.  If you have a preference, this may make the choice easier.

I live in Colorado.  School of Mines hosts our State MATHCOUNTS Competition every other year.  This is a very small school (~4000 students) and a very small campus.  The "back door" of the campus is literally a mountain.  The "front door" is the city of Golden, which is a suburb of Denver.  The setting is beautiful.  [But since I live in Colorado, I may be biased.]  This is one of the few schools that offer a degree in Mining Engineering.

My father was an engineering graduate of TAMU.  A&M'ers do stick together.  He always supported the school and he often talked about their traditions.  [I don't know if the modern A&M carries on quite as many traditions as in previous generations.  Remember the horrible bonfire accident a few years ago?]  This is a big school (~45000 students).  Would it be a surprise that Texas A&M is especially well respected in the field of Petroleum Engineering?


If x、y、z∈R，xy＋yz＋zx＝－1，
find x 2 +5 2 +8z 2
min value

(1) find x 2 +5y 2 +8z 2
question 2
  if a, b, c :varepsilon: R+,
  and k*a*b*c/(a + b +c)>=(a + b) 2  + (a + b + 4c) 2  for all a, b, c is ok.
find k's min value

f(x)= arctanx + arccosx + ax(a<0)
if f(x)'s min value is 0,
find its max value.

question 2 is easy
let a+b+c=1,and you will get it


1:[hide]
A and B are whole numbers.
A/11 + B/11 = 31/33
Find A and B
[/hide]

2:[hide]
The average of 5 weights is 13 grams. This set is increased by another weight which weighs 7 grams. What is the average of the 6 weights? [/hide]

3:[hide]
In a group of 30 high school students, 8 take French, 12 take Spanish, and 3 take both languages. How many students in the group take neither French nor Spanish? [/hide]

4:[hide]
X and Y are two different numbers selected from the first fifty counting numbers (including 1 and 50)
What is the largest value that  X+Y/X-Y can have? [/hide]

WARNING: 5:[hide]
A palimage of a natural number is the number that has the same digits but in reverse order. Ex: 659 and 956.
Add 261637 and its palimage. Call this sum X. Now add X and its palimage. Call this sum Y. Now add Y and its palimage. Call this sum Z. What is the value of 4Z?[/hide]

NEW: edited #1:
It's A/11 + B/3 = 31/33

[hide="1"]1. We must have $ \frac{A\plus{}B}{11}\equal{}\frac{31}{33}$, so $ A\plus{}B\equal{}\frac{31}{3}$.

Is there a typo in this? cuz this problem is impossible. You can't add two whole numbers to make a noninteger rational number.[/hide]

[hide="2"]2. Total weight of 5 weights=13*5=65 grams. The 6th weight is 7 grams, so the sum of the 6 weights is 65+7=72 grams. Average weight of the 6 weights is $ \frac{72}{6}\equal{}\boxed{12}$ grams.[/hide]

[hide="3"]3. Use the Principle of Inclusion-Exclusion. 3 Spanish kids also take French, so 12-3=9 kids take only Spanish. 8-3=5 students take only French. So the total number of kids who take a foreign language is 5+3+9=17 kids. Therefore the number of students who don't take any language is 30-17=$ \boxed{13}$ kids.[/hide]

[hide="4"]4. The two largest numbers that can be used are 50 and 49. We want to minimize X-Y, so it has to be 1. So X=50 and Y=49. (that must be done to also maximize X+Y) So the maximum possible value for $ \frac{X\plus{}Y}{X\minus{}Y}\equal{}50\plus{}49\equal{}\boxed{99}$.[/hide]

[quote="funtwo"]very funny. very, very amusing.
seriously post this in classroom math or something. moems is olympiad
as you wish:
1. multiply by 11 to see that you must have typoed
2. 12
3. 13
4. 99
5. 43806356[/quote]

Then I suggest you find out how many palindromes are smaller than 100000000000000

I also suggest you should not call people "amusing". We're not animals, are we?

 :(  :(  :(

[quote="gauss1181"][quote="funtwo"]very funny. very, very amusing.
seriously post this in classroom math or something. moems is olympiad
as you wish:
1. multiply by 11 to see that you must have typoed
2. 12
3. 13
4. 99
5. 43806356[/quote]
Eh, post your solution to #5 please. :roll:[/quote]

I haz solution:
[hide]Err we has 261637 + 736162 = 997799.
997799 + 997799 = 1995598
1995598 + 8955991 = 10951589
10951589*4 = 43806356[/hide]

I just realized you and funtwo joined EXACTLY one year apart. lolz :D :D :D

simple arithmetic bashing, cant take more than half a minute. what are the time constraints on moems anyway?


probability
number theory
relatively prime
algebra unsolved
algebra

1.) If one writes a proper fraction $\dfrac{A}{B}$ with both $A$ and $B$ selected at random, what is the probability that the fraction is in lowest terms ?

2.) If one eliminates even numbers and numbers ending in five or zero, what is the probability that numbers then selected are relatively prime ?

I think it was posted before (a looong time ago :)). If one selects $2$ numbers randomly, the probability of them not being both divisible by $p$ (a prime) is $1-\frac 1{p^2}$. We then multiply these expressions over the entire set $\mathcal P$ of primes and we get $\prod_{p\in\mathcal P}(1-\frac 1{p^2})$, and this is $\frac 1{\zeta(2)}=\frac 6{\pi^2}$.

As for the second part, eliminating numbers divisible by $2$ or $5$ (possibly both) is just the same as not considering divisions by $2,5$. The probability is the one obtained above, divided by $\frac 34\cdot \frac {24}{25}$. I'm not too sure about the second problem, though.


search
linear algebra
linear algebra unsolved

Suppose that $ n\geq2$, and $ T: \mathcal M_n(\mathbb C)\longrightarrow\mathcal M_n(\mathbb C)$ is a linear mapping.
a) Prove that if $ T(I)\equal{}I$ and $ T$ maps invertible matrices to invertible matrices. Prove that $ T$ is invertible itself, and \[ T(\mbox{Rank }A)\equal{}\mbox{Rank }A\]for each $ A\in\mathcal M_n(\mathbb C)$.
b) Suppose $ \det T(A)\equal{}\det A$ for each $ A\in \mathcal M_n(\mathbb C)$. Can we reach to first part's conclusions?


geometry
geometric transformation
reflection
inequalities
geometry theorems

Darij gave us a wealth of information about isogonal conjugates in a previous thread (thank you! :D), but I don't know much useful about isotomic conjugates. Does anyone know any cool properties of these?

So far I know:
1) Let the cevians through a point $P$ meet the $BC,CA,AB$ at $U,V,W$ respectively. Reflect $U,V,W$ in the midpoints of their sides to form $U',V',W'$. It follows (from flipping the ceva fraction over) from Ceva's theorem that $AU', BV', CW'$ are concurrent at a point $P'$, which we call the isotomic conjugate of $P$.

2) The isotomic conjugate of a point (in areals) $(x,y,z)$ has unnormalised co-ordinates $(1/x, 1/y, 1/z)$. This follows because a cevian through $(x,y,z)$ will meet BC at $(0,y,z)$, so its reflection is $(0,z,y)=(0, 1/y, 1/z)$, and projecting back up from all three sides gives $P'(1/x,1/y,1/z)$

[quote="Ilthigore"]So far I know:
1) Let the cevians through a point $P$ meet the $BC,CA,AB$ at $U,V,W$ respectively. Reflect $U,V,W$ in the midpoints of their sides to form $U',V',W'$. It follows (from flipping the ceva fraction over) from Ceva's theorem that $AU', BV', CW'$ are concurrent at a point $P'$, which we call the isotomic conjugate of $P$.[/quote]

An interesting result I know about two points $P,$ $P',$ isotomic conjugates with respect to a given triangle $\bigtriangleup ABC,$ inwardly to it, is as follows:

If $\bigtriangleup UVW,$ $\bigtriangleup U'V'W',$ are the cevian triangles of $P,$ $P'$ respectively, then the follows results may be proven.

$\frac{AP}{PU}+\frac{BP}{PV}+\frac{CP}{PW}= \frac{AP'}{P'U'}+\frac{BP'}{P'V'}+\frac{CP'}{P'W'}$ $,(1)$  and  $\frac{AP}{PU}\cdot \frac{BP}{PV}\cdot \frac{CP}{PW}= \frac{AP'}{P'U'}\cdot \frac{BP'}{P'V'}\cdot \frac{CP'}{P'W'}$ $,(2)$

It is easy to show these results as a simple application of the below, not well known, [b][size=109]Van Aubel's[/size][/b] equality, about an arbitrary point $P,$ inwardly to a given triangle $\bigtriangleup ABC,$ with the triangle $\bigtriangleup UVW,$ as it's cevian one.

$\left(\frac{AP}{PU}\cdot \frac{BP}{PV}\cdot \frac{CP}{PW}\right)-\left(\frac{AP}{PU}+\frac{BP}{PV}+\frac{CP}{PW}\right) = 2$ $,(3)$

They are well known the inequalities $\frac{AP}{PU}\cdot \frac{BP}{PV}\cdot \frac{CP}{PW}\geq 8$ $,(4)$ and $\frac{AP}{PU}+\frac{BP}{PV}+\frac{CP}{PW}\geq 6$ $,(5)$

But always is true the equality $(3),$ named as  [b][size=109]Van Aubel's[/size][/b] equality.

kostas Vittas.


[i]There are $p$ points bounded by an $n\times n$ square.  There is a line that goes through all points exactly once.  What is the maximum required distance of this line in terms of $n$ and $p$.[/i]

Masoud Zargar

As it looks from here, you should put some more constraints, since you can always increase the length of an arbitrary line infinitely.

Or I'm misunderstanding the problem. (Which is quite possible - I'm taking that the "line" is arbitrarily curved? Or is it?)

The so-called "line" is a union of small (straight) lines connecting points.  Also, I should say that we are to find the maximum length that would be [i]required[/i].  In other words, find an arrangement of the points in which the line that is required to connected the points is of maximum length.

If it is required, than it should be stated :)

So, we must have a POLYGONAL line with NO OTHER vertices than given points - and with that constraint we have to maximize its length: is that finally a correct and complete statement?

And as the source we have OWN, maybe you belong to Squareland

Stop posting them Masoud as i am sure you know the deadlines.

This problem is different.  The other one has more conditions. And besides, I already posted them.  I just wanted a solution for the pleasure of it.

This is a corollary to Problem463 of current issue of Olymon (Canadia Mathematical Olympiad Correspondence Program)

The deadline is November 30, 2006, So this is violation of competition rule.

[quote="boxedexe"]Source=own[/quote]
You are giving yourself full credit for this problem,With out mentioning the true source of problem.This is while, rephrasing a problem does not make you the author.

Such posts are either locked or deleted from Aops/mathlinks

Okay.  Sorry, I forgot to mention what it was inspired by.

EDIT: Tried to edit, but I can't.  Anyway, it was inspired by that problem.  However, I don't think it'll be of much use to solve that problem.  I'll wait for others to send in their solutions as well.  Then, hopefully, this discussion can continue.


If a certain negative number is multiplied by six, the result is the
same as 20 less than the original number.  What is the value of the
original number?


A sequence of 20 terms is formed by using the rule $ f(x)\equal{}4x\minus{}1$ for  $ x \equal{} 1, 2, 3, \ldots, 20$. What is the sum of the last eight terms?

The last 8 terms would occur for x=13,14,...20
Pair up the first and last terms, then the second and the second to last and so on, so that you get 4 pairs.  Observe that each pair adds up to $ 4(33)\minus{}2$.

Since there are 4 pairs adding up to this, we just have to compute $ 4(4(33)\minus{}2)\equal{}16(33)\minus{}8\equal{}528\minus{}8\equal{}520$.

This is just the same trick that Gauss used to add up the integers 1 to 100 very quickly!  You probably noticed that in one of the animated proofs on the left hand side of the page.


As people that live on this planet, we should know how the rest of the world lives.

Visit this site: [url=http://www.miniature-earth.com]The Miniature Earth[/url]

Let's discuss what you think here.

I liked it very much, and thought it was nicely done.

I agree with the "point" that we should all be more aware of the human race as a whole. We should all recognize that not everyone is as fortunate as us.

But I don't necessarily find those statistics shocking. I feel that to some degree, the movie was targeted at the small percentage of those who are "well-off". Obviously from that perspective, much of the world is behind. I don't feel that the world is behind. Merely the message I take away is that the difference between the wealthy and poor is greater than one would expect. [i]But there is nothing wrong with that.[/i] Although it would obviously be preferable for everyone to be "well-off", if that were indeed the case, then everyone would then simply be "average".

I highly doubt it, but people might walk to nearby Internet booths to go on the Internet. They might not be able to afford a personal computer, so they just pay some income to go on an hour or so.

That is a possibility.

[quote="banblink182"][quote="MysticTerminator"]uh...just by living in the US, I think that pretty much automatically qualifies you as wealthy, in some sense. The fact that you have internet access just underscores that.[/quote]

I completely disagree (unless you're referring to some very technical sense when you say "in some sense") - there are many destitute people living in the United States. Besides, I'm not just referring to wealth, I'm also referring to the other "privileges".

The fact that I have internet access is completely irrelevant if you don't assume your first statement to be true. I could very well be wealthy without the automatic qualification that everyone in the United States is wealthy "in some sense".[/quote]

There are indeed a high number of people living below the poverty line, but to my knowledge, they still live in relative comfort compared to the poor in other, less developed countries. To be sure, I do not envy the position of the poor of US and am not trying to downplay the struggle they face; I'm just saying that they may not represent the absolute rock bottom of survival. Also, although the US is not considered a welfare country like England or Sweden, etc., we still do have some welfare programs in place that I am sure are not replicated in some other countries.

I don't see what's wrong with my pointing out that you have internet access...how many people do you seriously think have internet access? As hwenterprise suggests, some people may be able to access it even when they don't own a computer themselves (e.g. at a library or something), but there are a lot of people who don't have any access whatsoever. Period.

These stats are FAKE. From snopes:

[quote]This
thing has been circulating for ages (in Internet time) now, and people seem to find the contrasts it highlights between the "haves" and "have-nots" of the world compelling. However, much of the information it presents is questionable, out of date, or poorly-defined:

    57 Asians
    21 Europeans
    14 from the Western Hemisphere, both north and south
    8 Africans

According to the United Nations World Population Prospects document, the world population in the year 2000 was about 6.1 billion, with this geographic distribution:

    * Africa: 794,000,000
    * Asia: 3,672,000,000
    * Latin America and Caribbean: 519,000,000
    * Europe: 727,000,000
    * North America: 314,000,000
    * Oceania: 51,000,000

If we calculate the corresponding percentages (and lump North America, Latin America, the Caribbean, and Oceania into the "western hemisphere" category), we get the following ratios for our population of 100:

    * Asians: 60
    * Europeans: 12
    * Western Hemisphereans: 15
      (9 Latin Americans/Caribbeans, 5 North Americans, 1 Oceanian)
    * Africans: 13

    52 would be female
    48 would be male

According to that same United Nations document, the world population in the year 2000 consisted of 3,051,099,000 men and 3,005,616,000 women, which (with a little rounding) breaks down to 50 men and 50 women in a population of 100.

    70 would be non-white
    30 would be white

Here we run into definitional problems trying to lump entire continents' worth of people into one class based on some nebulous concept of color. What makes a person "white" or "black"? If we say that Africans are considered "black," does that categorization apply equally to Nigerians, Egyptians, and South Africans? (Is the Middle East part of Africa or Asia?) Should the classification of Asians as "white" or "non-white" be based solely upon skin tone, or upon geographical and cultural factors as well?

The numbers given here seem to be based upon the classification of Europeans and Western Hemisphereans as "white" and Africans and Asians as "non-white" (and the assumption that those continents are homogeneous in racial composition). With those qualfications, a population of 100 (based on year 2000 numbers) would include 27 whites and 73 non-whites.

    70 would be non-Christian
    30 would be Christian

According to the chart at Adherents.com, which provides estimates of "the number of people who have at least a minimal level of self-identification as adherents" of a particular religion, the world's population in early 2001 was 33% Christian. So, our reduced population of 100 would be composed of 33 Christians and 67 non-Christians.

    89 would be heterosexual
    11 would be homosexual

This one is almost impossible to estimate with any reasonable degree of accuracy. The classification of people as being either "heterosexual" or "homosexual" is too dependent upon subjective criteria, and the answers of respondents to surveys about their sexual habits are too easily influenced by other social factors. The common figure of "10% of the population is homosexual" is often bandied about, but that number is derived from a misapplication of a Kinsey study which was not based upon a representative sample of the population. One can find estimates that place of the percentage of the population considered to be homosexual anywhere between 1-2% and 25%-35%, but a reasonable survey of the more controlled studies would put the figure in about the 2-3% range (for males, at least).

    6 people would possess 59% of the entire world's wealth, and all 6 would be from the United States.

This claim demonstrates the precariousness of trying to summarize a very large, diverse population in a few simple statistics. For starters, our miniature world of 100 people only includes 5 people from all of North America, so any statement involving 6 people from the United States just doesn't compute!

"Wealth" is a concept difficult to measure with any precision, but we can use Gross Domestic Product (GDP) as a reasonable approximation. If we take some figures from the CIA's World Factbook 2000, we find that the estimated GDP of the United States in 1999 was $9.255 trillion, out of a world total of $40.7 trillion. In other words, in 1999 the United States possessed about 23% of the world's wealth. If we assume that all 5 North Americans in our miniature world are from the United States, and that they have inherited an amount of wealth proportional to that held by the United States in the "real" world, together they'd still have only 23% of the world's wealth, not 59%. Even if you could find some combination of 6 people in our putative population of 100 who held 59% of the total wealth, they wouldn't all be from the United States.

    80 would live in substandard housing

This statement can't be assessed without knowing the definition of "substandard" being employed here. "Substandard" by whose standards? And if a full 80% of the world's population truly lives in "substandard" housing, doesn't that indicate whatever standard is being used must be too high?

Estimates for this figure are all over the map as well (some United Nations housing statistics are informative but don't really answer the question), but a 1999 article in International Wildlife puts the estimate at 33%, not the 80% figure offered here.

    70 would be unable to read

A 1998 UNICEF study put the world illiteracy rate at 16%, well short of the 70% claimed here.

    50 would suffer from malnutrition

The World Health Organization puts the malnutrition figure at about 33%.

    1 would be near death
    1 would be near birth

This statement is simply too vague to evaluate. At any given time, one person in a hundred is near death? Just how "near"? Is age a factor in this statistic?

    1 (yes, only 1) would have a college education

Again, we have to know whether our miniature world's inhabitants represent current population trends in age as well as other factors. According to the United Nations World Population Prospects document, the median age of the world's population was 26.5 in 2000, with that figure being lower in less developed (and more populous) areas of the world. (The median age of Africans, for example, was only 18.4 in 2000.) So, this statistic could be true simply because much of our miniature population would be too young to have finished college yet (assuming that "having a college education" means "graduated with the equivalent of a bachelor's degree"). However, if we assume everyone in our miniature world is of sufficient age and apply the current graduation rates of the USA (33%) to its share of the population (5), we'd have almost two college graduates from America alone. And other parts of the world (e.g., New Zealand, Netherlands, Britain, and Norway) have graduation rates equal to or higher than the USA's.

    1 would own a computer

Computer ownership rates in the USA now indicate that over 50% of American households have computers, so if we assume that "households" can be equated with "people" in our miniature world, our 5 Americans alone would have at least two computers between them.

On the other hand, if this is the kind of material having a computer gives one access to, the inhabitants of our miniature world just might opt to do without them. [/quote]

[quote="MysticTerminator"]There are indeed a high number of people living below the poverty line, but to my knowledge, they still live in relative comfort compared to the poor in other, less developed countries. To be sure, I do not envy the position of the poor of US and am not trying to downplay the struggle they face; I'm just saying that they may not represent the absolute rock bottom of survival. Also, although the US is not considered a welfare country like England or Sweden, etc., we still do have some welfare programs in place that I am sure are not replicated in some other countries.[/quote]

Not being the absolute rock bottom of survival and being wealthy are pretty different. I would not agree that being in the US automatically qualifies you as wealthy, but I would agree that the poor living in the US generally speaking live in relative comfort compared to the poor living in underdeveloped countries (such as Africa).

[quote="MysticTerminator"]I don't see what's wrong with my pointing out that you have internet access...how many people do you seriously think have internet access? As hwenterprise suggests, some people may be able to access it even when they don't own a computer themselves (e.g. at a library or something), but there are a lot of people who don't have any access whatsoever. Period.[/quote]

From a global perspective, I do not think many people have internet access. The fact that I have internet access, and therefore am one of the few does not say anything about the people in America.

My having internet access proves the fact that "being in America automatically qualifies you as wealthy" as much as the existence of a red balloon prove that all balloons are red.

[quote="banblink182"]as much as the existence of a red balloon prove that all balloons are red.[/quote]

darn, you mean proofs like that don't work? shucks...

anyway, enough of this silly argument; all of this is just missing the point


abstract algebra
invariant
Ring Theory
superior algebra
superior algebra unsolved

Let $ M$ be a module such that M has no submodule invariant under any automorphism of $ M$. Prove that $ M$ is semisimple homogeneous.

Thks in advance.

I think this is not true in general, but if the ring is Artinian then you can just use the radical.

Q is not a semisimple Z module.  The automorphisms of Q are the multiplications by nonzero elements of Q.  Any submodule invariant under all such automorphisms is divisible, so equal to 0 or Q.

If M is a module over an Artinian ring R, then consider JM where J is the Jacobson radical of R.  JM is clearly invariant under any automorphism, so JM=0, and M is a module over R/J, so is semisimple.  Each homogenous (=isotypic) component is characteristic, so M itself must be homogenous.


I think the answer is  $\frac{2004^3}{8} $. Since the expression is convex in each variable we may assume that all $ a_i$ are 0 or 1. Then the sum is exactly $\sum_{1\leq i<j\leq 2004} {(j-i)}-\sum_{ i<j,a_i=a_j} {(j-i)} $. Now, suppose k from the 2004 numbers are 0. It is clear that $\sum_{i<j,a_i=a_j=0}{(j-i)} $ is minimum when all indices of 0 values are consecutive. A similar argument for the sum when both numbers are 1 shows that the maximal value must be attained when the first k numbers are 0 and the last n-k are 1. But in this case we have some easy to compute sums and it is immediate after we compute them to show that the maximum value is attained when k=1002. Thus the maximal value must be the one stated in the begining of the solution.

[quote="arthur"]can you you rewrite it down with latex ?[/quote]

Suppose $a_{1},a_{2},\ldots, a_{2004} \in [0,1]$. Find the maximum of:
$\displaystyle \sum_{i>j} (i-j)|a_{i}-a_{j}|$.

Suppose 0<=a(1),a(2),...,a(2004)<=1.Find the maximum of
Sum(i>j) (i-j)*|a(i)-a(j)|


logarithms
limit
real analysis
real analysis unsolved

If $ x_n  \to 0$ does there exist p > 0 for which $ n^p x_n  \to 0$?

$ x_n \rightarrow 0$ is analogous to saying for an arbitrary $ \epsilon > 0$there exists $ N\in\mathbb{N}$ such that when $ n \geq N$ we have $ |x_n \minus{} 0| \equal{} |x_n| < \epsilon$.
So now let $ \epsilon' > 0$ be given and let $ \epsilon \equal{} n^p \epsilon'$. We have $ |x_n|<n^p|x_n \minus{} 0| \equal{} |n^p x_n \minus{} 0| < n^p \epsilon' \equal{} \epsilon$ and therefore $ n^p x_n \rightarrow 0$.

Thanks t0rajir0u.

Here is the problem that led me to this question:

If $ b_n  \equal{} o\left( {\frac{1}{n}} \right)$ show that there exists $ a_n$ for which $ \sum\limits_{n \equal{} 1}^\infty  {a_n }$converges and $ b_n  \equal{} o\left( {a_n } \right)$

[quote="beezees222"]We have $ |x_n| < n^p|x_n \minus{} 0| \equal{} |n^p x_n \minus{} 0| < n^p \epsilon' \equal{} \epsilon$ and therefore $ n^p x_n \rightarrow 0$.[/quote]
We don't have $ |x_n| < \epsilon'$ (you've mixed up your two epsilons).  In fact, as $ n \to \infty$ for any nonzero $ \epsilon$ we simply get $ \epsilon' \equal{} 0$.  Think carefully about what quantifiers you're using (this statement is "for all $ \epsilon$ there exists $ N$ such that for all $ n > N$...").  You'll notice that your proof puts no conditions on $ p$, which is clearly wrong.

i remember some questions like this in our section about Abels and Taubers ? perhaps those may be useful for your question

[quote="Kent Merryfield"]As for the full problem (§outh§tar's post #4), assuming that the intent is for all these sequences to be nonnegative, I think it's false.[/quote]

Well it doesn't say the sequences have to be all positive. Though if they were the series with nth term a(n) could be made to be absolutely convergent.

kenn4000, I looked up abel and tauber on google but here we require a sequence to be constructed and we don't know if anything is abel summable.

EDIT: I think I figured it out! It's a surprising answer, for sure. An enormous hint is contained in my response to Kent Merryfield.

If the sequences don't have to be non-negative then $ a_n \equal{} \frac{(\minus{}1)^n}{n}$ solves the problem pretty straightforwardly, so I (and Kent, apparently) doubt that was the intended question.

Works for me :p

It's very unlikely that's the answer the author was looking for since the keyword was "construct" and alternating series (or abel's theorem) hadn't been introduced. I initially thought of positive sequences, hence the initial post. I'll keep the problem in mind but I don't know how at all such a sequence would be found.

If it's any help, the first part of the question asked to show that if a(n) is a decreasing positive sequence and the series with nth term a(n) converges, then a(n) = o(1/n).

[quote="§outh§tar"]
If $ b_n \equal{} o\left( {\frac {1}{n}} \right)$ show that there exists $ a_n$ for which $ \sum\limits_{n \equal{} 1}^\infty {a_n }$converges and $ b_n \equal{} o\left( {a_n } \right)$[/quote]
If $ a_n$ and $ b_n$ are positive, this is a complete nonsense. Just take $ b_n\equal{}\frac 1{n\log n}$. What was meant is "if $ 0<b_n\equal{}o(1/n)$, show that there exists a decreasing sequence $ a_n>0$ for which $ \sum_n a_n<\plus{}\infty$ and $ a_n\ne o(b_n)$".

[quote="§outh§tar"]If it's any help, the first part of the question asked to show that if a(n) is a decreasing positive sequence and the series with nth term a(n) converges, then a(n) = o(1/n).[/quote]
That's not exactly the most obvious thing in the world. (And as an aside, I'm moving this out of the "computations and tutorials" forum, where I no longer think it fits.)

Here's a somewhat inelegant argument for the contrapositive:

Suppose that it is not true that $ a_n \equal{} o\left(\frac1n\right).$ That means that there exist a constant $ c > 0$ and an increasing sequence of indices $ n_k$ such that $ a_n > \frac {c}{n_k}.$ (For the sake of notation, let $ n_0 \equal{} 0.$) Then:

$ \sum_{n \equal{} 1}^{\infty}a_n \equal{} \sum_{k \equal{} 1}^{\infty}\sum_{n \equal{} n_{k \minus{} 1} \plus{} 1}^{n_k}a_n > \sum_{k \equal{} 1}^{\infty}(n_k \minus{} n_{k \minus{} 1})a_{n_k}$

$ > c\sum_{k \equal{} 1}^{\infty}\left(1 \minus{} \frac {n_{k \minus{} 1}}{n_k}\right)$

If we were to have that $ \sum a_n$ converges, then this last sum would converge; if it converges, then its terms tend to zero. Hence, if $ \sum a_n$ converges, then $ \lim_{k\to\infty}\frac {n_{k \minus{} 1}}{n_k} \equal{} 1$ and for sufficiently large $ k,$ $ \frac {n_{k \minus{} 1}}{n_k} > \frac12.$

Let $ n$ be sufficiently large but otherwise arbitrary. There exists $ k$ such that $ n_{k \minus{} 1} < n\le n_k.$ Then

$ na_n > n_{k \minus{} 1}a_{n_k} \equal{} n_ka_{n_k}\cdot \frac {n_{k \minus{} 1}}{n_k} > c\cdot\frac12.$

Hence for sufficiently large $ n,$ $ a_n > \frac c2\cdot\frac1n$ and $ \sum a_n$ diverges by comparison to the harmonic series.

So if we don't have $ a_n \equal{} 0\left(\frac1n\right),$ then $ \sum a_n$ diverges. (Assuming $ a_n$ is positive decreasing.)

Thanks KM.

Another way I remember is to use Cauchy's condensation test. The proposition follows immediately from the observations that the sequence is positive non-increasing and 2^n > n

Be careful! $ 2^n>n$ has nothing to do with it; you have $ 2^n$ both as the factor and as the index in the condensation test ;)


Let $ ABC$ be an arbitrary acute angled triangle. For any point $ P$ lying within the triangle, let
$ D$, $ E$, $ F$ denote the feet of the perpendiculars from $ P$ onto the sides $ AB$, $ BC$, $ CA$ respectively.
Determine the set of all possible positions of the point $ P$ for which the triangle $ DEF$ is isosceles.
For which position of $ P$ will the triangle $ DEF$ become equilateral?

An easy problem :
here goes the solution: Since B,D,E , P are concyclic, we have $ DE\equal{} PB \sin B$ and similarly due to concyclity of A,P,D and F we have $ DF \equal{} PA \sin A$. DEF to be isoscles, some two sides must be equal and so if DE= DF, we have $ \frac{PA}{PB} \equal{} \frac{\sin B}{\sin A} \equal{} \frac{b}{a}$(By sine law in triangle $ ABC$)

THis implies $ P$ lie son the Appolonius circle corresponding to $ A,B$. so if P lies on eithe rone of the appolonius circles corresponindg to $ A,B$ or $ B,C$ or $ C,A$, we have $ DEF$ to be isisceles. For equilateral, $ P$ must be a common point for two of the three above mentioned Appolonius circles.

THe problem can also be solved using the well known relations for the sidelengths of a pedal trinagle corresponding to a point in the plane of a triangle(of course a degenerate trinagle in case P lie son the circumcircle).


geometry
symmetry
angle bisector
perpendicular bisector
geometry solved

See figure below.

Let A and B be two points on a circle with center O such that < AOB = 90°.

P and Q are points on the arc AB such that PQ // AB. Let X and Y be the intersection between PQ and OA; PQ and OB. What's the value of $PX^2+ PY^2$ ?

I think we have [tex]PX^2+PY^2=2OA^2=AB^2[/tex].

Again a trivial problem with no solution posted...

[quote="wpolly"]$PX^2+PY^2=2OA^2=AB^2$[/quote]

Indeed, this is right. Here is a proof:

Because of the symmetry, the angle bisector g of the angle AOB is perpendicular to the line AB. Since PQ || AB, the same angle bisector g is also perpendicular to the line PQ, and since we know that the point O lies on the perpendicular bisector of the segment PQ (this is because the segment PQ is a chord of a circle with center O), we see that the line g is the perpendicular bisector of the segment PQ. Consequently, our whole configuration is symmetric with respect to the line g; to be precise, the points A and B are symmetric to each other with respect to g, the points P and Q are symmetric to each other with respect to g, and consequently, the points X and Y are symmetric to each other with respect to g. These symmetries yield PT = QT and XT = YT. Denote PT = QT = d and XT = YT = s.

Since the line g is the angle bisector of the angle AOB, we have < TOX = < BOA / 2 = 90 / 2 = 45. Since we also have < OTX = 90, the triangle OTX is a right isosceles triangle. Thus, OT = XT, so that OT = s.

We can WLOG assume that the point P lies inside the segment XQ; then, PX = XT - PT = s - d and PY = YT + PT = s + d. Then,

$PX^2 + PY^2 = \left(s-d\right)^2+\left(s+d\right)^2 = 2\cdot\left(s^2+d^2\right)$.

But $s^2+d^2 = OT^2 + PT^2$, and Pythagoras in the right-angled triangle OTP yields $OT^2 + PT^2 = OP^2$. Of course, OP = OA since the points P and A lie on one circle with center O. Thus, $s^2+d^2 = OP^2=OA^2$, and $PX^2+PY^2 = 2\cdot\left(s^2+d^2\right) = 2\cdot OA^2$.

Since the triangle AOB is a right isosceles triangle, we have $AB = \sqrt2 \cdot OA$, and thus $AB^2 = 2\cdot OA^2$. Thus, we can rewrite $PX^2+PY^2 = 2\cdot OA^2$ as $PX^2+PY^2 = AB^2$. And we are done.

I begin feeling like Myth posting in the Pre-Olympiad section...

  Darij


Let $ a,b,c\geq 0$.Prove that:
$ (a^2\minus{}bc)\sqrt{a^2\plus{}4bc}\plus{}(b^2\minus{}ca)\sqrt{b^2\plus{}4ca}\plus{}(c^2\minus{}ab)\sqrt{c^2\plus{}4ab}\geq 0$.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$.Prove that:
$ (a^2 \minus{} bc)\sqrt {a^2 \plus{} 4bc} \plus{} (b^2 \minus{} ca)\sqrt {b^2 \plus{} 4ca} \plus{} (c^2 \minus{} ab)\sqrt {c^2 \plus{} 4ab}\geq 0$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157776


geometry
perimeter
circumcircle
trigonometry
geometric transformation
rotation
angle bisector

Suppose $ABCD$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\ell_1$ and $\ell_2$, which are also $a$ units apart. The square $ABCD$ is placed on the plane so that sides $AB$ and $AD$ intersect $\ell_1$ at $E$ and $F$ respectively. Also, sides $CB$ and $CD$ intersect $\ell_2$ at $G$ and $H$ respectively. Let the perimeters of $\triangle AEF$ and $\triangle CGH$ be $m_1$ and $m_2$ respectively.

Prove that no matter how the square was placed, $m_1+m_2$ remains constant.

Let p be a line perpendicular to the lines $l_1 \parallel l_2$ intersecting these 2 parallel lines at $P_1, P_2$, so that their distance is $d = P_1P_2$. Let (O) be a circle with sufficiently large radius $r$ centered at an arbitrary point O on the segment $P_1P_2$. Let ABCD be a square with the side $a = r \sqrt 2$ inscribed in this circle intersecting both the lines $l_1, l_2$ in the way described in the problem. Let the angle $\phi = \angle P_1OA$ be fixed and let the square circumcenter O move on the perpendicular p. The triangles $\triangle AEF \sim \triangle CHG$ remain similar for all positions of the  circumcenter O and the sum of their A- and C-altitudes is constant, $h_a + h_c = 2r \cos \phi - d$, and the sum $m_1 + m_2$ of their perimeters (proportional to their altitudes) is then also constant. Consequently, we only have to consider the square circumcenter O being the midpoint of the segment $P_1P_2$, when the triangles $\triangle AEF \cong \triangle CHG$ are congruent, i.e., their perimeters $m_1 = m_2$ are equal, and we have to show that the perimeter $m_1$ of the triangle $\triangle AEF$ is constant for any rotation angle $\phi = \angle P_1OA$ iff a = d.

Denote e = AF, f = AE, h = EF the sides of the right triangle $\triangle AEF$. Let $A' \in EF$ be the foot of the bisector AO of the right angle $\angle FAE$. Using $h^2 = e^2 + f^2$, the angle bisector length b = AA' is

$b = \sqrt{ef\left(1 - \frac{h^2}{(e + f)^2}\right)} = \frac{ef \sqrt 2}{e + f}$

WLOG, assume $\angle AEF < \angle AFE$. Then $\angle EFA = \phi + 45^\circ$ and

$\frac f e = \tan \angle EFA = \frac{1 + \tan \phi}{1 - \tan \phi}$

Solving for e, f, h

$e = \frac{b \sqrt 2}{1 + \tan \phi},\ \ f = \frac{b \sqrt 2}{1 - \tan \phi}$

$h = \sqrt{e^2 + f^2} = \frac{2b \sqrt{1 + \tan^2 \phi}}{1 - \tan^2 \phi} = \frac{2b \cos \phi}{\cos^2 \phi - \sin^2 \phi}$

The perimeter $m_1$ of the triangle $\triangle AEF$ is then

$m_1 = e + f + h = \frac{2b \sqrt 2}{1 - \tan^2 \phi} + \frac{2b \cos \phi}{\cos^2 \phi - \sin^2 \phi} =$

$= \frac{2b \sqrt 2 \cos^2 \phi + 2b \cos \phi}{\cos^2 \phi - \sin^2 \phi} = \frac{2b \cos \phi (\sqrt 2 \cos \phi + 1)}{2 \cos^2 \phi - 1} =$

$= \frac{2b \cos \phi}{\sqrt 2 \cos \phi - 1}$

On the other hand, we have

$b = AA' = OA - OA' = r - \frac{d}{2 \cos \phi} = d\ \frac{\frac{2r}{d} \cos \phi - 1}{2 \cos \phi}$

$m_1 = d\ \frac{\frac{2r}{d} \cos \phi - 1}{\sqrt 2 \cos \phi - 1}$

It is clear that the perimeter $m_1 = d$ is constant, iff $\frac{2r}{d} = \sqrt 2$ or using $a = r \sqrt 2$, iff $a = d$.

Just a little bit of thinking before replying will, no doubt, produce in an olympiad

  caliber person the following observation:

    the ex-circles opposite the right angles of the triangles in question, having

    perimeters of m1 and m2, have centers located in the same point on the

    diagonal of the square connecting these right angles.

    From this point  the problem is just a pis cake.

    T.Y.

    M.T.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO2003Problem2

by Vo Duc Dien

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO2003Problem2

Vo Duc Dien

WLOG the square is a unit square. Employ Cartesian coordinates. Let $A=(0,0),B=(0,1),C=(1,1),D=(1,0).$ Then, suppose that $E=(0,b)$ and $F=(a,0)$ (sorry for using the variable $a$ again, it just means an arbitrary real between 0 and 1 now). Then, let line $\ell_2$ intersect line $AB$ again at $P$. We then have $$P=(0,b+\frac{\sqrt{a^2+b^2}}{a})$$ by e.g. dropping a perpendicular from $E$ to $\ell_2$ and using similar triangles. Thus, the equation of line $\ell_2$ is $$y=-\frac{b}{a}x+b+\frac{\sqrt{a^2+b^2}}{a}.$$ Thus, intersecting this with line $y=1$, we have $$G=(\frac{ab-a+\sqrt{a^2+b^2}}{b},1).$$ This means that $$GC=1-\frac{ab-a+\sqrt{a^2+b^2}}{b}=\frac{a+b-ab-\sqrt{a^2+b^2}}{b}.$$ Furthermore, note that $\triangle CGH\sim \triangle AFE$ due to the parallel lines, so the perimeter of $\triangle CGH$ is also equal to $$CG(\frac{a+b+\sqrt{a^2+b^2}}{a})$$ $$=(\frac{a+b-ab-\sqrt{a^2+b^2}}{b})(\frac{a+b+\sqrt{a^2+b^2}}{a})$$ $$=\frac{ab+b^2+b\sqrt{a^2+b^2}-a^2b-ab^2-ab\sqrt{a^2+b^2}+a^2+ab+a\sqrt{a^2+b^2}-a\sqrt{a^2+b^2}-b\sqrt{a^2+b^2}-a^2-b^2}{ab}$$ $$=\frac{ab-a^2b-ab^2-ab\sqrt{a^2+b^2}+ab}{ab}$$ $$=2-a-b-\sqrt{a^2+b^2}.$$ Since the perimeter of $\triangle AFE$ is $a+b+\sqrt{a^2+b^2}$, the sum of the two perimeters is just 2, so we are done.


[quote="kunny"]Write down one of figure by selecting among numbers $ 1,\ 2,\ 3,\ 4,\ 5$ at random. Doing the procedure repeatedly, let denote $ P_n\ (n\geq 5)$ the probability such that you can see the five kinds of figures are all presented in the $ n$ th round for the first time. Find the expected value $ E_n$ of $ n$.[/quote]

$ P_n\equal{}\frac {\binom 54\times\left(4^{n\minus{}1}\minus{}\binom 41 3^{n\minus{}1}\plus{}\binom 42 2^{n\minus{}1}\minus{}\binom 43 1^{n\minus{}1}\right)\times 1}{5^n}$

$ \implies \boxed{\boxed{P_n\equal{}\frac {4^{n\minus{}1}\minus{}4(3^{n\minus{}1})\plus{}6(2^{n\minus{}1})\minus{}4}{5^{n\minus{}1}}}}$

Could you explain how you got $ P_n$? Also, how does this determine $ E_n$?

I really like how that answer gives $ P_1 \equal{} \minus{}1$   :roll: .  (The correct formula for $ n \geq 5$ also happens to work for $ n \geq 1$.)

makar tried (but failed) to do a standard inclusion-exclusion argument; from there, you would need to sum an infinite series to get the expected value.  There's a much slicker proof based on linearity of expectation.  This is called the coupon collector problem, and a search on the forum should bring up plenty of past discussions.


If you release a ball from a certain height and time it with a stopwatch, how would you calculate the height (98-100% accurate) in which you released the ball (do not ignore air resistance)?

I would measure the time for known heights and plot a graph.

you need to know the coefficient of air friction, i forget exaclty how to do it, i found [url=http://www.karlscalculus.org/l12.1.html]this[/url] site which does it, it's different than the way i learned it, but the same idea just a different way


HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14720[/img]

 :(

The A-excircle of triangle ABC touches AB,AC at M,N respectively, AM = AN.
AM+AN=AB+AC+BC=AD+AE, so DM=EN. Let K be symmetric to E wrt MN.
Because triangle AMN is isosceles and DM=EN=KN, then DMNK is a parallelogram.
Hence DK // MN ⊥ KE, MN=DK ≤ DE.

[quote="jrrbc"] [color=darkred]Find points $ D$ and $ E$ on the extended sides $ AB$ & $ AC$ of a triangle $ ABC$ such that $ BD + CE = a$ and the length of $ DE$ is minimum.[/color] [/quote]

[color=darkblue][b][u]Proof (metric).[/u][/b] Exists the point  $ F\in (BC)$ for which $ \|\begin{array}{c} BF = BD = x \\
 \\
CF = CE = y \\
 \\
x + y = a\end{array}$ . Thus, $ \|\begin{array}{c} DF = 2x\cos\frac B2 \\
 \\
EF = 2y\cos \frac C2\end{array}$ and $ m(\widehat {DFE}) = 90^{\circ} + \frac A2$ .

Apply the [b]generalized Pythagoras' theorem[/b] to $ DE$ in $ \triangle\ DEF\ \ : \ \ DE^2 = 4x^2\cos^2\frac B2 + 4y^2\cos^2\frac C2 + 8xy\cos\frac B2\cos\frac C2\sin\frac A2$ .

Prove easily that $ \frac {\cos^2\frac B2}{b(p - b)} = \frac {\cos^2\frac C2}{c(p - c)} = \frac {\cos\frac B2\cos\frac C2\sin\frac A2}{(p - b)(p - c)} = \frac {p}{abc}$ . Therefore, the length of the segment $ [DE]$ is minimum $ \Longleftrightarrow$

$ b(p - b)x^2 + 2(p - b)(p - c)xy + c(p - c)y^2\ - \ \mathrm{min.}$ $ \Longleftrightarrow$ $ \frac {b(p - b)x^2 + 2(p - b)(p - c)xy + c(p - c)y^2}{(x + y)^2}\ - \ \mathrm{min.}$ . Denote

$ \frac xy = t > 0$ and $ f(t) = \frac {b(p - b)t^2 + 2(p - b)(p - c)t + c(p - c)}{(t + 1)^2}$ . Prove easily that $ f'(x)\mathrm{\ .s.s.}^*\ [(p - b)t - (p - c)]$ , i.e. $ t_{\mathrm{min}} = \frac {p - c}{p - b}$ .

Thus, $ \frac {x}{p - c} = \frac {y}{p - b}\implies \boxed {\begin{array}{c} x = p - c \\
\ y = p - c\end{array}}\implies$  $ D$ , $ E$ are the tangent points of the $ A$ - exincircle with the sidelines $ AB$ , $ AC$ respectively.

$ (*)$ [b]Notation.[/b] $ X\ .s.s.\ Y\ \Longleftrightarrow\ XY\ > \ 0\ \ \vee\ \ X = Y = 0$ .[/color]

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14737[/img]

 :thumbup: