This is from AoPS Vol I:

For every $ n$ the sum of $ n$ terms of an arithmetic progression is $ 3n^2 \plus{} 2n$.  What is the $ r^{th}$ term of the sequence in terms of $ r$?

First I set it equal to the sum of any sequence:

$ \frac{n}{2} (2a \plus{} (n \minus{} 1)d) \equal{} 3n^2 \plus{} 2n$

$ na \plus{}\frac{n^2d}{2} \minus{} \frac{nd}{2} \equal{} 3n^2 \plus{} 2n$

What do you do from here?  (hints only please!)

I've tried factoring out the $ n$ and or $ d$, but it doesn't seem to help at all.

Hint: Divide by $ n$, and $ a_r \equal{} a \plus{} (r \minus{} 1)d$.

EDIT: Yeah srry I see.

[quote="Stokes93"]:huh:

Well, $ a_n \equal{} a \plus{} (n \minus{} 1)d$

And $ S_n \equal{} \frac {n}{2}(a_1 \plus{} a_n)$ or $ S_n \equal{} \frac {n}{2}(a \plus{} a \plus{} (n \minus{} 1)d)$

EDIT: the $ r^{th}$ term of the sequence is equal to $ 6r \minus{} 1$

I just don't see how to get there....  I've also tried using the fact that the $ r^{th}$ term is $ S_r \minus{} S_{r \minus{} 1}$, but I didn't get to far with that either.

Solutions are okay now.[/quote]

Um try doing $ S_r \minus{} S_{r \minus{} 1}$ again because I am getting $ 6r \minus{} 1$.

I'll give you a hint and two solutions. Both solutions use the hint but in slightly different ways.

[hide="Hint"]
Play close attention to the part of the problem where it says, "for every $ n$".
[/hide]

[hide="Solution 1"]
Since it is true for every $ n$, we can plug in any value of $ n$ and the resulting equation has to be true. We conveniently plug in $ n \equal{} 1$ in $ na \plus{} \frac{n^2d}{2} \minus{} \frac{nd}{2} \equal{} 3n^2 \plus{} 2n$ to find that $ a \equal{} 5$. Therefore, our simplified equation is:

$ 5n \plus{} \frac{n^2d}{2} \minus{} \frac{nd}{2} \equal{} 3n^2 \plus{} 2n$

Now we can plug in any value of $ n$ to receive the value of $ d$. The next largest value of $ n$ we can choose is $ 2$ so we plug that it to find that $ d \equal{} 6$. The answer follows from here.
[/hide]

[hide="Solution 2"]
View both sides as polynomials in $ n$.

$ \left(\frac{d}{2}\right)n^2 \plus{} \left(a \minus{} \frac{d}{2}\right)n\equal{} 3n^2 \plus{} 2n$.

Because this equation is satisfied for all values of $ n$, both polynomials must have the same coefficients. Therefore,

$ \frac{d}{2} \equal{} 3\implies d \equal{} 6$

and

$ a \minus{} \frac{d}{2} \equal{} 2\implies a \minus{} 3 \equal{} 2\implies a \equal{} 5$.

Again, the answer follows from here.
[/hide]


Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \frac{a^2\plus{}b^2\plus{}c^2}{(a\plus{}b\plus{}c)^2} \plus{}\frac{2(a^3b\plus{}b^3c\plus{}c^3a)}{(ab\plus{}bc\plus{}ca)^2} \ge 1.\]

We have:
\[ LHS \minus{} RHS \equal{} 2\frac {\sum_{cyc}[(a^2 \minus{} bc)^2(ab \plus{} b^2)] \plus{} 2\sum_{cyc}abc(a^3 \minus{} abc)}{(a \plus{} b \plus{} c)^2(ab \plus{} bc \plus{} ca)^2}
\]
:)

EDIT: I think there shouldn't be any 2 in the denominator.
Can someone please check this identity for me?
I am sorry for my carelessness... I edited for the last time.

without brute forcing

$ \sum_{cyc}a^5b\geq \sum_{cyc} a^3b^2c$ (easy by cauchy)
which implies the weaker $ \left(\sum_{cyc}a^3b\right)\left(\sum_{cyc}a^2\right)\geq \left(\sum a^2b^2\right)\left(\sum_{cyc}ab\right)$
which implies the weaker $ \left(\sum_{cyc}a^3b\right)\left(\sum_{cyc}a\right)^2\geq \left(\sum_{cyc}ab\right)^3$
which is obviously equivalent to our inequality by the identity $ \left(\sum_{cyc}a\right)^2 \equal{} \sum_{cyc}a^2 \plus{} 2\sum_{cyc}ab$
 :)

[quote="Albanian Eagle"]
which implies the weaker $ \left(\sum_{cyc}a^3b\right)\left(\sum_{cyc}a\right)^2\geq \left(\sum_{cyc}ab\right)^3$
:)[/quote]
Actually, this inequality is just Holder Inequality:
\[ (a^3b\plus{}b^3c\plus{}c^3a)(b\plus{}c\plus{}a)(b\plus{}c\plus{}a) \ge (ab\plus{}bc\plus{}ca)^3\]

haha, OK I didn't notice that. It is indeed a simple, but nice inequality.

here is another simple one

$ 1\plus{}\frac{3\sum_{cyc} a^4}{(\sum_{cyc} ab)^2}\plus{}\frac{2\sum_{cyc} a^2b^2}{\sum_{sym}a^3b}\geq 3\frac{\sum_{cyc} a^3b}{\sum_{cyc}a^4}$


Show that there exist infinitely many positive integers $n$ with the following property: the sum of all the positive divisors of $n$, excluding $n$ itself, equals $n+12$.

Don't look at the solution, unless you tried really really hard:

[hide]Take $n = 6p$, where $p$ is prime.[/hide]


The first problem is directly taken from the meet we had today (The official title of the Math Team I forgot, but it was something like "Intermediate Eastern Massachustts Mathematical Competition" or something like that, and [b]it has nothing to do with MAML[/b]. I just wanted to clarify that since I may have said wrong things...)

Anyways, n number of lines are drawn through a circle, and each line goes through 2 points on the circle. No two lines are the same. Find the formula for the number of maximum sections that you can make by drawing n lines with the above rule. (By "section" they mean an enclosed space like a triangle, quadrilateral, two lines and an arc which surrounds an area, etc.)

Now find the formula for getting the least number of sections you can make with n lines.

Woaaah thats really cool.

[hide]

Most: It will be the n+2th Fibonacci number, where F_1 = F_2 = 1, and F_n+1 = F_n+F_n-1.

Least: 2n

[/hide]

Neither of you got the first one right, and TheSouthpaw is correct for the second one.
Also try to explain why a bit more. See if you can prove that your answer is correct (although I know this might be a bit above MATHCOUNTS level, but this concept is very important to middle-school leveled competitions)

I think Rep is right on the second one, not Southpaw, unless there's a restriction that the lines must intersect each other.

[quote] If at first you don't succeed, try another method( don't ever be thickheaded)   :lol:  [/quote]

The answer I think to the maximum number of sections created would be C(n,2) + n + 1. Tell me if I am wrong because it worked before.

Which can also be expressed as C(n, 2) + C(n, 1) + C(n, 0), (not-so) coincidentally.  That's right.  Now, why?  (That is, prove it!)

It means the number of sections that we can cut the circle into, does it not?  Or can I aim to make sections outside of the circle as well?  If that's the case, then I think Rep123, white_horse and I all had the same misconception.

But actually, I'm still getting the same answer -- am I doing something wrong?

no you're right. i missed the little spiel on sections meaning enclosing an area.

[quote="Tare"]Ah yes, I forgot about that...technically Rep123Max is right then.[/quote]

What do you mean, technically?  If there are multiple solutions depending on how you define something, then he needs both solutions to be right.  If there's only one solution, then either he is right or he's not.

- mathfanatic, who is just wondering about how one defines "technically" in the mathematical sense

I just forgot to write
[quote="JBL"]that the lines must intersect each other[/quote]
for the second question, making n+1 the correct answer (Rep123Max's answer). However what I wanted was 2n (TheSouthpaw's answer since I thought I made it clear that all lines must interect [b]at least once[/b] (hint hint :lol: )

Sections outside the circle does not count.

Also I forgot to mention n is a positive integer, but I'm pretty sure everyone assumed that :lol:


How would I go by starting a math circle in Oklahoma?  I'm sure there are enough students willing to come to a centralized place on Saturdays or Sundays for three or so hours. but how would i go by starting all of this. like finding the instructors and advertising and everything. suggestions?

edit: im sure teachers at Oklahoma School of Science and Math wouldnt mind helping out.  that school's in oklahoma city, and thats a centralized place.

Well then if you know any of those teachers, just talk to them about your idea some time (preferably in person).

hm, i dont know any of the teachers, but i do know students who go there. so i could tell them to ask about it and stuffs

[quote="Syntax Error"]How would I go by starting a math circle in Oklahoma?[/quote]

I'm trying to do the same in Minnesota.

http://mathforum.org/epigone/geometry-pre-college/woaphemzhin

 My model for what I would like to do is the [url=http://www.themathcircle.org/]Math Circle[/url] of Cambridge, MA, whose founder I have met on his book tour to my town. A parent of a participant in that circle has written to me by email about what a great program that is.

It is pretty easy to find interested parents. It is HARD, or at least it has been hard for me, to find an interested mathematician-mentor to make the program possible.

I would be interested in hearing from anyone who has ideas about how to grow more local math circles.

neal, if u really want to do this, can u please at least have it in tulsa?


What language do you normally program in? If its more than one, please enter one in the poll and list one here.

...is there one for calculator programs? as in ti 83? thats the only thing i know how to program with.

That's TI-Basic, I'd choose Basic.

Or assembly, but that's unlikely.

[quote="weird888"]sup...
im learning some c++[/quote]


probability
expected value
combinatorics unsolved
combinatorics

In an $ n\times n$ array each of the numbers $ 1, 2, \ldots, n$ appears exactly $ n$ times. Show that there is a row or a column in the array with at least $ \sqrt{n}$ distinct numbers.

I would be really interested to see a solution to this problem. Although I think there are few typo's in the previously mentioned post that needs to be cleared up.

Consider an arrangement of the numbers in $n \times n$ array. Pick up any particular row or column uniformly at random and call it $\mathcal{T}$.  Let $X$ denote the number of distinct integers appearing in $\mathcal{T}$ and  let $E(X)$ be the expected value of number of distinct integers appearing in $T$ . For all $1 \le i \le n $,let $P_i$ be the probability that the number $i$ appears, may be more than once, in $\mathcal{T}$.Since we are counting number of distinct numbers, if the probability of appearance of $i$ in $\mathcal{T}$ is $P_i$, then the expectation of $i$, in $\mathcal{T}$ is $1 \cdot P_i$. So by linearity of expectation $E(X) = \sum_{i=1}^{n} P_i$.

 Since all the numbers appears exactly $n$ times, for all $i$ ,$P_i$ will be minimum when all the same numbers (i.e $i$) are present in a $\sqrt{n} \times \sqrt{n}$ sub array. So the probability that $i \in \mathcal{T}$ , i.e $P_i$ is at least $\frac{2 \sqrt{n}}{2n} = \frac{1}{\sqrt{n}} $.

Therefore $E(X) = \sum_{i=1}^{n} P_i \ge \frac{n}{\sqrt{n}} = \sqrt{n}$. So the expectation of number of  distinct numbers in a row or column is at least $\sqrt{n}$. Hence there exists a row or column with at least $\sqrt{n}$ distinct numbers.

This has been posted and answered before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=412838


How many distinct terms are there when $ (1 \plus{} x^7 \plus{} x^{13})^{100}$ is multiplied out and simplified?

[hide="What I have figured out so far"]
We have to find all possible combination sums of up to 100 uses of either 7's or 13's. The leftover uses we will just assign the zero power to. It helps that 7 and 13 are coprime but...
[/hide]

Is there an easier way to do this than brute force?

[hide="Hint"]If you have a possible combination with more than thirteen 7's, you can replace these with seven 13's, which is still a possible combination since it has less terms.[/hide]


vector
superior algebra
superior algebra unsolved

Hello,

this problem is left as an exercise in a book but I keep thinking about it. :maybe:

Let $ V$ be a be a vector space, and $ U$ a nonsingular hermitian form on $ V$.  Suppose $ E$ is a totally isotropic subspace of $ V$ (meaning $ U$ simply vanishes on $ E$).  Show that every transformation in $ GL(E)$ can be extended to an element of $ SU(V)$ (the group of special linear isometries of $ U$).

If the problem would have been : prove that there is such an extension in $ U(V)$, it would have been the pretty well known theorem by Witt, but what to do now? Is this trivial or not? :huh:

Thank you very much,
fredbel6

I'm sorry, moderators, this should have been in "Superior Algebra". :blush:  Can you move it, please?


What is the largest possible median for the five number set
$ \{x, 2x, 3, 2, 5\}$ if $ x$ can be any integer?

The median is the middle number, in order from least to greatest. If we make x very large, say 100, then 2x will be, like 200. In this extreme case, we see the order is-

2, 3, 5, x, 2x

So we get $ \boxed{5}$


geometry
3D geometry
sphere
rotation
inequalities
trigonometry
area of a triangle

This isn't exactly a computation problem but is probably too easy for the other forums.

Find the radius and height of the right circular cone with the least volume that can be circumscribed about a sphere with radius R.

And, it seems reasonable to add, find the dimensions of the cone with largest volume that can be inscribed in the sphere.

Edit: post deleted, very erroneous assumption
[hide="lesson"]2D problems are not the same as 3D problems. :oops:[/hide]

radius- 0 height- 2r
radius- r  height- 0
either of these has the least possible volume

0714... : That's supposed to be a cone circumscribed around a sphere. Its volume has to be at least as large as the volume of the sphere.

quantum leap: There's no good reason to believe that you are correct. I'm afraid that you do need the actual computations.

On my first try, I get that for the cone of minimum volume, the radius of the base of the cone is ${r=(\frac{\sqrt{5}+1}2})R$ and the altitude is $h=2r=(\sqrt{5}+1)R.$

That's going to take some checking. Does anyone else get the same result?

I got $r = \sqrt2 R$ and $h = 4R$.  My volume is $\frac{8\pi}3R^3 \approx 8.377R^3$.  Yours is $\frac{2\pi(2+\sqrt5)}3r^3 \approx 8.872 R^3$.  So assuming the sphere fits inside my cone, you're wrong :)

[quote="quantum leap"]I know my assumptions were unjustified, I was just saying that's what i would expect before doing any calculations.  It's not unreasonable to have anexpectation of what an answer will be before starting a problem.  Things usually turn out to be regular though.[/quote]

Yes, but when you write such a light sketch, and especially when you finish it with, "Finding the actual dimensions are now technicalities. I can live without the actual computations," your work doesn't amount to much more than if I said, "I know the problem has an answer, but actually finding it would be tiresome." ;)

Also, your predictions are going to be wrong.  While the best way to solve the 3-D question is to look at the 2-D question, the 3-Dness of it means that their solutions don't correspond.  (That is, the smallest circumscribed triangle will not yield the smallest circumscribed cone, and ditto for inscribed.)  The reason is actually very intuitive: to get from the 2-D diagram to the 3-D image, I rotate the whole thing around.  The widest part of the triangle (the base) adds much more volume than the thinner parts.  Thus, it's not suprising that if I want to minimize volume, I need a triangle that's taller but thinner than equilateral, and if I want to maximize I should take one that's shorter and fatter.  Without doing any work at all, it should be clear that the obvious 2-D solution (equilateral) [i]won't be[/i] the solution for either 3-D question.

And by the way, JBL's answer is correct.

Yep. $r=\sqrt{2}R,\ h=4R.$

I never did find the mistake in my first attempt - I just did it over. I was using Lagrange multipliers: minimize $\pi r^2h$ subject to the constraint (the correct version, not my first version) $h(r^2-R^2)=2Rr^2.$

[quote="quantum leap"]I thought i did a problem similar to this a recently.  I guess it was 2-D :oops:
btw, i'm using a right triangle with sides $r, R, and (h-R)$ in the cone to rewrite the cone's radius in terms of its height and the sphere's radius, and since $R$ is constant, differentiating the volume with respect to $h$.  But i'm then getting that $h=\frac{4}{3}R$.  How did you get $h=4R$?[/quote]

$h=\frac{4}{3}R$ is correct for the inscribed cone.  The answers we mentioned were for the circumscribed cone.

Let $h$ be the hight of the circular cone  and $r$ be the radius of base\ (circle), remark the area of a triangle we have

$\frac{1}{2}\sqrt{r^2+h^2}\cdot R=\frac{1}{2}(h-R)r.$ Squaring both sides and rearranging we obtain

$r^2=\frac{R^2h}{h-2R}\ \cdots [*]$ Thus using the inequality A.M-G.M. the volume of the right circular cone:

$V=\frac{1}{3}\pi r^2h=\frac{\pi R^2}{3}\cdot \frac{h^2}{h-2R})=\frac{\pi}{3}R^2(h+2R+\frac{4R^2}{h-2R})=\frac{\pi}{3}R^2(h-2R+\frac{4R^2}{h-2R}+4R)$

$\geq \frac{\pi}{3}R^2(2\sqrt{(h-2R)\cdot \frac{4R^2}{h-2R}}+4R)\ \because h>2R.$

$=\frac{8}{3}\pi R^3.$

The equality holds when $h-2R=\frac{4R^2}{h-2R} \ (h>2R)\Longleftrightarrow h=4R.$ Therefore from $[*],$

the desired answer is $\boxed{h=4R,\ r=\sqrt{2}R}$

Oops I misread the problem. I thought the cone was inside the sphere.

One last method (don't know that it's any easier - I went through several incorrect versions before making it work).

Let $\phi$ be the angle between the the axis of the cone and a ray from the center of the sphere to a point on the ring of contact between the cone and the sphere. This is the standard co-latitude angle of spherical coordinates. We draw a cross-section and we see right triangles containing the angle $\phi.$ From the figure we derive that

$h=R+R\sec\phi=R(1+\sec\phi)$
$r=h\cot\phi=R\cot\phi(1+\sec\phi)=R(\cot\phi+\csc\phi)$

The volume of the cone is

$V=\pi r^2h=\pi R^3(\cot\phi+\csc\phi)^2(1+\sec\phi).$

This goes to $\infty$ as $\phi\to 0$ and $\phi\to\frac{\pi}2,$ so $V$ must have a minimum in the first quadrant. Looking for the critical point:

$\frac{dV}{d\phi}=\pi R^3\left[2(\cot\phi+\csc\phi)(-\csc^2\phi-\csc\phi\cot\phi) (1+\sec\phi)+(\cot\phi+\csc\phi)^2\sec\phi\tan\phi\right]$

$=\pi R^3(\cot\phi+\csc\phi)^2[\sec\phi\tan\phi-2\csc\phi(1+\sec\phi)]$

Setting this equal to zero, we get

$\sec\phi\tan\phi=2\csc\phi(1+\sec\phi)$

$\frac{\sin\phi}{\cos^2\phi}=\frac{2}{\sin\phi}\left(1+\frac1{\cos\phi}\right)$

$\sin^2\phi=2\cos\phi(\cos\phi+1)$

$1-\cos^2\phi=2\cos\phi(\cos\phi+1)$

$1-\cos\phi= 2\cos\phi$

$\cos\phi=\frac13.$

From that it follows that $\sec\phi=3$ and $\cot\phi=\frac1{2\sqrt{2}}$ and from that we get that $h=4R$ and $r=\sqrt{2}R.$

One last version: put the problem in $\mathbb{R}^n$. A "cone" is a figure whose base is an $n-1$-ball on an $n-1$-dimensional hyperplane; the cone is the convex hull of that set with a point whose orthogonal projection onto the hyperplane in of the base is the center of the base. We want to find the cone of minimum "volume" (meaning $n$-dimensional measure) that can be circumscribed around a sphere of radius $R.$

We use exactly the method of the previous post. The only difference is that this time the volume is $V=c_nr^{n-1}h.$ (We know what $c_n$ is but for the purposes of finding the minimum volume, all we need is some constant.) Everything works just as in the previous post, with $n-1$ in place of $2.$ Our minimum is located where

$\cos\phi=\frac1n,\ h=(n+1)R,\ r=\sqrt{\frac{n+1}{n-1}}\,R.$

Note that if $n=2$ we get the equilateral triangle that quantum leap wanted back at the beginning of this topic. For the reasons JBL pointed out, the cone gets narrower and more pointed as the number of dimensions increases.


3 ladies have each brought a child for admission to a school. The head of the school wishes to interview the 6 people one by one , taking care that no child is interviewed before its mother.The number of ways of doin this is......

This is a very very ugly solution that counts each possibility.

There are five different arrangements of the mothers, and for each possible arrangement the moms can be switched around in 3! ways. Then in each situation the children can be arranged differently...

Is $3!\cdot3! + 3!\cdot2^2 + 3!\cdot2 + 3!\cdot2 + 3! = 90$ the right answer?

I also got $90$ but I also don't have a nice solution for problems like these. I've encountered some problems similar to this one before and would like to know how to solve them properly.

Well if it wasn't for the order you would have $6!$ so divide it by $2^3$  to get the asnwer.
(Each case - mother/son and son/mother is counted twice)


inequalities
function
triangle inequality
real analysis
real analysis unsolved

As we know, the equality in the  triangle inequality can holds. For eg, $ |f| \plus{} |g| \equal{} |f \plus{} g|$ when $ f,g$ are non-negative functions.
However, does this happen in $ L^{\infty}(\mu)$.
i.e, does there exists $ f,g \in L^{\infty}(\mu)$, $ f \neq g$ such that $ \left\| f \right\|_{\infty} \plus{} \left\| g \right\|_{\infty} \equal{} \left\| f \plus{} g \right\|_{\infty}$ ?

Thanks :-)

And, what if we restrict it to $ \left\| f \right\|_{\infty} \equal{} \left\| g \right\|_{\infty}$ but $ f \neq g$ ?

Try letting $ f$ be the characteristic function of $ [0,2]$ and $ g$ the characteristic function of $ [1,3].$

So, in that case, we have $ \left\| f \right\| _{\infty} \equal{} \left\| g \right\| _{\infty} \equal{} 1$.

Is this right? :-)


Form 4 fractions by using each of the digits 1,2,3 and 4 twice; once as a numerator and once as a denominator. What is the greatest possible sum of these 4 fractions?

  :P

[quote="236factorial"][hide]biggest numbers on top, smaller numbers on bottom
4/1 + 4/1 + 3/2 + 3/2 = 11[/hide][/quote]

But it's asking for the numbers to be used once as a numerator and once as a denominator.

That makes it a little more complicated...
so what is the answer? I have 4/1 + 3/2 + 1/3 + 2/4 or 6 5/12 or something, and still needs the correct answer


http://armorgames.com/play/4987/obey-the-game

Fun game.

My highests:

Normal: 119
Hidenride: 128
Blind: 110
Reverse: 57
Rapidfire: 28

normal 93 - 2nd time
hidenride 68
blind 75
reverse 48
rapidfire 30  :D

first try each.
How do you beat break on reverse?

Normal : 77
Hidenride : 54
Blind : 80
Reverse : 74
Rapidfire : 48

[quote="Chompy"]first try each.
How do you beat break on reverse?[/quote]

Heh, nice point.

Ride and survive can really be a killer in the later levels. Thank goodness for walk.

My highest streak is 50, I believe. Hilariously, that was on blind.

174/128/134/74/41 now (played each 3-6 times)

Normal: 35
Hidenride:110   (streak of 60)
Blind: 37  (just to pass)
Reverse: 49  (after I reached forty I just let it go...)
Rapid Fire: 72

1st try on all.

I am proud of my rapid fire score.

I just started. Longest streak: 44 (on Hide 'n' Ride)

Normal: 135
Hide 'n' Ride: 116
Blind: 159
Reverse: 168
Rapidfire: 71

First time stats:

Normal -- 83
Hidenride -- 91
Blind -- 75
Reverse -- 49
Rapidfire -- 40

256 on normal 0.0
130 hidenride
180 on blind

I'll make the others later.

[quote="randomguy64
"]I hate you all.[/quote]

Why? What were your scores?
 :P

92
49
87
57
20 something

My top scores so far.

Best normal mode: 222 (at around the 200s, it really just becomes luck as to which ones you get)

Best rapidfire: Got a 114 around my 10th try, was really lucky though. I can usually get to 80 or so, but after that, the best strategy seems to be to just wait and do nothing...

Also, it seems that people with slower computers would have an advantage at this game, since it's dependent on frame rate and not on real time.

I found a similar ArmorGames game by way of ESPN.com

[url]http://espn.go.com/free-online-games/theocho/grid16[/url]

My scores:

Top Combo: 23
Best Multi (speed) 4.6x
Total time:299

Skill sets:

Prioritizing: 89%
Reflexes 80%
Timing 78%

the game in the link I provided is called grid 16.

116 on rapidfire now.  2nd on daily list.

137 on rapidfire with a hi streak of 42.

lol. still second.

oh no not another elephant game!
love it.
score on normal-45


我覺得世界上一大部份人(非基督徒/天主教徒)都不必要過聖誕節，因為這個節日對他們並無意義。所以中國的基督徒/天主教徒都應該過聖誕節，其他人則按個人喜好..

[quote="wuhao_jshj"]I've never celebrated Christmas Day.It's a western festival indeed.But we are Chinese.[/quote]
I would say it is a Catholic festival rather than a western one.

haha, 现在中国的圣诞节在一定程度上变成情人节了

well, I guess that's because the group of people who is now celebrating Xmas in China Mainland, is young people like us who loves romance at this age of our lives, haha
By the way, the Christmas I saw here in Canada is far more religious than what I expected years ago.

[quote="wuhao_jshj"]I've never celebrated Christmas Day.It's a western festival indeed.But we are Chinese.[/quote]
如果是这样的话，那元旦也不用过了，因为公元纪年是外国传来的，元旦也是。

元旦是放假的日子, 但基本上我不把它当作什么重要的节日.

促进消费。。。增添生活乐趣 :P

对我来说, 是补觉的日子...

无论在基督教国家还是中国,圣诞节对于非教徒都是胡闹.
这是教徒庆祝他们先知诞生的日子,其他人就HAPPY HOLIDAY就行了.

在中国很多贺卡上都写着'MERRY CHRISTMAS'.其实应该是HAPPY HOLIDAY.

why are we talking about christmas in a math website?

 我们为什么在这讨论这些东西呢？
 :?:

Porque LBs quiere earn los muchos money .  :roll:  And we have missed ourselves. :ninja:

因为老板们希望挣到更多的钱，而我们却已经迷失了自己。

[quote="liyi"]haha, 现在中国的圣诞节在一定程度上变成情人节了[/quote]

对你真无语了... :(

文化交流 最简单而且最重要的

[quote=wuhao_jshj]I've never celebrated Christmas Day.It's a western festival indeed.But we are Chinese.[/quote]

so。。。YOU DON‘T GET PRESENTS？？？
i am chinese, but we still celibrate chinese.
你好。


What mass of sodium bicarbonate and what volume of hydrochloric acid 7 M must react to completely fill a 600 mL plastic bag with carbon dioxide to a pressure equal to atmospheric pressure, at room temperature (298 K)?

Is it

[hide="Answers"] $ 2.06g$ of Sodium Bicarbonate and $ 3.51mL$ of Hydrochloric Acid.[/hide]


Given a sequence $\{ a_{n}\}$ such that $a_{n+1}=\frac{p_{1}a_{n}+p_{2}}{p_{3}a_{n}+p_{4}}$ and $a_{1}= p_{0}$. Find the general term of $a_{n}$.

See my post in "Sequence with frac{2+x_n}{1-2x_n}".


a, b   are complex numbers.
|a+b| ≥2   and |a+b|≥1+|ab|
Prove that:  |a^(n+1)  +b^(n+1) |≥|a^n  +b^n|.   for every n from N

You've already posted this here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131649[/url]

Do not double post.


calculus integration geometry logarithms limit ratio trigonometry

Can anyone please tell me how to determine whether the following 2 integrals converge or diverge as well as the related area problem? I have trouble with using the Direct and limit comparison tests.

1. Integral from (-1 to 1) ln |x| dx.
2. Integral from (0 to infinity) (1)/[(1+y 2 )(1 + Arctan y)] dy.
3. Find the area of the region in the first quadrant that lies under the given curve:

y = (ln x)/x.

Thanks for any help.

1. $\int_{-1}^1\ln|x|\,dx.$

Although $\ln|x|$ tends to $-\infty$ as $x\to0,$ it goes to infinity slower than any negative power of $x$. For instance, $|\ln|x||<\frac1{\sqrt{|x|}}$ for $x$ near 0.

Since $\int_0^1\frac{dx}{\sqrt{x}}$ converges, this integral converges by direct comparison.

Of course, in this case, you might as well just compute it:

$\int_{-1}^1\ln|x|\,dx=2\int_0^1\ln x\,dx=\left2(x\ln x-x)|_0^1 =2(0-1)-2(0-0)=-2.$ Note that we used $\lim_{x\to0^+}x\ln x=0$ because the logarithm goes to infinity slower than any power of $x.$

2. $\int_0^{\infty}\frac{dy}{(1+y^2)(1+\arctan y)}$

As $y\to\infty,\,1+\arctan y\to$ a constant. Hence, for large $y$, the integrand behaves like $\frac C{y^2}$ for some constant $C.$ Thus, it converges by limit comparison to $\int_1^{\infty}\frac{dy}{y^2}.$

That's what the limit comparison test is for: "de-cluttering" things and getting down to the essentials of size.

3. $\int_1^{\infty}\frac{\ln x}x\,dx$

For large enough $x,\, \frac{\ln x}x>\frac1x.$ Larger than a divergent integral implies divergence. This diverges by direct comparison to $\int_b^{\infty}\frac{dx}x.$

As Moubinool pointed out, you can find the exact antiderivative (primitive) for #3. You can also do that for #2. I suspect that the author of the textbook from which these problems came intends for them to be worked that way. I knew that when I made my post, but I was making a different point: that in many cases, the question of convergence or divergence is a matter of size, and calculus students should always think of size first. In all three instances of my post, it took me 10 times as long to write out what I meant as it took for me to think it; this line of reasoning is supposed to be quick.

Yeah, generally the inverval of convergence problems or determining convergence and divergence can be done more than one way (ratio test, limit comparison etc.) but it helps when you can actually deduce the antiderivative. Like you had, for example:

$\lim_{b\rightarrow \infty} \int^{b}_{0} \frac{dy}{(1+y^{2})(1 + \tan^{-1}y)}$

Substituting $u = \tan^{-1}y$ and seeing $du = \frac{1}{1+y^{2}}$ then doing the limit after you have your antiderivate yields that the integral converges for what I believe are all values of $y$.


Let $ S$ be the set of $ n\in\mathbb N$ such that $ 2n\plus{}1$ and $ 3n\plus{}1$ are complete squares. Find $ \gcd S$.

[quote="dgreenb801"]This is well known, and from Engel.

[hide]The answer is 40. Clearly n=40 works. We will show every other n is divisible by 40.
For some naturals a and b, we have
$ 2n \plus{} 1 \equal{} a^2$
$ 3n \plus{} 1 \equal{} b^2$, so that
$ 3a^2 \minus{} 2b^2 \equal{} 1$
We first take this mod 5. Quadratic residues mod 5 are 0,1, and 4. We find this can only be possible if $ a^2$ and $ b^2$ are 1 mod 5. So since $ 2n \plus{} 1 \equal{} a^2 \equal{} 5k \plus{} 1$, 5 divides n.
Now mod 8. Residues are 0,1,4. Again, only 1 for both $ a^2$ and $ b^2$ works. So $ 3n \plus{} 1 \equal{} b^2 \equal{} 8k \plus{} 1$, so 8 divides n.
So since 5 and 8 divide n, 40 divides n.[/hide][/quote]

Wow. I need serious help with that. I sort of got it but not really.

[quote="ifai"]Let $ S$ be the set of $ n\in\mathbb N$ such that $ 2n \plus{} 1$ and $ 3n \plus{} 1$ are complete squares. Find $ \gcd S$.[/quote]

In this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282100]post[/url] there are some other solutions that might help.


The area of a circle is $ 256\pi$. What is its circumference?

We know that the formula for the area of a circle is $ \pi r^2$.

Therefore, the radius of the circle must be $ \sqrt 256$, or $ 16$.

The circumference of a circle is $ \pi d$, or $ \boxed{32 \pi}$.


geometry
videos
3D geometry
algorithm
group theory

do u guys all think SAT is really important for applying a University?
because, my self, came to US almost 4 years, and my english is really bad, i can not get a good score on SAT english part, which really kills me! any suggestion?

Qiu - Which  grade are you?

For many immigrants this is a common problem. I do have  one  suggestiion. If you school offer Latin.  Take it.

im in 11th grade, and our school does not offer Latin,=/   why u suggest me to take Latin?

well, i took 4 ap in my sophmore year, and taking 2 in junior year, i am capable to take these classes, but well, english is just a major headache.
=(    what are the universities looking for? sigh, i think im screw

[quote="yinjiaqiu"]=(    what are the universities looking for? sigh, i think im screw[/quote]

Absolutely not...you just need to explain your situation on your application.

given your name, i'd guess that you're chinese...so go take the chinese SAT II (and you better get a perfect score!)...and if you don't do so well on the english ones, you can explain that chinese is your first language. If you have chinese speech or writing competition or the like in your area, do those.

SAT scores do count, but they're not everything...besides, the big-name colleges spend millions recruiting kids from other countries that don't know or know very little english at all...

[quote="cats..."]would taking the TOEFL help?

i wouldn't take SAT II chinese, i can't imagine how it could possibly help you.[/quote]

Actually, I hope that colleges do look at how long you've been in the country, so they can fairly judge how well you have learned english. It's not very fair to compare somebody who has lived in the US all their life to someone who just moved to the US when he or she was 16 and had lived in, say, Spain all their lives.

As for the Chinese SAT II, how could taking it hurt? That's kind of like saying, "Oh, I know all the material on this exam, but I'm not going to take it, even if it has a chance of helping me and can't hurt me at all."

I think the deal with SAT II Chinese is that colleges want to know what you've learned in high school, not what you've known since you were 7.

well,, i will have to explain it on the personal statement then,,
thanks you guys! love u all =)

and by the way, Toefl wont help me, because, i'm not an international student, so i do not need to take the Toefl exam..

[quote="MithsApprentice"]given your name, i'd guess that you're chinese...so go take the chinese SAT II[/quote]

Yes, that is really the main purpose of the SAT II Chinese test--to demonstrate that a person has essentially a native command of Chinese, which provides evidence a person hasn't had as much time to learn English as someone who grew up in the United States. MOST people who take the SAT II Chinese test are first-generation immigrants, for sure.

If you have just arrived in the United States in the last few years, the people you are competing against in college admission are the other people who arrived in the last few years. Turn off the video games, READ LOTS OF ENGLISH MAGAZINES AND BOOKS, and make sure you learn as much English as the most successful first-generation immigrants, but don't worry about what you haven't got time to learn if you put in your best effort.

Gyan's suggestion to learn Latin is learn about how many English words (which in many cases are derived from Latin) are put together, and build up your vocabulary. A useful English grammar course you can take online is EPGY's [url=http://epgy.stanford.edu/courses/english/EG20/]EG20 course[/url].

Good luck in your applications.

By the way, why do so many Chinese people move to Diamond Bar?

[quote="tokenadult"] Turn off the video games[/quote]

 :!: :!::!:  :cry: :cry:

hmmm...
i participated all the rubik's cube tournament that host in caltech, and i set 7 chinese national records, =.=!!! lol, will this special event help me for my college application?
by the way,, there is one cube tournament host this friday and saturday, so wish me luck=)))

yinjiaqiu - Good luck  Use some really nice lubricant so the cube moves smoothly  :)   (grahite powder works well ) .. what is your time around?

One  of my nephew routinely used to beat me ( but then he had small fingers, used  [b]my[/b]  algorithm .. and used butter to make the cube smooth) .. (The Rubik's cube were a craze in  early 1980's)  I once solved it using just my toes to impress some one .. (Well  I am happilly  marrid to that person for last 20+ years  - and my son  last year enjoyed his course in group theory  - so I think it was well worth  :) )

[quote="yinjiaqiu"]i,=/   why u suggest me to take Latin?[/quote]

Tokenadult has already answered that  better than I can.

Well I knew lot of kids (in  my extended family and  close friends)  who did extremely well in math but  did not score that high in verbal SAT.   When my son was in High School he had an excellent Latin teacher and he took Latin and he  did as well in verbal too.  He enjoyed latin and it helped  his vocabulary too. .. so those of you who can take latin, consider taking it... you may enjoy it.

On a side topic, if I may say so  Qiu - You may score much higher than you think ..so just don't worry so much about SAT  scores..they are just one of the  MANY things which good schools look for . .. Just do your best and enjoy and sat scores will take care of themselves...

Just my thoughts.

[quote="Gyan"]
On a side topic, if I may say so  Qiu - You may score much higher than you think ..so just don't worry so much about SAT  scores..they are just one of the  MANY things which good schools look for . .. Just do your best and enjoy and sat scores will take care of themselves...
[/quote]

agreed...*nods*.

Now, this is entirely off topic but anyone know where I could learn how to do the rubix cube?...since there's supposedly quite a few algorithms and stuffs that you could use to do it. I actually worked on it when I was little...most I ever got was 3 side...after sitting in my room for like 4 hours doing nothing back in 3rd grade...I kinda gave up after that but it'd still be nice to know a method to solve it...

thanks everyone, i am reading about the roots of the vacabulary by some sat books, it helps, lot of the roots are from Latin. thnx=))

my best average of 3x3 rubik's cube is 18.79seconds. hope ill stay calm during the competition..

for learning how to solve the cube, you can go to
http://www.speedcubing.com
it is the main website of the rubik's cube.. and u can find my records on that site too.=)


advanced fields
advanced fields unsolved

Let $m^*$ denote the Lebesque outer measure on $\mathbb{R}$. Prove that for any two sets $A,B$ we have
\[ m^*(A\cup B)+m^*(A\cap B)\leq m^*(A)+m^*(B)  \]

Lemma. For measurable sets $A,B$ we have $m(A\cup B)+m(A\cap B)=mA+mB$.
(Prove it using Caratheodory's condition)

Let $H_A$ and $H_B$ be two $G_\delta$ set containing $A$ and $B$ respectively and $mH_A=m^*A, mH_B = m^*B$.

Then $A\cap B\subseteq H_A\cap H_B$ and $A\cup B\subseteq H_A\cup H_B$.
$m^*(A\cap B)\leq m(H_A\cap H_B)$ and $m^*(A\cup B)\leq m(H_A\cup H_B)$. Apply the lemma.


[quote="justinahmann"]I don't understand problem 33 in the attached .pdf file.[/quote]

For an air column closed on one end, the length of the column is $ \frac{\lambda}{4}$, where $ \lambda$ is the wavelength of the resonant frequency (the frequency for which the sound is loudest.)


is it possible to make  Thread or forum for Primary School kids.
if not is there any other forum for them ?
can you please link me ?
caues i have a little Brother, he is 11 years old.

I think he would be best fit in the "Classroom Math" section under middle school.

I thought MOEMS was for middle schools...

Primary school forum would not be bad if we get enough people under age 10 to come to the site.


superior algebra
superior algebra unsolved

i encounter a problem as such:
Prove that a non empty set G,together with an associative binary operation * on G such that
a*x=b and y*a=b have solutions in G for all a,b in G
 is a group.

Can i start proving by stating that since the equations have solutions,then, it implies that inverses exist?Or do it have any other way to initialize my proof?

Can anyone help me???Thx!!

$ \mbox{If equations } a*x\equal{}e \mbox{ and } y*a\equal{}e \mbox{ have solutions }\Rightarrow x\equal{}y\equal{}a^{\minus{}1}.$

The problem is that we are not given the existence of an identity element.  You need to prove this first.

$ \mbox{If equations } a*x\equal{}a \mbox{ and } x*a\equal{}a \mbox{ have solution }\Rightarrow x\equal{}y\equal{}e.$

That's not enough.  It only tells you there is some element $ x$ that [i]depends on $ a$[/i] with that property (and the left and right identity might differ).  You need to show that the element you have identified is really an identity for everything in the set.

(Hint:  This result is false if you drop associativity.)

choose an element $ a\in G$, then there exist $ f\in G$ such that $ f*a\equal{}a$. Then try to show that this $ f$ is indeed the left identity for all element in $ G$. (i.e by showing that $ f*b\equal{}b$ for any $ b\in G$)

Once you prove that it has left identity, you can show that it too has left inverse and by similar argument you can show that its right identity and right inverse exist.

Finally just show that the left right identity and left right inverse are the same.


geometry
geometric transformation
reflection
circumcircle
ratio
trigonometry
symmetry

Dear Mathlinkers,
let ABC be a triangle, (1) the incircle of ABC, I the center of (1), DEF the contact triangle of ABC, E' the symmetric of E wrt BI,
B', C' the feet of the B, C-altitudes of ABC and L the meetpoint of B'C' and EF.
Prove: LE' is parallel to BC.
Sincerely
Jean-Louis

[quote="jayme"]E' the symmetric of I wrt BI, ...[/quote] Fixed.

$ AI, BI, CI$ cut $ BC, CA, AB$ at $ X, Y, Z.$ $ D', E', F'$ are reflections of $ D, E, F$ in $ AI, BI, CI.$ $ XD', YE', ZF'$ are tangents to the triangle incircle $ (I)$ (the common internal tangents of the incircle and the excircles other than the triangle sides). BY simple angle addition, $ E'F' \parallel BC,\ F'D' \parallel CA,\ D'E' \parallel AB$ and $ \triangle D'E'F' \sim \triangle ABC$ are centrally similar with similarity coefficient $ \minus{} \frac {r}{R}$ and similarity center at the internal similarity center of the incircle $ (I).$ and circumcircle $ (O).$

Let $ YE'$ meet $ AB$ at $ P$ and let $ ZF'$ meet $ CA$ at $ Q.$ The (concave) quadrilateral $ PYQZ$ is tangential and $ YZ, PQ$ are its diagonals. $ E'EF'F$ is its contact cyclic quadrilateral with diagonals $ E'F', EF.$ Since these 2 quadrilaterals have a common diagonal intersection, the lines $ PQ,\ YZ,\ EF,\ E'F'$ are concurrent at $ L.$

Easily calculated cross ratios $ \frac {\overline{YE}}{\overline{YA}} \cdot \frac {\overline{B'A}}{\overline{B'E}} \equal{} \minus{} \cos A \equal{} \frac {\overline{ZF}}{\overline{ZA}} \cdot \frac {\overline{C'A}}{\overline{C'F}}$ are equal. But then by Menelaus theorem for isosceles $ \triangle AEF,$ its transversals $ B'C', YZ$ intersect at the same point $ L \in EF.$ $ \square$
_________________________________________________
$ L$ is the A-Pelletier point of $ \triangle ABC.$
See [url]http://www.xtec.es/~qcastell/ttw/ttweng/definicions/d_Pelletier_p.html[/url]
If $ M_a$ is midpoint of $ BC$ and $ Fe$ the Feuerbach point, then $ M_aFe$ also goes through $ L.$

Let be the points $ Y\equiv AC\cap BI$ and $ Z\equiv AB\cap CI.$

The line segments $ B'C',\ EF,\ YZ,$ are concurrent at one point so be it $ L,$ as well ( this result is coming from the past and I will post here later the proof I have in mind, based in the [b][size=100]Double ratio theory[/size][/b] ).

$ \bullet$ Let be the point $ Q\equiv EF'\cap E'F,$ where $ F'$ is the symmetric point of $ F,$ with respect to $ CI.$ As [b][size=100]Vladimir[/size][/b] said, the line segments $ YE',\ ZF',$ tangent to $ (I),$ at points $ E'\ F',$ respectively.

Also, let be the point $ P\equiv YE'\cap ZF'$ and because of the circumscibed quadrilateral $ AYPZ,$ based on the [b][size=100]Newton theorem[/size][/b], we conclude that the line segments $ AP,\ YZ,\ EF',\ E'F,$ are concurrent at point $ Q\equiv EF'\cap E'F.$

$ \bullet$ Also, it is well known that the point $ L'\equiv EF\cap E'F',$ lies on the line segment $ YZ$ and so, we conclude that $ L'\equiv L.$

So, it is enough to prove that $ E'F'\parallel BC$ $ ,(1)$

But $ (1)$ is true, because of $ \angle PE'F' \equal{} \angle PF'E'$ and $ E'Y' \equal{} F'Z',$ where $ Y'\equiv BC\cap PE'$ and $ Z'\equiv BC\cap PF'$ ( from $ EE'\parallel AY'$ and $ YE \equal{} YE',$ we have $ AE \equal{} E'Y'$ and similarly, $ AF \equal{} F'Z'$ ).

Hence, we conclude that $ LE'\parallel BC$ and the proof is completed.

Kostas Vittas.

[quote="yetti"]...$ L$ is the A-Pelletier point of $ \triangle ABC.$
See [url]http://www.xtec.es/~qcastell/ttw/ttweng/definicions/d_Pelletier_p.html[/url]
If $ M_a$ is midpoint of $ BC$ and $ Fe$ the Feuerbach point, then $ M_aFe$ also goes through $ L.$[/quote]

Thank you dear [b][size=100]Vladimir[/size][/b] for the reference about the [b][size=100]Pelletier points[/size][/b].

Have you in mind any proof of the interesting collinearity you mentioned of the points $ M_{a},\ F_{e},\ L$?

Best regards, Kostas Vittas.

Dear Vladimir and Kostas
sorry for my typo, I will be more carefully next time… and thanks you for your interesting proofs.
A schème of my proof:

With the point E
1. let F' be the symmetric of F wrt CI ; then E'F' // BC. (I prove it with Pascal's theorem).
2. By symmetry, EF, BI, DE' are concurrent at M and MB is the inner M-bissector of triangle MDF.
3. Let B", C" be the feet of the B, C-altitudes of ABC and N the meetpoint of CI with EF.
4. B, C", N, B", M, C are cocyclic. (well-known)
5. NB = NB" ; in consequence, MN is the inner M-bissector of triangle MBB".
6. Let I' be the symmetric of I wrt EF ; I' is the incenter of triangle AB"C" ; then M, B" and I' are collinear.
7. Let E" be the symmetric of E' wrt B"M ; E" is on B"C".
8. We can prove that E' is the symmetric of E" wrt EF.

With the point F,
we conclude that  F' is the symmetric of F" wrt EF.

E'F' is the symmetric of B"C" wrt EF.
E'F', EF, B"C" are concurrent at L…

Sincerely
Jean-Louis

[quote="vittasko"] ... Have you in mind any proof of the interesting collinearity you mentioned of the points $ M_{a},\ F_{e},\ L$?[/quote]
I proved it before with a lot of trigonometry, but I am not able to get rid of trigonometry completely. Keeping notation of my previous post: $ P \equiv YE' \cap AB,\ Q \equiv ZF' \cap AC.$ In addition, $ S \equiv YE' \cap ZF'.$ By Menelaus theorem for isosceles $ \triangle SF'E'$ cut by the line $ YZL,$

$ \frac {\overline{LE'}}{\overline{LF'}} = \frac {\overline{YE'}}{\overline{YS}} \cdot \frac {\overline{ZS}}{\overline{ZF'}}.$

Using the line pencil $ L(Q, Z, S, F') \equiv L(P, Y, S, E'),$ cross ratios $ \frac {\overline{QS}}{\overline{QF'}} \cdot \frac {\overline{ZF'}}{\overline{ZS}} = \frac {\overline{PS}}{\overline{PE'}} \cdot \frac {\overline{YE'}}{\overline{YS}}$ are equal, hence

$ \frac {\overline{LE'}}{\overline{LF'}} = \frac {\overline{PE'}}{\overline{PS}} \cdot \frac {\overline{QS}}{\overline{QF'}} = \frac {\overline{PE'}}{\overline{PE'} - \overline{E'S}} \cdot \frac {\overline{QF'} - \overline{F'S}}{\overline{QF'}}.$

$ PBCY, QCBZ$ are both kites, $ \overline{PE'} = - \overline{CE},\ \overline{QF'} = - \overline{BF},$ hence

$ \frac {\overline{LE'}}{\overline{LF'}} = \frac {\overline{CE}}{\overline{CE} - \overline{SE'}} \cdot \frac {\overline{BF} - \overline{SF'}}{\overline{BF}}.$

Now comes the remaining little bit of trigonometry: By central similarity of $ \triangle D'E'F' \sim \triangle ABC$ with coefficient $ - \frac {r}{R},$

$ \frac {\overline{SE'}}{\overline{CE}} = \frac {r \tan A}{p - c} = \tan \frac {C}{2} \tan A = \frac {1 - \cos C}{\sin C} \cdot \frac {\sin A}{\cos A} = \frac {a}{c} \cdot \frac {1 - \cos C}{\cos A},$

$ 1 - \frac {\overline{SE'}}{\overline{CE}} = \frac {c \cos A - a(1 - \cos C)}{c \cos A} = \frac {b - a}{c \cos A}$ and similarly, $ 1 - \frac {\overline{SF'}}{\overline{BF}} = \frac {c - a}{b \cos A}$

$ \Longrightarrow\ \boxed{\frac {\overline{LE'}}{\overline{LF'}} = \frac {c}{b} \cdot \frac {c - a}{b - a}}$

Inversion with center $ M_a$ and power $ M_aD^2$ takes the incircle $ (I)$ and excircle $ (I_a)$ to themselves and the 9-point circle $ (N)$ passing through $ M_a$ into their common internal tangent $ XD'$ other than $ XD \equiv BC.$ Therefore, $ D'$ is the image of the Feuerbach point $ Fe$ and $ \overline{FeM_a} \cdot \overline{D'M_a} = M_aD^2.$ On the other hand, $ Fe$ is external similarity center of $ (I), (N)$ $ \Longrightarrow$ $ \frac {\overline{FeD'}}{\overline{FeM_a}} = \frac {2r}{R}.$ Combining,

$ FeM_a^2 = \frac {M_aD^2}{1 - \frac {2r}{R}} = \frac {R^2}{OI^2} \cdot \frac {(b - c)^2}{4}\ \Longrightarrow$

$ FeD'^2 = \frac {r^2}{OI^2} \cdot (b - c)^2$ and similarly, $ FeE'^2 = \frac {r^2}{OI^2} \cdot (c - a)^2,$ $ FeF'^2 = \frac {r^2}{OI^2} \cdot (a - b)^2.$ Signs of directed segments have to be chosen so that $ \overline{FeD'} + {\overline{FeE'} + \overline{FeF'} = 0,}$ for example

$ \boxed{\overline{FeD'} = \frac {r}{OI} \cdot (b - c),\ \overline{FeE'} = \frac {r}{OI} \cdot (c - a),\ \overline{FeF'} = \frac {r}{OI} \cdot (a - b)}$

$ \Longrightarrow\ \boxed{\frac {\overline{FeE'}}{\overline{FeF'}} = - \frac {c - a}{b - a}}$

Let the line $ M_aD'Fe$ cut $ E'F'$ at $ L'.$ Using directed triangle areas $ [\triangle...],$

$ \boxed{\frac {\overline{L'E'}}{\overline{L'F'}} = \frac {[\triangle FeD'E']}{[\triangle FeD'F']} = \frac {\overline{FeE'}}{\overline{FeF'}} \cdot \frac {\overline{D'E'}}{\overline{D'F'}} = \left( - \frac {c - a}{b - a}\right) \cdot \left( - \frac {c}{b}\right) = \frac {c}{b} \cdot \frac {c - a}{b - a}}.$

As a result, $ L' \equiv L$ and $ M_aFe$ goes through $ L.$ $ \square$
[hide]Hope I did not use the operator of identity with eigenvalue -1.[/hide]

Dear Kostas, Vladimir and Mathlinkers,
for your question, dear Kostas, you can see my synthetic proof on

http://perso.orange.fr/jl.ayme/   vol. 2  Les deux points de Schroeter p. 14-20.

Thank for Vladimir for his nice metric proof.
Sincerely
Jean-Louis

[quote]...$ L$ is the A-Pelletier point of $ \triangle ABC.$
See [url]http://www.xtec.es/~qcastell/ttw/ttweng/definicions/d_Pelletier_p.html[/url][/quote]

Here is my proof for the fact above. See the name of the points in attached file below this following proof.
[i]Proof[/i]
We have $ \frac {\overline {D_bN_a}}{\overline {D_bH_a}} \equal{} \frac {\overline {IN_a}}{\overline {IA}} \equal{} \frac {\overline {CN_a}}{\overline {CA}}$ $ \Longrightarrow$ $ (CD_bN_aH_a)$ $ \equal{} \frac {\overline {CH_a}}{\overline {CA}}$.

With the same argument, we also have $ (CD_aN_bH_b) \equal{} \frac {\overline {CH_b}}{\overline {CB}}$.
But $ \triangle {CBH_b}$ $ \cong$ $ \triangle {CAH_a}$ $ \Longrightarrow$ $ \frac {\overline {CH_a}}{\overline {CA}} \equal{} \frac {\overline {CH_b}}{\overline {CB}}$.

Then $ (CD_bN_aH_a) \equal{} (CD_aN_bH_b)$, which implies $ D_aD_b,N_aN_b,H_aH_b$ are concurrent. $ \square$

$ \bullet$ [b]REMARK-[/b] I also have discovered about $ 9$ or $ 10$ points are collinear from this problem, I will post it later.


If $ a,b,c>0$, then

$ \frac{a\plus{}3b}{b(b\plus{}c)^2}\plus{}\frac{b\plus{}3c}{c(c\plus{}a)^2}\plus{}\frac{c\plus{}3a}{a(a\plus{}b)^2}\geq\frac{9}{ab\plus{}bc\plus{}ca}$

[quote="filomen"]If $ a,b,c > 0$, then

$ \frac {a \plus{} 3b}{b(b \plus{} c)^2} \plus{} \frac {b \plus{} 3c}{c(c \plus{} a)^2} \plus{} \frac {c \plus{} 3a}{a(a \plus{} b)^2}\geq\frac {9}{ab \plus{} bc \plus{} ca}$[/quote]

only prove

$ \frac {a}{b(b \plus{} c)^2} \plus{} \frac {b}{c(c \plus{} a)^2} \plus{} \frac {c}{a(a \plus{} b)^2} \plus{} \frac {3}{(b \plus{} c)^2}$ $ \plus{} \frac {3}{(c \plus{} a)^2} \plus{} \frac {3}{(a \plus{} b)^2}\geq\frac {9}{ab \plus{} bc \plus{} ca}$

Note

$ (\frac {a}{b(b \plus{} c)^2} \plus{} \frac {b}{c(c \plus{} a)^2} \plus{} \frac {c}{a(a \plus{} b)^2})(ab \plus{} bc \plus{} ca)\geq$ $ (\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b})^2\geq (\frac {3}{2})^2$

we have

$ \frac {a}{b(b \plus{} c)^2} \plus{} \frac {b}{c(c \plus{} a)^2} \plus{} \frac {c}{a(a \plus{} b)^2}\geq \frac {9}{4(ab \plus{} bc \plus{} ca)}$

we only prove

$ \frac {9}{4(ab \plus{} bc \plus{} ca)} \plus{} \frac {3}{(b \plus{} c)^2} \plus{} \frac {3}{(c \plus{} a)^2} \plus{} \frac {3}{(a \plus{} b)^2}$ $ \geq\frac {9}{ab \plus{} bc \plus{} ca}$

$ \Longleftrightarrow 12\sum(a^3(b \plus{} c)(a \minus{} b)(a \minus{} c)) \plus{} 9(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \plus{}$ $ 48abc\sum(a(a \minus{} b)(a \minus{} c))\plus{} 12abc\sum((b \plus{} c)(a \minus{} b)(a \minus{} c))\geq 0$

[quote="filomen"]If $ a,b,c > 0$, then

$ \frac {a \plus{} 3b}{b(b \plus{} c)^2} \plus{} \frac {b \plus{} 3c}{c(c \plus{} a)^2} \plus{} \frac {c \plus{} 3a}{a(a \plus{} b)^2}\geq\frac {9}{ab \plus{} bc \plus{} ca}$[/quote]

It's obvious from Iran1996.

Is this correct?
[hide="Solution"]Note that using CS we have
\[ \begin{aligned}(ab\plus{}bc\plus{}ca)\left(\frac a{b(b\plus{}c)^2}\plus{}\frac b{c(c\plus{}a)^2}\plus{}\frac c{a(a\plus{}b)^2}\right) \\ \geq \left[ \frac a{b\plus{}c}\plus{}\frac b{c\plus{}a}\plus{}\frac c{a\plus{}b}\right]^2 \geq \frac 94 \ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}\]
Also from Iran 97 inequality we have
\[ (ab\plus{}bc\plus{}ca)\left(\frac {3b}{b(b\plus{}c)^2} \plus{} \frac {3c}{c(c\plus{}a)^2}\plus{}\frac{3a}{a(a\plus{}b)^2}\right)\geq 3\cdot \frac 94\]
Summing these up yields the desired result.[/hide]


[quote="mtms5467"]There are eight marbles. They all look exactly the same but one weighs slightly more than the others. How can you identify the heavier marble using two weighings?[/quote]

[hide]Weigh three and three.

Case 1.
They weigh the same. Use the second weigh to find which of the remaining two is heavier.

Case 2.
One side is heavier. Weigh two marbles from that group. You can then tell which one is heavier. If they're the same, it's the third marble.[/hide]

[hide] Weigh 3 marbles and 3 marbles with 2 marbles left.

1. One of the groups of 3 is heavier. Then, you take 2 of the 3 marbles and weigh them. If they're the same weight, the last of the 3 is the heavy one.

2. The 2 groups of 3 marbles is the same. Then weigh the 2 marbles left out. [/hide]
This can be extended to 9 marbles.

SplashD: Please hide your answer. And now I really can't quote you. Did you make a really really small hide?

Please refrain from identifying problems as "easy."  If it belongs in the forum you posted it, it's not easy or hard relative to the forum.  It's simply appropriately posted.

[quote="mtms5467"]There are eight marbles. They all look exactly the same but one weighs slightly more than the others. How can you identify the heavier marble using two weighings?[/quote]

[hide="Different Method"]
Every weighing will eliminate 2/3 of the marbles, leaving only 1/3. So:
$\lfloor \frac{1}{3}\cdot 8 = 3$
$\frac{1}{3}\cdot 3 = 1$

So it takes 2 weighings.

[/hide]

[quote="t0rajir0u"]Less easy:  Suppose we don't know whether the different marble is heavier or lighter.  How many weighings of $8, 9, n$ marbles do we need?[/quote]
[hide="8 marbles"] Umm...the fastest way I can think of is weigh pairs of 2 marbles at a time, until you find a difference. This can potentially take 4 weighings. Let's call the heavier marble $p$ and the lighter marble $q$. Then, we take marble $p$ and weigh it with another marble. If it is still heavier, $p$ is the correct marble. If they weigh the same, $q$ is the correct marble.

So, $\boxed{5}\text{weighings}$?[/hide]


este es un problema que me pusieron en unos entrenamientos a qui en méxico. No lo eh podido hacer, el problema dice:

Sea ABC un triángulo y sean D, E puntos sobre BC. Una cicunferencia por A corta las rectas  AB, AC, AD, AE en los puntos P,Q,R,S, respectivamente. Muestre que,

(AP·AB - AR·AD)/(AS·AE - AQ·AC) = BD/CE

La sug. que me dieron es usar potencia de un punto.

[quote="Tony2006"]
$((AP)(AB)-(AR)(AD))/((AS)(AE)-(AQ)(AC)) = BD/CE$

.[/quote] oe mi pregunta es k si esta bien $(AS)(AE)-(AQ)(AC))$ por k creo ke eso sale negativo. chaufas.


Question:
$ \text{A triangle with side lengths 6, 8, and 10 are inscribed in a circle.  What is the area of the circle?}$

Answer:
[hide]$ 25$[/hide]

My question:

How do you get the area of the circle, I know the area of the triangle is 24, and the area of the circle is 1 more.  How do I figure this out (or should I say you figure this out)?

Right triangle... Circumscribed circle has radius half of hypotenuse...

Answer is $ 25\pi$ not 25

[hide] The diameter is the side of length 10 since it's the longest. Now, the radius is then 5 and the area is 25pi.[/hide]

EDIT: Oh snap, Lyndon's too fast.

Thanks!  BTW: You guys are fast, the mathcrazed, you posted ~30-45 seconds after I posted the post!!  Huh. THe answer booklet has it wrong as 25 and not $ 25\pi$.  Thanks!

Mods: May be locked.

It's obvious the 6 and 8 wouldn't be the diameter of the circle, but how do we determine 10 is the diameter rather than another chord that simply appears to be the diameter?

[asy]path p = (0,0)--(8,0)--(8,6)--cycle;

draw(rotate(143.1)*p);
/* draw(p); */
draw(Circle((-5,0),5));
label("10", (-5, 0), S);
label("8", (-3, 3));
label("6", (-8.5, 3));[/asy]

any right triangle inscribed in a circle is also inscribed in the semicircle so 10 is the diameter of the semicircle

Ok, well, this can be proved but I'm too lazy.
Any [i]right triangle[/i] inscribed in a circle has its hypotenuse = diameter of the circle.

Consider the perpendicular bisectors of the sides with length 6 and 8.  This immediately gives us that the center is the midpoint of the hypotenuse.

The right angle is half the angle of the arc it intercepts. Therefore, the arc is $ 180^\circ$ and the diameter is the hypotenuse.


Nan has art class every sixth day of school.  Assume that school is held Mondayh through Friday for $ 36$ consecutive weeks.  School starts on a Monday and Nan has art the first the day of school.  How many times during the school year will Nan have art on Monday?

[hide]$ 6$[/hide]

**Does every sixth day mean multiples of $ 6$?**

[hide="hint"]well one thing that you could do to see if every sixth day is days with multiples of six is assume that it does, then how many times would she have art class on a monday!  I have a suspicion that it does in easier words i think that it means that if you took a calendar of 36 weeks from monday to friday and labeled the school days 1-6 and once you got to six you would start over, only days marked 6 will be art class[/hide]

[hide]
Because $ 6\equiv 1 \pmod{5}$, every week the art-class day changes from Monday to Tuesday, etc. (but every 6th week cannot have an art class!) Thus every $ 6$ weeks there is one art class on Monday. $ \dfrac{36}{6} \equal{} \fbox{6}$[/hide]

[quote="sbanerjee"][hide="really?"]Nan has art class on monday the first week, again on the sixth week Then again on the 11th, and the 16th and the 21st, 26th, 31st, and 36th. That's a total of $ 8$ weeks, not 6...[/hide][/quote]

Um, no. School days.

[hide]Art class: Day 1 (Monday, Week 1)
Art class: Day 7 (Tuesday, Week 2)
Art class: Day 13 (Wednesday, Week 3)
Art class: Day 19 (Thursday, Week 4)
Art class: Day 25 (Friday, Week 5)
Art class: Day 31 (Monday, Week 7)
and so on...[/hide]

[quote="sbanerjee"][hide="really?"]Nan has art class on monday the first week, again on the sixth week Then again on the 11th, and the 16th and the 21st, 26th, 31st, and 36th. That's a total of $ 8$ weeks, not 6...[/hide][/quote]
You can't have it on the 6th and the ones that are 5 after that. Week 1:Monday, 2:Tuesday,3:Thursday,4:Wednesday,5:Friday, but next week, you don't have it, so it goes 1st,7th,13th,19th,25th,31st.

Oh, heheheheheh heh....heh........heh. Ok, I wasn't paying attention. I'm not thinking it through enough.


If we are given n sides $a_1,a_2,...,a_n$ which are known that they can be the sides of a convex n-sided polygon, then does there always exist a cyclic n-sided polygon with these n sides? If yes, then is the circumcircle of that cyclic n-sided polygon unique? Moreover, does polygon which has fixed sides attain its max. area when it is cyclic? If no, then when does it attain its max. area?

I hope someone can help me to solve this problem, partial solution/opinion is also welcome.

Thx!

The fact that its maximal area is reached when it's cyclic was an Iranina problem posted recently by Omid (although I don't know if there always exists a cyclic polygon with the respective sides). Try searching for the problem. Meanwhile, I'll look for it and post the link if I find it.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16777]Here[/url]'s the topic. I don't know if the solution is correct, but it seems like it's correct to me.

hi, grobber, I think a friend of mine get a solution for the existence.
It's easy to know we can find a sufficient large circle to draw these n sides in this way (see the figure below). Then when the radius of the circle decreases, the distance between the two endpoints is strictly decreasing. If the radius continues to decrease, then the two endpoints will coincide with each other, which forms a cylic n-sided polygon. However, I don't know if such circumcircle is unique. My conjecture is that some special n-sided polygon with fixed sides have two circumcircles, but some only has one. Anyway, I'm still searching for a proof of it.

I have also found thr proof for the uniqueness, :)  actaully very easy!
Suppose there exists two circumcircles, then the polygon inscribed in it must be distinct. Let $A_1$ and $A_2$ be the area of the two polygons. From the proof by grobber, we know that $A_1\geq A_2$ and $A_2\geq A_1$, so $A_1=A_2$. The remaining is trivial.


conics
parabola
LaTeX
calculus
derivative
Pythagorean Theorem
geometry

Question: A point moves along the parabola y=X^2, it's x-ordinate is increasing uniformly at 2 units/min. At what rate is its distance from (1,-1) changin when it passes through (2,4)?

I just don't know where to start, what forumal(s) to begin with?

Thanks,
BB[/code]

all right, here is what i got (sorry i don't know Latex, maybe some one can rewrite this!!):
[hide]
the distance of a point on the graph of y=x^2 from the point (1,-1), is calculated by this equation:
distance=sqrt((x-1)^2+((x^2)+1)^2)
therefore the rate of change of distance is the derivative of the above:

rate of change:
[(x-1)+((x^2)-1)(2x)]/[sqrt((x-1)^2+((x^2)+1)^2)]
then all you need to do is to plug in the values for the point (2,4)
you get that the distance is changing at the rate of:
13sqrt(5)/5 :P  [/hide]

Sorry, thank you very much. Just need clarification on what you did.

distance=sqrt  (   (x-1)^2   + ((x^2)+1)^2       )  <-pythagorean theorem
      C^2      =           A^2     +        B^2

[(x-1)+((x^2)-1)(2x)]/[sqrt((x-1)^2+((x^2)+1)^2)]  <- what did you do here to find the derivative?

I tried deriving it myself and got approx 9.3 units/min and you got 5.81 units/min.

[(x-1)+((x^2)-1)(2x)]  <-product rule               / [sqrt((x-1)^2+((x^2)+1)^2)]

Thanks

all right.
i think you know this:
the distance of a point (a,b) from the poing (c,d) is calculated by this formula:
sqrt(((a-c)^2)+((b-d)^2))
now let (a,b) be (x,(x^2)). and (c,d) is (1,-1).
ok?

i might have made a mistake in the derivative. but i took the derivative of every thing that was under the sqrt, and devide it by 2*sqrt of the original,

I think i get it :) Our teacher has us use Libence (sp?) notation, just trying to follow what you're doing and rework it myself with the notation.

OK, here's what I did and got a reasonable answer.

Pythagorean Theorem: let  C = distance, A = x^2 - 1,  B = X + 1

C^2 = (X^2 - 1)^2  +  (X + 1)^2

Isolate C by square root of entire right side

Find the derivative of that using Chain Rule

Plugged in my given x value (2) and got an answer of approx 3.53 units/min.

Sound like i have the right idea? Thanks BB

you got the right idea, but i don't think the answer is but your pathagorean identitiy is wrong you wrote:
C^2 = (X^2 - 1)^2 + (X + 1)^2
the right one is
C^2 = (X^2 +1 1)^2 + (X- 1)^2 because your point is (1,-1)
and the final answer is (i had made a mistake before!):
21sqrt(26)/26=4.118 :lol:

Ack, right, thank you very much. I got it now.

BB


Let $ABC$ be a triangle with $AB=AC$. Let $D$ be the midpoint of $BC$, $M$ the midpoint of $AD$ and $N$ the foot of the perpendicular from $D$ to $BM$. Prove that $\angle ANC = 90^\circ$.

Since $MD \perp DB$, $DN \perp BM$, so $\angle{ADN}=\angle{DBM}$ and $MD^2 = MN \cdot MB$. Because $M$ is the midpoint of $AD$, so $AM^2=DM^2=MN \cdot BM$, thus $\angle{NAD}=\angle{ABM}$. Therefore we have:
\[ \angle{ACB}=\angle{ABC}=\angle{ADN}+\angle{DAN}=180^o-\angle{AND} \]
Hence $A, N, D, C$ are cyclic, thus $\angle{ANC}=\angle{ADC}=90^o$.

[hide="For V. Vornicu."]
[color=red]You placed correctly here this problem which is antiquated. Again I see $\mathrm {an\ acute\ absence\ of\ the\ imagination\ (inspiration)\ !}$[/color][/hide]
[hide="Here is a simple proof."]
[color=darkred]Define the point $E$ for which the quadrilateral $ADCE$ is a rectangle. Observe that $M\in BE$ and the points $D,E,N$ belong to the circle with the diameter $AC$, i.e. the points $A,C,D,E,N$ are cyclically.[/color][/hide]
[hide="Here is a nice generalization."]
[color=darkblue]Let $ABC$ be an acute triangle. Define the projection $D$ of the vertex $A$ on the side $[BC]$ , a point $M\in (AD)$ and the projection $N$ of the point $D$ on the line $BM$. Prove that $NA\perp NC\Longleftrightarrow \frac{MA}{MD}=\frac{DC}{DB}\Longleftrightarrow [ABM]=[CDM]$.
[b]Remark.[/b] It would be better if this last problem (the its proof is similarly with the above solution), which isn't evidently and which is more interesting, would be proposed at the Romania District Olympiad, 2006 ![/color][/hide]

This can also be done using vectors:
$\overrightarrow {AN} \cdot \overrightarrow {NC}=(\overrightarrow {AM} + \overrightarrow {MN})(\overrightarrow {ND}+\overrightarrow{ DC})=\overrightarrow{AM}\cdot \overrightarrow{ND} +\overrightarrow {MN} \cdot \overrightarrow {DC}$(Since $AM \perp DC$ and $MN\perp ND$) so now all we have to prove is that $\overrightarrow{AM}\cdot \overrightarrow{ND} +\overrightarrow {MN} \cdot \overrightarrow {DC} \iff \overrightarrow {AM} \cdot \overrightarrow{ND}= \overrightarrow{NM} \cdot \overrightarrow {DC}$ and this is true since $\angle{ADN}=\angle{MBC}$
End of story


find with proof all pairs of integers (x,y) satisfying 1+1996x+1998y=xy


In $\triangle{ABC}$, we have $\sin{A}\cdot\sin{B}=\cos^2{\frac{C}{2}}$.
Find what kind of property the triangle has.

[hide]
$\sin A \sin B = \sin^2 \left( 90 - \frac{C}{2} \right)$
$= \sin^2 \left( \frac{A+B}{2}\right)$
Using half angle formula
$= \frac{1+\cos(A+B)}{2}$
$= \frac{\sin A \sin B - \cos A \cos B + 1}{2}$
$\cos A \cos B + \sin A \sin B = 1$
$\cos(A - B) = 1$
$A - B = 0$
$A=B$
Isosceles.
[/hide]

Is there a way to show that $\cos{C} + \cos{(A+B)} = 0$?

cuz the way I was doing it
[hide]I got down to $\cos{C}+1 = \cos{(A-B)}-\cos{(A+B)}$ and looking at the second to last step in your solution, I wanted to make sure I didn't make a mistake.[/hide]

To show $\cos{C}+\cos{(A+B)}=0$, just note that since the points A, B, and C form a triangle, their sum must be 180. Thus, the equation becomes $\cos{C}+\cos{(180-C)}=0$. Since $\cos{(180-C)} = -\cos{C}$, the equation is true.

Why are you trying to show this, though? It is true for any triangle.

Because the way I went about it was this:
[hide]$\displaystyle \sin{A}\cdot\sin{B}=\cos^2{\frac{C}{2}}$
Using the half angle identity, we get
$\sin{A}\sin{B}=\frac{\cos{C}+1}{2}$
$2\sin{A}\sin{B}=\cos{C}+1$
$\cos{(A-B)}-\cos{(A+B)}=\cos{C}+1$
Noting that $\cos{C}+\cos{(A+B)}=0$, we have
$\cos{(A-B)}=1$
$A-B=0$
And finally $A=B$[/hide]


Let $ m\minus{}1\equal{}2^r x$ where $ x$ is odd less than or equal to $ 2^r \plus{}1.$ Show that if there is a $ b$ such that
$ 2^{2^{r\minus{}1}} \equiv \minus{}1\,  mod\, ( \,m)$
then $ m$ is a prime.
Morover show that otherway i.e., that if there is such $ m$ which is prime then $ b$ can be founded.

Sorry it should be

$ b^{2^{r\minus{}1}} \equiv \minus{}1\,  mod\, ( \,m)$

 and not

$ 2^{2^{r\minus{}1}} \equiv \minus{}1\,  mod\, ( \,m)$

Where can I foun the proof. Please help thanks. Can you please write the proof here, thanks.

what about the other way, if m is a prime then b can be founded? Thanks for the link.

Read the posts carefully before posting further questions. In the link I gave that was the word "iff".

Can somebody else please help. My question is not same as this theorem. Anyways I have show one part. Can somebody else help me to show that if m is prime then it is easy to find such a b such that the congurence holds.


I am having trouble finding information about certain rules for the Putnam, specifically what can and cannot be used.  I know calculators aren't allowed, but what about things like ruler, compass, and graph paper?

From the rules: (this is from several years ago, and I haven't opened this year's package yet, but I doubt they've changed anything.)

[quote]7. [b]CONTESTANTS ARE NOT PERMITTED TO BRING ANY PAPER, BOOKS, SLIDE RULES, RULERS, CALCULATORS, COMPUTERS, ETC., INTO THE EXAMINATION ROOM.[/b] They should bring sharpened pencils (or pens) and erasers. Watches are helpful.

....

10. The supervisor should provide an adequate supply (approximately 12 sheets per contestant) of 8 by 11 white paper that is [b]blank[/b] and unlined on both sides (such as copier paper) to be used, if necessary, as extra sheets or scratch paper.

...

[color=red]12. No copies of the competition, morning or afternoon session, may be taken out of the room in which the competition is being held until after the formal [b]END[/b] of the afternoon session. (This prohibition applies to both students and faculty.) The supervisor may lock up the competition in another location during the lunch break. Furthermore, the supervisor should insure that no written material of any kind (scratch paper, notes, etc.) leaves the examination room until after the [b]END[/b] of the afternoon session.[/color][/quote]
All of that boldface, all caps, and red ink was in the original. As I said, that's from a few years ago, but the rules shouldn't have changed.

I don't think you need to be told what to do with cell phones/Blackberrys/etc.

So, are we allow to use infinitely many scatch paper? Also, how many pages at most I can use for one solution?

Each question comes on its own folded double-page. Using both sides of that, you have four pages for each question. That's going to be enough 99.9% of the time. If you need more space than that, you are given instructions on how to add extra pages - but it almost never comes up.

You do understand that each question is totally independent of each other question, and each question will be graded by its own set of graders. You must not refer in question A4 to something you claimed in your work on question A2 - that wouldn't make any sense, as the A4 graders would have no access to your A2 work.

If you are a double major, and have finished one major can you still take the putnam exam?

I'm pretty sure you can take it as long as you haven't graduated, and you certainly can't take it more than $ 4$ times if you are going to take more than $ 4$ years to finish both degrees.


real analysis
integration
limit
real analysis unsolved

$f$ is differentiable on $\mathbb{R}$. Both $f$ and $f'$ are Lebesgue-integrable on $\mathbb{R}$. Show that $\int_\mathbb{R}f'(x)dx = 0$.

$f$ integrable on $R \Longrightarrow f(-\infty)=f(+\infty)=0 \Longrightarrow \int_{-\infty}^{+\infty}f'(x)dx = f(+\infty)-f(-\infty)=0$

:cool: Do you know the non-standard real numbers ? :cool:

[quote="Diogene"]$f$ integrable on $R \Longrightarrow f(-\infty)=f(+\infty)=0 \Longrightarrow \int_{-\infty}^{+\infty}f'(x)dx = f(+\infty)-f(-\infty)=0$[/quote]
Why do $f(+\infty)$ and $f(-\infty)$ exist?

[quote="Diogene"]:cool: Do you know the non-standard real numbers ? :cool:[/quote]
Yes, I do.

[quote="Kent Merryfield"][quote]Why do $f(+\infty)$ and $f(-\infty)$ exist? [/quote]
Because $f'$ is integrable.[/quote]
Could you please point out more?

$\int_\mathbb{R}f'(x)dx = \lim_{a\rightarrow -\infty;b\rightarrow +\infty}(f(b)-f(a))$
 :cool:

man, i know that....
i was asking why the integrability of $f'$ implies the existence of $f(+\infty)$ and $f(-\infty)$.

$f(x)=f(0)+\int_0^xf'(t)\,dt.$ Now take the limit as $x\to\infty.$ The limit exists and equals $f(0)+\int_0^{\infty}f'(t)\,dt$ by the Dominated Convergence Theorem.

[quote="Kent Merryfield"]$f(x)=f(0)+\int_0^xf'(t)\,dt.$ [/quote]
Why does it hold? We don't tell you that $f$ is absolutely countinuous.

$\int_\mathbb{R}f'(x)dx = \lim_{a\rightarrow -\infty;b\rightarrow +\infty}(f(b)-f(a))$
By definition, that means for ex.   $\lim_{b\rightarrow +\infty}(f(b)-f(a))= f(+\infty)-f(a)$  is convergent.

:cool:


vector
real analysis
real analysis unsolved

Let E be a vectorial space with inner product. Show that all isometric immersion f: R -> E can be write as f(t) = a + t.u, where a, u belong to E and |u| = 1.

pick $ a \equal{} f(0)$

Now, for arbitrary $ t$, we have that $ \|f(t) \minus{} f(0)\| \equal{} |t|$ this shows that $ f(t) \equal{} a \plus{} u(t) t$ where $ u(t)$ is a unit vector depending on $ t$.

Now, picking any two $ t$, $ t'$, we have that $ \|f(t) \minus{} f(t')\| \equal{} |t \minus{} t'|$, so $ \|u(t) t \minus{} u(t') t'\| \equal{} |t \minus{} t'|$ Squaring (allowed, because its an inner product space),

$ \|u(t)\|^2 t^2 \minus{} 2 \langle u(t)t,u'(t') \rangle \plus{} \|u(t')\|^2 t'^2 \equal{} t^2 \plus{} t'^2 \minus{} 2tt'$, so using $ \|u(t)\| \equal{} 1$ for all $ t$, we get that $ \langle u(t) t , u(t') t' \rangle \equal{} tt' \equal{} \|u(t')\|^2 tt' \equal{} \langle u(t')t, u(t')t' \rangle$. Thus subtracting

$ \langle u(t) t \minus{} u(t') t, u(t') t' \rangle \equal{} 0$, so as long as $ t,t'$ are nonzero, we get that $ u(t) \equal{} u(t')$ (because $ u(t)$ is a unit vector so the left hand side cant be zero.

Now all you have to show is that $ u(0) \equal{} u(t)$, which I'll leave to you.


we have m.cos(a+b)=sin(a-b).Prove that A dont depend on a,b
  A=1/(1-m.sin2a)+1/(1-m.sin2b) help me.............

I don't think so..  :maybe:
If $b = 0$, then $A = sec 2a$ which depends on the value of $a$..


病因 （一）

疯人院新任院长走到一个病人面前，问他为什么进了疯人院？
病人问答：＂医生，是这样的。我娶了一个已有成年女儿的寡妇，而我的父亲则
娶了她的女儿为妻。所以我太太成了她公公的岳母，她女儿成了我的继女和继母。
继 母生了个儿子，这个孩子成了我的弟弟和我太太的外孙。我也有了一个儿子，这个
儿子成了他祖父的内弟和他自己叔父的叔父。另一方面，我父亲提到他孙子的时候，
说是他的内弟，我的儿子叫他的姐姐做祖母。我现在认为我是我母亲的父亲，我外孙
的哥 哥，我太太是他女婿的儿媳，他外孙的姐姐。现在我不知道我是自己的祖父、我
弟 弟的父亲，还是我儿子的侄子，因为我的儿子是我父亲的内弟。院长，这就是我
来 这 里的原因。我觉得在这里比在家里清净些。”

英文字母的搞笑对话（二）
＄对S说：在俺们美国，有钱没钱看领带！
￥对Y说：在俺们中国，有钱没钱看裤带！
8对∞说：小样儿，躺地下打滚儿我就怕你啦？
n对∏说：戴个大盖帽就跟我玩派，是不？
a对A说：哥们，你的别墅好漂亮啊！
b对B说：姐们，在哪做的丰胸手术呀？
d对D说：姐姐，你看我梳“冲天辫”好看不？
e对E说：哥们，你在视力检查中心工作吧？
g对G说：你真的是我哥哥？为啥咱俩长得一点也不像呀？
i对I说：大哥，年轻时是不是也喜欢戴帽子呀？
j对J说：姐们，为什么玩牌的人都管你叫“钩儿”呢？
m对M说：哥们，这身肌肉块在哪练的呀？
o对O说：姐姐，为什么那帮小子一见咱俩就“呕呕”怪叫呢？
pq对bd说：夫妻感情好，就得对着咬！
qp对db说：看似般配的夫妻不一定就幸福！
pb对qd说：夫妻身高有差距，不等于感情有差距！
qb对dp说：身高差点不怕，要是感情也差，那就没法处了！[img]http://www.qqjia.com/mu/p906c/triangl.jpg[/img][img]http://www.qqjia.com/mu/p906c/115.jpg[/img]

《施氏食狮史》

石室诗士施氏，嗜狮，誓食十狮。氏时时适市视狮。十时，适十狮适市。是时，适施氏适市。氏视是十狮，恃矢势，使是十狮逝世。氏拾是十狮尸，适石室。石室湿，氏使侍拭室。石室拭，氏始试十狮尸，食时，始识是十狮尸，实十石狮尸。试释是事。

[quote="shobber"]Is the answer 7?[/quote]
你是不是没数就说这个答案的 :P

[quote="shobber"]我数了...很认真地数了. 从上到下:

1: 云.
2: 月亮上的黑影.
3: 山上的骑士.
4: 画面右边的那块石头.
5&6: 两匹活马.
7: 帐篷里的火影.[/quote]
我以为你是根据七匹马的画直接说答案的，误会了 :blush:

[quote="Soarer"][quote="shobber"]我数了...很认真地数了. 从上到下:

1: 云.
2: 月亮上的黑影.
3: 山上的骑士.
4: 画面右边的那块石头.
5&6: 两匹活马.
7: 帐篷里的火影.[/quote]
樹根位置的馬頭[/quote]
我看不出来， :blush:

[quote="Soarer"][quote="shobber"]我数了...很认真地数了. 从上到下:

1: 云.
2: 月亮上的黑影.
3: 山上的骑士.
4: 画面右边的那块石头.
5&6: 两匹活马.
7: 帐篷里的火影.[/quote]
樹根位置的馬頭[/quote]
我就觉得那根树上有文章......可我还是看不出来......

我也找到一些相似的圖。
(To Soarer: from Mr. Koo's personal web site)
[img]http://www.qc.edu.hk/it-school/homepage/KC/7Hearts.jpg[/img]
[img]http://www.qc.edu.hk/it-school/homepage/KC/7Dolphins.jpg[/img]

人，之所以被称为 homo sapiens.. 乃是因为 我们的创造力...

具体我也不清楚，但至少10匹以上。

萨达姆的《沁园春.雪》

美国风光,
千里兵疯,
万里血漂。
望纽约内外,
废墟莽莽；
白宫上下，
顿时逃逃。
你舞战斧，
我施冷枪，
真主上帝试比高。
石油如此稀少，
引无数枭雄竞折腰。
惜阿尔道夫，
功亏一篑；
东条英机，
两弹魂消。
一代天骄，
拉登老弟，
只识驾机撞世贸。
俱完矣，
数风流人物，
还看锦涛。

水调歌头之光棍篇

美眉几时有？把酒问室友。
不知隔壁姑娘，可有男朋友？
我欲凿墙看去，又恐墙壁太厚，
夯疼我的手。
改用望远镜，屋里人已走。
转楼梯，低头看，
那某某，果不单身走，
她正挎着俊男肘。
人有悲欢离合，月有阴晴圆缺，此事古来有。
但原没多久，他俩就分手

有两个修女，一个是叫做数学修女，另一个则是叫逻辑修女。现在已经快天黑了但她们离修道院还有很远的路程。

数学： 你有没有注意到，后面有个男人已经跟踪我们有三十八分钟三十秒了，
       不知道他想要做什么？

逻辑： 这很合理的，他想侵犯我们。

数学： 天哪！在这样的速度下，他会在十五分钟之内抓到我们的，我们该怎么办？

逻辑： 唯一合理的方法当然是走快一点。

数学： 好像没用呀！

逻辑： 当然没用，那个男人也很合理的越走越快。

数学： 那我们该怎么办？在这样的速度下，他还有一分钟就能抓到我们了。

逻辑： 唯一合理的方法就是我们分开逃，走那边，我走这边，他不可能两个都抓。

       那个男人继续跟踪逻辑修女。

数学修女平安地到达修道院，但很担心逻辑修女会不会出事，然后就看到逻辑修女进了门口。

数学： 逻辑修女你终于回来啦！感谢主！快告诉我发生什么事了？

逻辑： 发生了唯一合理的事情，那个男人不能两个都跟踪，所以他就来追我。

数学： 对对，但后来发生什么事？

逻辑： 发生了唯一合理的事情，我用尽全力地跑，他也用尽全力地在后面追。

数学： 然后呢？

逻辑： 发生了唯一合理的事情，他抓到我了。

数学： 天哪！那怎么办？

逻辑： 我做了唯一合理的事，把裙子拉起来。

数学： 天哪，逻辑修女！那个男人呢？

逻辑： 他做了唯一合理的事，他把裤子拉了下去。

数学： 我的天哪！那后来呢？

逻辑： 不是很合理吗，数学修女， 一个把裙子拉起来的修女，一定跑得比一个把裤子拉下去的男人快得多!!!

这个数学修女和这个逻辑修女。。。不过逻辑算不算是数学的一部分？？
我一直没太搞清楚。。。


combinatorial geometry
real analysis
real analysis unsolved

This is a nice result: if $n\geq 2$ and $f: R^{n}\to R$ continuous has the property that for all $x$ the set of solutions of $f(y)=x$ is bounded, then $f$ attains one of its extrema.

Call $S_{x}$ the compact subset $\{y : f(y)=x\}$ and $C_{x}$ the convex hull of $S_{x}$, that is also compact. Now, take an arbitrary value $x_{0}$ in the range of $f$. Then there are two cases: for all $x \not \in C_{x_{0}}$, $f(x)>x_{0}$, or for all $x \not \in C_{x_{0}}$, $f(x)<x_{0}$. (this is because $f$ is continuous, and if it were false, it would contradict the intermediate value theorem. Of course, since $n \geq 2$, $\mathbb{R}^{n}\setminus C_{x_{0}}$ is pathwise connected, so along a path between a point $f(y_{1})>x_{0}$ and another $f(y_{2})<x_{0}$ it would exists a point such that $f(y)=x_{0}$ which is not in $C_{x_{0}}$ : absurd).

  Suppose that the first case holds :  for all $x \not \in C_{x_{0}}$, $f(x)>x_{0}$. Then $f$ is continuous inside $C_{x_{0}}$, so achieved its minimum inside the compact set $C_{x_{0}}$. And this minimum is also a global minimum.


Find all positive integers x,y which satisfy:

$x^2-y^2=666$

Wow, great solution, I couldn't think that far.

[quote="Melissa"][hide="I find none :?"]
Since $666=2\times{3^2}\times{37}$, there is only one even factor. This means that $666$ will never be the product of two even numbers. Also, because $666$ is even, it will never be the product of two odd numbers. Therefore it has to be the product of one even and one odd number.
$x^2-y^2=(x+y)(x-y)$. Setting $\alpha{\beta}=666$,
$x+y=\alpha$ and
$x-y=\beta$
Solving, $2x=\alpha+\beta$. But $\alpha + \beta$ is odd, so $x$ cannot be an integer. Then there are no integer solutions.[/hide][/quote]

Same here :D


linear algebra
matrix
linear algebra unsolved

Let  $ A,B$ be a comlex $ 2x2$ matrix. Assume $ AB\neq BA$. Then show that $ AB, BA$and$ E$ are linearly independent, where$ E$ is $ 2x2$ unit matrix.

It's false:

Example: $ A\equal{}\begin{bmatrix}0&1\\0&0\end{bmatrix},\ B\equal{}\begin{bmatrix}1&0\\0&0\end{bmatrix}.$

Then $ AB\equal{}0$ and $ BA\equal{}A.$ But then, the set $ \{I,AB,BA\}$ is dependent, because it includes the zero element.

dear mathlinkers!
Excuse me for my stupid question!
My question was false!


function
calculus
linear algebra
matrix
AMC
AIME
MATHCOUNTS

What is the avatar above you thinking?

Rules:
--Remember, this is "What is the avatar above you thinking", not "What is a description of the avatar above you".
--People with no avatars are not allowed to post. However, if one shall do so we shall assume their avatar looks like this
[img]http://www.here-now.org/content/2004/04/29/0429bush50.jpg[/img]

The avatar above me is thinking...

"I don't understand human. What does all that text mean? WAIT A SEC... I think that pi symbol is talking about me..."

"I feel pretty special around here, but I could go for some apple pie right now..."

"I am [i]so[/i] cooler than the likes of lambda, omicron, or xi"

"i also get hurt by water attacks, dang it!"

"Be the ball, be the ball, be the ball..."

"Why does the cotangent function avoid me?"

I wish I could stop smiling, my lips really hurt...

What are we doing, playing Corrupt A Wish?

No, you stupid sailboat. We're playing "what is the avatar above you thinking?

and you can pretend mine is a surfing manatee if the sailboat is getting boring.

hey, babes, don't i look sexy with this surfboard?
(responding to manatee thingy)

Ahh... I wish this confetti did not have such sharp edges...
now I am Santa Claus stuck in a chimney.

i knew that beer wasn't gonna help... T.T

Moo.

I think this topic is going to the spam demons.

I would make an awesome surfboard...
now I am a cookie flying past the goal at a football stadium.
if you've posted a lot it makes it more interesting to change what your avatar is.

He He, these people are so dumb - they think they can guess what I'm actually thinking...
Well, I'd just gobble them up as punishment for their idiotness if I could...

- now my avatar is a bird about to fly into an airplane's engine. -

Hey! A match! Hey! Beer! I wonder what would happen if I -

10 minutes later: I DON'T KNOW WHO LIT THAT MATCH NEXT TO THAT BEER!!!

I wish 1=2 would give me a carpet... here I am, stuck in a little box with a vacuum. the least I could do is be useful...

-Now I am  :rotfl:  about to roll off a cliff.-

THAT IS SO FUNNY HAHAHAHHAHAHAHAHAHAHAHAHAHAHAAAAAAAAAAAAAAAAAAAAAaaaaaaaaaaaaaaa[size=75]aaaaaaaaaaaaaaaaaaaa[/size][size=59]aaaaaaaaaaaaaaaaaa[/size][size=9]aaaaaaaaaaaaaaaaaaaaaa.[/size]

*thud*

I wish I had another arm so I could give this one a rest...

now I am an airplane pilot who just ran out of gas over the atlantic.
please pretend you have a different avatar if you have posted a lot.

[quote="hunter34"]I think this topic is going to the spam demons.[/quote]

Yep. I don't think the creativeness is just there with the same user replying every other post.

Sigh...another marathon thread locked.


inequalities calculus integration logarithms induction

Prove that

$ \sum_{k\equal{}1}^{n}{\frac{1}{k}}\minus{}\ln(n)<\frac{5}{6}$ for $ n\ge 1$ integer.

 :)

Can we prove the problem without using Integration? :maybe:

Did you mean $ n>1$?
Unless I misinterpreted:
$ H_1\minus{}\ln(1)\equal{}1\minus{}0\not<\frac{5}{6}$

All you need is the inequality $ \ln\frac{n}{n\minus{}1}>\frac1{n}$ and induction. I'm not sure how you prove that or anything else about the natural logarithm without calculus, but it isn't very complicated

@Patterns_34: yes you are right it should b $ n > 1$

@kunny: yes it is possible, and not so hard, but you'll see that the inequalities are sort of an integration

@jmerry: something like that inequality is needed at least in my proof, but it is more like telescoping than induction

Try to work with the inequality jmerry provided and see what happens...

 :)


calculus
integration
function
analytic geometry
calculus computations

Let $ a,\ b$ be postive real numbers. For a real number $ t$, denote by $d(t)$ the distance between the origin and the line $ (ae^t)x \plus{} (be^{ \minus{} t})y \equal{} 1$.

Let $ a,\ b$ vary with $ ab \equal{} 1$, find the minimum value of $ \int_0^1 \frac {1}{d(t)^2}\ dt$.

Just need a little clarification.

With the defined properties of $ a,b,t$, we find $ \ell_t$ to be $ Ax\plus{}By\equal{}1$ for some $ A,B\in\mathbb{R}$, but by distance, do you mean the [i]smallest[/i] distance? Or do you mean simply the distance function for some coordinate lying on the line in question?

The min value of integral is $ e\minus{}\frac{1}{e}$.

Thus, $ \ell_t : ux+u^{-1}y=1\Rightarrow y=u-u^2x$

Then the distance from the origin to some point $ (x,y)$ is
$ D=\sqrt{x^2+y^2}=\sqrt{x^2+u^2-2u^3x+u^4x^2}$
$ \Leftrightarrow \frac{dD}{dx}=\frac{x-u^3+u^4x}{\sqrt{x^2+u^2-2u^3x+u^4x^2}}=0\Longrightarrow x=\frac{u^3}{1+u^4}$

Subsituting this for $ x$ in $ D$, we find
$ d(t)=\frac{u}{\sqrt{1+u^4}}\Rightarrow \frac{1}{d(t)^2}=u^{-2}+u^2$
And
$ \displaystyle \begin{eqnarray*}
\int_0^1 \frac{\,dt}{d(t)^2}&=&\frac{1}{n^2}\int^1_0 e^{-2t}\,dt+n^2\int^1_0 e^{2t}\,dt\\
&=&\left(-\frac{e^{-2t}}{2n^2}+\frac{n^2e^{2t}}{2}\right|^1_0\\
\end{eqnarray*}$

That as far as I can get. I tried differentiating with respect to $ n$ but that didn't seem to work.

Let the value of the integral be $ I(n)$ so that from igiul's work

\begin{eqnarray*} I^{\prime}(n) & = & \frac{\partial }{\partial n}\left[ - \frac {e^{ - 2t}}{2n^2} + \frac {n^2e^{2t}}{2}\right|^1_{t = 0} = \left[ \frac {e^{ - 2t}}{n^3} + ne^{2t}\right|^1_{t = 0}\\ & = & \frac {1}{n^3} \left( e^{ - 2}-1\right)+ n\left( e^{2}-1\right) = 0\\ & \Rightarrow & \boxed{ n  = \sqrt[4]{\frac {1-e^{-2}}{e^{2}-1}}}\\
\end{eqnarray*}

We also have $ I(n)= \frac {1}{2n^2} \left(1- e^{ - 2}\right)+ \frac{1}{2}n^2\left( e^{2}-1\right)$.

So the minimum value is

$ I\left(\sqrt[4]{\frac {1-e^{-2}}{e^{2}-1}}\right) =  e-\frac {1}{e}$

Here is my solution.

By the formula of the distance of a point and a line, $ d(t)\equal{}\frac{|\minus{}1|}{\sqrt{(ae^t)^2\plus{}(be^{\minus{}t})^2}}\equal{}\frac{1}{\sqrt{a^2e^{2t}\plus{}b^2e^{\minus{}2t}}}$.

$ \therefore I\equal{}\int_0^1 (a^2e^{2t}\plus{}b^2e^{\minus{}2t})\ dt$

$ \equal{}\left[\frac{1}{2}a^2e^{2t}\minus{}\frac{1}{2}b^2e^{\minus{}2t}\right]_0^1$

$ \equal{}\frac{e^2\minus{}1}{2}\left(a^2\plus{}\frac{b^2}{e^2}\right)$

$ \geq (e^2\minus{}1)\sqrt{a^2\cdot \frac{b^2}{e^2}}\ \text{by AM\minus{}GM}$

$ \geq e\minus{}\frac{1}{e}$
The equality holds when $ a^2\equal{}\frac{b^2}{e^2}$ and $ ab\equal{}1\ (a,\ b>0)\Longleftrightarrow a\equal{}\frac{1}{\sqrt{e}},\ b\equal{}\sqrt{e}$,
Therefore the desired minimum value is $ e\minus{}\frac{1}{e}$.


Let $a=\log_35,\ b=\log_57$. Express $\log_{63}105$ by using $a$ and $b$.


I have this homework problem that I can't solve!

help me please with the following inequality:
let x1,x2,..,xn be positive reals, n>=2. Prove that:
x1/(1*2)+x2/(2*3)+..+xn/(n*(n+1))>=x1^(1/(1*2))*x2^(1/(2*3))*..*xn^(1/(n*(n+1)))-1/(n+1).

thank you very much!

Tiger Sandre.
Germany, Stuttgart

Three successive hints in spoilers.  Try the problem again between hints.

Hint 1:

[hide]The 1/(1*2),1/(2*3),1/(3*4),...,1/(n+1) (I moved the last over) are awfully suggestive.  Do you recogize the telescoping sum?  It suggests that these are weights which conveniently sum to 1.[/hide]

Hint 2:

[hide]If a function is continuous and concave on an interval I and x1,x2,...,xn are in I, and 0<w1,w2,w3,...,wn<1 and w1+w2+w3+...+wn = 1, then

f(w1x1+w2x2+...+wnxn) >= w1 f(x1) + w2 f(x2) + ... + wn f(xn)

This is the weighted Jensen's inequality.  To see one use, let the weights all be 1/n, and let the f(x) be log x -> do you get AM-GM as a result?[/hide]

Hint 3:

[hide]We can use Jensen to prove the weighted AM-GM, in which the x's are positive and the weights are as above (positive, sum to 1):

w1x1 + w2x2 + ... + wnxn >= x1^w1*x2^w2*...*xn^wn[/hide]


Let three circle with center $ O$ are $ \omega_a, \omega_b, \omega_c$. The points $ A, B, C$ on $ \omega_a, \omega_b, \omega_c$. With any point $ P$ inside $ ABC$. Rays $ AP, BP, CP$ cut $ \omega_a, \omega_b, \omega_c$ at $ A_1, B_1, C_1$.
Rays $ B_1A_1, BA$ cut $ \omega_b$ at $ M, N$ and $ B_2\equiv MN\cap B_1B_2$;
Rays $ A_1C_1, AC$ cut $ \omega_a$ at $ P, Q$ and $ A_2\equiv PQ\cap A_1A_2$;
Rays $ C_1B_1, CB$ cut $ \omega_c$ at $ R, S$and $ C_2\equiv RS\cap C_1C_2$;
Prove that $ OP \equal{} OA_2 \equal{} OB_2 \equal{} OC_2$

Only a corollary of problem [url=http://www.mathlinks.ro/viewtopic.php?t=307672]geometry 1[/url], applying for triangles $ ABP, BCP, CAP$.


Find intervals containing solutions of: x - 3^(-x) = 0

Let f(x) = x - 3^(-x)
Then, f(0) = -1 and f(1) = 2/3. By Intermediate value theorem, there exists a  number c in (0,1) with f(c) = 0

By how can we show that the solution is unique? Can we use Rolle theorem here?

You can show the solution is unique by showing that the function in question is monotonic

$ f(x)\equal{} x \minus{} 3^{\minus{}x} \iff f'(x) \equal{} 1 \plus{} 3^{\minus{}x}\ln(3)$

It turns out that $ f'(x) \geq 0 \; \forall x$ thus $ f(x)$ increases $ \forall x$ as well.


inequalities
geometry
geometric transformation
analytic geometry
function
inequalities open

Hi everyone,

Nesbitt inequality is very familiar, aslo some extesion of it is well-known, too, such as \[\left(\frac{a}{b+c}\right)^{k}+\left(\frac{b}{c+a}\right)^{k}+\left(\frac{c}{a+b}\right)^{k}\ge \min\left(1,\frac{3}{2^{k}}\right).\] I want to ask the general-form of Nesbitt Inequality for $n$ numbers?

1. Sometime it's called $Shapiro$ Inequality, or

[i]Prove that for all even number $n\ (n\le 12)$ and for all odd number $n\ (n\le 23)$ then[/i] \[\frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+...+\frac{a_{n}}{a_{1}+a_{2}}\ge \frac{n}{2}.\] (a). Solution for $n\le 6$ is very simple by using Cauchy-Schwarz Inequality. How to solve it for [b]any[/b] other bigger number $n$?
(b). Do you know any conunter-example for $n=14$ and $n=25$? (truly, this thing is not very necessary).

2. For the general form

[i]Consider the following inequality (with positive real numbers $a_{1},a_{2},...,a_{n}$)[/i] \[\frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+...+\frac{a_{n}}{a_{1}+a_{2}}\ge \frac{kn}{2}.\] (a). Up to now, what is the best constant $k$ for the above inequality? or some formula?
(b). Any results in mathlinks or other about any values of $k$? (I remember $k=2(\sqrt{2}-1)$ is valid and solution is nice, if I'm not wrong).

3. For another form

[i]Prove or disprove that (for $a_{1},a_{2}...,a_{n}>0$)[/i] \[\sum\frac{a_{i}}{a_{i-1}+a_{i+1}}\ge \frac{n}{2}.\] (a). If it's wrong, for which value of $n$ then it's true?
(b). Could you change the number $\frac{n}{2}$ to another valid value? (of course, in form $\frac{kn}{2}$).

Above is quite important for many student, so please Mathlinks member help me complete it. Thank you.

[quote="puuhikki"]1a), 1b) I believe you should look

H. S. Shapiro, Problem 4603, Amer. Math. Monthly 61 (1954) 571.[/quote]

But the fact is that, I've never read any material about it, also i haven't the article you show as above. So could you please write something from it? Thank you,

[quote="spanferkel"]Here is one little link I could find:
http://www.math.niu.edu/~rusin/known-math/99/shapiro

it contains some more references  :)[/quote]

... but isn't it helpful for the questions above?  :(  I read it but I can't see any new things.

[quote="Flove"]
2. For the general form

[i]Consider the following inequality (with positive real numbers $a_{1},a_{2},...,a_{n}$)[/i] \[\frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+...+\frac{a_{n}}{a_{1}+a_{2}}\ge \frac{kn}{2}.\] (a). Up to now, what is the best constant $k$ for the above inequality? or some formula?
(b). Any results in mathlinks or other about any values of $k$? (I remember $k=2(\sqrt{2}-1)$ is valid and solution is nice, if I'm not wrong).
[/quote]

Approximate value of the best constant $k$ is 0.989. It's root of some equation, but I don't remember this equation. You can find details in original article of Drinfel'd "A certain cyclic inequality" Mat. Zametki, 9, 1971, P. 113--119 (english translation in Math. Notes 9 (1971), 68--71).

[quote="Flove"]
3. For another form

[i]Prove or disprove that (for $a_{1},a_{2}...,a_{n}>0$)[/i] \[\sum\frac{a_{i}}{a_{i-1}+a_{i+1}}\ge \frac{n}{2}.\] (a). If it's wrong, for which value of $n$ then it's true?
[/quote]

For $n \leq 4$ only.

[quote="Flove"]
(b). Could you change the number $\frac{n}{2}$ to another valid value? (of course, in form $\frac{kn}{2}$).
[/quote]

The best constant $k$ is zero  :lol:.

Actually for $x_{1}, x_{2}, \dots, x_{n}> 0$ and $n \geq 4$ \[\frac{x_{1}}{x_{n}+x_{2}}+\frac{x_{2}}{x_{1}+x_{3}}+\dots+\frac{x_{n-1}}{x_{n-2}+x_{n}}+\frac{x_{n}}{x_{n-1}+x_{1}}\geq 2 .\] The constant $2$ is the best. For example take $x_{k}= a^{k}$, $x_{n-1}= x_{n}= 1$ and tends $a$ to zero.

It's problem of Prokop'ev from Tournament of Towns, 1982 (or near).
See also Kvant, problem M749.

Thank you very much, Alexander Khrabrov. The result of part (3) is very amazing to me, although it's really not hard. But for some small value of $n$, could we find the minimum of the expression?

[b](3c).[/b] [i]Find the minimum value of[/i] \[\frac{b}{c+a}+\frac{c}{d+b}+\frac{d}{e+c}+\frac{e}{a+d}+\frac{a}{b+e}.\] [i]for all $a,b,c,d,e \ge 0$.[/i]

I think it's very nice and It definitely has a solution. I will try today.

Now, anyone help me to show some result for part (1)? It's about Shapiro Inequality.

Sorry, I just thought there is a full proof. There was only a conterexample for the case $n=20$:

From the article I referenced (not a direct copy to avoid copyright problems):

Let $n=20$ $a_{1},a_{2},\cdots a_{20}$ (in that order) to be $1+5\epsilon$, $6\epsilon$, $1+4\epsilon$, $5\epsilon$, $1+3\epsilon$, $4\epsilon$, $1+2\epsilon$, $3\epsilon$, $1+\epsilon$, $2\epsilon$, $1+2\epsilon$, $3\epsilon$, $1+\epsilon$, $2\epsilon$, $1+4\epsilon$, $3\epsilon$, $1+5\epsilon$, $4\epsilon$, $1+6\epsilon$, $5\epsilon$.

Then the expression is $10+\epsilon^{2}+O(\epsilon^{3})$ which is less than $n/2=10$ if $\epsilon$ is close enough to zero.

If I have time, I'll try to look some other references.

[quote="Flove"]Thank you very much, Alexander Khrabrov. The result of part (3) is very amazing to me, although it's really not hard. But for some small value of $n$, could we find the minimum of the expression?

[b](3c).[/b] [i]Find the minimum value of[/i] \[\frac{b}{c+a}+\frac{c}{d+b}+\frac{d}{e+c}+\frac{e}{a+d}+\frac{a}{b+e}.\] [i]for all $a,b,c,d,e \ge 0$.[/i]

I think it's very nice and It definitely has a solution. I will try today.
[/quote]

I missed $x_{n-1}=x_{n}=1$ in my post. So infimum is equal to 2 always for $n \geq 4$.

[quote="Flove"]
Now, anyone help me to show some result for part (1)? It's about Shapiro Inequality.[/quote]

Counterexamples:

For $n=14$: 50, 5, 48, 3, 48, 1, 50, 0, 52, 1, 54, 4, 53, 6 (Zulauf, 1958)

For $n=20$: 32, 0, 37, 0, 43, 0, 50, 0, 59, 8, 62, 21, 55, 29, 44, 32, 33, 31, 24, 30, 16, 29, 10, 29, 4 (from Kvant, 1991, No 4; but first example Daykin, 1971 or Malcolm, 1971)

I didn't check this counterexamples.

For $n \leq 6$ you can prove Sapiro's inequality using Cauchy-Bounyakovsky-Schwarz inequality. I heard that proof for $n=8$ is elementary. But I couldn't find the article (Djokovic D. Z. Sur une inegalite. Proc. Glasgow Math. Assoc., 6, 1963, P. 1-10).
For other $n$ proof is more complicated. I have no enthusiasm to read proofs. But I put references for person with strong spirit.

For $n=9$ and  $n=10$
Nowosad P. Isoperimetric eigenvalue problems in algebras.
Comm. Pure Appl. Math. 21, 1968, P. 401-465.

For $n = 11$ and $n=12$
Годунова Е. К., Левин В. И.,
Об одной циклической сумме двенадцати слагаемых.
Мат. Заметки 19, 1976, No 6, 873-885.

For $n=13$
Troesch B. A. On Shapiro's cyclic inequality for $N=13$.
Math. Comp. 45, 1985, No 171, P. 199-207.

For $n = 15$, 17, 19, 21 and 23.
Troesch B. A. The validity of Shapiro's cyclic inequality.
Math. Comp. 53, 1989, No 188, P. 657-664.

[quote="Alexander Khrabrov"][quote="Flove"]Thank you very much, Alexander Khrabrov. The result of part (3) is very amazing to me, although it's really not hard. But for some small value of $n$, could we find the minimum of the expression?

[b](3c).[/b] [i]Find the minimum value of[/i] \[\frac{b}{c+a}+\frac{c}{d+b}+\frac{d}{e+c}+\frac{e}{a+d}+\frac{a}{b+e}.\] [i]for all $a,b,c,d,e \ge 0$.[/i]

I think it's very nice and It definitely has a solution. I will try today.
[/quote]

I missed $x_{n-1}=x_{n}=1$ in my post. So infimum is equal to 2 always for $n \geq 4$.

[/quote]

Yes, after I post my opinion, I also see that the infimum of the expression for all number $n\ge 4$ is $2$. But now I'm intend some very nice inequalities which will posted in the future,

[quote="Alexander Khrabrov"]
[quote="Flove"]
Now, anyone help me to show some result for part (1)? It's about Shapiro Inequality.[/quote]

Counterexamples:

For $n=14$: 50, 5, 48, 3, 48, 1, 50, 0, 52, 1, 54, 4, 53, 6 (Zulauf, 1958)

For $n=20$: 32, 0, 37, 0, 43, 0, 50, 0, 59, 8, 62, 21, 55, 29, 44, 32, 33, 31, 24, 30, 16, 29, 10, 29, 4 (from Kvant, 1991, No 4; but first example Daykin, 1971 or Malcolm, 1971)

I didn't check this counterexamples.

For $n \leq 6$ you can prove Sapiro's inequality using Cauchy-Bounyakovsky-Schwarz inequality. I heard that proof for $n=8$ is elementary. But I couldn't find the article (Djokovic D. Z. Sur une inegalite. Proc. Glasgow Math. Assoc., 6, 1963, P. 1-10).
For other $n$ proof is more complicated. I have no enthusiasm to read proofs. But I put references for person with strong spirit.

For $n=9$ and  $n=10$
Nowosad P. Isoperimetric eigenvalue problems in algebras.
Comm. Pure Appl. Math. 21, 1968, P. 401-465.

For $n = 11$ and $n=12$
Годунова Е. К., Левин В. И.,
Об одной циклической сумме двенадцати слагаемых.
Мат. Заметки 19, 1976, No 6, 873-885.

For $n=13$
Troesch B. A. On Shapiro's cyclic inequality for $N=13$.
Math. Comp. 45, 1985, No 171, P. 199-207.

For $n = 15$, 17, 19, 21 and 23.
Troesch B. A. The validity of Shapiro's cyclic inequality.
Math. Comp. 53, 1989, No 188, P. 657-664.[/quote]

Thank you very much, although your reference helps me nothing  :(

[quote="Xevarion"]I have some of these articles in pdf format. If someone wants them, please PM me. :)[/quote]

Hi, so I will be the first one. Thank you very much,

[quote="silouan"]First of all Shapiro's is true for $n\leq 11$ for odd numbers $n$ and $n\leq 12$ when $n$ is even .But for even numers $n\geq 14$ and for odd numers $n\geq 27$ it is nt true .
Also
$\frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+....+\frac{a_{n}}{a_{1}+a_{2}}\geq c\frac{n}{2}$ holds for all $n$ where
$c=lim_{n\rightarrow\infty}c_{n}$ and $c$ is the y-coordinate of the point in which the common tangent of the functions
$f(x)=e^{-x}$  and $f(x)=\frac{2}{\sqrt{e^{x}}+e^{x}}$

intersect the axis $Oy$  .[/quote]

It's very good information, thank you. But can you show or find a clearer expression of $c$? Or... a approcimate value  :oops: I heard that someone can find it accurately with 100 decimal.

Above all, who can prove it for $n=7$? Certainly by elementaty mathematics.

[quote="spanferkel"]http://www.mathsoft.com/mathsoft_resources/mathsoft_constants/Analytic_Inequalities/2068.aspx
http://mathworld.wolfram.com/ShapirosCyclicSumConstant.html[/quote]

But anyway, I hope some result which can be proved by elementary mathematics. Is there a such proof when $n=7$?

I will off here for a long time, with my hope that questions here will be answered in a near future. Moreover

1, How to prove the Shapiro's inequality if $k=2(\sqrt{2}-1)$.
2, How to prove it for some bigger value of $k$, such as $k=5/6$.

All above means "proving with elementary mathematics".

Thanks anyway,

Hm, Flove. I have nothing to you. But my friend has a generation inequality about the following sum:
$\sum_{cyc}(\frac{a_{1}}{S-a_{1}})^\alpha$
where $a_{i}$are possitive and $S= \sum_{}a_{i}$
But i can't post it.


trigonometry
function
real analysis
real analysis unsolved

Find the maximum value of $a$ such that $f(x)=\frac{a-\cos x}{x^2}$ is increasing function at $0<x\leqq \frac{\pi}{2}$.


The pressure ($P$) of wind on a sail varies jointly as the area ($A$) of the sail and the square of the velocity ($V$) of the wind. The pressure on a square foot is 1 pound when the velocity is 16 miles per hour. The velocity of the wind when the pressure on a square yard is 36 pounds is:
(A) $10\frac{2}{3}$ mph
(B) $96$ mph
(C) $32$ mph
(D) $1\frac{2}{3}$ mph
(E) $16$ mph