calculus
derivative
integration
geometry
inequalities
complex numbers

Hi all

I am a university student majoring in business, and I've always wanted to learn maths but am not good at it.

I have only taken one maths course in university so far because I feel that I lack the knowledge necessary to fully understand what is taught there. And my university tends to teach you theorems without proofs. I learned (learn may not be an appropriate word) matrices in that course and understand virtually nothing. Computations are fairly mechanical in that course, so I scored quite well, but that's not the way I'd like to learn maths.

So please give me some advice on what books I should read to self-learn maths. I obviously aim not at competitions or anything like that, but to sharpen my mind a little.

Here's what I've learned in secondary school:-

Calculus (single variable, I remember things like first principle of differentiation, some basic techniques to optimise a function. I don't think I understand integration except that it is the opposite of differentation, although I can solve some simple questions.)

Geometry (understand poorly)

Inequality (this is by far the topic I understand the least. I know AM-GM, but can almost never apply it.)

Complex numbers (forgotten most already)

Counting (simple counting)

What are some topics that I should learn, or re-learn, now? What books do you commend?

My general goal about maths: I like what I can understand and love what I can intuitively understand, so I really don't mind progressing slowly! But then I'd really like to equip myself with enough maths knowledge so that I can take some maths courses and [b]actually benefit from them[/b] in my final year.

Thanks for you adivce  :lol:

Get these books:

Art and Craft of Problem Solving (this one is aimed for people just like you)

Calculus by Michael Spivak     (this one is pretty hard, I suggest you look into the next one before this)

Calculus Made Easy          (you'll fly through this, very well explained and nice problems)

And, if you aim to do advanced math, don't use those huge textbooks because most of the time they don't contain proofs (you could use them and the Internet in conjunction).
If you're trying to do problem solving, precalc math, then also get Art of Problem Solving Volume 2 by clicking the bookstore link at the left.


[quote="your .doc"]
$ 0 \le x \le y \le 1$
prove $ \frac {1}{{\left( {1 \plus{} 2x^2 } \right)^2 }} \plus{} \frac {2}{{\left( {2 \plus{} y^2 } \right)^2 }} \ge \frac {4}{{3\left( {1 \plus{} xy} \right)^2 }}$
[/quote]

try $ x \equal{} 0,y \equal{} 1$... :wink:


2.	The following quadratic formula can be used to approximate the distance in feet, d, that it takes to stop a car after the brakes are applied for a car traveling on a dry road at the rate of r miles per hour:
d = 0.045 r2 + 1.1r. To the nearest foot, how far will a car travel once the brakes are applied at 40mph, 55mph, and 65mph?

3.	The police measured the skid marks made by a car that crashed into a tree. If the measurement gave a braking distance of 250 feet, was the driver exceeding the legal speed limit of 55 miles per hour? To the nearest mph, what was the speed of the car before the brakes were applied? (Use the formula in #2.)

I think these belong in the High School Basics forum but...
[hide]For 2, simply subtitute $r$ for $40,55,65$ and in the equation and compute.

For 3, substitute in $250$ for $d$ in the equation and solve for $r$ using the quadratic formula.
[/hide]


Actually, they are all SOLVED inequalities. The method is not elementary. :blush:

Equality holds with $(0,0,\tfrac12,\tfrac12)$ in the first, and for the second it holds with $(\lambda,\gamma,0,0)$ where $\lambda,\gamma$ are roots of \[x^{2}-x+\frac15=0.\]

Unexpectedly, we have got another simple solutions for these two. All I need is AM-GM!!!!


[quote="Davron"]For x is element of integers. Show that if $f(x+1)=f(x)+1$ than we have $f(x)=x+1$.

davron[/quote]
Let g(x): $g: [0,1)\to R$ any function. If $f(x+1)=f(x)+1$ then $f(x)=[x]+g(\{x\})$, were g any function.


I've read an inequalities book online recently.
There are many problems in it,but I don't have the solution
brothers and sisters,who can help me?

Apollo_Mars,
If you want our helps, please post problems on forum, in box : Inequalities Unsolved Problems.  :)


what cond.s favour  $E1$ against $S_{N}1$..also arrange the halides according to increasing rate of their $E1$ reactivity and inc rate of $S_{N}1$ reactivity..

The rate determining step for both SN1 and E1 mechanisms is the same (carbocation formation), so the halides that are most reactive in SN1 are also the most reactive in E1, and these are the tertiary halides, because they can form the most stable carbocations.
In tertiary halides, E1 is greatly favoured over SN1 with an increase in temperature. This because eliminations reactions have greater activation energies, due to a greater change in bonding.
E1 is also favoured over SN1 in polar solvents that are weak nucleophiles, like polar aprotic solvents.


How to enter on the Mathlinks  site, not the AOPS?????
I mean, I know that they joined, but how can I see the old mathlinks interface???

Go to profiles and change the skin to Mathlinks

if you are not logged in typing http://www.mathlinks.ro will let you browse the forum with mathlinks skin, while at http://www.artofproblemsolving.com you will browse the forum with aops skin.

once you've logged in the profile settings come in place. you can change your style in the profile menu, as billzhao said :)


limit
trigonometry
logarithms
calculus
calculus computations

Compute
$ L\equal{} \mathop {\lim }\limits_{x \to 0 } (\cos x)^{cot2x}$

[quote="Victory.US"]Compute
$ L \equal{} \mathop {\lim }\limits_{x \to 0 } (\cos x)^{cot2x}$[/quote]

Taking log, we get $ \cos(2x)\frac{\ln(\cos(x))}{\sin(2x)}$ $ \sim$ $ \minus{}\frac x4$ whose limit is $ 0$ and so $ L\equal{}1$

$ L =\lim _{x \to 0 } (\cos x)^{cot2x}$

As the limit is of $ 1^{\infty}$ type,

$ L=e^{\lim_{x \to 0 }(cosx-1)cot(2x)}$

$ L=e^{\lim_{x \to 0 }\frac{-sinx}{2sec^2(2x)}}$      ...By L Hospital rule

$ L=e^0$

$ L=1$

You have some sort of a typo since there is no $ \ln$ in your answer, and there should be if you wish to use the exponential.. Your second line should be:

$ \lim_{x \to 0}e^{\cot (2x) \ln \cos x}$

How did you get $ \cos x \minus{} 1$?

Anyways, my solution after this is:

$ \lim_{x \to 0}\cot (2x) \ln \cos x \equal{} \lim_{x \to 0}\frac {\ln \cos x}{\tan (2x)} \equal{} ^{LH} \lim_{x \to 0}\frac { \minus{} \tan x \cos^2(2x)}{2} \equal{} 0$ so the limit is $ 1$.

Aashay THE ROKR is using this: if $ a\to 1$ and $ b\to\infty,$ then

$ \lim a^b \equal{} e^{\lim((a \minus{} 1)b)}.$

This comes from two things: first, taking the logarithm to get $ \lim b\ln a,$ then noting that as $ a\to 1,$ $ \frac {\ln a}{a \minus{} 1}\to 1,$ so we can replace $ \ln a$ by $ a \minus{} 1.$


LaTeX
number theory unsolved
number theory

suppose that [b]p[/b] is prime number. prove that:
[b]p.1 -1+p.3 -1+p.5 -1+...+p.(p-2) -1 -2 p-1+1 [/b] divisible by [b]p[/b].

hmm, isn't that trivial? are addends are trivially divisible by p, except $2^{p-1}-1$, but this is divisible by p by fermat...so where's the problem?

I think the problem he meant is in a topic called "congruence"

[b]no .. it is very very hard.........[/b]

do you mean divisible by $p^2$, because there is something similar, i found this days.

It doesn't seem to be true for $p^2$ though..

nop, the ing for $p^2$ is the following:

$(1+1)^p-2=\sum_{i=1}^{p-1}\binom{p}{i}$

and use $\binom{p}{i}==(-1)^{i-1}i^{p-2}p (modp^2)$ or something like that (I am not sure about the sign.

The question that irritates me: what do those dots mean? Do they stand for \cdot in LaTeX?


function
real analysis
real analysis unsolved

Suppose $f$ is absolutely continuous on $(a,b)$ and that $f'(x) = 0$ almost everywhere (in the sense of Lebesgue). Then $f$ is constant.

How is this done? This is a famous theorem I guess, so if you don't want to write something down for me feel free to reccomend a book which has the proof.

One approach is to define a signed measure on $(a,b)$ by $\mu(c,d)=f(d)-f(c)$, and apply the Radon-Nikodym theorem.

A direct proof: Replace the open interval $(a,b)$ with a smaller closed interval $[c,d]$. Choose $\epsilon>0$. Let $\delta>0$ be as in the definition of absolute continuity. Cover the exceptional set where $f'\ne 0$ or $f'$ does not exist by countably many open intervals $I_{k}$ with total length $<\delta$. Then $\sum_{k}\mathrm{osc}_{I_{k}}(f)<\epsilon$. (Here $\mathrm{osc}_{A}f=\sup_{A}f-\inf_{A}f$ is the oscillation of $f$ on set $A$).

The set $K=[c,d]\setminus\bigcup_{k}I_{k}$ is compact, and for each $x\in K$ there is an open interval $J_{x}$ containing $x$ such that the oscillation of $f$ on $J_{x}$ is at most $\epsilon |J_{x}|$. Choose a subcover of $\{J_{x}\}$ with finite multiplicity (at most $2$). Then the sum of oscillation of $f$ on the elements of the subcover is at most $2\epsilon |d-c|$ or something.

In conclusion, the oscillation of $f$ on $[c,d]$ was estimated to be at most $\epsilon$ times a constant.

Thanks mlok. I didn't respond to this before because I realized I was reading about this question before I had the appropriate theory to solve it. My textbook just says this is true and omits the proof because it is "too involved". Our lecturer showed us the Stieltjes Measure+Radon-Nikodym method to show that any absolutely continuous function can be represented with an integral. I felt that it was very uncomplicated and elegant, so I don't know why my book chooses to omit the complete proof of this.

It is nice that you can just brute force this problem though. I read in some paper that a proof of the Lebesgue Decomposition can also be given without the use of Radon-Nikodym, and it turns out much more general.


We need to prove :
  $10(a^3+b^3+c^3)(a+b+c)^2-9(a^5+b^5+c^5) \geq (a+b+c)^5$
$\iff 15\sum_{sym}(a^4b-a^2b^2c) \geq 0$ : obviously true by Muirhead

That is an obvious way, but what about other ways?

An elegant factorization [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=363017#p363017]here[/url]. :D

Can we prove that $abc\geq ab+bc+ca\ or\ not?$ :?

If it is possible, then I can show you another solution.

kunny

No, it's false.That one would be equivalent whith 1>=1/a+1/b+1/c.But because a+b+c=1, we have, for example,
a<1.Then 1/a>1.=>1/a+1/b+1/c>1, contradiction.

[quote=Skyward_Sea]If $a,b,c$ are positive reals such that $a+b+c=1$, prove that  \[ 10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\geq 1 . \][/quote]
Suppose $c = \max\{a,b,c\},$ set $t = \frac{a+b}{2}$ and
\[f(a,b,c) = 10(a^3+b^3+c^3)-9(a^5+b^5+c^5).\]
We have
\[f(a,b,c) - f(t,t,c) = \frac{15}{16}(a+b)[8 - 3(a+b)^2  -6(a^2+b^2)](a-b)^2.\]
Because $a + b \leqslant \frac{2}{3}$ so $3(a+b)^2 + 6(a^2+b^2) \leqslant 8,$ therefore $f(a,b,c) \geqslant f(t,t,c).$

Another, we have
\[f(t,t,c) = f(t,t,1-2t) = 1 + 30t(3t-1)^2(t-1)^2 \geqslant 1.\]
The proof is completed.

[quote=Skyward_Sea]If $a,b,c$ are positive reals such that $a+b+c=1$, prove that  \[ 10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\geq 1 . \][/quote]

By computer，we have
$$LHS-RHS=\frac{5}{2}\sum_{cyc}{(b+c)(bc+a)(3a-1)^2}\ge{0}$$

We need to prove $10(\sum x^3 - (\sum x)^3) \geq 9 (\sum x^5 - (\sum x)^5)$
Which is equivalent to $-30\prod(x+y) \geq -45\prod(x+y)(\sum x^2+\sum xy)$
$\leftrightarrow x^2+y^2+z^2+xy+yz+zx  \geq \frac{2}{3}$
$\leftrightarrow 1 - \sum xy \geq \frac{2}{3} \leftrightarrow \sum xy \leq \frac{1}{3}$ which is true because $1 = (x+y+z)^2 \geq 3\sum xy$

Just expand all
And get..
$a^2+b^2+c^2\ge (ab+bc+ac)$
Which is true


inequalities
integration
real analysis
real analysis unsolved

Integrate by parts to prove:
\[ \displaystyle\int_{U}|Du|^pdx\leq C\bigl(\displaystyle\int_{U}|u|^pdx\bigr)^{\frac {1}{2}}\bigl(\displaystyle\int_{U}|D^2U|^pdx\bigr)^{\frac {1}{2}}\]
for $ 2\leq p < \infty$ and all $ u\in W^{2,p}(U)\cap W_{0}^{1,p}(U)$.


probability
ratio
geometry
calculus
search
real analysis

Bob has never heard of it before. Define it for him so that Bob can understand it. (make it harder- fifty-five words or less).

Probabilty determines whether or not, and how often,  a given event would happen in a mathmatically perfect world. For example, in a perfect world, when flipping a coin, it would be heads exactly half the time and tails half the time. However, in the real world, the actual data is usually somewhere in that range.

Well... an exactly 55-word description on the first try! By this point, Bob either gets it or is so confused that he has retreated to his smelly den, never to get NEAR probability-or mathamatics- again. Depends. What are the odds or him getting it? Express to 2 decimal places.

.86 or so, depending on how smart Bob is. Throw in a  :pm:.14 to account for Bob's intelligence.

The probability of a discrete event (one that can only happen in a finite number of ways) is given by (The total number of "good ways" it could happen) divided by (The total number of ways it could happen.)  For continuous probability it gets a little bit more complicated and there isn't an elementary "nice" way of saying it.  However, in any given instance you can often find it as a ratio of areas or something.

Since bob = 3, I say that the answer is -e^pi*i.  Which is 1, by the way.

Although actually, bob = mysticterminator = 16 = 7 = 5 = 4, so the answer is sqrt(2).  Although that's not a real probability.

Hey, bleumoose is back!

my math teacher said that on the back of MIT's basketball jerseys or something it says "we're -e^(i pi)".  I thought it was funny; he thought they're jackasses.

about mit's jerseys- that is great.

about probability- bob is the average high school freshman. he's got absolutely no clue what you are talking about about. he needs layman's terms. as in, no "big words."

yes im intentionally trying to make this hard.

and if this post irritates you, blame richard.

i am a freshman,  but i wouldn't really call myself [i]average[/i]... ah well, no matter.

For discrete probability (that is cases where there are a finite number of outcomes - or even a countably infinite number of them), the probabilty of a set of outcomes is the total number of ways of obtaining those outcomes divided by the number of all possible outcomes.

So basically, it's the number of "good" ways over the total number of ways.

For non-discrete probability (cases where the number of outcomes is not finite), it becomes more difficult to define. To define it precisely actually requires calculus (or real analysis actually).

The basic idea though is that you just need some notion of measure (or size) of a set. For finite sets that is easy - it is just the number of members of the set. But for infinite sets there are lots of natural (and sometimes useful) ways to define a measure. The most natural way though is often to think of these infinite sets geometrically and then use the measure of area or volume on them.

There are a couple of good threads that have been posted here that really test your understanding of the definition of probability. They should turn up by doing a search for "probability".

ok that confused the heck out of me. dont even want to try to explain it to bob. *sigh*.... i tried to explan combinations in 15 seconds... never EVER do that.

I thought mine was simpler :).  Try reading it again, ignoring the word "discrete" and not bothering with the talk about continuous probability.  That should be all you need.

that worked out pretty well, thanks jbl....

How's -e^pi*i equal to 1?

-interesting_move

remember that e^ix is cos x + i*sin x...


geometry
circumcircle
angle bisector
geometry proposed

A triangle $ \bigtriangleup ABC$ is given and let $ O'$ be, the intersection point of the tangent lines of its circumcircle $ (O),$ at vertices $ B,$ $ C.$ The circle $ (O')$ centered at $ O'$ with radius $ O'B = O'C,$ intersects the angle bisector of $ \angle A,$ at a point so be it $ M.$ We denote as $ P,$ the intersection point of $ BC,$ from the segment line $ OM$ and as $ E,$ $ F,$ the orthogonal projections of $ M,$ on $ AC,$ $ AB$ respectively. Prove that $ PE\perp PF.$

Kostas Vittas.

Can you give a hint, Kostas Vittas? I like this problem very much but I can't get it  :blush:
I only can prove that $ AP,\,BE$ and $ CF$ are concurrent and I think it's nice, so I will write it here. Is it O.K?  :D

Denote $ S$ be the intersection of $ AB$ with $ (O').$

We have $ \angle BSC=\frac{1}{2}\cdot\angle BO'C=90^\circ-A\Rightarrow CA\perp CS\Rightarrow CS\parallel ME$
So $ \angle MBF=\angle MCS=\angle CME\Rightarrow\triangle MBF\sim\triangle CME$
$ \Rightarrow\frac{MB}{MC}=\frac{MF}{CE}=\frac{BF}{ME}\Rightarrow\left(\frac{MB}{MC}\right)^{2}=\frac{MF}{CE}\cdot\frac{BF}{ME}=\frac{BF}{CE}\quad (1)$

On the other hand, we have $ OM$ is $ M$ - symmedian line of triangle $ MBC,$ so $ \frac{PB}{PC}=\left(\frac{MB}{MC}\right)^{2}\quad (2)$

From $ (1)$ and $ (2)$ we have $ \frac{PB}{PC}\cdot\frac{EC}{FB}=1\Rightarrow\frac{PB}{PC}\cdot\frac{EC}{EA}\cdot\frac{FA}{FB}=1$
Hence $ AP,\,BE,\,CF$ are concurrent.

[quote="April"]Can you give a hint, Kostas Vittas? ....[/quote]
Thank you dear April for your interest and the nice result you proved $ ($ indeed the pencil $ P.BAFE$ is in harmonic conjugation and the segment line $ PE,$ is the angle bisector of $ \angle APB$ $ ).$

It is useful to see that the circle $ (M),$ centered at $ M$ with radius $ ME = MF,$ is a [b][size=100]special[/size][/b] circle with respect to our configuration.

Also, a helpful result is to show that the points $ B'\equiv (O)\cap BM$ and $ C'\equiv(O)\cap CM$ are antidiametric (=diametric opposite) each other, with respect to $ (O).$

Best regards, Kostas Vittas.

$ PE$ meets $ AB$ at $ T$
$ PE$ meets $ AM$ at $ K$, $ FP$ meets $ AM$ at $ Q$
$ EF$ meets $ AM$ at $ N$
because $ AP,BE,CF$ are congruent
then $ K(E,A,F,B)=-1$ and $ KA$ passes through the midpoint of $ EF$ then $ BK //EF$
thus $ B,F,M,K$are cyclic
then $ \angle MBF=\angle NKF=\angle NKE$
similarly $ \angle MCE=\angle MQE=\angle MQF$
From $ \angle NKE+\angle NEK=90$ and $ \angle MBF+\angle MCE=90$ then $ \angle NEK=\angle PQK$
thus $ E,N,P,Q$ are cyclic
Then $ \angle QPE=\angle QNE=90$ $ q.e.d$

Very good dear Huyen Vu. Thank you for a nice solution, completely different than the one I have in mind, which is based on a property in the configuration of a bicentric quadrilateral, the inspiration of this problem.

I posted here the problem, because I was interested in another approach irrespective of the bicentric configuration.

Many thanks to both of you [b][size=100]April[/size][/b] and [b][size=100]Huyen vu.[/size][/b]

Best regards, Kostas vittas.

PS. I will post here later the solution I have in mind.

Let $ H$ be the orthocenter of $ \triangle AEF$. $ E',F',M'$ is the projection of $ E,F,M$ to $ AB,AC,BC$.
Easy to see that $ \angle FM'E = 90^{o}$, so the circumcenter of $ \triangle EM'F$ is the midpoint of $ EF$, which is the midpoint of $ MH$, too.
In the other hands, $ \triangle EM'F$ is the pedal triangle of $ M$, so "the midpoint of $ MH$ is the circumcenter of $ \triangle EM'F$" means that $ H$ is the isogonal conjugate of $ M$ wrt $ \triangle BAC$.
Now $ \frac{BE'}{CF'}= \frac{HE' . cot \angle HBA}{HF' .cot \angle HCA}= \frac{ cot \angle HBA}{cot \angle HCA}= \frac{ cot \angle MBC}{cot \angle MCB}= \frac{BM'}{CM'}$
Prove $ \frac{BP}{CP}= \frac{BF}{CE}$ easily.
So $ \frac{BP.BM'}{CP.CM'}= \frac{BF.BE'}{CE.CF'}= \frac{\mathcal{P}_{B/(EM'F)}}{\mathcal{P}_{C/(EM'F)}}$
It means that $ P \in (EM'F)$, which ends the proof.

Thank you dear [b][size=100]PdatK40SP,[/size][/b] for another clever solution, new to me.

The result that the points $ F$ $ P,$ $ M',$ $ E$ are concyclic, is the wanted conclusion in the original problem I had in mind.

Let me to try some explanations, of the proof you presented and tell me please, if you thought an easier approach about them.

[b][size=100]1).[/size][/b] - I can only to show the result that $ \angle FM'E = 90^{o}$ based on the collinearity of $ B'\equiv (O)\cap BM,$ $ O,$ $ C'\equiv (O)\cap CM.$

Indeed, we have $ \angle FM'E = \angle FM'M+\angle EM'M = \angle FBM+\angle ECM = \angle ABB'+\angle ACC' =$ $ 90^{o}$

$ ($ because of the collinearity of $ B',$ $ O,$ $ C',$ from $ \angle CBB' = \angle CC'B'$ $ ,(1)$ and $ \angle CBB'\equiv CBM = \angle MCO = \angle MC'O$ $ ,(2)$

$ \Longrightarrow$ $ \angle CBB' = \angle MC'O$ $ \Longrightarrow$ $ C'O\equiv C'B'$ $ ).$

[b][size=100]2).[/size][/b] – We have the result that the points $ H,$ $ M,$ are isogonal conjugates with respect to $ \bigtriangleup ABC,$ based on the below useful Lemma, easy to prove.

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ D,$ $ E,$ $ F$ be, the orthogonal projections of an arbitrary point $ P$ inwardly to it, on its sidelines $ BC,$ $ AC,$ $ AB,$ respectively. Prove that the circumcircle of the triangle $ \bigtriangleup DEF,$ intersects again these sidelines at points $ D',$ $ E',$ $ F'$ respectively, as the orthogonal projections of the point $ P',$ isogonal conjugate of $ P,$ with respect to $ \bigtriangleup ABC.$[/color][/size][/b]

It is easy to show that the circumcircle of $ \bigtriangleup DEF,$ $ \bigtriangleup D'E'F',$ is centered at the midpoint of the segment $ PP'.$

[b][size=100]3).[/size][/b] – After that, we can also prove that $ \frac{BE'}{CF'}= \frac{BM'}{CM'},$ from the similarities of the right triangles $ \bigtriangleup E'BH,$ $ \bigtriangleup M'BM,$ and $ \bigtriangleup F'CH,$ $ \bigtriangleup M'CM.$

[b][size=100]4).[/size][/b] – About the result you say $ \frac{BP}{CP}= \frac{BF}{CE},$ I can only think the way of proof I learned, by [b][size=100]April's[/size][/b] presentation.

Thank you again and best regards, Kostas Vittas.

Dear Mathlinkers,
this result can be shown as a consequence in the following article which has been put on my site
http://perso.orange.fr/jl.ayme  vol. 4 "An unlikely concurrence, revisited and generalised"  p. 11-18.
Sincerely
Jean-Louis


linear algebra
matrix
trigonometry
algebra
linear equation

Two small bodies of masses $m_{1}$ and $m_{2}$ are connected by a spring of elasticity $k$. The spring is initially unstretched, and the system rests with the spring in the horizontal position. Suddenly, the body with mass $m_{2}$ is given an horizontal velocity $v$ away from body 1. Find the maximum elongation of the spring.

This problem was inspired in Djole's problem.

initially the speed of c.m is $v$ fsince no external force acts on the system $spring+m_{1}+m_{2}$ by conservation of linear momentum finally when the spring extenda to [b]max[/b] the speed is $v^{'}=\frac{m_{2}v}{m_{1}+m_{2}}$
by COE we have
$\Delta K_{1}+\Delta_{2}+\Delta U$ $=$ $0$
and with required manipulation we get the ans

I'm going to try working this purely as a differential equations problem with a minimum of physical insight. Let $y_{1}$ and $y_{2}$ be the positions of the two masses relative to their original rest position. After the motion has started, the spring is the only force acting, so we have:

$y_{1}''=-\frac{k}{m_{1}}(y_{2}-y_{1});\ y_{1}(0)=0,\,y_{1}'(0)=0$

$y_{2}''=-\frac{k}{m_{2}}(y_{1}-y_{2});\ y_{2}(0)=0,\,y_{2}'(0)=v$

Let's make this a matrix differential equation of the form $Y''=AY.$ In particular,

$\begin{bmatrix}y_{1}\\ y_{2}\end{bmatrix}''= \begin{bmatrix}-\frac{k}{m_{1}}&\frac{k}{m_{1}}\\ \frac{k}{m_{2}}&-\frac{k}{m_{2}}\end{bmatrix}\begin{bmatrix}y_{1}\\ y_{2}\end{bmatrix}$

with the appropriate initial conditions.

That matrix $A$ has eigenvalues $0$ and $-k\left(\frac1{m_{1}}+\frac1{m_{2}}\right).$

The eigenvector associated to the eigenvalue zero is $\begin{bmatrix}1\\1\end{bmatrix}.$

The eivenvector associated to the other eigenvalue is $\begin{bmatrix}-m_{2}\\ m_{1}\end{bmatrix}.$

We write $\begin{bmatrix}y_{1}\\ y_{2}\end{bmatrix}= u_{1}\begin{bmatrix}1\\1\end{bmatrix}+u_{2}\begin{bmatrix}-m_{2}\\ m_{1}\end{bmatrix}.$

Place this in the matrix equation and we have that $u_{1}''=0$ and $u_{2}''=-k\left(\frac1{m_{1}}+\frac1{m_{2}}\right)u_{2}.$

Hence $u_{1}$ satisfies a linear equation, and if we let $\omega^{2}=k\left(\frac1{m_{1}}+\frac1{m_{2}}\right),$ then $u_{2}$ satisfies $u_{2}=C\sin(\omega t).$

To determine the initial conditions, we note that

$\begin{bmatrix}1&-m_{2}\\ 1&m_{1}\end{bmatrix}^{-1}= \frac1{m_{1}+m_{2}}\,\begin{bmatrix}m_{1}&m_{2}\\-1&1\end{bmatrix}$

From this we get that $u_{1}=\frac{m_{1}y_{1}+m_{2}y_{2}}{m_{1}+m_{2}}$ and hence $u_{1}(0)=0,u'(0)=\frac{m_{2}v}{m_{1}+m_{2}}$ from which it follows that $u_{1}(t)=\frac{m_{2}vt}{m_{1}+m_{2}}.$ This is the motion of the center of mass.

The other side is that $u_{2}=\frac{y_{2}-y_{1}}{m_{1}+m_{2}}.$ This gives $u_{2}(0)=0,u_{2}'(0)=\frac{v}{m_{1}+m_{2}}.$

From $u_{2}=C\sin(\omega t)$ we get $u_{2}'(0)=C\omega$ so $C\omega=\frac{v}{m_{1}+m_{2}}$ or $C=\frac{v}{\omega(m_{1}+m_{2})}.$

${u_{2}=\frac{v}{\omega(m_{1}+m_{2})}\,\sin(\omega t}$ but then $u_{2}=\frac{y_{2}-y_{1}}{m_{1}+m_{2}}$ or $y_{2}-y_{1}=(m_{1}+m_{2})u_{2}.$

$y_{2}-y_{1}=\frac{v}{\omega}\sin(\omega t).$

That gives the maximum stretch of the spring as

$\frac{v}{\omega}=v\sqrt{\frac{m_{1}m_{2}}{k(m_{1}+m_{2})}}.$

[quote="Kent Merryfield"]Here's a much more physical solution - pardesi was trying for this but didn't get the details right.

Momentum is conserved, and the overall momentum of the system is given by the momentum of the center of mass. The momentum is initially $m_{2}v$ and does not change, so the velocity $v_{\text{cm}}$ will be given by $(m_{1}+m_{2})v_{\text{cm}}=m_{2}v$ or $v_{\text{cm}}=\frac{m_{2}v}{m_{1}+m_{2}).}$

The kinetic energy associated to the center of mass motion is then $\frac12(m_{1}+m_{2})v_{\text{cm}}^{2}= \frac12\cdot\frac{m_{2}^{2}v^{2}}{m_{1}+m_{2}}.$

But the intial kinetic energy is $\frac12m_{2}v^{2}.$ This leaves

$\frac12m_{2}v^{2}-\frac12\cdot\frac{m_{2}^{2}v^{2}}{m_{1}+m_{2}}= \frac12\cdot\frac{m_{1}m_{2}v^{2}}{m_{1}+m_{2}}$

as energy not otherwise accounted for. That extra energy has to go into oscillatory stretching of the spring. Let $\Delta y$ be the maximum stretch of the spring. At that point, the extra energy is all potential energy stored in the spring, in the amount $\frac12k(\Delta y)^{2}.$ Thus,

$\frac12k(\Delta y)^{2}=\frac12\cdot\frac{m_{1}m_{2}v^{2}}{m_{1}+m_{2}}$ or $\Delta y=v\sqrt{\frac{m_{1}m_{2}}{k(m_{1}+m_{2})}}.$[/quote]
well i didn't find any thing wrong in my approach well here's a solution
we have for
$\Delta K_{1}+\Delta K_{2}+\Delta K_{s}=W_{s_{1}}+W_{s_{2}}-W_{s_{1}}-W_{s_{2}}+W_{int}$
assuming the spring i s of negligible mass and attains negligible speed
this gives as i said before $\Delta K_{1}+\Delta K_{2}+\Delta U=0$
we get
$\Delta K_{1}=\frac{m_{1}m_{2}^{2}v^{2}}{2(m_{1}+m_{2})^{2}}$
$\Delta K_{2}=\frac{-m_{1}m_{2}v^{2}(m_{1}+2m_{2})}{2(m_{1}+m_{2}))^{2}}$
$W_{int}=-\Delta U=-\frac{kx^{2}}{2}$
so we have
 $\frac{m_{1}m_{2}v^{2}(m_{1}+2m_{2})}{2(m_{1}+m_{2})^{2}}-\frac{m_{1}m_{2}^{2}v^{2}}{2(m_{1}+m_{2})^{2}}=\frac{kx^{2}}{2}$
this gives ,
$x=v\sqrt{\frac{m_{1}m_{2}}{k(m_{1}+m_{2})}}$
so we get the same result. :wink:

Sorry, pardesi - I thought you had said one thing but when I read it again, I saw that you had said another, correct, thing.

it's ok ,that's very humble of you Dr.Kent :) .

Incidentally, this is the solution to the harder version of this problem I had posted a few days ago. Is there anyone willing to solve it? If not, I'll post the solution...

I'll post my solution.

I think the easiest way is to work in the reference frame of the center of mass.

So, the initial velocity of the body 2 relative to center of mass is:

$v_{2}=\frac{m_{1}}{m_{1}+m_{2}}v$

And the velocity of the body 1 relative to the c.o.m:

$v_{1}=\frac{m_{2}}{m_{1}+m_{2}}v$

By conservation of energy, no external forces are doing work on the system:

$\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}=\frac{1}{2}k\Delta x^{2}$

Which gives the same result:

$\Delta x=v\sqrt{\frac{m_{1}m_{2}}{k(m_{1}+m_{2})}}$


A straight line passes through the three points $ (2,5),(4,8),$ and $ (6,y)$. What is the value of $ y$?

hello, let $ y\equal{}mx\plus{}n$ we find the slope $ m$ as $ m\equal{}\frac{y_2\minus{}y_1}{x_2\minus{}x_1}\equal{}\frac{3}{2}$ and $ n$ as $ 2$ so we get the equation $ y\equal{}\frac{3}{2}x\plus{}2$ and $ y\equal{}\frac{3}{2} \cdot 6\plus{}2\equal{}9\plus{}2\equal{}11$.
Sonnhard.

Or simply note that everytime the $ x$ value increases by $ 2$, the y value increases by $ 3$ and $ 8\plus{}3\equal{}11$.  :)


linear algebra
matrix
vector
function
combinatorics unsolved
combinatorics

On Valentine's Day, $ n$ loving couples of chess players play chess in such a way that every pair of boys play a single game, and every pair of girls play a single game. (There are no games between boys and girls) No games ends in a draw, and each couple of lovers won a total of $ n\minus{}1$ game between the two.

Prove that the number of triples of boys $ (A,B,C)$ such that $ A$ beats $ B$,$ B$ beats $ C$ and $ C$ beats $ A$ is equal to the number of triples of girls $ (D,E,F)$ such that $ D$ beats $ E$,$ E$ beats $ F$ and $ F$ beats $ D$.

Classic, isn't it?  :)

Consider a graph $ G$ with $ n$ nodes for each of the couples.  We connect two couples with a blue directed edge for a boy victory and a red directed edge for a girl victory.  Let adjacency matrix $ \mathbf{B}$ denote the adjacency matrix for the blue edges and let $ \mathbf{A}$ denote the adjacency matrix for the red edges.  Let $ \mathbf{1}$ denote the vector with $ n$ ones and consider the (linear) function $ f(\mathbf{M}) \equal{} \mathbf{1} \mathbf{M} \mathbf{1}^T$, which gives the sum of the entries in $ \mathbf{M}$.

Now:  we are given that $ (\mathbf{A} \plus{} \mathbf{B}) \mathbf{1}^T \equal{} (n \minus{} 1) \mathbf{1}^T$ (in other words, $ \mathbf{1}$ is an eigenvector with eigenvalue $ n \minus{} 1$, which is important) and we wish to prove that $ f(\mathbf{A}^2) \equal{} f(\mathbf{B}^2)$, equivalently, that $ f(\mathbf{A}^2 \minus{} \mathbf{B}^2) \equal{} 0$.  But we know that each boy and each girl played $ n \minus{} 1$ games, so the matrices $ \mathbf{B} \plus{} \mathbf{B}^T$ and $ \mathbf{A} \plus{} \mathbf{A}^T$ consist of zero entries on the diagonal and ones elsewhere.  This implies that $ \mathbf{B} \minus{} \mathbf{A}$ is anti-symmetric, so the sum of its entries is zero and

$ f \left( ( \mathbf{A} \minus{} \mathbf{B})(\mathbf{A} \plus{} \mathbf{B} ) \right) \equal{} (n \minus{} 1) \mathbf{1} (\mathbf{A} \minus{} \mathbf{B}) \mathbf{1}^T \equal{} 0$.

It remains to prove that $ f(\mathbf{AB}) \equal{} f(\mathbf{BA})$; in other words, the number of games such that a boy beats a boy whose lover beat a girl is equal to the number of games such that a girl beats a girl whose lover beat a boy.  But the first given gives

$ f(\mathbf{AB} \plus{} \mathbf{B}^2) \equal{} \mathbf{1} (\mathbf{A} \plus{} \mathbf{B})(\mathbf{B}) \mathbf{1}^T \equal{} (n \minus{} 1) f(\mathbf{B})$
$ f(\mathbf{BA} \plus{} \mathbf{B}^2) \equal{} \mathbf{1} (\mathbf{B})(\mathbf{A} \plus{} \mathbf{B}) \mathbf{1}^T \equal{} (n \minus{} 1) f(\mathbf{B})$.

Edit:  A shorter solution.  The total number of wins is the same among the guys and the girls, so $ f(\mathbf{A}) \equal{} f(\mathbf{B}) \equal{} {n \choose 2}$.  Apply the same logic as above to $ f(\mathbf{AB} \plus{} \mathbf{A}^2) \equal{} (n \minus{} 1) f(\mathbf{A}) \equal{} f(\mathbf{BA} \plus{} \mathbf{A}^2)$ to conclude.

Edit 2:  One additional step.  $ (\mathbf{A} \plus{} \mathbf{B} \plus{} \mathbf{A}^T \plus{} \mathbf{B}^T)\mathbf{1}^T \equal{} 2(n\minus{}1) \mathbf{1}^T$, so $ (\mathbf{A}^T \plus{} \mathbf{B}^T) \mathbf{1}^T \equal{} (n\minus{}1) \mathbf{1}^T \implies \mathbf{1}(\mathbf{A} \plus{} \mathbf{B}) \equal{} (n\minus{}1) \mathbf{1}$, which I implicitly assumed above.


hi,

i was reading a book and i have arrived to a problem that i couldn't neither

solve  :( nor undestand the solution :blush: .i will be happy if you help me to

undestand the solution .

[u]problem :[/u]
a teacher plays at least one game of chess in the day but doesn't play more

than 10 games .prove that if that teacher continue playing chess in this way for

a long time, there would be a continu period of days where the teacher would

play exactly 21 games.

[u]solution :[/u]

in the solution we assume $ a_{i}$ is the number of games played until i days.

$ 1=<a_{1}<a_{2}<...<a_{k}<10\times[\frac{k}{7}]$

$ 22=<a_{1}+21<a_{2}+21<..<a_{k}+21<10\times[\frac{k}{7}]+21$

if k is bigger than a fixed value  ,we will have $ 2k>10\times[\frac{k}{7}]$

[u]there for there are a couple(i,j) such that : $ a_{j}= a_{i}+21$[/u]

please i didn't understand the  underlined sentence.
thanks so much for any help.

EDIT : the teacher doesn't play more than 10 games [u]in the week [/u].

Pigeon hole principle.
Out of the 2k numbers in the interval (with length less than 2k), there must be two which are the same.
But $ a_{1}, a_{2}, \ldots, a_{k}$ are all distinct
and $ a_{1}+21, a_{2}+21, \ldots, a_{k}+21$ are also distinct
Hence the result

that is good  :lol: ! thanks for your answer .


1/ Does exist $ B,A \in SL(3,Z)$ such that  $ diag(1;1;\minus{}1)\equal{}B\plus{}A$ ?

2/ Does exist $ B,A \in SL(3,Z)$ such that $ diag(a;b;c) \equal{} B\plus{}A$ ? where $ a,b,c$ are in $ Z$

Let's start with a simple answer
[quote="Moubinool"]2/ Does exist $ B,A \in SL(3,Z)$ such that $ diag(a;b;c) \equal{} B \plus{} A$ ? where $ a,b,c$ are in $ Z$[/quote]
In general, no. Take $ a \equal{} b \equal{} c \equal{} 0$, if there were $ B,A \in SL(3,Z)$ such that $ B \equal{} \minus{} A$, then $ \det A \equal{} \minus{} \det B$, which is impossible.

For $ n>2$ even, the answer is nicer, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289976 .


There are 8 houses along a street, 4 along each side. Each of them are painted brown, yellow or white. Each house is painted only one color and each color is used on at least one house. No two color is used on at least one house. No two colors are used to paint the same number of houses. In how many ways could the eight houses on the street be painted? one such way is shown here by my very skillful art skills  :lol: :P

[yellow] [yellow] [white] [brown]
___________________________
_  _  _  _  _  _  _  _  _  _  _  _  _  _  _  _  _
___________________________

[yellow] [yellow] [yellow] [white]

(^that's a pretty sweet drawing for using just symbols on ur keyboard =] )

I'm pretty sure I'm wrong, but... [hide]is 5 the answer?[/hide]

im pretty sure that 5 isn't the ans.  :o  :(  :mad:

I think davidlizeng's method is right.

Random weird thing: I was being an Ubemaya, so I ended with 1, 2, 6; and 1, 3, 5; which both add up to 9. Then, I did $6(\binom{8}{1}\binom{7}{2}\binom{6}{6}+\binom{8}{2}\binom{6}{3}\binom{5}{5})$, which got the same answer :rotfl:  :rotfl:  :rotfl:  :rotfl:

I deserve some pwnage points for that!

[quote="davidlizeng"][hide]
there are two choices for the number of painted houses
5,2,1 or 4,3,1 each has 6 ways to order, a total of 12 ways
then for the 5,2,1 way, there are $\frac{8!}{5!2!}$ for each ordering a total of $1008$
for the 4,3,1 way, there are $\frac{8!}{4!3!}$ for each ordering a total of $1680$
add them up to get $2688$
something tells me this is wrong.........
  [/hide][/quote]
I'm pretty sure that that is right actually.


Find positive solutions of the equation \[ x^3\plus{}x\minus{}a<0, a>0.\]

[quote="Hungarian_86"][quote="LeUyenHoc"]Find positive solutions of the equation
\[ x^3 \plus{} x \minus{} a < 0, a > 0.
\]
[/quote]

f(x)=x^3+x-a
After derivation:
f'(x)=3x^2+1
x^2+1 is always positive so your function is monotonously increasing therefore however big is 'a' there will always be a number after what f(x) is bigger than 'a'.[/quote]

Thank for your comments. But how is $ x$? Can we solve this?

[quote="Tyl"]So we assume $ x > 0$.

Fix $ a$.

Then, we solve $ x^3 \plus{} x \minus{} a \equal{} 0$ by using Cardano's formula:

$ x \equal{} \left( \minus{} \frac {1}{2}( \minus{} a) \plus{} \sqrt {\left(\frac { \minus{} a}{2}\right)^2 \plus{} \left(\frac {1}{3}\right)^3}\right)^{\frac {1}{3}} \plus{} \left( \minus{} \frac {1}{2}( \minus{} a) \minus{} \sqrt {\left(\frac { \minus{} a}{2}\right)^2 \plus{} \left(\frac {1}{3}\right)^3}\right)^{\frac {1}{3}}$

Or,

$ x \equal{} \left(\frac {a}{2} \plus{} \sqrt {\frac {a^2}{4} \plus{} \frac {1}{27}}\right)^{\frac {1}{3}} \plus{} \left(\frac {a}{2} \minus{} \sqrt {\frac {a^2}{4} \plus{} \frac {1}{27}}\right)^{\frac {1}{3}}$

As we have seen that $ x^3 \plus{} x$ is strictly monotonously increasing, all values of $ x$ smaller than the above will satisfy the inequality.[/quote]

Hi Tyl,

Thank you very much. :lol:


Έχουμε χ,y,z,w,a πραγματικοί αριθμοί , να αποδειχθεί ότι για δύο από αυτούς ισχύει :

0<(x-y)/(1+xy)<1

Συγγνώμη  εννοώ πραγματικοί και το code μπήκε μονο του :blush: .  Η άσκηση είναι 0<(x-y)/(1+xy)<1

Πρόσεξε οτι λέω για 2 από αυτούς και μπορώ να αντιστρέψω το χ με το y ώστε  η διαφορά τους να είναι θετικοί.Στη θέση του χ μπορώ να έχω οποιοδήποτε από τα x,y,z,w,a  και στη θέση του y ομοίως .

[quote="maniopas"]Έχουμε x,y,z,w,a παραγματικοί αριθμοί να αποδειχθεί ότι για δύο από αυτούς ισχύει :
$ 0 < \frac {x \minus{} y}{1 \plus{} xy} < 1$

Είσαι σίγουρος για την εκφώνηση;
Δε μου φαίνεται σωστή. Για $ x \equal{} y \equal{} z \equal{} w \equal{} a$ έχουμε $ 0 > \frac {a \minus{} a}{1 \plus{} 2a} \equal{} 0$[/quote]

Συγγνώμη ξέχασα να αναφέρω ότι οι αριθμoί ειναι διαφορετικοι μεταξυ τους αλλιώς θα ζητούσα (αν ισχυε να ειναι ισοι ) 0=<(x-y)/(1+xy)<1

Λοιπόν,

Αφού η συνάρτηση $ f(t) \equal{} \epsilon\phi t$ έχει σύνολο τιμών, το σύνολο των πραγματικών αριθμών, άρα υπάρχουν $ x',y',z',w',a' \neq k\pi \plus{} \frac {\pi}{2}, \ k\in\mathbb{Z}$ τέτοιοι ώστε $ \epsilon\phi x' \equal{} x, \epsilon\phi y' \equal{} y, \epsilon\phi z' \equal{} z, \epsilon\phi w' \equal{} w, \epsilon\phi a' \equal{} a$. Όμως τότε 2 τουλάχιστον από αυτούς θα βρίσκονται στο ίδιο τεταρτημόριο (από αρχή Dirichlet) ας πούμε οι z', w' και χωρίς βλάβη της γενικότητας υποθέτουμε ότι $ 0\leq z' < w'$ οπότε θα έχουμε

$ 0 < w' \minus{} z' < \frac {\pi}{2}, \ \ (1)$.

Όμως τότε η παράσταση $ \frac {w \minus{} z}{1 \plus{} wz}$ ισούται με $ \epsilon \phi (w' \minus{} z')$ που λόγω της (1) βρίσκεται στο επιθυμητό διάστημα.

Αλέξανδρος

Ωραία λύση . Παρόμοια λύση έκανα και εγώ

Δηλαδή η άσκηση γίνεται:
Έχουμε x,y,z,w,a διακριτούς παραγματικούς αριθμούς. Να αποδειχθεί ότι για δύο από αυτούς ισχύει :
$ 0 < \frac {x \minus{} y}{1 \plus{} xy} < 1$
Χωρίς βλάβη της γενικότητας, έστω $ x > y > z > w > a$.

Παρατηρούμε ότι τουλάχιστον 3 από τους αριθμούς αυτούς είναι ομόσημοι ($ x > y > z > w > a > 0$ ή $ x > y > z > w > 0 > a$ ή $ x > y > z > 0 > w > a$ ή $ x > y > 0 > z > w > a$ ή $ x > 0 > y > z > w > a$ ή $ 0 > x > y > z > w > a$).

Έστω, χωρίς βλάβη της γενικότητας, ότι ομόσημοι είναι οι $ x > y > z$ τότε

$ x \minus{} y > 0$ και $ xy > 0 < \equal{} > 1 \plus{} xy > 0$ άρα
$ \frac {x \minus{} y}{1 \plus{} xy} > 0$ (Πρώτο μέλος)

έστω ότι το δεύτερο μέλος δεν ισχύει για κανένα ζευγάρι τότε:
$ \frac {x \minus{} y}{1 \plus{} xy} > \equal{} 1 < \equal{} > (x \plus{} 1)(1 \minus{} y) > \equal{} 2$
και $ (y \plus{} 1)(1 \minus{} z) > \equal{} 2$
και $ (x \plus{} 1)(1 \minus{} z) > \equal{} 2$

άρα $ \frac {1 \minus{} y}{1 \minus{} z} > \equal{} 1 < \equal{} > ... < \equal{} > y < \equal{} z$ (χρησιμοποιώντας και τις παραπάνω σχέσεις και περιπτώσεις για το 1-z)
καταλήξαμε όμως σε άτοπο λόγω της αρχικής υπόθεσης

άρα ισχύει και το δεύτερο μέλος

Η εκφώνηση είναι πολύ ωραία, αλλά η λύση μου δε μου άρεσε. Πού βρήκες την άσκηση;

Είδα μια ασκηση IMO  που ήθελε αυτη την Ιδέα γι αυτο και την έβαλα

Ξέρεις, με αυτή την ιδέα μπορείς να βγάλεις τις μεταβλητές από τις ρίζες, αν για παράδειγμα θέσεις x=ημy στην παράσταση $ A \equal{} \sqrt {1 \minus{} x^{2}}, |x| <\equal{} 1$, παίρνεις $ A \equal{}$ συνy

[quote="thanhnam2902"]Why you don't post by English langues.[/quote]

 Nice that you read the greek community problems  :) . But here we write usually in Greek.If you like some problem , we can traslate it into english for you.

[u] Babis[/u]


:lol: good morning anyone, can someone please give me a brief definition,if any, of entropy vs gibbs free energy

Entropy is the degree of randomness ...like if you go from solid to gas.  And gibbs free energy is whether the reaction is spontaneous, not spontaneous, or at equlibrium.

[thanks for the reply.  I wish there were live chem chat rooms i could sit in on , do you know of any

nope but maybe other veteran forum members might!


algebra
polynomial
trigonometry
function
binomial theorem
algebra proposed

[quote="kunny"]$ P(x) \equal{} \sin x$[/quote]
 the problem asks to find [b]polynomials[/b]  and sin x is not a plynomial

I think P(x) = x is the only solution. We can find out a tedious solution by explicitly writing the standard polynomial form and then using binomial theorem on one side and seeing by comparing terms they do not occur on the other side. But hell, there will be a simpler way then that. I hope someone solves it in a smart way.

[quote="Svejk"]Find all polynomials $ P \in \mathbb{R}[X]$ satisfying
\[ (P(x))^2 \minus{} (P(y))^2 \equal{} P(x \plus{} y)\cdot P(x \minus{} y) , \forall x,y \in \mathbb R
\]
[/quote]

$ P(x)\equal{}c$ is solution iff $ c\equal{}0$
Consider now only non-constant solutions.

if $ P(x)$ is solution, then $ aP(x)$ is solution too, so we can consider only solutions whose greatest term has a coefficient of 1.
Let then n the degree of $ P(x)$ ($ P(x)\equal{}x^n\plus{}\cdots$).
Let $ y\equal{}ax$. We have then $ (P(x))^2 \minus{} (P(ax))^2 \equal{} P(x \plus{} ax)\cdot P(x \minus{} ax)$

Coefficient of $ x^{2n}$ in LHS is $ 1\minus{}a^{2n}$
Coefficient of $ x^{2n}$ in RHS is $ (1\plus{}a)^n(1\minus{}a)^n$

So $ 1\minus{}a^{2n}\equal{}(1\minus{}a^2)^n$ $ \forall a$ And so obviuously $ n\equal{}1$
So $ P(x)\equal{}x\plus{}a$ and so $ (x\plus{}a)^2\minus{}(y\plus{}a)^2\equal{}(x\plus{}y\plus{}a)(x\minus{}y\plus{}a)$ and so $ a\equal{}0$

So the solutions are $ P(x)\equal{}cx$ for any real c.

[quote="pco"][quote="Svejk"]Find all polynomials $ P \in \mathbb{R}[X]$ satisfying
\[ (P(x))^2 \minus{} (P(y))^2 \equal{} P(x \plus{} y)\cdot P(x \minus{} y) , \forall x,y \in \mathbb R
\]
[/quote]

$ P(x) \equal{} c$ is solution iff $ c \equal{} 0$
Consider now only non-constant solutions.

if $ P(x)$ is solution, then $ aP(x)$ is solution too, so we can consider only solutions whose greatest term has a coefficient of 1.
Let then n the degree of $ P(x)$ ($ P(x) \equal{} x^n \plus{} \cdots$).
Let $ y \equal{} ax$. We have then $ (P(x))^2 \minus{} (P(ax))^2 \equal{} P(x \plus{} ax)\cdot P(x \minus{} ax)$

Coefficient of $ x^{2n}$ in LHS is $ 1 \minus{} a^{2n}$
Coefficient of $ x^{2n}$ in RHS is $ (1 \plus{} a)^n(1 \minus{} a)^n$

So $ 1 \minus{} a^{2n} \equal{} (1 \minus{} a^2)^n$ $ \forall a$ And so obviuously $ n \equal{} 1$
So $ P(x) \equal{} x \plus{} a$ and so $ (x \plus{} a)^2 \minus{} (y \plus{} a)^2 \equal{} (x \plus{} y \plus{} a)(x \minus{} y \plus{} a)$ and so $ a \equal{} 0$

So the solutions are $ P(x) \equal{} cx$ for any real c.[/quote]

Awesome :lol:
I missed the constant term :oops:. I proved P(-y) = -P(y) and noticed it can have only odd powers, but was lost after that. "y = ax " was nice :)

we take $ Q_x\in R[X]: \ Q_x(y) \equal{} (P(x))^2 \minus{} (P(y))^2 \minus{} P(x \plus{} y)P(x \minus{} y)$
we have $ \forall y\in R: \ Q_x(y) \equal{} 0$ so $ \forall y\in C: \ Q_x(y) \equal{} 0$
if $ \exists a\{b\in C|P(b) \equal{} 0\}$ such that $ a\neq 0$ so $ 0 \equal{} Q_x(a) \equal{} (P(x))^2 \minus{} P(x \plus{} a)P(x \minus{} a)$
we take $ U_n \equal{} P(na)^{2^{n}}$ $ \forall n\ge 1\ U_n \equal{} 0$ because $ U_{n \plus{} 1} \equal{} U_nP((n \plus{} 2)a)^{2^{n \minus{} 1}}\ and\ U_1\equal{}0$ then $ P\equiv 0$
if $ \{b\in C|P(b) \equal{} 0\}\subset 0$ then $ P(X) \equal{} cX^n\ (c\neq 0)$ so $ {c(X^{2n} \minus{} Y^{2n}) \equal{} c^2(X^2 \minus{} Y^2)^n}$ gives $ n \equal{} 1$

finally $ P\equiv cX$ where $ c\in R$

And how come $ \sin{x}$ is not a polynomial? Try writing it as a Taylor series! Would this make it a solution?

What? It's not a polynomial since, for example, it has infinitely many zeroes.

and $ \sin x$ is a periodic function, but polynominal function isn't.

1)

It has an infinitive number of zeros because it has an infinitive number of powers.

2)

A constant is a polynomial and a periodic function!

A polynomial is defined to have finite degree.  We distinguish between polynomials, power series, and formal power series.

Ah..  :lol: There's an explanation I appreciate. Thanks for clearing that out t0rajir0u.

[quote=Svejk]Find all polynomials $ P(x) \in \mathbb{R}[X]$ satisfying
\[ (P(x))^2-(P(y))^2=P(x+y)\cdot P(x-y) ,\quad  \forall x,y \in \mathbb R.\][/quote]

The only constant polynomial which satisfies the given relation is $P(x)=0~\forall~x \in \mathbb{R}.$ Now let's find the polynomials with $\deg P=n \ge 1.$

We have: $$ \frac{d}{dx} \Big(P^2(x)-P^2(y) \Big)= \frac{d}{dx} \Big(P(x+y)P(x-y) \Big) ~\forall~x,y \in \mathbb{R}$$ or $$2P(x)P'(x)=P'(x+y)P(x-y)+P(x+y)P'(x-y)~\forall~x,y \in \mathbb{R}.$$

Now note that from the original assertion it follows that $P(0)=0.$ So, plugging $x=y$ in the last assertion, we get: $$2P(x)P'(x)=P(2x)P'(0) ~\forall~x \in \mathbb{R}.$$

Note that $LHS$ is a polynomial and its degree is $2n-1$ and $RHS$ is a polynomial of the degree $n.$ Thus $2n-1=n \Leftrightarrow n=1.$ So $P(x)=ax$ with $a \neq 0$ - which is indeed a solution.

Putting the solutions together: The polynomials are $P(x)=ax$ for every choice of $a.$

[quote=Kunihiko_Chikaya]$ P(x)\equal{}\sin x$[/quote]

In your oppinion this is polynomial function?  :what?:

[u]Other idea to the [b]Filipjack[/b] solution:[/u]
We have $\frac{P(x)-P(y)}{x-y}(P(x)+P(y))=P(x+y)\frac{P(x-y)-P(0)}{x-y-0}$ and if $y\rightarrow x$ we get $2P'(x)P(x)=P(2x)P'(0)$ and from here follow the [b]Filipjack[/b] solution.


Hi.. :)

I just passed my 10th boards from ICSE(92%), and now starting off with +1.

I have decided to take up Non-Med as my choice of subjects and aspire to go for IIT.  :roll:

I need ur advice guys as to which BOOKS to refer and which courses to join. I am from Chandigarh.

And also do advice if I should join a [b]correspondence course[/b]. Does it help. If yes, the which one should i go for?

[i][u]

Thanks a lot ...I shall be looking forward to get ur replies.

PS: IF anyone is from chd. plz do advice a suitable school too!!! ;) . I need help fast as time is running out.[/u][/i]

Chandigarh has awesome students who cleared INMO like rushil,ashwath,bubka and the iit topper as well....(the first three are a part of this forum).

yeah dats rite.......

but since there r thousands of coachings available in every nook and corner i m confused as to which to go for......so i wuld like u guys who belong to chandigarh to plz advice me.....

n plz do add ur thoughts about [b]correspondence courses[/b] too....does these help??

Thanks.

hi again

i was just wondering if i could get in touch wid someone from chandigarh or around like rushil,bubkar or aswath or others as shadysaysurspammed mentioned through pm or yahoo messenger........to clear all my doubts which i have in mind regarding preparations for the entrance.

if u guys r out there and can take out some time to help a beginner i shall be really honoured......

Thank You
take care

PS: my im id --> shampy89@yahoo.co.in

I'm very glad that people think of me as an "awesome student" but frankly I'm just entering the 10th in Chennai  and am far from "awesome" when the IIT is concerned.(I assume that you are refering to me as I'm the only Ashwath on this forum I think!)

well...yeah.

dats rite its u....dnt underestimate urself...there is a genius in everyone n ever body is awesome in one way or other.

welll......................was just wondering if u knew nething about how to work out things to get into iit or other good institues...books...correspondence courses ...nething at all...

do reply.......

Hello,
    Maybe,I could help you.You may PM me,if necessary.
Sathej

people say FIITJEE is very gud 4 IIT...three of my frenz got into IIT from there. They say u get max results if u join them in 11th and 12th....I may think of doing so!  I think they have a centre in Chandigarh.
(I am only in 9th so i dunno wat subjects and all... :) )
Congrats on ur awesome marks :D


For each question below, form a polynomial f(x) with real coefficients having the given degree and zeros.

[1] Degree 6; zeros: i, 4 - i, 2 + i

[2] Degree 5; zeros: 1, multiplicity 3; 1 + i

[hide="Solution"] [b]Conjugate Root Theorem:[/b]  Given a polynomial $f(x)$ with real coefficients such that $a+bi$ is a root of $f(x)$, we know that $a-bi$ must also be a root.

So, for [b]1[/b] the other roots must be $-i, 4+i, 2-i$, hence we have the polynomial

$(x-i)(x+i)(x-(4-i))(x-(4+i))(x-(2+i))(x-(2-i))$

And for [b]2[/b] the other root must be $1-i$ and we have the polynomial

$(x-1)^{3}(x-(1+i))(x-(1-i))$
[/hide]
[hide="Proof of the conjugate root theorem"] The simplest proof, in my opinion, is to consider the conjugation operator $\bar{z}$.

[b]Definition:[/b]  Given a complex number $z = a+bi$ define $\bar{z}= a-bi$.

Now, it is not hard to figure out that $\overline{z+w}= \bar{z}+\bar{w}$.  The really important thing is that $\overline{z \cdot w}= \bar{z}\cdot \bar{w}$.  In other words, conjugation distributes over multiplication and addition.

This is important because a polynomial $P(x) = a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}$ is just a series of multiplications and additions of $x$; hence

$\overline{ P(x) }= P ( \bar{x})$

This is [b]very important.[/b]  Now, if $z = a+bi$ is a complex root of $P(x)$, then $P(a+bi) = 0$.  But then

$\overline{ P(a+bi) }= 0 = P ( a-bi )$

Hence $a-bi$ is also a root.  QED. [/hide]
[hide="Remark"] The conjugate root theorem is actually a special case of the theorem of minimal polynomials, but that's not really high school level. [/hide]

you posted another problem just like this in the other thread.  either put them into the same topic or just choose one.


geometry
similar triangles
perpendicular bisector
power of a point
radical axis
angle bisector
geometry solved

Hi there,

Need some help with this problem

[img]http://dandamedia.com/hw3.12.jpg[/img]

In the picture, the circles $c_1$ and $c_2$ are tangent at P and have the centers $O_1$ and $O_2$ respectively. The point Q is on $c_2$. Use the rules of constructions to construct the circle through Q that is tangent to both $c_1$ and $c_2$. Explain steps and prove that it works

Hint: you might find the tangent lines at P and Q useful.

Ok so this is what I have done so far:

For those of you who do not know the rules of constructions, basically it is anything you can create with a straightedge and compass, not a ruler.

Lets say that this new circle is called $c_3$. If $c_3$ must be tangent to $c_2$ and it must go through Q, then wouldnt it follow that the point of tangency between $c_2$ and $c_3$ must be at Q?

If this is true, then that would mean that the center of $c_3$ must lie on the line $QO_2$ (if we extended $QO_2$ outwards towards infinity)

The other thing i noticed is that the lines p and q both intersect at a common point, lets call this point Z, then angle $QO_2Z$ is congruent to angle $PO_2Z$ so the line segment $O_2Z$ bisects angle $PO_2Q$.

From there I see a few similar triangles, but nothing I can really piece together to find $c_3$

Any suggestions?

thanks

There are 2 possible ways to construct the 3rd circle: expansion transformation and inversion. I will describe the former.

In an expansion transformation, radii of all circles are increased by the same quantity and all lines move sideways by the same quantity. If this quantity is negative, the circles contract. It looks simple, but actually, to put it on a firm basis, all lines and all circles have to be considered directed. Directed lines and circles are tangent only if they have the same directions at the tangency point. In the expansion by a positive quantity a, all directed lines move by the distance a when looking down their directions and radii of all circles directed counter-clockwise increase from by the quantity a. Circles directed clockwise are considered as circles with negative radius and points as circles of zero radius. The expansion transformation preserves tangencies and tangent distances.

In the problem at hand, assume that both circles are directed counter-clockwise, having positive radii $r_1 > 0,\ r_2 > 0$. These directed circles both pass through the point P, but they are not tangent, having different directions at P. This is OK, because the tangency of these 2 circles is not necessary for the solution. The line q is tangent to the circle $c_2$ at Q, which defines its direction. Expand the figure by the negative quantity $-r_1 < 0$ (i.e., compress it). Radius of the circle $c_1$ becomes $r_1' = r_1 - r_1 = 0$, i.e., it is transformed into the point $c_1' \equiv O_1$. Radius of the circle a point circle $c_2$ becomes $r_2' = r_2 - r_1 > 0$, i.e., it is transformed into a smaller circle $c_2'$ with the same center $O_2$ and with radius $r_2 - r_1$ (such circle is easy to construct with compass and straightedge). Line q moves into a line q', such that it remains parallel to q and tangent to $c_2'$ at a point Q', because expansion preserves tangencies. We transform the line q only to avoid confusion, somebody might incorrectly consider the point Q as a circle of zero radius and try to expand it into a circle, otherwise, we do not need the line q' at all. Now the problem is reduced to a trivial problem: draw a circle $c_3'$ tangent to the circle $c_2'$ at the point Q' and passing through the point $O_1$. Again, expansion preserves tangencies, so that the circle $c_3'$ has to be tangent to the circle $c_1'$, but this circle is the point $O_1$. The center $O_3$ of the circle $c_3'$ is the intersection of the line $O_2Q \equiv O_2Q'$ and the perpendicular bisector of the segment $O_1Q'$. Now the circles $c_1', c_2', c_3'$ are transformed back into the circles $c_1, c_2, c_3$ using an expansion by positive quantity $r_1 > 0$, but of course, their centers do not move. The circle $c_3'$ is directed clockwise, because it must be tangent to the circle $c_2'$, hence, its radius $-r_3'$ is negative. In the expansion by positive quantity $r_1$, it goes into a circle $c_3$ still directed clockwise, its radius increased by $r_1$, i.e., $-r_3 = -r_3' + r_1$. So the desired circle $c_3$ is the circle centered at the point $O_3$ and with radius $O_3Q$.

The "problem" with expansion transformations is that with directed lines and circles, things become "too" unique. Usually, you get only one solution, where you would normally expect more than one. Then you have to repeat the solution with circles assigned different directions. But in this case, reversing the direction of $c_2$ to clockwise makes it tangent to the directed circle $c_1$ at P and obviously tangent to itself at Q, hence, the other solution is trivial - a circle identical with the circle $c_2$.

[quote="yetti"]There are 2 possible ways to construct the 3rd circle: expansion transformation and inversion.[/quote]

This is surely right, but I think he is searching for a different way, which is much simpler but actually does require the circles $c_1$ and $c_2$ to be tangent to each other.

[color=blue][b]Theorem 1.[/b] Let a, b, c be three circles such that the circles b and c touch each other at the point X, the circles c and a touch each other at the point Y, and the circles a and b touch each other at the point Z. Then, the common tangent of the circles b and c at the point X, the common tangent of the circles c and a at the point Y, and the common tangent of the circles a and b at the point Z concur at one point.

In words: If three circles pairwisely touch each other, then their pairwise common tangents at their points of tangency concur at one point.[/color]

[i]Proof of Theorem 1.[/i] If you know the notion of radical axes, you can just note that if two circles touch each other, then their common tangent at their point of tangency is their radical axis; thus, Theorem 1 trivially follows from the fact that the pairwise radical axes of three circles concur at one point.

Another proof of Theorem 1 goes at follows: Since the points Y and Z lie on the circle a, whose center is A, we have AY = AZ. Similarly, BZ = BX and CX = CY. Hence,

$CA+AB-BC=\left(CY+AY\right)+\left(AZ+BZ\right)-\left(BX+CX\right)$
$=\left(CX+AY\right)+\left(AY+BX\right)-\left(BX+CX\right)=2\cdot AY$,

so that $AY=\frac{CA+AB-BC}{2}$. On the other hand, if Y' is the point where the incircle of triangle ABC touches its side CA, then a well-known formula yields $AY^{\prime}=\frac{CA+AB-BC}{2}$, so that AY' = AY. As both points Y and Y' lie on the segment CA, this yields that the points Y' and Y coincide. Thus, the point Y is the point where the incircle of triangle ABC touches its side CA. Hence, if I is the center of the incircle of triangle ABC, then $IY\perp CA$. Thus, the point I lies on the perpendicular to the line CA at the point Y. But this perpendicular to the line CA at the point Y is the common tangent of the circles c and a at the point Y (this is because C and A are the centers of the circles c and a); thus, the point I lies on the common tangent of the circles c and a at the point Y. Similarly, the point I also lies on the common tangent of the circles b and c at the point X and on the common tangent of the circles a and b at the point Z. Hence, these three common tangents concur at one point (namely, at the point I), and Theorem 1 is proven.

Now, how can we apply Theorem 1 to your problem?

Well, denote by p the common tangent of the circles $c_1$ and $c_2$ at their point of tangency P, and denote by q the tangent to the circle $c_2$ at the point Q. Denote the point of intersection of these lines p and q by I.

Since our three circles $c_1$, $c_2$ and $c_3$ pairwisely touch each other, by Theorem 1 it follows that their pairwise common tangents at their points of tangency concur at one point. Now, two of these common tangents are known to us: the common tangent of the circles $c_1$ and $c_2$ at their point of tangency P is the line p, and the common tangent of the circles $c_2$ and $c_3$ at their point of tangency Q is simply the tangent to the circle $c_2$ at the point Q, i. e. the line q. Thus, it follows that the point of intersection I of these two common tangents p and q must also lie on the third common tangent, i. e. on the common tangent r of the circles $c_3$ and $c_1$ at their point of tangency. Thus, we can construct this third common tangent r easily as the tangent from the point I to the circle $c_1$ different from the line p. Now, it is very easy to construct the circle $c_3$: Its center must lie on the angle bisector of the angle between the lines q and r (since the circle $c_3$ touches both lines q and r) and on the perpendicular to the line q at the point Q (since the circle $c_3$ touches the line q at the point Q); so we can easily construct its center, and since the circle $c_3$ passes through the known point Q, we can construct this circle.

So the construction of the circle $c_3$ is complete, and the problem is solved.

  Darij

[quote="darij grinberg"]
This is surely right, but I think he is searching for a different way, which is much simpler but actually does require the circles $c_1$ and $c_2$ to be tangent to each other.[/quote]

The way you proposed is certainly different from mine. It is usually a matter of opinion, which solution is simpler. In this case, they are both so simple that it is easy to determine, which one actually is simpler, by the generally accepted method of counting the construction steps.

My solution:
1. Draw the circle $(P, r_1)$.
2. Draw the circle $c_2' \equiv (O_2, r_2 - r_1)$.
3. Draw the line OQ intersecting the circle $c_2'$ at a point Q'.
4. Construct the the perpendicular bisector of the segment $O_1Q'$, intersecting the line $O_2Q$ at a point $O_3$.
5. Draw the circle $c_3 \equiv (O_3, r_3 = O_3Q)$.

Your solution:
1. Find the midpoint M of $IO_1$, where $I \equiv p \cap q$.
2. Draw the circle $(M, r = MO_1)$, intersecting the circle $c_1$ at the point P and at another point R.
3. Draw the line $O_2Q$.
4. Draw the line $O_1R$, intersecting the line $O_2Q$ at a point $O_3$.
5. Draw the circle $c_3 \equiv (O_3, r_3 = O_3Q)$.

Comparing the 2 constructions, the steps 2, 3, 5 are equivalent. The steps 4 and 1 are also equivalent, because to find the midpoint of a line segment, you have to construct its perpendicular bisector. Finally, the steps 1 and 4 are also equivalent, because drawing a circle with a known center passing through a given point is equivalent to drawing a line through 2 given points. Thus your construction is not much simpler, but exactly as simple as mine. If the lines p, q were not already drawn, your solution would actually be more complex. It is also less general, because it works only for 2 tangent circles, while mine works for any circles. Simplicity is only one factor in judging a solution, usually a subjective one, other factors (less subjective) might be presentation,  generality and extra results or consequences.

Yetti


This is related to my other post.

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0\;{\rm m}/{\rm s}^{2} . Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off and ignore effects of air resistance.

Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0m / sec^2. How far is Powers above the ground when the helicopter crashes into the ground?

I solved it like this...

10 seconds later, they are 250 m in the air.
Powers falls for 7 seconds.

Distance fallen = (9.8/2) * 7 = 240.1 m

Now, he only has 10 meters left. Did I read the problem wrong somewhere?

[quote="K81o7"]When he steps out of the helicopter, he is still traveling upwards due to inertia.[/quote]

 :?: Inertia?

Nevermind, I FINALLY figured it out.  :D Took a while to understand it.


Find real constants a,b,c such that sin3x−√3cos3x = asin(bx+c).

If you want to find $ a,b,c$ such that for all $ x$ is $ \sin 3x-\sqrt{3}\cos 3x=a\sin (bx+c)$ then:
[hide]
$ \sin 3x-\sqrt{3}\cos 3x=2\left(\frac{1}{2}\sin 3x-\frac{\sqrt{3}}{2}\cos 3x\right)=2\left(\cos\phi\sin 3x-\sin\phi\cos 3x\right)=$
$ =2\sin\left(3x-\phi\right)$ for $ \phi=\frac{\pi}{3}+2k\pi$ for $ k\in\mathbb{Z}$[/hide]


modular arithmetic
number theory unsolved
number theory


[code]
\decaheterov[af]{4==O}
  {1==CH#_3#;6==H#_3#C;9A==H;%
  {{10}A}==\lmoiety{HOCH#_2#}}
\hspace*{-15mm} \nonaheterov[bjge]{1==S;2==N}{3==Cl}

\cyclohexanev{2B==OH;3A==OH}
\cyclohexanev{2SA==H;2SB==OH;3SA==OH;3SB==H}
\cyclohexanev{2SA==H;2SB==OH;3Sd==OH;3Su==H}
[/code]

and got below image, replaced # for dollar sign due to $\LaTeX$

How can you make those diagrams?  Can you give me a step-by-step?


Let any number $x$ (no matter real or complex) satisfies the equation $x^{3}+x^{2}= 1$

Show, that it satisfies $x^{5}+x = 1$ as well.


How do the problems on BAMO (specifically this year's) compare to the ones on USAMO? I'm sure the first two (= problems 3 and 4) are much easier than USAMO problems, but what about the remaining three? Are they all easier than USAMO #1/4 ?

I don't think any of the BAMO problems are beyond USAMO 1/4 difficulty.  However, the BAMO's different time constraints should be taken into account.

No, there's usually 1-2 problems on the BAMO that are greater than USAMO 1/4 difficulty in my opinion. Last year's #6, this year's #7, for example.

[quote="archimedes1"]However, the BAMO's different time constraints should be taken into account.[/quote]

Yeah that's true, but I'm surprised that solving 3 problems got you third place, considering how easy the first two were.


[hide]the circle's radius is 6

line CD must be a diameter too

the formula to find the length of an arc is S=x/360 * 2 pi * r

S=arc length, x=degree measure, r=radius

S=90/360 * 12 pi

=3 pi

[/hide]

Silverfalcon wrote:Well, since I'm done with MATHCOUNTS, I usually don't bother with MATHCOUNTS problems. But since geometry is weak part of my math, I'll do this one.

[hide]First, in geometry, good picture is most important thing. We have AB, diameter, 12. Now AB is perpendicular to chord CD which also has length 12. Since diameter is longest chord that's possible, we can say CD is another diameter and they're perpendicular resulting arc AC has degrees of 90. (remember in the theorem, if there is two lines that are perpendicular line, they make up four congruent angles and as you know, they must be 90. And we use the formula: S(arc length) = theta * radius and that gives us:

S = 90 * 6 = 540.

This is a good question. I couldn't remember this formula so I had to look on my trig note.

 [/hide]

[hide]The circumference is 12pi, so how can u have a minor arc loength of 540?[/hide]

Oops, I forgot to change the theta into radians. My bad, good point.

Here it is:

[hide]90*6 = pi/2 * 6 = 3pi[/hide]

That should be an answer for sure.

[hide="Answer"]Since CD=AB, CD is a diameter, as well as being at a right angle from AB, these diameters must divide the circle into fourths. The circumference is 12*pi, and a quarter of that is [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/541c35006774db100fba341620781e41.gif[/img][/hide]

[quote="Treething"][quote="rrusczyk"]AB is a diameter of a circle with length 12.  Chord CD is perpendicular to AB, and CD = 12.  Find the length of minor arc AC.[/quote]

[hide="solution"]
Since chord CD is the same length as AB and AB is a diameter, CD must also be a diameter. Since it's perpendicular, you get a circle divided up into four equal portions, so the answer is 90 degrees.
[/hide][/quote]

But what's the answer?  ;)

It always happens to me, if the problem says solve for x, I somehow solve for y.  :D


projective geometry
geometry
circumcircle
geometry proposed

Suppose that the altitudes of the triangle from $A, B, C$ intersect the circumcircle at $A_{1}, B_{1}, C_{1}$ respectively. Let $X=B_{1}C_{1}\cap{BC}, Y=C_{1}A_{1}\cap{CA}, Z=A_{1}B_{1}\cap{AB}$. Prove that $X$, $Y$ and $Z$ are collinear.

Well, that's trivial, at least using the Desargues theorem: Since the lines $AA_1$, $BB_1$ and $CC_1$ are the altitudes of triangle ABC, they concur at one point, so that, after the Desargues theorem, the points $B_1C_1\cap BC$, $C_1A_1\cap CA$ and $A_1B_1\cap AB$ must be collinear. In other words, the points X, Y, Z are collinear. That's all...

  darij

It follows from the fact that $ABC,A_1B_1C_1$ are perspective, so we can apply Desargues' Theorem.

Edit: Darij beat me to it :)


Dot has a bag of apples and each apple weighs exactly 0.7 pounds. When the apples
are placed on the scale, the scale shows .3 pounds, because, while the tenths digit
lights up correctly, the lights recording the number of pounds are malfunctioning.
What is the fewest number of apples that could be in the bag?

[hide]
If I read the problem correctly, the tenths digit is OK but the units digit is bad.

So we want to find the lowest multiple of 7 that ends in 3. This is 63.
[/hide]

[hide]We want the smallest multiple of 7 such that the units digit is 3. 63=7*9.
There are 9 apples.[/hide]


Let $a,b,c,d$ be real positive numbers, can we find a restriction on these numbers such that the following is true?
\[ \frac ab+\frac bc+\frac cd+\frac da\geq\frac{a^2+b^2+c^2+d^2}{\sqrt{abcd}}. \]

oh, not true .
 i thinks that : this problem is :
                              $\sum \frac{a^2}{b^2} \geq \frac{a+b+c+d}{\sqrt{abcd}}$


AMC
AIME
USA(J)MO
USAMO
ARML
quadratics
number theory

Let $ a,\ b,\ p,\ q$ be positive integers such that $ \frac {p^2 \plus{} q^2}{a} \equal{} \frac {pq}{b}$ and $ gcd(a,\ b) \equal{} 1.$

(1) Show that $ pq$ is divisible by $ b.$

(2) Show that $ \sqrt {a \plus{} 2b}$ is positive integer.

1)  $ \frac{a}{b} \equal{} \frac{p^2 \plus{} q^2}{pq}$, so if $ \gcd(a, b) \equal{} 1$ then the LHS is the reduction of the RHS to lowest terms.  The conclusion follows.

2)  You mean natural number.  First, divide out common factors until $ \gcd(p, q) \equal{} 1$.  Then if a prime divides $ pq$, it divides either $ p$ or $ q$, and if additionally it divides $ p^2 \plus{} q^2$ then it divides both $ p$ and $ q$, so $ \gcd(p^2 \plus{} q^2, pq) \equal{} 1$ and we have $ a \equal{} p^2 \plus{} q^2, b \equal{} pq$ so that

$ \sqrt{a \plus{} 2b} \equal{} p \plus{} q$.

That's right, t0rajir0u. :) Kyoto university's problem related to number theory is difficult for Japanese high school students, but it seems a piece of cake for Math Linkers. By the way, do you like or know ''torasan'' in Japanese movie? What do you thinf how difficult the problme is compared to Competion Problem, say in U.S.A.?

Not sure.  This is a proof problem, so it's not exactly comparable with the AIME, but it's pretty far below USAMO level.  Perhaps a first or second question on an ARML power round?

I think part of the difficulty of the problem is its intimidating form; Olympiad-level Diophantine equations of this form tend to be very difficult (albeit with fewer variables).  By the way, here is a fun second way to do 2):  the quadratic equation

$ bx^2 \minus{} ax \plus{} b \equal{} 0$

has two rational roots $ \frac{p}{q}, \frac{q}{p}$, so by the Rational Root Theorem $ p, q | b$.  If we further assume $ \gcd(p, q) \equal{} 1$ then $ pq | b \implies b \equal{} pq$.


algebra
function
domain
geometry
3D geometry

I don't think there's any real solution... :?

There are no real solutions.

[hide="Solution"]
Rearrange the given equation:
\[ \sqrt{4x-8} - \sqrt{24-4x} = \sqrt[3]{2092 - 182x} - \sqrt[3]{14x-20} \]

Let $f(x)$ be the LHS and $g(x)$ be the RHS above.  We find that $f$ is only real on $[2,6]$.  We can also fairly simply show that $f$ is strictly increasing and $g$ is strictly decreasing on $[2,6]$.  Thus if $f(6)<g(6)$, then there are no real solutions.

Simple calculation gives:

$f(6) = \sqrt{24 - 8} - \sqrt{24 - 24} = 4$

$g(6) = \sqrt[3]{2092 - 1092} - \sqrt[3]{84 - 20} = 6$

Thus $f(6) < g(6)$, so for any $x$ in the domain of $f$, we must have $f(x) < g(x)$.

Suppose that $x$ is not in the domain of $f$, then we will have a complex number on the LHS above.  The cube root function, however, produces real values for all real numbers.  Thus if $x$ is real, the RHS above will be real.  Thus the two sides can never be equal, completing the proof that there are no real solutions.

QED[/hide]


function
modular arithmetic
search
advanced fields
advanced fields unsolved

I didn't know where to post this question, so I just decided to post in on the advanced fields section.
The key for an affine cipher consists of two integers $ k \equal{} (k_{1},k_{2})$ and encryption and decryption functions are given by:

$ e_{k}(m)\equiv k_{1}\cdot m\plus{}k_{2}\pmod{p}$
$ d_{k}(c)\equiv k'_{1}(c\minus{}k_{2})\pmod{p}$

where $ k'_{1}$ is the inverse of $ k_{1}$ modulo $ p$ and $ m$ is the plaintext message and $ c$ is the encrypted message.

Now assuming that $ p$ is public knowledge, explain why the affine cipher is vulnerable to a chosen plaintext attack(i.e. given a set of plaintext/ciphertext pairs, $ (m_{i}, c_{i})$, the attacker can find out the private key, $ k$.
Also, how many pairs of these plaintext/ciphertext are needed in order to retrieve the private key, $ k$?

You could've posted in Other Problem Solving Topics or even Number Theory.

The weakness of the affine cipher is that given a plaintext/ciphertext pair $ m_{1}, c_{1}$ there is a unique choice of $ k_{2}$ for any chosen $ k_{1}$ such that $ m_{1}\equiv k_{1}c_{1}\plus{}k_{2}\bmod p$.  With a [i]single[/i] plaintext/ciphertext pair, the search space now has size $ p\minus{}1$.

With a second plaintext/ciphertext pair, we can find $ k$.  That is, given the second equation

$ m_{2}\equiv k_{1}c_{2}\plus{}k_{2}\bmod p$

we can solve for $ k_{1}, k_{2}$ as follows:

$ (m_{1}\minus{}m_{2})\equiv k_{1}(c_{1}\minus{}c_{2})\bmod p\Leftrightarrow$
$ k_{1}\equiv\frac{m_{1}\minus{}m_{2}}{c_{1}\minus{}c_{2}}\bmod p\implies$
$ k_{2}\equiv m_{1}\minus{}k_{1}c_{1}\equiv m_{2}\minus{}k_{1}c_{2}\bmod p$

shouldn't that be the inverse of $ (c_{1}\minus{}c_{2})$?

What's "that"?  The notation $ \frac{m_{1}\minus{}m_{2}}{c_{1}\minus{}c_{2}}$ is understood to mean $ (m_{1}\minus{}m_{2})(c_{1}\minus{}c_{2})^{\minus{}1}$.


geometry
incenter
geometric transformation
reflection
geometry proposed

I is the incentre of triangle ABC.
C(I) denotes a circle centre I.
Lines through I perpendicular to  sides BC,CA and AB cut the circle C(I) in points D.E F respectively.
Prove AD,BE and CF are concurrent.

Now a frivolous question, but not entirely without some point .
The gorgons Euryale ,Medusa and Stheno are at the corners of a triangle.They start chasing each other in such a way that each moves in a clockwise  direction at the same fixed speed.
Perseus to save time and effort wants to dispatch them all at once.Where should he be in order to achieve this objective?(Mere mortals would flee to a point at infinity but not this intrepid hero!)
a.The Gergonne point? The generalized Gergonne point?
b.A Brocard point?
c.The Fermat point.?
d.Any reflection point relative to Medusa?

A generalization of the problem has been posted by Treegoner before.
For this problem, there is one additional property: the point of concurrence always lies on OI.


geometry
trigonometry
Pythagorean Theorem

Derive the formulas for  30-60-90 and 45-45-90 triangles Step by step given only the side lengths "s" to start.

[hide="First ‘Hint’"]Prove that the dashed lines divide the figures into two equal parts.[/hide]
[hide="Second ‘Hint’"]Use the Pythagorean theorem to find the lengths of the dashed lines.[/hide]

Thi[b]s[/b] [b]s[/b]eriou[b]s[/b]ly look[b]s[/b] like a textbook problem.

maybe this should be in HSB. its very easy with trigonometry but without trig 45-45-90 can be done bisecting squares.


The Greek army contained two types of soldiers: the upper class and the lower class soldiers. If there were a total of 5 upper class soldiers, and 10 lower class soldiers in a certain part of Athens, and the battle of Thermopylae demands a force of 4 upper class soldiers and 8 lower class soldiers, how many different battalions can be sent?

We can count this directly but it would involve large combinations so we could count it a diferent way.One way of doing it is counting how many ways there are to leave soldiers behind.The way to count this would be
$ \binom{5}{1}\cdot\binom{10}{2}\equal{} 225$.

The reason this works is because $ \binom{5}{1}\ \equal{} \binom{5}{4}$ and $ \binom{10}{2}\ \equal{} \binom{10}{8}$

It means choose. A choose is defined as $\binom{a}{b}=\frac{a!}{(b!)((a-b)!)}$


this inequality seems right to me but i cant prove it other than with the theorom like weierstraß:

for all $a,b,c\in\mathbb R: 0\le a,b,c\le 1$

\[\frac{a+b+c}{1+abc}\le 2\]

can anybody help here?

peeta

We have \[\begin{aligned} (1 - a)(1 - b) & \geq 0 \\ ab + 1 & \geq a + b \\ abc + c & \geq ac + bc. \end{aligned}\] Adding up the other inequalities, \[3abc + (a + b + c) \geq 2(ab + bc + ca).\] Thus \[\begin{aligned} & 2abc + 2 \\ & \geq 2(ab + bc + ca) - (a + b + c) - abc + 2 \\ & = 2(ab + bc + ca) - 2(a + b + c) - 2abc + 2 + abc + (a + b + c) \\ & = 2(1 - a)(1 - b)(1 - c) + abc + (a + b + c) \\ & \geq a + b + c \end{aligned}\] and we are done

nice ...

i really seldomly have such an idea, i always want to do everything with am-gm or cs etc. but i already assumed that it would be something like that because of the addtional condition.

thx though :)


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

lease help me solving this....finals come....so nervous....thank you!!!

Define T  ; P2(x)---->P3(x) by the linear transformation T(p)=p'+xp

a) Find a basis for imT.
b) Does T map P2(x) onto P3(x)? why or why not?

I don't know what did you mean by P2 and P3. I guess they are the space of polynomials of degree less than or equal to 2 and 3 resp.

If it's the case then the solution is like this.

P2(x) and P3(x) are isomorphic to $ \mathbb{R}^3$ and $ \mathbb{R}^4$. We then could think of the mapping $ T(ax^2 \plus{} bx \plus{} c) \equal{} ax^3 \plus{} bx^2 \plus{} (c\plus{}2a)x \plus{} b$ as the transformation $ T'$: $ \begin{pmatrix}a \\ b \\ c \end{pmatrix} \mapsto \begin{pmatrix} a \\ b \\ 2a\plus{}c \\ b \end{pmatrix}$.

Note that $ \begin{pmatrix} a \\ b \\ 2a\plus{}c \\ b \end{pmatrix} \equal{} \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \end{pmatrix} a \plus{} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix} b \plus{} \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} c$.

So you easily notice that $ \{ \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} \}$ forms the basis for $ Im(T')$. Then the equivalent for $ Im(T)$ must be $ \{ x^3 \plus{} 2x, x^2 \plus{} 1, x \}$.

For question b, intuitively, the dimension of $ P2$ is less than that of $ P3$ so the mapping could not be onto. You could easily verify this from part a: The basis for $ Im(T)$ does NOT span the entire $ P3$.

[hide="Advice"] Be not nervous before exam ! [/hide]

thank you...!!! it really helps a lot!!!


topology
real analysis
real analysis unsolved

Show that the closure of a connected set is connected.

More generally: if a metric space $ X$ has a dense connected subset $ D$, then $ X$ itself is connected. Indeed, if $ X$ could be partitioned into two nonempty open subsets $ A$ and $ B$, then $ A\cap D$ and $ B\cap D$ would form a partition of $ D$.

This does not work for path-connectedness. However, the following two properties are inherited by $ X$ from $ D$, and they obviously imply path-connectedness.

Definition 1. $ X$ has bounded turning if there is $ C > 0$ such that any two points $ a,b\in X$ can be connected by a curve of diameter at most $ Cd(a,b)$.

Definition 2. $ X$ is quasiconvex if there is $ C > 0$ such that any two points $ a,b\in X$ can be connected by a curve of length at most $ Cd(a,b)$.


Is there a nice way to find \[\sum_{k=1}^\infty \frac{1}{a^{k}+1}\] for $a > 1$ ? For $|a| > 1$?  We can tell it converges by the comparison test; but to what value?

If you mean an Intermediate- or Pre-Olympiad-level way, then I seriously doubt it.  The summation doesn't telescope or do anything nice.  If there is even a way to find a closed form, it will involve analysis.

Sam...random problems don't usually have nice solutions (especially summations). Of course, I cannot prove that there is not a nice solution, but I doubt that there is a nice solution. A funny coincidence, I was working with this summation for a=2 for a potential Mock ARML problem.

[quote="Phelpedo"]It does in fact converge to a relatively nice form, but the calculation is not simple. See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143231 for the $k=2$ case.[/quote]

$k$ is the index of the summation in this problem; your link contains a different problem.


There are several sets of three different numbers whose sum is 14 that can be chosen from the set $ \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. What fraction of these sets contains a $ 4$? Express your answer as a common fraction.

The number of successes is the number of ways you can choose two numbers from there other than 4 to sum to 10. There are 3 ways.

Here are the ways we can make 14:

Highest is 9:
1,4
2,3
Highest is 8:
1,5
2,4
Highest is 7:
1,6
2,5
3,4
Highest is 6:
2,6
3,5
Highest is 5:
No ways, and there are no ways for any lower numbers.

There are a total of 10 combinations, and 3 are success, so the probability is $ \frac{3}{10}$.

[quote="vallon22"]i know for a fact that the answer is $ \frac {3} {8}$

here's where you went wrong

[quote]Highest is 9:
1,4
2,3
Highest is 8:
1,5
2,4
Highest is 7:
1,6
2,5
3,4
Highest is [b]6:
2,6 [/b]
3,5
Highest is 5:
No ways, and there are no ways for any lower numbers. [/quote]

you can't use two $ 6$'s in the same number (in bold)

and for your answer, there are actually $ 9$ choices, not $ 10$

since we had nine, and one of them doesn't work, there are a total of $ 8$ choices, so our answer is $ \frac {3} {8}$[/quote]

Oh thanks for catching my error. I guess I was just listing ways to make 10. :D


* Does a positive integer $ n$ is determined uniquely by the sets ( without the multiplicities) of prime factors of $ n$ and of $ n\plus{}1$, that is by
yhe conductors $ N(n)$ and N(n+1)$ ($N(n) is the product of the prime factors of $ n$)?
NO, indeed there are infinitely many counterexamples, the smallest with natural numbers being $ n\equal{}2$ and $ n'\equal{}8$, which begins the infinite family $ {n,n'}\equal{}{2^{m}\minus{}2,2{m}(2^m\minus{}2)}$.

is the correponding statement for polynomial true?
[color=red]The polynomial $ P(x)$ and $ Q(x)$ with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true off ihe polynomials $ P(x)\plus{}1$ and $ Q(x\plus{}1)$.
Does we have $ P(x)\equal{}Q(x)$.[/color]

notes:
* it must be assumed that at least one of $ P$ and $ Q$ is not constant, else the claim is false.

* Does a positive integer $ n$ is determined uniquely by the sets ( without the multiplicities) of prime factors of $ n$ and of $ n\plus{}1$, that is by the conductors $ N(n)$ and $ N(n\plus{}1)$ ( $ N(n)$ is the product of the prime factors of $ n$)?
NO, indeed there are infinitely many counterexamples, the smallest with natural numbers being $ n\equal{}2$ and $ n'\equal{}8$, which begins the infinite family $ (n,n')\equal{}(2^{m}\minus{}2,2{m}(2^m\minus{}2))$.

is the correponding statement for polynomial true?
The polynomial $ P(x)$ and $ Q(x)$ with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true for the polynomials $ P(x)\plus{}1$ and $ Q(x)\plus{}1$.
Does we have $ P(x)\equal{}Q(x)$.

note:
* it must be assumed that at least one of $ P$ and $ Q$ is not constant, else the claim is false.

my solution is
I will use Mason's theorem
Let $ A(t)$, $ B(t)$ and  $ C(t)$ be polynomials whose coefficient belong to an algebraically closed field of characteristic 0. Suppose $ A(t)$, $ B(t)$ and  $ C(t)$ are relativetly prime and that $ A(t)\plus{}B(t)\equal{}C(t)$. Then
$ \max \deg(A(t),B(t),C(t)) \leq Z(A(t),B(t),C(t))\minus{}1$, where $ Z(P(t))$ is the number of distinct zeros of $ P(t)$.

without loss of generality, we suppoe $ m\equal{}\deg(P(X)) \geq \deg(Q(X))\equal{}n$.
Let $ F(X)\equal{}P(X)\plus{}1$. Obviously, we see that $ \gcd(F(X),P(X),1)\equal{}1$ then by Mason's theorem, we have $ \deg(P(X)) \leq Z(F(X)P(X))\minus{}1$.iIf we note $ r\equal{}Z(P(X))\equal{}Z(Q(X))$ and $ s\equal{}Z(P(X)\plus{}1)\equal{}Z(Q(X)\plus{}1)$, then
$ m < r\plus{}s$(*). If we note $ H\equal{}P\minus{}Q$ different from $ 0$, the roots of $ P$ and $ P\plus{}1$ are a roots of $ H$, hence it must be $ m \geq \deg(H(X)) \geq Z(H(X)) \geq Z(P(X))\plus{} Z(P(X)\plus{}1)\equal{}r\plus{}s$, it's imopossible with (*).Then $ H\equal{}0$ and $ P\equal{}Q$.

if you have an other solution post it.


Hello everyone, got another brain teaser for ya. This one's definitely a challenge so please show your work. Take care and enjoy!

"Bees"

A male bee is born from an unfertilized egg, a female bee from a fertilized one. In other words, a male bee has only a mother wheras a female bee has a mother and a father. How many total ancestors does a male bee have going back ten generations?

*Hint: Try drawing a diagram for this one...

[quote="samath"]This seems like computation, not really a brain teaser :P[/quote]
Fibonacci

how can you be born from an unfertalized egg? thereforeeeeeeeeee ZERO.

You can't be born from an unfertilized egg, can you? But assuming you could:
[hide]
A male bee has a mother. His generation has one bee. His mother's generation has one bee. His mother has a mother and a father. The bee's grandparent's generation has 2 bees. You can draw a diagram to see that each generation has a Fibonacci number of bees.
Then you compute:
1+1+2+3+5+8+13+21+34+55+89, to get 143.
[/hide]
Not sure if this is right.

Male bees actually are born from unfertilized eggs... however I think the queen bee gives birth to all of them anyway, so you can't really trace back 10 generations the way you'd normally do it (I guess the answer is 10?)


Is it possible to write  prime numbers at interval (0,100) such that neither the sum nor the substraction of neighbour numbers can be divided with 5?

Um... 3 and 7?

Do you mean "find the maximum number of primes between 1 and 100 such that no two consecutive ones have differences of anything that divides 5?"

Or do you mean "can all primes satisfy (that condition)?"

I think it means,

"Is it possible to write the prime numbers between 1 and 99 in some order so that neither the sum nor the difference of any two consecutive terms is divisible by 5?"

If this is the question, then the answer is [hide]no.  There are 25 primes between 1 and 99.  7 of them are congruent to 3 mod 5 and 7 of them are congruent to 2.  These 14 numbers must be pairwise separated by the remaining 11, but this is not possible, and we're done.

It becomes possible if the list of primes goes up to 2, 5, 11, 31 or 41, and then isn't possible again until you get to listing the primes up to 3581.[/hide]


Show that $ (ab)!$ is divisible by $ (a!)^b \cdot b!$.

How many ways are there to partition a set with $ ab$ elements into $ b$ indistinguishable disjoint subsets of $ a$ elements?

Oh, I see
You have, for example, the set
$ {1,1,1,1,2,2,2,2,3,3,3,3}$
There are $ \frac {(4 \cdot 3)!}{(4!)^3}$ possible ways to make a string of numbers using these.
But then for each string you could switch the places of each type of number in $ 3!$ different ways, so the number of strings must be divisible by $ 3!$. So $ (4!)^3 \cdot 3!$ divides $ (4 \cdot 3)!$

It said on Crux that this is a well known fact but I couldn't prove it, that's why I posted it.

Yes.  Another way to say this is that the [url=http://en.wikipedia.org/wiki/Wreath_product]wreath product[/url] $ S_b \wr S_a$ acts freely on a totally ordered set with $ ab$ elements.  The copies of $ S_a$ permute the elements of each partition and $ S_b$ permutes the partitions.


geometry
3D geometry
sphere
geometric transformation
vector
combinatorics solved
combinatorics

Each vertex of the 3-dimensional Euclidean space either is coloured red or blue. Prove that within those squares being possible in this space with edge length 1 there is at least one square either with three red vertices or four blue vertices !

First case : Suppose that there are no two red points A,B such that AB = 1.
- If all the points are blue the conclusion holds trivially.
- If there exists a red point, say A : Then, the sphere with centre A and radius 1 is all blue. It is easy to see that we may inscribe a square of side 1 in this sphere, thus the conclusion holds.

Second case : Suppose that there exist two red points, say A and B, such that AB = 1.
Let C(A) be the circle with center A and radius 1 contained in the plane orthogonal to the line (AB) through A. Define C(B) in the same way.

- If there exists a red point in C(A) \cup C(B), say M in C(A). Let P be the image of M by the translation with vector AB. Then, P is in C(b) and AMPB is a square of side 1 with at least three red vertices.

- If all the points in C(A) \cup C(B) then it is easy to see that we may find a square of side 1, say MNPQ where M,N are in C(A) and P,Q are in C(B), and then M,N,P,Q are blue.

Then, in each case the conclusion holds, and we are done.

Pierre.


This class started on the 19th of September, but I just enrolled tonight (the 23rd), and I can't see the Message Board. Is this because I only registered a few hours ago? I tried logging out and then logging back in again, and I have looked everywhere under the "Online School" forum and don't see it. I would like to get access to the first problems posted after the Monday night class. Could anyone tell me why I don't see a link to the Intermediate Counting board and exactly where on the screen to look if it does appear? Thanks!

I believe you have to give time for an administrator to put you in the usergroup which allows you access to that forum.  So wait a little bit and the problem should be solved.

joml88 is correct -- you wouldn't have been able to see the class forum until your enrollment was processed and I added you to the class.  That's now been done, so you should be able to see the class forum now.


$ ABCD$ is a convex quadrilateral with $ AB$ not parallel to $ CD$. A circle through $ A$ and $ B$ touches $ CD$ at $ X$, and a circle through $ C$ and $ D$ touches $ AB$ at $ Y$. The two circles meet at $ U$ and $ V$.
Show that $ UV$ bisects $ XY$ iff $ BC$ is parallel to $ DA$.

I know this is an old thread but I have found a solution,so I will post it here:

let $ XY$ intersect $ UV$ at $ K$,and the circumcircle of $ \triangle YCD$,for the second time, at $ R$ and the circumcircle of $ \triangle XAB$,for the second time,at $ S$.also let $ BA,CD$ meat at $ O$,now let:

$ OA\equal{}a,OB\equal{}b,OC\equal{}c,OD\equal{}d,OP\equal{}p,OQ\equal{}q$ then:

$ KX.KS\equal{}KU.KV\equal{}KY.KR$

hence:

$ \frac{XR}{KX}\equal{}\frac{KR}{KX}\minus{}1\equal{}\frac{KS}{KY}\minus{}1\equal{}\frac{YS}{KY}$

hence the equality $ KX\equal{}KY$ is equivalent to $ XR\equal{}YS$ which is equivalent to:

$ CX.DX\equal{}XR.XY\equal{}YS.XY\equal{}AY.BY$

now note that $ p^2\equal{}ab,q^2\equal{}cd$ hence the equality

$ (c\minus{}p)(p\minus{}d)\equal{}(q\minus{}a)(b\minus{}q)$

can be written as:

$ (ac\minus{}bd)(bc\minus{}ad)\equal{}0$

but $ b>a,c>d$ so we must have:

$ ac\minus{}bd\equal{}0\Rightarrow \frac ba\equal{}\frac cd$

which is true iff $ BC\parallel AD$.

you can see here[url]http://www.mathlinks.ro/viewtopic.php?t=212500[/url]


Very nice problem paolo :)

Here's my solution for other ppl to see:
[hide]Yes. It is true that $ a_1 > 500$.
Proof:
Let
$ X \equal{} \{a_i|0 < i\le 10000\}$
$ Y \equal{} \{x|10000\le x < 20000\text{ and }x\in \mathbb N\}$
Let For a number $ x < 20000$ there exists an unique non-negative integer $ t$ such that $ x\cdot 2^t\in Y$
Let $ f(x)$ denote the unique positive integer $ x\cdot 2^t$
For example $ f(12345) \equal{} 12345,f(4321) \equal{} 17284,f(700) \equal{} 11200,f(1) \equal{} 16384$ etc.
$ f(x) \equal{} f(y)\implies \frac {x}{y} \equal{} 2^k$ for some integer $ k$ (not necessarily positive) which implies  one of $ x,y$ divides the other.
So $ f(x)\neq f(y)\ \forall\ x,y\in X\text{ and }x\neq y$
This is equivalent to saying that $ f$ is an injective map from $ X\to Y$
also $ |X| \equal{} |Y|\implies f$ is a bijective map from $ X\to Y$
(For a set $ S$, $ |S|$ denotes the number of elements of $ S$)
For some odd $ x\in (0,20000)$ there exists unique $ t$ such that $ x\cdot 2^t\in Y$. Then There exists unique $ u$ such that $ x\cdot 2^u\in X$
For some odd $ x\in (0,20000)$ denote $ g(x)$ that unique $ u$
If $ h(x) \equal{} x\cdot 2^{g(x)}$ then similarly it can be showed that $ h$ is a bijective map between the set of odd integers $ < 20000$ and $ X$.
We claim to show that $ x\cdot 2^{g(x)} > 500\ \forall \text{ odd }x\in (0,20000)$
For odd $ x,y\in (0,20000)$ such that $ y\neq 1$, if $ g(x)\le g(xy)$ then $ x\cdot 2^{g(x)}|xy\cdot 2^{g(xy)}$.
But both $ x\cdot 2^{g(x)},xy\cdot 2^{g(xy)}\in X$ and are distinct. Contradiction.
So we have that for all $ x,y\in (0,20000)$ such that $ y\neq 1$, $ g(x) > g(xy)$ or $ g(x)\ge g(xy) \plus{} 1$
$ g(1)\ge g(3) \plus{} 1\ge \cdots \ge g(3^2) \plus{} 2\ge \cdots g(3^9) \plus{} 9\ge 9$
(note that $ 3^9 < 20000$)
$ 1\cdot 2^{g(1)}\ge 512 > 500$
$ g(3)\ge g(3^2) \plus{} 1\ge \cdots \ge g(3^9) \plus{} 8\ge 8$
$ 3\cdot 2^{g(3)}\ge 768 > 500$
$ g(5)\ge g(5\cdot 3) \plus{} 1\ge g(5\cdot 3^2) \plus{} 2\ge \cdots \ge g(5\cdot 3^7) \plus{} 7\ge 7$
(note that $ 5\cdot 3^7 < 20000$)
$ 5\cdot 2^{g(5)}\ge 640 > 500$
$ g(7)\ge g(7\cdot 3) \plus{} 1\ge \cdots \ge g(7\cdot 3^7) \plus{} 7\ge 7$
(note that $ 7\cdot 3^7 < 20000$)
$ 7\cdot 2^{g(7)}\ge 896 > 500$
Suppose $ t$ is an arbitrary odd integer $ \ge 9$ and $ < 20000$
We need to show that $ t\cdot 2^{g(t)} > 500$
Let $ m$ be the unique non-negative integer such that $ 3^{m \plus{} 2}\le t < 3^{m \plus{} 3}$
$ t\cdot 3^{6 \minus{} m} < 3^{m \plus{} 3}\cdot 3^{6 \minus{} m} \equal{} 3^9 < 20000$
$ g(t)\ge g(t\cdot 3) \plus{} 1\ge \cdots \ge g(t\cdot 3^{6 \minus{} m}) \plus{} 6 \minus{} m\ge 6 \minus{} m$
$ t\cdot 2^{g(t)}\ge 3^{m \plus{} 2}\cdot 2^{6 \minus{} m} \equal{} 9\cdot 3^m2^{6 \minus{} m}\ge 9\cdot 2^m2^{6 \minus{} m} \equal{} 9\cdot 2^6 \equal{} 576 > 500$
$ QED$[/hide]


Why this is hard? Choose $x,z\geqslant 0$ arbitrarily, and then $y=13-11\cdot 7^x+3\cdot 5^z$

And for $x,z<0$ it's very easy to prove that there's no integer solution for $y$.

[quote="akhrif"]solve  in integers:
                        $11\times7^x+y-3\times5^z=13$[/quote]
Del! :spam:


How hot is it there? We sometimes reach the 90's snd never use an air cooler other than fans.

It only gets to be like mid 80's in maryland but next week is supposed to get into mid 90s!!

Oh. I'm in NY in the southern regions. I'm only in a village.

[quote="Treething"][quote="shadysaysurspammed"]u bet i'm hot[/quote]

always the indian pimp

lols indian pimp[/quote]

 :rotfl:

The temperature in GA is in the 80's. My dad wants the thermostat at our house to be at 85.  :furious:

80's and 90's would be a relief here. in may we had 3 100 degree days in a row. in MAY!! yesterday i had to mow 2 lawns in 105 degree heat. luckily i have an in-ground swimming pool.

in 1998, we had 60 days of straight 100+ temperatures.
in maryland people gripe about 90 degrees. i wish i lived there.

Well in Texas its about 100 right now, which is about normal.  (Our thermostat is normally around 80-81?)

At least it's better than Arizona.  I lived by Phoenix for 8 years, it got up to 110-115 every summer.  Less humid than here though.

Edit: *cough*nomosquitosinPhoenix*cough*

I'm in the sacred land of Chocolate! :rotfl:

Ok, back on topic. It's getting cooler here....

[quote="mtms5467"][quote="akalra1"]new england is slowly getting rid of the heat wave

i live in ct[/quote]
Me too. By that logic, CT is atop an asteroid.  :P

Yeah, heat wave's declining. *checks weather* it is now 84 degrees.[/quote]
XD Wimp. That's how hot it is in my room. XDXDXDXD

over here, in the shade, it's..... BRB
*goes outside*
*comes back in*

86 steps to find out that it's 75 degrees in the shade.

You counted the number of steps you took?

in souther california, 100 people died because it was too hot

dang jonny you beat me

in my school slightly less than 70 people died because i was too hot

becuz [b]u[/b] were to hot?

y not the other guys?

[quote="nutz_for2.718281828"]:rotfl:  :rotfl:

It's cold in my room! *opens blinds**sun streams in*[/quote]
It's now cold in my town! It's like 78 degrees. This morning I walked 5 miles to a nearby town, went to the library (very air-conditioned) and felt no difference from the outside :P