geometry
geometric transformation
complex numbers
geometry proposed

Let quadrilateral $ ABCD$ with $ AD\equal{}BC$. Draw two equal triangles $ AED$ and $ BCF$ outside it. Prove the midpoints of $ AB, CD , EF$ are collinear :)

Uses lemes:
[b]1). [/b]$ \mbox{ If in quadrilateral ABCD we: } \angle{BAD} \equal{} \angle{BCD}\mbox{, then the inbisectors of } \angle{ABC} \mbox{ and }\angle{ADB}\mbox{ is parallels.}$
[b]2).[/b]  $ \mbox{ If in quadrilateral } ABCD \mbox{ we: }|AD| \equal{} |BC|; M, N \mbox{ is midpoints of sides }[AB], [CD]\mbox{ and }$
$ \{E\} \equal{} AD\cap BD\mbox{, then line }MN \mbox{ is parallel with inbisector of }\angle{AEB}.$

[quote="DangChienbn"]Let quadrilateral $ ABCD$ with $ AD \equal{} BC$. Draw two equal triangles $ AED$ and $ BCF$ outside it. Prove the midpoints of $ AB, CD , EF$ are collinear :)[/quote]
Use complex numbers, and let $ a \equal{} 0, b \equal{} y, d \equal{} r, c \equal{} y \plus{} s$, where $ |r| \equal{} |s|$. Let $ \omega \equal{} e^{i\frac {\pi}{3}}$. Then $ e \equal{} r\omega, f \equal{} y \plus{} s\overline{\omega}$.
Then $ n \equal{} \frac {a \plus{} b}{2} \equal{} \frac {y}{2}, m \equal{} \frac {c \plus{} d}{2} \equal{} \frac {y}{2} \plus{} \frac {r \plus{} s}{2}, p \equal{} \frac {e \plus{} f}{2} \equal{} \frac {y}{2} \plus{} \frac {r\omega \plus{} s\overline{\omega}}{2}$.
To prove that $ N,M,P$ are colinear we only have to prove that $ \frac {m \minus{} n}{p \minus{} n} \equal{} \frac {r \plus{} s}{\omega r \plus{} \overline{\omega}s} \in \mathbb{R}$.
But $ \overline{\frac {r \plus{} s}{\omega r \plus{} \overline{\omega}s}} \equal{} \frac {\frac {|r|^2}{r} \plus{} \frac {|s|^2}{s}}{\overline{\omega}\frac {|r|^2}{r} \plus{} \omega \frac {|s|^2}{s}} \equal{} \frac {r \plus{} s}{\omega r \plus{} \overline{\omega}s}$, so it is true. QED

Let M,N,P be the midpoints of $ AB, CD, EF$.
$ \vec{MP}\equal{}\frac{1}{2}(\vec{AE}\plus{}\vec{BF}), \vec{MN}\equal{}\frac{1}{2}(\vec{AD}\plus{}\vec{BC})$
It's easy to see that $ \vec{MN}\equal{}k\vec{MP}$ from two congruent triangles.

[quote="livetolove212"]Let M,N,P be the midpoints of $ AB, CD, EF$.
$ \vec{MP} \equal{} \frac {1}{2}(\vec{AE} \plus{} \vec{BF}), \vec{MN} \equal{} \frac {1}{2}(\vec{AD} \plus{} \vec{BC})$
It's easy to see that $ \vec{MN} \equal{} k\vec{MP}$ from two congruent triangles.[/quote]
Oh, I have a solution without Vecto :) Nice ! :)

[quote="livetolove212"]Let M,N,P be the midpoints of $ AB, CD, EF$.
$ \vec{MP} \equal{} \frac {1}{2}(\vec{AE} \plus{} \vec{BF}), \vec{MN} \equal{} \frac {1}{2}(\vec{AD} \plus{} \vec{BC})$
It's easy to see that $ \vec{MN} \equal{} k\vec{MP}$ from two congruent triangles.[/quote]
Can you please fill in the details?

A translation $ T_{\vec{BA}}: B\rightarrow A, C\rightarrow C', F\rightarrow F'$. We have $ AC'\equal{}AD, AE\equal{}AF', \angle C'AF\equal{}\angle DAE$ ...


function
advanced fields
advanced fields unsolved

If we define the continuous function $ \varphi : D \rightarrow \mathbb{R}^n$ s.t

$ D$ is connected set $ \Rightarrow \varphi{(D)}$ is also connected.

Is it necessary that the set $ T \equal{} \{ (x,\varphi{(x)}) : \; x\in D \}$ is connected in $ \mathbb{R}^2$

waiting your replay.

[quote="ZetaX"]What is $ D$ in the first line¿ And $ T$ isn't a subset of $ \mathbb R^2$ in general.[/quote]

D is any connected set ,this is what I read it from question

do you think there is something wrong in the question ?

Why should $ T$ even be a subset of $ \mathbb R^2$..
It's a subset of $ D \times \mathbb R^n$. As long as $ D$ isn't subset of a real space $ \mathbb R^m$ and $ m\plus{}n \leq 2$, this cannot happen.

[quote="ZetaX"]Why should $ T$ even be a subset of $ \mathbb R^2$..
It's a subset of $ D \times \mathbb R^n$. As long as $ D$ isn't subset of a real space $ \mathbb R^m$ and $ m \plus{} n \leq 2$, this cannot happen.[/quote]

yes ZetaX, that's warping me so much and I thought that but I didn't pretty sure about it.

So , if we change the question to be $ T$ subsets of $ \mathbb{R}^{m\plus{}n}$ where $ D$ subsets of $ \mathbb{R}^{m}$.

Would be true?

All you have to show is that the map $ d \mapsto (d, \varphi(d))$ is continuous (this holds for general spaces).


Prove or disprove:  Any possible real set solutions to the cubic equation

$x^3 - ax^2 + bx - a = 0$

Must be of the form $\tan \alpha, \tan \beta, \tan \gamma$ where $\alpha + \beta + \gamma = 180^{\circ}$.

Edit:  Follow-up question:  Prove or disprove that every possible real set of solutions corresponds bijectively to unique values of $\alpha, \beta, \gamma$ (not taking order into account).

The answer is partly true, since $\alpha+\beta+\gamma=n\pi\Leftrightarrow\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$ (there was a topic with the proof of that, I can't find it right now). Hence the sum of the angles must be $n\pi$.


calculus
integration
trigonometry
calculus computations

Calculate \[ \int_0^{2\pi } {x\sin 2xdx} .
\]
 :lol:


Am I wrong or it is quite trivial?
What is grade 7 in terms of age?

Pierre.

Nice problem -  Note that $2^2+3^2 = 13$ .

(Edited later:  a small typo )

[quote="Valentin Vornicu"]Let $a,b$ be two integers. Prove that

a) $13 \mid 2a+3b$ if and only if $13 \mid 2b-3a$;
b) If $13 \mid a^2+b^2$ then $13 \mid (2a+3b)(2b+3a)$.

[i]Mircea Fianu[/i][/quote]
Although this problem succumbs to much simpler argument, I will take the liberty of revealing its obvious connection with primes and sums of two squares (which, in turn, are connected to Fermat's Two Squares Theorem), through an important lemma. Call it an overkill, whatever... :P

[i]Solution.[/i] We start with a lemma, whose proof contains the essence of the connection mentioned earlier.

[color=#0000FF][b]Lemma.[/b] If a number that can be written as a sum of two squares is divisible by a prime that can also be so written, then the quotient is a sum of two squares.[/color]

[b]Proof.[/b] Suppose $a^2+b^2$ is divisible by $p^2+q^2$, and this latter number is a prime. Then \[p^2+q^2\mid (pb-aq)(pb+aq)=p^2b^2-a^2q^2=p^2(a^2+b^2)-a^2(p^2+q^2).\] As a result, since $p^2+q^2$ is prime, it must divide one of the factors, i.e. $p^2+q^2\mid pb-aq$ or $p^2+q^2\mid pb+aq$. Wlog, let $p^2+q^2\mid pb-aq$. Then by Diophantus' Identity, we have that \[(p^2+q^2)(a^2+b^2)=(pa+qb)^2+(pb-aq)^2\] and since $p^2+q^2\mid pb-aq$, it follows $p^2+q^2\mid pa+qb$. Thus, dividing through by $(p^2+q^2)^2$ yields \[\frac{a^2+b^2}{p^2+q^2}=\left(\frac{pa+qb}{p^2+q^2}\right)^2+\left(\frac{pb-aq}{p^2+q^2}\right)^2.\] Similarly for the case $p^2+q^2\mid pb+aq$, by instead using Diophantus' Identity in the form $(q^2+p^2)(a^2+b^2)=(aq+pb)^2+(ap-bq)^2$. The result follows. $\Box$

In fact, the proof of the lemma has already dispatched part (a) of the problem, so let us focus on the second part.

(b) Now, since $13=2^2+3^2\in \mathbb{P}$, and it divides $a^2+b^2$, then by the lemma, if $13\mid 2b-3a$ we have \[\frac{a^2+b^2}{2^2+3^2}=\left(\frac{2a+3b}{13}\right)^2+\left(\frac{2b-3a}{13}\right)^2\] and if $13\mid 2b+3a$ we have \[\frac{a^2+b^2}{2^2+3^2}=\left(\frac{2b+3a}{13}\right)^2+\left(\frac{2a-3b}{13}\right)^2.\] In the first case, $13\mid 2b-3a\implies 13\mid 2a+3b$, since both squares on $RHS$ are in $\mathbb{N}^{\star}$, and in the second case, we have by definition that $13\mid 2b+3a$.

Ultimately, it follows $13$ divides $(2a+3b)(2b+3a)$ since it has to divide one of the factors, and the proof is complete. $\blacksquare$

Arkan

[hide="a)"]\begin{align*}13 \mid 2a + 3b \Leftrightarrow 2a \equiv -3b \pmod{13} \Leftrightarrow 3a \equiv \frac{-9b}{2} \equiv \frac{4b}{2} \equiv 2b \pmod{13} \Leftrightarrow 13 \mid 2b - 3a \end{align*}[/hide]
[hide="b)"]\[(2a+3b)(2b+3a) = 6(a^2 + b^2) + 13ab \equiv 6 \cdot 0 + 0 \cdot ab \equiv 0 \pmod{13}\] Hence \[13 \mid (2a + 3b)(2b + 3a)\][/hide]

[quote="arkanm"]
Although this problem succumbs to much simpler argument, I will take the liberty of revealing its obvious connection with primes and sums of two squares (which, in turn, are connected to Fermat's Two Squares Theorem), through an important lemma. Call it an overkill, whatever... :P

[[/quote]
Thanks, That's precisely what I meant.. when I pointed that above
(Editing the typo now in the original...)
[quote="Gyan"]Nice problem -  Note that $2^2+3^2 = 13$ .[/quote]

\begin{align}
  & 2a+3b\overset{13}{\mathop{\equiv }}\,0\Leftrightarrow -8(2a+3b)\overset{13}{\mathop{\equiv }}\,0\Leftrightarrow -16a-24b\overset{13}{\mathop{\equiv }}\,0 \\
 & \Leftrightarrow -13a-3a-26b+2b\overset{13}{\mathop{\equiv }}\,0\Leftrightarrow 2b-3a\overset{13}{\mathop{\equiv }}\,0
\end{align}


function
algebra
domain
complex analysis

identity theorem states if two functions f and g are holomophic and f=g in the neibourhood of z then they are equal in whole domain . how can we apply in this theorem. we have only one function f?

There is a more suitable version of the identity theorem that states for holomorphic functions $ f$ and $ g$ that if the set $ S: \equal{} \{z\in\mathbb C|f(z) \equal{} g(z)\}$ has a limit point, then $ f \equal{} g$. The Wikipedia article on the identity theorem is incomplete!

how can we apply this theorem to evaluate $ f(\sqrt{2})$?

Probably, you want to find a $ g$ to compare to $ f$.

By the way, you appear to have improved your basic quality of posts, which is nice to see.  One next step you could take is better title-naming: you're posting in a complex analysis forum, so titling all your posts "mathematics" and "complex analysis" is not very helpful.

[quote="JAS"]Let  S be the disk │z│<3 in the complex plane and let f:S→C be an analytic function such that f((1+√2i/n)= -2/n^2 for each natural number n then what will be the value of f(√2)[/quote]

Just try some things alright?  How about $ f(z) \equal{} z$?  Then $ f(1 \plus{} i\sqrt {2}/2) \equal{} 1 \plus{} i\sqrt {2}/2$.  Won't do it.  How about $ f(z) \equal{} z^2$?  In that case:

$ f(1 \plus{} i\sqrt {2}/n) \equal{} 1 \plus{} 2i\sqrt {2}/n \minus{} \frac {2}{n^2}$.  Hey, that's close.  How about $ f(z) \equal{} z^2 \minus{} 1$?  That would give me $ 2i\sqrt {2}/n \minus{} 2/n^2$.  Closer.  Now, can you think of a way to change $ f(z) \equal{} 1 \minus{} z^2$ so that $ f(1 \plus{} i\sqrt {2}/n) \equal{} \minus{} 2/n^2$?

yes if we define $ f(z) \equal{} z^2 \plus{} 1 \minus{} 2z$ then $ f(\sqrt2) \equal{} 3 \minus{} 2\sqrt2$

thankyou

That is good hints but anyone could tell me how to measure similarly of two signal by Complex Analysis , this subject is low level for perception of human .

Thank for finding f(z) ,its hard for me for try all from one to one function to get solution in all such problems, Do you have another good method?

System of treatment is required    !!!!!!!!!!!!!!!!!!

kongp


integration
quadratics
Functional Analysis
probability and stats

came across a question telling me to prove for square integrable ito processes,

$ <aX\plus{}bY,Z>_t\equal{}a<X,Z>_t\plus{}b<Y,Z>_t$
$ <X,Y>_t\equal{}<Y,X>_t$

wondering what this inner product is meant to symbolize..
if its just $ <X,Y>_t\equal{}\int_0^t X(s)Y(s)ds$ it seems a bit too simple of a question.. does this notation usually mean something else?

Your definition of quadratic variation is wrong  here is what you should write in my opinion

if you have for example
$ dX_t\equal{}\mu_X(t)dt\plus{}\sigma_X(t)dW_t$
$ dY_t\equal{}\mu_Y(t)dt\plus{}\sigma_Y(t)dW_t$
$ dZ_t\equal{}\mu_Z(t)dt\plus{}\sigma_Z(t)dW_t$

then your covariation is equal to
$ <\alpha.X\plus{}\beta.Y,Z>_t\equal{}\int_0^t (\alpha.\sigma_X(s)\plus{}\beta.\sigma_Y(s)).\sigma_Z(s)ds$
(which is not equal to :
$ \int_0^t (\alpha.X(s)\plus{}\beta.Y(s)).Z(s)ds$)

From there it is easy to show what is asked ( the reasoning is the same that with the wrong definition of quadratic diffusion)

Is that what you wanted ?

I think that your are trying to show that Itô diffusion (are at least a subset of them) has an Hilbert Space structure


Determine a positive constant c such that

xy2 - y2 - x + y = c

has precisely three solutions (x, y) in positive integers.

What interests me is your strategy to find such a constant :)
I found one that works, but it was rather trial and error than strategy.

We have
xy 2  - y 2  - x + y = (y-1)[1+(y+1)(x-1)].

Note that, for each positive integer c such that c = 1 mod[3], we always have the solutions (x,y) = (1,c+1) and ( 1+(c - 1)/3 , 2).

Moreover, if x  \geq 2, then y-1 < 1 + (y+1)(x-1).
Thus, it suffices to choose (if possible) c such that c = 1 mod[3] and c = p*q   for some primes p,q, with p < q.
In that case, a solution of the equation which is not one of the two already founded is y-1 = p and 1+(y+1)(x-1) = q , which leads to y = p + 1 and x = 1 + (q - 1)/(p+2).

Then it suffices to choose primes p and q such that p*q= 1 mod[3] and p+2 divides q - 1.
Thus, it suffices to choose p,q = 1 mod[3] and p+2 divides q-1.
Let's choose p = 7. Then we want a prime q > 7 such that 9 divides q-1, say q = 19.
Then c = 7*19.

Pierre.


Hmm, just wondering, why is there are usergroup titled "VIP" with only Valentin in it?

Actually it's Valentin's Important Persons :D


Problem 4
Let $ ABCD$ be a trapezoid with $ AD\equal{}a$ ,$ AB\equal{}2a$ , $ BC\equal{}3a$ and $ \angle A\equal{}\angle B\equal{}90^{o}$. Let $ E$ and $ Z$ be the midpoints of the segments $ AB$ and $ CD$ respectively. Let $ ZI$ be a segment perpendicular to the base of the trapezoid(the point $ I$ is on the segment $ BC$ ). Prove that
$ a$) $ BDZ$ is an isosceles triangle,
$ b$) The midpoint $ O$ of the median $ EZ$ is the barycenter of the triangle $ BDZ$ ,
$ c$) The lines $ AZ$ and $ DI$ intersect on the line $ BO$.

[quote="Γιώργος"]Problem 4
Let $ ABCD$ be a trapezoid with $ AD \equal{} a$ ,$ AB \equal{} 2a$ , $ BC \equal{} 3a$ and $ \angle A \equal{} \angle B \equal{} 90^{o}$. Let $ E$ and $ Z$ be the midpoints of the segments $ AB$ and $ CD$ respectively. Let $ ZI$ be a segment perpendicular to the base of the trapezoid(the point $ I$ is on the segment $ BC$ ). Prove that
$ a$) $ BDZ$ is an isosceles triangle,
$ b$) The midpoint $ O$ of the median $ EZ$ is the barycenter of the triangle $ BDZ$ ,
$ c$) The lines $ AZ$ and $ DI$ intersect on the line $ BO$.[/quote]

[b]proof[/b]
 [hide="point a)"]Let $ D'$ be the perpendicular from $ D$ to $ BC$.
$ BD'\equal{}D'I\equal{}IC\equal{}a$
$ ZI\equal{}a\equal{}IC$
$ BD\equal{}BZ\equal{}a\sqrt{3}$, as diagonals in the equal rectangles $ ADD'B$,$ EZIB$.[/hide]

I don't know what is barycenter,but I supose that is smth with center of this triangle.
[hide="b)"]Let $ X\equal{}DD' \cap EZ$,$ O \in (DD')$
In the $ \Delta ABC$,$ EX\equal{}\frac{a}{2}\equal{}XO\equal{}\frac{OZ}{2}$
The same for $ DO\equal{}2OY$, where $ Y\equal{}BZ \cap DD'$
Results that $ O$ is the centroid for the triangle $ BZD$.[/hide]

[hide="c)"]Let $ M\equal{}DI\cap OZ$
$ N\equal{}AZ\cap OD$
$ DM$ and $ ZN$ are medians in the triangle $ DOZ$
$ BP$ is the median in the triangle $ BDZ$ $ \implies$ $ OP$ median in the triangle $ DOZ$.Therefore $ DM,ZN.OB$ are meeting in the same point[/hide]


inequalities
search
inequalities unsolved

Let $ x_1, x_2,\ldots ,x_n$ be non-negative real numbers such that
$ \sum_{i\equal{}1}^n\frac{1}{1\plus{}x_i}\equal{}1$.

Show that  $ \sum_{i\equal{}1}^n \sqrt{x_i}\geq (n\minus{}1)\sum_{i\equal{}1}^n\frac{1}{\sqrt{x_i}}$.

[quote="mkrnic"]Any solution  :maybe:[/quote]

Search before posting! This is Vojtech Jarnik 2002.

In "Problems from the Book" Titu Andreescu claims there exists a solution using Cauchy-Schwartz but doesnt show it. I failed to find such solution myself and also havent found it here (but I am still not very good at searching  :) ).

So any ideas?

So far I only saw solutions using Chebyshev, convexity and trigonometry.

See the third post on the link given by kuing

I know, perhaps this is what Titu Andreescu had in mind, but my opinion is that in this case the use of Cauchy-Schwarz is trivial and the main idea lies somewhere else. I was hoping to see some ingenious use of CS :). Thanks anyway.


$a,b,c \geq 0 and a^{2}+b^{2}+c^{2}=1 prove that: 1\leq \sum\frac{a}{1+bc}\leq \sqrt{2}$
the former one is easy ,and the latter one may be difficult

Prove $1\leq \sum{\frac{a}{1+bc}}$
if $abc=0$ then it is obvius true.
if $abc>0$, $\frac{a}{1+bc}=\frac{2a}{2+2bc}\geq\frac{2a}{3-a^{2}}\geq a^{2}$, the last step works with $(a-1)^{2}(a+2)\geq 0$.


5 red balls, 5 white balls, 5 blue balls, need to be placed into 3 indistinguishable boxes.
No box has more than 8 balls and No box has more than three balls of any one color.
How many ways can this be done?


Two [u]positive integers[/u] (call them M and N) differ by 90, and their arithmetic mean exceeds twice their positive geometric mean by 3.  Find the smaller of the two numbers.

[hide="hint..."]Let the numbers be x and x+90, and solve:
x+45=3+\sqrt{x(x+90)}[/hide]

[hide] plug in N+90 for M.  work your way down until you square both sides.  the quadratic equation factors nicely, and you are left with only one positive integer solution for the smaller term.  The answer i get is[/hide]  [hide]6[/hide]

Wlog $N < M$
$M - N = 90$
$\frac{M+N}{2} = 2\sqrt{M N} + 3$

Compute N.

This combines into:
$2N+90 = 4\sqrt{M N} +6$
Error here $\rightarrow 16 M N = (2N + 84)^2$
$16N^2 + 1440N = 4N^2 + 336N + 84^2$
$12N^2+1104N-84^2 = 0$
$N^2 + 92N - 588 = 0$
$(N-6)(N+98) = 0$

Thus, $N = 6$.

The resulting N [b]are no longer[/b] imaginary.  I know I made a mistake somewhere, please fix.

Edit: Thanks jhcreinardt

i dont know about all of ur steps, but the quadratic i got was x^2+92x-588=0.  this gives a positive integer solution.

oh i see it....in between step one and two you took square root of 4 when you should have squared it.


I tried to type the phrase:

Since 35 students can participate for 85 dollars (Mathcounts is 20 dollars per person, and a school is limited to 8 participants) ...

with dollar-signs, but I couldn't come up with any way of escaping the dollar signs to avoid the intervening part of my sentence being turned into a LaTeX math expression.  I tried putting backslashes before each of them, even tried making "85 dollars" into a math expression by typing <dollar> <backslash><dollar>85 <dollar> but that didn't work either.

I'm stumped!

This is an issue not with LaTeX, but with the way our forum works.

In regular LaTeX you simply put a backslash before the dollar sign, like so: \$

This doesn't work on our forum.

The easiest way to get dollar signs on our forum is to check the "Disable BBCode in this post" box when you compose your message.

[quote="Kent Merryfield"]It will cost $\$85.
dollar-slash-dollar
Weird, but it works.
How about $\$$85$? That took four dollar signs.[/quote]

I'm impressed ;-)

Can you explain why that works?  I tried dollar-slash-dollar-dollar which didn't work but makes more sense, IMO.

You can also click 'disable BBCode in this post', but then you won't be able to use any LaTeX or Bold, or whatever in your post.

[quote]Can you explain why that works? I tried dollar-slash-dollar-dollar which didn't work but makes more sense, IMO.[/quote]
I didn't say it makes sense. There is no "why" - just that trial-and-error found this. (The trial-and-error was actually jmerry's, not mine.)

[quote="henryyangrui"]wow, whats wrong? i typed "\[85\]"[/quote]Don't do that. The easiest and [b]SAFEST[/b] way to escape a dollar, besides disabling BBCode, is $\$ which is generated by the code: $\$\$\$

I've been looking for a way to type a dollar sign $ without disabling BBCode.  This seems to work: ampersand-poundsign-36-semicolon .  Now I can write [b]$1,000,000[/b].

edit: see below

Yes but you have to rechange that from the dollar if you edit your post, or someone quotes it. So it's not a good method ...

Hmm... I guess I can make do with the $\$ but the less ornate $ would be nicer.

I think I may have found a solution to the problem of entering dollar signs in TeX mode on these forums.  Rather than trying to escape the dollar sign, simply use the tag [color=red]\textdollar[/color].

The lottery jackpot was quoted as $\textdollar \hbox{180,000,000}$, but the winner chose the cash option, which was worth only $\hbox{40\% of \textdollar 180,000,000}=\textdollar72\thinspace 000\thinspace 000.$

I found this in [url=http://www.phil.cam.ac.uk/teaching_staff/Smith/LaTeX/resources/symbols-a4.pdf]The Comprehensive LaTeX Symbol List[/url]

$\textdollar$ :) Indeed a better idea!! :)


Just rewrite the definitions in terms of elements. "$ f(K)\subseteq W$" means "for all $ x\in K$, $ f(x)\in W$". "$ K\subseteq f^{\minus{}1}(W)$" means the same thing.

No, no, deleting questions, makes no sense.


During the "Manhattan Project,"there was a story about
a group of scientists who gather to tacle a very intricate mathematical problem.One of them is "Stanislaw Ulam,"a polish-born-american theoretical physicist.Each of the scientist recieved a copy of a problematic computation.The computer then was used.It is very noteworthy that before anyone give
their answer(Coming from computer.),Dr.Ulam solved the problem
single-handedly.

              Is there any formula to improve ones "counting ability,"thereby becoming good,and fast in counting any assigned
mathematics problem?(Ex.Arithmetic,algebra,etc.)

                                                       Respectfully,
                                                     A.Eddington

Practice practice practice! That's how I became fast at calculations, at least. :)

You can find techniques for fast arithmetic calculations. You can also find memory techniques. But solving a problem fast can only developed with highly developed intuition and the sharpness of a razor. And that comes, as Xevarion points out, from practice.

By the way, are you really called Arthur Eddington?


inequalities
logarithms
function
symmetry
inequalities unsolved

It seems to be a similar problem by Vasc, in P&O section. In that topic, Vasc's one had a relation with $\sum_{cyc}\frac{1}{1-ab}\le{0 \over 2}$.

[quote="zhaobin"]$a,b,c,d>0$,such that $a^{2}+b^{2}+c^{2}+d^{2}=1$
I wonder if we can prove:
$(1-ab)(1-bc)(1-cd)(1-da) \ge \left(\frac{3}{ 4}\right)^{4}$?[/quote]

Yes, we can.  :wink: By substitution it suffices to show the following:
For all $a,b,c,d \ge 0$,such that $a+b+c+d=1$ we have
$g(a,b,c,d) : = (1-\sqrt{ab})(1-\sqrt{bc})(1-\sqrt{cd})(1-\sqrt{da}) \ge \left(\frac{3}{ 4}\right)^{4}$.

For fixed $a,c$ let  $f(x) : = \log[(1-\sqrt{ax})(1-\sqrt{cx})]$, where $0 \le x \le 1-a-c$.
Both functions $\log(1-\sqrt{ax})$ and $\log(1-\sqrt{cx})$ can be seen to be convex.
(Differentiate w.r.t. $x$ and use $a+x \le 1$ which implies $2 \sqrt{ax}\le 1$ and the same for $c$.)
Hence $f$ is convex.

Thus by Jensen we have
$\log[(1-\sqrt{ab})(1-\sqrt{bc})(1-\sqrt{cd})(1-\sqrt{da})] = f(b)+f(d) \ge 2 f \Big( \frac{b+d}{2}\Big).$
So $g(a,b,c,d) \ge g\Big(a,\frac{b+d}{2},c,\frac{b+d}{2}\Big)$.
Note that we can do the same with the roles of $a,c$ and $b,d$ interchanged:
$g(a,b,c,d) \ge g\Big(\frac{a+c}{2},b,\frac{a+c}{2},d\Big)$

After all these observation the proof is easy:
$(1-\sqrt{ab})(1-\sqrt{bc})(1-\sqrt{cd})(1-\sqrt{da}) = g(a,b,c,d)$
$\ge g\Big(a,\frac{b+d}{2},c,\frac{b+d}{2}\Big)$
$\ge g\Big(\frac{a+c}{2},\frac{b+d}{2},\frac{a+c}{2},\frac{b+d}{2}\Big)$
$= \Big( 1-\sqrt{ \frac{a+c}{2}\frac{b+d}{2}}\Big)^{4}$
$\ge \Big( 1-\frac{1}{2}\Big( \frac{a+c}{2}+\frac{b+d}{2}\Big)\Big)^{4}= \left(\frac{3}{ 4}\right)^{4}$.

and we are done. :)

[quote="solyaris"]

For fixed $a,c$ let  $f(x) : = \log[(1-\sqrt{ax})(1-\sqrt{cx})]$, where $0 \le x \le 1-a-c$.
Both functions $\log(1-\sqrt{ax})$ and $\log(1-\sqrt{cx})$ can be seen to be convex.
(Differentiate w.r.t. $x$ and use $a+x \le 1$ which implies $2 \sqrt{ax}\le 1$ and the same for $c$.)
Hence $f$ is convex.

and we are done. :)[/quote]
Hi,Thank you.
But can you tell how you proved  $f$ is convex?
I can't understand :blush:

I hope I didn't make a stupid mistake. Anyhow, here we go:

We have that $f(x)=f_{a}(x)+f_{c}(x)$ where $f_{a}(x) = \log(1-\sqrt{ax})$ and  $f_{c}(x) = \log(1-\sqrt{cx})$.
In order to prove convexity for $f$ it suffices to prove convexity for $f_{a}$ and $f_{c}$.
But $f''_{a}(x) = \frac{ \sqrt{a}(1-2 \sqrt{ax})}{ 4 \sqrt{x}(\sqrt{x}-\sqrt{ax})^{2}}$ which is positive as $2\sqrt{ax}\le 1$.
By symmetry we can do the same for $f_{c}$. Thus $f_{a}$ and $f_{c}$ are convex.  :lol:


algebra
polynomial
modular arithmetic
algebra unsolved

Here is the first hint:
[url=http://ecademy.agnesscott.edu/~lriddle/ifs/siertri/Pascalmath.htm]Pascal's triangle mod 2[/url]

Hello

$P(x)=(x+1)^{n}=\sum_{k=0}^{n}{n\choose k}x^{k}$, so we have to determine the number of $n\choose k$ which are odd.
Let $\sum_{k=0}^{k_{m}}a_{i}2^{i}$ base $2$ representation of a fixed $k$. Then Lucas' Theorem states: \[{n\choose k}={\sum_{i=1}^{m}2^{k_{i}}\choose \sum_{k=0}^{k_{m}}a_{i}2^{i}}\equiv\prod_{i\notin\{k_{1},\ldots,k_{m}\}}{0\choose a_{i}}\prod_{i\in\{k_{1},\ldots,k_{m}\}}{1\choose a_{i}}\mod 2\] . The $a_{i}$ of the first factor have to be $0$, otherwise the product is $0$ and the binomial coefficient is even. The $a_{i}$ of the second factor can be chosen arbitrarily for which we have $2^{m}$ choices.

Finally $2^{m}$ coefficients of P(x) are odd.

Using Lucas Theorem if $n=\sum_{i=1}^{m}a_{i}2^{i},k=\sum_{i=1}^{m}b_{i}2^{i}$: \[\binom nk\equiv\prod_{i=1}^{m}\binom{a_{i}}{b_{i}}\pmod 2\] P.S. I didn't see Mash'Allah's post when I was writing this.


$ a,b,c \in \mathbb{R^{\plus{}*}}$, prove that,
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \geq \frac{a}{b\plus{}c} \plus{}\frac{b}{a\plus{}c} \plus{}\frac{c}{a\plus{}b} \plus{} \frac{3}{2}$

[quote="peine"]$ a,b,c \in \mathbb{R^{ \plus{} *}}$, prove that,
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {3}{2}$[/quote]
we have
$ \frac {a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge\ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}\plus{} \frac{3}{2}$
$ <\equal{}> \frac{ac}{b(b\plus{}c)}\plus{}\frac{ba}{c(c\plus{}a}\plus{}\frac{bc}{a(a\plus{}b)} \ge\ \frac{3}{2}$
By cauchy-schwarz inequality, we have :
 $ LHS\ge\ \frac{(ab\plus{}bc\plus{}ca)^2}{2abc(a\plus{}b\plus{}c)} \ge\ \frac{3}{2}$    (Q.E.D)
  :wink:  :)

Dear peine! Very nice problem, very nice your solution! Congratulations!
For $ 2p^2\ge 5q+p $ we used Schur's inequality:
$ q\le\frac{p^3+9}{4p}\le\frac{2p^2-p}{5}\Leftrightarrow (p-3)(3p^2+5p+15)\ge 0 $
which is true because $ p\ge 3 $!
________
Marin Sandu

[quote="peine"]yes good! there is my solution,
put $ x \equal{} \frac {a}{b}$   ,  $ y \equal{} \frac {b}{c}$   and   $ z \equal{} \frac {c}{a}$
we have $ xyz \equal{} 1$ and  $ \frac {a}{b \plus{} c} \equal{} \frac {xy}{y \plus{} 1} \equal{} x \minus{} \frac {x}{y \plus{} 1}$
then, the inequality becomes:
$ \sum\frac {x}{y \plus{} 1} \geq \frac {3}{2}$
$ \Leftrightarrow \sum\frac {2x}{y \plus{} 1} \geq 3$
$ \Leftrightarrow 2\sum x^2z \plus{} 2\sum x^2 \plus{} 2\sum xy \plus{} 2\sum x \geq 3\sum x \plus{} 3\sum xy \plus{} 3xyz \plus{} 3$
now let put $ x \plus{} y \plus{} z \equal{} p$   ,$ xy \plus{} yz \plus{} zx \equal{} q$ and we have $ xyz \equal{} 1$
therefore the inequality becomes,
$ \Leftrightarrow 2\sum x^2z \plus{} 2p^2\geq p \plus{} 5q \plus{} 6$ wich is true because;
By AM-GM $ 2\sum x^2z \geq 6$ and we have $ 2p^2 \equal{} \frac {5}{3} p^2 \plus{} \frac {1}{3}p^2 \geq p \plus{} 5q$.[/quote]

$ xyz \equal{} 1$ ,so $ x+y+z \geq  3 $
By cauchy-schwarz inequality,
$ \sum\frac{x}{y+1}\ge\ \frac{(x+y+z)^2}{x+y+z+xy+yz+zx}\ge\ \frac{(x+y+z)^2}{x+y+z+\frac{(x+y+z)^2}{3}} \ge\ \frac{3}{2} $

[quote="oldbeginner"]The following inequality is also true.

$ a,b,c \in \mathbb{R^{\plus{}*}}$, prove that,
\[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 6\cdot\left(\frac{a}{b+c}+\frac{b}{a+c}+ \frac{c}{a+b}\right)\cdot\frac{ab+bc+ca}{(a+b+c)^2} \][/quote]
An expanding gives $\sum_{cyc}(a^6c^2+a^5b^3+3a^5c^3+3a^4b^4+$
$+a^6bc-2a^5b^2c+3a^4b^3c+5a^4c^3b-4a^4b^2c^2-11a^3b^3c^2)\geq0$, which is obvious.


We want to divide {1, 2, $ \cdots$, 3n} into n subsets where no subset {x, y, x} satisfies x+y=3z.

find the minimum of n.

I don't understand. What are $ x,y,z$? Are they elements of the same subset?

Well, my instinct is to suppose the question is meant to say, "Find the minimal $ n$ such that no matter how $ \{1, \ldots, 3n\}$ is divided into $ n$ triplets, there must be some triplet $ \{x, y, z\}$ with $ x \plus{} y \equal{} 3z$."  But this is ridiculous, because there is no such minimal $ n$: we can always take $ \{1, 2, 3\}, \{4, 5, 6\}, \ldots$.  \

The answer to the question actually written is just $ n \equal{} 0$: the set $ \{\}$ can be divided into 0 triplets such that none of these triplets satisfies the given equation.

There are a couple other possible statements that might have been meant, but I don't really feel inspired to do the original poster's job and produce an actually meaningful question without typos and other errors.

Oh sorry , my mistake.

I meant that find the minimal of n where we can devide into n subsets, and every subset satisfies x+y=3z.

So, for example, all subsets needs to be like {1,2,5} (2*3=1+5).


Let $a,b,c$ be positive number such that $a^{2}+b^{2}+c^{2}=3$. Prove that
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}+\frac{3(ab+bc+ca)}{a+b+c}\ge 6 \]
[hide="Hint, just click when you have no idea!"]Use a nice and hard inequality of toanhocmuonmau
\[2\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)+3(a+b+c)\ge \frac{15(a^{2}+b^{2}+c^{2})}{a+b+c}\]
[/hide]

P/s:Edited!

[quote="The soul of rock"]Let $a,b,c$ be real number such that $a^{2}+b^{2}+c^{2}=3$. Prove that
\[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}+\frac{3(ab+bc+ca)}{a+b+c}\ge 6 \][/quote]

Something wrong ? [b] Consider $a=b=c=-1$ . [/b]

Yes, this is my mistake when typing! Sorry!
Are there others solution?

[quote="The soul of rock"]Let $ a,b,c$ be positive number such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3$. Prove that
\[ \frac{a^{2}}{b}\plus{}\frac{b^{2}}{c}\plus{}\frac{c^{2}}{a}\plus{}\frac{3(ab\plus{}bc\plus{}ca)}{a\plus{}b\plus{}c}\ge 6\]
[/quote]

I think I have solved it.
(All sums are cyclic)
We will prove that $ \frac{a^{2}}{b}\plus{}\frac{b^{2}}{c}\plus{}\frac{c^{2}}{a}\ge\frac{(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})}{ab\plus{}bc\plus{}ca}$

$ \iff (ab^{3}\plus{}bc^{3}\plus{}ca^{3})\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)\ge (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})$

$ \iff\sum a^{3}\plus{}\sum ab^{2}\plus{}\sum\frac{ab^{3}}{c}\ge\sum a^{3}\plus{}\sum a^{2}b\plus{}\sum ab^{2}$

$ \iff\sum\frac{ab^{3}}{c}\ge\sum a^{2}b$

$ \iff a^{2}b^{4}\plus{}b^{2}c^{4}\plus{}c^{2}a^{4}\ge a^{3}b^{2}c\plus{}ab^{3}c^{2}\plus{}a^{2}bc^{3}$

Let $ x \equal{} ab^{2},y \equal{} bc^{2},z \equal{} ca^{2}$. Then this is equivalent to $ x^{2}\plus{}y^{2}\plus{}z^{2}\ge xy\plus{}yz\plus{}zx$, which is true indeed. Thus we have proved
\[ \frac{a^{2}}{b}\plus{}\frac{b^{2}}{c}\plus{}\frac{c^{2}}{a}\ge\frac{(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})}{ab\plus{}bc\plus{}ca}\]
Now it remains to show that $ \frac{(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})}{ab\plus{}bc\plus{}ca}\plus{}\frac{9(ab\plus{}bc\plus{}ca)}{(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})}\ge 6$

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

Which is true by AM-GM. Equality holds when $ ab^{2}\equal{}bc^{2}\equal{}ca^{2}$ and $ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ca$ i.e. $ a \equal{} b \equal{} c \equal{} 1$.

[hide]Very sorry for the passing away of Mr Lupas(flip2004). :(  May he rest in peace.[/hide]


Mike purchased a pair of shoes at The Runner's Roost. As a member of the Breathe Better Runners Team, he received a $ 20\%$ discount. However, he still had to pay $ 6\%$ tax on the reduced price. Mike's final bill, including discount and tax was $ \$118.72$. How many dollars was the cost of the running shoes before discount and tax?


vector
analytic geometry
linear algebra
matrix
linear algebra unsolved

My teacher gave the following problem in class today:

Let $A,B \in \mathcal M_n \left( \mathbb R \right)$ such that $A \cdot A^T=I_n$, $B \cdot B^T=I_n$ and $n \in \mathbb N$, $n$ odd. Prove that $\det(A+B)=0$ or $\det(A-B)=0$.

Although I'm posting this in the "Unsolved" section, I also found a solution to it.

Let's suppose that $B=I_n$. We have to prove that $\det(A+I) \det(A-I) = 0 \ \ \Longleftrightarrow \ \ \det( A^2 - I) = 0$.

Multiplying $AA^T=I_n$ by $A$ we get $A^2 A^T=A \ \ \Longleftrightarrow \ \ A^2 A^T - I_n A^T= A - A^T \ \ \Longleftrightarrow \ \ (A^2-I_n) A^T = A-A^T$. Since $\det A^T = \pm 1$, it suffices to prove that $\det(A-A^T)=0$. Let $B=A-A^T$, $x=\det B$. Then $B^T = -B$, so $x = (-1)^n x$. $n$ is odd, so $x=-x$, i.e. $x=0$.

For the general case: the columns of $A,B$ represent mutually orthogonal vectors of length $1$, and the conclusion tells us that the sums or the differences of the vectors are linearly dependent. This can't change if we operate modifications on the coordinate system. (can it?) I propose to take coordinates s.t. $B=I_n$.

Well, is the last step correct?

If you find other solutions, you can go ahead and post them.

almost the same solution:
$(A+B)(A^t-B^t)=-AB^t+BA^t=-M+M^t$. So $det[(A+B)(A^t-B^t)]=det[(A+B)(A-B)]=0$ because $n$ is odd.

[quote="alekk"]$\det(M-M^t)=0$ ?[/quote]

$\begin{bmatrix}0&1\\0&0\end{bmatrix}-\begin{bmatrix}0&0\\1&0\end{bmatrix} =\begin{bmatrix}0&1\\-1&0\end{bmatrix},$ and that has determinant $1.$

I'm still not sure. Is it okay to suppose that $B=I_n$ in this context?

@alekk: Indeed, you found a "generalization" to my solution for the particular case. Very nice.

yes Kent, you're right : I wrote nonsense.

[quote="perfect_radio"]So, anybody cares to answer my stupid question?[/quote]

?


The digits of a positive integer $ n$ are four consecutive integers in decreasing order when read from left to right.  What is the sum of the possible remainders when $ n$ is divided by $ 37$?
How do you figure out this problem?
this is what I have so far [hide]$ 1000x\plus{}100(x\minus{}1)\plus{}10(x\minus{}2)\plus{}x\minus{}3 \equal{} 1111x\minus{}123$ not sure what to do after this.  [/hide]

thanks

[hide="Hint"]$ 111$ is divisible by $ 37$, and thus $ 1110$ is divisible by $ 37$.[/hide]

[hide]Simply take that mod 37.  That gives you $ x\minus{}12$. Since $ 3\le x \le9$, then you can easily find the corresponding remainders.[/hide]

brute force isnt the recommended way to go but its by far the easiest solution for this problem..

There are only 7 numbers in this situation

[quote="mathcrazed"][hide]Simply take that mod 37.  That gives you $ x \minus{} 12$. Since $ 3\le x \le9$, then you can easily find the corresponding remainders.[/hide][/quote]

But x-12 < 0
The remainder is x+25
Sum (x) for x between 3 and 9 = 42
Sum (25) for x between 3 and 9 = 175

It doesn't matter if $ x \minus{} 12 < 0$, just take $ 37 \plus{} x \minus{} 12$
Same as  yours.  Mod, at least in this problem, implies a positive number, and I'm assuming that anyone who takes AIME knows that.  It's common sense XD.

I'd say for this problem, you should just brute force it, since there are so few options, and simple division shouldn't take too long.  Also, since you've got about 12 minutes per problem, you should be fine.

You can just add the numbers in columns...

4321
5432
6543
7654
8765
9876


[i]How do you study math?[/i] Do you just do exercises or try to "memorize" the theory?
You shall reveal your tricks ... :P

I usually just read the theory than do some exercises. It is enough to have marks near 100% here in Portugal, altough I do not know half of you do about math.

I study the theory and "learn" it. Then I read the proofs and see if I can do them myself. After that I do the exercises and try some hard problems. For example, from a contest on the related topic.  :)

The "formal" way
1) Read topic.
2) Look through example of application of topic.
3) Do simple/repetitive problems with the "theory".
4) See how "topic" relates with other topics (proof, etc.)
5) Do Contest Math problems directly related to that topic
6) Do Contest Math problems that integrate other previously learned topics with this topic
7) Come back in a week and do harder problems with this topic.

I believe the Art Of Problem Solving books do an excellent job picking out the problems for you. :D


geometry
trigonometry
perpendicular bisector
extremal principle

Hi every body
ABC TRIANGLE
M midpoint of BC
such that: angle AMB is 45 degrees and angle B equal angle MAC
Show that B IS 30 DEGREE

[quote="stergiu"][quote="senouy"] $ABC$  is a triangle ,
$M$ is midpoint of  side $BC$ such that: $\angle AMB = 45$ and $\angle B = \angle MAC$ Show that $\angle B = 30$[/quote]

[b] HINT(A first approach )[/b]

 Triangles $ABC , AMC$ are similar. From this fact  we take $\angle A = 135$ and $CA^{2}= CM.CB$. Now take sinus law in triangle $ABC$ to find $\sin B = \frac{1}{2}$ ect.(Use  : $CA = \sqrt 2 BM , CB = 2 BM$ in the sinus law , so you can simplify  !!!)

[u] Babis[/u][/quote]

 [b]Synthetic Solution(HINT) ![/b]
[color=blue]
During a pause in the school I found this solution. I hope it is not bad ![/color]

 If we draw $AD \perp BC , CE \perp BA$ then we easily find that $ADCE$ is inscribed ,   $EA = EC$ and $DE$ is the perpendicular bisector of $AM$ , so $EM= EA$ .Thus $EC=EA=EM= \frac{BC}{2}= MC$.This gives $\angle BCE = 60$ , so $\angle B= 30$.

[u]Babis[/u]

There are $n > 3$ points in the plane such that any three of the points form a triangle of area $\le 1$. Show that all n points lie in a triangle of area $\le 4$.

I attempted a solution;

I'm not certain if the question is even correct.  Say that we choose three points, and these points are the vertecies of the main triagle that supposivley contains all the points, then we just found a triagnle who's area is 4 and not 1.

Could somebody verify this for me please?

[quote="stergiu"][quote="player"][quote="boxedexe"]The Extremal Principle solves the above problem.[/quote]

What is the Extremal Principle? I cannot seem to find it on wikipedia nor mathworld

 :([/quote]

[b] See in the book[/b] [color=red]: Englel, Problem Solving Strategies, Springer Verlag[/color][/quote]

Dang, that would take a while for me to get my hands on the book.

Could somebody please walk me through this problem?

Extend $\overline{BM}$, let $N$ be the ortogonal projection from $C$ to the extension of $\overline{BM}$. $\triangle CMN$ right isosceles. Seek similary triangles and you'll find that $\triangle CBN$ is...


calculus
derivative
function
algebra
domain
Functional Analysis
real analysis

Let V  and W be banach spaces.  Let H(V,W) be the space of all  linear and continuous  functions such that its inverse is also continuous.  Prove that the function f : H(V,W) to H(W,V) such that $ f(T)\equal{}T^^\minus{}1$   is
 k-th continuously differentiable for all  k (i.e. f is in C infinity, sorry, the infinity part should be an exponent).

First, which derivative are you using for function with infinite-dimensional space H(V,W) as domain?

Umm... H(V,W) is a (open)subspace of L(V,W) which is a Banach Space so think of the function as going from H(V,W) to L(W,V) so take the derivative there. does that answer your question?


In triangle $ABC$, prove that:
$\frac{a^{3}}{r_{a}}+\frac{b^{3}}{r_{b}}+\frac{c^{3}}{r_{c}}\ge\frac{abc}{r}$
where $r_{a}, r_{b},r_{c}$ are the radii escribed circle respective with angle $A, B, C$ of triangle $ABC$ and $r$ is the radius of inscribed circle

[hide]$R_{a}= \frac{rs}{s-a}$

RTP $a^{3}(s-a)+b^{3}(s-b)+c^{3}(s-c) \geq sabc$

is this true?[/hide]

I think sign reversed. \[\iff \sum a^{3}(b+c-a)\le \sum a^{2}bc \iff \sum-a^{3}(b+c)+a^{2}bc+a^{4}\ge 0 \\ \iff \sum a^{2}(a-b)(a-c) \ge 0\]
which is just Schur's. However, I think there is an easier proof since the similar inequality: \[abc\ge (a+b-c)(a-b+c)(-a+b+c) \] normally requires Schur's but in a triangle is an easy Am-Gm in terms of the substitution $x,y,z$.


Hey guys. I'm bad at Mathcounts. I was wondering if you guys could give me your MSN or AIM address/screename so you guys could help me.

ADD ME :D

MSN-wraithlordronin@hotmail.com
AIM-yaor890
plz say you are from the mathcounts forum. Thank you very much for your help.

[quote="kstan013"]Why don't you just post stuff on here that you're having trouble with?[/quote]

That's a good idea for RayRay.

Sorry, I'm new to MATHCOUNTS, so I couldn't help you much there. :oops:

[quote="anirudh"]I think he AIM'ed me. Anyway, there's [url=http://www.artofproblemsolving.com/Forum/index.php?f=132]this[/url] and [url=http://www.artofproblemsolving.com/Forum/index.php]this[/url] if you have questions.  :)[/quote]I think he knows what they are considering that's how he posted this :wink: . You know just a little hunch I had.

[quote="bpms"]I suck at math so I couldn't help you. However, this forum is the best tool for math that exists :)[/quote]

dude. how can you be bad at math and make nationals?

i think i saw you there  :P

[quote="bpms"]I suck at math so I couldn't help you. However, this forum is the best tool for math that exists :)[/quote]
Right....  :P

i think he was just being lazy :P

don't worry, i do it all the time :)
-jorian


geometry
perimeter
inequalities
triangle inequality

How many isosceles triangles exist which have a perimeter of 10 and integer side lengths?

Since the triangle must be isosceles, you can count the cases:

4, 4, 2
3, 3, 4

Note that 5, 5, 0 ;2, 2, 6 ;and 1, 1, 8 aren't in here.  These violate the triangle inequality theorem.  Therefore, the answer is $ \boxed{2}$ triangles.


topology
advanced fields
advanced fields unsolved

Hello, topological spaces are quite new to me, so forgive me if these questions are too easy or belong in theorems subsection..., if so please move them

A nonempty subset of a topological space is compact if and only if by definition all collections of open subsets of the space containing it together, can be reduced to be a finite collection of them, still containing the set

In analysis : $\mathbb{R}^n$ with the usual metric, my early uni definition was bounded and closed for compactness.  It was then proven that these subsets are exactly the ones for which every sequence in them has at least one subsequence that converges.

Then it was proven (as a theorem of Heine) that in the induced topology by the standard metric, these sets  are compact in the topological way

Now would you please help me check some generalisations in a general topological space:

subset A is compact $\rightarrow$ every sequence in it has a convergent subsequence

each countable collection of open sets containing A can be reduced to a finite collection $\Longleftrightarrow$ every sequence in A has a convergent subsequence

A is compact $\Longleftrightarrow$ every sequence in A has a convergent subspace

my guess is that only the latter is false?

Really, all comments are welcome, I really wanna get to the bottom of this.

In a metric space, all three of your statements are if and only if.

There are examples of topological spaces (not metric spaces) in which your second and third statements are forward implications but the converse fails. In other words, it is possible to have a set on which every sequence has a convergent subsequence that isn't compact by the open cover definition.

I would have expected that there was an if and only if when i talked about a countable covering?

Can you give such counterexamples or are they far from easy?


geometry
3D geometry
sphere
geometry unsolved

Vertices of a square $ABCD$ of side $\frac {25}{4}$ lie on a sphere . Parallel lines passing through points $A,B,C$ and $D$ intersect the sphere at points $A_1,B_1,C_1$ and $D_1$ , respectively. Given that $AA_1=2,BB_1=10,CC_1=6$, determine the length of the segment $DD_1$.

I think the right answer is DD'=2 and I think I can prove it (but I'm not sure) I don't know if the side lenght (of the square) has any importance


abstract algebra
number theory
number theory unsolved

Find all primes $ p,q,r,$ such that $ p^{3} \equal{} p^{2} \plus{}q^{2}\plus{}r^{2}$.

CASE(1): $ p\equal{}2$ obviously there is no solution in this case...

CASE(2):$ p\equal{}4k\plus{}1$

$ \Longrightarrow 4|p^2(p\minus{}1)\equal{}q^2\plus{}r^2 \Rightarrow 2|q , 2|r  \Rightarrow q\equal{}r\equal{}2 \Rightarrow p^2(p\minus{}1)\equal{}8$ which is impossible...

CASE(3):$ p\equal{}4k\plus{}3$

$ p|q^2\plus{}r^2  ,  p\equiv3 (\mod 4) \Longrightarrow p|q  ,  p|r \Rightarrow p\equal{}q\equal{}r \Rightarrow p\equal{}q\equal{}r\equal{}3$

It might just be me but the solution seems a bit vague. Can someone elaborate?

If $ p\equiv3(mod 4)$ ( $ p$ is a prime number) and $ p\mid a^2 \plus{} b^2$ so $ p \mid a$ and $ p \mid b$ because otherwise $ (\frac {a}{b})^2 \equiv \minus{} 1(mod p)$ hence for some integer $ x$ , $ x^2\equiv \minus{} 1(mod p)$ .By Fermat's Little Theorem $ x^{p \minus{} 1}\equiv 1(mod p)$ and $ x^4\equiv 1(mod p)$ Let $ d$ be degree of $ x$ on module $ p$.So $ d \mid p \minus{} 1$ and $ d \mid 4$.
$ d \mid 2 \Longrightarrow$ $ x^2\equiv 1(mod p)$ but $ x^2\equiv \minus{} 1( mod p)$ so $ p \equal{} 2$ but $ 2$ can't be. Hence $ d \equal{} 4$ and $ p\equiv 1(mod 4)$. :wink:


[hide="A somewhat elegant solution?"]
Graph it out. Then estimate the points and use guess and check.
[hide="An algebraic solution"]
$ x^3 \minus{} y^3 \equal{} 316\\
(x\minus{}y)(x^2\plus{}y^2\plus{}xy)\equal{}316\\
(a)(b\plus{}56)\equal{}316$

$ x^2 \plus{} y^2 \equal{} 56\\
x^2 \minus{} 2xy \plus{} y^2 \equal{} 56\minus{}2xy\\
a^2 \equal{} 56\minus{}2b$

$ b\equal{}\frac{316}a \minus{} 56\\
a^2 \plus{}\left(\frac{632}a \minus{} 112\right) \equal{} 56\\
a^3 \minus{} 168a \plus{} 632 \equal{} 0$

By RRT, this has no Rational Roots, so you go and use Cardano's method and find the three roots to this and then multiply each answer by two. Or you can check my work.
[/hide]
[/hide]

[quote="PhireKaLk6781"]If $ x^2 \plus{} y^2 \equal{} 56$ and $ x^3 \minus{} y^3 \equal{} 316$, find all possible values of $ 2x \minus{} 2y$.

Is there an "elegant" solution?[/quote]

We have $ x^3\minus{}y^3\equal{}(x\minus{}y)(x^2\plus{}xy\plus{}y^2)\equal{}(x\minus{}y)(56\plus{}xy)\equal{}316$.

Let $ x\minus{}y\equal{}k$. We have $ x^2\minus{}2xy\plus{}y^2\equal{}k^2 \implies 56\minus{}2xy\equal{}k^2$.

Thus, we have $ 56\minus{}k^2\equal{}2xy \implies xy\equal{}\frac{56\minus{}k^2}{2}$.

Plugging this in gives

$ k(56\plus{}\frac{56\minus{}k^2}{2})\equal{}316$.

Multiplying by 2 and expanding gives

$ k(168\minus{}k^2)\equal{}632$
$ 168k\minus{}k^3\equal{}632$.

Unfortunately, this does not have pretty solutions. Here are the [url=http://www.wolframalpha.com/input/?i=168k-k%5E3%3D632]approximations[/url].

Bleh, I just realized that I pretty much copied dragon's solution, but hopefully this is easier to understand.

[quote="Ihatepie"]Is there any reason that the negative of the sum of the two positive solutions is equal to the negative solution?[/quote]

$ k_1^3\minus{}168k_1\plus{}632\equal{}0$

$ k_2^3\minus{}168k_2\plus{}632\equal{}0$

We want to show that $ k_3\equal{}\minus{}(k_1\plus{}k_2)$.

Plugging this in gives

$ \minus{}(k_1\plus{}k_2)^3\plus{}168(k_1\plus{}k_2)\plus{}632\equal{}0$

Expanding:

$ \minus{}k_1^3\minus{}k_2^3\minus{}3k_1^2k_2\minus{}3k_1k_2^2\plus{}168k_1\plus{}168k\plus{}2\plus{}632\equal{}0$

Negating and rearranging gives

$ k_1^3\minus{}168k_1\plus{}k_2^3\minus{}168k^2\plus{}3k_1^2k_2\plus{}3k_1k_2^2\plus{}632\equal{}0$

We know that $ k_1^3\minus{}168k_1\equal{}k_2^3\minus{}168k_2\equal{}\minus{}632$, so we have

$ 3k_1^2k_2\plus{}3k_1k_2^2\equal{}632$.

This is clearly not a "nice" identity, so it appears to me to just be coincidential.

[quote="Ihatepie"]I was thinking that because maybe realizing it provides an easy way to solve the problem?[/quote]

I think it is just coincidential, since based on the coefficients, since the equation does not appear to be generally satisfied. For example, $ (x\minus{}1)(x\minus{}2)(x\minus{}5)\equal{}0$ clearly does not satisfy, etc.

[quote="tenniskidperson3"]Also, because the coefficient of $ x^2$ is 0, the result is true by Vieta's.  Not purely coincidental, but probably coincidence that the coefficient is 0.[/quote]

Oh of course, I didn't realize that.

Yes, the x^2 coefficient is the sum of the roots, so only if the coefficent is 0.


For any positive real numbers $a,b$ and $c$,
\[\frac{b+c}{a^{2}+bc}+\frac{c+a}{b^{2}+ca}+\frac{a+b}{c^{2}+ab}\le\frac{a+b+c}{\sqrt[3]{(abc)^{2}}}\]

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac {b \plus{} c}{a^{2} \plus{} bc} \plus{} \frac {c \plus{} a}{b^{2} \plus{} ca} \plus{} \frac {a \plus{} b}{c^{2} \plus{} ab}\le\frac {a \plus{} b \plus{} c}{\sqrt [3]{(abc)^{2}}}\]
[/quote]
See here: http://can-hang2007.blogspot.com/2010/03/inequality-139-p-h-duc.html

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac {b \plus{} c}{a^{2} \plus{} bc} \plus{} \frac {c \plus{} a}{b^{2} \plus{} ca} \plus{} \frac {a \plus{} b}{c^{2} \plus{} ab}\le\frac {a \plus{} b \plus{} c}{\sqrt [3]{(abc)^{2}}}\]
[/quote]
I think we have
\[ \frac{1}{a\plus{}b\plus{}c}\sum{\frac{b\plus{}c}{a^2\plus{}bc}}\leq \frac{1}{2a^2\plus{}bc}\plus{}\frac{1}{2b^2\plus{}ca}\plus{}\frac{1}{2c^2\plus{}ab}\leq \frac{1}{\sqrt[3]{a^2b^2c^2}}\]

[quote="Tourish"]I think we have
\[ \frac {1}{a \plus{} b \plus{} c}\sum{\frac {b \plus{} c}{a^2 \plus{} bc}}\leq \frac {1}{2a^2 \plus{} bc} \plus{} \frac {1}{2b^2 \plus{} ca} \plus{} \frac {1}{2c^2 \plus{} ab}\leq \frac {1}{\sqrt [3]{a^2b^2c^2}}\]
[/quote]
For the left inequality, try $ a\equal{}1.1,$ $ b\equal{}1$ and $ c\equal{}0.001.$

Sorry for my wrong caculation.... :blush:


The planet Shm'lort uses a different timing system from ours. After making contact, human astronauts tried to figure out how to convert between the two systems. They determined that there are 3 blarsfs and  18 crobs in a minute and 7 blarsfs and 2 crobs in two minutes. How much Earth time, in seconds, elapses in 9 blarsfs and 6 crobs?

Let $ x$= the number of seconds in a blarsf, and $ y$= the number of seconds in a crob. We then have a sytem of equations.
$ 3x\plus{}18y\equal{}60$
$ 7x\plus{}2y\equal{}120$

Multiplying the second equation by $ 9$, subtracting, and dividing through yields $ x\equal{}17$. Plugging this in, we find that $ y\equal{}\frac12$. So, $ 9x\plus{}6y\equal{}9(17)\plus{}6(\frac12)\equal{}153\plus{}3\equal{}\boxed{156}$.

you can also just use manipulation to solve it without actually finding values of variables


I forget what test this comes from, if you know please tell me.
I also can't remember the exact problem, but it is something like:
Some numbers can be expressed as the sum of consecutive squares, for example 1=1^2 and 50=3^2+4^2+5^2.  How many numbers between 1 and 1000 can be expressed in this way?

I think it was the difference of two squares, not the sum of two consecutive.

Yeah, it was #1 from the 1997 AIME:

http://www.kalva.demon.co.uk/aime/aime97.html


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a group if the following conditions hold, $ G$ becomes abelian group
i)  $ \forall a,b \in G, (ab)^5\equal{}a^5b^5$
ii) $ \forall a,b \in G, (ab)^3\equal{}a^3b^3$

$ (1): (ab)^3 \equal{} a^3 b^3$
$ (2): (ab)^5 \equal{} a^5 b^5$

We get immediately form this:
$ (1) \implies (3): (ba)^2 \equal{}a^2 b^2$
$ (3) \implies (4): (ab)^4 \equal{} ((ab)^2)^2 \overset{(3)}{\equal{}} (b^2 a^2)^2 \overset{(3)}{\equal{}} a^4 b^4$

From this we get:
$ ab \equal{} (ab)^5 (ab)^{\minus{}4} \overset{(2)}{\equal{}} (a^5 b^5)(b^{\minus{}1} a^{\minus{}1})^4 \overset{(4)}{\equal{}} a^5 b^5 b^{\minus{}4} a^{\minus{}4}$
$ \implies b \equal{} a^4 b a^{\minus{}4} \implies (5): a^4 \in Z(G)$
And similarly:
$ ab \equal{} (ab)^4 (ab)^{\minus{}3} \overset{(4)}{\equal{}} (a^4 b^4)(b^{\minus{}1} a^{\minus{}1})^3 \overset{(1)}{\equal{}} a^4 b^4 b^{\minus{}3} a^{\minus{}3}$
$ \implies b \equal{} a^3 b a^{\minus{}3} \implies (6): a^3 \in Z(G)$

And both (5) and (6) implie
$ a \equal{} a^4 a^{\minus{}3} \in Z(G)$
which means nothing but $ G$ is abelian.

The claim is a straightforward consequence of the fact that a group $ G$ is abelian [b]if and only if[/b] the identity $ (ab)^{n} \equal{} a^{n}b^{n}$ holds for three consecutive integers $ n$ and all $ a, b \in G$.

All the Best...

Notice that the identities are equivalent to $ (ba)^4\equal{}a^4b^4$ and $ (ba)^2\equal{}a^2b^2$.  So $ a^4b^4\equal{}a^2b^2a^2b^2\iff b^2a^2\equal{}a^2b^2\equal{}(ba)^2\equal{}baba\iff ab\equal{}ba$. $ \Box$


group theory
abstract algebra
number theory
advanced fields
advanced fields unsolved

Why is it that in some contexts, people will write a group quotient on the left, and in other cases on the right? I can't understand the pattern or what meaning is supposed to be conveyed by the order. Examples of common notations:

$ \text{SL}(n, \mathbb{Z}) \backslash \text{SL}(n, \mathbb{R})$
$ \Gamma$ a lattice; $ G$ a Lie group: $ \Gamma \backslash G$

$ \mathbb{Z}/n\mathbb{Z}$
$ G$ a finite group, $ H$ a normal subgroup: $ G/H$

etc...

Is this just a peculiarity of analytic number theory where people like to quotient Lie groups by lattices on the right? Or am I missing something?

if $ G$ acts on a set/space/graph/whatever $ X$ and this action is written on the left ($ (g,x) \mapsto gx$), the quotient set/space/graph is denoted by $ G\backslash X$. in the right case it's denoted by $ X \backslash G$. for example if $ H$ is a subgroup of $ G$, there are two natural operations from $ G$ on $ H$. $ G\backslash H$ is the set of right cosets and $ H\backslash G$ is the set of left cosets of $ H$ in $ G$. this is [i]not[/i] the same and isn't just a freedom of notation.

probably this is nothing new to you but I want to indicate that the whole thing is only a matter of fact how the action is written. also remark that you can always switch between left actions and right actions ($ xg \equal{} g^{\minus{}1} x$, or $ xg\equal{}gx$ if $ G$ is abelian), but I think in most cases there is a natural choice.

O.K. I think in the application where it confused me, it is just an idiosyncratic and arbitrary choice, but I see how it can be an important distinction in other situations. Thanks!

I guess the peculiarity started with, say, the upper half plane $ \mathbb H$ and its transitive group of isometries $ \mathrm{SL}_2(\mathbb R)$ operating on the left, so $ \mathbb H\equal{}\mathrm{SL}_2(\mathbb R)/\mathrm{SO}(2)$. Further factoring out $ \mathrm{SL}_2(\mathbb Z)$ gives $ \mathrm{SL}_2(\mathbb Z)\backslash\mathrm{SL}_2(\mathbb R)/\mathrm{SO}(2)$. So the compact subgroup is on the right, and the lattice is on the left. This also leads to the Iwasawa decomposition being written as $ NAK$ (as opposed to $ KAN$). owk


calculus
integration
algebra
polynomial
derivative
quadratics
partial fractions

[quote="cheerful coffin"]well I learn partial fraction when I am working on integral, so I guess this would be the right forum to ask.

say I want to partial fraction x/[(1+x)(1-4x)] into the form A/(1+x) + B/(1-4x)

so set them to be equal, eventually we have:

[A(1-4x) + B(1+x)]/[(1+x)(1-4x)] = x/[(1+x)(1-4x)]

so if we equate the numerator, it becomes:
A(1-4x) + B(1+x) = x

take limit of x approaching -1 on both sides, we have:
A(5) = -1
A = -1/5

B could be derived using the same way.

now magically this method of deriving a partial fraction would give you the exact same value of A and B as the normal method of partial fraction in which you need to solve system of 2 equations.

Unfortunately, this "new" method is suggested by my friend, and it hasn't been proven I think (none of my textbooks has mentioned it). So I don't know if it works all the time.

So could anyone please tell me if this "new" (less time-consuming too...) method is valid or not?[/quote]

It's supposed to hold for all values of $x$, so substituting one value of $x$ (in this case, $-1$) is perfectly legal.  It's a fine method.

It's completely accurate. After multiplying through by the denominators, we have a polynomial equality, which is true at all values of $x$. In particular, it's true at convenient values such as those which make some of the terms zero.

If there are higher powers of some factor, we won't get enough equations this way, and we'll have to modify the method (get equations based on the derivative at those points) or use some less convenient points. With irreducible quadratic factors, we'll either have to modify the method or use complex algebra.

This method isn't perfect, but it's usually very good. The "equate the coefficients" method you know is really a special case of this, with a fixed choice of points (0 and derivatives at zero).

Justifying why this method always works would require some knowledge of polynomial interpolation, and I'd rather not get into that here. On the other hand, you don't need that universal justification to use the method; the main theorem of partial fractions tells us that something works, and we find that only one possibility can work.


geometry
inradius
ratio
Pythagorean Theorem
algebra
system of equations

1. A 45* 45* 90* triangle has a circle inscribed in it. if the legs of the triangle have length of 50, what is the area of the region inside the triangle but outside the circle?
-round to 3 decimal places

2.find the length of the common internal tangent between two non-intersecting circles, if the first circle has a radius of 6, the second has the radius of 18, and the distance between the centers of the circle is 51.

Thanks in advance i think i might be done with the questions for now...

[quote="Pistachio"][quote="i_like_pie"][hide="1"]The area of the triangle is $\frac{50^{2}}{2}=1250.$

The inradius of any triangle is $\frac{a+b-c}{2}.$

In this case, $r=\frac{50+50-50\sqrt{2}}{2}=50-25\sqrt{2}.$

The area of the incircle is $\pi\left(50-25\sqrt{2}\right)^{2}.$

The answer is $1250-\pi\left(50-25\sqrt{2}\right)^{2}\approx576.235$[/hide][/quote]

thanks i like pie, you end up answering like all my questions  :)[/quote]
You're welcome. :)

[hide="2"][img]http://img134.imageshack.us/img134/293/tangentssg5.png[/img]

The first thing we see is that $\triangle ACE$ is similar to $\triangle BDE.$ This means their corresponding parts all have the same ratio. We know that $AC=18$ and $BD=6.$ This means the ratio of corresponding parts in the two triangles is $3: 1.$

Now that we know the ratio, we have a simple system of equations: $AE=3EB$ and $AE+EB=51.$

After solving the system, we know that $AE=\frac{153}{4}$ and $EB=\frac{51}{4}.$

We can apply the Pythagorean Theorem twice to find $CE$ and $ED.$

$18^{2}+CE^{2}=\left(\frac{153}{4}\right)^{2}\Longrightarrow CE=\frac{135}{4}$

$6^{2}+ED^{2}=\left(\frac{51}{4}\right)^{2}\Longrightarrow ED=\frac{45}{4}$

Adding those two segments, we get $CD=\boxed{45}.$[/hide]

I like pie - Which program do you use to create those diagrams?

Microsoft Visio. The full name is Microsoft Office Visio Professional 2003. I make a picture, then upload it via [url]http://imageshack.us/[/url].

2. d squared - (ra+r2) squared= int tangent squared

plug in this formula into the equation and get 45 for the answer


Is there a way to draw pictures and graph functions in LaTeX?

Yes.
1. You can use the picture environment. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122562]Clever ideas in LaTeX[/url] for examples and Chapter 5 of [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url] for an introduction to graphics.
2. You can also use external programs -see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url]


algebra
polynomial
quadratics
complex numbers

Don't triple post. Please. Especially three times the same day. Thank you.

And IMHO, I don't think it's factorable.

[quote="neelnal"]Can someone factor a^5 - b^5 for me please.

a^5 - (a^4)b + (a^4)b - (a^3)(b^2) + (a^3)(b^2)...
(a - b)(a^4 + (a^3)b + (a^2)(b^2) +  a(b^3) + b^4)

Can we get further than that while sticking to real numbers? It does not seem like we can but I am not sure. Unless we use the rational roots theorem, so can someone help me?[/quote]
Actually, it is factorable. To help with factoring this expression you might want to look at the equation $a^{5}-b^{5}=0$, which is clearly satisfied by $a=b$. This means that $a-b$ is a factor of $a^{5}-b^{5}$. (Try to convince yourself why this is true.) From there you can factor it out to get \[(a-b)(a^{4}+ab^{3}+a^{2}b^{2}+a^{3}b+b^{4}).\] In general, \[a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}).\]
EDIT: Sorry, I misunderstood your question.  :oops:

[quote="nebula42"][quote="neelnal"]Can someone factor a^5 - b^5 for me please.

a^5 - (a^4)b + (a^4)b - (a^3)(b^2) + (a^3)(b^2)...
(a - b)(a^4 + (a^3)b + (a^2)(b^2) +  a(b^3) + b^4)

Can we get further than that while sticking to real numbers? It does not seem like we can but I am not sure. Unless we use the rational roots theorem, so can someone help me?[/quote]
Actually, it is factorable. To help with factoring this expression you might want to look at the equation $a^{5}-b^{5}=0$, which is clearly satisfied by $a=b$. This means that $a-b$ is a factor of $a^{5}-b^{5}$. (Try to convince yourself why this is true.) From there you can factor it out to get \[(a-b)(a^{4}+ab^{3}+a^{2}b^{2}+a^{3}b+b^{4}).\] In general, \[a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}).\] [/quote]

...

repetition of wat i said. i'm asking if there is a way to factor:

a^4 + (a^3)b + (a^2)(b^2) + a(b^3) + b^4

Homogenous polynomials in two variables $a,b$ can always be factored into linear and quadratic factors by the fundamental theorem of algebra over the reals and into linear factors over the complex numbers (set $x=\frac{a}{b}$ and look what you get).
This will give ou also how to factor it.
It can in fact not be factored anymore over the rationals than given before.


function
inequalities
inequalities proposed

I contemplated on this question that suddenly popped in my head.

Prove or disprove:

$ (ab \plus{} bc \plus{} ca)(\frac {1}{a^2 \plus{} b^2} \plus{} \frac {1}{b^2 \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} a^2})\ge \frac{9}{2}$

[quote="earldbest"]I contemplated on this question that suddenly popped in my head.

Prove or disprove:

$ (ab \plus{} bc \plus{} ca)(\frac {1}{a^2 \plus{} b^2} \plus{} \frac {1}{b^2 \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} a^2})\ge \frac {9}{2}$[/quote]
I don't think it is true :wink:
Try $ a\equal{}1,b\equal{}2,c\equal{}\frac{1}{2}$

Thanks! for some reason, I couldn't find a counterexample (sigh) :lol:

I think want to find a function  $ f (a, b, c)$ satisfy:
 $ f(a,b,c).(\frac {1}{a^{2} \plus{} b^{2}} \plus{} \frac {1}{b^{2} \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} a^{2}})$ stronger than Iran 1996
is very difficult.
I have tried function $ f(a,b,c) \equal{} a^2 \plus{} b^2 \plus{} c^3 \plus{} ab \plus{} ac \plus{} bc$ but ineq not true(when $ a \equal{} 1,b \equal{} 8,c \equal{} 9$)
What $ f(a,b,c)$ :?:

[quote="math10"]I found a ineq.Prove that:
 $ (5a^2 \plus{} 5 b^2 \plus{} 5 c^2 \plus{} ab \plus{} ac \plus{} bc).(\frac {1}{a^{2} \plus{} b^{2}} \plus{} \frac {1}{b^{2} \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} a^{2}})\geq 27$[/quote]
$ (4a^2\plus{}4b^2\plus{}4c^2 \plus{} ab\plus{}bc\plus{}ca)\left ( \frac {1}{a^{2} \plus{} b^{2}} \plus{} \frac {1}{b^{2} \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} a^{2}} \right ) \ge \frac{35}{2}$ is the strongest :) There's equality when $ a\equal{}b\equal{}c\equal{}1$ and when $ a\equal{}b\equal{}1,c\equal{}0$.

I think ineq also true
$ (3a^{2}\plus{}3b^{2}\plus{}3c^{2}\plus{}ab\plus{}bc\plus{}ca)\left (\frac{1}{a^{2}\plus{}b^{2}}\plus{}\frac{1}{b^{2}\plus{}c^{2}}\plus{}\frac{1}{c^{2}\plus{}a^{2}}\right )\ge 18$
I suddenly feel the ineq following is incorrect:
$ (4a^{2}\plus{}4b^{2}\plus{}4c^{2}\plus{}(a\plus{}b\plus{}c)^2)\left (\frac{1}{a^{2}\plus{}b^{2}}\plus{}\frac{1}{b^{2}\plus{}c^{2}}\plus{}\frac{1}{c^{2}\plus{}a^{2}}\right )\ge \frac{63}{2}$

[quote="math10"][quote="Mathias_DK"][quote="math10"]I found a ineq.Prove that:
 $ (5a^2 \plus{} 5 b^2 \plus{} 5 c^2 \plus{} ab \plus{} ac \plus{} bc).(\frac {1}{a^{2} \plus{} b^{2}} \plus{} \frac {1}{b^{2} \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} a^{2}})\geq 27$[/quote]
$ (4a^2 \plus{} 4b^2 \plus{} 4c^2 \plus{} ab \plus{} bc \plus{} ca)\left ( \frac {1}{a^{2} \plus{} b^{2}} \plus{} \frac {1}{b^{2} \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} a^{2}} \right ) \ge \frac {35}{2}$ is the strongest :) There's equality when $ a \equal{} b \equal{} c \equal{} 1$ and when $ a \equal{} b \equal{} 1,c \equal{} 0$.[/quote]
I also think of this in yesterday. But very strange  :?: , when I get a wrong example.
But this is not found now :blush:
can you post solution?[/quote]
I made a mistake. It should be:
$ (4a^2 \plus{} 4b^2 \plus{} 4c^2 \plus{} ab \plus{} bc \plus{} ca)\left ( \frac {1}{a^{2} \plus{} b^{2}} \plus{} \frac {1}{b^{2} \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} a^{2}} \right ) \ge \frac {45}{2}$ of course. (else we wouldn't have equality :) )
$ (3a^{2}\plus{}3b^{2}\plus{}3c^{2}\plus{}ab\plus{}bc\plus{}ca)\left (\frac{1}{a^{2}\plus{}b^{2}}\plus{}\frac{1}{b^{2}\plus{}c^{2}}\plus{}\frac{1}{c^{2}\plus{}a^{2}}\right )\ge 18$ is not right.(Because I said that mine was the strongest :P ) Try $ a\equal{}b\equal{}1,c\equal{}0$ :)
The proof for my inequality is rather straightfoward with the uvw-method :) Do you want to try to solve it yourself, or do you want the proof? :)

[quote="arqady"][quote="Mathias_DK"]It should be:
$ (4a^2 \plus{} 4b^2 \plus{} 4c^2 \plus{} ab \plus{} bc \plus{} ca)\left ( \frac {1}{a^{2} \plus{} b^{2}} \plus{} \frac {1}{b^{2} \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} a^{2}} \right ) \ge \frac {45}{2}$ of course. [/quote]
SOS kills it easy enough.[/quote]
I think it is really complicated
can you post solution?

[quote="math10"]
can you post solution?[/quote]
Of cause!
$ (4a^2 \plus{} 4b^2 \plus{} 4c^2 \plus{} ab \plus{} bc \plus{} ca)\left ( \frac {1}{a^{2} \plus{} b^{2}} \plus{} \frac {1}{b^{2} \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} a^{2}} \right ) \ge \frac {45}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {4c^2 \plus{} ab \plus{} ac \plus{} bc}{a^2 \plus{} b^2} \minus{} \frac {7}{2}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {8c^2 \plus{} 2ab \plus{} 2ac \plus{} 2bc \minus{} 7a^2 \minus{} 7b^2}{a^2 \plus{} b^2}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(c \minus{} a)(7a \minus{} b \plus{} 4c) \minus{} (b \minus{} c)(7b \minus{} a \plus{} 4c)}{a^2 \plus{} b^2}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)\left(\frac {7b \minus{} c \plus{} 4a}{b^2 \plus{} c^2} \minus{} \frac {7a \minus{} c \plus{} 4b}{a^2 \plus{} c^2}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(a^2 \plus{} b^2)(4a^2 \plus{} 11ab \plus{} 4b^2 \minus{} ac \minus{} bc \minus{} 3c^2)\geq0.$
Let $ a\geq b\geq c$ and $ S_c \equal{} (a^2 \plus{} b^2)(4a^2 \plus{} 11ab \plus{} 4b^2 \minus{} ac \minus{} bc \minus{} 3c^2).$
Now it's easy to show that $ S_c\geq0,$ $ S_b\geq0$ and $ S_a \plus{} S_b\geq0.$ :wink:


function
integration
inequalities
LaTeX
real analysis
real analysis theorems

Can anyone help me the theorem:give me some advise or a link about it?
Let $f,g$ are the real functions satisfy the following conditions:
a,$f$ is monotonic on [a,b]
b,$g$ is Riemann integerable on [a,b]
So $f(x)g(x)$ is Riemann integrable on [a,b]
and there exists $\alpha \in [a,b]$ so that:
$\int^{b}_{a}f(x)g(x)dx=f(a)\int^{\alpha}_{a}g(x)dx+f(b)\int^{b}_{\alpha}g(x)dx$

Assume $f$ is increasing. Let $h(\alpha)=f(a)\int^{\alpha}_{a}g(x)dx+f(b)\int^{b}_{\alpha}g(x)dx$. This is a continuous function such that $h(a)=f(b)\int_{a}^{b}g(x)\,dx$ and $h(b)=f(a)\int_{a}^{b}g(x)\,dx$. Since $h(b)\le \int_{a}^{b}f(x)g(x)\,dx\le h(a)$, the conclusion follows from the intermediate value theorem.

[quote="mlok"] Since $h(b)\le \int_{a}^{b}f(x)g(x)\,dx\le h(a)$, the conclusion follows from the intermediate value theorem.[/quote]
This line isn't clear,can you post in detail?

The problem seem very easy as mlok think,but I think it isn't simple.
Can anyone help me this?
Thanks

It is because $f(a) \le f(x)$ for all $x \in [a,b]$ so

$h(b) = f(a) \int_{a}^{b}g(x) dx \le \int_{a}^{b}f(x)g(x)dx$.

Similarly, $f(b) \ge f(x)$ for all $x \in [a,b]$ so

$h(a) = f(b)\int_{a}^{b}g(x) dx \ge \int_{a}^{b}f(x)g(x)dx$.

I think the inequalities may reverse depending on the signs of $f$ and $g$, but either way the Intermediate Value Theorem solves it.

The problem is (conceptually) easier if you handle it in terms of Riemman [i]Stieltjes[/i] integrals. Unfortunately I don't know how to use Latex


probability
parameterization
probability and stats

Suppose that you continually collect coupons and that there are m different types. Suppose also that each time a new coupon is obtained, it is a type i coupon with probability p_i, i=1,...,m. Suppose that you have just collected your nth coupon. What is the probability that it is a new type?

suppose that the coupon you just collected is of type $ i$, which happens with probability $ p_i$: the probability that you do not have already a coupon of type $ i$ is
\[ (1\minus{}p_i)^{n\minus{}1}\]
Hence, the probability that the $ n$-th coupon is a new one is:
\[ \sum_i p_i (1\minus{}p_i)^{n\minus{}1}\]

question: estimate the time you have to wait until you get all the coupons.

[quote="alekk"]question: estimate the time you have to wait until you get all the coupons.[/quote]
If we let $ T_i$ to be the time taken to collect the $ i$th coupon, then $ T_i$ is geometric with success prob. $ p_i$. Therefore $ E(T_i)\equal{}\frac{1}{p_i}$
If $ T$ is the total time to wait to get all the coupons then $ T\equal{}T_1\plus{}T_2\plus{}\dots \plus{}T_m$
Therefore $ E(T)\equal{}\frac{1}{p_1}\plus{}\frac{1}{p_2}\plus{}\dots \plus{}\frac{1}{p_m}$

I do not think that $ T \equal{} T_1 \plus{} \ldots \plus{} T_m$: you are collecting all the coupons at the same time. (but you proved that $ E[T] \leq \sum \frac {1}{p_i}$)

ps: there is a neat expression for $ E[T]$ when $ p_i\equal{}\frac{1}{n}$ but I don't know the answer for the general case. It seems like it's getting nasty. Is it ?

[quote="alekk"]suppose that the coupon you just collected is of type $ i$, which happens with probability $ p_i$: the probability that you do not have already a coupon of type $ i$ is
\[ (1 \minus{} p_i)^{n \minus{} 1}\]
Hence, the probability that the $ n$-th coupon is a new one is:
\[ \sum_i p_i (1 \minus{} p_i)^{n \minus{} 1}\]
question: estimate the time you have to wait until you get all the coupons.[/quote]

Why do you need to multiply both probabilities? Does it have to do with independence of events?

exactly: the coupon you collect at time $ i$ does not depend on what you already have in hands.

[quote="alekk"]I do not think that $ T \equal{} T_1 \plus{} \ldots \plus{} T_m$[/quote]
Why?
Is the problem is to collect the coupons one by one with replacement (noting down each time we collect a coupon) till we collect all the $ m$ coupons?

[quote="alekk"]you are collecting all the coupons at the same time.[/quote]
I don't understand this.

[quote="alekk"]$ E[T] \leq \sum \frac {1}{p_i}$)[/quote]
I don't understand this also. I showed the equality, right?

[quote="limsup"][quote="alekk"]I do not think that $ T \equal{} T_1 \plus{} \ldots \plus{} T_m$[/quote]
Why?
Is the problem is to collect the coupons one by one with replacement (noting down each time we collect a coupon) till we collect all the $ m$ coupons?
[/quote]

Yes. And for example, the first coupon is collected at time $ \min(T_1, T_2, \ldots, T_m)\equal{}1$ and the last one at $ T\equal{}\max(T_1, T_2, \ldots, T_n)$ if $ T_i$ is the first time that a coupon of kind $ i$ is collected. Each $ T_i$ is a geometric random variable of parameter $ p_i$, but of course the dependence structure of $ (T_1, \ldots, T_m)$ seems quite intricate. Am I wrong ?

Yeah alekk. I m extremely sorry. You are absolutely right!  $ T\equal{}\max(T_{1}, T_{2},\ldots, T_{m})$
There is no closed form formula for $ E(T)$ as said [url=http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V1D-4NXHC5N-4&_user=512776&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_searchStrId=961147995&_rerunOrigin=google&_acct=C000025298&_version=1&_urlVersion=0&_userid=512776&md5=06957f3e9af7d35e4a64962ce22825e6]here[/url]

the article you mentioned deals with i.i.d. geometric random variables while in our problem they are highly correlated. There may be a closed form formula in this particular case :roll: (as I said, there is a closed form formula in the case $ p_i \equal{} \frac{1}{n}$: this is the classical "coupon collector" problem)


Euler
logarithms
limit
real analysis
real analysis unsolved

$ \gamma$ - Euler const.
$ H_n$ - harmonic number

let $ \gamma_n\equal{}H_n\minus{}\ln n$,
$ g_n\equal{}\gamma_n\minus{}\gamma$.

probably
\[ \lim_{n\to\infty}\log_{\frac{n}{n\plus{}1}}\frac{g_{n\plus{}1}}{g_n}\equal{}1\]
can anyone prove it?

I think limit is equal to $ 2$.
here is proof
:
$ ... \equal{} \frac{\ln (g_{n\plus{}1}/g_{n})}{\ln (n/(n\plus{}1))}\approx \frac{\ln (1\minus{}\frac{1}{n(n\plus{}1)(H_{n}\minus{}\ln n \minus{}\gamma)})}{\minus{}\frac{1}{n\plus{}1}}\approx \frac{1}{n(H_{n}\minus{}\ln n \minus{}\gamma )}\equal{}2$ when $ n\minus{}>\plus{}oo$
in my last step I use the fact that $ H_{n}\equal{}\ln n \plus{}\gamma \plus{}\frac{1}{2n}\plus{}O(\frac{1}{n^{2}})$ in last term we have $ \frac{1}{2n}$ therefore
$ n(H_{n}\minus{}\ln n \minus{}\gamma )\equal{}_{n\minus{}>\plus{}oo}2$


inequalities
Support
inequalities unsolved

[quote="Denbox"]${x,y,z \in R*.}$
a=$\sqrt{x^{2}+xy+y^{2}}$;
b=$\sqrt{y^{2}+yz+z^{2}}$;
b=$\sqrt{z^{2}+zx+x^{2}}$;
Prove that:${ab+bc+ca \geq x+y+z}$[/quote]
I think there should be $c=\sqrt{z^{2}+zx+x^{2}}$
And if I'm not mistaken there are two ways to solve this problem,with geometrical interpritation and with Caushy of course  :) .

[quote="RDeepMath91"]I have another inequality for...
$a = \sqrt{x^{2}+xy+y^{2}}$
$b = \sqrt{y^{2}+yz+z^{2}}$
$c = \sqrt{z^{2}+zx+x^{2}}$

By CS,
$\sum_{cyc}ab$
$= \sum_{cyc}\sqrt{[(x+\frac{y}{2})^{2}+\frac{3y^{2}}{4}][(z+\frac{y}{2})^{2}+\frac{3y^{2}}{4}]}$
$\geq \sum_{cyc}(x+\frac{y}{2})(z+\frac{y}{2})+\frac{3y^{2}}{4}$
$= (x+y+z)^{2}$

Thus, the initial inequality is right for $x+y+z \geq 1$, but if $x+y+z < 1$ ???
Who knows ??  :o[/quote]
I think that the initial inequality was just written with typos, in it!See the proposer's second post where he quote something :) .The way you provided is the second way I mentioned above.

P.S. Nice signature [b]RDeepMath91[/b], hope to see you in Vietnam and good luck with gold medal :wink: .

Thx for the support,  Tiks !!  :)

And yeah, that's because I only saw the first post ...  :blush:


1)If $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\equal{}1$ , prove that :

$ \frac{1}{2}(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})^2\ge (a\plus{}b\plus{}c)\sqrt{3}\plus{}\frac{a\plus{}b\plus{}c}{a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)}$

[quote="alex2008"]Thank you for finding a stronger one , quykhtn-qa1[/quote]
Thanks, alex2008 :lol:

Let $ a\plus{}b\plus{}c\equal{}x$ and $ abc\equal{}y$.

Inequality equivalents to $ \frac{1}{6y^2} \geq x\sqrt{3}\plus{}\frac{x}{x\minus{}3y}$

$ i) (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ac)\geq 9abc$

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

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

$ x\geq 9y , \frac{x}{x\minus{}3y}\leq \frac{3}{2} , xy\leq \frac{1}{3}, x \geq \sqrt{3}$

$ \frac{1}{6y^2} \geq \frac{3x^2}{2}$ and $ x\sqrt{3}\plus{}\frac{x}{x\minus{}3y} \leq x\sqrt{3}\plus{}\frac{3}{2}$

$ \frac{3x^2}{2} \geq x\sqrt{3}\plus{}\frac{3}{2} \Longleftrightarrow$

$ (x\minus{}\sqrt{3})(3x\plus{}\sqrt{3}) \geq 0$ which is obviously true.  :wink:

[quote="crazyfehmy"]Let $ a \plus{} b \plus{} c \equal{} x$ and $ abc \equal{} y$.

Inequality equivalents to $ \frac {1}{6y^2} \geq x\sqrt {3} \plus{} \frac {x}{x \minus{} 3y}$

$ i) (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ac)\geq 9abc$

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

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

$ x\geq 9y , \frac {x}{x \minus{} 3y}\leq \frac {3}{2} , xy\leq \frac {1}{3}, x \geq \sqrt {3}$

$ \frac {1}{6y^2} \geq \frac {3x^2}{2}$ and $ x\sqrt {3} \plus{} \frac {x}{x \minus{} 3y} \leq x\sqrt {3} \plus{} \frac {3}{2}$

$ \frac {3x^2}{2} \geq x\sqrt {3} \plus{} \frac {3}{2} \Longleftrightarrow$

$ (x \minus{} \sqrt {3})(3x \plus{} \sqrt {3}) \geq 0$ which is obviously true.  :wink:[/quote]
Nice solution, but it  similar my solution :oops:


I post this problem here but I don't know if it is really number theory and if it is elementary.

Let $ q$ be a rational number such that $ q>1$ and $ a_n \equal{} q^n \minus{} [q^n]$ for all positive integer $ n$.
Is it possible that $ \lim_{n\to\infty} a_n \equal{} 0$ ?
Can the sequence $ a_n$ be equireparted ? Is it always equireparted ?
Are there other theorems or conjectures about this sequence ?

Thank you

It seems it is quite a famous topic, with many results and many conjectures!

Try entering in Google things like "fractional parts of powers of rationals" (like I did), or similar.
I got these links: [url]http://oldweb.cecm.sfu.ca/Mahler/051.pdf[/url] and especially [url]http://www.ias.ac.in/mathsci/vol115/nov2005/PM2583.pdf[/url] which states that there are infinitely many limit points, so it is not true that $ \lim_{n\to\infty} a_n \equal{} 0$.
See also  [url]http://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number[/url] and [url]http://www.springerlink.com/content/90g48xwbvgb8a0qw/[/url] where they prove that if for a real algebraic number $ q$ one has $ \lim_{n\to\infty} a_n \equal{} 0$, then $ q \in (0, 1) \cup \mathbb{Z}$.


calculus
integration
linear algebra
matrix
algebra
polynomial
complex analysis

A fun problem in Conway, that I have no idea how to do: Prove the Cayley-Hamilton theorem over $ \mathbb{C}$, that every complex matrix satisfies its own characteristic polynomial, using the Cauchy integral formula.

I started with diagonalizable matrices, then used a density argument... but of course that doesn't use the Cauchy integral formula.  Does anybody know what proof is being referenced here?  What possible connection is there between these two results?

The connection is that
$ p(A) \equal{} \frac {1}{2\pi} \oint_{|z| \equal{} 1} \frac {p(z)}{zI \minus{} A} \, dz$
for matrices with norm less than one, at least.
(the division is multiplication by matrix inverse)


I sent my r2 copied w/ my r3. Which grader feedbacks will I see, and when will they be up??
Thanks,
mathemonster

also btw. what is AIME cutoff

[quote="Xantos C. Guin"]If we snail mailed our solutions, when should we receive a check mark?[/quote]All of the mailed solutions that we have received have been checked in.  If you mailed a solutions but it is not checked in, please write usamts@usamts.org with the details.

[quote="the future"]For people that did not turn in new Round 2 solutions (did not make copy), when will the recalculated scores be posted?

Right after round 3?  Or after round 4?  Because dont they have to be calculated before the AIME to give everybody a fair shot?[/quote]We will make an announcement about this when we are ready to do so.  We have not yet announced the AIME qualification process.

Were we supposed to have received an e-mail confirming that our solutions have been received?

Also, if my R2 (which was sent by mail) is checked, does that mean it was received for R2 or R3?

We are not "re-checking" Round 2: the Round 2 checkmark is for your original submission in November.

All of the Round 3 solutions that we have received have been checked in to the best of our knowledge.  If you mailed in Round 3 but it hasn't been checked in, please write usamts@usamts.org with the details.

Do we need to include any details besides our name and ID number?

[quote="mathpk"][quote="DPatrick"]The Round 2 scores/feedback will be available hopefully the week Feb 2-6, and the Round 3 scores/feedback a week or two after that.

We have not yet made any announcements about AIME qualification.[/quote]

Will round 2 scores surely be available in this week?[/quote]Unlikely: we just got them back from the graders today, and it'll take a while to scan them all in.

i kinda want to see how i did on r2 before i start r4, but if theyre [b]DEFINATELY[/b] going to be up next week i guess i can procrastinate some more

Better not procrastinate and start now. What does it matter whether you get a 25 on Round 2 or a 0? How is that going to affect how much effort you put into Round 4? I already finished three problems and I am working on my fourth now.

Hey atleast you guys will be getting a rd2 score before the rd4 deadline...

I wont even be getting a score because I did not make backup copies...and it sucks because just using estimates, i thought i got 22 that round and only 18 round 3 (so if they use the "calculation" thing to recalculate my score, that 18 will bring me down)

I know its my fault, but why did they have to lose it that round?? why not round 3?

As for me, I havent even looked at the problems yet.  I did not even know they had a new round out until I read this thread...and I still probably wont work on it until after AMC 12 B.

Actually, I think that if you get to see your score on round 2, you will know how hard you will have to work on round 4 to get the medal that you want (provided that your round 3 estimate is fairly close)...so its always an advantage...
like in sports, you have much of an advantage if you know what you need to win.

[quote="the future"]Actually, I think that if you get to see your score on round 2, you will know how hard you will have to work on round 4 to get the medal that you want (provided that your round 3 estimate is fairly close)...so its always an advantage...
like in sports, you have much of an advantage if you know what you need to win.[/quote]
Why wouldn't someone rather do the best they could on each round? The goal of the contest is to learn and do math, not to have someone tell you that you accumulated x points.

[quote="the future"]like in sports, you have much of an advantage if you know what you need to win.[/quote]I disagree. In sports, analyzing what I need to win generally just throws me off my game.

[quote="the future"]uhh sorry, i dont mean you shouldnt try your best.

You have to be time efficient...for example, if I know my total score is like 60 through 3 rounds, there is no way I can get a gold medal.  Also, I need only 15 points to get silver, so if you already know how to do 4 questions, it would be more efficient to check those questions than do a 5th question that you have no idea how to do.  Now of course, if you know how to do it or have some idea, then you should always do it.

Its similar parallel to the AIME...when you guys to the AIME, and your target is like 7 questions to pass, how many of you guys actually try to do all 15 questions (when the last 3-4 you know you have not learned the concepts to do)...for me I just focus on the first 10-12 and try to get all of them right instead of wasting time doing questions I will never be able to answer.

Im  not saying you should try only 4 if you know how to do 5...or try only the first 10-12 on the AIME if you know how to do 15...Im just saying that if you know what you want to get, its easier to create a plan out for yourself.  Of course, if you can get a perfect or want to, you should always go for it.  And yes I know for the

AIME has a time limit of 3 hours, far less than the USAMTS...but the USAMTS conflicts with school homework, activities, sleep...

I dont know, maybe this is a really bad philosophy...i do have to agree with you guys and say that sometimes you should freely go for it, without being restricted by the environment or any other factors.  I also agree that not everything should be about winning.  This contest is a great one for you to enhance your math abilities and have a fun time solving math problems.  However, winning still has to play some motivation, no matter if you believe it or not.  If you do not like winning, then you are not human.

I dont know...I was just trying to analyze Mathemonster's response...I dont know if this is what he meant when he said he wanted to know his rd2 score before starting rd4.[/quote]

Yes you are right; I wanted to know how much i have to do to get silver; prbly on 2-3 problems, and if I can get to AIME via USAMTS

USAMTS has essentially no time limit.

I'm going to say that again.

[i]USAMTS has essentially no time limit.[/i]

[quote="mathemonster"]what sport do you play wawb? because in tennis/squash you always know the score[/quote]Knowing the score is different from analyzing what you need to win. Whenever I play tennis (albeit badly), if I start thinking about what *could* happen (and especially what could go wrong), I do terribly. Maybe it's different for other people.

The only time limit of USAMTS is however much time is in the month (24 hours/day * # of days in the round). Of course, you're not actually going to stay up that long, are you?

[quote="worthawholebean"][quote="mathemonster"]what sport do you play wawb? because in tennis/squash you always know the score[/quote]Knowing the score is different from analyzing what you need to win. Whenever I play tennis (albeit badly), if I start thinking about what *could* happen (and especially what could go wrong), I do terribly. Maybe it's different for other people.[/quote]

That's why in tennis you play point by point and just focus on winning each point individually rather than focusing on winning the match...  That's how you save game, break, set, or even match points. :wink:

As for this discussion over knowing scores in advance, what does it really matter?  Just do your best and know that either way, your mathematical skill will stay just as high for you to do well on the final round. :)

[quote="Math Geek"]That's why in tennis you play point by point and just focus on winning each point individually rather than focusing on winning the match...  That's how you save game, break, set, or even match points. :wink: [/quote]My point exactly.

well this is a bit off topic...but I play tennis differently

I consider the whole match.  For example, if Im up 40-0 on my second serve, Ill sometimes hit my second serve like a first serve since its almost a win-win situation.  Also, if I am getting tired during the match, Ill go for winners every opportunity I get...if I am not, then Ill play a little more conservatively.  Also, if I am down like 5-2 and two breaks, I sometimes will just go for winners and not try to play long points (and thus conserve my energy for the next set)...thats why in professional tennis you get those lopsided scores like 6-1,1-6,6-3...

If I played the same on every point, I would defiently not be as successful.

By the way, I was wondering if a grader or someone else involved with USAMTS could tell me if points would be deducted from #4 for including the two endpoints of the major axis of the ellipse in the locus?

[quote="mathemonster"]No really?  :P
yes i have a good idea of what i got but its annoying they are taking so long[/quote]
try grading 12590214 papers faster than them.
patience is important.

Quite a heated debate going on here. I have to agree with both sides.

[hide]It is extremely important to put everything into every problem you encounter and every test you take. Math without passion may not only be unsuccessful, but more importantly, is meaningless.

However, although I am more inclined to the above argument, the opposing side has a minor point. Putting everything one has into every problem/test could include looking at the big picture. The tennis court examples a few posts ago were particularly illustrative. The sad thing is, we have a limited amount of time and energy in our hands, and we have to use it wisely.

For example, there is a limited amount of time until the upcoming AMC, AIME, and just today I had a local math competition. All three were different formats of tests with different types of questions. The week before was PUMaC. Oh, and there is also USAMTS, and some people I have know have even more upcoming math competitions (like Mu Alpha Theta or HMMT). Now, we could all improve our scores in most of these by practicing for them, but we have a limited amount of time.

So, for example, last week, I had to pick. I have AMC on Tuesday and the local math competition. On top of that, I was falling (am falling) very behind on schoolwork (more specifically, european history), and felt horribly sick such that some school workers told me I should be home. If I knew I would qualify for AIME through USAMTS, then I could take a few practice tests (3 hours long each; proof-writing contest) of my local math competition and get to know the format of it well since this is the first time I participated, and perform to the best of my abilities on it, quite possibly getting first place since it's a local test.

However, making AIME is undoubtedly much more important than this local math competition, and my AMC scores are not the best and have the ability to improve as I've seen the last few days, so I studied for the AMC instead, and only left the time of 9:30-12:30 on Friday night to take one practice test of the local math competition so I know what's going on (and the test was this (Saturday) morning at 9:00). I still think I did well, and maybe got first place, but I think I could have managed my time a little better, taken a calculator as was very necessary, and learned a little more from the types of problems in the competition. I probably would have done so had I made AIME through USAMTS, for I probably would not get higher than 130 (hopefully I'm underestimating myself), but instead I studied a lot for the AMC. Maybe it was for the best, who knows.[/hide]

I guess I've gone for too long here so I'll make my point. We have to treat every problem separate and try our best in each one, and that is more important than getting our results back (anyways, although not always, we can often grade ourselves). But we still have to be as smart as possible and manage our tragically not very long amount of time and energy as efficiently as possible to be the best we can be. Sometimes, knowing your USAMTS score helps to do the latter, but since life is not perfect I think we can manage without it all right.


Assume $ A \in M_{n,n}(\mathbb{F})$[color=red](corrected)[/color] and $ rank(A) < n$.

if for any $ m \in \mathbb{N^{ \plus{} }}$, $ A^{m} \not \equal{} A$

whether or not exists a number $ n \in \mathbb{N^{ \plus{} }}$ such that $ A^{n} \equal{} 0$ ?

 :o

I do not mean the martirx $ A$ is full rank. I think I have confused the notation :huh:  [color=red]thanks Kent Merryfield very much[/color]

Uh, what exactly do you mean by $ gl(n,\mathbb{F})?$

The only thing I can think of for that to mean would be the set of invertible $ n\times n$ matrices, which I would write as $ GL(n,\mathbb{F})$ (or maybe $ \text{GL}(n,\mathbb{F}).$)

But all such matrices have rank $ n,$ so saying $ \text{rank}\,A<n$ looks like nonsense.


integration
parameterization
calculus
derivative
absolute value

hi

i want to show that if

$ f ' \plus{} 2zf \equal{} 0$

then $ f \equal{} ce^{ \minus{} z^2}$, where c is some constant.

so far I have got this:

Assume $ f \equal{} \sum_{n \geq 0} a_nz^n$

So we need to show

$ \sum_{n \geq 1} na_nz^{n \minus{} 1} \plus{} \sum_{n \geq 0} 2a_nz^{n \plus{} 1} \equal{} 0$

So comparing coeffiencts of $ x^n$ we have $ (n \plus{} 1)a_{n \plus{} 1} \equal{} 2 a_{n \minus{} 1} \equal{} 0$

I.e., $ a_{n \plus{} 1} \equal{} \frac { \minus{} 2a_{n \minus{} 1}}{n \plus{} 1}$

Then I looked at odd and even values for $ n$ and got

$ a_{2n \plus{} 1} \equal{} \frac {( \minus{} 1)^n2^n}{(2n \plus{} 1)(2n \minus{} 1) \cdot ... \cdot 1}a_1$

and

$ a_{2n} \equal{} \frac {( \minus{} 1)^n}{n!}a_0$

So if I can show that $ a_{2n \plus{} 1} \equal{} 0$ then I think I'm done... but I get stuck here.

All help very appreciated.

H

$ \frac{df}{dz}\equal{}\minus{}2zf$

$ \Rightarrow \frac{df}{f}\equal{}\minus{}2z dz$

$ \Rightarrow \int\frac{df}{f}\equal{}\int\minus{}2z dz$

$ \Rightarrow \ln|f|\equal{}\minus{}z^2\plus{}C$

$ \Rightarrow f\equal{}ce^{\minus{}z^2}$

The derivation above uses a method commonly called "separation of variables".

[b]Note:[/b] The absolute value signs could be removed because if f>0, then we get $ f \equal{} e^C e^{ \minus{} z^2} \equal{} ce^{ \minus{} z^2}$, and thus, $ c \equal{} e^C > 0$. Similarly, if f<0, then we find that $ c \equal{} \minus{} e^C < 0$. Thus, both possibilities take the same form, and we can use an umbrella parameter for both: $ c \in \mathbb{R}$

Another useful method for these kinds of problems is the introduction of an [url=http://en.wikipedia.org/wiki/Integrating_factor]integrating factor[/url]. In this case:

$ f' \plus{} 2zf \equal{} 0$
$ e^{z^2} f' \plus{} 2ze^{z^2} f \equal{} 0$, where we multiplied through by $ e^{z^2}$, which is always positive.

By reversing the product rule, we can rewrite the left side as the derivative of a product: $ (e^{z^2} f)' \equal{} 0$

Integrating both sides: $ e^{z^2}f \equal{} c, \; \; c \in \mathbb{R} \implies \boxed{f \equal{} ce^{ \minus{} z^2}, \; \; c \in \mathbb{R}}$

In general, with differential equations of the form $ f' \plus{} a(z)f \equal{} b(z)$, it usually helps to multiply both sides by $ e^{\int a(z) dz}$.

[quote="nice_hat_harry"]Thanks to both of you - those methods are really slick. However, I'd REALLY like to know if it can be solved by finishing off the method I started.....[/quote]

When you set the coefficients of the left hand side of the equation to 0, the constant term on the left hand side is $ a_1$. So $ a_1\equal{}0$ and by your formula all odd coefficients are 0.

thanks - but just to clarify, when you say "the equation", which one do you mean?

Your differential equation written in terms of series was:
$ \sum_{n \geq 1} na_nz^{n \minus{} 1} \plus{} \sum_{n \geq 1} 2a_nz^{n \plus{} 1} \equal{} 0$

Separate the case of $ n \equal{} 1$ from the first summation, then factor out $ z$:
$ a_1 \plus{} \sum_{n \geq 2} na_nz^{n \minus{} 1} \plus{} \sum_{n \geq 1} 2a_nz^{n \plus{} 1} \equal{} 0$ $ \iff a_1 \plus{} z \left( \sum_{n \geq 2} na_nz^{n \minus{} 2} \plus{} \sum_{n \geq 1} 2a_nz^{n} \right) \equal{} 0$

The above must hold true when $ z \equal{} 0$; substitute $ z \equal{} 0$ to find that $ a_1 \equal{} 0$

Ahh.. The original equation was actually:
$ \sum_{n \geq 1} na_nz^{n \minus{} 1} \plus{} \sum_{n \geq 0} 2a_nz^{n \plus{} 1} \equal{} 0$
but the same concept carries over as the constant term on the RHS is 0, so the constant term on the LHS, which is $ a_1$ must also be zero!

The Zuton Force, caffeineboy - I love you both.

Thanks  :lol:


counting
derangement
combinatorics proposed
combinatorics

There are $n$ married couples on the dance floor. Determine the number $D(n)$  in which no wife dances with her husband.

Let the couples be ordered from $1$ to $n$ in any way.
Let $N_k$ be the number of constellations where the first $k$ (and also possibly some more) couples are together. Thus, you have $N_k=(n-k)!$. The number of constellations were some given $k$ (and also some more) couples are together is independent of these couples, only dependent on $k$ so also equal to $N_k$. By principle of inclusion-exclusion (used on the sets $E_i$ of those constellations in that the couple number $i$ is together) you get: $D(n)=\sum_{k=0}^n(-1)^k\cdot \frac {n!} {k! \cdot (n-k)!} \cdot N_k=\sum_{k=0}^n(-1)^k\cdot \frac {n!} {k! \cdot (n-k)!} \cdot (n-k)!=n! \cdot \sum_{k=0}^n(-1)^k\cdot \frac {1} {k!}$

(These numbers are the so-called "derangements")

I better understand the answer than the question  ;)

Pierre.