function
logarithms
Euler
algebra theorems
algebra

I've seen formulas for many sequences like this one but have never seen one for this particular one:

$ (i)$ Is there a formula for $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n}$? If so, what is it?

$ (ii)$ Is there a special limit for $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n}$ or does it just tend to infinity?

1)  That depends on what you mean by a formula.  All you really have is an expression in terms of the [url=http://mathworld.wolfram.com/HurwitzZetaFunction.html]Hurwitz zeta function[/url].  There's certainly no nice well-behaved exact expression, although there are pretty accurate asymptotics.

2)  That depends on what you mean by special limit.  It's known, for example, that $ \left| \sum_{k\equal{}1}^{n} \frac{1}{k} \minus{} \ln n \right|$ approaches a special constant known as the [url=http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant]Euler-Mascheroni constant[/url].

[quote="t0rajir0u"]1)  That depends on what you mean by a formula.  All you really have is an expression in terms of the [url=http://mathworld.wolfram.com/HurwitzZetaFunction.html]Hurwitz zeta function[/url].  There's certainly no nice well-behaved exact expression, although there are pretty accurate asymptotics.

2)  That depends on what you mean by special limit.  It's known, for example, that $ \left| \sum_{k \equal{} 1}^{n} \frac {1}{k} \minus{} \ln n \right|$ approaches a special constant known as the [url=http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant]Euler-Mascheroni constant[/url].[/quote]

Hmm... so you could say that if the Euler-Mascheroni constant is a known constant is the limit as $ n$ approaches $ \infty$ of $ \left| \sum_{k \equal{} 1}^{n} \frac {1}{k} \minus{} \ln n \right|$ and $ \ln n$ approaches $ \infty$ as $ n$ approaches $ \infty$, then the limit as $ n$ approaches $ \infty$ of $ \sum_{k \equal{} 1}^{n} \frac {1}{k}$ is $ \infty$

Whoops; for some reason I thought the wording of your question implied that you already knew that the limit was infinity.  That is indeed the correct answer, but the proof is much simpler than that:  one can estimate the difference between the two instead of proving that the limit exists or use a standard argument that is generalized by the [url=http://en.wikipedia.org/wiki/Cauchy_condensation_test]Cauchy condensation test[/url].


given a simple symmetric random walk $ S_k$ on $ Z^d$

where $ S_n\equal{}\sum_{k\equal{}1}^n X_k$,  $ P(X_k\equal{}e_j)\equal{}P(X_k\equal{}\minus{}e_j)\equal{}\frac{1}{2d}$

then for $ N\equal{}\min \{ n>0: S_n\equal{}0\}$

id like a decent look at $ f_n\equal{}P(N\equal{}n)$ thanks

are you looking for an exact value or for the behaviour of $ f(n)\equal{}P[N\equal{}n]$ for large $ n$. If this is just an equivalent for large $ n$ that you are looking for, try to do that with a Brownian motion (hitting time of $ B(0,1)$, starting from $ x\equal{}(2,0,\ldots,0)$, say): this is much easier I think, and should behave the same.


analytic geometry
complex numbers
geometry proposed
geometry

Given triangle $ABC$ centroid $G$, $P$ is an arbitrary point on plane, contruct the right isoceles triangle $APA'$(right at $P$ and follow positive direct) $G'$ is the centroid of triangle $A'BC$, $K$ is midpoint of $PG'$ contruct the right isoceles triangle $KG'J$ (right at $G'$ and follow positive direct).Prove that line $PJ$ is always through a fix point, find the barycentric of this point. :)

[quote="gemath"][color=darkred]Given triangle $ABC$[b][u]centroid $G$[/u][/b], $P$is an arbitrary point on plane, construct the right isoceles triangle $APA'$(right at $P$and follow positive direct) $G'$is the centroid of triangle $A'BC$, $K$is the midpoint of $PG'$construct the right isoceles triangle $KG'J$(right at $G'$and follow positive direct). Prove that the line $PJ$is always through a fix point, find the barycentric of this point.[/color][/quote][color=darkblue]What is the roll of the centroid $G$ ?
In the my proof with the complex numbers I found $6j=p+(1-2i)a+(2+i)(b+c)$.
I think is a little mistake in exercise (... with the letters !) or ... in the my calculus.[/color]

Dear Virgil! $G$ is centroid (the intersection of three median) of triangle $ABC$, I don't understand the term "roll" of you! Really the fix point is $G''$ such that $i\vec{G''A}+\vec{G''B}+\vec{G''C}=\vec{0}$ but it is quite hard to find the real barycentric coordinates of $G''$

[quote="Virgil Nicula"][quote="gemath"][color=darkred]Given triangle $ABC$[b][u]centroid $G$[/u][/b], $P$is an arbitrary point on plane, construct the right isoceles triangle $APA'$(right at $P$and follow positive direct) $G'$is the centroid of triangle $A'BC$, $K$is the midpoint of $PG'$construct the right isoceles triangle $KG'J$(right at $G'$and follow positive direct). Prove that the line $PJ$is always through a fix point, find the barycentric of this point.[/color][/quote][/quote][color=darkblue]The centroid $G$ has no any roll in the enunciation of this problem
because the definitions of the points $A'\ ,\ G'\ ,\ K\ ,\ J$ are independent of the point $G$ !
Denote $X(x)$- the point $X$ with the affix $x$. Then, $\left\{\begin{array}{ccccc}a-p=i(a'-p) & \Longrightarrow & a'=p-i(a-p) & \Longrightarrow & a'=(1+i)p-ia\\\ 3g'=a'+b+c & \Longrightarrow & & & 3g'=(1+i)p+(-ia+b+c)\\\ 2k=p+g' & \Longrightarrow & 6k=3p+3g' & \Longrightarrow & 6k=(4+1)p+(-ia+b+c)\\\ k-g'=i(j-g') & \Longrightarrow & j=(1+i)g'-ik & \Longrightarrow & \boxed{6j=1\cdot \boxed p+5\cdot\boxed{\frac{(2+i)(-ia+b+c)}{5}}}\end{array}\right\|$
Therefore, the point $J$ is dependent of the mobile point $P$ and
the fixed point $F(f)$ is $f=\frac{1}{5}\cdot (2+i)(-ia+b+c)$ because $f=\frac{1}{5}\cdot (6j-p)$.
The barycentrical coordinates of the fixed point $F$ are obtained immediately.
[/color]

Ah yes you are right, the point $G$ doesn't need for this problem! I'm sorry
My solution is quite simple
Let $I$ such that $i\vec{IA}+\vec{IB}+\vec{IC}=\vec{0}\Rightarrow I$ fix  we have for any point $P$ then
$i\vec{PA}+\vec{PB}+\vec{PC}=(i+2)\vec{PI}$ but easily seen
$i\vec{PA}=\vec{PA'}, (i+2)\vec{PI}=3\vec{PJ}$
thus $3\vec{PG'}=\vec{PA'}+\vec{PB}+\vec{PC}=3\vec{PJ}$
$\Rightarrow PJ$ always through $G'$.
Remark $I$ has complex barycentric coordinates $(i,1,1)$ and it fix and we can convert them to real coordinates , I found this problem base on it. I think it is quite new!


geometry
incenter
circumcircle
Euler
geometry proposed

Let ABC be a triangle with intouch triangle DEF.
I and I' be the incentre of triangle ABC and DEF.
ray II' intersect circumcircle (O) of triangle ABC at X.
Prove that simson line wrt. X and euler line are
concurrent on the circumcircle (O)
then, O lies on II".

Are you sure this problem is true? I drew in Geometer sketchpad and saw that it was wrong.


linear algebra
matrix
modular arithmetic
number theory unsolved
number theory

Let $ (a_{m,n})_{m,n\geq 1} $ be an infinite matrix of non-zero integers such that for all m the set of prime factors of the numbers in the m-th row is finite. Prove that the system of congruences $ x_{m+n}\equiv a_{m,n} (mod x_m) $ is solvable in primes.

I suppose we're not allowed to use Dirichlet's Theorem? Otherwise, unless I'm missing something here, I think we don't even need that condition about the set of prime factors of $(a_{m,n})_n$ being finite for all $m$.. :? We could simply define the primes inductively.

I'll try, and I hope it's not wrong.. First of all, I was probably wrong: you do need that condition.

Let $x_1$ be any prime which does not divide any of $a_{1,k}$. We need a prime $x_2=\equiv a_{1,1}\pmod {x_1}$. We can find such a prime which does not divide any of the $a_{2,k}$, by Dirichlet's Theorem. We now need $x_3$ s.t. $x_3\equiv a_{1,2}\pmod {x_1},\ x_3\equiv a_{2,1}\pmod {x_2}$, and such that $x_3$ does not divide any of the $a_{3,k}$. We can find such a prime, by Dirichlet. We continue like this, each time choosing $x_{m+n}$ s.t. it doesn't divide any of the $a_{m+n,k}$ and it satisfies the required congruences.

Doesn't it work?


A uniform steel chain 10m long and with a mass of 6kg per m, hangs vertically. Compute the work required to wind it up?

The same hint as in the previous problem.. :D

level of the center of mass... :wink:


geometry
perimeter
circumcircle
geometric transformation
reflection
rectangle
trigonometry

A triangle with acute angles $\alpha = \angle BAC$ and $\beta = \angle CBA$ is given. Externally, isosceles triangles $ACD$ and $BCE$ with bases $AC$, and $BC$, and vertex angles $\angle ADC = \beta$ and $\angle BEC = \alpha$, are drawn. Let $O$ be the center of the circle circumscribed to the triangle $ABC$. Prove that $|DO|+|EO|$ equals the perimeter of the triangle $ABC$ if and only if the angle $\angle ACB$ is the right.

Let $M$, $N$ be the midpoints of $BC, AC$; let $P,Q$ be the intersections of the perpendicular bisectors of $BC, AC$ with the circumcircle of $ABC$. Then $D,E$ are the reflections of $P,Q$ in $BC$, $AC$ respectively.
It follows that $DO=DM+MO=PO+MO=R+2MO$ and $EO=EN+NO=QO+NO=R+2NO$.
Then, the thesis becomes $2R+2MO+2NO=AB+BC+CA$ if and only if $ABC$ is rectangle in $C$.
If we write down those lenghts with a bit of trigonometry, we obtain the equivalent relation
\[ 1+\cos\alpha+\cos\beta=\sin\alpha+\sin\beta+\sin\gamma  \]
or, equivalently,
\[ \sin\alpha-\cos\alpha+\sin\beta-\cos\beta=1-\sin(\alpha+\beta)  \]
that is
\[ \sqrt{2}\left(\sin\left(\alpha-\frac{\pi}{4}\right)+\sin\left(\beta-\frac{\pi}{4}\right)\right)=1-\sin(\alpha+\beta)  \]
and, applying the sum->product and duplication formulas, we get to
\[ 2\sqrt{2}\sin\left(\frac{\alpha+\beta}{2}-\frac{\pi}{4}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=1-2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha+\beta}{2}\right)  \]
Now, let $x=\sin\left(\frac{\alpha+\beta}{2}\right)$ and $y=\cos\left(\frac{\alpha-\beta}{2}\right)$ and rewrite the last equation in terms of $x,y$ :
\[ 2(x-\sqrt{1-x^2})y+2x\sqrt{1-x^2}-1=0  \]
now, factoring, we get $(x-\sqrt{1-x^2})(2y-x+\sqrt{1-x^2})=0$.
Now, we note that if $0\leq \alpha,\beta\leq\frac{\pi}{2}$, necessarily we have $0\leq x\leq \frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}\leq y\leq 1$ so $2y-x+\sqrt{1-x^2}$ cannot be zero. Therefore
\[ (x-\sqrt{1-x^2})(2y-x+\sqrt{1-x^2})=0\Leftrightarrow x=\sqrt{1-x^2}\Leftrightarrow x=\frac{1}{\sqrt{2}}  \]
that is, if and only if $\alpha+\beta=\frac{\pi}{2}$.
So, the relation required holds if and only if $ABC$ is rectangle in $C$.


If $ a < b$ ,prove or disprove that $ a^{b} < b^{a}$

Edit:  ohh sorry ,It should be $ a^{b} > b^{a}$  ..sorry again  :oops:

All you need to do to disprove the statement is provide a counterexample, which he just did.

If, $ a>b>e \Rightarrow a^b<b^a$.
If, $ e>a>b \Rightarrow a^b>b^a$.

cf. http://www.mathlinks.ro/viewtopic.php?t=266800

$ 0 > 1$ but $ 0 \equal{} 0^1 < 1^0 \equal{} 1$.

Look up a graph of $ x^y \equal{} y^x$ in the first quadrant. You'll see that it consists of the line $ x\equal{}y$ and another curve, which together, divide the plane into four parts. If we assume $ y < x$, that is, we're under the $ x\equal{}y$ part of the graph of $ x^y \equal{} y^x$, this region is cut into two parts by the other part of the graph $ x^y \equal{} y^x$. Above this "other part", we have $ y^x > x^y$, and below it, we have $ x^y > y^x$. On it, of course, we have that they are equal.


[quote="Interval"]what exactly is 1 cubic foot?[/quote]

A cubic foot is the volume of a cube with dimensions $1 \text{ ft.}\times 1 \text{ ft.}\times 1 \text{ ft.}$

Now we we got that cleared up, can you help me with this question?

[quote="Interval"]A steel drum in the shape of a right circular cylinder is required to have a volume of 100 cubic feet.

This question has two parts:

(a) Express the amount A of material required to make the drum as a function of the radius r of the cylinder.

(b) How much material is required if the drum is of radius 3 feet?

NOTE: By the way, what exactly is 1 cubic foot?[/quote]

by the rate of your topics, there coul probably be a forum just for all your questions :rotfl:

[hide]the surface area of a right cylinder is $2\pi r^{2}+2\pi r h$.  PLEASE DONT MEMORIZE THAT!!!

and $v=\pi r^{2}h=100$.  you can solve for $h$ in terms of r and plug that into the surface area equation

when u figured tat out, just plug in 3 for r.[/hide]


trigonometry
geometry
trapezoid
Gauss
cyclic quadrilateral
congruent triangles
angle bisector

Given an acute triangle $ PBC$ with $ PB\neq PC.$ Points $ A,D$ lie on $ PB,PC,$ respectively. $ AC$ intersects $ BD$ at point $ O.$ Let $ E,F$ be the feet of perpendiculars from $ O$ to $ AB,CD,$ respectively. Denote by $ M,N$  the midpoints of $ BC,AD.$
$ (1)$: If four points $ A,B,C,D$ lie on one circle, then $ EM\cdot FN \equal{} EN\cdot FM.$
$ (2)$: Determine whether the converse of $ (1)$ is true or not, justify your answer.

Because of the cyclic quadrilateral $ ABCD,$ it is easy to see the configuration of the triangle $ \bigtriangleup OEF$ and the similar right triangles $ \bigtriangleup EBO\ \bigtriangleup FCO,$ erected ( outwardly to it ) on its side-segments $ OE,\ OF,$ respectively.

It is well known that the midpoint $ M$ of the side-segment $ BC,$ lies on the midperpendicular of the segment $ EF.$

Similarly the midpoint $ N$ of the side-segment $ AD,$ lies also on the same line ( we consider now, the similar right triangles $ \bigtriangleup EAO\ \bigtriangleup FDO,$ erected inwardly to it, on the same side-segments ).

That is we have that the line segment $ MN,$ is the midperpendicular of the segment $ EF$ and so, from the congruent triangles $ \bigtriangleup EMN,\ \bigtriangleup FMN,$

we have $ (EM)\cdot (FN) \equal{} (EN)\cdot (FM)$ $ ,(1)$ and the proof of the part [b][size=100](1)[/size][/b], is completed.

$ \bullet$ As for the part [b][size=100](2)[/size][/b] of the proposed problem, we can say that it is not true in general.

If we consider two points $ B',\ C',$ on $ AB,\ CD$ respectively, such that $ \angle EOB' \equal{} \angle FOC'$ then, the midpoint $ M'$ of the segment $ B'C',$ lies also on the line segment $ MN$ as the midperpendicular of $ EF$ and although the relation $ (EM')\cdot (FN) \equal{} (EN)\cdot (FM')$ is true, however the quadrilateral $ AB'C'D,$ is not cyclic $ ($ because $ B'C'$ is not parallel to $ BC$ $ ).$

But clearly, when the given triangle $ \bigtriangleup ABC$ is isosceles, the quadrilateral $ AB'C'D$ as before is cyclic and so, in this particular case, the part [b][size=100](2)[/size][/b] of the proposed problem is true.

Kostas Vittas.

as for (1)
Denote R,S,T,Q are midpoints of GD,GA,GB,GC , respectively
△NRF≌△ESN   △ETM≌△MQF
So EN=FN EM=FM
=> EM*FN=EN*FM

The first part of this problem is in fact a past olympiad problem.  :blush: I saw it few years ago, but we are required to prove $ EM\equal{}FM$ instead of $ (EM)\cdot (FN) \equal{} (EN)\cdot (FM)$.

For the second part, let $ K$ be the intersection of $ NS$ and $ MT$ (using the configuration in plane geometry' post), then $ KEST$ is a cyclic quadrilateral. Thus $ \angle NSE\equal{}\angle ETM$ no matter $ ABCD$ is cyclic or not. Then by cosine law, it can be shown that $ (OA)\cdot (OB) \equal{} (OC)\cdot (OD)$ also implys $ (EM)\cdot (FN) \equal{} (EN)\cdot (FM)$.

And it is easy for us to construct a triangle with this property  :lol:

[quote="thecool"]Can someone explain to me the solution of part b. In vittasko's solution O isn't the intersection of AC' and B'D, nor is it the intersection of AB' and DC'.[/quote]

I'm confused with vittasko's solution as well. The point $ O$ changes when you change $ B$ to $ B'$ and $ C$ to $ C'$...

I am sorry dear [b][size=100]thecool[/size][/b] and [b][size=100]jbmorgan[/size][/b] and all my friends, because of the confusion from all what I wrote about the solution of the part [b][size=100](2)[/size][/b] of the proposed problem.

Although we can say that it is not true in general, however I used a false argument ( as for the point $ O\equiv AC\cap BD$ ) in trying to prove it.

It is better to use the argument of an arbitrary (non isosceles) trapezium $ ABCD$ with $ AD\parallel BC,$ where is true the relation as the problem states (easy to prove) and clearly, it is not a cyclic quadrilateral.

I will post here later more details. Give please a little time.

Sorry again, Kostas Vittas.

[quote="vittasko"]…It is well known that the midpoint $ M$ of the side-segment $ BC,$ lies on the midperpendicular of the segment $ EF.$[/quote]
[b][size=100][color=DarkBlue]LEMMA  1. - A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABD,\ \bigtriangleup ACE$ be, two similar right triangles with $ \angle ABD \equal{} \angle ACE \equal{} 90^{o}$ and $ \angle BDA \equal{} \angle CEA \equal{} \angle \omega,$ erected outwardly (or inwardly) to it. Prove that the midperpendicular of the side-segment $ BC,$ also bisectw the segment $ DE.$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] - (we used here, the same argument as mentioned before, by [b][size=100]plane geometry[/size][/b] ).

Let $ N$ be, the midpoint of the segment $ DE$ and it is enough to prove that $ NB \equal{} NC$ $ ,(1)$

We denote as $ K,\ L,\ M,$ the midpoints of the segments $ AD,\ AE,\ BC$ respectively and is is easy to show that

$ KN \equal{} \frac {AE}{2} \equal{} LC$ $ ,(2)$ and $ LN \equal{} \frac {AD}{2} \equal{} KB$ $ ,(3)$

from $ KN\parallel AE,\ LN\parallel AD$ and the right triangles $ \bigtriangleup ADB,\ \bigtriangleup ACE.$

From$ (2),$ $ (3)$ and because of $ \angle BKN \equal{} \angle BKA \plus{} \angle AKN \equal{} \angle CLA \plus{} \angle ALN \equal{} \angle CLN,$

we conclude that the triangles $ \bigtriangleup KBN,\ \bigtriangleup LNC$ are congruent.

Hence, we have that $ NB \equal{} NC$ $ ,(4)$ and the proof of the Lemma is completed.

[b][size=100]NOTE :[/size][/b] As an interesting remark in the configuration of the above [b][size=100]Lemma 1[/size][/b], we can say that $ \angle BNC \equal{} \angle 2\omega.$

Kostas Vittas.

[quote="vittasko"]…It is better to use the argument of an arbitrary (non isosceles) trapezium $ ABCD$ with $ AD\parallel BC,$ where is true the relation as the problem states (easy to prove) and clearly, it is not a cyclic quadrilateral.[/quote]
[b][size=100][color=DarkBlue]LEMMA  2. - An arbitrary trapezium $ ABCD$ is given with $ AD\parallel BC$ and let be the point $ O\equiv AC\cap BD.$ We denote as $ M,\ N,$ the midpoints of its side-segments $ BC,\ AD$ respectively and let $ E,\ F$ be, the orthogonal projections of $ O,$ on $ AB,\ AC,$ respectively. Prove that $ (EN)\cdot (FM) \equal{} (EM)\cdot (FN).$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] - Let be the point $ P\equiv AB\cap CD$ and it is easy to show that the points $ M,\ N,\ P$ are collinear, applying the [b][size=100]Thales theorem.[/size][/b]

From the complete quadrilateral $ PAODBC,$ we have that the points $ P,\ N,\ O,\ M,$ are in harmonic conjugation as well.

So, we conclude that the pencil $ E.PNOM,$ is also in harmonic conjucation.

Because of now, $ EP\perp EO,$ we conclude that the line segment $ EO$ is the angle bisector of the angle $ \angle MEN$ and then, we have $ \frac {EM}{EN} \equal{} \frac {OM}{ON}$ $ ,(1)$

Similarly, the line segment $ FO$ is the angle bisector of the angle $ \angle MFN$ and then, we have also $ \frac {FM}{FN} \equal{} \frac {OM}{ON}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {EM}{EN} \equal{} \frac {FM}{FN}$ $ \Longrightarrow$ $ (EN)\cdot (FM) \equal{} (EM)\cdot (FN)$  and the proof is completed.

Kostas Vittas.

[quote="vittasko"]…[b][size=100]NOTE :[/size][/b] As an interesting remark in the configuration of the above [b][size=100]Lemma 1[/size][/b], we can say that $ \angle BNC \equal{} \angle 2\omega.$[/quote]
[b][size=100]PROOF.[/size][/b] - We draw the circles $ (K),\ (L),$ with diameters $ AD,\ AE$ respectively and let be the points $ B'\equiv (K)\cap BC$ and $ C'\equiv (L)\cap BC.$

We denote as $ D',\ K',\ A',\ L',\ E',$ the orthogonal projections of the points $ D,\ K,\ A,\ L,\ E$ respectively and it is easy to show that $ D'B' \equal{} BA'$ $ ,(1)$ because of the point $ K'$ is the midpoint of the segments $ BB'$ and $ A'D'.$

Similarly, we have that $ A'C \equal{} C'E'$ $ ,(2)$ because of the point $ L'$ is the midpoint of the segments $ A'E'$ and $ CC'.$

The right triangles $ \bigtriangleup D'B'D,\ \bigtriangleup E'C'E$ are similar, because they have $ \angle D'B'D \equal{} \angle BAD \equal{} 90^{o} \minus{} \angle \omega \equal{} \angle CAE \equal{} \angle E'C'E$ and so,

we conclude that $ \frac {D'B'}{E'C'} \equal{} \frac {D'D}{E'E}$ $ ,(3)$

From $ (3)$ $ \Longrightarrow$ $ \frac {D'B}{D'B \plus{} E'C'} \equal{} \frac {D'D}{D'D \plus{} E'E}$ $ \Longrightarrow$ $ \frac {D'B'}{2(MB)} \equal{} \frac {D'D}{2(MN)}$ $ ,(4)$

$ ($ from $ D'B' \plus{} C'E' \equal{} BA' \plus{} A'C \equal{} BC \equal{} 2(MB)$ and $ D'D \plus{} E'E \equal{} 2(MN)$ $ ).$

From $ (4)$ $ \Longrightarrow$ $ \frac {D'B'}{MB} \equal{} \frac {D'D}{MN}$ and so, we  conclude that the right triangles $ \bigtriangleup D'B'D,\ \bigtriangleup MBN,$ are similar.

Hence, we have that $ \angle MBN \equal{} \angle D'B'D \equal{} 90^{o} \minus{} \angle \omega$ $ ,(5)$

From $ (5)$ $ \Longrightarrow$ $ \angle BNC \equal{} \angle 2\omega$ and the proof is completed.

Kostas Vittas.

Here is my solution
Call P is midpoint of PO. By Gauss line theory, we have M, N, J is conlinear. Similar TST USA 2000, we get EF is perpendicular to MN, in the other hand, JE = JF. Hence, JMN is perpendicular bisector of MN……..


Find v(t) in the RLC circuit

[img]http://i39.tinypic.com/291dity.jpg[/img]

please tell me a guideline on how to proceed with this problem
soln. not necessary

I would first write out the complex impedances of all the circuit elements and calculate the total impedance to find the current as a function of time through the whole thing (which is just the current through the top resistor).  More specifically, you'll need to calculate both the phase angle and the current amplitude in terms of the AC voltage amplitude.  Then use this expression for the current to find the voltage across only the top resistor as a function of time (call this $ V_R(t)$).  Let the voltage of the AC source be $ V_S(t)$, so you can then solve the equation $ V_S(t) \minus{} V_R(t) \minus{} V(t) \equal{} 0$.

I haven't had too much practice with AC circuits, but that's the first thing that comes to mind.  As a side note, only a few months ago did I learn about complex impedances - previously, I had tried to deal with all AC circuits using phasors, which was quite obviously a nightmare.

thanks for the guide
ill see how it goes!!


Can someone please tell me how to download the National MATHCOUNTS Webcast to my computer?

I know that it is possible using Realplayer 11, but I dislike Realplayer as it is "badware" as listed on "Stopbadware.org". Does anyone know of an alternate way?

Badware? Why is it "badware" when it works?

Again, I must state that you have been very off recently. Unless that was sarcasm, badware is like malware. Kind of.

Realplayer 11 is labebed as "badware" because they secretly install advertising bots on your computer without telling you so in their privacy policy, and then fail to uninstall them if you remove Realplayer from your computer (which is why it is considered "badware").

So, anyone know how to download the National MATHCOUNTS webcast? That announcer guy said it'd be available for download, but I have no idea how to download it.

I think that's the only way to download it.  I have it, but it is like 100 megabytes.

Darn. Now I'll have to put the "badware" back on my computer, download the vid, convert it to AVI format, then remove the "badware" once more.

So, does anyone else know of a way to download the webcast? That announcer said it'd be up for download the entire year, but I have yet to see any sort of "download" button on the website.

Downloading is any type of data transferred from a remote site to your computer. In this case, it is streaming video, which is basically an non-savable (without the right software) download.

I was able to get a .wmx file of it. It's not the actual file but you can play it on your computer while not using internet explorer, firefox etc...

I put it in the .zip attachment file

Thanks, now that I know the URL of the site from which the strem originates, I can download the webcast to my computer using WM Recorder.

[quote="12markkram34"]I will post the webcast here if I am able to download it to my computer.[/quote]
No.  Unless you get written permission from MATHCOUNTS and an AoPS administrator, you won't post a 100MB, copyrighted Webcast on this site.

You may all view the streamed WebCast on your own computers.  You may attempt to capture that WebCast on your own computers.  You may even tell others how to capture the WebCast on their own computers.  But, please don't put AoPS in the position of hosting copyrighted material belonging to other organizations.  [Instead, why don't we all write to MATHCOUNTS, and tell them we would like to *purchase* a full-resolution DVD of the competition.  If enough people ask, maybe they will consider adding it to their product list.]

Acutally, you don't need RealPlayer.
Use Windows Media Player.  (Should be pre-installed for windows computers)
Although it shows your settings are saved, at the bottom click on preferences and choose Windows Media Player.
It will load from that perfectly fine.
No Badware.
:)

Several people have told me of this method, but none so far have explained to me exactly how to do so. Can you please elaborate?

Okay.  Here are my instructions more detailed:

Go to http://varicast.variview.net/getContent.aspx?Action=Factory&FileName=mathcounts08/frame.htm&WCID=d51efcc1-2aea-4488-bb57-47fd15f43812

As you said, it says it needs RealPlayer; well it doesn't.
Click on PREFERENCES which is at the bottom to the left of "Powered by Variview"

Now click on Windows Media Player and click save preferences.

Hopefully that works for you.
 :P

So by doing this you can download the webcast to your computer?

As long as Mathcounts leaves this up on their Variview thing, it will work if you have internet connection.
Mathcounts has a very abnormal way of doing things, making it very hard to save things.
If you go to this page:
http://wm.variview.net/Clients/WinterMedia/33087.asf?WCID=2245&ViewerID=f37f04d4-7b7b-4af0-bd0a-16df343bcca1
You will see a popup open with the Windows Media Player with the video.
Unfortunately, that is the farthest I can help you.
 :(

Thanks! Perhaps I will find a way to download it.


Let $ ABCD$ be a cyclic quadrilateral with no pair of opposite sides parallel. Assume that $ BD$ is a diameter of the circle. Let $ E=AB\cap CD$ and $ F=AD\cap BC$. Show that $ EF\perp BD$.

In the triangle $ DEF$, $ EA\perp FD, \ FC\perp DE$. Then $ B$ is orthocenter, so $ BD\perp EF$.


LaTeX algebra function domain search \/closed

Hi,

Whenever you make a $ \text{\LaTeX}$ code, for some reason, there are two tags that result as an HTML replacement.

For example, the one I just made above generates:

[code][/code]

On one of my sites, I transfer the information on a post to the site. If I transfer the code directly, a small blue hyperlink box surrounds this LaTeX code (on IE) and obviously leads me to a 404. (Since my domain is not AoPS) My suggestion is to change this HTML replacement to:
[code][/code]

I understand that this then in turn, does not allow the pretty little popup every time you click it, but my reasoning is that not many people use this function. If they really wanted to know the code for the LaTeX, they could just quote the post.

If this is absolutely not possible, please let me know if there is a way around this problem.

Thanks.

I'm not even sure how to be appropriately snarky about this ... if only one other person makes use of the pop-up feature, your comment can be summarized as "please make that person's life less convenient in order to make my life more convenient."  Oy vey.  Have you never come across a search-and-replace utility before?

Of course, I'm much more important than the other person. :D

I'm not sure if search and replace will work. Two codes may be similar, but each one is the slightest bit different.

Edit:

I thought of a solution. Since each one has the class of latex, I could change the coloring of the link to the color of the background for all tags of class latex. Does anyone know how I can do that?

Meet my good friend sed. cat thecontentyourerippingoffaops | sed "s/[<]a href=\"code.php?hash=[0-9a-f]*\" class=\"latex\"[>]//g" . Who cares about extra </a> tags?

Well... I actually got permission from the author... if that makes it any better?

What do you mean by:
[code]sed "s/[<]a href=\"code.php?hash=[0-9a-f]*\" class=\"latex\"[>]//g" .[/code]

If you apply a regex like that to your code, it'll take out the offending links.

[url=http://en.wikipedia.org/wiki/Sed]Sed[/url] is a UNIX command. It means "stream editor".

[quote="Dojo"]Well... I actually got permission from the author... if that makes it any better?[/quote]I don't think I ever gave you permission. What you are doing right now is creating extra http requests and basically stealing bandwidth from our servers. That is usually frowned upon ... we don't really care right now, but I am [b]definitely not[/b] changing the $ \text{\LaTeX}$ code to benefit your site  :!:

PS I actually thought the first post was a joke :)


geometry
parallelogram
vector
calculus
calculus computations

Its very easy:
$ A \equal{} \frac {1}{2} |d_1 \times d_2|$
 where $ d_1$ & $ d_2$ are the diagonal vectors respectively
$ d_1 \equal{} a \plus{} b
d_2 \equal{} a \minus{} b$
where a & b are the vector representing the sides.
I think u should revise vectors theory before sitting down to solve problems :)

this is solved..moderator can close my topic

@curioustoknow
thanks for ur interest but mods here are pretty active and it's not necessary to lock topics after just the answer has been found out....u never know someone may come up with some genralization or another good proof :)

And this is another reason why this user needs to know that AoPS/ML is not a [b]homework forum[/b].

AoPS'ers need to be aware that questions like these, ppl are recommended to post hints rather than solving it fully. Not only are you solving someone's supposedly homework/practice/whatever it is, the skills necessary to think, to grow ideas, and learn are destroyed by you (the problem solver), showing only a dull-procedure that 'hey now I see the solution, ok, hmm yea, this is how is done, ok, so boring'. Where is the sense of accomplishment of the student for putting the effort to try and solve a problem? These are only routine problems... Actually, what [u]is[/u] a [b]problem[/b]? If I tell you the answer to a puzzle, there is no mystery behind it anymore; the same applies here.

PS: Oh yes, I'm not a moderator and if you got any inquiries about me, please PM me. Thx.


A fair, six-sided die is tossed 12 times and the number that comes out is recorded each time, and after 12 tosses, the 12 numbers are summed. What number has the greatest probability of being this sum?

[hide]I'm not sure, but since the most likely sum of $2$ dice is $7$, the most likely for $12$ should be $42$.[/hide]

[hide="Solution"]
The sums range from 12 to 72, and with some funky probability casework, I seem to be getting 42 as the expected value.
That's because the expected value on any die in general is 3.5, and we see that $12 * 3.5 = 42$, which should be the expected value of the sum.

If you asked me what the probability is, though, I couldn't tell you.
 :lol:
[/hide]

EDIT: 500th post, and beaten by about 8 seconds by i_like_pie.
Woot I'm now a YangMills Theory! (I'll have to google them)

In my attempts to at least approximate the probability, I always get somewhere near a $16000\%$ chance of the sum being $42$.

[quote="illcrowflu"]I'm not sure if its right or not, but I got about $\%14.6$.
EDIT: That can't be right.  I think it should be about $\%50$ since each individual dice roll is uniformly distributed and by the central limit theorem, rolling 12 times approaches a normal distribution with $\mu = 42$.[/quote]
$50\%$? Isn't that a bit too high? There are $61$ possible sums, so if it were evenly distributed, it would be $\approx1.6\%$ for each sum. However, it won't be evenly distributed, since the probabilities would all be different. I would say somewhere between $2\%$ and $14\%$ would make sense, but I still can't figure out an exact value.

[quote="illcrowflu"]I'm not sure if its right or not, but I got about $\%14.6$.
EDIT: That can't be right.  I think it should be about $\%50$ since each individual dice roll is uniformly distributed and by the central limit theorem, rolling 12 times approaches a normal distribution with $\mu = 42$.[/quote]

I don't think the central limit theorem applies to this problem because it assumes a continuous variable and the sum of twelve dice rolls is discrete. Also, even if it did apply, the 50% wouldnt be the probability of getting a 42, but it would be the probability of getting less than or equal to a 42.

@whatthedeuce: Yea, you're right -  the $\%50$ would be the probability of less than or equal to $42$.

The central limit theorem does apply to discrete distributions.  The idea behind it is that if there are $n$ random variables that are identically distibuted, then as $n$ grows larger it approaches a normal distribution.  So any probabilities obtained through the central limit theorem are mearly approximations.  As $n\to \infty$ the approximations become more and more accurate.

I wrote a program to calculate the exact probability of rolling a sum of $42$ from $12$ rolls.  There are $144,840,476$ ways to roll a sum of $42$, and there are $6^{12}$ ways to roll $12$ dice.  The probability is about $\%6.65$.  Again as $n$ grows arbitrarily large, the probability of one "point" in the sample space occuring goes to zero.

I was wondering if it is possible to generalize this question.  In other words, if $n$ dice are rolled simultaneously, what is the probability of the expected value occuring?  Using my program I began to calculate the number of ways to roll the expected value.  For an odd number of dice the expected value is not an integer, so I was taking the floor in those cases.  Let $n$ be the number of dice and $f(n)$ be the number of ways to roll the expected value.  The first few values are:
$f(1) = 1$
$f(2) = 6$
$f(3) = 25$
$f(4) = 146$
$f(5) = 780$
$f(6) = 4332$
$\cdots$
$f(12) = 144,840,476$
Does anyone see a recursion in this? (as ugly as it might be)


What does the term "factor over the integers" mean? My math book does not have a definition of it in the index. Would $ (3x\plus{}1)(2x\plus{}1)$ be considered factored over the integers, although the end result is not an integer?

try this website
[url]http://en.wikipedia.org/wiki/Polynomial_factorization[/url]

Its also called Polynomial Factorization

When a polynomial is factored over the integers, the coefficients of the factors are integers. E.g. x^2+2x+1 can be factored over the integers, namely (x+1)^2.

[quote="bleuwiki"]sometimes the things that are posted on wikipedia is not accurate, but the reference that wz1118 referred isn't so bad...[/quote]
Actually, a recent British Magazine did an experiment with Wikipedia, and found that Wiki was actually [b]more[/b] accurate than a British Encyclopedia commonly used in British schools. [Citation needed]

yeah ure answer there is already factorised


linear algebra
matrix
linear algebra unsolved

$J$ is a matrix with $0$ everywhere except above the main diagonal where there are only $1$s.
Prove that $J$ is a product of idempotent matrices (do not use the fact that every non-invertible ...)

Now, could you prove that every non invertible complex matrix is a product of idempotent matrices ...

I think I know how to deduce from here the fact that each singular matrix is a product of idempotent matrices (I remember having posted this some time ago, topic "let's delirate" or something like this). We "know" (killer posted a very nice and complete proof  ;) ) that each non invertible matrix is a product of nilpotent matrices. Thus it suffices to prove the assertion for nilpotent matrices. This reduces to proving for their Jordan form and thus to this problem. So, all we need is to exhibit two idempotent whose product is the elementary Jordan matrix.


When i type in (2+ :rt3: )^x-(2- :rt3: )^x, it comes up with sinh(ln( :rt3: -2)*x)
Where does this come from?

If I type in (1/2+ :rt5: /2)^(x-1)*(1/2- :rt5: /2)^x, it spits out e^0*( :rt5: /2 - 1/2)*(-1)^x.  The answer makes sense, but why is it thinking in terms of e?

[quote="cats..."]The answer makes sense, but why is it thinking in terms of e?[/quote]

I'm not a calculator expert or anything, but since you're dealing with solving for x using exponents, your calculator is probably taking logs (e being the natural logarithm).  Switching from "exact" to "approximate" mode should give you a more understandable answer.


find all function $: Q^{+}\to Q^{+}$
and satisfying  $f(x+1)=f(x)+1$and$f(x)^n=f(x^n)$


limit
real analysis
real analysis unsolved

[quote="ozlem81"]a=[0,1]
b=[1,0]
p1 is any point
q1 is center of line between p1 and a
p2 is center of line between q1 and b
qn is center of line between pn and a
pn is center of line between qn-1 and b

find limit of pn by using epsilon delta way[/quote]That's not how you find the limit,it's how you [i]prove [/i]this is the limit.

Hint: $ \lim_{n \to \infty} p_n$ and $ \lim_{n \to \infty} q_n$ must be on the line segment that connects $ a$ and $ b$.


Let $f(n)$ be the sum of the first $n$ terms of the sequence \[ 0, 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, \ldots\, .  \] a) Give a formula for $f(n)$.
b) Prove that $f(s+t)-f(s-t)=st$ where $s$ and $t$ are positive integers and $s>t$.

The sum of the first $2k+1$ terms is just twice the sum of the first $k$ positive integers, thus $f(2k+1)=k(k+1)$.
Then it's easy to get $f(2k+1)=k^2$ by just subtracting the last number.

To see the last one: $s+t$ and $s-t$ have clearly the same parity, thus just two cases. But I'm foolish, thus...


trigonometry
function
complex numbers
calculus
calculus computations

i was curious, can one evaluate things like $\sin(i)$, or $\cos(i)$ by using tayor series, or euler's theorem? i am not sure if they are proven for imaginary arguments, (the taylor series), so i was thinking that

[hide]$\cos(i)= \frac{ e^{i*i}+e^{-i*i}}{2}=\frac{e^{-1}+e}{2}$

this seems correct, but is it?[/hide]

It is. And you'll get the same if you use Taylor series, though I don't know what Euler's theorem is.

I think by Euler's Theorem, Altheman was referring to:

$e^{iz}=\cos z + i\sin z$

This is true for all complex numbers $z,$ as is the closely related formula that Altheman used.

Altheman essentially calculated that $\cos(i)=\cosh(1).$

More generally, $\cos(iy)=\cosh(y)$ and $\sin(iy)=i\sinh(y).$

That shows that both the sine and the cosine grow exponentially fast as we move away from the real axis.

The identity $\cos^2z+\sin^2z=1$ remains true for all complex numbers $z.$ In fact, all of the important trigonometric identities you know remain true for all complex numbers.

The only thing ... well for real $x,\,\cos^2z+\sin^2z=1$ forces $\sin x$ and $\cos x$ to remain bounded and in the interval $[-1,1].$ But since the squares of complex numbers can be negative (or, in fact, to have any argument you please), this identity does not prevent the cosine and sine from being unbounded functions on the complex plane.


Diophantine equation
number theory
number theory proposed

maybe it is posted,(however i searched and found nothing)
let $ t_n$ denote the $ n \minus{} th$ triangular number,
prove that for each $ n$ there exists infinitely many integer $ m$ such that $ t_nt_m$ is a perfect square.
well i have proven it using pell equation but i wanted to know if there exists a construction giving all $ m$ which satisfies the condition.
thx in advance for your help. :)

From [L.Panaitopol and A.Gica - Number Theory Problems (in Romanian only)].

[quote]Denote $ t_n \equal{} \frac {n(n\plus{}1)} {2}$. $ t_1 \equal{} 1$ is a perfect square. We prove that if $ t_n$ is perfect square, then so is $ t_{4n^2 \plus{} 4n}$. Indeed, $ t_{4n^2 \plus{} 4n} \equal{} \frac {(4n^2 \plus{} 4n)(4n^2 \plus{} 4n \plus{} 1)} {2} \equal{} (2(2n\plus{}1))^2\frac {n(n\plus{}1)} {2} \equal{} (2(2n\plus{}1))^2t_n$.[/quote]

From [W. Sierpinski - 250 Problems of Elementary Number Theory, problem 6/238]. He first proves the almost trivial fact that $ t_{n}t_{n\plus{}1}$ is never a perfect square, then that given $ n$ there are infinitely many $ m$ such that $ t_nt_m$ is a perfect square.

[quote]The equation $ x^2 \minus{} n(n\plus{}1)y^2 \equal{} 1$ admits infinitely many solutions, since $ x\equal{}2n\plus{}1$ and $ y\equal{}2$ is one, and if $ (x,y)$ is a solution, then $ ((2n\plus{}1)x \plus{} 2n(n\plus{}1)y, 2x \plus{} (2n\plus{}1)y)$ is also a solution.

Now, if $ x^2 \minus{} n(n\plus{}1)y^2 \equal{} 1$, that means $ t_nt_{2t_ny^2} \equal{} t_nt_ny^2(2t_ny^2\plus{}1)^2 \equal{} t_n^2y^2x^2 \equal{} (t_nxy)^2$. For example, for $ n\equal{}2$, we have $ t_2t_{24} \equal{} 30^2$, and $ t_2t_{2400} \equal{} (3\cdot 20\cdot 49)^2$.[/quote]

(Quite equivalent to your use of Pell equations theory).

However, no one mentions finding all possible solutions.  :(


superior algebra
superior algebra unsolved

Very simple: For every n you have $ S_{n}\equal{}\langle(12),(12...n)\rangle$ because the transpositions $ (k,k\plus{}1)$ generate $ S_{n}$ and $ (k,k\plus{}1) \equal{} (12...n)^{k\minus{}1}(12)(12...n)^{\minus{}k\plus{}1}$


Prove that there exists extremely many natural numbers n for which:

n divides (2^n + 2)

Happy New Year!

Posted before .
Consider the sequence :
$ a_1\equal{}2$
$ a_{n\plus{}1}\equal{}2^{a_n}\plus{}2$
Easy to check it satisfy by induction.

[quote="TTsphn"]Posted before .
Consider the sequence :
$ a_1 \equal{} 2$
$ a_{n \plus{} 1} \equal{} 2^{a_n} \plus{} 2$
Easy to check it satisfy by induction.[/quote]
Nice answer friend!
Please tell where motivation for this could be made from.


calculus
integration
pigeonhole principle
number theory unsolved
number theory

Can someone please explain these two problems (I can't visualize that the question is asking):

1. What is the least number of weights required to weigh any integral number of pounds up to 63 pounds if one is allowed to put weights in only one pan of a balance?

2. Determine the least number of weights required to weigh any integral number of pounds up to 80 pounds if one is allowed to put weights in both pans of a balance.

These problems are supposed to encompass number base representation techniques.

I think the optimal situations are reached when you take powers of $2$ for the first case and powers of $3$ in the second case, until the sum of the weights is the maximal wight you have to obtain, but I'm not sure right now.

well, in both cases, powers of 2 are the best; that's a simple application of the pigeonhole principle.

Think about it man: in the second case you can put weights on both pans of the scales, so you can measure differences of weights as well. Try to see why you can do it with powers of $3$, which are fewer than the powers of $2$ required.

You can do it with $1,2,4,8,16,32,17$, but you can also do it with $1,3,9,27,40$. You can get $2$ by putting $1$ on one pan and $3$ on the other pan, for example. I'm not sure right now (I just woke up and I'm confused :D), but I think that works.


Let $M;N;P$ be three points outside a given triangle $ABC$ such that $\hat{CAN}=\hat{CBM}=30^{o}$;$\hat{ACN}=\hat{BCM}=20^{o}$:$\hat{PAB}=\hat{PBA}=40^{o}$.Show that $MNP$ is an equilateral angle

first denote $X_{\alpha}^{k}(S)(A)=B$ as "[b]$B$ is a point on the plane s.t $SB=k.SA$ and $\angle BSA=\alpha$ (counter clockwise)[/b]...

now with the notation above we get that:

$X_{30}^{\frac{BC}{BM}}(B)(M)=C$

$X_{30}^{\frac{AN}{AC}}(A)(C)=N$

now its easy to see that $X_{\alpha}^{k}(P) \circ X_{\beta}^{k'}(Q)=X_{\alpha+\beta}^{kk'}(S)$...

now we know that

$\triangle BMC \sim \triangle ANC \Rightarrow \frac{BC}{BM}=\frac{AC}{AN}\Rightarrow \frac{BC}{BM}.\frac{AN}{AC}=1$

hence we get that:

$X_{30}^{\frac{BC}{BM}}(B) \circ X_{30}^{\frac{AN}{AC}}(A)(M)=X_{60}^{1}(S)(M)=(N)$ so we get that $SM=SN$ and $\angle MSN=60$

so its sufficinet to how that $S=P$

now note that the only constant point of $X (O)$ is the $O$ itself...

so its sufficient to show that:

$X_{60}^{1}(S)(P)=P$

so its sufficient to show that:

$X_{30}^{\frac{BC}{BM}}(B) \circ X_{30}^{\frac{AN}{AC}}(A)(P)=P$...


Can we, as another poster already mentioned, use a method to verify our answer, something that is quite rote to do, and not include it in our solutions but merely mention that we verified that way?  I don't have the time nor the ability to tex this method since it uses some complicated things that are hard to do in Latex.

If it is a standard method, you are fine.  If it's something you're not sure is standard, you might cite a web page or text for the method.


\[ \left( {z \plus{} i} \right)^n  \minus{} \left( {z \minus{} i} \right)^n  \equal{} 0
\]

[quote="vazk"]
\[ \left( {z + i} \right)^n - \left( {z - i} \right)^n = 0
\]
[/quote]

Clearly $ z-i, z+i \neq 0$.

So the equation $ \Leftrightarrow (\frac{z+i}{z-i})^n=1$
$ \Rightarrow \frac{|z+i|}{|z-i|}|=1$
$ \Leftrightarrow |z+i|^2=|z-i|^2$
$ \Leftrightarrow (z+i)(\overline{z}-i)=(z-i)(\overline{z}+i)$
$ \Leftrightarrow z=\overline{z}$

So $ z$ is real.

Now let $ z=cot \theta$, where $ 0 <\theta<\pi$,  we get the equation

$ \Leftrightarrow \frac{z+i}{z-i}=e^{i\frac{2k \pi}{n}}$, where $ 1 \le k \le n-1$ is an integer.
$ \Leftrightarrow \frac{re^{i \theta}}{re^{i(-\theta)}}=e^{i (2\theta)}=e^{i\frac{2k \pi}{n}}$
$ \Leftrightarrow \theta=\frac{k \pi}{n}$

So $ z=cot \frac{k \pi}{n}$, where $ 1 \le k \le n-1$ , are the only solutions.


$1^{\circ}\blacktriangleright n\in N^*,\ \{a,b,c\}\subset (0,\infty)\Longrightarrow$$\sum \frac{a}{b+c}\le \frac{a^2+b^2+c^2}{a+b+c}\cdot \sum \frac{a}{b^2+c^2}\le \sum \frac{a^2}{b^2+c^2}\ .$
$2^{\circ}\blacktriangleright \sum \frac{a^n}{b^n+c^n}\le \sum \frac{a^{n+1}}{b^{n+1}+c^{n+1}}\ .$

Nice problem. Finally it’s very easy.
Let’s prove more generally \[ \sum_{cyc}\frac{a^n}{b^n+c^n}\le \frac{a^{n+1} +b^{n+1} +c^{n+1} }{a^n+b^n+c^n}\sum_{cyc}\frac{a^n}{b^{n+1} +c^{n+1} }\le\sum_{cyc}\frac{a^{n+1}}{ b^{n+1} +c^{n+1} }. \]
Multiplying with $a^n+b^n+c^n$ yields
\[ \sum_{cyc}\frac{a^{2n}}{b^n+c^n} + \sum_{cyc} {a^n}\le\sum_{cyc}\frac{a^{2n+1}}{ b^{n+1} +c^{n+1} }+ \sum_{cyc} {a^n} \le\sum_{cyc}\frac{a^{2n+1}}{ b^{n+1} +c^{n+1} }+ \sum_{cyc}\frac{a^{n+1}( b^n+c^n )}{ b^{n+1} +c^{n+1} } . \]

For the left inequality, let $N_a : = ( b^n+c^n )( b^{n+1} +c^{n+1})$ etc. Then it is equivalent to
$\sum_{cyc}\frac{a^{2n}b^n(a-b)+a^{2n}c^n(a-c)}{N_a}\ge0 \Leftrightarrow \sum_{cyc}a^nb^n(a-b)(\frac{a^n}{N_a}-\frac{b^n}{N_b}) \ge0$.
But this is clear because $a>b\Leftrightarrow \frac1{N_a}> \frac1{N_b}$ etc., so all terms are $\ge0$.

For the right inequality, let $N_a : = ( b^{n+1} +c^{n+1})$ etc. Same trick: it is equivalent to
$\sum_{cyc}\frac1{ N_a }(a^nb^n(a-b)+ a^nc^n(a-c))\ge0 \Leftrightarrow \sum_{cyc}a^nb^n(a-b)( \frac1{N_a}- \frac1{N_b})\ge0$.
This is clear because again $a>b\Leftrightarrow \frac1{N_a}>\frac1{N_b}$ etc.


calculus
integration
quadratics
algebra
number theory solved
number theory

Find all positive integer k such that the equation
     x 2 - (k 2 -4)y 2 = -24
has positive integral solutions

Source: Suggested by Namdung for School Team Selection Test, year ?

With the aid of a previous post regarding Markoff equations, the solution follows nicely.

x^2 - (k^2 -4)y^2 = -24      (*)

implies x^2 = k^2 y^2 - 4y^2 - 24 is a perfect square.

Consider the quadratic equation f(x) = x^2 - kyx + y^2 + 6.     (**)

This equation has integral roots iff its discriminant D is a perfect square.

D = k^2 y^2 - 4y^2 - 24, so we see that * has integral solutions <=> f(x) = 0 has integral solutions.

For f(x) = 0 to have integral solutions, xy > 0. If x, y are both negative, we can simply plug (-x) and (-y). Therefore, we only consider positive solutions to f(x) = 0. Now the equation is symmetric with respect to x and y, so there must exist a family of solutions such that x \geq y.

There also must then exist (x, y) that solves f(x) with x minimal. Let that x equal x_1.

By Viete's formula, x` = ky - x_1 is also a solution to f(x) = 0.

But x_1 is minimal, so x` \geq x_1 implies ky - x_1 \geq x_1.

Then y^2 + 6 = kyx - x^2 \geq x_1^2 implies

6 \geq (x_1 + y)(x_1 - y).

x_1 \geq y, so (x_1 + y)(x_1 - y) can only be one of {0, 1, 2, 3, 4, 5, 6}. Checking, we see that (x_1 + y)(x_1 - y) = 1, 2, 3, 4, 5, 6 yield contradictions.

So (x_1 + y)(x_1 - y) = 0. Then x_1 = y.

Thus, we solve for y^2 - ky^2 + y^2 + 6 = 0.
Then k = 2 + 6/y^2, which implies that k can only equal 8.

What we have just shown is that if * is solvable in integers, then k must equal 8.

Conversely, if k = 8, * indeed has a solution - namely, (x, y) = (6, 1). There may be more, but it is not required to find them. =)


Let $ A$ is a subset compact of metric space $ (X,d)$.$ A'\equal{}\{x | x
\text{ is accumulation point of } A\}$.Prove $ A'$ is compact

[quote="fedja"]Just show that $ A'$ is a closed subset of $ A$.[/quote]
Can you proof $ A'$ is a closed  set, please??

If $ x$ is not an accumulation point of $ A$, one can find an open ball $ B$ around $ x$ that contains only finitely many points of $ A$. No point in $ B$ can be an accumulation point of $ A$, whence $ B$ is contained in the complement of $ A'$.


geometry
3D geometry
sphere
geometry solved

How to prove 5 (or more than 5) points in space are spheric?

[quote="ashegh"]for each four points we can find a sphere wich goes through them.then i dont think that for each 5 are spheric :([/quote]
What I mean is that if 5 points are spheric, then what conditions does the points satisfy? Just like asking you how to prove 4 points are cyclic.

i thinks for each n points wich are spheric,if we project them in  a screen,they should be cyclic...

[quote="Arne"]Angles on a sphere are tricky. I guess you'll just need to show that there is a point equidistant from the 5 given points...[/quote]
That's okay but... how to find the centre?

ohh.. ure right Arne.i made a mistake :blush:

Ashegh to ...........

IFAI Do U wanna solve contest Problem? ay kalak, sheytoon

oh,oh,oh...

i forgot it...

its like one of the contests problem... :?


Hello. Since we are very likely to start this thread after today, I will go ahead and start it. Feel free to critique on the Bush and Kerry debate here! I really want to know other people's opinions on the debates! Does anyone else have to do a debate similar to this for school, because my history teacher is making us take notes on this debate.

There's a pretty big meeting at my school for boarders who are all watching it together. I know that JSRosen3 is there. I'm still stuck here doing homework at home though, so I'll just wait until the next one.

i only watched the second half of it..  kerry did much better, he didnt stop speaking and had lot to say. Bush seemed unknowlegable and always ended before his allocated time ran out.

but i heard the comentators later say that bush was stronger in the first half..

[quote="ZennyK"][quote="Rep123max"]He voted against the $87 (which is now over $200) billion after they wouldn't roll back the tax cuts on the rich to pay for it. See Kerry believes in not going into huge deficets.[/quote]

I'd rather have a leader who believes in stopping insane dictators who kill millions of their own people and terrorize the world by abusing their "authority," which they obtained out of backstabbing greed and lies in the first place. [/quote]

Then why doesn't Bush want to attack North Korea?  North Korea actually bragged about having nuclear weapons and repeatedly threatened to use them.

I thought it was kinda interesting when Kerry said that his main concern in foreign policy matters was the proliferation of nuclear weapons and George Bush responded that he agreed that nuclear proliferation, "in the hands of a terrorist enemy" was a bad thing.  North Koreans and Kim-Jong Il are by no means terrorists, so I guess it is alright for them to have nukes.

I thought Kerry definitely had the upper hand :) , and he made some good progress about clearing up his position on Iraq.

[quote="ZennyK"][quote="Rep123max"]He voted against the $87 (which is now over $200) billion after they wouldn't roll back the tax cuts on the rich to pay for it. See Kerry believes in not going into huge deficets.[/quote]

I'd rather have a leader who believes in stopping insane dictators who kill millions of their own people and terrorize the world by abusing their "authority," which they obtained out of backstabbing greed and lies in the first place.

That leader would be President Bush, not John Kerry, for whom the choice between voting to eliminate a crazy mass murderer and saving a money was a "complicated decision."[/quote]

And while we're at it, let's just say that since chickens lay eggs, the chicken came before the egg.  :maybe:

Things aren't always that simple, dude.

You can read the [url=http://news.bbc.co.uk/nol/shared/bsp/hi/pdfs/01_10_04_debate1.pdf]full debate transcript here.[/url]

Kerry may have been more fluent, but he also contradicted himself a good deal more frequently than Bush.  He simultaneously speaks of building strong alliances (with his "90% of casualties and 90% of the cost" statistic), and about how our troops outnumbering other troops in Iraq is a bad thing, then says he wants to increase the number of troops currently there, apparently without any say-so from the other countries about their contributions.  He also claimed to have laid out some sort of grand plan for America, but gives neither specifics nor hearsay any more than what can be stated in noncommittal terms.

[quote="white_horse_king88"]Kerry was way more fluent, but what he said was absurd sometimes. For example. He said that by the end of the year, they will have 125000 Iraqi personel trained. They have 50,000 so far. That is rediculous! How on earth do you train competent personel with little education in 2 months?[/quote]

You can do better than that. Clearly the training process has already begun for those other 75000 but has not finished yet.

[quote="white_horse_king88"]Kerry was way more fluent, but what he said was absurd sometimes. For example. He said that by the end of the year, they will have 125000 Iraqi personel trained. They have 50,000 so far. That is rediculous! How on earth do you train competent personel with little education in 2 months?[/quote]

I'm pretty sure Kerry was talking about the end of 2005, since he can't be president untill January.

I have a friend of mine who has some more political aptitude than I do who had something to say on the subject.  I can't vouch for credibility (he's an ber-conservative), but he makes an interesting point.

[quote]when kerry brought up the anecdote about parents buying kids body armor for christmas because the army wouldnt, one thing came to mind, he voted against that armor for troops, once I saw that, kerry lost what little semblence of competence he had to me. [/quote]

[quote="ZennyK"][quote="Rep123max"]He voted against the $87 (which is now over $200) billion after they wouldn't roll back the tax cuts on the rich to pay for it. See Kerry believes in not going into huge deficets.[/quote]

I'd rather have a leader who believes in stopping insane dictators who kill millions of their own people and terrorize the world by abusing their "authority," which they obtained out of backstabbing greed and lies in the first place.

That leader would be President Bush, not John Kerry, for whom the choice between voting to eliminate a crazy mass murderer and saving some money was a "complicated decision."[/quote]

1. i dont think Saddam killed millions of people.
2. There are many other dictators out there, are we gonna kill them all?
3. The war has caused over 1000 america death, even more injuries, and even more iraqi death..
4. if you want to save lives, why dont you save your food and give it to the 1 billion people in africa and asia that has nothing to eat? If americans saved the food they would have otherwise wasted, thats enough to cure world hunger..

[quote]when kerry brought up the anecdote about parents buying kids body armor for christmas because the army wouldnt, one thing came to mind, he voted against that armor for troops, once I saw that, kerry lost what little semblence of competence he had to me. [/quote]

but the 87 billion was still passed, yet they still didnt have armor.. that shows what bush really wanted the money for.. tell your friend that![/quote]

[quote]All right, then, I'll un-sugarcoat my position. [/quote].. Now be careful lest some one starts harping on [i]your[/i]  "flip-flop" :)
Just kidding ..Of course the snap polls are .. well just sanp polls.. One  can vote multiple  times (safe-guards  to prevent multiple voting is easy to bypass) and "one" need not be a human..just a computer program.

But some  experts analysis, on prediction of the polls,  I heard, IMO, was even more weird ...

Here's my scoop on the presidential debates:

More articulate and fluent: Kerry
Points more appealing to mainstream undecided voters: Tie
Who lied less: Tie (although I think that while Kerry has inserted little inaccuracies as all politicians do, Bush's entire criticism of Kerry waffling is based on distortion. Don't listen to me though, I'm biased here)

So both debates were even.

Now the vice presidential debate...(which took place in good ol' Cleveland!)

More articulate and fluent: Cheney
Points more appealing to mainstream undecided voters: Edwards
Who lied less: same thing as above - they both lied their rear ends off but I think that the lies of the Democratic candidate were less serious

So, naturally, being a Democrat, I think that Kerry and Edwards came off on the better side. I'm afraid though - the only way they'll win is a stellar performance on the last debate from Kerry and a dumb slip-up like "the illiteracy level of our children are appalling" from Bush.  :maybe:

[quote="EFuzzy"]When I voted on the cnn poll about 10 minutes after the debate ended, Kerry was leading 91% to 8. Then his lead gradually disappeared. I hope no-one's hacking  :)[/quote]

Meanwhile some one commenting in some other internet form: (Some part xxxx out to protect the innocents..  They were talking about election in Afganistan:

[quote]A young man who cast a ballot before voting was stopped showed CNN how he used water to remove the ink stain from his left thumb. While his voting card was punched, also to prevent fraud, he said many other voters have more than one voter card.

"I have cast my vote, but the ink was removable," he said. "This is ridiculous."

We should make this guy an honorary XXXX member. This is as good as the Procedure laid down   for Productive Internet Voting. Cool

My congratulations to President Dubya and Ambassador Khalilzad, and Of-Course-President_Elect Karzai. Afghanistan has certainly matured into a full democracy, if they have mastered Productive Voting to such an extent.

BTW, just out of curiosity: XXXXX  buddy, where were u between 11:30pm Oct. 8 and 6:30am Oct. 9, US EDT? Between those nocturnal hours, (the sun rises only at 7:30 around here these days) over 334,000 votes had been cast for Sen. Kerry on the CNN web site poll. Kerry was leading 80-20 then. In the many hours since, only 100,000 more votes for Kerry, and 400,000 more for Bush.....[/quote]

Looks like   these polls can be easily fooled by hackers ...

For those of you talking about the internet polls: Actually, the results have nothing to do with hackers.  Mostly thiey are a result of Bloggers (although the campaigns have gotten into the act as well).  For example, I read the blog [url=atrios.blotspot.com]Eschaton[/url].  After each debate, he posts a list of websites that are likely to have polls.  Since thousands of individuals read his website every day, he can quickly organize perhaps a hundred thousand people to vote on online polls in seconds, without really doing anything much.  This is the reason online polls are inaccurate -- it has nothing to do with hacking or insecurity at all.  Rather, a poll in which the pollees are self-selected is inherently unscientific and gives you no useful information at all.

Joel - true but I do  know some people who go  take it a little more seriously -(or have a little more what they call as fun) - I am told that its is as simple as deleting cookies  and let your script do the voting on some sites .. to vote multiple times..

Don't they have tracking devices that don't attach themselves to deletable cookies? Or are they not so concise about polling? Yeah. I will answer my question. Online polls are worthless, but they do represent a general picture of what is going on....

BUSH is getting cremated. He is very flustered. For some reason, he is not debating up to standards. Bush is stumbling over ideas, and the way he emphasizes them makes him sound very desperate.

For example, Bush just claimed a few seconds ago that crime with guns prosecution has increased by 68%. He said this was a good thing. What????? This isn't a good thing. Gun policies allow people to own AK-47's and not be prosecuted. Arresting people after crimes doesn't represent how safe a country is. It represents how incompetent police are with gun control in this country, but Bush believes that rising prosecution for gun crimes is a good thing. I think that is rediculous. You want prosecution to go down because that represents a decrease in crime.

Heard some place:[quote]During the second  debate, Bush was asked by a lady to name three mistakes he's made.  Looks like Bush has the answer now:, 'This debate, the last debate and the first  debate.'"[/quote]

I hated how Kerry, when asked what his plan was, told what his plan was intended to do.  Besides the fact that he needs to explain to the American public outright what on earth, and second of all, the public education system wasn't that terrific to begin with.

X - posting from the other thread -Enjoy:[url=http://www.badmash.org/dishoom.php]This from India[/url]

It's not fair to think that based on his arched back. What about Kerry's way-to-long face, it might be that his jaw has a communication device in it. imean, he got a botox injection right?

[quote="mathmonkey"]How about that hump on Bush's back? Anyone think it was for some sort of communication with someone else on the outside? He definitely seemed most-prepeared during the debate where the hump was noticed.[/quote]

I'm pretty sure that's been mentioned. Just read the whole topic.

[quote="RobinHood3000"]The "hump" was also noticed in the first debate.  Interesting to note that Bush seemed to do worst then, and his opponents still have to find something else to keep their twiddling thumbs busy.  You know, it COULD just be overly starched suits...[/quote]

Maybe the ironing lady took a sick day.


function
algebra
polynomial
quadratics
quadratic formula

Does anyone know any good methods for finding an explicit formula for a given sequence?
I know the thing about writing out the differences between every two terms, then doing it again until you get equal terms, but I don't know where to go from there.

For example, I want to find out if the sum of the first n perfect squares can ever be a perfect square. Show me how you would find that formula and solve the problem...

[quote="shobber"]Do you mean: $1 ^2+2^2+3^2+ \cdots + n^2= \frac {n(n+1)(2n+1)}{6}$?[/quote]
that is indeed the formula...

1 gives us a solution ;) (of course :P )
24 gives a perfect square as well (namely $70^2$)
I checked other numbers up to 400, but no solution  :?

This reminded me of a question of the BMO :)

BMO 1994 Round 2, question one.

Find the first integer $n>1$ such that the average of $1^2+2^2+3^2+...+n^2$ is itself a perfect square.

Can you solve this one? ;)

Proving of the identity: Use the fact: $(m+1)^3-m^3=3m^2+3m+1$
But i am not able to solve your problem since i knows nothing about number theory. I even do not know what perfect square is!

A perfect square is just a number of the form $n^2$ where $n\in \mathbb{Z}.$

I was looking for the METHOD to find an explicit formula for a sequence.

For example, I made this sequence today out of an explicit formula. Show me how you would find that formula!

1, 5, 15, 35, 70, 126, 210, 330...

[quote="Psycho"]Find the first integer $n>1$ such that the average of $1^2+2^2+3^2+...+n^2$ is itself a perfect square.

Can you solve this one? ;)[/quote]

Well, we have $\frac{1^2+2^2+3^2+...+n^2}{n}=k^2$, where $k\in \mathbb{Z}$. $1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$, so we have $\frac{\frac{n(n+1)(2n+1)}{6}}{n}=\frac{(n+1)(2n+1)}{6}=k^2 \Rightarrow 2n^2+3n+(1-6k^2)=0$. Using quadratic formula, we have $n=\frac{-3\pm \sqrt{9-4(2)(1-6k^2)}}{4}=\frac{-3\pm \sqrt{1+48k^2}}{4}$. Also, we know that $n>0$, so we can eliminate the negative root from the equation, so we essentially have $\frac{\sqrt{1+48k^2}}{4}=m-\frac{1}{4} \Rightarrow \sqrt{1+48k^2}=4m-1$, where $m\in \mathbb{Z}$. This first occurs for $k=1$, but that would also mean $n=1$, and the problem is asking for $n>1$.

Umm... I can't figure out where to go from here. I don't see any way of simplifying it further.

Take the polynomial $f(x)=a_7x^7 + a_6x^6 + ... + a_1x+a_0$. Plug in $x=0,1,2,...,7$ and the values for $f(0),f(1),...,f(7)$. Solve the system (matrices or calculator.) Now you're all set.

EDIT: or use Lagrange Interpolation.

[quote="JesusFreak197"]

Well, we have $\frac{1^2+2^2+3^2+...+n^2}{n}=k^2$, where $k\in \mathbb{Z}$. $1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$, so we have $\frac{\frac{n(n+1)(2n+1)}{6}}{n}=\frac{(n+1)(2n+1)}{6}=k^2 \Rightarrow 2n^2+3n+(1-6k^2)=0$. Using quadratic formula, we have $n=\frac{-3\pm \sqrt{9-4(2)(1-6k^2)}}{4}=\frac{-3\pm \sqrt{1+48k^2}}{4}$. Also, we know that $n>0$, so we can eliminate the negative root from the equation, so we essentially have $\frac{\sqrt{1+48k^2}}{4}=m-\frac{1}{4} \Rightarrow \sqrt{1+48k^2}=4m-1$, where $m\in \mathbb{Z}$. This first occurs for $k=1$, but that would also mean $n=1$, and the problem is asking for $n>1$.

Umm... I can't figure out where to go from here. I don't see any way of simplifying it further.[/quote]

This particular type of equation that you're trying to solve is called a Pell Equation. Its very famous so there's lots of information online about how to solve it. If you can't find anything useful though, maybe I'll write something up about it.

[hide]The next smallest integer n that works is n = 337.[/hide]

I guess you would have to say that it's the lowest degree polynomial possible.  I don't reemmber how to find it, though...though you can find the degree...
But you can see from your sequence that one way would be that the differences between the terms in your sequence are sums of triangular numbers.  Then you could derive it.  There is this way though...you can for instance find the sequence which goes 1, 2, 4, 8, 16.. and actually it after 16 goes 31, 58...

There's a wonderful general method for $k$ th pwers used Bernoulli nos. Anyone???

[quote="Hexaditidom"][quote="joml88"]There's not necessarily one set method to find a sequence given some elements.  In fact, there are infinitely many models that would fit your sequence.  Think about the first couple values that are given as points in the plane.  You can build infinitely many functions through these points.

If you are given some condition then it may be possible to narrow it down to one function.  For example, given that the sequence is modeled by a polynomial with integer coefficients would narrow down the possible functions immensely.[/quote]

Ok, then the problem in my second post is a polynomial with rational coefficients. It can be simplified into a single fraction.[/quote]

i guess u r looking for the difference thing...

take your sequence and start substracting

$1, 5, 15, 35, 70, 126, 210, 330 \dots$
$  4, 10, 20, 35, 56,  84,   120 \dots$
$     6,  10, 15, 21, 28,   36 \dots$
$        4,    5,   6,  7,    8 \dots$
$            1,   1,   1,   1   \dots$
$                0,   0,   0   \dots$

and the formula u look for is $U_n = (1 +  \Delta)^n * U_0$

expand and get:

$U_n = U_0 + \displaystyle\binom{n}{1} \Delta U_0 + \displaystyle\binom{n}{2} \Delta^2U_0  \dots $

where $1$ is $U_0$, $4$ is  $\Delta U_0$, etc.

then since everything after 6th row is 0, you only expand up to $\displaystyle\binom{n}{4} \Delta^4U_0$

and simplifying, ... ur done!

so your sequence is:

$\displaystyle \frac{n^4 + 10n^3 + 35n^2 + 50n + 24}{24}$

ps: u need to plug in $0$ to get the first term, $1$ to get the second, etc.
pps: this method assumes it's a polynomial equation, and each row of difference extends down infinitely in the same pattern, which means there will not be a $-1$ popping out of nowhere in the rows of $0$'s

p.s: $\Delta$ is only a symbol, but surprisingly its property can be act as an operator, thats why u can expand and attach the $U_0$. and $\Delta^n$ where $n$ is some power, just means that its the number of row of difference counting down

I found it in ARBELOS, the first issue.


probability
geometry
inradius
ratio
modular arithmetic
floor function
function

1. If a point is randomly chosen in the interior of a triangle with sides 7, 8 and 9, what is the probability that the point is 1 unit away from any side of the triangle?

2. 2009 is an example of a 4 digit number, exactly 2 digits of which are identical and adjacent. How many other 4 digit numbers have this property?

3. A student has a strange way of counting. He counts from 1 but only counts numbers that are relatively prime with 2009. If he had counted 2009 numbers, what is the last number counted?

4. Find the smallest n such that n+2 is divisible by 12, n+3 is divisible by 23, and n+4 is divisible by 34.

5, what is the smallest n such that n! is divisible by 2009^29?

6. IN a sequence of counting numbers, every odd term is doubled and every even term is increased by 1. If the new terms are arranged to form a sequence of increasing terms, find the 2009th term.

7. find all ordered pair (a, b) such that a^b +b^a = 17.

8. Find the 5th term of the sequence 2009, 2672, 3731, 5600.

[hide="Number Two"]
Case 1: The number is in the form aabc

There are $ 9 \cdot 1 \cdot 9 \cdot 9 \equal{} 729$ possible ways.

Case 2: The number is in the form baac

There are $ 10 \cdot 9 \cdot 1 \cdot 8 \equal{} 720$ possible ways.

Case 3: The number is in the form bcaa

There are $ 10 \cdot 9 \cdot 8 \cdot 1 \equal{} 720$ possible ways.

Thus our answer is $ 729 \plus{} 720 \plus{} 720 \equal{} 2169$.
[/hide]

EDIT: I think I see my mistake. Better?

Hi! Your answer is incorrect..One more try..

[hide="Number 2"]Case 1: aabc
$ 9 \times 1 \times 10 \times 10 = 900$

Case 2: abbc
$ 9 \times 10 \times 1 \times 10 = 900$

Case 3: abcc
$ 9 \times 10 \times 10 \times 1 = 900$

Cases 1 and 2 overlap: aaab
$ 9 \times 1 \times 1 \times 10 = 90$

Cases 1 and 3 overlap: aabb
$ 9 \times 1 \times 10 \times 1 = 90$

Cases 2 and 3 overlap: abbb
$ 9 \times 10 \times 1 \times 1 = 90$

Cases 1, 2, and 3 overlap: aaaa
$ 9 \times 1 \times 1 \times 1 = 9$

$ 900 + 900 + 900 - 90 - 90 - 90 + 9 = \boxed{2439}$[/hide]
[hide="Number 3"]In other words, he skips multiples of $ 7$ and $ 41$.  This means that out of every $ 287$ numbers, he skips $ 47$ of them and counts $ 240$.  After $ 8$ cycles of this, he has counted $ 1920$ numbers, the ending with (but not including) $ 2296$.  He has $ 89$ more numbers to go, which would land him at $ 2385$, but he must skip $ 12$ multiples of $ 7$ and $ 2$ multiples of $ 41$.  Thus the $ 2009^{th}$ number counted is $ \boxed{2404}$[/hide]
[hide="Number 5"]$ n$ must be the smallest number to have $ 29$ multiples of $ 41$ and $ 58$ multiples of $ 7$ less than it.  If the first restriction is fulfilled, however, the second must also be.  Therefore the answer is $ 29 \cdot 41 = \boxed{1189}$[/hide]
[hide="Number 6"]The odd numbers all become even numbers that aren't multiples of $ 4$, while the even numbers just become odd numbers, beginning with $ 3$.  Thus the new sequence is simply all natural numbers except for $ 1$ and all multiples of $ 4$.  Thus the $ 3n^{th}$ term is $ 4n+1$, so the $ 2007^{th}$ term is $ 2677$, which means the $ 2009^{th}$ term is $ \boxed{2679}$[/hide]
[hide="Number 7"]The intersection of the graphs
$ z = x^y + y^x$
$ z = 17$
has infinitely many points.  If you mean for integers, the only possibilities are $ \boxed{(1, 16), (2, 3), (3, 2), \text{and} (16, 1)}$[/hide]

[hide="For the others..."]In Number 1, if you mean exactly $ 1$, which I doubt, then the answer is clearly $ 0$.  If you mean $ \le 1$, I can't see any way other than brute force to do it.

For number 4, just use the Chinese Remainder Theorem.

Number 8 is not well-defined; there are infinitely many series beginning with those four terms.[/hide]

[hide="number 1"]
I'll assume that we want the point less than $ 1$ unit from any side of the triangle.

If a point $ P$ is exactly $ 1$ unit away from any side of the triangle, $ ABC$, then the locus of points $ P$ would be another triangle similar to $ \bigtriangleup ABC$. The smaller triangle would have an inradius $ 1$ unit less that triangle $ \bigtriangleup ABC$

Using herons formula $ [ABC] \equal{} \sqrt {12\cdot 5\cdot 4\cdot 3} \equal{} 12\sqrt {5}$
In radius of $ ABC \equal{} \frac {\bigtriangleup}{s} \equal{} \sqrt {5}$

So the ratio of the inradius for larger and smaller triangles is
$ \sqrt {5}: \sqrt {5} \minus{} 1$

So the ratio of area is
$ 5: 6 \minus{} 2\sqrt {5}$

So the probability that the pont is less than $ 1$ unit away is
$ \frac {6 \minus{} 2\sqrt {5}}{5}$
[/hide]

[hide="Number 4"]
I'm assuming that $ n\geq 0$ (otherwise there are infinitely many negative integer solutions).

$ n\plus{}2\equiv 0\implies n\equiv \minus{}2\equiv 10\pmod{12}\implies n\equal{}12k\plus{}10$.
$ 12k\plus{}13\equiv 0\implies 12k\equiv \minus{}13\equiv 10\pmod{23}$.
$ 12\cdot 2\equiv 1\pmod {23}$
$ 2\cdot 12k\equiv 2\cdot 10\implies k\equiv 20\pmod{23}\implies k\equal{}23j\plus{}20$.
$ 23j\plus{}24\equiv 0\implies 23j\equiv \minus{}24\equiv 10\pmod{34}$.
$ 3\cdot 23\equiv 1\pmod{34}$.
$ 3\cdot 23j\equiv 3\cdot 10\implies j\equiv 30\pmod{34}\implies j\equal{}34m\plus{}30$.
$ k\equal{}23(34m\plus{}30)\plus{}20$
$ n\equal{}12(23(34m\plus{}30)\plus{}20)\plus{}10\equal{}9384m\plus{}8530$.
And the smallest $ n$ is $ \boxed{8530}$.
[/hide]

Great solution by ocha for number 1

#2 1943 is the correct answer
#3 2402 is the correct answer
#4 3838 is the correct answer
#5 correct 1189
#6 correct 2679
#7 correct
#8 13,145 is the correct answer

For #$ 2$, would a number such as $ 2002$ be acceptable?

[hide="Possible answer."]If not, then then the answer would be $ (9 \cdot 9 \cdot 8) \cdot 3$.  For a given three-digit number with no two identical digits, you can duplicate any one of the digits to form a valid four-digit number.  The result would be $ 1944$, but that's off by $ 1$ from the answer you gave.[/hide]

[hide="number 2"]
$ 2009 \equal{} 7^2\cdot 41$
If we have skiped all the multiples of $ 7$ and $ 41$ then we would find that

$ t_{2009} \minus{} \left\lfloor \frac {t_{2009}}{7} \right\rfloor \minus{} \left\lfloor \frac {t_{2009}}{41} \right\rfloor \plus{} \left\lfloor \frac {t_{2009}}{7\cdot 41} \right\rfloor \equal{} 2009$

Solving without the floor function gives $ t_n \equal{} 2402.429... \approx 2402$
Since we solved without the floor function, this may be off by a couple digits If $ 7|2402$ or $ 41|2402$
However they dont, so plugging the answer back into the function yields

$ 2402 \minus{} \left\lfloor \frac {2402}{7} \right\rfloor \minus{} \left\lfloor \frac {2402}{41} \right\rfloor \plus{} \left\lfloor \frac {2402}{287} \right\rfloor \equal{} 2402 \minus{} 343 \minus{} 58 \plus{} 8 \equal{} 2009$

That works :) so thats our answer!
[/hide]

@[b]Begoner[/b]: haha, I was just about to post the exact same thing you did, now I'll have to answer number 2 instead...and $ 2002$ isn't acceptable because $ 7|2002$  :wink:


At a party, one is called [i]boring[/i] if he or she knows at most 3 other people. Prove that if everybody at the party has at least 3 [i]boring[/i] acquaintances, then everybody is [i]boring[/i]. How many participants may be there at the party in this case?

Isn't the statement obvious from the contrapositive?

In the case where everyone has 3 boring acquaintances, there are 4 people.

it is indeed quite obvious, but the answer for the second question is all even numbers greater equal $4$.
the construction of the graph would be e.g. a regular polygon in which also the longest diagonals are put in also.


[color=darkblue]Slove this equation:[/color] $ |x\minus{}2006|^{2008}\plus{}|x\minus{}2007|^{2009}\equal{}1$

[hide="answer"]$ x \equal{} 2006$[/hide]

[hide]We claim $ 2006,2007$ are the two solutions. Suppose there exists another solution $ x$, then $ 2006 < x < 2007$. Substituting in $ t \equal{} x \minus{} 2006$, we have
$ t^{2008} \plus{} (1 \minus{} t)^{2009} \equal{} 1$.
But since $ 0 < 1 \minus{} t < 1$, we have
$ t^{2008} \plus{} (1 \minus{} t)^{2009} < t^{2008} \plus{} (1 \minus{} t)^{2008} < (t \plus{} 1 \minus{} t)^{2008} \equal{} 1$
where the last inequality is due to Binomial Theorem. Contradiction.
[/hide]

What if $ x$ is not required to be real?

[color=darkblue]This equation have two root: $ x\equal{}2006$ and $ x\equal{}2007$[/color]


i took both alternate amc12 and regular amc 12

can you also to that for AIME?

I have another question. Did you try the search feature?

the answer to that one be a no as well :D

I dunno...I'd have to say it would be pretty hard to do a search and find what he was looking for in a reasonable amount of time for this particular question.  There's just so much stuff on AIME posted on the boards.

[quote="amirhtlusa"][quote="yif man12"]What if you have two different names and each name is registered at a different school and one of the schools takes the I and the other takes the alternate?[/quote]

what if you could move in forth dimention? ;)[/quote]

He should already  be moving in four dimensions :P

[quote="yif man12"]What if you have two different names and each name is registered at a different school and one of the schools takes the I and the other takes the alternate?[/quote]

Lol that would be Dae Ro. (Choate joke)


What are the formulas for:

1) $ \sin({\alpha\plus{}\beta})$
2) $ \cos({\alpha\plus{}\beta})$
and 3)$ \tan({\alpha\plus{}\beta})$

all in terms of $ \alpha$ and $ \beta$

Thanks. I am a noob at trig

[url=http://library.thinkquest.org/C0110248/trigonometry/formsine.htm]Take a look the link.[/url]

[hide="sin(a±b)"]
$ \forall \alpha,\beta \in R$:
$ \sin({\alpha\plus{}\beta}) \equal{} \sin(\alpha)\cos(\beta)\plus{}\sin(\beta)\cos(\alpha)$
$ \sin({\alpha\minus{}\beta}) \equal{} \sin(\alpha)\cos(\beta)\minus{}\sin(\beta)\cos(\alpha)$[/hide]
[hide="cos(a±b)"]
$ \forall \alpha,\beta \in R$
$ \cos({\alpha\plus{}\beta})\equal{} \cos\alpha \cos \beta \minus{}\sin \alpha \cos \beta$
$ \cos({\alpha\minus{}\beta})\equal{} \cos\alpha\cos\beta\plus{}\sin\alpha\cos\beta$[/hide]
[hide="tan(a±b)"]
$ \forall \alpha,\beta \in R, \alpha \neq\frac{\pi}{2}\plus{}k \pi,\beta \neq \frac{\pi}{2}\plus{}m \pi, \alpha\plus{} \beta\neq \frac{\pi}{2}\plus{}n \pi \neq \alpha\minus{} \beta, (k,m,n \in Z)$
$ \tan(\alpha\plus{}\beta)\equal{}\frac{\tan(\alpha)\plus{}\tan(\beta)}{1\minus{}\tan(\alpha)\tan(\beta)}$

 $ \tan(\alpha\minus{}\beta)\equal{}\frac{\tan(\alpha)\minus{}\tan(\beta)}{1\plus{}\tan(\alpha)\tan(\beta)}$[/hide]

hello, or here
http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html
Sonnhard.


Source: ARML 2001

Given $\frac{\log{xy}}{\log{\frac{x}{y}}}=\frac{1}{2}$, increasing $y$ by 50% decreases $x$ by a factor of $k$. Compute $k$.

[quote]Given $\frac{\log{xy}}{\log{\frac{x}{y}}}=\frac{1}{2}$, increasing $y$ by $50%
$ decreases $x$ by a factor of $k$. Compute $k$.[/quote]

Given $\frac{\log{xy}}{\log{\frac{x}{y}}}=\frac{1}{2}$, increasing $y$ by 50% decreases $x$ by a factor of $k$. Compute $k$.
 :D  :D

Cross-multiplying, we obtain $2\log xy=\log\frac{x}{y}$.  Using logarithm identities, we have $2\log x+2\log y=\log x-\log y$, or $\log x=-3\log y$.  Therefore, $\log x=\log y^{-3}$.  Since the logarithms are equal, $x=y^{-3}$, or $xy^{3}=1$.  We want the value of $k$ such that $(kx)\left(\frac{3y}{2}\right)^{3}=1$.  Expanding, we obtain $\frac{27kxy^{3}}{8}=1$, so $k=\frac{8}{27}$.


A compressed gas cylinder contains 1.00 x 10^3 g of argon gas.  The pressure inside the cylinder is 2050 psi at a temperature of 18 degrees Celsius.  How much gas remains in the cylinder if the pressure is decreased to 650 psi at a room temperature of 26 degrees Celsius?

I'm not sure which gas law(s) to use to find the number of moles of argon gas left in the cylinder.  Which laws should I use?  Please help.

The ideal gas equation seems good here, just keep the volume constant. So it's just $\frac{PV}{nT} = \textsc{Ideal Gas Constant}$. Since the ideal gas constant stays the same, you can just do $\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$

in this case, you can take out that V1 and V2 in the equation above and calculate n2...$\frac{P_1}{n_1T_1} = \frac{P_2}{n_2T_2}$

I set up the second equation, and found that the number of moles is 11.41.  Could someone please check to make sure I worked it right?

Converted to grams, the answer would be 456 grams.


geometry
3D geometry
sphere
analytic geometry
parameterization
inequalities
topology

any help on these would be great

[code]

Let $\alpha(s)$ be a regular curve

1) if all normal planes of $\alpha(s)$ pass thur a fixed point show that $\alpha(s)$ lies on a sphere.

2)
Let C be a regular curve on the unit sphere and S the surface
obtained by taking all lines thru the origin passing thur points on C. Find the first fundamental form (using convenient paramterization)? and also prove that S is locally isometric to the plane ?

3) The coordinate curves of a parametrization r(u,v) form a Chebyshev net if the lengths of the sides of any quadrilateral formed by them are equal. Prove that a necessary and sufficient condition for this is the $\frac{\partial E}{\partial v} $ = $\frac{\partial G}{\partial u}$ = 0

For problem 2) I have a parameteriztion in terms of spherical coordinates of the sphere and then used the equation of the line thru origin. But it seems that it is going to be a B to compute the first form in terms of these params.

Thanks for any help[/code]

Sorry for double post I forgot that the code tag was for  some thing else.
Thought you had to included it with the latex.
Anyway maybe this is more readable.

Let $\alpha(s)$ be a regular curve

1) if all normal planes of $\alpha(s)$ pass thur a fixed point show that $\alpha(s)$ lies on a sphere.

2)
Let C be a regular curve on the unit sphere and S the surface
obtained by taking all lines thru the origin passing thur points on C. Find the first fundamental form (using convenient paramterization)? and also prove that S is locally isometric to the plane ?

3) The coordinate curves of a parametrization r(u,v) form a Chebyshev net if the lengths of the sides of any quadrilateral formed by them are equal. Prove that a necessary and sufficient condition for this is the $\frac{\partial E}{\partial v}$ = $\frac{\partial G}{\partial u}$ = 0

For problem 2) I have a parameteriztion in terms of spherical coordinates of the sphere and then used the equation of the line thru origin. But it seems that it is going to be a B to compute the first form in terms of these params.

Thanks for any help

when is it due?  it's past 4am here and I can help but alas... it's 4am!  I"ll be back on later today.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Probably this is completely stupid, but I don't see an approach: what are those $n$ such that any group of order $n$ is the semi-direct product of two groups of smaller order?

The answer is none.
Related question: Find the positive integers $ n$, such that there exists a semidirect product of order $ n$.

Direct products are also semidirect.

Anything for which you can guarantee the existence of a normal subgroup and a complementary subgroup. The first two are 6 and 10.

I don't think there will be any explicit description of such n.  Partial answers are not too hard:

If n is a prime power, then the cyclic group of order n cannot be written as a non-trivial semidirect product.  If there is a quasi-simple group of order n (such as n in {60,120,168}), then it cannot be written as a non-trivial semidirect product.  If n in {48,144,192,240} then there is a group of order n that cannot be written as a non-trivial semi-direct product.  This handles n less than 320: all other n less than 320 are such that every group of order n can be written as a non-trivial semidirect product.

If n is not divisible by the square of any prime, then every group of order n is the semi-direct product of its derived subgroup (of course if the group is abelian, then it is cyclic, so as long as n is not prime, it is a non-trivial semi-direct product).

[quote="jmerry"]Direct products are also semidirect.
[/quote]
I took harazi's question to be about "non trivial semidirect products", i.e. not direct products.


Can you help me finding all solutions to a^b=b^a if a, b are integers?

Posted many times before.Search in PEN. :wink:

sry i'm a real noob here (meaning i registered 20 minutes ago). i could have guessed it's old but didn't know where to search, and how. it seems pointless to enter an equation in the search form, so could you give me a hint how to search?

See here:
http://www.mathlinks.ro/viewtopic.php?t=150682
But,unfortunately,you won't find a proof there.I'll post it,just give me some time.
Edit:
I've already posted it .

WLOG let $ a\leq b$. For every prime $ p$, if $ n$ is the highest power of $ p$ dividing $ a$, then $ b^a$ is divisible by $ p^{nb}$. In particular, $ p^n$ divides $ b$.

Since this was true for every prime, this means $ a$ divides $ b$; let $ b\equal{}ka$ so that the equation reduces to finding positive integers $ a,k$ such that $ a^{k\minus{}1}\equal{}k$.

If $ k\equal{}1$, then every $ a$ is a solution giving $ (a,a)$. If $ k\equal{}2$, then $ a\equal{}2$ giving $ (2,4)$ (also $ (4,2)$ if you take away the restriction $ a\leq b$).

If $ k\geq 3$, then it's easy to prove $ 2^{k\minus{}1}>k$ by induction. Since $ a\neq 1$, $ a^{k\minus{}1}\geq 2^{k\minus{}1}$ so no more solutions.

[quote="BaBaK Ghalebi"]Erken:this problem is different,it only asks for the integer solutions (the PEN version asks for all the rational solutions).[/quote]
I've already understood it.But both solutions are similar. :)


Find all nonnegative integer solutions $ (x,y,z,w)$ of the equation
$ (2^x)(3^y)\minus{}(5^z)(7^w)\equal{}1$

Well Ill need some help finishing this, but I got

(1,0,0,0), (1,1,1,0), (2,2,1,1), and (3,0,0,1).

I dont think there will be any bigger cases.

But I also dont know how to prove it....

I know that x>=1 because its gotta be even-odd on the left side.

Also, I do not think x will be greater than 3 because if you look at it in mod 8, you will see contradictions if x>=3...but thats all I really have.  I would appreciate any help.  Thanks!

This one took me a while, but think I got it.

[hide]Rearrange the equation as $ 2^{x}3^{y}-1 = 5^{z}7^{w}$.

I'll break this into 5 cases:
$ x \ge 3$ AND $ y \neq 0$
$ x = 0$
$ x = 1$ AND $ y \neq 0$
$ x = 2$ AND $ y \neq 0$
$ y = 0$ AND $ x \neq 0$

Case I: $ x \ge 3$ AND $ y \neq 0$
[hide]In $ \pmod{3}$ we have:
$ 2^{x}3^{y}-1 = 5^{z}7^{w}$
$ (-1)^x(0) - 1 \equiv (-1)^z(1)^w \pmod{3}$
$ -1 \equiv (-1)^z \pmod{3}$
Thus, $ z \equiv 1 \pmod{2}$

In $ \pmod{8}$ we have:
$ 2^{x}3^{y}-1 = 5^{z}7^{w}$
$ (0)(3)^y - 1 \equiv (-3)^z(-1)^w \pmod{3}$
$ -1 \equiv (-3)^z(-1)^w \pmod{3}$
Thus, $ z \equiv 0 \pmod{2}$ AND $ w \equiv 1\pmod{2}$

This is a contradiction. Therefore, there are no solutions for $ x \ge 3$ AND $ y \neq 0$. [/hide]

Case II: $ x=0$
[hide]The equation becomes: $ 3^y - 1 = 5^z7^w$
In $ \pmod{2}$, we have:
$ (1) - 1 = (1)(1) \pmod{2} \Rightarrow 0 \equiv 1 \pmod{2}$

This is a contradiction. Therefore, there are no solutions for $ x = 0$. [/hide]

Case III: $ x = 1$ AND $ y \neq 0$
[hide]The equation becomes: $ 2\cdot3^y - 1 = 5^z7^w$

Subcase IIIa: $ z,w \neq 0$:

In $ \pmod{5}$ we have:
$ 2\cdot(3)^y - 1  \equiv (0)(2)^w \pmod{5}$
$ 2\cdot(3)^y - 1  \equiv 0 \pmod{5}$
$ 3^y \equiv 3 \pmod{5}$
Thus, $ y \equiv 1 \pmod{4}$, and thus, $ y \equiv 1\pmod{2}$

In $ \pmod{7}$ we have:
$ 2\cdot(3)^y - 1  \equiv (-2)^z(0) \pmod{7}$
$ 2\cdot(3)^y - 1  \equiv 0 \pmod{7}$
$ 3^y \equiv 4 \pmod{7}$
Thus, $ y \equiv 4 \pmod{6}$, and thus, $ y \equiv 0\pmod{2}$
This is a contradiction. Thus, we have no solutions if $ z,w \neq 0$.

Subcase IIIb: $ z = 0$:
The equation becomes: $ 2\cdot3^y - 1 = 7^w$
In $ \pmod{6}$ we have:
$ (0) - 1 \equiv (1)^w \pmod{6} \Rightarrow -1 \equiv 1 \pmod{6}$
This is a contradiction. Thus, we have no solutions if $ z = 0$.

Subcase IIIc: $ w = 0$ AND $ z \neq 0$:
The equation becomes: $ 2\cdot3^y - 1 = 5^z$

$ y=1$ yields $ 2\cdot3-1 = 5 = 5^z \Rightarrow z=1$, which gives $ (x,y,z,w) = (1,1,1,0)$ as a solution.

For $ y \ge 2$, in $ \pmod{18}$ we have:
$ (0) - 1 \equiv 5^z \pmod{18}$
Thus, $ z \equiv 3 \pmod{6}$

In $ \pmod{7}$, we have:
$ 5^z \equiv 6 \pmod{7}$ if $ z \equiv 3 \pmod{6}$
$ 2\cdot3^y - 1 \equiv 5^z \equiv 6 \pmod{7}$
$ 3^y \equiv 0 \pmod{7}$
This is a contradiction, Thus we have no solutions in this subcase for $ y \ge 2$

Therefore, $ (x,y,z,w) = (1,1,1,0)$ is the only solution for $ x = 1$ AND $ y \neq 0$. [/hide]

Case IV: $ x = 2$ AND $ y \neq 0$
[hide]The equation becomes: $ 4\cdot3^y - 1 = 5^z7^w$

Subcase IVa: $ z \neq 0$:

In $ \pmod{5}$ we have:
$ 4\cdot(3)^y - 1  \equiv (0)(2)^w \pmod{5}$
$ -\cdot(3)^y - 1  \equiv 0 \pmod{5}$
$ 3^y \equiv 4 \pmod{5}$
Thus, $ y \equiv 2 \pmod{4}$

Since $ y$ is even, we have:
$ 4\cdot3^y - 1 = (A-1)(A+1) = 5^z7^w$ (where $ A = 2\cdot 3^\dfrac{y}{2}$).

Since $ A-1 \not \equiv A+1 \pmod{5 \text{ \ OR \ } 7}$, either:

i) $ A-1 = 1$, AND $ A+1 = 3 = 5^z7^w$ (which yields no solutions)
ii) $ A-1 = 5^z$ AND $ A+1 = 7^w$
iii) $ A-1 = 7^w$ AND $ A+1 = 5^z$

$ z=1$ yields $ A=6$, $ w=1$, $ y=2$, and $ (x,y,z,w) = (2,2,1,1)$ as a solution.

For $ z \ge 2$, we have: $ 7^w \equiv \pm 2 \pmod{25}$
However, $ 7^w \equiv \pm1, \pm7 \pmod{25}$.
This is a contradiction. Thus there are no solutions in this subcase of $ z \ge 2$.

Subcase IVb: $ z = 0$:
The equation becomes: $ 4\cdot3^y - 1 = 7^w$
In $ \pmod{6}$, we have:
$ (0) - 1 \equiv (1)^w \pmod{6} \Rightarrow -1 \equiv 1 \pmod{6}$
This is a contradiction. Thus there are no solutions in this subcase.

Therefore, $ (x,y,z,w) = (2,2,1,1)$ is the only solution for $ x = 2$ AND $ y \neq 0$.[/hide]

Case V: $ y = 0$ AND $ x \neq 0$
[hide]The equation becomes: $ 2^x - 1 = 5^z7^w$

Subcase Va: $ z \neq 0$:
In $ \pmod{5}$, we have:
$ 2^x - 1 \equiv (0) \pmod{5}$
$ 2^x \equiv 1 \pmod{5}$
Thus, $ x \equiv 0 \pmod{4}$

Since $ x$ is even, we have:
$ 2^x-1 = (2^\frac{x}{2} - 1)(2^\frac{x}{2} + 1) = 5^z7^w$

As shown earlier, the only solutions occur when $ 2^\frac{x}{2} = 6$ which has no solutions.
Thus, there are no solutions in this subcase.

Subcase Vb: $ z = 0$:
The equation becomes: $ 2^x - 1 = 7^w$
$ x=1$ yields $ w=0$, and $ (x,y,z,w)=(1,0,0,0)$ as a solution.
$ x=3$ yields $ w=1$, and $ (x,y,z,w)=(3,0,0,1)$ as a solution.

For $ x \ge 4$, we have in $ \pmod{16}$:
$ (0) - 1 \equiv 7^w \pmod{16}$
$ 7^w \equiv -1 \pmod{16}$
This is a contradiction since $ 7^w \equiv 1,7 \pmod{16}$. Thus, there are no solutions in this subcase for $ x \ge 4$.

Therefore, $ (x,y,z,w)=(1,0,0,0)$ and $ (x,y,z,w)=(3,0,0,1)$ are the only solutions for $ y = 0$ AND $ x \neq 0$.[/hide]

So, the four solutions $ (x,y,z,w)$ are:
$ (1,0,0,0)$
$ (1,1,1,0)$
$ (3,0,0,1)$
$ (2,2,1,1)$[/hide]


How does your school's team do at chapter? Also how hard is your chapter. My school had 1st, 6th, 7th, 9th written. We had four in countdown and two going to nats from the county so we are definitely the hardest county in Florida.

well, from our chapter, 3 are going to nats, 4 went last year.

Our school did okay, only me in countdown but decent scores of 37 36 36 from the teammates, and a 20 on the team round helped our team get 5th, advancing to state since our chapter's so large.

From my chapter:

My school has gotten first for as long as I can remember (i.e. just this year).

Only one person from my chapter is going to nationals (besides the coach) and that's meeeeeeeeeeeeeeeeeeeeeee.  :w00t:

[quote="Klebian"]well, from our chapter, 3 are going to nats, 4 went last year.
[/quote] You're from Sugarland. No surprises there :P.

Our school wasn't great. 45, 29, 20, and 17... Which means 1st, 49th, 90th, and 117th. but a perfect team round to make up for that got us 6th... We also got sixth last year with scores that weren't as great. I don't remember, but we had a 12 or something on the team round...

Last year, we sent 2 to nats, and this year it's just me (but the alternate is from our chapter, too)

Our school had no one in countdown at chapters, but 3 at state(10th, 4th, and 3rd), and 2 going to nationals. We barely even made states (we got 5th place team, which was the cutoff for states), and yet we got 2nd at states.

3 people from my chapter are going to nationals.

Individual scores at chapter: 39(me), 37, 30 something, 30(?). One of our individuals got 37(?). The cutoff for countdown was 40.

I have no clue what the Countdown cutoff for our chapter was... Let's see... If a 29 got you 49th, I'd have to say the cutoff was like 35-40 or something like that.

Our chapter sent 2nd, 3rd, and 4th, to nats as places 3rd, 1st, and 4th respectively. We sent 0 last year. Nobody else but me got to states. I am the second person ever from my school to go to states (and first repeat, as well as first to go to Nats as well as first to make countdown, etc.)

My school scores were
Team: 18:
28/14
20/8 <- Sev
12/6
12/4 <- Six
Individuals
12/10
12/0

[quote="Ignite168"]Our chapter sent 2nd, 3rd, and 4th, to nats as places 3rd, 1st, and 4th respectively. We sent 0 last year. Nobody else but me got to states. I am the second person ever from my school to go to states (and first repeat, as well as first to go to Nats as well as first to make countdown, etc.)[/quote]

Sounds like me, too. Second from our school to ever make State, first to ever win chapter, first to ever win chapter twice, first countdowner at state... Need I go on?

Yeah, our scores @ Chapter (I've said already but haven't done sprint/target):

29/16 -- 1st
17/12 -- 49th
12/8 -- 90th
9/8 -- 117th or so (7th grader)
That was the team, which had a 20 on team!

Individuals: (both 7th graders)
12/12 (Why didn't we put him on the team after one girl dropped?) - 70 somethingth
4/4 - 160?

Well, I think our team might be ok next year. The scores don't show it, but there are two sixth graders who I'm going to try to get and join next year. Plus, my coach wants me back to help coach the team, so that's always a good thing!

I just wish our team could have gotten past chapter this year...

My school has never and will never send a team to state.

our school is always in the top 3 at chapter. it's easy for me, at least. i don't know for the rest of the team. this year we got third. we had two eighth graders (me and my friend) and 2 sixth graders, who each got 2nd and 3rd sixth graders in the competition. so, this year was like a "training year". my chapter almost never sends anyone to nationals. we did maybe 5 years so far. we did have one state winner a few years back (from my school too.  :)  go duncan!) we also have this pretty smart seventh grader who didn't do it this year. she's getting in geometry next year. i got 1st and my friend got 5th. the sixth graders got around 40th out of 120 people. pretty good!
lawton Tomlinson beat us. oh well, we beat them at state.  :oops:

This year, our school was strong as ever. We had 6 ppl in countdown. We won as team and also had 1st and 2nd place individuals. When we went to state. we took 1st place as a team and had 2nd and 4th place individuals. Somebody else from our chapter also placed 3rd. so 3 ppl from our chapter made it to nats.

Our chapter had the top two teams and top 5 individuals at state.

One school in our chapter had 1st in both chapter in state and had individuals 1,2,4,5

Our team was really strong this year, but not nearly good enough as our inter-city rivals :(  :mad:

We had 44, 44, 41, and 40, and we got full score on the team round.
If we had gotten 0.75 more points, we would have gotten 1st.


geometry
geometric transformation
rotation
logarithms
trigonometry
reflection
function

The graph, $G,$ of $y = \log_{10} x$ is rotated $90^\circ$ counter-clockwise about the origin to obtain a new graph $G'$. Which of the following is an equation for $G'$?

$\text{(A)} \ y = \log_{10} \left(\frac{x+90}{9}\right) \qquad \text{(B)} \ y = \log_y {10} \qquad \text{(C)} \ y = \frac{1}{x+1}$
$\text{(D)} \ y = 10^{-x} \qquad \text{(E)} \ y = 10^x$

[hide="Not sure"]
Not sure: Putting it in terms of $x$, you get $x=10^y$.  Switching $x$ and $y$ to rotate it counterclockwise 90 degrees, you get $y=10^x$.  Now you need to flip it over the $y$ axis to get $y=10^{-x}$.
[/hide]

[quote="t0rajir0u"][quote="Silverfalcon"]Wow.. If that's correct way (the answer is correct), I feel so stupid..

[hide="View my ugly"]

Set $x = 10$ than $y = 1$. In trigonometric representation, $(10,1)$ is equal to $\sqrt{101} (\cos (\tan^{-1} (1/10)) + \sin (\tan^{-1} (1/10)))$. Rotating this by -90 degrees is same as multiplying by $-(\cos (-90) + \sin (-90))$. Hence, doing the multiplication by angles, we get $-\sqrt{101} (\cos (\tan^{-1} (1/10) + -90) + \sin (\tan^{-1} (1/10) + -90)) \rightarrow (-1,10)$

Clearly, $y = 10^{-x}$ has this point so $D$ is the answer. [/hide][/quote]

 :huh:

[hide="Solution"]
A $90^{\circ}$ counterclockwise rotation is given by the transformation $(x, y) \rightarrow (-y, x)$.

$y = \log_{10} x \implies x = 10^{y}$

After transformation, we wish any $(x_0, y_0)$ on the graph of the original equation to be $(-y_0, x_0)$ satisfying the new equation - we do this by substituting $x \rightarrow y$ and $y \rightarrow -x$.

$\boxed{y = 10^{-x}}$
[/hide][/quote]

I rotated the point first since this rotated point must also be in the equation. I don't know, I'm poor at rotation, reflection, adn whatever you like to do with flipping.  :D

[hide]
well if you visualize the curve, you'll think it's $y = 10^{-x}$ because of the nature of log curves.  You can graph other functions on TI to check it... (the TI rule is kinda bad...)

[/hide]