algebra
polynomial
function
algebra proposed

Construct a polynomial function in $ x$ , $ P(x)$, such that $ P(1), P(3), P(5), P(7)$ are positive while $ P(2), P(4), P(6), P(8)$ are negative.

This is a very particular application of Lagrange's interpolation polynomial: given reals $ x_1 < x_2 < \cdots x_n$, and values $ v_1, v_2, \ldots, v_n$, there exists a unique polynomial $ L(x)$ of degree $ \deg L \leq n\minus{}1$ such that $ L(x_i) \equal{} v_i$ for all $ 1 \leq i \leq n$. The formula is $ L(x) \equal{} \sum_{i\equal{}1}^n v_i \prod_{j\neq i} \frac {x\minus{}x_j} {x_i \minus{} x_j}$.

Just take $ n\equal{}8$, $ x_i \equal{} i$, $ v_1 \equal{} v_3 \equal{} v_5 \equal{} v_7 \equal{} 1$, $ v_2 \equal{} v_4 \equal{} v_6 \equal{} v_8 \equal{} \minus{}1$.

Mavropnevma, had the same polynomial but took some time to construct it as I did not know the Lagrange Interpolation 's Application.
Nice work.

Lagrange interpolation?  Really?  Take any polynomial of the right sign with simple roots between $ 1$ and $ 2$, $ 2$ and $ 3$, ...


[quote="SnowStorm"]Source: AHSME 1973 #13

The fraction $\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2+\sqrt{3}}}$ is equal to
$\text{(A)} \ \frac{2\sqrt{2}}{3} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ \frac{2\sqrt{3}}{3} \qquad \text{(D)} \ \frac{4}{3} \qquad \text{(E)} \ \frac{16}{9}$[/quote]

[hide] $4/3. \boxed {D}$.[/hide]

[hide="idea #1"]
\begin{eqnarray*}\sqrt{2} +\sqrt{6}=\sqrt{2}(1+\sqrt 3)\\=\sqrt{2}\sqrt{(1+\sqrt 3)^2}\\=\sqrt{2}\sqrt{4+2\sqrt 3}\\=2\sqrt{2+\sqrt 3}\end{eqnarray*}
[/hide]

[hide="idea #2"]
Apply [b]Lagrange's[/b] identity:
if $a^2\ge b\ge 0$ then $\sqrt{a+\sqrt b}=\sqrt\frac{a+\sqrt{a^2-b}}{2}+\sqrt\frac{a-\sqrt{a^2-b}}{2}$
[/hide]

I don't think idea number 2 is applicable in this forum (?)  ;)

Well, lets use it anyways!  :lol:

[hide]$\sqrt{a+\sqrt b}=\sqrt\frac{a+\sqrt{a^2-b}}{2}+\sqrt\frac{a-\sqrt{a^2-b}}{2}$ is our formula, so if $a=2, \ b=3$, then:

$\sqrt{2+\sqrt 3}=\sqrt\frac{2+\sqrt{1}}{2}+\sqrt\frac{2-\sqrt{1}}{2} =\sqrt\frac{3}{2}+\sqrt\frac{1}{2}= \frac{\sqrt{6}+\sqrt{2}}{2}$.

So then ${\frac{2(\sqrt{2} + \sqrt{6})}{3(\frac{\sqrt{6}+\sqrt{2}}{2}})}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\implies \boxed{D}$.[/hide]

Snowstorm, can you put the year inside the source field (i.e. not with the post)?

I'll appreciate it. Thanks.

The source field doesn't show up for me.. cause I am not a mod or something I guess.


calculus
algebra
polynomial
function
complex numbers

I plan to take the SAT Level 2 Math Subject Test soon, and I though it would be nice to have a few practice problems.  This is an easy problem by comparison, but it is an old SAT Math Subject Test problem, so I though I'd post it:

$\text{Problem 1}$
 If $f(x)=x^{4}-3x^{3}-9x^{2}+4$ for how many real numbers $k$ does $f(k)=2$?

I took away the answer choices to make it a bit harder...well, not really.

EDIT:  I realized that an allowed TI-83 makes this problem trivial.  So NO CALCULATORS, even though they are allowed!

I'm not very good at the SAT, but I'll try ...

If $f(k) = 2$, we have $k^{4}-3k^{3}-9k^{2}+4 = 2$. We can subtract two from both sides, and we get $k^{4}-3k^{3}-9k^{2}+2 = 0$. I'm not sure what to do from here though...any help?

Some calculus would make this trivial.  That's all I need to say...

Otherwise, plug in values and use the intermediate value thm (or is it mvt?)

[hide] $x^{4}-3x^{3}-9x^{2}+2=0$
By Descartes' Sign Rule, there are 2 or 0 positive roots, and 2 or 0 negative roots. We can evaluate the polynomial for a few simple values. $f(-1)=-3$, $f(0)=2$, and $f(1)=-9$. Since polynomials are continuous functions, there must be a root in the interval $(-1,0)$ and so there are 2 negative roots. Also, there is a root in the interval $(0,1)$, so there are 2 positive roots. $\boxed{4}$ [/hide]

I think there are two levels of SAT II math, one that has imaginary numbers, and one that does not...

[quote="Elemennop"][quote="Jimmy"][quote="Elemennop"]what

I don't think the SAT knows about copmlex numbers

so degree is 4 and there are four roots[/quote]

The SAT II, does have questions about complex numbers.  Also, by your logic how many real roots does the equation $x^{2}+1=0$ have?[/quote]

If it was on the SAT II, 2.[/quote]

What are the 2? I count 0.

[quote="mysmartmouth"]
$\text{Problem 1}$
 If $f(x)=x^{4}-3x^{3}-9x^{2}+4$ for how many real numbers $k$ does $f(k)=2$?
[/quote]

Um....... the SAT II may know about complex numbers, but it also knows what a [b]REAL[/b] number is! Honestly, people, read the question.


geometry
search
calculus
trigonometry
Galois Theory

hm maybe spherical geometry? usually the second most well known type of geometry (after euclidean, of course).

I'm honestly unsure.

I know AoPS Vol 1/2 and AoPS Intro. Geometry doesn't cover a lot of it, from what I recall.

I'm sure other people have suggestions... right?  :wink:

[EDIT]

Try this link...

[url]http://www.amazon.com/s/ref=nb_ss_b/105-1436726-4618814?url=search-alias%3Dstripbooks&field-keywords=spherical+geometry[/url]

[quote="Calculus"]I am not sure where to post this so move it to another forum if needed.... but anyways, I only have background of euclidean geometry and I am interested in learning some other type of geometry like hyperbolic, elliptical, projective, etc. What geometry is best to start off with and learn? What books would you recommend?[/quote]

It depends how you want to learn it.
There are two approaches.

[b]Classical Approach:[/b] This is simply all the theorems derivable without the Parallel Postulate. In fact, Euclid's first 31 Propositions do not use the Parallel Postulate. All this means that you prove theorems and results by just working with what you know. This type of geometry is called "neutral geometry".

[b]Analytic Approach:[/b] This approach is not a formal as the classical approach but it is more computational, i.e. you compute lengths, angles, ect... But in non-Euclidean geomeotry you are required to know some analytic methods. Those include a good understanding of Multivariable Calculus and Differential Equations. The book [u]Poincare Half-Plane[/u] provides an elementary transition into non-Euclidean geometry by using techiques of Calculus,Trigonometry and Algebra. So if you are strong in those area I would recommend this book (not that it is the best).

This is my viewpoint on non-Euclidean geometry. You asked about Projective Geometry. This geometry at its advanced level is a Graduate course, it requires the reader to be familar with something called Field Theory (Galois theory, splitting field, finite extensions ... ). So I would no recommend it as a first advanced geometry course.

But since you like geometry in general I would recommend the following: http://www.amazon.com/Geometry-Springer-Undergraduate-Mathematics-Roger/dp/1852330589/ref=sr_1_1/104-8166963-8466320?ie=UTF8&s=books&qid=1181777286&sr=8-1 . I never ordered the book so I honestly cant say how good it is. I just like Springer books and I think they are well written. This books seems to contain everything that interests you.


function
Divisibility
number theory
algebra
functional equation
IMO Shortlist
httpbit.ly3vjUcdg

Find all functions $ f: \mathbb{N^{*}}\to \mathbb{N^{*}}$ satisfying
\[ \left(f^{2}\left(m\right)+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\]
for any two positive integers $ m$ and $ n$.

[i]Remark.[/i] The abbreviation $ \mathbb{N^{*}}$ stands for the set of all positive integers:
$ \mathbb{N^{*}}=\left\{1,2,3,...\right\}$.
By $ f^{2}\left(m\right)$, we mean $ \left(f\left(m\right)\right)^{2}$ (and not $ f\left(f\left(m\right)\right)$).

[i]Proposed by Mohsen Jamali, Iran[/i]

oh, i think it is IMOSL-04
ANSWER: [tex]f(n)=n[/tex]
i think this pro is easy because very cleary to prove that f(p-1)=p-1 for all prime p
@pipfly: why are you post this pro in here and who are you?

How to do it after getting $f(p-1)=p-1$ for all primes $p$?

pigfly, can you please clear up the notation for me: What is [tex]\mathbb{N}^*[/tex], is it the set of positive integers? And, what is [tex]f^2(n)[/tex]? Is it a composition of [tex]f[/tex] on [tex]f[/tex] or it's [tex](f(n))^2[/tex]?

$\mathbb{N^{*}}$ is set of positive integers, $f^{2}(n)=(f(n))^{2}$.

Ok, let's solve it :

We know that $f^2(1)+f(1)$ divides $4$ and is greater than $1$, so that it is $2$ or $4$. Solving the quadratic equations in $f(1)$ we easily find that $f(1)=1.$
It follows that for each prime $p$ the number $1+f(p-1)$ divides $p^2$ and is greater than $1$ so that it is $p$ or $p^2$.

Suppose that for some prime $p$ we have $f(p-1)+1 = p^2.$
Then $p^4-2p^2+2 = (p^2-1)^2 + 1 = f^2(p-1)+f(1)$ divides $((p-1)^2+1)^2 = p^4-4p^3 + 8p^2 - 8p +4$.
But it is easy to verify that for $p \geq 2$ we have $p^4-4p^3 + 8p^2 - 8p +4 <2(p^4-2p^2+2)$, from which we deduce that we must have $p^4-4p^3 + 8p^2 - 8p +4 = p^4 - 2p^2 + 2$, that is $2p^3-5p^2+4p-1=0$. Thus $p$ divides $1$ which is absurd.

Then, for all prime $p$, we have $f(p-1)+1=p$ that is $f(p-1)=p-1.$

Now, for all positive integer $n$ and all prime $p$, we deduce that $f(n)+(p-1)^2$ divides $((p-1)^2+n)^2 =  ((p-1)^2+f(n))((p-1)^2 + 2n - f(n)) + (f(n) - n)^2$.
Thus $\frac {(f(n)-n)^2} {f(n) + (p-1)^2}$ is an integer.
Note that this integer is clearly non-negative. Choosing $p$ sufficientely large, the corresponding integer is less than $1$, so that it is $0$. Thus $f(n) = n$.

Conversely, $f(n)=n$ is clearly a solution of the problem.

Pierre.

Yes,it is Number Theory 3 from the IMO Shortlisted 2004.The solution from the jury is as similar as Pierre'solution.Nice solution!Sorry for posting this problem here :blush:
@[b]nguyenquockhanh[/b]:Pigfly not pipfly,grrrr....You asked me:"who are you?" but I don't know exactly!Who am I?By the way,I am the friend of you :)

[quote="pbornsztein"]Hmm... ;)  Btw, where did you find the IMO-SL 2004?

Pierre.[/quote]
I have known some problems from IMO-SL 2004(not all) but I think I don't post them here because:
[quote="orl"]
Some countries use them to select team members for the next IMO thus it should not get known before the team has to be announced to the IMO host. Otherwise it would be unfair to those students who did not know the problems in advance.  ;)[/quote]
When I posted this problem,I didn't know it from IMO-SL 2004.Sorry again  :blush: for posting this problem here!

[quote="K09"]This problem I known since excercise of Dang Hung Thang and my solution above.[/quote]
Really :? ?Maybe Mr Dang Hung Thang had the IMO-SL 2004( he was leader of VietNam team in IMO2004).I'm sure this problem from IMO-SL2004 ;).
@K09:the HaiDuong team is very strong,isn't it?

[quote="pbornsztein"][quote="pigfly"]Yes,it is Number Theory 3 from the IMO Shortlisted 2004. [/quote]

Wow! I solved an IMO-SL one  :)

[quote] The solution from the jury is as similar as Pierre'solution.[/quote]
:)  I could be part of it  :)  :)

Pierre.[/quote]

We are just waiting until you become the new France (deputy) IMO delegation leader finally. The time will come.  :)

You and Moubinool would form a cool team.  ;)

[quote="pbornsztein"][quote="orl"]The time will come.  :) [/quote]

Maybe, but not in this life  ;)

Pierre.[/quote]

I am sure it will be in this life.  :)

Find all functions $ f: \mathbb{N^{*}}\to \mathbb{N^{*}}$ satisfying
\[ \left(f(m)^2+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\]

Let $P(m,n)$ denote the given assertion.

$P(1,1): f(1)^2+1\mid 4\implies f(1)=1$.

$P(1,n): f(n)+1\mid (n+1)^2$.

Set $n=p-1$, where $p$ is a sufficiently large prime.

We will show $f(p-1)=p-1$.

Note that $f(p-1)+1\mid p^2$, so $f(p-1)\in \{0,p-1,p^2-1\}$. Clearly $f(p-1)>0$.

Suppose FTSOC $f(p-1)=p^2-1$.

Then $P(p-1,1): f(p-1)^2+1\mid ((p-1)^2+1)^2$.

So $p^4-2p^2+2\mid (p^2-2p+2)^2$.

Note that leading term of RHS is $p^4$, so $2\cdot LHS>RHS$ for sufficiently large $p$.

Thus, for sufficiently large $p$, $p^4-2p^2+2=(p^2-2p+2)^2$. We can check that this doesn't hold true for primes $p$.

Thus, $f(p-1)=p-1$.

Let $p-1=k$.

$P(n,k): f(n)^2+k\mid (n^2+k)^2=n^4+2n^2k+k^2$.

Thus, $f(n)^2+k\mid n^4+2n^2k+k^2-(2n^2+k)(f(n)^2+k)=n^4-2n^2f(n)^2-f(n)^2k$.

Now \[f(n)^2+k\mid n^4-2n^2f(n)^2-f(n)^2k+f(n)^2(f(n)^2+k)=n^4-2n^2f(n)^2+f(n)^4\]

So $f(n)^2+k\mid (n^2-f(n)^2)^2$. Set $k$ insanely large, which implies $n^2=f(n)^2\implies \boxed{f(n)=n}$, which works.

Let $P(m,n)$ be the assertion $f(m)^2+f(n)\mid(m^2+n)^2$.
$P(1,1)\Rightarrow f(1)^2+f(1)\mid4\Rightarrow f(1)=1$
$P(1,p-1)\Rightarrow 1+f(p-1)\mid p^2$ for large primes $p$.
Suppose $1+f(q-1)=q^2$ for some prime $q$.
$P(q-1,1)\Rightarrow (q^2-1)^2+1\mid(q^2-2q)^2$
but for $q\ge3$, $(q^2-2q)^2\le(q^2-1)^2+1$, a contradiction.
So $f(p-1)=p-1$.

For sufficiently large primes $p$:
$P(p-1,n)\Rightarrow (p-1)^2+f(n)\mid(p-1)^4+2n(p-1)^2+n^2\Rightarrow(p-1)^2+f(n)\mid(p-1)^4+2n(p-1)^2+n^2-((p-1)^2+f(n))((p-1)^2+2n-f(n))\Rightarrow(p-1)^2+f(n)\mid(n-f(n))^2\Rightarrow\boxed{f(n)=n}$
which is a solution.

Let $P(m,n)$ be assertion of $\left(f^{2}\left(m\right)+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}$. $P(1,1) \implies f(1)=1$. $P(m,1)\implies f(m)^2+1\le (m^2+1)^2 \implies f(m)\le m^2$. Let $p$ be prime. $P(1,p-1)\implies 1+f(p-1)\mid p^2$ and $1<1+f(p-1)\le 1+(p-1)^2<p^2$. So $1+f(p-1)=p  \implies f(p-1)=p-1$ for all prime $p$. $P(p-1,n) \implies (p-1)^2+f(n)\mid ((p-1)^2+n)^2\implies (p-1)^2+f(n)\mid (f(n)-n)^2$. Choosing $p$ extremely large prime gives that $LHS>|RHS|\implies RHS=0 \implies f(n)=n$ for all $n$, which indeed works. So we are done.

Let $P(x,y)$ denote the given assertion.

[b][color=red]Claim:[/color][/b] $f(p-1)=p-1$ for all primes $p.$
[i]Proof.[/i] Note $P(1,1)$ yields $$(f(1)+1)f(1)=f(1)^2+f(1)\mid (1^2+1)^2=4$$ so $f(1)=1.$ Also, $P(1,p-1)$ gives us $1+f(p-1)\mid p^2$ so $1+f(p-1)$ is equal to $p$ or $p^2.$ Suppose FTSOC the latter. Then, $P(p-1,1)$ yields $(p^2-1)^2+1\mid ((p-1)^2+1)^2.$ On the other hand, $$(p^2-1)^2+1>(p^2-1)^2>(p^2-2p+2)=((p-1)^2+1)^2,$$ a contradiction. $\blacksquare$

From $P(p-1,n)$ we have $(p-1)^2+f(n)\mid ((p-1)^2+n)^2.$ Notice if $a\mid b^2,$ then $a\mid a^2-2ab+b^2=(a-b)^2.$ Thus, $$(p-1)^2+f(n)\mid ((p-1)^2+f(n)-(p-1)^2-n)^2=(f(n)-n)^2.$$ Varying $p,$ we see $(f(n)-n)^2$ has an unlimited amount of divisors so $f(n)=n,$ which works. $\square$

Note that $f(1)=1,$ trivial.

[b][u]Lemma:[/u][/b] $f(p-1)=p-1~\forall$ primes $p.$
[i]Proof.[/i] We have that $1+f(p-1)\mid p^2.$ Thus $f(p-1)\in \{p-1,p^2-1\}.$
It is not hard to get a contradiction from the latter. $\blacksquare$

$m\mapsto p-1\implies f(n)+(p-1)^2\mid (n+(p-1)^2)^2 \implies f(n)+(p-1)^2\mid (n-f(n))^2.$ Now for large $n,$ the identity satisfies.

Solved with hints from [b]megarnie[/b]
-----
Let $P(m,n)$ be the assertion.

$P(1,1) \rightarrow f(1)^2+f(1) \mid 4 \rightarrow f(1) = 1$

Let $p$ be a prime number.
$P(1,p-1) \rightarrow f(p-1)+1 \mid p^2 \rightarrow f(p-1)=p-1 \text{ or } f(p-1)=p^2$
Assume $f(p-1)=p^2-1$.

$P(p-1, 1) \rightarrow \text{ contradiction }$, since $p \neq 1$ ($1$ isn't prime!).
Thus, $f(p-1)=p-1$.

$$P(n, p-1) \rightarrow (p-1)^2+f(n) \mid ((p-1)^2+n)^2 \rightarrow (p-1)^2+f(n) \mid (f(n)-n)^2,$$ which implies the only solution is $f(n)=n$. $\blacksquare$

Let $P(m,n)$ denote the assertion. We have
\begin{align*}
(a)~~P(1,1)&: f(1)^2+f(1) \mid 4 \implies f(1)=1 \\
(b)~~P(m,1)&: f(m)^2+1\mid (m^2+1)^2\\
(c)~~P(1,m)&: 1+f(m) \mid (1+m)^2
\end{align*}
Doing $m\to p-1$ for $(c)$ gives $1+f(p-1)\mid p^2$, so $f(p-1)$ is $p-1$ or $p^2-1$. If it is $p^2-1$, then putting in $b$ gives $p^4-2p^2+2\mid p^4-4p^3+8p^2-8p+4$, which implies that \[4p^3-10p^2+8p-2\mid p^4-4p^3+8p^2-8p+4.\]
Note that $p\ge 2$. Let $f(p)=4p^3-10p^2+8p-2$, then $f'(p)=4(2p-1)(p-2)>0$. Since $f(2)=6$, $f(p)>0$ for all $p$. Let \[g(p)=(p^4-4p^3+8p^2-8p+4)-(4p^3-10p^2+8p-2)=p^4-8p^3+18p^2-16p+6\] then $g'(p)=4(p-1)^2(p-4)$. Since $g(5)=1$, $g(p)>0$ for all $p\ge 5$. It is easy to check that the divisibility does not hold for $p=2$ or $3$, so that is a contradiction. Thus, $f(p-1)=p-1$ for all primes $p$. In particular, there are infinitely many values of $n$ such that $f(n)=n$. Let $y$ be one such value. Note that $P(y,m)$ gives $y^2+m\mid (y^2+f(m))^2.$ We have $(y^2+f(m))^2\equiv (f(m)-m)^2\pmod y^2+m$, so picking large $y$ lets us conclude that $f(m)=m$. This function is our only solution, and it works because $m^2+n\mid (m^2+n)^2$.

It felt nice doing doing math after a long time, and especially struggling on a problem and still persisting and completing it is something i had not done in a while so this was really nice...

I will write out all the claims that i found in the order i found them, even though i could shorten this solution by a lot, i don't see a reason to

Let $P(m,n)$ denote the assertion that $f(m)^2+f(n) \mid (m^2+n)^2$

[hide=f(1) is 1]$P(1,1) \implies f(1)^2+f(1) \mid 4$[/hide]

[hide=f(2) is 2]$P(2,2) \implies f(2) \in {2,3}$ and $P(1,2) \implies f(2)+1 \mid 9$ so combining gives $f(2)=2$[/hide]

[hide=f(p-1) is p-1 for odd primes p also]$P(1,p-1) \implies 1+f(p-1) \mid p^2$ and since $f: \mathbb N \to \mathbb N, f(x) \ge 1$, so $f(p-1) \in {p-1,p^2-1}$
Now let's prove the latter is impossible, if $f(p-1) = p^2 -1$,  $P(p-1,1) \implies (p^2-1)^2+1 \mid ((p-1)^2+1)^2
\implies (p^2-1)^2+1 \mid ((p-1)^2+1)^2$

Let's look at the right thing mod $(p^2-1)^2+1=p^4-2p^2+2$

$((p-1)^2+1)^2=(p^2-2p+2)^2=p^4+4p^2+4-4p^3-8p+4p^2 \equiv 10p^2+2-4p^3-8p$,

but $p^4-2p^2+2$ is odd, so this implies we can take out the factor of two from the above thing, so we get that it must divide $2p^3+4p-5p^2-1$

But note $p^4-2p^2+2 > 2p^3+4p-5p^2-1 \iff p^4+3p^2+3 \ge 2p^3+4p$ but by AM-GM ($p^4+p^2 \ge 2p^3$ and $2p^2+2 \ge 4p$), this is always true, so we get a contradiction (as $a \mid b \implies b \ge a$ unless $b=0$ and we can see that $2p^3+4p-5p^2-1$ has no integer solutions)
[/hide]

[hide=f is injective]Assume it was not injective, so there exists unequal integers $a$ and $b$ such that $f(a)=f(b)$
Let $c=f(m)^2+f(a)=f(m)^2+f(b) \ge 2$, from $P(m,a)$ and $P(m,b)$ we see that $c$ divides both $(m^2+a)^2$ and $(m^2+b)^2$, so $c$ divides their gcd,
which is equal to $(gcd(m^2+a,m^2+b))^2= (gcd(a-b,m^2+b))^2$
If $p_1,p_2,\cdots p_k$ divide both $a-b$ and $b$ and $q_1,q_2,\cdots q_r$ divide $a-b$ but not $b$, choosing $m=q_1q_2 \cdots q_r(p_1p_2 \cdots p_k+1)$, we see that any prime that divides $a-b$ does not divide $m^2+b$ so their $\gcd=1$ which in turn implies $c=1$ which is impossible.
[/hide]

[hide=f(m) divdes m^2 for all m]Choose $n=q-1$ where $q$ is a prime that is $1$ mod $f(m)^2$, it's existence is guaranteed by dirichlet. Then, $f(m)^2 \mid f(m)^2+f(n) \mid (m^2+n)^2$ but $n$ is divisible by $f(m)^2$ so we get that $f(m)^2 \mid m^4 \iff f(m) \mid m^2$[/hide]

[hide=f(p)=p for all primes]from above we get $f(p) \mid p^2 \implies f(p) \in {1,p,p^2}$ but $f(p)$ can't be $1$ due to injectivity, and $f(p) \neq p^2$ because otherwise, $P(1,p) \implies 1+p^2 \mid (1+p)^2 \implies 1+p^2 \mid 2p$ which is impossible due to size. [/hide]

[hide=f(n)=n for all n]
$P(p,n) \implies p^2+f(n) \mid (p^2+n)^2 \implies p^2+f(n) \mid (p^2+n)^2-(p^2+n)(p^2+f(n)) = (p^2+n)(f(n)-n)$ but $\mod p^2+f(n)$, this is just $-(f(n)-n)^2$ (as $p^2= -f(n) \mod p^2+f(n)$), but now taking very large $p$ gives contradiction unless $f(n)=n$
[/hide]

First plugging in $m = n = 1$ yields $$f(1)^2+f(1) \mid 4$$ but since $f(1)$ is a positive integer we must have $f(1) = 1$.

Now plug in $m=1$ to get $1+f(n) \mid (1+n)^2$ for all positive integers $n$. Plugging in $n = p-1$ here yields $1+f(p-1) \mid p^2$ which implies that $f(p)$ is either equal to $p-1$ or $p^2-1$. We now prove that we must actually have $f(p) = p-1$.

Plug in $m = n = p-1$ for a prime $p$ to get $$f(p-1)(1+f(p-1)) \mid p^2(p-1)^2$$

If $f(p-1) = p^2-1$ then this means $$(p^2-1)p^2 \mid p^2(p-1)^2$$ which implies $p^2-1 \mid (p-1)^2$ which implies $p^2-1 \mid 2-2p$ which is clearly false by size reasons. It follows that $f(p-1) = p-1$ as desired.

Now, plug in $m = p-1$ to get $$(p-1)^2+f(n) \mid ((p-1)^2+n)^2$$ which is equivalent to $$((p-1)^2+n)^2 \equiv 0 \pmod {(p-1)^2+f(n)}$$ This is also equivalent to $$(n-f(n))^2 \equiv 0 \pmod {(p-1)^2+f(n)}$$ and taking a large enough $p$ gives us $(n-f(n))^2 = 0$ or $f(n) = n$ for all positive integers $n$.

f(1)^2+f(1)|4 means f(1)=1 since the other values don't work; we're motivated to use primes p because dividing p^2 only gives three choices, so substitute (p-1,1) and (1,p-1) to get f(p-1)={p-1,p^2-1}, and f(p-1)^2+1|((p-1)^2+1)^2; the latter in the set doesn't work by plugging in (p^4-2p^2+2|p^4-4p^3+8p^2-8p+4, or p^4-2p^2+2|-4p^3+10p^2+2, contradiction for p\ne 2 since LHS>RHS\ne 0, and for p=2 we already know f(1)=1), so f(p-1)=p-1. The problem is now evident, because (p-1)^2+f(n)|((p-1)^2+n)^2\implies (p-1)^2+n|(n-f(n))^2; taking p large enough s.t. LHS>RHS, we must have RHS=0; in particular, n=f(n) for all n.

Denote the assertion with $P(m, n)$. Then by $P(1, 1)$ it follows that $f(1)^2 + f(1) \mid 4$, so $f(1) = 1$ must hold.

[b][color=pink]Claim:[/color][/b] $f(p-1) = p-1$ for primes $p$.
[i]Proof.[/i] By $P(1, p-1)$ it follows that \[ 1 + f(p-1) \mid p^2. \] so $f(p-1)$ is either $p-1$ or $p^2-1$.
Then, by $P(p-1, 1)$ it follows that \[ f(p-1)^2 + 1 \mid ((p-1)^2 + 1)^2 \] which implies $f(p-1) = p-1$. $\blacksquare$
By taking a sufficiently large prime $p$, it follows by $P(p-1, n)$ that \[ (p-1)^2 + f(n) \mid ((p-1)^2 + n)^2 \]
This simplifies as \[ (p-1)^2 + f(n) \mid (n - f(n))^2 \] which forces $n = f(n)$.

The idea is to just get $f(n)$ for $n=p-1$ for sufficiently large $p$. This is doable because $f(1)^2+f(1) \mid 4$, hence $f(1) = 1$, so in particular $$f(p-1) + 1 = f(1)^2 + f(p-1) \mid p^2.$$ This narrows down $f(p-1) \in \{p-1, p^2-1\}$, and furthermore $$f(p-1)^2 + 1 \mid (p^2-2p+2)^2$$ ensures $f(p-1) = p-1$. Now just writing $$f(n) + (p-1)^2 \mid (n+(p-1)^2)^2$$ for fixed $n$ and sufficiently large $p$ forces $f(n) = n$.

The only function is $f(n) = n$ for all $n \in \mathbb{N^{*}},$ which works. We will now show it is the only solution.

Let $P(m, n)$ denote the given assertion. From $P(1, 1)$ we have $f(1)^2 + f(1) \mid 4$, so $f(1) = 1$ since $f(1) \in  \mathbb{N^{*}}$. Now, we will show that for all primes $p$, $f(p - 1) = p - 1$. From $P(1, p - 1)$, we get
$$1 + f(p - 1) \mid p^2,$$
so $f(p - 1) \in \{p - 1, p^2 - 1\}$. But from $P(p - 1, p - 1),$ we have
$$f(p - 1)^2 + 1 \mid ((p - 1)^2 + 1)^2,$$
which is impossible if $f(p - 1) = p^2 - 1$ since $(p^2 - 1)^2 > (p^2 - 2p + 2)^2 = ((p - 1)^2 + 1)^2.$

So, we can now fix $n \in N$ and take $P(p - 1, n)$ for some prime $p$, which gives
$$(p - 1)^2 + f(n) \mid ((p - 1)^2 + n)^2.$$
Since $((p - 1)^2 + n)^2 - (n - f(n))^2 = ((p - 1)^2 + 2n - f(n))((p - 1)^2 + f(n)),$ this implies that $(p - 1)^2 + f(n) \mid (n - f(n))^2.$ Taking $p$ large enough, we get $f(n) = n$, as desired.

Pretty clearly, $f(x) = x$, and we'd like to prove this is the only such function.

We'll start with some basic substitutions. $P(1, 1)$ gives $f(1)^2 + f(1) \mid 4 \implies f(1) = 1$. To narrow down the function, we can substitute in something related to primes, which we can do by $P(1, p - 1)$, which gives:
$$1 + f(p - 1) \mid p^2 \implies f(p - 1) = p - 1, p^2 - 1$$since the only divisors of $p^2$ are $1, p, p^2$ and $f(p - 1) > 0$.

Say, for the sake of contradiction, $f(p - 1) = p^2 - 1$. $P(p - 1, p - 1)$ gives:
$$(p^2 - 1)^2 + p^2 - 1 \mid p^2(p - 3)^2$$$$\implies p^2(p^2 - 1) \mid p^2(p - 3)^2$$$$\implies p^2 - 1 \mid p^2 - 6p + 9$$which is impossible as $p^2 - 1 > p^2 - 6p + 9$.

So, we must have $f(p - 1) = p - 1$. To prove $f(n) = n$ for all $n$, we can substitute $P(n, p - 1)$.
$$f(n)^2 + f(p - 1) \mid (n^2 + (p - 1))^2 = n^4 + 2n^2(p - 1) + (p - 1)^2$$$$\implies f(n)^2 + p - 1 \mid n^4 + 2n^2(p - 1) + (p - 1)^2 - 2n^2(f(n)^2 + p - 1)$$$$\implies f(n)^2 + (p - 1) \mid n^4 - 2n^2f(n)^2 + (p - 1)^2$$This also means$f(n)^2 + (p - 1) \mid n^4 - 2n^2f(n)^2 + (p - 1)^2 - (p - 1)(f(n)^2 + (p - 1))$.

$$\implies f(n)^2 + (p - 1) \mid n^4 - 2n^2f(n)^2 - (p - 1)f(n)^2$$This implies $f(n)^2 + (p - 1) \mid n^4 - 2n^2 f(n)^2 - (p - 1)f(n)^2 + f(n)^2(f(n)^2 + (p - 1)) = n^2 - 2n^2f(n)^2 + f(n)^4 = (n^2 - f(n))^2$

For sufficiently large $p$, we must have $(n^2 - f(n))^2 = 0 \implies f(n) = n$.

[b]credits to my blog :)[/b]


In case you do have the results, please attach them.. I cannot see how was it possible for them not to appear anywhere.. Thanks!  :)

what do you mean by results? The final ranking was not hang out as promised, so no one knows.

I know 90 is in the middle of third prizes, since that was what I had, but much more I don't know.

By results I mean the place of everyone and the price he took. I'm sure that the final ranking was anounced but it wasn't posted on the website.

It will be posted there soon I think. No, I did not write down all the names on the ceremony :roll:

I've got the first prize. Congrats to all other mathlinkers who was in Bulgaria!!!

You got 'a' first prize or you got 'THE' (great grand) first prize? :)

(since that guy was also from bellorussia


geometry
circumcircle
trapezoid
analytic geometry
geometry unsolved

(1) Given a triangle ABC, find a construction by ruler and compass of three equal squares such that each square touches two sides and all three squares touch at a single common point (Kenmotu point)

(2) Prove that the edge lengths of inscribed squares are
\[
\ell  = \frac{{\sqrt 2 abc}}{{a^2  + b^2  + c^2  + 4\Delta }}
\]
where a,b,c,$\Delta$ are the sides and the area of triangle ABC.

Thanks in advance
Leon

Assume that  $\triangle KTS,\triangle KRQ,\triangle KPU$ are congruent isosceles right triangles with common apex $K.$ $(P,Q) \in BC,$ $(R,S) \in CA,$ $(T,U) \in AB$ (See the diagram below). Hexagon $PQRSTUV$ is obviously cyclic with circumcenter $K$ and $UPQR$ is an isosceles trapezoid with $UR \parallel PQ$ $\Longrightarrow$ $TS$ is antiparallel to $UR \parallel BC$ WRT $AB,AC.$ If $M_A,M_B,M_C$ denote the apices of the isosceles right triangles erected outside $\triangle ABC,$ then $ABM_AC \sim ASKT$ $\Longrightarrow$ $\angle BAM_A=\angle SAK,$ i.e. $AM_A$ and $AK$ are isogonals WRT $\angle A.$ Similarly, $BM_B$ and $CM_C$ are the isogonals of $BK$ and $CK$ WRT $\angle B$ and $\angle C.$ Hence, $K$ is the isogonal conjugate of the 1st Vecten point $V \equiv AM_A \cap BM_B \cap CM_C,$ i.e. the 1st Kenmotu point of $\triangle ABC.$

For the second part, we resort to calculations with barycentric coordinates with respect to $\triangle ABC.$ Using Conway's notation, the coordinates of $M_A,M_B$ and $M_C$ are

$M_A(-a^2:S_C+S:S_B+S)$

$M_B(S_C+S:-b^2:S_A+S)$

$M_C(S_B+S:S_A+S:-c^2)$

Therefore, $K(a^2(S_A+S):b^2(S_B+S):c^2(S_C+S))$

$\frac{|\triangle KAC|}{|\triangle ABC|}=\frac{b^2}{2S^2+S(a^2+b^2+c^2)} \Longrightarrow \text{d}(K,AC)=\frac{b(S_B+S)}{a^2+b^2+c^2}$

$\frac{|\triangle M_ABA|}{|\triangle ABC|}=\frac{S_B+S}{2S} \Longrightarrow \text{d}(M_A,AB)=\frac{S_B+S}{2c}$

From $ABM_AC \sim ASKT$ we deduce that $\frac{\ell}{BM_A}=\frac{\text{d}(K,AC)}{ \text{d}(M_A,AB)}.$ Then

$\ell=\frac{\sqrt{2}}{2}a \cdot \frac{2bc}{a^2+b^2+c^2+2S}=\frac{ \sqrt{2}abc}{a^2+b^2+c^2+4|\triangle ABC|}.$


geometry circumcircle perimeter trigonometry

How in the world do you find out the area of a dodecagon? I bumped into one of those questions once but I'm wondering how to solve it... :help:

The formula is 3 times the square of the radius of the circumcircle of the dodecagon.

Or, if r is the circumradius (radius of the circumcircle), then the area is
$3r^{2}$.

check the side bar, it's one of those visual proofs things  :)

I would use $Area=\frac{1}{2}\cdot Apothem \cdot Perimeter$.

Both of these can be found using trigonometry, using a central angle of $\frac{180}{12}=15^{\circ}$.

How do you prove (rigorously) the side bar one?

given the dodecagon's side length, how do you find the circumradius?

[hide="sexy solution"]
given the side length, you don't need the radius.
Extend the sides to make a hexagon, and subtract the extra area added[/hide]

Thanks, I'm glad you'll put your heads together, cuz now I understand, even though I'm a noob at math.  :D


A polynomial P has four roots 1/4, 1/2, 2, 4.  The product of the roots is 1, and P(1)=1.  Find P(0).

Why does it say Riemann Hypothesis under your name?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10831

contains the answers to all of life's greatest questions, and your question, too.


calculus
integration
trigonometry
algebra
polynomial
logarithms
calculus computations

Hi,
I think this might work:

1) Use the substitution $x=t^{12}$, it follows that

$\int\frac{dx}{\sqrt[3]{x}+\sqrt[4]{x}}=12\int\frac{t^{8}}{t+1}dt$, then simply divide the two  polynomials and the problem is solved.

2) $\int\frac{\sec x}{1+\sin x}dx=\int\frac{\cos x}{\cos^{2}x (1+\sin x)}dx$
Use the substitution $x=\sin t$ and  partial fraction decomposition to solve the problem.
Didi

[quote="sinajackson"]Hi, excuse me I don't understand the solution of $3)$, what do you mean $x^{2}+x+1=(x-z)^{2}$?!!! I solved it like this:
let $x=\frac{1}{t}$ so $dx=\frac{-dt}{t^{2}}$ and so:
$\int\frac{\frac{-dt}{t^{2}}}{\frac{1}{t}\sqrt{\frac{1}{t^{2}}+\frac{1}{t}+1}}=-\int\frac{dt}{\sqrt{t^{2}+t+1}}=-\int\frac{dt}{\sqrt{(t+\frac{1}{2})^{2}+\frac{3}{4}}}=-ln(t+\frac{1}{2}+\sqrt{t^{2}+t+1})+c$
where $t=\frac{1}{x}$. What's wrong with it?!!![/quote]

The question says $x^{2}+x+2$, not $x^{2}+x+1$.

[quote="greengrass"][quote="benorin"][quote="benorin"]3)$\int \frac{dx}{x\sqrt{x^{2}+x+2}}$ use the substitution $x^{2}+x+2=(z-x)^{2}$[/quote]

I saw this one all solved-out in some oldish (circa 1960) calc book [the other two came out of Stewart's Calculus (2e or 3e?) Test Item Book, Vol. 2, and were marked as "hard" with answer]:

$x^{2}+x+2=(z-x)^{2}\Rightarrow x=\frac{z^{2}-2}{1+2z}$ so that $dx=\frac{2(z^{2}+z+2)}{(1+2z)^{2}}dz$ and that $\sqrt{x^{2}+x+2}=\frac{z^{2}+z+2}{1+2z}$ hence \[\int \frac{dx}{x\sqrt{x^{2}+x+2}}= \int \frac{\frac{2(z^{2}+z+2)}{(1+2z)^{2}}dz}{\frac{z^{2}-2}{1+2z}\cdot\frac{z^{2}+z+2}{1+2z}}= 2\int \frac{dz}{z^{2}-2}=-\frac{1}{\sqrt{2}}\mbox{tanh}^{-1}\left(\frac{z}{\sqrt{2}}\right)+C\]  \[=-\frac{1}{\sqrt{2}}\mbox{tanh}^{-1}\left(\frac{\sqrt{x^{2}+x+2}+x}{\sqrt{2}}\right)+C\] [/quote]

Excuse me, benorin

$x^{2}+x+2\, =\, (z-x)^{2}\Rightarrow z\, = \, x \pm \sqrt{x^{2}+x+2}$

¿why you choose positive sign and not negative one?[/quote]

Well, I didn't, but rather the author of the text did and I hadn't really thought about it untill you asked: thank you. Here are my thoughts on thew matter: consider that $\mbox{tanh}^{-1}(z)=\frac{1}{2}\left( \log (1+z)-\log (1-z)\right)$ so that the branch cut is on $(-\infty ,-1]\cup [1,\infty )$ and is thus only defined for real arguements in the range $(-1,1)$ the following plot [made using http://www.quickmath.com ] will make clear when which sign ought to be chosen

[url=http://allyoucanupload.webshots.com/v/2004323843441873051][img]http://aycu12.webshots.com/image/3891/2004323843441873051_rs.jpg[/img][/url]

[quote="sinajackson"]Ok paladin8, i made a mistake , let $x=\frac{1}{t}$ so:
$\int\frac{\frac{-dt}{t^{2}}}{\frac{1}{t}\sqrt{\frac{1}{t^{2}}+\frac{1}{t}+2}}=-\int\frac{dt}{\sqrt{2t^{2}+t+1}}=-\frac{1}{\sqrt{2}}\int\frac{dt}{\sqrt{t^{2}+\frac{t}{2}+\frac{1}{2}}}=-\frac{1}{\sqrt{2}}\int\frac{dt}{\sqrt{(t+\frac{1}{4})^{2}+\frac{7}{16}}}$
$=-\frac{1}{\sqrt{2}}\ln \left(t+\frac{1}{4}+\sqrt{t^{2}+\frac{t}{2}+\frac{1}{2}}\right)$

now is it ok? :maybe:[/quote]

One mistake is that $\frac{7}{16}$ ought to be $\frac{3}{8}$ when you complete the square...

[quote="benorin"]Well, I didn't, but rather the author of the text did and I hadn't really thought about it untill you asked: thank you. Here are my thoughts on thew matter: consider that $\mbox{tanh}^{-1}(z)=\frac{1}{2}\left( \log (1+z)-\log (1-z)\right)$ so that the branch cut is on $(-\infty ,-1]\cup [1,\infty )$ and is thus only defined for real arguements in the range $(-1,1)$ the following plot [made using http://www.quickmath.com ] will make clear when which sign ought to be chosen [/quote]

Tks  :lol:

[quote="sinajackson"][quote="benorin"]One mistake is that $\frac{7}{16}$ ought to be $\frac{3}{8}$ when you complete the square...[/quote]

sorry but are you sure? :maybe:
$t^{2}+2\times\frac{t}{4}+\frac{1}{16}-\frac{1}{16}+\frac{1}{2}$

and $\frac{1}{2}-\frac{1}{16}=\frac{7}{16}$[/quote]

My bad  :blush: your absolutely correct.

[quote="sinajackson"]Ok paladin8, i made a mistake , let $x=\frac{1}{t}$ so:
$\int\frac{\frac{-dt}{t^{2}}}{\frac{1}{t}\sqrt{\frac{1}{t^{2}}+\frac{1}{t}+2}}=-\int\frac{dt}{\sqrt{2t^{2}+t+1}}=-\frac{1}{\sqrt{2}}\int\frac{dt}{\sqrt{t^{2}+\frac{t}{2}+\frac{1}{2}}}=-\frac{1}{\sqrt{2}}\int\frac{dt}{\sqrt{(t+\frac{1}{4})^{2}+\frac{7}{16}}}$
$=-\frac{1}{\sqrt{2}}\ln \left(t+\frac{1}{4}+\sqrt{t^{2}+\frac{t}{2}+\frac{1}{2}}\right)$

now is it ok? :maybe:[/quote]

I think your expression may be converted to benorin's by using $\tanh^{-1}{x}= \frac{1}{2}\ln{\left(\frac{1+x}{1-x}\right)}$.


Let $ a,b,c\ge 0$, s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that:

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

[quote="Inequalities Master"]Let $ a,b,c\ge 0$, s.t. $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:

$ 3(ab \plus{} bc \plus{} ca)\geq (a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2)^2$[/quote]
$ c \equal{} 0, a \equal{} b \equal{} \sqrt {\frac{3}{2}}$ then $ LHS \equal{} \frac {9}{2}$, $ RHS \equal{} \frac {243}{32} > \frac {9}{2}$


rotation
geometry
geometric transformation
AMC
AIME

In how many patterns can six identical L-shaped pieces, each consisting of three unit squares, be placed to perfectly cover a 3 by 6 board? One such pattern is shown.

[asy]draw((0,0)--(6,0),linewidth(2));
draw((0,3)--(6,3),linewidth(2));

draw((0,0)--(0,3),linewidth(2));
draw((2,0)--(2,3),linewidth(2));
draw((4,0)--(4,3),linewidth(2));
draw((6,0)--(6,3),linewidth(2));

draw((0,1)--(1,1)--(1,2)--(3,2)--(3,1)--(5,1)--(5,2)--(6,2),linewidth(2));[/asy]

I don't know how to explain it, but I think there are 3 ways.

I only found two, one is the one shown, and the other is each of the pieces rotated by 90 degrees.

yeah, i got $ \boxed{8}$ also by trial and errrrror

but i might have counted some twice

I know for a fact that the answer is $ 8$.  :wink:

[quote="Brut3Forc3"]You will only be able to tile the L-pieces in 3 separate 3 by 2 rectangles. Each one will have 2 possible ways of tiling it, so there will be 8 ways total.[/quote]

What he said.  :P

THis has something to do withFibonacci sequence, I can't remember what, except 8 is 6th fibonacci numb er (1,1,2,3,5,8)

[quote="myyellowducky82"][quote="herefishyfishy1"]I don't like this problem because it does not state whether rotations of a floor are distinguishable.[/quote]
I don't think it matters[/quote]

i think it does.....read the problem

Yeah, I agree it's a bad problem. Anyway, let's let this thread die.

why do you always say kill this thread?

im not a murderer  :(

lol

[quote="myyellowducky82"][quote="isabella2296"]...

That was the kind of spam I'm talking about. I'm serious, we need a moderator...I'd be happy to be one, if levans happens to be reading this! [Fat chance][/quote]
They choose you; you don't choose.
[did that make ANY sense??][/quote]

Um... does that mean that mods can choose to visit these threads or not?

I'm not the most proficient person at english.

guys, plz stop spamming
right now, i would rate of the recent posts in this spam, but unfortunately, i can't rate that many times in one hour
so i'll let these slide for now, but plz stop spamming
anyway, i'm pretty sure the answer to this is $ 8$

[quote="isabella2296"]A mod could but no mod does. But we NEED a mod. I told levans about this, and got no reply.  :([/quote]
Tell the other admins.

guys
i already pmed levans about this
he said to just mark the posts 1 star, as spam
so could we plz drop this discussion, especially in this post, which was created for answers to a question


Putnam
algorithm
induction
LaTeX
linear algebra
matrix
algebra

Let [tex]a[/tex] and [tex]b[/tex] be complex numbers with [tex]a \neq b[/tex], and let [tex]f(x) = (r_1 - x)(r_2 - x) \cdots (r_n - x)[/tex] for some fixed complex numbers [tex] r_1, r_2, \ldots, r_n [/tex].

Show that [tex] \displaystyle {det \left( \begin{array}{ccccc}
r_1 & a & a & \cdots & a \\
b & r_2 & a & \cdots & a \\
b & b & r_3 & \cdots & a \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
b & b & b & \cdots & r_n
\end{array} \right)} [/tex] [tex] \displaystyle {= \frac {bf(a) - af(b)}{b-a}.} [/tex]

What do you think happens to the determinant when [tex] a = b [/tex]?

Guess: [hide] Well obviously b/c of what you said the det equals, it would be undefined.  However I also know that there is no divison used in the algorithim for finding a determinant.[/hide]

You don't happen to own Larson's problem solving through problems, becasue I believe I've seen the last three problems you have posted in it.

In deed a do! That's where the last few problems have come from. This one was actually also on one of my midterms in college. I came up with a different solution than the one in the book which I like a lot better.

Guess at possible solution:

[hide]Expand the determinant along the top row to get some sort of recursive formula (hopefully).[/hide]

JBL wrote:I was just about to write how zabelman's method doesn't work, but [hide]I think it might if you use induction, too.[/hide]  Details to follow.

Induction looks very promising.

Okay, so the base case n = 2 is trivial, and the problem is pretty meaningless for n = 1.

Assume it's true for n = k.

Look at n = k + 1.
Expand the determinant along the top row.  I don't know how to latex matrices, so I'll just try and explain what the terms are.  The first term, by induction, is
[tex]\displaystyle r_1 \cdot \frac{b \cdot (r_2 - a)(r_3 - a) \cdots (r_{k+1}-a) - a \cdot (r_2 - b) \cdots (r_{k+1}-b)}{b-a}[/tex].

In all other terms, the left-most term of the "f(b)" product will be zero, so every other term will look like:
[tex]\displaystyle (-1)^{i+1} a\cdot\frac{b\cdot(r_1 - a)\cdots\hat{(r_i-a)}\cdots(r_{k+1} - a)}{b-a}[/tex], where that hat indicates the ith term is missing, for i = 2 to (k + 1).

I don't unfortunately have the texing skills to make it clear, I'm afraid.  I think that you can just group terms and everything will work out well, though.

To [tex]\TeX[/tex] matrices, use \begin{pmatrix} (or bmatrix or matrix or vmatrix or Vmatrix, but probably pmatrix) (1,1) & (1,2) & ... & (1,n) \\ (2,1) & (2,2) & ... & (2,n) \\ etc. \end{pmatrix}, where (i,j) is the ijth entry. The &s separate entries, and the \\s make new lines.

Sorry, forget that, it doesn't work anyhow ... the diagonal moves.

As for when a = b, if we set b = a + h and take the limit as h -> 0, I believe we get that the determinant is (f(a) - af'(a)).

Ok, I got it, but I'm not sure how to tex it, so I'll just walk through (most of) the steps.

[hide]Use induction (what a surprise!). Let the big determinant be X. First, factor out [tex]r_1/b[/tex] from the top row to get [tex]r_1/b[/tex] times the determinant of a matrix that has b in the (1,1) spot and [tex]ab/r_1[/tex] for the rest of the row. Then, subtract this row from all other rows ot get 0's down the left column, and then take the determinant along the left column. We end up with [tex]r_1[/tex] times the determinant of a matrix that has [tex]r_2-(ab/r_1)[/tex], [tex]r_3-(ab/r_1)[/tex] etc down the diagonal, all [tex]b-(ab/r_1)[/tex] below the diagonal, and all [tex]a-(ab/r_1)[/tex] above the diagonal. But this has the form of the inductive hypothesis! So, use the inductive hypothesis and some algebra to get the final result. [/hide]IM SORRY FOR THE UNCLARITY. WHEN I GET THE SKILL AND PATIENCE, I PROMISE I WILL TEX A FULL SOLUTION!

There is a sweet non-inductive proof for this problem. Hint: [hide]Call the matrix in question A = (aij). Let G(x) =  det (aij + x). Prove that G(x) is a polynomial in x of deg. :le: 1. Then use that information to determine the determinant.[/hide]

The ideas of gauss202 is very nice

With the multilinearity of the determinant we can see $g(x)$
is a polynomial of degree  less or equal to $1$

$g(x) = mx+n$

$g(-a) = f(a), g(-b)=f(b)$ gives a system of equation

$m = \frac{f(a)-f(b)}{b-a}, n = \frac{bf(a)-af(b)}{b-a}$

$g(0) = detA$

We fixe a and let b tend to a

$n = -a.f'(a) - f(a) $


calculus
integration
function
complex analysis
complex analysis unsolved

[quote="sylow_theory"]How would we show that $ \zeta\left(\frac{1}{2}\plus{}i\right) \equal{} 0$ for this specific number?

Is it hard? Has anyone been able to show that this has zeros of a certain form (I know Hardy proved infinitude).[/quote]

That's not zero.  I guess you mean showing zeta is zero at the zeros.  Call all of them $ \rho_{n}$.  Then how do we show:

$ \zeta[\rho_{1}]\equal{}0;\quad\rho_{1}\approx 1/2\plus{}14.1i$

I don't know about the official way to do that but I can suggest some ways:

1. Numerically integrate the contour integral function representation of Zeta at $ \rho_{1}$ (get an accurate value for it first)  with 25 decimal places of accuracy.  Nice exercise in Mathematica and see if you come up a decimal value with say 24 zeros.

2.  Use the $ \Xi$ function which has the same zeros as zeta but are pure real.  If it crosses the x-axis there, must be a zero there.

3.  Suppose could use the Argument principle with a contour around one of the zeros and the pole at s=1.  So if it comes out being zero then the contour must have encircled one zero since the equation is number of zeros - number of poles.


Let ABCD be a square of side 1. Suppose P lies on BC, Q lies on DC, and that AP = AQ . Show that

\[
AP + PQ + QA \le 2 + \sqrt 2
\]

This is easy. Denote $BP=x$. Then we have to find maximum of $f(x)=2\sqrt{1+x^2}+(1-x)\sqrt2$ if $x\in(0,1)$.
We have $f(0)=2+\sqrt2$ and $f^\prime(x)=2x/\sqrt{1+x^2}-\sqrt2\leq0$.

[img]http://img.photobucket.com/albums/v218/hudrea/prob01.jpg[/img]

$AP + PQ + QA \le 2 + \sqrt 2  \Leftrightarrow \sqrt {1 + x^2 }  + \sqrt {2\left( {1 - x} \right)^2 }  + \sqrt {1 + x^2 }  \le 2 + \sqrt 2  \Leftrightarrow $
$\mathop  \Leftrightarrow \limits_{\mathop {x \in \left[ {0,1} \right]}\limits^ \uparrow  } 2\sqrt {1 + x^2 }  + \sqrt 2 \left( {1 - x} \right) \le 2 + \sqrt 2  \Leftrightarrow 2\sqrt {1 + x^2 }  \le 2 + x\sqrt 2  \Leftrightarrow $
$ \Leftrightarrow 4\left( {1 + x^2 } \right) \le \left( {2 + x\sqrt 2 } \right)^2  \Leftrightarrow x\left( {x - 2\sqrt 2 } \right) \le 0$
that is true.


inequalities
function
inequalities proposed

Let $x,y,z$ be nonnegative integers, no two of which is $0$. Prove that \[\frac{3}{2}\le \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\]

By $\frac{x}{x+y}<\frac{x+z}{x+y+z},\frac{y}{y+z}<\frac{y+x}{x+y+z},\frac{z}{z+x}<\frac{z+y}{x+y+z}$, we have 2nd inequality  is proved!

[quote="arqady"][quote="nayel"]Let $x,y,z$ be nonnegative integers, no two of which is $0$. Prove that
\[\frac{3}{2}\le \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\]
[/quote]
It's wrong. Try $x=3,$ $y=1$ and $z=2.$ :wink:[/quote]

yes i found the same counterexample. but i have a proof of the inequality. i don't understand, its all mixed up! :o

If so, please post your proof here, We will check it!

[quote="nayel"]we have

$=\frac{\sum{x^{2}y}+xyz}{\sum{x^{2}y}+\sum{xy^{2}}+2xyz}+1$

$\ge \frac{\sum{x^{2}y}+xyz}{2(\sum{x^{2}y}+xyz)}+1=\frac{1}{2}+1=\frac{3}{2}$

this is wrong[/quote]

actually the minimum of the function sigma(x/x+y)is 1
and the maximum 2
try x,y--->0 andz---->infinity
or x---->0 and y,z---->infinity

<Edit: stupid typo in maximum 3---->2>

sorry this is wrong! just switch $z$ and $x$ and the inequality is reversed! :blush:

[quote="arqady"][quote="Zamfirmihai"][quote="nayel"]Let $x,y,z$ be nonnegative integers, no two of which is $0$. Prove that
\[\frac{3}{2}\le \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2 \]
[/quote]

the inequality holds ONLY if x,y,z are the sides of a triangle[/quote]
What if $x=4,$ $y=2$ and $z=3$ $?$ :wink:[/quote]
yes , i found that , when $x \geq z \geq y$ , the inequality is reversed (without the condition "lenght-side")

[quote="nayel"]Let $x,y,z$ be nonnegative integers, no two of which is $0$. Prove that
\[\frac{3}{2}\le \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\]
[/quote]
$\left(\frac{x}{x+y}\right)^{k}+\left(\frac{y}{y+z}\right)^{k}+\left(\frac{z}{z+x}\right)^{k}\ge \frac{3}{2^{k}}$
where $k \ge \frac{ln3}{ln2}$

I thought that is sum cyc from x/(y+z) :oops:  in my book by Arthur Engel this sum is also between 3/2 and 2 but only if a,b,c are the sides of a triangle. the solution, for case sum<2, is using x+y>p, p is the semiperimeter

finally from my post we can conclude that the right problem should be \[1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\]

[quote="Zamfirmihai"]I thought that is sum cyc from x/(y+z) :oops:  in my book by Arthur Engel this sum is also between 3/2 and 2 but only if a,b,c are the sides of a triangle. the solution [/quote]
but , here , we have sum cyc from x/(x+y) not x/(y+z) :D

about the LHS after changing ,
$\frac{x}{x+y}> \frac{x}{x+y+z}$
making two others similar ineq and adding them up

okey, my original inequality will be true if we restrict it to the condition that $x\ge y\ge z$. now prove it. :D
let $x\ge y\ge z>0$. prove that \[\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge \frac{3}{2}\]

first , let take a look at the sum
$B=\sum \frac{y}{x+y}$
and we may denote the above $LHS A= \frac{x}{x+y}$
consider $A-B = \sum\frac{x}{x+y}-\sum\frac{y}{x+y}$ and after a few caculations we obtain
$A-B=\frac{(x-y)(z-x)(z-y)}{(x+y)(x+z)(y+z)}\geq 0$
so we have $A \geq B$ which is equivalent to $2A \geq A+B=3$ , TCTP :D

I am reviving this thread as I want to see more solutions of this inequality  :oops:
[quote="nayel"]
let $x\ge y\ge z>0$. prove that \[\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge \frac{3}{2}\][/quote]
What a brilliant observation by Samin Riasat!!

[color=#0000FF][b]Problem.[/b] let x ≥ y ≥ z > 0. prove that

$\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge \frac{3}{2}$.[/color]

[i]Solution.[/i] Since $x\ge y\ge z$, we can apply the substitution $x = z + u + v$ and $y = z + u$ (with $u,v \ge 0$) to obtain

$\frac{z+u+v}{2z+2u+v}+\frac{z+u}{2z+u}+\frac{z}{2z+u+v}\ge \frac{3}{2}$.

Note that equality occurs if one of $u,v$ is $0$, so we can assume that $u,v>0$.
Making the substitution $a=z$, $b=u$ and $c=v$ (for visual niceness), the previous inequality transforms into

$\frac{a+b+c}{2a+2b+c}+\frac{a+b}{2a+b}+\frac{a}{2a+b+c}\ge \frac{3}{2}$.

Now we subtract 1 from both sides in the following manner:

$\left(\frac{a+b+c}{2a+2b+c}-\frac{1}{2}\right)+\left(\frac{a+b}{2a+b}-\frac{1}{2}\right)\ge \left(\frac{1}{2}-\frac{a}{2a+b+c}\right)$,

which is equivalent to

$\frac{c}{2a+2b+c}+\frac{b}{2a+b}\ge \frac{b+c}{2a+b+c}$.

By Cauchy-Schwarz,

$\frac{c}{2a+2b+c}+\frac{b}{2a+b}=\frac{c^2}{c^2+2bc+2ca}+\frac{b^2}{b^2+2ab}$
$\ge \frac{(b+c)^2}{2ab+(b+c)^2+2ca}=\frac{b+c}{2a+b+c}$,

and so it suffices to prove that

$\frac{b+c}{2a+b+c}\ge \frac{b+c}{2a+b+c}$,

which is obvious. This completes the proof. $\blacksquare$

Arkan


abstract algebra
Functional Analysis
linear algebra
linear algebra unsolved

Let $T$ be a compact operator on a Hilbert space $H$, and assume $\lambda \ne 0$.

(a) Show that the range of $\lambda I-T$ defined by
$\{g\in H : g = (\lambda I - T)f,$ for some $f \in H\}$
is closed. [Hint: Suppose $g_j \to g$ where $g_j = (\lambda I - T)f_j$. Let $V_\lambda$ denote the eigenspace of $T$ corresponding to $\lambda$, that is, the kernel of $\lambda I - T$. Why can one assume that $f_j \in V_\lambda^\bis$? Under this assumption prove that $\{f_j\}$ is a bounded sequence.]

(b) Show by example that this may fail when $\lambda = 0$.

(c) Show that the range of $\lambda I - T$ is all of $H$ if and only if the null-space of $\overline{\lambda} I - T^*$ is trivial.

-----------------------------

We thought we proved (a), but we couldn't figure out how our proof depended on $\lambda \ne 0$, so we think it is wrong. We also couldn't think of an example for (b). I think we had (c) but you are free to post your own proof of it if you want. Make sure your proofs work for infinite-dimensional spaces.

(c)

In general, if $A$ is an operator, the orthogonal complement of the range $\mathcal R(A)$ of $A$ is precisely the null space of $T^*$: $A^*y=0\iff(x,A^*y)=0,\ \forall x\iff(Ax,y)=0,\ \forall x\iff y\in\mathcal R(A)^\perp$. In particular, the range of $A$ is dense iff the null space of $A^*$ is trivial. When $A=\lambda-T$ we know that the range is closed (from (a)), so it's dense iff it's the entire space $H$, and that's that.

(a)

$\frac 1\lambda T$ is compact as well, so it suffices to show that $1-T$ has closed range. Suppose $(1-T)x_n\to y\ (*)$. This still holds if we replace each $x_n$ by its component which is orthogonal to $\ker(1-T)$, so we may assume that all $x_n$ are orthogonal to $\ker(1-T)$. Now if $\|x_n\|\to\infty$, we divide $(*)$ by $\|x_n\|$ and find $(1-T)\frac{x_n}{\|x_n\|}\to 0$. $\left(\frac{Tx_n}{\|x_n\|}\right)_n$ has a convergent subsequence, so assume WLOG that it's convergent to some $x$, but then $\left(\frac{x_n}{\|x_n\|}\right)_n$ converges to the same $x$, which must satisfy $x-Tx=0$, i.e. $x\in\ker(1-T)$. Since all $x_n$ were assumed to be orthogonal to $\ker(1-T)$, we get $x=0\Rightarrow\frac{x_n}{\|x_n\|}\to 0$, absurd.

All of the above shows that $(\|x_n\|)_n$ cannot go to $\infty$. This means that it has a bounded subsequence, and, in turn, this means that $(Tx_n)_n$ has a convergent subsequence. But then the corresponding subsequence of $(x_n)_n$ converges as well, to some $x$, and we get $x-Tx=y$, exactly what we wanted.

(b)
Any compact operator with infinite-dimensional range will do (I won’t prove this). Just take your Hilbert space to be $\ell^2$, the space of complex square-summable sequences, and the operator $T$ to be the diagonal operator sending the element $e_n,\ n\ge 1$ of an orthogonal basis to $\frac 1n$. It shouldn’t be too hard to check that it works. In fact, its range contains $\sum_{i=1}^n\frac 1ie_i\ (*)$ for every $n\ge 1$, but it does not contain the sequence $\left(1,\frac 12,\frac 13,\ldots\right)$, which belongs to the closed linear space spanned by $(*)$.


[hide]If we have $ n$ bits, then anything we receive could have actually been one of $ {n \choose 2} \plus{} {n \choose 1} \plus{} {n \choose 0}$ codes. We could have received $ 2^n$ things. So we need $ \frac {2^n}{{n \choose 2} \plus{} {n \choose 1} \plus{} {n \choose 0}} > 16$, and the smallest $ n$ for which this is true is $ n \equal{} 10$.

After playing around a bit, I am fairly sure $ 10$ does not actually suffice. Here's a way for $ 11$.

Suppose our 4 digit code is $ abcd$. Construct the 11 digit code

$ a, b, c, d, \\
a \plus{} b, a \plus{} c, \\
a \plus{} b \plus{} c, a \plus{} b \plus{} d, a \plus{} c \plus{} d, b \plus{} c \plus{} d, \\
a \plus{} b \plus{} c \plus{} d$

The reason this works is that to get from one valid code to another, you must flip at least 5 of the bits.
- if you flip one of $ a,b,c,d$, you flip three triple sums and the final bit.
- if you flip two of $ a,b,c,d$, you flip at least one of $ a \plus{} b, a \plus{} c$, and two of the triple sums.
- if you flip three of $ a,b,c,d$, you flip a triple sum and the final bit.
- if you flip everything, you flip all of the triple sums.
So if you think of the codes as having a metric, distance being how many bits must be flipped, then all valid strings are at least distance 5 from each other. Thus drawing balls of radius 2 around each one will give non-intersecting balls, exactly what we wanted.[/hide]

I believe that most mobiles ect. only use error correction for up to one switch. However, when NASA sends space crafts up into space they use error correction which compensates for up to two switches (because of all the radiation etc.)
So you could probably find the answer to this question if you googled: 'NASA communication error correction' or something along those lines

A code used as an error-correcting code in practice will almost certainly not be the densest possible code: efficient encoding and decoding is worth a lot, too, for large data streams.

[quote="miyomiyo"]What is the fewest number of bits (1 or 0) necessary to completely determine a four-bit code, if up to two bits in any code may be switched?  For example, the code 11101001 may become 11001011 or 00101001, etc.[/quote]
I don't really understand the question; by four-bit code, do you mean the code consisting of only four numerals?


Euler
geometry
geometric transformation
reflection
geometry proposed

A very well known problem. Just use complex numbers with the center in O.

[quote="silouan"]$ABCD$ is a quadrilateral inscribed ina circle .Prove that the six lines, each passing through a midpoint of one the sides of $ABCD$ and perpendicular to the opposite side, are concurrent .
Here , the diagonals are considered to be opposite sides.[/quote]

A well known fatc, the concurrency point is called the anticwenter of $\square ABCD$. (Called the Euler point by Darij)

[color=darkblue]$ABCD$ is a quadrilateral inscribed in a circle $C(O,R)$. Denote the centroid $G$ of the given quadrilateral. Then the six lines, each passing through a midpoint of one the sides of $ABCD$ and perpendicular to the opposite side, are concurrently in the reflection of the point $O$ w.r.t. the point $G$ (here , the diagonals are considered to be opposite sides).[/color]


Can we find a closed form expression for (1+3^n)/2^a where a represents the biggest power of 2 that can be factorised from the numerator.
Thanks.

[quote="Pascual2005"]Iff $n$ is even $a$ is $1$, iff $n$ is odd, $a$ is $2$[/quote]

Thanks
I am not looking for values for a.But for a closed expression when we divide by 2^a.May be sure i dont explain the things very well because of my english.

Why do you need a closed expression and what is allowed¿ (Pascual2005 gave you all you need to get it finished)

yes, just a simple evaluation mod $8$! ;)


calculus
algebra
polynomial
calculus computations

the question is simple,,
find a line which is tangent to the curve $ y \equal{} x^4 \minus{} 3x^2 \plus{} 2x$ in exactly 2 points of the curve..

I've used $ f'(x_1) \equal{} f'(x_2)\equal{}\frac {f(x_2) \minus{} f(x_1)}{x_2 \minus{} x_1}$
but,,I still can't find the answer, I might make some mistake..

thanks for your help..

Your solution doesn't reveal much insight. How did you know to modify the expression in that way Kunny?

[quote="kunny"]$ x^4 \minus{} 3x^2 \minus{} \left(2x \minus{} \frac {9}{4}\right) \equal{} \left(x \plus{} \sqrt {\frac {3}{2}}\right)^2\left(x \minus{} \sqrt {\frac {3}{2}}\right)^2$.

The tangent line is $ y \equal{} 2x \minus{} \frac {9}{4}$.[/quote]

your idea is so brilliant,,
but how could you conclude that's the tangent line,, without doing any calculus method?
please give me some explanation,,thx..:)

[quote="gauss202"]How did you know to modify the expression in that way Kunny?[/quote]  I assume that what Kunny did is exactly what I did, stripped of all the helpful explanation  :wink:

mhar_teens -- a polynomial with a double root at $ x \equal{} r$ is tangent to the $ x$-axis at that point.  Adding a linear term doesn't change tangency properties.  Go read the post I linked for more elaboration.

[quote="JBL"][quote="gauss202"]How did you know to modify the expression in that way Kunny?[/quote]  I assume that what Kunny did is exactly what I did, stripped of all the helpful explanation  :wink: [/quote]

No! My solution is not same as yours, JBL. :wink:

Anyway Here is a generalization.

[quote="kunny"][quote="JBL"]After you've solved it following my hint, how to reverse-engineer it should be pretty clear.  The result is that a quartic $ f(x) = x^4 + ax^3 + bx^2 + cx + d$ has such a tangent line if and only if there exist different real numbers $ r, s$ such that $ r + s = - \frac {a}{2}$ and $ r^2 + 4 r s + s^2 = b$.  Solving these equations for $ r, s$, one can show that this is equivalent to $ 3a^2 - 8b > 0$.[/quote]

${ x^4 + ax^3 + bx^2 + cx + d - (\frac {a(4b - a^2)}{8}x + \frac {(4b - a^2)^2}{64}})$

$ = (x + \frac {a + \sqrt {3a^2 - 8b}}{4})^2(x + \frac {a - \sqrt {3a^2 - 8b}}{4})^2$  with $ b < \frac {3}{8}a^2$.

The tangent line is $ y = \frac {a(4b - a^2)}{8}x + \frac {(4b - a^2)^2}{64}$.[/quote]


Julie is preparing a speech for her class. Her speech must last
between one-half hour and three-quarters of an hour. The ideal rate
of speech is 150 words per minute. If Julie speaks at the ideal
rate, which of the following number of words would be appropriate
length for her speech?\\
\\
A. 2250 \\
B. 3000 \\
C. 4200 \\
D. 4350 \\
E. 5650 \\
\\
Enter the correct letter as your response.

30*150=4500.
We see the letter it should be is E.


Show that there is no positive integer n such that n + k 2   is a perfect square for at least n positive values of k.

[quote="tetrahedr0n"]Assume there is such a n, so we have n+k^2 = a^2, where a>k,n. Then n=(a-k)(a+k) for at least n different possible values of k. Then the max(k) is at least n, so we have that n>=(a-n)(a+n), which is false, since a>n.[/quote]

I'm sorry but i didn't understand anything at all. :blush: (for example, why a>n, why the max (k) is at least n, why then n>=(a-n)(a+n), and why is t5his false?)
Wouldn't you mind writing the solution again with as much detailed as possible, please.


inequalities
algebra
polynomial
AMC
USA(J)MO
USAMO
rearrangement inequality

a) Prove that if $x,y>0$ then
\[ \frac x{y^2} + \frac y{x^2} \geq \frac 1x + \frac 1y. \]

b) Prove that if $a,b,c$ are positive real numbers, then
\[ \frac {a+b}{c^2} + \frac {b+c}{a^2} + \frac {c+a}{b^2} \geq 2 \left( \frac 1a + \frac 1b + \frac 1c \right). \]

[hide]a)  This follows from the best underused inequality, the rearrangement inequality.  The sequences x,y and 1/y^2, 1/x^2 are similarly sorted, so x*1/y^2+y*1/x^2>=x*1/x^2+y*1/y^2.

b)  This is a direct result of a.  The left side is equal to a/c^2+c/a^2+a/b^2+b/a^2+c/b^2+b/c^2, so just apply the first inequality 3 times.[/hide]

[hide="Proof for part A."]The degree of the inequality is everywhere 0, so we can scale it so that $y=1$.  Thus, it is sufficient to show that, for all $x>0$, we have
\[x+\frac{1}{x^2}\geq\frac{1}{x}+1\]

Consider the polynomial
\[(x-1)(x-1)(x+1)\]
Clearly, it is positive for $x=0$, and can only become negative when its graph crosses the $x$-axis.  As $x$ ranges among the positive real numbers, it touches the $x$-axis at $x=1$, but the root is of multiplicity 2; it doesn't really cross the $x$-axis.  Thus, for all positive $x$, we can say
\[(x-1)(x-1)(x+1)\geq0\]
or
\[x^3-x^2-x+1\geq0\]
\[x^3+1\geq x^2+x\]
Since $x$ is positive, we can safely divide both sides by $x^2$ and get
\[x+\frac{1}{x^2}\geq 1+\frac{1}{x}\]
and we are done.[/hide]

I'll add another solution to a, just because most people don't know the [hide]rearrangement inequality[/hide]  Though you should.  It rocks.

[hide]$(x-y)^2\geq{0}$
$x^2-xy+y^2\geq{xy}$
$\frac{x^2-xy+y^2}{xy}\geq{1}$
$\frac{x^2-xy+y^2}{x^2y^2}\geq{\frac{1}{xy}}$
$\frac{(x+y)(x^2-xy+y^2)}{x^2y^2}=\frac{x^3+y^3}{x^2y^2}\geq{\frac{x+y}{xy}}$
$\frac{x}{y^2}+\frac{y}{x^2}\geq{\frac{1}{x}+\frac{1}{y}}$[/hide]

[hide="a."]

Start with the trivial inequality

$(x-y)^2 \ge 0$
$x^2-xy+y^2 \ge xy$
$(x+y)(x^2-xy+y^2) \ge xy(x+y)$
$x^3+y^3 \ge xy(x+y)$

Divide through by $x^2y^2$ to get

$\frac{x}{y^2}+\frac{y}{x^2} \ge \frac{1}{x}+\frac{1}{y}$ as desired.[/hide]

[quote="dts"][quote="Magnara"]I'll add another solution to a, just because most people don't know the [hide]rearrangement inequality[/hide]  Though you should.  It rocks....[/quote]

*sigh*

I knew it, but didn't think to use it...[/quote]

 :)

I've developed a habit of thinking of rearrangment every time I see an inequality.  It works a lot.

[hide="Proof of a."] We can multiply both sides of the inequality by $x^2y^2$ to clear the fractions since $x, y > 0$.
\[x^3 + y^3 \geq xy^2 + yx^2\]
\[x^3 - yx^2 - xy^2 + y^3 \geq 0\]
\[x^2 (x - y) - y^2 (x - y) \geq 0\]
\[(x^2 - y^2)(x - y) \geq 0\]
\[(x + y)(x - y)(x -y) \geq 0\]
\[(x + y)(x - y)^2 \geq 0\]
Now since $A^2 \geq 0$ for any number $A$ and $x + y > 0$ since each $x$ and $y > 0$, we conclude that
$\frac{x}{y^2} + \frac{y}{x^2}\geq  \frac{1}{x} + \frac{1}{y}$[/hide]

Part a was pretty easy, since both x and y are greater than 0 you can multiply and divide terms across the inequality without worrying about the inequality changing directions.

Question: What did you do here?

[quote="dts"]The degree of the inequality is everywhere 0, so we can scale it so that $y=1$.[/quote]

Well, the degree of the equation is really -1, but that's a different story.  There's this idea of an inequality being "homogenous," which means that the degree of each term is the same.  Therefore, if you replaced every x with kx and every y with ky, the inequality wouldn't change.  Therefore, you can scale the inequality, because all of the k's cancel off.  That is, if you wanted to prove it for the numbers 5, 10, you could just as well prove it for 1/2, 1, because it's only the proportions of the numbers that affect the inequality.

That was horribly explained on my part.  Maybe one of you guys could do a better job....

Maybe you can explain like this,
replace x by x/y and y by 1 and you can get the same inequality.
so wlog u may assume y=1. This is a concept of normalization.

For a) use AM-GM:
$\frac{x}{y^2}+\frac{y}{x^2}+\frac{y}{x^2}+\geq 3\frac{1}{x}$
And sum with cyclic to this one and you'll got a.

for b)
$\because \frac12(\frac1a+\frac1b)\geq \frac2{a+b}$
by Cauchy-inequality
$(\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2})(\frac1{a+b}+\frac1{b+c}+\frac1{c+a})\geq
(\frac1a+\frac1b+\frac1c)^2\geq \\
(\frac1a+\frac1b+\frac1c)(\frac2{a+b}+\frac2{b+c}+\frac2{c+a})\Longrightarrow \frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2}\geq 2(\frac1a+\frac1b+\frac1c)$

Second one can be solved by using Cauchy-Schwarz :)

[hide="a"]
$\displaystyle \frac{x^3+y^3}{x^2y^2} \ge \frac{x+y}{xy}$
$\displaystyle\frac{x^3+y^3}{xy} \ge x+y$
$\displaystyle x^3+y^3 \ge (x+y)(xy)$
$\displaystyle(x+y)^3 - 3xy(x+y) \ge (x+y)(xy)$
$\displaystyle(x+y)^3 - 4xy(x+y) \ge 0$
$\displaystyle(x+y)((x+y)^2 - 4xy)) \ge 0$
$\displaystyle(x+y)(x^2 + y^2 +2xy - 4xy) \ge 0$
$\displaystyle(x+y)(x-y)^2 \ge 0$

since x and y are positive, and since any square is positive,

$x+y \text{ is positive } and (x-y)^2 \text { is positive }$

and QED.
[/hide]

[quote="xxreddevilzxx"][quote="paladin8"]
Start with the trivial inequality
[/quote]

in case we are writing this proof for USAMO, when he mentioned the trivial inequality, do we put that under lemma? and when we use lemma, can we just state the theorem or do we have to prove that too?[/quote]

From what I recall from last year's discussion, you can state pretty much anything that has a name (or if you forget the name, just call it a well-known inequality) but you have to prove anything that doesn't in a lemma. So, with the trivial inequality, I'm pretty sure you can just state it.

[quote="xxreddevilzxxin case we are writing this proof for USAMO, when he mentioned the trivial inequality, do we put that under lemma?[/quote]

NO.  A lemma is something that [i]you[/i] prove, not something that you know to already be true.

[quote] and when we use lemma, can we just state the theorem or do we have to prove that too?[/quote]

If you write the USAMO, there is no reason whatsoever for you to use lemmas.  Since you don't really understand how they work (which is completely normal for anyone who hasn't read a lot of "real" mathematics books), trying to use them will only waste your time and make your solution more difficult to read.

When you are using a "named theorem" such as the Trivial Inequality, Cauchy-Schwarz, or anything else, you just have to state it and show how it applies to your problem.

[quote=sqing]Let $a,b,c$ are positive real numbers. Prove that
$$\frac {a+kb}{c^2} + \frac {b+kc}{a^2} + \frac {c+ka}{b^2} \geq (k+1) \left( \frac 1a + \frac 1b + \frac 1c \right) $$
Where $k\ge 0.$[/quote]
$$\sum_{\text{cyc}}\frac{a+kb}{c^2}=(k+1)\left(\sum_{\text{cyc}}\frac1a\right)+\sum_{\text{cyc}}\frac{(ka+b)(a-b)^2}{b^2c^2}$$
[url=https://artofproblemsolving.com/community/c6h3045392p27431243]here[/url]

a) we wanna prove that
\[ \frac x{y^2} + \frac y{x^2} \geq \frac 1x + \frac 1y. \]
$\iff$ $\frac{x^3+y^3}{x^2y^2} \geq \frac{x+y}{xy}$
$\iff$ $ x^3+y^3 \geq xy(x+y)$
$\iff$ $ (x+y)(x^2+y^2-xy) \geq xy(x+y)$
$ x+y≠0 $ so we should prove that $x^2+y^2-2xy\geq 0$ that is obviously true

b) \[ \frac {a+b}{c^2} + \frac {b+c}{a^2} + \frac {c+a}{b^2} \geq 2 \left( \frac 1a + \frac 1b + \frac 1c \right). \]
$\iff$ $\Sigma \frac{b+c}{a^2} -\frac2a \geq 0$
By cauchy-schowrz we have
$(\Sigma\frac{b+c-2a}{a^2})(\Sigma a^2(b+c-2a)) \geq 0$
So we should prove that $\Sigma a^2(b+c-2a) \geq 0 \iff \Sigma a^2b +\Sigma ab^2 \geq 2\Sigma a^3$


I am going to attempt to mod a game of techonpoly.
The rules are located on [url=http://kevan.org/games/technopoly.html]this[/url] website. Since I know these are a bit complicated, I'd advise you to read through [url=http://www.mafiascum.net/forum/viewtopic.php?t=4089]this[/url] Technopoly game for an idea of how to play, and to develop some strategy as well.

Similar to that game, [list][*]Robots have the attributes Elaborate Worthless Automatic Manipulator Attacker
[*]Players start with 3 Robots.
[*]If a player has no robots at the beginning of their turn, they'll be airdropped three robots.[/list]

Robots have the attributes Elaborate Worthless Automatic Manipulator Attacker- This means: It takes 3 robots to create another robot. Robots count as 0 points on your endgame score. Robots can 'activate themselves' once per turn. Robots have the ability to invent a new Object, create a known Object,  activate a Machine, or Attack another Object. And, they can attack other objects.

[b]Please read the rules and the game provided before signing up. If you don't understand something, feel free to ask, but PLEASE don't sign up if you aren't going to read the rules or the game. Also, please don't sign up if you can check this game approximately ONCE EVERY TWENTY-FOUR HOURS except on special ocurrences. I do not want you to sign up if you can only access this forum on weekdays, or only on weekends, or something like that.[/b]

I'm taking 4 to 6 signups. This game will last more than a month, but shouldn't last more than 2 to 3. If you do not feel like you can commit to that amount of time, don't sign up.

[quote="Ignite168"]what happened to friendorfoe[/quote]
abandoned, I guess.

bah, never mind, it would be in my best interest not to host this.

Sorry.


Here is a math problem from a MathCounts 1983-1984 written round.  The answer that was given to me seems to be incorrect, and I just wanted to check to make sure I am not making a mistake.

Question: If an 8-inch square cake serves four, how many 12-inch square cakes are needed to provide equivalent servings for eighteen people?

[hide="Click Here to View My Work"]
[img]http://img.photobucket.com/albums/v221/Farazy/83_84_CW28.gif[/img]
Thus the final answer I got is two.
[/hide]
[hide="Answer I was Given"]
9
[/hide]

Tell me what you got.

This is a somewhat tricky problem, where one can be too smart for their own good. But, I'm not sure if the solution given is correct either.

[hide]The problem states there are 8 inch and 12 inch square pieces. This does not mean the sides of the squares are 8 and 12 inches! They are referring to the area. Thus, we have a simple proportion.

$\frac{8}{4}=\frac{12x}{18}$

Simplification gets us $2=\frac{2x}{3}$

Cross multiply to get $6=2x$ and finally $x=3$ [/hide]

I am curious how they got 9, though I may have made a mistake.

But is says 8 inch square not 8 inches squared...

Any more thoughts?

here is what I did-
(I don't know how to do the click to reveal content so please excuse)

Alright, we know that and 8 inch square cake serves 4 people.  SO from this we can determine that each people will get 16 square inches of cake.  (the area of the eight inch cake is 64) 64/4=16
From this we now see that a 12 inch cake is 144 in sq and 144/16=9  Therefore we see that one cake serves 9 people and the answer should come out as 2.  I think that the answer given was wrong.
My answer was 2

[hide]I can see how they got 9.  (8/4)^2=(12x/18)^2.  After squaring and solving, x=9.  However, this problem needs rewording because this problem could interpreted many ways.  When I first looked at the problem, I thought the cakes were cubes because you can't really have a flat cake.  So I thought that they were telling us there were cube cakes with square faces.  That didn't work.  However, if I did this problem, I would the answer 2.  Every person in the group of four consumes 8*8/4=16 squared inches of cake.  Multiply that by 18=17^2-1=288 and divide that by 12^2=144 to get 2.[/hide]

[quote]If an 8-inch square cake serves four,
how many 12-inch square cakes are needed to provide equivalent servings for 18 people?[/quote][hide]
Algebra is not required ... just common sense.

$\text{An }8\times8\text{ cake serves 4 people.}$
$\text{Each person gets a }4\times4\text{ piece of cake.}$[code]
    [ ] [ ]
    [ ] [ ][/code]

$\text{A }12\times12\text{ cake provides 9 such servings.}$[code]
    [ ] [ ] [ ]
    [ ] [ ] [ ]
    [ ] [ ] [ ][/code]
$\text{Therefore, }two\text{ cakes are required for 18 servings.}$[/hide]

[quote="dengmi"]Soroban, I cannot see what you wrote.[/quote]

This:

[code][ ][/code]

Represents a 4*4 piece of cake.


geometry trigonometry algebra polynomial AMC AIME

Evaluate the product \[(\sqrt 5+\sqrt6+\sqrt7)(-\sqrt 5+\sqrt6+\sqrt7)(\sqrt 5-\sqrt6+\sqrt7)(\sqrt 5+\sqrt6-\sqrt7).\]

[hide]Notice that we can apply the difference of squares here.

$(\sqrt 5+\sqrt6+\sqrt7)(-\sqrt 5+\sqrt6+\sqrt7)(\sqrt 5-\sqrt6+\sqrt7)(\sqrt 5+\sqrt6-\sqrt7)$
$=((\sqrt 5+\sqrt6)^2-7)((\sqrt 5-\sqrt6)^2+7)$
$=(4+2\sqrt{30})(4-2\sqrt{30})=120 - 16=\boxed{104}$.
[/hide]

[hide="Another Solution"]By Heron's formula, we see that
\[\sqrt{\frac{ \text{Product}} {16}} = \text{Area}\]
\[\text{Product} = 16\text{Area}^2.\]
Using laws of cosines, we have
\[7= 5 + 6 - 2 \sqrt{30} \cos \theta\]
\[\cos \theta = \frac{2}{\sqrt{30}}\]
Using the pythagorean theorem, we have $\sin\theta = sqrt{\frac{26}{30}}$. Now we have the sine of the angle between sides of $\sqrt5$ and $\sqrt6$. So the area is
\[\frac{1}{2} \cdot \sqrt{30}{\sqrt{\frac{26}{30}}} = \frac{\sqrt{26}}{2}\]. So we have
\[\frac{26}{4} \cdot 16 = \boxed{104}\][/hide]

[hide="Yet another"]
Let $x = \sqrt{5} + \sqrt{6} + \sqrt{7}$. Then $x^2 - 18 = 2\left(\sqrt{30} + \sqrt{42} + \sqrt{35}\right)$. Then $(x^2 - 18)^2 = 428 + 4\sqrt{210}\left(\sqrt{5} + \sqrt{6} + \sqrt{7}\right)$. So $(x^2 - 18)^2 = 428 + 4x\sqrt{210}$, so $((x^2 - 18)^2 - 428)^2 = 3360x^2$. Then the polynomial is $x^4 + \ldots + 10816 = 0$. So the product of the solutions is $10816$. The answer is $\sqrt{10816} = \boxed{104}$ using the minimal polynomial properties of $\sqrt{5} + \sqrt{6} + \sqrt{7}$.
[/hide]

[hide=Solution]Rewrite as $((\sqrt6+\sqrt7)+\sqrt5)((\sqrt6+\sqrt7)-\sqrt5) \cdot (\sqrt5+(\sqrt7-\sqrt6))(\sqrt5-(\sqrt7-\sqrt6))$. Difference of Squares gives $(8+2\sqrt42) \cdot (-8+2\sqrt42) = \boxed{104}$ from Difference of Squares again.[/hide]

[hide=Solution]Expand the first 2 terms and the last 2 to get $(8+2\sqrt{42})(-8+2\sqrt{42})$, which simplifies to $168-64=\boxed{104}$ through difference of squares.[/hide]

[hide=Sol]Consider $(\sqrt x+\sqrt y+\sqrt z)(-\sqrt x+\sqrt y+\sqrt z)(\sqrt x-\sqrt y+\sqrt z)(\sqrt x+\sqrt y-\sqrt z)$ Using difference of squares on the first two and on the last two, we get $((\sqrt y+\sqrt z)^2-x)(x-(\sqrt y -\sqrt z)^2)\Rightarrow -((y+z-x)+2\sqrt{yz})((y+z-x)-2\sqrt{yz})\Rightarrow (y+z-x)^2-4yz$. Plugging in $x=5$, $y=6$, and $z=7$, we have $-\left((6+7-5)^2-4\cdot6\cdot7\right)=-(8^2-4\cdot 6\cdot 7)=168-64=\boxed{104}$[/hide]

If you expand the first $2$ terms, and the last $2$ terms, to get $(8+2\sqrt{42})(-8+2\sqrt{42})$, by difference of squares, this is just $168-64=\boxed{104}.$


geometry
combinatorics unsolved
combinatorics

Suppose I draw 20 circles in the plane, all passing through the origin, but no two tangent at the origin. Also, except for the origin, no three circles pass through a common point. How many regions are created in the plane?

Please help, I am totally lost.

Thanks

Try an inversion centered at the origin ;)

That was my first reaction too :D
But, you may construct the configuration by adding circle by circle. If you already have $ n$ circles, a new circle will intersect each of the $ n$ circles at two points and, except from the origin, these points will be pairwise distinct (since we do not allow that three circles do not have another common point). Thus, there are exactly $ n\plus{}1$ intersection points on the new circle. The arc between two consecutive points divides exactly one of the regions into two parts. Thus, the new circle create $ n\plus{}1$ new regions.
If $ a_n$ is the number of regions for $ n$ circles, we then have $ a_1 \equal{} 2$ and $ a_{n\plus{}1} \equal{} a_n \plus{} n\plus{}1$.
Thus, $ a_n \equal{} 1\plus{}\frac{n(n\plus{}1)}2$ and $ a_{20} \equal{} 211$.

Pierre.

thanks a lot, I understand it fairly well.. but one thing puzzles me.. If we had a line or curve, what is the relation between regions and intersecting points? I thought a moment earlier that each intersection would represent a split of region, but that isn't true.. Also, if you cross two lines, you'll make 4 regions with only 1 intersection point.. but if you drew two circles, you'll still have 4 regions but you have TWO intersection points.. makes me wonder. Could anybody shed some light on this/

THanks

[quote="JBL"]Each arc cuts a region into two pieces.  When you have one line (two regions) and you throw down another line, the new line comes as two "arcs" (in this case, rays) each of which divide an old region into two parts.  If you then add a third line intersecting both of the other two at distinct points, it has three "arcs" (two rays and one segment) and thus adds three new regions.  If you have a line in the plane (two regions) and you draw a circle that intersects it in two points, the circle gets split into two arcs and so adds two new regions.  Etc.[/quote]

 ah I see, that's interesting. And instead of extending all the way to split the plane (as a line does), the arc just forms a circle to create an enclosed area for the new area.

Thanks!!

[quote="pbornsztein"] Thus, there are exactly $ n \plus{} 1$ intersection points on the new circle. The arc between two consecutive points divides exactly one of the regions into two parts.

Pierre.[/quote]

 I['m not sure what you mean by that, but you can make 3 circles such that one arc cuts two regions in a circle :S


geometry
calculus
articles
set theory
group theory

yes, u dont learn much from high school math anyways, till calc

[quote="cascadee"]If you know a good 95% of the math in your grade right now, would you skip a grade and take a higher level course to learn the topics and take an enrichment supplement for the previous grade, or would you stay in your current grade and hope to learn the extra bit?

Right now I'm a freshman taking sophmore math, should I skip 11th grade math and take BC Calc or just continue? (I know most of 10th and I'm learning 11th grade math at the same pace as they are being taught in school.)

Thanks[/quote]

You should really make sure that you know all of 11th grade math before proceeding to skip it. If you take the final exam and are not able to do every problem on sight (or after less than a minute's thought), you should take the time to learn it yourself, even if you do get out of 11th grade math.

I myself am in an identical situation. I am taking Geometry this year (I'm happy that I didn't get out of it, because Geometry is actually semi-non-boring and the textbook is surprisingly good-it has [rigorous] sections on the nine-point circle, ptolemy's theorem, and the simson line, to name a few), and will probably skip Precalculus (or Advanced Mathematics as my school calls it), but that class is relatively sophisticated, and there's some material I'm not too familiar with so I'm taking the time this year (by time meaning like a few hours total over 8 months) to familiarize myself with it.

[quote="13375P34K43V312"](I'm happy that I didn't get out of it, because Geometry is actually semi-non-boring and the textbook is surprisingly good-it has [rigorous] sections on the nine-point circle, ptolemy's theorem, and the simson line, to name a few)[/quote]Out of curiosity, what is the name and author of the book?

Skipping grades doesn't necessarily count as added learning. You can skip to multivariable calculus and still not learn a thing. I believe the one way you can learn something to the fullest extent is to take an AP course (one course you have not taken yet, for example chemistry).

depends. I know stuff but suck
I skipped Alg 1 but and I knew the geometry and stuff
STILL FAILED WTF

Depends on yourself really. I have skipped a few grades and despite being able to hang around people who think alike and have similar abilities you also sacrifice things like the opportunities to participate in IMO, IChO, IPhO etc.

It also relies largely on how socially mature you are I had guess because when I first skipped grades, everyone was 2 or 3 years older than me and sometimes it is hard to relate to the way they think and experiences they have. You may be a genius but inability to relate may leave you feeling quite alone (of course you can always hang around people in your own grade).

It is not enough to know the material in the course but you have to 'know it back-to-front' to put it roughly. You will be dealing with people who are not necessarily better than yourself but who have a lot more experience and have seen more types of questions. After all, most high school courses have very similar questions that constantly occur over and over again.

My advice would be to go with the flow unless you really feel that your intellectual capacity is limited greatly and that you would be unable to learn under these conditions. So in the end, it is still your decision.

Thanks,

I've talked to the others who have skipped in older grades, and they have informed me that skipping twice doesn't really do much unless you already know the stuff "back-to-front". Having off campus courses in junior year can really take it out of you.

Always skip if you can, it looks good on your collage resumé. I skipped two grades and I'm in eight grade, it's best if you can take collage courses during your senior year, that way you don't take math your last year in collage.

My school's supposedly one of the best

in Senior year everything is AP courses even if you're a normal person, and 2 AP's in Junior year. So realistically I would have 3 in Junior year. I think I will try to skip a grade

Thanks guys =)

I think [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]this[/url] article might be a good one to read.

it's ok to skip grades when u r in 7th or 8th but i seriously feel then on what u read will increase ur knowledge and skill not understanding like most people can 'complete' calculus without actually knowing the 'definition' of limits or even read set theory or complex number s without knowig group theory or so .for an 8th grader it's ok but as u progress rigor is imp. particularly in physics where people tend to accept things most blindly .Reading too quick without understanding and i mean that understanding not only the concepts in the book but also the assumptions proofs ....So skipping is good initially but it's the depth of uinderstanding that stands the test of time


summer program
Mathcamp
AwesomeMath
geometry
similar triangles

Hi,

I've gone to a mathcamp thingy before, that was 4 weeks long, and my parents paid for it. However, my parents are a little iffy about awesomemath (although they'll probably say yes in the end => but then I don't feel as good about their decision). Do you guys have any ideas on trying to persuade them ... I think they already know it's filled with math...but I just need ideas...because I don't want to do like a 1 week camp, even though I don't really think there are any. And I really would like to go to AMSP since I know a bunch of you guys are going to be there and it just seems like a lot more 'fun' than the other camps.

Thanks a lot, to whoever helps out

It's cool that you are interested in AMSP. I'm not going to lie, since I am a counselor there, I kinda have an bias to convince you to go. But even considering that, the reason that I am coming back as a counselor there is because I had a great experience as a student last summer.

Okay so like I think that the strengths of the camp are that you have a good community of kids who are interested in immersing themselves in math. Furthermore, there are a lot of really smart mentors and assistants that you can learn from.

--

This is a bit of a subtle point but the idea of talking of mentors and such should be to learn a new way of thinking. That is, you don't just want to learn techniques when you are there. You want to talk to mentors and learn how they approach problems.

How do you learn? When you see a new concept, what do you do? How do you find new materials to learn from? What do you get out of a problem that you cannot solve even after trying real hard? What do you focus on? How can I solve a problem more efficiently? How could I have come up with this solution myself? What is the motivation for this solution? What is important to look at when you read a solution? How do I write rigorous solutions?

Not to be arrogant or anything, but I consider myself on of the stronger math students. One of the reasons that I excel is because I constantly think about how I can improve my methods or learning. Its not just what you learn (like say similar triangles) but how do you actually solve a problem where you don't know how to start. I also force myself constantly to read materials to learn new things.

With that being said, you can get a lot from the camp by talking to the mentors and seeing how they learn. How do they reconcile the questions that I posed above? How do they organize things? I think that is the difference between learning along and learning with a community. You can get more general organizational things that you would not get from just learning on your own. You do have to be active in seeking this type of advice. I remember that Zach Abel was very helpful in this respect. Other teachers did not actively express these ideas. You had to ask them. It is important to learn to ask questions and seek out opportunities.

--

So beyond all of that, it is a lot of fun to be with kids who like math. I kind of get negative feedback from friends at school about how much math I do. However, everybody there is very positive about loving math. I also feel that I came out of the camp being more mature in terms of my ability to study well.

If you or your parents have more questions, I suggest that you e-mail Titu Andreescu at titu@awesomemath.org  (or I suppose that I can answer questions :D ).

Hey, thanks a lot for your advice - I'm sure that in the end I'll probably end up going...unless this family conflict gets in the way (but I've already talked to Titu about that). Anways, I hope I do go and get to see all of you there.

Thanks! - and I won't be here to post anything for a while, because of break


AMC
USA(J)MO
USAMO
geometry
angle bisector

Um... When is it?
I never heard of it.
Can you give me some information?

It is for high schoolers.  If you have had Algebra II it might be appropriate, however many of the problems are workable with less.  See [url]http://www.math.okstate.edu/~hsc/[/url] for some old tests.

You could sign up as an individual if you want to go.

[quote="danb80"]It is for high schoolers.  If you have had Algebra II it might be appropriate, however many of the problems are workable with less.

You could sign up as an individual if you want to go.[/quote]

Oh.. Then never mind....

[quote="Chinaboy"]Amy and her new OSSM gang's probably going.[/quote]
Oh my gosh! That is SCARY!

lol yeah and amy probably won. i did quite horrendous.

But OSSM is pretty sucky this year...

Only Amy got a perfect on the OML. (Now if only I didn't screw up #5. 2222 is the new 666!)

oh haha i got owned on the last one in that silly oml thing...tried to use cauchy, am-gm, etc. on it to prove that the answer was none...stupid me. someone else randomly got a 6 from our school though so that's cool.

In a sense, the last problem got me too, because I tried for most of the time to prove that the 5 answers I had were the only 5. Didn't think of doing #5 rigourously...

this last oml was really kind of harsh.....i got pwned... :(

I got lucky. Intuition and calculator pulled me through.

haha guessing saved me from utter disaster...i somehow guessed #6 right...lol almost randomly. somehow didn't see that incircle.
#5 was harsh..couldn't find an example for the case=failed.

I angle chased for so long on #6... (I got so frustrated I even banged the table like 3 times) then I gave up and used Angle Bisector theorem and solved for theta using my calculator. But I thought it looked like 80 all along. It just LOOKED like it haha. A lot of people in Jenks guessed that and got it right.

Math Leagues are a lot harder this year than years past.

lol so are you gonna win martin? :)
so far i've missed 2 already...sad...

wait until usamo.....

lol it doesn't end until April. I just cross my fingers and hope that ATai misses one.

ha....i ought to work on usamo then...DANG i got owned my sophomore year... u?

was that my junior year?

Yep, I got owned too.


geometry
3D geometry
HCSSiM
Euler
graph theory

Using Euler's Formula, F+V-E=2, where F= number of faces, V=number of vertices, and E= number of edges. F=32.

For an interesting generalization of Euler's Formula, take a look at a cube with a square hole punched through the center of one of it's faces to the other side. Count the number of Vertices, Edges, and Faces of this solid and compare it to Euler's Formula.

Yes -- another question, which I dont know how to solve offhand, is to find the minimum number of faces when the solid does not have to be convex.  I think this problem delves pretty deep into topological graph theory (which I took a course in over the summer as a part of the HCSSiM program.)

Limiting the problem to three dimensions:
The polyhedra can be represented as a cellular imbedding on a topological surface.  We want to imbed it on a surface with the smallest Euler charactaristic to minimize the number of faces.  In this case it is clear that we are bounded by the number of vertices.  I dont have a proof, but by messing around with polyhedra I think that the following equation holds:
7/2*(2 - euler charactaristic) + 2 = required number of vertices for a cellullarly imbedded polyhedron.

If this is correct it is easy to solve for the euler charactaristic of the desired surface (-2)  This means that the minimum number of faces the concave polyhedron could have is 28.

Note that I assumed you meant polyhedron when you said solid.  If you did not mean polyhedron then the answer I get is 2 (and I think the answer is 2 no matter what you chose for V and E, which is what leads me to believe you meant polyhedron).

Could someone with a better understanding of topological graph theory verify my answer, or tell me if I have done anything incorrectly?

the edges of a certain solid is 50 and the number of vertices is 20. what is the maximum number of faces the solid can have???
[hide]
if it's a simple polyhedra then the number of faces F+20-2=50, F=32
but if it's not simple, I can't work out.
[/hide]

paganinio, stop bumping up dead threads.  I looked in your post history and you've been doing it consistently on threads that are very, very old.


advanced fields
advanced fields unsolved

there is an immersion of the punctured torus $ S^{1}xS^{1}\minus{}\{ponit\}$ in $ \mathbb{R}^{2}$  :blush: .

Punctured torus is diffeomorphic to torus with a disk cut out. Stretch the hole until the torus becomes the union of two cylindrical strips. This space can be mapped by immersion onto the union of two intersecting annuli in $ \mathbb R^2$.


Aratati ca :

$ \sum_{cyc}\frac{a}{a\plus{}2b\plus{}3c\plus{}4d}\ge \frac{2}{5} , (\forall)a,b,c,d\in (0;\infty)$

[quote="alex2029"]Aratati ca :

$ \sum_{cyc}\frac {a}{a \plus{} 2b \plus{} 3c \plus{} 4d}\ge \frac {2}{5} , (\forall)a,b,c,d\in (0;\infty)$[/quote]
I think it is a simply application of Adreescu inequality... :huh:

$ {\begin{array}{*{20}c}
   {a \plus{} 2b \plus{} 3c \plus{} 4d \equal{} x}  \\
   {b \plus{} 2c \plus{} 3d \plus{} 4a \equal{} y}  \\
   {c \plus{} 2d \plus{} 3a \plus{} 4b \equal{} z}  \\
   {d \plus{} 2a \plus{} 3b \plus{} 4c \equal{} t}  \\
\end{array}}\[ \left\{ {\begin{array}{*{20}c}
   {a \plus{} 2b \plus{} 3c \plus{} 4d \equal{} x}  \\
   {b \plus{} 2c \plus{} 3d \plus{} 4a \equal{} y}  \\
   {c \plus{} 2d \plus{} 3a \plus{} 4b \equal{} z}  \\
   {d \plus{} 2a \plus{} 3b \plus{} 4c \equal{} t}  \\
\end{array} \Rightarrow a \equal{} \frac{{ \minus{} 9x \plus{} 11y \plus{} z \plus{} t}}{{40}}} \right. \Leftrightarrow \sum {\frac{{ \minus{} 9x \plus{} 11y \plus{} z \plus{} t}}{{40}} \ge \frac{2}{5} \Leftrightarrow }\] \Leftrightarrow \sum {\frac{{11y \plus{} z \plus{} t}}{x}}  \ge 52 \Leftrightarrow a \equal{} \frac{{11y}}{x} \plus{} \frac{z}{x} \plus{} \frac{t}{x} \plus{} \frac{{11z}}{y} \plus{} \frac{t}{y} \plus{} \frac{x}{y} \plus{} \frac{{11t}}{z} \plus{} \frac{x}{z} \plus{} \frac{y}{z} \plus{} \frac{{11x}}{t} \plus{} \frac{y}{t} \plus{} \frac{z}{t}$,$ \ge 52$$ a \equal{} \underbrace {\frac{y}{x} \plus{} ... \plus{} \frac{y}{x}}_{11ori} \plus{} \frac{z}{x} \plus{} \frac{t}{x} \plus{} \underbrace {\frac{z}{y} \plus{} ... \plus{} \frac{z}{y}}_{11} \plus{} ... \ge 52\sqrt[{52}]{1}.$


vector
invariant
algebra
polynomial
linear algebra

In the following questions, $ T$ is an operator over vector space $ V$.
(1) $ V$ is a vector space over $ R$. Prove if $ T$ is invariant upon all subspaces, of dimension $ dim V \minus{} 1$, of $ V$, then $ T$ is a scalar multiple of the identity.
(2) Suppose $ V$ is a vector space over $ C$. $ P$ is any polynomial, and $ a$ is any element of $ C$. Prove if $ a$ is an eigenvalue of $ P(T)$, then $ a \equal{} P(b)$ for some eigenvalue $ b$ of $ T$.
(3) Suppose $ V$ is a complex vector space. Prove that $ T$ has an invariant subspace of dimension $ k$, for all $ k \equal{} {1,2, 3 ... dim V}$.

1) suppose that $ Ta$ is not a multiple of $ a$. Consider an $ (n\minus{}1)$-dimensional subspace that contains $ a$ but does not contain $ Ta$.


[hide]Subtract the two equations to get $y+2z=25$.  Multiply by 4 to get $4(y+2z)=25\cdot4\Rightarrow4y+8z=100$.
$x+4y+7z=225$
$4y+8z=100$
$x-z=125$
Then, add $y+2z=25$
$x-z+2z+y=x+y+z=125+25=\boxed{150}$[/hide]


Let $ a, b, c$ be the [b]real[/b] numbers satisfying $ a\plus{}b\plus{}c\equal{}3$. Prove that:
\[ \frac{|a\plus{}b|}{4a\plus{}b^2\plus{}c^2}\plus{}\frac{|b\plus{}c|}{4b\plus{}c^2\plus{}a^2}\plus{}\frac{|c\plus{}a|}{4c\plus{}a^2\plus{}b^2} \le 1\]
:)

It can be solved by only Am-Gm and Cauchy Schwarts  :)

I still waiting nice solutions for this problem :)

[quote="nguoivn"]Let $ a, b, c$ be the [b]real[/b] numbers satisfying $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {|a \plus{} b|}{4a \plus{} b^2 \plus{} c^2} \plus{} \frac {|b \plus{} c|}{4b \plus{} c^2 \plus{} a^2} \plus{} \frac {|c \plus{} a|}{4c \plus{} a^2 \plus{} b^2} \le 1\]
:)[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-111-t-q-anh.html

[quote="nguoivn"]Very nice proof, Can_hang  :lol:
And we also have a nice general problem:

Let $ a, b, c$ be the [b]real[/b] numbers satisfying $ a \plus{} b \plus{} c \equal{} 3$ and $ 1 \le k \le 2$. Prove that:
\[ \frac {|a \plus{} b|^k}{4a \plus{} b^2 \plus{} c^2} \plus{} \frac {|b \plus{} c|^k}{4b \plus{} c^2 \plus{} a^2} \plus{} \frac {|c \plus{} a|^k}{4c \plus{} a^2 \plus{} b^2} \le 2^{k \minus{} 1}\]
[/quote]
Actually, it suffices to prove this inequality for $ k\equal{}1$ and $ k\equal{}2$ (and in these cases we can proceed in the same way as my proof above). This is the reason:

The inequality is equivalent to
$ f(k)\equal{}\sum \frac{\left| \frac{a\plus{}b}{2}\right|^k}{4a\plus{}b^2\plus{}c^2} \le \frac{1}{2}.$
Clearly, $ f(k)$ is a convex function in term of $ k,$ and hence $ f(k) \le \max\{f(1),\; f(2)\}.$ Thus, it suffices to prove that $ f(1) \le \frac{1}{2}$ and $ f(2) \le \frac{1}{2}.$

[quote="manlio"]Sorry, my dear friend, but I cannot do download from

http://public.bay.livefilestore.com/y1pltVILVNIBuqCmclJm4XDsOWijcUm6y2Y15GVAbwqkQdv-PP01glM5ZiCPOrkIKbyHBYbl70tP5zVryOlYQJLkQ/Inequality%20111.rar?download :(

Can you please tell me if you can do download from it?

Thank you very much. :)[/quote]
This is not what I have linked at my page, my dear friend. In my page, I linked to download the solution at: http://cid-b2deb869e6bb1093.skydrive.live.com/self.aspx/Inequality%20Problems/Inequality%20111.rar

[quote="manlio"]Sorry,  :( but also at this address it is impossible for me to download the file

 http://cid-b2deb869e6bb1093.skydrive.live.com/self.aspx/Inequality%20Problems/Inequality%20111.rar[/quote]
This is the last method - uploading directly here.


A poll shows that $ 70\%$ of all voters approve of the mayor's work. On three separate occasions a pollster selects a voter at random. What is the probability that on exactly one of these three occasions the voter approves of the mayor's work?

$ \textbf{(A)}\ 0.063 \qquad
\textbf{(B)}\ 0.189 \qquad
\textbf{(C)}\ 0.233 \qquad
\textbf{(D)}\ 0.333 \qquad
\textbf{(E)}\ 0.441$

it is b. could you post amc 10b problems 8 and 20?

[hide]
There are $ 3$ ways to order the responses, so the probability is $ 3(0.7)(0.3)^2\equal{}0.189$.  Answer is $ B$.
[/hide]

This problemnis totally out of place...in my opinion...

i thought this was easier than number 7 on the test.

[quote="E^(pi*i)=-1"][hide]
There are $ 3$ ways to order the responses, so the probability is $ 3(0.7)(0.3)^2 \equal{} 0.189$.  Answer is $ B$.
[/hide][/quote]

I forgot that order didn't matter...

Didn't you multiply by 3 because order DID matter though?

So since order doesn't matter you multiply by 3?

Order does matter.  You multiply by $ 3$ because there are $ 3$ ways to choose the person who responds favorably.

Is e^pi*i+1=0 here?
I have some interest about wootlings

PM me if you want to ask me something, so we don't clog up the forum.

I get [hide] B

there are three possibilities for getting exactly one voter approving

the first voter approves, the next two don't
The second voter approves, the first and last don't
the third one approves, the first one doesn't

the chances of each of the above possibilities happening is .3*.3*.7=.063.

multiply by three because of 3 possiblities and u get .189
[/hide]


Hi guys, I'm having trouble with this proof. Hopefully you guys can help me. Here's the problem:

Given trapezoid ROSE with diagonols RS and EO intersecting ar point I, prove that the diagnols of the trapezoid do no bisect each other.
O   __________S
    l                    l
   l           I          l
  l____________l
R                          E

Excuse the poor quality image but I just wanted to show what it looked like.
RS and OE intersect at I.

Hope that helps
Thanks

If you are doing a proof by contradiction, then you would assume the diagonals do bisect each other. With that piece of information, you'd go and prove facts one of which, should contradict your givens or information derived from them.

That's right... if the diagonals do bisect each other, try proving that in that case all 4 resulting triangles would have to be congruent... and that would mean there would be no trapezoid, there would be a square instead.


conics ellipse geometry perimeter invariant parameterization symmetry

[quote="zhaobin"]the triangel $ABC$ is inside connect with the ellipse $ \frac {x^2}{a^2}+\frac {y^2}{b^2}=1$
find out the maximum of perimeter of triangel $ABC$.
I think it is really hard[/quote]

Consider variation of $C$ with fixed $A, B$.   For $C$ to be optimal, the tangent line at $C$ must be parallel to $AB$.  This is an affine-invariant condition, and if one transforms the ellipse to a circle, it means that $C'$ is the midpoint of arc $A'B'$, and the same for the other vertices; that is, $A'B'C'$ is equilateral.   So we have a one-parameter family of affine images of equilateral triangles to test for maximality (I don't think they all have the same perimeter, it would be very surprising!).  Maybe the extrema occur when an axis of symmetry of the triangle is parallel to the eigenlines of the affine transformation, in this case the $x$ and $y$ axes, but I did not try to calculate what happens.

thanks fleeting_guest.
but does anyone do it in detial?
I have the answer.and the answer is $\frac{2 \sqrt{3}(a^2+b^2+D)}{\sqrt{a^2+b^2+2D}}$
where $D=\sqrt{a^4+b^2-a^2b^2}$
my friend told me it.and he said he get it in a book.but the proof is toooooo long.
have anyone want to give a short proof.?


If we send in our solutions before the deadline, will the green checkmark appear earlier, or are they put up in batches?

my dad just sent my handwritten solutions today, and i live in the NY.  Do you think it will reach CA before January 8th?

I think so. You have almost a week.

And even if it didn't, you can't really do much about it.  :|

I thought it only had to be postmarked by the due date.


geometry
combinatorics proposed
combinatorics

Please, give a solution to problem 5 of China 2002.