MATHCOUNTS
geometry
probability
LaTeX
FTW
abstract algebra
AMC

I'm currently working on both AoPS Vol. 1 and Intro to NT at the same time.

I'm using the book and taking the class for Intermediate Algebra

[quote="Dojo"]AoPS volume 1
Intro Geo (with class)
Intro Counting Prob
Intro Number Theory
Intro Alg (kinda sorta)
Inter Alg (with class)

Also:
MATHCOUNTS class[/quote]

Since u hav all the intro books..

Is it better to do all of he books (I meant he intro books) with Volume 1, or just Volume 1?

Hmm...the only aops book I have is Intermediate Algebra and I'm taking that class.  (I wish I had more).  I'm also taking the mathcounts class.

[quote="Urc"]
Is it better to do all of he books (I meant he intro books) with Volume 1, or just Volume 1?[/quote]

Well.. I don't have all the books.  But I've heard while volume 1 is a good review over everything the intro books go more into detail on the particular subjects.  So I would say you should do volume 1 and get the books for the subjects you're weak in.  Still if you can get them all that would be great, though not really necessary..

[quote="Urc"]Volume 1, tryin 2 finish it bu its ahrd for me...maybe im just stupid[/quote]

You're not stupid! I have it too, and I find it hard as well.

[quote="Poincare"]aaaaaaaahhhhhhh! THIS IS TORTURE! i made my school library get the intro books so i read them for a few weeks and now that school's over, i don't have them any more and now that i've had a taste of what Aops books are like i can't go back and i cant order the books for another 5 days (because we're moving)  :cursing:  :censored:  :wallbash_red:  :wallbash_red:[/quote]

Wait for 5 days then get the books then.

Patience =P hehe

If you want, they have excerpts of the AOPS books here so you can look at them. I'd always do that during my computer thing at school

Trying to pass out of Geometry and Algebra II so been doing a Geometry textbook so far. Is the Art of Problem Solving geometry any good? Does it have all the concepts you learn in school?

Dang, it sounds to me like you guys have all of the AoPS books... :o  :(  :( . I wish I did....

[quote="alton66"]This sounds stupid i know... what is AoPS i know nothing... :(  :(  :([/quote]

AoPS = Art of Problem Solving

and no ur not stupid you joined like 5 days ago

[quote="Poincare"]If only Aops books weren't sooooo  :o  expensive i would have bought all of them![/quote]

100% agree

i want to teach classes in 9th grade so i can earn enough to buy AoPs books on my own because my mom will probably not buy anymore
btw: i have vol 1/2 and c p

i love aops books they completely transformed me first in the 6th grade i got a b+  in math it was so boring i made stupid errors i mean who wants to do no thinking arithmetic all day so i hated math then i met aops and mathounts started  stufying the aops books with there CHALLENGING problems that REQUIRE THINKING as opposed to the problem repitition at school and NO JOKE in 7 th grade i was the best in my middle school
aops books are way better then school books when given the choice always choose aops :alien:  :starwars:  :spam:  :spidy:   :spider:  :first:  :rotfl:

same here i moved to the USA almost 3 years ago from India and when i did i used to do a lot of reading and writing(both in english). But since the school the math level here is slightly lower than india i used to do bonus long division for my fourth grade teacher. And then she put me and loads of other people from my into the MC program at the elementary school thinking it was just basic arithmetic and maybe a little of long division which was considered "challenging". :o  :rotfl: So i went there and within two days all the people that had entered vanished except for a kid called Colin. btw, colin is one of my best friends right now. So, i participated in the competition unofficially and did really badly not surprising. But in fifth grade i bought intro to algebra and i could do every mathcounts and AMC problem you could throw at me  :ninja:

Umm is it just me thats not taking any of the classes or have ANY of the books?  :blush:   :rotfl: (I think my parents think its useless but whatever)

What are the AoPS books like?

they're really good, in my opinion. (the books)

i'm reading intermediate c&p.

Aops books aren't like normal books. They have this "learn with discovery" thing that's really good. So, instead of just showing you the whole concept at once , they show it to you in parts (with examples) and you discover how do stuff your self. Some people might say there is this kind of writing style in some textbooks. Yes i agree but only in the kind of intermediate and more advanced sections of math like trig, precalc, calc, analysis etc. But no one really cares to put that stuff in the lower level book like algebra 1 or geometry. So all in all a Aops are awesome!!! :wow:

AoPS volume 1 and Intro to C and P. I have AoPS volume 2 but haven't started yet =/

AoPS books are different from ordinary textbooks. They start off with a set of problems in the beginning of each chapter, than teach you the key concepts, and show the solutions to each of the problems, and then end with a set of exercises. However, those problems are not similar to the ones at the beginning of the chapter - you have to figure out how to use the basic concepts you learned to solve challenging problems that use them.

Currently, Intro to CP. Want to get to Number Theory and AoPS Vol. 1.

Im doing aops Volume 2 already did Volume 1. I also have Intro to C&P but i just do bits and pieces of that cause i know most of it and im prepping aime with volume 2. Im also in the amc 10 and amc 12 classes.

holy crap that's a lot money for math!!!


Solve in nonnegative integers:

1) m^2 = n!
2) m^a = n!

I don't think any of them have any nontrivial solutions, but any ideas on how to prove/disprove that?

I believe you can prove them using the fact that [hide]there is always a prime number strictly between n and 2n for any n :ge: 2.[/hide]

I think I was just working on this.  0!=1!=1  This is m^0 for any m.  I think that I remember hearing about for each positive integer i, a prime number p exists where i<p :le: 2i.  You get one prime factor, and you can't get another before you run into another prime number.  An example is you have a factor of 5 in n!, like 5!, and to get another to make, for example, a perfect square out of m^2, you need it to be 10!, but you'll have a factor of 7 in it (10!), so it keeps on going.  I hope this helps you a bit.


vector
inequalities
inequalities proposed

Prove that : In any triangle $ ABC $ , we have :
 $ \sqrt3.cos2A + 2.cos2B + 2\sqrt3.cos2C \geq -4 $

[quote="Ghang Hwan"]We consider 3 vectors,

OA,OB,OC which satisfy OA=2,OB=3^0.5,OC=1

angleBOC=2angleA
angleCOA=2angleB
angleAOB=2angleC

then(OA+OB+OC)^2>=0
we get the question's inequality[/quote]

  More understandable : $ (2\vec{OA}+ \sqrt{3}.\vec{OB}+\vec{OC})^2 \geq 0 $


[quote="kheg_k"]but there are 3 negatives.

$-p^2=-(p^2)$[/quote]

ok...i guess i thought that it means (-p)^2


Can anyone prove the identity
sum i*nCi $=n*2^{n-1}$
using calculus...
the great H.S.K. taught me this and i forgot it
I think it involves expanding $(x+1)^n$ and differentiating both sides or somting....

Also, can we prove
$\displaystyle\sum_{i=1}^{n}nCi=2^n$ the same way

For the first identity, we can use the binomial formula
\[ (x + 1)^n = \sum_{i=0}^n \binom{n}{i} x^i \, .  \]
As you suggested, differentiate both sides, and then set $x = 1$.

Of course, there are other, more elementary, methods for proving the identity.  For example, we can interpret both expressions as the number of edges in an $n$-dimensional hypercube.

For the second identity, we can use the binomial formula again, but skip the differentiation.


Show that for any natural $p$ there is at least one real solution $x$ to the following equation!
$2^p cot2^px=cotx$

That $2^p$ isn't necessary, you can take $n$ instead (but you will need $n \geq 3$).
To prove: $n \cot (nx) = \cot(x)$ has a real solution.
The function $f(x)=n \cot (nx) - \cot(x)$ is defined everywhere except multiples of $\frac{\pi}{n}$.
Between any two of this poles, except iff one of them is also a pole of $cot(x)$ (then $f$ isn't defined yet where the roots would be), there is one and only one root of $f$. To see this, there are many possibilities. E.g. you could show that $f$ is monotonically decreasing or find the limits near to the poles.
So there will be a root between $\frac{\pi}{n}$ and $\frac{2\pi}{n}$; done.


group theory
abstract algebra
vector
LaTeX
number theory
relatively prime
superior algebra

Prove that any group of order 15 is cyclic (without quoting Sylow Theorem)

Remark : This question is in the topic before Sylow Theorem & Direct Product are being introduced , so possibly use other method to prove . Thanks

[hide]Can you use Cauchy's Theorem? If so, then let a be an element of order 5, and let b be an element of order 3, and let M and N be the cyclic subgroups generated by a and b respectively. Since 5 does not divide 3!=6, M is normal. Thus $ b a b^{ - 1} = a^n$ for some integer n not a multiple of 5. Then we have $ b^4 a b^{ - 4} = a^{n^4} = a = b a b^{ - 1}$ so a and b commute. Suppose that $ (ab)^k = e$ for some integer $ k$. Then we have $ a^k = b^{ - k} \in M %Error. "intersect" is a bad command.
N = \{e\}$, so $ a^k = b^k = e$, meaning that $ 3,5 | k$ or $ 15|k$, so the order of ab is 15, and our group is cyclic.

See problem 10 Section 2.9 for a generalization of this. It turns out that a group of order $ pq$, $ p > q$, is cyclic unless $ q|p - 1$, in which case there is a unique non-abelian group.[/hide]

If you want to do it without citing Cauchy's Theorem, you can show by elementary means that if all elements of a finite group have order q for some prime q, then the number of elements in the group is a power of q (and it is in fact $ C_q$ crossed with itself multiple times).
[hide="If you have seen them..."]think vector spaces.[/hide]

Thx for reply , some parts that im not sure how to obtain

(i) why $ b^4ab^{\minus{}4}\equal{}a^{n^4}$ ?
(ii) why $ MN\equal{}\{e\}$ ?

Can you further explain ? Thanks

Is it always true that if $ M,N$ are normal subgroup of $ G$ with $ |M|$ and $ |N|$ relatively prime , then $ M\cap N\equal{}\{e\}$ ?

[quote="shyong"]Is it always true that if $ M,N$ are normal subgroup of $ G$ with $ |M|$ and $ |N|$ relatively prime , then $ M\cap N \equal{} \{e\}$ ?[/quote]
It's true for all subgroups, not just normal ones.  The reason is that if $ M$ and $ N$ are subgroups, then $ M \cap N \equal{} H$ is a subgroup. (easy to prove)

But the order of $ |H|$ has to divide the $ |M|$ and $ |N|$. (basic property of subgroups)  Since $ |M|$ and $ |N|$ are relatively prime, the only way this can happen is if $ |H|\equal{}1$.

[quote="dphexen"][quote="shyong"](i) why $ b^4ab^{ \minus{} 4} \equal{} a^{n^4}$ ?

[/quote]

because you know that $ a^b \equal{} a^n$ so you conjugate $ a$ four times.

[quote](ii) why $ MN \equal{} \{e\}$ ?

Can you further explain ? Thanks[/quote]

It should be $ M \cap N \equal{} 1$, actually $ MN \equal{} G$ :)[/quote]

Yes, thank you, I didn't know how to do intersection in LaTeX  :blush:


The integers from 1 to 1000 are written in succession. Find the sum of all the digits.

[quote="Adam Shaffer"]13501..notice the phenominon that 999+000=27 sum of digits (9+9+9) 998+001 in terms of sum of digits is also 27..

so 27*1000/2+1 for 10000 to get that[/quote]

I don't understand what you are saying.

[hide="Here is my solution"]Between 000 and 999, each digit 0 1 2 3 4 5 6 7 8 9 will be used 100 times (each digit is used the same number of times). So the sum between 000 and 999 is $ 10(9+8+7+6+5+4+3+2+1) = 4500$. To add the $ 1000$, we just simply add $ 1$; $ \boxed{4501}$.[/hide]

EDIT: Shouldn't it be just 100 times?

Okay first they will not be used 10 times, they will be used 300 times

so 300*45+1=13501

or

pair them up

999 with 000 9+9+9=27
998 with 001 9+9+8+1=27...

you will notice this will continue so you do sum formula and get same answer (as i said above)

[quote="vishalarul"][quote="Adam Shaffer"]13501..notice the phenominon that 999+000=27 sum of digits (9+9+9) 998+001 in terms of sum of digits is also 27..
so 27*1000/2+1 for 10000 to get that[/quote]I don't understand what you are saying.[/quote]it's kinda like the proof for the formula $ \sum_{i=1}^{n}i = \frac{n(n+1)}{2}$.
Let's assume [ABC] is a 3 digit number, where A, B, C are digits. [ABC]=100A+10B+C.
So 999-[ABC] = 900-100A+90-10B+9-C = 100(9-A)+10(9-B)+9-C.
But A, B, and C are all less than or equal to 9, so the sum of digits of [ABC] and 999-[ABC] is always equal to 27.


integration
calculus
calculus computations

I have this problem,

Find the volume of the solid generated by revolving a loop of the curve $ r^2\equal{}a^2sin2\theta$ about the initial line.

To solve this I used the formula $ V\equal{}\frac{2}{3} \pi \int (f(\theta))^3 sin \theta d\theta$. And last I got the integration
$ \frac{4\sqrt{2}}{3}\pi a^3 \int sin^{5/2}\theta cos^{3/2}\theta d\theta.$
But I can't get the answer $ \frac{\pi^2 a^3}{8}$.

I believe that last integral was done in a post not too long ago, if that's your problem.  I can't completely remember it, but try looking for Kabi, maybe he solved it.  Again, I'm not completely sure.

Yes, this is the one I have posted earlier. But  can't get the answer. If u know pls tell me is this formula correct and also what is the limit?

I don't recognize that volume formula.  It very well could be correct but I've never seen it before so I couldn't tell you if you're right or wrong.  Are you sure you're evaluating the integral correctly (plugging in the right things, no arithmetic mistakes,...)?


abstract algebra
Ring Theory
superior algebra
superior algebra solved

I can't figure it out. The question is this:

Are there only a finite number of minimal prime ideals in a local noetherian ring?

This question arises in the dimension theory of such rings. An affirmative answer seems to be assumed in the proof I've seen of this proposition:

Let A be such a ring and assume that dim A = d. Let m be the maximal ideal of A. Then there is an m-primary ideal in A generated by d elements.

This is one of the steps towards the dimension theorem asserting the equivalence of three definitions of dimension for local noetherian rings.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=ideals+prime+minimal&t=70441[/url]

It's true for Noetherian rings in general, not just local ones.


Find the inverse, state the domain and range of f and f^-1.

1) f(x) = (x + 2)^2, x > or = to -2.

2) f(x) = 2(cuberoot{x})

Singular....I am not posting homework here.  I am not in high school.  I have not been in high school in over 20 years.  I am preparing for a state math test.  Get it?

1) $f(x)=(x+2)^2 \ \ x\ge -2.$

The domain of $f$ is all reals with the constraint given, i.e. $x\ge-2$ is the domain.  The range, all the values that $f(x)$ can assume, is $[0,\infty)$ (that means all reals greater than zero, including zero).

To find the inverse, we switch the $f(x)$'s and $x$'s and replace $f(x)$ with $f^{-1}(x)$:

\[x=(f^{-1}(x)+2)^2\Rightarrow f^{-1}(x)=\sqrt{x}-2.\]

Remember that the inverse of a function means that you switched the domain with the range of the original function and vice versa.  Therefore, the domain is $[0,\infty)$ and the range is $[-2,\infty).$  We check to see that all these values will work, and they do.

2) $f(x)=2\sqrt[3]{x}$

The domain is all reals.  The range is also all reals.

We again find the inverse:
\[x=2\sqrt[3]{f^{-1}(x)}\Rightarrow f^{-1}(x)=\frac 18 x^3.\]

The domain and range of this is, again, all reals.


Find all integral values of $x$ for which $\frac{12(x^2-4x+3)}{x^3-3x^2-x+3}$ has a positive integral value.

I got {0,1,2,3,5,11} but the answer is {0,2,5,11}can someone tell me what i did wrong?

Plugging in 1 you get 0, which is not a [b]positive[/b] integral value, and plugging in 3 the bottom becomes 0, which is undefined. This leaves you with your answer set without 1 and 3, which is {0,2,5,11}, and thus correct.


Find all complex numbers $ z,\,z\neq 0,$ such that

[list]$ \left|\begin{array}{ccr}i^8& z& i^2\\0&i^7&z\\i^5&0&\minus{}\overline{z}\end{array}\right|\equal{}1.$[/list]

Note: $ \overline{z}\equal{}a\minus{}bi\Longleftrightarrow z\equal{}a\plus{}bi$ and $ i\equal{}\sqrt{\minus{}1}.$

[quote="TZF"][hide]By computing the determinant, the equation becomes $ 1 \plus{} (z^2 \plus{} \overline{z})i \equal{} 1 \iff z^2 \plus{} \overline{z} \equal{} 0$

$ z : \equal{} (r, \theta ) \implies (r^2, 2 \theta ) \plus{} (r, \minus{} \theta) \equal{} 0 \implies (r^2, 2 \theta ) \equal{} (r, \pi \minus{} \theta)$

First coefficients: $ r^2 \equal{} r \implies r \equal{} 0,1$
Second coefficients: $ 2 \theta \equal{} (2n \plus{} 1) \pi \minus{} \theta \implies \theta \equal{} \frac {\pi}{3}, \; \pi, \; \frac {5 \pi}{3}$

$ \therefore z \in \left \{ 0, \; \minus{} 1, \; \text{cis} \frac {\pi}{3} , \; \text{cis} \frac {5 \pi}{3} \right \}$[/hide][/quote]

$ z\neq 0$   :roll:

Sigh.

$ z \in \left \{\minus{} 1, \; \text{cis} \frac {\pi}{3} , \; \text{cis} \frac {5 \pi}{3} \right \}$

Nitpickers (just kidding :P)

[quote="TZF"][hide="Solution"]By computing the determinant, the equation becomes $ 1 \plus{} (z^2 \plus{} \overline{z})i \equal{} 1 \iff z^2 \plus{} \overline{z} \equal{} 0$

$ z : \equal{} (r, \theta ) \implies (r^2, 2 \theta ) \plus{} (r, \minus{} \theta) \equal{} 0 \implies (r^2, 2 \theta ) \equal{} (r, \pi \minus{} \theta)$

First coefficients: $ r^2 \equal{} r, \; r \neq 0 \implies r \equal{} 1$
Second coefficients: $ 2 \theta \equal{} (2n \plus{} 1) \pi \minus{} \theta \implies \theta \equal{} \frac {\pi}{3}, \; \pi, \; \frac {5 \pi}{3}$

$ \therefore z \in \left \{ \minus{} 1, \; \text{cis} \frac {\pi}{3} , \; \text{cis} \frac {5 \pi}{3} \right \}$[/hide]

Edit: [i]Right, thanks.[/i][/quote]
[hide="another solution, not used in polar form"]
Also we get

$ 1 \plus{} (z^2 \plus{} \overline{z})i \equal{} 1 \iff z^2 \plus{} \overline{z} \equal{} 0$

$ z\equal{}a\plus{}bi ,z^2\equal{}a^2\minus{}b^2\plus{}2iab,\overline{z}\equal{}a\minus{}bi \\
z^2 \plus{} \overline{z}\equal{} a^2\minus{}b^2\plus{}2iab\plus{}a\minus{}bi\\
\implies (a^2\minus{}b^2\plus{}a)\plus{}(2ab\minus{}b)i\equal{}0\\
\implies a^2\minus{}b^2\plus{}a\equal{}0 \cdots (1),b(2a\minus{}1)\equal{}0\cdots (2)$

when $ b(2a\minus{}1)\equal{}0 \implies b\equal{}0$ or $ a\equal{}\frac{1}{2}$

when  $ b\equal{}0 \implies  z\equal{}Re(z)$ or $ a\equal{}0$ from $ (1)$which is rejected.

If $ z$ is real  then $ z^2 \plus{} \overline{z}\equal{}a^2\plus{}a\equal{}0 \implies a\equal{}\minus{}1$

when $ a\equal{}\frac{1}{2}$ then $ b\equal{}\frac{\pm\sqrt{3}}{2}$ from $ (1)$

$ \therefore z\in\left \{ \minus{}1, \frac{1\pm\sqrt 3}{2}\right\}$
[/hide] :P

$ z^2\plus{}\bar{z} \equal{} 0 \Rightarrow |z|^2 \equal{} |z| \Rightarrow |z| \equal{} 1 (\because \ z \ne 0)$

Now multiplying the equation by z since z = 0 is not a solution, we get $ z^3\plus{}z\bar{z} \equal{} 0$ or $ z^3\plus{}1 \equal{} 0$

Hence the solutions are the cube roots of -1


在这里能用letax语言吗？如何使用？请举例说明。
我发现有些帖子可以打印出类似于.pdf文件的文本。

Yes. Just type $ \$$ signs around your LaTeX math expressions, e.g. $ \$$ x^2 - 1 = 0 $ \$$ gives $ x^2 \minus{} 1 \equal{} 0$.

Try out in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=224]Test Forum[/url].

Also, here is a page containing some [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php]basic commands[/url]

Example: [code]$\sqrt[3]{5}$[/code]
gives
$ \sqrt[3]{5}$


This problem came from Olympiad competition but I think it's easy enough for this forum.

Assume $s_n = 1-2+3-4+5-6 \cdots + (-1)^{n-1} n$. Then $s_{1989} + s_{1990} =$?

A. negative
B. 0
C. 1
D. 2
E. greater than 2


We are to calculate $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_nF_{n+1}}$  with$ F_{n+2}=F_{n+1}+F_n,F_1=1,F_2=2$, which states like Fibonacci sequence.

[quote="kunny"]We are to calculate $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{F_nF_{n+1}}$  with$ F_{n+2}=F_{n+1}+F_n,F_1=1,F_2=1$, which states Fibonacci sequence.[/quote]
but notice in the question the in the first part, the fraction is $\frac{1}{1*2}$, but in the fibonacci sequence, $F(2)=1$ so the first term had to be $\frac{1}{1*1}$... ;)
also it will help to use the formula:
$F_n=\frac{(\displaystyle \frac{1+\sqrt{5}}{2})^n-(\displaystyle \frac{1-\sqrt{5}}{2})^n}{\sqrt{5}}$
so find this:
$\sum_{n=2}^{\infty}\frac{(\displaystyle \frac{1+\sqrt{5}}{2})^n-(\displaystyle \frac{1-\sqrt{5}}{2})^n}{\sqrt{5}}$

Yes. You are right, amirhtlusa.

I have just edited.


It's a poll,
I want to know YOUR point of view.... please?
 :D

Don't you think this is a little too "frivilous" even for the fun factory?

I think it should be locked.........

[quote="mathgirl98"]me thought this was a math site[/quote]
It IS a math site, notice the Fun Factory is 'For frivolous discussions.'

:what?:
magicbard what page was this on?
I have no idea who to be....

This right here is a fail of my used to be newbness, DON'T END UP LIKE THIS PLEASE  :(

I voted Einstein, yet my vote did not show. Why?

idk oprah??? just cus shes rich. though i was tempted with that numa numa guy ^.^


conics
ellipse
ratio
geometry proposed
geometry

Hi, everyone.

If AB, BC, CD, DA touch a ellipse and p, q, r, s be the perpendiculars from A, B, C, D on a variable tangent of the ellipse

prove that $ \frac {pr}{qs}$ is constant.

with synthetic proofs (no analytic, no trigonometric)

Thanks in advance.

May God bless you!

The proposition is true for any conic, but it can be proved by parallel projection only for an ellipse (similar to the proposition at [url]http://www.artofproblemsolving.com/viewtopic.php?t=262029[/url], where I used parallel projection). Later, in [url]http://www.artofproblemsolving.com/posting.php?p=1534636[/url],[quote="yetti"]Let $ a, b, c, d$ be 4 fixed tangents of a circle $ (O)$ at $ A, B, C, D$ and let $ p$ be a variable tangent of this circle at $ P.$ Let $ p$ cut $ a, b, c, d$ at $ X, Y, Z, T.$ Then $ OX \perp PA, OY \perp PB, OZ \perp PC, OT \perp PD.$ It follows that the line pencils $ O(X, Y, Z, T) \sim P(A, B, C, D)$ are similar. But angles of $ P(A, B, C, D)$ are independent of the variable point $ P \in (O)$ $ \Longrightarrow$ the cross ratio $ G(X, Y, Z, T) \equal{} k \equal{} \text{const}.$[/quote]

Since any conic can be projected to a circle by central projection and since the cross ratios are preserved, the lemma holds for any conic. Let $ a \equiv AB, b \equiv BC, c \equiv CD, d \equiv DA$ be tangent to the conic at $ E, F, G, H$ respectively. Let $ U \equiv AB \cap CD$ and $ V \equiv BC \cap DA.$ Let $ AB, BC, CD, DA$ cut the 5th arbitrary conic tangent $ t$ at $ X, Y, Z, T.$ Since $ p \parallel q \parallel r \parallel s$ (perpendicularity to $ t$ is not necessary),

$ \frac {p}{q} \cdot \frac {r}{s} \equal{} \frac {\overline{XA}}{\overline{XB}} \cdot \frac {\overline{ZC}}{\overline{ZD}}.$

Imagine tangents $ a, b, d, t$ fixed and tangent $ c \equiv CD$ moved to the position of $ a \equiv AB,$ which cuts itself at the tangency point $ E.$ By the above lemma, the following cross ratios are then equal:

$ \frac {\overline{ZC}}{\overline{ZD}} \cdot \frac {\overline{UD}}{\overline{UC}} \equal{} \frac {\overline{XB}}{\overline{XA}} \cdot \frac {\overline{EA}}{\overline{EB}}.$

Consequently,

$ \frac {\overline{XA}}{\overline{XB}} \cdot \frac {\overline{ZC}}{\overline{ZD}} \equal{} \frac {\overline{EA}}{\overline{EB}} \cdot \frac {\overline{UC}}{\overline{UD}}$ $ \Longrightarrow$ $ \boxed{\frac {p}{q} \cdot \frac {r}{s} \equal{} \frac {\overline{EA}}{\overline{EB}} \cdot \frac {\overline{UC}}{\overline{UD}}}$

The last expression does not contain any of the variable points $ X, Y, Z, T \in t$ on the 5th arbitrary tangent $ t,$ therefore, it is fixed.


A teacher writes a positive integer less than fifty thousand on the board. One student claims it is an exact multiple of 2; a second student says it is an exact multiple of 3; and so on, until the twelfth student says that it is an exact multiple of 13. The teacher observes that all but two of the students were right, and that the two students making incorrect statements spoke one after the other. What was the number written on the board?

[hide="My solution"]

Let $x$ be the number then $x<50,000$.

If $p$ is the first student who said wrong statement, then $p+1$ is the second student who said wrong statement. So, $p, p+1$ don't divde $x$. For $p$, we know that at least 2,3,4,5,6 are not the correct values because for instance, if $p$ was 6, then $6$ doesn't divide $x$ but this also means rest of the numbers (excluding 6 and 7) divides $x$ and since 12 is one of them, we have contradiction. Thus:

$p = 7,8,9,10,11,$ or $12$

Now, because 2,3,4,5,6 at least divides $x$, we can form the smallest number that will help us to define $x$. This is LCM of (2,3,4,5,6) and that is 60. So, $x$ is divisible by 60 so $p \neq 10, 12$. Since 11 and 13 are between them, they can't be $p$ or $p+1$ neither.Thus, 11 divides $x$ and 13 divides $x$. We need to multiply our numbers 60 by 11 and 13 which gives us 8580. Now, for $p$:

$p = 7,8$

(9 was excluded because since its $p+1$, 10, is divisible by $x$)

Now, if $p = 7$ then $p+1 = 8$. Also, 9 divides $x$ and this means $x = 8580 \cdot 3$ or $x=25740$.

If $p=8$ then $p+1 = 9$ so 7 divides $x$. We have no 7 at all so we need to multiply 8580 by 7, resulting 60,060. But this number is out of our problem's limit.

Thus, the number is 25740. [/hide]

Hope this is right..

i agree with your answer as well, i dont know if i did as much work though
i found the lcm assuming that all of them told the truth
$2^{3} * 3^{2} * 5 * 7 * 11 * 13 = 360360$
now if you think about it, if the number is not divisible by 2, it cannot be divisible by multiples of 2
since having no 2 would make 4,6,8,10,12 not possible, that is out
similar reasoning follows for 3,4,5,6 but then 7 and 8 come
since eliminating them would not messup a higher multiple, then they could be removed
so $\frac{2^{3} * 3^{2} * 5 * 7 * 11 * 13}{2 * 7} = 25740$ (divide by 2 since it still must be divisible by 4)
since it satisfies the given requirements, ${25740}$ is the correct answer

Let's say you were only throwing out one number.  The only numbers you could toss which would make a difference would be 13, 11, 7, 8 and 9.  Now if you're tossing two numbers, they have to be consecutive...And both of the numbers tossed out have to make a difference: otherwise, the one which didn't would have been correct.  Thus, it's (7, 8).


inequalities
calculus
integration
trigonometry
triangle inequality
Pythagorean Theorem
geometry

The sides of a triangle have lengths $11$,$15$, and $k$, where $k$ is an integer. For how many values of $k$ is the triangle obtuse?

$\textbf{(A)}\ 5 \qquad
\textbf{(B)}\ 7 \qquad
\textbf{(C)}\ 12 \qquad
\textbf{(D)}\ 13 \qquad
\textbf{(E)}\ 14$

[hide]by triangle inequality, 4<k<26

if the triangle was right, by pythagorean theorem, $k=\sqrt{346}$

for triangle to be obtuse, $k>\sqrt{346}$

$k^2>326$

19^2 is the first number that is greater than 326

k= all numbers from 19...25

7, B[/hide]

EDIT: oh yeah, :oops:

hm, 19 to 26 is 7?
(it can't be 26, so it's 19 to 25)

Are you sure that you have all the $k$ values?

Are you 100% sure?  :) In another word, that's not the correct answer.  :D

[hide="Answer"]Either $k$ is the longest side or $15$ is the longest side.

If $k$ is the longest side, we must have $k^2>11^2+15^2$. Since $k$ is integral, we have $7$ possible values.

If $15$ is the longest side, then $15^2>11^2+k^2\Rightarrow k^2<104$, so we have $6$ possible values.

Therefore, the answer is $6+7=13\Rightarrow \boxed{D}$.[/hide]

[hide="Mine"]First, with triangle inequality you know $4<k<26$

Now, Bust it out with a bit of law of cosines:
You know if $\theta$ is obtuse then $cos(\theta)$ has to be negative

The angle opposite the side 11:
$11^2 = 15^2 +k^2 -2*15*k*cos(\theta)$
$cos(\theta)=\frac{k^2+104}{330k}$
This is always positive so the angle opposite 11 is always acute

The angle opposite the side 15:
$15^2 = 11^2 +k^2 -2*11*k*cos(\theta)$
$cos(\theta)=\frac{k^2-104}{22k}$
Since x is always positive, cos($\theta$) is negative when
$k^2-104<0$
$k<\sqrt{104}$
Since $4<k$ there are 6 values that satisfy this

The angle opposite the side 15:
$k^2 = 11^2 +15^2 -2*11*15*cos(\theta)$
$cos(\theta)=\frac{346-k^2}{330}$
This is negative when
$346-k^2<0$
$\sqrt{346}<k$
There are 7 values that satisfy this

$7+6=13 \boxed{D}$[/hide]


function
number theory unsolved
number theory

Prove that $ \forall n\in\mathbb{N}: \;\frac{n^2}{2}<\phi (n) \theta (n) < n^2,$ where $ \phi (n)$ - Euler's function and $ \theta (n)$ - sum of positive divisors function.

$ \phi(n),\theta (n)$ are multiple and $ \phi(p^n)\theta(p^n)\equal{}p^{2n}\minus{}p^{n\minus{}1}<p^{2n}$.
Therefore $ \phi(n)\theta(n)<n^2 \ \forall n>1.$
Because $ \prod_p(1\minus{}\frac{1}{p^{n(p)\plus{}1}})>\prod_p(1\minus{}\frac{1}{p^2})\equal{}\frac{6}{\pi^2}$ we get
$ \frac{6}{\pi^2}n^2<\phi(n)\theta(n)<n^2 \ \forall n>1$.


integration
real analysis
real analysis solved

Prove:

As \[n \rightarrow \infty\]

\[\frac{1}{\sqrt{n}}(\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+...+\frac{1}{\sqrt{n+n}}) \rightarrow 2(\sqrt2-1)\]

It shouldn't be in the "Open Questions" section. This is for possibly unsolved questions.

Just take another $\frac 1{\sqrt n}$ out from each term of the sum to get $\frac 1n\left(\frac 1{\sqrt{1+\frac 1n}}+\ldots+\frac 1{\sqrt{1+\frac nn}}\right)$. This is a Riemann sum, and it's limit is $\int^1_0 \frac 1{\sqrt {1+x}}dx$. This is easy to compute.

[quote="Kent Merryfield"]I seem to have suddenly become a moderator. I was checking to see what powers that gives me - apparently I can do what grobber suggested and move this topic.[/quote]Sorry I understood that you'd want that ...  :blush:


Does $ \ce{H3 \plus{} }$ adopt a equilateral triangle structure or a linear structure? Why?

It has linear structure with +H - H- - H+ The central H has a negative charge.

That is wrong chemrock. Making the answer quite obvious. Question is why it adopts one structure over the other.

The answer has been proved to be Equilateral Triangle.

[quote]
APS Journals
Physical Review Online Archive Physical Review Online Archive
AMERICAN PHYSICAL SOCIETY

Three different measurements on the structure of the H3+ molecular ion are reported. The measurements all make use of a new technique—the foil-induced dissociation of a fast molecular-ion beam. It is shown that the structure is equilaterally triangular in shape. The most probable length of side of the triangle is determined by the three measurements to be 0.97 ± 0.03 Å, 0.95 ± 0.06 Å, and 1.2 ± 0.2 Å, respectively.

[/quote]

Is it because something like the Positive charge is kinda delocalized and so are the electrons, which is much easier to happen in case of a Triangle.


given is $w=-x^{2}-y^{2}+2xz$ find the extreme with using $x+y^{2}-z^{2}=1$

So $-x^{2}-y^{2}+2xz-\beta (x+y^{2}-z^{2})=0$

Then I derivate $\frac{\partial f}{\partial x}=-2x+2z-\beta$

$\frac{\partial f}{\partial y}=-2y-2\beta y$

$\frac{\partial f}{\partial z}=2x+2z\beta$

So I have try a lot to solove this system but every time I don't get a solotion What will the problem?

From the last equation, we have $\beta =-\frac{x}{z}$, blugging this into the first equation gives $x=\frac{2z^{2}}{2z-1}$, plugging both of these into the second equation gives $y=0$ or $z=0\Rightarrow x=0\Rightarrow\beta=0\Rightarrow y=0\Rightarrow x+y^{2}-z^{2}\not=1$, a contradiction so we must have $y=0$.  So, now in the constraint we have $\frac{2z}{2z-1}+0^{2}-z^{2}=1\Rightarrow z=1\Rightarrow x=2$.  And thus we have $(x,y,z)=(2,0,1)\Rightarrow w=0$.

[quote]The problem is that this function doesn't have an extremum on the set satisfying the constraint.
[/quote]

I know but I'mlocking for a description like jimmy gives thank you very well I will study your answer very well.

Actually, this is very odd, and someone who knows more about this than I do should explain what's going on:  if you just apply the lagrange multiplier method, you get a single real solution.  If you substitute the value of $x$ in terms of $y, z$ from the condition, then set both partials equal to zero, you get the same solution.  However, if you substitute in for $y^{2}$, you find that the partials cannot be simultaneously zero.  What does that mean?

What is the problem? $M = \{(x,y,z) \; | \; x+y^{2}-z^{2}= 1\}$ is not a compact set, so we don't know that the function $w$ has max/min on $M$. Remember that lagrange multipliers/stationary points give necessary conditions for a max/min point, not sufficient!

Be careful when doing the substitution $y^{2}= 1-x+z^{2}$. You will want to optimize the new function $f(x,z)$ on $1-x+z^{2}\ge 0$. So possible max/min points may also occur on the boundary, not just in stationary points. I missed this when I first made my post (I am too tired and still feel like I can't be responsible for this post. hopefully it is correct now).

[quote="Kalle"]So possible max/min points may also occur on the boundary, not just in stationary points. I missed this when I first made my post (I am too tired and still feel like I can't be responsible for this post. hopefully it is correct now).[/quote]

Ah, excellent, that's what I was missing as well.  Thanks.


[geogebra]cd1e5f684f1cb2b1a176488718d1fe1c891b52ec[/geogebra]

In the triangle $ \bigtriangleup ABC$ side $ AC \equal{} 2$ and $ BC \equal{} 1$ and $ AD \equal{} 1$, $ \bigtriangleup DBC$ is isosceles. What is $ AD * AB$?

[geogebra]ed61fb82fa3b64f91234b01b3c2e296a51d9afa6[/geogebra]

Draw a circle using $ C$ as center and $ CD$ as radius.
Then by intersecting chord theorem, $ AD \times AB \equal{} AE \times AF\equal{} 3$


algebra
polynomial
inequalities
algebra unsolved

Let $a_n,a_{n-1},\ldots,a_0$ be the decimal representation of a prime integer. Prove that the polynomial $P(X)=a_nX^n+a_{n-1}X^{n-1}+\cdots+a_0$ is irreducible in $\mathbb{Z}[X]$, if $a_n=1$. (For $a_n>1$, the problem was proposed at the Balkan Math Olympiad)

Hint:
Let $P(x)\in \mathbb{Z}[x]$.Suppose $k\in \mathbb{Z}$such that:$P(k-1)\neq 0,P(k)\in \mathbb{P}$ and if $c=a+ib$ be the root of $P(x)$ then :$a<k-\frac{1}{2}$.Prove that:$P(x)$ is irreducible in $\mathbb{Z}[X]$

You can paste tex directly in the posts (provided you change the command \operatorname to \textrm - has almost the same effect). I've done that so the solution is

We use the following
\textit{Lemma}. If $\alpha$ is a complex root of the polynomial
\[
P\left(  x\right)  =a_{n}x^{n}+\ldots+a_{1}x+a_{0},
\]
where $\frac{a_{n-1}}{a_{n}}\geq0,$ then the following inequality holds:
\[
\textrm{Re}\alpha<\frac{1+\sqrt{1+4M}}{2},
\]
where $M=\underset{0\leq k\leq n-2}{\max}\left|  \frac{a_{k}}{a_{n}}\right|  .$

\textsl{Proof}. Suppose that $\alpha$ is a root of $P$ such that
$\textrm{Re}\alpha\geq\frac{1+\sqrt{1+4M}}{2}.$ Then we also have
\[
\left|  \alpha\right|  \geq\textrm{Re}\alpha\geq\frac{1+\sqrt{1+4M}}%
{2}.
\]
Since $P\left(  \alpha\right)  =0,$ we can write
\[
a_{n}\alpha^{n}+a_{n-1}\alpha^{n-1}=-\left(  a_{n-2}\alpha^{n-2}+\ldots
+a_{1}\alpha+a_{0}\right)  ,
\]
which implies
\[
\alpha+\frac{a_{n-1}}{a_{n}}=-\left(  \frac{a_{n-2}}{a_{n}}\cdot\frac
{1}{\alpha}+\frac{a_{n-3}}{a_{n}}\cdot\frac{1}{\alpha^{2}}+\ldots+\frac{a_{0}%
}{a_{n}}\cdot\frac{1}{\alpha^{n-1}}\right)  .
\]
We obtain
\[
\left|  \alpha+\frac{a_{n-1}}{a_{n}}\right|  \leq M\left(  \frac{1}{\left|
\alpha\right|  }+\frac{1}{\left|  \alpha\right|  ^{2}}+\ldots+\frac{1}{\left|
\alpha\right|  ^{n-1}}\right)  <\frac{M}{\left|  \alpha\right|  -1}\leq
\frac{2M}{\sqrt{1+4M}-1}.
\]
On the other hand,
\[
\left|  \alpha+\frac{a_{n-1}}{a_{n}}\right|  \geq\textrm{Re}\left(
\alpha+\frac{a_{n-1}}{a_{n}}\right)  =\textrm{Re}\alpha+\frac{a_{n-1}%
}{a_{n}}\geq\textrm{Re}\alpha\geq\frac{1+\sqrt{1+4M}}{2}.
\]
We deduce that
\[
\frac{1+\sqrt{1+4M}}{2}<\frac{2M}{\sqrt{1+4M}-1},
\]
or
\[
4M<4M,
\]
a contradiction. The lemma is proved.

Returning to our problem, let us observe that if $\alpha$ is a complex number
such that $\textrm{Re}\alpha<\frac{19}{2},$ then
\[
\left|  10-\alpha\right|  >\left|  9-\alpha\right|  .
\]
Indeed, write $\alpha=x+yi,$ with $x,y\in\mathbb{R}.$ Then the above
inequality is equivalent to
\[
\left(  10-x\right)  ^{2}+y^{2}>\left(  9-x\right)  ^{2}+y^{2},
\]
or
\[
x<\frac{19}{2},
\]
as stated.

Now, assume, by way of contradiction, that $P\left(  x\right)  =Q\left(
x\right)  R\left(  x\right)  ,$ where $Q$ and $R$ are nonconstant polynomials
with integer coefficients. It follows that one of the numbers $\left|
Q\left(  10\right)  \right|  $ and $\left|  R\left(  10\right)  \right|  $ is
equal to $1.$ Suppose that $\left|  Q\left(  10\right)  \right|  =1$ and let
$x_{1},x_{2},\ldots,x_{k}$ be the roots of $Q.$ We then have
\[
Q\left(  x\right)  =a\left(  x-x_{1}\right)  \left(  x-x_{2}\right)
\ldots\left(  x-x_{k}\right)  ,
\]
for some integer $a$ and then
\[
1=\left|  a\right|  \left|  10-x_{1}\right|  \left|  10-x_{2}\right|
\ldots\left|  10-x_{k}\right|  .
\]
But $x_{1},x_{2},\ldots,x_{k}$ are also roots of $P$ and therefore, by the
lemma,
\[
\textrm{Re}x_{i}<\frac{1+\sqrt{37}}{2}<\frac{19}{2}.
\]
Hence
\[
\left|  10-x_{i}\right|  >\left|  9-x_{i}\right|  ,
\]
for all $i,$ $1\leq i\leq k.$ We deduce that
\[
1=\left|  a\right|  \left|  10-x_{1}\right|  \ldots\left|  10-x_{k}\right|
>\left|  a\right|  \left|  9-x_{1}\right|  \ldots\left|  9-x_{k}\right|
=\left|  Q\left(  9\right)  \right|  .
\]
Clearly $P\left(  9\right)  =Q\left(  9\right)  R\left(  9\right)  $ is a
positive integer, therefore $\left|  Q\left(  9\right)  \right|  \geq1,$ which
is a contradiction.

isn't this a weaker statement of cohn's irreductibility criterion???

Can someone throw it away if there is a more complete solution, I don’t understand these things, please


Does anyone know where i can get a pdf vversion of imo shorlist and longlists.

Mmm !
  This is an easy question with a difficult answer  :)
  You will find some lists in this forum in the [color=blue]Resources section,due to ORL and others.[/color].Probably some more lists and solutions you can find in the book << [color=red]IMO compendim [/color], Springer editions , [color=red]written by Dukic, Matic ect [/color]>>, which I warmly recomend you.

  [u] BAbis[/u]

Longlists have not been published recently (have they ever been published?), but there's certainly enough material in the shortlists to keep you busy for a long while. Shortlists can all be found reworded here: [url]http://www.kalva.demon.co.uk/short.html[/url], sometimes without solutions.

[quote="D.P.L"][quote="Boy Soprano II"]And do not forget the [url=http://www.artofproblemsolving.com/Forum/index.php?f=177]IMO Shortlist Forum[/url].  It is said that the longlists are generally of fairly low quality; I have found this to be true with the few longlists I have investigated (many are simply inelegant, many are too easy, etc.).  However, I think that there are supposed to be some exceptions, notably the longlist from 1992.[/quote]

If you have the longlist from 1992. Would you be kind enough to post it? Thanx in advance :)[/quote]
It's in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=177]IMO Shortlist Forum[/url] that I mentioned—about the fourteenth topic down, I believe. :wink:


[b]1.[/b][i]Skiing[/i]
You(tall rectangle, with mass $m$) stand on a wedge sliding down a frictionless plane, as shown in the figure. What weight does the scale (shaded rectangle) read? Use clearly labeled, well-separated free-body diagrams. (see attached figure)
[b]2.[/b][i]Virtual work[/i]
The mass $m_{1}$ slides down the plane with constant velocity, and $m_{2}$ rises with constant velocity (see the figure). Use the principle of virtual work to find the mass ratio $\frac{m_{1}}{m_{2}}$. We live as usual in the make believe world of physics the plane is frictionless, the string is massless, and the pulley is massless and frictionless. (see attached figure. please give for this a detailed solution since i know very few about the principle of virtual work and i would learn it)
[b]3.[/b][i]Pendulum[/i]
As a pendulum slowly loses energy, the amplitude of its swing decreases. How does the period change as the amplitude decreases?Is it constant, decreasing, or increasing? Justify your answer.

What does the source refer to?? Some website. Can you please specify??

I found these problems (and the other in the topic "Some Questions") in a website a long time ago. I can't remember the url of this site but i remember the name of the author of the collection: Sanjoy Mahajan.
Best regards  :)

I think this is it.... [url=http://www.inference.phy.cam.ac.uk/sanjoy/]www.inference.phy.cam.ac.uk/sanjoy/[/url]

Problem number 2 looks really familiar. Isn't it from Feynman Lectures on Physics?

Really, bus? Can you plz specify the page? i'm very interested in this.

Right in the beginning. I only have the Slovak version but it's in Volume 1, Chapter 4 - conservation of energy. Look for Figure 4.3.


geometry
inequalities
LaTeX
function
trigonometry
inequalities proposed

In the thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=3171 siuhochung mentioned the following inequality:

[quote="siuhochung"]how about this inequality?
mamb+mbmc+mcma>=3sqrt3 S
S is the area of triangle.
can it be used for this one?[/quote]

I have a proof. I would like to see different proofs. Thanks.

Here is the LaTeX version:

$m_am_b+m_bm_c+m_cm_a \geq 3\sqrt3 S$

We can even replace medians by bisectors. But the proof of this new problem is extremely difficult (see Crux for a solution).

For the medians problem:

Well, here is how I prove it:

Since the medians $m_a$, $m_b$, $m_c$ of triangle ABC form a triangle $D_m$ themselves, and the area of this triangle $D_m$ equals $\displaystyle S_m = \frac34 S$, where S is the area of the triangle ABC, your inequality $m_b m_c + m_c m_a + m_a m_b \geq 3 \sqrt3 S$ can be rewritten in the form $m_b m_c + m_c m_a + m_a m_b \geq 4 \sqrt3 S_m$. But this is just an application of the well-known inequality $bc + ca + ab \geq 4 \sqrt3 S$ to the triangle $D_m$.

Here is how I prove the inequality $bc + ca + ab \geq 4 \sqrt3 S$:

Since the function csc x is convex on the interval $\left[ 0;\;\pi\right]$, we can apply Jensen and get

  $\displaystyle \csc A + \csc B + \csc C \geq 3 \csc \frac{A+B+C}{3} = 3 \csc \frac{\pi}{3} = 3 \cdot \frac{2}{\sqrt3} = 2 \sqrt3$.

Now, since $\displaystyle S = \frac12 bc \sin A$, we have $\displaystyle bc = \frac{S}{\frac12 \sin A} = 2S \csc A$, and similarly $ca = 2S \csc B$ and $ab = 2S \csc C$, so that

  $bc + ca + ab = 2S \csc A + 2S \csc B + 2S \csc C = 2S \cdot \left( \csc A + \csc B + \csc C \right) \geq 2S \cdot 2 \sqrt3 = 4 \sqrt3 S$,

qed.. Note that the above inequality $bc + ca + ab \geq 4 \sqrt3 S$ is stronger than Weitzenbck.

  Darij

[quote="darij grinberg"][...] Note that the above inequality $bc + ca + ab \geq 4 \sqrt3 S$ is stronger than Weitzenbck.[/quote]

Nice solution. The first part is same as mine. I proved $bc + ca + ab \geq 4 \sqrt3 S$ by using that [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15558]"general inequality"[/url] again (taking x=ca, y=bc, z=ab and M the circumcenter). It's astonishing so many interesting inequalities can be derived from that "general inequality".

My BW139 is:
$\sum{bc} \geq \frac{4}{3}\sum{\frac{h_a}{m_a}(m_bm_c)} \geq 4\sqrt{3}S.$

<=>
$\sum{m_bm_c) \geq  \frac{3}{4}\sum{\frac{h_a}{m_a}(bc)} \geq 3\sqrt{3}S.}$

BQ


group theory
number theory theorems
number theory

for the discrete log problem:
given or known are the values a,b,c
number sought is x

     a^x = b ( mod c)

it is easy to generate the residue for  x +1 or x-1  (or for just about any difference in x)

The question: does being able to calculate the residues for a+1, a+2, or a-1 or a-2 provide any help in solving the problem?

If it were possible to calculate the residues for c+1 or c-1 then it appears as if the chinese remainder theorem would provide a solution.

restated:

   a1 ^ x  == b1 ( mod c)
   a2 ^ x  == b2 ( mod c)
   a3 ^ x  == b3 ( mod c)

x is not known
you know:  a1,b1,c, a2,b2, a3,b3
Does it help?

[quote="integer"]for the discrete log problem:
given or known are the values $a,b,c$
number sought is $x$

     $a^x \equiv b \textrm{  (mod c)}$

[/quote]

Partial solution when $c$ is a prime:

Let $z$ be a primitive root modulo $c$; i.e. $z$ is a generator of the cyclic group $(Z/cZ)^{*}$. Then your $a \equiv z^{u} \textrm{  (mod c)}$ and $b \equiv z^{v} \textrm{  (mod c)}$ for some $u,v \in Z$. Substitution yields $z^{ux} \equiv z^{v} \textrm{  (mod c)}$. This implies $ux \equiv v \textrm{  (mod }c-1)$. The rest is easy.

Best regards,
Murat Aygen


limit
calculus
integration
real analysis
function
algebra
domain

show that if $ \mu$ is a measure with dom. $ A$ and if also $ E_n\in A$ with $ E_{n+1}\subset E_n$ for every $ n$, and $ \mu (E_1)<\infty$, then we have $ \mu (\bigcap_{n=1}^{\infty} E_n)=\lim_{n\to\infty} \mu (E_n)$. Besides showing this, what is the importance of this statement?

Thanks!

I always refer to this as "measure theory lemma #2." Its counterpart, "measure theory lemma #1," says that for increasing sequences of measurable sets, the measure of the union is the limit of the measures. Lemma #2 - the decreasing sequence lemma - does require one additional hypothesis, that you're starting downward from finite measure.

Importance? Between them, these two lemmas are used so constantly in measure and integration that we stop noticing that we're using them. They're simply an indispensable part of the standard working vocabulary.

hm. i know that this will be used in lebesgue integration. but what is an example?

Also, how would you go about showing that statement?

Hint: let $ A_k\equal{}E_{k\minus{}1}\setminus E_{k}.$ These sets are disjoint. You have countable additivity for disjoint sets. What can you get from this?

An application:

Suppose $ f$ is a measurable function whose domain is a set of finite measure, and suppose that $ f$ is finite almost everywhere. Prove that for any $ \epsilon>0,$ there exists $ N$ such that the measure of the set $ \{x: f(x)>N\}$ is less than $ \epsilon.$

(Note that this would be false without the hypothesis that the domain has finite measure. Such a hypothesis is a hint that Measure Theory Lemma #2 may be pertinent.)


A+B+C =180 degrees and

Tan((B+C-A)/4) + Tan((C+A-B)/4) +Tan((A+B-C)/4) =1

Prove 1+ CosA +CosB +CosC = 0

Happy solving!Dont go off tangents!

hello, maybe $-1+\cos(\alpha)+\cos(\beta)+\cos(\gamma)=0$.
Regards,Sonnhard.

[quote="Michael Niland"]There is a major mistake in the statement of this problem>My most abject apologies to people who have tried to solve this question

The tangents of the the three angles are to be multipled together (not added) and their product is 1.[/quote]

Hmm, this doesn't appear correct to me. In my opinion, the right variant of the problem is the one by Dr Sonnhard Graubner in post #2:

[color=blue]Let A, B, C be three angles such that A + B + C = 180° and

$\tan\frac{B+C-A}{4}+\tan\frac{C+A-B}{4}+\tan\frac{A+B-C}{4}=1$.

Then, prove that cos A + cos B + cos C = 1.[/color]

[i]Proof.[/i] Let $x=\frac{B+C-A}{4}$, $y=\frac{C+A-B}{4}$ and $z=\frac{A+B-C}{4}$. Then, the equation $\tan\frac{B+C-A}{4}+\tan\frac{C+A-B}{4}+\tan\frac{A+B-C}{4}=1$ becomes tan x + tan y + tan z = 1. Also, A + B + C = 180° yields

$x+y+z=\frac{B+C-A}{4}+\frac{C+A-B}{4}+\frac{A+B-C}{4}=\frac{A+B+C}{4}=\frac{180^{\circ}}{4}=45^{\circ}$.

Since A + B + C = 180°, we have B + C = 180° - A, and thus B + C - A = (180° - A) - A = 180° - 2A = 2 (90° - A), and thus $x=\frac{B+C-A}{4}=\frac{2\left(90^{\circ}-A\right)}{4}=\frac{90^{\circ}-A}{2}$. Similarly, $y=\frac{90^{\circ}-B}{2}$ and $z=\frac{90^{\circ}-C}{2}$.

From $x=\frac{90^{\circ}-A}{2}$, it follows that 2x = 90° - A, and thus A = 90° - 2x, so that cos A = sin (2x). Similarly, cos B = sin (2y) and cos C = sin (2z). Hence, the equation we have to prove, cos A + cos B + cos C = 1, rewrites as sin (2x) + sin (2y) + sin (2z) = 1.

Now let X = tan x, Y = tan y and Z = tan z. Then, the equation tan x + tan y + tan z = 1 becomes X + Y + Z = 1, thus X + Y + Z - 1 = 0. On the other hand, since x + y + z = 45°, we have x + y = 45° - z, so that tan (x + y) = tan (45° - z). But by the addition formulas for the tangens function, $\tan\left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{X+Y}{1-XY}$ and $\tan\left(45^{\circ}-Z\right)=\frac{\tan 45^{\circ}-\tan z}{1+\tan 45^{\circ}\tan z}=\frac{1-Z}{1+Z}$ (since tan 45° = 1). Hence, $\frac{X+Y}{1-XY}=\frac{1-Z}{1+Z}$. Thus, (X + Y) (1 + Z) = (1 - XY) (1 - Z). Expanding yields X + XZ + Y + YZ = 1 - Z - XY + XYZ; in other words,

X + Y + Z + YZ + ZX + XY - XYZ - 1 = 0.

On the other hand, X + Y + Z - 1 = 0. Thus,

$\left(1-X\right)\left(1-Y\right)\left(1-Z\right)=1-X-Y-Z+YZ+ZX+XY-XYZ$
$=\left(X+Y+Z+YZ+ZX+XY-XYZ-1\right)-2\cdot\left(X+Y+Z-1\right)$
$=0-2\cdot 0=0$.

But a product of three factors is zero if and only if at least one of these factors is zero. Hence, at least one of the numbers 1 - X, 1 - Y and 1 - Z must be 0. We can WLOG assume that 1 - X = 0. Then, X = 1. Since we know that X + Y + Z = 1, we must thus have Y + Z = 0.

Since X = tan x, the equation X = 1 becomes tan x = 1. Thus, $x=45^{\circ}+k\cdot 180^{\circ}$ for some integer k. Hence, $2x=90^{\circ}+k\cdot 360^{\circ}$. It follows that sin (2x) = sin 90° = 1.

Also, since Y = tan y and Z = tan z, the equation Y + Z = 0 becomes tan y + tan z = 0, so that tan y = - tan z = tan (-z). Thus, $y=-z+m\cdot 180^{\circ}$ for some integer m. Consequently, $2y=-2z+m\cdot 360^{\circ}$. It follows that sin (2y) = sin (-2z) = - sin (2z).

Consequently,

sin (2x) + sin (2y) + sin (2z) = 1 + (- sin (2z)) + sin (2z) = 1,

and the proof is complete.

  Darij

This is the statement of the problem as it appears in the book by HALL &kNight.

A,B and C are three angles such that A + B + C =180 degrees.

IF  Tan[(B+C-A)/4] . Tan[(C+A-B)/4] . Tan[(A+B-C)/4]  = 1 ,

Prove that 1+ CosA  +CosB +CosC =0.

When I first submitted this question last year it contained a mistake.The correct form of the problem is as I have written it above.My solution used two pages of A4 paper.

The problem in its CORRECT form remains  unsolved .Does anybody want to solve this challenging question.?

Several other trigonometry questions submitted by me have also gone unanswered.


A * [(B+C) ^ (D-E) -F ^ (G*H)] / K = 10

Hints:

1) You know that each variable is a unique, single-digit, non-zero number.

2) C- B = 1

3) H - G = 3

What is ABCDEFGHK? (Note exactly what the letters are its not alphabetical)

[hide]256973148[/hide]

[quote="ProtestanT"]what did it mean by "it's not alphabetical"?[/quote]
$ \mathrm{ABCDEFGHI}\ne\mathrm{ABCDEFGHK}$

I'm sure I added some unnecessary parentheses but is this right?

$ A * \frac{(B\plus{}C)^{(D\minus{}E)}\minus{}F^{(G*H)}}{K} \equal{} 10$


a + b + c = 7
(1/b+c) + (1/a+b) + (1/a+c) = 10/7

what is the value of
(a/b+c) + (b/a+c) + (c/a+b) ?

a,b,c, are positive numbers.

Double-posted in Pre-Olympiad.

Just multiply the two expressions to get

${a\over b+c}+{b+c\over b+c}+{c\over a+b}+{a+b\over a+b}+{b\over a+c}+{a+c\over a+c}=10$. The result follows.


inequalities
limit
logarithms
LaTeX
real analysis
real analysis unsolved

For an positive sequence $ \{a_n\}$,we can prove:
$ \lim_{n \to \plus{} \infty} \sup \left(\frac {1 \plus{} a_{n \plus{} 1}}{a_n}\right)^n \ge e$
But can we prove:
$ \lim_{n \to \plus{} \infty} \sup\left( \frac {1 \plus{} a_{n \plus{} 1}}{a_n}\right)^n > e$
Or in the same meaning,does there exist a positive sequence such that:
$ \lim_{n \to \plus{} \infty} \sup\left( \frac {1 \plus{} a_{n \plus{} 1}}{a_n}\right)^n \equal{} e$?

Note that $ a_n\equal{}Kn$ gives limit $ e^{1\plus{}1/K}$. This suggests to try $ a_n\equal{}n\log n$.

just a note, $ \limsup$ (\limsup) is a latex command.

can some one give a proof of the first inequality

You can try to prove there is infinite $ n$ such that:$ \frac{1\plus{}a_{n\plus{}1}}{a_n} \ge \frac{n\plus{}1}{n}$.
Otherwise,it will contradict with the fact that all $ a_n$ is positive.


calculus
integration
limit
function
calculus computations

ok, so i'm a sophomore in high school and I just had a calc bc test on integration techniques. There was a problem with improper integrals...
\[\int\displaylimits_4^6 {1 \over {x-5}} dx\]
there is a discontinuity, but since the two portions are anti symmetric, is the answer 0 or divergent? classmates said different things and my teacher was absent

thanks in advance for answering!

i had a discussion about that too earlier

i believe the convention when integrating over [a,b] while f is unbounded in every ball around c in ]a,b[ is :

the integral (f,x=a..b) exists if both the limits int (f,x=a..c-e) and int(f,x=c+e..b) converge for e going to zero

however there is a weaker condition called cauchy principal value i believe which requires convergence

of int(f,x=a..c-e)+int(f,x=c+e..b) with e going to zero

so in short : i think your integral is not convergent

NOOOO.... :(

i said it was 0 :(

lol, oh well, if i get a B+ this quarter, i had an A first quarter and the it'll average out

This integral is divergent.

There's a standard here - it's a convention, really, but it is uniformly agreed upon.

If an integral is improper for more than one reason, then we accept it as convergent only if each of those reasons results in a convergent integral. Furthermore, if we have two pieces, and on one piece the integral converges to the real number and the other is (for example) $+\infty,$ then we are willing to say that (finite number) $+\infty=+\infty$ and the integral diverges to infinity. We can even say that $\infty+\infty=\infty.$ However, we declare that $\infty-\infty$ is meaningless for these purposes.

Your example is improper for two reasons - the left side of 5 and the right side of 5. The convention I was laying out above requires us to handle each side separately, and write it at

$\lim_{a\to5^-}\int_4^a\frac{dx}{x-5}+\lim_{b\to5^+}\int_b^6\frac{dx}{x-5}.$

The key is that $a$ and $b$ are independent of each other. The first limit is $-\infty$ and the second limit is $+\infty.$ Since $a$ and $b$ are independent of each other, we have no way of controlling which one grows faster, and hence can assign no limit.

Fredbel6 mentioned the Cauchy principal value. In this case, that would be

$\mathrm{p.v.}\int_4^6\frac{dx}{x-5}= \lim_{\delta\to0}\left(\int_4^{5-\delta}\frac{dx}{x-5} + \int_{5+\delta}^6\frac{dx}{x-5}\right) = 0$ (by symmetry.)

But the Cauchy principal value is not a default of the notation; if that's what you mean, you would have to specifically say that somewhere in your notation. If just the integral is presented, as Naib Stilgar did at the top, we must declare it divergent and assign no value.

$x-5=t$
$dx=dt$
$\int_4^6\frac{dx}{x-5}=\int_{-1}^1\frac{dx}{x}$.
Function $f(x)=\frac{1}{x}$ is odd. So, this integral equals $0$.

Did you even read the post immediately before yours?


How many solutions satisfy:

$\sin{x}=\frac{x}{100}$
[hide]
For some reason I'm getting $62$ and I don't understand why that's not the correct answer :( [/hide]

Looking, for now, only $\geq 0$ roots.

There are none for $x > 100$ and there are two in each $[2 \cdot k \cdot \pi, 2 \cdot (k+1) \cdot \pi]$, $k \leq 16$

So there are $32 + 31 = 63$ in total

I should be because that's one of the two intersections in the first loop (sorry if I'm calling them loops because I don't know what they're called)

[quote="Iversonfan2005"]Hmmm....yeah, the way I did it was that I drew the graphs of $y=\sin{x}$ and $y=\frac{1}{100}x$, and since each period of the $\sin$ graph is $\pi$, then at most loops it can have is $31$ (since $[\frac{100}{\pi}]=31$) and the $y=\frac{1}{100}x$ intersects each of the loops twice, so I got $62$....but I think I understand your explanation. :? I think I just left one of them out, hmm...let's see :|[/quote]

I don't understand what you did.  Isn't the period of a sine graph $2\pi?$

Yes, but it's the same thing since I'm also counting the negative values for $x$ ;), I'm sorry, I should've made that clear, I'll change it. It doesn't matter, I figured out how to do it :)


A 1 gm rifle bullet takes 0.5 seconds to reach the target.Regarding the bullet as a point mass, and neglecting the air resistance and the mottion of the earth,find the order of magnitude of the spread of successive shots at the target with optimum condition for aiming and firing.

I cannot understand what I am to find out in this problem.This problem appears in QM book of Schiff and possibly,it is a problem of uncertainty principle.

The book has no answer list at the back for which I am confused...whether to find the time interval or something else...
What is meant by: "the spread of successive shots"?

Can anyone please tell?

There was a suggestion that we are asked to find the spread in the position of the bullet in the problem.

Then,

∆E~[ћ/(2∆t)]~ћ

This uncertainty in KE of the bullet implies the bullet has some deviation from the target due to uncertainty in energy.

For ∆E~E and ∆p~p, we must have

(∆p)²~2mE=2mћ

=> (∆p)~√ћ
=> (∆x)~√ћ

I have not taken care with that 1 gm mass.Essentially the problem is like this... as I believe.

can anyone please tell me if it is correct?


Let $x, y, z$ non-zero real numbers such that $xy$, $yz$, $zx$ are rational. [list=a] [*] Show that the number $x^{2}+y^{2}+z^{2}$ is rational. [*] If the number $x^{3}+y^{3}+z^{3}$ is also rational, show that $x$, $y$, $z$ are rational. [/list]

Part a) is trivial:
$x^2 = \frac{xy \cdot zx}{yz}$

Then $x^2,y^2,z^2$ are rational numbers.

Let $k = x^3+y^3+z^3 \in \mathbb{Q}$.
Then $(k-x^3)^2 = (y^3+z^3)^2 = (y^2)^3 + (z^2)^3 + 2(yz)^3 \in \mathbb{Q}$
$k^2 + x^6 -2kx^3 \in \mathbb{Q}$, but $k, x^2 \in \mathbb{Q}$.
Therefore $2kx^3 \in \mathbb{Q} \Rightarrow x^3 \in \mathbb{Q} \Rightarrow x = \frac{x^3}{x^2} \in \mathbb{Q}$.

In the same way we prove that $y,z \in \mathbb{Q}$.

Hi,

Edriv

what about the case where k=0?
in fact this cannot occur because of Fermat's last Theorem (n=3) and the fact that x/z and y/z are both rational numbers.

$\frac{x}{y}$ (and same about other pairs) is rational, so as we can represent an irrartional number by $Aa$ such that $A$ is rational and $a$ is irrational,which it cannot devide into productions of an irrational and a rational, so let $x=Aa,y=Bb,Z=Cc;$ by what we have at the first we will get that $a=b=c$,the rest is obviouse.

From part a) it follows that the numbers  $x^2, y^2, z^2,\frac{x}{ y}, \frac{y}{ z}, \frac{z}{ x} $ are rational, and since

 $\frac{x^3+y^3+z^3}{y}=x^2(\frac{x}{ y})+y^2+z^2(\frac{z}{y})$,

 $y$ is rational and so are also $x, z$.


analytic geometry
graphing lines
slope
geometry
incenter
geometry proposed

[quote="mathematica"]$BC$ is the diameter of a circle and $XY$ is a chord perpendicular to $BC$. $P$ is on the line of $XY$ and $M$ is

on the line of $CY$ such that $CY\parallel PB$ and $CX\parallel MP$. The intersection point of $CX$ and $PB$ is

$K$. Prove that $PB\perp MK$.[/quote]

Let's see.
1) Triangle $CXY$ isosceles, hence triangle $KPY$ isosceles, hence $KP = KY$
2) $\angle CBK = \angle BCY = \angle BCK \Rightarrow CBK$ isosceles, hence $BK = KC = PM$
3) $\angle KPM = \angle KCM = \angle XKP$
4) From 1), 2) and 3) triangles $BKX$ and $MKP$ are equal, hence $\angle PKM = \angle BXK = 90$

Daniel

[quote="Davron"]Dear Daniel we can not say $BK=KC=PM$.

Davron[/quote]

Why not?? I'm assuming $XY$ nearer to $B$ than to $C$ if this is what you mean. This is clearly a flaw but wait let me check the other case...

Checked, It's the same, I made a mistake in the first line though, for some reason kknwon only to the Gods, I posted the whole message again.

Daniel

[quote="dhernandez"][quote="mathematica"]$BC$ is the diameter of a circle and $XY$ is a chord perpendicular to $BC$. $P$ is on the line of $XY$ and $M$ is

on the line of $CY$ such that $CY\parallel PB$ and $CX\parallel MP$. The intersection point of $CX$ and $PB$ is

$K$. Prove that $PB\perp MK$.[/quote][/quote]

Let's see.
1) Triangle $CXY$ isosceles, hence triangle $KPX$ isosceles, hence $KP = KX$
2) $\angle CBK = \angle BCY = \angle BCK \Rightarrow CBK$ isosceles, hence $BK = KC = PM$
3) $\angle KPM = \angle KCM = \angle XKB$
4) From 1), 2) and 3) triangles $BKX$ and $MKP$ are equal, hence $\angle PKM = \angle BXK = 90$

[color=red]NOTE (edit) :[/color] There was a mistake in the first line, said $X$ where it ought to be $Y$. Also in 3) $\angle XKB$ not $XKP$ for it to be valid in both cases.

Daniel

[quote="Davron"]That is not true.

We know that [b]XY[/b] is perpendicular to [b]BC[/b], now we have to show that [b]BK[/b] is perpendicular to [b]MK[/b]. The problem is similar to showing that the quadrilateral [b]KMEP[/b] is cyclic.

[color=red][u][b]Note:[/b][/u][/color]

[color=blue][b]1- E is the intersection point of XY and BC
2- L is the intersection point of MP and BX[/b][/color]

Since [b]BC[/b] is the diameter of the circumcribed circle [b]BXCY[/b], we have $<CXB=<MPB=90$. From here we can easily see that [b]P[/b] is the orthocentre of the triangle [b]BXM[/b] which means [b]BP[/b] is perpendicular to [b]XM[/b]. Which leads to contradiction of the fact.

Davron[/quote]

Are you sure you are having the problem right? $KMEP$, the way you're defining it, is NOT cyclic. Maybe you meant $KMLB$, or some other thing, there are many cyclic quadrilaterals out there. $P$ is not the orthocentre of triangle $BXM$, that would be the case if $XP$, or what is the same $XY$, would be orthogonal to BM. But for that $M$ would have to be on line $BC$ which actually, only happens if $X$ is the midpoint of arc $BC$ in which case $X,P,K$ are all the same point.

Daniel

[hide="Here are four proofs and a equivalent enunciation."]
[color=darkred][b]The first proof.[/b] Observe that $PK=XK\ ,\ PM=YM\ ;$ $BM^2=BY^2+YM^2\ ,$ $BK^2=BX^2+XK^2\ .$ Therefore, $PB\perp MK$ $\Longleftrightarrow$ $PM^2-PK^2=BM^2-BK^2$ $\Longleftrightarrow$ $YM^2-XK^2=(BY^2+YM^2)-(BX^2+XK^2)$ $\Longleftrightarrow$  $BY^2=BX^2\ ,$ what is truly. [u]A very nice and simple problem ![/u][/color]

[color=darkblue][b]The second proof.[/b] Observe that $PM=KC\ ,\ KP=KX\ .$ Denote the intersections $F\in BC\cap PM$ and $L\in BX\cap PM\ .$ Then $m(\widehat {BLM})=90^{\circ}$ and $PF=PB\ .$ Therefore, $\frac{PL}{PF}=\frac{KX}{KC}$ $\Longleftrightarrow$ $\frac{PL}{PB}=\frac{PK}{PM}$ $\Longleftrightarrow$ $PL\cdot PM=PB\cdot PK$ $\Longleftrightarrow$
The points $B\ ,$ $L\ ,$ $K\ ,$ $M$ are concyclically $\Longleftrightarrow$ $\widehat {BLM}\equiv\widehat {BKM}$ $\Longleftrightarrow$ $KB\perp KM\ .$[/color]

[color=blue][b]The third proof ([u]the best[/u]).[/b] $KB=CK\ ,$ $KX=CM\ (=KP)\ ,$ $\widehat {BKX}\equiv \widehat {KCM}\Longrightarrow$ $(S.A.S.)$ $\triangle BKX\equiv\triangle KCM$ $\Longrightarrow$ $\widehat {BXK}\equiv\widehat {KMC}$ $\Longrightarrow$ $MK\perp MC\parallel BK$ $\Longrightarrow$ $KB\perp KM\ .$[/color]

[color=brown][b]The fourth proof.[/b] We will use the analytical method.
Consider the points $C(c,0)$, $B(-b,0)$, $X(0,\sqrt{bc})$, $Y(0,-\sqrt{bc})\ .$
Then the equations of the lines which appear in the our figure are :
$\boxed {CX\ : \ \frac xc+\frac{y}{\sqrt{bc}}=1}\ \ ;\ \ \boxed {CY\ : \ \frac xc-\frac{y}{\sqrt{bc}}=1}\ \ ;\ \ \boxed {BP\ : \ y=\frac{\sqrt{bc}}{c}(x+b)}\ .$
The coordinates of the new points which appear in the our figure are :
$P\in XY\ ,\ BP\parallel CY\Longrightarrow \boxed {P\left(0,\frac bc\sqrt{bc}\right)}\ ;$
$K\in BP\cap CX\Longrightarrow \boxed {K\left(\frac{c-b}{2},\frac{b+c}{2c}\sqrt{bc}\right)}\ ;$
$M\in CY\ ,\ PM\parallel CX\Longrightarrow \boxed {M\left(\frac{b+c}{2},\frac{b-c}{2c}\sqrt{bc}\right)}\ .$
We came to the conclusion that the slopes of the lines $BP$ and $KM$ are :
$m_{BP}=\sqrt{\frac bc}$ and $m_{KM}=-\sqrt{\frac cb}$, i.e. $m_{BP}\cdot m_{KM}=-1\ .$
This means that $BP\perp KM\ .$ We observe that $\frac{\overline {PX}}{\overline {PY}}=\frac{b-c}{b+c}\ .$[/color]

[color=red][b]An equivalent enunciation.[/b] The incircle $w$ of the isosceles triangle $AUV\ (AU=AV)$ touches the its sides in the points $C\in UV\ ,$ $X\in AV$ and $Y\in AU\ .$ Denote: the second intersection $B\in AC\cap w\ ;$ $P\in (XY)\ ,\ PB\parallel CY\ ;$ $K\in BP\cap CX\ ;$ $M\in CY\ ,\ PM\parallel CX\ .$ Prove that $PB\perp MK\ .$
[u]Indication.[/u] The point $B$ is the incentre of the triangle $AXY\ !$[/color][/hide]

[color=blue][i]Davron,[/i] see in the [u]my first reply[/u] [b]the second my proof[/b] of this nice problem ![/color]
[b]Remark.[/b] You are very much alike with my grandson (two nice youngs).

[hide="Only for Davron."][quote="Davron"]Thanks you are like my teacher. Best teacher, one more time thanks.[/quote][color=darkblue]Thank you very much !. My dear ... grandson Davron, see the my first reply where appeared and ... the third proof which is the best ![/color][/hide]

[quote="Davron"]Sorry everybody it is really easy, but i thought that point M is on the line CB .

Davron[/quote]

Nothing to be sorry about  :) These things happen to me all the time :)

Daniel

[hide][color=darkblue]Dear my grandsan Mathematica, DHernandez and Davron, in the my first reply appeared an interesting equivalent enunciation ![/color][/hide]

[color=darkblue]In the my first reply I added and a fourth proof (an analytical method).[/color]


function
logarithms
calculus
integration
algebra unsolved
algebra

Hello!

I'm not so good at explaining math with english but here goes a try.

Basically, I got a equation looking something like this:

1400 x e^0,03x = 15x
(this is a primitive function from 42*e^0,03x/0,03 =15)

Now, I divide with 15 on both sides to get x alone on the right side:

1400 x e^0,03x / 15 = x

Now, my problem is that I can't solve out the x in the "new" equation(the one in e^0,03X). Anyone got any ideas or am I thinking wrong.

Thanks for any advice!

[quote="puuhikki"]We see easily that $x=0$ is a solution. If $x\ne 0$, we can divide both sidex by $x$ and we get $1400e^{0.03x}=15$ or equivalently $e^{0.03x}=\frac{3}{280}$. Taking a natural logarithm this is equivalent to $0.03x=\ln\frac{3}{280}$ and thus $x=\frac{\ln\frac{3}{280}}{0.03}$.[/quote]

Thanks alot, it made it alot more clear!

I'm going to post the full solution for this task in the solved problems forum and see if I did right, I got some more work to do.

Again, thank you.

I found a different approach to this task and that was turning it into a integral.
 I'm going to write the whole task here so it will be easier to understand and solve i:

 Two veichles move against the same direction on a linear field. Va:s speed is 42 * e^0,03t and Vb:s speed is 15. t is the time counted from the beginning of the observation. How long will it take before A catches up with B , if A starts 7200 meters behind B.

Now, the two tasks is to: 1. Create a equation to decide the time and 2. Solve the approximate value for the time in seconds.

I did a wrong approach with my first post and equation so I decided to do a integral of it. I just can't find the right method.

[i]Sorry for not using the miktex(?) I can't install it on my computer[/i]


A book of tickets for 30 games of a local university's baseball
team sold for $ \$$120, and a book of tickets for 7 games of the
same university's football team sold for $ \$$161. By how many
dollars did the average cost of one football game ticket exceed
that of one baseball game ticket?

baseball ticket price $ \frac{120}{30} \equal{} 4$

football ticket price $ \frac{161}{7} \equal{} 23$

$ 23 \minus{} 4 \equal{} 19$ dollars.


[quote="jagdish"][b](1) If x>0, Then Find Min value of the expression y = ((x^2 +2) - (x^4 +4)^1/2 ) / x[/b][/quote]

$ y\equal{}\frac{x^2\plus{}2\minus{}\sqrt{x^4\plus{}4}}{x}$ $ \equal{}\frac{4x}{x^2\plus{}2\plus{}\sqrt{x^4\plus{}4}}$ and so $ y>0$ and $ \lim_{x\to 0^\plus{}}y\equal{}\lim_{x\to\plus{}\infty}y\equal{}0$

Hence the answer : $ 0$, value never reached over $ \mathbb R^{\plus{}*}$

[quote="jagdish"][b]If x is a real no. then  Min value is [/b][/quote]

Huh ? Do you mean there was (again) an error in your problem statement and that the min value is to be searched for $ x\in\mathbb R^*$ and not $ x\in\mathbb R^{  \plus{} *}$ ?
:(

[quote="jagdish"][b]  yes you are saying right.[/b][b] I think there is some mistake in this question.[/b][/quote]

:(  :(

$ f(x) \equal{} \frac {x^2 \plus{} 2 \minus{} \sqrt {x^4 \plus{} 4}}{x}$

$ f'(x)\equal{}\frac{2\minus{}x^2}{(x^2\plus{}2)\sqrt{x^4\plus{}4}\plus{}x^4\plus{}4}$

And so minimum over $ \mathbb R^{*}$ is $ f(\minus{}\sqrt 2)\equal{}2\minus{}2\sqrt 2$


inequalities
geometry
rectangle
inequalities unsolved

Prove that any finite set of closed squares with total area 1/2 can be fitted inside a unit square without overlapping.

k, i think this is the main idea:

Fit the squares (s_i) with edge (x_i) one by one in descending order (so s_1 is the largest, s_2 second largest, so on) into the unit square ABCD. Put s_1 in one of the corners of ABCD (so one vertex coinciding, two on respective edges of ABCD, and last one inside the square), and extend the edges to divide ABCD into 4 smaller rectangles. First show it's possible to put each of s_2, s_3 and s_4 into any of the other three rectangles. But deliberately put s_4 in the opposite corner (so in the (1-x_1)x(1-x_1) square) Now extend edges of s_4 to divide that square into 4 smaller rectangles, prove it's possible to put each of s_5, s_6, s_7 in the other 3 rectangles, but be sure to put s_7 in the (1-x_1-x_4)x(1-x_1-x_4) square (opposite to s_4 in the (1-x_1)x(1-x_1) square). And continue in this fashion... :D (hope that's clear... :lol: )


geometry
analytic geometry
linear algebra
matrix
geometry unsolved

I've got three questions

$\bigstar$ I have found in a Book [b]College Geometry[/b] the following theorem: [i][color=blue]"The isotomic points of three collinear points are collinear"[/color][/i] but I have proved it's not true... What do you think???

$\bigstar$ What is the proof for [i][color=blue]"The isodynamic points divide the Brocard Diameter harmonically[/color][/i]".

$\bigstar$ What is the proof for [i][color=blue]"An Apollonian circle of a triangle makes an angle of 120º with each of the remaining two circles"[/color][/i].

Could anybody help me??

If $P(x,y,z)$ is our point (in barycentric coordinates), then its isodynamic is $P^\prime \left( \frac1{x}, \frac1{y}, \frac1{z}\right)$.

So the statement says that \[\left| \begin{matrix}x_{1}& y_{1}& z_{1}\\ x_{2}& y_{2}& z_{2}\\ x_{3}& y_{3}& z_{3}\end{matrix}\right| = 0 \, \, \Longrightarrow \, \, \left| \begin{matrix}\frac1{x_{1}}& \frac1{y_{1}}& \frac1{z_{1}}\\ \frac1{x_{2}}& \frac1{y_{2}}& \frac1{z_{2}}\\ \frac1{x_{3}}& \frac1{y_{3}}& \frac1{z_{3}}\end{matrix}\right| = 0 .\] But that statement seems totally unreasonable. I took some numerical example and I obtained $584 = 600$, so it's false.

I think the third one follows from the following (I think) fact: inverting one apollonian circle about another gives the third one. I think the "fact" is not hard to prove once you note that the centers are collinear, which should follow from the circles concurring at two points.


function
Vieta
algebra
polynomial
sum of roots

OH YES. SORRY!!! my bad
The quartic function
           [size=150]x^4-16x^3+94x^2+px+q[/size]
has four total roots(2 double roots).

What is p+q?

i think there is missing information
do you know its roots or factorization?

Let the roots be a,a,b,b. By Vieta's
$ 16\equal{}2a\plus{}2b$
$ 94\equal{}a^2\plus{}b^2\plus{}4ab$
$ p\equal{}\minus{}2a^2b\minus{}2ab^2$
$ q\equal{}a^2b^2$

Divide the first by 2, square it, and then subtract from the second to get ab=15. Since a+b=8, we find a=5,b=3.
So p=-240, q=225, p+q=-15

but how did you end up with  16=2a+2b and 94=a^2+b^2+4ab ???
we haven't learned any vieta's formulas or patterns yet. so. i'm having difficulty understanding the process.

is there another approach to this problem?

[b]viete's formula[/b]
$ s_k\equal{}(\minus{}1)^k\frac{a_{n\minus{}k}}{a_n}$
here, $ k$ stands for the sum of the products. i.e if sum of roots is taken one at a time, then $ k\equal{}1$ and if in pairs $ k\equal{}2$. e.g. if roots are a, b, c. then $ a\plus{}b\plus{}c\equal{}\frac{\minus{}b}{a}$


inequalities
function
real analysis
real analysis unsolved

Hi

I have this exercise, and Ididn't manage to complete it:

There are 2 sequences, An and Bn, defined as:
[url=http://img517.imageshack.us/i/yyyx.jpg/][img]http://img517.imageshack.us/img517/8638/yyyx.jpg[/img][/url]
And: a>b>0

I need to prove that the sequences converge, and to calculate thier limit.

Now, I proved that they converge:
Because of the inequality of means, a>=An>Bn>=b, so An and Bn are bounded between a and b.
An monotonically decreases and Bn monotonically increases, so they both converge.
An=(An+Bn)/2
So when I calculate the limit of that when n goes to infinity:
La=(La+Lb)/2
2La=La+Lb
La=Lb
And I get the same result for the definition of Bn, so I can't find a way to calculate this limit (as a function of a and b).

Any ideas?

It has been studied by Gauss.
See [url]http://en.wikipedia.org/wiki/Arithmetic-geometric_mean[/url]


geometry
3D geometry
quadratics
trigonometry
analytic geometry
graphing lines
slope

These are all from past USC math contests.

1. A cube measuring 100 units on each side is painted only on the outside and cut into unit cubes.  The number of cubes with paint only on two sides is?

2. The largest constant $ C$ such that $ sin x \ge Cx$ for all $ x$ in $ [0, \frac{\pi}{2} ]$ is?

I solved this by plugging in some basic values of x, and noticing that $ C$ must get smaller as $ x$ gets larger.  But I would like to know how to solve this problem more "rigorously".

3. For what values of $ a$ does the system of equations:

$ \{          x^2 \equal{} y^2$
             $ (x \minus{} a)^2 \plus{} y^2 \equal{} 1$

Have exactly three solutions?

a) for all $ a \ge 0$    b) for $ \minus{}2 \le  a \le 2$    c) for $ a \equal{} \pm 1$    d) for all $ a \epsilon  \{\minus{}1, 0, 1 \}$    e) none of the above

First, I notice that $ a \equal{} 0$ doesn't work which leaves us with choices c) and e).  When I plugged $ a\equal{} 1$ into the system I got:

$ x^2 \equal{} y^2$
$ x^2 \minus{}2x \plus{} 1 \plus{} y^2 \equal{} 1 \Rightarrow 2x^2 \minus{} 2x \equal{} 0$  which only has two solutions.

However, I messed up somewhere and the answer is c) :)

[quote="Stokes93"]These are all from past USC math contests.

1. A cube measuring 100 units on each side is painted only on the outside and cut into unit cubes.  The number of cubes with paint only on two sides is?

2. The largest constant $ C$ such that $ sin x \ge Cx$ for all $ x$ in $ [0, \frac {\pi}{2} ]$ is?

I solved this by plugging in some basic values of x, and noticing that $ C$ must get smaller as $ x$ gets larger.  But I would like to know how to solve this problem more "rigorously".

3. For what values of $ a$ does the system of equations:

$ \{ x^2 \equal{} y^2$
             $ (x \minus{} a)^2 \plus{} y^2 \equal{} 1$

Have exactly three solutions?

a) for all $ a \ge 0$    b) for $ \minus{} 2 \le a \le 2$    c) for $ a \equal{} \pm 1$    d) for all $ a \epsilon \{ \minus{} 1, 0, 1 \}$    e) none of the above

First, I notice that $ a \equal{} 0$ doesn't work which leaves us with choices c) and e).  When I plugged $ a \equal{} 1$ into the system I got:

$ x^2 \equal{} y^2$
$ x^2 \minus{} 2x \plus{} 1 \plus{} y^2 \equal{} 1 \Rightarrow 2x^2 \minus{} 2x \equal{} 0$  which only has two solutions.

However, I messed up somewhere and the answer is c) :)[/quote]
[hide="1"]Unit cubes with paint on only two sides means that it is an edge piece but NOT a corner piece. Since there are 12 edges, and each edge has 8 such unit cubes, we have 96 such unit cubes with two painted sides.[/hide]

[hide="3"]Since $ x^{2}\equal{}y^{2}$ we can substitute this into the 2nd equation to get $ (x\minus{}a)^{2}\plus{}x^{2}\equal{}1$.
Simplifying...
$ 2x^{2}\minus{}2ax\plus{}a^{2}\minus{}1\equal{}0$. Since this results in a quadratic, the only way to obtain three solutions is if x=0 and x=some other number. If x=0 we get that $ a^{2}\equal{}1$, getting $ a\equal{}\pm 1$. For x=some other number, this value doesn't rely on $ a$ because for any $ a$ we get a quadratic which results in some $ x$. The intersection of these two is when $ a\equal{}\pm 1$. [/hide]

2) $ \sin x$ over that interval is concave down and goes from $ (0,0)$ to $ \left(\frac{\pi}{2}\right)$

Therefore, the largest possible value of $ C$ is the slope between those points which is

$ \frac{\pi}{2}$

When I understand #2 and #3 now, but I'm pretty sure your solution to #1 is wrong.  Wouldn't there be $ 96$ cubes per side of side?

Now both of you are wrong. There are $ 98$ such cubes on each side.

$ 98\times12\equal{}1176$

The question is ambiguous. It could mean either that there are 100 units along each edge, or that each face has area 100 square units.

In each case, there are 98 and 8 such cubes on each edge respectively.

Well, it said "100 units", not "100 units square" or "100 square units".

If you're measuring the side like in the problem, I would assume it's the side length (the edge), but maybe not.

Stokes93, what do you have to say?

And fishythefish, your answer to #2 is wrong. Just substitute your answer into the question and clearly the inequality does not hold at $ x\equal{}\frac{\pi}{2}$.

Shouldn't the slope between $ (0,0)$ and $ (\frac {\pi}{2},1)$ be:

$ \frac {2}{\pi}$

Yeah, I figured that out. I messed up my $ \text{\LaTeX}$.

The slope between those two points (and I was missing my second ordinate) is

$ \frac{1-0}{\frac{\pi}{2}-0}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}$

Well you're right in that it is 100 units, not 100 unit cubes, and the answer is $ 1176$, however, I'm still having problems visualizing it; why is it 98 cubes per side?

You know that each side has a length of 100 units. But the cubes on the ends are painted on 3 sides, not 2, so we only want to count the 98 in the middle, then multiply by 12 to account for all the sides.


Co bac nao biet thong tin ve ket qua lam bai thi cua doi tuyen Viet Nam khong ? :?

[size=150]$46^{th}$ International Mathematical Olympiad[/size]
Mérida, Mexico, Thursday 14 July 2005
Language: Vietnamese

[b]Bài 1.[/b] Trên các cạnh của tam giác đều
$ABC$ chọn sáu điểm: $A_1, A_2$ trên $BC$; $B_1, B_2$ trên $CA$;
$C_1, C_2$ trên $AB$. Các điểm này là đỉnh của một lục giác lồi
$A_1A_2B_1B_2C_1C_2$ có sáu cạnh bằng nhau. Chứng minh rằng các
đường thẳng $A_1B_2$, $B_1C_2$ và $C_1A_2$ đồng qui.

[b]Bài 2.[/b] Cho $a_1, a_2, \ldots$ là một dãy số nguyên
với vô hạn số hạng dương và vô hạn số hạng âm. Biết rằng với mỗi số
nguyên dương $n$, các số $a_1, a_2, \ldots, a_n$ cho $n$ số dư khác
nhau trong phép chia cho $n$. Chứng minh rằng mỗi số nguyên xuất
hiện đúng một lần trong dãy số này.

[b]Bài 3.[/b] Cho $x, y$ và $z$ là các số thực dương sao
cho $xyz \geq 1$. Chứng minh rằng
\[
\frac{x^5 - x^2}
{x^5 + y^2 + z^2}
+
\frac{y^5 - y^2}
{y^5 + z^2 + x^2}
+
\frac{z^5 - z^2}
{z^5 + x^2 + y^2}
\geq 0.\]
[b]
Bài 4.[/b] Xét dãy số $a_1, a_2, \ldots$ được xác định
như sau
\[a_n = 2^n + 3^n + 6^n -1\ \ (n=1,2,\ldots).\]
Tìm tất cả các số nguyên dương nguyên tố cùng nhau với mọi số hạng
của dãy số trên.

[b]Bài 5.[/b]  Cho $ABCD$ là một tứ giác lồi có các cạnh
$BC$ và $AD$ bằng nhau và không song song với nhau. Xét $E$ và $F$
lần lượt là các điểm nằm ở phần trong các cạnh $BC$ và $AD$ sao cho
$BE=DF$. Các đường thẳng $AC$ và $BD$ cắt nhau tại điểm $P$, các
đường thẳng $BD$ và $EF$ cắt nhau tại điểm $Q$, các đường thẳng $EF$
và $AC$ cắt nhau tại điểm $R$. Chứng minh rằng khi $E$ và $F$ thay
đổi, đường tròn ngoại tiếp tam giác $PQR$ luôn đi qua một điểm cố
định thứ hai khác $P$.

[b]Bài 6.[/b] Trong một kỳ thi học sinh giỏi, các thí sinh
phải giải 6 bài toán. Biết rằng với hai bài toán bất kỳ luôn có
nhiều hơn $\frac 25$ số thí sinh dự thi giải được cả hai bài này.
Ngoài ra, không có thí sinh nào giải được cả 6 bài toán. Chứng minh
rằng có ít nhất 2 thí sinh sao cho mỗi người trong họ giải được đúng
5 bài toán.

[i]Thời gian làm bài: 4 giờ 30 phút
Mỗi bài được 7 điểm[/i]


algebra
binomial theorem
number theory
Perfect Square
IMO Shortlist

A positive integer is called a [b]double number[/b] if its decimal representation consists of a block of digits, not commencing with 0, followed immediately by an identical block. So, for instance, 360360 is a double number, but 36036 is not. Show that there are infinitely many double numbers which are perfect squares.

An idea may be to find $A$ and $B$ prime such as :

$A \mid 10^n+1$
$A^2 \nmid 10^n+1$
$B^2 \mid 10^n+1$
$A^3 < B^2$

Then let $N = A^2 * (10^n+1) / B^2$ then $N * (10^n+1)$ is a double number.

After some tries, it happens that $A=7$ and $B = 19$ works and that (using modulo, Euler's theorem, ... to prove it) with $n = 171*(1+14*k)$

Using Euler's theorem and/or brute force:-
$10^{21} \equiv -1 (mod 49) \\$ and
$10^{42} \equiv 1 (mod 49) \\$
so I think that if $A = 16.(10^{42k+21} + 1)/49$ , then $A.(10^{42k+21} + 1)$ is square, and $A$ is also large enough to prevent leading zeroes (i.e. $A > 10^{42k+20}$ )

Observe that: $ 1001=7\cdot143\Rightarrow 10^3=-1+7a$, $ (a=143).$ Using [color=blue]binomial theorem[/color] we get that : $ 10^{21}=(-1+7a)^7=-1+7b^2$ for some integer $ b$, so that we also have $ 10^{21n}\equiv-1(\mod 49)$ for any odd integer $ n>0.$

Hence $ N=\frac{9}{49}(10^{21n}+1)$ is an integer of $ 21n$ digits, and ${ N(10^{21n}+1)=(\frac{3}{7}\cdot(10^{21n}+1}))^2$ is a double number that is a perfect square!


In long-distance races, speed is referred to as pace, a rate measure in minutes per mile rounded to the nearest hundredth. If a runner completes a marathon, 26.2 miles, in 4 hours, 24 minutes, 30 seconds, what is the runner’s pace?

A) 10.10 min/mi B) 0.10 min/mi C) 10.9 min/mi D) 5.52 min/mi E) 10 min/mi

This is definitely not geometry...
[hide]
4 hrs 24 min 30 sec = 264.5 min
264.5 / 26.2 = ?
Now the only problem is arithmetic; cheating with a computer 10.09542 rounds to 10.10, or A
[/hide]


Let $ a,b,c \in R^\plus{}$ and $ abc\equal{}1$
Prove that  \[ \sum_{cyc}\frac{4}{a^5(b\plus{}c)^2} \geq \frac{3\sqrt{3}}{\sqrt{a^2\plus{}b^2\plus{}c^2}}\]

I have a feeling this will heavily involve AM-GM for some reason

[quote="arqady"]Yes! It's AM-GM only. :wink:[/quote]
can you please post your solution if possible.  :omighty:
I use Cauchy-Schwarz.

I am sorry, apollo. :(  I have meant the following easiest inequality:
$ \sum_{cyc}\frac {2}{a^5(b \plus{} c)} \geq \frac {3\sqrt {3}}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}}.$

[quote="unlimitd3K"]Just a question.. I've never seen that "cyc" sum before, what does it mean? Thanks![/quote]

I'm guessing it means cyclic, or:

$ \frac{4}{a^5(b\plus{}c)^2}\plus{}\frac{4}{b^5(a\plus{}c)^2}\plus{}\frac{4}{c^5(a\plus{}b)^2}$

Basically, $ \sum_{cyc} f(x_1,x_2,x_3,\cdots,x_n) = f(x_1,x_2,\cdots,x_n) + f(x_2,x_3,\cdots,x_n,x_1) + f(x_3,x_4,\cdots,x_n,x_1,x_2) + \cdots + f(x_n,x_1,x_2,\cdots,x_{n-1})$.

In this case we can rewrite the inequality as

$ \sum_{cyc} \frac{(a+b)(c+a)}{a^5} \ge \frac{3\sqrt{3}}{\sqrt{a^2+b^2+c^2}$.

But this becomes $ \sum_{cyc} a^2b^5c^5+ab^5c^6+ab^6c^5+b^6c^6 \ge \frac{3\sqrt{3}}{\sqrt{a^2+b^2+c^2}}$

Now we have $ \sqrt{\frac{a^2+b^2+c^2}{3}}\sum_{cyc} a^2b^5c^5+ab^5c^6+ab^6c^5+b^6c^6 \ge 3$

After QM-AM-GM, this becomes equivalent to $ \sum_{cyc} a^2b^5c^5+ab^5c^6+ab^6c^5+b^6c^6 \ge 3$, which probably follows pretty easily from Muirhead or whatever.

[quote="apollo"]Let $ a,b,c \in R^ \plus{}$ and $ abc \equal{} 1$
Prove that
\[ \sum_{cyc}\frac {4}{a^5(b \plus{} c)^2} \geq \frac {3\sqrt {3}}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}}
\]
[/quote]
$ \sum_{cyc}\frac {1}{a^5(b \plus{} c)^2}\cdot\sum_{cyc}(b \plus{} c)^2a\geq\left(\sum_{cyc}\frac {1}{a^2}\right)^2 \equal{} \left(\sum_{cyc}a^2b^2\right)^2\geq(a \plus{} b \plus{} c)^2.$
$ \sum_{cyc}(b \plus{} c)^2a \equal{} \sum_{cyc}(a^2b \plus{} a^2c \plus{} abc) \plus{} 3abc\leq(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc) \plus{}$
$ \plus{} \frac {(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}{3} \equal{} \frac {4}{3}(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc).$
Thus, it's remains to prove that
$ 4(a \plus{} b \plus{} c)^2\geq\frac {4}{3}(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)\cdot\frac {3\sqrt {3}}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}},$
which equivalent to $ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)\geq3(ab \plus{} ac \plus{} bc)^2,$ which  true because $ (a \plus{} b \plus{} c)^2\geq3(ab \plus{} ac \plus{} bc)$ and $ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} ac \plus{} bc.$
Your proof is better, apollo! :D

This is my proof
By Powermean ; $ \sqrt{{a^2\plus{}b^2\plus{}c^2}} \geq \frac{a\plus{}b\plus{}c}{\sqrt{3}}$
It's remain to prove \[ \sum_{cyclic}\frac{4}{a^5(b\plus{}c)^2} \geq \frac{9}{a\plus{}b\plus{}c}\]
\[ \leftrightarrow (a\plus{}b\plus{}c)(\sum_{cyclic}\frac{1}{a^5(b\plus{}c)^2}) \geq \frac{9}{4}\]
By Cauchy-Schwarz ;\[ (a\plus{}b\plus{}c)(\sum_{cyclic}\frac{1}{a^5(b\plus{}c)^2}) \geq (\sum_{cyclic}\frac{1}{a^2(b\plus{}c)})\]
It's equivalent to prove that  $ (\sum_{cyclic}\frac{1}{a^2(b\plus{}c)}) \geq \frac{3}{2}$
Substitute \[ a \equal{} \frac{1}{x},b \equal{} \frac{1}{y},c \equal{} \frac{1}{z}\]
$ \because abc\equal{}1 \rightarrow xyz\equal{}1$
It's equivalent to prove $ \frac{x}{y\plus{}z} \geq \frac{3}{2}$ which is nessbit


[hide="hint"]Break 2002 down into it's prime factors.[/hide]
[hide="solution"]x = 7[/hide]


You are right Markelov :blush: , but why my solution is wrong :maybe: ?Do you know the answer?

[quote="Erken"]I don't know maybe my solution is wrong :maybe:  so please check it.
Solution:
$ \frac{bc}{a^{2}+3}=\frac{abc}{a^{3}+3a}$.It is easy to see that $ abc\leq1$,so $ S\leq\sum\frac{1}{a^{3}+3a}$ but  $ \sum\frac{1}{a^{3}+3a}\leq\frac{9}{\sum a^{3}+\sum3a}$. It is easy to understand that $ \sum a^{3}+\sum3a\geq12$ so the answer is $ \frac{3}{4}$ and equality holds for  $ a=b=c=1$[/quote]

you 're solution is wrong because
\[ \sum{\frac{1}{{a^{3}+3a}}}\ge \sum{\frac{9}{{\sum{a^{3}+\sum{3a}}}}}\]
from cauchy

[quote="Markelov"]Try $ a=6-\sqrt{33}$, $ b=c=\frac{\sqrt{33}-3}{2}$ and you will find $ \frac{ab}{c^{2}+3}+\frac{ac}{b^{2}+3}+\frac{bc}{a^{2}+3}= \frac{11\sqrt{33}-45}{24}$ which is about $ 0.7579 > \frac{3}{4}$

Very nice problem, thank you, Techno ![/quote]

 :wink: you're right ! are you continue ....answer ?  :?:

[quote="zaizai-hoang"][quote="quangpbc"][quote="techno"]Let $ a,b,c\geq 0$ and $ a+b+c=3$. Find maximum :

$ \text{S}=\frac{bc}{a^{2}+3}+\frac{ca}{b^{2}+3}+\frac{ab}{c^{2}+3}$[/quote]

You can using ABC theorem of Anhcuong on [img]mathnfriend.org[/img][/img][/hide][/quote]
Can you post yours? Are you sure that ABC theorem kills it?  :mad:  :?:[/quote]
Sorry because i have seen it can be solved by ABC ,but it is wrong .
[quote="ducpbc"][quote="techno"]Let $ a,b,c\geq 0$ and $ a+b+c=3$. Find maximum :

$ \text{S}=\frac{bc}{a^{2}+3}+\frac{ca}{b^{2}+3}+\frac{ab}{c^{2}+3}$[/quote]
the similar inequalities:$ \text{S}=\frac{bc}{\sqrt{a^{2}+3}}+\frac{ca}{\sqrt{b^{2}+3}}+\frac{ab}{\sqrt{c^{2}+3}}$
It was proposed by Can Vo (THMM).But it is easier than the first ineq.It can be solved by AM-GM  :lol:  :D[/quote]

can you post this solution , duc?

[quote="que ga"]
[quote="ducpbc"][quote="techno"]Let $ a,b,c\geq 0$ and $ a+b+c=3$. Find maximum :

$ \text{S}=\frac{bc}{a^{2}+3}+\frac{ca}{b^{2}+3}+\frac{ab}{c^{2}+3}$[/quote]
the similar inequalities:$ \text{S}=\frac{bc}{\sqrt{a^{2}+3}}+\frac{ca}{\sqrt{b^{2}+3}}+\frac{ab}{\sqrt{c^{2}+3}}$
It was proposed by Can Vo (THMM).But it is easier than the first ineq.It can be solved by AM-GM  :lol:  :D[/quote]

can you post this solution , duc?[/quote]

By Cauchy-Schwarz,we have:$ 4(a^{2}+3)=4(a^{2}+(\frac{a+b+c}{3})^{2}+(\frac{a+b+c}{3})^{2}+(\frac{a+b+c}{3})^{2}) \ge (2a+b+c)^{2}$
Hence,$ S \le \sum \frac{2bc}{2a+b+c}$.
But, $ \sum \frac{2bc}{a+b+a+c}\le \sum (\frac{bc}{2(a+b)}+\frac{bc}{2(a+c)})=\sum (\frac{bc}{2(a+b)}+\frac{ac}{2(b+a)})=\sum (\frac{c}{2})=\frac{3}{2}$


Show that if $3 \leq d \leq 2^{n+1}$, then $d \nmid (a^{2^{n}}+1)$ for all positive integers $a$.

Suppose d divedes the expression (for contrary). Then a and d are obviously coprime. Order of a mod d divides $2^{n+1}$ but doesn't divide $2^{n}$, so the order is $2^{n+1}$ and it must divide the totient of d, so $2^{n+1}<d$, contrary.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group and $ N$ its normal subgroup of index $ n$. Prove that $ \forall{g \in G} \ g^{n} \in N$.


This is something I made a few months ago... I took about 2000 still pictures and assembled it into an animation.  It took an insane amount of work, and I'm pretty happy with the results... click the link below to download it.  It's about 3 minutes long, AVI file in MPEG format.

[url=http://s47.yousendit.com/d.aspx?id=1EBD2RP901F8E1V2BH0BXMP0GO]http://s47.yousendit.com/d.aspx?id=1EBD2RP901F8E1V2BH0BXMP0GO[/url]

yousendit.com will only hang onto the file for about a week or so, so you've got a limited time to DL it!  The version I posted is pretty bad quality, but any better and the filesize would get huge.

Let me know what you guys think!  I'd say it's pretty unique... there's lots of model animation done with things like clay, but paper is pretty unusual. ;)

wow thats really amazing. Yeah I can tell it took a lot of work. The music goes with it good.  :)


The diagonals AC and BD of a convex quadrilateral ABCD are perpendicular. Through the midpoints
of AB and AD the lines perpendicular to the corresponding opposite sides CD and BC are drawn. Prove
that these lines intersect at a point on AC.

[b]HINT:[/b]
$ \mbox{If M, N, P and Q is the midpoint of sides [AB], [BC], [CD] and [CD], respectively; then MNPQ \minus{} rectangle.}$
$ \mbox{ Let } \{G\} \equal{} MP\cap NQ \mbox{ and O the circumcircle center of }\triangle{BCD}.$
$ \mbox{Let } \{R\} \equal{} OG\cap AC; \mbox{ the point R is symmetrically of O respect to point G}$
$ (|OG| \equal{} |GR|), \mbox{ then R \minus{} is the point of intersection of two lines.}$


Let's assume it converges, which it should if your guess is anywhere close to correct. So suppose it converges. Then we need to have x_i and x_(i+1) be very close to each other after a certain point, for example, within epsilon. That is, for any epsilon > 0, there will exist some a such that for all i>a, |x_i-x_{i+1}|<epsilon.

Let x_i be close to b for large values of i. Then we need (n+b^2)/2b=b, or, solving for b, b=sqrt(n). So that's how it works, informally. You can prove it better by formalizing all these concepts and actually proving that it converges, which isn't that easy.

You can see it better if you rewrite that formula (I'll just use x cos I hate subscripts): each term equals 1/2 (x + N/x) where x is the previous term. Now we know that the square root of N is somewhere between x and N/x, and so we try halfway in between. Then we repeat with that, etc, getting closer and closer each time. Hard to "prove" but easy to see.

There's not LaTeX rendering here Rep123max
, but you can write in LaTeX language anyway.

No LaTeX, but there's BBCode (I think that's what it's called).  Anyway, click on the "View more Emoticons" to view some symbols.  Or use the shortcuts (type ^2, ^3, ^4, ^5, pi, sqrt, le, ge, equiv with colons on both sides, for instance)


Alright, I have some ideas for some games, but I want to see how they will work out.

We will have sign-ups, then we will play one game, and then the other one.

The first game;

All the players randomly vote off each other and persuade each other to keep them in the game. Majority votes will decide who loses each round. When there are two people left, the people that already lost, vote on who will win.

The second game;

We will imagine all the players standing on stilts. Some of the people's stilts are less stable than others. Half of the people are standing on unstable stilts, but it will take a little bit of wind to blow them off. Nobody knows who is standing on unstable stilts. There is a fan nearby the stilts. I will randomly pick a person. This person has control of the fan for the round. This person will choose who will be blown down. They will blow down the person that is the
-est of something. For instance they might say "I will blow down the person with the longest username", then the fan will be pointed at the person with the longest username and they will be blown down if they are standing on unstable stilts. Otherwise, it moves onto the person with the second longest username, and it continues until somebody with unstable stilts finally gets blown down. Once this happens, the next round begins, and they all get new stilts. This whole thing continues until it is at the top 2. When it is at the top 2, I will choose a person and they will say whether they want to keep their stilts or trade with the other person's. The person that remains with the stable stilts will be the winner. (kinda lame, but I am just trying)

Sign-ups begin.

apparently, i have won the first game

GAH I JUST LOST

[quote="chenhsi"]The first one is just like EtE, or mafia.

/in for second[/quote]

Not really, because nobody really has a motive for who they are going to vote off. It is pretty much random, but I think it would be fun.

First Game;
ernie
rt_08

Second Game;
ak909090
aleph0
chenhsi
ernie
pacman2812
eggylv999
dragon96
kimmystar94
rt_08

Guys, please join game 1, you guys aren't really seeing what it is like by only having 2 people. This game isn't really even official. I just want to test them out. Please join.

First Game;
ernie
rt_08
Poincare
mathelete
evilhamster

Second Game;
ak909090
aleph0
chenhsi
ernie
pacman2812
eggylv999
dragon96
kimmystar94
rt_08
evilhamster
pythag011

We will begin in a few days.
Poincare

can we pleeeeeeeeaaaaaaaaaase start???

 :(

Oh, fine. We are doing game 1 first though. We can save the better one for last. It will probably be longer anyway.

Players;
ernie
rt_08
Poincare
mathelete
evilhamster
jjx1

Okay, your goal is to be the last person. People get eliminated by getting four votes. (That is majority at the moment.) The votes don't even have to be for a reason. The way to make it far is by persuading others to keep you in.

Should we PM you or put our picks in our posts?

All discussion must be on the thread, and inactivity will result in me taking people out of the game.

Hmph. I don't know who I want to get rid of, as I don't know everyone too well.

If you don't know somebody too well, look at their posts on other threads.

First Game;
ernie
rt_08
Poincare
mathelete
evilhamster
jjx1

Second Game;
ak909090
aleph0
chenhsi
ernie
pacman2812
eggylv999
dragon96
kimmystar94
rt_08
evilhamster
pythag011
Poincare
jjx1
erabel


Do anyone of you have any idea of the type of questions that appear in the Singapore mathematics Olympiad { Junior level } ??
Even a web link will be fine.I tried searching , but I could get only the papers for the senior level. Sample question papers , or even a few examples of the type of questions will be really useful...
Thanks a real lot in advance.

may I know what's the website of the senior papers?

thank you :)

senior stuff can be found at http://goodlymaths.0catch.com/


Let S = {$ p \minus{} 1^2, p \minus{} 2^2, p \minus{} 3^2, p \minus{} 4^2,..., p \minus{} n^2$}, where p is a prime number and n is a natural numbers such that $ p \minus{} n^2> 1$ and $ p \minus{} (n \plus{} 1)^2< 0$. Prove that exist $ i,j \in S$  such that i divides j.

Locked due to bad title and being alrady posted (see e.g. PEN).


calculus integration algebra function domain geometry rectangle

I'm having trouble with integrating this to get a expression in terms of t:

$\displaystyle\int^{\infty}_{-\infty} e^{xt }\displaystyle\frac{e^\frac{x-a}{b}}{b(1 + e^\frac{x-a}{b})^2} dx$

Already knowing that the second term integrates to $\frac{-1}{1+e^\frac{x-a}{b}}$, I tried integration by parts, and then tried to integrate the second part of the answer, but everything cancelled out...sigh. Help me, please!

Beware of convergence issues. If you try that integration by parts, you lose half of the domain on which this converges.

You need a completely different method.
The first one that comes to mind for me is complex analysis; integrate $e^{zt}\frac{e^{\frac{z-a}b}}{b(1+e^{\frac{z-a}b})^2}$ around a rectangle $K$ with corners $-R,R,R+2\pi ib,-R+2\pi ib$ by the residue theorem.

We should assume $|t|<\frac1{|b|}$ for convergence. The following calculations also assume $b>0$, but could be easily modified for $b<0$.

The integrals on the horizontal segments are constant multiples of each other, and there is a double pole at $a+\pi ib$ to calculate the residue at. I don't have time for the details now- I might come back later.
[Edit- I'm back]

[hide="Calculation"]
$\int_{-R}^R e^{(x+2\pi ib)t}\frac{e^{\frac{x+2\pi ib-a}b}}{b(1+e^{\frac{x+2\pi ib-a}b})^2}\,dx=\int_{-R}^R e^{xt}e^{2\pi ibt}\frac{e^{\frac{x-a}b}e^{2\pi i}}{b(1+e^{\frac{x-a}b}e^{2\pi i})^2}\,dx$
$=e^{2\pi ibt}\int_{-R}^Re^{xt} \frac{e^{\frac{x-a}b}}{b(1+e^{\frac{x-a}b})^2}\,dx$
The integrals over the vertical bars at $\pm R$ tend to zero as $R\to\infty$, so $\lim_{R\to\infty}\int_{\partial K} e^{zt}\frac{e^{\frac{z-a}b}}{b(1+e^{\frac{z-a}b})^2}\,dz=(1-e^{2\pi ibt})\int_{-\infty}^\infty e^{xt}\frac{e^{\frac{x-a}b}}{b(1+e^{\frac{x-a}b})^2}\,dx$.

On the other hand, the integral around $\partial K$ is $2\pi i$ times the residue at the pole $a+\pi ib$. Set $w=z-a-\pi ib$, and calculate the first two terms of the Laurent series for the function at $w=0$. $e^{zt}\frac{e^{\frac{z-a}b}}{b(1+e^{\frac{z-a}b})^2}= e^{at}e^{\pi ibt}e^{wt}\frac{-e^{\frac{w}b}}{b(1-e^{\frac{w}b})^2}=e^{at}e^{\pi ibt}(1+w+O(w^2))\frac{-1-\frac wb+O(w^2)}{(-w-\frac{w^2}{2b}+O(w^3))^2}$
$=-e^{at}e^{\pi ibt}\frac{1+(1+\frac1b)w+O(w^2)}{w^2(1+\frac 1bw+O(w^2))}= -e^{at}e^{\pi ibt}(\frac1{w^2}+\frac1w+O(1))$.

Taking the $\frac1w$ term, the residue is $-e^{at}e^{\pi ibt}$. This gives
$2\pi ie^{at}e^{\pi ibt}=(e^{2\pi ibt}-1)\int_{-\infty}^\infty e^{xt}\frac{e^{\frac{x-a}b}}{b(1+e^{\frac{x-a}b})^2}\,dx$
$\int_{-\infty}^\infty e^{xt}\frac{e^{\frac{x-a}b}}{b(1+e^{\frac{x-a}b})^2}\,dx= \pi e^{at}\frac{2ie^{\pi ibt}}{e^{2\pi ibt}-1}=\pi e^{at}\cdot\frac1{\sin \pi bt}$[/hide]