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if anyone have anymore suggestions u can posted here....

Each week I post a few problems.  They might as well be called "Problems of the Week."

[quote="math92"]well the people who solve them have to be awarded or recognized so people will know that they solved the "problem of the week".[/quote]

Managing this site already takes an enormous amount of time and resources.  That's just not something we have time to do, but thank you for the suggestion.  If there were more of us, it's something we might do.

i realized that you do alot to run this website.

I was wondering if we can enter html codes in the message text!!
It didn't work on my previous message.
And what about malicious code? do u check the text before storing it??

By the way great site specially the smileys :starwars: it's amazing.
Thx


[quote="chavez"]Find all functions (1,2,...) --> (1,2,...) such that
f(m+1).f(m-1) = f(m^2) for all m,n.[/quote]
For all $m,n$? I don't see $n$ here. And I guess $m\geq 2$?


Determine all functions $f: \mathbb R \to \mathbb R$ such that for all reals $x$ and $y$,
\[ f(xf(x) \plus{} f(y))\equal{}(f(x))^2 \plus{} y.\]

[quote="mahanmath"]Hi !  :)
Find all $ f: R\rightarrow R$ such that :

$ f(xf(x) \plus{} f(y)) \equal{} (f(x))^2 \plus{} y$[/quote]

Let $ G(x,y)$ be the assertion of $ f(xf(x) \plus{} f(y)) \equal{} (f(x))^2 \plus{} y$

$ G(0,0) \implies f(a) \equal{} a^2$ with $ f(0) \equal{} a$

$ G(0,y) \implies f(f(y)) \equal{} a^2 \plus{} y \implies f(c) \equal{} 0$ with $ c \equal{} f( \minus{} a^2)$
$ G(c,y) \implies ff(y) \equal{} y$
$ G(f(x),y) \implies f(xf(x) \plus{} f(y)) \equal{} x^2 \plus{} y$

which implies that $ f(x)^2 \equal{} x^2$

and hence the solution are $ f(x) \equal{} \pm x$

[quote="mahanmath"]Hi !  :)
Find all $ f: R\rightarrow R$ such that :

$ f(xf(x) \plus{} f(y)) \equal{} (f(x))^2 \plus{} y$[/quote]

Let $ P(x,y)$ be the assertion $ f(xf(x)\plus{}f(y))\equal{}f(x)^2\plus{}y$
Let $ a\equal{}f(0)$

$ P(0,x)$ $ \implies$ $ f(f(x))\equal{}x\plus{}a^2$ and so $ f(x)$ is a bijection and $ \exists u$ such that $ f(u)\equal{}0$
$ P(u,x)$ $ \implies$ $ f(f(x))\equal{}x$ and so $ a\equal{}0$
$ P(x,0)$ $ \implies$ $ f(xf(x))\equal{}f(x)^2$
$ P(f(x),0)$ $ \implies$ $ f(xf(x))\equal{}x^2$

And so $ f(x)^2\equal{}x^2$

If $ f(1)\equal{}1$, if $ \exists x$ such that $ f(x)\equal{}\minus{}x$, then $ P(1,f(x))$ $ \implies$ $ f(x\plus{}1)\equal{}\minus{}x\plus{}1$ and so either $ x\plus{}1\equal{}\minus{}x\plus{}1$, either $ x\plus{}1\equal{}x\minus{}1$, and so $ x\equal{}0$ and so $ f(x)\equal{}x$ $ \forall x$
If $ f(1)\equal{}\minus{}1$, if $ \exists x$ such that $ f(x)\equal{}x$, then $ P(1,f(x))$ $ \implies$ $ f(x\minus{}1)\equal{}x\plus{}1$ and so either $ x\minus{}1\equal{}x\plus{}1$, either $ \minus{}x\plus{}1\equal{}x\plus{}1$, and so $ x\equal{}0$ and so $ f(x)\equal{}x$ $ \forall x$

And it is easy to check back that these two functions indeed are solutions.

Hence the two solutions $ f(x)\equal{}x$ $ \forall x$ and $ f(x)\equal{}\minus{}x$ $ \forall x$

@abdek : $ f(x)^2\equal{}x^2$ need some more proof to deduce the two solutions. You must show that there are no solutions in which $ f(2)\equal{}2$ while $ f(3)\equal{}\minus{}3$, for example.

Let $ P(x,y)$ the assertion of $ f(xf(x)\plus{}f(y))\equal{}(f(x))^2\plus{}y$.

Let $ f(0)\equal{}s$.

Let $ y_1,y_2 \in \mathbb{R}$ so that
$ f(y_1)\equal{}f(y_2) \Longrightarrow xf(x)\plus{}f(y_1)\equal{}xf(x)\plus{}f(y_2) \Longrightarrow f(xf(x)\plus{}f(y_1))\equal{}f(xf(x)\plus{}f(y_2)) \Longrightarrow (f(x))^2\plus{}y_1\equal{}(f(x))^2\plus{}y_2 \Longrightarrow y_1\equal{}y_2$

$ P(0,0) \Longrightarrow f(s)\equal{}s^2$

$ P(s,0) \Longrightarrow f(sf(s)\plus{}s)\equal{} s^2$.

But $ s^2\equal{}f(s)$ so
$ f(s^3\plus{}s)\equal{}f(s)$ and because f is 1-1 function we take $ s^3\plus{}s\equal{}s \Longleftrightarrow s\equal{}0$.

$ P(x,0) \Longrightarrow f(xf(x))\equal{}(f(x))^2$

$ P(f(x),0) \Longrightarrow f(xf(x))\equal{}x^2$.

So $ (f(x))^2\equal{}x^2$....

Now look at pco's post.... :)


trigonometry
induction
algebra
polynomial
number theory proposed
number theory

Prove that  \[ \sin (1^\circ )\] is algebraic.

Use trigonometric identities to prove by induction that $ \sin 2nx \equal{} \cos x P_{2n}(\sin x)$ where $ P_{2n}$ is some polynomial for each $ n\in \mathbb{N}$.  In particular, this implies that there is a polynomial $ P_{180}$ so that
\[ 0 \equal{} \sin^2 180^\circ \equal{} \left(1 \minus{} \sin^2 1^\circ \right) \left[P_{180}\left(\sin 1^\circ\right)\right]^2
\]
and hence $ \sin 1^\circ$ is algebraic by definition.  The same argument can be used to prove $ \sin p\pi$ is algebraic for each $ p\in \mathbb{Q}$.
[hide="remark"]In particular, you can prove $ \sin 2nx \equal{} ( \minus{} 1)^{n \minus{} 1}\cos x U_{2n}(\sin x)$, where $ U_k$ are [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html]Chebsyhev Polynomials of the Second Kind[/url][/hide]

This seems similar to a Fourier Series (or atleast I think it was that I came across it awhile ago but didn't understand it much).

Me,too.It's very hard to understand it.  :oops:


I posted this question in the mathematical books and journal board but I haven't got a response in ages. Now that I think about it, It's probably better to ask questions about Canadian books in the Canada forum. >_>

Are the Problems Problems Problems books from Waterloo good books to study from? Or is it better to read an actual text rather than just problems and their solutions?

The only way to learn problem solving is to solve problems. There is no substitute for that.

Of course, you'll need to know some techniques, but Problems*3 is self-contained enough for you to learn as you do the problems. I've used the Waterloo books myself in the past, and I highly recommend them.

Simply reading some text will not give you the skills required to solve contest problems. Sure, you can read through "Introduction to Number Theory" (or whatever), and by the end of it, I can give you number theory problem, and you still won't know how to solve it.

On the other hand, if you learn by solving problems, then the theory comes naturally and firmly.

go for volume 8, i thought volume 3 had some nice problems in it.

[quote="Farnak"]Thanks Yufei  :) . I just wasn't sure if a pure problems book was comparable to a text and problems book.

About the Problems Problems Problems series, aren't there around 8? I scored 142 on Pascal and 115 on AMC 12, so which volumes would I actually benefit from?[/quote]

Go for the ones for Senior High School Students.

Also, you may find [i]Problems and How to Solve Them Vol 2[/i] beneficial. It's also published by Waterloo.

I like Problems Problems Problems when I'm extremely bored and I have absolutely nothing to do. Or when I'm absolutely desperate for math. But otherwise, I don't think it's that great a practice book.


In a set of four consecutive even positive integers the difference of twice the smallest integer and the cube root of the largest integer is 112. What is the smallest integer?

Thanks in advance

General Idea:
Use an algebraic representation of the integers and perform all of the arithmetic that the problem asks. Then its basically guess and check using the most reasonable numbers.

I believe the smallest is then 58.

-If you are working on handbook problems from this year, then you can simply view the official solutions rather than asking on the forum each time.


integration
trigonometry
calculus
calculus computations

Compute $ \int \frac{1}{\sin^6x \plus{} \cos^6x}\,dx$.

We have:

$ \sin^{6}(x) \plus{} \cos^{6}(x)$

$ \equal{} [\sin^{2}(x) \plus{} \cos^{2}(x)]^{3} \minus{} \sin^{2}(x)\cos^{2}(x)[\sin^{2}(x) \plus{} \cos^{2}(x)]$

$ \equal{}1 \minus{} \frac{\sin^{2}(2x)}{4} \equal{} \frac{7 \plus{} \cos4x}{8}$

So that:

$ \int \frac{1}{\sin^{6}(x) \plus{} \cos^{6}(x) }dx$

$ \equal{} 8\int \frac{1}{7 \plus{} \cos(4x)}dx \equal{} 2\int \frac{1}{7 \plus{} cosu}du$ (Use Weierstrass'  to get the next line)

$ \equal{} 2\int \frac{x}{4 \plus{} 3x^{2}}dx \equal{} \frac{\ln(4 \plus{} 3x^{2})}{3} \plus{} C.$

[quote="Carcul"][color=darkred]Ascertain[/color] $ I\equiv\int \frac {1}{\sin^6x + \cos^6x}\ \mathrm {dx}$.[/quote]
[color=darkblue]$ \frac {1}{\sin^6x + \cos^6x} =$ $ \frac {(\sin^2x + \cos^2x)^3}{\sin^6x + \cos^6x} =$ $ \frac {(\sin^2x + \cos^2x)^2}{\sin^4x - \sin^2x\cos^2x + \cos^4x} =$

$ \frac {(\tan^2x + 1)^2}{\tan^4x - \tan^2x + 1}$ $ \implies$ $ I = \int\frac {\tan^2x + 1}{\tan^4x - \tan^2x + 1}\cdot(\tan\ x)'\ \mathrm {dt}$ $ \implies$

$ \boxed {I = F(\tan\ x) + \mathcal C}$, where $ F(t) = \int\frac {t^2 + 1}{t^2 - t + 1}\ \mathrm {dt} =$ $ \int\frac {1 + \frac {1}{t^2}}{t^2 + \frac {1}{t^2} - 1}\ \mathrm {dt} =$

 $ \int\frac {1}{(t - \frac 1t)^2 + 1}\cdot(t - \frac 1t)'\ \mathrm {dt}$ $ \implies$ $ \boxed {F = \arctan\ \frac {t^2 - 1}{t} + \mathcal C}$.

Thus,  $ I = \arctan\ \frac {\tan^2 - 1}{\tan\ x} + \mathcal C$, i.e. $ \boxed {I = - \arctan\ [2\cot\ (2x)] + \mathcal C}$.

[b][u]Remark.[/u][/b] Using the well-known identity

$ \arctan\ x + \arctan\ \frac 1x = \{\begin{array}{ccc} - \frac {\pi}{2} & \mathrm {,} & x < 0 \\
 \\
\frac {\pi}{2} & \mathrm {,} & x > 0\end{array}$

obtain finally $ \boxed {I = \arctan[\frac 12\cdot\tan\ (2x)] + \mathcal C}\ .$

We can use directly the substitution $ t = \phi (x) = \frac 12\cdot\tan\ (2x)\ !$[/color]


analytic geometry
geometry
geometric transformation
reflection
trapezoid

A triangle $\triangle ABC$ with sides $AB=3$, $AC=4$, and $BC=5$, is placed on the coordinate plane with $BC$ on the $x$-axis and the midpoint of $BC$ is at the origin. $A$ is reflected across the $y$-axis to get $A'$. Find $AA'$.

[hide]The area of the triangle is $6$.  Let $h$ be the altitude to side $BC$.  We have $\frac{1}{2}(5)h=6$, so $h=\frac{12}{5}$.

Reflect the entire triangle across the $y$-axis and connect points $A$ and $A'$ to get an isosceles trapezoid with larger base $5$, height $\frac{12}{5}$, legs $3$, and diagonals $4$.  We want to find the length of the other base.

Draw an altitude from one endpoint of the smaller base, so that we have a right triangle with hypotenuse $3$ and leg $\frac{12}{5}$.  The other leg, which forms part of the larger base of the trapezoid, has length $\frac{9}{5}$.  The length of the smaller base is $5-2\left(\frac{9}{5}\right)=\boxed{\frac{7}{5}}$.
[/hide]

[quote="matt276eagles"]...segment $AA'$ is twice the altitude to side $BC$...[/quote]

How? BTW, I got a different answer.

My mistake, it's reflected over the $y$-axis, not the $x$-axis.  I'll edit my other post.

It's correct now.

[hide="hint for another solution"]
quadrilateral...
[hide]
cyclic
[/hide]
[/hide]

[quote="LordoftheMorons"]I'm getting.... $\frac{7}{5}$?[/quote]

Like I said, that's correct...

[quote="matt276eagles"][hide="Ptolemy Theorem Solution"]Reflect the entire triangle across the $y$-axis and connect points $A$ and $A'$ to get an isosceles trapezoid with larger base $5$, legs $3$, and diagonals $4$.  Let $x$ be the length of the shorter base.  Isosceles trapezoids are cyclic, so we can use Ptolemy's Theorem.

$3(3)+5(x)=4(4)\Rightarrow x=\boxed{\frac{7}{5}}$[/hide][/quote]

Yup, that was the solution I had in mind! :yup:

Yeah, I didn't think of it until you posted that hint. :D


Find all positive integers whose decimal expansion is $ \overline{a_ka_{k\minus{}1}\cdots a_1a_0}$ such that $ \overline{a_ka_{k\minus{}1}\cdots a_1a_0}\equal{}2\cdot\overline{a_0a_1\cdots a_{k\minus{}1}a_k}\minus{}1$.

Sorry if the problem is already here, but I haven't found it.

That's a known problem João.

The only numbers that work are $ 79...93$ with a number of $ 9$'s $ \geq 0$

Try to prove first the first and last digits.

You missed $ 1$ and $ 73$, but that's a good ideia. I started solving it that away.

Actually, $ 73$ has zero $ 9$'s. I have clearly said $ \geq 0$.

And $ 1$ is a trivial case.

I've noticed you're Portuguese too, please register on[url=http://www.mat.uc.pt/~delfos/phpBB2/index.php]this forum[/url]  where the problem was already posted :wink:

[quote="rmfgcampos"]Actually, $ 73$ has zero $ 9$'s. I have clearly said $ \geq 0$.[/quote]
Yes, you're right... Sorry.
Ricardo, you are talking like if you didn't know me... :P


irrational number
algebra unsolved
algebra

I'm sure this is a standard result...

[b]Prove that every irrational number has a non-terminating, non-recurring decimal representation and every rational has a non-terminating/terminating and recurring representation.[/b]

[quote="Rushil"]I'm sure this is a standard result...  [/quote]

Sure, it is ! :D
:cool:


Find all natural numbers with 3 digits, which are equal to the third power of the sum of their digits.

(as simple and as short as possible of course)

Equation to solve is N=100a+10b+c=(a+b+c)^3 (*)
100<=N<1000 so 5<=a+b+c<=9. i.e N € {125, 216, 343, 512, 729}
Now (*) shows that a+b must be even. So only possible solution is 512=8^3 which is checked to be correct.

It's a matter of maturity. I didn't do better in my starting days.

[quote="fatihk"][quote="Peter"]indeed, that's the easy way :D

I gave it to my fellow pupils here, and they all started working immediately:

 (100a+10b+c) = (a+b+c)³

well guess what, they never found it :)[/quote]

is there any way without guessing.. as you say my pupils have the same problem at the same level..  :)[/quote]

I am not sure if you are asking for a clarification of the solution...I will give one anyway. The core of the problem lies in the fact that the integer is equal to a cube of some integer. That is to say $ N\equal{}q^3$. Since $ N$ is a 3 digit number, we have very few cases to check. We have $ N\equal{}5^3,6^3,...,9^3$. From this, we see that the only solution is $ 8^3$.

I addressed this issue in the first week of my high school's math club meetings.  The key here is that many students are used to the idea of solving equations directly using algebra, rather than making use of the expedient of [b]casework[/b] (this was the title of the first week's lecture).  I will be emphasizing this point throughout the year :)


[b]In how many ways can 5 different rings be worn in 4 fingers such that exactly 1 finger has no ring?[/b]

If the two rings on a finger are switched in order, is it different?

Order does matter because it says "5 [b]different[/b] rings." There 5 options for the 1st ring, 4 for the 2nd, and so on, so you get $ 5!\equal{}120$ possible ways.

@pascal12; sorry if I am incorrect but I thought that one finger has no ring.
Here is my [hide="answer"]
I get $ 4 \cdot 5!$
[/hide]

One finger does have no ring, but it doesn't affect the answer, because once you have 4 rings on, there's only 1 possibility for the last ring. Technically, you would do a permutation resulting in $ \frac{5!}{1!}$

wait are they identical rings

and does it matter which fingers you where them on

If you read the posts before yours, this would save a lot of confusion and repetition of questions.

[quote="pascal12"]Order does matter because it says "5 [b]different[/b] rings." There 5 options for the 1st ring, 4 for the 2nd, and so on, so you get $ 5! \equal{} 120$ possible ways.[/quote]

[hide="Another way to think about it"]
There are $ \binom{5}{1}$ ways to select the finger that is not going to have the ring. Then for each of them there are $ 4!$ ways to arrange the other 4 rings on the four fingers as finger 1 having ring A, finger 2 having ring B, finger 3 having C, finger 4 having D is not the same as finger 1 having ring B, finger 2 having A, finger 3 having D, and finger 4 having C. Thus, we have a total of $ 5 \cdot 4! \equal{} \boxed{120}.$ [/hide]

[hide=" !!!!different answer!!!!"]

ok, if you notice, in the question it says "[b]5[/b] different rings be worn in [b]4 [/b]fingers such that exactly 1 finger has no ring?"

what does this mean? this means that one finger has 2 rings, and one finger has 0 rings.

there is 5C1 = 5 ways to pick the finger that does not have a ring

then for the 4 fingers remaining, you have 4! ways to arrange 4 rings. then for the 5th ring you have 4 options, it can go on any of the 4 fingers.

so one would think that the answer is $ 5\cdot 4!\cdot 4\equal{} 480$ but that is wrong.

because when you have the finger with 2 rings, each arrangement in counted twice. therefore the answer is 480/2= $ \boxed {240}$[/hide]

the answer would be 5! if all the 5 fingers were alike

[quote="dragon96"][hide="My Thoughts"]I agree with dafjbomb above. I do not get something in all the previous solutions

limac:

So that means you chose a finger NOT to have a ring, that part I agree. But when you get to the 4! part, this is counting only 4 of the rings. So where does the 5th go?

pascal12:

So once you have 4 rings on, then the 5th has ... 1 finger to go on? Wait, there are actually 4 other fingers.

harekrishna:

You are close, but don't forget if your thumb has a red ring and a blue ring, then it can be arranged in two ways.[/hide][/quote]

Whoops, sorry, I misread the problem. I thought it was 4 rings on 5 fingers.

the answer is 2400. because there are 5 ways in which 5 different fingers can be paired without a ring.

then the 4 fingers that are paired with a ring can have A, B, C, D, E so the first finger has 5 choices, second finger has 4 choices, third finger has 3 choices, fourth finger has 2 choices.

now there's 1 more ring left over and there is 4 ways to distribute the last ring.

so (5)(5 x 4 x 3 x 2)(4)= 2400.

itzmiko, wouldn't you have to then divide by 2 because the order of the rings on the finger with 2 rings would not matter.

can someone clearify what the answer is the[b] correct answer[/b]

[quote="tenniskidperson3"][hide="correct solution"]If there is one finger with no ring, then there must be 5 rings for 3 fingers, with one at least on each.  The only ways to do this is 3 on one and 1 on the other 2, or 2 on 2 of them and 1 on the third.

Case 1: 3 on one finger, 1 on other 2

We pick which finger has 3, which has 0, which rings go on the first finger, which ring goes on the second.

$ 4\cdot3\cdot\binom{5}{3}\cdot2 \equal{} 240$

Case 2: 2 on 2 fingers, 1 on third

We pick the finger without any rings, then which finger has 1, then the two rings for the first of the 2, then the two rings for the second.

$ 4\cdot3\cdot\binom{5}{2}\cdot\binom{3}{2} \equal{} 360$

The final answer is 360+240=600.[/hide]

Everyone else is assuming that the question says 5 fingers, not 4.[/quote]

Well, dafjbomb actually interpreted the question correctly. Since we now read the problem correctly, it's a matter of deciding if dafjbomb is correct in which you have 5 fingers and wear rings among 4, leaving one empty, or have 4 fingers and leave one of those empty and therefore wear it on 3. However, the problem does say "in 4 fingers" and not "within 4 fingers" so I think dafjbomb is correct.

let the rings be different
1 finger has no ring
$ \implies$5 rings in 3 fingers
let 1 finger contain 3 rings
$ \implies3\binom{3}{2}$ ways
let 2 fingers contain 2 rings
$ \implies3\binom{3}{2}$ ways
total number of ways
$ \equal{}6\binom{3}{2}*\binom{4}{3}$ ways as any one of the four fingers can be without a ring

am i correct?


Prove that :

$ \frac {1}{bc} \minus{} \frac {1}{b(a \plus{} b)} \minus{} \frac {1}{c(a \plus{} c)} \equal{} \frac {2}{(a \plus{} b \plus{} c)^2}$ , where $ a,b,c$ are the side lenghts of a right-angled triangle and $ a$ is the hypothenuse .

Try $ a\equal{}b\equal{}c\equal{}1.$ :wink:

Sorry , Kunny . I forgot to mention that $ a,b,c$ are the side lenghts of a triangle with $ \sphericalangle A\equal{}90^{o}$

[quote="PowerOfPi"]Are $ a$ and $ b$ the legs and is $ c$ the hypotenuse?[/quote]

$ a$ is the side which oppose to $ \sphericalangle A (\equal{}90^{o})$ so $ a$ is the hypothenuse and $ b\ ,\ c$ are the legs .

$ \frac {1}{bc} \minus{} \frac {1}{b(a \plus{} b)} \minus{} \frac {1}{c(a \plus{} c)} \equal{} \frac {2}{(a \plus{} b \plus{} c)^2}$

$ \Longleftrightarrow (a\plus{}b\plus{}c)^2\equal{}2(a\plus{}b)(a\plus{}c)$

$ \Longleftrightarrow a^2\equal{}b^2\plus{}c^2.$

I don't see how you went from the second step to the third step.

$ (a \plus{} b \plus{} c)^2 \equal{} 2(a \plus{} b)(a \plus{} c)$
$ \Longleftrightarrow a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca) \equal{} 2\{a^2 \plus{} (b \plus{} c)a \plus{} bc\}$
$ \Longleftrightarrow b^2 \plus{} c^2 \equal{} a^2$ by Pythagorean theorem, which shows $ \angle{A} \equal{}90^\circ$.


If a circle is to be cut by $n$ straight lines , what is the maximum possible section that can be formed ? i.e. the most section can be formed by three cuts is $7$ .

[quote="Gabriel(fr)"][hide]$\frac{(n+1)n + 2}{2}$[/hide][/quote]

It would help if you posted the solution as well... :P

Well, for the nth line...there are at maximum (n-1) lines it can cut.  Without cutting any other lines, it divides a section into two parts, and for each line it hits, it cuts a new section.  Thus, the nth line creates at maximum n new sections.  Thus, $1+\frac{n(n+1)}{2}$.

That is one cool formula that I [b]will[/b] remember.


Each outcome on the spinner below has equal probability. If
you spin the spinner three times and form a three-digit number
from the three outcomes, such that the first outcome is the
hundreds digit, the second outcome is the tens digit and the
third outcome is the units digit, what is the probability that you
will end up with a three-digit number that is divisible by 4?
Express your answer as a common fraction.

[asy]draw(Circle((0,0),10));
draw((0,0)--(8.7,-5));
draw((0,0)--(-8.7,-5));
draw((0,0)--(0,10));
label("1",(7,7),SW);
label("3",(-7,7),SE);
label("2",(0,-2),S);
draw((0,0)--(3,5),Arrow);[/asy]

I asked this.  Becuase the divisibility rule states that the last two digits need to be divisible by 4, and all the combos with 2 as the last digit work, wouldn't it just be 1/3?  The solution states 2/9.

The only way you could have this is with $ \overline{x12}$ or $ \overline{x32}$, but [i]not[/i] with $ \overline{x22}$.  Thus, the answer is $ \frac13 \times \frac23$, or $ \boxed{\frac29}$, just as the answer stated.

NOOOO I SILLIED
ALERT ALERT THE SKY IS FALLING THE SKY IS FALLING THE SKY IS FALLING
I FORGOT ABOUT THE HUNDREDS DIGIT NOOOOOOO


geometry
circumcircle
vector
geometry unsolved

In triangle ABC, O is the circumcenter, D is the midpoint of AB, E is the centroid of triangle ACD. Prove that EO is perpendicular to CD iff AB = AC.

The problem comes from Honsberger's [i]From Erdos to Kiev[/i] (pp. 230-234). There are two analytic solutions and a mention of alternative solutions at Crux Mathematicorum, 1991, 105 and 228, to which I have no access.

I am interested in a synthetic solution.

Thank you,
A. Bogomolny

Drawing the altitude $AH$ that passes through $O$ (now $H$ is midpoint of $BC$), $DF//BC$($F$ is midpoint of $AC$), $AM$ is median of $ADC$. Now $D,E,F$ and $H,M,F$ are collinear, due to $DF$ is median of $ADC$ and $HF$ is median of $AHC$,so $HF//AB$, $BF$ and $CD$ concur at $P$ on $AH$($P$ is centroid of $ABC$). Besides: $DF$ cut $AH$ at $N$ and $DF$ is perpendicular to $AH$ then $PN$ is altitude of $DPE$, also: $OD$ is perpendicular to $AB$ and $PE//AB$, due to $AP=2PH$, $AE=2EM$, so: $PE//HM//HF//AB$, then $O$ is the orthocenter of $DPE$, hence $OE$ is perpendicular to $CD$ ($DP$).

Let me see.

In your notations, PE||AB because P is on the median CD of triangle ABC while E is on the median of triangle ACD through C. DF||BC as a midline.

When P lies on AOH, O is the orthocenter of triangle PDE, because DO is perpendicular to PE and OP is perpendicular to DE. It follows that EO is the third altitude and therefore is perpendicular to the third side, DP.

Conversely, if EO is perpendicular to CD (DP), then since OD is perpendicular to PE, O is again the orthocenter of PDE. Hence, the altitude from P passes through O, such that in triangle ABC the circumcenter O and the centroid P are collinear with the midpoint of BC. Thus the median from A is also the altitude and ABC is bound to be isosceles.

Very good.

Thank you,
AB

hmm this is in PSS's vector section (just did it today :rotfl: ) so here's the vector solution.

It is best here to let $O$ be the origin.  We have $D = \frac{A+B}{2}$ and $E = \frac{3A+2C+B}{6}$.  $EO$ perpendicular to $CD$ implies $(3A+2C+B)(A+B-2C) = 0$, and $|B-A| = |C-A|$ implies $(B-A)^{2}= (C-A)^{2}$. (All multiplication stands for dot product here).  expanding and taking into account that $A^{2}= B^{2}= C^{2}$ since $O$ is our origin, we can see that one implies the other.

[color=darkred][b]Proposed problem.[/b] In the triangle $ABC$, $O$ is the circumcenter, $D$ is the midpoint of $AB$,
$E$ is the centroid of triangle $ACD$. Prove that $EO$ is perpendicular to $CD$ iff $AB = AC$.[/color]

[color=darkblue][b]The first method (metric).[/b] Denote the middlepoints $F$, $L$ of the segments $[AC]$, $[AD]$ respectively. Then,
$4\cdot (CL^{2}-DF^{2})=2\cdot (CD^{2}+CA^{2})-AD^{2}-2\cdot (DA^{2}+DC^{2})+AC^{2}\Longleftrightarrow$ $4\cdot (CL^{2}-DF^{2})=3\cdot (AC^{2}-AD^{2})\ .$
$EO\perp CD$ $\Longleftrightarrow$ $EC^{2}-ED^{2}=OC^{2}-OD^{2}$ $\Longleftrightarrow$ $EC^{2}-ED^{2}=OB^{2}-OD^{2}$ $\Longleftrightarrow$ $EC^{2}-ED^{2}=BD^{2}$ $\Longleftrightarrow$
$4\cdot (CL^{2}-DF^{2})=9\cdot BD^{2}$ $\Longleftrightarrow$ $3\cdot (AC^{2}-AD^{2})=9\cdot AD^{2}$ $\Longleftrightarrow$ $AC=2\cdot AD$ $\Longleftrightarrow$ $AC=AB\ .$

[b]The second method (vectorial).[/b] Let $O$ be the origin of the vectorial system.
Thus, $(X,X)=OX^{2}$ and $(X,Y)=\frac{1}{2}\cdot\left(OX^{2}+OY^{2}-XY^{2}\right)\ .$
$2D=A+B$, $2F=A+C$ ; $3E=A+D+C\Longleftrightarrow$ $6E=2A+(A+B)+2C$ $\Longleftrightarrow$ $6E=3A+B+2C\ .$
Therefore, $EO\perp DC$ $\Longleftrightarrow$ $(E,C-D)=0$ $\Longleftrightarrow$ $(3A+B+2C,2C-A-B)=0$ $\Longleftrightarrow$
$6(A,C)-3(A,A)-3(A,B)+2(B,C)-$ $(A,B)-(B,B)+4(C,C)-2(A,C)-2(B,C)=0$ $\Longleftrightarrow$
$(*)\ \ \--4(A,B)+4(C,A)-3(A,A)-(B,B)+4(C,C)=0$ $\Longleftrightarrow$ $(A,B)=(A,C)$, i.e. $AB=AC\ .$
Otherwise, the relation $(*)$ is equivalently with $-2(2R^{2}-c^{2})+2(2R^{2}-b^{2})=0$ $\Longleftrightarrow$ $AB=AC\ .$

[b]Remark.[/b] $AB=AC$ $\Longleftrightarrow$ $AO\perp BC$ $\Longleftrightarrow$ $(A,B-C)=0$ $\Longleftrightarrow$ $(A,B)=(A,C)\ .$[/color]

[color=darkred][b]Equivalent enunciation.[/b] Let $ADOF$ be a quadrilateral inscribed in the circle with the diameter $[AO]$.
Let $E\in (DF)$ be the point for which $DE=2\cdot EF$. Denote the middlepoint $L$ of the side $[AD]$.
Prove that $OE\perp FL\Longleftrightarrow AD=AF\ .$[/color]

[size=134][color=darkblue][b]Remark.[/b] An extension of the proposed problem you find to the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=107793 [/size][/color]


Hello,

I am trying to find the value of $ tan\frac{pi}{2}$ and $ cot\frac{pi}{2}$ but my calculator is giving an ERROR for the closing parenthesis of both trigonometric functions. I am in RADIAN mode and am using a TI-83 plus.

Thank you.

*Note: Moderators, please move topic if this isn't the right forum for graphing calculator discussions.

$ \tan x\equal{}\frac{\sin x}{\cos x}$ and $ \cos\frac{\pi}{2}\equal{}0$, so the denominator of $ \tan x$ is 0.

$ \cot\frac{\pi}{2}\equal{}0$


Pascal's Pizza Parlor is changing the size of its circular pizza from 12 inches to 16 inches and increasing the number of slices per pizza from 8 to 10 equal pieces. What is the percent of increase to the nearest whole percent of the size of each new slice?

You are increasing the area of the pizza by 16/9, but decrease area by 8/10. Multiplying, 128/90=1.4222...

Or, [b]42[/b] percent increase.

The area of the pizza increases by (16/12)^2 = 16/9. The portion for one slice decreases by 8/10=4/5. So the area of a slice increases by 64/45=1.42 repeating so our answer is 42%.

*when I say increases/decreases I mean "is multiplied by"

Note: It may be easier not to simplify 8/10 as you can just divide 128/9 and then move down the decimal point


function
integration
calculus
limit
number theory
prime factorization
calculus computations

Suppose $ F_{a}(x) \equal{} \int_{a}^{x} f(t) \ dt$.

What are some examples of functions $ F_a$ which are continuous but not differentiable?

Well if f is a step function, then F is clearly not differentiable where the step is. The integral of any function, when it exists, is continuous.

but is the step function Riemann integrable?

[quote]Well if f is a step function, then F is clearly not differentiable where the step is. [/quote]

It's also not continuous, so it doesn't satisfy the premise of the problem.

The simplest example I can think of is $ x^{1/3}$. It is continuous on $ \mathbb{R}$ though not differentiable at $ 0$. And like Fibonnaci said, all functions that are continuous on some interval $ [a,b]$ (or have finite jump discontinuities, but that's not relevant to this case) will be integrable on the same interval.

Edit: What's the difference between $ F(x)$ and $ F_{a}(x)$?

Also, did you mean a function that is continuous everywhere but differentiable nowhere?

Let $ a \in \mathbb{R}$, let $ I$ be an interval containing $ a$, and let $ f: I \to \mathbb{R}$ be a function. Suppose that $ \int_{a}^{x} f$ exists for all $ x \in I$. Define the function $ F_a: I \to \mathbb{R}$ by $ F_{a}(x) \equal{} \int_{a}^{x} f(t) \ dt$.

So we want $ F_{a}$ to be continuous without being differentiable.

No, Prime factorization, you misunderstand. There is no requirement that $ f$ be continuous, and step functions are Riemann integrable. FibonacciFan is right.

In fact, that's almost exactly the condition; if this is the proper Riemann integral and we assume $ f$ is integrable on a large enough interval, $ F_a$ is not differentiable at $ x$ if and only if $ \lim_{t\to x}f(t)$ does not exist.
[Edit- fixed a mistake]
[Edit 2- Ack, I wasn't thinking straight; there are other possible discontinuities. Fixed now.]

So the step function is an example where $ F_a$ is continuous but not differentiable?

Hi :) !!!

I didn't understand what you wand exactly but i think that:

$ F_a(x)\equal{}\int_{a}^{b|x\minus{}c|\plus{}a}dt\equal{} b|x\minus{}c|$ with $ a,b,c \in I$

than $ F_a$ are continue in $ c$ but not differentiable in $ c$

....
thx
Ps: ????
__________________________
lahoucine

[quote="jmerry"]No. That can't be an antiderivative, and it's continuous so it can't be $ f$ either. It's all about non-removable discontinuities, and step functions are the easiest way to get there.[/quote]

But the step function is not continuous. We want functions $ F_a$ that are continuous but not differentiable.

[quote="jmerry"]Read carefully. There are [b]two[/b] functions here.[/quote]

yep. $ f$ is the step function. and $ F_a$ is continuous but not differentiable.


[hide]
Multiply both sides by $x^2-3x+2$ to get

$35x-29=N_1(x-2)+N_2(x-1)\\ 35x-29=N_1x-2N_1+N_2x-N_2$

Now let the terms of degree one be equal to each other and the terms of degree 0 be equal. Then we have

$N_1+N_2=35\\ -2N_1-N_2=-29\\ \text { Adding them up, we find}\\ -N_1=6\\ N_1=-6, N_2=41, N_1N_2=-246 \implies A$[/hide]


conics
LaTeX
percent
analytic geometry
graphing lines
slope
geometry

This is old and all, but I want to know what careless mistakes everyone else made in the 8th grade, because that is essentially the whole point of the IML: not making dumb mistakes.

Wait, Honey Creek didn't make mistakes.  NVM ignore this. :D  Maybe that was just insulting to us.  Who knows...

[quote="unimpossible"]what the heck is indiana math league?  :P[/quote]

lol its a(n) *easy* math contest that i failed on.

Yes, it's a lower-than-prealgebra math contest that 99% of students at my school would get below 20/40.  It's so sad, yet I don't pity them at all.

maybe you're one of those 99%? heh jaykayyy

did anyone beat my/mihika's score of 40? :O

um yeah i failed. 38 shut up. one of the questions i missed

find the reciprocal of $ \frac{1}{2 \plus{} \frac{1}{2}}$.

my answer: 2/5
correct answer: 5/2

those are the kinds of mistakes i make. i believe someone in our school beat me. she got a 39. subsy isn't she quite a beast i mean she got like 41 on chapter.  :o

oh i didn't even know indiana had that.

we had the same questions in virginia, except called Virginia Math league.

40. yay. 3 perfects in my school, a 39, 2 38's. bunch of 37's

The whole country actually takes the same test, you just compete against the people in your state or region.  There is a New England Math League for the 6 states for example.  Youcan view past results on mathleague.com.  They have a high school contest too that is very different and monthly.

The score that I received is equal to either latus rectum of the conic whose transverse axis is a line parallel to the x-axis and has a length of 6.  Furthermore, this conic's conjugate axis has a length of 2 times the square root of 108.

Good luck finding that out to you all, except cognos and mustitalwaysbeme, who already know.

BTW, do I have to download things to use LaTeX on AoPS?

$ \mbox{No, you don't need to download anything to use }$. $ \mbox{Btw mathandknowledge}$, $ \mbox{i solved that problem and i figured out your score but i won't tell anyone i will say that it is \ge 90 percent}$. $ \mbox{you don't want people thinking you got less}$  :wink: $ \mbox{ I will teach you about  \text{\LaTeX}\ at school sometime wonny}.$

guess what my score was

$ \frac{x}{100} \equal{} \frac{2}{5}$

solve THAT :wow:

se-hyoon, we all know you got a 40...and we all now know I got a 37.  And my name isn't subsy.

I missed 5, 29, and 39.

5 was something like: $ \mbox{? makes 6 points every 12 minutes. In how many hours will he make 36 points.}$
The right answer was $ 1.2$, I put $ 1$.

29: $ \mbox{Starting with 2, counting by 3, what is the 50th number in the sequence?}$
The right answer was 149, I put 152.

39: $ \mbox{How many factors of (270?) are multiples of 9.}$
I forgot to include $ 9$ as a multiple.

So there you have it.  My stupidity.

Edit: @MathAndKnowledge: The $ \text{\LaTeX}$ software is installed on the AoPS servers, so you don't have to download anything.  You simply put your $ \text{\LaTeX}$ code in 2 $ \$$ signs.

haha. very funny. mine was

$ 39 \plus{} i^{1\plus{}2\plus{}3\plus{}\ldots\plus{}64}$

lol  $ \frac{64(65)}{2}$ is obviously a multiple of $ 4$.

mwpl should shut up.

or the side length of a square with area 1600. :D

[quote="cognos599"][quote="undefined117"]and subsy has 64 posts (:

mine's the slope of this line:

$ y \equal{} 40x$[/quote]

shut up sooey, my mathcounts chapter score was like twice ur's plus mihika's combined. OHHHHH. what now??!??!?!?!?! :wow:[/quote]

huh, sarcasm is usually used towards those who are [i]inferior[/i]. this just doesn't make sense.

[quote="cognos599"]
find the reciprocal of $ \frac{1}{2 \plus{} \frac{1}{2}}$
[/quote]
I missed the reciprocal question too. :(

[quote="cognos599"][quote="undefined117"]and subsy has 64 posts (:

mine's the slope of this line:

$ y \equal{} 40x$[/quote]

shut up sooey, my mathcounts chapter score was like twice ur's plus mihika's combined. OHHHHH. what now??!??!?!?!?! :wow:[/quote]

chapter this year was easier than 1990 chapter.

um, 1990? well yeah i guess this test was really really easy.

[quote="MustItAlwaysBeMe"][quote="mathcountsfreak"]
I missed the reciprocal question too. :([/quote]
I thoughted you gotsded a perfect.  :D[/quote]

We only had 2 perfects. The others in our top 5 were 39s.

[quote="mathcountsfreak"][quote="MustItAlwaysBeMe"]
I thoughted you gotsded a perfect.  :D[/quote]

We only had 2 perfects. The others in our top 5 were 39s.[/quote]

Yeah, I heard (cognos).  So we pretty much got pwned... :(

well the results are out online here.
http://www.mathleague.com/reports/leagues.htm

Wow... I got the only 40 in my state.
And my school won! Yay!

[quote="MustItAlwaysBeMe"]1.  Mark Sellke took it and got a perfect.
2.  Me and MathAndKnowledge aren't even on it by region because of him.
3.  I got 13th?!  Is this a new low?  Probably not.
4.  Do I get to regrade Mark's test?
5.  Do I get to regrade my test?
6.  Will they take off for (the thing that happened that shouldn't be discussed here) on my test?
7.  We got pwned by Honey Creek.
8.  We got pwned by Mark Sellke.
9.  We got pwned by Andrew Keiser.
10. Our school even got pwned by Mayfawny.

That's a lot of reasons that I am mad.[/quote]

mad because you lost to people who are better than you?

i lost to person(s) who are worse than me. for example a person in my school beat my and i have beaten her in every other math contest. i just make too many stupid mistakes. that is why mathleague is such a rough contest for me. like on AMC 8 where i missed #4 or 3 or something.  :rotfl:

[quote="academicsrule"]I'm not from Indiana, but...
I got a 40 on the same test! Yay no stupid mistakes!
We have GML, which is the exact same test, so... yeah.[/quote]
GML must stand for the nonexistant Connecticut Math League.  Is your name Jared? Or Qian?  (Georgia Math League)


If $ a\in\mathbb{N}$, prove that $ \gcd(2a\plus{}1,3a\plus{}1)\equal{}1.$

Um...$ \gcd(2a \plus{} 1,3a \plus{} 1) \equal{} \gcd(2a \plus{} 1,a) \equal{} \gcd(1,a) \equal{} 1$...HSB level?

consider $ \left( 2a\plus{}1,3a\plus{}1\right)\equal{}d$
$ \Rightarrow  3a\plus{}1\equal{}\left(2a\plus{}1 \right)\plus{}a \vdots d$
$ \Rightarrow a \vdots d$
we have $ 2a\plus{}1 \vdots d \Rightarrow  1 \vdots d$
$ \Rightarrow d\equal{}1$


logarithms quadratics function algebra domain

[quote="D.P.L"][hide="6"]No solution :D [/hide][/quote]

Dude, please [hide="reconsider"]Try $x = 1$.  Then, $1^{1}= 1 = 10^{1-1^{2}}= 10^{0}= 1$. [/hide]

[hide="3"]
Let $a=\log x$, then $\frac{a}{2}= \log_{100}x$. Rewriting the equation in terms of $a$, \[3a^{2}+6(\frac{a}{2})a-2 \left(\frac{a}{2}\right)^{2}= 0 \Longrightarrow \frac{11a^{2}}{2}= 0 \Longrightarrow a = 0\] then $x$ equals $1$.
[/hide]

[hide=" Number 2"]

Substitute: $a= (\frac{1}{2})^{x}$

Solve the quadratic and you get: $a=\frac{3}{4}$ or $a=-1$

So: $(\frac{1}{2})^{x}=\frac{3}{4}$ $\Rightarrow$ $\boxed{x=\log_{2}{\frac{4}{3}}}$

[/hide]

[hide=" Number 4"]
$x-y=\ln 3$
$3x+4y=\ln 8$

Solving the system: $x=\ln \sqrt[7]{648}$  and  $y=\ln \sqrt[7]{\frac{8}{27}}$[/hide]

[hide=" NUmber 5"]
$b^{2x}a^{-2x-2}=1$
$(\frac{b}{a})^{2x}=a^{2}$  $\Rightarrow$  $(\frac{b}{a})^{x}=a$
$\boxed{x=\log_\frac{b}{a}a}$

$x \log\frac{b}{a}= \log_\frac{b}{a}a \log\frac{b}{a}=\boxed{\log a}$[/hide]

1b =)[hide]
For A(t), the function is 1 to 1 when it is both x vs. y and y vs. x. I forget the term used to describe something like that. well anyway,

A(t): Domain: Based on common sense, t[u]>[/u]0
                  since you don't get negative time.
         Range: When t=0, A(0)=P. t=infinity, A(inf) = infitny
         And based on the exponential form, the function is ever increasing.
              Range is [P,inf)

For
           A -1  (t), Domain is [P,inf)
                     Range is [0,inf)[/hide]

[quote="D.P.L"][hide="6"]No solution other than $1$ [/hide]

Thanx  :oops:   :roll:[/quote]

Please don't edit it if someone else already points out the mistake. Just move on.
Anyways,
#6:

x^x = 10^x/(10^(x^2))
x= 10/(10^x)
x = 10^(1-x)
logx + x - 1 = 0
x=1 since x cannot be greater than 1, or else, logx+x>1 and if x is less than 1, logx + x<1.

let A-1(t) = g

t = p(1-r/m)^(mg)
Take the log of both sides.

logt = mg log[p(1-r/m)]
g = (logt)/(log(p(1-r/m))

Voila!

I wrote up solutions to all these in some other forum, I think it was getting started...


ratio
algebra
polynomial
number theory solved
number theory

Find out all the natural numbers n such that ,for every aritmethic sequence  a_1,a_2,...a_2n (with 2n natural terms) the number:

a_1*a_2*...*a_2n + r^2n is a perfect square. (r is the ratio)

I found a solution, but I'm convinced that there exists a far less complicated solution for this problem. Anyway, I will post it.
  It is based on two famous results. The first one says that if P is a polynomial with integer coefficients such that for all natural n  P(n) is a square, then P is the square of a polynomial with integer coefficients. Since for all natural k the number k(k+1)...(k+2n-1)+1 is a perfect square, it follows that one can find a polynomial f with integer coefficients such that f^2(X)=1+X(X+1)...(X+2n-1). It follows that degf=n and that
(f(X)-1)(f(X)+1)=X(X+1)...(X+2n-1). Hence we can say that f(X)-1=(X+a_1)...(X+a_n) and f(X)+1=(X+a_n+1)...(X+a_n), where a_1,...,a_2n are the numbers 0,1,...,2n-1 in a certain order.So, (X+a_n+1)...(X+a_2n)-(X+a_1)...(X+a_n)=2. Hence (a_1-a_n+1)*...*(a_n-a_n+1)=-2 and from here it follows easily that n<=2. Obviously, n=1 is not a solution and one can easily check that n=2 is a solution.


Explain how to locate point P given that

$ PA: PB: PC : : 1: 2: 4$ in an equilateral triangle $ ABC$

Two conditions are enough, say $ \frac {PA}{PB} \equal{} \frac {1}{2}$ and $ \frac {PA}{PC} \equal{} \frac {1}{4}$ This implies intersecting two locus of P , the locus of P whose ratio of distances to A,B is 1/2 and the locus of P whose ratio of distances to A,C is 1/4,  The locus are the Apollonian circumferences , which is the locus of the point whose ratio of distances to two fixed points is a constant number . Draw the two Apollonian circumferences with respect the segments AB,AC with ratios 1/2,1/4 .
the meetpoint of both circumferences satisfies the wanted condition.

[quote="sunken rock"]It seems the problem is impossible, since $PA, PB, PC$ cannot form a triangle, under given conditions!

Best regards,
sunken rock[/quote]

Don't know, exactly what you mean..
P is   at the intersection of two circles. (There are two possible location's of P). One can easily solve (and get the equation of required circles as suggested above in Nebraska boy's post

For example, if A is at (0,0) and B is at (3,0), the point P has to lie on a circle, with radius of 2 and center at (-1,0).
(Or two points of the diameter are at (1,0) and (-3,0), easy to see that PA:PB ratio is observed)

One can also see this by simply solving:
$2^2 (x^2+y^2) = (x-3)^2+y^2$
or $x^2+y^2+2x =3 $ (or $(x+1)^2+y^2 = 4$

(Similarly, the other circle, will have two points of diameter on line AC such that PA:PC::1:4 - In other words   A point  X between A and C where XA = AC/5 and a point Y on the other side, such that AY = 4*AC)
(Of course, one can solve it analytically easily enough ... Point P need not (and is not) inside the triangle)

I think he meant this: http://en.wikipedia.org/wiki/Pompeiu's_theorem


algebra
polynomial
function
induction
floor function
algebra unsolved

Determine all polynomials $ p$ with real coefficients such that

$ p(0)\equal{}0$ and $ f(f(n))\plus{}n\equal{}4f(n)$ for all natural numbers $ n$, where

$ f(n)$ is the floor function of $ p(n)$ for all $ n$.

$ x^2-4x+1=0$ has roots $ 2\pm\sqrt{3}<4$.
So for each $ n\in\mathbb{N}$ we find $ \alpha,\beta\in\mathbb{R}$
with $ f^{(k)}(n): =\underbrace{f\circ f\circ \cdots \circ f}_{k\text{ times}}(n)=\alpha (2+\sqrt{3})^k+\beta (2-\sqrt{3})^k$ for all $ k\in\mathbb{N}$.
This gives $ {f^{(k)}(n)\over 4^k}\mathop{\longrightarrow}_{k\to\infty} 0$ for all $ n\in\mathbb{N}$.

But if $ \deg p\ge 2$ we find $ N\in\mathbb{N}$ with $ |p(x)|>4|x|+1$ for all $ x\in\mathbb{R},|x|>N$.
Then $ 1>|p(x)-f(x)|\ge |p(x)|-|f(x)|$ and $ |f(x)|>|p(x)|-1>4|x|$ for all $ x\in\mathbb{R},|x|>N$.
This gives by induction $ |f^{(k)}(x)|>4^k|x|$ for all $ k\in\mathbb{N}$ and $ x\in\mathbb{R},|x|>N$.
So $ n<\left|{f^{(k)}(n)\over 4^k}\right|\mathop{\longrightarrow}_{k\to\infty} 0$ for all $ n\in\mathbb{N},n>N$. Contradiction!

So $ p(x)=ax+b$ for some $ a,b\in\mathbb{R}$ (forget $ p(0)=0$)
and $ f(n)=[an+b]=an+b-\{an+b\}$

Then
$ 0=f(f(n))-4f(n)+n\\
\hspace*{0.07in}=[g(f(n))]-4f(n)+n\\
\hspace*{0.07in}=[af(n)+b-4f(n)+n]\\
\hspace*{0.07in}=[(a-4)(an+b-\{an+b\})+b+n]\\
\hspace*{0.07in}=[(a^2-4a+1)n+ (a-3)b+(4-a)\{an+b\}]\\
\Leftrightarrow (a^2-4a+1)n+ (a-3)b+(4-a)\{an+b\}\in[0,1)$
for all $ n\in\mathbb{N}$.

Then $ \{(a^2-4a+1)n\}_{n\in\mathbb{N}}$ is bounded, as $ \{an+b\}\in[0,1)$ for all $ n\in\mathbb{N}$.
So $ a^2-4a+1=0$ and $ 4>a=2\pm\sqrt{3}\not\in\mathbb{Q}$
and therefore $ \{\{an+b\}: n\in\mathbb{N}\}$ is dense in $ [0,1)$.

This gives
$ f(f(n))-4f(n)+n=0$ for all $ n\in\mathbb{N}$
$ \Leftrightarrow (a-3)b+(4-a)\{an+b\}\in[0,1)$ for all $ n\in\mathbb{N}$
$ \Leftrightarrow (a-3)b=\inf_{n\in\mathbb{N}}( (a-3)b+(4-a)\{an+b\})\ge 0$
and $ (a-3)b+4-a=\sup_{n\in\mathbb{N}}( (a-3)b+(4-a)\{an+b\})\le 1$
$ \Leftrightarrow 4-a\le 1$ and $ a=2+\sqrt{3}>3$ and $ 0\le b\le 1$.

This gives the only solutions $ p(x)=(2+\sqrt{3})x+b$ for $ 0\le b\le 1$.


algebra
difference of squares
special factorizations

How much greater than $ 19^2$ is $ 31^2$ ?

Factor a as a difference of squares: $ a^2 \minus{} b^2 \equal{} (a\minus{}b)(a\plus{}b)$

$ 31^2 \minus{} 19^2 \equal{} (31 \minus{} 19)(31\plus{}19) \equal{} (12)(50) \equal{} \boxed{600}$


$ ABC$ is an acute angled triangle.Let $ D$ be a point on $ BC$ s.t. $ AD$ is the bisector of the angle $ \angle BAC$.
Let $ E$,$ F$ be the feet of perpendiculars from $ D$ to $ AC$,$ AB$,respectively.Suppose $ BE$ and $ CF$ meet at $ H$.
Prove that $ AH$ is perpendicular on $ BC$.

Call J the feet of the altitude from A and $ K: AH \cap BC$. For the Ceva theorem we have:

$ \frac{BF}{FA} \cdot \frac{AE}{EC} \cdot \frac{CK}{KB}\equal{}1$ and then $ \frac{CK}{KB}\equal{}\frac{EC}{BF}\equal{}\frac{\tan{\beta}}{\tan{\gamma}}\equal{}\frac{CJ}{JB} \ \Longleftrightarrow K \equiv J$

Dear Mathlinkers,
the problem can be reformulated in this way:
let ABC be a triangle, D a point on BC, E and F the feet of perpendiculars from D to AC, AB,respectively, and H the point of intersection of BE and CF.
Prove that  D is the foot of the A-bisector if, and only if, AH is perpendicular on BC.
Sincerely
Jean-Louis

Call J the point on BC such that AJ and BC are perpendicular, see that A, F, J, D and E are cyclic, hence <BJF = <CJE ( 1 ) and <BFJ + <JEC = 180 ( 2 ). Applying sine theorem, from the triangles BFJ and ECJ we get, with (1) and (2),  BF/CE = BJ/CJ  ( 3 ). Further, with AE = EF, from Ceva we get AJ, BE and CF concurrent.

Best regards,
sunken rock


geometry
circumcircle
parallelogram
power of a point
radical axis
geometry solved

We have 2 circles which intersect at A and B. Let PQ be the common tangent of the 2 circles. The tangents at P and Q to the circumcircle of PAQ intersect at S. The symmetric of B with respect to PQ is T.

Show that T, A and S are collinear.

Just a thought: I think that the Proposed problems - Discussed Topics is redundant. After someone solves one of the problems, it ( the problem) can be moved to Algebra Solved Problems, Geometry SP, NT SP,...

Now to the problem.

Let C1(O1,r1) and C2(O2,r2) be the 2 circles. Suppose P is on C1 and Q is on C2. Suppose A is closer to PQ than B. (this is if you want to get the same picture as mine).
Let O be the circumcenter of PAQ, and let r=OA=OP=OQ.
Let R=AB \cap PQ

Let <PO1A=2a, <QO2A=2b. It is easy to see that PBA=QPA=a and <QBA=<PQA=2b.
AO1PO is a kyte, so <OO1P=a. Also< OO2Q=b.
O is the circumcenter of PAQ, so <POA=2b and QOA=2a. It is easy to see that <POO1=b and QOO2=a.
So the triangles POO1 and QO2O are similar, so PO/r2=r1/QO, so r 2 =r1*r2.

R lies on AB, the radical axis of C1 and C2, so R has equal powers with respect to C1 and C2. But RP and RQ are the tangents from R to C1 and C2, so R is the midpoint of PQ. It is now clear that OS \cap PQ=R.

Let A' be the symmetric of A with respect to R. Then A'PAQ is a paralelogram and A'PQ=A'BQ=b, so A'PBQ can be inscribed in a circle. This circle is circumscribed to PA'Q which is congruent to PAQ, so the circle has radius r.
From these it should be easy to see that TPAQ is inscribed in the circle of center O and radius r.

Lemma: TS is the symedian (symmetric of the median with respect to the bisector) from T of triangle TPQ.
Proof of Lemma: <PTQ=1/2*<POQ=a+b, so <SPQ=<SQP=a+b.
Let M=TS \cap PQ. PM/MQ=area[TPS]/area[TQS]=(TP*PS*sin(TPS))/(TQ*QS*sin(TQS))=TP*sin(TQP)/TQ*sin(TPQ)=TP 2 /TQ 2, so TS is the symedian.
End of proof of Lemma.

So all we need to prove now is that A lies on the symedian from T.
Let N=PQ \cap TA. Then PN/NQ=area[PAT]/area[QAT]=TP*PA/TQ*QA [b](1)[/b](remember TPAQ is inscribed in a circle). We want to peove that TA is the symedian. This is equivalent to proving PN/NQ=(TP/TQ) 2 , which using (1) is equivalent to PA/AQ=TP/TQ.
Looking back at good old inscribed quadrilateral A'PBQ, the last is equivalent to A'Q/A'P=PB/QB whis is equivalent to A'P*PB=A'Q*QB, which because the quadrilateral is inscribed is equivalent to area[A'PB]=area[A'QB].
The last is true because BA' is the median of BPQ.

The problem is now solved.

Let $U\equiv AB \cap PQ ,V$ is the symmetric of $B $  in $U$.
first $U$ is on the radical axis  thus $U$ is midpoint of $PQ ,\angle PBQ=\angle PBA +\angle ABQ=\angle APQ +\angle QPA =\pi - \angle PAQ $ so $T,V$ are in $\odot (APQ)$ but $BPVQ$ is parallelogram then $QV=BP=PT \implies \angle VAQ = \angle  TAP $ which means that $ AT$ is the symmedian therefore it pass through   $S$


The set $ S$ is a finite subset of $ [0,1]$ with the following property: for all $ s\in S$, there exist $ a,b\in S\cup\{0,1\}$ with $ a, b\neq s$ such that $ s \equal{}\frac{a\plus{}b}{2}$. Prove that all the numbers in $ S$ are rational.

Assume there was an irrational number in $ S$, only finitely much, and number them $ x_{1}\le\ldots\le x_{n}$. Now $ x_{1}\equal{}\frac{a\plus{}x_{k}}{2}$ with $ a<x_{1}$ rational. So, we can express $ x_{1}$ as a (nonnegative) linear combination of rationals and irrationals with higher indices (over $ \mathbb{Q}$).

Now let $ t<n$ be the smallest index for which this is impossible, then $ x_{t}\equal{}\frac{a\plus{}b}{2}$ with $ a>x_{t}>b$ and necessarily $ b$ irrational (else $ a$ irrational but must have higher index). But then $ b\in\{x_{1},...,x_{t\minus{}1}$, which is a nonnegative linear combination of $ x_{t},...,x_{n}$. Now, if the coefficient in that combination of $ x_{t}$ is not $ 1$, then we can solve it for $ x_{t}$ as a (nonnegative) linear combination of $ x_{t\plus{}1},\ldots,x_{n}$, contradiction. So $ x_{t}$ appears in coefficient $ 1$ and, by nonnegativity, the rest appears in coefficient $ 0$, so $ b\equal{}x_{t}$, contradiction.

So $ t\equal{}n$ and thus every $ x_{i}$ is a rational multiple of $ x_{n}$. But then $ x_{n}\equal{}\frac{a\plus{}b}{2}$ with $ b>x_{n}$ rational and $ a<x_{n}$ irratiional. But then $ a\equal{}qx_{n}$, and thus $ b\equal{}(2\minus{}q)x_{n}$ would be irrational, contradiction. So there can be no irrational number in $ S$.

Another solution:

We consider $\mathbb R$ as  a vectorial space over $\mathbb Q$. Let $V$ be the subspace generated by $S$. If $V\neq\mathbb Q$, take $v_1,v_2,\ldots$ in $S$ such that $\{1,v_1,v_2,\ldots\}$ is a basis for $V$.

Each $x\in S$ can be written as $a_0+a_1v_1+...$, with $a_i\in\mathbb Q$.

Consider the number $x$ whose coefficient of $v_1$ is maximal. It is the average o two other numbers. Clearly, both of them have the same $v_1$ coefficient as $x$.

We find a contradiction by choosing $x$ as large as possible.$\square$


We say that an integer $m \ge 1$ is super-abundant if \[\frac{\sigma(m)}{m}>\frac{\sigma(k)}{k}\] for all $k \in \{1, 2,\cdots, m-1 \}$. Prove that there exists an infinite number of super-abundant numbers.

[quote="Robert Gerbicz"]Suppose indirectly that there are only finite number of super-abundant numbers and let $ N$ is the largest. Let $ p>N$ prime number, then $ gcd(N,p)=1$ so $ {\sigma(N*p)}={\sigma(N)}*(p+1)$ from this \[ \frac{{\sigma(N*p)}}{N*p}=\frac{{\sigma(N)}*(p+1)}{N*p}>\frac{\sigma(N)}{N}\]

So there is at least one $ m$ number greater than N for that \[ \frac{\sigma(m)}{m}>\frac{\sigma(N)}{N}\], if K is the smallest among them, then K is trivially a super-abundant number.[/quote]

In fact, it is a trivial corollary to this simple (derived from the above)

[b]Theorem.[/b] [color=#0000FF]Let $(a_n)_{n\geq 1}$ be a sequence of positive real numbers. Call a term $a_m$, $m\geq 2$, to be a [i]gigant[/i] if $a_m > a_k$ for all $1\leq k<m$. Then, if there exists no largest element, there are infinitely many gigants.[/color]

[i]Proof.[/i] Assume $a_N$ is the largest gigant. Since there exists no largest element, the set $\{m \mid a_m > a_N\}$ is non-empty. Its least element $M$ then yields a gigant $a_M$, since $a_k \geq a_M$ for some $1\leq k < M$ would force $a_k > a_N$, thus $N<k<M$, contradicting the minimality of $M$.

All that was left then was to show that the sequence $\left (\dfrac {\sigma(n)} {n}\right )_{n\geq 1}$ has no largest element; in fact it is even upper unbounded. Indeed, for $n= \prod_{k=1}^s p_k$ we have $\dfrac {\sigma(n)} {n} = \prod_{k=1}^s \left (1+ \frac {1} {p_k}\right ) > 1+   \sum_{k=1}^s \frac {1} {p_k} \to \infty$ when $s\to \infty$.


[i]NOT[/i] The kind that you order to go at McDonalds. The mathematics kind! What is it?

If you have boxes of $n$ chicken McNuggets and boxes of $m$ Chicken McNuggets, the least amount of chicken mcnuggets that you cannot buy (assume your rich) is $mn-m-n$ or $mn-(m+n)$ or product - sum.  Note: They MUST be relatively prime.

Please don't start unnecessary threads.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=97888[/url]

[color=blue][size=75]kyyuanmathcount: oops, didn't see that, locked.[/size][/color]


algebra
polynomial
number theory open
number theory

Let r(n) be the number of distinct prime divisors of n
r(50)=2 (as 50=2*5*5 and has 2 distinct prime factors 2 and 5)

Let f(n)=n^2+1
Then, my conjecture is that:

The minimum k (integer) satisfying
r(f(k))=i (i >=2 is an integer)
is a prime number.

To explain, lets take an example.
Take i=3.
f(1)=2
f(2)=5
f(3)=10
f(4)=17
f(5)=26
f(6)=37
f(7)=50
.......
f(13)=170=2*5*17..................therefore, 13 is the min. number for which r(f(k))=3 and 13 is a prime number!!

for i=2, we can see that 3 is the required min number, which is again prime.

Is this true for all i ?? (where i=positive integer >=2)
And, is this true for any non constant polynomials ??(except f(n)=n)

It's false; the smallest $ k$ for which $ r(f(k))\equal{}5$ is $ 447\equal{}3\cdot 149$.


algebra
polynomial
linear algebra
linear algebra unsolved

T:V->V is a linear transformation and V is a unitary space, and:
1) |c|=1 for every eigen value c (c is complex).
2) for every v in V, ||Tv||<=||v||

prove: T is a unitary linear transformation

and another question:

does exist a linear transformation T:R^n->R^n
such that:
1) for every v, ||Tv||=2001||v||
and: 2) the minial polynomial of T is : m(t) = t^2+2t+3           ?


A 1$ \%$ late charge was added to Jenna's bill on the $ 30^{\text{th}}$ day
past its due date. The resulting total was then increased by 1$ \%$
because she did not pay the bill in the next 30 days either. Her
original bill was $ \$400$. Exactly how much is the bill now?


http://ocw.mit.edu/OcwWeb/Mathematics/18-S66The-Art-of-CountingSpring2003/CourseHome/index.htm

Great problem collection.

[color=cyan]Might you post some of them here?[/color]


topology
real analysis
real analysis unsolved

Let $E$ be a separable space and $F$ be an uncountable family of open sets of $E$. Prove that there exists $x$ in $E$ that belongs to uncountable many elements of $F$.

What do you mean by separable space, exactly? Is it a metric space, in which case separability and having a countable basis are equivalent? Or is it just a topological space with a countable basis? Or a topological space which has a dense countable subset? (this is what is usually known as separability; it's a weaker condition that having a countable basis, in general)

You have to specify these "details", man :).

Oh, sorry, in "taupe" we work only with metric spaces, thus here the space is a separable metric space. Sorry...  :blush:  :blush:

Ok, thanks for clarifying :).

Let $(x_n)_{n\ge 1}$ be a dense countable subset of $F$. Each open subset of $F$ contains one of the $x_n$, so our collection os open sets is the union of $\mathcal A_n,\ n\ge 1$, where $\mathcal A_n$ consists of those sets which contain $x_n$. Since $\bigcup_{n\ge 1}\mathcal A_n$ is uncountable, at least one $\mathcal A_n$ will have to be uncountable, and $x_n$ is the point we're looking for.

P.S.

I guess it didn't really matter anyway; this works in the weakest hypotheses possible: a topological space with a dense countable sybset :).


Prove or disprove that any rational number in the interval $(0,1)$ with odd denominator can be written in form of the fraction part of $\frac{xyz}{x^2+y^2+z^2}$ for some integers $x,y,z$

i thought if x^2+y^2+z^2 = nxyz has solutions in integers x,y,z, n can only be 1 or 3. :?

Well, the problem asks to prove for the fraction part of $\frac{xyz}{x^2+y^2+z^2}$


function
Functional inequality
algebra
functional equation
IMO
IMO 1977

Let $\mathbb{N}$ be the set of positive integers. Let $f$ be a function defined on $\mathbb{N}$, which satisfies the inequality $f(n + 1) > f(f(n))$ for all $n \in \mathbb{N}$. Prove that for any $n$ we have $f(n) = n.$


Princeton college Stanford Duke MIT Harvard geometry

Heya all, just wondering, for the tons of grade 12's out there, where are you going for university?  Where have you applied?

I, for te record, was accepted into the best university ever.....lol, going to Waterloo for Bioinformatics this fall.  Sooo excited.  Hope to see some of you there!  Hehehe, and of course, meet old friends there too.   :D

Congratulations! :clap2:

Haha...you guys call it University.  :D

Betchu 5 bucks Elyot will respond to this thread.

Yea.  Go Waterloo.  Bioinformatics seems like such a perfect fit for you, Sarah.  I know a guy who's doing some grad studies in that... and it's very interesting stuff.  There's a bio-informatics lab in the Davis Centre too.  Can't wait to see you there.  I'll likely be on co-op in the fall though... hopefully we can plan some stuff.  I remember the first few days at Waterloo... so exciting... so much math... so awesome.

Yea... for CS/math, Waterloo is definitely the best university EVAR!  W00T!

[quote="The Master"]You already got accepted??? How so early?[/quote]

No idea. lol.  I'm quite thrilled....I thought they'd wait to see how I'd do in the Euclid.  Oh well, i'm not complaining.

[quote="The Master"]wow, well, congratulations

n here I am preparing for the Euclid, filling out my 'Admission Information Form' tryin to impress them.......[/quote]

Hehehehe good luck!  It might have helped that I've written Euclid for a long time....

hmmmm....... I think I should have spoken to you sooner.. like 4 or 5 years ago.....

But hopefully I'll see you at Waterloo

I'm tryin to get into Mathematics, by the way.

Already? Wow...congratulations! And I'm still trying to make up some nicer things for the AIF...

[quote="YY"]Already? Wow...congratulations! And I'm still trying to [b]make up[/b] some nicer things for the AIF...[/quote]

 :rotfl:

Good job!!!!!!!!!!

I've just filled out my AIF on Wednesday, I think. So I guess I'll have to wait a bit longer.. But hopefully I'll get accepted soon..  :D
See you in September!!!!

[quote="Sunny"][quote="Elyot"]I gave Ian Vanderburgh the news too.  I think he's horrified (his worst nightmare come true?).  But he still has your seminar button in his office.  I'm gonna laugh if he's one of your profs.[/quote]

Hehehe that's awesome.  But I want my button back.  I'm gonna steal it back eventually.[/quote]

lol, good luck with that.. ;)

[quote="jpark"]
I've just filled out my AIF on Wednesday, I think. [/quote]
How did you get your Fermat score? No one ever told me the score, so I have no idea how I can get the score. :(

[quote="Sunny"]
In that case, awesomeness.

Only need to survive 3 more IB tests....then I'm free!! Yay!  If only alberta didn't have these stupid diploma exams too!!!!![/quote]

I lucked out in that regard. My Chemistry 30 class is piloting a new curriculum, so the entire class is excused from the diploma. That means that I only have 1 diploma exam to write (physics), not that it really makes a difference anyways, since there's really nothing that hinges on your final grades that come out in the middle of summer

It just kind of hit me that I might be seeing some of you in person in September, which is pretty bizarre for me.

[quote="Sisyphus"]

I lucked out in that regard. My Chemistry 30 class is piloting a new curriculum, so the entire class is excused from the diploma. That means that I only have 1 diploma exam to write (physics), not that it really makes a difference anyways, since there's really nothing that hinges on your final grades that come out in the middle of summer

It just kind of hit me that I might be seeing some of you in person in September, which is pretty bizarre for me.[/quote]
WHAT?!  NOOOOOOO SOOOO NOT FAIR!  Humph.  What kinda new program?  bleh, lucky!

[quote="Sunny"]
WHAT?!  NOOOOOOO SOOOO NOT FAIR!  Humph.  What kinda new program?  bleh, lucky![/quote]

From what my teacher tells me, they've updated a lot of the symbols according to today's conventiosn...rearranged some units...added organic chemistry...and are testing out a new text book. The diploma thing's cool, but there are tons of mistakes in the textbook, and there are a lot of labs (which I'm not too crazy about).

[quote="Elyot"][quote="Sunny"]
WHAT?!  NOOOOOOO SOOOO NOT FAIR!  Humph.  What kinda new program?  bleh, lucky![/quote]

Don't get upset... just take comfort in knowing that when you get to Waterloo, you'll just remember high school as "that bothersome thing you did before you got to Waterloo".

Good luck on your Alberta finals (although, they should be easy after IB, right?!)[/quote]

they're usually not that bad, but the fact that they account for 50% of your overall grade is very disconcerting. I know that before I was accepted to Waterloo and was awaiting my diploman marks for math, english, and social studies, I was chewing my nails. It's all good now, though.

[quote="Sisyphus"]
they're usually not that bad, but the fact that they account for 50% of your overall grade is very disconcerting.[/quote]

That's really typical of any final exam.  All the finals at Waterloo seem to be worth about that much... some even more.  Maybe that's a bit tough for high school, but back when Ontario had it's old curriculum, all the math courses had 40% finals.

Oh no, I'm pissed off that if it weren't for this Alberta Diploma exam, I'd be done high school tomorrow.  When I'm done my IB exams.  But no.....we gotta study for these extra tests, which are always last two weeks of june.  So another 5 weeks of school left.

[quote="Sisyphus"][quote="Elyot"] All the finals at Waterloo seem to be worth about that much... some even more.  .[/quote]

Are there a lot of assignments for math courses?[/quote]

Yes, there are a lot of assignments.  But if you're good, they don't take too long.  Expect to spend maybe 2-5 hours a week per course on assignments.  Less on easy courses.

Of course, they're usually worth like 1% each (sometimes more in CS), so if you neglect to do one, you won't be too screwed.  But it's easy to fall behind if you start skipping over assignments.

The hardest thing for me has been to actually get stuff done ahead of time.  When you have 4 assignments due on one day, and each will take you multiple hours, you've gotta work on them ahead of time, or you'll get screwed.

In my first term, we had weekly assignments for Algebra, which took maybe 4 hours, and biweekly assignments for calculus, which took about 10 hours (our prof gave us some INSANE questions).  But all the assignments in the advanced classes are marked out of less than they're worth.  So you can get like 125/100 if you finish everything on the calc assignment.  A lot of people don't get the hard questions (which can be tricky... even contest-level sometimes), so you've got that buffer to fall back upon.

[quote="J-I-M-B-O"][quote="claralex"]I'm getting incredibly excited about university reading all of this.
Which is dumb because . . . . . . I'm in Grade 11. :P So I won't be going to university for another year and 4 months. Hahahaha.

But I have my heart set on going to U of Waterloo, most likely for pure math :) So hopefully, if all goes well I'll see some of you guys there.

But in the meantime I have seminar to look forward to :)[/quote]

don't worry, i was excited about university in gr 11 too![/quote]

everyone that was bored with highschool was in grade 11, ie, all of aops.

so ah, ditto.

Have you guys picked your M-Sections?

I ended up going with the Physics-based CS 135 one. I'm going to have to switch into the advanced classes if I will want to during orientation week.

I ended up going with what was being said here and went with stream 8 with the intentions of switching back into stream 4 when I declare my major. It looks like I'll have to have two consecutive workterms during summer and fall of 07, but I guess it all kind of worked out in the end.

[quote="Sunny"]I went with ADV-BIO CS 133, stream 4....I really wanted a stream 8, but I don't get a choice in this matter.  Oh well.  I'm excited though!!  hehehe.[/quote]

Aww stream 4... we'll be in different streams :( :(  But I'm working in Waterloo at Maple again, so I'll still be around often in the evenings.

I think they just slot you in there for bioinformatics.  However, if you really want stream 8, you can plead your case with the co-op people and they'll easily allow you to make a switch.  But, they generally want a decent reason.  A buddy of mine switched streams so he could be in the same co-op stream as his girlfriend, and he didn't really have any trouble with it.  You will have much smaller classes during the 1b term in the first summer (which can be good, or bad... in my case, most of my friends were gone for the term and I had to make new ones.)

And, if you don't get a job first term (which happens to a good number of first years, although probably not ones of your calibre) then you can always just do another academic term in the winter (which is kinda like an automatic stream switch).  Co-op doesn't really actually keep track of streams to the full extent that they could.  Generally, they just want to know what you're planning on doing next term, and whether or not you want your name in the online job application system.


Using a hose, Eugene is able to fill his pool with water in 2 hours and 12 minutes. With help from his neighbor's hose, Eugene can fill his pool in 1 hour 34 minutes. How long would it take to fill the pool using only his neighbor's hose?

Rate of Eugene's pump: 1/132 poolfuls of water per minute
Rate of hoses combinded: 1/94 poolfuls of water per minute

Just subtract rates.

38/12408=19/6204 poolfuls of water per minute, or 6204/19 minutes = 326 and 10/19 minutes = 5 hours 26 10/19 minutes


Assume that a set S has n distinct elements.

How many subsets does S have?

I know the answer is going to be 2^n . But I don't know how....

Please any one help me...

You don't post this kind of problems in olympic section.
There are $ { n \choose k}$ subsets of $ S$ which have $ k$ elements. $ k$ can be every number from 0 to n, so there are $ \sum_{k\equal{}0}^n {n \choose k}$ subsets, and this sum is equal to $ 2^n$ because $ (1\plus{}1)^n$ by binom formula is exactly $ \sum_{k\equal{}0}^n {n \choose k}$.

Or you can prove it like this: one n-tuple of $ 0$'s and $ 1$'s uniqely determins a subset of $ S$ (you write $ 1$ if that element is in the subset, else you write $ 0$), and there are $ 2^n$ different n-tuples.

Bye


AMC
AMC 10
AIME
USA(J)MO
USAMO
AwesomeMath
summer program

Quick background:
I've scored a 120 on AMC10 (ehhh....)
a 3 on the AIME (ugh...)
I really want to go on to the USAMO next year. I plan on doing two sessions, should I take all four intermediate courses or two intermediate the first time and the follow up advanced the second time? Which would improve my chances further of going on?

It depends on which fields of mathematics you are taking, and also, you really don't know until you go to at least one session.

Since your AIME score is a 3, (which is near average, so feel proud that you qualified for it, and don't feel bad) intermediate would probably be the better choice over advanced for at least the first session, like you said. On the awesomemath site, it says if your score is under 8, you should go for intermediate, and if it is over 8, you should go for advanced.

Now, if you thought intermediate was easy in your first session, take advanced. But I highly doubt that will happen, because these courses are very hard and designed for real top-notch students, which you are for qualifying AIME, but I mean, really really hard.

You should also consider which fields of math you like, or are good at. For example, if you're really square with number theory and algebra, but need some work in geometry and combinatorics, take geometry and combinatorics, definitely, for your first session. Then, you could take either number theory and algebra advanced for next session (and if it's too hard, they let you change classes, so don't worry, really..) or step up geometry and combinatorics.

It's really up to you. But seeing from your AIME score, (which is not bad, trust me.. and please don't fell offended) I'd take intermediate.. And just to let you know, the cutoff score is around 210 ~ 220 usually.. So that means you have to get a score of around 9 or 10 for the AIME to qualify, saying that you get a 120 again.. But you will DEFINITELY improve from AwesomeMath.. Good luck, and hope you qualify for USAMO!


let f is a function from R to R such that:

for all (x in R)  3f(x)+f(-x)= -4x^3-2x

is this function bijective?

obviously if you plug $ x\rightarrow\minus{}x$ you find $ 8f(x) \equal{}\minus{}16x^{3}\minus{}8x$ so it is bijective :)

but f(x) = -2x^3-x
is it bijectif how do you found it please

I meant it is surjective (clearly because takes arbitrarily big and small values, and is continous)
to see if it is injective see that $ f(x) = f(y)$ and $ x\doteq y$ imply $ x^{2}+xy+y^{2}=-\frac{1}{2}$ which is imposible

And yes your titles (and post contents in general) are quite meaningless.


Given that $ a_1\equal{}1$, $ a_2\equal{}5$, $ \displaystyle a_{n\plus{}1} \equal{} \frac{a_n \cdot a_{n\minus{}1}}{\sqrt{a_n^2 \plus{} a_{n\minus{}1}^2 \plus{} 1}}$. Find a expression of the general term of $ \{ a_n \}$.


algebra
polynomial
number theory proposed
number theory

Prove that the polynomial $P(x)=x^{2n}+x^n+1$ for $n \in \mathbb{N}$ is divisible by the polynomial $Q(x)=x^2+x+1$ iff $n$ isn't a multiple of $3$.

A polynomial $p$ is divisible by a polynomial $q$ iff all (counted with multiplicity) the roots of $q$ are roots of $p$.
Here, the roots of $q$ are $\zeta , \zeta ^2$ with $\zeta = e^\frac{2 \pi}{3} = \frac{-1 +i \sqrt{3}}{2}$. If $3|n$, then $\zeta$ is not a root of the first one; but in all other cases it is.


geometry
similar triangles
calculus
calculus computations

A tight rope is stretched 30 feet above the ground between building J and building T, which are 50 feet apart.  A tightrope walker, walking at a constant rate of 2 feet per second from point $ A$ to point $ B$, is illuminated by a spotlight 70 feet above point $ A$.  ($ A$ is a point on building J, 30 feet above the ground; $ B$ is directly across, on building T, also 30 feet above the ground)

a) How fast is the shadow of the tightrope walker's feet moving along the ground when she is midway between the buildings?

b) How far from point $ A$ is the tightrope walker when the shadow of her feet reaches the base of building T?

c) How fast is the shadow of the tightrope walker's feet moving up the wall of building Y when she is 10 feet from point $ B$?

x= distance travelled by tightrope walker from A,
y= distance from base of J to shadow on ground

by similar triangles

$ \frac{70}{x}\equal{}\frac{100}{y}$ or $ 7y\equal{}10x$, so $ 7\frac{dy}{dt}\equal{}10\frac{dx}{dt}$

a) since $ \frac{dx}{dt}\equal{}2$ feet per second,  $ \frac{dy}{dt}\equal{}\frac{20}{7}$ feet per second

b) $ 7(50)\equal{}10x \to x\equal{}35$

when the walker is x<15 feet from B, and her shadow is on the building, new similar triangles give
for z= feet above the ground of the shadow on building T

$ \frac{70}{50\minus{}x}\equal{}\frac{30\minus{}z}{x}$ or $ 70x\equal{}(30\minus{}z)(50\minus{}x)$
or $ 70x\equal{}1500\minus{}30x\minus{}50z\plus{}xz \to z\equal{}100\frac{15\minus{}x}{50\minus{}x}$
and $ \frac{dz}{dt}\equal{} \minus{}100\frac{1}{50\minus{}x}\frac{dx}{dt}\plus{}100\frac{15\minus{}x}{(50\minus{}x)^2}\frac{dx}{dt}\equal{}\frac{\minus{}3500\frac{dx}{dt}}{(50\minus{}x)^2}$

c) so for $ x\equal{}10, \frac{dx}{dt}\equal{}\minus{}2$ we get $ \frac{dy}{dt}\equal{} \frac{7000}{40^2}\equal{}\frac{35}{8}$ ft per second


In the triangle $ ABC$ , $ M$ is midpoint of $ AC$ and $ BM \equal{} \frac {1}{2}BC$ . Prove that $ AC > AB$

[hide="solution (click me!)"]Let $ BM\equal{}x$, $ AB\equal{}y$, and $ AM\equal{}MC\equal{}a$. We then have $ BC\equal{}2x$. From Stewart's Theorem,
\[ 4x^2a\plus{}y^2a\equal{}2ax^2\plus{}2a^3\]
\[ 4x^2\plus{}y^2\equal{}2x^2\plus{}2a^2\]
\[ y^2\equal{}2(a^2\minus{}x^2)\]
\[ y\equal{}\sqrt{2(a^2\minus{}x^2)}\]
We must prove that $ y<2a$, or
\[ \sqrt{2(a^2\minus{}x^2)}<2a\]
\[ 2a^2\minus{}2x^2<4a^2\]
\[ \minus{}2x^2<2a^2\]
This is true, since the lesser amount is negative and the greater amount is positive. Therefore $ AC>AB$.[/hide]
I have a feeling that I've done something wrong; I solved this quicker than most Olympiad Geometry problems.

hello, we have $ \overline{BM}^2 \equal{} \frac {a^2 \plus{} c^2}{2} \minus{} \frac {b^2}{4}$, with $ \overline{BM} \equal{} \frac {a}{2}$ we get $ b^2 \equal{} 2c^2 \plus{} a^2$ and $ b > c$ is equivalent with $ \sqrt {2c^2 \plus{} a^2} > c$ or $ a^2 \plus{} c^2 > 0$ which is obviously true.
Sonnhard.

A solution without metric relations:

Extend the BM during part MB' = MB.
 So  the    ABCB'   is  parallelogram  and we have:

$AB' = CB = BB' \Rightarrow B\hat CB' = C\hat B'B = B'\hat BA = B\hat AB'$

So  in triangle ABC is    $\hat A < \hat B$$ \Rightarrow CB < CA$

and in triangle ABM:    $AB < MB + MA \Rightarrow $

 $2AB < 2MB + 2AM \Rightarrow 2AB < BC + AC < 2AC \Rightarrow AB < AC$


What summer camps have you attended or attending?
I did tennis camp, creative wrinting camp, architechture camp, and math puzzle camp.

SPRI (summer program for research interns), running camp, hmm...that's about it

running camp???
do you just run all day?
that would be torcher to me.

I've been camping for a week in Canada, where we just canoed the whole time, two camps in our temple, and 2 Boy Scout summer camps.  And I also went to a camp a couple weeks ago run by a mission with no missionaries, that meets in a hotel.

i went to "work-a-part-time-job-that-makes-you-do-over-30-hours-a-week-camp" and "study-for-AP-classes-for-senior-year-during-the-summer" camp....ooh don't forget "sit-at-home-and-play-diablo II exp-camp"...that was definitely the most fun part  :D

[quote="mwahahahaha"]ooh don't forget "sit-at-home-and-play-diablo II exp-camp"...that was definitely the most fun part  :D[/quote]

whoa. that's what i did all summer too! :D

[quote="nr1337"]I went to MathPath.[/quote]

I also went to mathpath. There were many AoPSters there.

[quote="h78reg"]no camps for me this summer :([/quote]whay do you do then?

after I came to US..

Nothing 2002
CCNY 2003
Mathcamp 2004
CCNY as TA 2005

Germany/California 2006 hopefully

[quote]whay do you do then?[/quote]

er....interesting question. basically AOPS and cycling. :D

Hmm...no MathCampers have posted here yet. Come on folks...

*prods*

Alright Miths, I'm here.

What's MathCamper...?!

http://mathcamp.org/

i went to mathcamp for couple weeks Duke TIP then space camp it was fun in that order most fun first. food was the same.

wow...mathcamp sounds very very fun
hmmmmm too bad i've never heard of it till now lol

blue lake, located in muskegon, michigan (west coast) is a fine arts camp. I went there for violin and got into the top orchestra! Drunken_Math is in blue lake right now too... :D

you guys dont play instuments or something?[url=http://bluelake.org]bluelake website![/url]

OMG my friend went to blue lake for choir She's their right now and couldn't get her schedule yesterday and I won't know for another week if she has any classes with me!!! :(


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $S$ be the group of permutations of the integers and let $G$ the subgroup of those permutations such that the set of $x$ with $f(x)\ne x$ is finite. Prove that the subgroup of $G$ generated by the $3$-cycles is simple.

Sounds a bit to simple:
We can see $G$ as union of all $S_{n}$. We know that $A_{n}$ is generated by the $3$-cycles in $S_{n}$ for big $n$, thus the desired subgroup is the union of the $A_{n}$, and as a result it has index $2$ and is normal in $G$. Since conjugation preserves finitness of the non-fixed elements, it's also normal in $S$.

An alternative method.

Let $G_{0}$ be the subgroup generated by 3-cycles in $G$ and suppose that $0 \neq N$ is a normal subgroup of $G_{0}$. Note that if $N$ contains a 3-cycle then $N$ contains all 3-cycles and hence $N=G_{0}$. Let $(a,b,c) \in N$ and $(x,y,z)$ a general 3-cycle. Examining some cases one can find an element $g \in G_{0}$ so that $g(a)=x$, $g(b)=y$ and $g(c)=z$ thus $g(a,b,c)g^{-1}=(x,y,z)$. The hardest case is when $a,b,c,x,y,z$ are all distinct elements. Let $n=\max\{x,y,z,a,b,c\}+1$. Then $g=(a, x, n)(b, y, n+1)(c,z,n+2)$ will work.

For each permutation $g$ define $mov(g)=\{ x \in \mathbb{Z}\mid g(x)\neq x\}$. One can easily prove that if $|mov(g) \cap mov(h)|=1$ then $[g,h]=ghg^{-1}h^{-1}$ is a 3-cycle. Let $h \in N$ not be the identity. Since $mov(h)$ finite pick $x,y \not\in mov(h)$ and $z \in mov(h)$ such that $x,y$ and $z$ are distinct. Then $g=(x,y,z)$ has $mov(h)\cap mov(g)=\{z\}$ which yields $ghg^{-1}h^{-1}$ is a 3-cycle. Since $ghg^{-1}$ and $h^{-1}$ are both in $N$ so is $[g,h]$.


There's a game of luck in which a lucky (or unlucky) man has to place $ 3$ shots accidentally in the shot places of a pistol (which contains $ 7$ shot places) and rolls it once then puts the gun on his head and shoots twice. What is the probability that the man is lucky?!!! that means he remains alive after the second shoot!

$ \text{a)}\;\;\frac {1}{7}\;\;\;\;\;\;\;\text{b)}\;\;\frac {12}{49}\;\;\;\;\;\;\;\text{c)}\;\;\frac {4}{7}\;\;\;\;\;\;\;\text{d)}\;\;\frac {16}{49}\;\;\;\;\;\;\;\text{e)}\;\;\frac {2}{7}$

Wow. What a violent problem.  :|

The probability that the man will not be shot the first time is $ \frac{4}{7}$, and the probability the man is lucky for the second shot is $ \frac{3}{6} \equal{} \frac{1}{2}$, multiplying these together, we have $ \frac{2}{7}$, or $ \boxed{E}$.

Correct!
Yeah it is, but it's just a problem! :D I hope and I know such a game doesn't exist!


analytic geometry
geometry open
geometry

It is given a triangle ABC. Inside triangle are chosen two points M and N. If <MAC=<NBC, <MCA=<NCB and AN=BM is it true that AC=BC?

If I understood the problem correctly, from the three given conditions follows that triangles $ MAC$ and $ NBC$ are congruent (it's clear that they are similar, and also we know $ AM\equal{}BN$, which are opposite to the equal angles $ MCA$ and $ NCB$ respectively).

I'm sorry the last equality should be AN=BM

I'm surprised this problem remain so long time unsolved. It seems to be not so difficult.

A man from another forum gave me the following answer.
1) Can someone give me an explanation how the equation for BM is obtained
and
2) Why from 2 * (cos(a))^2 * (tg(a)+tg(b)) = 1 follows <A=<B?

Let's <MAC=a, <MCA=b. If true 2*(cos(a))^2*(tg(a)+tg(b))=1

More detailed description:

If we consider rectangular coordinate system with axe OX == AC (right-handed system) and OX == BC (left-handed system), we can get that

$ BM^2 \equal{} AB^2 * ((cos(A) \minus{} (sin(B)/sin(C)) / (tg(a) \plus{} tg(b)))^2 \plus{} (sin(A) \minus{} (sin(B)/sin(C)) * (tg(a)/ (tg(a) \plus{} tg(b)))^2)$

$ AN^2 \equal{} AB^2 * ((cos(B) \minus{} (sin(A)/sin(C)) / (tg(a) \plus{} tg(b)))^2 \plus{} (sin(B) \minus{} (sin(A)/sin(C)) * (tg(a)/ (tg(a) \plus{} tg(b)))^2)$

If we transform AN = BM, we can get that 2 * (cos(a))^2 * (tg(a)+tg(b)) = 1 or <A = <B for execution equality.

I'm sorry for loosing your time. The problem is very easy. I solved it in my own very simple way. It requires more concentration.


Anyone know any good chemistry contests that students from other countries can sign up for and take at their high school? I live in Canada and there isn't many chemistry contests to do locally and nationally.

Victor

I found a [url=http://www.cheminst.ca/index.cfm/ci_id/2762/la_id/1.htm]Canadian[/url] one for you on the Web

Thanks for the site.  :D  But do you know other chemistry contests that students from other countries can participate in such as US chemistry contests?

I know that there is the chemistry olympiad. You can go to their website to find out more info. Chemistry olympiad is when students around the world take a rigorous exam dealing with MANY chem topics; including kinetics, equilibrium, etc.

If you qualify after taking the exam, you can compete in the international competition against students from other countries.

If you aren't interested in an exam, but want to enter a science competition or fair, you could compete in the INTEL International Science and Engineering Fair. Before you can reach this level, however, you must win a top prize in your region and then your country (i.e. Canada).

The Intel Science Fair allows for high school students to do graduate level research in Bio, Chem, and Physics. The competition takes months of hard work and preparation before you can actually create a research plan, poster, and present in the science fair. I know that in order to participate in one of these competitions, a student must work in a university lab and must be thoroughly knowledgable in their topic. In past years, some students have developed drugs for controlling infections or even determined the evolution path of humans. :)


[quote="zaizai-hoang"]Let $a,b,c\ge 0$ such that: $a+b+c=3$. Prove that:
\[\frac{a}{b\sqrt{c^{2}+3}}+\frac{b}{c\sqrt{a^{2}+3}}+\frac{c}{a\sqrt{b^{2}+3}}\ge \frac{a^{2}+b^{2}+c^{2}}{2}\][/quote]
By Holder $\left(\sum_{cyc}\frac{a}{b\sqrt{c^{2}+3}}\right)^2\sum_{cyc}ab^2(c^2+3)(a+c)^3\geq\left(\sum_{cyc}(a^2+ab)\right)^3$.
Hence, it remains to prove that
$4\left(\sum_{cyc}(a^2+ab)\right)^3\geq(a^2+b^2+c^2)^2\sum_{cyc}ab^2(c^2+3)(a+c)^3$, which is true after some work.


[quote="The soul of rock"]This is my 250-th post
Let $a,b,c>0$ prove that
\[\frac{1}{a^{2}+b^{2}+c(a+b)}+\frac{1}{b^{2}+c^{2}+a(b+c)}+\frac{1}{c^{2}+a^{2}+b(b+c)}\le \frac{9}{4(ab+bc+ca)}\]
[/quote]

As you can see we need to prove that $\sum{3-\frac{4(ab+bc+ac)}{a^{2}+b^{2}+c(a+b)}}\geq 0$ which is equivalent to $\sum{\frac{2(a-b)^{2}+a^{2}+b^{2}-c(a+b)}{a^{2}+b^{2}+c(a+b)}}\geq 0$ hense it is sufficent to show that $\sum{\frac{a^{2}+b^{2}-c(a+b)}{a^{2}+b^{2}+c(a+b)}}\geq 0$ which is equivalen to $\sum{\frac{2(a^{2}+b^{2})}{a^{2}+b^{2}+c(a+b)}}\geq 3$. We have that $2(a^{2}+b^{2})\geq (a+b)^{2}$ thus \[\frac{2(a^{2}+b^{2})}{a^{2}+b^{2}+c(a+b)}=\frac{2}{1+\frac{c(a+b)}{a^{2}+b^{2}}}\geq \frac{2}{1+\frac{2c}{a+b}}=\frac{2(a+b)}{(a+c)+(b+c)}\]
and we need to show that $\sum{\frac{(a+b)}{(a+c)+(b+c)}}\geq \frac{3}{2}$, but this(as you can recognize it) is the famous Nasbitt's inequality for $x=a+b,y=a+c$ and $z=b+c$. :wink:


[quote="white_horse_king88"]Find the exact value of the infinite continued fraction:

$2+\frac{1}{3+\frac{1}{2+\frac{1}{3+\ldots}}}$[/quote][hide="hint"]If \[\alpha=2+\frac{1}{3+\frac{1}{2+\frac{1}{3+\ldots}}},\]
then $\alpha$ satisfies \[\alpha=\frac{1}{\frac{1}{\alpha-2}-3}.\][/hide]

I'm probably wrong, but oh well.

[hide]You could make $x = 2+\frac{1}{3+\frac{1}{2+\frac{1}{3+\ldots}}}$, which makes $x = 2 + \frac{1}{3+\frac{1}{x}}$
When you solve for x, you get $x = \frac{6+4\sqrt{3}}{3} \text{ or } x = \frac{6-4\sqrt{3}}{3}$[/hide]

the trick to doing these infinity questions are to let the infinite sum/diff/etc equal to a variable. and then sub it back in.

E.g. if the questions was to find the value of sqrt{3+sqrt[3+sqrt(3+...)]}, then let the whole thing equal to x. You get:

x=sqrt(3+x), solving u can find the value of x.

It applies to all sorts of infinity questions, not just fractions or square roots.

Hope this helped. Cheers!

Hey, you've just disproved math as we know it! The fraction has two values!

The expression is 2+(something). The (something) is positive, so the expression must be greater than two. Therefore we take the positive squareroot, not the negative one.

[quote="h_s_potter2002"][hide]You could make $x = 2+\frac{1}{3+\frac{1}{2+\frac{1}{3+\ldots}}}$, which makes $x = 2 + \frac{1}{3+\frac{1}{x}}$
When you solve for x, you get $x = \frac{6+4\sqrt{3}}{3} \text{ or } x = \frac{6-4\sqrt{3}}{3}$[/hide][/quote]

I got the same equation but a different value.  Here is my solution:
[hide]$x = 2 + \frac{1}{3+\frac{1}{x}}$
$(x-2)(3+\frac{1}{x})=1$
$3x-6+1-\frac{2}{x}=1$
$3x^2-6x-2=0$
$x=1+\sqrt{5/3}=2.290994$ approximately
[/hide]

A fraction with 2 values, base ten: $\frac{9}{9}=0.999999$ repeating and $\frac{9}{9}=1$  Although I guess you could say they are the same number, which is true.

[quote="Wumbate"]A fraction with 2 values, base ten: $\frac{9}{9}=0.999999$ repeating and $\frac{9}{9}=1$[/quote]Given that $\frac{9}{9}=0.999999\ldots$ and $\frac{9}{9}=1$, either (a) $\frac{9}{9}\ne\frac{9}{9}$ and math as we know it falls apart, or (b) $.9999\ldots=1$. Fortunately case (b) is correct.

I think I just plugged the numbers into quadratic formula wrong, because I got the same final equation as Wumbate.


[quote="nut_sk129"]use the cauchy schwarz inequality
and use x^2+y^2+z^2>=xy+yz+zx[/quote]
Show the full proof, please. Thank you!

I have already used chachy-schwars but it doesn't lead the answer. :(

[quote="nut_sk129"]use the cauchy schwarz inequality
and use x^2+y^2+z^2>=xy+yz+zx[/quote]

i don't think u can CS this

if u wanna do CS, the natural thing to try to show is that

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

which is patently false (in fact, the reverse is true)

although i might be following the wrong train of thought :\

[quote="TRAN THAI HUNG"]:lol:I have found the answer of this problem by using S.O.S[/quote]

Please wait!!!! :?:

This topic was very old.Did anyone have any idea? :)

[quote="TRAN THAI HUNG"]$ (ab \plus{} bc \plus{} ac)(\frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }}) \ge \frac {9}{4}$
a,b,c>0[/quote]
some algebra:
$ (ab \plus{} bc \plus{} ac)(\frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }}) \\
\equal{} (a^2 \plus{} b^2 \plus{} c^2 \minus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{2 })( \frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }}) \\
\equal{} (a^2 \plus{} b^2 \plus{} c^2)( \frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }}) \\
\minus{} \frac {1}{2}((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2)( \frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }})$
Cauchy
$ \\
\ge (\frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{{c \plus{} a}})^2 \minus{} \frac {1}{2}(\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a })^2 \\
\ge (\frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a})^2 \minus{} \frac {1}{2}(\frac {2a}{a \plus{} b} \plus{} \frac {2b}{b \plus{} c} \plus{} \frac {2c}{c \plus{} a} \minus{} 3 )^2$
let $ x \equal{} \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{{c \plus{} a}}$

thus the above is $ x^2 \minus{} \frac {1}{2}(2x \minus{} 3)^2 \equal{} x^2 \minus{} \frac {1}{2}(4x^2 \minus{} 12x \plus{} 9) \equal{} \minus{} 9/2 \minus{} 6x \minus{} x^2 \equal{} 9/2 \minus{} (3 \minus{} x)^2$
to minimize this we require x to be as close to 3 as possible  as x is Nesbitt's it minimizes at 3/2 thus thus minimal value of the inequality is $ 9/2 \minus{} (3 \minus{} 3/2)^2 \equal{} 9/2 \minus{} 9/4 \equal{} 9/4$ as required

I must be missing the easy solution though.

[quote="Dadeyemi"]$ (a^2 \plus{} b^2 \plus{} c^2)( \frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }})\\
\minus{} \frac {1}{2}((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2)( \frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }})$
Cauchy
$ \\
\ge (\frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{{c \plus{} a}})^2 \minus{} \frac {1}{2}(\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a })^2$[/quote]Is this true? I think you applied Cauchy as follows: \[ (a^2 \plus{} b^2 \plus{} c^2)\left( \frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }}\right)\ge \left(\frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{{c \plus{} a}}\right)^2\]and\[ \minus{} \frac {1}{2}((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2)\left( \frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }}\right)\ge\minus{}\frac {1}{2}\left(\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a }\right)^2\]and then you added them up. But the second application of Cauchy is the other way, since both sides are negative.

[quote="TRAN THAI HUNG"][quote="mathking123"][quote="TRAN THAI HUNG"]:lol:I have found the answer of this problem by using S.O.S[/quote] What is S.O.S??[/quote]
S.O.S means sum of square you can find this method in many math books
Yeah, I think  Dadeyemi has made some mistake.I think we can't achieve the solution by B.C.S[/quote]
@Tran Thai Hung: Your inequality is actually very famous. It's a problem in IRAN TST 1996.Your method is from Pham Kim Hung(Hungkhtn)'s "Secrets in Inequality" book. However, there is a proof just by Cauchy-Schwartz and its author is: can_hang2007(Vo Quoc Ba Can) if I'm not mistaken. Using "Search" bar to find out it in the Olympiad Section, Inequality. :)

[quote="ghjk"][/quote][quote="TRAN THAI HUNG"][quote="mathking123"][quote="TRAN THAI HUNG"]:lol:I have found the answer of this problem by using S.O.S[/quote] What is S.O.S??[/quote]
S.O.S means sum of square you can find this method in many math books
Yeah, I think  Dadeyemi has made some mistake.I think we can't achieve the solution by B.C.S[/quote][quote="ghjk"]
@Tran Thai Hung: Your inequality is actually very famous. It's a problem in IRAN TST 1996.Your method is from Pham Kim Hung(Hungkhtn)'s "Secrets in Inequality" book. However, there is a proof just by Cauchy-Schwartz and its author is: can_hang2007(Vo Quoc Ba Can) if I'm not mistaken. Using "Search" bar to find out it in the Olympiad Section, Inequality. :)[/quote]
Can you give me the link of the solution by using Cauchy Schwarz?I :) That would be nice because I used to think we can't use B.C.S for this problem.Thanks!

completly wrong solution was posted here ... :oops:

[quote="Wichking"]I solved it like this:

By Cauchy

$ (ab + bc + ac)(\frac {1}{{(a + b)^2 }} + \frac {1}{{(a + c)^2 }} + \frac {1}{{(c + b)^2 }}) \ge ( \frac {\sqrt {ab}}{a + b} + \frac {\sqrt {bc}}{b + c} + \frac {\sqrt {ac}}{a + c})^2$

by A-G

$ \boxed{\frac {\sqrt {ab}}{a + b} + \frac {\sqrt {bc}}{b + c} + \frac {\sqrt {ac}}{a + c} \ge 3 \sqrt [3]{\frac {abc}{(a + b)(b + c)(a + c)}}}$

We use A-G once again

$ (a + b)(b + c)(a + c) = ab^2 + a^2b + abc + a^2c + b^2c + abc + bc^2 + ac^2 \ge 8 \sqrt [8]{a^8b^8c^8} = 8abc$

$ \Rightarrow$

$ \boxed{\frac {abc}{(a + b)(b + c)(a + c)} \le \frac {1}{8}}$

$ \Rightarrow$

$ \boxed{\sqrt [3]{\frac {abc}{(a + b)(b + c)(a + c)}} \le \frac {1}{2}}$

$ \Rightarrow$

$ \boxed{\frac {\sqrt {ab}}{a + b} + \frac {\sqrt {bc}}{b + c} + \frac {\sqrt {ac}}{a + c} \ge \frac {3}{2}}$

$ \Rightarrow$

$ \( \frac {\sqrt {ab}}{a + b} + \frac {\sqrt {bc}}{b + c} + \frac {\sqrt {ac}}{a + c})^2 \ge \frac {9}{4}$

$ \Rightarrow$

$ (ab + bc + ac)(\frac {1}{{(a + b)^2 }} + \frac {1}{{(a + c)^2 }} + \frac {1}{{(c + b)^2 }}) \ge \frac {9}{4}$ what we wanted to show[/quote]

Sorry, my guy.I don't agree , because

you would use this agrument :

$ A\ge B , C\ge B, \implies A\ge C$

It is illogical. :play_ball:

ooooooooooooooooooooooooooooooooops :blush:  ... yes you are correct .... and it is illgical  :lol:

[quote="Aravind Srinivas L"]Just use Cauchy lemma and finish off.[/quote]
Can you post the full solution? :lol:

There are many solutions in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3547]this topic[/url] and also links to other discussions at the bottom of darij grinberg's [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204800#204800]post[/url].

See my solution. [hide="Click to reveal hidden content"]

[/hide]

EDIT: I am really very sorry :blush:
Hence I have deleted this post(my proof was completely mad one and I am shocked at myself :o )

Vandermonde: Of course,
$ (ab \plus{} bc \plus{} ac)(\frac {1}{{(a \plus{} b)^2 }} \plus{} \frac {1}{{(a \plus{} c)^2 }} \plus{} \frac {1}{{(c \plus{} b)^2 }})\geq (ab \plus{} bc \plus{} ca)(\frac {1}{4ab} \plus{} \frac {1}{4ca} \plus{} \frac {1}{4bc})\ge \frac {9}{4}$! :D
As $ (a \plus{} b)^2\geq 4ab, (b \plus{} c)^2\ge 4bc, (c \plus{} a)^2\ge 4ca$ :P

EDIT: I am very sorry. Thanks. :oops:

[color=#666666]Hi.........[/color]

[quote]The inequality is the wrong direction $ (a \plus{} b)^2\ge 4ab\implies \frac {1}{4ab}\ge \frac {1}{(a \plus{} b)^2}$[/quote]
Yeah. Potla answer is wrong.

[color=#666666]By.......[/color]

I have deleted my solution. I get my flaw. :blush:  Thank you for your coordination, Altheman.

Here is a nice solution:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=105728512&t=239240

[quote="TRAN THAI HUNG"][quote="mathking123"][quote="TRAN THAI HUNG"]:lol:I have found the answer of this problem by using S.O.S[/quote] What is S.O.S??[/quote]
S.O.S means sum of square you can find this method in many math books
Yeah, I think  Dadeyemi has made some mistake.I think we can't achieve the solution by [u]B.C.S[/u][/quote]
What is B.C.S?

B.C.S is just Bunhiaxcopski-Cauchy-Schwarz :)