Show that there are infinitely many integer solutions to

a3 + b5 + c7 + d11 = e13.

[i]France 2001 - 2002, IMO training program, Sheet 1, Problem 4[/i]

(a^13)^3 +0+0+0 =  (a^3)^13

Is there something missing in the terms of the problem (abcde != 0 ?) ?

Sorry, positive integer solutions :) and positive integers don't include 0..

It is really quite trivial.
But the problem seems "connected" with the Belarussian problem solved here:

Show that infinitely many positive integers can't be written as

a3 + b5 + c7 + d9 + e11 or, alternatively,
a3 + b5 + c7 + d11 + e13.

can anyone tell me what is the meaning of the bracket of 2^(5.7.11.k)

(i am sorry to reply to an old post..)

Wow... that's a very old post.  :huh:
I'm pretty sure it just means multiplication: 2^(5.7.11.k) = $ 2^{5\cdot7\cdot11\cdot k}$.


linear algebra
matrix
induction
algebra unsolved
algebra

Given $n$ integer $x_i$ satysfying the folow system :
$2x_1-x_2=1$
$-x_1+2x_2-x_3=1$
$-x_2+2x_3-x_4=1$
$...$
$-x_{n-2}+2x_{n-1}-x_n=1$
$-x_{n-1}+2x_n=1$
Prove that $2|n$
I don't think this problem is too tough but I didn't find the right way

Does this make sense:
Consider the nxn tridiagonal matrix with $2$ on the diagonal and $-1$ on the upper and lower diagonal (the matrix of the system!), and name $\Delta_{n}$ its determinant. We have $\Delta_{1}=2$,  $\Delta_{2}=3$, and the relation $\Delta_{n+1}=2\Delta_{n}-\Delta_{n-1}$ so an easy induction gives: $\Delta_{n}=n+1$. This proves that the matrix is invertible, and the solution to that system in real numbers exists and is unique. But we note that if $(x_{1},...,x_{n})$ is solution of the system, then $(x_{n},...,x_{1})$ is also solution. This means that $x_{1}=x_{n}$. Now add all the equations together and you will find $x_{1}+x_{n}=n$ so $n=2x_{1}$ is even.


Possibly follows may become the hint.
http://www.mathlinks.ro/viewtopic.php?t=222221


function
logarithms
algebra unsolved
algebra

Find all functions $ f: R\rightarrow R$ such that
$ (i) f$ is injective
$ (ii) f(\frac{x\plus{}y}{x\minus{}y})\equal{}\frac{f(x)\plus{}f(y)}{f(x)\minus{}f(y)}\forall x\neq y$

First of all $ f$ cannot be constant. Now $ y = 0$ gives $ f(1) = \frac {f(x) + f(0)}{f(x) - f(0)}$ for all $ x$. Suppose that $ f(0)\neq 0$ and choose real numbers $ a\neq b$. Then $ f(1) = \frac {f(a) + f(0)}{f(a) - f(0)} = \frac {f(b) + f(0)}{f(b) - f(0)}$ implies $ f(a) = f(b)$. Thus by injectivity $ a = b$, a contradiction. Hence $ f(0) = 0$. Thus $ y = 0$ yields $ f(1) = 1$ and $ x = 0$ yields $ f( - 1) = - 1$.

Therefore $ y = 1$ implies $ f\left(\frac {x + 1}{x - 1}\right) = \frac {f(x) + 1}{f(x) - 1}$. On the other hand, $ y = - 1$ implies $ f\left(\frac {x - 1}{x + 1}\right) = \frac {f(x) - 1}{f(x) + 1}$. Thus $ \frac {x + 1}{x - 1}\mapsto x$ implies $ f\left(\frac 1x\right) = \frac {1}{f(x)}$ for all $ x\neq 0$. Hence
\begin{align*} & f\left(\frac {x + y}{x - y\right) = \frac {f(x) + f(y)}{f(x) - f(y)} \\
\Rightarrow & f\left(\frac {\frac xy + 1}{\frac xy - 1}}\right) = \frac {\frac {f(x)}{f(y)} + 1}{\frac {f(x)}{f(y)} - 1} \\
\Rightarrow & \frac {f\left(\frac xy\right) + 1}{f\left(\frac xy\right) - 1} = \frac {\frac {f(x)}{f(y)} + 1}{\frac {f(x)}{f(y)} - 1} \\
\Rightarrow & f\left(\frac xy\right) = \frac {f(x)}{f(y)} \\
\Rightarrow & f(xy) = f(x)f(y)\end{align*}
Hence $ f(x) = x^c$ for some constant $ c$. But then $ f\left(\frac {x + 1}{x - 1}\right) = \frac {f(x) + 1}{f(x) - 1}$ implies $ \frac {(x + 1)^c}{(x - 1)^c} = \frac {x^c + 1}{x^c - 1}$ for all $ x$. Thus we conclude that $ c = 1$ and $ f(x) = x$.

$ f$ is not continuous so $ f(xy)\equal{}f(x)f(y)\Rightarrow f(x)\equal{}x^c$ is wrong.
I have idea that we prove $ f(xy)\equal{}f(x)f(y)$ and$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ then we conclude that $ f(x)\equal{}x$ but i didn't prove $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$
nayel,can you help me?

[quote="toanIneq"]$ f$ is not continuous so $ f(xy) \equal{} f(x)f(y)\Rightarrow f(x) \equal{} x^c$ is wrong.
I[/quote]
nayel's proof is true,toanineq,note $ f(x)$ is injective ,my solution is same to his

[hide="to toanineq"]Bạn giở mấy trang đầu của cuốn mà bạn lấy đề xem lại định lí mà nayel dùng nhé[/hide]

oh,allnames.I don't agree with you. as i know,if $ f: R^\plus{}\rightarrow R^\plus{}$ and f is monotone then $ f(x)\equal{}x^c$ but f is injective so we didn't conclude that $ f(x)\equal{}x^c$

That $ f$ is injective and $ f(xy) \equal{} f(x)f(y)$ isn't enough. Let $ g(x) \equal{} \ln f(e^x)$, then $ g(x\plus{}y) \equal{} g(x) \plus{} g(y)$. Construct an injective, noncontinuous solution $ g$ to this equation. This defines $ f$ for positive reals, mapping it to positive reals. Let $ f(\minus{}x) \equal{} \minus{}f(x)$ and $ f(0) \equal{} 0$, and we get a counterexample.


abstract algebra
superior algebra
superior algebra theorems

Let H and K be groups and let G be isomorphic to HxK (cross-product).
Prove that G is solvable iff both H and K are solvable.

Regards,

Hello,

First assume $H$ and $K$ to be solvable. We then have two abelian towers $\{e\}=H_{0} \subseteq \dots \subseteq H_{n}=H$ and $\{e\}=K_{0} \subseteq \dots \subseteq K_{n}=K$. It is clear that for each $i$, $H_{i} \times K_{i}$ is normal in $H_{i+1} \times K_{i+1}$, and the factor product is also abelian (since it is isomorphic to $\frac {H_{i+1}}{H_{i}} \times \frac {K_{i+1}}{K_{i}}$), whence $G_{i}=H_{i} \times K_{i}$ is an abelian tower for $G$.

Reciprocally, let $G$ be solvable and let $\{e\}=G_{0} \subseteq \dots \subseteq G_{n}=H \times K$ be an abelian tower for $H \times K$ (that is, up to an isomorphism, for $G$). Then I let you check that $H_{i}=p_{H}(G_{i})$ is an abelian tower for $H$ (where $p_{H}$ is the projection on $H$), and similarly $K_{i}=p_{K}(G_{i})$ is an abelian tower for $K$. Whence $H$ and $K$ are solvable.


vector
invariant
linear algebra
linear algebra unsolved

I need help proving the answer to this problem.

Let n be a positive integer, let A, B be square symmetric n x n matrices with real entries.  Suppose there are n x n matrices X, Y (with complex entries) such that det(AX + BY) does not equal 0.  Prove that det(A^2 + B^2) does not equal 0.

I started by supposing that det(A^2 + B^2) = 0.  That means there is a n x 1 column vector, c, such that

(A^2 + B^2)c = 0.

I don't know where to go from here.  Any suggestions?

Assume $ A^2+B^2$ is not invertible; let $ c$ be a non zero vector s.t. $ 0=c^T(A^2+B^2)c=\parallel{}Ac\parallel{}^2+\parallel{}Bc\parallel{}^2$; therefore $ Ac=Bc=0$. Let $ E$ be the orthogonal vector space to $ c$. Then $ E$ is $ A,B$-invariant and, for all $ X,Y$, $ Im(AX+BY)\subset{Im(A)+Im(B)\subset{E}}$, a contradiction.

Nice proof, loup blanc.

$ Ac\equal{}Bc\equal{}0$ gives $ (AX\plus{}BY)^Tc\equal{}(X^TA\plus{}Y^TB)c\equal{}0$,
and therefore $ 0\equal{}\det (AX\plus{}BY)^T\equal{}\det (AX\plus{}BY)$.


search calculus derivative algebra polynomial function integration

i know there was a thread that had this before, but we are not supposed to revive threads no one has posted on in a month, so here it is again...

Search fun things to do at walmart on google. Here are some of the stuff they say.

[quote]13. Walk up to complete strangers and say, "Hi. I haven't seen
you in so long." etc. See if they play along.  Insist on calling them 'Bob', and if they protest, get angry about it.[/quote]i am going to do that someday

[code]10. Walk up to an employee and tell him in an official tone, "I
think we've got a code 3 in housewares," and see what happens.[/code]creative

[quote]19. As the cashier runs your purchase over the scanner say "BEEP" in a loud voice.  Repeat this for every item, and for other customers items. [/quote]tacky

[code]46. When someone steps away from his or her cart to look at
something, quickly place random combinations of items in their cart, such as 'A Large Cucumber and a Tub of Vasceline'.[/code]clever

interesting...
y are 2 of them codes, and the other 2 quotes? couldnt decide on just 1?

i swear that was accidental. funny isn't it? i didn't even notice.

oh well, didya like it?

haha i am going to do them all when i go to wal mart next time!  :D

take a box of dog food to a employee and ask "how many calories per serving in this? im on a diet"

lmao, I actually tried this once; but it was like, a fusion.

I ran up to a walmart dude looking distressed and did the code 3 thing, and then I was like, "Mr Gomez?? Is that you??" and tried to tell him how I met him at his nephew's house, but he just looked utterly clueless, it was pretty good, though it couldve gone better

My fav one is the one about making up your own product. Didn't try this yet, but i could imagine it :D

aanchal, do u really get the last one. I really don't think u would get it unless u were more mature or a man...

Oh yeah,

When old people and anyone else u want to, drop packs of extra large condoms in their cart or something somewhere where they can't really see it. U can tail them around and watch their surprisement when they take it out!

I had this really odd moment last year, when I thought I saw a guy from school, and then bowed to him in greeting, and then it turns out it wasn't him. It looked just like him from the back!!! Then I just like half-ran away. EMBARASSMENT.

I think you all should try that.

By the way, I think there are cameras there, so if you did that, and then got reported, then I dunno. :)

1. If you ever see a shopping cart with no hands on it, Move it one isle over.

2. If you see a nice, clean isle, and there's no one looking, switch the products around.

3. If you are under 21, complain that the beer is too expensive.

4. Go to the toys section, and find the balls. Get one, and run around bouncing it. Doing it infront of toddlers is best.

5. If you can find one, open the package and scramble the rubiks cube. Make sure that the package is closed properly.

6. Go to the greeting card isle, and scramble the cards. Just don't do it with any other store, ok?

7. If you can do it without being seen, rearrange the digits on the price boards so that they are the most expensive possible. The prices 1.89 and 4.99 are best.

8. If you see a customer eyeing a product, tell them that yours broke in 3 days. Repeat until sales decrease.

good ways to get kicked out with style:

1. Play dodgeball near the jewelry section.

2. Chuck a frisbee around. If asked, just tell them that you are testing it.

3. Repeat 2 except with plastic plates. Say that you are testing their endurancy.

4. Repeat 3 with books.

5. Repeat 1, just near the electronics section.

6. Open a package of scissors and cut all of the paper you can find. Again, say that you are testing the scissors.

7. Repeat 4 with computer disks.

8. Dance near the fish tanks. Preferably a dance that requires that your feet should be raised.   :diablo:

another thing i do at clothing stores is switch the sizes, so the XL have an XS on top, etc. Nobody really readss those things, though, so it might not work

[quote="azhang"]aanchal, do u really get the last one. I really don't think u would get it unless u were more mature or a man...[/quote]

I'm not quite sure what being a man has anything to do with getting the joke. Given that only 1.1% of all men are homosexual...

#whatever

Look for tv's which display what security cameras pick up. Then dance in front of it until you have a group dancing.

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl: vv

These are hilarious. Especially 1=2's.

during halloween, put on a custume, and then stand where they sell the custume, and then when ppl walk by, go boo, or just randomly follow someone, or go to the men's health section, and then ask about stuff and say do they have this for ...

Get a shopping cart and fill it with junk food and drive it around Wal-mart. Then fill it with fruits, vegetables, grains, dairy, you know what I'm talking about. Then drive that around Wal-mart. Do it with a hypocritical style.

take a pack of ONE A DAY vitamins and ask the cashier how many u should take each day.

Get some vitamins A, B6, C, D, and if they have it, E. Complain to the pharmacy that there is no F.

Take two full carts of items into the express checkout lane. :huh:

Be one of those charade people who get people to play charades. Make hard ones for people who seem in a hurry.

ask the cashiers for B Batteries. (i don't think they exist)

Ask for d batteries in a foriegn acsent

i think many people will agree with this one. While getting your recipt, start talking in a very high level mathematical knoledge on how they overcharged you.

i like ashpotter's last one, although if that worked, would that really be legal?

If that trick worked, I highly doubt that it would be legal (especially if you purposely tricked them into thinking that they overcharged you when you knew that they really didn't). I'm pretty sure there are at least some people out there who would consider that stealing or attempted theft. If you're actually going to do ashpotter's last idea, make sure you're ready/smart enough to know when to draw the line (aka, if they offer you the extra money, just tell them it was a joke), and don't say that they overcharged you by anything more than 50 cents. If you're not smart enough, then I don't particularly recommend following through on ashpotter's last idea.

i know, i wasnt going to try it, or any of the other ones as a matter of fact, i was just saying i thought it was funny

Yeah, it was meant to be a message to any AoPSer in general that reads this thread. :)

Tear off or otherwise damage all the barcodes on the items, and then take them to the register.

go to the dental care aisle and use their products to brush your teeth.

Go get a fishin' pole and try to get the fishies in the fishie isle.

Ask if the milk comes from first grade cows, and is milked by first grade milkers, and is processed by first grade processors, and is handled by first grade handlers.

Run up to a group of customers and endorse your own personal product: A bottle of lotion that burns your skin.

Streak through the store holding a big mac and yelling "MC FLURRY!!!!!!!"

Bring a friend to play a tennis match in the electronic section.

Tell a cash registrer that you are conspiring to blow up wal-marts and that you love trade unions.

Burn the American flag in the food department and scream "Treason and futility!!!!!!!!!!"

write in all of the notebooks

Point at the XBOX 360 and say "Hey, that looks just like mine! You stole it!"

Call security and bring them to the plastic toy guns

Backwash into the water bottles and put them back

Open a huge pack of coca-cola and spill it all over the floor and when someone asks, say "I'm testing it to see how durable it is" in a snappy voice. and then say, oops

Get some speakers and find an AC outlet.  plug in your mp3 player and boost it up really loud and say "I'm seeing how loud before it blows out so I know how long I can play a song this loud" in a smarty voice.

Open a kiddie bowling set and a real bowling ball.  Set it up in the Jewelry department.  When an employee asks you why you chose there say, "I have already proved mathematically that this is the best place to test this."  Then ask "Wanna Join me?" Hopefully they say yes. ;)

NOTE: Do [b][i][u]NOT[/u][/i][/b] try this at a store.  They will hate you for the rest of your life.

well, this isnt for wallmart persay, but

go into the camping section and bring a bunch of pillows and blankets into a tent and take a nap. when someone asks, say "im testing the comfort of this tent."

go into a changing room and wait there. after like 5 minutes say "hey! wheres the toilet paper!"

[quote="hunter34"]Go to the Mcdonalds part of the store and ask for a whopper and a frostie.[/quote]
lol

[quote="1=2"]Go to a cash register, slap the belt with our hands, and shout "I wanna beer."[/quote]
WOW THAT'S FUNNY

Shout "FIRE!!!!!!!!!!!!!" and shoot everyone with toy water guns.

Shout "AVALANCHE!!!!!" adn run away from the balls that were just set loose.

Open a bowling ball and a 12 pack and keep bowling until one of the following occurs:

I. You get a strike
II. The store security comes.
III. I and II happen.

Then it's your show!!! :lol:

Go and try on some pants that are WAY too large. Then complain to the person there that the *insert your size pants here* are way to large now a days and stomp off before he/she can read the size.

Set up a stand where you teach people the derivatives of polynomial functions of degree 5 or higher.

Say that the cashiers should integrate the sum of all the stuff you buy. If they ask huh?, then just start opening up everything and show them that integration is not necessary any more :D

Yodel. Really loud. Until someone cuts you off on the loudspeaker.

Something else that's cool:

Get fat:

[hide=" :maybe: "]As you see the ladies that are passing out samples for food, start grabbing all of them and stuff them up down your throat when they are busy or something.[/hide]

 :D

Shout "RUN!!! IT'S CHUCK NORRIS!!!!!". That should leave you with enough space to play frisbee in the store.


inequalities
geometry
inequalities solved

a,b,c>=0 . Show  that
(a+b+c)/3- cuberoot(abc) <= max{ (sqrta-sqrtb)^2 , (sqrtb-sqrtc)^2 , (sqrtc-sqrta)^2 }

thanks

We can obtain a^3 + b^3 + c^3 + 3abc  >= a^2 b + a b^2 + b^2 c + b c^2 + c^2 a + c a^2 >= 2  \sum (ab)^(3/2)

a^3=x, b^3=y, c^3=z

then x + y + z + 3 (xyz)^(1/3) >= 2 \sum  \sqrt xy

<=>  2(x + y + z) - 2 \sum  \sqrt xy >= x + y + z - 3 (xyz)^(1/3)

<=>   \sum ( \sqrt x -  \sqrt y)^2 >= 3 ( (x+y+z)/3 - (xyz)^(1/3) )

If all of  (\sqrt x - \sqrt y)^2 is smaller than (x+y+z)/3 - (xyz)^(1/3) ,

Adding all the inequality, \sum ( \sqrt x -  \sqrt y)^2 < 3 ( (x+y+z)/3 - (xyz)^(1/3) ), contradiction.

Therefore (x+y+z)/3 - (xyz)^(1/3) <= max { (\sqrt x - \sqrt y)^2 }.

Another way to solve this problem :
     We only need to solve this problem in the case : 0  \leq  a  \leq  b  \leq  c
      The inequality becomes :
            a + b + c - 3(abc)^(1/3)  \leq  3( \sqrt a -  \sqrt c) 2
      <=> 2a + 2c - b + 3(abc)^(1/3) - 6 \sqrt (ac)  \geq  0
      Now we fix a,c and denote
         f(b) = 2a + 2c - b + 3(abc)^(1/3) - 6 \sqrt (ac)   ( a  \leq  b  \leq  c )
         f'(b) = (ac)^(1/3)b^(-2/3) - 1
         f'(b) = 0 <=> b =  \sqrt (ac)
         f'(b) > 0 when a  \leq  <  \sqrt (ac)
         f'(b) < 0 when  \sqrt (ac) < b  \leq  c
       Hence : f(b)  \geq  Min { f(a) , f(c) }
           f(a) = a + 2c + 3(a 2 c)^(1/3) - 6 \sqrt (ac)
       We note that : c + 3(a 2 c)^(1/3)  \geq  4 \sqrt (ac) ( Cauchy )
        So : f(a)  \geq  a + c - 2 \sqrt (ac) = ( \sqrt a -  \sqrt c)^2  \geq  0
        Similarity : f(c)  \geq  0
      => f(b)  \geq 0 when a  \leq  b  \leq  c
      => qed

In fact, we can give a geometric proof for 2sum sqrt(xy)-sum x<=3 (xyz)^1/3. Let x=a^4 and so on. Then LHS is (a+b+c)(a+b-c)(b+c-a)(c+a-b). Clearly, we can assume that a,b,c are sides of a triangle. The(a+b+c)(a+b-c)(b+c-a)(c+a-b) is 16S^2 (S is the area)=a^2b^2c^2/R^2<=9a^2b^2c^2/(a^2+b^2+c^2)<=3(abc)^4/3.

[quote="nickolas"]
$\frac{x+y+z}{3}-\sqrt[3]{xyz}\leq\max\{(\sqrt{y}-\sqrt{z})^{2}, (\sqrt{z}-\sqrt{x})^{2}, (\sqrt{x}-\sqrt{y})^{2}\}$
[/quote]

you also can find a short solution in my article here : [url]http://reflections.awesomemath.org/2007_1/2007_1_nice.pdf[/url], problem A5.

[quote="pohoatza"][quote="nickolas"]
$\frac{x+y+z}{3}-\sqrt[3]{xyz}\leq\max\{(\sqrt{y}-\sqrt{z})^{2}, (\sqrt{z}-\sqrt{x})^{2}, (\sqrt{x}-\sqrt{y})^{2}\}$
[/quote][/quote]

I think it exists more easier solutions;)

it`s enough to prove that x+y+z/3 - sqrt[3]{xyz} is greater or equal than one of them, because they are symmetrical.

here it is the generalization:

(x1+x2+...+xn)/n - sqrt[n]{x1*x1*...*xn} is greater or equal than max from :  [sqrt(x1) - sqrt(x2)]^2 /n and the others. the prove is simple: use AM-GM


function
integration
probability and stats

Assume we want to choose n random numbers all of which is between 0 and sigma (by a program).how can I find the sum  i*y(i) over 1 to n , in terms of n and sigma, or the sum y(i)^k where y(i) is the ith number chosen and k is a positive integer. Of course I ask the average values af these sums(or assume n is very large; for example we can say that the sum of y(i) is n*sigma/2 in average). Is there a more general answer for some other sums? I estimate the sum i^m*y(i) to be n^(m+1)*sigma/2(m+1) by experience; but I am neither sure of that nor know how to actually derive it.

Three things to know:

1) $ E[cX]\equal{}c EX$

2) $ E[X_1\plus{}\dots\plus{}X_n]\equal{}EX_1\plus{}\dots\plus{}EX_n$

3) If $ f$ is the density function of a random variable $ X$ and $ g$ is any function, then $ E[g(X)]\equal{}\int_\mathbb R g(x)f(x)\,dx$.

These three rules plus the standard formulae for the sums like $ \sum_{i\equal{}1}^n i^k$ (see [url=http://mathpages.com/home/kmath279/kmath279.htm]here[/url] or [url=http://planetmath.org/encyclopedia/SumOfKthPowersOfTheFirstNPositiveIntegers.html]there[/url])should be enough to figure the answer to all your questions out :).


We denote 2 polynoms $f,g\in\mathbb{R}[X]$ like $f(X)=X(X+1)^{8n+1}+(m-1)^{4n}$ and $g(X)=X^{2}+X+1$. Find all real $m$ such that $g/f$. Since I'm not so good at solving with mods this kind of pb. I could need some help(I got stucked at some point). Or any other solution/hind I would apreciate :). Thanks!

If $g/f$ iff $f(\epsilon)=0,f(\epsilon^2)=0$ .... cheers :P

[hide]Let $\omega$ be a root of $g$; then $\omega$ must be a root of $f$. Since $\omega +1=-\omega^2$, $f(\omega )=\omega (-\omega^2)^{8n+1} +(m-1)^{4n}$.

Also note that $\omega^3=1$. Then $f(\omega )=-(\omega^2)^{2n}+(m-1)^{4n}=(m-1)^{4n}-\omega^n$.

If $f(\omega )=0$ then it must be real, so given $m$ real, $n$ is a multiple of $3$. For such $n$, $f(\omega )=(m-1)^{4n}-1\Rightarrow (m-1)^{4n}=1\Rightarrow \boxed{m=0,2}$.[/hide]


Let $ A\equal{}\{A_1,\dots,A_m\}$ be a family distinct subsets of $ \{1,2,\dots,n\}$ with at most $ \frac n2$ elements. Assume that $ A_i\not\subset A_j$ and $ A_i\cap A_j\neq\emptyset$ for each $ i,j$. Prove that:
\[ \sum_{i\equal{}1}^m\frac1{\binom{n\minus{}1}{|A_i|\minus{}1}}\leq1\]

[hide="Hint"]Do the same as in the proof of the Erdos-Ko-Rado theorem.[/hide]

By a [i]chain[/i], we denote a sequence of $n$ subsets $C_1, \cdots, C_n \subset [n]$ such that $C_i\subset C_{i+1}$ for all $1\le i < n$ and $|C_i| = i$. We double count pairs $(A_i, C)$, where $C$ is a chain that contains $A_i$. On one hand, we have that for each $A_i$, there are $(|A_i| !)(n - |A_i|)!$ chains that contain it. This gives
$$\sum_i|A_i|!(n-|A_i|)!$$ pairs.

We say that a permutation $\sigma$ of $[n]$ [i]induces[/i] a chain $C$ if for all $1\le i < n$, $~C_{i+1} \setminus C_i = \{\sigma(i)\}$. Say a chain $C$ contains $A_k$. Let $(c_1, c_2, ..., c_n)$ be the sequence that induces $C$. The key observation is that a chain induced by any sequence of the form $(d_1, d_2, ... d_{n-|A_k|}, c_1, c_2, ..., c_{|A_k|})$ does not contain any element of $A$. If it contained $A_i$, then we would have that $|A_i| + |A_k| > n$, since $|A_i|\cap |A_k| \neq \emptyset$, which contradicts the problem statement. Therefore, any chain containing $A_k$ also generates at least $\big [ \frac{n}{2}\big ] !$ distinct chains not containing any $A_i$.

Now, any chain contains at most one $A_i$. Associated with each such chain is a unique set of at least $\big [ \frac{n}{2}\big ] !$ chains not containing any $A_j$. There are a total of $n!$ chains.Therefore, there are at most $n! / \big [ \frac{n}{2}\big ] !$ pairs. This is at most $(n-1)!$ for $n \ge 6$. In this case, we have that
$$(n-1)! \ge \sum_i|A_i|!(n-|A_i|)! > \sum_i(|A_i|-1)!(n-|A_i|)! $$
Dividing by $(n-1)!$ gives the desired inequality. it remains to prove this when $n < 6$, which is easily done by exhaustion of cases.


[quote="Multiple"]Suppose I want to write just "Hello" in one page.

How can I center the word?

And how can I delete the first page number?[/quote]

To center stuff, try
[code]\begin{center}
Hello
\end{center}[/code]

[url]http://web.engr.oregonstate.edu/~kaj/ltxtips.html#others[/url]
[quote="That site"]How to eliminate page numbers
Here is the word from Péter. (Thanks Péter)

To ignore page numbers:

[code]\pagestyle{empty}[/code]

but which will still keeps the page number for the first page. To ignore
that one also you should add to the first page (only):

[code]\thispagestyle{empty}[/code]
[/quote]

Yeah, thanks

But also center vertically, how do I do that?

[code]\thispagestyle{empty}
\null
\vfill
\centerline{Hello}
\vfill
\null
\newpage [/code]

Thank you

But I can't set this for example

[img]http://img503.imageshack.us/img503/8351/psvv2.png[/img]

How could be there?

But you changed the question! You only wanted one word. For more lines you need the center environment. It's not clear what you want.
1. Your name etc centered below the A<B text
[code]\thispagestyle{empty}
\null
\vfill
\begin{center}
$A<B$\\
$A$ is less than B
\vspace{6ex}
Kiran Kedlaya\\
based on the notes for the Math Olympiad Program (MOP)\\
Version 1.0, last revised August 2,1999
\end{center}
\vfill
\null
\newpage [/code]Change \vspace{6ex} to give the vertical spacing you want.

2. Your name etc at the bottom of the page
[code]\thispagestyle{empty}
\null
\vfill
\begin{center}
$A<B$\\
$A$ is less than B
\vfill
Kiran Kedlaya\\
based on the notes for the Math Olympiad Program (MOP)\\
Version 1.0, last revised August 2,1999
\end{center}
\newpage [/code]

Sorry!

I thought that it'd work as the same way.

Now works fine!

Thanks stevem.


geometry
3D geometry
sphere
calculus
integration
LaTeX
calculus computations

I am having some trouble with this problem and the final is in 2 days.

Find an equation for the plane through the origin such that the circulation of the flow field F=zi+xj+yk around the circle of intersection of the plane with the sphere x^2+y^2+z^2=4 is a maximum.

Thanks

Use Stokes' theorem: the circulation is equal to the flux of $ \nabla\times\vec F$ through the disk that is the intersection of the plane with the solid sphere $ x^2+y^2+z^2\le 4$. It's important to realize that $ \nabla\times\vec F$ is a constant field (it does not depend on $ x,y,z$). I'll leave the rest to you.

Sorry, but i am still not making the connection.  I have been trying to use stokes, but do not see how to integrate over a surface which the problem is asking that I find.

You want to find $ \iint_S (\nabla\times\vec F)\cdot\,d\vec S$. This integral can also be written as $ \iint_S ((\nabla\times\vec F)\cdot \vec n)\,dS$ where $ \vec n$ is the unit normal to $ S$. The crucial observation is that the expression that you integrate is a constant. Therefore, the integral is equal to the value of that constant multiplied by the area of $ S$. You can easily find the area of $ S$, which is a disk of radius $ 2$. The problem boils down to choosing $ \vec n$ (which amounts to rotating the plane) so that $ ((\nabla\times\vec F)\cdot \vec n)$ is as large as possible.

Thank you, that's basically what I was trying, but did not realize that i could just multiply the SA by the product of F and n.  I was trying to come up with some limits that would yeild the SA of the circle dependent on n, which I defined as ai+bj+ck, but as you can see, I was not having much success.

Btw, how do you put equations into your texts?  I tried to copy from Microsoft Equation, but was without success.

The pretty math on this forum is $ \text{\LaTeX}$. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261]here[/url] for more.


[hide="hint"]A regular hexagon can be seperated into 6 equilateral triangles.[/hide]

[quote="DaReaper"]if ABCDEF is a regular hexagon inscribed in a circle with a radius of 5, find the area of triangle ACE.[/quote]
[hide="solution"]
Imagine the hexagon as 6 equilateral triangles, each with a side length of 5.
ACE would cut all 6 triangles in half, so it's the area of 3 of those equilateral triangles.
$ A\equal{}\frac{(3)(5)(\frac{5\sqrt{3}}{2})}{2}$
$ A\equal{}\frac{75\sqrt{3}}{4}$
[/hide]

why are you multiplying 3 and 5 together, and how did you find the length of the base?

EDIT**** NEVER MIND, i figured it out*****  THANK YOU !!

actually never mind... how do you find the length of the base? :huh:

Since one of the sides of the equilateral triangle is one of the sides of the hexagon, all the sides have length 5.


Let be $ f: N\minus{}\minus{}\minus{}> N$  a function such that :  (N-natural numbers)

a) For each natural $ a$ and $ b$ we have $ f(ab)\equal{}f(a)f(b)$

b)For each natural $ a$ and $ b$ if $ a<b$ imply $ f(a)<f(b)$

c) $ f(3)\ge 7$

Find the minimal value of $ f(3)$

$ f(n)\equal{}n^2$ satisfies the conditions and $ f(3)\equal{}9$
Now we have to prove that $ f(3)>8$
 If $ f(3)\equal{}7$ or $ f(3)\equal{}8$:
Clearly $ f(1)\equal{}1$ and then $ f(2)>1$
$ f(2)^2\equal{}f(4)>f(3)$ And this implies $ f(2)>2$
$ f(3)^2\equal{}f(9)>f(8)\equal{}f(2)^3$ so $ f(2)<4$
Then we get $ f(2)\equal{}3$

But also $ 2400\equal{}7^4\minus{}1<f(3)^4\equal{}f(81)<f(128)\equal{}f(2)^7\equal{}3^7\equal{}2187$ contradiction
So $ f(3)\equal{}9$ is the minimal value

Nice problem

It is well-known (from Cauchy equation) that for $ f(xy)\equal{}f(x)f(y)$ the only solutions are $ f(x)\equal{}x^p$. Because of the $ \text{b} )$, $ p \geq 1$. We have that $ f(3) \equal{} 3^p \geq 7$, so, because $ 3^p$ is natural, $ p \geq 2$, $ p \equal{} 2$ and $ f(3)\equal{}9$, $ q.e.d.$

[quote="jevgeniy"]It is well-known (from Cauchy equation) that for $ f(xy) \equal{} f(x)f(y)$ the only solutions are $ f(x) \equal{} x^p$[/quote]

Maybe it's well known, but it's unfortunately wrong.

It exists some functions from $ \mathbb N\to\mathbb N$, different from $ x^p$ and such that $ f(xy) \equal{} f(x)f(y)$

In order to obtain $ f(x) \equal{} x^p$, you need to be in $ \mathbb Q$


calculus integration trigonometry logarithms Putnam function

Let $I = \int^{1}_{0}\frac{\arctan x}{x+1}dx$.  We use the substitution $y = \tan^{-1}x$ to get

$I = \int_{0}^{\frac{\pi}{4}}\frac{y\sec^{2}y}{1+\tan y}\, dy = \int_{0}^{\frac{\pi}{4}}y\, d(\ln (1+\tan y))$

$= y\ln (1+\tan y)\Big|_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\ln (1+\tan y)\, dy = \frac{\pi}{4}\ln 2-I_{1}$ (say)

Now, $I_{1}= \int_{0}^{\frac{\pi}{4}}\ln (1+\tan y)\, dy = \int_{0}^{\frac{\pi}{4}}\ln (1+\tan \left( \frac{\pi}{4}-y \right))\, dy$
...[Using the identity $\int_{0}^{a}f(x)\, dx = \int_{0}^{a}f(a-x)\, dx$]

$\therefore I_{1}= \int_{0}^{\frac{\pi}{4}}\ln (1+\frac{1-\tan y}{1+\tan y})\, dy = \int_{0}^{\frac{\pi}{4}}\ln(\frac{2}{1+\tan y})\, dy = \ln 2 \int_{0}^{\frac{\pi}{4}}\, dy-I_{1}$

$\Rightarrow 2I_{1}= \frac{\pi}{4}\ln 2 \Rightarrow I_{1}= \boxed{\frac{\pi}{8}\ln 2}$

So, $I = \frac{\pi}{4}\ln 2-\frac{\pi}{8}\ln 2 = \boxed{\frac{\pi}{8}\ln 2}$

My guess is that the idea for this integral is drawn from [b]Putnam A5/5[/b]. Here is the discussion on that problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436

Thanks, [b]Peter VDD[/b], for sharing the problem.

the first test of the brazilian undergraduated competition from 2005 i guess, has this problem, but instead of beggining from the beggining of this problem, it begins from $\int_{0}^{\pi/4}ln(1+tgx)dx$

ugh... so they copy from the putnam competition FROM THE SAME YEAR?  :|

hehehe maybe they were inspired by the putnam competition, this problem was proposed in 2005, but i don´t think that this is a real copy, i´ve just seen the 2005 putnam test, but about the brazilian competition, the always like to take some integrals that we may notice that the function is odd, so we could use some tricks to find it´s values. But.... who knows?!?!?!  :roll:

[quote="Brambila"]so we could use some tricks to find it´s values[/quote]In fact, it's not just an alike trick. I found 4 equivalent versions (only posted the best looking one), one being the $\ln(1+\tan x)$ which FieryHydra even used as step in his proof. They're completely equivalent, so that would be unfair advantage for pupils knowing that question.

$J\equiv\int_{0}^{1}\frac{\ln (1+x)}{1+x^{2}}\ dx=\int_{0}^{\frac{\pi}{4}}\ln (1+\tan x)\ dx=$ $\int_{0}^{\frac{\pi}{4}}\ln\left[1+\tan\left(\frac{\pi}{4}-x\right)\right]\ dx=$ $\int_{0}^{\frac{\pi}{4}}\ln\left(1+\frac{1-\tan x}{1+\tan x}\right)\ dx=$ $\int_{0}^{\frac{\pi}{4}}\ln\frac{2}{1+\tan x}\ dx=$ $\frac{\pi}{4}\cdot \ln 2-J$ $\Longrightarrow$ $\boxed{\ J=\frac{\pi}{8}\cdot\ln 2\ }$ $\Longrightarrow$ $I\equiv\int_{0}^{1}\frac{\arctan x}{1+x}\ dx=$ $\left[\arctan x\cdot ln (1+x)\right]_{0}^{1}-\int_{0}^{1}\frac{\ln (1+x)}{1+x^{2}}\ dx=$ $\frac{\pi}{4}\cdot\ln 2-J$ $\Longrightarrow$ $\boxed{\ I=\frac{\pi}{8}\cdot \ln 2\ }\ .$
$\boxed{\ \begin{array}{ccc}u(x)=\arctan x & \Longrightarrow & u'(x)=\frac{1}{1+x^{2}}\\ v'(x)=\frac{1}{1+x}& \Longrightarrow & v(x)=\ln (1+x)\end{array}\ }\ .$

[b]Remark.[/b] This problem is very old. I know it from 1975 ! In the our gazette G.M.B. appeared and an article on this theme.

[quote][b]Remark.[/b] This problem is very old. I know it from 1975 ! In the our gazette G.M.B. appeared and an article on this theme.[/quote]
Hehe, it is much older. It is an example in Fihtengolz's mathematical analysis textbook, which is published before 1950, I am afraid.

[quote="Peter VDD"][quote="Virgil Nicula"]$1+\tan\left(\frac{\pi}{4}-x\right)=\frac{2}{1+\tan x}$[/quote]Interesting. Is this a known identity, or did you come up with this somehow? I have tried precisely that way, but I thought their integrals were just equal without the functions being equal.[/quote]
We have
$\tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$.

in fact, i was checking some dates, and found that while the brazilian contest proposed this problem on september/2005, the putnam contest proposed it on december/2005


Duke college MATHCOUNTS email percent Harvard

On the [url=http://www.wcu.edu/2575.asp]Western Carolina University Math Contest website[/url] there is the following tidbit:

[quote]Students scoring in the top 7 percent in the Comprehensive division qualify for the State Mathematics Contest April 19 at the School of Science and Mathematics in Durham.

[color=blue]The top 20 finishers in the state are offered full scholarships to attend a North Carolina college or university of their choice, including WCU.[/color] [/quote]

Anyone know if there is any truth to the full scholarships?  Seems like an incredibly good deal.  Might this be an April Fool's joke that was left on the site for a few extra days?  Ralph Willis, who runs the WCU contest, is well connected and would know about prizes, if any exist.

The Duke scholarship to the highest scoring senior is the only one I remember from past NC state math contests.

I thought the #1 person got a full scholarship iff they were a senior... :maybe:

The full scholarship to Duke is for the highest scoring senior in the top 10.

That is pretty amazing, but remember that everybody at NCSSM gets full scholarships as well, so they're really pretty free with their scholarships. And no offense to the state schools, but just looking at the facts, most of the top 20 do not take those scholarships anyway.

Even if that's not true, I imagine any of those people could get full scholarships or close to it to most state schools anyway.

I'd agree that many of the scholarships would go unused if given to the top 20 at state contest.  However, I was talking to Drew last month and he seemed inclined to take the Park scholarship to NC State.  Other top math students might be interested in the UNC system scholarships if they were available.

The scholarships would be a good recruiting tool to increase serious contest participation.  We've seen rather significant declines in math contest participation in the last few years.  The number of schools is decreasing as well as the number of students.  AMC participation is off at least 30% from 5 years ago.  Some of the local state qualifying events are down as well.  I strongly suspect that less than half of the top math students are being identified through math contest success.

Purdue gives full scholarships to the top 10 Indiana MathCounts students each year.  I don't know how many actually go to Purdue 4 years later - probably not many.  However, it starts a competitive math culture.  Considering it's size, Indiana is one of the most competitive states in mathematics - especially in middle school.  From the perspective of the state, the scholarships are a relatively low-cost way to produce much better top students (assuming that not many take up the scholarship offers.)

It would be interesting to know the percentage of NCSSM students who take advantage of the scholarships to UNC system schools.  I'd suspect less than 1/3.

what is WCU? edit 2 - just saw top line

20 scholarships to top math students seems pretty reasonable - I mean, look, maybe a fourth of the kids at NCSSM are from out west and I could cream them in math if I was a comatose paraplegic, man.  they all get full rides to UNC schools. (if they get accepted - if you look closely at the stats, there was one kid who applied to UNC-CH who did not get accepted  :rotfl: )

edit - yay, this brought me up to 100 posts and two whole stars.  now I'm a millenium problem that HAS been solved! (wow that almost rhymes.  very sad)

According to John Goebel, state math contest coordinator, the Western Carolina University web reference to the top 20 getting full scholarships is not entirely correct.

[quote="John Goebel"]Many of the colleges in NC will offer some or in the case of Duke, full tuition scholarships to the top students at the State Math Contest.  You can see our webpage for the list of schools that have offered such scholarships.  To qualify, a student must be admitted to the school and win in their senior year.  If you are interested, you should contact the individual school and get more details.
[/quote]

Looks like Duke is still the only school to offer a [color=blue]full[/color] scholarship based solely on performance at State Math Contest.  To receive the scholarship, you must have applied and been accepted to Duke and be the highest scoring senior on the NC State Comprehensive Math Contest.

Perhaps with a bit of work, we could sweeten the prize pool at state math contest.

There's been a recent trend toward prizes for specific problem solving activities and technological achievements.  The logic is that those chasing the prize money will get the problem solved faster than other means.  The same logic might apply to contest prizes.

In response to the question about how many NCSSM students use the tuition grant, I just got these statistics in an email:

-Since 2004, an average of 82 percent of NCSSM graduates have attended UNC system schools, up from an average of 55 percent prior to 2004.
-NCSSM currently has alumni at 15 of the 16 University of North Carolina constituent institutions.
-All but six members of NCSSM’s Class of 2007 applied to a UNC system school.

That sounds about right.
Tuition to out-of-state schools (Ivy League, other state schools, and others) is a lot, and costs are rising. Some would still want to go out-of-state, but a lot either can't afford it, or just want to stay near home. And since students at NCSSM seem to be pretty good, they might have gone to a prestigious boarding school if they could afford it and were fine about being very far away from home. Plus they can always go out-of-state for graduate school...

I think it's more like the average S&M student looks at UNC as their best school they can get into. Mainly because the ones above them won't accept 85% of any high school, and sm isn't as accomplished as TJ or exeter.

Actually, it's funny what has happened as a result of the tuition grant. Normal students are choosing UNC or NC State because that's where they were likely to go anyways, and now this grant has removed all doubt. My friend quoted some stats saying that the average SM GPA at UNC is around 3.00 or 3.05, which indicates that the school doesn't do well in college at all, compared with enloe or east. But those schools don't send 85% of their students to unc/state, so it is a bit biased, and a greater percentage of SM top students go to duke or ivy league.

So overall, I don't think much changed; the difference between 85% and 55% is the non-top-tier group of students at SM picking UNC instead of Wake or UVA or other random schools like that, not top-tier students choosing UNC instead of say Harvard or MIT.


Find all second degree polynomial $d(x)=x^{2}+ax+b$ with integer coefficients, so that there exists an integer coefficient polynomial $p(x)$ and a non-zero integer coefficient polynomial $q(x)$ that satisfy: \[\left( p(x) \right)^{2}-d(x) \left( q(x) \right)^{2}=1, \quad \forall x \in \mathbb R.\]

By "nonzero" polynomial do you mean a polynomial which never equals zero? If so...

$d(x) = \frac{(p(x))^2 - 1}{(q(x))^2} = x^2 + ax + b$.

$deg(p(x)) - deg(q(x)) = 2$ which yields:

$deg(p(x)) = 2, \ deg(q(x)) = 1$

or

$deg(p(x)) = 1, \ deg(q(x)) = 0$.

We must discard the first possibility since q(x) would have a zero.
Thus, $p(x) = ax+b, \ p(x) = a^2x^2 + 2abx + b^2, \ q(x) = a \ne 0$

$d(x) = \frac{a^2x^2 + 2abx + b^2-1}{a^2} = x^2 + \frac{2b}{a}x + \frac{b^2-1}{a^2}$

The third term yields $b^2 = na^2+1$ which I daresay has only the solutions $b = 1, \ -1$ corresponding to n = 0 (please correct me if I err).

It follows from the second term that $a = \pm 1, \ \pm 2$. Thus,
$d(x) = x^2 \pm x$ and $d(x) = x^2 \pm 2x$.

If "nonzero" means that q is nontrivial, then we have some additional work to do.

I think he mean that non of coefficients of $q(x)$ is zero. :D

[quote="mathisfun1"]By "nonzero" polynomial do you mean a polynomial which never equals zero? If so...

$d(x) = \frac{(p(x))^{2}-1}{(q(x))^{2}}= x^{2}+ax+b$.
$deg(p(x))-deg(q(x)) = 2$ which yields:

$deg(p(x)) = 2, \ deg(q(x)) = 1$
or
$deg(p(x)) = 1, \ deg(q(x)) = 0$.
[/quote]
Why?

[quote="mathisfun1"]By "nonzero" polynomial do you mean a polynomial which never equals zero? If so...

$ d(x) \equal{} \frac {(p(x))^2 \minus{} 1}{(q(x))^2} \equal{} x^2 \plus{} ax \plus{} b$.

$ deg(p(x)) \minus{} deg(q(x)) \equal{} 2$ which yields:

$ deg(p(x)) \equal{} 2, \ deg(q(x)) \equal{} 1$

or

$ deg(p(x)) \equal{} 1, \ deg(q(x)) \equal{} 0$.

We must discard the first possibility since q(x) would have a zero.
Thus, $ p(x) \equal{} ax \plus{} b, \ p(x) \equal{} a^2x^2 \plus{} 2abx \plus{} b^2, \ q(x) \equal{} a \ne 0$

$ d(x) \equal{} \frac {a^2x^2 \plus{} 2abx \plus{} b^2 \minus{} 1}{a^2} \equal{} x^2 \plus{} \frac {2b}{a}x \plus{} \frac {b^2 \minus{} 1}{a^2}$

The third term yields $ b^2 \equal{} na^2 \plus{} 1$ which I daresay has only the solutions $ b \equal{} 1, \ \minus{} 1$ corresponding to n = 0 (please correct me if I err).

It follows from the second term that $ a \equal{} \pm 1, \ \pm 2$. Thus,
$ d(x) \equal{} x^2 \pm x$ and $ d(x) \equal{} x^2 \pm 2x$.

If "nonzero" means that q is nontrivial, then we have some additional work to do.[/quote]
Sorry,but I don't understand a few things.
We get $ degP\minus{}degQ\equal{}1$,why did you just consider the case $ degP\equal{}2,degQ\equal{}1$,what about the other cases?Like $ degP\equal{}3,degQ\equal{}2$?Thanks.

That is my solution. Maybe too long
If a is odd then $4b=a^{2}-1$ and if a is even then $b=\frac{a^{2}}{4}+k$ with $k\mid 2$
First:
if $a=2r$
$d(x)= (a+r)^{2}+k$ here $k=b-r^{2}$
Let $f(x)=P(x+r), g(x)=Q(x+r)$ then we have: $(P(x))^{2}-(x^{2}+k)(Q(x))^{2})=1$
$R(x)=P(x)-xQ(x)$ we have $R(x).(2xQ(x)+R(x))=1+k(Q(x))^{2}$(*)
Easy to see that: If $deg(P)=2$ then only root for (*) in $Q_{[x]}$ is $Q(x)=\frac{2}{k}.x, P(x)=1+2/k.x^{2}$
                                 we have root in $Q_{[x]}$ with deg(P)=1 if and only if $k=-1/m^{2}$ and $m\in Q^{+}$ and only root in that case is;   $P(x)=mx.Q(x)=m$
Let $P_{1},Q_{1}$ is root of (*) in $Q_{[x]}$ and $deg P$ is min and $Q_{1}$ not equal 0
 If $deg(P_{1})=n\geq 3$
    $R_{1}(x).(2xQ_{1}(x)+R_{1}(x))=1+k(Q_{1}(x))^{2}$(**)
Let $Q^{*}(x)=\frac{2xR(x)}{k}-Q(x)$ with (**)then $deq Q^{*}=n-2>0$ and $R(x), Q^{*}$ is a root of (**) hence $Q^{*}, P^{*}$ is a root of (*)
$P^{*}= xQ^{*}+R(x)$
And $deg(P^{*})=n-1$ .That is not true.

Then $deqP_{1} \leq 2$
    If k not equal $-m^{2}$ , $m \in Z^{+}$
Hence  $Q_{1}(x)=\frac{2}{k}.x, P_{1}(x)=1+2/k.x^{2}$ is minnimal root
Introduce with deq P we have that:

All roots for (*) in $Q_{[x]}$

$P_{n}(x)=2P_{0}P_{n-1}(x)-P_{n-2}(x), Q_{n}(x)=2P_{0}Q_{n-1}(x)-Q_{n-2}(x)$ and $P_{0}=1, Q_{0}=0,$  $Q_{1}(x)=\frac{2}{k}.x, P_{1}(x)=1+2/k.x^{2}$
and Easy too see max degree coefficient of $P_n$ if $\frac{2^{2n-1}}{k^{n}}$ then (*) have root in $Z_{[x]}$ iff $k\mid 2$
      if $k=-m^{2}$ similar we have
All roots for (*) in $Q_{[x]}$

$P_{n}(x)=2P_{0}P_{n-1}(x)-P_{n-2}(x), Q_{n}(x)=2P_{0}Q_{n-1}(x)-Q_{n-2}(x)$ and $P_{0}=1, Q_{0}=0,$ $P(x)=\frac{1}{m}x, Q(x)=\frac{1}{m}$

 max degree coefficient of $P_n$ if $\frac{2^{n-1}}{m^{n}}$ then (*) have root in $Z_{[x]}$ iff $m\mid 2$ and $\mid m \mid <2$
Hence $m=1$ then $k=-1$
End of proof with a is even.
Case of a is odd similar, note more $k=4b-a^{2}=4r-1$

 If a is odd then $4b=a^{2}-1$ and if a is even then $b=\frac{a^{2}}{4}+k$ with $k\mid 2$
Easy too see a solution: $Q(x)=\frac{2}{k}.x, P(x)=1+2/k.x^{2}$ if a even
                                      $P(x)=mx+n$ with a odd
QED :)


pigeonhole principle
function
modular arithmetic
geometry
3D geometry
number theory
totient function

[hide="hint"]Use [url=http://en.wikipedia.org/wiki/Euler%27s_theorem]Euler's Theorem[/url] to prove more generally that $ 3^{4\cdot 10^{k \minus{} 1}} \equiv {1 \mod 10^k}$ for any integer $ k$.[/hide]

[quote="JoeBlow"][hide="hint"]Use [url=http://en.wikipedia.org/wiki/Euler%27s_theorem]Euler's Theorem[/url] to prove more generally that $ 3^{4\cdot 10^{k \minus{} 1}} \equal{} {1 \mod 10^k}$ for any integer $ k$.[/hide][/quote]

Oh yes ,Definitely , that would solve the problem . !!! THANKS FOR THE HINT !!

Bye the way , Can you please Tell me , how could one use Pigeonhole to Solve this Problem !!!?
 :?:

[hide="Step 1"] [b]Claim:[/b]  The sequence $ 3^n \bmod 1000$ is periodic.

[i]Proof.[/i]  The set $ \{ 3^0, 3^1, ... 3^{1000} \}$ contains at least two members which are the same by Pigeonhole, hence there exist $ i, j$ less than or equal to $ 1000$ such that $ 3^i \equiv 3^j \bmod 1000$.[/hide]
[hide="Step 2"][b]Claim:[/b]  If $ 3^i \equiv 3^j \bmod 1000$ then $ 3^{i \minus{} j} \equiv 1 \bmod 1000$.

[i]Proof.[/i]  Since $ 3 \cdot 333 \equal{} 999 \equiv \minus{} 1 \bmod 1000$ it follows that $ 3 \cdot \minus{} 333 \equiv 1 \bmod 1000$, so the multiplicative inverse exists.  (Generally, this is possible because $ \gcd(3, 1000) \equal{} 1$.) [/hide]

[quote="amangupta2"]Can anyone Please Help me out with this Questions:->

Q) Prove that there exists a Power of 3 that ends with the digits 001 ( in Base 10)

** I just know that pigeonhole principle can be applied here but how to apply it ?????[/quote]

[hide="Another way"]
In fact,
Use Carmichaei function $ \lambda(1000)\equal{}100$

$ 3^{100}\equiv 1 \pmod {1000}$
[url]http://en.wikipedia.org/wiki/Carmichael_function[/url][/hide]

[hide="Another way"]
In fact,
Use Carmichaei function $ \lambda(1000) \equal{} 100$

$ 3^{100}\equiv 1 \pmod {1000}$
[url]http://en.wikipedia.org/wiki/Carmichael_function[/url][/hide][/quote]

If I am Not Mistaken than isn't this Carmicheal Function Same as Order of an Integer Modulo N....???
If this is correct, then please can you tell me how did you calculate the order of 3 modulo 1000 , I mean is there any shortcut Formula

But Hey !!! A Big Thanks to all of you for Helping me , THANK YOU VERY MUCH !!!

[quote="amangupta2"]If I am Not Mistaken than isn't this Carmicheal Function Same as Order of an Integer Modulo N....???[/quote]
No.  The Carmichael function describes the [i]exponent[/i] of the group of units $ \bmod N$, which is the smallest positive integer $ \lambda(N)$ such that the order of every element divides $ \lambda(N)$.  (The totient function just describes the order of the group, which is usually strictly larger.)  It is true that there always exists an element with order the Carmichael function, but in general there isn't a shortcut calculation you can do to compute the order of a residue $ \bmod N$.

Thanks very much !! Your reply was really helpful !!   :)

I think there's another easy method:

Assume that there exists n^3=1000k+1
Thus (n-1)(n^2+n+1)=1000k
Thus we can let n=1001

A cube is not the same thing as a power of three; you want $ 3^n \equal{} 1000k \plus{} 1$.


calculus
geometric sequence
algebra proposed
algebra

Hello to everyone, i solved the following problem with computer programming, but in fact , i want to learn how it can be solved in purely mathematical ways:

We are given a  sequence a1, a2, ..., an  with    a1 = 1, and  an+1 = an   +  (1/an) for n>=2. Find the integer part of a1000( That is, find the number m  which satisfies   m< = a1000< m+1).
I'll be waiting for your comments/answers.

[quote="outsider"]

We are given a geometric sequence $a_1, a_2, ..., a_n$ with $a_1 = 1$, and  $a_{n+1} = a_n + \frac{1}{a_n}$ for $n \ge 2$..[/quote]

TeXed.  That's not a geometric sequence at all, though  :huh:

Well, I can certainly imagine that it could be done with some very ugly calculus to put bounds on $a_{1000}$...

Actually, it's quite simple. Here's a hint: we have $a_{k+1}^2=a_k^2+\frac{1}{a_k^2}+2>a_k^2+2$. Add this for $k=2,\ldots ,n$ to obtain $a_n>\sqrt{2n-1}$. Next, use  $a_{k+1}^2=a_k^2+\frac{1}{a_k^2}+2<a_k^2+2+\frac{1}{2k-1}$ to obtain an upper bound.
Evaluating $\sum{\frac{1}{2k-1}}$ can be done without calculus.
I'll post details later, if needed :)


linear algebra
matrix
SFFT
special factorizations

Let $A \in R^{n\times n}$, resolve : \[A X = A+X \] with $X \in R^{n\times n}$.

We could rewrite this as $(A-I)(X-I)=I.$

(Long-time AoPS users may recognize this as "Simon's favorite factoring trick", applied to matrices.)

If $A-I$ is singular (that is, if 1 is an eigenvalue of $A$,) then this has no solutions, as the determinant of the left hand side would be zero, which is not the determinant of $I.$

Assume $A-I$ is nonsingular. Then

$X=I+(A-I)^{-1}.$

Well trick.

$(A-I)X = A$, unique solution is obvious.

Changing basis we can use instead A its jordan matrix, seen other solution it's impossible (noncompatible: if a-1=a, then -1=0). The same but trick looks better (but it's trick :)).


Is the following statement true or false-
 The radius of a  spherical air bubble ,situated at the bottom of a lake of depth $8.5 m$ doubles if it rises to the top of the lake.

The bubble will inflate because the hidrostatic pressure (and thus the pressure inside the bubble) decreases. Assuming an isothermic process, the ratio of the volumes of the bubble at the bottom and at the top is $V'/V = p/p' = (p_{0}+\rho gh) / p_{0}= 1+\rho gh/p_{0}$, where $p_{0}$ is the atmospheric pressure. Assuming standard values for the constants, we get $V'/V \approx 1.8$, so the radius hasn't doubled.


Euler
geometry
incenter
circumcircle
geometric transformation
reflection
radical axis

Let $ H$ be orthocenter of triangle $ ABC$ and $ A_1B_1C_1$ is pedal triangle of $ H$ wrt $ ABC$, $ A_2B_2C_2$ is pedal triangle of $ H$ wrt $ A_1B_1C_1$. Take $ A_3,B_3,C_3$ on ray $ HA_2,HB_2,HC_2$ such that $ HA_3 \equal{} HB_3 \equal{} HC_3$.

a) Prove that radical center of circles $ (A_1,A_1A_3),(B_1,B_1B_3),(C_1,C_1C_3)$ lie on Euler line of triangle $ ABC$.

b) Prove that radical center of circles $ (A,AA_3),(B,BB_3),(C,CC_3)$ lie on Euler line of triangle $ ABC$.

Note that, compare with problems [url=http://www.mathlinks.ro/viewtopic.php?t=298310]Radical center again![/url] and [url=http://www.mathlinks.ro/viewtopic.php?t=297000]Radical center and line OI[/url]

Dear [b]buratinogigle[/b],
Thanks you for your nice generalization.
a, Similar the solution of [b]Luis[/b] in topic [url=http://www.mathlinks.ro/viewtopic.php?t=297000]Radical center and line OI[/url].Construct lines through $ A_3,B_3,C_3$ and perpendicular to $ HA_3,HB_3,HC_3$, respectively, they intersect and make triangle $ A'B'C'$. We can show that the radical center of three circles is the orthocenter $ P$ of triangle $ A'B'C'. P$ lies on a line which connects incenter $ H$ and circumcenter $ O'$ of triangle $ A'B'C'$, it's also Euler line of triangle $ ABC$.
b, We will show problem in case $ A_3\equiv A_2, B_3\equiv B_2, C_3\equiv C_2$. The generalization could be solved similarly if we construct triangle $ A'B'C'$.
[b]An equivalent problem:[/b]
Given triangle $ ABC$ with incenter $ I$, excenters $ I_a, I_b, I_c$, circumcenter $ O. (I)$ is tangent to $ BC, CA, AB$ at $ A',B',C'$, respectively. Prove that the radical center of $ (I_a,I_aA'), (I_b,I_bB'), (I_c,I_cC')$ lies on the line $ OI$.
[b]Proof:[/b]
Let $ A_1$ be the intersection of  $ (I_a,I_aA')$ and $ (I)$, similar for $ B_1,C_1. A'A_1, B'B_1, C'C_1$ intersect each other and make triangle $ A_2B_2C_2$.
Since $ A_2C_1.A_2C' \equal{} A_2B_1.A_2B'$ we get $ A_2$ lies on the radical axis of $ (I_b,I_bB')$ and $ (I_c,I_cC')$
But $ A'A_1\perp IA, IA\perp B'C'$ then $ B_2C_2//B'C'$, similarly we obtain $ (A'B'C')$ or $ (I)$ is the Nine-point circle of triangle $ A_2B_2C_2 \Rightarrow A_2A_1\perp B_2C_2$
$ \Rightarrow A_2A_1\perp I_bI_c$ or $ A_2A_1$ is the radical axis of $ (I_b,I_bB')$ and $ (I_c,I_cC')$
Therefore the orthocenter $ L'$ of triangle $ A_2B_2C_2$ is the radical center of three circles. $ L'$ lies on Euler line of triangle $ A_2B_2C_2$ or line $ OI$.
We are done.
[b]Remark:[/b]
If $ L$ is the radical center of $ (A,AA'),(B,BB'),(C,CC')$ we can easily show that $ L'$ is the reflection of $ L$ across point $ I$.

Dear [b]livetolove212[/b]
Thank you for your proof. Some my problems, I created them by myself but in my thinking so I don't have solution, this is an example. I will post my solution for the problems which I know solution.


[quote="arqady"]Solve in real numbers the following equation:
  $ x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}\frac{2}{5}\equal{} 0.$
Exists nice solution. :wink:[/quote]

[hide="One classical, but rather simple, solution"]Multipliying by $ (x\plus{}1)$, we have $ 5x^{7}\minus{}3x\plus{}2\equal{}0$. Let then $ f(x)\equal{}5x^{7}\minus{}3x\plus{}2$. $ f'(x)\equal{}35x^{6}\minus{}3$ has 2 zeroes $ \minus{}a$ and $ \plus{}a$ with $ a\equal{}(\frac{3}{35})^{\frac{1}{6}}$.
So $ f(x)\equal{}0$ has one negative root (obviously $ \minus{}1$) and zero, one or two positive roots depending on sign of $ f(a)$ :

$ f(a)\equal{}5a^{7}\minus{}3a\plus{}2\equal{}2\minus{}\frac{18}{7}a$ (since $ a^{6}\equal{}\frac{3}{35}$). And then it is easy to check that $ 2>\frac{18}{7}a$ : $ 7^{6}.35>3.9^{6}$ $ \implies$ $ (\frac{7}{9})^{6}>\frac{3}{35}$ $ \implies$ $ \frac{7}{9}>a^{6}$ $ \implies$ $ 2>\frac{18}{7}a$.

So $ f(x)$ has only one real root ($ \minus{}1$) which appears not to be a root of original equation and so original equation has no real roots.[/hide]

OK, pco. But exists elementary solution without derivatives. :wink:

[hide="Casework"]
For $ |x| > 1, x^{2n}> x^{2n\minus{}1}\implies x^{2n}\minus{}x^{2n\minus{}1}> 0$
$ \frac{2}{5}> 0\implies (x^{6}\minus{}x^{5})\plus{}(x^{4}\minus{}x^{3})\plus{}(x^{2}\minus{}x)\plus{}\frac{2}{5}> 0$
Hence no solution here.

For $ \minus{}1 < x\geq 0, x^{2n}> 0, x^{2n\minus{}1}< 0\implies x^{2n}\minus{}x^{2n\minus{}1}> 0$
Again, $ \frac{2}{5}> 0\implies (x^{6}\minus{}x^{5})\plus{}(x^{4}\minus{}x^{3})\plus{}(x^{2}\minus{}x)\plus{}\frac{2}{5}> 0$

Hence, $ 0 < x < 1$. We can now turn to $ (x\plus{}1)f(x) \equal{} 5x^{7}\minus{}3x\plus{}2 \equal{} 2x^{7}\plus{}3(x^{7}\minus{}x)\plus{}2$
It suffices to show that $ x\minus{}x^{7}<\frac{2}{3}$. Since the maximum occurs at $ x \equal{}\sqrt [6]{\frac{1}{6}}$, we can quickly evaluate this to be true. So, $ 2x^{7}\plus{}3(x^{7}\minus{}x)\plus{}2 > 3(x^{7}\minus{}x)\plus{}2 > 3\left(\minus{}\frac{ 2}{3}\right)\plus{}2 > 0$
$ \implies (x\plus{}1)f(x) > 0\implies f(x) > 0$
No solutions here.

Therefore, there are no real solutions to the equation.
[/hide]

[quote="A M Y"]
It suffices to show that $ x\minus{}x^{7}<\frac{2}{3}$. Since the maximum occurs at $ x \equal{}\sqrt [6]{\frac{1}{6}}$[/quote]
Maybe when $ x \equal{}\sqrt [6]{\frac{1}{7}}$ $ ?$ :wink:
And what if without differential calculus... :wink:

Notice that the expression is the same with :
$ x^{4}\left(x\minus{}\frac{1}{2}\right)^{2}\plus{}\frac{3}{4}x^{2}\left(x\minus{}\frac{2}{3}\right)^{2}\plus{}\frac{2}{3}\left(x\minus{}\frac{3}{4}\right)^{2}\plus{}\frac{1}{40}>0$
Immediately,it gives a slightly stronger result:
 $ x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}\frac{3}{8}> 0.$
Arqady: is this what you meant? :maybe: Or do you have another solution???

[quote="stephen38"]
Immediately,it gives a slightly stronger result:
 $ x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}\frac{3}{8}> 0.$
Arqady: is this what you meant? :maybe: Or do you have another solution???[/quote]
Nice result, stephen38!  :lol:
My solution:
if $ x\geq1$ or $ x\leq0$ then $ x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}\frac{2}{5}> 0$ obviously true.
Let $ 0 < x < 1.$ We obtain:
$ x^{6}\minus{}x^{5}\plus{}x^{4}\minus{}x^{3}\plus{}x^{2}\minus{}x\plus{}\frac{2}{5}\equal{}\frac{2}{5}\minus{}x(1\minus{}x)\minus{}x^{3}(1\minus{}x)\minus{}x^{5}(1\minus{}x) \equal{}$
$ \equal{}\frac{2}{5}\minus{}x(1\minus{}x)(1\plus{}x^{2}\plus{}x^{4}) \equal{}\frac{2}{5}\minus{}x(1\minus{}x)(1\plus{}x\plus{}x^{2})(1\minus{}x\plus{}x^{2}) \equal{}$
$ \equal{}\frac{2}{5}\minus{}(x\minus{}x^{2}\plus{}x^{3})(1\minus{}x^{3})\geq\frac{2}{5}\minus{}\left(\frac{x\minus{}x^{2}\plus{}x^{3}\plus{}1\minus{}x^{3}}{2}\right)^{2}\equal{}$
$ \equal{}\frac{2}{5}\minus{}\left(\frac{1\plus{}x\minus{}x^{2}}{2}\right)^{2}> 0$ because $ \frac{2}{5}\minus{}\left(\frac{1\plus{}x\minus{}x^{2}}{2}\right)^{2}> 0\Leftrightarrow$
$ \Leftrightarrow x^{2}\minus{}x\minus{}1\plus{}2\sqrt{\frac{2}{5}}> 0\Leftrightarrow\left(x\minus{}\frac{1}{2}\right)^{2}\plus{}2\sqrt{\frac{2}{5}}\minus{}\frac{5}{4}> 0.$ :)


search
function
superior algebra
superior algebra unsolved

Let G be a finite group such each element of it (except the unity) has order 2. Show that the order of G is a power of 2.

Another trivial solution: Use the search function  ;)
[url]http://www.mathlinks.ro/Forum/topic-63237[/url]


Let be $ a,b,c>0$ such that $ abc\equal{}1$ . Prove that :

$ a^2\plus{}b^2\plus{}c^2\plus{}3\ge a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca$

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ abc \equal{} 1$ . Prove that :

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3\ge a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca$[/quote]
Nice inequality,alex2008 :)
I will must prove that:
i) $ a^2\plus{}b^2\plus{}c^2\plus{}3 \ge 2(a\plus{}b\plus{}c)$
We have $ a^2\plus{}b^2\plus{}c^2 \plus{}3 \minus{}2(a\plus{}b\plus{}c)\equal{}(a\minus{}1)^2\plus{}(b\minus{}1)^2\plus{}(c\minus{}1)^2 \ge 0$,then,we have Q.E.D
ii) $ a^2\plus{}b^2\plus{}c^2\plus{}3 \ge 2(ab\plus{}bc\plus{}ca)$
We can Assume $ (b\minus{}1)(c\minus{}1) \ge 0$,then
$ a^2\plus{}b^2\plus{}c^2\plus{}3\minus{}2(ab\plus{}bc\plus{}ca)\equal{}(b\minus{}c)^2\plus{}(a\minus{}1)^2\plus{}2a(b\minus{}1)(c\minus{}1) \ge 0$
Then,we have Q.E.D :)

This ineq is weaker than old result of canhang2007(with same condition)
$ a^2 \plus{} b^2 \plus{} c^2 \ge \frac {3}{2} (a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca ) \minus{} 6$


Score distribution, percentile, and basic statistics in a nice format  :lol:

Thanks for posting, I guess.

hm... there are less people-who-scored-a-5 than USAMO winners... does that mean we're more special?

Lol it's just that a 5 is pretty hard to get on a problem.

Aw man, I'm part of the most common group of people who scored a 2.

And I thought I had a good chance of getting 4...

Oh, well, I'll prepare better next year. And then I'll overestimate again.

By the way, 220 593 AMC test-takers, 10 029 AIME takers, and 525 USAMO takers.

Congrats to everyone!


inequalities
real analysis
real analysis unsolved

Suppose $A$ and $B$ are open sets in $\mathbb{R}^{d}$ of finite positive measure. Then the Brunn-Minkowski inequality
$m(A+B)^{1/d}\ge m(A)^{1/d}+m(B)^{1/d}$
obtains equality if and only if $A$ and $B$ are convex and similar, i.e., $\exists \delta > 0$ and $h \in \mathbb{R}^{d}$ st
$A = \delta B+h$
($m$ is Lebesgue measure on $\mathbb{R}^{d}$)

[quote="Yury"]I like the proof presented in the lecture notes of Keith Ball:
http://www.msri.org/publications/books/Book31/files/ball.pdf[/quote]

It's a very nice proof. More about this approach and many similar things you can find at thesis of Franck Barthe. Thesis and summary (Franck Barthe "Autour de l'inégalité de Brunn-Minkowski" Annales de la faculté des sciences de Toulouse Sér. 6, 12 no. 2 (2003), p. 127-178) is available at Internet, but I lost the reference.


[quote="ElChapin"]Let $ a$, $ b$ and $ c$ be real numbers.

Prove that

$ \sqrt {\dfrac{a^2 \plus{} b^2}{2} \plus{} (c \minus{} a)(c \minus{} b)}\ge\dfrac{a \plus{} b}{2}$

:)[/quote]

$ \sqrt {\dfrac{a^2 \plus{} b^2}{2} \plus{} (c \minus{} a)(c \minus{} b)}\equal{}\sqrt {\dfrac{a^2 \plus{} b^2}{2} \plus{} (c\minus{}\frac{a\plus{}b}{2})^2\minus{}(\frac{a\minus{}b}{2})^2}$
$ \ge \sqrt {\dfrac{a^2 \plus{} b^2}{2} \minus{}(\frac{a\minus{}b}{2})^2}\equal{}\sqrt{\frac{a^2\plus{}2ab\plus{}b^2}{4}} \ge \frac{a\plus{}b}{2}$

Alternatively

$ \sqrt{\frac{a^{2}\plus{}b^{2}}{2}\plus{}(c\minus{}a)(c\minus{}b)} \equal{}  \sqrt{\frac{a^{2}\plus{}2ab \plus{} b^{2}}{2}\minus{}c(a\plus{}b) \plus{}c^2} \equal{}  \sqrt{\frac{(a\plus{}b\minus{}c)^2\plus{}c^2}{2}} \ge \frac{a\plus{}b}{2}$

I like stephenchengs better though  :P

Alternative:
$ 2(a^2\plus{}b^2)\plus{}4c^2\minus{}4c(a\plus{}b)\plus{}4ab \ge (a\plus{}b)^2 \Leftrightarrow a^2\plus{}b^2\plus{}4c^2\minus{}4c(a\plus{}b)\plus{}2ab\equal{}(a\plus{}b\minus{}2c)^2 \ge 0$,which is true

Nice proofs. My solution is the same as ocha's one.

 :)


How many $ 3$-digit positive integers have digits whose product equals $ 24$?

$ \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24$

They are basically asking us for three one-digit numbers whose product is 24.
Such numbers are:

1,3,8
1,4,6
2,3,4
2,2,6

The first three triples have six possible orderings. The last one has 3. 6 + 6 + 6 + 3 = 21.

Therefore, there are 21 such numbers.

[hide="a better way"]The prime factorization of $24$ is $2^3\cdot3$. So if we add two dividers and find different combinations of $1112223||$, we can get numbers like $112|22|13$ and if we multiply all the numbers in each slots, we have $2|4|3$, a number where the product of its digit is $24$. So if we order the dividers, we have $\binom{7}{2}=21$.[/hide]

Actually the answer I got was coincidental. I don't know why.

Just under two years too late. :wink:
And your method is not sound. The best way in this problem would be to just list them out.


[quote="Interval"]A ribbon 56 centimeters long is cut into two pieces. One of the
pieces is three times longer than the other. Find the lengths, in
centimeters, of both pieces of ribbon.[/quote]
[hide]So one piece is x cm long. The other one is 3x long. So solve 4x=56. You get that x=14. So one is 14 cm and the other one is 42 cm.[/hide]


[quote="Wikipedian1337"]On my USAMTS profile, page, it says:
[quote]If any of the information below is incorrect, please correct it and click Submit Changes. Your Grade should reflect your grade during the 2006-7 school year. Past participants and those who registered before June 4, 2007, must set their Grade for the 2007-8 school year below (click Submit Changes after setting your grade).[/quote]
I'm a sophomore this year, and this is my first year doing USAMTS. Why should I put 9 as my grade? (Or am I misreading the directions in some way?)[/quote]
put this year's grade
they forgot to update that page

I had the same problem. Just don't mind it, and it'll be fine.

Yes, I also noticed that people I know are putting last year's grades.

Maybe USAMTS should send out an e-mail announcement to clarify things?


geometry
perimeter
inequalities unsolved
inequalities

ABC is triangle , x,y,z>0  with  x+y+z=1  show that
[tex] (xbc+yac+zab)-s(yza+xzb+xyc)\geq3Rr  [/tex]
Thnaks!

i just  found that this one  is  crux 1771


algebra
polynomial
quadratics
Rational Root Theorem

[hide] I got $\frac{1}{12}$:

$r+s+t=\frac{1}{2}$
$rst=3$

So you have $\frac{4rst(r+s+t)}{8r^2s^2t^2}$
So you just substitute the values in and if i did everything correctly you should get $\frac{1}{12}$.[/hide]

[quote="ch1n353ch3s54a1l"][quote="weatherbottle"]Source: Massachusetts Mathematics League March 2003

The roots of $-2x^3+x^2+13x+6=0$ are $r, s,$ and $t$. Compute the value of

$\frac 1{2rs} + \frac 1{2rt} + \frac 1{2st}$[/quote]

[hide]$-2$ is a root. Factor the polynomial into $(2x+1)(3-x)(x+2)$. The roots are $-2$, $3$, and $-\frac{1}{2}$. $\frac 1{2rs} + \frac 1{2rt} + \frac 1{2st}=\frac{1}{12}$[/hide][/quote] Dont mean to be rude or anything but why would you ever find the roots?

[hide] $-2x^3+x^2+13x+6=0$

Use Descartes Rule of Sign [two negative, one positive], Rational Root Theorem, and Synthetic Division [to simplify to quadratic]. [if you can't use a calculator].

$r=3$
$s=-1/2$
$t=-2$

$-1/3 + -1/12 + 1/2= \boxed {1/12}$

I didn't bother to use a calculator so i may have made some computational errors, and i'm too lazy to factor. [/hide]

Well, this problem is not that hard just solve the equation and substitute in the values of its zeros into the three fractions.

By synthetic division and factorization we could do the following:

$-2x^3+x^2+13x+6=(x+2)(-2x^2+5x+3)=(x+2)(2x+1)(x-3)$
$So$ $r=-2,s=-\frac{1}{2}, t=3$
Now let's just substitute this into the fractions
$\frac 1{2rs} + \frac 1{2rt} + \frac 1{2st}=$
$\frac{1}{2(-2)(-\frac{1}{2})} + \frac 1{2(-2)(3)} + \frac 1{2(-\frac{1}{2})(3)} =$
$\frac 1 {2}-\frac 1 {12}-\frac 1 {3}=$
$\frac 6 {12}-\frac 1 {12} -\frac 4 {12}=$
$\frac 1 {12}$

[quote="lkryptonicl"][hide] I got $-\frac{1}{12}$:

$r+s+t=\frac{1}{2}$
$rst=-3$

So you have $\frac{4rst(r+s+t)}{8r^2s^2t^2}$
So you just substitute the values in and if i did everything correctly you should get $-\frac{1}{12}$.[/hide][/quote]

What's the formula for the sum and product of the roots of a [b]cubic [/b]equation?

Also, when I did that common denominator of those three expressions... I got:

$\frac{r+s+t}{2rst}$, making it even simpler if I knew the formula for sum/product of a cubic equation.

i just did it |kryptonic|'s way and im not sure about the product and sum formulas, but i think sum is $-b/a$ and product is $-d/a$...particularly the product one...but if it is $-d/a$ instead of $d/a$, you get 1/12.

and then I did it the root-finding way and got 1/12 and double checked it (foolproof).

Think about it.

Say you have $(x-1)(x-2)(x-3)=0$ roots are obviously $1,2,3$ so 1+2+3=6 and 1*2*3=6 so sum = -b/a and product = -d/a.

[hide]
You don't really need to find the roots.

Expanding $\frac 1{2rs} + \frac 1{2rt} + \frac 1{2st}$ we have $\frac12\left(\frac 1{rs}+\frac 1{rt}+\frac 1{st}\right)=\frac{rst^2+rs^2t+r^3st}{2r^2s^2t^2}=\frac{(r+s+t)(rst)}{2r^2s^2t^2}=\frac{r+s+t}{2rst}$

Now look at the polynomial $-2x^3+x^2+13x+6=0$.  The roots $(r,s,t)$ of it are the same as the roots of $x^3-\frac{x^2}{2}-\frac{13}{2}x-3=0$.

$rst$ would then be $-(-3)$ and $r+s+t$ would be $-(-\frac12)$  Substituting, we have $\frac{r+s+t}{2rst}=\frac{1}{12}$.
[/hide]

[quote="lkryptonicl"]Think about it.

Say you have $(x-1)(x-2)(x-3)=0$ roots are obviously $1,2,3$ so 1+2+3=6 and 1*2*3=6 so sum = -b/a and product = -c/a.[/quote]

You kind of contradicted yourself. the product is the free term (6) divided by the x^3 coefficient (1), but in your solution, you said it was -3 when -d/a=3.

Edit: never mind.

Right, it's $-d/a$;

So the product of the roots is $3$.

Yes, the best way to solve this one is to ignore the actual roots themselves and compute the value by using the sum and products of the roots... :)


1. A positive 6 digit number is split into two parts. The first consists of the first three digits, while the second consists of the last three digits. Whne the two parts are added and the sum is squared, the result is the original six digit number. Find all six digit numbers which have this property.

2. Find all positive integers $(x,y)$ such that $y^{2}+2=x^{3}$.

1]
Let the parts be a,b.
$(a+b)^{2}=1000a+b$, leting $a+b=c$
$c^{2}-c=999a$,
$c(c-1)=27.37a$.
So we have c either 0 or 1 mod both 27 and 37. That gives 4 possibilities mod 999: 0, 1,  297 or 703 (simple linear congruences). So c may be
0 - a=0
1 - a=0
297 - a=088
703 - a=494, b=209.
999 - a=998, b=001
1000 - too much (a=1000*999/999)
1296 - too much
1961 - too much

2]
Well known problem, solved by Fermat.Answer is y=5, x=3, reasoning can be found somwhere in forum (ring $\mathbb{Z}[\sqrt{-2}]$ works).

[quote="boxedexe"]It is very easy to solve this problem; the reasoning is also simple.  Factor in complex numbers, or the ring (you may not be familiar with rings) mentioned above, and then use the same standard logic for solving a normal problem using unique factorization.  You forgot the solution $(x,y)=(3,-5)$.

Masoud Zargar[/quote]
The problem mentions $x$ and $y$ are positive integers.

Positive is no restriction here, just a useless spam to make it to have only one solution.
It makes it not a single bit easier.

Oh, sorry. I missed that part.  However, you can use the unique factorization method to find all solutions, including negative solutions.


function
algebra
polynomial
floor function

Prove that

$P_{n, r}(x) = \frac{ (1-x^{n+1})(1-x^{n+2})...(1-x^{n+r}) }{(1-x)(1-x^{2})...(1-x^{r})}$

Is a polynomial in $x$ of degree $nr$, where $n$ and $r$ are nonnegative integers.  (When $r = 0$ the empty product is understood to be $1$ and we have $P_{n, 0}= 1$ for all $n \ge 0$. )

Nothing?

[hide="Ideas"] Recall that $(1-x^{n}) | (1-x^{nm})$.  Recall also that

$\frac{(n+1)(n+2)...(n+r)}{1(2)...(r)}={n+r \choose r}$

Which is an integer. [/hide]

Still no takers?

[hide="Proof 1"] We will show that the roots of the denominator are a subset of the roots of the numerator.  Now, the roots of the denominator are simply the $k^{th}$ roots of unity, $k = 1, 2, 3, ... r$.  With what multiplicity do they occur?

A particular (primitive) $k^{th}$ root of unity occurs as a root of $(1-x^{m})$ if and only if $k | m$.  Hence each $k^{th}$ root occurs $\lfloor \frac{r}{k}\rfloor$ times.

Applying the same logic to the numerator, each $k^{th}$ root occurs $\lfloor \frac{ (n+r)-(n+1)+1}{k}\rfloor$ times, which is, of course, the same number.  QED. [/hide]


Show that $ \sqrt{p}$ is irrational for any prime $ p$.

[hide="Solution"]
[i]Proof by Contradiction:[/i]

Assume that $ \sqrt{p}$ is an irreducible rational number,
\[ \sqrt{p}\equal{}\dfrac{a}{b}\]
Square both sides,
\[ p\equal{}\dfrac{a^2}{b^2}\implies pb^2\equal{}a^2\implies p|a\]
Therefore we can write $ a$ as $ pk$, then we get
\[ pb^2\equal{}p^2k^2\implies b^2\equal{}pk^2\implies p|b\]
From which we see a contradiction because we said that $ \dfrac{a}{b}$ was irreducible, hence $ \sqrt{p}$ is irrational.
[/hide]


Let $O$ be the center of the excircle of $\triangle ABC$ opposite $A$. Let $M$ be the midpoint of $AC$ and let $P$ be the intersection of lines $MO$ and $BC$. Prove that if $\angle BAC = 2\angle ACB$, then $AB = BP$

Errr...not too hard...
First, We need to prove that AP bisects $\angle CAO$.
Since $\frac{AO}{CO}=\frac{\sin (\frac{\pi}{2}+\alpha)}{\sin 2\alpha}=\frac{1}{2\sin \alpha}$
and  $\frac{MP}{PO}=\frac{S_{MPC}}{S_{POC}}=\frac{CM \sin 2\alpha}{CO \sin (\frac{\pi}{2}-\alpha)}=\frac{AM \sin 2\alpha}{CO}=\frac{AM}{AO}$
We get that AP is a bisector.
so $\angle CAP=\angle OAP=\alpha$, $\angle PAB=3\alpha=\angle APB$, and this completes the proof.□


analytic geometry
geometry
geometric transformation
reflection
geometry solved

Hi!

Considerer two circles, $K_{1}$
with center $A$ passing by $B$ and
$K_{2}$ with center $C$ passing by $D$.
Suppose $K_{1}$ is in the interior of $K_{2}$.

Now, let $X$ a point on the circle $K_{1}$.
Draw the ray $\overrightarrow{AX}$, the intersects the circle $K_{2}$
at the point $Y$. Let be $M$ the midpoint of the
segment $XY$.

What is the locus of $M$ when $X$ "walks" over
the circle $K_{1}$?

What happens when $K_{1}$ is not in the interior
of the circle $K_{2}$?

Thanks!

The locus of the point M is an algebraic curve of degree 6. This can be best seen using a Cartesian coordinate system with center A. Then, the point X is the point where the ray AM meets the circle $K_1$, and the point M lies on the required locus if and only if the reflection of the point X in the point M lies on the circle $K_2$. These things can be easily computed and we get an equation of degree 6 for M which doesn't factor.

  Darij


quadrilateral ABCD is inscribed in a circle whose radius is 6.5

BD = 13, AB=5, BC=2. find AC.

[quote="worthawholebean"][hide="solution"]Obviously, $BD$ is a diameter of the circle. Since now $\angle BCD$ is inscribed in a semicircle, it is a right triangle and thus $169-4=CD^{2}$ and $CD=\sqrt{165}$. Similar logic can be used to find that $AD=12$. By Ptolemy's theorem, $5\sqrt{165}+2(12)=13(AC)$ and thus $AC=\boxed{\frac{5\sqrt{165}+24}{13}}$.[/hide]

edit: fixed hide[/quote]
ahhhhh...i missed the whole BD is diameter part. man im slow.
good call


Does anyone have an idea of what the AMC 10 cutoff will be this year based on the difficulty?

A lot less people passed the 120 this year.. so, they're probably going to drop the 1% thing.. I got a 118, so hopefully, I will still qualify.

you cant really tell yet, but you can always hope the AMC 10B will be easy and you can get 120 :D


The quantity inside the parentheses is 100 so the total is $ 100^{2}\equal{}10000$.

hello, one should use the formula $ a^2 \minus{} b^2 \equal{} (a \minus{} b)(a \plus{} b)$ and we have $ (26^2 \minus{} 24^2)^2 \equal{} \left((26 \minus{} 24)(26 \plus{} 24)\right)^2 \equal{} 2^2\cdot50^2 \equal{} 10000$.
Sonnhard.


Beginning at point $ A$ in the diagram below, Dora selects one of the four possible directions with equal probability. Each time she comes to an intersection, she again randomly selects one of the possible directions. What is the probability that, in her first four steps, she will walk completely around the gray square? Express your answer as a common fraction.
[asy]size(100);
fill((1,1)--(1,2)--(2,2)--(2,1)--cycle, gray(.6));
draw((0,0)--(0,3)--(3,3)--(3,0)--cycle, linewidth(1.5));
draw((0,1)--(3,1), linewidth(1.5));
draw((0,2)--(3,2), linewidth(1.5));
draw((1,0)--(1,3), linewidth(1.5));
draw((2,0)--(2,3), linewidth(1.5));
dot(MP("A", (1,2), NW));[/asy]

There is a total of two possible ways for Dora to walk completely around the gray square in her first four steps if she starts at point A.  These are:
\[ \text{Down\minus{}Right\minus{}Up\minus{}Left}\\
\text{and}\\
\text{Right\minus{}Down\minus{}Left\minus{}Up}\]
At each vertex of the gray square, she has four different places she can go:  Up, Down, Left, or Right.  Therefore, the probability that Dora goes Down-Right-Up-Left is $ \left(\frac{1}{4}\right)^4\equal{}\frac{1}{256}$, as is the probability that Dora goes Right-Down-Left-Up.  Therefore, the probability that Dora goes completely around the gray square in her first four steps is $ \frac{1}{256}\plus{}\frac{1}{256}\equal{}\boxed{\frac{1}{128}}$


Prove that

$(A'\cap B\cap C)\cup [A\cap (B'\cup C)]=(A\cap B')\cup (B\cap C)$

using venn diagram is not hard but i would be appreciate if it is worked out algrebically :)

[quote="10000th User"][quote="shyong"]Prove that

$(A'\cap B\cap C)\cup [A'\cap (B'\cup C)]=(A\cap B')\cup (B\cap C)$[/quote]
The way this is written originally by shyong is wrong. Please double check.[/quote]

opps I copy the question wrongly , it should be $A\cap (B'\cup C)$ instead . Editted . :blush:

Sorry, but I don't know what $A'$ means in terms of sets. :blush:

$A'$ is the complement of $A$, everything in the universal set but not in $A$. Another notation that is used is $A^C$.

A Venn diagram would be easiest and any algebraic solution could be summed up with a Venn diagram.
If you assign values to everything in one set but not in the other two, everything in two of the three sets, everything in all three sets and everything not in any of the sets, then the intersections and unions become algebraic expressions that can be manipulated to produce the other intersection-union combination.
I'm sorry for being so vague.

What am I doing wrong?

Prove $(A'\cap B\cap C)\cup(A\cap(B'\cup C))=(A\cap B')\cup(B\cap C)$

$A\cap(B'\cup C)=(A\cap B')\cup(A\cap C)$

$(A'\cap B\cap C)\cup(A\cap B')\cup(A\cap C)=(A\cap B')\cup(A'\cap B\cap C)\cup(A\cap C)$

$(A'\cap B\cap C)\cup(A\cap C)=((A'\cap B)\cup A)\cap C$

$(A'\cap B)\cup A=(A'\cup A)\cap(B\cup A)$

$A'\cup A=U$, $U$ is universal set

$U\cap(B\cup A)=B\cup A$

$(A'\cap B\cap C)\cup(A\cap(B'\cup C))=(A\cap B')\cup((B\cup A)\cap C)$

10000th User , you didnt make any mistake there but somehow when you keep working , you will get

$(B\cup A)\cap C=(B\cap C)\cup (A\cap C)$

And $A\cap C$ is the subset of $(B\cap C)\cup (A\cap B')$ . This is the part that Im stuck . Could it be proved algebrically (well vern diagram works ...) ?

[quote="peeta"]how about this:
$(A'\cap B\cap C)\cup(A\cap (B'\cup C))\\ =(A'\cap B\cap C)\cup(A\cap B')\cup(A\cap C)\\ =(A\cap B')\cup (C\cap(A\cup(A'\cap B)))\\ =(A\cap B')\cup (C\cap((A\cup A')\cap (A\cup B)))\\ =(A\cap B')\cup (C\cap(A\cup B))\\ =(A\cap B')\cup (C\cap A)\cup(C\cap B)\\ =(A\cap B'\cap U)\cup (C\cap A\cap B')\cup (C\cap A\cap B)\cup(C\cap B\cap U) \\ = (A\cap B'\cap(U\cup C))\cup (B\cap C\cap (A\cup U))\\ = (A\cap B') \cup(B\cap C)$[/quote]

Wow peeta , thats cool . Thanks  :)


Find all poltnomials $ P(x)$ with real coefficients satisfying: For all $ a>1995$, the number of real roots of $ P(x)\equal{}a$ (incuding multiplicity of each root) is greater than 1995, and every roots are greater than 1995.

I think the problem is wrong. This is the problem statement from "The Vietnamese Mathematical Olympiad" book:

Determine all polynomials $ P(x)$ satisfying the following condition: for every $ a>1995$ the number of real roots (counted with the multiplicities) of the equation $ P(x)\equal{}a$ is equal to the degree of $ P(x)$ and all roots of this equation is greater than 1995.

I had a solution for it but I can't understand. Who can explain for me? Thanks a lots !


AMC USA(J)MO USAMO inequalities geometry circumcircle incenter

Let $ABC$ be an acute-angled triangle whose side lengths satisfy the inequalities $AB < AC < BC$. If point $I$ is the center of the inscribed circle of triangle $ABC$ and point $O$ is the center of the circumscribed circle, prove that line $IO$ intersects segments $AB$ and $BC$.

Heres my solution:

We already know that angle A > angle B > angle C. Let <OBC = <OCB = x, <OAB = <OBA = y, and <OAC = <OCA = z. Then since BC > AC > AB, z + y > x + y > x + z.

However, for OI to intersect AC, <OAC must be greater than <OAB (since AI bisects <CAB). This means that z > y, a contradiction.

Thus OI intersects only AB and BC.

[quote="froufrou4189"]Heres my solution:

However, for OI to intersect AC, <OAC must be greater than <OAB (since AI bisects <CAB). This means that z > y, a contradiction.

[/quote]

I don't quite understand this part. Why must <OAC be greater than <OAB?
I believe it is possible to construct such a "contradiction"

For OI to intersect AC, O must be on the same side of AI as B. Since AI bisects angle A, angle OAB must be smaller than angle OAC (this means that O is on the same side of AI as B).

A similar way to prove this would be to use the above reasoning to conclude that O is on the same side of AI as C and on the same side of CI as B. So if AI intersects BC at D, then O must lie inside triangle CDI, which means it cannot intersect AC.

Here's my solution, very different from the above two:

First we show that if given a triangle $ ABC$ with circumcenter $ O$ and incenter $ I$, it is true that $ OI$ intersects a vertex, then the triangle is isosceles.

WLOG let that vertex is $ A$. Then $ AI$ is the bisector of $ \angle BAC$ and $ AI$ is the same as $ AO$, so $ \angle BAO\cong\angle CAO$. Since $ \triangle BAO$ and $ \triangle CAO$ are isosceles with $ AO \equal{} BO \equal{} CO$, it follows that they are congruent and that $ AB \equal{} AC$ and $ \triangle ABC$ is isosceles as desired.

Say we have $ \triangle ABC$ with circumcenter $ O$ and incenter $ I$. If we move $ A$, $ B$, and $ C$ continuously without allowing the triangle to become isosceles, then $ O$ and $ I$ move continuously because they are solutions to quadratic equations in terms of $ A$, $ B$, and $ C$, and thus the intersection of $ OI$ with the boundary of $ \triangle ABC$ will move continuously. If one of the intersections moves from one side to another, it must pass a vertex, but this would make the triangle become isosceles, so the intersections of $ OI$ with the triangle must remain on the same segments. Also note that if the triangle is scalene then $ O$ and $ I$ must both lie inside the triangle and not coincide, and their line's intersection with the interior and boundary of the triangle should be a line segment with two endpoints because the triangle is convex.

We note that in a triangle with sides $ AB \equal{} 5$, $ AC \equal{} 7$, and $ BC \equal{} 8$, line $ OI$ intersects the smallest and largest sides. This is trivial to check and is left as an exercise to the reader.

We now show that we can continuously move $ A$, $ B$, and $ C$ through the plane while keeping the triangle acute and scalene to make $ AB$ and $ AC$ have any positive real lengths $ c < b < 8$. To increase or decrease any side, we rotate one of its vertices about the opposite vertex while keeping the other vertices stationary. This moves one of the vertices along the arc of a circle, which is a continuous motion, and we can achieve any desired length assuming that keeps the triangle non-degenerate. We then proceed by casework to avoid making the triangle isosceles:
If $ c > 5$ and $ b > 7$ we increase $ AC$ to $ b$ and then $ AB$ to $ c$.
If $ c > 5$ and $ b < 7$ we increase $ AB$ and then decrease $ AC$.
If $ c < 5$ and $ b > 7$ we increase $ AC$ and then decrease $ AB$.
If $ c < 5$ and $ b < 7$ we decrease $ AB$ and then decrease $ AC$.

This will keep $ BC \equal{} 8$ as the largest side since $ b,c,5,7 < 8$ and will thus cause the triangle to remain acute since we increase $ AB$ and $ AC$ before doing any decreasing. It also avoids any isosceles triangles in the path, as one can check in each of the case. It thus holds that in any acute scalene triangle with greatest side $ 8$, $ OI$ intersects the smallest and largest side. We can then appropriately scale the triangle to account for all possible scalene triangles, and the result is proved.

QED

I don't think that continuous transformation stuff is that rigirous? What do you mean by it exactly?

We can also easily use barycentric coordinates to solve this problem.

By continuous transformation I just mean the definition from topology, using standard real topology. I guess by "continuously move" the vertices I should say move them along a connected path.

But yeah what we're basically trying to show is that if we move it continuously and change the sides that $ OI$ intersects then $ OI$ would have to intersect one of the vertices, so using the geometric fact I proved at the beginning, we can just prove it for one triangle, show we can move $ A$, $ B$, and $ C$ to any scalene triangle and thus prove it for all of them.

[hide]Consider the lines that pass through the circumcenter O. Extend AO, BO, CO to D,E,F on a,b,c, respectively.

We notice that IO passes through sides a and c if and only if I belongs to either regions AOF or COD.

Since AO = BO = CO = R, we let $ \alpha \equal{} \angle OAC \equal{} \angle OCA$, $ \beta \equal{} \angle BAO \equal{} \angle ABO$, $ \gamma \equal{} \angle BCO \equal{} \angle CBO$.

We have $ c < b < a \implies C < B < A \implies \gamma \plus{} \alpha < \beta \plus{} \gamma < \alpha \plus{} \beta \implies \gamma < \alpha < \beta$.

Since IA divides angle A into two equal parts, it must be in the region marked by the $ \beta$ of angle A, so I is in ABD.

Similarly, I is in ACF and ABE. Thus, I is in their [intersection], AOF. From above, we have IO passes through a and c. $ \blacksquare$[/hide]

Edit: Is essentially the same as froufrou4189. Meh, I'm still happy.

Edit2: Oops, meant intersection not union.

[quote="#H34N1"]Is there a way to do this without topology? :maybe: I'm confused.[/quote]

Uh, the solutions don't directly use topology, he just called upon a definition to make it more rigorous.

[quote="#H34N1"]Is there a way to do this without topology? :maybe: I'm confused.[/quote]
Look here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=261335187&t=125578

[quote="#H34N1"]Let $ ABC$ be an acute-angled triangle whose side lengths satisfy the inequalities $ AB < AC < BC$. If point $ I$ is the center of the inscribed circle of triangle $ ABC$ and point $ O$ is the center of the circumscribed circle, prove that line $ IO$ intersects segments $ AB$ and $ BC$.

PostPosted: Sun Dec 24, 2006 9:45 pm[/quote]
There is a certain amount of irony here  :rotfl:

I do not understand pianoforte's solution that I must be within some region for OI to intersect BC. Can someone provide a clear solution(I am horrible at geometry).


Solve an equation

$ y^{2} \equal{} x^{3} \plus{} 16$ , $ x,y \in Z$

My idea :
[hide] I expect that the equation has only one solution : $ (x,y) \equal{} (0,4)$ ,
$ x^{3} \equal{} y^{2} \minus{} 4^{2} \equal{} (y \minus{} 4)(y \plus{} 4)$
Lemma : The equation hasn't any solution for $ x\neq 0$
Proof :
Let d=(y-4, y+4). Let $ p$ be prime number ($ p\neq 2$) which divide $ d$. Also it must be :
$ p|x$,  $ p|y \minus{} 4$, $ p|y \plus{} 4$ => $ p|(y \plus{} 4) \plus{} (y \minus{} 4) \equal{} 2y$ -> $ p|y$ => $ p|4$ what is impossible, therefore $ d$ hasn't any prime number divider ($ p\neq 2$) so $ d$ is a power of 2 with index >=0
It can't be 16|d, because$ d|((y \plus{} 4) \minus{} (y \minus{} 4) \equal{} 8$ and it would be$ 16|8,$ ,
It can't be d=4, becasue it'd be $ y \plus{} 4 \equal{} 4k$, $ y \minus{} 4 \equal{} 4l$ , (k,l)=1
$ x^{3} \equal{} 16kl$ hence 4|x =>4|kl  It may be only 4|k, l=2a+1 (because of (k,l)=1 )
But $ 4 \equal{} d|4 \plus{} y$=> $ 4|y \equal{} 2(k \plus{} l)$ what is impossible !!! so $ d\neq 4$
There are options else :d=1, d=8. I expect i should eliminate one of option in similar way like above and then prove that last possibility is wrong, and it'd be end of my proof :) But how can i do that ? Or maybe part of proof above is completely wrong ?
[/hide]

Here is D=1.
$ y\minus{}4\equal{}a^3, y\plus{}4\equal{}b^3$ for some a,b.
Then $ 8\equal{}b^3\minus{}a^3\equal{}(b\minus{}a)(b^2\plus{}ab\plus{}a^2)$.
If b-a=1, then $ b^2\plus{}ab\plus{}a^2\equal{}8$, subbing in b=a+1, we find no integer solution.
If b-a=2, then $ b^2\plus{}ab\plus{}a^2\equal{}4$, subbing in b=a+2, we find a=0, so y=4, x=0  or a=-2, b=0, hence y=-4, x=0 (we already know these solutions).
If b-a=4, then $ b^2\plus{}ab\plus{}a^2\equal{}2$, subbing in b=a+4 we find no integer solution.
If b-a=8, then $ b^2\plus{}ab\plus{}a^2\equal{}1$, subbing in b=a+8, we find no integer solution.

Great idea to look on the divisors:)

You use a lot of explanation, and it could be done shorter. Indeed let $ d \equal{} (y\minus{}4,y\plus{}4) \equal{} (\minus{}8, y\plus{}4) \Rightarrow d \equal{} 2^n, n \in \{0,1,2,3\}$ And $ y\minus{}4 \equiv y\plus{}4 \bmod 8$, so if $ d\equal{}4$ we have $ x^3 \equiv 16 \bmod 32$. Contradiction.
I think you forgot $ d\equal{}2$? It's the same thing anyways.  $ y\minus{}4 \equiv y\plus{}4 \bmod 4 \Rightarrow x^3 \equiv 4 \bmod 8$. Contradiction again.

For $ d\equal{}8$ we have $ y \equal{} 8k\plus{}4 \Rightarrow x^3 \equal{} 2^6(k)(k\plus{}1) \Rightarrow x \equal{} 4x_1 \Rightarrow x_1^3 \equal{} k(k\plus{}1)$.  Then we have $ k \equal{} a^3, k\plus{}1 \equal{} b^3 \Rightarrow a^3\minus{}b^3 \equal{} 1 \Rightarrow a\equal{}1, b\equal{}0$. So the only solution for $ d\equal{}8$ is $ (x,y) \equal{} (0,4)$

Then we have $ d\equal{}1$, but then we have $ y\minus{}4 \equal{} k^3$ and $ y\plus{}4\equal{}j^3$, which implies $ j^3\minus{}k^3 \equal{} 8$ which have no solution.

So we have $ \boxed{(x,y) \equal{} (0,4)}$

Yeah well done guys thank you :D, it was easy isn't it ? :P (2 solution $ (x,y)\equal{}(0,4)$, $ (x,y)\equal{}(0,\minus{}4)$ ) Only one question else :

Mathias wrote
[quote]$ d\equal{}(y\minus{}4,y\plus{}4)\equal{}(\minus{}8,y\plus{}4)$ [/quote]

How do you know that ? or maybe it's after i proved $ 16\nmid d$ ? :)

[quote="Pablo09"]Yeah well done guys thank you :D, it was easy isn't it ? :P (2 solution $ (x,y) \equal{} (0,4)$, $ (x,y) \equal{} (0, \minus{} 4)$ ) Only one question else :

Mathias wrote
[quote]$ d \equal{} (y \minus{} 4,y \plus{} 4) \equal{} ( \minus{} 8,y \plus{} 4)$ [/quote]

How do you know that ? or maybe it's after i proved $ 16\nmid d$ ? :)[/quote]
$ (a,b) \equal{} (a,b \minus{} a) \Rightarrow d \equal{} (y \minus{} 4, y \plus{} 4) \equal{} (y \minus{} 4 \minus{} y \minus{} 4,y \plus{} 4) \equal{} ( \minus{} 8,y \plus{} 4). \Rightarrow \quad d \in \{1,2,4,8\}$.
I was actually just providing a shorter proof:)

Haha Mati you're brillant :D thanks again

This is a [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url] with $ n \equal{} 16$.  Some cases can be solved with elementary methods (such as this one) whereas others require more technical machinery.


Couldn't you also treat a^3 as an x and solve for x through the quadratic formula.Then you could cube root both sides and plug it in the expression. This would however require a calculator.

Yeah, you can, but you will use calculator machine, and generaly, in olympiad competition, the calculator machine aren't allowed  :P


Find the point T in a triangle, such that the ekspresion xyz is maksimal, where x y and z are the distances from T to a b and c.

the answer is the point of inter section of the medians,
it is easy to check,

I know the answer, but i would like to see a nice solution


Determine a two-digit number knowing that if we divide the sum of its digits, the quotient is 7 and the rest is 3.If we reverse the digits and divide the resulting number by the sum of its digits, the quotient is 3 and the remainder is 7. What's the number ?

[b][color=red]Please stop posting your questions in the Preolympiad forum; they are far below that level. - Moderator [/color][/b]

[hide="Solution"]
Let the number be 10a+b. From the first sentence, we know that:
\[ 10a\plus{}b \equal{} 7(a\plus{}b) \plus{} 3\]

From the second sentence, we know that:
\[ 10b\plus{}a \equal{} 3(a\plus{}b) \plus{} 7\]

We have 2 linear equations, and if we solve we see that:
a = 7 and b = 3, so the number is 73.

[/hide]

Ok, nice method. It's a interested question. Thanks ;)


integration
calculus
calculus computations


calculus
derivative
combinatorics proposed
combinatorics

You can use differentiation
In $ n \choose k $ shouldn't $k$ be replaced be $ r$.If yes, I define (standard)

$ C_{r}$ = $ {n}\choose {r} $
Thus we have to find $ \sum _{r} r ( C_r ) ^2 $
Consider
$ (1+x)^n = C_0 + C_1 x + .... C_r x^r +.... \\
\\ n(1+x)^{n-1} = \sum r C_r x^ {r-1} $

Now we consider the product of the above series and $ (1+x)^n = C_0 x^n + C_1 x^ {n-1} +.... C_r x^{n-r} +.... $
Clearly, the coefficient of $ x^ {n-1) }$ in the product $ n (1+x)^{n-1} \times (1+x)^n $ is the required answer (compare!!)

Thus the answer is coefficeint of $ x^{n-1} $ in $ n (1+x)^{2n-1} $. Thus, using binomial theorm, the answer is

$n \times $ $ {2n-1} \choose {n-1} $

Another way of looking at it:

Given two sets of $n$ elements, how many ways are there to pick the same number of elements from each set and to mark one of the elements you picked from the first set?

On one hand, we can view this as choosing a size $r$, choosing a size $r$ subset of each set, then picking the element at the end.  This gives the left hand side.

On the other hand, we can view it as first picking the marked element ($n$ choices), then coloring $n-1$ of the remaining $2n-1$ total elements blue.  One set will consist of the blue elements in the first set and the marked element, the other set will consist of the red elements in the second set.   It's easy to check that the two sets are the same size, and that any pair of sets+marked element can be realized by precisely one coloring.  This gives the right hand side.


$ a,b,c,d$ positive real number . Prove:

$ \frac{a\plus{}b\plus{}c\plus{}d}{abcd}\leq \frac{1}{a^{3}}\plus{}\frac{1}{b^{3}}\plus{}\frac{1}{c^{3}}\plus{}\frac{1}{d^{3}}$

It's obviously true by AM-GM ; $ \frac{1}{a^3}\plus{}\frac{1}{b^3}\plus{}\frac{1}{c^3} \geq \frac{3}{abc}$


Source: ACoPS
Prove that $n!<\left(\frac{n+1}{2}\right)^{2}$, for $n=2,3,4,....$

Here's how i solved:
[hide]
$n!=n\cdot (n-1)\cdot (n-2)\cdot\cdot\cdot2 \cdot 1$
[/hide]

Are you sure that $\left(\frac{n+1}{2}\right)^{2}?$

Please stop responding to this topic. kimby accidentally double posted. There is already a discussion on that other topic.

Mods: please lock this topic.


abstract algebra
group theory
superior algebra
superior algebra solved

Let $G=\left\{ A_{(}a,b)=\left(\begin{array}{cc}a+\hat{3}b & b \\ \hat{4}b & a+\hat{3}b \end{array}\right) | \ a,b\in\mathbb{Z}_{6}, \ \det(A)=\hat{1}\right \}$. Show that $(G,\cdot)$ is an abelian group isomrphic to Klein's group.

From $\det(A)=\hat{1}$ we get $a^{2}+\hat{5}b^{2}=\hat{1}$. Then the solutions gives directly the solutions... I don't know how they did this  :| The rest was pretty clear to me.

Despite the fact that I have seen only 5-6 problems with isomorphism all this problems seem kind of odd for me. :fool: Untill now every solution that I saw was different, [b]not[/b] obivous at all. I'm not freaking out because this is one of next year's subject matter and I belive I'll get used with problems involving isomprphisms.

What didn't you understand? How to solve $a^{2}+5 b^{2}= 1$?

Just note that $5b^{2}=-b^{2}$. Write the map $a \to a^{2}$ in a table, then just find the solutions by hand.

Next, you can just do computations to check that it is isomorphic to Klein's Group.

Thanks! Indeed $-\hat{1}=\hat{5}$ in $\mathbb{Z}_{6}$..  so :blush:


geometry
circumcircle
geometric transformation
rotation
symmetry
geometry solved

Given an equilateral triangle $ABC$ and a point $M$ in the plane ($ABC$). Let $A', B', C'$ be respectively the symmetric through $M$ of $A, B, C$.

[b]I.[/b] Prove that there exists a unique point $P$ equidistant from $A$ and $B'$, from $B$ and $C'$ and from $C$ and $A'$.
[b]II.[/b] Let $D$ be the midpoint of the side $AB$. When $M$ varies ($M$ does not coincide with $D$), prove that the circumcircle of triangle $MNP$ ($N$ is the intersection of the line $DM$ and $AP$) pass through a fixed point.

You can get from the triangle $ABC$ to $B'C'A'$ by composing a rotation of angle $\frac{2\pi}3$ centered at $O'$, the center of $A'B'C'$, with the symmetry wrt $M$. This yields a rotation of angle $-\frac\pi 3$, and $P$ is the center of this rotation.

If $O$ is the center of $ABC$, then $POO'$ is an equilateral triangle, and $M$ is the midpoint of $OO'$, which means that $\angle POM=\frac\pi 3$. At the same time, $PAB'$ is an equilateral triangle, and $DM\|AB'$, so $\angle PNM=\frac\pi 3$. From these two we find that $P,O,M,N$ are concyclic, i.e. the circumcircle of $PMN$ always passes through $O$.

Rocco Lupoi (he is a member of ML, too) posted a solution of the problem in the Hyacinthos newsgroup, but it was deleted because of the use of non-ASCII signs. Here is the post:

[quote="Hyacinthos message #9560"][b]From:[/b] Rocco Lupoi
[b]Subject:[/b] RE: [EMHL] equilateral triangle ABC

> Given an equilateral triangle ABC and a point M in the plane (ABC).
> Let A', B', C' be respectively the symmetric through M of A, B, C.
>
> a) Prove that there exists a unique point P equidistant from A and
> B', from Band C' and from C and A'.
>
> b) Let D be the midpoint of the side AB. When M varies (M does not
> coincide with D), prove that the circumcircle of triagle MNP (N is
> the intersection of the line DM and AP) pass through a fixed point.
>

a) If G is the centroid of ABC and G' is the symmetric through M of G,
then P is the third vertex of equilateral triangle GG'P (or
equilateral triangle AB'P, and similar). One can see A'B'C' as a
rotation of 60° through P of ABC.

b) given that MD is parallel to AB' and given that APB' is
equilateral, then <MNP = 60° = <MGP. This means that as M varies
c(MNP) pass through G.

Ciao

Rocco[/quote]


pigeonhole principle
invariant
combinatorics solved
combinatorics

there are 70 numbers smaller than  200 . prove that two of them is such that:
a_b=4 or 5 or 9

[quote="ooiler"]a_b=4 or 5 or 9[/quote]

Do you mean $a - b \in \{4,5,9\}$?  :?

the following makes this almost trivial.
  by repeated application of pigeonhole the question becomes equivalent to the statement that among the integers from $1$ to $13$ any set of $5$ integers will necessarily contain two whose difference is either 4,5 or 9.
suppose not and $S$ is such a set of $5$ elements.since this property is invariant under subtraction of all elements by a constant, we can WLOG assume that $1\in S$. Then $5,6,10\notin S$.among the remaining integers consider the following array created by them:
$\begin{array}{rrr} 2&7&11\\3&8&12\\4&9&13\end{array}$.
note that since there are $4$ more elements in $S$ atleast two of them are in the same row.but the differences of any two in the same row is prohibited.contradiction.

[quote="seshadri"]the following makes this almost trivial.
  by repeated application of pigeonhole the question becomes equivalent to the statement that among the integers from $1$ to $13$ any set of $5$ integers will necessarily contain two whose difference is either 4,5 or 9. [/quote]

seshadri,

Can you elaborate on how you reduced the original problem to the simpler problem? ("the repeated application of pigeonhole part")? Thanks.

that is because if you have 70 numbers(positive integers) less than 200, then you have atleast 35 numbers in one of the two halves(1-100 or 100-200). Again reducing by half, you have atleast 18 numbers in a set of 50 consecutive integers, atleast 9 numbers in a set of 25 consecutive integers and atleast 5 in a set of 13 consecutive integers.Note that since we finally only care about the difference between the integers, we can as well assume the integers to be 1-13.