What is the value of the arithmetic series $ 28 \plus{} 30 \plus{} 32 \plus{} \ldots \plus{} 86$?

Consider the arithmetic series:
\[ a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_n
\]
The value of the arithmetic series is:
\[ \frac {a_1 \plus{} a_n}{2} \times \left (\frac {a_n \minus{} a_1}{a_2 \minus{} a_1} \plus{} 1 \right)
\]
Thus, the value of that series is $ \boxed{1710}$.

Or, if you don't know the formula

Notice that we can pair the numbers as follows: \[ (28\plus{}86)\plus{}(30\plus{}84)\plus{}(32\plus{}82)\plus{}\cdots \plus{} (56\plus{}58)\]  So there are a bunch of pairs with a sum of $ 114$, so we just need the number of pairs.

The number of numbers in the list \[ 28,30,32,...,86\] doesn't change if we divide by $ 2$, so the number of numbers is the same as the number of numbers in the list \[ 14,15,16,...,43\]  Subtracting $ 13$ from all the numbers then gives us \[ 1,2,3,...,30\] There are clearly $ 30$ numbers in this list, so there are $ \frac{30}{2}\equal{}15$ pairs.

Finally, our answer is $ 114\cdot 15\equal{}\boxed{1710}$


geometry
circumcircle
incenter
power of a point
radical axis
geometry unsolved

Let $ A_0,B_0,C_0$ respectively be the midpoints of $ B_1C_1,C_1A_1,A_1B_1$. Denote $ r$ by the radii of $ \triangle ABC$. Consider the inversion through pole $ I$, power $ r^2$, we have $ \mathcal {I}(I,r): A\mapsto A_0$, $ B\mapsto B_0$, $ C\mapsto C_0$, $ A_1,B_1,C_1$ is mapped to itself. Hence $ \mathcal {I}(I,r): (AIA_1)\mapsto A_1A_0$, $ (BIB_1)\mapsto B_1B_0$, $ (CIC_1)\mapsto C_1C_0$. But $ B_0B_1,C_0C_1,A_0A_1$ are the medians of $ \triangle A_1B_1C_1\Longrightarrow B_0B_1,C_0C_1,A_0A_1$ are concurrent. Therefore $ (AIA_1),(BIB_1),(CIC_1)$ are coaxal. The result is lead as follow.
Our proof is completed then.

:D Nice proof...
And Inversion still keeps on amazing me...

Dear Mathlinkers, the circumcircles are concurrent in a second point named "the Schröder's point".
Sincerely
Jean-Louis

[quote="Xaenir"]$ I$ is the incenter of a non-isosceles triangle $ \Delta ABC$ .If the incircle touches $ BC,CA,AB$ at $ A_1,B_1,C_1$ respectively, prove that the circumcenter of the triangles $ \Delta AIA_1$,$ \Delta BIB_1$,$ \Delta CIC_1$ are collinear.[/quote]
$ \frac{O_aM_b}{O_aM_c}\equal{}\frac{A_2B}{A_2C}\equal{}\frac{AB}{AC}$

By Menelaus's theorem for $ \triangle M_aM_bM_c$...we have done.

Posted many times before, e.g; see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=221942]An old problem[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=181827]Incenter and 3 circles[/url] and elsewhere.

[quote="jayme"]Dear Mathlinkers, the circumcircles are concurrent in a second point named "the Schröder's point". [/quote]
I found three more concurrent circles at the Schröder's point. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1489741#1489741]Circles through the Schroder point[/url].

Let G be the centroid of triangle $ A_1B_1C_1. A_1G, B_1G, C_1G$ cut $ (I)$ at $ A_3, B_3, C_3$, respectively.
We have $ A_2A.A_2I\equal{}A_2C_1.A_2B_1\equal{}A_2A_1.A_2A_3 \Rightarrow AA_1IA_3$ is cyclic.
Similarly $ BB_1IB_3$ is cyclic
Moreover, $ GA_1.GA_3\equal{}GB_1.GB_3$ then G lies on the radical axis of $ (I_a)$ and $ (I_b)$
$ \Rightarrow I_aI_b\perp IG$, similarly $ I_bI_c\perp IG$ so $ I_a,I_b,I_c$ are collinear.


A motorboat traveling across a wide river is headed 50 degrees north of
 west at a constant speed of 10 mi/hr.  The river current is 7 mi/hr
 due west.  How far west does the motorboat travel in 3 minutes?  How far
 north does it travel in 3 minutes?

Do you have a solution to this, or are you trying to get someone else to do your work? I've seen countless of these routine problems come from you.

Post a solution (hidden preferably), then I'll work on it.


geometry
circumcircle
cyclic quadrilateral
power of a point

Let ABCD be a cyclic quadrilateral with AC perpendicular to BD. As well, let E denote
the intersection of AC and BD. Given that AE, BE, CE, and DE are the roots of $ (x^2 \minus{} 4x \plus{} 2)(x^2 \minus{} 6x \plus{} 2) \equal{} 0$ (which turns out to be $ 2\pm\sqrt2,3\pm\sqrt7$), what is the square of the circumradius of ABCD?

[hide="Sketch"]
By power of a point, we can assume WLOG $ AE \equal{} 2 \minus{} \sqrt {2}$, $ CE \equal{} 2 \plus{} \sqrt {2}$, $ BE \equal{} 3 \plus{} \sqrt {7}$, $ DE \equal{} 3 \minus{} \sqrt {7}$
By Pythag, we can find $ AD$ and $ AB$
We can easily find the area of $ \triangle ABD$ using the formula $ area \equal{} \frac {base \cdot height}{2}$
Now we have in any triangle area$ \equal{} \frac {abc}{4R}$, where $ a$,$ b$, and $ c$ are the sidelengths and $ R$ is the circumradius. Apply this to $ \triangle ABD$ to get $ R$.[/hide]
[hide="Answer (After a ton of algebra)"]$ 44$[/hide]

Well, I think the answer above is wrong, namely the value of $4R^2$!
Indeed, as above we get $AC=4, BD=6$ and, if $M, N$ are midpoints of $AC, BD$ respectively and $O$ - the circumcenter, we easily get $EM=\sqrt{2}, EN=\sqrt{7}$; as $AC\bot BD\implies OE^2=EM^2+EN^2=9$.
Next, see that $R^2=AE\cdot CE+OE^2=2+9=11$.

Best regards,
sunken rock


when i put $\boxed{\sqrt{3}}$, i get "Unparseable or potentially dangerous latex formula" :? (lol @ the "potentially dangerous")

Try \fbox{$\sqrt{3}$} (this works for me).

But can you get it to work on the forums, jhaussmann?

No, but it does work when making your own document.  :(  You'd have to ask the admins why.

$\boxed{\sqrt{3}}$

Use \boxed in math mode and \fbox in text mode.

$\boxed{\sqrt{3}}$

Cool deal. I like boxes.

so wait, in TexnicCenter, use /fbox, but here use /boxed?

Everywhere. \fbox is only for text mode. \boxed is for math mode.

Woo hoo.  I finally know how to emphasize which part of my solution is the answer! :D

\[
2 + 2 = \boxed{5}
\]

You'll get the hang of this addition stuff in no time. I have faith in you.


calculus
derivative
vector
analytic geometry
function
topology
linear algebra

I noticed in the linear algebra book that I'm teaching from right now (Larson, Edwards, and Falvo) a question asking students to prove that there are no 2x2 matrices A,B such that AB - BA = I.

This is not so hard to do with brute force.  It becomes a wee bit trickier to prove it for the 3x3 case and the 4x4 case, but those cases also work out nice and pat with an easy contradiction along the main diagonal.

To generalize this, however, it seems it *should* be fairly direct to attack it with the Lie bracket, and interpret it somehow as a Lie derivative of two vector fields.  The statement would then say that the Lie derivative of one vector field with respect to another cannot be the identity.  Proving this by going into the differential coordinate functions and whatnot seems to me to be not so far off from a brute force approach.  It will probably work, but frankly (my dear), I don't give a damn.  I want to understand how this is true from more of a differential *topology* perspective.  What is it about the change of one vector field along another in this "particular" topological space (R^n, I believe, if we interpret a GL(n) matrix as a vector field on R^n) that restricts it from being the identity?

Of course, I'm having enough difficulty visualizing why the Lie derivative of X along I (or vice versa) is trivial (0).  It seems to me that, for example, if we shove any vector field X straight outward from the origin (along the flow of the vector field I), we're getting scalar multiples of X.  So are we actually normalizing the scaling that I is providing, thus keeping X stable, and that's why the Lie derivative is 0?

Partly I'm looking for a more elegant solution to nonexistence of solutions to AB-BA=I for matrices A,B in GL(n) (or M(n,n) if you can find a particuarly sneaky approach)----but partly I'm also looking for a better understanding of how the Lie derivative is futzing with basic vector fields.

I'd say the "tricky" solution would be to note that $ tr(AB)\equal{}tr(BA)$ and also that $ tr(A\plus{}B)\equal{}tr(A)\plus{}tr(B)$, so $ tr(AB\minus{}BA)\equal{}tr(AB)\plus{}tr(\minus{}BA)\equal{}tr(AB)\minus{}tr(BA)\equal{}0$, but on the RHS $ tr(I)\neq 0$.

The statement is true for $ A, B: E\rightarrow E$ be bounded linear operators.  The proof is via the identity $ A^nB\minus{}BA^n\equal{}nA^{n\minus{}1}$ provided $ AB\minus{}BA\equal{}id$ holds.

[quote="HilbertThm90"]I'd say the "tricky" solution would be to note that $ tr(AB) \equal{} tr(BA)$ and also that $ tr(A \plus{} B) \equal{} tr(A) \plus{} tr(B)$, so $ tr(AB \minus{} BA) \equal{} tr(AB) \plus{} tr( \minus{} BA) \equal{} tr(AB) \minus{} tr(BA) \equal{} 0$, but on the RHS $ tr(I)\neq 0$.[/quote]

There are some cases when $ Tr I\equal{}n\equal{}0$. Something like positive characteristic. But it seems that characteristic 0 is assumed in the hypothesis.


integration
trigonometry
calculus
calculus computations

If $ I_{m,n}$ denotes $ \int_{0}^{\frac{\pi}{2}} \sin^{m}\theta \cos^{n}\theta\ d\theta$, prove that $ I_{m,n} \equal{} \frac{n\minus{}1}{m\plus{}n}I_{m,n\minus{}2}$. By using a suitable substitution, show also that $ I_{m,n} \equal{} I_{n,m}$, and deduce that $ I_{m,n} \equal{} \frac{m\minus{}1}{m\plus{}n}I_{m\minus{}2,n}$.

Find (a) $ \int_{0}^{\frac{\pi}{2}} \sin^{8}\theta \cos^{9}\theta\ d\theta$, (b) $ \int_{0}^{\frac{\pi}{2}} \sin^{9}\theta \cos^{10}\theta \ d\theta$, (c) $ \int_{0}^{\frac{\pi}{2}}\sin^{8}\theta \cos^{10}\theta \ d\theta$.

hello
$ I_{m,n} \equal{}\int_{0}^{\frac{\pi}{2}}\sin^{m}\theta\cos^{n}\theta\ d\theta$
Let $ P\equal{}\sin^{m\minus{}1}\theta\cos^{n\plus{}1}\theta$
$ \frac{dP}{d\theta}\equal{}(m\minus{}1)\sin^{m\minus{}2}\theta\cos^{n}\theta\minus{}(m\plus{}n)\sin^{m}\theta\cos^{n}\theta.......(1)$
so
$ (m\plus{}n)\int_{0}^{\frac{\pi}{2}}\sin^{m}\theta\cos^{n}\theta d\theta\equal{}(m\minus{}1)\int_{0}^{\frac{\pi}{2}}\sin^{m\minus{}2}\theta\cos^{n}\theta d\theta\minus{}[\sin^{m\minus{}1}\theta\cos^{n\plus{}1}\theta]^{\frac{\pi}{2}}_{0}$
$ I_{m,n}\equal{}\frac{m\minus{}1}{m\plus{}n}I_{m\minus{}2,n}$
for second proof ::
use property
$ \int_{0}^{a}f(x)dx\equal{}\int_{0}^{a}f(a\minus{}x)dx$
for proving first part Let $ P\equal{}\sin^{m\plus{}1}\theta\cos^{n\minus{}1}\theta$
better to Use
[u][b]WALLI'S FORMULA[/b][/u] for these kind of problem .
thank u


The question is:
Let $ B\equal{}[b_{ij}]$ with $ b_{ij}\equal{}0$ if $ i>j$ or $ i<j\minus{}1$, $ b_{ii}\equal{}a_{ii}>0$ and $ b_{i,i\plus{}1}\equal{}\minus{}a_{i,i\plus{}1}<0$.
Show that the elements of $ (B^TB)^{\minus{}1}$ are positive.
Proof: $ (B^TB)^{\minus{}1}\equal{}B^{\minus{}1}(B^T)^{\minus{}1}$. $ B^{\minus{}1}\equal{}[c_{ij}]$ where $ c_{ij}\equal{}0$ if $ i>j$ and else $ c_{ij}\equal{}\dfrac{a_{i,i\plus{}1}a_{i\plus{}1,i\plus{}2}\cdots{a}_{j\minus{}1,j}}{a_{ii}a_{i\plus{}1,i\plus{}1}\cdots{a}_{jj}}$. $ B^{\minus{}1}$ is a full upper triangular non negative matrix and by the same way $ (B^T)^{\minus{}1}$ is a full lower triangular non negative matrix. Their product  is a positive matrix.


At the mid-point of line segment AB which is p units long, a perpendicular MR is erected with length q units. An arc is described from R with radius equal to half of AB, meeting AB at T. Then, AT and TB are the roots of which quadratic equation?


Find the biggest $ m$ such that $ a^3 \plus{} b^3 \plus{} c^3 \plus{} m(a^2b \plus{} b^2c \plus{} c^2a) \ge (m \plus{} 1)(ab^2 \plus{} bc^2 \plus{} ca^2)$  $ \forall a,b,c \in [0;\infty)$

[quote="Mathias_DK"]Find the biggest $ m$ such that $ a^3 \plus{} b^3 \plus{} c^3 \plus{} m(a^2b \plus{} b^2c \plus{} c^2a) \ge (m \plus{} 1)(a^2b \plus{} b^2c \plus{} c^2a)$  $ \forall a,b,c \in [0;\infty)$[/quote]
but this is equivalent to $ a^3\plus{}b^3\plus{}c^3\geq a^2b\plus{}b^2c\plus{}c^2a$... maybe there is a typo somewhere?

[quote="limes123"]
but this is equivalent to $ a^3 \plus{} b^3 \plus{} c^3\geq a^2b \plus{} b^2c \plus{} c^2a$... maybe there is a typo somewhere?[/quote]
Thanks for pointing out the mistake :)

The number is
[hide="This"]The positive root in $ \minus{}27m^4 \minus{} 54m^3 \plus{} 135m^2 \plus{} 162m \plus{} 621$ which is $ 2.48444$![/hide]


algebra
polynomial
Vieta
algebra proposed

f(x) is a polynomial of degree 10, and g(x) is a polynomial of degree 2. Can the roots of f(g(x)) be 1, 4, 9, 16, 25, ..., 361, 400?

the new degree is $10.2$

So its correct

A simple Case

${\{n,k\}\in \mathbb{Z}}$
${\{n,k\}>0}$
${f(x)=a x^{n}}$
${g(x)=\beta x^{k}}$

The degree of $f$ is $n$ and the degree of $g$ is $k$

${f(g(x))=a \left(x^{k}\beta \right)^{n}=a x^{k n}\beta^{n}}$
${g(f(x))=\left(a x^{n}\right)^{k}\beta =a^{k}x^{k n}\beta }$

There exist no such $f,g\in\mathbb{C}[x]$.
Let $\lambda\in\mathbb{C}$ be some root of $f(x)$, $0\not =c\in\mathbb{C}$ the leading coefficient of $g(x)$
and set $\tilde{f}(x): =f(cx+\lambda)$ and $\tilde{g}(x): ={1\over c}(g(x)-\lambda)$.
Then $\tilde{f}(\tilde{g}(x))=f(g(x))$ with $\tilde{f}(0)=0$ and $\tilde{g}(x)$ monic.
So after replacing $f,g$ by $\tilde{f},\tilde{g}$ we may assume $f(0)=0$ and $g(x)$ monic.

Now if $\lambda_{1}=0,\lambda_{2},\ldots,\lambda_{r}$ are all the roots of $f(x)$, then $r\leq \deg f=10$
and each equation $g(x)=\lambda_{i}$ has at most $2$ solutions, as $\deg g\leq 2$.
But $f(g(x))=0$ has $20=10\cdot 2$ different integer solutions. So we must have $r=10$
and each equation $g(x)=\lambda_{i}$ has exactly $2$ solutions in the set $\{1^{2},2^{2},\ldots,20^{2}\}$.
In particular there exists $a,b\in\{1,2,\ldots,20\},a\not =b$ with $g(a^{2})=g(b^{2})=\lambda_{1}=0$.
So $g(x)=(x-a^{2})(x-b^{2})\in\mathbb{Z}[x]$, as $g$ monic. Set $t: ={a^{2}+b^{2}\over 2}$.
Then $g$ is strictly decreasing in $(-\infty,t]$ and strictly increasing in $[t,\infty)$.
So in case $t\geq 11^{2}$ we have $g(1^{2})>g(2^{2})>\cdots>g(11^{2})$
and in case $t\leq 10^{2}$ we have $g(10^{2})<g(11^{2})<\cdots<g(20^{2})$,
in both cases $11$ different roots of $f$, which contradicts $\deg f=10$.
So $10^{2}<t<11^{2}$ and $g(1^{2})>g(2^{2})>\cdots>g(10^{2})$ and
$g(11^{2})<g(11^{2})<\cdots<g(20^{2})$ both run through all the different roots of $f$.

This now gives $g(1^{2})=g(20^{2}),g(2^{2})=g(19^{2}),\ldots,g(10^{2})=g(11^{2})$,
and by Vieta $1^{2}+20^{2}=2^{2}+19^{2}=\cdots=10^{2}+11^{2}$. Contradiction!

This is my solution:

That the roots of f(g(x)) are 1, 4, 9, ..., 400 means that g(1), g(4), g(9), ..., g(400) are only 10 distinct numbers - the roots of f(x). This means that g(1) = g(400), g(4) = g(361), g(9) = g(324), ..., g(100) = g(121). Define h(x) = g(x)-g(1). The roots of h(x) are 1 and 400, so h(x) = c(x-1)(x-400) where c is not 0. g(4)=g(361), so h(4) must equal h(361); but h(4)=-1188c, h(361)=-14040c. Contradiction.
Is my solution OK?
Based on your idea, I can simplify my solution - g(1)=g(400), g(4)=g(361). But this can be true only if 1+400=4+361, which is false.


Just from thinking about world affairs and societal or even familial issues, is it?

Strictly speaking: no. I suggest that we apply common sense: if you burn yourself while making coffee and no one else learns of it, a net decrease of happiness results. This contradicts the notion that happiness is a zero-sum game.

Alternatively, think of romantic relationships: if feelings are mutual, then there will be a net increase of happiness as both parties gain happiness simultaneously. Again, this contradicts the notion that happiness is a zero-sum game. Keep in mind that happiness is not a resource; it can be generated and depleted at will without "flowing" from one person to another.

I can provide a satirical response demonstrating the opposite, if needed.

In fact, I'm quite sure happiness is more likely to be loss-loss or win-win than zero-sum. That's because what different people consider happiness is usually different and likely to be complementary.

I would imagine it is very independent (assuming you could quantify happiness. what would the unit even be?) although if you want to be sketchy you could look at it from a "what-could-have-been" thing.

Example of that: I find a penny on the ground. Yay I got +X happies. However, each person had an expected value of gaining X/N (where n is the total number of people in the world), so my gain in happies is actually X(N-1)/N and everyone else gets -X/N so its 0.

BUt happiness can't be quantified so its a moot point.

When Buddha achieved enlightenment, I'm sure nobody felt sad about it.

[quote="Mewto55555"]I would imagine it is very independent (assuming you could quantify happiness. what would the unit even be?) although if you want to be sketchy you could look at it from a "what-could-have-been" thing.[/quote]In my economic class my teacher said that sometimes the term "utils" is used to measure satisfaction (so sorta like happiness.) But no one really uses it so I guess it doesn't matter much.


Let ABC be a triangle. Prove that:

 \sqrt a *la+ \sqrt b *lb+ \sqrt c *lc \leq 2p \sqrt p

Use the fact that l_a<=  \sqrt p(p-a). Thus,  \sum  \sqrt a*l_a \leq  \sum  \sqrt a*a*(p-a) \leq  \sqrt  \sum a * \sum a(p-a) \leq 2p \sqrt p


Prve that this equation has finite roots$ \in N*$
$ a^b\plus{}b^c\plus{}c^d\plus{}d^a\equal{}a^d$

No one can solve it,ZeTax,where are you :o

None can? :maybe: ,Really,I think it is very interseting(but my proof is very long) and I wait a nice soluion for it


I stumbled upon this link in Gmail:

http://www.relativitychallenge.com/index.htm

This person thinks they've found an error in Einstein's theory of Special Relativity.

I haven't carefully perused everything on the site, but it seems like pseudoscience to me.

Comments?

:huh:
Theories like this pop up every so often, someone just wants to be "The One" that disproved Einstein.
Though this one seem quite similar to SRT
[quote]A Complete Coordinate System is one in which the phenomena under observation (or its associated travel medium) is contained within the coordinate system and moves with it at velocity v.

An Incomplete Coordinate System is one in which the phenomena under observation (or its associated travel medium) is not completely contained within the coordinate system and does not move with it at velocity v.
[/quote]
That seems to match with Einstein's theory, in it being that events occur on different "time slices" unless travelling the same speed.
It is also funny how they attempt to find flaws in experiments that were "proved" Relativity, instead of using new experiments.
Like one, for example, the Michelson Morley, which has been repeated many times, to prove there does not exist an ether through which events occur.

In science it is still important to be open to new theories, like String Theory ( :rotfl: ), for it to progress, BUT like fraud lawsuits in district courts, fraud theories just disrupt the system.

Looks like a load of BS to me. On this page, http://www.relativitychallenge.com/math1905Mistake.htm, it seems that the author ignores the fact that there are some substitutions that can be made so that both sides are actually equal. It appears that $x = Vt$ is the necessary equation to resolve the "error." I don't know what the variables actually are, but probably distance = velocity * time. And there's obviously no way Einstein and millions of physicists/mathematicians have missed an error so trivial.

Didn't someone famous disprove small portions of the Theory of Special Relativity years ago?

You may want to look up "Doubly special relativity" on Wikipedia or somewhere else. Lee Smolin seems to be famous enough, indeed.

Lee Smolin.  :)  Nice guy that writes great books.


What is $ \dbinom{n}{0}$ for any positive integer $ n$?

$ \binom{n}{0}$ is just $ \frac{n!}{0!n!}\equal{}\frac{1}{0!}\equal{}\boxed{1}$ for any positive integer $ n$.

Alternatively, there is only $ \boxed{1}$ to choose $ 0$ objects from a group of $ n$.


limit
integration
real analysis
real analysis unsolved

Prove that there is only one solution, $\psi_{0}(.)$ to the following equation

                                           \[ x'=\frac{t^{2}+(1+2t^{2})x}{t} \]

  such that there is $\lim_{t\to\infty}\psi_{0}(t)\in\mathbb{R}$.

Subtitute $x$ by $ty$  gives  $y'=1+2ty$ (E)
Solving $y'=2ty$ gives $y(t)=Ae^{t^2}$
Plug $y_{0}(t)=A(t)e^{t^2}$ in (E) gives $A(t)=\int_{a}^{t}e^{-z^2}dz$
General solution of (E) are :$y(t)=(A+\int_{a}^{t}e^{-z^2}dz)e^{t^2}$
Thus $x(t)=(A+\int_{a}^{t}e^{-z^2}dz)te^{t^2}$
If $A\neq0$ it's clear that $\lim_{t\to\infty}x(t)\not\in\mathbb{R}$
So $x(t)=(\int_{a}^{t}e^{-z^2}dz)te^{t^2}$
Now how to choose $a$ such that :$\lim_{t\to\infty}x(t)\in\mathbb{R}$ :?:


(IMO) Let ABC be a triangle with circumcentre O. The points P and Q are interior points
of the sides CA and AB, respectively. Let K, L and M be the midpoints of the segments BP, CQ
and PQ, respectively, and let TT be the circle passing through K, L and M. Suppose that the line
PQ is tangent to the circle TT. Prove that OP = OQ.

IMO Germany 2009 http://www.mathlinks.ro/viewtopic.php?t=288839


Hallo!

   Have you seen the new book by Titu Andreescu and Andrica on Complex numbers(from A to Z)? It has many good problems and it is  very good for an introduction to this area( school students - teachers -math competitions and  first year students).
                Enjoy it !!!

 babis

Could you guys give me a review? Always interested in adding books to my collection.

Wow! :lol:
for all I know the authors are distinguished mathematicians!
So their work must be an excellent one!
which is the editor/publishing house??

I think the editor is [b]Birkhauser[/b].

  The book is very nice for high school students. I have also the Romanian book for this area by the same autors, but this one , written in english  and with Titu Andreescu as one more autor , is more rich. All problems are solved or have good hints. Some problems are of IMO level and are solved with complex Geometry.

     [u]babis[/u]


The Wake Forest University regional contests were held today; here are the results that I can remember (official results aren't released for about 1 month):

[hide="Algebra I"]1)  Richard Sun - Hanes Middle School (39/40)
2)  Vipul Vachharajani - Hanes Middle School (36/40)
3)  Alice Wong - Guy B. Phillips Middle School (33/40)
4)  An Li - Hanes Middle School(31/40)

School Winner:  Hanes Middle School[/hide]

[hide="Geometry"]1)  Daniel Tong (33/40)
2)  Taylor Anderson - North Davie Middle School (31/40)
3)  Daniel Nees - North Davie Middle School(29/40)
4)  Brad Neal - North Davie Middle School(28/40)
4)  Jimmy Jong - Phillips Middle School(28/40)

School Winner:  North Davie Middle School[/hide]

[hide="Algebra II"]1)  Bryce Taylor - Hanes Middle School (36/40)
2)  Michelle Phillips - Davie High (30/40)
3)  Powwel Upman (29/40)
4)  Austin March - R. J. Reynolds High School(23/40)

The person who wrote the Algebra II contest said that he had made it too hard (it was difficult).

School Winner:  Reynolds High School[/hide]

EDIT:  Filled out schools & some more names

yay U WON!

go bryce

better than you did at mathcounts :|

Yay! me getted 2nd place in Alg I! Me not so bad! GOOOOOOO BRYCE AND THE DRAGONS!!


Linspire Linux is a fairly famous distro, highly recommended for ease of use. They have decided to show everyone how great Linspire is by letting people get it free until midnight, September 6th. Hurry and get it! (Due to the extreme popularity of this offer, it is recommended that you be persistent and try to get it between 11PM-5AM PST - it took me several tries before I succeeded.)

Here is the link to the page: http://info.linspire.com/freespire/index.html

It's a great way to start your Linux Adventure!

Billy

Better read [url=http://www.linux.org/lessons/beginner/toc.html]this[/url] also.

[quote="solafidefarms"]But really all you need to know is that Linux will overwrite everything on the partition you install it on. Other then that, its pretty self-explanitory; all you need is a internet connection to figure out what to do when it isn't perfectly clear.[/quote]

Do you want to pull out the link of the partition-resizer, or should I?

EDIT:[url=http://www.softpedia.com/get/System/Hard-Disk-Utils/Partition-Resizer.shtml]I will[/url]

But Linspire has its own partitioner...!

Who in their right mind woudl use lispire?

If you want something user-friendly, use Ubuntu or something. Comes 100% free and GPL'd every single day of the year.

Personally, I use Gentoo, although unless you can someone to maintain the system, it's not recommended to newbies.


integration
logarithms
function
calculus
algebra
rational function
partial fractions

I am currently writing an essay about numerical method

I have covered all the topics except differential equation which Korean syllabus does not cover

but now I am stuck with differential eqaution that describes the free fall of the body:

$ \frac {dv}{dt} = - g + cv^{r}$

my teacher told that this equation cannot be solved exactly

I don't get what 'cannot be solved exactly' means..

I solved it anyway which might be wrong..

$ \frac {dv}{dt} = - g + cv^{r}$

$ \frac {1}{ - g + cv^r} \frac {dv}{dt} = 1$

$ \int \frac {1}{ - g + cv^r} \frac {dv}{dt}\, dt = \int \, dt$

$ \ln {\left| - g + cv^r\right|} = t + C$

$ - g + cv^r = \pm e^{t + C}$

$ cv^r = \pm e^{t + C} + g$

$ v = \sqrt [r]{\frac {\pm e^{t + C} + g}{c}}$

I sense some incorrectness

That is not a correct antiderivative, and the antiderivative there is not an elementary function in general.

Yes you have made a mistake in the evaluation of the integral:  $ \int \frac{1}{\minus{}g \plus{} cv^r} \,dv$

Remember that: $ \int \frac{f'(x)}{f(x)} \,dx \equal{} log|f(x)| \plus{} C$

So $ \int \frac{rcv^{r\minus{}1}}{\minus{}g \plus{} cv^r} \,dv \equal{} log|\minus{}g \plus{} cv^r|$.

I think you have just become too familiar with applying the case of the linear denominator too often (and even then in the case $ r\equal{}1$, the logarithm should be multiplied by a constant factor $ \frac{1}{c}$.)

$ \int \frac {1}{ \minus{} g \plus{} cv^r} \frac {dv}{dt}\, dt \equal{} \int \, dt$

$ \ln {\left| \minus{} g \plus{} cv^r\right|} \equal{} t \plus{} C$

I think this is the mistake. Since you're integrating with respect to v, it isn't just $ \ln{x}$. This is by the Chain Rule,
since differentiating what you would not give you what you want which is
$ \frac {1}{ \minus{} g \plus{} cv^r}$

but you want $ \frac {1}{ \minus{} g \plus{} cv^r}\frac {dv}{dt}$ which I can't find an explicit formula on my paper...

Anyways, I think I remember my lectures on this were about $ r \equal{} 2$. This is based on experimental evidence.

For that, $ \minus{} {arctanh} \left( {\frac {cv}{\sqrt {gc}}} \right) {\frac {1}{ \sqrt {gc}}}$.

[quote="Fermat's Little Turtle"] This is by the Chain Rule,
since differentiating what you have would yield
$ \frac {1}{ \minus{} g \plus{} cv^r}$[/quote]
Differentiating what he has would yield the expression sludgethrower gave.

When $ r$ is an integer, the integrand is a rational function and does, in fact, have an elementary antiderivative (partial fractions), but it is difficult to concisely describe as a function of $ r$.

Ahh my apologies, typing mistake. I fixed that :P
Thanks t0rajir0u :)

If $ r$ is a positive integer then we can use partial fractions as follows. I have redefined the constants, to make the problem easier to read.

$ \frac {1}{v^r - c} = \prod_{n = 0}^{r - 1} \frac {1}{v - ce^{i\frac {2 \pi n}{r}}}$.

We can now express this in partial fractions as:

$ \prod_{n = 0}^{r - 1} \frac {1}{v - ce^{i\frac {2 \pi n}{r}}} = \sum_{n = 0}^{r - 1} \frac {A_n}{v - ce^{i\frac {2 \pi n} {r}}}$

Multiplying through by the denominator of the left hand side:

$ 1 = \sum_{n = 1}^{r - 1} A_n \prod_{j = 0, j \neq n}^{r - 1} \left(v - ce^{i\frac {2 \pi j}{r} \right)}$

Setting $ v = ce^{i\frac {2 \pi k}{r}}$ where $ k \in \left(0,1,...,r - 1 \right)$ simplifies this expression to:

$ 1 = \prod_{j = 0, j \neq k}^{r - 1} A_k \left(ce^{i\frac {2 \pi k}{r}} - ce^{i\frac {2 \pi j}{r} \right)}$

Thus $ A_k = \prod_{j = 0, j \neq k}^{r - 1} \frac {1}{ce^{i\frac {2 \pi k}{r}} - ce^{i\frac {2 \pi j}{r}}}$

Now we have the partial fraction decomposition of $ \frac {1}{v^r - c}$

$ \int \frac {1}{v^r - c} = \sum_{n = 0}^{r - 1} \int \frac {A_n}{v - ce^{i\frac {2 \pi n}{r}}} = \sum_{n = 0}^{r - 1} A_n \log(v - ce^{i\frac {2 \pi n}{r}}) + C$

The complex logarithm is defined up to an additive integer multiple of $ 2 \pi i$. but as long as we have a consistant additive multiple for each logarithmic term, this causes no problems. They just "add" to the complex constant of integration. So we may assume, without loss of generality, that our integer multiple is 0.

We can probably express this enitrely in terms of real functions, by attempting to pair off our logarithms according to conjugate arguments. Also the fact the logarithms in the sum that have conjugate arguments, also have conjugate coefficients would probably be useful i.e. $ A_n^* = A_{r - n}$. But I am way to lazy to try this myself.


Let be given a semicircle with the diameter $ AB$, and points $ C,D$ on it such that $ AC\equal{}CD$. The tangent at $ C$ intersects the line $ BD$ at $ E$. The line $ AE$ intersects the arc of the semicircle at $ F$. Prove that $ CF<FD$.

$ OC\perp AD$ and $ CE\perp CO$ then $ AD//CE$ we get $ \angle CED\equal{}90^o$
$ \angle FCD\equal{}\angle FAD\equal{}90^o\minus{}\angle AED\equal{}\angle CEF$
$ \angle CDF\equal{}\angle CAF$
We need to prove that $ \angle CEF>\angle CAF \Leftrightarrow CA>CE$
$ \Leftrightarrow CD>CE$
It's true because $ \angle CDE<90^o\equal{}\angle CED$


[b]Suppose 4 friends go to a carnival and each plays a game in which they can win $ 2, win$1, break even (win or lose nothing), lose $ 1, or lose$2. In how many different ways can the friends collectively break even? [/b]

Basically you want the coefficient of $ x^0$ (constant term) in the expansion of
$ ( x^{ \minus{} 2} \plus{} x^{ \minus{} 1} \plus{} x^{0} \plus{} x^{1} \plus{} x^{2} )^4$ :)

Just an explanation how the above expression comes: What we want is the number of ways you can choose these variables such that

$ x\plus{}y\plus{}z\plus{}w \equal{} 0$ where each of these variables range between [-2,2].

However, since this question involves only 4 variables. just find all unordered pairs satisfying the condition and multiply by $ 4!$. You can probably do this faster than calculating binomial co-efficients.

[hide="Solution."]
You can also do it the following way; for simplicity, just add $ 2$ to the amount of money won or lost and make the desired target $ 8$.  Then you can create the following chart where the number in a particular position is the sum of the five numbers above it and to the left.  That is, define $ f(n, x)$ as the amount of ways for $ x$ friends to win $ n$ money.  Then $ f(n, x)$ = $ f(n, x \minus{} 1)$ $ \plus{}$ $ f(n \minus{} 1, x \minus{} 1)$ $ \plus{}$ $ f(n \minus{} 2, x \minus{} 1)$ $ \plus{}$ $ f(n \minus{} 3, x \minus{} 1)$ $ \plus{}$ $ f(n \minus{} 4, x \minus{} 1)$.

Friend 1: 1	1	1	1	1	0	0	0	0
Friend 2: 1	2	3	4	5	4	3	2	1
Friend 3: 1	3	6	10     15     18      19     18      15
Friend 4: 1	4	10	20	35	52	68	80	85

So, from the chart, there are $ \boxed{85}$ ways to break even.
[/hide]


inequalities
trigonometry
geometry
cyclic quadrilateral
trig identities
Law of Sines
inequalities unsolved

In a cyclic quadrilateral of sides $a,b,c,d$ and diagonals $p,q$ prove that

$ \frac{a}{c} + \frac{c}{a}  > \frac{p}{q} + \frac{q}{p} $

unless $ a=c$

I don't know if it is easy or difficult to solve but it was published by Durell in1950 and I have never seen a solution.

Let [tex]ABCD[/tex] be a cyclic quadrilateral, and let [tex]AB = a, BC = b, CD = c, AC = p,[/tex] and [tex]BD = q[/tex].  Pick some point [tex]P[/tex] on arc [tex]DA[/tex], and label accordingly [tex]m\angle{BPA} = \alpha, m\angle{CPB} = \beta, m\angle{DPC} = \gamma[/tex]. Then, by the Law of Sines what we are trying to prove is equivalent to showing

[tex]
\displaystyle{
\frac{\sin{\alpha}}{\sin{\gamma}} + \frac{\sin{\gamma}}{\sin{\alpha}} \ge
}[/tex] [tex]\displaystyle{
\frac{\sin{(\alpha+\beta)}}{\sin{(\gamma+\beta)}} + \frac{\sin{(\gamma+\beta)}}{\sin{(\alpha+\beta)}}
}[/tex]

[tex]\forall \alpha, \beta, \gamma \in (0, \pi)[/tex] [tex]|[/tex] [tex]0 < \alpha + \beta + \gamma < \pi[/tex] with equality iff [tex]\alpha = \gamma[/tex].

We may assume WLOG that [tex]\alpha \le \gamma[/tex]. Then

[tex]\displaystyle{
\frac{\sin{\alpha}}{\sin{\gamma}} \le \frac{\sin{(\alpha+\beta)}}{\sin{(\gamma+\beta)}}
}[/tex] and [tex]\displaystyle{
\frac{\sin{(\alpha+\beta)}}{\sin{(\gamma+\beta)}} \le \frac{\sin{\gamma}}{\sin{\alpha}}}[/tex]

since quick algebra yields [tex]0 \le \sin{(\gamma - \alpha)}[/tex] and [tex]\cos{(2\gamma + \beta)} \le \cos{(2\alpha + \beta)}[/tex] respectively.  The former is obviously true. The latter is also true:  Suppose it isn't.  Since [tex]\alpha \le \gamma[/tex], then we must have [tex]2\gamma + \beta \ge 2\alpha + \beta[/tex].  By the shape of the graph of cosine, we must have [tex]2\alpha + \beta = \pi - x[/tex] and [tex]2\gamma + \beta = \pi + y[/tex] for some [tex]y > x > 0[/tex].  This, however, is a contradiction since [tex]2\pi > 2\alpha + 2\beta + 2\gamma = 2\pi + y - x > 2\pi[/tex].  It is clear that equality in either inequality holds iff [tex]\alpha = \gamma[/tex].

However, the desired result follows from these two inequalities, since it is known that [tex]\forall x, y \in \mathbb{R^{+}}[/tex] [tex]|[/tex] [tex]\frac{1}{x} \le y \le x[/tex], we have [tex]x + \frac{1}{x} \ge y + \frac{1}{y}[/tex], and we are done.


It might be a bit of a stretch, but I think as long as you plan on doing a lot of work and studying you'll be fine. I'll also be a junior, taking AP: US History, English Language, Physics C, Chemistry. Doesn't look too bad, since I got all my math APs out of the way.

the only struggle might be with taking the us history and government together, but if you work hard i'm sure you can manage.  the rest of it is somewhat typical in my school for a junior.

Yeah. 5 AP's aren't too bad... if that's what you want, and if you're willing to commit yourself, I'd say, go ahead.
I'm taking 3 classes and 5 tests this year, and I'll probably be taking 6 and 6 next year (Physics C, Bio, both English, USH, Art H).

I'm taking 4 and independently studying 4. Shouldn't be too big a deal, especially since all I do is stay home and

I'm in 5 AP classes

And takin 2 tests

Just about every AP test I've taken in high school is useless

Don't be fooled, depending on your major and what you wanna do in the future, AP tests may be useless

The classes just look good on paper...that's about all I've benefitted thus far....

i'm taking 2 next yr and 3or4 in my junior yr

Ask towersfreak2006 for advice. That's kid's like crazy. (7 APs *mutters*)

I've got 5 this year.

[quote="furious"]hmm....i wish i had the problem of having too many ap classes. my school randomly doesn't offer certain ap classes. for example, next year, ap economics and english aren't going to be offered. and since i wouldn't commit myself to independent studying, i'm not able to take as many ap classes.  :(
but i'll end up with 7: us(got a 5), stats(3 days away), euro(5 days away), bio(8 days away), calc, chem, and gov.[/quote]

Like you would take English if it were offered... :P

the amount of work required by an AP course varies a lot from school to school and from teacher to teacher. You will get the best advice talking to seniors at your school rather than relying on the experience of people at other schools.

[quote="anhnguyen"]I am going to be junior next year and course selection is coming soon. I really want to take a lot of AP classes. My friends told me that I shouldn't take more than 3 because more would be too much to handle. I plan to take AP US History, US government, Computer Science, Physics C, English. Do you guys think this is too much?

Thanks for any suggestion.

Anh.

P.S. One more week till AP Exams  :?[/quote]

Depends on your goals and how much work the classes are.  Find out from other students how much work the classes are.

Personally, I think APs are wildly overrated and are largely responsible for avid students not being able to really explore what they are interested in due to putting too much time demands on students.  Y'all would be way better off with a few less APs and more time spent pursuing studies that really interest you.

Too bad getting into college seems to be all about doing what they want you to do and not what you actually care about and what is actually meaningful.

btw Mr. Rusczyk what college did you go to? (edit--wow I got it right somehow)

[quote="MithsApprentice"][quote="joml88"]Like you would take English if it were offered... :P[/quote]

Argh note of advice to people here:

Don't try taking Literature and Language APs in the same year...*sigh*...it's like psychological suicide. *mutters* (Furthermore, don't try taking both in the same year without studying for either...)[/quote]

Lol...I might actually be taking AP English and AP Latin next year :D .  I think that I'm only going to take 3 AP's so I have more time for taking a math class at a college and just more time for math in general :P .

[quote="joml88"]Like you would take English if it were offered... :P[/quote]

no i wouldn't, but that's not the point. the point is, now i have an excuse to colleges why i didn't take an english my senior year. i could say i had planned to take ap english, the class didn't make, i burst into tears, and then signed up for physics.  :)

You're going to Clemson...it hardly makes a difference if you have 7 or 8 AP's there

[quote="MithsApprentice"][quote="joml88"]Like you would take English if it were offered... :P[/quote]

Argh note of advice to people here:

Don't try taking Literature and Language APs in the same year...*sigh*...it's like psychological suicide. *mutters* (Furthermore, don't try taking both in the same year without studying for either...)[/quote]

here is even a better note:

dont take 2 of language/literature/psychology at the same time. now that is called a psychological suicide.

[quote="xxreddevilzxx"]here is even a better note:

dont take 2 of language/literature/psychology at the same time. now that is called a psychological suicide.[/quote]

I thought psychology was supposed to be one of the easiest...

Ap might be, bu thte class it self i heard was hard.


analytic geometry
geometry open
geometry

Given triangle ABC and unknown point X.
The 3D coordinates of A, B and C are known.
The following differences are known:
XA-XB, XA-XC and XB-XC (Xi is the distance from X to i)

Can the coordinates of X be found from the above information?
If so, what are the coordinates of X?

One of the relations, say $ XB \minus{} XC$ is redundant, since $ XB \minus{} XC \equal{} (XA \minus{} XC) \minus{} (XA \minus{} XB)$.
Now, relation $ XA \minus{} XB \equal{} v$ forces $ X$ on a hyperboloid of foci $ A$ and $ B$, and similarly, relation $ XA \minus{} XC \equal{} w$ forces $ X$ on a hyperboloid of foci $ A$ and $ C$. These two hyperboloids may not intersect, be tangent, or intersect (in infinitely many points), so in general the position of $ X$ cannot be determined.


Solve in naturals $ x,\;y,\;z,\;t$ equation
\[ 520(xyzt\plus{}xy\plus{}xz\plus{}zt\plus{}1)\equal{}577(yzt\plus{}y\plus{}z)\]

[quote="Yuriy Solovyov"]Solve in naturals $ x,\;y,\;z,\;t$ equation
\[ 520(xyzt \plus{} xy \plus{} xz \plus{} zt \plus{} 1) \equal{} 577(yzt \plus{} y \plus{} z)
\]
[/quote]

$ \frac{xyzt \plus{} xy \plus{} xz \plus{} zt \plus{} 1}{yzt \plus{} y \plus{} z}\equal{}\frac{577}{520}$ ---> $ x\plus{}\frac{zt\plus{}1}{yzt \plus{} y \plus{} z}\equal{}1\plus{}\frac{57}{520}$ ---> $ 0<\frac{zt\plus{}1}{yzt \plus{} y \plus{} z}<1$ --> $ x\equal{}1$  --> $ \frac{zt\plus{}1}{yzt \plus{} y \plus{} z}\equal{}\frac{57}{520}$ -->

$ \frac{yzt\plus{}y\plus{}z}{zt\plus{}1}\equal{}\frac{520}{57}$  --> $ y\plus{}\frac{z}{zt\plus{}1}\equal{}9\plus{}\frac{7}{57}$ ---> $ y\plus{}\frac{1}{t\plus{}\frac{1}{z}}\equal{}9\plus{}\frac{1}{8\plus{}\frac{1}{7}}$

$ x\equal{}1;y\equal{}9;z\equal{}7;t\equal{}8$

It's same my post. :lol:

http://www.mathlinks.ro/viewtopic.php?t=227861


limit
function
search
geometry
geometric transformation
reflection

Why is the sum of all positive integers -1/12?

this seems crazy... but apparently it has been "proven" by physics experiments (caismir effect . sp?)
sorry if this is in the wrong level... not actually knowing the answer, I wouldn't know how difficult it is.

this is by any means NOT true!
(It's even very easy to prove that the sum of it is not defined.)

I don't know who told you this non-sence, but anyway, don't believe it! :D

Yea, that makes no sense that all POSITIVE integers would add up to -1/12.

If you mean what is:

[tex]1+2+3+4+5+...+n+...\to\infty[/tex]

IE

[tex]\displaystyle\lim_{n\to\infty}{S_n}=\infty[/tex] where [tex]a_n=n[/tex]

heh it's even of all positive integers :D

in that case, I don't even get it, even if you don't know it's $+\infty$, it's just logical that by adding positive terms to something above zero, you can impossibly drop below zero.. :?

there is actually a way to "proof" it: consider the zeta function
$\displaystyle \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}$
for $s>1$(or $Re[s]>1$ if you know complex numbers). it is well-known that this function can be extended to a meromorphic function to R(or C) everywhere except at 1 where it has a simple pole. furthermore, one can prove that $\zeta(-1)=-\frac{1}{12}$. but "$\displaystyle \zeta(-1)=\sum_{n=1}^{\infty} \frac{1}{n^{-1}}=\sum_{n=1}^{\infty} n$".
this is the connection.

look at http://mathworld.wolfram.com/RiemannZetaFunction.html

Peter

However, in one of the first letters Ramanujan wrote to Hardy, trying to explain his way of thinking, the Indian genius stated just that! He wrote that, although it might sound weird, he believed the sum of the integers was -1/12, and this led him to important results. It was sort of a "shortcut" that he didn't formally prove, but that he successfully applied to derive "standard" and true formulae.

yeah, i believe that as well. we even have $1+1+1+...=-\frac{1}{2}$, using the same idea ($\zeta(0)=-\frac{1}{2}$) and some similar results(considering the sum of all squares etc). i guess every of ramanujan's proofs may be translated into formally correct proofs using true identities like the ones for the zeta function.

Peter

Peter, a proof giving such results is simply a fool's proof.

It just means there's a little hole in it, using a theorem that doesnt work for the field your working in, for example

Like as you can't use $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ in complex numbers.

Search for the Caismir effect. If you put two metal plates facing each other very close, only a few wavelengths will be able to reflect between the two plates, so the energy on the outside of the plates is higher than that on the inside... so the two plates are attracted for no apparent reason (its been proven... and the maths shows that this sum works!!?!). quantum fluctuations in a vacuum!

I don't know how, because it's very complicated, but surely if it is proven in the real world it is a valid answer.

very interesting.... :?

I know about the plates and about the vacuum thingie, and I can understand it.
I really don't believe it means something about the sum of integers though.

peter, you recognized i put the word proof in hyphons(took me long time to remember the word :D )? i am sure aware of the fact that this thing will ALWAYS be nonsense looking strictly at it. but maybe, you want to proof some formula where there is a convergent sum which we can split into the sum of integers and some other divergent sum and now use those formulas to proof it. strictly speaking this is nonsense. but maybe you can proof a more general equation (e.g. an equation for the zeta function and a silar function for the other divergent sum) which is true in the region where these functions converge. now, since both sides are maybe meromorphic on the complex plane, that equation holds for the whole complex plane, thus also for our special case where both sums diverge.
thus, one has to be careful about using such identities like the sum of all integers being $-\frac{1}{12}$(one can derive many nonsense out of it, because the connection to the zeta function gets lost during these arguments), but doing it the right way, one can modify the proof to get a mathematical correct proof.

Peter


[size=150]If Susie Susan has 6 keys and 4 locks and each key only goes to one lock. And each lock has only one key that opens it how many times must she try the keys to the locks if she wants to unlock all 4 locks?[/size]

Oh, I've seen this problem before  :)  but I might have forgotten how to do it :blush:. But, is this how:

Since Susan has 6 keys and 4 locks, she has a 1/2 chance that each key goes into a lock. Also, you can do combinations, 6 C 4, which gets you 15. So 1/2X=15. X=30. Or is that wrong how I'm doing it?

Hmm i am terrible at math but i think 30 is rather large since even if you try all 6 keys in each lock each time, you will get 24...

Well i will approch this the conventional way:

6+5+4+3=18

Assuming that Susie is the unluckiest person in the world, she will try all 6 on the first lock, finds one, then use the 5 she has left for the next one...etc.

This will GUARANTEE that she will open all locks assuming she knows when locks are open

If not: 6*4=24..

I think Im wrong but oh well.

yeah Dunedain's first method is correct. usually in these type of problems you usually wanna do a worst case scenario thing. So if she is the "unluckiest" person in the world, she'd have to try 6+5+4+3 times so 18 tries should be right


$ x^3\plus{}y^3\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2)$ substitute $ 3r$ for x and $ 4$ for y and get $ (3r\plus{}4)(9r^2\minus{}12r\plus{}16)$. doing a quick check tells us $ 9r^2\minus{}12\plus{}16$ is unfactorable so we are done here


Let $ ABCD$ be a convex quadrilateral with $ AB\plus{}CD\equal{}BC\plus{}AD$. Let the diagonals $ AC,BD$ cut at $ K$. Let $ I_1,I_2,I_3,I_4$ be incenters of triangles $ ABK,BCK,CDK,DAK$ respectively. Prove that $ I_1I_2I_3I_4$ is cyclic.

This problem was posted in many times.
You can find a solution in [i]Toshio Seimiya, Peter Y. Woo, Solution of problem 2338, Crux Mathematicorum 4 / 25 (1999) pages 243-245.[/i]

[url]http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv25n4/body/HTML/243?template=CRUX[/url] The proof is very long and hard.

Yes, I have found a new proof but darij showed that my proof had posted in a Russian magazine (1995), and I forgot the link related to it...


Ten women sit in 10 seats in a line. All of the 10 get up and then reseat themselves using all 10 seats, each sitting in the seat she was in before or a seat next to the one she occupied before In how many ways can the women be reseated?

 Is there a way to solve this question without a recurrence?

well, the answer is a Fibonacci number, and Fibonacci numbers are defined recursively, so even if there is some direct counting argument, it wouldn't be obvious. there are formulas relating Fibonacci numbers to binomial coefficients (which count things), like $ \sum_{k \equal{} 0}^n\binom{n \minus{} k}{k} \equal{} F_{n \plus{} 1}$, and they may help you get an alternate solution to this problem, but all in all, the best and most natural way to do it is by recurrence :)

[quote="007math"]What is recurrence? (sorry if this is a dumb question)[/quote]

http://www.lmgtfy.com/?q=recurrence :D

It's a recursive relation. For example, $ F_{n\plus{}2}\equal{}F_{n\plus{}1}\plus{}F_{n}$ is a recurrence relation, because the definition of $ F_n$ depends on itself.


A recent market research survey of $ 100$ people found that $ 60$ people own a VCR, $ 40$ people own a CD player, and $ 20$ people own both a VCR and a CD player. How many in the survey own neither a VCR or a CD player?

Using PIE, we have 60 + 40 -20 = 80 people who own a VCR or CD player. 100 - 80 = $ \boxed{20}$ own neither.


Solve in real numbers the following system:
$tan^{2}x+2cot^{2}2y=1$
$tan^{2}y+2cot^{2}2z=1$
$tan^{2}z+2cot^{2}2x=1$


It was already posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=fibonacci&t=103431]here[/url] but not yet a complete proof  :blush:


This might sound stupid, but...

How do you ask a teacher if he has sent out a recommendation form without sounding pushy?

I asked a teacher of mine to give me a recommendation to a summer program, but he seems to have forgotten. I want to remind him without being to bothersome, but the due date is coming soon.

The same thing happened to me last year with Mathcamp.. I just approached the teacher and said something along the lines of, "Hey, I don't want to bug you too much, but I was just wondering if you sent out my reccommendation for Mathcamp."  He actually had forgotten, and thanked me for reminding him.

Be sure to thank them whether or not they sent it out.  =)

Just go up to him and ask politely and casually if he has the rec finished. I'm sure he will understand.


The form can be simplified as :
1/2 cos 40 + 4 cos 20 cos 40 - 1/2 cos 20
=2 cos 20 - 2 cos 40 + 16 (cos 20.cos 40)^2/(4 cos 20. cos 40)
Notice also that cos 40 = 2 (cos 20)^2-1
May someone can continue...


probability
geometry
3D geometry
algebra
polynomial
advanced fields
advanced fields unsolved

Suppose you have a few invertible chemical reactions with just 2 reagents, i.e., a few equations like $ 2x\plus{}3y\leftrightarrow x\plus{}5y$. With each such equation, one associates two speed coefficients: $ f(t)$ for the forward reaction and $ b(t)$ for the backward reaction. Consider the corresponding differential system for concentrations. In the case of just one reaction above, it will be
\[ \begin{aligned}
\frac{dx}{dt}\equal{}&\minus{}f(t)x^2y^3\plus{}b(t)xy^5
\\
\frac{dy}{dt}\equal{}&2f(t)x^2y^3\minus{}2b(t)xy^5
\end{aligned}\]
the logic being that the probability that 2 molecules of $ x$ and 3 molecules of $ y$ will meet to cause the forward reaction is proportional to the square of the concentration of $ x$ times the cube of the concentration of $ y$ and one extra $ x$ is produced in the forward reaction and lost in the backward one.

The problem is to show that if all the speed coefficients are separated from $ 0$ and $ \infty$, then the closure of every trajectory is a compact subset of the open first octant $ \{x,y>0\}$, so neither a blow up, nor running out of any reagent is possible.

There is a rather neat and short solution to this, so I decided to propose this problem. If you can do something for more than 2 reagents, I'll be happy to see it too. I spent a few days trying to do the case of 3 reagents and I believe I have a couple of workable ideas but my approach is rather complicated, so I hope somebody will put me to shame by demonstrating a simple way to do the general case :)

[quote="Fedja"]invertible chemical reactions with just 2 reagents[/quote]

1. "Reversible" chemical reactions would be more correct.

2. Such reactions, as you have written, do not exist. Perhaps you want to analyse something like

$ 2w \plus{} 3x \rightleftharpoons y \plus{} 5z$

@Carcul:
It is meant as a mathematical problem, thus being realistic is irrelevant  :wink:
By the way, why should such a reaction not exist¿ Simplifying some things, such reactions are possible.

[quote="ZetaX"]By the way, why should such a reaction not exist[/quote]

Because there is no mass conservation in that reaction.

[quote="ZetaX"]Simplifying some things, such reactions are possible.[/quote]

Then give us an example.

they don't exist because the  matter doesn't get produced or destroyed in a chemical reaction...just converrted from one form to the other...
so no. of $ x$ and $ y$ shud be conserved :wink:

EDIT:Didn't see carcul posting :(

Where does the matter change/get destroyed/get created¿
For one of the most simplest examples, take $ x \equal{} O_2$ and $ y \equal{} O$ (oxygen building radicals).

This would then simplify to $ x \leftrightarrow 2y$. If we want some conditions more realistic to the original equation, we could e.g. take some step from cell respiration.

Cell respiration? Can you then give an example?

I already gave an example  :wink:
I mentioned cell respiration because there sometimes a product is needed before the reaction (catalysing it).

Folks, you know yourself that you argue over irrelevant things :).

First $ 2x \plus{} 3y\leftrightarrow x \plus{} 5y$ is not entirely impossible for 2 reasons:
a) $ x \equal{} A_2$, $ x \equal{} A$ where $ A$ is some element or combination of elements.
b) The problem came from a biologist who said that they usually care only about complicated organic molecules and ignore all simple junk like water and carbon dioxide that gets produced or consumed, so even $ x\leftrightarrow 2x$ is nothing unusual. If you really want to, I can ask him to tell me some real reactions.

Second, when you do a mathematical model, you try to get rid of all assumptions that you deem unnecessary for the problem. The point is that if you can prove something for all reactions, including non-realistic, nobody is going to argue that your result does not apply to the realistic ones but if you start using certain realistic assumptions in your solution, somebody will certainly argue that you haven't reached full generality and there may be interesting cases to which your method does not apply.

Third, it was posted to the Advanced Fields subforum, not to the Chemistry one. This is the case that is the simplest from the mathematical point of view. It is a perfectly well posed mathematical problem and it has a solution. The motivation was included just for the fun of it. I could just say: consider a system of polynomial differential equations obtained in the following way: blah blah blah... By the way, by saying "2 reagents", I mean that every reaction is of the form $ ax\plus{}by\leftrightarrow cx\plus{}dy$ with the same $ x,y$ but different $ a,b,c,d$, so we consider dynamics on the plane. I'm not sure I made it clear in my original post :blush:

Fourth, how about solving the problem rather than arguing about it ;)


integration
calculus
calculus computations


geometry
circumcircle
geometric transformation
reflection
symmetry
analytic geometry
power of a point

Dear Mathlinkers,
prove this conjecture:
for a triangle, the radical axis of the K-Hagge circle and the Brocard circle goes through the Brocard midpoint,
K being the symmedian point of that triangle.
Sincerely
Jean-Louis

Dear Mathlinkers,
Paul Yiu, chief redactor of Forum Geometricorum has confirmed this conjecture.
How can we prove it synthetically.
Sincerely
Jean-Louis

[b]General lemma:[/b] Center of the P-Hagge circle WRT $\triangle ABC$ is the midpoint of the segment connecting  its orthocenter $H$ with the anticomplement of the isogonal conjugate $P^{+}$ of $P.$

Let $X,Y,Z$ be the 2nd intersections of $AP,BP,CP$ with the circumcircle $(O)$ of $\triangle ABC$ and $X',Y',Z'$ the reflections of $X,Y,Z$ about $BC,CA,AB.$ If $\triangle A_0B_0C_0$ is the anticomplementary triangle of $\triangle ABC,$ then $B,C$ lie on a circle $\omega_a$ with diameter $HA_0.$

Let $P^{-1}$ be the isogonal conjugate of $P$ WRT  $\triangle ABC$ and $P_1 \equiv AP^{-1} \cap (O),$ different from $A.$ Since $\angle (AB,AP^{-1})=\angle (AC,AP),$ it follows that $P_1X \parallel BC$ $\Longrightarrow$  $P_1,X$ are symmetric about the perpendicular bisector $\ell_a$ of $BC.$ Congruent circles $(O) \cong \omega_a$ are both symmetric about $BC$ and its midpoint $M_a.$  Since the composition of the axial symmetries across the orthogonal axes $BC$ and $\ell_a$ is a central symmetry about the intersection $\ell_a \cap BC,$ it follows that $X',$ $M_a$ and $P_1$ are collinear, thus $A_0X' \parallel AP^{-1}$ $\Longrightarrow$ anticomplement $P^{+}$ of $P^{-1}$ lies on $A_0X',$ but $X' \in \odot(A_0BC)$ $\Longrightarrow$ $\angle HX'P^{+}=90^{\circ}$ $\Longrightarrow$ $X',Y',Z'$ lie on a circle with diameter $HP^{+}.$
____________________________________________

The anticomplement of the isogonal conjugate of the symmedian point $K$ is obviously the centroid $G$ of $\triangle ABC$ $\Longrightarrow$ K-Hagge circle is the Orthocentroidal circle of $\triangle ABC.$

We use barycentric coordinates WRT $\triangle ABC.$ Brocard circle $\mathcal{K}$ passing through $K (a^2:b^2:c^2)$ and Brocard points $\Omega_1 \left ( \frac{1}{b^2}:\frac{1}{c^2}:\frac{1}{a^2} \right),$  $\Omega_2 \left ( \frac{1}{c^2}:\frac{1}{a^2}:\frac{1}{b^2} \right)$ has equation:

$\mathcal{K} \equiv a^2yz+b^2zx+c^2xy-\frac{a^2b^2c^2}{a^2+b^2+c^2}(x+y+z) \left ( \frac{x}{a^2}+\frac{y}{b^2}+\frac{z}{c^2} \right)=0$

Equation of the Orthocentroidal circle $\mathcal{G}$ whose diameter is $\overline{HG}$ is given by:

$\mathcal{G} \equiv a^2yz+b^2zx+c^2xy-\frac{_2}{^3}(x+y+z)(S_Ax+S_By+S_Cz)=0$

Radical axis $\tau$ of $\mathcal{K}$ and $\mathcal{G}$ is then:

$ (b^4+c^4-a^4-b^2c^2)x+(a^4+c^4-b^4-a^2c^2)y+(a^4+b^4-c^4-a^2b^2)z=0$

$\tau$ passes through the Brocard midpoint $X_{39} \ (a^2(b^2+c^2):b^2(a^2+c^2):c^2(a^2+b^2))$

Dear Luis and Mathlinkers,
thanks for your approach of a synthetical proof of my conjecture. I will it investigate when I have more time...
For the center, I prove it on my site
http://perso.orange.fr/jl.ayme   vol. 5  Le P-cercle de Hagge p.4.
Sincerely
Jean-Louis


Find the equation of the line that traverses the vertex of a right angle whose edges are constantly tangent with the parabola $ y^2\equal{}2px$.

[hide="My solution"]
We search for the equation of the tangent in this form:
$ y \minus{} y_0  \equal{} k(x \minus{} x_0 )$
$ y \equal{} kx \plus{} (y_0 \minus{} kx_0 )$(1)
From the condition of the tangent: $ 2kt\equal{}p$ we find $ t \equal{} \frac{p}{{2k}}$(2)
From (1) and (2) we have: $ y_0 \minus{} kx_0 \equal{} \frac{p}{{2k}}$
$ 2ky_0 \minus{} 2k^2x_0 \minus{} p \equal{} 0$
$ \minus{}2x_0k^2 \plus{} 2y_0k \minus{} p \equal{} 0$
$ D \equal{} (2y_0)^2 \minus{} 4(\minus{}2x_0)(\minus{}p)$
$ D \equal{} 4y_0 ^2 \minus{} 8x_0p$
There are two solutions $ k_1  \equal{} \frac{{ \minus{} b \plus{} \sqrt D }}{{2a}}$ and $ k_2  \equal{} \frac{{ \minus{} b \minus{} \sqrt D }}{{2a}}$.
Since the tangents are perpendicular $ k_1  \cdot k_2  \equal{}  \minus{} 1 \equal{} \frac{{4y_0 ^2  \minus{} D}}{{16x_0 ^2 }}$
$ \frac{{4y_0 ^2  \minus{} 4y_0 ^2  \minus{} 8x_0 p}}{{16x_0 ^2 }} \equal{}  \minus{} 1$
$ x_0 \equal{} \frac{p}{2}$
So the geometric place is a straight line parallel with OY. Now analytically everything seems fine to me, but my intuition says it can't be a line. Especially i can't see how can this solution be right for the points of the segment of the line among the parabola's arms.
[/hide]
Do you find this solution correct?

May be this helps you.

http://www.mathlinks.ro/viewtopic.php?t=169849

 :)


Let $ x,y,z>0$ and $ xyz\equal{}1$. Prove that

$ \frac{x}{x^{\minus{}8}\plus{}y\plus{}z}\plus{}\frac{y}{x\plus{}y^{\minus{}8}\plus{}z}\plus{}\frac{z}{x\plus{}y\plus{}z^{\minus{}8}} \ge 1$

[quote="aktyw19"]Let $ x,y,z > 0$ and $ xyz = 1$. Prove that

$ \frac {x}{x^{ - 8} + y + z} + \frac {y}{x + y^{ - 8} + z} + \frac {z}{x + y + z^{ - 8}} \ge 1$[/quote]

Its equivalent with
\begin{align*} & \sum\frac {x^6y^5z^5}{y^8z^8 + y^6x^5z^5 + z^6x^5y^5}\geq1 \\
\iff & \sum\frac {x^6}{y^3z^3 + y^6 + z^6}\geq1 \end{align*}

Its obvious, that $ y^3z^3\leq\frac {y^6 + z^6}2$

So it remains to prove
\begin{align*} & \sum\frac {x^6}{\frac {y^6 + z^6}2 + y^6 + z^6}\geq1 \\
\iff & \frac 2 3 \sum\frac {x^6}{y^6 + z^6}\geq 1 \end{align*}
The proof is completed since this ineq is true considering Nesbitt  :!:


parameterization
integration
trigonometry
calculus
calculus computations

$a$ real parameter

Is $f_{a}:\mathbb {R}^2\to\mathbb R$
\[ f_{a}(x,y)={\begin{cases}
\displaystyle\int_{\frac{\pi}6}^{\frac{\pi}3}\dfrac{2|xy|^{a}}{x^2\sin^2t+y^2\cos^2t}\,dt\qquad\text{if}\quad(x,y)\not=(0,0)\\
\quad\\
0\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{if}\quad(x,y)=(0,0)
\end{cases}}\]
continuous in $\mathbb{R}^2$?

2/ study the differentiability ?

Néné's original example was postively homogeneous of degree 0. This variant is postively homogeneous of degree $2a-2.$

[hide]Continuous if $a>1,$ differentiable if $a>\frac32.$[/hide]


calculus
derivative
analytic geometry
graphing lines
slope
algebra
polynomial

Given line: $ 2x^2\plus{}4x\minus{}x^\minus{}1$

Find the line tangent at $ x\equal{} \minus{}2$

[hide="Solution"]Taking the derivative of $ f(x) \equal{} 2x^2 \plus{} 4x \minus{} x^{ \minus{} 1}$, we get $ f'(x) \equal{} 4x \plus{} \frac {1}{x^2} \plus{} 4$.

The slope of $ f(x)$ at $ x \equal{} \minus{} 2$ is $ f'( \minus{} 2) \equal{} 4( \minus{} 2) \plus{} \frac {1}{( \minus{} 2)^2} \plus{} 4 \equal{} \minus{} 8 \minus{} \frac {1}{4} \plus{} 4 \equal{} \minus{} \frac {17}{4}$. So far, we know that the equation of the line is $ y \equal{} \minus{} 2x \plus{} b$. To find $ b$, we find $ f( \minus{} 2)$, and then plug $ \left( \minus{} 2,f( \minus{} 2)\right)$ into $ y \equal{} \minus{} 2x \plus{} b$.

$ f( \minus{} 2) \equal{} 2( \minus{} 2)^2 \plus{} 4( \minus{} 2) \minus{} \frac {1}{ \minus{} 2} \equal{} 8 \minus{} 8 \plus{} \frac {1}{2} \equal{} \frac {1}{2}$, so we can now solve for $ b$:
\[ \frac {1}{2} \equal{} \minus{} 2( \minus{} 2) \plus{} b\quad\Longrightarrow\quad b \equal{} \minus{} \frac {7}{2}
\]
Thus, the equation of the line tangent to $ f(x) \equal{} 2x^2 \plus{} 4x \minus{} x^{ \minus{} 1}$ at $ x \equal{} \minus{} 2$ is $ y \equal{} \minus{} 2x \minus{} \frac {7}{2}$.[/hide]

Please read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438]this[/url].
 :wink:

Anyway,

[hide="Outline"]We can just take the derivative, solve for $ x$ in $ f'(x) \equal{} 0$. We then find $ (x,f(x))$, and then find the $ y$ intercept. We can now put it into form $ ax\plus{}by\equal{}c$ and find our answer.[/hide]

Why are you setting the derivative equal to zero? We have a specific x value that we are given... and also $ f'(a) \equal{} 0 \implies y_{tangent} (x) \equal{} f(a)$

I get the impression we're supposed to solve this without calculus. This problem is trivial otherwise. Finding tangent lines to polynomials is easy enough but the $ x^{\minus{}1}$ term is a little bit difficult to deal with. Maybe try to see if finding tangent lines of $ xf(x)$ says anything about tangent lines of $ f(x)$.

[b]moogra[/b] — $ 2x^{2} \plus{} 4x \minus{} x^{\minus{}1}$ is a quadratic, so the graph of $ f(x)$ is actually a [very weird] curve.

[hide="To: i_like_pie"]Wouldn't the line be $ y \equal{} \minus{}\left(\frac {17}{4}\right)x \minus{} 8$, since the derivative is equal to the [i]slope[/i] of the line tangent to the function at a certain point $ x$, and the line $ y \equal{} \minus{}2x \minus{} \frac {7}{2}$ is actually [i]secant[/i] to $ f(x)$?[/hide]

[quote="metafor"][b]moogra[/b] — $ 2x^{2} \plus{} 4x \minus{} x^{ \minus{} 1}$ is a quadratic, so the graph of $ f(x)$ is actually a [very weird] curve.

[hide="To: i_like_pie"]Wouldn't the line be $ y \equal{} \minus{} \left(\frac {17}{4}\right)x \minus{} 8$, since the derivative is equal to the [i]slope[/i] of the line tangent to the function at a certain point $ x$, and the line $ y \equal{} \minus{} 2x \minus{} \frac {7}{2}$ is actually [i]secant[/i] to $ f(x)$?[/hide][/quote]

Yes that is the correct answer.

@#H34N1 I did not post this problem in this forum. It got moved here (from High SCHOOL BASICS  :rotfl: ). And btw I don't think this is a terribly hard problem because this matches the material I do at school, so I thought HSB was appropriate (FOR SCHOOL STUFF IN HIGH SCHOOL).

[quote="metafor"][hide="To: i_like_pie"]Wouldn't the line be $ y \equal{} \minus{} \left(\frac {17}{4}\right)x \minus{} 8$, since the derivative is equal to the [i]slope[/i] of the line tangent to the function at a certain point $ x$, and the line $ y \equal{} \minus{} 2x \minus{} \frac {7}{2}$ is actually [i]secant[/i] to $ f(x)$?[/hide][/quote]
You're absolutely correct; I completely ignored that number and decided that the slope was $ \minus{}2$ for some reason. :D

I'm so full of myself, so I like linking others to my previous posts :wink:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194721&start=7]Proof of the power rule of differentiation via AM-GM[/url].

If you use weighted AM-GM, it's even cleaner and more easily applicable, but I feel like unweighted AM-GM places greater emphasis on the fact that finding the slope of the tangent line to a polynomial-type curve can be approached by elementary means. I swear, I'm going to give this rant to everyone that complains about differentiating polynomials in this forum.

[quote="mathemonster"]this problem is very trivial with calculus.........what contest is it from?[/quote]

This isn't from a contest. It was merely a school problem my friend wanted me to put up (we were arguing on what to plug values into after the differentiation). It is really easy since this is school material.


How many whole numbers between 1000 and 2000 have exactly 3 positive whole numbers divisors?

It must be of the form $ p^2$ where $ p$ is a prime.  Thus, $ 32 \le p \le 44 \implies p \in \{37,41,43\}$ so our answer is $ \boxed{3}$.


function
induction
algebra
polynomial
algebra solved

define function F_0=1 and F_1=x, F_(k+1)=2xF_k - F_(k-1)
find all the roots of F_(n+1)

Fk (cos [i]x[/i]) = cos [i]kx[/i]

I think, it's enough...

well, it is quite simple to show that fk(cosx)=coskx.
Now suppose fn=0. we shall prove that  x=cos t for some t and therefore we will be done.

suppose abs(x)>1. we shall show that absfk>absfk-1 for all k.
note that fk+1=2xfk-fk-1.now use induction:
hence absfk+1>=2abs(x)abs(fk)-abs(fk-1) >=absfk. where the last one follows by induction.
but then fk can not be zero.
hence abs(x)<1.

F_k(x) are called Chebyshev's polynomials.

BTW you don't have to prove that x=cos t for some t, because
Fk has n real roots lying in [-1;1] so it can't have more,

because
it's a polynomial with degree n.


Hello too!
But do you know that there is a forum dedicated to inequalities?... ;)

Pierre.


superior algebra
superior algebra unsolved

Let $ R$ be a commutative ring and $ I$ an ideal of $ R$. Show that every element in $ I^n/I^{n\plus{}1}$ can be written as linear combination of products of $ n$ elements in $ I/I^2$

you seem to use some embedding $ I/I^2 \to I^n / I^{n \plus{} 1}$? (which one?)

More like a surjective map $ (I/I^2)^{\otimes n}\to I^n/I^{n\plus{}1}$.

"Products" most likely refers to multiplication in the associated graded ring $ \mathrm{gr}_I\,R\equal{}\bigoplus_{n\geq0}I^n/I^{n\plus{}1}$. owk

For completeness I shall confirm that I was indeed working in graded rings.


Αυτή είναι μια όμορφη μικρή ασκησούλα με δυνατή σκέψη, αλλά εύκολη εκτέλεση:

Έστω $ a_na_{n\minus{}1}...a_0$ πρώτος αριθμός με $ n$ άρτιο και έστω πολυώνυμο
$ P(x) \equal{} a_nx^{n} \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} ... a_0$ που αν έχει πραγματικές ρίζες,
από αυτές μόνο μία δεν είναι ακέραια.
Να δείξετε ότι το $ P(x)$ δεν έχει πραγματικές ρίζες.

Θα δώσω μια λύση κρυφή, αλλά η άσκηση πραγματικά αξίζει.

[hide]

Έστω ότι το $ P(x)$ έχει ρίζα $ m \in \mathbb{Z}$,
τότε από τη διαίρεση πολυωνύμου προκύπτει ότι υπάρχει $ U(x)$ με ακέραιους συντελεστές τέτοιο ώστε:
$ \frac{P(x)}{x \minus{} m} \equal{} U(x) < \equal{} > P(x) \equal{} (x \minus{} m)U(x) \equal{} > P(10) \equal{} (10 \minus{} m)U(10)$
$ < \equal{} > a_na_{n\minus{}1}...a_0 \equal{} (10 \minus{} m)U(10)$ άρα ο πρώτος αριθμός $ a_na_{n\minus{}1}...a_0$
γράφεται ως γινόμενο δύο άλλων ακεραίων $ \equal{} >$ άτοπο.
Άρα το $ P(x)$ δεν έχει ακέραιες ρίζες.

Άρα το $ P(x)$ έχει το πολύ μία πραγματική ρίζα. $ (I)$
Έστω ότι το $ P(x)$ έχει πραγματική ρίζα, τότε
επειδή είναι άρτιου βαθμού (γιατί $ n$ άρτιος), θα έχει και δεύτερη πραγματική ρίζα. $ (II)$
$ (I), (II) \equal{} >$ άτοπο

Άρα το $ P(x)$ δεν έχει πραγματικές ρίζες.

[/hide]

Αναφέρετέ μου οποιαδήποτε ένσταση έχετε.

[quote="maniopas"]
$ < \equal{} > a_na_{n \minus{} 1}...a_0 \equal{} (10 \minus{} m)U(10)$ άρα ο πρώτος αριθμός $ a_na_{n \minus{} 1}...a_0$
γράφεται ως γινόμενο δύο άλλων ακεραίων $ \equal{} >$ άτοπο.
Άρα το $ P(x)$ δεν έχει ακέραιες ρίζες.
[/quote]

To U(10) δεν μπορεί να είναι μονάδα ? :wink:   . Το δεν δεν μπορεί να είναι 9 ? Νομίζω ότι αυτές οι περιπτώσεις δεν είναι τόσο προφανείς για να παραλειφθούν (αν και βγαίνει  :wink: )

Όντως! Δίκιο έχεις. Απλώς θεώρησα δεδομένη τη σκέψη και δεν την ανέφερα ότι:
αν $ m > 0$ τότε $ P(x) > 0$ οπότε δεν είναι ρίζα, οπότε $ m <\equal{} 0$ άρα το $ m$ δε μπορεί ν είναι 1 ή 9

[quote="maniopas"]Όντως! Δίκιο έχεις. Απλώς θεώρησα δεδομένη τη σκέψη και δεν την ανέφερα ότι:
αν $ m > 0$ τότε $ P(x) > 0$ οπότε δεν είναι ρίζα, οπότε $ m < \equal{} 0$ άρα το $ m$ δε μπορεί ν είναι 1 ή 9[/quote]

Συμφωνώ με τον Silouan ότι βγαίνει, αλλά μπορείς να δώσεις μερικές επεξηγήσεις για το τί εννοείς παραπάνω?

Cheerio,

Durandal 1707

Επειδή οι συντελεστές στο πολυώνυμο είναι θετικοί και ο σταθερός όρος θετικός διάφορος του 0 (οι αριθμοί που τελειώνουν σε 0 δεν είναι πρώτοι), για μη αρνητικά x, ο κάθε όρος έίναι $ > \equal{} 0$ και ο σταθερός όρος $ > 0$. Προσθέτοντας κατά μέλη βγαίνει ότι αν $ x > \equal{} 0$ είναι $ P(x) > 0$. Άρα το πολυώνυμο δεν έχει θετικές ρίζες. Άρα απορίπτονται οι και οι θετικοί 1, 9 από τις ρίζες.

[quote="maniopas"]Όντως! Δίκιο έχεις. Απλώς θεώρησα δεδομένη τη σκέψη και δεν την ανέφερα ότι:
αν $ m > 0$ τότε $ P(x) > 0$ οπότε δεν είναι ρίζα, οπότε $ m < \equal{} 0$ άρα το $ m$ δε μπορεί ν είναι 1 ή 9[/quote]

Ok ;) εννοείς ότι "αν $ m>0$ τότε $ P(\mathbf{m})>0$ οπότε...". Ήθελα να βεβαιωθώ ότι το παραπάνω ήταν τυπογραφικό και δεν εννοούσες κάτι διαφορετικό.

Τα υπόλοιπα προφανώς κατανοητά, απλώς είχα μπερδευτεί αν ήταν typo ή όχι.

Cheerio,

Durandal 1707

Ναι, αυτό εννοώ. Από μια παρόμοια άσκηση έχω βγάλει και τον άλλο τρόπο λύσης, αλλά μου φαίνεται πιο μεγάλος.


Hello,

 Find all functions $ f : \mathbb{R}^{ \plus{} } \to \mathbb{R}^{ \plus{} }$ satisfying the conditions:
\[ \forall x \in \mathbb{R}^{ \plus{} } ,\quad 4f(x) \geq 3x \: \text{ and } \: f\Big(4f(x) \minus{} 3x\Big) \equal{} x
\]

This is posted before. [url]http://www.mathlinks.ro/viewtopic.php?t=268067[/url]


limit
function
calculus
calculus computations

Let $(F_n)_{n\in{N^*}}$ be the Fibonacci sequence defined by
$F_1=1$, $F_2=1$, $F_{n+1}=F_n+F_{n-1}$ for every $n\geq{2}$. Find
the limit: \[ \lim_{n \to \infty}(\sum_{i=1}^n{\frac{F_i}{2^i}}) \]

Write $F_n$ in the canonical form :$F_n=a x^n+by^n$ for some $a,b,x,y$.
The limit becomes: $\lim a \sum \left(\frac{y}{2}\right)^n+b\left(\frac{x}{2}\right)^n= a\frac{1}{1-\frac{x}{2}} + b\frac{1}{1-\frac y 2}$. First prove that $x,y<2$.

Using generating function we have $\sum_{i=1}^{\infty} F_n x^n = \frac{x}{1-x-x^2}$.  for $-\frac{\sqrt{5}+1}{2} < x < \frac{\sqrt{5}-1}{2}$.

what is the generating function? :blush:

It some kind of method to find the closed formula of reccurence relation. Using power series $\sum_{k=1}^{\infty} a_kx^k$. Where $a_k$ is the sequence of reccurence.

i solved it using the general form of fibonacci like xirti said( forgive me if i mispelled your name)...


limit
integration
vector
calculus
linear algebra
matrix
function

I found this equation in an electromagnetics book.  It doesn't seem to make sense to me.

$curl \vec {F} = \lim_{V \to \ 0} \frac{1}{V} \int_S \vec {n} \times \vec {F} da$

As I understand it, the curl of a vector field is orthogonal to the surface around which the line integral applies.  (I don't know how to get the circle on the integral sign)

This equation suggests that the curl F is a vector that is within the plane of the surface, rather than projecting orthogonal to it.  Is something wrong here?

I think you want \oint: $\oint_S$.

I see the curl as really belonging to the dual space- comparing it to an ordinary vector is dangerous. This deals with that principle by having a cross product on both sides.
In this case, there is no one direction perpendicular to the surface; the surface points in all directions around the point we're looking at. $V$ is the volume enclosed by $S$, and if you're calculating, $S$ will probably be a sphere. We're going to have to use something like the divergence theorem to prove it:

We're looking at $\oint_S dA\times F=\oint_S -F\times dA$. Write $F=\begin{bmatrix}F_x\\F_y\\F_z\end{bmatrix}$ and $dA=\begin{bmatrix}dA_x\\dA_y\\dA_z\end{bmatrix}$. The integral becomes
$\oint_S \begin{bmatrix}0&-F_z&F_y\\F_z&0&-F_x\\-F_y&F_x&0\end{bmatrix}\cdot \begin{bmatrix}dA_x\\dA_y\\dA_z\end{bmatrix}$, which we can deal with (I might normally write the cross product as a row vector, but I would have to introduce transposes here to do it). Now we apply the divergence theorem. What's the divergence of a matrix? Take the divergence of each row independently. We have
$\oint_S \begin{bmatrix}0&-F_z&F_y\\F_z&0&-F_x\\-F_y&F_x&0\end{bmatrix}\cdot \begin{bmatrix}dA_x\\dA_y\\dA_z\end{bmatrix}=\int_V\begin{bmatrix} 0-\frac{\partial}{\partial y}F_z+\frac{\partial}{\partial z}F_y\\ \frac{\partial}{\partial x}F_z+0- \frac{\partial}{\partial z}{F_x}\\-\frac{\partial}{\partial x}{F_y}+\frac{\partial}{\partial z}{F_y}+0\end{bmatrix}dV=\int_V \text{curl }F\,dV$

wow. . . It's really been too long.   :blush:

I'm afraid I don't quite understand all of your posting, but I'll study up.  Thank you for your explanation.

The trouble I'm having is trying to picture exactly what my function means physically.  I take it to mean that you have an infinitesimal number of tiny surfaces (essentially planes), such that you add n x F da for all of them.  However, since n is orthogonal to the tiny surface (any given surface), the resulting vector (n x F) lies within the plane of the surface in question, whereas the curl (if I remember) points orthogonal to the plane. . . On the other hand, if you add them all up, you'll end up with a vector that is, by default, orthogonal to the volume's "center" point (such as the center of gravity).  Correct?

Does curl F of a volume represent the axis around which the volume might rotate? (assume F is a vector field representing forces, say).

I suppose the most important question is whether or not the equation is true.  If so, I can continue to trust the book I am using.

Thanks again for your time.

The curl isn't orthogonal to anything in particular. I have no idea what the physical interpretation is, but this is right.


Αυτή την άσκηση την έχω ξαναστείλει κάπου : ή εδώ ή στο αγγλικό φόρουμ. Έχω την επίσημη λύση. Εϊχα κάνει και παλιά μία τριγωνομετρικά.
 Μια μέρα, καθώς ξεφύλλιζα ένα από τα δεκάδες βιβλία πάνω στο γραφείο, είδα για το πρόβλημα αυτό  μια λύση χωρίς τριγωνομετρία.Δεν την σημείωσα αμέσως  και άντε τώρα να βρω που την είδα. Έφαγα μια μέρα ψάχνοντας , αλλά τίποτα. Λοιπόν , την ξαναβάζω , μήπως και θυμηθείτε ή βρείτε τίποτα.Θα με βοηθήσετε να ξεκολίσω και αυτό θα μου δώσει ... χαρά!

 [color=red][b]ΑΣΚΗΣΗ[/b][/color]

[color=blue] Σε τρίγωνο [b]ΑΒΓ[/b] φέρουμε τη διχοτόμο [b]ΑΔ[/b] , τις προβολές [b]Ε, Ζ[/b] του [b]Δ [/b]στις [b]ΑΒ,ΑΓ [/b]και το σημείο τομής [b]Η[/b] των [b]ΒΖ,ΓΕ.[/b] Να αποδειχθεί ότι η [b]ΑΗ [/b]είναι κάθετη στην [b]ΒΓ[/b][/color]

ΣΧΟΛΙΑ
 Νομίζω ότι προέκτεινε το ΑΗ  μέχρι το Η (στην ΒΓ) και με κάτι ομοιότητες τριγώνων έβγαιναν κάτι σχέσεις. Νομίζω ότι η συνέχεια πρέπει να γίνονταν με Ceva.
 Ευχαριστώ - Μπάμπης

[quote="silouan"]Λύσεις για όλα τα γούστα εδώ (just search for AH,BC sto ellinika) http://www.mathlinks.ro/viewtopic.php?search_id=680352741&t=200801[/quote]

Σιλουανέ, δεν αποκλείεται καθόλου να είδα αυτή τη λύση στις τυπωμένες σελίδες από το φόρουμ και σίγουρος ότι θα την ξαναδώ, την άφησα για αργότερα. Έτσι την έχασα και μετά παιδευόμουνα. Αυτά παθαίνεις , αν ανοίγεις πολλά μέτωπα.
 Σε ευχαριστώ πολύ !

 Μπάμπης


[quote="felipesa"][quote="nat mc"]just another way to write it
$\prod_{a=0}^{n}{(7^{2a}+2)}$[/quote]

Did you find a value for P?[/quote]
no, i'm just writing it in a way to make it look a little nicer.

You can of course expand it and by paaterns collect terms and find sum using GP.

Anu "beautiful " soln???


algorithm
greatest common divisor
algebra unsolved
algebra

Theorem:The number of divisions necessary to find the greatest common divisor of two natural numbers by means of the Enclidean algorithm is not greater than 5 multiplied by the number of digits in the decimal expansion of the smaller of the numbers

Can you write my theorem by vietnamese? I am vietnamese

I think it is an Number Theory problem,isn't it?

$ A\equal{}$The number of divisions necessary to find the greatest common divisor of $ a\ge b>1$ by means of the Enclidean
$ A\le \frac{ln(b)}{ln(2)}\minus{}1$
and if $ 10^{n}> b\ge 10^{n\minus{}1}$ then $ 5n\ge 5\frac{ln(b)}{ln(10)}$
and we have $ 5\frac{ln(b)}{ln(10)}\minus{}\frac{ln(b)}{ln(2)}\plus{}1>0$

This theorem is called Lame's theorem. see for example http://mathworld.wolfram.com/LamesTheorem.html


These are the problems from the book I was practicing Permutation and Combination. I tried them but was unable to solve few of them. Try them and have a good PROBLEM SOLVING...
Important- Please post your answers as hidden.

1. How many subsets of the subset {1,2,...,10} contain tleast one odd integer?

2. Prove that the number of isosceles triangles with integer sides, no side exceeding n is $\frac{1}{4}(3n^2+1)$ or $\frac{3}{4}n^2$ according as n is odd or even.

3. What is the number of distinct terms in the expansion of $(x_1+x_2+\cdot\cdot\cdot+x_m)^n$?

*4. If three distinct integers are chosen at random show that there will exist two among tham, say a and b such that 30 divides $(a^3b-ab^3)$.

*5. n objects are arranged in a row. A subset of these objects is called [i]unfriendly[/i] if no two of its elements are consecutive. Show that the number of unfriendly subsets of a [i]k[/i]-element set is\[ \binom{n-k+1}{k}  \]

6. Four numbers are chosen from 1 to 20. If $1\le {k}\le17$,in how many ways is the difference between the smallest and largest equal to [i]k[/i]?

Note:- '*' marked questions I was unable to solve.

[quote="varun"]These are the problems from the book I was practicing Permutation and Combination. I tried them but was unable to solve few of them. Try them and have a good PROBLEM SOLVING...
Important- Please post your answers as hidden.

1. How many subsets of the subset {1,2,...,10} contain tleast one odd integer?

*2. Prove that the number of isosceles triangles with integer sides, no side exceeding n is $\frac{1}{4}(3n^2+1)$ or $\frac{3}{4}n^2$ according as n is odd or even.

3. What is the number of distinct terms in the expansion of $(x_1+x_2+\cdot\cdot\cdot+x_m)^n$?

*4. If three distinct integers are chosen at random show that there will exist two among tham, say a and b such that 30 divides $(a^3b-ab^3)$.

*5. n objects are arranged in a row. A subset of these objects is called [i]unfriendly[/i] if no two of its elements are consecutive. Show that the number of unfriendly subsets of a [i]k[/i]-element set is\[ \binom{n-k+1}{k}  \]

6. Four numbers are chosen from 1 to 20. If $1\lek\le17$,in how many ways is the difference between the smallest and largest equal to [i]k[/i]?

Note:- '*' marked questions I was unable to solve.[/quote]

[hide="problem 1"]
there are $2^{10}$ total subsets of ${1,2,3,\ldots,10}$ we subtract the subsets that have only evens to find the subsets that contain at least one odds. $2^5$ are purely even. so $2^{10} - 2^5$ is the answer.[/hide]

i think you have a typo in the last problem. if 1<17?

Yes there is a TYPO and I have corrected it.

4-
[hide]
If one of the number is divisible by 30 it's finished.
Else separate the numbers depending on their remainder modulo 30 :

$1, 4, 6, 9, 11, 14, 16, 19, 21, 24, 26, 29$
$2, 3, 5, 7, 8, 10, 12, 13, 15, 17, 18, 20, 22, 23, 25, 27, 28$

By pigeon hole there are two numbers in the same group. As you can check these two numbers work (if I built my groups correctly :|).
 [/hide]

5-
[hide]
If u(n,k) is the number of unfriendly k-subset of $1 ... n$ then we have the induction relation :

u(n,k) = u(n-1,k) + u(n-2,k-1)

An unfriendly k-subset in a n-set may be built from the unfriendly k-subsets in a (n-1)-set or the unfriendly (k-1)-subsets in a (n-2)-set, depending whether (n-1) is used or not in the k-subset.

since u(n,1) = 1 and u(k,k) = 0 we have the same relation as binomial coefficients[/hide]


In a district of my city there exist $ 420$ houses, each house contain either a small car or a van or both. If the number of the small cars is $ 320$ and the number of vans in $ 280$. How many houses has a small car without a van?

$ \textbf{(A)}\ 90\qquad\textbf{(B)}\ 140\qquad\textbf{(C)}\ 180\qquad\textbf{(D)}\ 320$

[hide]Let $ x$ be the number of houses with both a small car and a van. Then, by PIE we have $ 420\equal{}320\plus{}280\minus{}x \implies x\equal{}180$. Subtracting this from the total number of small cars, we have $ 320\minus{}180\equal{}140$. Thus, the answer is B.[/hide]


A,B,C unghiurile unui triunghiu; unul dintre acestea este mai mic decat 30.
Aratati ca
$2sin \frac{A}{2}sin \frac{B}{2}sin \frac{C}{2}\le sin15-sin^{2}15$

Rescriem concluzia în forma $\sum \ln \sin \frac{A}2 \leq \ln \left( \sin \frac{\pi}{12}-\sin^{2}\frac{\pi}{12}\right)$.

Fie $f : \left( 0, \frac{\pi}2 \right) \to \mathbb R, \, f(x) = \ln \sin x$. Avem $f^{\prime \prime}(x) =-\frac1{\sin^{2}x}< 0$, deci $f$ e concavă.

WLOG, $\frac{A}2 \leq \frac{\pi}{12}$, iar $\frac{B}2 \geq \frac{5 \pi}{24}$.
Din Karamata avem $\ln \sin \frac{A}2+\ln \sin \frac{B}2 \leq \ln \sin \frac{\pi}{12}+\ln \sin \left( \frac{A+B}2-\frac{\pi}{12}\right)$, iar din Jensen
\[\ln \sin \left( \frac{A+B}2-\frac{\pi}{12}\right)+\ln \sin \frac{C}2 \leq 2 \ln \sin \left( \frac{5 \pi}{24}\right) .\]
Adunându-le, obţinem \[\sum \ln \sin \frac{A}2 \leq \ln \left( \sin \frac{\pi}{12}\cdot \sin^{2}\frac{5 \pi}{24}\right) = \ln \left( \frac{\sin \frac{\pi}{12}-\sin^{2}\frac{\pi}{12}}2 \right) ,\]
deci \[\prod \sin \frac{A}2 \leq \frac{\sin \frac{\pi}{12}-\sin^{2}\frac{\pi}{12}}2 .\]

Am greşit ceva?

o singura chestie ar fi :D  de unde stii ca stie derivate? daca problema a fost data la clasa a X-a? si chiar acolo cred ca a fost data, am eu o banuiala

Cred ca concavitatea lui $f$ iese şi fără derivate, dar nu prea am chef să verific definiţia  :noo:

Ceea ce mi se pare ciudat e că mi-a ieşit un $\frac12$ în faţă.

imi dau si eu cu parerea, dar nu am pus problema pe foaie deci e probabil sa gresesc eu la calcul
daca google imi zice bine, atunci $\sin \frac{B}{2}\sin \frac{C}{2}= \frac{1}{2}\Bigg(\cos\left(\frac{B-C}{2}\right)-\cos\left(\frac{B+C}{2}\right)\Bigg)$.
acum, $B+C=\pi-A$, deci $\cos\left(\frac{B+C}{2}\right)=\sin \frac{A}{2}$.
notez $t=\sin \frac{A}{2}$.
expresia devine $E=\frac{t}{2}\cdot\Bigg(\cos\left(\frac{B-C}{2}\right)-t\Bigg) \leq \frac{t(1-t)}{2}$.
acuma presupun ca $A\leq 30^{\circ}$, si rezulta $t\leq \sin 15^{\circ}<\frac{1}{2}$, si de aici
$E\leq \frac{\sin 15^{\circ}-\sin^{2}15^{\circ}}{2}$.
($f(x)=x(1-x)$ e crescatoare pe $[0,1/2]$.)
se pare ca si mie imi iese un $1/2$ in fata :)


Nathan will roll two six-sided dice. What is the probability that he will roll a number
less than three on the first die and a number greater than three on the second die?
Express your answer as a common fraction.

[b]First Die[/b]
1, 2 for a total of 2 successes out of 6, so the probability is $ \frac{2}{6}\equal{}\frac{1}{3}.$

[b]Second Die[/b]
4, 5, 6 for a total of 3 successes out of 6, so the probability is $ \frac{3}{6}\equal{}\frac{1}{2}.$

Since these events are independent, we multiply the probabilities to get $ \boxed{\frac{1}{6}}.$

Oh wow! You wrote this like less than a year after I was born!!!

On that note, I wasn't born when that was written.


Does MathCounts allow state teams to have assistant coaches? Furthermore, are there certain requirements that all coaches have to meet?

Don't tell me you want to coach the CA team...

I'm not sure... but we have an assistant coach who is in 11th grade. :|

I think mathcounts posted somewhere to have absolutely no more than one coach per state team.

Mathcounts will only pay for one coach, and only one coach can attend the coaches' dinner on Thursday, but several teams bring more than one coach.  At our coaches' dinner they always ask us to give the names of any assitant coaches they want recognized in the event one of our kids places high.

Ubemaya, you REALLY make me laugh (Trying to get a FREE trip to Nationals - haha)! :D

What about assistant [color=red][size=59]to[/size][/color] the coach? (credit: The Office :rotfl: )

Can you come to MathPath maybe?

I need to meet >1 aops people from different states (and due to the no from my dad towards mc nats, I need to look elsewhere).

jorian

[quote="jhredsox"]Ubemaya, you REALLY make me laugh (Trying to get a FREE trip to Nationals - haha)! :D [/quote]
Some people actually enjoy helping out the community that helped them.
Some people also enjoy helping their coaches that helped them get so far.
Some people enjoy helping other talented students suceed with math.

[quote="Ubemaya"]Wait, are the coaches/team members fees to Nats completely paid for?[/quote]
Yes, the coach and team members fees to nats are completely paid.
I don't know if assistant coaches are paid for as well.

Hi XPMath,

I believe you are in North CA, right since you know Miller?

Did your assistant coach go to Nationals? Did he take care members of your team when your official coach is away?

When I was in 8th grade, my team had two coaches.  Mathcounts will only pay for one, but a second is allowed to attend.  I don't know how it works with things like the coaches' dinner and all that.  I do know, however, that the school district covered at least part if not all of the cost of sending the second coach.  So if your state team ends up with two coaches, and you want to send both, you might look into external sources of funding like that.

Yeah. Last year we had 3 coaches come. Our state coach, the coach of two other team members, and my coach. This year, we'll probably have 3 coming. Our other unofficial coaches all of the other years came to some of the breakfasts (but just stopped in to say hi (too expensive)). So I guess who the national assistants are is up to your state coach.


search
calculus
derivative
calculus computations

For real numbers $ p,\ q$, find the necessary and sufficient condition such that a cubic eqiation $ x^3\minus{}3px\plus{}q\equal{}0$ has three distinct real roots in terms of $ p,\ q$.

The condition has been known since the 16th century. One hint: if there are three real roots, then the Cardano-Tartaglia formulas work very strangely, as they involve the square root of a negative number. Here are two loosely related links:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=675567764&t=187053]Link 1.[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=675567764&t=57856]Link 2.[/url]

Can anyone solve the problem with differentiation?

Let $ f(x) \equal{} x^3 \minus{} 3px \plus{} q.$ Assume $ p > 0.$ Else, $ f$ is strictly increasing and can only have one real root.

$ f'(x) \equal{} 3x^2 \minus{} 3p,$ so $ f'(x) \equal{} 0$ for $ x \equal{} \pm\sqrt {p}.$ Further examination of this derivative shows that $ f$ has a local maximum at $ \minus{} \sqrt {p}$ and a local minimum at $ \sqrt {p}.$ In order for $ f$ to have three distinct roots, we must have both $ f\left( \minus{} \sqrt {p}\right) > 0$ and $ f\left(\sqrt {p}\right) < 0.$

That is, $ 2p^{3/2} \plus{} q > 0$ and $ \minus{} 2p^{3/2} \plus{} q < 0.$

Rewrite this as $ 2p^{3/2} > \minus{} q$ and $ 2p^{3/2} > q.$ We can consolidate that as $ 2p^{3/2} > |q|.$

Squaring, we get that $ 4p^3 > q^2$ or $ 4p^3 \minus{} q^2 > 0.$ Phrasing it this way, we don't have to make $ p>0$ a separate condition, as it is included automatically.

Of course, this is identical to the conclusion reached by the algebraic means of the Cardano-Tartaglia method.


Putnam
algebra
polynomial
college contests

Prove that, for any integers $ a, b, c$, there exists a positive integer $ n$ such that $ \sqrt{n^3 \plus{} an^2 \plus{} bn \plus{} c}$ is not an integer.

Query:  would it have been at all reasonable to cite Siegel's theorem here, since it sort of trivializes the problem?

I don't know about quoting the theorem; it's not something I'm familiar with.

On the other hand, there is a reasonably simple way to do it directly. We split into cases:
Case 1: $ c\neq 0$. Consider $ n\equal{}9k^2$. We have that $ (27k^3\plus{}ak)^2\equal{}3^6k^6\plus{}3^4ak^4\plus{}9bk^2\plus{}c\equal{}n^3\plus{}an^2\plus{}\frac19a^2n$, and the difference between this and our hypothetical square is $ (9b\minus{}a^2)k^2\plus{}c$. For large enough $ k$, this is nonzero because $ c\neq 0$, and also strictly less than $ 2(27k^3\plus{}ak)\minus{}1$, the smallest difference between $ (27k^3\plus{}ak)^2$ and any other perfect square.
Case 2: $ c\equal{}0,b\neq 0$. Here, let $ n$ be a prime $ p$ not dividing $ b$. $ p$ divides $ n^3\plus{}an^2\plus{}bn$, but $ p^2$ does not.
Case 3: $ b\equal{}c\equal{}0$. $ \sqrt{n^3\plus{}an^2}\equal{}n\sqrt{n\plus{}a}$. Choose $ n$ so that $ n\plus{}a$ is not a perfect square.

In all cases, we have constructed a value of $ n$ making $ n^3\plus{}an^2\plus{}bn\plus{}c$ not a square, and we are done.

The approach generalizes. Any polynomial which is not the square of a polynomial takes nonsquare values infinitely often.

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]Finite differences[/url] to the rescue!

[quote="WOOT forum, where this problem was posted,"]Assume the contrary. Define $ k_n \equal{} \sqrt {n^3 \plus{} an^2 \plus{} bn \plus{} c}$. Taking finite differences on $ k_n^2$, we have:
\[ k_n^2 \minus{} 3k_{n \minus{} 1}^2 \plus{} 3k_{n \minus{} 2}^2 \minus{} k_{n \minus{} 3}^2 \equal{} 6
\]
Take both $ \text{mod }4$.
\[ k_n^2 \plus{} k_{n \minus{} 1}^2 \minus{} k_{n \minus{} 2}^2 \minus{} k_{n \minus{} 3}^2\equiv2\pmod4
\]
Note that the only squares $ \mod 4$ are $ 0,1$. If $ k_n$ is odd, we cannot subtract enough to get to $ \minus{} 2$, so $ k_{n \minus{} 1}$ must be odd and $ k_{n \minus{} 2}$ and $ k_{n \minus{} 3}$ even. Similarly, if $ k_n$ is even then $ k_{n \minus{} 1}$ is even and $ k_{n \minus{} 2}$ odd. This identity also holds true for $ k_{n \minus{} 1}$, so $ k_{n \minus{} 1}$ and $ k_{n \minus{} 2}$ have the same sign, a contradiction.[/quote]

Designate the expression $ n^3\plus{}an^2\plus{}bn\plus{}c$ as $ P(n)$. The statement claims that regardless of $ a,b,c$, there exists $ n$, such that $ P(n)$ is not a complete square. Assume the contrary:

Thus, $ P(b\plus{}1)\minus{}P(b\minus{}1)\equal{} 6b^2\plus{}2\plus{}4a\cdot b\plus{}2b$, which means that $ P(b\minus{}1)$ and $ P(b\plus{}1)$ have the same parity. Therefore, their difference has to be divisible by $ 4$, which is not the case, as long as $ 6b^2\plus{}2\plus{}2b$ is not divisible by $ 4$.

Finally, either of $ b\minus{}1$ and $ b\plus{}1$ must be suitable as $ n$.


Let $\mathcal{S}$ be the set of rational numbers such that for every $a,b\in \mathcal{S}$ the numbers $a+b$ and $ab$ belong to $\mathcal{S}$ and for every $r\in\mathbb{Q}$ just one of the statements $r\in \mathcal{S},-r\in \mathcal{S}, r=0$ is true. Prove that $\mathcal{S}=\mathbb{Q}\bigcap(0,+\infty).$

Assuming those rules are still meant to hold when $a = b$,

[hide]If $-1 \in S$ then $-1\cdot-1 \in S$ so $1 \in S$ and we're dead.  Thus $1 \in S$, and so $1+1, 1+1+1, \ldots \in S$ so ${\bf Z}_{>0}\in S$ and ${\bf Z}_{<0}\not\in S$.  Take any $r \in{\bf Q}_{< 0}$.  Then we can write $r =-\frac{p}{q}, p, q \in{\bf Z}$, and since $q \in S$, $-p \in S$, a contradiction.  Thus no negative rationals are in $S$, so all the positive rationals must be.  $0 = 0$ so $0 \in S$ is false, and we're done.[/hide]

Actually, I can get it without ever applying the rules to $a$ and itself:
[hide]If $-1$ and $2$ are in $S$, so is $-1+2 = 1$ and we lose.  But if $-1$ and $-2$ are in $S$, so is $-1 \cdot-2$, so we lose again and $-1 \not\in S, 1 \in S$.  If $-2 \in S$, so is $-2+1 =-1$, so we lose, so $-2 \not \in S, 2 \in S$.  This gives us the positive integers, as before, which is enough to rule out all negative rationals, which gives us all positive rationals.[/hide]


Σήμερα μετά από μια παρατήρηση έφτιαξα την ακόλουθη άσκηση που είναι εύκολη αλλά ωραία νομίζω .

Βρείτε δύο λύσεις της εξίσωσης
$ 16x^{4} \plus{} 8x^{3} \minus{} 12x^{2} \minus{} 4x \plus{} 1 \equal{} 0$

(όχι προσσεγγιστικά)

$-\frac{1}{2}$ και $\eta\mu(10^{o})$

P.S. Το ίδιο πρόβλημα είχα φτιάξει κι εγώ όταν πρωτοπήρα το βιβλίο άλγεβρας της Β' λυκείου.  :P  :P   :P

Σωστός ........... :)
Νομίζω ότι ήταν πολύ εύκολο ε??

παιδιά η εξίσωση αυτή λέγεται ειδικά κάπως θα ήθελα να μου πείτε τον όρο της εξίσωσης η οποία ειναι "συμμετρικές" ως πρός τις δυνάμεις
ευχαριστώ


The incircle of a triangle $ ABC$ touches the side $ AB$ at $ X$ and $ AC$ at $ Y$ . Let $ K$ be the midpoint of the arc $ AB$ of the circumcircle of triangle $ ABC$. Assume that the line $ XY$ bisects the segment $ AK$. What are the possible measures of angle $ BAC$?

see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=205955]here[/url]  :wink:


That's right. The problem is somewhat tricky, I don't think any team solved it.

I know this solution is not a nice one,e.lopes's solution is really good and enough...but me and my friend solved it this way:
we proved that for each $n\geq1$ we have $a_{n}\geq\frac{n-1}{2}$ then:

$\sum^{98}_{k=1}\frac{1}{a_{k}a_{k+2}}=\frac{1}{a_{1}a_{3}}+\sum^{98}_{k=2}\frac{1}{a_{k}a_{k+2}}\leq\frac{4}{7}+\sum^{98}_{k=2}\frac{4}{(k-1)(k+1)}=\frac{4}{7}+2\sum^{98}_{k=2}\left(\frac{1}{k-1}-\frac{1}{k}+\frac{1}{k}-\frac{1}{k+1}\right) =\frac{4}{7}+2\left(1-\frac{1}{98}+\frac{1}{2}-\frac{1}{99}\right)<4$ :)


[quote="ellysekwan"]There are 25 people in a room. It is given that any group of 3 people in the room 2 of them are foes. Prove that there is a person who has at least 12 foes.[/quote]

Take person A. Either person A has 24 foes or he doesn't. If he has 24, then he has atleast 12. If he doesn't then there is someone who isn't his foe. Call that Person B.
There are 23 other people. Because A and B aren't foes, each of these 23 people must be a foe of either A or B or both. A and B have 23 foes, so one of them must have at least 12.


Two circles intersect in 2 different points.  If their radii are 2000 and 6, respectively, how many integer values are possible for the distance between their centers?

[hide]Consider the two circles externally tangent to each other.  The two circles are now intersecting in 1 point and the distance between their centers is $2000+6=2006$.

Keeping the circle of radius 2000 stationary, moving the circle of radius 6 closer to the center of the other circle makes both circles intersect in two points.  Moving the circle of radius 6 one unit closer to the center of the other, the distance between their centers is $2000+6-1=2005$

Intuitively, if the circle of radius 6 moves closer to the center of the other circle a distance $x$, then the distance between their centers is $2000+6-x=2006-x$

However, we are only interested in values where the circles intersect in 2 points.  When $x=12$, the two circles will be internally tangent and only intersect in 1 point.  As a result, we can obtain integer values of distance between the centers when $0<x<12$, which yields $\boxed{11}$ different values.[/hide]


can someone please explain to me the chicken mcnugget therom?

Let's say we have an unlimited number of two kinds of boxes of chicken mcnuggets. One kind of box has m chicken nuggets, and the other kind has n, where m and n are relatively prime. The largest finite amount of chicken nuggets that we CANNOT have by using these boxes only is mn-m-n. Note: this only works if m and n are relatively prime.

[quote="lotrgreengrapes7926"]What if $\text{gcf}(m,n)=d\ne1$? Is there still a general formula?[/quote]

No because any integer that's not a multiple of d cannot be attained.

Yeah, but if we only considered multiples of d, then we can m and n by d, then use chicken McNugget on that, and then multiply by d, to get the largest multiple of d impossible to be achieved.


Let a segment of length L varie its positions across the first quartile, as in the scheme. Find the movement of its points.
Note: I don't need just a simple equation for its points; I've already found  one, but I need a certain curve that delimits the  surface  which the segment "scans" while moving .

I don't understand completly your enuntiation  :oops:  but  I think that maybe the curve described by the segment L is the tractrix

http://xahlee.org/SpecialPlaneCurves_dir/Tractrix_dir/tractrix.html


quadratics
analytic geometry
function
algebra
polynomial

If you know any three points and that it is a function (as opposed to a quadratic in y) you can determine the quadratic.

[quote="4everwise"]Is it possible to find the formula for a quadratic equation that does not have real solutions by looking at it in a coordinate plane?

For example, if the graph has real roots 3, 8, 9, and 2, you can find the equation by multiplying

(x-3)(x-8)(x-9)(x-2)

Is there a way to do this without seing the zeroes of the function?[/quote]

I think you mean polynomial, since quadratic implies the highest power of the variable is to the second degree.
You can determine a polynomial of degree $n$ if you know $n+1$ points using some tedious algebra. Compare this with solving a system of $n+1$ equations with $n+1$ variables.


From experience with Sudoku or other stuffs, we see that the answer is $ \boxed5$ (or by noting that the design must be pretty in order to be correct).  :P


Just out of curiosity, can anyone decipher this grader comment, specifically the underlined word?
[img]http://img.photobucket.com/albums/v189/bigsister/math/y18r4p5comment.png[/img]
It looks like "uzlea". The only word I can think of that looks similar is "unclear", but there is no way the letter between the u and the l is an n and a c put together. Plus the r is missing.

A Mathematical Test of 1976

Complete the sentence:

1)

A Few of you _____ we ____, But ___ rigth

I'd say "unclear" too, though the n and the c seem to have combined into one squiggly. I can only assume the tail of the a is the r. I write sort of like that sometimes.

I think it says, "A few of your claims are unclear, but overall right."

Ironic, isn't it?  He criticizes your work for being too "unclear", yet his handwriting is so unclear you can't really tell that the word he's using is "unclear".

[quote="bookaholic"]Just out of curiosity, can anyone decipher this grader comment, specifically the underlined word?
It looks like "uzlea". The only word I can think of that looks similar is "unclear", but there is no way the letter between the u and the l is an n and a c put together. Plus the r is missing.[/quote]
I can't read the handwriting any better than you, but as a grader, I can try to guess what that grader said. (No, that is not my handwriting.) For problem 5/4/18, four points was typically given for a workable proof with holes. "Unclear" fits that context. It would mean that you had a valid proof, but your write-up of that proof skipped some steps or poorly explained some steps.

We graders write our comments off the tops of our heads, in haste and in pen. I often look at the comment I just wrote and think that I phrased it awkwardly. Unfortunately, changing a comment would leave a mess.

Erin Schram
USAMTS grader