Let a circle with diameter $  BC $ and a point $  A $ on his plane , such that  $\triangle ABC $ is regular. On side  $ BC $  we take a point $ D $ with $ CD = 2DB $.Line
 $ AD $  meets the circle [b]for second time [/b]at $  E $ . Prove that $  BE / / AC $ .

 Babis

[i](  It could be preolympiad , but some times easier problems encourange the solvers ! So , to this problem I' d like to see as many as possible solutions)[/i]

are you sure the wording of the problem is correct?? i just cant get AD  to intersect the circle

Well , A is an exterior point of the circle , triangle$\triangle ABC  $is equilateral and D is on  -cirle's diameter - side BC. Line AD intersects the circle first in a point interior to the triangle and then, for second time , at E. Does this help , or I must write it in an other way?
Babis

Stergiu, you frisk with this problem. See (High School, Pre-Olimpiad, Shobber)
http://www.mathlinks.ro/Forum/viewtopic.php?p=331011#p331011

Levi ,yes ! It took me more than an hour to find a geometric solution( you know that I' m not so young !!!) for Shobber ' s problem . So I changed the wording  and posted it to solvers with more experience , - you are beetween them of course.I ' m sure that I have seen a good solution in the past , but I did not write it. Believe me , this is the only reason for reposting the problem.
   Cu respekt - Babis
(By the way , Tsitsifas has also an excellent geometry book in greek. )

It is O.K.  Try to prove the fourth criterion of equivalence and of similarity
(the enunciations are in reply at Grobber's problem).

Consider the omotety $f$ with center D and ratio -1/2; then
$f(C)=B$
$f(B)=O$
where O is the midpoint of BC.
Now, $F=f(A)$ is the vertex of the equilateral triangle on BO, in the halfplane that doesn't cointain A, therefore is on the circle with diameter BC, because $BF=BC/2$ and $\angle BOF=\pi/3$.
So, A,D and F are collinear and F is on the circle with diameter BC, then $F\equiv E$, but $BF\parallel AC$ (the second is the image of the first under omotety) and this is the thesis.


geometry
geometric transformation
rotation

Any suggestions on how  to teach a five year old mathematics? I have tried, but I don't know how to explain it to her. I want my little sister to be a math nerd like me, but its kind of hard when she still does not know addition or subtraction.  :|

Addition and subtraction? Use Cheerios. So easy to count and sort, and when you're done, you can eat them.

This is meant as a serious suggestion.

When you want to teach multiplication, first teach grids, rows, and columns.  Then you can visually represent it to her.  Another concept that helps multiplication is counting by intervals.  Counting by fives also greatly helps in money and time.  Normal multiplication might be beyond her now, but you never know how soon this might become helpful...

I do pattern recognition with my little four year-old brother. Sometimes it's the simple ones like <square> <triangle> <square> to rotations to number progressions. They're all pretty much what comes next. It's really tough trying to get a feel for what he can understand and what he fooled you into thinking he can understand.

THE questions for testing understanding are "How?" and "Why?"

Occasionally, when the situation allows, review topics, and get the kid to explain the topic to YOU.  If he/she can clearly phrase his/her thoughts, I, at least, think that's worth half a cookie or a shiny nickel.

Slight tangent?: Totally serious about the nickel.  Wash it and it becomes shiny and clean.  Like alot of kids, I still like shiny nickels.  You should have seen me at the nearby museum's gem exhibition.   :wink:

[quote="alexhhmun"]Totally serious about the nickel. [/quote]
Only for a child that's old enough not to automatically put everything into his or her mouth.

true.  bad idea.   :(   I should have left while I was ahead.

Start with word problems and real life situations instead of abstractions. Say "How many are there?" and "How long will it take?" instead of "What comes after eighty-seven?" and "What is $ 20 \minus{} 4$?". Make it amusing. Not "How many pencils does the teacher have?", but "How many chocolate bars did Mr. Fat eat?" Encourage thinking and early exposure to geometric ideas: "What is a circle?" "...A round thingy?" Draw an extremely huge elongated and twisted but round figure. Ask if it is a circle. "...NooooooOOOO... like this..." Pretend you don't know what a circle is. "Oh, so smaller!" Draw a small twisted and loopy thing. Etc.  :)


Find all real solutions to the following inequality:

$ \frac {8}{\sqrt {x \plus{} 6} \minus{} \sqrt {x \minus{} 2}} \leq 6 \minus{} \sqrt {x \plus{} 1}$

Reminds me quite a bit of an IMO problem from the mid-1960s.

[hide="First step"]
$ x\minus{}2\geq0\implies x\geq 2$, giving us a lower bound.

$ \text{LHS}\cdot\frac{\sqrt{x\plus{}6}\plus{}\sqrt{x\minus{}2}}{\sqrt{x\plus{}6}\plus{}\sqrt{x\minus{}2}}\equal{}\frac{8(\sqrt{x\plus{}6}\plus{}\sqrt{x\minus{}2})}{4}\equal{}2\sqrt{x\plus{}6}\plus{}2\sqrt{x\minus{}2}$

$ 2\sqrt{x\plus{}6}\plus{}2\sqrt{x\minus{}2}<6\minus{}\sqrt{x\plus{}1}$

Now it seems kind of ugly, unless there's a substitution or something that makes things easier.
[/hide]

hint:

[hide]Think about what values of $ x$ will make the radical values real, you can find an interval for $ x$.[/hide]

[quote="grn_trtle"]Reminds me quite a bit of an IMO problem from the mid-1960s.

[hide="First step"]
$ x \minus{} 2\geq0\implies x\geq 2$, giving us a lower bound.

$ \text{LHS}\cdot\frac {\sqrt {x \plus{} 6} \plus{} \sqrt {x \minus{} 2}}{\sqrt {x \plus{} 6} \plus{} \sqrt {x \minus{} 2}} \equal{} \frac {8(\sqrt {x \plus{} 6} \plus{} \sqrt {x \minus{} 2})}{4} \equal{} 2\sqrt {x \plus{} 6} \plus{} 2\sqrt {x \minus{} 2}$

$ 2\sqrt {x \plus{} 6} \plus{} 2\sqrt {x \minus{} 2} < 6 \minus{} \sqrt {x \plus{} 1}$

Now it seems kind of ugly, unless there's a substitution or something that makes things easier.
[/hide][/quote]

Are you sure that $ \text{LHS}\cdot\frac {\sqrt {x \plus{} 6} \plus{} \sqrt {x \minus{} 2}}{\sqrt {x \plus{} 6} \plus{} \sqrt {x \minus{} 2}} \equal{} \frac {8(\sqrt {x \plus{} 6} \plus{} \sqrt {x \minus{} 2})}{4}$?

I think once you fix it, it should be easier.

[hide="Solution"]It is apparent that $ x \geq 2$. Observe that:

(1) $ \sqrt {x \plus{} 1}$ is strictly increasing, thus $ 6 \minus{} \sqrt {x \plus{} 1}$ is strictly decreasing.

(2) $ \sqrt {x \plus{} 6} \minus{} \sqrt {x \minus{} 2}$ is strictly decreasing, thus $ \frac {8}{\sqrt {x \plus{} 6} \minus{} \sqrt {x \minus{} 2}}$ is strictly increasing.

(3) At our lower bound $ x \equal{} 2$, the former is greater than the latter.

(4) The LHS is strictly positive, but the RHS is eventually negative, so eventually, the LHS exceeds the RHS

Considering the above, we realize there must be a unique number $ r>2$ such that the solution set is $ 2 \leq x \leq r$.

By inspection, this number is $ r\equal{}3$, and the solution set is $ \boxed{2 \leq x \leq 3 }$[/hide]


Is there supposed to be another parenthesis anywhere?

There is not supposed to be a ) at the end of the 128s

Uh... in mixed fractions... that's equal to $ 1\plus{}\frac{256}{(x^2\minus{}16)\sqrt{x^2\minus{}64}\plus{}x^2\minus{}48x\minus{}128}$ and $ x^2\minus{}48x\pm128$ are both not reducible in $ \mathbb{Z}$. Do you want to rationalize the numerator/denominator? It's a mess  :maybe:

It's supposed to simplify to something, I don't know what.


I don't know if this is true or not.  I am not aware of any proof of it, but it would certainly be neat if the so-called [tex]p[/tex]-norm were a descreasing function of [tex]p[/tex].

Let [tex]n[/tex] be a positive integer.  Prove (or disprove) that [tex]\forall p, q \in \mathbb{R^{+}} \,| \,p \ge q; \forall\mathbf{x} \in \mathbb{R}^{n}[/tex], we have

[tex]
\displaymode{
||\mathbf{x}||_{p} \le ||\mathbf{x}||_{q}
}[/tex]

Where we define the [tex]p[/tex]-norm of [tex]\mathbf{x} = \,\,<x_{1}, \dots , x_{n}>[/tex] by

[tex]
\displaymode{
||\mathbf{x}||_{p} = \sqrt[p]{|x_{1}|^{p} + \cdots + |x_{n}|^{p}}
}[/tex]

Shouldn't the inequality go the other way? I think you can prove it with Holder's Inequality.

I think the above is the correct statement, or the following is a quick counterexample

[tex]\sqrt[3]{91} = \sqrt[3]{3^3 + 4^3} <  \sqrt{3^2 + 4^2} = 5[/tex].

Does it follow from the expansion of the [tex]pq[/tex]-th power of both sides?

Interesting. Then I think you are right. I'm not sure if you can do it by raising each side to the power of $pq$. But geometrically it might be ``obvious'' by looking at unit ``circles'' with respect to various $l_p$-norms and considering them as distance formulae. (At least that convinces me that it's true...)

Ok, this is actually quite a low-tech result.   :D   Since the entire inequality is homogenous in the first degree, we assume [tex]x_{1}^{p} + \cdots + x_{n}^{p} = 1[/tex]. Let [tex]a_{i} = x_{i}^{p}[/tex].  Substituting [tex]x_{i}[/tex] with [tex]a_{i}[/tex], we want

[tex]\displaymode{
1 = \sqrt[p]{a_{1} + \cdots + a_{n}} \le \sqrt[q]{a_{1}^{\frac{q}{p}} + \cdots + a_{n}^{\frac{q}{p}}}[/tex]

But we have [tex]0 \le a_{i} \le 1[/tex], where the upper bound results from the sum of the [tex]a_{i}[/tex] being 1.  Hence, [tex]a_{i}^{\frac{q}{p}} \ge a_{i}[/tex] (*), and it follows immediately that

[tex]\displaymode{
\sqrt[q]{a_{1}^{\frac{q}{p}} + \cdots + a_{n}^{\frac{q}{p}}} \ge \sqrt[q]{a_{1} + \cdots + a_{n}} = 1}[/tex]

which is sufficient.  As for the equality condition, equality in (*) holds iff [tex]a_{j} = 0[/tex] or [tex]a_{j} = 1[/tex].  It is now clear that equality in the [tex]p[/tex]-norm inequality holds iff [tex]x_{1} = \cdots = x_{j-1} = x_{j+1} = \cdots = x_{n} = 0[/tex] and [tex]x_{j} \in \mathbb{R}[/tex].  Moreover, [tex]||\mathbf{x}||_{p}[/tex] is a strictly decreasing function of [tex]p[/tex] iff [tex]\exists[/tex] at least two non-zero components of [tex]\mathbf{x}.[/tex]


1.A body falling from a given height H hits an inclined plane in its path at a height

   'h',as a result of its impact the direction of the body becomes horizontal.For what

    value of  (h/H),the body will take maximum time to reach the ground

2.A police inspector in a jeep is chasing a pickpocketer in a straight road.The

  jeep is going at its maximum speed v.The pickpocketer rides on the motorcycle

  of a waiting friend.When the jeep is at a distance 'd' away and the motorcycle starts

  with a constant acceleration 'a'.Show that pickpocketer will be caught if

  $v\ge\sqrt{2ad}$

[quote="madatmath"][size=150][b]$1$[/b][/size].whatever happens, the net time taken has to be [b]$\sqrt{\frac{2(H - h)}{g}} + \sqrt{\frac{2h}{g}} \geq 2 \sqrt{\frac{2}{g} \sqrt{(H - h)(h)}} \geq 2\sqrt{\frac{h}{g}}$[/b]
 the required ratio of [b]$\frac{h}{H} = \frac{1}{2}$[/b]

 [size=150][b]$2$[/b][/size]. Consider the frame of reference of the pickpocketer.
 w.r.t him, the inspector is slowing down with a decelration of [b]$a$[/b] ,
 having started with initial velocity[b] $v$.[/b]
 then, we require to find the minimum [b]$v$[/b] such that the inspector covers [b]$d$[/b].
 using third eq^n of kinematics,
 [b]$v^2 = 2ad$[/b],
 hence the result.[/quote]

1.your answer is right but why is $\sqrt{\frac{2}{g} \sqrt{(H - h)(h)}} \geq 2\sqrt{\frac{h}{g}}$
how did you conclude $\frac{h}{H} = \frac{1}{2}$

2.if the inspector covers the distance d with the minimum v then the pickpocketer could have covered some distance .then how could he catch the p.p

[quote="bkunchanapalli"][quote="madatmath"]$1$. The minimum value of $(H - h)(h)$ is $h^2/4$and occurs at
        $h = H/2$. this is easily verifiable by differentiation.

$2$. Not with minimum $v$.
        you see, we are talking in the pickpocketer's frame of reference,
        and the pickpocketer does not move w.r.t himself. He watches the inspector being the distance $d$ away, and  then actually decelerating as he moves towards the p.p. this is becoz' in the ground frame of ref. , the p.p. is acclerating away frm the inspector.[/quote]

yes.thanks for the solution[/quote]

wait a minute
i think after differentiation it should be $H^2$/4 instead of $h^2$/4

[quote="madatmath"]true, true.
 my mistake. i think i forgot to hit the shift key   :oops:
 the correct answer is $2 \sqrt{\frac{H}{g}}$[/quote]

it's ok once again thanks for the solution


[hide]That is the general form for all perfect numbers that we know of (all even ones): http://en.wikipedia.org/wiki/Perfect_number.

There are $ (k-1+1)2=2k$ factors.

The sum of the factors is:

$ (1+2+4...+2^{k-1})(1+2^{k}-1)$
$ (2^{k}-1)2^{k}$

And if we subtract the number itself we get the original number, which is the definition of a perfect number.[/hide]

[quote="jonysatie"]Let $ n=2^{k-1}(2^{k}-1)$ be a natural, where $ k$ is a positive integer such that $ 2^{k}-1$ is prime.
 Compute the number of divisors of $ n$ and the sum of them.[/quote]

The number theoretic function $ \sigma (n)$ which is the sum of the divisors is [b]weakly multiplicative[/b]. Meaning $ \sigma (ab) = \sigma(a) \sigma(b)$ whenever $ \gcd(a,b)=1$.

So, $ \sigma \left( 2^{k-1}(2^{k}-1) \right) = \sigma (2^{k-1}) \cdot \sigma (2^{k}-1)$

Now, $ \sigma (2^{k}-1)=2^{k}$ because it is prime.

Und, $ \sigma (2^{k-1}) = 1+2+2^{2}+...+2^{k-1}= 2^{k}-1$

So their product is,
$ 2^{k}(2^{k}-1)$

Which is TWICE as much as the number itself. So the number is perfect.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Compute:
$ \int_{ \minus{} \infty}^{ \plus{} \infty} \frac {e^{ix t} }{1 \plus{} i x} \, \mathbf{d}x$

${ \textbf I\;=\;\int_{-\infty}^\infty}\;\dfrac{e^{\imath \,x\,t}}{1+\imath\,x}\;\textbf dx$

${ =\; \int_{-\infty}^\infty} \dfrac {\left[\cos\,(x\,t)+\imath\,\sin\,(x\,t)\right]\cdot\left(1-\imath\,x\right) } {1+x^2}\;\textbf dx$

${{ \;=\int_{-\infty}^\infty} \left[ \dfrac{\cos\,(x\,t)+x\,\sin\,(x\,t)}{1+x^2}\right]\textbf dx\;+\;\imath\;\int_{-\infty}^\infty} \left[ \dfrac{\sin\,(x\,t)-x\,\cos\,(x\,t)}{1+x^2}\right]\textbf dx$

${{ \;=\int_{-\infty}^\infty} \left[ \dfrac{\cos\,(x\,t)-\dfrac{d}{dt}\cos\,(x\,t)}{1+x^2}\right]\textbf dx\;+\;\imath\;\int_{-\infty}^\infty} \left[ \dfrac{\sin\,(x\,t)-\dfrac{d}{dt}\sin\,(x\,t)}{1+x^2}\right]\textbf dx$

$ =\;\left(\pi\,e^{-t} -\dfrac{d}{dt}\left(\pi\,e^{-t}\right)\right)+\;\imath\left( 0 -\dfrac{d}{dt} (0)\right)$

$ =\;\boxed{2\pi\,e^{-t}}$

[quote="@petko"]Very nice :D Thank you!
But isn't it true only for t>0 ? :huh:[/quote]
When t<0, the answer is zero. When t=0, the integral does not converge, however it is reasonable to set its value $ \pi$.
You can see the conclusion easily from the following (rough) derivation
$ \int_{ \minus{} \infty}^{ \plus{} \infty}\frac {e^{ix t} }{1 \plus{} i x}dx \equal{} \int_{ \minus{} \infty}^{ \plus{} \infty}\int_{ \minus{} \infty}^t e^{i x u \plus{} u \minus{} t}du dx \equal{} \int_{ \minus{} \infty}^t 2\pi\delta(u)e^{u \minus{} t} du$

[b]Buffalo[/b] could you explain in more detail the very last part of your equation, please?
I mean:

$ \int\limits_{ \minus{} \infty}^{ \plus{} \infty}\int\limits_{ \minus{} \infty}^t e^{i x u \plus{} u \minus{} t}du dx \equal{} \int\limits_{ \minus{} \infty}^t 2\pi\delta(u)e^{u \minus{} t} du \stackrel{\text{is it true that...?}}{\Rightarrow} \int\limits_{ \minus{} \infty}^{ \plus{} \infty} e^{i x u} dx \equal{} 2 \pi \delta(u)$

:?: :huh:

I think that if t=0 the integral is $ 2\pi$.


Is there a subset of $ E\subset [0,1]\times [0,1] \times [0,1]$ such that $ m(E)>0$ and $ E$ is connected, compact, and nowhere dense?

Let $ A$ be a "fat" Cantor set in $ [0,1]$ - construct it by deleting middle intervals, but arrange the sizes of the deleted intervals so that $ m(A)>0.$ Now let ${ E=\{(x,y,z}: x\in A \text{ or }y\in A\text{ or }z\in A\}.$ This should be path-connected, hence connected, and compact and nowhere dense should be true as well.

Oh. This was what I was fiddling with, but I couldn't quite tweak it properly. I didn't think about getting connected from allowing the other two dimensions to vary at each point.


[quote="Christine"]Show that if a is an integer and d is a positive integer then when a divides d the quotient is floor (a/d) and remainder is a - d * floor (a/d)[/quote]

[hide="Not sure"]I guess this really depends on how you define quotient. If you define it as the largest multiple of $ d$ that divides $ a$, then it is obviously $ \lfloor a/d \rfloor$. So:

$ a/d \equal{} \lfloor a/d \rfloor \plus{} r/d$

Solve for $ r$ to get $ r\equal{}a\minus{}d\cdot\lfloor a/d \rfloor$.[/hide]

[hide="Part 2"]The remainder is $ a \minus{} n$, where $ n$ is the largest multiple of $ d$ less than $ a$. This is $ d\left\lfloor\frac {a}{d}\right\rfloor$, so the remainder is $ a \minus{} d\left\lfloor\frac {a}{d}\right\rfloor$.[/hide]


[quote="fuzzylogic"]I can't see quote for the messages in the Sentbox, Outbox and Savebox, Why?[/quote]

I don't know about savebox, but outbox and sentbox is stuff that you sent, so usually you ain't quoting yourself :)

The thing doesn't think of latex apparently :)

fuzzylogic, I tried it out, and in my box there are quote buttons there. Just check it out, they're at the place where reply/newtopic buttons usually are..

[quote="fuzzylogic"][quote="Valentin Vornicu"][quote="fuzzylogic"]I would like to copy some LaTeX code in the private message that I sent. But I can't see quote for the messages in the Sentbox, Outbox and Savebox, Why?[/quote]if you are really lazy, you can send yourself that pm (works!) and then quote yourself :D[/quote]

I don't understand what you mean. I sent a message before. So there is a saved copy in the Sentbox. How do I reply that pm to myself?

I am pretty sure it's very easy for the administrator to add a "post reply" or "quote" button, but they are not there right now.[/quote]I am one of the Administrators, and I was there :) actually I added the quote buttons today, but I didn't yet get them to work. Have a little patience :)

If you really need it you can just go to [u]V[/u]iew -> Sour[u]c[/u]e in your menu

there you should find the latex code somewhere ( ... alt="$here_the_latex_code$"...)

[quote="darij grinberg"]
[quote="fuzzylogic"]I strongly suggest admin to let us see the quote of our own PMs.[/quote]

I agree. Note that the "quote message" button is there, but strangely enough it just leads back to the inbox.

  Darij[/quote]

Admin, could you fix the problem?


Show that the triangle whose angles satisfy the equality \[\frac{\sin^{2}{A}+\sin^{2}{B}+\sin^{2}{C}}{\cos^{2}{A}+\cos^{2}{B}+\cos^{2}{C}}=2\]
is a right-angled triangle .

Trivial by the well-known identity $\cos^{2}A+\cos^{2}B+\cos^{2}C = 1-2\cos A\cos B\cos C$.

[hide]
$\frac{3-(\cos^{2}A+\cos^{2}B+\cos^{2}C)}{\cos^{2}A+\cos^{2}B+\cos^{2}C}=2$
$3-(\cos^{2}A+\cos^{2}B+\cos^{2}C)=2(\cos^{2}A+\cos^{2}B+\cos^{2}C)$
$3=3(\cos^{2}A+\cos^{2}B+\cos^{2}C)$
$\cos^{2}A+\cos^{2}B+\cos^{2}C=1$
$1-2\cos A\cos B\cos C=1$
$\cos A\cos B\cos C=0$
So one of those cosines has to be 0 meaning one of the angles is right.
[/hide]


geometry
circumcircle
geometric transformation
homothety
cyclic quadrilateral
power of a point
radical axis

Quadrilateral ABCD is inscribed in circle k centered at O. The bisector of angle ABD meets AD and k at points K and M, respectively. The bisector of angle CBD meets CD and k at points L and N, respectively. Suppose that KL ll MN.  Denote midpoint of BD with S. Prove that fopur points M, O, N and S are concyclic.

What you need to prove is to show that $MBND$ is harmonic.

Let $(P)$ be the circumcircle of the triangle $\triangle MON$ intersecting the center line $OP$ of the circles $(O), (P)$ at another point $Q$ different from $O.$  Since $OQ$ is a diameter of $(P),$ $OM \perp QM,\ ON \perp QN$ and $QM, QN$ are tangents to $(O)$ from $Q.$ Hence

$\angle QMN = \angle QNM = \angle MAN = \frac{\angle ABC}{2}$

$\angle QMN = \angle QNM = \frac{\angle MON}{2}$

It follows that $\angle MON = \angle ABC$ and from the cyclic quadrilateral $OMQN,\ ABCD,$ $\angle MQN = \angle ADC \equiv \angle KDL.$ Let $(O)$ meet the center line $OP$ at $U, V$ outside and inside of $(P),$ respectively, and let a parallel to $UV \equiv OP$ through $D$ meet $(O)$ again at $T.$ Then

$\angle AOT = 180^\circ-(2 \angle MOV+\angle VOD)+\angle UOT = 180^\circ-2 \angle MOV$

$\angle COT = 180^\circ-(2 \angle NOV-\angle VOD)-\angle UOT = 180^\circ-2 \angle NOV$

It follows that $T$ is the midpoint of the arc $AC$ opposite to $D,$ so that $DT \parallel OP$ bisects the angle $\angle ADC \equiv \angle KDL.$  The radical axis of $(O), (P)$ is perpendicular to their center line, $MN \perp OP.$ By the problem condition, $KL \parallel MN.$ Combining, we get $DT \perp KL,$ which means that the triangle $\triangle KDL$ is isosceles with $DK = DL.$ As a result, the isosceles triangles $\triangle KDL \sim \triangle MQN$ with $\angle KDL = \angle MQN$ are centrally similar with the similarity center $B \equiv KM \cap LN.$ Thus the line $QD$ connecting their remaining corresponding vertices $Q, D$ passes through the similarity center $B.$ Let $QD \equiv BD$ meet $(P)$ again at $S.$ Since $OQ$ is a diameter of $(P),$ the angle $\angle QSO \equiv \angle DSO = 90^\circ$ is right, which means that $S$ is the midpoint of the chord $BD.$

Consider the homotety centered at $B$ which sends $KL$ into $MN$.
It will map the point $D$ into a point $X$ which is the intersection of the tangents to the circumcircle of $ABCD$ at $M$ and $N$. It's easy to see that $MOSNX$ is cyclic and we are done.


[quote=MellowMelon]$ x \equal{} y \equal{} 0$ gives $ f(0)^2 \equal{} 0$, so $ f(0) \equal{} 0$.

$ x \equal{} 1, y \equal{} \minus{}1$ gives $ f(1)f(\minus{}1) \equal{} \minus{}f(1)$. Either $ f(1) \equal{} 0$ or $ f(\minus{}1) \equal{} \minus{}1$.

Suppose $ f(1) \equal{} 0$. Then $ x \equal{} 1$ gives $ f(1\plus{}y) \equal{} 0$, or $ f \equiv 0$. This is a solution. Now assume it away. $ f(\minus{}1) \equal{} \minus{}1$.

$ y \equal{} \minus{}1$ gives $ f(x^2) \equal{} xf(x)$. Put $ x \equal{} \minus{}1$ in to get $ f(1) \equal{} 1$.

$ x \equal{} 1$ gives $ f(y\plus{}1) \equal{} f(y)\plus{}1$.

Substituting $ f(x^2)$ for $ xf(x)$ gives $ xf(x\plus{}xy) \equal{} xf(x) \plus{} xf(x)f(y)$. Let's only allow $ x$ to be nonzero, and then the equation reduces to $ f(x(y\plus{}1)) \equal{} f(x) \plus{} f(x)f(y)$, which reduces to $ f(xy) \equal{} f(x)f(y)$ when $ y$ is shifted by $ 1$.

Put in $ x \equal{} y$ in this new equation to get $ f(x^2) \equal{} f(x)^2$, or $ xf(x) \equal{} f(x)^2$. So for all $ x$, $ f(x) \equal{} 0$ or $ f(x) \equal{} x$. If $ f(a) \equal{} 0$ for some $ a \neq 0$, then $ f(xa) \equal{} f(x)f(a) \equal{} 0$. $ xa$ can take any nonzero real, so $ f \equiv 0$ in this case.

We end up with $ f(x) \equal{} x$ or $ f(x) \equal{} 0$ as our solutions.[/quote]

Can you pleas explain " which reduces to $ f(xy) = f(x)f(y)$ when y is shifted by 1"
I just cant figure out what you did there. Help please ;)

$f(x(y+1)) = f(x) + f(x)f(y)$ is equivalent to $f(xy) = f(x) + f(x)f(y-1) = f(x)(1 + f(y-1) = f(x)f(y)$ when you replace $y+1$ with $y$.

Let $P(x,y)$ be the assertion into the problem statement.
$P(0,y)\implies f(y)f(0)=0.$  Either $\boxed{f(x)=0}$ for all $x$ or we have $f(0)=0$.  Assume that there is some $x$ such that $f(x)\neq 0$; thus, we have $f(0)=0$.
$P(1,-1)\implies 0=f(1)+f(1)f(-1)\implies f(1)(1+f(-1))=0.$  Either $f(1)=0$ or $f(-1)=-1.$

[b]Case 1:[/b] $f(1)=0$.
$P(1,y)\implies f(y+1)=0$ but this doesn't satisfy the assumption that there exists some $x$ such that $f(x)\neq 0$.  We already counted this solution before the assumption of $f(0)=0$.

[b]Case 2:[/b] $f(-1)=-1$
$P(-1,-1)\implies 0=1-f(1)$ so $f(1)=1$.

Thus, $P(1,y)\implies f(y+1)=f(y)+1$.
$P(x,-1)\implies xf(x)=f(x^2)$.
 Substituting these two identities into the problem yields $xf(x+xy)=xf(x)+f(x^2)f(y)=xf(x)(1+f(y))=xf(x)(f(y+1))$ so $$f(x+xy)=f(x)(f(y+1))$$ given that $x\neq 0.$
Let $Q(x,y)$ be the assertion into this new function.
$Q(x,x-1)\implies f(x^2)=(f(x))^2$.
Hence, $(f(x))^2=f(x^2)=xf(x)$ so $$f(x)(f(x)-x)=0.$$
Now, we've run into the Pointwise Values Trap. We claim that the only possible solution from this if $\boxed{f(x)=x}$ and pathological solutions do not exist. Suppose for the sake of contradiction that there exists distinct values $s$ and $t$ such that $f(s)=s$ and $f(t)=0$ for $s,t\neq 0$(since they still can be in the same function).  Then, $f(s+st)=s+st$ or $f(s+st)=0$.  We consider the revised function $$xf(x+xy)=xf(x)(f(y)+1).$$

[b]Case 2.1:[/b] $f(s)=s, f(t)=0$, and $f(s+st)=s+st$
Let $x=s$ and $y=t$ in the revised function yields $sf(s+st)=sf(s)(f(t)+1)\implies s^2+s^2t=s^2$ or $s^2t=0$ so we must have either $s=0$ or $t=0$, contradiction.

[b]Case 2.2:[/b] $f(s)=s, f(t)=0$, and $f(s+st)=0$.
Let $x=s$ and $y=t$ in the revised function yields $sf(s+st)=sf(s)(f(t)+1)\implies 0=s^2$ or $s=0$, contradiction.

Hence, we have proven that $f(x)=x$ is the only solution that is not identically zero and our two solutions are $f(x)=x$ and $f(x)=0.$ $\blacksquare$

all the solutions are beautiful but i found an easier solution

[quote=orl]Determine all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that
\[ x f(x \plus{} xy) \equal{} x f(x) \plus{} f \left( x^2 \right) f(y) \quad  \forall  x,y \in \mathbb{R}.\][/quote]

Can anyone check this proof?
The only solutions are $f\equiv \text{Id}$ and $f\equiv 0$ which clearly works.We will now prove that indeed these are the only solutions.

Let $P(x,y)$ denote the assertion $x f(x + xy) = x f(x) + f \left( x^2 \right) f(y)$.$\qquad$.           We proceed in the following steps:-
              Note that $P(0,0)$ gives $f(0)=0$.From $P(x,-1)$ we get $f(-1)f(x^2)+xf(x)=0$.       Now split into two cases; If $f(-1)=0$ then we get one of the solutions $f\equiv 0$.Otherwise Compare $P(-1,x)$ and $P(-1,-x)$ to see $xf(x)=-xf(-x)$ which implies $f$ is odd.

Now $x=1$ in the above para easily gives $f(1)=1$.Thus $f(-1)=-1$ and hence $f(x^2)=xf(x)$ and  $P(1,y)$ gives $f(y+1)=f(y)+1$.To finish up $$P(x,x-1)\implies x^2f(x)=xf(x^2)=xf(x)+f(x^2)f(x-1)=xf(x)+f(x^2)f(x-1)=xf(x)+xf(x)(f(x)-1) \implies f(x)=x \forall x\neq 0 $$ but since $f(0)=0$ hence we conclude $f(x)=x$ for all $x\in \mathbb{R}$.$\square$

[quote=Pluto1708]$x^2f(x)=......=xf(x)+xf(x)(f(x)-1) \implies f(x)=x \forall x\neq 0 $
No
$x^2f(x)=xf(x)^2$ implies
$\forall x\ne 0$, either $f(x)=0$, either $f(x)=x$

[quote=pco][quote=MathLuis]$(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) \implies x+1=1+f(x) \implies f(x)=x$ for all $x \in \mathbb R$[/quote]
No.
$(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) $ implies :
$\forall x\in\mathbb R$, either $f((x+1)^2)=0$, either $f(x)=x$[/quote]

Yes @pco i made a mistake in that part i have done yet the case when $f(x)=0$ soo that doesnt make a problem i think

$xf(x+xy)=xf(x)+f(x^2)f(y)$
$P(0,0)$ $\implies$ $f(0)=0$
$P(-1,-1)$ $\implies$ $f(-1)=0$ or $f(1)=1$
If $f(-1)=0$ then $P(x,-1)$ $\implies$ $f(x)=0$ which is a solution.
If $f(1)=1$
$P(1,x)$ $\implies$ $f(x+1)=1+f(x)$
$P(x,x-1)$
$xf(x^2)=xf(x)+f(x^2)f(x-1)=xf(x)+f(x^2)(f(x)-1)$
$xf(x^2)=xf(x)+f(x^2)f(x)-f(x^2)$
$f(x^2)=\frac{xf(x)}{x+1-f(x)}$
it is obvious that $f(x) \ne x+1$
Let's substitute this with original equation
$xf(x+xy)=xf(x)+\frac{xf(x)f(y)}{x+1-f(x)}$
Call it's assertion $Q(x,y)$
$Q(1,-1)$  $\implies$ $f(-1)=-1$
$Q(-1,x)$ $\implies$ $f(x)=-f(-x)$
$Q(-x,-1)$ $\implies$ $x+1-f(x)=1$
so that, $f(x)=x$
We have pointwise trap which is easy to get contradicition.
Sols:$f(x)=0,x$


Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.5×10^4 m/s when at a distance of 2.3×10^11 m from the center of the sun, what is its speed when at a distance of 5.9×10^10 m .

[hide="HINT"]Conversation of angular momentum.[/hide]


linear algebra
matrix
vector
LaTeX
linear algebra unsolved

Hello, Can someone help me on any of these problems ?

1) Let A be a matrix with n columns. Prove that Null(A)=R^n if and only if A=0

2) Let A and B be matrices with n columns. Prove that A=B if and only if Ax=Bx, for all x in R^n

3)Let V=R^2={(a) (b)* | a,b is contain in R } be the vector space of column vectors of length 2.
                                                                                Show that W={(a) (b)* | a,b is contain/element of Z}is not a vector subspace of V

I would appreciate any help and thank you in advance

* the (b) should be written below the letter "a" but I don't know how to use the computer to do it like that. I am not very good at computers  :blush:

for the first and second questions ,start plugging in special vectors in $\mathbb{R}^n$ :

like these : $v_{i}=\left[\begin{matrix} 0\\ \vdots\\ 1\\ \vdots \\0 \\ \end{matrix}\right]$ with zeroes everywhere except on the i th position

I am terrible with computers yet these people here helped me  :oops:  to write in a much neater way :

Press the button "Latex Help" on top of your screen, it will take you to this link:
[url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]

Thanks for your reply fredbel6, I'll try to learn and use Latex next time as only Question 3 require the use of Latex anyway.

Meanwhile, I really want to know what the solution is for Question 1 and Question 2, but I am more interested to know what the solution is for Question 1  than Question 2 as I can't figure it out. However, if anyone who can solve Question 2 but not Question 1, feel free to post it here as I am dying to know the solution.

Can you fredbel6 or anyone  help me on any of the first two questions ? I reallly really want to know it  :)

(Question 1 and Question 2 doesn't need to use Latex for it , so the question is as it is :)

1) The dimension of $N(A)$ (the nullspace of $A$) is $n-r$ ($r$ is the rank) so $N(A) = \mathbb{R}^n$ iff $r = 0$, i.e. $A = 0$.

2) If $A = B$, then clearly $Ax = Bx$ for all $x \in\mathbb{R}^n$.  If $Ax = Bx$ for all $x \in\mathbb{R}^n$ then $(A-B)x = 0$ for all $x \in\mathbb{R}^n$, i.e. $A-B$ is the zero matrix, and this is exactly what we want.

3) So $V = \{(a,b) | a,b\in\mathbb{R}\}$ and $W = \{(a,b) | a,b\in\mathbb{Z}\}$.  If $W$ is a subspace of $V$, then for instance, $c(a,b) \in W$ for any $c \in\mathbb{R}$ and $(a,b) \in V$ which is absurd.


Princeton college geometry email 2007 function videos

In this thread you answer a question that's posted, and then you ask a generalized question to someone on this forum (whoever chooses to answer it), and so on. ( No math questions)
 ( and if you're on geobeecentral -yes i did copy this from there)

I'll start---

Do you like chocolate?

No, I'm allergic to it.

(BTW, geobeecentral.com is a geography site with a forum. There are a few GBC members who are also AoPS members. )

What is the most useless emoticon?

:arrow: I think it's this one....I don't know why..

Do you have a sibling?

[img]http://www.venganza.org/forum/images/smiles/fsm7.gif[/img]

[url=http://www.venganza.org/]That^[/url]
What is the largest 20 digit prime?

[quote="nonie"]

Do you have a sibling?[/quote]
yes, she's going into 6th grade.
 Do you have an i-pod?

yeah i do  :D

What country do you live in?

The USA. What is your favorite kind of rodent?

[quote="nonie"]Do you have a sibling?[/quote]

[quote="PenguinIntegral"][img]http://www.venganza.org/forum/images/smiles/fsm7.gif[/img]

[url=http://www.venganza.org/]That^[/url][/quote]

Is that your brother or your sister? :D

chinchillas

if you are to be turned into a dessert of your choice(dont ask why), what would you be?

ice cream, I think..

If you could be any of the emoticons on this forum, which one would you be?

It sounds familiar.

Do you own a pet reptile?

nope

do you know what Japanese guy is nicknamed "Tsunami"?

Because you PROBABLY are.. but you're never sure...

why is  :P 's tongue pentagonular?

idk, my bff rose?

Why are GEICO's commercials so stupid?

Stupidity attracts money.

Is money 1337?

[quote="Brut3Forc3"]It is a language where you talk like this.

If pro is the opposite of con, what is the opposite of Congress?[/quote]
It should be progress by logic, but by common sense it isn't.
 :huh:
\/\/l-lY 4lVl 1 5711_1_ 741_1<1l\lG 1_11<3 7l-l15?

Because you think its cool for some odd reason.

How much wood would a woodchuck chuck if a woodchuck could chuck would?

The Woodchuck International Theological Society (TWITS) does.

What question answers itself?

Is this a question?

What does x = if 1=2 and 1=x?

Square both sides of the second equation to get x=4.

Mirror mirror, on t3h wall,
Who's t3h 1337est of them all?

xralZ backwards.

:ddr:  :ninja:  :rotfl: ?

This thread has pretty much degenerated into spam as well.
Locking this.


This was posted before, but I haven't seen a solution yet.

Prove that the Generalized Binomial Coefficients defined as:

[tex]\displaystyle {n\choose k}_C = \frac {\prod^n_{i=1}C_i}{\left(\prod^k_{i=1}C_i\right)\left( \prod^{n-k}_{i=1}C_i\right)}[/tex] for [tex]1\le k\le n[/tex] are all integers,

where [tex]\{C_n\}^{\infty}_{n=1}[/tex] is a sequence of positive integers such that [tex]\gcd(C_m,C_n)=C_{\gcd(m,n)}[/tex].

let $p$ be an arbitrary prime.  for each $ i \geq 1 $ let $ m_i $ (if it exists) be the smallest positive integer such that $ p^i | C_{m_i} $.  then if $ p^i | C_k $, where $ k = qm_i+r $, then $ p^i | (C_{m_i}, C_{qm_i+r}) = C_{(m_i, qm_i+r)} = C_{(m_i,r)} $, so $ r = 0 $ (else $(m_i, r)$ contradicts the minimality of $m_i$.  hence the only $ k $ for which $ p^i | C_k $ are the multiples of $ m_i $, and, in general, the number of $ C_j $ with $ j \leq N $ for which $ p^i | C_j $ is $ [\frac{N}{m_i}] $.

this means that, in general, the highest power of $ p $ dividing $ \prod^N_{i=1}C_i $ is $ \sum^{\infty}_{j = 1}[\frac{N}{m_j}] $.

so we need to show $ \sum^{\infty}_{j = 1}[\frac{n}{m_j}] \geq \sum^{\infty}_{j = 1}[\frac{k}{m_j}] + \sum^{\infty}_{j = 1}[\frac{n-k}{m_j}] $

this is evident from the general fact that

${ [\frac{n}{r}}] \geq [\frac{k}{r}] + [\frac{n-k}{r}] $

[quote="mna851"]Yikes, I don't think this is pre-olympiad.[/quote]

I disagree.  His techniques are just a fancier version of the ones you use to solve a problem like: "Find how many zeroes are at the end of 100!"


integration
real analysis
real analysis unsolved

$\sum_{n\geq 0}a_nz^n$ power serie with radius convergence 1
write
$f(z)=\sum_{n\geq 0}a_nz^n$ when $|z|<1$

suppose

$|f(z)|\leq \frac{1}{1-|z|}$ for any $|z|<1$

Prove that $|a_n|<(n+1)e$ for any $n\geq 0$

Just write that $a_n=\frac{1}{2\cdot\pi\cdot r^n}{\int_{0}^{2\pi}{ f(re^{it})e^{-int}dt}}$ and then integrate and take $r=n/(n+1)$. I hope it's correct.


integration
abstract algebra
algebra
polynomial
calculus
derivative
real analysis

Let $n , (n> 2) ,$ be a given integer and  $K\in C[0,1]$ such that :
[b]i)[/b]  $\int_{0}^{1}K(t)\; dt = 1$,
[b]ii)[/b] there exists a system $c_{1},c_{2},...,c_{n}$ of real numbers such that $\int_{0}^{1}K(t)f^{(n-1)}(t)\; dt = \sum_{j=1}^{n}c_{j}f\left(\frac{j}{n}\right)\; \; ,\; \; \forall \; f\in C^{n-1}[0,1] .$
[i]Questions : [/i]
[b]1.[/b] Prove that there exists at least one such continuous ,,[i]kernel[/i]"  $K$   having above properties [b]i)[/b]--[b]ii)[/b] ;
[b]2.[/b] Suppose that $S(n,j) , 0\le j \le n,$ are Stirling numbers of second kind, i.e. the coefficients from  $x^{n}=\sum_{j=0}^{n}S(n,j)x(x-1)\cdots (x-j+1) .$
Prove that for  $p \in{\mathbb N}$ one has \[\int_{0}^{1}K(t)t^{p}\; dt = \frac{S(n+p,n)}{n^{p}{n-1+p\choose p}}\; .\] [b]3.[/b] Prove or disprove that $\int_{\frac{1}{n}}^{1}K(t)\frac{dt}{t^{n}}=\frac{n^{n}}{n!}$ and $\int_{\frac{1}{n}}^{1}K(t)\frac{dt}{t^{n+1}}=\frac{n^{n}}{n!}H_{n}$ where  $H_{n}: = \sum_{j=1}^{n}\frac{1}{j}\; .$

[quote="Myth"]I think something is wrong with the condition [b]ii)[/b].[/quote]
Hi Myth, many thanks ! Alex

I don't like meaningless discussions!

We can smoothly change values of $f$ inside of intervals $(i/n,(i+1)/n)$, then LHS changes and RHS doesn't changes, therefore the equlity doesn't hold for all $f\in C^{n-1}[0,1]$!

Not quite. It means only that $K$ is a polynomial of small degree between the points and satisfies some conditions at the points imposed on low order left and right derivatives (like in the formula $\int_0^1 1\cdot f'(t)\,dt=f(1)-f(0)$ or in $\int_{-1}^1 (1-|t|)f''(t)\,dt=f(1)+f(-1)-2f(0)$).  ;)

I see. Indeed, I forgot about this trick, though it was in the course of numerical methods.

[quote="flip2004"][quote="Myth"]Anyway, I can't believe it is true, since we can smoothly change $f$ on $[0,1/n]$, then LHS will change and RHS will remain unchanged. Am I missing something?[/quote]
Further I consider the problem to be OK  ! Likewise, my opinion is that 0 instead 1/n it was  also good.
Perhaps the reason  ist that the kernel K statisfies (after me)  the condition $ K(t)=0 $ , $ \forall t \in   [0,1/n] $.
HINT : consider the linear functional  $ A[f] =\frac{1}{(n-1)!}\int\limits_{0}^{1}K(t)f^{(n-1)}(t)\equiv \frac{1}{(n-1)!}\sum\limits_{j=1}^{n}c_j f\left(\frac{j}{n}\right)  .
$ Observe that $ A[h]=0 $ for all polynomials of degree $\le n-2 . $
Moreover  $ A[e_{n-1}]=1 $ where   $e_m(t)=t^m . $  Further find the explicit form of coefficients  $ c_1,c_2 ,...,c_n $ . It's  possible !  Recognize the functional $ A $,.... .
Alex.[/quote]
For $ k\in \{1,2,...,n\} $, $x_j=\frac{j}{n} , 1\le j\le n ,$ denote ${  \omega(x)=\prod\limits_{\nu=1}^{n}(x-x_{\nu})\; \; ,\; \; \;
l_k(x)=\frac{\omega(x)}{(x-x_k)\omega^{\prime}(x_k)}} \; .
$
Consider the polynomial  $ H_k $ of degree  $\le n-2$ defined as  $ H_k(x)= l_k(x) - \frac{1}{\omega^{\prime}(x_k)}x^{n-1} . $ Because   $ l_k(x_j)=\left\{\begin{array}{lcl}
1 & , & j =k\\
0 & , & j\ne k
\end{array}\right. , $ and $A[e_{n-1}]=1, $ we obtain $ 0 = A[H_k]= c_k -\frac{1}{\omega{\prime}(x_k)}, $ that is    $ c_k= \frac{1}{\omega{\prime}(x_k)}. $ In other words  \[  A[f]=\frac{1}{(n-1)!}\int\limits_{0}^{1}K(t)f^{(n-1)}(t)\; dt = \frac{1}{(n-1)!}\sum\limits_{j=1}^{n}\frac{f(x_j)}{\omega^{\prime}(x_j)} .
\]

.........


conics ellipse hyperbola LaTeX parameterization

For the hyperbola $frac{x^{2}}{4}-\frac{y^{2}}{12}$= 1

a)eccentricity
b)coordinates of foci
c)eqn of directrices

$\LaTeX$ correction $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$
[hide="HINT"]
Just use the definitions of the parameters of the hyperbola equation.
[/hide]

Well the titles say Ellipse Equation so:

[hide="Solutions"]

$E: \frac{x^{2}}{4}+\frac{y^{2}}{12}=1$

a) $e= \sqrt{1-\frac{4}{12}}= \sqrt{\frac{2}{3}}$

b) $\text{Foci}= (0, \pm \sqrt12 e)= (0, \pm \sqrt{8})$

c) $y= \pm \sqrt{18}$

[/hide]


How many zeros are there at the end of the base 2 representation of $132!$?

please detail your  solution

thanks!!!

[hide]If you multiply something in base 2 by $10_2$ (= $2_{10}$) then it adds another 0 to the end of the base 2 representation.

So, we only have to count how many powers of 2 are in 132!.

132/2 = 66
132/4 = 33
132/8 = 16.5 (so 16 powers of 2)
132/16 = 8.25 (so 8 powers of 2)
132/32 = 4.125 (so 4 powers of 2)
132/64 = 2.0625 (so 2 powers of 2)
132/128 = 1.3125 (so 1 power of 2)

So there are 66 + 33 + 16 + 8 + 4 + 2 + 1 = 130 powers of 2 in 132!. That means that there are [b]130[/b] zeros at the end when written in base 2.

Is that right...?[/hide]

This is Legendre formula see http://mathworld.wolfram.com/Factorial.html for details

When you say written in base two, do you mean in like computer code.
Then the answer would be 2.  132 is divisble by 2. which correponds to 0.
66 is divisble by 2. which correponds to 0.   33 is not divisble by 2. which correponds to 1.  16 is divisble by 2. which correponds to 0.  8 is divisble by 2. which correponds to 0.   4 is divisble by 2. which correponds to 0.  2 is divisble by 2. which correponds to 0.   1 is not divisble by 2. which correponds to 1.

so we get 0010001.  We have to flip the numbers to put to binary.

So 132 =   1000100  in binary.  There are two zeros at the end.

[size=24][/size]2

The problem asks for the number of trailing 0's of $132!$ in binary, not 132 ;)

thanks MaThWhIz2004,tµtµ,zacharytmitchell for the help!!

Anyone has a different solution?

thanks!!

I was thinking about the ninth ciphering question at this year's UGA Math Meet.

How many zeroes are at the end of the base three decimal for 27!?

So, does the solution follow like...

[hide="This"]

Count the number of threes.

27 / 3 = 9

27 / 9 = 3

27 / 27 = 1.

9 + 3 + 1 = 13.

[/hide]

?

[hide]I don't really think there are many alternate ways to do this type of problem.

Basically what you're do when you're trying to find the number of zeroes at the end of a factorial is you think:  What would make a zero in this base system?

In Base 2, 2s
In Base 3, 3s
In Base 4, 2 2s
In Base 5, 5
In Base 6, 2*3
In Base 7, 7
In Base 8, 2*2*2
In Base 9, 3*3
In Base 10, 2*5
In Base 11, 11.
...
In Base 14, 7*2

But now you start to think, which one is the limiting factor here (like in 2*5), because there is a 2 in every other number, there are many more twos than 5s.  So we're trying to count the number of 5s in base 10.  But we're talking about base 2 here so we're trying to count the number of 2s which occur.  132/2+132/4+132/8+132/16+132/32+132/64+132/128

Round each individual part down

66+33+16+8+4+2+1=130[/hide]

To find the number of terminal zeroes of a number $n!$ in base $b$, just divide powers of the largest prime factor of $b$ into $n$ until you start to get zeroes in the division.

[quote="erik-the-red"]
So, does the solution follow like...
This
?[/quote]

Yes, it does ;)
Pakman2012 explained why + tµtµ's link and Elemennop's post.


$300$ people must elect one of three persons $A , B, C$. The first $200$ of them gave $90$ votes to $A$ , 60 to the second , $40$ votes to the third and  $5+5 = 10$ votes were invalid or white. To be elected from the first round, one person must have at least $127$ votes.Ih how many ways can  vote the other $100$ people , in order to elect person $A$ in the first round ?

 [u]Babis[/u]

 [i]The problem was given to students in a Greek technology Institute.[/i]

[hide]

127-90 = 37

So, 63 of the votes are up in the air.

63x4x3x2 = 1512 ways? Not sure.[/hide]


Prove that:
\[ 1+\frac12+\frac13+\cdots+\frac1{2^n-1} > \frac{n}{2}\ ,\ \forall n \in \mathbb{N^*}  \]

[quote="luimichael"]Do you mean:
$\frac 1 2 + (\frac1 3 + \frac 1 4 ) + (\frac 1 5 + \frac 1 6 + \frac 17 + \frac 18) + ..+(\frac 1{2^{n-1}+1}+...+\frac 1{2^n}) \\ > \frac 1 2 + 2 \times \frac 1 4 + 4 \times \frac 1 8 + ....+ 2^{n-1} \times \frac 1{2^n}= \frac n 2$[/quote]

Well that proves the given problem too... since $1 > \frac{1}{2^n}$... it is a pretty weak inequality as stated, though.


algebra
polynomial
number theory proposed
number theory

Consider a finite sequence of integers $ a_1, a_2, \ldots, a_n$. We say a polynomial $ W$ fits the sequence
if $ W(i) \equal{} a_i$ for $ i \equal{} 1, 2,\ldots, n$:
(a) Does there necessarily (that is, for every finite sequence $ a_i$) exist a polynomial $ W$ with
integer coefficients which fits this sequence?
(b) Does there necessarily exist a polynomial W which fits the sequence and such that $ W(n)$ is an integer for all integers $ n$?

a)  No.  Polynomials with integer coefficients are "locally obstructed" by the fact that $ m \minus{} n | W(m) \minus{} W(n)$.  For example, $ W(0), W(2), W(4), ...$ must all have the same parity.  (I believe this is the only such obstruction; in other words, if for every prime $ p$ the sequence $ a_i$ satisfies $ a_{i\plus{}p} \equiv a_i \bmod p$, then $ W$ should exist.)

b)  Yes.  If you require that the polynomial has degree $ n \minus{} 1$, then the polynomial is unique and the fact that it only takes integer values is a corollary of some basic theorems about [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]finite differences[/url].  Such a polynomial takes the form

$ W(x) \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} b_i {x \choose i}$

where the $ b_i$ are certain integers that appear in the finite difference table of $ W$.


Let $f: [0,1]\rightarrow[0,1]$ a strictly increasing function such that:
    * $f(0)=0$
    * $\forall x\in[0,1]: f\left(\frac{x}{3}\right)=\frac{f(x)}{2}$
    * $\forall x\in[0,1]: f(1-x)=1-f(x)$

Find $f\left(\frac{18}{1991}\right)$


I have a question.

Say, when scientists do experiments, record data, and come up wtih a formula, how do they prove that the formula is correct? There will inevitably be some error so how do they prove, besides just looking at the graph, that the formula still stands in spite of the errors affecting the data? Is there some staticistical analytic way to prove the relationship they found (such as using software)? What are some ways?

This gets into statistics, and such formulas cannot be proven to be completely correct, only approximated very well. Using staistics, you can show that it is very very very unlikely that your formula is incorrect, but you cannot prove that it's absolutely true, and there will sometimes be slight differences between your output and the actual observations.

Thank you.

Can you tell me (or offer me a site) how to use statistics to show that the formula is unlikely to be untrue?

you can do a best fit line on your graphing calculator!
but thats only more linear data
for the other ones you'll have to transfer your data
like if exponential , you log y
if power log both X and Y
and it goes to very complicated stuff

Yes. My question is, however, how can you prove mathematically or show using statistics that the formula that you come up with is unlikely to be wrong. That is, the line of best fit really exists, and the data is not completely random. The data will inevitably have errors, so you can't say that the formula stands just by finding a line of best fit.

So how do you show that the formula that you come up with, with the line of best fit and everything, is true or likely to be true?

For some physical laws, like Snell's Law, there is a mathematical proof of a natural phenomenon. If you are in this forum, you can probably understand the link I'll post here. This is a nicely laid out proof of Snell's Law. [url]http://scienceworld.wolfram.com/physics/SnellsLaw.html[/url]
This is an example of a mathematically derived proof of a physics law. There are many more of these, to be sure, this is just an excellent example that is laid out well.

Thanks a bunch.

What about for laws that involve an experimental factor? Like, say, $F=uF_{N}$?

Which law is that? I might be able to explain that if I had a proper name...

That's the Coulomb friction equation. I'm talking about laws in which there is a constant that must be determined experimentally for each kind of material. Another example is Hooke's Law.

So for these formulae, is there a MATHEMATICAL way using statistical analysis to show that the formula that you've derived is unlikely to be false? Like, sure, I can look on the graph and see most of my plotted points are in a line, some are off, but most are on a line. But can I prove, not just by looking at the graph, but by using statistics (like standard deviation and whatever), that my formula is unlikely to be untrue?

Thanks.

Think of the probabilty definiton.
If you do an experiment correctly (assuming very slight human error) infinity times, then your result should be infinitely accurate.
Have you done limits in math yet? Take the graph y = 1 / x.
As X values increase without bound, Y values decrease to the horizontal asymptote (the X-axis).
Because the more times we do an experiment, the more accurate it is, we can say that if we do an experiment an infinite amount of times, we will be infinitely accurate.

Thus let the previous graphs' Y values represent units of innacuracy, and let the X values represent repetitions of experiment. This way, you can statistically say how innaccurate you are after doing X amount of trials.

I *think* this is how they derive laws from experimental data.

Thanks.

Say I've been experimenting on Hooke's Law $F=-kx$ with a spring, a ruler, and some masses. There will be imprecision and random errors from the ruler. And if I collect the data, plot it, it will not assume the shape of a perfect line. So my question is: how do I know that I should fit a linear regression to it? How can I prove that it is very likely to be a linear relationship and not some other, like an exponential relationship? I look at the graph, yes, it look slinear, but how do I prove on paper or with a software mathematically that it is highly likely that this is a linear relationship?

Thanks a lot.

does your calculator give you a value for 'r'?
'r' has not been well explained to me but i have been told it indicates the accuracy with which you line describes your data. I am not sure how this value is calculated... can someone explain??
What I do know is that one way that you can show that your data is most likely linear as opposed to exponential is that the 'r' value for the linear regression is closer to one (with 1 being perfect accuracy) than the 'r' value for exponential regression. again, this is only what I have learned in my math class and an explanation of what 'r' is and how it is found woul be useful... thanks


algebra
polynomial
special factorizations

For which c real numbers ,there can be found a line that intersects  $y=x^4+9x^3+cx^2+9x+4$ curve at four  distinct points?

$x^4+9x^3+cx^2+9x+4=mx+b$ has four real solutions for some m, b. $x^4+9x^3+cx^2+(9-m)x+(4-b)=0$ I can't remember the rules about numbers of real solutions for polynomials though. I think there should be an easy enough way of telling how many solutions there are. Change of signs shows how many negative/positive solutions there are? Can't remember.

[hide]
We want two distinct points of inflection.
$y''=12x^2+54x+2c$ $y'''=24x+54$, so points of inflection
occur at x values such that $12x^2+54x+2c=0$ and $24x+54$ is not 0
So, by completing the square, $x^2+\frac{9x}{2}+\frac{81}{16}=\frac{81}{16}-\frac{c}{6}$ and and x does not equal $-\frac{9}{4}$
In order to have two distinct real solutions, $\frac{81}{16}-\frac{c}{6}>0$ and $c<\frac{243}{8}$[/hide]

[hide]If $r_k$ denote the roots of the polynomial $x^4+9x^3+cx^2+(9-a)x+(4-b)$, then the $r_k$ are all real and distinct. Then $\sum_{j<k} (r_j-r_k)^2> 0$.

Expanding each, this is $3\sum_{k} r_k^2> 2\sum_{j<k} r_jr_k=2c$.

Adding on $6c=6\sum_{j<k} r_jr_k$ to both sides to complete the square, $3\left( \sum_k r_k \right)^2>8c$. But $\sum_k r_k=-9\Rightarrow c<\frac{243}{8}$.

So if $c\geq \frac{243}{8}$ it certainly won't work. But if $c<\frac{243}{8}$ then the sum $\sum_{j<k} (r_j-r_k)^2$, or $243-8c$, is positive.

Just suppose that $r_1=r_2+\alpha=r_3+2\alpha=r_4+3\alpha$. Then this sum is $(3\alpha)^2+2(2\alpha)^2+3\alpha^2=243-8c$.

Then $20\alpha^2=243-8c$, so $\alpha=\sqrt{\frac{243-8c}{20}}$.

Then from $\sum_k r_k=4r_4+6\alpha=-9$, $r_4=-\frac{9+6\alpha}{4}$. So a set of $r_k$ can be constructed, which then determine the coefficients.

Therefore there is a line if and only if $\boxed{c<\frac{243}{8}}$.[/hide]


gcd(2^{p}-1,2^{q}-1)=1  equivalent with gcd(p,q)=1 ?  :?: : i don't work it  ,,,,please help me !
thanks you !everyone

Stronger:  $ \gcd(2^p \minus{} 1, 2^q \minus{} 1) \equal{} 2^{\gcd(p, q)} \minus{} 1$.

Let $ d \equal{} \gcd(2^p \minus{} 1, 2^q \minus{} 1)$; then $ 2^p \equiv 2^q \equiv 1 \bmod d$.  Let $ a, b$ be integers such that $ ap \plus{} bq \equal{} \gcd(a, b)$ (by [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity]Bezout's lemma[/url]); then $ 2^{ap \plus{} bq} \equiv 1 \bmod d$.

Hence $ d | 2^{\gcd(p, q)} \minus{} 1$.  And it's obvious that $ 2^{\gcd(p, q)} \minus{} 1 | d$.


Hmmm.
I'd say nice video except lately all of the videos posted in G&FF have only played for 2 seconds then stopped but said they were playing...

Awesome video  :) .

If it was choreographed then that would seriously be on Youtube's favorites.

[quote="chickendude"][youtube]2bpKiwVDZP8[/youtube]

A parody of Crank that Soulja boy... about math =).
There was very little choreographing done, so bear with it.
I hope you like it. The lyrics are in the "about this video". I'll post them here too.

[hide="lyrics"]
Sigma boy add-em
Hey I got this new algorithm fo yall. It's called tha Sigma Boy

You gotta partition and add n times from i to infinity

(Chorus){
Sigma boi up in Big O
Watch me max the bound with Rolle
Watch me crank that sigma boi and
Substitute dat ho

now sub with UUUUUUUU crank that sigma boi
now sub with UUUUUUUU crank that sigma boi
now sub with UUUUUUUU crank that sigma boi
now sub with UUUUUUUU crank that sigma boi
}

Sigma boi up in big O
Watch me work it with dat chalk
Simplify and show
That Riemann cranks dat Robocop
Plug and chug, now watch me calc
Calckin' on dem hatahs mang
When I do dat sigma boy
I punch it in tha calc and watch it summate
I'm calckin on you (calckin on you)
And if we get to graphin'
I'll be plottin' on ya (plottin on ya)
You catch me at yo local party
Integrating with yo date
Haterz get mad cuz
They got F's and I got A

Chorus

Spherical C.O.s
I roll with Theta, phi, and rho
Ima pass it to Leibniz
And he gon' calc it up fo sho
Haterz wanna be me Sigma boy, I'm the man
They be lookin' at my index
Iterate through the summand (maan) (maan)
Add it through (Add it through)
Dance (Dance)
Lie tangent to (tangent to)
Nope, you can't add it like me
Hoe, so don't add it like me
Folk, I see you tryin to add it like me
Man that sum was ugly

Chorus
[/hide][/quote]

lol :lol:


How is your GPA this year, university students?
Here's mine:
Semester 1:
Overall 3.29
Major 3.59
Semester 2:
Overall 3.07  :(
Major 3.59

Though I think I deserve my low major GPA for my utter laziness...
No one believes I was participating IMO last year  :D

uh, let's not. we hate talking about our GPA

Well, a lot of 8th graders don't have GPA's yet.

But it said university students.

senior in high school with 4.93 weighted gpa (3.94 unweighted)

[quote="bleumoose"]*cough* university students, furious *cough*[/quote]

Yeah, but to be fair, The Round Table isn't exactly swarming with university students...

this is true. aight, im 3.93 unweighted, 4.5something weighted.

4.0 uw. We don't weigh or rank (I wish!)

Well, I don't technically have a gpa yet, since I just started high school, but the high school classes that we took in 7th/8th grade count for it.  So only counting the high school classes, I have a 4.0 (w), and am probably ranked 1/615, since only 6 people are ranked.

I'm a Freshman, but considering school started today, my unweighed GPA is probably either 0 or 4.  :D

My unweighted is 4, my weighted is 4.5. Im Freshman, and school started 3 weeks ago.

Community colleges count, right?

4.0 @ Bellevue Community College (I'm still "technically" a high school student though lol) last year. :)

4.0, 4.5

it would be higher but my high school only has 2 ap classes for freshmen...and honors english doesn't even count as a 5...

Junior in High School:
Cumulative GPA: 3.98 unweighted (#^@! A-'s), 4.75 weighted

I am a freshman in high school. I'm taking 1 AP class so um... I dunno.  :)

Wow I'd expect much higher GPA's from USAMO qualifiers... Maybe you guys spend all your time studying math? Probably becuase you school doesn't offer that many AP classes. Yes that must be it.

My GPA is low, I'm ashamed.


Let $ ABCD$ be a cyclic quadrilateral. And let $ E \equal{} AB\cap CD$, $ F \equal{} BC\cap DA$ and $ G \equal{} AC\cap BD$. The circumcircles of triangles $ DAE$ and $ DCF$ intersect each other at $ D$ and $ H$. The angle bisectors of the angles $ AHB$, $ DHC$ intersect $ AB$, $ CD$ at $ I$, $ J$, respectively. Prove that $ I$, $ J$ and $ G$ are collinear.

Use Miguel's lemma with triangle $ BAF$ we have $ H \in (BCE)$. From this we have $ \Delta HAD \sim \Delta HBC \Rightarrow \frac{HA}{HB} \equal{} \frac{HD}{HC} \equal{} \frac{AD}{BC} \Rightarrow \frac{IA}{IB} \equal{} \frac{GA}{GB} , \frac{JD}{JC} \equal{} \frac{GD}{GC}$
It means that $ GI,GJ$ is the bisectors of the angles $ \angle BGA , \angle DGC$, so $ G,I,J$ are collinear

Thank you dear [b]PDatK40SP[/b] for your interest and for your solution. We can easily to prove two other nice properties of this configure that $ E$, $ H$, $ F$ are collinear and $ \angle AHB\equal{}\angle CHD$  :wink:


linear algebra
matrix
vector
arithmetic sequence

(sorry for posting so many questions)

 How can one know what the reduced row echelon form will look like? I know that you can row reduce any matrix into something, but is there a way to tell how it will look like? (i.e. how many independent columns will it have?) I know that if a matrix A is m x n then it can only have at most m independent columns (obviously)

 thanks

for example, I thought

1 2 3
4 5 6
7 8 9

 could eventually get you

1 0 0
0 1 0
0 0 1

 but it doesn't. How could I have known this without just "experimenting"?

thanks a lot

I think I found the answer for this one. There's a theorem that states, for scalars a1,...,an and columns c1,....,cn, if column K = a1c1 + a2c2 +..... + ancn, then in the row reduced form, it will equal to a1e1 + .... + anen where e1, .... , en are columns for the standard F^n identity matrix

[quote="jmerry"]That doesn't make sense, and can't possibly be right.

There is no simple answer; the easiest way to find the row-reduced form in general is to actually do the row reduction.[/quote]

  it's from my textbook which states:

 "Theorem 3.16: Let A be an m x n matrix of rank r, where r > 0, and let B be the reduced row echelon form of A. Then

 ... For each k = 1,2,....,n if column k of B is d1e1 + d2e2 +.... + drer then column K of A is  d1a1 + d2a2 +.... + drar"

 Oops, I messed up the notation in my previous post, I said d1e1 + d2e2 +.... + dnen

thanks for your time

Here's the thing I can see at a glance: if the columns (or rows) of a matrix form an arithmetic progression (which is to say, each pair of adjacent columns has the same difference), then the rank of the matrix is at most 2. I see that right away, and see that it applies to this matrix, meaning its rank is at most 2. And since the columns are not all multiples of the same thing, the rank is exactly 2. That means that your reduced row echelon form will be
\[ \begin{bmatrix}1&0&*\\0&1&*\\0&0&0\end{bmatrix}.\]
(And its determinant is zero, and it's not invertible, and its null space has dimension 1, and so on.)

For more details: we can see that the difference of adjacent columns is $ \begin{bmatrix}1\\1\\1\end{bmatrix}.$ That means that all three columns can be written as linear combinations of the first column and the difference I just stated - and if all three columns lie in the span of a set of two vectors, then the dimension of the column space (also known as the rank) of this matrix is at most 2.


trigonometry geometry incenter LaTeX inradius circumcircle

An equilateral triangle is inscribed in a circle of radius 6cm. Find the length of each side of the triangle...
I just wanna check if everyone else does it the same way I have done it so please show ur working..

[hide="with trig"]the incenter of the triangle = the center of the circle

thus, the distance from each vertex of the triangle to the incenter = 6cm

this splits our equilateral triangle into 3 isosceles triangles ABC with:

<A = 30 degrees
<B = 30 degrees
<C = 120 degrees
AC = 6cm
BC = 6cm
AB = ?? (this is one edge of the original triangle)

AB / sin120 = 6 / Sin30
thus
AB = 6sin120 / sin30 = 10.39cm

so each side of the triangle is 10.39cm[/hide]

sorry I dont know how to use latex just yet..

[hide="Without trig"]Call the triangle ABC. The center of the triangle is also the middle of the circle since the triangle is equilateral. Draw the radius to any vertice, lets do it to A. Then from the center draw a perpendicular line to AB. Call the point where the perpendicular line X. Call the center of the circle O. Angle OAB is 30 degrees, and that means that we have a 30-60-90 triangle. Since OA equals 6, we have that AX equals $3\sqrt{3}$. But that is half of the side of the equilateral triangle so the side length is $6\sqrt{3}$.[/hide]

[b] :rotfl:

Inscribe triangles in a circle or in another triangle are easy to solve if you'll be given the givens you need! :lol:
Am I right?

And being stupid adds up a lot because the more stupid you are the least problems you'll have? because you're to stupid to know that you don't even know that you don't. :P [/b]

[quote="7h3.D3m0n.117"][quote="bikturya"][b]

Inscribe triangles in a circle or in another triangle are easy to solve if you'll be given the givens you need!
Am I right?

And being stupid adds up a lot because the more stupid you are the least problems you'll have? because you're to stupid to know that you don't even know that you don't.  [/b][/quote]

What the...?

 :spam:  :spam:  :spam:[/quote]
Yeah, I agree.
 I know you're new here, but some people don't know how to do it that fast. So be nice to those people.

[quote="13375P34K43V312"]you could be really cool and use A=rs...

so $\frac{\sqrt{3}}{4}x^{2}=\frac{3x}{2}\cdot6$
$\frac{\sqrt{3}}{4}x=9$
so whazam $x=36/sqrt{3}=12\sqrt{3}$

wait wtf did I do something wrong?[/quote]
What does the x represent, the radius?

[quote="ckck"]
What does the x represent, the radius?[/quote]

Ideally, it should represent the side of the equilateral triangle !!!

[quote="13375P34K43V312"]you could be really cool and use A=rs...

so $\frac{\sqrt{3}}{4}x^{2}=\frac{3x}{2}\cdot6$
$\frac{\sqrt{3}}{4}x=9$
so whazam $x=36/sqrt{3}=12\sqrt{3}$

wait wtf did I do something wrong?[/quote]
I know what you did wrong. That formula is when the circle is inscribed in the triangle. Not the other way around.

I did it with trig  and I got the same answer...thanx!

[quote="13375P34K43V312"]you could be really cool and use A=rs...

so $\frac{\sqrt{3}}{4}x^{2}=\frac{3x}{2}\cdot6$
$\frac{\sqrt{3}}{4}x=9$
so whazam $x=36/sqrt{3}=12\sqrt{3}$

wait wtf did I do something wrong?[/quote]

yes. r stands for the inradius of the triangle, which is 3, not the circumradius, 6. But once you've found the inradius, it's easier to just find the side lengths using that 30-60-90 triangle.


Prove that for $a,b,c>0$ is valid:

$\sum_{cyclic}{\frac{a}{\sqrt{b+c}}}\geq \sqrt{\frac{3}2(a+b+c)}$.

Denoting by $LHS$ by $S$ and applying chebyshev inequality we get
$3S \geq (a+b+c)(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{a+c}}+\frac{1}{\sqrt{b+a}}) \geq (a+b+c)(\frac{9}{\sqrt{b+c}+\sqrt{a+c}+\sqrt{b+a}}) \geq (a+b+c)(\frac{9}{\sqrt{6(a+b+c)}}) = 3\sqrt{\frac{3}{2}(a+b+c)}$
 So $S \geq \sqrt{\frac{3}{2}(a+b+c)}$

--------------------
By [b]Holder Inequality[/b] \[\left(\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \right ) ^2\left( a(b+c)+b(a+c)+c(a+b) \right ) \geq (a+b+c)^3\] $\longrightarrow$ \[\left(\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \right) \geq \sqrt{\frac{(a+b+c)^3}{2(ab+bc+ca)}}\] So it remains to prove that \[\sqrt{\frac{(a+b+c)^3}{2(ab+bc+ca)}}\geq \sqrt{\frac{3}2(a+b+c)}\] $\longrightarrow$ $$(a+b+c)^2\geq 3(ab+bc+ca)$$ which is obvious.
------------

$\sum_{cyc}{\frac{a}{\sqrt{b+c}}}\geq \sqrt{\frac{3}2(a+b+c)}$
assume:a+b+c=6,now we can prove
$\sum_{cyc}{\frac{a}{\sqrt{b+c}}}\geq 3$
M=$\sum_{cyc}{\frac {a} {\sqrt{b+c}}}$,N=$\sum_{cyc}{\frac {\sqrt{b+c}} {a}}$
then M+N≥6
and M-N=$\sum_{cyc}{\frac {a^2-b-c} {a\sqrt{b+c}}}$
so $\sum_{cyc}{\frac{a}{\sqrt{b+c}}}\geq \sqrt{\frac{3}2(a+b+c)}$


An Isoceles triangle ABC, with base AC, has a point P that lies in BC, and a point Q that lies in AB. Given that AC=AP=QP=BQ, find the measure in degrees of the angle B.

 :)

I hand-wrote this on a tablet computer, sorry if my handwriting is a little messy.

From my algebraic solution, I got an answer of $ \angle B\equal{}\boxed{\frac{180}{7}^{\circ}}$.

Thank you gauss1181; I understood  :lol:


trigonometry
inequalities
inequalities proposed

prove that for any triangle ABC
[2r/R] 2  < cos{(A-B)/2}cos{(B-C)/2}cos{(C-A)/2}
in my opinion this one's really sweet ;)

I think I have something for this one:

using Mollweide formula we have: $\cos (\frac{A-B}{2})=\frac{\sin( \frac{C}{2})(a+b)}{c}$ we rewrite the inequality like so: $64 \prod (a) \prod \sin (\frac{A}{2}) < \prod(a+b)$, but $\prod \sin (\frac{A}{2}) = \frac{r}{4R}$. Thus the inequality is equivalent to:$ R \prod(a+b) \geq 16r \prod(a)$ but this is the product of $R\geq 2r$ and $ (a+b) \geq 2\sqrt{ab}$ (and the simmilar relations).

cheers! :D  I hope it is okay. What is your sol galois?

nice sol,lagrangia(tho i've to go thru it fully but it seems impressive)
i converted everyting in terms of half angles and then converted it into an obvious inequality in homogenous polynomials(with the variables being cot(A/2) and similar terms) using the fact that
cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)


I'm told that the answer to:

T - xg = xg((y-x)/(x + y)) (Find T)

is:

T = g(2xy/(x + y))

Can anyone tell me the intermediate steps please?

Thanks very much for the quick answer! :)


If I have a homework assignment that has 25 questions and I finish 10% of the remaining homework on any given day, how many days will it take me to finish the all but the last question?

(You can finish a little bit of a question at a time)

[hide="answer"]I multiplied 25 by powers of $\frac{9}{10}$ until I got an answer smaller than $1$, which was $31$[/hide]

Since 10% of the homework is completed, 90% always remains. Thus we can write the exponential equation 25(9/10)^t. We want to know when all but the last question is answered, so set this equation equal to 1. Graph is equation on a graphing calculator and the answer should be 31 days.

P.S. Could someone check this?

I got the same answer, but shouldn't it be $25(\frac{9}{10})^{t}\le 1$?  The answer would have to be a fraction for it to equal $1$ exactly

You really don't need to guess.  :P There will always be 9/10 of the homework remaining (you can never actually finish it), and there are 25 problems:
[hide]
((9/10)^n)25 = 1
(9/10)^n = 0.04
n = (log0.04)/log(0.9)
Hitting this value onto your calculator...
n=30.5510637...
Rounding up,
n=31[/hide]

[hide]Set up the equation $25\left(\frac{9}{10}\right)^{x}\le 1$

Rounding $x$ to an integer, we get $31$.[/hide]

[hide]
Even though you aren't allowed to use logs,

You can easily find out that you could find out
that because this is true:
$\frac{9}{10}^{4}= 0.04$, that the answer of $n$ would be:
$\frac{\log 0.04}{\log 0.9}$
Then, you use Log tables or a calculator to find out that the answer to that would be $30.5520\dots$. I think you round up to get $31$.[/hide]

[hide]25*.9 all the way until it is lessthan or equal to one (im too lazy to do it)[/hide]

you can use logs after that, which is what I did


1)find f:R-->R such that
$ f(f(x)\plus{}y))\equal{}2f(x)\plus{}f(y)$
2)find g:R-->R such that
$ g(xy)\minus{}g(x)g(y)\plus{}g(x\plus{}y)\equal{}g(x)(y\plus{}1)\plus{}g(y)(x\plus{}1)$
3)find f:R->R such that
$ f(x^2\minus{}y^2)\equal{}xf(x)\minus{}yf(y)$
4)find h:R->R such that
$ h(x\plus{}y)\minus{}2h(x\minus{}y)\equal{}2h(y)\minus{}h(x)$
Can you help me? :(

[quote="tronghieu"]

3)find f:R->R such that
$ f(x^2 \minus{} y^2) \equal{} xf(x) \minus{} yf(y)$
[/quote]
[url]http://www.mathlinks.ro/viewtopic.php?search_id=901152022&t=138850[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=337857#337857[/url]


Prove that there doesn't exist any positive integer $n$ such that $2n^2+1,3n^2+1$ and $6n^2+1$ are perfect squares.

Very nice problem,I tried to solve it with Pell's equations till I came up with this
(6n^3-3n) 2  <= (2n 2 -1)(3n 2-1 )(6n 2 -1) <= (6n ^3-3n+1) 2

very nice key.thank you.I also want to solve it by Pell's equations but failed
but is that be (6n^3+3n) 2 <(2n 2+1)(3n 2 +1 )(6n2 +1)<(6n ^3+3n+1)  2

$(3n)^2 (2n^2+1)+1=18n^4+9n^2+1=(3n^2+1)(6n^2+1)$, so you would have two squares with difference 1, so its only possible for $n=0$ which is not positive.

[quote=aThousandWords][quote=darij grinberg]The problem does not only come from the Japanese mathematical olympiad, it was also used as problem 1 in the 1st German TST 2005:

[color=blue]Prove that there is no positive integer n such that all three numbers $2n^2+1$, $3n^2+1$ and $6n^2+1$ are perfect squares.[/color]

My solution uses the fact that $\left(2n^2+1\right)\left(6n^2+1\right)-\left(2n\right)^2\left(3n^2+1\right)=4n^2+1$ (now I think you can complete it for yourself). All other solutions were easier ;) .

  darij[/quote]

How would you advance from this, for a dumb person like me?[/quote]

Difference of two squares couldn't be of kind $4k+1$.


Let $ x,y,z \ge  0$.Prove that $ \frac{\sqrt{x}}{1\plus{}x}\plus{}\frac{\sqrt{y}}{1\plus{}x\plus{}y}\plus{}\frac{\sqrt{z}}{1\plus{}x\plus{}y\plus{}z}<\sqrt{3}$

_______________________________________
Azerbaijan Land of the Fire  :P

[quote="Pirkuliyev Rovsen"]Let $ x,y,z \ge 0$.Prove that $ \frac {\sqrt {x}}{1 \plus{} x} \plus{} \frac {\sqrt {y}}{1 \plus{} x \plus{} y} \plus{} \frac {\sqrt {z}}{1 \plus{} x \plus{} y \plus{} z} < \sqrt {3}$

_______________________________________
Azerbaijan Land of the Fire  :P[/quote]

$ \frac {\sqrt {x}}{1 \plus{} x} \plus{} \frac {\sqrt {y}}{1 \plus{} x \plus{} y} \plus{} \frac {\sqrt {z}}{1 \plus{} x \plus{} y \plus{} z} < \sqrt {3}$

we just need to prove

$ \frac {\sqrt {x}}{1 \plus{} x} \plus{} \frac {\sqrt {y}}{1 \plus{} y} \plus{} \frac {\sqrt {z}}{1 \plus{} z} \le \sqrt {3}$

$ \frac {\sqrt {x}}{1 \plus{} x} \plus{} \frac {\sqrt {y}}{1 \plus{} y} \plus{} \frac {\sqrt {z}}{1 \plus{} z} \le\frac {\sqrt {x}}{2\sqrt{x}} \plus{}$ $ \frac {\sqrt {y}}{2\sqrt{y}} \plus{} \frac {\sqrt {z}}{2\sqrt{z}} \equal{}\frac{3}{2}\le\sqrt {3}$

 :lol:


Let $ ABCD$ be cyclic quadrilateral.$ P$ $ Q$ $ R$ are orthogonal projections of $ D$ on $ BC$ $ CA$ $ AB$ respectively. Prove that $ PQ \equal{} QR$ if and only if bisectors of angles $ ABC and ADC$ intersect on $ AC$.

thanks for Your help

$  \begin{array}{c} |PQ|=|CD|.sin{C}=|CD|.\frac{|AB|}{2R}\\
|RQ|=|AD|.sin{A}=|AD|.\frac{BC|}{2R}\end{array}\}\Rightarrow \frac{|PQ|}{|RQ|}=\frac{|AB|.|CD|}{|AD|.|BC|}\Leftrightarrow\\
\Leftrightarrow [|PQ|=|QR|\Leftrightarrow |AB|.|CD|=|AD|.|BC|\Leftrightarrow \mbox{the bisectors of angles ABC and ADC intersect in (AC).}]$

Why Your last conclusion is equivalent to the statement from the problem( why then this bisectors intersect each other on $ AC$) thanks


FIND AREA OF IIGM AND TRIANGLE.BD =15

ANGLE( D )OF IIGM=35

FIND AREA OF IIGM AND TRIANGLE.BD =15

ANGLE( D )OF IIGM=35

SORRY FOR DIAGRAM

SORRY FOR DIAGRAM

[url][/url][youtube]hide[/youtube]

sorry i am unable to understand the question,so please post what is in diagram in words
it may help

[hide]though even then i'ld not be able to answer[/hide]


induction
number theory
relatively prime
number theory proposed

Prove that the sequence $ a_n$ with

$ a_n \equal{} 6a_{n \minus{} 1} \minus{} a_{n \minus{} 2}$ with  $ a_1 \equal{} 4$ and $ a_2 \equal{} 24$

contains no perfect squares

Dunno, but I think this could potentially help:
$ a_{n}\equal{}\frac{\sqrt{2}}{2}((3\plus{}2\sqrt{2})^{n}\minus{}(3\minus{}2\sqrt{2})^{n})$
Maybe this will help...  :wink:

[quote]Can somebody prove this statement?I have a proof for this...Nobody? [/quote]

Can you give hint for solution?

[quote="Αρχιμήδης 6"]There is no help....Its only a transformation... :wink:[/quote]
Every proof consists of transformations...

[quote="Αρχιμήδης 6"]Can somebody prove this statement?I have a proof for this...Nobody? :wink:[/quote]
Stop posing. Either post a solution, a hint, etc., or don't post.

$ a_1\equal{}4$ is a perfect square  :rotfl:
A hint is to divide every term of the sequence by 4. Now you have smaller numbers and the same problem.

maybe $ a_i a_{i \plus{} 1} \plus{} 4$ would be a perfect square,,,,
I just guess, not prove yet :oops:

divide the sequence by 4.
then a_1=1,a_2=6,a_3=35  etc
and a_n=6a_n-1-a_n-2
we need to show that the sequence a_n contains no squares.
Lemma.1 a_n=a_k*a_(n-k+1)-a_(k-1)*a_(n-k)(easy proof by induction)
(in general this holds for a_n=a_2*a_n-1-a_1*a_n-2)....
Lemma 2. gcd(a_n,a_n-1)=1 (easy proof by induction)
using this, we get a_2n=a_n*(a_n+1-a_n-1) and a_(2n+1)=a_(n+1)*(a_n+1-a_n)
if a_2n is a square, since a_n and a_n+1-a_n-1 are coprime (transform a_n+1 into 6a_n-a_n-1 and apply lemma)
a_n is a square
if a_2n+1 is a square, similar argument leads to a_n+1 is a square.
since a_1 is square, a_2, a_3,a_4 is not,
choosing the smallest square that is not a_1 we get contradiction

sorry for not latexing..

Let $ (1\plus{}\sqrt 2 )^n\equal{}x_n\plus{}y_n\sqrt 2$, then $ a_n\equal{}y_{2n}$.
Because $ (1\minus{}\sqrt 2 )^n\equal{}x_n\minus{}y_n\sqrt 2$ we get $ x_n^2\minus{}2y_n^2\equal{}(\minus{}1)^n\to (x_n,y_n)\equal{}1.$
If $ n\equal{}2^km$, m - odd, then $ y_{2n}\equal{}2^{m\plus{}1}x_nx_{2^{k\minus{}1}m}...x_my_m$, were $ x_{2^i}m,i\equal{}0,..k$ are relatively prime odd numbers and $ (x_{2^im},y_m)\equal{}1$. If $ y_{2n}$ is square any $ x_{2^im}$ must be square. But $ x_m\equal{}3\mod 8$ is not square.

SORRY :blush:
$ 1)$If $ (b_n \minus{} 1)^2 < a_n < b_n^2$ and $ n > 6$ then:
$ a)$for $ n \equal{} 0,2,4mod5 \minus{} > b_n \equal{} 2b_{n \minus{} 1} \plus{} b_{n \minus{} 2} \minus{} 1$
$ b)$for $ n \equal{} 1,3mod5 \minus{} > b_n \equal{} 2b_{n \minus{} 1} \plus{} b_{n \minus{} 2}$
$ 2)lim(b_{n \plus{} 1}/b_n)^2 \equal{} lim(a_{n \plus{} 1}/a_n) \equal{} c$ and hold that $ c^2 \minus{} 6c \plus{} 1 \equal{} 0$  and  $ c \equal{} 3 \plus{} 2\sqrt {2}$


Given a triangle $ABC$ such that $AB=2BC$ and $\angle ABC=60^{o}$. A point $P$ lies within the triangle and satisfy the condition: $\angle APB= \angle BPC= \angle CPA$. Find the area of triangle $APC$ in term of the area of triangle $ABC$

If you solved the above problem, try this:
Given a convex quadrangle $ABCD$ such that $AD=AB$, $\angle DAB= \angle DCB= \angle AEC=90^{o}$ and $AE=5$. Find the area of quadrangle $ABCD$


Show  the next equality:
$ \begin{vmatrix} \frac{1}{p\plus{}1}   & \frac{1}{p\plus{}2}  & \ldots   &\frac{1}{p\plus{}n}\\ \frac{1}{p\plus{}2}   &    \frac{1}{p\plus{}3}       & \ldots &     \frac{1}{p\plus{}n\plus{}1}            \\ \vdots &\vdots &\ddots &\vdots\\    \frac{1}{p\plus{}n}            &   \frac{1}{p\plus{}n\plus{}1}   & \ldots &     \frac{1}{p\plus{}2n\minus{}1} \end{vmatrix}\equal{}\frac{[1! 2! \ldots (n\minus{}1)! (p\plus{}1)! (p\plus{}2)!\ldots (p\plus{}n\minus{}1)!  ] ^2 }{(p\plus{}1)! (p\plus{}2)!\ldots (p\plus{}2n\minus{}1)!  }$.

Read Determinant Cauchy : http://vi.wikipedia.org/wiki/%C4%90%E1%BB%8Bnh_th%E1%BB%A9c_Cauchy

A Romanian probably won't get much out of a Vietnamese-language link.

My best idea here is to find an $ LL^T$ or maybe $ LDL^T$ factorization by exploiting the inner product structure on polynomials. Unfortunately, that $ p$ means we've got something meaner and not just the Legendre polynomials.

He probable meant the Cauchy deteminant.
http://www.mathlinks.ro/viewtopic.php?t=290874 contains the statement and proofs.

First apply column operations, col i -> col i - col 1 for i = 2,...,n
Then for each row, we take out the common factors and get $ \frac{1}{(p\plus{}1)...(p\plus{}n)}$
For each column, we take out the common factors and get $ (\minus{}1)^{n\minus{}1}(n\minus{}1)!$
We are then left with a matrix with the first column becomes all 1s.
Now, apply row operations row i -> row i - row 1 for i = 2,...,n
and use similar method, then proceed by induction.


when f(x)= 3sin(4x) the max turning point on f(x) should be (pie/2,3) right?

how would you figure out the domain and range of f(x)=-3sec(x)?

[hide="not sure if it needs to be hidden"]
considering the unit circle, the angle with the greatest sine value is $\frac{\pi}{2}$. However, we are looking at sin(4x). so we want $4x=\frac{\pi}{2}$, so $x=\frac{\pi}{8}$. plug it in and you get your point to be
$(\frac{\pi}{8},3)$

[/hide]


At what velocity $v_{0}$ should a bullet of mass $m$ hit(and get stuck in) a block of mass $M$ resting on a flat surface for it to move a distance $d$ if the coefficient of static and kinetic friction are $\mu_{s}$ and $\mu_{s}$, respectively and air resistance is negligible in terms of the given variables.

By conservation of momentum, $mv_{0}= (M+m)v$.

Since the collision is virtually instantaneous, the static friction is negligible.

Thus, $(M+m) \mu g d = \frac{1}{2}(M+m)v^{2}= \frac{m^{2}v_{0}^{2}}{2(M+m)}$.

Therefore, $v_{0}^{2}= \frac{2 \mu (M+m)^{2}g d}{m^{2}}$.

So we have $v_{0}= \frac{M+m}{m}\sqrt{2 \mu gd}$.

r u sure about static friction, cuz thats y i asked this question, consider a speed not large enough to move the block, its definately greater than zero, if a marble hits an 18 wheeler, it won't move

The momentum lost in the collision due to static friction is $\mu (M+m)g t$, where $t$ is the time of the collision.

Thus, since this collision is assumed to be nearly instantaneous, $t$, and therefore the momentum lost due to static friction, is negligible.

The reason is that these are assumed to be two solid incompressible blocks.  In an actual collision, all objects involved undergo some compression, however small, during which the impulse of the collision is applied to them.

If the problem were changed to include a spring on one of the blocks, say with spring constant $k$, then the static friction could be taken into account, since it takes a nonzero time to compress the spring.  If you wanted the blocks to still stick, you could say that after the spring was compressed as far as it would be, a latch holds the blocks together.

fine how would one solve the problem with a spring on the 2nd block with spring constant k


function
induction
group theory
abstract algebra
superior algebra
superior algebra unsolved

Here's a problem I just composed; hope you'll find it interesting!  :)

[b]Problem. [/b] Let $f(n)$ denote the number of groups of order $n$.
(a) Find all pairs $(m,n)$ of positive integers such that $(m,n)=1$ and $f(mn) = f(m)f(n)$.
(b) Find all pairs $(m,n)$ of positive integers such that $(m,n) = 1$ and $f(mn) = f(m)f(n)  + 1$.
(c)* Is it possible to characterize the pairs $(m,n)$ with $f(mn)=f(m)f(n)$ without the assumption of $m$ and $n$ being co-prime?

--Vesselin

I think I can solve (a), but this is going to be rather long.

In [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=automorphisms&t=45849]another thread[/url], vess introduced the multiplicative function $g$ defined by $g(p^{k})=(p^{k}-1)(p^{k}-p)\ldots(p^{k}-p^{k-1})$. I’ll (try to :)) prove that if $(m,n)=1$, then $(*)\ f(mn)=f(m)f(n)$ is equivalent to $(**)\ (m,g(n))=(n,g(m))=1$.

First, notice that $(*)$ is equivalent to saying that every group of order $mn$ is a direct product of a group of order $m$ and one of order $n$. Next, notice that $(**)$ implies that an element $x$ of order dividing $n$ which normalizes a group $H$ of prime power order $p^{a}|m$ actually centralizes $H$: I showed in that other thread that every prime dividing $|\mbox{Aut}~H|$ divides $|H|\cdot g(p^{a})$, and since $x$ acts by conjugation on $H$ and the order of $x$ is coprime to $g(p^{a})$ and hence to $|\mbox{Aut}~H|$, the conclusion follows. I’ll use this observation below.

$(*)\Rightarrow (**)$

Assume $(**)$ doesn’t hold because, say, a prime divisor $p$ of $m$ divides $g(q^{k})$ for a prime power $q^{k}|n$. There is then an automorphism of $(\mathbb Z/q\mathbb Z)^{k}$ of order $p$, so $\mathbb Z/p\mathbb Z$ can be made to act non-trivially on $(\mathbb Z/q\mathbb Z)^{k}$. This means that there is a semidirect product $H=(\mathbb Z/q\mathbb Z)^{k}\rtimes\mathbb Z/p\mathbb Z$ which is not direct. We can take our counterexample $G$ to be the direct product between $H$ and, for example, a cyclic group of order $\frac{mn}{pq^{k}}$. $G$ is a group of order $mn$ which is not a direct product of one of order $m$ and one of order $n$, i.e. $(*)$ does not hold.

$(**)\Rightarrow (*)$

Assume first that we’ve managed to prove this for the case when $m,n$ are prime powers. Then we could solve the general case by induction on the order $mn$ of $G$ as follows:

Assume that $m$ has at least two prime factors, and let $p^{a}\|m$ for some prime $p$. Clearly, $mn$ is odd, so $G$ is solvable (the Feit-Thompson theorem). I’ll also use Hall’s theorem, stating that if $G$ is solvable, and $u\ |\ |G|,\ \left(u,\frac{|G|}u\right)=1$, then $G$ has subgroups of order $u$ which are all conjugate. By Hall’s theorem, $G$ has a subgroup $N$ of order $n$, and all such subgroups of $G$ are conjugates of $N$. This means that we can find subgroups $N\le N_{1}\le G,\ N\le N_{2}\le G$ such that $|N_{1}|=\frac{mn}{p^{a}},\ |N_{2}|=np^{a}$. By the induction hypothesis, we can find subgroups $M_{1}\le N_{1},\ M_{2}\le N_{2}$ with $|M_{1}|=\frac m{p^{a}},\ |M_{2}|=p^{a}$ which centralize $N$. This implies that the centralizer of $N$ contains a subgroup $M$ of order $m$, and it’s clear from here that $G=M\times N$.

All that's left now is to prove that $(**)\Rightarrow(*)$ when $m=p^{a},n=q^{b}$ are prime powers.

Let $P$ be a Sylow $p$-subgroup of $G$. Another result due to Hall states that if for every subgroup $H$ of $P$, every element $x\in G$ of order coprime to $p$ which normalizes $H$ also centralizes $H$, then $G$ has a normal subgroup $Q$ of order $q^{b}$. This condition holds thanks to $(**)$, so we do have such a group $Q$. $(**)$ now tells us that $P$ centralizes $Q$, so we get $G=P\times Q$, as desired.

I hope it’s finally ok. I’ve posted several incorrect versions up until now :).


If I find the formula/method/theorem in a book/journal, how do I cite it???  Say for example,  I find in a journal call "Mathematics" issue number 2, in November 1950, how to find the roots to a cubic polynomial, how do I cite it???

There are various formats, but a standard citation might be:

J. Smith, "An interesting equation", [i]Mathematics[/i] [b]5[/b] (1950), 129-131.

Author, "Title", [i]Journal Name[/i] [b]Volume[/b] (year), pages.

If citing a theorem or result from a paper, state the result in full; do not simply say something like "By Theorem 3 in [1], f(x) is convex".

So like we have a page of "works cited" and we put that line in there???or we put the line in the middle of our solutions???

As long as it's clear to the reader, it doesn't matter.


What is the minimum number of squares that is necessary to draw on a white sheet to obtain a square grid of side $n$?

I think it is n^2-(n-2)^2/2 if n is even and n^2-((n-2)^2+1)/2 if n is odd,
because every exterior square must be drawn and for the rest (n-2)*(n-2)
just notice that in any 2*2 piece at least 2 squares must be drawn.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=square&t=1403

Pierre.


does anyone know here an excellent math forum site (intended for math undergraduates and graduates) aside from mathlinks???please tell me the link..thanks

you can try the mathcounts website
its like mathcounts.saab.com
but im not sure so search on google

Is MathCounts website for undergraduates/graduates?

Anyway, I would suggest [url=http://www.mathoverflow.net]Math Overflow[/url]. I think many undergraduates/graduates can learn a lot by seeing how professional mathematicians/other graduate students think.


Express the following as a simplified common fraction: \\

$ \frac{1^2\plus{}2^2\plus{}3^2\plus{}4^2\plus{}5^2}{
\sqrt{1^2} \cdot
\sqrt{2^2} \cdot
\sqrt{3^2} \cdot
\sqrt{4^2} \cdot
\sqrt{5^2}}$

The formula for sum of squares is $ \frac{n(n \plus{} 1)(2n \plus{} 1)}{6}$ where $ n$ is the last term (this only works when starting from one). Plug in five to get fifty-five. Everything on the bottom is just $ 1\times 2 \times 3 \times 4 \times 5$. This obviously equals $ 120$. $ \frac{55}{120} \implies \boxed{\frac{11}{24}}$.


A line dance requires two steps forward and then one step back. If a wall is ten steps away, how many steps must the dancers take to reach the wall?

Should go into the middle school forum ;)


Two equal circles in the same plane cannot have the following number of common tangents:

$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
Two equal circles in the same plane cannot have only one common tangent (draw a diagram with them intersecting, with them tangent, or with them separate).  Thus, the answer is $ \boxed{\textbf{(A)}}$.
[/hide]


greatest common divisor
quadratics
number theory proposed
number theory

Multiplying both sides by 64 and adding $16x^2$ yields the equation
\[ 16x^2+64xy+64y^2=64x^3-432x^2+704x-192  \] that is,
\[ (4x+8y)^2=(4x)^3-27(4x)^2+176(4x)-192  \] and so we can deal instead with the equation $A^2=B^3-27B^2+176B-192$ where $A=4(x+2y)$ and $B=4x$ are also integers.

Note $B=8$ is a root of the cubic on the RHS, so $A^2=(B-8)(B^2-19B+24)=(B-8)[(B-8)^2-3(B-8)-64]$.  If $B=8$ the LHS is 0 and we have the solution $(x, y)=(2, -1)$ to the original equation.  The second factor has no integer roots.  Otherwise, the product of $B-8$ and $B^2-19B+24$ is a nonzero square.  As we know that the gcf of these two numbers is a factor of 64, $B^2-19B+24$ must be either a square or twice a square.

If $B^2-19B+24=k^2$, $(2B-19)^2-(2k)^2=265=1\times 265=5\times 53$, so $2B-19$ is one of $\pm\frac{265+1}{2}=\pm 133$ and $\pm\frac{53+5}{2}=\pm 29$, and $B\in\{76,-57,24,-5\}$.  In fact, as $A^2=(B-8)(B^2-19B+24)$, $B-8$ is square also, and this happens only for $B=24$.  In this case $A^2=16\times 144$ so $A=\pm 48$, and the two resulting solutions $(x, y)$ to the original equation are $(6,3)$ and $(6,-9)$.

If $B^2-19B+24=2k^2$, $(2B-19)^2-2(2k)^2=265$, which, checking quadratic residues mod 5, implies $2B-19$ and $2k$ are multiples of 5, a contradiction since then the RHS should be a multiple of 25, which it is not.  So there are no solutions in this case.

Nice solution $Agrippina$ :clap2:  :clap:


abstract algebra
superior algebra
superior algebra theorems

I just have a general question about showing a group G is Abelian. In trying to show that G is Abelian, does it suffice to show that the inverses of two arbitrary elements in the group commute? That is (a^-1)(b^-1) = (b^-1)(a^-1) for any a and b in G. Or would I have to take it one step further and show that this implies that ab=ba? Thanks for your help.

The statement with inverses is precisely equivalent to the original statement. You can prove this by manipulation, or see that inversion is a bijection from the set to itself.


[hide]24

the reason i didn't show my work is because i had i a whole big page of work.  then my annoying little sister deleted it all and typed all this weird stuff.  now i can't get it back. :mad: [/hide]

Note that this is asking the same as the question, "What are the number of zeros at the end of $100!$?"

[quote="Silverfalcon"]The largest integer $k$ such that $5^k$ is a factor of $100! = (1)(2)(3)(4)...(100)$ is what?[/quote]

[hide="Solution"]
20 + 4
[b]24[/b]
[/hide]

you can generalise this: (i had this problem in one of my correspondence programmes)

ummm...i give u a rigourous proof (yeh im aussie)

Prove: find an expression for the greatest power of a prime number $k$ that divides $n$!

Proof:
let's imagine each of the integers from 1 to n strecthed out on a line as in a continued product:
n!=1.2.3.4.5....n
CLEARLY each term of the sequence p, 2p, 3p, ..., kp will contribute to at least one factor of p, kp being the largest multiple of p not exceeding n.

let the greatest power of p which divides n! be x
if 1 <= n <= p-1 then x=0
if p <= n <= $p^2$, n! will contain one factor of p from each of the corresponding multiples of p; that is x=[$n/p$] (integer part function)

if $p^2$ <= n <= $p^3$-1, n! will contain one factor for each corresponding multiple of p that is not a multiple of $p^2$ and an additional one for each factor that is a multiple of $p^2$.
that is x=[$n/p$]+[$n/p^2$]

this pattern will continue:

that is x=[$n/p$]+[$n/p^2$]+[$n/p^3$]+...+[$n/p^j$] where $p^j$ is the largest power of p that does not exceed n.

plug in 100 into n, 5 into p and then you get the answer: 24

That about covers it all. Nice proof random_mathematician.

I always did it that way, random_mathematician, only I used a simple proof based on doublecounting :P .

[hide="Answer"]
20 + 20/5
20 + 4
[b]24[/b]
[/hide]


vector
induction
superior algebra
superior algebra solved

Let $K$ a corp of caracteristic $0$ and $E$ a vectorial space over $K$.Let $E1,....En$ vectorial subspaces in $E$ with their reunion a vectorial subspace.Then there is one which contains all the others.

it suffices to prove that if K is an infinite field then a vector space E over K is not a finite union of subspaces strictly included in E. You can prove it by induction on n, the number of subspace:
It's clear for n=1,n=2.
suppose n>=3 and the result proved for k<n.
 [code]  :cup:  [/code]

it suffices to prove that if K is an infinite field then a vector space E over K is not a finite union of subspaces strictly included in E. You can prove this result by induction on n=dim E, the number of subspace:
It's clear for n=1,n=2.
suppose that E=F1 U F2 U .. U Fk where Fi is a strict subspace of E. Then take an arbitrary hyperplan H of E. Then H=F'1 U F'2 U .. U F'k where F'i=(H inter Fi). But , because of the inductive assumption, H is equal to one of the Fi. -contradiction because there are infinitely many hyperplans in E ( K is infinite)-

I've just realized I've only proved the result for vector spaces of finite dimension  :(
(for the case where K=R you can prove the result thanks to Baire's theorem and the fact that a subspace have void interior)

it also works in the general case. Assume that E=(F1 U F2 U..U Fn) where n is minimal. Take A in E\(F1 U F2 U..U F(n-1) ) and B in E\Fn and consider the line L that goes through A and B. Then this line is not included in Fi. Thus L cuts Fi in only one point. -contradiction-


For the sequence $ a_n$ we have

   $ a_{n \plus{} 1} \minus{} a_n \equal{} 2n \plus{} 1 , \forall n \in \mathbb N*$ and $ a_1 \in \mathbb Z$

Prove that the product of any two consecutive  terms of the sequence is also a term of the sequense.

[quote="stergiu"]For the sequence $ a_n$ we have

   $ a_{n \plus{} 1} \minus{} a_n \equal{} 2n \plus{} 1 , \forall n \in \mathbb N*$

Prove that the product of any two consecutive  terms of the sequence is also a term of the sequense.[/quote]

We have $ (a_{n\plus{}1}\minus{}(n\plus{}1)^2)\minus{}(a_{n}\minus{}n^2)\equal{}0$ and so $ a_n\equal{}n^2\plus{}c$

Then $ a_na_{n\plus{}1}\equal{}(n^2\plus{}c)((n\plus{}1)^2\plus{}c)$ $ \equal{}(n^2\plus{}n\plus{}c)^2\plus{}c$  $ \equal{}a_{n^2\plus{}n\plus{}c}$

Q.E.D.

Call $ D_n\equal{}a_{n\plus{}1}\minus{}a_{n}$. Call $ a_1\equal{}k$

So, $ \sum_{i\equal{}1}^nD_i\plus{}k\equal{}a_{n\plus{}1}\equal{}n(n\plus{}2)\plus{}k$

So, $ a_n\equal{}(n\minus{}1)(n\plus{}1)\plus{}k$

So,
$ a_na_{n\plus{}1}\equal{}[(n\minus{}1)(n\plus{}1)\plus{}k][n(n\plus{}2)\plus{}k]\equal{}n^4\plus{}2n^3\plus{}(2k\minus{}1)n^2\plus{}(2k\minus{}2)n\minus{}k\plus{}k^2\equal{}(n^2\plus{}n\plus{}k\minus{}2)(n^2\plus{}n\plus{}k)\plus{}k\equal{}a_{n^2\plus{}n\plus{}k\minus{}1}$


geometry calculus derivative perimeter conics parabola

Uhm...Yeah I'm in Pre-Calc. and have a test Monday. I've been absent for a week and have no idea how to do these, there are two problems:

ENCLOSING THE MOST AREA WITH A FENCE: A farmer with 4000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed?

yet another one like it...

A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed

Please help me...
AIM: KomyunisutoNeko
YIM: floopyjellytinkiefishie

Ok, optimization.

I've never taken pre-calc, so I'm not sure if you want to find the answer by derivatives, or basic inequalities.  :huh:

this can be solved using getting started techniques though (since its halfway through the year in precalc, i highly doubt theyd ask you do differentiate anything)

[quote="4everwise"][quote="Scrambled"]I dont think this exactly qualifies for preolympiad (maybe getting started?)[/quote]Anything with calculus goes in other problem solving topics. Definately NOT getting started.  :P[/quote]

(cough)...yah I'm new so i didn't know where the hell to put this sorry ^^;

Ok, here's to start you off.

If he wants to enclose 3 sides, that means the perimeter is $2l+w=4000$ instead of $2l+2w$, where $l$ and $w$ are length and width. The area is $A=lw$.

Because $2l+w=4000$, $4000-2l=w$ and $A=l(4000-2l)$. Then just take the derivative and find the maximum and minimum values.  :)

When working on optimization, try to get everything in terms of one variable so that you can solve it easily.

First of all, I also don't know where "-" is, so yeah. But anyways this probably belongs in intermediate? Pre-calculus $\ne$ pre-olympiad :D.

Oh yeah ignore 4everwise I'm AIMing you with a much beastlier method.

[quote="4everwise"][quote="probability1.01"]Oh yeah ignore 4everwise I'm AIMing you with a much beastlier method.[/quote]What's that supposed to mean?  :mad:  :roll:[/quote]

Errmm... it means I told her* to be all like $l(4000-2l) = 1,000,000 - 2(l-1000)^2 \le 1,000,000$.

*oops I said him darn oh well

[quote="probability1.01"][quote="4everwise"][quote="probability1.01"]Oh yeah ignore 4everwise I'm AIMing you with a much beastlier method.[/quote]What's that supposed to mean?  :mad:  :roll:[/quote]

Errmm... it means I told him to be all like $l(4000-2l) = 1,000,000 - 2(l-1000)^2 \le 1,000,000$.[/quote]

thank youu-u-u! But it was all about parabolas (in that section of the book anyways)...

((sends you jelly filled donuts in the mail))


A single large banquet table is created by joining the sides of several square tables, each of which seats one person for each open side. What is the maximum number of people that could be seated at a table which uses 10 small tables.

I think that's correct. The initial count is $ 4\cdot10\equal{}40$, but there are at least $ 9$ places where two tables join, so we take away $ 2$ from each of those places to get $ 40\minus{}2\cdot9\equal{}\boxed{22}$.

[quote="math154"]I think that's correct. The initial count is $ 4\cdot10 \equal{} 40$, but there are at least $ 9$ places where two tables join, so we take away $ 2$ from each of those places to get $ 40 \minus{} 2\cdot9 \equal{} \boxed{22}$.[/quote]

Which means that every single formation seats $ 22$?

You wish to maximize the perimeter of a rectangle with area 10, which clearly occurs with a 10x1 rectangle. So it is 2x(10+1)=22.

As for math154's solution, he says there are at LEAST 9 junctions.


calculus
derivative
limit
function
real analysis
real analysis unsolved

If f: R --->R is continous at 0. Determine a necessary and sufficient condition for ( f(2x) - f(x) ) / x having a limit at 0.

I guess the answer should be that f is differentiable at 0, and I manage to do the forward direction. However, I am stuck in doing the reverse direction.  :oops:

$ \lim_{x\to 0}\frac{f(2x)\minus{}f(x)}{x}\equal{}\lim_{x\to 0} \frac{f(2x)\minus{}f(0)\plus{}f(0)\minus{}f(x)}{x}\equal{}$

$ \lim_{x\to 0} \left( 2\frac{f(2x)\minus{}f(0)}{2x}\minus{}\frac{f(x)\minus{}f(0)}{x}\right)\equal{}f'(0)$

That has to be what distantstar called the "forward direction," which is the easy direction. You assumed differentiability and derived that the limit in question equals the derivative. What if you start merely by assuming that the limit in question exists?

Note the extra hypothesis that $ f$ is continuous at zero. Without that we could simply have an example like $ f(x) \equal{} \text{sgn}(x),$ in which case the limit is zero but the function is not even continuous, much less differentiable.

Actually, $ f$ is Lipschitz is some neighborhood of zero: indeed, for $ x$ small enough $ |f(2x) \minus{} f(x)|\leq C|x|$ - follows from the existence of a limit. Let $ f(0) \equal{} 0$. We can write $ |f(x)| \equal{} |f(x) \minus{} f(\frac x2) \plus{} f(\frac x2)|\leq |f(\frac x2)| \plus{} c|\frac x2|\leq \dots\leq f(0) \plus{} c\sum \frac {x}{2^k} \equal{} c|x|$.
Now it's easy to prove the existence of right and left limits and conclude that they are equal.

Yeah, you are right. However, we can only show that f(x) / x is bounded by some positive constant c, how do we actually know that it admits a limit?

If $ M \equal{} \lim \frac {f(2x) \minus{} f(x)}{x}$, apply my previous post to $ g(x) \equal{} f(x) \minus{} Mx$ and see what you get.


At Pigskins High School $ 20\%$ of the boys and $ 80\%$ of the girls attended a football game. If $ 45\%$ of the student body is female, what percent of the school population attended the game?

If $ 45\%$ of the student body is female, $ 55\%$ is male. To solve, we find the percentages of each gender percent that attended:

$ 0.8(45\%)\plus{}0.2(55\%)\equal{}\fbox{47}\%$.


[hide]4x + 3y/xy = 44
12x - 2y/x = 44
so
4x + 3y/xy = 12x - 2y/xy
so
4x + 3y = 12x - 2y
5y = 8x

the ratio of x to y is 8:5 or 8/5 or 8 to 5...[/hide]

[color=indigo][hide] first you use substitution to combine the two equations and from there you can get x/y=5/8 :lol:  [/color][/hide]

My method would be ...

[hide="My answer"]
Because the two equations are equal to 44, set the variable expressions equal to each other, so that:

4/y + 3/x = 12/y - 2/x

Multiply by x*y.

4x + 3y = 12x - 2y

5y = 8x

y = 8/5 * x

There's your answer.
[/hide]

[quote="Silverfalcon"]Given that:

\[\frac {4}{y} + \frac {3}{x} = 44\\
\frac {12}{y} - \frac {2}{x} = 44\]

Find the ratio of x to y.[/quote]

[hide="click here"]$\[\frac {4}{y} + \frac {3}{x} =\frac {12}{y} - \frac {2}{x}$

$4x+3y=12x-2y$

$8x=5y$

$x:y=5:8$[/hide][/hide]