geometry
geometric transformation
homothety
reflection
cyclic quadrilateral
geometry proposed

let $A$ be one of the intersection point of two circles,let $\omega$ be a circle that is tangent to these 2 circles externaly,prove that the polars of $A$ wrt $\omega$s  are concurrent.

hint:
[hide]the tangency points are anti-congruent[/hide]
(sorry for the bad english)

[quote="BaBaK Ghalebi"]hint:

the tangency points are anti-congruent
(sorry for the bad english)[/quote]

What is anti-congruent?
if you meant [b]Padmotajanes[/b] (this a persian word).
Note that Padmotajanes points are not on polar of $A$.

yeah I meant [b]padmotajanes[/b] thats not the whole solution,that was just a hint...
I`ll post the complete solution if no one else solved it...

finally I find another solution , that shows problem is more easier than what my first solution shows.
after a month I will post it.

and I`ll post my solution as soon as possible before you do...

Let $A,$ $B,$ be, the intersection points of two given circles $(O_{1},$ $(O_{2})$ and let $P$ be, an arbitrary point on the segment line $AB,$ as their radical axis.

Through $P,$ we draw the tangent lines of $(O_{1}),$ $(O_{2}),$ at points, so be it $C,$ $D,$ respectively and we denote as $O,$ the intersection point of $O_{1}C,$ $O_{2}D.$

It is easy to show that $OC = OD,$ from the right triangles $\bigtriangleup PCO,$ $\bigtriangleup PDO,$ because they have $PC = PD.$

Also, it is easy to show that the segment line $CD,$ intersects the line $O_{1}O_{2},$ connecting the centers $O_{1},$ $O_{2},$ of $(O_{1}),$ $(O_{2})$ res[ectively, at a point so be it $M,$ as the homothety center of $(O_{1}),$ $(O_{2}).$

 $($ Apply the [b][size=100]Menelaus’s theorem,[/size][/b] in triangle $\bigtriangleup OO_{1}O_{2},$ taken the segment line $CD,$ as its transversal,

we have that $\frac{MO_{1}}{MO_{2}}\cdot \frac{DO_{2}}{DO}\cdot \frac{CO}{CO_{1}}= 1$ $\Longrightarrow$ $\frac{MO_{1}}{MO_{2}}= \frac{CO_{1}}{DO_{2}}$ $).$

Through the point $A,$ we draw the tangent lines of $(O),$ at points so be it $E,$ $F$ and we denote the points $N\equiv AM\cap EF$ and $K\equiv CD\cap EF.$

So, we have now, the configuration as the problem states and it is enough to prove that the point $N,$ is a constant point lies on the constant segment line $AM.$

We will prove that $AM = MN.$

$\bullet$ Because of the segment lines $CD,$ $EF,$ as the polars of the points $P,$ $A$ respectively, with respect to the circle $(O),$ pass through the point $K,$ we conclude that the segment line $AP\equiv AB,$ is the polar of $K,$ wrt $(O)$ $($ $\Longrightarrow$ $AP\perp KO$ $\Longrightarrow$ $KO\parallel O_{1}O_{2}$ $).$

We denote as $C',$ $D',$ the intersection points of $(O),$ from the segment lines $AC,$ $AD$ respectively and as $L,$ the one of $CD',$ $C'D.$

The points $K',$ $L,$ where $K'\equiv CD\cap C'D',$ lie on the segment line $EF,$ as the polar of $A,$ with respect to $(O),$ as well known.

So, we conclude that $K'\equiv K.$ That is the segment line $C'D',$ passes through the point $K.$

The point $L,$ also lies on the segment line $AP,$ as the polar of $K,$ wrt $(O)$ and hence, we conclude that $L\equiv EF\cap CD'\cap C'D.$

$\bullet$ From the complete quadrilateral $ACLDC'D',$ we conclude that the pencil $K.ACLC',$ is in harmonic conjugation.

From the cyclic quadrilateral $CDD'C',$ we have that $\angle C'D'D = \angle DCA$ $,(1)$

Based on the below Lemma, we conclude that $\angle DCA = \angle DAM$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $\angle C'D'D = \angle DAM$ $\Longrightarrow$ $KD'\parallel AM$ $,(3)$

Hence, because of $(3),$ we conclude that the segment line$AM,$ as parallel to $KD'\equiv KC',$ the one of the rays of the pencil $K.ACLC'$ in harmonic conjugation, intersects the other three its rays at two points $M$ and $N,$ formed two congruent segments. That is we have that $AM =MN.$

So, the polar of $A,$ with respect to the arbitrary circle $(O),$ as the problem states ( externally tangents to the given circles $(O_{1},$ $(O_{2}$ ), passes through the constant point $N,$ on the segment line $AM$ and the proof is completed.

Kostas Vittas.

PS. I will post here later, the simple Lemma I used, easy to prove.

[quote="BaBaK Ghalebi"]this problem was supposed to be solved by me (or amir.s),we both solved it,and we were waitng for the time to post our solutions,so it would be great if no one else would send any more solutions until we post ours,tnx...  :)
anyway I didnt read your solution vittasko but it seems pretty difficult,it has an easier solution,and as far as I remember the constant point is the reflection of $A$ respect to point $S$,where $S$ is the homothecy center of THE two circles...
(as far as I remember...)
note that if the tangency point are $P,Q$ then we have $SA^{2}=SP.SQ$...[/quote]

Sorry dear [b][size=100]BaBak Ghalebi[/size][/b] for the disturbance ( out of my purpose ), but I think that when a problem has already been posted in the Forum, then any solution different or probably not different than our, is desirable from the proposer without limits of time.

By the way , I enjoyed this [b][size=100]Amir.S’s[/size][/b] nice problem and thank him very much.

Sorry again and best regards from hot Athens.

Kostas Vittas.

PS. As you said, the constant point is the reflexion of $A,$ with respect to the homothety center of the given circles and this is exactly the result which is proved in the ( difficult I agree ) solution I presented.

:)
yeah you`re right sorry...
thank you anyway...  :lol:

actually I find some free times,so I will post my solution not in details.

[quote="vittasko"]... By the way , I enjoyed this [b][size=100]Amir.S’s[/size][/b] nice problem and thank him very much...   [/quote]

you're welcome.
at first I find a dificult solutoin like yours dear [b]vittasko[/b],but when I thought again on it , I found the following:

let the tangency points be $C,D$ and $S$ is homothecy center of two circles($\not = \omega$) .

[b][color=blue]Lemma 1.[/color] [/b] $SA^{2}=SD.SC$

hence circle $C(S,SA)$ is perpendicular to $\omega$

[b][color=blue]Lemma 2.[/color] [/b] if we have two perpendicular circles , the polar of a point on one circle wrt to another circle passes though oppsite point of a diameter wich passes through that point.

now we are done.

[quote="Amir.S"]...
[b][color=blue]Lemma 1.[/color] [/b] $SA^{2}=SD.SC$

hence circle $C(S,SA)$ is perpendicular to $\omega$

[b][color=blue]Lemma 2.[/color] [/b] if we have two perpendicular circles , the polar of a point on one circle wrt to another circle passes though oppsite point of a diameter wich passes through that point.

now we are done.[/quote]

Dear [b][size=100]Amir.S[/size][/b] and all my friends.

About the [b][size=100]Lemma 1,[/size][/b] it is a consequent of the [b][size=100]Lemma[/size][/b] I mentioned, because of the similarity of the triangles $\bigtriangleup MAC,$ $\bigtriangleup MDA,$ where $M,$ is the homothety center of the circles $(O_{1}),$ $(O_{2}),$ in my wording, instead of the point $S.$

About the [b][size=100]Lemma 2,[/size][/b] although I can prove it, however it is a new result for me, because of I have not much experience in the polar theory.

[b][size=100][color=DarkBlue]Our knowledge is a procedure of learning, mainly from the other men and specially from the younger ones ![/color][/size][/b]

Thanks a lot, Kostas Vittas.

PS. I am looking forward to the  [b][size=100]BaBak Ghalebi's[/size][/b] solution.

[quote="vittasko"]....
Based on the below Lemma, we conclude that $\angle DCA = \angle DAM$ $,(2)$
....
PS. I will post here later, the simple Lemma I used, easy to prove.[/quote]

[quote="Amir.S"]....
[b][color=blue]Lemma 1.[/color] [/b] $SA^{2}=SD.SC$

hence circle $C(S,SA)$ is perpendicular to $\omega$

[b][color=blue]Lemma 2.[/color] [/b] if we have two perpendicular circles , the polar of a point on one circle wrt to another circle passes though opposite point of a diameter which passes through that point.

now we are done.[/quote]

[b][size=100][color=DarkBlue]LEMMA 1. - Two circles $(O_{1},)$ $(O_{2}),$ intersect each other are given and let $A,$ $B$ be their intersection points.  Through the point $S,$ as the external homothetic center of $(O_{1}),$ $(O_{2}),$ we draw an arbitrary line which intersects the circle $(O_{1})$ at points $C,$ $X$ and the circle $(O_{2}$ at points $D,$ $Y$ $($ $C,$ $D,$ adjoined points to $B$ $).$  Prove that $\angle DCA = \angle DAS$ $\Longrightarrow$ $(SA)^{2}= (SD)\cdot(SC).$ [/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – Because of $S,$ is the external homothetic center of $(O_{1}),$ $(O_{2}),$ we have that $O_{1}X\parallel O_{2}D$ $,(1)$ $($ also $O_{1}C\parallel O_{2}Y$ $),$ as well known.

We denote as $Z,$ the intersection point of $(O_{1}),$ from the segment line $SA$ and so, we have also $O_{1}Z\parallel O_{2}A$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $DA\parallel XZ$ $,(3)$ because of $O_{1}X = O_{1}Z$ and $O_{2}D = O_{2}A.$

From $(3)$ $\angle DAS = \angle XZA$ $,(4)$

From cyclic quadrilateral $CXZA$ $\Longrightarrow$ $\angle XZA = \angle DCA$ $,(5)$

From $(4),$ $(5)$ $\Longrightarrow$ $\angle DCA = \angle DAS$ $,(6)$

From $(6),$ we conclude that the triangles $\bigtriangleup SDA,$ $\bigtriangleup SAC,$ are similar.

Hence, we have that $\frac{SA}{SC}= \frac{SD}{SA}$ $\Longrightarrow$ $(SA)^{2}= (SD)\cdot(SC)$ $,(7)$ and the proof is completed.

[b][size=100]REMARK.[/size][/b] – It is easy to show that $OC = OD,$ where $O\equiv O_{1}C\cap O_{2}D$ $($ applying the [b][size=100]Menelaus’s theorem[/size][/b] in the triangle $\bigtriangleup OO_{1}O_{2},$ taken as its transversal the segment line $CD$ $)$ and so, the circle $(O),$ with radius $OC = OD,$ tangents externally to $(O_{1}),$ $(O_{2}),$ at points $C,$ $D,$ respectively.

From $(7),$ we conclude that the circle $(O),$ and the circle centered at $S,$ with radius $SA,$ are orthogonal each other.

[b][size=100][color=DarkBlue]LEMMA 2. – The polar of an arbitrary point which lies on the one of two given orthogonal circles, with respect to the other circle, passes through its antidiametric ( = diametric opposite ) point. [/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – Let $(O_{1)}),$ $(O{2})$ be, the given orthogonal circles, intersect each other at points so be it, $A,$ $B$ $($ $\Longrightarrow$ $O_{1}A\perp O_{2}A$ $)$ and also let $P$ be an arbitrary point lies on $(O_{2})$ and we denote as $P',$ its antidiametric point.

Through the point $P,$ we draw the tangent lines of $(O_{1}),$ at points so be it $C,$ $D$ and we will prove that the segment line $CD,$ as the polar of $P,$ with respect to $(O_{1}),$ passes through the point $P'.$

We denote as $M,$ the intersection point of $(O_{2}),$ from the segment line $O_{1}P'$ and as $E,$ $F,$ the tangent points of $(O_{1}),$ from the lines through $M.$

So, we have that $(O_{1}A)^{2}= (O_{1}P')\cdot(O_{1}M$ $\Longrightarrow$ $(O_{1}E)^{2}= (O_{1}P')\cdot(O_{1}M)$ and hence, in the right triangle $\bigtriangleup EO_{1}M$ we conclude that $EP'\perp O_{1}M$ $\Longrightarrow$ $EP'\equiv EF.$

That is the segment line $EF,$ as the polar of $M,$ with respect to $(O_{1}),$ passes through the point $P'.$

As well known we conclude now, that the polars of all the points lie on the segment line $PM\perp O_{1}M,$ pass through the point $P'$  as the harmonic conjugation of $M,$ with respect to the intersection points of $(O_{1}),$ from the segment line $MP'.$ $(*)$

Hence, the segment line $CD$ as the polar of $P,$ with respect to $(O_{1}),$ passes through the point $P'$ and the proof is completed.

$(*)$ - An elementary approach about the proof of this result, has already been posted at  http://www.mathlinks.ro/Forum/viewtopic.php?t=139564

Kostas Vittas.

[quote="BaBaK Ghalebi"]nice sol,but mine is even easier I think...
I`ll post it next week[/quote]

waiting for solution from BaBaK GhAlEbI ....

I think the easiest way to prove this is inversion.


geometry
3D geometry
algebra unsolved
algebra

Find all integer numbers $(a,b,c)$ such that $a^3+6ab+20, b^3+6bc+20, c^3+6ca+20$ are  all cubes of integer numbers.

I assume $\ a'^3 = a^3+6ab+20, b'^3=b^3+6bc+20 , c'^3 = c^3+6ca+20$
Since $\ a^3+6ab+20+ b^3+6bc+20+ c^3+6ca+20 \geq 3*a'b'c'$
And I think $\ \sum(a+1) \leq a'+b'+c' \leq \sum(a+4)$ or $\ a+5$ which I can't find a proof of it.
If that's true, by the two ineq above,we could find a range for a,b,c.
If someone can give out an answer, i'll really apreciate him.

$a',b',c'$ are integers, we cannot use AM-GM. Maybe we should divide the problem into cases.

Finally get some time to work on math.  :)
 $\ \ a'^3 = a^3+6ab+20, b'^3=b^3+6bc+20 , c'^3 = c^3+6ca+20$ :arrow: $\ a'^3 - a^3 = 6ab+20$ which is an even number, so that $\ a'$ must be the same as $\ a$odd or even, $\ a' = a+2,a+4$, or even $\ a+6$ The same as $\ b'$ and $\ c'$.
Case 1:(discussing a' value first)
$\ a' = a+2 , 6ab +20 = 2[(a+2)^2+a(a+2)+a^2] \stackrel{to}{\rightarrow} 3ab +10 = 3a^2 +6a+4 \rightarrow a^2-2ab+2a-2 = 0, a(a-2b+2)=2$ Since both $\ a$& $\ b$ are integers, $\ a= 1,2$, which gives us no integer solution for b.

Case 2:
$\ a' = a+4, 6ab+20 = 4[(a+4)^2 +a(a+4) +a^2] \rightarrow ...$

Case 3:
$\ a' = a+2m, 6ab +20 =m[(a+m)^2 +a(a+m)+a^2] \rightarrow ... (m \in N)$
And $\ m$ cannot be too big, like $\ 10$ or something.
I hope this will work out the problem.
Frank Morgan

$a'=a+2\stackrel{to}{\rightarrow}3ab+10=3a^2+6a+4\stackrel{to}{\rightarrow} ab+2=a^2+2a, a(a-b+2a)=2$.
$a=1,-1,2,-2\stackrel{to}{\rightarrow}b=1,-1,5,-5.$ Similary when $a'=a+4,a+6,..$
$a'=a+2m, m\in Z^,$so we shoul also see the cases when $a'=a-2,a-4..$
Maybe the only solution is $a=b=c=1,a'=b'=c'=3$


limit
trigonometry
geometry
calculus
function
integration
geometric transformation

I want to see a proof of
\[\mathop{\lim }_{x \to 0}\frac{{\sin x}}{x}= 1 \]
which does not use Euclidean Geometry. I read plenty of books, all prove it using the unit circle definition of sine and properties of Euclidean Geometry which are of course based on the axioms of Euclidean Geometry.

Now, I may be out of the loop here, but I was taught that in constructing Calculus, we only base ourselves on set theory axioms (ZPC). Using Euclidean axioms in order to prove a Calculus limit sounds like one huge mess to me.

However, the task becomes a hard one. The fact that the arc AB is larger than AB is a simple fact in Euclidean geometry but not so intuitive and obvious via ZPC and Calculus tools.

In short, I would like to see a rigorous proof of the above limit, using no Euclidean Geometry Axioms whatsoever. The proof of the above limit in Moise's Calculus 1 is close to being what I want to see but he only shows the first few steps of the proof.

Of course, Sine can be defined by an infinite series, therefore making the above limit extremely easy to find. I've been told that due to the geometric nature of the traditional definition of the sine function, it is impossible to prove a fundamental limit such as the above without using axioms of Euclidean Geometry. Is this true?

[quote="perfect_radio"]Well, you could define $\sin z = \frac{e^{iz}-e^{-iz}}{2i}, \, \forall z \in \mathbb C$, if you really want to use just calculus.

[quote]
However, the task becomes a hard one. The fact that the arc AB is larger than AB is a simple fact in Euclidean geometry but not so intuitive and obvious via ZPC and Calculus tools.
[/quote]
How are you going to define the length of an arc or the area of a sector without calculus?[/quote]

The above definition for sine is based on Euler's formula which is basiclly like defining sine as a Taylor Series (actually, one of the proofs of Euler's formula uses the sine Taylor Series).

How am I going to define the length of an arc? How about defining it as an infinite series of straight lines converging to the arc and have N (the number of straight lines) approach to infinity? A Limit-based approach does not seem surprising to me. This IS Calculus after all. This is how we basiclly define the length of arc of a function via Integrals, no? Divide to N segments and have N approach infinity. And I suppose it should be possible to prove that the limit of the above series converging to the arc length is is larger than the straight line between the two points (the beginning and end of the arc).

As for the area of a sector, this can be done via Integrals as well, no?

Yes, you can define arclength as the limit (supremum) of the length of a piecewise linear curve with all vertices on the circle (or other curve we want to find the arclength of). In order to get anything useful about a circle, you need rotation invariance of length.
By this method, an arc of the unit circle with central angle $\theta$ has length $\lim_{n\to\infty}2n\sin\frac{\theta}{2n}$. If we don't know much of anything about $\sin$, we're in trouble.

We could also develop the general form for the arclength of a curve $y=f(x)$ between $x=a$ and $x=b$: the arclength is $\int_{a}^{b}\sqrt{1+(f'(x))^{2}}\,dx$. For a semicircle of radius 1 $y=\sqrt{1-x^{2}}$, the arclength between $x=0$ and $x=\sin\theta$ is $\int_{0}^{\sin\theta}\sqrt{1+\left(\frac{-x}{\sqrt{1-x^{2}}}\right)^{2}}\,dx= \int_{0}^{\sin\theta}\sqrt{1+\frac{x^{2}}{1-x^{2}}}\,dx=\int_{0}^{\sin\theta}\frac1{\sqrt{1-x^{2}}}\,dx$

We want this arclength to be $\theta$, of course. To do this, define $\arcsin x$ to be $\int_{0}^{x}\frac1{\sqrt{1-t^{2}}}\,dt$, and let $\sin$ be the inverse function. We can get all of the familiar properties this way.

Why don't calculus textbooks do this (or the power series/ differential equations) definition? We want to be able to differentiate $\sin$ and $\cos$ before developing the machinery of integration. They make very good examples, so we use a less satisfying but more elementary definition just to get them out there.

Hmmm...some Calculus books do go through integration before differentiation (Apostol's Calculus, for example, no?) and yet they do not give a satisfying argument of why the above limit is as it is. I mean, they give the geometric argument which is based on Euclidean Geometry (otherwise, we would not know things like "the hypotenuse of a right triangle is always larger than any of the other sides of the triangle", unless we find another way), with the unit circle etc. , but this argument seems week because naturally Calculus is based upon Set Theory axioms and adding up the Euclidean axioms (which seem to me fairly weak) seems like a mess. The book usually mentions "The above procedure was only to show the geometric interpretation of the limit and the properties of sine. A more formal definition will be given via infinite series in chapter X.Y".

So sure, defining sine as an infinite series solves all problems. But it does seem a bit artificial. I mean, sine is a geometric creature. It seems natural to prove it by geometric means. But can it be proven via geometric means of Calculus only (as area, arc length, etc. are defined in Calculus and not in Euclidean Geometry)? The above post gives me the feeling that yes. However, I have yet to see such a proof in any book, mathematical article or journal, which is strange, to say the least.

In my opinion proving things from the "ZPC" is an idea of the 1800s, when scientists believed that they knew everything and they were just in the process of finishing the last few steps (when in fact natural science began at the time). Even if you want to be a 'formalist' Gödel's theorems show that it's futile to try to implement completely 'rigourous' mathematics anyway. You have to 'dare' to take that first step and begin with math even if you do not have mathematical (or should I say philosophical) logic behind it.

In fact, proving that $\lim_{x\to0}\frac{\sin x}{x}= 1$ from the ZPC is illusory rigour. Because who will want to read that? The standards for what is called a proof has drastically increased the past 100 years, but it still is the ideas that matter, and it will always be that way. ZPC won't help if you don't have an idea.

Note that I don't object to defining arc-length formally and what not, rectifiable curves are important, but to believe that you should prove something from the set theory axioms is absurd and pointless.

Ok, sorry to derail  :P

EDIT: It isn't really artifical to define $\sin z$ with power series, because if you define length properly the functions that describe the relationship between length of the unit circle and the corresponding coordinates will be exactly the power series that we define $\sin$ and $\cos$ with.

Well, I didn't mean that you should prove directly from the axioms. I only said that any proof of the mentioned limit that uses tools or theorems that are eventually justified by axioms that are not a part of ZPC, should not be considered rigour. Why? It doesn't seem formal or nice or "clean", it seems like a mess, like I said. I don't know of anything else in Calculus that uses Euclidean Geometry (or for that matter, any tool that can't be traced back to the ZPC) to justify A CALCULUS ARGUMENT. Why should this? Just because it's the most intuitive way? Huh! Forget about Euclidean Geometry, define the trigonometric functions as infinite Maclurin series and that's that!

I would still like to see a proof of the limit based on geometric arguments via Calculus (arc length definition, etc.) and not Euclidean Geometry. It seems possible. So if anyone could refer me to a book or an article in which such a proof is given, I would be most grateful.

Well, after consulting countless Calculus books, I've arrived to the following conclusion: The limit Sin(x)/x as x approaches zero cannot be evaluated by geometric means, never, due to the geometric nature of the definition of the sine function. A correct proof of the limit in question will always require that sine will be defined purely by an analytic expression (such as an infinite series, via Euler's formula, etc.) and never geometric in nature.

The problem with the geometric proof of this limit (beyond the fact that most Calculus books base it on Euclidean Geometry axioms, which seems not in place for a Calculus book) is that a major part of the proof is based on comparison of areas, using the fact that the area of the unit circle is pi. However, other than strictly defining the area as pi (but that's the same trick as defining sine to be an infinite series or something), you cannot possibly arrive to the conclusion that the area of a circle is pi without using the sin(x)/x limit! Sometimes, the limit is used in a very subtle way, like if you use integrals to calculate the area, you will inevitably have to use a trigonometric substitution in the process, therefore requiring you to know the anti derivative of cosine, which you cannot possibly know without knowing the limit sin(x)/x. Another kind of proof is bound a circle by regular polygons with n sides and let n approach infinity. This would be the intuitive way of finding what the area of a circle is. However, it is inevitable that you will require the limit of sin(x)/x somewhere in the process. Which you obviously don't know unless you know what the area of a circle is.

To make long things short, the trigo. functions should be defined analytically and innocent Calculus students should not be taught such circular and meaningless proofs.

The combination of calculus and Euclidean geometry goes back to the Greeks. Have you ever heard of the "method of exhaustion"? Mathematicians such as Eudoxus and Archimedes were using limit arguments to derive results about areas. Euclid's definition of a ratio is essentially a Dedekind cut; that gives you the real numbers, and a way to rigorously prove equality or inequality of limits.
Using these ancient methods, what do we know about the area of a circle with radius $r$? We know that this area is half the product of the radius and the circumference. Since all circles are similar, the ratio of circumference to radius in a circle is the same for all circles. We can't find an exact numerical value for this ratio, but we can approximate it; Archimedes proved that the ratio of a circle's circumference to its diameter is between $3\frac{10}{71}$ and $3\frac17$. The only step that eluded the Greeks is giving this ratio a standard name; many centuries later, this ratio was named $\pi$.
Trigonometry came later, and was initially developed as an aid to astronomy and navigation. For these applications, angles in degrees were sufficient. Radians came much later; their invention is contemporary with Newton and Leibniz. Radians exist to make the calculus of trigonometric functions nice.

You can't avoid defining $\pi$ somehow, and the geometric definition is historically correct. While a strictly analytic cluster of definitions could be used, the hybrid geometric/analytic approach works and has history behind it. Students don't come into a calculus class ignorant of what $\sin$ and $\cos$ are; why ignore that knowledge?

We have
\[\lim_{z\to 0}\frac{e^{iz}-1}{iz}= \frac{1}{i}\frac{d}{dz}e^{iz}\Big|_{z=0}= 1 \]
and, similarly,
\[\lim_{z\to 0}\frac{e^{-iz}-1}{iz}= \frac{1}{i}\frac{d}{dz}e^{-iz}\Big|_{z=0}=-1 \]
Subtracting these shows that
\[\lim_{z\to 0}\frac{e^{iz}-e^{-iz}}{iz}= 2 \]
and hence
\[\lim_{z\to 0}\frac{\sin z}{z}= 1 \]
as desired.  In 'higher math' we usually use the definition
\[e^{z}= \sum_{n\geq 0}\frac{z^{n}}{n!}\]
anyways, and define $\sin z$ and $\cos z$ to be the imaginary and real parts of $e^{iz}$ (note the order! $e$ comes first!).

The fact that a purely analytical approach based on power series (such as what blahblahblah just outlined) can be logically supported does not mean I ever intend to start a Calculus I or Calculus II course with that.

In the end, the average student is going to think that you can prove that $\lim_{x\to0}\frac{\sin x}x=1$ by using L'Hôpital's rule. It they can swallow such a blatant logical circularity as that, how much negative pedagogical impact do you think there can be to a little geometric hand-waving about arclengths and the areas of circular wedges?

I agree with the main thrust of Kalle's post, especially his use of the phrase "illusory rigor."

[And towards the end of a second semester calculus course, I will introduce the complex exponential - but this is resting on the framework of already having gotten the students used to a number of things, including the sine, cosine, and exponential, and including power series.]


quadratics
algebra
polynomial
quadratic formula

Here's a problem.

Prove that in $ax^2+bx+c=0$,
\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]

[hide=" My Proof"]
We use the completing squares method and generalize it into a formula, as follows.
$ax^2+bx+c=0$
$\\x^2+\frac{b}{a}x=-\frac{c}{a}\\ x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}{a}\\ \left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac{c}{a}=\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}=\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm\sqrt{b^2-4ac}}{2a}\\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \mathbb{QED}$
[/hide]

Does this belong in intermediate? I learned it in middle school and wasn't that great at math in middle school.  ;)

Just complete the square.  :)

lol, that's what I did, however, since proofs are not supposed to be in anything below intermediate forums, I placed it here. :)

Masoud Zargar

even though it is obvious to complete the square, another way to prove the quadratic formula is to input those values into the equation, and show it equals 0, and you have found all the solutions because a second degree polynomial has 2 roots (not necessarily distinct)


http://www.manilatimes.net/national/2005/may/05/yehey/top_stories/20050505top4.html

So...axiom of choice and axiom of trichotomy are wrong.  Number theory is wrong...so FLT is wrong.  That's essentially his proof.

Is my post above the previous post or below the previous post?? We don't know!!! Maybe its skew, because Escultura seems to think so.

:-p

**EDIT** This is a repost, but talk anyway :D

umm....the poll is kind of one sided, don't you think?  All three choices are "yes"

Yes, eryaman.  That's the point. The guy's bogus.

Wow. This was a really tough choice. I ended up voting "Yes."

I skimmed his website.  On the article where he disproves FLT, he apparently manages to construct a countably infinite counterexample.  When he declared that he was going to do that, I stopped reading and vomited for hours.

Edit:  I just found this quotation at http://mathforge.net/index.jsp?page=seeReplies&messageNum=919.  It is this guy describing his own work.

[quote]The most advanced development of analysis is the construction of the real numbers without the false axioms – completeness and trichotomy axioms – using only three simple axioms and built on the basic integers 0, 1, …, 9.[/quote]

Yes.  That's right.  The basic integers 0-9.  Time to go vomit some more.

Now, what I wonder is...how does he even manage to establish the basic relationships between the "basic" 10 integers with only 3 axioms. I mean..."1 + 1 = 2" is one already...

I imagine that that question is one of the few where you would actually become [i]less[/i] intelligent upon finding out the answer.

C'mon, I don't think that someone really took seriusly Escultura since the begining!

Now that's not nice! We should give everyone the opportunity to present their ideas without a prejudice audience. :-)

[quote="MithsApprentice"]We should give everyone the opportunity to present their ideas without a prejudice audience. :)[/quote]
well, we have to keep in mind that this site is very prejudiced, not everyone reading this article would have the same reaction as the poeple on this thread.

[quote="polymorphic"]i don't get it...is that article some kind of joke?[/quote]

To most of us, yes. To him, evidently not, which is probably why it's such a joke to us :P

hmm, i had to go with "yes" but apparently my yes was as popular as the other answer.

[quote="polymorphic"]i don't get it...is that article some kind of joke?[/quote]Somehow it does, somehow it does not. Remember that mathematics is a science based on axioms. Mathematics does not build knowledge from nothing, because nothing can create knowledge from nothing (it sounds recursive :D ) hence, there are theories where the proof by contradiction are not acceptable, because there is a third option (I think it is called [i]the third excluded[/i] theory, or something like that) Basically, most of us work under a math where everything is either true or false, but you can have a theory where everything is true, false or something else (not necessarly in the middle). In the same way, you can change the axioms of a theory and develope one similar but different (example: Geometry from Euclid to Lobachebsky or Riemann). Then, I didn't read the article totally (actually, I didn't spend more than a few minutes half reading it), but it looks like he is kind of denying some of the axioms (axiom of choice for example) that are fundamental on modern mathematics. That last part is what most of the people here consider a joke, that he is basically saying that all the modern mathematic is incorrect!


Let $ F: R^2\rightarrow R^2$ be defined by $ F(x, y) \equal{} (3x \plus{} 5y, 2x \plus{} 3y)$, and let $ S$ be the unit circle in $ R^2$.

Find
a) the image $ F(S)$
b) the preimage $ F^{ \minus{} 1}(S)$.

Can someone give me a hint?

Hint: this is a linear transformation, and $ S$ is a circle. The image is an ellipse, and the preimage is an ellipse.

I can definitely picture the image and preimage being ellipses, but I'm without a clue as to how I can show that on paper.

Would the equation of $ F(S)$ just be $ (3x \plus{} 5y)^2 \plus{} (2x \plus{} 3y)^2 \equal{} 1$?

No; that's actually $ F^{\minus{}1}(S)$. For $ F(S)$, put the inverse matrix there.

Is the matrix  you speak of  $ \left[\begin{array}{cc}3&5\\2&3\end{array}\right]$?

Yes. Its determinant is $ \minus{}1$, so the inverse also has integer coefficients.


geometry
circumcircle
perpendicular bisector
geometry unsolved

$ ABCD$ is a convex quadrilateral and $ \odot O$ is its circumcircle. $ BA\cap CD \equal{}H$, $ AC \cap BD\equal{}G$

$ O_1, O_2$ are the circumcenter of $ \triangle AGD$ and $ \triangle BGC$ respectively. $ O_1O_2\cap OG\equal{}N$

$ HG$ meets $ \odot O_1,\odot O_2$ at $ P,Q$ respectively. $ M$ is the midpoint of $ PQ$

Prove that $ NO\equal{}NM$

[hide="Solution"]Denote K be the second intersection of $ (O_1),(O_2)$.
We have two well know lemmas:
Lemma 1: O,K,H are collinear and GK is perpendicular to OH.
Lemma 2: An arbitrary line through G meets $ (O_1),(O_2)$ at P,Q respectively. Then, OP=OQ
According to the lemma 1, because of the right angle OKG, the perpendicular bisector $ O_1O_2$ of KG cuts OG at the midpoint N of OG. According to the second lemma, OM is perp to GM, so NM=ON.  [/hide]


Let ABC be a triangle.$ BB_1$ and $ CC_1$ are bisector angle of angle $ \hat{ABC}$and $ \hat{ACB}$.And$ B_1C_1$ intersects circumfrence of triangle at $ M,N$.$ I$ be intercentre of trangle $ ABC$ .Provethat :$ R$=2$ r$ where $ R,r$ are radius of ($ MNI$) and ($ ABC$) respectively.

http://www.mathlinks.ro/viewtopic.php?t=86561

  darij


geometry
inradius
3D geometry
tetrahedron
geometry unsolved

[quote="pankajsinha"]Find the radius of incircle of the regular tetrahedron the length of each side being $ k$ units.[/quote]
A tetrahedron is a $ 3\minus{}D$ object, how can it have an incircle :maybe: ? Do you mean in"sphere"?

True.
I worked it out.
If I be the centre of the insphere of tetrahedron OABC,then
$ Vol(OABC)\equal{}Vol(IOAB)\plus{}Vol(IOBC)\plus{}Vol(IOCA)\plus{}Vol(IABC)$
$ Vol(OABC)\equal{}\frac{1}{6}{[\vec a  \vec b  \vec c]}$ and the volume of the other tetrahedrons can be calculated using
$ Vol\equal{}\frac{1}{3} (Area of \triangle ABC) (height)$
and this height will be the required radius

[quote="pankajsinha"]True.
I worked it out.
If I be the centre of the insphere of tetrahedron OABC,then
$ Vol(OABC) \equal{} Vol(IOAB) \plus{} Vol(IOBC) \plus{} Vol(IOCA) \plus{} Vol(IABC)$
$ Vol(OABC) \equal{} \frac {1}{6}{[\vec a \vec b \vec c]}$ and the volume of the other tetrahedrons can be calculated using
$ Vol \equal{} \frac {1}{3} (Area of \triangle ABC) (height)$
and this height will be the required radius[/quote]
Ok, I got it.


trigonometry
geometry
trapezoid
cyclic quadrilateral

I was reading the solution to this problem;

The sides of a quadrilateral are 3, 3, 4, and 8 (in some order). Two of its angles have equal sines but unequal cosines, yet the quadrilateral cannot be inscribed in a circle. Compute the area of the quadrilateral.

And there were parts of the solution that I didn't understand. We know that $ sin A \equal{} sin (180 \minus{} A)$, so the two angles with equal sines add to 180, so they are supplementary. We know that the quadrilateral isn't cyclic because it can't be  inscribed in a circle. It is not a cyclic quadrilateral, so the opposite angles don't add to 180 degrees. That means the two supplementary angles are the adjacent ones. I understand all of this so far.

Then somehow we come to the conclusion that the quadrilateral is a trapezoid based on the knowledge that the adjacent angles are supplementary. I DO NOT understand this part. Could somebody please explain that please?

Next the solution says that an isosceles trapezoid is a cyclic quadrilateral. Can somebody prove this for me?

After concluding this, the solution says that the bases have lengths of 3 and 8. Why can't the bases have lengths of 3 and 4?

After constructing the quadrilateral, I understand the solution, but those parts I need some explanations about. Can somebody please help? Thanks.

If we label the quadrilaeral $ ABCD$, then since the internal angle at $ B$ is $ 180^\circ \minus{} A$, the external angle at $ B$ is $ \equal{}A$.
So $ BC$ is parallel to $ AD$.

Now if we havean isosceles trapezium $ ABCD$ with the angles at the base $ AB$ = $ \alpha$, say, then the angles at the top are $ 90^\circ \plus{} (90^\circ \minus{} \alpha) \equal{} 180^\circ$ so we have a cyclic quadrilaeral.

If the bases were $ BC\equal{}3$ and $ AD\equal{}4$, then drawing a line $ DE$ parallel to $ AB$, we would have a triagle at the top with sides $ 4, 1$ and $ 8$ which is impossible.

I hope this is helpful. Unfortunately I don't know how to post a diagram as this would make it much easier for me to explain.

I'm sorry. I don't understand any of those. You can make diagrams using Geogebra BTW. When you are making a post it is in a small box when you are making your post.

Edit: I understand why eight has to be one of the bases. I don't understand the other two though.

If two adjacent angles are supplementary, then the sides that the angles do not share must be parallel. Suppose they aren't. Extend them until they meet. The triangle that forms has an angle sum greater than $ 180^\circ$.

I don't know why they mention isosceles trapezoids, but this one certainly is not isosceles. Isosceles trapezoids are cyclic since their base angles are congruent, and one base angle is therefore supplementary to the angle opposite from it (the one that the other base angle is supplementary to).

Okay, I understand why an isosceles trapezoid is cyclic, but I still don't understand why the quadrilateral has to be a trapezoid. Sorry.


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

[1] $\int \frac{x}{\sqrt{5-x}}dx$

[2] $\int \frac{\sin x \cos ^2 x}{1+\cos x}dx$

[3] $\int (\sin x+\cos x)^2dx$

[4] $\int \frac{x-\cos ^2 x}{x\cos^ 2 x}dx$

[5]$\int (\sin x+\sin 2x)^2 dx$

[quote="amirhtlusa"]
4
$\tan{x}-\ln{x}$

You forgot something.

5
$x-\frac{1}{2}(\frac{2\sin{2x}+\sin{(4x)}}{4})-\frac{\sin{(3x)}}{3}$

You are wrong.
[/quote]

To Japanese High School Students:

$d(\cos x)$ means that $(\cos x)'dx$.

I attempted to solve the first integral by integration by parts and arrived at a different answer. Could someone please point out to me where my line of reasoning went wrong?

$\int \frac{x}{\sqrt{5-x}} \,dx=\int x\left(\frac{1}{\sqrt{5-x}}\right) \,dx=-2x\sqrt{5-x}+2\int \sqrt{5-x} \,dx=-2x\sqrt{5-x}+\frac{4}{3}(\sqrt{5-x})^3$.

Oh whoops, so I was actually correct up to a sign flip.

Kunny,

In the first integral $\int \frac{x}{\sqrt{5-x}}dx=-\int \frac{1}{\sqrt{5-x}}d(5-x)= -2\sqrt{5-x} + C$

You put a $x$ more

[quote=kunny]Here is my solution.

[hide]

[1] $\int \frac{x}{\sqrt{5-x}}dx=-\int \frac{x}{\sqrt{5-x}}d(5-x)=\int \frac{(5-x-5)}{\sqrt{5-x}}d(5-x)$

$=\int \left(\sqrt{5-x}-\frac{5}{\sqrt{5-x}}\right)d(5-x)=\boxed{\frac{2}{3}(5-x)^{\frac{3}{2}}-10\sqrt{5-x}+C}$

[2] $\int \frac{\sin x \cos ^2 x}{1+\cos x}dx=-\int \frac{(\cos ^2 x -1+1)\sin x}{\cos x +1}dx$

$=\int \left(1-\cos x-\frac{1}{1+\cos x}\right)d(\cos x)=\boxed{\cos x-\frac{1}{2}\cos ^ 2 x -\ln (1+\cos x)+C}$

[3] $\int (\sin x+\cos x)^2dx=\int (1+\sin 2x )dx=\boxed{x-\frac{1}{2}\cos 2x +C}$

[4] $\int \frac{x-\cos ^2 x}{x\cos^ 2 x}dx=\int \left(\frac{1}{\cos ^ 2 x}-\frac{1}{x}\right)dx=\boxed{\tan x-\ln |x|+C}$

[5]$\int (\sin x+\sin 2x)^2 dx=\int (\sin ^ 2 x+2\sin x \sin 2x +\sin ^ 2 2x)dx$

$=\int (\sin ^ 2 x+4\sin ^ 2 x \cos x +\sin ^ 2 2x)dx$

$=\int \left(\frac{1-\cos 2x}{2}+4\sin ^ 2 x (\sin x)' +\frac{1-\cos 4x}{2}\right)dx$

$=\boxed{x-\frac{1}{4}\sin 2x -\frac{1}{8}\sin 4x +\frac{4}{3}\sin ^ 3 x +C}$

[/hide][/quote]


MATHCOUNTS
trigonometry
geometry
circumcircle
trig identities
Law of Sines
Pythagorean Theorem

Let ABC be a triangle such that AB=10 cm, the measure of angle CAB is 30 degrees, and the measure of angle ACB is 45 degrees. What is the number of centimeters in the length of BC ?

[hide]

Draw perpendicular to AC.

5:rt2:

[/hide]

Mmm, i got a different answer. Please show your process, and try to always show your process. please.

And I correct myself. It was called [hide]Laws of Sines[/hide].

[hide]

I got 5 :rt3:  + 5

edit:

oops... that was AC...

Ok then....

BD(altitude from B to AC) = sin(30) x 10 = 5

so then

BC =  :rt2:  * 5 = 5 :rt2: [/hide]

sigh methods please. and i don't think that's right...

Yah, I edited it. Figured what I did wrong.

yeah that's probably right...now use tare's hint, and while you're at it, find the circumradius of the triangle.

Hmm...I thought the Laws of Sines (a/SinA=c/SinC) because you know one pair of an angle and the length of the opposing segment, and you know the angle opposing the length of the segment you're trying to figure out.

Oh, and try to figure out the radius of the inscribed circle too.

A/sinA=C/sinC is true. also equal to B/sinB

Yup, also leads to MysticTerminator's problem (Hint:[hide]the circumradius[/hide] is also involved in the Laws of Sines)

yeah, it would lead to [hide]2r, or the diameter[/hide]

Aaaarg! You're the comod :) , you have to remember to use Spoilers. (and yes, I edited yours in case someone wanted to prove the Laws of Sines themselves to find out; it's not that hard to prove, you know.)

yes, tare, the law of sines would be easier. anyways, someone try that. you [i]could[/i]find the radius of the inscribed circle, but I would not suggest trying to do so. (it's not worth it)

However, go ahead and find the circumradius.

Why do you people (tare) post the same things as me just after I post?

Well [b]right[/b] when I'm typing people post short messages. I tend to write long, so at most a couple of people post the things I was going to say before me.
I'm not doing it on purpose, fyi :)

and where did I post the same thing in this thread?

Um, is it just me or is trig not needed for this?

[hide]

Draw perpendicular from B to AC, we'll call the point where the height is H. Clearly ABH is a 30-60-90 triangle. Using that, BH is 5. BHC is an iscosoles right triangle, so HC is 5. Using the pythagorean theorem, we get BC = 5:rt2:

[/hide]

no, you do not need trig, but it is so much easier if you just use that one simple formula...

I mean, not all triangles are 30-60-90 or equilatteral-right. This is one of the easy cases where the triangle is a combination of these two special triangles, but just keep in mind that not all triangles are as simple as these.

As for the circumcircle, any takers?

correct, but seeing as to how this is a mathcounts question, trig usually isnt needed. actually, never is needed, just may be preferable

Yes, there's always that easy way around the thorn bush...(but I don't usually find it :) )

make 2 right triangles.  now you have a 45-45-90 and a 30-60-90.  5sqrt(2) is the answer.

[quote="TheSouthpaw"]Um, is it just me or is trig not needed for this?
[/quote]

The second one.

Draw a perpendicular.
Then it is just knowledge of 30-60-90 and 45-45-90 triangles.

Answer:
[hide]$5\sqrt{2}$[/hide]


function
combinatorics proposed
combinatorics

Show that the number $s_{m,n}$ of surjective functins $f: X \rightarrow Y$ whit
$\mid X\mid =n$ and $\mid Y\mid =m$ is given by the expression
$s_{m,n}= m^n - {m \choose 1}(m-1)^n+{m \choose 2}(m-2)^n+...+(-1)^{m-1}m$

this is a straight application of the I-E principle. just note that the $k$th term is just the number of functions whose image is contained in a set of size $m-k+1$.


On Monday a school library was attended by 5 students, on Tuesday, by 6, on Wednesday, by 4, on Thursday, by 8, and on Friday, by 7. None of them have attended the library two days running. What is the least possible number of students who visited the library during a week?

[hide]$ \boxed{15}$

Less than 15 is impossible since between the distinct individuals on Thursday and Friday we have 15. It is also possible by having those people attend on alternate days.[/hide]

[hide]easy problem but hard to express your thought(especially if english is not your first language)...
well from monday and tuesday we have at least 11 differents students.
on wednesday we have 4 students visiting the library. To minimize the whole number of students we guess that these 4 students were 4 of the 5 that have visited the library on monday.
so our number remains 11.
on thursday though we have 8 people.They are differents than the 4 of the wednesday and of the monday.11-4=7students remaining to visit the library this day.
because they were 8 we have a new student here.
so our number is now 12.
finally at friday we have 7 students completely irrelevant with the 8 of the last day. So 11-8=3
so 3 new students.
and we finished.
the least possible number of the students is 15. :) (i analysed it as more as i could)[/hide]


geometry
3D geometry
sphere
polyhedron
IMO Shortlist
IMO Longlist

Faces of a convex polyhedron are six squares and 8 equilateral triangles and each edge is a common side for one triangle and one square. All dihedral angles obtained from the triangle and square with a common edge, are equal. Prove that it is possible to circumscribe a sphere around the polyhedron, and compute the ratio of the squares of volumes of that polyhedron and of the ball whose boundary is the circumscribed sphere.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


Prove that $\cos \frac{\pi}7+\cos \frac{3\pi}7+\cos \frac{5\pi}7=\frac{1}{2}$.

$2 \sin \frac{\pi}{7}(\cos \frac{\pi}7+\cos \frac{3\pi}7+\cos \frac{5\pi}7)=\sin \frac{6\pi}{7}-\sin \frac{4\pi}{7}+\sin \frac{4\pi}{7}-\sin \frac{2\pi}{7}+\sin \frac{2\pi}{7}= \sin \frac{\pi}{7}$
$\Rightarrow \boxed{ cos \frac{\pi}{7}+cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}= \frac{1}{2}}$   $\mathbb{QED}$

Look at the general case:

$cos\alpha+cos(\alpha+r)+cos (\alpha+2r)+....+\cos (\alpha+(n-1)r)$


What is the largest integer k such that  3/2 . 2/1 . 1/2 . 2/3 . 3/4 .  .....  k / k+1   ≥  1/8

$ \frac {3}{2} \cdot \frac {2}{1} \times \left(\frac {1}{2} \cdot \frac {2}{3} \cdot \cdots \cdot \frac {k}{k \plus{} 1} \right) \equal{} 3 \times \frac {1}{k \plus{} 1} \ge \frac {1}{8} \iff k \le \boxed{23}$


[hide="this can Help!"]take $ a=\sin\alpha$ and $ b=\cos\alpha$ where $ 0\leq\alpha\leq\frac{\pi}{2}$ :) [/hide]

let a+b=A, so ab=(A square -1)/2,  and  -2^0.5<= A <= 2^0.5
then substitute A into the equation and eliminate all forms of a,b.
the rest is just normal.

[quote="chien than"]Let $ a;b$ be real numbers satisfying $ -1 \leq a;b \leq 1$ and $ a^{2}+b^{2}=1$
Find min of $ A=|a+b+\frac{1}{a}+\frac{1}{b}+\frac{a}{b}+\frac{b}{a}|$[/quote]

This reminds of me the [url=http://www.kalva.demon.co.uk/putnam/psoln/psol033.html]Putnam 2003 A3[/url] if you use sinajackon's nice hint. :)


modular arithmetic
quadratics
number theory unsolved
number theory

Prove, that $\forall k\in\mathbb{N}$ congruence: \[x^{2}\equiv{-7}\pmod{2^{k}}\] has solution.

$-7=1(mod 8)$, therefore by Gensels lemm we have 2-adic solution $x^{2}=-7$.

I don't know what is "2-adic solution" and "Gensels lemma" :( I guess it is not in the elemetry number theory.
I have a elementry soluiton here:
for $k=3$ ,obvious.
asumme that $\forall k\leq l$ ,$x_{k}^{2}\equiv-7(\mod 2^{k})$
suppose $x_{l}^{2}=2^{l}A-7$ if $A$ be even then let $x_{l+1}=x_{l}$
if $A$ is odd,then $(x_{l}+2^{l-1})^{2}=x_{l}^{2}+2^{k}x_{l}+2^{2k-2}=2^{k}(x_{l}+A+2^{k-2})-7$and notice that$x_{l}+A+2^{k-2}$ is even.
thus $x_{l+1}=x_{l}+2^{k-1}$ satisfies $x_{l+1}^{2}\equiv-7(\mod 2^{l+1})$

@Rust: it's called "Hensels lemma".
PS: it is elementary.

Can you tell me what it is,Zetax?Thanks.

In this case, you only need the following fact:

If $k\geq 3$ is an integer and $a$ is an odd integers, then $a$ is a quadratic residue modulo $2^{k}$ iff $a$ is a quadratic residue modulo $8$.

For the sake of completeness, here is the general form for odd primes:
If $p$ is an odd prime, $k$ is a positive integer and $a$ is an integer not divisible by $p$, then $a$ is a quadratic residue modulo $p^{k}$ iff $a$ is a quadratic residue modulo $p$.

I think these well known facts have been proven many times here (see eg http://www.mathlinks.ro/Forum/viewtopic.php?&p=530704#530704 post 4) :)

In addition we can prove that the equation $x^{2}+7y^{2}=2^{k}$ has solution with $x,y\in\mathbb{Z}$ for all $k\in\mathbb{N}.$

he most probably meant $x^{2}+7y^{2}=2^{k}$ which is equivalent to the poblem and is well known. :)


probability
geometry
number theory
\/closed

i'm interested in the APoS  books but I'm don't excel in math (I'm average level probably but my school's curriculum is kinda lame- currently in gr.9, in integrated math class that teaches basic alg I/geo).

so i've never done any math competition but I really want to try. really.
would the books help get my skills up and prepare for comp? if so, which books? if not, what should I do first?

[quote="JRav"]The books are really helpful if you're willing to sit down and work with them.  If you don't consider yourself great at math yet, get Volume 1, and some of the other lower-level books.[/quote]

I specifically would recommend Introduction to Counting & Probability and Introduction to Number Theory, as these will expose you deeply to important areas of math that are overlooked in your "lame" school curriculum.  These are also areas that show up quite a bit on contests, since they are interesting, important areas of math.

To answer your 'other question' is it reasonable for any grade:

To a certain level.
Any kid with pre-algebra to college and above who is interested in math is good for this site.

Its so cute because my little 7 yr old cousin asked if he could make an account too...


algebra
polynomial
calculus
derivative
algebra unsolved

The expansion of
            $ f(x) \equal{} (1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^{1000})^{1000}$
is the polynomial
            $ f(x) \equal{} a_0 \plus{} a_1x \plus{} a_2 x^2 \plus{} ... \plus{} a_{10^6} x^{10^6}$
Calculate
            $ S \equal{} a_0 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_{1000}$

Follow Rust's argument, it is easy to get
$ a_i\equal{}C_{n\plus{}i\minus{}1}^{n\minus{}1}$ for $ i<\equal{}n$, where $ n\equal{}1000$.
We observe $ S$ is the coefficient of $ x^{n\minus{}1}$ in $ f(x)\equal{}\sum_{i\equal{}0}^n(1\plus{}x)^{n\plus{}i\minus{}1}$,
$ f(x)\equal{}\frac{(1\plus{}x)^{2n}\minus{}(1\plus{}x)^{n\minus{}1}}x$.
So $ S\equal{}C_{2n}^n$.

I didn´t understand Rust´s argument. Can somebody explain me it?

Ah, I observe that we can write

$ f(x) \equal{} (1 \plus{} x \plus{} ... \plus{} x^n)(1 \plus{} x \plus{} ... \plus{} x^n)...(1 \plus{} x \plus{} ... \plus{} x^n)$

and that $ x^k$ appears if and only if we take $ x^{a_1}$ from the first factor, $ x^{a_2}$ from the second factor, ... and $ x^{a_{1000}}$ from the last factor and $ \Sigma a_i \equal{} k$... So all we need is to count the number of solutions of the last equations, and this is a known problem.

If my understanding is correct. Rust is using Taylor expansion of $ (1\minus{}x)^{\minus{}m}\equal{}\sum_{i\equal{}0}\frac{f^{(i)}(0)}{i!}x^i$, where $ f(x)\equal{}(1\minus{}x)^{\minus{}m}$ and it is not hard to verify the i-th derivative of $ f(x)$ is $ m(m\plus{}1)...(m\plus{}i\minus{}1)(1\minus{}x)^{\minus{}m\minus{}i}$.


linear algebra
matrix
combinatorics unsolved
combinatorics

In an $ n\times n$ matrix, a generalized diagonal is a set of $ n$ entries, one from each row and column. Let $ A$ be a $ 0, 1$-matrix having exactly one generalized diagonal all of whose entries are $ 1$. Prove that one can permute the rows and columns of $ A$ to obtain an upper triangular matrix.

It's a nice question -- it's also been posted before: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19558


Very neat (but I guess once you start thinking about bases it is hard not to miss it)

Has anyone heard of the look - and -say sequence?

[quote="me@home"]Has anyone heard of the look - and -say sequence?[/quote]
$1, 11, 21, 1211, 111221, 312211,\ldots$

Do you know of anybody who guessed what that sequence was on their first try? (I know 2 people, out of about 60 i've seen who tried to guess it...)

Kev-Eile, is that your guess for the sequence? I'm a little confused...


Wow, Moubi, where did you take this from?  ;)

After drinking japanese green tea, I wrote this

Are you sure there was no vodca or something else around? :D


Circles $A_{1},A_{2}$ intersect at $P,Q$. A common external tangent touches $A_{1}$ at $X_{1}$ and $A_{2}$ at $X_{2}$. Another common external tangent touches $A_{1}$ at $Y_{1}$ and $A_{2}$ at $Y_{2}$. Show that $PQ$ is parallel to $X_{1}Y_{1}$

[hide="Solution"]
Let the centres of the circles be $C_{1}$ and $C_{2}$. $C_{1}C_{2}\perp PQ$. Let $R$ be the point where $X_{1}Y_{1}$ crosses $C_{1}C_{2}$. The triangle $C_{1}X_{1}R$ is congruent to $C_{2}X_{2}R$ as the whole diagram is symmetric about $C_{1}C_{2}$. Hence $C_{1}C_{2}\perp X_{1}Y_{1}$, and we're done.
[/hide]


I'm trying to prepare for the USABO, especially the first level.
However, I was not successful in finding any of the past USABO competitions.
Besides the CEE website, which only has IBO and foreign competitions, does anyone know where I can get the past USABO exams(first level)?

http://www.ibo-info.org/

go to the NATIONAL OLYMPIADS section on the left.

if you aren't lazy, you can go to each country and see which countries have old exams... US is relatively stingy in this regard :)

Well, that site is nice, but as you know there is no USABO.
ANYONE know a site with all the past USABOs???
is there any way to order them?
...

I'm leaving on vacation for about a week and a half but drop me your e-mail address and I'll send you some old semifinal exams that I have.

I am still in shock that I didn't make national finals. I needed a 125 to go on and I missed it by two points. That is terrible. Next year I'll be at finals though. My e-mail is adidasty@aol.com


inequalities
LaTeX
inequalities unsolved

[quote="navid"]not very hard! :arrow:[/quote]
The same as in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=112522[/url]:
[quote="Rzeszut"][quote="navid"]Please send the solution of this:[/quote]
Please, use $\LaTeX$. You can learn it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5261]here[/url]. Using Word documents is extremely inconvenient.
[/quote]
After writing in $\LaTeX$, the Word document you attached is
[quote="navid"]Let $x_{1},\ldots,x_{n}$ be real numbers satisfying $\sum_{i=1}^{n}\frac{1}{x_{i}}=1$. Find the minimum of the expression $\sum_{i=1}^{n}\frac{x_{i}^{i}}{i}$. [/quote]
While rewriting the problem, you made two mistakes: $x_{1},\ldots,x_{n}$ should be positive numbers and the sum of their reciprocals should be $n$. The correct version is
[color=blue][i]Let $x_{1},\ldots,x_{n}$ be positive numbers satisfying $\sum_{i=1}^{n}\frac{1}{x_{i}}=n$. Find the minimum of the expression $\sum_{i=1}^{n}\frac{x_{i}^{i}}{i}$. [/i][/color]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=57179[/url].


superior algebra
superior algebra unsolved

Are there extensions of degree $4$ of $\mathbb{R}$ and $\mathbb{Q}(i)$ ?

No for the first since $\mathbb{C}$ has degree $2$ over $\mathbb{R}$ and is algebraically closed, so every algebraic extension can be assume to lie between them, not letting the degree be higher than $2$ (and non-algberaic=transcendental extensions have degree $\infty$).

Yes for the second one, take for example $\mathbb{Q}[\zeta_{16}]$ with $\zeta_n$ a primitive $16$-th root of unity: the degree of this extension over $\mathbb{Q}$ is $\varphi(16)=8$, so (it clearly contains $i=\zeta_2$) it has degree $4$ over $\mathbb{Q}[i]$. To see the $\varphi$-part, either some stuff about cyclotomic extensions or again calculation is necassary.
Another one: $\sqrt{1+\sqrt{3}}$ is real and has degree $4$ (see some of the other threads). Thus adjoining $i \not \in \mathbb{R}$ to $\mathbb{Q}[\sqrt{1+\sqrt{3}}] \subset \mathbb{R}$, we get something of degree $8$, thus what we wanted.


trigonometry
calculus
function
algebra
polynomial
Euler
complex analysis

Evaluate:

$\left( 1 + \frac{1}{1^2} \right) \times \left( 1 + \frac{1}{2^2} \right) \times \left( 1 + \frac{1}{3^2} \right) \times \left( 1 + \frac{1}{4^2} \right) \times \left( 1 + \frac{1}{5^2} \right) \times ...$

(With proof of course  :) )

Using :

$\sin z = z \cdot \prod_{n=1}^\infty \left(1 - \frac{z^2}{n^2 \cdot \pi^2} \right)$ which is true $\forall z \in \mathbb{C}$ (which is a good exercice in itself ;) ) with $z = i \cdot \pi$  we get

$\frac{\sinh \pi}{\pi}$

Doesn't that formula need complex analysis to prove?

I think there is supposed to be a more specific but simpler way to solve this problem.
Although, I'm not sure, because this type of question lends itself to calculus techniques.

No, I'm pretty sure that's the right way to do this problem, and I think I [hide="know the derivation of the formula"]
We know that $\sin x = 0$ for $x \equiv 0 \bmod \pi$, so we can write $\sin x = x(1 - \frac{x}{\pi})(1 + \frac{x}{\pi})(1 - \frac{x}{2\pi})(1 + \frac{x}{2\pi}) ... = x \prod_{k=1}^{\infty} (1 - \frac{x^2}{k^2 \pi^2} )$, and I think it can be shown that this function gives the correct values of $\sin x$ up to multiplication by a constant simply by assuming that a polynomial (specifically, the Taylor polynomial of $\sin x$) is uniquely described by its roots (again, up to multiplication by a constant).  Yes?

Or does it require more rigor than that?  :|
[/hide]

It requires a lot more rigor than that :!:
That is not a polynomial, so it's not described by it's roots.

The whole problem seems inadequate for this forum since there should be no truly elementary prove.

[quote="t0rajir0u"]assuming that a polynomial (specifically, the Taylor polynomial of $\sin x$) is uniquely described by its roots (again, up to multiplication by a constant).  Yes?[/quote]

I don't think it's that simple. After all, functions like $1/x$ and $e^x$ don't have any complex roots.

[quote="t0rajir0u"]No, I'm pretty sure that's the right way to do this problem, and I think I
We know that $\sin x = 0$ for $x \equiv 0 \bmod \pi$, so we can write $\sin x = x(1 - \frac{x}{\pi})(1 + \frac{x}{\pi})(1 - \frac{x}{2\pi})(1 + \frac{x}{2\pi}) ... = x \prod_{k=1}^{\infty} (1 - \frac{x^2}{k^2 \pi^2} )$, and I think it can be shown that this function gives the correct values of $\sin x$ up to multiplication by a constant simply by assuming that a polynomial (specifically, the Taylor polynomial of $\sin x$) is uniquely described by its roots (again, up to multiplication by a constant).  Yes?

Or does it require more rigor than that?  :|
[/quote]

This is exactly how Euler did it ! But nowadays we require perhaps a little more rigor.

The idea you want is a common one in complex analysis - you take some sort of function which is meromorphic, subtract out all the poles, and apply Liouville's theorem to the remaining part, which should be holomorphic and bounded in the plane.

Or you use the residue theorem...
Or you use some properties of the $\Gamma$-function
Or...
Complex analysis offers a lot of nice ways (which often rely on the methode mentioned by blahblahblah) to kill such problems.

[quote="joml88"]I moved it to Complex Analysis.  Hopefully this is the right place...[/quote]
There are elementary proofs of this, all we need is $(1+n/z)^n \rightarrow e^z$ $\forall z \in \mathbb{R}$ and some polynomials tweaking.

[quote="tµtµ"] all we need is $(1+n/z)^n \rightarrow e^z$ $\forall z \in \mathbb{R}$ [/quote]
Little typos here, it should be : $(1+z/n)^n$ and $z \in \mathbb{C}$ ;)


inequalities
function
inequalities proposed

If three positive numbers $a,b,c$ satisfy $a+b+c=1,$ then show that

\[\sqrt{9a+1}+\sqrt{9b+1}\plus{}\sqrt{9c+1} \le 6 \]

the function $ f(x)\equal{}\sqrt{x}$is concave then by Jensen
$ \sum \sqrt{9a\plus{}1} \leq 3\sqrt{\frac{9(a\plus{}b\plus{}c)\plus{}3}{3}}\equal{}6$

By AMGM we have that
$ \sum{\sqrt{9a\plus{}1}}\le\dfrac{1}{2}\sum{\dfrac{4\plus{}(9a\plus{}1)}{2}}\equal{}\dfrac{9}{4}(a\plus{}b\plus{}c)\plus{}\dfrac{15}{4}\equal{}6$

:)

dear friend,don't post old problem on box.

We have,
$ (\sum\sqrt{9a\plus{}1})^2\le 3(9(a\plus{}b\plus{}c)\plus{}3)\equal{}36$ (from Cauchy.) :wink:
Hence $ \sum\sqrt{9a\plus{}1}\le6$

Another solution to your problem: :D
From the RMS-AM inequality or the QM-AM inequality, we have:.
$ \frac{\sum\sqrt{9a\plus{}1}}{3}\le\sqrt{\frac{9\plus{}3}{3}}$ (as a+b+c=1)
$ \implies \sum\sqrt{9a\plus{}1}\le3\sqrt{4}\equal{}6$.
$ QED$ :wink:  :lol:


vector
function
superior algebra
superior algebra theorems

hi

let R be a noncommutative ring with identity

and suppose u*v=1
can we be sure now that there is a q such that q*u=1
and is in particular v*u=1

thx

fred

Simple answer: no.

Full answer: Let $V$ be a vector space of infinite sequences $(a_1,a_2,a_3,\dots)$. Let $A$ be the ring of endomorphisms of this vector space (multiplication is composition).
Define $L(a)=(a_2,a_3,a_4,\dots)$ and $R(a)=(0,a_1,a_2,a_3,\dots)$. Then $L\cdot R=1$ but $R\cdot L (a)=(0,a_2,a_3,a_4,\dots)\neq a$ and in fact $L$ has no left inverse.

Any of the standard sequence spaces will allow this example; I don't care which one you interpret this in.
Whenever you're looking for noncommutative rings and sometimes groups, look for matrices. In this case, finite-dimensional ones are too well-behaved, so we went to an infinite-dimensional example, where functions that are onto but not one-one are possible.

:D  thx jmerry
that's a very interesting example!


Is there any support available if I want to convert my AoPS blog to, say, Wordpress?

No.  We are interested to hear what features of Word Press you'd like added to the blogs, as we'll be doing a major overhaul of some parts of the site in the not-too-distant future.


probability
geometry
geometric transformation
reflection
algebra
polynomial
Putnam

a) Year 1872 Texas
3 gold miners found a peice of gold. They have a coin that with possibility of $\frac 12$ it will come each side, and they want to give the piece of gold to one of themselves depending on how the coin will come. Design a fair method (It means that each of the 3 miners will win the piece of gold with possibility of $\frac 13$) for the miners.

b) Year 2005, faculty of Mathematics, Sharif university of Technolgy
Suppose $0<\alpha<1$ and we want to find a way for people name $A$ and $B$ that the possibity of winning of $A$ is $\alpha$. Is it possible to find this way?

c) Year 2005 Ahvaz, Takhti Stadium
Two soccer teams have a contest. And we want to choose each player's side with the coin, But we don't know that our coin is fair or not. Find a way to find that coin is fair or not?

d) Year 2005,summer
In the National mathematical Oympiad in Iran. Each student has a coin and must find a way that the possibility of coin being TAIL is $\alpha$ or no. Find a way for the student.

For (c) you must mean "find a way to choose the team fairly, even if the coin is unfair".

There are infinitely many ways to do that, but is there a solution which is optimal in some precise sense?

No you are wrong Fleeting guess,
Let me clear the meaning of problem with a example:
Let the coint has the side $+,-$.
and we have $22$  players. and two teams with the names of $A , B$(In the exam the names where [b]SAYPA[/b]&[b]FOLAD[/b],I think the designer(Master ALISHAHI) is one of the fans of these teams :D ).
For each player ,we drop the coint and $+$ will give the player to $A$ and $-$ will give the player to $B$.
Find a way to claim that coin is fair or not?

Since I don't know Farsi and don't have the exam, let me just discuss the mathematical part:

If the question is "how to test whether a given coin has $p = 1/2$", I think that there is no procedure guaranteed to do that (with probability 1) after some finite number of steps, let alone a simple procedure that can be done at the Takhti Stadium (probably the locations reflect the mathematical sophistication of the solution).

If the question is "how to simulate a fair coin using an unfair coin with unknown constant $p \in (0,1)$", there is a simple procedure that with probability 1 will terminate in a finite number of steps.  It is a nice problem whether or not it was part (c) of this olympiad.

First the designer was not Mr. Alishahi the n he is not the fan of these 2 teams.

First the question was for the 2 teams Real Madrid And Barcelona

Come on man,why are you saying barcelona & real madrid,There were Iranian teams(I have misstyped Saba battery with saypa  :D ),
But As you wish ,Mr.hatami ,you are one of the coordinators
but who on the earth can design such a problems beside of Mr.alishahi ?? he designed the problem ,doesnt him?

Fleeting guess:I completely agree with you,but This is beauty of the problem ,It seems impossible but how can you
prove your  [quote]think that there is no procedure guaranteed to do that (with probability 1) after some finite number of steps, let alone a simple procedure that can be done at the Takhti Stadium[/quote]

First I didn't say that it was these 2 teams in the exam. I told that they changed these 2 teams to 2 Iranian teams for the exam.

And also this problem was designed by all of the members of the National Mathematical olympiad Committee. Not by a person, However some problems were given by one person.

re: impossiblity of a finite-with-probability-1 test for $p = 1/2$:
Consider any testing procedure that stops (with probability 1) after a finite number of coin tosses and declares YES or NO to the question "is $p = 1/2$?".
For any sequence of tosses where the procedure stops, those tosses are consistent with any $p \in (0,1)$; they do not determine $p$, so any declaration by the testing procedure is not necessarily correct.

What can be guaranteed is to answer the question "is $|p - 1/2| < \delta$" with a probability greater than $1 - \epsilon$ that the answer is correct.  To do this (for a given $\delta, \epsilon > 0$) one can just flip the coin  some large number of times and answer YES/NO according to whether the proportion of Heads/Tails is close to 1/2.

reply of (a) :
let the persons be $A,B,C$ . consider this sets : {$A,B$},{$B,C$},{$C,A$}. in the first step, we ballot by the coin, in each set , and in each set one person can win . then if one person didn't win in any sets (second step) we delete him and ballot between two other.
if each of persons won in one of his sets , we ballot in each of three sets again . we know finaly have the above manner(the above manner : one of persons don't win in any sets)  ( we  have that manner in finitely ballots )so we do the second step and finally one lucky person wins the coin fairly.

[quote="cm"](a) i think there is no finite procedure. Caz the probability for each person to win is a polynomial of 1/2^k.
Let T denotes tail and H denotes head. At each step toss the coin twice, if it's TT, then the first person win, TH for the second one and HT for the third. If it's HH, redo the procedure.
(b) similar to (a), just write a  in base 2, then it's a polynomial of 1/2^k. Define A to win at each step according to each k.[/quote]

if you read the above solution , you will understand that you were rong  ;) .

[quote="cm"](a) i think there is no finite procedure. Caz the probability for each person to win is a polynomial of 1/2^k. [/quote]

This shows that there is no procedure that stops within $k$ steps, where $k$ is given in advance.
It is possible that, with probability 1, the procedure will stop after some finite (but not limitable in advance) number of steps.   For example, this is true of the procedure "flip coin until a Heads appears".

Question (b) is actually Putnam 1989 A4.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1868731525&t=75385 or http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1868731525&t=106683


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let G be a simple group  and $f: S_{n}\to G$  be an epimorphism for some positive integer n. Prove that G and

$S_{k}$ are isomorphic for some $k\leq n$

please help me :?:

ZetaX, could it be that Rustam's proposition is correct though? :maybe:
I won't deny the weirdness of the question, but here goes :

We are given that $G$  is simple and that there is an epimorphism from an $S_{n}$.  The latter means, because of the third isomorphism system, precisely that $G$ is isomorphic to a quotient of $S_{n}$.
So the question as it is written there now is equivalent to :
prove that every simple quotient of a $S_{n}$ is isomorphic to a $S_{k},k\leq n$

If $n\geq 5$ , the only normal subgroups of $S_{n}$ are $A_{n}$, the trivial and the full one.  That means the quotients can only be trivial, have two elements, or  $S_{n}$.  As $G$ was simple, only the second option is possible : $G$ is isomorphic to $\mathbb{Z}_{2}$ or thus $S_{k}$ with $k=2$

When $n=2$, the only simple quotient is the group itself which is obviously isomorphic to $S_{k}$ with $k=2$.

When $n=3$, the only simple quotient again is isomorphic to $S_{2}$

When $n=4$, the only simple quotient again is isomorphic to $S_{2}$

But I have to be extremely cautious seeing you disagree :wink:

[quote="ZetaX"]Are you sure with the statement¿ Because $S_{n}$ is never simple for $n \geq 3$, thus they can't be isomorphic to a simple group $G$.[/quote]That's right, but that just means $G\cong S_{2}$.

And if we take $f$ the epimorphism $S_{n}\rightarrow S_{2}$ that tells wether a permutation is even or odd, you have an example that his problem (even though totally trivial) makes sense for all $n$.


Two distinct positive real numbers A and B are chosen at random.  B>A.  What is the expected value of B/A?

Three distinct positive real numbers A, B, and C are chosen at random.  C>B>A.  What is the expected value of (A+B)/C?  (A*B)/C?

What if the numbers must be integers?

Are there generalizations for when there are n numbers chosen?

I'm not sure how many of these questions are even possible to solve without being trivial...

It depends on what distribution one is choosing from.  There is no uniform distribution over the reals or integers, so the phrase "a random real" doesn't mean anything if you don't specify a distribution.


Centsports is basically a free online betting thingy. They start out by giving you ten cents, and you can use it to bet on baseball or basketball or whatever you want, and try to get more money. If you get enough money, you can send them your address and they will send you a check. If you lose all you money, they will give you another 10 cents to bet with. This is paid for by the advertisers, which is how they can run this without losing money. You can sign up for an account here [url]http://www.centsports.com/?opcode=97296[/url] , you don't need to give them anything but an email address until you want to cash out, and I do know somebody who has gotten money from them so this is legit!

*note* this isn't an ad, just a recommendation. I'm pretty much just doing this because i think it's cool, and if you do sign up, I would get 5% of anything that you won (you would get the same amount, I would get an extra 5%).

Hey Mustafa,

My name is Mustafa Said and I also like to gamble but I play poker....


Anyone knowing much abt it?
The official site said that ICSE 10th standard physics is required,and any idea abt how to participate? :D

err.... dont worry , 1st level is easy ( are you 95 born)?
Just do the 11th Physics portion from ncert , mechanics and optics should be enough . , if youll see , the 1st level is much tougher than the 2nd
I do not know much of icse , but ncert is recommended , i noticed , few questions were directly from ncert last year  :P

sorry mate , but wouldnt that imply u have to give senior level? ... i am 94 born too

Oooooooooops! :oops_sign:
I thought NSEA(Jr) was meant for 10thies and below.
So what's the syllabus?I hope u know what I mean.

ahh , for senior it would be the whole of 11th and 12th physics , specifically mechanics and optics  :rotfl:

The syllabus of NSEA Senior is 11th and 12th NCERT.Its level is much easy though.

Then this year is probably not for me,may be next year I'll try. :D

give it nevertheless , it is the easiest compared to other olympiads.....
for physics - u need to go much higher than jee physics , same for chem ...
For Bio - ask ritu
For math - Its a completely diff. genre  :D  (just love it )
Astro - Well .... Tough , but comparitively easy  :P
In astro , the real stuff starts in the camp ( ... well thats what ive heard  :( )

How tough is it for a 12thie to make it to INAO Sr.? Is it even worth a try or should I conc. on IPhO/INOI?

@nuclear
It is not very tough for a 12thie to crack NSEA Senior.But INAO Senior is tough to crack.

I mean I know nothing of 11-12 physics.For the time being,let me remain satisfied with math. :D

@scary
Don't lose heart.You can crack NSEA.Its not that difficult.Just make an attempt.Cutoffs are not that high as in case of NSEP/NSEC.So you can make it :thumbup:

@aashay
Yah,that's okay,but I dont think I will be getting much time to start off with 11-12 physics right now,cause u see,my parents and teachers are worried that even my math olympiad may end up nowhere,and may instead end up ruining my ICSE. :(
Moreover,it's serious time now,and boards do matter a lot,when it comes to admissions,job interviews and all!
I just thought that 10thies are supposed to answer NSEA(Jr) and the official site was saying that ICSE level physics is required,so I kind of got interested,u see.And another thing is that right now,my only tutor(ie my dad  :D )is reluctant to teach me calculus,and without that,I can go nowhere,as far as physics is concerned :(

@scary
r u in class 10. then i feel u shudn't try for nsep/c/a and focus on ur rmo instead.
i myself participated in nsep/c/a for the first time in 12th.
rmo is fine because they look at ur class too, and those in lower classes are preferred (this is evident from the fact that 12th class students have a tough time cracking it)

but since nsep/c/a are from ncert syllabus most of them who crack it are 12thies

[quote="sahilsharma94"]woa woa woa!!!!abhi tak toh 1st round bhi nahi hua!!!!!camp!!2nd phase!! u sure sound pretty confident of cracking this eh??[/quote]
He's been there, done that already :P

[quote="sahilsharma94"][quote]aadil
what sort of astronomy questions might be asked in nsea?
 [/quote]

Ya sure i think he has given NSEA b4 n also qualified..that explains his confidence...
but then y did he ask this question i've quoted....mayb he gave JER last YR n is giving SR this YR....

n btw who is an alchemist(if u can answer nuclear??)[/quote]
He did the INJSO instead maybe though...
And an alchemist- It was a term used in medieval times to denote a person who transformed one substance to another(The predecessor of the modern chemist) Often, alchemy was associated with the discovery of a method to make the fabled philosopher's stone which would turn everything into gold.

well! guyz let me clear ur doubts!!!!!!!!!!!

i went to both ASTRO(jr.)  and  JUNIOR SCI  camp prev year but failed 2 qualify for international in any!!!

this time there's no GRoup 1 and Group 2 in Senior so, X students(like me) have 2 compete wid Kota Toppers of XII in 1st phase!!!!!!!!!!!!!!!!!

so i am nervous and i am anxious to again goto camp as i am missing  camp experience a lot !!!!
but 1st phase will doom me!!!!

@ nuclear alchemist

i have very lean chance of making 2 2nd phase(leave alone camp!!!)

so, i declare MY WILLINGNESS TO JOIN UR KMOG CAMPAIGN!!!!!!!!!!!!!!!!!

KMOG - KILL MULTIPLE OLYMPIAD GUYS

let's start recruitment!!!! who wanna join???

PS :- Strong punishment for TRAITORS(WHo join and are later MOGs themselves!)

So join only if u have the guts!!!

hard time for kota guys starting 2011 i predict .........coz den NCERT CBSE weightage will be included to make Merit List n only 1s who have >85% marks in XII boards will be eligible to sit for IITJEE....n btw fbaggins ....if dat kinda training is given to some1 who just has the basics..he sure will also be at par...mostly all d stu.s are at par WRT their talent(n i Use MOSTLY coz dere are exceptions)...its just Xtra practice..experience which makes 1 triamphant n oders no quite :wink: ..so they are no geniuses...n if u start thinking b4hand dat chances are bleak...they sure will be....n believe me!!!kota n stuff are not proportional to ppl going for inter. olympiads..theres a guy in our school namely RAJAT..who dis just narayana(from 1 of the weakest branches of delhi, dwarka)..n still represented india at IJSO..got us a silver...he had raw intelligence...so be upbeat n trust ur gray matter..... :D   :lol: ..n btw can some1 plz tell me what all are the topics covered in XII optics(i'm doing mechanics form HC but dunno about optics XII)

@ sahil

thanks for encouraging me, i sure will try hard dis year!!!

well is that Rajat Kulshestra???

abt XII optics , i think it has lens and mirror equations, fermat's principle,lens maker formula, telescopes,microscopes,combination of lenses,etc.etc.

Yup..he was rajat kulshreth....how was he in the camp..was he among the top guys who qualified..ior did he just qualify...anyways..nobody can doubt his genius  :D thats certain :lol:

well actually everybody has raw intelligence but only some americans and a lot of south asians have actually found it in them.

if you feel racist don't be

I'm indian too.

hey!!!OFC i feel bad when u say that..its just that our system AKA r school concentrates on mugging up everthing..but believe me!!N students try not to do it..at least dellites do..i do..rajat does...n sure we will be faaaar ahead of amercians by end of century..the asian century is here  :lol:

When the result would be declared?
Swati


linear algebra
matrix
integration
vector

If A is an n x n Hilbert matrix over a field F of characteristic zero, how can we show that A is invertible without resorting to determinants? This was asked in my book several chapters before determinants were introduced.

Also, the same question arises for Vandermonde matrices.

Well, you show that the rows (or columns) are linearly independent, which is usually done by reducing the matrix to the row echelon form and checking that you do not have any 0 on the diagonal.

Hi fedja, I know that is the general method but it is not clear how to do this in the present case for the general n x n matrix.

There are some calculations of the inverse on the forum- it's easy to find by searching.

The simplest proof I know: we prove that the Hilbert matrix is invertible by showing that it's positive definite.
Using the inner product $ \langle f,g\rangle\equal{}\int_0^1 f(x)g(x)\,dx$ and vectors $ f_i(x)\equal{}x^i$ for $ 0\le i\le n\minus{}1$, the $ ij$-entry of the matrix is $ \langle f_i,f_j\rangle$. Since the $ f_i$ are linearly independent and our inner product is positive definite, this Gramian matrix must be positive definite.


show 2aCa doesnt divide 2nCn where a is natural no. bet. n/2 and n.nCr- n choose r.

I can't understand your words,who. :(
Could you express more clearly?

i think that:
show $C_{2a}^{a}$ doesn't divide $C_{2n}^{n}$

He also wants $n/2\leq a\leq n$, if I understood corectly his words...

statement reads 2aCa for e.g. 4C2=6.IS IT CLEAR NOW?


probability
combinatorics unsolved
combinatorics

A machine generates 0's and 1's with the same probability 1/2, beginning at time t=0. Which string is more likely to appear first: 011 or 110? A rigorous solution is preferred.

011 is more likely to occur first.

the only sequences that produce 110 before 011 are those of the form 11...110, wheareas there are many more types of sequences that produce 011 first.

In particular, if you start with 0 then you're guaranteed 011 first, and likewise if you start with 10, so the probability that you get 011 first is at least 3/4.  These sorts of things have been discussed on the forum before -- look particularly for threads that include the words "state diagram" and those on which Ravi B. has posted, to find some of them.

The best way to solve this problem is via a coupling (basically a rigorous version of the solution outlined by JBL), but you can compute the expected amount of time until you see a 011 or 110 string without too much difficulty.


I can't find the solution to this:
Solve for $x$. $\sin(\sin(\sin(x)))=\cos(\cos(\cos(x)))$.

Im not to sure if it will have a nice form for us.... anyhoo, slapping it into mathematica gives us an approximation of 0.960246

I would first reduce the outermost sines and cosines, then convert the rest of the cosines into sines.

[hide]
$\cos{\sin{^-1 (x)}=\sqrt{1-x^2}}$

$\sin{(\sin{(\sin{(x)}))}}=\cos{(\cos{(\cos{(x)})})}$

Not sure how to proceed from here...

[/hide]


We are given $ n$ positive numbers $ x_{1}, ..., x_{n}$. Form the fraction

$ \frac{1}{(x_{1})(x_{1} \plus{} x_{2})(x_{1}\plus{}x_{2}\plus{}x_{3})...(x_{1}\plus{}...\plus{} x_{n})}$.

By permuting thed $ x_{i}$, we can form $ n!$ such fractions. What is their sum?

We show by induction that the answer is $ \frac{1}{x_1x_2\cdots{x_n}}$. For n=1 it is trivial.  If it is true for n=k, then for n=k+1 we can factor out $ \frac{1}{x_1\plus{}x_2\plus{}\ldots\plus{}x_{k\plus{}1}}$ and grouping the terms by what x term is not in the sum, we obtain that the answer is $ \frac{\frac{1}{x_1x_2\cdots{x_k}}\plus{}\frac{1}{x_1x_2\cdots{x_{k\minus{}1}}x_{k\plus{}1}}\plus{}...\plus{}\frac{1}{x_2x_3\cdots{x_{k\plus{}1}}}}{x_1\plus{}x_2\plus{}\ldots\plus{}x_{k\plus{}1}}\equal{}\frac{1}{x_1x_2\cdots{x_{k\plus{}1}}}$, as desired.


limit
real analysis
real analysis unsolved

[quote="arp"]Let $P(x) = x^{n}+\sum_{k=0}^{n-1}a_{k}x^{k}$ and compute
$\lim_{x \rightarrow+\infty}\left( [P(x)]^{1/n}-x \right)$[/quote]

$\sqrt[n]{a}-\sqrt[n]{b}=\frac{a-b}{\sum_{k=1}^{n}(\sqrt[n]{a})^{n-k}\sqrt[n]{b}^{k-1}}$ so
$\lim_{x \rightarrow+\infty}\left( [P(x)]^{1/n}-x \right)=\lim_{x \rightarrow+\infty}\frac{\sum_{k=0}^{n-1}a_{k}x^{k}}{\sum_{k=1}^{n}\left(\sqrt[n]{P(x)}\right)^{n-k}x^{k-1}}=\frac{a_{n-1}}{n}$


For how many values of $k$, $0\leq k \leq 2006$, is $\binom{2006}{k}$ even?

[b]Solution.[/b]
By Lucas' Theorem, $\binom{n}{j}\equiv\prod_{i=0}^{m}\binom{n_{i}}{j_{i}}\pmod{p}$, where $p$ is a prime, $n=(\overline{n_{m}n_{m-1}...n_{0}})_{p}$, and $j=j_{0}+j_{1}p+...+j_{m}p^{m}$.  Therefore, $\binom{n}{j}\equiv 0\pmod p$ if and only if $j_{i}>n_{i}$ for some $i\in [0,m]\cap\mathbb{Z}$.  As a results, $\binom{n}{j}$ is not divisible by $p$ for all $j=j_{0}+j_{1}p+...+j_{m}p^{m}$ such that $j_{i}\in [0,n_{i}]\cap\mathbb{Z}$ for $i\in [0,m]\cap\mathbb{Z}$. Clearly, the number of numbers in $\left\{\binom{n}{0},...,\binom{n}{n}\right\}$ divisible by $p$ is $n+1-\prod_{k=0}^{m}(n_{k}+1)$.

In the case of $p=2$, we can find the number of even elements in $\left\{\binom{n}{0},...,\binom{n}{n}\right\}$ by just finding $n+1-\prod_{k=0}^{m}(n_{k}+1)$, where $n=(\overline{n_{m}n_{m-1}...n_{0}})_{2}$.

When $n=2006=11111010110_{2}$, we have the desired answer $2007-256=1751.\blacksquare$

Masoud Zargar

can you specify what the notation with the bar above the numbers means?

It just differentiates between multiplication and the digits of a number.

But with the expressions for $n$ and $j$ that you introduced, aren't you implying that $n$ and $j$ both have $m$ digits in base $p$? Or are you allowing some of the $j_{i}$'s to be 0?

It's understood that if $j$ has less than $m$ digits than all digits to the left of the left most nonzero digit is counted as zero.

[quote="drunner2007"]ok. that's a very nice solution. so is the first congruence actually lucas' thm., or is that a more of a consequence.
how do you prove lucas' thm?[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=theorem+lucas&t=121963]Search[/url] far and wide!

wow, how did you know about the theorem, it is crazy!!! :|


trigonometry
function
calculus
calculus computations

Hi everybody I wanna know how many different proofs do you know to show that the sequence $ x_n\equal{}\sin (n)$ is divergent. I still only have the one which computes $ \sin (n\plus{}2) \minus{} \sin (n)$ and then $ \cos (n\plus{}2) \minus{} \cos (n)$ but it's not so interesting...

So thanks :)

By Jacobi's Theorem $ n$ becomes arbitrarily close to $ \pi k$ infinitely often but also becomes arbitrarily close to $ \pi k \plus{} \frac{\pi}{2}$ infinitely often.

You may think that since $ \sin(x)$ is a period function then the sequence is divergent.

i probably know just this one, and i keep it really simple, $ \sin$ is a [b]periodic [/b]function with period $ 2\pi$, so as $ n$ approaches $ \infty$, the terms of the sequence in question just move around in the interval $ [\minus{}1,1]$ --- they don’t go anywhere! and since the terms do not converge to $ 0$ as $ n\rightarrow\infty$, the sequence must diverge by the $ n^{th}$ term test.

why it don't covnerge to 0?
because
 suppose it convergenc to $ 0$, then also $ \sin^{2} n$ converge to $ 0$. then also $ \sin (n \plus{} 1)$ converge to $ 0$ or $ \sin (n \plus{} 1) \equal{} \sin n \cos 1 \plus{} \sin 1 \cos n$ if $ n \minus{} > \plus{} oo$ then $ \sin n \cos 1 \plus{} \sin 1 \cos n \minus{} > \sin 1 \cos n$ which must converge to $ 0$ or we have that $ \cos n \minus{} > 0$
therefore
$ \cos^{2} n \minus{} > 0$
in finnaly we have that
$ \sin^{2}n \minus{} > 0$
and
$ \cos^{2}n \minus{} > 0$
or $ 1 \equal{} \sin^{2}n \plus{} \cos^{2}n \minus{} > 0$ or $ 1 \equal{} 0$  :lol:


$I$ is the incenter of the $\triangle{ABC}$ and the incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $AD$ meets the incircle again at $P$. $M$ is the midpoint of $EF$. Show that $PMID$ is cyclic (or the points are collinear).
I think that this is only possible in an [u][b]isosceles[/b][/u] triangle, but this is from a contest.Any solutions?

isoscele is only the collinear case, i think.

As I said I agree but this is from a competition. Look at this [url]http://www.kalva.demon.co.uk/ibero/ib90.html[/url]

I have showed that this problem is to prove the two coloured arc are equal, but i can't proceed any more...


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 12
AMC 8

The AMC now has a newly created Compact Disk with the:
AMC 8  -- 2001-2005
2001 AMC 10 and AMC 10 A and AMC 10 B -- 2002 - 2006
2001 AMC 12 and AMC 12 A and AMC 12 B -- 2002 - 2006
AIME I and AIME II -- 2001 - 2006
USAMO -- 2001 - 2006
all on one disk.  Each contest has two PDF files, one with the problems, the other with the solutions.  These are the images that we use to print the contests.   This disk is an effective way to obtain copies of all the 21st Centuy Contests, so we call it the 21st Century Contests Disk, Volume I.

You can order it with the AMC order form at:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/2007-Pub,Revision,9-22.pdf
or go to Publications  on our website and follow the links.

Steve Dunbar
AMC Director

where?

I followed the link and it gave me an error.

if you go to the AMC site and go to "publications" you should find it!

[quote="AMCDirector"]We are overhauling our web pages.  The new address is:
http://www.unl.edu/amc/d-publication/publication.shtml

Steve Dunbar
AMC Director[/quote]
If we order it from now and onwards will it have the AMC 8 of 07, and so on?

[quote="AMCDirector"]Well, it obviously won't have the 2007 AMC 8, but we are re-mastering an update that will inlcude the 2006 AMC 8, and some additional materials, intended primarily for teachers and coaches.   We expect to take it to the manufacturer on Monday, Dec 11 should hae it back in a week.  If you can wait to order Dec 18, we should have the new update in then.

Steve Dunbar
AMC Director[/quote]
Sorry, I meant the 06 I get mixed up a lot with the competitions...

I got the CD with teh math club book. Are they the same CD's

The disks sold individually, and included with the Math Club book are absolutely identical.
The intended audience for the Math Club Package is Math Club advisers and Math Team coaches as well as teachers.

-- Steve Dunbar
AMC Director

Actually I purchased the CD, but i didin't see the AMC10 and AMC12 2006 problems and solutions?

If we bought a disk with, for example, 2001-2006 but we want to practice with, for example, the 2007, is there any way to get that without buying another disk?

The easiest thing to do would be to order single copies of the most year's contests.  Use the Publications Ordering Instructions at :
http://www.unl.edu/amc/d-publication/publication.shtml
and be clear about what years and editions (e.g. 2007 AMC 12 B) that you want.

Steve Dunbar
MAA Director, American Mathematics Competitions


let $E$ be a vetor space of finite dimension $n \geq 1$. and $f \in L(E)$
let $\Psi : L(E) \rightarrow L(E)$ such that : $(\forall u \in L(E)): \Psi (u)=fou$
Determine $rk(\Psi)$ in funtion of $n$ and $rk(f)$

Salam,
Well here is my solution even if it isn't a rigorous one though and might even be stupid  as i'm so sleepy :D .
well let's precise what is Ker(psi) then :
u belongs to Ker(psi) if and only if f*u=0(L(E))
so u belongs to Ker(psi) if and only if Im(u) is included in Ker(f)
then the Ker(psi) is the set of linear maps from E to Ker(f) the dimension of that vectorial space of maps is exactly dim(E)*dim(ker(f)) (this is a classic lemma ;))
then dim(ker(psi))=n*dim(ker(f))
we have  rk(psi)=n²-dim(ker(psi)) and dim(ker(f))=n-dim(E).
Thus  rk(psi)=n²-n(n-rk(f))
So rk(psi)=n*rk(f)

i think $ rk(\psi) \plus{} dim(ker(\psi)) \equal{} n^4$ ,because $ \psi$ belong $ \L(L(E),L(E))$,and $ dim(L(L(E),L(E))) \equal{} (dim(L(E)))^2$


abstract algebra
search
ratio
probability
LaTeX
AoPS Books

In this game you post words that have a property that no words above have.

You post the word and the property that that word has that no other word above it has. In addition, the word posted must not satisfy any of the previous words' unique properties; that would make them not unique!

If you get stuck, clear the list and start over...
Also, there is no need to post a property on the first word posted in a set.

I think scoring will be +1 for each word you make and then +5 if you are the last one to post before a clear.

For example, if the first word is grapefruit:

grapefruit
[b]banana[/b]: Only word with only one unique vowel.

[hide]scores
randomdragoon: 1[/hide]

Note: Copy the entire word list and the scoreboard for each post!
(The next person needs to post a word that has a property that "grapefruit" and "banana" do not have, while [i]not [/i]having the property of only having one unique vowel)

grapefruit
banana: Only word with only one unique vowel
[b]sunglasses[/b]: only word with unique letter "l"

grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
[b]potato[/b]: Only word with two "o"s

Scores:
randomdragoon: 1
Flame: 1
cf249: 1

I think this is right...

grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
potato: Only word with two "o"s
cheese: Only word with three "e"s.

Scores:
randomdragoon: 1
Flame: 1
cf249: 1
iLord: 1

[quote="iLord"]cheese: Only word with three "e"s.[/quote]

Um... "Cheese" does not leave "Banana" as a unique word (Cheese only has 1 vowel) so.. do I just not count it, or do I end the round?

[quote="erabel"]Um... "Cheese" does not leave "Banana" as a unique word (Cheese only has 1 vowel) so.. do I just not count it, or do I end the round?
[/quote]

/me slaps himself.

I knew that.
 :blush:

So, if the round isn't (hopefully) going to be cleared:

grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
potato: Only word with two "o"s
cheesecake: Only word with three "e"s.

Scores:
randomdragoon: 1
Flame: 1
cf249: 1
iLord: 1

Ch[b]ee[/b]s[b]e[/b]cak[b]e[/b] has 4 e's. This has been a public shaming of iLord. Thank you. Try again.

[quote]Cheesecake has 4 e's. This has been a public shaming of iLord. Thank you. Try again.
[/quote]

Today's just not my day.

grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
potato: Only word with two "o"s
cheesecake: Only word with four "e"s.

Scores:
randomdragoon: 1
Flame: 1
cf249: 1
iLord: 1
[/quote]

grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
potato: Only word with two "o"s
cheesecake: Only word with four "e"s.
humour: Only word with British spelling

Scores:
randomdragoon: 1
Flame: 1
cf249: 1
iLord: 1
james4l: 1

grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
potato: Only word with two "o"s
cheesecake: Only word with four "e"s.
humour: Only word with British spelling
apprehension: only work using 4 different vowels

Scores:
randomdragoon: 1
Flame: 1
cf249: 1
iLord: 1
james4l: 1
NoTA: 1

#11
grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
potato: Only word with two "o"s
cheesecake: Only word with four "e"s.
humour: Only word with British spelling
apprehension: only work using 4 different vowels
[b]Default: only word with the letter "d" in the beginning (or elsewhere!)[/b]

grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
potato: Only word with two "o"s
cheesecake: Only word with four "e"s.
humour: Only word with British spelling
apprehension: only word using 4 different vowels
default: Only word with "d"
grape: only word with 5 letters

Scores:
randomdragoon: 1
Flame: 1
cf249: 1
iLord: 1
james4l: 1
NoTA: 2
D3m0n Shad0w: 1

Since you didn't put the score, I added your point for you.

grapefruit
banana: Only word with only one unique vowel
sunglasses: only word with unique letter "l"
potato: Only word with two "o"s
cheesecake: Only word with four "e"s.
humour: Only word with British spelling
apprehension: only work using 4 different vowels
default: only word with the letter "d" in the beginning
American: only proper noun

crunch: the only word with 6 or more letters
a: the only word with a ratio of vowels to letters greater than 1/4
bash: the only word with an 's' in it
bleh: the only reaction to someone else
Cramp: the only word ending in "amp"
click: the on word with the letter i in it.
PPP: the only 3-lettered acronym
about: the only word with the letter 'o' in it.

apparently 3/5<1/4. THBASOW, TY

feet: the only real word. :P

feet: The only word which there are an equal number of vowels and consonants.

feet: The only word which there are an equal number of vowels and consonants.
super: The only word that has more consonants than vowels.
deepa: The only misspelled word.
awe: The only word that begins with "a".
your: the only word with the letter y.

y counts as a consanent in "your".

so your and feet both have an equal number of vowels and consanents.

game: The only word spelled correctly

game: the only word spelled correctly
gaimes: the only word that is the plural of another word on the list.
mxi: the only 3-lettered word
foor: the only word that is supposed to be a number.
shick: the only word describing a piece of wood
emotticon: the only 9 letter word (ok, 8)
sooper: the only adjective other than shick.


geometry
3D geometry
sphere
logarithms
floor function
puzzles

There are 9 identical spheres , of them one is slightly lighter, can you find the lighter one .Try using the pan the lowest number of times. :rotfl:  :starwars:  :play_ball:

[hide="very well known..."]
Separate into 3 piles of 3. Weigh two piles - if equal, third pile has light ball, if not equal, lighter pile has light ball. Now we have 3 balls remaining. Weigh two balls - if equal, third ball is light ball, if not equal, lighter ball is light ball. In general, we can find the lighter (or heavier, as long as it is given) ball among $n$ balls in $\log_3{n}$ weighings.[/hide]

[hide="you mean"] $\lfloor\log_3(n-\frac1{2})\rfloor+1$
[/hide]


By how much do the values of six dozen dozen and a half-dozen dozen differ?


I have found the following as subsets of $ \mathbb{Z}_{15}\times\mathbb{Z}_{5}$:

$ \left\{\left(0,0\right)\right\},$

$ \left\{\left(x,0\right)\ |\ x\in\left<1\right>\subseteq\mathbb{Z}_{15}\right\},$

$ \left\{\left(x,0\right)\ |\ x\in\left<3\right>\subseteq\mathbb{Z}_{15}\right\},$

$ \left\{\left(x,0\right)\ |\ x\in\left<5\right>\subseteq\mathbb{Z}_{15}\right\},$

$ \left\{\left(0,y\right)\ |\ y\in\mathbb{Z}_{5}\right\},$

$ \mathbb{Z}_{5}\times\mathbb{Z}_{5},$

and of course, $ \mathbb{Z}_{15}\times\mathbb{Z}_{5}$.

Are these all of the subgroups?  By the Second Sylow Theorem, the number of Sylow $ 3$-subgroups is $ 1$ and the number of Sylow $ 5$-subgroups is also $ 1$ so I think that covers them all.

$ \left< 5\right> \times \mathbb{Z}_5$ is the one you're missing.

To make it easier, the group's isomorphic to $ \mathbb{Z}_3\times \mathbb{Z}_5\times \mathbb{Z}_5$. The subgroups of a direct product are exactly the direct products of all possible subgroups. But the only proper subgroup of $ \mathbb{Z}_p$ for $ p$ prime is trivial, so the number of total subgroups is $ 2^3\equal{}8$. This also gives a way of describing the subgroups (for example, the subgroup shown above is $ \mathbb{Z}_3\times \{ 0\} \times \mathbb{Z}_5$).

[quote="scorpius119"]The subgroups of a direct product are exactly the direct products of all possible subgroups. [/quote]

this is not true in general, but holds here in the case of finite abelian groups.

Isn't it true that $ \mathbb{Z}_{3}$ is not an actual subgroup of $ G\equal{}\mathbb{Z}_{3}\oplus\mathbb{Z}_{5}\oplus\mathbb{Z}_{5}$ but rather [i]isomorphic[/i] to a subgroup of $ G$ (namely $ \left\{\left(x,0,0\right)\ |\ x\in\mathbb{Z}_{3}\right\}$)?

This being the case, is it proper to simply say that $ \mathbb{Z}_{3}$ is a subgroup of $ \mathbb{Z}_{3}\oplus\mathbb{Z}_{5}\oplus\mathbb{Z}_{5}$ or do we need to explicitly write out the subgroup as above?

Thanks for your help :)

if there is a "canonical" monomorphism U -> G, you may say that U is a subgroup of G.

You are in fact missing a lot of subgroups (it is not true that the subgroups of a direct product are the direct products of subgroups; it's wrong for abelian groups, too):
E.g., all of you miss $ <(3,5)>$ (isomorphic to $ (\mathbb Z / 5 \mathbb Z)^2$).

Oh, right. If $ H$ is a subgroup of $ A\times B$, then $ H\cap A,H\cap B$ are normal in $ H$ and intersect only in the identity - but their product may not cover all of $ H$. Even in $ \mathbb{Z}_p\times \mathbb{Z}_p$ this fails miserably.  :blush:

In that case:

Using the form $ \mathbb{Z}_3\times \mathbb{Z}_5\times \mathbb{Z}_5$, there is a unique subgroup of order 3 and a unique subgroup of order 25. All groups of order 5 are subgroups of $ \mathbb{Z}_5\times \mathbb{Z}_5$ and there are 6 of them (24 elements of order 5, 4 per subgroup, distinct subgroups have trivial intersection). All groups of order 15 have the form $ \mathbb{Z}_3\times H$, where $ H$ has order 5. Add on the trivial group and the whole group and you get 16 subgroups. (Specifically, the groups of order 5 are given by $ <(0,a,1)>,<(0,1,0)>$.)

The subgroups of the direct product of tho finite abelians groups are the direct products of the subgroups only when the two groups have prime order.
For instance, let's just consider $ \mathbb{Z}_5\times\mathbb{Z}_5$, it has the trivial subgroup, the full group, and the subgroups generated by (1,x) for x = 0,1,..., 5, and by (0,1), that are all isomorphic to $ \mathbb{Z}_5$.